# Does set of finitely additive probability measures embed linearly into a strictly convex dual Banach space?

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 $$f(\mu_n) for some $$\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)) contradicting $$f(\alpha_0)=c.$$
• 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. – Pietro Majer Nov 13 '20 at 20:05
• @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. – Harry West Nov 14 '20 at 7:16