Introduction
A boundary-to-boundary walk can be weighted critically without fixing its length. In that law the separation of the endpoints supplies the spatial scale. Our question is whether prescribing the terminal boundary point, rather than summing over an interval, changes the power relating length to that scale. The distinction is substantial: a macroscopic terminal interval contains order R R R
We use the regular honeycomb lattice, embedded as the centers of the triangles of an equilateral triangular tiling of side one. Neighboring centers are joined across a common triangle edge. A port is the midpoint of a boundary edge of a domain made of whole triangles. A path enters and leaves normally at its ports, visits only centers in the interior, and visits no center twice. Its length L ( γ ) L(\gamma) L ( γ )
w ( γ ) = κ L ( γ ) , κ = ( 2 cos ( π / 8 ) ) − 1 = ( 2 + 2 ) − 1 / 2 . w(\gamma)=\kappa^{L(\gamma)},\qquad\kappa=(2\cos(\pi/8))^{-1}=(2+\sqrt{2})^{-1/2}. w ( γ ) = κ L ( γ ) , κ = ( 2 cos ( π /8 ) ) − 1 = ( 2 + 2 ) − 1/2 . Ports carry no weight. For a domain D D D a , b a,b a , b
K D ( a , b ) = ∑ γ : a → b in D w ( γ ) . K_D(a,b)=\sum_{\gamma:a\to b\text{ in }D}w(\gamma). K D ( a , b ) = γ : a → b in D ∑ w ( γ ) . Whenever the sum is positive and finite, dividing by it defines the critical chord law. The path has no other boundary exit. A pure cut follows one of the three tiling-edge directions; its admissible translates bound whole triangular rows. Distances refer to this fixed embedding.
Theorem 1.1 (Prescribed boundary endpoints). Let D R D_R D R R R R a R , b R a_R,b_R a R , b R ε > 0 \varepsilon>0 ε > 0
P D R ; a R , b R { R 4 / 3 − ε ≤ L ( γ ) ≤ R 4 / 3 + ε } ⟶ 1. \mathbb{P}_{D_R;a_R,b_R}\left\{R^{4/3-\varepsilon}\le L(\gamma)\le R^{4/3+\varepsilon}\right\}\longrightarrow1. P D R ; a R , b R { R 4/3 − ε ≤ L ( γ ) ≤ R 4/3 + ε } ⟶ 1. *The same conclusion holds in a full half-plane for prescribed boundary ports at distance R R R R R R O ( R ) O(R) O ( R ) π / 3 \pi/3 π /3 2 π / 3 2\pi/3 2 π /3 R R R O ( R ) O(R) O ( R ) c R − 5 / 4 cR^{-5/4} c R − 5/4
The theorem concerns each admissible choice of the two ports. A first step, proved in Section 3, fixes the source and sums the terminal port over a separated macroscopic interval. Its normalizing mass is of order R − 1 / 4 R^{-1/4} R − 1/4
For half-plane arches we additionally prove E L = R 4 / 3 + o ( 1 ) \mathbb{E}L=R^{4/3+o(1)} E L = R 4/3 + o ( 1 ) [15 ] , Theorem 8.1, where the gap ranges over [ R , 2 R ] [R,2R] [ R , 2 R ] R R R [ H , 2 H ] [H,2H] [ H , 2 H ]
Historical context. Nienhuis’s analysis of the dilute O ( n ) O(n) O ( n ) 4 / 3 4/3 4/3 [11 ] . Duminil-Copin and Smirnov proved the honeycomb connective constant using a local winding relation and a boundary identity [3 ] . Their mid-edge convention is the port convention used here. For critical strip mass, Beaton, Bousquet-Mélou, de Gier, Duminil-Copin and Guttmann proved decay to zero in their work on surface adsorption [2 ] . Glazman and Manolescu gave a shorter proof and a logarithmic subsequence bound, and proved boundary two-point invariance for columnwise rhombic half-planes with angles in [ π / 3 , 2 π / 3 ] [\pi/3,2\pi/3] [ π /3 , 2 π /3 ] [6 ] . Krachun and Panagiotis subsequently proved a polynomial upper bound for strip-crossing mass and a quantitative sub-ballistic estimate for uniform honeycomb walks [9 ] . These results distinguish critical activity, decay of crossing mass, and length–distance exponents; the finite inputs below specify the stronger estimates used here.
The irreducible-bridge construction goes back to Kesten [8 ] ; its honeycomb mid-edge form appears in [2 ] , Appendix]. On the square lattice, the infinite critical law and its relation to strip-spanning laws are developed by Lawler–Schramm–Werner [10 ] , Appendix] and Dhy–Gilbert–Kennedy–Lawler–Passon [4 ] , Section 2.4]. Gilbert’s appendix records Kennedy’s heuristic connecting strip mass of order h − 1 / 4 h^{-1/4} h − 1/4 h − 3 / 4 h^{-3/4} h − 3/4 n 4 / 3 n^{4/3} n 4/3 [5 ] , Appendix]. Here whole irreducible bridges are independent increments; height, length and lateral displacement within one increment remain dependent.
The quantitative starting point is the sharp finite marked-polygon and bridge estimates of [16 ] , together with the renewal estimates of [15 ] ; Section 2 states exactly what we use. These are stronger inputs than the preceding critical and sub-ballistic results. The terminal-free fixed-height probability law and the prescribed half-plane arch probability law already appear in [15 ] , Propositions 5.7 and 5.8]. We give a different proof of the latter, through common-level pair renewals and arch-kernel regularity, and develop the local estimates needed to assemble the prescribed-endpoint laws in the other geometries. The mean estimates require additional length-weighted control and retain separate finite inputs.
The three endpoint mechanisms. For two ports on one line, grow independent bridge sequences upward from the two ports and stop at their common renewal levels. Their common height transform has exponent 1 / 2 1/2 1/2 O ( m − 2 ) O(m^{-2}) O ( m − 2 ) m m m
For parallel cuts one must match height and lateral displacement at once. A height-tail bound alone does not pay for one prescribed height. Section 5 bounds all fixed-order differences of the strip mass, starting from the finite projection representation in [16 ] . This local regularity gives the one-piece height atom bound in Section 6. Together with two-dimensional concentration and a sum of heightwise maximal atoms, it pays the parallel endpoint restriction in Section 7. For nonparallel cuts, small normal prefixes initially exhaust only a sum over terminal ports. A local visit estimate and an artificial-arc boundary identity then control the exceptional mass at the prescribed terminal port. Each case thus pays for its actual endpoint restriction before dividing by the chord mass; Section 8 completes the nonparallel case.
The means use two further arguments. In Section 9, the calibrated finite estimates of [12 ] and their renewal consequences control marked arms at both ends of an arch. The resulting length-weighted contributions are summable over arch heights at each prescribed gap. In Section 10, the finite strip moments of [14 ] feed a renewal-reward identity. A two-bridge boundary comparison controls variation between terminal ports after averaging over heights, leading to the density-one and aligned-mixture mean laws rather than an all-height pointwise claim.
Finite estimates and the irreducible law
We first identify the inputs that do not depend on prescribed endpoints. A strip of integral height h h h d h dh d h d = 3 / 2 d = \sqrt{3}/2 d = 3 /2 B h B_h B h B 0 = 1 B_0 = 1 B 0 = 1 Z + h / 2 \mathbb{Z} + h/2 Z + h /2 B h B_h B h
Proposition 2.1 (Finite boundary and length inputs). The following estimates hold in the port convention above.
For every integer h ≥ 0 h \ge0 h ≥ 0 B h ≲ ( 1 + h ) − 1 / 4 B_h \lesssim(1+h)^{-1/4} B h ≲ ( 1 + h ) − 1/4 h h h C h 13 / 12 C h^{13/12} C h 13/12 B h { L ≥ c h 4 / 3 } ≥ c h − 1 / 4 B_h\{L \ge c h^{4/3}\} \ge c h^{-1/4} B h { L ≥ c h 4/3 } ≥ c h − 1/4 h ≥ 1 h \ge1 h ≥ 1 r ≥ 0 r \ge0 r ≥ 0 r r r C ( 1 + r ) − 1 / 4 C(1+r)^{-1/4} C ( 1 + r ) − 1/4
In a convex pure domain, the mass from one boundary source of completed paths of diameter at least r r r C ( 1 + r ) − 1 / 4 C(1+r)^{-1/4} C ( 1 + r ) − 1/4 r ≥ 0 r \ge0 r ≥ 0 l ≥ 1 l \ge1 l ≥ 1 K l ≍ l − 5 / 4 K_l \asymp l^{-5/4} K l ≍ l − 5/4 K l + 1 ≤ K l K_{l+1} \le K_l K l + 1 ≤ K l A l Al A l c l − 5 / 4 cl^{-5/4} c l − 5/4 A A A
From a prescribed port a a a n n n m ⋅ ( z − a ) = s m \cdot(z-a) = s m ⋅ ( z − a ) = s m m m n n n m m m π / 3 \pi/3 π /3 s > 0 s > 0 s > 0 a a a s − 1 / 4 s^{-1/4} s − 1/4 s s s s s s a + [ 0 , s ] m a + [0,s]m a + [ 0 , s ] m
In a convex pure domain of diameter O ( R ) O(R) O ( R )
These are the finite results of [16 ] , Theorem 1.1, Propositions 7.1, 7.4, 10.1 and Corollary 8.1. The statement records the input before use; the full analytic and finite geometric proofs are in that companion. In particular its free-boundary mean is a finite input, not a consequence of the endpoint theorem here. The disk-transfer construction [13 ] , Theorem 1.1 provides an independent fixed-corridor first moment. We use the specified finite method at each application below; in particular the logarithmic confinement in the polynomial-vacuum route retains its own precision.
We use also the local contour identity with phase e i σ W e^{i\sigma W} e iσ W σ = 3 / 8 \sigma= 3/8 σ = 3/8 W W W e − 5 i W / 8 e^{-5iW/8} e − 5 iW /8 [3 ] , Section 2, Lemma 1. At an unvisited center the two outgoing weights sum to the incoming weight, since 2 κ cos ( π / 8 ) = 1 2\kappa\cos(\pi/8) = 1 2 κ cos ( π /8 ) = 1 8 π / 3 8\pi/3 8 π /3 [ cos ( 3 π / 8 ) , 1 ] [\cos(3\pi/8),1] [ cos ( 3 π /8 ) , 1 ] [3 ] ; below we explain the additional arc and spectator constructions when they are used.
An irreducible strict bridge has positive integral height and no intermediate singly crossed level. Every bridge has a unique ordered factorization into such pieces. The honeycomb mid-edge version of the irreducible construction is given in [2 ] , Appendix]. Let p ( γ ) = κ L ( γ ) p(\gamma) = \kappa^{L(\gamma)} p ( γ ) = κ L ( γ ) H , Z , L , D H,Z,L,D H , Z , L , D
Proposition 2.2 (Renewal inputs). The mass p p p p p p s > 0 s > 0 s > 0
∑ h ≥ 0 B h e − s h = ( 1 − E p e − s H ) − 1 ≍ s − 3 / 4 ( s ↓ 0 ) . \sum_{h\ge0} B_h e^{-sh} = (1-\mathbb{E}_p e^{-sH})^{-1} \asymp s^{-3/4}\qquad(s\downarrow0). h ≥ 0 ∑ B h e − s h = ( 1 − E p e − sH ) − 1 ≍ s − 3/4 ( s ↓ 0 ) . Moreover,
p ( H > x ) + p ( D > x ) ≤ C x − 3 / 4 , E p [ L ; H ≤ x ] ≤ x 7 / 12 + o ( 1 ) , p(H>x)+p(D>x)\le Cx^{-3/4},\qquad\mathbb{E}_p[L;H\le x]\le x^{7/12+o(1)}, p ( H > x ) + p ( D > x ) ≤ C x − 3/4 , E p [ L ; H ≤ x ] ≤ x 7/12 + o ( 1 ) , p ( L > n ) ≤ n − 9 / 16 + o ( 1 ) , 1 − E p e − u L = u 9 / 16 + o ( 1 ) . p(L>n)\le n^{-9/16+o(1)},\qquad1-\mathbb{E}_p e^{-uL}=u^{9/16+o(1)}. p ( L > n ) ≤ n − 9/16 + o ( 1 ) , 1 − E p e − uL = u 9/16 + o ( 1 ) . Almost surely, for independent pieces, their first k k k k 4 / 3 + o ( 1 ) k^{4/3+o(1)} k 4/3 + o ( 1 ) k 16 / 9 + o ( 1 ) k^{16/9+o(1)} k 16/9 + o ( 1 ) k 4 / 3 + o ( 1 ) k^{4/3+o(1)} k 4/3 + o ( 1 )
The proofs and the full joint-law conventions are [15 ] , Propositions 2.2, 2.4, Remark 2.5 and Lemma 2.6. Only the independent pieces are asserted here; their coordinates are not independent. B h B_h B h
Throughout, f ( R ) = R α + o ( 1 ) f(R)=R^{\alpha+o(1)} f ( R ) = R α + o ( 1 ) R α − ϵ ≤ f ( R ) ≤ R α + ϵ R^{\alpha-\epsilon}\le f(R)\le R^{\alpha+\epsilon} R α − ϵ ≤ f ( R ) ≤ R α + ϵ ϵ > 0 \epsilon>0 ϵ > 0
A prescribed source and a terminal interval
Prescribing one endpoint and summing the other over a separated interval gives normalizing mass of order R − 1 / 4 R^{-1/4} R − 1/4
Interior connectors
Proposition 3.1 (Interior connectors). For two macroscopically separated side-interior ports in the hexagons of Theorem 1.1, the critical mass of connecting paths inside the hexagon is at least c R − 5 / 4 cR^{-5/4} c R − 5/4 R R R O ( R ) O(R) O ( R )
Proof. We adapt the cut-averaging construction in the proof of [16 ] , placing the routes in the interior rather than the exterior. The geometric choices below ensure that the decomposition can be recovered for two different lists of cuts.
Routes and junction cuts. Work first in a hexagon and scale distances by R R R
Approximate the arcs by finitely many segments in pure normal directions, retaining the initial and terminal straight portions and making successive direction changes at most π / 3 \pi/3 π /3 R R R
Index a branch’s segments by their guiding directions n j n_j n j j j j n j n_j n j arriving direction. The next leg starts with inward normal n j n_j n j n j + 1 n_{j+1} n j + 1
The geometric choices have positive clearances. They persist in a neighborhood of the scaled endpoint data, including the small support perturbations allowed in Theorem 1.1. Compactness gives uniform choices of the number of segments and their relative tolerances.
One list and two lists of levels. Vary each junction cut in a fixed small window of admissible levels of width comparable to R R R k k k Λ R \Lambda_R Λ R k − 2 k-2 k − 2
For a list λ \lambda λ I λ ( γ ) I_\lambda(\gamma) I λ ( γ )
F R ( γ ) = 1 ∣ Λ R ∣ ∑ λ ∈ Λ R I λ ( γ ) , ∫ G d μ = ∑ γ : a → b κ L ( γ ) G ( γ ) . F_R(\gamma)=\frac{1}{|\Lambda_R|}\sum_{\lambda\in\Lambda_R} I_\lambda(\gamma), \qquad\int G\,d\mu=\sum_{\gamma:a\to b}\kappa^{L(\gamma)}G(\gamma). F R ( γ ) = ∣ Λ R ∣ 1 λ ∈ Λ R ∑ I λ ( γ ) , ∫ G d μ = γ : a → b ∑ κ L ( γ ) G ( γ ) . Multiplication of the k k k
∫ F R d μ ≥ c R − k / 4 − 5 / 4 . \int F_R\,\mathrm{d}\mu\ge cR^{-k/4-5/4}. ∫ F R d μ ≥ c R − k /4 − 5/4 . For two lists, split each branch at both crossings in each junction neighborhood. The two crossings occur in their normal-level order: the incoming portions start below both levels, and the outgoing portions end above both. The intervening portion is consequently a strict bridge of the level difference. The portions between successive neighborhoods still have separation comparable to R R R O ( R − 1 / 4 ) O(R^{-1/4}) O ( R − 1/4 ) k k k O ( R − 5 / 4 ) O(R^{-5/4}) O ( R − 5/4 )
For two independent uniform levels u , u ′ u,u' u , u ′ R R R
E ( 1 + ∣ u − u ′ ∣ ) − α ≤ C α R − α ( 0 < α < 1 ) . \mathbb{E}(1+|u-u'|)^{-\alpha}\le C_\alpha R^{-\alpha}\qquad(0<\alpha<1). E ( 1 + ∣ u − u ′ ∣ ) − α ≤ C α R − α ( 0 < α < 1 ) . Every ordinary junction gives one difference bridge and uses this bound with α = 1 / 4 \alpha=1/4 α = 1/4 α = 1 / 2 \alpha=1/2 α = 1/2 B 0 = 1 B_0=1 B 0 = 1
∫ F R 2 d μ ≤ C R − k / 4 ⏟ long portions R − ( k − 2 ) / 4 R − 1 / 2 ⏟ difference bridges R − 5 / 4 ⏟ terminal arch = C R − 2 k / 4 − 5 / 4 . \int F_R^2\,\mathrm{d}\mu\le C\underbrace{R^{-k/4}}_{\text{long portions}}\underbrace{R^{-(k-2)/4}R^{-1/2}}_{\text{difference bridges}}\underbrace{R^{-5/4}}_{\text{terminal arch}}=CR^{-2k/4-5/4}. ∫ F R 2 d μ ≤ C long portions R − k /4 difference bridges R − ( k − 2 ) /4 R − 1/2 terminal arch R − 5/4 = C R − 2 k /4 − 5/4 . Cauchy–Schwarz now yields
μ ( F R > 0 ) ≥ ( ∫ F R d μ ) 2 ∫ F R 2 d μ ≥ c R − 5 / 4 . \mu(F_R>0)\ge\frac{\left(\int F_R\,\mathrm{d}\mu\right)^2}{\int F_R^2\,\mathrm{d}\mu}\ge cR^{-5/4}. μ ( F R > 0 ) ≥ ∫ F R 2 d μ ( ∫ F R d μ ) 2 ≥ c R − 5/4 . For a wedge, add distant pure cuts at distance A R AR A R A A A c R cR c R c ′ R c'R c ′ R
Theorem 3.2 (A source and a macroscopic terminal interval). In the hexagons of Theorem 1.1, fix a source port tending after scaling to a side-interior point. Let the terminal ports form a macroscopic interval converging to a compact side-interior interval at positive distance from the source limit. Under critical weights, normalized over those chords, length is R 4 / 3 + o ( 1 ) R^{4/3+o(1)} R 4/3 + o ( 1 ) R − 1 / 4 R^{-1/4} R − 1/4
The diameter bound gives the upper normalization, and summing Proposition 3.1 over order-R R R
Bridge length and displacement windows
Lemma 3.3 (Fixed-height bridge length). Under critical weights on strict bridges of integral height h h h δ > 0 \delta>0 δ > 0
Pr { h 4 / 3 − δ ≤ L ≤ h 4 / 3 + δ } ⟶ 1. \Pr\{h^{4/3-\delta}\le L\le h^{4/3+\delta}\}\longrightarrow1. Pr { h 4/3 − δ ≤ L ≤ h 4/3 + δ } ⟶ 1. *The same conclusion holds after any restriction retaining a fixed positive fraction of B h B_h B h
The fixed-height conclusion also appears in [15 ] , Proposition 5.7. We give the argument from the finite and renewal inputs selected in Section 2.
Proof. It suffices to treat 0 < δ < 1 0<\delta<1 0 < δ < 1 B h − 1 ∑ L w ≤ C h 4 / 3 B_h^{-1}\sum Lw\le Ch^{4/3} B h − 1 ∑ L w ≤ C h 4/3 p p p M , A > 1 M,A>1 M , A > 1
p { x ≤ H ≤ M x , D ≤ A x } ≥ c x − 3 / 4 p\{x\le H\le Mx,\ D\le Ax\}\ge cx^{-3/4} p { x ≤ H ≤ M x , D ≤ A x } ≥ c x − 3/4 for all sufficiently large x x x E p [ 1 − e − H / ( K x ) ] ≍ ( K x ) − 3 / 4 \mathbb{E}_p[1-e^{-H/(Kx)}]\asymp(Kx)^{-3/4} E p [ 1 − e − H / ( K x ) ] ≍ ( K x ) − 3/4 H ≤ x H\le x H ≤ x O ( x − 3 / 4 / K ) O(x^{-3/4}/K) O ( x − 3/4 / K ) H > M x H>Mx H > M x O ( ( M x ) − 3 / 4 ) O((Mx)^{-3/4}) O (( M x ) − 3/4 ) K K K M M M A ≥ M A\ge M A ≥ M
This window gives the concentration estimate
sup j Pr ( H 1 + ⋯ + H k = j ) ≤ C k − 4 / 3 . \sup_j\Pr(H_1+\cdots+H_k=j)\le Ck^{-4/3}. j sup Pr ( H 1 + ⋯ + H k = j ) ≤ C k − 4/3 . To see this, symmetrize the characteristic function using an independent copy of H H H (3.1) , take x x x ∣ θ ∣ − 1 |\theta|^{-1} ∣ θ ∣ − 1 c ∣ θ ∣ 3 / 4 c|\theta|^{3/4} c ∣ θ ∣ 3/4 (3.2) .
