Question

Let $G$ be a countable group and $\lambda \colon G \to U(\ell^2 G)$ its left-regular representation. Suppose that there exists a constant $C>0$ such that for all $T \in B(\ell^2 G)$

$$\inf \lbrace\|T-S\| \mid S \in \lambda(G)' \rbrace \leq C \cdot \sup\lbrace \|\lambda(g)T- T \lambda(g) \| \mid g \in G\rbrace.$$

(Here $\lambda(G)'$ denotes the commutant of $G$ in $B(\ell^2 G)$.)

Question: Is $G$ amenable?

It is fairly easy to see that amenability of $G$ implies the existence of such a constant. Indeed, one may take $S$ to be some fixed point for the conjugation action on $\overline{\rm conv}\lbrace \lambda(g)T\lambda(g)^* \mid g \in G\rbrace$. I am asking for the converse of this statement.

EDIT: Since the derivation problem came up in Kate's comment, I want to clarify to what version of it my question is related. The inequality above holds for some $C$ if and only if the first bounded cohomology of $G$ with coefficients in $B(\ell^2 G)$ (with the conjugation action induced by $\lambda$) is reduced. This is a straightforward application of the open mapping theorem. Now, two things are unclear:

Question: Can $H^1_b(G,B(\ell^2 G))$ be reduced without being zero?

and

Question: Can $H^1_b(G,B(\ell^2 G))$ be zero without $G$ being amenable?

Answer 1

Let $M=C^*_\lambda(G)''$ be group von Neumann algebra of $G$. The the condition above implies:

$d(T, M')\leq C ||ad(T)|_{M}||$ for every $T\in B(l^2 G)$. The last inequality is equivalent to saying that every derivation of $M$ into $B(l^2 G)$ is inner.

Edit: the above inequality is satisfied automatically (was clarified to me by Stuart White).

It is known that if $M\subset B(H)$ has a cyclic vector, then every bounded derivation from $M$ into $B(H)$ is inner [E. Christensen, Extensions of derivations II, Math. Scand., 1982]. Thus [Christensen, op cit, Cor 5.4] we have

$d(T,M')\leq 3/2\|(\mathrm{ad}| T)_{M}\|$

for every $T\in B(l^2 G)$.