Proof. We first expose one sign of each radial edge with uniform slack in all other constraints. Write \(x=v+ke_1\), \(y=v+(k+1)e_1\) after applying a lattice symmetry. If \(k\ge k_0\), prescribe \(a(x)=h(x)\) and \(a(y)=h(x)+1\). Both points are at distance at least \(k_0\) from the shell, and the change at \(y\) is at most two. Hence, for sufficiently small \(\beta\le\delta/2\), \[|a(z)-h(b)|\le(1-\delta_0)\operatorname{dist}(z,b)+2 \le(1-\beta)\operatorname{dist}(z,b) \quad(z=x,y,\ b\in\partial Q).\] If \(k<k_0\), put \(b_l=v+le_2\) on the compressed ray and \(f_l=h(b_l)-h(v)\). Its distances are \[\operatorname{dist}(x,b_l)=\max(k,l),\qquad \operatorname{dist}(y,b_l)=\max(k+1,l).\] Choose \(\tau=\operatorname{sign}f_1\), and prescribe \(a(x)=h(v)\) and \(a(y)=h(v)+\tau\). For \(l\ge1\), \[s_0-(1-\delta)(l-1)\le\tau f_l\le(1-\delta)l,\] so \[ |\tau-f_l|\le l-\min\{s_0+\delta(l-1),1+\delta l\} \le(1-\min\{s_0,\delta\})l. \tag{9}\] The difference at \(x\) is bounded by \((1-\delta)l\). On the opposite ray, the distances are \(k+l,k+1+l\). The estimate \(1+(1-\delta)l\le(1-\delta/2)(k+1+l)\) holds when \(k+l\ge1\). The omitted case is the designated edge itself. For another face, put \(l=\operatorname{dist}(v,b)\ge L/2\). Both prescribed differences are at most \(1+(1-\delta_0)l\) and both distances are at least \(l-k-1\). The choice of \(L\) gives the required relative slack there as well.

Choose \(\beta<\min\{\delta/2,s_0/2,1/4\}\), decreasing it if necessary. Give the designated edge length one and all other edges length \(1-\beta\). All prescribed pairs are Lipschitz for this weighted metric. Its distance across the designated edge is one, since every alternative path has at least two edges and \(2(1-\beta)>1\). Extending by the minimum of the upper distance cones gives a filling with designated gradient \(\tau\) and all other gradients at most \(1-\beta\) in absolute value.

We need the following quantitative finite-dimensional consequence of slice log-concavity, and give its proof. If positive-volume convex bodies \(K_-,K_+\subset\mathbb R^m\) have projection widths at most \(A\) in a unit direction \(u\), their top facet areas divided by volume are at least \(c_0>0\), and \(K\supset(K_-+K_+)/2\), then \[ \log\frac{|K|}{\sqrt{|K_-||K_+|}} \ge\kappa\bigl[(b_+-u\cdot m_+)-(b_--u\cdot m_-)\bigr]^2. \tag{10}\] Here \(m_\pm\) are the centroids, \(b_\pm\) the top projection endpoints, and \(\kappa>0\) depends only on \(A,c_0\). This is a finite-dimensional lemma; it has no restriction on the spatial dimension of the lattice.

To prove (10), let \(f_i\) be the normalized density of the projection of uniform measure on \(K_i\), and let \(q_i(p)\) be its quantile, \(0<p<1\), for \(i\in\{-,+\}\). Parallel-slice Brunn–Minkowski makes \(f_i\) log-concave, with \(f_i(b_i)\ge c_0\). If \(f_i(q_i(p))<c_0\), every density value to its left is at most \(f_i(q_i(p))\): otherwise log-concavity between that value and the top slice gives a contradiction. Integration over an interval of length at most \(A\) then gives \(p\le A f_i(q_i(p))\). In the other case \(q_i'(p)=1/f_i(q_i(p))\le c_0^{-1}\). Thus, with \(C=\max\{A,c_0^{-1}\}\), \[0<q_i'(p)\le C/p.\] The midpoint slice of \(K\) at \((q_-(p)+q_+(p))/2\) contains the midpoint sum of the two corresponding slices. Slice Brunn–Minkowski and change of variable along these increasing levels yield \[\frac{|K|}{\sqrt{|K_-||K_+|}} \ge\int_0^1\frac{q_-'+q_+'}{2\sqrt{q_-'q_+'}}\,\mathrm dp \ge1+\frac1{8C^2}\int_0^1p^2(q_-'-q_+')^2\,\mathrm dp.\] For the second inequality, subtract one from the integrand; its numerator is \((q_-'-q_+')^2\) and its denominator is \(2\sqrt{q_-'q_+'}(\sqrt{q_-'}+\sqrt{q_+'})^2\le8C^2/p^2\). Tonelli gives \(b_i-u\cdot m_i=\int_0^1p q_i'(p)\,\mathrm dp\). Cauchy–Schwarz therefore bounds the last integral below by the squared difference \(d^2\) of the centroid gaps. Since \(|d|\le A\), \(\log(1+x)\ge x/(1+x)\) proves (10) with \(\kappa=(8C^2+A^2)^{-1}\). All slice integrals may first be taken over compact subintervals of \((0,1)\) and then by monotone limits.

Apply it to \(K_\pm=K_Q(s\pm\alpha\chi)\) with \(\alpha\le\beta/(8L)\). Endpoint feasibility implies \(|\chi(b)-\chi(c)|\le\operatorname{dist}(b,c)\le2L\) on the shell. Shift every interior coordinate in the exposed filling by \(\pm\alpha\chi(v)\) and use the new shell data. Every undesignated edge retains slack at least \(\beta/2\). Thus each selected facet contains a Euclidean ball of radius \(\beta/8\) in its hyperplane. Body volumes are at most \((4L)^{(L-1)^2}\), and the projection widths are at most two. The hypotheses of (10) therefore hold uniformly.

Define \(U=m_+-m_-\) on the interior and \(U=2\alpha\chi\) on the shell. For an interior radial edge use the direction \(\tau(e_y-e_x)/\sqrt2\). The top projection endpoints are both \(1/\sqrt2\), so the difference of the two centroid gaps in (10) is \(-\tau(U(y)-U(x))/\sqrt2\). For an edge from a face center \(v\) to an interior vertex \(y\), use the direction \(\tau e_y\). Its top endpoints are \(1+\tau(s(v)\pm\alpha\chi(v))\), so the difference of the centroid gaps is \(-\tau(U(y)-U(v))\). Thus (10) bounds every \(|U(y)-U(x)|^2\) by a fixed multiple of the common midpoint gap \(g_\alpha\). The bodies and their centroids are the same for all radial edges; varying \(\tau\) changes no squared increment. Telescoping at most \(L\) radial edges between \(v,w\) and using \(g_\alpha\le g_Q\) proves the lemma. ◻

Proof. The controlled-gradient theorem [CAP], on a graph of maximum degree six with an arbitrary nonempty zero fixed set, states that for distinct edges \(e_1,\ldots,e_j\) and \(0<u\le1/8\), \[\mathbb P\bigl(|\nabla_{e_i}h|\ge1-u\text{ for all }i\bigr) \le(C_6u)^{j/C_6}.\] Fix one shortest path for every pair in an enlarged block. A path of length \(m\) violating linear slack with \(\delta_0=\varepsilon/2\) has at least \(m/2\) gradients of magnitude at least \(1-\varepsilon\). Choose \(\varepsilon\) sufficiently small to dominate the \(2^m\) subsets of this path, then \(k_0\) large enough to dominate the polynomial number of pairs. Choose \(L\) at a fixed sufficiently large multiple of \(k_0/\delta_0\), and finally choose \(\delta\) small enough to dominate the finite edge count. For disjoint enlargements, the edge witnesses are distinct. Apply the controlled-gradient bound to the short-edge and long-path witnesses separately and bound the intersection by the geometric mean of these two bounds. This gives an arbitrarily small product bound for failures of the first two tests.

For a non-tie test, condition on its enlarged shell. If there is a filling with the stated long slack, fix one such filling and write \(x,y\) for the tested edge. Prescribe its values to be \(h(x),h(x)+1/2\), or \(h(x),h(x)-1/2\). The distance of the edge to the shell is a sufficiently large fixed multiple of \(1/\delta_0\); long slack and the triangle inequality make both prescriptions compatible with the shell. Upper-cone extension produces two feasible fillings. Contracting the entire conditional body toward either filling with ratio \(1/4\) gives a volume fraction \(4^{-m}\) in each of the difference intervals \([1/8,5/8]\) and \([-5/8,-1/8]\), where \(m\) is the fixed number of free coordinates. The difference marginal is log-concave. A one-dimensional log-concave density placing a fixed positive mass in these separated intervals has bounded supremum: its half-height points lie within \(2/\sup f\) of a mode, and concavity of \(\log f\) bounds the farther tails exponentially. Consequently \[\mathbb P\bigl(|h(x)-h(y)|<s_0,\ \text{long slack}\mid\text{shell}\bigr) \le C_Ls_0.\] If no long-slack filling exists the event is empty. Conditional laws in disjoint enlarged interiors are independent given their complement. Choosing \(s_0\) small, combining the possible witnesses, and coloring the block grid by finitely many residue classes proves (11). For a fixed thickening, extract a separated positive fraction of witnesses in the original set; the same counting proves the last assertion. ◻

Proof. Normal reflection has a forward Brownian martingale part and a finite-variation part consisting of boundary reflection and any constant drift. Reversibility gives the same decomposition backwards from \(T\). Adding the forward increment and the negative reversed increment cancels the finite-variation terms, so each coordinate displacement is half the sum of the corresponding forward and reversed martingale increments. Each of the two martingale vectors has independent variance-two Brownian coordinates, although the two vectors need not be independent. If all \(j\) coordinate displacements exceed \(r\), at least \(j/2\) are witnessed in one of the two directions. A union over subsets and the Brownian maximal inequality prove (12). A failed block persistence test is witnessed by a coordinate displacement exceeding a fixed constant. Separated blocks have distinct witnesses; their finite coordinate count is absorbed by taking \(T\) smaller. Edge slack, long-distance slack, and non-ties all persist once their two endpoint displacements are below the appropriate fixed fraction of their original margins. ◻

Proof. Follow the filled outer grid contour of a hole and its exterior good rim. After a fixed thickening this gives nearest-neighbor rim paths. A hole of diameter \(d\) has at most \(C(d+1)^2\) vertices and boundary edges, and any required rim path has at most that length. Choose the lexicographically first rim site as the filling value. For an edge entering a hole, telescope along its rim; the inequality \(|\sum_{i=1}^m t_i|^p\le m^{p-1}\sum_i|t_i|^p\), followed by the boundary-edge count, gives \[\sum|\nabla v|^p\le\sum_{e\text{ good}}W_e|\nabla u(e)|^p, \qquad W_e\le C_p(1+d_e)^{2p}.\] Outer rims have bounded overlap. Holder with exponents \(2/p\) and \(2/(2-p)\) gives (14). For unweighted energy the required hole exponent is \(4p/(2-p)\); in the weighted case another Holder inequality separates the hole and reciprocal-weight moments. A hole filling value lies within distance \(C(d+1)\) of each of its sites, proving the pointwise error.

If a deterministic site belongs to a hull of diameter at least \(r\), a bad star-animal of size \(m\ge cr\) has a vertex within \(Cm\) of it. There are \(O(m^2)\) root choices and at most \(C_0^m\) such animals. Applying (11), or extracting a fixed-fraction separated subset before doing so, proves (13). In particular all fixed moments have uniformly bounded expectations.

Here is the stronger spatial statement. Clusters with size in \([k,2k)\) contribute at most \(Ck^{q+2}\) each to the sum of hole moments. For \(r\) disjoint such cluster witnesses in an \(O(N)\) box, root and animal counting and (11) bound the probability by \[\frac{(CN^2)^r}{r!}C_0^{2kr}\rho^{ckr} \le\left(C\frac{N^2}{r}e^{-c'k}\right)^r.\] For all sufficiently large fixed \(k\), take \(r=\lceil N^2k^{-(q+4)}\rceil\). For \(k\le A\log N\) and large \(N\), the resulting bound is \[\exp(-cN^2/k^{q+3}).\] Larger clusters occur with probability at most \(CN^2e^{-cA\log N}\), which is \(O(N^{-D})\) by choosing \(A\) large. On all dyadic size ranges the resulting contribution to \(N^{-2}\sum_z(1+d_z)^q\) is summable; clusters below the fixed cutoff give a fixed deterministic bound. This proves the assertion. The same argument excludes holes crossing the macroscopic buffer, and a union over the polynomially many subboxes proves the stated simultaneous version. Lexicographic selection commutes with translation; a random grid phase can be adjoined when invariance under single lattice steps is desired. ◻

Proof. Discrete \(L^1\) Poincare on successive dyadic squares, clipped to the ambient square at its boundary, gives \[|v(x)-\overline v|\le C\sum_e \frac{|\nabla v(e)|}{1+\operatorname{dist}(x,e)}.\] For \(p'=p/(p-1)\), a dyadic contribution to the averaged kernel is \(F_r(e)=Cr^{-1}\nu(B(e,Cr))\). Its supremum is at most \(CN^{-\alpha}r^{\alpha-1}\) and its \(\ell^1\) norm at most \(Cr\). Thus \[\|F_r\|_{\ell^{p'}}\le C N^{-\alpha/p}r^{1+(\alpha-2)/p}.\] The dyadic series is bounded by \(CN^{1-2/p}\) precisely under the stated strict inequality on \(\alpha\). Holder finishes the proof. ◻

Proof. Use Lemma 6 with \(p=3/2\). A Markov bound on the twelfth spatial hole moment, and exclusion of holes crossing a fixed buffer, give \(\|\nabla v\|_{3/2}\le CN^{1/3}\sqrt E\). The filling error tested against \(|w|\) is bounded by the same spatial moment. In the anchored case choose the deterministic pin measure \(\nu\) just constructed. Although filling may overwrite a pin value, the Lipschitz bound gives \[|\nu(v)|\le CK\sum_z\nu(z)(1+d_z).\] The right side has bounded expectation by (13); another Markov bound makes it a fixed constant. Applying Lemma 7 to \(\nu\) and \(|w|\) proves (16). In the zero-mass case the ordinary mean cancels and no pin measure is needed. The truncation error is at most \(M\sum|w|\). Choose the two moment thresholds so their failure probabilities sum to at most \(1/4\); the buffer event has failure at most \(1/4\) for large \(N\). Bounded \(N\) is absorbed by deterministic Lipschitzness and the anchor or zero-mass condition. All error bounds depend on hole sizes rather than on \(X\), which proves the simultaneous assertion. ◻

Proof of Proposition 2 for zero pins. For \(R/L\) large integrate the interiors of pin-free blocks in a fixed large box around the statistic and its anchor. Retain all other vertices, including every vertex of pin blocks, and write \(s\) for their heights. The unnormalized marginal density is exactly \[ p(s)=\mathbf1_{\mathcal A}(s)\prod_Q e^{F_Q(s|_{\partial Q})}, \tag{17}\] where \(\mathcal A\) imposes the remaining edge constraints. This density is even, log-concave, compactly supported, and upper semicontinuous. Let \(Z=\int p>0\).

For feasible \(s\pm\chi\) with positive density, put \(X_z=\chi(v_z)\) at a fixed face-center representative of each block. The difference of any two full endpoint fillings, divided by two, extends \(\chi\) as a 1-Lipschitz field vanishing on pins. Consequently neighboring \(X\) values differ by at most \(CL\). Choose \(M\) above the distance of every representative and shared face center in or next to a pin block to a pin, and use \(u=T_M(X)\). It vanishes on pin blocks. For neighboring ordinary good blocks \(z,w\), their common face center and Lemma 3 give \[|u_z-u_w|^2\le2c_*^{-1}(g_z+g_w).\] For a good block next to a pin block, the common face center \(f\) satisfies \(|\chi(f)|\le M\), so \(|u_z|^2\le|\chi(v_z)-\chi(f)|^2\le g_z/c_*\). Treating pin blocks as good zero blocks therefore gives \[ \sum_{e\text{ good}}|\nabla u(e)|^2\le C\sum_Q g_Q. \tag{18}\]

Partition vertices into half-open blocks and replace the statistic by the retained linear functional \(\ell(s)=\sum_z w_zs(v_z)\), where \(w_z=\sum_{x\text{ in block }z}q_x\). On every original configuration, \[ \left|\sum_xq_xh(x)-\ell(s)\right|\le CL\sum_x|q_x|. \tag{19}\] The coarse weights satisfy the hypotheses of Lemma 8, and exact zero mass is preserved. That lemma and (18) show, on a full-configuration event of probability at least \(1/2\), simultaneously for every feasible \(\chi\), \[|\ell(\chi)|\le C_0+C_2\sqrt{\sum_Qg_Q}.\] Take fixed \(t>C_0\) and \(D=((t-C_0)/C_2)^2>0\). Since the logarithmic midpoint deficit of (17) is \(\sum_Qg_Q\), it follows that \[ \sqrt{p(s+\chi)p(s-\chi)}\le e^{-D}p(s) \quad\text{whenever }\ell(\chi)=t. \tag{20}\]

Define \(A_t(s)=\sup_{\ell(\chi)=t}\sqrt{p(s+\chi)p(s-\chi)}\). Compactness of the support restricts the supremum to a compact set; upper semicontinuity gives measurability. Log-concavity implies \(0\le A_t\le p\). The good event may depend on discarded midpoint coordinates: the simultaneous assertion (20) still holds for its retained vector. Thus \[Z^{-1}\int A_t\le1-\rho_0,\qquad \rho_0=(1-e^{-D})/2>0.\] If \(\ell\) is zero on the free coordinates there is nothing to prove. Otherwise choose linear coordinates \(s=(r,\sigma)\) with \(r=\ell(s)\), absorbing their constant Jacobian into the complementary measure. For the density \(a(r)\) of \(\ell(S)\), Prékopa–Leindler in \(\sigma\) [Prekopa] gives \[\sqrt{a(r-t)a(r+t)}\le Z^{-1}\int A_t(r,\sigma)\,\mathrm d\sigma.\] Integrating yields overlap at most \(1-\rho_0\). Since \(a\) is even and log-concave, this overlap is at least \(\int\min\{a(r-t),a(r+t)\}\,\mathrm dr=1-\int_{-t}^t a\). Hence \(\mathbb P(|\ell(S)|\le t)\ge\rho_0\). Its survival function is log-concave, again by Prekopa–Leindler, so \[\mathbb P(|\ell(S)|>jt)\le(1-\rho_0)^j\quad(j\ge1).\] This gives a positive uniform exponential moment. Equation (19) transfers it to the original statistic. For bounded \(R/L\), distance to the anchor, or cancellation of \(q\), gives the required deterministic bound. ◻

Proof. Run monotone single-site uniform heat baths with common update variables for pin values \(-B,b,+B\). The two extreme chains can be maintained as exact translates, since their conditional intervals are translates. The middle chain remains between them: both endpoints of a conditional interval are increasing functions of the neighboring heights. A stationary subsequential coupling on the finite compact polytopes gives the asserted coupling with the uniform marginals. Each tested statistic changes by at most \(B\sum|q_x|\). Condition first on mixed pin values and integrate the same uniform bound. ◻

Proof. Apply (5) to \(Y/c\) to bound \(\mathbb Ee^{2Y}\), and use Jensen together with the uniform bound on \(|\mathbb EY|\) to bound \(\mathbb Ee^Y\) away from zero. Cauchy–Schwarz gives the assertions; for moments apply it with the square of the desired moment. Multiple bounded tests can be combined by Holder and a further reduction of \(c\). ◻

Proof. Recall the interval extension criterion. Data \([\alpha_z,\beta_z]\) at specified vertices admit a 1-Lipschitz selection exactly when \(\alpha_z\le\beta_w+\operatorname{dist}(z,w)\) for every pair. Sufficiency follows from the lower-cone function \(g(y)=\sup_z\{\alpha_z-\operatorname{dist}(z,y)\}\), which is Lipschitz, lies above every lower endpoint and below every upper endpoint. Infinite interval endpoints are omitted from the corresponding bound.

