# Subgradient in a predual under weak* continuity

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?

• If $Y$ is a non-reflexive Banach space and $X = Y^*$, then the unit ball $B_Y$ of $Y$ is closed, bounded and convex, but not weak*-closed in $X^* = Y^{**}$ (due to Goldstine theorem). Thus, your first parenthesis might fail and, similarly, $f$ might fail to be weak*-lower semicontinuous.
– gerw
Sep 30, 2021 at 8:30
• Whoops, that's embarrassing. Thank you! I'm now amending the question to not have this error. Oct 1, 2021 at 9:59

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 $$1 \in \ell^\infty \setminus c_0$$.