Uniform Interior \(C^{1,\alpha}\) Estimates for Infinity-Harmonic Functions

Proof. For each x_0\in\Omega, choose r>0 with \overline B_{2r}(x_0)\subset\Omega. Continuity bounds u on this closed ball, and y\mapsto u(x_0+2ry) is infinity-harmonic on B_1. Theorem 1.1 applies and gives the conclusion near x_0.□

Proof. Suppose that \phi touches U from above at (y_0,b_0). Restriction to the vertical line gives \phi_b(y_0,b_0)=1. Put p=\nabla_y\phi(y_0,b_0). If p\ne0, choose an affine function \ell with \ell(y_0)=b_0 and \nabla\ell\cdot p=1. The function

touches u from above at y_0. At that point \nabla\psi=p, because \phi_b=1, and

The viscosity inequality for u gives the required sign. If p=0, the vertical upper contact gives \phi_{bb}\ge0, and this is \Dinf\phi because \nabla\phi=(0,1). Lower tests are identical with the inequalities reversed.□

Proof. Fix \overline B_r(z)\subset D and L>L_r^+(w;z). Choose \gamma>0 so small that

On \partial B_r(z) the function w lies strictly below

It lies below this function throughout the ball. Otherwise the positive maximum of their difference is attained at an interior point different from z, and the displayed function, shifted by a constant, is an upper test there. Writing \rho=|Z-z|, its infinity Laplacian is

contrary to the viscosity inequality. Letting \gamma\downarrow0 and then L\downarrow L_r^+(w;z) gives

Applying the same argument to -w gives the lower bound with L_r^-(w;z). These bounds prove monotonicity in the radius.

For centers in a compact neighborhood, choose one small radius whose balls remain in D. The two slopes at that radius are uniformly bounded by the oscillation on a larger compact set divided by the radius. The preceding inequalities then give a common local Lipschitz bound. Continuity in the center at fixed radius follows from uniform continuity of w on a compact neighborhood, taking the maximum over the fixed set \{|h|\le r\}.□

Proof. The local Lipschitz bound and the exhaustion of \R^n by the rescaled domains give subsequential convergence on compact sets. By Lemma 2.2, the finite limits

exist. They are equal. Indeed, if h_r attains the positive maximum, then for each fixed small R>r, once r is sufficiently small,

First let r\downarrow0, using continuity at fixed R and |h_r|\le r, and then let R\downarrow0. This gives L_*^+\le L_*^-. Reversing signs gives equality; write the common value as L_*.

Let V be a limit. Uniform convergence on balls shows that

and that the slopes at every other center retain monotonicity in their radius. In particular |V(Y)|\le L_*|Y|. For any center q,

For a fixed radius at q, bound its slope by the slope at T and send T\to\infty. Thus every slope of V is at most L_*, so V is globally L_*-Lipschitz.

If L_*=0, then V=0. Otherwise let x_R and y_S attain the positive and negative extrema on the balls of radii R and S centered at zero. The equality of the origin slopes gives

Equality throughout forces |x_R|=R, |y_S|=S, and these points to lie on opposite rays of one line. Comparing all choices of R,S gives a unit vector e such that V(te)=L_*t for every real t. The Lipschitz bound then yields

Letting t\to+\infty in the lower bound and t\to-\infty in the upper bound gives V(Y)=L_*e\cdot Y.□

Proof. Put w_i=U_i-\ell_i and \varepsilon_i=\|w_i\|_{L^\infty(B_1)}. The slopes of \ell_i are bounded: evaluating at e/2 and -e/2 for every unit vector e gives |\nabla\ell_i\cdot e|\le\Lip(U_i)+2\varepsilon_i. Thus w_i have a common Lipschitz bound C.

Suppose that (2.1) fails. Passing to a subsequence and replacing U_i,\ell_i by their negatives if necessary, there are endpoint pairs in B_{1/4} with positive midpoint defect greater than \delta|x-y| for some fixed \delta>0. For triples x,y,z\in\overline B_{3/4} with x\ne y, write

Here 0<r\le3/2, so \nu r\le f(r)\le r with \nu=1-\sqrt{3/2}/8>0. Choose a smooth function B(c) that equals \delta/2 on \overline B_{1/4}, is at least \delta/2 everywhere, and equals a sufficiently large constant B_{\mathrm{out}} when |c|\ge1/2. Consider

Allowing the third point to move permits the use of Lemma 2.4, while the quadratic penalty keeps it close to the midpoint. The factor B(c) localizes the maximum, and the concavity of f will supply the dominant negative second variation. The violating pair with z=c makes this expression positive. On the other hand it is at most

Its upper limit as r\downarrow0 is therefore at most zero, uniformly in the other variables. A positive maximum is attained at some triple with r>0.

At these maximizing triples, r\to0 as i\to\infty: positivity and Bf\ge\delta\nu r/2 give r\le8\varepsilon_i/(\delta\nu). Comparing the maximizing value with the admissible choice z=c also gives

Choose B_{\mathrm{out}}>(C+\tfrac32C^2)/\nu. The preceding upper bound for (2.2) is then nonpositive when |c|\ge1/2. Hence |c|<1/2 at a positive maximum. Since r\to0 and h=O(r^2), the maximizing triples are interior to (B_{3/4})^3 for all large i.

At one such triple set q=\nabla\ell_i and e=(x-y)/r. Apply Lemma 2.4 to the functions U_i,U_i,-2U_i and the test

This test is C^2 near the triple. Define p_1=D_x\Psi, p_2=D_y\Psi, and p_0=-D_z\Psi/2. Direct differentiation gives

These vectors and a are bounded. Since f'(r)=1-3\sqrt r/16 and |h|=O(r^2), we also have a\ge\delta/4 for all sufficiently large i.

Let \sigma=(p_1,p_2,p_0). The third jet supplied by Lemma 2.4 has gradient -2p_0. The function -2U_i is infinity-harmonic, so the viscosity inequalities for the three superjets give

In particular X_3[p_0]\ge0. If A=D^2\Psi at the maximum, the matrix inequality evaluated at \sigma gives

The maximum, A, and \sigma are fixed while \eta\downarrow0, although the jets may vary. It follows that A[\sigma]\ge0.

We compute the same quadratic form by moving (x,y,z) along the straight line with velocity \sigma and parameter \tau, with \tau=0 at the maximum. The endpoint relative velocity p_1-p_2=2ae is parallel to x-y, and the third velocity differs from the mean endpoint velocity by O(r). Write c(\tau),h(\tau),r(\tau) for the induced quantities, and use dots for their derivatives at zero. The formulas above give

Since h=O(r^2),

Also f''(r)=-3/(32\sqrt r), and therefore

The affine part of \Psi has zero Hessian. Thus A[\sigma]\to-\infty, contradicting its nonnegativity and proving (2.1).□

Proof. The cylinder contains \overline B_{\sqrt\lambda r}(z). Evaluating the fit on this ball and using the Lipschitz bound gives the first inequality in (3.6). The exact increment of U in direction e_b equals the increment length, which gives the second.

Suppose that (3.5) fails. There are s_i\to0, functions U_i\in\mathcal U, centers z_i\in G, radii 0<r_i\le s_iD(z_i), and parameters \lambda_i that are not good at these triples, with \lambda_i\ge\delta>0. Passing to a subsequence, the common bounds give

The limit U_\circ is infinity-harmonic on B_{7/8}^n by the usual viscosity stability argument [8, Section 6]: add a positive fourth-order term to an upper test to make its contact strict, take nearby interior maxima for the approximating functions minus that test, and pass their inequalities to the limit. The added term leaves the first two derivatives at contact unchanged. A negative fourth-order term and minima treat lower tests. This argument also applies to the varying continuous jet operators below, since they converge uniformly on compact jet sets. Choose scales R_i\downarrow0 so slowly that

Such a decreasing sequence can be obtained after passage to a subsequence, since all three terms tend to zero. The point z_* is interior to B_{7/8}^n. Proposition 2.3, applied to the one function U_\circ at z_*, gives a further subsequence on which its R_i blow-ups converge locally uniformly to a linear function. By (3.7), the functions

converge to that same linear function uniformly on B_1^n. They are infinity-harmonic and uniformly Lipschitz there. Proposition 2.5 therefore gives numbers \varepsilon_i\to0 such that

Return now to the smaller radii r_i. The functions

are uniformly bounded and Lipschitz on \overline B_2^n, so a further subsequence converges uniformly there to W. The slow-scale choice ensures that z_i+r_i\overline B_2^n\subset B_{R_i/4}(z_*) for all large i. Substituting

into (3.8) shows that the midpoint defect of W_i is at most \varepsilon_i|H-K|. Thus W satisfies the midpoint identity on \overline B_2^n. Continuity and dyadic subdivision on segments make W affine on this ball. Since W(0)=0 and W(e_b)=1, it has the form W(H)=P\cdot H with P\cdot e_b=1.

