R Monneau

Strictly convex Hamilton-Jacobi equations: strong trace of the derivatives in codimension ≥ 2

Keywords: 35F21 Strong trace, Hamilton-Jacobi equations, codimensional boundaries, Liouville-type result, regularity, viscosity solutions

We consider Lipschitz continuous viscosity solutions to an evolutive Hamilton-Jacobi equation. The equation arises outside a closed set Γ. Under a condition of strict convexity of the Hamiltonian, we show that there exists a notion of strong trace of the derivatives of the solution on the Lipschitz boundary Γ of codimension d ≥ 2. The very special case d = 1 is done in a separated work.

This result is based on a Liouville-type result of classification of global solutions with zero Dirichlet condition on the boundary Γ, where Γ is an affine subspace. We show in particular that such solutions only depend on the normal variable to Γ. As a consequence, we show more generally that the existence of a pointwise tangential gradient along Γ implies the existence of pointwise directional derivatives in all directions.

This result also holds true for Hamiltonians depending on the time-space variables, under an additional Dini condition involving certain moduli of continuity. We also give a counter-example for d = 2 in the stationary case, where the Hamiltonian is only continuous in the space variable, and where the solution has no directional derivatives in any directions normal to Γ. Such phenomenon does not hold for d = 1.

Introduction 1.Main results

Let m ≥ 0 and let us consider globally Lipschitz continuous solutions u = u(t, x ′ , x) of (1.1)

u t + H(Du) = 0 on R × R m × Ω u = 0 on R × R m × ∂Ω
where the Dirichlet condition is satisfied in the strong sense (i.e. pointwisely). Here we consider the open set whose boundary has codimension bigger or equal to 2

Ω := R d \ {0} , ∂Ω = {0 R d } , with d ≥ 2.
The case of dimension d = 1 with a half line Ω = (0, +∞) has been studied in [START_REF] Monneau | Strictly convex Hamilton-Jacobi equations: strong trace of the gradient[END_REF]. Here for d ≥ 2, we find a new method and get different results. Assume where we recall that H strictly convex means H(λP + (1 -λ)Q) < λH(P ) + (1 -λ)H(Q) for all λ ∈ (0, 1), P, Q ∈ R m+d , P ̸ = Q.

We consider the coordinates X := (t, x) with t ∈ R and x = (x ′ , x) ∈ R m × R d .

Theorem 1.1 (Liouville-type result) Assume that the strictly convex function H satisfies (1.2). Then every globally Lipschitz continuous viscosity solution u of (1.1) only depends on the normal variable, i.e. u(t, x ′ , x) = u(0, 0, x) for all (t, x ′ , x) ∈ R × R m × Ω.

For local problems, we will need to describe directional derivatives with a single object. To this end, we introduce the following definition.

Definition 1.2 (Bouligand derivative)

We say that a function f : R n ⊃ B 1 (0) → R has a Bouligand derivative at the origin if there exists a (positively) 1-homogeneous function Bf (0) : R n → R (i.e. Bf (0)(λy) = λBf (0)(y) for all λ ≥ 0) such that

f (y) -f (0) = Bf (0)(y) + o(|y|)
It is easy to see that any Lipschitz function has directional derivatives in any directions if and only if it has a Bouligand derivative. The formalism of Bouligand derivative has just the practical advantage to deal with all directional derivatives at the same time.

We now consider the following problem localized on a cylinder Q 0 ⊂ R 1+m+d . We denote the boundary Γ := R 1+m × {0 R d } and the open ball B r = B r (0) of center the origin and radious r > 0. Then we consider the problem:

(1.3) u t + H(Du) = 0 on Q 0 \Γ ⊂ R 1+m × Ω, u = g(t, x ′ , 0) on Q 0 ∩ Γ ⊂ R 1+m × ∂Ω.
We have the following result (at least surprising for the author).

Theorem 1.3 (Tangential gradient implies full directional derivatives) Let Q 0 := (-1, 1) × B 1 ⊂ R × R m+d . Assume that H : R m+d → R satisfies (1.2). Consider a Lipschitz continuous function u : Q 0 → R which is a viscosity solution of (1.3) with u(0) = 0. Assume that u has a tangential gradient at the origin, i.e. there exists (λ, p ′ ) ∈ R × R m such that

(1.4) u(t, x ′ , 0) = λt + p ′ • x ′ + o(|(t, x ′ )|) for all (t, x ′ , 0) ∈ R × R m × ∂Ω.
Then u has Bouligand time-space derivative at the origin, i.e. there exists a (positively) 1-homogeneous function Bu(0) : R 1+m+d → R such that for X = (t, x ′ , x) ∈ Q 0

(1.5) u(X) = Bu(0)(X) + o(|X|)
with the splitting Bu(0)(X ′ , x) = X ′ • D ′ u(0) + Bu(0)(0, x) where X ′ • D ′ u(0) := λt + p ′ • x ′ for X ′ = (t, x ′ ).

Theorem 1.4 (A notion of strong trace of directional derivatives)

Let Q 0 := B 1 × B 1 ⊂ R 1+m × R d .
Assume that H : R m+d → R satisfies (1.2). Consider a Lipschitz continuous function u : Q 0 → R which is a viscosity solution of (1.3), with g :

Q 0 ∩ Γ → R Lipschitz continuous.
Then for Y ∈ R 1+m+d with X + Y ∈ Q 0 , we have

(1.6) u(X + Y ) -u(X) = Bu(X)(Y ) + o(|Y |) for a.e. X = (t, x ′ , 0) ∈ Γ
where Bu(X) : R 1+m+d → R is (positively) 1-homogeneous and Lipschitz. Moreover for a.e. X ∈ Γ, the quantity Bu(X) satisfies the splitting Bu(X)(Y ′ , y) = Y ′ •D ′ u(X)+Bu(X)(0, y) for all Y ′ = (s, y ′ ) ∈ R×R m and y ∈ R d , with Y ′ • D ′ u(X) := λs + p ′ • y ′ where λ := u t (X), and the tangential space gradient is p ′ := D(u |Γ )(X). Moreover, we have the following limit for the time-space derivatives Du := (u t , Du) with X = (X ′ , x)

(1.7) lim

ε→0 + B1×B1
| Du(X ′ , εx) -D(Bu(X ′ , 0))(0, x)| dX = 0.

This convergence is our definition of the strong trace of the time-space derivatives of the solution u on the boundary Γ of codimension d ≥ 2.

Remark 1. [START_REF] Barles | An Illustrated Guide of the Modern Approaches of Hamilton-Jacobi Equations and Control Problems with Discontinuities[END_REF] We see here that the strong trace of the gradient has to be a Bouligand derivative in general, and not a standard gradient. In other words, we see that taking the strong trace of the gradient, we may loose the linearity along the normal variable for the first order approximation map when the boundary Γ has codimension d ≥ 2. In particular, for a.e. point X of the boundary Γ, we get that u has directional derivatives at X.

Remark 1.6 Notice that it is straightforward to adapt the proof of Theorem 1.4 to the case of Lipschitz continuous boundary Γ.

Again for a cylinder Q 0 ⊂ R 1+m+d , we consider now the X-dependence in the Hamiltonian H, for X = (t, x ′ , x), and the problem (1.8) u t + H(X, Du) = 0 on Q 0 \Γ u = g(X)

on Q 0 ∩ Γ.

Recall that we say that a function u is semiconcave on an open set Q 0 if there exists a modulus of continuity ω : (0, +∞) → (0, +∞), with ω(0 + ) = 0 such that (1.9) λu(X)

+ (1 -λ)u(Y ) -u(λX + (1 -λ)Y ) ≤ λ(1 -λ)|X -Y |ω(|X -Y |) for all λ ∈ [0, 1], X, Y ∈ Q 0 .
Recall also that semiconcavity implies the existence of directional derivatives everywhere (see [START_REF] Cannarsa | Semiconcave Functions, Hamilton-Jacobi Equations, and Optimal Control[END_REF]). Here we show that the lack of regularity in X can break semiconcavity property of the solution, even if the Hamiltonian is strictly convex in the gradient.

Proposition 1.7 (A counter-example to semiconcavity)

We assume that m = 0, ∂ t = 0 and d = 2. Let h : [0, +∞) → (0, +∞) be C 2 with h(1) = 1 and h ′ (0) = 0 and h ′ > 0 on (0, +∞) and h ′′ > 0 on [0, +∞). Then there exists some continuous function a : R 2 ⊃ B 1 → R and H(x, P ) = a(x)h(|P |) such that there exists a Lipschitz continuous solution u = u(x) of

H(x, Du) = 1 on B 1 ⊂ R 2 .
We can choose the continuous function a(•) such that our solution u has no directional derivatives at x = 0. In particular u is not semiconcave (i.e. does not satisfy (1.9) with Q 0 := B 1 2 ). Still, there exists a modulus of continuity ω 0 such that we have but ω 0 is not Dini integrable, i.e. that r 0 ds s ω 0 (s) = +∞ for all r > 0.

We refer the reader to Theorem 2.1.10 on page 35 in Cannarsa, Sinestrari [START_REF] Cannarsa | Semiconcave Functions, Hamilton-Jacobi Equations, and Optimal Control[END_REF], which says that (1.10) implies semiconcavity estimate (1.9) with ω(r) := C r 0 ds s ω 0 (s). For our ptoblem, and under a certain Dini condition on the X-dependence of H(X, P ) which is continuous in both variables and C 1 strictly convex and superlinear in P , it is possible to recover the results of the homogeneous case. This is the result below.

Theorem 1.8 (Generalization to variable coefficients X) Theorems 1.3 and 1.4 hold for equation (1.8), under the condition that H : Q 0 × R m+d → R is continuous and that P → H(X, P ) satisfies (1.2) for all X ∈ Q 0 , under the additional Dini condition (1.13) on a certain modulus given below.

Assume that the solution u to (1.8) has Lipschitz constant L > 0. Let L be the Legendre-Fenchel transform of H defined by (1.11) L(X, ξ) := sup

P ∈R m+d ξ • P -H(X, P ).
Assume also that there exists a map L 0 : R m+d → R such that L(X, ξ) ≥ L 0 ( ξ) for all X ∈ Q 0 , ξ ∈ R m+d , and for L > 0, we consider R L ≥ 1 such that

(1.12) ξ ∈ R m+d , L 0 ( ξ) ≤ L|(1, ξ)| ⊂ B R L .
i) (modulus of strict convexity of the map ξ → L(X, ξ))

For every R > 0, we assume that there exists a (monotone) modulus of continuity ωR : (0, +∞) → (0, +∞) with ωR (0 + ) = 0 such that for r > 0, we set

ΩR (r) := r 0 ωR (s) ds with ωR (r) := inf | ξ1-ξ2|≥r, ξi∈BR, X∈Q0 ξ1 -ξ2 | ξ1 -ξ2 | , DL(X, ξ1 ) -DL(X, ξ2 )
where ΩR is convex increasing.

ii) (modulus of continuity of the map X → L(X, ξ))

We assume the existence of the following monotone modulus of continuity

ω R (r) := sup X,Y ∈Q0, |X-Y |≤r, ξ∈B R L(X, ξ) -L(Y, ξ) .
Then we require the modulus Ω-1 R • ω R to satisfy the following Dini condition

(1.13) 1 0 ds s ( Ω-1 R • ω R )(s) < +∞ for R := 1 + R L .
Remark 1.9 It is easy to check that Dini condition (1.13) is satisfied for instance if P → H(X, P ) is C 2 with δ -1 ≥ D 2 P P H ≥ δ > 0 and X → H(X, P ) is β-Hölder continuous for some β ∈ (0, 1]. Such result can be compared to the precise interior semiconcavity results in Cannarsa, Cardaliaguet [START_REF] Cannarsa | Regularity results for eikonal-type equations with nonsmooth coefficients[END_REF] for Hölder continuous Hamiltonians in the space variable, in the case of stationary solutions where the Hamiltonian is 1-homogeneous in the gradient.

Remark 1.10 It is straightforward to adapt Theorem 1.8 to the case of a Lipschitz continuous boundary Γ. It would be also interesting (and not quite difficult) to develop an interior theory (i.e. with Γ = ∅) along the same lines as in the present paper. For instance, the analogue of the Liouville-type result claims that Lipschitz global time-space solutions are concave. Here we refrain us to go further in that direction.

Brief review of the literature

We refer to the pioneering work of Lions [START_REF] Lions | Generalized solutions of Hamilton-Jacobi equations[END_REF] on viscosity solutions of Hamilton-Jacobi equations and their properties. The reader can also consult the reference books Cannarsa, Sinestrari [START_REF] Cannarsa | Semiconcave Functions, Hamilton-Jacobi Equations, and Optimal Control[END_REF] on semiconcave functions and Bardi, Capuzo-Dolcetta [START_REF] Bardi | Capuzzo-Dolcetta Optimal Control and Viscosity Solutions of Hamilton-Jacobi-Bellman Equations[END_REF] for Hamilton-Jacobi equations related to control problems. In Cannarsa, Soner [START_REF] Cannarsa | Generalized one-sided estimates for solutions of Hamilton-Jacobi equations and applications[END_REF] (see also Theorem 5.3.8 on page 118 in [START_REF] Cannarsa | Semiconcave Functions, Hamilton-Jacobi Equations, and Optimal Control[END_REF]), it is proven that for Hamiltonians H(X, P ) which are locally Lipschitz in all variables and strictly convex in P , the locally Lipschitz continuous solutions are semiconcave. This result has been extended to stationary solutions for 1-homogeneous Hamiltonians which are Hölder in the space variable in Cannarsa, Cardaliaguet [START_REF] Cannarsa | Regularity results for eikonal-type equations with nonsmooth coefficients[END_REF].