For k ≥ 2 k\ge2 k ≥ 2 h h h h / 2 h/2 h /2 C k h − 3 / 4 Ckh^{-3/4} C k h − 3/4 C k − 4 / 3 Ck^{-4/3} C k − 4/3 k = 1 k=1 k = 1 C h − 3 / 4 Ch^{-3/4} C h − 3/4 η > 0 \eta>0 η > 0
∑ 1 ≤ k ≤ h 3 / 4 − η Pr { ∑ j ≤ k H j = h } ≤ C h − 3 / 4 ( 1 + ∑ 2 ≤ k ≤ h 3 / 4 − η k − 1 / 3 ) = o ( B h ) . \sum_{1\le k\le h^{3/4-\eta}}\Pr\left\{\sum_{j\le k}H_j=h\right\}
\le Ch^{-3/4}\left(1+\sum_{2\le k\le h^{3/4-\eta}}k^{-1/3}\right)=o(B_h). 1 ≤ k ≤ h 3/4 − η ∑ Pr ⎩ ⎨ ⎧ j ≤ k ∑ H j = h ⎭ ⎬ ⎫ ≤ C h − 3/4 1 + 2 ≤ k ≤ h 3/4 − η ∑ k − 1/3 = o ( B h ) . For the remaining counts put t = h 4 / 3 − δ t=h^{4/3-\delta} t = h 4/3 − δ
Pr { ∑ j ≤ k L j < t } ≤ ( E p e − L / t ) k ≤ exp { − k t − 9 / 16 + o ( 1 ) } . \Pr\left\{\sum_{j\le k}L_j<t\right\}\le\left(\mathbb{E}_p e^{-L/t}\right)^k\le\exp\{-kt^{-9/16+o(1)}\}. Pr ⎩ ⎨ ⎧ j ≤ k ∑ L j < t ⎭ ⎬ ⎫ ≤ ( E p e − L / t ) k ≤ exp { − k t − 9/16 + o ( 1 ) } . Choose η < 9 δ / 16 \eta<9\delta/16 η < 9 δ /16 k > h 3 / 4 − η k>h^{3/4-\eta} k > h 3/4 − η B h B_h B h B h B_h B h
We next arrange a displacement that points inward at both chord ends. This is needed when a translated chord is attached back to its original supports. Distances in the following geometric statement are physical; the bridge height h h h
Lemma 3.4 (Positive projections at two supports). Let n , m n,m n , m m ≠ − n m\ne-n m = − n b , B , A , c 0 > 0 b,B,A,c_0>0 b , B , A , c 0 > 0 δ > 0 \delta>0 δ > 0 h h h n n n v v v
b h ≤ m ⋅ v ≤ B h , D ≤ A h , h 4 / 3 − δ ≤ L ≤ h 4 / 3 + δ , bh\le m\cdot v\le Bh,\qquad D\le Ah,\qquad h^{4/3-\delta}\le L\le h^{4/3+\delta}, bh ≤ m ⋅ v ≤ B h , D ≤ A h , h 4/3 − δ ≤ L ≤ h 4/3 + δ , is at least c 0 h − 1 / 4 c_0h^{-1/4} c 0 h − 1/4 n , m n,m n , m δ \delta δ δ \delta δ
Proof. If m ⋅ n > 0 m \cdot n > 0 m ⋅ n > 0 n n n 2 π / 3 2\pi/3 2 π /3 n ′ n' n ′ π / 3 \pi/3 π /3 T = d h T = dh T = d h n ′ n' n ′ 3 T / 2 3T/2 3 T /2 n n n n n n n ′ n' n ′ n n n n n n m m m 3 T / 4 3T/4 3 T /4 T / 4 T/4 T /4 n n n m m m 5 T / 8 5T/8 5 T /8 n n n O ( h ) O(h) O ( h )
Average that cut over a small window of order h h h c h − 1 / 2 ch^{-1/2} c h − 1/2 C h − 3 / 4 Ch^{-3/4} C h − 3/4 c h − 1 / 4 ch^{-1/4} c h − 1/4 B n B_n B n h h h
The cost of two translation choices
Fix pure normals n , m n,m n , m m ≠ − n m \ne-n m = − n z z z n ⋅ z > 0 n \cdot z > 0 n ⋅ z > 0 D n ( z ) D_n(z) D n ( z ) n n n z z z n ⋅ z < 0 n \cdot z < 0 n ⋅ z < 0 D n ( z ) = D n ( − z ) D_n(z) = D_n(-z) D n ( z ) = D n ( − z ) n n n D n ( 0 ) = 1 D_n(0) = 1 D n ( 0 ) = 1 D n ( z ) = 0 D_n(z) = 0 D n ( z ) = 0 z z z n ⋅ z = 0 n \cdot z = 0 n ⋅ z = 0
Lemma 3.5 (Mixed displacement estimate). For every fixed C 0 < ∞ C_0 < \infty C 0 < ∞ C < ∞ C < \infty C < ∞ s ≥ 1 s \ge1 s ≥ 1
∑ ∣ z ∣ ≤ C 0 s D n ( z ) ( 1 + ∣ m ⋅ z ∣ ) − 1 / 4 ≤ C s 1 / 2 . \sum_{|z|\le C_0s} D_n(z)(1+|m\cdot z|)^{-1/4} \le Cs^{1/2}. ∣ z ∣ ≤ C 0 s ∑ D n ( z ) ( 1 + ∣ m ⋅ z ∣ ) − 1/4 ≤ C s 1/2 . If Y Y Y m m m n n n
sup y Pr { ∑ j ≤ k Y j = y } ≤ C k − 4 / 3 , sup w E ( 1 + ∣ w + ∑ j ≤ k Y j ∣ ) − 1 / 4 ≤ C k − 1 / 3 . (1) \sup_y \Pr\left\{\sum_{j\le k}Y_j=y\right\}\le Ck^{-4/3},\qquad
\sup_w \mathbb{E}\left(1+\left|w+\sum_{j\le k}Y_j\right|\right)^{-1/4}\le Ck^{-1/3}. \tag*{(1)} y sup Pr ⎩ ⎨ ⎧ j ≤ k ∑ Y j = y ⎭ ⎬ ⎫ ≤ C k − 4/3 , w sup E 1 + w + j ≤ k ∑ Y j − 1/4 ≤ C k − 1/3 . (1) Proof. The height-and-diameter window (3.1) and lateral reflection symmetry imply, for fixed B > b > 0 B>b>0 B > b > 0
p { b x ≤ ∣ Y ∣ ≤ B x } ≥ c x − 3 / 4 . p\{bx\le|Y|\le Bx\}\ge cx^{-3/4}. p { b x ≤ ∣ Y ∣ ≤ B x } ≥ c x − 3/4 . Indeed the two reflected displacements have m m m 2 ( m ⋅ n ) d H 2(m\cdot n)dH 2 ( m ⋅ n ) d H m ⋅ n ≠ 0 m\cdot n\ne0 m ⋅ n = 0 H H H (3.2) gives the first bound in (1) on the projection’s actual lattice. For the second, sum this atom bound with weight ( 1 + ∣ y ∣ ) − 1 / 4 (1+|y|)^{-1/4} ( 1 + ∣ y ∣ ) − 1/4 ∣ y ∣ ≤ k 4 / 3 |y|\le k^{4/3} ∣ y ∣ ≤ k 4/3 k − 1 / 3 k^{-1/3} k − 1/3 w w w
It suffices to sum over n ⋅ z ≥ 0 n\cdot z\ge0 n ⋅ z ≥ 0 (3.3) . Use renewal factorization, splitting k ≥ 2 k\ge2 k ≥ 2 C 1 s C_1s C 1 s
C exp { − c k / s 3 / 4 } . C\exp\{-ck/s^{3/4}\}. C exp { − c k / s 3/4 } . Conditioning on that half, the second bound in (1) applies to the other half, with a shift given by the first half’s projection. Thus the sum is at most
C + C ∑ k ≥ 2 e − c k / s 3 / 4 k − 1 / 3 ≤ C s 1 / 2 . C+C\sum_{k\ge2}e^{-ck/s^{3/4}}k^{-1/3}\le Cs^{1/2}. C + C k ≥ 2 ∑ e − c k / s 3/4 k − 1/3 ≤ C s 1/2 . The constant covers one piece and the zero-height match. No coordinates within one piece have been made independent.
Recoverable attachments
Consider the two inward support normals n n n m m m a a a a a a R R R R R R m ≠ − n m \ne-n m = − n c > 0 c > 0 c > 0 s ≤ c R s \le cR s ≤ c R v v v ∣ v ∣ ≤ C s \lvert v\rvert\le Cs ∣ v ∣ ≤ C s
g 1 = n ⋅ v / d , g 2 = m ⋅ v / d g_1 = n \cdot v/d, \qquad g_2 = m \cdot v/d g 1 = n ⋅ v / d , g 2 = m ⋅ v / d between fixed positive multiples of s s s g i g_i g i C ′ s C's C ′ s s 4 / 3 − δ s^{4/3-\delta} s 4/3 − δ s 4 / 3 + δ s^{4/3+\delta} s 4/3 + δ c c c
Proposition 3.6 (Attachment moments). Let E \mathcal{E} E a a a E E E N s N_s N s μ a \mu_a μ a
∫ N s d μ a ≥ c s 1 / 2 E , ∫ N s 2 d μ a ≤ C s R − 1 / 4 . (2) \int N_s\,\mathrm{d}\mu_a \ge c s^{1/2} E, \qquad\int N_s^2\,\mathrm{d}\mu_a \le C s R^{-1/4}. \tag*{(2)} ∫ N s d μ a ≥ c s 1/2 E , ∫ N s 2 d μ a ≤ C s R − 1/4 . (2) Here the side-1 outer landing is exactly a a a
If the input chords lie in a convex pure domain of diameter O ( R ) O(R) O ( R ) s = c R s=cR s = c R O ( R ) O(R) O ( R )
∫ N R f r e e d μ f r e e ≥ c R 3 / 2 E , ∫ ( N R f r e e ) 2 d μ f r e e ≤ C R 7 / 4 . (3) \int N_R^{\mathrm{free}}\,\mathrm{d}\mu_{\mathrm{free}} \ge c R^{3/2} E, \qquad\int\left(N_R^{\mathrm{free}}\right)^2\,\mathrm{d}\mu_{\mathrm{free}} \le C R^{7/4}. \tag*{(3)} ∫ N R free d μ free ≥ c R 3/2 E , ∫ ( N R free ) 2 d μ free ≤ C R 7/4 . (3) The measure μ f r e e \mu_{\mathrm{free}} μ free δ \delta δ s s s δ \delta δ
Proof. For fixed source a a a a a a v v v V s V_s V s s s s c 0 s − 1 / 4 c_0s^{-1/4} c 0 s − 1/4 v v v c s 3 / 4 cs^{3/4} c s 3/4 c s − 1 / 4 cs^{-1/4} c s − 1/4 v v v
Let A v ( E ) \mathcal{A}_v(\mathcal{E}) A v ( E ) v v v
N s ( Γ ) = ∑ v ∈ V s 1 { Γ ∈ A v ( E ) } . N_s(\Gamma)=\sum_{v\in V_s}\mathbf{1}_{\{\Gamma\in\mathcal{A}_v(\mathcal{E})\}}. N s ( Γ ) = v ∈ V s ∑ 1 { Γ ∈ A v ( E )} . For a fixed v v v (2) . Recovering two translations. For v , v ′ v,v' v , v ′
The central portion is between the two innermost crossings. It stays inside both corresponding supports, because each construction remains beyond its own translated support after crossing, except for its localized branch at the other end when the supports coincide. Those other-end branches lie outside the central portion. Its initial point is prescribed once the two translations are fixed, its terminal point is free, and their separation is of order R R R C R − 1 / 4 CR^{-1/4} C R − 1/4
Put z = v ′ − v z=v'-v z = v ′ − v z z z D n ( z ) D_n(z) D n ( z ) C ( 1 + ∣ m ⋅ z ∣ ) − 1 / 4 C(1+|m\cdot z|)^{-1/4} C ( 1 + ∣ m ⋅ z ∣ ) − 1/4 v = v ′ v=v' v = v ′ D n D_n D n C s − 1 / 4 Cs^{-1/4} C s − 1/4 v o u t v_{\mathrm{out}} v out D n ( v o u t ) D_n(v_{\mathrm{out}}) D n ( v out )
∑ v o u t ∈ V s D n ( v o u t ) ≤ C ∑ h ≤ C s B h ≤ C s 3 / 4 . \sum_{v_{\mathrm{out}}\in V_s}D_n(v_{\mathrm{out}})\leq C\sum_{h\leq Cs}B_h\leq Cs^{3/4}. v out ∈ V s ∑ D n ( v out ) ≤ C h ≤ C s ∑ B h ≤ C s 3/4 . These upper bounds may be multiplied after dropping avoidance constraints between the five portions. Lemma 3.5 sums the two difference factors. The complete accounting is
∫ N s 2 d μ a ≤ C s 3 / 4 ⏟ side-1 outer sum s 1 / 2 ⏟ two differences R − 1 / 4 ⏟ central chord s − 1 / 4 ⏟ side-2 outer = C s R − 1 / 4 . \int N_s^2\,d\mu_a\leq C\underbrace{s^{3/4}}_{\text{side-1 outer sum}}\underbrace{s^{1/2}}_{\text{two differences}}\underbrace{R^{-1/4}}_{\text{central chord}}\underbrace{s^{-1/4}}_{\text{side-2 outer}}=CsR^{-1/4}. ∫ N s 2 d μ a ≤ C side-1 outer sum s 3/4 two differences s 1/2 central chord R − 1/4 side-2 outer s − 1/4 = C s R − 1/4 . The argument used the two supports and endpoint clearance, not a bound on the diameter of the input chord.
Both outer landings free. Take s = c R s=cR s = c R R 2 R^2 R 2 c R 2 R − 1 / 2 E cR^2R^{-1/2}E c R 2 R − 1/2 E O ( R 2 R − 1 / 4 ) O(R^2R^{-1/4}) O ( R 2 R − 1/4 )
∫ ( N R f r e e ) 2 d μ f r e e ≤ C R 2 − 1 / 4 R 1 / 2 R − 1 / 4 R − 1 / 4 = C R 7 / 4 . \int(N_R^{\mathrm{free}})^2\,d\mu_{\mathrm{free}}\leq CR^{2-1/4}R^{1/2}R^{-1/4}R^{-1/4}=CR^{7/4}. ∫ ( N R free ) 2 d μ free ≤ C R 2 − 1/4 R 1/2 R − 1/4 R − 1/4 = C R 7/4 . The localized attachments and O ( R ) O(R) O ( R ) O ( R ) O(R) O ( R ) □ \square □
Proof of the interval length law
Proof of Theorem 3.2. It suffices to prove the length bounds for 0 < ϵ < 1 0<\epsilon<1 0 < ϵ < 1 R − 1 / 4 R^{-1/4} R − 1/4 h h h h ≍ R h\asymp R h ≍ R B h B_h B h
For an input event of mass E = p 0 R − 1 / 4 E=p_0R^{-1/4} E = p 0 R − 1/4
μ a ( N s > 0 ) ≥ ( ∫ N s d μ a ) 2 ∫ N s 2 d μ a ≥ c p 0 2 R − 1 / 4 . (4) \mu_a(N_s>0)\geq\frac{\left(\int N_s\,d\mu_a\right)^2}{\int N_s^2\,d\mu_a}\geq c p_0^2R^{-1/4}. \tag*{(4)} μ a ( N s > 0 ) ≥ ∫ N s 2 d μ a ( ∫ N s d μ a ) 2 ≥ c p 0 2 R − 1/4 . (4) All these outputs have endpoints separated by order R R R C R − 1 / 4 CR^{-1/4} C R − 1/4
Apply this to input lengths below R 4 / 3 − ε R^{4/3-\varepsilon} R 4/3 − ε M M M γ j \gamma_j γ j ( 1 − ε / 3 , 1 ) (1-\varepsilon/3,1) ( 1 − ε /3 , 1 ) s j = R γ j s_j=R^{\gamma_j} s j = R γ j δ > 0 \delta>0 δ > 0 M M M δ \delta δ M c p 0 2 ≤ C + o ( 1 ) Mc p_0^2\leq C+o(1) M c p 0 2 ≤ C + o ( 1 ) R → ∞ R\to\infty R → ∞ M → ∞ M\to\infty M → ∞ p 0 → 0 p_0\to0 p 0 → 0
For input lengths above R 4 / 3 + ε R^{4/3+\varepsilon} R 4/3 + ε (4) gives output mass at least
c R 3 E 2 R 7 / 4 = c p 0 2 R 3 / 4 . c\frac{R^3E^2}{R^{7/4}}=cp_0^2R^{3/4}. c R 7/4 R 3 E 2 = c p 0 2 R 3/4 . Every output remains long. Proposition 2.1(iv) bounds its mass in the encompassing convex domain by
R − 4 / 3 − ε R 25 / 12 + o ( 1 ) = o ( R 3 / 4 ) . R^{-4/3-\varepsilon}R^{25/12+o(1)}=o(R^{3/4}). R − 4/3 − ε R 25/12 + o ( 1 ) = o ( R 3/4 ) . Thus p 0 → 0 p_0\to0 p 0 → 0
Two prescribed ports on one line
Fix two boundary ports at distance l l l [15 ] . Here the common-level argument supplies local regularity of the arch kernel as well as the length law, which we then transfer to the same-side hexagon law.
Theorem 4.1 (Same-side probability law). Under the critical half-plane law w ( γ ) / K l w(\gamma)/K_l w ( γ ) / K l for every ε > 0 \varepsilon>0 ε > 0
P { l 4 / 3 − ε ≤ L ( γ ) ≤ l 4 / 3 + ε } ⟶ 1. \mathbb{P}\left\{l^{4/3-\varepsilon}\leq L(\gamma)\leq l^{4/3+\varepsilon}\right\}\longrightarrow1. P { l 4/3 − ε ≤ L ( γ ) ≤ l 4/3 + ε } ⟶ 1. The same conclusion holds in the hexagons of Theorem 1.1 when both scaled endpoints converge to distinct points in one side interior.
Common blocks and lateral concentration
We use the intersection-renewal construction discussed in [1 ] , proving the required identities directly. Distances along the base are physical distances, with port spacing one; heights are in row units. Run two independent sequences of upward irreducibles of law p p p r r r T T T
A pair block contains one complete bridge segment from each sequence, possibly made of several irreducibles. Let D ∗ D_* D ∗ Y Y Y ∣ Y ∣ ≤ 2 D ∗ |Y|\leq2D_* ∣ Y ∣ ≤ 2 D ∗
Lemma 4.2 (Common-block estimates). The pair-block mass r r r For x ≥ 1 x \ge1 x ≥ 1
r ( T > x ) + r ( D ∗ > x ) ≤ C x − 1 / 2 . r(T > x) + r(D_* > x) \le Cx^{-1/2}. r ( T > x ) + r ( D ∗ > x ) ≤ C x − 1/2 . If Y ( m ) Y^{(m)} Y ( m ) m m m
sup j Pr ( Y ( m ) = j ) ≤ C m − 2 ( m ≥ 1 ) . \sup_j \operatorname{Pr}(Y^{(m)} = j) \le Cm^{-2} \qquad(m \ge1). j sup Pr ( Y ( m ) = j ) ≤ C m − 2 ( m ≥ 1 ) . Proof. The common renewal masses are B h 2 B_h^2 B h 2
C ( s ) = ∑ h ≥ 0 e − s h B h 2 = ( 1 − r ^ ( s ) ) − 1 ≍ s − 1 / 2 , r ^ ( s ) = ∑ r e − s T . C(s) = \sum_{h \ge0} e^{-sh}B_h^2 = (1-\hat r(s))^{-1} \asymp s^{-1/2}, \qquad\hat r(s) = \sum r e^{-sT}. C ( s ) = h ≥ 0 ∑ e − s h B h 2 = ( 1 − r ^ ( s ) ) − 1 ≍ s − 1/2 , r ^ ( s ) = ∑ r e − s T . Sending s s s r r r 1 − r ^ ( s ) ≍ s 1 / 2 1-\hat r(s) \asymp s^{1/2} 1 − r ^ ( s ) ≍ s 1/2 s = D / x s=D/x s = D / x D D D C ( s ) C(s) C ( s ) D ∗ > x D_* > x D ∗ > x C x − 1 / 4 s − 3 / 4 / C ( s ) < 1 / 2 Cx^{-1/4}s^{-3/4}/C(s) < 1/2 C x − 1/4 s − 3/4 / C ( s ) < 1/2 B h B_h B h d s = 1 − r ^ ( s ) d_s=1-\hat r(s) d s = 1 − r ^ ( s ) q s = E r [ e − s T ; D ∗ > x ] q_s=\mathbb{E}_r[e^{-sT};D_* > x] q s = E r [ e − s T ; D ∗ > x ] q s / ( d s + q s ) q_s/(d_s+q_s) q s / ( d s + q s ) 1 / 2 1/2 1/2 q s ≤ d s = O ( x − 1 / 2 ) q_s \le d_s=O(x^{-1/2}) q s ≤ d s = O ( x − 1/2 ) x x x 1 − r ^ ( s ) 1-\hat r(s) 1 − r ^ ( s )
A lateral fluctuation in one common block. The common-height transform identifies the time scale. We next show that the transverse difference fluctuates on the same scale; otherwise it could not smooth a prescribed separation. We claim there are fixed D 1 > D 0 > 0 D_1>D_0>0 D 1 > D 0 > 0
r ( D 0 x ≤ ∣ Y ∣ ≤ D 1 x ) ≥ c x − 1 / 2 r(D_0x \le|Y| \le D_1x) \ge cx^{-1/2} r ( D 0 x ≤ ∣ Y ∣ ≤ D 1 x ) ≥ c x − 1/2 for large x x x h h h ≳ h − 1 / 4 \gtrsim h^{-1/4} ≳ h − 1/4 h h h d h dh d h 60 ∘ 60^\circ 6 0 ∘ ( 3 / 2 ) d h (3/2)dh ( 3/2 ) d h c h − 1 / 2 ch^{-1/2} c h − 1/2 t , t ′ t,t' t , t ′
C h − 1 / 2 ( 1 + ∣ t − t ′ ∣ ) − 1 / 4 . Ch^{-1/2}(1+|t-t'|)^{-1/4}. C h − 1/2 ( 1 + ∣ t − t ′ ∣ ) − 1/4 . The outer crossings are still paid by the two-cut upper estimate. Averaging the gap factor over the order-h h h h − 3 / 4 h^{-3/4} h − 3/4
Normalize C ( 1 / x ) C(1/x) C ( 1/ x ) x ≤ h ≤ 2 x x \le h \le2x x ≤ h ≤ 2 x a x ax a x a > 0 a>0 a > 0 0 , 1 , … , 0,1,\ldots, 0 , 1 , … , O ( x 1 / 2 ) O(x^{1/2}) O ( x 1/2 ) e − T / x / r ^ ( 1 / x ) e^{-T/x}/\hat r(1/x) e − T / x / r ^ ( 1/ x ) ∣ Y ∣ ≤ 2 D ∗ |Y| \le2D_* ∣ Y ∣ ≤ 2 D ∗ ∣ Y ∣ > D 1 x |Y|>D_1x ∣ Y ∣ > D 1 x O ( D 1 − 1 / 2 ) O(D_1^{-1/2}) O ( D 1 − 1/2 ) ∣ Y ∣ < D 0 x |Y|<D_0x ∣ Y ∣ < D 0 x O ( D 0 x ) O(\sqrt{D_0x}) O ( D 0 x ) r r r
Y Y Y Y ( m ) Y^{(m)} Y ( m ) m m m
sup j Pr ( Y ( m ) = j ) ≤ C m − 2 . \sup_j \Pr(Y^{(m)}=j) \leq C m^{-2}. j sup Pr ( Y ( m ) = j ) ≤ C m − 2 . Indeed symmetrizing with an independent copy (bounded with positive probability), the window bound yields 1 − ∣ E e i θ Y ∣ 2 ≳ ∣ θ ∣ 1 / 2 1-\lvert\mathbb{E}e^{i\theta Y}\rvert^2 \gtrsim\lvert\theta\rvert^{1/2} 1 − ∣ E e i θ Y ∣ 2 ≳ ∣ θ ∣ 1/2 x x x 1 / ∣ θ ∣ 1/\lvert\theta\rvert 1/ ∣ θ ∣ m m m
Regularity of the joining kernel
We next attach short prefixes and recover them from the completed arch. Figure 1 shows this construction.
