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?


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. $\endgroup$ – Pietro Majer Nov 13 '20 at 20:05
  • $\begingroup$ Thanks, this is really helpful. $\endgroup$ – aduh Nov 13 '20 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 '20 at 7:16
  • $\begingroup$ Thank you.. I should have said "continuous" indeed $\endgroup$ – Pietro Majer Nov 14 '20 at 8:29

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.