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Let $(X, \lVert \cdot \rVert)$ be a normed space over $\mathbb{R}$ and $A = \{ a_1,a_2 \ldots \} \subseteq X$ be a closed bounded set.

Let $\overline{\mathrm{co}}(A)$ denote the closed convex hull of $A$, i.e. the intersection of all closed convex subsets of $X$ which contain $A$.


Question:

Does it follow that for every $x \in \overline{\mathrm{co}}(A)$ there is a sequence $(t_n)_n$ of elements of $[0,1]$ such that \begin{equation} x = \sum_{n=1}^\infty t_na_n \quad \text{ and } \quad \sum_{n=1}^\infty t_n=1, \end{equation} so $x$ is a sum of convex series of elements of $A$?

If the above property does not hold in the mentioned general setting, I would still be interested if there are some additional properties than one can impose on either $X$ or $A$ so that the property would still hold.

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    $\begingroup$ Let $X$ be the reals, let $A=\{\,0,.9,.99,.999.\dots\,\}$. Then $1$ is in the closed convex hull of $A$, but it can't be a convex combination of elements of $A$. $\endgroup$ May 15, 2022 at 0:44
  • $\begingroup$ I edited the question so that now we require for $A$ to also be closed. Previously, I mentioned the linear independence, however, I found that I might want for some of the elements of $A$ to be linearly dependent. Hence, I changed the property to closedness. $\endgroup$ May 15, 2022 at 1:16
  • $\begingroup$ The set $C=\{\sum_{n=1}^{\infty} t_na_n: t_n\geq 0, \sum_{n=1}^{\infty} t_n=1 \}$ is convex and contains $A$. Thus, its closure must be the closed convex hull of $A$. Do you think $C$ is closed? $\endgroup$
    – Onur Oktay
    May 15, 2022 at 5:45

1 Answer 1

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In a Banach space $X$, the representation $$\overline{co}(A)=\left\{\sum_{n=1}^\infty t_n a_n: t_n\ge 0, \sum_{n=1}^\infty t_n\le 1\right\}$$ holds for $A=\{a_n:n\in\mathbb N\}$ if $a_n\to 0$ (if $X$ is only normed, you can apply this to the completion).

To prove this, consider the sequence space $\ell^1$ as the dual of the space $c_0$ of all null sequences and $T:\ell^1\to X$, $(t_n)_{n\in\mathbb N}\mapsto \sum_{n=1}^\infty t_n a_n$ (which clearly converges because $X$ is complete). Then $T$ is linear and $\sigma(\ell^1,c_0)-\sigma(X,X')$-continuous. Indeed, for every $f\in X'$ we need the continuity of $f\circ T$ which is the map $(t_n)_n\mapsto \sum_n t_n f(x_n)$ which is $\sigma(\ell^1,c_0)$ continuous because $f(x_n)\to 0$.

The dual unit ball of $\ell^1$ is $\sigma(\ell^1,c_0)$-compact by Alaoglu and hence so is $B=\{(t_n)_n: t_n\ge 0, \sum_n t_n\le 1\}$ because of the continuity of the projections $(t_k)_k \mapsto t_n$. Therefore, the image $T(B)$ (which is the right hand side in the displayed formula) is weakly compact and hence norm closed in $X$. This proves the inclusion $\overline{co}(\{a_n:n\in\mathbb N\})\subseteq T(B)$ whereas the reverse one is elementary (truncating the series).


As in Gerry Myerson's comment, $a_n=1/n$ shows that one cannot replace the condition $\sum_n t_n\le 1$ by $\sum_n t_n=1$ (unless, of course, one term $a_k=0$) . In the proof above, the problem is that $\tilde B=\{(t_n)_n: t_n\ge 0,\, \sum_nt_n=1\}$ is not $\sigma(\ell^1,c_0)$-compact because $(t_n)_n\mapsto \sum_n t_n$ is not $\sigma(\ell^1,c_0)$-continuous. The idea to replace $c_0$ by the space $c$ of all convergent sequences fails because then the duality between $c$ and $\ell^1$ is not of the form $\langle (s_n)_n,(t_n)_n \rangle=\sum_n s_nt_n$ because the functional $(s_n)_n\mapsto \lim s_n$ is not representable in that way.


There are more sophisticated representations of elements of the closed convex hull of more general sets $A$ considered in Choquet theory. The question when the closed convex hull of a compact or weakly compact set is again (weakly) compact is also very relevant. The compact case is due to Mazur and the weakly compact case is a theorem of Krein.

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