Introduction and main results
Let k ≥ 2 k\geq2 k ≥ 2 T T T k k k x 0 x_0 x 0 k k k
Throughout the paper, we denote A ≲ θ B A\lesssim_\theta B A ≲ θ B c θ c_\theta c θ θ \theta θ A ≤ c θ B A\leq c_\theta B A ≤ c θ B A ≃ θ B A\simeq_\theta B A ≃ θ B A ≲ θ B A\lesssim_\theta B A ≲ θ B B ≲ θ A B\lesssim_\theta A B ≲ θ A
It is possible to equip T T T d d d μ \mu μ d ( x , y ) d(x,y) d ( x , y ) x x x y y y μ \mu μ A ⊂ T A\subset T A ⊂ T ∣ A ∣ = μ ( A ) |A|=\mu(A) ∣ A ∣ = μ ( A )
f ( A ) = ∫ A f ( x ) d x = ∑ x ∈ A f ( x ) . f(A)=\int_A f(x)\,dx=\sum_{x\in A}f(x). f ( A ) = ∫ A f ( x ) d x = x ∈ A ∑ f ( x ) . All radii are nonnegative integers. By a weight we mean a function w : T → [ 0 , ∞ ) w:T\to[0,\infty) w : T → [ 0 , ∞ ) T T T w ( E ) = ∑ x ∈ E w ( x ) w(E)=\sum_{x\in E}w(x) w ( E ) = ∑ x ∈ E w ( x )
Let S ( x , r ) S(x,r) S ( x , r ) r r r S ( x , r ) = { y ∈ T : d ( x , y ) = r } S(x,r)=\{y\in T:d(x,y)=r\} S ( x , r ) = { y ∈ T : d ( x , y ) = r } S ( x , r ) S(x,r) S ( x , r )
∣ S ( x , r ) ∣ ≃ k r , |S(x,r)|\simeq k^r, ∣ S ( x , r ) ∣ ≃ k r , and μ \mu μ T T T
The study of classical operators in the context of k k k [7 ] and Ombrosi, Rivera-Ríos and Safe [9 ] . We also refer to the articles [12 , 13 ] in which the authors studied connections between the geometrical properties of finite and infinite graphs and the boundedness of the Hardy-Littlewood maximal operator. For weighted estimates on trees, we refer to [4 , 5 , 8 , 15 ] .
In [7 ] , Naor and Tao proved the unweighted weak type ( 1 , 1 ) (1,1) ( 1 , 1 ) k k k T T T
∣ { x ∈ T : M f ( x ) > λ } ∣ ≤ c λ ∑ x ∈ T ∣ f ( x ) ∣ \left|\{x \in T : Mf(x) > \lambda\}\right| \leq\frac{c}{\lambda}\sum_{x\in T}|f(x)| ∣ { x ∈ T : M f ( x ) > λ } ∣ ≤ λ c x ∈ T ∑ ∣ f ( x ) ∣ where M M M
M f ( x ) = sup r ≥ 0 1 ∣ B ( x , r ) ∣ ∑ y ∈ B ( x , r ) ∣ f ( y ) ∣ . Mf(x)=\sup_{r\geq0}\frac{1}{|B(x,r)|}\sum_{y\in B(x,r)}|f(y)|. M f ( x ) = r ≥ 0 sup ∣ B ( x , r ) ∣ 1 y ∈ B ( x , r ) ∑ ∣ f ( y ) ∣. Recently, in [9 ] Ombrosi, Rivera-Ríos and Safe proved a weighted Fefferman–Stein inequality for the Hardy–Littlewood maximal operator on T T T
w ( { M f > λ } ) ≲ 1 λ ∑ x ∈ T ∣ f ( x ) ∣ M s w ( x ) , s > 1 , w(\{Mf>\lambda\})\lesssim\frac{1}{\lambda}\sum_{x\in T}|f(x)|M_sw(x), \qquad s>1, w ({ M f > λ }) ≲ λ 1 x ∈ T ∑ ∣ f ( x ) ∣ M s w ( x ) , s > 1 , where M s ( w ) = M ( w 1 / s ) M_s(w)=M(w^{1/s}) M s ( w ) = M ( w 1/ s ) s = 1 s=1 s = 1 2 n / k r 2^n/k^r 2 n / k r
The Fefferman-Stein estimate was originally proved in [3 ] , later these estimates and related variants have been studied in a variety of contexts [1 , 6 , 10 , 11 , 14 ] . In the fractional setting, the Fefferman-Stein for the fractional maximal operator is a consequence of classical covering lemmas [3 ] . In [2 ] , the authors proved that the Fefferman-Stein estimate for the fractional integral holds with a fractional L log L L\log L L log L
The purpose of the present paper is to develop the corresponding theory for the fractional maximal operator with 0 < α < 1 0<\alpha<1 0 < α < 1
M α f ( x ) = sup r ≥ 0 1 ∣ B ( x , r ) ∣ 1 − α ∑ y ∈ B ( x , r ) ∣ f ( y ) ∣ , M_\alpha f(x)=\sup_{r\geq0}\frac{1}{|B(x,r)|^{1-\alpha}}\sum_{y\in B(x,r)}|f(y)|, M α f ( x ) = r ≥ 0 sup ∣ B ( x , r ) ∣ 1 − α 1 y ∈ B ( x , r ) ∑ ∣ f ( y ) ∣ , and its spherical counterpart
M α ∘ f ( x ) = sup r ≥ 0 1 ∣ S ( x , r ) ∣ 1 − α ∑ y ∈ S ( x , r ) ∣ f ( y ) ∣ . M_\alpha^\circ f(x)=\sup_{r\geq0}\frac{1}{|S(x,r)|^{1-\alpha}}\sum_{y\in S(x,r)}|f(y)|. M α ∘ f ( x ) = r ≥ 0 sup ∣ S ( x , r ) ∣ 1 − α 1 y ∈ S ( x , r ) ∑ ∣ f ( y ) ∣. Note that, in contrast to the standard Euclidean setting, here it makes sense to consider this kind of maximal function because S ( x , r ) S(x,r) S ( x , r ) k ≥ 2 k\geq2 k ≥ 2 M α ≃ α M α ∘ M_\alpha\simeq_\alpha M_\alpha^\circ M α ≃ α M α ∘
The fractional problem is not obtained by a formal substitution in the proof of [9 ] . The factor ∣ S ( x , r ) ∣ α ≃ k r α |S(x,r)|^\alpha\simeq k^{r\alpha} ∣ S ( x , r ) ∣ α ≃ k r α
We consider the fractional L s L^s L s
M α , s w ( x ) = sup r ≥ 0 ∣ B ( x , r ) ∣ α ( 1 ∣ B ( x , r ) ∣ ∑ y ∈ B ( x , r ) w ( y ) s ) 1 / s . M_{\alpha,s}w(x)=\sup_{r\geq0}|B(x,r)|^\alpha\left(\frac{1}{|B(x,r)|}\sum_{y\in B(x,r)}w(y)^s\right)^{1/s}. M α , s w ( x ) = r ≥ 0 sup ∣ B ( x , r ) ∣ α ∣ B ( x , r ) ∣ 1 y ∈ B ( x , r ) ∑ w ( y ) s 1/ s . The first result gives a weighted endpoint estimate that survives fractional normalization in a space with exponential volume growth. The following theorem provides a Fefferman-Stein estimate with M α , s w M_{\alpha,s}w M α , s w
Theorem 1.1. Let k > 2 k>2 k > 2 0 < α < 1 0<\alpha<1 0 < α < 1 s > 1 s>1 s > 1 w ≥ 0 w\geq0 w ≥ 0 T T T
w ( { x ∈ T : M α f ( x ) > λ } ) ≤ C α , s 1 λ ∑ x ∈ T ∣ f ( x ) ∣ M α , s w ( x ) , (1) w(\{x\in T:M_\alpha f(x)>\lambda\})\leq C_{\alpha,s}\frac{1}{\lambda}\sum_{x\in T}|f(x)|M_{\alpha,s}w(x), \tag*{(1)} w ({ x ∈ T : M α f ( x ) > λ }) ≤ C α , s λ 1 x ∈ T ∑ ∣ f ( x ) ∣ M α , s w ( x ) , (1) where C α , s C_{\alpha,s} C α , s k k k s ↓ 1 s\downarrow1 s ↓ 1
Theorem 1.1 is proved in § 2 . The weighted sphere-incidence lemma, Lemma 2.3, is a fractional variant of the combinatorial estimate in [9 ] , but its power of k k k
A central issue is whether the assumption s > 1 s>1 s > 1 0 < α < 1 / 2 0<\alpha<1/2 0 < α < 1/2 s = 1 s=1 s = 1 M α M_\alpha M α
Theorem 1.2. For any n ∈ N n\in\mathbb{N} n ∈ N 0 < α < 1 / 2 0<\alpha<1/2 0 < α < 1/2 w w w M α w ≲ w M_\alpha w\lesssim w M α w ≲ w { f j } \{f_j\} { f j }
w ( { M α f j > 1 } ) ∑ x ∈ T ∣ f j ( x ) ∣ M α ( n ) w ( x ) → j → ∞ ∞ . \frac{w(\{M_\alpha f_j>1\})}{\sum_{x\in T}|f_j(x)|M_\alpha^{(n)}w(x)}\xrightarrow[j\to\infty]{}\infty. ∑ x ∈ T ∣ f j ( x ) ∣ M α ( n ) w ( x ) w ({ M α f j > 1 }) j → ∞ ∞. Consequently, no estimate of the form
w ( { M α f > λ } ) ≤ C λ ∑ x ∈ T ∣ f ( x ) ∣ M α ( n ) w ( x ) w(\{M_\alpha f>\lambda\})\leq\frac{C}{\lambda}\sum_{x\in T}|f(x)|M_\alpha^{(n)}w(x) w ({ M α f > λ }) ≤ λ C x ∈ T ∑ ∣ f ( x ) ∣ M α ( n ) w ( x ) can hold uniformly for all finitely supported f f f λ > 0 \lambda>0 λ > 0
Although Theorem 1.2 is the fractional counterpart of the result of [9 ] , the construction of the counterexample requires a different approach. In their example, the weight is chosen to compensate the average of the ball. In the fractional setting, however, this normalization becomes ∣ B ( x , r ) ∣ 1 − α |B(x,r)|^{1-\alpha} ∣ B ( x , r ) ∣ 1 − α
We next obtain a necessary condition for the weighted weak-type ( 1 , 1 ) (1,1) ( 1 , 1 ) M α w ≲ w M_\alpha w\lesssim w M α w ≲ w
Theorem 1.3. Let 0 < α < 1 0<\alpha<1 0 < α < 1 w w w T T T C w > 0 C_w>0 C w > 0
w ( { x ∈ T : M α f ( x ) > λ } ) ≤ C w λ ∑ x ∈ T ∣ f ( x ) ∣ w ( x ) (2) w(\{x\in T:M_\alpha f(x)>\lambda\})\leq\frac{C_w}{\lambda}\sum_{x\in T}|f(x)|w(x) \tag*{(2)} w ({ x ∈ T : M α f ( x ) > λ }) ≤ λ C w x ∈ T ∑ ∣ f ( x ) ∣ w ( x ) (2) for every finitely supported f f f λ > 0 \lambda>0 λ > 0
M α w ( x ) ≲ α , k , C w w ( x ) , x ∈ T . M_\alpha w(x)\lesssim_{\alpha,k,C_w}w(x),\qquad x\in T. M α w ( x ) ≲ α , k , C w w ( x ) , x ∈ T . Consequently, for 0 < α < 1 / 2 0<\alpha<1/2 0 < α < 1/2 1 / 2 ≤ α < 1 1/2\leq\alpha<1 1/2 ≤ α < 1 M α w ≲ w M_\alpha w\lesssim w M α w ≲ w Proposition 3.11 .
