Strong convergence of subgradients of a convex Fréchet differentiable function on a normed space 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 $X, Y$ be normed spaces, $A$ a subset of $X$, $f:A\to \mathbb R$, $\Phi: A \to \mathcal P(Y)$ a multivalued mapping, and $a \in A$.

*

*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 image by $\Phi$ of a subset $E$ of $A$ is defined as $\Phi(E) := \bigcup_{x \in E} \Phi(x)$. We say that $\Phi$ is single-valued and continuous at $a$ if $\Phi(a)=\left\{y\right\}$ and
$$
\forall \varepsilon>0, \exists \delta>0:\left[x \in A, |x-a|<\delta \implies \Phi(x) \subset B \left(y, \varepsilon\right)\right] .
$$
It can be proved that $\Phi$ is single-valued and continuous at $a$ if and only if $\Phi(a) = \{y\}$ and $[A \ni x_{n} \rightarrow a, y_n \in \Phi (x_n) \implies y_{n} \to y]$.

Theorem:  Let $A$ be open and $f$ convex continuous. Then $f$ is Fréchet differentiable at $a$ if and only if $\partial f$ is single-valued and continuous at $a$.


A direct corollary is as follows:

Let $A$ be open and $f$ convex continuous. If $f$ is Fréchet differentiable at each point of $A$, then $f \in \mathcal{C}^{1}(A)$.

 A: *

*Let $\partial f$ be single-valued and continuous at $a$.

Let $x^* \in \partial f (a)$. Let $r>0$ such that $B(a, r) \subset A$. Let $h_n \in B(0, r)$ such that $h_n \to 0$. By this result, $\partial f (x) \neq \emptyset$ for all $x\in A$. Let $x^*_n \in \partial f (a+h_n)$. Then
$$
0 \le \frac{f(a+h_n)-f(a) - \langle x^*, h_n \rangle}{|h_n|} \le \frac{\langle x_n^*, h_n  \rangle - \langle x^*, h_n \rangle}{|h_n|} = \frac{\langle x_n^* -  x^*, h_n  \rangle }{|h_n|} \le \| x_n^* -  x^*\|.
$$
It follows from $x_n^* \to  x^*$ that
$$
\lim_n \frac{f(a+h_n)-f(a) - \langle x^*, h_n \rangle}{|h_n|} = 0.
$$
Hence $f$ is Fréchet differentiable at $a$.

*

*Let $f$ is Fréchet differentiable at $a$.

Then $f$ is Gâteaux differentiable at $a$. By this result, $\partial f (a)$ is a singleton. Let $x^* \in \partial f (a)$.  Let $x_n \to a$ and $x_n^* \in \partial f (x_n)$. It suffices to show that $x_n^* \to x^*$. Let
$$
\varphi(v) := \frac{f(a+v)-f(a) - \langle x^*, v \rangle}{|v|}.
$$
We have

*

*$-\langle x^*, v \rangle = |v| \varphi (v) +f(a)-f(a+v)$.

*$f(x_n+v)-f(x_n) \ge \langle x_n^*, v \rangle$.

By this result, there are $L,h>0$ such that $B(a,h) \subset A$ and that $f$ is $L$-Lipschitz on $B(a, h)$. For all $0<r<h/2$, we have
$$
\begin{align}
\|x_n^*-x^*\|  &=  \sup_{|v|=r}  \frac{\langle x_n^* - x^*, v \rangle}{r} \\
&\le \sup_{|v|=r} \frac{f(x_n+v)-f (x_n)  +|v| \varphi (v) +f(a)-f(a+v)}{|v|} \\
&= \sup_{|v|=r} \left [\frac{[f(x_n+v) - f(a+v)]+ [f(a) - f (x_n)]}{|v|}  +\varphi (v) \right ]\\
&\le \sup_{|v|=r} \left [\frac{2L|x_n-a|}{|v|}  +\varphi (v) \right ] \\
&= \frac{2L|x_n-a|}{r}  + \sup_{|v|=r}  \varphi (v).
\end{align}
$$
For all $0<r<h/2$, we have
$$
\lim_n \|x_n^*-x^*\| \le \frac{2L}{r} \lim_n |x_n-a| +  \sup_{|v|=r}  \varphi (v) = \sup_{|v|=r}  \varphi (v).
$$
Notice that
$$
\lim_{r \to 0^+} \sup_{|v|=r}  \varphi (v) = 0.
$$
So $\lim_n \|x_n^*-x^*\| = 0$. This completes the proof.
