$1 \leq p \leq \infty$. Let $\phi \in {M_{\mathcal{CB}}A}(G)$. Then, $$\Vert M_{\phi}^p \Vert_{{\mathcal{CB}}({\mathcal{S}}_p(L^2G))} \leq \Vert T_{\phi}^p \Vert_{{\mathcal{CB}}(L^p({\mathcal{L}}G))}.$$

Let $q$ be the conjugate exponent of $p$. The values of $p$ for which $p/q \in \Q$ form a dense subset of $[1,\infty]$. It suffices to prove the theorem for such a $p$. Indeed, since $M_\phi^p$ is defined by means of interpolation, it follows from [@BerghLof Theorem 4.1.2 and Section 2.4 (6)] and the re-iteration theorem that the logarithm of $\Vert M_{\phi}^p \Vert_{{\mathcal{CB}}({\mathcal{S}}_p(L^2G))}$ is a convex, hence continuous function in $p$. Similarly, $\Vert T_{\phi}^p \Vert_{{\mathcal{CB}}({\mathcal{S}}_p(L^2G))}$ is continuous in $p$.

Let $F = \{ s_1, \ldots, s_k \} \subseteq G$ be a finite subset, and consider two matrices $a,b \in M_k(\C)$. We claim that $$|\sum_{i,j} \phi(s_i s_j^{-1}) a_{i,j}  b_{j,i}| \leq \Vert T_{\phi}^p \Vert_{{\mathcal{CB}}(L^p({\mathcal{L}}G))} \|a\|_{{\mathcal{S}}_p^k} \|b\|_{{\mathcal{S}}_q^k}.$$ By Theorem \[thm=mult\_de\_symb\_continu\] this implies that $\Vert M_{\phi}^p \Vert_{B({\mathcal{S}}_p(L^2G))} \leq \Vert T^p_\phi  \Vert_{{\mathcal{CB}}(L^p({\mathcal{L}}G)) }$. The inequality $\Vert M_\phi^{p} \Vert_{{\mathcal{CB}}({\mathcal{S}}_p(L^2G))} \leq \Vert T^p_\phi  \Vert_{{\mathcal{CB}}(L^p({\mathcal{L}}G)) }$ follows similarly, by taking $a_{i,j} \in {\mathcal{S}}_p^n$, $b_{i,j} \in {\mathcal{S}}_q^n$ instead of $\mathbb{C}$.

Take $x_\alpha,y_\alpha$ given by Lemma \[lem=recover\_phi\_from\_Iphi\], and consider the element $A_\alpha \in {\mathcal{S}}_p^k \otimes L^p({\mathcal{L}}G)$ and $B_\alpha \in {\mathcal{S}}_q^k \otimes L^q({\mathcal{L}}G)$ defined by $$A_\alpha = (a_{i,j} \lambda(s_i) x_\alpha \lambda(s_j^{-1}))_{i,j \leq k}, B_\alpha 