Strict convexity and differentiability of the planar exponential first-passage limit shape

Proof. Integrating the Gamma density with the substitution y=(\lambda-\theta)t gives the moment identity. At \theta=\lambda/2 it reads \E e^{\lambda\tau/2}=2^\kappa. Independence and exponential Markov inequality give

□

Proof. Each of a_- and a_+ is a subgradient at zero of the convex function t\mapsto\mu(u+te_2). If x(z)>0, write z=x(z)(u+r(z)x(z)^{-1}e_2) and use positive homogeneity to get f_\sigma(z)\le\mu(z). If this inequality failed at any other point z, then a sufficiently small positive convex combination of z with u would still violate it, by convexity, and would have positive first coordinate. This is impossible. The average f is also bounded above by \mu. Equation (2.5) is the respective one-sided derivative expansion.□

Proof. Uniform shape estimates. Use the constants \alpha,\beta from (2.2). Equation (2.1) implies, for each fixed C>0,

Indeed, outside any sufficiently large fixed ball the error is bounded by an arbitrarily small multiple of |z|, while the finitely many passage times inside that ball contribute o(s). The same shape theorem gives

Translation invariance gives the same probability bounds at every deterministic lattice origin. This uniformity will allow the origins to depend on the endpoints and on s.

Short pieces cannot escape. Choose a fixed integer B so large that \beta D/B<\alpha\delta/24. Round the B+1 equally spaced points of [p,q] to vertices p_0=p,p_1,\ldots,p_B=q. Uniformly in the endpoints,

Apply (3.1) and (3.2) at these finitely many deterministic vertices. With probability tending to one, uniformly in p,q, every short passage time is at most

and escape from the ball of radius \delta s/3 about its starting vertex costs at least \alpha\delta s/6. For each j=0,\ldots,B-1, choose a path from p_j to p_{j+1} with cost at most T(p_j,p_{j+1})+1. For all sufficiently large s, each chosen path has cost strictly below the corresponding escape bound. Nonnegativity of the weights forces it to stay inside the ball about p_j: its initial portion up to a vertex outside would already cost at least the escape bound.

Concatenation. Concatenating the B paths gives total cost at most \mu(q-p)+\eta s/2+B\le\mu(q-p)+\eta s. Each ball is centered within bounded distance of the original segment, so it lies in its \delta s-neighborhood for large s. All probability bounds used above are uniform in the deterministic vertices, proving the stated uniformity.□

Proof. The assertion is vacuous if either interval contains no vertex.

Finite nets and vertical attachments. Let \beta be the upper norm bound in (2.2), and choose a fixed \gamma>0 so small that

In each interval, choose a net of lattice vertices containing its extreme lattice vertices and having consecutive gaps at most \lfloor\gamma s\rfloor. For all large s, each net needs at most 2D/\gamma+2 vertices. Assign every interval vertex to a net vertex at vertical distance at most \gamma s, using only edges of its intervening gap.

For any such gap, the total weight V is a sum of at most \lfloor\gamma s\rfloor independent edge weights. Lemma 2.1, with A=\gamma and v=s, gives

There are only boundedly many gaps. Thus, with probability tending to one, all vertical connections from interval vertices to their chosen net vertices cost at most 2\kappa\gamma s/\lambda.

Connections between net vertices. Apply Lemma 3.1, with errors \eta/2 and \delta/2, to every pair of net vertices. Its uniformity and the bounded number of pairs give these connections simultaneously.

All endpoint pairs on the same event. On the intersection of these high-probability events, take arbitrary p\in I_1\cap\Z^2 and q\in I_2\cap\Z^2. Concatenate their vertical attachments with the path between the assigned net vertices \widehat p,\widehat q. The resulting cost is at most

The segments [p,q] and [\widehat p,\widehat q] have Hausdorff distance at most \gamma s. The choice of \gamma therefore gives both the cost bound and the required localization, including the vertical attachments. The same event works for every p,q, and its probability bound is uniform in the deterministic locations of the intervals.□

Proof. A finite reachable set if there is no infinite path. Call a site good when G_v occurs. Let S consist of the endpoints of finite oriented paths from 0 for which every site except possibly the last is good. In particular 0\in S. If S were infinite, the finitely branching tree of these paths would have an infinite branch; every site on that branch would be good. Hence, on the event of no infinite good path, S is finite.

A surrounding contour forces bad sites. For such a finite S, orient the boundaries of the unit squares centered at its sites counterclockwise and cancel shared sides. The remaining edges form a balanced directed graph and hence decompose into closed directed trails, with no edge repeated in a trail. Repeated vertices are allowed, so corner contacts cause no difficulty. Every edge has an S-square on its left and a complementary square on its right. Winding number is additive under cancellation and decomposition. Its total about 0 is one, contributed by the square centered at 0; therefore some trail has nonzero winding.

Let m be the length of this trail. Closedness implies that exactly m/2 of its steps are upward or leftward. For these steps, the site on the right is respectively an east or north neighbor of the site on the left. The left-hand site must be bad: if it were good, extending a path that reaches it would put this neighbor in S as well.

Independent bad sites along a fixed contour. A site borders at most four edges of the trail, so at least m/8 distinct sites are forced to be bad. For any fixed candidate trail these sites are deterministic. Greedy selection in a fixed order gives at least

of them at pairwise sup distance greater than k_0: each selected site removes at most (2k_0+1)^2 candidates. Their tests are independent, so the probability of all the badness conditions forced by this particular trail is at most \zeta^{m/[8(2k_0+1)^2]}.

Summing over contours. Nonzero winding forces 0 into the coordinate bounding box of the trail. A length-m trail has coordinate spans at most m, so it lies in a square of side O(m) about 0. Choosing its starting vertex and then its step sequence bounds the number of candidates by C(m+1)^2 4^m. Consequently the probability of no infinite good path is at most

This is less than one when \zeta is sufficiently small. The argument uses no stationarity assumption on the tests.□

Proof. Suppose that the probability in (3.3) is at least p_0>0 along a sequence satisfying the stated limits. At each auxiliary site we will try several disjoint translates of this crossing event and require connections to both forward neighbors. Enough trials make a successful crossing likely; the one-sided contact estimate makes the connections likely without spending the whole crossing deficit.

1. Parameters and site geometry. The dependence radius must be fixed before we choose how many trials to make. Set

and let \zeta be supplied by Lemma 3.3. Choose a fixed integer M so large that (1-p_0)^M<\zeta/2. For each member of the sequence put

where the fixed constant D_0 will be chosen below, after M.

Identify the signs + and - with 1 and -1 when they multiply coordinates. For a test site v=(v_1,v_2)\in\Z^2, write

An oriented step sends i to i+1 and changes Y_v by \sigma S_0 or 2\sigma S_0. Thus every step advances to the next longitudinal block and moves transversely to the same prescribed side.

2. Independent crossing trials within a site. At site v, make M trials using exact lattice translates of the event in (3.3), all with initial column X_i, and with their strips disjoint and contained in |r-Y_v|\le S_0/4. Here is an explicit choice, which also accounts for lattice phases. Put

The translated strip has center r(t_{v,j}), within 1/2 of Y_v+h_j. For s\ge1, consecutive centers are separated by more than 2Rs, and every strip of half-width Rs lies inside the designated S_0/4 window. Thus the trials use disjoint internal edge sets. Translation leaves both displacement and its supporting functional unchanged, so each trial has probability at least p_0, exactly as in the original event. Their probability of all failing is at most (1-p_0)^M.

3. Connection tests for every endpoint pair. For each of the two forward neighbors v' of v, impose an additional test. Every vertex on

must be connectible to every vertex on

by a path of cost at most f_\sigma of its displacement plus \varepsilon s/2, lying in

These are existence tests using only the internal edges of this region. Requiring every endpoint pair will permit the successful trials to be chosen later, after the environment has been observed.

4. Choosing the connector length. We next choose D_0 and verify that both connection tests pass with probability tending to one uniformly in v. A connector displacement z satisfies

Set C_0=200M(R+1), so that S_0/s\le C_0 for s\ge1. The one-sided expansion (2.5) supplies h_0>0 such that

Choose D_0\ge1 with 3C_0/D_0\le h_0. For every possible connector, h=r(z)/d is on the required side and has absolute value at most h_0. Homogeneity therefore gives

This leaves another \varepsilon s/4 for the travel error.

The two endpoint intervals have union diameter at most Ds for a fixed D depending only on M,R,D_0. Their joining segments have longitudinal coordinates between the two endpoint columns and satisfy |r-Y_v|\le9S_0/4. Since |b|\le1, moving Euclidean distance \delta s changes x by at most \delta s and r by at most \sqrt2\delta s. Thus a tube with any fixed 0<\delta\le1/2 fits inside (3.4). Lemma 3.2, with cost error \varepsilon s/4, proves the required connection assertions simultaneously for all endpoint pairs. Its probability estimate is uniform in the intervals’ locations, including their lattice phases.

5. A dependence radius independent of the later choices. Call v good when at least one of its trials succeeds and both connection tests pass. Its bad probability is at most (1-p_0)^M+o(1), uniformly in v. We verify that the initially chosen k_0 is a dependence radius. Take a sufficiently large member of the sequence that n\ge d+s. The entire test at v is supported on internal edges in

To compare two sites, use the integer coordinates i=v_1+v_2 and j=i+v_2. If their support regions share a vertex, the longitudinal and transverse bounds give, respectively,

where \Delta denotes the difference between the two sites. Since v_2=j-i and v_1=2i-j, this implies

Tests at sites at pairwise sup distance greater than k_0 therefore use pairwise disjoint edge sets and are jointly independent. In particular, the dependence radius does not increase when M or D_0 is chosen. The order of choices was consequently legitimate: k_0 and \zeta first, then M, then D_0, and finally a sufficiently large member of the sequence. For that member the bad probability is at most \zeta.

6. Concatenation at one fixed scale. By Lemma 3.3, with positive probability there is an infinite oriented path of good sites starting at 0. Fix this one scale s,n,d,S_0 and a configuration with such a path. At each visited site choose a successful trial and join it to the next by a tested connector. Prepend a vertical path from the lattice origin to the initial trial’s starting vertex; its defect H_\sigma is a finite random quantity C. If z_t is the endpoint of the trial at the tth site, with the initial site numbered 0, additivity gives a path from 0 to z_t of defect at most

The trials and connectors need not be disjoint: costs are counted with edge multiplicities, so their concatenation still bounds passage time from above.

7. The shape-theorem contradiction. At that site, i=t and 0\le v_2\le t. Hence

Thus |z_t|\to\infty and |z_t|\le C'(t+1) for a deterministic constant C' at this fixed scale. The limit here is t\to\infty; the parameters s,n,d,S_0 remain fixed. On the probability-one event of the origin shape theorem, its uniform outside-ball estimate applies also to these adaptively chosen endpoints and yields

where the supporting inequality is from Lemma 2.2. This contradicts (3.5), whose negative part grows at least as t\varepsilon s/2. The infinite-path event consequently has probability zero, contradicting its construction. No sequence with crossing probability bounded below by p_0 exists, proving the asserted joint limit. Exact lattice-translation invariance proves the last assertion.□

Proof. Turning an allowed crossing into a fast crossing. Put n=s^3. Choose a fixed \eta>0 such that

and set \delta=\eta s/n and \alpha=1-\delta. Take s sufficiently large that 0<\delta\le1/2. Multiply every internal edge weight of the region defining E(s,n) by \alpha, leaving the other weights unchanged. If P witnesses E(s,n) before this operation, its transverse displacement obeys \abs{r(\Delta P)}\le16s, and in the modified environment

This computation includes every traversal if P repeats an edge. Let F(s,n) be the crossing event H_+(P)<-s in the same region, and let Q be the law of the modified environment. The deterministic implication gives

The cost of the change of measure. Under Q, a modified edge has distribution \operatorname{Gamma}(\kappa,\lambda/\alpha). On t>0 its likelihood ratio with respect to the original edge law is

Both densities are positive on (0,\infty), including when 0<\kappa<1. Direct integration gives

The region has at most (3n+1)(16s+2) vertices and at most twice that many internal unoriented edges. Thus the number M of modified coordinates is at most 144ns. The likelihood ratio is a product over these distinct coordinates, regardless of path multiplicity. By independence,

Transfer back to the original law. Lemma 3.4, applied under the original law with A=1, R=8, and \varepsilon=1, gives \PP(F(s,n))\to0 because n/s=s^2\to\infty. Cauchy–Schwarz therefore gives

All supports and the norm in F(s,n) are those of the original fixed law; no shape theorem for the modified environment is used.□

Proof. Short crossings are likely. First suppose n\to\infty with n\leq Cs, and put q=(n,\lfloor bn\rfloor). Then

Apply Lemma 3.1 at scale n, with endpoints 0,q, with cost error \eta n where \eta=K/(4C), and with tube radius \delta_0n, where

The endpoints have Euclidean distance at most 3n. Since \abs{b}\leq1, changing a point by Euclidean distance t changes its r-coordinate by at most \sqrt2t. The resulting tube is inside -n\leq x\leq2n, and inside \abs{r}\leq8s for all sufficiently large s: its transverse extent is at most 1+\sqrt2\delta_0n. With probability tending to one its path cost is at most

This proves (4.2) when n\to\infty.

