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.


1 Answer 1


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 $y$.

  • $\begingroup$ 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. $\endgroup$ Jul 22, 2020 at 14:35
  • $\begingroup$ 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. $\endgroup$
    – s.harp
    Jul 22, 2020 at 16:37
  • 1
    $\begingroup$ 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. $\endgroup$ Jul 22, 2020 at 17:42

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