A Short Introduction to Viscosity Solutions and the Large Time Behavior of Solutions of Hamilton–Jacobi Equations

Abstract

We present an introduction to the theory of viscosity solutions of first-order partial differential equations and a review on the optimal control/dynamical approach to the large time behavior of solutions of Hamilton–Jacobi equations, with the Neumann boundary condition. This article also includes some of basics of mathematical analysis related to the optimal control/dynamical approach for easy accessibility to the topics.

In memory of Riichi Iino, my former adviser at Waseda University.

Appendix

1.1 A.1 Local maxima to global maxima

We recall a proposition from [56] which is about partition of unity.

Proposition A.1.

Let $\mathcal{O}$ be a collection of open subsets of ${\mathbb{R}}^{n}$ . Set $W :=\bigcup _{U\in \mathcal{O}}U$ . Then there is a collection $\mathcal{F}$ of ${C}^{\infty }$ functions in ${\mathbb{R}}^{n}$ having the following properties:

• (i) 0 ≤ f(x) ≤ 1 for all x ∈ W and $f \in \mathcal{F}$.

• (ii) For each x ∈ W there is a neighborhood V of x such that all but finitely many $f \in \mathcal{F}$ vanish in V.

• (iii) $\sum _{f\in \mathcal{F}}f(x) = 1$ for all x ∈ W.

• (iv) For each $f \in \mathcal{F}$ there is a set $U \in \mathcal{O}$ such that supp f ⊂ U.

Proposition A.2.

Let Ω be any subset of ${\mathbb{R}}^{n}$ ,$u \in \mathrm{USC}(\Omega, \mathbb{R})$ and ϕ ∈ C ¹ (Ω). Assume that u −ϕ attains a local maximum at y ∈ Ω. Then there is a function ψ ∈ C ¹ (Ω) such that u −ψ attains a global maximum at y and ψ = ϕ in a neighborhood of y.

Proof.

As usual it is enough to prove the above proposition in the case when (u − ϕ)(y) = 0.

By the definition of the space C ¹(Ω), there is an open neighborhood W ₀ of Ω such that ϕ is defined in W ₀ and $\phi \in {C}^{1}(W_{0})$.

There is an open subset U _y  ⊂ W ₀ of ${\mathbb{R}}^{n}$ containing y such that $\max _{U_{y}\cap \Omega }(u\,-\,\phi ) = (u-\phi )(y)$. Since $u \in \mathrm{USC}(\Omega, \mathbb{R})$, for each x ∈ Ω ∖ { y} we may choose an open subset U _x of ${\mathbb{R}}^{n}$ so that x ∈ U _x , y ∉ U _x and $\sup _{U_{x}\cap \Omega }u < \infty$. Set $a_{x} =\sup _{U_{x}\cap \Omega }u$ for every x ∈ Ω ∖ { y}.

We set $\mathcal{O} =\{ U_{z}\, :\, z \in \Omega \}$ and $W =\bigcup _{U\in \mathcal{O}}U$. Note that W is an open neighborhood of Ω. By Proposition A.1, there exists a collection $\mathcal{F}$ of functions $f \in {C}^{\infty }({\mathbb{R}}^{n})$ satisfying the conditions (i)–(iv) of the proposition. According to the condition (iv), for each $f \in \mathcal{F}$ there is a point z ∈ Ω such that $\mathrm{supp}f \subset U_{z}$. For each $f \in \mathcal{F}$ we fix such a point z ∈ Ω and define the mapping $p\, :\, \mathcal{F}\rightarrow \Omega$ by p(f) = z. We set

$$\displaystyle{\psi (x) =\sum _{f\in \mathcal{F},\ p(f)\not =y}a_{p(f)}f(x) +\sum _{f\in \mathcal{F},\ p(f)=y}\phi (x)f(x)\ \ \ \text{ for }x \in W.}$$

By the condition (ii), we see that ψ ∈ C ¹(W). Fix any x ∈ Ω and $f \in \mathcal{F}$, and observe that if f(x) > 0 and p(f) ≠ y, then we have $x \in \mathrm{supp}f \subset U_{p(f)}$ and, therefore, $a_{p(f)} =\sup _{U_{p(f)}\cap \Omega }u \geq u(x)$. Observe also that if f(x) > 0 and p(f) = y, then we have x ∈ U _y and ϕ(x) ≥ u(x). Thus we see that for all x ∈ Ω,

$$\displaystyle{\psi (x) \geq \sum _{f\in \mathcal{F},\ p(f)\not =y}u(x)f(x) +\sum _{f\in \mathcal{F},\ p(f)=y}u(x)f(x) = u(x)\sum _{f\in \mathcal{F}}f(x) = u(x).}$$

Thanks to the condition (ii), we may choose a neighborhood V ⊂ W of y and a finite subset $\{f_{j}\}_{j=1}^{N}$ of $\mathcal{F}$ so that

$$\displaystyle{\sum _{j=1}^{N}f_{ j}(x) = 1\ \ \ \text{ for all }x \in V.}$$

If p(f _j ) ≠ y for some j = 1, …, N, then $U_{p(f_{j})} \cap \{ y\} = \varnothing$ and hence $y\not\in \mathrm{supp}f_{j}$. Therefore, by replacing V by a smaller one we may assume that p(f _j ) = y for all j = 1, …, N. Since f = 0 in V for all $f \in \mathcal{F}\setminus \{ f_{1},\ldots,f_{N}\}$, we see that

$$\displaystyle{\psi (x) =\sum _{ j=1}^{N}\phi (x)f_{ j}(x) =\phi (x)\ \ \ \text{ for all }x \in V.}$$

It is now easy to see that u − ψ has a global maximum at y. □ 

1.2 A.2 A Quick Review of Convex Analysis

We discuss here basic properties of convex functions on ${\mathbb{R}}^{n}$.

By definition, a subset C of ${\mathbb{R}}^{n}$ is convex if and only if

$$\displaystyle{(1 - t)x + ty \in C\ \ \ \text{ for all }\ x,y \in C,\,0 < t < 1.}$$

For a given function $f\, :\, U \subset {\mathbb{R}}^{n} \rightarrow [-\infty,\,\infty ]$, its epigraph epi(f) is defined as

$$\displaystyle{\mathrm{epi}(f) =\{ (x,y) \in U \times \mathbb{R}\, :\, y \geq f(x)\}.}$$

A function f : U → [ − ∞, ∞] is said to be convex if epi(f) is a convex subset of ${\mathbb{R}}^{n+1}$.

We are henceforth concerned with functions defined on ${\mathbb{R}}^{n}$. When we are given a function f on U with U being a proper subset of ${\mathbb{R}}^{n}$, we may think of f as a function defined on ${\mathbb{R}}^{n}$ having value ∞ on the set ${\mathbb{R}}^{n} \setminus U$.

It is easily checked that a function $f\, :\, {\mathbb{R}}^{n} \rightarrow [-\infty,\,\infty ]$ is convex if and only if for all $x,y \in {\mathbb{R}}^{n}$, $t,s \in \mathbb{R}$ and λ ∈ [0, 1],

$$\displaystyle{f((1-\lambda )x +\lambda y) \leq (1-\lambda )t +\lambda s\ \ \ \mathrm{if}\ \ t \geq f(x)\ \text{ and }\ s \geq f(y).}$$

From this, we see that a function $f\, :\, {\mathbb{R}}^{n} \rightarrow (-\infty,\,\infty ]$ is convex if and only if for all $x,y \in {\mathbb{R}}^{n}$ and λ ∈ [0, 1],

$$\displaystyle{f((1-\lambda )x +\lambda y) \leq (1-\lambda )f(x) +\lambda f(y).}$$

Here we use the convention for extended real numbers, i.e., for any $x \in \mathbb{R}$, − ∞ < x < ∞, x ± ∞ =  ± ∞, x ⋅( ± ∞) =  ± ∞ if x > 0, 0 ⋅( ± ∞) = 0, etc.

Any affine function f(x) = a ⋅x + b, where $a \in {\mathbb{R}}^{n}$ and $b \in \mathbb{R}$, is a convex function on ${\mathbb{R}}^{n}$. Moreover, if $A \subset {\mathbb{R}}^{n}$ and $B \subset \mathbb{R}$ are nonempty sets, then the function on ${\mathbb{R}}^{n}$ given by

$$\displaystyle{f(x) =\sup \{ a \cdot x + b\, :\, (a,b) \in A \times B\}}$$

is a convex function. Note that this function f is lower semicontinuous on ${\mathbb{R}}^{n}$. We restrict our attention to those functions which take values only in ( − ∞, ∞].

Proposition B.1.