If n remains bounded, direct lattice paths from 0 to (n,\lfloor bn\rfloor) use a bounded number of edges and remain in a bounded transverse region with 0\leq x\leq n. Their costs are bounded in probability, whereas Ks\to\infty. This proves the same assertion for bounded n. The two cases give the joint limit in (4.2) by the subsequence criterion.

The first dyadic threshold. Lemma 4.1 shows that the dyadic integer 2^{3k} satisfies (4.3) for all sufficiently large k. Thus L_k exists and has the stated upper bound. If L_k/2^k failed to tend to infinity, a subsequence would satisfy L_k\leq C2^k for some fixed C, contradicting (4.2) and (4.3). The last assertion follows from the definition of the first dyadic length; it uses no monotonicity in n of the events E(s,n).□

Proof. The positive sequence L_k2^{-4k} is at most 2^{-k}, so it tends to zero and has infinitely many strict record minima. Take these indices for \mathcal K. At a record index k, comparison with k-j gives

which is (4.5).□

Proof. Independent crossing trials. Bin the values of r_0 in intervals [hw_j,(h+1)w_j], taking those bins with h\in\Z that meet [-Rs,Rs]. Fix j,X,h, and write w=w_j, l=l_j. Make j^2 trials, using the lattice translates of E(w,l) by

Here w is an integer, so each v_t is a lattice vector. The transverse center of its strip is

Thus the lattice phase contributes the same offset, of absolute value less than one, to every trial. Consecutive centers differ by exactly 30w, and each strip has half-width 8w, so their vertex sets and internal edge sets are disjoint. Each trial is an exact lattice translate of E(w,l) and has the same probability as that event. By independence and (5.2), all trials fail with probability at most

Vertical attachments for every endpoint. On each column x=X,X+l, consider the deterministic vertical interval whose vertices satisfy

It contains every vertex whose transverse coordinate is within one unit of the chosen bin, as well as every possible trial endpoint. There are at most 50j^2w internal vertical edges. If V is their total weight, Lemma 2.1, with A=50 and v=j^2w, gives

Except with probability 2e^{-c_0j^2w}, vertical travel between any vertices in either of these intervals costs at most 100\kappa j^2w/\lambda.

Cost and localization. If a trial succeeds and the vertical bounds hold, join p vertically to that trial, follow its crossing, and join its other endpoint vertically to q. The total cost is at most

The path stays in the longitudinal range [X-l,X+2l], and all its vertices are within 50j^2w transversely of every r_0 in the chosen bin. Since K=100(1+\abs{a_-}+\abs{a_+})\ge100, C_1=K+200\kappa/\lambda supplies both bounds. This argument does not require independence between a successful trial and its attachments.

One event for all grid connections. Let G_{k,J} be the event that, for every tested triple j,X,h, some crossing trial succeeds and both vertical bounds hold. There are at most C_{A,R}2^{5j} pairs X,h to test at level j: the grid interval count is bounded by a constant times L/l_j\le2^{4j+1} and the bin count by a constant times s/w_j=2^j. The first bound also holds at the terminal mesh l_j=1, because then m\le4j+1. Taking the union over these pairs and the levels, and using w_j\ge1, gives

For every fixed q_0>0 and \kappa>0 this series is convergent, and its tail tends to zero as J\to\infty, independently of k. Whole vertical intervals were tested, so the event gives every allowed endpoint pair simultaneously.□

Proof. Fix A,R,\varepsilon, and let J\geq1 be a fixed integer, to be chosen large. We first work on the event G_{k,J} and construct extensions for all vertices in the region. Set l=l_J.

Connections to the coarse columns. For a vertex p in

define coarse columns

Each of x(p)-X_p^- and X_p^+-x(p) lies in [l,2l).

Decompose [X_p^-,x(p)] into grid intervals by taking one full interval of length l_J, then filling the remaining interval greedily with successively finer meshes. There are at most 16 intervals of each level: at finer levels this follows from the ratio bound 16, and the mesh of one reaches the integer endpoint exactly. At every integer column X that occurs, use the vertex

It satisfies \abs{r(z_p(X))-r(p)}<1, and z_p(x(p))=p. Apply Lemma 5.1 with the same transverse target r_0=r(p) to every interval, and concatenate its paths. This gives an entrance path P_p^- from the column X_p^- to p. Decomposing [x(p),X_p^+] from its right endpoint in the same way, and then concatenating the resulting intervals in left-to-right order, gives an exit path P_p^+ from p to the column X_p^+.

The cost and region of the extensions. Each path costs at most m_0 times its longitudinal displacement plus

The r-coordinate of every intermediate vertex used to join successive paths differs from r(p) by less than 1. In particular the whole entrance and the whole exit each have transverse displacement of absolute value less than one. Thus, for either sign,

Only the displacement of the whole extension enters this bound; no rounding error is accumulated over the levels.

The grid intervals used here lie inside [-(A+4)L,(A+4)L]. Their paths have longitudinal range within 3l of x(p), so they also lie in this enlarged longitudinal interval. Since \sup_{j\geq1}j^2 2^{-j}<2, their transverse localization in (5.4) gives

All these statements hold simultaneously for every p\in\mathcal D.

Preserving a negative defect. Now suppose a path from p to z in \mathcal D satisfies H_\sigma(P)<-\varepsilon s. Put a_* =\max\{\abs{a_-},\abs{a_+}\}. Nonnegativity of its cost and \abs{r(z)-r(p)}\leq2Rs imply

In particular

Thus the final endpoint can lie at most C_3s to the left of the initial endpoint. This bound lets the extensions produce a left-to-right crossing even when the original endpoints occur in reverse order.

Choose J so large that

Then take k\in\mathcal K large enough that 2a_*\leq\varepsilon s/4. By (5.8), the concatenation

satisfies H_\sigma(\widetilde P)<-\varepsilon s/2. It starts on the coarse column X_p^-, ends on the coarse column X_z^+, and stays in (5.9).

A long crossing between coarse columns. For this fixed J, eventually

Using (5.10) and the two attachments of longitudinal length at least l_J, the new span satisfies

for all sufficiently large k. Thus d' is a positive integer and d'/s\to\infty, uniformly over the possible endpoint choices.

Reduction to finitely many crossing tests. Both coarse columns lie in [-(A+2)L,(A+2)L] and are multiples of l_J. There are at most a constant depending on A,J possible pairs, because L/l_J=2^{4J+1}. For each pair obeying (5.11), translate its first column to zero by the lattice vector (X_p^-,\lfloor bX_p^-\rfloor). The transverse center changes by less than one. Also L/d'\leq2^{4J+1}. It follows that the translated region in (5.9) is contained in a region of the form

where, for example, A_J=(2A+8)2^{4J+1} is independent of k. Lemma 3.4 therefore shows that the probability of such an extended crossing tends to zero. This application is uniform over the bounded number of pairs: otherwise a sequence of violating pairs would contradict that lemma’s joint limit, since every admissible pair has d'/s\geq l_J/s\to\infty. Taking the union over the pairs and both signs still gives a probability tending to zero.

Order of limits and translations. Write B_k for the event in (5.6). We have proved, for every fixed sufficiently large J,

All constants in the crossing tests above were fixed before taking the limit in k. We may now let J\to\infty, so that \beta_J\to0, proving the assertion. Every exact lattice translate of B_k has exactly the same probability by translation invariance of the edge-weight law and of path displacements, giving the stated uniformity.□

Proof. Let F_{\kappa,\lambda} be the Gamma distribution function and define its quantile function by

Here \Phi(0)=0. The density is positive for every t>0, so the distribution function is continuous and strictly increasing from zero to one. Consequently \Phi is increasing and continuous on [0,1). Represent the independent weights as

where all B_{e,h} are independent Bernoulli variables of parameter 1/2. At precision t put

Then U_e^{(t)}\downarrow U_e; because U_e<1 almost surely, continuity of \Phi gives \tau_e^{(t)}\downarrow\tau_e almost surely on the finite union of the edge sets under consideration. Define A_j^{(t)} and D^{(t)} by using these rounded weights.

At each fixed precision, the events A_j^{(t)} are decreasing events of finitely many independent bits. Fixing all precision-t bits of the edges of a witnessing path forces its strict cost inequality regardless of the other bits. Edge-disjoint path witnesses thus give disjoint bit certificates. Complement the bits so that the events are increasing. For each certificate, retain its realized one-bits and set every other bit to zero. This configuration still belongs to the event, because it agrees with the realized configuration on the certificate, and its one-set lies inside that certificate. Disjoint certificates therefore give disjoint positive witnesses in the convention of [32]; any unused one-bits can be assigned by monotonicity. The finite Bernoulli BK inequality [32, Theorem 3.3, Equation (3.4)], iterated over the list, gives

Monotone disjoint occurrence is associative, so iteration allows repeated events. The second inequality follows from \tau_e^{(t)}\ge\tau_e.

The events D^{(t)} increase with t. A fixed finite list of original witnessing paths involves finitely many traversals. Its rounded path costs decrease to the original costs, so every strict inequality holds at all sufficiently fine precisions. Conversely, every rounded witness remains a witness with the original weights. Therefore, up to a null set,

Continuity of probability from below proves (6.1). No bound on the Gamma density or on the derivative of its quantile function is used, so the proof includes 0<\kappa<1.□

Proof. There are three possible exit behaviors. A transverse exit uses the slope gap, a backward exit has large defect by nonnegativity, and a forward exit reduces to the crossing event after controlling its outer pieces.

If \abs{r(\Delta Q)}\ge s, the support identity gives

The whole chunk is then a witness for the first defining event.

Otherwise a longitudinal exit occurred, and x(\Delta Q)\in\{-2L,2L\}. In the negative case, nonnegative weights and (6.4) imply

contrary to the hypothesis.

It remains to consider x(\Delta Q)=2L. The initial column is strictly to the left of (i+1)L, and the final column is at or to the right of (i+2)L. Let z be the first vertex of Q on column (i+2)L, and let p be the last vertex on column (i+1)L preceding z. The subpath Q' from p to z stays between these two columns. Its transverse offsets from v_{i+1,h} have absolute value at most 2s+2\le8s. Consequently Q' fits inside the translate of the block defining E(s,L) at v_{i+1,h}.

Write Q=Q_{\rm in}*Q'*Q_{\rm out}, allowing an empty outer part. If either outer part has H<-\eta s, it witnesses the first defining event, because it lies in \mathcal B_{i,h} and \min_\sigma H_\sigma\le H. Otherwise additivity gives H(Q')<3\eta s. Since \abs{r(\Delta Q')}\le2(s+1), we obtain

Thus Q' witnesses the second defining event.□

Proof. For each oriented nearest-neighbor step e,

Starting at the beginning of Q, split off the first prefix with H<-\eta s, if one exists, and repeat with the remaining path. At the last step of each split prefix, its defect decreases from a value at least -\eta s to a value below -\eta s. By (6.4), the defect of this prefix is at least -\eta s-C_f\ge-2\eta s. Every split prefix is a witness for the original chunk’s event \mathcal A_{i,h}: it remains in the same box, and \min_\sigma H_\sigma\le H<-\eta s.

If this procedure splits off t prefixes, simplicity of P makes their edge sets disjoint, so t\le M_Q. The unsplit final part has H\ge-\eta s. Additivity now gives

which proves (6.6).

For a full chunk with M_Q=0, Lemma 6.2 gives H(Q)\ge\eta s. If M_Q\ge1, then

This proves (6.7). Apply that bound to all full chunks and use (6.6), weakened to -\eta s-4\eta sM_Q, on the remainder. Their sum is (6.8).

Finally, choose a family attaining M_Q in each chunk. Distinct chunks of a simple path have disjoint edge sets, so the union of these families is still edge-disjoint. The assertion does not require their associated events or their boxes to be distinct.□

Proof. First quantify how many witnesses a path of small defect must supply. Let B_N be the event that there exists such a simple path with N full chunks and H(P)<N\eta s/2. If it occurs, (6.8) implies

For all sufficiently large N, we may therefore select exactly t_N=\lfloor N/16\rfloor of the edge-disjoint witnesses supplied by Lemma 6.3.

Next count the possible event lists. List the chunk indices in order, including the index of the remainder. The first is (0,0). Each coordinate of consecutive indices differs by at most 3, by the displacement bounds for a full chunk and (6.5). There are consequently at most 49^N possible index itineraries. For a fixed itinerary, the numbers of selected witnesses in its N+1 positions form a weak composition of t_N. The number of these compositions is

The allocation is indexed by positions along the itinerary, so repeated visits to one coarse cell retain separate positions. Each composition specifies a list of t_N events \mathcal A_{i,h}, with repetitions when appropriate. Its selected witnesses certify disjoint occurrence of that list. Lemma 6.1 and (6.3) bound its probability by (2q_0)^{t_N}. There is no additional count of witness paths: that lemma bounds the existence of all choices of disjoint witnessing paths at once. Thus the path-dependent selection is covered by a union over deterministic itineraries and compositions. Taking this union yields

for all sufficiently large N. All the events involved are measurable; in particular, finite lattice paths form a countable family.

