Deterministic Selection in Empirical Wasserstein-One Matching

1. Introduction

Optimal transport with the Euclidean distance cost can have many minimizers, even for compactly supported measures with densities. We ask whether independent sampling selects a definite transport plan. For samples from the uniform distributions on ${[0,1]}^2$ and $[2,3]\times [0,1]$, this question is listed as an open problem in the survey of del Barrio, González-Sanz, Loubes, and Rodríguez-Vítores [2].

We prove that the optimal empirical matchings converge weakly in probability to a deterministic plan as the sample size grows. We characterize this plan through the source-Palm distribution of the unique stationary matching obtained at the scale of nearby sample points.

1.1. Model and Main Result

For each positive integer n, let

$$X_i=(U_i,T_i)\in {[0,1]}^2,Y_j=(2+V_j,S_j)\in [2,3]\times [0,1],1\le i,j\le n,$$

where the $4n$ coordinates are independent and uniform on $[0,1]$. Let ${\sigma }_n$ minimize

$${\displaystyle \frac{1}{n}}{\displaystyle \sum _{i=1}^n}\sqrt{{(2+V_{{\sigma }_n\left(i\right)}-U_i)}^2+{(S_{{\sigma }_n\left(i\right)}-T_i)}^2}$$

(1.1)

over permutations of $\{1,\dots ,n\}$. A fixed ordering of the permutations gives a measurable choice whenever there is a tie. In the coordinate order $(t,u,s,v)$, define

$${\pi }_n:={\displaystyle \frac{1}{n}}{\displaystyle \sum _{i=1}^n}{\delta }_{(T_i,U_i,S_{{\sigma }_n\left(i\right)},V_{{\sigma }_n\left(i\right)})}.$$

(1.2)

The physical transport plan is its pushforward by $(t,u,s,v)\mapsto \left(\right(u,t),(2+v,s\left)\right)$. For a coupling $\kappa$ of two uniform $[0,1]$ laws, write

$${\Pi }_{\kappa }(dt,du,ds,dv):=dt{\delta }_t\left(ds\right)\kappa (du,dv).$$

Thus ${\Pi }_{\kappa }$ preserves height, and its paired horizontal marks have law $\kappa$, independently of height.

Theorem 1.1

(Deterministic empirical selection). There is a deterministic coupling ${\kappa }_{\infty }$ of two uniform $[0,1]$ laws such that

$${\pi }_n\stackrel{\mathbb{P}}{\to }{\Pi }_{{\kappa }_{\infty }}\mathit{in}\mathit{the}\mathit{weak}\mathit{topology}\mathit{on}\mathit{probability}\mathit{measures}\mathit{on}{[0,1]}^4.$$

(1.3)

Equivalently, for every continuous $F:{[0,1]}^4\to \mathbb{R}$,

$${\displaystyle \frac{1}{n}}{\displaystyle \sum _{i=1}^n}F(T_i,U_i,S_{{\sigma }_n\left(i\right)},V_{{\sigma }_n\left(i\right)})\stackrel{\mathbb{P}}{\to }{\int }_0^1{\int }_{{[0,1]}^2}F(t,u,t,v){\kappa }_{\infty }(du,dv)dt.$$

The theorem uses the iid law at each sample size; samples of different sizes may have arbitrary dependence. The selected coupling is the source-Palm distribution of paired horizontal marks in the stationary microscopic matching. This is the joint law of the horizontal coordinates seen from a typical source point and its partner.

We can describe all optimal plans for the limiting problem. For $x=(u,t)$ and $y=(2+v,s)$,

$$|x-y|\ge 2+v-u,$$

with equality precisely when $s=t$. Every admissible continuum coupling integrates the right-hand side to 2, and a height-preserving coupling attains this value. Consequently the optimal plans are exactly

$$dt{\delta }_t\left(ds\right){\kappa }_t(du,dv),$$

(1.4)

where $t\mapsto {\kappa }_t$ is a measurable family of couplings of two uniform laws, up to equality for almost every t. This description follows by disintegrating over the common height. Theorem 1.1 therefore identifies both a deterministic horizontal coupling and its constancy across macroscopic heights.

1.2. Selection and Stationary Matching

Secondary variational selection for the distance cost minimizes an additional energy over the continuum optimal set; its relationship with transport rays and variational asymptotic expansions is developed by Ambrosio and Pratelli [1]. In our model, quadratic secondary selection is explicit: on the optimal set,

$${\int |x-y|}^{2}d\pi =4+\int {(v-u)}^{2}d\pi .$$

It therefore selects the horizontal translation $v=u$. The approximation scheme matters for selection. Gwóźdź [3] constructs weakly convergent absolutely continuous marginals whose ray-monotone selectors converge to a different optimal plan from the selector of the limiting marginals. Our result establishes selection for the iid empirical approximation in (1.1).

The local matching problem connects the proof to invariant minimal matchings. Holroyd, Janson and Wästlund [4] study point-process matchings that minimize cost under every finite rematching, including their uniqueness and translation-invariant realizations. Here the points carry horizontal marks, and a feedback change of coordinates gives a quadratic local cost. We prove uniqueness for stationary locally minimizing matchings with zero relative flux and finite source-Palm second moment in this environment.

Optimal transport between stationary random measures provides another point of comparison. Huesmann [5] studies mean transport cost per unit volume and approximation by problems on bounded regions, with uniqueness for an absolutely continuous source. We also pass from finite problems to a stationary matching. For the atomic source measure in our setting, regeneration and stationary mass transport establish uniqueness within the locally minimizing class.

The selection result also has a stochastic homogenization interpretation. The continuum cost determines the optimal set, while the rescaled sample geometry determines which plan is selected. Variational asymptotic expansions study this type of selection by resolving finer scales within a minimizing set [1].

In the present proof, subtracting terms that depend only on a source or a target exposes a quadratic local cost. Optimality passes to the local matching limits, and their unique distribution of paired marks determines the transport plan at every macroscopic height.

1.3. Proof Strategy

Fix an auxiliary parameter $0<\lambda <1/8$. A predictable feedback graph compensates the source–target counting imbalance on the microscopic rank scale. Section 2 constructs this graph and controls its current and a comparison matching. Section 3 transfers the comparison estimate to the exact Euclidean optimizer by a row–column cost gauge, obtaining a quadratic bound on feedback edge lengths.

Section 4 identifies the stationary marked environment through its martingale problem. Section 5 then shows that every selected local matching limit is perfect, locally minimizing, of zero relative flux, and has finite source-Palm second moment. Its source-Palm law means the law seen from a typical source event; the paired source and target marks give the candidate horizontal coupling.

Section 6 and Section 7 prove uniqueness of this stationary matching. A long zero-current hold, together with a deterministic energy iteration, produces a common empty cut in any joining of two candidate matchings. Stationary mass transport and the absence of finite alternating ties then force the matchings to coincide. Finally, Section 8 uses vertical concentration and blockwise Palm readout to reconstruct ${\pi }_n$ and prove Theorem 1.1.

The feedback graph is built at a fixed outer error tolerance $\epsilon >0$, and its estimates are derived before intersecting the required high-probability events. The limit $\epsilon \downarrow 0$ is taken only after returning to the original empirical quantities. This order of limits is used throughout.

2. Finite Feedback Estimates

We construct a graph that keeps the source and target crossing counts close. Matching the crossings in order will give a comparison matching with controlled squared edge lengths.

Set

$$a_i:=1-U_i\in [0,1],b_j:=1+V_j\in [1,2],D_{ij}:=a_i+b_j\in [1,3],$$

and use the rank-scale phase lines

$${\widehat{A}}_i\left(p\right):=n{T}_i+a_{i}p,{\widehat{B}}_j\left(p\right):=n{S}_j-b_{j}p.$$

(2.1)

Fix once and for all

$$0<\lambda <{\displaystyle \frac{1}{8}},0<\eta <{\displaystyle \frac{\lambda }{4(2+\lambda )}}.$$

(2.2)

Auxiliary graphs and outer limits.

For every $\epsilon >0$, choose a fixed $A=A\left(\epsilon \right)$, set $B_n=A\sqrt{n}$, and construct the auxiliary graph $g_{n,\epsilon }$. There is a good event ${\mathcal{G}}_{n,\epsilon }$ such that

$$\underset{n\to \infty }{\mathrm{lim\; inf}}\mathbb{P}\left({\mathcal{G}}_{n,\epsilon }\right)\ge 1-\epsilon ,$$

(2.3)

and all graph-dependent estimates below hold on ${\mathcal{G}}_{n,\epsilon }$, with constants allowed to depend on $\epsilon$ but not on n. To reduce notation, the superscript $\epsilon$ is suppressed inside a proof carried out at fixed $\epsilon$.

Allocation of error probabilities.

Whenever a proof uses finitely many high-probability events, we first apply the corresponding estimates with tolerances $\epsilon /m$, where m is a fixed deterministic integer, and then intersect the resulting events. We relabel the total tolerance as $\epsilon$. Throughout, “on the good event” refers to this final intersection. We first prove the stopped conditional estimates with respect to the filtration at the stated stopping coordinate, and then form the intersection.

For

$${\mathsf{J}}_n(p,x):=\#\{i:{\widehat{A}}_i\left(p\right)\le x\}-\#\{j:{\widehat{B}}_j\left(p\right)\le x\},$$

(2.4)

the unclipped half-space volumes in the bulk are

$$V^{-}(p,x)=x-{\displaystyle \frac{p}{2}},V^{+}(p,x)=x+{\displaystyle \frac{3p}{2}},$$

(2.5)

and we write

$${\Delta }_n^{-}(p,x)=\#\{i:{\widehat{A}}_i\left(p\right)\le x\}-V^{-}(p,x),$$

$${\Delta }_n^{+}(p,x)=\#\{j:{\widehat{B}}_j\left(p\right)\le x\}-V^{+}(p,x).$$

(2.6)

Thus, whenever (2.5) is unclipped,

$$\begin{array}{|c|}\hline {\mathsf{J}}_n(p,x)={\Delta }_n^{-}(p,x)-{\Delta }_n^{+}(p,x)-2p.\\ \hline\end{array}$$

(2.7)

2.1. Predictable Deferred Decisions and Adaptive Rates

Let

$${\mathcal{X}}_n=[0,n]\times [0,1],P_i^{-}=(n{T}_i,U_i),P_j^{+}=(n{S}_j,V_j).$$

Lemma 2.1

(Predictable-frontier deferred decisions). Let $p=g\left(x\right)$ be a left-to-right, piecewise $C^1$, predictable graph with $\mathrm{Lip}\left(g\right)\le \lambda <1/2$. Define the swept sets

$$H_x^{-}\left(g\right)=\{(t,u)\in {\mathcal{X}}_n:t+g\left(x\right)(1-u)\le x\},$$

$$H_x^{+}\left(g\right)=\{(s,v)\in {\mathcal{X}}_n:s-g\left(x\right)(1+v)\le x\}.$$

(2.8)

Let ${\mathcal{F}}_x^{g,0}$ be generated by the graph up to x and by all labeled source and target points whose trajectories have crossed it by coordinate x. In stopping-coordinate statements we use its usual augmentation

$${\mathcal{F}}_x^g:=\bigcap _{y>x}\left({\mathcal{F}}_y^{g,0}\vee \mathcal{N}\right)=\bigcap _{q>x,q\in \mathbb{Q}}\left({\mathcal{F}}_q^{g,0}\vee \mathcal{N}\right),$$

where $\mathcal{N}$ is the family of ambient terminal null events. Conditional on ${\mathcal{F}}_x^g$, the unrevealed source labels are independent uniforms on ${\mathcal{X}}_n\setminus H_x^{-}\left(g\right)$, the unrevealed target labels are independent uniforms on ${\mathcal{X}}_n\setminus H_x^{+}\left(g\right)$, and the two families are conditionally independent. The same assertion holds at every ${\mathcal{F}}^g$-stopping coordinate τ with $0\le \tau \le n$ almost surely.

Proof. 

Spacelikeness makes both swept families increasing. Indeed, for $y>x$,

$$[y-g(y\left)\right(1-u\left)\right]-[x-g(x\left)\right(1-u\left)\right]\ge (1-\lambda )(y-x)>0,$$

and

$$[y+g(y\left)\right(1+v\left)\right]-[x+g(x\left)\right(1+v\left)\right]\ge (1-2\lambda )(y-x)>0.$$

(2.9)

Suppose at a crossing coordinate $x_0$ that the unrevealed points have the stated product-uniform law on the current complements. Until the next crossing the continuation of the graph is a deterministic function of the exposed filtration. If $C_x$ is either decreasing complement and Z is one unrevealed label, then

$$\mathbb{P}\{Z\in C_x\mid Z\in C_{x_0},{\mathcal{F}}_{x_0}^g\}={\displaystyle \frac{|C_x|}{|C_{x_0}|}},$$

and conditional on $Z\in C_x$, Z is uniform on $C_x$. The conditioning factorizes over unrevealed labels, so absence of a crossing up to x preserves the product-uniform law.

Simultaneous first crossings have probability zero. To see this, freeze the two initial revealed-label sets and a finite history of the previously exposed labels. On each such history cell the graph continuation is a measurable function of the coordinates in that history. If two unused labels were to cross next at the same coordinate, then, after fixing every coordinate except the height of the first label, spacelikeness makes the crossing equation injective in that height. Thus the exceptional height section contains at most one point. The height law is nonatomic, and there are only finitely many initial patterns, histories, steps, and label pairs, so the union of all simultaneous-crossing events is null. The same one-coordinate argument shows that a label lies on a fixed frontier with probability zero.

At the next crossing exactly one label is exposed almost surely. Condition first on entrance through a boundary slab of thickness h, reveal that label and its position, and let $h\downarrow 0$; the remaining labels retain the product-uniform law on the new complement. There are at most $2n$ crossings.

To construct the conditional law, fix a finite labeled history h and the tuple z of revealed coordinates. On that history cell define $Q_{h,z,x}$ by fixing the revealed coordinates at z and assigning to every unrevealed source or target label the normalized restriction of Lebesgue measure to its current strict complement; take the product over the unrevealed labels. The boundary-slab induction gives, first for a future cylinder E and a generator D of the raw history sigma-field,

$${\int }_{D\cap \left\{h\right\}}{Q}_{h,z\left(\omega \right),x}\left(E\right)d\mathbb{P}\left(\omega \right)=\mathbb{P}(D\cap E\cap \left\{h\right\}).$$

(2.10)

The kernel is measurable in the revealed tuple: all its cylinder values are finite products of measurable section areas divided by the corresponding positive complement areas. (On terminal zero-area cells its value may be chosen arbitrarily.) Summing over the finitely many histories and applying the pi-lambda theorem first in D and then in E extends (2.10) to the whole raw history sigma-field and all Borel future events. Thus the assembled kernel is a regular conditional law at each deterministic coordinate. The identity also holds almost surely after adding ambient terminal null events.

The two definitions of ${\mathcal{F}}_x^g$ above agree: monotonicity of the raw filtration and density of $\mathbb{Q}$ make the rational right-neighbourhoods cofinal among all right-neighbourhoods. For rational $x_m\downarrow x$, frontier nullity and dominated convergence give $Q_{x_m}\left(E\right)\to Q_x\left(E\right)$ first for cylinders, while reverse-martingale convergence identifies the limit with conditioning on ${\mathcal{F}}_x^g$. A second monotone-class argument extends this identification to every Borel E.

For a stopping coordinate $0\le \tau \le n$, let ${\tau }_m=2^{-m}\lceil 2^m\tau \rceil$. Apply the deterministic-coordinate conditional identity on the finitely many range slices $\{{\tau }_m=k{2}^{-m}\}$. Then ${\tau }_m\downarrow \tau$, right continuity gives ${\mathcal{F}}_{\tau }^g={\bigcap }_m{\mathcal{F}}_{{\tau }_m}^g$, and the reverse martingale convergence theorem passes the conditional expectations to ${\mathcal{F}}_{\tau }^g$.

Continuity of the graph and the frontier-null statement above give monotone convergence of the remaining-set indicators and their normalizing areas. Applying this argument to finite products of bounded Borel test functions identifies the asserted product-uniform conditional law at $\tau$. □

For fixed $A>10$, put $B_n=A\sqrt{n}$ and

$${\mathcal{R}}_{A,n}=\{(p,x):B_n\le x\le n-B_n,\left|p\right|\le B_n/8\}.$$

(2.11)

For every $n\ge 1$, $(p,x)\in {\mathcal{R}}_{A,n}$, and $u,v\in [0,1]$,

$${\displaystyle \frac{7B_n}{8}}\le x-p(1-u)\le n-{\displaystyle \frac{7B_n}{8}},{\displaystyle \frac{3B_n}{4}}\le x+p(1+v)\le n-{\displaystyle \frac{3B_n}{4}}.$$

Thus all source and target sections lie strictly inside the rank strip before the curve exits ${\mathcal{R}}_{A,n}$. We use the unclipped area formulas on this interval. Let

$$m_{-}(p,x)=\min \{V^{-}(p,x),n-V^{-}(p,x)\},m_{+}(p,x)=\min \{V^{+}(p,x),n-V^{+}(p,x)\}.$$

Lemma 2.2

(Half-space regularity and adaptive crossing rates). There is an event ${\mathcal{E}}_{A,n}$ with $\mathbb{P}\left({\mathcal{E}}_{A,n}\right)\to 1$ on which, uniformly on ${\mathcal{R}}_{A,n}$,

$$|{\Delta }_n^{\pm }(p,x)|\le C\left(\sqrt{m_{\pm }(p,x)\log n}+\log n\right).$$

(2.12)

Moreover

$$\underset{{\mathcal{R}}_{A,n}}{\sup }\left(\right|{\Delta }_n^{-}|+|{\Delta }_n^{+}\left|\right)=O_{\mathbb{P}}\left(\sqrt{n}\right),$$

(2.13)

and at $x_R=n-B_n$,

$$\underset{\left|p\right|\le B_n/8}{\sup }\left(\right|{\Delta }_n^{-}(p,x_R)|+|{\Delta }_n^{+}(p,x_R)\left|\right)=O_{\mathbb{P}}\left(\sqrt{B_n}\right).$$

(2.14)

Let now $p=g\left(x\right)$ be any predictable curve in ${\mathcal{R}}_{A,n}$ with $|g^{\prime }|\le \lambda$. Write ${\tau }_i^{-}$ and ${\tau }_j^{+}$ for the source and target crossing coordinates and define the pre-crossing counts

$$D_{-}(x-):=\#\{i:{\tau }_i^{-}<x\},D_{+}(x-):=\#\{j:{\tau }_j^{+}<x\},$$

and the remaining counts

$$M_{-}(x-):=n-D_{-}(x-),M_{+}(x-):=n-D_{+}(x-).$$

Put $K(x-)=D_{-}(x-)-D_{+}(x-)$ and let $w_{-}\left(x\right)=g_{-}^{\prime }\left(x\right)$ be the predictable left slope. Away from crossing coordinates, $K(x-)={\mathsf{J}}_n(g\left(x\right),x)$ and $w_{-}\left(x\right)=g^{\prime }\left(x\right)$. The predictable marked crossing intensities are

$$r_{-}\left(x\right)[1-w_{-}\left(x\right)(1-u)]du,r_{+}\left(x\right)[1+w_{-}\left(x\right)(1+v)]dv,$$

(2.15)

where

$$r_{-}\left(x\right)={\displaystyle \frac{M_{-}(x-)}{n-V^{-}(g\left(x\right),x)}},r_{+}\left(x\right)={\displaystyle \frac{M_{+}(x-)}{n-V^{+}(g\left(x\right),x)}}.$$

(2.16)

Consequently the total source and target jump rates are

$${\lambda }_{-}\left(x\right)=r_{-}\left(x\right)\left(1-{\displaystyle \frac{w_{-}\left(x\right)}{2}}\right),{\lambda }_{+}\left(x\right)=r_{+}\left(x\right)\left(1+{\displaystyle \frac{3w_{-}\left(x\right)}{2}}\right),$$

(2.17)

and, on ${\mathcal{E}}_{A,n}$,

$$\underset{x}{\sup }|r_{\pm }\left(x\right)-1|=o\left(1\right)$$

(2.18)

uniformly along every such curve before it exits ${\mathcal{R}}_{A,n}$.