At a movable vertex \(x\), the intervals propagated from all non-shell data contain both \(h(x)\) and zero. Indeed feasibility gives the first assertion. For a nonfixed interval datum the second follows because its interval contains zero. For a fixed pin \(p\), its distance to \(x\) is at least one, whereas \(|h(p)|\le1/2\), so its propagated interval contains zero as well. The datum at \(x\) itself, if present, also contains zero.

If both endpoints \(u,v\) are movable, put \(c=\operatorname{median}(0,h(u),h(v))\) and prescribe \(g(u)=g(v)=c\). The value \(c\) lies between zero and each old endpoint value; thus it respects every propagated non-shell interval at both endpoints. Moreover \(|c-h(u)|,|c-h(v)|\le|h(u)-h(v)|\le1\). Equation (21) preserves all endpoint–shell inequalities. The new endpoint pair has difference zero, and every pair among the old data was already consistent. The interval criterion therefore extends the new prescriptions to a feasible filling.

If one endpoint is a fixed pin, prescribe zero at the other endpoint. Its original height has absolute value at most \(3/2\), so (21) again preserves all shell inequalities. The propagated non-shell intervals contain zero, and the resulting edge gradient has absolute value at most \(1/2\). Apply the same extension criterion. ◻

Proof. First freeze all string values, making them pins in \([-1,1]\). Lemma 9 couples the conditional field within coordinatewise distance one of the corresponding zero-pinned field. The controlled-gradient witness argument gives arbitrary-parameter long-distance slack, with the additive comparison loss absorbed by increasing its fixed threshold. This bound is uniform over the conditioned string values, so integrating preserves its joint form.

Now condition only on the exterior of each enlarged patch; its interior strings remain variable. On a shell for which a filling has the preceding long slack, central endpoints satisfy (21). Apply Lemma 11. If the conditional body has dimension \(m\), contraction toward its strict filling with ratio \(1-2s\) sends the entire body into \(\{|\nabla h|\le1-s\}\). Therefore the conditional relative volume of the near-contact region is at most \(1-(1-2s)^m\le2ms\). This remains true for arbitrarily thin sign intervals; their effect on the body’s total volume cancels in the relative-volume estimate. Degenerate prescribed intervals are treated as fixed coordinates.

For disjoint enlarged patches the conditional laws are independent. For a specified collection of bad boxes, either at least half have long-slack failures, or at least half have near contacts with good long slack. The joint bounds just proved apply to these two alternatives; the subset count is absorbed by choosing the preliminary parameters and then \(s\) small enough. A finite coloring removes the separation restriction. Lemma 5 gives the short-step statement. The tilt transfer, when its statistic meets the anchored or zero-mass hypotheses, is Lemma 10. ◻

Proof. Embed the lattice cells in a dyadic coordinate square with \(2^J\) cells on each side, where \(2^J\) is comparable to \(n\). Its physical size remains bounded above and below. Expand the piecewise constant field in the orthonormal two-dimensional Haar basis; the expansion terminates at the cell scale. A Haar coefficient on a cell of physical side \(r=R/n\) is \(r\) times a statistic satisfying (4) and having exactly zero total mass. For zero pins the tests include all ambient lattice vertices, including the zero extension, so this cancellation also holds across the boundary. Bounded data differ from the same zero-pinned sample by a uniformly bounded field. Proposition 2 consequently gives a second moment at most \(Cr^2\) for each centered coefficient. The coarse scaling coefficient is bounded by the anchored assertion.

Write \(F_j\) for the sum of wavelets at level \(j\). There are \(O(r_j^{-2})\) coefficients there, so \(\mathbb E\|F_j\|_2^2\le C\). Each parent-cell integral of \(F_j\) vanishes. For \(0<t\le1\), fractional Poincare on the parent cells and duality give \[\|F_j\|_{H^{-t}}\le Cr_j^t\|F_j\|_2.\] Minkowski in \(L^2(\mathbb P;H^{-t})\) and the geometric sum of \(r_j^t\) give a uniform second moment in \(H^{-t}\). Choose \(0<t<s\) and use compactness of its embedding into \(H^{-s}\) on the fixed torus. ◻

Proof. For a fixed finite query history, every decision indicator is a function of recorded values and auxiliary randomness. Disintegrate the original product density over these values. The indicators introduce no factor depending on an unqueried coordinate, so the remaining density is the product of the two conditional Gibbs densities. Induction on the query count, followed by partitioning over the stopping count, proves the claim. Any coupling of these two kernels has the same individual conditional marginals. Integration over the transcript recovers the original marginals. A conditional invariant joining, obtained by time averaging the synchronous semigroup on its compact state space, is stationary. One may first take this time average jointly with the transcript; disintegration of a weak limit supplies a measurable choice of conditional invariant joinings. For the boundary-data version, disintegrate directly over the joint boundary vector and any recorded fixed set or component. Integrating either marginal of a coupling of the two conditional Gibbs kernels restores its original annealed law. The subsequent sparse-event bounds use only these individual marginals; independence of the two boundary vectors is unnecessary. ◻

Proof. On the first event let \(N=\sum_{i\in I}\mathbf1_{A_i}\) and \(k=|I|\). Then \(\mathbb E[N\mid\mathcal I]>qk\) and \(N\le k\), so \(\mathbb P(N\ge qk/2\mid\mathcal I)\ge q/(2-q)\ge q/2\). The unconditional probability is at most \(2^k\rho^{\lceil qk/2\rceil}\) by choosing a witnessing subset. For the second assertion use conditional Markov on \(\mathbb E[X_n\mathbf1_{X_n>M}\mid\mathcal I]\). ◻

Proof. Use the finite-dimensional construction by projecting after each common Brownian increment. Interpolate the two preprojection inputs, and also their pinned values. Except at finitely many parameter breakpoints, the derivative of projection is computed on a fixed active face. An active edge makes its two free derivatives equal. A free component therefore receives the average of its input derivatives, unless it touches a pin or individual wall, when it receives the prescribed derivative. Several pins active on a parameter interval have equal derivatives. Integrating over the interpolation parameter gives the asserted convex mixture of orthogonal averaging projections and zero blocks. All intermediate contacts belong to the union of endpoint near-contact clusters: an infinitesimal projection cannot contact an edge whose slacks in both endpoint fields are bounded below.

For \(u=(z-b)_+\), convexity and the pin bound \(z|_P\le b\) give \(u'\le Au\) coordinatewise. Write this as \(u'=RAu\) with a diagonal matrix \(0\le R\le I\). The lower tail is identical after changing sign. Each elementary matrix \(A\) or \(R\) is a positive self-adjoint contraction, hence \[\|v-Av\|_2^2\le\|v\|_2^2-\|Av\|_2^2.\] Add zero as a possible initial level. Every intermediate coordinate is a substochastic mixture of the initial levels, with lost mass put at zero. Jensen’s identity and the column deficits show that its variance about that mixture is at most the total loss \(D\). It is therefore within \(\sqrt D\) of some initial level. Each elementary jump is at most \(\sqrt D\). A gap greater than \(4\sqrt D\) between successive initial levels cannot be crossed. There are at most \(H+1\) levels, which proves (29) uniformly in the projection mesh. Passing to the reflected limit proves the assertion. ◻

Proof. Write the inequalities in residual form \(y_i-y_j\le c_{ij}\), with individual inequalities using the ground vertex. All costs are nonnegative because \(y=0\) is feasible. Delete redundant inequalities. The strongest bound on a difference in a finite system of difference constraints is the shortest directed path cost in its constraint graph: summing a path proves one direction, and shortest-path potentials give the converse. The original central bound has cost at most two, so the facet graph contains a path of cost at most two. Remove cycles. If it uses ground, split the path there. Neither part can use an artificial collar bound, since its segment from or to the tested endpoint would already cost more than \(\Lambda\). For a wall use the same argument with budget one. A visit to a vertex carrying a wall is not itself a visit to ground; only its individual-constraint edge gives ground input.

We justify uniformity of the forcing construction. Move every signed coordinate a distance \(\tau\le\min(s/8,r_0/4)\) toward its allowed side, leaving other coordinates unchanged. All edge constraints retain slack at least \(s-2\tau\), and the artificial collar retains slack at least \(r_0-\tau\). A fixed-radius Euclidean ball about the new point lies in the auxiliary polytope. Every feasible displacement satisfies \(|y_x|\le2N+r_0\), by following a path to a pin or collar. Individual bounds with greater slack than \(2N+r_0+1\) may be capped at that value: they remain redundant. Thus, after fixing one of finitely many graph and sign patterns, the effective parameters form a compact family with a uniform interior ball.

Fix a parameter value and take the facet paths just obtained for its polytope. For one facet choose a point in its relative interior where all other distinct-normal inequalities are strict. The artificial collar inequalities are among the strict ones. Drive the coordinates from the initial point along a feasible segment to this point, and then a fixed distance in the outward normal direction. The reflected trajectory stays at the facet during the outward part and acquires a positive impulse. An initial segment toward the interior-ball center can be inserted. A sufficiently small uniform tube around this piecewise linear driver preserves the collar inequality and at least half of the final impulse.

For a fixed polytope, uniform-driver continuity and convergence of the projection recursion are the deterministic Skorokhod assertions in [Slominski]. The geometric hypotheses hold here: if \(B(a,r)\subset K\) and \(|a-x|\le R\) on \(K\), every inward unit supporting normal at \(y\) has scalar product at least \(r\) with \(a-y\); within distance \(r/2\) of \(x\) its scalar product with \((a-x)/|a-x|\) is at least \(r/(2R)\). Convexity supplies the exterior sphere condition. On the final interval only one distinct facet normal is active, and the one-dimensional running-minimum formula gives impulse stability.

Parameter stability does not require interpreting this fixed-body reference as a changing-body theorem. Fix the graph and the free coordinates, and perturb initial heights and individual pin or wall levels by at most \(\delta\), in absolute height coordinates. Apply the projection derivative argument of Lemma 18, now allowing these individual inputs to vary. Every output difference is a subconvex combination of the previous free differences and the individual-input differences. Under the same driver its supremum norm is therefore at most \(\delta\) at every projection step and in the reflected limit. The original edge bounds remain exactly one. Combining this parameter bound with fixed-body driver continuity preserves the chosen tube in a parameter neighborhood. In the final single-normal interval, normal balance also preserves the positive impulse. Graph or free-coordinate patterns belong to a finite list and are treated separately.

Brownian motion has positive probability of belonging to the chosen tube: its piecewise linear center has square-integrable derivative, so translation by that center preserves positivity of a Brownian small-ball event. The construction persists in a neighborhood of the parameter value. Merge identical free normals by retaining the strongest bound. The selected relative-interior facet point then moves slightly along its normal, with every other inequality still strict. Use facets of each limiting polytope rather than following a facet that disappears in a limit. A finite cover of the compact parameter family supplies positive minimum probability and impulse. ◻

Proof. Cut the edges across an initially strictly slack shell. For the cut exterior process let \(A_B\) be the event that an exterior endpoint moves by more than \(s/4\). If in the original stationary process every endpoint of a cut edge moves by less than \(s/4\), none of these edges is contacted. Until a first cut contact the cut and uncut solutions agree. Uniqueness therefore makes them agree throughout this quiet event. In particular \(A_B\) implies a shell-displacement event of the original process. Equation (28) gives joint sparse bounds for these events on separated boxes. Lemma 17, with \(q=1/2\), shows that boxes with \(\mathbb P(A_B\mid\mathcal I)>1/2\) are still arbitrarily sparse. On the remaining initial configurations exterior quietness has probability at least \(1/2\) and depends only on exterior noises. The controlled interior noises are independent of them. Their intersection has probability at least \(p_0/2\), and no cut edge is contacted there. Thus the cut controlled path is a path of the original process. ◻

Proof. Use the original lattice vertices for the extension. Bad boxes are first enlarged by the fixed amount needed to control every original edge on a surrounding good rim. Their outer filled hulls give the holes of 6; passing from boxes to their unions of lattice sites changes diameter and counting constants only by the fixed box size. Enlarge the rim by a fixed layer if needed to include an exit by a triangular-lattice diagonal edge. The same square-coordinate contour construction supplies connected good rim paths of length at most \(C(d+1)^2\) for a hole of diameter \(d\). No extra probabilistic claim for independently marked individual sites is needed.

The zero set \(E_n\) is connected to infinity in the lattice graph. Indeed the polygon has one connected boundary cycle, and its exterior is the unbounded complementary component; including boundary vertices removes any vertex-level corner separation. For a hole \(H\) which meets \(E_n\), follow an \(E_n\)-path from a vertex in \(H\) to infinity. Since \(H\) is finite, the path exits it. Its first exterior rim vertex lies in \(E_n\) and has value zero. Choose this vertex as the anchor of \(H\), and fill all of \(H\) with zero. If \(H\) does not meet \(E_n\), choose any surrounding good anchor as before. Outside the holes retain the original values. The resulting field is zero on every vertex of \(E_n\).

We verify the energy bound for this choice, rather than relying on preservation by an arbitrary anchor. If an edge joins a good vertex \(x\) to a hole, its new jump is the difference of \(u(x)\) and the value at that hole’s chosen rim anchor. Join these two rim vertices by the good rim path just described and telescope. The path length, the number of incident edges, and the repeated uses of a rim edge have the same polynomial diameter bounds for every choice of anchor. Applying the path inequality and then Holder gives exactly the weighted bound in 6, with possibly a changed fixed exponent of \(1+d\). Edges with both endpoints in a filled hole have zero jump. Edges outside the holes retain their old jump; exterior zero–zero edges may be given weight one and contribute no energy. In particular the weighted gradient estimate holds across the actual polygon boundary. The spatial moment estimates and expected energy bound then give its squared \(W^{1,p}\) gradient bound. The local weighted square bound already proved controls its \(L^p\) norm.

The new anchor also preserves the vanishing filling error. For holes of diameter at most \(M\), join each vertex to its selected anchor by a full-lattice path of length at most \(CM\). Such paths lie in a fixed enlargement of the hole’s bounding box, and their congestion is bounded by a fixed polynomial of \(M\). The all-edge estimate (39) applies to the zero-extended original field, including boundary-incident edges. It makes the expected error from these holes tend to zero for each fixed \(M\). For larger holes, \(|U(x)-u(x)|\le C(1+d_x)\) by full-lattice Lipschitzness; high diameter moments make their error tend to zero as \(M\to\infty\). These estimates also cover the exceptional event by setting \(U=0\) there and using the established moment bounds.

For a comparison with identical boundary heights, \(z=h-\widetilde h=0\) on the boundary. Its extension by zero is \(2\)-Lipschitz on the full lattice: edges crossing the boundary first meet a fixed boundary vertex, and exterior edges have zero difference. Apply the construction to \(z\), or to both of its tails and subtract. All triangle vertices outside the lattice polygon have value zero, so affine interpolation vanishes there as well. Norm equivalence on the fixed lattice triangles gives the displayed global Sobolev bound. This proof uses the unbounded connected exterior, and does not apply to a hole containing an entire isolated internal component of observed vertices. ◻

Proof. First consider bounded continuous cylinder functions of finitely many gradients and Brownian increments in a fixed time window. Tile an interior region by disjoint boxes of side \(L\). In each box run a zero-boundary equilibrium bath using the original Brownian drivers at its free sites. The simultaneous stationary joining exists by time averaging the finite reflected semigroup. The reference baths have independent path laws: their drivers are independent, and the unique invariant marginal of their product dynamics is the product of their equilibrium measures.

Discard a boundary layer of relative width \(\eta\) in each box. The centered free comparison, uniform in the ambient domain size, makes the difference of each fixed local gradient tend to zero in volume probability as \(L\to\infty\). Stationarity gives the same assertion at any fixed finite list of times. The drivers are identical. Consequently cylinder averages for the original bath differ in \(L^1\) from those for the references by a quantity tending to zero, followed by an error \(O(\eta)\) from discarded sites. Comparing two original systems through reference boxes of the same size proves that their volume-sampled cylinder laws are Cauchy.

For a spatial observation box of side \(R\gg L\), its reference-box averages are independent and bounded. Their average has variance at most \(C(L/R)^2\). Choose \(L\to\infty\) slowly with \(L/R\to0\), and then let \(\eta\downarrow0\). This proves concentration of spatial cylinder averages. The estimates allow \(R\) itself to grow arbitrarily slowly relative to the original domain, so the concentration passes to the infinite-volume local limit. Cylinder approximation then proves spatial ergodicity of the full gradient-and-driver law. Spatial stationarity follows by shifting the uniform root, and time stationarity passes from the finite stationary laws. The same comparison includes the small smooth tilts and proves their independence in the unscaled limit.

We give the path and reflection details separately from this static comparison. For a stationary finite-volume bath, Lemma 5 and the forward and reversed martingale decompositions express each coordinate increment as half the sum of two variance-two Brownian increments. The usual Brownian modulus bound therefore gives tightness of each coordinate path after subtracting its initial value. Relative initial heights in a fixed neighborhood are bounded by graph distance. This gives tightness of all local continuous paths, jointly with their drivers. A bounded drift of order \(n^{-2}\) contributes a vanishing term. Independence of future Brownian increments from the forward past passes first for finite cylinder tests and then, by a monotone-class argument, for the full forward sigma-field.

Fix one short step and a finite set of vertices. Outside an event of arbitrarily small probability, its incident transmission clusters lie in a fixed finite neighborhood, and all edges crossing their boundary have positive slack throughout the step. This is the short-time isolation input. There are only finitely many possible clusters under this restriction. Include their markings in the joint extraction. On the event corresponding to a particular finite cluster \(C\), the trajectory coincides with the normally reflected motion using only the internal edge constraints of \(C\): removed edges exert no force. The finite-dimensional Skorokhod map is continuous in the initial point and the common continuous drive on this isolated problem. Use the orthonormal constant vector \(|C|^{-1/2}\mathbf1_C\) and an orthonormal basis of its orthogonal complement. The constant coordinate is unconstrained variance-two Brownian motion; every edge normal lies in the orthogonal complement, where the Brownian covariance remains \(2I\) and the edge polytope is compact. The arithmetic mean, in contrast, has variance parameter \(2/|C|\). Thus the limiting trajectory has the same normal reflection. Letting the cluster restriction increase identifies the dynamics everywhere locally.

The improving initial strict-slack bounds imply that each fixed initial edge is strict almost surely in the limit. On an isolated finite cluster, a union of intersections of two distinct facets is polar for normal reflection. One can verify this from the Neumann Dirichlet form: a logarithmic cutoff in the two transverse coordinates has energy tending to zero. Starting at an interior point, such a polar set is not reached. This applies before restricting to the isolation event, so conditioning on that future event is unnecessary. The countable collection of finite clusters and time intervals proves absence of simultaneous contacts. The one-dimensional normal coordinate and the strong Markov property at its first hit show that a visited face has positive regulator increment on an interval containing that hit in its interior. Applying this assertion to rational enclosing intervals makes it simultaneous for all episodes used below. Stationarity and strict initial slack also exclude contact at either fixed step endpoint. ◻

Proof. For each facet \(e\), its contact times form a compact subset of \([0,T]\). Different nonempty contact sets are disjoint and thus have positive distance. Since the number of facets is finite, there can be only finitely many switches of contact label. Repetitions of the same label can be grouped. Choose a finite partition with endpoints off all facets so that each contact interval allows only its designated facet, and each remaining interval has no contact. All undesignated inequalities on each such interval have positive minimum slack.