Let $f\, :\, {\mathbb{R}}^{n} \rightarrow (-\infty,\,\infty ]$ be a convex function. Assume that p ∈ D ⁻ f(y) for some $y,p \in {\mathbb{R}}^{n}$ . Then

$$\displaystyle{f(x) \geq f(y) + p \cdot (x - y)\ \ \ \text{ for all }\ x \in {\mathbb{R}}^{n}.}$$

Proof.

By the definition of D ⁻ f(y), we have

$$\displaystyle{f(x) \geq f(y) + p \cdot (x - y) + o(\vert x - y\vert )\ \ \ \text{ as }\ x \rightarrow y.}$$

Hence, fixing $x \in {\mathbb{R}}^{n}$, we get

$$\displaystyle{f(y) \leq f(tx + (1 - t)y) - tp \cdot (x - y) + o(t)\ \ \ \text{ as }\ t \rightarrow 0 +.}$$

Using the convexity of f, we rearrange the above inequality and divide by t > 0, to get

$$\displaystyle{f(y) \leq f(x) - p \cdot (x - y) + o(1)\ \ \ \text{ as }\ t \rightarrow 0 +.}$$

Sending t → 0 + yields

$$\displaystyle{f(x) \geq f(y) + p \cdot (x - y)\ \ \ \text{ for all }\,\,x \in {\mathbb{R}}^{n}.}$$

 □ 

Proposition B.2.

Let $\mathcal{F}$ be a nonempty set of convex functions on ${\mathbb{R}}^{n}$ with values in (−∞, ∞]. Then $\sup \mathcal{F}$ is a convex function on ${\mathbb{R}}^{n}$ having values in (−∞, ∞].

Proof.

It is clear that $(\sup \mathcal{F})(x) \in (-\infty,\,\infty ]$ for all $x \in {\mathbb{R}}^{n}$. If $f \in \mathcal{F}$, $x,y \in {\mathbb{R}}^{n}$ and t ∈ [0, 1], then we have

$$\displaystyle{f((1 - t)x + ty) \leq (1 - t)f(x) + tf(y) \leq (1 - t)(\sup \mathcal{F})(x) + t(\sup \mathcal{F})(y)}$$

and hence

$$\displaystyle{(\sup \mathcal{F})((1 - t)x + ty) \leq (1 - t)(\sup \mathcal{F})(x) + t(\sup \mathcal{F})(y),}$$

which proves the convexity of $\sup \mathcal{F}$. □ 

We call a function $f\, :\, {\mathbb{R}}^{n} \rightarrow (-\infty,\,\infty ]$ proper convex if the following three conditions hold:

• (a) f is convex on ${\mathbb{R}}^{n}$.

• (b) $f \in \mathrm{LSC}({\mathbb{R}}^{n})$.

• (c) f(x) ≢ ∞.

Let $f\, :\, {\mathbb{R}}^{n} \rightarrow [-\infty,\,\infty ]$. The conjugate convex function (or the Legendre–Fenchel transform) of f is the function ${f}^{\star }\, :\, {\mathbb{R}}^{n} \rightarrow [-\infty,\,\infty ]$ given by

$$\displaystyle{{f}^{\star }(x) =\sup _{ y\in {\mathbb{R}}^{n}}(x \cdot y - f(y)).}$$

Proposition B.3.

If f is a proper convex function, then so is f ^⋆ .

Lemma B.1.

If f is a proper convex function on ${\mathbb{R}}^{n}$ , then D ⁻ f(y)≠∅ for some $y \in {\mathbb{R}}^{n}$.

Proof.

We choose a point $x_{0} \in {\mathbb{R}}^{n}$ so that $f(x_{0}) \in \mathbb{R}$. Let $k \in \mathbb{N}$, and define the function g _k on $\bar{B}_{1}(x_{0})$ by the formula g _k (x) = f(x) + k | x − x ₀ | ². Since $g_{k} \in \mathrm{LSC}(\overline{B}_{1}(x_{0}))$, and $g_{k}(x_{0}) = g(x_{0}) \in \mathbb{R}$, the function g _k has a finite minimum at a point $x_{k} \in \overline{B}_{1}(x_{0})$. Note that if k is sufficiently large, then

$$\displaystyle{\min _{\partial B_{1}(x_{0})}g_{k} =\min _{\partial B_{1}(x_{0})}f + k > f(x_{0}).}$$

Fix such a large k, and observe that x _k  ∈ B ₁(x ₀) and, therefore, $-2k(x_{k} - x_{0}) \in {D}^{-}f(x_{k})$. □ 

Proof (Proposition B.3). 

The function $x\mapsto x \cdot y - f(y)$ is an affine function for any $y \in {\mathbb{R}}^{n}$. By Proposition B.2, the function f ^⋆ is convex on ${\mathbb{R}}^{n}$. Also, since the function $x\mapsto x \cdot y - f(y)$ is continuous on ${\mathbb{R}}^{n}$ for any $y \in {\mathbb{R}}^{n}$, as stated in Proposition 1.5, the function f ^⋆ is lower semicontinuous on ${\mathbb{R}}^{n}$.

Since f is proper convex on ${\mathbb{R}}^{n}$, there is a point $x_{0} \in {\mathbb{R}}^{n}$ such that $f(x_{0}) \in \mathbb{R}$. Hence, we have

$$\displaystyle{{f}^{\star }(y) \geq y \cdot x_{ 0} - f(x_{0}) > -\infty \ \ \ \text{ for all }\ y \in {\mathbb{R}}^{n}.}$$

By Lemma B.1, there exist points $y,p \in {\mathbb{R}}^{n}$ such that p ∈ D ⁻ f(y). By Proposition B.1, we have

$$\displaystyle{f(x) \geq f(y) + p \cdot (x - y)\ \ \ \text{ for all }\ x \in {\mathbb{R}}^{n}.}$$

That is,

$$\displaystyle{p \cdot y - f(y) \geq p \cdot x - f(x)\ \ \ \text{ for all }\ x \in {\mathbb{R}}^{n},}$$

which implies that ${f}^{\star }(p) = p \cdot y - f(y) \in \mathbb{R}$. Thus, we conclude that ${f}^{\star }\, :\, {\mathbb{R}}^{n} \rightarrow (-\infty,\,\infty ]$, f ^⋆ is convex on ${\mathbb{R}}^{n}$, ${f}^{\star } \in \mathrm{LSC}({\mathbb{R}}^{n})$ and f ^⋆(x) ≢ ∞. □ 

The following duality (called convex duality or Legendre–Fenchel duality) holds.

Theorem B.1.

Let $f\, :\, {\mathbb{R}}^{n} \rightarrow (-\infty,\,\infty ]$ be a proper convex function. Then

$$\displaystyle{{f}^{\star \star } = f.}$$

Proof.

By the definition of f ^⋆, we have

$$\displaystyle{{f}^{\star }(x) \geq x \cdot y - f(y)\ \ \ \text{ for all }\ x,y \in {\mathbb{R}}^{n},}$$

which reads

$$\displaystyle{f(y) \geq y \cdot x - {f}^{\star }(x)\ \ \ \text{ for all }\ x,y \in {\mathbb{R}}^{n}.}$$

Hence,

$$\displaystyle{f(y) \geq {f}^{\star \star }(y)\ \ \ \text{ for all }\ y \in {\mathbb{R}}^{n}.}$$

Next, we show that

$$\displaystyle{{f}^{\star \star }(x) \geq f(x)\ \ \ \text{ for all }\ x \in {\mathbb{R}}^{n}.}$$

We fix any $a \in {\mathbb{R}}^{n}$ and choose a point $y \in {\mathbb{R}}^{n}$ so that $f(y) \in \mathbb{R}$. We fix a number R > 0 so that | y − a |  < R. Let $k \in \mathbb{N}$, and consider the function $g_{k} \in \mathrm{LSC}(\overline{B}_{R}(a))$ defined by g _k (x) = f(x) + k | x − a | ². Let $x_{k} \in \overline{B}_{R}(a)$ be a minimum point of the function g _k . Noting that if k is sufficiently large, then

$$\displaystyle{g_{k}(x_{k}) \leq f(y) + k\vert y - a{\vert }^{2} <\min _{ \partial B_{R}(a)}f + k{R}^{2} =\min _{ \partial B_{R}(a)}g_{k},}$$

we see that x _k  ∈ B _R (a) for k sufficiently large. We henceforth assume that k is large enough so that x _k  ∈ B _R (a). We have

$$\displaystyle{{D}^{-}g_{ k}(x_{k}) = {D}^{-}f(x_{ k}) + 2k(x_{k} - a) \ni 0.}$$

Accordingly, if we set ξ _k  =  − 2k(x _k  − a), then we have $\xi _{k} \in {D}^{-}f(x_{k})$. By Proposition B.1, we get

$$\displaystyle{f(x) \geq f(x_{k}) +\xi _{k} \cdot (x - x_{k})\ \ \ \text{ for all }\ x \in {\mathbb{R}}^{n},}$$

or, equivalently,

$$\displaystyle{\xi _{k} \cdot x_{k} - f(x_{k}) \geq \xi _{k} \cdot x - f(x)\ \ \ \text{ for all }\ x \in {\mathbb{R}}^{n}.}$$

Hence,

$$\displaystyle{\xi _{k} \cdot x_{k} - f(x_{k}) = {f}^{\star }(\xi _{ k}).}$$

Using this, we compute that

$$\displaystyle\begin{array}{rcl}{ f}^{\star \star }(a)& \geq & a \cdot \xi _{ k} - {f}^{\star }(\xi _{ k}) =\xi _{k} \cdot a -\xi _{k} \cdot x_{k} + f(x_{k}) {}\\ & =& 2k\vert x_{k} - a{\vert }^{2} + f(x_{ k}). {}\\ \end{array}$$

We divide our argument into the following cases, (a) and (b).

