Introduction
In the late 1990s problems connected to quasiconformal and quasisymmetric maps led to the development of a Sobolev spaces theory on metric measure spaces (mms). In their foundational work [26 ] , Heinonen and Koskela introduced the notion of upper gradient, meant to replace the norm of gradients of smooth functions. To define it, let ( X , μ ) (X,\mu) ( X , μ ) be a mms. A nonnegative Borel function g : X → [ 0 , ∞ ] g : X \to[0,\infty] g : X → [ 0 , ∞ ] is said to be an upper gradient of a real valued function u u u if
M.H. is supported by the Research Council of Finland via the project Quantitative differentiability and rectifiability in metric spaces , grant no. 363800.
M.V. was supported by the European Union MSCA IF QuReViMal, grant no. 101108515, and by an IKERBASQUE starting grant. M.V. was also supported by the Academy of Finland via the project “Higher dimensional Analyst’s Traveling Salesman theorems and Dorronsoro estimates on non-smooth sets”, grant No. 347828.
I.Y.V. was supported by the European Union (ERC, Con-Fine, 101078057) and the MIUR Excellence Department Project awarded to the Department of Mathematics, University of Pisa, CUP I57G22000700001. the inequality
∣ u ( x ) − u ( y ) ∣ ≤ ∫ γ g |u(x)-u(y)| \leq\int_{\gamma} g ∣ u ( x ) − u ( y ) ∣ ≤ ∫ γ g
is satisfied for all rectifiable curves γ \gamma γ joining x x x to y y y in X X X . The path integral is computed with respect to arc length along γ \gamma γ . Coupling this notion with that of Poincaré inequality, Shanmugalingam developed a theory of what she termed Newton-Sobolev spaces [Sha00], denoted by N 1 , p ( X ) N^{1,p}(X) N 1 , p ( X ) (see also [11 ] ).
Definition 1.1. Let p ∈ [ 1 , ∞ ) p \in[1,\infty) p ∈ [ 1 , ∞ ) . A metric measure space ( X , μ ) (X,\mu) ( X , μ ) is said to support a weak ( 1 , p ) (1,p) ( 1 , p ) -Poincaré inequality (or ( 1 , p ) (1,p) ( 1 , p ) -PI) if there exist constants C > 0 , Λ ≥ 1 C>0,\Lambda\geq1 C > 0 , Λ ≥ 1 such that for all open balls B B B in X X X and all pairs of functions u u u and g g g defined on B B B , whenever g g g is an upper gradient of u u u on B B B and u u u is integrable on B B B the following inequality holds:
∫ B ∣ u − u B ∣ d μ ≤ C diam ( B ) ( 1 μ ( B ) ∫ B g p d μ ) 1 / p . \int_B |u-u_B|\,d\mu\leq C\operatorname{diam}(B)\left(\frac{1}{\mu(B)}\int_B g^p\,d\mu\right)^{1/p}. ∫ B ∣ u − u B ∣ d μ ≤ C diam ( B ) ( μ ( B ) 1 ∫ B g p d μ ) 1/ p . Any such ( X , μ ) (X,\mu) ( X , μ ) is called a PI space.
The importance of (1.1) for a first-order Sobolev theory lies in the fact that it ensures the presence of ‘thick’ families of rectifiable curves in X X X . When coupled with the doubling property of μ \mu μ , this ensures that upper gradients reflect appropriately local changes of the associated functions. Indeed, Cheeger proved that if ( X , μ ) (X,\mu) ( X , μ ) supports a Poincaré inequality and is doubling, then it possesses an infinitesimal linear structure sufficiently rich to permit upper gradients of Lipschitz functions to be controlled by their point-wise Lipschitz constant. In other words, Cheeger proved in this setting a version of Rademacher’s differentiation theorem about Euclidean Lipschitz functions. Spaces ( X , μ ) (X,\mu) ( X , μ ) which support this infinitesimal linear structure needed for Cheeger’s Rademacher’s result to apply were termed Lipschitz differentiability spaces . A converse of Cheeger’s theorem was proved in the remarkable works by Bate and Li, and Eriksson-Bique [10 , 21 ] , where they showed that any Lipschitz differentiability space (with the Radon-Nikodym Property) is PI-rectifiable – it can be ‘covered’ by a countable number of PI-spaces.
The property of supporting a Poincaré inequality is quantitative and multiscale. As such, one would expect it to have quantitative and multiscale consequences. Cheeger’s theorem, however, is qualitative: it says that Lipschitz functions on X X X are infinitesimally linear in an appropriate generalised sense. In Euclidean setting, there is a quantitative version of Rademacher’s theorem stating that, in fact, Lipschitz functions are close to being affine at the typical location and scale – this is Dorronsoro’s theorem [18 ] . This begs the question: does an appropriate version of this theorem hold on PI spaces? And, conversely, do spaces with a Dorronsoro theorem also have PI?
To proceed, recall that a metric measure space ( X , μ ) (X,\mu) ( X , μ ) is Ahlfors d d d -regular if X = supp ( μ ) X=\operatorname{supp}(\mu) X = supp ( μ ) and μ \mu μ is Ahlfors regular, meaning that there is A ≥ 1 A\geq1 A ≥ 1 such that
A − 1 r d ≤ μ ( B ( x , r ) ) ≤ A r d for all x ∈ X , 0 < r < 2 diam ( X ) . A^{-1}r^d \leq\mu(B(x,r)) \leq Ar^d \qquad\text{for all } x\in X,\ 0<r<2\operatorname{diam}(X). A − 1 r d ≤ μ ( B ( x , r )) ≤ A r d for all x ∈ X , 0 < r < 2 diam ( X ) .
When ( X , μ ) (X,\mu) ( X , μ ) is Ahlfors d d d -regular, and X ⊂ R n X\subset\mathbb{R}^n X ⊂ R n , we now know precisely the condition for which ( X , μ ) (X,\mu) ( X , μ ) supports a Dorronsoro type theorem: uniform d d d -rectifiability is that condition. This is due to David and Semmes [20 ] and the proof of the WALA conjecture by Merlo [Mer26]. We recall the definition of uniform rectifiability (introduced in [19 ] ) below.
Definition 1.2. An Ahlfors d d d -regular set X ⊂ R n X \subset\mathbb{R}^n X ⊂ R n is said to be uniformly d d d -rectifiable if there exist constants 0 < θ < 1 0 < \theta< 1 0 < θ < 1 and L ≥ 1 L \ge1 L ≥ 1 such that, for each ball B B B centered on X X X , there exists a L L L -Lipschitz map g : R d ⊃ B ( 0 , r ( B ) ) → R n g : \mathbb{R}^d \supset B(0,r(B)) \to\mathbb{R}^n g : R d ⊃ B ( 0 , r ( B )) → R n such that
H d ( X ∩ B ∩ g ( B ( 0 , r ( B ) ) ) ) ≥ θ r ( B ) d . \mathcal{H}^d\left(X \cap B \cap g(B(0,r(B)))\right) \ge\theta r(B)^d. H d ( X ∩ B ∩ g ( B ( 0 , r ( B ))) ) ≥ θ r ( B ) d .
Returning to the first question above, Azzam [5 ] proved that if X ⊂ R n X \subset\mathbb{R}^n X ⊂ R n (equipped with the Euclidean metric) is Ahlfors d d d -regular and supports a ( 1 , d ) (1,d) ( 1 , d ) -Poincaré, then it is uniformly d d d -rectifiable. In this article, we prove an appropriate converse. Namely:
Theorem 1.3. If E ⊆ B ( 0 , 1 ) ⊆ R n E \subseteq B(0,1) \subseteq\mathbb{R}^n E ⊆ B ( 0 , 1 ) ⊆ R n is uniformly d d d -rectifiable, then there is an Ahlfors d d d -regular set Z ⊆ R n + 1 Z \subseteq\mathbb{R}^{n+1} Z ⊆ R n + 1 with E ⊆ Z E \subseteq Z E ⊆ Z such that ( Z , H d ∣ Z ) (Z,\mathcal{H}^d|_Z) ( Z , H d ∣ Z ) supports a ( 1 , 1 ) (1,1) ( 1 , 1 ) -Poincaré inequality. Moreover, the Ahlfors regularity constants and the constants in the Poincaré inequality depending only on n n n , d d d and the uniform rectifiability constants for E E E .
For Azzam, it was enough to assume a ( 1 , d ) (1,d) ( 1 , d ) -PI to guarantee uniform rectifiability. Since we construct a space which supports a ( 1 , 1 ) (1,1) ( 1 , 1 ) -PI, and since a ( 1 , p ) (1,p) ( 1 , p ) -PI is stronger than a ( 1 , q ) (1,q) ( 1 , q ) -PI whenever p < q p<q p < q , it follows that the space we construct is also d d d -UR. One observes also that Theorem 1.3 and Azzam’s result imply that any d d d -ADR ( 1 , d ) (1,d) ( 1 , d ) -PI space in R n \mathbb{R}^n R n is a subset of a d d d -ADR ( 1 , 1 ) (1,1) ( 1 , 1 ) -PI space in R n + 1 \mathbb{R}^{n+1} R n + 1 .
Remarks on the literature
The literature on Lipschitz differentiability, PI spaces and Sobolev spaces on mms is very broad and we only scratch the surface here. First, we should remark that, aside from those already mentioned, two other definition of Sobolev spaces in metric measure spaces where proposed: that of Hajłasz [Haj96] and of Ambrosio-Gigli-Savaré [1 ] . All the definitions coincide for a general mms, except for the one of Hajłasz which coincides with the others in the presence of a PI. Cheeger, after his seminal result [11 ] , proved quantitative analogues [12 , 13 ] . While a notion of uniform rectifiability for general metric measure spaces has been introduced in [8 ] by Bate, Schul and the first author, it is unclear under which conditions Cheeger’s quantitative result may imply this notion. For geometric quantified flatness estimates across scales related to uniform rectifiability in metric spaces see also [23 ] . As we mentioned above, if X ⊂ R n X \subset\mathbb{R}^n X ⊂ R n , quantitative affine approximation of Lipschitz function does imply uniform rectifiability. This is the content of [Mer26]. In fact, if X ⊂ R n X \subset\mathbb{R}^n X ⊂ R n is d d d -ADR, one may recover the full Donoroso characterisation of Sobolev spaces [3 ] . See also [29 ] for an analogous result on RCD spaces.
In general, the qualitative theory of Lipschitz differentiability is much better understood for general metric measure spaces, thanks to the collective efforts of several mathematicians, see e.g. the work of Bate [7 , 6 ] , of Bate-Li [9 ] , of Cheeger-Kleiner-Schioppa [15 ] , of Schioppa [34 ] , of Keith [30 , 31 ] , of Alberti-Marchese [2 ] , and of David, Eriksson-Bique and Schul [17 ] (in no particular order).
The theory of Sobolev spaces has much progressed, too: we refer the reader to the monograph [27 ] . Remarkable, and connected to the progress on Lipschitz differentiability and directions present in X X X mentioned above, is the construction of Sobolev differential by Eriksson-Bique and Soultanis [22 ] .
Outline of the proof. Consider E E E as in the statement of Theorem 1.3 . Since E E E is uniformly rectifiable, it admits a corona decomposition, i.e., it is well-approximated in most locations and scales by a controlled ‘number’ of Lipschitz graphs (and contained in their union up to a set of measure zero). It is well-known that Lipschitz graphs admit ( 1 , 1 ) (1,1) ( 1 , 1 ) -Poincaré inequalities. Thus, the basic idea is build bridges between the graphs from the corona decomposition in such a way that generates a global Poincaré inequality.
There are several problems we need to overcome. First, we only have access to graphs in the good scales and locations for E E E but we need to describe the coarse behaviour of the enveloping set in all cubes. This begs the question, what does the set look like in the bad locations and scales and how do you glue this to the good graphs in a way that is consistent with a Poincaré inequality? We solve this by considering approximations by d d d -dimensional dyadic skelata, i.e., d d d -dimensional boundary of dyadic cubes. Instead of using the graphs directly, our approximations in the good regions will be unions of dyadic skelata which are ‘aligned’ along the original graphs and appear at the scale of the minimal cubes. In the bad locations we have no geometry to deal with, so we just look at a fine-mesh approximation of ambient n n n -dimensional ball. See § 6 . The benefit of considering such a approximations is the following: if two dyadic skelata with comparable size have large intersection, then their union supports a Poincaré inequality (see § 5 ). Since the good and bad regions are built using dyadic skelata, this makes the gluing procedure far easier. One caveat of using skeletal approximations is that it is not immediate that they support a Poincaré inequality. Thus, we must check that the skeletal approximations admit a Poincaré inequality. We check this condition in Proposition 5.7 . Note that, since we want to make use of the above good intersection properties of nearby skelata, we will want to consider smooth trees where nearby minimal cubes have comparable sizes. These are defined in § 4 .
The next problem is to avoid bad intersections of the skeletal approximations and the bridges that we build between each of the skeletal approximations. For this we add an additional dimension and ‘lift’ the approximations in a controlled way which depends on how well it approximates E E E , see (69) . To connect one skeletal approximation to the next layer of skeletal approximations, we pick a single d d d -dimensional skeleton for each minimal cube and ‘drag’ one of its ( d − 1 ) (d-1) ( d − 1 ) -dimensional faces down to the approximation below it (here, down and below refer the height in this additional dimension). One should have in mind connecting the two sets with a bridge which is bi-Lipschitz to a d d d -dimensional cube, although the reality is a little more complicated that. These bridges are built in § 8 .
Finally, we have to verify a Poincaré inequality for the set we construct as above. We start in § 9 by showing a local Poincaré inequality, i.e., we check that a skeletal approximation, together with all its connecting bridges, supports a Poincaré inequality. To check a global Poincaré inequality we find, for each pair of points, a connecting ‘Harnack chain’ of balls such that consecutive balls lie in one of the spaces that already have Poincaré estimates. This gives us a way to estimate the oscillation of functions in consecutive balls and, thus, this type of connectivity with local Poincaré estimates is enough to conclude a global Poincaré inequality. This is carried out in Section 10 and relies on the results in § 3 .
Statement on AI. No AI tools were used for this article.
Preliminaries
Notation
Throughout the paper C C C will denote some absolute constant (depending perhaps on n n n ) which may change from line to line. If there exists C ≥ 1 C \ge1 C ≥ 1 such that a ≤ C b a \le Cb a ≤ C b , then we will write a ≲ b a \lesssim b a ≲ b . If C C C depends on parameters ( t 1 , … , t m ) (t_1,\ldots,t_m) ( t 1 , … , t m ) , then we will write a ≲ t 1 , … , t m b a \lesssim_{t_1,\ldots,t_m} b a ≲ t 1 , … , t m b . We will write a ∼ b a \sim b a ∼ b if a ≲ b a \lesssim b a ≲ b and b ≲ a b \lesssim a b ≲ a . The notation a ∼ t 1 , … , t m b a \sim_{t_1,\ldots,t_m} b a ∼ t 1 , … , t m b is defined similarly.
For a metric space ( X , d ) (X,d) ( X , d ) , we let B ( x , r ) B(x,r) B ( x , r ) denote the open in X X X centred at x x x with radius r r r , that is,
B ( x , r ) = { y ∈ X : d ( y , x ) < r } . B(x,r) = \{y \in X : d(y,x) < r\}. B ( x , r ) = { y ∈ X : d ( y , x ) < r } .
If B B B is a ball in X X X , we will often write x B x_B x B and r B r_B r B to denote its centre and radius, respectively (although such objects are not necessarily uniquely defined). For sets E , F ⊆ X E,F \subseteq X E , F ⊆ X , let
dist ( E , F ) : = inf { d ( x , y ) : x ∈ E and y ∈ F } ; \operatorname{dist}(E,F) := \inf\{d(x,y): x \in E \text{ and } y \in F\}; dist ( E , F ) := inf { d ( x , y ) : x ∈ E and y ∈ F } ;
diam ( E ) : = sup { d ( x , y ) : x , y ∈ E } . \operatorname{diam}(E) := \sup\{d(x,y): x,y \in E\}. diam ( E ) := sup { d ( x , y ) : x , y ∈ E } .
The k k k -dimensional Hausdorff measure of E E E is defined as
H k ( E ) : = lim δ → 0 { ∑ i diam ( U i ) k : E ⊆ ⋃ i U i and diam ( U i ) < δ } . \mathcal{H}^k(E) := \lim_{\delta\to0} \left\{\sum_i \operatorname{diam}(U_i)^k : E \subseteq\bigcup_i U_i \text{ and } \operatorname{diam}(U_i) < \delta\right\}. H k ( E ) := δ → 0 lim { i ∑ diam ( U i ) k : E ⊆ i ⋃ U i and diam ( U i ) < δ } .
For a set A ⊆ R n × { 0 } A \subseteq\mathbb{R}^n \times\{0\} A ⊆ R n × { 0 } and a function f : A → R f:A \to\mathbb{R} f : A → R , denote the graph of f f f over A A A by Graph A ( f ) \operatorname{Graph}_A(f) Graph A ( f ) , that is,
Graph A ( f ) = { ( x , f ( x ) ) : x ∈ A } ⊆ R n + 1 . (1) \operatorname{Graph}_A(f) = \{(x,f(x)): x \in A\} \subseteq\mathbb{R}^{n+1}. \tag*{(1)} Graph A ( f ) = {( x , f ( x )) : x ∈ A } ⊆ R n + 1 . (1)
Dyadic cubes and their skeleta
Let Δ \Delta Δ denote the collection of dyadic cubes on R n \mathbb{R}^n R n . For I ∈ Δ I \in\Delta I ∈ Δ , let x I x_I x I denote its centre and ℓ ( I ) \ell(I) ℓ ( I ) denote its side-length. For c > 0 c > 0 c > 0 , let c I cI c I denote the cube concentric to I I I with side-length equal ℓ ( c I ) = c ℓ ( I ) \ell(cI) = c\ell(I) ℓ ( c I ) = c ℓ ( I ) . Let
B I = B ( x I , diam ( I ) ) . B_I = B(x_I,\operatorname{diam}(I)). B I = B ( x I , diam ( I )) .
We let ∂ d I \partial_d I ∂ d I denote its d d d -dimensional skeleton, i.e., if I = ∏ j = 1 n I j I = \prod_{j=1}^{n} I_j I = ∏ j = 1 n I j for some (half open, half closed) intervals I j ⊆ R I_j \subseteq\mathbb{R} I j ⊆ R , then
∂ d I = { ( x 1 , … , x n ) ∈ I : x j ∈ ∂ I j for all least n − d indices j ∈ { 1 , … , n } } . \partial_d I = \{(x_1,\ldots,x_n) \in I: x_j \in\partial I_j \text{ for all least } n-d \text{ indices } j \in\{1,\ldots,n\}\}. ∂ d I = {( x 1 , … , x n ) ∈ I : x j ∈ ∂ I j for all least n − d indices j ∈ { 1 , … , n }} .
These dyadic skeleta will be the basic building block from which we construct the set Z Z Z from Theorem 1.3 . We work with these sets because it is easy to describe how nearby skeleta interact with each other. The properties we need are proven below. First, we show that the gluing metric between nearby cubes respect the ambient Euclidean distance (a sort of orthogonality). Second, we show that skeleta corresponding to nested cubes have large intersections. Both of these properties will be useful in showing that unions of such skeleta support a Poincaré inequality.
Lemma 2.1. Let I , J , K ∈ Δ I,J,K \in\Delta I , J , K ∈ Δ be such that I ⊆ K I \subseteq K I ⊆ K , ∂ d I ∩ ∂ d K ≠ ∅ \partial_d I \cap\partial_d K \ne\varnothing ∂ d I ∩ ∂ d K = ∅ , ℓ ( I ) = 2 − m ℓ ( K ) \ell(I) = 2^{-m}\ell(K) ℓ ( I ) = 2 − m ℓ ( K ) and ℓ ( J ) = ℓ ( K ) \ell(J) = \ell(K) ℓ ( J ) = ℓ ( K ) . Then, for each x ∈ ∂ d I x \in\partial_d I x ∈ ∂ d I and y ∈ ∂ d J y \in\partial_d J y ∈ ∂ d J , we have
∣ x − y ∣ ∼ m inf z ∈ ∂ d I ∩ ∂ d K ∣ x − z ∣ + ∣ y − z ∣ . (2) |x-y| \sim_m \inf_{z \in\partial_d I \cap\partial_d K} |x-z| + |y-z|. \tag*{(2)} ∣ x − y ∣ ∼ m z ∈ ∂ d I ∩ ∂ d K inf ∣ x − z ∣ + ∣ y − z ∣. (2)
Proof. By scaling and translating, we may assume that K = [ 0 , 1 ] n K=[0,1]^n K = [ 0 , 1 ] n . Translating and rotating again, we may also assume I ⊆ [ 0 , 1 / 2 ] n I\subseteq[0,1/2]^n I ⊆ [ 0 , 1/2 ] n . Let us write x = ( x 1 , … , x n ) x=(x_1,\ldots,x_n) x = ( x 1 , … , x n ) and y = ( y 1 , … , y n ) y=(y_1,\ldots,y_n) y = ( y 1 , … , y n ) . Since x ∈ ∂ d I x\in\partial_d I x ∈ ∂ d I , there is a set A x A_x A x with ∣ A x ∣ = n − d |A_x|=n-d ∣ A x ∣ = n − d such that
x k ∈ { j 2 − m : j ∈ { 0 , 1 , … , 2 m − 1 } } for all k ∈ A x . x_k\in\{j2^{-m}:j\in\{0,1,\ldots,2^{m-1}\}\}\quad\text{for all }k\in A_x. x k ∈ { j 2 − m : j ∈ { 0 , 1 , … , 2 m − 1 }} for all k ∈ A x .
Similarly, there exists a set A y A_y A y with ∣ A y ∣ = n − d |A_y|=n-d ∣ A y ∣ = n − d such that
y k ∈ Z for all k ∈ A y . y_k\in\mathbb{Z}\quad\text{for all }k\in A_y. y k ∈ Z for all k ∈ A y .
Suppose to begin with that there exists k ∈ A x k\in A_x k ∈ A x such that x k ≠ 0 x_k\ne0 x k = 0 . In this case, we have x k ∈ [ 2 − m , 1 / 2 ] x_k\in[2^{-m},1/2] x k ∈ [ 2 − m , 1/2 ] . Thus, since y k ∈ Z y_k\in\mathbb{Z} y k ∈ Z , this implies ∣ x k − y k ∣ ≥ 2 − m |x_k-y_k|\ge2^{-m} ∣ x k − y k ∣ ≥ 2 − m . By taking an arbitrary point z ∈ ∂ d I ∩ ∂ d K z\in\partial_d I\cap\partial_d K z ∈ ∂ d I ∩ ∂ d K , since x ∈ K x\in K x ∈ K , we get
∣ x − z ∣ + ∣ y − z ∣ ≤ 2 ∣ x − z ∣ + ∣ x − y ∣ ≲ m 2 − m + ∣ x − y ∣ ≲ ∣ x − y ∣ . |x-z|+|y-z|\le2|x-z|+|x-y|\lesssim_m 2^{-m}+|x-y|\lesssim|x-y|. ∣ x − z ∣ + ∣ y − z ∣ ≤ 2∣ x − z ∣ + ∣ x − y ∣ ≲ m 2 − m + ∣ x − y ∣ ≲ ∣ x − y ∣.
Now assume that x k = 0 x_k=0 x k = 0 for all k ∈ A x k\in A_x k ∈ A x . Since ∣ A x ∣ = n − d |A_x|=n-d ∣ A x ∣ = n − d and I ⊆ K I\subseteq K I ⊆ K we have x ∈ ∂ d I ∩ ∂ d K x\in\partial_d I\cap\partial_d K x ∈ ∂ d I ∩ ∂ d K . The estimate (2) is now immediate by taking z = x z=x z = x as the infimising point. □
Lemma 2.2. Let I ⊆ J I\subseteq J I ⊆ J and suppose that x ∈ ∂ d I ∩ ∂ d J x\in\partial_d I\cap\partial_d J x ∈ ∂ d I ∩ ∂ d J . Then, there are d d d -dimensional faces F I ⊆ ∂ d I F_I\subseteq\partial_d I F I ⊆ ∂ d I and F J ⊆ ∂ d J F_J\subseteq\partial_d J F J ⊆ ∂ d J such that x ∈ F I ⊆ F J x\in F_I\subseteq F_J x ∈ F I ⊆ F J .
Proof. Let us write I = ∏ k = 1 n I k I=\prod_{k=1}^n I_k I = ∏ k = 1 n I k and J = ∏ k = 1 n J k J=\prod_{k=1}^n J_k J = ∏ k = 1 n J k , where I k ⊆ J k I_k\subseteq J_k I k ⊆ J k are half-open half-closed intervals in R \mathbb{R} R . Let x = ( x 1 , … , x n ) ∈ ∂ d I ∩ ∂ d J x=(x_1,\ldots,x_n)\in\partial_d I\cap\partial_d J x = ( x 1 , … , x n ) ∈ ∂ d I ∩ ∂ d J and let
K I : = { k : x k ∈ ∂ I k } and K J : = { k : x k ∈ ∂ J k } . K_I:=\{k:x_k\in\partial I_k\}\quad\text{and}\quad K_J:=\{k:x_k\in\partial J_k\}. K I := { k : x k ∈ ∂ I k } and K J := { k : x k ∈ ∂ J k } . Since I k ⊆ J k I_k\subseteq J_k I k ⊆ J k for each $k, it is clear that K J ⊆ K I K_J\subseteq K_I K J ⊆ K I . Since x ∈ ∂ d J x\in\partial_d J x ∈ ∂ d J , we have that ∣ K J ∣ ≥ n − d |K_J|\ge n-d ∣ K J ∣ ≥ n − d . Thus, we can find a set K K K such that
∣ K ∣ = n − d and K ⊆ K I ∩ K J . |K|=n-d\quad\text{and}\quad K\subseteq K_I\cap K_J. ∣ K ∣ = n − d and K ⊆ K I ∩ K J . We define the faces F I F_I F I and F J F_J F J as
F I : = { z ∈ I : z k = x k for all k ∈ K } ; F_I:=\{z\in I:z_k=x_k\text{ for all }k\in K\}; F I := { z ∈ I : z k = x k for all k ∈ K } ; F J : = { z ∈ J : z k = x k for all k ∈ K } . F_J:=\{z\in J:z_k=x_k\text{ for all }k\in K\}. F J := { z ∈ J : z k = x k for all k ∈ K } . If z = ( z 1 , … , z n ) ∈ F I z=(z_1,\ldots,z_n)\in F_I z = ( z 1 , … , z n ) ∈ F I , then we have z k = x k z_k=x_k z k = x k for all k ∈ K k\in K k ∈ K and z k ∈ I k ⊆ J k z_k\in I_k\subseteq J_k z k ∈ I k ⊆ J k for all k ∈ { 1 , … , n } ∖ K k\in\{1,\ldots,n\}\setminus K k ∈ { 1 , … , n } ∖ K . That is, z ∈ F J z\in F_J z ∈ F J , as required. □
Christ-David cubes
Throughout the paper, it will be convenient for us to work with a version of dyadic cubes tailored to a set X X X . These are the so-called Christ-David cubes, which were first introduced by David [16 ] and generalized in [14 ] . The following formulation is due to [28 ] .
Lemma 2.3. Let X X X be a doubling metric space and { X k } k ∈ Z \{X_k\}_{k\in\mathbb{Z}} { X k } k ∈ Z be a sequence of maximal ϱ k \varrho^k ϱ k -separated nets such that X k + 1 ⊆ X k X_{k+1}\subseteq X_k X k + 1 ⊆ X k , where ϱ = 1 / 1000 \varrho=1/1000 ϱ = 1/1000 and let c 0 = 1 / 500 c_0=1/500 c 0 = 1/500 . Then, for each k ∈ Z k\in\mathbb{Z} k ∈ Z , there is a collection D k \mathcal{D}_k D k of “cubes” such that the following conditions hold.
For each k ∈ Z k\in\mathbb{Z} k ∈ Z , X = ⋃ Q ∈ D k Q X=\bigcup_{Q\in\mathcal{D}_k}Q X = ⋃ Q ∈ D k Q .
If Q 1 , Q 2 ∈ D : = ⋃ k D k Q_1,Q_2\in\mathcal{D}:=\bigcup_k\mathcal{D}_k Q 1 , Q 2 ∈ D := ⋃ k D k and Q 1 ∩ Q 2 ≠ ∅ Q_1\cap Q_2\ne\varnothing Q 1 ∩ Q 2 = ∅ , then Q 1 ⊆ Q 2 Q_1\subseteq Q_2 Q 1 ⊆ Q 2 or Q 2 ⊆ Q 1 Q_2\subseteq Q_1 Q 2 ⊆ Q 1 .
For Q ∈ D k Q\in\mathcal{D}_k Q ∈ D k set ℓ ( Q ) = 5 ϱ k \ell(Q)=5\varrho^k ℓ ( Q ) = 5 ϱ k . Then there is x Q ∈ X k x_Q\in X_k x Q ∈ X k such that
B X ( x Q , c 0 ℓ ( Q ) ) ⊆ Q ⊆ B X ( x Q , ℓ ( Q ) ) . B_X(x_Q,c_0\ell(Q))\subseteq Q\subseteq B_X(x_Q,\ell(Q)). B X ( x Q , c 0 ℓ ( Q )) ⊆ Q ⊆ B X ( x Q , ℓ ( Q )) .
For each Q ∈ D Q \in\mathcal{D} Q ∈ D we set B Q : = B X ( x Q , ℓ ( Q ) ) B_Q := B_X(x_Q,\ell(Q)) B Q := B X ( x Q , ℓ ( Q )) .
Given a collection of cubes D \mathcal{D} D and Q ∈ D Q \in\mathcal{D} Q ∈ D , define
D ( Q ) = { R ∈ D : R ⊆ Q } . \mathcal{D}(Q)=\{R\in\mathcal{D}:R\subseteq Q\}. D ( Q ) = { R ∈ D : R ⊆ Q } .
Let Parent ( Q ) \operatorname{Parent}(Q) Parent ( Q ) be the unique cube R ∈ D R\in\mathcal{D} R ∈ D such that R ⊇ Q R\supseteq Q R ⊇ Q and ℓ ( Q ) = ϱ ℓ ( Q ) \ell(Q)=\varrho\ell(Q) ℓ ( Q ) = ϱ ℓ ( Q ) . Let Child ( Q ) \operatorname{Child}(Q) Child ( Q ) denote the collection of cubes R ⊆ Q R\subseteq Q R ⊆ Q such that Q = Parent ( R ) Q=\operatorname{Parent}(R) Q = Parent ( R ) . Let Sibling ( Q ) \operatorname{Sibling}(Q) Sibling ( Q ) be the collection of cubes R R R such that Parent ( R ) = Parent ( Q ) \operatorname{Parent}(R)=\operatorname{Parent}(Q) Parent ( R ) = Parent ( Q ) .
Fix for the moment a doubling metric space and a system of Christ-David cubes D \mathcal{D} D on X X X .
Definition 2.5. A collection of cubes S ⊆ D \mathcal{S}\subseteq\mathcal{D} S ⊆ D is called a stopping-time region if the following conditions holds.
(1) There is a top cube Q ( S ) ∈ S Q(\mathcal{S})\in\mathcal{S} Q ( S ) ∈ S such that, if Q ∈ S Q\in\mathcal{S} Q ∈ S , then Q ⊆ Q ( S ) Q\subseteq Q(\mathcal{S}) Q ⊆ Q ( S ) .
(2) If Q ∈ S Q\in\mathcal{S} Q ∈ S and Q ⊆ R ⊆ Q ( S ) Q\subseteq R\subseteq Q(\mathcal{S}) Q ⊆ R ⊆ Q ( S ) , then R ∈ S R\in\mathcal{S} R ∈ S .
(3) If Q ∈ S Q\in\mathcal{S} Q ∈ S , then R ∈ S R\in\mathcal{S} R ∈ S for all R ∈ Sibling ( Q ) R\in\operatorname{Sibling}(Q) R ∈ Sibling ( Q ) .
Definition 2.6. A collection of cubes B ⊆ D \mathcal{B}\subseteq\mathcal{D} B ⊆ D satisfies a Carleson packing condition if there exists C > 0 C>0 C > 0 such that
∑ Q ∈ B Q ⊆ Q 0 ℓ ( Q ) n ≤ C ℓ ( Q 0 ) n \sum_{\substack{Q\in\mathcal{B}\\Q\subseteq Q_0}}\ell(Q)^n\leq C\ell(Q_0)^n Q ∈ B Q ⊆ Q 0 ∑ ℓ ( Q ) n ≤ C ℓ ( Q 0 ) n for all Q 0 ∈ D Q_0\in\mathcal{D} Q 0 ∈ D .
Definition 2.7. For a collection of cubes C ⊆ D \mathcal{C}\subseteq\mathcal{D} C ⊆ D , define the distance function to C \mathcal{C} C by
d C ( x ) = inf { ℓ ( R ) + dist ( x , R ) : R ∈ C } . d_{\mathcal{C}}(x)=\inf\{\ell(R)+\operatorname{dist}(x,R):R\in\mathcal{C}\}. d C ( x ) = inf { ℓ ( R ) + dist ( x , R ) : R ∈ C } . For Q ∈ D Q\in\mathcal{D} Q ∈ D , set
d C ( Q ) = inf { d C ( x ) : x ∈ Q } = inf { ℓ ( R ) + dist ( Q , R ) : R ∈ C } . d_{\mathcal{C}}(Q)=\inf\{d_{\mathcal{C}}(x):x\in Q\}=\inf\{\ell(R)+\operatorname{dist}(Q,R):R\in\mathcal{C}\}. d C ( Q ) = inf { d C ( x ) : x ∈ Q } = inf { ℓ ( R ) + dist ( Q , R ) : R ∈ C } .
The following can be easily deduced from the definitions.
Lemma 2.8. Let A , C ≥ 1 A,C\geq1 A , C ≥ 1 and c ∈ ( 0 , 1 ) c\in(0,1) c ∈ ( 0 , 1 ) . Suppose { B i } i ∈ I \{B_i\}_{i\in I} { B i } i ∈ I is a collection of balls such that c B i ∩ c B j = ∅ cB_i\cap cB_j=\varnothing c B i ∩ c B j = ∅ and C − 1 r i ≤ r j ≤ C r i C^{-1}r_i\leq r_j\leq Cr_i C − 1 r i ≤ r j ≤ C r i for all i , j ∈ I i,j\in I i , j ∈ I such that A B i ∩ A B j ≠ ∅ AB_i\cap AB_j\neq\varnothing A B i ∩ A B j = ∅ , then { A B i } i ∈ I \{AB_i\}_{i\in I} { A B i } i ∈ I has bounded overlap depending on d , c , A d,c,A d , c , A and C C C . Let { Q i } i ∈ I \{Q_i\}_{i\in I} { Q i } i ∈ I be a collection of disjoint Christ-David cubes. If C − 1 ℓ ( Q j ) ≤ ℓ ( Q i ) ≤ C ℓ ( Q j ) C^{-1}\ell(Q_j)\leq\ell(Q_i)\leq C\ell(Q_j) C − 1 ℓ ( Q j ) ≤ ℓ ( Q i ) ≤ C ℓ ( Q j ) for all i , j ∈ I i,j\in I i , j ∈ I such that A B Q i ∩ A B Q j ≠ ∅ AB_{Q_i}\cap AB_{Q_j}\neq\varnothing A B Q i ∩ A B Q j = ∅ , then { A B Q i } i ∈ I \{AB_{Q_i}\}_{i\in I} { A B Q i } i ∈ I has bounded overlap with constant depending on d , c 0 , A d,c_0,A d , c 0 , A and C C C .
Gluing Ahlfors regular sets
At many points in the paper we will need to glue various configurations of (upper, lower or upper and lower) regular sets. Here, we gather some useful result that will allow us to preserves these regularity properties after the gluing. We start with the following easy lemma regarding gluing finitely many spaces with non-empty intersection.
Lemma 2.9. Let Z Z Z be a metric space and { X i } i = 1 N ⊆ Z \{X_i\}_{i=1}^{N}\subseteq Z { X i } i = 1 N ⊆ Z be a finite collection of Ahlfors d d d -regular subsets with uniform regularity constant C 0 C_0 C 0 . If X i ∩ X i + 1 ≠ ∅ X_i\cap X_{i+1}\neq\varnothing X i ∩ X i + 1 = ∅ for each i ∈ 1 , … , N − 1 i\in1,\ldots,N-1 i ∈ 1 , … , N − 1 , then ⋃ i = 1 N X i \bigcup_{i=1}^{N}X_i ⋃ i = 1 N X i is Ahlfors d d d -regular with regularity constant depending only on C 0 C_0 C 0 and N N N .
Proof. By induction, it is enough to show that X ∪ Y X \cup Y X ∪ Y is Ahlfors d d d -regular whenever X X X and Y Y Y are Ahlfors d d d -regular with uniform constant C 0 C_0 C 0 and satisfy X ∩ Y ≠ ∅ X \cap Y \ne\emptyset X ∩ Y = ∅ . Upper regularity with constant 2 C 0 2C_0 2 C 0 follows directly from the upper regularity of X X X and Y Y Y . For the lower regularity, fix a ball B B B centred on X ∪ Y X \cup Y X ∪ Y with r B ∈ ( 0 , 2 diam ( X ∪ Y ) ) r_B \in(0, 2\operatorname{diam}(X \cup Y)) r B ∈ ( 0 , 2 diam ( X ∪ Y )) and assume (without loss of generality) that B B B is centred at x B ∈ X x_B \in X x B ∈ X . If r B ≤ 4 diam ( X ) r_B \le4\operatorname{diam}(X) r B ≤ 4 diam ( X ) , then the lower regularity follows from the lower regularity of X X X . Suppose instead that r B > 4 diam ( X ) r_B > 4\operatorname{diam}(X) r B > 4 diam ( X ) . Observe first that
diam ( Y ) > r B / 4. (3) \operatorname{diam}(Y) > r_B/4. \tag*{(3)} diam ( Y ) > r B /4. (3)
Indeed, if this were not true, then we would have diam ( X ∪ Y ) < 2 r B \operatorname{diam}(X \cup Y) < 2r_B diam ( X ∪ Y ) < 2 r B , which is a contradiction. Now, take any point z ∈ X ∩ Y z \in X \cap Y z ∈ X ∩ Y and consider the ball B ′ = B ( z , r B / 2 ) B' = B(z, r_B/2) B ′ = B ( z , r B /2 ) . Since d ( x B , z ) < r B / 4 d(x_B,z) < r_B/4 d ( x B , z ) < r B /4 , we have that B ′ ⊆ B B' \subseteq B B ′ ⊆ B , so the lower regularity of B B B follows from the lower regularity of the space Y Y Y and the diameter estimate (3) . □ \square □
If the spaces satisfy certain properties, we are able to obtain a gluing where the regularity constant is independent of the number of spaces in the collection.
Lemma 2.10. Let { C i } i = 0 3 ⊆ [ 1 , ∞ ) \{C_i\}_{i=0}^{3} \subseteq[1,\infty) { C i } i = 0 3 ⊆ [ 1 , ∞ ) . Let ( Z , d ) (Z,d) ( Z , d ) be a metric space and X , { Y i } i ∈ I ⊆ Z X,\{Y_i\}_{i\in I} \subseteq Z X , { Y i } i ∈ I ⊆ Z be Ahlfors d d d -regular subsets with regularity constant C 0 C_0 C 0 . Let { B i } i ∈ I \{B_i\}_{i\in I} { B i } i ∈ I be a collection of balls such that the following conditions hold.
(1) The set { 5 C 1 B i } i ∈ I \{5C_1B_i\}_{i\in I} { 5 C 1 B i } i ∈ I has overlap bounded by C 2 C_2 C 2 .
(2) If B j = B ( x j , r j ) B_j=B(x_j,r_j) B j = B ( x j , r j ) , then
x j ∈ Y j , r j ≤ max { diam ( Y j ) , C 1 diam ( X ) } and Y i ⊆ C 1 B i . x_j \in Y_j,\qquad r_j \le\max\{\operatorname{diam}(Y_j),C_1\operatorname{diam}(X)\}\quad\text{and}\quad Y_i \subseteq C_1B_i. x j ∈ Y j , r j ≤ max { diam ( Y j ) , C 1 diam ( X )} and Y i ⊆ C 1 B i . (3) We have
dist ( x j , X ) ≤ C 1 r i . \operatorname{dist}(x_j,X) \le C_1r_i. dist ( x j , X ) ≤ C 1 r i . Then, the set
Y : = ⋃ i ∈ I Y i Y := \bigcup_{i\in I}Y_i Y := i ∈ I ⋃ Y i is upper Ahlfors d d d -regular, with constant depending on d d d , C 0 C_0 C 0 , C 1 C_1 C 1 and C 2 C_2 C 2 . Furthermore, if
(4) X ⊆ ⋃ i ∈ I C 1 B i X \subseteq\bigcup_{i\in I}C_1B_i X ⊆ ⋃ i ∈ I C 1 B i ,
and for all all i , j ∈ I i,j \in I i , j ∈ I such that 3 C 1 B i ∩ 3 C 1 B j ≠ ∅ 3C_1B_i \cap3C_1B_j \ne\emptyset 3 C 1 B i ∩ 3 C 1 B j = ∅ we have
C 3 − 1 ≤ r i / r j ≤ C 3 , C_3^{-1} \le r_i/r_j \le C_3, C 3 − 1 ≤ r i / r j ≤ C 3 , then Y Y Y is also lower Ahlfors d d d -regular with diam ( Y ) ∼ C 1 diam ( X ) \operatorname{diam}(Y) \sim_{C_1} \operatorname{diam}(X) diam ( Y ) ∼ C 1 diam ( X ) and constant depending on d d d , C 0 C_0 C 0 , C 1 C_1 C 1 , C 2 C_2 C 2 and C 3 C_3 C 3 .
Proof. Let x ∈ Y x \in Y x ∈ Y and r ∈ ( 0 , 2 diam ( Y ) ) r \in(0,2\operatorname{diam}(Y)) r ∈ ( 0 , 2 diam ( Y )) . We start with the lower regularity estimate. Let i ( x ) ∈ I i(x) \in I i ( x ) ∈ I be such that x ∈ Y i ( x ) ⊆ C 1 B i ( x ) x \in Y_{i(x)} \subseteq C_1B_{i(x)} x ∈ Y i ( x ) ⊆ C 1 B i ( x ) .
If C 1 r i ( x ) ≥ r / ( 100 C 3 ) C_1r_{i(x)} \ge r/(100C_3) C 1 r i ( x ) ≥ r / ( 100 C 3 ) , then the lower regularity estimate in Y Y Y for B ( x , r ) B(x,r) B ( x , r ) (with constant depending on C 1 C_1 C 1 ) follows by considering a ball B ⊆ B ( x , r ) B \subseteq B(x,r) B ⊆ B ( x , r ) centred on Y i ( x ) Y_{i(x)} Y i ( x ) , with radius comparable to r r r , and using the lower regularity of Y i ( x ) Y_{i(x)} Y i ( x ) (which we have by assumption). Suppose instead that
C 1 r i ( x ) < r / ( 100 C 3 ) . C_1r_{i(x)} < r/(100C_3). C 1 r i ( x ) < r / ( 100 C 3 ) . We also note that
diam ( X ) ≥ r / 10. \operatorname{diam}(X) \ge r/10. diam ( X ) ≥ r /10.
Suppose indeed that diam ( X ) < r / 10 \operatorname{diam}(X)<r/10 diam ( X ) < r /10 . Then C 1 r i < r / 10 C_{1}r_{i}<r/10 C 1 r i < r /10 for all i i i , because otherwise X ⊂ 3 C 1 B i X\subset3C_{1}B_{i} X ⊂ 3 C 1 B i by (3) and so 3 C 1 B i ∩ 3 C 1 B i ( x ) ≠ ∅ 3C_{1}B_{i}\cap3C_{1}B_{i}(x)\ne\varnothing 3 C 1 B i ∩ 3 C 1 B i ( x ) = ∅ and so C 1 r i ≤ C 1 C 3 r i ( x ) < C 1 r / 100 C_{1}r_{i}\le C_{1}C_{3}r_{i}(x)<C_{1}r/100 C 1 r i ≤ C 1 C 3 r i ( x ) < C 1 r /100 which is a contradiction.
diam ( Y ) ≤ diam ( X ) + 2 C 1 r i ≤ r / 10 + 2 / 5 r = r / 2 , \operatorname{diam}(Y)\le\operatorname{diam}(X)+2C_{1}r_{i}\le r/10+2/5r=r/2, diam ( Y ) ≤ diam ( X ) + 2 C 1 r i ≤ r /10 + 2/5 r = r /2 ,
which contradicts our choice of r r r . Let x ∗ ∈ X x^{*}\in X x ∗ ∈ X be any point such that d ( x , x ∗ ) ≤ 2 dist ( x , X ) d(x,x^{*})\le2\operatorname{dist}(x,X) d ( x , x ∗ ) ≤ 2 dist ( x , X ) . By assumption (3), we have ∣ x − x ∗ ∣ ≤ 3 C 1 r i ( x ) ≤ r / 50 |x-x^{*}|\le3C_{1}r_{i}(x)\le r/50 ∣ x − x ∗ ∣ ≤ 3 C 1 r i ( x ) ≤ r /50 . Let
I = { i ∈ I : X ∩ B ( x ∗ , r / 4 ) ∩ C 1 B i ≠ ∅ } . \mathcal{I}=\{i\in I:X\cap B(x^{*},r/4)\cap C_{1}B_{i}\ne\varnothing\}. I = { i ∈ I : X ∩ B ( x ∗ , r /4 ) ∩ C 1 B i = ∅ } .
By assumption (4), { C 1 B i } i ∈ I \{C_{1}B_{i}\}_{i\in I} { C 1 B i } i ∈ I form a cover of X ∩ B ( x ∗ , r / 4 ) X\cap B(x^{*},r/4) X ∩ B ( x ∗ , r /4 ) . Thus, since X X X is Ahlfors d d d -regular with diam ( X ) ≥ r / 10 \operatorname{diam}(X)\ge r/10 diam ( X ) ≥ r /10 and each Y i Y_{i} Y i is Ahlfors d d d -regular with diam ( Y i ) ≥ r i \operatorname{diam}(Y_{i})\ge r_{i} diam ( Y i ) ≥ r i , we have
r d ≲ d , C 0 H d ( X ∩ B ( x ∗ , r / 4 ) ) ≲ d , C 0 , C 1 ∑ i ∈ I r i d ≲ d , C 0 ∑ i ∈ I H d ( Y i ) . r^{d}\lesssim_{d,C_{0}}\mathcal{H}^{d}(X\cap B(x^{*},r/4))\lesssim_{d,C_{0},C_{1}}\sum_{i\in I}r_{i}^{d}\lesssim_{d,C_{0}}\sum_{i\in I}\mathcal{H}^{d}(Y_{i}). r d ≲ d , C 0 H d ( X ∩ B ( x ∗ , r /4 )) ≲ d , C 0 , C 1 i ∈ I ∑ r i d ≲ d , C 0 i ∈ I ∑ H d ( Y i ) .
We claim that Y i ⊂ B ( x , r ) Y_{i}\subset B(x,r) Y i ⊂ B ( x , r ) for all i ∈ I i\in I i ∈ I . This would be enough to conclude the lower regularity estimate. Indeed the balls { C 1 B i } \{C_{1}B_{i}\} { C 1 B i } have bounded overlap and the last term in the above inequality is at most C 2 H d ( Y ∩ B ( x , r ) ) C_{2}\mathcal{H}^{d}(Y\cap B(x,r)) C 2 H d ( Y ∩ B ( x , r )) .
Suppose by contradiction that there exists i ∈ I i\in I i ∈ I and y ∈ Y i ∖ B ( x , r ) y\in Y_{i}\setminus B(x,r) y ∈ Y i ∖ B ( x , r ) . Then, since B ( x ∗ , r / 4 ) ∩ C 1 B i ≠ ∅ B(x^{*},r/4)\cap C_{1}B_{i}\ne\varnothing B ( x ∗ , r /4 ) ∩ C 1 B i = ∅ ,
r ≤ ∣ x − y ∣ ≤ ∣ x − x ∗ ∣ + ∣ x ∗ − y ∣ < r / 50 + r / 4 + 2 C 1 r i ≤ r / 2 + 2 C 1 r i . r\le|x-y|\le|x-x^{*}|+|x^{*}-y|<r/50+r/4+2C_{1}r_{i}\le r/2+2C_{1}r_{i}. r ≤ ∣ x − y ∣ ≤ ∣ x − x ∗ ∣ + ∣ x ∗ − y ∣ < r /50 + r /4 + 2 C 1 r i ≤ r /2 + 2 C 1 r i .
Hence r < 4 C 1 r i r<4C_{1}r_{i} r < 4 C 1 r i and thus also ∣ x − x i ∣ < r / 2 + C 1 r i ≤ 3 C 1 r i |x-x_{i}|<r/2+C_{1}r_{i}\le3C_{1}r_{i} ∣ x − x i ∣ < r /2 + C 1 r i ≤ 3 C 1 r i , which gives x ∈ 3 C 1 B i x\in3C_{1}B_{i} x ∈ 3 C 1 B i . Therefore, 3 C 1 B i ( x ) ∩ 3 C 1 B i ≠ ∅ 3C_{1}B_{i}(x)\cap3C_{1}B_{i}\ne\varnothing 3 C 1 B i ( x ) ∩ 3 C 1 B i = ∅ and by assumption r i ≤ C 3 r i ( x ) ≤ r 100 C 1 r_{i}\le C_{3}r_{i}(x)\le\frac{r}{100C_{1}} r i ≤ C 3 r i ( x ) ≤ 100 C 1 r , which is in contradiction with r i > r / ( 4 C 1 ) r_{i}>r/(4C_{1}) r i > r / ( 4 C 1 ) obtained above.
We now prove the upper regularity. Let
J : = { i ∈ I : Y i ∩ B ( x , r ) ≠ ∅ } . \mathcal{J}:=\{i\in I:Y_{i}\cap B(x,r)\ne\varnothing\}. J := { i ∈ I : Y i ∩ B ( x , r ) = ∅ } .
By definition, the set { Y i } i ∈ J \{Y_{i}\}_{i\in\mathcal{J}} { Y i } i ∈ J form a cover of Y ∩ B ( x , r ) Y\cap B(x,r) Y ∩ B ( x , r ) . Thus,
H d ( Y ∩ B ( x , r ) ) ≤ ∑ i ∈ J H d ( Y i ∩ B ( x , r ) ) . \mathcal{H}^{d}(Y\cap B(x,r))\le\sum_{i\in\mathcal{J}}\mathcal{H}^{d}(Y_{i}\cap B(x,r)). H d ( Y ∩ B ( x , r )) ≤ i ∈ J ∑ H d ( Y i ∩ B ( x , r )) .
To estimate the right-hand side, we split J \mathcal{J} J in two sub-families, those whose corresponding ball C 1 B i C_{1}B_{i} C 1 B i is large relative to B ( x , r ) B(x,r) B ( x , r ) and those whose corresponding ball is relatively small. Indeed, let
J l a r g e : = { i ∈ J : C 1 r i ≥ r / 2 } and J s m a l l : = { i ∈ J : C 1 r i < r / 2 } . \mathcal{J}_{\mathrm{large}}:=\{i\in\mathcal{J}:C_{1}r_{i}\ge r/2\}\quad\text{and}\quad\mathcal{J}_{\mathrm{small}}:=\{i\in\mathcal{J}:C_{1}r_{i}<r/2\}. J large := { i ∈ J : C 1 r i ≥ r /2 } and J small := { i ∈ J : C 1 r i < r /2 } .
We start by showing that
∑ i ∈ J l a r g e H d ( Y i ∩ B ( x , r ) ) ≲ C 0 , C 2 r d . \sum_{i\in\mathcal{J}_{\mathrm{large}}}\mathcal{H}^{d}(Y_{i}\cap B(x,r))\lesssim_{C_{0},C_{2}}r^{d}. i ∈ J large ∑ H d ( Y i ∩ B ( x , r )) ≲ C 0 , C 2 r d .
Since each Y i Y_{i} Y i is upper regular with constant C 0 C_{0} C 0 , it suffices to show that
∣ J l a r g e ∣ ≤ C 2 . (4) |\mathcal{J}_{\mathrm{large}}|\le C_{2}. \tag*{(4)} ∣ J large ∣ ≤ C 2 . (4)
This follows from Lemma 2.10 (1). Indeed, let i ∗ ∈ J l a r g e i_{*}\in\mathcal{J}_{\mathrm{large}} i ∗ ∈ J large be the index whose corresponding ball has the largest radius. Since C 1 B i ∩ B ( x , r ) ≠ ∅ C_{1}B_{i}\cap B(x,r)\ne\varnothing C 1 B i ∩ B ( x , r ) = ∅ and C 1 r i > r / 2 C_{1}r_{i}>r/2 C 1 r i > r /2 for all i ∈ J l a r g e i\in\mathcal{J}_{\mathrm{large}} i ∈ J large , it follows that C 1 B i ⊂ 5 C 1 B i ∗ C_{1}B_{i}\subset5C_{1}B_{i_{*}} C 1 B i ⊂ 5 C 1 B i ∗ for all i ∈ J l a r g e i\in\mathcal{J}_{\mathrm{large}} i ∈ J large . The estimate (4) now follows directly from the bounded overlap of { 5 C 1 B i } i ∈ I \{5C_{1}B_{i}\}_{i\in I} { 5 C 1 B i } i ∈ I .
We finish by showing that
∑ i ∈ J s m a l l H d ( Y i ∩ B ( x , r ) ) ≲ r d . \sum_{i\in\mathcal{J}_{\mathrm{small}}} \mathcal{H}^{d}(Y_i\cap B(x,r)) \lesssim r^{d}. i ∈ J small ∑ H d ( Y i ∩ B ( x , r )) ≲ r d .
We further divide J s m a l l 1 = { j ∈ J : diam ( X ) > r i } \mathcal{J}_{\mathrm{small}}^{1}=\{j\in\mathcal{J}:\operatorname{diam}(X)>r_i\} J small 1 = { j ∈ J : diam ( X ) > r i } and J s m a l l 2 = J ∖ J s m a l l 1 \mathcal{J}_{\mathrm{small}}^{2}=\mathcal{J}\setminus\mathcal{J}_{\mathrm{small}}^{1} J small 2 = J ∖ J small 1 . For each i ∈ J s m a l l 1 i\in\mathcal{J}_{\mathrm{small}}^{1} i ∈ J small 1 , let z i ∈ X ∩ 2 C 1 B i z_i\in X\cap2C_1B_i z i ∈ X ∩ 2 C 1 B i (which exists by Lemma 2.10 (3)) and let B i ′ = B ( z i , r i ) B'_i=B(z_i,r_i) B i ′ = B ( z i , r i ) . Note that B i ′ ⊆ 2 C 1 B i B'_i\subseteq2C_1B_i B i ′ ⊆ 2 C 1 B i which, by Lemma 2.10 (1), implies
{ B i ′ } i ∈ J s m a l l \{B'_i\}_{i\in\mathcal{J}_{\mathrm{small}}} { B i ′ } i ∈ J small
has overlap bounded by C 2 C_2 C 2 .
Note also, since C 1 B i ∩ B ( x , r ) ≠ ∅ C_1B_i\cap B(x,r)\ne\varnothing C 1 B i ∩ B ( x , r ) = ∅ and C 1 r i < r / 2 C_1r_i<r/2 C 1 r i < r /2 , we have B i ′ ⊆ 2 C 1 B i ⊆ B ( x , 3 r ) B'_i\subseteq2C_1B_i\subseteq B(x,3r) B i ′ ⊆ 2 C 1 B i ⊆ B ( x , 3 r ) . Moreover, since B i ′ ∩ X ≠ ∅ B'_i\cap X\ne\varnothing B i ′ ∩ X = ∅ we have B i ′ ⊆ B ( x ∗ , 6 r ) B'_i\subseteq B(x^*,6r) B i ′ ⊆ B ( x ∗ , 6 r ) for some x ∗ ∈ X x^*\in X x ∗ ∈ X . Putting this together, and using the Ahlfors regularity of X X X and { Y i } \{Y_i\} { Y i } , diam ( X ) > r i \operatorname{diam}(X)>r_i diam ( X ) > r i , we have
∑ i ∈ J s m a l l H d ( Y i ∩ B ( x , r ) ) ≲ d , C 0 , C 1 ∑ i ∈ J s m a l l H d ( X ∩ B i ′ ) \sum_{i\in\mathcal{J}_{\mathrm{small}}}\mathcal{H}^{d}(Y_i\cap B(x,r))
\lesssim_{d,C_0,C_1}
\sum_{i\in\mathcal{J}_{\mathrm{small}}}\mathcal{H}^{d}(X\cap B'_i) i ∈ J small ∑ H d ( Y i ∩ B ( x , r )) ≲ d , C 0 , C 1 i ∈ J small ∑ H d ( X ∩ B i ′ )
≲ d , C 0 , C 1 C 2 H d ( X ∩ B ( x ∗ , 4 r ) ) ≲ d , C 0 , C 1 , C 2 r d . \lesssim_{d,C_0,C_1} C_2\mathcal{H}^{d}(X\cap B(x^*,4r))
\lesssim_{d,C_0,C_1,C_2}r^d. ≲ d , C 0 , C 1 C 2 H d ( X ∩ B ( x ∗ , 4 r )) ≲ d , C 0 , C 1 , C 2 r d .
To conclude the proof it suffices to show ∣ J s m a l l 2 ∣ ≤ C 2 |\mathcal{J}_{\mathrm{small}}^{2}|\le C_2 ∣ J small 2 ∣ ≤ C 2 and then argue as for J l a r g e \mathcal{J}_{\mathrm{large}} J large . Since diam ( X ) ≤ r i \operatorname{diam}(X)\le r_i diam ( X ) ≤ r i and d ( X , Y i ) ≤ C 1 r i d(X,Y_i)\le C_1r_i d ( X , Y i ) ≤ C 1 r i , it holds X ⊆ 2 C 1 B i X\subseteq2C_1B_i X ⊆ 2 C 1 B i for all i ∈ J s m a l l 2 i\in\mathcal{J}_{\mathrm{small}}^{2} i ∈ J small 2 . From this ∣ J s m a l l 2 ∣ ≤ C 2 |\mathcal{J}_{\mathrm{small}}^{2}|\le C_2 ∣ J small 2 ∣ ≤ C 2 follows from the bounded overlap assumption.
Finally, we prove the diameter estimate diam ( Y ) ∼ C 1 diam ( X ) \operatorname{diam}(Y)\sim_{C_1}\operatorname{diam}(X) diam ( Y ) ∼ C 1 diam ( X ) under the assumption (4). The estimate diam ( Y ) ≲ C 1 diam ( X ) \operatorname{diam}(Y)\lesssim_{C_1}\operatorname{diam}(X) diam ( Y ) ≲ C 1 diam ( X ) follows from the triangle inequality, using the facts that diam ( Y i ) ≤ 2 C 1 r i \operatorname{diam}(Y_i)\le2C_1r_i diam ( Y i ) ≤ 2 C 1 r i , dist ( x j , X ) ≤ C 1 r i \operatorname{dist}(x_j,X)\le C_1r_i dist ( x j , X ) ≤ C 1 r i and r i ≤ C 1 diam ( X ) r_i\le C_1\operatorname{diam}(X) r i ≤ C 1 diam ( X ) for all i ∈ I i\in I i ∈ I . For the estimate diam ( X ) ≲ C 1 diam ( Y ) \operatorname{diam}(X)\lesssim_{C_1}\operatorname{diam}(Y) diam ( X ) ≲ C 1 diam ( Y ) , consider two point x , x ′ ∈ X x,x'\in X x , x ′ ∈ X such that d ( x , x ′ ) ≥ diam ( X ) / 2 d(x,x')\ge\operatorname{diam}(X)/2 d ( x , x ′ ) ≥ diam ( X ) /2 . By assumption, there are indices i , j ∈ I i,j\in I i , j ∈ I such that x ∈ C 1 B i x\in C_1B_i x ∈ C 1 B i and x ′ ∈ C 1 B j x'\in C_1B_j x ′ ∈ C 1 B j . If one of r i r_i r i or r j r_j r j is larger than diam ( X ) / ( 8 C 1 ) \operatorname{diam}(X)/(8C_1) diam ( X ) / ( 8 C 1 ) , then the diameter estimate follows directly from the fact that diam ( Y k ) ≥ r k \operatorname{diam}(Y_k)\ge r_k diam ( Y k ) ≥ r k . In the other case that r i , r j ≤ diam ( X ) / ( 8 C 1 ) r_i,r_j\le\operatorname{diam}(X)/(8C_1) r i , r j ≤ diam ( X ) / ( 8 C 1 ) , we have
d ( x i , x j ) ≥ d ( x , x ′ ) − d ( x i , x ) − d ( x j , x ′ ) ≥ diam ( X ) / 2 − C 1 r i − C 1 r j ≥ diam ( X ) / 4 , \begin{aligned}
d(x_i,x_j)\ge d(x,x')-d(x_i,x)-d(x_j,x') \\
&\ge\operatorname{diam}(X)/2-C_1r_i-C_1r_j\ge\operatorname{diam}(X)/4,
\end{aligned} d ( x i , x j ) ≥ d ( x , x ′ ) − d ( x i , x ) − d ( x j , x ′ ) ≥ diam ( X ) /2 − C 1 r i − C 1 r j ≥ diam ( X ) /4 ,
which finishes the estimate. □ \square □
Poincaré inequalities
There are a number of key results we will use to check a Poincaré inequality in the set Z Z Z that we construct for Theorem 1.3 . The first is a well-known characterisation of Poincaré inequalities via maximal function estimates. To state a more precise result, we will introduce some notation.
Notation 3.1. Let C , λ ≥ 1 C,\lambda\ge1 C , λ ≥ 1 , ( X , d , μ ) (X,d,\mu) ( X , d , μ ) be a metric measure space, B B B a ball in X X X and u : X → R u:X\to\mathbb{R} u : X → R Lipschitz.
(1) We say ( u , B ) ∈ PI ( C , λ ) (u,B)\in\operatorname{PI}(C,\lambda) ( u , B ) ∈ PI ( C , λ ) if the estimate
\fint_{B'}|u-u_B|\,d\mu\le C\fint_{\lambda B'}\operatorname{Lip}u\,d\mu
holds for all B ′ ⊆ B B'\subseteq B B ′ ⊆ B .
(2) We say ( u , B ) ∈ MPI ( C , λ ) (u,B) \in\operatorname{MPI}(C,\lambda) ( u , B ) ∈ MPI ( C , λ ) if the estimate
∣ u ( x ) − u ( y ) ∣ ≤ C d ( x , y ) ( M λ d ( x , y ) [ Lip u ] p ( x ) + M λ d ( x , y ) [ Lip u ] p ( y ) ) 1 p |u(x)-u(y)| \leq Cd(x,y)\left(M_{\lambda d(x,y)}[\operatorname{Lip}u]^p(x)+M_{\lambda d(x,y)}[\operatorname{Lip}u]^p(y)\right)^{\frac{1}{p}} ∣ u ( x ) − u ( y ) ∣ ≤ C d ( x , y ) ( M λ d ( x , y ) [ Lip u ] p ( x ) + M λ d ( x , y ) [ Lip u ] p ( y ) ) p 1
holds for all x , y ∈ B x,y \in B x , y ∈ B .
Above we used the maximal function notation M_{R}g(x):=\sup_{r<R}\fint_{B(x,r)}|g|\,d\mu . The following is proved in [26 ] , Lemma 5.15. Although not stated as such in [26 ] , it can be deduced by examining the proof there.
Lemma 3.2. Let C , λ > 1 C,\lambda>1 C , λ > 1 , ( X , d , μ ) (X,d,\mu) ( X , d , μ ) be a locally compact doubling metric space with doubling constant C d b C_{\mathrm{db}} C db , B B B a ball in X X X and u : X → R u:X\to\mathbb{R} u : X → R . Then,
( u , B ) ∈ PI ( C , λ ) ⟹ ( u , 1 10 B ) ∈ MPI ( C ′ , 10 λ ) . (u,B)\in\operatorname{PI}(C,\lambda)\Longrightarrow\left(u,\frac{1}{10}B\right)\in\operatorname{MPI}(C',10\lambda). ( u , B ) ∈ PI ( C , λ ) ⟹ ( u , 10 1 B ) ∈ MPI ( C ′ , 10 λ ) . and
( u , B ) ∈ MPI ( C , λ ) ⟹ ( u , B ) ∈ PI ( C ′ , 2 λ ) . (u,B)\in\operatorname{MPI}(C,\lambda)\Longrightarrow(u,B)\in\operatorname{PI}(C',2\lambda). ( u , B ) ∈ MPI ( C , λ ) ⟹ ( u , B ) ∈ PI ( C ′ , 2 λ ) . In both cases C ′ C' C ′ depends quantitatively on C C C and C d b C_{\mathrm{db}} C db . In particular, X X X admits a weak ( 1 , p ) (1,p) ( 1 , p ) -Poincaré inequality if and only if there is a constant C , λ ≥ 1 C,\lambda\geq1 C , λ ≥ 1 such that
∣ u ( x ) − u ( y ) ∣ ≤ C d ( x , y ) ( M λ d ( x , y ) [ Lip u ] p ( x ) + M λ d ( x , y ) [ Lip u ] p ( y ) ) 1 p (5) |u(x)-u(y)|\leq Cd(x,y)\left(M_{\lambda d(x,y)}[\operatorname{Lip}u]^p(x)+M_{\lambda d(x,y)}[\operatorname{Lip}u]^p(y)\right)^{\frac{1}{p}} \tag*{(5)} ∣ u ( x ) − u ( y ) ∣ ≤ C d ( x , y ) ( M λ d ( x , y ) [ Lip u ] p ( x ) + M λ d ( x , y ) [ Lip u ] p ( y ) ) p 1 (5) for all x , y ∈ X x,y\in X x , y ∈ X and u : X → R u:X\to\mathbb{R} u : X → R Lipschitz.
As a consequence of Lemma 3.2 one can show that the pointwise estimate from the MPI ( C , λ ) \operatorname{MPI}(C,\lambda) MPI ( C , λ ) implies control of averages of Lipschitz functions on nearby balls. The proof (with slightly different constants) can be found in [25 ] . Below we denote M_{R',R}g(x):=\sup_{R'<r<R}\fint_{B(x,r)}|g|\,d\mu .
Lemma 3.3. Let C , λ > 1 C,\lambda>1 C , λ > 1 , ( X , d , μ ) (X,d,\mu) ( X , d , μ ) be a locally compact doubling metric space with doubling constant C d b C_{\mathrm{db}} C db , B 0 B_0 B 0 a ball in X X X with radius r 0 r_0 r 0 and u : X → R u:X\to\mathbb{R} u : X → R Lipschitz. If ( u , B 0 ) ∈ MPI ( C , λ ) (u,B_0)\in\operatorname{MPI}(C,\lambda) ( u , B 0 ) ∈ MPI ( C , λ ) then
∣ u B − u B ′ ∣ ≲ C ′ , C d b r B 0 ( M r B , 4 λ r B 0 Lip ( x B ) + M r B ′ , 4 λ r B 0 Lip ( x B ′ ) ) |u_B-u_{B'}|\lesssim_{C',C_{\mathrm{db}}}r_{B_0}\left(M_{r_B,4\lambda r_{B_0}}\operatorname{Lip}(x_B)+M_{r_{B'},4\lambda r_{B_0}}\operatorname{Lip}(x_{B'})\right) ∣ u B − u B ′ ∣ ≲ C ′ , C db r B 0 ( M r B , 4 λ r B 0 Lip ( x B ) + M r B ′ , 4 λ r B 0 Lip ( x B ′ ) ) for all balls B , B ′ ⊆ 1 4 B 0 B,B'\subseteq\frac{1}{4}B_0 B , B ′ ⊆ 4 1 B 0 .
The construction of the set Z Z Z from Theorem 1.3 also relies on gluing PI spaces together. The following tells us when a gluing preserves a Poincaré inequality and is proved in [26 ] .