Put p=2q_0. Condition (6.2) implies 0<p<1 and 196p^{1/16}<1, while the right-hand side of (6.10) is at most

Thus \sum_N\PP(B_N)<\infty. The first Borel–Cantelli lemma proves the simultaneous eventual bound (6.9) for all the paths in the proposition.□

Proof. Fix an arbitrary deterministic b\in[-1,1] and suppose that c>0. Make all the preceding choices for this b, including the single fixed scale k. For positive integers t, let

A path from 0 to z_t with N full chunks has N+1 chunks when the remainder is included. Each changes the x-coordinate by at most 2L, so

Work on the probability-one event where the shape theorem and Proposition 6.4 both hold. For each t, choose a finite path from 0 to z_t with cost at most T(0,z_t)+1 and erase its loops. The resulting simple path P_t still has cost between T(0,z_t) and T(0,z_t)+1, since the edge weights are nonnegative. The shape theorem gives T(0,z_t)/t\to m_0, and

It follows that

On the other hand, the numbers N_t of full chunks of P_t tend to infinity by (6.11). Proposition 6.4 therefore applies for all sufficiently large t and gives

This contradicts (6.12). Hence c>0 is impossible, and convexity gives a_-=a_+.

The choice of b was arbitrary and deterministic. The corner gap of the deterministic norm is itself deterministic, and the argument excludes a positive gap for each such b. Accordingly the conclusion holds for every b\in[-1,1]; no intersection of uncountably many probability-one events is being asserted or needed.□

Proof. Differentiability and a unique support. Fix a nonzero vector u and choose a vector e linearly independent of it. The function t\mapsto\mu(u+te) has finite one-sided derivatives a_-\le a_+. As in Lemma 2.2, the linear functionals taking the value \mu(u) on u and the values a_- and a_+ on e support \mu globally. Any other supporting functional at u must have its value on e between these two numbers, by the inequalities for positive and negative t.

If the endpoints agree with value a, the two-sided expansion gives, as xu+re\to u with x>0,

Since x\to1 and |r|\le C|xu+re-u|, this is Fréchet differentiability at u. The expression is exact when r=0. Conversely, Fréchet differentiability makes the two one-sided derivatives equal. Thus differentiability is equivalent to a unique linear functional g\le\mu satisfying g(u)=\mu(u).

Supports of the norm and of its unit ball. Because zero lies in the interior of the unit ball, every supporting line at u/\mu(u) has a unique normalization g=1 with g\le\mu and g(u)=\mu(u). This identifies uniqueness of the functional with uniqueness of the line.

Continuity of the derivative and the boundary. Suppose now that this uniqueness holds everywhere off zero. All supporting functionals lie in the compact dual unit ball. If u_j\to u\ne0 and a subsequence of the supports at u_j converges to g, passing to the limit shows g\le\mu and g(u)=\mu(u). Uniqueness determines g, so the whole sequence of supports converges to the support at u. The derivative of \mu is therefore continuous off zero. At a unit vector for this norm it is nonzero, since homogeneity gives

The implicit function theorem shows that the level set \mu=1 is a C^1 curve.

Conversely, at any point of a regular C^1 convex boundary a supporting functional has a local maximum along a local boundary parametrization. It therefore annihilates the tangent direction. Its supporting line is the tangent line, which is unique. This proves the converse and all the equivalences.□

Proof of Theorem 1.2. Proposition 6.5 gives a_-=a_+ for every fixed b\in[-1,1]. The expansion in the proof of Lemma 7.1, applied to u=(1,b) and e=e_2, gives Fréchet differentiability at u. Signed coordinate permutations and positive homogeneity cover every nonzero vector. Lemma 7.1 gives the remaining conclusions. The parameters \kappa,\lambda>0 were arbitrary and fixed throughout, so the conclusion holds for each Gamma law in the theorem.□

Proof. Equation (8.1) is the nonrandom-fluctuation estimate of Damron and Kubota [12, Proposition 1.1]. The two-sided concentration theorem of Damron, Hanson, and Sosoe [11, Theorem 1.1] gives the stronger scale \sqrt{R/\log R}; enlarging constants gives (8.2). Its hypotheses hold since \PP(\tau_e=0)=0 and \E e^{a\tau_e}=(1-a)^{-1}<\infty for a<1. The finitely many small separations can be bounded by a coordinate path and the exponential tail of its cost.

There are polynomially many pairs in the displayed box. Since \sqrt n\log n=o(n^\alpha), a union bound in (8.2) proves the simultaneous assertion. For the lower deficit, use \mu(y-x)\le\E T(x,y) and integrate the bound \min\{1,C n^{C_p}e^{-ct/\sqrt n}\}. For interior-edge endpoints, attach the endpoint of each incident edge, at a cost bounded by twice the maximum edge weight in a slightly larger box; changes in \mu are bounded by a fixed constant. The expected maximum of polynomially many exponential variables is O(\log n). This proves the remaining statements.□

Proof. By the joint stationarity of the edge and private inputs, it is enough to consider i=0. Combining the centered cocycle identity with (9.5) gives

The first term has mean zero; the second has the law of P_n(X_0). All terms are integrable because each cut lies on one of finitely many deterministic edges, where domination bounds its calibration value by a finite sum of weights and a deterministic linear term. Taking expectations proves the first identity, and (9.2) proves the second. The mean of P_n(X_0) itself need not vanish.□

Proof. A fixed vertex starts at most 4\cdot3^{m-1} self-avoiding paths of length m. For each path, independence and the identity \E e^{-5\tau_e}=1/6 give

Sum over paths and over m\ge R to obtain (9.7).

For the geodesic estimate, compare with a deterministic coordinate path of length \ell=|z|_1 and cost S. Its exponential moment gives \PP (S>t)\le2^\ell e^{-t/2}. Since weights are positive, a geodesic is self-avoiding. If it has at least R edges and S\le t, it is counted by (9.7) with V=\{0\}. Taking t=(\log2)R/10 makes both probabilities exponentially small in R when R\ge40\ell. The same estimate excludes arbitrarily long paths of bounded cost. Comparing with S therefore also proves existence of a finite minimizing path.□

Proof. We concatenate geodesics between closely spaced deterministic anchors, restricting each geodesic to a small ball to make the rule local. Choose an integer m large enough that, with \delta=1/m,

Increase the fixed divisibility requirement so that n\delta A and n\delta B are lattice vectors. For d=A or B, join successive deterministic anchors k\delta d by a minimizing path restricted to the lattice portion of the \|\cdot\|_*-ball of scaled radius r/4 about the first anchor. Each microscopic edge has \|\cdot\|_*-length at most K_{\mathrm{lat}}/n, so a coordinate path between the anchors stays within K_{\mathrm{lat}}\delta|d|_1\le r/1000 of the first anchor. The restricted minimum therefore exists. Use a translation-covariant tie rule.

Restriction changes the minimum only if the unrestricted geodesic leaves the ball. Such a geodesic has at least rn/(4K_{\mathrm{lat}}) edges, so Lemma 9.3, with integer rounding, bounds this event by Ce^{-c_s n} for large n. On that event the increase is at most the coordinate-path cost S_k. Since (\E S_k^2)^{1/2}\le C_s n, Cauchy–Schwarz bounds the expected increase by C_s n e^{-c_s n/2}.

For each patch, Lemma 8.1 bounds the mean unrestricted cost above its norm value by C_s\sqrt n\log(2+n). The latter value is \mu(n\delta d)=n\delta h(d) because d lies in the cone of the face. Summing over the m patches in a unit period proves the first inequality in (9.8); increasing C'_s gives the second.

Parametrize each patch monotonically from k\delta to (k+1)\delta, for example proportionally to its number of edge occurrences. Its \|\cdot\|_*-distance from its nominal location is at most r/4+\delta\|d\|_*\le r/4+K_{\mathrm{lat}}\delta|d|_1<r. The inputs for each patch lie within this same distance of every nominal point in the patch. Thus the radius-ten locality condition holds. The rule commutes with translation by d, and each period has finite expected cost. This proves admissibility.□

Proof. Two routes around a large square have small expected defects if \beta is small. We decrease weights near the middle of one route, leaving the other route and their common endpoints fixed. The first route then becomes a cheaper competitor, while the change in law is small enough to transfer that saving back to the original environment.

The two routes. Fix an admissible pair with common bound \beta. Write (\xi,\eta)=[A\ B]^{-1}x for its oblique coordinates, and put \rho=2r. Choose the fixed even integer

Use four translated roads, with nominal lines

They use the same physical edge weights and independent private inputs. Their strips have oblique half-width at most \rho.

At each nominal corner, determine the intersection from parameter windows of half-width 4\rho on the two incident roads. One road crosses the square of coordinate half-width 2\rho from left to right inside its horizontal strip, and the other crosses from bottom to top inside its vertical strip. The elementary planar crossing property for a rectangle gives an intersection. Nearest- neighbor lattice edges can meet only at a common vertex or along a common edge. If K_{\mathrm{lat}}/n\le r, a common vertex can be chosen with coordinate distance at most 2\rho from the corner and with both nominal parameters within 3\rho of their corner values: moving from an interior intersection to an endpoint of its edge costs at most 2K_{\mathrm{lat}}/n\le\rho in coordinates, after which tracking costs at most another \rho in nominal parameter. The chosen vertex remains in the inspected windows. Select the lexicographically first eligible vertex and then the first pair of eligible occurrences. These finite choices give a measurable local corner rule.

The bottom and right sides, traversed forward between these selected corners, form R_1; the left and top sides form R_2. Their initial and terminal vertices agree. Since 3\rho<1, every side used is a subpath of the periods with indices -1,0,\ldots,L. Nonnegative defects and Lemma 9.2 give, with C_R=2(L+2),

A region traversed by only one route. Consider the deterministic middle-side parallelogram

and let \mathcal E_n(S) be the edges having both endpoints in S\cap\mathcal G_n. The entire input support of R_2, including its selected endpoints and the middle corner, is outside S. Indeed that support is contained in the \|\cdot\|_*-neighborhood of radius 10 of the deterministic union \mathcal C of the left and top nominal segments and all incident nominal corner windows of parameter half-width 4\rho. The coordinate distance from S to \mathcal C is at least L/2-2-4\rho. By (9.3), its \|\cdot\|_*-distance is at least \tfrac34(L/2-2-4\rho)>10. The same separation holds for the unused corner window. Thus changing only weights in \mathcal E_n(S) leaves R_2 and its endpoints unchanged, even when its rule and corner selections are recomputed.

The bottom-road portion with nominal parameters between L/2-1 and L/2+1 belongs to R_1. For sufficiently large n, every edge meeting this portion has both endpoints in S. Discarding the first and last partial edges leaves endpoints with coordinate separation at least 2-2\rho-4K_{\mathrm{lat}}/n\ge1. Each microscopic edge has (A,B)-coordinate length at most 2K_{\mathrm{lat}}/n, so the resulting lattice path has at least n/(2K_{\mathrm{lat}}) edges, up to integer rounding. Loop erasure preserves the endpoints and their separation, so the resulting self-avoiding path obeys the same length bound and has no greater cost.

There are at most C_V n^2 vertices in S\cap\mathcal G_n, where one may take C_V=(8M+3)^2. Apply Lemma 9.3 with

For n\ge4K_{\mathrm{lat}} the resulting bound is

The constants c,C>0 depend only on K_{\mathrm{lat}},M. Uniformity in the road rule follows because the cheap-path event includes every self-avoiding path starting at every vertex of S.

Perturbation and transfer of the saving. Keep the private inputs fixed. For 0<\varepsilon\le1/2, put a_n=1-\varepsilon/n and multiply every weight in \mathcal E_n(S) by a_n. Let \PP be the original joint law of weights and private inputs, and \mathbb Q its pushforward under this map. Write m_n=|\mathcal E_n(S)|. Since each vertex is charged at most two undirected edges, m_n\le C_E n^2, where C_E=2C_V. The relative entropy of a scaled exponential variable with respect to an unscaled one is a_n-1-\log a_n. Product additivity, followed by Pinsker’s inequality [33, Theorem 31, order one], with total variation taken as one half of the L^1 distance, gives

Here we used -x-\log(1-x)\le x^2 for 0\le x\le1/2.