In Bianchini, De Lellis, Robyr [START_REF] Bianchini | SBV regularity for Hamilton-Jacobi equations in R n[END_REF], the authors show that for a uniformly C 2 Hamiltonian H, the timespace gradient of the solution is in SBV loc , which can be seen as a refinement of semiconcavity estimates. This result has been extended to the case of C 3 Hamiltonians depending also on (t, x) in Bianchini, Tonon [START_REF] Bianchini | SBV Regularity for Hamilton-Jacobi Equations with Hamiltonian Depending on (t, x)[END_REF]. We also refer the reader to Rifford [START_REF] Rifford | On Viscosity Solutions of Certain Hamilton-Jacobi Equations: Regularity Results and Generalized Sard's Theorems[END_REF] for further interesting regularity results of solutions.

In the context of homogeneous scalar conservation laws, a notion of strong trace on a Lipschitz boundary of a domain (i.e. for d = 1) has been introduced by Vasseur [START_REF] Vasseur | Strong Traces for Solutions of Multidimensional Scalar Conservation Laws[END_REF] under a condition of genuine nonlinearity of the C 3 flux function. This result has been generalized by Panov [START_REF] Yu | Existence of strong traces for quasi-solutions of multidimensional conservation laws[END_REF] to the case of C 0 homogeneous fluxes, and C 1 boundary (the case of Lipschitz boundary is also claimed to remain valid with the same proof).

In Monneau [START_REF] Monneau | Strictly convex Hamilton-Jacobi equations: strong trace of the gradient[END_REF], the existence of strong traces of the gradient of the solution has been obtained for strictly convex Hamilton-Jacobi equations with boundary Γ of codimension d = 1. For d = 1, no Dini condition is required: only the continuity of the map (X, P ) → H(X, P ) is sufficient.

Actual researches try to understand convex Hamilton-Jacobi equations either on stratified domains (see for instance Barles, Chasseigne [START_REF] Barles | Almost) everything you always wanted to know about deterministic control problems in stratified domains[END_REF][START_REF] Barles | An Illustrated Guide of the Modern Approaches of Hamilton-Jacobi Equations and Control Problems with Discontinuities[END_REF]), or with a point defect (see Achdou, Le Bris [START_REF] Achdou | Homogenization of some periodic Hamilton-Jacobi equations with defects[END_REF]). Here we hope that our present work will help to understand better these probems where codimensions play a key role.

Organization of the paper

In Section 2, we recall the representation formula of the solutions to convex Hamilton-Jacobi equations.

In Section 3, we construct a fundamental solution û associated to the Hamiltonian H(P ), which is the analogue of the distance to the origin for the eikonal equation. Then we show that u + := û and u -= -û(-x) are both barriers, and that every global solution with zero value on the boundary Γ is sandwiched in between u + ≥ u -. This is a key step towards the Liouville-type result (Theorem 1.1) whose proof is done in Section 4. This proof is in particular based on a key equality (Lemma 4.2) satisfied by the solution on a characteristic trajectory.

In Section 5, we study in details the characteristics of 1-homogeneous concave global solutions. This section is of independent interest and uses the notion of extreme points and exposed points of a compact convex set. This section is also a key step towards the proof of Theorem 1.3 on the existence of full directional derivatives for local solutions.

Section 6 is a technical result which localizes the result of Lemma 4.2, and which is necessary for the proof of Theorem 1.3.

Section 7 is fully devoted to the proof of Theorem 1.3 (on the existence of directional derivatives) using ingredients of Sections 5 and 6.

Section 8 is devoted to the proof of Theorem 1.4 on the strong traces of the directional derivatives. It starts with a building block result, which shows the strong L 1 loc convergence of the gradient of the blow-up of the solution (see Proposition 8.1). The remaining part of the section proves Theorem 1.4, using covering arguments and identification of the limits through Young measures.

Section 9 gives the proof of the counter-example to semiconcavity, namely Proposition 1.7. Section 10 gives the proof of Theorem 1.8 which generalizes our results to Hamiltonians with time-space dependence.

Section 11 is an appendix where we collect results on exposed points of convex sets, which are used in the main part of the paper.

Main notations

Ω = R d \ {0} = reduced space domain R m × Ω = space domain Γ = R 1+m × {0 R d } = time-space boundary Γ = R m × {0 R d } = space boundary x ∈ R d = normal coordinates X ′ = (t, x ′ ) ∈ R × R m = tangential coordinates x = (x ′ , x) ∈ R m × R d = space coordinates X = (t, x ′ , x) = (t, x) = (X ′ , x)
= time-space coordinates

Q 0 = (-1, 1) × B 1 or B 1 × B 1 = cylinder in R × R m+d or R 1+m × R d Q τ,ρ = (-τ, 0) × B ρ = local cylinder in R × R m+d Γ τ,ρ = Q τ,ρ ∩ Γ = (closed) tangential boundary of the local cylinder P, Q ∈ R m+d = space gradient Du = (u t , Du) = time-space gradient Bu = Bouligand derivative of u H = the Hamiltonian L = H * = the Legendre-Fenchel transform of H ξ = (ξ ′ , ξ) ∈ R m × R d = velocity ℓ(ξ) = inf ξ ′ ∈R m L(ξ ′ , ξ) = the Legendre-Fenchel transform of H(0, •) û = fundamental convex 1-homogeneous solution u ± = barriers.

Representation formula

In this section, we recall the representation formula of the solutions to convex Hamilton-Jacobi equations.

Given X = (t, x) ∈ R × R m+d and ξ(•) ∈ L 1 loc ((-∞, t]; R m+d ), we consider the following backward trajectory d dσ γ ξ X (σ) = ξ(σ) for σ ≤ t, with terminal data γ ξ X (t) = x and call for all t 0 < t

E t0 X :=    (s, ξ) ∈ [t 0 , t) × L 1 loc ((-∞, t]; R m+d ), γ ξ X (σ) ∈ R m × Ω, for all σ ∈ (s, t), with γ ξ X (s) ∈ R m × ∂Ω if s ∈ (t 0 , t) γ ξ X (s) ∈ R m × Ω if s = t 0   
which is the set of parameters such that the backward trajectory stays in the set R m × Ω and in a time interval contained in [t 0 , t].

We recall the following standard result for convex Hamiltonians (which can be seen as a generalization of Lax-Hopf formula).

Lemma 2.1 (Representation formula)

Assume that H : R 1+m+d × R m+d → R be continuous such that P → H(X, P ) satisfies (1.2) for all X, and let L be the Legendre-Fenchel transform of H given in (1.11). Assume that u : R 1+m+d → R is a globally Lipschitz continuous function satisfying

u = g on Γ = R 1+m × {0 R d }.
Then u satisfies for all X = (t, x) ∈ R 1+m+d and all t 0 ∈ (-∞, t)

           u(X) = inf (s,ξ)∈E t 0 X G(s, t; γ ξ X ) with G(s, t; γ ξ X ) - t s L σ, γ ξ X (σ), d dσ γ ξ X (σ) dσ := u(s, γ ξ X (s)) if γ ξ X (s) ∈ R m × Ω g(s, γ ξ X (s)) if γ ξ X (s) ∈ R m × ∂Ω if and only if u solves (1.8) with Q 0 = R 1+m+d . (2.1)
Representation formula (2.1) means that u(t, x) is the infimum of some cost function over all trajectories with terminal point (t, x) and initial point on the part of the boundary ([t 0 , t)×R m ×∂Ω)∪({t 0 }×R m ×Ω).

Sketch of the proof

The standard proof first shows the dynamic programming principle which implies (by variations/comparison) the viscosity inequalities on the time interval (t 0 , +∞) (see for instance [START_REF] Imbert | Flux-limited solutions for quasi-convex Hamilton-Jacobi equations on networks[END_REF], or [START_REF] Cannarsa | Semiconcave Functions, Hamilton-Jacobi Equations, and Optimal Control[END_REF] for results of the same flavour). Conversely, the comparison principle implies that every solution of (1.8) on the time interval [t 0 , +∞) coincides with the unique solution given by the representation formula (2.1). Notice that the comparison principle is valid here because u is globally Lipschitz continuous. This ends the sketch of the proof.

Existence of barriers

In this section, we show that inf R d H(0, •) ≤ 0, and that when the inequality is strict, then we can describe the maximal solution u + and the minimal solution u -to equation (1.1). We show that u + = û and u -= -û(-x), where û is a convex 1-homogeneous solution, whose we study the rich properties. The existence of the two barriers u ± is a key step towards the Liouville-type result which will be developed in the next section.

Lemma 3.1 (Dichotomy)

Assume that H : R m+d → R satisfies (1.2) and that u is a globally Lipschitz continuous solution of (1.1). Then either inf R d H(0, •) = 0 and there exists a unique p 0 ∈ R d such that H(0, p 0 ) = 0 with u(t, x ′ , x) = p 0 •x, or

(3.1) inf R d H(0, •) < 0.
Proof of Lemma 3.1

We define u(x) := sup

(t,x ′ )∈R×R m u(t, x ′ , x), u(x) := inf (t,x ′ )∈R×R m u(t, x ′ , x).
Because u is globally Lipschitz continuous this is also the case of u and u. Moreover u is a subsolution, and by classical Barron-Jensen argument (see [START_REF] Barron | Semicontinuous viscosity solutions of Hamilton-Jacobi Equations with convex hamiltonians[END_REF]) for convex Hamiltonians and Lipschitz solutions, the minimum (here infimum) of solutions is still a solution. Hence H(0, Du) ≤ 0 and H(0, Du) = 0 on Ω. Using a test function from above, we see that µ := inf R d H(0, •) ≤ 0. In case µ = 0, the strict convexity of H implies the uniqueness of some p 0 such that H(0, p 0 ) = 0. Then any test function φ touching from above either u or u at some point x 0 ∈ Ω, satisfies Dφ(x 0 ) = p 0 . It is then easy to see that Du = p 0 = Du which gives the result. This ends the proof of the lemma.

In the remaining part of the paper, we assume the negativity of the infimum of H as in (

(3.3) û(x) = x • Dû(x) with Dû(x) ∈ ∂K for all x ∈ Ω
where the convex K is defined in (3.2). In particular û is a (viscosity) solution of

(3.4) H(0, Dû) = 0 on Ω û = 0 on ∂Ω = {0 R d } .
We have also for all x ∈ R d

(3.5) û(x) = inf τ >0 τ ℓ( x τ
) where ℓ(ξ) := sup

p∈R d {ξ • p -h(p)} for all ξ ∈ R d , with h(p) := H(0, p) and for x ̸ = 0 (3.6) inf τ >0 τ ℓ( x τ ) = τ * ℓ( x τ * ) for some unique τ * = τ * (x) > 0 and      ξ * := x τ * (x) , p * := Dû(x) ∈ ∂K, ξ * = D x H(0, p * ), p * = Dℓ (ξ * ) .
Moreover the two Legendre-Fenchel transforms ℓ and L satisfy

(3.7) ℓ(ξ) = inf ξ ′ ∈R m L(ξ ′ , ξ) with L(ξ ′ , ξ) = sup P ∈R m+d {(ξ ′ , ξ) • P -H(P )} , (ξ ′ , ξ) ∈ R m × R d .
Finally, û enjoys the following strict convexity property (not along the radials):

(3.8) û(λx + (1 -λ)y) < λû(x) + (1 -λ)û(y) for all λ ∈ (0, 1), if [0, +∞) • x ̸ = [0, +∞) • y, x, y ̸ = 0. Proof of Lemma 3.2 Step 1: proof that u ∈ C 1 (R d \ {0})
We first notice that by construction, û is convex, (positively) homogeneous of degree one and Lipschitz continuous. We now check that the strict convexity of

K implies û ∈ C 1 (R d \ {0}).
To this end, we introduce the lower semi-

continuous convex function v = (+∞) • 1 R d \K which satisfies v = 0
on K and v = +∞ outside K. Then the Legendre-Fenchel transform v * of v satisfies v * = û, and by convex duality, we have û * = v. Now assume by contradiction that for some x 0 ̸ = 0

p 1 ̸ = p 2 with p 1 , p 2 ∈ ∂ û(x 0 ) := p ∈ R d , û(x 0 + y) -û(x 0 ) ≥ p • y, for all y ∈ R d where ∂ û(x 0 ) is the subdifferential of û at x 0 . Then the full segment satisfies [p 1 , p 2 ] ⊂ ∂ û(x 0 ), which (by convex duality) implies x 0 ∈ ∂ û * (p) = ∂v(p) for all p ∈ [p 1 , p 2 ].
In particular, we get v(p) < +∞, and then

v(p) = 0, i.e. [p 1 , p 2 ] ⊂ K. If there exists p ∈ [p 1 , p 2 ] ∩ Int(K), then we have x 0 ∈ ∂v(p) = {0} which is impossible. Hence [p 1 , p 2 ] ⊂ ∂K which is also impossible because K is strictly convex. Hence we deduce that ∂ û(x) = singleton = {Dû(x)} for all x ̸ = 0 which implies û ∈ C 1 (R d \ {0}).
Step 2: proof of (3.3)

We notice that û satisfies û

(x) = sup p∈∂K p • x.
In particular for every x ̸ = 0, there exists some p ∈ ∂K such that û(x) = p • x and then

p ∈ ∂ û(x). Because û ∈ C 1 (R d \ {0}
), we deduce that p = Dû(x) is unique. This shows (3.3) and consequently (3.4).

Step 3: proof of (3.5)

We now define û+ (x) := inf

τ >0 τ ℓ( x τ
). By definition of ℓ, we have

(3.9) û+ (x) = inf τ >0 τ sup p∈R d x τ • p -H(0, p) ≥ inf τ >0 τ sup p∈K x τ • p -H(0, p) ≥ inf τ >0 τ sup p∈K x τ • p = sup p∈K x • p = û(x).
From assumption (3.1), we have

(3.10) ℓ(0) = -inf p∈R d H(0, p) < 0. Hence (3.11) û+ (0) = 0.
Now for x ̸ = 0, (3.10) shows on the one hand that the infimum defining û+ (x) is not reached as τ → +∞.