Figure 1. Two independent prefixes meet a common renewal height and are joined above it. In the restricted small-prefix construction, the prefixes remain near their respective starting ports. The first m m m
Lemma 4.3 (Sublinear changes of the gap). For positive integers l , l ′ l,l' l , l ′ l → ∞ l\to\infty l → ∞ l ′ / l → 1 l'/l\to1 l ′ / l → 1
K l ′ K l ⟶ 1. \frac{K_{l'}}{K_l}\longrightarrow1. K l K l ′ ⟶ 1. Proof. For initial separation j j j m m m ∑ i ≤ m D ∗ , i < j / 4 \sum_{i\leq m}D_{*,i}<j/4 ∑ i ≤ m D ∗ , i < j /4 O ( m / j ) O(m/\sqrt{j}) O ( m / j )
K j ≥ E [ 1 g o o d K j + Y ( m ) ] . K_j \geq\mathbb{E}\left[\mathbf{1}_{\mathrm{good}}K_{j+Y^{(m)}}\right]. K j ≥ E [ 1 good K j + Y ( m ) ] . Indeed product probabilities on the pair sequences are ordinary critical weights. The first m m m m m m Excluding a jump on a sublinear scale. Use monotonicity and the two-sided bounds of order l − 5 / 4 l^{-5/4} l − 5/4 ϵ l − 5 / 4 \epsilon l^{-5/4} ϵ l − 5/4 l − w l-w l − w l + w l+w l + w 1 ≤ w = o ( l ) 1 \le w=o(l) 1 ≤ w = o ( l ) q q q m = ⌊ q ⌋ m=\lfloor\sqrt{q}\rfloor m = ⌊ q ⌋ Pr ( Y ( m ) ≤ − b q ) ≥ 1 / 4 \Pr(Y^{(m)}\le-bq)\ge1/4 Pr ( Y ( m ) ≤ − b q ) ≥ 1/4 b > 0 b>0 b > 0 Pr ( ∣ Y ( m ) ∣ > A q ) = O ( A − 1 / 2 ) \Pr(|Y^{(m)}|>Aq)=O(A^{-1/2}) Pr ( ∣ Y ( m ) ∣ > A q ) = O ( A − 1/2 ) A A A w ≪ q ≪ l w\ll q\ll l w ≪ q ≪ l [ l + b q / 4 , l + ( A + b + 1 ) q ] [l+bq/4,l+(A+b+1)q] [ l + b q /4 , l + ( A + b + 1 ) q ] o ( l − 5 / 4 ) o(l^{-5/4}) o ( l − 5/4 ) O ( l − 5 / 4 ) O(l^{-5/4}) O ( l − 5/4 ) j = l + ⌊ b q / 2 ⌋ j=l+\lfloor bq/2\rfloor j = l + ⌊ b q /2 ⌋ ∣ Y ( m ) ∣ ≤ A q |Y^{(m)}|\le Aq ∣ Y ( m ) ∣ ≤ A q K j − o ( l − 5 / 4 ) K_j-o(l^{-5/4}) K j − o ( l − 5/4 ) Y ( m ) ≤ − b q Y^{(m)}\le-bq Y ( m ) ≤ − b q l − w l-w l − w ϵ l − 5 / 4 \epsilon l^{-5/4} ϵ l − 5/4 K j K_j K j m / j = O ( q / l ) m/\sqrt{j}=O(\sqrt{q/l}) m / j = O ( q / l ) O ( A − 1 / 2 ) K j O(A^{-1/2})K_j O ( A − 1/2 ) K j 1 / 4 − O ( A − 1 / 2 ) − o ( 1 ) 1/4-O(A^{-1/2})-o(1) 1/4 − O ( A − 1/2 ) − o ( 1 ) A A A ϵ \epsilon ϵ
Concentration under the prescribed-gap law
Proof of Theorem 4.1. Lemma 4.3 shows that sublinear perturbations preserve the joining mass. It remains to transfer the independent-prefix law and bound an unusually long joining arch. Take m = ⌊ l 1 / 2 − η ⌋ m=\lfloor l^{1/2-\eta}\rfloor m = ⌊ l 1/2 − η ⌋ η > 0 \eta>0 η > 0 1 − o ( 1 ) 1-o(1) 1 − o ( 1 ) ∑ D ∗ , i / l → 0 \sum D_{*,i}/l\to0 ∑ D ∗ , i / l → 0 K l K_l K l 1 1 1 m m m m 2 + o ( 1 ) m^{2+o(1)} m 2 + o ( 1 ) ( m 2 ) 4 / 3 + o ( 1 ) (m^2)^{4/3+o(1)} ( m 2 ) 4/3 + o ( 1 ) l l l 4 / 3 4/3 4/3 η \eta η
Finally bound the contribution among these paths with joining length above 1 2 l 4 / 3 + ϵ \frac{1}{2}l^{4/3+\epsilon} 2 1 l 4/3 + ϵ
C m − 2 ( C R − 1 / 4 + l − 4 / 3 − ϵ R 13 / 12 + o ( 1 ) ) , R = l 1 + γ . Cm^{-2}\left(CR^{-1/4}+l^{-4/3-\epsilon}R^{13/12+o(1)}\right), \qquad R=l^{1+\gamma}. C m − 2 ( C R − 1/4 + l − 4/3 − ϵ R 13/12 + o ( 1 ) ) , R = l 1 + γ . Indeed after summing over the other joining endpoint this uses the half-plane diameter bound above R R R R R R O ( R ) O(R) O ( R ) R 25 / 12 + o ( 1 ) R^{25/12+o(1)} R 25/12 + o ( 1 ) γ > 0 \gamma>0 γ > 0 ϵ \epsilon ϵ η > 0 \eta>0 η > 0
2 η < γ / 4 , 2 η + 13 γ / 12 < ϵ . 2\eta<\gamma/4,\qquad2\eta+13\gamma/12<\epsilon. 2 η < γ /4 , 2 η + 13 γ /12 < ϵ . Both displayed contributions are then o ( l − 5 / 4 ) o(l^{-5/4}) o ( l − 5/4 ) l ≻ R l\succ R l ≻ R c l − 5 / 4 cl^{-5/4} c l − 5/4 o ( l − 5 / 4 ) o(l^{-5/4}) o ( l − 5/4 )
Regularity of the strip partition function
The renewal identity alone gives a tail estimate for the height of one irreducible bridge. Prescribing the terminal port requires an estimate for a single height. Our first step is to show that the strip partition function varies smoothly on its natural scale: every fixed forward difference gains the corresponding power of the strip width.
Proposition 5.1. Let B N B_N B N critical mass of strict bridges from one fixed port to all ports on the opposite pure cut at integral height N N N Δ f N = f N + 1 − f N \Delta f_N=f_{N+1}-f_N Δ f N = f N + 1 − f N ℓ ≥ 0 \ell\ge0 ℓ ≥ 0 η > 0 \eta>0 η > 0 C ℓ , η < ∞ C_{\ell,\eta}<\infty C ℓ , η < ∞
∣ Δ ℓ B N ∣ ≤ C ℓ , η N − 1 / 4 − ℓ + η ( N ≥ 1 ) . \lvert\Delta^\ell B_N\rvert\le C_{\ell,\eta}N^{-1/4-\ell+\eta}\qquad(N\ge1). ∣ Δ ℓ B N ∣ ≤ C ℓ , η N − 1/4 − ℓ + η ( N ≥ 1 ) . The probabilistic application needs precisely Proposition 5.1: its passage to an exact height atom begins in Section 6. We now prove the proposition from the exact finite strip representation and its positive interpolation spaces. We state this input before introducing the holomorphic kernels used to estimate it. The spectral parameter λ = π / 8 \lambda=\pi/8 λ = π /8 0 0 0 − 4 -4 − 4 N N N B N B_N B N
Put b = 3 λ b=3\lambda b = 3 λ C t ( x ) = cosh ( t x ) C_t(x)=\cosh(t\sqrt{x}) C t ( x ) = cosh ( t x ) D ( x ) = 2 C λ ( x ) − 1 \mathcal{D}(x)=2C_\lambda(x)-1 D ( x ) = 2 C λ ( x ) − 1 a ∗ = D ( − 4 ) = 2 cos ( 2 λ ) − 1 > 0 a_*=\mathcal{D}(-4)=2\cos(2\lambda)-1>0 a ∗ = D ( − 4 ) = 2 cos ( 2 λ ) − 1 > 0 μ \mu μ ( 0 , ∞ ) (0,\infty) ( 0 , ∞ )
d μ ( x ) = z D ( z 2 ) sinh ( 4 λ z ) d z , x = z 2 , z > 0. (5) \mathrm{d}\mu(x)=\frac{z\mathcal{D}(z^2)}{\sinh(4\lambda z)}\,\mathrm{d}z,\qquad x=z^2,\quad z>0. \tag*{(5)} d μ ( x ) = sinh ( 4 λ z ) z D ( z 2 ) d z , x = z 2 , z > 0. (5) For N ≥ 1 N\ge1 N ≥ 1
M N = span { 1 , x , … , x N − 1 , ∂ t j C t ∣ t = λ : 0 ≤ j < N } , M_N=\operatorname{span}\{1,x,\ldots,x^{N-1},\partial_t^j C_t|_{t=\lambda}:0\le j<N\}, M N = span { 1 , x , … , x N − 1 , ∂ t j C t ∣ t = λ : 0 ≤ j < N } , c 1 = C λ / 2 , c 3 = C 3 λ / 2 , h 0 = D − 1 , h 1 = c 1 / D , c_1=C_{\lambda/2},\qquad c_3=C_{3\lambda/2},\qquad h_0=\mathcal{D}^{-1},\qquad h_1=c_1/\mathcal{D}, c 1 = C λ /2 , c 3 = C 3 λ /2 , h 0 = D − 1 , h 1 = c 1 / D , g = ∫ 0 λ C t d t , F 0 , N = M N + R c 1 , G N = M N + R g . g=\int_0^\lambda C_t\,\mathrm{d}t,\qquad F_{0,N}=M_N+\mathbb{R}c_1,\qquad G_N=M_N+\mathbb{R}g. g = ∫ 0 λ C t d t , F 0 , N = M N + R c 1 , G N = M N + R g . Let Π N h ∈ F 0 , N \Pi_Nh\in F_{0,N} Π N h ∈ F 0 , N G N G_N G N L 2 ( μ ) L^2(\mu) L 2 ( μ ) r N h = h − Π N h r_Nh=h-\Pi_Nh r N h = h − Π N h c 3 c_3 c 3 G N G_N G N c 3 ∉ L 2 ( μ ) c_3\notin L^2(\mu) c 3 ∈ / L 2 ( μ )
We use the following finite facts from the strip analysis in [16 ] .
The above projection exists. Whenever the denominator is nonzero,
B N = 2 a ∗ [ r N h 0 ( − 4 ) − r N ( c 3 − h 1 ) ( − 4 ) r N h 0 ( 0 ) r N ( c 3 − h 1 ) ( 0 ) ] . B_N=2a_*\left[r_Nh_0(-4)-\frac{r_N(c_3-h_1)(-4)\,r_Nh_0(0)}{r_N(c_3-h_1)(0)}\right]. B N = 2 a ∗ [ r N h 0 ( − 4 ) − r N ( c 3 − h 1 ) ( 0 ) r N ( c 3 − h 1 ) ( − 4 ) r N h 0 ( 0 ) ] . The corresponding undivided statement is that there is a scalar s N s_N s N 2 B N = 4 a ∗ [ r N h 0 ( − 4 ) + s N r N h 0 ( 0 ) ] 2B_N=4a_*[r_Nh_0(-4)+s_Nr_Nh_0(0)] 2 B N = 4 a ∗ [ r N h 0 ( − 4 ) + s N r N h 0 ( 0 )] r N ( c 3 − h 1 ) ( − 4 ) + s N r N ( c 3 − h 1 ) ( 0 ) = 0 r_N(c_3-h_1)(-4)+s_Nr_N(c_3-h_1)(0)=0 r N ( c 3 − h 1 ) ( − 4 ) + s N r N ( c 3 − h 1 ) ( 0 ) = 0
For 0 < t < λ 0<t<\lambda 0 < t < λ G N , t = M N + R C t G_{N,t}=M_N+\mathbb{R}C_t G N , t = M N + R C t X = ( x 1 < ⋯ < x 2 N + 1 ) X=(x_1<\cdots<x_{2N+1}) X = ( x 1 < ⋯ < x 2 N + 1 ) ( F 0 , N , G N , t ) (F_{0,N},G_{N,t}) ( F 0 , N , G N , t ) G N , t G_{N,t} G N , t e 0 c 3 e_0c_3 e 0 c 3 − e 0 c 3 -e_0c_3 − e 0 c 3 G N , t G_{N,t} G N , t G N G_N G N
Let F N = F 0 , N + R c 3 F_N = F_{0,N} + \mathbb{R}c_3 F N = F 0 , N + R c 3 A N = { f ∈ F N : f ( 0 ) = 0 } A_N = \{f \in F_N : f(0) = 0\} A N = { f ∈ F N : f ( 0 ) = 0 } U N = { f ∈ M N : f ( 0 ) = 0 } U_N = \{f \in M_N : f(0) = 0\} U N = { f ∈ M N : f ( 0 ) = 0 } ( F 0 , N , G N , t ) (F_{0,N},G_{N,t}) ( F 0 , N , G N , t ) ( A N , G N , t ) (A_N,G_{N,t}) ( A N , G N , t ) ∏ i r ( x i ) \prod_i r(x_i) ∏ i r ( x i ) r ( x ) = ( x + α ) / ( x + β ) r(x) = (x + \alpha)/(x + \beta) r ( x ) = ( x + α ) / ( x + β ) α , β > 0 \alpha,\beta> 0 α , β > 0 ( M N , M N ) (M_N,M_N) ( M N , M N ) ( U N , U N ) (U_N,U_N) ( U N , U N ) N , t N,t N , t [16 ] : its positive weight and exponential-domination hypotheses persist under each such change, and the bound uses only the range of log r \log r log r t t t log r \log r log r
For the probability measure μ c \mu_c μ c d z / cosh ( c z ) dz/\cosh(cz) d z / cosh ( cz ) x = z 2 x = z^2 x = z 2 α c = π / ( 2 c ) \alpha_c = \pi/(2c) α c = π / ( 2 c ) w t = tan 2 ( α c t ) w_t = \tan^2(\alpha_c t) w t = tan 2 ( α c t )
A t ( x ) = cos ( α c t ) C t ( x ) , ∫ A t A s d μ c = 1 1 − w t w s . (6) A_t(x) = \cos(\alpha_c t)C_t(x), \qquad\int A_t A_s\,d\mu_c = \frac{1}{1-w_t w_s}. \tag*{(6)} A t ( x ) = cos ( α c t ) C t ( x ) , ∫ A t A s d μ c = 1 − w t w s 1 . (6) Thus A t = ∑ j ≥ 0 P j ( x ) w t j A_t = \sum_{j\geq0} P_j(x)w_t^j A t = ∑ j ≥ 0 P j ( x ) w t j P j P_j P j j j j ∣ t ∣ + ∣ s ∣ < c |t| + |s| < c ∣ t ∣ + ∣ s ∣ < c
All base weights below have the required exponential domination. In particular these inputs apply to μ \mu μ d W = z d z / sinh ( b z ) dW = z\,dz/\sinh(bz) d W = z d z / sinh ( b z ) r j W r^jW r j W j j j r r r (5.2) .
Set
v ( ω ) = tan π ω 2 b , J ( ω ) = v ′ ( ω ) , Q ( v ) = v 2 ( v 2 − v ( λ ) 2 ) 1 − v 2 v ( λ ) 2 . v(\omega) = \tan\frac{\pi\omega}{2b}, \qquad J(\omega) = v'(\omega), \qquad Q(v) = \frac{v^2(v^2-v(\lambda)^2)}{1-v^2v(\lambda)^2}. v ( ω ) = tan 2 b π ω , J ( ω ) = v ′ ( ω ) , Q ( v ) = 1 − v 2 v ( λ ) 2 v 2 ( v 2 − v ( λ ) 2 ) . The strip ∣ Re ω ∣ < b / 2 |\operatorname{Re}\omega| < b/2 ∣ Re ω ∣ < b /2 Q Q Q 0 0 0 ± v ( λ ) \pm v(\lambda) ± v ( λ ) A A A
( T N A ) ( ω ) = J ( ω ) Q ( v ( ω ) ) N A ( v ( ω ) ) . (T_N A)(\omega) = J(\omega)Q(v(\omega))^N A(v(\omega)). ( T N A ) ( ω ) = J ( ω ) Q ( v ( ω ) ) N A ( v ( ω )) . The factor Q N Q^N Q N M N M_N M N h ∈ M N ⊥ ⊂ L 2 ( μ ) h \in M_N^\perp\subset L^2(\mu) h ∈ M N ⊥ ⊂ L 2 ( μ )
∫ h ( x ) C ω ( x ) d μ ( x ) = T N A ( ω ) . \int h(x)C_\omega(x)\,d\mu(x) = T_NA(\omega). ∫ h ( x ) C ω ( x ) d μ ( x ) = T N A ( ω ) . The transform is even, has a zero of order 2 N 2N 2 N N N N ± λ \pm\lambda ± λ J Q N JQ^N J Q N A A A
Here is the Hilbert-space content of this identification. Let w μ ( z ) w_\mu(z) w μ ( z ) w W ( z ) w_W(z) w W ( z ) z z z μ \mu μ W W W F = T N A F = T_NA F = T N A f ( z ) = h ( z 2 ) w μ ( z ) f(z) = h(z^2)w_\mu(z) f ( z ) = h ( z 2 ) w μ ( z )
F ( ω ) = 1 2 ∫ R f ( z ) e ω z d z , ∥ h ∥ L 2 ( μ ) 2 = 1 2 ∫ R ∣ f ( z ) ∣ 2 w μ ( z ) d z . F(\omega) = \frac{1}{2}\int_{\mathbb{R}} f(z)e^{\omega z}\,dz, \qquad\|h\|_{L^2(\mu)}^2 = \frac{1}{2}\int_{\mathbb{R}}\frac{|f(z)|^2}{w_\mu(z)}\,dz. F ( ω ) = 2 1 ∫ R f ( z ) e ω z d z , ∥ h ∥ L 2 ( μ ) 2 = 2 1 ∫ R w μ ( z ) ∣ f ( z ) ∣ 2 d z . We compare norms while holding F F F f f f W W W h W = h d μ / d W h_W=h\,d\mu/dW h W = h d μ / d W w μ w_\mu w μ w W w_W w W e z Re ω e^{z\operatorname{Re}\omega} e z Re ω sinh ( b z ) / z = 1 / w W ( z ) \sinh(bz)/z=1/w_W(z) sinh ( b z ) / z = 1/ w W ( z ) J J J
G N 0 ( A , A ) = c b ∫ ∣ v ∣ < 1 ∣ Q ( v ) ∣ 2 N ∣ A ( v ) ∣ 2 d 2 v , c b > 0. (7) \mathcal{G}^{0}_{N}(A,A)=c_b\int_{\lvert v\rvert<1}\lvert Q(v)\rvert^{2N}\lvert A(v)\rvert^2\,d^2v,\qquad c_b>0. \tag*{(7)} G N 0 ( A , A ) = c b ∫ ∣ v ∣ < 1 ∣ Q ( v ) ∣ 2 N ∣ A ( v ) ∣ 2 d 2 v , c b > 0. (7) For completeness, the converse Fourier representation follows by Fourier transforming the Cauchy–Riemann equation locally in frequency: the data on parallel lines must have the same exponential multiplier. Strip square integrability gives the stated weighted L 2 L^2 L 2 h h h
Write G N ( A , A ) = ∥ h ∥ L 2 ( μ ) 2 \mathcal{G}_N(A,A)=\lVert h\rVert_{L^2(\mu)}^2 G N ( A , A ) = ∥ h ∥ L 2 ( μ ) 2 (5.4) . The density ratio of μ \mu μ W W W G N ≍ G N 0 \mathcal{G}_N\asymp\mathcal{G}^{0}_N G N ≍ G N 0 N N N T N A ( ± d t r + i y ) T_NA(\pm d_{\mathrm{tr}}+iy) T N A ( ± d tr + i y ) d t r < b / 2 d_{\mathrm{tr}}<b/2 d tr < b /2 b / 2 b/2 b /2
z − 1 { sinh ( 4 λ z ) D ( z 2 ) − sinh ( b z ) } . z^{-1}\left\{\frac{\sinh(4\lambda z)}{\mathcal{D}(z^2)}-\sinh(bz)\right\}. z − 1 { D ( z 2 ) sinh ( 4 λ z ) − sinh ( b z ) } . It is bounded at zero and is O ( e 2 d t r ∣ z ∣ ) O(e^{2d_{\mathrm{tr}}\lvert z\rvert}) O ( e 2 d tr ∣ z ∣ ) L 2 L^2 L 2
If A ( v ) = ∑ j ≥ 0 a j v j A(v)=\sum_{j\geq0}a_jv^j A ( v ) = ∑ j ≥ 0 a j v j
∥ A ∥ N 2 = ∑ j ≥ 0 ∣ a j ∣ 2 N + j . (8) \lVert A\rVert_N^2=\sum_{j\geq0}\frac{\lvert a_j\rvert^2}{N+j}. \tag*{(8)} ∥ A ∥ N 2 = j ≥ 0 ∑ N + j ∣ a j ∣ 2 . (8) The boundary phase speed of the finite Blaschke product Q Q Q ∣ Q ( r e i θ ) ∣ 2 N \lvert Q(re^{i\theta})\rvert^{2N} ∣ Q ( r e i θ ) ∣ 2 N e − c N ( 1 − r ) e^{-cN(1-r)} e − c N ( 1 − r ) e − C N ( 1 − r ) e^{-CN(1-r)} e − C N ( 1 − r ) r r r
Every fixed ℓ \ell ℓ G N 0 \mathcal{G}^{0}_N G N 0 O ℓ ( N − ℓ ) O_\ell(N^{-\ell}) O ℓ ( N − ℓ ) (8) , since it inserts ( ∣ Q ∣ 2 − 1 ) ℓ (\lvert Q\rvert^2-1)^\ell (∣ Q ∣ 2 − 1 ) ℓ G N \mathcal{G}_N G N v → ± i v\to\pm i v → ± i Q ( ± i ) = 1 Q(\pm i)=1 Q ( ± i ) = 1 ∣ 1 − Q ∣ ≤ C ( 1 − ∣ Q ∣ ) \lvert1-Q\rvert\leq C(1-\lvert Q\rvert) ∣ 1 − Q ∣ ≤ C ( 1 − ∣ Q ∣) T N T_N T N Q − 1 Q-1 Q − 1 T ⌊ N / 2 ⌋ T_{\lfloor N/2\rfloor} T ⌊ N /2 ⌋ O ( N − 1 ) O(N^{-1}) O ( N − 1 ) G N − 1 G N = I \mathcal{G}^{-1}_N\mathcal{G}_N=I G N − 1 G N = I ℓ \ell ℓ O ℓ ( N − ℓ ) O_\ell(N^{-\ell}) O ℓ ( N − ℓ )
The model evaluation kernel
Denote by K N 0 ( v , u ) K^0_N(v,u) K N 0 ( v , u ) G N 0 \mathcal{G}^{0}_N G N 0 v v v u u u A ( v ) = G N 0 ( A , K N 0 ( ⋅ , u ) ) A(v)=\mathcal{G}^{0}_N(A,K^0_N(\cdot,u)) A ( v ) = G N 0 ( A , K N 0 ( ⋅ , u )) ∣ u ∣ < 1 \lvert u\rvert<1 ∣ u ∣ < 1
Lemma 5.2. On every compact subset of the closed bidisk on which ∣ 1 − v 2 u ‾ 2 ∣ > 0 \lvert1-v^2\overline{u}^2\rvert>0 ∣ 1 − v 2 u 2 ∣ > 0
∣ Δ ℓ K N 0 ( v , u ) ∣ ≤ C ℓ N 1 − ℓ , K N 0 ( v , u ) N ⟶ O ( v ) O ( u ) ‾ 1 − v 2 u ‾ 2 . \lvert\Delta^\ell K^0_N(v,u)\rvert\leq C_\ell N^{1-\ell},\qquad\frac{K^0_N(v,u)}{N}\longrightarrow\frac{O(v)\overline{O(u)}}{1-v^2\overline{u}^2}. ∣ Δ ℓ K N 0 ( v , u )∣ ≤ C ℓ N 1 − ℓ , N K N 0 ( v , u ) ⟶ 1 − v 2 u 2 O ( v ) O ( u ) . Here O O O
Proof. Let P j k P_{jk} P j k v 2 j v^{2j} v 2 j (7) , and R = P − 1 R=P^{-1} R = P − 1 K N 0 ( v , u ) = ∑ j , k ≥ 0 v 2 j R j k u ‾ 2 k K_N^0(v,u)=\sum_{j,k\geq0}v^{2j}R_{jk}\overline{u}^{2k} K N 0 ( v , u ) = ∑ j , k ≥ 0 v 2 j R j k u 2 k δ \delta δ j , k j,k j , k ℓ , s \ell,s ℓ , s
∣ Δ ℓ δ s P j k ∣ ≤ C ℓ , s e − c ∣ j − k ∣ ( N + j + k ) − 1 − ℓ − s . (9) \left|\Delta^\ell\delta^s P_{jk}\right|\leq C_{\ell,s}e^{-c|j-k|}(N+j+k)^{-1-\ell-s}. \tag*{(9)} Δ ℓ δ s P j k ≤ C ℓ , s e − c ∣ j − k ∣ ( N + j + k ) − 1 − ℓ − s . (9) In the radial integral the differences insert ( ∣ Q ∣ 2 − 1 ) ℓ ( r 4 − 1 ) s (|Q|^2-1)^\ell(r^4-1)^s ( ∣ Q ∣ 2 − 1 ) ℓ ( r 4 − 1 ) s θ \theta θ ∣ j − k ∣ |j-k| ∣ j − k ∣ Q ( r e i θ ) Q ( r e − i θ ) Q(re^{i\theta})Q(re^{-i\theta}) Q ( r e i θ ) Q ( r e − i θ ) r = 1 r=1 r = 1 e − c ( 1 − r ) e^{-c(1-r)} e − c ( 1 − r ) O ( 1 − r ) O(1-r) O ( 1 − r ) (9) .
