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I am stuck on the following question for quite sometime now. Please help, any comment is welcome.

Let $S$ be a topological semigroup and $\mathcal{F}$ be a translation invariant, conjugate closed linear subspace of $B(S)$ containing constant functions. For any functional $\mu\in \mathcal{F}^*$ and $f\in \mathcal{F}$, the left introversion operator is defined as $T_\mu : \mathcal{F} \rightarrow B(S)$ where

$T_\mu f(x) := \mu(L_xf)$ for any $f\in \mathcal{F}$, $x\in S$.

We need to show that the set $\{T_\mu f : ||\mu|| \leq 1\}$ is the closure of $cco(\mathcal{O}_r(f))$ in $B(S)$ with respect to the pointwise topology.

Here $cco(E)$ denotes the convex circled hull of $E$ and $\mathcal{O}_r(f)$ denotes the right orbit of $f$, i.e, $\mathcal{O}_r(f) = \{R_xf: x\in S\}$ where $R_xf \in B(S)$ is defined as $R_xf(y) = f(yx) = L_yf(x)$.

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  • $\begingroup$ Does S carry the discrete topology? If not, what continuity assumptions are you making about the map S \times S \to S ? $\endgroup$
    – Yemon Choi
    Commented Nov 16, 2015 at 17:32
  • $\begingroup$ Do you at least know how to prove this when $\mathcal{F}=B(S)$? $\endgroup$
    – Yemon Choi
    Commented Nov 16, 2015 at 17:36
  • $\begingroup$ @Yemon Choi: No, $S$ may not carry the discrete topology. $S$ is any topological semigroup here. So the multiplication map is jointly continuous. I do not see how to approach this problem for any $\mathcal{F}$. $\endgroup$
    – Carl
    Commented Nov 16, 2015 at 17:52
  • $\begingroup$ OK, thanks. Can you at least do the case $S={\bf Z}_+$ and $\mathcal{F}=c_0({\bf Z}_+)$? $\endgroup$
    – Yemon Choi
    Commented Nov 16, 2015 at 20:56
  • $\begingroup$ What is $B(S)$? $\endgroup$
    – Jochen Wengenroth
    Commented Nov 17, 2015 at 15:25

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