I am looking at the proof of Proposition 2.5. from Convex Functions, Monotone Operators and Differentiability, which asserts that the subdifferntial map $x \to \partial f(x)$ of a continous convex function $f: D\to \mathbb R$ on Banach ($D \subset E,$ convex open)space is $||\cdot||-w^*$ upper semicontinous.
The proof goes as follows: the usc property here means that for $x \in D$ and $W$ any $w^*-$ open set in $E^{*}$ containing $\partial f(x)$ and any sequence $\{x_n\} \subset D$ with $||x_n -x|| \to 0$ we have $\partial f(x_n) \subset W$ for all sufficiently large $n.$ If we assume to the contrary to derive a contradiction, passing to a subseqeunce and relabeling it again $\{x_n\}$, s.t. there is $x_n^* \in \partial f(x_n) \setminus W.$ Local boundedness of $partial f$ at $x$ implies that for sufficiently larce $n$ all the sets $\partial f (x_n) \subset M.B^*$ are contained a closed bounded ball in $E^*.$ Which is weak^* compact from Alaoglu. Also $\partial f(x_n)$ are weak^* closed so for sufficiently large $n,$ they are weak star compacts. And so the bounded sequence $\{x_n^*\}$ has s cluster point $x^*.$ Now I need to prove that $x^* \in \partial f(x) \setminus W,$ which would be obvious contradiciton.
Now. This is in some sense closedness of $\operatorname{Gr}(\partial f).$ But this is no good because I know that for a multivalued maps on a well behaved topological spaces uppersemicontinuity can be made equivalent to closedness of the graph (plus extra assumpions for closed images for example).
Can somebody provide a proof of closedness of $\operatorname{Gr}(\partial f)?$
In this particular case just to prove $x^* \in \partial f(x)$ we need to take limit of a subnet in $\left<x_{n_\alpha}^*,y-x_{n_{\alpha}}\right> \leq f(y)-f(x_{n_{\alpha}}), \forall y \in D, \alpha \in I.$ But here i need to use closedness of the graph. Also why should $x^*$ not be in $W?$
I am well aware of this post.