Case (a): $\lim _{k\rightarrow \infty }k\vert x_{k} - a{\vert }^{2} = \infty$. In this case, if we set $m =\min _{\bar{B}_{R}(a)}f$, then we have

$$\displaystyle{{f}^{\star \star }(a) \geq \liminf _{ k\rightarrow \infty }2k\vert x_{k} - a{\vert }^{2} + m = \infty,}$$

and, therefore, f ^⋆ ⋆(a) ≥ f(a).

Case (b): $\liminf _{k\rightarrow \infty }k\vert x_{k} - a{\vert }^{2} < \infty$. We may choose a subsequence $\{x_{k_{j}}\}_{j\in \mathbb{N}}$ of {x _k } so that $\lim _{j\rightarrow \infty }x_{k_{j}} = a$. Then we have

$$\displaystyle{{f}^{\star \star }(a) \geq \liminf _{ j\rightarrow \infty }\left (2k_{j}\vert x_{k_{j}} - a{\vert }^{2} + f(x_{ k_{j}})\right ) \geq \liminf _{j\rightarrow \infty }f(x_{k_{j}}) \geq f(a).}$$

Thus, in both cases we have f ^⋆ ⋆(a) ≥ f(a), which completes the proof. □ 

Theorem B.2.

Let $f\, :\, {\mathbb{R}}^{n} \rightarrow (-\infty,\,\infty ]$ be proper convex and $x,\xi \in {\mathbb{R}}^{n}$ . Then the following three conditions are equivalent each other.

• (i) ξ ∈ D ⁻ f(x).

• (ii) $x \in {D}^{-}{f}^{\star }(\xi )$.

• (iii) x ⋅ξ = f(x) + f ^⋆ (ξ).

Proof.

Assume first that (i) holds. By Proposition B.1, we have

$$\displaystyle{f(y) \geq f(x) +\xi \cdot (y - x)\ \ \ \text{ for all }\ y \in {\mathbb{R}}^{n},}$$

which reads

$$\displaystyle{\xi \cdot x - f(x) \geq \xi \cdot y - f(y)\ \ \ \text{ for all }y \in {\mathbb{R}}^{n}.}$$

Hence,

$$\displaystyle{\xi \cdot x - f(x) =\max _{y\in {\mathbb{R}}^{n}}(\xi \cdot y - f(y)) = {f}^{\star }(\xi ).}$$

Thus, (iii) is valid.

Next, we assume that (iii) holds. Then the function $y\mapsto \xi \cdot y - f(y)$ attains a maximum at x. Therefore, ξ ∈ D ⁻ f(x). That is, (i) is valid.

Now, by the convex duality (Theorem B.1), (iii) reads

$$\displaystyle{x\cdot \xi = {f}^{\star \star }(x) + {f}^{\star }(\xi ).}$$

The equivalence between (i) and (iii), with f replaced by f ^⋆, is exactly the equivalence between (ii) and (iii). The proof is complete. □ 

Finally, we give a Lipschitz regularity estimate for convex functions.

Theorem B.3.

Let $f\, :\, {\mathbb{R}}^{n} \rightarrow (-\infty,\,\infty ]$ be a convex function. Assume that there are constants M > 0 and R > 0 such that

$$\displaystyle{\vert f(x)\vert \leq M\ \ \ \text{ for all }\ x \in B_{3R}.}$$

Then

$$\displaystyle{\vert f(x) - f(y)\vert \leq \frac{M} {R} \vert x - y\vert \ \ \ \text{ for all }\ x,y \in B_{R}.}$$

Proof.

Let x, y ∈ B _R and note that | x − y |  < 2R. We may assume that x ≠ y. Setting ξ = (x − y) ∕ | x − y | and z = y + 2Rξ and noting that z ∈ B _3R ,

$$\displaystyle{ x - y = \frac{\vert x - y\vert } {2R} (z - y), }$$

and

$$\displaystyle{ x = y + \frac{\vert x - y\vert } {2R} (z - y) = \frac{\vert x - y\vert } {2R} z + \left (1 -\frac{\vert x - y\vert } {2R} \right )y, }$$

we obtain

$$\displaystyle{f(x) \leq \frac{\vert x - y\vert } {2R} f(z) + \left (1 -\frac{\vert x - y\vert } {2R} \right )f(y),}$$

and

$$\displaystyle{f(x) - f(y) \leq \frac{\vert x - y\vert } {2R} (f(z) - f(y)) \leq \frac{\vert x - y\vert } {2R} (\vert f(z)\vert + \vert f(y)\vert ) \leq \frac{M\vert x - y\vert } {R}.}$$

In view of the symmetry in x and y, we see that

$$\displaystyle{\vert f(x) - f(y)\vert \leq \frac{M} {R} \vert x - y\vert \ \ \ \text{ for all }\ x,y \in B_{R}.}$$

 □ 

1.3 A.3 Global Lipschitz Regularity

We give here a proof of Lemmas 2.1 and 2.2.

Proof (Lemma 2.1). 

We first show that there is a constant C > 0, for each $z \in \overline{\Omega }$ a ball B _r (z) centered at z, and for each $x,y \in B_{r}(z) \cap \overline{\Omega }$, a curve $\eta \in \mathrm{AC}([0,T], {\mathbb{R}}^{n})$, with $T \in \overline{\mathbb{R}}_{+}$, such that η(s) ∈ Ω for all s ∈ (0, T), $\vert \dot{\eta }(s)\vert \leq 1$ for a.e. s ∈ (0, T) and T ≤ C | x − y | .

Let ρ be a defining function of Ω. We may assume that $\|D\rho \|_{\infty,{\mathbb{R}}^{n}} \leq 1$ and | Dρ(x) | ≥ δ for all $x \in {(\partial \Omega )}^{\delta } :=\{ y \in {\mathbb{R}}^{n}\, :\, \mathrm{dist}(y,\partial \Omega ) <\delta \}$ and some constant δ ∈ (0, 1).

Let z ∈ Ω. We can choose r > 0 so that B _r (z) ⊂ Ω. Then, for each x, y ∈ B _r (z), with x ≠ y, the line η(s) = x + s(y − x) ∕ | y − x | , with s ∈ [0,  | x − y | ], connects two points x and y and lies inside Ω. Note as well that $\dot{\eta }(s) = (y - x)/\vert y - x\vert \in \partial B_{1}$ for all s ∈ [0,  | x − y | ].

