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Let $F$ be a continuous convex function on $\mathbb{R}^n$.

If the subdifferential $\partial F(x)$ of $F(x)$ admits a continuous selection, for every $x \in \mathbb{R}^n$, does it mean that $F$ is differentiable on $ \mathbb{R}^n$ ?

I was trying to use theorem 25.5 and 25.6 (Rockafellar: Convex Analysis) but the normal cone $K(x)$ in theorem 25.6 gives me some problems.

With continuous selection I mean that there exists an element $f(x) \in \partial F(x)$ continuous for every $x \in \mathbb{R}^n$.

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    $\begingroup$ I sketched the proof earlier in this answer, but judging from your question, your issue is basically why, in that sketch, I can ignore the set $K(x)$: the answer is that for $x$ an interior point (in your case since the domain is all of $\mathbb{R}^n$ this applied to all $x$), the normal cone $K(x) = \{0\}$. $\endgroup$ Commented May 26, 2023 at 16:09

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$\newcommand\p\partial\newcommand\R{\mathbb R}\newcommand\cl{\operatorname{cl}}\newcommand\conv{\operatorname{conv}}$The answer is yes, and you were almost there.

Indeed, suppose the contrary: that we we have a continuous function $f\colon\R^n\to\R^n$ such that $f(z)\in\p F(z)$ for all $z\in\R^n$ and yet $F$ is not differentiable at some $x\in\R^n$.

By Theorem 25.1 in Rockafellar's book, the non-differentiability of $F$ at $x$ means that the cardinality $|\p F(x)|$ of the subdifferential $\p F(x)$ of $F$ at $x$ is $>1$. The subdifferential $\p F(x)$ is a closed convex set. Moreover, in this case $\p F(x)$ is bounded, since the function $F$ is continuous and hence locally bounded. So, the subdifferential $\p F(x)$ contains two distinct extreme points, say $u$ and $v$.

By Theorem 25.6 in Rockafellar's book, $\p F(x)=\cl\conv S(x)+K(x)$, where $\cl\conv S(x)$ is the closed convex hull of the set $S(x)$ of all limits of the sequences of the form $(\nabla F(x_k))$ such that $F$ is differentiable at all $x_k$'s and $x_k\to x$ (as $k\to\infty$), and $K(x)$ is the normal cone to $\operatorname{dom}F$ at $x$. In this case, $K=\{0\}$, since $\operatorname{dom}F=\R^n$. Also, the set $S(x)$ is closed and bounded, again because the function $F$ is locally bounded. So, $\p F(x)=\cl\conv S(x)$ and hence any extreme point of $\p F(x)$ is in $S(x)$. So, the two distinct points $u$ and $v$ in $\p F(x)$ are in $S(x)$. So, $u=\lim_k \nabla F(y_k)$ and $v=\lim_k \nabla F(z_k)$ for some sequences $(y_k)$ and $(z_k)$ converging to $x$ such that $F$ is differentiable at all $y_k$'s and at all $z_k$'s.

But, again by Theorem 25.1 in Rockafellar's book, $\nabla F(y_k)=f(y_k)$ and $\nabla F(z_k)=f(z_k)$. So, $u=\lim_k\nabla F(y_k)=\lim_k f(y_k)=f(x)$ and $v=\lim_k\nabla F(z_k)=\lim_k f(z_k)=f(x)$, which contradicts the condition that $u$ and $v$ are distinct. $\quad\Box$

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Can I recommend Proposition 2.8 in Robert Phelps’ book “Convex Functions, Monotone Operators and Differentiability” second edition, vol 1364 of Lecture Notes in Mathematics, Springer 1993.

It exactly answers your question. Namely, a continuous convex function is Gateaux differentiable at a point $x$ if and only if there is a selection of the subdifferential mapping that is norm-to-weak$^*$ continuous at $x$. Of course in finite dimensions weak$^*$ continuous is just norm continuous (which one would normally just call continuous), and Gateaux differentiable is the same as Frechet differentiable, which one would just call differentiable.

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