At the critical value α = 1 / 2 \alpha=1/2 α = 1/2
We present the power-type radial weights of the form
w β ( x ) = k β ∣ x ∣ , w_\beta(x)=k^{\beta|x|}, w β ( x ) = k β ∣ x ∣ , where ∣ x ∣ |x| ∣ x ∣ x x x β ∈ R \beta\in\mathbb{R} β ∈ R
These weights allow us to compare the necessary condition with the sufficient condition involving M α , s M_{\alpha,s} M α , s β \beta β α = 1 / 2 \alpha=1/2 α = 1/2
Theorem 1.4. If 0 < α < 1 / 2 0<\alpha<1/2 0 < α < 1/2
M α w β ≈ w β ⟺ β ∈ [ α − 1 , − α ] . M_\alpha w_\beta\approx w_\beta\qquad\Longleftrightarrow\qquad\beta\in[\alpha-1,-\alpha]. M α w β ≈ w β ⟺ β ∈ [ α − 1 , − α ] . If 1 / 2 ≤ α < 1 1/2 \le\alpha< 1 1/2 ≤ α < 1 s > 1 s > 1 s > 1
M α , s w β ≃ w β M_{\alpha,s}w_\beta\simeq w_\beta M α , s w β ≃ w β holds precisely in the range
s < 1 2 α , β ∈ [ α − 1 s , − α ] . s < \frac{1}{2\alpha}, \qquad\beta\in\left[\alpha-\frac{1}{s},-\alpha\right]. s < 2 α 1 , β ∈ [ α − s 1 , − α ] . The paper is organized as follows. Section 2 contains the comparison between ball and spherical fractional maximal operators, the weighted incidence estimate, the new fractional distributional lemma, and the proof of Theorem 1.1. Section 3 is devoted to radial weights, sharpness, counterexamples, necessary conditions, and the phase transition at α = 1 / 2 \alpha= 1/2 α = 1/2
Proof of Theorem 1.1
To prove Theorem 1.1 we need several weighted estimates. We first compare the ball and spherical fractional maximal operators. Spheres are convenient for the argument because pairs of vertices at a fixed distance can be classified by their common ancestor. We keep track of the constants in this comparison so that the final estimate is uniform in k k k
We write
B ( x , r ) = { y ∈ T : d ( x , y ) ≤ r } , S ( x , r ) = { y ∈ T : d ( x , y ) = r } . B(x,r)=\{y\in T:d(x,y)\le r\},\qquad S(x,r)=\{y\in T:d(x,y)=r\}. B ( x , r ) = { y ∈ T : d ( x , y ) ≤ r } , S ( x , r ) = { y ∈ T : d ( x , y ) = r } . The geometry of the rooted tree gives, uniformly in x x x r r r k ≥ 2 k\ge2 k ≥ 2
k r ≤ ∣ S ( x , r ) ∣ ≤ 2 k r , ∣ B ( x , r ) ∣ ≤ 2 ∣ S ( x , r ) ∣ . (3) k^r\le|S(x,r)|\le2k^r,\qquad|B(x,r)|\le2|S(x,r)|. \tag*{(3)} k r ≤ ∣ S ( x , r ) ∣ ≤ 2 k r , ∣ B ( x , r ) ∣ ≤ 2∣ S ( x , r ) ∣. (3) For 0 < α < 1 0<\alpha<1 0 < α < 1 1 ≤ s < ∞ 1\le s<\infty 1 ≤ s < ∞
M α , s f ( x ) = sup r ≥ 0 ∣ B ( x , r ) ∣ α ( 1 ∣ B ( x , r ) ∣ ∑ y ∈ B ( x , r ) ∣ f ( y ) ∣ s ) 1 / s M_{\alpha,s}f(x)=\sup_{r\ge0}|B(x,r)|^\alpha\left(\frac{1}{|B(x,r)|}\sum_{y\in B(x,r)}|f(y)|^s\right)^{1/s} M α , s f ( x ) = r ≥ 0 sup ∣ B ( x , r ) ∣ α ∣ B ( x , r ) ∣ 1 y ∈ B ( x , r ) ∑ ∣ f ( y ) ∣ s 1/ s and
M α , s ∘ f ( x ) = sup r ≥ 0 ∣ S ( x , r ) ∣ α ( 1 ∣ S ( x , r ) ∣ ∑ y ∈ S ( x , r ) ∣ f ( y ) ∣ s ) 1 / s . M^\circ_{\alpha,s}f(x)=\sup_{r\ge0}|S(x,r)|^\alpha\left(\frac{1}{|S(x,r)|}\sum_{y\in S(x,r)}|f(y)|^s\right)^{1/s}. M α , s ∘ f ( x ) = r ≥ 0 sup ∣ S ( x , r ) ∣ α ∣ S ( x , r ) ∣ 1 y ∈ S ( x , r ) ∑ ∣ f ( y ) ∣ s 1/ s . Thus M α , 1 = M α M_{\alpha,1}=M_\alpha M α , 1 = M α M α , 1 ∘ = M α ∘ M^\circ_{\alpha,1}=M^\circ_\alpha M α , 1 ∘ = M α ∘
Proposition 2.1. Let 0 < α < 1 0<\alpha<1 0 < α < 1
M α f ( x ) ≤ C α M α ∘ f ( x ) , M α ∘ f ( x ) ≤ 2 1 − α M α f ( x ) , M_\alpha f(x)\le C_\alpha M^\circ_\alpha f(x),\qquad M^\circ_\alpha f(x)\le2^{1-\alpha}M_\alpha f(x), M α f ( x ) ≤ C α M α ∘ f ( x ) , M α ∘ f ( x ) ≤ 2 1 − α M α f ( x ) , where C α C_\alpha C α k k k 1 ≤ s < ∞ 1\le s<\infty 1 ≤ s < ∞
M α , s ∘ f ( x ) ≤ 2 ( 1 / s − α ) + M α , s f ( x ) . (4) M^\circ_{\alpha,s}f(x)\le2^{(1/s-\alpha)_+}M_{\alpha,s}f(x). \tag*{(4)} M α , s ∘ f ( x ) ≤ 2 ( 1/ s − α ) + M α , s f ( x ) . (4) If, in addition, α s < 1 \alpha s<1 α s < 1
M α , s f ( x ) ≤ C α , s M α , s ∘ f ( x ) , (5) M_{\alpha,s}f(x)\le C_{\alpha,s}M^\circ_{\alpha,s}f(x), \tag*{(5)} M α , s f ( x ) ≤ C α , s M α , s ∘ f ( x ) , (5) where C α , s C_{\alpha,s} C α , s k k k
Proof. By (2.1), for every r ≥ 0 r\ge0 r ≥ 0
1 ∣ S ( x , r ) ∣ 1 − α ∑ S ( x , r ) ∣ f ∣ ≤ ( ∣ B ( x , r ) ∣ ∣ S ( x , r ) ∣ ) 1 − α 1 ∣ B ( x , r ) ∣ 1 − α ∑ B ( x , r ) ∣ f ∣ ≤ 2 1 − α M α f ( x ) . \frac{1}{|S(x,r)|^{1-\alpha}}\sum_{S(x,r)}|f|
\le\left(\frac{|B(x,r)|}{|S(x,r)|}\right)^{1-\alpha}
\frac{1}{|B(x,r)|^{1-\alpha}}\sum_{B(x,r)}|f|
\le2^{1-\alpha}M_\alpha f(x). ∣ S ( x , r ) ∣ 1 − α 1 S ( x , r ) ∑ ∣ f ∣ ≤ ( ∣ S ( x , r ) ∣ ∣ B ( x , r ) ∣ ) 1 − α ∣ B ( x , r ) ∣ 1 − α 1 B ( x , r ) ∑ ∣ f ∣ ≤ 2 1 − α M α f ( x ) . Hence M α ∘ f ( x ) ≤ 2 1 − α M α f ( x ) M^\circ_\alpha f(x)\le2^{1-\alpha}M_\alpha f(x) M α ∘ f ( x ) ≤ 2 1 − α M α f ( x )
Conversely,
1 ∣ B ( x , r ) ∣ 1 − α ∑ B ( x , r ) ∣ f ∣ = 1 ∣ B ( x , r ) ∣ 1 − α ∑ ℓ = 0 r ∑ S ( x , ℓ ) ∣ f ∣ ≤ M α ∘ f ( x ) ∑ ℓ = 0 r ∣ S ( x , ℓ ) ∣ 1 − α ∣ B ( x , r ) ∣ 1 − α ≤ M α o f ( x ) 2 1 − α ∑ j = 0 r k − j ( 1 − α ) ≤ 2 1 − α 1 − 2 α − 1 M α o f ( x ) . (6) \begin{aligned}
\frac{1}{|B(x,r)|^{1-\alpha}}\sum_{B(x,r)}|f|
=\frac{1}{|B(x,r)|^{1-\alpha}}\sum_{\ell=0}^{r}\sum_{S(x,\ell)}|f|
\le M^\circ_\alpha f(x)\frac{\sum_{\ell=0}^{r}|S(x,\ell)|^{1-\alpha}}{|B(x,r)|^{1-\alpha}} \\
&\le M_{\alpha}^{o}f(x)2^{1-\alpha}\sum_{j=0}^{r}k^{-j(1-\alpha)} \\
&\le\frac{2^{1-\alpha}}{1-2^{\alpha-1}}M_{\alpha}^{o}f(x). \tag*{(6)}
\end{aligned} ∣ B ( x , r ) ∣ 1 − α 1 B ( x , r ) ∑ ∣ f ∣ = ∣ B ( x , r ) ∣ 1 − α 1 ℓ = 0 ∑ r S ( x , ℓ ) ∑ ∣ f ∣ ≤ M α ∘ f ( x ) ∣ B ( x , r ) ∣ 1 − α ∑ ℓ = 0 r ∣ S ( x , ℓ ) ∣ 1 − α ≤ M α o f ( x ) 2 1 − α j = 0 ∑ r k − j ( 1 − α ) ≤ 1 − 2 α − 1 2 1 − α M α o f ( x ) . (6) This proves the first assertion.
Finally, for each r r r
∣ S ( x , r ) ∣ α − 1 / s ( ∑ S ( x , r ) ∣ f ∣ s ) 1 / s ≤ ( ∣ S ( x , r ) ∣ ∣ B ( x , r ) ∣ ) α − 1 / s ∣ B ( x , r ) ∣ α − 1 / s ( ∑ B ( x , r ) ∣ f ∣ s ) 1 / s . |S(x,r)|^{\alpha-1/s}\left(\sum_{S(x,r)}|f|^{s}\right)^{1/s}\le\left(\frac{|S(x,r)|}{|B(x,r)|}\right)^{\alpha-1/s}|B(x,r)|^{\alpha-1/s}\left(\sum_{B(x,r)}|f|^{s}\right)^{1/s}. ∣ S ( x , r ) ∣ α − 1/ s S ( x , r ) ∑ ∣ f ∣ s 1/ s ≤ ( ∣ B ( x , r ) ∣ ∣ S ( x , r ) ∣ ) α − 1/ s ∣ B ( x , r ) ∣ α − 1/ s B ( x , r ) ∑ ∣ f ∣ s 1/ s . If α − 1 / s ≥ 0 \alpha-1/s\ge0 α − 1/ s ≥ 0 α − 1 / s < 0 \alpha-1/s<0 α − 1/ s < 0 2 1 / s − α 2^{1/s-\alpha} 2 1/ s − α
Assume now that α s < 1 \alpha s<1 α s < 1 0 ≤ ℓ ≤ r 0\le\ell\le r 0 ≤ ℓ ≤ r
( ∑ S ( x , ℓ ) ∣ f ∣ s ) 1 / s ≤ M α , s o f ( x ) ∣ S ( x , ℓ ) ∣ 1 / s − α . \left(\sum_{S(x,\ell)}|f|^{s}\right)^{1/s}\le M_{\alpha,s}^{o}f(x)|S(x,\ell)|^{1/s-\alpha}. S ( x , ℓ ) ∑ ∣ f ∣ s 1/ s ≤ M α , s o f ( x ) ∣ S ( x , ℓ ) ∣ 1/ s − α . Therefore, by (2.1),
∣ B ( x , r ) ∣ α − 1 / s ( ∑ B ( x , r ) ∣ f ∣ s ) 1 / s ≤ M α , s o f ( x ) ∣ B ( x , r ) ∣ α − 1 / s ( ∑ ℓ = 0 r ∣ S ( x , ℓ ) ∣ 1 − α s ) 1 / s |B(x,r)|^{\alpha-1/s}\left(\sum_{B(x,r)}|f|^{s}\right)^{1/s}\le M_{\alpha,s}^{o}f(x)|B(x,r)|^{\alpha-1/s}\left(\sum_{\ell=0}^{r}|S(x,\ell)|^{1-\alpha s}\right)^{1/s} ∣ B ( x , r ) ∣ α − 1/ s B ( x , r ) ∑ ∣ f ∣ s 1/ s ≤ M α , s o f ( x ) ∣ B ( x , r ) ∣ α − 1/ s ( ℓ = 0 ∑ r ∣ S ( x , ℓ ) ∣ 1 − α s ) 1/ s ≲ α , s M α , s o f ( x ) k r ( α − 1 / s ) k r ( 1 / s − α ) ≲ α , s M α , s o f ( x ) . \lesssim_{\alpha,s}M_{\alpha,s}^{o}f(x)k^{r(\alpha-1/s)}k^{r(1/s-\alpha)}\lesssim_{\alpha,s}M_{\alpha,s}^{o}f(x). ≲ α , s M α , s o f ( x ) k r ( α − 1/ s ) k r ( 1/ s − α ) ≲ α , s M α , s o f ( x ) . For the geometric sum to converge, we need α s < 1 \alpha s<1 α s < 1
Remark 2.2. At α = 0 \alpha=0 α = 0 ∣ S ( x , ℓ ) ∣ 1 − α |S(x,\ell)|^{1-\alpha} ∣ S ( x , ℓ ) ∣ 1 − α
We will denote by 1 ⊗ w 1\otimes w 1 ⊗ w
1 ⊗ w ( A × B ) = ∣ A ∣ w ( B ) = ∑ ( x , y ) ∈ A × B w ( y ) . 1\otimes w(A\times B)=|A|w(B)=\sum_{(x,y)\in A\times B}w(y). 1 ⊗ w ( A × B ) = ∣ A ∣ w ( B ) = ( x , y ) ∈ A × B ∑ w ( y ) . Lemma 2.3. Let E , F E,F E , F T T T s > 1 s>1 s > 1 r ∈ N 0 r\in\mathbb{N}_{0} r ∈ N 0
1 ⊗ w ( { ( x , y ) ∈ E × F : d ( x , y ) = r } ) ≤ C s k s ′ s ′ + 1 ( 1 − α ) w ( F ) 1 s ′ + 1 ( M α , s o w ( E ) ) s ′ s ′ + 1 , 1\otimes w\left(\{(x,y)\in E\times F:d(x,y)=r\}\right)\le C_{s}k^{\frac{s'}{s'+1}(1-\alpha)}w(F)^{\frac{1}{s'+1}}\left(M_{\alpha,s}^{o}w(E)\right)^{\frac{s'}{s'+1}}, 1 ⊗ w ( {( x , y ) ∈ E × F : d ( x , y ) = r } ) ≤ C s k s ′ + 1 s ′ ( 1 − α ) w ( F ) s ′ + 1 1 ( M α , s o w ( E ) ) s ′ + 1 s ′ , where s ′ = s / ( s − 1 ) s'=s/(s-1) s ′ = s / ( s − 1 ) C s C_s C s k k k
Proof. Write T = ⋃ j ≥ 0 T j T=\bigcup_{j\ge0}T^j T = ⋃ j ≥ 0 T j T j T^j T j j j j i , j ≥ 0 i,j\ge0 i , j ≥ 0 E j = E ∩ T j E_j=E\cap T^j E j = E ∩ T j F i = F ∩ T i F_i=F\cap T^i F i = F ∩ T i
I i , j = 1 ⊗ w ( { ( x , y ) ∈ E j × F i : d ( x , y ) = r } ) I_{i,j}=1\otimes w\left(\{(x,y)\in E_j\times F_i:d(x,y)=r\}\right) I i , j = 1 ⊗ w ( {( x , y ) ∈ E j × F i : d ( x , y ) = r } ) If x ∈ E j x\in E_j x ∈ E j y ∈ F i y\in F_i y ∈ F i d ( x , y ) = r d(x,y)=r d ( x , y ) = r
for some m ∈ { 0 , … , r } m\in\{0,\ldots,r\} m ∈ { 0 , … , r } I i , j = 0 I_{i,j}=0 I i , j = 0 i ≠ j + r − 2 m i\ne j+r-2m i = j + r − 2 m m ∈ { 0 , … , r } m\in\{0,\ldots,r\} m ∈ { 0 , … , r }
1 ⊗ w ( { ( x , y ) ∈ E × F : d ( x , y ) = r } ) = ∑ m = 0 r ∑ i , j ∈ N 0 i = j + r − 2 m I i , j . 1\otimes w\left(\{(x,y)\in E\times F:d(x,y)=r\}\right)=\sum_{m=0}^{r}\sum_{\substack{i,j\in\mathbb{N}_{0}\\i=j+r-2m}}I_{i,j}. 1 ⊗ w ( {( x , y ) ∈ E × F : d ( x , y ) = r } ) = m = 0 ∑ r i , j ∈ N 0 i = j + r − 2 m ∑ I i , j . For fixed y ∈ T i y\in T^i y ∈ T i k m k^m k m x ∈ T j x\in T^j x ∈ T j r r r
I i , j ≤ k m w ( F i ) . (7) I_{i,j}\le k^m w(F_i). \tag*{(7)} I i , j ≤ k m w ( F i ) . (7) On the other hand, for fixed x ∈ T j x \in T^j x ∈ T j k r − m k^{r-m} k r − m T i T^i T i r r r
I i , j = ∑ x ∈ E j ∑ y ∈ F i d ( x , y ) = r w ( y ) = ∑ x ∈ E j w ( F i ∩ S ( x , r ) ) ≤ ∑ x ∈ E j ∣ F i ∩ S ( x , r ) ∣ 1 s ′ w s ( F i ∩ S ( x , r ) ) 1 s = ∑ x ∈ E j ∣ F i ∩ S ( x , r ) ∣ 1 s ′ k r ( 1 s − α ) k r ( 1 s − α ) w s ( F i ∩ S ( x , r ) ) 1 s ≤ C α , s ∑ x ∈ E j ∣ F i ∩ S ( x , r ) ∣ 1 s ′ k r ( 1 s − α ) ∣ S ( x , r ) ∣ α w s ( F i ∩ S ( x , r ) ) 1 s ∣ S ( x , r ) ∣ 1 s ≤ C α , s ∑ x ∈ E j ∣ F i ∩ S ( x , r ) ∣ 1 s ′ k r ( 1 s − α ) M α , s ∘ ( w ) ( x ) ≤ C α , s k r − m s ′ k r ( 1 s − α ) M α , s ∘ ( w ) ( E j ) . \begin{aligned}
I_{i,j} &= \sum_{x\in E_j}\sum_{\substack{y\in F_i\\d(x,y)=r}} w(y) \\
&= \sum_{x\in E_j} w(F_i\cap S(x,r)) \\
&\le\sum_{x\in E_j} |F_i\cap S(x,r)|^{\frac{1}{s'}}w^s(F_i\cap S(x,r))^{\frac{1}{s}} \\
&= \sum_{x\in E_j} |F_i\cap S(x,r)|^{\frac{1}{s'}}\frac{k^{r(\frac{1}{s}-\alpha)}}{k^{r(\frac{1}{s}-\alpha)}}w^s(F_i\cap S(x,r))^{\frac{1}{s}} \\
&\le C_{\alpha,s}\sum_{x\in E_j}|F_i\cap S(x,r)|^{\frac{1}{s'}}k^{r(\frac{1}{s}-\alpha)}|S(x,r)|^{\alpha}\frac{w^s(F_i\cap S(x,r))^{\frac{1}{s}}}{|S(x,r)|^{\frac{1}{s}}} \\
&\le C_{\alpha,s}\sum_{x\in E_j}|F_i\cap S(x,r)|^{\frac{1}{s'}}k^{r(\frac{1}{s}-\alpha)}M_{\alpha,s}^{\circ}(w)(x) \\
&\le C_{\alpha,s}k^{\frac{r-m}{s'}}k^{r(\frac{1}{s}-\alpha)}M_{\alpha,s}^{\circ}(w)(E_j).