Lemma 3.4. Suppose E , F ⊆ R n E,F\subseteq\mathbb{R}^n E , F ⊆ R n are Ahlfors d d d -regular, each admitting a weak ( 1 , 1 ) (1,1) ( 1 , 1 ) -Poincaré inequality. Suppose the set A = E ∩ F A=E\cap F A = E ∩ F satisfies
H d ( A ∩ B ) ≥ C − 1 r B d \mathcal{H}^{d}(A\cap B)\geq C^{-1}r_B^d H d ( A ∩ B ) ≥ C − 1 r B d for all balls B B B centred on A A A with radius at most min { diam ( E ) , diam ( F ) } \min\{\operatorname{diam}(E),\operatorname{diam}(F)\} min { diam ( E ) , diam ( F )} and that, for any x ∈ E x\in E x ∈ E and y ∈ F y\in F y ∈ F , we have
inf a ∈ A ∣ x − a ∣ + ∣ a − y ∣ ≲ ∣ x − y ∣ . (6) \inf_{a\in A}|x-a|+|a-y|\lesssim|x-y|. \tag*{(6)} a ∈ A inf ∣ x − a ∣ + ∣ a − y ∣ ≲ ∣ x − y ∣. (6) Then, E ∪ F E\cup F E ∪ F admits a weak ( 1 , 1 ) (1,1) ( 1 , 1 ) -Poincaré inequality.
Lemmas Lemmas 3.5 and 3.6 below provide us with some convenient ways to check the condition (6) .
Lemma 3.5. Let ( Z , d ) (Z,d) ( Z , d ) be a metric space and X , Y ⊆ Z X,Y \subseteq Z X , Y ⊆ Z such that A : = X ∩ Y ≠ ∅ A := X \cap Y \ne\varnothing A := X ∩ Y = ∅ . If x ∈ X x \in X x ∈ X and y ∈ Y y \in Y y ∈ Y are such that
d ( x , y ) ≳ min { diam ( X ) , diam ( Y ) } d(x,y) \gtrsim\min\{\operatorname{diam}(X),\operatorname{diam}(Y)\} d ( x , y ) ≳ min { diam ( X ) , diam ( Y )} then (6) holds.
Proof. Suppose, without loss of generality, that diam ( X ) ≤ diam ( Y ) \operatorname{diam}(X) \le\operatorname{diam}(Y) diam ( X ) ≤ diam ( Y ) . Choose an arbitrary point a ∈ A a \in A a ∈ A . Since A ⊆ X A \subseteq X A ⊆ X , we have d ( x , a ) ≤ diam ( X ) d(x,a) \le\operatorname{diam}(X) d ( x , a ) ≤ diam ( X ) . Then, by the triangle inequality,
d ( x , a ) + d ( a , y ) ≤ 2 d ( x , a ) + d ( x , y ) ≲ d ( x , y ) , d(x,a) + d(a,y) \le2d(x,a) + d(x,y) \lesssim d(x,y), d ( x , a ) + d ( a , y ) ≤ 2 d ( x , a ) + d ( x , y ) ≲ d ( x , y ) , as required. □
Lemma 3.6. For each c 1 ∈ ( 0 , 1 ) c_1 \in(0,1) c 1 ∈ ( 0 , 1 ) there is τ 1 ∈ ( 0 , 1 ) \tau_1 \in(0,1) τ 1 ∈ ( 0 , 1 ) such that the following holds. Let τ ∈ ( 0 , τ 1 ) \tau\in(0,\tau_1) τ ∈ ( 0 , τ 1 ) , g : R n → R g:\mathbb{R}^n \to\mathbb{R} g : R n → R be τ \tau τ -Lipschitz and Γ \Gamma Γ be its graph. Let x , y ∈ R n + 1 x,y \in\mathbb{R}^{n+1} x , y ∈ R n + 1 such that π 1 ( x ) ≥ g ( π n ( x ) ) \pi_1(x) \ge g(\pi_n(x)) π 1 ( x ) ≥ g ( π n ( x )) and π 1 ( y ) ≤ g ( π n ( y ) ) \pi_1(y) \le g(\pi_n(y)) π 1 ( y ) ≤ g ( π n ( y )) . If a ∈ Γ a \in\Gamma a ∈ Γ is such that either
c 1 ∣ a − x ∣ + g ( π n ( a ) ) − g ( π n ( x ) ) ≤ ∣ π 1 ( a − x ) ∣ , (7) c_1|a-x| + g(\pi_n(a)) - g(\pi_n(x)) \le|\pi_1(a-x)|, \tag*{(7)} c 1 ∣ a − x ∣ + g ( π n ( a )) − g ( π n ( x )) ≤ ∣ π 1 ( a − x ) ∣ , (7) or
c 1 ∣ a − y ∣ + g ( π n ( a ) ) − g ( π n ( y ) ) ≤ ∣ π 1 ( a − y ) ∣ , c_1|a-y| + g(\pi_n(a)) - g(\pi_n(y)) \le|\pi_1(a-y)|, c 1 ∣ a − y ∣ + g ( π n ( a )) − g ( π n ( y )) ≤ ∣ π 1 ( a − y ) ∣ , then (6) holds with this choice of a a a .
Proof. We will prove (6) in the case that (7) holds. The other case is similar, and we will not include the details. We begin by showing how (7) implies the more natural condition
c 1 ∣ a − x ∣ ≤ 2 ∣ π 1 ( a − x ) ∣ . (8) c_1|a-x| \le2|\pi_1(a-x)|. \tag*{(8)} c 1 ∣ a − x ∣ ≤ 2∣ π 1 ( a − x ) ∣. (8) Indeed, taking g ( π n ( a ) ) − g ( π n ( x ) ) g(\pi_n(a)) - g(\pi_n(x)) g ( π n ( a )) − g ( π n ( x )) to the right-hand side of (7) and using that g g g is τ \tau τ -Lipschitz and π n \pi_n π n is 1 1 1 -Lipschitz, we get
c 1 ∣ a − x ∣ ≤ ∣ π 1 ( a − x ) ∣ + τ ∣ a − x ∣ . c_1|a-x| \le|\pi_1(a-x)| + \tau|a-x|. c 1 ∣ a − x ∣ ≤ ∣ π 1 ( a − x ) ∣ + τ ∣ a − x ∣. If τ 1 = τ 1 ( c 1 ) \tau_1 = \tau_1(c_1) τ 1 = τ 1 ( c 1 ) is sufficiently small, this implies (8) . We now use (8) to prove (6) . By the triangle inequality, we have ∣ x − a ∣ + ∣ a − y ∣ ≤ ∣ x − y ∣ + 2 ∣ a − x ∣ |x-a| + |a-y| \le|x-y| + 2|a-x| ∣ x − a ∣ + ∣ a − y ∣ ≤ ∣ x − y ∣ + 2∣ a − x ∣ . Thus, it suffices to show that
∣ a − x ∣ ≲ ∣ x − y ∣ . (9) |a-x| \lesssim|x-y|. \tag*{(9)} ∣ a − x ∣ ≲ ∣ x − y ∣. (9) Assume to begin with that
π n ( y ) ∉ B : = B n ( π n ( x ) , ∣ π 1 ( a − x ) ∣ ) ⊆ R n . \pi_n(y) \notin B := B_n(\pi_n(x),|\pi_1(a-x)|) \subseteq\mathbb{R}^n. π n ( y ) ∈ / B := B n ( π n ( x ) , ∣ π 1 ( a − x ) ∣ ) ⊆ R n . In this case, we have by (8) that
∣ x − y ∣ ≥ ∣ π n ( x − y ) ∣ ≥ ∣ π 1 ( a − x ) ∣ ≳ ∣ a − x ∣ , |x-y| \ge|\pi_n(x-y)| \ge|\pi_1(a-x)| \gtrsim|a-x|, ∣ x − y ∣ ≥ ∣ π n ( x − y ) ∣ ≥ ∣ π 1 ( a − x ) ∣ ≳ ∣ a − x ∣ , which proves (9) . Suppose instead that
π n ( y ) ∈ B . \pi_n(y) \in B. π n ( y ) ∈ B . Recalling (8) , we also have π n ( a ) ∈ 2 C 1 B \pi_n(a) \in2C_1B π n ( a ) ∈ 2 C 1 B , where C 1 : = c 1 − 1 C_1 := c_1^{-1} C 1 := c 1 − 1 . In particular, ∣ π n ( y ) − π n ( a ) ∣ ≤ 4 C 1 ∣ π 1 ( a − x ) ∣ |\pi_n(y)-\pi_n(a)| \le4C_1|\pi_1(a-x)| ∣ π n ( y ) − π n ( a ) ∣ ≤ 4 C 1 ∣ π 1 ( a − x ) ∣ . Since g g g is τ \tau τ -Lipschitz, setting y ′ = ( π n ( y ) , g ( π n ( y ) ) ) y'=(\pi_n(y),g(\pi_n(y))) y ′ = ( π n ( y ) , g ( π n ( y ))) we
have ∣ x − y ′ ∣ ≤ 1 + τ 2 ∣ x − y ∣ |x-y'| \le\sqrt{1+\tau^2}|x-y| ∣ x − y ′ ∣ ≤ 1 + τ 2 ∣ x − y ∣ . Using these facts, and the fact that g g g is τ \tau τ -Lipschitz, if τ 1 = τ 1 ( c 1 ) \tau_1=\tau_1(c_1) τ 1 = τ 1 ( c 1 ) is small enough, we have
1 + τ 2 ∣ x − y ∣ ≥ ∣ x − y ′ ∣ ≥ ∣ π 1 ( x − y ′ ) ∣ ≥ ∣ π 1 ( a − x ) ∣ − ∣ π 1 ( y ′ − a ) ∣ = ∣ π 1 ( a − x ) ∣ − ∣ g ( π n ( y ) ) − g ( π n ( a ) ) ∣ ≥ ∣ π 1 ( a − x ) ∣ − 4 C 1 τ ∣ π 1 ( a − x ) ∣ ≥ ( 1 / 2 ) ∣ π 1 ( a − x ) ∣ ≳ ∣ a − x ∣ , \sqrt{1+\tau^2}|x-y| \ge|x-y'| \ge|\pi_1(x-y')| \ge|\pi_1(a-x)|-|\pi_1(y'-a)|
=|\pi_1(a-x)|-|g(\pi_n(y))-g(\pi_n(a))|
\ge|\pi_1(a-x)|-4C_1\tau|\pi_1(a-x)| \ge(1/2)|\pi_1(a-x)|\gtrsim|a-x|, 1 + τ 2 ∣ x − y ∣ ≥ ∣ x − y ′ ∣ ≥ ∣ π 1 ( x − y ′ ) ∣ ≥ ∣ π 1 ( a − x ) ∣ − ∣ π 1 ( y ′ − a ) ∣ = ∣ π 1 ( a − x ) ∣ − ∣ g ( π n ( y )) − g ( π n ( a )) ∣ ≥ ∣ π 1 ( a − x ) ∣ − 4 C 1 τ ∣ π 1 ( a − x ) ∣ ≥ ( 1/2 ) ∣ π 1 ( a − x ) ∣ ≳ ∣ a − x ∣ ,
which gives the required estimate provided τ 1 \tau_1 τ 1 is small enough. □
Poincaré inequalities via Harnack chains
In this and the next subsection we collected two results which give us ways of transferring local Poincaré inequalities to global inequalities. The result below states, roughly speaking, that if every two points is connected by a Harnack chain of balls, where consecutive balls lie in a common PI space, then the whole space has PI.
Lemma 3.7. Let A > 1 A>1 A > 1 , ( X , d , μ ) (X,d,\mu) ( X , d , μ ) be a metric measure space where μ \mu μ is Ahlfors d d d -regular with constant C d b C_{\mathrm{db}} C db , u : X → R u:X\to\mathbb{R} u : X → R be Lipschitz, and x , y ∈ X x,y\in X x , y ∈ X . Suppose there exists a sequence of balls { B i } i = 1 N \{B_i\}_{i=1}^{N} { B i } i = 1 N in X X X and subsets { X i } i = 1 N \{X_i\}_{i=1}^{N} { X i } i = 1 N of X X X such that the following conditions holds.
(1) Each X i X_i X i is Ahlfors d d d -regular with constant C d b C_{\mathrm{db}} C db and supports a ( 1 , p ) (1,p) ( 1 , p ) -Poincaré inequality with constants ( C P I , λ P I ) (C_{\mathrm{PI}},\lambda_{\mathrm{PI}}) ( C PI , λ PI ) .
(2) For each i ∈ { 1 , … , N − 1 } i\in\{1,\ldots,N-1\} i ∈ { 1 , … , N − 1 } , the space X i X_i X i completely contains the balls B i B_i B i and B i + 1 B_{i+1} B i + 1 i.e.,
B i ⊆ X i and B i + 1 ⊆ X i . B_i\subseteq X_i \qquad\text{and}\qquad B_{i+1}\subseteq X_i. B i ⊆ X i and B i + 1 ⊆ X i . (3) The end spaces contain the end points and balls i.e.,
{ x } ∪ B 1 ⊆ X 1 and { y } ∪ B N ⊆ X N . \{x\}\cup B_1\subseteq X_1 \qquad\text{and}\qquad\{y\}\cup B_N\subseteq X_N. { x } ∪ B 1 ⊆ X 1 and { y } ∪ B N ⊆ X N . (4) If z i z_i z i and r i r_i r i denote the centre and radius, respectively, of the ball B i B_i B i , then there exists K ∈ { 1 , … , N } K\in\{1,\ldots,N\} K ∈ { 1 , … , N } such that
x ∈ B ( z i , A r i ) for all i ∈ { 1 , … , K } ; x\in B(z_i,Ar_i)\quad\text{for all }i\in\{1,\ldots,K\}; x ∈ B ( z i , A r i ) for all i ∈ { 1 , … , K } ; y ∈ B ( z i , A r i ) for all i ∈ { K , … , N } . y\in B(z_i,Ar_i)\quad\text{for all }i\in\{K,\ldots,N\}. y ∈ B ( z i , A r i ) for all i ∈ { K , … , N } . (5) We have
∑ i = 1 N r i ≤ A d ( x , y ) . \sum_{i=1}^{N}r_i\leq Ad(x,y). i = 1 ∑ N r i ≤ A d ( x , y ) . Then, there is C P I ′ = C P I ′ ( A , C d b , C P I , d , p ) C'_{\mathrm{PI}}=C'_{\mathrm{PI}}(A,C_{\mathrm{db}},C_{\mathrm{PI}},d,p) C PI ′ = C PI ′ ( A , C db , C PI , d , p ) and λ P I ′ = λ P I ′ ( A , λ P I ) \lambda'_{\mathrm{PI}}=\lambda'_{\mathrm{PI}}(A,\lambda_{\mathrm{PI}}) λ PI ′ = λ PI ′ ( A , λ PI ) such that
∣ u ( x ) − u ( y ) ∣ ≤ C P I ′ d ( x , y ) ( M λ P I ′ d ( x , y ) [ Lip u ] p ( x ) + M λ P I ′ d ( x , y ) [ Lip u ] p ( y ) ) 1 p . (10) |u(x)-u(y)|\leq C'_{\mathrm{PI}}d(x,y)\left(M_{\lambda'_{\mathrm{PI}}d(x,y)}[\operatorname{Lip}u]^p(x)+M_{\lambda'_{\mathrm{PI}}d(x,y)}[\operatorname{Lip}u]^p(y)\right)^{\frac{1}{p}}. \tag*{(10)} ∣ u ( x ) − u ( y ) ∣ ≤ C PI ′ d ( x , y ) ( M λ PI ′ d ( x , y ) [ Lip u ] p ( x ) + M λ PI ′ d ( x , y ) [ Lip u ] p ( y ) ) p 1 . (10) Proof. We begin by showing that
∣ u B i − u B i + 1 ∣ ≲ A , C d b , C P I max { r i , r i + 1 } ( M λ P I ′ d ( x , y ) [ Lip u ] p ( x ) ) 1 p ; (11) |u_{B_i}-u_{B_{i+1}}|\lesssim_{A,C_{\mathrm{db}},C_{\mathrm{PI}}}\max\{r_i,r_{i+1}\}\left(M_{\lambda'_{\mathrm{PI}}d(x,y)}[\operatorname{Lip}u]^p(x)\right)^{\frac{1}{p}}; \tag*{(11)} ∣ u B i − u B i + 1 ∣ ≲ A , C db , C PI max { r i , r i + 1 } ( M λ PI ′ d ( x , y ) [ Lip u ] p ( x ) ) p 1 ; (11) ∣ u B j − u B j + 1 ∣ ≲ A , C d b , C P I max { r i , r i + 1 } ( M λ P I ′ d ( x , y ) [ Lip u ] p ( y ) ) 1 p , (12) |u_{B_j}-u_{B_{j+1}}|\lesssim_{A,C_{\mathrm{db}},C_{\mathrm{PI}}}\max\{r_i,r_{i+1}\}\left(M_{\lambda'_{\mathrm{PI}}d(x,y)}[\operatorname{Lip}u]^p(y)\right)^{\frac{1}{p}}, \tag*{(12)} ∣ u B j − u B j + 1 ∣ ≲ A , C db , C PI max { r i , r i + 1 } ( M λ PI ′ d ( x , y ) [ Lip u ] p ( y ) ) p 1 , (12) for each i ∈ { 1 , … , K − 1 } i\in\{1,\ldots,K-1\} i ∈ { 1 , … , K − 1 } and j ∈ { K , … , N − 1 } j\in\{K,\ldots,N-1\} j ∈ { K , … , N − 1 } , where
λ P I ′ : = 960 λ P I A 3 . \lambda'_{\mathrm{PI}}:=960\lambda_{\mathrm{PI}}A^3. λ PI ′ := 960 λ PI A 3 .
We shall prove (11) only (the proof for (12) is the same, up to notational changes). Fix some i ∈ { 1 , … , K − 1 } i \in\{1,\ldots,K-1\} i ∈ { 1 , … , K − 1 } and assume, without loss of generality, that r i + 1 ≥ r i r_{i+1} \ge r_i r i + 1 ≥ r i . Under this assumption, we have by (4) that B i ⊆ 3 A B i + 1 B_i \subseteq3AB_{i+1} B i ⊆ 3 A B i + 1 . Hence B i , B i + 1 ⊂ 1 4 12 A B i + 1 B_i,B_{i+1} \subset\frac{1}{4}12AB_{i+1} B i , B i + 1 ⊂ 4 1 12 A B i + 1 . Since X i X_i X i contains B i ∪ B i + 1 B_i \cup B_{i+1} B i ∪ B i + 1 and supports a ( 1 , p ) (1,p) ( 1 , p ) -Poincaré inequality, it follows from Lemma 3.3 and Lemma 3.2 that
∣ u B i − u B i + 1 ∣ ≲ C d b , C P I A r i + 1 ( M r i , 480 λ P I A r i + 1 X i [ Lipu ] p ( z i ) ) 1 p + A r i + 1 ( M r i + 1 , 480 λ P I A r i + 1 X i [ Lipu ] p ( z i + 1 ) ) 1 p (13) \lvert u_{B_i}-u_{B_{i+1}}\rvert\lesssim_{C_{d_b},C_{\mathrm{PI}}} Ar_{i+1}\left(M^{X_i}_{r_i,480\lambda_{\mathrm{PI}}Ar_{i+1}}[\operatorname{Lipu}]^p(z_i)\right)^{\frac{1}{p}}+Ar_{i+1}\left(M^{X_i}_{r_{i+1},480\lambda_{\mathrm{PI}}Ar_{i+1}}[\operatorname{Lipu}]^p(z_{i+1})\right)^{\frac{1}{p}} \tag*{(13)} ∣ u B i − u B i + 1 ∣ ≲ C d b , C PI A r i + 1 ( M r i , 480 λ PI A r i + 1 X i [ Lipu ] p ( z i ) ) p 1 + A r i + 1 ( M r i + 1 , 480 λ PI A r i + 1 X i [ Lipu ] p ( z i + 1 ) ) p 1 (13)
(recall that z i z_i z i is the centre of B i B_i B i ), where M X i M^{X_i} M X i denotes the maximal function with distance and measure restricted to X i X_i X i . For r ∈ [ r i , 480 λ P I A r i + 1 ] r \in[r_i,480\lambda_{\mathrm{PI}}Ar_{i+1}] r ∈ [ r i , 480 λ PI A r i + 1 ] , it follows from (4) and (26) that
B ( z i , r ) ⊆ B ( x , ( A + 1 ) r ) ⊆ B ( x , 480 ( A + 1 ) λ P I A r i + 1 ) ⊆ B ( x , λ P I ′ d ( x , y ) ) . B(z_i,r) \subseteq B(x,(A+1)r) \subseteq B(x,480(A+1)\lambda_{\mathrm{PI}}Ar_{i+1}) \subseteq B(x,\lambda'_{\mathrm{PI}}d(x,y)). B ( z i , r ) ⊆ B ( x , ( A + 1 ) r ) ⊆ B ( x , 480 ( A + 1 ) λ PI A r i + 1 ) ⊆ B ( x , λ PI ′ d ( x , y )) .
Thus, since B ( z i , r i ) ⊂ X i B(z_i,r_i) \subset X_i B ( z i , r i ) ⊂ X i and by the Ahlfors d d d -regularity of X i X_i X i and X X X ,
\fint_{B(z_i,r)\cap X_i}[\operatorname{Lipu}]^p\,d\mu\lesssim_{A,C_{d_b}} \fint_{B(x,(A+1)r)}[\operatorname{Lipu}]^p\,d\mu\le M_{\lambda'_{\mathrm{PI}}d(x,y)}[\operatorname{Lipu}]^p(x). \tag*{(14)}
Similarly, for r ∈ [ r i + 1 , 480 λ P I A r i + 1 ] r \in[r_{i+1},480\lambda_{\mathrm{PI}}Ar_{i+1}] r ∈ [ r i + 1 , 480 λ PI A r i + 1 ] ,
\fint_{B(z_{i+1},r)\cap X_i}[\operatorname{Lipu}]^p\,d\mu\lesssim_{A,C_{d_b}} M_{\lambda'_{\mathrm{PI}}d(x,y)}[\operatorname{Lipu}]^p(x). \tag*{(15)}
Inequality (11) now follows from (13) , (14) and (15) .
Since x ∈ B ( x 1 , A r 1 ) ⊆ B ( x 1 , A 2 d ( x , y ) ) x \in B(x_1,Ar_1) \subseteq B(x_1,A^2d(x,y)) x ∈ B ( x 1 , A r 1 ) ⊆ B ( x 1 , A 2 d ( x , y )) and y ∈ B ( x N , A r N ) ⊆ B ( x N , A 2 d ( x , y ) ) y \in B(x_N,Ar_N) \subseteq B(x_N,A^2d(x,y)) y ∈ B ( x N , A r N ) ⊆ B ( x N , A 2 d ( x , y )) , it follows from Lemma 3.3 that
∣ u ( x ) − u B 1 ∣ ≲ A , C d b , C P I d ( x , y ) ( M 4 λ P I A 2 d ( x , y ) [ Lipu ] p ( x ) ) 1 p ; (16) \lvert u(x)-u_{B_1}\rvert\lesssim_{A,C_{d_b},C_{\mathrm{PI}}}d(x,y)\left(M_{4\lambda_{\mathrm{PI}}A^2d(x,y)}[\operatorname{Lipu}]^p(x)\right)^{\frac{1}{p}}; \tag*{(16)} ∣ u ( x ) − u B 1 ∣ ≲ A , C d b , C PI d ( x , y ) ( M 4 λ PI A 2 d ( x , y ) [ Lipu ] p ( x ) ) p 1 ; (16)
∣ u ( y ) − u B N ∣ ≲ A , C d b , C P I d ( x , y ) ( M 4 λ P I A 2 d ( x , y ) [ Lipu ] p ( y ) ) 1 p . (17) \lvert u(y)-u_{B_N}\rvert\lesssim_{A,C_{d_b},C_{\mathrm{PI}}}d(x,y)\left(M_{4\lambda_{\mathrm{PI}}A^2d(x,y)}[\operatorname{Lipu}]^p(y)\right)^{\frac{1}{p}}. \tag*{(17)} ∣ u ( y ) − u B N ∣ ≲ A , C d b , C PI d ( x , y ) ( M 4 λ PI A 2 d ( x , y ) [ Lipu ] p ( y ) ) p 1 . (17)
Now, putting (11) , (12) together with (16) , (17) , and applying (26) , we get
∣ u ( x ) − u ( y ) ∣ ≤ ∣ u ( x ) − u B 1 ∣ + ∑ i = 1 N − 1 ∣ u B i − u B i + 1 ∣ + ∣ u B N − u ( y ) ∣ ≲ A , C d b , C P I ( d ( x , y ) + ∑ i = 1 N r i ) [ ( M λ P I ′ d ( x , y ) [ Lipu ] p ( x ) ) 1 p + ( M λ P I ′ d ( x , y ) [ Lipu ] p ( y ) ) 1 p ] ≲ A d ( x , y ) [ M λ P I ′ d ( x , y ) [ Lipu ] p ( x ) + M λ P I ′ d ( x , y ) [ Lipu ] p ( y ) ] 1 p . \begin{aligned}
\lvert u(x)-u(y)\rvert&\le\lvert u(x)-u_{B_1}\rvert+\sum_{i=1}^{N-1}\lvert u_{B_i}-u_{B_{i+1}}\rvert+\lvert u_{B_N}-u(y)\rvert\\
&\lesssim_{A,C_{d_b},C_{\mathrm{PI}}}\left(d(x,y)+\sum_{i=1}^{N}r_i\right)\left[\left(M_{\lambda'_{\mathrm{PI}}d(x,y)}[\operatorname{Lipu}]^p(x)\right)^{\frac{1}{p}}+\left(M_{\lambda'_{\mathrm{PI}}d(x,y)}[\operatorname{Lipu}]^p(y)\right)^{\frac{1}{p}}\right]\\
&\lesssim_A d(x,y)\left[M_{\lambda'_{\mathrm{PI}}d(x,y)}[\operatorname{Lipu}]^p(x)+M_{\lambda'_{\mathrm{PI}}d(x,y)}[\operatorname{Lipu}]^p(y)\right]^{\frac{1}{p}}.
\end{aligned} ∣ u ( x ) − u ( y )∣ ≤ ∣ u ( x ) − u B 1 ∣ + i = 1 ∑ N − 1 ∣ u B i − u B i + 1 ∣ + ∣ u B N − u ( y )∣ ≲ A , C d b , C PI ( d ( x , y ) + i = 1 ∑ N r i ) [ ( M λ PI ′ d ( x , y ) [ Lipu ] p ( x ) ) p 1 + ( M λ PI ′ d ( x , y ) [ Lipu ] p ( y ) ) p 1 ] ≲ A d ( x , y ) [ M λ PI ′ d ( x , y ) [ Lipu ] p ( x ) + M λ PI ′ d ( x , y ) [ Lipu ] p ( y ) ] p 1 .
This completes the proof of (10) .
Poincaré inequalities via Whitney balls
Let Z Z Z be an ambient metric space. The next result says that if a space X ⊆ Z X \subseteq Z X ⊆ Z is covered by a collection of balls resembling a Whitney decomposition (i.e., well-separated with nearby balls having comparable size), which have PI estimates locally around each ball, and cover nearby PI space Y ⊆ Z Y \subseteq Z Y ⊆ Z , then X X X also has PI.
Proposition 3.8. Let { C i } i = 0 5 ⊆ [ 1 , ∞ ) \{C_i\}_{i=0}^{5} \subseteq[1,\infty) { C i } i = 0 5 ⊆ [ 1 , ∞ ) such that
C 3 ≥ 5 C 1 and C 5 ≥ 50 C 1 . C_3 \ge5C_1 \qquad\text{and}\qquad C_5 \ge50C_1. C 3 ≥ 5 C 1 and C 5 ≥ 50 C 1 . Let ( Z , d ) (Z,d) ( Z , d ) , X X X , { Y } i ∈ I ⊆ Z \{Y\}_{i\in I} \subseteq Z { Y } i ∈ I ⊆ Z and { B i } i ∈ I \{B_i\}_{i\in I} { B i } i ∈ I satisfy the hypotheses Lemma 2.10 (1)–(4) with constants C 0 , C 1 C_0,C_1 C 0 , C 1 and C 2 C_2 C 2 . Suppose further that X X X support a ( 1 , 1 ) (1,1) ( 1 , 1 ) -Poincaré inequality with constant ( C P I , λ P I ) (C_{\mathrm{PI}},\lambda_{\mathrm{PI}}) ( C PI , λ PI ) and that the following condition holds
(1) The exists c 4 > 0 c_4>0 c 4 > 0 such that, whenever i ≠ j ∈ I i\ne j\in I i = j ∈ I , we have
c 4 B i ∩ c 4 B j = ∅ . c_4B_i\cap c_4B_j=\varnothing. c 4 B i ∩ c 4 B j = ∅ . (2) For all i , j ∈ I i,j\in I i , j ∈ I such that C 3 B i ∩ C 3 B j ≠ ∅ C_3B_i\cap C_3B_j\ne\varnothing C 3 B i ∩ C 3 B j = ∅ we have
C 4 − 1 ≤ r i / r j ≤ C 4 C_4^{-1}\le r_i/r_j\le C_4 C 4 − 1 ≤ r i / r j ≤ C 4 (3) For all i ∈ I i\in I i ∈ I and u : X → R u:X\to\mathbb{R} u : X → R we have ( u , C 5 B i ) ∈ MPI ( C 6 , C 6 ) (u,C_5B_i)\in\operatorname{MPI}(C_6,C_6) ( u , C 5 B i ) ∈ MPI ( C 6 , C 6 ) .
Then, the set
Y = ⋃ i ∈ I Y i Y=\bigcup_{i\in I}Y_i Y = i ∈ I ⋃ Y i is Ahlfors d d d -regular, with regularity constant depending on d , C 0 , C 1 d,C_0,C_1 d , C 0 , C 1 and C 2 C_2 C 2 , and supports a ( 1 , 1 ) (1,1) ( 1 , 1 ) -Poincaré inequality, with constant C P I ′ C'_{\mathrm{PI}} C PI ′ depending on all the above constants (with the exception of C 3 C_3 C 3 ) and d d d , and λ P I ′ \lambda'_{\mathrm{PI}} λ PI ′ depending on λ P I , C 1 , C 5 \lambda_{\mathrm{PI}},C_1,C_5 λ PI , C 1 , C 5 and C 6 C_6 C 6 .
The Ahlfors regularity follows directly from Lemma 2.10. We will let C 0 ( Y ) C_0(Y) C 0 ( Y ) denote the regularity constant of Y Y Y for remainder of the section. Our goal now it to prove a Poincaré inequality, which will be done over a number of lemmas. By Lemma 3.2 , it suffices to find C 7 ≥ 1 C_7\ge1 C 7 ≥ 1 such that
∣ u ( x 0 ) − u ( y 0 ) ∣ ≤ C 7 ∣ x 0 − y 0 ∣ ( M C 7 ∣ x 0 − y 0 ∣ Lip u ( x 0 ) + M C 7 ∣ x 0 − y 0 ∣ Lip u ( y 0 ) ) (18) |u(x_0)-u(y_0)|\le C_7|x_0-y_0|\left(M_{C_7|x_0-y_0|}\operatorname{Lip}u(x_0)+M_{C_7|x_0-y_0|}\operatorname{Lip}u(y_0)\right) \tag*{(18)} ∣ u ( x 0 ) − u ( y 0 ) ∣ ≤ C 7 ∣ x 0 − y 0 ∣ ( M C 7 ∣ x 0 − y 0 ∣ Lip u ( x 0 ) + M C 7 ∣ x 0 − y 0 ∣ Lip u ( y 0 ) ) (18)
for any u : Y → R u:Y\to\mathbb{R} u : Y → R Lipschitz and x 0 , y 0 ∈ Y x_0,y_0\in Y x 0 , y 0 ∈ Y , which we now fix. If x 0 , y 0 ∈ C 5 B i x_0,y_0\in C_5B_i x 0 , y 0 ∈ C 5 B i for some i ∈ I i\in I i ∈ I , then this follows by assumption, so let us suppose for the moment that there is no such i ∈ I i\in I i ∈ I . Note that, by Lemma 3.3 and the hypothesis (3) in Proposition 3.8 ,
∣ u B − u B ′ ∣ ≲ C 0 ( Y ) , C 5 , C 6 ( M r B , λ r i Lip u ( x B ) + M r B ′ , λ r i Lip u ( x B ′ ) ) (19) |u_B-u_{B'}|\lesssim_{C_0(Y),C_5,C_6} \left(M_{r_B,\lambda r_i}\operatorname{Lip}u(x_B)+M_{r_{B'},\lambda r_i}\operatorname{Lip}u(x_{B'})\right) \tag*{(19)} ∣ u B − u B ′ ∣ ≲ C 0 ( Y ) , C 5 , C 6 ( M r B , λ r i Lip u ( x B ) + M r B ′ , λ r i Lip u ( x B ′ ) ) (19)
for all indices i ∈ I i\in I i ∈ I and balls B , B ′ ⊆ ( C 5 / 4 ) B i B,B'\subseteq(C_5/4)B_i B , B ′ ⊆ ( C 5 /4 ) B i , where
λ : = 4 C 5 C 6 . \lambda:=4C_5C_6. λ := 4 C 5 C 6 .
We define functions u ~ , ρ ~ : X → R \widetilde{u},\widetilde{\rho}:X\to\mathbb{R} u , ρ : X → R as follows. First, let { ϕ i } i ∈ I \{\phi_i\}_{i\in I} { ϕ i } i ∈ I be a collection of Lipschitz functions such that
supp ( ϕ i ) ⊆ 2 C 1 B i , Lip ( ϕ i ) ≲ r i − 1 and ∑ i ∈ I ϕ i = 1 on ⋃ i ∈ I C 1 B i ⊇ X , (20) \operatorname{supp}(\phi_i)\subseteq2C_1B_i,\quad\operatorname{Lip}(\phi_i)\lesssim r_i^{-1}\quad\text{and}\quad\sum_{i\in I}\phi_i=1\text{ on }\bigcup_{i\in I}C_1B_i\supseteq X, \tag*{(20)} supp ( ϕ i ) ⊆ 2 C 1 B i , Lip ( ϕ i ) ≲ r i − 1 and i ∈ I ∑ ϕ i = 1 on i ∈ I ⋃ C 1 B i ⊇ X , (20)
where the final inclusion follows from hypothesis (2) in Proposition 3.8 . The existence of such a collection follows from the fact that Y Y Y is Ahlfors regular. Similarly, let { φ i } i ∈ I \{\varphi_i\}_{i\in I} { φ i } i ∈ I be a collection of Lipschitz functions such that
supp ( φ i ) ⊆ 3 C 1 B i , Lip ( φ i ) ≲ r i − 1 and φ i ≳ 1 on ⋃ i ∈ I 2 C 1 B i . \operatorname{supp}(\varphi_i)\subseteq3C_1B_i,\quad\operatorname{Lip}(\varphi_i)\lesssim r_i^{-1}\quad\text{and}\quad\varphi_i\gtrsim1\text{ on }\bigcup_{i\in I}2C_1B_i. supp ( φ i ) ⊆ 3 C 1 B i , Lip ( φ i ) ≲ r i − 1 and φ i ≳ 1 on i ∈ I ⋃ 2 C 1 B i .
Then, set
u ~ : = ∑ i ∈ I ⟨ u ⟩ B i φ i and ρ ~ : = ∑ i ∈ I ⟨ Lip u ⟩ 2 λ B i φ i . \widetilde{u}:=\sum_{i\in I}\langle u\rangle_{B_i}\varphi_i \qquad\text{and}\qquad\widetilde{\rho}:=\sum_{i\in I}\langle\operatorname{Lip}u\rangle_{2\lambda B_i}\varphi_i. u := i ∈ I ∑ ⟨ u ⟩ B i φ i and ρ := i ∈ I ∑ ⟨ Lip u ⟩ 2 λ B i φ i .
We will show via several lemmas the following.
Lemma 3.9. The function u ~ \widetilde{u} u is Lipschitz and satisfies
Lip u ~ ≲ C 0 ( Y ) , C 2 , C 4 , C 5 , C 6 ρ ~ . \operatorname{Lip}\widetilde{u}\lesssim_{C_0(Y),C_2,C_4,C_5,C_6}\widetilde{\rho}. Lip u ≲ C 0 ( Y ) , C 2 , C 4 , C 5 , C 6 ρ .
To help with the proof of Lemma 3.9 , we will need some notation. For each y ∈ Y y\in Y y ∈ Y , choose a point p X ( y ) ∈ X p_X(y)\in X p X ( y ) ∈ X and an index i ( y ) ∈ I i(y)\in I i ( y ) ∈ I such that
d ( y , p X ( y ) ) = dist ( y , X ) and y ∈ C 1 B i ( y ) . d(y,p_X(y))=\operatorname{dist}(y,X) \qquad\text{and}\qquad y\in C_1B_{i(y)}. d ( y , p X ( y )) = dist ( y , X ) and y ∈ C 1 B i ( y ) .
For each z ∈ Z z\in Z z ∈ Z , let
I ( z ) = { i ∈ I : z ∈ 4 C 1 B i } I(z)=\{i\in I:z\in4C_1B_i\} I ( z ) = { i ∈ I : z ∈ 4 C 1 B i }
and
j ( z ) : = arg max i ∈ I ( z ) r i . (21) j(z):=\underset{i\in I(z)}{\operatorname{arg\,max}}\ r_i. \tag*{(21)} j ( z ) := i ∈ I ( z ) arg max r i . (21)
Lemma 3.10. For each z ∈ Z z\in Z z ∈ Z and i ∈ I ( z ) i\in I(z) i ∈ I ( z ) , we have
r i ∼ C 4 r j ( z ) , ⋃ i ∈ I ( z ) 2 B i ⊆ ( C 5 / 4 ) B j ( z ) and ∣ I ( z ) ∣ ≤ C 2 , (22) r_i\sim_{C_4}r_{j(z)},\qquad\bigcup_{i\in I(z)}2B_i\subseteq(C_5/4)B_{j(z)}\qquad\text{and}\qquad|I(z)|\leq C_2, \tag*{(22)} r i ∼ C 4 r j ( z ) , i ∈ I ( z ) ⋃ 2 B i ⊆ ( C 5 /4 ) B j ( z ) and ∣ I ( z ) ∣ ≤ C 2 , (22) where ∣ ⋅ ∣ |\cdot| ∣ ⋅ ∣ denotes the cardinality of a set.
Proof. The right-most property follows from Lemma 2.10 (1) and the definition of I ( z ) I(z) I ( z ) . For the remaining properties, fix some z ∈ Z z\in Z z ∈ Z and i ∈ I ( z ) i\in I(z) i ∈ I ( z ) . Since i , j ( z ) ∈ I ( z ) i,j(z)\in I(z) i , j ( z ) ∈ I ( z ) , the left-most property follows directly from hypothesis (2) in Proposition 3.8 (2) since C 3 ≥ 5 C 1 C_3\geq5C_1 C 3 ≥ 5 C 1 . Since 4 C 1 B i ∩ 4 C 1 B j ( z ) ≠ ∅ 4C_1B_i\cap4C_1B_{j(z)}\neq\varnothing 4 C 1 B i ∩ 4 C 1 B j ( z ) = ∅ and r j ( z ) ≥ r i r_{j(z)}\geq r_i r j ( z ) ≥ r i , we have 4 C 1 B i ⊆ 12 C 1 B j ( z ) 4C_1B_i\subseteq12C_1B_{j(z)} 4 C 1 B i ⊆ 12 C 1 B j ( z ) . Thus, the middle claim in (22) follows from the fact that C 5 ≥ 50 C 1 C_5\geq50C_1 C 5 ≥ 50 C 1 . □ \square □
Lemma 3.11. For each y ∈ Y y\in Y y ∈ Y , w ∈ B ( p X ( y ) , r i ( y ) ) w\in B(p_X(y),r_{i(y)}) w ∈ B ( p X ( y ) , r i ( y ) ) and i ∈ I ( w ) i\in I(w) i ∈ I ( w ) , we have
B ( y , r i ( y ) ) ∪ B i ⊆ ( C 5 / 4 ) B j ( w ) . B(y,r_{i(y)})\cup B_i\subseteq(C_5/4)B_{j(w)}. B ( y , r i ( y ) ) ∪ B i ⊆ ( C 5 /4 ) B j ( w ) .
Proof. Fix y , w y,w y , w and i ∈ I ( w ) i\in I(w) i ∈ I ( w ) as above. By Lemma 3.10 , we immediately have B i ⊆ ( C 5 / 4 ) B j ( w ) B_i\subseteq(C_5/4)B_{j(w)} B i ⊆ ( C 5 /4 ) B j ( w ) . Since y ∈ C 1 B i ( y ) y\in C_1B_{i(y)} y ∈ C 1 B i ( y ) and dist ( x i ( y ) , X ) ≤ C 1 r i ( y ) \operatorname{dist}(x_{i(y)},X)\leq C_1r_{i(y)} dist ( x i ( y ) , X ) ≤ C 1 r i ( y ) , it follows that
∣ y − p X ( y ) ∣ ≤ 2 C 1 r i ( y ) . (23) |y-p_X(y)|\leq2C_1r_{i(y)}. \tag*{(23)} ∣ y − p X ( y ) ∣ ≤ 2 C 1 r i ( y ) . (23)
Since y ∈ C 1 B i ( y ) y\in C_1B_{i(y)} y ∈ C 1 B i ( y ) and w ∈ B ( p X ( y ) , r i ( y ) ) w\in B(p_X(y),r_{i(y)}) w ∈ B ( p X ( y ) , r i ( y ) ) , this gives ∣ w − x i ( y ) ∣ ≤ ( 1 + 3 C 1 ) r i ( y ) ≤ 4 C 1 r i ( y ) |w-x_{i(y)}|\leq(1+3C_1)r_{i(y)}\leq4C_1r_{i(y)} ∣ w − x i ( y ) ∣ ≤ ( 1 + 3 C 1 ) r i ( y ) ≤ 4 C 1 r i ( y ) , i.e., i ( y ) ∈ I ( w ) i(y)\in I(w) i ( y ) ∈ I ( w ) . Since B ( y , r i ( y ) ) ⊆ 2 B i ( y ) B(y,r_{i(y)})\subseteq2B_{i(y)} B ( y , r i ( y ) ) ⊆ 2 B i ( y ) , the inclusion B ( y , r i ( y ) ) ⊆ ( C 5 / 4 ) B j ( y ) B(y,r_{i(y)})\subseteq(C_5/4)B_{j(y)} B ( y , r i ( y ) ) ⊆ ( C 5 /4 ) B j ( y ) follows from Lemma 3.10 . □ \square □
Proof of Lemma 3.9 . That u ~ \widetilde{u} u is Lipschitz is easy to see since it is the sum of Lipschitz functions whose supports have bounded overlap by C 2 C_2 C 2 . To prove the estimate on Lip u ~ \operatorname{Lip}\widetilde{u} Lip u , fix some x ∈ X x \in X x ∈ X and let j : = j ( x ) j := j(x) j := j ( x ) as in (21) . Recalling the properties of ϕ i \phi_i ϕ i in (20) , and applying Lemma 3.3 , Lemma 3.10 and (19) , we have
Lip u ~ ( x ) = Lip ( u ~ − ⟨ u ⟩ B j ) ( x ) = Lip ( ∑ i ∈ I ( ⟨ u ⟩ B i − ⟨ u ⟩ B j ) ϕ i ) ( x ) ≲ ∑ i ∈ I ( x ) r i − 1 ∣ ⟨ u ⟩ B i − ⟨ u ⟩ B j ∣ ≲ C 2 , C 4 sup i ∈ I ( x ) r j − 1 ∣ ⟨ u ⟩ B i − ⟨ u ⟩ B j ∣ ≲ C 0 ( Y ) , C 5 , C 6 sup i ∈ I ( x ) ( M r i , λ r j Lip u ( x i ) + M r j , λ r j Lip u ( x j ) ) . \begin{aligned}
\operatorname{Lip}\widetilde{u}(x)
&=\operatorname{Lip}\left(\widetilde{u}-\langle u\rangle_{B_j}\right)(x)
=\operatorname{Lip}\left(\sum_{i\in I}\left(\langle u\rangle_{B_i}-\langle u\rangle_{B_j}\right)\phi_i\right)(x)\\
&\lesssim\sum_{i\in I(x)}r_i^{-1}\left|\langle u\rangle_{B_i}-\langle u\rangle_{B_j}\right|\\
&\lesssim_{C_2,C_4}\sup_{i\in I(x)}r_j^{-1}\left|\langle u\rangle_{B_i}-\langle u\rangle_{B_j}\right|\\
&\lesssim_{C_0(Y),C_5,C_6}\sup_{i\in I(x)}\left(M_{r_i,\lambda r_j}\operatorname{Lip}u(x_i)+M_{r_j,\lambda r_j}\operatorname{Lip}u(x_j)\right).
\end{aligned} Lip u ( x ) = Lip ( u − ⟨ u ⟩ B j ) ( x ) = Lip ( i ∈ I ∑ ( ⟨ u ⟩ B i − ⟨ u ⟩ B j ) ϕ i ) ( x ) ≲ i ∈ I ( x ) ∑ r i − 1 ⟨ u ⟩ B i − ⟨ u ⟩ B j ≲ C 2 , C 4 i ∈ I ( x ) sup r j − 1 ⟨ u ⟩ B i − ⟨ u ⟩ B j ≲ C 0 ( Y ) , C 5 , C 6 i ∈ I ( x ) sup ( M r i , λ r j Lip u ( x i ) + M r j , λ r j Lip u ( x j ) ) .
(recall that x i x_i x i is the centre of B i B_i B i ). Fix an arbitrary i ∈ I ( x ) i\in I(x) i ∈ I ( x ) . Since 2 B i ⊆ ( C 5 / 4 ) B j 2B_i\subseteq(C_5/4)B_j 2 B i ⊆ ( C 5 /4 ) B j , any ball B ( x i , r ) B(x_i,r) B ( x i , r ) with r ∈ [ r i , λ r j ] r\in[r_i,\lambda r_j] r ∈ [ r i , λ r j ] must satisfy B ( x i , r ) ⊆ 2 λ B j B(x_i,r)\subseteq2\lambda B_j B ( x i , r ) ⊆ 2 λ B j and r ∼ C 4 , λ r j r\sim_{C_4,\lambda}r_j r ∼ C 4 , λ r j . On the other hand, any ball B ( x j , r ) B(x_j,r) B ( x j , r ) with r ∈ [ r j , λ r j ] r\in[r_j,\lambda r_j] r ∈ [ r j , λ r j ] must satisfy B ( x j , r ) ⊆ B ( x j , 2 λ r j ) B(x_j,r)\subseteq B(x_j,2\lambda r_j) B ( x j , r ) ⊆ B ( x j , 2 λ r j ) and r ∼ λ r j r\sim_\lambda r_j r ∼ λ r j . By comparing the averages of Lip u \operatorname{Lip}u Lip u in these balls, it follows from this that
M r i , λ r j Lip u ( x i ) + M r j , λ r j Lip u ( x j ) ≲ C 0 ( Y ) , C 4 , λ ⟨ Lip u ⟩ 2 λ B j . M_{r_i,\lambda r_j}\operatorname{Lip}u(x_i)+M_{r_j,\lambda r_j}\operatorname{Lip}u(x_j)\lesssim_{C_0(Y),C_4,\lambda}\langle\operatorname{Lip}u\rangle_{2\lambda B_j}. M r i , λ r j Lip u ( x i ) + M r j , λ r j Lip u ( x j ) ≲ C 0 ( Y ) , C 4 , λ ⟨ Lip u ⟩ 2 λ B j .
Recalling that x ∈ 2 C 1 B j x\in2C_1B_j x ∈ 2 C 1 B j and ϕ j ( x ) ≳ 1 \phi_j(x)\gtrsim1 ϕ j ( x ) ≳ 1 on 2 C 1 B j 2C_1B_j 2 C 1 B j , we now have
Lip u ~ ( x ) ≲ C 0 ( Y ) , C 2 , C 4 , C 5 , C 6 ⟨ Lip u ⟩ 2 λ B j ϕ j ( x ) ≤ ρ ~ ( x ) , \operatorname{Lip}\widetilde{u}(x)\lesssim_{C_0(Y),C_2,C_4,C_5,C_6}\langle\operatorname{Lip}u\rangle_{2\lambda B_j}\phi_j(x)\leq\widetilde{\rho}(x), Lip u ( x ) ≲ C 0 ( Y ) , C 2 , C 4 , C 5 , C 6 ⟨ Lip u ⟩ 2 λ B j ϕ j ( x ) ≤ ρ ( x ) ,
as required. □
For the remainder of this section, we will denote ν : = H d ∣ X \nu:=\mathcal{H}^d|_X ν := H d ∣ X . To avoid confusion over which measure various average integrals are being computed with respect to, for a function f : Z → R f:Z\to\mathbb{R} f : Z → R and a ball B ⊆ Z B\subseteq Z B ⊆ Z , we will write
\langle f\rangle_{B,\nu}:=\fint_B f\,d\nu.
Lemma 3.9 and the assumptions on X X X tell us that we can estimate the oscillations of u ~ \widetilde{u} u by the integral of ρ ~ \widetilde{\rho} ρ with respect to ν \nu ν . To prove the Poincaré inequality on Y Y Y , we must now prove estimates which related u u u to u ~ \widetilde{u} u and the integral of Lip u \operatorname{Lip}u Lip u with respect to μ \mu μ to the integral of ρ ~ \widetilde{\rho} ρ with respect to ν \nu ν .
Lemma 3.12. We have
∣ u ( y ) − ⟨ u ~ ⟩ B ( p X ( y ) , r i ( y ) ) , ν ∣ ≲ C 0 ( Y ) , C 4 , C 5 , C 6 r i ( y ) M C 8 r i ( y ) Lip u ( y ) \left|u(y)-\langle\widetilde{u}\rangle_{B(p_X(y),r_i(y)),\nu}\right|\lesssim_{C_0(Y),C_4,C_5,C_6}r_i(y)M_{C_8r_i(y)}\operatorname{Lip}u(y) u ( y ) − ⟨ u ⟩ B ( p X ( y ) , r i ( y )) , ν ≲ C 0 ( Y ) , C 4 , C 5 , C 6 r i ( y ) M C 8 r i ( y ) Lip u ( y ) for all y ∈ Y y\in Y y ∈ Y , where C 8 : = 2 C 4 λ C_8:=2C_4\lambda C 8 := 2 C 4 λ .
Proof. Fix some w ∈ B ( p X ( y ) , r i ( w ) ) w\in B(p_X(y),r_i(w)) w ∈ B ( p X ( y ) , r i ( w )) and i ∈ I ( w ) i\in I(w) i ∈ I ( w ) for the moment. Recalling from Lemma 3.11 that B ( y , r i ( y ) ) ∪ B i ⊆ ( C 5 / 4 ) B j ( w ) B(y,r_i(y))\cup B_i\subseteq(C_5/4)B_{j(w)} B ( y , r i ( y )) ∪ B i ⊆ ( C 5 /4 ) B j ( w ) , by applying (19) , we get
∣ u ( y ) − ⟨ u ⟩ B i ∣ ≤ ∣ u ( y ) − ⟨ u ⟩ B ( y , r i ( y ) ) ∣ + ∣ ⟨ u ⟩ B ( y , r i ( y ) ) − ⟨ u ⟩ B i ∣ ≲ C 0 ( Y ) , C 5 , C 6 r i ( y ) M λ r i ( y ) Lip u ( y ) + r j ( w ) M r i ( y ) , λ r j ( w ) Lip u ( y ) + r j ( w ) M r i , λ r j ( w ) Lip u ( x i ) . \begin{aligned}
|u(y)-\langle u\rangle_{B_i}|
&\leq|u(y)-\langle u\rangle_{B(y,r_i(y))}|+|\langle u\rangle_{B(y,r_i(y))}-\langle u\rangle_{B_i}|\\
&\lesssim_{C_0(Y),C_5,C_6}r_i(y)M_{\lambda r_i(y)}\operatorname{Lip}u(y)+r_j(w)M_{r_i(y),\lambda r_j(w)}\operatorname{Lip}u(y)\\
&\qquad+r_j(w)M_{r_i,\lambda r_j(w)}\operatorname{Lip}u(x_i).
\end{aligned} ∣ u ( y ) − ⟨ u ⟩ B i ∣ ≤ ∣ u ( y ) − ⟨ u ⟩ B ( y , r i ( y )) ∣ + ∣ ⟨ u ⟩ B ( y , r i ( y )) − ⟨ u ⟩ B i ∣ ≲ C 0 ( Y ) , C 5 , C 6 r i ( y ) M λ r i ( y ) Lip u ( y ) + r j ( w ) M r i ( y ) , λ r j ( w ) Lip u ( y ) + r j ( w ) M r i , λ r j ( w ) Lip u ( x i ) . Arguing as in the proof of Lemma 3.9 , any ball B ( x j ( w ) , r ) B(x_{j(w)},r) B ( x j ( w ) , r ) which satisfies r ∈ [ r i ( y ) , λ r j ( w ) ] r\in[r_i(y),\lambda r_{j(w)}] r ∈ [ r i ( y ) , λ r j ( w ) ] is contained in B ( y , 2 λ r j ( w ) ) ⊆ B ( y , 2 λ C 4 r i ( y ) ) B(y,2\lambda r_{j(w)})\subseteq B(y,2\lambda C_4r_i(y)) B ( y , 2 λ r j ( w ) ) ⊆ B ( y , 2 λ C 4 r i ( y )) and satisfies r ∼ C 4 , λ r i ( y ) r\sim_{C_4,\lambda}r_i(y) r ∼ C 4 , λ r i ( y ) . Using also the fact that r j ( w ) ≤ C 4 r i ( y ) r_{j(w)}\leq C_4r_i(y) r j ( w ) ≤ C 4 r i ( y ) , we can now estimate
∣ u ( y ) − ⟨ u ⟩ B i ∣ ≲ C 0 ( Y ) , C 4 , C 5 , C 6 r i ( y ) M 2 C 4 λ r i ( y ) Lip u ( y ) . |u(y)-\langle u\rangle_{B_i}|\lesssim_{C_0(Y),C_4,C_5,C_6}r_i(y)M_{2C_4\lambda r_i(y)}\operatorname{Lip}u(y). ∣ u ( y ) − ⟨ u ⟩ B i ∣ ≲ C 0 ( Y ) , C 4 , C 5 , C 6 r i ( y ) M 2 C 4 λ r i ( y ) Lip u ( y ) .
Letting C 8 : = 2 C 4 λ C_8 := 2C_4\lambda C 8 := 2 C 4 λ , the above estimate implies
∣ u ( y ) − ⟨ u ~ ⟩ B ( p X ( y ) , r i ( y ) ) , ν ∣ ≤ ∫ B ( p X ( y ) , r i ( y ) ) ∣ u ( y ) − u ~ ( w ) ∣ d ν ( w ) = ∫ B ( p X ( y ) , r i ( y ) ) ∣ u ( y ) − ∑ i ∈ I ( w ) ⟨ u ⟩ B i ϕ i ( w ) ∣ d ν ( w ) ≤ ∫ B ( p X ( y ) , r i ( y ) ) ∑ i ∈ I ( w ) ∣ u ( y ) − ⟨ u ⟩ B i ∣ ϕ i ( w ) d ν ( w ) ≲ ∫ B ( p X ( y ) , r i ( y ) ) ∑ i ∈ I ( w ) r i ( y ) M C 8 r i ( y ) Lip u ( y ) ϕ i ( w ) d ν ( w ) = r i ( y ) M C 8 r i ( y ) Lip u ( y ) , \begin{aligned}
\left|u(y)-\langle\widetilde{u}\rangle_{B(p_X(y),r_i(y)),\nu}\right|
&\le\int_{B(p_X(y),r_i(y))}|u(y)-\widetilde{u}(w)|\,d\nu(w) \\
&= \int_{B(p_X(y),r_i(y))}\left|u(y)-\sum_{i\in I(w)}\langle u\rangle_{B_i}\phi_i(w)\right|\,d\nu(w) \\
&\le\int_{B(p_X(y),r_i(y))}\sum_{i\in I(w)}\left|u(y)-\langle u\rangle_{B_i}\right|\phi_i(w)\,d\nu(w) \\
&\lesssim\int_{B(p_X(y),r_i(y))}\sum_{i\in I(w)}r_i(y)M C_{8r_i(y)}\operatorname{Lip}u(y)\phi_i(w)\,d\nu(w) \\
&=r_i(y)M C_{8r_i(y)}\operatorname{Lip}u(y),
\end{aligned} u ( y ) − ⟨ u ⟩ B ( p X ( y ) , r i ( y )) , ν ≤ ∫ B ( p X ( y ) , r i ( y )) ∣ u ( y ) − u ( w ) ∣ d ν ( w ) = ∫ B ( p X ( y ) , r i ( y )) u ( y ) − i ∈ I ( w ) ∑ ⟨ u ⟩ B i ϕ i ( w ) d ν ( w ) ≤ ∫ B ( p X ( y ) , r i ( y )) i ∈ I ( w ) ∑ ∣ u ( y ) − ⟨ u ⟩ B i ∣ ϕ i ( w ) d ν ( w ) ≲ ∫ B ( p X ( y ) , r i ( y )) i ∈ I ( w ) ∑ r i ( y ) M C 8 r i ( y ) Lip u ( y ) ϕ i ( w ) d ν ( w ) = r i ( y ) M C 8 r i ( y ) Lip u ( y ) ,
where the implicit constant depends on C 0 ( Y ) C_0(Y) C 0 ( Y ) , C 4 C_4 C 4 , C 5 C_5 C 5 and C 6 C_6 C 6 . This finishes the proof.
Lemma 3.13. Let y ∈ Y y\in Y y ∈ Y and r ≥ r i ( y ) r\ge r_i(y) r ≥ r i ( y ) . If i ∈ I i\in I i ∈ I is such that B ( p X ( y ) , r ) ∩ B i ≠ ∅ B(p_X(y),r)\cap B_i\ne\varnothing B ( p X ( y ) , r ) ∩ B i = ∅ , then we have r i ≤ max { 3 + 2 C 1 , C 4 } r r_i\le\max\{3+2C_1,C_4\}r r i ≤ max { 3 + 2 C 1 , C 4 } r .
Proof. Suppose towards a contradiction that there exists i ∈ I i\in I i ∈ I such that B ( p X ( y ) , r ) ∩ B i ≠ ∅ B(p_X(y),r)\cap B_i\ne\varnothing B ( p X ( y ) , r ) ∩ B i = ∅ and r i > A r r_i>Ar r i > A r , where A = max { 3 + 2 C 1 , C 4 } A=\max\{3+2C_1,C_4\} A = max { 3 + 2 C 1 , C 4 } . Recalling that y ∈ B i ( y ) y\in B_{i(y)} y ∈ B i ( y ) and from (23) that ∣ p X ( y ) − y ∣ ≤ 2 C 1 r i ( y ) ≤ 2 C 1 r |p_X(y)-y|\le2C_1r_i(y)\le2C_1r ∣ p X ( y ) − y ∣ ≤ 2 C 1 r i ( y ) ≤ 2 C 1 r , it follows from that triangle inequality that
B i ( y ) ⊆ B ( y , 2 r i ( y ) ) ⊆ B ( p X ( y ) , ( 2 + 2 C 1 ) r i ( y ) ) ⊆ B ( p X ( y ) , ( 2 + 2 C 1 ) r ) . B_{i(y)}\subseteq B(y,2r_i(y))\subseteq B(p_X(y),(2+2C_1)r_i(y))\subseteq B(p_X(y),(2+2C_1)r). B i ( y ) ⊆ B ( y , 2 r i ( y )) ⊆ B ( p X ( y ) , ( 2 + 2 C 1 ) r i ( y )) ⊆ B ( p X ( y ) , ( 2 + 2 C 1 ) r ) .
By our choice of A A A and the triangle inequality, we have B i ( y ) ⊆ 2 B i B_{i(y)}\subseteq2B_i B i ( y ) ⊆ 2 B i . Thus, it follows from Proposition 3.8 (2) that
C 4 r ≤ A r < r i ≤ C 4 r i ( y ) ≤ C 4 r , C_4r\le Ar<r_i\le C_4r_i(y)\le C_4r, C 4 r ≤ A r < r i ≤ C 4 r i ( y ) ≤ C 4 r ,
which is a contradiction.
Lemma 3.14. For each y ∈ Y y\in Y y ∈ Y and r ≥ r i ( y ) r\ge r_i(y) r ≥ r i ( y ) we have
∫ B ( p X ( y ) , r ) ρ ~ d ν ≲ C 0 , C 0 ( Y ) , C 1 , C 4 , c 4 , λ ∫ B ( y , C 9 r ) Lip u d μ , \int_{B(p_X(y),r)}\widetilde{\rho}\,d\nu\lesssim_{C_0,C_0(Y),C_1,C_4,c_4,\lambda}\int_{B(y,C_9r)}\operatorname{Lip}u\,d\mu, ∫ B ( p X ( y ) , r ) ρ d ν ≲ C 0 , C 0 ( Y ) , C 1 , C 4 , c 4 , λ ∫ B ( y , C 9 r ) Lip u d μ , where C 9 C_9 C 9 depends only on λ \lambda λ , C 1 C_1 C 1 and C 5 C_5 C 5 .
Proof. By a standard volume argument, using Ahlfors regularity of Y Y Y , Proposition 3.8 (1) and (2), we know that the balls { 2 λ B i } i ∈ I \{2\lambda B_i\}_{i\in I} { 2 λ B i } i ∈ I have bounded overlap with constant depending only on C 0 ( Y ) C_0(Y) C 0 ( Y ) , C 4 C_4 C 4 , c 4 c_4 c 4 and λ \lambda λ . By Lemma 3.13 and the estimate ∣ y − p X ( y ) ∣ ≤ 2 C 1 r i ( y ) ≤ 2 C 1 r |y-p_X(y)|\le2C_1r_i(y)\le2C_1r ∣ y − p X ( y ) ∣ ≤ 2 C 1 r i ( y ) ≤ 2 C 1 r (see (23) for the first estimates), if i ∈ I i\in I i ∈ I is such that B i ∩ B ( p X ( y ) , r ) ≠ ∅ B_i\cap B(p_X(y),r)\ne\varnothing B i ∩ B ( p X ( y ) , r ) = ∅ then it is easy to check from the triangle inequality that
2 λ B i ⊆ C 9 B ( y , r ) 2\lambda B_i\subseteq C_9B(y,r) 2 λ B i ⊆ C 9 B ( y , r )
for some C 9 C_9 C 9 depending only on λ \lambda λ , C 1 C_1 C 1 and C 4 C_4 C 4 . Combining these two facts we can estimate
∫ B ( p X ( y ) , r ) ρ ~ d ν = ∫ B ( p X ( y ) , r ) ∑ i ∈ I ⟨ Lip u ⟩ 2 λ B i φ i d ν ≤ ∑ i ∈ I B i ∩ B ( p X ( y ) , r ) ≠ ∅ ∫ B i ⟨ Lip u ⟩ 2 λ B i d ν ≲ C 0 , C 0 ( Y ) , λ ∑ i ∈ I 2 λ B i ⊂ B ( y , C 9 r ) ≠ ∅ ∫ 2 λ B i Lip u d μ ≲ C 0 ( Y ) , C 4 , c 4 , λ ∫ B ( y , C 9 r ) Lip u d μ . \begin{aligned}
\int_{B(p_X(y),r)} \widetilde{\rho}\,d\nu
&=\int_{B(p_X(y),r)}\sum_{i\in I}\langle\operatorname{Lip}u\rangle_{2\lambda B_i}\varphi_i\,d\nu\\
&\le\sum_{\substack{i\in I\\B_i\cap B(p_X(y),r)\ne\varnothing}}\int_{B_i}\langle\operatorname{Lip}u\rangle_{2\lambda B_i}\,d\nu\\
&\lesssim_{C_0,C_0(Y),\lambda}\sum_{\substack{i\in I\\2\lambda B_i\subset B(y,C_9r)\ne\varnothing}}\int_{2\lambda B_i}\operatorname{Lip}u\,d\mu\\
&\lesssim_{C_0(Y),C_4,c_4,\lambda}\int_{B(y,C_9r)}\operatorname{Lip}u\,d\mu.
\end{aligned} ∫ B ( p X ( y ) , r ) ρ d ν = ∫ B ( p X ( y ) , r ) i ∈ I ∑ ⟨ Lip u ⟩ 2 λ B i φ i d ν ≤ i ∈ I B i ∩ B ( p X ( y ) , r ) = ∅ ∑ ∫ B i ⟨ Lip u ⟩ 2 λ B i d ν ≲ C 0 , C 0 ( Y ) , λ i ∈ I 2 λ B i ⊂ B ( y , C 9 r ) = ∅ ∑ ∫ 2 λ B i Lip u d μ ≲ C 0 ( Y ) , C 4 , c 4 , λ ∫ B ( y , C 9 r ) Lip u d μ .
The desired estimate follows this since, by Ahlfors regularity of both X X X and Y Y Y , μ ( B ( y , C 9 r ) ) ∼ C 0 , C 0 ( Y ) , C 9 ν ( B ( p X ( y ) , r ) ) \mu(B(y,C_9r))\sim_{C_0,C_0(Y),C_9}\nu(B(p_X(y),r)) μ ( B ( y , C 9 r )) ∼ C 0 , C 0 ( Y ) , C 9 ν ( B ( p X ( y ) , r )) .
Lemma 3.15. The estimate (18) holds for the points x 0 , y 0 x_0,y_0 x 0 , y 0 with C 7 C_7 C 7 depending on all constants in the statement of Proposition 3.8 , except for C 3 C_3 C 3 , and C 8 C_8 C 8 depending on λ P I \lambda_{\mathrm{PI}} λ PI , C 1 C_1 C 1 , C 5 C_5 C 5 and C 6 C_6 C 6 .
