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Let $u$ be a real valued function defined on the open set $\Omega \subset \mathbb{R}^n$. Assume that $u$ is continuous, for any $x\in \Omega$, the sets \begin{align*} D^-u(x) &= \left\lbrace p\in \mathbb{R}^n: \liminf_{y\longrightarrow x} \frac{u(y) - u(x) - \langle p,y-x\rangle}{\Vert y-x\Vert} \geq 0 \right\rbrace \\ D^+u(x) &= \left\lbrace p\in \mathbb{R}^n: \limsup_{y\longrightarrow x} \frac{u(y) - u(x) - \langle p,y-x\rangle}{\Vert y-x\Vert} \leq 0 \right\rbrace \end{align*} are called, respectively the subdifferential and superdifferential of $u$ at $x$.

One can prove that for a $C^1$ function $\phi$ then

\begin{align*} u-\phi \;\text{has a strict max at}\;x_0 &\Longleftrightarrow u\;\text{is touched from above by}\;\phi\;\text{at}\;x_0\\ &\Longleftrightarrow D\phi(x_0) \in D^+u(x_0),\\ u-\phi \;\text{has a strict min at}\;x_0 &\Longleftrightarrow u\;\text{is touched from below by}\;\phi\;\text{at}\;x_0\\ &\Longleftrightarrow D\phi(x_0) \in D^-u(x_0). \end{align*} And $u$ is differentiable at $x$ if and only if $D^+u(x) = D^-u(x) = \{\nabla u(x)\}$.

MY QUESTION: If I have $u$ is continuous on the whole space $\mathbb{R}^n$ and there exists a constant $C>0$ such that

\begin{align*} \Vert p\Vert \leq C \qquad \text{for all}\; p\in D^+u(x)\;\text{or}\; p\in D^-u(x)\;\text{for all}\;x. \end{align*} Could I have $u$ is Lipschitz globally? This question pops out when I study the notion of viscosity solution for Hamilton-Jacobi equations. We know that if $u$ is differentiable and $\Vert Du\Vert$ is bounded then $u$ is Lipschitz, I just wonder that maybe my statement is true, but I failed to prove it, though I can prove that it is locally Lipschitz.

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Yes. The proof is similar to how one proves that viscosity solutions are Lipschitz when the Hamiltonian is coercive. You will have to assume some growth conditions on $u$ at $\infty$ (though this could be relaxed with some effort). I'll give the proof for $u$ bounded.

Assume $u$ is bounded and let $y \in \mathbb{R}^n$. Define $\phi(x) = (C+\epsilon)|x-y|$. Since $u$ is bounded, $u-\phi$ attains its maximum at some point $x_0 \in \mathbb{R}^n$. If $x_0\neq y$ then $\phi$ is smooth in a neighborhood of $x_0$ and so $p := D\phi(x_0) \in D^+u(x_0)$. Computing $|p| = C+\epsilon>C$, we get a contradiction.

Therefore $x_0=y$ and we get

$$u(x) - (C+\epsilon)|x-y| \leq u(y)$$

for all $\epsilon>0$ and $x \in \mathbb{R}^n$. It follows that

$$u(x) - u(y) \leq C|x-y|$$

for all $x,y \in \mathbb{R}^n$. The basic idea of the proof is that we showed we can touch the graph of $u$ with a cone $\phi(x)$.

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  • $\begingroup$ Thank you very much, could you give me some instruction how to deal with the case $u$ is not bounded with some growth condition of $u$? I really appreciate. $\endgroup$ – Sean Nov 25 '16 at 19:40
  • $\begingroup$ You have to cut off $u$ so that it is bounded. Take a smooth function $\Phi:\mathbb{R}\to (-2,2)$ such that $0 < \Phi'(s) \leq 1$ for all $s$ and $\Phi(s) = s$ for $s \in (-1,1)$. Then set $v(x) = \Phi(u(x)/R)R$ for large $R>0$. Then show that $v$ is bounded and satisfies the same hypotheses as $u$ (regarding super and sub differentials being bounded). $\endgroup$ – Jeff Nov 25 '16 at 21:15
  • $\begingroup$ But then what is the growth condition that $u$ must satisfy? $\endgroup$ – Sean Nov 26 '16 at 0:06
  • $\begingroup$ There is none! My initial thoughts were wrong. $\endgroup$ – Jeff Nov 26 '16 at 3:27
  • $\begingroup$ Looks like we don't need to use the fact that $D^-u(x)$ is bounded, cause from $$ u(x) - u(y) \leq C|x-y|$$ we can redo the same argument to obtain $$ u(y) - u(x) \leq C|y-x|$$ I think it is reasonable. $\endgroup$ – Sean Jul 3 '17 at 4:31

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