On the other hand, we know that H satisfies (1.2), and then this is also the case of h, and by duality of ℓ. Therefore ℓ is superlinear and the infimum defining û+ (x) is not reached as τ → 0. Hence it is reached for some τ * ∈ (0, +∞). The first variation in τ of the map τ → τ ℓ( x τ ) shows that

(3.12) ℓ(ξ * ) = ξ * • p * with ξ * := x τ * and p * := Dℓ(ξ * ). Hence (3.13) h(p * ) = sup ξ∈R d {p * • ξ -ℓ(ξ)} = p * • ξ * -ℓ(ξ * ) = 0 and p * ∈ ∂K.
We get for

x ̸ = 0 û+ (x) = τ * ℓ(ξ * ) = τ * ξ * • Dℓ(ξ * ) = x • p * ≤ û(x).
Together with (3.9) for x ̸ = 0 and (3.11) for x = 0, we deduce that û+ = û, which shows (3.5).

Step 4: proof of (3.6)

Let us consider the function f (τ, x) := τ ℓ( x τ ), for all (τ, x) ∈ (0, +∞) × R d . For τ 1 , τ 2 > 0 and x 1 τ 1 ̸ = x 2 τ 2 and λ ∈ (0, 1) and τ := λτ 1 + (1 -λ)τ 2 , we get τ -1 {λf (τ 1 , x 1 ) + (1 -λ)f (τ 2 , x 2 )} = τ -1 λτ 1 ℓ( x 1 τ 1 ) + (1 -λ)τ 2 ℓ( x 2 τ 2 ) > ℓ τ -1 {λx 1 + (1 -λ)x 2 } = τ -1 f (τ, λx 1 + (1 -λ)x 2 ).
In particular for x 1 = x 2 = x ̸ = 0 and τ 1 ̸ = τ 2 , we see that the map τ → f (τ, x) is strictly convex on (0, +∞). This shows the uniqueness of the minimizer τ * = τ * (x) which is the first part of (3.6).

On the other hand, we have û(x) = τ * ℓ(ξ * ) with ξ * = x τ * , and (3.12), (3.13) show that

û(x) = x • p * with Dℓ(ξ * ) = p * ∈ ∂K
and Step 2 shows that p * = Dû(x). Therefore

x τ * = ξ * = Dh(p * ) = D x H(0, p * ) = D x H(0, D x û(x)) and also Dû(x) = Dℓ (ξ * ) with ξ := x τ * (x)
which shows the last part of (3.6).

Step 5: proof of (3.7) From convex duality, we have sup

q {p • q -ℓ(q)} = H(0, p) = sup (q ′ ,q) {(0, p) • (q ′ , q) -L(q ′ , q)} = sup q p • q -inf q ′ L(q ′ , q)
which shows by reverse convex duality that ℓ(q) = inf q ′ L(q ′ , q) which is (3.7).

Step 6: proof of (3.8) Assume by contradiction that û is affine on the segment [

x 1 , x 2 ] with [0, +∞)•x 1 ̸ = [0, +∞)•x 2 and x 1 , x 2 ̸ = 0.
From the positive 1-homogeneity of û, we deduce that û is linear on the cone Λ := [0, +∞)

x 1 + [0, +∞)x 2 .
Moreover, by property (3.6) of û, we have

û(x k ) = τ k ℓ(ξ k ), ξ k := x k τ k , τ k = τ * (x k ), k = 1, 2 i.e. û(ξ k ) = ℓ(ξ k ).
Hence for λ ∈ (0, 1) and

ξ λ := λξ 1 + (1 -λ)ξ 2 , we get ℓ(ξ λ ) ≥ inf τ >0 τ ℓ( ξ λ τ ) = û(ξ λ ) = λû(ξ 1 ) + (1 -λ)û(ξ 2 ) = λℓ(ξ 1 ) + (1 -λ)ℓ(ξ 2 ) > ℓ(ξ λ )
where the last inequality follows from the strict convexity of ℓ. Contradiction. Hence we deduce that û is convex not affine on [x 1 , x 2 ] as desired, which proves (3.8). This ends the proof of the lemma.

u + (t, x ′ , x) := û(x) for all (t, x ′ , x) ∈ R × R m × R d .
Then u + is a globally Lipschitz continuous viscosity solution of (1.1). Moreover any globally Lipschitz continuous viscosity solution u of (1.1) satisfies u + ≥ u.

Proof of Lemma 3.3

Recall that inf H(0, •) = H(0, p 0 ) < 0 with p 0 ∈ Int(K). Up to redefine H, we can assume that p 0 = 0. Then we have u + (0 R 1+m , •) > 0 on Ω. The graph of the solution u + is a cone that we will shrink like an umbrella λu + for λ > 1 to get a strict supersolution above u. Precisely, for any ε > 0, we have for large λ > 1

u ε,λ + := ε + λu + > u on R 1+m × Ω.
This is always possible, because u is globally Lipschitz. Then we now continuously unfold the umbrella (i.e. decrease λ). Either we reach the value λ = 1, or we have to stop at a heigher level λ * > 1 and get

u ε,λ * + ≥ u on R 1+m × Ω
and for any δ > 0, there exists

X δ = (X ′ δ , x δ ) ∈ R 1+m × R d such that (u ε,λ * -δ + -u)(X δ ) < 0. Case 1: x δ stays bounded Then we set u δ := u(• + (X ′ δ , 0
)) and up to extract a subsequence, we have x δ → x 0 , u δ → u 0 , and we get

u ε,λ * + ≥ u 0 with equality at (0, x 0 ) ∈ R 1+m × Ω. Because u ε,λ * + -u 0 = ε on R 1+m × ∂Ω,
and because u 0 is globally Lipschitz continuous, we deduce that (0, x 0 ) ∈ R 1+m × Ω. Then u ε,λ * + is a test function for u at (0, x 0 ), and this gives a contradiction because

λ * > 1. Case 2: |x δ | → +∞ Then we set u δ := u(• + X δ ) -u(X δ ) and u ε,λ * +,δ = u ε,λ * + (• + X δ ) -u(X δ )
, and up to extract a subsequence, we have u δ → u 0 and u ε,λ * +,δ → u ε,λ * +,0 , which gives u ε,λ * +,0 ≥ u 0 with equality at 0.

Here u ε,λ * +,0 is a strict supersolution of u t + H(Du) = 0 on R 1+m+d , because λ * > 1. Again this leads to a contradiction because u ε,λ * +,0 is affine.

Conclusion

We conclude that λ * > 1 is impossible, and then we always reach the value λ = 1. We get ε+u + ≥ u. Because this is true for every ε > 0, we can pass to the limit ε → 0 and get u + ≥ u. This ends the proof of the lemma.

u -(t, x ′ , x) := -û(-x) for all (t, x ′ , x) ∈ R × R m × R d .
Then u -is a globally Lipschitz continuous viscosity solution of (1.1). Moreover any globally Lipschitz continuous viscosity solution u of (1.1) satisfies u ≥ u -.

Proof of Lemma 3.4

The fact that û ∈ C 1 (Ω) and that H(0, Dû) = 0 implies that u -is also a viscosity solution of (1.1). We now want to show that

(3.14) u ≥ u -on R 1+m × Ω.
We start with short preliminaries and then proceed to a proof by contradiction.

Step 1: preliminaries

Step 1.1: representation formula for any solution of (1.1) From Lemma 2.1, we have for all t > s (3.15)

u(t, x) = min inf s∈[s,t), ỹ∈R m ×∂Ω 0 + t s L x - ỹ t -s dσ , inf ỹ∈R m ×Ω u(s, ỹ) + t 0 L x - ỹ t -s dσ .
Step 1.2: reformulation of û From Lemma 3.2, we have for

x ̸ = 0 û(x) = τ * ℓ(ξ * ) with ξ * := x τ * = τ * L(ξ ′ * , ξ * ) with L(ξ ′ * , ξ * ) = inf ξ ′ ∈R m L(ξ ′ , ξ * )
where ξ ′ * is uniquely defined from the strict convexity of L, inherited from (1.2) satisfied by H.

Step 2: comparison with u - We proceed by contradiction.

Step 2.1: first consequences (of the contrapositive) If (3.14) is false, then there exists some ε > 0 and some X0 ∈ R 1+m × Ω such that (3.16) ε + u( X0 ) < u -( X0 ).

From the global Lipschitz continuity of u and u -and their common zero value on R 1+m × ∂Ω, we deduce that there exists r > 0 and some ball Br = Br (0) ⊂ R d such that

(3.17) ε + u ≥ u -on R 1+m × Br .
Because of (3.16), we can increase the size of the ball up to some finite r * > 0 such that (3.17) holds true for r = r * , but not for any r > r * . This implies that for any δ > 0, there exists

X δ ∈ R 1+m × ( Br * +δ \ Br * ) such that ε + u(X δ ) < u -(X δ ). For X δ = (t δ , x ′ δ , x δ
) with x δ ∈ Br * +δ \ Br * , we define the points Xδ := (t δ , x ′ δ , 0) and Y δ := (0, 0, x δ ) and get

(3.18) ε + u δ (Y δ ) < u -(Y δ ) with u δ (X) := u(X + Xδ ).
Up to extract a subsequence, we have

u δ → u 0 , Y δ = (0, 0, x δ ) → (0, 0, y 0 ) = Y 0 ∈ {0 R 1+m } × ∂B r * .
Passing to the limit in (3.18), we get ε + u 0 (Y 0 ) ≤ u -(Y 0 ), and then passing to the limit in (3.17) for r = r * , we get

(3.19) ε + u 0 ≥ u - on R 1+m × Br * ε + u 0 = u - at Y 0 = (0, y 0 ) ∈ R 1+m × ∂B r * .
Step 2.2: getting a contradiction From Step 2.1, we have (3.20)

u -(Y 0 ) = -û(-y 0 ) = -τ * L( ξ * ) with ξ * = (ξ ′ * , ξ * ), ξ * = -y 0 τ * , τ * = τ * (-y 0 ), ℓ(ξ * ) = L(ξ ′ * , ξ * ).
We also have 0

= u 0 (τ * , τ * ξ ′ * , 0) = u 0 (Y 0 + τ * (1, ξ * )) ≤ u 0 (Y 0 ) + 0 -τ * L( ξ * ) dσ = (3.20) u 0 (Y 0 ) -u -(Y 0 ) = (3.19) -ε
where we have used the representation formula (3.15) in the third line. This gives a contradiction and ends the proof of the lemma.

Proof of Theorem 1.1: a Liouville-type result

The proof of the Liouville-type result is based on the barriers identified in the previous section. We make full use of the representation formula along characteristics trajectories, for which we show a key equality (Lemma 4.2), that is probably standard in other contexts. From this equality we deduce that the global solutions u are concave on {u < u + }, which will very soon imply the Liouville-type result.

Lemma 4.1 (Solution along an optimal trajectory)

Assume that H satisfies (1.2) and (3.1). Let u be a globally Lipschitz continuous solution of (1.1). Let

X 0 := (t 0 , x0 ) ∈ R 1+m × Ω be such that u(X 0 ) < u + (X 0 ).
Then for any τ > 0, there exists

X 1 := (t 0 -τ, ỹτ ) ∈ R 1+m × Ω such that u(X 0 ) = u(X 1 ) + τ L( ξτ ) with ξ := x0 -ỹτ τ and u(X 1 ) < u + (X 1 ).
Proof of Lemma 4.1

Step 1: splitting the representation formula in two parts Recall that from the representation formula, we have for X 0 = (t 0 , x0 ) and t < t 0

(4.1) u(X 0 ) = min(u t b (X 0 ), u t d (X 0 ))
with the boundary term

u t b (X 0 ) := inf s0∈[t,t0) ūs0 b (X 0 ) with ūs0 b (X 0 ) := inf ỹ∈R m ×∂Ω 0 + t s L x0 - ỹ t 0 -s 0 dσ
and the domain term

(4.2) u t d (X 0 ) := inf ỹ∈R m ×Ω u(t, ỹ) + t 0 L x0 - ỹ t 0 -t dσ .
Step 2: The boundary contribution We write x0 = (x ′ 0 , x 0 ) and define (τ 0 , ξ+ ) ∈ (0, +∞) × R m+d such that

u + (X 0 ) = û(x 0 ) = inf τ >0 τ ℓ( x 0 τ ) = τ 0 ℓ( x 0 τ 0 ) = τ 0 L( ξ+ ), ξ+ = (ξ ′ + , ξ + ), ξ + := x 0 τ 0
where we have used Lemma 3.2. Now for τ > 0 and s 0 := t 0 -τ , we get

ūs0 b (X 0 ) = inf ỹ∈R m ×∂Ω τ L x0 - ỹ τ = τ L( ξτ ) for ξτ := x0 -ỹτ τ for some ỹτ ∈ R m × ∂Ω.
Recall that by convexity, we have for any ξ ∈ R m+d with P + := DL( ξ+ ) = (0, p + ) with p + ∈ ∂K

(4.3) L( ξ) ≥ L( ξ+ ) + ( ξ -ξ+ ) • DL( ξ+ ) ≥ L( ξ+ ) + ( ξ -ξ+ ) • P + = ξ • P + -H(P + ) = ξ • P +
where we have used in the third line the fact that

P + = DL( ξ+ ) is equivalent to L( ξ+ ) = ξ+ • P + -H(P + ). Hence τ -1 ūs0 b (X 0 ) = L( ξτ ) ≥ ξτ • P + = x0 τ • P + = τ -1 u + (X 0 ) and then ūs0 b (X 0 ) ≥ u + (X 0 ), which implies (4.4) u t b (X 0 ) ≥ u + (X 0 ).
Step 3: The domain contribution Because u(X 0 ) < u + (X 0 ), we deduce from (4.1) and (4.4) that

(4.5) u(X 0 ) = u t d (X 0 ) for all t < t 0 .
Setting τ := t 0 -t > 0 and G(ỹ

) := u(t, ỹ) + τ L x0 - ỹ τ , we get u + (X 0 ) > u(X 0 ) = u t d (X 0 ) = inf ỹ∈R m ×Ω G(ỹ) = inf ỹ∈R m ×Ω G(ỹ)
where the infimum is reached for some ỹτ ∈ R m × Ω. We have used the superlinearity of L and the global Lipschitz regularity of u. Notice that (4.2) implies that ỹτ ̸ ∈ R m × ∂Ω, i.e. ỹτ ∈ R m × Ω. We get

P + • x0 = u + (x 0 ) > u(X 0 ) = u(t, ỹτ ) + τ L( ξτ ) ≥ u(t, ỹτ ) + τ P + • ξτ with ξτ := x0 - ỹ τ
where we have used (4.3) in the last inequality. For X 1 := (t, ỹτ ), this implies u + (X 1 ) > u(X 1 ), which is the desired inequality. This ends the proof of the lemma.