Proof. Since we have assumed that there is no i ∈ I i\in I i ∈ I such that x 0 , y 0 ∈ C 5 B i x_0,y_0\in C_5B_i x 0 , y 0 ∈ C 5 B i and C 5 ≥ 2 C_5\ge2 C 5 ≥ 2 , it follows that ( C 5 / 2 ) max { r i ( x 0 ) , r i ( y 0 ) } ≤ ∣ x 0 − y 0 ∣ (C_5/2)\max\{r_i(x_0),r_i(y_0)\}\le|x_0-y_0| ( C 5 /2 ) max { r i ( x 0 ) , r i ( y 0 )} ≤ ∣ x 0 − y 0 ∣ . Furthermore, recalling from (23) that ∣ y 0 − p X ( y 0 ) ∣ ≤ 2 C 1 r i ( y 0 ) |y_0-p_X(y_0)|\le2C_1r_i(y_0) ∣ y 0 − p X ( y 0 ) ∣ ≤ 2 C 1 r i ( y 0 ) and ∣ x 0 − p X ( x 0 ) ∣ ≤ 2 C 1 r i ( x 0 ) |x_0-p_X(x_0)|\le2C_1r_i(x_0) ∣ x 0 − p X ( x 0 ) ∣ ≤ 2 C 1 r i ( x 0 ) , and from the statement that C 5 ≥ 50 C 1 C_5\ge50C_1 C 5 ≥ 50 C 1 , we have by the triangle inequality that
max { r i ( x 0 ) , r i ( y 0 ) } ≤ ∣ p X ( x 0 ) − p X ( y 0 ) ∣ ≲ ∣ x 0 − y 0 ∣ . (24) \max\{r_i(x_0),r_i(y_0)\}\le|p_X(x_0)-p_X(y_0)|\lesssim|x_0-y_0|. \tag*{(24)} max { r i ( x 0 ) , r i ( y 0 )} ≤ ∣ p X ( x 0 ) − p X ( y 0 ) ∣ ≲ ∣ x 0 − y 0 ∣. (24)
We use this, the triangle inequality and Lemma 3.12 to get
∣ u ( x 0 ) − u ( y 0 ) ∣ ≤ ∣ u ( x 0 ) − ⟨ u ~ ⟩ B ( x ~ 0 , r i ( x 0 ) ) , ν ∣ + ∣ ⟨ u ~ ⟩ B ( x ~ 0 , r i ( x 0 ) ) , ν − ⟨ u ~ ⟩ B ( p X ( y 0 ) , r i ( y 0 ) ) , ν ∣ + ∣ ⟨ u ~ ⟩ B ( p X ( y 0 ) , r i ( y 0 ) ) , ν − u ( y 0 ) ∣ ≲ r i ( x 0 ) M C 8 r i ( x 0 ) Lip u ( x 0 ) + ∣ ⟨ u ~ ⟩ B ( p X ( x 0 ) , r i ( x 0 ) ) , ν − ⟨ u ~ ⟩ B ( p X ( y 0 ) , r i ( y 0 ) ) , ν ∣ + r i ( y 0 ) M C 8 r i ( y 0 ) Lip u ( y 0 ) ≤ ∣ x 0 − y 0 ∣ ( M C 8 ∣ x 0 − y 0 ∣ Lip u ( x 0 ) + M C 8 ∣ x 0 − y 0 ∣ Lip u ( y 0 ) ) + ∣ ⟨ u ~ ⟩ B ( p X ( x 0 ) , r i ( x 0 ) ) , ν − ⟨ u ~ ⟩ B ( p X ( y 0 ) , r i ( y 0 ) ) , ν ∣ , \begin{aligned}
|u(x_0)-u(y_0)|
&\le|u(x_0)-\langle\widetilde{u}\rangle_{B(\widetilde{x}_0,r_i(x_0)),\nu}|\\
&\quad+|\langle\widetilde{u}\rangle_{B(\widetilde{x}_0,r_i(x_0)),\nu}-\langle\widetilde{u}\rangle_{B(p_X(y_0),r_i(y_0)),\nu}|\\
&\quad+|\langle\widetilde{u}\rangle_{B(p_X(y_0),r_i(y_0)),\nu}-u(y_0)|\\
&\lesssim r_i(x_0)M_{C_8r_i(x_0)}\operatorname{Lip}u(x_0)
+|\langle\widetilde{u}\rangle_{B(p_X(x_0),r_i(x_0)),\nu}-\langle\widetilde{u}\rangle_{B(p_X(y_0),r_i(y_0)),\nu}|\\
&\quad+r_i(y_0)M_{C_8r_i(y_0)}\operatorname{Lip}u(y_0)\\
&\le|x_0-y_0|\left(M_{C_8|x_0-y_0|}\operatorname{Lip}u(x_0)+M_{C_8|x_0-y_0|}\operatorname{Lip}u(y_0)\right)\\
&\quad+|\langle\widetilde{u}\rangle_{B(p_X(x_0),r_i(x_0)),\nu}-\langle\widetilde{u}\rangle_{B(p_X(y_0),r_i(y_0)),\nu}|,
\end{aligned} ∣ u ( x 0 ) − u ( y 0 ) ∣ ≤ ∣ u ( x 0 ) − ⟨ u ⟩ B ( x 0 , r i ( x 0 )) , ν ∣ + ∣ ⟨ u ⟩ B ( x 0 , r i ( x 0 )) , ν − ⟨ u ⟩ B ( p X ( y 0 ) , r i ( y 0 )) , ν ∣ + ∣ ⟨ u ⟩ B ( p X ( y 0 ) , r i ( y 0 )) , ν − u ( y 0 ) ∣ ≲ r i ( x 0 ) M C 8 r i ( x 0 ) Lip u ( x 0 ) + ∣ ⟨ u ⟩ B ( p X ( x 0 ) , r i ( x 0 )) , ν − ⟨ u ⟩ B ( p X ( y 0 ) , r i ( y 0 )) , ν ∣ + r i ( y 0 ) M C 8 r i ( y 0 ) Lip u ( y 0 ) ≤ ∣ x 0 − y 0 ∣ ( M C 8 ∣ x 0 − y 0 ∣ Lip u ( x 0 ) + M C 8 ∣ x 0 − y 0 ∣ Lip u ( y 0 ) ) + ∣ ⟨ u ⟩ B ( p X ( x 0 ) , r i ( x 0 )) , ν − ⟨ u ⟩ B ( p X ( y 0 ) , r i ( y 0 )) , ν ∣ ,
where the implicit constant depends on C 0 ( Y ) C_0(Y) C 0 ( Y ) , C 4 C_4 C 4 , C 5 C_5 C 5 and C 6 C_6 C 6 . By applying the Poincaré inequality and Ahlfors regularity on X X X , the final term in the above inequality is bounded above by
∣ ⟨ u ~ ⟩ B ( p X ( x 0 ) , r i ( x 0 ) ) , ν − ⟨ u ~ ⟩ B ( p X ( x 0 ) , ∣ p X ( x 0 ) − p X ( y 0 ) ∣ ) , ν ∣ + ∣ ⟨ u ~ ⟩ B ( p X ( x 0 ) , ∣ p X ( x 0 ) − p X ( y 0 ) ∣ ) , ν − ⟨ u ~ ⟩ B ( p X ( y 0 ) , ∣ p X ( x 0 ) − p X ( y 0 ) ∣ ) , ν ∣ + ∣ ⟨ u ~ ⟩ B ( p X ( y 0 ) , ∣ p X ( x 0 ) − p X ( y 0 ) ∣ ) , ν − ⟨ u ~ ⟩ B ( p X ( y 0 ) , r i ( y 0 ) ) , ν ∣ ≲ C 0 , C P I ∣ p X ( x 0 ) − p X ( y 0 ) ∣ M r i ( x 0 ) , 3 λ P I ∣ p X ( x 0 ) − p X ( y 0 ) ∣ Lip u ~ ( p X ( x 0 ) ) + ∣ p X ( x 0 ) − p X ( y 0 ) ∣ M r i ( y 0 ) , 3 λ P I ∣ p X ( x 0 ) − p X ( y 0 ) ∣ Lip u ~ ( p X ( y 0 ) ) . \begin{aligned}
&|\langle\widetilde{u}\rangle_{B(p_X(x_0),r_i(x_0)),\nu}-\langle\widetilde{u}\rangle_{B(p_X(x_0),|p_X(x_0)-p_X(y_0)|),\nu}|\\
&\quad+|\langle\widetilde{u}\rangle_{B(p_X(x_0),|p_X(x_0)-p_X(y_0)|),\nu}
-\langle\widetilde{u}\rangle_{B(p_X(y_0),|p_X(x_0)-p_X(y_0)|),\nu}|\\
&\quad+|\langle\widetilde{u}\rangle_{B(p_X(y_0),|p_X(x_0)-p_X(y_0)|),\nu}
-\langle\widetilde{u}\rangle_{B(p_X(y_0),r_i(y_0)),\nu}|\\
&\lesssim_{C_0,C_{\mathrm{PI}}}|p_X(x_0)-p_X(y_0)|
M_{r_i(x_0),3\lambda_{\mathrm{PI}}|p_X(x_0)-p_X(y_0)|}
\operatorname{Lip}\widetilde{u}(p_X(x_0))\\
&\quad+|p_X(x_0)-p_X(y_0)|
M_{r_i(y_0),3\lambda_{\mathrm{PI}}|p_X(x_0)-p_X(y_0)|}
\operatorname{Lip}\widetilde{u}(p_X(y_0)).
\end{aligned} ∣ ⟨ u ⟩ B ( p X ( x 0 ) , r i ( x 0 )) , ν − ⟨ u ⟩ B ( p X ( x 0 ) , ∣ p X ( x 0 ) − p X ( y 0 ) ∣ ) , ν ∣ + ∣ ⟨ u ⟩ B ( p X ( x 0 ) , ∣ p X ( x 0 ) − p X ( y 0 ) ∣ ) , ν − ⟨ u ⟩ B ( p X ( y 0 ) , ∣ p X ( x 0 ) − p X ( y 0 ) ∣ ) , ν ∣ + ∣ ⟨ u ⟩ B ( p X ( y 0 ) , ∣ p X ( x 0 ) − p X ( y 0 ) ∣ ) , ν − ⟨ u ⟩ B ( p X ( y 0 ) , r i ( y 0 )) , ν ∣ ≲ C 0 , C PI ∣ p X ( x 0 ) − p X ( y 0 ) ∣ M r i ( x 0 ) , 3 λ PI ∣ p X ( x 0 ) − p X ( y 0 ) ∣ Lip u ( p X ( x 0 )) + ∣ p X ( x 0 ) − p X ( y 0 ) ∣ M r i ( y 0 ) , 3 λ PI ∣ p X ( x 0 ) − p X ( y 0 ) ∣ Lip u ( p X ( y 0 )) .
By Lemma 3.9, Lemma 3.14 and (24) , these final terms are
≲ C 0 ( Y ) , C 2 , C 4 , C 5 , C 6 ∣ x 0 − y 0 ∣ M r i ( y 0 ) , 3 λ P I ∣ p X ( x 0 ) − p X ( y 0 ) ∣ Lip ρ ~ ( p X ( y 0 ) ) + ∣ p X ( x 0 ) − p X ( y 0 ) ∣ M r i ( y 0 ) , 3 λ P I ∣ p X ( x 0 ) − p X ( y 0 ) ∣ Lip ρ ~ ( p X ( y 0 ) ) \begin{aligned}
\lesssim C_0(Y),C_2,C_4,C_5,C_6\left|x_0-y_0\right|M_{r_i(y_0),3\lambda_{\mathrm{PI}}}\left|p_X(x_0)-p_X(y_0)\right|\operatorname{Lip}\widetilde{\rho}(p_X(y_0)) \\
&+\left|p_X(x_0)-p_X(y_0)\right|M_{r_i(y_0),3\lambda_{\mathrm{PI}}\left|p_X(x_0)-p_X(y_0)\right|}\operatorname{Lip}\widetilde{\rho}(p_X(y_0))
\end{aligned} ≲ C 0 ( Y ) , C 2 , C 4 , C 5 , C 6 ∣ x 0 − y 0 ∣ M r i ( y 0 ) , 3 λ PI ∣ p X ( x 0 ) − p X ( y 0 ) ∣ Lip ρ ( p X ( y 0 )) + ∣ p X ( x 0 ) − p X ( y 0 ) ∣ M r i ( y 0 ) , 3 λ PI ∣ p X ( x 0 ) − p X ( y 0 ) ∣ Lip ρ ( p X ( y 0 ))
≲ C 0 , C 0 ( Y ) , C 1 , C 4 , c 4 , λ ∣ x 0 − y 0 ∣ ( M 24 λ P I C 9 ∣ x 0 − y 0 ∣ Lip u ( x 0 ) + M 24 λ P I C 9 ∣ x 0 − y 0 ∣ Lip u ( y 0 ) ) . \lesssim C_0,C_0(Y),C_1,C_4,c_4,\lambda\left|x_0-y_0\right|\left(M_{24\lambda_{\mathrm{PI}}C_9\left|x_0-y_0\right|}\operatorname{Lip}u(x_0)+M_{24\lambda_{\mathrm{PI}}C_9\left|x_0-y_0\right|}\operatorname{Lip}u(y_0)\right). ≲ C 0 , C 0 ( Y ) , C 1 , C 4 , c 4 , λ ∣ x 0 − y 0 ∣ ( M 24 λ PI C 9 ∣ x 0 − y 0 ∣ Lip u ( x 0 ) + M 24 λ PI C 9 ∣ x 0 − y 0 ∣ Lip u ( y 0 ) ) .
Recalling that C 9 = C 9 ( λ , C 5 , C 6 ) C_9=C_9(\lambda,C_5,C_6) C 9 = C 9 ( λ , C 5 , C 6 ) , this finishes the proof. □
The stopping-time regions
With the preliminaries out of the way, we can start making some inroads towards Theorem 1.3. Thus, for the rest of the paper, E ⊆ R n E\subseteq\mathbb{R}^n E ⊆ R n will be a fixed set satisfying the hypotheses of that theorem, i.e., a uniformly d d d -rectifiable subset of B ( 0 , 1 ) ⊆ R n B(0,1)\subseteq\mathbb{R}^n B ( 0 , 1 ) ⊆ R n . We will also fix system of Christ-David cubes D \mathcal{D} D on E E E . By scaling and translating, we can assume there is Q 0 ∈ D Q_0\in\mathcal{D} Q 0 ∈ D such that
Q 0 = E and B Q 0 = B ( 0 , 5 ) . (25) Q_0=E \qquad\text{and}\qquad B_{Q_0}=B(0,5). \tag*{(25)} Q 0 = E and B Q 0 = B ( 0 , 5 ) . (25)
Our goal in this section is to define a relevant collection of stopping-time regions which will be used in the construction of the approximating surface.
The cubes Layer(N), Stop(N) and Top(N)
Before defining the stopping-time regions we need to introduce several collection of cubes in D ( Q 0 ) \mathcal{D}(Q_0) D ( Q 0 ) which we call Layer(N), Stop(N) and Top(N). We also gather various useful properties that will be used throughout the remainder of the construction.
Constants 4.1. Let α 0 , ε 0 , θ 0 , τ 0 ∈ ( 0 , 1 ) \alpha_0,\varepsilon_0,\theta_0,\tau_0\in(0,1) α 0 , ε 0 , θ 0 , τ 0 ∈ ( 0 , 1 ) , A 0 > 1 A_0>1 A 0 > 1 and M 0 = 20 A 0 M_0=20A_0 M 0 = 20 A 0 . Each of α 0 , ε 0 , θ 0 , τ 0 \alpha_0,\varepsilon_0,\theta_0,\tau_0 α 0 , ε 0 , θ 0 , τ 0 will be chosen sufficiently small with ε 0 = ε 0 ( A 0 , α 0 , θ 0 , τ 0 ) \varepsilon_0=\varepsilon_0(A_0,\alpha_0,\theta_0,\tau_0) ε 0 = ε 0 ( A 0 , α 0 , θ 0 , τ 0 ) . We will choose A 0 A_0 A 0 sufficiently large.
Notation 4.2. Set
G = { Q ∈ D : b β E ( M 0 B Q ) ≤ ε 0 } . \mathcal{G}=\left\{Q\in\mathcal{D}:b\beta_E(M_0B_Q)\leq\varepsilon_0\right\}. G = { Q ∈ D : b β E ( M 0 B Q ) ≤ ε 0 } .
and
B = D ∖ G . \mathcal{B}=\mathcal{D}\setminus\mathcal{G}. B = D ∖ G .
For Q ∈ G Q\in\mathcal{G} Q ∈ G , let P Q P_Q P Q be a d d d -plane such that
b β E ( M 0 B Q , P Q ) ≤ 2 b β E ( M 0 B Q ) ≤ 2 ε 0 . b\beta_E(M_0B_Q,P_Q)\leq2b\beta_E(M_0B_Q)\leq2\varepsilon_0. b β E ( M 0 B Q , P Q ) ≤ 2 b β E ( M 0 B Q ) ≤ 2 ε 0 .
The starting point is the following result due to David and Semmes, which is proven in Sections 7 - 11 of [19 ] .
Lemma 4.3. For A 0 A_0 A 0 sufficiently large, and α 0 \alpha_0 α 0 and ε 0 = ε 0 ( A 0 , α 0 ) \varepsilon_0=\varepsilon_0(A_0,\alpha_0) ε 0 = ε 0 ( A 0 , α 0 ) sufficiently small, there exists a collection of disjoint stopping-time regions F \mathcal{F} F such that the following holds.
(1) The family F \mathcal{F} F partitions D ( Q 0 ) \mathcal{D}(Q_0) D ( Q 0 ) .
(2) The collection of cubes { Q ( S ′ ) } S ′ ∈ F \{Q(S')\}_{S'\in\mathcal{F}} { Q ( S ′ ) } S ′ ∈ F satisfy a Carleson packing condition.
(3) If Q ( S ′ ) ∈ B Q(S')\in\mathcal{B} Q ( S ′ ) ∈ B then S ′ = { Q ( S ′ ) } S'=\{Q(S')\} S ′ = { Q ( S ′ )} .
(4) We have
⋃ S ′ ∈ F Q ( S ′ ) ∈ G S ′ ⊆ G . \bigcup_{\substack{S'\in\mathcal{F}\\Q(S')\in\mathcal{G}}}S'\subseteq\mathcal{G}. S ′ ∈ F Q ( S ′ ) ∈ G ⋃ S ′ ⊆ G .
For each S ′ ∈ F such that Q ( S ′ ) ∈ G and each Q ∈ S ′ we have (26) \text{For each } S' \in\mathcal{F} \text{ such that } Q(S') \in\mathcal{G} \text{ and each } Q \in S' \text{ we have} \tag*{(26)} For each S ′ ∈ F such that Q ( S ′ ) ∈ G and each Q ∈ S ′ we have (26)
∠ ( P Q ( S ′ ) , P Q ) ≤ α 0 . \angle(P_{Q(S')},P_Q) \le\alpha_0. ∠ ( P Q ( S ′ ) , P Q ) ≤ α 0 .
Remark 4.4. The above formulation differs slightly from the one in [19 ] since we are treating bad cubes as stopping-time regions with a single cube. For clarity, we have chosen to denote stopping-time regions in F \mathcal{F} F with a prime and we will denote stopping-time regions in T \mathcal{T} T (to be defined) without a prime.
Constants 4.5. For the remainder of the paper we will suppose the A 0 A_0 A 0 , α 0 \alpha_0 α 0 and ε 0 \varepsilon_0 ε 0 are chosen to satisfy the conditions of Lemma 4.3 . We will fix F \mathcal{F} F as in the conclusion of Lemma 4.3 .
Notation 4.6. For Q ∈ D ( Q 0 ) Q \in\mathcal{D}(Q_0) Q ∈ D ( Q 0 ) , let
S Q ′ = { R ∈ S ′ : R ⊆ Q } , S'_Q = \{R \in S' : R \subseteq Q\}, S Q ′ = { R ∈ S ′ : R ⊆ Q } ,
where S ′ ∈ F S' \in\mathcal{F} S ′ ∈ F is such that Q ∈ S ′ Q \in S' Q ∈ S ′ .
Notation 4.7. For each integer N ≥ 0 N \ge0 N ≥ 0 , define collections of cubes Layer ( N ) \operatorname{Layer}(N) Layer ( N ) , Stop ( N ) \operatorname{Stop}(N) Stop ( N ) and Up ( N ) \operatorname{Up}(N) Up ( N ) as follows. First, set
Top ( 0 ) = { Q 0 } . \operatorname{Top}(0) = \{Q_0\}. Top ( 0 ) = { Q 0 } .
Then, assuming Top ( N ) \operatorname{Top}(N) Top ( N ) has been defined for some N ≥ 0 N \ge0 N ≥ 0 , let
Layer ( N ) = ⋃ Q ∈ Top ( N ) { S Q ′ : Q ∈ Top ( N ) } ; \operatorname{Layer}(N) = \bigcup_{Q \in\operatorname{Top}(N)} \{S'_Q : Q \in\operatorname{Top}(N)\}; Layer ( N ) = Q ∈ Top ( N ) ⋃ { S Q ′ : Q ∈ Top ( N )} ;
Stop ( N ) = { Q ∈ D : Q max. such that there is Q ′ ∈ Sibling ( Q ) with ℓ ( Q ′ ) < τ 0 d Layer ( N ) ( Q ′ ) } ; \operatorname{Stop}(N) = \{Q \in\mathcal{D} : Q\ \text{max. such that there is } Q' \in\operatorname{Sibling}(Q)
\text{ with } \ell(Q') < \tau_0 d_{\operatorname{Layer}(N)}(Q')\}; Stop ( N ) = { Q ∈ D : Q max. such that there is Q ′ ∈ Sibling ( Q ) with ℓ ( Q ′ ) < τ 0 d Layer ( N ) ( Q ′ )} ;
Top ( N + 1 ) = ⋃ Q ∈ Stop ( N ) Child ( Q ) . \operatorname{Top}(N+1) = \bigcup_{Q \in\operatorname{Stop}(N)} \operatorname{Child}(Q). Top ( N + 1 ) = Q ∈ Stop ( N ) ⋃ Child ( Q ) .
For brevity, in the remainder of the paper we will denote
d N : = d Layer ( N ) . d_N := d_{\operatorname{Layer}(N)}. d N := d Layer ( N ) .
The following lemma says that the size of cubes in Stop ( N ) \operatorname{Stop}(N) Stop ( N ) is determined by d N d_N d N , similar to a Whitney decomposition.
Lemma 4.8 ([4 ] ). For τ 0 \tau_0 τ 0 sufficiently small, if R ∈ Stop ( N ) R \in\operatorname{Stop}(N) R ∈ Stop ( N ) then
1 2 ℓ ( R ) ≤ τ 0 d N ( R ) ≤ 2 ρ ℓ ( R ) . \frac{1}{2}\ell(R) \le\tau_0 d_N(R) \le\frac{2}{\rho}\ell(R). 2 1 ℓ ( R ) ≤ τ 0 d N ( R ) ≤ ρ 2 ℓ ( R ) .
Moreover, if C > 1 C > 1 C > 1 and τ 0 = τ 0 ( C ) \tau_0 = \tau_0(C) τ 0 = τ 0 ( C ) is sufficiently small then
1 4 ℓ ( R ) ≤ τ 0 d N ( x ) ≤ 4 ρ ℓ ( R ) \frac{1}{4}\ell(R) \le\tau_0 d_N(x) \le\frac{4}{\rho}\ell(R) 4 1 ℓ ( R ) ≤ τ 0 d N ( x ) ≤ ρ 4 ℓ ( R )
for all R ∈ Stop ( N ) R \in\operatorname{Stop}(N) R ∈ Stop ( N ) and x ∈ C B R x \in C B_R x ∈ C B R .
The following result says that nearby cubes in Stop ( N ) \operatorname{Stop}(N) Stop ( N ) have comparable sizes.
Lemma 4.9 ([4 ] ). Let C > 1 C > 1 C > 1 . If τ 0 = τ 0 ( C ) \tau_0 = \tau_0(C) τ 0 = τ 0 ( C ) is sufficiently small, then
C 5 − 1 ℓ ( R ) ≤ ℓ ( R ′ ) ≤ C 5 ℓ ( R ) C_5^{-1}\ell(R) \le\ell(R') \le C_5\ell(R) C 5 − 1 ℓ ( R ) ≤ ℓ ( R ′ ) ≤ C 5 ℓ ( R )
for all R , R ′ ∈ Stop ( N ) R,R' \in\operatorname{Stop}(N) R , R ′ ∈ Stop ( N ) satisfying C B R ∩ C B R ′ ≠ ∅ C B_R \cap C B_{R'} \ne\varnothing C B R ∩ C B R ′ = ∅ . Here C 5 > 1 C_5 > 1 C 5 > 1 is an absolute constant independent of both C C C and τ 0 \tau_0 τ 0 .
Remark 4.10. The above statement remains true with Top ( N ) \operatorname{Top}(N) Top ( N ) in place of Stop ( N ) \operatorname{Stop}(N) Stop ( N ) . Indeed, if R , R ′ ∈ Top ( N ) R,R' \in\operatorname{Top}(N) R , R ′ ∈ Top ( N ) satisfy C B R ∩ C B R ′ ≠ ∅ C B_R \cap C B_{R'} \ne\varnothing C B R ∩ C B R ′ = ∅ , then Parent ( R ) , Parent ( R ′ ) ∈ Stop ( N − 1 ) \operatorname{Parent}(R),\operatorname{Parent}(R') \in\operatorname{Stop}(N-1) Parent ( R ) , Parent ( R ′ ) ∈ Stop ( N − 1 ) and C B Parent ( R ) ∩ C B Parent ( R ′ ) ≠ ∅ C B_{\operatorname{Parent}(R)} \cap C B_{\operatorname{Parent}(R')} \ne\varnothing C B Parent ( R ) ∩ C B Parent ( R ′ ) = ∅ . One then applies Lemma 4.8 for cubes in Stop ( N − 1 ) \operatorname{Stop}(N-1) Stop ( N − 1 ) and uses the fact that ℓ ( Q ) = ρ ℓ ( Parent ( Q ) ) \ell(Q) = \rho\ell(\operatorname{Parent}(Q)) ℓ ( Q ) = ρ ℓ ( Parent ( Q )) for any Q ∈ D Q \in\mathcal{D} Q ∈ D .
The extended stopping-time regions
We now introduce the stopping-time regions S Q S_Q S Q for each Q ∈ Top ( N ) Q \in\operatorname{Top}(N) Q ∈ Top ( N ) , which is obtained from S Q ′ S'_Q S Q ′ by adding all those cubes between min ( S Q ′ ) \min(S'_Q) min ( S Q ′ ) and Stop ( N ) \operatorname{Stop}(N) Stop ( N ) . More precisely,
S Q = { R ⊆ Q : R ⊇ T for some T ∈ Stop ( N ) } ⊇ S Q ′ . S_Q = \{R \subseteq Q : R \supseteq T \text{ for some } T \in\operatorname{Stop}(N)\} \supseteq S'_Q. S Q = { R ⊆ Q : R ⊇ T for some T ∈ Stop ( N )} ⊇ S Q ′ .
For N ≥ 0 N \geq0 N ≥ 0 , let
T N = { S Q : Q ∈ Top ( N ) } and T = ⋃ N ≥ 0 T N . \mathcal{T}^{N} = \{S_Q : Q \in\operatorname{Top}(N)\}
\qquad\text{and}\qquad
\mathcal{T} = \bigcup_{N \geq0} \mathcal{T}^{N}. T N = { S Q : Q ∈ Top ( N )} and T = N ≥ 0 ⋃ T N .
Lemma Lemma 4.11 and Lemma Lemma 4.12 below tell us that the collection of stopping-time regions T \mathcal{T} T satisfy a Carleson packing condition and that there is a graph which well approximates the stopping-time regions near a good stopping-time regions (i.e., a region in which the top cube is flat). Moreover, the approximation between the graph and the set is bi-lateral . For the reader familiar with corona decompositions, this formulation implies that usual corona decomposition. The point is that (in this formulation) nearby top cubes have comparable sizes and a graph approximations nearby stopping-time regions as well. Since the proofs are essentially standard (and implicit in [4 ] ), we have decided not to include them.
Lemma 4.11. The family T \mathcal{T} T partitions D ( Q 0 ) \mathcal{D}(Q_0) D ( Q 0 ) and the collection of cubes { Q ( S ) } S ∈ T \{Q(S)\}_{S \in\mathcal{T}} { Q ( S ) } S ∈ T satisfies a Carleson packing condition.
Lemma 4.12. Let M ≥ 1 M \geq1 M ≥ 1 and { Q k } k = 0 K ⊆ Top ( N ) \{Q_k\}_{k=0}^{K} \subseteq\operatorname{Top}(N) { Q k } k = 0 K ⊆ Top ( N ) such that Q 0 ∈ G Q_0 \in\mathcal{G} Q 0 ∈ G and M B Q i ∩ M B Q 0 ≠ ∅ MB_{Q_i} \cap MB_{Q_0} \ne\varnothing M B Q i ∩ M B Q 0 = ∅ . If A 0 = A 0 ( M ) A_0 = A_0(M) A 0 = A 0 ( M ) is large enough and τ 0 = τ 0 ( M ) \tau_0 = \tau_0(M) τ 0 = τ 0 ( M ) , ε 0 = ε 0 ( A 0 , τ 0 , θ 0 ) \varepsilon_0 = \varepsilon_0(A_0,\tau_0,\theta_0) ε 0 = ε 0 ( A 0 , τ 0 , θ 0 ) are small enough, there exists a C α 0 C\alpha_0 C α 0 -Lipschitz map g 0 : P Q 0 → P Q 0 ⊥ g_0 : P_{Q_0} \to P_{Q_0}^{\perp} g 0 : P Q 0 → P Q 0 ⊥ with ∥ g ∥ ∞ ≤ θ 0 ℓ ( Q 0 ) \|g\|_{\infty} \leq\theta_0\ell(Q_0) ∥ g ∥ ∞ ≤ θ 0 ℓ ( Q 0 ) such that the following conditions hold.
(1) If Q ∈ S i Q \in S_i Q ∈ S i and x ∈ E ∩ A 0 B Q x \in E \cap A_0B_Q x ∈ E ∩ A 0 B Q , then
∣ x − ( π 0 ( x ) + g 0 ( π 0 ( x ) ) ) ∣ ≤ θ 0 ℓ ( Q ) , \left|x-\left(\pi_0(x)+g_0(\pi_0(x))\right)\right| \leq\theta_0\ell(Q), ∣ x − ( π 0 ( x ) + g 0 ( π 0 ( x )) ) ∣ ≤ θ 0 ℓ ( Q ) , where π 0 \pi_0 π 0 is the orthogonal projection onto P Q 0 P_{Q_0} P Q 0 . In particular, if Γ 0 \Gamma_0 Γ 0 is the graph of g 0 g_0 g 0 over P Q 0 P_{Q_0} P Q 0 , then
dist ( x , Γ 0 ) ≤ θ 0 ℓ ( Q ) . \operatorname{dist}(x,\Gamma_0) \leq\theta_0\ell(Q). dist ( x , Γ 0 ) ≤ θ 0 ℓ ( Q ) . (2) If Q ∈ S i Q \in S_i Q ∈ S i and x ∈ Γ 0 ∩ A 0 B Q x \in\Gamma_0 \cap A_0B_Q x ∈ Γ 0 ∩ A 0 B Q , then
dist ( x , E ) ≤ θ 0 ℓ ( Q ) . \operatorname{dist}(x,E) \leq\theta_0\ell(Q). dist ( x , E ) ≤ θ 0 ℓ ( Q ) .
The truncated (extended) stopping-time regions
In later sections it will be convenient to truncate the stopping-time regions from the previous section at some specified scale. Associated to these will be a truncated distance function.
Notation 4.13. For k 0 ∈ N k_0 \in\mathbb{N} k 0 ∈ N , let
T N ( k 0 ) : = { S ∈ T N : Q ( T ) ∈ D ≤ k 0 } . \mathcal{T}^{N}(k_0) := \{S \in\mathcal{T}^{N} : Q(\mathcal{T}) \in\mathcal{D}_{\leq k_0}\}. T N ( k 0 ) := { S ∈ T N : Q ( T ) ∈ D ≤ k 0 } .
For S ∈ T N ( k 0 ) S \in\mathcal{T}^{N}(k_0) S ∈ T N ( k 0 ) , let
S ( k 0 ) : = S ∩ D ≤ k 0 S(k_0) := S \cap\mathcal{D}_{\leq k_0} S ( k 0 ) := S ∩ D ≤ k 0
and
Stop ( N , k 0 ) : = ⋃ S ∈ T N ( k 0 ) min ( S ( k 0 ) ) . \operatorname{Stop}(N,k_0) := \bigcup_{S \in\mathcal{T}^{N}(k_0)} \min(S(k_0)). Stop ( N , k 0 ) := S ∈ T N ( k 0 ) ⋃ min ( S ( k 0 )) .
Also, let
Top ( N , k 0 ) : = { Q ( T ) : T ∈ T N ( k 0 ) } . \operatorname{Top}(N,k_0):=\{Q(T):T\in\mathcal{T}^{N}(k_0)\}. Top ( N , k 0 ) := { Q ( T ) : T ∈ T N ( k 0 )} .
Finally, define
d N , k 0 : = max { d N , 5 ρ k 0 / τ 0 } . d_{N,k_0}:=\max\{d_N,5^{\rho k_0}/\tau_0\}. d N , k 0 := max { d N , 5 ρ k 0 / τ 0 } .
We shall now prove variants of Lemma 4.8 and Lemma 4.9 for these truncated regions and the truncated distance function.
Lemma 4.14. Let S ∈ T N ( k 0 ) S\in\mathcal{T}^{N}(k_0) S ∈ T N ( k 0 ) and Q ∈ Stop ( N , k 0 ) Q\in\operatorname{Stop}(N,k_0) Q ∈ Stop ( N , k 0 ) such that Q ⊆ Q ( S ) Q\subseteq Q(S) Q ⊆ Q ( S ) . Then,
ℓ ( Q ) ∼ τ 0 d N , k 0 ( Q ) : = max { τ 0 d N ( Q ) , 5 ρ k 0 } . (27) \ell(Q)\sim\tau_0d_{N,k_0}(Q):=\max\{\tau_0d_N(Q),5^{\rho k_0}\}. \tag*{(27)} ℓ ( Q ) ∼ τ 0 d N , k 0 ( Q ) := max { τ 0 d N ( Q ) , 5 ρ k 0 } . (27) More generally, if C > 1 C>1 C > 1 and τ 0 = τ 0 ( C ) \tau_0=\tau_0(C) τ 0 = τ 0 ( C ) is sufficiently small, then
τ 0 d N , k 0 ( x ) ∼ ℓ ( Q ) (28) \tau_0d_{N,k_0}(x)\sim\ell(Q) \tag*{(28)} τ 0 d N , k 0 ( x ) ∼ ℓ ( Q ) (28) for all x ∈ C B Q x\in CB_Q x ∈ C B Q .
Proof. We start with (27) . Since Q Q Q contains some element of Stop ( N ) \operatorname{Stop}(N) Stop ( N ) , it follows from maximality that ℓ ( Q ) ≥ τ 0 d N ( Q ) \ell(Q)\geq\tau_0d_N(Q) ℓ ( Q ) ≥ τ 0 d N ( Q ) . Since Q ∈ D ≤ k 0 Q\in\mathcal{D}_{\leq k_0} Q ∈ D ≤ k 0 , we also have ℓ ( Q ) ≥ 5 ρ k 0 \ell(Q)\geq5^{\rho k_0} ℓ ( Q ) ≥ 5 ρ k 0 . These two inequalities imply ℓ ( Q ) ≥ d N , k 0 ( Q ) \ell(Q)\geq d_{N,k_0}(Q) ℓ ( Q ) ≥ d N , k 0 ( Q ) . For the converse, note that if ℓ ( Q ) = 5 ρ k 0 \ell(Q)=5^{\rho k_0} ℓ ( Q ) = 5 ρ k 0 then the inequality ℓ ( Q ) ≤ τ 0 d N , k 0 ( Q ) \ell(Q)\leq\tau_0d_{N,k_0}(Q) ℓ ( Q ) ≤ τ 0 d N , k 0 ( Q ) is immediate. Now, let us suppose that ℓ ( Q ) > 5 ρ k 0 \ell(Q)>5^{\rho k_0} ℓ ( Q ) > 5 ρ k 0 . In this case, Q ∈ Stop ( N ) Q\in\operatorname{Stop}(N) Q ∈ Stop ( N ) . Thus, ℓ ( Q ) ≲ τ 0 d N ( Q ) ≤ τ 0 d N , k 0 ( Q ) \ell(Q)\lesssim\tau_0d_N(Q)\leq\tau_0d_{N,k_0}(Q) ℓ ( Q ) ≲ τ 0 d N ( Q ) ≤ τ 0 d N , k 0 ( Q ) by Lemma 4.8.
We now prove (28) . Let C > 1 C>1 C > 1 , x ∈ C B Q x\in CB_Q x ∈ C B Q and suppose initially that τ 0 \tau_0 τ 0 small enough for (27) . Let z ∈ Q z\in Q z ∈ Q such that d N , k 0 ( z ) = d N , k 0 ( Q ) d_{N,k_0}(z)=d_{N,k_0}(Q) d N , k 0 ( z ) = d N , k 0 ( Q ) , so that
c ℓ ( Q ) ≤ d N , k 0 ( z ) ≤ C ′ ℓ ( Q ) . (29) c\ell(Q)\leq d_{N,k_0}(z)\leq C'\ell(Q). \tag*{(29)} c ℓ ( Q ) ≤ d N , k 0 ( z ) ≤ C ′ ℓ ( Q ) . (29)
Since τ 0 d N , k 0 \tau_0d_{N,k_0} τ 0 d N , k 0 is τ 0 \tau_0 τ 0 -Lipschitz, we have
∣ τ 0 d N , k 0 ( x ) − τ 0 d N , k 0 ( z ) ∣ ≤ 2 C τ 0 ℓ ( Q ) < min { c , C ′ } ℓ ( Q ) / 2 (30) \lvert\tau_0d_{N,k_0}(x)-\tau_0d_{N,k_0}(z)\rvert\leq2C\tau_0\ell(Q)<\min\{c,C'\}\ell(Q)/2 \tag*{(30)} ∣ τ 0 d N , k 0 ( x ) − τ 0 d N , k 0 ( z )∣ ≤ 2 C τ 0 ℓ ( Q ) < min { c , C ′ } ℓ ( Q ) /2 (30)
(assuming τ 0 < min { c , C ′ } / ( 4 C ) \tau_0<\min\{c,C'\}/(4C) τ 0 < min { c , C ′ } / ( 4 C ) ). The estimate (28) then follows directly from (29) and (30) after applying the triangle inequality. □
Lemma 4.15. Given C > 1 C>1 C > 1 , if τ 0 = τ 0 ( C ) \tau_0=\tau_0(C) τ 0 = τ 0 ( C ) is sufficiently small, then C 5 − 1 ℓ ( R ) ≤ ℓ ( R ′ ) ≤ C 5 ℓ ( R ) C_5^{-1}\ell(R)\leq\ell(R')\leq C_5\ell(R) C 5 − 1 ℓ ( R ) ≤ ℓ ( R ′ ) ≤ C 5 ℓ ( R ) for all R , R ′ ∈ Stop ( N , k 0 ) R,R'\in\operatorname{Stop}(N,k_0) R , R ′ ∈ Stop ( N , k 0 ) satisfying C B R ∩ C B R ′ ≠ ∅ CB_R\cap CB_{R'}\neq\varnothing C B R ∩ C B R ′ = ∅ . Here C 5 > 1 C_5>1 C 5 > 1 is an absolute constant independent of both C C C and τ 0 \tau_0 τ 0 . In particular, the set { C B R } R ∈ Stop ( N , k 0 ) \{CB_R\}_{R\in\operatorname{Stop}(N,k_0)} { C B R } R ∈ Stop ( N , k 0 ) has bounded overlap with constant depending on d d d , c 0 c_0 c 0 , C 5 C_5 C 5 and C C C .
Proof. The second statement (regarding bounded overlap) is an immediate consequence of the first and Lemma 2.8 . For the first, let R , R ′ ∈ Stop ( N , k 0 ) R,R'\in\operatorname{Stop}(N,k_0) R , R ′ ∈ Stop ( N , k 0 ) satisfy the above hypotheses. If R , R ′ ∈ D < k 0 R,R'\in\mathcal{D}_{<k_0} R , R ′ ∈ D < k 0 (so that R , R ′ ∈ Stop ( N ) R,R'\in\operatorname{Stop}(N) R , R ′ ∈ Stop ( N ) ) then the result follows directly from Lemma 4.9. If R , R ′ ∈ D k 0 R,R'\in\mathcal{D}_{k_0} R , R ′ ∈ D k 0 then ℓ ( R ) = ℓ ( R ′ ) \ell(R)=\ell(R') ℓ ( R ) = ℓ ( R ′ ) and the result is obvious. Suppose instead that R ∈ D < k 0 R\in\mathcal{D}_{<k_0} R ∈ D < k 0 (so that R ∈ Stop ( N ) R\in\operatorname{Stop}(N) R ∈ Stop ( N ) ) and R ′ ∈ D k 0 R'\in\mathcal{D}_{k_0} R ′ ∈ D k 0 . By symmetry, this is the only remaining case that we need to consider. Since ℓ ( R ) > ℓ ( R ′ ) \ell(R)>\ell(R') ℓ ( R ) > ℓ ( R ′ ) , we only need to prove the reverse inequality. Since C B R ∩ C B R ′ ≠ ∅ CB_R\cap CB_{R'}\neq\varnothing C B R ∩ C B R ′ = ∅ (and ℓ ( R ) > ℓ ( R ′ ) \ell(R)>\ell(R') ℓ ( R ) > ℓ ( R ′ ) ), it follows from the triangle inequality that R ′ ⊆ 3 C B R R'\subseteq3CB_R R ′ ⊆ 3 C B R . Let R ′ ′ ∈ Stop ( N ) R''\in\operatorname{Stop}(N) R ′′ ∈ Stop ( N ) such that R ′ ′ ⊆ R ′ ⊆ 3 C B R R''\subseteq R'\subseteq3CB_R R ′′ ⊆ R ′ ⊆ 3 C B R . Choosing τ 0 = τ 0 ( 3 C ) \tau_0=\tau_0(3C) τ 0 = τ 0 ( 3 C ) as in Lemma 4.9, it now follows that
ℓ ( R ′ ) ≥ ℓ ( R ′ ′ ) ≥ C 5 − 1 ℓ ( R ) , \ell(R')\geq\ell(R'')\geq C_5^{-1}\ell(R), ℓ ( R ′ ) ≥ ℓ ( R ′′ ) ≥ C 5 − 1 ℓ ( R ) ,
as required. □
It turns out that graph constructed in Lemma 4.12 are (locally) covered by the corresponding minimal cubes. More precisely, we have the following.
Lemma 4.16. Let M ≥ 1 M \ge1 M ≥ 1 and { Q k } k = 0 K \{Q_k\}_{k=0}^{K} { Q k } k = 0 K as in Lemma 4.12 , and choose A 0 = A 0 ( M ) A_0=A_0(M) A 0 = A 0 ( M ) and τ 0 \tau_0 τ 0 such that the conclusions there hold. If B = B ( x Q 0 , r ) B=B(x_{Q_0},r) B = B ( x Q 0 , r ) is a ball such that r ∈ ( θ 0 ℓ ( Q 0 ) , A 0 ℓ ( Q 0 ) ) r \in(\theta_0\ell(Q_0),A_0\ell(Q_0)) r ∈ ( θ 0 ℓ ( Q 0 ) , A 0 ℓ ( Q 0 )) and
E ∩ 2 B ⊆ ⋃ k = 0 K Q k , E \cap2B \subseteq\bigcup_{k=0}^{K} Q_k, E ∩ 2 B ⊆ k = 0 ⋃ K Q k , then
Γ 0 ∩ B ⊆ ⋃ k = 0 K ⋃ R ∈ min ( S Q k ( k 0 ) ) Γ 0 ∩ B ∩ 2 B R ≠ ∅ 2 B R . \Gamma_0 \cap B \subseteq\bigcup_{k=0}^{K}\ \bigcup_{\substack{R\in\min(S_{Q_k}(k_0))\\ \Gamma_0\cap B\cap2B_R\ne\varnothing}} 2B_R. Γ 0 ∩ B ⊆ k = 0 ⋃ K R ∈ m i n ( S Q k ( k 0 )) Γ 0 ∩ B ∩ 2 B R = ∅ ⋃ 2 B R .
Proof. Fix a ball B B B (as in the statement) and a point z ∈ Γ 0 ∩ B z \in\Gamma_0 \cap B z ∈ Γ 0 ∩ B . Let m ≥ 0 m \ge0 m ≥ 0 be the largest integer such that
z ∈ ⋃ k = 0 K ⋃ Q ∈ S Q k A 0 B Q . z \in\bigcup_{k=0}^{K}\ \bigcup_{Q\in S_{Q_k}} A_0B_Q. z ∈ k = 0 ⋃ K Q ∈ S Q k ⋃ A 0 B Q .
which exists since each S Q k ( k 0 ) S_{Q_k}(k_0) S Q k ( k 0 ) is a finite collection of cubes. It satisfies 5 ρ 0 m ≤ ℓ ( Q 0 ) 5\rho_0^m \le\ell(Q_0) 5 ρ 0 m ≤ ℓ ( Q 0 ) , since z ∈ A 0 B Q 0 z \in A_0B_{Q_0} z ∈ A 0 B Q 0 . By Lemma 4.12 (2), there exists y ∈ E y \in E y ∈ E such that ∣ y − z ∣ ≤ 5 θ 0 ρ 0 m ≤ θ 0 ℓ ( Q 0 ) |y-z| \le5\theta_0\rho_0^m \le\theta_0\ell(Q_0) ∣ y − z ∣ ≤ 5 θ 0 ρ 0 m ≤ θ 0 ℓ ( Q 0 ) . Since z ∈ B z \in B z ∈ B and r B ≥ θ 0 ℓ ( Q 0 ) r_B \ge\theta_0\ell(Q_0) r B ≥ θ 0 ℓ ( Q 0 ) , this implies that y ∈ E ∩ 2 B ⊆ ⋃ k = 0 K Q k y \in E \cap2B \subseteq\bigcup_{k=0}^{K}Q_k y ∈ E ∩ 2 B ⊆ ⋃ k = 0 K Q k . Let T ∈ D m + 1 T \in\mathcal{D}_{m+1} T ∈ D m + 1 such that y ∈ T ⊆ Q k y \in T \subseteq Q_k y ∈ T ⊆ Q k for some k k k . For θ 0 \theta_0 θ 0 small enough, it follows from the estimate ∣ y − z ∣ ≤ 5 θ 0 ρ 0 m |y-z| \le5\theta_0\rho_0^m ∣ y − z ∣ ≤ 5 θ 0 ρ 0 m that z ∈ 2 B T ⊆ A 0 B T z \in2B_T \subseteq A_0B_T z ∈ 2 B T ⊆ A 0 B T . So, by maximality of m m m , we must have T ∉ S Q k T \notin S_{Q_k} T ∈ / S Q k . In particular, there exists R ∈ min ( S Q k ) R \in\min(S_{Q_k}) R ∈ min ( S Q k ) such that T ⊆ R T \subseteq R T ⊆ R . This then satisfies z ∈ Γ 0 ∩ B ∩ 2 B R z \in\Gamma_0 \cap B \cap2B_R z ∈ Γ 0 ∩ B ∩ 2 B R , as required. □
Below we introduce a slightly variant of the cubes Stop ( N , k ) \operatorname{Stop}(N,k) Stop ( N , k ) . Indeed, we need to identify those cubes in Stop ( N , k ) \operatorname{Stop}(N,k) Stop ( N , k ) which have an element of Top ( N , k ) \operatorname{Top}(N,k) Top ( N , k ) as a child. It is from nearby these cubes that we will build bridges between the approximating spaces.
Notation 4.17. Let
Stop ′ ( N − 1 , k ) : = { Q ∈ Stop ( N − 1 , k ) : Child ( Q ) ∩ Top ( N , k ) ≠ ∅ } = Stop ( N − 1 , k ) ∩ D < k . \operatorname{Stop}'(N-1,k):=\{Q\in\operatorname{Stop}(N-1,k):\operatorname{Child}(Q)\cap\operatorname{Top}(N,k)\ne\varnothing\}
=\operatorname{Stop}(N-1,k)\cap\mathcal{D}_{<k}. Stop ′ ( N − 1 , k ) := { Q ∈ Stop ( N − 1 , k ) : Child ( Q ) ∩ Top ( N , k ) = ∅ } = Stop ( N − 1 , k ) ∩ D < k .
Nearby cubes from Stop ′ ( N , k ) \operatorname{Stop}'(N,k) Stop ′ ( N , k ) , it turns out that the distance functions d N d_N d N and d N , k d_{N,k} d N , k are comparable.
Lemma 4.18. Let C > 1 C>1 C > 1 , Q ∈ Stop ′ ( N , k ) Q\in\operatorname{Stop}'(N,k) Q ∈ Stop ′ ( N , k ) and x ∈ C B Q x\in CB_Q x ∈ C B Q . Then, for τ 0 = τ 0 ( C ) \tau_0=\tau_0(C) τ 0 = τ 0 ( C ) sufficiently small, we have
d N ( x ) = d N , k ( x ) . (31) d_N(x)=d_{N,k}(x). \tag*{(31)} d N ( x ) = d N , k ( x ) . (31) In particular, if R ∈ Top ( N , k ) R\in\operatorname{Top}(N,k) R ∈ Top ( N , k ) and T ∈ Stop ′ ( N − 1 , k ) T\in\operatorname{Stop}'(N-1,k) T ∈ Stop ′ ( N − 1 , k ) are such that Q ⊆ R Q\subseteq R Q ⊆ R , then
ℓ ( Q ) ≲ τ 0 ℓ ( R ) ≤ τ 0 ℓ ( T ) . (32) \ell(Q)\lesssim\tau_0\ell(R)\le\tau_0\ell(T). \tag*{(32)} ℓ ( Q ) ≲ τ 0 ℓ ( R ) ≤ τ 0 ℓ ( T ) . (32)
Proof. Since d N ≤ d N , k d_N\le d_{N,k} d N ≤ d N , k , the only inequality we need to prove is d N ( x ) ≥ d N , k ( x ) d_N(x)\ge d_{N,k}(x) d N ( x ) ≥ d N , k ( x ) . If d N , k ( x ) = d N ( x ) d_{N,k}(x)=d_N(x) d N , k ( x ) = d N ( x ) , then this is trivial. Suppose instead that d N , k ( x ) = 5 ρ 0 k / τ 0 d_{N,k}(x)=5\rho_0^k/\tau_0 d N , k ( x ) = 5 ρ 0 k / τ 0 . Let S ∈ T N ( k ) S \in\mathcal{T}^{N}(k) S ∈ T N ( k ) such that Q ∈ S ( k ) Q \in S(k) Q ∈ S ( k ) . Since Q ∈ Stop ′ ( N , k ) Q \in\operatorname{Stop}'(N,k) Q ∈ Stop ′ ( N , k ) (so that ℓ ( Q ) ≥ 5 ρ k − 1 \ell(Q) \ge5\rho^{k-1} ℓ ( Q ) ≥ 5 ρ k − 1 ), we actually have Q ∈ min ( S ) Q \in\min(S) Q ∈ min ( S ) , which implies Q ∈ Stop ( N ) Q \in\operatorname{Stop}(N) Q ∈ Stop ( N ) . Thus, by Lemma 4.8,
d N ( x ) ≥ ℓ ( Q ) 4 τ 0 ≥ 5 ρ k τ 0 = d N , k ( x ) , d_N(x) \ge\frac{\ell(Q)}{4\tau_0} \ge\frac{5\rho^k}{\tau_0} = d_{N,k}(x), d N ( x ) ≥ 4 τ 0 ℓ ( Q ) ≥ τ 0 5 ρ k = d N , k ( x ) ,
as required for (31) . Lemma 4.14 and (31) imply (32) since
ℓ ( Q ) ≲ τ 0 d N , k ( Q ) ≲ τ 0 d N ( Q ) ≤ τ 0 ℓ ( R ) ≤ τ 0 ℓ ( T ) . \ell(Q) \lesssim\tau_0 d_{N,k}(Q) \lesssim\tau_0 d_N(Q) \le\tau_0\ell(R) \le\tau_0\ell(T). ℓ ( Q ) ≲ τ 0 d N , k ( Q ) ≲ τ 0 d N ( Q ) ≤ τ 0 ℓ ( R ) ≤ τ 0 ℓ ( T ) .
Nearby cubes in Stop ′ ( N − 1 , k ) \operatorname{Stop}'(N-1,k) Stop ′ ( N − 1 , k ) , we can also guarantee that the distance function is d N , k d_{N,k} d N , k is smaller than d N − 1 , k d_{N-1,k} d N − 1 , k .
Lemma 4.19. Let C > 1 C > 1 C > 1 , N ∈ N N \in\mathbb{N} N ∈ N and Q ∈ Stop ′ ( N − 1 , k ) Q \in\operatorname{Stop}'(N-1,k) Q ∈ Stop ′ ( N − 1 , k ) . Then,
d N , k ( x ) < d N − 1 , k ( x ) d_{N,k}(x) < d_{N-1,k}(x) d N , k ( x ) < d N − 1 , k ( x ) for all x ∈ C B Q x \in CB_Q x ∈ C B Q .
Proof. By Lemma 4.8 and Lemma 4.18, we have
d N − 1 , k ( x ) = d N ( x ) ≥ ℓ ( Q ) / 4. d_{N-1,k}(x) = d_N(x) \ge\ell(Q)/4. d N − 1 , k ( x ) = d N ( x ) ≥ ℓ ( Q ) /4.
On the other hand, we first note that 5 ρ k ≤ ρ ℓ ( Q ) 5\rho^k \le\rho\ell(Q) 5 ρ k ≤ ρ ℓ ( Q ) . To estimate d N ( x ) d_N(x) d N ( x ) from above, consider an arbitrary cube R ∈ Top ( N , k ) ⊆ Top ( N ) R \in\operatorname{Top}(N,k) \subseteq\operatorname{Top}(N) R ∈ Top ( N , k ) ⊆ Top ( N ) such that R ⊆ Q R \subseteq Q R ⊆ Q . Since { x } ∪ B R ⊆ C B Q \{x\} \cup B_R \subseteq CB_Q { x } ∪ B R ⊆ C B Q , we have
τ 0 d N ( x ) ≤ τ 0 ( ℓ ( R ) + dist ( x , R ) ) ≤ τ 0 ( 1 + 2 C ) ℓ ( Q ) ≤ ρ ℓ ( Q ) . \tau_0d_N(x) \le\tau_0\bigl(\ell(R) + \operatorname{dist}(x,R)\bigr) \le\tau_0(1+2C)\ell(Q) \le\rho\ell(Q). τ 0 d N ( x ) ≤ τ 0 ( ℓ ( R ) + dist ( x , R ) ) ≤ τ 0 ( 1 + 2 C ) ℓ ( Q ) ≤ ρ ℓ ( Q ) .
Thus, we have d N , k ( x ) ≤ ρ ℓ ( Q ) < d N − 1 ( x ) d_{N,k}(x) \le\rho\ell(Q) < d_{N-1}(x) d N , k ( x ) ≤ ρ ℓ ( Q ) < d N − 1 ( x ) , as required.
In fact, we can prove a more quantitative estimate regarding the distance between the distance functions at different generations. The following result ensure that the graphs we construct are well-separated after pushing them into a further dimension.
Lemma 4.20. Let M , N ∈ N M,N \in\mathbb{N} M , N ∈ N such that M > N M > N M > N and C > 1 C > 1 C > 1 . For each Q ∈ Stop ′ ( N , k ) Q \in\operatorname{Stop}'(N,k) Q ∈ Stop ′ ( N , k ) and x , y ∈ C B Q x,y \in CB_Q x , y ∈ C B Q we have
τ 0 d N , k ( x ) − 2 τ 0 d M , k ( y ) ≳ τ 0 d N , k ( x ) ∼ ℓ ( Q ) . (33) \tau_0d_{N,k}(x) - 2\tau_0d_{M,k}(y) \gtrsim\tau_0d_{N,k}(x) \sim\ell(Q). \tag*{(33)} τ 0 d N , k ( x ) − 2 τ 0 d M , k ( y ) ≳ τ 0 d N , k ( x ) ∼ ℓ ( Q ) . (33) Proof. The right-most inequality is just Lemma 4.14, so the main point is to prove the left-most inequality. We start by giving a lower bound for τ 0 d N , k ( x ) \tau_0d_{N,k}(x) τ 0 d N , k ( x ) . Indeed, since Stop ′ ( N , k ) ⊆ Stop ( N ) \operatorname{Stop}'(N,k) \subseteq\operatorname{Stop}(N) Stop ′ ( N , k ) ⊆ Stop ( N ) and x ∈ C B Q x \in CB_Q x ∈ C B Q , we have by Lemma 4.8 and Lemma 4.14 that
ℓ ( Q ) / 4 ≤ τ 0 d N ( x ) ≤ τ 0 d N , k ( x ) . (34) \ell(Q)/4 \le\tau_0d_N(x) \le\tau_0d_{N,k}(x). \tag*{(34)} ℓ ( Q ) /4 ≤ τ 0 d N ( x ) ≤ τ 0 d N , k ( x ) . (34) We now show that
2 τ 0 d M , k ( y ) ≤ ℓ ( Q ) / 10. (35) 2\tau_0d_{M,k}(y) \le\ell(Q)/10. \tag*{(35)} 2 τ 0 d M , k ( y ) ≤ ℓ ( Q ) /10. (35) By Lemma 4.19, it suffices to check the above estimate with N + 1 N+1 N + 1 in place of M M M . Since Q ∈ Stop ′ ( N , k ) Q \in\operatorname{Stop}'(N,k) Q ∈ Stop ′ ( N , k ) we have ℓ ( Q ) ≥ 5 ρ k 0 − 1 \ell(Q) \ge5\rho^{k_0-1} ℓ ( Q ) ≥ 5 ρ k 0 − 1 (otherwise then can be no child of Q Q Q in Top ( N + 1 , k 0 ) \operatorname{Top}(N+1,k_0) Top ( N + 1 , k 0 ) ). Thus,
10 ρ k 0 ≤ 2 ρ ℓ ( Q ) ≤ ℓ ( Q ) / 10. (36) 10\rho^{k_0} \le2\rho\ell(Q) \le\ell(Q)/10. \tag*{(36)} 10 ρ k 0 ≤ 2 ρ ℓ ( Q ) ≤ ℓ ( Q ) /10. (36) On the other hand, if R ∈ Top ( N + 1 , k ) ⊆ Top ( N ) R \in\operatorname{Top}(N+1,k) \subseteq\operatorname{Top}(N) R ∈ Top ( N + 1 , k ) ⊆ Top ( N ) is such that R ⊆ Q R \subseteq Q R ⊆ Q , we by the definition of d N + 1 d_{N+1} d N + 1 that
2 τ 0 d N + 1 ( y ) ≤ τ 0 ( ℓ ( R ) + dist ( y , R ) ) ≤ τ 0 ( ρ ℓ ( Q ) + C ρ − 1 ℓ ( Q ) ) ≤ ℓ ( Q ) / 10. (37) 2\tau_0d_{N+1}(y) \le\tau_0\bigl(\ell(R) + \operatorname{dist}(y,R)\bigr) \le\tau_0\bigl(\rho\ell(Q) + C\rho^{-1}\ell(Q)\bigr) \le\ell(Q)/10. \tag*{(37)} 2 τ 0 d N + 1 ( y ) ≤ τ 0 ( ℓ ( R ) + dist ( y , R ) ) ≤ τ 0 ( ρ ℓ ( Q ) + C ρ − 1 ℓ ( Q ) ) ≤ ℓ ( Q ) /10. (37)
The inequality (36) and (37) imply that d N + 1 , k ( y ) ≤ ℓ ( Q ) / 10 d_{N+1,k}(y) \le\ell(Q)/10 d N + 1 , k ( y ) ≤ ℓ ( Q ) /10 , as required. Combing (34) with (35) and the right-most inequality in (33) , we get
τ 0 d N , k ( x ) − 2 τ 0 d M , k ( y ) ≥ ℓ ( Q ) / 10 ≳ τ 0 d N , k ( x ) , \tau_0 d_{N,k}(x)-2\tau_0 d_{M,k}(y)\ge\ell(Q)/10 \gtrsim\tau_0 d_{N,k}(x), τ 0 d N , k ( x ) − 2 τ 0 d M , k ( y ) ≥ ℓ ( Q ) /10 ≳ τ 0 d N , k ( x ) ,
as required.
Good and bad stopping-time regions
In order to build the set Z Z Z from Theorem 1.3 , we will need to construct approximating sets for each S ∈ T N ( k 0 ) S\in\mathcal{T}^{N}(k_0) S ∈ T N ( k 0 ) . The construction will depend on whether S S S lies in a certain good class G 2 N ( k 0 ) \mathcal{G}^{N}_{2}(k_0) G 2 N ( k 0 ) of stopping-time regions (in which the top cube is near another flat top cube which is not too small) or its complement B 2 N ( k 0 ) \mathcal{B}^{N}_{2}(k_0) B 2 N ( k 0 ) (where all flat top cubes are far away or very small). It will also be convenient to introduce here some slight variants of these classes G 1 N ( k 0 ) ⊆ G 2 N ( k 0 ) \mathcal{G}^{N}_{1}(k_0)\subseteq\mathcal{G}^{N}_{2}(k_0) G 1 N ( k 0 ) ⊆ G 2 N ( k 0 ) and B 1 N ( k 0 ) ⊇ B 2 N ( k 0 ) \mathcal{B}^{N}_{1}(k_0)\supseteq\mathcal{B}^{N}_{2}(k_0) B 1 N ( k 0 ) ⊇ B 2 N ( k 0 ) .
Constants 4.21. Let M 1 ∈ N M_1\in\mathbb{N} M 1 ∈ N to be chosen sufficiently large. We will suppose for the remainder of the section that τ 0 = τ 0 ( M 1 ) \tau_0=\tau_0(M_1) τ 0 = τ 0 ( M 1 ) is small enough such that
C 5 − 1 ℓ ( R ) ≤ ℓ ( Q ) ≤ C 5 ℓ ( R ) C_5^{-1}\ell(R)\le\ell(Q)\le C_5\ell(R) C 5 − 1 ℓ ( R ) ≤ ℓ ( Q ) ≤ C 5 ℓ ( R )
for any Q , R ∈ Stop ( N , k 0 ) Q,R\in\operatorname{Stop}(N,k_0) Q , R ∈ Stop ( N , k 0 ) (resp. Top ( N , k 0 ) \operatorname{Top}(N,k_0) Top ( N , k 0 ) ) such that 2 M 1 B Q ∩ 2 M 1 B R ≠ ∅ 2M_1B_Q\cap2M_1B_R\ne\varnothing 2 M 1 B Q ∩ 2 M 1 B R = ∅ , see Lemma 4.9.
Constants 4.22. Let k 2 = k 2 ( C 5 ) ∈ N k_2=k_2(C_5)\in\mathbb{N} k 2 = k 2 ( C 5 ) ∈ N be the small integer such that the following property holds: if Q , R ∈ Stop ( N , k 0 ) Q,R\in\operatorname{Stop}(N,k_0) Q , R ∈ Stop ( N , k 0 ) (resp. Top ( N , k 0 ) \operatorname{Top}(N,k_0) Top ( N , k 0 ) ) are such that Q ∈ D < k 2 Q\in\mathcal{D}_{<k_2} Q ∈ D < k 2 and 2 M 1 B Q ∩ 2 M 1 B R ≠ ∅ 2M_1B_Q\cap2M_1B_R\ne\varnothing 2 M 1 B Q ∩ 2 M 1 B R = ∅ , then R ∈ D ≤ k 0 R\in\mathcal{D}_{\le k_0} R ∈ D ≤ k 0 .
Notation 4.23. Let
G ( k 0 ) : = G ∩ D < k 2 and B ( k 0 ) : = G ( k 0 ) c = B ∪ D ≥ k 2 . (38) \mathcal{G}(k_0):=\mathcal{G}\cap\mathcal{D}_{<k_2}\qquad\text{and}\qquad\mathcal{B}(k_0):=\mathcal{G}(k_0)^{c}=\mathcal{B}\cup\mathcal{D}_{\ge k_2}. \tag*{(38)} G ( k 0 ) := G ∩ D < k 2 and B ( k 0 ) := G ( k 0 ) c = B ∪ D ≥ k 2 . (38)
Notation 4.24. For i ∈ { 1 , 2 } i\in\{1,2\} i ∈ { 1 , 2 } , let
\mathcal{G}^{N}_{i}(k_0):=\left\{S\in\mathcal{T}^{N}(k_0)\mathrel{\left|}\begin{gathered}\text{there exists }T\in\mathcal{T}^{N}(k_0)\text{ satisfying}\\ Q(T)\in\mathcal{G}(k_0)\text{ and }iM_1B_{Q(T)}\cap iM_1B_{Q(S)}\ne\varnothing\end{gathered}\right.\right\}. \tag*{(39)}
B i N ( k 0 ) : = T N ∖ G i N ( k 0 ) . (40) \mathcal{B}^{N}_{i}(k_0):=\mathcal{T}^{N}\setminus\mathcal{G}^{N}_{i}(k_0). \tag*{(40)} B i N ( k 0 ) := T N ∖ G i N ( k 0 ) . (40)
We show below that any cube from Stop ( N , k ) \operatorname{Stop}(N,k) Stop ( N , k ) which is sufficiently close to a good stopping-time region must be relatively small in comparison to the top cube. On the other hand, any cube in Stop ( N , k ) \operatorname{Stop}(N,k) Stop ( N , k ) which is relatively close to some stopping-time region from B 1 N ( k ) \mathcal{B}^{N}_{1}(k) B 1 N ( k ) must be relatively large compared to the top cube.
Lemma 4.25. If S ∈ G 1 N ( k 0 ) S\in\mathcal{G}^{N}_{1}(k_0) S ∈ G 1 N ( k 0 ) and Q ∈ Stop ( N , k 0 ) Q\in\operatorname{Stop}(N,k_0) Q ∈ Stop ( N , k 0 ) satisfy 3 B Q ∩ 2 ρ − 1 B Q ( S ) ≠ ∅ 3B_Q\cap2\rho^{-1}B_{Q(S)}\ne\varnothing 3 B Q ∩ 2 ρ − 1 B Q ( S ) = ∅ , then ℓ ( Q ) ≤ ρ ℓ ( Q ( S ) ) \ell(Q)\le\rho\ell(Q(S)) ℓ ( Q ) ≤ ρ ℓ ( Q ( S )) .
Proof. If ℓ ( Q ) = 5 ρ k 0 \ell(Q)=5\rho^{k_0} ℓ ( Q ) = 5 ρ k 0 then the estimate follows from the fact that ℓ ( Q ( S ) ) > 5 ρ k 0 \ell(Q(S))>5\rho^{k_0} ℓ ( Q ( S )) > 5 ρ k 0 . Suppose instead that ℓ ( Q ) > 5 ρ k 0 \ell(Q)>5\rho^{k_0} ℓ ( Q ) > 5 ρ k 0 , which implies Q ∈ Stop ′ ( N , k 0 ) Q\in\operatorname{Stop}'(N,k_0) Q ∈ Stop ′ ( N , k 0 ) . In this case, Lemma 4.18 implies ℓ ( Q ) ≲ τ 0 ℓ ( Q ( S ) ) \ell(Q)\lesssim\tau_0\ell(Q(S)) ℓ ( Q ) ≲ τ 0 ℓ ( Q ( S )) , which gives the desired estimate for τ 0 \tau_0 τ 0 small enough.
Proof. If Q ∈ D k 0 Q \in\mathcal{D}_{k_0} Q ∈ D k 0 then it is clear that ℓ ( Q ) ≤ ℓ ( Q ( S ) ) \ell(Q) \le\ell(Q(S)) ℓ ( Q ) ≤ ℓ ( Q ( S )) since Q ( S ) ∈ D ≤ k 0 Q(S) \in\mathcal{D}_{\le k_0} Q ( S ) ∈ D ≤ k 0 . Suppose instead that Q ∈ D ≤ k 0 − 1 Q \in\mathcal{D}_{\le k_0-1} Q ∈ D ≤ k 0 − 1 . In this case we know that Q ∈ Stop ( N ) Q \in\operatorname{Stop}(N) Q ∈ Stop ( N ) . By Lemma 4.8, and the fact that Q ( S ) ∈ Layer ( N ) Q(S) \in\operatorname{Layer}(N) Q ( S ) ∈ Layer ( N ) and 3 B Q ∩ 2 ρ − 1 B Q ( S ) ≠ ∅ 3B_Q \cap2\rho^{-1}B_{Q(S)} \ne\varnothing 3 B Q ∩ 2 ρ − 1 B Q ( S ) = ∅ , we have
ℓ ( Q ) ≲ τ 0 d N ( Q ) ≤ τ 0 ( ℓ ( Q ( S ) ) + 3 ℓ ( Q ) + 2 ρ − 1 ℓ ( Q ( S ) ) ) . \ell(Q) \lesssim\tau_0 d_N(Q) \le\tau_0\left(\ell(Q(S)) + 3\ell(Q) + 2\rho^{-1}\ell(Q(S))\right). ℓ ( Q ) ≲ τ 0 d N ( Q ) ≤ τ 0 ( ℓ ( Q ( S )) + 3 ℓ ( Q ) + 2 ρ − 1 ℓ ( Q ( S )) ) .
Thus, for τ 0 \tau_0 τ 0 sufficiently small (after taking 3 τ 0 ℓ ( Q ) 3\tau_0\ell(Q) 3 τ 0 ℓ ( Q ) to the left-hand side), the above inequality implies ℓ ( Q ) ≤ ℓ ( Q ( S ) ) \ell(Q) \le\ell(Q(S)) ℓ ( Q ) ≤ ℓ ( Q ( S )) .
Lemma 4.26. There is c 6 ∈ ( 0 , 1 ) c_6 \in(0,1) c 6 ∈ ( 0 , 1 ) such that the following holds. Suppose S ∈ B 1 N ( k 0 ) S \in\mathcal{B}_1^N(k_0) S ∈ B 1 N ( k 0 ) and Q ∈ Stop ( N , k 0 ) Q \in\operatorname{Stop}(N,k_0) Q ∈ Stop ( N , k 0 ) are such that Q ⊆ ( M 1 / 2 ) B Q ( S ) Q \subseteq(M_1/2)B_{Q(S)} Q ⊆ ( M 1 /2 ) B Q ( S ) . Then, ℓ ( Q ) ≥ c 6 τ 0 ℓ ( Q ( S ) ) \ell(Q) \ge c_6\tau_0\ell(Q(S)) ℓ ( Q ) ≥ c 6 τ 0 ℓ ( Q ( S )) .