Proof. 

The sets in (2.8), with deterministic $p,x$, form two finite-VC classes. The standard relative VC Bernstein inequality (obtained by dyadic peeling according to the smaller of a set and its complement, followed by Sauer’s lemma and Bernstein’s inequality) gives (2.12); the ordinary VC maximal inequality gives (2.13).

For (2.14), keep $x=x_R$ fixed. As p varies, each complement is a nested one-parameter family contained in a deterministic set of area $O\left(B_n\right)$. If Z is the number of sample points in the largest member, then $Z=O_{\mathbb{P}}\left(B_n\right)$ and $Z-\left|C\right|=O_{\mathbb{P}}\left(\sqrt{B_n}\right)$. Conditional on Z, the nested counts are an empirical distribution function of Z iid points; its maximal fluctuation is $O_{\mathbb{P}}\left(\sqrt{Z}\right)$. This proves (2.14).

For the rates, invoke Lemma 2.1. Conditional on the explored filtration, the unrevealed source labels are uniform on a region of area $n-V^{-}$. Advancing the graph by $dx$ sweeps source area

$${\int }_0^1[1-w(1-u)]dudx=\left(1-{\displaystyle \frac{w}{2}}\right)dx,$$

and marked u-slice area $[1-w(1-u\left)\right]dudx$. Multiplication by the current unexplored density gives the first formula in (2.15) and the first rate in (2.17). The target swept area is

$${\int }_0^1[1+w(1+v)]dvdx=\left(1+{\displaystyle \frac{3w}{2}}\right)dx,$$

giving the second formulas.

To identify the predictable intensities, fix deterministic $s<t$ before exit from ${\mathcal{R}}_{A,n}$ and $B\in {\mathcal{F}}_s^g$. On each of the finitely many revealed-label histories, Fubini’s theorem and the conditional product law give the exact identities

$$\mathbb{E}\left[{\mathbf{1}}_B(D_{-}\left(t\right)-D_{-}\left(s\right))\right]=\mathbb{E}\left[{\mathbf{1}}_B{\int }_{(s,t]}{\lambda }_{-}\left(x\right)dx\right],$$

$$\mathbb{E}\left[{\mathbf{1}}_B(D_{+}\left(t\right)-D_{+}\left(s\right))\right]=\mathbb{E}\left[{\mathbf{1}}_B{\int }_{(s,t]}{\lambda }_{+}\left(x\right)dx\right].$$

(2.19)

The same calculation with bounded Borel mark tests gives the marked versions of these identities. Frontier-nullity removes the endpoints, while the strictly positive remaining areas in ${\mathcal{R}}_{A,n}$ justify every normalization. A monotone-class extension from the predictable rectangles $B\times (s,t]$, followed by finite linear combinations and bounded approximation, proves that the counting processes compensated by $\int {\lambda }_{\pm }$ are martingales. This establishes the predictable intensity formulas (2.15)–(2.17).

Finally, on ${\mathcal{R}}_{A,n}$ the unexplored volumes are at least $c{B}_n$; (2.12) divided by these volumes gives (2.18). □

2.2. Inward Feedback, Current Energy, and Spacelikeness

Starting at $x=B_n$ with $g_n\left(B_n\right)=0$, construct the graph recursively by

$$g_n^{\prime }\left(x\right)=\lambda \mathrm{sgn}{K}_n\left(x\right),K_n\left(x\right):={\mathsf{J}}_n(g_n\left(x\right),x),$$

(2.20)

between crossings, using the new value of $K_n$ immediately after a crossing and $\mathrm{sgn}0=0$. Define

$${\tau }_{\mathrm{sh}}=\inf \{x\ge B_n:|g_n\left(x\right)|=B_n/8\},$$

$${\tau }_{\mathrm{reg}}=\inf \{x\ge B_n:|r_{-}\left(x\right)-1|\vee |r_{+}\left(x\right)-1|\ge \eta \},$$

$${\tau }_n=(n-B_n)\wedge {\tau }_{\mathrm{sh}}\wedge {\tau }_{\mathrm{reg}}.$$

(2.21)

Lemma 2.3

(Lower occupation from predictable rates). Let N be a simple counting process with predictable intensity $\Lambda \left(x\right)$, and assume on a stochastic interval $[\sigma ,\sigma +r]$ that $0<c_0\le \Lambda \le C_0<\infty$. Then there are $a_0,b_0>0$ depending only on $c_0,C_0$ such that

$$\mathbb{P}\{N(\sigma +r)-N\left(\sigma \right)\le a_{0}r\mid {\mathcal{F}}_{\sigma }\}\le e^{-b_{0}r}.$$

(2.22)

The upper rate bound also gives exponential upper tails for the number of jumps on length-r intervals.

Proof. 

For $\theta >0$, the exponential counting-process martingale

$$\exp \left\{-\theta N\left(x\right)+(1-e^{-\theta }){\int }^x\Lambda \left(y\right)dy\right\}$$

and optional sampling between $\sigma$ and $\sigma +r$ give

$$\mathbb{P}\{N(\sigma +r)-N\left(\sigma \right)\le ar\mid {\mathcal{F}}_{\sigma }\}\le \exp \{\theta ar-(1-e^{-\theta })c_{0}r\}.$$

Take, for example, $\theta =1$ and $a_0=c_0(1-e^{-1})/2$. The upper tail is analogous. □

For the endpoint estimate, we separate the failure time from the deterministic right endpoint. Put

$${\widehat{\tau }}_n:={\tau }_{\mathrm{sh}}\wedge {\tau }_{\mathrm{reg}},x_R:=n-B_n,$$

(2.23)

and define the modified killing time

$${\tau }_n^{†}:=\left\{\begin{array}{cc}{\widehat{\tau }}_n,\hfill & {\widehat{\tau }}_n<x_R,\hfill \\ +\infty ,\hfill & {\widehat{\tau }}_n\ge x_R.\hfill \end{array}\right.$$

(2.24)

Then ${\tau }_n^{†}$ is a stopping coordinate and the process is alive at $x_R$ exactly when there has been no premature bulk failure.

Theorem 2.4

(Feedback current estimates). Before ${\tau }_n$, the feedback current has a uniform drift toward zero. There are deterministic $c,C>0$ such that, for all large n, whenever $K_n\left(x\right)\ne 0$,

$$\mathrm{sgn}\left(K_n\left(x\right)\right)({\lambda }_{-}\left(x\right)-{\lambda }_{+}\left(x\right))\le -c,c\le {\lambda }_{-}\left(x\right),{\lambda }_{+}\left(x\right)\le C.$$

(2.25)

Consequently

$$\mathbb{E}{\int }_{B_n}^{{\tau }_n}{K}_n{\left(x\right)}^{2}dx\le Cn+C\mathbb{E}{\left|K_n\left(B_n\right)\right|}^3=O\left(n\right),$$

(2.26)

$$\underset{B_n\le x\le {\tau }_n}{\sup }\left|K_n\left(x\right)\right|=O_{\mathbb{P}}(n^{1/4}+\log n),$$

(2.27)

and

$$K_n(n-B_n){\mathbf{1}}_{\{{\tau }_n=n-B_n\}}=O_{\mathbb{P}}\left(1\right).$$

(2.28)

Proof. 

Before ${\tau }_n$, $1-\eta <r_{\pm }<1+\eta$. If $K_n>0$, then $g_n^{\prime }=\lambda$, and

$${\lambda }_{-}-{\lambda }_{+}\le (1+\eta )\left(1-{\displaystyle \frac{\lambda }{2}}\right)-(1-\eta )\left(1+{\displaystyle \frac{3\lambda }{2}}\right)=-2\lambda +\eta (2+\lambda )\le -c.$$

If $K_n<0$, then $g_n^{\prime }=-\lambda$, and

$${\lambda }_{-}-{\lambda }_{+}\ge (1-\eta )\left(1+{\displaystyle \frac{\lambda }{2}}\right)-(1+\eta )\left(1-{\displaystyle \frac{3\lambda }{2}}\right)=2\lambda -\eta (2-\lambda )\ge c.$$

The individual rate bounds follow from $\lambda <1/8$.

For $V\left(k\right)={\left|k\right|}^3$, the jump generator satisfies

$$\mathcal{L}V\left(k\right)\le -c_1{k}^2+C_1.$$

(2.29)

Dynkin’s formula up to ${\tau }_n$ gives (2.26). At $x=B_n$ the shear is zero, so $K_n\left(B_n\right)$ is the difference of two independent $\mathrm{Bin}(n,B_n/n)$ variables and

$$\mathbb{E}|K_n\left(B_n\right){|}^3=O\left(B_n^{3/2}\right)=O\left(n^{3/4}\right)$$

for fixed A.

For $W\left(k\right)=e^{\theta \left|k\right|}$, sufficiently small $\theta >0$ gives

$$\mathcal{L}W\left(k\right)\le -aW\left(k\right)+b.$$

(2.30)

We use a stopped excursion estimate to control the maximum current. For $R\ge 1$, put

$$W_R\left(k\right):={\exp \left\{\theta \right(\left|k\right|-R)}_{+}\}.$$

The same one-step calculation, including the flat region $\left|k\right|\le R$, gives

$$\mathcal{L}{W}_R\left(k\right)\le -a{W}_R\left(k\right)+b$$

(2.31)

with the same positive constants after increasing b, if necessary. Let $G_R=\left\{\right|K_n\left(B_n\right)|\le R\}$, which is ${\mathcal{F}}_{B_n}$-measurable, and put $T_n=x_R-B_n\le n$. Multiplying the stopped Dynkin identity by ${\mathbf{1}}_{G_R}$, using $W_R\left(K_n\left(B_n\right)\right)=1$ on $G_R$, and discarding the nonnegative terminal term gives

$$\mathbb{E}\left[{\mathbf{1}}_{G_R}{\int }_{B_n}^{{\tau }_n}{W}_R\left(K_n(x-)\right)dx\right]\le {\displaystyle \frac{1+b{T}_n}{a}}.$$

(2.32)

For $q\ge 1$, let $J_{R,q}$ count outward jumps, through the right-closed stopping coordinate ${\tau }_n$, which enter $\left\{\right|K_n|\ge R+q\}$ from its complement. On $G_R$, attainment of height $R+q$ forces $J_{R,q}\ge 1$. Immediately before every such jump,

$$W_R\left(K_n(x-)\right)\ge e^{\theta (q-1)}.$$

The two outward directions are disjoint, and their predictable rates are bounded by a deterministic constant C. The compensator identity and (2.32) therefore yield

$$\mathbb{P}\left(G_R\cap \left\{\underset{B_n\le x\le {\tau }_n}{\sup }\left|K_n\left(x\right)\right|\ge R+q\right\}\right)\le {\displaystyle \frac{C}{a}}(1+b{T}_n)e^{-\theta (q-1)}.$$

(2.33)

Together with the preceding cubic-moment estimate and Markov’s inequality,

$$\mathbb{P}\left\{\underset{B_n\le x\le {\tau }_n}{\sup }\left|K_n\left(x\right)\right|\ge R+q\right\}\le {\displaystyle \frac{\mathbb{E}|K_n\left(B_n\right){|}^3}{R^3}}+{\displaystyle \frac{C}{a}}(1+b{T}_n)e^{-\theta (q-1)}.$$

(2.34)

Choose $R=D\sqrt{B_n}$ and $q=2{\theta }^{-1}\log (n+1)$. The first term in (2.34) is $O\left(D^{-3}\right)$, uniformly in n, and the second is $O\left(n^{-1}\right)$. First choosing D and then letting $n\to \infty$ proves $\sup |K_n|=O_{\mathbb{P}}(\sqrt{B_n}+\log n)$, which is (2.27).

For (2.28), kill the process at ${\tau }_n^{†}$, send the killed state to a cemetery point ∂, and set $W(\partial )=0$. Killing adds only a nonpositive jump to the generator, hence

$$\tilde{\mathcal{L}}W\le -aW+b.$$

Dynkin’s formula and Gronwall give, for $B_n\le x\le x_R$,

$$\mathbb{E}\left[e^{\theta |K_n\left(x\right)|};{\tau }_n^{†}>x\right]\le e^{-a(x-B_n)}\mathbb{E}{e}^{\theta |K_n\left(B_n\right)|}+{\displaystyle \frac{b}{a}}.$$

(2.35)

At $x=B_n$,

$$K_n\left(B_n\right)=X_n-Y_n,X_n,Y_n\stackrel{\mathrm{iid}}{\sim }\mathrm{Bin}\left(n,{\displaystyle \frac{B_n}{n}}\right).$$

Using $e^{\theta \left|z\right|}\le e^{\theta z}+e^{-\theta z}$ and the binomial mgf,

$$\mathbb{E}{e}^{\theta |K_n\left(B_n\right)|}\le 2\exp \left\{2B_n(\cosh \theta -1)+O(B_n^2/n)\right\}\le e^{C_{\theta }{B}_n}.$$

(2.36)

At $x=x_R$, the event $\{{\tau }_n^{†}>x_R\}$ is exactly $\{{\widehat{\tau }}_n\ge x_R\}$, equivalently $\{{\tau }_n=x_R\}$. The first term in (2.35) is

$$\exp \{-a(n-2B_n)+C_{\theta }{B}_n\}=o\left(1\right),$$

because $B_n=A\sqrt{n}$. Therefore

$$\mathbb{P}\left\{\right|K_n\left(x_R\right)|>M,{\tau }_n=x_R\}\le e^{-\theta M}\left({\displaystyle \frac{b}{a}}+o\left(1\right)\right),$$

uniformly in n, proving (2.28). □

Theorem 2.5

(Spacelike shear control). For every $\epsilon >0$ there is a finite $A=A\left(\epsilon \right)>10$ such that, with $B_n=A\sqrt{n}$,

$$\mathbb{P}\{{\tau }_n=n-B_n\}\ge 1-\epsilon -o\left(1\right).$$

(2.37)

On this event

$$\mathrm{Lip}\left(g_n\right)\le \lambda <{\displaystyle \frac{1}{2}},\underset{B_n\le x\le n-B_n}{\sup }\left|g_n\left(x\right)\right|=O_{\mathbb{P}}\left(\sqrt{n}\right),$$

(2.38)

and at the right junction

$$q_n:=g_n(n-B_n)=O_{\mathbb{P}}\left(\sqrt{B_n}\right)=O_{\mathbb{P}}\left(n^{1/4}\right).$$

(2.39)

Extend $g_n$ by the constant 0 to the left of $B_n$ and by the constant $q_n$ to the right of $n-B_n$. The extension is globally λ-Lipschitz.

Proof. 

On ${\mathcal{E}}_{A,n}$, (2.18) is $<\eta$ for all large n, so ${\tau }_{\mathrm{reg}}$ cannot occur before ${\tau }_{\mathrm{sh}}\wedge (n-B_n)$. While the graph remains in the bulk rectangle, (2.7) gives

$$2g_n\left(x\right)={\Delta }_n^{-}(g_n\left(x\right),x)-{\Delta }_n^{+}(g_n\left(x\right),x)-K_n\left(x\right).$$

(2.40)

By (2.13), the bridge supremum divided by $\sqrt{n}$ is tight, while (2.27) is $o_{\mathbb{P}}\left(\sqrt{n}\right)$. At the first shear hit, $2|g_n|=A\sqrt{n}/4$. Therefore

$$\underset{n}{\mathrm{lim\; sup}}\mathbb{P}\{{\tau }_{\mathrm{sh}}<n-B_n\}\le \mathbb{P}\{Z>A/8\}+o\left(1\right)$$

for a tight random variable Z. Choose A so the right side is $<\epsilon$, proving (2.37).

On survival, (2.28), (2.40), and the nested endpoint bound (2.14) give

$$2|q_n|\le |{\Delta }_n^{-}(q_n,x_R)|+|{\Delta }_n^{+}(q_n,x_R)|+|K_n\left(x_R\right)|=O_{\mathbb{P}}\left(\sqrt{B_n}\right),$$

which is (2.39). The constant extensions preserve the Lipschitz bound. □

Predictable continuation on all samples.

