In a paper that I am reading there is a following step:



> 
> Let $X$ be a Banach space and let $(x\_k) \subset X$ be a normalized sequence that converges weakly to $0$.
>  Then $\overline{co}(x\_k)$ is a weakly compact set.
> 
> 
> 


(notice that $\overline{co}(x\_k)$ denotes the norm-closure of the convex hull of $(x\_k)$.)


I think that I managed to prove the claim, but I had to do a lot of manual checking. My line of thought is given below.


My question is: Can this be proved more directly than I did (assuming that my proof is without error)? For example, is closed convex hull of a weakly compact set always weakly compact? (similar question was asked here, but in a bit different context, so it does not seem to apply to this situation: [Convex hulls of compact sets](https://mathoverflow.net/questions/156321/convex-hulls-of-compact-sets)).


I reasoned as follows:


* $\{x\_k \mid k \in \mathbb{N} \}$ is an weakly compact set.
* I checked that given a family $(y\_{\alpha}) \subset co(x\_k)$, it contains an weakly convergent subfamily, which weakly converges to an element in $\{ \sum\_{k=1}^{\infty} {\alpha\_k x\_k \mid (\alpha\_k) \in B\_{\ell\_1} }\}$.
* I observed that given any family $(y\_{\alpha})\_{\alpha \in I} \subset \overline{co}(x\_k)$, I can construct a family $(z\_{\alpha, \epsilon})\_{\alpha \in I, \epsilon > 0} \subset {co}(x\_k)$ such that $\forall \alpha \in I, \epsilon > 0$ we have that $\lVert y\_{\alpha} - z\_{\alpha, \epsilon} \rVert < \epsilon$. By defining order for family $z\_{\alpha, \epsilon}$ in such a way that $(\alpha\_1, \epsilon\_1) \leq (\alpha\_2, \epsilon\_2) \Leftrightarrow \alpha\_1 \leq \alpha\_2$ and $\epsilon\_1 \geq \epsilon\_2$, I can verify that (assuming that I did not make a mistake):
\begin{equation\*}
(y\_{\alpha}) \text { converges weakly to } w \Leftrightarrow (z\_{\alpha, \epsilon}) \text{ converges weakly to } w
\end{equation\*}
* Since $(y\_{\alpha}) \subset \overline{co}(x\_k)$, then $(z\_{\alpha, \epsilon})\_{\alpha \in I, \epsilon > 0} \subset {co}(x\_k)$. The latter family contains an weakly convergent subfamily $(z\_{\beta}')$, which converges to an element $c \in \{ \sum\_{k=1}^{\infty} {\alpha\_k x\_k \mid (\alpha\_k) \in B\_{\ell\_1} }\} \subset \overline{co}(x\_k)$. Therefore the original family $(y\_{\alpha}) \subset \overline{co}(x\_k)$ can also be shown to have a subfamily $(y\_{\gamma}')$ which converges weakly to the same element $c$.
* Therefore $\overline{co}(x\_k)$ is a weakly compact set.



Is there an easier proof to show that the closed convex hull of a normalized weakly null sequence is weakly compact?
You can find in many functional analysis text books the theorem that the closed convex hull of a weakly compact subset of a Banach space is weakly compact. But what you want is simpler than the general theorem. Here is a simple conceptual proof: Let $(y\_n)$ be a weakly null sequence in $X$ and consider the bounded linear operator $T:\ell\_1 \to X$ that maps the $n$th unit vector in $\ell\_1$ to $y\_n$. By Banach-Alaoglu, to show that the closed convex hull of $(y\_n)$ is weakly compact it is sufficient to verify that $T$ is weak$^\*$ to weak continuous. But that is the same as saying that $T^\*$ maps $X^\*$ into the predual $c\_0$ of $\ell\_1$, which in turn is the same as saying that $(y\_n)$ converges weakly to zero.