Proof. If ℓ ( Q ( S ) ) ≤ 5 C 5 ρ k 2 \ell(Q(S)) \le5C_5\rho^{k_2} ℓ ( Q ( S )) ≤ 5 C 5 ρ k 2 then the estimate is immediate since ℓ ( Q ) ≥ 5 ρ k 0 ≳ 5 C 5 ρ k 2 \ell(Q) \ge5\rho^{k_0} \gtrsim5C_5\rho^{k_2} ℓ ( Q ) ≥ 5 ρ k 0 ≳ 5 C 5 ρ k 2 . Assume instead that
ℓ ( Q ( S ) ) ≥ 5 C 5 ρ k 2 − 1 . (41) \ell(Q(S)) \ge5C_5\rho^{k_2-1}. \tag*{(41)} ℓ ( Q ( S )) ≥ 5 C 5 ρ k 2 − 1 . (41)
Let R ∈ Stop ( N ) R \in\operatorname{Stop}(N) R ∈ Stop ( N ) such that R ⊆ Q ⊆ ( M 1 / 2 ) B Q ( S ) R \subseteq Q \subseteq(M_1/2)B_{Q(S)} R ⊆ Q ⊆ ( M 1 /2 ) B Q ( S ) . By Lemma 4.8, it is sufficient to show
d N ( R ) = inf x ∈ R d N ( x ) ≳ ℓ ( Q ( S ) ) . (42) d_N(R) = \inf_{x\in R} d_N(x) \gtrsim\ell(Q(S)). \tag*{(42)} d N ( R ) = x ∈ R inf d N ( x ) ≳ ℓ ( Q ( S )) . (42)
To get (42) , fix an arbitrary x ∈ R x \in R x ∈ R , an arbitrary cube P ∈ Layer ( N ) P \in\operatorname{Layer}(N) P ∈ Layer ( N ) . It suffices to show
ℓ ( P ) + dist ( x , P ) ≳ ℓ ( Q ( S ) ) . (43) \ell(P) + \operatorname{dist}(x,P) \gtrsim\ell(Q(S)). \tag*{(43)} ℓ ( P ) + dist ( x , P ) ≳ ℓ ( Q ( S )) . (43)
Since x ∈ R ⊆ ( M 1 / 2 ) B Q ( S ) x \in R \subseteq(M_1/2)B_{Q(S)} x ∈ R ⊆ ( M 1 /2 ) B Q ( S ) and P ⊆ B P P \subseteq B_P P ⊆ B P , if B P ∩ M 1 B Q ( S ) = ∅ B_P \cap M_1B_{Q(S)} = \varnothing B P ∩ M 1 B Q ( S ) = ∅ , then (43) is obvious. Suppose instead that B P ∩ ( M 1 / 2 ) B Q ( S ) ≠ ∅ B_P \cap(M_1/2)B_{Q(S)} \ne\varnothing B P ∩ ( M 1 /2 ) B Q ( S ) = ∅ . Let T ∈ T N ( k 0 ) T \in\mathcal{T}^N(k_0) T ∈ T N ( k 0 ) such that P ∈ S T ′ P \in S'_T P ∈ S T ′ (recall Notation 4.6). This region satisfies
B Q ( T ) ∩ ( M 1 / 2 ) B Q ( S ) ≠ ∅ . B_{Q(T)} \cap(M_1/2)B_{Q(S)} \ne\varnothing. B Q ( T ) ∩ ( M 1 /2 ) B Q ( S ) = ∅ .
Thus, by Lemma 4.15 and (41) , we have
ℓ ( Q ( T ) ) ≥ C 5 − 1 ℓ ( Q ( S ) ) ≥ 5 ρ k 2 − 1 ⟹ Q ( T ) ∉ D ≥ k 2 . (44) \ell(Q(T)) \ge C_5^{-1}\ell(Q(S)) \ge5\rho^{k_2-1} \Longrightarrow Q(T) \notin\mathcal{D}_{\ge k_2}. \tag*{(44)} ℓ ( Q ( T )) ≥ C 5 − 1 ℓ ( Q ( S )) ≥ 5 ρ k 2 − 1 ⟹ Q ( T ) ∈ / D ≥ k 2 . (44)
On the other case, by the definition of B 1 N ( k 0 ) \mathcal{B}_1^N(k_0) B 1 N ( k 0 ) ,
Q ( T ) ∉ G ( k 0 ) ⟹ Q ( T ) ∈ B ( k 0 ) . (45) Q(T) \notin\mathcal{G}(k_0) \Longrightarrow Q(T) \in\mathcal{B}(k_0). \tag*{(45)} Q ( T ) ∈ / G ( k 0 ) ⟹ Q ( T ) ∈ B ( k 0 ) . (45)
Recalling (38) , (44) and (45) imply Q ( T ) ∈ B Q(T) \in\mathcal{B} Q ( T ) ∈ B . In particular, S T ′ = { Q ( T ) } = { P } S'_T = \{Q(T)\} = \{P\} S T ′ = { Q ( T )} = { P } . Thus, by (44) ,
ℓ ( P ) = ℓ ( Q ( T ) ) ≳ ℓ ( Q ( S ) ) , \ell(P) = \ell(Q(T)) \gtrsim\ell(Q(S)), ℓ ( P ) = ℓ ( Q ( T )) ≳ ℓ ( Q ( S )) ,
as required.
Approximations for individual cubes
Our goal for § 5 –§ 9 it to gather the necessary ingredients for defining a certain set Z ( k 0 ) Z(k_0) Z ( k 0 ) . The set Z Z Z from Theorem 1.3 will then be obtained from Z ( k 0 ) Z(k_0) Z ( k 0 ) after sending k 0 → ∞ k_0 \to\infty k 0 → ∞ . For now, let us fix some k 0 ∈ N k_0 \in\mathbb{N} k 0 ∈ N . The basic building blocks of Z k 0 Z_{k_0} Z k 0 are element of { G Q } Q ∈ D \{G_Q\}_{Q\in\mathcal{D}} { G Q } Q ∈ D , a collection d d d -dimensional boundaries of dyadic meshes associated to each Christ-David cube, which we describe in this section.
Constants 5.1. Let k 1 ∈ N k_1 \in\mathbb{N} k 1 ∈ N . Eventually, we will choose k 1 = k 1 ( τ 0 ) k_1 = k_1(\tau_0) k 1 = k 1 ( τ 0 ) sufficiently small.
Definitions and key properties
For a cube Q ∈ D Q \in\mathcal{D} Q ∈ D , we construct a set G Q G_Q G Q as follows. First, let J Q J_Q J Q be the dyadic cube with x Q ∈ J Q x_Q \in J_Q x Q ∈ J Q and ℓ ( J Q ) / 2 < ℓ ( Q ) ≤ ℓ ( J Q ) \ell(J_Q)/2 < \ell(Q) \leq\ell(J_Q) ℓ ( J Q ) /2 < ℓ ( Q ) ≤ ℓ ( J Q ) . Note, in particular, that
4 ρ − 1 B Q ⊆ λ 1 J Q ⊆ 2 n λ 1 B Q , where λ 1 : = 4 ρ − 1 + 1. (46) 4\rho^{-1}B_Q \subseteq\lambda_1J_Q \subseteq2\sqrt{n}\lambda_1B_Q,\qquad\text{where }\lambda_1:=4\rho^{-1}+1. \tag*{(46)} 4 ρ − 1 B Q ⊆ λ 1 J Q ⊆ 2 n λ 1 B Q , where λ 1 := 4 ρ − 1 + 1. (46)
Note also (since λ 1 \lambda_1 λ 1 is odd) that, for any k ∈ N k \in\mathbb{N} k ∈ N with 2 − k ≤ ℓ ( J Q ) 2^{-k} \leq\ell(J_Q) 2 − k ≤ ℓ ( J Q ) , the cubes in Δ k \Delta_k Δ k which intersect λ 1 J Q \lambda_1J_Q λ 1 J Q also form a partition of it. Let
J Q ′ : = { I ∈ Δ : I ⊆ λ 1 J Q and ℓ ( I ) = 2 − k 1 ℓ ( J Q ) } , \mathcal{J}'_Q:=\{I\in\Delta:I\subseteq\lambda_1J_Q\text{ and }\ell(I)=2^{-k_1}\ell(J_Q)\}, J Q ′ := { I ∈ Δ : I ⊆ λ 1 J Q and ℓ ( I ) = 2 − k 1 ℓ ( J Q )} ,
J Q : = J Q ′ ∩ { I ∈ Δ : ∂ I ∩ ∂ ( λ 1 J Q ) = ∅ } , \mathcal{J}_Q:=\mathcal{J}'_Q\cap\{I\in\Delta:\partial I\cap\partial(\lambda_1J_Q)=\varnothing\}, J Q := J Q ′ ∩ { I ∈ Δ : ∂ I ∩ ∂ ( λ 1 J Q ) = ∅ } ,
and
G Q : = ⋃ I ∈ J Q ∂ d I . G_Q:=\bigcup_{I\in\mathcal{J}_Q}\partial_d I. G Q := I ∈ J Q ⋃ ∂ d I .
It will be useful to be able to find d d d -dimensional cubes in G Q G_Q G Q that are close to the centres of sub-cubes of Q Q Q and relatively large compared to Q Q Q . If we allow the cube to be relatively large with some dependency on τ 0 \tau_0 τ 0 , then this is just immediate from the definition. However, it will be useful to have the relative size independent of τ \tau τ .
Lemma 5.2. Let κ ∈ ( 0 , 1 ) \kappa\in(0,1) κ ∈ ( 0 , 1 ) , R ⊆ Q ∈ D R\subseteq Q\in\mathcal{D} R ⊆ Q ∈ D and I ∈ Δ I\in\Delta I ∈ Δ , and suppose that κ ℓ ( Q ) ≤ ℓ ( I ) ≤ ℓ ( Q ) \kappa\ell(Q)\leq\ell(I)\leq\ell(Q) κ ℓ ( Q ) ≤ ℓ ( I ) ≤ ℓ ( Q ) and x R ∈ I x_R\in I x R ∈ I . Then, provided k 1 = k 1 ( κ ) k_1=k_1(\kappa) k 1 = k 1 ( κ ) is sufficiently large, we have
∂ d I ⊆ G Q . \partial_d I\subseteq G_Q. ∂ d I ⊆ G Q .
Proof. Since ∣ x R − x Q ∣ ≤ ℓ ( Q ) ≤ ℓ ( J Q ) |x_R-x_Q|\leq\ell(Q)\leq\ell(J_Q) ∣ x R − x Q ∣ ≤ ℓ ( Q ) ≤ ℓ ( J Q ) and x Q ∈ J Q x_Q\in J_Q x Q ∈ J Q there is a cube J ∈ Δ J\in\Delta J ∈ Δ such that
x R ∈ J ⊆ 3 J Q and ℓ ( J ) = ℓ ( J Q ) . x_R\in J\subseteq3J_Q\qquad\text{and}\qquad\ell(J)=\ell(J_Q). x R ∈ J ⊆ 3 J Q and ℓ ( J ) = ℓ ( J Q ) .
Since x R ∈ I x_R\in I x R ∈ I and ℓ ( I ) ≤ ℓ ( J ) \ell(I)\leq\ell(J) ℓ ( I ) ≤ ℓ ( J ) , it follows from the properties of dyadic cubes that I ⊆ J ⊆ 3 J Q I\subseteq J\subseteq3J_Q I ⊆ J ⊆ 3 J Q . Since the cube J Q \mathcal{J}_Q J Q partition 3 J Q 3J_Q 3 J Q , if k 1 ( κ ) k_1(\kappa) k 1 ( κ ) is chosen sufficiently large such that 2 − k 1 ℓ ( J Q ) ≤ ℓ ( I ) 2^{-k_1}\ell(J_Q)\leq\ell(I) 2 − k 1 ℓ ( J Q ) ≤ ℓ ( I ) , then we can write
I = ⋃ K ∈ J Q K ⊆ I K . I=\bigcup_{\substack{K\in\mathcal{J}_Q\\K\subseteq I}}K. I = K ∈ J Q K ⊆ I ⋃ K .
This implies
∂ d I = ⋃ K ∈ J Q K ⊆ I ∂ d K ⊆ G Q , \partial_d I=\bigcup_{\substack{K\in\mathcal{J}_Q\\K\subseteq I}}\partial_d K\subseteq G_Q, ∂ d I = K ∈ J Q K ⊆ I ⋃ ∂ d K ⊆ G Q ,
as required. □ \square □
In a moment we will prove that the G Q G_Q G Q , and various configurations of them, support a Poincaré inequality. The following lemma will be useful for this. It shows that the grids corresponding to cubes with comparable sizes are either very disjoint or they have large intersection. Moreover, the gluing metric between grids with large intersection is bi-Lipschitz equivalent (with respect to the identity) to the Euclidean metric.
Lemma 5.3. Let C ≥ 1 C \ge1 C ≥ 1 and suppose Q , R ∈ D Q,R \in\mathcal{D} Q , R ∈ D are such that C − 1 ≤ ℓ ( R ) / ℓ ( Q ) ≤ C C^{-1} \le\ell(R)/\ell(Q) \le C C − 1 ≤ ℓ ( R ) / ℓ ( Q ) ≤ C . If k 1 k_1 k 1 large enough (depending only on C C C ) then we have either λ 1 J Q ∩ λ 1 J R = ∅ \lambda_1 J_Q \cap\lambda_1 J_R = \varnothing λ 1 J Q ∩ λ 1 J R = ∅ , in which case
dist ( G Q , G R ) ≥ 2 − k 1 max { ℓ ( Q ) , ℓ ( R ) } , \operatorname{dist}(G_Q,G_R) \ge2^{-k_1}\max\{\ell(Q),\ell(R)\}, dist ( G Q , G R ) ≥ 2 − k 1 max { ℓ ( Q ) , ℓ ( R )} , or λ 1 J Q ∩ λ 1 J R ≠ ∅ \lambda_1 J_Q \cap\lambda_1 J_R \ne\varnothing λ 1 J Q ∩ λ 1 J R = ∅ , in which G Q ∩ G R G_Q \cap G_R G Q ∩ G R is non-empty, lower Ahlfors d d d -regular and
∣ x − y ∣ ∼ inf z ∈ G Q ∩ G R ∣ x − z ∣ + ∣ y − z ∣ (47) |x-y| \sim\inf_{z \in G_Q \cap G_R} |x-z|+|y-z| \tag*{(47)} ∣ x − y ∣ ∼ z ∈ G Q ∩ G R inf ∣ x − z ∣ + ∣ y − z ∣ (47) for all x ∈ G Q x \in G_Q x ∈ G Q and y ∈ G R y \in G_R y ∈ G R .
Proof. Since G Q ⊆ λ 1 J Q G_Q \subseteq\lambda_1 J_Q G Q ⊆ λ 1 J Q and dist ( G Q , ∂ ( λ 1 J Q ) ) ≥ 2 − k 1 ℓ ( Q ) \operatorname{dist}(G_Q,\partial(\lambda_1 J_Q)) \ge2^{-k_1}\ell(Q) dist ( G Q , ∂ ( λ 1 J Q )) ≥ 2 − k 1 ℓ ( Q ) (with the same properties holding for the cube R R R ), if λ 1 J Q ∩ λ 1 J R = ∅ \lambda_1 J_Q \cap\lambda_1 J_R = \varnothing λ 1 J Q ∩ λ 1 J R = ∅ then the second conclusion holds. So, let us suppose that
λ 1 J Q ∩ λ 1 J R ≠ ∅ , \lambda_1 J_Q \cap\lambda_1 J_R \ne\varnothing, λ 1 J Q ∩ λ 1 J R = ∅ ,
and let us assume (without loss of generality) that ℓ ( Q ) ≤ ℓ ( R ) ≤ C ℓ ( Q ) \ell(Q) \le\ell(R) \le C\ell(Q) ℓ ( Q ) ≤ ℓ ( R ) ≤ C ℓ ( Q ) .
To see G Q ∩ G R G_Q \cap G_R G Q ∩ G R is non-empty, note to begin that there exists a dyadic cube I ⊆ λ 1 J Q ∩ λ 1 J R I \subseteq\lambda_1 J_Q \cap\lambda_1 J_R I ⊆ λ 1 J Q ∩ λ 1 J R with side length ℓ ( I ) = ℓ ( J Q ) \ell(I)=\ell(J_Q) ℓ ( I ) = ℓ ( J Q ) . Consider the cube I ′ I' I ′ such that x I ∈ I ′ x_I \in I' x I ∈ I ′ and ℓ ( I ′ ) = 2 − k 1 ℓ ( J R ) ≤ C 2 − k 1 ℓ ( J Q ) \ell(I')=2^{-k_1}\ell(J_R) \le C2^{-k_1}\ell(J_Q) ℓ ( I ′ ) = 2 − k 1 ℓ ( J R ) ≤ C 2 − k 1 ℓ ( J Q ) . As long as k 1 = k 1 ( C ) k_1=k_1(C) k 1 = k 1 ( C ) is sufficiently small, it follows that
I ′ ∩ ( ∂ ( λ 1 J Q ) ∪ ∂ ( λ 1 J R ) ) = ∅ , (48) I' \cap\left(\partial(\lambda_1 J_Q)\cup\partial(\lambda_1 J_R)\right)=\varnothing, \tag*{(48)} I ′ ∩ ( ∂ ( λ 1 J Q ) ∪ ∂ ( λ 1 J R ) ) = ∅ , (48)
i.e., I ′ ∈ I R I' \in\mathcal{I}_R I ′ ∈ I R . Consider also an arbitrary dyadic cube I ′ ′ ⊆ I ′ I'' \subseteq I' I ′′ ⊆ I ′ with side length ℓ ( I ′ ′ ) = 2 − k 1 ℓ ( J Q ) \ell(I'')=2^{-k_1}\ell(J_Q) ℓ ( I ′′ ) = 2 − k 1 ℓ ( J Q ) such that ∂ d I ′ ∩ ∂ d I ′ ′ ≠ ∅ \partial_d I' \cap\partial_d I'' \ne\varnothing ∂ d I ′ ∩ ∂ d I ′′ = ∅ . Since I ′ ′ ⊆ I ′ I'' \subseteq I' I ′′ ⊆ I ′ , (5.3) also holds with I ′ ′ I'' I ′′ in place of I ′ I' I ′ . Thus, I ′ ′ ∈ I Q I'' \in\mathcal{I}_Q I ′′ ∈ I Q . Since the d d d -dimensional boundaries of I ′ I' I ′ and I ′ ′ I'' I ′′ have non-empty intersection, it now follows that G Q ∩ G R ≠ ∅ G_Q \cap G_R \ne\varnothing G Q ∩ G R = ∅ .
Our next goal is to show that G Q ∩ G R G_Q \cap G_R G Q ∩ G R is lower Ahlfors d d d -regular. We start by proving the following.
Claim 1. For each x ∈ G Q ∩ G R x \in G_Q \cap G_R x ∈ G Q ∩ G R there are cubes I ∈ I Q I \in\mathcal{I}_Q I ∈ I Q and J ∈ I R J \in\mathcal{I}_R J ∈ I R such that I ⊆ J I \subseteq J I ⊆ J and x ∈ ∂ d I ∩ ∂ d J x \in\partial_d I \cap\partial_d J x ∈ ∂ d I ∩ ∂ d J .
Let T Q = ⋃ I ∈ I Q I T_Q=\bigcup_{I\in\mathcal{I}_Q} I T Q = ⋃ I ∈ I Q I and T R = ⋃ I ∈ I R I T_R=\bigcup_{I\in\mathcal{I}_R} I T R = ⋃ I ∈ I R I . Since T Q T_Q T Q and T R T_R T R are cubes in R n \mathbb{R}^n R n (so that intersections and closures commute), we have
G Q ∩ G R ⊆ T Q ‾ ∩ T R ‾ = T Q ∩ T R ‾ . G_Q \cap G_R \subseteq\overline{T_Q}\cap\overline{T_R}=\overline{T_Q\cap T_R}. G Q ∩ G R ⊆ T Q ∩ T R = T Q ∩ T R .
Since every cube in I Q \mathcal{I}_Q I Q is smaller than every cube in I R \mathcal{I}_R I R , we can write
T Q ∩ T R ‾ = ⋃ I ∈ I Q I ⊆ T R I = ⋃ I ∈ I Q I ⊆ T R I ‾ . \overline{T_Q\cap T_R}
=
\bigcup_{\substack{I\in\mathcal{I}_Q\\ I\subseteq T_R}} I
=
\bigcup_{\substack{I\in\mathcal{I}_Q\\ I\subseteq T_R}} \overline{I}. T Q ∩ T R = I ∈ I Q I ⊆ T R ⋃ I = I ∈ I Q I ⊆ T R ⋃ I .
In particular, for each x ∈ G Q ∩ G R x\in G_Q\cap G_R x ∈ G Q ∩ G R there are cubes I ∈ I Q I\in\mathcal{I}_Q I ∈ I Q and J ∈ I R J\in\mathcal{I}_R J ∈ I R such that x ∈ I ‾ ⊆ J ‾ x\in\overline{I}\subseteq\overline{J} x ∈ I ⊆ J (so that, I ⊆ J I\subseteq J I ⊆ J ). Since x ∈ G Q x\in G_Q x ∈ G Q (i.e., it is in the d d d -dimensional boundary of some cube I ′ I' I ′ with ℓ ( I ′ ) = ℓ ( I ) \ell(I')=\ell(I) ℓ ( I ′ ) = ℓ ( I ) ), we conclude that x ∈ ∂ d I x\in\partial_d I x ∈ ∂ d I . By similar reasoning, we also conclude that x ∈ ∂ d J x\in\partial_d J x ∈ ∂ d J , as required.
:::
Fix some x ∈ G Q ∩ G R x \in G_Q \cap G_R x ∈ G Q ∩ G R and 0 < r < ℓ ( R ) 0 < r < \ell(R) 0 < r < ℓ ( R ) . By Claim 1, there are cubes I ∈ J Q I \in\mathcal{J}_Q I ∈ J Q and J ∈ J R J \in\mathcal{J}_R J ∈ J R such that I ⊆ J I \subseteq J I ⊆ J and x ∈ ∂ d I ∩ ∂ d J x \in\partial_d I \cap\partial_d J x ∈ ∂ d I ∩ ∂ d J . By Lemma 2.2, there are d d d -dimensional sub-faces F I ⊆ ∂ d I F_I \subseteq\partial_d I F I ⊆ ∂ d I and F J ⊆ ∂ d J F_J \subseteq\partial_d J F J ⊆ ∂ d J such that x ∈ F I ⊆ F J x \in F_I \subseteq F_J x ∈ F I ⊆ F J . Thus, by the definition of G Q G_Q G Q and G R G_R G R , we have
F I ⊆ G Q ∩ G R . F_I \subseteq G_Q \cap G_R. F I ⊆ G Q ∩ G R .
Since ℓ ( F I ) = ℓ ( I ) ∼ ℓ ( R ) \ell(F_I) = \ell(I) \sim\ell(R) ℓ ( F I ) = ℓ ( I ) ∼ ℓ ( R ) and F I F_I F I is lower Ahlfors d d d -regular, it follows that
H d ( ( G Q ∩ G R ) ∩ B ( x , r ) ) ≥ H d ( F I ∩ B ( x , r ) ) ≳ r d , \mathcal{H}^d((G_Q \cap G_R) \cap B(x,r)) \geq\mathcal{H}^d(F_I \cap B(x,r)) \gtrsim r^d, H d (( G Q ∩ G R ) ∩ B ( x , r )) ≥ H d ( F I ∩ B ( x , r )) ≳ r d ,
as required.
Finally, let us check (47) . Fix some x ∈ G Q x \in G_Q x ∈ G Q and y ∈ G R y \in G_R y ∈ G R . Let I ∈ J Q I \in\mathcal{J}_Q I ∈ J Q such that x ∈ ∂ d I x \in\partial_d I x ∈ ∂ d I and let J ∈ J R J \in\mathcal{J}_R J ∈ J R such that y ∈ ∂ d J y \in\partial_d J y ∈ ∂ d J . If ∂ d I ∩ ∂ d K ′ = ∅ \partial_d I \cap\partial_d K' = \varnothing ∂ d I ∩ ∂ d K ′ = ∅ for all K ′ ∈ J R K' \in\mathcal{J}_R K ′ ∈ J R , then ∣ x − y ∣ ≥ ℓ ( I ) |x-y| \geq\ell(I) ∣ x − y ∣ ≥ ℓ ( I ) . Thus, by taking an arbitrary z ∈ G Q ∩ G R z \in G_Q \cap G_R z ∈ G Q ∩ G R and using the fact that diam ( G Q ∪ G R ) ≲ ℓ ( R ) ≲ ℓ ( I ) \operatorname{diam}(G_Q \cup G_R) \lesssim\ell(R) \lesssim\ell(I) diam ( G Q ∪ G R ) ≲ ℓ ( R ) ≲ ℓ ( I ) , we get
∣ x − z ∣ + ∣ z − y ∣ ≤ 2 ∣ x − z ∣ + ∣ x − y ∣ ≲ ℓ ( I ) + ∣ x − y ∣ ≲ ∣ x − y ∣ . |x-z| + |z-y| \leq2|x-z| + |x-y| \lesssim\ell(I) + |x-y| \lesssim|x-y|. ∣ x − z ∣ + ∣ z − y ∣ ≤ 2∣ x − z ∣ + ∣ x − y ∣ ≲ ℓ ( I ) + ∣ x − y ∣ ≲ ∣ x − y ∣.
Thus, we can assume that there exists some K ′ ∈ J R K' \in\mathcal{J}_R K ′ ∈ J R such that ∂ d I ∩ ∂ d K ′ ≠ ∅ \partial_d I \cap\partial_d K' \ne\varnothing ∂ d I ∩ ∂ d K ′ = ∅ . Applying Claim 1, we find another K ∈ J R K \in\mathcal{J}_R K ∈ J R such that I ⊆ K I \subseteq K I ⊆ K . The estimate (47) now follows from Lemma 2.1 (recalling that ∂ d I ∩ ∂ d K ⊆ G Q ∩ G R \partial_d I \cap\partial_d K \subseteq G_Q \cap G_R ∂ d I ∩ ∂ d K ⊆ G Q ∩ G R ). □
Poincaré inequalities for the G Q G_Q G Q
We now collect various properties of the set we constructed above in terms of Poincaré inequalities. We start with the most basic, that each G Q G_Q G Q supports a Poincaré inequality.
Lemma 5.4. The set G Q G_Q G Q is Ahlfors d d d -regular and supports a ( 1 , 1 ) (1,1) ( 1 , 1 ) -Poincaré inequality with constants depending on k 1 k_1 k 1 and n n n .
Proof. We first observe that for any cube I ⊂ R n I \subset\mathbb{R}^n I ⊂ R n it holds that ∂ d I \partial_d I ∂ d I is Ahlfors d d d -regular with regularity constant depending only on n n n . Up to rescaling I I I , we can assume that ℓ ( I ) = 1 \ell(I) = 1 ℓ ( I ) = 1 . The d d d -regularity follows e.g. from Lemma 2.9, indeed ∂ d I \partial_d I ∂ d I is the union of less than 3 n 3^n 3 n closed d d d -dimensional cubes Q i Q_i Q i of side-length one and such that F i : = Q i ∩ Q i + 1 ≠ ∅ F_i := Q_i \cap Q_{i+1} \ne\varnothing F i := Q i ∩ Q i + 1 = ∅ .
Recall that G Q G_Q G Q is the union of the d d d -skeleton of all the dyadic cubes of side-length 2 − k 1 ℓ ( J Q ) 2^{-k_1}\ell(J_Q) 2 − k 1 ℓ ( J Q ) contained in the dyadic cube λ 1 J Q ∈ Δ \lambda_1J_Q \in\Delta λ 1 J Q ∈ Δ minus the outermost layer (recall that λ 1 \lambda_1 λ 1 is odd). Hence G Q = ⋃ i = 1 N ∂ d I i G_Q = \bigcup_{i=1}^N \partial_d I_i G Q = ⋃ i = 1 N ∂ d I i , with I i I_i I i disjoint dyadic cubes such that ℓ ( I i ) = 2 − k 1 ℓ ( J Q ) \ell(I_i) = 2^{-k_1}\ell(J_Q) ℓ ( I i ) = 2 − k 1 ℓ ( J Q ) with N ≤ ( 2 k 1 λ 1 ) n ≲ n ( 2 k 1 ) n N \leq(2^{k_1}\lambda_1)^n \lesssim_n (2^{k_1})^n N ≤ ( 2 k 1 λ 1 ) n ≲ n ( 2 k 1 ) n (in fact N = L n N = L^n N = L n with L = 2 k 1 λ 1 − 2 L = 2^{k_1}\lambda_1 - 2 L = 2 k 1 λ 1 − 2 ) because λ 1 = 4 ρ − 1 + 1 = 4001 \lambda_1 = 4\rho^{-1} + 1 = 4001 λ 1 = 4 ρ − 1 + 1 = 4001 . Up to a rescaling, we can assume that ℓ ( J Q ) = 2 k 1 \ell(J_Q) = 2^{k_1} ℓ ( J Q ) = 2 k 1 and so G Q G_Q G Q coincides with the d d d -skeleton of a cubical one-by-one grid in R n \mathbb{R}^n R n of side-length L L L . In other words, up to translation, G Q = { ( x 1 , … , x n ) ∈ [ 0 , L ] n : x i is a non-negative integer for at least n − d choices of i } G_Q = \{(x_1,\ldots,x_n) \in[0,L]^n : x_i \text{ is a non-negative integer for at least } n-d \text{ choices of } i\} G Q = {( x 1 , … , x n ) ∈ [ 0 , L ] n : x i is a non-negative integer for at least n − d choices of i } . We can also choose an enumeration so that ∂ d I i ∩ ∂ d I i + 1 \partial_d I_i \cap\partial_d I_{i+1} ∂ d I i ∩ ∂ d I i + 1 is non-empty. Applying Lemma 2.9 N − 1 N-1 N − 1 times follows that G Q G_Q G Q is Ahlfors d d d -regular with regularity constant depending only on k 1 k_1 k 1 .
To show the validity of the Poincaré inequality we aim to apply the results in [33 ] . We need to check that G Q G_Q G Q satisfies (1)-(5) in [33 ] , Assumptions 1.10]. Note that in the terminology of [33 ] the set G Q G_Q G Q equipped with the intrinsic length distance d l d_l d l is a connected Euclidean homogeneous polyhedral complex of dimension d d d .
Since G Q G_Q G Q is homogeneous (the maximal polyhedra have have dimension d d d ), to verify (1) we need only to check that if F , E F,E F , E are two closed d d d -dimensional faces of
G Q G_Q G Q such that F ∩ E F \cap E F ∩ E is a k k k -dimensional face P P P , k ≤ d − 1 k \le d-1 k ≤ d − 1 , then there is a chain of closed d d d -dimensional faces E = G 1 , … , G m = F E = G_1, \ldots, G_m = F E = G 1 , … , G m = F such that G i ∩ G i + 1 G_i \cap G_{i+1} G i ∩ G i + 1 is a ( d − 1 ) (d-1) ( d − 1 ) -dimensional face and P ⊂ G i P \subset G_i P ⊂ G i . We have that E = ∏ i = 1 n E i E = \prod_{i=1}^n E_i E = ∏ i = 1 n E i and F = ∏ i = 1 n F i F = \prod_{i=1}^n F_i F = ∏ i = 1 n F i , with E i , F i E_i, F_i E i , F i either closed unit intervals or singletons and E i ∩ F i ≠ ∅ E_i \cap F_i \ne\varnothing E i ∩ F i = ∅ for all i i i . Set G 1 : = E G_1 := E G 1 := E and proceed inductively as follows. Take the first i i i such that E i ≠ F i E_i \ne F_i E i = F i . If both E i E_i E i and F i F_i F i are intervals replace E i E_i E i in E E E by F i F_i F i . If E i E_i E i is an interval and F i F_i F i is a point, since the E i E_i E i ’s and F i F_i F i ’s are intervals exactly d d d times, there must be j j j such that E j E_j E j is a point while F j F_j F j is not. Then we replace E i E_i E i with F i F_i F i and E j E_j E j by F j F_j F j . If E i E_i E i is a point and F i F_i F i is an interval we do the same but reversed. This produces a d d d -dimensional face (still in G Q G_Q G Q ) G 2 G_2 G 2 such that E ∩ G 2 E \cap G_2 E ∩ G 2 is a ( d − 1 ) (d-1) ( d − 1 ) -dimensional face and P ⊂ G 2 P \subset G_2 P ⊂ G 2 , indeed P ⊂ ∏ i = 1 n E i ∩ F i ⊂ G 2 P \subset\prod_{i=1}^n E_i \cap F_i \subset G_2 P ⊂ ∏ i = 1 n E i ∩ F i ⊂ G 2 by construction because G 2 G_2 G 2 is obtained by swapping some of the E i E_i E i ’s with some of the F i F_i F i ’s. After finitely many step we end up with G m = F G_m = F G m = F as desired.
Assumption (2) is trivially true with Λ = 1 \Lambda= 1 Λ = 1 because each polyhedron is equipped with the Euclidean metric.
For (3) we have that for each j < k ≤ d j < k \le d j < k ≤ d and any j j j -dimensional face F F F of G Q G_Q G Q , the maximal number of k k k -dimensional faces that contain F F F is clearly bounded above by M = 3 n M = 3^n M = 3 n . Indeed F = ∏ i = 1 n F i F = \prod_{i=1}^n F_i F = ∏ i = 1 n F i where each F i F_i F i is either a unit intervals with integer endpoints or an integer. If a k k k -dimensional face E E E of G Q G_Q G Q contains F F F then E = ∏ i = 1 n E i E = \prod_{i=1}^n E_i E = ∏ i = 1 n E i , where E i E_i E i are again either a unit intervals with integer endpoints or an integer and F i ⊂ E i F_i \subset E_i F i ⊂ E i . However, independently of F i F_i F i there are clearly at most three choices of E i E_i E i .
For (4) we can immediately check that the distance between two disjoint k k k -dimensional faces of G Q G_Q G Q is at least ℓ = 1 \ell= 1 ℓ = 1 .
Assumption (5) requires a uniform solid angle lower bound α \alpha α . We recall the meaning of this. Take any 0 ≤ k ≤ d 0 \le k \le d 0 ≤ k ≤ d and any ball B ⊂ G Q B \subset G_Q B ⊂ G Q with respect to the length distance with center x B x_B x B lying on a k k k -dimensional face F F F and radius r < d l ( x B , G Q k − 1 ) r < d_l(x_B, G_Q^{k-1}) r < d l ( x B , G Q k − 1 ) (where G Q k − 1 G_Q^{k-1} G Q k − 1 is the ( k − 1 ) (k-1) ( k − 1 ) -skeleton of G Q G_Q G Q ), assuming simply r < 1 r < 1 r < 1 if k = 0 k = 0 k = 0 . A wedge W W W is the closure of a connected component of B ∖ G Q d − 1 B \setminus G_Q^{d-1} B ∖ G Q d − 1 . Having a uniform solid angle lower bound α \alpha α means that H d ( W ) / ( ω d r d ) ≥ α \mathcal{H}^d(W)/(\omega_d r^d) \ge\alpha H d ( W ) / ( ω d r d ) ≥ α for all wedges W W W (note that this ratio is always ≤ 1 \le1 ≤ 1 ). In our case, for any B B B as above, we have that B B B is isometric to the union of copies of B R d ( 0 , r ) ∩ R k × [ 0 , ∞ ) d − k B^{\mathbb{R}^d}(0,r) \cap\mathbb{R}^k \times[0,\infty)^{d-k} B R d ( 0 , r ) ∩ R k × [ 0 , ∞ ) d − k . Therefore each wedge W W W is a copy of B R d ( 0 , r ) ∩ R k × [ 0 , ∞ ) d − k B^{\mathbb{R}^d}(0,r) \cap\mathbb{R}^k \times[0,\infty)^{d-k} B R d ( 0 , r ) ∩ R k × [ 0 , ∞ ) d − k and so H d ( W ) / ( ω d r d ) ≥ 2 − ( d − k ) ≥ 2 − d ≥ 2 − n \mathcal{H}^d(W)/(\omega_d r^d) \ge2^{-(d-k)} \ge2^{-d} \ge2^{-n} H d ( W ) / ( ω d r d ) ≥ 2 − ( d − k ) ≥ 2 − d ≥ 2 − n . Hence we can choose α = 2 − n \alpha= 2^{-n} α = 2 − n .
Therefore we can apply [33 Corollary 2.10] choosing R 0 ′ : = 2 diam d l ( G Q ) R'_0 := 2\operatorname{diam}_{d_l}(G_Q) R 0 ′ := 2 diam d l ( G Q ) (see also [33 Theorem 2.7] ) to deduce that G Q G_Q G Q , endowed with the length metric d l d_l d l , supports a ( 1 , 1 ) (1,1) ( 1 , 1 ) -Poincaré inequality with constants depending only on k 1 k_1 k 1 , d d d and n n n . Indeed note that their notion of gradient for a Lipschitz function f f f in G Q G_Q G Q satisfies ∣ ∇ f ∣ ≤ Lip ( f ) H d |\nabla f| \le\operatorname{Lip}(f) \mathcal{H}^d ∣∇ f ∣ ≤ Lip ( f ) H d -a.e. To pass to G Q G_Q G Q with the Euclidean distance d e d_e d e we only need to observe that d e ≤ d l ≲ n d e d_e \le d_l \lesssim_n d_e d e ≤ d l ≲ n d e . The first inequality is obvious. For the second fix x = ( x 1 , … , x n ) , y = ( y 1 , … , y n ) ∈ G Q x = (x_1, \ldots, x_n), y = (y_1, \ldots, y_n) \in G_Q x = ( x 1 , … , x n ) , y = ( y 1 , … , y n ) ∈ G Q . Recall that at least n − d n-d n − d entries for both x x x and y y y must be integers. Suppose first that ∣ x − y ∣ ≤ 1 |x-y| \le1 ∣ x − y ∣ ≤ 1 . Then if x i x_i x i and y i y_i y i are both integers they must coincide. We join x x x and y y y by the curve which is the union of the segments formed by subsequent linear interpolation from x i x_i x i to y i y_i y i , starting first from all the non-integer entries x i x_i x i . This is contained in G Q G_Q G Q , since at all times there are at least n − d n-d n − d integer coordinates. Moreover this curve has length ∑ i = 1 n ∣ x i − y i ∣ ≤ n ∣ x − y ∣ \sum_{i=1}^n |x_i-y_i| \le\sqrt{n}|x-y| ∑ i = 1 n ∣ x i − y i ∣ ≤ n ∣ x − y ∣ . Suppose now that ∣ x − y ∣ ≥ 1 |x-y| \ge1 ∣ x − y ∣ ≥ 1 . Since both x x x and y y y lie in d d d -dimensional faces of G Q G_Q G Q we can take vertices of such faces x ˉ , y ˉ \bar{x}, \bar{y} x ˉ , y ˉ having all integer coordinates such that ∣ x − x ˉ ∣ , ∣ y − y ˉ ∣ ≤ n |x-\bar{x}|, |y-\bar{y}| \le\sqrt{n} ∣ x − x ˉ ∣ , ∣ y − y ˉ ∣ ≤ n . Moreover the segments x , x ˉ x,\bar{x} x , x ˉ and y , y ˉ y,\bar{y} y , y ˉ are contained in G Q G_Q G Q (because d d d -dimensional faces are convex). As above we can build a curve from x ˉ \bar{x} x ˉ to y ˉ \bar{y} y ˉ contained in G Q G_Q G Q of length at most n ∣ x ˉ − y ˉ ∣ \sqrt{n}|\bar{x}-\bar{y}| n ∣ x ˉ − y ˉ ∣ which, joined with the two segments x , x ˉ x,\bar{x} x , x ˉ and y , y ˉ y,\bar{y} y , y ˉ , produces a curve from x x x to y y y of length at most n ∣ x ˉ − y ˉ ∣ + 2 n ≤ ( 3 n + 2 n ) ∣ x − y ∣ \sqrt{n}|\bar{x}-\bar{y}|+2\sqrt{n} \le(3\sqrt{n}+2n)|x-y| n ∣ x ˉ − y ˉ ∣ + 2 n ≤ ( 3 n + 2 n ) ∣ x − y ∣ , as desired. □
More generally, we would like to show that the union of sets of this form support a Poincaré inequality (under certain hypotheses). This will be our goal for the remainder of the section. In Lemma 5.5 below we show this to be true provided the union is connected (however, we allow the constant to depend on the number of elements in the union).
Lemma 5.5. Let C ≥ 1 C \ge1 C ≥ 1 , Q \mathcal{Q} Q be a finite collection of cubes such that C ≤ ℓ ( Q ) / ℓ ( R ) ≤ V C \le\ell(Q)/\ell(R) \le V C ≤ ℓ ( Q ) / ℓ ( R ) ≤ V for all Q , R ∈ Q Q,R \in\mathcal{Q} Q , R ∈ Q , and suppose that G : = ⋃ Q ∈ Q G Q G := \bigcup_{Q\in\mathcal{Q}}G_Q G := ⋃ Q ∈ Q G Q is connected. If k 1 k_1 k 1 large enough (depending only on C C C ) then G G G is Ahlfors d d d -regular and supports a ( 1 , 1 ) (1,1) ( 1 , 1 ) -Poincaré inequality with constant depending only on k 1 k_1 k 1 , n n n and ∣ Q ∣ |\mathcal{Q}| ∣ Q ∣ .
Proof. Ahlfors regularity follows directly from Lemma 2.9 . Thus, our main goal is to prove prove a Poincaré inequality. Let K : = ∣ Q ∣ K := |\mathcal{Q}| K := ∣ Q ∣ and let k 1 = k 1 ( C ) k_1 = k_1(C) k 1 = k 1 ( C ) be as in Lemma 5.3 . Inductively, we will define a sequence of cubes Q = { Q i } i = 1 K \mathcal{Q} = \{Q_i\}_{i=1}^K Q = { Q i } i = 1 K such that
G k : = ⋃ i = 1 k G Q k supports a ( 1 , 1 ) -Poincar e ˊ inequality (49) G_k := \bigcup_{i=1}^{k} G_{Q_k}\text{ supports a }(1,1)\text{-Poincaré inequality} \tag*{(49)} G k := i = 1 ⋃ k G Q k supports a ( 1 , 1 ) -Poincar e ˊ inequality (49)
for each k ∈ { 1 , … , K } k \in\{1,\ldots,K\} k ∈ { 1 , … , K } with constant depending on k k k . Indeed, to begin with we let Q 1 Q_1 Q 1 be an arbitrary cube in Q \mathcal{Q} Q . Then, (49) follows immediately for G 1 G_1 G 1 by Lemma 5.4 . Inductively, suppose that we have constructed cubes { Q i } i = 1 k \{Q_i\}_{i=1}^k { Q i } i = 1 k with k < K k<K k < K such that G k G_k G k satisfies (49) . Since G G G is connected, there exists some cube in Q \mathcal{Q} Q , which we label Q k + 1 Q_{k+1} Q k + 1 , such that G k ∩ G Q k + 1 ≠ ∅ G_k \cap G_{Q_{k+1}} \ne\varnothing G k ∩ G Q k + 1 = ∅ . Let
I 1 : = { i ∈ { 1 , … , k } : G Q i ∩ G Q k + 1 ≠ ∅ } I_1 := \{i \in\{1,\ldots,k\}: G_{Q_i} \cap G_{Q_{k+1}} \ne\varnothing\} I 1 := { i ∈ { 1 , … , k } : G Q i ∩ G Q k + 1 = ∅ } and I 2 : = { 1 , … , k } ∖ I 1 I_2 := \{1,\ldots,k\}\setminus I_1 I 2 := { 1 , … , k } ∖ I 1 .
By Lemma 5.3 , we know that
A : = G k ∩ G Q k + 1 = ⋃ i ∈ I 1 G Q i ∩ G Q k + 1 A := G_k \cap G_{Q_{k+1}} = \bigcup_{i\in I_1} G_{Q_i} \cap G_{Q_{k+1}} A := G k ∩ G Q k + 1 = i ∈ I 1 ⋃ G Q i ∩ G Q k + 1
is lower Ahlfors d d d -regular. By Lemma 3.4 , the only thing left to show is that
∣ x − y ∣ ∼ inf z ∈ A ∣ x − z ∣ + ∣ y − z ∣ (50) |x-y| \sim\inf_{z\in A}|x-z|+|y-z| \tag*{(50)} ∣ x − y ∣ ∼ z ∈ A inf ∣ x − z ∣ + ∣ y − z ∣ (50)
for any pair of points x ∈ G k x \in G_k x ∈ G k and y ∈ G Q k + 1 y \in G_{Q_{k+1}} y ∈ G Q k + 1 , which we now fix.
Let i ∈ { 1 , … , k } i \in\{1,\ldots,k\} i ∈ { 1 , … , k } such that x ∈ G Q i x \in G_{Q_i} x ∈ G Q i . If i ∈ I 1 i \in I_1 i ∈ I 1 , then (50) follows immediately from Lemma 5.3 since A ⊇ G Q i ∩ G Q k + 1 A \supseteq G_{Q_i} \cap G_{Q_{k+1}} A ⊇ G Q i ∩ G Q k + 1 . Suppose now that i ∈ I 2 i \in I_2 i ∈ I 2 . In this case, it follows from Lemma 5.3 that ∣ x − y ∣ ≳ ℓ ( Q k + 1 ) |x-y| \gtrsim\ell(Q_{k+1}) ∣ x − y ∣ ≳ ℓ ( Q k + 1 ) . Furthermore, since A ∩ G Q k + 1 ≠ ∅ A \cap G_{Q_{k+1}} \ne\varnothing A ∩ G Q k + 1 = ∅ and diam ( G Q k + 1 ) ≲ ℓ ( Q k + 1 ) \operatorname{diam}(G_{Q_{k+1}}) \lesssim\ell(Q_{k+1}) diam ( G Q k + 1 ) ≲ ℓ ( Q k + 1 ) , there exists z ∈ A z \in A z ∈ A such that ∣ y − z ∣ ≲ ℓ ( Q k + 1 ) |y-z| \lesssim\ell(Q_{k+1}) ∣ y − z ∣ ≲ ℓ ( Q k + 1 ) . Combing these two facts, we get
∣ x − z ∣ + ∣ y − z ∣ ≤ ∣ x − y ∣ + 2 ∣ y − z ∣ ≲ ∣ x − y ∣ + ℓ ( Q k + 1 ) ≲ ∣ x − y ∣ , |x-z|+|y-z| \le|x-y|+2|y-z| \lesssim|x-y|+\ell(Q_{k+1}) \lesssim|x-y|, ∣ x − z ∣ + ∣ y − z ∣ ≤ ∣ x − y ∣ + 2∣ y − z ∣ ≲ ∣ x − y ∣ + ℓ ( Q k + 1 ) ≲ ∣ x − y ∣ ,
as required. □
We now show how to remove this dependency on the number of elements in the union provided they lie close to a graph generated by the class G 2 N ( k 0 ) \mathcal{G}_2^N(k_0) G 2 N ( k 0 ) of good stopping-time regions from (39) . In order to fix constants, we introduce the following.
Constants 5.6. Let A 1 : = 3000 n λ 1 C 5 A_1 := 3000\sqrt{n}\lambda_1C_5 A 1 := 3000 n λ 1 C 5 and C 7 = ( c 6 τ 0 ) − 1 C_7 = (c_6\tau_0)^{-1} C 7 = ( c 6 τ 0 ) − 1 .
Proposition 5.7. Let
k ∈ N , A ∈ [ A 1 , 2 A 1 2 ] and C ∈ [ 1 , C 7 ] . k \in\mathbb{N}, \qquad A \in[A_1,2A_1^2] \qquad\text{and} \qquad C \in[1,C_7]. k ∈ N , A ∈ [ A 1 , 2 A 1 2 ] and C ∈ [ 1 , C 7 ] . Let S ∈ G 2 N ( k ) S \in\mathcal{G}_2^N(k) S ∈ G 2 N ( k ) and { R j } j ∈ J ⊆ ⋃ T ∈ T N ( k 0 ) T ( k 0 ) \{R_j\}_{j\in J} \subseteq\bigcup_{T\in\mathcal{T}^N(k_0)}T(k_0) { R j } j ∈ J ⊆ ⋃ T ∈ T N ( k 0 ) T ( k 0 ) be a disjoint collection of cubes such that
E ∩ A B Q ( S ) ⊆ ⋃ j ∈ J R j E \cap AB_{Q(S)} \subseteq\bigcup_{j\in J} R_j E ∩ A B Q ( S ) ⊆ ⋃ j ∈ J R j ;
E ∩ 3 B R j ∩ A B Q ( S ) ≠ ∅ E \cap3B_{R_j} \cap AB_{Q(S)} \ne\varnothing E ∩ 3 B R j ∩ A B Q ( S ) = ∅ for all j ∈ J j \in J j ∈ J ;
C − 1 ≤ ℓ ( R i ) / ℓ ( R j ) ≤ C C^{-1} \le\ell(R_i)/\ell(R_j) \le C C − 1 ≤ ℓ ( R i ) / ℓ ( R j ) ≤ C for all i , j ∈ J i,j \in J i , j ∈ J such that A B R i ∩ A B R j ≠ ∅ AB_{R_i} \cap AB_{R_j} \ne\varnothing A B R i ∩ A B R j = ∅ .
Then, provided
τ 0 = τ 0 ( C 5 , M 1 ) , θ 0 = θ 0 ( τ 0 ) and α 0 \tau_0=\tau_0(C_5,M_1),\quad\theta_0=\theta_0(\tau_0)\quad\text{and}\quad\alpha_0 τ 0 = τ 0 ( C 5 , M 1 ) , θ 0 = θ 0 ( τ 0 ) and α 0 are sufficiently small and
M 1 = M 1 ( A 1 ) , A 0 = A 0 ( C 5 , M 1 ) and k 1 = k 1 ( τ 0 ) M_1=M_1(A_1),\quad A_0=A_0(C_5,M_1)\quad\text{and}\quad k_1=k_1(\tau_0) M 1 = M 1 ( A 1 ) , A 0 = A 0 ( C 5 , M 1 ) and k 1 = k 1 ( τ 0 ) are sufficiently large, we have that ⋃ j ∈ J G R j \bigcup_{j\in J}G_{R_j} ⋃ j ∈ J G R j is Ahlfors d d d -regular with constants depending on d d d , n n n , C 0 C_0 C 0 , C C C , and supports a ( 1 , 1 ) (1,1) ( 1 , 1 ) -Poincaré inequality with constant C P I = C P I ( d , n , A , C , C g r i d ) C_{\mathrm{PI}}=C_{\mathrm{PI}}(d,n,A,C,C_{\mathrm{grid}}) C PI = C PI ( d , n , A , C , C grid ) and λ P I = λ P I ( n , C , λ g r i d ) \lambda_{\mathrm{PI}}=\lambda_{\mathrm{PI}}(n,C,\lambda_{\mathrm{grid}}) λ PI = λ PI ( n , C , λ grid ) .
We will prove Proposition 5.7 by applying Proposition 3.8 with the collection of balls { B j } j ∈ J : = { B R j } j ∈ J \{B_j\}_{j\in J}:=\{B_{R_j}\}_{j\in J} { B j } j ∈ J := { B R j } j ∈ J , the space
Y = ⋃ j ∈ J G R j , Y=\bigcup_{j\in J}G_{R_j}, Y = j ∈ J ⋃ G R j ,
and X X X a (truncated) Lipschitz graph associated to the nearby stopping-time with flat root. To be more precise regarding X X X , let T ∈ T N ( k 0 ) T\in\mathcal{T}^N(k_0) T ∈ T N ( k 0 ) such that Q ( T ) ∈ G ( k 0 ) Q(T)\in\mathcal{G}(k_0) Q ( T ) ∈ G ( k 0 ) and 2 M 1 B Q ( T ) ∩ 2 M 1 B Q ( S ) ≠ ∅ 2M_1B_{Q(T)}\cap2M_1B_{Q(S)}\ne\varnothing 2 M 1 B Q ( T ) ∩ 2 M 1 B Q ( S ) = ∅ (which exists by the definition of G 2 N ( k 0 ) \mathcal{G}_2^N(k_0) G 2 N ( k 0 ) , recall (39) ). Let
{ Q i } i ∈ I : = { Q ∈ Top ( N , k 0 ) : Q ∩ 5 C 5 M 1 B Q ( T ) ≠ ∅ } , \{Q_i\}_{i\in I}:=\{Q\in\operatorname{Top}(N,k_0):Q\cap5C_5M_1B_{Q(T)}\ne\varnothing\}, { Q i } i ∈ I := { Q ∈ Top ( N , k 0 ) : Q ∩ 5 C 5 M 1 B Q ( T ) = ∅ } ,
where C 5 C_5 C 5 is as in Lemma 4.15. Assume that A 0 = A 0 ( C 5 , M 1 ) A_0=A_0(C_5,M_1) A 0 = A 0 ( C 5 , M 1 ) is large enough and τ 0 = τ 0 ( C 5 , M 1 ) \tau_0=\tau_0(C_5,M_1) τ 0 = τ 0 ( C 5 , M 1 ) is small enough so that we can apply Lemma 4.12 with the collection { Q i } i ∈ I \{Q_i\}_{i\in I} { Q i } i ∈ I and Q ( T ) Q(T) Q ( T ) . Let Γ \Gamma Γ be the resulting graph, then set
X : = Γ ∩ π T − 1 ( P Q ( T ) ∩ A B Q ( S ) ) . X:=\Gamma\cap\pi_T^{-1}\left(P_{Q(T)}\cap AB_{Q(S)}\right). X := Γ ∩ π T − 1 ( P Q ( T ) ∩ A B Q ( S ) ) .
It only remains to check the hypotheses of Proposition 3.8 with this data. We will check it with
C 1 : = 6 max { n λ 1 , C 5 } C_1:=6\max\{\sqrt{n}\lambda_1,C_5\} C 1 := 6 max { n λ 1 , C 5 }
(the remaining constants shall be found shortly). This is done in Lemmas 5.8–5.12 below, recall that some of these hypotheses are stated in Lemma 2.10.
Lemma 5.8. Proposition 3.8 {(1)} and {(2)} with constants c 4 = c 0 c_4=c_0 c 4 = c 0 , C 3 = A C_3=A C 3 = A and C 4 = C C_4=C C 4 = C , and Lemma 2.10 {(1)} and {(2)} hold with constants C 1 C_1 C 1 (as above) and C 2 C_2 C 2 depending on d d d , C C C , C 0 C_0 C 0 , C 1 C_1 C 1 and c 0 c_0 c 0 .
Proof. Proposition 3.8 (1) holds with constant c 4 = c 0 c_4=c_0 c 4 = c 0 since { R j } j ∈ J \{R_j\}_{j\in J} { R j } j ∈ J is a disjoint collection of cubes. Proposition 3.8 (2) holds with C 3 = A C_3=A C 3 = A and C 4 = C C_4=C C 4 = C by assumption Lemma 5.7 (3). Lemma 2.10 (2) follows from Lemma 6.1. Lemma 2.10 (1) follows from Proposition 3.8 (1) and (2) by a volume argument. □
Lemma 5.9. Lemma 2.10 (3) holds with C 1 C_1 C 1 .
Proof. Fix some j ∈ J j\in J j ∈ J and x ∈ E ∩ 3 B R j ∩ A B Q ( S ) x\in E\cap3B_{R_j}\cap AB_{Q(S)} x ∈ E ∩ 3 B R j ∩ A B Q ( S ) , which exists by assumption (Lemma 5.7 (2)). By Lemma 4.12 , there exists some y ∈ Γ ∩ 4 B R j y\in\Gamma\cap4B_{R_j} y ∈ Γ ∩ 4 B R j such that ∣ π T ( y ) − π T ( x Q ( S ) ) ∣ ≤ ℓ ( R j ) \lvert\pi_T(y)-\pi_T(x_{Q(S)})\rvert\leq\ell(R_j) ∣ π T ( y ) − π T ( x Q ( S ) )∣ ≤ ℓ ( R j ) . By sliding along Γ \Gamma Γ , we can find a further point z ∈ Γ ∩ 5 B R j z\in\Gamma\cap5B_{R_j} z ∈ Γ ∩ 5 B R j such that π T ( z ) ∈ P Q ( T ) ∩ A B Q ( S ) \pi_T(z)\in P_{Q(T)}\cap AB_{Q(S)} π T ( z ) ∈ P Q ( T ) ∩ A B Q ( S ) i.e., z ∈ X z\in X z ∈ X . □
Lemma 5.10. Lemma 2.10 (4) holds with constant C 1 C_1 C 1 .
Proof. Observe to begin with that
E ∩ 2 A B Q ( S ) ⊆ E ∩ 5 C 5 M 1 B Q ( T ) ⊆ ⋃ i ∈ I Q i . (51) E\cap2AB_{Q(S)}\subseteq E\cap5C_5M_1B_{Q(T)}\subseteq\bigcup_{i\in I}Q_i. \tag*{(51)} E ∩ 2 A B Q ( S ) ⊆ E ∩ 5 C 5 M 1 B Q ( T ) ⊆ i ∈ I ⋃ Q i . (51)
Indeed, the first inclusion uses Lemma 4.15 along with the fact that 2 M 1 B Q ( S ) ∩ 2 M 1 B Q ( T ) ≠ ∅ 2M_1B_{Q(S)}\cap2M_1B_{Q(T)}\ne\varnothing 2 M 1 B Q ( S ) ∩ 2 M 1 B Q ( T ) = ∅ . The second inclusion follows since E ∩ 5 C 5 M 1 B Q ( T ) E\cap5C_5M_1B_{Q(T)} E ∩ 5 C 5 M 1 B Q ( T ) is covered by the set of Q ∈ Top ( N ) Q\in\operatorname{Top}(N) Q ∈ Top ( N ) such that Q ∩ 5 C 5 M 1 B Q ( T ) ≠ ∅ Q\cap5C_5M_1B_{Q(T)}\ne\varnothing Q ∩ 5 C 5 M 1 B Q ( T ) = ∅ . Since Q ( T ) ∈ G ( k 0 ) Q(T)\in\mathcal{G}(k_0) Q ( T ) ∈ G ( k 0 ) , our choice of k 1 k_1 k 1 (see Constants 5.1) implies that any such cube lies in D ≤ k 0 \mathcal{D}_{\leq k_0} D ≤ k 0 i.e., Q ∈ { Q i } i ∈ I Q\in\{Q_i\}_{i\in I} Q ∈ { Q i } i ∈ I .
By (5.6) and Lemma 4.16 , we have
X ⊆ ⋃ i ∈ I ⋃ R ∈ min ( S Q i ( k 0 ) ) Γ ∩ A B Q ( S ) ∩ 2 B R ≠ ∅ 2 B R . X\subseteq\bigcup_{i\in I}\ \bigcup_{\substack{R\in\min(S_{Q_i}(k_0))\\ \Gamma\cap AB_{Q(S)}\cap2B_R\ne\varnothing}}2B_R. X ⊆ i ∈ I ⋃ R ∈ m i n ( S Q i ( k 0 )) Γ ∩ A B Q ( S ) ∩ 2 B R = ∅ ⋃ 2 B R .
Fix some R ∈ min ( S Q i ( k 0 ) ) R\in\min(S_{Q_i}(k_0)) R ∈ min ( S Q i ( k 0 )) such that Γ ∩ A B Q ( S ) ∩ 2 B R ≠ ∅ \Gamma\cap AB_{Q(S)}\cap2B_R\ne\varnothing Γ ∩ A B Q ( S ) ∩ 2 B R = ∅ . We claim that there exists j ∈ J j\in J j ∈ J such that
2 B R ⊆ 6 C 5 B R j , (52) 2B_R\subseteq6C_5B_{R_j}, \tag*{(52)} 2 B R ⊆ 6 C 5 B R j , (52)
which is sufficient for Lemma 2.10 (4). Indeed, by travelling along Γ \Gamma Γ slightly, we can find a point y ∈ Γ ∩ 3 B R y\in\Gamma\cap3B_R y ∈ Γ ∩ 3 B R such that B ( y , 5 θ 0 ℓ ( R ) ) ⊆ A B Q ( S ) B(y,5\theta_0\ell(R))\subseteq AB_{Q(S)} B ( y , 5 θ 0 ℓ ( R )) ⊆ A B Q ( S ) . By Lemma 4.12 there exists a point z ∈ E ∩ A B Q ( S ) ∩ 4 B R z\in E\cap AB_{Q(S)}\cap4B_R z ∈ E ∩ A B Q ( S ) ∩ 4 B R such that ∣ y − z ∣ ≤ θ 0 ℓ ( R ) \lvert y-z\rvert\leq\theta_0\ell(R) ∣ y − z ∣ ≤ θ 0 ℓ ( R ) . By assumption (Lemma 5.7 (1)), there exists j ∈ J j\in J j ∈ J such that z ∈ R j ∩ 4 B R z\in R_j\cap4B_R z ∈ R j ∩ 4 B R . It now follows from Lemma 4.15 that C 5 ℓ ( R j ) ≥ ℓ ( R ) C_5\ell(R_j)\geq\ell(R) C 5 ℓ ( R j ) ≥ ℓ ( R ) . Then, (5.7) follows from the triangle inequality. □
It only remains to check that Proposition 3.8 (3) holds with constant C 5 = 50 C 1 C_5=50C_1 C 5 = 50 C 1 . Note that C 5 ≤ A 1 ≤ A C_5\leq A_1\leq A C 5 ≤ A 1 ≤ A .
Fix an index j ∈ J j\in J j ∈ J and a pair of points x , y ∈ C 5 B j x,y\in C_5B_j x , y ∈ C 5 B j . Let Q , R ∈ { R i } i ∈ J Q,R\in\{R_i\}_{i\in J} Q , R ∈ { R i } i ∈ J such that x ∈ G Q x\in G_Q x ∈ G Q and y ∈ G R y\in G_R y ∈ G R . Before proving Proposition 3.8 (3), we will prove the following useful lemma.
Lemma 5.11. If A 0 = A 0 ( C 5 ) A_0=A_0(C_5) A 0 = A 0 ( C 5 ) is sufficiently large, and θ 0 = θ 0 ( τ 0 ) \theta_0=\theta_0(\tau_0) θ 0 = θ 0 ( τ 0 ) is sufficiently small, there is a finite sequence of cubes { T k } k = 0 K \{T_k\}_{k=0}^K { T k } k = 0 K , with K ≲ C , C ∗ 1 K\lesssim_{C,C_*}1 K ≲ C , C ∗ 1 , such that T 0 = Q T_0=Q T 0 = Q , T K = R T_K=R T K = R and 6 B T k ∩ 6 B T k + 1 ≠ ∅ 6B_{T_k}\cap6B_{T_{k+1}}\ne\varnothing 6 B T k ∩ 6 B T k + 1 = ∅ for each k ∈ { 0 , … , K − 1 } k\in\{0,\ldots,K-1\} k ∈ { 0 , … , K − 1 } .
Proof. Since G T ⊆ 2 n λ 1 B T G_T\subseteq2\sqrt{n}\lambda_1B_T G T ⊆ 2 n λ 1 B T for each T ∈ D T\in\mathcal{D} T ∈ D , we have
x ∈ 2 n λ 1 B Q ∩ C 5 B R j and y ∈ 2 n λ 1 B R ∩ C 5 B R j , (53) x\in2\sqrt{n}\lambda_1B_Q\cap C_5B_{R_j}
\quad\text{and}\quad
y\in2\sqrt{n}\lambda_1B_R\cap C_5B_{R_j}, \tag*{(53)} x ∈ 2 n λ 1 B Q ∩ C 5 B R j and y ∈ 2 n λ 1 B R ∩ C 5 B R j , (53)
Since A ≥ C 5 A \ge C_5 A ≥ C 5 , it follows by assumption (Lemma 5.7 (3)) that
ℓ ( Q ) ∼ C ℓ ( R j ) ∼ C ℓ ( R ) . \ell(Q) \sim_C \ell(R_j) \sim_C \ell(R). ℓ ( Q ) ∼ C ℓ ( R j ) ∼ C ℓ ( R ) .
Let Q ∗ ∈ { Q , R , R j } Q^* \in\{Q,R,R_j\} Q ∗ ∈ { Q , R , R j } have the maximal side-length.
One can check (from (53) and the triangle inequality) that
5 B Q ∪ 5 B R ⊆ λ B Q ∗ , (54) 5B_Q \cup5B_R \subseteq\lambda B_{Q^*}, \tag*{(54)} 5 B Q ∪ 5 B R ⊆ λ B Q ∗ , (54)
where
λ = 5 + 4 n λ 1 + 600 max { n λ 1 , C 5 } ≤ 100 n λ 1 C 5 ≤ A 1 / 3 . \lambda= 5 + 4\sqrt{n}\lambda_1 + 600\max\{\sqrt{n}\lambda_1,C_5\} \le100\sqrt{n}\lambda_1C_5 \le A^{1/3}. λ = 5 + 4 n λ 1 + 600 max { n λ 1 , C 5 } ≤ 100 n λ 1 C 5 ≤ A 1/3 .
By arguing as in the proof of Lemma 2.10 (3), there are points z Q ∈ X ∩ 5 B Q z_Q \in X \cap5B_Q z Q ∈ X ∩ 5 B Q and z R ∈ X ∩ 5 B R z_R \in X \cap5B_R z R ∈ X ∩ 5 B R . Let L L L be the line segment from z Q z_Q z Q to z R z_R z R such that π T ( Γ L ) = π T ( L ) \pi_T(\Gamma_L)=\pi_T(L) π T ( Γ L ) = π T ( L ) . Let β > 0 \beta>0 β > 0 (to be chosen small enough, depending on C C C ) and { w k } k = 0 K \{w_k\}_{k=0}^K { w k } k = 0 K be a collection of points in L L L such that w 0 = π T ( z Q ) w_0=\pi_T(z_Q) w 0 = π T ( z Q ) , w K = π T ( z R ) w_K=\pi_T(z_R) w K = π T ( z R ) and
β ℓ ( Q ∗ ) / 2 ≤ ∣ w i + 1 − w i ∣ ≤ β ℓ ( Q ∗ ) . (55) \beta\ell(Q^*)/2 \le|w_{i+1}-w_i| \le\beta\ell(Q^*). \tag*{(55)} β ℓ ( Q ∗ ) /2 ≤ ∣ w i + 1 − w i ∣ ≤ β ℓ ( Q ∗ ) . (55)
By (54) and (55) , we have
K ≲ C ∗ , β 1. K \lesssim_{C^*,\beta} 1. K ≲ C ∗ , β 1.
Let Γ L ⊆ Γ \Gamma_L \subseteq\Gamma Γ L ⊆ Γ such that π T ( Γ L ) = π T ( L ) \pi_T(\Gamma_L)=\pi_T(L) π T ( Γ L ) = π T ( L ) , and let { y k } k = 0 K \{y_k\}_{k=0}^K { y k } k = 0 K be the corresponding points in Γ L \Gamma_L Γ L , i.e., π T ( y k ) = w k \pi_T(y_k)=w_k π T ( y k ) = w k . By convexity, it follows from (54) that L ⊆ λ B Q ∗ L\subseteq\lambda B_{Q^*} L ⊆ λ B Q ∗ . Since Γ \Gamma Γ is a C α 0 C\alpha_0 C α 0 -Lipschitz graph, we have
dist ( y , L ) ≤ ( 1 + C α 0 ) ∣ z Q − z R ∣ for all y ∈ Γ L . \operatorname{dist}(y,L) \le(1+C\alpha_0)|z_Q-z_R| \quad\text{for all } y\in\Gamma_L. dist ( y , L ) ≤ ( 1 + C α 0 ) ∣ z Q − z R ∣ for all y ∈ Γ L .
These two facts imply
{ y k } k = 0 K ⊆ 2 λ B Q ∗ . \{y_k\}_{k=0}^K \subseteq2\lambda B_{Q^*}. { y k } k = 0 K ⊆ 2 λ B Q ∗ .
If A 0 = A 0 ( λ ) A_0=A_0(\lambda) A 0 = A 0 ( λ ) is large enough, it follows from Lemma 4.12 that for each y i y_i y i there is z i ∈ E ∩ A B Q ( S ) ∩ 3 λ B Q ∗ z_i\in E\cap AB_{Q^{(S)}}\cap3\lambda B_{Q^*} z i ∈ E ∩ A B Q ( S ) ∩ 3 λ B Q ∗ such that ∣ y i − z i ∣ ≤ θ 0 ℓ ( Q ∗ ) |y_i-z_i|\le\theta_0\ell(Q^*) ∣ y i − z i ∣ ≤ θ 0 ℓ ( Q ∗ ) . Let T 0 = Q T_0=Q T 0 = Q and T K = R T_K=R T K = R . For the remaining indices, let T i ∈ { R j } j ∈ J T_i\in\{R_j\}_{j\in J} T i ∈ { R j } j ∈ J such that z i ∈ T i z_i\in T_i z i ∈ T i . Since z i ∈ 3 λ B Q ∗ z_i\in3\lambda B_{Q^*} z i ∈ 3 λ B Q ∗ , it follows by assumption that ℓ ( T i ) ∼ C ℓ ( Q ∗ ) \ell(T_i)\sim_C\ell(Q^*) ℓ ( T i ) ∼ C ℓ ( Q ∗ ) (recall that A ≥ A 1 ≥ 3 λ A\ge A_1\ge3\lambda A ≥ A 1 ≥ 3 λ ). Thus, using (55) and the fact that X X X is the (truncated) graph of an C α 0 C\alpha_0 C α 0 -Lipschitz function, we have
∣ z i + 1 − z i ∣ ≤ 2 θ 0 ℓ ( Q ∗ ) + ∣ y i + 1 − y i ∣ ≤ 2 θ 0 ℓ ( Q ∗ ) + 2 ∣ w i + 1 − w i ∣ ≤ C ( θ 0 + β ) ℓ ( T i ) . |z_{i+1}-z_i| \le2\theta_0\ell(Q^*)+|y_{i+1}-y_i| \le2\theta_0\ell(Q^*)+2|w_{i+1}-w_i|
\le_C(\theta_0+\beta)\ell(T_i). ∣ z i + 1 − z i ∣ ≤ 2 θ 0 ℓ ( Q ∗ ) + ∣ y i + 1 − y i ∣ ≤ 2 θ 0 ℓ ( Q ∗ ) + 2∣ w i + 1 − w i ∣ ≤ C ( θ 0 + β ) ℓ ( T i ) .
