\in {M_{\mathcal{CB}}A}(G)$. There is a unique completely bounded linear map $T_\phi^p: L^p({\mathcal{L}}G) \to L^p({\mathcal{L}}G)$ that satisfies $P_F T_\phi^p(x) P_F = M_\phi (P_F x P_F)$. It has completely bounded norm less than $\|\phi\|_{{M_{\mathcal{CB}}A}(G)}$. This map is called the Fourier multiplier with symbol $\phi$.

Let us first prove that $T_\phi^p$ exists and is bounded. For $p=\infty$ this is obvious. Assume $p=1$. Consider $\widetilde \phi(s) = \phi(s^{-1})$. Then $\|\widetilde \phi\|_{{M_{\mathcal{CB}}A}(G)}=\|\phi\|_{{M_{\mathcal{CB}}A}(G)}$. Take $\varphi \in A(G) \to x_\varphi \in L^1({\mathcal{L}}G)$ the isometry described in Lemma \[lemma=isom\_AG\_L1\]. Define $T_\phi^1$ by $T_\phi^1(x_\varphi)= x_{\widetilde \phi \varphi}$. It is a map of norm at most $\|\phi\|_{{M_{\mathcal{CB}}A}(G)}$, and by Lemma \[lemma=isom\_AG\_L1\] it satisfies $P_F T_\phi^1(x) P_F = M_\phi (P_F x P_F)$.

The general case follows by interpolation. We claim that the maps $T_\phi^1$ and $T_\phi^\infty$ are compatible with respect to the pair $(A_0,A_1)$ given by Theorem \[thm=interpolation\_BB\]. Indeed, let $a \in A_0 \cap A_1$. We have to show that $j_1^{-1}(T_\phi^1(j_1(a)) = j_\infty^{-1}(T_\phi^\infty(j_\infty(a))$. By it suffices to have that $u_F(j_1^{-1}(T_\phi^1(j_1(a))) = u_F (j_\infty^{-1}(T_\phi^\infty(j_\infty(a)))$ for every relatively compact $F \subseteq G$. But by , this equality is equivalent to $j_{1,F}^{-1}(M_\phi(j_{1,F}(u_F(a))) = j_{\infty,F}^{-1}(M_\phi(j_{\infty,F}(u_F(a)))$, which holds by .

By interpolation, there is therefore a map $T_\phi^p:L^p({\mathcal{L}}G) \to L^p({\mathcal{L}}G)$ of norm at most $\|\phi\|_{{M_{\mathcal{CB}}A}(G)}$. It 