To form the marked rooted laws in Section 4, we define a predictable graph on every sample. The compactness estimates above will then apply on the final good event. Follow (2.20) until

$${\overline{\tau }}_n:={\widehat{\tau }}_n\wedge (n-B_n),$$

and, if ${\overline{\tau }}_n<n-B_n$, keep the graph height constant from ${\overline{\tau }}_n$ onward until all source and target trajectories have crossed; use the same constant extensions at the two outer tails. Denote this everywhere-defined predictable spacelike graph by ${\overline{g}}_n$. On $\{{\tau }_n=n-B_n\}$, it agrees with the graph already used above.

This continuation supplies an unconditioned finite crossing path and its regular conditional likelihood. We use ${\overline{g}}_n$ in the globally marked constructions below and suppress the bar. On the good event, all previous estimates apply to this graph.

We next control the boundary contribution to the energy of the increasing matching.

Lemma 2.6

(Boundary current energy). Write $B=B_n$, $R=n-B$, and $q_n=g_n\left(R\right)$. On the left,

$$\mathbb{E}{\int }_0^B{\mathsf{J}}_n{(0,x)}^{2}dx\le C{B}^2.$$

(2.41)

For fixed L, let the ${\mathcal{F}}_R$-measurable localization be

$${\mathcal{A}}_{n,L}:=\{{\tau }_n=R,|q_n|\le L\sqrt{B},|q_n|\le B/8,|K_n\left(R\right)|\le L,B/2\le m_{\pm }\left(R\right)\le 2B\}.$$

(2.42)

Here $m_{\pm }\left(R\right)$ are the two remaining label counts. On this event, for every deterministic $x\in [R,n+2B]$,

$$\mathbb{E}[{\mathsf{J}}_n{(q_n,x)}^2\mid {\mathcal{F}}_R]\le C_{L}B,\mathbb{E}\left[\left.{\int }_R^{n+2B}{\mathsf{J}}_n{(q_n,x)}^{2}dx\right|{\mathcal{F}}_R\right]\le C_L{B}^2.$$

(2.43)

For every prescribed tolerance, first choose A using Theorem 2.5 and then L large enough that the failure probability of ${\mathcal{A}}_{n,L}$ is below that tolerance for all sufficiently large n.

Proof. 

At zero shear the two counts are independent binomials with equal mean x, so $\mathbb{E}{\mathsf{J}}_n{(0,x)}^2\le 2x$, proving (2.41). Conditional on ${\mathcal{F}}_R$, Lemma 2.1 makes the unseen labels independent uniforms on their remaining complements. If these complements have areas $C_{\pm }\left(x\right)$, put ${\rho }_{\pm }=m_{\pm }\left(R\right)/C_{\pm }\left(R\right)$. The constant shear gives

$$C_{\pm }\left(R\right)=B+O\left(\right|q_n\left|\right),|C_{\pm }\left(x\right)-{(n-x)}_{+}|\le 2|q_n|,|{\rho }_{+}-{\rho }_{-}|\le C{\displaystyle \frac{|q_n|+|K_n\left(R\right)|}{B}}.$$

(2.44)

The two remaining populations at x are conditionally independent binomials. Their difference consequently has conditional mean at most $C\left(\right|q_n|+|K_n\left(R\right)\left|\right)$ in absolute value and variance at most $4B$. This proves the first bound in (2.43); conditional Fubini over an interval of length $3B$ proves the second. The localization probability follows from Theorem 2.5, endpoint-current tightness, and the remaining-count bounds. □

Lemma 2.7

(Right interval density). For fixed L, set

$$H=n-(8L\sqrt{B}+1),e_n=60\log n+\frac{1}{4}.$$

(2.45)

On ${\mathcal{A}}_{n,L}$, except for conditional probability at most $16n^{-3}$, every $R\le s\le t\le H$ satisfies

$$N_{\pm }\left((s,t)\right)\ge \frac{1}{8}(t-s)-e_n.$$

(2.46)

Here $N_{\pm }$ count the physical graph intersections of each color.

Proof. 

The strip $[R,H]$ is unclipped by the constant shear. Conditional on ${\mathcal{F}}_R$, each color’s unseen intersections have expected count at least $(b-a)/4$ in every interval $(a,b)\subset [R,H]$: its swept area is $b-a$, its remaining area is at most $2B$, and its remaining population is at least $B/2$. The binomial lower-tail bound therefore gives

$$\mathbb{P}\{N_{\pm }\left((a,b)\right)<(b-a)/8-u\mid {\mathcal{F}}_R\}\le 2e^{-u/12}.$$

(2.47)

Apply this to integer endpoints in $[0,n]$, to both colors, with $u=60\log n$. The union bound is at most $4{(n+1)}^2{n}^{-5}\le 16n^{-3}$. Round the endpoints of an arbitrary $(s,t)$ inward to integers. The lost length is at most two, costing $2/8=1/4$; intervals shorter than two satisfy (2.46) trivially. Every unseen crossing in this strip is an actual graph intersection, so the resulting lower bound also holds for the full graph counts. □

Theorems 2.4 and 2.5 imply, on the joint good event,

$${\int }_B^{n-B}{K}_n{\left(x\right)}^{2}dx=O_{\mathbb{P}}\left(n\right),$$

(2.48)

and the constant extensions make the graph globally spacelike.

2.3. Energy of the Increasing Graph Matching

Let ${\widehat{\xi }}_i,{\widehat{\eta }}_j$ be the unique intersections

$${\widehat{A}}_i\left(g_n\left({\widehat{\xi }}_i\right)\right)={\widehat{\xi }}_i,{\widehat{B}}_j\left(g_n\left({\widehat{\eta }}_j\right)\right)={\widehat{\eta }}_j.$$

(2.49)

Uniqueness follows from $\mathrm{Lip}\left(g_n\right)<1/2$. Match the two intersection lists increasingly, with permutation ${\tau }_n^g$, and set

$$d_i^g=|{\widehat{\eta }}_{{\tau }_n^g\left(i\right)}-{\widehat{\xi }}_i|.$$

(2.50)

Their rank discrepancy is exactly the graph current:

$$\#\{i:{\widehat{\xi }}_i\le x\}-\#\{j:{\widehat{\eta }}_j\le x\}=K_n\left(x\right).$$

(2.51)

Write $l\left(e\right)\le r\left(e\right)$ for an edge’s two endpoints.

Lemma 2.8

(Stopped lower occupation). If a simple counting process N has predictable intensity $c_0\le \Lambda \left(x\right)\le C_0$ before a stopping coordinate τ, then there are $a_0,b_0>0$, depending only on $c_0,C_0$, such that for every stopping coordinate σ and deterministic $u\ge 0$,

$$\mathbb{P}\{N(\sigma +u)-N\left(\sigma \right)\le a_{0}u,\tau \ge \sigma +u\mid {\mathcal{F}}_{\sigma }\}\le e^{-b_{0}u}.$$

(2.52)

Proof. 

Apply the exponential counting-process martingale as in Lemma 2.3, stopping at $\tau$. On the event displayed, the compensator between $\sigma$ and $\sigma +u$ is at least $c_{0}u$. Optional sampling gives the bound with survival inside the probability. □

Lemma 2.9

(Stopped graph rank-gap energy). For two simple point processes started with no unmatched points, suppose both predictable intensities remain in $[c_0,C_0]$ before τ. Their increasing matching satisfies

$$\mathbb{E}\sum _e{\left(\min \left\{r\right(e),\tau \}-l\left(e\right)\right)}_{+}^2\le C\mathbb{E}{\int }_0^{\tau }(1+K{\left(x\right)}^2)dx.$$

(2.53)

The same bound holds when only edges completed before τ are retained.

Proof. 

Charge an edge at its actual left endpoint x. A censored gap longer than u requires $x+u<\tau$, and either $\left|K\left(x\right)\right|>c_{1}u$ or fewer than $c_{2}u$ opposite-color points in $(x,x+u]$. Thus Lemma 2.8 gives

$$\mathbb{P}\{\mathrm{censored}\mathrm{gap}>u\mid {\mathcal{F}}_x\}\le {\mathbf{1}}_{\left\{\right|K\left(x\right)|>c_{1}u\}}+C{e}^{-c_{2}u}.$$

(2.54)

Integration against $2udu$ and summation over genuine left endpoints using the stopped compensators prove (2.53). Censoring endpoints creates no new arrivals at $\tau$. □

Energy before the right boundary.

Stop the zero-shear left process at its first rate exit from $[1/2,2]$, capped at B. This stop reaches B except on an $o\left(1\right)$ event: $N_{\pm }\left(B\right)\le 2B$ has exponentially high probability and $B=o\left(n\right)$. Let $P_n$ be its censored prefix energy, $I_n$ the sum of squared suffixes after B of initial-backlog edges completed by ${\tau }_n$, and $U_n$ the energy of edges starting after B and completed by ${\tau }_n$. These are stopped random variables on the original sample space. Put

$$Z_n^{\mathrm{pre}}:=2P_n+2I_n+U_n.$$

(2.55)

The left bound (2.41) and Lemma 2.9 give $\mathbb{E}{P}_n\le C(B+B^2)$.

For the initial backlog, condition on ${\mathcal{F}}_B$: its ranks have backlogs at most $1,\dots ,|K_n\left(B\right)|$. The same occupation argument gives a squared suffix bound $C(k^2+1)$ for backlog k. Consequently $\mathbb{E}{I}_n\le C(n+\mathbb{E}|K_n\left(B\right){|}^3)$.

For bulk edges the endpoint charging argument in (2.54), now starting at their actual arrivals after B, and Theorem 2.4 give $\mathbb{E}{U}_n\le C(n+\mathbb{E}|K_n\left(B\right){|}^3)$. Since $\mathbb{E}|K_n{\left(B\right)|}^3=O\left(B^{3/2}\right)$ and $B=A\sqrt{n}$,

$$\begin{array}{|c|}\hline \mathbb{E}{Z}_n^{\mathrm{pre}}\le C_{A}n.\\ \hline\end{array}$$

(2.56)

On left and bulk survival, every edge completed by R is controlled by this majorant: an edge crossing B is split there using ${(a+b)}^2\le 2a^2+2b^2$.

Two-cut energy accounting.

For an increasing edge with $R<r\left(e\right)\le H$, let $d_R\left(e\right)=r\left(e\right)-\max \{R,l\left(e\right)\}$. Under (2.46), rank order implies $\frac{1}{8}(r\left(e\right)-x)-e_n\le \left|K_n\left(x\right)\right|$ on this suffix. Integrating and using Young’s inequality yields

$$d_R{\left(e\right)}^2\le 32{\int }_{\max \{R,l(e\left)\right\}}^{r\left(e\right)}\left|K_n\left(x\right)\right|dx+256e_n^2.$$

At each cut the increasing edges all cross in the same direction, and their number is $|K_n\left(x\right)|$. If $N_R$ counts edges with $r\left(e\right)>R$, summing the preceding inequality gives

$$2\sum _{R<r\left(e\right)\le H}{d}_R{\left(e\right)}^2\le 64{\int }_R^H{K}_n{\left(x\right)}^{2}dx+512N_R{e}_n^2.$$

(2.57)

We next bound the left prefixes of edges crossing R. Choose a fixed W so large that each color has at least L arrivals in $[R-W,R]$, outside a probability smaller than any prescribed tolerance on bulk survival. Lemma 2.8, applied to the deterministic subinterval $[R-W,R-W/2]$, proves this assertion. Since $|K_n\left(R\right)|\le L$, rank order then bounds the entire crossing-prefix contribution by

$$2\sum _{l\left(e\right)\le R<r\left(e\right)}{(R-l\left(e\right))}^2\le 2L{W}^2.$$

(2.58)

Splitting at B and R, and keeping full lengths for $r\left(e\right)>H$, gives

$$\sum _e{\left|e\right|}^2\le Z_n^{\mathrm{pre}}+2L{W}^2+64{\int }_R^H{K}_n{\left(x\right)}^{2}dx+512N_R{e}_n^2+E_n^{\mathrm{term}},$$

(2.59)

where $E_n^{\mathrm{term}}:={\sum }_{r\left(e\right)>H}{\left|e\right|}^2$. In particular, (2.59) includes the prefixes of edges crossing the first right cut.

Lemma 2.10

(Terminal localization). Fix $a>0$. On ${\mathcal{A}}_{n,L}$ and the right-density event, also require $|K_n\left(H\right)|\le a\sqrt{B}$. For all sufficiently large n,

$$N_{\mathrm{term}}\le N_R\le 4B,E_n^{\mathrm{term}}\le 4{(10L+8a+9)}^2{B}^2.$$

(2.60)

The added current event fails on ${\mathcal{A}}_{n,L}$ with probability at most $C_L/a^2$.

Proof. 

Each edge counted by $N_R$ uses a remaining endpoint at R, so $N_R\le m_{-}\left(R\right)+m_{+}\left(R\right)\le 4B$. Put $V=8(a\sqrt{B}+e_n)$. For large n, $R\le H-V$ and $e_n\le \sqrt{B}$. The interval bound (2.46) gives at least $a\sqrt{B}\ge \left|K_n\left(H\right)\right|$ points of each color in $(H-V,H)$. Thus every edge crossing H starts after $H-V$; all right endpoints are at most $n+2|q_n|$. Every terminal length is consequently at most

$$n+2|q_n|-H+V\le (10L+8a+9)\sqrt{B}.$$

This proves (2.60). The failure bound for $|K_n\left(H\right)|$ is Chebyshev’s inequality applied to the pointwise conditional second moment in (2.43). □

Theorem 2.11

(Energy of the comparison matching). For every $\epsilon >0$ there are finite $A=A\left(\epsilon \right)$ and $C=C\left(\epsilon \right)$ such that

$$\begin{array}{|c|}\hline \mathbb{P}\left\{{\displaystyle \sum _{i=1}^n}{|{\widehat{\eta }}_{{\tau }_n^g\left(i\right)}-{\widehat{\xi }}_i|}^2\le Cn\right\}\ge 1-\epsilon -o\left(1\right).\\ \hline\end{array}$$

(2.61)

Proof. 

Allocate fixed fractions of $\epsilon$ to Theorem 2.5, the boundary localization, and the backward window in (2.58), choosing $A,L,W$ in that order. Then choose a large enough for Lemma 2.10. The expectation estimates (2.56) and (2.43), with ${\mathcal{A}}_{n,L}\in {\mathcal{F}}_R$, allow deterministic $K_1,K_2$ such that

$$\mathbb{P}\{Z_n^{\mathrm{pre}}>K_{1}n\}<\epsilon /8,\mathbb{P}\left({\mathcal{A}}_{n,L}\cap \left\{{\int }_R^{n+2B}{K}_n{\left(x\right)}^{2}dx>K_2{B}^2\right\}\right)<\epsilon /8.$$

(2.62)

The left-rate and right-grid failures are $o\left(1\right)$. On the remaining events, (2.59)–(2.62), $N_R\le 4B$, and $e_n^2\le B$ give

$$\sum _e{\left|e\right|}^2\le K_{1}n+2L{W}^2+\left[64K_2+2048+4{(10L+8a+9)}^2\right]B^2\le Cn.$$

(2.63)

Here $B^2=A^{2}n$; the terminal $O\left(B^2\right)=O\left(n\right)$ bound suffices. Adding the allocated failure probabilities proves (2.61). □

2.4. Vanishing Normalized Graph Height

Corollary 2.12

(Vanishing normalized graph height). For the feedback graph define

$${\gamma }_n\left(z\right):={\displaystyle \frac{1}{n}}{g}_n\left(nz\right).$$

Then

$$\mathrm{Lip}\left({\gamma }_n\right)\le \lambda$$

(2.64)

deterministically, and for every $\delta >0$,

$$\mathbb{P}\left({\mathcal{G}}_{n,\epsilon }\cap \{\parallel {\gamma }_n{\parallel }_{\infty }>\delta \}\right)⟶0.$$

(2.65)

Proof. 

Rescaling preserves the Lipschitz constant. The graph starts at zero and is stopped no later than its first shear exit. Continuity at that stop and constant extension give the deterministic bound

$$\parallel g_n{\parallel }_{\infty }\le B_n/8,{\parallel {\gamma }_n\parallel }_{\infty }\le {\displaystyle \frac{A}{8\sqrt{n}}}⟶0.$$

On the survival event this stopped graph agrees with the horizon graph used above. This proves (2.65). The same normalized-height bound holds without conditioning for the everywhere-defined stopped continuation. □

The Euclidean assignment is equivalently the minimizer of the residual

$$c_{ij}:=D_{ij}L\left({\displaystyle \frac{S_j-T_i}{D_{ij}}}\right),L\left(s\right):=\sqrt{1+s^2}-1,$$

(2.66)

because the subtracted horizontal term is permutation invariant.

3. Quadratic Localization of the Optimizer

We now transfer the squared-length bound from the comparison matching to the Euclidean optimizer. Subtracting row and column potentials leaves the optimizer unchanged and makes the cost comparable to squared distance along the graph.

Put

$${\xi }_i:={\displaystyle \frac{{\widehat{\xi }}_i}{n}},{\eta }_j:={\displaystyle \frac{{\widehat{\eta }}_j}{n}}.$$

Then

$$T_i+a_i{\gamma }_n\left({\xi }_i\right)={\xi }_i,S_j-b_j{\gamma }_n\left({\eta }_j\right)={\eta }_j.$$

(3.1)

3.1. Exact Feedback Gauge

For

$${\Phi }_n\left(x\right):={\int }_0^x{L}^{\prime }\left({\gamma }_n\left(s\right)\right)ds$$

define row and column potentials

$${\alpha }_i:=a_{i}L\left({\gamma }_n\left({\xi }_i\right)\right)-{\Phi }_n\left({\xi }_i\right),{\beta }_j:=b_{j}L\left({\gamma }_n\left({\eta }_j\right)\right)+{\Phi }_n\left({\eta }_j\right),$$

and the gauged edge cost

$${\Gamma }_{ij}^{\left(n\right)}:=c_{ij}-{\alpha }_i-{\beta }_j.$$

(3.2)

Since ${\sum }_i{\alpha }_i+{\sum }_j{\beta }_j$ is independent of the permutation, the Euclidean optimizer ${\sigma }_n$ also minimizes ${\sum }_i{\Gamma }_{i,\sigma \left(i\right)}^{\left(n\right)}$.

