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
    Commented Sep 30, 2021 at 8:30
  • $\begingroup$ Whoops, that's embarrassing. Thank you! I'm now amending the question to not have this error. $\endgroup$ Commented 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. $\endgroup$ Commented Jan 27, 2022 at 17:17

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