Continuity of the finite Skorokhod map preserves these inactive margins under small perturbations. Positive impulse supplies a strict running-minimum witness in the normal coordinate. Therefore both sufficiently close synchronous paths reach the designated face within that interval. The one-dimensional reflection formula then removes their normal difference exactly; their tangential difference stays unchanged. Their difference at the end of the interval is \(P_e\) times its input difference. Contact-free intervals preserve the difference. Composition proves the exact identity.

The conclusion is stronger than an error \(o(1)\), which would not survive multiplication by \(n\). On each fixed regular path neighborhood the identity is exact. Normal monotonicity bounds a finite-dimensional perturbation from a differential drift by its integrated Euclidean magnitude, at most \(C(C)Tn^{-2}\). Its product with \(n\) tends to zero.

For the random application, first bound the cluster size and the scaled initial oscillation. Next bound the positive inactive margins, temporal separations, and impulse witnesses away from zero. Each restriction has probability tending to one when its bound is relaxed. The deterministic argument applies on the restricted event. Remove the restrictions last. The union of the two replicas’ near-contact clusters provides the needed isolation. Subtracting a common spatial constant is harmless because all free edge constraints are invariant under that translation. ◻

Proof. The comparison estimates give tightness on every finite neighborhood and every finite set of step times. A diagonal extraction preserves the exact additivity of the differences. Shifting the root by a fixed lattice vector changes its macroscopic position by \(o(1)\) and changes its sampling law only by a vanishing boundary error. The conditional local law is therefore spatially stationary. Time stationarity passes directly. Its environmental marginal is the same ergodic law at every macroscopic position by Proposition 23. Lemma 24 identifies the update.

For integrability, distribute \(n^2 e_C^2\) equally among the vertices of \(C\). Its expectation at a volume-sampled site is bounded by (40). Cluster displacement is bounded by a fixed polynomial in cluster size times the square root of cluster loss. For any \(1<r<2\), Holder’s inequality consequently bounds the \(r\)-th moment of each scaled displacement multiplied by any fixed cluster power. Lower semicontinuity and finite-cluster testing give the corresponding integrability for the limit.

Include a marking for one fixed sparse-hole construction in the extraction. Randomize the block-grid phase, and choose a rim representative by a translation-equivariant rule. By 6, the filled limiting potential \(\widehat w\) has stationary gradients in \(L^p\) for some \(p>1\). Let \(\mathcal I\) be the invariant sigma-field of this joint spatial law. The spatial ergodic theorem makes the rescaled gradients converge weakly to their conditional mean \(q\) given \(\mathcal I\). To see that the potential has affine growth, subtract its mean on each box and rescale by \(R\). Poincare’s inequality and the spatial \(L^p\) bound give weak \(W^{1,p}\) compactness and strong \(L^p\) compactness. Every compact limit has distributional gradient \(q\), and is therefore affine. This proves the asserted volume growth for \(\widehat w\).

The correction \(w_x-\widehat w_x\) is a finite stationary random variable: the hole has finite diameter and every finite collection of original scaled differences is tight. No integrability of this correction is required to transfer volume growth. For fixed \(\delta>0\), stationarity and bounded convergence show that the expected fraction of sites where \(|w_x-\widehat w_x|>\delta R\) tends to zero. The original cocycle thus has the same slope. The identical argument for its stationary step displacement shows that \(q\) is time invariant.

For any Borel slope event \(B\), the function \(\mathbb P(q\in B\mid\mathcal E)\) is spatially invariant. Ergodicity of \(\mathcal E\) makes it constant. Conditioning on \(q\) therefore leaves the environmental marginal unchanged. Stationarity and the controlled bounds disintegrate for almost every slope value, proving the claimed deterministic-slope solutions.

Finally, the filled gradients are uniformly integrable in volume sampling. Discrete summation by parts against a smooth macroscopic test identifies their expected weak limit with \(\nabla m\). The expected filled and original potentials have the same limit by their vanishing \(L^1\) difference. The conditional spatial mean of the extracted filled gradients is \(q\), so its expectation is precisely the stated derivative. ◻

Proof. Use a time grid whose spacing accommodates a force attempt and is a multiple of \(T\). Choose measurably one class at an overlap. Maintain a separate visit counter for each class and cycle through its assigned facets. On infinitely many visits to that class, each facet is attempted infinitely often. Conditional Borel–Cantelli, applied to the forward filtration, makes each such facet occur infinitely often: its conditional success probability at an attempted step is at least \(p_j\).

Stationarity and Poincare recurrence imply \[\mathbb P(\text{infinitely many visits to }G_j)\ge\mathbb P(G_j).\] Some class is visited infinitely often on this event, and its contact path connects the prescribed endpoints. Let \(j\to\infty\), and then intersect the resulting probability-one events over the countably many edges. Neither temporal ergodicity nor conditioning on a temporal invariant event is used. ◻

Proof. Fix a lattice edge and a quality index \(j\). Use the finite cover in Lemma 19, choosing the first neighborhood at an overlap. It has finitely many labels \(1,\ldots,M_j\), each carrying a fixed path of genuine free facets joining the edge endpoints. Artificial grounding is excluded by the collar slack, and there are no genuine pins or walls in this bulk test box. Let \(G_{j,n}\) mark the configurations satisfying the initial slack and insulation conditions, and let \(L_{j,n}\) be their cover label. The estimates give \(\mathbb P(G_{j,n}=0)\le\delta_j+o(1)\) with \(\delta_j\downarrow0\). The force duration and its positive probability \(p_j\) may depend on \(j\).

At each time of a grid accommodating the force duration, these marks are functions of the full finite-volume initial configuration, before future drives are sampled. For an assigned facet \(e\) write \(C_e\) for the event of visiting that facet during the next force interval. The force and insulation proof, conditional on the full initial data, and independence of subsequent Brownian increments from the past give \[ \mathbb E\bigl[\Psi\mathbf1_{\{G_{j,n}=1,L_{j,n}=\ell\}} \mathbf1_{C_e}\bigr] \ge p_j\mathbb E\bigl[\Psi\mathbf1_{\{G_{j,n}=1,L_{j,n}=\ell\}}\bigr] \tag{41}\] for every bounded nonnegative function \(\Psi\) of the forward past. Here the initial bad-insulation event was discarded before forcing; no conditioning on a future cluster or future success event is used.

Retain the finite-valued marks in the local extraction constructing the bulk law, for every fixed quality, spatial translate, and time \(kT\). For attempts of a particular quality use a coarser subgrid whose spacing accommodates its force duration. A diagonal extraction is possible because these index sets are countable. The resulting marked extension is stationary. Its unmarked marginal is still the unique bulk environment of Proposition 23. The limiting marks need not be functions of the limiting initial state, and we do not assert that they are.

Take \(\Psi\) to be a bounded nonnegative continuous cylinder function of the unmarked past. Visiting a specified facet in a fixed compact time interval is closed in the uniform local path topology: the maximum of its oriented edge gradient equals the facet bound. The discrete mark event is clopen. Thus the left integrand in (41) is upper semicontinuous, and Portmanteau gives \[\mathbb E\bigl[\Psi\mathbf1_{\{G_j=1,L_j=\ell\}}\mathbf1_{C_e}\bigr] \ge \limsup_n \mathbb E\bigl[\Psi\mathbf1_{\{G_{j,n}=1,L_{j,n}=\ell\}} \mathbf1_{C_e}\bigr] \ge p_j\mathbb E\bigl[\Psi\mathbf1_{\{G_j=1,L_j=\ell\}}\bigr].\] The last expectation passes exactly by continuity. Approximation of finite-dimensional past laws by continuous functions, followed by a monotone-class argument over past cylinders, extends the inequality to every bounded nonnegative function of the full unmarked forward past \(\mathcal F_0\). Contact at a deterministic interval endpoint is a null event by Proposition 23, so the closed-event passage does not create an endpoint-only contact for the later tangent argument.

Define \[a_\ell=\mathbb P(G_j=1,L_j=\ell\mid\mathcal F_0), \qquad a=\sum_{\ell=1}^{M_j}a_\ell.\] Dropping the mark from the left side of the preceding inequality gives \(\mathbb P(C_e\mid\mathcal F_0)\ge p_j a_\ell\) for every facet of the path with label \(\ell\). Since \(\mathbb E(1-a)\le\delta_j\), \(\mathbb P(a<1/2)\le2\delta_j\). On \(\{a\ge1/2\}\) choose a maximizing label, with a fixed tie rule. Every facet of its path then has conditional contact probability at least \(p_j/(2M_j)\). These are the required forward-past measurable opportunity classes. Choose conditional-expectation versions equivariantly under the countable spatial and time shifts; their joint law is stationary. Their probabilities approach one, so Lemma 27 applies. The auxiliary marks served only in the transfer; the resulting contact-connectivity assertion concerns the original environment alone. ◻

Proof. Let \(a\) be a nonnegative compactly supported Lipschitz function, equal to one on the unit ball. For \(M=\varepsilon R\), let \[\Phi_M(s)= \begin{cases}s^2,&|s|\le M,\\2M|s|-M^2,&|s|>M,\end{cases} \qquad F_R(w)=\min_c\sum_x a(x/R)\Phi_M(w_x-c).\] Its Lipschitz constant in each input coordinate is bounded by \(2M a(x/R)\). Hence \(F_R(Jw)-F_R(w)\) is integrable, although the two terms need not separately have finite expectations. They have the same distribution by time stationarity. Truncate each scalar minimum at level \(K\) and use \(|(s\wedge K)-(t\wedge K)|\le|s-t|\); dominated convergence gives \[ \mathbb E[F_R(Jw)-F_R(w)]=0. \tag{42}\]

Choose a minimizing center \(c_R\). On a cluster \(C\) set \(a_C=\min_{x\in C}a(x/R)\) and \[D_C^\Phi=\sum_{x\in C} \{\Phi_M(w_x-c_R)-\Phi_M((Jw)_x-c_R)\}\ge0.\] Clusterwise Jensen and evaluation of the new minimum at \(c_R\) give \[ F_R(Jw)-F_R(w) \le-\sum_C a_CD_C^\Phi +C\varepsilon\sum_{C:C\cap\operatorname{supp}(a(\cdot/R))\ne\varnothing} \sum_{x\in C}H_x^k|(Jw-w)_x|. \tag{43}\] Indeed the cutoff oscillation on \(C\) is at most \(C\operatorname{diam}(C)/R\), while \(\Phi_M\) is \(2\varepsilon R\)-Lipschitz. The controlled displacement moments bound the expectation of the last sum by \(C R^2\); clusters reaching the support from outside are covered by a further fixed power of \(H_x\). Equations (42)–(43) therefore bound the normalized expected convex loss by \(C\varepsilon\).

For fixed \(\varepsilon>0\), volume sublinearity puts all but a vanishing fraction of the weighted sites within \(o(R)\) of a common center. The Huber minimizer is itself \(o(R)\) from that center. To verify this, the derivative from an exceptional fraction \(\delta\) of sites is at most \(2\varepsilon R\delta\) times total weight; the derivative from the remaining sites has a fixed sign and larger magnitude if the center is displaced by more than their oscillation plus \(C\varepsilon R\delta\). Thus such a displacement cannot minimize the loss.

First bound cluster size. The entire input of a typical interior cluster then lies in the quadratic region, with probability tending to one. Stochastic averaging keeps its output there. On that cluster \(D_C^\Phi=\sum_C(w_x^2-(Jw)_x^2)\), independently of the center. The stationary integrable quadratic loss allows the exceptional fraction to be discarded by truncating that loss before taking \(R\to\infty\). The cluster-size restriction is then removed using its moments. It follows that \(\mathbb Ed_0\le C\varepsilon\). Send \(\varepsilon\downarrow0\).

Each factor of \(J\) is an orthogonal averaging projection. Its quadratic loss is nonnegative, and the losses telescope. Zero total loss therefore makes every constituent projection fix its input. The relative heights are constant in time, and each actual contact equates its incident values. Lemma 27 connects every pair of neighboring sites and proves spatial constancy. This argument never conditions on the tangent solution, which may depend on future drives. ◻

Proof. Join two solutions of the same slope relatively independently over the full two-sided environment. The joining is invariant under space and time shifts. Their difference has zero slope; its dissipation and controlled displacements are bounded by constant multiples of the sums for the two solutions. Lemma 29 makes their relative heights identical. This proves uniqueness of flux, including conditional uniqueness given the environment. Addition of joined solutions and scalar multiplication preserve the controlled class. Hence the attained slopes form a vector subspace and the flux is linear on it.

The bulk law is invariant under lattice rotations and reflections, since it is the unique limit of the symmetric zero-boundary model. A nonzero attained slope and its sixty-degree rotation span \(\mathbb R^2\). The resulting linear flux map commutes with these rotations and reflections. A real two-dimensional map commuting with a nontrivial sixty-degree rotation is a scalar matrix plus a scalar multiple of the quarter-turn matrix; a reflection eliminates the latter. Thus \(j(q)=Aq\). Its coefficient is finite by the integrability of (44).

We justify passage of fluxes to expectations. At a volume-sampled site let \(D_n=n^2e_C^2/|C|\) and \(H_n=|C|\). A scaled step displacement, or the scaled cluster first moment per site, obeys \[|F_n|\le C H_n^a\sqrt{D_n}\] for a fixed \(a\). For \(1<r<2\), \[ \mathbb E|F_n|^r\le C(\mathbb ED_n)^{r/2} (\mathbb EH_n^{2ar/(2-r)})^{(2-r)/2}. \tag{46}\] The right side is bounded. Thus these fluxes are uniformly integrable; a moment of \(D_n\) larger than one is not needed. Exceptional localization events are removed using the polynomial error bounds in the comparison estimate before applying this calculation.

Test stationarity of \(z\) over one step against \(f(x/n)\). On each complete free cluster, the constant Taylor term sees only the added mass \(Tn^{-2}\sum_{x\in C}b(x/n)\), because reflections conserve the sum. The first-order Taylor term is the negative of the cluster first moment in (44), with the height differences multiplied by \(n\) and with an overall factor \(Tn^{-2}\). The second-order remainder has expected magnitude bounded by \[C n^{-3}\mathbb E\sum_{x\in n\operatorname{supp}(f)+O(H_x)} H_x^a|n(z_T(x)-z_0(x))|=O(n^{-1}).\] The same moment estimate handles clusters touching the edge of the testing region. The contribution of the added drift to the cluster first moment is also \(o(1)\). Lemmas 24 and 26, finite-cluster truncation, and (46) identify the limiting transport term as the conditional mean of \(j(q)\). The source and flux Riemann sums both carry the density \(v\), which cancels. Linearity and Lemma 26 give \(\mathbb E[j(q)\mid y]=j(\nabla m(y))\), proving (45). If the attained subspace is zero, this argument still applies and its flux is identically zero. ◻

Proof. Begin with zero fixed boundary values, and compare the untilted law with its tilt by a nonzero nonnegative smooth source. The filled coupled difference, extended by zero across its common pinned boundary, is bounded in \(W^{1,p}(\mathbb R^2)\) for some \(p>1\), by Lemma 21. To recall its geometric content, put the original difference equal to zero on the boundary and exterior, and retain the coercive edges incident to the common pins. Whenever a hole meets this connected zero set, a path to infinity exits through a zero rim vertex; choose that vertex as the hole’s anchor. The usual rim energy estimate holds for this anchor, so the extension is still bounded in \(W^{1,p}\) and remains exactly zero outside the polygon. Its expected value is therefore weakly compact in that space and strongly compact locally in \(L^p\). The original and filled expectations have the same limit.

Uniform convergence of the polygonal boundaries to the smooth boundary implies that this zero extension vanishes outside an arbitrarily small neighborhood of \(\overline D\). Its limiting extension consequently vanishes outside \(D\). The Sobolev trace theorem for a smooth domain gives \(m\in W^{1,p}_0(D)\). Thus no assertion about random slit covariance is needed for this ordinary boundary identification.

If only slope zero were attained, (45) would give \(\int b f=0\) for every compact test, a contradiction. A nonzero slope is therefore attained, so the cell flux is \(Aq\) on all of \(\mathbb R^2\). The same identity excludes \(A=0\).

For completeness, the zero-trace weak equation has a unique \(W^{1,p}_0(D)\) solution even before an energy-space conclusion is known. If \(u\) is the difference of two such solutions, take a smooth compact test \(g\) and solve \(-\Delta\phi=g\) with zero boundary data on smooth \(D\). The solution is smooth up to the boundary. Approximate \(\phi\) in \(W^{1,p'}_0(D)\) by compactly supported smooth functions in the weak harmonic equation, and integrate by parts using the zero trace of \(u\). This gives \(\int u g=0\). Hence \(m=A^{-1}(-\Delta_D)^{-1}b\). Monotonicity under the nonnegative tilt gives \(m\ge0\). The Dirichlet Green solution of a nonzero nonnegative source is positive, so \(A<0\) is impossible. This proves \(A>0\).

For arbitrary identical bounded boundary data, the same zero-extension argument and weak identity identify every mean-response subsequence with \(A^{-1}(-\Delta_D)^{-1}b\). Compactness therefore gives full convergence.

For the untilted mean, compare the prescribed-boundary bath with the zero-boundary bath on the same domain. The latter has mean zero by height reflection symmetry. Proposition 30, with source zero, makes every limiting prescribed mean harmonic. Monotone comparison with the zero-boundary law shifted by \(\pm\lambda\) bounds this mean between \(-\lambda\) and \(\lambda\). Near a compact subarc, compare instead with the zero-boundary law shifted by that arc’s prescribed value. The coupled difference has exact zero boundary on that subarc. A cutoff supported away from the marks permits the same local zero-extension argument, giving the stated trace. A bounded harmonic function in smooth \(D\) is uniquely determined by these two open-arc values; the two marked points have zero harmonic measure. This identifies the mean and every subsequence. ◻

Proof. For real \(f\in C_c^\infty(D)\) define \[X_n(f)=\frac1{vn^2}\sum_{x\in V_n} f(x/n)(h_x-\mathbb Eh_x), \qquad L_n(t)=\log\mathbb Ee^{tX_n(f)}.\] By 2, there is \(a_f>0\) such that \(\sup_n\mathbb Ee^{a_f|X_n(f)|}<\infty\). Tilting by \(e^{tX_n(f)}\) adds drift \(n^{-2}tf/v\) to the variance-two reflected diffusion. Proposition 31 therefore gives, for sufficiently small \(t\), \[L_n'(t)=\mathbb E_{n,t}X_n(f) \longrightarrow \frac{t}{vA}\int_D f(-\Delta_D)^{-1}f.\] For \(|t|\le a_f/2\), Jensen’s inequality bounds the denominator of the tilted expectation below by one, while \[|L_n'(t)|\le\mathbb E[|X_n(f)|e^{a_f|X_n(f)|/2}] \le C_{a_f}\mathbb Ee^{a_f|X_n(f)|}.\] Integration from zero and dominated convergence yield the Gaussian log moment-generating function near zero. The exponential bound gives tightness and uniform integrability of these moment-generating functions; uniqueness identifies every subsequential scalar law. Apply the same argument to every finite real linear combination of tests to obtain Gaussian finite-dimensional laws with covariance \((vA)^{-1}(-\Delta_D)^{-1}\).

For distributional tightness, enclose the physical domains in a fixed flat torus and extend the centered affine field by zero. A Fourier coefficient has vertex weights obtained by integrating a bounded mode against the nodal hat functions. Uniformly over the mode, their absolute sum is bounded, their maximum is \(O(n^{-2})\), and their support has lattice diameter \(O(n)\). The original boundary supplies a connected pin set of diameter comparable to \(n\). Proposition 2, applied at scale \(R=n\) to real and imaginary parts, therefore bounds all centered Fourier-coefficient second moments uniformly. Summing with Fourier weights gives a uniform \(H^{-s}\) second moment for every \(s>1\). Compact embedding into a more negative Sobolev space, followed by restriction to \(D\), proves tightness as random distributions.