For a fixed pair write

$$d:={\eta }_j-{\xi }_i,u:={\gamma }_n\left({\xi }_i\right),v:={\gamma }_n\left({\eta }_j\right),D:=a_i+b_j,q:={\displaystyle \frac{S_j-T_i}{D}}.$$

By (3.1),

$$q={\displaystyle \frac{d+a_{i}u+b_{j}v}{D}}.$$

(3.3)

Lemma 3.1

(Peano–hinge representation). Assume $d>0$. For $t\in \mathbb{R}$ set

$$\begin{array}{cc}\hfill H_{ij}\left(t\right):=& D{(q-t)}_{+}-a_i{(u-t)}_{+}-b_j{(v-t)}_{+}\hfill \\ \hfill & -\left|\{x\in [{\xi }_i,{\eta }_j]:{\gamma }_n\left(x\right)>t\}\right|.\hfill \end{array}$$

(3.4)

Then

$${\Gamma }_{ij}^{\left(n\right)}={\int }_{\mathbb{R}}{L}^{\prime \prime }\left(t\right)H_{ij}\left(t\right)dt,$$

(3.5)

and

$$H_{ij}\left(t\right)\ge 0\mathit{for}\mathit{all}t.$$

(3.6)

For $d<0$, set

$${\xi }^{\prime }=-{\xi }_i,{\eta }^{\prime }=-{\eta }_j,\tilde{\gamma }\left(x\right)=-{\gamma }_n(-x).$$

Keep the source coefficient $a_i$ attached to ${\xi }^{\prime }$ and the target coefficient $b_j$ attached to ${\eta }^{\prime }$. Then

$$d^{\prime }=-d>0,u^{\prime }=-u,v^{\prime }=-v,q^{\prime }=-q.$$

The reflected gauged functional equals the original one because L is even and $L^{\prime }$ is odd.

Proof. 

The linear functional in L defined by the right side of (3.2) annihilates affine functions. The standard Peano representation modulo affine functions therefore gives

$$\mathcal{T}\left(L\right)={\int }_{\mathbb{R}}{L}^{\prime \prime }\left(t\right)\mathcal{T}\left({(·-t)}_{+}\right)dt$$

for this functional $\mathcal{T}$. Evaluating $\mathcal{T}\left({(·-t)}_{+}\right)$ with (3.3) gives exactly $H_{ij}\left(t\right)$, and hence (3.5).

It remains to prove (3.6). Since

$$|v-u|\le \lambda d,$$

and $a_i\le 1,b_j\le 2,\lambda <1/2$,

$$q-u={\displaystyle \frac{d+b_j(v-u)}{D}}>0,q-v={\displaystyle \frac{d-a_i(v-u)}{D}}>0.$$

(3.7)

Moreover q lies above the whole graph segment. Indeed, if for some $x={\xi }_i+s$ both $u+\lambda s>q$ and $v+\lambda (d-s)>q$, then

$$d>{\displaystyle \frac{(q-u)+(q-v)}{\lambda }}.$$

But, with $r=(v-u)/d$,

$${\displaystyle \frac{(q-u)+(q-v)}{d}}={\displaystyle \frac{2+(b_j-a_i)r}{a_i+b_j}}\ge {\displaystyle \frac{2-\lambda }{3}}>\lambda ,$$

a contradiction.

We prove nonnegativity by considering the possible positions of t. If $t<\min (u,v)$, the first three terms of (3.4) equal d, while the last term is at most d.

If $u\le t<v$, they equal $d-a_i(t-u)$, whereas the $\lambda$-Lipschitz constraint gives

$$|\{{\gamma }_n>t\}|\le d-{\displaystyle \frac{t-u}{\lambda }},$$

and $a_i<1/\lambda$. The case $v\le t<u$ is identical, using $b_j<1/\lambda$.

If $\max (u,v)\le t<q$, then

$$D{(q-t)}_{+}=d-a_i(t-u)-b_j(t-v).$$

If the graph rises above t, it spends at least $(t-u)/\lambda$ before the first up-crossing and at least $(t-v)/\lambda$ after the last down-crossing. Hence

$$|\{{\gamma }_n>t\}|\le {\left[d-{\displaystyle \frac{t-u}{\lambda }}-{\displaystyle \frac{t-v}{\lambda }}\right]}_{+},$$

and again $a_i,b_j<1/\lambda$.

Finally, if $t\ge q$, (3.7) and the fact that q lies above the graph make every term in (3.4) zero. Thus $H_{ij}\ge 0$. □

Let $Q\left(s\right):=s^2/2$, and write ${\Gamma }_{ij}^Q$ for the same gauge functional with L replaced by Q. Lemma 3.1 gives

$${\Gamma }_{ij}^Q={\int }_{\mathbb{R}}{H}_{ij}\left(t\right)dt.$$

(3.8)

Lemma 3.2

(Quadratic spacelike coercivity). For every $\lambda <1/2$ there are $0<c_{\lambda }<C_{\lambda }<\infty$ such that

$$c_{\lambda }{d}^2\le {\Gamma }_{ij}^Q\le C_{\lambda }{d}^2.$$

(3.9)

One may take

$$c_{\lambda }={\displaystyle \frac{(1-\lambda )(1-2\lambda )}{2(3-4\lambda )}}.$$

(3.10)

Proof. 

For $d>0$, put $\Delta \gamma =v-u$ and $r=\Delta \gamma /d$. A direct calculation yields

$${\Gamma }_{ij}^Q={\displaystyle \frac{(d-a_i\Delta \gamma )(d+b_j\Delta \gamma )}{2D}}+{\int }_{{\xi }_i}^{{\eta }_j}\left({\mathrm{Chord}}_{{\xi }_i,{\eta }_j}{\gamma }_n-{\gamma }_n\right)ds.$$

(3.11)

Among $\lambda$-Lipschitz functions with endpoint slope r, the largest possible area above the chord is

$${\displaystyle \frac{{\lambda }^2-r^2}{4\lambda }}{d}^2.$$

(3.12)

Indeed, after subtracting the chord, the deviation h satisfies $h\left(0\right)=h\left(d\right)=0$ and its a.e. derivative lies in $[-\lambda -r,\lambda -r]$. The maximal area is attained by the bang–bang tent with slope $\lambda -r$ up to its peak and $-(\lambda +r)$ afterwards. Its peak height is $({\lambda }^2-r^2)d/\left(2\lambda \right)$, and its triangular area is exactly (3.12).

Next,

$${\displaystyle \frac{(1-ar)(1+br)}{2(a+b)}}$$

is decreasing in both a and b, so its minimum over $a\in [0,1],b\in [1,2]$ occurs at $a=1,b=2$. Therefore

$${\displaystyle \frac{{\Gamma }_{ij}^Q}{d^2}}\ge {\displaystyle \frac{(1-r)(1+2r)}{6}}-{\displaystyle \frac{{\lambda }^2-r^2}{4\lambda }}.$$

The right side is minimized on $[-\lambda ,\lambda ]$ at $r=-\lambda /(3-4\lambda )$, giving (3.10). The upper bound follows immediately from (3.11), $\left|r\right|\le \lambda$, and $|\int (\mathrm{Chord}{\gamma }_n-{\gamma }_n)|\le \lambda d^2/2$.

For $d<0$, the same estimate follows from $x\mapsto -x$, $\tilde{\gamma }\left(x\right)=-{\gamma }_n(-x)$, and the evenness of the quadratic cost. □

Lemma 3.3

(Euclidean coercivity). On the event $\parallel {\gamma }_n{\parallel }_{\infty }\le 1$,

$$2^{-3/2}{c}_{\lambda }{d}^2\le {\Gamma }_{ij}^{\left(n\right)}\le C_{\lambda }{d}^2\mathit{for}\mathit{every}i,j.$$

(3.13)

Proof. 

For the Euclidean L,

$$L^{\prime \prime }\left(t\right)={(1+t^2)}^{-3/2}.$$

The support of the nonnegative kernel in Lemma 3.1 lies between the graph values and q. Here

$$\left|q\right|=\left|{\displaystyle \frac{S_j-T_i}{D}}\right|\le 1,\left|u\right|,\left|v\right|\le 1.$$

Hence $2^{-3/2}\le L^{\prime \prime }\le 1$ on the support. Combining (3.5), (3.8), and Lemma 3.2 proves (3.13). □

Theorem 3.4

(Localization of the Euclidean optimizer). Let ${\sigma }_n$ be the exact Euclidean optimal permutation. Then, for every $\epsilon >0$, the corresponding auxiliary feedback graph satisfies

$$\begin{array}{|c|}\hline \mathbb{P}\left\{{\displaystyle \sum _{i=1}^n}{\left|{\widehat{\eta }}_{{\sigma }_n\left(i\right)}-{\widehat{\xi }}_i\right|}^2\le C_{\epsilon }n\right\}\ge 1-\epsilon -o\left(1\right).\\ \hline\end{array}$$

(3.14)

Proof. 

Fix $\epsilon >0$ and allocate fixed fractions of this tolerance to Theorem 2.11 and Corollary 2.12. On the intersection of their good events, $\parallel {\gamma }_n{\parallel }_{\infty }\le 1$ for all large n, and the optimizer property together with the row/column gauge gives

$$\sum _i{\Gamma }_{i,{\sigma }_n\left(i\right)}^{\left(n\right)}\le \sum _i{\Gamma }_{i,{\tau }_n^g\left(i\right)}^{\left(n\right)}.$$

Apply Lemma 3.3 and multiply by $n^2$:

$$2^{-3/2}{c}_{\lambda }\sum _i|{\widehat{\eta }}_{{\sigma }_n\left(i\right)}-{\widehat{\xi }}_i{|}^2\le C_{\lambda }\sum _i{|{\widehat{\eta }}_{{\tau }_n^g\left(i\right)}-{\widehat{\xi }}_i|}^2.$$

Theorem 2.11 then yields (3.14) after adding the allocated failure probabilities.

□

Joint high-probability event.

For each outer tolerance $\epsilon$, apply the preceding estimates with a fixed finite subdivision of $\epsilon$, and let ${\mathcal{G}}_{n,\epsilon }$ denote their intersection, including the energy event

$${\displaystyle \frac{1}{n}}\sum _i{|{\widehat{\eta }}_{{\sigma }_n\left(i\right)}-{\widehat{\xi }}_i|}^2\le C_{\epsilon }.$$

(3.15)

After relabeling the total tolerance,

$$\mathbb{P}\left({\mathcal{G}}_{n,\epsilon }\right)\ge 1-\epsilon -o\left(1\right).$$

(3.16)

4. Stationary Feedback Environment

We identify the local environment seen from a uniformly chosen point in the bulk. Spatial averaging turns the finite martingale identities into a stationary marked process.

For fixed $\epsilon >0$, choose a deterministic sequence $h_{n,\epsilon }\to \infty$, $h_{n,\epsilon }=o\left(n\right)$, with $h_{n,\epsilon }\ge 2B_n$, such that

$$\mathbb{P}\left({\mathcal{G}}_{n,\epsilon }\cap \left\{\parallel g_{n,\epsilon }{\parallel }_{\infty }>{\displaystyle \frac{h_{n,\epsilon }}{4}}\right\}\right)⟶0.$$

(4.1)

Corollary 2.12 ensures this choice. Set

$$I_{n,\epsilon }:=[h_{n,\epsilon },n-h_{n,\epsilon }].$$

Let $Z_n$ be uniform on $I_{n,\epsilon }$, independently conditional on the finite sample. Recenter the feedback scene at $Z_n$, and normalize the graph height by

$$\left({\theta }_{Z_n}{g}_n\right)\left(x\right):=g_n(Z_n+x)-g_n\left(Z_n\right).$$

At fixed $\epsilon$, write $I_n:=I_{n,\epsilon }$ and ${\mathbf{Q}}_n^{\mathrm{env}}$ temporarily for ${\mathbf{Q}}_{n,\epsilon }^{\mathrm{env}}$ inside local calculations. Define the random empirical environment law

$${\mathbf{Q}}_n^{\mathrm{env}}:={\displaystyle \frac{1}{|I_n|}}{\int }_{I_n}{\delta }_{{\theta }_x{\omega }_n^{\mathrm{env}}}dx.$$

(4.2)

Lemma 4.1

(Spatial martingale closure). Let G be a bounded local functional depending only on the rooted past window $[-R,0]$, let $f:\mathbb{Z}\to \mathbb{R}$ have finite support, and let $t>0$. Write

$${\Psi }_{f,t}\left(\omega \right)=f\left(K\left(t\right)\right)-f\left(K\left(0\right)\right)-{\int }_0^t\mathcal{L}f\left(K(s-)\right)ds,$$

where

$$\mathcal{L}f\left(k\right)=b_k[f(k+1)-f\left(k\right)]+d_k[f(k-1)-f\left(k\right)].$$

Then, for every $\delta >0$,

$$\mathbb{P}\left({\mathcal{G}}_{n,\epsilon }\cap \left\{\left|\int G{\Psi }_{f,t}d{\mathbf{Q}}_{n,\epsilon }^{\mathrm{env}}\right|>\delta \right\}\right)⟶0.$$

(4.3)

The analogous assertions hold for the two marked martingale residuals with bounded mark tests $h_{-}$ and $h_{+}$.

Proof. 

Fix $\epsilon >0$, and let ${\tau }_{n,\epsilon }$ be the stopping coordinate in the feedback construction. Use the compensated current martingale stopped at ${\tau }_{n,\epsilon }$, denoted $M_{n,\tau }^f$. Since $h_{n,\epsilon }\ge 2B_n$, every fixed past and future observation window rooted in $I_n$ lies inside the bulk for all sufficiently large n. For these roots put

$$X_x:=G\left({\theta }_x{\omega }_n\right)[M_{n,\tau }^f(x+t)-M_{n,\tau }^f\left(x\right)].$$

(4.4)

If $x+t\le y$, then $X_{x}G\left({\theta }_y{\omega }_n\right)$ is measurable at time y, so the martingale property gives $\mathbb{E}{X}_x{X}_y=0$. The same holds when $y+t\le x$. Bounded $f,G$ and bounded stopped rates give $\mathbb{E}{X}_x^2\le C_{f,G,t}$ uniformly in $x,n$. Ordinary Fubini and Cauchy–Schwarz therefore give

$$\mathbb{E}{\left|{\displaystyle \frac{1}{|I_n|}}\int X_{x}dx\right|}^2\le {\displaystyle \frac{C_{f,G,t}}{|I_n{|}^2}}\underset{I_n^2}{\int \int }{\mathbf{1}}_{\left\{\right|x-y|<t\}}dxdy\le {\displaystyle \frac{2t{C}_{f,G,t}}{|I_n|}}⟶0.$$

(4.5)

On ${\mathcal{G}}_{n,\epsilon }$, the stopping coordinate is the right bulk endpoint, so the stopped increment equals the desired feedback martingale increment on every admissible rooted window.

The finite generator also differs from $\mathcal{L}$ by uniformly $o\left(1\right)$ on this event. To obtain this estimate at the fixed stopping tolerance, put $q_n=\log n/(13B_n/16)$ and ${\eta }_n=2·33280(\sqrt{q_n}+q_n)$. Then ${\eta }_n\to 0$, and

$$4{\left(33280\right)}^2\log n\le {\eta }_n^2(13B_n/16),2·33280\log n\le {\eta }_n(13B_n/16).$$

The same VC good event therefore controls residual densities by $|r_{\pm }-1|\le {\eta }_n$ throughout the surviving bulk. A mark test bounded by C has drift-integral error at most $2Ct{\eta }_n$. Chebyshev’s inequality gives (4.3).

The marked assertions use the same orthogonality argument; bounded marks and stopped rates bound the second moments of their fixed-window compensated jump counts. All martingale estimates are taken under the original sample law, and then restricted to ${\mathcal{G}}_{n,\epsilon }$. □

Lemma 4.2

(Marked feedback martingale problem). Consider a marked càdlàg process whose current takes values in $\mathbb{Z}$. For every finitely supported $f:\mathbb{Z}\to \mathbb{R}$ and bounded continuous $h_{-},h_{+}:[0,1]\to \mathbb{R}$, require the processes

$$\begin{array}{cc}\hfill M_f\left(t\right):=& f\left(K\right(t\left)\right)-f\left(K\right(0\left)\right)\hfill \\ \hfill & -{\int }_0^t[b_{K(s-)}(f(K(s-)+1)-f\left(K(s-)\right))\hfill \\ \hfill & +d_{K(s-)}(f(K(s-)-1)-f\left(K(s-)\right))]ds\hfill \end{array}$$

(4.6)

and

$$M_{-}\left(t\right):=\sum _{\begin{array}{c}0<s\le t\\ \Delta K\left(s\right)=+1\end{array}}{h}_{-}\left(U_s\right)-{\int }_0^t{\int }_0^1{h}_{-}\left(u\right)[1-\lambda \mathrm{sgn}K(s-)(1-u)]duds,$$

(4.7)

$$M_{+}\left(t\right):=\sum _{\begin{array}{c}0<s\le t\\ \Delta K\left(s\right)=-1\end{array}}{h}_{+}\left(V_s\right)-{\int }_0^t{\int }_0^1{h}_{+}\left(v\right)[1+\lambda \mathrm{sgn}K(s-)(1+v)]dvds$$

(4.8)

are martingales, where

$$b_k=1-{\displaystyle \frac{\lambda }{2}}\mathrm{sgn}k,d_k=1+{\displaystyle \frac{3\lambda }{2}}\mathrm{sgn}k.$$

This marked martingale problem is nonexplosive and uniquely determines the marked feedback path once the law of $K\left(0\right)$ is specified.

Proof. 

Let $\tau$ be the first future event. Conditional on the entire initial past and $K\left(0\right)=k$, stop the compensators before $\tau$. The survival equation and the two marked identities give, with $q_k=b_k+d_k$,

$$\begin{array}{cc}\hfill \mathbb{P}\{\tau \in dt,-\mathrm{event},U\in du\mid {\mathcal{F}}_0\}& =e^{-q_{k}t}[1-\lambda \mathrm{sgn}\left(k\right)(1-u)]dtdu,\hfill \\ \hfill \mathbb{P}\{\tau \in dt,+\mathrm{event},V\in dv\mid {\mathcal{F}}_0\}& =e^{-q_{k}t}[1+\lambda \mathrm{sgn}\left(k\right)(1+v)]dtdv.\hfill \end{array}$$

Optional sampling gives the same conditional kernel after each event. The bound $0<2-\lambda \le q_k\le 2+\lambda$ excludes explosion and ensures each next waiting time is finite. Iterating these conditional holding, colour and mark kernels determines all finite histories and hence the entire future path, given the initial current law. □

Lemma 4.3

(Deterministic sample selection on good events). Fix $\epsilon >0$. Suppose events $E_n\subset {\mathcal{G}}_{n,\epsilon }$ have probability at least $p>0$ along a subsequence. One can select samples from these events and a further increasing subsequence so that all countably many fixed-window spatial martingale residuals vanish, local count second moments are uniformly bounded on each fixed window, and every weak limit has no simultaneous two-colour events. The root-current moment bound and every additional finite collection of errors tending to zero in good-event probability may be imposed simultaneously.