The inverse matrix satisfies
∣ Δ ℓ δ s R j k ∣ ≤ C ℓ , s e − c ∣ j − k ∣ ( N + j + k ) − ℓ ( 1 + min { j , k } ) − s . (10) \left|\Delta^\ell\delta^s R_{jk}\right|\leq C_{\ell,s}e^{-c|j-k|}(N+j+k)^{- \ell}(1+\min\{j,k\})^{-s}. \tag*{(10)} Δ ℓ δ s R j k ≤ C ℓ , s e − c ∣ j − k ∣ ( N + j + k ) − ℓ ( 1 + min { j , k } ) − s . (10) First take ℓ = s = 0 \ell=s=0 ℓ = s = 0 [7 ] , Proposition 2. The parameter differences will be retained explicitly below. Conjugating P P P N + j \sqrt{N+j} N + j (8) . Conjugation of this normalized operator by weights of sufficiently small exponential logarithmic slope changes it by an arbitrarily small operator norm, by row and column sums in (9) . Its inverse remains bounded. Truncate these weights and pass to the limit to obtain exponential decay of inverse entries, proving the undifferenced case.
Here are details for the differences. Write ( S A ) j k = A j + 1 , k + 1 (SA)_{jk}=A_{j+1,k+1} ( S A ) j k = A j + 1 , k + 1 δ A = S A − A \delta A=SA-A δ A = S A − A
S ( A B ) = ( S A ) ( S B ) + E ( A , B ) , E ( A , B ) j k = A j + 1 , 0 B 0 , k + 1 . S(AB)=(SA)(SB)+E(A,B),\qquad E(A,B)_{jk}=A_{j+1,0}B_{0,k+1}. S ( A B ) = ( S A ) ( S B ) + E ( A , B ) , E ( A , B ) j k = A j + 1 , 0 B 0 , k + 1 . In particular P R = I PR=I P R = I N N N
δ R = − R ( δ P ) S R − R E ( P , R ) , Δ R N = − R N ( Δ P N ) R N + 1 . \delta R=-R(\delta P)SR-RE(P,R),\qquad\Delta R_N=-R_N(\Delta P_N)R_{N+1}. δ R = − R ( δ P ) S R − R E ( P , R ) , Δ R N = − R N ( Δ P N ) R N + 1 . The first identity displays the boundary contribution that is absent from a difference in N N N (10) , with ( N + j + k ) a (N+j+k)^a ( N + j + k ) a a a a E ( P , R ) E(P,R) E ( P , R ) j + k j+k j + k N N N s s s ℓ \ell ℓ R R R P P P (9) , these bounds prove (10) .
These estimates also give convergence at the boundary off the even diagonal. On each matrix diagonal j − k j-k j − k n = min { j , k } n=\min\{j,k\} n = min { j , k } n n n ( v 2 u ‾ 2 ) n (v^2\overline{u}^2)^n ( v 2 u 2 ) n s s s n ∼ m n\sim m n ∼ m N ℓ − 1 Δ ℓ K N 0 N^{\ell-1}\Delta^\ell K_N^0 N ℓ − 1 Δ ℓ K N 0 C s m 2 − s C_s m^{2-s} C s m 2 − s ∣ 1 − v 2 u ‾ 2 ∣ |1-v^2\overline{u}^2| ∣1 − v 2 u 2 ∣ s > 2 s>2 s > 2
For fixed j , k j,k j , k N P j k NP_{jk} N P j k Q Q Q R j k / N R_{jk}/N R j k / N 1 / ∣ O ∣ 2 1/|O|^2 1/∣ O ∣ 2 O ( 0 ) > 0 O(0)>0 O ( 0 ) > 0 [17 ] , Section 5. Real analyticity, evenness, and reflection symmetry of the symbol give the asserted analytic continuation, nonvanishing, and symmetries of O O O O O O 1 / O 1/O 1/ O ( 1 − v 2 v ˉ 2 ) − 1 (1-v^2\bar v^2)^{-1} ( 1 − v 2 v ˉ 2 ) − 1
The model has now isolated the only large part of the evaluation kernel. The remaining change of measure contributes a bounded vector, which is small enough for every residual estimate below. Let K N K_N K N G N \mathcal{G}_N G N u = v ( t ) u=v(t) u = v ( t ) 0 ≤ t ≤ λ 0\le t\le\lambda 0 ≤ t ≤ λ t ↑ b / 2 t\uparrow b/2 t ↑ b /2
∥ Δ ℓ ( K N ( ⋅ , u ) − K N 0 ( ⋅ , u ) ) ∥ N ≤ C ℓ N − ℓ . (11) \lVert\Delta^\ell(K_N(\mathord{\cdot},u)-K_N^0(\mathord{\cdot},u))\rVert_N\le C_\ell N^{-\ell}. \tag*{(11)} ∥ Δ ℓ ( K N ( ⋅ , u ) − K N 0 ( ⋅ , u )) ∥ N ≤ C ℓ N − ℓ . (11) At t = b / 2 t=b/2 t = b /2 u = 1 u=1 u = 1 ∥ ⋅ ∥ N \lVert\cdot\rVert_N ∥ ⋅ ∥ N K N ( ⋅ , 1 ) K_N(\mathord{\cdot},1) K N ( ⋅ , 1 ) u u u
∣ J ∣ ≍ e − π ∣ y ∣ / b , 1 − ∣ Q ∣ ≳ e − π ∣ y ∣ / b . |J|\asymp e^{-\pi|y|/b},\qquad1-|Q|\gtrsim e^{-\pi|y|/b}. ∣ J ∣ ≍ e − π ∣ y ∣/ b , 1 − ∣ Q ∣ ≳ e − π ∣ y ∣/ b . Lemma 5.2 and discrete product rules therefore give
∥ Δ ℓ ( T N K N 0 ( ⋅ , u ) ) ∥ L 2 ( { ± d t r + i y } ) ≤ C ℓ N − ℓ . \lVert\Delta^\ell(T_NK_N^0(\mathord{\cdot},u))\rVert_{L^2(\{\pm d_{\mathrm{tr}}+iy\})}\le C_\ell N^{-\ell}. ∥ Δ ℓ ( T N K N 0 ( ⋅ , u )) ∥ L 2 ({ ± d tr + i y }) ≤ C ℓ N − ℓ . For example the undifferenced integrand is bounded by
C N e − π ∣ y ∣ / b exp { − c N e − π ∣ y ∣ / b } . CNe^{-\pi|y|/b}\exp\{-cNe^{-\pi|y|/b}\}. C N e − π ∣ y ∣/ b exp { − c N e − π ∣ y ∣/ b } . Its squared integral is bounded; each difference supplies its stated additional factor. Apply the bounded correction form to this column and solve with G N − 1 \mathcal{G}_N^{-1} G N − 1 O ( N ) O(\sqrt{N}) O ( N ) v , u v,u v , u 0 ≤ t ≤ λ 0\le t\le\lambda 0 ≤ t ≤ λ N N N ℓ \ell ℓ O ( N 1 − ℓ ) O(N^{1-\ell}) O ( N 1 − ℓ )
Residual estimates at the two evaluation points
We prove the following bounds, which are the full analytic input to the quotient in (5.2):
∣ Δ ℓ r N h j ( − a ) ∣ ≤ N − 1 + b a / π − ℓ + o ( 1 ) ( j = 0 , 1 ) , ∣ Δ ℓ r N c 3 ( − a ) ∣ ≤ N b a / π − ℓ + o ( 1 ) ( a = 0 , 4 ) , − r N c 3 ( 0 ) ≥ c > 0. (12) \begin{aligned}
|\Delta^\ell r_N h_j(-a)|&\le N^{-1+b\sqrt{a/\pi}-\ell+o(1)} &&(j=0,1),\\
|\Delta^\ell r_N c_3(-a)|&\le N^{b\sqrt{a/\pi}-\ell+o(1)} &&(a=0,4),\\
-r_Nc_3(0)&\ge c>0. \tag*{(12)}
\end{aligned} ∣ Δ ℓ r N h j ( − a ) ∣ ∣ Δ ℓ r N c 3 ( − a ) ∣ − r N c 3 ( 0 ) ≤ N − 1 + b a / π − ℓ + o ( 1 ) ≤ N b a / π − ℓ + o ( 1 ) ≥ c > 0. ( j = 0 , 1 ) , ( a = 0 , 4 ) , (12) All orders ℓ \ell ℓ o ( 1 ) o(1) o ( 1 )
First take the orthogonal residual against M N M_N M N h = h j h=h_j h = h j T N A h , N T_NA_{h,N} T N A h , N A h , N A_{h,N} A h , N h h h
h ( z 2 ) = ∫ R f h ( y ) cos ( y z ) d y . h(z^2)=\int_{\mathbb{R}} f_h(y)\cos(yz)\,dy. h ( z 2 ) = ∫ R f h ( y ) cos ( y z ) d y . The function f h f_h f h e u ∣ y ∣ e^{u|y|} e u ∣ y ∣ u < π / ( 3 λ ) = π / b u<\pi/(3\lambda)=\pi/b u < π / ( 3 λ ) = π / b D ( z 2 ) D(z^2) D ( z 2 ) π / ( 3 λ ) \pi/(3\lambda) π / ( 3 λ ) A A A ∫ f h ( y ) ( T N A ) ( i y ) d y \int f_h(y)(T_NA)(iy)\,dy ∫ f h ( y ) ( T N A ) ( i y ) d y
On ∥ A ∥ N ≤ 1 \lVert A\rVert_N\le1 ∥ A ∥ N ≤ 1 T ⌊ N / 2 ⌋ T_{\lfloor N/2\rfloor} T ⌊ N /2 ⌋
∣ ( Δ ℓ T N A ) ( i y ) ∣ ≤ C ℓ N − ℓ exp { − c N e − π ∣ y ∣ / b } . |(\Delta^\ell T_NA)(iy)|\le C_\ell N^{-\ell}\exp\{-cNe^{-\pi|y|/b}\}. ∣ ( Δ ℓ T N A ) ( i y ) ∣ ≤ C ℓ N − ℓ exp { − c N e − π ∣ y ∣/ b } . Indeed 1 − ∣ Q ( v ( i y ) ) ∣ ≳ e − π ∣ y ∣ / b 1-|Q(v(iy))|\gtrsim e^{-\pi|y|/b} 1 − ∣ Q ( v ( i y )) ∣ ≳ e − π ∣ y ∣/ b ∣ 1 − Q ∣ ≲ 1 − ∣ Q ∣ |1-Q|\lesssim1-|Q| ∣1 − Q ∣ ≲ 1 − ∣ Q ∣ f h f_h f h
∥ Δ ℓ A h , N ∥ N ≤ N − 1 − ℓ + o ( 1 ) . (13) \lVert\Delta^\ell A_{h,N}\rVert_N\le N^{-1-\ell+o(1)}. \tag*{(13)} ∥ Δ ℓ A h , N ∥ N ≤ N − 1 − ℓ + o ( 1 ) . (13) For an input C s C_s C s 0 ≤ s < b / 2 0\le s<b/2 0 ≤ s < b /2 J ( s ) Q ( v ( s ) ) N K N ( ⋅ , v ( s ) ) J(s)Q(v(s))^NK_N(\cdot,v(s)) J ( s ) Q ( v ( s ) ) N K N ( ⋅ , v ( s )) s ↑ b / 2 s\uparrow b/2 s ↑ b /2 c 3 c_3 c 3 Q ( 1 ) = 1 Q(1)=1 Q ( 1 ) = 1 u = 1 u=1 u = 1
We now impose the one remaining oblique constraint. Put u 0 = v ( λ / 2 ) u_0=v(\lambda/2) u 0 = v ( λ /2 ) r N h r_Nh r N h T N T_N T N
A h , N − K N ( ⋅ , u 0 ) χ N ( A h , N ) χ N ( K N ( ⋅ , u 0 ) ) , A_{h,N}-K_N(\cdot,u_0)\frac{\chi_N(A_{h,N})}{\chi_N(K_N(\cdot,u_0))}, A h , N − K N ( ⋅ , u 0 ) χ N ( K N ( ⋅ , u 0 )) χ N ( A h , N ) , where χ N \chi_N χ N v ( t ) v(t) v ( t ) 0 ≤ t ≤ λ 0\le t\le\lambda 0 ≤ t ≤ λ J ( t ) ∣ Q ( v ( t ) ) ∣ N d t J(t)|Q(v(t))|^Ndt J ( t ) ∣ Q ( v ( t )) ∣ N d t M N M_N M N T N T_N T N ∫ 0 λ C t d t = g \int_0^\lambda C_tdt=g ∫ 0 λ C t d t = g Q Q Q χ N = 0 \chi_N=0 χ N = 0
Every fixed ℓ \ell ℓ χ N \chi_N χ N O ℓ ( N − ℓ ) O_\ell(N^{-\ell}) O ℓ ( N − ℓ ) N / 2 N/2 N /2 N N N
χ N ( K N ( ⋅ , u 0 ) ) ≥ c N , ∣ Δ ℓ χ N ( K N ( ⋅ , u 0 ) ) ∣ ≤ C ℓ N 1 − ℓ . \chi_N(K_N(\cdot,u_0))\ge cN,\qquad|\Delta^\ell\chi_N(K_N(\cdot,u_0))|\le C_\ell N^{1-\ell}. χ N ( K N ( ⋅ , u 0 )) ≥ c N , ∣ Δ ℓ χ N ( K N ( ⋅ , u 0 )) ∣ ≤ C ℓ N 1 − ℓ . For h j h_j h j (5.12) has the same bound as (13) : use the O ( N ) O(\sqrt{N}) O ( N ) O ( N ) O(\sqrt{N}) O ( N ) N N N c 3 c_3 c 3 ℓ \ell ℓ O ℓ ( N − ℓ ) O_\ell(N^{-\ell}) O ℓ ( N − ℓ ) 1 1 1 u 0 u_0 u 0 T N T_N T N O ( N − ℓ ) O(N^{-\ell}) O ( N − ℓ ) ℓ \ell ℓ F h , N ( ω ) = ∫ ( r N h ) C ω d μ F_{h,N}(\omega)=\int(r_Nh)C_\omega\,d\mu F h , N ( ω ) = ∫ ( r N h ) C ω d μ
∣ Δ ℓ F h , N ( i y ) ∣ ≤ N a h − ℓ + o ( 1 ) , a h 0 = a h 1 = − 1 , a c 3 = 0 , |\Delta^\ell F_{h,N}(iy)|\le N^{a_h-\ell+o(1)},\qquad a_{h_0}=a_{h_1}=-1,\qquad a_{c_3}=0, ∣ Δ ℓ F h , N ( i y ) ∣ ≤ N a h − ℓ + o ( 1 ) , a h 0 = a h 1 = − 1 , a c 3 = 0 , uniformly for y ∈ R y\in\mathbb{R} y ∈ R
Growth on a shifted contour. The bound at − 4 -4 − 4 h = h 0 , h 1 , c 3 h=h_0,h_1,c_3 h = h 0 , h 1 , c 3
ρ h , N ( z ) = ( r N h ) ( z 2 ) z D ( z 2 ) sinh ( 4 λ z ) . \rho_{h,N}(z)=(r_Nh)(z^2)\frac{zD(z^2)}{\sinh(4\lambda z)}. ρ h , N ( z ) = ( r N h ) ( z 2 ) sinh ( 4 λ z ) z D ( z 2 ) . We first prove that, for every fixed Y > 0 Y>0 Y > 0
∫ R ∣ ρ h , N ( z + i Y ) ∣ d z ≤ C Y N b Y / π + C , (14) \int_{\mathbb{R}}|\rho_{h,N}(z+iY)|\,dz\le C_YN^{bY/\pi+C}, \tag*{(14)} ∫ R ∣ ρ h , N ( z + iY ) ∣ d z ≤ C Y N bY / π + C , (14) where the additive exponent C C C Y Y Y Y Y Y D D D h 0 , h 1 h_0,h_1 h 0 , h 1
In the orthogonal polynomial expansion (6) with c = b c=b c = b F 0 , N F_{0,N} F 0 , N N − 1 N-1 N − 1
q j p ( j ) + d 0 q 1 j , deg p < N , 0 < q 1 = w λ / 2 < q = w λ < 1. q^jp(j)+d_0q_1^j,\qquad\deg p<N,\qquad0<q_1=w_{\lambda/2}<q=w_\lambda<1. q j p ( j ) + d 0 q 1 j , deg p < N , 0 < q 1 = w λ /2 < q = w λ < 1. Divide by q j q^j q j N N N d 0 d_0 d 0 e O ( N ) e^{O(N)} e O ( N ) N N N C ′ N C'N C ′ N C ′ C' C ′ 2 λ < c < b 2\lambda<c<b 2 λ < c < b μ c \mu_c μ c f ∈ F 0 , N f\in F_{0,N} f ∈ F 0 , N O ( N ) O(N) O ( N ) f f f μ c \mu_c μ c ∣ P j ( c ) ( z 2 ) ∣ ≤ C c j + 1 e c z / 4 |P_j^{(c)}(z^2)|\le C_c^{j+1}e^{cz/4} ∣ P j ( c ) ( z 2 ) ∣ ≤ C c j + 1 e cz /4 C ′ C' C ′ μ c \mu_c μ c z > C ′ N z>C'N z > C ′ N ∥ f ∥ μ c \|f\|_{\mu_c} ∥ f ∥ μ c μ c \mu_c μ c f f f z ≤ C ′ N z\le C'N z ≤ C ′ N d μ c / d μ b d\mu_c/d\mu_b d μ c / d μ b e O ( N ) e^{O(N)} e O ( N )
∥ f ∥ μ c ≤ e O ( N ) ∥ f ∥ μ b , f ∈ F 0 , N . \|f\|_{\mu_c}\le e^{O(N)}\|f\|_{\mu_b},\qquad f\in F_{0,N}. ∥ f ∥ μ c ≤ e O ( N ) ∥ f ∥ μ b , f ∈ F 0 , N . All constants here may depend on the fixed c c c N N N
Pairing against c 3 c_3 c 3 μ \mu μ N N N z ≤ N 2 z\le N^2 z ≤ N 2 c 3 ( z 2 ) 2 d μ / d z = O ( 1 + z ) c_3(z^2)^2d\mu/dz=O(1+z) c 3 ( z 2 ) 2 d μ / d z = O ( 1 + z ) N N N N 2 N^2 N 2 μ c \mu_c μ c e − ( b − c ) z e^{-(b-c)z} e − ( b − c ) z e O ( N ) e^{O(N)} e O ( N ) c 3 c_3 c 3 M N M_N M N (5.12) ; for h j h_j h j (13) .
Finally the generating formula on ∣ w t ∣ = exp ( − 1 / ( j + 1 ) ) |w_t|=\exp(-1/(j+1)) ∣ w t ∣ = exp ( − 1/ ( j + 1 ))
∣ P j ( ( z + i Y ) 2 ) ∣ ≤ C Y ( j + 1 ) 1 / 2 + b Y / π e b z ∣ z ∣ / 2 . |P_j((z+iY)^2)|\le C_Y(j+1)^{1/2+bY/\pi}e^{bz|z|/2}. ∣ P j (( z + iY ) 2 ) ∣ ≤ C Y ( j + 1 ) 1/2 + bY / π e b z ∣ z ∣/2 . Here the corresponding parameter satisfies ∣ Re t ∣ < b / 2 |\operatorname{Re}t|<b/2 ∣ Re t ∣ < b /2 ∣ Im t ∣ ≤ ( b / π ) log ( j + 1 ) + O ( 1 ) |\operatorname{Im}t|\le(b/\pi)\log(j+1)+O(1) ∣ Im t ∣ ≤ ( b / π ) log ( j + 1 ) + O ( 1 ) (14) : multiplication by the density leaves integrable exponential decay in ∣ z ∣ |z| ∣ z ∣ b Y / π bY/\pi bY / π Y Y Y
Recovering the two evaluations. Fix the difference order ℓ \ell ℓ η > 0 \eta>0 η > 0 S N = ( b / π + η ) log N S_N=(b/\pi+\eta)\log N S N = ( b / π + η ) log N F h , N ( i y ) F_{h,N}(iy) F h , N ( i y ) z z z Im z = Y \operatorname{Im}z=Y Im z = Y 1 / sinh ( 4 λ z ) 1/\sinh(4\lambda z) 1/ sinh ( 4 λ z ) z = 2 i n z=2in z = 2 in n ≥ 1 n\ge1 n ≥ 1 y ≥ S N y\ge S_N y ≥ S N
F h , N ( i y ) = ∑ 0 < 2 n < Y b n , N e − 2 n y + O Y ( e − Y y N b Y / π + C ) . (15) F_{h,N}(iy)=\sum_{0<2n<Y}b_{n,N}e^{-2ny}+O_Y(e^{-Yy}N^{bY/\pi+C}). \tag*{(15)} F h , N ( i y ) = 0 < 2 n < Y ∑ b n , N e − 2 n y + O Y ( e − Y y N bY / π + C ) . (15) where b 1 , N b_{1,N} b 1 , N r N h ( − 4 ) r_N h(-4) r N h ( − 4 ) (15) also at N + 1 , … , N + ℓ N+1,\ldots,N+\ell N + 1 , … , N + ℓ S N S_N S N y = S N + j y=S_N+j y = S N + j j ≥ 0 j\geq0 j ≥ 0 Y > 2 Y>2 Y > 2 ℓ \ell ℓ η \eta η N N N C C C Y Y Y N − η Y + C N^{-\eta Y+C} N − η Y + C (5.13) then bounds every coefficient e − 2 n S N Δ ℓ b n , N e^{-2nS_N}\Delta^\ell b_{n,N} e − 2 n S N Δ ℓ b n , N N a h − ℓ + o ( 1 ) N^{ah-\ell+o(1)} N ah − ℓ + o ( 1 ) n = 1 n=1 n = 1
∣ Δ ℓ r N h ( − 4 ) ∣ ≤ N a h + 2 b / π + 2 η − ℓ + o ( 1 ) . \left|\Delta^\ell r_N h(-4)\right| \leq N^{ah+2b/\pi+2\eta-\ell+o(1)}. Δ ℓ r N h ( − 4 ) ≤ N ah + 2 b / π + 2 η − ℓ + o ( 1 ) . Letting η \eta η y ≥ S N y\geq S_N y ≥ S N N a h − ℓ + o ( 1 ) N^{ah-\ell+o(1)} N ah − ℓ + o ( 1 ) S N S_N S N (5.13) ; its logarithmic factor is absorbed in o ( 1 ) o(1) o ( 1 ) z = 0 z=0 z = 0 μ \mu μ (12) also at a = 0 a=0 a = 0
It remains to prove − r N c 3 ( 0 ) ≥ c -r_Nc_3(0)\geq c − r N c 3 ( 0 ) ≥ c G N G_N G N G N , t G_{N,t} G N , t 0 < t < λ 0<t<\lambda 0 < t < λ W W W K N 0 K_N^0 K N 0 χ N \chi_N χ N v ( t ) v(t) v ( t ) J ( b / 2 ) J(b/2) J ( b /2 ) J ( i y ) Q ( v ( i y ) ) N J(iy)Q(v(iy))^N J ( i y ) Q ( v ( i y ) ) N
K N 0 ( v , 1 ) − K N 0 ( v , u 0 ) K N 0 ( v ( t ) , 1 ) K N 0 ( v ( t ) , u 0 ) , v = v ( i y ) . K_N^0(v,1)-K_N^0(v,u_0)\frac{K_N^0(v(t),1)}{K_N^0(v(t),u_0)},\qquad v=v(iy). K N 0 ( v , 1 ) − K N 0 ( v , u 0 ) K N 0 ( v ( t ) , u 0 ) K N 0 ( v ( t ) , 1 ) , v = v ( i y ) . After division by N N N v = ± i v=\pm i v = ± i
1 2 − 1 − v ( t ) 2 u 0 2 ( 1 + u 0 2 ) ( 1 − v ( t ) 2 ) , \frac{1}{2}-\frac{1-v(t)^2u_0^2}{(1+u_0^2)(1-v(t)^2)}, 2 1 − ( 1 + u 0 2 ) ( 1 − v ( t ) 2 ) 1 − v ( t ) 2 u 0 2 , which is strictly negative, uniformly for 0 ≤ t ≤ λ 0\leq t\leq\lambda 0 ≤ t ≤ λ 1 / ( 1 + u 0 2 ) > 1 / 2 1/(1+u_0^2)>1/2 1/ ( 1 + u 0 2 ) > 1/2 O O O
Fourier inversion at zero integrates the bracket divided by N N N
c N J ( i y ) Q ( v ( i y ) ) N d y , c > 0. cNJ(iy)Q(v(iy))^N\,dy,\qquad c>0. c N J ( i y ) Q ( v ( i y ) ) N d y , c > 0. This positive measure has mass bounded above and below and concentrates at y → ± ∞ y\to\pm\infty y → ± ∞ Q ≥ 0 Q\geq0 Q ≥ 0 1 − Q ≍ e − π ∣ y ∣ / b 1-Q\asymp e^{-\pi|y|/b} 1 − Q ≍ e − π ∣ y ∣/ b J ≍ e − π ∣ y ∣ / b J\asymp e^{-\pi|y|/b} J ≍ e − π ∣ y ∣/ b W W W t t t
To pass to μ \mu μ W W W
d μ d W = ( 2 u − 1 ) 2 ( 2 u + 1 ) 4 u ( 2 u 2 − 1 ) , u = cosh ( λ x ) . \frac{d\mu}{dW}=\frac{(2u-1)^2(2u+1)}{4u(2u^2-1)},\qquad u=\cosh(\lambda\sqrt{x}). d W d μ = 4 u ( 2 u 2 − 1 ) ( 2 u − 1 ) 2 ( 2 u + 1 ) , u = cosh ( λ x ) . It is positive and smooth on [ 0 , ∞ ) [0,\infty) [ 0 , ∞ ) r ( x ) = ( x + 2 ) / ( x + 1 ) r(x)=(x+2)/(x+1) r ( x ) = ( x + 2 ) / ( x + 1 ) j > 0 j>0 j > 0 μ / ( r j W ) \mu/(r^jW) μ / ( r j W ) − ( log r ) ′ -(\log r)' − ( log r ) ′ − e 0 c 3 -e_0c_3 − e 0 c 3 μ \mu μ r j W r^jW r j W
Each multiplication of the base weight by r r r ∏ i r ( x i ) \prod_i r(x_i) ∏ i r ( x i ) ( A N , G N , t ) (A_N,G_{N,t}) ( A N , G N , t ) ( F 0 , N , G N , t ) (F_{0,N},G_{N,t}) ( F 0 , N , G N , t ) ( U N , U N ) (U_N,U_N) ( U N , U N ) ( M N , M N ) (M_N,M_N) ( M N , M N ) log r \log r log r N , t N,t N , t j j j G N , t G_{N,t} G N , t G N G_N G N (12) .
