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It’s relatively easy to show that if $J$ is a closed subgroup of a finite-dimensional real Banach space, $B$, then it is a vector subspace iff for all bounded linear functionals $\sigma$ of $B$, $\sigma(J)$ is not rank one (i.e., not of the form $\delta\, {\bf Z}$ for some nonzero real $\delta$); restated, iff either $\sigma(J)$ is zero or ${\bf R}$.

Is the same true for infinite-dimensional real Banach spaces? I am primarily thinking of $B = {\rm Aff\,} K$ where $K$ is a Choquet simplex. Explicitly, the question (posed in terms of subgroups, rather than the equivalent form in terms of closed subgroups) is,

Suppose $H$ is a subgroup of the real Banach space $B$, and for every $\sigma \in B^*$, either $\sigma(H)$ is zero or a dense subgroup of ${\bf R}$. Then the closure of $H$ is a real vector space.

Of course, there is an immediate reduction to the case that $\sigma(H)$ is dense for all nonzero $\sigma \in B^*$, which then boils down to showing density (unfortunately, this reduction does not preserve the type of Banach space, in my case, ${\rm Aff\,}K$). But I suspect it’s not true (or have I missed something completely obvious?).

The question comes from the study of good or refinable and related measures (following Akin et al) on Cantor sets, specifically those probability measures invariant under a minimal self-homeomorphism. These measures constitute the $K$ of ${\rm Aff\,} K$; $H$ is a subgroup of a countable dense subgroup $G$ of ${\rm Aff\,} K$, and of a special form, $H = G \cap {\rm ker\,} \tau$, where $\tau \in K$. The $G$ is the ordered Grothendieck group of the crossed product arising from the minimal action; alternatively, $G$ is the ordered cohomology group arising from the minimal action.

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    $\begingroup$ In mathoverflow.net/questions/45322/… you have an example of a closed connected subgroup of a infinite-dimensional separable real Hilbert space $B$ that spans a dense subspace of $B$ and contains no real line. Clearly the image of this subgroup by any continuous linear form is all of $\mathbf{R}$. $\endgroup$
    – YCor
    Apr 26, 2015 at 20:28
  • $\begingroup$ Thanks very much! I suppose it is known whether real Hilbert space can be embedded in ${\rm Aff\,}K$? $\endgroup$ Apr 26, 2015 at 23:08
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    $\begingroup$ They can. In fact, in separable $L_1$ predual (in particular, the universal space $C[0,1]$) can be isometrically embedded as a norm one complemented subspace of Aff $K$ for some $K$. This is an old result of Lazar and Lindenstrauss. $\endgroup$ Apr 27, 2015 at 0:22
  • $\begingroup$ Thanks again. Does the result of Lazar & Lindenstrauss appear in the same paper as the one in which they prove that every metrizable Choquet simplex is an inverse limit of finite-dimensional simplices? (If so, I should have known it ....) $\endgroup$ Apr 27, 2015 at 0:32
  • $\begingroup$ Yes, their Acta paper. $\endgroup$ Apr 27, 2015 at 0:49

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