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 $x \in \mathbb{R}^n$.



Subdifferential of a convex function admits a continuous selection
$\newcommand\p\partial\newcommand\R{\mathbb R}\newcommand\cl{\operatorname{cl}}\newcommand\conv{\operatorname{conv}}$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. $\quad\Box$


