s}} {x_{\widetilde \gamma}} x_{\widetilde \gamma s} \lim_{\alpha, {\mathcal{U}}} {\mbox{Re\,}}\: {\mathrm{Tr}}\left(s^{-1}.(\xi_\gamma^{(\alpha)} \xi_{\widetilde \gamma}^{(\alpha)*}) \xi_{\widetilde \gamma s}^{(\alpha)}\xi_{\gamma s}^{(\alpha)*}\right)\\
& \leq & \sum_{s,\gamma,\widetilde \gamma} x_\gamma {x_{\gamma s}} {x_{\widetilde \gamma}} x_{\widetilde \gamma s} 1= \|x\|_{L^4({\mathcal{L}}G)}^4.\end{aligned}$$ Hence if $\gamma,\widetilde \gamma,\gamma s,\widetilde \gamma s$ belong to the support of $x$, $$\lim_{\alpha,{\mathcal{U}}} {\mbox{Re\,}}\: {\mathrm{Tr}}\left(s^{-1}.(\xi_\gamma^{(\alpha)} \xi_{\widetilde \gamma}^{(\alpha)*}) \xi_{\widetilde \gamma s}^{(\alpha)}\xi_{\gamma s}^{(\alpha)*}\right) = 1,$$ so that by the uniform convexity Lemma \[lemma=uniform\_convexity\] below $$\label{eq=limit_terme_a_terme}
\lim_{\alpha, \mathcal U}\|\lambda_s (\eta_{\widetilde \gamma s}^{(\alpha)})^2 - (\eta_{\gamma}^{(\alpha)})^2\|_{\ell^2(\Gamma)} = 0.$$ Since $x$ was arbitrary, holds for all $s,\gamma,\widetilde \gamma \in \Gamma$. This proves , and hence that $\Gamma$ is amenable.

The converse was proved in [@NeuRic].

\[lemma=uniform\_convexity\] Let $\Gamma$ be a discrete group and let ${\mathcal{H}}$ be a Hilbert space. For all $\epsilon >0$, there exists $\delta>0$ such that the following holds: Let $s \in \Gamma$ and $\xi_i \in \ell^4(\Gamma;{\mathcal{S}}_4({\mathcal{H}}))$, $i=1\dots 4$ be unit vectors, and denote by $\eta_i$ the unit vector $(\|\xi_i(t)\|_{{\mathcal{S}}_4({\mathcal{H}})})_{t \in \Gamma}$ in $\ell^4 \Gamma$. If $${\rm Re}\: {\mathrm{Tr}}\left(s^{-1}.(\xi_1 \xi_2) \xi_3 \xi_4\right) \geq 1-\delta,$$ then $\|\lambda_s \eta_3^2 - \eta_1^2\|_{\ell^2\Gamma} \leq \epsilon$.

In the proof we write $o(1)$ for a quantity that goes to $0$ as $\delta$ goes to $0$. By Hölder’s inequality in ${\mathcal{S}}_4({\mathcal{H}})$, $$\begin{aligned}
1-\delta\leq{\mbox{Re\,}}\: {\mathrm{Tr}}\left(s^{-1}.(\xi_1 \xi_2) \xi_3
  \xi_4\right)&\leq& \langle \lambda_{s^{-1}}(\eta_1 