the case when $\varphi$ is positive definite, so that the corresponding element of ${\mathcal{L}}G_*$ is positive. By the definition of $L^1({\mathcal{L}}G)$ and the spatial derivative, $x_\varphi$ is the positive self-adjoint operator characterized by $\|x_\varphi^{1/2} \xi\|^2 =\langle \varphi, \lambda(\xi)\lambda(\xi)^*\rangle$ for all $\xi \in L^2G$ that are $\psi$-bounded. By definition of $P_F x_\varphi P_F$, and the fact that $L^2 F \subseteq L^1 G$, we get $$\langle P_F x_\varphi P_F \xi,\xi\rangle = \iint_{F \times F} \varphi(t s^{-1}) \xi(t) \overline{\xi(s)} ds dt,$$ and by polarization $$\langle P_F x_\varphi P_F \xi,\eta\rangle = \iint_{F \times F} \varphi(t s^{-1}) \xi(t) \overline{\eta(s)} ds dt.$$ This proves the lemma when $\varphi$ is positive definite. The general case follows by the linearity of $\varphi \mapsto P_F x_\varphi P_F$.

Fourier multipliers
-------------------

Given a function $\phi \in {M_{\mathcal{CB}}A}(G)$, we would like to define the Fourier multiplier on $L^p({\mathcal{L}}G)$ as the restriction to $L^p({\mathcal{L}}G)$ of $M_\phi$. This is possible when $G$ is unimodular because $L^p({\mathcal{L}}G) \cap B(L^2 G)$ is dense in $L^p({\mathcal{L}}G)$. When $G$ is not unimodular (and $p \neq \infty$) this is more problematic: $L^p({\mathcal{L}}G) \cap B(L^2 G) = \{0\}$, whereas $M_\phi$ is only defined on $B(L^2G)$. The definition is therefore done with the help of Proposition \[prop=def\_of\_PFxPF\].

\[dfprop=def\_\_Tphip\] Let $\phi 