\end{aligned} I i , j = x ∈ E j ∑ y ∈ F i d ( x , y ) = r ∑ w ( y ) = x ∈ E j ∑ w ( F i ∩ S ( x , r )) ≤ x ∈ E j ∑ ∣ F i ∩ S ( x , r ) ∣ s ′ 1 w s ( F i ∩ S ( x , r ) ) s 1 = x ∈ E j ∑ ∣ F i ∩ S ( x , r ) ∣ s ′ 1 k r ( s 1 − α ) k r ( s 1 − α ) w s ( F i ∩ S ( x , r ) ) s 1 ≤ C α , s x ∈ E j ∑ ∣ F i ∩ S ( x , r ) ∣ s ′ 1 k r ( s 1 − α ) ∣ S ( x , r ) ∣ α ∣ S ( x , r ) ∣ s 1 w s ( F i ∩ S ( x , r ) ) s 1 ≤ C α , s x ∈ E j ∑ ∣ F i ∩ S ( x , r ) ∣ s ′ 1 k r ( s 1 − α ) M α , s ∘ ( w ) ( x ) ≤ C α , s k s ′ r − m k r ( s 1 − α ) M α , s ∘ ( w ) ( E j ) . We therefore have
I i , j ≤ C α , s min { k r − m s ′ k r ( 1 s − α ) M α , s ∘ w ( E j ) , k m w ( F i ) } . (8) I_{i,j}\le C_{\alpha,s}\min\left\{k^{\frac{r-m}{s'}}k^{r(\frac{1}{s}-\alpha)}M_{\alpha,s}^{\circ}w(E_j),k^m w(F_i)\right\}. \tag*{(8)} I i , j ≤ C α , s min { k s ′ r − m k r ( s 1 − α ) M α , s ∘ w ( E j ) , k m w ( F i ) } . (8) Set
c j = k − j / s ′ M α , s ∘ w ( E j ) , d i = k − i w ( F i ) , i , j ≥ 0. c_j=k^{-j/s'}M_{\alpha,s}^{\circ}w(E_j),\qquad d_i=k^{-i}w(F_i),\qquad i,j\ge0. c j = k − j / s ′ M α , s ∘ w ( E j ) , d i = k − i w ( F i ) , i , j ≥ 0. Then
∑ j ≥ 0 k j / s ′ c j = M α , s ∘ w ( E ) , ∑ i ≥ 0 k i d i = w ( F ) . \sum_{j\ge0}k^{j/s'}c_j=M_{\alpha,s}^{\circ}w(E),\qquad\sum_{i\ge0}k^i d_i=w(F). j ≥ 0 ∑ k j / s ′ c j = M α , s ∘ w ( E ) , i ≥ 0 ∑ k i d i = w ( F ) . Since i = j + r − 2 m i=j+r-2m i = j + r − 2 m
r − m + j = r + i + j 2 , m + i = r + i + j 2 . r-m+j=\frac{r+i+j}{2},\qquad m+i=\frac{r+i+j}{2}. r − m + j = 2 r + i + j , m + i = 2 r + i + j . Since i = j + r − 2 m i=j+r-2m i = j + r − 2 m
min { k r − m s ′ k r ( 1 s − α ) M α , s ∘ w ( E j ) , k m w ( F i ) } = min { k r − m s ′ k r ( 1 s − α ) k j / s ′ c j , k m + i d i } = min { k r − m + j s ′ k r ( 1 s − α ) c j , k m + i d i } . (9) \begin{aligned}
\min\left\{k^{\frac{r-m}{s'}}k^{r(\frac{1}{s}-\alpha)}M_{\alpha,s}^{\circ}w(E_j),k^m w(F_i)\right\}
&=\min\left\{k^{\frac{r-m}{s'}}k^{r(\frac{1}{s}-\alpha)}k^{j/s'}c_j,k^{m+i}d_i\right\}\\
&=\min\left\{k^{\frac{r-m+j}{s'}}k^{r(\frac{1}{s}-\alpha)}c_j,k^{m+i}d_i\right\}. \tag*{(9)}
\end{aligned} min { k s ′ r − m k r ( s 1 − α ) M α , s ∘ w ( E j ) , k m w ( F i ) } = min { k s ′ r − m k r ( s 1 − α ) k j / s ′ c j , k m + i d i } = min { k s ′ r − m + j k r ( s 1 − α ) c j , k m + i d i } . (9) Since i = j + r − 2 m i=j+r-2m i = j + r − 2 m m = 1 2 ( j − i + r ) m=\frac{1}{2}(j-i+r) m = 2 1 ( j − i + r )
r − m + j s ′ = 1 s ′ [ r − 1 2 ( j − i + r ) + j ] = r + j + i 2 s ′ . \frac{r-m+j}{s'}=\frac{1}{s'}\left[r-\frac{1}{2}(j-i+r)+j\right]=\frac{r+j+i}{2s'}. s ′ r − m + j = s ′ 1 [ r − 2 1 ( j − i + r ) + j ] = 2 s ′ r + j + i . Then, for a θ \theta θ
∑ m = 0 r ∑ i , j ∈ N ∪ { 0 } i = j + r − 2 m min { k r − m + j s ′ k r ( 1 s − α ) c j , k m + i d i } ≤ ∑ m = 0 r ∑ i , j ∈ N ∪ { 0 } i = j + r − 2 m min { k r 2 s ′ k i + j 2 s ′ k r ( 1 s − α ) c j , k r 2 k i + j 2 d i } \sum_{m=0}^{r}\sum_{\substack{i,j\in\mathbb{N}\cup\{0\}\\i=j+r-2m}}\min\left\{k^{\frac{r-m+j}{s'}}k^{r(\frac{1}{s}-\alpha)}c_j,k^{m+i}d_i\right\}
\le\sum_{m=0}^{r}\sum_{\substack{i,j\in\mathbb{N}\cup\{0\}\\i=j+r-2m}}\min\left\{k^{\frac{r}{2s'}}k^{\frac{i+j}{2s'}}k^{r(\frac{1}{s}-\alpha)}c_j,k^{\frac{r}{2}}k^{\frac{i+j}{2}}d_i\right\} m = 0 ∑ r i , j ∈ N ∪ { 0 } i = j + r − 2 m ∑ min { k s ′ r − m + j k r ( s 1 − α ) c j , k m + i d i } ≤ m = 0 ∑ r i , j ∈ N ∪ { 0 } i = j + r − 2 m ∑ min { k 2 s ′ r k 2 s ′ i + j k r ( s 1 − α ) c j , k 2 r k 2 i + j d i } ≤ ∑ i , j ∈ N 0 i < j + θ k r 2 s ′ k i + j 2 s ′ k r ( 1 s − α ) c j + ∑ i , j ∈ N 0 i ≥ j + θ k r 2 k i + j 2 d i = k r 2 s ′ k r ( 1 s − α ) ∑ i , j ∈ N 0 i < j + θ k i + j 2 s ′ c j + k r 2 ∑ i , j ∈ N 0 i ≥ j + θ k i + j 2 d i = k r 2 s ′ k r ( 1 s − α ) ∑ j = 0 ∞ k j 2 s ′ c j ∑ i ∈ N 0 i < j + θ k i 2 s ′ + k r 2 ∑ i = 0 ∞ k i 2 d i ∑ j ∈ N 0 j ≤ i − θ k j 2 ≤ C s [ k r 2 s ′ k r ( 1 s − α ) ∑ j = 0 ∞ k j 2 s ′ c j k j + θ 2 s ′ + k r 2 ∑ i = 0 ∞ k i 2 d i k i − θ 2 ] ≤ C s [ k r 2 s ′ k r ( 1 s − α ) k θ 2 s ′ ∑ j = 0 ∞ k j s ′ c j + k r 2 k − θ 2 ∑ i = 0 ∞ k i d i ] ≤ C s [ k r 2 s ′ k r ( 1 s − α ) k θ 2 s ′ M α , s o w ( E ) + k r 2 k − θ 2 w ( F ) ] . \begin{aligned}
&\le\sum_{\substack{i,j\in\mathbb{N}_0\\i<j+\theta}} k^{\frac{r}{2s'}}k^{\frac{i+j}{2s'}}k^{r(\frac{1}{s}-\alpha)}c_j+\sum_{\substack{i,j\in\mathbb{N}_0\\i\ge j+\theta}}k^{\frac{r}{2}}k^{\frac{i+j}{2}}d_i\\
&=k^{\frac{r}{2s'}}k^{r(\frac{1}{s}-\alpha)}\sum_{\substack{i,j\in\mathbb{N}_0\\i<j+\theta}}k^{\frac{i+j}{2s'}}c_j+k^{\frac{r}{2}}\sum_{\substack{i,j\in\mathbb{N}_0\\i\ge j+\theta}}k^{\frac{i+j}{2}}d_i\\
&=k^{\frac{r}{2s'}}k^{r(\frac{1}{s}-\alpha)}\sum_{j=0}^{\infty}k^{\frac{j}{2s'}}c_j\sum_{\substack{i\in\mathbb{N}_0\\i<j+\theta}}k^{\frac{i}{2s'}}+k^{\frac{r}{2}}\sum_{i=0}^{\infty}k^{\frac{i}{2}}d_i\sum_{\substack{j\in\mathbb{N}_0\\j\le i-\theta}}k^{\frac{j}{2}}\\
&\le C_s\left[k^{\frac{r}{2s'}}k^{r(\frac{1}{s}-\alpha)}\sum_{j=0}^{\infty}k^{\frac{j}{2s'}}c_jk^{\frac{j+\theta}{2s'}}+k^{\frac{r}{2}}\sum_{i=0}^{\infty}k^{\frac{i}{2}}d_ik^{\frac{i-\theta}{2}}\right]\\
&\le C_s\left[k^{\frac{r}{2s'}}k^{r(\frac{1}{s}-\alpha)}k^{\frac{\theta}{2s'}}\sum_{j=0}^{\infty}k^{\frac{j}{s'}}c_j+k^{\frac{r}{2}}k^{-\frac{\theta}{2}}\sum_{i=0}^{\infty}k^id_i\right]\\
&\le C_s\left[k^{\frac{r}{2s'}}k^{r(\frac{1}{s}-\alpha)}k^{\frac{\theta}{2s'}}M_{\alpha,s}^{o}w(E)+k^{\frac{r}{2}}k^{-\frac{\theta}{2}}w(F)\right].
\end{aligned} ≤ i , j ∈ N 0 i < j + θ ∑ k 2 s ′ r k 2 s ′ i + j k r ( s 1 − α ) c j + i , j ∈ N 0 i ≥ j + θ ∑ k 2 r k 2 i + j d i = k 2 s ′ r k r ( s 1 − α ) i , j ∈ N 0 i < j + θ ∑ k 2 s ′ i + j c j + k 2 r i , j ∈ N 0 i ≥ j + θ ∑ k 2 i + j d i = k 2 s ′ r k r ( s 1 − α ) j = 0 ∑ ∞ k 2 s ′ j c j i ∈ N 0 i < j + θ ∑ k 2 s ′ i + k 2 r i = 0 ∑ ∞ k 2 i d i j ∈ N 0 j ≤ i − θ ∑ k 2 j ≤ C s [ k 2 s ′ r k r ( s 1 − α ) j = 0 ∑ ∞ k 2 s ′ j c j k 2 s ′ j + θ + k 2 r i = 0 ∑ ∞ k 2 i d i k 2 i − θ ] ≤ C s [ k 2 s ′ r k r ( s 1 − α ) k 2 s ′ θ j = 0 ∑ ∞ k s ′ j c j + k 2 r k − 2 θ i = 0 ∑ ∞ k i d i ] ≤ C s [ k 2 s ′ r k r ( s 1 − α ) k 2 s ′ θ M α , s o w ( E ) + k 2 r k − 2 θ w ( F ) ] . where C s = 1 2 1 2 s ′ − 1 C_s=\dfrac{1}{2^{\frac{1}{2s'}}-1} C s = 2 2 s ′ 1 − 1 1
If both terms are nonzero, choose θ \theta θ
A = k r 2 s ′ k r ( 1 s − α ) M α , s o w ( E ) , B = k r 2 w ( F ) , A=k^{\frac{r}{2s'}}k^{r(\frac{1}{s}-\alpha)}M_{\alpha,s}^{o}w(E),\qquad B=k^{\frac{r}{2}}w(F), A = k 2 s ′ r k r ( s 1 − α ) M α , s o w ( E ) , B = k 2 r w ( F ) , choose θ \theta θ A k θ / ( 2 s ′ ) = B k − θ / 2 Ak^{\theta/(2s')}=Bk^{-\theta/2} A k θ / ( 2 s ′ ) = B k − θ /2
A k θ / ( 2 s ′ ) + B k − θ / 2 = 2 A s ′ s ′ + 1 B 1 s ′ + 1 . Ak^{\theta/(2s')}+Bk^{-\theta/2}=2A^{\frac{s'}{s'+1}}B^{\frac{1}{s'+1}}. A k θ / ( 2 s ′ ) + B k − θ /2 = 2 A s ′ + 1 s ′ B s ′ + 1 1 . The cases A = 0 A=0 A = 0 B = 0 B=0 B = 0
1 ⊗ w ( { ( x , y ) ∈ E × F : d ( x , y ) = r } ) ≤ C s k r s ′ + 1 k r ( 1 s − α ) s ′ s ′ + 1 ( M α , s o w ( E ) ) s ′ s ′ + 1 w ( F ) 1 s ′ + 1 . \mathbf{1}\otimes w\bigl(\{(x,y)\in E\times F:d(x,y)=r\}\bigr)
\le C_sk^{\frac{r}{s'+1}}k^{r(\frac{1}{s}-\alpha)\frac{s'}{s'+1}}\bigl(M_{\alpha,s}^{o}w(E)\bigr)^{\frac{s'}{s'+1}}w(F)^{\frac{1}{s'+1}}. 1 ⊗ w ( {( x , y ) ∈ E × F : d ( x , y ) = r } ) ≤ C s k s ′ + 1 r k r ( s 1 − α ) s ′ + 1 s ′ ( M α , s o w ( E ) ) s ′ + 1 s ′ w ( F ) s ′ + 1 1 . Since
1 s ′ + 1 + ( 1 s − α ) s ′ s ′ + 1 = s ′ s ′ + 1 ( 1 − α ) , \frac{1}{s'+1}+\left(\frac{1}{s}-\alpha\right)\frac{s'}{s'+1}
=\frac{s'}{s'+1}(1-\alpha), s ′ + 1 1 + ( s 1 − α ) s ′ + 1 s ′ = s ′ + 1 s ′ ( 1 − α ) , we obtain (2.4). □
Remark 2.4. The exponent r s ′ s ′ + 1 ( 1 − α ) r^{\frac{s'}{s'+1}}(1-\alpha) r s ′ + 1 s ′ ( 1 − α ) [9 ] . It comes from the factor k r ( 1 / s − α ) k^{r(1/s-\alpha)} k r ( 1/ s − α ) L s L^s L s M α , s o w M_{\alpha,s}^{o}w M α , s o w k r α k^{r\alpha} k r α
For 0 ≤ α < 1 0\le\alpha<1 0 ≤ α < 1 r ≥ 0 r\ge0 r ≥ 0
A r , α o h ( x ) = 1 ∣ S ( x , r ) ∣ 1 − α ∑ y ∈ S ( x , r ) ∣ h ( y ) ∣ . A_{r,\alpha}^{o}h(x)=\frac{1}{|S(x,r)|^{1-\alpha}}\sum_{y\in S(x,r)}|h(y)|. A r , α o h ( x ) = ∣ S ( x , r ) ∣ 1 − α 1 y ∈ S ( x , r ) ∑ ∣ h ( y ) ∣. Thus M α o h = sup r ≥ 0 A r , α o h M_{\alpha}^{o}h=\sup_{r\ge0}A_{r,\alpha}^{o}h M α o h = sup r ≥ 0 A r , α o h α = 0 \alpha=0 α = 0 A r o h ( x ) = A r , 0 o h ( x ) A_r^{o}h(x)=A_{r,0}^{o}h(x) A r o h ( x ) = A r , 0 o h ( x )
Lemma 2.5. Let 0 < α < 1 0<\alpha<1 0 < α < 1 s > 1 s>1 s > 1 r ≥ 1 r\ge1 r ≥ 1 λ > 0 \lambda>0 λ > 0 f f f w w w
w ( { x : A r , α o f ( x ) > λ } ) ≤ C α , s ∑ n ∈ N 0 1 ≤ 2 n < k r ( 2 n k r ) 1 + 1 2 s ′ k r ( 1 − α ) M α , s o w ( { x : ∣ f ( x ) ∣ k r α ≥ c α 2 n − 1 λ } ) , (10) w\bigl(\{x:A_{r,\alpha}^{o}f(x)>\lambda\}\bigr)
\le C_{\alpha,s}\sum_{\substack{n\in\mathbb{N}_0\\1\le2^n<k^r}}\left(\frac{2^n}{k^r}\right)^{1+\frac{1}{2s'}}k^{r(1-\alpha)}M_{\alpha,s}^{o}w\bigl(\{x:|f(x)|k^{r\alpha}\ge c_\alpha2^{n-1}\lambda\}\bigr), \tag*{(10)} w ( { x : A r , α o f ( x ) > λ } ) ≤ C α , s n ∈ N 0 1 ≤ 2 n < k r ∑ ( k r 2 n ) 1 + 2 s ′ 1 k r ( 1 − α ) M α , s o w ( { x : ∣ f ( x ) ∣ k r α ≥ c α 2 n − 1 λ } ) , (10) where c α c_\alpha c α α \alpha α C α , s C_{\alpha,s} C α , s α \alpha α s s s C α , s → ∞ C_{\alpha,s}\to\infty C α , s → ∞ s → 1 s\to1 s → 1
Proof. By (3) ,
A r , α o f ( x ) = ∣ S ( x , r ) ∣ α A r o ( ∣ f ∣ ) ( x ) . A_{r,\alpha}^{o}f(x)=|S(x,r)|^{\alpha}A_{r}^{o}(|f|)(x). A r , α o f ( x ) = ∣ S ( x , r ) ∣ α A r o ( ∣ f ∣ ) ( x ) . Hence
A r , α o f ( x ) > λ ⟹ A r o g ( x ) > 1 , g = 2 α k r α λ ∣ f ∣ . A_{r,\alpha}^{o}f(x)>\lambda\quad\Longrightarrow\quad A_{r}^{o}g(x)>1,\qquad g=\frac{2^{\alpha}k^{r\alpha}}{\lambda}|f|. A r , α o f ( x ) > λ ⟹ A r o g ( x ) > 1 , g = λ 2 α k r α ∣ f ∣. We estimate the latter level set.
