# A question about properties of convex subdifferential

Let $X$ be a reflexive Banach space. Let $$\mathcal{P}_{fc}(X)=\{A\subset X\, |\, A\,\text{is nonempty, closed, convex\}}.$$ Let $F:X\to \mathcal{P}_{fc}(X^*)$ be an operator. Consider a convex and lsc function $f:X\to \mathbb{R}$. Can we set $\partial f=F$, where $\partial f$ denotes convex subdifferential of $f$?

In other words, is $\partial f(x)$ nonempty, closed and convex?

Look at my answer in Continuity of subdifferential mapping . Since $X$ is reflexive, then $w^*$-$X^*$ topology and $w$-$X^*$ topology coincide and if I show that $\partial f(x)$ is convex, then from Mazur lemma $w-X^*$ topology coincides with strong topology, which implies the closedness. And the fact that $\partial f(x)$ is convex follows directly from the definition of convex subdifferential. Do you find my considerations correct?

• What precisely is your question? I see two possibilities: 1. Given $F$, is there $f$ with $\partial f = F$? 2. Given $f$ does $\partial f(x) \in \mathcal P_{fc}(X^*)$ for all $x \in X$? – gerw Oct 6 '17 at 6:41
• Given $f$ does $\partial f(x)\in\mathcal{P}_{fc}(X^*)$ for all $x\in X$. I think that this can be easily proved setting $x_n=x$ in closed graph theorem. – zorro47 Oct 6 '17 at 17:22

A very simple and quick reason to reject your claim is that $\partial f$ is always monotone operator so what if $F$ is not monotone !
• What about a case when $\partial f$ is a special case of $F$ - in this particular situation $F$ inherits properties of $\partial f$? – zorro47 Oct 5 '17 at 19:34
• I'm not sure if the closed graph theorem $(X,w^*-X^*)$ for convex subdifferential (which was in fact used above) is a suitable argumentaion for the proof. – zorro47 Oct 5 '17 at 19:48
• yes $\partial f(x)$ is convex and $w^* -$closed. in any topological spaces.... note that always $w^* -$closeness is stronger than topology closeness – Red shoes Oct 5 '17 at 19:54