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?
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 !