Proof. 

For integer $R\ge 1$, let $N_R$ count both colours in $[-R,R]$ and put $V_{n,R}=\int N_R^{2}d{\mathbf{Q}}_n^{\mathrm{env}}$. Integrating the stopped count estimates over the root gives $\mathbb{E}\left[{\mathbf{1}}_{{\mathcal{G}}_{n,\epsilon }}{V}_{n,R}\right]\le C_{\epsilon ,R}$ for each fixed R and all sufficiently large n. Choose $K_R$ so that $C_{\epsilon ,R}/K_R\le p{2}^{-R-4}$. Markov’s inequality then bounds the probability of each good-event failure $V_{n,R}>K_R$ by $p{2}^{-R-4}$.

For the collision constraints, cover $[-R,R]$ by overlapping open windows of length $O\left({\delta }_{R,l}\right)$. The two-colour stopped factorial moment bound gives probability at most $36{\Delta }^2$ for two events in a window of length $\Delta$. The expected sum of empirical collision masses over the grid is consequently at most $C_R{\delta }_{R,l}$. Require that sum to be at most $\sqrt{{\delta }_{R,l}}$, choosing the meshes to decrease to zero and satisfy ${\sum }_{R,l}{C}_R\sqrt{{\delta }_{R,l}}<p/8$. The same good-event indicator controls the sum over all windows.

For each grid, the condition that every supported continuous test $0\le f\le 1$ has total two-colour mass at most one is closed in the vague topology. Portmanteau passes these grid bounds to a weak limit. Letting $l\to \infty$, and then taking the countable union over R, excludes both multiplicities and cross-colour collisions.

Enumerate the spatial residuals as $R_{n,j}$. At step m, choose $n_m$ sufficiently large that the preceding estimates for the first m constraints apply and the total failure probability of $|R_{n_m,j}|\le 1/m$, $j\le m$, is below $p/8$. Impose also the first m requested vanishing-error bounds. Their combined loss can be made smaller than p, leaving a sample in $E_{n_m}$. Each step imposes a finite prefix of the constraints, and every required observation window eventually fits inside the bulk. For every fixed constraint the resulting sequence satisfies it on a tail.

The root-current bound follows directly from the localization estimate. For a finite permutation, current equals signed crossing count, so with $d_i=|{\widehat{\eta }}_{{\sigma }_n\left(i\right)}-{\widehat{\xi }}_i|$,

$$\int \left|K\left(0\right)\right|d{\mathbf{Q}}_n^{\mathrm{env}}\le {\displaystyle \frac{1}{|I_n|}}\sum _i{d}_i\le {\displaystyle \frac{n+{\sum }_i{d}_i^2}{2|I_n|}}\le {\displaystyle \frac{n(1+C_{\epsilon })}{2|I_n|}}.$$

This gives tightness of the integer root current on every selected sample, directly from (3.15). Together with the local count bounds, compact marks and the $\lambda$-Lipschitz normalized graphs, it proves tightness of the environment laws. The local second-moment bounds give uniform integrability of all fixed-window counts. □

Theorem 4.4

(The stationary feedback limit). Let ${\mathbf{Q}}_{n,\epsilon }^{\mathrm{env}}$ be the empirical environment law generated by the feedback graph associated with the outer tolerance ε. Then, for every compatible metric $d_{\mathrm{loc}}$ on local environment laws and every $\delta >0$,

$$\begin{array}{|c|}\hline \underset{\epsilon \downarrow 0}{\lim }\underset{n\to \infty }{\mathrm{lim\; sup}}\mathbb{P}\left\{d_{\mathrm{loc}}\left({\mathbf{Q}}_{n,\epsilon }^{\mathrm{env}},{\mathfrak{F}}_{\lambda }\right)>\delta \right\}=0.\\ \hline\end{array}$$

(4.9)

where ${\mathfrak{F}}_{\lambda }$ is the unique two-sided stationary marked feedback law described below. Its current $K\left(x\right)\in \mathbb{Z}$ is the birth–death chain

$$k\to k+1\mathrm{at}\mathrm{rate}{b}_k:=1-{\displaystyle \frac{\lambda }{2}}\mathrm{sgn}k,$$

$$k\to k-1\mathrm{at}\mathrm{rate}{d}_k:=1+{\displaystyle \frac{3\lambda }{2}}\mathrm{sgn}k.$$

(4.10)

Conditional on $K(x-)=k$, a $+1$ jump carries a source mark with intensity

$$[1-\lambda \mathrm{sgn}(k\left)\right(1-u\left)\right]du,$$

(4.11)

and a $-1$ jump carries a target mark with intensity

$$[1+\lambda \mathrm{sgn}(k\left)\right(1+v\left)\right]dv.$$

(4.12)

The graph cocycle is

$$g\left(y\right)-g\left(x\right)={\int }_x^y\lambda \mathrm{sgn}K\left(s\right)ds.$$

(4.13)

The invariant current law is

$${\varpi }_0={\displaystyle \frac{\lambda }{1+\lambda }},$$

(4.14)

$${\varpi }_k={\displaystyle \frac{{\varpi }_0}{1+3\lambda /2}}{\left({\displaystyle \frac{1-\lambda /2}{1+3\lambda /2}}\right)}^{k-1},k\ge 1,$$

(4.15)

$${\varpi }_{-k}={\displaystyle \frac{{\varpi }_0}{1+\lambda /2}}{\left({\displaystyle \frac{1-3\lambda /2}{1+\lambda /2}}\right)}^{k-1},k\ge 1.$$

(4.16)

In particular,

$${\mathbb{E}}_{\varpi }\left[\mathrm{sgn}K\right]=0,$$

(4.17)

and the stationary source and target mark intensities are exactly

$$du,dv.$$

(4.18)

Proof. 

Fix $\epsilon >0$. If good-event convergence to ${\mathfrak{F}}_{\lambda }$ failed, there would be a subsequence and events

$$E_n={\mathcal{G}}_{n,\epsilon }\cap \{d_{\mathrm{loc}}({\mathbf{Q}}_{n,\epsilon }^{\mathrm{env}},{\mathfrak{F}}_{\lambda })>\delta \}$$

of probability at least $p>0$. Apply Lemma 4.3 inside these events. It gives a deterministic sequence of empirical laws with a weakly convergent subsequence, denoted $Q_n\Rightarrow Q$, while retaining the deviation and all the residual, count-moment and collision bounds. Here $\epsilon ,p$ are fixed before any moment cutoff is chosen.

Fixed cuts and stationarity. Write $L_n=\left|I_n\right|$. Uniform rooting and Fubini give, for either colour, any fixed $x\in \mathbb{R}$ and $r>0$,

$$\int N\left([x-r,x+r]\right)d{Q}_n\le 2r{\displaystyle \frac{n}{L_n}}.$$

Take a continuous compactly supported test equal to one on the marked fibre above x, supported in this interval, and bounded by one. Its counting integral is a nonnegative continuous vague observable. Bounded truncation and Portmanteau pass the upper moment bound to Q. Since $n/L_n$ is bounded, letting $r\downarrow 0$ proves that the fixed cut x has no event Q-almost surely. This conclusion holds separately for each fixed x.

For a fixed shift h, rerooting is continuous when cuts $0,h$ are vacant. The exact finite-root estimate

$$|Q_n(F\circ {\theta }_h)-Q_n\left(F\right){|\le 2\parallel F\parallel }_{\infty }\left|h\right|/L_n$$

therefore passes to the limit and proves stationarity of Q.

The full marked martingale problem. Choose a countable algebra of bounded environment-past cylinders, generated by continuous marked-count tests in rational past windows and the integer root current. The observation filtration consists of the initial current and the subsequently revealed two-colour counts. Use a countable uniformly dense family of continuous mark tests, current tests ${\mathbf{1}}_{\left\{k\right\}}$, and rational time intervals. These cylinders depend on the revealed environment past and are measurable in the original filtration. The selected samples satisfy all their residual identities asymptotically by Lemma 4.1.

Simplicity and vacancy of deterministic endpoints imply continuity of each residual observable Q-almost everywhere. A fixed residual is bounded by $C(1+N_R)$ for some finite R. The selected bounds $\int N_R^{2}d{Q}_n\le K_R$ give uniform integrability, so its expectation converges and the limiting residual has expectation zero.

For a fixed rational interval, the residual defines a finite signed measure on the past sigma-field. It vanishes against the generating cylinder algebra, hence against every bounded past-measurable function by the monotone-class theorem. Increasing the past window recovers the entire past.

The error from uniformly approximating a mark test is bounded in $L^1$ by the uniform approximation error times ${\mathbb{E}}_Q{N}_R+CR$. Right continuity and the same second-moment bound extend rational times to all real times. Thus the marked compensators are martingales; their $L^1$ right continuity also permits usual augmentation.

The invariant current and two-sided points. Stationarity and the current indicator identities give the balance equations for $p_k=Q\{K\left(0\right)=k\}$. The probability flux $p_k{b}_k-p_{k+1}{d}_{k+1}$ is constant in k. Since $p_k\to 0$ and the rates are bounded, this constant is zero. The resulting detailed balance gives (4.15)–(4.16), and normalization gives (4.14) and (4.17).

Lemma 4.2 now identifies the joint law of $K\left(0\right)$ and both complete marked configurations on $(0,\infty )$. By stationarity, shift any finite window into this positive half-line and recover its current by signed event counts. Increasing the windows identifies the joint law of the root current and both full two-sided configurations with that of ${\mathfrak{F}}_{\lambda }$.

The original graph. The finite recursive curve has right derivative $\lambda \mathrm{sgn}{K}_n$ on every surviving bulk segment, including the post-event right neighbourhood. The fundamental theorem of calculus therefore gives the exact original-graph identity

$$g_n(B_n+t)-g_n(B_n+s)={\int }_s^t\lambda \mathrm{sgn}{K}_n(B_n+u)du,0\le s\le t\le n-2B_n.$$

Every fixed rooted window is eventually in this range. For each rational q, the bounded residual

$$\min \left\{1,\left|g\left(q\right)-{\int }_0^q\lambda \mathrm{sgn}K\left(s\right)ds\right|\right\}$$

is continuous at every scene with vacant root cut. Indeed, on a compact interval the current stabilizes locally away from its event coordinates; a small open neighbourhood of those coordinates has arbitrarily small Lebesgue measure, while the velocity is bounded by $\lambda$. This proves continuity of the integral observable.

The finite residual is identically zero for all sufficiently large n; the almost-everywhere continuous mapping theorem therefore makes it zero Q-almost surely. Intersect the resulting full-measure sets over rational q, then use continuity of both primitives to obtain (4.13) for all endpoints. Thus the normalized graph is a measurable function of the identified current and point configurations. The whole law Q equals ${\mathfrak{F}}_{\lambda }$, contradicting the retained deviation.

We have proved, for each fixed $\epsilon ,\delta >0$,

$$\mathbb{P}\left({\mathcal{G}}_{n,\epsilon }\cap \{d_{\mathrm{loc}}({\mathbf{Q}}_{n,\epsilon }^{\mathrm{env}},{\mathfrak{F}}_{\lambda })>\delta \}\right)⟶0.$$

(4.19)

Adding the probability of ${\mathcal{G}}_{n,\epsilon }^c$ yields a limiting failure probability at most $\epsilon$. We then let $\epsilon \downarrow 0$. Averaging the identified marked intensities and using (4.17) gives (4.18). □

Corollary 4.5

(Local limiting feedback cost). Normalize $g\left(0\right)=0$. If a source event is $(\xi ,u)$, a target event is $(\eta ,v)$, and

$$a=1-u,b=1+v,D=a+b,d=\eta -\xi ,$$

then the rescaled finite gauged costs converge uniformly on bounded rooted configurations to

$$\begin{array}{|c|}\hline \begin{array}{cc}\hfill {\mathfrak{c}}_{\lambda }((\xi ,u),(\eta ,v))=& {\displaystyle \frac{d^2}{2D}}-{\displaystyle \frac{ab}{2D}}{\left(g\left(\eta \right)-g\left(\xi \right)\right)}^2\hfill \\ \hfill & +d{\displaystyle \frac{ag\left(\xi \right)+bg\left(\eta \right)}{D}}-{\int }_{\xi }^{\eta }g\left(s\right)ds.\hfill \end{array}\\ \hline\end{array}$$

(4.20)

Moreover

$$c_{\lambda }^{\prime }{d}^2\le {\mathfrak{c}}_{\lambda }\le C_{\lambda }^{\prime }{d}^2.$$

(4.21)

Proof. 

On a fixed rank-scale rooted window, along every good-event subsequential limit, the normalized graph height at the root tends to zero by Corollary 2.12, while graph increments and rank displacements are $O\left(1\right)$.

Apply the exact gauge decomposition of Section 3 and Taylor-expand only the curvature coefficient: $L^{\prime \prime }\left(o\left(1\right)\right)\to 1$. After multiplication by $n^2$, the affine and common-height terms cancel through the gauge, and the remaining quadratic expression is exactly (4.20). Uniformity on compact rooted configurations follows from the uniform continuity of $L^{\prime \prime }$ near zero and the $\lambda$-Lipschitz bound. The coercivity is the rank-scale form of Lemmas 3.2–3.3. □

5. Compactness of Selected Matchings

Let ${\mathcal{G}}_{\lambda ,2}^0$ be the class of all stationary perfect locally minimizing zero-relative-flux matching laws $\Sigma$ over ${\mathfrak{F}}_{\lambda }$ whose source Palm law has finite feedback-edge second moment,

$$\int d_0^{2}d{\Sigma }^{-}<\infty .$$

(5.1)

This is exactly the class reached by the compactness theorem below, because Lemma 5.4 transfers the localization bound (3.15).

Throughout the local lemmas in this section, $\epsilon >0$ is fixed. Whenever a good-event failure has probability bounded away from zero, Lemma 4.3 selects deterministic samples with the required moment and residual bounds. The local limits below are taken along these selected sequences. We use the same selection to recover mark averages over the full sample and over macroscopic blocks.

For $\Sigma \in {\mathcal{G}}_{\lambda ,2}^0$, let

$$\kappa [\Sigma ]:={\mathrm{Law}}_{\mathrm{source}\mathrm{Palm}}(U_0,V_{\Sigma \left(0\right)}).$$

Lemma 5.1

(Bulk intersection count). At fixed ε, with $I_n:=I_{n,\epsilon }$, for every $\delta >0$,

$$\mathbb{P}\left({\mathcal{G}}_{n,\epsilon }\cap \left\{{\displaystyle \frac{1}{n}}\#\{i:{\widehat{\xi }}_i\notin I_n\}>\delta \right\}\right)\to 0,$$

(5.2)

and the same statement holds for targets. Hence the source- and target-intersection intensities converge to one along every positive-probability good-event subsequence.

Proof. 

On ${\mathcal{G}}_{n,\epsilon }$ outside the $o\left(1\right)$ event in (4.1),

$$|{\widehat{\xi }}_i-n{T}_i|=a_i\left|g_n\left({\widehat{\xi }}_i\right)\right|\le {\displaystyle \frac{h_{n,\epsilon }}{4}},$$

and

$$|{\widehat{\eta }}_j-n{S}_j|=b_j\left|g_n\left({\widehat{\eta }}_j\right)\right|\le {\displaystyle \frac{h_{n,\epsilon }}{2}}.$$

Hence a source with

$$T_i\in \left[{\displaystyle \frac{2h_{n,\epsilon }}{n}},1-{\displaystyle \frac{2h_{n,\epsilon }}{n}}\right]$$

must satisfy ${\widehat{\xi }}_i\in I_n$, and the analogous statement holds for targets. Therefore the excluded source and target counts are bounded by the numbers of $T_i$’s and $S_j$’s in physical boundary layers of total width $4h_{n,\epsilon }/n=o\left(1\right)$. Their expectations are $O\left(h_{n,\epsilon }\right)=o\left(n\right)$; Markov’s inequality gives (5.2). □

Let ${\omega }_n$ denote the finite feedback environment together with the optimal matching ${\sigma }_n$. Define its empirical Lebesgue-rooted scene law

$${\mathbf{Q}}_n:={\displaystyle \frac{1}{|I_n|}}{\int }_{I_n}{\delta }_{{\theta }_x{\omega }_n}dx.$$

(5.3)

At fixed $\epsilon$, this is the abbreviation ${\mathbf{Q}}_n={\mathbf{Q}}_{n,\epsilon }$; similarly ${\omega }_n={\omega }_{n,\epsilon }$ and $I_n=I_{n,\epsilon }$ in the local lemmas below. A rooted scene records the two marked point measures, the marked edge measure, the graph cocycle normalized by $g\left(0\right)=0$, and the integer root current $K\left(0\right)$. The three point measures carry their vague topologies, the graph carries the local uniform topology, and $K\left(0\right)\in \mathbb{Z}$ carries the discrete topology. The current at other coordinates is recovered from $K\left(0\right)$ and the signed counts of source and target events between the root and that coordinate.

For a source event i, write

$$d_i:=|{\widehat{\eta }}_{{\sigma }_n\left(i\right)}-{\widehat{\xi }}_i|.$$

Lemma 5.2

(No loss of partners). Fix the outer tolerance $\epsilon >0$. For every $\delta >0$,

$$\underset{R\to \infty }{\lim }\underset{n\to \infty }{\mathrm{lim\; sup}}\mathbb{P}\left({\mathcal{G}}_{n,\epsilon }\cap \left\{{\displaystyle \frac{1}{n}}\#\{i:d_i>R\}>\delta \right\}\right)=0.$$

(5.4)

The same holds when rooted from target endpoints. Consequently no partner is lost in any positive-probability good-event subsequential limit.

