I am trying to better understand a condition that appears in Theorem 1 of [this paper](https://arxiv.org/abs/1510.07921).


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



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



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> 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?
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Does set of finitely additive probability measures embed linearly into a strictly convex dual Banach space?
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 $f(\alpha\_0)=c.$