Let $z \in \partial \Omega$. Since | Dρ(z) | ² ≥ δ ², by continuity, we may choose r ∈ (0, δ ³ ∕ 4) so that Dρ(x) ⋅Dρ(z) ≥ δ ² ∕ 2 for all $x \in B_{{4\delta }^{-2}r}(z)$. Fix any $x,y \in B_{r}(z) \cap \overline{\Omega }$. Consider the curve ξ(t) = x + t(y − x) − t(1 − t)6δ ^− 2 | x − y | Dρ(z), with t ∈ [0, 1], which connects the points x and y. Note that

$$\displaystyle\begin{array}{rcl} \vert \xi (t) - z\vert & \leq & \,(1 - t)\vert x - z\vert + t\vert y - z\vert + 6t{(1 - t)\delta }^{-2}\vert x - y\vert \vert D\rho (z)\vert {}\\ & <& \,(1 + {3\delta }^{-2})r < {4\delta }^{-2}r {}\\ \end{array}$$

and 4δ ^− 2 r < δ. Hence, we have $\xi (t) \in B_{{4\delta }^{-2}r}(z) \cap {(\partial \Omega )}^{\delta }$ for all t ∈ [0, 1]. If t ∈ (0, 1 ∕ 2], then we have

$$\displaystyle\begin{array}{rcl} \rho (\xi (t))& \leq & \,\rho (x) + tD\rho (\theta \xi (t) + (1-\theta )x) \cdot (y - x - 6{(1 - t)\delta }^{-2}\vert x - y\vert D\rho (z)) {}\\ & \leq & \,t\vert x - y\vert (1 - 3(1 - t)) < 0 {}\\ \end{array}$$

for some θ ∈ (0, 1). Similarly, if t ∈ [1 ∕ 2, 1), we have

$$\displaystyle{\rho (\xi (t)) \leq \rho (y) + (1 - t)\vert x - y\vert (1 - 3t) < 0.}$$

Hence, ξ(t) ∈ Ω for all t ∈ (0, 1). Note that

$$\displaystyle{\vert \dot{\xi }(t)\vert \leq \vert y - x\vert (1 + {6\delta }^{-2}).}$$

If x = y, then we just set η(s) = x = y for s = 0 and the curve $\eta \,:\, [0,\,0] \rightarrow {\mathbb{R}}^{n}$ has the required properties. Now let x ≠ y. We set t(x, y) = (1 + 6δ ^− 2) | x − y | and η(s) = ξ(s ∕ t(x, y)) for s ∈ [0, t(x, y)]. Then the curve $\eta : [0,\,t(x,y)] \rightarrow {\mathbb{R}}^{n}$ has the required properties with C = 1 + 6δ ^− 2.

Thus, by the compactness of $\overline{\Omega }$, we may choose a constant C > 0 and a finite covering $\{{B}^{i}\}_{i=1}^{N}$ of $\overline{\Omega }$ consisting of open balls with the properties: for each $x,y \in \hat{ B}_{i} \cap \overline{\Omega }$, where $\hat{B}_{i}$ denotes the concentric open ball of B _i with radius twice that of B _i , there exists a curve $\eta \in \mathrm{AC}([0,\,t(x,y)], {\mathbb{R}}^{n})$ such that η(s) ∈ Ω for all s ∈ (0, t(x, y)), $\vert \dot{\eta }(s)\vert \leq 1$ for a.e. s ∈ [0, t(x, y)] and t(x, y) ≤ C | x − y | .

Let r _i be the radius of the ball B _i and set r = minr _i and R = ∑r _i , where i ranges all over i = 1, …, N.

Let $x,y \in \overline{\Omega }$. If | x − y |  < r, then $x,y \in \hat{ B}_{i}$ for some i and there is a curve $\eta \in \mathrm{AC}([0,t(x,y)], {\mathbb{R}}^{n})$ such that η(s) ∈ Ω for all s ∈ (0, t(x, y)), $\vert \dot{\eta }(s)\vert \leq 1$ for a.e. s ∈ [0, t(x, y)] and t(x, y) ≤ C | x − y | . Next, we assume that | x − y | ≥ r. By the connectedness of Ω, we infer that there is a sequence $\{B_{i_{j}}\, :\, j = 1,\ldots,J\} \subset \{ B_{i}\, :\, i = 1,\ldots,N\}$ such that $x \in B_{i_{1}}$, $y \in B_{i_{J}}$, $B_{i_{j}} \cap B_{i_{j+1}} \cap \Omega \not =\varnothing$ for all 1 ≤ j < J, and $B_{i_{j}}\not =B_{i_{k}}$ if j ≠ k. It is clear that J ≤ N. If J = 1, then we may choose a curve η with the required properties as in the case where | x − y |  < r. If J > 1, then we may choose a curve $\eta \in \mathrm{AC}([0,\,t(x,y)],\, {\mathbb{R}}^{n})$ joining x and y as follows. First, we choose a sequence {x _j  : j = 1, …, J − 1} of points in Ω so that $x_{j} \in B_{i_{j}} \cap B_{i_{j+1}} \cap \Omega$ for all 1 ≤ j < J. Next, setting x ₀ = x, x _J  = y and t ₀ = 0, since $x_{j-1},x_{i_{j}} \in B_{j} \cap \overline{\Omega }$ for all 1 ≤ j ≤ J, we may select $\eta _{j} \in \mathrm{AC}([t_{j-1},\,t_{j}],\, {\mathbb{R}}^{n})$, with 1 ≤ j ≤ J, inductively so that $\eta _{j}(t_{j-1}) = x_{j-1}$, $\eta _{j}(t_{j}) = x_{j}$, η _j (s) ∈ Ω for all s ∈ (t _j − 1, t _j ) and $t_{j} \leq t_{j-1} + C\vert x_{j} - x_{j-1}\vert$. Finally, we define $\eta \in \mathrm{AC}([0,\,t(x,y)], {\mathbb{R}}^{n})$, with t(x, y) = t _J , by setting η(s) = η _i (s) for s ∈ [t _j  − 1, t _j ] and 1 ≤ j ≤ J. Noting that

$$\displaystyle{T \leq C\sum _{j=1}^{J}\vert x_{ j} - x_{j-1}\vert \leq C\sum _{j=1}^{J}r_{ i_{j}} \leq CR \leq CR{r}^{-1}\vert x - y\vert,}$$

we see that the curve $\eta \in \mathrm{AC}([0,\,t(x,y)],\, {\mathbb{R}}^{n})$ has all the required properties with C replaced by CRr ^− 1. □ 

Remark C.1.

(i) A standard argument, different from the above one, to prove the local Lipschitz continuity near the boundary points is to flatten the boundary by a local change of variables. (ii) One can easily modify the above proof to prove the proposition same as Lemma 2.1, except that Ω is a Lipschitz domain.

Proof (Lemma 2.2). 

Let C > 0 be the constant from Lemma 2.1. We show that | u(x) − u(y) | ≤ CM | x − y | for all x, y ∈ Ω.

To show this, we fix any x, y ∈ Ω such that x ≠ y. By Lemma 2.1, there is a curve $\eta \in \mathrm{AC}([0,\,t(x,y)],\, {\mathbb{R}}^{n})$ such that  η(0) = x,  η(t(x, y)) = y,  t(x, y) ≤ C | x − y | ,  η(s) ∈ Ω  for all s ∈ [0, t(x, y)]  and  $\vert \dot{\eta }(s)\vert \leq 1$  for a.e. s ∈ [0, t(x, y)].

By the compactness of the image η([0, t(x, y)]) of interval [0, t(x, y)] by η, we may choose a finite sequence $\{B_{i}\}_{i=1}^{N}$ of open balls contained in Ω which covers η([0, t(x, y)]). We may assume by rearranging the label i if needed that x ∈ B ₁, y ∈ B _N and $B_{i} \cap B_{i+1}\not =\varnothing$ for all 1 ≤ i < N. We may choose a sequence $0 = t_{0} < t_{1} < \cdots < t_{N} = t(x,y)$ of real numbers so that the line segment $[\eta (t_{i-1}),\,\eta (t_{i})]$ joining η(t _i − 1) and η(t _i ) lies in B _i for any i = 1, …, N.

Thanks to Proposition 1.14, we have

$$\displaystyle{\vert u(\eta (t_{i})) - u(\eta (t_{i-1}))\vert \leq M\vert \eta (t_{i}) -\eta (t_{i-1})\vert \ \ \ \text{ for all }i = 1,\ldots,N.}$$

Using this, we compute that

$$\displaystyle\begin{array}{rcl} \vert u(y) - u(x)\vert & =& \,\vert u(\eta (t_{N})) - u(\eta (t_{0}))\vert \leq \sum _{i=1}^{N}\vert u(\eta (t_{ i})) - u(\eta (t_{i-1}))\vert {}\\ &\leq & \,M\sum _{i=1}^{N}\vert \eta (t_{ i}) -\eta (t_{i-1})\vert \leq M\sum _{i=1}^{N}\int _{ t_{i-1}}^{t_{i} }\vert \dot{\eta }(s)\vert \mathrm{d}s {}\\ & =& \,M\int _{t_{0}}^{t_{N} }\vert \dot{\eta }(s)\vert \mathrm{d}s \leq M(t_{N} - t_{0}) = \mathit{Mt}(x,y) \leq CM\vert x - y\vert. {}\\ \end{array}$$

This completes the proof. □ 

1.4 A.4 Localized Versions of Lemma 4.2

Theorem D.1.