Set \lambda=a_*\varepsilon/2. Let E be the event that the cost of the current R_2 exceeds passage time between its current endpoints by at least \lambda. This event uses only the current R_2, its endpoints and the passage metric. In the perturbation coupling, outside the event in (9.11), the old R_1 saves at least a_*\varepsilon. It still joins the endpoints of the unchanged R_2. If its old cost exceeded that of R_2 by at most \lambda, it is therefore a competitor cheaper by at least \lambda after perturbation. Equation (9.10) and Markov’s inequality give

Only the old R_1 is used as a competitor; its recomputed rule may produce a different path. Transferring E to the original law by (9.12) yields

In the original environment domination implies D_n(R_2)\ge\lambda on E, also for its random endpoints. Hence

Fix \varepsilon>0 with C_{\mathrm{TV}}\varepsilon\le1/8 and then take n large enough that q_n\le1/8. Rearranging yields

The constants in c_0 depend only on the uniform geometry bounds and the fixed locality radius. Taking the infimum over admissible pairs proves (9.9). The transfer used the event E, so no comparison of calibrations in the two environments was needed.□

Proof. We build a grid from the two road families, then attach any additional anchor to that grid by a bounded signed trip. Put an A-road on each line jB+\R A and a B-road on each line iA+\R B. Write physical coordinates in the basis (A,B). The coordinate-conversion bound implies that the first road lies in |x_2-j|\leq\delta and the second in |x_1-i|\leq\delta, where \delta=2r. Take r_0=1/100, so 8\delta<1/2.

The portions whose nominal parameters are within 4\delta of the nominal crossing connect opposite sides of a small rectangle. The planar crossing property gives an intersection of the two portions. An intersection lies in both strips and hence within coordinate distance \delta of the nominal crossing. Nearest-neighbor paths meet at a common vertex or along a common edge. Each scaled edge has (A,B) coordinate length at most 2K_{\rm lat}/n\leq\delta; a vertex of that edge therefore has coordinate distance at most 2\delta from the crossing and nominal parameter distance at most 3\delta on either road. It lies within the inspected portions. Choose the first such vertex occurrence along the A portion, breaking any remaining tie by its occurrence on the B portion. This choice uses only those two bounded portions, is covariant under lattice translations, and specifies X_{(i,j)} together with its occurrence label on each road. Loops and repeated visits cause no ambiguity because occurrences, rather than only vertex names, are used.

The disjoint parameter windows order neighboring intersections along the relevant road. Define E_{z,1} as the Q-cost of the intervening forward A portion, and define E_{z,2} using the B portion. Each portion lies in a fixed number of marked periods, so additivity and nonnegativity bound its expected defect by C\beta. This proves (10.1).

Tracking and locality put all inputs in a fixed number of unit parallelograms. Assign boundary weights by a fixed half-open convention and private inputs to the cell of their nominal location. Since each microscopic input is assigned to one cell, the cell variables are independent.

To attach an additional anchor, follow its road with either sign to a nearby transverse network road, then follow that road to a network intersection. A bounded number of network intervals reaches any other anchor under consideration. All nominal distances are bounded, and crossings use the same local selection rule. The error on each signed interval is its signed defect, whose absolute value is bounded by complete-period defects. The triangle inequality gives the stated comparison. Its constant depends on the allowed anchor distance, independently of the number of other labels.□

Proof. We decompose a local increment according to the finite sets of cells on which it depends. Each component can be integrated locally; an exterior detour controls the error where that integration is cut off.

For a nonempty finite S\subset\Z^2, define

These are finite-product ANOVA (Hoeffding) projections; see Efron and Stein [15, Section 2] for the classical square-integrable decomposition. Here we use only finite inclusion–exclusion identities in L^1. Each projection has mean zero and L^1 operator norm at most 2^{|S|}. If F is measurable with respect to a finite cell set N, then \Pi_SF=0 unless S\subset N, and

Cancellation and independence prove both identities directly in L^1; no infinite or L^2 expansion is involved.

A comparison supported on Q_H(z) has components only on sets of at most m=(2H+1)^2 cells and diameter at most 2H. Accordingly, let \mathscr S be the collection of nonempty finite sets S with |S|\leq m and \operatorname{diam}_\infty S\leq2H. Let k(S) be the lexicographically least point of S, put

and use z_S as a reference beyond the cutoff U_S. Define

There are only a bounded number N_H of summands: k(S) must lie in Q_R(z) and S\subset Q_{2H}(k(S)). For example N_H\leq(2R+1)^2 2^{(4H+1)^2} suffices. Every summand is measurable with respect to cells in Q_{R+2H}(z) and has mean zero. This proves locality and (10.3).

The cutoff error. Suppose z\in U_S but a neighbor z' lies outside. There is a lattice path from z to z_S of length at most 20(R+4H+10) lying outside Q_{R-2}(k(S)): follow the boundary of Q_R(k(S)) to its eastern side and then move east. Every local support Q_H(u) along this path is disjoint from S\subset Q_{2H}(k(S)). If necessary first switch from label \lambda to label 0 at z. By (10.2), telescoping and the projection norm,

Each local comparison has zero \Pi_S projection because its support misses S. The connected exterior in two dimensions is used exactly at this step.

Recovery of an increment. Consider an edge (z,\lambda),(z',\lambda') from the hypothesis and write W=V_{z',\lambda'}-V_{z,\lambda} and E for its local comparison. The centered variables satisfy \|W-\E W-E+\E E\|_1\leq2\varepsilon. Every nonempty subset of Q_H(z) contributes to both endpoint gauges, since R>H+1. A set contributing at both endpoints gives \Pi_SW, within 2^m\varepsilon of \Pi_SE; this latter projection vanishes when the set is not contained in Q_H(z). A set contributing at only one endpoint is controlled by (10.7). Thus (10.5), the centered error bound above and at most 2N_H component errors prove (10.4). For a same-index label change there are no cutoff terms.

Finally, the choices k(S), U_S and z_S commute with index translations. The random summands in (10.6) contain only differences of V, so a random common additive constant under translation cancels. Covariance of the increments and the deterministic means therefore passes to the gauge.□

Proof. Define U as a fixed multiple of the sum of the suprema of the listed corrected costs and switch gauge differences, large enough for the allowed number of switches. It is local because the road windows and their gauges are local.

For its expectation, compare with an auxiliary sum over the finitely many relevant periods and anchors: include |C_i|, the defects D(i,i+1), and the absolute increment discrepancies in (10.4). At a switch the two calibration values refer to the same physical vertex and cancel. Equations (10.10) and (10.11) therefore bound U by a constant times this auxiliary sum. Its expectation is O(\beta), regardless of how many microscopic occurrences lie in the windows.

For physical competitors, let U' be the negative part of the minimum endpoint-adjusted corrected cost in the prescribed subgraph. This is again a local measurable variable by the preceding endpoint-selection argument. Domination gives Q\ge\Delta P for every competitor. After endpoint corrections, only calibration-error increments and signed corrected anchor adjustments remain in the lower bound. The same auxiliary sum controls their suprema in expectation by O(\beta).

The auxiliary estimates may be nonlocal; U and U' were defined from the local costs themselves. Thus the argument does not require local pointwise dominators for the absolute errors P-g.□

Proof. We bound the expectations of the local suprema appearing in the statement, then take their sum as V. The passage-time deficits used to estimate these expectations need not be local.

Choose a fixed macroscopic box containing the road portions in question, the bounded comparison trips used below and every lattice edge meeting them. Let K_n^{\rm vert} be the maximum of (\mu(n(y-x))-T(nx,ny))_+ over lattice vertices in that box, and let M_n be its maximum edge weight. Put

The lower concentration estimate, summed over polynomially many vertex pairs and then integrated, gives \E K_n^{\rm vert}\le C n^{\alpha'}. The exponential tail and O(n^2) edges give \E M_n=O(\log(2+n)). Hence \E K_n\le Cn^{\alpha'}.

The enlargement handles interior-edge endpoints as well. Given a road subinterval from x to y, attach partial edges to endpoints x',y' of the incident lattice edges. The resulting vertex-to-vertex path costs at most the subinterval cost plus 2M_n, while

The definition of K_n^{\rm vert} therefore shows that every forward road subinterval in the box satisfies Q\geq-K_n, because \mu\geq h. For a complete period q=Q(i,i+1) we therefore have \E |q|\leq\E q+2\E K_n \leq\beta+2\E K_n. If q=q_1+q_2+q_3 is split at any two occurrences, then q_2\geq-K_n and q_2\leq q+2K_n. Thus the supremum of absolute raw subinterval costs is bounded by |q|+2K_n, including for intervals selected after inspecting the road. The same conclusion holds for a bounded signed trip by addition.

For the gauges, each summand in (10.6) is a projection of a potential difference. Lemma 10.1 approximates that difference by a bounded signed trip with L^1 error O(\beta). The projection norm and the fixed number of terms therefore give an O_s(n^{\alpha'}) bound. Taking expectations of the same comparison bounds the differences of deterministic label means. Interpolation by (10.9) extends the bound to the windows. After the common deterministic centering, the sum of these local absolute suprema supplies V. Markov’s inequality then gives the asserted tail.

For random phases, take the common centering constant to be the mean of a specified fixed-phase reference anchor, and attach its translated copy as an extra label near each site. The signed-trip comparison bounds the difference between any phase-conditioned anchor mean and this reference by C_s n^{\alpha'}, uniformly in the phase. The centered cocycle identity makes the reference mean independent of the site. Thus centering remains deterministic and introduces no shared random phase.□

Proof. Use the same deterministic coarse cell partition in every world. Within each cell, group the common tube coordinates, common private coordinates, and additional coordinates of that world. These cells are independent. Until we establish that a projection is common to all worlds, \Pi_S denotes the finite-cell projection in the product space of the world under consideration.

We first replace each calibration difference in the gauge formula by a common baseline cost. Let x be the anchor’s coarse index, and choose a marked anchor y on the same baseline, a fixed integer number L_0 of periods earlier. Take L_0 large enough that y lies outside every cutoff U_S contributing at x in (10.6), with an additional margin of 4H+20. The coordinate-conversion bound allows one choice of L_0 for all 0<s<1/4.

Let Q_{y,x} be the forward baseline cost from y to x. The traveled edges lie in the baseline’s own preserved tube, and the path and its marked occurrences depend only on (\eta_0,\xi). Thus Q_{y,x} is the same random variable in every world. Adding the period defects along this fixed number of periods gives

Fix a contributing set S. Attach y to the comparison network locally near y, and connect that network anchor to the reference anchor at z_S by signed comparisons along the boundary of Q_R(k(S)) and its exterior. Such a route can reach the square’s boundary, follow it to the eastern side, and then move east to z_S, just as in the cutoff argument leading to (10.7). The initial attachment is separated from S by the margin at y. Along the remaining route, the local supports Q_H(u) miss S\subset Q_{2H}(k(S)), since R=8H+20. The route has bounded length: L_0 is fixed, k(S)\in Q_R(x), and z_S-k(S) is fixed by the gauge construction.

Every local comparison on this route therefore has zero \Pi_S projection. Telescoping their calibration differences, using their O(\beta) errors and the bounded projection norm, yields

Combining this estimate with the preceding baseline estimate shows that each summand of g_x^j differs by O(\beta) from \Pi_SQ_{y,x}.

It remains to check that these projected baseline costs, including the deterministic mean term, are common to all worlds. The joint marginal law of the world weights and the shared baseline is the same in every world: the baseline uses its own tube weights together with private inputs independent of that world’s actual exterior. Since the calibration map is also the same, the mean m_x=\E P^j(X_x) is independent of j. Define

For any cell subset, conditioning Q_{y,x} on its independent \eta_{\rm other} or \zeta^j coordinates adds no information to its (\eta_0,\xi) coordinates. Consequently, every conditional expectation in \Pi_SQ_{y,x} is a function only of that subset’s (\eta_0,\xi) coordinates. These conditional expectations, and hence the projections themselves, can be chosen as the same functions in every world. They have no dependence on weights in the other preserved tubes.

Only a fixed number of sets S contributes, all within the fixed support radius of (10.6). Summing the projection errors proves (10.12) and locality, with constants independent of the number of worlds. The displayed construction also proves the independence assertion: disjoint baseline tubes supply disjoint sets of product edge weights, their private input collections are independent, and m_x is deterministic.

Finally, interpolate the common anchor gauges using the common raw baseline cost. In (10.9), the raw Q terms cancel when the two interpolations are subtracted. For t\in[i,i+1], their discrepancy is therefore the linear interpolation of the discrepancies at i and i+1. Its supremum is bounded by the sum of those two anchor discrepancies. A bounded interval meets only a bounded number of periods, which proves the stated expected supremum bound.□

Proof. We first check phase dependence, including that of the conditional expectations used to construct a gauge. An anchor is determined by local weights, local private inputs, and the phases of the roads in a fixed neighborhood. The calibration itself depends on the weights alone. Each projected variable in (10.6) is the difference of calibration values at two anchors: the anchor carrying the gauge label and the reference anchor indexed by z_S. Both anchors lie within a fixed distance of the gauge’s location. Indeed, for a gauge indexed by z, one has k(S)\in Q_R(z) and z_S=k(S)+(R+4H+10)e_1.