Finally the Riemann pairing and affine pairing have vanishing difference. An interior nodal hat function has integral \((vn^2)^{-1}\) and, by the symmetry of its triangular star, zero first moment about its vertex. Taylor expansion of \(f\) therefore gives \[\int f h^{\mathrm{aff}}-\frac1{vn^2}\sum_xf(x/n)h_x =O\!\left(n^{-2}\frac1{n^2} \sum_{x\text{ near }n\operatorname{supp}(f)}|h_x|\right).\] Deterministic Lipschitzness and bounded boundary values give \(|h_x|=O(n)\) in the bounded rescaled domain, so this error is \(O(n^{-1})\). The same is true after centering. The scalar limits and tightness thus identify the asserted affinely interpolated distributional limit. ◻

Proof. Fix a deterministic sequence as in the statement. For fixed \(t\), [t:input-extension] and Proposition 38 give subsequential compactness of the mean filled response in \(L^p\), for a fixed \(3/2<p<2\), and a zero \(1/2\)-Frostman trace on \(K\). The original response has the same distributional limit. By (55) and Theorem 40, that limit is uniquely \[ m_t=\frac{t}{vA}G_{Q\setminus K}f. \tag{58}\] It is zero almost everywhere on \(K\), including its positive-area parts. This proves convergence against general smooth \(f\), rather than only against tests separated from the strings. Uniqueness removes the need to choose a further subsequence.

Set \(L_n(t)=\log\mathbb E_{n,s_n}e^{tX_{n,s_n}(f)}\). Differentiation of the finite-volume integral gives \[L_n'(t) =\frac1{vn^2}\sum_xf(x) \bigl(\mathbb E_{n,s_n,t}h_n(x)-\mathbb E_{n,s_n}h_n(x)\bigr) \longrightarrow t\sigma^2\langle f,G_{Q\setminus K}f\rangle.\] The factor \(1/v\) in the source is necessary: the tilt adds the microscopic drift \(t f(x)/(vn^2)\) to the reflected bath. No boundary term is added, since the walls and the exterior values are common to both laws.

The exponential moment in [t:input-moments] makes these derivatives uniformly bounded for \(|t|\le a/4\). Indeed the denominator of a centered moment generating function is at least one, and \[|L_n'(t)|\le \mathbb E_{n,s_n}\bigl[|X_{n,s_n}(f)|e^{a|X_{n,s_n}(f)|/4}\bigr]\le C_a.\] Integrating from zero proves (56). The same exponential bound gives tightness and uniform integrability of every fixed moment. Each limiting law is the Gaussian determined by this moment generating function near zero; hence the second moments converge. Applying this to linear combinations and polarizing gives cross covariances and the finite-dimensional conclusion.

The proof also works with \(f_n\) in a compact family converging to \(f\). Choose \(d_n\downarrow0\) sufficiently slowly that \((f_n-f)/d_n\) belongs to a compact family in \(C_c^\infty(Q)\). Apply [t:input-moments] to this additional compact family. Linearity then makes the change of the pairing tend to zero in every fixed moment. This uses the stated estimates for each fixed compact family, with its own constants, and requires no uniform constant over all such families. To obtain (57), suppose its supremum did not tend to zero and select a deterministic sequence realizing a fixed positive error. Hausdorff compactness and compactness of the test family give a convergent subsequence. The convergence already proved contradicts that error, once the continuity of the Green form is checked. That continuity is proved in the following lemma and is purely analytic. ◻

Proof. Put \(v_n=G_{Q\setminus K_n}g_n\) and extend by zero. Their variational identities and the Poincare inequality on \(Q\) give a common \(H^1\) bound. After taking a subsequence they converge strongly in \(L^2\) and weakly in \(H^1\) to \(v\). Every compact subset of \(Q\setminus K\) is eventually in \(Q\setminus K_n\), so \(-\Delta v=g\) there.

The zero traces pass to \(K\) by the deterministic version of the proof of Proposition 38: use the continuous, uniformly Frostman approximating measures from Lemma 36, before its optional projection onto lattice vertices. The \(H^1_0(Q\setminus K_n)\) approximation by compactly supported smooth functions and Lemma 35 show that \(v_n\) has zero trace on these measures. Atomic vertex traces of \(v_n\) are neither asserted nor used. There are no wall errors or filling maps in this case. The \(H^1\) bound implies the required \(W^{1,p}\) bound, and the averaging estimate passes the traces. Theorem 40 identifies \(v=G_{Q\setminus K}g\) on all of \(Q\). Thus every subsequence has this same limit. Strong \(L^2\) convergence and uniform convergence of \(f_n\) prove the assertion. For mesh-thick sets perform the measure construction only at scales larger than \(c/n\); its errors tend to zero before the fixed averaging scale is removed. ◻

Proof. Condition on \(S_n\) and apply the deterministic uniform error (57); its expectation is no larger than that error. Lemma 43 gives convergence of the random Green forms. They are uniformly bounded for bounded tests, because the Dirichlet Green quadratic form of every subdomain is bounded above by that of \(Q\). This also gives convergence of expectations and the stated \(L^1\) assertion under a common coupling. ◻

Proof. The Green function of \(V\) is bounded above by that of a fixed disk containing \(Q\), hence by \(C(1+\log^+(1/|x-y|))\). Dyadic summation with the Frostman bound gives \[\iint_{|x-y|<\rho}G_V(x,y)\,\mathrm d\mu_\varepsilon(x)\,\mathrm d\mu_\varepsilon(y) \le C\rho^{1/2}(1+|\log\rho|).\] For \(|x-y|\ge\rho\), a Brownian path started at \(x\) must leave a \(c\rho\)-neighborhood of a nearest point of \(K\) before it contributes to the Green function at \(y\). Beurling bounds that probability by \(C(\varepsilon/\rho)^{1/2}\). On the exit circle the Green function is at most \(C(1+|\log\rho|)\). Thus the remaining contribution is at most \(C(\varepsilon/\rho)^{1/2}(1+|\log\rho|)\). First send \(\varepsilon\) to zero and then \(\rho\) to zero. Restricting the measure preserves all these bounds.

For the last assertion, the inclusion \(H^1_0(V')\subset H^1_0(V)\) and the variational formula \[\langle f,G_Vf\rangle =\sup_{u\in H^1_0(V)} \left\{2\langle f,u\rangle-\int_V|\nabla u|^2\right\}\] give the quadratic-form inequality, also for signed tests. Apply it first to smooth measures and then use the finite logarithmic energy to pass to the stated measures. ◻

Proof. Construct an auxiliary height law in the portion of the original domain inside \(Q_3\). Retain all original fixed data and all sign constraints on its free vertices, and put zero values on the new artificial boundary. At a vertex already fixed by the original problem retain that original value. Drop sign walls on vertices newly fixed at the artificial rim; their values are now prescribed to be zero. The two models have exactly the same graph, fixed data, and sign walls in \(Q_2\) once \(n\) is large. A strict feasible filling for the auxiliary free coordinates is obtained by assigning small signed values at the sign sites and small values at the other free sites: each original fixed value has absolute value at most \(1/2\). Thus the auxiliary conditional polytope has positive volume. At a cut string endpoint, a free sign site whose opposite neighbor was removed is adjacent to the zero artificial boundary; it therefore remains bounded. Every truncated observed component reaching the artificial boundary joins the exterior observed set. The stated thickness assumption consequently puts this auxiliary law within the bounded domain class of Theorem 42.

Let \(h_n^{\rm loc}\) denote the auxiliary field. Put it and the original field \(h_n\) in the stationary synchronous joining supplied by 15, using the common walls in \(Q_2\). The artificial outer data may differ, as permitted in that local comparison theorem. The original marginal has uniform local average bounds from the nearby observed piece. The auxiliary marginal has them from its artificial boundary, or the retained observed piece. If \(U_n\) is the filled extension of \(z_n=h_n-h_n^{\rm loc}\) in \(Q_1\), its energy and squared Sobolev norm satisfy \[ \mathbb EE_n+\mathbb E\|U_n\|_{W^{1,p}(Q_1)}^2\le C_p, \qquad 3/2<p<2, \tag{62}\] with constants independent of the original ambient diameter and of \(\eta\). The coupling is constructed before choosing the test.

Transport \(\mu_{n,\eta}\) to the nearby common observed continuum, then to its mesh vertices and past the holes to retained vertices. The transport in Lemma 36 moves points by \(O(\eta)\); the last two movements are \(O(\ell_n/n)\) and give the coarse Frostman bound (48). The squared anchor estimate is \[\left(\sum_x\nu_n(x)|U_n(x)|\right)^2 \le E_n\sum_x\frac{\nu_n(x)^2}{w_x} \le CE_n n^{-1/2+3/(2q)}(\log n)^C, \qquad q>3.\] Its expectation tends to zero. Compare each testing value with its \(\eta\) average as in Lemmas 35 and 36. The bottom mesh scales contribute \(o(1)\) at fixed \(\eta\). Squaring these inequalities and using (62) gives \[ \limsup_{n\to\infty} \mathbb E\left|\langle\mu_{n,\eta},U_n\rangle\right|^2 \le C_p\eta^{2\beta},\qquad \beta=1-\frac{3}{2p}>0. \tag{63}\] All these inequalities are samplewise before expectation; the moved measure may depend on the holes. No independence between that measure and \(U_n\) is required.

It remains to transfer the upper bound to the original difference. At fixed \(\eta\), put \[Z_n=\langle\mu_{n,\eta},h_n-h_n^{\rm loc}\rangle, \qquad W_n=\langle\mu_{n,\eta},U_n\rangle.\] The original–filled \(L^1\) estimate in 15, together with the bounded test density at this fixed scale, gives \(\mathbb E|Z_n-W_n|\to0\). The marginal fourth-moment estimates in 2 make \(Z_n^2\) uniformly integrable. Their constants may depend on \(\eta\), which is harmless for this order of limits. For \(M>0\), the function \(x\mapsto\min\{x^2,M\}\) is \(2\sqrt M\)-Lipschitz, so \[\limsup_n\mathbb E\min\{Z_n^2,M\} \le\limsup_n\mathbb EW_n^2\le C_p\eta^{2\beta}.\] Letting \(M\to\infty\) proves \(\limsup_n\mathbb EZ_n^2\le C_p\eta^{2\beta}\). No \(L^2\) estimate for the original–filled error has been assumed.

The auxiliary domain has uniformly bounded diameter in these scaled coordinates. At each fixed \(\eta\), Theorem 42 applies uniformly to its compact family of tests, and Proposition 45 makes its limiting tested variance tend to zero as \(\eta\downarrow0\). Finally the coupling inequality \[\sqrt{\operatorname{Var}\langle\mu_{n,\eta},h_n\rangle} \le \sqrt{\operatorname{Var}\langle\mu_{n,\eta},h_n^{\rm loc}\rangle} +\|Z_n\|_2\] and (63) prove (61). The proof identifies no Poisson solution on an unbounded domain and imposes no condition at infinity. ◻

Proof. Write \(\ell_i\geq0\) for the normalized affine slack of facet \(F_i\). There is a finite constant \(C_i\), depending here on the polytope, such that \(\operatorname{dist}(x,F_i)\leq C_i\ell_i(x)\) for \(x\in P\). Indeed, express \(x\) as a convex combination of vertices, fix \(v_0\in F_i\), and replace each vertex \(v\notin F_i\) in that combination by \(v_0\); use the maximum of \(|v-v_0|/\ell_i(v)\) over these finitely many vertices. Tangency and Lipschitz continuity give \[|\nabla\ell_i\cdot u(x)| \leq C_i\operatorname{Lip}(u)\ell_i(x).\] Consequently, for sufficiently small positive or negative \(t\), \(T_t=I+tu\) maps \(P\) into itself and maps \(\partial P\) into \(\partial P\). It is injective when \(|t|\operatorname{Lip}(u)<1\). To prove surjectivity, fix \(y\in P\). The map \(x\mapsto\Pi_P(y-tu(x))\), where \(\Pi_P\) is Euclidean projection, is a contraction of \(P\) and has a fixed point \(x\). Put \(n=y-tu(x)-x\). The projection inequality says that \(n\) is an outward normal to \(P\) at \(x\), so \(n\cdot(y-x)\le0\). Tangency to every facet containing \(x\) gives \(n\cdot u(x)=0\), because the normal cone is generated by those facet normals. Consequently \(0\ge n\cdot(y-x)=|n|^2\), so \(n=0\) and \(T_t(x)=y\).

The Lipschitz area formula now gives \[\int_P \det(I+tDu(x))\,dx=|P|.\] For the piecewise affine field constructed below, this identity is elementary: subdivide \(P\) into its finitely many affine regions, apply the linear determinant formula on each region, and sum. Injectivity makes the image interiors disjoint, while their boundaries have zero volume; surjectivity makes their union all of \(P\). The determinant is positive for these \(t\). Comparing the linear and quadratic coefficients of this polynomial identity proves (66), since the quadratic coefficient is \(((\operatorname{div}u)^2-\operatorname{tr}((Du)^2))/2\). Finally \(|\operatorname{tr}(A^2)|\leq\|A\|_{\mathrm{HS}}^2\). Thus no integration by parts across corners or nonsmooth interfaces is required. The polytope-dependent restriction on \(t\) has disappeared from the resulting identity. ◻

Proof. Take \(\alpha=1/8\) and \(G_n=\{H_{\max}\leq n^\alpha\}\), using the enlarged threshold \(s_2\). By (65), \(\mathbb P(G_n^c)=o(1)\) uniformly. If \(h\in G_n\), then (68) gives \(\|u-f\|_\infty\leq Ln^{\alpha-1}\). For an edge with \(t_e\leq s_0\), its two shifts agree exactly. For every other edge the original slack is greater than \(s_0\), whereas \[|u_x-u_y|\leq L(2n^\alpha+1)/n<s_0\] for large \(n\). Thus \(h+u(h)\) is feasible. Walls are preserved because their sites are grounded.

To prove injectivity, suppose \(h,k\in G_n\) and \(h+u(h)=k+u(k)\). The prescribed \(f\) is the same for both fields, so \[ \|h-k\|_\infty\leq 2Ln^{\alpha-1}. \tag{72}\] The whole segment \(h_t=(1-t)h+tk\) belongs to the convex polytope. If an edge has \(t_e(h_t)\leq s_1\), then \[t_e(h)\leq s_1+2\|h_t-h\|_\infty \leq s_1+4Ln^{\alpha-1}<s_2\] for large \(n\). Every \(s_1\)-component anywhere along this segment is therefore contained in an \(s_2\)-component of \(h\) and has size at most \(n^\alpha\). Apply the component proof of (70) along the segment to get \[\|u(k)-u(h)\|_2 \leq C L n^{2\alpha-1}\|k-h\|_2.\] For a segment lying on a subdivision boundary, the same conclusion follows by restricting the continuous piecewise-affine formula to that segment; the tangential derivatives agree with limits from adjacent pieces. Since the coefficient tends to zero, the collision equation forces \(h=k\). Notice that this argument does not presume that \(G_n\) is convex.

On \(G_n\), \(\|Du\|_{\mathrm{op}}=o(1)\), so for large \(n\) the determinant of \(I+Du\) is positive and \[ \log\det(I+Du)\geq \operatorname{div}u-\|Du\|_{\mathrm{HS}}^2. \tag{73}\] Indeed, use the power series for \(\log(I+A)\) and \(|\operatorname{tr}(A^j)| \leq\|A\|_{\mathrm{HS}}^2\|A\|_{\mathrm{op}}^{j-2}\) for \(j\geq2\); the remainder is at most \(\|A\|_{\mathrm{HS}}^2\) when \(\|A\|_{\mathrm{op}}\leq1/2\). By (71), choose a fixed \(K=K(p_0,L)\) sufficiently large that \[\mathbb P\bigl[ |\operatorname{div}u|>K \ \hbox{or}\ \|Du\|_{\mathrm{HS}}^2>K\bigr]<p_0/4.\] For sufficiently large \(n\), \(\mathbb P(G_n^c)<p_0/4\) also. Intersecting \(E\) with the complementary good events gives \(E'\) of probability at least \(p_0/2\), and (73) gives \(\det DT\geq e^{-2K}\) almost everywhere on \(E'\). The area formula and injectivity now imply \[\mathbb P(T(E')) =\frac1{|P|}\int_{E'}\det DT(h)\,dh \geq e^{-2K}p_0/2.\] This proves the claim. For finitely many independent replicas use the product polytope and the block-diagonal map, or intersect the corresponding good events and multiply the determinant lower bounds. ◻

Proof. Every string site has an oppositely sign-constrained nearest neighbor, so Lipschitzness on their edge bounds both heights in absolute value by one. This supplies the local anchor for the stated patches and also for a singleton patch at any string site, including one far from the tips. Lipschitzness therefore bounds heights in the patch by a deterministic \(M=M(L)\). Enclose it in a larger fixed box and use Corollary 12 with a long-distance deficit greater than \(2M+20\); the small linear-slack version gives this by enlarging the fixed box. The good-shell event has probability at least \(1/2\), uniformly in the strand conditioning and the lengths of the additional upper intervals.

Fix such a shell and give the prescribed vertices, including the tested positive string site if present, their requested values \(\pm b\), where \(0<b<1/4\). Clip any feasible filling between the lower and upper distance cones of these prescriptions. At each other sign-constrained vertex the clipping interval contains zero, since that vertex has graph distance at least one from the prescribed vertices. Consequently clipping preserves its sign. Since the clipping interval contains zero, it also preserves every old interval \([0,d]\), without making \(b\) depend on \(d\), and preserves fixed exterior values of absolute value at most \(1/2\). The shell remains unchanged because of the long-distance deficit. This is the local feasible-filling construction underlying Lemma 11.

Let \(g\) be the resulting filling. For every original filling \(h\), the map \(h\mapsto t h+(1-t)g\), with \(0<t<b/(2(M+b))\), gives the required signs on the patch and makes every prescribed positive height greater than \(b/2\). Take \(b_*=b/2\) for the singleton assertion. Convexity preserves all the old constraints. If the number of free box coordinates is \(N\), its Jacobian is \(t^N\); both \(N\) and \(t\) depend only on the fixed patch and box. Integrating the conditional volume bound over good shells proves the result. For an interface extension prescribe the signs on the two sides of its finitely many edges. ◻

Proof. Inside \(Q_{\rm out}\) retain the union \(C\) of components of \(G\cap Q_{\rm out}\) meeting \(T\). A fixed pin incident to \(C\) before the outer rim is controlled. Otherwise a path from one of its tests to that pin, contained in \(Q_{\rm out}\) apart from one terminal lattice step, would have diameter less than \(c_0R\) for large \(R\), contrary to (SC1). Every edge from \(C\) to a removed component inside the square therefore passes through a controlled fixed pin. Edges at the outer rim are not re-pinned or deleted from the conditional baths: that rim is only an artificial boundary for the local estimate. It is a fixed positive ambient distance from \(Q_{\rm in}\). The theorem is applied once to this union, not separately with a constant for each of its possibly many components.

These are precisely the local one-sided hypotheses of Theorem 15. To see the localization in its proof, a transmission cluster which brings any uncontrolled input to an inner cutoff must reach the outer rim. Its diameter is then a fixed multiple of \(R\), so it is covered by the large-cluster estimate after (33). A fixed forcing box away from the rim has no uncontrolled incident pin; each of its free components reaches its artificial collar or a genuine controlled pin. Thus Lemma 19 and the allocation argument use the original annealed good marks. No estimate for a newly selected quenched pin vector is being asserted.

