# Subdifferential of a convex function admits a continuous selection

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

• 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 $K(x) = \{0\}$. Commented May 26, 2023 at 16:09

$$\newcommand\p\partial\newcommand\R{\mathbb R}\newcommand\cl{\operatorname{cl}}\newcommand\conv{\operatorname{conv}}$$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. $$\quad\Box$$

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.