Uniform convergence now gives, for all large i,

The nonzero slope P therefore gives a fit with error at most \lambda_i r_i on Z_{\lambda_i}(z_i,r_i;P)\subset B_{2r_i}(z_i). This contradicts the choice of \lambda_i and proves (3.5).□

Proof. Choose any sequence \gamma_i\downarrow0. The function b(s)s^{-\gamma_i} is bounded on intervals separated from zero and is unbounded near zero. Taking a factor-two approximate maximum on a sufficiently long truncated interval [h_i,1/8] gives a point s_i such that s_i\to0 and

For example, make the approximate maximum larger than the supremum on [1/i,1/8] to ensure s_i<1/i. Pass to a subsequence to make s_i decreasing. We have b(s_i)>0. Choose U_i\in\mathcal U, z_i\in G, and r_i\le s_iD(z_i) with \beta_{U_i}(z_i,r_i)>b(s_i)/2, and then choose a parameter \lambda_i>\beta_{U_i}(z_i,r_i)/2 that is not good for U_i at this center and radius. This uses only approximate suprema and gives

In particular \lambda_i\to0; pass to a further subsequence if needed to make it decreasing.

Fix M\ge2. For large i, we have Mr_i\le Ms_iD(z_i)<D(z_i), so every z with |z-z_i|\le Mr_i belongs to G. The distance to the boundary then satisfies

For r\le Mr_i we have r/D(z)\le2Ms_i\le1/8, so the center and radius are admissible. By (3.10) and (3.11), uniformly over all these choices,

for all sufficiently large i. Here (2M)^{\gamma_i}\to1 for each fixed M, so the same \Lambda=32 works for every fixed M; only the threshold in i depends on M. The parameter \Lambda\lambda_i is strictly greater than every bad parameter and is therefore good.□

Proof. In the common-center case both cylinders contain B_{\theta H}(z). Evaluation of the difference of the fits on this ball gives |P-P'|\le3\theta. Normalization of the vectors, using the lower norm bound, gives |n-n'|\le C\theta. Choose a fixed small c>0. The point z+cHn belongs to both cylinders: relative to n' its transverse displacement is at most cH|n-n'|\le Cc\theta H, and its axial displacement is at most cH. Taking c sufficiently small, the fit errors at that point give

Since 1-n'\cdot n=|n-n'|^2/2,

For displaced centers, a ball B_{c\theta H}(z') with fixed small c belongs to both cylinders by the quarter-radius margins. Subtracting the first fit’s value at z' makes it an estimate for increments from z' with error 2\theta^2H; the second fit has error \theta^2H. Evaluation on the common ball again gives the vector and direction bounds. The segment z'+s n' for |s|\le c'H lies in both cylinders when c'>0 is sufficiently small: the first cylinder’s initial offsets use a quarter of its radii, and the additional transverse offset is at most c'H|n-n'|. Evaluation on this segment and the same norm calculation give the last bound in (3.12).□

Proof. Set \theta_i=\sqrt{\Lambda\lambda_i}. A good fit of radius Rr_i has error \theta_i^2Rr_i and transverse radius \theta_i Rr_i. Its slope obeys uniform upper and lower norm bounds by (3.6), once i is large enough.

For each fixed R\ge2, Lemma 3.4, applied with any fixed M>R, provides fits at z_i through successive doublings from r_i, followed if necessary by a final step to Rr_i of ratio at most two, for all sufficiently large i. Comparing them by Lemma 3.5 gives, for a slope P_R at the latter radius and its unit vector n_R,

Fix a box |x|,|t|\le K in the coordinates of (3.14) and a radius factor S>0. Choose R\ge2 so large that R>16K and

Such a choice is possible because R/\log R\to\infty. Next fix M>\max\{2,2K,R,S\}. The physical centers in the box are within 2Kr_i of z_i, and all radii in the comparisons below are at most \max\{R,S\}r_i. Lemma 3.4 therefore supplies these fits uniformly once i is sufficiently large. The physical image of the box then lies inside the cylinder at z_i for P_R, with axial and transverse offsets at most one quarter of the respective radii. For the axial offset use |\sqrt{\lambda_i}x|+|t|\le2K. For the transverse offset, (3.16) gives the bound

in units of r_i.

Compare a fit centered at any point of this box, first at radius Rr_i with the enclosing fit, and then through successive doublings or halvings, with a final ratio at most two, to radius Sr_i. The number of comparisons depends on K,S but not on i. If P=(P_x,P_t) is any such slope in the coordinates of (3.14), Lemma 3.5 gives

Since P_i/c_i is the last coordinate vector,

It follows that the normalized coefficients

In particular, using the enclosing fit at z_i in (3.14) gives on the box

Thus v_i are uniformly bounded on each fixed box.

We next express the domain of a fit in these coordinates. Suppress the index i temporarily, write \lambda=\lambda_i, and set

For fixed K,S, (3.17) gives A>0, N/A=1+O_{K,S}(\lambda), and |n_x|=O_{K,S}(\sqrt\lambda) for large i. The physical axis has ratio \sqrt\lambda\,p_i'/A of transverse to axial displacement, which becomes the finite shear p_i'/A after the transverse stretch. Figure 1 illustrates this change of geometry. The next calculation identifies a sheared box contained in the cylinder’s image in normalized coordinates, with a fixed margin at the cylinder’s axial ends. The exact decomposition

shows that the mapped increment lies in the radius-Sr_i cylinder whenever

for all large i. Indeed the residual in (3.18) has transverse projection at most \sqrt\lambda S\le\theta_i S, and its axial projection is O_{K,S}(\lambda S). The first term has axial length at most (1+O_{K,S}(\lambda))S/2, leaving a strict axial margin. Dividing (3.3) by r_i c_i\lambda_i therefore gives

on (3.19). The coefficient of S is independent of K and S.

This proves equicontinuity on compact sets. Given an accuracy \varepsilon>0, first fix S>0 so small that 2\Lambda S<\varepsilon/3. For a fixed box, (3.17) now bounds its coefficients by some finite C_{K,S}. Choose the increments sufficiently small, depending on this bound and S, that (3.19) holds and the affine increment in (3.20) is less than \varepsilon/3. For all sufficiently large i, (3.13) and (3.20) then bound the value increment by 2\varepsilon/3. Discard an initial segment so that all domains contain the compact box. The threshold in i may depend on \varepsilon; the finitely many remaining preceding continuous functions can be included by decreasing the increment size. The local bounds and domain exhaustion now give a locally uniformly convergent subsequence, with limit v and v(0,0)=0.

The jet operators X[(p_x,1+\lambda_i p_t)] converge locally uniformly to X[(p_x,1)]. Viscosity stability gives (3.1); equivalently, strictify a smooth upper or lower test for v, choose nearby touching points for v_i, and pass to the limit in (3.15). To pass the fits, fix a center and a target radius R>0, and take S=4R in (3.20). After a further subsequence its bounded coefficients converge to p,q. The shears p_i'/(1+\lambda_i q_i') converge to p. Every increment with |\ell|\le R and |h-p\ell|\le R then satisfies (3.19), uniformly, for all sufficiently large i. Passing to the limit in (3.20) yields (3.2) with

The additional subsequence may depend on the chosen center and radius: the limit function v stays fixed, and (3.2) asserts existence separately for each choice.

