Assume $A$ is open convex and $f$ convex continuous. Then $f$ is $L$-lipschitz on $A$ if and only if $\partial f(A) \subset L B_{X^*}$ This thread is meant to record a question that I feel interesting during my self-study. I'm very happy to receive your suggestion and comments. See: SE blog: Answer own Question and MSE meta: Answer own Question.

Let $A$ be a subset of a normed space $X$ and $f:A \to \mathbb R$. The subdifferential of $f$ at $a \in A$ is the set
$$
\partial f(a)=\left\{x^* \in X^* \mid f(x) - f(a) \ge \langle x^*, x-a \rangle \text { for each } x \in A\right\}.
$$
The elements of $\partial f(a)$ are called subgradients of $f$ at $a$. The multivalued mapping $\partial f: A \to \mathcal P(X^*), x \mapsto \partial f(x)$, is called the subdifferential of $f$. The image $\partial f(E)$ of a set $E \subset A$ is the set
$$
\partial f(E)=\bigcup_{x \in A} \partial f(x) .
$$

Theorem: Assume $A$ is open convex and $f$ convex continuous. Then $f$ is $L$-lipschitz on $A$ if and only if $\partial f(A) \subset L B_{X^*}$.

 A: Let $f$ be $L$-lipschitz on $A$. Let $a \in A$ and $x^* \in \partial f (a)$. Let $r>0$ such that $B(a, r) \subset A$. Then $\overline B(a, r) \subset A$.  We have
$$
\begin{align}
\|x^*\| &= \frac{1}{r} \sup_{|x|=r} \langle x^*, x \rangle \\
&= \frac{1}{r} \sup_{|x-a|=r} \langle x^*, x-a \rangle \\
&\le \frac{1}{r} \sup_{|x-a|=r} [f(x)-f(a)] \\
&\le \frac{1}{r} \sup_{|x-a|=r} L|x-a| =L. \\
\end{align}
$$
Now we assume $\partial f(A) \subset L B_{X^*}$. We need the following lemma

Lemma: Assume $A$ is open convex and $f$ convex continuous. Then $\partial f (a) \neq \emptyset$ for all $a \in A$.

Let $a,b \in A$ and $x^* \in \partial f (a), y^* \in \partial f (b)$. Then
$$
f(a)-f(b) \le \langle x^*, a-b \rangle \le \|x^*\||a-b| \le L|a-b|.
$$
Similarly,
$$
f(b)-f(a) \le \langle y^*, b-a \rangle \le \|y^*\||b-a| \le L|b-a|.
$$
Then
$$
\frac{|f(a)-f(b)|}{|a-b|} \le L.
$$
This completes the proof.
Corollary: Assume $A$ is open convex and $f$ convex continuous. Then

*

*$\partial f$ is locally bounded on $A$. Indeed, $f$ is locally Lipschitz on $A$.

*$\partial f(a)$ is convex, bounded and $w^{*}$-closed (and hence $w^{*}$-compact).

