Let $G$ be a locally compact group. For $1<p<\infty$ let $\lambda\_p:G\to\mathcal{B}(L^p(G))$ (resp. $\rho\_p:G\to\mathcal{B}(L^p(G))$) be the left (resp. right) regular representation. $CV\_p(G)$ is defined as the commutant of $\{\rho\_p(s)\ :\ s\in G\}$ and $PM\_p(G)$ is the weak\*-closed linear span of $\{\lambda\_p(s)\ :\ s\in G\}$.


In this paper <http://arxiv.org/abs/1308.1073> it is stated that $CV\_p(G)$ is the bicommutant of $PM\_p(G)$. If I'm not mistaken, the inclusion $PM\_p(G)''\subseteq CV\_p(G)$ follows directly from the fact that $\lambda\_p(s)$ and $\rho\_p(t)$ commute for all $s,t\in G$, but I don't see why the other inclusion holds. Any ideas or references?



Why are the convolvers in the bicommutant of the pseudo-measures? ($CV_p(G)\subseteq PM_p(G)''$)
Yes, I'm sorry, this is definitely a mistake. I don't know where that claim came from.


As $X''' = X'$ for any set $X$, it follows that $CV\_p = CV\_p''$, and so also that $PM\_p'' = CV\_p$ if and only if $PM\_p' = CV\_p' = \rho\_p(G)''$. As trivially $PM\_p \subseteq \rho\_p(G)'$ (as you note) we need only show that $PM\_p' \subseteq \rho\_p(G)''$.


Spelling this out, if $T\lambda\_p(g) = \lambda\_p(g)T$ for all $g$, and $S\rho\_p(g) = \rho\_p(g)S$ for all $g$, then we want $ST=TS$.


Like Yemon, I can prove this claim if $G$ is discrete.


**Edit:** You can now find a full proof here: <https://arxiv.org/abs/1308.1073>


