real analysis - Sub-gradient and super-gradient are bounded implies globally Lipschitz.

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.

Answer

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 $\phi(x)$.