Boundary points in $\overline{\operatorname{conv}\{z_i\}_{i\in I}}$

Question

Let $X$ be an infinitely-dimensional Banach space and $\{z_i\}_{i\in I}$ be a set of linearly independent points in $X_{\leq 1}$, the closed unit ball of $X$. $I$ the index set is not necessarily countable. Define $Z = \overline{\operatorname{conv}\{z_i\}_{i\in I}}$. Fix a countable subset $J\subseteq I$ and suppose $\{n_j\}_{j\in\mathbb{N}}$ is an enumeration of $J$. Observe that for each sequence $\{\lambda_j\}_{j\in\mathbb{N}}\subseteq (0, 1)$, which sums to $1$, we have for each $N\in\mathbb{N}$:

$$ \Big( \sum_{j\leq N} \lambda_j z_{n_j} \Big) + \Big( 1 - \sum_{j \leq N}\lambda_j\Big) z_{n_{N+1}} \in \operatorname{conv}\{z_i\}_{i\in I} $$

Therefore $\sum_{j\in\mathbb{N}} \lambda_j z_{n_j} \in Z$. My question is: is there a sufficient (and ideally necessary) condition such that for each $z\in Z$, there exists a countable subset $J\subseteq I$ and a summable sequence $\{\lambda_j\}_{j\in J}$ which sums up to $1$, such that $z = \sum_{j\in J}\lambda_j z_j$.

I start from the case where $I=\mathbb{N}$. From the paper Weak compactness and reflexivity written by James. R. C. (link), the following result proves the existence of such a sequence of points $\{z_i\}_{i\in\mathbb{N}} \subseteq X_{\leq 1}$, such that each point in $\overline{\operatorname{conv}\{z_i\}}$ has the form $\sum_{i\in\mathbb{N}} \lambda_i z_i$ where $\sum_{i\in\mathbb{N}} \lambda_i=1$.

$X_{\leq 1}$ is not weakly compact iff there exists a sequence $\{z_i\}\subseteq X_{\leq 1}$, $\{g_n\}_{n\in\mathbb{N}} \subseteq X^*$ and $\theta>0$ such that for each $i, n\in\mathbb{N}$:

$$ g_n(z_i) = \begin{cases} \theta, \hspace{1cm} n\leq i\\ 0, \hspace{1cm} n>i \end{cases} $$

For each $n\in\mathbb{N}$, define $\alpha_n = \dfrac{1}{\theta}\big( g_n - g_{n+1} \big)$. Then it is proved in the paper that for each $z\in \overline{\operatorname{conv}\{z_i\}}$, $\sum_{n\in\mathbb{N}} \alpha_n(z)=1$ and $z = \sum_{n\in\mathbb{N}}\alpha_n(z) z_n$.Hence, $X_{\leq1}$ being weakly compact implies the existence of such a sequence $\{z_i\}_{i\in\mathbb{N}}$.

Above is an example where such a sequence exists so I am afraid for a general (linealy independent) sequence the answer could be negative but feel free to add conditions if needed. Also any other relevant references are much appreciated.

Answer 1

If $(z_n)_{n\in\mathbb N_0}$ is a weakly convergent sequence with limit $z_0$ then $$ \overline{conv\{z_n:n\in\mathbb N_0\}}= \left\{\sum_{n=0}^\infty\lambda_nz_n: \lambda_n\ge 0, \sum_{n=0}^\infty \lambda_n=1\right\}. $$ By translating, we may assume $z_0=0$ but the example of Alex Ravsky in the comments shows that it is essential that the limit is an element of the sequence (even norm convergence does not help, e.g., for $z_n=1/(n+1)$ times the standard unit vector in $\ell^2$).

For the proof we consider the usual duality between $c_0=c_0(\mathbb N)$ and $\ell^1$ so that the unit ball $B$ of $\ell^1$ is $\sigma(\ell^1,c_0)$-compact. The weak$^*$-continuity of the evaluations $\lambda\mapsto \lambda_n$ implies that also $C=\{\lambda\in B: \lambda_n\ge 0$ for all $n\}$ is compact (note however, that $\{\lambda\in C: \sum_{n=1}^\infty\lambda_n =1\}$ is not $\sigma(\ell^1,c_0)$-closed which is the technical reason that one needs the limit $z_0$ as an element of the sequence). Since $z_0=0$, the right hand side of the claimed formula is the image $F(C)$ for the linear map $$ F:\ell^1\to X, \quad \lambda\mapsto \sum_{n=1}^\infty \lambda_nz_n $$ which is well defined by the completeness of $X$. Moreover, $F$ is $\sigma(\ell^1,c_0)$-$\sigma(X,X^*)$ continuous: The universal property of the weak topology yields that it is enough to show the continuity of $x^*\circ F$ for every $x^*\in X^*$ and this follows from $$x^*\circ F(\lambda)=\sum_{n=1}^\infty \lambda_nx^*(z_n)$$ because $(x^*(z_n))_{n\in\mathbb N}\in c_0$.

We obtain that $F(C)$ is $\sigma(X,X^*)$-compact and hence norm closed. This implies the inclusion $\subseteq$ in the claimed equality whereas the other one is obvious.