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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 $f \in \mathcal{C}^{1}(A)$.

Analyst
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1 Answers1

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  • 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.

Analyst
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