Convex and Symmetric subset of a Banach space

Question

Let X be a Banach space and A be a convex and symmetric subset of X. Is it true then that the closure of A will be a subset of 2A=A+A?

I doubt that this always holds, but can't seem to find a counter-example

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Answer 1

If $X$ is infinite dimensional, you can simply take $A$ to be any non-closed linear subspace, such as the polynomials in $C([0,1])$. Then $2A=A$, but $A$, not being closed, does not contain its closure.

If $X$ is finite dimensional the claim is true.

First, note it is trivially true if $A$ is empty. If $A$ is nonempty then by symmetry and convexity, $0 \in A$.

Suppose first that $A$ spans $X$. Then we may find a set $e_1, \dots, e_n \subset A$ which is a basis for $X$. Let $f_1, \dots, f_n \in X^*$ be the dual basis. Let $M = \max_i \|f_i\|_{X^*}$ and choose $0 < \epsilon < 1/(nM)$. I claim that $B(0, \epsilon) \subset A$, which implies the desired statement.

Suppose $x \in X$ with $\|x\| < \epsilon$. We already argued that $0 \in A$ so suppose $x \ne 0$. Then we may write $x = \sum_{i=1}^n a_i u_i$ where $a_i = |f_i(x)| \ge 0$ and $u_i = \pm e_i \in A$. Let $\lambda = \sum_{i=1}^n a_i > 0$. Note that $a_i = |f_i(x)| \le M \epsilon < 1/n$, so $\lambda < 1$. Thus if we set $b_i = a_i / \lambda$, we have $\sum_{i=1}^n b_i = 1$. Setting $y = \sum_{i=1}^n b_i u_i$, which is in $A$ by convexity, we have $x = \lambda y + (1-\lambda) 0 \in A$ by convexity again.

If $A$ does not span $X$ then let $E$ be the span of $A$. By the previous argument, $\bar{A} \cap E \subset 2A$. But as we are in finite dimensions, $E$ is closed, so $\bar{A} \cap E = \bar{A}$.