Let N r N_r N r 2 N r ≤ k r 2^{N_r}\le k^r 2 N r ≤ k r
E n = { x : 2 n − 1 ≤ g ( x ) < 2 n } , 0 ≤ n ≤ N r . E_n=\{x:2^{n-1}\le g(x)<2^n\},\qquad0\le n\le N_r. E n = { x : 2 n − 1 ≤ g ( x ) < 2 n } , 0 ≤ n ≤ N r . Then
g ≤ 1 2 + ∑ n = 0 N r 2 n χ E n + g χ { g ≥ k r / 2 } . (11) g\le\frac{1}{2}+\sum_{n=0}^{N_r}2^n\chi_{E_n}+g\chi_{\{g\ge k^r/2\}}. \tag*{(11)} g ≤ 2 1 + n = 0 ∑ N r 2 n χ E n + g χ { g ≥ k r /2 } . (11) Consequently,
{ A r o g > 1 } ⊂ { ∑ n = 0 N r 2 n A r o ( χ E n ) ≥ 1 2 } ∪ { A r o ( g χ { g ≥ k r / 2 } ) ≠ 0 } . (12) \{A_r^og>1\}\subset\left\{\sum_{n=0}^{N_r}2^nA_r^o(\chi_{E_n})\ge\frac{1}{2}\right\}\cup\left\{A_r^o(g\chi_{\{g\ge k^r/2\}})\ne0\right\}. \tag*{(12)} { A r o g > 1 } ⊂ { n = 0 ∑ N r 2 n A r o ( χ E n ) ≥ 2 1 } ∪ { A r o ( g χ { g ≥ k r /2 } ) = 0 } . (12) Let 0 < β < 1 0<\beta<1 0 < β < 1
∑ n = 0 N r 2 n A r o ( χ E n ) ( x ) ≥ 1 2 , \sum_{n=0}^{N_r}2^nA_r^o(\chi_{E_n})(x)\ge\frac{1}{2}, n = 0 ∑ N r 2 n A r o ( χ E n ) ( x ) ≥ 2 1 , then for some 0 ≤ n ≤ N r 0\le n\le N_r 0 ≤ n ≤ N r
A r o ( χ E n ) ( x ) ≥ 2 β − 1 2 n + 2 ( 2 n k r ) β . (13) A_r^o(\chi_{E_n})(x)\ge\frac{2^\beta-1}{2^{n+2}}\left(\frac{2^n}{k^r}\right)^\beta. \tag*{(13)} A r o ( χ E n ) ( x ) ≥ 2 n + 2 2 β − 1 ( k r 2 n ) β . (13) Indeed, if (13) failed for all n n n
∑ n = 0 N r 2 n A r o ( χ E n ) ( x ) < 2 β − 1 4 k r β ∑ n = 0 N r 2 β n < 2 β ( N r + 1 ) 4 k r β ≤ 2 β 4 < 1 2 , \begin{aligned}
\sum_{n=0}^{N_r}2^nA_r^o(\chi_{E_n})(x)<\frac{2^\beta-1}{4k^{r\beta}}\sum_{n=0}^{N_r}2^{\beta n} \\
&<\frac{2^{\beta(N_r+1)}}{4k^{r\beta}}\le\frac{2^\beta}{4}<\frac{1}{2},
\end{aligned} n = 0 ∑ N r 2 n A r o ( χ E n ) ( x ) < 4 k r β 2 β − 1 n = 0 ∑ N r 2 β n < 4 k r β 2 β ( N r + 1 ) ≤ 4 2 β < 2 1 , a contradiction.
Define
F n = { x : A r o ( χ E n ) ( x ) ≥ τ n } , τ n = 2 β − 1 2 n + 2 ( 2 n k r ) β . F_n=\{x:A_r^o(\chi_{E_n})(x)\ge\tau_n\},\qquad\tau_n=\frac{2^\beta-1}{2^{n+2}}\left(\frac{2^n}{k^r}\right)^\beta. F n = { x : A r o ( χ E n ) ( x ) ≥ τ n } , τ n = 2 n + 2 2 β − 1 ( k r 2 n ) β . Since f f f E n E_n E n F n F_n F n y ∈ F n y\in F_n y ∈ F n
∣ E n ∩ S ( y , r ) ∣ ≥ τ n ∣ S ( y , r ) ∣ ≥ τ n k r . |E_n\cap S(y,r)|\ge\tau_n|S(y,r)|\ge\tau_n k^r. ∣ E n ∩ S ( y , r ) ∣ ≥ τ n ∣ S ( y , r ) ∣ ≥ τ n k r . Therefore
1 ⊗ w ( { ( x , y ) ∈ E n × F n : d ( x , y ) = r } ) ≥ τ n k r w ( F n ) . (14) \mathbb{1}\otimes w(\{(x,y)\in E_n\times F_n:d(x,y)=r\})\ge\tau_nk^rw(F_n). \tag*{(14)} 1 ⊗ w ({( x , y ) ∈ E n × F n : d ( x , y ) = r }) ≥ τ n k r w ( F n ) . (14) By Lemma 2.3,
1 ⊗ w ( { ( x , y ) ∈ E n × F n : d ( x , y ) = r } ) ≤ C s k r s ′ s ′ + 1 ( 1 − α ) w ( F n ) 1 s ′ + 1 ( M α , s o w ( E n ) ) s ′ s ′ + 1 . (15) \mathbb{1}\otimes w(\{(x,y)\in E_n\times F_n:d(x,y)=r\})
\le C_sk^{\frac{rs'}{s'+1}}(1-\alpha)w(F_n)^{\frac{1}{s'+1}}(M_{\alpha,s}^{o}w(E_n))^{\frac{s'}{s'+1}}. \tag*{(15)} 1 ⊗ w ({( x , y ) ∈ E n × F n : d ( x , y ) = r }) ≤ C s k s ′ + 1 r s ′ ( 1 − α ) w ( F n ) s ′ + 1 1 ( M α , s o w ( E n ) ) s ′ + 1 s ′ . (15) Combining (14) and (2.13, and raising to the power ( s ′ + 1 ) / s ′ (s'+1)/s' ( s ′ + 1 ) / s ′
w ( F n ) ≤ C s , β ( 2 n k r ) ( 1 − β ) s ′ s ′ + 1 k r ( 1 − α ) M α , s o w ( E n ) . (16) w(F_n)\le C_{s,\beta}\left(\frac{2^n}{k^r}\right)^{\frac{(1-\beta)s'}{s'+1}}k^{r(1-\alpha)}M_{\alpha,s}^{o}w(E_n). \tag*{(16)} w ( F n ) ≤ C s , β ( k r 2 n ) s ′ + 1 ( 1 − β ) s ′ k r ( 1 − α ) M α , s o w ( E n ) . (16) Choose
β = 1 2 ( s ′ + 1 ) . \beta=\frac{1}{2(s'+1)}. β = 2 ( s ′ + 1 ) 1 . Then
( 1 − β ) s ′ + 1 s ′ = 1 + 1 2 s ′ , (1-\beta)\frac{s'+1}{s'}=1+\frac{1}{2s'}, ( 1 − β ) s ′ s ′ + 1 = 1 + 2 s ′ 1 , so
w ( F n ) ≤ C s ( 2 n k r ) 1 + 1 2 s ′ k r ( 1 − α ) M α , s , w o ( E n ) ≤ C s ( 2 n k r ) 1 + 1 2 s ′ k r ( 1 − α ) M α , s , w o ( { g ≥ 2 n − 1 } ) (17) w(F_n) \le C_s\left(\frac{2^n}{k^r}\right)^{1+\frac{1}{2s'}} k^{r(1-\alpha)}M_{\alpha,s,w}^o(E_n)
\le C_s\left(\frac{2^n}{k^r}\right)^{1+\frac{1}{2s'}} k^{r(1-\alpha)}M_{\alpha,s,w}^o(\{g\ge2^{n-1}\}) \tag*{(17)} w ( F n ) ≤ C s ( k r 2 n ) 1 + 2 s ′ 1 k r ( 1 − α ) M α , s , w o ( E n ) ≤ C s ( k r 2 n ) 1 + 2 s ′ 1 k r ( 1 − α ) M α , s , w o ({ g ≥ 2 n − 1 }) (17) It remains to control of { A r o ( g χ { g ≥ k r / 2 } ) ≠ 0 } \{A_r^o(g\chi_{\{g\ge k^r/2\}})\ne0\} { A r o ( g χ { g ≥ k r /2 } ) = 0 }
{ A r o ( g χ { g ≥ k r / 2 } ) ≠ 0 } ⊂ ⋃ y ∈ { g ≥ k r / 2 } S ( y , r ) . \{A_r^o(g\chi_{\{g\ge k^r/2\}})\ne0\}\subset\bigcup_{y\in\{g\ge k^r/2\}}S(y,r). { A r o ( g χ { g ≥ k r /2 } ) = 0 } ⊂ y ∈ { g ≥ k r /2 } ⋃ S ( y , r ) . By Hölder’s inequality,
w ( S ( y , r ) ) ≤ ∣ S ( y , r ) ∣ 1 / s ′ ( ∑ z ∈ S ( y , r ) w ( z ) s ) 1 / s = ∣ S ( y , r ) ∣ 1 − α [ ∣ S ( y , r ) ∣ α ( 1 ∣ S ( y , r ) ∣ ∑ z ∈ S ( y , r ) w ( z ) s ) 1 / s ] ≤ C α k r ( 1 − α ) M α , s , w o ( y ) . \begin{aligned}
w(S(y,r))\le|S(y,r)|^{1/s'}\left(\sum_{z\in S(y,r)}w(z)^s\right)^{1/s} \\
&=|S(y,r)|^{1-\alpha}\left[|S(y,r)|^\alpha\left(\frac{1}{|S(y,r)|}\sum_{z\in S(y,r)}w(z)^s\right)^{1/s}\right]
\le C_\alpha k^{r(1-\alpha)}M_{\alpha,s,w}^o(y).
\end{aligned} w ( S ( y , r )) ≤ ∣ S ( y , r ) ∣ 1/ s ′ z ∈ S ( y , r ) ∑ w ( z ) s 1/ s = ∣ S ( y , r ) ∣ 1 − α ∣ S ( y , r ) ∣ α ∣ S ( y , r ) ∣ 1 z ∈ S ( y , r ) ∑ w ( z ) s 1/ s ≤ C α k r ( 1 − α ) M α , s , w o ( y ) . Thus
w ( { A r o ( g χ { g ≥ k r / 2 } ) ≠ 0 } ) ≤ C α k r ( 1 − α ) M α , s , w o ( { g ≥ k r / 2 } ) . (18) w(\{A_r^o(g\chi_{\{g\ge k^r/2\}})\ne0\})\le C_\alpha k^{r(1-\alpha)}M_{\alpha,s,w}^o(\{g\ge k^r/2\}). \tag*{(18)} w ({ A r o ( g χ { g ≥ k r /2 } ) = 0 }) ≤ C α k r ( 1 − α ) M α , s , w o ({ g ≥ k r /2 }) . (18) Now 2 N r ≤ k r < 2 N r + 1 2^{N_r}\le k^r<2^{N_r+1} 2 N r ≤ k r < 2 N r + 1
{ g ≥ k r / 2 } ⊂ { g ≥ 2 N r − 1 } , 2 N r k r > 1 2 . \{g\ge k^r/2\}\subset\{g\ge2^{N_r-1}\},\qquad\frac{2^{N_r}}{k^r}>\frac{1}{2}. { g ≥ k r /2 } ⊂ { g ≥ 2 N r − 1 } , k r 2 N r > 2 1 . Therefore (18) is dominated, up to a constant, by the term n = N r n=N_r n = N r (17) . This is the point where the cutoff 2 n ≤ k r 2^n\le k^r 2 n ≤ k r
Combining this observation with (12) ,
w ( { A r o g > 1 } ) ≤ C α , s ∑ n = 0 N r ( 2 n k r ) 1 + 1 2 s ′ k r ( 1 − α ) M α , s , w o ( { g ≥ 2 n − 1 } ) . w(\{A_r^o g>1\})\le C_{\alpha,s}\sum_{n=0}^{N_r}\left(\frac{2^n}{k^r}\right)^{1+\frac{1}{2s'}}k^{r(1-\alpha)}M_{\alpha,s,w}^o(\{g\ge2^{n-1}\}). w ({ A r o g > 1 }) ≤ C α , s n = 0 ∑ N r ( k r 2 n ) 1 + 2 s ′ 1 k r ( 1 − α ) M α , s , w o ({ g ≥ 2 n − 1 }) . Since g = 2 α k r α ∣ f ∣ / λ g=2^\alpha k^{r\alpha}|f|/\lambda g = 2 α k r α ∣ f ∣/ λ c α = 2 − α c_\alpha=2^{-\alpha} c α = 2 − α (16) , together with β ↓ 0 \beta\downarrow0 β ↓ 0 s ↓ 1 s\downarrow1 s ↓ 1 C α , s → ∞ C_{\alpha,s}\to\infty C α , s → ∞ s ↓ 1 s\downarrow1 s ↓ 1
Remark 2.6. Following the account in [9 ] , the first attempt is to cut the dyadic levels in k r α k^{r\alpha} k r α k r k^r k r k r k^r k r f f f k r α k^{r\alpha} k r α k r k^r k r α = 0 \alpha=0 α = 0
Now, we proceed to the proof of the main result, Theorem 1.1 .
