# On a result of Mazur about convergence in locally convex spaces

My question is about the following result from Simon (2011) (Theorem 5.3).

Let $$X$$ be a locally convex space and $$Y$$ its space of continuous [linear] functionals. Let $$\{x_n\}$$ be a sequence in $$X$$ with $$x_n \to x_\infty$$ in the $$\sigma(X,Y)$$-topology. Then, $$x_\infty = \bigcap_n \mathrm{cch}(\{x_m\}_{m\ge n}).$$

Here, $$\sigma(X,Y)$$ is the weak topology on $$X$$ with respect to $$Y$$. For a set $$A$$, $$\mathrm{cch}(A)$$ is the closed convex hull of $$A$$. Note that $$X$$ may be endowed with a topology stronger than the weak topology (subject to the requirement that this topology gives $$Y$$ as its dual).

The proof establishes fairly quickly that $$x_\infty \in \cap_n \mathrm{cch}(\{x_m\}_{m\ge n})=:A$$. It is, however, silent on the (non-)existence of other points in $$A$$. Thus, my question. How does one show that $$x_\infty$$ is the only member of $$A$$?

For reference, here is the proof from the text.

Let $$C_n=\mathrm{cch}(\{x_m\}_{m\ge n})$$. If $$x_\infty \notin C_n$$, there exists $$y\in Y$$ such that $$\langle y,x_\infty \rangle > \sup_{x\in C_n} \langle y,x\rangle \ge \sup_{m \ge n} \langle y, x_m \rangle$$. This is incompatible with $$\langle y, x_n \rangle \to \langle y, x_\infty \rangle$$.

If $$X$$ were a Fréchet space, then the result might follow from a version of Cantor's Intersection Theorem. However, $$X$$ may not be metrisable. Moreover, showing that the sequence $$\{C_n\}$$ has vanishing diameter might require that $$x_n \to x_\infty$$ in the original topology, and this need not be the case.

Suppose $$X$$ is Hausdorff so that $$Y$$ separates points, otherwise this statement is not true (give $$X$$ the indiscrete topology for a counter-example). Let $$C=\bigcap_n cch (\{ x_m\}_{m≥n})$$. If $$x\in C$$ lets show that for any $$y\in Y$$ that $$y(x)=y(x_\infty)$$ holds. So since $$Y$$ separates points you get that $$x=x_\infty$$.
Now suppose $$x\in C$$, let $$y\in Y$$ be some fixed dual element. There must be some sequence $$v_n\in \langle \{ x_m\}_{m≥n}\rangle$$ with $$y(v_n)\to y(x)$$, here $$\langle\cdot\rangle$$ denotes the convex hull. Write $$v_n=\sum_{k≥n}t_{k}(n)\, x_k$$ where $$t_{k}(n)≥0$$, $$\sum_k t_{k}(n)=1$$ and for each $$n$$ only finitely many $$t_k(n)$$ are non-zero.
Remember that $$x_n\to x_\infty$$, so for every $$\epsilon$$ there is some $$N$$ so that if $$n>N$$ you have $$|y(x_n)-y(x_\infty)|<\epsilon$$. Then if $$n>N$$: $$|y(v_n)-y(x_\infty)| = \left|\sum_{k≥n}t_{k}(n)\, y(x_k) -y(x_\infty)\right|=\left|\sum_k{t_k}(n) (y(x_k)-y(x_\infty))\right|\\ ≤\sum_{k}t_k(n)|y(x_k)-y(x_\infty)| ≤\epsilon$$ implying that $$y(v_n)\to y(x_\infty)$$ also, hence $$y(x)=y(x_\infty)$$ for every $$y$$.
• Typo: You forgot the second 'e' in "denotes". The reason I'm not just fixing that typo myself: a) It's not obvious (to me) that there must be such a sequence converging to $x$, but of course we can work with nets. b) I'm not sure whether you just accidentally used $n$ both as the index of $v_n$ and as the cutoff index for the tail of $\{x_m\}$ or that's deliberate. Jul 22, 2020 at 14:35
• You are right, it was an oversight to use a sequence $v_n$ rather than a net. Maybe it can be corrected to work with a net, but I think by shifting the definition of $v_n$ so that one simply has $y(v_n)\to y(x)$ for an arbitrary fixed $y$ (and $v_n\in \langle \{x_m\}_{m≥n}\rangle$) rather than $v_n\to x$ is simpler and one can retain the proof. Having $v_n$ be in the convex set generated by the $x_m$ with $m≥n$ was deliberate but I don't think its necessary. I would need a sequence of sequences $v_{n,k}$ otherwise. Jul 22, 2020 at 16:37
• Right. You'd need a sequence of sequences/nets, at least implicitly, then. We could of course phrase the argument without any sequences/nets by noting that $$\sup \{ \lvert y(x) - y(x_{\infty})\rvert : x \in \operatorname{cch}(\{x_m\}_{m \geqslant n})\} = \sup \{ \lvert y(x_m) - y(x_{infty})\rvert : m \geqslant n\}.$$ But six of this, half a dozen of that. Jul 22, 2020 at 17:42