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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?

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  • $\begingroup$ 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 $x \in X$? $\endgroup$
    – gerw
    Commented Oct 6, 2017 at 6:41
  • $\begingroup$ 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. $\endgroup$
    – zorro47
    Commented Oct 6, 2017 at 17:22

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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 !

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  • $\begingroup$ What about a case when $\partial f$ is a special case of $F$ - in this particular situation $F$ inherits properties of $\partial f$? $\endgroup$
    – zorro47
    Commented Oct 5, 2017 at 19:34
  • $\begingroup$ I don't know !. $\endgroup$
    – Red shoes
    Commented Oct 5, 2017 at 19:41
  • $\begingroup$ 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. $\endgroup$
    – zorro47
    Commented Oct 5, 2017 at 19:48
  • $\begingroup$ yes $\partial f(x)$ is convex and $w^* -$closed. in any topological spaces.... note that always $ w^* -$closeness is stronger than topology closeness $\endgroup$
    – Red shoes
    Commented Oct 5, 2017 at 19:54

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