$G$ contains $S_\alpha$ for all $n$ large enough. By in Theorem \[thm=interpolation\_BB\], $j_\infty^{-1}(\lambda(h_\alpha))$ belongs to $A_0 \cap A_1$, we can therefore define $a_\alpha =j_{2r}(j_\infty^{-1}(\lambda(h_\alpha))))\in L_{2r}({\mathcal{L}}G)$. Normalise $h_\alpha$ so that $\|a_\alpha\|_{2r}=1$. Consider the unit vectors $x_\alpha \in L_p({\mathcal{L}}G)$ and $y_\alpha \in L_q({\mathcal{L}}G)$ defined by $x_\alpha = |a_\alpha|^{2k}$ and $y_\alpha = |a_\alpha|^{2k'}$. By in Theorem \[thm=interpolation\_BB\], we can write $x_\alpha = j_p( j_\infty^{-1}(\lambda(f_\alpha)))$ for a non-negative function $f_\alpha \in A_c(G)$ with support contained in $(S_\alpha^{-1}S_\alpha)^k$. Similarly $j_q^{-1}(y_\alpha)$ belongs to $A_0 \cap A_1$, and $(j_1(j_q^{-1}(y_\alpha)))^* \in L_1({\mathcal{L}}G)$ corresponds to $g_\alpha \in A(G)$ satsifying $g_\alpha(s) \geq 0$ for every $s \in S$.

For every $\phi \in {M_{\mathcal{CB}}A}(G)$, since the maps $T_\phi^p$ are compatible with respect to the interpolation given by Theorem \[thm=interpolation\_BB\], we can write by , $$\langle T_\phi^p x_\alpha,y_\alpha^*\rangle_{L^p, L^q} = \langle T_\phi^\infty(\lambda(f_\alpha)),(j_1(j_q^{-1}(y_\alpha)))^*\rangle_{L^\infty, L^1} = \int_G \phi(s) f_\alpha(s) g_\alpha(s) ds.$$ Taking $\phi=1$ we get $\int f_\alpha(s) g_\alpha(s) ds = \langle x_\alpha,y_\alpha\rangle = \|a_\alpha\|_{2r}^{2r} = 1$. Hence, $$\begin{aligned}
|\langle T_\phi^p x_\alpha,y_\alpha^*\rangle - \phi(e)| &=& |\int (\phi(s)-\phi(e)) f_\alpha(s) g_\alpha(s) ds|\\ & \leq & \sup_{s \in \textrm{support}(f_\alpha)} |\phi(s) - \phi(e)|.\end{aligned}$$ But $\phi$ is continuous and the support of $f_\alpha$ converges to $\{e\}$. This proves the lemma.

\[Thm-TransferenceEstimate\] Let $G$ be a locally compact group and let 