Lemma 4.2 (Key equality along a characteristic ξ-)

Assume that H satisfies (1.2) and (3.1). Let u be a globally Lipschitz continuous solution of (1.1). Let X 0 := (t 0 , x0 ) ∈ R 1+m × Ω be such that u(X 0 ) < u + (X 0 ). Then there exists ξ-∈ R m+d such that (4.6)

P -:= DL( ξ-) = (0, p -) with p -∈ ∂K such that for all τ > 0, we have

u(X 0 ) = u(X 0 -τ (1, ξ-)) + τ L( ξ-).
Remark 4.3 Notice that Lemma 4.2 does not exclude the existence of possibly different admissible values of ξ-, in particular at points X 0 where u is not C 1 .

Proof of Lemma 4.2

Step 1: the direction ξ2 remains fixed for τ 2 > τ 0

We apply two times Lemma 4.1. For any τ 0 , the first time applied from the point X 0 gives the existence of some direction ξτ0 X0 such that

u(X 0 ) = u(X 1 ) + τ 0 L( ξτ0 X0 ), X 1 := X 0 -τ 0 (1, ξτ0 X0 ), u(X 1 ) < u + (X 1 ).
For any τ 1 > 0, the second time applied from the point X 1 , gives the direction ξτ1 X1 such that

u(X 1 ) = u(X 2 ) + τ 1 L( ξτ1 X1 ), X 2 := X 1 -τ 1 (1, ξτ1 X1 ), u(X 2 ) < u + (X 2 ).
Hence for τ 2 := τ 0 + τ 1 , we get

(4.7) u(X 0 ) = u(X 2 ) + τ 1 L( ξτ1 X1 ) + τ 0 L( ξτ0 X0 ) ≥ u(X 2 ) + τ 2 ( ξ2 ) with ξ2 := τ -1 2 (τ 0 ξτ0 X0 + τ 1 ξτ1 X1 )
where the convex inequality remains strict if ξτ1

X1 ̸ = ξτ0 X0 . Now for X 2 := (t 2 , x2 ), we have t 2 = t 0 -τ 2 and ξ2 = x0 -x2 τ 2 with x2 ∈ R m
× Ω, and we get

u(X 2 ) + τ 2 L( ξ2 ) ≤ u(X 0 ) = u t2 d (X 0 ) = inf ỹ∈R m ×Ω u(t 0 -τ 2 , ỹ) + τ 2 L( ξ) with ξ := x 0 - ỹ τ 2
where we have used (4.5) for the first equality. Hence the infimum is reached for ξ = ξ2 and we have equality in (4.7). This implies ξτ1 X1 = ξτ0 X0 = ξ2 . This also shows that we can choose ξτ2 X0 = ξ2 and ỹτ2 := x2 , i.e. for all τ 2 > τ 0 , there exists

x2 ∈ R m × Ω such that X 2 = (t 0 -τ 2 , x2 ) satisfies (4.8) u(X 0 ) = u(X 2 ) + τ 2 L( ξ2 ) with ξ2 = x2 -x 2 τ 2 = ξτ2 X0 = ξτ0 X0 .
Because this is true for arbitrary τ 2 > τ 0 , this shows that we can find a map τ → ξτ X0 which is constant equal to ξ 2 .

Step 2: proof that ξ2 = ξsatisfies (4.6) We now want to show that ξ2 = ξis indeed specific, i.e. satisfies (4.6). By assumption, we have with

X 2 = X 0 -τ 2 (1, ξ2 ) u + (X 0 ) > u(X 0 ) = u(X 2 ) + τ 2 L( ξ2 ) ≥ u -(X 2 ) + τ 2 L( ξ2 ) and then u + (X 0 ) -u -(X 0 ) > u -(X 2 ) -u -(X 0 ) + τ 2 L( ξ2 ), i.e. for P τ2 -:= Du -(X 2 ) with H(P τ2 -) = 0 L( ξ2 ) ≤ A τ 2 + τ -1 2 {u -(X 0 ) -u -(X 2 )} with A := u + (X 0 ) -u -(X 0 ) > 0 ≤ A τ 2 + Du -(X 2 ) • X 0 -X 2 τ 2 ≤ A τ 2 + P τ2 -• ξ2
where in the second line we have used the concavity of u -. For any P -∈ R m+d such that H(P -) = 0, we set

S A τ (P -) := ξ ∈ R m+d , L( ξ) < P -• ξ + A τ .
Setting ξ-:= DH(P -), we have P -= DL( ξ-) and

L( ξ-) = ξ-• P --H(P -) = ξ-• P -. Therefore L( ξ) ≥ L( ξ-) + ( ξ -ξ-) • DL( ξ-) = ξ • P -, i.e.
L( ξ) ≥ P -• ξ with equality only at ξ = ξ-:= DH(P -) with H(P -) = 0.

Hence from the strict convexity of L, we deduce that for a fixed P -∈ R m+d , we have dist( ξ-, S A τ (P -)) → 0 as τ → +∞, if ξ-:= DH(P -),

H(P -) = 0.
Indeed this convergence is event true and uniform if P -stays in some compact set. Hence we have also dist( ξτ2

- , S A τ (P τ2 -)) → 0 as τ 2 → +∞, with        ξτ2 -:= DH(P τ2 -), H(P τ2 -) = 0, P τ2 -= DL( ξτ2 -), P τ2 -:= Du -(X 2 ) = Du -(X 0 -τ 2 (1, ξ2 )).
Because ξ2 ∈ S A τ2 (P τ2 -) with ξ2 fixed, we deduce that

| ξ2 -ξτ2 -| → 0, P τ2 -→ P -= DL( ξ2 ) = Du -(-(1, ξ2 )) = (0, p -), p -∈ ∂K.
This shows that ξτ2 -→ ξ-= ξ2 satisfies (4.6). This ends the proof of the lemma.

Lemma 4.4 (Property of global solutions)

Assume that H satisfies (1.2) and (3.1). Let u be a globally Lipschitz continuous solution of (1.1) and let

X 0 := (t 0 , x0 ) ∈ R 1+m × Ω.
If u(X 0 ) < u + (X 0 ), then there exists P -= (0, p -) with p -∈ ∂K such that we have

u(t, x) ≤ u(t 0 , x0 ) + P -• (x -x0 ) for all X := (t, x) ∈ R × R m+d .
Proof of Lemma 4.4 From Lemma 4.2 applied to the point X 0 , we get the existence of some characteristic ξ-, such that P -= DL( ξ-) = (0, p -) with p -∈ ∂K, and for all τ > 0

u(X 0 ) = u(Y 0 ) + τ L( ξ-) with Y 1 := (t 1 , ỹ1 ) = (t 0 -τ, x0 -τ ξ-).
From the representation formula, we also have for

X ∈ R 1+m+d u(X) ≤ u(Y 0 ) + τ L( ξ1 ) with ξ1 := x -ỹ1 τ = ξ-+ ξ, ξ := x -x0 τ . Hence we get u(X 0 ) -u(X) = u(Y 0 ) + τ L( ξ-) -u(X) ≥ τ L( ξ-) -τ L( ξ1 ) ≥ -τ 1 0 dσ ξ • DL( ξ-+ σ ξ) → -(x -x0 ) • DL( ξ-) as τ → +∞ = -(x -x0 ) • P -
which gives the desired result. This ends the proof of the lemma.

Corollary 4.5 (Characterization of global solutions)

Assume that H satisfies (1.2) and (3.1). Let u be a globally Lipschitz continuous solution of (1.1). Then u(t, x ′ , x) = u(0, 0, x). Moreover either u is concave, or u is the minimum of the convex function û and of a concave function and K u := {û = u} is strictly convex, possibly unbounded, and contains some small ball B r (0). Moreover we have

(4.9) û(y) ≥ u(0, 0, x + y) -u(0, 0, x) ≥ -û(-y) for all x, y ∈ R d .
Proof of Corollary 4.5

Step 1: main result Because u is Lipschitz continuous in time-space, its time-space gradient Du := (u t , Du) is defined almost everywhere on R 1+m × Ω. If X 0 is such a point where Du is defined, then we have two cases. Case 1: u(X 0 ) = u + (X 0 ) Because we know that u ≤ u + , and

∂ t u + = 0 = D x ′ u + , we deduce also that ∂ t u(X 0 ) = 0 = D x ′ u(X 0 ). Case 2: u(X 0 ) < u + (X 0 )
Then from Corollary 4.5, we have for all

X = (t, x) ∈ R 1+m × Ω u(t, x) ≤ u(t 0 , x0 ) + P -• (x -x0 ), P -= (0, p -), p -∈ ∂K.
This implies that Du(X 0 ) = (0, 0, p -).

Conclusion

From both cases, we then conclude that ∂ t u = 0 = D x ′ u for a.e. X 0 ∈ R 1+m × Ω. Hence the Lipschitz continuity of u implies

u(t, x ′ , x) = u(0, 0, x) for all (t, x ′ , x) ∈ R × R m × Ω.
Step 2: further properties of the solution

Step 2.1: first properties In this step, we make abuse of notation, and simply note u(x) = u(t, x ′ , x) = u(0, 0, x) and u

+ (x) = u + (t, x ′ , x) = u + (0, 0, x). From Lemma 4.4, if x 0 ∈ {u < u + } ∩ Ω, then there exists p -= p -(x 0 ) ∈ ∂K such that u(x) ≤ u(x 0 ) + (x -x 0 ) • p -(x 0 ) for all x ∈ Ω. We define for x ∈ Ω w(x) := inf x0∈{u<u+}∩Ω {u(x 0 ) + (x -x 0 ) • p -(x 0 )}
which is concave by construction, and also viscosity solution of H(0, Dw) = 0 on R d . By construction of w, we also have for all x, y ∈ Ω (4.10) û(y) := sup

p∈∂K p • y ≥ w(x + y) -w(x) ≥ inf p∈∂K p • y ≥ -û(-y).
By construction of w, we also have u ≤ w and u = w on {u < û}, i.e. and then w(0) ≥ u(0) = 0. Notice that (4.10) implies (4.9). We now distinguish two cases. Case 1: w(0) = 0 Then w ≤ û and u = w and u is concave. Case 2: w(0) > 0 Then K u := {u = û} = {w ≥ û} = {û -w ≤ 0} ⊂ Ω. Because û -w is convex, we deduce that K u is also convex. Moreover the condition w(0) > 0 = û(0) implies that the origin 0 is in the interior of K u .

Step 2.2: strict convexity of K u when 0 ∈ Int(K u ) Assume by contradiction that K u is not strictly concave. Hence there exists [

x 1 , x 2 ] ⊂ ∂K u with x 1 ̸ = x 2 .
Because ∂K u ⊂ {û = w}, we see that the equality of a convex function û and a concave function w, implies that û = w = affine on [x 1 , x 2 ]. Because û is known to be strictly convex (see (3.8)), except along lines Re for some e ∈ S d-1 , we deduce that there exists such e such that [x 1 , x 2 ] ⊂ Re. Moreover, the case 0 ∈ (x 1 , x 2 ) is impossible, because we know that p 0 = 0 ∈ Int(K), and then û is not linear on (x 1 , x 2 ) ∋ 0. We conclude that [x 1 , x 2 ] ⊂ (0, +∞)e for such e. Then there exists an hyperplan Π tangential to the convex K u at x 2 ∈ ∂K u which contains the vector e. Hence 0 ∈ Π, which is impossible because 0 ∈ Int(K u ). We conclude that K u is strictly convex.

Step 2.3: conclusion Hence u is convex on the the strictly convex set K u ⊃ B r (0) for some small r > 0, and concave outside K u . This ends the proof of the corollary.

Proof of Theorem 1.1 From Dichotomy Lemma 3.1, we have either u(t, x ′ , x) = p 0 • x with 0 = H(0, p 0 ) = inf H(0, •), or inf H(0, •) < 0, and we get u(t, x ′ , x) = u(0, 0, x) from Corollary 4.5. This ends the proof of the theorem.

Explicit characteristics for concave 1-homogeneous solutions

We now provide a refinement of Lemma 4.2, where we characterize explicitely the set of characteristics ending at a point x 0 where the key equality holds. This result is of independent interest and is done for positively 1-homogeneous concave solutions. A large part of this section will be also used in the proof of Theorem 1.3 later in Section 7.

Lemma 5.1 (Explicit characteristics for concave 1-homogeneous solutions) Assume that H satisfies (1.2) and (3.1), and let K := {H(0, •) = 0}. Let w be a globally Lipschitz continuous solution of (1.1) which is (positively) 1-homogeneous. Let x 0 ∈ Ω be such that w(x 0 ) < û(x 0 ) = u + (0, 0, x 0 ). i) (General result) Then w is concave.