Regard the phases as fixed parameters when taking expectations over weights and private inputs. The variable being integrated depends on the phases only through the two anchors. Its conditional expectations therefore depend on the same finite list of phases. This applies to every projection and to the deterministic anchor mean in the gauge formula. The number of potentially involved B-wall indices is bounded by their spatial range, independently of the number of possible phase values. The spatial support of each projection is the one given by Lemma 10.2.

All mesh offsets lie in one bounded fundamental cell. Thus the supports of the gauges being averaged lie in one fixed larger neighborhood, uniformly in |\mathcal H| and \sigma. The triangle inequality for the normalized average preserves the L^1 estimates without a factor |\mathcal H|. At a fixed physical anchor, each mesh gauge has the same mean, namely the expected calibration value at that anchor, so averaging preserves this mean. For a switch between two labels, the triangle inequality applied to the average of their within-mesh gauge differences preserves the switch estimate without a factor |\mathcal H|. Interpolation is affine in the anchor gauges and uses the same physical road cost on every mesh. It therefore commutes with averaging. The construction from the preserved incoming road can be averaged in the same way; its common-tube comparison and its dependence only on the baseline’s own tube inputs and private field are preserved.

Translation by A advances the wall index by one and the absolute row height by s. Since \theta_j=js+\phi_j, its action on relative phases is the ordinary shift of the sequence (\phi_j)_{j\in\Z}. The road rules, spatial inputs, and gauge construction are covariant under this joint translation, giving covariance under A. Translating the decorated network by aB, a\in\mathcal H, adds a modulo one to every wall phase and permutes the offset meshes. Integer carries are absorbed by unit-period covariance. The independent uniform phase law is invariant under this addition, and the spatial product law is translation invariant. This proves the asserted covariance of the row laws.

Finally, the non-phase inputs for an entire column lie in a strip of fixed width about that column. Sufficiently separated columns have disjoint strips and disjoint finite sets of wall-phase indices. The auxiliary A-roads contribute only private coordinates at the spatial locations being inspected. Hence the corresponding column sigma-fields are generated by disjoint independent inputs. The strip width and the range of wall indices are independent of |\mathcal H| and \sigma, as required.□

Proof. First suppose all indicators are independent. If a path visits \ell distinct marked sites, list them in their order of first encounter as z_1,\ldots,z_\ell. With z_0=0, \sum_{i=1}^{\ell}|z_i-z_{i-1}|_1\le L. There are at most 4(d+1) displacements of \ell^1 length d. The number of possible lists is therefore at most

The identity follows from \sum_{d\ge0}(d+1)x^d=(1-x)^{-2}, followed by summing coefficients through degree L. The count includes lists that cannot occur along a path, which only increases the bound.

Distinctness makes the probability that a specified list is marked at most p^\ell. If M_L is the maximum in the statement and a=C(L+1)\sqrt p, the union bound gives

For a\le1/2, summing over \ell\ge1 gives at most \sum_{\ell\ge1}a^{2\ell}\le2a^2. For a>1/2, split at \lceil2a\rceil: the initial sum is O(a) and the remaining sum is geometric. Thus \E M_L\le C\min\{a,a^2\}, after adjusting constants also for 1/2<a<1. Combining this with M_L\le L+1 proves (12.1).

For a partition into independence classes, apply the same list count to sites in each class and sum the resulting q maxima. No independence between classes is needed. For example, when I_z depends on independent inputs in z+[-R,R]^2, residue classes modulo an integer greater than 2R give the required partition. The same partition argument works when sigma-fields generated by sets at distance greater than a fixed range are independent.

Only the budget for successive distances entered the counting argument, so it also proves the assertion about lists. An invertible linear identification of a fixed full-rank lattice with \Z^2 changes this budget by a fixed factor and proves the lattice version.□

Proof. For u\ge0,

Take the supremum over paths, move it inside the nonnegative integral as an upper bound, and apply Lemma 12.1 at every level. Tonelli’s theorem gives (12.3), possibly with an infinite right-hand side. Integrating \sqrt A\,t^{-2} proves the fourth-power assertion.

The sharper minimum of (L+1)\sqrt{p(t)} and (L+1)^2p(t) in (12.1) can also be used inside the integral. In particular, the estimate has no additive constant independent of t to be integrated over an infinite interval.□

Proof. Before selecting an object, form

where the maximum ranges over all tuples in the region with pairwise disjoint supports, including those that no admissible object supplies. Set V=0 if there are no such tuples. This is a local variable and gives the required simultaneous domination. For each fixed tuple, independence and Markov’s inequality bound the probability that all m values exceed t by \min\{1,(A/t)^m\}. There are at most J^m tuples, so a union bound proves the tail estimate. Finally,

The conditional proof is identical. The object itself may depend on the whole environment: independence is used only for each fixed tuple, before taking the finite maximum.□

Proof. Compare the tilings (T(z+Q))_{z\in\Lambda} and (w+Q)_{w\in\Lambda}. Join z to w when the corresponding tiles intersect in positive area. This bipartite graph is countable and locally finite. All tiles have the same area v_Q>0. Any union of k tiles on either side has area kv_Q and is covered, up to a null set, by its neighboring tiles on the other side. It therefore has at least k neighbors. Both finite Hall conditions hold.

Here is the compactness step giving a perfect matching. By finite Hall’s theorem [18, Theorem 1], any prescribed finite set of left vertices is covered by a finite matching M_1, and any prescribed finite set of right vertices by a finite matching M_2. The union of the two matchings consists of alternating paths and cycles, together with common edges. On each component one can choose an alternating matching covering all prescribed vertices. Internal vertices are covered by either choice. On an odd-length path, one choice covers both endpoints. On an even-length path, the endpoints lie on the same side, and the matching that covered the prescribed vertices on that side already covers every prescribed endpoint. Thus one matching covers both prescribed finite sets.

Exhaust the two vertex sets by finite sets and choose such a matching at each stage. A diagonal subsequence converges on every edge. Since each vertex has finite degree and is covered at all sufficiently large stages, exactly one of its incident edges belongs to the limit. The limit is a perfect matching.

Let \psi(z)=w for the matched pair. An intersection point has the form Tz+Tq=w+q', with q,q'\in Q, giving the stated displacement bound. Translate the first tiling by a to obtain the affine statement, with the same error relative to Tz+a.□

Proof. The scalar prefactor makes the determinant one. In E-coordinates, the map contracts the first extent of the box by F and expands the second by F. The prefactor and U have uniform bounds, so the transformed diameter is at most C(N/F+F\delta N)+C_F. The fundamental-tile bound in Lemma 12.4 bounds the matching error uniformly in E for fixed F. Adding this error proves (12.6). Its N-dependent coefficient C is independent of F; the deformation of the fixed margins and the matching error are included in C_F.

For v=z_j-z_i, condition (12.4) gives v=Ea+e with a\ge0 coordinatewise and |e|\le CH. The vector T_FEa is in the cone generated by u_1,u_2, strictly inside \mathcal K. Consequently its distance to the complement of \mathcal K, and each of its positive coordinates, are at least \rho_{\mathcal K}|T_FEa| for a fixed \rho_{\mathcal K}>0. For fixed F,

with c_F>0. Replacing T_Fv by \psi_F(z_j)-\psi_F(z_i) adds at most twice the matching error. Choose R_F so that the cone margin and the two coordinate margins exceed these errors plus r_0 whenever |v|>R_F. Compactness of the allowed matrices makes this choice uniform in E.□

Proof. Fix a law and t>0, and set p=\PP(Y_z>t). There is nothing to prove if p=0. Otherwise choose L comparable to p^{-1/2}, with lattice-dependent comparison constants, so that L\ge1 and [-L,L]^2 contains at least p^{-1} lattice sites. Independence implies that some value in the square exceeds t with probability at least 1-e^{-1}. Since singletons are allowed in G,

Rearrange and combine with p\le1. The constants do not depend on the law.□

Proof. Fix F, and take R_F from Lemma 12.5. We divide the selected values into nearby collisions, which have a lighter tail, and one separated set, to which the collection bound applies.

Collision values. Define the local field

If D_z>t, both Y_z and one distinct neighboring value exceed t. Independence and Lemma 12.6 give

This argument includes laws with atoms. The dependence range of D is fixed once F is fixed. The path budget and Corollary 12.2 therefore bound the expected supremum of its selected tail sums by C_F(N+1)/T.

Separated values. For any admissible sequence, keep one occurrence of each site in its original order and pay a factor m_0 for multiplicity. Among these sites with Y_z>T, process values in decreasing order, breaking ties by a fixed deterministic order. Retain a site if its distance from every previously retained site is greater than R_F. Every discarded site has a distinct retained neighbor with at least its value, so its value equals D_z. The retained sites are pairwise separated by more than R_F.

Put the retained sites back in their original order. The all-pairs order condition still holds, so their images under the deterministic bijection \psi_F form an admissible directed chain. They lie in a deterministic square with radius

The transported field \widetilde Y_w=Y_{\psi_F^{-1}(w)} has exactly the original iid law. Thus the retained reward is at most the corresponding G_n(L_F,x_F) in this transported field, for a deterministic center x_F.

Crucially, the same square and the same field permutation work for every original admissible sequence. Hence this domination holds after taking the supremum over those sequences, even though the pruning and the retained chain depend on the observations. By (12.7), its expectation is at most

Add the collision estimate and absorb the fixed factor m_0 to obtain (12.8).

Choose F first so that C/F<\varepsilon/3, then choose T_0 so that C_F/T_0<\varepsilon/3, and set \delta_0=F^{-2}. The remaining terms vanish after division by N(n)\to\infty, uniformly over the laws. For finitely many types, bound the supremum of a sum by the sum of the suprema. For a finite-range field, first split into finitely many sublattices on which each fixed template has an iid law, provided the collection hypothesis holds for every such law. Apply the argument separately and sum.□

Proof. The case B=0 is immediate. Otherwise fix a deterministic row and omission set. Split the centered sum into its q residue classes. Within each class the summands are independent, have support in intervals of length at most 2B, and number at most N/q+1. For the total sum to have absolute value greater than \varepsilon N, some class sum must have absolute value greater than \varepsilon N/q. Hoeffding’s inequality [20, Theorem 2, Equation (2.6)] and a union bound therefore give, for N\ge q,

as an upper bound on this probability.

It remains to count the choices over which the supremum is taken. Put V=\lfloor\delta N\rfloor. For a fixed initial row, stars and bars gives \binom{N-1+V}{V} choices of the absolute increments with sum at most V. Their signs have at most 2^V choices. The total number of rows is therefore at most

where

For 0<\eta\le1/2, there are at most (N+1)\exp\{Nh_{\mathrm{bin}}(\eta)\} omission sets, where h_{\mathrm{bin}}(\eta)=-\eta\log\eta-(1-\eta)\log(1-\eta). Choose \delta,\eta so that the two entropy terms sum to less than \varepsilon^2/(8qB^2). Multiplying the counts by (12.11) gives a polynomial factor times 2q\exp\{-\varepsilon^2N/(8qB^2)\}, uniformly in n.

If rows are specified only at nonomitted columns, assume the same bound on their first specified row and on their successive jumps. Extend to the full interval by holding that first row on the initial gap and, at each later gap, holding the old row until the next specified column. Hold the last row on the final gap. This does not increase the total variation and gives a row counted above. Indicators have B=1. For a fixed finite sum, apply the result to each summand with a divided tolerance, then use the triangle inequality and a finite union bound. Alternatively, the sum itself has a fixed bound and can be treated directly when its combined supports satisfy the column independence condition.□

Proof. For every u\ge0, a geometric sum gives

Take expectations and sum over a. At least one of the J resulting nonnegative summands is no larger than their average, which is at most 2A/J.□

Proof. For I\subseteq\mathcal N(c) put \mathcal F_I=\sigma(\eta_d:d\in I), and define

Independence of the input cells and measurability of U_c give

This is one of the nonnegative summands defining W_c. There are at most 2^{m_0} summands, each with expectation \E U_c. Since only the finitely many intersections with \mathcal N(c) matter, versions can be chosen for which all the inequalities hold simultaneously.□

Proof. The nominal h-coordinate increases along either road at a fixed positive rate, and tracking changes it by a bounded additive amount. The slab condition therefore lets us pass the full observation halo by moving a fixed number of periods before the incoming window or after the outgoing window. The relevant transverse-road index is chosen from the deterministic test type. For example, on an incoming B-wall of nominal first coordinate i, take the A-road of index

where the fixed M_0 exceeds the final slab padding b_0, its input halo, and the support radius of a connector by a strict margin. Then use the canonical intersection of those two prescribed roads. On an incoming A-wall interchange the coordinates; for exits use the analogous ceiling above the upper face. Tracking changes the h-level by only O(\rho), absorbed by the margin. The chosen parameter differs by a bounded amount from the cut window. Thus the index choice uses no unrevealed input, and all possible attachment occurrences remain in fixed local windows.