For clarity, the tail is zero-extended only across the controlled boundary inside \(Q_{\rm out}\), where every new edge has a zero pin between its retained and removed sides. This extension is uniformly lattice-Lipschitz there. Do not set it to zero across the artificial outer rim. The hole construction and the test cutoffs are taken in nested inner squares with a fixed buffer before that rim. A hole reaching from those squares to the rim has diameter comparable to \(R\) and is in the arbitrarily small exceptional event of Lemma 6. Off that event, a positive copied value \(U(x)=u(y)\) comes from a retained rim vertex \(y\) in \(C\). The full-window Lipschitz bound on the unmasked \(z\) gives \[U(x)^2\le(a-z(x))U(x)+C(1+d_x)U(x)\] for the lower tail, and the reflected inequality for the upper tail. A copied zero satisfies it automatically, even on a removed component. Pairing with the deterministic ambient cutoff is exactly (37)–(38); it never uses a randomly masked original field. The energy, Sobolev, and original–filled estimates now give (SC2). Retained controlled pins still have value zero. Conditioning on an event of probability at least \(p_*\) costs only its reciprocal in these annealed and exceptional-event estimates. ◻

Proof. The conditional law is log-concave and monotone in its boundary data and in its one-sided constraints. Lemma 50 gives a uniform positive chance that a tested positive string height exceeds some fixed \(b>0\), even after other positive sites have been restricted to \((0,d]\). A one-dimensional log-concave density on \([0,1]\) with this tail bound assigns at most \(C d\) mass to \([0,d]\). One proof integrates the chord lower bounds for its logarithm between a point in \([0,d]\) and a point carrying the prescribed tail mass. Thus, for small fixed \(0<d\le1/2\), the tested height exceeds \(d\) with conditional probability at least \(1/2\). Monotonicity makes the conditioning that all earlier indicators vanish the worst one. Successive uniform couplings show that the positive string heights dominate \(d\) times independent fair Bernoulli variables.

Write \(A\) for the full original string-height vector, and freeze all its coordinates. Retain the actual free component or union of components on the tested side. The positive portion in the preceding domination may be all positive string pins incident to this retained graph: the successive bound is uniform in their number, and the newly tested site is never one of the earlier upper-bounded sites. Put \(c_*=d/2\) for a string-only pure sector, and put \(c_*=(d/2)\wedge\lambda_*\) in a mixed sector. Using the same fair indicators, couple pins \(B_x\in\{0,2c_*\}\) below all these positive coordinates of \(A\), and extend to the remaining coordinates of \(A\) by disintegration. Conditional on \((A,B)\), compare the original Gibbs kernel on the retained graph with the kernel pinned to \(B\) at these positive string pins and to \(c_*\) at every remaining incident pin. The geometry of the retained graph is fixed by the strand observation, not by \(A\) or \(B\). The comparison mixture is invariant under \(h\mapsto2c_*-h\), so its mean is \(c_*\). Its pins lie in \([0,2c_*]\subset[-1/2,1/2]\) and have the uniform comparison estimates.

Rejoin the conditional kernels synchronously, keeping the conditional baths on every other original component. Lemma 16 then preserves the full original annealed marginal. Extend the comparison field by \(c_*\) off the retained graph, except at its \(B\) pins; this is a uniformly Lipschitz bounded-pin extension. Thus deterministic ambient tests of both full fields have their required anchored and zero-mass moments. For the lower tail \(u=(\widetilde h-h)_+\), all selected string pins are controlled because \(A_x\ge B_x\). Every incident positive exterior pin in the mixed sector is controlled because it is at least \(\lambda_*\ge c_*\). All remaining incident pins failing this inequality are included in \(P_{\rm u}\) of Lemma 51; none is silently discarded.

The access hypothesis permits a fixed finite cover of a testing neighborhood by inner squares whose outer diameters are smaller than that margin. In each square take \(T\) to include every free vertex under the smooth tests in the screened neighborhood. Lemma 51 gives the filled lower tail with uniformly bounded Sobolev norm on this cover, and (SC2) identifies its pairings with the original tail there. Exhausting such test supports identifies the distributional response only on the open screened side, which is all that is needed.

For the trace, choose a component \(C_0\) of \(G\cap Q_{\rm out}\) meeting \(T\), and a compact accessible pure subarc of the boundary of its free-vertex hexagon union, contained in a smaller square compactly inside \(Q_{\rm in}\), of diameter at least a fixed multiple of \(R\). Trace that hexagon boundary along the subarc. The neighboring fixed hexagons are incident pins; consecutive ones meet at a tiling vertex and their triangular-lattice centers are adjacent. The local access condition makes all these pins controlled. They therefore give a connected controlled chain of diameter comparable to \(R\). This chain has the required local thickness, and its vertices can be moved past the polylogarithmic holes to retained controlled pins. Apply the localized proof of Proposition 38, with \(3/2<p<2\), only to measures on a smaller compact part of this chain. The Sobolev square is its buffer; the artificial rim is not used as an anchor. It gives zero trace for every limit of the filled tail mean there. Together with the interior identification above this proves the needed tail inequality. No harmonic equation for the tail is used. Jensen’s inequality bounds the negative part of the limiting mean difference by this nonnegative zero-trace limit. The two field means are bounded, harmonic away from the observed sets, and their remote contribution is bounded by a constant times the harmonic probability of reaching an endpoint or an incompatible side. This probability tends to zero along Brownian approaches to the pure portion, by Beurling. The lower trace is therefore at least \(c_*\). All bounds used are uniform for each deterministic admissible conditioning sequence, which gives the claimed sequential uniformity. Reflect the field for the negative statement.

The graph-access condition has a concrete meaning for the uses below. After freezing all strand-adjacent heights, every primal edge crossed by a dual strand has both endpoints fixed. A free path cannot cross that strand without first ending at one of those pins. A nearby negative pin on the other side may still be incident to the global component of an open prefix, but it is not reachable from the tested side by crossing the slit; any remaining route to it must go around an endpoint or reach an incompatible boundary, and the asserted access margin is exactly what excludes such a short route. In the endpoint collars this margin is verified below. For the completed reference component with exterior values \(\pm1/2\), take all incident positive string pins in the coupling. Every other incident pin on its positive component is \(+1/2\ge c_*\), and no negative pin is incident: the crossed-edge incidence also closes the two marked ends, as detailed in Lemma 71. Here \(P_{\rm u}\) is empty and its access diameter is infinite. This completed-component application, not an open-tip comparison, is the strict positivity input for the reference height gap. ◻

Proof. Condition on all heights on the negative side. At a positive string vertex, a crossing edge gives an upper bound at most one. Delete the crossing edges and replace these bounds by independent walls \([0,1]\). This raises the conditional field by monotonicity. Raising the exterior positive data to \(1/2\) raises it again. The resulting positive-component law is invariant under \(h\mapsto1-h\), so its mean equals \(1/2\). Average the comparison over the conditioned negative heights. The negative statement follows by reflection. Disconnection of the two components by the full crosscut justifies deletion of precisely these crossing constraints. ◻

Proof. We give the two-field proof. Explore every cluster of \(h_1-h_2\geq K\) connected to a deterministic transverse seed. Every queried vertex away from the seed has difference at least \(K-2\): positive queries satisfy the threshold, and failed queries are adjacent to a positive query. Freeze both samples on the common queried set, then use the stationary synchronous joining of their conditional laws. The exploration decisions used only recorded values; hence this replacement preserves each annealed marginal. All statements below are annealed over this transcript, not asserted for every possible conditional pin value. Use a middle testing subcorridor a fixed positive distance from the transverse seed. The free-buffer hypothesis removes all original pins and sign walls from its larger comparison window; the newly queried common pins there satisfy the displayed threshold inequality. Thus the one-sided comparison has no uncontrolled original boundary in that window.

Away from the seed, set \[u_K=\left(1-\frac{h_1-h_2}{K-2}\right)_+ .\] It has zero data on the revealed set. The local truncated comparison estimate in Theorem 15, applied after division by \(K-2\), gives compact filled extensions with uniformly bounded local Sobolev energy. Here the cutoff energy grows at most quadratically with the truncation level: after division it is bounded by the squared cutoff norm and the macroscopic second moments of \(h_1/(K-2)\) and \(h_2/(K-2)\). Proposition 2 controls these moments uniformly. The same bounds hold after conditioning on an event of probability bounded below, with its reciprocal as a constant.

Suppose a crossing had probability bounded below along \(K\to\infty\) and \(n\to\infty\). On that event the revealed set contains a connected continuum of fixed positive diameter away from the seed. Pass to a subsequence jointly with this continuum and the filled extensions. Macroscopic averaged field tightness gives \((h_1-h_2)/(K-2)\to0\) as a distribution in this iterated limit. Since \(u_K\geq1-(h_1-h_2)/(K-2)\), its limit is at least one almost everywhere. The varying-set anchor argument of Proposition 38 gives zero trace for that same limit. Here one uses its samplewise inequalities, without a Poisson equation for the truncated tail. Choose a probability measure on the crossing continuum by coordinate projection, with ball masses bounded by \(Cr\) in physical units. Move it along the revealed set past each hole to a retained zero vertex. The maximum hole diameter is polylogarithmic outside the negligible exceptional event, and the crossing component cannot be contained in such a hole. Lemma 35 bounds the integral of a radius-\(r\) average of the filled tail by \(C(r^\beta+o(1))\|U_n\|_{W^{1,p}}\), for some \(\beta>0\). These inequalities hold for the random measure before expectation. Joint compactness and strong local \(L^p\) convergence permit the mesh limit at fixed \(r\); the bounded conditional Sobolev norm permits \(r\downarrow0\). This proves the zero trace. A \(W^{1,p}\) function which is at least one almost everywhere cannot have zero trace on a continuum carrying a fixed positive-diameter Frostman test with exponent greater than \(2-p\). This is a contradiction.

For one field retain the entire original field on the deterministic testing window. Given its transcript, compare its unvisited Gibbs kernel with a symmetric field zero-pinned on the revealed set and original exterior, and extend this comparison field by zero on its fixed coordinates. Keep the original field, not its restriction to a random unvisited component, in the full-window difference. For each recorded set, deterministic ambient tests of the comparison field have the bounds of Proposition 2; the revealed crossing component provides a macroscopic anchor on the crossing event. The original full field keeps its annealed bounds by disintegration. Queried nonseed differences are at least \(K-1\), and the same lower tail with \(K-1\) has zero input there. The free buffer excludes all original walls from this comparison window. The preceding argument therefore applies without a random mask on the original field. ◻

Proof. The field part of \(\Phi_{R'}\) constrains only sites adjacent to the external arms, hence outside radius \(R'-O(1)\). The sites specifying \(\beta_R^\sigma\) lie inside radius \(R+O(1)\). Choose comparison disks between these two radii, with outer radius \((R+R')/2\) and inner radius \(R+(R'-R)/4\). For sufficiently large \(R\), they have a free intervening annulus of fixed relative width. The original base observations are all outside the outer disk, and the external arms give the required anchors. Thus Lemma 56 applies to the two orientations of the inner strand, comparing the exterior data with their global sign reversal. Together with sign symmetry it gives a uniformly positive probability that the two observed unoriented configurations have compatible signs on the side containing the walk. In that case join corresponding interface tips and the two walk tips by three polar centerlines. We may assume \(a\le1/100\), since otherwise the lower-bound hypothesis is empty. After rounding their same-radius angular gaps are at least \(a/2\). Put \(s=aR/(800\pi)\) and \(w=s/100\), reserve straight radial stems of length \(20w\) at both ends, and interpolate the angles linearly between the stems. Since \(R'-R\ge R/8\) and \(40w\le R/(200000\pi)\), the angular slopes are at most \(25\pi/R\). At a common radius the distance between centerlines is at least \(.49aR\). If two radii differ by at least \(s\), that gives the required separation directly; otherwise moving along one centerline to the common radius costs at most \((1+125\pi)s<.158aR\). Thus the mutual separation is at least \(s\). The observed external arms, the two free interface centerlines, and the inner strand complete to one prospective simple oriented chord \(\Gamma_0=K\cup J\), where \(K\) is the observed field system and \(J\) its free completion. The compatible orientation gives its two sides \(\Omega_+\) and \(\Omega_-\).

Offset a lane by \(w\) on each side of each interface centerline. At a tip the two offset lanes meet its radius-\(4w\) circle at transverse offsets \(\pm w\) and radial depth \(\sqrt{15}w\), up to bounded lattice error. Follow from the chosen lane the circular arc away from the radial stem. The long arc between the two lane points stays at distance at least \(w-O(1)\) from \(J\); the other connector is farther by the centerline separation. Its endpoints are on opposite ideal sides, so this arc must meet the old boundary arc at that end. At its start the full-boundary clearance is greater than \(3w\), since the old record is on the old side of its stopping circle and the other record is remote. Stop at the first point of full-boundary distance \(w/10\) and append a globally shortest segment to \(b\in\partial D(K)\). This segment remains in the selected side; its endpoint is on the appropriate old pure bank, even if that bank shadows the tip. Moreover \[\begin{aligned} d(b,J)&\ge .89w-O(1),\\ d(b,\hbox{all observed tips})&\ge3.8w-O(1),\\ d(b+tu,\partial D(K))&=t\quad(0\le t\le w/10). \end{aligned}\]

Take \(d=w/(200A)\), with one fixed sufficiently large \(A\). To check the endpoint pocket, in each ideal side join free collars of the two connectors \(J_1,J_2\) by the same-side arc on the middle-radius circle. This arc is free and has macroscopic clearance, since the annulus has no old observations. Delete the triangles meeting \(K\) and use the access-resolved component containing this connected reference set. The deleted triangles attach to its old boundary arcs, so its resolved completion is a disk. The radius-\(d\) crosscut met by the nearest prefix misses the reference set. The retained cap arc has clearance \(w/10>d\), and a short transverse segment at its lane end reaches the local radial stem with clearance greater than \(d\). Thus the tail and both connector collars are on the reference side of the crosscut. The other side is bounded by that crosscut and one pure old boundary arc, and contains the initial nearest prefix. This is the required pocket; the prefix crosses each smaller circle once. After the cap, a stem has radial depth at least \(\sqrt{15}w\), and the interpolating portion is at radial distance at least \(20w\) from either stopping layer before the offset. Subtracting the offset and curvature/lattice errors leaves clearance greater than \(w/10\); the retained cap itself has that clearance by its first-stop definition. Choose the tube width so that its enlargements are at most \(w/100\). Use barrier scale \(\bar R=12R\). All four barriers lie in \(B_{6R}\) and have radial span at least \(R/8-O(w)>d\), so their diameters are less than \(\bar R\) and greater than \(d\). Thus \[\theta=\frac{d}{\bar R} =\frac{a}{192\,000\,000\pi A}.\] The remaining lanes have clearance \(w/10\), so (UB1) holds with clearance constant \(1/2\) after lattice error. The full endpoint modules have radius at most \(d+L\ell_c\le w/100\). Hence the sign-\(\pm\) confined domains and prescription supports lie in \(\Omega_\pm\) and have distance at least \[w-w/10-w/100-w/100-O(1)\ge .85w\] from \(J\), including lattice error. A free path between opposite sides cannot cross the fixed \(K\), so it must cross \(J\). This verifies (UB2) at scale \(\bar R\) with \(\kappa=.85w/(12R)>0\). Proposition 55 applied to these four signed barriers forces the interface joins within the disjoint corridors with uniformly positive probability. Planarity forces the outer arms to merge with the corresponding inner arms; there is no other exit through the fences.

The remaining walk bridge must reach the joining vertex of the reversed inner piece. Its conditioning to reach this vertex before zero has probability bounded below. Harnack gives a positive probability to follow the free third corridor to a fixed relative neighborhood of that vertex; the two-dimensional point-hitting estimate gives probability comparable to \(1/\log R\) to hit it there. The loop portion between its first and last visits stays in that corridor with probability bounded below, by the last-exit decomposition. This proves the lower bound. Conditional on any completed field configuration, hitting the same vertex before a nearby interface continuum has probability at most \(C/\log R\); this proves the upper bound. These are precisely the walk hypotheses of [SSdiscrete]: radii in the specified compact ratio range, observer outside the larger disk, and two prescribed first/last walk pieces. All field estimates used in its corridor construction have been supplied above.

We record the two augmented upper bounds used below. First, the upper bound still holds after the whole field, hence \(\gamma\), is fixed; the middle walk bridge remains unexposed. Second, put \(H=(\Phi_{3r},\Theta_{2r})\), and expose that middle walk only until its first exit from a circle of radius \(\rho/3>10r\), if this occurs before its prescribed terminal piece. Call this stopped prefix \(S_{**}\). On this escape event, \[\mathbb P(Z^\sigma\mid H,S_{**},\hbox{escape}) \leq C/\log r. \tag{GK}\] For verification, let \(\hat q\) be the joining vertex of the fixed reversed inner piece. The first-hit/last-exit decomposition leaves a walk from the current exit vertex to \(\hat q\), conditioned to hit \(\hat q\) before zero, followed by its last-exit loop. Harnack bounds the conditioning probability below by a fixed constant, since the exit vertex is outside \(B_{10r}\) while \(|\hat q|=2r+O(1)\). To realize \(Z^\sigma\), this first part must hit \(\hat q\) before the connected inner interface. The same point-versus-continuum estimate as above is at most \(C/\log r\); the remaining loop can only reduce the probability. This argument holds for every completed field consistent with the records, and averaging gives (GK). It does not expose the later successful bridge. ◻

Proof. Write \({\cal F}_{r,s}=\sigma(\Phi_r,\Theta_s)\), and write \(\widehat q_{\rm o},\widehat q_{\rm i}\) for the two numerical quality formulas above, before the following common live test. An exact pair is called open if both pairs of strand tips exist, neither observed interface has closed or prematurely merged, and the observed walk has made no forbidden contact with any observed strand or with the original exterior; the prescribed terminal contact at zero is the sole exception. The raw qualities are zero if these conditions fail. A pair is called live when \(\mathbb P(Z^\sigma\mid{\cal F}_{r,s})>0\); its two qualities are the raw qualities when live and are set to zero otherwise. A live pair is open. We shall update only one of the two records at a time. For the outer record put \(\rho=r\) and \(e=-1\), and for the inner record put \(\rho=s\) and \(e=1\). If the selected record has quality \(q>0\), put \[L=q\rho,\qquad \rho_1=(1+10eq)\rho .\] Let \(q_1\) denote its quality after this update, including the live-pair test, with the other record held fixed. We first prove, uniformly for \[3.3R<r\le4R,\qquad R\le s<1.7R,\qquad 0<L<R/200,\qquad L\le\rho_{\rm other}q_{\rm other},\] where \(\rho_{\rm other}\) and \(q_{\rm other}\) are the radius and live quality of the unchanged record, that \[\mathbb P\bigl(q_1\ge2q\mid{\cal F}_{r,s}\bigr)\ge c_0>0. \tag{G3}\] Here \(q=L/\rho<1/200\), so the cap is inactive. The geometric, barrier, and walk construction will be proved in the wider layout buffer \[2.8R<r\le4R,\qquad R\le s<1.84R,\qquad 0<\widehat q\le1/100,\] for any jointly admissible open pair, using its selected raw quality \(\widehat q\), again denoted by \(q\) there. Only the positive-support step will use the smaller- physical-separation premise in (G3). For the low-quality updates used later we also prove the following joint form. Put \(F=\{q_1<2q\}\), and, on \(F\cap\{q_1>0\}\), make one further update of the same record to \(\rho_2=(1+10e q_1)\rho_1\), leaving the other record fixed. Put \(q_2=0\) on \(\{q_1=0\}\). If \(q<1/500\), \(r>3.25R\), and \(s<1.61R\), then \[\mathbb P(F,q_2=0\mid{\cal F}_{r,s}) \ge c_1\mathbb P(F\mid{\cal F}_{r,s}). \tag{G4}\] Both subsequent radii in (G4) remain in the wider layout buffer just displayed. The estimates condition on the stopped records, not on \(Z^\sigma\).

