Is there an easier proof to show that the closed convex hull of a normalized weakly null sequence is weakly compact?

Question

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).

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.

Answer 1

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.

Answer 2

Rauni’s link to the Freeman et al paper perhaps provides sufficient excuse for me to give here a simple proof (which, however, uses some theory) of the theorem in that paper. The background can more or less be found in Diestel’s book “Sequences and Series in Banach Spaces”, Springer GTM 92. The theorem says that if every weakly compact subset of the Banach space $X$ is in the closed convex hull of a weakly null sequence, then $X$ is a Schur space (which means that every weakly convergent sequence in $X$ is norm convergent). If $X$ has the property but is not Schur, then there is a normalized weakly null sequence $(y_n)$ in $X$ that, by passing to a subsequence, can assumed to be basic. As in my other answer to this MO problem, let $T:\ell_1 \to X$ be the weak$^*$ to weak continuous bounded linear operator that maps the $n$th unit vector in $\ell_1$ to $y_n$. The operator $T$ is weakly compact but not compact, and $T$ is injective because $(y_n)$ is basic. By the factorization theorem for weakly compact operators, $T= JS$ where $S:\ell_1 \to R$, $J:R\to X$ are injective bounded linear operators and $R$ is reflexive. We can now forget about $T$ and focus on $J$. Necessarily $J$ is not compact. But the image $JB_R$ of the closed unit ball $B_R$ of $R$ is weakly compact, so by hypothesis is in the closed convex hull of a weakly null sequence $(x_n)$ in $X$. Again, the sequence $(x_n)$ is the image under a weak$^*$ to weak continuous bounded linear operator $U: \ell_1 \to X$ that maps the $n$th unit vector in $\ell_1$ to $x_n$. If $U$ were injective we would be done since $U^{-1}J$ would be a necessarily non compact bounded linear operator from a reflexive space into $\ell_1$, which is impossible because $\ell_1$ has the Schur property. The same contradiction is available as long as $\ell_1/U^{-1}(0)$ has the Schur property. But since $U$ is weak$^*$ to weak continuous, the kernel $U^{-1}(0)$ of $U$ is weak$^*$ closed in $\ell_1$, which is to say that $\ell_1/U^{-1}(0)$ is the dual of the subspace $U^{-1}(0)_\perp$ of $c_0$. But the dual of every subspace of $c_0$ has the Schur property (proofs that $\ell_1$ has the Schur property also give this).