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?

  • 1
    $\begingroup$ 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. $\endgroup$
    – gerw
    Sep 30, 2021 at 8:30
  • $\begingroup$ Whoops, that's embarrassing. Thank you! I'm now amending the question to not have this error. $\endgroup$ Oct 1, 2021 at 9:59

1 Answer 1


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

  • $\begingroup$ Awesome, thanks! Very clean. $\endgroup$ Jan 27, 2022 at 17:17

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.