Hulanicki's theorem

Remarks

Condition 2 says that the trivial representation is weakly contained in $\lambda$. From Reiter's condition one passes to condition 2 by taking $\xi = f^{1/2}$, using $\norm{a^{1/2} - b^{1/2}}_2^2 \le \norm{a - b}_1$ for nonnegative $a, b$. Condition 2 gives condition 3, because $\ip{\lambda(f)\xi}{\xi}$ is close to $\int f$ for almost invariant $\xi$, while $\norm{\lambda(f)} \le \norm{f}_1$ always.

Example: for $G = \Z$, take $\xi = n^{-1/2} \mathbf{1}_{\set{0, \ldots, n-1}}$; then $\norm{\lambda(k)\xi - \xi}_2^2 = 2\abs{k}/n$ for $\abs{k} \le n$. For the free group $F_2$ with generators $a, b$ and $f = \frac14(\delta_a + \delta_{a^{-1}} + \delta_b + \delta_{b^{-1}})$, one has $\norm{\lambda(f)} = \sqrt{3}/2 < 1$, so $F_2$ is not amenable.

A consequence is that, for amenable $G$, the full and the reduced group C*-algebras coincide.

Going deeper

For a student

The theorem puts amenability into the language of unitary representations: the trivial representation lies in the reduced dual, which is the support of $\lambda$ for the Fell topology. For a discrete group, condition 3 applied to a symmetric probability measure gives Kesten's criterion. Non-amenability is a weak relative of Kazhdan's property (T). A non-amenable group has no almost invariant vectors in $L^2(G)$. Property (T) forbids them in every representation without invariant vectors.