Since z i ∈ 5 B T i z_i\in5B_{T_i} z i ∈ 5 B T i , if θ 0 \theta_0 θ 0 and β \beta β are small enough (depending on C C C ), this implies z i + 1 ∈ 6 B T i z_{i+1}\in6B_{T_i} z i + 1 ∈ 6 B T i . Since z i + 1 ∈ 5 B T i + 1 z_{i+1}\in5B_{T_{i+1}} z i + 1 ∈ 5 B T i + 1 , this finishes the proof. □
Lemma 5.12. Proposition 3.8 (3) holds with constant C 6 C_6 C 6 depending only on C C C , C 5 C_5 C 5 and n n n .
Proof. Let { T k } k = 0 K \{T_k\}_{k=0}^K { T k } k = 0 K be the sequence obtain by applying Lemma 5.11 and consider the set
G : = ⋃ k = 0 K G T k . G := \bigcup_{k=0}^K G_{T_k}. G := k = 0 ⋃ K G T k .
By (46) and Lemma 5.3 , we see that G T k + 1 ∩ G T k i ≠ ∅ G_{T_{k+1}}\cap G_{T_k}^i\ne\varnothing G T k + 1 ∩ G T k i = ∅ for each k ∈ { 0 , … , K − 1 } k\in\{0,\ldots,K-1\} k ∈ { 0 , … , K − 1 } . In particular, G T k + 1 ∪ G T k G_{T_{k+1}}\cup G_{T_k} G T k + 1 ∪ G T k is connected. Since this hold for arbitrary k k k , we have that G G G is connected. Thus, by Lemma 5.5 , G G G supports a (1,1)-Poincaré inequality with constant depending on K = K ( C , C 5 ) K=K(C,C_5) K = K ( C , C 5 ) and n n n . Since x , y ∈ G ⊆ Y x,y\in G\subseteq Y x , y ∈ G ⊆ Y , this completes the proof. □
Constants 5.13. We will assume for the remainder of the paper that the constants ( τ 0 , θ 0 , M 1 , A 0 , κ 1 ) (\tau_0,\theta_0,M_1,A_0,\kappa_1) ( τ 0 , θ 0 , M 1 , A 0 , κ 1 ) are chosen such that the Proposition 5.7 holds.
Approximations for layers of cubes
In this short section we define the sets Σ S , k \Sigma_{S,k} Σ S , k and Σ N , k \Sigma_{N,k} Σ N , k . The first set will be a skeletal approximation for a full stopping-time region S ∈ T N ( k ) S\in\mathcal{T}^N(k) S ∈ T N ( k ) (as constructed in § 5 ). The second set will simply be the union each of the skeletal approximation at layer N N N i.e.,
Σ N , k : = ⋃ S ∈ T N ( k ) Σ S , k . \Sigma_{N,k}:=\bigcup_{S\in\mathcal{T}^N(k)}\Sigma_{S,k}. Σ N , k := S ∈ T N ( k ) ⋃ Σ S , k .
These will be constructed from the collection { G Q } Q ∈ D \{G_Q\}_{Q\in\mathcal{D}} { G Q } Q ∈ D . However, the definition will depend on whether S S S lies G 1 N ( k ) \mathcal{G}_1^N(k) G 1 N ( k ) or B 1 N ( k ) \mathcal{B}_1^N(k) B 1 N ( k ) (recall (39) and (40) ). Indeed, for S ∈ G 1 N ( k ) S\in\mathcal{G}_1^N(k) S ∈ G 1 N ( k ) , let
Σ S , k : = ⋃ Q ∈ Stop ( N , k ) 3 B Q ∩ 2 ρ − 1 B Q ( S ) ≠ ∅ G Q . \Sigma_{S,k}:=\bigcup_{\substack{Q\in\operatorname{Stop}(N,k)\\3B_Q\cap2\rho^{-1}B_{Q(S)}\ne\varnothing}}G_Q. Σ S , k := Q ∈ Stop ( N , k ) 3 B Q ∩ 2 ρ − 1 B Q ( S ) = ∅ ⋃ G Q .
For S ∈ B 1 N ( k ) S\in\mathcal{B}_1^N(k) S ∈ B 1 N ( k ) , let
Σ S , k : = G Q ( S ) . \Sigma_{S,k}:=G_{Q(S)}. Σ S , k := G Q ( S ) .
Before proving more serious properties of these skeletal approximations in the next section, we end this section by showing two relatively easy properties. First, we show that each set Σ S , k \Sigma_{S,k} Σ S , k is upper regular and contained in a sufficiently large dilate of the ball B Q ( S ) B_{Q(S)} B Q ( S ) .
Lemma 6.1. For each S ∈ T N ( k ) S\in\mathcal{T}^N(k) S ∈ T N ( k ) , we have that Σ S , k \Sigma_{S,k} Σ S , k is upper Ahlfors d d d -regular and
Σ S , k ⊆ 2 n λ 1 B Q ( S ) . (56) \Sigma_{S,k}\subseteq2\sqrt{n}\lambda_1B_{Q(S)}. \tag*{(56)} Σ S , k ⊆ 2 n λ 1 B Q ( S ) . (56) Proof. The upper regularity of Σ S , k \Sigma_{S,k} Σ S , k follows directly from Proposition 5.7 . For (56) , if S ∈ B 1 N ( k ) S\in\mathcal{B}_1^N(k) S ∈ B 1 N ( k ) , this follows directly from (46) . If S ∈ G 1 N ( k ) S\in\mathcal{G}_1^N(k) S ∈ G 1 N ( k ) and Q ∈ Stop ( N , k ) Q\in\operatorname{Stop}(N,k) Q ∈ Stop ( N , k ) are such that 3 B Q ∩ 2 ρ − 1 B Q ( S ) ≠ ∅ 3B_Q\cap2\rho^{-1}B_{Q(S)}\ne\varnothing 3 B Q ∩ 2 ρ − 1 B Q ( S ) = ∅ , it follows from Lemma 4.25 that ℓ ( Q ) ≤ ρ ℓ ( Q ( S ) ) \ell(Q)\leq\rho\ell(Q(S)) ℓ ( Q ) ≤ ρ ℓ ( Q ( S )) . Thus, for each x ∈ G Q ⊆ 2 n λ 1 B Q x\in G_Q\subseteq2\sqrt{n}\lambda_1B_Q x ∈ G Q ⊆ 2 n λ 1 B Q , we have
∣ x − x Q ( S ) ∣ ≤ ( 3 + 2 n λ 1 ) ℓ ( Q ) + 2 ρ − 1 ℓ ( Q ( S ) ) \lvert x-x_{Q(S)}\rvert\leq(3+2\sqrt{n}\lambda_1)\ell(Q)+2\rho^{-1}\ell(Q(S)) ∣ x − x Q ( S ) ∣ ≤ ( 3 + 2 n λ 1 ) ℓ ( Q ) + 2 ρ − 1 ℓ ( Q ( S )) ≤ ( 5 n ( 1 + 4 ρ − 1 ) ρ + 2 ρ − 1 ) ℓ ( Q ( S ) ) \leq(5\sqrt{n}(1+4\rho^{-1})\rho+2\rho^{-1})\ell(Q(S)) ≤ ( 5 n ( 1 + 4 ρ − 1 ) ρ + 2 ρ − 1 ) ℓ ( Q ( S )) ≤ ( 25 n + 2 ρ − 1 ) ℓ ( Q ( S ) ) ≤ 2 n λ 1 ℓ ( Q ( S ) ) , \leq(25\sqrt{n}+2\rho^{-1})\ell(Q(S))\leq2\sqrt{n}\lambda_1\ell(Q(S)), ≤ ( 25 n + 2 ρ − 1 ) ℓ ( Q ( S )) ≤ 2 n λ 1 ℓ ( Q ( S )) , as required. □
Lemma 6.2. Let S ∈ T N ( k ) S\in\mathcal{T}^N(k) S ∈ T N ( k ) ,
Q ∈ { min ( S ( k ) ) if S ∈ G 1 N ( k ) ; min ( S ( k ) ) ∪ { Q ( S ) } if S ∈ B 1 N ( k ) . Q\in
\begin{cases}
\min(S(k)) & \text{if }S\in\mathcal{G}_1^N(k);\\
\min(S(k))\cup\{Q(S)\} & \text{if }S\in\mathcal{B}_1^N(k).
\end{cases} Q ∈ { min ( S ( k )) min ( S ( k )) ∪ { Q ( S )} if S ∈ G 1 N ( k ) ; if S ∈ B 1 N ( k ) .
and R ∈ Child ( Q ) R \in\operatorname{Child}(Q) R ∈ Child ( Q ) such that x R = x Q x_R=x_Q x R = x Q . Then, supposing k 1 = k 1 ( τ 0 ) k_1=k_1(\tau_0) k 1 = k 1 ( τ 0 ) is sufficiently large, there is a d d d -dimensional I Q I_Q I Q such that
I Q ⊆ Σ S , k ∩ ( c 0 / 8 ) B R and ℓ ( I Q ) ≥ c 0 n − 1 / 2 ℓ ( R ) / 10. I_Q \subseteq\Sigma_{S,k} \cap(c_0/8)B_R \qquad\text{and} \qquad\ell(I_Q) \ge c_0 n^{-1/2}\ell(R)/10. I Q ⊆ Σ S , k ∩ ( c 0 /8 ) B R and ℓ ( I Q ) ≥ c 0 n − 1/2 ℓ ( R ) /10.
Proof. Let k 3 ∈ N k_3 \in\mathbb{N} k 3 ∈ N such that
ρ c 0 n − 1 / 2 ℓ ( Q ) / 20 < 2 − k 3 ≤ ρ c 0 n − 1 / 2 ℓ ( Q ) / 10. \rho c_0 n^{-1/2}\ell(Q)/20 < 2^{-k_3} \le\rho c_0 n^{-1/2}\ell(Q)/10. ρ c 0 n − 1/2 ℓ ( Q ) /20 < 2 − k 3 ≤ ρ c 0 n − 1/2 ℓ ( Q ) /10.
Let K K K be the dyadic cube such that K ∈ Δ k 3 K \in\Delta_{k_3} K ∈ Δ k 3 and x Q ∈ K x_Q \in K x Q ∈ K , and let I Q I_Q I Q be an arbitrary d d d -dimensional subface. By our choice of k 3 k_3 k 3 , we have diam ( K ) ≤ ( ρ c 0 / 10 ) ℓ ( Q ) \operatorname{diam}(K) \le(\rho c_0/10)\ell(Q) diam ( K ) ≤ ( ρ c 0 /10 ) ℓ ( Q ) . Since x Q ∈ K x_Q \in K x Q ∈ K and I Q ⊆ K I_Q \subseteq K I Q ⊆ K , this implies
I Q ⊆ ( ρ c 0 / 8 ) B Q = ( c 0 / 8 ) B R I_Q \subseteq(\rho c_0/8)B_Q=(c_0/8)B_R I Q ⊆ ( ρ c 0 /8 ) B Q = ( c 0 /8 ) B R
ℓ ( I Q ) ≥ ρ c 0 n − 1 / 2 ℓ ( Q ) / 10 = c 0 n − 1 / 2 ℓ ( R ) / 10. (57) \ell(I_Q) \ge\rho c_0 n^{-1/2}\ell(Q)/10=c_0 n^{-1/2}\ell(R)/10. \tag*{(57)} ℓ ( I Q ) ≥ ρ c 0 n − 1/2 ℓ ( Q ) /10 = c 0 n − 1/2 ℓ ( R ) /10. (57)
It only remains to show that
I Q ⊆ Σ S , k . (58) I_Q \subseteq\Sigma_{S,k}. \tag*{(58)} I Q ⊆ Σ S , k . (58)
Suppose to begin with that S ∈ G 1 N ( k ) S \in\mathcal{G}_1^N(k) S ∈ G 1 N ( k ) and Q ∈ min ( S ( k ) ) Q \in\min(S(k)) Q ∈ min ( S ( k )) . By Lemma 5.2 and (57) we have I Q ⊆ G Q I_Q \subseteq G_Q I Q ⊆ G Q for k 1 k_1 k 1 sufficiently large. Since, G Q ⊆ Σ S , k G_Q \subseteq\Sigma_{S,k} G Q ⊆ Σ S , k in this case, this implies (58) . Suppose now that S ∈ B 1 N ( k ) S \in\mathcal{B}_1^N(k) S ∈ B 1 N ( k ) . If Q = Q ( S ) Q=Q(S) Q = Q ( S ) then the same argument as above give (58) , so let us suppose Q ∈ min ( S ( k ) ) Q \in\min(S(k)) Q ∈ min ( S ( k )) . By Lemma 4.26 and (57) , we have
ℓ ( I Q ) ≳ ℓ ( Q ) ≳ τ 0 ℓ ( Q ( S ) ) . \ell(I_Q) \gtrsim\ell(Q) \gtrsim\tau_0\ell(Q(S)). ℓ ( I Q ) ≳ ℓ ( Q ) ≳ τ 0 ℓ ( Q ( S )) .
Thus, for k 1 = k 1 ( τ 0 ) k_1=k_1(\tau_0) k 1 = k 1 ( τ 0 ) sufficiently small, Lemma 5.2 implies I Q ⊆ G Q ( S ) = Σ S , k I_Q \subseteq G_{Q(S)}=\Sigma_{S,k} I Q ⊆ G Q ( S ) = Σ S , k , as required. □
Next, we show every point in Q ∈ S ( k ) Q \in S(k) Q ∈ S ( k ) , for some S ∈ T N ( k ) S \in\mathcal{T}^N(k) S ∈ T N ( k ) , lies close to Σ S , k \Sigma_{S,k} Σ S , k .
Lemma 6.3. Let S ∈ T N ( k ) S \in\mathcal{T}^N(k) S ∈ T N ( k ) , Q ∈ S ( k ) Q \in S(k) Q ∈ S ( k ) and x ∈ Q x \in Q x ∈ Q . Then,
dist ( x , Σ S , k ) ≤ 2 ℓ ( Q ) . \operatorname{dist}(x,\Sigma_{S,k}) \le2\ell(Q). dist ( x , Σ S , k ) ≤ 2 ℓ ( Q ) .
Proof. By definition, there exists R ∈ min ( S ( k ) ) R \in\min(S(k)) R ∈ min ( S ( k )) such that R ⊆ Q R \subseteq Q R ⊆ Q . By Lemma 6.2, there is a point z ∈ Σ S , k ∩ ( c 0 / 8 ) B R ⊆ Σ S , k ∩ B Q z \in\Sigma_{S,k} \cap(c_0/8)B_R \subseteq\Sigma_{S,k} \cap B_Q z ∈ Σ S , k ∩ ( c 0 /8 ) B R ⊆ Σ S , k ∩ B Q . Since x ∈ Q ⊆ B Q x \in Q \subseteq B_Q x ∈ Q ⊆ B Q , the lemma follows by the triangle inequality. □
Local Poincaré inequalities for a single layer
In later sections we will want to glue the sets { Σ N , k } N ∈ N \{\Sigma_{N,k}\}_{N\in\mathbb{N}} { Σ N , k } N ∈ N that we defined in the previous section. We do this by first lifting these sets into R n + 1 \mathbb{R}^{n+1} R n + 1 , and then building bridges between consecutive layers. It will be important to know that any point in Σ S , k \Sigma_{S,k} Σ S , k (more accurately, a lifted version of this set) can be connected to one of these bridges within some larger subspace of Σ N , k \Sigma_{N,k} Σ N , k which supports a Poincaré inequality. We call this set Γ S , k \Gamma_{S,k} Γ S , k . There will be at least one neighbour of S S S (which we call U S U_S U S ) which is connected to a bridge, so we only need to know that Γ S , k \Gamma_{S,k} Γ S , k connects Σ S , k \Sigma_{S,k} Σ S , k and Σ U S , k \Sigma_{U_S,k} Σ U S , k in a nice way. It will also be useful to know how these sets { Γ S , k } S ∈ T N ( k ) \{\Gamma_{S,k}\}_{S\in\mathcal{T}^N(k)} { Γ S , k } S ∈ T N ( k ) interact with each other. It turns our that we can build these sets in such a ways that any pair is either contained in an even larger space supporting a Poincaré inequality or is well-separated. Let us be more precise.
Notation 7.1. For S ∈ T N ( k ) S \in\mathcal{T}^N(k) S ∈ T N ( k ) , let U S ∈ T N ( k ) U_S \in\mathcal{T}^N(k) U S ∈ T N ( k ) (possibly equal to S S S ) be the stopping-time region such that x Q ( U S ) = x Q x_{Q(U_S)}=x_Q x Q ( U S ) = x Q , where Q ∈ Stop ( N − 1 ) Q \in\operatorname{Stop}(N-1) Q ∈ Stop ( N − 1 ) is the parent of Q ( S ) Q(S) Q ( S ) .
Constants 7.2. Let A 2 = A 1 5 A_2=A_1^5 A 2 = A 1 5 .
Our main goal in this section is to prove the following.
Proposition 7.3. Let k ∈ N k\in\mathbb{N} k ∈ N . There is a collection of sets { Γ S , k } S ∈ T N ( k ) \{\Gamma_{S,k}\}_{S\in\mathcal{T}^{N}(k)} { Γ S , k } S ∈ T N ( k ) in Σ N , k \Sigma_{N,k} Σ N , k , each of which is Ahlfors d d d -regular and supports a ( 1 , 1 ) (1,1) ( 1 , 1 ) -Poincaré inequality with uniform constants such that
Σ S , k ∪ Σ U S , k ⊆ Γ S , k (59) \Sigma_{S,k}\cup\Sigma_{U_S,k}\subseteq\Gamma_{S,k} \tag*{(59)} Σ S , k ∪ Σ U S , k ⊆ Γ S , k (59) for each S ∈ T N ( k ) S\in\mathcal{T}^{N}(k) S ∈ T N ( k ) . If S , T ∈ T N ( k ) S,T\in\mathcal{T}^{N}(k) S , T ∈ T N ( k ) are such that A 1 B Q ( S ) ∩ A 1 B Q ( T ) ≠ ∅ A_1B_{Q(S)}\cap A_1B_{Q(T)}\ne\varnothing A 1 B Q ( S ) ∩ A 1 B Q ( T ) = ∅ , either there exists a set
Γ S , T , k ⊆ Σ N , k ∩ A 2 B Q ( S ) \Gamma_{S,T,k}\subseteq\Sigma_{N,k}\cap A_2B_{Q(S)} Γ S , T , k ⊆ Σ N , k ∩ A 2 B Q ( S ) which contains Γ S , k ∪ Γ T , k \Gamma_{S,k}\cup\Gamma_{T,k} Γ S , k ∪ Γ T , k and supports a ( 1 , 1 ) (1,1) ( 1 , 1 ) -Poincaré inequality, or
dist ( Γ S , k , Γ T , k ) ≳ ℓ ( Q ( S ) ) + ℓ ( Q ( T ) ) . (60) \operatorname{dist}(\Gamma_{S,k},\Gamma_{T,k})\gtrsim\ell(Q(S))+\ell(Q(T)). \tag*{(60)} dist ( Γ S , k , Γ T , k ) ≳ ℓ ( Q ( S )) + ℓ ( Q ( T )) . (60) In the case that 2 ρ − 1 B Q ( S ) ∩ 2 ρ − 1 B Q ( T ) ≠ ∅ 2\rho^{-1}B_{Q(S)}\cap2\rho^{-1}B_{Q(T)}\ne\varnothing 2 ρ − 1 B Q ( S ) ∩ 2 ρ − 1 B Q ( T ) = ∅ , the first alternative always holds.
Before defining the sets Γ S , k \Gamma_{S,k} Γ S , k and Γ S , T , k \Gamma_{S,T,k} Γ S , T , k , we show that, for any given S ∈ G 2 N ( k ) S\in\mathcal{G}^{N}_{2}(k) S ∈ G 2 N ( k ) , there exists a family of sets satisfying something like the conclusion of Proposition 7.3 (and, indeed, Γ S \Gamma_S Γ S will be chosen to be one of these sets for some particular choice of parameter).
Lemma 7.4. For each A ∈ [ A 1 , A 1 2 ] A\in[A_1,A_1^2] A ∈ [ A 1 , A 1 2 ] and S ∈ G 2 N ( k ) S\in\mathcal{G}^{N}_{2}(k) S ∈ G 2 N ( k ) there is a set
Γ S , k A ⊆ Σ N , k ∩ 10 A 2 B Q ( S ) \Gamma^{A}_{S,k}\subseteq\Sigma_{N,k}\cap10A^2B_{Q(S)} Γ S , k A ⊆ Σ N , k ∩ 10 A 2 B Q ( S ) which is Ahlfors d d d -regular, supports a ( 1 , 1 ) (1,1) ( 1 , 1 ) -Poincaré inequality and satisfies
Σ T , k ⊆ Γ S , k A (61) \Sigma_{T,k}\subseteq\Gamma^{A}_{S,k} \tag*{(61)} Σ T , k ⊆ Γ S , k A (61) for all T ∈ T N ( k ) T\in\mathcal{T}^{N}(k) T ∈ T N ( k ) such that Σ T ∩ A B Q ( S ) ≠ ∅ \Sigma_T\cap AB_{Q(S)}\ne\varnothing Σ T ∩ A B Q ( S ) = ∅ . Moreover, if A , A ′ ∈ [ A 1 , A 1 2 ] A,A'\in[A_1,A_1^2] A , A ′ ∈ [ A 1 , A 1 2 ] and S , S ′ ∈ G 2 N ( k ) S,S'\in\mathcal{G}^{N}_{2}(k) S , S ′ ∈ G 2 N ( k ) are such that A B Q ( S ) ⊆ A ′ B Q ( S ′ ) AB_{Q(S)}\subseteq A'B_{Q(S')} A B Q ( S ) ⊆ A ′ B Q ( S ′ ) , then
Γ S , k A ⊆ Γ S ′ , k A ′ . (62) \Gamma^{A}_{S,k}\subseteq\Gamma^{A'}_{S',k}. \tag*{(62)} Γ S , k A ⊆ Γ S ′ , k A ′ . (62)
Proof. Fix some A ∈ [ A 1 , A 1 2 ] A\in[A_1,A_1^2] A ∈ [ A 1 , A 1 2 ] and S ∈ G 2 N ( k ) S\in\mathcal{G}^{N}_{2}(k) S ∈ G 2 N ( k ) . Let A ∗ : = 4 n λ 1 C 5 + A ∈ [ A 1 , 2 A 1 2 ] A^*:=4\sqrt{n}\lambda_1C_5+A\in[A_1,2A_1^2] A ∗ := 4 n λ 1 C 5 + A ∈ [ A 1 , 2 A 1 2 ] and set
S : = { T ∈ T N ( k ) : E ∩ A ∗ B Q ( S ) ∩ Q ( T ) ≠ ∅ } ; \mathcal{S}:=\{T\in\mathcal{T}^{N}(k):E\cap A^*B_{Q(S)}\cap Q(T)\ne\varnothing\}; S := { T ∈ T N ( k ) : E ∩ A ∗ B Q ( S ) ∩ Q ( T ) = ∅ } ;
T : = { T ∈ T N ( k ) : Σ T , k ∩ A B Q ( S ) ≠ ∅ } . \mathcal{T}:=\{T\in\mathcal{T}^{N}(k):\Sigma_{T,k}\cap AB_{Q(S)}\ne\varnothing\}. T := { T ∈ T N ( k ) : Σ T , k ∩ A B Q ( S ) = ∅ } .
Recalling from Lemma 6.1 that Σ T ⊆ 2 n λ 1 B Q ( T ) \Sigma_T\subseteq2\sqrt{n}\lambda_1B_{Q(T)} Σ T ⊆ 2 n λ 1 B Q ( T ) , by applying Lemma 4.9 and the triangle inequality, one has
4 ρ − 1 B Q ( T ) ⊆ 2 n λ 1 B Q ( T ) ⊆ A ∗ B Q ( S ) (63) 4\rho^{-1}B_{Q(T)}\subseteq2\sqrt{n}\lambda_1B_{Q(T)}\subseteq A^*B_{Q(S)} \tag*{(63)} 4 ρ − 1 B Q ( T ) ⊆ 2 n λ 1 B Q ( T ) ⊆ A ∗ B Q ( S ) (63)
for each T ∈ T T\in\mathcal{T} T ∈ T . In particular,
T ⊆ S . \mathcal{T}\subseteq\mathcal{S}. T ⊆ S .
The same argument which is used to establish (63) can also be used to show that
2 n λ 1 B Q ( T ) ⊆ ( 4 n λ 1 C 5 + A ∗ ) B Q ( S ) ⊆ 3 A 1 2 B Q ( S ) (64) 2\sqrt{n}\lambda_1B_{Q(T)}\subseteq(4\sqrt{n}\lambda_1C_5+A^*)B_{Q(S)}\subseteq3A_1^2B_{Q(S)} \tag*{(64)} 2 n λ 1 B Q ( T ) ⊆ ( 4 n λ 1 C 5 + A ∗ ) B Q ( S ) ⊆ 3 A 1 2 B Q ( S ) (64)
for each T ∈ S T\in\mathcal{S} T ∈ S . Let
R T : = { R ∈ min ( T ( k ) ) : E ∩ A ∗ B Q ( S ) ∩ R ≠ ∅ } for T ∈ G 1 N ( k ) , \mathcal{R}_T:=\{R\in\min(\mathcal{T}(k)):E\cap A^*B_{Q(S)}\cap R\ne\varnothing\}\quad\text{for }T\in\mathcal{G}^{N}_{1}(k), R T := { R ∈ min ( T ( k )) : E ∩ A ∗ B Q ( S ) ∩ R = ∅ } for T ∈ G 1 N ( k ) ,
R T : = { Q ( T ) } for T ∈ B 1 N ( k ) , \mathcal{R}_T:=\{Q(T)\}\quad\text{for }T\in\mathcal{B}^{N}_{1}(k), R T := { Q ( T )} for T ∈ B 1 N ( k ) ,
and set
{ R j } j ∈ J : = ⋃ T ∈ J R T and Γ : = Γ S , k A : = ⋃ j ∈ J G R j . \{R_j\}_{j\in J}:=\bigcup_{T\in\mathcal{J}}\mathcal{R}_T
\quad\text{and}\quad
\Gamma:=\Gamma^A_{S,k}:=\bigcup_{j\in J}G_{R_j}. { R j } j ∈ J := T ∈ J ⋃ R T and Γ := Γ S , k A := j ∈ J ⋃ G R j .
It follows directly from (64) that Γ ⊆ 10 A 2 B Q ( S ) \Gamma\subseteq10A^2B_{Q(S)} Γ ⊆ 10 A 2 B Q ( S ) . Property (62) is also easy to see since the definitions of S \mathcal{S} S and R T \mathcal{R}_T R T are monotone with respect to set inclusions of balls.
To prove that Γ \Gamma Γ is Ahlfors regular and supports a Poincaré inequality we check that { R j } j ∈ J \{R_j\}_{j\in J} { R j } j ∈ J satisfies the hypotheses of Lemma 5.7 with constants A ∗ A^* A ∗ and C ≤ C 7 C\leq C_7 C ≤ C 7 . Lemma 5.7 (1) and (2) are immediate from the above definitions. Lemma 5.7 (3) follows from Lemma 4.15 whenever R i ∈ R T R_i\in\mathcal{R}_T R i ∈ R T and R j ∈ R T ′ R_j\in\mathcal{R}_{T'} R j ∈ R T ′ for some T , T ′ ∈ G 1 N ( k ) T,T'\in\mathcal{G}^N_1(k) T , T ′ ∈ G 1 N ( k ) and from Lemma 4.9 (recall also Remark 4.10) whenever T , T ′ ∈ B 1 N ( k ) T,T'\in\mathcal{B}^N_1(k) T , T ′ ∈ B 1 N ( k ) . Thus, it only remains to check Lemma 5.7 (3) in the case that R i ∈ min ( T ( k ) ) ⊆ Stop ( N , k ) R_i\in\min(\mathcal{T}(k))\subseteq\operatorname{Stop}(N,k) R i ∈ min ( T ( k )) ⊆ Stop ( N , k ) for some T ∈ G 1 N ( k ) T\in\mathcal{G}^N_1(k) T ∈ G 1 N ( k ) and R j = { Q ( T ′ ) } R_j=\{Q(T')\} R j = { Q ( T ′ )} for some T ′ ∈ B 1 N ( k ) T'\in\mathcal{B}^N_1(k) T ′ ∈ B 1 N ( k ) . Since A ∗ B Q ( T ) ∩ A ∗ B Q ( T ′ ) ≠ ∅ A^*B_{Q(T)}\cap A^*B_{Q(T')}\neq\varnothing A ∗ B Q ( T ) ∩ A ∗ B Q ( T ′ ) = ∅ , Lemma 4.9 implies
ℓ ( Q ( T ) ) ∼ ℓ ( Q ( T ′ ) ) and B Q ( T ) ⊆ ( M / 2 ) B Q ( T ′ ) \ell(Q(T))\sim\ell(Q(T'))
\quad\text{and}\quad
B_{Q(T)}\subseteq(M/2)B_{Q(T')} ℓ ( Q ( T )) ∼ ℓ ( Q ( T ′ )) and B Q ( T ) ⊆ ( M /2 ) B Q ( T ′ )
for M M M sufficiently large. In particular, R i ⊆ ( M / 2 ) B Q ( T ′ ) R_i\subseteq(M/2)B_{Q(T')} R i ⊆ ( M /2 ) B Q ( T ′ ) . Lemma 4.26 now implies that ℓ ( R i ) ≥ c 6 τ 0 ℓ ( Q ( T ′ ) ) = C 7 ℓ ( Q ( T ′ ) ) \ell(R_i)\geq c_6\tau_0\ell(Q(T'))=C_7\ell(Q(T')) ℓ ( R i ) ≥ c 6 τ 0 ℓ ( Q ( T ′ )) = C 7 ℓ ( Q ( T ′ )) , as required.
We finish the proof of Lemma 7.4 by checking (61) . Fix some T ∈ T T\in\mathcal{T} T ∈ T . If T ∈ B 1 N ( k ) T\in\mathcal{B}^N_1(k) T ∈ B 1 N ( k ) , then { Q ( S ) } = R T ⊆ { R j } j ∈ J \{Q(S)\}=\mathcal{R}_T\subseteq\{R_j\}_{j\in J} { Q ( S )} = R T ⊆ { R j } j ∈ J . Since Σ T , k = G Q ( T ) \Sigma_{T,k}=G_{Q(T)} Σ T , k = G Q ( T ) , the inclusion (61) now follows. Suppose instead that T ∈ G 1 N ( k ) T\in\mathcal{G}^N_1(k) T ∈ G 1 N ( k ) . Recalling the definition of Σ T , k \Sigma_{T,k} Σ T , k in this case, it suffices to show that
{ Q ∈ Stop ( N , k ) : 3 B Q ∩ 2 ρ − 1 B Q ( T ) ≠ ∅ } ⊆ ⋃ T ∈ S R T . (65) \{Q\in\operatorname{Stop}(N,k):3B_Q\cap2\rho^{-1}B_{Q(T)}\neq\varnothing\}
\subseteq
\bigcup_{T\in\mathcal{S}}\mathcal{R}_T. \tag*{(65)} { Q ∈ Stop ( N , k ) : 3 B Q ∩ 2 ρ − 1 B Q ( T ) = ∅ } ⊆ T ∈ S ⋃ R T . (65)
Fix some Q ∈ Stop ( N , k ) Q\in\operatorname{Stop}(N,k) Q ∈ Stop ( N , k ) as above. It follows from Lemma 4.25 that Q ⊆ 4 ρ − 1 B Q ( T ) Q\subseteq4\rho^{-1}B_{Q(T)} Q ⊆ 4 ρ − 1 B Q ( T ) . Recalling (63) , this implies Q ⊆ A ∗ B Q ( S ) Q\subseteq A^*B_{Q(S)} Q ⊆ A ∗ B Q ( S ) . Thus, if T ′ ∈ T N ( k ) T'\in\mathcal{T}^N(k) T ′ ∈ T N ( k ) is such that Q ∈ min ( T ′ ( k ) ) Q\in\min(T'(k)) Q ∈ min ( T ′ ( k )) , then E ∩ A ∗ B Q ( S ) ∩ Q ( T ′ ) ≠ ∅ E\cap A^*B_{Q(S)}\cap Q(T')\neq\varnothing E ∩ A ∗ B Q ( S ) ∩ Q ( T ′ ) = ∅ i.e., T ′ ∈ S T'\in\mathcal{S} T ′ ∈ S . The property Q ⊆ A ∗ B Q ( S ) Q\subseteq A^*B_{Q(S)} Q ⊆ A ∗ B Q ( S ) also implies that Q ∈ R T ′ Q\in\mathcal{R}_{T'} Q ∈ R T ′ , which finishes the proof of (65) . □
We now define
Γ S , k : = { Γ S , k A 1 if S ∈ G 2 N ( k ) ; Σ S , k ∪ Σ U S , k if S ∈ B 2 N ( k ) . \Gamma_{S,k}:=
\begin{cases}
\Gamma^{A_1}_{S,k} & \text{if } S\in\mathcal{G}^N_2(k);\\
\Sigma_{S,k}\cup\Sigma_{U_S,k} & \text{if } S\in\mathcal{B}^N_2(k).
\end{cases} Γ S , k := { Γ S , k A 1 Σ S , k ∪ Σ U S , k if S ∈ G 2 N ( k ) ; if S ∈ B 2 N ( k ) .
Proof of Proposition 7.3 for S ∈ G 2 N ( k ) S\in\mathcal{G}^N_2(k) S ∈ G 2 N ( k ) . The inclusion (59) and the Poincaré inequality follow directly from Lemma 7.4. Fix now some some T ∈ T N ( k ) T\in\mathcal{T}^N(k) T ∈ T N ( k ) such that
A 1 B Q ( S ) ∩ A 1 B Q ( T ) ≠ ∅ . A_1B_{Q(S)}\cap A_1B_{Q(T)}\neq\varnothing. A 1 B Q ( S ) ∩ A 1 B Q ( T ) = ∅ .
It follows Lemma 4.15 and the triangle inequality that
2 ρ − 1 B Q ( T ) ⊆ A 1 B Q ( T ) ⊆ ( 1 + 2 C 5 ) A 1 B Q ( S ) ⊆ A 1 2 B Q ( S ) . (66) 2\rho^{-1}B_{Q(T)}\subseteq A_1B_{Q(T)}
\subseteq(1+2C_5)A_1B_{Q(S)}
\subseteq A_1^2B_{Q(S)}. \tag*{(66)} 2 ρ − 1 B Q ( T ) ⊆ A 1 B Q ( T ) ⊆ ( 1 + 2 C 5 ) A 1 B Q ( S ) ⊆ A 1 2 B Q ( S ) . (66)
We set Γ S , T , k : = Γ S , k A 1 \Gamma_{S,T,k}:=\Gamma^{A_1}_{S,k} Γ S , T , k := Γ S , k A 1 (as in Lemma 7.4). It is clear that Γ S , T , k \Gamma_{S,T,k} Γ S , T , k is Ahlfors d d d -regular, supports a ( 1 , 1 ) (1,1) ( 1 , 1 ) -Poincaré inequality, contains Γ S \Gamma_S Γ S and that
Γ S , T , k ⊆ 10 A 1 4 B Q ( S ) ⊆ A 2 B Q ( S ) . (67) \Gamma_{S,T,k}\subseteq10A_1^4B_{Q(S)}\subseteq A_2B_{Q(S)}. \tag*{(67)} Γ S , T , k ⊆ 10 A 1 4 B Q ( S ) ⊆ A 2 B Q ( S ) . (67)
(we note the slightly better first inclusion for later on). If T ∈ G 2 N ( k ) T \in\mathcal{G}_2^N(k) T ∈ G 2 N ( k ) , then Γ T , k ⊆ Γ S , T , k \Gamma_{T,k} \subseteq\Gamma_{S,T,k} Γ T , k ⊆ Γ S , T , k by (62) and (66) . Since Σ T , k \Sigma_{T,k} Σ T , k and Σ U T , k \Sigma_{U_T,k} Σ U T , k have non-empty intersection with 2 ρ − 1 B Q ( T ) 2\rho^{-1}B_{Q(T)} 2 ρ − 1 B Q ( T ) , if T ∈ B 2 N ( k ) T \in\mathcal{B}_2^N(k) T ∈ B 2 N ( k ) , then the inclusion Γ T , k ⊆ Γ S , T , k \Gamma_{T,k} \subseteq\Gamma_{S,T,k} Γ T , k ⊆ Γ S , T , k follows by (61) and (66) .
Proof of Proposition 7.3 for S ∈ G 2 N ( k ) S \in\mathcal{G}_2^N(k) S ∈ G 2 N ( k ) . The inclusion (59) follows from the definition. Since 2 n λ 1 J Q ( S ) ∩ 2 n λ 1 J Q ( U S ) ≠ ∅ 2\sqrt{n}\lambda_1J_{Q(S)} \cap2\sqrt{n}\lambda_1J_{Q(U_S)} \ne\varnothing 2 n λ 1 J Q ( S ) ∩ 2 n λ 1 J Q ( U S ) = ∅ , the Poincaré inequality and Ahlfors regularity for Γ S \Gamma_S Γ S follows from Lemma 5.3, Lemma 5.5 and Lemma 2.9 once we show that
U S ∈ B 1 N ( k ) . (68) U_S \in\mathcal{B}_1^N(k). \tag*{(68)} U S ∈ B 1 N ( k ) . (68)
To see (68) , note that M 1 B Q ( U S ) ⊆ 2 M 1 B Q ( S ) M_1B_{Q(U_S)} \subseteq2M_1B_{Q(S)} M 1 B Q ( U S ) ⊆ 2 M 1 B Q ( S ) for M 1 = M 1 ( C 5 ) M_1=M_1(C_5) M 1 = M 1 ( C 5 ) sufficiently large, by Lemma 4.9. The inclusion (68) now follows since S ∈ B 2 N ( k ) S \in\mathcal{B}_2^N(k) S ∈ B 2 N ( k ) .
Now, fix some T ∈ T N ( k ) T \in\mathcal{T}^N(k) T ∈ T N ( k ) such that A 1 B Q ( S ) ∩ A 1 B Q ( T ) ≠ ∅ A_1B_{Q(S)} \cap A_1B_{Q(T)} \ne\varnothing A 1 B Q ( S ) ∩ A 1 B Q ( T ) = ∅ .
If T ∈ G 2 N ( k ) T \in\mathcal{G}_2^N(k) T ∈ G 2 N ( k ) then we set Γ S , T , k : = Γ T , S , k \Gamma_{S,T,k}:=\Gamma_{T,S,k} Γ S , T , k := Γ T , S , k , where Γ T , S , k \Gamma_{T,S,k} Γ T , S , k is the set obtained by applying Proposition 7.3 for T T T . It follows from Lemma 4.9 and the stricter inclusion in (67) that
Γ S , T , k ⊆ 10 A 1 4 B Q ( T ) ⊆ A 1 5 B Q ( S ) . \Gamma_{S,T,k} \subseteq10A_1^4B_{Q(T)} \subseteq A_1^5B_{Q(S)}. Γ S , T , k ⊆ 10 A 1 4 B Q ( T ) ⊆ A 1 5 B Q ( S ) .
Suppose instead that T ∈ B 2 N ( k ) T \in\mathcal{B}_2^N(k) T ∈ B 2 N ( k ) . In this case, we have
Γ S , k ∪ Γ T , k = Σ S , k ∪ Σ U S , k ∪ Σ T , k ∪ Σ U T , k = G S ∪ G U S ∪ G T ∪ G U T . \begin{aligned}
\Gamma_{S,k} \cup\Gamma_{T,k}
&= \Sigma_{S,k} \cup\Sigma_{U_S,k} \cup\Sigma_{T,k} \cup\Sigma_{U_T,k} \\
&= G_S \cup G_{U_S} \cup G_T \cup G_{U_T}.
\end{aligned} Γ S , k ∪ Γ T , k = Σ S , k ∪ Σ U S , k ∪ Σ T , k ∪ Σ U T , k = G S ∪ G U S ∪ G T ∪ G U T .
(note that (68) also holds for T T T ). Thus, if Γ S , k ∪ Γ T , k \Gamma_{S,k} \cup\Gamma_{T,k} Γ S , k ∪ Γ T , k is connected, we have by Lemma 5.5 that it is Ahlfors regular and supports a Poincaré inequality. Note that Γ S , k ∪ Γ T , k \Gamma_{S,k} \cup\Gamma_{T,k} Γ S , k ∪ Γ T , k is always connected in the case that 2 ρ − 1 B Q ( S ) ∩ 2 ρ − 1 B Q ( T ) ≠ ∅ 2\rho^{-1}B_{Q(S)} \cap2\rho^{-1}B_{Q(T)} \ne\varnothing 2 ρ − 1 B Q ( S ) ∩ 2 ρ − 1 B Q ( T ) = ∅ . If it is not connected, then Lemma 5.3 implies (60) .
Bridges between consecutive layers
For N ∈ N N \in\mathbb{N} N ∈ N and S ∈ T N ( k 0 ) S \in\mathcal{T}^N(k_0) S ∈ T N ( k 0 ) , consider the sets
Σ S , k 0 ∗ = Graph Σ S , k 0 ( τ 0 d N , k 0 ) and Σ N , k 0 ∗ = Graph Σ N , k 0 ( τ 0 d N , k 0 ) . (69) \Sigma^*_{S,k_0}=\operatorname{Graph}_{\Sigma_{S,k_0}}(\tau_0d_{N,k_0}) \quad\text{and}\quad\Sigma^*_{N,k_0}=\operatorname{Graph}_{\Sigma_{N,k_0}}(\tau_0d_{N,k_0}). \tag*{(69)} Σ S , k 0 ∗ = Graph Σ S , k 0 ( τ 0 d N , k 0 ) and Σ N , k 0 ∗ = Graph Σ N , k 0 ( τ 0 d N , k 0 ) . (69)
(recall the notation Graph A ( f ) \operatorname{Graph}_A(f) Graph A ( f ) from (1) ). In this section we describe a collection of bridges { b Q } Q ∈ Stop ( N − 1 , k 0 ) \{b_Q\}_{Q\in\operatorname{Stop}(N-1,k_0)} { b Q } Q ∈ Stop ( N − 1 , k 0 ) between consecutive layer Σ N − 1 , k 0 ∗ \Sigma^*_{N-1,k_0} Σ N − 1 , k 0 ∗ and Σ N , k 0 ∗ \Sigma^*_{N,k_0} Σ N , k 0 ∗ . In particular, we will prove the following.
Proposition 8.1. For each k , N ∈ N k,N \in\mathbb{N} k , N ∈ N there is a collection of set { b Q , k } k ∈ Stop ′ ( N − 1 , k ) \{b_{Q,k}\}_{k\in\operatorname{Stop}'(N-1,k)} { b Q , k } k ∈ Stop ′ ( N − 1 , k ) in R n + 1 \mathbb{R}^{n+1} R n + 1 satisfying the following conditions.
Each b Q , k b_{Q,k} b Q , k is Ahlfors d d d -regular with
diam ( b Q , k ) ∼ ℓ ( Q ) , \operatorname{diam}(b_{Q,k}) \sim\ell(Q), diam ( b Q , k ) ∼ ℓ ( Q ) ,
and supports a ( 1 , 1 ) (1,1) ( 1 , 1 ) -Poincaré inequality with uniform constants, and satisfies
We have
π n ( b Q , k ) ⊆ ( c 0 / 2 ) B Q \pi_n(b_{Q,k}) \subseteq(c_0/2)B_Q π n ( b Q , k ) ⊆ ( c 0 /2 ) B Q
and
b Q , k ⊆ { z ∈ R n + 1 : π 1 ( z ) ∈ [ τ 0 d N , k ( π n ( z ) ) , τ 0 d N − 1 ( π n ( z ) ) ] } . b_{Q,k} \subseteq\{z \in\mathbb{R}^{n+1}: \pi_1(z) \in[\tau_0d_{N,k}(\pi_n(z)),\tau_0d_{N-1}(\pi_n(z))]\}. b Q , k ⊆ { z ∈ R n + 1 : π 1 ( z ) ∈ [ τ 0 d N , k ( π n ( z )) , τ 0 d N − 1 ( π n ( z ))]} .
(3) The set Σ N − 1 , k ∗ ∩ b Q , k \Sigma^{*}_{N-1,k} \cap b_{Q,k} Σ N − 1 , k ∗ ∩ b Q , k is lower Ahlfors d d d -regular with diameter comparable to ℓ ( Q ) \ell(Q) ℓ ( Q ) . Moreover, if S ∈ T N − 1 ( k ) S \in\mathcal{T}^{N-1}(k) S ∈ T N − 1 ( k ) is such that Q ∈ S Q \in S Q ∈ S , then
Σ N − 1 , k ∗ ∩ b Q , k = Σ S , k ∗ ∩ b Q , k . \Sigma^{*}_{N-1,k} \cap b_{Q,k} = \Sigma^{*}_{S,k} \cap b_{Q,k}. Σ N − 1 , k ∗ ∩ b Q , k = Σ S , k ∗ ∩ b Q , k .
(4) The set Σ N , k ∗ ∩ b Q , k \Sigma^{*}_{N,k} \cap b_{Q,k} Σ N , k ∗ ∩ b Q , k is lower Ahlfors d d d -regular with diameter comparable to ℓ ( Q ) \ell(Q) ℓ ( Q ) . Moreover, if T ∈ T N ( k ) T \in\mathcal{T}^{N}(k) T ∈ T N ( k ) is such that x Q ( T ) = x Q x_Q(T) = x_Q x Q ( T ) = x Q , then
Σ N , k ∗ ∩ b Q , k = Σ T , k ∗ ∩ b Q , k . \Sigma^{*}_{N,k} \cap b_{Q,k} = \Sigma^{*}_{T,k} \cap b_{Q,k}. Σ N , k ∗ ∩ b Q , k = Σ T , k ∗ ∩ b Q , k .
(5) Denoting
A 1 : = b Q , k ∩ Σ N − 1 , k ∗ and A 2 : = b Q , k ∩ Σ N , k ∗ , A_1 := b_{Q,k} \cap\Sigma^{*}_{N-1,k} \qquad\text{and} \qquad A_2 := b_{Q,k} \cap\Sigma^{*}_{N,k}, A 1 := b Q , k ∩ Σ N − 1 , k ∗ and A 2 := b Q , k ∩ Σ N , k ∗ ,
for each x ∈ b Q , k x \in b_{Q,k} x ∈ b Q , k ,
y 1 ∈ { z ∈ R n + 1 : π 1 ( z ) ≥ τ 0 d N − 1 , k ( π n ( z ) ) } ⊇ Σ N − 1 , k ∗ , y_1 \in\{z \in\mathbb{R}^{n+1} : \pi_1(z) \ge\tau_0 d_{N-1,k}(\pi_n(z))\} \supseteq\Sigma^{*}_{N-1,k}, y 1 ∈ { z ∈ R n + 1 : π 1 ( z ) ≥ τ 0 d N − 1 , k ( π n ( z ))} ⊇ Σ N − 1 , k ∗ ,
y 2 ∈ { z ∈ R n + 1 : π 1 ( z ) ≤ τ 0 d N , k ( π n ( z ) ) } ⊇ Σ N , k ∗ , y_2 \in\{z \in\mathbb{R}^{n+1} : \pi_1(z) \le\tau_0 d_{N,k}(\pi_n(z))\} \supseteq\Sigma^{*}_{N,k}, y 2 ∈ { z ∈ R n + 1 : π 1 ( z ) ≤ τ 0 d N , k ( π n ( z ))} ⊇ Σ N , k ∗ ,
we have
inf a ∈ A i ∣ x − a ∣ + ∣ a − y i ∣ ≲ ∣ x − y i ∣ . \inf_{a \in A_i} |x-a| + |a-y_i| \lesssim|x-y_i|. a ∈ A i inf ∣ x − a ∣ + ∣ a − y i ∣ ≲ ∣ x − y i ∣.
Preliminary results
Before detailing on the construction of the b Q b_Q b Q , we state a useful technical preliminaries. Lemma 8.2 below give us a way to construct a bi-Lipschitz image of d d d -dimensional cube which connects the graph (over R n \mathbb{R}^n R n ) of a ( d − 1 ) (d-1) ( d − 1 ) -segment in R n \mathbb{R}^n R n to a different graph (over R n \mathbb{R}^n R n ) of a different ( d − 1 ) (d-1) ( d − 1 ) -segment in R n \mathbb{R}^n R n . Assuming the graphs are sufficiently well-separated with small enough Lipschitz constant, we are able to do this in such a way that these bi-Lipschitz images interact well (keeping in mind Lemma 3.4) with each of the graphs. This will be important for proving a Poincaré inequality later. The proofs are technical, but rather easy. As such, we have decided to include them in Appendix A.
Lemma 8.2. Let 0 ≤ η , θ , λ < 1 0 \le\eta, \theta, \lambda< 1 0 ≤ η , θ , λ < 1 , 0 < γ , δ < 1 0 < \gamma, \delta< 1 0 < γ , δ < 1 and C 1 > 1 C_1 > 1 C 1 > 1 . Let Q , R ⊆ R n Q, R \subseteq\mathbb{R}^n Q , R ⊆ R n be d d d -dimensional cubes with ℓ ( Q ) ≥ ℓ ( R ) \ell(Q) \ge\ell(R) ℓ ( Q ) ≥ ℓ ( R ) , I ⊆ Q I \subseteq Q I ⊆ Q and J ⊆ R J \subseteq R J ⊆ R be ( d − 1 ) (d-1) ( d − 1 ) -dimensional faces, P ⊆ R n P \subseteq\mathbb{R}^n P ⊆ R n be the d d d -dimensional plane containing R R R , u : P → P ⊥ u : P \to P^\perp u : P → P ⊥ be 1-Lipschitz and K K K be the graph of u u u over R R R . Let g 1 : R n → [ 0 , ∞ ) g_1 : \mathbb{R}^n \to[0,\infty) g 1 : R n → [ 0 , ∞ ) be η \eta η -Lipschitz and g 2 : R n → [ 0 , ∞ ) g_2 : \mathbb{R}^n \to[0,\infty) g 2 : R n → [ 0 , ∞ ) be θ \theta θ -Lipschitz such that
γ ℓ ( I ) ≤ inf x ∈ I g 1 ( x ) − sup y ∈ u ( J ) g 2 ( y ) and sup x ∈ I g 1 ( x ) ≤ C 1 ℓ ( I ) . (70) \gamma\ell(I) \le\inf_{x \in I} g_1(x) - \sup_{y \in u(J)} g_2(y)
\qquad\text{and}\qquad
\sup_{x \in I} g_1(x) \le C_1\ell(I). \tag*{(70)} γ ℓ ( I ) ≤ x ∈ I inf g 1 ( x ) − y ∈ u ( J ) sup g 2 ( y ) and x ∈ I sup g 1 ( x ) ≤ C 1 ℓ ( I ) . (70) Suppose there exists a map h : I × [ 0 , ℓ ( I ) ] → R n h : I \times[0,\ell(I)] \to\mathbb{R}^n h : I × [ 0 , ℓ ( I )] → R n which takes I I I to J J J such that
δ ∣ x − y ∣ ≤ ∣ h ( x , t ) − h ( y , t ) ∣ ≤ ∣ x − y ∣ (71) \delta|x-y| \le|h(x,t)-h(y,t)| \le|x-y| \tag*{(71)} δ ∣ x − y ∣ ≤ ∣ h ( x , t ) − h ( y , t ) ∣ ≤ ∣ x − y ∣ (71) for all x , y ∈ I x,y \in I x , y ∈ I and t ∈ [ 0 , ℓ ( I ) ] t \in[0,\ell(I)] t ∈ [ 0 , ℓ ( I )] ,
∣ h ( x , ℓ ( I ) ) − h ( y , ℓ ( I ) ) ∣ ≤ C 1 δ ∣ x − y ∣ |h(x,\ell(I))-h(y,\ell(I))| \le C_1\delta|x-y| ∣ h ( x , ℓ ( I )) − h ( y , ℓ ( I )) ∣ ≤ C 1 δ ∣ x − y ∣ and
∣ h ( x , t ) − h ( x , s ) ∣ ≤ C 1 ∣ t − s ∣ (72) |h(x,t)-h(x,s)| \le C_1|t-s| \tag*{(72)} ∣ h ( x , t ) − h ( x , s ) ∣ ≤ C 1 ∣ t − s ∣ (72) for all x ∈ I x \in I x ∈ I and t , s ∈ [ 0 , ℓ ( I ) ] t,s \in[0,\ell(I)] t , s ∈ [ 0 , ℓ ( I )] . Finally, suppose that is a ball B ⊆ R n B \subseteq\mathbb{R}^n B ⊆ R n such that
Q , u ( R ) , h ( I × [ 0 , ℓ ( I ) ] ) ⊆ B . (73) Q,u(R),h(I \times[0,\ell(I)]) \subseteq B. \tag*{(73)} Q , u ( R ) , h ( I × [ 0 , ℓ ( I )]) ⊆ B . (73)
Then, for η = η ( γ , δ , C 1 ) \eta=\eta(\gamma,\delta,C_1) η = η ( γ , δ , C 1 ) and θ = θ ( γ , C 1 ) \theta=\theta(\gamma,C_1) θ = θ ( γ , C 1 ) small enough, there exists a bi-Lipschitz map H : I × [ − ℓ ( I ) , 2 ℓ ( I ) ] → B × R H:I\times[-\ell(I),2\ell(I)]\to B\times\mathbb{R} H : I × [ − ℓ ( I ) , 2 ℓ ( I )] → B × R such that
H ( I , [ − ℓ ( I ) , 0 ] ) = Graph Q ( g 1 ) ⊆ R n + 1 , (74) H(I,[-\ell(I),0])=\operatorname{Graph}_{Q}(g_1)\subseteq\mathbb{R}^{n+1}, \tag*{(74)} H ( I , [ − ℓ ( I ) , 0 ]) = Graph Q ( g 1 ) ⊆ R n + 1 , (74)
and
H ( I , [ ℓ ( I ) , 2 ℓ ( I ) ] ) = Graph K ( g 2 ) ⊆ R n + 1 . (75) H(I,[\ell(I),2\ell(I)])=\operatorname{Graph}_{K}(g_2)\subseteq\mathbb{R}^{n+1}. \tag*{(75)} H ( I , [ ℓ ( I ) , 2 ℓ ( I )]) = Graph K ( g 2 ) ⊆ R n + 1 . (75)
Furthermore, for each x ∈ I x\in I x ∈ I and t ∈ [ 0 , ℓ ( I ) ] t\in[0,\ell(I)] t ∈ [ 0 , ℓ ( I )] ,
g 2 ( π R n ( H ( x , t ) ) ) + γ ⋅ ( ℓ ( I ) − t ) 2 ≤ π R ( H ( x , t ) ) ≤ g 1 ( π R n ( H ( x , t ) ) ) − γ t 2 . (76) g_2(\pi_{\mathbb{R}^n}(H(x,t)))+\frac{\gamma\cdot(\ell(I)-t)}{2}\leq\pi_{\mathbb{R}}(H(x,t))\leq g_1(\pi_{\mathbb{R}^n}(H(x,t)))-\frac{\gamma t}{2}. \tag*{(76)} g 2 ( π R n ( H ( x , t ))) + 2 γ ⋅ ( ℓ ( I ) − t ) ≤ π R ( H ( x , t )) ≤ g 1 ( π R n ( H ( x , t ))) − 2 γ t . (76)
Finally, denoting X : = H ( I , [ − ℓ ( I ) , 2 ℓ ( I ) ] ) X:=H(I,[-\ell(I),2\ell(I)]) X := H ( I , [ − ℓ ( I ) , 2 ℓ ( I )]) , X u p : = H ( I , [ − ℓ ( I ) , 0 ] ) X_{\mathrm{up}}:=H(I,[-\ell(I),0]) X up := H ( I , [ − ℓ ( I ) , 0 ]) , X d o w n : = H ( I , [ ℓ ( I ) , 2 ℓ ( I ) ] ) X_{\mathrm{down}}:=H(I,[\ell(I),2\ell(I)]) X down := H ( I , [ ℓ ( I ) , 2 ℓ ( I )]) , if η = η ( γ ) \eta=\eta(\gamma) η = η ( γ ) and θ = θ ( γ ) \theta=\theta(\gamma) θ = θ ( γ ) are sufficiently small, we have
inf a ∈ X u p ∣ x − a ∣ + ∣ a − y ∣ ≲ ∣ x − y ∣ (77) \inf_{a\in X_{\mathrm{up}}}|x-a|+|a-y|\lesssim|x-y| \tag*{(77)} a ∈ X up inf ∣ x − a ∣ + ∣ a − y ∣ ≲ ∣ x − y ∣ (77)
for all x ∈ X x\in X x ∈ X and y ∈ R n + 1 y\in\mathbb{R}^{n+1} y ∈ R n + 1 such that π 1 ( y ) ≥ g 1 ( π n ( y ) ) \pi_1(y)\geq g_1(\pi_n(y)) π 1 ( y ) ≥ g 1 ( π n ( y )) , and
inf a ∈ X d o w n ∣ x − a ∣ + ∣ a − y ∣ ≲ ∣ x − y ∣ (78) \inf_{a\in X_{\mathrm{down}}}|x-a|+|a-y|\lesssim|x-y| \tag*{(78)} a ∈ X down inf ∣ x − a ∣ + ∣ a − y ∣ ≲ ∣ x − y ∣ (78)
for all x ∈ X x\in X x ∈ X and y ∈ R n + 1 y\in\mathbb{R}^{n+1} y ∈ R n + 1 such that π 1 ( y ) ≤ λ g 2 ( π n ( y ) ) \pi_1(y)\leq\lambda g_2(\pi_n(y)) π 1 ( y ) ≤ λ g 2 ( π n ( y )) .
The part of the bridge in Σ N − 1 , k 0 ∗ \Sigma^{*}_{N-1,k_0} Σ N − 1 , k 0 ∗
Fix some N ∈ N N\in\mathbb{N} N ∈ N and Q ∈ Stop ′ ( N − 1 , k 0 ) Q\in\operatorname{Stop}'(N-1,k_0) Q ∈ Stop ′ ( N − 1 , k 0 ) for the remainder of the section. Let S ∈ T N − 1 ( k 0 ) S\in\mathcal{T}^{N-1}(k_0) S ∈ T N − 1 ( k 0 ) such that Q ∈ S Q\in S Q ∈ S ; T ∈ T N ( k 0 ) T\in\mathcal{T}^{N}(k_0) T ∈ T N ( k 0 ) such that x Q ( T ) = x Q x_Q(T)=x_Q x Q ( T ) = x Q .
The construction of the bridge b Q b_Q b Q will depend on whether T ∈ G 1 N ( k 0 ) T\in\mathcal{G}^{N}_{1}(k_0) T ∈ G 1 N ( k 0 ) or T ∈ B 1 N ( k 0 ) T\in\mathcal{B}^{N}_{1}(k_0) T ∈ B 1 N ( k 0 ) . However, in either case, the intersection of b Q b_Q b Q with Σ N − 1 , k 0 ∗ \Sigma^{*}_{N-1,k_0} Σ N − 1 , k 0 ∗ will be the same. In this subsection, we identify this set.
Throughout this section, I R I_R I R will denote the denote the d d d -dimensional cube from Lemma 6.2, and F R F_R F R will denote an arbitrary ( d − 1 ) (d-1) ( d − 1 ) -dimensional sub-face, which we fix. In this way, we have
F R ⊆ I R ⊆ ( ρ c 0 / 8 ) B R and ℓ ( F R ) = ℓ ( I R ) ∼ ℓ ( R ) . (79) F_R\subseteq I_R\subseteq(\rho c_0/8)B_R\quad\text{and}\quad\ell(F_R)=\ell(I_R)\sim\ell(R). \tag*{(79)} F R ⊆ I R ⊆ ( ρ c 0 /8 ) B R and ℓ ( F R ) = ℓ ( I R ) ∼ ℓ ( R ) . (79)
By scaling the metric, we will assume without loss of generality that
ℓ ( I Q ) = ℓ ( F Q ) = 1. \ell(I_Q)=\ell(F_Q)=1. ℓ ( I Q ) = ℓ ( F Q ) = 1.
For the cube Q Q Q (and only Q Q Q ), we set
I Q ∗ : = Graph I Q ( τ 0 d N − 1 , k 0 ) and F Q ∗ : = Graph F Q ( τ 0 d N − 1 , k 0 ) , I_Q^*:=\operatorname{Graph}_{I_Q}(\tau_0d_{N-1,k_0})\quad\text{and}\quad F_Q^*:=\operatorname{Graph}_{F_Q}(\tau_0d_{N-1,k_0}), I Q ∗ := Graph I Q ( τ 0 d N − 1 , k 0 ) and F Q ∗ := Graph F Q ( τ 0 d N − 1 , k 0 ) ,
which are both subsets of Σ S , k 0 ∗ \Sigma^{*}_{S,k_0} Σ S , k 0 ∗ . As we will see later, I Q ∗ I_Q^* I Q ∗ will be the intersection of b Q b_Q b Q with Σ N − 1 , k 0 ∗ \Sigma^{*}_{N-1,k_0} Σ N − 1 , k 0 ∗ . For now, let us collect some basic facts about these sets.
Lemma 8.3. The set I Q ∗ I_Q^* I Q ∗ is Ahlfors d d d -regular with diameter comparable to ℓ ( Q ) \ell(Q) ℓ ( Q ) . For each x , y ∈ ( c 0 / 2 ) B Q x,y\in(c_0/2)B_Q x , y ∈ ( c 0 /2 ) B Q , we have
τ 0 d N − 1 , k 0 ( x ) ≲ ℓ ( F Q ) (80) \tau_0d_{N-1,k_0}(x)\lesssim\ell(F_Q) \tag*{(80)} τ 0 d N − 1 , k 0 ( x ) ≲ ℓ ( F Q ) (80) and
τ 0 d N − 1 , k 0 ( x ) − 2 τ 0 d N , k 0 ( y ) ≳ ℓ ( F Q ) (81) \tau_0d_{N-1,k_0}(x)-2\tau_0d_{N,k_0}(y)\gtrsim\ell(F_Q) \tag*{(81)} τ 0 d N − 1 , k 0 ( x ) − 2 τ 0 d N , k 0 ( y ) ≳ ℓ ( F Q ) (81)
Proof. The Ahlfors regularity of I Q ∗ I_Q^* I Q ∗ follows since it is bi-Lipschitz equivalent to a d d d -dimensional cube with side-length comparable to ℓ ( Q ) \ell(Q) ℓ ( Q ) . Inequality (80) follows from Lemma 4.14 and (79) . Inequality (81) follows from Lemma 4.20 and (79) .
We now split to construction into our two cases.
Construction of b Q b_Q b Q when T ∈ B 1 N ( k 0 ) T \in\mathcal{B}^{N}_{1}(k_0) T ∈ B 1 N ( k 0 )
For brevity, let us denote I T : = I Q ( T ) I_T := I_{Q(T)} I T := I Q ( T ) and F T : = I Q ( T ) F_T := I_{Q(T)} F T := I Q ( T ) . Then, set
I T ∗ = Graph I T ( τ 0 d N , k 0 ) and F T ∗ = Graph F T ( τ 0 d N , k 0 ) , I_T^*=\operatorname{Graph}_{I_T}(\tau_0d_{N,k_0}) \quad\text{and}\quad F_T^*=\operatorname{Graph}_{F_T}(\tau_0d_{N,k_0}), I T ∗ = Graph I T ( τ 0 d N , k 0 ) and F T ∗ = Graph F T ( τ 0 d N , k 0 ) ,
which are both subsets of Σ T , k 0 ∗ ⊆ Σ N , k 0 ∗ \Sigma^*_{T,k_0} \subseteq\Sigma^*_{N,k_0} Σ T , k 0 ∗ ⊆ Σ N , k 0 ∗ (recall Lemma 6.2).
Remark 8.4. We note that I T ∗ I_T^* I T ∗ is the graph of the function τ 0 d N , k 0 \tau_0d_{N,k_0} τ 0 d N , k 0 as opposed to I Q ∗ I_Q^* I Q ∗ which is the graph of the function τ 0 d N − 1 , k 0 \tau_0d_{N-1,k_0} τ 0 d N − 1 , k 0 (similar for F T F_T F T versus F Q F_Q F Q ). We hope that this slight abuse of notation does not cause significant misunderstanding.
Since Q ( T ) Q(T) Q ( T ) is a child of Q Q Q , we have by (79) that
ℓ ( F T ) ∼ ℓ ( F Q ) = 1. (82) \ell(F_T)\sim\ell(F_Q)=1. \tag*{(82)} ℓ ( F T ) ∼ ℓ ( F Q ) = 1. (82)
Since I R ⊆ ( c 0 / 8 ) B R I_R \subseteq(c_0/8)B_R I R ⊆ ( c 0 /8 ) B R for each R ∈ D R \in\mathcal{D} R ∈ D , and ( c 0 / 8 ) B Q ( T ) ⊆ ( c 0 / 8 ) B Q (c_0/8)B_{Q(T)} \subseteq(c_0/8)B_Q ( c 0 /8 ) B Q ( T ) ⊆ ( c 0 /8 ) B Q , there is a rigid motion h : F Q × [ 0 , 1 ] → ( c 0 / 8 ) B Q h:F_Q\times[0,1]\to(c_0/8)B_Q h : F Q × [ 0 , 1 ] → ( c 0 /8 ) B Q taking F Q F_Q F Q to F T F_T F T such that
ℓ ( F T ) ℓ ( F Q ) ∣ x − y ∣ ≤ ∣ h ( x , t ) − h ( y , t ) ∣ ≤ ∣ x − y ∣ \frac{\ell(F_T)}{\ell(F_Q)}|x-y|\leq|h(x,t)-h(y,t)|\leq|x-y| ℓ ( F Q ) ℓ ( F T ) ∣ x − y ∣ ≤ ∣ h ( x , t ) − h ( y , t ) ∣ ≤ ∣ x − y ∣
for all x , y ∈ F Q x,y\in F_Q x , y ∈ F Q and t ∈ [ 0 , 1 ] t\in[0,1] t ∈ [ 0 , 1 ] ,
∣ h ( x , 1 ) − h ( y , 1 ) ∣ = ℓ ( F T ) ℓ ( F Q ) ∣ x − y ∣ |h(x,1)-h(y,1)|=\frac{\ell(F_T)}{\ell(F_Q)}|x-y| ∣ h ( x , 1 ) − h ( y , 1 ) ∣ = ℓ ( F Q ) ℓ ( F T ) ∣ x − y ∣
for all x , y ∈ F Q x,y\in F_Q x , y ∈ F Q , and
∣ h ( x , t ) − h ( x , s ) ∣ ≲ ∣ t − s ∣ |h(x,t)-h(x,s)|\lesssim|t-s| ∣ h ( x , t ) − h ( x , s ) ∣ ≲ ∣ t − s ∣
for all x , y ∈ F Q x,y\in F_Q x , y ∈ F Q and t , s ∈ [ 0 , 1 ] t,s\in[0,1] t , s ∈ [ 0 , 1 ] .
Lemma 8.5. The hypotheses of Lemma 8.2 are satisfied with the data
( Q , R ) = ( I Q , I T ) , ( I , J ) = ( F Q , F T ) , u ≡ 0 , (Q,R)=(I_Q,I_T),\quad(I,J)=(F_Q,F_T),\quad u\equiv0, ( Q , R ) = ( I Q , I T ) , ( I , J ) = ( F Q , F T ) , u ≡ 0 , ( g 1 , g 2 ) = ( τ 0 d N − 1 , k 0 , τ 0 d N , k ) , h = h and B = ( c 0 / 8 ) B Q , (g_1,g_2)=(\tau_0d_{N-1,k_0},\tau_0d_{N,k}),\quad h=h\text{ and }B=(c_0/8)B_Q, ( g 1 , g 2 ) = ( τ 0 d N − 1 , k 0 , τ 0 d N , k ) , h = h and B = ( c 0 /8 ) B Q , and parameters
η = θ = τ 0 , γ ∼ δ ∼ C 1 ∼ 1. \eta=\theta=\tau_0,\quad\gamma\sim\delta\sim C_1\sim1. η = θ = τ 0 , γ ∼ δ ∼ C 1 ∼ 1.