Let U, V be open subsets of ${\mathbb{R}}^{n}$ with the properties: $\overline{V } \subset U$ and $V \cap \Omega \not =\varnothing$ . Let $u \in C(U \cap \overline{\Omega })$ be a viscosity solution of

$$\displaystyle{ \left \{\begin{array}{llll} &H(x,\mathit{Du}(x)) \leq 0&&\mathrm{in}\ U \cap \Omega, \\ &\frac{\partial u} {\partial \gamma } (x) \leq g(x) &&\mathrm{on}\ U \cap \partial \Omega. \end{array} \right. }$$

(131)

Then, for each $\varepsilon \in (0,\,1)$ , there exists a function ${u}^{\varepsilon } \in {C}^{1}(V \cap \overline{\Omega })$ such that

$$\displaystyle{ \left \{\begin{array}{llll} &H(x,{\mathit{Du}}^{\varepsilon }(x)) \leq \varepsilon &&\mathrm{in}\ V \cap \Omega, \\ &\frac{\partial {u}^{\varepsilon }} {\partial \gamma } (x) \leq g(x) &&\mathrm{on}\ V \cap \partial \Omega,\\ &\|{u}^{\varepsilon } - u\|_{ \infty,V \cap \Omega } \leq \varepsilon. \end{array} \right. }$$

Proof.

We choose functions $\zeta,\,\eta \in {C}^{1}({\mathbb{R}}^{n})$ so that 0 ≤ ζ(x) ≤ η(x) ≤ 1 for all $x \in {\mathbb{R}}^{n}$, ζ(x) = 1 for all x ∈ V, η(x) = 1 for all $x \in \mathrm{supp}\zeta$ and $\mathrm{supp}\eta \subset U$.

We define the function $v \in C(\overline{\Omega })$ by setting v(x) = η(x)u(x) for $x \in U \cap \overline{\Omega }$ and v(x) = 0 otherwise. By the coercivity of H, u is locally Lipschitz continuous in $U \cap \overline{\Omega }$, and hence, v is Lipschitz continuous in $\overline{\Omega }$. Let L > 0 be a Lipschitz bound of v in $\overline{\Omega }$. Then v is a viscosity solution of

$$\displaystyle{ \left \{\begin{array}{llll} &\vert Dv(x)\vert \leq L&&\mathrm{in}\ \Omega, \\ &\frac{\partial v} {\partial \gamma } (x) \leq M &&\mathrm{in}\ \ \partial \Omega, \end{array} \right. }$$

where $M := L\|\gamma \|_{\infty,\partial \Omega }$. In fact, we have a stronger assertion that for any $x \in \overline{\Omega }$ and any p ∈ D ⁺ v(x),

$$\displaystyle{ \left \{\begin{array}{llll} &\vert p\vert \leq L &&\mathrm{if}\ \ x \in \Omega,\\ &\gamma (x) \cdot p \leq M&&\mathrm{if}\ \ x \in \partial \Omega. \end{array} \right. }$$

(132)

To check this, let $\phi \in {C}^{1}(\overline{\Omega })$ and assume that v − ϕ attains a maximum at $x \in \overline{\Omega }$. Observe that if x ∈ Ω, then | Dϕ(x) | ≤ L and that if x ∈ ∂Ω, then

$$\displaystyle\begin{array}{rcl} 0& \leq & \,\liminf _{t\rightarrow 0+}\frac{(v-\phi )(x - t\gamma (x)) - (v-\phi )(x)} {-t} {}\\ & =& \,\liminf _{t\rightarrow 0+}\frac{v(x - t\gamma (x)) - v(x)} {-t} -\frac{\partial \phi } {\partial \gamma }(x), {}\\ \end{array}$$

which yields

$$\displaystyle{\gamma (x) \cdot D\phi (x) \leq L\vert \gamma (x)\vert \leq M.}$$

Thus, (132) is valid.

We set

$$\displaystyle\begin{array}{rcl} h(x)& =& \,\zeta (x)g(x) + (1 -\zeta (x))M\ \ \text{ for }\ x \in \partial \Omega, {}\\ G(x,p)& =& \,\zeta (x)H(x,p) + (1 -\zeta (x))(\vert p\vert - L)\ \ \text{ for }\ (x,p) \in \overline{\Omega } \times {\mathbb{R}}^{n}. {}\\ \end{array}$$

It is clear that h ∈ C(∂Ω) and G satisfies (A5)–(A7), with H replaced by G

In view of the coercivity of H, we may assume by reselecting L if necessary that for all $(x,p) \in \overline{\Omega } \times {\mathbb{R}}^{n}$, if | p |  > L, then H(x, p) > 0. We now show that v is a viscosity solution of

$$\displaystyle{ \left \{\begin{array}{llll} &G(x,Dv(x)) \leq 0&&\mathrm{in}\ \Omega, \\ &\frac{\partial v} {\partial \gamma } (x) \leq h(x) &&\mathrm{on}\ \partial \Omega. \end{array} \right. }$$

(133)

To do this, let $\hat{x} \in \overline{\Omega }$ and $\hat{p} \in {D}^{+}v(\hat{x})$. Consider the case where $\zeta (\hat{x}) > 0$, which implies that $\hat{x} \in U$. We have η(x) = 1 near the point $\hat{x}$, which implies that $\hat{p} \in {D}^{+}u(\hat{x})$. As u is a viscosity subsolution of (131), we have  $H(\hat{x},\hat{p}) \leq 0$ if $\hat{x} \in \Omega$ and $\min \{H(\hat{x},\hat{p}),\,\gamma (\hat{x}) \cdot \hat{ p} - h(\hat{x})\} \leq 0$ if $\hat{x} \in \partial \Omega$. Assume in addition that $\hat{x} \in \partial \Omega$. By (132), we have $\gamma (\hat{x}) \cdot \hat{ p} \leq M$. If $\vert \hat{p}\vert > L$, we have both

$$\displaystyle{\gamma (\hat{x}) \cdot \hat{ p} \leq g(\hat{x})\ \ \text{ and }\ \ \gamma (\hat{x}) \cdot \hat{ p} \leq M.}$$

Hence, if $\vert \hat{p}\vert > L$, then $\gamma (\hat{x}) \cdot \hat{ p} \leq h(\hat{x})$. On the other hand, if $\vert \hat{p}\vert \leq L$, we have two cases: in one case we have $H(\hat{x},\hat{p}) \leq 0$ and hence, $G(\hat{x},\hat{p}) \leq 0$. In the other case, we have $\gamma (\hat{x}) \cdot \hat{ p} \leq g(\hat{x})$ and then $\gamma (\hat{x}) \cdot \hat{ p} \leq h(\hat{x})$. These observations together show that

$$\displaystyle{\min \{G(\hat{x},\hat{p}),\,\gamma (\hat{x}) \cdot \hat{ p} - h(\hat{x})\} \leq 0.}$$

We next assume that $\hat{x} \in \Omega$. In this case, we easily see that $G(\hat{x},\hat{p}) \leq 0$.

Next, consider the case where $\zeta (\hat{x}) = 0$, which implies that $G(\hat{x},\hat{p}) = \vert \hat{p}\vert - L$ and $h(\hat{x}) = M$. By (132), we immediately see that $G(\hat{x},\hat{p}) \leq 0$ if $\hat{x} \in \Omega$ and $\min \{G(\hat{x},\hat{p}),\gamma (\hat{x}) \cdot \hat{ p} - h(\hat{x})\} \leq 0$ if $\hat{x} \in \partial \Omega$. We thus conclude that v is a viscosity solution of (133).

We may invoke Theorem 4.2, to find a collection $\{{v}^{\varepsilon }\}_{\varepsilon \in (0,1)} \subset {C}^{1}(\overline{\Omega })$ such that

$$\displaystyle{\left \{\begin{array}{llll} &G(x,D{v}^{\varepsilon }(x)) \leq \varepsilon &&\text{ for all }x \in \Omega, \\ &\frac{\partial {v}^{\varepsilon }} {\partial \gamma } (x) \leq h(x) &&\text{ for all }x \in \partial \Omega,\\ &\|{v}^{\varepsilon } - v\|_{ \infty,\Omega } \leq \varepsilon. \end{array} \right.}$$

But, this yields

$$\displaystyle{\left \{\begin{array}{llll} &H(x,{v}^{\varepsilon }(x)) \leq \varepsilon &&\text{ for all }\ x \in V \cap \Omega, \\ &\frac{\partial {v}^{\varepsilon }} {\partial \gamma } (x) \leq g(x) &&\text{ for all }\ x \in V \cap \partial \Omega,\\ &\|{v}^{\varepsilon } - u\|_{ \infty,V \cap \Omega } \leq \varepsilon. \end{array} \right.}$$

The functions ${v}^{\varepsilon }$ have all the required properties. □ 

The above theorem has a version for Hamilton–Jacobi equations of evolution type.