Proof. Before hookup conditioning, the field observations in each \(\Phi_{3R}\) are only the signs adjacent to its two external arms, supported outside radius \(3R-O(1)\). The event specifying an oriented \(\beta_{2R}^\sigma\) tests vertices inside radius \(2R+O(1)\). Apply Lemma 56 with inner disk of radius \(9R/4\) and outer disk of radius \(5R/2\). Both base laws have no observations in this outer disk, their graphs agree there, and their external arms connect scale \(3R\) to the original boundary outside scale \(6R\), providing the anchors. Thus the inner oriented-strand laws are comparable by a fixed factor under these different external records. Summing this comparison over the two possible orientation lifts gives the same bound for the unoriented inner record; no landing side is added to that record or conditioned on. At this step no \(\Theta\)-record has already been imposed in either base law: the inner record is precisely the additional event being compared. The reversed-walk law is also comparable by a fixed factor: a walk from either entrance point has probability bounded below to visit the other before reaching zero. Apply the strong Markov property at that visit. The field and walk factors are independent before hookup. Thus there is a reference weight \(p_{2R}(\theta)\) for inner records such that their probabilities given a record at radius \(3R\) are comparable to this weight.

Choose the quality threshold in Lemma 58 with error \(1/2\), and let \(X\) be the inner records at \(2R\) exceeding it. Let \(\mathcal Q\) be the event that both the \(2R\) inner and \(3R\) outer records exceed it. Separation holds even conditional on a specified \(\Theta_R=\vartheta\). Hence conditioning further on \(\mathcal Q\) changes the probability of this specified inner record by at most a fixed factor. On \(\mathcal Q\), apply Lemma 57 with inner radius \(2R\) and outer radius \(3R\). Its ratio condition holds, and its clearance requirement is precisely \(\mathfrak B_{12R}\), as assumed. Bayes’ formula and the reference-weight comparison give \[\mathbb P(\Theta_{2R}=\theta\mid\mathcal Q,Z^\sigma,\Phi_{3R}) \asymp \frac{p_{2R}(\theta)}{\sum_{\xi\in X}p_{2R}(\xi)} \quad(\theta\in X).\] Indeed the numerator before normalization is comparable to \(p_{2R}(\theta)/\log R\), and summing determines its normalization. Average over \(\Phi_{3R}\), then sum over all \(\theta\in X\) whose restriction to radius \(R\) is \(\vartheta\). The resulting reference ratio is independent of the external experiment, proving (G7) first for single records, then for events. Conditioning on data farther out and averaging over the radius \(4R\) records preserves the same constant. ◻

Proof. Apply Lemma 58 in the outer buffer with base radius \(R/4\); its clearance requirement is \(3R<4R\). In the inner buffer the base radius is \(r\), and its requirement \(12r<4R\) holds after increasing \(A(\varepsilon)\) so that \(A(\varepsilon)>4\). Separation in these buffers makes all four records at scales \(2r,3r,R/2,3R/4\) have quality at least \(c(\varepsilon)\), except with probability \(O(\varepsilon)\). Put \(\rho\asymp\sqrt{rR}\). Let \(H=(\Phi_{3r},\Theta_{2r})\). The walk portion \(S_*\) from its first radius-\(3r\) entrance to the last visit, before zero, to the endpoint of its radius-\(r\) reversed piece is unlikely to leave \(\mathfrak B_{\rho/3}\) without hitting the external arms. The ordinary Beurling estimate and first-hit/last-exit decomposition bound this probability, conditional on \(H\), by \(b(R/r)\to0\). If it escapes, reveal only its prefix \(S_{**}\) up to first exit. The augmented bound (GK), not a bound conditioned on the completed walk, gives \[\mathbb P(S_*\not\subset\mathfrak B_{\rho/3},Z^\sigma\mid H) \leq C b(R/r)/\log r.\] Divide by the lower bound \(c(\varepsilon)/\log r\) on separated records and average over them. This proves confinement under the landing law.

For the outer construction, use only \(\operatorname{ext}_{3r}S\). Let \(S'\) be its terminal path from its last exit of \(\mathfrak B_{R/3}\) to its endpoint at radius \(3r\). Choose measurably from \(S'\) an arc on the circle of radius \(\rho\) whose union with \(S'\) separates the inner and outer circles. Intersect this open circular arc with \(D(\operatorname{ext}_{3r}\gamma)\). Its two components next to \(S'\) have other endpoints on the boundary of this deleted-triangle domain, not on the interface itself; call them \(\alpha_1,\alpha_2\). Thus an endpoint attachment does not already count as an interface intersection. They are \(\Phi_{3r}\)-measurable. For each \(i=1,2\) separately, in the component on that side place nested barriers separating \(\alpha_i\) from the inner circle. The deterministic construction is the clearance construction in [SSdiscrete]: take a separating circular arc; if its middle has adequate boundary clearance, attach shortest endpoint segments, and otherwise repeat at the smaller distance between its two facing pure boundary pieces. The procedure terminates because the two boundary pieces are disjoint. Use the shortest endpoint segments and fixed relative endpoint radius of the circular-clearance layout above. Thus each output satisfies (UB1), although its lattice diameter may be small. Barriers selected in distinct dyadic annuli remain separated.

For a fixed \(i\), after exposing the two interface arms to one barrier, the next single-sign barrier satisfies Proposition 55, including its bounded-scale case. Thus crossing \(N\) barriers in that access has probability at most \((1-c)^N\), conditional on \(H\) but not on the landing. Apply this separately to the two accesses, without conditioning on the other shielding event; their union costs a factor of two. For \(F=\{\gamma\cap(\alpha_1\cup\alpha_2)\ne\varnothing\}\), fix the completed field while retaining the unexposed walk bridge. The first augmented upper bound following (G1) gives \[\mathbb P(F,Z^\sigma\mid H) =\mathbb E[1_F\mathbb P(Z^\sigma\mid H,\gamma)\mid H] \leq \frac{C}{\log r}\mathbb P(F\mid H).\] The same lower normalization gives \(C(\varepsilon)(1-c)^N\) under the landing law, where \(N\to\infty\) with \(R/r\). Repeat the whole construction from the inner reversed record \(\Theta_{R/3}\), without also revealing the forward intermediate bridge.

Outside these exceptional events, confinement identifies the terminal forward path \(S'\) with the corresponding portion of the reversed inner record. The same deterministic circular-arc selection then makes the two crosscuts from the separate descriptions coincide. Their boundary endpoints therefore attach to lattice edges incident to both the inner incident strand and the two external arms. These strand attachments are on opposite sides of the incident edge: the other possibility contradicts the winding number of the circular separator completed along the common walk path. This is the deterministic Jordan-curve conclusion in [SSdiscrete]; its hypotheses are the two coinciding crosscuts, a simple interface, and the common walk avoiding it before zero, all obtained above. Thus each external arm has merged with a distinct inner arm, and there can be no additional annular interface crossing.

The total error is \(C\varepsilon+o_{R/r}(1)+C(\varepsilon)(1-c)^N\). Choose the buffer error first and then \(R/r\) large. Average over \(\Theta_r\) and the six possible incident edges for the external-only statement. ◻

Proof. We adapt the full-walk argument of [SSdiscrete]. We first establish pure-arc repulsion. For \(y\in\partial_+\), choose any deterministic lower bound \(b\leq d_D(y,\partial_-)\). Start the fixed-ratio iteration at \(\rho\le b/8\), losing only \(O(1)\) outer scales. Take the component of \(B(y,\rho)\cap D\) accessible from \(y\), and its circular crosscut separating \(y\) from \(\partial_-\). Use the clearance construction of Lemma 60, with \(\delta\le1/100\) and \(c_A\le\delta/100\) in the circular-clearance layout. Write \(\beta_1,\beta_2\) for the two components of \(\partial_+\setminus\{y\}\). Its endpoint \(z_j\) is accessible from \(y\) by a path of diameter at most \(2(1+2\delta)\rho+O(1)\). Its internal separation \(\sigma_j\) from \(\beta_{3-j}\) is greater than \(\ell_{\rm att}\) in the wide case, and in the narrow case is at least \(\delta\,\delta^k\rho\), with \(\ell_{\rm att}\le\delta^k\rho\). For the endpoint radius \(d_{\rm e}=c_A\ell_{\rm att}\le\delta\rho/100\), in both cases \(\sigma_j>3d_{\rm e}\).

Let \(\chi_j\) be the first circular crosscut at radius \(d_{\rm e}\) met by the initial segment from \(z_j\). Each endpoint of \(\chi_j\) joins \(z_j\) along that segment and \(\chi_j\) by a route of diameter at most \(2d_{\rm e}\). It cannot lie on \(\beta_{3-j}\), by \(\sigma_j>3d_{\rm e}\). It cannot lie on \(\partial_-\), since concatenating with the preceding access from \(y\) gives diameter at most \[2(1+2\delta)\rho+2d_{\rm e}+O(1)<b\] at diverging scales, contrary to \(b\le d_D(y,\partial_-)\). Prior exposed arms stay outside by the annular separation. Thus both endpoints of \(\chi_j\) lie on \(\beta_j\). The barrier tail avoids \(\chi_j\): its own nearest segment crosses the circle once, and the wide middle has clearance \(\ell_{\rm att}>d_{\rm e}\). If the other nearest segment met \(\chi_j\), its remaining length to \(\beta_{3-j}\) would be at most \(d_{\rm e}\); together with the preceding \(2d_{\rm e}\) route this would contradict \(\sigma_j>3d_{\rm e}\). Hence \(\chi_j\) and the intervening \(\beta_j\) subarc cut off the pure side containing the initial prefix. This is its endpoint pocket \(A_{z_j}\). At each stage this is one positive barrier, so no mixed layout is used. Explore the interface only up to the next outer crosscut. Its exact sign transcript leaves the inner barrier compatible, so Proposition 55 gives a fixed conditional chance of avoiding this crosscut and every smaller access to \(y\). Iterating, then reflecting for \(y\in\partial_-\), gives constants \(C,\zeta>0\) such that \[\mathbb P\{d_D(y,\gamma)<r\} \leq C(r/b)^\zeta+o_d(1), \tag{GB2}\] when \(r,b\) are fixed positive multiples of \(d\) and \(r<b\). The lattice error is uniform for these fixed multiples. This is the deterministic crosscut argument of [SSdiscrete], with its field input replaced by our exterior version of Proposition 55.

Fix \(r=\eta d\), and condition on \(\gamma\). At the first walk visit to \(M\) at distance less than \(r\) from \(\partial D\), the strong Markov property and the planar walk estimate give a fixed probability of exiting \(D\) in the ball of radius \(2r+O(1)\) about a closest boundary point. One can force the walk to surround that point inside this ball, which forces the exit. Its continuation and the adjacent interface edge then give an internal path of diameter at most \(4r+O(1)\) from the exit \(Y=S_{\tau_D}\) to \(\gamma\). Consequently \[\mathbb P\{\operatorname{dist}(M,\partial D)<r\} \leq C\mathbb P\{d_D(Y,\gamma)<4r+O(1)\}. \tag{GB3}\] Here \(Y\), unlike the interface landing, is independent of \(\gamma\). Conditional on \(Y\), apply (GB2) whenever its internal distance to the opposite arc exceeds \(s=\sqrt\eta\,d\), using the deterministic lower bound \(b=s\) in that estimate. This gives \(C\eta^{\zeta/2}+o_d(1)\).

It remains to bound exits within internal distance \(s\) of both sign arcs. Delete all paths of diameter at most \(2s\) joining the two arcs and retain the component of \(v\). Its new boundary lies in two balls of radius \(4s\). To verify this elementary fact, classify each deleted crosscut by which marked sign-change endpoint it separates from \(v\). Any two crosscuts in the same class which touch the retained component intersect: otherwise their separation order prevents one of them from touching that component. Their diameter bound puts the whole class in one ball. There are two classes. An exit near both arcs is beyond this new boundary. The walk hitting estimate for these two balls gives \(C(s/d)^{1/2}=C\eta^{1/4}\). Combining with (GB3) proves (GB1).

Every nonexterior old-prefix landing is a vertex of \(M\), pathwise, because the prefix is contained in \(\gamma\) and the full walk has not yet exited \(D\). This proves the simultaneous assertion without conditioning on any landing event or on any stopping time. ◻

Proof. The nested shielding argument follows [SSdiscrete], using Proposition 55. Expose the first interface approach to the circles centered at \(v\) of radii \(2^{-j-1}d\). After the first approach to radius \(\rho\), use the free closed circle of radius \(3\rho/4\) as one single-sign barrier. It has no endpoints, and has fixed relative clearance from both the observed strands and the original boundary. Thus (UB1) holds without an attachment module. Proposition 55 gives a fixed chance of protecting the next smaller disk. Failure through \(N\) scales has probability at most \((1-p)^N\). Fix \(N\) first and let \(d\to\infty\), so every used scale diverges; then let \(N\to\infty\). This proves (GB4). It also excludes all terminal and old-prefix landings within \(\eta d\) of \(v\), up to one lattice spacing. ◻

Proof. Conditional on an exact stopped prefix \(K\), the field has the ordinary strand-conditioned law: whether the rule stopped at this prefix is already decided by that prefix. The independent walk adds no field conditioning. Put \(U=D(K)\). For small fixed \(\delta>0\), except on an event tending uniformly to zero with \(\delta\), the following four conditions hold: \[\begin{split} &d_U(X_T,\hbox{opposite-sign side})\geq2\delta d,\\ &\operatorname{diam} S[0,\tau_T]<\delta^{-1}d,\\ &\operatorname{diam} S[t_*,\tau_T]<\delta d/4, \quad t_*=\inf\{t:d(S_t,\partial U)\leq\delta^2d\},\\ &d(v,K)\geq\delta^{1/2}d. \end{split} \tag{G10}\] The middle two statements are the planar walk hitting estimate uniform in \(U\). The last statement follows from Lemma 62 and \(K\subset\gamma\). For the observer-separated stopping times needed in the driving argument, the last condition can instead be imposed deterministically.

For the first condition delete all crosscuts of diameter at most \(4\delta d\) joining the two sign sides. The new boundary of the component containing \(v\) lies in two balls of radius \(8\delta d\). Indeed each deleted crosscut separates \(v\) from one of the two sign-change points; any two crosscuts for the same point which touch the retained component must intersect. All such crosscuts are consequently within twice their diameter of a fixed one. Apply the walk hitting estimate to those two balls, using the last condition. This also explains why arbitrary narrowing of \(U\) does not invalidate (G10).

Fix records satisfying (G10), and suppose the landing is positive. The deterministic clearance construction used in Lemma 60 gives a compatible crosscut \(\Upsilon_0\) in \(B(X_T,\delta d)\setminus B(X_T,\delta d/2)\), joining the positive boundary pieces on either side of \(X_T\) and separating it from the negative side. If no hexagon centered at a vertex of the stopped walk meets this crosscut, avoidance of this crosscut protects the entire walk. Otherwise add the boundaries of the walk hexagons on its unprotected side and take the boundary facing the negative side. Its diameter is at most \(2\delta^{-1}d\); its new part is at distance at least \(\delta^2d/2\) from \(\partial U\), by the terminal-walk condition. Indeed a contacting walk hexagon occurs before \(t_*\), since the terminal walk piece lies in \(B(X_T,\delta d/4)\), whereas the crosscut lies outside \(B(X_T,\delta d/2)\). That contact is at distance at least \(\delta^2d-O(1)\) from \(\partial U\); therefore the boundary-attached original crosscut has diameter at least \(\delta^2d-O(1)\) as well. The old endpoints retain their pure pockets and their shortest initial segments. In the contacting case choose their endpoint radius \(d_{\rm e}\) also below \(\delta^2d/(100A)\). The fixed relative circular construction and the preceding diameter bound make this a fixed further shrink depending only on \(\delta\). The added part cannot enter \(B(z,d_{\rm e})\) and has clearance much larger than \(d_{\rm e}\). Together with the circular-clearance qualification this proves (UB1) for the modified crosscut. It is one landing-sign barrier with parameter \(c\delta^3\) after a fixed shrink, at scale \(2\delta^{-1}d\).

Proposition 55 gives probability at least \(p_\delta>0\) that the future interface avoids the resulting barrier. On this event it is adjacent to no vertex of the stopped walk, so \(Y=X_T\). In the first branch \(\Upsilon_0\) can have bounded lattice diameter; this is covered by the all-scale part of that proposition. Reflect signs for a negative landing. Choose \(\delta\) to make the failure probability of (G10) smaller than \(\varepsilon\), and then take \(d\) sufficiently large. ◻

Proof. Let \(\mathcal G=\sigma(K,S)\), \(A=\{Y=X_T\}\), and \(q=\mathbb P(A\mid\mathcal G)\). Lemma 63 implies, for every \(\mathcal G\)-measurable event \(E\), \[\mathbb P(E)\leq\varepsilon_1+p^{-1}\mathbb P(E\cap A). \tag{G12}\] Take \(E\) to be the event that the old prefix has more than two crossings of an annulus about its walk landing. On \(A\), this implies the corresponding event for the completed interface at its completed harmonic landing. Lemma 60, with error smaller than \(p\varepsilon_1\), therefore removes all additional old excursions from a small comparison window.

Exclude landings close to the old tip using the walk hitting bound. Lemmas 61 and 62 exclude landings close to the original boundary or observer, uniformly over all prefixes. Choose the largest comparison window smaller than one tenth of this retained clearance. This verifies the original boundary and observer exclusions when Lemma 60 is applied after recentering at the landing.

In the remaining window exactly two incident old arms reach the outer circle, and the old tip lies outside it. There are two relevant complementary components. For each component separately, under the same conditioning \((K,S)\), select nested single-sign circular-clearance barriers attached to its pure old side, with the shortest endpoint segments and fixed relative endpoint radii specified above. After each failed shield, reveal the future arms up to that barrier; the conditional field law is again an exact-strand law, so the next shield succeeds with probability at least a fixed \(c>0\). Failure through \(N\) scales in that access has probability at most \((1-c)^N\). We do not condition one access estimate on the other shielding event; the union bound costs two. When neither complementary component is traversed, the future cannot enter the innermost Euclidean ball. The removal of additional old excursions is what converts this internal-access conclusion to Euclidean nonreturn.

Choose \(\varepsilon_1\), obtain \(p\) from survival, choose the completed excursion error below \(p\varepsilon_1\), choose \(N\), and then let all the finitely many retained scales tend to infinity in lattice units. Finally choose their innermost relative radius \(s\). The sum of the discarded probabilities can be made less than \(\varepsilon\), proving (G11). ◻

Proof. We first recall the required deterministic screening argument. For a simply connected polygonal slit \(U\), two complementary prime-end arcs, and \(R>0\), let \(L_R\) be the union of internal paths of diameter at most \(R\) joining those arcs. The frontier of the observer component of \(U\setminus L_R\) is contained in two balls of radius \(2R\). Indeed, classify a crosscut by which of the two arc endpoints it separates from the observer. Two crosscuts in one class touching the observer component must intersect, and their diameter bound then puts that entire class in one ball. The argument uses prime ends on a slit, not an identification of its two sides. This is the planar argument in [SSdiscrete]. Beurling’s estimate therefore gives \[\mathbb P_z\{\text{Brownian motion enters }L_R\} \le C(R/\rho)^\alpha .\] Apply this to each arc against its complementary arc. For a partition into \(m\) cyclic arcs, entrance into the union of short paths joining different labels has probability at most \(Cm(R/\rho)^\alpha\), for a fixed \(\alpha>0\).

The lattice walk has the required physical incidence. Every primal edge crossed by an explored dual edge has both endpoints observed and absorbing. Thus the interpolated walk cannot cross the dual strand before it is killed. An interior killing vertex has a connector of diameter \(C/n\) through its own dual hexagon to an incident strand edge, on the side bearing its sign. If several strand edges are incident, every such access has that same sign. At original boundary the connector is trivial and the incoming edge specifies its access. Consecutive explored dual edges cross primal edges belonging to a common triangle; their absorbing endpoints form a connected set attached to the original absorbing boundary. This set lies within \(C/n\) of the physical boundary and reaches across the annuli used below. It supplies the connected obstacle for the discrete Beurling estimate, including at the last explored dual vertex.

