Is there a convex function $F$ that is not differentiable, but whose subdifferential admits a continuous selection, i.e. a continuous $g$ with $g(x) \in \partial F(x)$ for all $x$ in the domain?


In one dimension I think I can prove there is not: if $|\partial F(x)| > 1$ then it contains an open set, whose inverse image under $g$ is an open set, so $g$ is not a selection after all - I think it maps some $y \neq x$ to a point in the interior of $\partial F(x)$, which cannot be a subgradient at $y$.



Nondifferentiable convex function whose subdifferential admits a continuous selection
The answer is no.


See Rockafellar's *Convex Analysis*, part V.


First, let $D$ be the set of points where $F$ is differentiable. *Theorem 25.5* proves that $D$ is dense in the interior of the domain of $F$, with measure zero complement. And that $\nabla F$ is a continuous mapping on $D$.


Next, if the domain of $F$ has non-empty interior, then at every interior point the subdifferential has the property (*Theorem 25.6*):


$$ \partial F(x) = \mathrm{cl} ( \mathrm{conv}(S(x))) $$


where $S(x)$ is the set of limits of all sequences of the form $\nabla F(x\_i)$ where $x\_i \to x$ and $x\_i \in D$.


Now if $\partial F$ admits a continuous section $g$, then for every $x$ in the interior of the domain of $F$, we have that for every sequence $x\_i \in D$ such that $x\_i \to x$ it holds that $g(x\_i) \to g(x)$ (continuity of $g$); but since $\partial F(x\_i) = g(x\_i)$ is unique, this implies that $S(x) = \{g(x)\}$.


And hence $F$ is differentiable at $x$.


