Let $X := \mathbb R^d$ and $h: X \to \mathbb R$ be convex. Then
- $h$ is locally Lipschitz, and
- $\partial h (x)$ is non-empty, convex, and compact for all $x \in X$.
Let $d_H$ be the Hausdorff distance on the collection of compact subsets of $X$. We have
Theorem Let $h$ be strictly convex. If $x_n, x \in X$ and $d_H(\partial h(x_n),\partial h(x)) \to 0$, then $x_n \to x$.
I would like to ask if the following converse is true, i.e.,
If $x_n, x \in X$ and $x_n \to x$, then $d_H(\partial h(x_n),\partial h(x)) \to 0$.
I tried to prove it, but failed in the second case below.
My attempt WLOG we assume there is $r>0$ such that $(x_n) \subset B(x, r)$. There is $L>0$ such that $h$ is $L$-Lipschitz on $B(x, r)$. By this result, we get $$ \bigcup_{y \in B(x,r)} \partial h(y) \subset B(0, L). $$
Let $A_n := \partial h(x_n)$ and $A := \partial h(x)$. We need to prove that $d_H (A_n, A) \to 0$. Assume the contrary that $d_H (A_n, A) \not\to 0$. WLOG, there is $\varepsilon > 0$ such that $d_H (A_n, A) \ge \varepsilon$ for all $n$. By definition of $d_H$, we have $2$ cases.
WLOG, there exist $p_n \in A_n$ such that $d (p_n, A) \ge \varepsilon$ for all $n$. Then $|p_n|\le L$ for all $n$. Wlog, we assume $p_n \to p$ for some $p \in X$. For all $y\in X$, $$ \begin{align} h(y) - h(x_n) &\ge \langle p_n, y-x_n \rangle \\ &= \langle p_n, y-x \rangle + \langle p_n, x-x_n \rangle \\ &\ge \langle p_n, y-x \rangle - L |x-x_n|. \end{align} $$ Taking the limit, we get $h(y)-h(x) \ge \langle p, y-x \rangle$ for all $y\in X$. This implies $p \in A$ and thus $d(p_n, A) \to 0$. This is a contradiction.
WLOG, there exist $p_n \in A$ such that $d (p_n, A_n) \ge \varepsilon$ for all $n$. Wlog, we assume $p_n \to p$ for some $p\in A$. There is $N \in \mathbb N$ such that $$ |p-y| \ge \varepsilon/2 \quad \forall y \in \bigcup_{n \ge N} A_n. $$