Fix \(n^{-1}\ll r\ll R\ll\rho\). Couple the unrestricted interpolated walk, with its diffusive clock, and Brownian motion so that their distance up to a fixed time cutoff is less than \(r/4\), except with probability \(o_n(1)\). The ordinary planar invariance principle gives this coupling independently of the slit; the triangular walk is isotropic. Both exit-time tails at the cutoff are bounded by those for a fixed enclosing ball. For each process use its own first entrance into the \(2r\) boundary layer. Beurling’s estimate at these two marginal stopping times confines each remaining tail to diameter \(R\), except with total probability \(C((r+n^{-1})/R)^\alpha\). No Markov assertion about the joint coupling filtration is made.

Let \(\theta\) be the first time either coupled path is within \(r\) of the physical boundary. On successful coupling both paths are still alive, both of their \(2r\) layer times have occurred, and their positions at \(\theta\) can be joined by a segment inside \(U_n\). If the exit labels differ, their confined tails, this segment, and the lattice terminal connector give an internal path of diameter \(CR\) joining different labels. The Brownian path has therefore entered the corresponding mixed-access set. The preceding screening bound gives \[\mathbb P\{\text{label mismatch or exit distance}>CR\} \le o_n(1)+C(r/R)^\alpha+Cm(R/\rho)^\alpha +\text{cutoff error}. \tag{H3}\] Take the time cutoff large, then \(R\) small, then \(r/R\) small, and finally \(n\) large. This proves (H2). In particular the proof compares the walk directly with the physical dual slit, not with an unmarked Hausdorff approximation or a thickened domain with possibly many label changes. ◻

Proof. The protected observer and the common enclosing ball make the normalized inverse maps a nondegenerate normal family. Kernel convergence identifies their limit with the observer component. Extract also all transition angles on the unit circle, preserving their cyclic order. The associated bounded step functions converge at every angle except the finitely many limiting transitions. Poisson integration proves compact convergence; normalized arclength gives zero mass to those transitions. Colliding transitions merely delete an arc of vanishing harmonic mass. This retains the boundary marks rather than reconstructing them from the limiting closed set.

For approach samples, use a free Brownian path independent of all domain and field records. At fixed positive distance from the boundary, kernel convergence and compact harmonic convergence give convergence of its smoothed readings. The boundary-layer tail estimate in the preceding proof lets that distance decrease to zero after taking the mesh limit. Equivalently, the normalized conformal maps give the limiting Brownian exit angle.

Finally suppose the two finite domains and their labels agree on the component of a ball of radius \(R\) next to a common oriented strand. Their bounded label-Poisson extensions differ, at a point in the concentric ball of radius \(R/2\) and at distance \(r\ll R\) from that strand, by at most \(CM(r/R)^\alpha\): the difference is zero on the common local boundary, and Beurling bounds escape from the ball. The same bound holds if the common boundary is an original arc. It is uniform in the domains, so it persists in marked kernel limits. The mixed-access screening in the preceding proof excludes conflating distinct sides in this procedure. This proves the asserted local identification, without claiming continuity of prime-end labels from closed sets alone. ◻

Proof. Pass to a subsequence on which \(\nu_n\) and the laws of all \(U^i_n\) converge. Conditional integration gives \[\mathbb E|U^i_n-V^i_n| \ge c\,\mathbb E\left[\mathbf1_{G^i_n} \int|U^i_n-y|\,\nu_n(\,\mathrm dy)\right].\] The lost contribution outside \(G^i_n\) is at most \(2M\mathbb P((G^i_n)^c)\). If \(\mu_i\) and \(\nu\) are the limiting laws, bounded continuity yields \(\int|x-y|\,\mu_i(\,\mathrm dx)\nu(\,\mathrm dy)=0\). A product of two probability measures supported on the diagonal must consist of the same point mass in each coordinate. The point is common to all \(i\) because \(\nu\) is common. Agreement of the readings completes the proof. ◻

Proof. For the first assertion compare the two fields with common reflecting sign walls. On shared strands their limiting mean difference has zero Brownian-tested trace by the comparison and trace estimates. The common original fixed data give zero differences at the corresponding pins, so the same trace argument gives zero trace on the common original boundary pieces. Away from these observed boundaries the difference is harmonic by the weak equation and \(A>0\). Its absolute value on the other boundary pieces is at most \(2M\). The maximum principle therefore bounds it by \(2M\) times the probability of escaping the common ball before hitting the connected common strand. Beurling bounds that probability by \(C(r/R)^{1/2}\). For arbitrary sequences with \(r\to\infty\) and \(r/R\to0\), rescale by \(r\) and first restrict to a common ball of radius \(Lr\) about a nearest strand point, with \(L\) fixed. Centrality puts this ball inside the shared region for all sufficiently large scales. Local \(L^1\) compactness transfers its \(CM L^{-1/2}\) bound to the fixed-cutoff averages. First take the mesh limit, then \(L\to\infty\), and finally remove the averaging cutoff. Sequential uniformity gives the assertion for every admissible conditioning sequence.

For the second assertion apply the identical argument in the common exterior region, with the small changed window as the only uncontrolled boundary. The probability of reaching that window from distance \(r\) before hitting a connected common arm is at most \(C(\rho/r)^{1/2}\). The bounds persist under averaging in a sufficiently small ball about the observation point. In both assertions the zero trace is a conclusion of the common-wall comparison, not an equality of the sampled wall heights. ◻

Proof. Regard each nonvanishing event \(E_{n,\sigma}\) as one member of the finite collection; independence between these members is not needed. If none remain, the assertion about strand samples is vacuous. Work under each branch’s conditional law, disintegrated over the landing location and translated as above. We use the resulting \(Z^\sigma\) notation, always with the landing sign still averaged. The actual orientation remains part of each signed conditional mean. We construct auxiliary signed inner and outer readings \(\widetilde V_n,\widetilde U_n\) which approximate the same signed boundary sample. Their absolute values will be the readings \(V_n,U_n\). The inner absolute reading, unlike its signed auxiliary version, will be a function of the unoriented record and so will admit the common lower comparison from (G7).

For a target error, first use (74) to discard strand landings within distance \(\eta d_n\) of the original boundary or observer, choosing \(\eta>0\) so that their probability is small. Recenter a retained landing at zero. Choose the lattice radii \[1\ll r_{\rm in}\ll R_{\rm core}\ll R_{\rm in}\ll R_{\rm mid} \ll R_{\rm conn}\ll r_{\rm out}\ll n. \tag{H1}\] Require also \(r_{\rm out}<\eta d_n/20\). Thus the disk \(\mathfrak B_{12R_{\rm in}}\) used by (G7) and the disks \(\mathfrak B_{4R_{\rm in}}\), \(\mathfrak B_{4R_{\rm conn}}\) used by (G8) lie inside the original domain and exclude the observer. Clearance is measurable in the outer record, which includes the domain and observer; retaining it imposes no additional condition on the inner strand. Write \(\mathcal A_n=\sigma(\Phi_{R_{\rm mid}})\) under this side-unconditioned \(Z^\sigma\) mixture, and let \(G_n\in\mathcal A_n\) be the retained clearance event. The diagonal choices below make \(\mathbb P(G_n)\to1\).

Agreement of complete-profile readings. First take mesh limits at fixed positive physical approach distances. In each limiting component the bounded harmonic martingale has a boundary limit. Its values at first entrances into balls of radii tending to zero about the eventual landing have that same limit: these entrance times approach the exit time pathwise. They need not be stopping times after the future-dependent center is specified.

At an interior approach point use a radial smooth average in a ball smaller than one tenth of its distance to the strand and of the approach radius, with the fixed probability kernel defining \(P_s\) in Section 6. The limiting harmonic average equals the value at its center. Impose a positive lower cutoff on the averaging radius before taking the mesh limit, then remove it. At each fixed positive physical approach scale the sampling point is interior almost surely, so this cutoff loses arbitrarily little probability. The comparison compactness and the vanishing original–filled \(L^1\) error give convergence of the conditional means in \(L^1\) on every compact subset of the limiting free regions, on a joint subsequential realization. At a fixed radius cutoff \(a>0\), the sampling supports have distance at least \(9a\) from the strand; the retained landing clearance and (H1) also keep them a fixed positive distance from the original exterior. Hausdorff convergence therefore places all these supports in one compact subset \(C_a\) of the limiting free regions. Since the kernels have supremum norm at most \(\|\varphi\|_\infty a^{-2}\), the discrepancy of their readings is bounded uniformly by \(\|\varphi\|_\infty a^{-2}\|m_n-m\|_{L^1(C_a)}\), which tends to zero. This transfers both readings to the lattice, including for kernels selected from the records. Denote the signed complete-profile readings at \(r_{\rm in}\) and \(r_{\rm out}\) by \(B_n^{\rm in}\) and \(B_n^{\rm out}\), assigning zero when their kernel cutoff fails. Such failures have vanishing probability. A diagonal choice of the radii in (H1) makes both readings converge in \(L^1\) to \(Y\) on the joint subsequential realization, and in particular \(\mathbb E|B_n^{\rm in}-B_n^{\rm out}|\to0\). These statements hold under the side-unconditioned branch law: the bounded errors under the original sampling law are divided only by the branch’s positive probability lower bound. The scales are chosen slowly; no convergence rate is asserted.

The reading determined by the inner record. Write \(\theta=(\beta,w)\) for the unoriented record \(\Theta_{R_{\rm in}}\). An admissible connected incident strand through \(e_\sigma^*\) has exactly two lifts to adjacent sign observations: choosing the sign at the landing vertex \(0\) propagates the signs along the nonbranching strand. Denote by \(o_+(\theta)\) the lift for which \(0\) is positive, and by \(o_-(\theta)\) its sign reversal. This deterministic choice of a lift only defines an auxiliary function of \(\theta\); it does not condition the sampled field on that sign.

For \(o\in\{o_+,o_-\}\), let \(m^{\rm can}_{\theta,o}\) be the conditional mean in the centered rounded disk of radius \(2R_{\rm in}\), with zero exterior heights and precisely the strand signs in \(o\). Choose the rounding equivariantly under the six lattice rotations. These polytopes have positive volume: assign small heights \(+b\) and \(-b\) to their respective string sites and zero to the other free sites, with \(0<b<1/2\). All free edge inequalities and sign inequalities are then strict. The volume-preserving map \(h\mapsto-h\) exchanges the two canonical polytopes and fixes the zero exterior data. Consequently \[m^{\rm can}_{\theta,o_-}(z)=-m^{\rm can}_{\theta,o_+}(z) \quad\text{at every }z. \tag{H1a}\]

Apply (G8) at \((R_{\rm core},R_{\rm in})\) and average its conditional bound. Except on an event of vanishing probability under \(Z^\sigma\), the walk does not leave \(R_{\rm in}\) after its first entrance at \(R_{\rm core}\). Its first entrance at \(r_{\rm in}\) therefore belongs to the recorded terminal piece, and reversing that piece determines the sampling point. Any complete-chord piece in \(B(0,R_{\rm core})\) other than the incident strand \(\beta_{R_{\rm in}}^\sigma\) would give two additional crossings of the same annulus. Outside the same type of exceptional event, the canonical conditioning with the actual oriented lift and the complete conditioning thus agree throughout \(B(0,R_{\rm core})\).

Choose the inner radial kernel using the distance to the recorded geometric strand and the cutoff just described; denote it by \(\kappa_\theta\). Both its center and its radius are functions only of \(\theta\). For a valid approach point and cutoff put \[v_{n,o}(\theta) =\int\kappa_\theta(z)m^{\rm can}_{\theta,o}(z)\,\mathrm dz, \qquad F_n(\theta)=|v_{n,o_+}(\theta)|.\] If the record has no admissible incident strand, does not contain the required approach point, or fails the kernel cutoff, set both \(v_{n,o}\) and \(F_n\) equal to zero. These fallback rules are geometric. Equation (H1a) gives \(v_{n,o_-}=-v_{n,o_+}\), including the fallback cases. Thus \[V_n:=F_n(\Theta_{R_{\rm in}}) =|\widetilde V_n|, \qquad \widetilde V_n:=v_{n,o_{\rm actual}}(\Theta_{R_{\rm in}}). \tag{H1b}\] The inner record space is countable, and the lift, kernel, and fallback choices are deterministic. Hence \(V_n\) is one bounded measurable function of the unoriented record, the same in every experiment, and does not use \(G_n\). The absolute value is taken after testing an oriented canonical mean; averaging the two orientations before taking the absolute value would instead give zero and is not the construction.

Apply Lemma 69 to each actual oriented conditioning, with \(r_{\rm in}/R_{\rm core}\to0\). Its sequential uniformity allows these errors to be averaged over the actual orientations under \(Z^\sigma\). The extra-excursion estimate supplies both the common strand neighborhood and the presence of the chosen approach point in the recorded walk. Boundedness absorbs its exceptional events and the cutoff failures, so \[\mathbb E|B_n^{\rm in}-\widetilde V_n|\longrightarrow0.\] No landing sign has been conditioned on in this expectation.

The reading determined by the outer record. Retain the external arms and initial walk in \(\Phi_{R_{\rm mid}}\). Delete every oriented dual-edge observation whose adjacent primal sign pair meets \(B(0,R_{\rm conn})\); retain the remaining external-arm edge observations and both vertices of each pair. Every retained sign vertex still has an oppositely constrained neighbor.

Apply (G8) at \((R_{\rm mid},R_{\rm conn})\) and average its bound. Except on an event of vanishing probability under \(Z^\sigma\), the unrecorded connection of the two external arms stays inside \(B(0,R_{\rm conn})\): the arms already supply two disjoint annular crossings, and an excursion of their connection would supply two more. Hence the modified and complete conditionings agree outside \(B(0,R_{\rm conn}+C)\), with an absolute lattice constant \(C\). All retained observations are determined by \(\Phi_{R_{\rm mid}}\). The first walk entrance at \(r_{\rm out}\) is also in that record, since it precedes the first entrance at \(R_{\rm mid}\).

Use a radial kernel smaller than one tenth of \(r_{\rm out}\) and of the distance to the recorded external strands, with a positive lower cutoff as above. Its support stays outside the changed window. The mean under the modified conditioning, tested with this kernel, is \(\mathcal A_n\)-measurable, since that record includes the oriented external-arm observations; denote this signed reading by \(\widetilde U_n\). Set \(\widetilde U_n=0\) when its required outer clearance or kernel cutoff fails, and put \(U_n=|\widetilde U_n|\). In the common exterior component, limiting mean differences are bounded harmonic functions with zero trace on the unchanged strands and original boundary. The only uncontrolled boundary is the deleted window, where the difference has absolute value at most \(2M\). A connected common arm crosses the intervening annulus. The maximum principle and Beurling’s estimate therefore bound the influence of this window at the sampling ball by \[C M(R_{\rm conn}/r_{\rm out})^{1/2}.\] The bounded enlargement by \(C\) is negligible at these scales. Sequentially uniform compactness and trace comparison transfer the bound to the lattice with an error tending to zero. After including the bounded exceptional contributions, this gives \(\mathbb E|\widetilde U_n-B_n^{\rm out}|\to0\). The inequality \(\bigl||a|-|b|\bigr|\le |a-b|\) now gives \[\mathbb E|U_n-V_n| \le \mathbb E|\widetilde U_n-B_n^{\rm out}| +\mathbb E|B_n^{\rm out}-B_n^{\rm in}| +\mathbb E|B_n^{\rm in}-\widetilde V_n| \longrightarrow0. \tag{H1c}\] All three errors are integrated under the side-unconditioned branch law. In particular the absolute value is taken only after the signed locality and harmonic-mean arguments.

One common lower kernel and the order of limits. Fix a geometric edge label \(\sigma_0\). Write \(F_{n,\sigma}\) for the function just constructed in branch \(\sigma\), and let \(T_\sigma\) be the lattice rotation fixing \(0\) and carrying \(e_\sigma^*\) to \(e_{\sigma_0}^*\). The zero exterior canonical disk, the lift making \(0\) positive, the radial kernel, and every fallback rule commute with this rotation. A spatial change of variables therefore gives \[F_{n,\sigma}(\theta) =F_{n,\sigma_0}(T_\sigma\theta). \tag{H1d}\] For each sufficiently large \(n\), choose one admissible reference external record \(\phi_n^*\) at radius \(4R_{\rm in}\) with label \(\sigma_0\) and positive hookup probability. Such a record can be taken from any nonvanishing branch after its outer clearance and a rotation: the clearance probability tends to one, and the discrete record space is countable. Let \(\nu_n\) be the law on \([0,M]\) obtained by pushing \[\operatorname{Law}_*(\Theta_{R_{\rm in}}\mid \Phi_{4R_{\rm in}}=\phi_n^*,Z^{\sigma_0})\] through \(F_{n,\sigma_0}\).

For any fine external record \(\phi\) in branch \(\sigma\) with positive hookup probability and satisfying the disk hypotheses of (G7), rotate the experiment by \(T_\sigma\) and apply the orientation-averaged (G7) to the event \(\{F_{n,\sigma_0}(\Theta_{R_{\rm in}})\in B\}\). Equation (H1d) yields \[\mathbb P(F_{n,\sigma}(\Theta_{R_{\rm in}})\in B \mid\Phi_{4R_{\rm in}}=\phi,Z^\sigma) \ge C^{-1}\nu_n(B) \tag{H1e}\] for every Borel set \(B\). No landing-side sign is present in this conditioning. On \(G_n\), the translated domain and observer meet the disk hypotheses of (G7) for every compatible fine record, because that clearance is already in the outer record.

Since \(R_{\rm mid}>4R_{\rm in}\), the record at \(R_{\rm mid}\) is a restriction of the record at \(4R_{\rm in}\). Disintegrate also over the original landing location in the branch mixture. Conditional integration of (H1e) over that location and the fine record, given \(\mathcal A_n\), gives on \(G_n\) \[\mathbb P(V_n\in B\mid\mathcal A_n)\ge C^{-1}\nu_n(B).\] The reference pushforward includes all geometric fallback records. Neither the extra-excursion events nor successful kernel cutoffs restrict this lower comparison. Edge-branch normalization was used only for the averaged errors, and no factor is paid here for labels or unused annuli.

Here is the scale order for a finite collection of limiting Brownian boundary readings and a requested error less than \(\varepsilon>0\). First extract the edge-branch probabilities; the finitely many nonvanishing branches then have a common positive probability lower bound. Choose the clearance exclusion before any physical buffers. Choose the outer physical approach scale for both the boundary-reading error and the retained clearance. Next choose \(R_{\rm conn}/r_{\rm out}\) for exterior locality and \(R_{\rm mid}/R_{\rm conn}\) for its excursion error. Choose \(R_{\rm in}\ll R_{\rm mid}\), then \(R_{\rm core}/R_{\rm in}\) for the inner excursion error, and finally \(r_{\rm in}/R_{\rm core}\) for inner locality and the inner boundary reading. These choices are in physical units before the mesh limit, so every retained lattice scale diverges. Hold the scales and kernel cutoffs fixed while taking that limit, and then remove the cutoffs. Repeat for \(\varepsilon\downarrow0\) and take a diagonal subsequence. The comparison constant in (G7) is independent of all these unused buffers. Lemma 68 now applies to \(U_n,V_n\). Equation (H1c) and the convergence of the complete readings identify their common limiting variable with \(|Y|\), not with a side-conditioned signed value. ◻

Proof. Freeze the full vector \(A\) of strand-adjacent heights in the exact complete-strand kernel. On almost every positive-volume fiber the remaining conditional laws are the pinned Gibbs kernels of their free primal components, as in Section 3. Every primal edge crossed by the dual crosscut has both endpoints among the frozen string vertices. Removing these vertices therefore disconnects the positive and negative free graphs. A pin incident to a positive free component is either a positive string pin or a pin on the positive original arc: an edge from a positive free vertex to a negative pin would cross the dual strand and would make that free vertex a frozen string vertex. At a mark the two endpoints of the marked boundary edge are fixed, and the crossed edge in its incident triangle has frozen string endpoints. The exterior is fixed as well, so there is no free route around either end of the crosscut. This also verifies the asserted incidence at the marks.