The first and last road portions in the asserted concatenation are physical forward portions. Additivity of signed corrected costs gives

Adding C_\pi(x,y) proves (13.6), whatever the relative orders of x,x_0 and of y,y_0. In particular, neither signed center arm has to be traversed backward. A label change at an exterior intersection adds the corresponding difference of the two local gauges. All extra signed travels and switches lie in the bounded windows of Section 13.1. Their absolute costs are therefore bounded by U_{c_-}+U_{c_+} after a fixed enlargement of those envelopes.□

Proof. Normalize the two costs together:

By (13.1) and Lemma 13.1, \E S_c\le C uniformly. Increase the fixed support radius R_0 to include the cell diameter.

Separated candidates. Choose m nested annuli about a potential exit mark, with annulus widths larger than the diameter bound for a component of the road complement, and with gaps larger than 4R_0. Also choose an integer J>2R_0+1+2\rho. Take H_m large enough that all annuli fit inside its ball and H_m>(m-1)J+2\rho+1, then take K>20H_m. On the path portion preceding an exit mark, use the last passage from the outer boundary of this ball to the mark. Its successive crossings of the annuli meet the road network; choose one occurrence at such a meeting in each annulus. If the mark is not at a cell center, allow every unit cell meeting the annulus enlarged by two cell diameters. The larger gaps ensure disjoint input supports for any candidate tuple having one cell in each enlarged annulus. The finite number of such tuples depends on K,m,R_0, but not on n.

One bound for every path. For the mark cell z, let \mathcal T_z contain every possible annular tuple and every horizontal tuple described below, allowing all center offsets within z, and set

A finite list of discretized annular and horizontal patterns suffices, because the allowed candidate sets are subsets of a fixed finite list of unit cells. For each fixed tuple, independence, Markov’s inequality, and the uniform mean bound give

Taking the union over \mathcal T_z proves (13.8). The actual path supplies a tuple in this list. Choose the candidate minimizing S_c, breaking ties by the deterministic cell and occurrence order. This gives (13.7) for every path on the same event, regardless of how that path was selected.

Geometry and gauges. The selected cut lies within H_m<K/10 of its mark. A preliminary exit has displacement exactly K from the preceding mark, in the continuous max norm, so the triangle inequality gives the asserted displacement bounds. Each portion between preliminary marks stays in its K-ball. The two adjacent short terminal portions give the O(K) diameter bound. The selected cut occurs after the preceding preliminary mark, since that mark lies outside the small terminal ball. Thus the cuts preserve path order.

Finally,

A union bound over O(n^q) mark cells proves the fourth assertion. The single objective S_c is essential here: it selects both bounds at once, so the same product estimate controls the charge and the gauge.

Horizontal terminal portions. Let [t_-,t_+] be the terminal occurrence-time interval and let P linearly interpolate the fine cuts on their time intervals. The diameter bound gives |P(t)-\gamma(t)|_\infty\le\delta_f in the present coordinates. For t_-\le a\le b\le t_+, the increment P(b)-P(a) is a sum of nonnegative multiples of the good fine increments. Hence, for \Delta=\gamma(b)-\gamma(a) in these coordinates,

This includes both partial fine segments at the ends. At a=t_-,b=t_+ we have |\Delta|_\infty=H_m. The strict inequality in (13.9) excludes \Delta_1=-H_m and |\Delta_2|=H_m, so \Delta_1=H_m. Thus the last boundary hit is on the left face. By lastness and continuity, all later points of the terminal arc are in the open ball.

Put p_1=\xi_1(\gamma(t_-)) and choose the wall levels

Our choice of H_m gives p_1<i_j-\rho and \xi_1(\gamma(t_+))>i_j+\rho. A proper B-road in each such vertical strip separates these two sides: apply the planar rectangle-crossing property to a bounded vertical road portion and a left-to-right subarc. Select the first meeting along the terminal arc. It is a vertex occurrence, since an interior meeting of two square-lattice paths lies on a common edge whose entering vertex would already be a meeting; the initial arc point is outside the wall. At repeated road visits choose the first eligible vertex occurrence in the bounded window determined by the cut cell. The first meetings with i_1,\ldots,i_m occur in that order, because reaching a later wall requires crossing every earlier wall. All of them lie strictly inside the terminal ball.

The center of a cell meeting W_{i_j} has first coordinate within \rho+1/2 of i_j. Distinct candidate centers are therefore separated by at least J-1-2\rho>2R_0, so their input supports are disjoint. For a fixed mark cell all candidates lie in a fixed finite (H_m+1)-neighborhood. The horizontal tuples were included in \mathcal T_z before the common objective was maximized, so the same product estimate applies without conditioning on the path or on its terminal test. The selected point stays on the terminal arc, preserving the preceding order and displacement bounds.□

Proof. Minimality gives c\ge-u, and a_{\mathcal D}\ge0 gives

Lemma 13.1 bounds a_{\mathcal D} by the two W/\beta terms in (13.12).

On \{Y>0\}, Lemma 13.2 realizes the internal minimizer with normalized corrected cost at most -u+(U_{c_-}+U_{c_+})/\beta. Taking its conditional mean yields -u+a_{\mathcal D}=-Y. The envelopes already bound every admissible attachment occurrence, so this estimate holds for the selected minimizer.□

Proof. Choose the chain and realize its route. Observe the inputs of every test in the sparse array. In the given finite square, choose a reward-maximizing chain, resolve ties lexicographically, and omit zero-reward sites. This is a function of the observations. Fix an external southwest and an external northeast corridor intersection, outside the square by 3S in both coordinates, with the same prescribed intersection-label convention. These endpoints depend on the square, not on the observed chain.

Route monotonically on the corridor grid from the start to the southwest portal of the first selected cell, use that tunnel and its private access, then continue from its northeast portal to the next southwest portal. The assumed 2S gaps make both coordinates of each latter displacement nonnegative, with a full corridor spacing to spare. Continue similarly to the fixed end. Every portion is a physical forward road portion or a physical internal tunnel path. Thus the result is an actual path.

Its total ordinary-corridor length is O(L+S), by telescoping the two positive coordinate increments. A chain has at most C(L/S+1) sites. The private uncharged connector lengths are O(S) per site, so their total length is also O(L+S). The number of road switches has the same bound when measured in unit road spacings. The two short charged attachments at each site are excluded from this count.

All selected route costs are integrable. A used tunnel has Y>0 and therefore has negative center-label cost. Rearranging that cost identity bounds its positive physical passage cost by a bounded nh displacement and the integrable local gauge and raw-arm envelopes. Charged attachments and ordinary finite road portions are integrable by the same window estimates. The chain is finite, and its maximum reward is integrable by the bound following (13.11). Thus the conditional costs and the expectation rearrangements below are finite.

Take conditional expected costs. For every prescribed ordinary connector, its full input support misses all observations. Its inputs therefore retain their product law under the conditioning. Subdivide the connector into unit road periods and apply the local envelopes to those periods and its switches. Its expected normalized corrected cost is at most a constant times its nominal length plus its number of switches. Summing uses only linearity of conditional expectation; connectors may share unobserved inputs.

The support of a charged access envelope meets no observation other than its own test. Independence gives

Thus each selected tunnel together with its charged attachments has conditional expected normalized corrected cost at most -Y_p, by Lemma 13.4. Consequently the completed path \Pi satisfies

Apply the calibration at the fixed endpoint windows. Corrected costs use only differences of gauges. For this endpoint calculation, undo any common deterministic centering and use the original mean-matching gauge representatives; the corrected cost of \Pi is unchanged. The constructed path has endpoint labels prescribed by translated copies of the exterior anchor convention. Calibration domination gives, pathwise,

Although an endpoint intersection is random, it lies in a prescribed bounded road window. Put R_W=P_n-g_W on that exterior road and choose a fixed marked anchor k before the entire window of possible endpoint occurrences t. Equation (10.3) gives \E R_W(k)=0, and the defect and interpolation identities give

The interval is covered by a fixed number of complete road periods. Additivity and nonnegativity bound its defect by the defects of those periods; Lemma 10.4 controls its corrected cost. Thus the expected supremum of the absolute right side is at most C\beta. In particular,

The estimate covers every possible intersection occurrence and allowed endpoint label. Consequently the expected normalized corrected cost of \Pi is at least -C. Taking expectations in (13.14) yields \E G_n(L)\le C(L+S)+C, and the fixed S is absorbed into C_0. The bound holds at each finite L, uniformly over the tunnel law; no law-dependent limiting passage is needed.□

Proof. Put F=S_\mu\setminus U. If the first assertion failed, there would be t_j\downarrow0 and probability measures \nu_j with barycenters within t_j of a and \nu_j(F)\ge e>0. A weakly convergent subsequence exists on the compact sphere. Its limit \nu has barycenter a, and the closed-set inequality gives \nu(F)\ge\limsup_j\nu_j(F)\ge e.

The convex hull of F is compact and omits a. Indeed, Carathéodory’s theorem would otherwise express a as a finite nontrivial convex combination of points different from a, contrary to extremality. If \nu(F)=1, its barycenter lies in that convex hull. If 0<\nu(F)<1, the conditional barycenters on F and its complement lie in the norm ball and express a as a nontrivial convex combination. Extremality makes both equal to a, whereas the first lies in the convex hull of F. Both cases are impossible, proving \omega_U(t)\to0.

For the second assertion put S=\sum_i\mu(z_i)>0 and give z_i/\mu(z_i) mass \mu(z_i)/S. Its barycenter is (\sum_i z_i)/S, at distance at most

from a. The mass on F, multiplied by S\le(1+d_0)N, is the left side of (14.5).□

Proof. Let B_0 be the bad set. For \theta>0, put

A one-dimensional maximal-interval bound gives |R|\le3|B_0|/\theta: select disjoint witness intervals in decreasing order of length. Their triple enlargements cover all discarded witness intervals, and their lengths sum to at most |B_0|/\theta. Every interval containing a retained index therefore has bad density at most \theta.

Let \ell_1,\ell_2 be the coordinate functionals of \mathcal C_2. Compactness and the strict cone inclusion give \ell_j(x_{i+1}-x_i)\ge\kappa_0K at good indices, whereas at bad indices they are at least -CK. Choose \theta<\kappa_0/(2(C+\kappa_0)). Every interval containing a retained index then satisfies

Apply this to the interval between two retained cut indices, then absorb the bounded errors in choosing their centers. To include the initial endpoint and a retained starting-cut representative x_i, use the prefix through increment i and subtract that increment, whose size is at most CK. The suffix from x_i already contains the retained index. Center rounding adds another bounded error. The coordinate matrices range in a fixed compact set, so these errors remain uniform as the cone enlargement shrinks.

For multiplicity, apply the same density argument to h. After decreasing \theta if necessary, it gives h(x_j-x_i)\ge cK(j-i) on the relevant intervals, where c is independent of the cone enlargement. Only boundedly many retained centers can therefore lie in any fixed-radius ball.

For the cost assertion, bound each removed gauge difference separately:

Choose \lambda<c_*\theta/24 and then n large. The bracket is at least c_*/2, retaining a positive penalty per bad index. The set R depends only on the bad indices, so this conclusion holds for every assignment of effective gauges satisfying the stated bound.□

Proof. An edge-only event and fine directions. Choose the nearest endpoint and the geodesic by fixed tie conventions. The metric and cheap-path estimates give an event \mathsf M_n depending only on the edge weights, with \rho_n=\mathbb P_{\rm edge}(\mathsf M_n^c)\to0, on which

Here L>99 makes (Nn)^{0.99}=o(N); endpoint rounding contributes only O(1) to Q_n. Write T_n(x,y)=T(nx,ny) for scaled metric-graph endpoints. On the same event, every such pair in a fixed enlargement of this box and at the bounded scaled distances used below satisfies

The graph-endpoint assertion includes the maximum incident-edge bound from Lemma 8.1. Thus it holds simultaneously before any cuts are chosen.

The fine choices in the stated parameter order can be made as follows. Choose the rational approximation to a, a neighborhood U, and \varepsilon_f>0 so that every increment directed in U satisfies

Indeed u(a)>0, y(a)\to0 as v\to a, and u(z)\le C\mu(z) by fixed norm comparison. Take \varepsilon_f small enough and choose the fixed radius \varrho so that

These conditions are compatible because s<1/4 and 2r<b'.

Use the fine exit partition of radius \varrho in the (A,B) maximum norm. Each fine trace has diameter at most 2\varrho, including the final incomplete trace. The edge budget bounds the number of fine segments by CN/\varrho+1. For their increments z_i, put S_n=\sum_i\mu(z_i). Summing the metric lower bounds gives

while the triangle inequality and endpoint rounding give S_n\ge N-C/n and |\sum_i z_i-Na|\le C/n. Thus both errors in Lemma 14.1, divided by N, are at most

Define

The total norm mass outside U is at most \epsilon_n^{\rm fine}N. Every full fine increment has norm length at least c\varrho, so there are at most C\epsilon_n^{\rm fine}N/\varrho+1 bad increments, including a possible bad incomplete one. A zero final segment is omitted. The good transverse variation is at most \tfrac12\eta\sigma S_n\le\eta\sigma N for large n, and the bad variation is at most C\epsilon_n^{\rm fine}N. All these bounds hold on the same edge event for every road law.