Finally suppose that v is affine. Since v(0,0)=0, write v(x,t)=p\cdot x+qt. In the physical coordinates consider

These slopes are nonzero for large i. The cylinders Z_{\lambda_i}(z_i,r_i;\widetilde P_i) lie in one fixed box of the rescaled coordinates. In fact, if

a point in such a cylinder, in units of r_i, is s(\sqrt{\lambda_i}p,1+\lambda_i q)/N_i+\sqrt{\lambda_i}\xi with |s|\le1, |\xi|\le1, and \xi perpendicular to the axis. Its rescaled coordinates satisfy |x|\le |p|/N_i+1 and |t|\le1+\sqrt{\lambda_i}. On this fixed box the residual for the proposed fit is exactly

uniformly, by local uniform convergence and the upper bound for c_i. It is eventually at most \lambda_i r_i on the whole cylinder. This makes \lambda_i good for U_i at (z_i,r_i), contrary to Lemma 3.4. Hence v is nonaffine.□

Proof of Proposition 3.3. If no power bound held, Lemma 3.4 and Proposition 3.6 would produce a nonaffine entire solution satisfying all hypotheses of Theorem 3.1, with m=d and C_0=256. The theorem makes that solution affine, a contradiction. Hence a power bound holds. Reducing its exponent if necessary gives \gamma\le1. Since the family \mathcal U and the function b were fixed solely by d, the constants in (3.9) depend only on d.□

Proof of Theorem 1.1, assuming Theorem 3.1. Take B,\gamma,s_0 from Proposition 3.3. At every center z\in\overline B_{1/2}^n one has D(z)\ge d_0:=1/4. Choose A_0>Bd_0^{-\gamma}. There is r_0>0, depending only on d, such that for all 0<r\le r_0,

For every U\in\mathcal U and each such center and radius,

Thus \lambda=A_0r^\gamma is good, with the same constants for all functions and centers.

Fix U\in\mathcal U. Set

and at each center z\in\overline B_{1/2}^n choose a fitted slope P_j(z) for the parameter A_0r_j^\gamma. The fit holds on the inscribed ball \overline B_{\rho_j}(z), with error at most A_0r_j^{1+\gamma}. Comparing consecutive fits on their common ball \overline B_{\rho_{j+1}}(z) gives

The geometric tail converges uniformly in U and z. Consequently there is a vector P_\infty(z) with

If 0<|h|\le\rho_0, choose j so that \rho_{j+1}<|h|\le\rho_j. The fit at radius r_j and (3.21) give

where

The final inequality follows from \rho_{j+1}=\sqrt{A_0}\,2^{-(1+\gamma/2)}r_j^{1+\gamma/2} and (1+\gamma/2)\alpha_d=\gamma/2. Thus U is differentiable at every such z, with gradient P_\infty(z), and

All increments here remain in G, since each lies in one of the admissible inscribed balls. The original Lipschitz bound gives |\nabla U(z)|\le L.

To compare gradients, take z,z'\in\overline B_{1/2}^n with \ell=|z-z'|\le\rho_0/2 and \ell>0, and fix any unit vector e\in\R^n. The evaluation point z'+\ell e need not lie in B_{1/2}^n; the Taylor estimates centered at z and z' still apply. Applying (3.23) at z' to the increment \ell e gives

Applying the same estimate at z to the increments z'-z and z'+\ell e-z, whose lengths are at most 2\ell, gives

The constants are independent of e,U,z,z'. Subtracting and taking the supremum over unit vectors proves

For larger distances the same estimate follows, after increasing C, from |\nabla U|\le L. Restriction to points (y,0) therefore gives the claimed uniform gradient norm and Hölder seminorm for every normalized u.

Finally, if u is as in Theorem 1.1 and its oscillation is M>0, then (u-u(0))/M is a normalized solution: positive scaling and addition of constants preserve the viscosity equation by transformation of tests. Multiplying the normalized estimates by M gives the theorem. If M=0, the function is constant and the conclusion is immediate.□

Proof. Set

For smooth functions, at corresponding points, \nabla_y\bar v=\nabla_xv-p_0 and D^2\bar v=R B^{\mathsf T}D^2vB. Since B(\nabla_y\bar v,1)=(\nabla_xv,1), this gives F[\bar v]=R F[v]. Pulling back an upper or lower smooth test gives the same implication in the viscosity sense. For the fits, an increment (h,\ell) in the new variables corresponds to R(h+p_0\ell,\ell). The old affine increment, divided by R and with the affine part of (4.1) subtracted, is

The old box conditions become |\ell|\le\rho and |h-(p-p_0)\ell|\le\rho, and the error is C_0\rho.□

Proof. We prove the estimates for centering at the origin. Translation by one unit in time gives the other case after enlarging the boxes by a fixed factor.

First compare two fits (p,a),(p',a') at the same center, with radii R and R'\in[R,2R]. On the horizontal slice \ell=0, both estimates hold for |h|\le R, so |p-p'|\le 3C_0\le4C_0. Write

With c=(2+8C_0)^{-1}, the point (p cR,cR) lies in both sheared boxes. At this point

The two fitting errors sum to at most 3C_0R. It follows that |a-a'|\le 3C_0/c+8C_0^2. Thus the differences of both parameters are bounded by one constant whenever the radii differ by at most a factor of two. Comparing along successive dyadic radii with the centered fit of radius 1 gives, for every fit at the origin of radius r,

Fix K\ge2. Choose once and for all a sufficiently large constant A=A(m,C_0) and set R=A K L(K). If (p_R,a_R) is any fit of radius R at the origin, then \mathcal B_K lies in its sheared box with

Indeed, (4.7) bounds the latter offset by K(1+C(1+\log R)), while \log R\le\log A+2L(K). Since L(K)\ge1, the asserted two inequalities hold for all K\ge2 once A is chosen larger than a fixed multiple of 1+\log A. Such a choice is possible because \log A/A\to0. The fit at radius R now gives

which is (4.2).

Let z=(x,t)\in\mathcal B_K, and compare a fit (p_z,a_z) of radius R at z with the enclosing fit (p_R,a_R). For horizontal increments of length at most R/2 from z, the enclosing estimate can be subtracted at the two endpoints, with error at most 2C_0R. Comparison with the fit at z therefore gives |p_z-p_R|\le6C_0. Set c'=(4(1+6C_0))^{-1}. The increments (p_z c'R,c'R) from z remain in both boxes: in the enclosing box the transverse offset is at most R/4+6C_0c'R\le R/2, and the time offset is at most R/2. The same affine-increment identity used above gives a bound on |a_z-a_R| depending only on C_0. Hence |p_z|+|a_z|\le C L(K). Dyadic comparisons at z, from radius R to an arbitrary radius r, give

as claimed. These comparisons apply to every choice of fit.

Using a fit of radius r on the horizontal slice proves (4.4). For a time increment \tau with 0<|\tau|\le1/2, put B=L(K)+|\log|\tau|| and choose the fit radius r=A'|\tau|B. Here A' is a constant to be fixed, independent of K and \tau. Since B\ge1,

The fit-parameter bound consequently gives |p|\le C(3+\log A')B. Choose A'\ge1 sufficiently large that A'\ge C(3+\log A'). This choice is possible and is made before K and \tau vary. Then |\tau|\le r and |p\tau|\le r, so the vertical increment lies in the fitted box. In that estimate, the affine time coefficient and error are bounded by

This proves (4.5).

The two moduli, applied successively to a spatial and a time increment in a slightly larger box, give equicontinuity on compact sets. The growth bound gives local uniform boundedness. Arzelà–Ascoli and a diagonal choice of subsequence therefore give local uniform convergence. To pass an upper viscosity test to the limit, add a positive fourth-order term at its contact, preserving its first and second derivatives and making the contact strict on a small closed ball. Maxima of the sequence minus this modified test on that ball converge to the contact, so they are interior for large indices. The viscosity inequalities there pass to the limit by continuity of F. Subtracting the fourth-order term and using minima treats a lower test. To pass the fits, fix one center and one radius r. By (4.3), pass to a further subsequence of the corresponding fits with (p_j,a_j)\to(p,a). For a point with |\ell|\le r and |h-p\ell|\le r, use in the jth fitted box the increment h_j=p_j\ell+(h-p\ell). It converges to h and belongs to that closed box. Local uniform convergence gives the limiting inequality. This subsequence may depend on the fixed center and radius, which is sufficient for existence of a fit at each such pair. The centered fit passes directly to the limit. Finally, for t in a fixed compact interval, (4.2) gives a bound of the form C_J(1+|x|)(1+\log(2+|x|))^2, uniformly in the centered class. Dividing by |x|^2 proves (4.6).□

Proof. Fix a compact range of centers and of allowed positive lengths. Uniform subquadratic growth and the quadratic cost force all maximizers and minimizers into a common bounded spatial set. They exist there, and the maximum and minimum of a continuous function over this fixed compact set depend continuously on the parameters. The same bounded-set conclusion holds when the length varies in (0,\ell_0]. On that set the values of f are bounded. Comparison with the competitor y=x then gives |y-x|^2\le C\ell for an optimizer as \ell\downarrow0. Uniform continuity of f on the compact set gives convergence to f(x,t) locally uniformly, as well as y\to x locally uniformly. This proves continuity at zero. The composition identities follow by optimizing first over the intermediate point and using

For the convexity assertion, let g(\ell)=\Penv_\ell f(x(\ell),t), and let a smooth function \psi touch g from above at \ell_0>0. Choose an optimizer y_0 at that length. After a constant adjustment,

touches (y,\ell)\mapsto f(y,t+\ell) from above at (y_0,\ell_0). Put q=(y_0-x(\ell_0))/\ell_0, which is the horizontal gradient of \Psi there. Along the direction (q,1), the Hessian of the quadratic term has value |d|^2/\ell_0. For example, writing r=y-x(\ell), its value along a direction with r-velocity q-d is \ell_0^{-1}|q-d-r/\ell_0|^2=|d|^2/\ell_0. The viscosity inequality gives \psi''(\ell_0)+|d|^2/\ell_0\ge0. Thus every upper test for g(\ell)+|d|^2\ell\log\ell has nonnegative second derivative at positive lengths. This is the one-dimensional viscosity criterion for convexity. To see it directly, a failure of convexity on [a,b] lets one place a strictly concave quadratic above the chord but below the function somewhere; the positive maximum of their difference yields an upper test with negative second derivative. Continuity at zero extends the convexity to that endpoint.