Proof of Theorem 1.1. We first assume that f f f M α o M_\alpha^o M α o M α , s , w o M_{\alpha,s,w}^o M α , s , w o
w ( { M α o f > λ } ) ≤ ∑ r = 0 ∞ w ( { A r , α o f > λ } ) . w(\{M_\alpha^o f>\lambda\})\le\sum_{r=0}^{\infty}w(\{A_{r,\alpha}^o f>\lambda\}). w ({ M α o f > λ }) ≤ r = 0 ∑ ∞ w ({ A r , α o f > λ }) . The radius r = 0 r=0 r = 0
w ( { ∣ f ∣ > λ } ) ≤ 1 λ ∑ x ∈ T ∣ f ( x ) ∣ w ( x ) ≤ 1 λ ∑ x ∈ T ∣ f ( x ) ∣ M α , s , w o ( x ) . w(\{|f|>\lambda\})\le\frac{1}{\lambda}\sum_{x\in T}|f(x)|w(x)\le\frac{1}{\lambda}\sum_{x\in T}|f(x)|M_{\alpha,s,w}^o(x). w ({ ∣ f ∣ > λ }) ≤ λ 1 x ∈ T ∑ ∣ f ( x ) ∣ w ( x ) ≤ λ 1 x ∈ T ∑ ∣ f ( x ) ∣ M α , s , w o ( x ) . For r ≥ 1 r \ge1 r ≥ 1 Lemma 2.5 gives,
w ( { M α o f > λ } ) ≤ C α , s ∑ r ≥ 1 ∑ n ≥ 0 2 n ≤ k r ( 2 n k r ) 1 + 1 2 s ′ k r ( 1 − α ) M α , s o w ( { ∣ f ∣ k r α ≥ c α 2 n − 1 λ } ) + C α λ ∑ ∣ f ∣ M α , s o w . w(\{M_{\alpha}^{o}f > \lambda\}) \le C_{\alpha,s}\sum_{r \ge1}\sum_{\substack{n \ge0 \\ 2^n \le k^r}}\left(\frac{2^n}{k^r}\right)^{1+\frac{1}{2s'}} k^{r(1-\alpha)}M_{\alpha,s}^{o}w(\{|f|k^{r\alpha} \ge c_{\alpha}2^{n-1}\lambda\})+\frac{C_{\alpha}}{\lambda}\sum|f|M_{\alpha,s}^{o}w. w ({ M α o f > λ }) ≤ C α , s r ≥ 1 ∑ n ≥ 0 2 n ≤ k r ∑ ( k r 2 n ) 1 + 2 s ′ 1 k r ( 1 − α ) M α , s o w ({ ∣ f ∣ k r α ≥ c α 2 n − 1 λ }) + λ C α ∑ ∣ f ∣ M α , s o w . Interchanging the sums, fix x ∈ T x \in T x ∈ T
A = C α ∣ f ( x ) ∣ λ , a = 1 + 1 2 s ′ > 1. A=C_{\alpha}\frac{|f(x)|}{\lambda}, \qquad a=1+\frac{1}{2s'}>1. A = C α λ ∣ f ( x ) ∣ , a = 1 + 2 s ′ 1 > 1. The contribution associated with x x x
Φ ( A ) : = ∑ r ≥ 1 k r ( 1 − α ) ∑ n ≥ 0 2 n ≤ k r 2 n ≤ A k r α ( 2 n k r ) a \Phi(A):=\sum_{r \ge1}k^{r(1-\alpha)}\sum_{\substack{n \ge0 \\ 2^n \le k^r \\ 2^n \le Ak^{r\alpha}}}\left(\frac{2^n}{k^r}\right)^a Φ ( A ) := r ≥ 1 ∑ k r ( 1 − α ) n ≥ 0 2 n ≤ k r 2 n ≤ A k r α ∑ ( k r 2 n ) a ≲ s ∑ r ≥ 1 k r ( 1 − α − a ) min { k r , A k r α } a = ∑ r ≥ 1 k r ( 1 − α ) min { 1 , A k − r ( 1 − α ) } a . \begin{aligned}
\lesssim_s\sum_{r \ge1}k^{r(1-\alpha-a)}\min\{k^r,Ak^{r\alpha}\}^a \\
&=\sum_{r \ge1}k^{r(1-\alpha)}\min\{1,Ak^{-r(1-\alpha)}\}^a.
\end{aligned} ≲ s r ≥ 1 ∑ k r ( 1 − α − a ) min { k r , A k r α } a = r ≥ 1 ∑ k r ( 1 − α ) min { 1 , A k − r ( 1 − α ) } a . We claim that
Φ ( A ) ≲ α , s A , A ≥ 0. (19) \Phi(A)\lesssim_{\alpha,s} A,\qquad A\ge0. \tag*{(19)} Φ ( A ) ≲ α , s A , A ≥ 0. (19) with a constant independent of k ≥ 2 k \ge2 k ≥ 2 0 ≤ A ≤ 1 0 \le A \le1 0 ≤ A ≤ 1
Φ ( A ) ≤ A a ∑ r ≥ 1 k − r ( 1 − α ) ( a − 1 ) ≲ α , s A a ≤ A . \Phi(A)\le A^a\sum_{r \ge1}k^{-r(1-\alpha)(a-1)}\lesssim_{\alpha,s} A^a\le A. Φ ( A ) ≤ A a r ≥ 1 ∑ k − r ( 1 − α ) ( a − 1 ) ≲ α , s A a ≤ A . If A > 1 A>1 A > 1 R ∈ N 0 R\in\mathbb{N}_0 R ∈ N 0
k R ( 1 − α ) ≤ A < k ( R + 1 ) ( 1 − α ) . k^{R(1-\alpha)}\le A<k^{(R+1)(1-\alpha)}. k R ( 1 − α ) ≤ A < k ( R + 1 ) ( 1 − α ) . Then
Φ ( A ) ≤ ∑ r ≤ R k r ( 1 − α ) + A a ∑ r > R k − r ( 1 − α ) ( a − 1 ) ≲ α , s A . \Phi(A)\le\sum_{r\le R}k^{r(1-\alpha)}+A^a\sum_{r>R}k^{-r(1-\alpha)(a-1)}\lesssim_{\alpha,s} A. Φ ( A ) ≤ r ≤ R ∑ k r ( 1 − α ) + A a r > R ∑ k − r ( 1 − α ) ( a − 1 ) ≲ α , s A . Since k ≥ 2 k\ge2 k ≥ 2 k k k
It follows that
w ( { M α o f > λ } ) ≤ C α , s 1 λ ∑ x ∈ T ∣ f ( x ) ∣ M α , s o w ( x ) . w(\{M_{\alpha}^{o}f>\lambda\})\le C_{\alpha,s}\frac{1}{\lambda}\sum_{x\in T}|f(x)|M_{\alpha,s}^{o}w(x). w ({ M α o f > λ }) ≤ C α , s λ 1 x ∈ T ∑ ∣ f ( x ) ∣ M α , s o w ( x ) . Using Proposition 2.1 once more, namely M α ≲ α M α o M_{\alpha}\lesssim_{\alpha}M_{\alpha}^{o} M α ≲ α M α o M α , s o w ≤ 2 ( 1 / s − α ) + M α , s w M_{\alpha,s}^{o}w\le2^{(1/s-\alpha)+}M_{\alpha,s}w M α , s o w ≤ 2 ( 1/ s − α ) + M α , s w
For a general f f f ∣ f ∣ |f| ∣ f ∣
As an immediate result of Theorem 1.1 we obtain the following sufficient condition for the weighted weak-type ( 1 , 1 ) (1,1) ( 1 , 1 )
Corollary 2.8. Let k ≥ 2 k \ge2 k ≥ 2 0 < α < 1 0 < \alpha< 1 0 < α < 1 s > 1 s > 1 s > 1 C w > 0 C_w > 0 C w > 0
M α , s w ( x ) ≤ C w w ( x ) , x ∈ T . M_{\alpha,s}w(x) \le C_w w(x), \qquad x \in T. M α , s w ( x ) ≤ C w w ( x ) , x ∈ T . Then
w ( { x ∈ T : M α f ( x ) > λ } ) ≲ α , s , C w 1 λ ∑ x ∈ T ∣ f ( x ) ∣ w ( x ) , w\left(\{x \in T : M_\alpha f(x) > \lambda\}\right) \lesssim_{\alpha,s,C_w} \frac{1}{\lambda}\sum_{x\in T}|f(x)|w(x), w ( { x ∈ T : M α f ( x ) > λ } ) ≲ α , s , C w λ 1 x ∈ T ∑ ∣ f ( x ) ∣ w ( x ) , with a constant independent of k k k
Radial weights, sharpness, and necessary conditions
Let T T T k k k
T = ⋃ j = 0 ∞ T j , T=\bigcup_{j=0}^{\infty}T^j, T = j = 0 ⋃ ∞ T j , where T 0 T^0 T 0 T j T^j T j j j j
w ( x ) = ∑ j = 0 ∞ c j χ T j ( x ) , c j ≥ 0. w(x)=\sum_{j=0}^{\infty}c_j\chi_{T^j}(x), \qquad c_j\ge0. w ( x ) = j = 0 ∑ ∞ c j χ T j ( x ) , c j ≥ 0. Radial weights and the subcritical range
A natural first problem is to identify explicit weights satisfying the fractional A 1 A_1 A 1
M α w ≲ w . M_\alpha w \lesssim w. M α w ≲ w . One would expect the natural response to be that for each 0 ≤ α < 1 0 \le\alpha< 1 0 ≤ α < 1 α \alpha α α = 0 \alpha=0 α = 0 [9 ] . However, the admissible interval of radial powers shrinks as α \alpha α α = 1 / 2 \alpha=1/2 α = 1/2
For β ∈ R \beta\in\mathbb{R} β ∈ R
w β ( x ) = ∑ j = 0 ∞ k j β χ T j ( x ) . w_\beta(x)=\sum_{j=0}^{\infty}k^{j\beta}\chi_{T^j}(x). w β ( x ) = j = 0 ∑ ∞ k j β χ T j ( x ) . We first record the geometric computation that will be used later.
Lemma 3.1. Let x ∈ T j x \in T^j x ∈ T j r ≥ 0 r \ge0 r ≥ 0 L = min { j , r } L=\min\{j,r\} L = min { j , r } k ≥ 2 k \ge2 k ≥ 2 j j j r r r
1 ∣ S ( x , r ) ∣ 1 − α ∑ y ∈ S ( x , r ) w β ( y ) ≃ α w β ( x ) k r ( α + β ) ∑ m = 0 L k − m ( 1 + 2 β ) . (20) \frac{1}{|S(x,r)|^{1-\alpha}}\sum_{y\in S(x,r)}w_\beta(y)\simeq_\alpha w_\beta(x)k^{r(\alpha+\beta)}\sum_{m=0}^{L}k^{-m(1+2\beta)}. \tag*{(20)} ∣ S ( x , r ) ∣ 1 − α 1 y ∈ S ( x , r ) ∑ w β ( y ) ≃ α w β ( x ) k r ( α + β ) m = 0 ∑ L k − m ( 1 + 2 β ) . (20) More generally, for 1 ≤ s < ∞ 1 \le s < \infty 1 ≤ s < ∞
∣ S ( x , r ) ∣ α ( 1 ∣ S ( x , r ) ∣ ∑ y ∈ S ( x , r ) w β s ( y ) ) 1 / s ≃ α , s w β ( x ) k r ( α + β ) ( ∑ m = 0 L k − m ( 1 + 2 s β ) ) 1 / s , (21) |S(x,r)|^\alpha\left(\frac{1}{|S(x,r)|}\sum_{y\in S(x,r)}w_\beta^s(y)\right)^{1/s}\simeq_{\alpha,s}w_\beta(x)k^{r(\alpha+\beta)}\left(\sum_{m=0}^{L}k^{-m(1+2s\beta)}\right)^{1/s}, \tag*{(21)} ∣ S ( x , r ) ∣ α ∣ S ( x , r ) ∣ 1 y ∈ S ( x , r ) ∑ w β s ( y ) 1/ s ≃ α , s w β ( x ) k r ( α + β ) ( m = 0 ∑ L k − m ( 1 + 2 s β ) ) 1/ s , (21) Proof. Let x ∈ T j x \in T^j x ∈ T j y ∈ S ( x , r ) ∩ T i y \in S(x,r)\cap T^i y ∈ S ( x , r ) ∩ T i i = j + r − 2 m i=j+r-2m i = j + r − 2 m 0 ≤ m ≤ L 0\le m\le L 0 ≤ m ≤ L
S ( x , r ) = ⋃ m = 0 L { y ∈ S ( x , r ) : y ∈ T i , i = j + r − 2 m } . S(x,r)=\bigcup_{m=0}^{L}\{y\in S(x,r):y\in T^i,\ i=j+r-2m\}. S ( x , r ) = m = 0 ⋃ L { y ∈ S ( x , r ) : y ∈ T i , i = j + r − 2 m } . The number of vertices in the corresponding class is comparable, uniformly in k ≥ 2 k\ge2 k ≥ 2 k r − m k^{r-m} k r − m m = 0 m=0 m = 0 k r k^r k r ( k − 1 ) k r − m − 1 (k-1)k^{r-m-1} ( k − 1 ) k r − m − 1
∑ y ∈ S ( x , r ) w β ( y ) ≃ ∑ m = 0 L k r − m k β ( j + r − 2 m ) = k j β k r ( 1 + β ) ∑ m = 0 L k − m ( 1 + 2 β ) . (22) \sum_{y\in S(x,r)}w_\beta(y)\simeq\sum_{m=0}^{L}k^{r-m}k^{\beta(j+r-2m)}
=k^{j\beta}k^{r(1+\beta)}\sum_{m=0}^{L}k^{-m(1+2\beta)}. \tag*{(22)} y ∈ S ( x , r ) ∑ w β ( y ) ≃ m = 0 ∑ L k r − m k β ( j + r − 2 m ) = k j β k r ( 1 + β ) m = 0 ∑ L k − m ( 1 + 2 β ) . (22) Since ∣ S ( x , r ) ∣ ≃ k r |S(x,r)| \simeq k^r ∣ S ( x , r ) ∣ ≃ k r (20) . Replacing w β w_\beta w β w β s w_\beta^s w β s s s s ∣ S ( x , r ) ∣ α − 1 / s |S(x,r)|^{\alpha-1/s} ∣ S ( x , r ) ∣ α − 1/ s (21) . □ \square □
We can now characterize the radial powers completely.
