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.


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 $\phi(x)$.

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.