Finally, the map \widetilde f(x,t)=-f(x,-t) exchanges subsolutions and supersolutions of (3.1), as pullback reverses the sign of F. It also gives \Penv_\ell\widetilde f(x,t)=-\Menv_\ell f(x,-t). The assertion for \Menv follows.

For the last convergence assertion, fix a compact range of allowed centers and positive lengths. The common subquadratic bound and the local uniform bounds confine every optimizer for f_j and for f to one compact spatial set, by the first argument of the proof. On that set the functions inside the supremum and infimum converge uniformly. Their extrema therefore converge uniformly on the chosen parameter range.□

Proof. Convexity of \Penv_\ell f(z), with value f(z) at zero, gives the first monotonicity. Let \widetilde z_\ell=(y_\ell,t-\ell) be a backward optimizing point for \Menv_\ell f(z). Using z as a forward competitor from \widetilde z_\ell gives

The optimizer localization at length zero in Lemma 4.3 shows \widetilde z_\ell\to z. Fix a small h>0 whose forward time argument is allowed near z. For all sufficiently small \ell<h, monotonicity bounds the right side by \Senv_h^+f(\widetilde z_\ell). Continuity for fixed h gives an upper limit at most \Senv_h^+f(z). Now let h\downarrow0 to obtain (4.12). This order of limits also covers the value -\infty.

Apply this conclusion to \widetilde f(x,t)=-f(x,-t) to obtain the supersolution statements. For a solution both monotone limits exist and the two inequalities identify them. Each is bounded above by the finite slope at any fixed positive allowed length, proving the range in (4.13). Locally in z, choose a common small range of lengths. The common limit is the infimum of the continuous functions \Senv_\ell^+f(z) over that range, so it is upper semicontinuous.□

Proof. By Lemma 4.4, it suffices to bound the two slopes at length K. Suppose an optimizing displacement has length BK with B\ge2. Comparing its value with the competitor having no displacement, and using (4.2) in a box of size (B+2)K, gives

This estimate holds for the supremum and for the infimum. Consequently B\le C(L(K)+\log B)^2\le2C L(K)^2+2C(\log B)^2. Choose a fixed B_0 so large that 2C(\log B)^2\le B/2 for B\ge B_0. The last inequality then yields B\le4C L(K)^2 for B\ge B_0, and the same type of bound holds when B<B_0. This proves the displacement claim. The optimizer and its time argument now lie in a box of size C K L(K)^2. Applying (4.2) there and observing that L(C K L(K)^2)\le C L(K) bounds the absolute values of v there by C K L(K)^4. Dropping the nonpositive cost in the expressions for the two slopes gives (4.14).□

Proof. For fixed (t,s) put g(y)=f(y,t)-|y|^2/(2s) and

The expression on the left of (5.4) is the supremum of \sum_i\theta_i g(y_i) over configurations of mean x, because \sum_i\theta_i|y_i-x|^2=\sum_i\theta_i|y_i|^2-|x|^2. It is therefore at most \inf_A\Gamma(A;x,t,s).

Fix a compact range of (x,t,s) with s>0. For A\ne0, use y=x-A/|A| in the supremum to get \Gamma(A;x,t,s)\ge |A|-C on this range. The value at A=0 is uniformly bounded above, since the negative quadratic part of g dominates the subquadratic growth of f. Thus every minimizing sequence whose values are at most \Gamma(0)+1 lies in a fixed bounded set of slopes. For slopes in that set, the quantity g(y)-A\cdot(y-x) tends uniformly to -\infty as |y|\to\infty. Its maximizing positions therefore lie in one compact set. It follows that \Gamma is continuous on the relevant set, the minimum and maxima are attained, and their value depends continuously on (x,t,s).

Let A_0 be a minimizing slope, and let E be its nonempty compact set of maximizing positions. Then x belongs to the convex hull of E. Otherwise strict separation gives a vector e and \delta>0 with e\cdot(y-x)\ge\delta for all y\in E. If y_h maximizes \Gamma(A_0+he;x,t,s) for h>0, the uniform compactness of maximizers and continuity show that every limit of y_h as h\downarrow0 belongs to E. For small h, e\cdot(y_h-x)\ge\delta/2, and hence

contradicting minimality. The finite-dimensional convex hull theorem now represents x as a convex combination of finitely many points of E. Discard any zero weights. Their weighted values of g equal \Gamma(A_0;x,t,s), proving the reverse inequality in (5.4) and finite attainment in (5.1).

The affine majorant g(y)\le\Gamma(A_0;x,t,s)+A_0\cdot(y-x) can also be integrated against any probability measure of mean x and finite second moment. All terms are integrable by the growth hypothesis. Thus such measures cannot increase the value. The right side of (5.4) is an infimum of affine functions of x, so it is concave in x. This is (5.5).

For the convergence assertion, on each compact range of (x,t,s) the same preceding bounds on slopes and maximizing positions hold uniformly for f_j: local uniform convergence gives common bounds near the compact range, and the common subquadratic control handles large y. Both extrema in (5.4) can therefore be restricted to fixed compact sets. On these sets the expressions converge uniformly, proving the asserted local uniform convergence.□

Proof. Replace \Phi by \Phi^\sharp=\Phi+|Y-Y_0|^4+|t-t_0|^4. This makes the local maximum strict while preserving the derivatives through order two at the contact. Choose a small closed product neighborhood of ((y_i^0,t_0))_{i=1}^N, contained in the domain, such that the diagonal points (Y,t) lie in this strict neighborhood. For independent time variables T=(t_1,\ldots,t_N), put

Maximize \sum_i\theta_i v(y_i,t_i)-\Psi_\lambda(Y,T) on the closed product. Comparison with the original diagonal point and boundedness on this product show \lambda\chi(T_\lambda)\le C at any maximum. Thus all time deviations tend to zero as \lambda\to\infty. Every limit of such maximum points is diagonal; dropping the nonnegative penalty in the maximum inequality shows that its diagonal limit is a maximum of V-\Phi^\sharp. Strictness forces this limit to be (Y_0,t_0). Hence all the maximum points converge to the original point and are interior for sufficiently large \lambda.

