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Let $G$ be a group and let $c > 1$ be a constant. We denote by $B_c(G)$ the space of all coefficients of the representations of $G$ which are uniformly bounded by $c$; more precisely, a function $f: G \rightarrow \mathbb{C}$ belongs to $B_c(G)$ if there is a uniformly bounded representation $\pi: G \rightarrow \mathcal{B}(\mathcal{H}_\pi)$ such that $\sup_{x \in G} \| \pi(x)\| \leq c$ and $$ f(x) = \langle \pi(x) \xi, \eta\rangle $$ for some $\xi, \eta \in \mathcal{H}_\pi$.

Further, define $$ \| f \|_{B_c} := \inf \{ \|\xi\| \; \|\eta\| \} $$ where the infimum runs over all possibilities of $\pi$ as well as $\xi,\eta$. One can show that $B_c(G)$ is a Banach space.

Apparently it was conjectured for a while that $$ \bigcup_{c>1} B_c(G) = M_{cb}A(G) $$ where $M_{cb}A(G)$ is the space of completely bounded multipliers of the Fourier algebra of $G$ (which is isometrically isomorphic to the space of Herz-Schur multipliers of $G$). It is easy to show that $\bigcup_{c>1} B_c(G) \subseteq M_{cb}A(G)$ based on a complete characterization of the later space (by Bozejko-Fendler 91 and Jolissaint 92).

But Haagerup, in a never published manuscript 85, proved that this is not the case for $\Bbb{F}_N$, non-commutative free group with $N$ generators.

Question: Are there classes/examples of non-amenable groups for which the equation holds? What if we look at a locally compact group $G$ (instead of a discrete one)?

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  • $\begingroup$ Small correction to 1st paragraph: I assume you wish to say that $\Vert \pi(g)\Vert\leq c$ for all $g\in G$? $\endgroup$
    – Yemon Choi
    Commented Feb 5, 2018 at 16:18
  • $\begingroup$ @YemonChoi: Thanks I definitely meant that! $\endgroup$
    – Mahmood Al
    Commented Feb 5, 2018 at 16:23
  • $\begingroup$ Mahmood, do you know what these "exotic" cb-multipliers look like for the case of free groups? Is this another trick with Leinert sets or similar? $\endgroup$
    – Yemon Choi
    Commented Feb 5, 2018 at 16:28
  • $\begingroup$ @YemonChoi: The proof is using radial cb-multipliers. I have a copy of the paper which I will send you via email. (I personally do not see any indication of Leinert sets there, but I might be naive.) $\endgroup$
    – Mahmood Al
    Commented Feb 5, 2018 at 16:33
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    $\begingroup$ @AdriánGonzález-Pérez So based on your comment, if $G$ is unitarizable then $B(G)=M_{cb}A(G)$. But this implies amenability! Ruan (in an unpublished manuscript generalizing the argument of Losert in [Proc. Amer. Math. Soc. 92 (1984), no. 3, 347–354]) proved that $G$ is amenable if and only if $B(G)=M_{cb}A(G)$. All of this can be found in Pisier's book on similarity on page 54 (2nd edition). So your remark would solve the conjecture. Would you tell us where you have seen the proof? $\endgroup$
    – Mahmood Al
    Commented Feb 6, 2018 at 16:36

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