Theorem D.2.

Let U, V be bounded open subsets of ${\mathbb{R}}^{n} \times \mathbb{R}_{+}$ with the properties:   $\overline{V } \subset U$ ,  $\overline{U} \subset {\mathbb{R}}^{n} \times \mathbb{R}_{+}$  and   $V \cap Q\not =\varnothing$ . Let $u \in \mathrm{Lip}(U \cap Q)$ be a viscosity solution of

$$\displaystyle{ \left \{\begin{array}{llll} &u_{t}(x,t) + H(x,D_{x}u(x,t)) \leq 0&&\mathrm{in}\ U \cap (\Omega \times \mathbb{R}_{+}), \\ &\frac{\partial u} {\partial \gamma } (x,t) \leq g(x) &&\mathrm{on}\ U \cap (\partial \Omega \times \mathbb{R}_{+}). \end{array} \right. }$$

Then, for each $\varepsilon \in (0,\,1)$ , there exists a function ${u}^{\varepsilon } \in {C}^{1}(V \cap Q)$ such that

$$\displaystyle{ \left \{\begin{array}{llll} &u_{t}^{\varepsilon }(x,t) + H(x,D_{x}{u}^{\varepsilon }(x,t)) \leq \varepsilon &&\mathrm{in}\ V \cap (\Omega \times \mathbb{R}_{+}), \\ &\frac{\partial {u}^{\varepsilon }} {\partial \gamma } (x,t) \leq g(x) &&\mathrm{on}\ V \cap (\partial \Omega \times \mathbb{R}_{+}),\\ &\|{u}^{\varepsilon } - u\|_{ \infty,V \cap Q} \leq \varepsilon. \end{array} \right. }$$

(134)

Proof.

Choose constants $a,b \in \mathbb{R}_{+}$ so that $U \subset {\mathbb{R}}^{n} \times (a,b)$ and let ρ be a defining function of Ω. We may assume that ρ is bounded in ${\mathbb{R}}^{n}$. We choose a function $\zeta \in {C}^{1}(\mathbb{R})$ so that ζ(t) = 0 for all t ∈ [a, b], ζ′(t) > 0 for all t > b, ζ′(t) < 0 for all t < a and $\min \{\zeta (a/2),\zeta (2b)\} >\|\rho \| _{\infty,\Omega }$.

We set

$$\displaystyle\begin{array}{rcl} \tilde{\rho }(x,t)& =& \,\rho (x) +\zeta (t)\ \ \text{ for }\ (x,t) \in {\mathbb{R}}^{n+1}, {}\\ \tilde{\Omega }& =& \,\{(x,t) \in {\mathbb{R}}^{n+1}\, :\,\tilde{\rho } (x,t) < 0\}. {}\\ \end{array}$$

It is easily seen that

$$\displaystyle{\tilde{\Omega } \subset \Omega \times (a/2,\,2b)\ \ \ \text{ and }\ \ \ \tilde{\Omega } \cap \left ({\mathbb{R}}^{n} \times [a,\,b]\right ) = \Omega \times [a,\,b].}$$

Let $(x,t) \in {\mathbb{R}}^{n+1}$ be such that $\tilde{\rho }(x,t) = 0$. It is obvious that $(x,t) \in \overline{\Omega } \times [a/2,\,2b]$. If a ≤ t ≤ b, then ρ(x) = 0 and thus Dρ(x) ≠ 0. If either t > b or t < a, then | ζ′(t) |  > 0. Hence, we have $D\tilde{\rho }(x,t)\not =0$. Thus, $\tilde{\rho }$ is a defining function of $\tilde{\Omega }$.

Let M > 0 and define $\tilde{\gamma }\in C(\partial \tilde{\Omega }, {\mathbb{R}}^{n+1})$ by

$$\displaystyle{\tilde{\gamma }(x,t) = \left ((1 + M\rho (x))_{+}\gamma (x),\,\zeta ^{\prime}(t)\right ),}$$

where we may assume that γ is defined and continuous in $\overline{\Omega }$. We note that for any $(x,t) \in \partial \tilde{\Omega }$,

$$\displaystyle{\tilde{\gamma }(x,t) \cdot D\tilde{\rho }(x,t) = (1 + M\rho (x))_{+}\gamma (x) \cdot D\rho (x) +\zeta ^{\prime}{(t)}^{2}.}$$

Note as well that (1 + Mρ(x))₊ = 1 for all x ∈ ∂Ω and

$$\displaystyle{\lim _{M\rightarrow \infty }(1 + M\rho (x))_{+} = 0\ \ \ \text{ locally uniformly in }\Omega.}$$

Thus we can fix M > 0 so that for all $(x,t) \in \partial \tilde{\Omega }$,

$$\displaystyle{\tilde{\gamma }(x,t) \cdot D\tilde{\rho }(x,t) = (1 + M\rho (x))_{+}\gamma (x) \cdot D\rho (x) +\zeta ^{\prime}{(t)}^{2} > 0.}$$

Noting that for each x ∈ Ω, the x-section $\{t \in \mathbb{R}\, :\, (x,t) \in \tilde{ \Omega }\}$ of $\tilde{\Omega }$ is an open interval (or, line segment), we deduce that $\tilde{\Omega }$ is a connected set. We may assume that g is defined and continuous in $\overline{\Omega }$. We define $\tilde{g} \in C(\partial \tilde{\Omega })$ by $\tilde{g}(x,t) = g(x)$. Thus, assumptions (A1)–(A4) hold with n + 1, $\tilde{\Omega }$, $\tilde{\gamma }$ and $\tilde{g}$ in place of n, Ω, γ and g.

Let L > 0 be a Lipschitz bound of the function u in $U \cap Q$. Set

$$\displaystyle{\tilde{H}(x,t,p,q) = H(x,p) + q + 2(\vert q\vert - L)_{+}\ \ \ \text{ for }(x,t,p,q) \in \overline{\tilde{\Omega }\,} \times {\mathbb{R}}^{n+1},}$$

and note that $\tilde{H} \in C(\overline{\tilde{\Omega }\,} \times {\mathbb{R}}^{n+1})$ satisfies (A5)–(A7), with Ω replaced by $\tilde{\Omega }$.

We now claim that u is a viscosity solution of

$$\displaystyle{\left \{\begin{array}{llll} &\tilde{H}(x,t,\mathit{Du}(x,t)) \leq 0 &&\mathrm{in}\ U \cap \tilde{ \Omega },\\ &\tilde{\gamma }(x, t) \cdot \mathit{Du } (x, t) \leq \tilde{ g}(x, t) & &\mathrm{on } \ U \cap \partial \tilde{\Omega }. \end{array} \right.}$$

Indeed, since $U \cap \tilde{ \Omega } = U \cap Q$ and $U \cap \partial \tilde{\Omega } = U \cap \partial Q$, if $(x,t) \in U \cap \overline{\tilde{\Omega }}$ and (p, q) ∈ D ⁺ u(x, t), then we get | q | ≤ L by the cylindrical geometry of Q and, by the viscosity property of u,

$$\displaystyle{\left \{\begin{array}{llll} &q + H(x,p) + 2(\vert q\vert - L)_{+} \leq 0 &&\mathrm{if}\ \ (x,t) \in \tilde{ \Omega }, \\ &\min \{q + H(x,p) + 2(\vert q\vert - L)_{+},\gamma (x) \cdot p - g(x)\} \leq 0&&\mathrm{if}\ (x,t) \in \partial \tilde{\Omega }.\end{array} \right.}$$

We apply Theorem D.1, to find a collection $\{{u}^{\varepsilon }\}_{\varepsilon \in (0,1)} \subset {C}^{1}(V \cap \overline{\tilde{\Omega }\,})$ such that

$$\displaystyle{\left \{\begin{array}{llll} &\tilde{H}(x,t,{\mathit{Du}}^{\varepsilon }(x,t)) \leq \varepsilon &&\mathrm{in}\ V \cap \tilde{ \Omega }, \\ &\tilde{\gamma }(x,t) \cdot {\mathit{Du}}^{\varepsilon }(x,t) \leq \tilde{ g}(x,t)&&\mathrm{on}\ U \cap \tilde{ \Omega }, \\ &\|{u}^{\varepsilon } - u\|_{\infty,V \cap \tilde{\Omega }} \leq \varepsilon.\end{array} \right.}$$

It is straightforward to see that the collection $\{{u}^{\varepsilon }\}_{\varepsilon \in (0,1)} \subset {C}^{1}(V \cap Q)$ satisfies (134). □ 

1.5 A.5 A Proof of Lemma 5.4

This subsection is mostly devoted to the proof of Lemma 5.4, a version of the Dunford–Pettis theorem. We also give a proof of the weak-star compactness of bounded sequences in ${L}^{\infty }(J, {\mathbb{R}}^{m})$, where J = [a, b] is a finite interval in $\mathbb{R}$.