Proof. The properties in the preamble of Lemma 8.2 are immediate from the definitions with the stated parameters. The inequalities in (70) with parameters γ ∼ C 1 ∼ 1 \gamma\sim C_1\sim1 γ ∼ C 1 ∼ 1 follow from (80) and (81) since F Q , F T ⊆ ( c 0 / 8 ) B Q F_Q,F_T\subseteq(c_0/8)B_Q F Q , F T ⊆ ( c 0 /8 ) B Q . That h h h satisfies (71) –(72) with parameters δ ∼ C 1 ∼ 1 \delta\sim C_1\sim1 δ ∼ C 1 ∼ 1 , follows from the definition above and (82) . The inclusions in (73) also follows directly from the construction since ( c 0 / 8 ) B Q ( T ) ⊆ ( c 0 / 8 ) B Q (c_0/8)B_{Q(T)}\subseteq(c_0/8)B_Q ( c 0 /8 ) B Q ( T ) ⊆ ( c 0 /8 ) B Q . □ \square □
Proof of Proposition 8.1 when T ∈ B 1 N ( k 0 ) T\in\mathcal{B}^{N}_{1}(k_0) T ∈ B 1 N ( k 0 ) . By Lemma 8.5, if we choose τ 0 = τ 0 ( 1 ) \tau_0=\tau_0(1) τ 0 = τ 0 ( 1 ) small enough there is a bi-Lipschitz map H : F Q × [ − 1 , 2 ] → ( c 0 / 8 ) H:F_Q\times[-1,2]\to(c_0/8) H : F Q × [ − 1 , 2 ] → ( c 0 /8 ) satisfying the conclusion of Lemma 8.2 with that data. We set
b Q : = H ( F Q × [ − 1 , 2 ] ) . b_Q:=H(F_Q\times[-1,2]). b Q := H ( F Q × [ − 1 , 2 ]) .
Proposition 8.1 (1) holds since b Q b_Q b Q is the bi-Lipschitz image of a d d d -dimensional rectangle with each side-lengths comparable to 1 ∼ ℓ ( Q ) 1 \sim\ell(Q) 1 ∼ ℓ ( Q ) . Proposition 8.1 (2) holds by (76) and the fact that H H H maps into ( c 0 / 8 ) B Q (c_0/8)B_Q ( c 0 /8 ) B Q . Proposition 8.1 (3) follows from (74) and (76) since these imply
Σ N − 1 , k 0 ∗ ∩ b Q = H ( F Q × [ − 1 , 0 ] ) = Graph I Q ( τ 0 d N − 1 , k 0 ) = I Q ∗ ⊆ Σ S , k 0 ∗ . \Sigma^*_{N-1,k_0}\cap b_Q=H(F_Q\times[-1,0])=\operatorname{Graph}_{I_Q}(\tau_0d_{N-1,k_0})=I^*_Q\subseteq\Sigma^*_{S,k_0}. Σ N − 1 , k 0 ∗ ∩ b Q = H ( F Q × [ − 1 , 0 ]) = Graph I Q ( τ 0 d N − 1 , k 0 ) = I Q ∗ ⊆ Σ S , k 0 ∗ .
Proposition 8.1 (4) follows from (75) and (76) since these imply
Σ N , k 0 ∗ ∩ b Q = H ( F Q × [ 1 , 2 ] ) = Graph I T ( τ 0 d N , k 0 ) = I T ∗ ⊆ Σ T , k 0 ∗ . \Sigma^*_{N,k_0}\cap b_Q=H(F_Q\times[1,2])=\operatorname{Graph}_{I_T}(\tau_0d_{N,k_0})=I^*_T\subseteq\Sigma^*_{T,k_0}. Σ N , k 0 ∗ ∩ b Q = H ( F Q × [ 1 , 2 ]) = Graph I T ( τ 0 d N , k 0 ) = I T ∗ ⊆ Σ T , k 0 ∗ .
Proposition 8.1 (5) follows directly from (77) and (78) .
Construction of b Q b_Q b Q when T ∈ G 1 N ( k 0 ) T\in\mathcal{G}^N_1(k_0) T ∈ G 1 N ( k 0 )
We start by setting some notation.
Notation 8.6. Let Γ \Gamma Γ denote the graph constructed from Lemma 4.12 with M = 10 M=10 M = 10 and { Q k } = { Q ( T ) } \{Q_k\}=\{Q(T)\} { Q k } = { Q ( T )} . Let P ⊆ R n P\subseteq\mathbb{R}^n P ⊆ R n and g : P → P ⊥ g:P\to P^\perp g : P → P ⊥ be the corresponding d d d -plane and map, respectively, such that Γ \Gamma Γ is the graph of g g g over P P P . Through this section π \pi π will denote the orthogonal projection onto P P P . Since E ∩ c 0 B Q ( T ) ⊆ Q ( T ) E\cap c_0B_{Q(T)}\subseteq Q(T) E ∩ c 0 B Q ( T ) ⊆ Q ( T ) note, by Lemma 4.16, that
Γ ∩ ( c 0 / 2 ) B Q ( T ) ⊆ ⋃ R ∈ min ( T ( k 0 ) ) Γ ∩ ( c 0 / 2 ) B Q ( T ) ∩ 2 B R ≠ ∅ 2 B R . (83) \Gamma\cap(c_0/2)B_{Q(T)}\subseteq\bigcup_{\substack{R\in\min(T(k_0))\\ \Gamma\cap(c_0/2)B_{Q(T)}\cap2B_R\ne\varnothing}}2B_R. \tag*{(83)} Γ ∩ ( c 0 /2 ) B Q ( T ) ⊆ R ∈ m i n ( T ( k 0 )) Γ ∩ ( c 0 /2 ) B Q ( T ) ∩ 2 B R = ∅ ⋃ 2 B R . (83)
Notation 8.7. Let I 0 ⊆ P I_0\subseteq P I 0 ⊆ P be the d d d -dimensional cube in P P P such that
ℓ ( I 0 ) = ℓ ( I Q ) and π ( x Q ) = x I 0 , \ell(I_0)=\ell(I_Q)\quad\text{and}\quad\pi(x_Q)=x_{I_0}, ℓ ( I 0 ) = ℓ ( I Q ) and π ( x Q ) = x I 0 ,
and let F 0 F_0 F 0 be an arbitrary ( d − 1 ) (d-1) ( d − 1 ) -dimensional face.
Lemma 8.8. We have
F 0 ⊆ I 0 ⊆ ( c 0 / 4 ) B Q ( T ) . F_0\subseteq I_0\subseteq(c_0/4)B_{Q(T)}. F 0 ⊆ I 0 ⊆ ( c 0 /4 ) B Q ( T ) .
In particular, there exists a rigid motion h : F Q × [ 0 , 1 ] → ( c 0 / 4 ) B Q ( T ) h:F_Q\times[0,1]\to(c_0/4)B_{Q(T)} h : F Q × [ 0 , 1 ] → ( c 0 /4 ) B Q ( T ) which takes F Q F_Q F Q to F 0 F_0 F 0 and satisfies (71) –(72) with parameters δ ∼ C 1 ∼ 1 \delta\sim C_1\sim1 δ ∼ C 1 ∼ 1 .
Proof. Recall from Lemma 6.2 that I Q ⊆ ( c 0 / 8 ) B Q ( T ) I_Q\subseteq(c_0/8)B_{Q(T)} I Q ⊆ ( c 0 /8 ) B Q ( T ) . Since I 0 I_0 I 0 is isometric to I Q I_Q I Q , this implies I 0 ⊆ B ( x I 0 , c 0 ℓ ( Q ( T ) ) / 8 ) I_0\subseteq B(x_{I_0},c_0\ell(Q(T))/8) I 0 ⊆ B ( x I 0 , c 0 ℓ ( Q ( T )) /8 ) . Since ∥ g ∥ ∞ ≤ θ 0 ℓ ( Q ( T ) ) \lVert g\rVert_\infty\leq\theta_0\ell(Q(T)) ∥ g ∥ ∞ ≤ θ 0 ℓ ( Q ( T )) and dist ( x Q , Γ ) = dist ( x Q ( T ) , Γ ) ≤ θ 0 ℓ ( Q ( T ) ) \operatorname{dist}(x_Q,\Gamma)=\operatorname{dist}(x_{Q(T)},\Gamma)\leq\theta_0\ell(Q(T)) dist ( x Q , Γ ) = dist ( x Q ( T ) , Γ ) ≤ θ 0 ℓ ( Q ( T )) (see Lemma 4.12), we have ∣ x Q − x I 0 ∣ ≤ 2 θ 0 ℓ ( Q ( T ) ) \lvert x_Q-x_{I_0}\rvert\leq2\theta_0\ell(Q(T)) ∣ x Q − x I 0 ∣ ≤ 2 θ 0 ℓ ( Q ( T )) . The triangle inequality now implies the desired inclusion, so long as θ 0 \theta_0 θ 0 is sufficiently small.
Notation 8.9. Let
Q : = { R ∈ min ( T ( k 0 ) ) : Γ ∩ π − 1 ( I 0 ) ∩ ( c 0 / 8 ) B R ≠ ∅ } . \mathcal{Q}:=\left\{R\in\min(T(k_0)):\Gamma\cap\pi^{-1}(I_0)\cap(c_0/8)B_R\ne\varnothing\right\}. Q := { R ∈ min ( T ( k 0 )) : Γ ∩ π − 1 ( I 0 ) ∩ ( c 0 /8 ) B R = ∅ } .
Lemma 8.10. We have
Γ ∩ π − 1 ( I 0 ) ⊆ ( c 0 / 4 ) B Q ( T ) ∩ ⋃ R ∈ Q C 8 B R , \Gamma\cap\pi^{-1}(I_0)\subseteq(c_0/4)B_{Q(T)}\cap\bigcup_{R\in\mathcal{Q}}C_8B_R, Γ ∩ π − 1 ( I 0 ) ⊆ ( c 0 /4 ) B Q ( T ) ∩ R ∈ Q ⋃ C 8 B R ,
where C 8 : = 17 d C 0 2 ρ − 1 C_8:=17\sqrt{d}C_0^2\rho^{-1} C 8 := 17 d C 0 2 ρ − 1 .
Proof. Let x ∈ Γ ∩ π − 1 ( I 0 ) x\in\Gamma\cap\pi^{-1}(I_0) x ∈ Γ ∩ π − 1 ( I 0 ) . Since ∥ g ∥ ∞ ≤ θ 0 ℓ ( Q ( T ) ) \lVert g\rVert_\infty\leq\theta_0\ell(Q(T)) ∥ g ∥ ∞ ≤ θ 0 ℓ ( Q ( T )) and I 0 ⊆ ( c 0 / 4 ) B Q ( T ) I_0\subseteq(c_0/4)B_{Q(T)} I 0 ⊆ ( c 0 /4 ) B Q ( T ) , we already have
x ∈ ( c 0 / 4 ) B Q ( T ) x\in(c_0/4)B_{Q(T)} x ∈ ( c 0 /4 ) B Q ( T )
for θ 0 \theta_0 θ 0 sufficiently small. Thus, by (83) , there is R ∈ min ( T ( k 0 ) ) R \in\min(T(k_0)) R ∈ min ( T ( k 0 )) such that x ∈ 2 B R x \in2B_R x ∈ 2 B R . Let m ∈ N m \in\mathbb{N} m ∈ N such that R ∈ D m R \in\mathcal{D}_m R ∈ D m and let l ∈ N l \in\mathbb{N} l ∈ N be the large integer less than m m m such that ρ l ≥ 2 C 5 ρ m \rho^l \ge2C_5\rho^m ρ l ≥ 2 C 5 ρ m , with C 5 C_5 C 5 as in Lemma 4.15 . By maximality,
2 C 5 ρ m ≤ ρ l < 2 C 5 ρ m − 1 . 2C_5\rho^m \le\rho^l < 2C_5\rho^{m-1}. 2 C 5 ρ m ≤ ρ l < 2 C 5 ρ m − 1 .
Consider a point x ′ ∈ Γ ∩ π − 1 ( I 0 ) x' \in\Gamma\cap\pi^{-1}(I_0) x ′ ∈ Γ ∩ π − 1 ( I 0 ) such that
∣ x − x ′ ∣ ≤ 10 d C 5 ℓ ( R ) and dist ( π ( x ′ ) , ∂ I 0 ) ≥ 4 C 5 ℓ ( R ) . |x-x'| \le10\sqrt{d}C_5\ell(R) \quad\text{and} \quad\operatorname{dist}(\pi(x'),\partial I_0) \ge4C_5\ell(R). ∣ x − x ′ ∣ ≤ 10 d C 5 ℓ ( R ) and dist ( π ( x ′ ) , ∂ I 0 ) ≥ 4 C 5 ℓ ( R ) .
(imagine sliding π ( x ) \pi(x) π ( x ) towards the centre of I 0 I_0 I 0 and letting x ′ x' x ′ be the associated graph point). By Lemma 4.12 , there exists a point y ∈ E ′ y \in E' y ∈ E ′ such that ∣ x ′ − y ∣ ≤ θ 0 ℓ ( R ) |x'-y| \le\theta_0\ell(R) ∣ x ′ − y ∣ ≤ θ 0 ℓ ( R ) . Let R ′ ∈ D l R' \in\mathcal{D}_l R ′ ∈ D l such that y ∈ R ′ y \in R' y ∈ R ′ and note that
π ( c 0 B R ′ ) ⊆ I 0 and B R ′ ⊆ 15 d C 5 ρ − 1 B R . (84) \pi(c_0B_{R'}) \subseteq I_0 \quad\text{and} \quad B_{R'} \subseteq15\sqrt{d}C_5\rho^{-1}B_R. \tag*{(84)} π ( c 0 B R ′ ) ⊆ I 0 and B R ′ ⊆ 15 d C 5 ρ − 1 B R . (84)
Indeed, for each z ∈ B R ′ z \in B_{R'} z ∈ B R ′ we have (by our choice of l l l ) that
∣ z − x ′ ∣ ≤ ∣ z − x ′ ∣ + ∣ x ′ − y ∣ ≤ ( 2 + θ 0 ) ℓ ( R ′ ) ≤ 3 C 5 ρ − 1 ℓ ( R ) ; |z-x'| \le|z-x'|+|x'-y| \le(2+\theta_0)\ell(R') \le3C_5\rho^{-1}\ell(R); ∣ z − x ′ ∣ ≤ ∣ z − x ′ ∣ + ∣ x ′ − y ∣ ≤ ( 2 + θ 0 ) ℓ ( R ′ ) ≤ 3 C 5 ρ − 1 ℓ ( R ) ;
∣ z − x R ∣ ≤ ∣ z − x ′ ∣ + ∣ x ′ − x ∣ + ∣ x − x R ∣ ≤ ( 3 C 5 ρ − 1 + 10 d C 5 + 2 ) ℓ ( R ) . |z-x_R| \le|z-x'|+|x'-x|+|x-x_R| \le(3C_5\rho^{-1}+10\sqrt{d}C_5+2)\ell(R). ∣ z − x R ∣ ≤ ∣ z − x ′ ∣ + ∣ x ′ − x ∣ + ∣ x − x R ∣ ≤ ( 3 C 5 ρ − 1 + 10 d C 5 + 2 ) ℓ ( R ) .
The second inclusion in (84) is immediate from the second estimate above, the first inclusion follows since dist ( π ( x ′ ) , ∂ I 0 ) ≥ 4 C 5 ℓ ( R ) \operatorname{dist}(\pi(x'),\partial I_0) \ge4C_5\ell(R) dist ( π ( x ′ ) , ∂ I 0 ) ≥ 4 C 5 ℓ ( R ) . We note also that there exists R ′ ′ ∈ min ( T ( k 0 ) ) R'' \in\min(T(k_0)) R ′′ ∈ min ( T ( k 0 )) such that
c 0 B R ′ ′ ⊆ c 0 B R ′ and C 5 ℓ ( R ′ ′ ) ≥ ℓ ( R ) . (85) c_0B_{R''} \subseteq c_0B_{R'} \quad\text{and} \quad C_5\ell(R'') \ge\ell(R). \tag*{(85)} c 0 B R ′′ ⊆ c 0 B R ′ and C 5 ℓ ( R ′′ ) ≥ ℓ ( R ) . (85)
Indeed, by the first inclusion in (84) and Lemma 8.8, we must have R ′ ⊂ Q ( T ) R' \subset Q(T) R ′ ⊂ Q ( T ) . However, R ′ R' R ′ cannot be contained in any minimal cube from min ( T ( k 0 ) ) \min(T(k_0)) min ( T ( k 0 )) since ℓ ( R ′ ) ≥ 2 C 5 ℓ ( R ) \ell(R') \ge2C_5\ell(R) ℓ ( R ′ ) ≥ 2 C 5 ℓ ( R ) (by our choice of l l l ) and B R ′ ⊆ 15 d C 5 ρ − 1 B R B_{R'} \subseteq15\sqrt{d}C_5\rho^{-1}B_R B R ′ ⊆ 15 d C 5 ρ − 1 B R (otherwise we contradict Lemma 4.15 ). The cube R ′ ′ R'' R ′′ from (85) is then just the cube in min ( T ( k 0 ) ) \min(T(k_0)) min ( T ( k 0 )) with the same centre as R ′ R' R ′ . Finally, we note that
R ′ ′ ∈ Q . R'' \in\mathcal{Q}. R ′′ ∈ Q .
Indeed, Lemma 4.12 implies Γ ∩ ( c 0 / 8 ) B R ′ ′ ≠ ∅ \Gamma\cap(c_0/8)B_{R''} \ne\varnothing Γ ∩ ( c 0 /8 ) B R ′′ = ∅ which, when combined with (84) and (85) , give Γ ∩ π − 1 ( I 0 ) ∩ ( c 0 / 8 ) B R ′ ′ ≠ ∅ \Gamma\cap\pi^{-1}(I_0) \cap(c_0/8)B_{R''} \ne\varnothing Γ ∩ π − 1 ( I 0 ) ∩ ( c 0 /8 ) B R ′′ = ∅ .
The proof of Lemma 8.10 will be complete once we show
x ∈ 17 d C 5 ρ − 1 c B R ′ ′ . x \in17\sqrt{d}C_5\rho^{-1}cB_{R''}. x ∈ 17 d C 5 ρ − 1 c B R ′′ .
This follows from the second inclusion in (84) and (85) since
∣ x − x R ′ ′ ∣ ≤ ∣ x − x R ∣ + ∣ x R − x R ′ ′ ∣ ≤ ( 2 + 15 d C 5 ρ − 1 ) ℓ ( R ) |x-x_{R''}| \le|x-x_R|+|x_R-x_{R''}| \le(2+15\sqrt{d}C_5\rho^{-1})\ell(R) ∣ x − x R ′′ ∣ ≤ ∣ x − x R ∣ + ∣ x R − x R ′′ ∣ ≤ ( 2 + 15 d C 5 ρ − 1 ) ℓ ( R )
≤ 17 d C 5 2 ρ − 1 ℓ ( R ′ ′ ) , \le17\sqrt{d}C_5^2\rho^{-1}\ell(R''), ≤ 17 d C 5 2 ρ − 1 ℓ ( R ′′ ) ,
as required. □
Lemma 8.11. For each R ∈ Q R \in\mathcal{Q} R ∈ Q , we have
( c 0 / 4 ) B R ⊆ ( c 0 / 2 ) B Q . (c_0/4)B_R \subseteq(c_0/2)B_Q. ( c 0 /4 ) B R ⊆ ( c 0 /2 ) B Q . Proof. Let x ∈ ( c 0 / 4 ) B R x \in(c_0/4)B_R x ∈ ( c 0 /4 ) B R and y ∈ Γ ∩ π − 1 ( I 0 ) ∩ ( c 0 / 8 ) B R y \in\Gamma\cap\pi^{-1}(I_0) \cap(c_0/8)B_R y ∈ Γ ∩ π − 1 ( I 0 ) ∩ ( c 0 /8 ) B R (which exists by definition). Then, by applying Lemma 8.10 and the fact that ℓ ( R ) ≤ ℓ ( Q ( T ′ ) ) = ρ ℓ ( Q ) \ell(R) \le\ell(Q(T')) = \rho\ell(Q) ℓ ( R ) ≤ ℓ ( Q ( T ′ )) = ρ ℓ ( Q ) , we have
∣ x − x Q ∣ ≤ ∣ x − y ∣ + ∣ y − x Q ∣ ≤ c 0 ℓ ( R ) / 2 + c 0 ℓ ( Q ( T ′ ) ) / 4 ≤ ℓ ( Q ) / 2 , |x-x_Q| \le|x-y|+|y-x_Q| \le c_0\ell(R)/2+c_0\ell(Q(T'))/4 \le\ell(Q)/2, ∣ x − x Q ∣ ≤ ∣ x − y ∣ + ∣ y − x Q ∣ ≤ c 0 ℓ ( R ) /2 + c 0 ℓ ( Q ( T ′ )) /4 ≤ ℓ ( Q ) /2 , which implies the state inclusion. □
Notation 8.12. For each R ∈ Q R \in\mathcal{Q} R ∈ Q , let
z R ∈ Γ ∩ π − 1 ( I 0 ) ∩ ( c 0 / 8 ) B R . z_R \in\Gamma\cap\pi^{-1}(I_0) \cap(c_0/8)B_R. z R ∈ Γ ∩ π − 1 ( I 0 ) ∩ ( c 0 /8 ) B R .
Let D R D_R D R be a d d d -dimensional cube in P P P such that
y R : = π ( z R ) ∈ D R ⊆ I 0 and diam ( D R ) = ( c 0 / 8 ) ℓ ( R ) . (86) y_R := \pi(z_R) \in D_R \subseteq I_0 \qquad\text{and} \qquad\operatorname{diam}(D_R) = (c_0/8)\ell(R). \tag*{(86)} y R := π ( z R ) ∈ D R ⊆ I 0 and diam ( D R ) = ( c 0 /8 ) ℓ ( R ) . (86)
Let
σ R = { x + z R − y R : x ∈ D R } \sigma_R = \{x + z_R - y_R : x \in D_R\} σ R = { x + z R − y R : x ∈ D R }
be the translate of D R D_R D R through z R z_R z R , and let ζ R \zeta_R ζ R be an arbitrary ( d − 1 ) (d-1) ( d − 1 ) -dimensional sub-face. Finally, let
D = ⋃ R ∈ Q D R ⊆ I 0 . D = \bigcup_{R\in\mathcal{Q}} D_R \subseteq I_0. D = R ∈ Q ⋃ D R ⊆ I 0 .
Lemma 8.13. For each R ∈ Q R \in\mathcal{Q} R ∈ Q , we have
ζ R ⊆ σ R ⊆ ( c 0 / 4 ) B R and ℓ R : = ℓ ( ζ R ) ∼ ℓ ( R ) . (87) \zeta_R \subseteq\sigma_R \subseteq(c_0/4)B_R \qquad\text{and} \qquad\ell_R := \ell(\zeta_R) \sim\ell(R). \tag*{(87)} ζ R ⊆ σ R ⊆ ( c 0 /4 ) B R and ℓ R := ℓ ( ζ R ) ∼ ℓ ( R ) . (87) In particular, there exists a rigid motion h R : ζ R × [ 0 , ℓ R ] → ( c 0 / 4 ) B R h_R : \zeta_R \times[0,\ell_R] \to(c_0/4)B_R h R : ζ R × [ 0 , ℓ R ] → ( c 0 /4 ) B R which takes ζ R \zeta_R ζ R to F R F_R F R and satisfies (8.3)–(8.4) with parameters δ ∼ C 1 ∼ 1 \delta\sim C_1 \sim1 δ ∼ C 1 ∼ 1 .
Proof. Since σ R \sigma_R σ R is isometric to D R D_R D R , it follows that diam ( σ R ) = ( c 0 / 8 ) ℓ ( R ) \operatorname{diam}(\sigma_R) = (c_0/8)\ell(R) diam ( σ R ) = ( c 0 /8 ) ℓ ( R ) (in particular, this implies ℓ ( σ R ) ∼ ℓ ( R ) \ell(\sigma_R) \sim\ell(R) ℓ ( σ R ) ∼ ℓ ( R ) ). Since z R ∈ σ R z_R \in\sigma_R z R ∈ σ R and z R ∈ ( c 0 / 8 ) B R z_R \in(c_0/8)B_R z R ∈ ( c 0 /8 ) B R , we deduce the left-most inclusion in (87) after applying the triangle inequality. □
Lemma 8.14. The set σ : = ⋃ R ∈ Q σ R \sigma:= \bigcup_{R\in\mathcal{Q}}\sigma_R σ := ⋃ R ∈ Q σ R is the graph of a C α 0 C_{\alpha_0} C α 0 -Lipschitz map u ~ : D → P ⊥ \widetilde{u}: D \to P^\perp u : D → P ⊥ which satisfies
∣ u ~ ( x ) ∣ ≤ θ 0 ℓ ( Q ) (88) |\widetilde{u}(x)| \le\theta_0\ell(Q) \tag*{(88)} ∣ u ( x ) ∣ ≤ θ 0 ℓ ( Q ) (88) for all $x \in D.
Proof. We define u ~ : D → P ⊥ \widetilde{u}: D \to P^\perp u : D → P ⊥ by the formula u ~ ( x ) = z R − y R \widetilde{u}(x) = z_R - y_R u ( x ) = z R − y R whenever x ∈ D R x \in D_R x ∈ D R (i.e., constant on each D R D_R D R ). In this way, we have
u ~ ( y R ) = g ( y R ) for each R ∈ Q . (89) \widetilde{u}(y_R) = g(y_R) \qquad\text{for each } R \in\mathcal{Q}. \tag*{(89)} u ( y R ) = g ( y R ) for each R ∈ Q . (89)
By construction, σ \sigma σ is the graph of u ~ \widetilde{u} u over D D D . Recall from Lemma 4.12 that ∥ g ∥ ∞ ≤ θ 0 ℓ ( Q ( T ) ) ≤ θ 0 ℓ ( Q ) \|g\|_\infty\le\theta_0\ell(Q(T)) \le\theta_0\ell(Q) ∥ g ∥ ∞ ≤ θ 0 ℓ ( Q ( T )) ≤ θ 0 ℓ ( Q ) . Since z R ∈ Γ z_R \in\Gamma z R ∈ Γ , it follows that
∣ u ~ ( x ) ∣ = ∣ u ~ ( y R ) ∣ = ∣ g ( y R ) ∣ ≤ θ 0 ℓ ( Q ) |\widetilde{u}(x)| = |\widetilde{u}(y_R)| = |g(y_R)| \le\theta_0\ell(Q) ∣ u ( x ) ∣ = ∣ u ( y R ) ∣ = ∣ g ( y R ) ∣ ≤ θ 0 ℓ ( Q )
for each x ∈ D R x \in D_R x ∈ D R . Thus, it only remains to check that it is C α 0 C_{\alpha_0} C α 0 -Lipschitz. Fix w , x ∈ D w,x \in D w , x ∈ D and let R , R ′ ∈ Q R,R' \in\mathcal{Q} R , R ′ ∈ Q such that w ∈ D R w \in D_R w ∈ D R and x ∈ D R ′ x \in D_{R'} x ∈ D R ′ . If R = R ′ R=R' R = R ′ then u ~ ( w ) = u ~ ( x ) \widetilde{u}(w)=\widetilde{u}(x) u ( w ) = u ( x ) and there is nothing to show. Suppose instead that R ≠ R ′ R \ne R' R = R ′ . First, by (89) and that fact that g ( y R ) , g ( y R ′ ) ∈ Γ g(y_R),g(y_{R'}) \in\Gamma g ( y R ) , g ( y R ′ ) ∈ Γ (which is C α 0 C_{\alpha_0} C α 0 -Lipschitz), we have
∣ u ~ ( w ) − u ~ ( x ) ∣ = ∣ u ~ ( y R ) − u ~ ( y R ′ ) ∣ = ∣ g ( y R ) − g ( y R ′ ) ∣ ≤ C α 0 ∣ y R − y R ′ ∣ . |\widetilde{u}(w)-\widetilde{u}(x)| = |\widetilde{u}(y_R)-\widetilde{u}(y_{R'})| = |g(y_R)-g(y_{R'})| \le C_{\alpha_0}|y_R-y_{R'}|. ∣ u ( w ) − u ( x ) ∣ = ∣ u ( y R ) − u ( y R ′ ) ∣ = ∣ g ( y R ) − g ( y R ′ ) ∣ ≤ C α 0 ∣ y R − y R ′ ∣.
Thus, it suffices to show that
∣ y R − y R ′ ∣ ≲ ∣ w − x ∣ . |y_R-y_{R'}| \lesssim|w-x|. ∣ y R − y R ′ ∣ ≲ ∣ w − x ∣.
For this, note to begin with that, since z R ∈ σ R ⊆ ( c 0 / 4 ) B R z_R \in\sigma_R \subseteq(c_0/4)B_R z R ∈ σ R ⊆ ( c 0 /4 ) B R , z R ′ ∈ σ R ′ ⊆ ( c 0 / 4 ) B R ′ z_{R'} \in\sigma_{R'} \subseteq(c_0/4)B_{R'} z R ′ ∈ σ R ′ ⊆ ( c 0 /4 ) B R ′ and c 0 B R ∩ c 0 B R ′ = ∅ c_0B_R \cap c_0B_{R'}=\varnothing c 0 B R ∩ c 0 B R ′ = ∅ , we have
∣ z R − z R ′ ∣ ≥ ( 3 c 0 / 4 ) ( ℓ ( R ) + ℓ ( R ′ ) ) . |z_R-z_{R'}| \ge(3c_0/4)\bigl(\ell(R)+\ell(R')\bigr). ∣ z R − z R ′ ∣ ≥ ( 3 c 0 /4 ) ( ℓ ( R ) + ℓ ( R ′ ) ) .
Since y R , y R ′ y_R,y_{R'} y R , y R ′ are the points in Γ \Gamma Γ above z R , z R ′ z_R,z_{R'} z R , z R ′ , respectively, and Γ \Gamma Γ is C α 0 C_{\alpha_0} C α 0 -Lipschitz, this implies that
∣ y R − y R ′ ∣ ≥ ( c 0 / 2 ) ( ℓ ( R ) + ℓ ( R ′ ) ) . \lvert y_R-y_{R'}\rvert\geq(c_0/2)\bigl(\ell(R)+\ell(R')\bigr). ∣ y R − y R ′ ∣ ≥ ( c 0 /2 ) ( ℓ ( R ) + ℓ ( R ′ ) ) .
Thus, after recalling the choice of diameter for D R D_R D R and D R ′ D_{R'} D R ′ from (86) , we have by the triangle inequality that
∣ w − x ∣ ≥ ∣ y R − y R ′ ∣ − ∣ w − y R ∣ − ∣ x − y R ′ ∣ ≥ ∣ y R − y R ′ ∣ − ( c 0 / 8 ) ℓ ( R ) − ( c 0 / 8 ) ℓ ( R ′ ) ≥ ∣ y R − y R ′ ∣ − 1 2 ∣ y R − y R ′ ∣ ≥ 1 2 ∣ y R − y R ′ ∣ , \begin{aligned}
\lvert w-x\rvert&\geq\lvert y_R-y_{R'}\rvert-\lvert w-y_R\rvert-\lvert x-y_{R'}\rvert\\
&\geq\lvert y_R-y_{R'}\rvert-(c_0/8)\ell(R)-(c_0/8)\ell(R')\\
&\geq\lvert y_R-y_{R'}\rvert-\frac{1}{2}\lvert y_R-y_{R'}\rvert\geq\frac{1}{2}\lvert y_R-y_{R'}\rvert,
\end{aligned} ∣ w − x ∣ ≥ ∣ y R − y R ′ ∣ − ∣ w − y R ∣ − ∣ x − y R ′ ∣ ≥ ∣ y R − y R ′ ∣ − ( c 0 /8 ) ℓ ( R ) − ( c 0 /8 ) ℓ ( R ′ ) ≥ ∣ y R − y R ′ ∣ − 2 1 ∣ y R − y R ′ ∣ ≥ 2 1 ∣ y R − y R ′ ∣ ,
as required.
Notation 8.15. Let u : P → P ⊥ u:P\to P^\perp u : P → P ⊥ be a C α 0 C_{\alpha_0} C α 0 -Lipschitz extension of u ~ \tilde{u} u ~ to the whole of P P P . Let
G Q = Graph I 0 ( u ) and G Q ∗ = Graph G Q ( 2 τ 0 d N , k 0 ) . G_Q=\operatorname{Graph}_{I_0}(u)\qquad\text{and}\qquad G_Q^*=\operatorname{Graph}_{G_Q}(2\tau_0d_{N,k_0}). G Q = Graph I 0 ( u ) and G Q ∗ = Graph G Q ( 2 τ 0 d N , k 0 ) . We emphasise here the factor of 2 appearing on the right-hand side. This will be important for us as it gives a little wiggle room between the sets G Q ∗ G_Q^* G Q ∗ and Σ N , k 0 ∗ \Sigma_{N,k_0}^* Σ N , k 0 ∗ .
Lemma 8.16. We have
∥ u ∥ L ∞ ( I 0 ) ≲ ( α 0 + θ 0 ) ℓ ( Q ) . (90) \lVert u\rVert_{L^\infty(I_0)}\lesssim(\alpha_0+\theta_0)\ell(Q). \tag*{(90)} ∥ u ∥ L ∞ ( I 0 ) ≲ ( α 0 + θ 0 ) ℓ ( Q ) . (90) Furthermore, for each x ∈ G Q ∗ x\in G_Q^* x ∈ G Q ∗ there exists R ∈ Q R\in\mathcal{Q} R ∈ Q such that
dist ( x , σ R ∗ ) ≲ ℓ ( ζ R ) . (91) \operatorname{dist}(x,\sigma_R^*)\lesssim\ell(\zeta_R). \tag*{(91)} dist ( x , σ R ∗ ) ≲ ℓ ( ζ R ) . (91) Proof. The estimate (90) follows from (88) and the fact that diam ( I 0 ) ≤ ℓ ( Q ) \operatorname{diam}(I_0)\leq\ell(Q) diam ( I 0 ) ≤ ℓ ( Q ) and u u u is C α 0 C_{\alpha_0} C α 0 -Lipschitz. To see (91) , fix some x ∈ σ x\in\sigma x ∈ σ , let y = π n ( x ) ∈ G Q y=\pi_n(x)\in G_Q y = π n ( x ) ∈ G Q and z = π ( y ) ∈ I 0 z=\pi(y)\in I_0 z = π ( y ) ∈ I 0 . By Lemma 8.10, there exists R ∈ Q R\in\mathcal{Q} R ∈ Q such that z ∈ π ( C 8 B R ) z\in\pi(C_8B_R) z ∈ π ( C 8 B R ) . Since σ R ⊆ ( c 0 / 4 ) B R \sigma_R\subseteq(c_0/4)B_R σ R ⊆ ( c 0 /4 ) B R , this implies dist ( z , π ( σ R ) ) ≲ ℓ ( R ) \operatorname{dist}(z,\pi(\sigma_R))\lesssim\ell(R) dist ( z , π ( σ R )) ≲ ℓ ( R ) . Since { y } ∪ σ R ⊆ G Q \{y\}\cup\sigma_R\subseteq G_Q { y } ∪ σ R ⊆ G Q and G Q G_Q G Q is a Lipschitz graph with small constant, this implies dist ( y , σ R ) ≲ ℓ ( R ) \operatorname{dist}(y,\sigma_R)\lesssim\ell(R) dist ( y , σ R ) ≲ ℓ ( R ) . For similar reasons, we also have (91) .
Lemma 8.17. The hypotheses of Lemma 8.2 are satisfied with the data
( Q , R ) = ( I Q , I 0 ) , ( I , J ) = ( F Q , F 0 ) , u = u , (Q,R)=(I_Q,I_0),\qquad(I,J)=(F_Q,F_0),\qquad u=u, ( Q , R ) = ( I Q , I 0 ) , ( I , J ) = ( F Q , F 0 ) , u = u , ( g 1 , g 2 ) = ( τ 0 d N − 1 , k 0 , 2 τ 0 d N , k 0 ) , h = h and B = ( c 0 / 2 ) B Q , (g_1,g_2)=(\tau_0d_{N-1,k_0},2\tau_0d_{N,k_0}),\qquad h=h\text{ and }B=(c_0/2)B_Q, ( g 1 , g 2 ) = ( τ 0 d N − 1 , k 0 , 2 τ 0 d N , k 0 ) , h = h and B = ( c 0 /2 ) B Q , and parameters
η = τ 0 , θ = 2 τ 0 , γ ∼ δ ∼ C 1 ∼ 1. \eta=\tau_0,\qquad\theta=2\tau_0,\qquad\gamma\sim\delta\sim C_1\sim1. η = τ 0 , θ = 2 τ 0 , γ ∼ δ ∼ C 1 ∼ 1. In particular, for τ 0 \tau_0 τ 0 small enough, there exists a bi-Lipschitz map H : F Q × [ − 1 , 2 ] → ( c 0 / 2 ) B Q × R H:F_Q\times[-1,2]\to(c_0/2)B_Q\times\mathbb{R} H : F Q × [ − 1 , 2 ] → ( c 0 /2 ) B Q × R such that, if we denote
C Q : = H ( F Q × [ − 1 , 2 ] ) , C_Q:=H(F_Q\times[-1,2]), C Q := H ( F Q × [ − 1 , 2 ]) , then the following conditions hold. First,
I Q ∗ = H ( F Q × [ − 1 , 0 ] ) = C Q ∩ Σ N − 1 , k 0 ∗ ; (92) I_Q^*=H(F_Q\times[-1,0])=C_Q\cap\Sigma_{N-1,k_0}^*; \tag*{(92)} I Q ∗ = H ( F Q × [ − 1 , 0 ]) = C Q ∩ Σ N − 1 , k 0 ∗ ; (92) G Q ∗ = H ( F Q × [ 1 , 2 ] ) = C Q ∩ Graph R n ( 2 τ 0 d N , k 0 ) . (93) G_Q^*=H(F_Q\times[1,2])=C_Q\cap\operatorname{Graph}_{\mathbb{R}^n}(2\tau_0d_{N,k_0}). \tag*{(93)} G Q ∗ = H ( F Q × [ 1 , 2 ]) = C Q ∩ Graph R n ( 2 τ 0 d N , k 0 ) . (93) Second, for each x ∈ F Q x\in F_Q x ∈ F Q and t ∈ [ 0 , 1 ] t\in[0,1] t ∈ [ 0 , 1 ] ,
2 τ 0 d N , k 0 ( π n ( H ( x , t ) ) ) + c ( 1 − t ) ≤ π 1 ( H ( x , t ) ) (94) 2\tau_0d_{N,k_0}\bigl(\pi_n(H(x,t))\bigr)+c(1-t)\leq\pi_1(H(x,t)) \tag*{(94)} 2 τ 0 d N , k 0 ( π n ( H ( x , t )) ) + c ( 1 − t ) ≤ π 1 ( H ( x , t )) (94) ≤ τ 0 d N − 1 , k 0 ( π n ( H ( x , t ) ) ) − c t . (95) \leq\tau_0d_{N-1,k_0}\bigl(\pi_n(H(x,t))\bigr)-ct. \tag*{(95)} ≤ τ 0 d N − 1 , k 0 ( π n ( H ( x , t )) ) − c t . (95)
Finally, we have
inf a ∈ I Q ∗ ∣ x − a ∣ + ∣ a − y ∣ ≲ ∣ x − y ∣ (96) \inf_{a\in I_Q^*} |x-a|+|a-y| \lesssim|x-y| \tag*{(96)} a ∈ I Q ∗ inf ∣ x − a ∣ + ∣ a − y ∣ ≲ ∣ x − y ∣ (96)
for all x ∈ C Q x\in C_Q x ∈ C Q and y ∈ R n + 1 y\in\mathbb{R}^{n+1} y ∈ R n + 1 such that π 1 ( y ) ≥ τ 0 d N , k 0 ( π n ( y ) ) \pi_1(y)\geq\tau_0d_{N,k_0}(\pi_n(y)) π 1 ( y ) ≥ τ 0 d N , k 0 ( π n ( y )) .
Proof. Once we verify the hypotheses of Lemma 8.2 with the stated data and parameters, properties (92) –(96) follow directly from Lemma 8.2 after some definition chasing. The conditions in the preamble of Lemma 8.2 with the stated parameters are immediate from the definitions above. By (79) , Lemma 8.8 and (90) , we have
F Q , u ( F 0 ) , h ( F Q × [ 0 , 1 ] ) ⊆ ( c 0 / 2 ) B Q . (97) F_Q,u(F_0),h(F_Q\times[0,1])\subseteq(c_0/2)B_Q. \tag*{(97)} F Q , u ( F 0 ) , h ( F Q × [ 0 , 1 ]) ⊆ ( c 0 /2 ) B Q . (97)
In particular, (73) holds with B = ( c 0 / 2 ) B Q B=(c_0/2)B_Q B = ( c 0 /2 ) B Q . The inequalities in (70) with parameters γ ∼ C 1 ∼ 1 \gamma\sim C_1\sim1 γ ∼ C 1 ∼ 1 follow from (80) , (81) and (97) . Finally, properties (71) –(72) follow from Lemma 8.8 with parameters δ ∼ C 1 ∼ 1 \delta\sim C_1\sim1 δ ∼ C 1 ∼ 1 . □
Notation 8.18. For each R ∈ Q R\in Q R ∈ Q , let
σ R ∗ : = Graph σ R ( 2 τ 0 d N , k 0 ) and ζ R ∗ : = Graph ζ R ( 2 τ 0 d N , k 0 ) , \sigma_R^*:=\operatorname{Graph}_{\sigma_R}(2\tau_0d_{N,k_0})\quad\text{and}\quad\zeta_R^*:=\operatorname{Graph}_{\zeta_R}(2\tau_0d_{N,k_0}), σ R ∗ := Graph σ R ( 2 τ 0 d N , k 0 ) and ζ R ∗ := Graph ζ R ( 2 τ 0 d N , k 0 ) , and
I R ∗ : = Graph I R ( τ 0 d N , k 0 ) and F R ∗ : = Graph F R ( τ 0 d N , k 0 ) . I_R^*:=\operatorname{Graph}_{I_R}(\tau_0d_{N,k_0})\quad\text{and}\quad F_R^*:=\operatorname{Graph}_{F_R}(\tau_0d_{N,k_0}). I R ∗ := Graph I R ( τ 0 d N , k 0 ) and F R ∗ := Graph F R ( τ 0 d N , k 0 ) .
Lemma 8.19. For each R ∈ Q R\in Q R ∈ Q , the hypotheses of Lemma 8.2 are satisfied with the data
( Q , R ) = ( σ R , I R ) , ( I , J ) = ( ζ R , F R ) , u = 0 , (Q,R)=(\sigma_R,I_R),\quad(I,J)=(\zeta_R,F_R),\quad u=0, ( Q , R ) = ( σ R , I R ) , ( I , J ) = ( ζ R , F R ) , u = 0 , ( g 1 , g 2 ) = ( 2 τ 0 d N , k 0 , τ 0 d N , k 0 ) , h = h R and B = ( c 0 / 4 ) B R , (g_1,g_2)=(2\tau_0d_{N,k_0},\tau_0d_{N,k_0}),\quad h=h_R\quad\text{and}\quad B=(c_0/4)B_R, ( g 1 , g 2 ) = ( 2 τ 0 d N , k 0 , τ 0 d N , k 0 ) , h = h R and B = ( c 0 /4 ) B R , and parameters
η = 2 τ 0 , θ = τ 0 , γ ∼ δ ∼ C 1 ∼ 1. \eta=2\tau_0,\quad\theta=\tau_0,\quad\gamma\sim\delta\sim C_1\sim1. η = 2 τ 0 , θ = τ 0 , γ ∼ δ ∼ C 1 ∼ 1. In particular, for τ 0 \tau_0 τ 0 small enough, there exists a bi-Lipschitz map H R : ζ R × [ − ℓ R , 2 ℓ R ] → ( c 0 / 4 ) B R × R H_R:\zeta_R\times[-\ell_R,2\ell_R]\to(c_0/4)B_R\times\mathbb{R} H R : ζ R × [ − ℓ R , 2 ℓ R ] → ( c 0 /4 ) B R × R such that, if we denote
L R : = H R ( ζ R × [ 0 , ℓ R ] ) , L_R:=H_R(\zeta_R\times[0,\ell_R]), L R := H R ( ζ R × [ 0 , ℓ R ]) , then the following conditions holds. First,
σ R ∗ = H R ( ζ R × [ − ℓ R , 0 ] ) = L R ∩ Graph R n ( 2 τ 0 d N , k 0 ) ; (98) \sigma_R^*=H_R(\zeta_R\times[-\ell_R,0])=L_R\cap\operatorname{Graph}_{R_n}(2\tau_0d_{N,k_0}); \tag*{(98)} σ R ∗ = H R ( ζ R × [ − ℓ R , 0 ]) = L R ∩ Graph R n ( 2 τ 0 d N , k 0 ) ; (98) I R ∗ = H R ( ζ R × [ ℓ R , 2 ℓ R ] ) = L R ∩ Σ N , k 0 ∗ . (99) I_R^*=H_R(\zeta_R\times[\ell_R,2\ell_R])=L_R\cap\Sigma^*_{N,k_0}. \tag*{(99)} I R ∗ = H R ( ζ R × [ ℓ R , 2 ℓ R ]) = L R ∩ Σ N , k 0 ∗ . (99) Second, for each x ∈ ζ R x\in\zeta_R x ∈ ζ R and t ∈ [ 0 , ℓ R ] t\in[0,\ell_R] t ∈ [ 0 , ℓ R ] ,
τ 0 d N , k 0 ( π n ( H R ( x , t ) ) ) + c ( ℓ R − t ) ≤ π 1 ( H R ( x , t ) ) ≤ 2 τ 0 d N , k 0 ( π n ( H R ( x , t ) ) ) − c t . (100) \tau_0d_{N,k_0}(\pi_n(H_R(x,t)))+c(\ell_R-t)\leq\pi_1(H_R(x,t))
\leq2\tau_0d_{N,k_0}(\pi_n(H_R(x,t)))-ct. \tag*{(100)} τ 0 d N , k 0 ( π n ( H R ( x , t ))) + c ( ℓ R − t ) ≤ π 1 ( H R ( x , t )) ≤ 2 τ 0 d N , k 0 ( π n ( H R ( x , t ))) − c t . (100) Finally, we have
inf a ∈ I R ∗ ∣ x − a ∣ + ∣ a − y ∣ ≲ ∣ x − y ∣ . (101) \inf_{a\in I_R^*}|x-a|+|a-y|\lesssim|x-y|. \tag*{(101)} a ∈ I R ∗ inf ∣ x − a ∣ + ∣ a − y ∣ ≲ ∣ x − y ∣. (101) x ∈ L R x\in L_R x ∈ L R and y ∈ R n + 1 y\in\mathbb{R}^{n+1} y ∈ R n + 1 with π 1 ( y ) ≤ τ 0 d N , k 0 ( π n ( y ) ) \pi_1(y)\leq\tau_0d_{N,k_0}(\pi_n(y)) π 1 ( y ) ≤ τ 0 d N , k 0 ( π n ( y )) .
Proof. Fix R ∈ Q R \in\mathcal{Q} R ∈ Q . Once we verify the hypotheses of Lemma 8.2 with the stated data and parameters, properties (98) –(101) follow directly from Lemma 8.2 after some definition chasing. The conditions in the preamble of Lemma 8.2 with the stated parameters are immediate from the definitions above. To check (70) , let us fix x ∈ ζ R x \in\zeta_R x ∈ ζ R and y ∈ F R y \in F_R y ∈ F R . By (79) and (87) , we know
x , y ∈ ( c 0 / 4 ) B R ⟹ ∣ x − y ∣ ≲ ℓ ( R ) ∼ ℓ ( ζ R ) . x,y \in(c_0/4)B_R \mathrel{\Longrightarrow} \lvert x-y\rvert\lesssim\ell(R) \sim\ell(\zeta_R). x , y ∈ ( c 0 /4 ) B R ⟹ ∣ x − y ∣ ≲ ℓ ( R ) ∼ ℓ ( ζ R ) .
Thus, by applying Lemma 4.14, we have
2 τ 0 d N , k 0 ( x ) − τ 0 d N , k 0 ( y ) = τ 0 d N , k 0 ( x ) − τ 0 ( d N , k 0 ( y ) − d N , k 0 ( x ) ) ≥ c ℓ ( ζ R ) − C τ 0 ℓ ( ζ R ) ≳ ℓ ( ζ R ) \begin{aligned}
2\tau_0 d_{N,k_0}(x)-\tau_0 d_{N,k_0}(y)=\tau_0 d_{N,k_0}(x)-\tau_0\left(d_{N,k_0}(y)-d_{N,k_0}(x)\right) \\
&\ge c\ell(\zeta_R)-C\tau_0\ell(\zeta_R)\gtrsim\ell(\zeta_R)
\end{aligned} 2 τ 0 d N , k 0 ( x ) − τ 0 d N , k 0 ( y ) = τ 0 d N , k 0 ( x ) − τ 0 ( d N , k 0 ( y ) − d N , k 0 ( x ) ) ≥ c ℓ ( ζ R ) − C τ 0 ℓ ( ζ R ) ≳ ℓ ( ζ R )
and
2 τ 0 d N , k 0 ( x ) ≲ ℓ ( ζ R ) . 2\tau_0 d_{N,k_0}(x)\lesssim\ell(\zeta_R). 2 τ 0 d N , k 0 ( x ) ≲ ℓ ( ζ R ) .
The above two inequalities imply (70) with parameters γ ∼ C 1 ∼ 1 \gamma\sim C_1 \sim1 γ ∼ C 1 ∼ 1 . Properties (71) –(72) follows from Lemma 8.13 with parameters δ ∼ C 1 ∼ 1 \delta\sim C_1 \sim1 δ ∼ C 1 ∼ 1 . The inclusions in (74) with B = ( c 0 / 4 ) B R B=(c_0/4)B_R B = ( c 0 /4 ) B R follow from (79) and (87) . □
We define
b Q : = C Q ∪ ⋃ R ∈ Q L R . b_Q := C_Q \cup\bigcup_{R\in\mathcal{Q}} L_R. b Q := C Q ∪ R ∈ Q ⋃ L R .
Proof of Proposition 8.1 when T ∈ G 1 N ( k 0 ) T \in\mathcal{G}_1^N(k_0) T ∈ G 1 N ( k 0 )
Proposition 8.1 (1) is a consequence of Lemma 8.20 and Lemma 8.22 below.
Lemma 8.20. The bridge b Q b_Q b Q is Ahlfors d d d -regular with diam ( b Q ) ∼ ℓ ( Q ) \operatorname{diam}(b_Q)\sim\ell(Q) diam ( b Q ) ∼ ℓ ( Q ) .
Proof. By Lemma 2.9, it suffices to check that ⋃ R ∈ Q L R \bigcup_{R\in\mathcal{Q}} L_R ⋃ R ∈ Q L R is Ahlfors regular, which we will check this Lemma 2.10. For each R ∈ Q R\in\mathcal{Q} R ∈ Q , let w R ∈ L R w_R\in L_R w R ∈ L R and r R : = ℓ R r_R:=\ell_R r R := ℓ R . By Lemma 8.19, ζ R ⊆ L R \zeta_R\subseteq L_R ζ R ⊆ L R , which implies r R ≤ diam ( L R ) ≲ r R r_R\le\operatorname{diam}(L_R)\lesssim r_R r R ≤ diam ( L R ) ≲ r R . In particular, { L R } \{L_R\} { L R } and { B ( w R , r R ) } \{B(w_R,r_R)\} { B ( w R , r R )} satisfy Lemma 2.10 (2) with C 1 ′ ∼ 1 C'_1\sim1 C 1 ′ ∼ 1 . By Lemma 8.10, Γ ∩ π − 1 ( I 0 ) \Gamma\cap\pi^{-1}(I_0) Γ ∩ π − 1 ( I 0 ) and { B ( w R , r R ) } \{B(w_R,r_R)\} { B ( w R , r R )} satisfy Lemma 2.10 (3) and (4) with C 1 ′ ′ ∼ 1 C''_1\sim1 C 1 ′′ ∼ 1 . Then, for C 1 : = max { C 1 ′ , C 1 ′ ′ } C_1:=\max\{C'_1,C''_1\} C 1 := max { C 1 ′ , C 1 ′′ } , it follows from Lemma 2.8 that Lemma 2.10 (1) is satisfied with constant C 1 C_1 C 1 and C 2 C_2 C 2 depending on d d d and C 1 C_1 C 1 . This finishes the proof. □
:::
Our next goal is the check that b Q b_Q b Q supports a Poincaré inequality. The following lemma is key.
Lemma 8.21. For each distinct pair R , R ′ ∈ Q R,R'\in\mathcal{Q} R , R ′ ∈ Q , each of the sets C Q C_Q C Q , C Q ∪ L R C_Q\cup L_R C Q ∪ L R and C Q ∪ L R ∪ L R ′ C_Q\cup L_R\cup L_{R'} C Q ∪ L R ∪ L R ′ support a (1, 1)-Poincaré inequality with uniform constants.
Proof. By Lemma 8.17 and Lemma 8.19, each of C Q C_Q C Q , L R L_R L R and L R ′ L_{R'} L R ′ support a (1, 1)-Poincaré inequality since they are bi-Lipschitz to d d d -dimensional cubes. To verify the other cases we appeal to Lemma 3.4.
We start with C Q ∪ L R C_Q\cup L_R C Q ∪ L R . Observe to begin with that
C Q ∩ L R = σ R ∗ . (102) C_Q\cap L_R=\sigma_R^*. \tag*{(102)} C Q ∩ L R = σ R ∗ . (102)
Indeed, by (8.26) and (100) we have C Q ∩ L R ⊆ Graph R n ( 2 τ 0 d N , k 0 ) C_Q\cap L_R\subseteq\operatorname{Graph}_{\mathbb{R}^n}(2\tau_0d_{N,k_0}) C Q ∩ L R ⊆ Graph R n ( 2 τ 0 d N , k 0 ) . Recalling from (98) that σ R ∗ = L R ∩ Graph R n ( 2 τ 0 d N , k 0 ) \sigma_R^*=L_R\cap\operatorname{Graph}_{\mathbb{R}^n}(2\tau_0d_{N,k_0}) σ R ∗ = L R ∩ Graph R n ( 2 τ 0 d N , k 0 ) , the forward inclusion now follows. For the reverse inclusion, note first that σ R ∗ ⊆ G Q ∗ \sigma_R^* \subseteq G_Q^* σ R ∗ ⊆ G Q ∗ . The reverse inclusion then follows by (93) and (99) . The identity (102) implies
C Q ∩ L R is Ahlfors d -regular with diam ( C Q ∩ L R ) ∼ diam ( L R ) . (103) C_Q \cap L_R \text{ is Ahlfors } d\text{-regular with } \operatorname{diam}(C_Q \cap L_R) \sim\operatorname{diam}(L_R). \tag*{(103)} C Q ∩ L R is Ahlfors d -regular with diam ( C Q ∩ L R ) ∼ diam ( L R ) . (103)
We now show that
inf w ∈ C Q ∩ L R ∣ y − w ∣ + ∣ w − z ∣ ≲ ∣ y − z ∣ . (104) \inf_{w \in C_Q \cap L_R} |y-w|+|w-z| \lesssim|y-z|. \tag*{(104)} w ∈ C Q ∩ L R inf ∣ y − w ∣ + ∣ w − z ∣ ≲ ∣ y − z ∣. (104)
for points y ∈ C Q y \in C_Q y ∈ C Q and z ∈ L R z \in L_R z ∈ L R . Let y ′ ∈ F Q y' \in F_Q y ′ ∈ F Q and t ∈ [ 0 , ℓ ] t \in[0,\ell] t ∈ [ 0 , ℓ ] such that y = H ( y ′ , t ) y=H(y',t) y = H ( y ′ , t ) , and let z ′ ∈ σ R z' \in\sigma_R z ′ ∈ σ R and s ∈ [ 0 , ℓ R ] s \in[0,\ell_R] s ∈ [ 0 , ℓ R ] such that z = H R ( z ′ , s ) z=H_R(z',s) z = H R ( z ′ , s ) . Furthermore, let w : = H R ( z ′ , 0 ) w:=H_R(z',0) w := H R ( z ′ , 0 ) . Since H R H_R H R is bi-Lipschitz, we have
∣ w − z ∣ ≤ C s . |w-z| \le Cs. ∣ w − z ∣ ≤ C s .
On the other hand, by (100) ,
∣ π 1 ( w − z ) ∣ ≥ 2 τ 0 d N , k 0 ( π n ( w ) ) − ( 2 τ 0 d N , k 0 ( π n ( x ) ) − c s ) ≥ 2 τ 0 d N , k 0 ( π n ( w ) ) − 2 τ 0 d N , k 0 ( π n ( x ) ) + c ∣ w − z ∣ . \begin{aligned}
|\pi_1(w-z)| &\ge2\tau_0 d_{N,k_0}(\pi_n(w))-\left(2\tau_0 d_{N,k_0}(\pi_n(x))-cs\right) \\
&\ge2\tau_0 d_{N,k_0}(\pi_n(w))-2\tau_0 d_{N,k_0}(\pi_n(x))+c|w-z|.
\end{aligned} ∣ π 1 ( w − z ) ∣ ≥ 2 τ 0 d N , k 0 ( π n ( w )) − ( 2 τ 0 d N , k 0 ( π n ( x )) − cs ) ≥ 2 τ 0 d N , k 0 ( π n ( w )) − 2 τ 0 d N , k 0 ( π n ( x )) + c ∣ w − z ∣.
The estimate (104) now follows directly from Lemma 3.6 so long as τ 0 \tau_0 τ 0 is sufficiently small. Properties (103) , (104) and Lemma 3.4 imply that C Q ∪ L R C_Q \cup L_R C Q ∪ L R supports a ( 1 , 1 ) (1,1) ( 1 , 1 ) -Poincaré inequality.
We finish by checking that C Q ∪ L R ∪ L R ′ C_Q \cup L_R \cup L_{R'} C Q ∪ L R ∪ L R ′ supports a ( 1 , 1 ) (1,1) ( 1 , 1 ) -Poincaré inequality. Let A : = ( C Q ∪ L R ) ∩ L R ′ A:=(C_Q \cup L_R)\cap L_{R'} A := ( C Q ∪ L R ) ∩ L R ′ . By Lemma 8.19,
π R n ( L R ) ⊆ ( c 0 / 4 ) B R and π R n ( L R ′ ) ⊆ ( c 0 / 4 ) B R ′ . (105) \pi_{\mathbb{R}^n}(L_R) \subseteq(c_0/4)B_R \qquad\text{and}\qquad\pi_{\mathbb{R}^n}(L_{R'}) \subseteq(c_0/4)B_{R'}. \tag*{(105)} π R n ( L R ) ⊆ ( c 0 /4 ) B R and π R n ( L R ′ ) ⊆ ( c 0 /4 ) B R ′ . (105)
In particular, L R L_R L R and L R ′ L_{R'} L R ′ are disjoint, which implies A = C Q ∩ L R ′ A=C_Q\cap L_{R'} A = C Q ∩ L R ′ . Thus, by the same argument used to establish (103) , we get
A A A is Ahlfors d d d -regular with diam ( A ) ∼ diam ( L R ′ ) \operatorname{diam}(A)\sim\operatorname{diam}(L_{R'}) diam ( A ) ∼ diam ( L R ′ ) .
We now fix y ∈ C Q ∪ L R y\in C_Q\cup L_R y ∈ C Q ∪ L R , z ∈ L R ′ z\in L_{R'} z ∈ L R ′ and check that
inf w ∈ A ∣ y − w ∣ + ∣ w − z ∣ ≲ ∣ y − z ∣ . (106) \inf_{w\in A}|y-w|+|w-z|\lesssim|y-z|. \tag*{(106)} w ∈ A inf ∣ y − w ∣ + ∣ w − z ∣ ≲ ∣ y − z ∣. (106)
If y ∈ C Q y\in C_Q y ∈ C Q , this follows by the same argument used to establish (104) . Suppose instead that y ∈ L R ′ y\in L_{R'} y ∈ L R ′ . In this case we have ∣ x − y ∣ ≳ ℓ ( R ′ ) ∼ diam ( L R ′ ) |x-y|\gtrsim\ell(R')\sim\operatorname{diam}(L_{R'}) ∣ x − y ∣ ≳ ℓ ( R ′ ) ∼ diam ( L R ′ ) by (105) . Thus, Lemma 3.5 implies the estimate (106) . □
Lemma 8.22. The bridge b Q b_Q b Q supports a ( 1 , 1 ) (1,1) ( 1 , 1 ) -Poincaré inequality.
Proof. By Lemma 8.21, any pair of points in b Q b_Q b Q are contained in some Ahlfors regular subset of b Q b_Q b Q which supports a ( 1 , 1 ) (1,1) ( 1 , 1 ) -Poincaré inequality with uniform constants. This property implies that b Q b_Q b Q supports a ( 1 , 1 ) (1,1) ( 1 , 1 ) -Poincaré inequality. □
With Proposition 8.1 (1) out of the way, we continue with the remaining points.
Lemma 8.23. Proposition 8.1 (2) holds.
Proof. The bounds on π 1 ( x ) \pi_1(x) π 1 ( x ) for each x ∈ b Q x\in b_Q x ∈ b Q follow from (8.26) and (100) . Since H H H takes values in ( c 0 / 2 ) B Q × R (c_0/2)B_Q\times\mathbb{R} ( c 0 /2 ) B Q × R , we know that π n ( C Q ) ⊆ ( c 0 / 2 ) B Q \pi_n(C_Q)\subseteq(c_0/2)B_Q π n ( C Q ) ⊆ ( c 0 /2 ) B Q . Since H R H_R H R takes values in ( c 0 / 4 ) B R × R (c_0/4)B_R\times\mathbb{R} ( c 0 /4 ) B R × R , we know that π n ( L R ) ⊆ ( c 0 / 4 ) B R ⊆ ( c 0 / 2 ) B Q \pi_n(L_R)\subseteq(c_0/4)B_R\subseteq(c_0/2)B_Q π n ( L R ) ⊆ ( c 0 /4 ) B R ⊆ ( c 0 /2 ) B Q for each R ∈ Q R\in\mathcal{Q} R ∈ Q (recall Lemma 8.11 for this final inclusion). □
Lemma 8.24. Proposition 8.1 (3) holds.
Proof. By Lemma 4.20 and (100) , we have L R ∩ Σ N − 1 , k 0 ∗ = ∅ L_R \cap\Sigma^*_{N-1,k_0} = \emptyset L R ∩ Σ N − 1 , k 0 ∗ = ∅ for each R ∈ Q R \in\mathcal{Q} R ∈ Q . Thus, by (92) ,
Σ N − 1 , k 0 ∗ ∩ b Q = I Q ∗ ⊆ Σ S , k 0 ∗ . \Sigma^*_{N-1,k_0} \cap b_Q = I^*_Q \subseteq\Sigma^*_{S,k_0}. Σ N − 1 , k 0 ∗ ∩ b Q = I Q ∗ ⊆ Σ S , k 0 ∗ .
Since I Q I_Q I Q is bi-Lipschitz to a d d d -dimensional cube with side length comparable to ℓ ( Q ) \ell(Q) ℓ ( Q ) , we conclude the lower regularity and diameter properties of Σ N − 1 , k 0 ∗ ∩ b Q \Sigma^*_{N-1,k_0} \cap b_Q Σ N − 1 , k 0 ∗ ∩ b Q .
Lemma 8.25. Proposition 8.1 (4) holds.
Proof. By Lemma 4.20 and (8.26), we have C Q ∩ Σ N , k 0 ∗ = ∅ C_Q \cap\Sigma^*_{N,k_0} = \emptyset C Q ∩ Σ N , k 0 ∗ = ∅ . If R ∈ Q R \in\mathcal{Q} R ∈ Q , it then follows from (99) and (100) that L R ∩ Σ N , k 0 ∗ = I R ⊆ Σ T , k 0 ∗ L_R \cap\Sigma^*_{N,k_0} = I_R \subseteq\Sigma^*_{T,k_0} L R ∩ Σ N , k 0 ∗ = I R ⊆ Σ T , k 0 ∗ . In particular,
A ∗ : = Σ N , k 0 ∗ ∩ b Q = ⋃ R ∈ Q I R ∗ ⊆ Σ T , k 0 ∗ . A^* := \Sigma^*_{N,k_0} \cap b_Q = \bigcup_{R \in\mathcal{Q}} I^*_R \subseteq\Sigma^*_{T,k_0}. A ∗ := Σ N , k 0 ∗ ∩ b Q = R ∈ Q ⋃ I R ∗ ⊆ Σ T , k 0 ∗ .
It now only remains to check the lower regularity and diameter estimates for A ∗ A^* A ∗ . Since they are bi-Lipschitz equivalent, it suffices to check the same properties for the set
A : = ⋃ R ∈ Q I R . A := \bigcup_{R \in\mathcal{Q}} I_R. A := R ∈ Q ⋃ I R .
We will check the hypotheses of Lemma 2.10 with
X = Γ ∩ π − 1 ( I 0 ) , { Y i } = { I R } and { B i } = { B R } . X = \Gamma\cap\pi^{-1}(I_0), \qquad\{Y_i\} = \{I_R\} \qquad\text{and}\qquad\{B_i\} = \{B_R\}. X = Γ ∩ π − 1 ( I 0 ) , { Y i } = { I R } and { B i } = { B R } .
First, the upper regularity of X X X is immediate since Γ \Gamma Γ is a graph and the upper regularity of each Y i Y_i Y i follows from since they are d d d -dimensional cubes. Lemma 2.10 (2) and (3) follow directly from Lemma 6.2 and the definition of Q \mathcal{Q} Q with constant C 1 = 1 C_1 = 1 C 1 = 1 . The bounded overlap of { 5 B R } \{5B_R\} { 5 B R } (Lemma 2.10 (1)) follows directly from Lemma 2.8 .
Proposition 8.1 (5) is a consequence of Lemma 8.26 and Lemma 8.27 below.
Lemma 8.26. For each x ∈ b Q x \in b_Q x ∈ b Q and y ∈ R n + 1 y \in\mathbb{R}^{n+1} y ∈ R n + 1 such that π 1 ( y ) ≥ τ 0 d N − 1 , k 0 ( π n ( y ) ) \pi_1(y) \ge\tau_0 d_{N-1,k_0}(\pi_n(y)) π 1 ( y ) ≥ τ 0 d N − 1 , k 0 ( π n ( y )) we have
inf a ∈ I Q ∗ ∣ x − a ∣ + ∣ a − y ∣ ≲ ∣ x − y ∣ . (107) \inf_{a \in I^*_Q} |x-a| + |a-y| \lesssim|x-y|. \tag*{(107)} a ∈ I Q ∗ inf ∣ x − a ∣ + ∣ a − y ∣ ≲ ∣ x − y ∣. (107)
Proof. If x ∈ C Q x \in C_Q x ∈ C Q , then this already follows from (96) , so let us suppose that x ∈ L R x \in L_R x ∈ L R for some R ∈ Q R \in\mathcal{Q} R ∈ Q . Since π n ( x ) ∈ ( c 0 / 2 ) B Q \pi_n(x) \in(c_0/2)B_Q π n ( x ) ∈ ( c 0 /2 ) B Q , if π n ( y ) ∉ c 0 B Q \pi_n(y) \notin c_0B_Q π n ( y ) ∈ / c 0 B Q then ∣ x − y ∣ ≳ ℓ ( Q ) ∼ diam ( b Q ) |x-y| \gtrsim\ell(Q) \sim\operatorname{diam}(b_Q) ∣ x − y ∣ ≳ ℓ ( Q ) ∼ diam ( b Q ) and the estimate follows from Lemma 3.5 . Suppose now that x , y ∈ c 0 B Q x,y \in c_0B_Q x , y ∈ c 0 B Q . In this case, (100) and Lemma 4.20 imply
π 1 ( y ) − π 1 ( x ) ≥ τ 0 d N − 1 , k 0 ( π n ( y ) ) − 2 τ 0 d N , k 0 ( π n ( x ) ) ≳ ℓ ( Q ) ∼ diam ( b Q ) . \pi_1(y) - \pi_1(x) \ge\tau_0 d_{N-1,k_0}(\pi_n(y)) - 2\tau_0 d_{N,k_0}(\pi_n(x)) \gtrsim\ell(Q) \sim\operatorname{diam}(b_Q). π 1 ( y ) − π 1 ( x ) ≥ τ 0 d N − 1 , k 0 ( π n ( y )) − 2 τ 0 d N , k 0 ( π n ( x )) ≳ ℓ ( Q ) ∼ diam ( b Q ) .
