I am trying to better understand a condition that appears in Theorem 1 of this paper.

Let $K$ be a convex and compact subset of a locally convex tvs. The condition is:

$K$ embeds linearly into a strictly convex dual Banach space endowed with its weak* topology. Call this the embedding condition.

In particular, I would like to know whether the embedding condition holds when $K$ is taken to be the set of all finitely additive probability measures on some measurable space, equipped with the weak* topology. In other words,

Is there a measurable space $(\Omega, \mathcal A)$ such that, with $K$ the set of finitely additive probability measures on $(\Omega, \mathcal A)$ equppied with the weak* topology, the embedding condition fails?


1 Answer 1


The space $M$ of finitely additive probability measures on $\omega_1,$ with the weak topology defined by the seminorms $\mu\mapsto|\mu(S)|$ for $S\subseteq\omega_1,$ does not admit a strictly convex lower semicontinuous function. This answers your question because of the easy direction of Theorem 1.1 of that paper - compose the embedding with the norm. If you're interested, $\omega_1$ can be replaced by $\omega$: pick a strictly increasing sequence $\langle A_\alpha:\alpha\in\omega_1\rangle$ in $\mathcal{P}(\omega)/fin,$ and replace the conditions "$\mu([\alpha,\beta))=1$" below by "$\mu(A_\beta\setminus A_\alpha)=1$ and $\mu(\{n\})=0$ for all $n$". This means the embedding condition fails iff your $\mathcal A$ is infinite.

Suppose for contradiction that there is a strictly convex lower semicontinuous function $f:M\to\mathbb R.$ For all $\alpha<\omega_1$ let $g(\alpha)$ be the infimum of $f(\mu)$ where $\mu$ is restricted to measures $\mu\in M$ with $\mu([\alpha,\beta))=1$ for some countable $\beta>\alpha.$ The infimum is actually attained, because if $f(\mu_n)<g(\alpha)+1/n$ for some $\mu_n$ with $\mu_n([\alpha,\beta_n))=1,$ then by lower semicontinuity $f$ attains a minimum $\leq g(\alpha)$ on the compact space of $\mu\in M$ with $\mu([\alpha,\cup_n\beta_n))=1.$ Since $g$ is non-decreasing, it is eventually constant: there is a real $c$ and a $\alpha_0<\omega_1$ such that $g(\alpha)=c$ for $\alpha\geq \alpha_0.$ There exists $\alpha_1>\alpha_0$ and a measure $\mu_1$ with $\mu_1([\alpha_0,\alpha_1))=1$ and $f(\mu_1)=g(\alpha_0)=c,$ and there exists $\alpha_2>\alpha_1$ and a measure $\mu_2$ with $\mu_2([\alpha_1,\alpha_2))=1$ and $f(\mu_2)=g(\alpha_1)=c.$ But by strict convexity $f(\tfrac12(\mu_1+\mu_2))<c,$ contradicting $f(\alpha_0)=c.$

  • $\begingroup$ Isn't this true more generally for any infinite dimensional real vector space $E$, endowed with the weak topology $\sigma(E,F)$ induced by a family $F$ of linear functionals on $E$? Any non-empty open set in this topology contains a finite co-dimensional affine space. So it has no non-empty, strictly convex open sets. Therefore no strictly convex lower semicontinuous functions either. $\endgroup$ Nov 13, 2020 at 20:05
  • $\begingroup$ Thanks, this is really helpful. $\endgroup$
    – aduh
    Nov 13, 2020 at 21:39
  • 2
    $\begingroup$ @PietroMajer: it's not true for a the usual weak topology on $\ell^2$ for example, where the norm function $f(x)=\|x\|$ is strictly convex on any closed affine subspace not including the origin, and weakly lower semicontinuous. In this example the sublevelsets $\{x:f(x)\leq t\}$ are weakly closed and contain no nonempty weakly open set. $\endgroup$
    – Harry West
    Nov 14, 2020 at 7:16
  • $\begingroup$ Thank you.. I should have said "continuous" indeed $\endgroup$ Nov 14, 2020 at 8:29

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