The quotient. We can now finish Proposition 5.1. The denominator r N ( c 3 − h 1 ) ( 0 ) r_N(c_3-h_1)(0) r N ( c 3 − h 1 ) ( 0 ) (5.2) is bounded away from zero for all sufficiently large N N N − r N c 3 ( 0 ) ≥ c -r_Nc_3(0)\ge c − r N c 3 ( 0 ) ≥ c ∣ r N h 1 ( 0 ) ∣ ≤ N − 1 + o ( 1 ) |r_Nh_1(0)|\le N^{-1+o(1)} ∣ r N h 1 ( 0 ) ∣ ≤ N − 1 + o ( 1 ) ℓ \ell ℓ N − ℓ + o ( 1 ) N^{-\ell+o(1)} N − ℓ + o ( 1 ) (12) . In the numerator, r N h 0 ( − 4 ) r_Nh_0(-4) r N h 0 ( − 4 ) N − 1 + 2 b / π − ℓ + o ( 1 ) N^{-1+2b/\pi-\ell+o(1)} N − 1 + 2 b / π − ℓ + o ( 1 ) r N ( c 3 − h 1 ) ( − 4 ) r_N(c_3-h_1)(-4) r N ( c 3 − h 1 ) ( − 4 ) r N h 0 ( 0 ) r_Nh_0(0) r N h 0 ( 0 ) 2 b / π = 3 / 4 2b/\pi=3/4 2 b / π = 3/4 − 1 / 4 − ℓ + o ( 1 ) -1/4-\ell+o(1) − 1/4 − ℓ + o ( 1 ) ℓ \ell ℓ N N N C ℓ , η C_{\ell,\eta} C ℓ , η
Exact height and displacement atoms
We now turn strip regularity into estimates for a bridge with a prescribed terminal port. Throughout this section, p p p H H H Z Z Z L L L D D D X = ( H , Z ) X=(H,Z) X = ( H , Z )
S k = X 1 + ⋯ + X k , T k = L 1 + ⋯ + L k . S_k=X_1+\cdots+X_k,\qquad T_k=L_1+\cdots+L_k. S k = X 1 + ⋯ + X k , T k = L 1 + ⋯ + L k . Independence is used between whole pieces, never between coordinates of one piece. We use the renewal identity and one-piece estimates stated earlier:
∑ h ≥ 0 B h z h = 1 1 − E p z H , B 0 = 1 , B h ≲ ( 1 + h ) − 1 / 4 . (16) \sum_{h\ge0}B_hz^h=\frac{1}{1-\mathbb{E}_p z^H},\qquad B_0=1,\qquad B_h\lesssim(1+h)^{-1/4}. \tag*{(16)} h ≥ 0 ∑ B h z h = 1 − E p z H 1 , B 0 = 1 , B h ≲ ( 1 + h ) − 1/4 . (16) and, for x , t ≥ 1 x,t\ge1 x , t ≥ 1 u ↓ 0 u\downarrow0 u ↓ 0
p ( H > x ) + p ( D > x ) ≤ C x − 3 / 4 , p(H>x)+p(D>x)\le Cx^{-3/4}, p ( H > x ) + p ( D > x ) ≤ C x − 3/4 , p ( L > t ) ≤ t − 9 / 16 + o ( 1 ) , 1 − E p e − u L = u 9 / 16 + o ( 1 ) . (17) p(L>t)\le t^{-9/16+o(1)},\qquad1-\mathbb{E}_p e^{-uL}=u^{9/16+o(1)}. \tag*{(17)} p ( L > t ) ≤ t − 9/16 + o ( 1 ) , 1 − E p e − uL = u 9/16 + o ( 1 ) . (17) The first identity implies 1 − E p e − s H ≍ s 3 / 4 1-\mathbb{E}_p e^{-sH}\asymp s^{3/4} 1 − E p e − sH ≍ s 3/4 p p p
An atom of the one-piece height
The extra information supplied by strip regularity is one full power of height decay beyond the tail bound. This is the estimate that rules out bridges made from too few irreducibles at one prescribed endpoint.
Proposition 6.1. For every η > 0 \eta>0 η > 0 C η < ∞ C_\eta<\infty C η < ∞
p ( H = n ) ≤ C η n − 7 / 4 + η ( n ≥ 1 ) . p(H=n)\le C_\eta n^{-7/4+\eta}\qquad(n\ge1). p ( H = n ) ≤ C η n − 7/4 + η ( n ≥ 1 ) . Proof. Let b ( z ) = ∑ m ≥ 0 B m z m b(z)=\sum_{m\geq0}B_m z^m b ( z ) = ∑ m ≥ 0 B m z m n ≥ 2 n\geq2 n ≥ 2 r = e − 1 / n r=e^{-1/n} r = e − 1/ n s = min { 1 , n − 1 + ∣ θ ∣ } s=\min\{1,n^{-1}+|\theta|\} s = min { 1 , n − 1 + ∣ θ ∣ } − π ≤ θ ≤ π -\pi\leq\theta\leq\pi − π ≤ θ ≤ π
∣ 1 − E p ( r e i θ ) H ∣ ≤ C E p min { 1 , s H } ≤ C s 3 / 4 . \left|1-\mathbb{E}_p(re^{i\theta})^H\right|\leq C\mathbb{E}_p\min\{1,sH\}\leq Cs^{3/4}. 1 − E p ( r e i θ ) H ≤ C E p min { 1 , sH } ≤ C s 3/4 . Thus ∣ b ( r e i θ ) ∣ ≥ c s − 3 / 4 |b(re^{i\theta})|\geq cs^{-3/4} ∣ b ( r e i θ ) ∣ ≥ c s − 3/4 b b b
For every fixed integer k ≥ 0 k\geq0 k ≥ 0 δ > 0 \delta>0 δ > 0
∣ ∂ θ k b ( r e i θ ) ∣ ≤ C k , δ s − 3 / 4 − k − δ . (18) \left|\partial_\theta^k b(re^{i\theta})\right|\leq C_{k,\delta}s^{-3/4-k-\delta}. \tag*{(18)} ∂ θ k b ( r e i θ ) ≤ C k , δ s − 3/4 − k − δ . (18) Here are the details of the summation estimate. Choose a smooth dyadic partition of unity on the positive integers. On m ≍ M m\asymp M m ≍ M j j j m k B m r m m^kB_mr^m m k B m r m C j , k , δ M k − 1 / 4 − j + δ e − M / ( C n ) C_{j,k,\delta}M^{k-1/4-j+\delta}e^{-M/(Cn)} C j , k , δ M k − 1/4 − j + δ e − M / ( C n ) j j j
C j , k , δ M 3 / 4 + k + δ e − M / ( C n ) ( 1 + M ∣ θ ∣ ) − j . C_{j,k,\delta}M^{3/4+k+\delta}e^{-M/(Cn)}(1+M|\theta|)^{-j}. C j , k , δ M 3/4 + k + δ e − M / ( C n ) ( 1 + M ∣ θ ∣ ) − j . Choose j > 3 / 4 + k + δ j>3/4+k+\delta j > 3/4 + k + δ M M M M ≤ s − 1 M\leq s^{-1} M ≤ s − 1 (18) .
Differentiating the reciprocal twice now gives
∣ ∂ θ 2 b ( r e i θ ) − 1 ∣ ≤ C δ s − 5 / 4 − 2 δ . \left|\partial_\theta^2b(re^{i\theta})^{-1}\right|\leq C_\delta s^{-5/4-2\delta}. ∂ θ 2 b ( r e i θ ) − 1 ≤ C δ s − 5/4 − 2 δ . Consequently its integral over [ − π , π ] [-\pi,\pi] [ − π , π ] C δ n 1 / 4 + 2 δ C_\delta n^{1/4+2\delta} C δ n 1/4 + 2 δ E p z H = 1 − b ( z ) − 1 \mathbb{E}_p z^H=1-b(z)^{-1} E p z H = 1 − b ( z ) − 1 n n n
r n p ( H = n ) ≤ 1 2 π n 2 ∫ − π π ∣ ∂ θ 2 b ( r e i θ ) − 1 ∣ d θ ≤ C δ n − 7 / 4 + 2 δ . r^n p(H=n)\leq\frac{1}{2\pi n^2}\int_{-\pi}^{\pi}\left|\partial_\theta^2b(re^{i\theta})^{-1}\right|\,d\theta\leq C_\delta n^{-7/4+2\delta}. r n p ( H = n ) ≤ 2 π n 2 1 ∫ − π π ∂ θ 2 b ( r e i θ ) − 1 d θ ≤ C δ n − 7/4 + 2 δ . The boundary terms vanish by periodicity, r n = e − 1 r^n=e^{-1} r n = e − 1 δ \delta δ
We need both a planar atom bound and a bound summed over possible heights. The latter is stronger than merely summing the planar bound over all heights, since the height distribution has an infinite tail.
Proposition 6.2. There is C < ∞ C<\infty C < ∞ k ≥ 1 k\geq1 k ≥ 1
sup w Pr ( S k = w ) ≤ C k − 8 / 3 , ∑ m ∈ Z sup H ( w ) = m Pr ( S k = w ) ≤ C k − 4 / 3 . (19) \sup_w\Pr(S_k=w)\leq Ck^{-8/3},\qquad\sum_{m\in\mathbb{Z}}\sup_{H(w)=m}\Pr(S_k=w)\leq Ck^{-4/3}. \tag*{(19)} w sup Pr ( S k = w ) ≤ C k − 8/3 , m ∈ Z ∑ H ( w ) = m sup Pr ( S k = w ) ≤ C k − 4/3 . (19) Proof. We first establish a directional window for one piece. There are constants 0 < c 0 < C 0 < ∞ 0<c_0<C_0<\infty 0 < c 0 < C 0 < ∞ c 1 > 0 c_1>0 c 1 > 0 e e e x x x
p { c 0 x ≤ ∣ e ⋅ X ∣ ≤ C 0 x } ≥ c 1 x − 3 / 4 . (20) p\{c_0x\leq|e\cdot X|\leq C_0x\}\geq c_1x^{-3/4}. \tag*{(20)} p { c 0 x ≤ ∣ e ⋅ X ∣ ≤ C 0 x } ≥ c 1 x − 3/4 . (20) Use bridges of heights in [ x , 2 x ] [x,2x] [ x , 2 x ] x x x c x − 5 / 4 cx^{-5/4} c x − 5/4 x x x c x 3 / 4 cx^{3/4} c x 3/4
By the renewal factorization, this is a sum of probabilities for S k S_k S k k > A x 3 / 4 k>Ax^{3/4} k > A x 3/4
∑ k > A x 3 / 4 e 2 ( E p e − H / x ) k ≤ C x 3 / 4 e − c A . \sum_{k>Ax^{3/4}} e^2\left(\mathbb{E}_p e^{-H/x}\right)^k \le Cx^{3/4}e^{-cA}. k > A x 3/4 ∑ e 2 ( E p e − H / x ) k ≤ C x 3/4 e − c A . Fix A A A k ≤ A x 3 / 4 k\le Ax^{3/4} k ≤ A x 3/4 x x x D > C 0 x D>C_0x D > C 0 x C A C 0 − 3 / 4 CA C_0^{-3/4} C A C 0 − 3/4
E ∑ i ≤ k ∣ e ⋅ X i ∣ 1 { ∣ e ⋅ X i ∣ < c 0 x } ≤ C A c 0 1 / 4 x . \mathbb{E}\sum_{i\le k}|e\cdot X_i|\mathbf{1}_{\{|e\cdot X_i|<c_0x\}} \le CA c_0^{1/4}x. E i ≤ k ∑ ∣ e ⋅ X i ∣ 1 { ∣ e ⋅ X i ∣ < c 0 x } ≤ C A c 0 1/4 x . The second estimate follows by integrating the diameter tail; fixed lattice conversion constants are absorbed in C C C C 0 C_0 C 0 c 0 c_0 c 0 (20) . A union bound and k ≤ A x 3 / 4 k\le Ax^{3/4} k ≤ A x 3/4
Let φ ( θ ) = E e i θ ⋅ X \varphi(\theta)=\mathbb{E}e^{i\theta\cdot X} φ ( θ ) = E e i θ ⋅ X X ′ X' X ′ (20) in direction θ / ∣ θ ∣ \theta/|\theta| θ /∣ θ ∣ x x x ∣ θ ∣ − 1 |\theta|^{-1} ∣ θ ∣ − 1 ∣ θ ⋅ ( X − X ′ ) ∣ |\theta\cdot(X-X')| ∣ θ ⋅ ( X − X ′ ) ∣ π \pi π
1 − ∣ φ ( θ ) ∣ 2 = E [ 1 − cos ( θ ⋅ ( X − X ′ ) ) ] ≥ c ∣ θ ∣ 3 / 4 . 1-|\varphi(\theta)|^2=\mathbb{E}[1-\cos(\theta\cdot(X-X'))]\ge c|\theta|^{3/4}. 1 − ∣ φ ( θ ) ∣ 2 = E [ 1 − cos ( θ ⋅ ( X − X ′ ))] ≥ c ∣ θ ∣ 3/4 . near zero. The directional windows also show that the affine span of the support is two-dimensional. Its difference group is therefore a full-rank lattice subgroup. There are only finitely many points of the lattice torus where ∣ φ ∣ = 1 |\varphi|=1 ∣ φ ∣ = 1
sup w Pr ( S k = w ) ≤ C ∫ 0 1 e − c k r 3 / 4 r d r + C e − c k ≤ C k − 8 / 3 . \sup_w\Pr(S_k=w)\le C\int_0^1e^{-ckr^{3/4}}r\,dr+Ce^{-ck}\le Ck^{-8/3}. w sup Pr ( S k = w ) ≤ C ∫ 0 1 e − c k r 3/4 r d r + C e − c k ≤ C k − 8/3 . This also explains why no aperiodicity assumption beyond the actual support lattice is needed.
It remains to sum the heightwise supremums. For a nonnegative measure q q q
∥ q ∥ 1 , ∞ = ∑ m sup H ( w ) = m q ( w ) . \|q\|_{1,\infty}=\sum_m\sup_{H(w)=m}q(w). ∥ q ∥ 1 , ∞ = m ∑ H ( w ) = m sup q ( w ) . Convolution satisfies ∥ q ∗ r ∥ 1 , ∞ ≤ ∥ q ∥ 1 , ∞ ∥ r ∥ 1 \|q*r\|_{1,\infty}\le\|q\|_{1,\infty}\|r\|_1 ∥ q ∗ r ∥ 1 , ∞ ≤ ∥ q ∥ 1 , ∞ ∥ r ∥ 1 (19) immediately bounds the sum over m ≤ 2 k 4 / 3 m\le2k^{4/3} m ≤ 2 k 4/3 C k − 4 / 3 Ck^{-4/3} C k − 4/3
For a dyadic shell m ≍ Λ k 4 / 3 m\asymp\Lambda k^{4/3} m ≍ Λ k 4/3 Λ ≥ 2 \Lambda\ge2 Λ ≥ 2 k k k k k k k k k C k − 8 / 3 Ck^{-8/3} C k − 8/3 t ≥ k 4 / 3 t\ge k^{4/3} t ≥ k 4/3 C k t − 3 / 4 Ckt^{-3/4} C k t − 3/4 t t t
Suppose first that one block has height at most k 4 / 3 Λ k^{4/3}\sqrt{\Lambda} k 4/3 Λ Λ \Lambda Λ c Λ k 4 / 3 c\Lambda k^{4/3} c Λ k 4/3 ∥ ⋅ ∥ 1 , ∞ \|\cdot\|_{1,\infty} ∥ ⋅ ∥ 1 , ∞ C Λ k − 4 / 3 C\sqrt{\Lambda}k^{-4/3} C Λ k − 4/3 C Λ − 3 / 4 C\Lambda^{-3/4} C Λ − 3/4 C Λ − 1 / 4 k − 4 / 3 C\Lambda^{-1/4}k^{-4/3} C Λ − 1/4 k − 4/3 k 4 / 3 Λ k^{4/3}\sqrt{\Lambda} k 4/3 Λ C Λ − 3 / 8 C\Lambda^{-3/8} C Λ − 3/8 O ( Λ k 4 / 3 ) O(\Lambda k^{4/3}) O ( Λ k 4/3 ) C Λ − 37 / 8 k − 4 / 3 C\Lambda^{-37/8}k^{-4/3} C Λ − 37/8 k − 4/3 Λ \Lambda Λ Λ \Lambda Λ k k k
Prescribed ports on parallel sides
The strip estimates above concern a fixed initial port and all terminal ports. We now prescribe the second port as well. The connector construction gives the required normalization: when the height is h h h O ( h ) O(h) O ( h ) c h − 5 / 4 ch^{-5/4} c h − 5/4 h h h o ( h − 5 / 4 ) o(h^{-5/4}) o ( h − 5/4 )
Theorem 7.1. Fix a bound on the ratio of transverse displacement to height. For compatible prescribed ports on two parallel pure cuts at integral height h → ∞ h \to\infty h → ∞ ϵ > 0 \epsilon> 0 ϵ > 0
Pr { h 4 / 3 − ϵ ≤ L ≤ h 4 / 3 + ϵ } ⟶ 1. \Pr\{h^{4/3-\epsilon} \le L \le h^{4/3+\epsilon}\} \longrightarrow1. Pr { h 4/3 − ϵ ≤ L ≤ h 4/3 + ϵ } ⟶ 1. The convergence is uniform under the fixed displacement bound. It also holds in convex pure-sided domains in a fixed compact family of nondegenerate scaled geometries, with the ports on opposite parallel sides and with order-h h h
Proof. It is enough to treat 0 < ϵ < 1 0 < \epsilon< 1 0 < ϵ < 1 w w w H ( w ) = h H(w) = h H ( w ) = h ∑ k ≥ 1 Pr ( S k = w ) \sum_{k \ge1} \Pr(S_k = w) ∑ k ≥ 1 Pr ( S k = w )
Fix a small ξ > 0 \xi> 0 ξ > 0 K = h 3 / 4 − ξ K = h^{3/4-\xi} K = h 3/4 − ξ a = h 1 − τ a = h^{1-\tau} a = h 1 − τ 0 < τ < 8 ξ / 21 0 < \tau< 8\xi/21 0 < τ < 8 ξ /21 k ≤ K k \le K k ≤ K ∑ i ≤ k H i = h \sum_{i \le k} H_i = h ∑ i ≤ k H i = h H i < a H_i < a H i < a
e − h / a ( E e H / a ; H < a ) k ≤ exp { − h τ + C k a − 3 / 4 } . e^{-h/a}\bigl(\mathbb{E}e^{H/a}; H<a\bigr)^k \le\exp\{-h^\tau+ Cka^{-3/4}\}. e − h / a ( E e H / a ; H < a ) k ≤ exp { − h τ + C k a − 3/4 } . We used e t − 1 ≤ e t e^t - 1 \le et e t − 1 ≤ e t 0 ≤ t ≤ 1 0 \le t \le1 0 ≤ t ≤ 1 E [ H ; H < a ] ≤ C a 1 / 4 \mathbb{E}[H;H<a] \le Ca^{1/4} E [ H ; H < a ] ≤ C a 1/4 k a − 3 / 4 ≤ h − ξ + 3 τ / 4 ka^{-3/4} \le h^{-\xi+3\tau/4} k a − 3/4 ≤ h − ξ + 3 τ /4 k ≤ K k \le K k ≤ K h h h
If some H i ≥ a H_i \ge a H i ≥ a
Pr { S k = w , H i ≥ a } ≤ sup n ≥ a p ( H = n ) ∑ H ( v ) = m sup Pr ( S k − 1 = v ) ≤ a − 7 / 4 + o ( 1 ) C ( 1 + k ) − 4 / 3 . \Pr\{S_k = w,\ H_i \ge a\}
\le\sup_{n \ge a} p(H=n)\sum_{H(v)=m}\sup\Pr(S_{k-1}=v)
\le a^{-7/4+o(1)}C(1+k)^{-4/3}. Pr { S k = w , H i ≥ a } ≤ n ≥ a sup p ( H = n ) H ( v ) = m ∑ sup Pr ( S k − 1 = v ) ≤ a − 7/4 + o ( 1 ) C ( 1 + k ) − 4/3 . For k = 1 k=1 k = 1 k ≤ K k \le K k ≤ K
C a − 7 / 4 + o ( 1 ) K 2 / 3 = h − 5 / 4 − 2 ξ / 3 + 7 τ / 4 + o ( 1 ) = o ( h − 5 / 4 ) . Ca^{-7/4+o(1)}K^{2/3}
= h^{-5/4-2\xi/3+7\tau/4+o(1)}
= o(h^{-5/4}). C a − 7/4 + o ( 1 ) K 2/3 = h − 5/4 − 2 ξ /3 + 7 τ /4 + o ( 1 ) = o ( h − 5/4 ) . This is the step that needs both the one-piece height atom and the heightwise supremum estimate.