Proposition 3.2. Let 0 < α < 1 0 < \alpha< 1 0 < α < 1 0 < α < 1 / 2 0 < \alpha< 1/2 0 < α < 1/2
M α w β ≃ α , β w β M_\alpha w_\beta\simeq_{\alpha,\beta} w_\beta M α w β ≃ α , β w β uniformly on T T T k ≥ 2 k \ge2 k ≥ 2
α − 1 ≤ β ≤ − α . \alpha- 1 \le\beta\le-\alpha. α − 1 ≤ β ≤ − α . If 1 / 2 ≤ α < 1 1/2 \le\alpha< 1 1/2 ≤ α < 1 β ∈ R \beta\in\mathbb{R} β ∈ R M α w β ≲ w β M_\alpha w_\beta\lesssim w_\beta M α w β ≲ w β
Proof. Since the radius r = 0 r = 0 r = 0 M α w β ≥ w β M_\alpha w_\beta\ge w_\beta M α w β ≥ w β M α o M_\alpha^o M α o Lemma 3.1 , for x ∈ T j x \in T^j x ∈ T j
A r , α o ( w β ) ( x ) = 1 ∣ S ( x , r ) ∣ 1 − α ∑ y ∈ S ( x , r ) w β ( y ) ≃ α w β ( x ) k r ( α + β ) ∑ m = 0 L k − m ( 1 + 2 β ) . A^o_{r,\alpha}(w_\beta)(x)=\frac{1}{|S(x,r)|^{1-\alpha}}\sum_{y\in S(x,r)}w_\beta(y)\simeq_\alpha w_\beta(x)k^{r(\alpha+\beta)}\sum_{m=0}^{L}k^{-m(1+2\beta)}. A r , α o ( w β ) ( x ) = ∣ S ( x , r ) ∣ 1 − α 1 y ∈ S ( x , r ) ∑ w β ( y ) ≃ α w β ( x ) k r ( α + β ) m = 0 ∑ L k − m ( 1 + 2 β ) . First assume j ≥ r j \ge r j ≥ r L = r L=r L = r Lemma 3.1 . If β < − 1 / 2 \beta< -1/2 β < − 1/2
∑ m = 0 r k − m ( 1 + 2 β ) ≃ β k r ( − 1 − 2 β ) , \sum_{m=0}^{r}k^{-m(1+2\beta)}\simeq_\beta k^{r(-1-2\beta)}, m = 0 ∑ r k − m ( 1 + 2 β ) ≃ β k r ( − 1 − 2 β ) , and therefore
A r , α o ( w β ) ( x ) w β ( x ) ≃ k r ( α − β − 1 ) . \frac{A^o_{r,\alpha}(w_\beta)(x)}{w_\beta(x)}\simeq k^{r(\alpha-\beta-1)}. w β ( x ) A r , α o ( w β ) ( x ) ≃ k r ( α − β − 1 ) . Uniform boundedness forces β ≥ α − 1 \beta\ge\alpha- 1 β ≥ α − 1 β = − 1 / 2 \beta=-1/2 β = − 1/2
A r , α o ( w β ) ( x ) w β ( x ) ≃ ( r + 1 ) k r ( α − 1 / 2 ) , \frac{A^o_{r,\alpha}(w_\beta)(x)}{w_\beta(x)}\simeq(r+1)k^{r(\alpha-1/2)}, w β ( x ) A r , α o ( w β ) ( x ) ≃ ( r + 1 ) k r ( α − 1/2 ) , which is uniformly bounded exactly when α < 1 / 2 \alpha< 1/2 α < 1/2 β > − 1 / 2 \beta> -1/2 β > − 1/2
A r , α o ( w β ) ( x ) w β ( x ) ≃ k r ( α + β ) , \frac{A^o_{r,\alpha}(w_\beta)(x)}{w_\beta(x)}\simeq k^{r(\alpha+\beta)}, w β ( x ) A r , α o ( w β ) ( x ) ≃ k r ( α + β ) , so necessarily β ≤ − α \beta\le-\alpha β ≤ − α
These conditions are also sufficient. In the first case, the conditions β < − 1 / 2 \beta< -1/2 β < − 1/2 β ≥ α − 1 \beta\ge\alpha-1 β ≥ α − 1 α < 1 / 2 \alpha< 1/2 α < 1/2 α + β < 0 \alpha+\beta< 0 α + β < 0 r ≤ j r \le j r ≤ j
k r ( α − β − 1 ) ≤ 1. k^{r(\alpha-\beta-1)}\le1. k r ( α − β − 1 ) ≤ 1. If r > j r>j r > j L = j L=j L = j
k r ( α + β ) k j ( − 1 − 2 β ) ≤ k j ( α − β − 1 ) ≤ 1. k^{r(\alpha+\beta)}k^{j(-1-2\beta)}\le k^{j(\alpha-\beta-1)}\le1. k r ( α + β ) k j ( − 1 − 2 β ) ≤ k j ( α − β − 1 ) ≤ 1. At β = − 1 / 2 \beta=-1/2 β = − 1/2 α < 1 / 2 \alpha< 1/2 α < 1/2
( L + 1 ) k r ( α − 1 / 2 ) ≲ α 1 (L+1)k^{r(\alpha-1/2)}\lesssim_\alpha1 ( L + 1 ) k r ( α − 1/2 ) ≲ α 1 for both r ≤ j r\le j r ≤ j r > j r>j r > j α + β ≤ 0 \alpha+\beta\le0 α + β ≤ 0 M α o w β ≲ w β M_\alpha^o w_\beta\lesssim w_\beta M α o w β ≲ w β
Combining the three regimes gives exactly [ α − 1 , − α ] [\alpha-1,-\alpha] [ α − 1 , − α ] 0 < α < 1 / 2 0<\alpha<1/2 0 < α < 1/2 α = 1 / 2 \alpha=1/2 α = 1/2 β = − 1 / 2 \beta=-1/2 β = − 1/2 r + 1 r+1 r + 1 α > 1 / 2 \alpha>1/2 α > 1/2 □ \square □
The behavior of L s L^s L s M α , s M_{\alpha,s} M α , s s > 1 s>1 s > 1
Proposition 3.3. Let 0 < α < 1 0 < \alpha< 1 0 < α < 1 s > 1 s > 1 s > 1
M α , s w β ≃ α , s , β w β M_{\alpha,s}w_\beta\simeq_{\alpha,s,\beta} w_\beta M α , s w β ≃ α , s , β w β uniformly on T T T k ≥ 2 k \ge2 k ≥ 2
s < 1 2 α , α − 1 s ≤ β ≤ − α . s < \frac{1}{2\alpha}, \qquad\alpha-\frac{1}{s} \le\beta\le-\alpha. s < 2 α 1 , α − s 1 ≤ β ≤ − α . In particular, nontrivial radial weights satisfying M α , s w ≲ w M_{\alpha,s}w \lesssim w M α , s w ≲ w 0 < α < 1 / 2 0 < \alpha< 1/2 0 < α < 1/2 1 < s < 1 / ( 2 α ) 1 < s < 1/(2\alpha) 1 < s < 1/ ( 2 α )
Proof. The proof is analogous to the case s = 1 s=1 s = 1
Necessity follows already from the spherical operator, because M α , s ∘ ≲ M α , s M_{\alpha,s}^{\circ} \lesssim M_{\alpha,s} M α , s ∘ ≲ M α , s (4) . Take j ≥ r j \ge r j ≥ r β < − 1 / ( 2 s ) \beta< -1/(2s) β < − 1/ ( 2 s )
( ∑ m = 0 r k − m ( 1 + 2 s β ) ) 1 / s ≃ s , β k r ( − 1 / s − 2 β ) , \left(\sum_{m=0}^{r} k^{-m(1+2s\beta)}\right)^{1/s} \simeq_{s,\beta} k^{r(-1/s-2\beta)}, ( m = 0 ∑ r k − m ( 1 + 2 s β ) ) 1/ s ≃ s , β k r ( − 1/ s − 2 β ) , so the quotient is comparable to k r ( α − β − 1 / s ) k^{r(\alpha-\beta-1/s)} k r ( α − β − 1/ s ) β ≥ α − 1 / s \beta\ge\alpha-1/s β ≥ α − 1/ s β = − 1 / ( 2 s ) \beta=-1/(2s) β = − 1/ ( 2 s )
( r + 1 ) 1 / s k r ( α − 1 / ( 2 s ) ) , (r+1)^{1/s}k^{r(\alpha-1/(2s))}, ( r + 1 ) 1/ s k r ( α − 1/ ( 2 s )) , and is bounded only if α < 1 / ( 2 s ) \alpha< 1/(2s) α < 1/ ( 2 s ) β > − 1 / ( 2 s ) \beta> -1/(2s) β > − 1/ ( 2 s ) β ≤ − α \beta\le-\alpha β ≤ − α α ≤ 1 / ( 2 s ) \alpha\le1/(2s) α ≤ 1/ ( 2 s ) s ≤ 1 / ( 2 α ) s \le1/(2\alpha) s ≤ 1/ ( 2 α ) β = − 1 2 s \beta=-\frac{1}{2s} β = − 2 s 1 α < 1 2 s \alpha< \frac{1}{2s} α < 2 s 1
Conversely, assume these conditions. In particular α s < 1 / 2 < 1 \alpha s < 1/2 < 1 α s < 1/2 < 1 M α , s w β M_{\alpha,s}w_\beta M α , s w β β < − 1 / ( 2 s ) \beta< -1/(2s) β < − 1/ ( 2 s ) r > j r>j r > j
k r ( α + β ) k j ( − 1 / s − 2 β ) . k^{r(\alpha+\beta)}k^{j(-1/s-2\beta)}. k r ( α + β ) k j ( − 1/ s − 2 β ) . Here α + β < 0 \alpha+\beta<0 α + β < 0
k j ( α − β − 1 / s ) ≤ 1. k^{j(\alpha-\beta-1/s)} \le1. k j ( α − β − 1/ s ) ≤ 1. For r ≤ j r \le j r ≤ j α − β − 1 / s ≤ 0 \alpha-\beta-1/s \le0 α − β − 1/ s ≤ 0 β = − 1 / ( 2 s ) \beta=-1/(2s) β = − 1/ ( 2 s ) ( L + 1 ) 1 / s (L+1)^{1/s} ( L + 1 ) 1/ s α − 1 / ( 2 s ) < 0 \alpha-1/(2s)<0 α − 1/ ( 2 s ) < 0 β > − 1 / ( 2 s ) \beta>-1/(2s) β > − 1/ ( 2 s ) α + β ≤ 0 \alpha+\beta\le0 α + β ≤ 0 r = 0 r=0 r = 0 □ \square □
Remark 3.4. These formulas recover the setting α = 0 \alpha=0 α = 0 [ − 1 , 0 ] [-1,0] [ − 1 , 0 ] [9 Theorem 3.1] . Similarly, Proposition 3.3 becomes − 1 s ≤ β ≤ 0 -\frac{1}{s}\le\beta\le0 − s 1 ≤ β ≤ 0
Thus, every β ∈ ( − 1 , 0 ] \beta\in(-1,0] β ∈ ( − 1 , 0 ] s > 1 s>1 s > 1 1 1 1 β = − 1 \beta=-1 β = − 1 s > 1 s>1 s > 1 [9 Remarks 3.2–3.3] .
For α > 0 \alpha>0 α > 0 [ α − 1 , − α ] [\alpha-1,-\alpha] [ α − 1 , − α ] 1 − 2 α 1-2\alpha 1 − 2 α α = 1 / 2 \alpha=1/2 α = 1/2 M α , s M_{\alpha,s} M α , s 1 / s − 2 α 1/s-2\alpha 1/ s − 2 α s < 1 / ( 2 α ) s<1/(2\alpha) s < 1/ ( 2 α ) α = 1 2 \alpha=\frac{1}{2} α = 2 1
Remark 3.5. If 0 < α < 1 / 2 0<\alpha<1/2 0 < α < 1/2 1 < s < 1 / ( 2 α ) 1<s<1/(2\alpha) 1 < s < 1/ ( 2 α ) β ∈ [ α − 1 / s , − α ] \beta\in[\alpha-1/s,-\alpha] β ∈ [ α − 1/ s , − α ]
w β ( { M α f > λ } ) ≲ α , s , β 1 λ ∑ x ∈ T ∣ f ( x ) ∣ w β ( x ) . w_\beta(\{M_\alpha f>\lambda\}) \lesssim_{\alpha,s,\beta} \frac{1}{\lambda}\sum_{x\in T}|f(x)|w_\beta(x). w β ({ M α f > λ }) ≲ α , s , β λ 1 x ∈ T ∑ ∣ f ( x ) ∣ w β ( x ) . Thus the positive Fefferman–Stein estimate is non-vacuous in a genuinely subcritical weighted regime.
Failure of the classical fractional Fefferman–Stein estimate
We now show that the estimate obtained above with M α , s w M_{\alpha,\mathrm{sw}} M α , sw s > 1 s>1 s > 1 s = 1 s=1 s = 1 w w w { f j } \{f_j\} { f j }
w ( { x ∈ T : M α f j ( x ) > 1 } ) ⟶ ∞ , w\left(\left\{x\in T:M_\alpha f_j(x)>1\right\}\right)\longrightarrow\infty, w ( { x ∈ T : M α f j ( x ) > 1 } ) ⟶ ∞ , while
sup j ∑ x ∈ T ∣ f j ( x ) ∣ M α w ( x ) < ∞ . \sup_j\sum_{x\in T}|f_j(x)|M_\alpha w(x)<\infty. j sup x ∈ T ∑ ∣ f j ( x ) ∣ M α w ( x ) < ∞. The main idea is to construct a subtree whose growth is of order k ( 1 − α ) r k^{(1-\alpha)r} k ( 1 − α ) r ∣ B ( x , r ) ∣ 1 − α |B(x,r)|^{1-\alpha} ∣ B ( x , r ) ∣ 1 − α
A subtree of fractional growth. For ℓ ≥ 0 \ell\geq0 ℓ ≥ 0
ε ℓ = ⌊ ( ℓ + 1 ) ( 1 − α ) ⌋ − ⌊ ℓ ( 1 − α ) ⌋ . \varepsilon_\ell=\lfloor(\ell+1)(1-\alpha)\rfloor-\lfloor\ell(1-\alpha)\rfloor. ε ℓ = ⌊( ℓ + 1 ) ( 1 − α )⌋ − ⌊ ℓ ( 1 − α )⌋ . Since 0 < α < 1 0<\alpha<1 0 < α < 1 ε ℓ ∈ { 0 , 1 } \varepsilon_\ell\in\{0,1\} ε ℓ ∈ { 0 , 1 } S ⊂ T S\subset T S ⊂ T T T T S S S ℓ \ell ℓ k ε ℓ k^{\varepsilon_\ell} k ε ℓ ε ℓ = 1 \varepsilon_\ell=1 ε ℓ = 1 k k k ε ℓ = 0 \varepsilon_\ell=0 ε ℓ = 0
Let
S j = S ∩ T j , N j = ∣ S j ∣ . S_j=S\cap T^j,\qquad N_j=|S_j|. S j = S ∩ T j , N j = ∣ S j ∣. Then
N j = ∏ ℓ = 0 j − 1 k ε ℓ = k ⌊ j ( 1 − α ) ⌋ . (23) N_j=\prod_{\ell=0}^{j-1}k^{\varepsilon_\ell}=k^{\lfloor j(1-\alpha)\rfloor}. \tag*{(23)} N j = ℓ = 0 ∏ j − 1 k ε ℓ = k ⌊ j ( 1 − α )⌋ . (23) In particular,
k − 1 k ( 1 − α ) j ≤ N j ≤ k ( 1 − α ) j . (24) k^{-1}k^{(1-\alpha)j}\leq N_j\leq k^{(1-\alpha)j}. \tag*{(24)} k − 1 k ( 1 − α ) j ≤ N j ≤ k ( 1 − α ) j . (24) Moreover, if x ∈ S i x\in S_i x ∈ S i x x x S i + r S_{i+r} S i + r
D ( i , r ) = N i + r N i = k ⌊ ( i + r ) ( 1 − α ) ⌋ − ⌊ i ( 1 − α ) ⌋ , (25) D(i,r)=\frac{N_{i+r}}{N_i}=k^{\lfloor(i+r)(1-\alpha)\rfloor-\lfloor i(1-\alpha)\rfloor}, \tag*{(25)} D ( i , r ) = N i N i + r = k ⌊( i + r ) ( 1 − α )⌋ − ⌊ i ( 1 − α )⌋ , (25) and therefore
k − 1 k ( 1 − α ) r ≤ D ( i , r ) ≤ k k ( 1 − α ) r . (26) k^{-1}k^{(1-\alpha)r}\leq D(i,r)\leq kk^{(1-\alpha)r}. \tag*{(26)} k − 1 k ( 1 − α ) r ≤ D ( i , r ) ≤ k k ( 1 − α ) r . (26) We define the weight
w ( x ) = ∑ j = 0 ∞ 1 N j χ S j ( x ) . (27) w(x)=\sum_{j=0}^{\infty}\frac{1}{N_j}\chi_{S_j}(x). \tag*{(27)} w ( x ) = j = 0 ∑ ∞ N j 1 χ S j ( x ) . (27) Thus w = 0 w=0 w = 0 S S S
w ( S j ) = 1 for every j ≥ 0. (28) w(S_j)=1\qquad\text{for every }j\geq0. \tag*{(28)} w ( S j ) = 1 for every j ≥ 0. (28) The next lemma is the key estimate for the right-hand side of the counterexample.