Proof. 

Put

$$Z_n:={\displaystyle \frac{1}{n}}\sum _i{d}_i^2.$$

By the energy bound (3.15), on ${\mathcal{G}}_{n,\epsilon }$ one has $Z_n\le C_{\epsilon }$ outside an $o\left(1\right)$ event. Since

$${\displaystyle \frac{1}{n}}\#\{i:d_i>R\}\le {\displaystyle \frac{Z_n}{R^2}},$$

(5.4) follows. The same edge set, viewed from target endpoints, gives the target statement. □

Lemma 5.3

(Projective Palm readout). Let fixed-sample rooted matching laws $Q_m$, with root intervals $A_m=(a_m,b_m]$ and lengths $L_m$, converge weakly to a stationary perfect matching law Q with environment ${\mathfrak{F}}_{\lambda }$. Suppose the source count in $A_m$, divided by $L_m$, tends to one, and the source count within distance one of its two endpoints, divided by $L_m$, tends to zero. Then for every continuous mark test $f:{[0,1]}^2\to \mathbb{R}$,

$${\displaystyle \frac{1}{L_m}}\sum _{{\widehat{\xi }}_i\in A_m}f(U_i,V_{{\sigma }_m\left(i\right)})⟶\int fd\kappa \left[Q\right].$$

(5.5)

These two count assumptions follow from Corollary 2.12 and the uniform height empirical-distribution bound for both $A_m=I_m$ and fixed macroscopic blocks $A_m=n_{m}J⋐(0,n_m)$.

Proof. 

Choose $\phi \ge 0$, continuous, supported in $[-1,1]$, with $\int \phi =1$, and set $W_m\left(x\right)={\int }_{A_m}\phi (x-z)dz$. Fubini gives the exact identity

$$\int \sum _{(x,u;y,v)\in \mathrm{edges}}\phi \left(x\right)f(u,v)d{Q}_m={\displaystyle \frac{1}{L_m}}\sum _i{W}_m\left({\widehat{\xi }}_i\right)f(U_i,V_{{\sigma }_m\left(i\right)}).$$

Since $0\le W_m\le 1$, its difference from the block sum is bounded by ${\parallel f\parallel }_{\infty }$ times the normalized endpoint-strip source count.

First take $0\le f\le 1$. Multiply the edge test by continuous compact cutoffs in both endpoint positions, increasing to one. The corresponding nonnegative continuous vague observables and Portmanteau, followed by monotone convergence, give

$$\int fd\kappa \left[Q\right]\le \underset{m}{\mathrm{lim\; inf}}{\displaystyle \frac{1}{L_m}}\sum _i{W}_m\left({\widehat{\xi }}_i\right)f(U_i,V_{{\sigma }_m\left(i\right)}).$$

Campbell’s formula identifies the limiting integral. Perfectness identifies the test $f=1$ with the source statistic, whose finite mass tends to one by the two count assumptions. Applying the same lower bound to $1-f$ gives the reverse upper bound. Shifting and scaling proves the assertion for every continuous real f.

To verify the two count assumptions, put ${\zeta }_m={\parallel g_m\parallel }_{\infty }/n_m$ and ${\Delta }_m={\sup }_{t\in [0,1]}|F_m\left(t\right)-t|$, where $F_m$ is the empirical source-height distribution. The original intersection equation gives $|{\widehat{\xi }}_i/n_m-T_i|\le {\zeta }_m$. Consequently each block count differs from its physical interval mass by $O(n_m{\zeta }_m+n_m{\Delta }_m+1)$, and the two endpoint strips contain at most $O(n_m{\zeta }_m+n_m{\Delta }_m+1)$ sources. Corollary 2.12 and the VC bound give ${\zeta }_m,{\Delta }_m\to 0$ on the selected samples. Since $L_m/n_m$ tends to a positive constant, both required normalized conclusions follow. □

Lemma 5.4

(Inherited source-Palm second moment). Let $\mathbf{Q}$ be any positive-probability good-event subsequential scene-law limit at fixed ε. Then its source Palm law satisfies

$$\begin{array}{|c|}\hline \int d_0^{2}d{\mathbf{Q}}^{-}<\infty .\\ \hline\end{array}$$

(5.6)

More precisely, the deterministic localization bound on the selected samples gives

$$\int d_0^{2}d{\mathbf{Q}}^{-}\le C_{\epsilon }.$$

(5.7)

Proof. 

Work along the selected deterministic subsequence, where $Z_n=n^{-1}{\sum }_i{d}_i^2\le C_{\epsilon }$. Choose a nonnegative continuous $\phi$, supported in $[-1,1]$, with integral one, and continuous cutoffs ${\chi }_R\left(d\right)\uparrow 1$, supported where $d\le R+1$. The edge statistic

$$H_R\left(\omega \right)=\sum _{(x,u;y,v)\in \mathrm{edges}}{\phi \left(x\right)|y-x|}^2{\chi }_R\left(\right|y-x\left|\right)$$

is a nonnegative continuous vague observable. Exact Fubini and $0\le {\int }_{I_n}\phi ({\widehat{\xi }}_i-z)dz\le 1$ give $\int H_{R}d{Q}_n\le (n/|I_n\left|\right)Z_n$. Portmanteau and Campbell’s formula imply

$$\int d_0^2{\chi }_R\left(d_0\right)d{Q}^{-}=\int H_{R}dQ\le C_{\epsilon }.$$

Monotone convergence as $R\to \infty$ proves (5.7). For roots in $nJ$, the bound is $C_{\epsilon }/\left|J\right|$, still finite. □

Lemma 5.5

(Finite crossing number by Campbell mass transport). Let

$$N_{\mathrm{cross}}\left(0\right)=\#\left\{\mathit{selected}\mathit{edges}\mathit{crossing}0\right\}.$$

Then

$$\int N_{\mathrm{cross}}\left(0\right)d\mathbf{Q}=\int \left|d_0\right|d{\mathbf{Q}}^{-}<\infty ,$$

(5.8)

hence $N_{\mathrm{cross}}\left(0\right)<\infty$ almost surely. Moreover

$$\int N_{>R,\mathrm{cross}}\left(0\right)d\mathbf{Q}=\int \left|d_0\right|{\mathbf{1}}_{\left\{\right|d_0|>R\}}d{\mathbf{Q}}^{-}\le {\displaystyle \frac{1}{R}}\int d_0^{2}d{\mathbf{Q}}^{-}.$$

(5.9)

Proof. 

For each selected edge, send unit density along the interval joining its two feedback endpoints. Stationary Campbell/mass transport gives

$$\int N_{\mathrm{cross}}\left(0\right)d\mathbf{Q}=\int \left|d_0\right|d{\mathbf{Q}}^{-}.$$

Cauchy–Schwarz and Lemma 5.4 make the right side finite. Restricting the same transport to $|d_0|>R$ gives the first equality in (5.9), and

$$|d_0|{\mathbf{1}}_{\left\{\right|d_0|>R\}}\le d_0^2/R$$

gives the estimate. □

Lemma 5.6

(Zero-relative-flux closure). Every selected good-event local limit of ${\mathbf{Q}}_n$ has zero relative flux with respect to its feedback current.

Proof. 

At every non-event cut x, define

$$C_n\left(x\right)=\#\{i:{\widehat{\xi }}_i\le x<{\widehat{\eta }}_{{\sigma }_n\left(i\right)}\}-\#\{i:{\widehat{\eta }}_{{\sigma }_n\left(i\right)}\le x<{\widehat{\xi }}_i\}.$$

For every finite permutation,

$$\begin{array}{|c|}\hline C_n\left(x\right)=K_n\left(x\right)\\ \hline\end{array}$$

(5.10)

identically.

For $R<\infty$, let $C_n^{\left(R\right)}\left(x\right)$ count only selected crossing edges of feedback length at most R. This is a local observable. Since $C_n=K_n$,

$$\begin{array}{cc}\hfill {\mathbf{Q}}_n\{C_n^{\left(R\right)}\left(0\right)\ne K_n\left(0\right)\}& \le {\displaystyle \frac{1}{|I_n|}}\sum _{d_i>R}{d}_i\hfill \\ \hfill & \le {\displaystyle \frac{n}{|I_n|}}{\displaystyle \frac{C_{\epsilon }}{R}}.\hfill \end{array}$$

(5.11)

Along the selected deterministic subsequence, take $Q_n\Rightarrow Q$. For every continuity radius R, vacancy of the root cut and local edge convergence make the truncated integer-current discrepancy an almost-everywhere continuous observable. Passing (5.11) to the limit gives

$$\mathbf{Q}\{C^{\left(R\right)}\left(0\right)\ne K\left(0\right)\}\le {\displaystyle \frac{C_{\epsilon }}{R}}.$$

(5.12)

By Lemma 5.5,

$$N_{>R,\mathrm{cross}}\left(0\right)⟶0\mathrm{in}{L}^1\left(\mathbf{Q}\right),$$

and therefore

$$C^{\left(R\right)}\left(0\right)⟶C\left(0\right)$$

in probability (indeed eventually almost surely for the finite realized set of crossing edges). Hence

$$\begin{array}{cc}\hfill \mathbf{Q}\left\{C\right(0)\ne K(0\left)\right\}& \le \mathbf{Q}\{C\left(0\right)\ne C^{\left(R\right)}\left(0\right)\}\hfill \\ & +\mathbf{Q}\{C^{\left(R\right)}\left(0\right)\ne K\left(0\right)\}⟶0\hfill \end{array}$$

along continuity radii $R\to \infty$. It follows that

$$C\left(0\right)=K\left(0\right)\mathbf{Q}-\mathrm{a}.\mathrm{s}.$$

(5.13)

At a source event both C and K jump by $+1$, at a target event both jump by $-1$, and between events both are constant. Thus $C-K$ is a spatially constant integer, and (5.13) forces it to be zero. □

Theorem 5.7

(Compactness of locally minimizing matchings). Let ${\mathbf{Q}}_{n,\epsilon }$ be the empirical matching-scene law built from the feedback graph associated with outer tolerance ε. For every compatible metric $d_{\mathrm{scene}}$ on local matching-scene laws and every $\delta >0$,

$$\underset{\epsilon \downarrow 0}{\lim }\underset{n\to \infty }{\mathrm{lim\; sup}}\mathbb{P}\left\{d_{\mathrm{scene}}\left({\mathbf{Q}}_{n,\epsilon },{\mathcal{G}}_{\lambda ,2}^0\right)>\delta \right\}=0.$$

(5.14)

Every good-event subsequential limit $\mathbf{Q}$ has the following properties:

(i)

$\mathbf{Q}$ is translation invariant and its environment marginal is ${\mathfrak{F}}_{\lambda }$;

(ii)

the matching is a perfect matching of the source and target events;

(iii)

the matching is locally minimizing for ${\mathfrak{c}}_{\lambda }$ against every finite alternating modification;

(iv)

the matching has zero relative flux;

(v)

if ${\mathbf{Q}}^{-}$ denotes its source Palm law, then the pushforward

$$(\xi =0,u;\eta ,v=\mathit{mark}\mathit{of}\mathit{the}\mathit{matched}\mathit{target})⟼(u,v)$$

is exactly a subsequential weak limit of

$${\kappa }_n:={\displaystyle \frac{1}{n}}{\displaystyle \sum _{i=1}^n}{\delta }_{(U_i,V_{{\sigma }_n\left(i\right)})}.$$

(5.15)

In particular both marginals of that projective law are uniform.

Proof. 

Tightness and stationarity. Fix $\epsilon >0$ and work on ${\mathcal{G}}_{n,\epsilon }$. The environment marginal is tight by Theorem 4.4. Theorem 3.4 gives a bound $Z_n\le C_{\epsilon }$ outside an additional $o\left(1\right)$ event. On this event, Lemma 5.2 implies that, in any fixed rooted window, the probability of seeing a source whose partner lies farther than R is $O_{\epsilon }\left(R^{-2}\right)$; the same holds for targets. More explicitly, for every fixed observation radius A,

$${\mathbf{Q}}_n\{\exists \mathrm{a}\mathrm{source}\mathrm{in}[-A,A]\mathrm{with}{d}_i>R\}\le {\displaystyle \frac{(2A+o(1\left)\right)}{|I_n|}}\#\{i:d_i>R\}\le (2A+o\left(1\right)){\displaystyle \frac{Z_n}{R^2}}.$$

The same estimate holds for targets. Together with local finiteness of the point processes this gives tightness of the matching relation.

Approximate translation invariance follows directly from the averaging definition (5.3): for a bounded local F and fixed h,

$$\left|\int F\circ {\theta }_{h}d{\mathbf{Q}}_n-\int Fd{\mathbf{Q}}_n\right|\le 2{\parallel F\parallel }_{\infty }{\displaystyle \frac{\left|h\right|}{|I_n|}}+o\left(1\right).$$

Hence every good-event limit is translation invariant. Its environment marginal is ${\mathfrak{F}}_{\lambda }$ by Theorem 4.4 on the good event.

Perfectness. Lemma 5.2 rules out escape of a partner to infinity under either source or target Palm rooting. Since every finite state is a bijection, the local limit is therefore a bijection as well.

Palm law, inherited second moment, and zero relative flux. Lemma 5.4 transfers the finite-sample second-moment bound directly through compact edge tests, and Lemmas 5.5–5.6 give a finite crossing current with zero relative-flux constant.

Local optimality. For each m, let $F_m$ be the closure of the union of all rooted optimizer scenes from the selected samples with index at least m. Each finite law has mass one on $F_m$ eventually; closed-set Portmanteau gives $Q\left(F_m\right)=1$, hence $Q\left({\bigcap }_m{F}_m\right)=1$.

Suppose a simple perfect scene in this intersection admitted a strict finite improvement by $\delta >0$. All involved endpoints lie in a bounded window. Simplicity and strictness give a neighbourhood of that scene in which the same finite modification remains improving. The compact-window cost convergence of Corollary 4.5 then excludes every sufficiently late optimizer scene from that neighbourhood: its gauged cost improvement would also improve the original Euclidean assignment. This contradicts membership in the corresponding $F_m$. Thus local minimality holds Q-almost surely.

Projective Palm identification. The mass-conservation readout in Lemma 5.3 proves (v). The first marginal is uniform by the identified source intensity. For the second, send each target-mark indicator along the matched edge. Stationary mass transport identifies its outgoing source-window mean with the target-window intensity, which is also uniform.

We now prove the outer statement. If, for some fixed $\epsilon ,\delta >0$, the event

$$E_n:={\mathcal{G}}_{n,\epsilon }\cap \left\{d_{\mathrm{scene}}({\mathbf{Q}}_{n,\epsilon },{\mathcal{G}}_{\lambda ,2}^0)>\delta \right\}$$

had probability bounded below along a subsequence, apply Lemma 4.3 inside $E_n$, retaining the localization bound and all the fixed-window constraints used above.

The resulting deterministic scene laws are tight. A convergent subsequence has the identified environment, perfectness, finite Palm second moment, local minimality and zero relative flux, so its limit belongs to ${\mathcal{G}}_{\lambda ,2}^0$, contradicting the retained distance $>\delta$. Thus $\mathbb{P}\left(E_n\right)\to 0$. Since $\mathbb{P}\left({\mathcal{G}}_{n,\epsilon }^c\right)\le \epsilon +o\left(1\right)$,

$$\underset{n\to \infty }{\mathrm{lim\; sup}}\mathbb{P}\left\{d_{\mathrm{scene}}\left({\mathbf{Q}}_{n,\epsilon },{\mathcal{G}}_{\lambda ,2}^0\right)>\delta \right\}\le \epsilon ,$$

and $\epsilon \downarrow 0$ gives (5.14). □

6. Kinetic Gauge and Finite Genericity

Let $(K,g,{\mathcal{N}}^{-},{\mathcal{N}}^{+})$ have the stationary feedback law ${\mathfrak{F}}_{\lambda }$. The graph g is specified up to an additive constant by

$$g\left(y\right)-g\left(x\right)={\int }_x^y\lambda \mathrm{sgn}K\left(s\right)ds.$$

(6.1)

Write source marks as $a=1-u\in [0,1]$ and target marks as $b=1+v\in [1,2]$. With respect to the natural left-to-right filtration, their predictable marked intensities are

$$[1-a\lambda \mathrm{sgn}K(x-\left)\right]dadx,[1+b\lambda \mathrm{sgn}K(x-\left)\right]dbdx.$$

(6.2)

6.1. Exact Kinetic Action

For any representative of the graph cocycle, let

$$X=\xi -ag\left(\xi \right),Y=\eta +bg\left(\eta \right),\mathsf{h}((X,a),(Y,b)):={\displaystyle \frac{{(Y-X)}^2}{2(a+b)}}.$$

(6.3)

Choose an absolutely continuous primitive $G^{\prime }=g$.

Proposition 6.1

(Exact kinetic gauge). The limiting feedback cost (4.20) satisfies

$$\begin{array}{|c|}\hline {\mathfrak{c}}_{\lambda }((\xi ,a),(\eta ,b))=\mathsf{h}((X,a),(Y,b))+R(\xi ,a)+C(\eta ,b),\\ \hline\end{array}$$

(6.4)

where

$$R(\xi ,a)=G\left(\xi \right)-{\displaystyle \frac{a}{2}}g{\left(\xi \right)}^2,C(\eta ,b)=-G\left(\eta \right)-{\displaystyle \frac{b}{2}}g{\left(\eta \right)}^2.$$

(6.5)

Thus finite alternating comparisons for ${\mathfrak{c}}_{\lambda }$ are exactly the finite alternating comparisons for $\mathsf{h}$.

Proof. 

Put $p=g\left(\xi \right)$, $q=g\left(\eta \right)$, $d=\eta -\xi$, and $D=a+b$. Since $Y-X=d+ap+bq$, direct expansion gives

$${\mathfrak{c}}_{\lambda }-\mathsf{h}=-{\displaystyle \frac{ab{(q-p)}^2+{(ap+bq)}^2}{2D}}-{\int }_{\xi }^{\eta }g\left(s\right)ds.$$

The identity

$$ab{(q-p)}^2+{(ap+bq)}^2=D(a{p}^2+b{q}^2)$$

reduces this to (6.5). Row and column terms cancel under every finite alternating rematching. □

Lemma 6.2

(No finite alternating ties). Almost surely under the feedback environment law, no finite alternating cycle has equal costs for its two pairings.