Let us consider the following set

Ξ x0 -:= ξ -∈ R d , w(x 0 ) = w(x 0 -τ ξ -) + τ ℓ(ξ -) for all τ > 0
with ℓ defined in (3.5). Then we have

Ξ x0 -= D x H(0, A 0 ) with A 0 := D + w(x 0 ) ∩ ∂K ̸ = ∅, D + w(x 0 ) ⊂ K w ⊂ K
where there exists a unique compact convex set K w (only depending on w) such that

w(x) = inf p∈Kw p • x and D + w(x 0 ) = {p ∈ K w , p • x 0 = w(x 0 )} . ii) (Properties of E w ) Let E w := K w ∩ ∂K.
Then E w ̸ = ∅ and we also have the convex hull reconstruction Moreover, for any p -∈ E w , and ξ -:= D x H(0, p -), we have (5.2) Ξ x0 -= {ξ -} for all x 0 = -λξ -with λ > 0, and w(-ξ -) = -ℓ(ξ -).

We also have

(5.3) E w = {p -∈ ∂K, w(-ξ -) = -û(ξ -) for ξ -:= D x H(0, p -)} . Remark 5.2
The notation E w is chosen to think to it as the set of exposed points of K w .

Proof of Lemma 5.1

Step 1: preliminaries on w

Step 1.1: general preliminaries Notice that w is in particular a solution to the Liouville-type problem. Hence by Corollary 4.5, either w is concave, or the set {w = û} has the origin in its interior, which implies w = û. This is impossible because w(x 0 ) < û(x 0 ). We conclude that w is concave, and then w ≥ -û(-•). Therefore

(5.4) v(x) := -w(-x) ≤ û(x) = sup p∈K p • x
is convex, and its the Legendre-Fenchel conjugate is with

K v := D -v(0) v * (p) := sup x∈R d {p • x -v(x)} = 0 if p ∈ K v +∞ otherwise = (+∞) • 1 R d \Kv (p)
where (5.4)

implies v * ≥ û * = (+∞)•1 R d \K and then K v ⊂ K. This means that v(x) = sup p∈R d {p • x -v * (p)} = sup p∈Kv p • x, i.e.
(5.5)

w(x) = inf p∈Kw p • x with the compact convex set K w := D + w(0) = K v ⊂ K.
Step 2: properties of every characteristics ending at x 0 We now choose x 0 ̸ = 0 and consider some ξ -∈ R d such that (5.6) w(x 0 ) = w(x 0 -τ ξ -) + τ ℓ(ξ -) for all τ > 0.

The fact that ℓ(0) = -inf H(0, •) > 0 excludes the case ξ -= 0.

Step 2.1: far away consequences From (5.6), we have ℓ(ξ -) = w(x 0 ) -w(x 0 -τ ξ -) τ → -w(-ξ -). Recall that w(x) ≥ -û(-x), and then

ℓ(ξ -) = -w(-ξ -) ≤ û(ξ -) = inf τ >0 τ ℓ( ξ - τ ) ≤ ℓ(ξ -)
which shows that we have equality in each inequality. Hence w(x) ≥ -û(-x) with equality at x := -ξ -, and τ * (ξ -) = 1.

The fact that û ∈ C 1 (Ω) implies that w has a derivative at -ξ -̸ = 0, which satisfies (5.7)

p -:= Dw(-ξ -) with K w ∋ Dw(-ξ -) = Dû(ξ -) ∈ ∂K and p -= Dℓ(ξ -)
where we have identified the gradient in K w using (5.5), and in ∂K using (3.3). Finally, we have used (3.6) in the last equality. Hence (5.8)

ξ -= D x H(0, p -) with p -∈ K w ∩ ∂K.
Step 2.2: close consequences

We also have -w(-ξ -) = û(ξ -) = ℓ(ξ -) = w(x 0 ) -w(x 0 -τ ξ -) τ = w( x 0 τ ) -w( x 0 τ -ξ -), i.e. w( x 0 τ ) = w( x 0 τ -ξ -) -w(-ξ -) = x 0 τ • p -+ o x 0 τ
where we have used (5.7) to identify p -. This gives w(x 0 ) = x 0 • p -+ τ • o x 0 τ , which implies in the limit τ → +∞ w(x 0 ) = x 0 • p -with p -∈ K w ∩ ∂K.

From (5.5), we also deduce that p -∈ D + w(x 0 ), and then (5.9) p -∈ A 0 := D + w(x 0 ) ∩ ∂K.

Step 2.3: a property of D + w(x 0 ) We first recall that for p ∈ D + w(x 0 ), we have w(x 0 + τ h) -w(x 0 ) ≤ p • (τ h), and in the limit τ → +∞, we get w(h) ≤ p • h which shows that p ∈ D + w(0) = K w . Hence D + w(x 0 ) ⊂ K w . We want to show that (5.10)

D + w(x 0 ) = K x0 w with K x0 w := {p ∈ K w , p • x 0 = w(x 0 )} .
Let p ∈ D + w(x 0 ). Then we have w(x 0 + h) -w(x 0 ) ≤ p • h. For h = ±εx 0 with ε > 0, we get in the limit ε → 0 + the two inequalities ±w(x 0 ) ≤ ±p • x 0 , which shows that

D + w(x 0 ) ⊂ K x0 w .
Conversely, for any p 0 ∈ K x0 w , we have

w(x 0 + h) -w(x 0 ) = inf p∈Kw p • (x 0 + h) -p 0 • x 0 ≤ p 0 • (x 0 + h) -p 0 • x 0 = p 0 • h
and then p 0 ∈ D + w(x 0 ), i.e. K x0 w ⊂ D + w(x 0 ) which is the reverse inequality. We conclude to (5.10).

Step 3: reaching the set DH(0, A 0 ) with A 0 = D + w(x 0 ) ∩ ∂K First notice that Lemma 4.2 applies to w and shows that ∅ ̸ = Ξ x0 -, while we know from Step 2 that Ξ x0

-⊂ D x H(0, A 0 ). Therefore A 0 ̸ = ∅. Because D + w(0) = K w ⊂ K, we also know that w is a subsolution at the origin. Moreover if D -w(0) ̸ = ∅, then w(x) = p • x with p ∈ K w has to be a solution for x ̸ = 0, and then is also a supersolution at x = 0. Hence w is a solution on the whole space of

H(0, Dw) = 0 on R d .
From the representation formula, we have for any τ > 0

w(x 0 ) = inf ξ∈R d {w(x 0 -τ ξ) + τ ℓ(ξ)} = inf ξ∈R d inf p∈Kw p • (x 0 -τ ξ) + τ ℓ(ξ) = inf p∈Kw p • x 0 -τ sup ξ∈R d {p • ξ -ℓ(ξ)} = inf p∈Kw {p • x 0 -τ H(0, p)} ≤ inf p∈A0 {p • x 0 -τ H(0, p)} = w(x 0 )
where using (5.10), we have

A 0 = K x0 w ∩ ∂K ̸ = ∅.
This shows that the infimum in p is reached for any p -∈ A 0 . Then in the third line the suppremum in ξ is reached for ξ = ξ -= Dℓ(p -) uniquely associated to each p -. Hence the infimum in the first line is reached for such ξ = ξ -. This shows that for all τ > 0, we have

w(x 0 ) = w(x 0 -τ ξ -) + τ ℓ(ξ -) for all ξ -:= D x H(0, p -) with p -∈ A 0 .
Step 4: proof of (5.2) For any x 1 ̸ = 0, we know from Step 3, that

A 1 := D + w(x 1 ) ∩ ∂K ̸ = ∅, and A 1 ⊂ E w := K w ∩ ∂K. Hence E w ̸ = ∅.
and we can consider any p -∈ E w and ξ -:= D x H(0, p -). Then p -= Dℓ(ξ -) and ℓ(ξ -)+H(0, p -) = p -•ξ -. Because inf H(0, •) < 0 and H is strictly convex, we first deduce that ξ -̸ = 0. We then notice that for all λ > 0 and x 0 := -λξ -, we have (5.11) w

(x 0 ) ≥ -û(-x 0 ) = -λû(ξ -) = -λℓ(ξ -) = -λp -• ξ -= p -• x 0 ≥ inf p∈Kw p • x 0 = w(x 0 )
which shows that we have equality in all inequalities. Because û ∈ C 1 (Ω), we deduce that w is C 1 at x 0 and we get

D + w(x 0 ) = {p * } , p * = Dû(-x 0 ) = Dû(ξ -). Moreover, we have ℓ(ξ -) = û(ξ -) = inf τ >0 τ ℓ( ξ - τ
), where we have used (3.6) which also shows that τ * (ξ -) = 1 and then that p * = Dû(ξ -) = Dℓ(ξ -) = p -. Therefore Ξ x0 -= {ξ -}. Moreover we have w(x 0 ) = w(x 0τ ξ -) + τ ℓ(ξ -), and then in the limit τ → +∞, we get w(-ξ -) = -ℓ(ξ -).

Step 5: proof of (5.1) Consider now some exposed point p 0 ∈ exp K w . Hence there exists a unit vector n ∈ S d-1 such that for Π ± p0,n := p ∈ R d , (p -p 0 , ±n) ≥ 0 , we have

K w ⊂ Π - p0,n , K w ∩ Π + p0,n = {p 0 } . Then Lemma 11.3 shows that v(x) = -w(-x) satisfies ∂v(n) = {p 0 }
i.e. that v is C 1 at n. This means that w is C 1 at -n with Dw(-n) = p 0 . Because the PDE is satisfied at -n, this shows that p 0 ∈ ∂K, and therefore, all exposed points of K w are on ∂K. From ii) of Lemma 11.2 in the Appendix, we deduce that we have the following closure of the convex hull property

K w = co(exp K w ) with exp K w ⊂ ∂K
which also implies (using the boundedness of A 2 to assert that co(A 2 ) = co(A 2 ), see for instance Theorem 1.4.3 on page 31 in [START_REF] Hiriart-Urruty | Fundamentals of Convex Analysis[END_REF])

(5.12)

K w = co(exp K w ) with exp K w ⊂ (K w ∩ ∂K) = E w .
Hence K w = co(E w ). Furthermore point iii) of Lemma 11.2 shows that

w(x) = inf p∈exp Kw p • x.
From the inclusion in (5.12), we deduce that

w(x) = inf p∈Ew p • x.
Step 6: proof of (5.3)

We set (5.11) for λ = 1 shows that (5.13)

E ′ w := {p -∈ ∂K, w(-ξ -) = -û(ξ -) for ξ -:= D x H(0, p -)} . Notice that relation
K w ∩ ∂K = E w ⊂ E ′ w .
Conversely, choose any p -∈ ∂K and ξ -:= D x H(0, p -). Because K is strictly convex and ξ -is orthogonal to K at p -, we get

ξ -• (p -p -) < 0 for all p ∈ K\ {p -} . Now if p -̸ ∈ K w ⊂ K, then we have on the compact set K w , the strict inequality sup p∈Kw ξ -• (p -p -) < 0, i.e. w(-ξ -) = inf p∈Kw p • (-ξ -) > p -• (-ξ -) = -û(ξ -)
where we have used (3.3) and (3.6) for the last inequality with p -= Dû(ξ -) and τ * (ξ -) = 1. This shows that p -̸ ∈ E ′ w , and then (∂K\K w ) ∩ E ′ w = ∅, i.e.

E w ⊃ E ′ w .
With (5.13), this shows that E w = E ′ w , which is (5.3). This ends the proof of the lemma.

Key equality for localized solutions

In this section we present a localization of Lemma 4.2. For a cylinder Q 0 ⊂ R 1+m+d , we consider solutions u(t, x ′ , x) of (6.1)

u t + H(Du) = 0 on Q 0 \Γ u(t, x ′ , 0) = g(t, x ′ , 0) on Q 0 ∩ Γ.
We also set for τ, ρ > 0

(6.2) Q τ,2ρ := (-τ, 0) × B 2ρ , Γ τ,2ρ := Q τ,2ρ ∩ Γ.

Lemma 6.1 (Key equality for localized solutions)

We use notation introduced in (6.2). Assume that H satisfies (1.2) and (3.1). For τ, ρ > 0, let

u : Q 0 → R with Q 0 := (-τ, 0) × B 2ρ ⊂ R × R m+d
be a Lipschitz continuous solution of (6.1), of Lipschitz constant

L > 0. Define R L ≥ 1 such that ξ ∈ R m+d , L( ξ) ≤ L|(1, ξ)| ⊂ B R L
and assume that

(6.3) ρ ≥ τ R L and set κ := sup x∈∂B1 τ * (x) > 0, α 0 := inf n∈S d-1 , ξ0=(2κ) -1 n {ℓ(ξ 0 ) -ξ 0 • Dℓ(ξ 0 )} > 0
for τ * defined in (3.6). For this local problem, we define the Dirichlet boundary (including the initial data)

Γ D τ,2ρ := Γ τ,2ρ ∪ Γ init τ,2ρ
with Γ init τ,2ρ := {-τ } × B 2ρ . Then we have the representation formula

(6.4) u(X) = inf s∈(0,τ ], ξ∈R m+d , Y :=X-s(1, ξ)∈Γ D τ,2ρ u(Y ) + sL( ξ) , for all X = (0, x) ∈ {0} × B ρ . Moreover for r ∈ (0, ρ], if X 0 = (0, x0 ) ̸ ∈ Γ with x0 = (x ′ 0 , x 0 ) ∈ B r , and
(6.5) u(X 0 ) < u + (X 0 ) + inf s∈(0,τ ] inf Γ s,r+sR L u + α 0 max {0, s -2κ|x 0 |}
then there exists ξ-∈ B R L such that for all s ∈ [0, τ ], we have

(6.6) u(X 0 ) = u(X 0 -s(1, ξ-)) + sL( ξ-).
Remark 6.2 Lemma 6.1 allows to connect the behaviour of the solution at short distances (for small s) to large distances (large s). This result is crucial to show the uniqueness of the blow-up limit of the solutions at the boundary Γ (in Theorem 1.3).