Fix one such \lambda and one maximum. Set

where A_\lambda is written in the separate variables ((y_i,t_i))_i. Apply Lemma 2.4 to the functions \theta_i v(y_i,t_i) and the test \Psi_\lambda. For every \eta>0 it supplies closed superjets with Hessians X_i and first derivatives D_{(y_i,t_i)}\Psi_\lambda, with \operatorname{diag}(X_i)\le A_\lambda+\eta A_\lambda^2. Dividing each jet by the positive number \theta_i gives a closed superjet of v with horizontal gradient q_i. The viscosity inequalities therefore give X_i[(q_i,1)]\ge0. Evaluating the matrix inequality on \zeta_\lambda yields

First let \eta\downarrow0, keeping \lambda and its maximum fixed. The quantities A_\lambda and \zeta_\lambda do not depend on \eta, so A_\lambda[\zeta_\lambda]\ge0. The time component of \zeta_\lambda is (1,\ldots,1), which is in the kernel of D^2\chi. The linear map (Y,T)\mapsto(Y,\bar t) sends \zeta_\lambda to ((q_i)_i,1). We conclude that

Now let \lambda\to\infty. The maximum points converge to the contact, and the horizontal derivatives converge there. The derivatives of \Phi^\sharp at that point are those of \Phi, giving (5.6). The time components of the first derivatives of \Psi_\lambda need no bound: they do not enter the direction (\nabla_xv,1) in the individual viscosity inequalities.□

Proof. Continuity follows from Lemma 5.1. Fix c\in\R, and let \phi touch (x,t)\mapsto\Cenv_t f(x,t)-ct from above at (x_0,t_0) with t_0>0. Choose an attaining finite configuration (\theta_i,y_i^0) for \Cenv_{t_0}f(x_0,t_0), and keep its positive weights fixed. For variable Y=(y_i)_i let X(Y)=\sum_i\theta_i y_i and define

The defining supremum for \Cenv_t shows that \sum_i\theta_i f(y_i,t)-\Phi(Y,t) has a local maximum at (Y_0,t_0): for nearby (Y,t) it is at most \Cenv_t f(X(Y),t)-ct-\phi(X(Y),t), and equality holds at the chosen configuration. Set p=\nabla_x\phi(x_0,t_0). The individual horizontal directions in Lemma 5.2 are

Their weighted mean is p. Along the straight motion with velocities y_i'=q_i and t'=1, the mean has velocity p and each offset y_i-X has velocity (y_i^0-x_0)/t_0. Consequently, along this motion the variance term is

which is affine in h. The term ct is also affine, and the Hessian of the remaining term along this motion is D^2\phi[(p,1)]. Lemma 5.2, applied to the subsolution inequality for f, therefore gives F[\phi](x_0,t_0)\ge0. This proves the claim for every c, including c=0.□

Proof. A positive finite horizontal defect. Start with a nonaffine member v^{(0)} and center it at the origin using Lemma 4.1. Some horizontal slice of v^{(0)} is nonaffine. To verify this, suppose instead that v^{(0)}(x,t)=p(t)\cdot x+c(t) for every t. The coefficients p,c are continuous, as follows by evaluating the function at 0 and at the coordinate vectors. Direct optimization gives

For every fixed x,t this is convex in \ell. In each convexity inequality, the coefficient of x must vanish, since the inequality holds for all x\in\R^m. Hence p satisfies the affine interpolation identity on every interval, and so is affine in t. On the other hand, (4.2), applied at x=\pm Ke and |t|\le K, implies |p(t)|\le C L(K)^2. An affine function of t with this bound as K\to\infty is constant. For constant p, the displayed forward formula and its backward counterpart

show that c is both convex and concave. It is therefore affine, contradicting the choice of v^{(0)}.

A continuous nonaffine spatial slice has a nonzero midpoint Jensen defect, since the midpoint identity everywhere would make that slice affine. The transformation f(x,t)\mapsto-f(x,-t) preserves \Aclass_{m,C_0}: it exchanges the two viscosity inequalities, and it sends a fit (p,a) to (-p,a). It also changes the sign of a horizontal Jensen defect. Apply this transformation if needed, so that v^{(0)} has a positive defect. Define

Then \kappa>0. It is finite as well. For any configuration in this supremum, choose a fit at its mean (x,t) of radius r and use (4.1) with these data. At time zero the new positions are (y_i-x)/r, with mean zero and second moment one. The defect divided by r is unchanged, since the affine terms cancel in the weighted difference. The new function is centered. Its growth bound implies |\bar v(y,0)|\le C(1+|y|^2), with one constant for all these normalizations. Integrating this inequality over a configuration of second moment one bounds the quotient by a fixed constant. Thus 0<\kappa<\infty. More generally, (4.1) preserves every such quotient, at every time: spatial differences and their root mean square radius both scale by R, and the subtracted affine terms cancel.

Attainment after compactness. Choose configurations whose quotients in (5.11) tend to \kappa, and for each one make the normalization just described. Denote the resulting functions by v_j and the normalized finite probability measures by \mu_j. Then

Every v_j is centered at the origin, and every finite configuration for it satisfies the Jensen bound with constant \kappa. First use Lemma 4.2 to pass to a locally uniform limit \widehat v\in\Aclass_{m,C_0}, centered at the origin. The Jensen bound passes to this limit for each fixed finite configuration.

The moment bound gives \mu_j(\{|y|>M\})\le M^{-2}, so the measures are tight. After a further subsequence they converge weakly to a probability measure \mu. Lower semicontinuity of the second moment gives \int|y|^2\,d\mu\le1. The mean remains zero: the bound \int_{|y|>M}|y|\,d\mu_j\le M^{-1} makes the first moments uniformly integrable, and the same bound holds for the limit. To retain the extremal defect, we also need convergence of the value integrals despite the varying functions and measures. For a compactly supported continuous cutoff \chi_M, local uniform convergence and weak convergence give

Choose \chi_M=1 on \{|y|\le M\} and zero outside \{|y|<2M\}. By (4.6),

uniformly in j; the analogous tail bound holds for \widehat v and \mu. Letting first j\to\infty and then M\to\infty proves

Apply (5.3) to \widehat v. By Lemma 5.1, the measure in (5.12) is admissible for the value of \Cenv_{1/\kappa}\widehat v(0,0). It gives

This is equality by (5.3). In particular, loss of second moment in the weak limit causes no loss of contact; the equality also forces \int|y|^2\,d\mu=1.

Set R=1/\kappa, and choose a fit (p_0,a_0) for \widehat v at (0,0) of radius R. Define the final function by the single change of frame and time translation

It is centered at (0,1) and still has the global Jensen bound with constant \kappa. Under the change of frame (4.1), at corresponding points Z and (y,s) the envelope gaps obey

This follows by scaling the spatial variance by R^2 in (5.1); the affine terms again cancel. The preceding equality at parameter 1/\kappa=R therefore becomes

The subsolution below v. Define W by (5.7). Equations (5.3) and (5.13) give (5.8). At any contact, (5.7) also gives (5.9). Lemma 5.1 gives continuity and the stated spatial semiconcavity. The sandwich and (4.6) give subquadratic spatial growth on every compact positive-time interval. Since v is a solution with the growth in (4.6), Corollary 5.3, with f=v and c=\kappa^2/2, gives the subsolution property of W.

A finite slope at the contact. Lemma 4.4 already bounds H_v above at every point. We first bound it below at the finitely many points supporting the extremal configuration, using one common forward paraboloid, and then transfer this bound to their barycenter through equality in the Jensen bound. Choose a finite attaining configuration in (5.13), with mean x=0, radius r, and positive weights. If D=\sum_i\theta_i v(y_i,1)-v(0,1), its attainment and the Jensen bound imply

All inequalities are equalities, so

Let A be a minimizing slope in (5.4) for (x,t,s)=(0,1,1). The weighted attaining value equals its affine majorant. Since all weights are positive, each y_i must attain that majorant. The identity

therefore shows that all y_i are optimizers for \Penv_1v(b,0) with b=-A.

Put h(s)=\Penv_s v(b,0) for 0\le s\le1. It is finite and convex by Lemma 4.3. For 0<\ell<1, use the point y_{i,\ell}^-=b+(1-\ell)(y_i-b) at time 1-\ell as a competitor for \Menv_\ell v(y_i,1). Its value is bounded above using \Penv_{1-\ell}v(b,0). The two quadratic costs on this linear path add, giving explicitly

It follows that

The last inequality is the chord inequality for the convex function h. Taking the slope limit shows H_v(y_i,1)\ge h(1)-h(0)>-\infty. These finitely many values are also finite above by Lemma 4.4.

For each \ell>0, choose an optimizer \widehat y_i for \Penv_\ell v(y_i,1) and let \widehat x=\sum_i\theta_i\widehat y_i. Write \xi_i=y_i-x, \widehat\xi_i=\widehat y_i-\widehat x, and r'=(\sum_i\theta_i|\widehat\xi_i|^2)^{1/2}. The exact orthogonal decomposition and the reverse triangle inequality in the weighted Euclidean space give

Use the Jensen bound at time 1+\ell, (5.14) at time 1, and \widehat x as a forward competitor from x. The decomposition yields

Divide by \ell and take \ell\downarrow0. There are only finitely many endpoints, all with finite limits, so

Its upper finiteness is part of Lemma 4.4; hence H_v(z_*)\in\R.