The estimate (107) now follows by Lemma 3.5 .
Lemma 8.27. For each x ∈ b Q x \in b_Q x ∈ b Q and y ∈ R n + 1 y \in\mathbb{R}^{n+1} y ∈ R n + 1 such that π 1 ( y ) ≤ τ 0 d N , k 0 ( π n ( y ) ) \pi_1(y) \le\tau_0 d_{N,k_0}(\pi_n(y)) π 1 ( y ) ≤ τ 0 d N , k 0 ( π n ( y )) there exists R ∈ Q R \in\mathcal{Q} R ∈ Q such that
inf a ∈ I R ∗ ∣ x − a ∣ + ∣ a − y ∣ ≲ ∣ x − y ∣ . (108) \inf_{a \in I^*_R} |x-a| + |a-y| \lesssim|x-y|. \tag*{(108)} a ∈ I R ∗ inf ∣ x − a ∣ + ∣ a − y ∣ ≲ ∣ x − y ∣. (108)
Proof. If x ∈ L R x \in L_R x ∈ L R for some R ∈ Q R \in\mathcal{Q} R ∈ Q then this already follows from (101) . Suppose that x ∈ C Q x \in C_Q x ∈ C Q . If x ∈ Σ N − 1 , k 0 ∗ x \in\Sigma^*_{N-1,k_0} x ∈ Σ N − 1 , k 0 ∗ , then dist ( x , Σ N , k ∗ ) ≳ ℓ ( Q ) ∼ diam ( b Q ) \operatorname{dist}(x,\Sigma^*_{N,k}) \gtrsim\ell(Q) \sim\operatorname{diam}(b_Q) dist ( x , Σ N , k ∗ ) ≳ ℓ ( Q ) ∼ diam ( b Q ) by Lemma 4.20 . 4.20, and the estimate follows from Lemma 3.5 . Thus, we will assume also that x ∉ Σ N − 1 , k 0 ∗ x \notin\Sigma^*_{N-1,k_0} x ∈ / Σ N − 1 , k 0 ∗ . Define a point
w ∈ C Q ∩ Graph R n ( 2 τ 0 d N , k 0 ) w \in C_Q \cap\operatorname{Graph}_{\mathbb{R}^n}(2\tau_0d_{N,k_0}) w ∈ C Q ∩ Graph R n ( 2 τ 0 d N , k 0 )
as follows. If x ∈ Graph R n ( 2 τ 0 d N , k 0 ) x \in\operatorname{Graph}_{\mathbb{R}^n}(2\tau_0d_{N,k_0}) x ∈ Graph R n ( 2 τ 0 d N , k 0 ) , let w = x w=x w = x . Otherwise, let x ′ ∈ F Q x' \in F_Q x ′ ∈ F Q and t ∈ [ 0 , 1 ] t \in[0,1] t ∈ [ 0 , 1 ] such that x = H ( x ′ , t ) x=H(x',t) x = H ( x ′ , t ) , and set w = H ( x ′ , 1 ) ∈ G Q ∗ w=H(x',1)\in G_Q^* w = H ( x ′ , 1 ) ∈ G Q ∗ . In either case, using the fact that H H H is bi-Lipschitz, we have
∣ w − x ∣ ≲ 1 − t . |w-x|\lesssim1-t. ∣ w − x ∣ ≲ 1 − t .
Let R ∈ Q R\in Q R ∈ Q such that σ R ∗ \sigma_R^* σ R ∗ is closest to w w w , and let y ∈ ζ R ∗ y\in\zeta_R^* y ∈ ζ R ∗ . In this way, it follows from (91) that
∣ w − y ∣ ≲ ℓ R . |w-y|\lesssim\ell_R. ∣ w − y ∣ ≲ ℓ R .
Let y ′ ∈ ζ R y'\in\zeta_R y ′ ∈ ζ R such that y = H R ( y ′ , 0 ) y=H_R(y',0) y = H R ( y ′ , 0 ) and let z = H R ( y ′ , ℓ R ) z=H_R(y',\ell_R) z = H R ( y ′ , ℓ R ) . Since H R H_R H R is bi-Lipschitz, we have
∣ y − z ∣ ≲ ℓ R . |y-z|\lesssim\ell_R. ∣ y − z ∣ ≲ ℓ R .
Combing the previous estimates, we conclude that
∣ x − z ∣ ≲ 1 − t + ℓ R . (109) |x-z|\lesssim1-t+\ell_R. \tag*{(109)} ∣ x − z ∣ ≲ 1 − t + ℓ R . (109)
On the other hand, by (8.26),
∣ π 1 ( x − z ) ∣ ≥ 2 τ 0 d N , k 0 ( π n ( x ) ) + c ( 1 − t ) − τ 0 d N , k 0 ( π n ( z ) ) ≥ τ 0 d N , k 0 ( π n ( x ) ) − τ 0 d N , k 0 ( π n ( z ) ) + c ( 1 − t ) + d N , k 0 ( π n ( x ) ) . (110) \begin{aligned}
|\pi_1(x-z)| &\ge2\tau_0d_{N,k_0}(\pi_n(x))+c(1-t)-\tau_0d_{N,k_0}(\pi_n(z))\\
&\ge\tau_0d_{N,k_0}(\pi_n(x))-\tau_0d_{N,k_0}(\pi_n(z))+c(1-t)+d_{N,k_0}(\pi_n(x)). \tag*{(110)}
\end{aligned} ∣ π 1 ( x − z ) ∣ ≥ 2 τ 0 d N , k 0 ( π n ( x )) + c ( 1 − t ) − τ 0 d N , k 0 ( π n ( z )) ≥ τ 0 d N , k 0 ( π n ( x )) − τ 0 d N , k 0 ( π n ( z )) + c ( 1 − t ) + d N , k 0 ( π n ( x )) . (110)
Lemma 8.19 implies that π n ( z ) ∈ ( c 0 / 4 ) B R \pi_n(z)\in(c_0/4)B_R π n ( z ) ∈ ( c 0 /4 ) B R . Thus, if 1 − t ≤ ℓ R 1-t\le\ell_R 1 − t ≤ ℓ R , then it follows from (109) that π n ( x ) ∈ C B R \pi_n(x)\in CB_R π n ( x ) ∈ C B R for some C > 1 C>1 C > 1 , and from Lemma 4.14 that d N , k 0 ( π n ( x ) ) ∼ ℓ ( R ) ∼ ℓ R d_{N,k_0}(\pi_n(x))\sim\ell(R)\sim\ell_R d N , k 0 ( π n ( x )) ∼ ℓ ( R ) ∼ ℓ R . If 1 − t > ℓ R 1-t>\ell_R 1 − t > ℓ R , then 1 − t ≳ 1 − t + ℓ R 1-t\gtrsim1-t+\ell_R 1 − t ≳ 1 − t + ℓ R . In either case, we have
1 − t + d N , k 0 ( π n ( x ) ) ≳ 1 − t + ℓ R . 1-t+d_{N,k_0}(\pi_n(x))\gtrsim1-t+\ell_R. 1 − t + d N , k 0 ( π n ( x )) ≳ 1 − t + ℓ R .
Thus, carrying on from (110) and applying (109) , we get
∣ π 1 ( x − z ) ∣ ≥ τ 0 d N , k 0 ( π n ( x ) ) − τ 0 d N , k 0 ( π n ( z ) ) + c ∣ x − z ∣ . |\pi_1(x-z)|\ge\tau_0d_{N,k_0}(\pi_n(x))-\tau_0d_{N,k_0}(\pi_n(z))+c|x-z|. ∣ π 1 ( x − z ) ∣ ≥ τ 0 d N , k 0 ( π n ( x )) − τ 0 d N , k 0 ( π n ( z )) + c ∣ x − z ∣.
The estimate (108) now follows directly from Lemma 3.6 . □
Local Poincaré inequalities for two consecutive layers
As mentioned at the beginning of § 5 , the set Z Z Z from Theorem 1.3 will be obtained as a limit of (a subsequence of) a sequence of sets { Z ( k ) } k ∈ N \{Z(k)\}_{k\in\mathbb{N}} { Z ( k ) } k ∈ N . We now have all the ingredients need to define this sequence. Indeed, for k ∈ N k\in\mathbb{N} k ∈ N , the set Z ( k ) Z(k) Z ( k ) is defined as
Z ( k ) : = ⋃ N ∈ N ( ⋃ S ∈ T N ( k ) Σ S , k ∗ ∪ ⋃ Q ∈ Stop ′ ( N , k ) b Q , k ) . Z(k):=\bigcup_{N\in\mathbb{N}}\left(\bigcup_{S\in\mathcal{T}^N(k)}\Sigma^*_{S,k}\cup\bigcup_{Q\in\operatorname{Stop}'(N,k)}b_{Q,k}\right). Z ( k ) := N ∈ N ⋃ S ∈ T N ( k ) ⋃ Σ S , k ∗ ∪ Q ∈ Stop ′ ( N , k ) ⋃ b Q , k .
We will show in the next section that each Z ( k ) Z(k) Z ( k ) is Ahlfors regular and supports a Poincaré inequality. In this section, we prove two useful preliminary results in this direction. First, we prove upper regularity of the union of a Σ S , k ∗ \Sigma^*_{S,k} Σ S , k ∗ together with all its bridges (Proposition 9.1 ). Second, we show that the union of the sets and bridge corresponding to nearby stopping-time regions in consecutive layers are contained in a subset of Z ( k ) Z(k) Z ( k ) which supports a Poincaré inequality (Proposition 9.2 ).
Proposition 9.1. For each S ∈ T N ( k ) S \in\mathcal{T}^{N}(k) S ∈ T N ( k ) , the set
F S : = Σ S , k ∗ ∪ ⋃ Q ∈ Stop ′ ( N , k ) Q ⊆ Q ( S ) b Q , k F_{S}:=\Sigma^{*}_{S,k}\cup\bigcup_{\substack{Q\in\operatorname{Stop}'(N,k)\\Q\subseteq Q(S)}} b_{Q,k} F S := Σ S , k ∗ ∪ Q ∈ Stop ′ ( N , k ) Q ⊆ Q ( S ) ⋃ b Q , k is upper Ahlfors d d d -regular with diameter at most a constant multiple of ℓ ( Q ( S ) ) \ell(Q(S)) ℓ ( Q ( S )) and satisfies π n ( F S ) ⊆ 2 n λ 1 B Q ( S ) \pi_{n}(F_{S})\subseteq2\sqrt{n}\lambda_{1}B_{Q(S)} π n ( F S ) ⊆ 2 n λ 1 B Q ( S ) .
Proof. The diameter estimate and the inclusion π n ( F S ) ⊆ 2 n λ 1 B Q ( S ) \pi_{n}(F_{S})\subseteq2\sqrt{n}\lambda_{1}B_{Q(S)} π n ( F S ) ⊆ 2 n λ 1 B Q ( S ) follow directly from Lemma 6.1 and Proposition 8.1 (2). By Lemma 6.1 we need only check need only check up regularity of
⋃ Q ∈ Stop ′ ( N , k ) Q ⊆ Q ( S ) b Q , k \bigcup_{\substack{Q\in\operatorname{Stop}'(N,k)\\Q\subseteq Q(S)}} b_{Q,k} Q ∈ Stop ′ ( N , k ) Q ⊆ Q ( S ) ⋃ b Q , k to conclude upper regularity of F S F_{S} F S . For this, we we will check the hypotheses of Lemma 2.10 with data
X = Σ N , k ∗ , { Y i } = { b Q } Q ∈ Stop ′ ( N , k ) and { B i } = { B Q } Q ∈ Stop ′ ( N , k ) X=\Sigma^{*}_{N,k},\qquad\{Y_{i}\}=\{b_{Q}\}_{Q\in\operatorname{Stop}'(N,k)}\qquad\text{and}\qquad\{B_{i}\}=\{B_{Q}\}_{Q\in\operatorname{Stop}'(N,k)} X = Σ N , k ∗ , { Y i } = { b Q } Q ∈ Stop ′ ( N , k ) and { B i } = { B Q } Q ∈ Stop ′ ( N , k ) for some collection { B Q } Q ∈ Stop ′ ( N , k ) \{B_{Q}\}_{Q\in\operatorname{Stop}'(N,k)} { B Q } Q ∈ Stop ′ ( N , k ) to be define momentarily. To begin with, the regularity properties with some uniform constant C 0 C_{0} C 0 follow immediately from Lemma 6.1 and Proposition 8.1 (1). To define the collection of balls, for each Q ∈ Stop ′ ( N , k ) Q\in\operatorname{Stop}'(N,k) Q ∈ Stop ′ ( N , k ) choose a point ζ Q ∈ b Q ∩ Σ N , k ∗ \zeta_{Q}\in b_{Q}\cap\Sigma^{*}_{N,k} ζ Q ∈ b Q ∩ Σ N , k ∗ and consider
B Q : = B ( ζ Q , ( c 0 / 2 ) ℓ ( Q ) ) . \mathcal{B}_{Q}:=B(\zeta_{Q},(c_{0}/2)\ell(Q)). B Q := B ( ζ Q , ( c 0 /2 ) ℓ ( Q )) . It follows from Proposition 8.1 (1) and (2) that Lemma 2.10 (2) holds with absolute constant C 1 ∼ 1 C_{1}\sim1 C 1 ∼ 1 . It is easy to check that π n ( 5 C 1 B Q ) ⊆ 6 C 1 B Q \pi_{n}(5C_{1}\mathcal{B}_{Q})\subseteq6C_{1}B_{Q} π n ( 5 C 1 B Q ) ⊆ 6 C 1 B Q . Thus, by a volume argument in E E E we conclude that { 5 C 1 B Q } Q ∈ Stop ′ ( N , k ) \{5C_{1}\mathcal{B}_{Q}\}_{Q\in\operatorname{Stop}'(N,k)} { 5 C 1 B Q } Q ∈ Stop ′ ( N , k ) has bounded overlap with some constant C 2 ∼ 1 C_{2}\sim1 C 2 ∼ 1 . Finally, Lemma 2.10 is trivial since we have chosen the balls B Q \mathcal{B}_{Q} B Q to be centred on Σ N , k ∗ \Sigma^{*}_{N,k} Σ N , k ∗ . □
Proposition 9.2. Let k ∈ N k\in\mathbb{N} k ∈ N , L ≥ 1 L\geq1 L ≥ 1 and S L , T L ∈ T L ( k ) S_{L},T_{L}\in\mathcal{T}^{L}(k) S L , T L ∈ T L ( k ) such that A 1 B Q ( S L ) ∩ A 1 B Q ( T L ) ≠ ∅ A_{1}B_{Q(S_{L})}\cap A_{1}B_{Q(T_{L})}\neq\varnothing A 1 B Q ( S L ) ∩ A 1 B Q ( T L ) = ∅ . For any Q L ∈ min ( S L ) ∩ Stop ′ ( L , k ) Q_{L}\in\min(S_{L})\cap\operatorname{Stop}'(L,k) Q L ∈ min ( S L ) ∩ Stop ′ ( L , k ) and R L ∈ min ( T L ) ∩ Stop ′ ( L , k ) R_{L}\in\min(T_{L})\cap\operatorname{Stop}'(L,k) R L ∈ min ( T L ) ∩ Stop ′ ( L , k ) there is a set X ⊆ Z ( k ) X\subseteq Z(k) X ⊆ Z ( k ) which is Ahlfors d d d -regular, supports a ( 1 , 1 ) (1,1) ( 1 , 1 ) -Poincaré inequality and contains
⋃ i = L − 1 L ( Σ S i , k ∗ ∪ Σ T i , k ∗ ∪ b Q i , k ∪ b R i , k ) , (111) \bigcup_{i=L-1}^{L}\left(\Sigma^{*}_{S_{i,k}}\cup\Sigma^{*}_{T_{i,k}}\cup b_{Q_{i,k}}\cup b_{R_{i,k}}\right), \tag*{(111)} i = L − 1 ⋃ L ( Σ S i , k ∗ ∪ Σ T i , k ∗ ∪ b Q i , k ∪ b R i , k ) , (111) where Q L − 1 , R L − 1 ∈ Stop ( L − 1 ) Q_{L-1},R_{L-1}\in\operatorname{Stop}(L-1) Q L − 1 , R L − 1 ∈ Stop ( L − 1 ) and S L − 1 , T L − 1 ∈ T L − 1 ( k ) S_{L-1},T_{L-1}\in\mathcal{T}^{L-1}(k) S L − 1 , T L − 1 ∈ T L − 1 ( k ) are such that
Q ( S L ) ⊆ Q L − 1 ⊆ Q ( S L − 1 ) and Q ( T L ) ⊆ R L − 1 ⊆ Q ( T L − 1 ) . Q(S_{L})\subseteq Q_{L-1}\subseteq Q(S_{L-1})\qquad\text{and}\qquad Q(T_{L})\subseteq R_{L-1}\subseteq Q(T_{L-1}). Q ( S L ) ⊆ Q L − 1 ⊆ Q ( S L − 1 ) and Q ( T L ) ⊆ R L − 1 ⊆ Q ( T L − 1 ) . We will often apply Proposition 9.2 via the following easy corollary.
Corollary 9.3. Let k ∈ N k\in\mathbb{N} k ∈ N , L ≥ 1 L\geq1 L ≥ 1 , Q ∈ Stop ( L − 1 ) Q\in\operatorname{Stop}(L-1) Q ∈ Stop ( L − 1 ) and R ∈ Stop ′ ( L , k ) R\in\operatorname{Stop}'(L,k) R ∈ Stop ′ ( L , k ) such that R ⊆ Q R\subseteq Q R ⊆ Q . Then, there exists a set X ⊆ Z ( k ) X\subseteq Z(k) X ⊆ Z ( k ) which is Ahlfors d d d -regular, supports a ( 1 , 1 ) (1,1) ( 1 , 1 ) -Poincaré inequality, and contains b Q , k ∪ b R , k b_{Q,k}\cup b_{R,k} b Q , k ∪ b R , k .
Proof of Corollary 9.3 assuming Proposition 9.2. Let S = S L = T L ∈ T L ( k ) S=S_{L}=T_{L}\in\mathcal{T}^{L}(k) S = S L = T L ∈ T L ( k ) be the stopping-time regions such that R ⊆ Q ( S ) ⊆ Q R\subseteq Q(S)\subseteq Q R ⊆ Q ( S ) ⊆ Q . The set X X X is obtained by applying Proposition 9.2 with this data (note the the intersection property is trivial). □
We begin the proof of Proposition 9.2 in earnest. Let k ∈ N k \in\mathbb{N} k ∈ N , L ≥ 1 L \ge1 L ≥ 1 and S L , T L ∈ T L ( k ) S_L,T_L \in\mathcal{T}^{L}(k) S L , T L ∈ T L ( k ) as in the statement of Proposition 9.2 , and fix some arbitrary Q L ∈ min ( S L ) ∩ Stop ′ ( L , k ) Q_L \in\min(S_L) \cap\operatorname{Stop}^{\prime}(L,k) Q L ∈ min ( S L ) ∩ Stop ′ ( L , k ) and R L ∈ min ( T L ) ∩ Stop ′ ( L , k ) R_L \in\min(T_L) \cap\operatorname{Stop}^{\prime}(L,k) R L ∈ min ( T L ) ∩ Stop ′ ( L , k ) . To ease notation, we will drop the dependency of k k k on our notation for the subset. For example, we will write
Σ S = Σ S , k , Σ S ∗ = Σ S , k ∗ and b Q = b Q , k . \Sigma_S=\Sigma_{S,k},\quad\Sigma_S^*=\Sigma_{S,k}^*\quad\text{and}\quad b_Q=b_{Q,k}. Σ S = Σ S , k , Σ S ∗ = Σ S , k ∗ and b Q = b Q , k .
The first thing we show is that Q ( S L − 1 ) Q(S_{L-1}) Q ( S L − 1 ) and Q ( T L − 1 ) Q(T_{L-1}) Q ( T L − 1 ) are sufficiently close (in terms of Proposition 7.3 ).
Lemma 9.4. We have
2 B Q ( S L − 1 ) ∩ 2 B Q ( T L − 1 ) ≠ ∅ . 2B_{Q(S_{L-1})}\cap2B_{Q(T_{L-1})}\ne\varnothing. 2 B Q ( S L − 1 ) ∩ 2 B Q ( T L − 1 ) = ∅ . In particular, there exists Γ L − 1 ⊆ Σ L − 1 ∩ A 2 B Q ( S L − 1 ) \Gamma_{L-1}\subseteq\Sigma_{L-1}\cap A_2B_{Q(S_{L-1})} Γ L − 1 ⊆ Σ L − 1 ∩ A 2 B Q ( S L − 1 ) with
diam ( Γ L − 1 ) ≲ min { ℓ ( Q ( S L − 1 ) ) , ℓ ( Q ( T L − 1 ) ) } \operatorname{diam}(\Gamma_{L-1})\lesssim\min\{\ell(Q(S_{L-1})),\ell(Q(T_{L-1}))\} diam ( Γ L − 1 ) ≲ min { ℓ ( Q ( S L − 1 )) , ℓ ( Q ( T L − 1 ))} which is Ahlfors d d d -regular, supports a ( 1 , 1 ) (1,1) ( 1 , 1 ) -Poincaré inequality and satisfies
Σ S L − 1 ∪ Σ T L − 1 ∪ ⊆ Γ L − 1 . \Sigma_{S_{L-1}}\cup\Sigma_{T_{L-1}}\cup\subseteq\Gamma_{L-1}. Σ S L − 1 ∪ Σ T L − 1 ∪ ⊆ Γ L − 1 .
Proof. The existence of the set Γ L − 1 ⊆ Σ L − 1 ∩ A 2 B Q ( S L − 1 ) \Gamma_{L-1}\subseteq\Sigma_{L-1}\cap A_2B_{Q(S_{L-1})} Γ L − 1 ⊆ Σ L − 1 ∩ A 2 B Q ( S L − 1 ) (which implies the diameter estimate by Lemma 4.9) satisfying the above properties follows directly from Proposition 7.3 once we prove the stated intersection property. Since Q L − 1 ∈ Stop ′ ( L − 1 , k ) Q_{L-1}\in\operatorname{Stop}^{\prime}(L-1,k) Q L − 1 ∈ Stop ′ ( L − 1 , k ) and Q ( S L ) ⊆ Q L − 1 ⊆ Q ( S L − 1 ) Q(S_L)\subseteq Q_{L-1}\subseteq Q(S_{L-1}) Q ( S L ) ⊆ Q L − 1 ⊆ Q ( S L − 1 ) , it follows from Lemma 4.18 that
ℓ ( Q ( S L ) ) ≤ ℓ ( Q L − 1 ) ≲ τ 0 ℓ ( Q ( S L − 1 ) ) . \ell(Q(S_L))\leq\ell(Q_{L-1})\lesssim\tau_0\ell(Q(S_{L-1})). ℓ ( Q ( S L )) ≤ ℓ ( Q L − 1 ) ≲ τ 0 ℓ ( Q ( S L − 1 )) .
Thus, by the triangle inequality and the fact that Q ( S L ) ⊆ Q ( S L − 1 ) Q(S_L)\subseteq Q(S_{L-1}) Q ( S L ) ⊆ Q ( S L − 1 ) , we have
A 1 B Q ( S L ) ⊆ B ( x Q ( S L − 1 ) , ℓ ( Q ( S L − 1 ) ) + A 1 ℓ ( Q ( S L ) ) ) ⊆ ( 1 + C τ 0 ) B Q ( S L − 1 ) A_1B_{Q(S_L)}\subseteq B(x_{Q(S_{L-1})},\ell(Q(S_{L-1}))+A_1\ell(Q(S_L)))\subseteq(1+C\tau_0)B_{Q(S_{L-1})} A 1 B Q ( S L ) ⊆ B ( x Q ( S L − 1 ) , ℓ ( Q ( S L − 1 )) + A 1 ℓ ( Q ( S L ))) ⊆ ( 1 + C τ 0 ) B Q ( S L − 1 )
for some C = C ( A 1 ) C=C(A_1) C = C ( A 1 ) . For τ 0 \tau_0 τ 0 sufficiently small, depending on A 1 A_1 A 1 , this implies that
A 1 B Q ( S L ) ⊆ 2 B Q ( S L − 1 ) . A_1B_{Q(S_L)}\subseteq2B_{Q(S_{L-1})}. A 1 B Q ( S L ) ⊆ 2 B Q ( S L − 1 ) .
Similarly,
A 1 B Q ( T L ) ⊆ 2 B Q ( T L − 1 ) . A_1B_{Q(T_L)}\subseteq2B_{Q(T_{L-1})}. A 1 B Q ( T L ) ⊆ 2 B Q ( T L − 1 ) .
Since A 1 B Q ( S L ) A_1B_{Q(S_L)} A 1 B Q ( S L ) and A 1 B Q ( T L ) A_1B_{Q(T_L)} A 1 B Q ( T L ) have non-empty intersection, this proves the lemma. □
By Lemma 9.4 , we are able to apply Proposition 7.3 to construct an Ahlfors regular PI space which covers the set corresponding to level L − 1 L-1 L − 1 . More precisely, we have the following.
Lemma 9.5. If Γ L − 1 ∗ \Gamma_{L-1}^* Γ L − 1 ∗ denotes the graph of τ 0 d L − 1 , k \tau_0d_{L-1,k} τ 0 d L − 1 , k over Γ L − 1 \Gamma_{L-1} Γ L − 1 , then the set
X L − 1 : = Γ L − 1 ∗ ∪ b Q L − 1 ∪ b R L − 1 X_{L-1}:=\Gamma_{L-1}^*\cup b_{Q_{L-1}}\cup b_{R_{L-1}} X L − 1 := Γ L − 1 ∗ ∪ b Q L − 1 ∪ b R L − 1 is Ahlfors d d d -regular, supports a ( 1 , 1 ) (1,1) ( 1 , 1 ) -Poincaré inequality and has diameter at most some constant multiple of min { ℓ ( Q ( S L − 1 ) ) , ℓ ( Q ( T L − 1 ) ) } \min\{\ell(Q(S_{L-1})),\ell(Q(T_{L-1}))\} min { ℓ ( Q ( S L − 1 )) , ℓ ( Q ( T L − 1 ))} .
Proof. To simplify notation, let us drop the dependency on L L L . For example, we will denote Γ = Γ L − 1 \Gamma=\Gamma_{L-1} Γ = Γ L − 1 and b Q = b Q L − 1 b_Q=b_{Q_{L-1}} b Q = b Q L − 1 . The Ahlfors regularity of X L − 1 X_{L-1} X L − 1 follows directly from Lemma 2.9 since each b Q ′ b_{Q'} b Q ′ and Γ L − 1 ∗ \Gamma_{L-1}^* Γ L − 1 ∗ is Ahlfors d d d -regular. We begin with the diameter estimate. Since Γ ∗ \Gamma^* Γ ∗ is the graph a Lipschitz function with small constant over Γ \Gamma Γ , it follows from Lemma 9.4 that diam ( Γ ∗ ) ≲ min { ℓ ( Q ( S L − 1 ) ) , ℓ ( Q ( T L − 1 ) ) } \operatorname{diam}(\Gamma^*)\lesssim\min\{\ell(Q(S_{L-1})),\ell(Q(T_{L-1}))\} diam ( Γ ∗ ) ≲ min { ℓ ( Q ( S L − 1 )) , ℓ ( Q ( T L − 1 ))} . By Proposition 8.1 (1), we have diam ( b Q ) ≲ ℓ ( Q ) ≤ ℓ ( Q ( S ) ) \operatorname{diam}(b_Q)\lesssim\ell(Q)\leq\ell(Q(S)) diam ( b Q ) ≲ ℓ ( Q ) ≤ ℓ ( Q ( S )) and, similarly, diam ( b R ) ≲ ℓ ( Q ( T ) ) \operatorname{diam}(b_R) \lesssim\ell(Q(T)) diam ( b R ) ≲ ℓ ( Q ( T )) . Since Q ( S ) Q(S) Q ( S ) and Q ( T ) Q(T) Q ( T ) have comparable sizes (Lemma 4.9) and X L − 1 X_{L-1} X L − 1 is connected, these estimates imply the desired diameter estimate.
We now prove the Poincaré inequality. Since Γ \Gamma Γ supports a ( 1 , 1 ) (1,1) ( 1 , 1 ) -Poincaré inequality, the same is true of Γ ∗ \Gamma^* Γ ∗ with comparable constants (since they are bi-Lipschitz equivalent with constant close to 1). Recall from Proposition 8.1 (1) that both b Q b_Q b Q and b R b_R b R support a ( 1 , 1 ) (1,1) ( 1 , 1 ) -Poincaré inequality. We start by showing that
X ~ : = Γ ∗ ∪ b Q supports a ( 1 , 1 ) -Poincar e ˊ inequality. (112) \widetilde{X}:=\Gamma^* \cup b_Q\text{ supports a }(1,1)\text{-Poincaré inequality.} \tag*{(112)} X := Γ ∗ ∪ b Q supports a ( 1 , 1 ) -Poincar e ˊ inequality. (112)
Indeed, since Γ ∗ ⊇ Σ S ∗ \Gamma^* \supseteq\Sigma^*_S Γ ∗ ⊇ Σ S ∗ , it follows from Proposition 8.1 (1) and Proposition 8.1 (3) that Γ ∗ ∩ b Q \Gamma^* \cap b_Q Γ ∗ ∩ b Q is lower Ahlfors d d d -regular on all balls with radius at most ℓ ( Q ) ∼ diam ( b Q ) \ell(Q) \sim\operatorname{diam}(b_Q) ℓ ( Q ) ∼ diam ( b Q ) . Moreover, by Proposition 8.1 (3) and Proposition 8.1 (5),
inf a ∈ Γ ∗ ∩ b Q ∣ x − a ∣ + ∣ a − y ∣ ≲ ∣ x − y ∣ \inf_{a\in\Gamma^*\cap b_Q}|x-a|+|a-y|\lesssim|x-y| a ∈ Γ ∗ ∩ b Q inf ∣ x − a ∣ + ∣ a − y ∣ ≲ ∣ x − y ∣
for all x ∈ b Q x\in b_Q x ∈ b Q and y ∈ Γ ∗ y\in\Gamma^* y ∈ Γ ∗ . Thus, (112) follows immediately from Lemma 3.4 .
We now check that
X = X ~ ∪ b R supports a ( 1 , 1 ) -Poincar e ˊ inequality. X=\widetilde{X}\cup b_R\text{ supports a }(1,1)\text{-Poincaré inequality.} X = X ∪ b R supports a ( 1 , 1 ) -Poincar e ˊ inequality.
Again, we will use Lemma 3.4 . We begin by noting that, by Proposition 8.1 (2), the bridges satisfy
dist ( b Q , b R ) ≳ ℓ ( R ) ∼ diam ( b R ) . (113) \operatorname{dist}(b_Q,b_R)\gtrsim\ell(R)\sim\operatorname{diam}(b_R). \tag*{(113)} dist ( b Q , b R ) ≳ ℓ ( R ) ∼ diam ( b R ) . (113)
In particular, the bridges are disjoint and (for the same reasons as above) this implies that X ~ ∩ b R \widetilde{X}\cap b_R X ∩ b R is lower Ahlfors d d d -regular on all balls with radius at most ℓ ( R ) ∼ diam ( b R ) \ell(R)\sim\operatorname{diam}(b_R) ℓ ( R ) ∼ diam ( b R ) . The only thing left to show is that
inf a ∈ Γ ∗ ∣ x − a ∣ + ∣ a − y ∣ ≲ ∣ x − y ∣ \inf_{a\in\Gamma^*}|x-a|+|a-y|\lesssim|x-y| a ∈ Γ ∗ inf ∣ x − a ∣ + ∣ a − y ∣ ≲ ∣ x − y ∣
for all x ∈ b R x\in b_R x ∈ b R and y ∈ X ~ y\in\widetilde{X} y ∈ X . As in the proof of (112) , if y ∈ Γ ∗ y\in\Gamma^* y ∈ Γ ∗ then this follows directly from Proposition 8.1 (3) and Proposition 8.1 (5). If y ∈ b Q y\in b_Q y ∈ b Q , this follows from Lemma 3.5 . Indeed, by (113) , ∣ x − y ∣ ≳ diam ( b R ) |x-y|\gtrsim\operatorname{diam}(b_R) ∣ x − y ∣ ≳ diam ( b R ) . □
We now want to find a set which covers the sets corresponding to level L L L and which interacts well (in terms of a Poincaré inequality) with the set Γ L − 1 ∗ \Gamma^*_{L-1} Γ L − 1 ∗ . The construction of this set (and the proof of Proposition 9.2 ) splits into two cases. Recall from Proposition 7.3 that there are two possibilities for the relationship between Γ S L \Gamma_{S_L} Γ S L and Γ T L \Gamma_{T_L} Γ T L . Either they are contained in some set which supports a Poincaré inequality (within the same layer) or they are well-separated. We deal now with the first case.
Case 1. Suppose that there exists a set Γ L ⊆ Σ L ∩ A 2 B Q ( S L ) \Gamma_L\subseteq\Sigma_L\cap A_2B_{Q(S_L)} Γ L ⊆ Σ L ∩ A 2 B Q ( S L ) which contains Γ S L ∪ Γ T L \Gamma_{S_L}\cup\Gamma_{T_L} Γ S L ∪ Γ T L , is Ahlfors d d d -regular and supports a ( 1 , 1 ) (1,1) ( 1 , 1 ) -Poincaré inequality.
We start by noting the following, whose proof is the same (up to notational changes) as the proof of Lemma 9.5 .
Lemma 9.6. If Γ L ∗ \Gamma^*_L Γ L ∗ denotes the graph of τ d L , k \tau d_{L,k} τ d L , k over Γ L \Gamma_L Γ L , then the set
X L : = Γ L ∗ ∪ b Q L ∪ b R L X_L:=\Gamma^*_L\cup b_{Q_L}\cup b_{R_L} X L := Γ L ∗ ∪ b Q L ∪ b R L is Ahlfors d d d -regular, supports a ( 1 , 1 ) (1,1) ( 1 , 1 ) -Poincaré inequality and satisfies diam ( X L ) ≲ min { ℓ ( Q ( S L ) ) , ℓ ( Q ( T L ) ) } \operatorname{diam}(X_L)\lesssim\min\{\ell(Q(S_L)),\ell(Q(T_L))\} diam ( X L ) ≲ min { ℓ ( Q ( S L )) , ℓ ( Q ( T L ))} .
:::`
Proof of Proposition 9.2 in Case 1. We set
X : = X L − 1 ∪ X L . X := X_{L-1} \cup X_L. X := X L − 1 ∪ X L .
The inclusion (111) is immediate from the definition of X L − 1 X_{L-1} X L − 1 and X L X_L X L . The Ahlfors regularity of X X X follows from Lemma 2.9 since each of X L − 1 X_{L-1} X L − 1 and X L X_L X L are Ahlfors regular with non-trivial intersection. To check the Poincaré inequality we will (as usual) apply Lemma 3.4. We show first that
X L − 1 ∩ X L is lower Ahlfors d -regular . (114) X_{L-1} \cap X_L \text{ is lower Ahlfors } d\text{-regular}. \tag*{(114)} X L − 1 ∩ X L is lower Ahlfors d -regular . (114)
Indeed, recall from Proposition 7.3 and Lemma 9.6 that
Γ L ∗ ⊇ Γ S L ∗ ⊇ Σ S L ∗ . \Gamma_L^* \supseteq\Gamma_{S_L}^* \supseteq\Sigma_{S_L}^*. Γ L ∗ ⊇ Γ S L ∗ ⊇ Σ S L ∗ .
Thus, it follows from Proposition 8.1 (1) and (4) that X L ∩ b Q L − 1 X_L \cap b_{Q_{L-1}} X L ∩ b Q L − 1 is lower Ahlfors d d d -regular on all balls with radius at most ℓ ( Q L − 1 ) ∼ ℓ ( Q ( S L ) ) \ell(Q_{L-1}) \sim\ell(Q(S_L)) ℓ ( Q L − 1 ) ∼ ℓ ( Q ( S L )) . By similar reasoning, X L ∩ b R L − 1 X_L \cap b_{R_{L-1}} X L ∩ b R L − 1 is lower Ahlfors d d d -regular on all balls with radius at most ℓ ( R L − 1 ) ∼ ℓ ( Q ( T L ) ) \ell(R_{L-1}) \sim\ell(Q(T_L)) ℓ ( R L − 1 ) ∼ ℓ ( Q ( T L )) . Since diam ( X L ) ≲ min { ℓ ( Q L − 1 ) , ℓ ( R L − 1 ) } \operatorname{diam}(X_L) \lesssim\min\{\ell(Q_{L-1}),\ell(R_{L-1})\} diam ( X L ) ≲ min { ℓ ( Q L − 1 ) , ℓ ( R L − 1 )} (by Lemma 9.6) and X L − 1 ∩ X L X_{L-1} \cap X_L X L − 1 ∩ X L is contained in b Q L − 1 ∪ b R L − 1 b_{Q_{L-1}} \cup b_{R_{L-1}} b Q L − 1 ∪ b R L − 1 , the lower regularity of this set on balls with radius at most diam ( X L ) \operatorname{diam}(X_L) diam ( X L ) is implied by the lower regularity estimates we established above. Now we check that
inf a ∈ X L − 1 ∩ X L ∣ x − a ∣ + ∣ a − y ∣ ≲ ∣ x − y ∣ (115) \inf_{a\in X_{L-1}\cap X_L} |x-a|+|a-y| \lesssim|x-y| \tag*{(115)} a ∈ X L − 1 ∩ X L inf ∣ x − a ∣ + ∣ a − y ∣ ≲ ∣ x − y ∣ (115)
for all x ∈ X L − 1 x \in X_{L-1} x ∈ X L − 1 and y ∈ X L y \in X_L y ∈ X L . If x ∈ b Q L − 1 x \in b_{Q_{L-1}} x ∈ b Q L − 1 or x ∈ b R L − 1 x \in b_{R_{L-1}} x ∈ b R L − 1 , then this estimate follows from Proposition 8.1 (5) since, by Proposition 8.1 (2) and the definition of Γ L ∗ \Gamma_L^* Γ L ∗ , we have π 1 ( y ) ≤ τ 0 d L , k ( π n ( x ) ) \pi_1(y) \le\tau_0 d_{L,k}(\pi_n(x)) π 1 ( y ) ≤ τ 0 d L , k ( π n ( x )) . Suppose instead that x ∈ Γ L − 1 ∗ x \in\Gamma_{L-1}^* x ∈ Γ L − 1 ∗ . Since
π n ( y ) ∈ A 2 B Q ( S L ) ⊆ A 2 B Q L − 1 , (116) \pi_n(y) \in A_2 B_{Q(S_L)} \subseteq A_2 B_{Q_{L-1}}, \tag*{(116)} π n ( y ) ∈ A 2 B Q ( S L ) ⊆ A 2 B Q L − 1 , (116)
if π n ( x ) ∉ 2 A 2 B Q L − 1 \pi_n(x) \notin2A_2B_{Q_{L-1}} π n ( x ) ∈ / 2 A 2 B Q L − 1 , then we have ∣ x − y ∣ ≳ ℓ ( Q ( S L ) ) ∼ diam ( X L ) |x-y| \gtrsim\ell(Q(S_L)) \sim\operatorname{diam}(X_L) ∣ x − y ∣ ≳ ℓ ( Q ( S L )) ∼ diam ( X L ) and the estimate follows from Lemma 3.5. Suppose instead that π n ( x ) ∈ 2 A 2 B Q L − 1 \pi_n(x) \in2A_2B_{Q_{L-1}} π n ( x ) ∈ 2 A 2 B Q L − 1 . Then, recalling (116) and applying Lemma 4.20, we have
π 1 ( x ) − π 1 ( y ) ≥ τ 0 d L − 1 , k ( π n ( x ) ) − τ 0 d L , k ( π n ( y ) ) ≳ ℓ ( Q L − 1 ) ∼ diam ( X L ) . \begin{aligned}
\pi_1(x)-\pi_1(y) &\ge\tau_0d_{L-1,k}(\pi_n(x))-\tau_0d_{L,k}(\pi_n(y))\\
&\gtrsim\ell(Q_{L-1})\sim\operatorname{diam}(X_L).
\end{aligned} π 1 ( x ) − π 1 ( y ) ≥ τ 0 d L − 1 , k ( π n ( x )) − τ 0 d L , k ( π n ( y )) ≳ ℓ ( Q L − 1 ) ∼ diam ( X L ) .
and the estimate again follows from Lemma 3.5. Properties (114) and (115) imply that X L − 1 X_{L-1} X L − 1 and X L X_L X L satisfies the hypotheses of Lemma 3.4. Thus, we reach our desired conclusion. □
Case 2. Suppose that there is no set satisfying the hypotheses of Proposition 7.3. In which case,
dist ( Γ S L , Γ T L ) ≳ ℓ ( Q ( S ) ) + ℓ ( Q ( T ) ) . (117) \operatorname{dist}(\Gamma_{S_L},\Gamma_{T_L}) \gtrsim\ell(Q(S))+\ell(Q(T)). \tag*{(117)} dist ( Γ S L , Γ T L ) ≳ ℓ ( Q ( S )) + ℓ ( Q ( T )) . (117)
We define
X L : = Γ S L ∗ ∪ Γ T L ∗ ∪ b Q L ∪ b R L . X_L := \Gamma_{S_L}^* \cup\Gamma_{T_L}^* \cup b_{Q_L} \cup b_{R_L}. X L := Γ S L ∗ ∪ Γ T L ∗ ∪ b Q L ∪ b R L .
Then, the following lemma implies Proposition 9.2 in this case.
Lemma 9.7. The set X : = X L − 1 ∪ X L X := X_{L-1} \cup X_L X := X L − 1 ∪ X L is Ahlfors d d d -regular and supports a ( 1 , 1 ) (1,1) ( 1 , 1 ) -Poincaré inequality.
Proof. The inclusion (111) is immediate from the definitions of X L − 1 X_{L-1} X L − 1 and X L X_L X L . It follows from the definitions that X X X is the union of a sequence of Ahlfors regular spaces with non-trivial intersections. Thus, by Lemma 2.9 , it is Ahlfors regular. Since X L X_L X L does not support a Poincaré inequality, we cannot directly apply the gluing lemma (Lemma 3.4 ). As in the proof of Lemma 9.5 , we will glue X L X_L X L to X L − 1 X_{L-1} X L − 1 in two steps. We begin by showing that
X ~ : = X L − 1 ∪ Γ S L ∗ ∪ b Q L supports a ( 1 , 1 ) -Poincar e ˊ inequality . (118) \widetilde{X} := X_{L-1} \cup\Gamma_{S_L}^{*} \cup bQ_L \text{ supports a } (1,1)\text{-Poincaré inequality}. \tag*{(118)} X := X L − 1 ∪ Γ S L ∗ ∪ b Q L supports a ( 1 , 1 ) -Poincar e ˊ inequality . (118)
We start with the lower regularity of X L − 1 ∩ ( Γ S L ∗ ∪ b Q L ) X_{L-1} \cap(\Gamma_{S_L}^{*} \cup bQ_L) X L − 1 ∩ ( Γ S L ∗ ∪ b Q L ) . Note to begin with that
X L − 1 ∩ ( Γ S L ∗ ∪ b Q L ) ⊆ Σ L ∗ . X_{L-1} \cap(\Gamma_{S_L}^{*} \cup bQ_L) \subseteq\Sigma_L^{*}. X L − 1 ∩ ( Γ S L ∗ ∪ b Q L ) ⊆ Σ L ∗ .
Since the only parts of X L − 1 X_{L-1} X L − 1 which have non-empty intersection with Σ L ∗ \Sigma_L^{*} Σ L ∗ are b Q L − 1 bQ_{L-1} b Q L − 1 and b R L − 1 bR_{L-1} b R L − 1 , and since (by Lemma 7.3 and Proposition 8.1 (3)) we have b Q L ∩ Σ L ∗ ⊆ Γ S L ∗ bQ_L \cap\Sigma_L^{*} \subseteq\Gamma_{S_L}^{*} b Q L ∩ Σ L ∗ ⊆ Γ S L ∗ , it follows that
X L − 1 ∩ ( Γ S L ∗ ∪ b Q L ) ⊆ Γ S L ∗ ∩ ( b Q L − 1 ∪ b R L − 1 ) . X_{L-1} \cap(\Gamma_{S_L}^{*} \cup bQ_L) \subseteq\Gamma_{S_L}^{*} \cap(bQ_{L-1} \cup bR_{L-1}). X L − 1 ∩ ( Γ S L ∗ ∪ b Q L ) ⊆ Γ S L ∗ ∩ ( b Q L − 1 ∪ b R L − 1 ) .
By Proposition 7.3 and Proposition 8.1 (4), we have b R L − 1 ∩ Σ L ∗ ⊆ Σ U T L ∗ ⊆ Γ T L ∗ bR_{L-1} \cap\Sigma_L^{*} \subseteq\Sigma_{U_TL}^{*} \subseteq\Gamma_{T_L}^{*} b R L − 1 ∩ Σ L ∗ ⊆ Σ U T L ∗ ⊆ Γ T L ∗ , which, since Γ S L ∗ \Gamma_{S_L}^{*} Γ S L ∗ and Γ T L ∗ \Gamma_{T_L}^{*} Γ T L ∗ are well-separated, implies that
X L − 1 ∩ ( Γ S L ∗ ∪ b Q L ) = Γ S L ∗ ∩ b Q L − 1 . X_{L-1} \cap(\Gamma_{S_L}^{*} \cup bQ_L) = \Gamma_{S_L}^{*} \cap bQ_{L-1}. X L − 1 ∩ ( Γ S L ∗ ∪ b Q L ) = Γ S L ∗ ∩ b Q L − 1 .
The lower regularity now follows directly from Proposition 8.1 (4). To establish (118) , the only other thing one needs to check is that
inf a ∈ X L − 1 ∩ ( Γ S L ∗ ∪ b Q L ) ∣ x − a ∣ + ∣ a − y ∣ ≲ ∣ x − y ∣ . (119) \inf_{a \in X_{L-1} \cap(\Gamma_{S_L}^{*} \cup bQ_L)} |x-a|+|a-y| \lesssim|x-y|. \tag*{(119)} a ∈ X L − 1 ∩ ( Γ S L ∗ ∪ b Q L ) inf ∣ x − a ∣ + ∣ a − y ∣ ≲ ∣ x − y ∣. (119)
for all x ∈ X L − 1 x \in X_{L-1} x ∈ X L − 1 and y ∈ Γ S L ∗ ∪ b Q L y \in\Gamma_{S_L}^{*} \cup bQ_L y ∈ Γ S L ∗ ∪ b Q L . The proof of (119) (up to notational changes) identical to the proof of (115) , and we will not repeat the argument.
We now finish by checking that
X = X ~ ∪ Γ T L ∗ ∪ b R L supports a ( 1 , 1 ) -Poincar e ˊ inequality . X = \widetilde{X} \cup\Gamma_{T_L}^{*} \cup bR_L \text{ supports a } (1,1)\text{-Poincaré inequality}. X = X ∪ Γ T L ∗ ∪ b R L supports a ( 1 , 1 ) -Poincar e ˊ inequality .
By the same argument as above, we have that
X L − 1 ∩ ( Γ T L ∗ ∪ b R L ) = Γ T L ∗ ∩ b R L − 1 . X_{L-1} \cap(\Gamma_{T_L}^{*} \cup bR_L) = \Gamma_{T_L}^{*} \cap bR_{L-1}. X L − 1 ∩ ( Γ T L ∗ ∪ b R L ) = Γ T L ∗ ∩ b R L − 1 .
By Proposition 8.1 (4), this set is lower Ahlfors d d d -regular. Thus, the only thing left to show is that
inf a ∈ X L − 1 ∩ ( Γ S L ∗ ∪ b Q L ) ∣ x − a ∣ + ∣ a − y ∣ ≲ ∣ x − y ∣ . \inf_{a \in X_{L-1} \cap(\Gamma_{S_L}^{*} \cup bQ_L)} |x-a|+|a-y| \lesssim|x-y|. a ∈ X L − 1 ∩ ( Γ S L ∗ ∪ b Q L ) inf ∣ x − a ∣ + ∣ a − y ∣ ≲ ∣ x − y ∣.
for all x ∈ X ~ x \in\widetilde{X} x ∈ X and y ∈ Γ T L ∗ ∪ b R L y \in\Gamma_{T_L}^{*} \cup bR_L y ∈ Γ T L ∗ ∪ b R L . If x ∈ X ~ x \in\widetilde{X} x ∈ X then the same proof as in (115) gives the desired estimate. So, let us suppose that x ∈ Γ S L ∗ ∪ b Q L x \in\Gamma_{S_L}^{*} \cup bQ_L x ∈ Γ S L ∗ ∪ b Q L . In this case, it follows from (117) that ∣ x − y ∣ ≳ ℓ ( Q ( S L ) ) ∼ diam ( Γ S L ∪ b Q L ) |x-y| \gtrsim\ell(Q(S_L)) \sim\operatorname{diam}(\Gamma_{S_L} \cup bQ_L) ∣ x − y ∣ ≳ ℓ ( Q ( S L )) ∼ diam ( Γ S L ∪ b Q L ) , and the estimate follows from Lemma 3.5 . □ \square □
Global Poincaré inequalities across all layers
We have now reached the point where we can prove the desired properties of the set Z ( k ) Z(k) Z ( k ) (Ahlfors regularity, Poincaré inequality, approximation of E E E ). The precise statement that we will prove is the following.
Proposition 10.1. For each k ∈ N k \in\mathbb{N} k ∈ N , the set Z ( k ) Z(k) Z ( k ) is closed, contained in the closed ball B ‾ ( 0 , 15 n λ 1 ) \overline{B}(0,15\sqrt{n}\lambda_1) B ( 0 , 15 n λ 1 ) in R n + 1 \mathbb{R}^{n+1} R n + 1 , Ahlfors d d d -regular and supports a ( 1 , 1 ) (1,1) ( 1 , 1 ) -Poincaré inequality with constant that depend only on n n n , d d d and the uniformly rectifiability constants i.e., independent of k k k . Moreover, we have
dist ( x , Z ( k ) ) ≲ ρ k (120) \operatorname{dist}(x,Z(k)) \lesssim\rho^k \tag*{(120)} dist ( x , Z ( k )) ≲ ρ k (120) for each x ∈ E x \in E x ∈ E .
Before proving Proposition 10.1, let us see why it implies Theorem 1.3.
Proof of Theorem 1.3 assuming Proposition 10.1. Since the space of non-empty compact subsets of B ‾ ( 0 , 15 n λ 1 ) \overline{B}(0,15\sqrt{n}\lambda_1) B ( 0 , 15 n λ 1 ) is compact with respect to Hausdorff convergence, there is a subsequence { Z ( k i ) } i ∈ N \{Z(k_i)\}_{i\in\mathbb{N}} { Z ( k i ) } i ∈ N converging to a compact set Z ⊆ B ‾ ( 0 , 15 n λ 1 ) Z \subseteq\overline{B}(0,15\sqrt{n}\lambda_1) Z ⊆ B ( 0 , 15 n λ 1 ) . Since each Z ( k i ) Z(k_i) Z ( k i ) is Ahlfors d d d -regular and supports a ( 1 , 1 ) (1,1) ( 1 , 1 ) -Poincaré inequality with uniform constants, the same is true of Z Z Z . Finally, (120) implies E ⊆ Z E \subseteq Z E ⊆ Z . □
Fix some k ∈ N k \in\mathbb{N} k ∈ N . We now prove Proposition 10.1.
Lemma 10.2. The set Z ( k ) Z(k) Z ( k ) is closed and contained in the closed ball B ‾ ( 0 , 15 n λ 1 ) \overline{B}(0,15\sqrt{n}\lambda_1) B ( 0 , 15 n λ 1 ) in R n + 1 \mathbb{R}^{n+1} R n + 1 .
Proof. It is clear that Z ( k ) Z(k) Z ( k ) is closed since it is the union of finitely many closed sets. To prove the inclusion, fix some the moment some N ∈ N N \in\mathbb{N} N ∈ N and S ∈ T N ( k ) S \in\mathcal{T}^{N}(k) S ∈ T N ( k ) . By Proposition 9.1, we know that π n ( F S ) ⊆ 2 n λ B Q ( S ) \pi_n(F_S) \subseteq2\sqrt{n}\lambda B_{Q(S)} π n ( F S ) ⊆ 2 n λ B Q ( S ) . Since, 0 ≤ π 1 ( x ) ≤ τ 0 d N , k ( π n ( x ) ) 0 \leq\pi_1(x) \leq\tau_0 d_{N,k}(\pi_n(x)) 0 ≤ π 1 ( x ) ≤ τ 0 d N , k ( π n ( x )) for each x ∈ F S x \in F_S x ∈ F S , this implies that F S ⊆ 3 n λ B Q ( S ) F_S \subseteq3\sqrt{n}\lambda B_{Q(S)} F S ⊆ 3 n λ B Q ( S ) (as a ball in R n + 1 \mathbb{R}^{n+1} R n + 1 ). Recalling (25) , this implies
F S ⊆ 3 n λ 1 B Q ( S ) ⊆ 3 n λ 1 B Q 0 ⊆ B ( 0 , 15 n λ 1 ) . F_S \subseteq3\sqrt{n}\lambda_1 B_{Q(S)} \subseteq3\sqrt{n}\lambda_1 B_{Q_0} \subseteq B(0,15\sqrt{n}\lambda_1). F S ⊆ 3 n λ 1 B Q ( S ) ⊆ 3 n λ 1 B Q 0 ⊆ B ( 0 , 15 n λ 1 ) .
Since Z ( k ) Z(k) Z ( k ) is just the union over the F S F_S F S , this finishes the proof. □
Lemma 10.3. The set Z ( k ) Z(k) Z ( k ) is lower Ahlfors d d d -regular with regularity constant depending only on n n n , d d d and the uniform rectifiability constants for E E E .
Proof. Fix x ∈ Z ( k ) x \in Z(k) x ∈ Z ( k ) and r ∈ ( 0 , diam ( Z ( k ) ) ) r \in(0,\operatorname{diam}(Z(k))) r ∈ ( 0 , diam ( Z ( k ))) . Moreover, suppose that
x ∈ Σ S , k ∗ ∪ b Q , k x \in\Sigma^*_{S,k} \cup b_{Q,k} x ∈ Σ S , k ∗ ∪ b Q , k
for some N ∈ N N \in\mathbb{N} N ∈ N , S ∈ T N ( k ) S \in\mathcal{T}^{N}(k) S ∈ T N ( k ) and Q ′ ∈ Stop ′ ( N , k ) Q' \in\operatorname{Stop}'(N,k) Q ′ ∈ Stop ′ ( N , k ) . We can suppose without loss of generality that Q ′ ⊆ Q ( S ) Q' \subseteq Q(S) Q ′ ⊆ Q ( S ) . Let Q ∗ ∈ Stop ( N − 1 ) Q^* \in\operatorname{Stop}(N-1) Q ∗ ∈ Stop ( N − 1 ) such that Q ( S ) ⊆ Q ∗ Q(S) \subseteq Q^* Q ( S ) ⊆ Q ∗ . Let λ > 1 \lambda> 1 λ > 1 (to be chosen sufficiently large). If r ≤ λ ℓ ( Q ∗ ) ∼ λ diam ( Γ S , k ∗ ∪ b Q , k ) r \leq\lambda\ell(Q^*) \sim\lambda\operatorname{diam}(\Gamma^*_{S,k} \cup b_{Q,k}) r ≤ λ ℓ ( Q ∗ ) ∼ λ diam ( Γ S , k ∗ ∪ b Q , k ) , then the lower regularity estimates follow directly from the Ahlfors regularity of this space (recall that Γ S , k ∗ \Gamma^*_{S,k} Γ S , k ∗ and b Q , k b_{Q,k} b Q , k are Ahlfors regular by Propositions 7.3 and 8.1, respectively). So, we suppose instead that
r > 10 λ ℓ ( Q ∗ ) . (121) r > 10\lambda\ell(Q^*). \tag*{(121)} r > 10 λ ℓ ( Q ∗ ) . (121)
Let Q ∗ ∗ Q^{**} Q ∗∗ be the largest cube in the set
{ R ∈ ⋃ L = 0 N − 1 Stop ( L ) : R ⊇ Q and λ ℓ ( R ) ≤ π 1 ( x ) + r / 10 } . \left\{R \in\bigcup_{L=0}^{N-1}\operatorname{Stop}(L): R \supseteq Q \text{ and } \lambda\ell(R) \leq\pi_1(x)+r/10\right\}. { R ∈ L = 0 ⋃ N − 1 Stop ( L ) : R ⊇ Q and λ ℓ ( R ) ≤ π 1 ( x ) + r /10 } .
(which is non-empty by (121) ). Let L ∈ { 0 , … , N − 1 } L \in\{0,\ldots,N-1\} L ∈ { 0 , … , N − 1 } and T ∈ T L ( k ) T \in\mathcal{T}^{L}(k) T ∈ T L ( k ) such that Q ∗ ∗ ∈ T Q^{**} \in T Q ∗∗ ∈ T . By maximality, the parent R R R of Q ( T ) Q(T) Q ( T ) satisfies λ ℓ ( R ) > π 1 ( x ) + r / 10 \lambda\ell(R) > \pi_1(x)+r/10 λ ℓ ( R ) > π 1 ( x ) + r /10 . In particular,
diam ( Γ T , k ∗ ) ∼ diam ( Σ T , k ∗ ) ∼ ℓ ( Q ( T ) ) ∼ ℓ ( R ) ≳ r . (122) \operatorname{diam}(\Gamma^*_{T,k}) \sim\operatorname{diam}(\Sigma^*_{T,k}) \sim\ell(Q(T)) \sim\ell(R) \gtrsim r. \tag*{(122)} diam ( Γ T , k ∗ ) ∼ diam ( Σ T , k ∗ ) ∼ ℓ ( Q ( T )) ∼ ℓ ( R ) ≳ r . (122)
Let x ∗ ∈ Σ T , k ∗ x^* \in\Sigma^*_{T,k} x ∗ ∈ Σ T , k ∗ such that
∣ π n ( x Q ∗ ∗ ) − π n ( x ∗ ) ∣ ≤ 2 ℓ ( Q ∗ ∗ ) , (123) \left|\pi_n(x_{Q^{**}})-\pi_n(x^*)\right| \le2\ell(Q^{**}), \tag*{(123)} ∣ π n ( x Q ∗∗ ) − π n ( x ∗ ) ∣ ≤ 2 ℓ ( Q ∗∗ ) , (123)
which exists by Lemma 6.3 . We claim that this point satisfies
B ( x ∗ , r / 2 ) ⊆ B ( x , r ) . (124) B(x^*,r/2) \subseteq B(x,r). \tag*{(124)} B ( x ∗ , r /2 ) ⊆ B ( x , r ) . (124)
Since Γ T , k ∗ \Gamma^*_{T,k} Γ T , k ∗ is Ahlfors regular (Proposition 7.3 ) with diameter at least some constant multiple of r r r (by (122) ), this is sufficient for the lower regularity estimates. We commence with the proof of (124) in earnest. First, by Lemma 6.1 and Proposition 8.1 (2) we know that
π n ( Σ S , k ∗ ∪ b Q , k ) ⊆ 2 n λ 1 B Q ( S ) ⊆ 2 n λ 1 B Q ∗ ∗ (125) \pi_n(\Sigma^*_{S,k}\cup b_{Q,k})\subseteq2\sqrt{n}\lambda_1B_{Q(S)}\subseteq2\sqrt{n}\lambda_1B_{Q^{**}} \tag*{(125)} π n ( Σ S , k ∗ ∪ b Q , k ) ⊆ 2 n λ 1 B Q ( S ) ⊆ 2 n λ 1 B Q ∗∗ (125)
and π 1 ( x ) ≤ τ 0 d N , k ( π n ( x ) ) \pi_1(x)\le\tau_0d_{N,k}(\pi_n(x)) π 1 ( x ) ≤ τ 0 d N , k ( π n ( x )) . Thus, by (28) and Lemma 4.19 , we have
0 ≤ π 1 ( x ) ≤ λ ℓ ( Q ∗ ∗ ) ≤ r / 10 (126) 0\le\pi_1(x)\le\lambda\ell(Q^{**})\le r/10 \tag*{(126)} 0 ≤ π 1 ( x ) ≤ λ ℓ ( Q ∗∗ ) ≤ r /10 (126)
for λ \lambda λ sufficiently large. On the other hand, it follows from (123) and Lemma 4.18 that we have π 1 ( x ∗ ) ∼ τ 0 d L , k ( π n ( x ∗ ) ) ∼ ℓ ( Q ∗ ∗ ) \pi_1(x^*)\sim\tau_0d_{L,k}(\pi_n(x^*))\sim\ell(Q^{**}) π 1 ( x ∗ ) ∼ τ 0 d L , k ( π n ( x ∗ )) ∼ ℓ ( Q ∗∗ ) . Thus, for λ \lambda λ sufficiently large,
0 ≤ π 1 ( x ∗ ) ≤ λ ℓ ( Q ∗ ∗ ) ≤ π 1 ( x ) + r / 10. (127) 0\le\pi_1(x^*)\le\lambda\ell(Q^{**})\le\pi_1(x)+r/10. \tag*{(127)} 0 ≤ π 1 ( x ∗ ) ≤ λ ℓ ( Q ∗∗ ) ≤ π 1 ( x ) + r /10. (127)
Inequalities (126) and (127) give
∣ π 1 ( x ) − π 1 ( x ∗ ) ∣ ≤ r / 10. (128) \left|\pi_1(x)-\pi_1(x^*)\right|\le r/10. \tag*{(128)} ∣ π 1 ( x ) − π 1 ( x ∗ ) ∣ ≤ r /10. (128)
Furthermore, by (123) and (125) ,
∣ π n ( x ) − π n ( x ∗ ) ∣ ≤ ∣ π n ( x ) − π n ( x Q ∗ ∗ ) ∣ + ∣ π n ( x Q ∗ ∗ ) − π n ( x ∗ ) ∣ ≤ ( 2 n λ 1 + 2 ) ℓ ( Q ∗ ∗ ) ≤ r / 10 (129) \begin{aligned}
\left|\pi_n(x)-\pi_n(x^*)\right|
&\le\left|\pi_n(x)-\pi_n(x_{Q^{**}})\right|+\left|\pi_n(x_{Q^{**}})-\pi_n(x^*)\right|\\
&\le(2\sqrt{n}\lambda_1+2)\ell(Q^{**})\le r/10 \tag*{(129)}
\end{aligned} ∣ π n ( x ) − π n ( x ∗ ) ∣ ≤ ∣ π n ( x ) − π n ( x Q ∗∗ ) ∣ + ∣ π n ( x Q ∗∗ ) − π n ( x ∗ ) ∣ ≤ ( 2 n λ 1 + 2 ) ℓ ( Q ∗∗ ) ≤ r /10 (129)
after choosing λ \lambda λ sufficiently large. Inequalities (128) and (129) imply ∣ x − x ∗ ∣ ≤ r / 5 |x-x^*|\le r/5 ∣ x − x ∗ ∣ ≤ r /5 . The inclusion (124) is an easy consequence of this estimate. □
Lemma 10.4. The set Z ( k ) Z(k) Z ( k ) is upper Ahlfors d d d -regular with regularity constant depending only on n n n , d d d and the uniform rectifiability constants for E E E .
Proof. Fix x ∈ Z ( k ) x\in Z(k) x ∈ Z ( k ) and r ∈ ( 0 , diam ( Z ( k ) ) ) r\in(0,\operatorname{diam}(Z(k))) r ∈ ( 0 , diam ( Z ( k ))) , and let N ∈ N N\in\mathbb{N} N ∈ N , S ∈ T N ( k ) S\in\mathcal{T}^N(k) S ∈ T N ( k ) and Q ′ ∈ Stop ′ ( N , k ) Q'\in\operatorname{Stop}'(N,k) Q ′ ∈ Stop ′ ( N , k ) as in the proof of Lemma 10.3 . For each T ∈ T ( k ) T\in\mathcal{T}(k) T ∈ T ( k ) , consider the set
F T : = Σ T , k ∗ ∪ ⋃ Q ∈ Stop ′ ( N , k ) Q ⊆ Q ( T ) b Q , k . F_T:=\Sigma^*_{T,k}\cup\bigcup_{\substack{Q\in\operatorname{Stop}'(N,k)\\Q\subseteq Q(T)}}b_{Q,k}. F T := Σ T , k ∗ ∪ Q ∈ Stop ′ ( N , k ) Q ⊆ Q ( T ) ⋃ b Q , k .
Then, let
T r ( k ) : = { T ∈ T ( k ) : F T ∩ B ( x , r ) ≠ ∅ } . \mathcal{T}_r(k):=\{T\in\mathcal{T}(k):F_T\cap B(x,r)\ne\varnothing\}. T r ( k ) := { T ∈ T ( k ) : F T ∩ B ( x , r ) = ∅ } .
By Proposition 9.1 , each F T F_T F T is upper Ahlfors d d d -regular with diameter at most some constant multiple of ℓ ( Q ( T ) ) \ell(Q(T)) ℓ ( Q ( T )) . In particular,
H d ( Z ( k ) ∩ B ( x , r ) ) ≤ ∑ T ∈ T r ( k ) H d ( F T ∩ B ( x , r ) ) ≲ ∑ T ∈ T r ( k ) min { ℓ ( Q ( T ) ) , r } d . (130) \mathcal{H}^d(Z(k)\cap B(x,r))\le\sum_{T\in\mathcal{T}_r(k)}\mathcal{H}^d(F_T\cap B(x,r))
\lesssim\sum_{T\in\mathcal{T}_r(k)}\min\{\ell(Q(T)),r\}^d. \tag*{(130)} H d ( Z ( k ) ∩ B ( x , r )) ≤ T ∈ T r ( k ) ∑ H d ( F T ∩ B ( x , r )) ≲ T ∈ T r ( k ) ∑ min { ℓ ( Q ( T )) , r } d . (130)
To deal with this final sum, we split it into two pieces. Indeed, let λ > 1 \lambda> 1 λ > 1 (to be chosen sufficiently large) and consider the sets
T s m a l l ( k ) : = { T ∈ T r ( k ) : ℓ ( Q ( T ) ) ≤ λ r } ; T l a r g e ( k ) : = { T ∈ T r ( k ) : ℓ ( Q ( T ) ) > λ r } . \mathcal{T}_{\mathrm{small}}(k) := \{T \in\mathcal{T}_r(k): \ell(Q(T)) \le\lambda r\};
\qquad
\mathcal{T}_{\mathrm{large}}(k) := \{T \in\mathcal{T}_r(k): \ell(Q(T)) > \lambda r\}. T small ( k ) := { T ∈ T r ( k ) : ℓ ( Q ( T )) ≤ λ r } ; T large ( k ) := { T ∈ T r ( k ) : ℓ ( Q ( T )) > λ r } .
To deal with the sum over cubes in T s m a l l N ( k ) \mathcal{T}_{\mathrm{small}}^N(k) T small N ( k ) , first recall from Proposition 9.1 that π n ( F T ) ⊆ 2 n λ 1 B Q ( T ) \pi_n(F_T) \subseteq2\sqrt{n}\lambda_1 B_{Q(T)} π n ( F T ) ⊆ 2 n λ 1 B Q ( T ) . Thus, 2 n λ 1 B Q ( T ) ∩ π n ( B ( x , r ) ) ≠ ∅ 2\sqrt{n}\lambda_1 B_{Q(T)} \cap\pi_n(B(x,r)) \ne\varnothing 2 n λ 1 B Q ( T ) ∩ π n ( B ( x , r )) = ∅ . For each T ∈ T l a r g e N ( k ) T \in\mathcal{T}_{\mathrm{large}}^N(k) T ∈ T large N ( k ) , this means that B Q ( T ) ⊆ B ( π n ( x ) , R ) B_{Q(T)} \subseteq B(\pi_n(x),R) B Q ( T ) ⊆ B ( π n ( x ) , R ) for some R R R depending on λ \lambda λ . Thus, since T T T satisfies a Carleson packing condition (Lemma 4.11 ), we have
∑ T ∈ T s m a l l ( k ) ℓ ( Q ( T ) ) d ≲ r d . (131) \sum_{T \in\mathcal{T}_{\mathrm{small}}(k)} \ell(Q(T))^d \lesssim r^d. \tag*{(131)} T ∈ T small ( k ) ∑ ℓ ( Q ( T ) ) d ≲ r d . (131)
The upper regularity of Z ( k ) Z(k) Z ( k ) in B ( x , r ) B(x,r) B ( x , r ) follows from (130) and (131) once we show that
∑ T ∈ T l a r g e ( k ) r d ≲ r d . (132) \sum_{T \in\mathcal{T}_{\mathrm{large}}(k)} r^d \lesssim r^d. \tag*{(132)} T ∈ T large ( k ) ∑ r d ≲ r d . (132)
To prove (132) , consider first the set N \mathcal{N} N of integers N ∈ N N \in\mathbb{N} N ∈ N such that
T l a r g e N ( k ) : = T N ( k ) ∩ T l a r g e ( k ) ≠ ∅ . \mathcal{T}_{\mathrm{large}}^N(k) := \mathcal{T}^N(k) \cap\mathcal{T}_{\mathrm{large}}(k) \ne\varnothing. T large N ( k ) := T N ( k ) ∩ T large ( k ) = ∅ .