Cuts and the uniform good event. Place the actual preliminary K-exit marks in the (A,B) maximum norm. At each mark test every fine interval whose occurrence-time interval meets the terminal arc, including at either endpoint. The coordinate identities

put every good fine displacement in \mathcal G_\kappa of (13.9). With \delta_f=2\varrho, \kappa\le1/2 and 12\varrho<H_m imply its exact strict condition

If every fine interval meeting the terminal arc is good, use the B-wall candidates of Lemma 13.3; otherwise use its annulus candidates. Both tuple families and all future datum choices were included before this test in the single objective W_c/\beta+V_c/a_n.

With a_n=C_1n^{1-\vartheta}, that lemma supplies

The inequalities cover every allowed effective gauge at the chosen cut. Let \mathcal Z_n be the deterministic set of possible mark cells in the box of (14.14), with its fixed boundary enlargement. It has at most Cn^{2L+10} cells. Define

The union bound in the selected-cuts lemma gives

uniformly in \mathfrak r. No independence between the metric event and the cut fields is needed. A selected cut lies within H_m of its mark. Since successive preliminary displacements equal K, selected displacements lie between K-2H_m and K+2H_m, in particular between K/2 and 2K. The intervening pieces and the two terminal fragments have diameter O(K).

The halo and its cost. Each terminal fragment has raw cost at least -e_n, so the selected pieces have total raw cost at most q_nN+2e_n. Every such piece has raw cost at least -e_n. For an off-\mathcal C_1 displacement in the fixed annulus, compactness gives a positive lower bound for \mu(z)-h(z). For a unit backward h-excursion, split before, during, and after the excursion; \mu\ge|h| gives raw cost at least 2n-3e_n. For some fixed c_*>0, bad pieces therefore cost at least c_*n for large n. Let B_0 be their index set. The edge budget gives at most CN/K+C pieces, whence

Use the fixed \lambda<c_*\theta/24 from the halo lemma and \epsilon_n^{\rm raw}=e_n/n. The halo set R depends only on B_0, and |R|\le3|B_0|/\theta. On \Omega_{n,\mathfrak r} the gauge bound holds simultaneously for every later datum, so the removed cost is nonnegative for the eventual shared choices. They telescope on the full consecutive list, giving

The halo’s endpoint order puts the retained centers in (14.10): the second E-coordinate of a tends to zero as \mathcal C_2 shrinks to \mathcal C. The endpoint and center errors stay bounded because E ranges in a compact family. Each preliminary exit uses at least cKn edges, so there are O(N/K) centers. Their successive-distance budget, including the initial O(K) distance, is O(N). The halo lemma also gives the stated multiplicity bound.

Rows and regular links. Track the fine polygon by hysteresis: keep its current row until it leaves the central sub-band of half-width b/4, then choose the nearest row. Since \sigma<b/10, each change except possibly the first consumes at least b/10 of transverse variation. The row-index variation is bounded by a constant times that variation divided by \sigma, plus a constant times the number of changes. It is at most

Let \mathcal I_0 contain exactly the pieces whose closed occurrence-time intervals meet a row change or a bad fine interval, or which are incident to a failed terminal test. Define this set on the full selected list, regardless of membership in R. A row-change time lies in at most two closed coarse intervals. A fine trace has diameter at most 2\varrho. Its closed occurrence-time interval can meet at most two closed coarse intervals, since otherwise it contains two consecutive cuts separated by K/2. Each terminal arc occurs after its preceding preliminary mark. A fine interval meeting three terminal arcs would contain two consecutive preliminary marks, separated by K. Each failed test is incident to at most two pieces and is charged to a bad fine interval meeting its terminal arc. Thus

On a piece outside \mathcal I_0, every fine interval meeting it chronologically is good. The polygon calculation in (13.10), now applied to any submotion of this piece with \delta_f=2\varrho, gives horizontal drawdown at most 4d\varrho. The path is within transverse distance 2(1+s)\varrho of its fine polygon. The hysteresis and (14.15) therefore keep the whole piece in one full row band, with drawdown less than b'. Its two successful terminal tests supply B-wall vertex occurrences. Write k_i for this constant hysteresis row. Two adjacent pieces outside \mathcal I_0 have the same row: their closed time intervals share a cut, and a change there would put the incident pieces in \mathcal I_0.

The first (A,B) coordinate increases across this piece. Its maximum norm displacement is at least K-2H_m. If the negative first coordinate realized that norm, its size would be at most 4\varrho by (13.10); if the second coordinate realized it, its size would be at most 4(1+\kappa)\varrho/(1-\kappa)\le12\varrho. Both contradict 12\varrho<H_m<K/20. Hence the first-coordinate increase is at least K-2H_m. Since \rho=2r is the wall tracking width in these coordinates, the terminal wall index exceeds the initial one by at least K-2H_m-2\rho>0.

Split the piece at successive first meetings of the next B-wall. At each intermediate meeting, use one occurrence of that wall for both adjoining subpaths. Planarity forces a meeting before passage beyond that wall’s far padded side, and the drawdown bound for every submotion supplies the backward allowance. End the final subpath at the original terminal cut, even if it met that wall earlier. Its overshoot is at most 2r+4d\varrho<b', so the entire trailing portion is retained. These are admissible links in (14.2); the piece diameter bounds their number by a fixed q_K. Write q_i\le q_K for this number.

Figure 3 shows the terminal-wall candidates and the first-meeting decomposition of a regular piece, including its final subpath after an earlier visit to the terminal wall.

Distinct columns. At every original cut put

At a cut on W_j, tracking gives \lvert \xi_1(x_h)-j\rvert\le\rho+O(1/n)<1/2 for large n, so v_h=j. Form one integer walk by expanding each piece outside \mathcal I_0\cup R into its q_i positive unit wall steps from v_i to v_{i+1}, and replacing each piece in \mathcal I_0\cup R by the jump between these endpoints. Include the two terminal jumps 0\to v_0 and v_{m_N}\to j_n. The walk begins at 0, ends at j_n, and is fixed before any whole piece is rejected by the following scan. Each jump has magnitude O(K), including the terminal jumps. If q_{\rm all} is the number of unit steps and P,D are the sums of the positive and negative jump magnitudes, then

Scan this fixed walk once, updating its running maximum at every visited vertex. Call a unit step j\to j+1 a candidate if its endpoint is a new strict maximum and is at most j_n, and let q_{\rm rec} be the number of candidates. For each integer level \ell\in\{1,\ldots,j_n\}, the first step whose endpoint is at least \ell is either a candidate unit step ending at \ell or a positive jump. A positive jump is charged at most its magnitude. Hence

This counts every rejected unit step, including record steps beyond j_n. Let \mathcal C_{\rm rec} be the pieces outside \mathcal I_0\cup R containing at least one rejected step, and set

Distinct pieces in \mathcal C_{\rm rec} contain distinct rejected steps, so \lvert\mathcal C_{\rm rec}\rvert\le D. Keep all links of the pieces in \mathcal G, without rescanning the walk. They are candidate steps, so their columns are distinct, increasing, and in \mathcal J_n. Discarding a whole piece loses at most q_K candidates, and therefore

The bounds on P,D prove the stated irregular and omission bounds. Since u(a)/d is bounded above and below, j_n is comparable to N; for sufficiently small \eta, the omission bound also gives \lvert\mathcal L\rvert\ge cN for large n.

The selected rows form a chronological subsequence of the hysteresis rows, so deleting terms cannot increase variation. Hold the old row across each omitted gap, as in Lemma 12.8. The box in (14.14) bounds the first selected row by CN/\sigma\le j_n^2 for large n. This proves (14.12).

Shared data and the tunnel charge. After these masks are fixed, assign one datum to every original cut x_h, 0\le h\le m_N, retaining its selected road label and occurrence. If at least one incident original piece belongs to \mathcal G, use that piece’s row center on the cut’s B-wall. When both incident pieces belong to \mathcal G, their rows agree by the argument for closed time intervals above, and they use the same selected wall occurrence, so they request the same center. If neither incident piece belongs to \mathcal G, use the prescribed reference center for the cut cell and road label. At h=0,m_N, only the one existing incident piece is considered. Use this datum for both incident pieces even if one belongs to R or \mathcal I; all incidence here is incidence in the original list.

The masks R,\mathcal I_0,\mathcal C_{\rm rec} did not use the eventual data, and the simultaneous V_{\rm cut} bound covers all the prescribed choices. Thus these are the shared data used in the telescope over the full list and the halo estimate above, which holds for every allowed datum choice. Every piece in \mathcal G receives its own row center at both endpoints. Any retained nonregular piece accepting a neighboring center is already in \mathcal I, while any removed incident piece is in R; the datum choice requires no further irregular or omission charge. For i\in\mathcal G, let \gamma_{i,\nu} be its consecutive physical subpaths from the decomposition at first meetings. Intermediate center arms cancel, including across the retained terminal subpath, and give

This proves (14.11).

For each retained piece, choose the access type with its cut cells, road labels, the two prescribed center conventions just used, and its rounded padded domain. This type belongs to the fixed finite family, and H(\gamma_i)=\beta c_i^{\rm cost} in (13.5). For the tunnel comparison, choose an exterior wall occurrence a_i before both \bar x_i,x_i, and b_i after both x_{i+1},\bar x_{i+1}, as in Lemma 13.2. Additivity gives

The right side is the corrected cost of the physical forward concatenation in Lemma 13.2. Those attachments enter the access charge; a negative signed arm is never treated as a negative physical traversal. Every retained piece lies in its padded h-slab because it has no unit backward h-excursion. Lemma 13.4 therefore gives (14.13). At each coarse center, sum all cut error fields in its fixed O(K) neighborhood and over the finite cut-type list. This fixed local field dominates the two selected endpoint errors for every allowed piece, keeps a fixed range, and preserves the tail exponent. The finite type and phase-template lists are those fixed in Section 13.6.

Finally, one common deterministic error sequence is

The fixed constant absorbs 1/\varrho,1/\sigma,1/\theta,1/c_0 and the bounded endpoint adjustments. The preceding cost, halo, irregular, omission, and row estimates then have exactly the stated bounds. Together with the uniform \pi_n above, this proves the proposition.□

Proof. Fix an error \varepsilon>0, to be chosen relative to \zeta. Choose F,T in (14.17), reserving the corresponding upper bound for the final \delta. By (14.19), choose J after T,M so that (q_KM+2T)C_KJ^{-1/2}<\varepsilon, and then choose \eta so that its contribution has the same bound.

Apply Lemma 12.8 to f_{J,M}(X_{j,k}/\beta), using the original column independence from Lemma 11.1. Its fixed bound is \max(J,M). The interval length j_n is comparable to N; hence (14.12) meets the lemma’s row and omission tolerances after decreasing \eta and taking n large. Uniformly over the road law, with probability tending to one,

on the extraction event. The supremum in the row lemma is taken before the rows and omissions are selected.

If (14.20) failed uniformly, choose an offending sequence of scales and road laws. On the geometric and row events the capped sum is at least c\zeta N-\varepsilon N, while (14.9) bounds the left side of (14.18) by \epsilon_nN. Since \mathcal E_J is empty off the good event, (14.19) and (14.17) bound the expected nonnegative losses in (14.18) by C\varepsilon N for all large n. By Markov’s inequality, they are at most 4C\varepsilon N with probability at least 3/4. This event intersects the geometric and row events. Choosing \varepsilon<c\zeta/(10C+10) contradicts (14.18). Every threshold and geometric tolerance was fixed before n, proving the asserted uniformity.□

Proof. Fix an error \varepsilon>0. Choose F,T as in (14.17), uniformly over the finite type family, and reserve the corresponding upper bound for the final \delta. Choose M_0\ge M_{\rm prescribed} so large that 2TC_s/M_0<\varepsilon. Let A bound the sum of the first moments of (X^{(a)}/\beta)^- and the U_b. Choose H so large that 2A/H<\varepsilon. Lemma 12.9 ensures that for each law some level in the fixed menu satisfies

We fix all concentration and geometric tolerances for the whole menu before selecting this law-dependent level.

For equal caps, the square-root probability in (14.19) would be multiplied by the cap. Instead use the original row-link fields and their column independence. For every M\in\mathcal M and every link template, apply Lemma 12.8 separately to \mathbf1_{\{X_{j,k}^{(a)}<-M\beta\}} and to the capped costs. Take a finite union of these conclusions, using M_{\max}=2^{H-1}M_0 as a common bound. The resulting event has probability tending to one uniformly over the law, and controls all candidate levels before one is selected.

On this event and the corresponding extraction events, the selected negative-link indicators have frequency at most their marginal probability plus a fixed error \nu. Columns are distinct, so each exceptional regular piece is counted by at least one such indicator. For any candidate M, the regular exceptions in (14.18) with J=M therefore cost at most

The irregular contribution is at most (q_KM+2T)(C_K\eta+\epsilon_n)N. After fixing M_{\max}, choose \nu, the row and omission tolerances, and \eta so that their contributions are at most \varepsilon N. The term containing \epsilon_n then has the same bound for large n. Select a level satisfying (14.24). Its M-weighted term is O(\varepsilon N); its T-weighted term is small by the same bound or by \mathbb P(X<-M\beta)\le C_s/M_0.