It remains to identify the forward slope of W. At the contact, the sandwich gives for every \ell>0 \Senv_\ell^+W(z_*)\le\Senv_\ell^+v(z_*). For 0<\ell<1 it also gives \Senv_\ell^-W(z_*)\ge\Senv_\ell^-v(z_*), since the backward time is positive. The subsolution property and growth of W allow the one-sided inequality (4.12) to be applied to it. Consequently

This proves (5.10).□

Proof. Put s=t+\ell. The function f(\cdot,s)-|\cdot|^2/(2s) is concave, because the same is true of \Cenv_s v(\cdot,s)-|\cdot|^2/(2s). Thus

is strictly concave: after subtracting the concave function f(y,s)-|y|^2/(2s), its quadratic part has Hessian (s^{-1}-\ell^{-1})I=-t(\ell s)^{-1}I. It has a maximizer by the subquadratic growth and the quadratic cost, and strict concavity makes the maximizer unique. Keeping this maximizer fixed when x is replaced by x+h gives

For the upper bound, the identity

shows that

is concave in x. Indeed, the expression in braces is jointly concave in (x,y); applying that concavity to a pair of maximizing points proves concavity of its partial supremum. A finite concave function on \R^m has an upper supporting affine function at each point. Adding back the quadratic therefore gives a vector q such that

Comparison with (6.7), first for h=\rho e and then for h=-\rho e, and passage to \rho\downarrow0, shows that q=(y-x)/\ell. The two quadratic remainders now prove differentiability and (6.5)–(6.6).□

Proof. Convexity in the length of \Penv_\ell f(z) gives D\ge0, and concavity in the length of \Menv_\ell v(z) gives \widetilde D\ge0. The point z is a forward competitor from z^- and a backward competitor from z^+, which gives e\ge0 and \widetilde e\ge0. The identities in (6.8) and the definitions of the competitor gaps yield

The same difference of the two half-step costs occurs in both lines, so it cancels on subtraction. Dividing by 4\epsilon proves the identity and its first inequality.

For the final inequality let b,d\in\R^m be arbitrary. Apply the path-convexity property of the forward envelope to

For this path the function \Penv_\ell f(x(\ell),t)+|b-d|^2\ell\log\ell/(4\epsilon^2) is convex, with \ell\log\ell assigned its continuous value zero at \ell=0. The midpoint inequality, and 2\epsilon\log(2\epsilon)-2\epsilon\log\epsilon=2\epsilon\log2, give

Using x+d as a forward competitor from z^- and expanding its cost gives

Use the upper supporting inequality in (6.6) for f^{2\epsilon}(x+b,t), and the lower one for f^\epsilon(x+(b+d)/2,t). The left side of (6.11) is then at most

Combining this upper bound with (6.11), using t>\epsilon and |b\pm d|^2\le2(|b|^2+|d|^2), proves

Choose

The result is D+e\ge\epsilon(|Q_+-Q_0|^2+|Q_--Q_0|^2)/(4(1+\log2)), as required.□

Proof. Lemma 4.1, followed by the time translation that puts the fitted point at (0,1), proves the class and centering assertions. At each fixed time the spatial offsets are multiplied by r, and values are multiplied by r up to an affine function. The affine terms cancel from a barycentric defect, so its quotient by the root mean square radius is unchanged. This proves the Jensen assertion.

For the remaining calculations put

Scaling the positions of a barycentric configuration and cancelling its weighted affine terms gives

Subtracting the displayed formula for v/r proves the first identity in (6.13). For s>0, take \sigma=s. The old time and the old barycentric parameter in W are both t=rs, while the correction \kappa^2t/(2r) is \kappa^2s/2. Hence

which proves the second gap identity. This is why the frame uses s=t/r: it preserves the time-zero boundary in the definition of W.

The quadratic envelopes obey the following value identities for \ell>0, with s>0 in the identity involving W:

To check the first identity, parameterize an old forward competitor by a new position y' at time s+\ell. Its old displacement divided by r is y'-y+p_0\ell. The cost divided by r is

The affine change in A from (y,s) to (y',s+\ell) is p_0\cdot(y'-y)+(a_0+|p_0|^2/2)\ell, leaving a_0\ell after the cost is subtracted. For a backward competitor the old displacement divided by r is y'-y-p_0\ell; the same expansion with the cost added leaves -a_0\ell. These prove the first two identities. The last follows by the same forward calculation using W/r=\bar W+A at every positive time. Subtract the formula for v/r from the forward identity, and subtract the backward identity from that formula. Dividing each result by \ell gives (6.14).

The competitor calculations also show that old optimizers map to optimizers for the transformed envelopes. The affine coordinate map sends their straight intermediate points to the new straight intermediate points. Thus the mapped steps are precisely the steps of (6.8) for the transformed pair. The first two velocity equalities in (6.15) follow from (6.12) and t^\pm-t=\pm\epsilon. For the reference velocity, differentiate the last identity of (6.16) in y with r\ell=\epsilon. The derivative of the left side is \nabla_x\Penv_\epsilon W=Q_0, since D_yx=rI, and the derivative of A is p_0. This proves the equality for Q'_0 and the invariance of the velocity discrepancies, with no additional scale factor.□

Proof. Let s=t+\epsilon, so 2\epsilon<s\le K+1. At the point x, choose any upper supporting slope p for the semiconcave function f(\cdot,s):

Such a slope exists because f(\cdot,s)-|\cdot|^2/(2s) is finite and concave on all of \R^m. If p\ne0, insert h=-sp/|p| to obtain

The sandwich bound for W gives, in the other direction,

The spatial increment estimate from the fits, applied with increment length s at the center (x,s) in the box of size 2K, bounds the right side below by

The resulting estimate for p also holds when p=0. Since 1<s/t<2, we have |\log s|\le|\log t|+\log2, and hence

Let h be the optimizing displacement for f^\epsilon(x,t). Its value is at least the value of the competitor h=0. In conjunction with (6.18), this gives

For h\ne0, using s>2\epsilon yields |h|/\epsilon\le4|p|; the same inequality is immediate when h=0. By (6.5), Q_0=h/\epsilon, so (6.19) proves the lemma.□

Proof. We first construct finite stopped walks with common expectation and error bounds, and then localize their terminal points in the frames of their marks. Finally, one joint limiting law will retain the terminal point and all marks, allowing us to select infinitely many localized contacts for the same endpoint.

The stopped walks and their expectation bounds. Fix for the moment an integer T\ge2 and a radius R\ge2. For each N\ge1 choose \epsilon_N to be an inverse power of two with

In particular, \epsilon_N divides 1,T, and every \delta_j with j\le N. We will make each \epsilon_N smaller if necessary, retaining these conditions. Write \epsilon=\epsilon_N while constructing the N-th walk.

Start at z_0=z_*. At a state z_n=(x_n,t_n) with \delta_N<t_n<T and |x_n|<R, choose the two steps of (6.8) and take z_{n+1}=z_n^+ or z_n^- with equal probabilities. Stop on first reaching t_n=\delta_N, t_n=T, or |x_n|\ge R; call this stopping index \tau_N. The choices of optimizers can be made separately at every node of the countable tree of finite histories. Taking independent fair signs \xi_n\in\{-1,1\} to select the steps then defines the walk and the filtration \mathcal F_n=\sigma(\xi_0,\ldots,\xi_{n-1}); all velocities at step n are \mathcal F_n-measurable. In expressions such as Q_-(z_n) below, the velocity denotes the choice made at that history. Before stopping, t_n>\delta_N\ge4\epsilon, so the inequality (6.10) applies.

We choose \epsilon_N sufficiently small that every position up to and including the stop lies in \overline B_{R+1}(0). Here is the uniform localization of optimizers that permits this choice. The sandwich for W extends it continuously to t=0 by W(x,0)=v(x,0). The growth bounds for v and the sandwich give uniform subquadratic growth for v and W on compact time intervals in \R and [0,\infty), respectively. For base points in a fixed compact box and lengths \ell\le1, a quadratic cost first confines all optimizing positions for \Penv_\ell W and \Menv_\ell v to a fixed spatial compact set: outside a sufficiently large ball the cost dominates the uniform subquadratic bound, even with the weaker coefficient corresponding to \ell=1. On that compact set the functions are bounded. Comparing an optimizer with the zero displacement competitor then gives |y-x|^2/(2\ell)\le C there, uniformly as \ell\downarrow0. Thus the optimizing displacements tend to zero uniformly. Apply this to length 2\epsilon and base points with |x|\le R, 0\le t\le T, in a slightly larger time interval. The intermediate displacements used for the steps are half the full displacements, so they can all be made at most one. This proves the asserted bound at stopping. We also choose \epsilon_N small enough that

which is possible by (6.4).

