Let $X$ be a Banach space. Suppose $f:X^\*\to\mathbb R\cup\{\infty\}$ is convex, has weak\*-compact effective domain, and is weak\*-continuous on its effective domain. In particular, $f$ is weak\*-lower semicontinuous on $X^\*$.


Suppose I know $f$ is subdifferentiable at $x^\*\in \text{dom}(f)$, i.e. the subdifferential $\partial f(x^\*)\subseteq X^{\*\*}$ is nonempty. Does this necessarily imply that $X\cap \partial f(x^\*)$ is nonempty? If not, are there known sufficient conditions for $X\cap \partial f(x^\*)$ to be nonempty?



Subgradient in a predual under weak* continuity
Finally, I was able to cook up a counterexample. We choose $X = c\_0$ (zero sequences equipped with supremum norm). Thus, the dual spaces are (isometric to) $X^\* = \ell^1$ and $X^{\*\*} = \ell^\infty$.


We define
$$
C := \{ x \in \ell^1 \mid \forall n \in \mathbb N : |x\_n| \le 1/n^2 \}$$
and $f \colon \ell^1 \to \mathbb R \cup \{\infty\}$ via
$$
f(x) = \sum\_{n=1}^\infty x\_n \in \mathbb R
$$
for all $x \in C$ and $f(x) = \infty$ for all $x \in \ell^1 \setminus C$.


Let us check, that the assumptions are satisfied. The set $C$ is bounded due to $\sum\_{n = 1}^\infty 1/n^2 < \infty$ and weak-$\star$ closed since it is the intersection of the weak-$\star$ closed "stripes"
$$ \{x \in \ell^1 \mid |x\_n| \le 1/n^2\} \qquad\forall n \in \mathbb N.$$
Thus, it is weak-$\star$ compact.
The function $f$ is convex and it remains to check weak-$\star$ continuity on $C$. Let $x\_0 \in C$ be given and consider a net $(x\_i)\_{i\in I} \subset C$ with $x\_i \to x\_0$.
For an arbitrary $\varepsilon > 0$, there is $N \in \mathbb N$ with
$\sum\_{n = N+1}^\infty 1/n^2 < \varepsilon$.
Next, there is $i \in I$ with
$$
\left| \sum\_{n = 1}^N (x\_{j,n} - x\_{0,n}) \right| < \varepsilon
\qquad\forall j \ge i$$
since $y \mapsto \sum\_{n = 1}^N y\_n$ is weak-$\star$ continuous.
Thus,
$$
|f(x\_j) - f(x\_0)|
\le 
\left| \sum\_{n = 1}^N (x\_{j,n} - x\_{0,n}) \right|
+
\sum\_{n = N+1}^\infty |x\_{j,n}|
+
\sum\_{n = N+1}^\infty |x\_{0,n}|
<
3 \varepsilon
\qquad\forall j \ge i.$$
Since $\varepsilon > 0$ was arbitrary, this shows weak-$\star$ continuity on $C$.


Finally, it is easy to check that $\partial f(0) = \{1\}$, but $1 \in \ell^\infty \setminus c\_0$.