For k > h 3 / 4 + ξ k > h^{3/4+\xi} k > h 3/4 + ξ
∑ k > h 3 / 4 + ξ Pr { ∑ i ≤ k H i = h } ≤ e ∑ k > h 3 / 4 + ξ ( E e − H / h ) k ≤ C h 3 / 4 e − c h ξ . \sum_{k>h^{3/4+\xi}}\Pr\left\{\sum_{i \le k}H_i=h\right\}
\le e\sum_{k>h^{3/4+\xi}}\left(\mathbb{E}e^{-H/h}\right)^k
\le Ch^{3/4}e^{-ch^\xi}. k > h 3/4 + ξ ∑ Pr { i ≤ k ∑ H i = h } ≤ e k > h 3/4 + ξ ∑ ( E e − H / h ) k ≤ C h 3/4 e − c h ξ . We are left with h 3 / 4 − ξ < k ≤ h 3 / 4 + ξ h^{3/4-\xi}<k\le h^{3/4+\xi} h 3/4 − ξ < k ≤ h 3/4 + ξ ξ \xi ξ ϵ \epsilon ϵ (17) , with u = h − 4 / 3 + ϵ u=h^{-4/3+\epsilon} u = h − 4/3 + ϵ k k k
Pr { T k < h 4 / 3 − ϵ } ≤ e ( E e − u L ) k ≤ exp { − h 9 ϵ / 16 − ξ − o ( 1 ) } . \Pr\{T_k<h^{4/3-\epsilon}\}
\le e\left(\mathbb{E}e^{-uL}\right)^k
\le\exp\{-h^{9\epsilon/16-\xi-o(1)}\}. Pr { T k < h 4/3 − ϵ } ≤ e ( E e − uL ) k ≤ exp { − h 9 ϵ /16 − ξ − o ( 1 ) } . Summing in k k k
For the upper tail, split the pieces into two independent batches with sizes comparable to k k k T k > h 4 / 3 + ϵ T_k > h^{4/3+\epsilon} T k > h 4/3 + ϵ t = 1 2 h 4 / 3 + ϵ t = \frac{1}{2}h^{4/3+\epsilon} t = 2 1 h 4/3 + ϵ η > 0 \eta> 0 η > 0
Pr { that batch has length > t } ≤ C η k t − 9 / 16 + η ≤ h − δ \Pr\{\text{that batch has length}>t\}\le C_{\eta}kt^{-9/16+\eta}\le h^{-\delta} Pr { that batch has length > t } ≤ C η k t − 9/16 + η ≤ h − δ for some δ > 0 \delta> 0 δ > 0 ξ , η \xi,\eta ξ , η ϵ \epsilon ϵ w w w C k − 8 / 3 Ck^{-8/3} C k − 8/3 k k k
C h − δ ∑ k > h 3 / 4 − ξ k − 8 / 3 ≤ C h − 5 / 4 + 5 ξ / 3 − δ = o ( h − 5 / 4 ) , Ch^{-\delta}\sum_{k>h^{3/4-\xi}} k^{-8/3}\le Ch^{-5/4+5\xi/3-\delta}=o(h^{-5/4}), C h − δ k > h 3/4 − ξ ∑ k − 8/3 ≤ C h − 5/4 + 5 ξ /3 − δ = o ( h − 5/4 ) , after decreasing ξ \xi ξ
Both length exceptions therefore have absolute strip mass o ( h − 5 / 4 ) o(h^{-5/4}) o ( h − 5/4 )
Prescribed ports on nonparallel cuts
The difference from the preceding two geometries is that a normal bridge prefix changes the distance to the opposite support. Summing over that support first gives an almost exhaustive prefix construction. We must then show that its small exceptional mass stays small at a specified port. All estimates below are uniform on a fixed compact set of the nondegenerate wedge geometries in Theorem 1.1. We continue to use d = 3 / 2 d=\sqrt{3}/2 d = 3 /2 σ = 3 / 8 \sigma=3/8 σ = 3/8
Theorem 8.1 (Tilted prescribed endpoints). In a pure wedge of angle π / 3 \pi/3 π /3 2 π / 3 2\pi/3 2 π /3 R R R O ( R ) O(R) O ( R ) R 4 / 3 + o ( 1 ) R^{4/3+o(1)} R 4/3 + o ( 1 )
We now treat the case when the supporting sides at the two marked midpoints have distinct, nonparallel pure directions. Write W W W n , m n,m n , m a , Q a,Q a , Q R R R O ( R ) O(R) O ( R ) R → ∞ R\to\infty R → ∞ W W W U U U K ( a , U ; W ) K(a,U;W) K ( a , U ; W ) K U K_U K U
First we explain exhaustion by normal bridge prefixes when summing over the target ray. Put r r r a a a F ( r ) F(r) F ( r ) ∑ U K U \sum_U K_U ∑ U K U r r r
1 = cos ( σ π ) A ( r ) + cos ( σ θ ) F ( r ) 1=\cos(\sigma\pi)A(r)+\cos(\sigma\theta)F(r) 1 = cos ( σ π ) A ( r ) + cos ( σ θ ) F ( r ) where A A A θ \theta θ π / 3 \pi/3 π /3 2 π / 3 2\pi/3 2 π /3 θ \theta θ A A A F ( r ) ≲ ( 1 + r ) − 1 / 4 F(r) \lesssim(1+r)^{-1/4} F ( r ) ≲ ( 1 + r ) − 1/4
Use the ordinary iid irreducible pieces directed along n n n k k k k k k J J J w w w
F ( r ) ≥ E [ 1 s m a l l F ( r + m ⋅ w / d ) ] . F(r) \ge\mathbb{E}\bigl[\mathbf{1}_{\mathrm{small}}F(r+m\cdot w/d)\bigr]. F ( r ) ≥ E [ 1 small F ( r + m ⋅ w / d ) ] . by using the smaller wedge W ′ W' W ′ a ′ = a + w a'=a+w a ′ = a + w a ′ a' a ′ k k k
We record why F ( r + o ( r ) ) − F ( r ) = o ( r − 1 / 4 ) F(r+o(r))-F(r)=o(r^{-1/4}) F ( r + o ( r )) − F ( r ) = o ( r − 1/4 ) q q q k ≤ C q 3 / 4 k\le Cq^{3/4} k ≤ C q 3/4 m ⋅ w / d ≤ − b 0 q m\cdot w/d\le-b_0q m ⋅ w / d ≤ − b 0 q b 0 > 0 b_0>0 b 0 > 0 [ q , 2 q ] [q,2q] [ q , 2 q ] ≳ q 3 / 4 \gtrsim q^{3/4} ≳ q 3/4 − m -m − m C q 3 / 4 Cq^{3/4} C q 3/4 O ( q 3 / 4 e − c C ) O(q^{3/4}e^{-cC}) O ( q 3/4 e − c C ) k k k T q Tq T q O ( T − 3 / 4 ) O(T^{-3/4}) O ( T − 3/4 ) T ≥ 1 T\ge1 T ≥ 1
If there is a drop at least ϵ r − 1 / 4 \epsilon r^{-1/4} ϵ r − 1/4 r − l , r + l r-l,r+l r − l , r + l 1 ≤ l = o ( r ) 1\le l=o(r) 1 ≤ l = o ( r ) A 0 A_0 A 0 l ≪ q ≪ r l\ll q\ll r l ≪ q ≪ r [ r + b 0 q / 4 , r + ( A 0 + b 0 + 1 ) q ] [r+b_0q/4,r+(A_0+b_0+1)q] [ r + b 0 q /4 , r + ( A 0 + b 0 + 1 ) q ] o ( r − 1 / 4 ) o(r^{-1/4}) o ( r − 1/4 ) j j j r + b 0 q / 2 r+b_0q/2 r + b 0 q /2 k k k ∣ m ⋅ w / d ∣ ≤ A 0 q |m\cdot w/d|\le A_0q ∣ m ⋅ w / d ∣ ≤ A 0 q F ( j ) − o ( r − 1 / 4 ) F(j)-o(r^{-1/4}) F ( j ) − o ( r − 1/4 ) ≤ − b 0 q \le-b_0q ≤ − b 0 q ϵ r − 1 / 4 \epsilon r^{-1/4} ϵ r − 1/4 A 0 A_0 A 0 F ( j ) ≲ r − 1 / 4 F(j)\lesssim r^{-1/4} F ( j ) ≲ r − 1/4
In particular take k = ⌊ R 3 / 4 − η ⌋ k=\lfloor R^{3/4-\eta}\rfloor k = ⌊ R 3/4 − η ⌋ η > 0 \eta>0 η > 0 ∑ J \sum J ∑ J R R R W ′ W' W ′ U U U E U E_U E U r ≲ R r\lesssim R r ≲ R F ( r ) F(r) F ( r ) o ( R − 1 / 4 ) o(R^{-1/4}) o ( R − 1/4 ) ∑ U E U = o ( R − 1 / 4 ) \sum_U E_U=o(R^{-1/4}) ∑ U E U = o ( R − 1/4 )
Visits near the prescribed target
The prefix construction has so far controlled only the sum of exceptional masses. To pass to a single target, we need a bound for paths that approach that target before ending a specified distance away. We give a local visit estimate. For q ≤ c R q\le cR q ≤ c R a a a P P P q q q 2 q 2q 2 q Q Q Q t t t Q Q Q 1 ≤ t ≤ q 1\le t\le q 1 ≤ t ≤ q
H q ( t ) ≲ R − 1 / 4 ( t / q ) 3 / 8 . (21) H_q(t)\lesssim R^{-1/4}(t/q)^{3/8}. \tag*{(21)} H q ( t ) ≲ R − 1/4 ( t / q ) 3/8 . (21) Here and below c > 0 c>0 c > 0 a a a q q q q / 20 q/20 q /20 q q q q q q c 1 q 1 / 2 H q ( t ) c_1q^{1/2}H_q(t) c 1 q 1/2 H q ( t ) ≳ q 3 / 4 \gtrsim q^{3/4} ≳ q 3/4 a a a ≳ q − 1 / 4 \gtrsim q^{-1/4} ≳ q − 1/4 a a a O ( R − 1 / 4 ) O(R^{-1/4}) O ( R − 1/4 )
For simultaneous translations v , v ′ v,v' v , v ′ P v , P v ′ P_v,P_{v'} P v , P v ′ q q q 2 q 2q 2 q Q Q Q
For t ≤ q / 10 t\le q/10 t ≤ q /10 Q + v Q+v Q + v Q + v ′ Q+v' Q + v ′ R R R q / 20 q/20 q /20 P v + v P_v+v P v + v P v ′ + v ′ P_{v'}+v' P v ′ + v ′ q q q
This common central portion stays above both target cuts. Its visit within t t t ∣ m ⋅ ( v − v ′ ) ∣ ≤ C t |m\cdot(v-v')|\le Ct ∣ m ⋅ ( v − v ′ ) ∣ ≤ C t P P P O ( q ) O(q) O ( q ) α \alpha α C R − 1 / 4 ( t / q ) α CR^{-1/4}(t/q)^\alpha C R − 1/4 ( t / q ) α α = 0 \alpha=0 α = 0
The mixed difference factor now sums to O ( t 1 / 2 ) O(t^{1/2}) O ( t 1/2 )
∑ z : ∣ m ⋅ z ∣ ≤ C t D n ( z ) ( 1 + ∣ m ⋅ z ∣ ) − 1 / 4 ≤ C ′ t 1 / 2 . \sum_{z:\,|m\cdot z|\le Ct}D_n(z)(1+|m\cdot z|)^{-1/4}\le C't^{1/2}. z : ∣ m ⋅ z ∣ ≤ C t ∑ D n ( z ) ( 1 + ∣ m ⋅ z ∣ ) − 1/4 ≤ C ′ t 1/2 . with D n D_n D n [3 ] . The projected concentration and negative-quarter-moment bounds in (3.4) bound the contribution from j j j C min ( j − 1 / 3 , ( 1 + t ) 3 / 4 j − 4 / 3 ) C\min(j^{-1/3},(1+t)^{3/4}j^{-4/3}) C min ( j − 1/3 , ( 1 + t ) 3/4 j − 4/3 ) j j j O ( q − 1 / 4 ) O(q^{-1/4}) O ( q − 1/4 ) D n ( v o u t ) D_n(v_{\mathrm{out}}) D n ( v out ) O ( q 3 / 4 ) O(q^{3/4}) O ( q 3/4 ) v o u t v_{\mathrm{out}} v out
C q 1 / 2 t 1 / 2 R − 1 / 4 ( t / q ) α . Cq^{1/2}t^{1/2}R^{-1/4}(t/q)^\alpha. C q 1/2 t 1/2 R − 1/4 ( t / q ) α . Cauchy–Schwarz proves the assertion with exponent 1 / 4 + α / 2 1/4+\alpha/2 1/4 + α /2 t ≥ q / 10 t\ge q/10 t ≥ q /10 O ( q ) O(q) O ( q ) c c c
An endpointwise comparison
We adapt the exterior-arc comparison from [16 ] , proof of Proposition 7.1 to the nonparallel supports. Take two individual endpoints P , Q P,Q P , Q P P P ∣ P − Q ∣ ∈ [ q , 2 q ] |P-Q|\in[q,2q] ∣ P − Q ∣ ∈ [ q , 2 q ] P P P Q Q Q e i θ Δ W e^{i\theta\Delta W} e i θ Δ W W W W Figure 2 records its orientation. Arrange the coordinates so the initial ray points right, and the target ray lies counterclockwise at angle θ \theta θ a → P , a → Q a \to P, a \to Q a → P , a → Q θ \theta θ P → Q P \to Q P → Q − π -\pi − π ω : a → Q \omega: a \to Q ω : a → Q
Figure 2. Local orientation at the target support, rotated to horizontal. The blue semicircle is a formal exterior connection, not an added lattice path. The orange path arrives at Q Q Q
f W ( P ; ω ) = 1 2 cos ( σ π ) − ∑ b ′ ∈ ( P , Q ) ∑ ζ : P → b ′ in W ζ disjoint from ω κ L ( ζ ) . f_W(P;\omega)=\frac{1}{2\cos(\sigma\pi)}-\sum_{b'\in(P,Q)}\ \sum_{\substack{\zeta:P\to b' \text{ in } W\\ \zeta\text{ disjoint from }\omega}} \kappa^{L(\zeta)}. f W ( P ; ω ) = 2 cos ( σ π ) 1 − b ′ ∈ ( P , Q ) ∑ ζ : P → b ′ in W ζ disjoint from ω ∑ κ L ( ζ ) . For paths to P P P f W ( Q ; ω ) f_W(Q;\omega) f W ( Q ; ω ) Q Q Q
∣ ∫ a → Q f W ( P ; ω ) − ∫ a → P f W ( Q ; ω ) ∣ ≤ C R − 1 / 4 ( K Q + K P ) . \left|\int_{a\to Q}f_W(P;\omega)-\int_{a\to P}f_W(Q;\omega)\right|\le CR^{-1/4}(K_Q+K_P). ∫ a → Q f W ( P ; ω ) − ∫ a → P f W ( Q ; ω ) ≤ C R − 1/4 ( K Q + K P ) . Indeed the flux delivered by arc traversals before further continuation equals the outgoing flux using the arc once. Prefix cancellations at all ordinary vertices preserve the count of arc uses under unfinished-loop reversal; the whole arc is well after the initial cut so the source cannot be inside a loop. Use first a large convex truncation; distant contributions vanish by ordinary diameter bounds. After rotation by e − i σ θ e^{-i\sigma\theta} e − iσ θ sin ( σ π ) ( K Q − K P ) \sin(\sigma\pi)(K_Q-K_P) sin ( σ π ) ( K Q − K P ) C R − 1 / 4 ( K Q + K P ) CR^{-1/4}(K_Q+K_P) C R − 1/4 ( K Q + K P ) a → Q a\to Q a → Q P P P Q Q Q P P P 1 1 1 sin ( 2 σ π ) \sin(2\sigma\pi) sin ( 2 σ π ) π \pi π a → P a\to P a → P ( P , Q ) (P,Q) ( P , Q ) sin ( 2 σ π ) \sin(2\sigma\pi) sin ( 2 σ π )
Reflection and the half-plane contour identity give total arch mass 1 / ( 2 cos ( σ π ) ) 1/(2\cos(\sigma\pi)) 1/ ( 2 cos ( σ π )) Q Q Q
c ′ q − 1 / 4 ≤ f W ( P ; ω ) ≤ C ( 1 + min ( q , dist ( P , ω ) ) ) − 1 / 4 c'q^{-1/4}\le f_W(P;\omega)\le C(1+\min(q,\operatorname{dist}(P,\omega)))^{-1/4} c ′ q − 1/4 ≤ f W ( P ; ω ) ≤ C ( 1 + min ( q , dist ( P , ω )) ) − 1/4 and analogously after exchanging the endpoints. Indeed all completions counted there are among half-plane arches in one direction, but omitting at least the long-offset tail of mass ≳ q − 1 / 4 \gtrsim q^{-1/4} ≳ q − 1/4
With q q q R R R O ( R − 1 / 4 ) K Q O(R^{-1/4})K_Q O ( R − 1/4 ) K Q q q q P P P H q ( t ) H_q(t) H q ( t ) > 1 / 4 >1/4 > 1/4 f W ( Q ; ω ) f_W(Q;\omega) f W ( Q ; ω ) P P P O ( R − 1 / 4 q − 1 / 4 ) O(R^{-1/4}q^{-1/4}) O ( R − 1/4 q − 1/4 ) > q >q > q K Q ≲ R − 5 / 4 K_Q\lesssim R^{-5/4} K Q ≲ R − 5/4
Removing the exceptional mass
The bound for each individual kernel is now available. Apply the same comparison to the exceptional set left by the prefix construction. The same absolute comparison holds with the two left integrals restricted to exceptional paths and the right error still C R − 1 / 4 ( K Q + K P ) CR^{-1/4}(K_Q+K_P) C R − 1/4 ( K Q + K P ) C C C q ≤ c R q\le cR q ≤ c R W ′ W' W ′ a ′ a' a ′ f f f O ( R − 1 / 4 ) O(R^{-1/4}) O ( R − 1/4 ) R R R
P → Q : Δ W = − π , modulus 1 P \mathbin{\to} Q:\quad\Delta W=-\pi,\quad\text{modulus }1 P → Q : Δ W = − π , modulus 1 Fix q q q ρ R \rho R ρR ρ > 0 \rho>0 ρ > 0
e R = R 1 / 4 ∑ U E U ⟶ 0. e_R=R^{1/4}\sum_U E_U\longrightarrow0. e R = R 1/4 U ∑ E U ⟶ 0. Let I R ( q ) I_R(q) I R ( q ) q q q P P P J a → P J_{a\to P} J a → P f W ( Q ; ω ) f_W(Q;\omega) f W ( Q ; ω ) 0 < δ < 1 / 2 0<\delta<1/2 0 < δ < 1/2 dist ( Q , ω ) ≥ δ q \operatorname{dist}(Q,\omega)\ge\delta q dist ( Q , ω ) ≥ δ q C ( δ q ) − 1 / 4 ∑ P E P C(\delta q)^{-1/4}\sum_P E_P C ( δ q ) − 1/4 ∑ P E P (8.2) . For the remaining paths, discard the exceptional restriction and sum (21) over dyadic distance bins below δ q \delta q δ q 3 / 8 − 1 / 4 = 1 / 8 3/8-1/4=1/8 3/8 − 1/4 = 1/8 R R R ρ , δ \rho,\delta ρ , δ
I R ( q ) ≤ C R − 1 / 4 q − 1 / 4 ( δ − 1 / 4 e R + δ 1 / 8 ) . I_R(q)\le CR^{-1/4}q^{-1/4}\left(\delta^{-1/4}e_R+\delta^{1/8}\right). I R ( q ) ≤ C R − 1/4 q − 1/4 ( δ − 1/4 e R + δ 1/8 ) . First let R → ∞ R\to\infty R → ∞ ρ , δ \rho,\delta ρ , δ δ ↓ 0 \delta\downarrow0 δ ↓ 0 I R ( q ) = o ( R − 1 / 4 q − 1 / 4 ) I_R(q)=o(R^{-1/4}q^{-1/4}) I R ( q ) = o ( R − 1/4 q − 1/4 ) ρ \rho ρ q q q (8.2) and the individual kernel upper bound, now yields
q − 1 / 4 E Q ≤ o ( R − 1 / 4 q − 5 / 4 ) + C R − 1 / 4 R − 5 / 4 . q^{-1/4}E_Q\le o(R^{-1/4}q^{-5/4})+CR^{-1/4}R^{-5/4}. q − 1/4 E Q ≤ o ( R − 1/4 q − 5/4 ) + C R − 1/4 R − 5/4 . Here the little-oh may depend on ρ \rho ρ C C C
lim sup R → ∞ R 5 / 4 E Q ≤ C ρ 1 / 4 . \limsup_{R\to\infty}R^{5/4}E_Q\le C\rho^{1/4}. R → ∞ lim sup R 5/4 E Q ≤ C ρ 1/4 . Letting ρ ↓ 0 \rho\downarrow0 ρ ↓ 0 E Q = o ( R − 5 / 4 ) E_Q=o(R^{-5/4}) E Q = o ( R − 5/4 )
The prescribed length law
Finally count lengths violating either power bound R 4 / 3 ± ϵ R^{4/3\pm\epsilon} R 4/3 ± ϵ o ( R − 5 / 4 ) o(R^{-5/4}) o ( R − 5/4 ) Q Q Q
3 / 4 − η 9 / 16 = 4 3 − 16 η 9 \frac{3/4-\eta}{9/16}=\frac{4}{3}-\frac{16\eta}{9} 9/16 3/4 − η = 3 4 − 9 16 η in probability. On small prefixes the center factor is at most C R − 5 / 4 CR^{-5/4} C R − 5/4 o ( 1 ) o(1) o ( 1 ) o ( R − 5 / 4 ) o(R^{-5/4}) o ( R − 5/4 ) Q Q Q η < 9 ϵ / 16 \eta< 9\epsilon/16 η < 9 ϵ /16 1 2 R 4 / 3 + ϵ \frac{1}{2}R^{4/3+\epsilon} 2 1 R 4/3 + ϵ
For the appended chord, any specific prefix displacement costs O ( k − 8 / 3 ) O(k^{-8/3}) O ( k − 8/3 ) τ w \tau_w τ w
x w = a + w + τ w , y w = Q + τ w , w = ( x w − y w ) − ( a − Q ) . x_w = a + w + \tau_w,\qquad y_w = Q + \tau_w,\qquad w = (x_w-y_w) - (a-Q). x w = a + w + τ w , y w = Q + τ w , w = ( x w − y w ) − ( a − Q ) . Thus each endpoint pair determines its prefix displacement. Since the angles are fixed and ∣ w ∣ = O ( R ) |w| = O(R) ∣ w ∣ = O ( R ) τ w \tau_w τ w O ( R ) O(R) O ( R ) O ( R ) O(R) O ( R ) O ( R ) O(R) O ( R )
We may now sum freely over the endpoint pairs. For appended chords of diameter above S ∗ = R 1 + λ S_* = R^{1+\lambda} S ∗ = R 1 + λ C R S ∗ − 1 / 4 CRS_*^{-1/4} C R S ∗ − 1/4 O ( S ∗ ) O(S_*) O ( S ∗ ) 1 2 R 4 / 3 + ϵ \frac{1}{2}R^{4/3+\epsilon} 2 1 R 4/3 + ϵ C R − 4 / 3 − ϵ S ∗ 25 / 12 + o ( 1 ) CR^{-4/3-\epsilon}S_*^{25/12+o(1)} C R − 4/3 − ϵ S ∗ 25/12 + o ( 1 ) k − 8 / 3 k^{-8/3} k − 8/3
C R − 5 / 4 − λ / 4 + 8 η / 3 , R − 5 / 4 − ϵ + 25 λ / 12 + 8 η / 3 + o ( 1 ) . CR^{-5/4-\lambda/4+8\eta/3},\qquad R^{-5/4-\epsilon+25\lambda/12+8\eta/3+o(1)}. C R − 5/4 − λ /4 + 8 η /3 , R − 5/4 − ϵ + 25 λ /12 + 8 η /3 + o ( 1 ) . Choose first λ > 0 \lambda> 0 λ > 0 25 λ / 12 < ϵ 25\lambda/12 < \epsilon 25 λ /12 < ϵ η > 0 \eta> 0 η > 0 − 5 / 4 -5/4 − 5/4 η < 9 ϵ / 16 \eta< 9\epsilon/16 η < 9 ϵ /16 o ( R − 5 / 4 ) o(R^{-5/4}) o ( R − 5/4 )
Restriction to a hexagon in Theorem 1.1 can only decrease these absolute exceptional masses. Its own normalizer is at least c R − 5 / 4 cR^{-5/4} c R − 5/4
This proves Theorem 8.1. Together with Theorem 4.1 and Theorem 7.1, it proves Theorem 1.1: the two side supports are identical, parallel, or nonparallel, and the connector lower bound supplies the normalizer in each finite hexagon.