The time coordinate before stopping is the simple symmetric walk with step \epsilon starting at 1. It is stopped no later than this time walk hits \delta_N or T. The latter hitting time is almost surely finite and has finite expectation, for example by the elementary gambler’s ruin formula. The same is therefore true of \tau_N. Put

The upper slope bound for v on boxes and (6.3) give a constant U_B=C L(B)^4, independent of N, such that

For any stopping time \sigma\le\tau_N, summing the conditional form of (6.10) up to \sigma\wedge h gives

For each fixed N, J_\epsilon is continuous and bounded on the compact box \overline B_{R+1}(0)\times[\delta_N,T] containing the stopped walk. Thus dominated convergence applies to its term as h\to\infty. Monotone convergence applies to the nonnegative sum. We obtain, for all these stopping times,

Taking \sigma=\tau_N in (7.5) and using (7.3)–(7.4) also gives

Here C_* is fixed for v and independent of N,R,T; it absorbs the finite number H_* and the absolute constant c in (6.10). Both (7.6) and (7.7) are unconditional expectations. Together with the upper slope bound, the lower expectation bound will give contact and a finite terminal slope in the limit. We first use the discrepancy bound to control horizontal-exit probabilities, both in the original coordinates and in fitted frames.

Reaching the lower level with positive probability. We next show how to choose the outer cutoffs so that the walks reach t=\delta_N with probability at least 3/4, uniformly in N. Couple the time signs with the simple symmetric walk starting at 1 and stopped only at 0 or T. Its probability of reaching T is 1/T, which also bounds that probability for our walk, since it may stop sooner. To control horizontal exit, we will bound the expected maximal displacement by C_T L(R+T+2)^2, with C_T independent of R and N. Markov’s inequality will then make the probability of horizontal exit tend to zero as R\to\infty for fixed T.

The expected number of visits of the unrestricted time walk to any fixed interior grid level is at most CT/\epsilon. To verify this directly, write q=T/\epsilon and identify the level with k\in\{1,\ldots,q-1\}. After a visit to k, the probability of reaching an endpoint before the next return is

This follows from the linear hitting probability for a symmetric walk on each of the intervals [0,k] and [k,q]. Starting from the first visit, the expected total number of visits is at most q/2, by the geometric return calculation; the probability of that first visit is at most one. Stopping the actual walk earlier can only decrease its visit counts. Summing this bound over levels gives

For the second inequality we used Lemma 6.4 before each step. More explicitly, it bounds the summand velocity squared by C(L(B)^2+|\log t_n|^2), while the visit estimate gives at most

This has the asserted bound. Indeed, on (0,1] the function |\log s|^2 is decreasing and integrable, so its right-endpoint grid sum times \epsilon is at most its integral there; on [1,T] it is bounded by (\log T)^2. The constants C_T in this proof can depend on T and the fixed data, but not on R or N.

For n<\tau_N the spatial increment can be written exactly as

where Q_{1}=Q_+ and Q_{-1}=Q_-. Extend both kinds of increments by zero after stopping. The first summands are martingale differences, since the active-step indicator and Q_0(z_n) are known before the fair sign. For a deterministic truncation in the number of steps, Doob’s L^2 maximal inequality and orthogonality of these differences give an upper bound

for the expected supremum of the norm of their partial sums. Letting the truncation increase preserves this bound by monotone convergence of the suprema. On each path, the sum of the norms of the other increments is at most

Cauchy–Schwarz in expectation gives the travel inequality

This calculation uses only predictability of the coefficients of the fair signs in the reference increments, including the active-step indicators, and the pathwise bound on error lengths. It therefore applies as well to the displacement from the starting point of a stopped portion after a stopping time, with its increments extended by zero outside that portion. Equations (7.7) and (7.8) now show that

The probability of a horizontal stop is at most the right side of (7.10) divided by R. For fixed T this tends to zero as R\to\infty. Choose an integer T\ge8, and then choose R large enough that the horizontal-stop probability is at most 1/8, uniformly in N. Fix these T,R for the rest of the proof, so B and B_0 are now fixed finite constants. If

the two preceding bounds and the definition of the stop imply

This remains true if the lower time level and the horizontal cutoff are reached simultaneously, since that case has \eta_N=1.

Marks and localization in fitted frames. For 1\le j\le N, let

Thus a mark is the first point on its time level, or the terminal point if the walk stops before that level. The divisibility of the grid ensures t_j\ge\delta_j, with equality on \{\eta_N=1\}. At the time of this mark choose a fit (p_j,a_j) at z_j of radius \delta_j. These choices can again be made on the countable tree of histories, so the parameters are \mathcal F_{\sigma_{j,N}}-measurable. The bound on fit parameters on boxes gives, for a fixed constant independent of N,

increasing C to make every \Gamma_j\ge1.

The cutoffs T,R and the error bound B_0 are now fixed. We will choose two further cutoffs, uniformly in j,N, to control the terminal point in each fitted frame. Fix 1\le j\le N, put r=\delta_j, and let \mathcal R_{j,N}=\{t_j=r\} be the event that the level r is reached, including the case where it is reached at the original stop. This event is \mathcal F_{\sigma_{j,N}}-measurable. On it z_j=(x_j,r), so we may apply Lemma 6.3 pathwise with the chosen fit (p_j,a_j). In its coordinates,

The transformed solution \bar v is centered at (0,1) and has the same Jensen bound. Its \bar W therefore has the standing properties of Section 6. The mapped steps are optimizing steps, and (6.15) gives identical squared velocity discrepancies. Contact at this finite mark is not needed. The fit and frame are known at the marking time and will be held fixed afterward.

On \mathcal R_{j,N} consider the portion of the original walk from its j-th mark to its stop, with the additional rule that this portion stops on first reaching s=M or |y|\ge K, where M\ge2 is an integer and K\ge2. It includes the endpoint of this additional stopping rule. If the original walk stopped at the mark, the portion has no steps. A hit of an additional cutoff simultaneous with the original stop is counted as a hit of that additional cutoff. On \mathcal R_{j,N}^{\rm c} define the portion to be empty as well, and assign zero to all its sums and spatial suprema. Let n' be its number of steps, and write \sum' for a sum over its pre-step states.

The total squared velocity error on this portion, in the new frame, has the unconditional bound

Indeed, on each path the sum is a subsum of the original sum in (7.7), with identical summands by (6.15); on an empty portion it is zero. This argument gives an unconditional estimate. It makes no assertion that the same bound holds after conditioning on a particular history at the mark.

The time-walk estimates, in contrast, do hold conditionally on that history. On \mathcal R_{j,N} the new time starts at s=1 and has step \epsilon'=\epsilon_N/r\le1/4, which divides 1 and M. Given \mathcal F_{\sigma_{j,N}}, the subsequent signs remain fair. The portion is stopped no later than the unrestricted new time walk first hits 0 or M. Its conditional expected visits to each interior level are therefore at most CM/\epsilon' by the same visit argument used above. Before any step of the portion, the original walk is still live, so s=t/r>\delta_N/r\ge4\epsilon'. Also |y|<K and s<M before the additional stop. Lemma 6.4 applies to the centered \bar v and its \bar W uniformly, and bounds |Q'_0| by C(L(K+M+2)+|\log s|). The integrable logarithmic grid sum gives

The frame is known at the conditioning time, and the estimates of the centered class have constants independent of that frame. This explains the uniformity of these conditional bounds. Integrating them gives the same bounds unconditionally without the indicator.

Extend the new increments by zero outside the portion. Their reference part is \epsilon'\xi_nQ'_0 and their error part is \epsilon'\xi_n(Q'_{\xi_n}-Q'_0). The reference increments are martingale differences after the random starting mark: the fit and reached event are known at that mark, and on an active step the indicator and Q'_0=Q_0-p_j are \mathcal F_n-measurable while \xi_n is still fair given \mathcal F_n. Thus the calculation of (7.9) applies in the primed variables. Using the integrated bounds from (7.13) and the unconditional bound (7.12), it gives

The supremum is understood to be zero for an empty portion. In particular, the probability that this portion reaches |y|\ge K is at most the right side of (7.14) divided by K. The probability it reaches s=M is at most 1/M: conditionally on reaching the mark, the unrestricted symmetric time walk starting from 1 hits M before 0 with probability 1/M, and an earlier stop only removes this event.

