Convexity of open balls in a metrizable subset of a locally convex t.v.s.

Question

The following result is well known (Rudin, Functional Analysis, Theorem 1.24): If $X$ is a locally convex topological vector space with a countable local base then there is a compatible metric $d$ such that all open balls are convex. My question is: Assume that $X$ is a locally convex t.v.s. which is not metrizable and $C$ is a compact, convex and metrizable subset of $X$. Is there a metric compatible with the topology of $C$ such that all open balls are convex?

Answer 1

Let $ s: X\times X\to X $ be the subtraction map, namely $$ s(x,y) = x-y, \quad \forall x,y\in X, $$ and put $$ D:= s(C\times C) = \{x-y: x, y\in C\}. $$

We next claim that $D$ is homeomorphic to a quotient of $C\times C$. Indeed, consider the equivalence relation $R$ on $C\times C$ according to which two points are equivalent iff they have the same image under $s$. Since $s$ obviously respects $R$, there is a unique continuous map $$ \tilde s:(C\times C)/{R}\ \to \ D, $$ such that $s=\tilde s\circ q$, where $q$ is que quotient map.

It is not hard to see that $\tilde s$ is a bijective map, and since $(C\times C)/{R}$ is compact and $D$ is Hausdorff, we conclude that $\tilde s$ is a homeomorphism, thus proving the claim.

Every Hausdorff quotient of a compact metrizable space is itself metrizable, so $D$ is metrizable and hence also first countable.

Noticing that $0$ lies in $D$, we may then find a countable open neighborhood base for $0$ in the relative topology of $D$, say $\{V_n\}_n$. Write each $V_n$ as $W_n\cap D$, where $W_n$ is an open subset of $X$, and choose an open, convex, balanced set $U_n$, with $$ 0\in U_n\subseteq W_n. $$

It is then evident that $$ V'_n := U_n\cap D \subseteq W_n\cap D = V_n, $$ so the $V'_n$ form an open neighborhood base for $0$ in the relative topology of $D$.

For each $n$, let $p_n$ be the Minkowsky functional associated to $U_n$ and consider the (not necessarily Hausdorff), locally convex topology $\tau'$ on $X$ generated by the $p_n$.

We next claim that $(C, \tau ')$ is Hausdorff. To see this, let $x$ and $y$ be distinct points in $C$, and notice that $$ z:= y-x\in C-C = D. $$ Choose an open set $A\subseteq X$, such that $0\in A$ and $z\notin A$, and let $n$ be such that $V'_n\subseteq A\cap D$.

Notice that $z\notin U_n$ since otherwise $$ z\in U_n\cap D = V'_n\subseteq A\cap D\subseteq A. $$ It follows that $$ 1\leq p_n(z)=p_n(y-x), $$ so $x$ and $y$ may be separated in the topology generated by $p_n$, and hence also in $\tau '$, proving our claim.

Clearly $\tau '$ is coarser than the default topology $\tau $ of $X$, so the identity map $$ \text{id} : (C, \tau ) \to (C, \tau ') $$ is continuous.

Given that $(C, \tau)$ is compact, and that we now know that $(C, \tau ')$ is Hausdorff, we have that the above map is a homeomorphism.

In case $(X,\tau ')$ happens to be Hausdorff, an application of the result by Rudin mentioned by the OP would provide the desired metric, but a slight adaptation of it also works even if $\tau '$ is not Hausdorff. By that I mean that Rudin's proof carries through without the assumption that $(X,\tau ')$ is Hausdorff, as long as we drop the requirement that $d$ satisfies $$ d(x,y)=0\Rightarrow x=y, $$ that is, as long as we satisfy ourselves with a pseudo-metric, rather than a metric.

The resulting pseudometric $d$ would of course be compatible with the topological space $(C, \tau ')$, and hence also $(C, \tau )$, but since this is a Hausdorff space, as seen above, $d$ must be a metric when restricted to $C$.

Answer 2

$Y=\operatorname{span}(C)$ is a locally convex tvs which is still metrisable I think (using that $C$ has a countable base, being compact metrisable). So the first result would apply there.