Mean length of a prescribed half-plane arch
The same-side probability theorem does not control mean length. We give a separate marked-arm proof, whose length-weighted contributions are summable over arch heights at each prescribed gap. It uses the calibrated finite method and its renewal consequences.
Theorem 9.1 (Every-gap half-plane mean length). For two half-plane boundary ports separated by each integer g → ∞ g \to\infty g → ∞ κ L ( γ ) / K g \kappa^{L(\gamma)}/K_g κ L ( γ ) / K g
L ( γ ) = g 4 / 3 + o ( 1 ) in probability , E L ( γ ) = g 4 / 3 + o ( 1 ) . L(\gamma) = g^{4/3+o(1)} \quad\text{in probability},\qquad\mathbb{E}L(\gamma) = g^{4/3+o(1)}. L ( γ ) = g 4/3 + o ( 1 ) in probability , E L ( γ ) = g 4/3 + o ( 1 ) . The finite calibrated estimates in [12 ] , Theorem 1.2 and Proposition 11.1 give B h ≲ h − 1 / 4 B_h \lesssim h^{-1/4} B h ≲ h − 1/4 K g ≳ g − 5 / 4 K_g \gtrsim g^{-5/4} K g ≳ g − 5/4 A g Ag A g A A A C R 25 / 12 CR^{25/12} C R 25/12 R R R [15 ] , Proposition 4.1, Lemma 4.2, Lemmas 7.2 and 7.4, Section 7.3, and Theorem 7.1. Proposition 4.1 and Lemma 4.2 give the opening height, diameter, and length-deficit estimates; Lemma 7.2 gives the atom bounds; Lemma 7.4 and Section 7.3 give the censored survival test and its shared-turn application; Theorem 7.1 gives the absolute short-path bound.
Proposition 9.2 (Calibrated renewal inputs). For the port-rooted irreducible law, write Y , Z , L , D Y,Z,L,D Y , Z , L , D 2 Z 2Z 2 Z Pr ( D > r ) ≤ C r − 3 / 4 \Pr(D>r)\le Cr^{-3/4} Pr ( D > r ) ≤ C r − 3/4 1 − E e − t Y ≍ t 3 / 4 1-\mathbb{E}e^{-tY}\asymp t^{3/4} 1 − E e − t Y ≍ t 3/4 1 − E e − t L = t 9 / 16 + o ( 1 ) 1-\mathbb{E}e^{-tL}=t^{9/16+o(1)} 1 − E e − t L = t 9/16 + o ( 1 )
(i) For k k k sup j Pr ( S Y ( k ) = j ) ≤ C k − 4 / 3 \sup_j\Pr(S_Y(k)=j)\le Ck^{-4/3} sup j Pr ( S Y ( k ) = j ) ≤ C k − 4/3 sup w Pr ( S ( Y , Z ) ( k ) = w ) ≤ C k − 8 / 3 \sup_w\Pr(S_{(Y,Z)}(k)=w)\le Ck^{-8/3} sup w Pr ( S ( Y , Z ) ( k ) = w ) ≤ C k − 8/3
(ii) Start two independent sequences at the same bottom port through the same first center v v v s s s s s s 0 < d 1 < d 2 < s 0<d_1<d_2<s 0 < d 1 < d 2 < s v v v v v v d 1 d_1 d 1 C η ( 1 + s ) − 3 / 4 + η C_\eta(1+s)^{-3/4+\eta} C η ( 1 + s ) − 3/4 + η η > 0 \eta>0 η > 0 s = 1 s=1 s = 1 s s s κ \kappa κ
(iii) For every sufficiently small fixed ξ > 0 \xi>0 ξ > 0 c , C > 0 c,C>0 c , C > 0
∑ ω rooted at a fixed center R ≤ diam ω ≤ 2 R , L ( ω ) ≤ R 4 / 3 − ξ κ L ( ω ) ≤ C e − R c . \sum_{\substack{\omega\ \text{rooted at a fixed center}\\ R\le\operatorname{diam}\omega\le2R,\ L(\omega)\le R^{4/3-\xi}}}\kappa^{L(\omega)}\le Ce^{-R^c}. ω rooted at a fixed center R ≤ diam ω ≤ 2 R , L ( ω ) ≤ R 4/3 − ξ ∑ κ L ( ω ) ≤ C e − R c . The two arms and the dyadic target
Split an arch at a chosen highest vertex. Its two real incident edges go downwards. Extending both arms to the directly adjacent top wall port makes them strict downward bridges, duplicating just that vertex. Their critical product weight therefore differs from the arch weight by a fixed factor. We count the two shapes from a reference top port, impose equal heights and transverse displacement difference g g g
The two ends of this decomposition impose different constraints. At the top the arms share a turn, so Proposition 9.2(ii) applies. At the bottom they start at distinct ports a distance g g g
Put d ∗ = 4 / 3 d_* = 4/3 d ∗ = 4/3 a = 3 / 4 a=3/4 a = 3/4 H H H
M g ( H ) = ∑ γ : 0 → g a half-plane arch H ≤ h ( γ ) < 2 H L ( γ ) κ L ( γ ) . \mathcal{M}_g(H)=\sum_{\substack{\gamma:0\to g\ \text{a half-plane arch}\\ H\le h(\gamma)<2H}}L(\gamma)\kappa^{L(\gamma)}. M g ( H ) = γ : 0 → g a half-plane arch H ≤ h ( γ ) < 2 H ∑ L ( γ ) κ L ( γ ) . where h ( γ ) h(\gamma) h ( γ )
M g ( H ) ≤ C η g 1 / 4 H d ∗ − 2 a + O ( η ) ( H ≥ g / log 2 g ) . (22) \mathcal{M}_g(H)\le C_\eta g^{1/4}H^{d_*-2a+O(\eta)}
\qquad(H\ge g/\log^2 g). \tag*{(22)} M g ( H ) ≤ C η g 1/4 H d ∗ − 2 a + O ( η ) ( H ≥ g / log 2 g ) . (22) The smaller height scales will be negligible by strip confinement. Since d ∗ − 2 a = − 1 / 6 d_*-2a=-1/6 d ∗ − 2 a = − 1/6 η \eta η g 1 / 12 + o ( 1 ) g^{1/12+o(1)} g 1/12 + o ( 1 ) K g ≳ g − 5 / 4 K_g\gtrsim g^{-5/4} K g ≳ g − 5/4
A local truncated reward
The first ingredient bounds the cost of the piece carrying the length weight.
Lemma 9.3 (Truncated length reward). For the strict irreducible law, as H → ∞ H \to\infty H → ∞
E [ L ; Y ≤ 2 H ] ≤ H d ∗ − a + o ( 1 ) . \mathbb{E}[L; Y \le2H] \le H^{d_* - a + o(1)}. E [ L ; Y ≤ 2 H ] ≤ H d ∗ − a + o ( 1 ) . Proof. Write L h \mathcal{L}_h L h h h h 2 H 2H 2 H H H H
( ∑ 0 ≤ u ≤ H B u ) 2 E [ L ; Y ≤ 2 H ] ≤ ∑ 0 ≤ h ≤ 4 H L h . \left(\sum_{0 \le u \le H} B_u\right)^2 \mathbb{E}[L; Y \le2H] \le\sum_{0 \le h \le4H} \mathcal{L}_h. ( 0 ≤ u ≤ H ∑ B u ) 2 E [ L ; Y ≤ 2 H ] ≤ 0 ≤ h ≤ 4 H ∑ L h . Indeed the same output can have several eligible marked pieces, but their lengths sum to at most its full length. The bridge sum on the left is of order H a H^a H a
For height h h h h log 2 h h\log^2 h h log 2 h h h h O ( h log 2 h ) O(h\log^2 h) O ( h log 2 h ) L h ≤ h 13 / 12 + o ( 1 ) \mathcal{L}_h \le h^{13/12+o(1)} L h ≤ h 13/12 + o ( 1 )
∑ h ≤ 4 H L h ≤ H a + d ∗ + o ( 1 ) . \sum_{h \le4H} \mathcal{L}_h \le H^{a+d_*+o(1)}. h ≤ 4 H ∑ L h ≤ H a + d ∗ + o ( 1 ) . The insertion inequality now proves the claim.
Survival of the order at separated bottom ports
The shared-turn estimate does not apply at two different starting ports. We therefore prove the separate weak-order estimate required at the bottom ends of the arch. This estimate uses the calibrated finite inputs throughout, independently of the off-axis construction in Section 4.
Lemma 9.4 (Weak order from separated ports). Start two independent upward irreducible sequences at bottom ports a distance g ≥ 1 g \ge1 g ≥ 1 Π g ( S ) \Pi_g(S) Π g ( S ) S S S S ≥ 2 S \ge2 S ≥ 2
Π g ( S ) ≤ C ( g / S ) 1 / 4 log ( 2 + S ) . \Pi_g(S) \le C(g/S)^{1/4}\sqrt{\log(2+S)}. Π g ( S ) ≤ C ( g / S ) 1/4 log ( 2 + S ) . Proof. Their strictly positive common renewal levels themselves form renewals; write ( T , W ) (T,W) ( T , W ) j j j V j = B j 2 ≍ ( 1 + j ) − 1 / 2 V_j = B_j^2 \asymp(1+j)^{-1/2} V j = B j 2 ≍ ( 1 + j ) − 1/2 b = 1 / 2 b=1/2 b = 1/2 1 − E e − T / r ≍ r − b 1-\mathbb{E}e^{-T/r}\asymp r^{-b} 1 − E e − T / r ≍ r − b ( T , W ) (T,W) ( T , W ) W W W − W -W − W
Pr ( ∣ W ∣ > r ) ≤ C r − b , E min ( 1 , ∣ W ∣ / r ) ≳ r − b . \Pr(|W| > r) \le Cr^{-b}, \qquad\mathbb{E}\min(1,|W|/r) \gtrsim r^{-b}. Pr ( ∣ W ∣ > r ) ≤ C r − b , E min ( 1 , ∣ W ∣/ r ) ≳ r − b . Pairs in a height bin around c 0 r c_0r c 0 r c 0 c_0 c 0 ≳ r b \gtrsim r^b ≳ r b r / 3 r/3 r /3 W W W r r r h ≍ r h \asymp r h ≍ r h h h c h ch c h ≳ h − 1 / 4 \gtrsim h^{-1/4} ≳ h − 1/4 [12 ] , Section 11, Recoverable half-plane growth uses two pure inward normals with dot product 1 / 2 1/2 1/2 r r r S ≥ K r S \ge Kr S ≥ K r c S − 1 / 4 ( S / r ) − C cS^{-1/4}(S/r)^{-C} c S − 1/4 ( S / r ) − C C ′ S C'S C ′ S r r r
Apply it first with normals y , u y,u y , u π / 3 \pi/3 π /3 c ∗ h c_*h c ∗ h u u u [ ( 2 − 2 δ ∗ ) h , ( 2 − δ ∗ ) h ] [(2-2\delta_*)h,(2-\delta_*)h] [( 2 − 2 δ ∗ ) h , ( 2 − δ ∗ ) h ] u , y u,y u , y δ ∗ \delta_* δ ∗ δ ∗ \delta_* δ ∗ c ∗ c_* c ∗
The first continuation gains y y y 1 / 2 1/2 1/2 0 < y < h 0<y<h 0 < y < h δ ∗ h / 4 \delta_*h/4 δ ∗ h /4 2 δ ∗ h 2\delta_*h 2 δ ∗ h c h ch c h ≳ h − 1 / 4 \gtrsim h^{-1/4} ≳ h − 1/4 u u u ≳ h 3 / 4 \gtrsim h^{3/4} ≳ h 3/4 ∑ j ≤ C h B j ≤ C h 3 / 4 \sum_{j\le Ch}B_j\le Ch^{3/4} ∑ j ≤ C h B j ≤ C h 3/4 ≳ 1 \gtrsim1 ≳ 1
From each prefix apply the u , y u,y u , y y = h y=h y = h u u u O ( δ ∗ h ) O(\delta_*h) O ( δ ∗ h ) ≳ h − 1 / 4 \gtrsim h^{-1/4} ≳ h − 1/4 δ ∗ \delta_* δ ∗ y = 0 y=0 y = 0
Reflect the construction for the second bridge. The critical product mass with total transverse difference ≥ c h \ge ch ≥ c h ≳ h − 1 / 2 = h b − 1 \gtrsim h^{-1/2}=h^{b-1} ≳ h − 1/2 = h b − 1 min ( 1 , ∣ ⋅ ∣ / r ) \min(1,|\cdot|/r) min ( 1 , ∣ ⋅ ∣/ r ) max s ≤ h ∑ i + j = s V i V j \max_{s\le h}\sum_{i+j=s}V_iV_j max s ≤ h ∑ i + j = s V i V j Pr ( c ≤ ∣ W ∣ ≤ r ) ≳ r − b \Pr(c\le|W|\le r)\gtrsim r^{-b} Pr ( c ≤ ∣ W ∣ ≤ r ) ≳ r − b 1 − ∣ φ W ( u ) ∣ 2 ≳ ∣ u ∣ b 1-|\varphi_W(u)|^2\gtrsim|u|^b 1 − ∣ φ W ( u ) ∣ 2 ≳ ∣ u ∣ b k k k ≤ C k − 1 / b \le Ck^{-1/b} ≤ C k − 1/ b
Counting near-minimum times. The distinction between weak and strict survival for lattice increments is essential; compare [18 ] , Section 2.2. We use the following direct factorization, which counts all near-minimum times. Kill at an independent exponential height E E E S S S
Q S ( w ) = E [ e − T / S ; W = w ] , q 0 = 1 − ∑ w Q S ( w ) ≍ S − b . Q_S(w)=\mathbb{E}[e^{-T/S};W=w],\qquad q_0=1-\sum_w Q_S(w)\asymp S^{-b}. Q S ( w ) = E [ e − T / S ; W = w ] , q 0 = 1 − w ∑ Q S ( w ) ≍ S − b . Let p p p − g -g − g N g N_g N g g g g E N g = p 2 / q 0 \mathbb{E}N_g=p^2/q_0 E N g = p 2 / q 0 a j ( g ) a_j(g) a j ( g ) j j j − g -g − g p = q 0 ∑ j ≥ 0 a j ( g ) p=q_0\sum_{j\ge0}a_j(g) p = q 0 ∑ j ≥ 0 a j ( g ) i i i j j j q 0 a i ( g ) a j ( g ) q_0a_i(g)a_j(g) q 0 a i ( g ) a j ( g ) g g g i , j i,j i , j p 2 / q 0 p^2/q_0 p 2 / q 0
Bounding the occupation count. The killed kernel is dominated by the original displacement law, including after convolution:
Q S ∗ k ( w ) ≤ Pr ( W 1 + ⋯ + W k = w ) . Q_S^{*k}(w) \le\Pr(W_1+\cdots+W_k=w). Q S ∗ k ( w ) ≤ Pr ( W 1 + ⋯ + W k = w ) . Thus the atom estimate above bounds the expected number L A L_A L A A A A 2 g 2g 2 g
E L A ≤ 1 + ∑ k ≥ 1 min { 1 , C g k − 1 / b } ≤ C g b . \mathbb{E}L_A \le1+\sum_{k\ge1}\min\{1,Cgk^{-1/b}\}\le Cg^b. E L A ≤ 1 + k ≥ 1 ∑ min { 1 , C g k − 1/ b } ≤ C g b . Markov’s inequality and restarting after successive visit-count thresholds give an exponential tail for L A L_A L A g b g^b g b g g g N g N_g N g 1 / q 0 1/q_0 1/ q 0
Pr ( N g ≥ t ) ≤ min { 1 , ( C / q 0 ) e − c t / g b } , E N g ≤ C g b log ( 2 + 1 / q 0 ) . \Pr(N_g\ge t)\le\min\{1,(C/q_0)e^{-ct/g^b}\},\qquad\mathbb{E}N_g\le Cg^b\log(2+1/q_0). Pr ( N g ≥ t ) ≤ min { 1 , ( C / q 0 ) e − c t / g b } , E N g ≤ C g b log ( 2 + 1/ q 0 ) . Together with the factorization this bounds p p p C ( g / S ) b / 2 log ( 2 + S ) C(g/S)^{b/2}\sqrt{\log(2+S)} C ( g / S ) b /2 log ( 2 + S ) S S S E E E { E < S } \{E<S\} { E < S } E E E
( 1 − e − 1 ) Π g ( S ) ≤ p . (1-e^{-1})\Pi_g(S)\le p. ( 1 − e − 1 ) Π g ( S ) ≤ p . Since b / 2 = 1 / 4 b/2=1/4 b /2 = 1/4
Summing the marked-arm decomposition
Proof of Theorem 9.1. We now have the two different survival tests needed for an arch: the shared-turn test at its top and Lemma 9.4 at its bottom. Write b = 1 / 2 b=1/2 b = 1/2 (22) by allocating independent groups of pieces to these tests, the reward, and the two-coordinate match.
Use dyads H ≤ h < 2 H H\le h<2H H ≤ h < 2 H H < g / log 2 g H<g/\log^2g H < g / log 2 g g g g C H log 2 H CH\log^2H C H log 2 H O A ( H − A ) O_A(H^{-A}) O A ( H − A ) A A A
Let k i k_i k i K i K_i K i K = max K i K=\max K_i K = max K i k i > H a + η k_i>H^{a+\eta} k i > H a + η η > 0 \eta>0 η > 0 k i ≤ 2 H k_i\le2H k i ≤ 2 H > H 1 − η / ( 2 a ) >H^{1-\eta/(2a)} > H 1 − η / ( 2 a ) k i ≤ H a − η k_i\le H^{a-\eta} k i ≤ H a − η Pr ( Y > t ) ≤ C t − a \Pr(Y>t)\le Ct^{-a} Pr ( Y > t ) ≤ C t − a
Sort the index distances (counting from 1 1 1 I i , J i I_i,J_i I i , J i max ( I i , J i ) ≍ K i \max(I_i,J_i)\asymp K_i max ( I i , J i ) ≍ K i
C η H d ∗ − 2 a + O ( η ) ∏ i = 1 2 min ( I i , J i ) . C_\eta H^{d_*-2a+O(\eta)}\prod_{i=1}^{2}\min(I_i,J_i). C η H d ∗ − 2 a + O ( η ) i = 1 ∏ 2 min ( I i , J i ) . Indeed for a given length piece use Lemma 9.3; the jump costs C H − a + O ( η ) CH^{-a+O(\eta)} C H − a + O ( η ) min ( I i , J i ) H − a + η ≳ 1 \min(I_i,J_i)H^{-a+\eta}\gtrsim1 min ( I i , J i ) H − a + η ≳ 1
Put I = min i I i I=\min_i I_i I = min i I i J = min i J i J=\min_i J_i J = min i J i r i r_i r i i i i k i k_i k i
u i = r i − 1 , v i = k i − r i , u = min i u i , v = min i v i . u_i=r_i-1,\quad v_i=k_i-r_i,\quad u=\min_i u_i,\quad v=\min_i v_i. u i = r i − 1 , v i = k i − r i , u = i min u i , v = i min v i . Take the first ⌊ u / 16 ⌋ \lfloor u/16\rfloor ⌊ u /16 ⌋ ⌊ v / 16 ⌋ \lfloor v/16\rfloor ⌊ v /16 ⌋ u + v ≤ k i − 1 u+v\leq k_i-1 u + v ≤ k i − 1 k max k_{\max} k m a x 15 k max / 16 − 1 15k_{\max}/16-1 15 k m a x /16 − 1 ⌊ k max / 4 ⌋ \lfloor k_{\max}/4\rfloor ⌊ k m a x /4 ⌋
If u < 16 u<16 u < 16 I I I v < 16 v<16 v < 16 J J J I I I J J J k max k_{\max} k m a x p ′ = b / ( 2 a ) = 1 / 3 p'=b/(2a)=1/3 p ′ = b / ( 2 a ) = 1/3 m i = min ( I i , J i ) m_i=\min(I_i,J_i) m i = min ( I i , J i ) H H H η \eta η
Table 1.
The second row includes the designated-index counting allowance. For a large-jump mark this is the actual sum of index choices. In the deterministic middle-index case its cost one is bounded by m i H − a + η ≳ 1 m_iH^{-a+\eta}\gtrsim1 m i H − a + η ≳ 1 H − a + η H^{-a+\eta} H − a + η C K 1 K 2 CK_1K_2 C K 1 K 2
With S I = I 1 / a H − C 0 η S_I=I^{1/a}H^{-C_0\eta} S I = I 1/ a H − C 0 η C 0 C_0 C 0 S I S_I S I S I ≥ 2 S_I\geq2 S I ≥ 2 S J = J 1 / a H − C 0 η S_J=J^{1/a}H^{-C_0\eta} S J = J 1/ a H − C 0 η h h h S I S_I S I [15 ] , Section 7.3, with the common top-port convention stated in Proposition 9.2(ii), at probability cost ≤ C η I − 1 H O ( η ) \leq C_\eta I^{-1}H^{O(\eta)} ≤ C η I − 1 H O ( η ) S J S_J S J ≤ C g b / 2 J − b / ( 2 a ) H O ( η ) \leq Cg^{b/2}J^{-b/(2a)}H^{O(\eta)} ≤ C g b /2 J − b / ( 2 a ) H O ( η ) < 2 <2 < 2 O ( η ) O(\eta) O ( η )
In addition the required exact height equality and transverse difference together cost ≤ C K − 2 / a \leq CK^{-2/a} ≤ C K − 2/ a ( Y , Z ) (Y,Z) ( Y , Z ) K K K K K K C K 1 K 2 CK_1K_2 C K 1 K 2 p ′ = b / ( 2 a ) = 1 / 3 p'=b/(2a)=1/3 p ′ = b / ( 2 a ) = 1/3
K 1 K 2 K − 2 / a ∏ i min ( I i , J i ) I − 1 J − p ′ ≤ C K − 2 / a max ( I 1 , I 2 ) max ( J 1 , J 2 ) J 1 − p ′ ≤ C . K_1K_2K^{-2/a}\prod_i \min(I_i,J_i)I^{-1}J^{-p'}\leq CK^{-2/a}\max(I_1,I_2)\max(J_1,J_2)J^{1-p'}\leq C. K 1 K 2 K − 2/ a i ∏ min ( I i , J i ) I − 1 J − p ′ ≤ C K − 2/ a max ( I 1 , I 2 ) max ( J 1 , J 2 ) J 1 − p ′ ≤ C . Thus including polylogarithmic bin sums proves (22) . Since d ∗ − 2 a < 0 d_* - 2a < 0 d ∗ − 2 a < 0 η \eta η g 1 / 12 + o ( 1 ) g^{1/12+o(1)} g 1/12 + o ( 1 )
The length-weighted bound is K g E L ≤ g 1 / 12 + o ( 1 ) K_g\mathbb{E}L \le g^{1/12+o(1)} K g E L ≤ g 1/12 + o ( 1 ) K g ≥ c g − 5 / 4 K_g \ge cg^{-5/4} K g ≥ c g − 5/4 g g g K g K_g K g E L ≥ g 4 / 3 − o ( 1 ) \mathbb{E}L \ge g^{4/3-o(1)} E L ≥ g 4/3 − o ( 1 )
The following terminal-summed consequence also appears in the proof of [15 ] . We include its normalization and first-moment argument to distinguish this confined mixture from the individual-gap law.
Corollary 9.5 (Confined macroscopic-gap arches). Fix a half-plane boundary source and choose a sufficiently large fixed C C C [ h , 2 h ] [h,2h] [ h , 2 h ] C h Ch C h h → ∞ h \to\infty h → ∞
L ( γ ) = h 4 / 3 + o ( 1 ) in probability , E L ( γ ) = h 4 / 3 + o ( 1 ) . L(\gamma) = h^{4/3+o(1)} \quad\text{in probability}, \qquad\mathbb{E}L(\gamma) = h^{4/3+o(1)}. L ( γ ) = h 4/3 + o ( 1 ) in probability , E L ( γ ) = h 4/3 + o ( 1 ) . Proof. The confined pointwise arch lower bound from the calibrated finite input above, summed over the order h h h c h − 1 / 4 ch^{-1/4} c h − 1/4 C C C h h h O C ( h ) O_C(h) O C ( h ) [12 ] , divided by the number of sources, gives first length mass at most C C h 13 / 12 C_C h^{13/12} C C h 13/12 E L ≤ C C h 4 / 3 \mathbb{E}L \le C_C h^{4/3} E L ≤ C C h 4/3 h h h C h Ch C h
Endpoint means from polynomial vacuum estimates
We now estimate mean lengths for bridges to prescribed top ports. The polynomial-vacuum representation controls endpoint variation after summing over heights. This gives individual-height estimates on one density-one set, and estimates at every large scale after critical averaging over even heights.
