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.

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.

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 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.

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.