Proof of Lemma 6.1

We first notice that because the solution u is Lipschitz, and DH is continuous, we have finite propagation of information. Then it is possible to justify representation formula (6.4), which itself implies that minimizers ξs have to satisfy ξs ∈ B R L . Moreover for

X 0 = (0, x0 ) ̸ ∈ Γ with x0 = (x ′ 0 , x 0 ) ∈ B r and r ∈ (0, ρ], if Y = (-s, y ′ , 0) ∈ Γ τ,2ρ is such that u(X 0 ) = u(Y ) + sL( ξs ), Y = X 0 -s(1, ξs ), s ξs = (x ′ 0 -y ′ , x 0 ).
then we have s ∈ (0, τ ]. On the one hand, we have for f (s, x 0 ) := sℓ( x 0 s )

sL( ξs ) ≥ sℓ( x 0 s ) = f (s, x 0 ) ≥ inf s ′ >0 f (s ′ , x 0 ) = f (τ * (x 0 ), x 0 ) = û(x 0 ) = u + (X 0 )
where the map τ * is positively 1-homogeneous. From Step 4 of the proof of Lemma 3.2, we know that the map f (•, x 0 ) : (0, +∞) → R is strictly convex. If 2κ|x 0 | ≥ 2τ * (x 0 ), then we have for

α := f ′ s (2κ|x 0 |, x 0 ) = ℓ(ξ 0 ) -ξ 0 • Dℓ(ξ 0 ) ≥ α 0 > 0 for ξ 0 := (2κ) -1 x 0 |x 0 |
with α 0 defined in (6.3). Hence

f (s, x 0 ) ≥ f (2κ|x 0 |, x 0 ) + α(s -2κ|x 0 |) ≥ f (τ * (x 0 ), x 0 ) + α(s -2κ|x 0 |) = u + (X 0 ) + α(s -2κ|x 0 |) and then sL( ξs ) ≥ f (s, x 0 ) ≥ u + (X 0 ) + α 0 max {0, s -2κ|x 0 |} . Hence u(X 0 ) = u(Y ) + sL( ξs ) ≥ inf ([-s,0)×B sR L (X0))∩Γτ,2ρ u + u + (X 0 ) + α 0 max {0, s -2κ|x 0 |} ≥ u + (X 0 ) + inf Γ s,r+sR L u + α 0 max {0, s -2κ|x 0 |} i.e. u(X 0 ) ≥ u + (X 0 ) + inf s∈(0,τ ] inf Γ s,r+sR L u + α 0 max {0, s -2κ|x 0 |} .
Then condition (6.5) prevents to have

Y ∈ Γ τ,2ρ . Hence Y ∈ Γ init τ,2ρ \({-τ } × Γ).
As in the proof of Lemma 4.2, we show that we can choose ξs independent on s for s ∈ (0, τ ], and this shows (6.6). This ends the proof of the lemma. and we can similarly find sequences b ε i → 0 + which go to zero sufficiently slowly such that

u b ε i ε i (t, x ′ , x) → ûi 0 (x) with ûi 0 := lim µ→0 + (u i ) µ .
Hence up to redefine the sequences ε i (by a ε i ε i → 0 or b ε i ε i → 0), and redefine the limit u i , we can assume that for i = 1, 2 (7.4)

u ε i (t, x ′ , x) → ûi (x) as ε i → 0
with ûi solution of (1.1), where ûi is 1-homogeneous (i.e. ûi (αx) = αû i (x) for all α ≥ 0), and either ûi = û, or ûi is concave and then from (5.1), we see that there exists a compact set E i ⊂ ∂K such that

(7.5) ûi (x) = inf p∈E i p • x = inf p∈K i p • x and K i = co(E i ), E i = K i ∩ ∂K.
Moreover, we have the following property. For any η > 0, there exists ε η > 0 such that for all ε i < ε η , we have for X = (X ′ , x) and X ′ := (t, x ′ ) (7.6)

|u ε i (X ′ , x) -ûi (x)| ≤ η for all (X ′ , x) ∈ [-1, 1] × B 1 .
Step 3: framework for a proof by contradiction

Step 3.1: position of the problem We want to show that û1 = û2 . Assume by contradiction that û1 ̸ ≡ û2 , i.e.

(

< û2 (e) for some unit vector e ∈ S d-1 .

In particular this forces û1 to be concave as in (7.5), and û2 is either equal to û or also concave as in (7.5). We distinguish two cases.

Case A: û2 is concave Then, with notation in (7.5), we have

E 1 ̸ = E 2 . If E 1 ⊂ E 2
, then (7.5) implies û1 ≥ û2 which is not the case by assumption (7.7). Hence E 1 \E 2 ̸ = ∅. We now choose

p -∈ E 1 \E 2 , ξ -:= D x H(0, p -) ̸ = 0.
Then from (5.3), we get (using

-û(ξ -) ≤ û2 (-ξ -)) û1 (-ξ -) = -û(ξ -) < û2 (-ξ -) and û1 (-ξ -) = -û(ξ -) < 0 < û(-ξ -).
Case B: û2 = û Then consider any

p -∈ E 1 , ξ -:= D x H(0, p -) ̸ = 0. Again from (5.3), we get û1 (-ξ -) = -û(ξ -) < 0 < û(-ξ -) = û2 (-ξ -).

Conclusion

Hence in both cases, there exists p -∈ E 1 with ξ -= D x H(0, p -) ̸ = 0 such that for x 1 := -λ 1 ξ -for λ 1 > 0, we have from ii) of Lemma 5.1

(7.8) Ξ x1 -= Ξ x1 -(û 1 )
= {ξ -} and û1 (x 1 ) < min û(x 1 ), û2 (x 1 ) .

Step 3.2: framework for ε 1 < ε 2 We assume (7.8). The idea of the proof consists to use the key equality along a characteristic of velocity ξ - for û1 approached by u ε1 , and to propagate the information far away where now u ε1 behaves like u ε2 , i.e. like û2 . This will lead to a contradiction because û1 (-ξ -) ̸ = û2 (-ξ -).

We set ε = ε 2 , and get from (7.6) that

|u ε (X ′ , x) -û2 (x)| ≤ η for all (X ′ , x) ∈ [-1, 1] × B 1 .
For

µ := ε 1 ε 2 ∈ (0, 1), we have |u µε (X ′ , x) -û1 (x)| ≤ η for all (X ′ , x) ∈ [-1, 1] × B 1 , i.e. (7.9) |u ε (X ′ , x) -û1 (x)| ≤ µη for all (X ′ , x) ∈ [-µ, µ] × B µ .
The limit σ → +∞ gives û1 (-ξ 0 -) + L( ξ0 -) = 0 with ξ0 -:= (ξ 0′ -, ξ 0 -). Step 4.5: identification of the limit characteristic ξ 0

- Now we have ℓ(ξ 0 -) = inf ξ ′ ∈R m L(ξ ′ , ξ 0 -) ≤ L(ξ 0′ -, ξ 0 -) = -û 1 (-ξ 0 -) ≤ û(ξ 0 -) = inf τ >0 τ ℓ( ξ 0 - τ )
≤ ℓ(ξ 0 -). Hence we have equality in all inequalities and

(7.15) -û 1 (-ξ 0 -) = L( ξ0 -) = ℓ(ξ 0 -). Now for x 1 = -λ 1 ξ -, relation (7.14) shows that ξ 0 -∈ Ξ x1 -= Ξ x1 -(û 1 ) = {ξ -}, i.e. ξ 0 -= ξ -= D x H(0, p -) with p -∈ E 1 = K 1 ∩ ∂K.
Step 4.6: contradiction Now (7.13) also shows that û2 (-ξ 0 -) = -L( ξ0 -) = û1 (-ξ 0 -) where the last equality follows for instance from (7.15). For

x 1 = -λ 1 ξ -= -λ 1 ξ 0 -, we deduce that û2 (x 1 ) = û1 (x 1 )
which is in contradiction with (7.8).

Step 5: conclusion We conclude that û1 = û2 , and then 1-homogeneous blow-up limits coincide. This implies the uniqueness of the blow-up limit, which also has to be 1-homogeneous. This means that for X = (t, x ′ , x) ε -1 u(εX) → û1 (x) locally on compact set of R 1+m+d as ε → 0 with û1 (αx) = αû 1 (x) for all α ≥ 0. This shows (1.5) and ends the proof of the theorem.

Proof of Theorem 1.4: strong traces of directional derivatives

This section is devoted to the proof of Theorem 1.4. We start with the following building block result.

Proposition 8.1 (Strong convergence of the blow-up gradient at the boundary)

We work under the assumptions of Theorem 1.3 with (λ, p ′ ) replaced by (λ 0 , p ′ 0 ). In particular, there exists (λ 0 , p ′ 0 ) ∈ R × R m and a 1-homogeneous function u 0 : R 1+m+d → R such that for X = (t, x ′ , x), we have

u(X) = u 0 (X) + o(|X|) as X → 0 in R 1+m+d , with u 0 (X) = λ 0 t + p ′ 0 • x ′ + u 0 (0, 0, x).
Recall that u is a Lipschitz continuous viscosity solution of u t + H(Du) = 0 in a neighborhood of 0 in R 1+m × Ω, with H strictly convex. Then for ε > 0, the blow-up u ε (X) := ε -1 {u(εX) -u(0)} enjoys the following strong convergence of its time-space gradient towards a 0-homogeneous function

(8.1) (u ε t , Du ε ) → (λ 0 , Du 0 ) in L 1 loc (R 1+m+d ; R 1+m+d ) as ε → 0.
Proof of Proposition 8.1

Step 1: preliminaries Because Du ε is uniformly bounded, we can use standard compactness of (Young) measures, and extract a subsequence (still denoted by ε) and find a family of probability measures ν X on R m+d for X ∈ R 1+m+d such that for any continuous function F : R m+d → R, we have

F (Du ε ) → F := ⟨ν X , F ⟩ = R m+d F (P )dν X (P ) in L ∞ loc (R 1+m+d ) weak - * .
Notice that we also have for any measurable function G : R 1+m+d × R m+d → R, such that P → G(X, P ) is continuous on compact sets uniformly in X, and X → G(X, P ) is bounded uniformly for bounded P , we also have (see for instance [START_REF] Ball | Partial differential equations and continuum models of phase transitions[END_REF])

(8.2) G(X, Du ε ) → Ḡ(X) := ⟨ν X , G(X, •)⟩ = R m+d G(X, P )dν X (P ) in L ∞ loc (R 1+m+d ) weak - * .
Because u is Lipschitz continuous, we have in particular almost everywhere

u ε t + H(Du ε ) = 0, u 0 t + H(Du 0 ) = 0.
Step 2: limit of a nonnegative integral We set (8.3) 0 ≤ Ψ(X, P ) := H(P ) -H(P 0 ) -(P -P 0 ) • DH(P 0 ) with

P 0 = Du 0 (X)
where the nonnegativity of Ψ (a.e. in X) follows from the convexity of H. Now for any test function 0 ≤ φ ∈ C ∞ c (R 1+m+d ), we consider the following integral

0 ≤ I ε := R 1+m+d φ(X) Ψ(X, Du ε (X)) dX.
On the other hand, setting B ε := -(Du ε -Du 0 ) • DH(Du 0 ), we get (8.4)

I ε = R 1+m+d φ B ε + H(Du ε ) -H(Du 0 ) dX = R 1+m+d φB ε + φ t (u ε -u 0 ) dX
where we have used the PDE for the last line. From the strong uniform convergence of u ε towards u 0 , we also get φ t (u ε -u 0 ) dX → 0. On the other hand, we split M := DH(Du 0 ) ∈ L ∞ in two parts M δ := ρ δ ⋆ M and Mδ := M -M δ , where ρ δ is a standard mollifier. We get M δ ∈ C 1 with bounded gradient, and Mδ → 0 in L 1 loc (R 1+m+d ). Hence we write

B ε = -(Du ε -Du 0 ) • M = -(Du ε -Du 0 ) • M δ -(Du ε -Du 0 ) • Mδ =: B ε δ + Bε δ and get R 1+m+d φB ε δ dX = R 1+m+d (u ε -u 0 ) div(φM δ ) dX → 0 as ε → 0. We also have | R 1+m+d φ Bε δ dX| ≤ 2Lip(Du) R 1+m+d φ| Mδ | dX → 0 as δ → 0.
This shows that we also get φB ε dX → 0. Therefore we get

I ε → 0 = I 0 := R 1+m+d
φ(X)Ψ(X) dX with 0 ≤ Ψ(X) = R m+d Ψ(X, P ) dν X (P ) for a.e. X ∈ R 1+m+d , where the nonnegativity of Ψ follows again from the convexity of H.

Step 3: consequence

Step 1 implies φ Ψ = 0 a.e. for all test function φ ≥ 0. Therefore we get Ψ = 0 a.e. on R 1+m+d . Now the strict convexity of H implies that supp(ν X ) ⊂ P 0 (X) and then ν X (P ) = δ 0 (P -P 0 (X)) for a.e. X ∈ R 1+m+d .

We then deduce that Du ε → P 0 = Du 0 in L 1 loc (R 1+m+d ; R d ) not only for the subsequence, but also for the full sequence ε → 0, because any limit Young measure is a unique Dirac mass. Finally, writing again u ε t -u 0 t = H(Du ε ) -H(Du 0 ) and using the fact that H is locally Lipschitz, we get the convergence u ε t → u 0 t = λ 0 in L 1 loc . This ends the proof of the lemma.