Proof. 

Condition on the jump coordinates, jump types, and all marks except one source mark a on a given cycle. The feedback path g is fixed by the coordinates and jump types. The two targets competing at that source have distinct marks $b_1,b_2$ almost surely.

By the exact kinetic gauge, the cycle cost difference, as a function of a, is a fixed constant plus the difference of the two rational edge costs. The latter have distinct poles at $-b_1,-b_2$. The numerator at $-b_j$ is the square of

$${\eta }_j-\xi +b_j\left(g\left({\eta }_j\right)-g\left(\xi \right)\right),$$

which is nonzero: the map $x\mapsto x+b_j(g\left(x\right)-g\left(\xi \right))$ is strictly increasing and equals $\xi$ only at $x=\xi$, while event coordinates are distinct. Hence the cost difference is not identically zero. Its zero set in the admissible interval for a is finite.

Conditional on the unmarked jump path, the source mark at pre-jump state k has density

$${\displaystyle \frac{1-a\lambda \mathrm{sgn}k}{1-(\lambda /2)\mathrm{sgn}k}},0<a<1,$$

and marks at different jumps are conditionally independent. This density is bounded and strictly positive, so the tie has probability zero. Taking the union over the countably many finite label cycles proves the claim. □

7. Regenerative Finite-Energy Uniqueness

The compactness theorem supplies stationary, locally minimizing, zero-relative-flux matching laws with finite source-Palm second moment. We prove uniqueness by finding a cut that two such matchings leave empty. To obtain this cut, we condition on a long remaining zero-current hold at a fixed spatial origin. Regeneration describes the conditional environment, and an energy iteration controls the edges that could cross the hold. The finite source-Palm second moment closes the iteration.

For a stationary phase $\Sigma$, write $d_e$ for the feedback-coordinate length of a selected edge e. For a cut x and bounded interval I, put

$$T_{\Sigma }\left(x\right):=\#\{e:x\in I_e\},S_{\Sigma }\left(x\right):=\sum _{e:x\in I_e}\left|d_e\right|,$$

(7.1)

$${\mathcal{H}}_{\Sigma }\left(I\right):={\int }_I\left(T_{\Sigma }\left(x\right)+S_{\Sigma }\left(x\right)\right)dx,$$

(7.2)

where $I_e$ is the closed interval joining the two feedback endpoints of e.

Lemma 7.1

(Mean local crossing energy). For every $\Sigma \in {\mathcal{G}}_{\lambda ,2}^0$, there is a finite constant $C_{\Sigma }$ such that

$$\mathbb{E}{\mathcal{H}}_{\Sigma }\left(I\right)=\left|I\right|\left({\mathbb{E}}_{{\Sigma }^{-}}\left|d_0\right|+{\mathbb{E}}_{{\Sigma }^{-}}{d}_0^2\right)\le C_{\Sigma }\left|I\right|.$$

(7.3)

Proof. 

For each selected edge send unit density along its feedback interval. Stationary Campbell mass transport gives

$$\mathbb{E}{T}_{\Sigma }\left(0\right)={\mathbb{E}}_{{\Sigma }^{-}}\left|d_0\right|.$$

If an edge of length $\left|d\right|$ sends mass density $\left|d\right|$ along the same interval, the mass through the origin is $S_{\Sigma }\left(0\right)$, and Campbell gives

$$\mathbb{E}{S}_{\Sigma }\left(0\right)={\mathbb{E}}_{{\Sigma }^{-}}{d}_0^2.$$

Integrate over I and use Cauchy–Schwarz together with the definition of the finite source-Palm second moment. □

7.1. Return-Cell Regeneration

Let $\mathcal{R}$ be the point process of jump coordinates at which the stationary feedback current jumps from a nonzero state into zero. Under the Palm law ${\mathbb{P}}^{\mathcal{R}}$, enumerate

$$\cdots <R_{-1}<R_0=0<R_1<R_2<\cdots .$$

(7.4)

For event-count purposes, the j-th return cell is the path segment $(R_j,R_{j+1}]$: it starts immediately after the return jump at $R_j$, includes the zero holding time following $R_j$, all subsequent excursion events, and the return jump at $R_{j+1}$. Let $L_j=R_{j+1}-R_j$, let $N_j$ count the source and target jumps in $(R_j,R_{j+1}]$, and write

$$Z_j:=\mathrm{its}\mathrm{initial}\mathrm{zero}\mathrm{hold},E_j:=L_j-Z_j.$$

The marked excursion consists of all subsequent holding increments, jump types and marks, starting with the departure from zero and ending at the next return. Its duration is $E_j$.

Lemma 7.2

(IID return cells with exponential moments). Under ${\mathbb{P}}^{\mathcal{R}}$, the marked return cells are iid. The field ${\left(Z_j\right)}_{j\in \mathbb{Z}}$ consists of iid $\mathrm{Exp}\left(2\right)$ variables independent of the entire marked excursion field. There exists ${\theta }_0>0$ such that

$${\mathbb{E}}^{\mathcal{R}}{e}^{{\theta }_0{N}_0}<\infty ,{\mathbb{E}}^{\mathcal{R}}{e}^{{\theta }_0{L}_0}<\infty .$$

(7.5)

Every return cell contains equally many source and target events.

Proof. 

At state zero the total jump rate is two, so its waiting time is exponential of rate two and independent of the departure type and mark. The strong Markov property at the departure makes the subsequent marked excursion independent of that waiting time.

After leaving zero, the sign of the current cannot change before returning to zero. On the positive half-line the embedded jump chain moves away from zero with probability

$$p_{+}={\displaystyle \frac{1-\lambda /2}{(1-\lambda /2)+(1+3\lambda /2)}}={\displaystyle \frac{1-\lambda /2}{2+\lambda }}<{\displaystyle \frac{1}{2}}.$$

On the negative half-line it moves away from zero with probability

$$p_{-}={\displaystyle \frac{1-3\lambda /2}{(1+\lambda /2)+(1-3\lambda /2)}}={\displaystyle \frac{1-3\lambda /2}{2-\lambda }}<{\displaystyle \frac{1}{2}}.$$

Thus the number of jumps in an excursion from zero back to zero has an exponential moment. The holding rate is always one of $2,2+\lambda ,2-\lambda$, hence is bounded below by $2-\lambda >0$. Conditional on the jump count, the cell length is dominated by a sum of that many iid exponentials with rate $2-\lambda$, plus the initial zero holding time. Choosing ${\theta }_0>0$ sufficiently small gives the second exponential moment in (7.5).

The strong Markov property at each jump into zero gives independence and identical distribution of the forward cells. The stationary Palm construction gives the same law in both directions.

Immediately after $R_j$ the current is zero, and after the included return jump at $R_{j+1}$ it is zero again. Hence the net number of $+1$ minus $-1$ jumps in $(R_j,R_{j+1}]$ is zero, exactly equality of the source and target counts. □

For a return cell, match its source events increasingly to its target events in feedback coordinate, and let $W_j$ be the resulting feedback cost.

Lemma 7.3

(Comparison rewards excluding the zero hold). There are iid nonnegative random variables

$$V_j=C_{\lambda }{N}_j{E}_j^2$$

which dominate the cellwise increasing comparison rewards $W_j$, are independent of the entire field ${\left(Z_j\right)}_{j\in \mathbb{Z}}$, and satisfy

$${\mathbb{P}}^{\mathcal{R}}\{V_j>t\}\le C{e}^{-c{t}^{1/3}}.$$

(7.6)

Proof. 

The cell convention is $(R_j,R_{j+1}]$, so the return event at $R_j$ is not one of the events matched inside cell j. Every matched vertex in that cell lies in $[R_j+Z_j,R_{j+1}]$. Each comparison edge therefore has length at most $E_j$. The upper bound in (4.21) gives $W_j\le C_{\lambda }{N}_j{E}_j^2$. The variables on the right are functions only of the marked excursions. They are iid and independent of all zero holds.

Both $N_j$ and $E_j$ have exponential moments by the proof of Lemma 7.2. If $N_j{E}_j^2>t/C_{\lambda }$, either $N_j$ or $E_j$ exceeds a constant multiple of $t^{1/3}$, proving (7.6). □

Lemma 7.4

(The zero hold seen from a fixed spatial origin). Under the stationary environment law ${\mathfrak{F}}_{\lambda }$, conditional on $K\left(0\right)=0$, let $A_0$ be the time since the last return to zero and $B_0$ the time until the next departure. Then $A_0,B_0$ are independent $\mathrm{Exp}\left(2\right)$ variables. Reindex the cells so this zero hold belongs to cell zero. The entire marked excursion field and all other zero holds have their ordinary return-Palm joint law, independently of $(A_0,B_0)$.

Proof. 

Let ${\rho }_{\mathcal{R}}>0$ be the return intensity. The intervals $[R_j,R_{j+1})$ partition the line. Campbell’s formula applied to this partition gives the stationary renewal inversion

$$\mathbb{E}F\left(\omega \right)={\rho }_{\mathcal{R}}{\mathbb{E}}^{\mathcal{R}}{\int }_0^{L_0}F\left({\theta }_t\omega \right)dt$$

(7.7)

for every nonnegative measurable environment function F. Indeed, integrate $F\left({\theta }_x\omega \right)$ over $[0,1]$, split the integral over the return intervals, and use stationarity and Campbell; Tonelli removes the spatial unit-window indicator. In particular, ${\rho }_{\mathcal{R}}{\mathbb{E}}^{\mathcal{R}}{L}_0=1$.

Write $\Xi$ for the marked excursions and the holds ${\left(Z_j\right)}_{j\ne 0}$, all indexed relative to the return starting the distinguished cell. Lemma 7.2 makes $\Xi$ independent of $Z_0$. Except at event coordinates, $K\left(t\right)=0$ in cell zero exactly when $0<t<Z_0$. For any nonnegative measurable $\varphi$, inversion and the density $2e^{-2z}$ of $Z_0$ therefore give

$$\begin{array}{cc}\hfill \mathbb{E}\left[{\mathbf{1}}_{\left\{K\right(0)=0\}}\varphi (A_0,B_0,\Xi )\right]& ={\rho }_{\mathcal{R}}{\mathbb{E}}^{\mathcal{R}}{\int }_0^{Z_0}\varphi (t,Z_0-t,\Xi )dt\hfill \\ \hfill & ={\rho }_{\mathcal{R}}{\int }_0^{\infty }{\int }_0^{\infty }2e^{-2(a+b)}{\mathbb{E}}^{\mathcal{R}}\varphi (a,b,\Xi )dadb.\hfill \end{array}$$

(7.8)

Here $\Xi$ on the left is read after reindexing at the last return; it does not depend on the position within the zero hold. Taking $\varphi =1$ yields $\mathbb{P}\{K\left(0\right)=0\}={\rho }_{\mathcal{R}}/2=\varpi \left(0\right)>0$. Dividing (7.8) by this probability gives the product density $4e^{-2a}{e}^{-2b}$ times the original law of $\Xi$, proving the full joint assertion for the environment law. □

7.2. Conditional Environmental Regularity

For $h\ge 1$, define an event in the original spatial stationary law by

$${\mathsf{A}}_h:=\{K\left(0\right)=0,h\le B_0\le h+1\},p_h:=\mathbb{P}\left({\mathsf{A}}_h\right)=\varpi \left(0\right)e^{-2h}(1-e^{-2})>0.$$

(7.9)

For any interval I of length L and center c, let $I^{*}$ be its concentric length-$4L$ enlargement and set

$$J^{-}=[c-2L,c-L],J^{+}=[c+L,c+2L].$$

Given constants $C_0,c_0>0$, let $\mathsf{R}\left(I\right)$ mean that

(i)

between every pair of non-event zero-current cuts $x^{-}\in J^{-}$, $x^{+}\in J^{+}$, there are at most $C_{0}L$ source events;

(ii)

the sum of $V_j$ over the complete return-cell event sets between any such pair is at most $C_{0}L$;

(iii)

each of $J^{-},J^{+}$ has zero-current occupation at least $c_{0}L$.

The counts of the two colours between these cuts agree, because those events form a union of complete return-cell event sets.

Lemma 7.5

(Environmental regularity conditional on a long residual hold). There are constants $C_0,c_0,C,L_{*}$, depending only on the feedback law, such that, if I has center $h/2$ and length $L\ge \max \{L_{*},16(h+1)\}$, then

$$\mathbb{P}\{\mathsf{R}{\left(I\right)}^c\mid {\mathsf{A}}_h\}\le {\displaystyle \frac{C}{L}}.$$

(7.10)

The constants do not depend on h or on any matching phase.

Proof. 

By Lemma 7.4, conditional on ${\mathsf{A}}_h$, the age $A_0$ remains a rate-two exponential, independent of $B_0$ and of the ordinary excursion and neighboring-cell data.

Collapse $[0,B_0]$, leaving the negative half-line fixed and translating the subsequent environment left by $B_0$. In the auxiliary environment the distinguished return is at $-A_0$, its departure is at zero, and its next return is at $E_0$. The distinguished cell now has zero hold $A_0$, so the entire indexed cell array has the ordinary iid Palm law, independently of h and $B_0$. The original matching and its conditional law are retained throughout.

It suffices to control the partial sums of four nonnegative cell observables: length $L_j$, zero hold $Z_j$, event count $N_j$, and reward $V_j$. Write ${\mu }_X=\mathbb{E}{X}_0$ for any of these observables. All have finite second moments, and ${\mu }_L,{\mu }_Z>0$. Put $k=\lfloor L/M\rfloor$. For each of the two index directions, let $S_X\left(n\right)$ be the sum of the first n cell observables in that direction. For fixed integers $M,H$, Chebyshev’s inequality and a finite union bound give

$$\mathbb{P}\left\{\exists X,m:|S_X\left(mk\right)-{\mu }_{X}mk|>k,0\le m\le H\right\}\le C_H/k.$$

On the complementary event, monotonicity between consecutive grid indices gives

$$|S_X\left(n\right)-{\mu }_{X}n|\le (1+{\mu }_X)k,0\le n\le Hk,$$

in both directions. This interpolation constant is independent of H.

Differences of these bounds also control individual boundary cells by $Ck$. In particular $A_0\le L_0\le Ck$, so the auxiliary return coordinates differ from their linear means by at most $Ck$.

Choose M first, large enough that these errors are a small fraction of L, and then choose H, depending on M, so that the controlled returns extend beyond $[-3L,3L]$. Increase $L_{*}$ so that $k\ge L/\left(2M\right)$. The event-count and reward sums then bound the total counts and rewards of cells meeting this window by $CL$.

For a subinterval of length $\ell \ge L/2$ in the window, discarding its two boundary cells loses at most $Ck$ in length. Subtracting the length and zero-hold partial sums over the remaining complete cells gives zero occupation at least

$${\displaystyle \frac{{\mu }_Z}{{\mu }_L}}\ell -Ck\ge c_{0}L,$$

after the preceding choice of M, with constants depending only on the cell law. Event coordinates have zero Lebesgue measure. The exceptional probability is at most $C_H/k\le C^{\prime }/L$.

To transfer these estimates back, put $c=h/2$. Then $c\le L/32$ and $B_0\le L/16$. A pair of original search cuts corresponds in the auxiliary environment to $x^{-}$ and $x^{+}-B_0$, both lying in $[-3L,3L]$. The event sets and their cell rewards are unchanged, so the preceding upper bounds give (i) and (ii).

For the lower occupation bounds, use the fixed auxiliary intervals

$$K^{-}=J^{-},K^{+}=[c+L,c+2L-L/16].$$

Both lie in $[-3L,3L]$ and have length at least $L/2$. The negative interval is unchanged when the hold is restored, while $K^{+}+B_0\subset J^{+}$ for every $0\le B_0\le L/16$. Translation preserves its zero occupation, proving (iii) in the original environment. Thus all three estimates are uniform over the conditioned values $B_0\in [h,h+1]$. □

7.3. The Deterministic Energy Iteration

Lemma 7.6

(Zero-current surgery). Let x be a cut which is not an event coordinate, with $K\left(x\right)=0$, and suppose

$$T_{\Sigma }\left(x\right)\le \Theta ,S_{\Sigma }\left(x\right)\le \Theta .$$

(7.11)

Delete the selected edges crossing x. The unmatched vertices on either side of the cut can be paired separately on that side, and the total feedback cost of all new repair edges is at most

$$C_{\lambda }{\Theta }^3.$$

(7.12)

Proof. 

Zero relative flux gives oriented crossing current $K\left(x\right)=0$, so the unmatched vertices on each side have equal source and target counts. There are at most $T_{\Sigma }\left(x\right)\le \Theta$ crossing edges.

Every endpoint of a crossing edge is at distance at most the length of that edge from the cut, and every individual crossing length is at most $S_{\Sigma }\left(x\right)\le \Theta$. Hence all unmatched endpoints on a fixed side lie within distance $\Theta$ of the cut. Pairing them arbitrarily produces at most $\Theta$ new edges, each of length at most $2\Theta$. The upper coercivity bound in (4.21) gives (7.12). □

Lemma 7.7

(One-step bootstrap). For fixed $C_0,c_0>0$, there are constants $C_B,r_{*},L_{*}$ such that the following holds. Let $L\ge L_{*}$, $r\ge r_{*}$, and choose $\Theta >0$ satisfying

$$\Theta <L/4,C_B{\Theta }^3\le rL/2.$$

(7.13)

On the regular environment event $\mathsf{R}\left(I\right)$,

$${\mathcal{H}}_{\Sigma }\left(I\right)>rL⟹{\mathcal{H}}_{\Sigma }\left(I^{*}\right)>{\displaystyle \frac{c_0}{8}}\Theta \left|I^{*}\right|.$$

(7.14)

Proof. 

Suppose first that each search interval contains a zero-current cut $x^{-}\in J^{-}$, $x^{+}\in J^{+}$, chosen away from the countable event coordinates, satisfying

$$T_{\Sigma }\left(x^{\pm }\right)\le \Theta ,S_{\Sigma }\left(x^{\pm }\right)\le \Theta .$$

(7.15)

Because the cuts are at distance at least $L/2>2\Theta$ from I, an edge crossing one of the cuts cannot meet I. Also no edge can cross both cuts.