Choose a fixed integer M\ge8, and then choose K so large that

This is possible because B_0 is already fixed and L(K+M+2)/K\to0. On \{\eta_N=1\} the j-th level is reached. If neither additional cutoff intervenes, the terminal point is in the portion and has |y|\le K. Combining this observation with (7.11) and the two additional-cutoff probabilities proves, uniformly for j\le N,

One limiting law and the contact properties. We use one fixed compact space for all these laws. Define

The coordinates are the terminal point, the flag, and the marked point and fit at each index. This countable product of compact metric spaces is compact and metrizable. For indices j>N, fill the coordinates by any fixed point of \mathcal K_j; these dummy entries need not be fits. Let \mu_N be the law on \mathcal X of the terminal point z_{\tau_N}, the flag \eta_N, and the marked points and fits. Compactness of the space of probability measures on a compact metric space gives a subsequence converging weakly to a probability law \mu; all subsequent limits in N are along this subsequence. Denote its coordinates by z_{\rm end}=(x_{\rm end},t_{\rm end}), \eta, and (z_j,p_j,a_j).

Here are first the properties that pass directly to this law. The flag is a coordinate in a discrete space, so its probability converges and \mu(\eta=1)\ge3/4. Also the continuous nonnegative function \eta t_{\rm end} on \mathcal X has integral at most \delta_N under \mu_N. Its integral under \mu is zero. It follows that t_{\rm end}=0 almost surely on \{\eta=1\}. For each fixed j, the sets \{t_j\ge\delta_j\} and \{\eta=0\}\cup\{t_j=\delta_j\} are closed and have \mu_N-measure one for all N\ge j. The closed-set direction of the Portmanteau Theorem, \limsup_N\mu_N(F)\le\mu(F), makes their \mu-measures one. Fits at the specified radius form a closed subset of \mathcal K_j. To see this without varying the sheared domain, write its increments in (3.2) as

For each (\theta,\zeta) the fitting inequality is closed in (z_j,p_j,a_j) by continuity of v, and their intersection over this fixed compact set of increments is closed. Portmanteau therefore also passes the fitting property to \mu. Taking the intersection of the resulting full-measure sets for all j, we have

almost surely.

Two further properties come from the expectation bounds:

For the first, fix \ell>0 and set

This is a finite continuous function on \mathcal B. For all sufficiently large N, \epsilon_N\le\ell, and monotonicity of both slopes with respect to their length and (6.3) give J_{\epsilon_N}\le\Phi_\ell. Apply (7.6) at \tau_N and pass to the weak limit using this fixed continuous function and (6.4). The result is

Take a decreasing sequence \ell_h\downarrow0 with \ell_h\le1. The functions \Phi_{\ell_h} decrease to H_v pointwise, because the two slope limits agree for v. Their uniform upper bound on \mathcal B is U_B after increasing its fixed constant. Hence U_B-\Phi_{\ell_h}(z_{\rm end}) are nonnegative and increase to U_B-H_v(z_{\rm end}). Monotone convergence and (7.18) give

The integrand would be infinite where H_v=-\infty, proving the first claim of (7.17).

For the contact claim fix j and let \nu_N^j and \nu^j be the marginals of \mu_N and \mu for the marked point z_j. They converge weakly. Suppose that \nu^j gives positive mass to \{v-W>0\}. All these limit marks have t\ge\delta_j. Since v-W is continuous and nonnegative for t>0, there is \alpha>0 such that the relatively open set

has \nu^j(U)>0 and its closure is contained in \{t\ge\delta_j/2,\ v-W\ge\alpha\}. For example, first choose \alpha with positive mass where v-W>2\alpha. The zero-length continuity of the quadratic envelopes is uniform on this compact closure. Thus (6.2) gives

on U for all sufficiently large N. In particular, J_{\epsilon_N}\le-\alpha/(4\epsilon_N) on U. By the open-set direction of Portmanteau, \nu^j(U)\le\liminf_N\nu_N^j(U), so these probabilities are bounded below by a fixed positive number for large N. The uniform upper bound (7.4) on all marks and (7.6) at \sigma_{j,N} would then give

This contradicts (6.4). Thus W(z_j)=v(z_j) almost surely. A countable intersection proves this simultaneously for all j. The derivation of (7.17) used unconditional expectations; it does not require conditioning on the flag.

For each fixed j define the subset of \mathcal X

This event retains the actual terminal time in the prelimit laws. It is closed: the flag condition is closed and the norm is a continuous function of the terminal point, the j-th marked point, and p_j. Equation (7.15) gives \mu_N(E_j)\ge1/2 whenever N\ge j. The closed-set Portmanteau inequality has the direction

These events need not be independent. For every h their tail union has measure at least 1/2, since it contains each E_j with j\ge h. Continuity of probability from above therefore gives

Choose one outcome in this limsup event and in the full-measure sets of (7.16)–(7.17). Every E_j includes \eta=1, so this outcome has t_{\rm end}=0 and t_j=\delta_j for every j. It belongs to E_j for an infinite set of indices, at each of which its localization inequality is

The same outcome has a finite common slope at the endpoint, contact at all the marks, and the indicated fits. Its endpoint and the infinite set of these marks establish (7.1).□

Proof. Translate (b,t_0) to (0,0) and subtract the value there; write V for the resulting function. These changes preserve the class and the envelope slopes. The forward equality in (8.1) gives \Penv_tV(0,0)=Bt for t>0, and therefore

Fix any (x,t)\in\R^m\times\R. At a large length \ell, a maximizing position y_\ell for \Penv_\ell V(x,t) satisfies

This is the optimizer estimate from Section 4, applied on a box of size comparable to \ell; the translated function has the same growth estimates with constants depending on a fixed value and fit. Explicitly, if the displacement is b_\ell\ell with b_\ell\ge2, comparison with zero displacement and the bound |V(y,s)|\le C K(1+\log K)^2 on boxes of size K give

Absorbing the logarithmic term in b_\ell gives b_\ell\le C_{x,t}'(1+\log\ell)^2, which proves (8.3).

For \ell sufficiently large, t+\ell>0, so (8.2) applies at the optimizer. The difference between the two quadratic costs is

by (8.3). Consequently

The forward slopes are nondecreasing in \ell by Lemma 4.4. Thus \Senv_\ell^+V(x,t)\le B for every \ell>0. Evaluating the envelope at any competitor gives the global directed comparison

The envelopes are attained by Lemma 4.3. At the origin, (8.1) therefore supplies, for every \ell,r>0, a forward equality point (\ell q_+,\ell) and a backward equality point (-r q_-,-r), with

Apply (8.4) directly from the backward point to the forward point. After canceling B(\ell+r), it gives

Hence every forward and backward equality velocity is one vector q, independent of all lengths and choices. We have

Let A(x,t)=q\cdot x+(B-|q|^2/2)t. Comparing (x,t) to a point (qs,s) on this line in (8.4) yields

Letting s\to-\infty and s\to+\infty proves V=A.□

Proof of Theorem 3.1. Suppose that the class contains a nonaffine function. Proposition 5.4 supplies a centered solution v, a number \kappa>0, and its subsolution envelope

with the stated contact and finite-slope properties. Proposition 7.1 supplies a point x_{\mathrm{end}}, infinitely many indices j with r_j=2^{-j}, contacts (x_j,r_j), and fits (p_j,a_j) there of radius r_j, such that

for one finite K.

Apply Lemma 6.3 at each of these marks, and denote the transformed functions by \bar v_j. Their images of (x_{\mathrm{end}},0) are (b_j,0), where

These vectors are bounded by (8.5). Every \bar v_j has value zero and the zero fit of radius one at (0,1). Lemma 4.2 gives a subsequence for which \bar v_j\to v_\infty locally uniformly and b_j\to b. The uniform growth bounds also give convergence of the forward and backward envelopes at each fixed positive length, by Lemma 4.3.

The slope covariance (6.14) gives

Let B=\Senv_1^+v_\infty(b,0), which is finite. For each fixed \ell>0, subtract the equality with the forward slope at length one:

The limit is zero because both unshifted slopes tend to the same finite value H_v(x_{\mathrm{end}},0). Passing to the limit on the left gives

This argument uses each fixed length along the same locally convergent subsequence; no convergence of the individual shifts a_j is required. Lemma 8.1 now makes v_\infty affine.

The contacts retain a positive envelope gap. By (5.9) at the contacts in (8.5) and the gap covariance (6.13),

Lemma 5.1 passes this identity to the locally uniform limit with its uniform growth bound. Thus

For an affine function, the average of its values at any configuration is its value at the barycenter, so the nonnegative variance penalty makes \Cenv_1v_\infty=v_\infty. This contradiction proves Theorem 3.1. The reduction in Section 3 then proves Theorem 1.1.□