Proof of Theorem 1.4

We precise the notation with the ball B ′ 1 ⊂ R 1+m that we distinguish from the ball

B 1 ⊂ R d . Step 1: preliminaries For X ′ = (t, x ′ ) ∈ B ′
1 and (X ′ , 0) ∈ Γ, we consider the tangential gradient (u t , D x ′ u)(X ′ , 0) of the Lipschitz continuous function X ′ → u(X ′ , 0). From Rademacher's theorem, we know that the tangential gradient exists a.e.. Now from Theorem 1.3 (up to a rescaling for each X ′ = (t, x ′ ) by a factor µ > 0 depending on

X ′ such that (X ′ , 0) + (-µ, µ) × B µ ⊂ B ′ 1 × B 1 ), we deduce that u(X + Y ) -u(X) = Bu(X)(Y ) + o(|Y |) for a.e. X = (t, x ′ , 0) ∈ Γ
where Bu(X) : R 1+m+d → R is 1-homogeneous and Lipschitz such that for Y = (s, y ′ , y) Bu(X)(Y ) = λs + P ′ • y ′ + Bu(X)(0, y) with λ := u t (X ′ , 0), P ′ := D x ′ u(X ′ , 0). This shows (1.6). We now set the gradients for a.e. X ′ ∈ B ′ 1 and all x ∈ B 1 P 0 (X ′ , x) := D(Bu(X ′ , 0))(0, x), P (X ′ , x) := Du(X ′ , x).

Step 2: rescaling and extraction of some Young measure For ε > 0, we consider the anisotropic rescaling

P ε (X ′ , x) := P (X ′ , εx).
Because Du is bounded, we can use standard compactness of (Young) measures, and extract a subsequence (still denoted by ε) and find a family of probability measures ν X on R m+d for X = (X ′ , x) ∈ Q 0 = B ′ 1 × B 1 such that for any continuous function F : R m+d → R, we have

F (P ε ) → F := ⟨ν X , F ⟩ = R d F (P )dν X (P ) in L ∞ loc (Q 0 ) weak - * .
Step 3: characterization of the Young measure Our goal is to show that the limit Young measure ν X is a Dirac mass of center P 0 (X).

Let us consider a test function 0 ≤ φ ∈ C ∞ c (B 1 (0)), and let us consider the following integral which is well defined for ε > 0 small enough (because φ has compact support in the unit ball

) for Y = (Y ′ , y) ∈ B ′ 1 × B 1 J ε := |B ′ 1 | -1 B ′ 1 φ(X ′ ) B ′ 1 ×B1
|P ((X ′ , 0) + εY ) -P 0 (X ′ , y)| dY dX ′ .

From Proposition 8.1, we have for the special case X = (X ′ , 0) for almost every

X ′ ∈ B ′ 1 u ε X (Y ) := ε -1 {u(X + εY ) -u(X)} → u 0 X (Y )
and for fixed X = (X ′ , 0)

P (X + εY ) = Du ε X (Y ) → Du 0 X (Y ) = (D X ′ u 0 X (Y ′ , 0), D x u 0 X (0, y)) = P 0 (X ′ , y) in L 1 loc (R 1+m+d Y ).
Hence on the one hand, from the Lebesgue dominated convergence theorem, we get J ε → 0. On the other hand, consider the change of variable Z ′ = X ′ + εY ′ . We get

J ε = |B ′ 1 | -1 B ′ 1 ×B1 B ′ 1 φ(Z ′ -εY ′ ) |P ε (Z ′ , y) -P 0 (Z ′ -εY ′ , y)| dZ ′ dY.
We now introduce

Ĵε := |B ′ 1 | -1 B ′ 1 ×B1 B ′ 1 φ(Z ′ ) |P ε (Z ′ , y) -P 0 (Z ′ , y)| dZ ′ dY which satisfies
Ĵε -J ε → 0 from the continuity of translations in L 1 for the term P 0 and from the uniform continuity for the factor φ.

Hence Ĵε → 0 with (for z = y and Z = (Z ′ , z))

Ĵε = B ′ 1 ×B1 φ(Z ′ ) |P ε (Z) -P 0 (Z)| dZ.
By density of continuous functions in L 1 (B ′ 1 × B 1 ), it is easy to justify by aproximations (of P 0 ) that we have (for a subsequence still denoted by ε) the following limit (as in (8.2))

Ĵε → 0 = Ĵ0 := B ′ 1 ×B1 φ(Z ′ ) R d |P -P 0 (Z)| dν Z (P ) dZ.
Because φ ≥ 0, this implies supp(ν Z ) ⊂ P 0 (Z) , and then ν Z (P ) = δ 0 (P -P 0 (Z)) for a.e. Z ∈ B ′ 1 × B 1 .

Step 4: conclusion From the uniqueness and the expression of the Young measure ν Z , we deduce that we have

P ε → P 0 in L 1 (B ′ 1 × B 1 )
not only for the extracted subsequence, but also for the whole sequence ε (even for a continuous parameter ε → 0). Finally, the convergence of u t (X ′ , εx) follows from the PDE, the uniform bounds on the gradient, the L 1 convergence of the gradient P ε , and the continuity of H. This shows convergence (1.7) of the time-space gradient. This ends the proof of the theorem.

9 Proof of Proposition 1.7: a counter-example

Proof of Proposition 1.7

Step 1: the rotation of the kink For

x = (x 1 , x 2 ) ∈ R 2 ,
= 1 + (rθ ′ 0 ) 2 + rθ ′ 0 sin(2α) > 0. Assume that rθ ′ 0 (r) → 0 as r → 0 + . Then we have h( 1 + (rθ ′ 0 ) 2 + rθ ′ 0 sin(2α)) → h(1) = 1 as r → 0 + . Defining a(x) := 1 h( 1 + (rθ ′ 0 ) 2 + rθ ′ 0 sin(2α))
for α := θ -θ 0 (r) and (r, θ) := Φ -1 (x).

we see that a ∈ C(B 1 ) and u is Lipschitz continuous. Moreover u is also C 1 except on the curve

Γ 0 := {Φ(r, θ), (r, θ) ∈ (0, 1] × R, θ = θ(r) mod π, r > 0} ∪ {0 R 2 }
We have D + u(x) ̸ = ∅ for all x ∈ Γ 0 , while D -u(x) = ∅. Because H(x, P ) is convex in P , we can easily show that u is a Lipschitz continuous viscosity solution of

a(x)h(|Du|) = 1 on B 1 ⊂ R 2 .
On the other hand, if we choose θ 0 such that θ 0 (r) → +∞ as r → 0 then we get a counter-example. We can for instance choose θ 0 (r) := -ln r 2 .

Step 2: checking strict convexity of g 0 We set g 0 (p) := h(|p|)

which is C 1 because h ′ (0) = 0. For p ̸ = 0, we have Dg 0 (p) = h ′ (|p|) p |p|
, while Dg 0 (0) = 0. We also have for

p ̸ = 0 with h ∈ C 2 D 2 g 0 (p) = h ′′ (|p|)p ⊗ p + h ′ (|p|) |p| • {Id -p ⊗ p} with p := p |p|
which is continuous as p → 0 with limit D 2 g 0 (0) = h ′′ (0) • Id. Hence g 0 ∈ C 2 with D 2 g 0 > 0.

Step 3: proof of (1. Then Liouville-type Theorem 1.1 (precisely (4.11)) implies that u 0 is concave which is in contradiction with (9.1). Finally the fact that ω 0 is not Dini integrable follows from Theorem 2.1.10 on page 35 in Cannarsa, Sinestrari [START_REF] Cannarsa | Semiconcave Functions, Hamilton-Jacobi Equations, and Optimal Control[END_REF], and from the fact that semiconcavity implies the existence of directional derivatives (see Theorem 3.2.1 on page 55 in [START_REF] Cannarsa | Semiconcave Functions, Hamilton-Jacobi Equations, and Optimal Control[END_REF]). This ends the proof of the proposition.

10 Proof of Theorem 1.8: generalization to variable coefficients

We start with the following result.

Lemma 10.1 (Modulus of strict convexity) Assume that L : R m+d → R is strictly convex and C 1 . Moreover assume that for some R > 0, there exists a continuous modulus ωR : (0, +∞) → (0, +∞) with ωR (0 + ) = 0, such that

(10.1) (P -Q, DL(P ) -DL(Q)) ≥ |P -Q|ω R (|P -Q|) for all P, Q ∈ B R .
Then we have

(10.2) λL(P ) + (1 -λ)L(Q) -L(λP + (1 -λ)Q) ≥ λ(1 -λ) ΩR (|P -Q|), for all λ ∈ [0, 1], P, Q ∈ B R with ΩR (r) := r 0 ωR (s) ds.
Proof of Lemma 10.1 For λ ∈ [0, 1], for P, Q ∈ B R , and P λ := λP + (1 -λ)Q, and P := P -Q, we have Setting p s := P λ + s(1 -λ) P , q s := P λ -sλ P , with p s -q s = s P we get p s , q s ∈ B R and

λL(P ) + (1 -λ)L(Q) -L(P λ ) = λ(1 -λ)(P -Q) • 1 0 ds DL(P λ + s(P -P λ )) + (1 -λ)λ(Q -P ) • 1 0 ds DL(P λ + s(Q -P λ )) = λ(1 -λ)
λL(P ) + (1 -λ)L(Q) -L(P λ ) = λ(1 -λ) 1 0 ds s (p s -q s ) • {DL(p s ) -DL(q s )} ≥ λ(1 -λ) 1 0 ds| P |ω R (s| P |) = λ(1 -λ) ΩR (| P |)
where we have applied (10.1) in the third line. This shows (10.2), which ends the proof of the lemma. where in (7.2) the Hamiltonian H(P ) has been replaced by

H 0 (P ) = H(0 R 1+m+d , P ) for P ∈ R m+d .
Again u 0 is unique and linear if inf

R d H 0 (0 R m , •) = 0.
We then assume that H 0 satisfies (3.1), and set

K := p ∈ R d , H 0 (0 R m , p) ≤ 0 .
We recall the obtained framework. For any η > 0, there exists ε η > 0 such that for all i = 1, 2 and ε i < ε η , we have for X = (X ′ , x) and X ′ := (t, x ′ )

|u ε i (X ′ , x) -ûi (x)| ≤ η for all X ∈ [-1, 1] × B 1 . For ε = ε 2 > ε 1 and µ = ε 1 ε 2 ∈ (0, 1), we have in particular |u ε (X ′ , x) -û2 (x)| ≤ η for all X ∈ [-1, 1] × B 1 , |u ε (X ′ , x) -û1 (x)| ≤ µη for all X ∈ [-µ, µ] × B µ .
Moreover, we have û1 (x) = inf p∈E 1 p • x with the compact set E 1 ⊂ ∂K. There exists p -∈ E 1 with ξ -= D x H(0, p -) such that for x 1 := -λ 1 ξ -for λ 1 > 0 and with notation of Lemma 5.1, we have Ξ x1 -= Ξ x1 -(û 1 ) = {ξ -} and û1 (x 1 ) < min û(x 1 ), û2 (x 1 ) .

Step 4: core of the proof by contradiction

Step 4.1: first bound from above on u ε (X µ )

We proceed as in Step 4.1 of the proof of Theorem 1.3. Recall that u has Lipschitz constant L > 0, and fix τ 0 > 0 such that 2τ 0 R L = 1 with R L ≥ 1 given in (1.12). Then for ρ = τ 0 R L = 1/2, and for µ > 0 small enough, we get

(10.4) u ε (X µ ) ≤ -µη + u + (X µ )
where X µ := (0, xµ ) = (0, 0, x µ ) and x µ = µx 1 with x1 = (0, x 1 ) = (0, -λ 1 ξ -) ∈ B 1 and xµ := µx 1 ∈ B µ .

Starting from now, the proof differs from the one of Theorem 1.3.

Step 4.2: strict convexity quantified From Lemma 10.1 (applied for frozen X), we have with P λ = λP + (1 -λ)Q (10.5) λL(X, P )+(1-λ)L(X, Q)-L(X, P λ ) ≥ λ(1-λ) ΩR (|P -Q|), for all λ ∈ [0, 1], P, Q ∈ B R , X ∈ Q 0 .

Step 4.3: minimization and dyadic estimate Then we still have the following representation formula (identified with the solution, because the comparison principle still arises for Lipschitz continuous solutions, even for low regularity in X, here continuity). Now for ρ = 1 2 and all τ ∈ (0, τ 0 ), and X = (0, x) ̸ ∈ Γ with x ∈ B ρ , we have u(X) = inf From the convexity of L in the variable γ(σ) and its continuity in X = (σ, γ(σ)), the existence of a minimizer γ * with "optimal exit time" s * is known. where we have used (10.6) in the last inequality.

Setting the approximate characteristic velocity ξτ x :=

x -γ * (-τ ) τ , we get by convexity of ξ → L(X, ξ) that u(X) -u(-τ, γ * (-τ )) ≥ τ L(X, ξτ x ) -τ ω(τ R). Similarly for 2τ ∈ (0, τ 0 ), we have (by monotonicity of ω R ) When Dini condition (1.13) holds true, this implies that the characteristic velocity converges ξ2 -j τ x → ξ0

x as j → +∞. This is then sufficient to imply the non rotation of the blow-up limit, and then the convergence of the blow-up to a unique limit, along the same lines as the remaining part of the proof of Theorem 1.3, using equality along the characteristic curve γ * : u(X) = u(-τ, γ * (-τ )) + 0 -τ dσ L(σ, γ * (σ), γ * (σ)) for all τ ∈ (0, τ 0 ).

We skip the details.

Part B: Proof of generalization of Theorem 1.4

Step 1: in the adaptation of the proof of Proposition 8.1

Once we know the convergence of the blow-up u ε → u 0 locally uniformly, the proof is very similar to the one of Theorem 1.4. We just have to replace everywhere H(P ) by H(0, P ) (and then in particular in the also theorem 1.4.7 on page 18 in Schneider [START_REF] Straszewicz | Über exponierte Punkte abgeschlossener Punktmengen[END_REF], or theorem 18.6 on page 167 in Rockafeller [START_REF] Rockafeller | Convex analysis[END_REF]).

Step 2: proof of ii)

The classical Minkowski theorem (see Theorem 2.3.4 on page 42 in [START_REF] Hiriart-Urruty | Fundamentals of Convex Analysis[END_REF]) claims that if K is a compact convex set, then K = co(ext K). Then (11.1) implies K = co(ext K) ⊂ co(exp K) ⊂ K, which shows (11.2).