The capped row sums at this level are compared with their means by the same simultaneous concentration event. The extraction gives |\mathcal L|\ge cN and retained cost at most \epsilon_nN for each template. Markov applied to the finite sum of their H_N’s supplies a loss event of fixed positive probability, which intersects all the events whose probabilities tend to one. The contradiction in the asymmetric proof, with \varepsilon sufficiently small relative to \zeta, therefore proves (14.22) for every template. Equation (14.24) gives (14.23). The same cap is used for both conclusions; no expectation bound on a discarded heavy negative tail is required.□

Proof. The local rule. Consecutive flagged indices belong to the same cluster when their difference is at most G_*. A cluster with extremes a\leq b is accepted exactly when b-a\leq L_* and

For such a cluster replace the primary slots shown in Figure 4:

Use the incoming transition, then the spare baseline, then the outgoing transition. The spare portion runs from parameter a-B_*+T_* to parameter b+B_*+1-T_*. This interval has nonnegative length because B_*>T_*. The endpoints of the whole replacement are precisely c_{a-B_*} and c_{b+B_*+1}. Its normalized corrected cost C(a,b) is in their primary gauge convention and satisfies

This follows by exact telescoping through the two transitions and the spare gauge. Different accepted intervals are disjoint because distinct clusters have gap greater than G_*>2B_*. All long or infinite clusters are left unchanged.

The latter instruction has an exact finite implementation. A proposed leftmost flag a and rightmost flag b\in[a,a+L_*] form a complete short cluster if and only if they are flagged, all consecutive flag gaps in [a,b] are at most G_*, and there is no flag in [a-G_*,a-1] or [b+1,b+G_*]. These tests read only [a-G_*,a+L_*+G_*]. They can never accept a portion of a longer or infinite cluster. Let O_i indicate replacement of slot i. A cluster covering i has

so O_i is determined by flags within distance 13B_* of i and transition tests within distance 10B_* of i. It follows in particular that

Uncovered positive costs. If a flagged slot is left uncovered because its cluster is long or infinite, it contains a flag at distance strictly greater than R. Following its chain from i until the first such flag gives an index j with

There are 2G_* possible witnesses. If the cluster is short but not accepted, one of the possible incoming tests indexed in [i-L_*-B_*,i-B_*], or outgoing tests indexed in [i+B_*,i+L_*+B_*], fails. Each of these at most 2(L_*+1) tests is independent of Z_i, because its index is more than R away. Applying independence to these deterministic possible witnesses, before any random choice of an extreme, gives

For example each remote-flag term is \mathbb E[Z_i^+F_iF_j]=\mathbb E[Z_i^+F_i]p\leq C_0p; the failed-test terms are analogous. Thus the expectation of an uncovered large cost is bounded directly, even if rare values carry a substantial part of the first moment.

Retained and discarded primary costs. The retained costs satisfy

Indeed on a covered slot the right side is at least -M+M=0; on an uncovered slot only the portion of Z_i above M can increase its clipped value. In particular discarding negative primary costs loses at most M times the covered density, whether or not those costs are correlated with nearby flags.

Replacement costs. By (15.15) and independence of the entire spare and primary fields,

We have bounded the success indicator by the event of a nearby primary flag; success itself need not be independent of V_i.

Accepted clusters have density at most p, so their two successful transitions contribute at most 2K_\delta p to the mean upper bound. Charge the remaining bound in (15.14) to the leftmost flag of its cluster. By stationarity, its mean per slot is bounded by \mathbb E[V_0O_0]; unused spare slots in that display are simply overcharged. Summing these bounds with (15.16)–(15.18) proves (15.12). Each charge and each reassignment involves a uniformly bounded number of slots. Hence all charges are integrable, and the expectation calculation requires only first moments.□

Proof. The structural constants C_{\rm tr},R_{\rm out},C_1 are uniform in sufficiently small s and the nearby rational directions. First choose an integer

Next fix rational 0<s<1/10 and \epsilon>0 so small that DC_1s<1/4 and D\epsilon<1/4. The geometry and the constants R,B_*,G_*,L_*,H_*,C_0,C_V have already been fixed uniformly. If C_0=0, the primary costs vanish almost surely and no replacements are necessary. Otherwise choose

then the corresponding transition cap K_\delta. Set

Choose the finite matched-cap range from (L) with this M_0,\eta, and then fix the geometric comparison tolerances and the rational directions so that it applies uniformly over that range. At every sufficiently large admissible n, choose its deterministic good cap. Equation (15.6) gives Mp\leq\eta, while p\leq C_0/M_0 by Markov’s inequality. In (15.12) the failed-test error, the terms A_*p, and the terms (H_*+1)\eta are each at most \epsilon/3. We obtain an actual stationary local road with

Regard D old slots as one new period and use space units Dn. The tracking error is then at most s/1000, the input radius at most 10, and the rule commutes with translation by DA in old units, that is by A in new units. Reindexing the private product field by this rescaling preserves its shift-invariant law: a new lattice shift corresponds to the old lattice shift multiplied by D. Thus it is in the original road class at scale Dn. The same construction in the other direction gives a pair with common mean bound

All the rational meshes have fixed denominators; multiplication of an admissible n by the integer D preserves admissibility.

Let b(n) be the infimum of the common mean bound over that road class. Invoke the uniform input estimates with the fixed enlarged prefactor 2C_s; this changes only the threshold for n. That threshold is therefore uniform over members with \beta\leq2C_s n^{1-c}. Since b(n)\leq C_s n^{1-c}, for each sufficiently small t>0 one can choose a pair with \beta\leq b(n)+t\leq2C_s n^{1-c}. Apply (15.21) and then let t\downarrow0; existence of a minimizing road is unnecessary. This proves

All directions, meshes, and tolerances are fixed in the stated order, and the choices give \rho=D(C_1s+\epsilon)<1/2. This proves the assertion for all admissible n above the common threshold.□

Proof of Theorem 1.1. Suppose \partial\Bcal contains a nontrivial segment, and fix the maximal supporting face with extreme endpoints a,b described in Section 8. The same admissible road class has the bounds

from Propositions 9.5 and 9.4.

Proposition 15.2 supplies fixed rational directions, s>0, an integer D\ge1, and \rho<1 such that b(Dn)\le\rho b(n) for every sufficiently large admissible n. Choose an admissible n_0 above the common threshold. Multiplication by D preserves admissibility, so iteration gives

a contradiction. Thus every supporting face is a single point.

To express this as strict convexity, take distinct x,y\in\partial\Bcal and 0<t<1. Convexity gives \mu((1-t)x+ty)\le1. Equality would provide a supporting functional \ell\le\mu at this point, and therefore

Both endpoint values would be one, so the entire segment [x,y] would lie in a nontrivial supporting face. This has been excluded. Therefore \mu((1-t)x+ty)<1 for the rate-one exponential law.

Theorem 1.2 with shape \kappa=1 gives Fréchet differentiability of this norm away from the origin and a C^1 unit sphere, with a unique supporting line at every point. To pass to any rate \lambda>0, couple the weights by \tau_e^{(\lambda)}=\lambda^{-1}\tau_e^{(1)}. Then every path cost, and hence every passage time, is multiplied by \lambda^{-1}. Thus

Multiplication by a positive scalar preserves differentiability, strict convexity, and the C^1 boundary property. This proves every assertion of the theorem.□

Proof. Damron–Hanson’s product-measure hypothesis A1' holds: the common law is continuous, and the minimum Y of the four weights incident to the origin is exponential of rate 4\lambda, whence \E Y^2=1/(8\lambda^2)<\infty. Their geometric assumptions (2.1)–(2.2) require differentiability at the chosen boundary point and at both endpoints of its supporting face. The preceding supporting-line calculation identifies that face with \{v_u\}. Thus their supporting sector is exactly the direction u. Theorems 1 and 2 of [10] give the convergence, coalescence, bigeodesic exclusion, and Busemann statements on one probability-one event for this fixed u. Their normalized supporting vector is \rho_u=\nabla\mu(u) by that calculation.

For completeness, uniqueness follows from the convergence assertion. If (x=x_0,x_1,\ldots) is any infinite geodesic of direction u, then its segment ending at x_n is \Gamma(x,x_n) by uniqueness of finite geodesics. These segments converge both to that infinite geodesic and to \Gamma_x^u, so the paths agree.□

Proof. Set p_n=a_n-z_n and q_n=b_n-z_n. Translation invariance gives \PP(z_n\in\Gamma(a_n,b_n))= \PP(0\in\Gamma(p_n,q_n)). If the asserted probabilities had positive limsup, the event that 0\in\Gamma(p_n,q_n) for infinitely many n would have positive probability. Intersect this event with \Omega_u from Corollary 17.1. Restrict to the infinite set of indices for which 0\in\Gamma(p_n,q_n). On the intersection, the arm from 0 to q_n converges locally to \Gamma_0^u. Both arms have lengths tending to infinity, since their endpoints tend to infinity. Local finiteness of the lattice permits a diagonal subsequence along which the other arm also converges to an infinite path. Every finite segment of the resulting two-sided path is a segment of one of the finite geodesics \Gamma(p_n,q_n), and hence is geodesic. The two arms have no vertex in common except 0, because each finite geodesic is simple. This constructs a bigeodesic with an end of direction u, contradicting Corollary 17.1. The first assertion uses a_n=0, b_n=nv, and z_n=\lfloor n/2\rfloor v.□

Proof. Theorem 1.2 gives a C^1 boundary. We verify that such a compact planar convex body has infinitely many extreme points. At a boundary point choose a supporting line. Its intersection with the body is a point or a compact segment, and its endpoints are extreme points of the body: any convex decomposition of an endpoint must stay in the supporting line, where extremality is immediate. Thus every boundary point is a convex combination of extreme points. Every interior point lies between two boundary points, so the whole body is the convex hull of its extreme points. If there were only finitely many, the body would be a polygon with nonempty interior. At a vertex of that polygon the two adjacent sides give distinct supporting lines, contradicting differentiability. Hence the number of extreme points exceeds forty.

The Gamma density is absolutely continuous, has no atom at zero, and has a finite exponential moment by Lemma 2.1. All the hypotheses of [13, Theorem 1.2] are therefore satisfied. Apply that theorem at rate one to obtain C_\kappa. The coupling t_e^{(\kappa,\lambda)}= t_e^{(\kappa,1)}/\lambda leaves every geodesic unchanged and proves the assertion for all positive rates with the same constant.□

Proof. The Gamma law is atomless for every \kappa>0, including \kappa<1. For four independent samples,

Thus the continuity and minimum-second-moment hypotheses of [1, Theorem 1] hold.

The infinite seed set causes no ambiguity in the arrival-time comparison. If I_1 is nonempty, choose one seed as a finite upper bound for T(I_1,z). Only finitely many seeds lie in the corresponding passage ball about z, so the infimum is attained. Simple paths from distinct seeds to z have different edge sets. Conditioning on all edge weights except one in their symmetric difference and using atomlessness rules out equal costs; countability gives this simultaneously for all such paths and sites. Thus the shared-metric competition is described by the displayed strict comparison defining \mathcal E_2.

Theorem 1.2 gives a C^1 boundary. For the fixed \theta, put h(\theta)=\max_{x\in\Bcal}x\cdot\theta>0. The supporting line

is perpendicular to the positive \theta-ray and meets that ray at h(\theta)\theta. At any contact point with \Bcal, the C^1 boundary makes L_\theta the unique supporting line, hence a tangent in the sense of [1]; the contact point need not lie on the ray. For a seed set I\subset\Z^2\setminus\{0\}, write \mathcal E_2(I)=\{z\in\Z^2:T(0,z)<T(I,z)\}, with the same empty-set convention for T(I,z). If I\subset J, then T(J,z)\le T(I,z) and hence \mathcal E_2(J)\subset\mathcal E_2(I). If \alpha<\pi/2, choose \alpha<\alpha'<\pi/2. The source’s cone of half-angle \alpha' contains the closed \alpha-cone, regardless of its boundary convention, so its Theorem 1 gives positive survival for the closed \alpha-cone as well. If \alpha\ge\pi/2, the closed \alpha-cone contains the closed critical half-plane. Theorem 5 gives almost-sure finite type-2 growth for that half-plane, and monotonicity gives the same conclusion for the larger cone. This proves the stated closed-cone convention throughout [0,\pi], without strict convexity, for every prescribed orientation.

When \kappa=1, the shared exponential edge field gives the equal-rate two-type Richardson model. The coupling \tau_e^{(\lambda)}=\tau_e^{(1)}/\lambda multiplies both arrival times by the same positive constant and leaves every comparison defining \mathcal E_2 unchanged. This proves the stated specialization.□