Let B h ( e ) \mathcal{B}_h(e) B h ( e ) 0 0 0 e ∈ Z + h / 2 e \in\mathbb{Z}+h/2 e ∈ Z + h /2 h h h
b h ( e ) = ∑ γ ∈ B h ( e ) κ L ( γ ) , b h ( L ) ( e ) = ∑ γ ∈ B h ( e ) L ( γ ) κ L ( γ ) . b_h(e) = \sum_{\gamma\in\mathcal{B}_h(e)} \kappa^{L(\gamma)}, \qquad b_h^{(L)}(e) = \sum_{\gamma\in\mathcal{B}_h(e)} L(\gamma)\kappa^{L(\gamma)}. b h ( e ) = γ ∈ B h ( e ) ∑ κ L ( γ ) , b h ( L ) ( e ) = γ ∈ B h ( e ) ∑ L ( γ ) κ L ( γ ) . Thus b h = ∑ e b h ( e ) = B h b_h = \sum_e b_h(e) = B_h b h = ∑ e b h ( e ) = B h b 0 ( e ) = 1 e = 0 b_0(e) = \mathbf{1}_{e=0} b 0 ( e ) = 1 e = 0
Theorem 10.1 (One set of density-one heights). There is one set G ⊆ N G \subseteq\mathbb{N} G ⊆ N ∣ G ∩ [ 1 , N ] ∣ / N → 1 |G \cap[1,N]|/N \to1 ∣ G ∩ [ 1 , N ] ∣/ N → 1 δ , ε > 0 \delta,\varepsilon> 0 δ , ε > 0 h ∈ G h \in G h ∈ G
h 4 / 3 − ε ≤ b h ( L ) ( e ) b h ( e ) ≤ h 4 / 3 + ε ( e ∈ Z + h / 2 , ∣ e ∣ ≤ h 1 − δ ) . h^{4/3-\varepsilon} \le\frac{b_h^{(L)}(e)}{b_h(e)} \le h^{4/3+\varepsilon}
\qquad(e \in\mathbb{Z}+h/2,\ |e| \le h^{1-\delta}). h 4/3 − ε ≤ b h ( e ) b h ( L ) ( e ) ≤ h 4/3 + ε ( e ∈ Z + h /2 , ∣ e ∣ ≤ h 1 − δ ) . For E H = 2 N ∩ [ H , 2 H ] E_H = 2\mathbb{N} \cap[H,2H] E H = 2 N ∩ [ H , 2 H ] Z H = ∑ h ∈ E H b h ( 0 ) Z_H = \sum_{h\in E_H} b_h(0) Z H = ∑ h ∈ E H b h ( 0 ) M H = ∑ h ∈ E H b h ( L ) ( 0 ) M_H = \sum_{h\in E_H} b_h^{(L)}(0) M H = ∑ h ∈ E H b h ( L ) ( 0 )
Theorem 10.2 (All-even aligned height mixture). Give each pair ( h , γ ) (h,\gamma) ( h , γ ) h ∈ E H h \in E_H h ∈ E H γ ∈ B h ( 0 ) \gamma\in\mathcal{B}_h(0) γ ∈ B h ( 0 ) κ L ( γ ) / Z H \kappa^{L(\gamma)}/Z_H κ L ( γ ) / Z H H → ∞ H \to\infty H → ∞
Z H = H − 1 / 4 + o ( 1 ) , M H = H 13 / 12 + o ( 1 ) , E H L = M H / Z H = H 4 / 3 + o ( 1 ) . Z_H = H^{-1/4+o(1)}, \qquad M_H = H^{13/12+o(1)}, \qquad\mathbb{E}_{H}L = M_H/Z_H = H^{4/3+o(1)}. Z H = H − 1/4 + o ( 1 ) , M H = H 13/12 + o ( 1 ) , E H L = M H / Z H = H 4/3 + o ( 1 ) . The height marginal is b h ( 0 ) / Z H b_h(0)/Z_H b h ( 0 ) / Z H
The complete finite construction in [14 ] gives b h = h − 1 / 4 + o ( 1 ) b_h = h^{-1/4+o(1)} b h = h − 1/4 + o ( 1 ) ∑ e b h ( L ) ( e ) = h 13 / 12 + o ( 1 ) \sum_e b_h^{(L)}(e) = h^{13/12+o(1)} ∑ e b h ( L ) ( e ) = h 13/12 + o ( 1 ) [14 ] is b h + cos ( 3 π / 8 ) A h = 1 b_h + \cos(3\pi/8)A_h = 1 b h + cos ( 3 π /8 ) A h = 1 A h A_h A h h h h i i i i ( L ) ( h , e ) i^{(L)}(h,e) i ( L ) ( h , e ) L ( γ ) κ L ( γ ) L(\gamma)\kappa^{L(\gamma)} L ( γ ) κ L ( γ ) [15 ] .
Proposition 10.3 (Polynomial-vacuum rewards). Write J J J i i i L L L α = 1 / 4 \alpha= 1/4 α = 1/4 β = 3 / 4 \beta= 3/4 β = 3/4 d ∗ = 4 / 3 d_* = 4/3 d ∗ = 4/3
b = ∑ k ≥ 0 i ∗ k , b ( L ) = b ∗ i ( L ) ∗ b , b = \sum_{k\geq0} i^{*k}, \qquad b^{(L)} = b * i^{(L)} * b, b = k ≥ 0 ∑ i ∗ k , b ( L ) = b ∗ i ( L ) ∗ b , i ( J > r ) = r − β + o ( 1 ) , E i [ L ; J ≤ r ] ≤ r d ∗ − β + o ( 1 ) , E i [ L e − J / r ] = r d ∗ − β + o ( 1 ) . i(J > r) = r^{-\beta+o(1)}, \qquad\mathbb{E}_i[L; J \leq r] \leq r^{d_*-\beta+o(1)}, \qquad\mathbb{E}_i[Le^{-J/r}] = r^{d_*-\beta+o(1)}. i ( J > r ) = r − β + o ( 1 ) , E i [ L ; J ≤ r ] ≤ r d ∗ − β + o ( 1 ) , E i [ L e − J / r ] = r d ∗ − β + o ( 1 ) . The discounted lower reward is retained on pieces with height at most r log 2 r r\log^2 r r log 2 r r log 4 r r\log^4 r r log 4 r
Only in this section, J J J σ = 3 / 8 \sigma= 3/8 σ = 3/8 θ = π / 8 \theta= \pi/8 θ = π /8
Reflection makes b h ( e ) b_h(e) b h ( e ) e e e
Endpoint variation from two disjoint bridges
First put t h t_h t h 0 , 1 0,1 0 , 1 J h ( u , v ) J_h(u,v) J h ( u , v ) u < v u < v u < v
b h ( y ) = ∑ u ≤ y < v J h ( u , v ) . b_h(y) = \sum_{u\leq y<v} J_h(u,v). b h ( y ) = u ≤ y < v ∑ J h ( u , v ) . Indeed take a bridge spectator from 1 1 1 y ′ y' y ′ 0 0 0 e i σ W e^{i\sigma W} e iσ W 0 0 0 1 1 1 0 0 0 y ′ y' y ′ s 0 = e 6 i θ s_0 = e^{6i\theta} s 0 = e 6 i θ θ = π / 8 \theta= \pi/8 θ = π /8
At a first contact the three arms from 0 , 1 , y ′ 0,1,y' 0 , 1 , y ′ e i σ W y ′ = e i θ e i σ W 0 = e − i θ e i σ W 1 e^{i\sigma W_{y'}}=e^{i\theta}e^{i\sigma W_0}=e^{-i\theta}e^{i\sigma W_1} e iσ W y ′ = e i θ e iσ W 0 = e − i θ e iσ W 1 s 0 = − e − 2 i θ s_0=-e^{-2i\theta} s 0 = − e − 2 i θ
The remaining active arches in the first balance can only go left, separated by the bridge from exits beyond the adjacent port (also no exit at it); in the second only right. Their phases including the combination factor are thus both ℓ = e 3 i θ \ell=e^{3i\theta} ℓ = e 3 i θ X ↦ ℑ ( ℓ − 1 X ) / ℑ ( ℓ − 1 ) X\mapsto\Im(\ell^{-1}X)/\Im(\ell^{-1}) X ↦ ℑ ( ℓ − 1 X ) /ℑ ( ℓ − 1 ) 1 , − 1 1,-1 1 , − 1
b h ( y ′ − 1 ) − b h ( y ′ ) = ∑ u < y ′ J h ( u , y ′ ) − ∑ v > y ′ J h ( y ′ , v ) . b_h(y'-1)-b_h(y')=\sum_{u<y'}J_h(u,y')-\sum_{v>y'}J_h(y',v). b h ( y ′ − 1 ) − b h ( y ′ ) = u < y ′ ∑ J h ( u , y ′ ) − v > y ′ ∑ J h ( y ′ , v ) . Sum up to y ′ = y y'=y y ′ = y
Consequently sup e b h ( e ) ≤ t h \sup_e b_h(e)\leq t_h sup e b h ( e ) ≤ t h b h ( ⋅ ) b_h(\cdot) b h ( ⋅ ) 2 t h 2t_h 2 t h t h ≤ C ( A h + 1 − A h ) t_h\leq C(A_{h+1}-A_h) t h ≤ C ( A h + 1 − A h ) 0 < c 1 < c 2 < ∞ 0<c_1<c_2<\infty 0 < c 1 < c 2 < ∞
∑ c 1 H ≤ j ≤ c 2 H t j ≤ C b ⌊ c 1 H ⌋ = H − α + o ( 1 ) \sum_{c_1H\leq j\leq c_2H}t_j\leq Cb_{\lfloor c_1H\rfloor}=H^{-\alpha+o(1)} c 1 H ≤ j ≤ c 2 H ∑ t j ≤ C b ⌊ c 1 H ⌋ = H − α + o ( 1 ) (as an upper bound).
A pointwise lower bound at every height
The variation bound is available only after summing heights. Nevertheless it will yield a lower bound at each height by averaging internal renewal cuts. Here is a uniform pointwise lower bound at all large heights for any fixed δ > 0 \delta>0 δ > 0
b h ( e ) ≥ h − 1 − α − o ( 1 ) ( ∣ e ∣ ≤ h 1 − δ , e ∈ Z + h / 2 ) . b_h(e)\geq h^{-1-\alpha-o(1)}\qquad(|e|\leq h^{1-\delta},\ e\in\mathbb{Z}+h/2). b h ( e ) ≥ h − 1 − α − o ( 1 ) ( ∣ e ∣ ≤ h 1 − δ , e ∈ Z + h /2 ) . First take even h h h γ ∈ B h ( e ) \gamma\in B_h(e) γ ∈ B h ( e ) N h ( γ ) N_h(\gamma) N h ( γ ) j ∈ [ h / 8 , h / 7 ] j\in[h/8,h/7] j ∈ [ h /8 , h /7 ] j , 2 j , h / 2 + j j,2j,h/2+j j , 2 j , h /2 + j k = h / 2 − j k=h/2-j k = h /2 − j j , j , k , k j,j,k,k j , j , k , k
∑ γ ∈ B h ( e ) N h ( γ ) κ L ( γ ) = ∑ h / 8 ≤ j ≤ h / 7 ( b j ∗ b j ∗ b k ∗ b k ) ( e ) . \sum_{\gamma\in B_h(e)}N_h(\gamma)\kappa^{L(\gamma)}
=
\sum_{h/8\leq j\leq h/7}(b_j*b_j*b_k*b_k)(e). γ ∈ B h ( e ) ∑ N h ( γ ) κ L ( γ ) = h /8 ≤ j ≤ h /7 ∑ ( b j ∗ b j ∗ b k ∗ b k ) ( e ) . For all but o ( h ) o(h) o ( h ) j j j t j ≤ h − 1 − α + δ 0 / 8 t_j\leq h^{-1-\alpha+\delta_0/8} t j ≤ h − 1 − α + δ 0 /8 δ 0 = min ( δ , 1 / 2 ) \delta_0=\min(\delta,1/2) δ 0 = min ( δ , 1/2 ) h h h A = b j ( ⋅ ) ∗ b k ( ⋅ ) A=b_j(\cdot)*b_k(\cdot) A = b j ( ⋅ ) ∗ b k ( ⋅ ) b j b k b_jb_k b j b k ∣ x ∣ ≤ C h log 2 h |x|\leq Ch\log^2h ∣ x ∣ ≤ C h log 2 h t j b k , 2 t j b k t_jb_k,2t_jb_k t j b k , 2 t j b k
∑ x A ( x ) A ( e − x ) = ∑ x A ( x ) A ( x − e ) ≥ ∥ A ∥ 2 2 − 2 ∣ e ∣ t j 2 b k 2 , ∥ A ∥ 2 2 ≥ h − 1 − o ( 1 ) ( b j b k ) 2 . \sum_x A(x)A(e-x)=\sum_x A(x)A(x-e)\geq\lVert A\rVert_2^2-2|e|t_j^2b_k^2,\qquad\lVert A\rVert_2^2\geq h^{-1-o(1)}(b_jb_k)^2. x ∑ A ( x ) A ( e − x ) = x ∑ A ( x ) A ( x − e ) ≥ ∥ A ∥ 2 2 − 2∣ e ∣ t j 2 b k 2 , ∥ A ∥ 2 2 ≥ h − 1 − o ( 1 ) ( b j b k ) 2 . For good j j j h − 4 α − o ( 1 ) h^{-4\alpha-o(1)} h − 4 α − o ( 1 )
For the second moment, diagonal terms cost at most h − 4 α + o ( 1 ) h^{-4\alpha+o(1)} h − 4 α + o ( 1 ) sup b j ( ⋅ ) ≤ t j \sup b_j(\cdot)\leq t_j sup b j ( ⋅ ) ≤ t j j < j ′ j<j' j < j ′ h h h j j j Δ \Delta Δ 2 Δ 2\Delta 2Δ Δ \Delta Δ Δ = j ′ − j \Delta= j' - j Δ = j ′ − j j j j ∑ Δ ≤ h b Δ 2 b 2 Δ ≤ h 1 − 3 α + o ( 1 ) \sum_{\Delta\le h} b_{\Delta}^{2} b_{2\Delta} \le h^{1-3\alpha+o(1)} ∑ Δ ≤ h b Δ 2 b 2Δ ≤ h 1 − 3 α + o ( 1 ) t j t_j t j
∑ γ ∈ B h ( e ) N h ( γ ) 2 κ L ( γ ) ≤ h 1 − 7 α + o ( 1 ) . \sum_{\gamma\in\mathcal{B}_h(e)} N_h(\gamma)^2 \kappa^{L(\gamma)} \le h^{1-7\alpha+o(1)}. γ ∈ B h ( e ) ∑ N h ( γ ) 2 κ L ( γ ) ≤ h 1 − 7 α + o ( 1 ) . Weighted Cauchy–Schwarz therefore gives
b h ( e ) ≥ ( ∑ γ ∈ B h ( e ) N h ( γ ) κ L ( γ ) ) 2 ∑ γ ∈ B h ( e ) N h ( γ ) 2 κ L ( γ ) ≥ h − 1 − α − o ( 1 ) . b_h(e) \ge\frac{\left(\sum_{\gamma\in\mathcal{B}_h(e)} N_h(\gamma)\kappa^{L(\gamma)}\right)^2}{\sum_{\gamma\in\mathcal{B}_h(e)} N_h(\gamma)^2\kappa^{L(\gamma)}} \ge h^{-1-\alpha-o(1)}. b h ( e ) ≥ ∑ γ ∈ B h ( e ) N h ( γ ) 2 κ L ( γ ) ( ∑ γ ∈ B h ( e ) N h ( γ ) κ L ( γ ) ) 2 ≥ h − 1 − α − o ( 1 ) . Odd heights follow by a bounded one-layer extension from even height with the appropriate half-step endpoint shift, using a lower-bound window with a smaller positive δ \delta δ
Inserting length between two bridge kernels
Write K = b ∗ b K=b*b K = b ∗ b
K h ( e ) ≥ h − 2 α − o ( 1 ) ( ∣ e ∣ ≤ h 1 − δ ) K_h(e) \ge h^{-2\alpha-o(1)} \qquad(|e| \le h^{1-\delta}) K h ( e ) ≥ h − 2 α − o ( 1 ) ( ∣ e ∣ ≤ h 1 − δ ) again for admissible offsets. Indeed sum over heights h / 3 ≤ l ≤ 2 h / 3 h/3 \le l \le2h/3 h /3 ≤ l ≤ 2 h /3 ∣ x ∣ ≤ h 1 − η |x| \le h^{1-\eta} ∣ x ∣ ≤ h 1 − η 0 < η < δ / 2 0 < \eta< \delta/2 0 < η < δ /2 η \eta η
∑ j ≤ 2 H sup x K j ( x ) ≤ H 2 β − 1 + o ( 1 ) . \sum_{j \le2H} \sup_x K_j(x) \le H^{2\beta-1+o(1)}. j ≤ 2 H ∑ x sup K j ( x ) ≤ H 2 β − 1 + o ( 1 ) . Order the two factor heights by size, larger l l l r r r sup b l ( ⋅ ) ≤ t l \sup b_l(\cdot) \le t_l sup b l ( ⋅ ) ≤ t l ∑ r ≤ l b r ≤ l β + o ( 1 ) \sum_{r \le l} b_r \le l^{\beta+o(1)} ∑ r ≤ l b r ≤ l β + o ( 1 ) β − α > 0 \beta-\alpha>0 β − α > 0
The length insertion is exactly b ( L ) = K ∗ i ( L ) b^{(L)}=K*i^{(L)} b ( L ) = K ∗ i ( L )
b h ( L ) ( e ) ≥ h d ∗ + β − 2 − o ( 1 ) . b_h^{(L)}(e) \ge h^{d_*+\beta-2-o(1)}. b h ( L ) ( e ) ≥ h d ∗ + β − 2 − o ( 1 ) . In fact take the middle piece from i ( L ) i^{(L)} i ( L ) ≤ m log 2 m \le m\log^2 m ≤ m log 2 m ≤ m log 4 m \le m\log^4 m ≤ m log 4 m m = h 1 − ε m=h^{1-\varepsilon} m = h 1 − ε ε > 0 \varepsilon>0 ε > 0 m d ∗ − β − o ( 1 ) m^{d_*-\beta-o(1)} m d ∗ − β − o ( 1 ) K K K h h h m log 4 m ≤ h 1 − ε / 2 m\log^4 m \le h^{1-\varepsilon/2} m log 4 m ≤ h 1 − ε /2 h h h δ , ε > 0 \delta,\varepsilon>0 δ , ε > 0 K K K ε \varepsilon ε
∑ H ≤ h ≤ 2 H sup e b h ( L ) ( e ) ≤ H d ∗ + β − 1 + o ( 1 ) \sum_{H \le h \le2H} \sup_e b_h^{(L)}(e) \le H^{d_*+\beta-1+o(1)} H ≤ h ≤ 2 H ∑ e sup b h ( L ) ( e ) ≤ H d ∗ + β − 1 + o ( 1 ) by the upper K K K i ( L ) i^{(L)} i ( L ) 2 H 2H 2 H
One density-one set and the aligned mixture
Put U h = sup e b h ( e ) U_h=\sup_e b_h(e) U h = sup e b h ( e ) V h = sup e b h ( L ) ( e ) V_h=\sup_e b_h^{(L)}(e) V h = sup e b h ( L ) ( e ) H H H ρ H → ∞ \rho_H \to\infty ρ H → ∞ ρ H = H o ( 1 ) \rho_H=H^{o(1)} ρ H = H o ( 1 )
∑ H ≤ h < 2 H U h ≤ ρ H H − α , ∑ H ≤ h < 2 H V h ≤ ρ H H d ∗ + β − 1 . \sum_{H \le h < 2H} U_h \le\rho_H H^{-\alpha}, \qquad\sum_{H \le h < 2H} V_h \le\rho_H H^{d_*+\beta-1}. H ≤ h < 2 H ∑ U h ≤ ρ H H − α , H ≤ h < 2 H ∑ V h ≤ ρ H H d ∗ + β − 1 . In each such interval retain precisely the heights satisfying
U h ≤ ρ H 2 H − 1 − α , V h ≤ ρ H 2 H d ∗ + β − 2 , U_h \le\rho_H^2 H^{-1-\alpha}, \qquad V_h \le\rho_H^2 H^{d_\ast+\beta-2}, U h ≤ ρ H 2 H − 1 − α , V h ≤ ρ H 2 H d ∗ + β − 2 , and let G G G 2 H / ρ H = o ( H ) 2H/\rho_H=o(H) 2 H / ρ H = o ( H ) G G G G G G
sup e b h ( e ) ≤ h − 5 / 4 + o ( 1 ) , sup e b h ( L ) ( e ) ≤ h 1 / 12 + o ( 1 ) . \sup_e b_h(e) \le h^{-5/4+o(1)}, \qquad\sup_e b_h^{(L)}(e) \le h^{1/12+o(1)}. e sup b h ( e ) ≤ h − 5/4 + o ( 1 ) , e sup b h ( L ) ( e ) ≤ h 1/12 + o ( 1 ) . The construction of G G G δ \delta δ ε \varepsilon ε δ > 0 \delta>0 δ > 0 b h ( e ) ≥ h − 5 / 4 − o ( 1 ) b_h(e) \ge h^{-5/4-o(1)} b h ( e ) ≥ h − 5/4 − o ( 1 ) b h ( L ) ( e ) ≥ h 1 / 12 − o ( 1 ) b_h^{(L)}(e) \ge h^{1/12-o(1)} b h ( L ) ( e ) ≥ h 1/12 − o ( 1 )
h 4 / 3 − o ( 1 ) ≤ b h ( L ) ( e ) b h ( e ) ≤ h 4 / 3 + o ( 1 ) ( h ∈ G , ∣ e ∣ ≤ h 1 − δ ) . h^{4/3-o(1)} \le\frac{b_h^{(L)}(e)}{b_h(e)} \le h^{4/3+o(1)} \qquad(h \in G,\ |e| \le h^{1-\delta}). h 4/3 − o ( 1 ) ≤ b h ( e ) b h ( L ) ( e ) ≤ h 4/3 + o ( 1 ) ( h ∈ G , ∣ e ∣ ≤ h 1 − δ ) . This proves Theorem 10.1 with its stated quantifiers.
For completeness, the mixture in Theorem 10.2 follows directly from the estimates before that deletion of heights. At each large even height, the aligned lower bounds are b h ( 0 ) ≥ h − 5 / 4 − o ( 1 ) b_h(0) \ge h^{-5/4-o(1)} b h ( 0 ) ≥ h − 5/4 − o ( 1 ) b h ( L ) ( 0 ) ≥ h 1 / 12 − o ( 1 ) b_h^{(L)}(0) \ge h^{1/12-o(1)} b h ( L ) ( 0 ) ≥ h 1/12 − o ( 1 ) H H H [ H , 2 H ] [H,2H] [ H , 2 H ] Z H ≥ H − 1 / 4 − o ( 1 ) Z_H \ge H^{-1/4-o(1)} Z H ≥ H − 1/4 − o ( 1 ) M H ≥ H 13 / 12 − o ( 1 ) M_H \ge H^{13/12-o(1)} M H ≥ H 13/12 − o ( 1 ) Z H ≤ H − 1 / 4 + o ( 1 ) Z_H \le H^{-1/4+o(1)} Z H ≤ H − 1/4 + o ( 1 ) M H ≤ H 13 / 12 + o ( 1 ) M_H \le H^{13/12+o(1)} M H ≤ H 13/12 + o ( 1 )
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