Proof (Lemma 5.4). 

We define the functions $F_{j} \in C(J, {\mathbb{R}}^{m})$ by

$$\displaystyle{F_{j}(x) =\int _{ a}^{x}f_{ j}(t)\mathrm{d}t.}$$

By the uniform integrability of {f _j }, the sequence $\{F_{j}\}_{j\in \mathbb{N}}$ is uniformly bounded and equi-continuous in J. Hence, the Ascoli–Arzela theorem ensures that it has a subsequence converging to a function F uniformly in J. We fix such a subsequence and denote it again by the same symbol {F _j }. Because of the uniform integrability assumption, the sequence {F _j } is equi-absolutely continuous in J. That is, for any $\varepsilon > 0$ there exists δ > 0 such that

$$\displaystyle\begin{array}{rcl} & & a \leq a_{1} < b_{1} < a_{2} < b_{2} < \cdots < a_{n} < b_{n} \leq b,\ \quad \ \sum _{i=1}^{n}(b_{ i} - a_{i}) <\delta,\ {}\\ & & \quad \Rightarrow\ \sum _{i=1}^{n}\vert f_{ j}(b_{i}) - f_{j}(a_{i})\vert <\varepsilon \ \ \ \text{ for all }\ j \in \mathbb{N}. {}\\ \end{array}$$

An immediate consequence of this is that $F \in \mathrm{AC}(J, {\mathbb{R}}^{m})$. Hence, for some $f \in {L}^{1}(J, {\mathbb{R}}^{m})$, we have

$$\displaystyle{F(x) =\int _{ a}^{x}f(t)\,\mathrm{d}t\ \ \ \text{ for all }\ x \in J.}$$

Next, let ϕ ∈ C ¹(J), and we show that

$$\displaystyle{ \lim _{j\rightarrow \infty }\int _{a}^{b}f_{ j}(x)\phi (x)\,\mathrm{d}x =\int _{ a}^{b}f(x)\phi (x)\,\mathrm{d}x. }$$

(135)

Integrating by parts, we observe that as j → ∞,

$$\displaystyle\begin{array}{rcl} \int _{a}^{b}f_{ j}(x)\phi (x)\,\mathrm{d}x& & \,=\big [F_{j}\phi \big]_{a}^{b} -\int _{ a}^{b}F_{ j}(x)\phi ^{\prime}(x)\,\mathrm{d}x {}\\ & & \,\rightarrow \big [F\phi \big]_{a}^{b} -\int _{ a}^{b}F(x)\phi ^{\prime}(x)\,\mathrm{d}x =\int _{ a}^{b}f(x)\phi (x)\,\mathrm{d}x. {}\\ \end{array}$$

Hence, (135) is valid.

Now, let ϕ ∈ L ^∞(J). We regard the functions f _j , f, ϕ as functions defined in $\mathbb{R}$ by setting f _j (x) = f(x) = ϕ(x) = 0 for x < a or x > b. Let $\{k_{\varepsilon }\}_{\varepsilon >0}$ be a collection of standard mollification kernels. We recall that

$$\displaystyle{ \lim _{\varepsilon \rightarrow 0}\|k_{\varepsilon } {\ast}\phi -\phi \|_{{L}^{1}(J)} = 0, }$$

(136)

$$\displaystyle{ \vert k_{\varepsilon } {\ast}\phi (x)\vert \leq \|\phi \|_{{L}^{\infty }(J)}\ \ \ \text{ for all }\ x \in J,\ \varepsilon > 0. }$$

(137)

Fix any δ > 0. By the uniform integrability assumption, we have

$$\displaystyle{M :=\sup _{j\in \mathbb{N}}\|f_{j} - f\|_{{L}^{1}(J)} < \infty.}$$

Let α > 0 and set

$$\displaystyle{E_{j} :=\{ x \in J\, :\, \vert (f_{j} - f)(x)\vert >\alpha \}.}$$

By the Chebychev inequality, we get

$$\displaystyle{\vert E_{j}\vert \leq \frac{M} {\alpha }.}$$

By the uniform integrability assumption, if α > 0 is sufficiently large, then

$$\displaystyle{ \int _{E_{j}}\vert (f_{j} - f)(x)\vert \,\mathrm{d}x <\delta. }$$

(138)

In what follows we fix α > 0 large enough so that (138) holds. We write f _j  − f = g _j  + b _j , where  $g_{j} = (f_{j} - f)(1 -\mathbf{1}_{E_{j}})$  and  $b_{j} = (f_{j} - f)\mathbf{1}_{E_{j}}$. Then,

$$\displaystyle{\vert g_{j}(x)\vert \leq \alpha \ \ \ \text{ for all }\ x \in J\quad \text{ and }\quad \|b_{j}\|_{{L}^{1}(J)} <\delta.}$$

Observe that

$$\displaystyle\begin{array}{rcl} I_{j} :& =& \,\int _{J}f_{j}(x)\phi (x)\,\mathrm{d}x -\int _{J}f(x)\phi (x)\,\mathrm{d}x {}\\ & =& \,\int _{J}(f_{j} - f)(x)\,k_{\varepsilon } {\ast}\phi (x)\,\mathrm{d}x +\int _{J}(f_{j} - f)(x)(\phi -k_{\varepsilon }{\ast}\phi )(x)\,\mathrm{d}x {}\\ \end{array}$$

and

$$\displaystyle\begin{array}{rcl} & & \Big\vert \int _{J}(f_{j} - f)(x)(\phi -k_{\varepsilon }{\ast}\phi )(x)\,\mathrm{d}x\Big\vert {}\\ & &\quad \leq \Big\vert \int _{J}g_{j}(x)(\phi -k_{\varepsilon }{\ast}\phi )(x)\,\mathrm{d}x\Big\vert +\Big \vert \int _{J}b_{j}(x)(\phi -k_{\varepsilon }{\ast}\phi )(x)\,\mathrm{d}x\Big\vert {}\\ & &\quad \leq \alpha \| k_{\varepsilon } {\ast}\phi -\phi \|_{{L}^{1}(J)} + 2\delta \|\phi \|_{{L}^{\infty }(J)}. {}\\ \end{array}$$

Hence, in view of (135) and (136), we get  $\limsup _{j\rightarrow \infty }\vert I_{j}\vert \leq 2\delta \|\phi \|_{{L}^{\infty }(J)}.$ As δ > 0 is arbitrary, we get  $\lim _{j\rightarrow \infty }I_{j} = 0,$ which completes the proof. □ 

As a corollary of Lemma 5.4, we deduce that the weak-star compactness of bounded sequences in ${L}^{\infty }(J, {\mathbb{R}}^{m})$:

Lemma E.1.

Let J = [a, b], with −∞ < a < b < ∞. Let $\{f_{k}\}_{k\in \mathbb{N}}$ be a bounded sequence of functions in ${L}^{\infty }(J, {\mathbb{R}}^{m})$ . Then {f _k } has a subsequence which converges weakly-star in ${L}^{\infty }(J, {\mathbb{R}}^{m})$.

Proof.

Set $M =\sup _{k\in \mathbb{N}}\|f_{k}\|_{{L}^{\infty }(J)}$. Let E ⊂ J be a measurable set, and observe that

$$\displaystyle{\int _{E}\vert f_{k}(t)\vert \mathrm{d}t \leq M\vert E\vert \ \ \ \text{ for all }k \in \mathbb{N},}$$

which shows that the sequence {f _k } is uniformly integrable in J. Thanks to Lemma 5.4, there exists a subsequence $\{f_{k_{j}}\}_{j\in \mathbb{N}}$ of {f _k } which converges to a function f weakly in L ¹(J, R ^m).

Let $i \in \mathbb{N}$ and set E _i  = { t ∈ J :  | f(t) |  > M + 1 ∕ i} and $g_{i}(t) = \mathbf{1}_{E_{i}}(t)f(t)/\vert f(t)\vert$ for t ∈ J. Since $g_{i} \in {L}^{\infty }(J, {\mathbb{R}}^{m})$, we get

$$\displaystyle{\int _{J}f_{k_{j}}(t) \cdot g_{i}(t)\mathrm{d}t \rightarrow \int _{J}\vert f(t)\vert \mathbf{1}_{E_{i}}(t)\mathrm{d}t\ \ \text{ as }\ j \rightarrow \infty.}$$

Hence, using the Chebychev inequality, we obtain

$$\displaystyle{\Big(M + \frac{1} {i} \big)\vert E_{i}\vert \leq \int _{J}\vert f(t)\vert \mathbf{1}_{E_{i}}(t)\mathrm{d}t \leq \int _{J}M\mathbf{1}_{E_{i}}(t)\mathrm{d}t = M\vert E_{i}\vert,}$$

which ensures that | E _i  |  = 0. Thus, we find that | f(t) | ≤ M a.e. in J.

