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I am reading here that in a Banach space, the sum of the closed unit ball with a closed bounded convex set might fail to be closed itself. It seems there is a counterexample if and only if the space fails to be reflexive.

This was a surprise, because I thought I had proved the sum is always closed. Suppose $A$ is closed, bounded and convex. Then the sum with the unit ball $B$ consists of the points $A+B = \{ x \in V: d(x,A) \le 1\}$. Here $d(x,A) = \inf\{d(x,a): a \in A\}$ is the point-to-set distance function. Since the point-to-set function is continuous (convex too) the set $\{ x \in V: d(x,A) \le 1\}$ is just the preimage of $[0,r]$ a continuous map. Hence it is closed.

I feel like there is some silly reason by proof doesn't work. But I am not getting it at the moment. Has anyone thought about this before?

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Your proof fails because $A+B \ne \{x \in V: d(x,A) \le 1\}$. Indeed, $d(x,A) = \inf_{y \in A} \|x - y\|$, but there is no reason the infimum has to be attained. It could be that there is a sequence $y_n \in A$ with $\|x - y_n\| \to 1$ as $n \to \infty$, but the sequence has no limit so there is no $y \in A$ with $|x - y| = 1$, so $x \notin A+B$.

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  • $\begingroup$ Thanks very much. I was under the impression that the minimum exists, essentially because every closed bounded set is weakly compact, hence your sequence $y_n$ admits a weakly convergent subsequence and yadda yadda yadda . . . the weak limit minimises the point-to-set function. But now I see that result also only holds under reflexivity. $\endgroup$
    – Daron
    Commented May 31, 2023 at 6:58

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