\lim_{\alpha, \mathcal U} \|\lambda_s 
(\eta_\gamma^{(\alpha)})^2 - (\eta_\gamma^{(\alpha)})^2\|_{\ell^2\Gamma} = 0.$$ In particular, $\lambda$ has almost invariant vectors in $\ell^2\Gamma$, which implies that $\Gamma$ is amenable.

To prove the claim, take $x = \sum_\gamma x_\gamma \lambda_\gamma$ arbitrary in the group algebra $\C[\Gamma]$ with $x_\gamma \geq 0$. Then, we find $x^* x = \sum_s (\sum_\gamma
\overline{x_\gamma} x_{\gamma s}) \lambda_s$, so that $$\|x\|_{L^4({\mathcal{L}}G)}^4 = \sum_{s,\gamma,\widetilde \gamma} x_\gamma \overline{x_{\gamma s}} \overline{x_{\widetilde \gamma}} x_{\widetilde \gamma s}.$$ To compute $\|j(x)\|^4_{\prod_{\mathcal U} {\mathcal{S}}_4(\ell^2 \Gamma \otimes
  {\mathcal{H}})} $ and to simplify the notation, we shall denote, for $s \in\Gamma$ and $\xi \in \ell^p(\Gamma;{\mathcal{S}}_p({\mathcal{H}}))$, $s \cdot \xi=
(\xi(s^{-1}t))_{t \in \Gamma}$. We shall also use the relation $s \cdot
\xi = (\lambda_s \otimes id) \xi (\lambda_s^{\ast} \otimes id)$. Hence, $$j(x)^* j(x) = \left( \sum_s (\lambda_s \otimes id) (\sum_\gamma \overline{x_\gamma}x_{\gamma s}(s^{-1} \cdot \xi_\gamma^{\alpha})^* \xi_{\gamma s}^\alpha)\right)_{\alpha},$$ so that $$\begin{aligned}
  \|j(x)\|_{\prod_{\mathcal U} {\mathcal{S}}_4(\ell^2 \Gamma \otimes
  {\mathcal{H}})} ^4 &=& \lim_{\alpha, \mathcal U} \sum_{s,\gamma,\widetilde \gamma} x_\gamma \overline{x_{\gamma s}} \overline{x_{\widetilde \gamma}} x_{\widetilde \gamma s} {\mathrm{Tr}}\left(\xi_{\gamma s}^{(\alpha)*} s^{-1}.(\xi_\gamma^{(\alpha)} \xi_{\widetilde \gamma}^{(\alpha)*}) \xi_{\widetilde \gamma s}^{(\alpha)} \right).
$$ Here, ${\mathrm{Tr}}$ is the natural (semi-finite) trace on $\ell^\infty(\Gamma; B({\mathcal{H}}))$. By $x_\gamma\geq 0$ and Hölder’s inequality, we have $$\begin{aligned}
  \|j(x)\|_{\prod_{\mathcal U} {\mathcal{S}}_4(\ell^2 \Gamma \otimes
  {\mathcal{H}})} ^4 & = & \sum_{s,\gamma,\widetilde \gamma} x_\gamma {x_{\gamma 