$A$ relatively weakly compact then $\overline{\text{aco}} (A)$ is relatively weakly compact.

Question

Let $(X,\|.\|) $ be a separable Banach space.

Let $A$ be a subset of $X$, the absolutely convex hull of the set $A$ is : $$ \text{aco}(A)=\{\sum_{i=1}^{n}\lambda_{i}x_{i}:x_{i}\in A, \lambda_{i}\in \mathbb{R}, \sum_{i=1}^{n}|\lambda_{i}|\leq 1\} $$

Theorem. ("Krein's theorem")

$A$ be a weakly compact subset of X. Then $\overline{\text{aco}} (A)$ the closure of $\text{aco(A)}$ is weakly compact.

Show that :

$A$ relatively weakly compact then $\overline{\text{aco}} (A)$ is relatively weakly compact.

Answer 1

Hint
I suppose "relatively compact" means "closure is compact". I use overbar for weak closure.

Suppose $A$ is relatively weakly compact. Then $\overline{A}$ is weakly compact. So by Krein, $\overline{\text{aco}(\overline{A})}$ is weakly compact. Is it true that $\overline{\text{aco}({A})}=\overline{\text{aco}(\overline{A})}$ ?

Answer 2

Note that the map $\mathbb{C}\times X \rightarrow X$, $(\alpha, x)\mapsto \alpha x$ is continuous if $X$ is endowed with the weak topology. Hence the set $A'=[-1,1]A$ is weakly compact since it is the image of $[-1,1]\times A$ under this map. Now we claim that $\text{acot}(A)=\text{conv}(A')$. Indeed, if $\sum\limits_{i=1}^n \lambda_i x_i\in \text{acot}(A)$ with $0<\sum\limits_{i=1}^n |\lambda_i|\leq 1$ and $x_i\in A$ we can rewrite $\sum\limits_{i=1}^n \lambda_i x_i$ as $\sum\limits_{i=1}^n\frac{|\lambda_i|}{\sum\limits_{j=1}^n|\lambda_j|}(\text{sign}(\lambda_i)\sum\limits_{j=1}^n|\lambda_j|)$ to see that it in fact lies in $\text{conv}(A')$. Conversely, it is easy to see that $\text{conv}(A')\subseteq \text{acot}(A)$ (and actually not needed for this result). Now, the Eberlin-Smulian theorem states that $\text{conv}(A')$ is weakly precompact, i.e. $\text{acot}(A)$is weakly precompact.