Since 2 n λ 1 B Q ( T ) ∩ π n ( B ( x , r ) ) ≠ ∅ 2\sqrt{n}\lambda_1 B_{Q(T)} \cap\pi_n(B(x,r)) \ne\varnothing 2 n λ 1 B Q ( T ) ∩ π n ( B ( x , r )) = ∅ and ℓ ( Q ) > λ r \ell(Q) > \lambda r ℓ ( Q ) > λ r for each T ∈ T l a r g e ( k ) T \in\mathcal{T}_{\mathrm{large}}(k) T ∈ T large ( k ) , it follows from Lemma 4.9 and a simple volume argument that
∣ T l a r g e N ( k ) ∣ ≲ d , n , C 0 1 for all N ∈ N . (133) |\mathcal{T}_{\mathrm{large}}^N(k)| \lesssim_{d,n,C_0} 1 \quad\text{for all } N \in\mathcal{N}. \tag*{(133)} ∣ T large N ( k ) ∣ ≲ d , n , C 0 1 for all N ∈ N . (133)
We claim that
∣ N ∣ ≤ 2 , (134) |\mathcal{N}| \le2, \tag*{(134)} ∣ N ∣ ≤ 2 , (134)
which, along with (133) , is enough to imply (132) . Let N N N be the largest integer in N \mathcal{N} N , T ∈ T l a r g e N ( k ) T \in\mathcal{T}_{\mathrm{large}}^N(k) T ∈ T large N ( k ) and Q ∈ Stop ′ ( N − 1 , k ) Q \in\operatorname{Stop}'(N-1,k) Q ∈ Stop ′ ( N − 1 , k ) such that Q ( T ) ⊆ Q Q(T) \subseteq Q Q ( T ) ⊆ Q . Moreover, fix a point
z ∈ F T ∩ B ( x , r ) . z \in F_T \cap B(x,r). z ∈ F T ∩ B ( x , r ) .
Note that, since π n ( F T ) ⊆ 2 n λ 1 B Q ( T ) \pi_n(F_T) \subseteq2\sqrt{n}\lambda_1 B_{Q(T)} π n ( F T ) ⊆ 2 n λ 1 B Q ( T ) and ℓ ( Q ( T ) ) > λ r \ell(Q(T)) > \lambda r ℓ ( Q ( T )) > λ r , we have
π n ( B ( x , r ) ) ⊆ 6 n λ 1 B Q ( T ) ⊆ 6 n λ 1 B Q . (135) \pi_n(B(x,r)) \subseteq6\sqrt{n}\lambda_1 B_{Q(T)} \subseteq6\sqrt{n}\lambda_1 B_Q. \tag*{(135)} π n ( B ( x , r )) ⊆ 6 n λ 1 B Q ( T ) ⊆ 6 n λ 1 B Q . (135)
Suppose towards a contradiction that there exists M ≤ N − 2 M \le N-2 M ≤ N − 2 , S ∈ T l a r g e M ( k ) S \in\mathcal{T}_{\mathrm{large}}^M(k) S ∈ T large M ( k ) and w ∈ F S ∩ B ( x , r ) w \in F_S \cap B(x,r) w ∈ F S ∩ B ( x , r ) . By (135) , π n ( w ) , π n ( z ) ∈ 6 n λ B Q \pi_n(w),\pi_n(z) \in6\sqrt{n}\lambda B_Q π n ( w ) , π n ( z ) ∈ 6 n λ B Q . Then, since π 1 ( z ) ≤ τ 0 d N , k ( π n ( z ) ) \pi_1(z) \le\tau_0 d_{N,k}(\pi_n(z)) π 1 ( z ) ≤ τ 0 d N , k ( π n ( z )) and π 1 ( w ) ≥ τ 0 d M + 1 ( π n ( w ) ) \pi_1(w) \ge\tau_0 d_{M+1}(\pi_n(w)) π 1 ( w ) ≥ τ 0 d M + 1 ( π n ( w )) (and M + 1 ≤ N − 2 M+1 \le N-2 M + 1 ≤ N − 2 ), it follows from Lemma 4.20 that
π 1 ( w ) − π 1 ( z ) ≥ c ℓ ( Q ) ≥ c λ r > r \pi_1(w)-\pi_1(z) \ge c\ell(Q) \ge c\lambda r > r π 1 ( w ) − π 1 ( z ) ≥ c ℓ ( Q ) ≥ c λ r > r
for λ \lambda λ sufficiently large, which is a contradiction. We have just argued that N ⊆ { N , N − 1 } \mathcal{N} \subseteq\{N,N-1\} N ⊆ { N , N − 1 } , which clearly implies (134) . □ \square □
Lemma 10.5. Inequality (120) holds.
Proof. Let x ∈ E x \in E x ∈ E and let Q ∈ D k Q \in\mathcal{D}_k Q ∈ D k such that x ∈ Q x \in Q x ∈ Q . By construction, there exists N ∈ N N \in\mathbb{N} N ∈ N and S ∈ T N ( k ) S \in\mathcal{T}^{N}(k) S ∈ T N ( k ) such that Q ∈ min ( S ( k ) ) Q \in\min(S(k)) Q ∈ min ( S ( k )) . Thus, by Lemma 6.3 , there exists z ∈ Σ S , k z \in\Sigma_{S,k} z ∈ Σ S , k such that ∣ x − z ∣ ≤ 2 ℓ ( Q ) |x-z| \leq2\ell(Q) ∣ x − z ∣ ≤ 2 ℓ ( Q ) . Moreover, this point satisfies τ 0 d N , k ( z ) ≲ ℓ ( Q ) \tau_{0}d_{N,k}(z) \lesssim\ell(Q) τ 0 d N , k ( z ) ≲ ℓ ( Q ) (see Lemma 4.14 ). Letting z ∗ : = ( z , τ d N , k ( z ) ) ∈ Σ S ∗ z^{*}:=(z,\tau d_{N,k}(z)) \in\Sigma^{*}_{S} z ∗ := ( z , τ d N , k ( z )) ∈ Σ S ∗ , we now have
∣ x − z ∗ ∣ ≤ ∣ x − z ∣ + ∣ z − z ∗ ∣ ≲ ℓ ( Q ) ≲ ρ k , |x-z^{*}| \leq|x-z|+|z-z^{*}| \lesssim\ell(Q) \lesssim\rho^{k}, ∣ x − z ∗ ∣ ≤ ∣ x − z ∣ + ∣ z − z ∗ ∣ ≲ ℓ ( Q ) ≲ ρ k ,
as required.
The only thing left to show is that Z ( k ) Z(k) Z ( k ) supports a ( 1 , 1 ) (1,1) ( 1 , 1 ) -Poincaré inequality. Fix some x , y ∈ Z ( k ) x,y \in Z(k) x , y ∈ Z ( k ) and u : Z ( k ) → R u:Z(k)\to\mathbb{R} u : Z ( k ) → R Lipschitz. Our goal is to prove (5) with C 1 C_{1} C 1 independent of x , y x,y x , y and u u u . Let N ≥ 0 N \geq0 N ≥ 0 , S ∈ T N ( k ) S \in\mathcal{T}^{N}(k) S ∈ T N ( k ) and Q N ∈ Stop ′ ( N , k ) Q_{N} \in\operatorname{Stop}^{\prime}(N,k) Q N ∈ Stop ′ ( N , k ) such that Q N ⊆ Q ( S ) Q_{N} \subseteq Q(S) Q N ⊆ Q ( S ) and
x ∈ Σ S ∗ ∪ b Q N , (136) x \in\Sigma^{*}_{S}\cup b_{Q_{N}}, \tag*{(136)} x ∈ Σ S ∗ ∪ b Q N , (136)
and let M ≥ 0 M \geq0 M ≥ 0 , T ∈ T N ( k ) T \in\mathcal{T}^{N}(k) T ∈ T N ( k ) and R M ∈ Stop ′ ( M , k ) R_{M} \in\operatorname{Stop}^{\prime}(M,k) R M ∈ Stop ′ ( M , k ) such that R M ⊆ Q ( T ) R_{M} \subseteq Q(T) R M ⊆ Q ( T ) and
y ∈ Σ T ∗ ∪ b R M . (137) y \in\Sigma^{*}_{T}\cup b_{R_{M}}. \tag*{(137)} y ∈ Σ T ∗ ∪ b R M . (137)
We can assume, without loss of generality, that
Note that
π 1 ( x ) ≤ τ 0 d N , k ( π n ( x ) ) and π 1 ( y ) ≥ τ 0 d M + 1 , k ( π n ( y ) ) . (138) \pi_{1}(x) \leq\tau_{0}d_{N,k}(\pi_{n}(x)) \quad\text{and}\quad\pi_{1}(y) \geq\tau_{0}d_{M+1,k}(\pi_{n}(y)). \tag*{(138)} π 1 ( x ) ≤ τ 0 d N , k ( π n ( x )) and π 1 ( y ) ≥ τ 0 d M + 1 , k ( π n ( y )) . (138)
Let Q 0 ∈ D Q_{0} \in\mathcal{D} Q 0 ∈ D with smallest cube such that π n ( x ) , π n ( y ) ∈ 3 n λ 1 B Q 0 \pi_{n}(x),\pi_{n}(y) \in3\sqrt{n}\lambda_{1}B_{Q_{0}} π n ( x ) , π n ( y ) ∈ 3 n λ 1 B Q 0 (recall the definition of λ 1 \lambda_{1} λ 1 from (46) ). Let K ≥ 0 K \geq0 K ≥ 0 and U ∈ T K U \in\mathcal{T}^{K} U ∈ T K such that Q 0 ∈ U Q_{0} \in U Q 0 ∈ U .
The proof of (5) will be split into a number of sub-cases. In two of these cases, we check (5) by constructing a pencil of balls from x x x to y y y satisfying the hypotheses of Lemma 3.7 . The balls will a subset of the following set.
Lemma 10.6. For each Q ∈ Stop ( N , k ) Q \in\operatorname{Stop}(N,k) Q ∈ Stop ( N , k ) there is a ball B Q \mathbb{B}_{Q} B Q centred on b Q b_{Q} b Q with radius comparable to ℓ ( Q ) \ell(Q) ℓ ( Q ) and such that
Z ( k ) ∩ B Q = b Q ∩ B Q . Z(k)\cap\mathbb{B}_{Q}=b_{Q}\cap\mathbb{B}_{Q}. Z ( k ) ∩ B Q = b Q ∩ B Q .
The sets in Lemma 3.7 (which contain these balls) will be constructed by Proposition 9.2 . The following will be useful in verifying the hypotheses of that lemma.
Lemma 10.7. Let L ≤ K L \leq K L ≤ K and V , W ∈ T L ( k ) V,W \in\mathcal{T}^{L}(k) V , W ∈ T L ( k ) such that Q N ⊆ Q ( V ) Q_{N} \subseteq Q(V) Q N ⊆ Q ( V ) and R M ⊆ Q ( W ) R_{M} \subseteq Q(W) R M ⊆ Q ( W ) . Then, A 1 B Q ( V ) ∩ A 1 B Q ( W ) ≠ ∅ A_{1}B_{Q(V)}\cap A_{1}B_{Q(W)}\neq\varnothing A 1 B Q ( V ) ∩ A 1 B Q ( W ) = ∅ .
Proof. Recall from Lemma 6.1 that π n ( x ) ∈ 2 n λ 1 B Q ( S ) \pi_{n}(x)\in2\sqrt{n}\lambda_{1}B_{Q(S)} π n ( x ) ∈ 2 n λ 1 B Q ( S ) . By definition, we have Q ( S ) ⊆ Q ( V ) Q(S)\subseteq Q(V) Q ( S ) ⊆ Q ( V ) . Thus, it follows from the triangle inequality that π n ( x ) ∈ 3 n λ 1 B Q ( V ) \pi_{n}(x)\in3\sqrt{n}\lambda_{1}B_{Q(V)} π n ( x ) ∈ 3 n λ 1 B Q ( V ) . In particular, 3 n λ 1 B Q ( V ) ∩ 3 n λ 1 B Q 0 ≠ ∅ 3\sqrt{n}\lambda_{1}B_{Q(V)}\cap3\sqrt{n}\lambda_{1}B_{Q_{0}}\neq\varnothing 3 n λ 1 B Q ( V ) ∩ 3 n λ 1 B Q 0 = ∅ . Applying Lemma 4.9 and the triangle inequality, it is easy to check that
3 n λ 1 B Q 0 ⊆ 3 ( 2 + C 5 ) n λ 1 B Q ( V ) ⊆ A 1 B Q ( V ) . 3\sqrt{n}\lambda_{1}B_{Q_{0}}\subseteq3(2+C_{5})\sqrt{n}\lambda_{1}B_{Q(V)}\subseteq A_{1}B_{Q(V)}. 3 n λ 1 B Q 0 ⊆ 3 ( 2 + C 5 ) n λ 1 B Q ( V ) ⊆ A 1 B Q ( V ) . Since 3 n λ 1 B Q ( W ) ⊆ A 1 B Q ( W ) 3\sqrt{n}\lambda_{1}B_{Q(W)}\subseteq A_{1}B_{Q(W)} 3 n λ 1 B Q ( W ) ⊆ A 1 B Q ( W ) and 3 n λ 1 B Q ( W ) ∩ 3 n λ 1 B Q 0 ≠ ∅ 3\sqrt{n}\lambda_{1}B_{Q(W)}\cap3\sqrt{n}\lambda_{1}B_{Q_{0}}\neq\varnothing 3 n λ 1 B Q ( W ) ∩ 3 n λ 1 B Q 0 = ∅ , the result now follows.
We commence with the proof of (5) .
Case 1. Suppose that K ≤ M K \leq M K ≤ M .
We construct a sequence of balls from x x x to y y y and subsets of Z ( k ) Z^{(k)} Z ( k ) satisfying the hypotheses of Lemma 3.7 . We start with the balls. Let { Q i } i = K N − 1 ⊆ D \{Q_i\}_{i=K}^{N-1} \subseteq\mathcal{D} { Q i } i = K N − 1 ⊆ D be such that Q i ∈ Stop ( i , k ) Q_i \in\operatorname{Stop}(i,k) Q i ∈ Stop ( i , k ) satisfies Q ( S ) ⊆ Q i Q(S) \subseteq Q_i Q ( S ) ⊆ Q i , and let { R j } j = K M − 1 ⊆ D \{R_j\}_{j=K}^{M-1} \subseteq\mathcal{D} { R j } j = K M − 1 ⊆ D be such that R j ∈ Stop ( j , k ) R_j \in\operatorname{Stop}(j,k) R j ∈ Stop ( j , k ) satisfies Q ( S ) ⊆ R j Q(S) \subseteq R_j Q ( S ) ⊆ R j .
Lemma 10.8. The collection of balls
B Q : Q ∈ { Q i } i = K N − 1 ∪ { R j } j = K M − 1 \mathcal{B}_{\mathcal{Q}}: Q \in\{Q_i\}_{i=K}^{N-1} \cup\{R_j\}_{j=K}^{M-1} B Q : Q ∈ { Q i } i = K N − 1 ∪ { R j } j = K M − 1 satisfy Lemma 3.7 (4) and (5). That is, there exists A ≥ 1 A \ge1 A ≥ 1 such that x ∈ A B Q i x \in A\mathcal{B}_{Q_i} x ∈ A B Q i and y ∈ A B R j y \in A\mathcal{B}_{R_j} y ∈ A B R j for each i ∈ { K , … , N − 1 } i \in\{K,\ldots,N-1\} i ∈ { K , … , N − 1 } and j ∈ { K , … , M − 1 } j \in\{K,\ldots,M-1\} j ∈ { K , … , M − 1 } , and
∑ j = K N − 1 r B Q i + ∑ j = K M − 1 r B R j ≲ ∣ x − y ∣ . (139) \sum_{j=K}^{N-1} r_{\mathcal{B}_{Q_i}}+\sum_{j=K}^{M-1} r_{\mathcal{B}_{R_j}}\lesssim|x-y|. \tag*{(139)} j = K ∑ N − 1 r B Q i + j = K ∑ M − 1 r B R j ≲ ∣ x − y ∣. (139)
Proof. We only deal with properties related to { Q i } \{Q_i\} { Q i } and x x x . The properties for { R i } \{R_i\} { R i } and y y y are identical up to notational changes. Let us begin by showing that
x ∈ A B Q i x \in A\mathcal{B}_{Q_i} x ∈ A B Q i
for some suitable choice of A A A . {#eq:10.21}
Fix some i ∈ { K , … , N − 1 } i \in\{K,\ldots,N-1\} i ∈ { K , … , N − 1 } . For brevity, let us define x i : = x B Q i x_i := x_{B_{Q_i}} x i := x B Q i . First, by Lemma 6.1, we have π n ( x ) ∈ 2 n λ 1 B Q ( S ) ⊆ 3 n λ 1 B Q i \pi_n(x) \in2\sqrt{n}\lambda_1 B_{Q(S)} \subseteq3\sqrt{n}\lambda_1 B_{Q_i} π n ( x ) ∈ 2 n λ 1 B Q ( S ) ⊆ 3 n λ 1 B Q i . On the other hand, since B Q i ⊆ b Q i B_{Q_i} \subseteq b_{Q_i} B Q i ⊆ b Q i , we have by Proposition 8.1 (2) that π n ( x i ) ∈ ( c 0 / 2 ) B Q i \pi_n(x_i) \in(c_0/2)B_{Q_i} π n ( x i ) ∈ ( c 0 /2 ) B Q i . In particular,
∣ π n ( x − x i ) ∣ ≲ ℓ ( Q i ) . |\pi_n(x-x_i)|\lesssim\ell(Q_i). ∣ π n ( x − x i ) ∣ ≲ ℓ ( Q i ) .
Since π 1 ( x ) ≤ τ 0 d N , k ( x ) \pi_1(x) \le\tau_0 d_{N,k}(x) π 1 ( x ) ≤ τ 0 d N , k ( x ) , (4.3) and Lemma 4.19 implies π 1 ( x ) ≲ ℓ ( Q i ) \pi_1(x)\lesssim\ell(Q_i) π 1 ( x ) ≲ ℓ ( Q i ) . On the other hand, by Proposition 8.1 (2) and Lemma 4.18 , we have π 1 ( x i ) ≲ τ 0 d N , k ( π n ( x i ) ) ≲ ℓ ( Q i ) \pi_1(x_i)\lesssim\tau_0 d_{N,k}(\pi_n(x_i))\lesssim\ell(Q_i) π 1 ( x i ) ≲ τ 0 d N , k ( π n ( x i )) ≲ ℓ ( Q i ) . In particular,
∣ π 1 ( x − x i ) ∣ ≲ ℓ ( Q i ) . |\pi_1(x-x_i)|\lesssim\ell(Q_i). ∣ π 1 ( x − x i ) ∣ ≲ ℓ ( Q i ) .
Recalling from Lemma 10.6 that r B Q i ≳ ℓ ( Q i ) r_{B_{Q_i}}\gtrsim\ell(Q_i) r B Q i ≳ ℓ ( Q i ) , the above inequalities imply (10.21).
We now turn our attention to (139) . Recall from Lemma 4.14 that ℓ ( Q i ) ∼ τ 0 d i , k ( Q i ) \ell(Q_i)\sim\tau_0 d_{i,k}(Q_i) ℓ ( Q i ) ∼ τ 0 d i , k ( Q i ) for each i i i . Since Q i + 1 ⊆ Q i Q_{i+1}\subseteq Q_i Q i + 1 ⊆ Q i , it follows from Lemma 4.18 that ℓ ( Q i + 1 ) ≲ τ 0 ℓ ( Q i ) \ell(Q_{i+1})\lesssim\tau_0\ell(Q_i) ℓ ( Q i + 1 ) ≲ τ 0 ℓ ( Q i ) . In particular, by summing over a geometric series, it suffices to show that
ℓ ( Q K ) ≲ ∣ x − y ∣ . \ell(Q_K)\lesssim|x-y|. ℓ ( Q K ) ≲ ∣ x − y ∣.
This estimate follows by minimality. Indeed, let Q ∈ Child ( Q 0 ) Q\in\operatorname{Child}(Q_0) Q ∈ Child ( Q 0 ) such that π n ( x ) ∈ 2 n λ 1 B Q \pi_n(x)\in2\sqrt{n}\lambda_1 B_Q π n ( x ) ∈ 2 n λ 1 B Q (note that such a cube exists since π n ( x ) ∈ 2 n λ 1 B Q ( S ) \pi_n(x)\in2\sqrt{n}\lambda_1 B_{Q(S)} π n ( x ) ∈ 2 n λ 1 B Q ( S ) by Lemma 6.1). By minimality, we know that π n ( y ) ∉ 3 n λ 1 B Q \pi_n(y)\notin3\sqrt{n}\lambda_1 B_Q π n ( y ) ∈ / 3 n λ 1 B Q . Since ℓ ( Q K ) ≲ ℓ ( Q ) \ell(Q_K)\lesssim\ell(Q) ℓ ( Q K ) ≲ ℓ ( Q ) , the estimate now easily follows.
We now construct the sets satisfying Lemma 3.7 (1)–(3). These are constructed by applying Proposition 9.2 . The properties Lemma 3.7 (1) and (3) are explicit in the statement below, we note that (10.22) and (10.23) are sufficient for Lemma 3.7 (2) by Lemma 10.6 .
Lemma 10.9. There are sets { X i } i = K − 1 N − 1 , { Y j } j = K − 1 M − 1 ⊆ Z ( k ) \{X_i\}_{i=K-1}^{N-1}, \{Y_j\}_{j=K-1}^{M-1}\subseteq Z^{(k)} { X i } i = K − 1 N − 1 , { Y j } j = K − 1 M − 1 ⊆ Z ( k ) , each of which is Ahlfors d d d -regular and supports a ( 1 , 1 ) (1,1) ( 1 , 1 ) -Poincaré inequality, such that x ∈ X N − 1 x\in X_{N-1} x ∈ X N − 1 , y ∈ Y M − 1 y\in Y_{M-1} y ∈ Y M − 1 and
b Q i ∪ b Q i + 1 ⊆ X i and b R j ∪ b R j + 1 ⊆ Y j b_{Q_i}\cup b_{Q_{i+1}}\subseteq X_i \qquad\text{and}\qquad b_{R_j}\cup b_{R_{j+1}}\subseteq Y_j b Q i ∪ b Q i + 1 ⊆ X i and b R j ∪ b R j + 1 ⊆ Y j for each i ∈ { K , … , N − 1 } i\in\{K,\ldots,N-1\} i ∈ { K , … , N − 1 } and j ∈ { K , … , M − 1 } j\in\{K,\ldots,M-1\} j ∈ { K , … , M − 1 } , and (10.23):
b Q K ∪ b R K ⊆ X K − 1 . (140) b_{Q_K}\cup b_{R_K}\subseteq X_{K-1}. \tag*{(140)} b Q K ∪ b R K ⊆ X K − 1 . (140)
Proof. Except for the end points i ∈ { K − 1 , N − 1 } i \in\{K-1,N-1\} i ∈ { K − 1 , N − 1 } and j ∈ { K − 1 , M − 1 } j \in\{K-1,M-1\} j ∈ { K − 1 , M − 1 } , we let X i X_i X i and Y j Y_j Y j be the sets constructed by Corollary 9.3 with the cubes Q i Q_i Q i , Q i + 1 Q_{i+1} Q i + 1 and R j R_j R j , R j + 1 R_{j+1} R j + 1 , respectively. In this range of indices, (140) is then immediate. For i = N − 1 i=N-1 i = N − 1 (resp. j = M − 1 j=M-1 j = M − 1 ), we let X N − 1 X_{N-1} X N − 1 (resp. Y M − 1 Y_{M-1} Y M − 1 ) be the set construct by applying Proposition 9.2 with data S S S , U S U_S U S and Q M Q_M Q M (resp. T T T , U T U_T U T and R M R_M R M ). Then,
Σ S ∗ ∪ Q ^ N ⊆ X N − 1 and Σ T ∗ ∪ R ^ M ⊆ Y M − 1 , \Sigma^*_S \cup\widehat{Q}_N \subseteq X_{N-1} \qquad\text{and} \qquad\Sigma^*_T \cup\widehat{R}_M \subseteq Y_{M-1}, Σ S ∗ ∪ Q N ⊆ X N − 1 and Σ T ∗ ∪ R M ⊆ Y M − 1 ,
which implies (140) for this choice of indices and also implies x ∈ X N − 1 x \in X_{N-1} x ∈ X N − 1 and y ∈ Y M − 1 y \in Y_{M-1} y ∈ Y M − 1 (recall (136) and (137) ). For (10.23), we also apply Proposition 9.2. Indeed, let V , W ∈ T K ( k ) V,W \in\mathcal{T}^K(k) V , W ∈ T K ( k ) such that Q K ∈ V Q_K \in V Q K ∈ V and R K ∈ W R_K \in W R K ∈ W . It follows from Lemma 10.7 that A 1 B Q ( V ) ∩ A 1 B Q ( W ) ≠ ∅ A_1B_Q(V) \cap A_1B_Q(W) \ne\varnothing A 1 B Q ( V ) ∩ A 1 B Q ( W ) = ∅ , and now from Proposition 9.2 that such a set exists. □ \square □
Case 2. Suppose that N ≥ M + 2 N \ge M+2 N ≥ M + 2 and K ≥ M + 1 K \ge M+1 K ≥ M + 1 .
As in Case 1, we construct a sequence of balls from x x x to y y y and subsets of Z ( k ) Z(k) Z ( k ) satisfying the hypotheses of Lemma 3.7. Let
I = { M , … , N − 1 } and I ′ = I ∖ { M } . I=\{M,\ldots,N-1\} \qquad\text{and} \qquad I'=I\setminus\{M\}. I = { M , … , N − 1 } and I ′ = I ∖ { M } .
For each i ∈ I i \in I i ∈ I , let Q i ∈ Stop ( i ) Q_i \in\operatorname{Stop}(i) Q i ∈ Stop ( i ) such that Q ( S ) ⊆ Q i Q(S) \subseteq Q_i Q ( S ) ⊆ Q i .
Lemma 10.10. The collection of balls { B Q i : i ∈ I ′ } \{\mathcal{B}_{Q_i}:i\in I'\} { B Q i : i ∈ I ′ } satisfy Lemma 3.7 (4) and (5) i.e., there is A ≥ 1 A \ge1 A ≥ 1 such that x ∈ A B Q i x \in A\mathcal{B}_{Q_i} x ∈ A B Q i for each i ∈ I ′ i \in I' i ∈ I ′ and
∑ i ∈ I ′ ℓ ( Q i ) ≲ ∣ x − y ∣ . (141) \sum_{i\in I'}\ell(Q_i)\lesssim|x-y|. \tag*{(141)} i ∈ I ′ ∑ ℓ ( Q i ) ≲ ∣ x − y ∣. (141) Proof. Since I ′ = ∅ I'=\varnothing I ′ = ∅ in the case that N = M + 1 N=M+1 N = M + 1 , we can assume that N ≥ M + 2 N \ge M+2 N ≥ M + 2 . The proof that x ∈ A B Q i x \in A\mathcal{B}_{Q_i} x ∈ A B Q i is the same as in the proof of Lemma 10.8 and we will not repeat it. Furthermore, as in the proof of Lemma 10.8, (141) follows once we show that
ℓ ( Q M + 1 ) ≲ ∣ x − y ∣ . (142) \ell(Q_{M+1})\lesssim|x-y|. \tag*{(142)} ℓ ( Q M + 1 ) ≲ ∣ x − y ∣. (142) It suffice to prove (142) in the case that ∣ π n ( x ) − π n ( y ) ∣ ≤ ℓ ( Q M + 1 ) |\pi_n(x)-\pi_n(y)|\le\ell(Q_{M+1}) ∣ π n ( x ) − π n ( y ) ∣ ≤ ℓ ( Q M + 1 ) (otherwise there is nothing to show). In this case, we have π n ( x ) , π n ( y ) ∈ 3 n λ 1 B Q M + 1 \pi_n(x),\pi_n(y)\in3^n\lambda_1\mathcal{B}_{Q_{M+1}} π n ( x ) , π n ( y ) ∈ 3 n λ 1 B Q M + 1 . Thus, by Lemma 4.20 and (138) , we have
∣ x − y ∣ ≥ π 1 ( y ) − π 1 ( x ) ≳ ℓ ( Q M + 1 ) , |x-y|\ge\pi_1(y)-\pi_1(x)\gtrsim\ell(Q_{M+1}), ∣ x − y ∣ ≥ π 1 ( y ) − π 1 ( x ) ≳ ℓ ( Q M + 1 ) , as required. □ \square □
The properties Lemma 3.7 (1)–(3) follow directly from the next two lemmas.
Lemma 10.11. There are sets { X i } i ∈ I ′ ⊆ Z ( k ) \{X_i\}_{i\in I'}\subseteq Z(k) { X i } i ∈ I ′ ⊆ Z ( k ) , each of which is Ahlfors d d d -regular and supports a ( 1 , 1 ) (1,1) ( 1 , 1 ) -Poincaré inequality, such that x ∈ X N − 1 x\in X_{N-1} x ∈ X N − 1 and
Q ^ i ∪ Q ^ i + 1 ⊆ X i (143) \widehat{Q}_i\cup\widehat{Q}_{i+1}\subseteq X_i \tag*{(143)} Q i ∪ Q i + 1 ⊆ X i (143) for each i ∈ I ′ i\in I' i ∈ I ′ .
Proof. Except for the endpoint i = N − 1 i=N-1 i = N − 1 , we let X i X_i X i be the set constructed by Corollary 9.3 with the cubes Q i Q_i Q i and Q i + 1 Q_{i+1} Q i + 1 . In this range of indices, (143) is then immediate. For i = N − 1 i=N-1 i = N − 1 , we let X N − 1 X_{N-1} X N − 1 be the set constructed by applying Proposition 9.2 with the data S S S , U S U_S U S and Q N Q_N Q N . The remaining properties hold in the same way as in Lemma 10.9. □ \square □
Lemma 10.12. We have ∣ y − x B Q M + 1 ∣ ≲ ∣ x − y ∣ |y-x_{B Q_{M+1}}| \lesssim|x-y| ∣ y − x B Q M + 1 ∣ ≲ ∣ x − y ∣ . Moreover, there exists a space X ⊂ Z ( k ) X \subset Z(k) X ⊂ Z ( k ) which is Ahlfors d d d -regular, supports a ( 1 , 1 ) (1,1) ( 1 , 1 ) -Poincaré inequality and satisfies { y } ∪ b Q M + 1 ⊆ X \{y\} \cup b_{Q_{M+1}} \subseteq X { y } ∪ b Q M + 1 ⊆ X .
Proof. To simplify our notation, we will denote z : = x B Q M + 1 z:=x_{B Q_{M+1}} z := x B Q M + 1 . We start by estimating
∣ y − z ∣ ≤ ∣ y − x ∣ + ∣ π 1 ( x ) − π 1 ( z ) ∣ + ∣ π n ( x ) − π n ( z ) ∣ ≤ ∣ y − x ∣ + ∣ π 1 ( z ) ∣ + ∣ π n ( x ) − π n ( z ) ∣ . \begin{aligned}
|y-z| \leq|y-x|+|\pi_1(x)-\pi_1(z)|+|\pi_n(x)-\pi_n(z)| \\
&\leq|y-x|+|\pi_1(z)|+|\pi_n(x)-\pi_n(z)|.
\end{aligned} ∣ y − z ∣ ≤ ∣ y − x ∣ + ∣ π 1 ( x ) − π 1 ( z ) ∣ + ∣ π n ( x ) − π n ( z ) ∣ ≤ ∣ y − x ∣ + ∣ π 1 ( z ) ∣ + ∣ π n ( x ) − π n ( z ) ∣.
Since Q ( S ) ⊆ Q M + 1 Q(S) \subseteq Q_{M+1} Q ( S ) ⊆ Q M + 1 , it follows immediately that ∣ π n ( x ) − π n ( z ) ∣ ≲ ℓ ( Q M + 1 ) |\pi_n(x)-\pi_n(z)| \lesssim\ell(Q_{M+1}) ∣ π n ( x ) − π n ( z ) ∣ ≲ ℓ ( Q M + 1 ) . Also, since z ∈ b Q M + 1 z \in b_{Q_{M+1}} z ∈ b Q M + 1 , it follows from Proposition 8.1 (2) and Lemma 4.14 that
π 1 ( z ) ≤ τ 0 d M + 1 , k ( π n ( z ) ) ∼ ℓ ( Q M + 1 ) . \pi_1(z) \leq\tau_0 d_{M+1,k}(\pi_n(z)) \sim\ell(Q_{M+1}). π 1 ( z ) ≤ τ 0 d M + 1 , k ( π n ( z )) ∼ ℓ ( Q M + 1 ) .
As we proved in Lemma 10.10 , we have ℓ ( Q M + 1 ) ≲ ∣ x − y ∣ \ell(Q_{M+1}) \lesssim|x-y| ℓ ( Q M + 1 ) ≲ ∣ x − y ∣ , which finishes the proof of the first assertion.
The set X X X is constructed as follows. Let V ∈ T M + 1 ( k ) V \in\mathcal{T}^{M+1}(k) V ∈ T M + 1 ( k ) such that Q M + 1 ∈ V Q_{M+1} \in V Q M + 1 ∈ V and let W ∈ T M + 1 ( k ) W \in\mathcal{T}^{M+1}(k) W ∈ T M + 1 ( k ) such that Q ( W ) ⊆ R M Q(W) \subseteq R_M Q ( W ) ⊆ R M . By Lemma 10.7 , we have that A 1 B Q ( V ) ∩ A 1 B Q ( W ) ≠ ∅ A_1B_{Q(V)} \cap A_1B_{Q(W)} \neq\varnothing A 1 B Q ( V ) ∩ A 1 B Q ( W ) = ∅ . Thus, we let X X X be the set constructed by applying Proposition 9.2 with this data. It is easy to see that this set satisfies the desired conclusion. □
Case 3. Suppose that N = M + 1 N=M+1 N = M + 1 or N = M N=M N = M .
In this final case, we simply find an Ahlfors regular PI space in Z ( k ) Z(k) Z ( k ) which contains the endpoints x x x and y y y .
Lemma 10.13. There is a set X ⊆ Z ( k ) X \subseteq Z(k) X ⊆ Z ( k ) which is Ahlfors d d d -regular and supports a ( 1 , 1 ) (1,1) ( 1 , 1 ) -Poincaré inequality and satisfies x , y ∈ X x,y \in X x , y ∈ X . In particular, (5) holds.
Proof. If N = M + 1 N=M+1 N = M + 1 , let W ∈ T N ( k ) W \in\mathcal{T}^{N}(k) W ∈ T N ( k ) such that Q ( W ) ⊆ R M Q(W) \subseteq R_M Q ( W ) ⊆ R M and π ( y ) ∈ 3 n λ 1 B Q ( W ) \pi(y) \in3\sqrt{n}\lambda_1B_{Q(W)} π ( y ) ∈ 3 n λ 1 B Q ( W ) (such a cube exists by the minimality of Q 0 Q_0 Q 0 ). If N = M N=M N = M , let W = T W=T W = T . In either case, we have by Lemma 10.7 that
A 1 B Q ( S ) ∩ A 1 B Q ( W ) ≠ ∅ . A_1B_{Q(S)} \cap A_1B_{Q(W)} \neq\varnothing. A 1 B Q ( S ) ∩ A 1 B Q ( W ) = ∅ .
Let X X X be the set constructed by applying Proposition 9.2 with this data. The conclusion is then immediate. □
Appendix A. Bi-Lipschitz maps associated to the bridges
In this section, we fill in the details for the proofs of Lemma 8.2 . We start by explicitly defining the map H H H that we will work with. By scaling, we will suppose that ℓ ( I ) = 1 \ell(I)=1 ℓ ( I ) = 1 . Define a map h ~ : [ − 1 , 2 ] → R n \widetilde{h}:[-1,2]\to\mathbb{R}^{n} h : [ − 1 , 2 ] → R n (piecewise) as follows. First, let h ~ : I × [ − 1 , 0 ] → Q \widetilde{h}:I\times[-1,0]\to Q h : I × [ − 1 , 0 ] → Q be an isometry with h ~ ( x , 0 ) = h ( x , 0 ) = x \widetilde{h}(x,0)=h(x,0)=x h ( x , 0 ) = h ( x , 0 ) = x . For x ∈ I x\in I x ∈ I and t ∈ [ 0 , 1 ] t\in[0,1] t ∈ [ 0 , 1 ] let
h ~ ( x , t ) = { h ( x , 2 t ) if t ∈ [ 0 , 1 / 2 ] ; h ( x , 1 ) + 2 ( t − 1 / 2 ) u ( h ( x , 1 ) ) if t ∈ [ 1 / 2 , 1 ] . \widetilde{h}(x,t)=
\begin{cases}
h(x,2t) & \text{if }t\in[0,1/2];\\
h(x,1)+2(t-1/2)u(h(x,1)) & \text{if }t\in[1/2,1].
\end{cases} h ( x , t ) = { h ( x , 2 t ) h ( x , 1 ) + 2 ( t − 1/2 ) u ( h ( x , 1 )) if t ∈ [ 0 , 1/2 ] ; if t ∈ [ 1/2 , 1 ] .
Finally, let h ′ : I × [ 1 , 2 ] → R h':I\times[1,2]\to\mathbb{R} h ′ : I × [ 1 , 2 ] → R be a surjective map such that h ′ ( x , 1 ) = h ( x , 1 ) h'(x,1)=h(x,1) h ′ ( x , 1 ) = h ( x , 1 ) and
∣ h ′ ( x , t ) − h ′ ( y , s ) ∣ = ℓ ( R ) ℓ ( Q ) ∣ ( x , t ) − ( y , s ) ∣ |h'(x,t)-h'(y,s)|=\frac{\ell(R)}{\ell(Q)}|(x,t)-(y,s)| ∣ h ′ ( x , t ) − h ′ ( y , s ) ∣ = ℓ ( Q ) ℓ ( R ) ∣ ( x , t ) − ( y , s ) ∣
for all $(x,t),(y,s)\in I\times[1,2], and set
h ~ : = u ∘ h ′ . \widetilde{h} := u \circ h'. h := u ∘ h ′ .
We have that h ~ \widetilde{h} h is continuous since it is continuous on each component and agrees at the boundaries. Also, h ~ \widetilde{h} h is an isometry (1-bi-Lipschitz) on I × [ − 1 , 0 ] I \times[-1,0] I × [ − 1 , 0 ] and ℓ ( R ) / ℓ ( Q ) \ell(R)/\ell(Q) ℓ ( R ) / ℓ ( Q ) -bi-Lipschitz on I × [ 1 , 2 ] I \times[1,2] I × [ 1 , 2 ] . It follows from the definitions and (73) that
h ~ ( I × [ − 1 , 0 ] ) = Q , h ~ ( I × [ 1 , 2 ] ) = u ( R ) = K and h ~ ( I × [ − 1 , 2 ] ) ⊆ B . (144) \widetilde{h}(I \times[-1,0]) = Q,\quad\widetilde{h}(I \times[1,2]) = u(R) = K\quad\text{and}\quad\widetilde{h}(I \times[-1,2]) \subseteq B. \tag*{(144)} h ( I × [ − 1 , 0 ]) = Q , h ( I × [ 1 , 2 ]) = u ( R ) = K and h ( I × [ − 1 , 2 ]) ⊆ B . (144)
Finally, one can also check, using (70) –(72) and the fact that u u u is 1-Lipschitz, that if x , y ∈ I x,y \in I x , y ∈ I and t , s ∈ [ 0 , 1 ] t,s \in[0,1] t , s ∈ [ 0 , 1 ] then
δ ∣ x − y ∣ ≤ ∣ h ~ ( x , t ) − h ~ ( y , t ) ∣ ≤ 2 ∣ x − y ∣ ; (145) \delta|x-y| \leq|\widetilde{h}(x,t)-\widetilde{h}(y,t)| \leq2|x-y|; \tag*{(145)} δ ∣ x − y ∣ ≤ ∣ h ( x , t ) − h ( y , t ) ∣ ≤ 2∣ x − y ∣ ; (145)
∣ h ~ ( x , 1 ) − h ~ ( y , 1 ) ∣ ≤ 2 C 1 δ ∣ x − y ∣ ; (146) |\widetilde{h}(x,1)-\widetilde{h}(y,1)| \leq2C_1\delta|x-y|; \tag*{(146)} ∣ h ( x , 1 ) − h ( y , 1 ) ∣ ≤ 2 C 1 δ ∣ x − y ∣ ; (146)
∣ h ~ ( x , t ) − h ~ ( x , s ) ∣ ≤ C 1 ∣ t − s ∣ . (147) |\widetilde{h}(x,t)-\widetilde{h}(x,s)| \leq C_1|t-s|. \tag*{(147)} ∣ h ( x , t ) − h ( x , s ) ∣ ≤ C 1 ∣ t − s ∣. (147)
We define H : I × [ − 1 , 2 ] → R n + 1 H : I \times[-1,2] \to\mathbb{R}^{n+1} H : I × [ − 1 , 2 ] → R n + 1 by
H ( x , t ) : = { ( h ~ ( x , t ) , g 1 ( h ~ ( x , t ) ) ) for t ∈ [ − 1 , 0 ] ; ( h ~ ( x , t ) , ( 1 − t ) g 1 ( h ~ ( x , 0 ) ) + t g 2 ( h ~ ( x , 1 ) ) ) for t ∈ [ 0 , 1 ] ; ( h ~ ( x , t ) , g 2 ( h ~ ( x , t ) ) ) for t ∈ [ 1 , 2 ] . H(x,t) :=
\begin{cases}
(\widetilde{h}(x,t),g_1(\widetilde{h}(x,t))) & \text{for } t \in[-1,0];\\
(\widetilde{h}(x,t),(1-t)g_1(\widetilde{h}(x,0))+tg_2(\widetilde{h}(x,1))) & \text{for } t \in[0,1];\\
(\widetilde{h}(x,t),g_2(\widetilde{h}(x,t))) & \text{for } t \in[1,2].
\end{cases} H ( x , t ) := ⎩ ⎨ ⎧ ( h ( x , t ) , g 1 ( h ( x , t ))) ( h ( x , t ) , ( 1 − t ) g 1 ( h ( x , 0 )) + t g 2 ( h ( x , 1 ))) ( h ( x , t ) , g 2 ( h ( x , t ))) for t ∈ [ − 1 , 0 ] ; for t ∈ [ 0 , 1 ] ; for t ∈ [ 1 , 2 ] .
Lemma A.1. The map H H H takes values in B × R B \times\mathbb{R} B × R and satisfies (74) –(8.8) .
Proof. That H H H takes values in B × R B \times\mathbb{R} B × R and satisfies (74) and (75) follows directly from (A.1). Let us check (76) . Fix x ∈ I x \in I x ∈ I and t ∈ [ 0 , 1 ] t \in[0,1] t ∈ [ 0 , 1 ] . Recalling that h ~ ( x , 0 ) ∈ I \widetilde{h}(x,0) \in I h ( x , 0 ) ∈ I and h ~ ( x , 1 ) ∈ u ( J ) \widetilde{h}(x,1) \in u(J) h ( x , 1 ) ∈ u ( J ) , by applying (70) we have
g 1 ( h ~ ( x , 0 ) ) − π R ( H ( x , t ) ) ≥ t γ . (148) g_1(\widetilde{h}(x,0))-\pi_{\mathbb{R}}(H(x,t)) \geq t\gamma. \tag*{(148)} g 1 ( h ( x , 0 )) − π R ( H ( x , t )) ≥ t γ . (148)
Similarly,
π R ( H ( x , t ) ) − g 2 ( h ~ ( x , 1 ) ) = ( 1 − t ) ( g 1 ( x ) − g 2 ( h ~ ( x , 1 ) ) ) ≥ ( 1 − t ) γ . (149) \pi_{\mathbb{R}}(H(x,t))-g_2(\widetilde{h}(x,1))=(1-t)\left(g_1(x)-g_2(\widetilde{h}(x,1))\right)\geq(1-t)\gamma. \tag*{(149)} π R ( H ( x , t )) − g 2 ( h ( x , 1 )) = ( 1 − t ) ( g 1 ( x ) − g 2 ( h ( x , 1 )) ) ≥ ( 1 − t ) γ . (149)
Applying (A.4), (A.5), and using the fact that g 1 g_1 g 1 is η \eta η -Lipschitz, if η = η ( C 1 , γ ) \eta=\eta(C_1,\gamma) η = η ( C 1 , γ ) is chosen small enough, we have
π R ( H ( x , t ) ) ≤ g 1 ( h ~ ( x , 0 ) ) − t γ ≤ g 1 ( h ~ ( x , t ) ) + ∣ g 1 ( h ~ ( x , t ) ) − g 1 ( h ~ ( x , 0 ) ) ∣ − t γ ≤ g 1 ( h ~ ( x , t ) ) + ( C 1 η − γ ) t ≤ g 1 ( h ~ ( x , t ) ) − γ t / 2. (150) \begin{aligned}
\pi_{\mathbb{R}}(H(x,t)) &\leq g_1(\widetilde{h}(x,0))-t\gamma\\
&\leq g_1(\widetilde{h}(x,t))+|g_1(\widetilde{h}(x,t))-g_1(\widetilde{h}(x,0))|-t\gamma\\
&\leq g_1(\widetilde{h}(x,t))+(C_1\eta-\gamma)t \leq g_1(\widetilde{h}(x,t))-\gamma t/2. \tag*{(150)}
\end{aligned} π R ( H ( x , t )) ≤ g 1 ( h ( x , 0 )) − t γ ≤ g 1 ( h ( x , t )) + ∣ g 1 ( h ( x , t )) − g 1 ( h ( x , 0 )) ∣ − t γ ≤ g 1 ( h ( x , t )) + ( C 1 η − γ ) t ≤ g 1 ( h ( x , t )) − γ t /2. (150)
Similarly, applying (A.4), (A.6), and using the fact that g 2 g_2 g 2 is θ \theta θ -Lipschitz, if θ = θ ( C 1 , γ ) \theta=\theta(C_1,\gamma) θ = θ ( C 1 , γ ) is chosen small enough, we have
π R ( H ( x , t ) ) ≥ g 2 ( h ~ ( x , t ) ) + ( 1 − t ) γ / 2. (151) \pi_{\mathbb{R}}(H(x,t)) \geq g_2(\widetilde{h}(x,t))+(1-t)\gamma/2. \tag*{(151)} π R ( H ( x , t )) ≥ g 2 ( h ( x , t )) + ( 1 − t ) γ /2. (151)
Inequalities (A.7) and (A.8) imply (76) . □
Next, we shall check (77) and (78) , which will be important for checking that H H H is bi-Lipschitz.
Lemma A.2. The estimates (77) and (78) hold.
Proof. The proofs of (77) and (78) are similar, and so we will only show (78) . Fix x ∈ X x \in X x ∈ X . If x ∈ X d o w n x \in X_{\mathrm{down}} x ∈ X down then (78) is trivial (just take a = x a=x a = x ). Suppose instead that x ∈ H ( I × [ − 1 , 1 ] ) x \in H(I \times[-1,1]) x ∈ H ( I × [ − 1 , 1 ]) . Let ( w , t ) ∈ I × [ − 1 , 1 ] (w,t) \in I \times[-1,1] ( w , t ) ∈ I × [ − 1 , 1 ] such that x = H ( w , t ) x=H(w,t) x = H ( w , t ) . Then, let a 2 = H ( w , 1 ) a_2=H(w,1) a 2 = H ( w , 1 ) . The proof split into two cases.
Suppose first that t ∈ [ 0 , 1 ] t \in[0,1] t ∈ [ 0 , 1 ] . Since H H H is Lipschitz, we have
∣ x − a 2 ∣ ≲ 1 − t . |x-a_2| \lesssim1-t. ∣ x − a 2 ∣ ≲ 1 − t .
On the other hand, it follows from (76) that
∣ π 1 ( x ) − π 1 ( a 2 ) ∣ ≥ λ g 2 ( π n ( x ) ) + c γ ⋅ ( 1 − t ) − λ g 2 ( π n ( a ) ) ≥ c γ ∣ x − a 2 ∣ + λ g 2 ( π n ( x ) ) − λ g 2 ( π n ( y ) ) . \begin{aligned}
|\pi_1(x)-\pi_1(a_2)| &\ge\lambda g_2(\pi_n(x))+c\gamma\cdot(1-t)-\lambda g_2(\pi_n(a))\\
&\ge c\gamma|x-a_2|+\lambda g_2(\pi_n(x))-\lambda g_2(\pi_n(y)).
\end{aligned} ∣ π 1 ( x ) − π 1 ( a 2 ) ∣ ≥ λ g 2 ( π n ( x )) + c γ ⋅ ( 1 − t ) − λ g 2 ( π n ( a )) ≥ c γ ∣ x − a 2 ∣ + λ g 2 ( π n ( x )) − λ g 2 ( π n ( y )) .
The estimate (78) now follows directly from Lemma 3.6 so long as θ = θ ( γ ) \theta=\theta(\gamma) θ = θ ( γ ) is sufficiently small.
Suppose now that t ∈ [ − 1 , 0 ] t \in[-1,0] t ∈ [ − 1 , 0 ] . Let x ′ = H ( w , 0 ) x'=H(w,0) x ′ = H ( w , 0 ) , so that
∣ x − x ′ ∣ ≲ 1 ∼ ∣ x − a 2 ∣ . (152) |x-x'| \lesssim1 \sim|x-a_2|. \tag*{(152)} ∣ x − x ′ ∣ ≲ 1 ∼ ∣ x − a 2 ∣. (152)
Repeating the estimates above for the point x ′ x' x ′ , we have
∣ π 1 ( x ′ ) − π 1 ( a 2 ) ∣ ≥ λ g 2 ( π n ( x ′ ) ) + c γ − λ g 2 ( π n ( y ) ) . (153) |\pi_1(x')-\pi_1(a_2)|\ge\lambda g_2(\pi_n(x'))+c\gamma-\lambda g_2(\pi_n(y)). \tag*{(153)} ∣ π 1 ( x ′ ) − π 1 ( a 2 ) ∣ ≥ λ g 2 ( π n ( x ′ )) + c γ − λ g 2 ( π n ( y )) . (153)
Since x x x and x ′ x' x ′ lie in the graph of g 1 g_1 g 1 , we can use (A.9), (A.10) and the fact that g 1 g_1 g 1 is η \eta η -Lipschitz and λ g 2 \lambda g_2 λ g 2 is λ θ \lambda\theta λ θ -Lipschitz, to get
∣ π 1 ( x ) − π 1 ( a 2 ) ∣ ≥ ∣ π 1 ( x ′ ) − π 1 ( a 2 ) ∣ − C η ≥ ( c γ − C η − C λ θ ) + λ g 2 ( π n ( x ) ) − λ g 2 ( π n ( y ) ) . \begin{aligned}
|\pi_1(x)-\pi_1(a_2)| &\ge|\pi_1(x')-\pi_1(a_2)|-C\eta\\
&\ge(c\gamma-C\eta-C\lambda\theta)+\lambda g_2(\pi_n(x))-\lambda g_2(\pi_n(y)).
\end{aligned} ∣ π 1 ( x ) − π 1 ( a 2 ) ∣ ≥ ∣ π 1 ( x ′ ) − π 1 ( a 2 ) ∣ − C η ≥ ( c γ − C η − C λ θ ) + λ g 2 ( π n ( x )) − λ g 2 ( π n ( y )) .
Assuming η = η ( γ ) \eta=\eta(\gamma) η = η ( γ ) and θ = θ ( γ ) \theta=\theta(\gamma) θ = θ ( γ ) are small enough, this implies
∣ π 1 ( x ) − π 1 ( a 2 ) ∣ ≥ c γ ∣ x − a 2 ∣ + λ g 2 ( π n ( x ) ) − λ g 2 ( π n ( y ) ) . |\pi_1(x)-\pi_1(a_2)|\ge c\gamma|x-a_2|+\lambda g_2(\pi_n(x))-\lambda g_2(\pi_n(y)). ∣ π 1 ( x ) − π 1 ( a 2 ) ∣ ≥ c γ ∣ x − a 2 ∣ + λ g 2 ( π n ( x )) − λ g 2 ( π n ( y )) .
The estimate (78) now follows directly from Lemma 3.6 so long as θ = θ ( γ ) \theta=\theta(\gamma) θ = θ ( γ ) is sufficiently small. □ \square □
The only thing left to check is that H H H is bi-Lipschitz. We start with the following preliminary result.
Lemma A.3. The map H H H is bi-Lipschitz when restricted to either of I × [ − 1 , 0 ] I \times[-1,0] I × [ − 1 , 0 ] , I × [ 0 , 1 ] I \times[0,1] I × [ 0 , 1 ] or I × [ 1 , 2 ] I \times[1,2] I × [ 1 , 2 ] .
Proof. That H H H is bi-Lipschitz on when restricted to I × [ − 1 , 0 ] I \times[-1,0] I × [ − 1 , 0 ] and I × [ 1 , 2 ] I \times[1,2] I × [ 1 , 2 ] (since it is Lipschitz in the second component and bi-Lipschitz in the first). To check the last case, fix x , y ∈ I x,y \in I x , y ∈ I , t , s ∈ [ 0 , 1 ] t,s \in[0,1] t , s ∈ [ 0 , 1 ] . Denote
w = h ~ ( x , 1 ) and z = h ~ ( y , 1 ) w=\widetilde{h}(x,1)\quad\text{and}\quad z=\widetilde{h}(y,1) w = h ( x , 1 ) and z = h ( y , 1 )
and observe for later that, after applying (A.3) and the fact that g 2 g_2 g 2 is θ \theta θ -Lipschitz and u u u is 1-Lipschitz, we have
∣ g 2 ( w ) − g 2 ( z ) ∣ ≤ θ ∣ w − z ∣ ≤ 2 C 1 θ δ . (154) |g_2(w)-g_2(z)|\le\theta|w-z|\le2C_1\theta\delta. \tag*{(154)} ∣ g 2 ( w ) − g 2 ( z ) ∣ ≤ θ ∣ w − z ∣ ≤ 2 C 1 θ δ . (154)
By (A.2) and (A.4), we have
∣ h ~ ( x , t ) − h ~ ( y , s ) ∣ ≤ ∣ h ~ ( x , t ) − h ~ ( y , t ) ∣ + ∣ h ~ ( y , t ) − h ~ ( y , s ) ∣ ≲ C 1 ∣ x − y ∣ + ∣ t − s ∣ . |\widetilde{h}(x,t)-\widetilde{h}(y,s)|\le|\widetilde{h}(x,t)-\widetilde{h}(y,t)|+|\widetilde{h}(y,t)-\widetilde{h}(y,s)|\lesssim_{C_1}|x-y|+|t-s|. ∣ h ( x , t ) − h ( y , s ) ∣ ≤ ∣ h ( x , t ) − h ( y , t ) ∣ + ∣ h ( y , t ) − h ( y , s ) ∣ ≲ C 1 ∣ x − y ∣ + ∣ t − s ∣.
Furthermore, using the second estimate in (70) , (A.11), and the fact that g 1 g_1 g 1 is η \eta η -Lipschitz, it follows that
∣ ( 1 − t ) g 1 ( x ) + g 2 ( w ) − ( 1 − s ) g 1 ( y ) − s g 2 ( z ) ∣ ≤ ( 1 − t ) ∣ g 1 ( x ) − g 1 ( y ) ∣ + t ∣ g 2 ( w ) − g 2 ( z ) ∣ + ∣ s − t ∣ ( ∣ g 1 ( y ) ∣ + ∣ g 2 ( z ) ∣ ) ≤ ( η + 2 C 1 θ δ ) ∣ x − y ∣ + 2 C 1 ∣ t − s ∣ . \left|(1-t)g_1(x)+g_2(w)-(1-s)g_1(y)-sg_2(z)\right|
\leq(1-t)\left|g_1(x)-g_1(y)\right|+t\left|g_2(w)-g_2(z)\right|+|s-t|\left(\left|g_1(y)\right|+\left|g_2(z)\right|\right)
\leq(\eta+2C_1\theta\delta)|x-y|+2C_1|t-s|. ∣ ( 1 − t ) g 1 ( x ) + g 2 ( w ) − ( 1 − s ) g 1 ( y ) − s g 2 ( z ) ∣ ≤ ( 1 − t ) ∣ g 1 ( x ) − g 1 ( y ) ∣ + t ∣ g 2 ( w ) − g 2 ( z ) ∣ + ∣ s − t ∣ ( ∣ g 1 ( y ) ∣ + ∣ g 2 ( z ) ∣ ) ≤ ( η + 2 C 1 θ δ ) ∣ x − y ∣ + 2 C 1 ∣ t − s ∣.
Combining both previous sets of inequality, we deduce that H H H is Lipschitz. To prove that H H H is bi-Lipschitz, we first choose η = η ( γ , δ , C 1 ) \eta=\eta(\gamma,\delta,C_1) η = η ( γ , δ , C 1 ) and θ = θ ( γ , C 1 ) \theta=\theta(\gamma,C_1) θ = θ ( γ , C 1 ) small enough such that
η δ + 2 θ C 1 ≤ γ 4 C 1 , (155) \frac{\eta}{\delta}+2\theta C_1\leq\frac{\gamma}{4C_1}, \tag*{(155)} δ η + 2 θ C 1 ≤ 4 C 1 γ , (155)
then split into two cases. Suppose first that 2 C 1 ∣ t − s ∣ ≥ δ ∣ x − y ∣ 2C_1|t-s|\geq\delta|x-y| 2 C 1 ∣ t − s ∣ ≥ δ ∣ x − y ∣ . By the first inequality in (70) ,
∣ g 1 ( x ) − g 2 ( w ) ∣ ≥ γ . |g_1(x)-g_2(w)|\geq\gamma. ∣ g 1 ( x ) − g 2 ( w ) ∣ ≥ γ .
Applying this, in conjunction with the first estimate in (70) , (A.11), (A.12), the fact that g 1 g_1 g 1 is η \eta η -Lipschitz, and our assumption that 2 C 1 ∣ t − s ∣ ≥ δ ∣ x − y ∣ 2C_1|t-s|\geq\delta|x-y| 2 C 1 ∣ t − s ∣ ≥ δ ∣ x − y ∣ , it follows that
∣ ( 1 − t ) g 1 ( x ) + t g 2 ( w ) − ( 1 − s ) g 1 ( y ) − s g 2 ( z ) ∣ ≥ ∣ t − s ∣ ∣ g 1 ( x ) − g 2 ( w ) ∣ − ( 1 − s ) ∣ g 1 ( x ) − g 1 ( y ) ∣ − s ∣ g 2 ( w ) − g 2 ( z ) ∣ ≥ γ ∣ t − s ∣ − ( η + 2 C 1 θ δ ) ∣ x − y ∣ ≥ γ ∣ t − s ∣ − 2 C 1 ( η δ + 2 θ C 1 ) ∣ t − s ∣ ≳ γ ∣ t − s ∣ . \begin{aligned}
\left|(1-t)g_1(x)+tg_2(w)-(1-s)g_1(y)-sg_2(z)\right|
&\geq|t-s|\left|g_1(x)-g_2(w)\right|-(1-s)\left|g_1(x)-g_1(y)\right|-s\left|g_2(w)-g_2(z)\right|\\
&\geq\gamma|t-s|-(\eta+2C_1\theta\delta)|x-y|\\
&\geq\gamma|t-s|-2C_1\left(\frac{\eta}{\delta}+2\theta C_1\right)|t-s|\\
&\gtrsim\gamma|t-s|.
\end{aligned} ∣ ( 1 − t ) g 1 ( x ) + t g 2 ( w ) − ( 1 − s ) g 1 ( y ) − s g 2 ( z ) ∣ ≥ ∣ t − s ∣ ∣ g 1 ( x ) − g 2 ( w ) ∣ − ( 1 − s ) ∣ g 1 ( x ) − g 1 ( y ) ∣ − s ∣ g 2 ( w ) − g 2 ( z ) ∣ ≥ γ ∣ t − s ∣ − ( η + 2 C 1 θ δ ) ∣ x − y ∣ ≥ γ ∣ t − s ∣ − 2 C 1 ( δ η + 2 θ C 1 ) ∣ t − s ∣ ≳ γ ∣ t − s ∣.
Suppose instead that 2 C 1 ∣ t − s ∣ < δ ∣ x − y ∣ 2C_1|t-s|<\delta|x-y| 2 C 1 ∣ t − s ∣ < δ ∣ x − y ∣ . Then, using (71) , we have
∣ h ( x , t ) − h ( y , s ) ∣ ≥ ∣ h ( x , t ) − h ( y , t ) ∣ − ∣ h ( y , t ) − h ( y , s ) ∣ ≥ δ ∣ x − y ∣ − C 1 ∣ t − s ∣ ≳ δ ∣ x − y ∣ . \begin{aligned}
|h(x,t)-h(y,s)|
&\geq|h(x,t)-h(y,t)|-|h(y,t)-h(y,s)|\\
&\geq\delta|x-y|-C_1|t-s|\\
&\gtrsim\delta|x-y|.
\end{aligned} ∣ h ( x , t ) − h ( y , s ) ∣ ≥ ∣ h ( x , t ) − h ( y , t ) ∣ − ∣ h ( y , t ) − h ( y , s ) ∣ ≥ δ ∣ x − y ∣ − C 1 ∣ t − s ∣ ≳ δ ∣ x − y ∣.
The completes the proof. □
The following finishes the proof of Lemma 8.2.
Lemma A.4. The map H H H is bi-Lipschitz on I × [ − 1 , 2 ] I\times[-1,2] I × [ − 1 , 2 ] . Proof. Note that H H H is globally Lipschitz since it is Lipschitz on each component, so we only need to check the lower bounds. Fix x , y ∈ I x,y\in I x , y ∈ I and s , t ∈ [ − 1 , 2 ] s,t\in[-1,2] s , t ∈ [ − 1 , 2 ] . For brevity, let us denote w = H ( x , t ) w=H(x,t) w = H ( x , t ) and z = H ( y , s ) z=H(y,s) z = H ( y , s ) .
Case 1. Suppose that t ∈ [ − 1 , 0 ] t\in[-1,0] t ∈ [ − 1 , 0 ] and s ∈ [ 0 , 1 ] s\in[0,1] s ∈ [ 0 , 1 ] .
Consider the point a 1 = H ( y , 0 ) a_1=H(y,0) a 1 = H ( y , 0 ) . By (77) , we know that ∣ w − z ∣ ≳ ∣ w − a ∣ + ∣ a − z ∣ |w-z|\gtrsim|w-a|+|a-z| ∣ w − z ∣ ≳ ∣ w − a ∣ + ∣ a − z ∣ . This gives the lower bound ∣ H ( x , t ) − H ( y , s ) ∣ ≳ ∣ x − y ∣ + ∣ t − s ∣ |H(x,t)-H(y,s)|\gtrsim|x-y|+|t-s| ∣ H ( x , t ) − H ( y , s ) ∣ ≳ ∣ x − y ∣ + ∣ t − s ∣ since ( y , 0 ) ∈ I × { 0 } (y,0)\in I\times\{0\} ( y , 0 ) ∈ I × { 0 } and H H H is bi-Lipschitz on each of I × [ − 1 , 0 ] I\times[-1,0] I × [ − 1 , 0 ] and I × [ 0 , 1 ] I\times[0,1] I × [ 0 , 1 ] by Lemma A.3.
Case 2. Suppose that t ∈ [ − 1 , 0 ] t\in[-1,0] t ∈ [ − 1 , 0 ] and s ∈ [ 1 , 2 ] s\in[1,2] s ∈ [ 1 , 2 ] .
Consider the points w ′ = H ( x , 0 ) ∈ g 1 ( I ) w' = H(x,0) \in g_1(I) w ′ = H ( x , 0 ) ∈ g 1 ( I ) and z ′ = H ( y , 1 ) ∈ g 2 ( u ( J ) ) z' = H(y,1) \in g_2(u(J)) z ′ = H ( y , 1 ) ∈ g 2 ( u ( J )) . By (70) , these points satisfy ∣ w ′ − z ′ ∣ ≥ γ ℓ ( I ) |w' - z'| \ge\gamma\ell(I) ∣ w ′ − z ′ ∣ ≥ γ ℓ ( I ) . Since g 1 g_1 g 1 is η \eta η -Lipschitz and H H H is Lipschitz (so that ∣ w − w ′ ∣ ≲ ℓ ( I ) |w-w'| \lesssim\ell(I) ∣ w − w ′ ∣ ≲ ℓ ( I ) ), we have
∣ π 1 ( w − w ′ ) ∣ = ∣ g 1 ( π n ( w ) ) − g 1 ( π n ( w ′ ) ) ∣ ≲ η ℓ ( I ) . |\pi_1(w-w')| = |g_1(\pi_n(w))-g_1(\pi_n(w'))| \lesssim\eta\ell(I). ∣ π 1 ( w − w ′ ) ∣ = ∣ g 1 ( π n ( w )) − g 1 ( π n ( w ′ )) ∣ ≲ η ℓ ( I ) .
Similarly, ∣ π 1 ( z − z ′ ) ∣ ≲ θ ℓ ( I ) |\pi_1(z-z')| \lesssim\theta\ell(I) ∣ π 1 ( z − z ′ ) ∣ ≲ θ ℓ ( I ) . Putting these estimates together, we get
∣ w − z ∣ ≥ ∣ π 1 ( w − z ) ∣ ≥ ∣ π 1 ( w ′ − z ′ ) ∣ − ∣ π 1 ( w − w ′ ) ∣ − ∣ π 1 ( z − z ′ ) ∣ ≥ γ ℓ ( I ) − C η ℓ ( I ) − C θ ℓ ( I ) ≳ γ ℓ ( I ) \begin{aligned}
|w-z| \ge|\pi_1(w-z)| \ge|\pi_1(w'-z')| - |\pi_1(w-w')| - |\pi_1(z-z')| \\
&\ge\gamma\ell(I)-C\eta\ell(I)-C\theta\ell(I) \gtrsim\gamma\ell(I)
\end{aligned} ∣ w − z ∣ ≥ ∣ π 1 ( w − z ) ∣ ≥ ∣ π 1 ( w ′ − z ′ ) ∣ − ∣ π 1 ( w − w ′ ) ∣ − ∣ π 1 ( z − z ′ ) ∣ ≥ γ ℓ ( I ) − C η ℓ ( I ) − C θ ℓ ( I ) ≳ γ ℓ ( I )
≳ γ ( ∣ x − y ∣ + ∣ t − s ∣ ) . \gtrsim\gamma\left(|x-y|+|t-s|\right). ≳ γ ( ∣ x − y ∣ + ∣ t − s ∣ ) .
Case 3. Suppose that t ∈ [ 0 , 1 ] t \in[0,1] t ∈ [ 0 , 1 ] and s ∈ [ 1 , 2 ] s \in[1,2] s ∈ [ 1 , 2 ] .
This has is similar to Case 1, and the proof passes via Lemma 3.6 . In this case, one considers the point a = H ( x , 1 ) a=H(x,1) a = H ( x , 1 ) . We will leave the details to the reader. □ Matthew Hyde.
University of Jyväskylä, Department of Mathematics and Statistics, P.O. Box 35 (MaD), FI-40014 University of Jyväskylä, Finland
Email address: matthew.j.hyde@jyu.fi
Michele Villa,
Ikerbasque, Basque Foundation for Science Plaza Euskadi 5, 48009 Bilbao, Spain; Euskal Herriko Unibertsitatea/Universidad del País Vasco Enparantza Torres Quevedo Ingeniaritzaren 1C, Basurtu-Zorrotza, 48013 Bilbao, Bizkaia, Spain
Email address: michele.villa@ehu.eus
Ivan Yuri Violo,
La Sapienza Università di Roma, Dipartimento di Matematica Guido Castelnuovo, Piazzale Aldo Moro, 5, 00185 Roma RM, Italy
Email address: ivanyuri.violo@uniroma1.it