# Convex series and closed convex hulls in normed spaces

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 $$$$x = \sum_{n=1}^\infty t_na_n \quad \text{ and } \quad \sum_{n=1}^\infty t_n=1,$$$$ so $$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.

• 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 $A$. May 15, 2022 at 0:44
• 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. May 15, 2022 at 1:16
• 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? May 15, 2022 at 5:45

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.