Delete all selected edges meeting either cut and every selected edge with both endpoints between the cuts. Match all source and target events between the cuts by the cellwise increasing environmental matching. Each non-event zero cut lies in an initial zero hold, so the events between the cuts are a union of complete return-cell event sets $(R_j,R_{j+1}]$, each with equal colour counts. Repair the unmatched exterior endpoints separately at the two cuts using Lemma 7.6. The resulting perfect matching differs from $\Sigma$ on finitely many edges.

Local minimality, the lower coercivity in (4.21), the environmental reward bound, and the two repair estimates give

$$\sum _{e:I_e\cap I\ne ⌀}{d}_e^2\le C\left(L+{\Theta }^3\right).$$

(7.16)

At most $CL$ selected edges meet I, since their source endpoints lie between the cuts. Using $\left|d\right|\le (1+d^2)/2$, we obtain

$${\mathcal{H}}_{\Sigma }\left(I\right)\le \sum _{e:I_e\cap I\ne ⌀}\left(\right|d_e|+d_e^2)\le {\displaystyle \frac{CL}{2}}+{\displaystyle \frac{3}{2}}\sum _{e:I_e\cap I\ne ⌀}{d}_e^2\le C^{\prime }(L+{\Theta }^3).$$

(7.17)

Take $C_B\ge C^{\prime }$ and $r_{*}\ge 2C_B$. The cubic condition gives

$${\mathcal{H}}_{\Sigma }\left(I\right)\le C_{B}L+C_B{\Theta }^3\le rL.$$

(7.18)

Thus ${\mathcal{H}}_{\Sigma }\left(I\right)>rL$ is impossible if both good cuts exist.

Hence on that event at least one search interval contains no zero-current cut satisfying (7.15). On the regular environment event, the zero set in that search interval has Lebesgue measure at least $c_{0}L$. At almost every point of that zero set, $T_{\Sigma }+S_{\Sigma }>\Theta$. Therefore

$${\mathcal{H}}_{\Sigma }\left(I^{*}\right)\ge c_0\Theta L={\displaystyle \frac{c_0}{4}}\Theta \left|I^{*}\right|.$$

This is strictly larger than $(c_0/8)\Theta \left|I^{*}\right|$, proving (7.14). □

Theorem 7.8

(Local energy conditional on a long zero hold). There are constants $A,C,L_{*}$ such that, for every $\Sigma \in {\mathcal{G}}_{\lambda ,2}^0$, every $h\ge 1$, and every $L\ge \max \{L_{*},16(h+1)\}$, if I has center $h/2$ and length L, then

$$\mathbb{P}\left\{{\mathcal{H}}_{\Sigma }\left(I\right)>AL\mid {\mathsf{A}}_h\right\}\le {\displaystyle \frac{C}{L}}.$$

(7.19)

Proof. 

Use the constants from Lemmas 7.5 and 7.7. Choose $A\ge r_{*}$ and increase $L_{*}$ so that, for every $L\ge L_{*}$,

$${\displaystyle \frac{16A}{c_0}}<{\displaystyle \frac{L}{4}},C_B{\left({\displaystyle \frac{16A}{c_0}}\right)}^3\le {\displaystyle \frac{AL}{2}}.$$

Let $I_n$ be concentric intervals centered at $h/2$, with

$$L_n=4^{n}L,r_n=2^{n}A,{\Theta }_n={\displaystyle \frac{16r_n}{c_0}}.$$

The cubic inequality persists at every scale, since $r_n^2/L_n=A^2/L$; also ${\Theta }_n/L_n$ decreases. The one-step conclusion has exactly the next threshold, because $(c_0/8){\Theta }_n=r_{n+1}$. Hence

$$\{{\mathcal{H}}_{\Sigma }\left(I_n\right)>r_n{L}_n\}\subset \{{\mathcal{H}}_{\Sigma }\left(I_{n+1}\right)>r_{n+1}{L}_{n+1}\}\cup \left\{\mathsf{R}{\left(I_n\right)}^c\right\}.$$

This deterministic inclusion holds under the conditional joint law. Iteration gives

$$\begin{array}{cc}\hfill \mathbb{P}\{{\mathcal{H}}_{\Sigma }\left(I_0\right)>AL\mid {\mathsf{A}}_h\}\le & \mathbb{P}\{{\mathcal{H}}_{\Sigma }\left(I_N\right)>r_N{L}_N\mid {\mathsf{A}}_h\}\hfill \\ \hfill & +C{\displaystyle \sum _{n=0}^{N-1}}{\displaystyle \frac{1}{4^{n}L}}.\hfill \end{array}$$

(7.20)

By Lemma 7.1 and Markov’s inequality, the first term is at most

$${\displaystyle \frac{C_{\Sigma }}{p_h{r}_N}}⟶0.$$

Here h is fixed, so $p_h>0$ is fixed. The remaining series is $4/\left(3L\right)$, proving (7.19) after changing C. Thus the conditioning factor $p_h^{-1}$ affects only the vanishing terminal term. □

Corollary 7.9

(A common empty cut in any stationary joining). Let $({\Sigma }^{+},{\Sigma }^{-})$ be a stationary common-environment joining of two elements of ${\mathcal{G}}_{\lambda ,2}^0$. In its original spatial stationary law, there is positive probability of a cut crossed by neither matching.

Proof. 

Take $L=16(h+1)$ and I centered at $h/2$. On ${\mathsf{A}}_h$, there is no event in $(0,h)$. Any edge crossing $h/2$ therefore has length at least h, crosses all of $(0,h)$, and contributes at least $h^2$ to ${\int }_0^h{S}_{\Sigma }\left(x\right)dx$.

For sufficiently large fixed h, one has $h^2>AL$ and the sum of the two bounds in (7.19) is less than $1/2$. Conditional on ${\mathsf{A}}_h$, the union bound then gives probability at least $1/2$ that neither matching crosses $h/2$. Its unconditional spatial probability is at least $p_h/2>0$. □

7.4. Disagreement Components and Uniqueness

Lemma 7.10

(Every stationary disagreement component spans both infinities). In a stationary common-environment joining of two elements of ${\mathcal{G}}_{\lambda ,2}^0$, almost surely every component of the symmetric difference has one end tending to $-\infty$ and one end tending to $+\infty$. Consequently, a common empty cut implies equality of the entire matchings.

Proof. 

After common edges are removed, every remaining vertex has degree two. Thus each component is a finite alternating cycle or a double ray. The first possibility is excluded by local minimality and Lemma 6.2: switching the cycle in each matching would otherwise force its two costs to be equal.

Lemma 5.5 holds simultaneously at all rational cuts. Every end of a double ray therefore tends to one of the two spatial infinities. Otherwise local finiteness would fail in a bounded interval, or that ray would traverse a rational cut infinitely often.

If both ends tended right, the component would have a unique leftmost vertex. Send one unit of mass from each vertex of every such component to its leftmost vertex. This transport is translation covariant and has outgoing intensity at most the finite total vertex intensity. Each recipient, however, receives infinite mass. The stationary mass-transport identity excludes every such recipient almost surely. The analogous transport to a rightmost vertex excludes two left ends.

Every remaining component thus crosses every cut. If both matchings have no edge crossing one cut, no disagreement component can exist. □

Theorem 7.11

(Regenerative uniqueness at finite energy). Any stationary common-environment joining of two matchings in ${\mathcal{G}}_{\lambda ,2}^0$ is supported on equal matchings.

Proof. 

Suppose the invariant event $D=\{{\Sigma }^{+}\ne {\Sigma }^{-}\}$ has positive probability, and condition the stationary joining on D. The feedback environment is ergodic, as follows from its iid return cells. Conditioning an extension over an ergodic environment on a translation-invariant event leaves its environmental marginal unchanged. Indeed, for a bounded environmental test function f, invariance gives

$$\mathbb{E}\left[{\mathbf{1}}_{D}f\right]=\mathbb{E}\left[{\mathbf{1}}_D{\displaystyle \frac{1}{2T}}{\int }_{-T}^{T}f\circ {\theta }_{x}dx\right]⟶\mathbb{P}\left(D\right)\mathbb{E}f$$

by the environmental ergodic theorem and bounded convergence.

The conditioned matching marginals remain stationary, locally minimizing and of zero relative flux. Their source-Palm second moments remain finite: the corresponding unnormalized stationary Palm integrals are bounded by the original integrals divided by $\mathbb{P}\left(D\right)$, and the environmental source intensity is unchanged.

Corollary 7.9 therefore applies to this conditioned joining and gives positive spatial probability of a common empty cut. Lemma 7.10 then forces the matchings to coincide, contradicting D. Thus $\mathbb{P}\left(D\right)=0$. □

Corollary 7.12

(A unique matching kernel over the environment). There is at most one stationary matching law in ${\mathcal{G}}_{\lambda ,2}^0$. When such a law exists, its conditional matching kernel given the feedback environment is almost surely a point mass.

Proof. 

Join any two regular conditional matching kernels independently over their common feedback environment. Stationarity of the joint laws implies stationarity of this relative independent joining. Theorem 7.11 makes the two samples equal almost surely. Applying the same assertion to two conditional copies of one law shows that its kernel is a point mass. Applying it to two different laws shows that their kernels coincide almost surely. □

Theorem 7.13

(Projective uniqueness). Every $\Sigma \in {\mathcal{G}}_{\lambda ,2}^0$ has the same source-Palm projective law $\kappa [\Sigma ]={\kappa }_{\infty }$.

Proof. 

The class is nonempty by the good-event compactness theorem. Corollary 7.12 identifies its stationary matching law, and hence its source-Palm projective law. Its first mark marginal is uniform by the source intensity (4.18).

For the second, send ${\mathbf{1}}_B\left(v\right)$ along every matched edge to its target. Stationary mass transport and perfectness identify the source-window outgoing mass with the target-window count carrying marks in B. Its expectation is $\left|B\right|$ by the target intensity (4.18). Thus the second marginal is also uniform. □

8. Macroscopic Reconstruction

8.1. Vertical Concentration

Proposition 8.1

(Vertical diagonal concentration). For the Euclidean optimizer,

$$\begin{array}{|c|}\hline {\displaystyle \frac{1}{n}}{\displaystyle \sum _{i=1}^n}|S_{{\sigma }_n\left(i\right)}-T_i|\stackrel{\mathbb{P}}{\to }0.\\ \hline\end{array}$$

(8.1)

Proof. 

For a permutation $\sigma$, set

$${\mathcal{E}}_n\left(\sigma \right)={\displaystyle \frac{1}{n}}\sum _i{D}_{i,\sigma \left(i\right)}L\left({\displaystyle \frac{S_{\sigma \left(i\right)}-T_i}{D_{i,\sigma \left(i\right)}}}\right).$$

The Euclidean objective differs from ${\mathcal{E}}_n$ by the permutation-independent average of $D_{ij}=2+V_j-U_i$. Let ${\rho }_n$ match the order statistics of T and S. By DKW,

$$\underset{i}{\max }|T_{\left(i\right)}-S_{\left(i\right)}|=O_{\mathbb{P}}\left(n^{-1/2}\right).$$

Since $L\left(z\right)\le z^2/2$,

$${\mathcal{E}}_n\left({\sigma }_n\right)\le {\mathcal{E}}_n\left({\rho }_n\right)\le {\displaystyle \frac{1}{2n}}\sum _i{|T_{\left(i\right)}-S_{\left(i\right)}|}^2=O_{\mathbb{P}}\left(n^{-1}\right).$$

For $\left|z\right|\le 1$,

$$L\left(z\right)\ge {\displaystyle \frac{z^2}{1+\sqrt{2}}}.$$

Because $1\le D_{ij}\le 3$,

$${\mathcal{E}}_n\left({\sigma }_n\right)\ge {\displaystyle \frac{1}{3(1+\sqrt{2})}}{\displaystyle \frac{1}{n}}\sum _i{|S_{{\sigma }_n\left(i\right)}-T_i|}^2.$$

Hence the average squared vertical displacement is $O_{\mathbb{P}}\left(n^{-1}\right)$, and Cauchy–Schwarz gives (8.1). □

8.2. Block Identification and Reconstruction

Let ${\kappa }_{\infty }$ be the common projective law from Theorem 7.13.

Lemma 8.2

(Macroscopic block identification). Let $J=[a,b]⋐(0,1)$ be fixed. For every bounded continuous $f:{[0,1]}^2\to \mathbb{R}$,

$${\displaystyle \frac{1}{n}}{\displaystyle \sum _{i=1}^n}{\mathbf{1}}_{\{T_i\in J\}}f(U_i,V_{{\sigma }_n\left(i\right)})\stackrel{\mathbb{P}}{\to }\left|J\right|\int fd{\kappa }_{\infty }.$$

(8.2)

Proof. 

Fix $\epsilon >0$ and use its original feedback graph. Suppose for some $\delta >0$ the events

$${\mathcal{G}}_{n,\epsilon }\cap \left\{\left|{\displaystyle \frac{1}{n}}\sum _{{\widehat{\xi }}_i\in nJ}f(U_i,V_{{\sigma }_n\left(i\right)})-\left|J\right|\int fd{\kappa }_{\infty }\right|>\delta \right\}$$

have probability bounded below along a subsequence. Apply Lemma 4.3 within these events, now with roots in $nJ$. The same stopped estimates apply: all fixed windows lie inside the bulk eventually, and the global current and energy bounds acquire the fixed factor ${\left|J\right|}^{-1}$.

The proof of Theorem 5.7 gives a further deterministic weak limit $Q\in {\mathcal{G}}_{\lambda ,2}^0$. Corollary 2.12 and the global VC event give the two count inputs of Lemma 5.3. Its Palm readout and Theorem 7.13 show that the displayed block statistic tends to $\left|J\right|\int fd{\kappa }_{\infty }$, a contradiction. Thus its good-event failure probability tends to zero.

To replace intersection coordinates by physical heights, put ${\zeta }_{n,\epsilon }={\parallel g_{n,\epsilon }\parallel }_{\infty }/n$ and ${\Delta }_n={\sup }_t|F_n\left(t\right)-t|$. The symmetric difference between $\{i:{\widehat{\xi }}_i/n\in J\}$ and $\{i:T_i\in J\}$ is contained in height strips of width ${\zeta }_{n,\epsilon }$ around the two endpoints. Its normalized size is at most

$$4{\zeta }_{n,\epsilon }+4{\Delta }_n+O(1/n)=o_{\mathbb{P}}\left(1\right)\mathrm{on}{\mathcal{G}}_{n,\epsilon }.$$

(8.3)

Multiplication by ${\parallel f\parallel }_{\infty }$ bounds the statistic’s change. The physical block failure probability therefore has limsup at most $\epsilon$. This statistic and its asserted limit are independent of the auxiliary tolerance; letting $\epsilon \downarrow 0$ proves convergence in probability. □

Theorem 8.3

(Blockwise reconstruction of the empirical plan). The empirical plans satisfy

$$\begin{array}{|c|}\hline {\pi }_n\stackrel{\mathbb{P}}{\to }dt{\delta }_t\left(ds\right){\kappa }_{\infty }(du,dv).\\ \hline\end{array}$$

(8.4)

Proof. 

By Lemma 8.2, for every interval $J⋐(0,1)$ with rational endpoints and every bounded continuous $f(u,v)$,

$${\displaystyle \frac{1}{n}}\sum _i{\mathbf{1}}_{\{T_i\in J\}}f(U_i,V_{{\sigma }_n\left(i\right)})\stackrel{\mathbb{P}}{\to }\left|J\right|\int fd{\kappa }_{\infty }.$$

(8.5)

Finite linear combinations of such interval indicators approximate every continuous $\varphi :[0,1]\to \mathbb{R}$ uniformly after controlling arbitrarily small endpoint intervals. Hence

$${\displaystyle \frac{1}{n}}\sum _i\varphi \left(T_i\right)f(U_i,V_{{\sigma }_n\left(i\right)})\stackrel{\mathbb{P}}{\to }\left({\int }_0^1\varphi \left(t\right)dt\right)\left(\int fd{\kappa }_{\infty }\right).$$

(8.6)

By Stone–Weierstrass, finite sums of products $\varphi \left(t\right)f(u,v)$ are uniformly dense in $C\left({[0,1]}^3\right)$. Therefore

$${\displaystyle \frac{1}{n}}\sum _i{\delta }_{(T_i,U_i,V_{{\sigma }_n\left(i\right)})}\stackrel{\mathbb{P}}{\to }dt{\kappa }_{\infty }(du,dv).$$

(8.7)

Finally, for any Lipschitz test $F(t,u,s,v)$,

$$\left|{\displaystyle \frac{1}{n}}\sum _{i}F(T_i,U_i,S_{{\sigma }_n\left(i\right)},V_{{\sigma }_n\left(i\right)})-{\displaystyle \frac{1}{n}}\sum _{i}F(T_i,U_i,T_i,V_{{\sigma }_n\left(i\right)})\right|\le \mathrm{Lip}\left(F\right){\displaystyle \frac{1}{n}}\sum _i|S_{{\sigma }_n\left(i\right)}-T_i|,$$

which tends to zero in probability by Proposition 8.1. Approximation of continuous tests by Lipschitz tests on the compact state space then gives (8.4). □

Remark 8.4 (The feedback parameter is auxiliary).

The empirical plans in Theorem 8.3 depend only on the original samples. Running the argument with two admissible feedback parameters therefore gives two weak limits of the same sequence, which must coincide. Thus ${\kappa }_{\infty }$ is independent of $\lambda$.

8.3. Conclusion of the Selection Proof

Proof of Theorem 1.1.

Theorem 7.13 supplies the unique source-Palm horizontal coupling. Theorem 8.3 combines physical block identification and vertical concentration to give the asserted convergence. The two marginals of ${\kappa }_{\infty }$ are uniform because the finite empirical mark marginals are the laws of all $U_i$’s and all $V_j$’s, respectively. □

Corollary 8.5

(Deterministic projective mark). The original full empirical mark coupling ${\kappa }_n=n^{-1}{\sum }_i{\delta }_{(U_i,V_{{\sigma }_n\left(i\right)})}$ converges weakly in probability to ${\kappa }_{\infty }$.

Proof. 

Apply Theorem 1.1 to tests of the form $F(t,u,s,v)=f(u,v)$, where f is continuous on ${[0,1]}^2$. □