Lemma 3.6. For every x ∈ S x\in S x ∈ S
M α ∘ w ( x ) ≲ α , k w ( x ) and M α w ( x ) ≲ α , k w ( x ) . M_\alpha^\circ w(x)\lesssim_{\alpha,k}w(x)\qquad\text{and}\qquad M_\alpha w(x)\lesssim_{\alpha,k}w(x). M α ∘ w ( x ) ≲ α , k w ( x ) and M α w ( x ) ≲ α , k w ( x ) . Moreover,
M α ∘ w ( x ) ≃ α , k w ( x ) , M α w ( x ) ≃ α , k w ( x ) , x ∈ S . M_\alpha^\circ w(x)\simeq_{\alpha,k}w(x),\qquad M_\alpha w(x)\simeq_{\alpha,k}w(x),\qquad x\in S. M α ∘ w ( x ) ≃ α , k w ( x ) , M α w ( x ) ≃ α , k w ( x ) , x ∈ S . Proof. Fix x ∈ S i x\in S_i x ∈ S i r ≥ 0 r\geq0 r ≥ 0 0 ≤ m ≤ min { r , i } 0\leq m\leq\min\{r,i\} 0 ≤ m ≤ min { r , i } E m E_m E m y ∈ S ∩ S ( x , r ) y\in S\cap S(x,r) y ∈ S ∩ S ( x , r ) x x x y y y x x x i − m i-m i − m
If y ∈ E m y\in E_m y ∈ E m y y y i + r − 2 m i+r-2m i + r − 2 m z z z x x x i − m i-m i − m z z z S i + r − 2 m S_{i+r-2m} S i + r − 2 m
N i + r − 2 m N i − m . \frac{N_{i+r-2m}}{N_{i-m}}. N i − m N i + r − 2 m . Hence
∣ E m ∣ ≤ N i + r − 2 m N i − m . |E_m|\leq\frac{N_{i+r-2m}}{N_{i-m}}. ∣ E m ∣ ≤ N i − m N i + r − 2 m . Since every vertex of S i + r − 2 m S_{i+r-2m} S i + r − 2 m N i + r − 2 m − 1 N_{i+r-2m}^{-1} N i + r − 2 m − 1
∑ y ∈ E m w ( y ) ≤ 1 N i − m . \sum_{y\in E_m} w(y) \le\frac{1}{N_{i-m}}. y ∈ E m ∑ w ( y ) ≤ N i − m 1 . Therefore,
∑ y ∈ S ( x , r ) w ( y ) ≤ ∑ m = 0 min { r , i } 1 N i − m . (29) \sum_{y\in S(x,r)} w(y) \le\sum_{m=0}^{\min\{r,i\}} \frac{1}{N_{i-m}}. \tag*{(29)} y ∈ S ( x , r ) ∑ w ( y ) ≤ m = 0 ∑ m i n { r , i } N i − m 1 . (29) Using (3.3),
N i N i − m = k ⌊ ( 1 − α ) i ⌋ − ⌊ ( 1 − α ) ( i − m ) ⌋ ≤ k k ( 1 − α ) m , \frac{N_i}{N_{i-m}} = k^{\lfloor(1-\alpha)i\rfloor-\lfloor(1-\alpha)(i-m)\rfloor} \le k k^{(1-\alpha)m}, N i − m N i = k ⌊( 1 − α ) i ⌋ − ⌊( 1 − α ) ( i − m )⌋ ≤ k k ( 1 − α ) m , and consequently
1 N i − m ≤ k k ( 1 − α ) m 1 N i . \frac{1}{N_{i-m}} \le k k^{(1-\alpha)m}\frac{1}{N_i}. N i − m 1 ≤ k k ( 1 − α ) m N i 1 . It follows from (29) that
∑ y ∈ S ( x , r ) w ( y ) ≲ α , k k ( 1 − α ) r N i . (30) \sum_{y\in S(x,r)} w(y) \lesssim_{\alpha,k} \frac{k^{(1-\alpha)r}}{N_i}. \tag*{(30)} y ∈ S ( x , r ) ∑ w ( y ) ≲ α , k N i k ( 1 − α ) r . (30) Since x x x k r k^r k r r r r ∣ S ( x , r ) ∣ ≥ k r |S(x,r)|\ge k^r ∣ S ( x , r ) ∣ ≥ k r
1 ∣ S ( x , r ) ∣ 1 − α ∑ y ∈ S ( x , r ) w ( y ) ≲ α , k 1 k ( 1 − α ) r k ( 1 − α ) r N i = 1 N i = w ( x ) . \frac{1}{|S(x,r)|^{1-\alpha}}\sum_{y\in S(x,r)} w(y) \lesssim_{\alpha,k} \frac{1}{k^{(1-\alpha)r}}\frac{k^{(1-\alpha)r}}{N_i}=\frac{1}{N_i}=w(x). ∣ S ( x , r ) ∣ 1 − α 1 y ∈ S ( x , r ) ∑ w ( y ) ≲ α , k k ( 1 − α ) r 1 N i k ( 1 − α ) r = N i 1 = w ( x ) . Taking the supremum in r r r
M α O w ( x ) ≲ α , k w ( x ) . M_\alpha^O w(x)\lesssim_{\alpha,k}w(x). M α O w ( x ) ≲ α , k w ( x ) . For the maximal function over balls, using (30) ,
∑ y ∈ B ( x , r ) w ( y ) = ∑ ℓ = 0 r ∑ y ∈ S ( x , ℓ ) w ( y ) ≲ α , k 1 N i ∑ ℓ = 0 r k ( 1 − α ) ℓ ≲ α , k k ( 1 − α ) r N i . \sum_{y\in B(x,r)}w(y)=\sum_{\ell=0}^r\sum_{y\in S(x,\ell)}w(y)\lesssim_{\alpha,k}\frac{1}{N_i}\sum_{\ell=0}^r k^{(1-\alpha)\ell}\lesssim_{\alpha,k}\frac{k^{(1-\alpha)r}}{N_i}. y ∈ B ( x , r ) ∑ w ( y ) = ℓ = 0 ∑ r y ∈ S ( x , ℓ ) ∑ w ( y ) ≲ α , k N i 1 ℓ = 0 ∑ r k ( 1 − α ) ℓ ≲ α , k N i k ( 1 − α ) r . Since ∣ B ( x , r ) ∣ ≥ k r |B(x,r)|\ge k^r ∣ B ( x , r ) ∣ ≥ k r
1 ∣ B ( x , r ) ∣ 1 − α ∑ y ∈ B ( x , r ) w ( y ) ≲ α , k 1 N i = w ( x ) . \frac{1}{|B(x,r)|^{1-\alpha}}\sum_{y\in B(x,r)}w(y)\lesssim_{\alpha,k}\frac{1}{N_i}=w(x). ∣ B ( x , r ) ∣ 1 − α 1 y ∈ B ( x , r ) ∑ w ( y ) ≲ α , k N i 1 = w ( x ) . Thus
M α w ( x ) ≲ α , k w ( x ) . M_\alpha w(x)\lesssim_{\alpha,k}w(x). M α w ( x ) ≲ α , k w ( x ) . Finally, taking r = 0 r=0 r = 0 M α O w ( x ) ≥ w ( x ) M_\alpha^O w(x)\ge w(x) M α O w ( x ) ≥ w ( x ) M α w ( x ) ≥ w ( x ) M_\alpha w(x)\ge w(x) M α w ( x ) ≥ w ( x )
We now prove the counterexample. A function supported on a single generation of the subtree has uniformly bounded pairing with M α w M_\alpha w M α w 0 < α < 1 0<\alpha<1 0 < α < 1
Theorem 3.7. Let k ≥ 2 k\ge2 k ≥ 2 0 < α < 1 0<\alpha<1 0 < α < 1 w ≥ 0 w\ge0 w ≥ 0 T T T { f j } j ≥ 1 \{f_j\}_{j\ge1} { f j } j ≥ 1
sup j ≥ 1 ∑ x ∈ T ∣ f j ( x ) ∣ M α w ( x ) < ∞ , (31) \sup_{j\ge1}\sum_{x\in T}|f_j(x)|M_\alpha w(x)<\infty, \tag*{(31)} j ≥ 1 sup x ∈ T ∑ ∣ f j ( x ) ∣ M α w ( x ) < ∞ , (31) whereas
w ( { x ∈ T : M α f j ( x ) > 1 } ) ⟶ ∞ as j → ∞ . (32) w(\{x\in T:M_\alpha f_j(x)>1\})\longrightarrow\infty\qquad\text{as }j\to\infty. \tag*{(32)} w ({ x ∈ T : M α f j ( x ) > 1 }) ⟶ ∞ as j → ∞. (32) Consequently, there is no constant C > 0 C>0 C > 0
w ( { x ∈ T : M α f ( x ) > λ } ) ≤ C λ ∑ x ∈ T ∣ f ( x ) ∣ M α w ( x ) w(\{x\in T:M_\alpha f(x)>\lambda\})\le\frac{C}{\lambda}\sum_{x\in T}|f(x)|M_\alpha w(x) w ({ x ∈ T : M α f ( x ) > λ }) ≤ λ C x ∈ T ∑ ∣ f ( x ) ∣ M α w ( x ) holds for every finitely supported f f f λ > 0 \lambda>0 λ > 0
Proof. Let w w w (27) . For every j ≥ 1 j \ge1 j ≥ 1
f j = C 0 χ S j , f_j=C_0\chi_{S_j}, f j = C 0 χ S j , where C 0 > 4 1 − α k C_0>4^{1-\alpha}k C 0 > 4 1 − α k
⋃ i = 0 j S i ⊂ { x ∈ T : M α f j ( x ) > 1 } . (33) \bigcup_{i=0}^{j}S_i\subset\{x\in T:M_\alpha f_j(x)>1\}. \tag*{(33)} i = 0 ⋃ j S i ⊂ { x ∈ T : M α f j ( x ) > 1 } . (33) Let x ∈ S i x\in S_i x ∈ S i 0 ≤ i ≤ j 0\le i\le j 0 ≤ i ≤ j r = j − i r=j-i r = j − i x x x S j S_j S j D ( i , r ) = N j / N i D(i,r)=N_j/N_i D ( i , r ) = N j / N i (26) ,
D ( i , r ) ≥ k − 1 k ( 1 − α ) r . D(i,r)\ge k^{-1}k^{(1-\alpha)r}. D ( i , r ) ≥ k − 1 k ( 1 − α ) r . All these descendants belong to B ( x , r ) B(x,r) B ( x , r )
∑ y ∈ B ( x , r ) f j ( y ) ≥ C 0 D ( i , r ) ≥ C 0 k − 1 k ( 1 − α ) r . (34) \sum_{y\in B(x,r)}f_j(y)\ge C_0D(i,r)\ge C_0k^{-1}k^{(1-\alpha)r}. \tag*{(34)} y ∈ B ( x , r ) ∑ f j ( y ) ≥ C 0 D ( i , r ) ≥ C 0 k − 1 k ( 1 − α ) r . (34) Recall that ∣ S ( x , ℓ ) ∣ ≤ 2 k ℓ |S(x,\ell)|\le2k^\ell ∣ S ( x , ℓ ) ∣ ≤ 2 k ℓ ∣ B ( x , ℓ ) ∣ ≤ 4 k ℓ |B(x,\ell)|\le4k^\ell ∣ B ( x , ℓ ) ∣ ≤ 4 k ℓ δ = 1 − α \delta=1-\alpha δ = 1 − α (34) ,
M α f j ( x ) ≥ 1 ∣ B ( x , r ) ∣ δ ∑ y ∈ B ( x , r ) f j ( y ) ≥ C 0 k − 1 k δ r 4 δ k δ r = C 0 4 1 − α k > 1. M_\alpha f_j(x)\ge\frac{1}{|B(x,r)|^\delta}\sum_{y\in B(x,r)}f_j(y)\ge\frac{C_0k^{-1}k^{\delta r}}{4^\delta k^{\delta r}}=\frac{C_0}{4^{1-\alpha}k}>1. M α f j ( x ) ≥ ∣ B ( x , r ) ∣ δ 1 y ∈ B ( x , r ) ∑ f j ( y ) ≥ 4 δ k δ r C 0 k − 1 k δ r = 4 1 − α k C 0 > 1. This proves (33) . Using w ( S i ) = 1 w(S_i)=1 w ( S i ) = 1 i ≥ 0 i\ge0 i ≥ 0
w ( { M α f j > 1 } ) ≥ w ( ⋃ i = 0 j S i ) = ∑ i = 0 j w ( S i ) = j + 1. w(\{M_\alpha f_j>1\})\ge w\left(\bigcup_{i=0}^{j}S_i\right)=\sum_{i=0}^{j}w(S_i)=j+1. w ({ M α f j > 1 }) ≥ w ( i = 0 ⋃ j S i ) = i = 0 ∑ j w ( S i ) = j + 1. Hence, the left-hand side tends to infinity. On the other hand, since f j f_j f j S j S_j S j Lemma 3.6 yields
∑ x ∈ T ∣ f j ( x ) ∣ M α w ( x ) = C 0 ∑ x ∈ S j M α w ( x ) ≲ α , k C 0 ∑ x ∈ S j w ( x ) = C 0 w ( S j ) = C 0 . \sum_{x\in T}|f_j(x)|M_\alpha w(x)=C_0\sum_{x\in S_j}M_\alpha w(x)\lesssim_{\alpha,k}C_0\sum_{x\in S_j}w(x)=C_0w(S_j)=C_0. x ∈ T ∑ ∣ f j ( x ) ∣ M α w ( x ) = C 0 x ∈ S j ∑ M α w ( x ) ≲ α , k C 0 x ∈ S j ∑ w ( x ) = C 0 w ( S j ) = C 0 . This bound is uniform in j j j
Remark 3.8. The essential feature of the construction is that, for x ∈ S i x\in S_i x ∈ S i x x x S i + r S_{i+r} S i + r
D ( i , r ) ≃ k k ( 1 − α ) r . D(i,r)\simeq_k k^{(1-\alpha)r}. D ( i , r ) ≃ k k ( 1 − α ) r . This growth exactly compensates the denominator ∣ B ( x , r ) ∣ 1 − α ≃ k ( 1 − α ) r |B(x,r)|^{1-\alpha}\simeq k^{(1-\alpha)r} ∣ B ( x , r ) ∣ 1 − α ≃ k ( 1 − α ) r [9 ] .
