Closure of balanced convex set and its dilation

Question

While reading a proof, one of the steps was showing that for a (specific) balanced and convex set in a Banach space we had the inclusion $\overline{B} \subseteq 2B$. The proof continued using specific properties of the set in question. This seems like a very reasonable inclusion geometrically (for my untrained intuition, at least), so it made me curious as to whether it is true in general or only for specific sets. I tried it myself, but couldn't find a proof nor counterexample. More precisely, I would like to know if for any set $B$ in a normed space that's balanced and convex the inclusion $\overline{B} \subseteq 2B$ holds, if the hypothesis that the space is Banach is needed or if there is a counterexample such that the inclusion is not generally true. Thanks!

Answer 1

Partial answer. It is not true in $l_2$. The following counterexample in $l_2$ was inspired by Example 2.14 here.

Let $ I_n = \{ t e_n \mid |t| \le {1 \over n} \}$ and $B = \operatorname{co} \cup_n I_n$. Note that $B$ is convex, balanced and any element has a finite number of non zero components (and hence the same is true of $rB$ for any $r>0$).

Let $x_n = { \sum_{k=1}^n {1 \over 2^k} {1 \over k}e_k \over \sum_{k=1}^n {1 \over 2^k}}$, note that $x_n \in B$ and $x_n$ is Cauchy, hence converges to some $x \in l_2$ and all components of $x$ are non zero, for which we see that $x \in \overline{B} \setminus B$. In particular, $\overline{B} \not \subset rB$ for any $r>0$.