All pins incident to a positive free component are consequently nonnegative. Compare its conditional Gibbs kernel with the kernel on the same component having all these pins set to zero. The monotone single-site heat-bath coupling used in Lemma 9 orders the two kernels, and the zero-pinned kernel has mean zero by \(h\mapsto-h\). Thus \(\mathbb E[h(x)\mid\gamma_n,A]\ge0\) at each positive free vertex. Integrating over \(A\) preserves this inequality; it already holds at positive string vertices by their observed sign. Reflection gives the negative inequality.

These assertions are literal at primal vertices. Affine triangles mixing the two sides lie within \(O(n^{-1})\) of the dual slit. Every compact subset of a limiting component avoids them for large \(n\), so the compact-interior convergence of the means preserves the sign throughout that harmonic component. Its Brownian martingale and its boundary limit preserve the same weak inequality. Lemmas 66 and 67 identify that component sign with the incident-side label of the sampled access. ◻

Proof. Include a fixed reference Dobrushin experiment with original boundary heights \(\pm1/2\) and a fixed interior observer. Choose compact subarcs \(J_+\Subset A_+\) and \(J_-\Subset A_-\) of positive length. Let \(p_n^\pm\) be the probabilities that the independent full walk, stopped only on the original boundary, exits on their lattice approximations. Full-domain walk convergence gives \(p_n^\pm\to\omega_D(v,J_\pm)>0\). Conditional on any completed chord, an observer not already adjacent to it lies in one of its two components. A full walk exiting on the opposite \(J_\pm\) must cross the chord first. The absorbing-incidence fact in Lemma 66 makes it visit a strand-adjacent vertex before that exit. Since \(J_\pm\) avoid the marks, this forced crossing cannot be the marked boundary edge; if a crossed edge has a boundary endpoint, its interior endpoint is already absorbing before boundary priority can apply. An already adjacent observer has a strand landing at time zero. Hence, for some fixed \(\delta>0\) and all large \(n\), \[\mathbb P(\xi_n\hbox{ is a strand landing}\mid\gamma_n) \ge \min(p_n^+,p_n^-)\ge\delta. \tag{H1f}\] This is a deterministic separation argument for each chord and does not use a barrier estimate.

Compactness supplies joint mean profiles and harmonic samples. Proposition 70 and Lemma 68 make their absolute strand-boundary samples equal to one deterministic constant \(l\ge0\) on every nonvanishing geometric edge branch, shared by all included experiments and labels. To pass back to all strand landings, for every error event \(D_n\) use the finite-union bound \[\mathbb P(D_n,\ \xi_n\hbox{ is a strand landing}) \le\sum_{\sigma=1}^6\mathbb P(D_n\cap E_{n,\sigma}).\] Zero-limit branches have vanishing unnormalized mass, and on every other branch the magnitude error tends to zero. The overlaps between the \(E_{n,\sigma}\) cause no problem. Lemma 71 now gives \(Y=s|Y|=sl\) on a strand access with label \(s\).

For strict positivity, (H1f) and the finite cover give a reference edge branch with positive limiting probability. Choose the unnormalized landing-exclusion errors smaller than a fixed fraction of that probability, and then choose the conditional (G8) errors small. A positive limiting mass of accesses remains away from the original boundary and observer, with only the incident strand in the inner core and with the recorded walk approaching its accessed side. In particular the marks, which lie on the original boundary, and the ends of this local strand sector are outside the smaller testing scale.

Here is the access condition for applying Lemma 52 on these reference sectors. In a positive completed component, freeze all strand-adjacent heights \(A\) and take the positive portion in that lemma’s Bernoulli coupling to be all positive string pins incident to the component. For this mixed sector take \(\lambda_*=1/2\) and \(c_*=(d/2)\wedge1/2\) as in that lemma. Its pins \(B_x\in\{0,2c_*\}\) satisfy \(B_x\le A_x\), and every original-arc pin there is \(+1/2\ge c_*\), the remaining comparison boundary value. All incident pins therefore satisfy the lower-tail inequality. By the free-graph separation in Lemma 71, the retained graph \(G\) is a union of full common unqueried components: no free–free edge is deleted, all string walls have been frozen, and no negative pin is incident, including at the marks. Thus \(P_{\rm u}=\varnothing\) in Lemma 51. With fixed vertices terminal as in (SC1), every test \(x\in G\) satisfies \(\delta_G(x,P_{\rm u})=\infty\), which exceeds every required fixed-multiple access margin. The negative case follows by reflection. This is a graph-access assertion, not a Euclidean separation from the nearby opposite string pins.

At fixed kernel cutoff the support stays a positive fraction of the diverging testing scale from the dual strand, so it avoids the bounded lattice layer of frozen string vertices and lies in this unqueried component. The exclusions and (G8) give the isolated local access at diverging scales. Fix the oriented complete strand and the independent walk record defining the test before sampling the auxiliary pin transcript \((A,B)\). Given the strand, this walk record carries no additional height information. The square cover in Lemma 51 can therefore be fixed before that transcript, as required. Apply the pure-side trace bound of Lemma 52 directly to the completed-component original/comparison pair just described, for each actual orientation. Its original marginal is the complete-strand kernel after averaging the frozen heights, so this bounds the complete-profile trace itself; the canonical open-strand pair is not used in this positivity step. The sequentially uniform bound may then be averaged over the two signs; it gives \(sY\ge c_*>0\) on the retained positive-mass reference accesses. This pure-access trace bound is the strict positivity input. Since \(|Y|=l\), it yields \(l\ge c_*>0\). Finally Lemma 53, applied to the completed reference profile together with Lemma 71, gives \(|Y|\le1/2\) and hence \(l\le1/2\). No upper bound for the canonical open-strand system is used.

Apply this argument to a countable collection of interior balls, with a diagonal subsequence. If a terminal profile had a set of incorrect signed strand-boundary values of positive harmonic measure from any ball in that collection, its sampled value would have positive probability of differing from \(sl\) on its label-\(s\) access. The finite edge cover would detect that discrepancy on a nonvanishing branch, contrary to the preceding conclusion. Thus the statement holds almost surely in every component seen from that exhaustion. Original boundary arcs retain their prescribed values by the boundary comparison, rather than by the height-gap argument. ◻

Proof. It suffices to treat an arbitrary sequence \(T_n\), since failure of the supremum would select a counterexample sequence. Use one independent full walk \(S\). Write \(\xi\) for its old-prefix landing, \(\upsilon\) for its completed-strand landing, and set \[\mathcal G_n=\sigma(\gamma_n[0,T_n],S),\qquad A_n=\{\xi=\upsilon\},\qquad q_n=\mathbb P(A_n\mid\mathcal G_n).\] For any \(\mathcal G_n\)-measurable \(Z_n\in[0,2M]\), \[ \mathbb EZ_n\le 2M\mathbb P(q_n<p)+p^{-1}\mathbb E[Z_n\mathbf1_{A_n}]. \tag{77}\] Indeed, on \(\{q_n\ge p\}\) integrate the identity \(\mathbb E[Z_n\mathbf1_{A_n}\mid\mathcal G_n]=Z_nq_n\). Lemma 63 makes the first term arbitrarily small with a fixed \(p>0\), uniformly in \(T_n\). This is the required change of sampling law; mere absolute continuity is not used.

At physical radius \(r\) let \(V^{\rm old}_{n,r}\) be a small radial average of the stopped conditional mean at the first walk entrance into \(B(\xi,r)\). Its averaging radius is less than one tenth of the distance to the observed strand and one tenth of \(r\). Define \(V^{\rm end}_{n,r}\) in the same way with the complete strand and \(\upsilon\). Positive cutoffs on the averaging radius are removed after the mesh limit, as in Proposition 70. Let \(b(\xi)\) be \(+l\) or \(-l\) according to the old oriented side, and put \[Z_n=|V^{\rm old}_{n,r}-b(\xi)| \mathbf1_{\{\xi\notin\partial D_n\}} .\] This is measurable with respect to \(\mathcal G_n\). By Proposition 72 and Lemmas 66–67, \[\lim_{r\downarrow0}\limsup_n \mathbb E\!\left[|V^{\rm end}_{n,r}-b(\upsilon)| \mathbf1_{\{\upsilon\notin\partial D_n\}}\right]=0. \tag{H4}\] This is the bounded harmonic martingale boundary limit in the completed component, sampled by the lattice walk through (H2). First discard arbitrarily small completed-observer clearances using Lemma 62; (H2) then applies with a fixed protected observer. The first entrance around its eventual landing need not itself be a stopping time: these approach points tend to the exit pathwise, which is all that this boundary-limit assertion uses.

Discard old-side landings within distance \(2R\) of the original boundary and those within distance \(2R\) of the future strand. Their total probability, denoted \(e_n(R)\), satisfies \(\lim_{R\downarrow0}\limsup_n e_n(R)=0\) by (74) and (75). On the remaining event, and on \(A_n\), the two readings use the same walk segment, the same label, and the same averaging kernel when \(r<R/10\). The prefix extends across this common ball: its starting point is on the distant original boundary and its tip belongs to the distant future, or to the original boundary if exploration is complete. The stopped and completed conditionings have exactly the same sign observations there. Lemma 69 therefore bounds their mean difference by \(CM(r/R)^\alpha+o_n(1)\) at fixed \(r,R\). Consequently \[\mathbb E[Z_n\mathbf1_{A_n}] \le \mathbb E\!\left[|V^{\rm end}_{n,r}-b(\upsilon)| \mathbf1_{\{\upsilon\notin\partial D_n\}}\right] +CM(r/R)^\alpha+2M e_n(R)+o_n(1). \tag{H5}\] First choose the survival error in (77), obtaining \(p>0\). Choose \(R\) so that the discarded mass is small compared with \(p\) times the desired final error. Then choose \(r\ll R\) to make both (H4) and the locality error equally small. Keep these physical scales and the kernel cutoffs fixed while taking \(n\to\infty\). This order proves that the old-side boundary reading equals its prescribed label in \(L^1\), uniformly in the chosen sequence of stopping rules.

Original boundary accesses retain their original values. For clarity, the argument proving (GB2)–(GB3) also yields \[\lim_{R\downarrow0}\limsup_n \mathbb P\{d_{D_n}(S_{\tau_{D_n}},\gamma_n)<R\}=0 \tag{H6}\] in physical units: condition on the independent original-boundary exit, apply pure-arc repulsion away from the opposite arc, and use the two-ball bound at the two sign-change accesses. On the complementary event the final approach has a neighborhood, in its original-domain component, free of the completed strand. The original-boundary comparison gives the prescribed trace there. The pathwise coupling used for (H2), including confinement of the terminal tails, transfers this internal-access assertion to Brownian motion. Thus (H6) applies also on the event that the old landing lies on the original boundary. No boundary avoidance deduced from SLE is used.

Finally extract jointly the stopped closed sets, bounded conditional profiles, normalized conformal maps, and their boundary marks. The limiting stopped profile is harmonic by the weak equation. The preceding readings give its boundary values at Brownian-almost every marked access. Bounded harmonic uniqueness and Lemma 67 identify this profile with the limit of \(H^l_{n,T_n}\). The comparison compactness and original–filled \(L^1\) estimate give convergence of the lattice conditional means against the fixed smooth test \(f\). The harmonic functions \(H^l_{n,T_n}\) converge uniformly on its protected support. Integration against \(f\) and boundedness therefore prove (76). Proposition 2 supplies the moments of the underlying field averages if those averages are retained in the joint limit.

Reverse exploration reveals the same kind of sign prefixes and satisfies the same comparisons and nonreturn estimate. Repeating the proof gives the reverse assertion. ◻

Proof. We specify which part of the harmonic-observable criterion is used. Let \(x_t<y_t\) be the two images of the starting point under \(g_t\). For \(t>0\) introduce the bounded-driver coordinates \[s=\log(y_t-x_t),\qquad G_s(z)=\frac{2g_t(z)-x_t-y_t}{y_t-x_t},\qquad \widetilde U_s=\frac{2U_t-x_t-y_t}{y_t-x_t}\in[-1,1].\] Let \(\mathcal F_s\) be the sign-prefix filtration at the corresponding time, equivalently the filtration of the explored dual curve after this deterministic coordinate change. In these coordinates, as in [SSdiscrete], fix a finite interval \([-S,S]\) and the two deterministic interior probes protected throughout that interval. Replace each point probe by a radial smooth average in a sufficiently small protected disk. The harmonic mean-value property makes the average of the continuum observable equal to its value at the center. Thus (76) supplies the same two approximate martingales as the point probes.

For stopping times \(s_0\le s_1\) with \(s_1-s_0\le\varepsilon^2\) and driver oscillation at most \(\varepsilon\), the two-probe Taylor calculation gives, outside an event of arbitrarily small prescribed probability, \[\begin{align*} \left|\mathbb E[\Delta\widetilde U+ 2\widetilde U_{s_0}\Delta s\mid\mathcal F_{s_0}]\right| &\le C_S\varepsilon^3, \tag{79}\\ \left|\mathbb E[(\Delta\widetilde U)^2- 2(1-\widetilde U_{s_0}^{\,2})\Delta s \mid\mathcal F_{s_0}]\right| &\le C_S\varepsilon^3. \tag{80}\end{align*}\] Here \(\Delta s=s_1-s_0\) and \(\Delta\widetilde U=\widetilde U_{s_1}-\widetilde U_{s_0}\). To obtain these estimates from qualitative convergence, fix \(S,\varepsilon\) and the exceptional probability first, and only then take the mesh sufficiently fine. Conditional Markov inequality converts the \(L^1\) observable errors into the needed conditional errors. No predetermined rate is required.

To apply the approximate-diffusion criterion, start the comparison diffusion at the same bounded-driver value as the discrete process. For a fixed comparison horizon \(S\), choose the exceptional tolerance \(\eta<\delta^5/S^2\), where \(\delta\) is the criterion’s approximation parameter, before choosing the mesh; Proposition 4.5 of [SSdiscrete] then gives its comparison on \([0,S-1]\). For a limiting window \([-S,S]\), start instead at \(-S'\) and let \(S'\to\infty\) after the mesh limit. The coupling argument in the proof of Theorem 4.1 of [SSdiscrete], pp. 115–116, removes the initial value and identifies the stationary whole-line bounded-coordinate diffusion. Thus an arbitrary initial Jacobi law is not substituted for the chordal limit.

The deterministic Taylor argument and the approximate-diffusion argument of [SSdiscrete] use precisely these martingale and harmonic-observable inputs; no Gaussian conditional law enters these steps. They identify the bounded-coordinate diffusion with squared diffusion coefficient \(2(1-u^2)\) and drift \(-2u\). The inverse coordinate change [SSdiscrete] gives the chordal driver \(2B\). The discrete harmonic-measure comparison in that source is replaced here by the explicit continuum comparison assumed in (76). ◻

Proof. The argument uses more than convergence of the two endpoint drivers. The stopped observable identifies their limiting side labels in the same coupling. Harmonic-mass nonreturn then excludes macroscopic backtracking. All the estimates invoked here precede this proposition.

Proof. Use the matched experiment with original heights \(\pm l\) and the coupled finite-prefix convergence (81). Fix a nonzero nonnegative smooth test \(f\) supported in an interior ball. Choose a positive rational \(t\) such that a fixed neighborhood of this ball is deterministically protected through capacity time \(t\). This is possible: its preimages under \(\Phi_n\) have imaginary part bounded below, and a hull of capacity \(2t\) has height at most \(2\sqrt t\). Leave a fixed clearance so that the one-edge physical overshoot in \(T_n(t)\) also preserves protection.

Conditioning on \(\mathcal F^n_{T_n(t)}\) gives exactly the sign-string kernel of the realized oriented prefix. Whether the rule stopped there is already decided by that prefix; no unobserved height values are frozen. The absorbing vertices along crossed primal edges connect to each other and to the exterior. The observed set, including the exterior in a fixed ambient box, therefore has a connected component of uniformly positive macroscopic diameter and satisfies the common thickness condition in Theorem 42. It converges in Hausdorff distance to the exterior with \(K_t\) adjoined. That theorem and Corollary 44 consequently give \[ \mathbb E\operatorname{Var}(X_n(f)\mid\mathcal F^n_{T_n(t)}) \longrightarrow\sigma^2\mathbb E\langle f,G_{D\setminus K_t}f\rangle. \tag{88}\] Here brackets denote Euclidean area pairing. The Riemann-to-affine pairing error from Theorem 32 tends deterministically to zero; the fixed-test moment bounds transfer both second moments and conditional-mean second moments between the two pairings.

Proposition 73 approximates \(M^n_{T_n(t)}(f)\) in \(L^1\) by the harmonic average. ODE continuity and the protected support identify its limit as \(\int fH_t\). The exponential moments of Proposition 2, and conditional Jensen’s inequality, make the squares of these conditional means uniformly integrable. Thus their variances converge to \(\operatorname{Var}(\int fH_t)\). The unconditional field limit also has convergent second moments. The law of total variance and (88) yield \[ \operatorname{Var}\left(\int fH_t\right) =\sigma^2\mathbb E\langle f,(G_D-G_{D\setminus K_t})f\rangle. \tag{89}\]

For \(0\le s\le t\) set \(Z_s(z)=g_s(\Phi^{-1}(z))-2B_s\). The imaginary part of \(Z_s\) on the test support is bounded below, since its square decreases by at most \(4s\). Itô’s formula gives \[\,\mathrm d\log Z_s(z)=-\frac{2}{Z_s(z)}\,\mathrm dB_s.\] The cross variation of the harmonic observables is therefore \[ \frac{16l^2}{\pi^2} \operatorname{Im}\frac1{Z_s(z)} \operatorname{Im}\frac1{Z_s(w)}\,\mathrm ds. \tag{90}\] For covariance \((-\Delta)^{-1}\) the half-plane Green function is \[G_{\mathbb H}(z,w)=\frac1{2\pi} \log\left|\frac{z-\overline w}{z-w}\right|.\] Differentiating its conformal pullback gives \[ -\partial_sG_{D\setminus K_s}(z,w)=\frac2\pi \operatorname{Im}\frac1{Z_s(z)} \operatorname{Im}\frac1{Z_s(w)}. \tag{91}\] These normalizations agree with [SScontinuum]. Only this calculation is used; the stronger conditional coupling in that work is not an input.

Integrate both identities against \(f\) twice and over \(0\le s\le t\). The integrands are bounded on the protected support, and the Green singularity is integrable for area pairing. No pointwise value of the limiting field is used. Equation (89) becomes \[\frac{16l^2}{\pi^2}\, \mathbb E\int_0^t\left(\int f(z)\operatorname{Im}(1/Z_s(z))\,\mathrm dz\right)^2\,\mathrm ds = \frac{2\sigma^2}{\pi}\, \mathbb E\int_0^t\left(\int f(z)\operatorname{Im}(1/Z_s(z))\,\mathrm dz\right)^2\,\mathrm ds.\] The common factor is finite and strictly positive because \(f\ge0\) is nonzero and \(Z_s(z)\in\mathbb H\). Cancelling it proves the formula. ◻

Proof. Apply the reference height-gap extraction to every subsequence. Proposition 77 gives the same value \(\lambda_*\) on each extraction; its membership in \((0,1/2]\) comes from the reference complete-crosscut bounds. Include, in each finite comparison, the fixed experiment with exterior heights \(\pm\lambda_*\). Every subsequence then has a further subsequence to which the driving and uniform-path conclusions apply with the same matched value. This proves full convergence. The field assertions are the identifications in Theorem 32 and Proposition 31. ◻