A global counterexample and failure under finite iterations
We now focus on Theorem 1.2 in the subcritical range 0 < α < 1 / 2 0<\alpha<1/2 0 < α < 1/2 n ≥ 1 n\ge1 n ≥ 1 M α ( n ) M_\alpha^{(n)} M α ( n ) n n n M α M_\alpha M α
Proof of Theorem 1.2. Consider the weight
w ∗ ( x ) = ∑ j = 0 ∞ k ( α − 1 ) j χ T j ( x ) . w_*(x)=\sum_{j=0}^{\infty}k^{(\alpha-1)j}\chi_{T_j}(x). w ∗ ( x ) = j = 0 ∑ ∞ k ( α − 1 ) j χ T j ( x ) . The first assertion follows from Proposition 3.2 , since w ∗ = w α − 1 w_*=w_{\alpha-1} w ∗ = w α − 1 α − 1 ∈ [ α − 1 , − α ] \alpha-1\in[\alpha-1,-\alpha] α − 1 ∈ [ α − 1 , − α ] 0 < α < 1 / 2 0<\alpha<1/2 0 < α < 1/2 n n n
M α ( n ) w ∗ ≲ α , k , n w ∗ . (35) M_\alpha^{(n)}w_*\lesssim_{\alpha,k,n}w_*. \tag*{(35)} M α ( n ) w ∗ ≲ α , k , n w ∗ . (35) Let S S S
f j = C 0 χ S j , f_j=C_0\chi_{S_j}, f j = C 0 χ S j , with the same fixed constant C 0 > 4 1 − α k C_0>4^{1-\alpha}k C 0 > 4 1 − α k Theorem 3.7 gives
⋃ i = 0 j S i ⊂ { M α f j > 1 } . \bigcup_{i=0}^{j} S_i \subset\{M_\alpha f_j>1\}. i = 0 ⋃ j S i ⊂ { M α f j > 1 } . Since ∣ S i ∣ = k ⌊ ( 1 − α ) i ⌋ |S_i|=k^{\lfloor(1-\alpha)i\rfloor} ∣ S i ∣ = k ⌊( 1 − α ) i ⌋
w ∗ ( S i ) = k ⌊ ( 1 − α ) i ⌋ k − ( 1 − α ) i , w_*(S_i)=k^{\lfloor(1-\alpha)i\rfloor}k^{-(1-\alpha)i}, w ∗ ( S i ) = k ⌊( 1 − α ) i ⌋ k − ( 1 − α ) i , and therefore
k − 1 ≤ w ∗ ( S i ) ≤ 1. (36) k^{-1}\le w_*(S_i)\le1. \tag*{(36)} k − 1 ≤ w ∗ ( S i ) ≤ 1. (36) It follows that
w ∗ ( { M α f j > 1 } ) ≥ ∑ i = 0 j w ∗ ( S i ) ≥ j + 1 k . w_*(\{M_\alpha f_j>1\})\ge\sum_{i=0}^{j}w_*(S_i)\ge\frac{j+1}{k}. w ∗ ({ M α f j > 1 }) ≥ i = 0 ∑ j w ∗ ( S i ) ≥ k j + 1 . On the other hand, by (3.15) and (3.16),
∑ x ∈ T ∣ f j ( x ) ∣ M α ( n ) w ∗ ( x ) ≲ α , k , n C 0 ∑ x ∈ S j w ∗ ( x ) = C 0 w ∗ ( S j ) ≤ C 0 , \sum_{x\in T}|f_j(x)|M_\alpha^{(n)}w_*(x)\lesssim_{\alpha,k,n}C_0\sum_{x\in S_j}w_*(x)=C_0w_*(S_j)\le C_0, x ∈ T ∑ ∣ f j ( x ) ∣ M α ( n ) w ∗ ( x ) ≲ α , k , n C 0 x ∈ S j ∑ w ∗ ( x ) = C 0 w ∗ ( S j ) ≤ C 0 , uniformly in j j j □ \square □
Remark 3.9. The extreme weight w ∗ = w α − 1 w_*=w_{\alpha-1} w ∗ = w α − 1 M α w ∗ ≃ w ∗ M_\alpha w_*\simeq w_* M α w ∗ ≃ w ∗ M α , s w ∗ ≲ w ∗ M_{\alpha,s}w_*\lesssim w_* M α , s w ∗ ≲ w ∗ s > 1 s>1 s > 1 j ≥ 1 j\ge1 j ≥ 1 x j ∈ T j x_j\in T^j x j ∈ T j x 0 x_0 x 0 j j j x 0 ∈ S ( x j , j ) x_0\in S(x_j,j) x 0 ∈ S ( x j , j )
M α , s w ∗ ( x j ) ≥ ∣ S ( x j , j ) ∣ α − 1 / s w ∗ ( x 0 ) ≳ α , s w ∗ ( x j ) k j ( 1 − 1 / s ) . M_{\alpha,s}w_*(x_j)\ge|S(x_j,j)|^{\alpha-1/s}w_*(x_0)\gtrsim_{\alpha,s}w_*(x_j)k^{j(1-1/s)}. M α , s w ∗ ( x j ) ≥ ∣ S ( x j , j ) ∣ α − 1/ s w ∗ ( x 0 ) ≳ α , s w ∗ ( x j ) k j ( 1 − 1/ s ) . Since 1 − 1 / s > 0 1-1/s>0 1 − 1/ s > 0 j → ∞ j\to\infty j → ∞ M α , s M_{\alpha,s} M α , s [9 Remark 3.3] .
A necessary condition for weighted weak type (1, 1).
Proof of Theorem 1.3. Fix x 0 ∈ T x_0\in T x 0 ∈ T r ≥ 0 r\ge0 r ≥ 0 f = δ x 0 f=\delta_{x_0} f = δ x 0 y ∈ B ( x 0 , r ) y\in B(x_0,r) y ∈ B ( x 0 , r ) d = d ( x 0 , y ) d=d(x_0,y) d = d ( x 0 , y ) d ≤ r d\le r d ≤ r
M α ( δ x 0 ) ( y ) ≥ 1 ∣ B ( y , d ) ∣ 1 − α . M_\alpha(\delta_{x_0})(y)\ge\frac{1}{|B(y,d)|^{1-\alpha}}. M α ( δ x 0 ) ( y ) ≥ ∣ B ( y , d ) ∣ 1 − α 1 . Since ∣ B ( y , d ) ∣ ≤ 4 k d ≤ 4 k r |B(y,d)|\le4k^d\le4k^r ∣ B ( y , d ) ∣ ≤ 4 k d ≤ 4 k r
M α ( δ x 0 ) ( y ) ≥ 4 − ( 1 − α ) k − ( 1 − α ) r . M_\alpha(\delta_{x_0})(y)\ge4^{-(1-\alpha)}k^{-(1-\alpha)r}. M α ( δ x 0 ) ( y ) ≥ 4 − ( 1 − α ) k − ( 1 − α ) r . Thus, with
λ = 1 2 4 − ( 1 − α ) k − ( 1 − α ) r , \lambda=\frac{1}{2}4^{-(1-\alpha)}k^{-(1-\alpha)r}, λ = 2 1 4 − ( 1 − α ) k − ( 1 − α ) r , we have
B ( x 0 , r ) ⊂ { M α ( δ x 0 ) > λ } . B(x_0,r)\subset\{M_\alpha(\delta_{x_0})>\lambda\}. B ( x 0 , r ) ⊂ { M α ( δ x 0 ) > λ } . Applying (1.2),
w ( B ( x 0 , r ) ) ≲ α , k , C w k ( 1 − α ) r w ( x 0 ) . w(B(x_0,r))\lesssim_{\alpha,k,C_w}k^{(1-\alpha)r}w(x_0). w ( B ( x 0 , r )) ≲ α , k , C w k ( 1 − α ) r w ( x 0 ) . Since ∣ B ( x 0 , r ) ∣ ≥ k r |B(x_0,r)|\ge k^r ∣ B ( x 0 , r ) ∣ ≥ k r
w ( B ( x 0 , r ) ) ∣ B ( x 0 , r ) ∣ 1 − α ≲ α , k , C w w ( x 0 ) . \frac{w(B(x_0,r))}{|B(x_0,r)|^{1-\alpha}}\lesssim_{\alpha,k,C_w}w(x_0). ∣ B ( x 0 , r ) ∣ 1 − α w ( B ( x 0 , r )) ≲ α , k , C w w ( x 0 ) . Taking the supremum over r r r □ \square □
Using Theorem 1.3 and Theorem 1.2 with n = 1 n=1 n = 1
Corollary 3.10. If 0 < α < 1 / 2 0<\alpha<1/2 0 < α < 1/2
M α w ≲ w M_\alpha w\lesssim w M α w ≲ w is necessary but not sufficient for the weighted weak type estimate (1.2).
The phase transition at α = 1 / 2 \alpha= 1/2 α = 1/2
The radial classification suggests that the threshold α = 1 / 2 \alpha= 1/2 α = 1/2 Lemma 3.1 .
Proposition 3.11. Let 1 / 2 ≤ α < 1 1/2 \le\alpha< 1 1/2 ≤ α < 1 There is no nonzero weight w w w T T T
M α w ( x ) ≤ C w ( x ) , x ∈ T , M_\alpha w(x) \le Cw(x), \qquad x \in T, M α w ( x ) ≤ C w ( x ) , x ∈ T , for some finite constant C C C
Proof. First assume that α > 1 / 2 \alpha> 1/2 α > 1/2 x 0 x_0 x 0 w ( x 0 ) > 0 w(x_0) > 0 w ( x 0 ) > 0 D r ( x 0 ) D_r(x_0) D r ( x 0 ) x 0 x_0 x 0 r r r ∣ D r ( x 0 ) ∣ = k r |D_r(x_0)| = k^r ∣ D r ( x 0 ) ∣ = k r
For y ∈ D r ( x 0 ) y \in D_r(x_0) y ∈ D r ( x 0 ) x 0 ∈ B ( y , r ) x_0 \in B(y,r) x 0 ∈ B ( y , r ) ∣ B ( y , r ) ∣ ≤ 4 k r |B(y,r)| \le4k^r ∣ B ( y , r ) ∣ ≤ 4 k r
C w ( y ) ≥ M α w ( y ) ≥ w ( x 0 ) ∣ B ( y , r ) ∣ 1 − α ≳ α , k k − ( 1 − α ) r w ( x 0 ) . Cw(y) \ge M_\alpha w(y) \ge\frac{w(x_0)}{|B(y,r)|^{1-\alpha}} \gtrsim_{\alpha,k} k^{-(1-\alpha)r}w(x_0). C w ( y ) ≥ M α w ( y ) ≥ ∣ B ( y , r ) ∣ 1 − α w ( x 0 ) ≳ α , k k − ( 1 − α ) r w ( x 0 ) . Summing over D r ( x 0 ) D_r(x_0) D r ( x 0 )
w ( B ( x 0 , r ) ) ≥ w ( D r ( x 0 ) ) ≳ α , k , C k α r w ( x 0 ) . (37) w(B(x_0,r)) \ge w(D_r(x_0)) \gtrsim_{\alpha,k,C} k^{\alpha r}w(x_0). \tag*{(37)} w ( B ( x 0 , r )) ≥ w ( D r ( x 0 )) ≳ α , k , C k α r w ( x 0 ) . (37) On the other hand, applying M α w ( x 0 ) ≤ C w ( x 0 ) M_\alpha w(x_0) \le Cw(x_0) M α w ( x 0 ) ≤ C w ( x 0 ) B ( x 0 , r ) B(x_0,r) B ( x 0 , r ) ∣ B ( x 0 , r ) ∣ ≤ 4 k r |B(x_0,r)| \le4k^r ∣ B ( x 0 , r ) ∣ ≤ 4 k r
w ( B ( x 0 , r ) ) ≲ α , k , C k ( 1 − α ) r w ( x 0 ) . (38) w(B(x_0,r)) \lesssim_{\alpha,k,C} k^{(1-\alpha)r}w(x_0). \tag*{(38)} w ( B ( x 0 , r )) ≲ α , k , C k ( 1 − α ) r w ( x 0 ) . (38) Combining (37) and (38) gives k ( 2 α − 1 ) r ≲ 1 k^{(2\alpha-1)r} \lesssim1 k ( 2 α − 1 ) r ≲ 1 r r r α > 1 / 2 \alpha> 1/2 α > 1/2
We now consider α = 1 / 2 \alpha= 1/2 α = 1/2 M 1 / 2 w ≤ C w M_{1/2}w \le Cw M 1/2 w ≤ C w w w w x 0 x_0 x 0 w ( x 0 ) > 0 w(x_0) > 0 w ( x 0 ) > 0 x 0 x_0 x 0 w w w x ∈ T j x \in T^j x ∈ T j B ( x , j ) B(x,j) B ( x , j )
C w ( x ) ≥ M 1 / 2 w ( x ) ≥ w ( x 0 ) ∣ B ( x , j ) ∣ 1 / 2 ≥ w ( x 0 ) 2 k − j / 2 . Cw(x) \ge M_{1/2}w(x) \ge\frac{w(x_0)}{|B(x,j)|^{1/2}} \ge\frac{w(x_0)}{2} k^{-j/2}. C w ( x ) ≥ M 1/2 w ( x ) ≥ ∣ B ( x , j ) ∣ 1/2 w ( x 0 ) ≥ 2 w ( x 0 ) k − j /2 . Thus, writing v ( x ) = k − ∣ x ∣ / 2 v(x) = k^{-|x|/2} v ( x ) = k − ∣ x ∣/2
w ( x ) ≥ w ( x 0 ) 2 C v ( x ) , x ∈ T . (39) w(x) \ge\frac{w(x_0)}{2C}v(x), \qquad x \in T. \tag*{(39)} w ( x ) ≥ 2 C w ( x 0 ) v ( x ) , x ∈ T . (39) For x ∈ T j x \in T^j x ∈ T j (20) with α = 1 / 2 \alpha= 1/2 α = 1/2 β = − 1 / 2 \beta= -1/2 β = − 1/2 r = j r = j r = j
1 ∣ S ( x , j ) ∣ 1 / 2 ∑ y ∈ S ( x , j ) v ( y ) ≃ ( j + 1 ) k − j / 2 . \frac{1}{|S(x,j)|^{1/2}}\sum_{y\in S(x,j)}v(y) \simeq(j+1)k^{-j/2}. ∣ S ( x , j ) ∣ 1/2 1 y ∈ S ( x , j ) ∑ v ( y ) ≃ ( j + 1 ) k − j /2 . Since ∣ B ( x , j ) ∣ ≤ 2 ∣ S ( x , j ) ∣ |B(x,j)| \le2|S(x,j)| ∣ B ( x , j ) ∣ ≤ 2∣ S ( x , j ) ∣
M 1 / 2 v ( x ) ≥ 1 ∣ B ( x , j ) ∣ 1 / 2 ∑ y ∈ S ( x , j ) v ( y ) ≳ ( j + 1 ) k − j / 2 . M_{1/2}v(x) \ge\frac{1}{|B(x,j)|^{1/2}}\sum_{y\in S(x,j)}v(y) \gtrsim(j+1)k^{-j/2}. M 1/2 v ( x ) ≥ ∣ B ( x , j ) ∣ 1/2 1 y ∈ S ( x , j ) ∑ v ( y ) ≳ ( j + 1 ) k − j /2 . Combining this with (39) and the monotonicity of the maximal operator, we obtain
C w ( x ) ≥ M 1 / 2 w ( x ) ≥ w ( x 0 ) 2 C M 1 / 2 v ( x ) ≳ w ( x 0 ) C ( j + 1 ) k − j / 2 . Cw(x) \ge M_{1/2}w(x) \ge\frac{w(x_0)}{2C}M_{1/2}v(x) \gtrsim\frac{w(x_0)}{C}(j+1)k^{-j/2}. C w ( x ) ≥ M 1/2 w ( x ) ≥ 2 C w ( x 0 ) M 1/2 v ( x ) ≳ C w ( x 0 ) ( j + 1 ) k − j /2 . Summing over the k j k^j k j T j T^j T j
w ( T j ) ≳ C − 2 w ( x 0 ) ( j + 1 ) k j / 2 . (40) w(T^j) \gtrsim C^{-2}w(x_0)(j+1)k^{j/2}. \tag*{(40)} w ( T j ) ≳ C − 2 w ( x 0 ) ( j + 1 ) k j /2 . (40) On the other hand, applying the assumed inequality at the root gives
w ( T j ) ≤ w ( B ( x 0 , j ) ) ≤ C w ( x 0 ) ∣ B ( x 0 , j ) ∣ 1 / 2 ≤ 2 C w ( x 0 ) k j / 2 . w(T^j) \le w(B(x_0,j)) \le Cw(x_0)|B(x_0,j)|^{1/2} \le2Cw(x_0)k^{j/2}. w ( T j ) ≤ w ( B ( x 0 , j )) ≤ C w ( x 0 ) ∣ B ( x 0 , j ) ∣ 1/2 ≤ 2 C w ( x 0 ) k j /2 . Together with (40) , this implies j + 1 ≲ C 3 j+1 \lesssim C^3 j + 1 ≲ C 3 j ≥ 0 j \ge0 j ≥ 0
Combining Theorem 1.3 with Proposition 3.11, we obtain the following result.
Corollary 3.12. Let 1 / 2 ≤ α < 1 1/2 \le\alpha< 1 1/2 ≤ α < 1 (2) .
AI Disclosure
ChatGPT was used for language editing and proofreading and also helped suggest considering subtrees for the example demonstrating the failure of the Fefferman–Stein estimate in Section 3. All mathematical results, arguments, proofs, and conclusions were developed and established by the authors.
G. Ibañez Firnkorn. Instituto de Matemática (INMABB), Departamento de Matemática, Universidad Nacional del Sur (UNS)-CONICET, Bahía Blanca, Argentina Email address: gonzalo.ibanez@uns.edu.ar
Jorgelina Recchi. Instituto de Matemática (INMABB), Departamento de Matemática, Universidad Nacional del Sur (UNS)-CONICET, Bahía Blanca, Argentina Email address: drecchi@uns.edu.ar