Now, fix any $\phi \in {L}^{1}(J, {\mathbb{R}}^{m})$. We select a sequence $\{\phi _{i}\}_{i\in \mathbb{N}} \subset {L}^{\infty }(J, {\mathbb{R}}^{m})$ so that, as i → ∞, ϕ _i  → ϕ in ${L}^{1}(J, {\mathbb{R}}^{m})$. For each $i \in \mathbb{N}$, we have

$$\displaystyle{\lim _{j\rightarrow \infty }\int _{J}f_{k_{j}}(t) \cdot \phi _{i}(t)\mathrm{d}t =\int _{J}f(t) \cdot \phi _{i}(t)\mathrm{d}t.}$$

On the other hand, we have

$$\displaystyle{\Big\vert \int _{J}f_{k_{j}}(t) \cdot \phi (t)\mathrm{d}t -\int _{J}f_{k_{j}}(t) \cdot \phi _{i}(t)\mathrm{d}t\Big\vert \leq M\|\phi -\phi _{i}\|_{{L}^{1}(J)}\ \ \ \text{ for all }\ j \in \mathbb{N}}$$

and

$$\displaystyle{\Big\vert \int _{J}f(t) \cdot \phi (t)\mathrm{d}t -\int _{J}f(t) \cdot \phi _{i}(t)\mathrm{d}t\Big\vert \leq M\|\phi -\phi _{i}\|_{{L}^{1}(J)}.}$$

These together yield

$$\displaystyle{\lim _{j\rightarrow \infty }\int _{J}f_{k_{j}}(t) \cdot \phi (t)\mathrm{d}t =\int _{J}f(t) \cdot \phi (t)\mathrm{d}t.}$$

 □ 

1.6 A.6 Rademacher’s Theorem

We give here a proof of Rademacher’s theorem.

Theorem F.1 (Rademacher). 

Let $B = B_{1} \subset {\mathbb{R}}^{n}$ and $f \in \mathrm{Lip}(B)$ . Then f is differentiable almost everywhere in B.

To prove the above theorem, we mainly follow the proof given in [1].

Proof.

We first show that f has a distributional gradient Df ∈ L ^∞(B).

Let L > 0 be a Lipschitz bound of the function f. Let i ∈ { 1, 2, …, n} and e _i denote the unit vector in ${\mathbb{R}}^{n}$ with unity as the i-th entry. Fix any ϕ ∈ C ₀ ¹(B) and observe that

$$\displaystyle\begin{array}{rcl} \int _{B}f(x)\phi _{x_{i}}(x)\mathrm{d}x& =& \,\lim _{r\rightarrow 0+}\int _{B}f(x)\frac{\phi (x + re_{i}) -\phi (x)} {r} \mathrm{d}x {}\\ & =& \,\lim _{r\rightarrow 0+}\int _{B}\frac{f(x - re_{i}) - f(x)} {r} \phi (x)\mathrm{d}x {}\\ \end{array}$$

and

$$\displaystyle{\Big\vert \int _{B}f(x)\phi _{x_{i}}(x)\mathrm{d}x\Big\vert \leq L\int _{B}\vert \phi (x)\vert \mathrm{d}x \leq L\vert B{\vert }^{1/2}\|\phi \|_{{ L}^{2}(B)}.}$$

Thus, the map

$$\displaystyle{C_{0}^{1}(B) \ni \phi \mapsto -\int _{ B}f(x)\phi _{x_{i}}(x)\mathrm{d}x \in \mathbb{R}}$$

extends uniquely to a bounded linear functional G _i on L ²(B). By the Riesz representation theorem, there is a function g _i  ∈ L ²(B) such that

$$\displaystyle{G_{i}(\phi ) =\int _{B}g_{i}(x)\phi (x)\mathrm{d}x\ \ \ \text{ for all }\ \phi \in {L}^{2}(B).}$$

This shows that g = (g ₁, …, g _n ) is the distributional gradient of f.

We plug the function ϕ ∈ L ²(B) given by $\phi (x) = (g_{i}(x)/\vert g_{i}(x)\vert )\mathbf{1}_{E_{k}}(x)$, where $k \in \mathbb{N}$ and E _k  = { x ∈ B :  | g _i (x) |  > L + 1 ∕ k}, into the inequality $\vert G_{i}(\phi )\vert \leq L\|\phi \|_{{L}^{1}(B)}$, to obtain

$$\displaystyle{\int _{B}\vert g_{i}(x)\vert \mathbf{1}_{E_{k}}(x)\mathrm{d}x \leq L\int _{B}\mathbf{1}_{E_{k}}(x)\mathrm{d}x = L\vert E_{k}\vert,}$$

which yields

$$\displaystyle{(L + 1/k)\vert E_{k}\vert \leq L\vert E_{k}\vert.}$$

Hence, we get | E _k  |  = 0 for all $k \in \mathbb{N}$ and | {x ∈ B :  | g _i (x) |  > L} |  = 0. That is, g _i  ∈ L ^∞(B) and | g _i (x) | ≤ L a.e. in B.

The Lebesgue differentiation theorem (see [57]) states that for a.e. x ∈ B, we have $g(x) \in {\mathbb{R}}^{n}$ and

$$\displaystyle{ \lim _{r\rightarrow 0+} \frac{1} {{r}^{n}}\int _{B_{r}}\vert g(x + y) - g(x)\vert \mathrm{d}y = 0. }$$

(139)

Now, we fix such a point x ∈ B and show that f is differentiable at x. Fix an r > 0 so that B _r (x) ⊂ B. For δ ∈ (0, r), consider the function $h_{\delta } \in C(\overline{B})$ given by

$$\displaystyle{h_{\delta }(y) = \frac{f(x +\delta y) - f(x)} {\delta }.}$$

We claim that

$$\displaystyle{ \lim _{\delta \rightarrow 0}h_{\delta }(y) = g(x) \cdot y\ \ \ \text{ uniformly for }\ y \in \overline{B}. }$$

(140)

Note that h _δ (0) = 0 and h _δ is Lipschitz continuous with Lipschitz bound L. By the Ascoli–Arzela theorem, for any sequence {δ _j } ⊂ (0, r) converging to zero, there exist a subsequence $\{\delta _{j_{k}}\}_{k\in \mathbb{N}}$ of {δ _j } and a function $h_{0} \in C(\overline{B})$ such that

$$\displaystyle{\lim _{k\rightarrow \infty }h_{\delta _{j_{ k}}}(x) = h_{0}(y)\ \ \ \text{ uniformly for }\ y \in \overline{B}.}$$

In order to prove (140), we need only to show that h ₀(y) = g(x) ⋅y for all y ∈ B.

Since h _δ (0) = 0 for all δ ∈ (0, r), we have h ₀(0) = 0. We observe from (139) that

$$\displaystyle{\int _{B}\vert g(x +\delta y) - g(x)\vert \mathrm{d}y =\int _{B_{\delta }}\vert g(x + y) - g(x){\vert \delta }^{-n}\mathrm{d}y\ \rightarrow \ 0\ \ \text{ as }\ \delta \rightarrow 0.}$$

Using this, we compute that for all ϕ ∈ C ₀ ¹(B),

$$\displaystyle\begin{array}{rcl} \int _{B}h_{0}(y)\phi _{y_{i}}(y)\mathrm{d}y& =& \,\lim _{k\rightarrow \infty }\int _{B}h_{\delta _{j_{ k}}}(y)\phi _{y_{i}}(y)\mathrm{d}y {}\\ & =& \,-\lim _{k\rightarrow \infty }\int _{B}g_{i}(x +\delta _{j_{k}}y)\phi (y)\mathrm{d}y {}\\ & =& \,-\int _{B}g_{i}(x)\phi (y)\mathrm{d}y =\int _{B}g(x) \cdot y\phi _{y_{i}}(y)\mathrm{d}y. {}\\ \end{array}$$

This guarantees that h ₀(y) − g(x) ⋅y is constant for all y ∈ B while h ₀(0) = 0. Thus, we see that h ₀(y) = g(x) ⋅y for all y ∈ B, which proves (140).

Finally, we note that (140) yields

$$\displaystyle{f(x + y) = f(x) + g(x) \cdot y + o(\vert y\vert )\ \ \ \text{ as }\ y \rightarrow 0.}$$

 □