Step 3: proof of iii) Let φ(p) := p • x. From Proposition 2.4.6 on page 46 in [START_REF] Hiriart-Urruty | Fundamentals of Convex Analysis[END_REF], we know that point iii) is true for the set exp K replaced by ext K, i.e. φ and (11.5) show (11.4). This ends the proof of the lemma. Then (11.6) implies that for all p ∈ R d \ {p 0 }, we have sup

x∈R d {p • x -v(x)} = v * (p) > v * (p 0 ) + n • (p -p 0 ).
Hence there exists x p ∈ R d such that

p • x p -v(x p ) > v * (p 0 ) + n • (p -p 0 ) = n • p -v(n)
where we have used (11.8) in the equality. This means v(x p ) < v(n) + p • (x p -n) and then p ̸ ∈ ∂v(n). Hence (11.8) implies ∂v(n) = {p 0 }, which ends the proof of the lemma.

Aknowledgements

The author thanks P. Cardaliaguet for very stimulating discussions. He also thanks P. Cardaliaguet and A. Chambolle for indication of references. The author thanks C. Imbert for his advices for publications. Finally the author thanks J. Dolbeault, C. Imbert and T. Lelièvre for providing him good working conditions.

(1. 2 )

 2 H : R m+d → R is C 1 , strictly convex, and superlinear (i.e. lim |P |→+∞ H(P ) |P | = +∞)

( 1

 1 .10) u(x + b) + u(x -b) -2u(x) ≤ |b|ω 0 (|b|) for all x + b, x -b ∈ B 1 2

Lemma 3 . 3 (

 33 Barrier u + ) Under assumptions (1.2) and (3.1) on H, for û defined in (3.2), we set

Lemma 3 . 4 (

 34 Barrier u -) Under assumptions (1.2) and (3.1) on H, for û defined in (3.2), we set

( 4 .

 4 11) u = min(w, û)

(5. 1 )

 1 K w = co(E w ) and w(x) = inf p∈Ew p • x.

  let us consider the kink function U (x) := -|x 1 |, which is a viscosity solution of h(|DU |) = 1 because h(1) = 1. We now introduce polar coordinates x = (x 1 , x 2 ) = Φ(r, θ) := (r cos θ, r sin θ). Given a C 1 function θ 0 : (0, 1] → R, we define u(x) = (U • Φ)(r, θ -θ 0 (r)) = -r| cos(α)|, for α := θ -θ 0 (r). For e r = x |x| , e θ := e ⊥ r , we have -Du(x) = {| cos α| + rθ ′ 0 (r) • sign(cos α) sin α} e r -r {sign(cos α) sin α} e θ r a.e. and |Du| 2

10 )

 10 Assume by contradiction that (1.10) is false. Then for any δ > 0, there exists sequencesR 2 \ {0} ∋ b → 0, and (x b ) b with x b ∈ B 1 2 , such that u(x b + b) + u(x b -b) -2u(x b ) > δ|b|.Defining the blow-up u b with moving center x b , and the normal vector n b u b (x) := |b| -1 {u(x b + x) -u(x b )} , n b := b |b| up to extraction of a subsequence, we get u b → u 0 and n b → n 0 and (9.1) H(x 0 , Du 0 ) = 0 on R 2 , and u 0 (n 0 ) + u 0 (-n 0 ) -2u 0 (0) ≥ δ > 0

  ) • DL(P λ + s(1 -λ) P ) -DL(P λ -sλ P ) .

Proof of Theorem 1. 8

 8 Part A: Proof of generalization of Theorem 1.3 Up to redefine H, we can assume that λ = 0 = p ′ . Steps 1 to 3: as in the proof of Theorem 1.3 These steps are identical to the ones of the proof of Theorem 1.3, except that after the first blow-up, equation (7.2) has to be replaced by the following (10.3) u 0 t + H 0 (Du 0 ) = 0 on R 1+m × Ω (in the viscosity sense) u 0 = 0 on R 1+m × ∂Ω (in the strong sense).

  s∈[-τ,0), γ(s)∈Γ D τ,2ρ , γ(0)=X, γ((s,0))⊂Bρ\ Γ u(γ(s)) + 0 s dσ L(σ, γ(σ), γ(σ))whereΓ := R m × {0 R d }.Notice in particular that the L-Lipschitz continuity of u imposes (for R L given in (1.12))(10.6) | γ(σ)| ≤ R Land to have short hand notation, we set ω := ω R , Ω := ΩR , R := 1 + R L .

  Case A: γ * (s * ) ∈ Γ τ,2ρ Recall that X = (0, x) with x = (x ′ , x) and x ̸ = 0. Then we have|x| ≤ |x -γ * (s * )| = 0 s * γ * (σ) dσ ≤ R L |s * | with s * ∈ [-τ, 0).This shows that for τ > 0 small enough, Case A is ruled out.Case B: γ * (s * ) ̸ ∈ Γ τ,2ρ Because γ * (s * ) ̸ ∈ Γ τ,2ρ , then s * = τ and u(X) -u(-τ, γ * (-τ )) = 0 -τ dσ L(σ, γ * (σ), γ * (σ)) ≥ 0 -τ dσ L(0, x, γ * (σ)) -τ ω(τ R)

≥ 1 4 Ω 4 Ω| ≤ 4 (

 444 u(-τ, γ * (-τ )) -u(-2τ, γ * (-2τ )) ≥ τ L(X, ξτ γ * (-τ ) ) -τ ω(2τ R), ξτ γ * (-τ) := γ * (-τ ) -γ * (-2τ ) τ and then u(X) -u(-2τ, γ * (-2τ )) ≥ τ L(X, ξτ x ) + τ L(X, ξτ γ * (-τ ) ) -2τ ω(2τ R). Now notice also that u(X) -u(-2τ, γ * (-2τ )) = 0 -2τ dσ L(σ, γ * (σ), γ * (σ)) = inf γ(0)=x, γ(-2τ )=γ * (-2τ ), | γ(σ)|≤R 0 -2τ dσ L(σ, γ(σ), γ(σ)) ≤ 2τ ω(2τ R) + inf γ(0)=x, γ(-2τ )=γ * (-2τ ), | γ(σγ * (-τ ) .Hence we get τ L(X, ξτx ) + τ L(X, ξτ γ * (-τ ) ) -2τ ω(2τ R) ≤ 2τ ω(2τ R) + 2τ L(X, ξ2τ x ). Therefore 2ω(2τ R) ≥ 1 2 L(X, ξτ x ) + L(X, ξτ γ * (-τ ) ) -L(X, ξ2τ x ) (| ξτ x -ξτ γ * (-τ ) |)where the last inequality follows from (10.5) for λ = 1 2 . Notice that ξτx -ξ2τ x = 1 2 ξτ x -ξτ γ * (-τ ) . Hence 2ω(2τ R) ≥ 1 (2| ξτ x -ξ2τ x |).Because Ω is convex, C 1 with Ω′ (0) = 0 = Ω(0), we have the chord inequality Ω(θr) ≤ θ Ω(r) for all θ ∈ [0Ω-1 • ω)(τ R) for all τ ∈ (0, τ 0 ).Step 4.4: conclusion by dyadic sums If f : [0, 1) → [0, +∞) is non-decreasing, then we have

  follows from the continuity of φ. This shows(11.3). Now co Argmax ext K φ = co Argmax exp K

Lemma 11 . 3 (

 113 Exposed point and pointwise C 1 support function) Let K ⊂ R d be a convex compact set and its support function v(x) := sup p∈K p • x. Assume that p 0 is an exposed point of K with admissible unit normal n, i.e for Π ± p0,n := p ∈ R d , (p -p 0 , ±n) ≥ 0 , we have(11.6) K ⊂ Π - p0,n , K ∩ Π + p0,n = {p 0 } .Then the subdifferential of the convex function v satisfies (11.7) ∂v(n) = {p 0 } i.e. v is C 1 at n. Proof of Lemma 11.3 Notice that the Legendre-Fenchel transform of v is v * = (+∞)1 R d \K . Moreover for p 0 as in (11.6), we have n ∈ ∂v * (p 0 ) and then (11.8) v * (p 0 ) + v(n) = p 0 • n, v * (p 0 ) = 0, p 0 ∈ ∂v(n).

Proof of Theorem 1.3: full directional derivatives

This section is fully devoted to the proof of Theorem 1.3.

Proof of Theorem 1.3

Up to redefine H, we can assume that λ = 0 = p ′ .

Step 1: blow-up limits and double blow-up limits For ε > 0, we consider the blow-up functions (7.1) u ε (t, x) = ε -1 u(εt, εx)

which are Lipschitz continuous, uniformly with respect to ε, with the same Lipschitz constant. By Ascoli-Arzela theorem, from any sequence ε → 0, we can extract a subsequence (still denoted by ε) such that u ε → u 0 locally uniformly on compact sets of R 1+m+d . Moreover by stability of viscosity solutions, the limit u 0 solves the whole space problem (7.2)

From Theorem 1.1, we know that u 0 = u 0 (x) with x ∈ Ω = R d and from Corollary 4.5, we even know that

and convex û defined in (3.2). We can now consider the rescaling for µ > 0

Because w is globally Lipschitz and concave, on the one hand, we know that we have the blow-up convergence

On the other hand, for the same reason, we have the blow-down convergence

Here by construction, both w 0 and w ∞ are globally Lipschitz continuous, concave, and moreover 1homogeneous and solve H(Dw) = 0. As a consequence, we get for u 0 = u 0 (x) that

Notice that the limit of (u 0 ) µ is then either equal to û, or is concave and 1-homogeneous.

Step 2: setting of the problem Our goal is to show later the uniqueness of the blow-up limit. If inf R d H(0, •) = 0 = H(0, p 0 ), then we know that the blow-up limit is u 0 (X) = p 0 • x and then is unique. We then assume from Lemma 3.1 that (3.1) holds true. We start as follows. Consider now two sequences ε i = ε i k → 0 for i = 1, 2, such that for rescaling (7.1), we have u ε i → u i locally uniformly on compact sets of R 1+m+d . Notice that each limit u i has a shape as in (7.3). Then by a diagonal extraction argument, we can always find sequences a ε i → +∞ which go to infinity sufficiently slowly such that a ε i ε i → 0 and

Step 4: core of the proof by contradiction

Step 4.1: first bound from above on u ε (X µ ) Recall that u has Lipschitz constant L > 0, and fix τ 0 > 0 such that 2τ 0 R L = 1 with R L ≥ 1 given in Lemma 6.1. Then there exists some fixed λ 1 > 0 small enough such that x1 = (0,

Then for ρ = τ 0 R L = 1/2, we have for µ > 0 small enough B µ ⊂ B ρ ⊂ B 2ρ ⊂ B 1 , and from (7.8), we have û1 (x 1 ) < û(x 1 ) and then for X µ := (0, xµ ) = (0, 0, x µ ) and

for η small enough such that 0 < 2η < û(x 1 ) -û1 (x 1 ). We get

Step 4.2: effective bound from above on u ε (X µ ) Now from (1.4) with λ = p ′ = 0, we deduce that for s > 0 and µ > 0 (and using

With notation of Lemma 6.1, we have for X 0 := X µ , x 0 := x µ = µx 1 ∈ B µ , and r := µ (and

Hence from (7.10) for µ > 0 small enough, we get

Step 4.3: properties along characteristics Now from Lemma 6.1, there exists ξµ

Now at the scale µε with X 1 = (0, x1 ), we get by change of scales s = µσ

i.e.

(7.12)

Step 4.4: passing to the limit Now in the limit ε → 0 with µ → 0, we have (up to extraction of subsequences)

and passing to the limit in (7.11) for ûi (t, x ′ , x) = ûi (x) (7.13) û2 (0) = û2 (-s(1, ξ0 -)) + sL( ξ0 -) for all s ∈ [0, τ 0 ] and in the limit in (7.12)

definition (8.3) of Ψ(X, P ), and in the expression of B ε ). The definition of the integral I ε is unchanged. Because of the relations u ε + H(εX, Du ε ) = 0 and u 0 t + H(0, Du 0 ) = 0 we just have in the second line of (8.4), to introduce the error term A ε := -{H(εX, Du ε ) -H(0, Du ε )} which converges uniformly to zero, and then does not affect the reasoning.

Step 2: in the adaptation of the proof of Theorem 1.4

The proof is unchanged (notice that to show the convergence of u t (X ′ , εx) we have to use the PDE with the X-dependence, but the same argument applies). This ends the proof of the theorem.

Appendix

In this appendix, we grasp together some results on exposed points of closed convex sets, that are useful in the main part of the paper.

We recall the following definitions. Definition 11.1 (Extreme points and exposed points) Let K ⊂ R d be a convex compact set. i) (Extreme point) We say that p 0 ∈ K is an extreme point K and denote it by p 0 ∈ ext K, if there are no two different points p 1 , p 2 ∈ K such that p 0 = λp 1 + (1 -λ)p 2 for some λ ∈ (0, 1). ii) (Exposed point) We say that p 0 ∈ K is an exposed point of K and denote it by p 0 ∈ exp K, if there exists a closed half space Π ± p0,n := p ∈ R d , (p -p 0 , ±n) ≥ 0 for some non-zero vector n (we can in particular choose a unit vector n ∈ S d-1 ) such that K ⊂ Π - p0,n , Π + p0,n ∩ K = {p 0 } .

We get immediately exp K ⊂ ext K, and have the following complementary result.

Lemma 11.2 (Exposed points of convex compact sets) Let K ⊂ R d be a convex compact set. i)(Extreme-exposed relation)

We have

ii) (Property of exposed points)

We have (11.2) K = co(exp K).

iii) (Support functions) For every x ∈ R d , we have The results are more or less classical. For completness of the argument we give some details.

Step 1: proof of i)

The first inclusion in (11.1) is straightforward, and the second inclusion is Straszewicz's theorem [START_REF] Straszewicz | Über exponierte Punkte abgeschlossener Punktmengen[END_REF] (see