---
abstract: |
    Consider a completely bounded Fourier multiplier $\phi$ of a locally compact group $G$, and take $1 \leq p \leq \infty$. One can associate to $\phi$ a Schur multiplier on the Schatten classes ${\mathcal{S}}_p(L^2 G)$, as well as a Fourier multiplier on $L^p({\mathcal{L}}G)$, the non-commutative $L^p$-space of the group von Neumann algebra of $G$. We prove that the completely bounded norm of the Schur multiplier is not greater than the completely bounded norm of the $L^p$-Fourier multiplier. When $G$ is amenable we show that equality holds, extending a result by Neuwirth and Ricard to non-discrete groups.

    For a discrete group $G$ and in the special case when $p\neq 2$ is an even integer, we show the following. If there exists a map between $L^p({\mathcal{L}}G)$ and an ultraproduct of $L^p({\mathcal{M}}) \otimes {\mathcal{S}}_p(L^2G)$ that intertwines the Fourier multiplier with the Schur multiplier, then $G$ must be amenable. This is an obstruction to extend the Neuwirth-Ricard result to non-amenable groups.
address:
- 'M. Caspers, Laboratoire de Mathématiques, Université de Franche-Comté, 16 Route de Gray, 25030 Besançon, France'
- 'M. de la Salle, CNRS, Laboratoire de Mathématiques, Université de Franche-Comté, 16 Route de Gray, 25030 Besançon, France'
author:
- Martijn Caspers
- Mikael de la Salle
date: |
    . [*MSC2010*]{}: 43A15, 46B08, 46B28, 46B70. [*Keywords*]{}: non-commutative $L^p$-spaces, Schur multiplier, Fourier multipliers, amenability.\
    \
    The first author was supported by the ANR project: ANR-2011-BS01-008-01.\
    The scond author was partially supported by the ANR projects NEUMANN and OSQPI. 
title: 'Schur and Fourier multipliers of an amenable group acting on non-commutative $L^p$-spaces'
---

Introduction
============

This paper studies the close connection between Schur and Fourier multipliers of locally compact groups $G$. In particular, we are interested in the relation between the completely bounded norms of such multipliers.

Given a bounded function $\phi:G \to \C$, the associated Fourier multiplier $T_\phi$, when it exists, is the unique weak-\* continuous map on the von Neumann algebra ${\mathcal{L}}G$ of $G$ extending $\lambda_g \mapsto \phi(g)\lambda_g$. The associated Schur multiplier $M_\phi$, when it exists, is the unique weak-\* continuous map on $B(L^2G)$ extending $(a_{s,t})_{s,t \in G} \in {\mathcal{S}}_2(L^2 G) \mapsto (\phi(s t^{-1}) a_{s,t})_{s,t \in G}$. Bo[ż]{}ejko and Fendler proved [@BozFen84] that $M_\phi$ indeed determines a bounded map if and only if $T_\phi$ defines a completely bounded map, and in this case the completely bounded norms coincide. The function $\phi$ is then called a completely bounded Fourier multiplier and the space of such $\phi$ is denoted ${M_{\mathcal{CB}}A}(G)$. Note that $T_\phi$ is the restriction of $M_\phi$ to ${\mathcal{L}}G \subset B(L^2G)$.

In this paper we investigate the Bo[ż]{}ejko-Fendler result for $L^p$-multipliers. When $G$ is discrete and $\phi \in {M_{\mathcal{CB}}A}(G)$, there is no technical difficulty for defining the Fourier multiplier $T_\phi^p:L^p({\mathcal{L}}G) \to L^p({\mathcal{L}}G)$ and the Schur multiplier $M_\phi^p:{\mathcal{S}}_p(\ell^2 G) \to {\mathcal{S}}_p(\ell^2 G)$ (they are just the extension of $T_\phi$/ restriction of $M_\phi$). When $G$ is not discrete, the definition of the Schur multiplier $M_\phi^p:{\mathcal{S}}_p(L^2 G) \to {\mathcal{S}}_p(L^2 G)$ is known (see Subsection \[subsection=Schur\]), and we define in Subsection \[subsection=Fourier\] the $L^p$-Fourier multiplier $T_\phi^p:L^p({\mathcal{L}}G) \to L^p({\mathcal{L}}G)$ spatially [@Con], [@Hilsum].

We are interested in the following question.

\[Que-One\] Let $G$ be a locally compact group. Is it true that, $$\Vert T_\phi^p \Vert_{{\mathcal{CB}}(L^p({\mathcal{L}}G))} = \Vert M_\phi^p \Vert_{{\mathcal{CB}}({\mathcal{S}}_p(L^2G))}.$$ for all completely bounded Fourier multipliers $\phi$ of $G$?

There are several motivations to study this problem. One is that a positive answer would imply that for a discrete group $G$, the property ${\rm AP}^{Schur}_{p,cb}$ considered in [@LafSal], see also [@Laa], is an invariant of the group von Neumann algebra of $G$. Furthermore, one is often interested in strict estimates of the norm of the transfered Fourier multiplier, see for example [@ChenXuYin Section 7].

In [@NeuRic], Neuwirth and Ricard studied Question \[Que-One\] for discrete groups. They noted that the inequality $\geq$ always holds, and they proved the other inequality in the case $G$ is amenable. This is an $L^p$-version of the Bo[ż]{}ejko-Fendler result. Their proof relies on the fact that the amenability of $G$ allows them to construct a completely isometric embedding of $L^p({\mathcal{L}}G)$ into an ultrapower of ${\mathcal{S}}_p(\ell^2 G)$ that intertwines Fourier and Schur multipliers.

Marius Junge (personal communication) pointed out to the second-named author that a positive answer to Question \[Que-One\] would also follow from the existence of a completely isometric embedding of $L^p({\mathcal{L}}G)$ into an ultraproduct of $L^p({\mathcal{M}}_n) \otimes {\mathcal{S}}_p(L^2 G)$ for some net of von Neumann algebras ${\mathcal{M}}_n$, that intertwines Fourier and Schur multipliers. He suggested that some weaker approximation property (as exactness) might provide such an embedding. Our first result (Theorem \[Thm-CounterExample\]) partially answers his suggestion negatively: if $p\neq 2$ is an even integer, $G$ is a discrete group, and such an embedding exists, then $G$ is amenable. This bad news was the motivation to study Question \[Que-One\] for general locally compact groups.

The main part of the present paper is to extend the Neuwirth-Ricard result to arbitrary locally compact groups: we first prove in Theorem \[Thm-TransferenceEstimate\] by a transference technique that the inequality $\geq$ always holds. We then prove the inequality $\leq$ for $G$ amenable in Theorem \[thm=embedding\_in\_ultraproduct\] and its Corollary \[Cor-MainEquality\]. Our proof is close to the proof given in [@NeuRic Section 3], but we encounter several technical issues that we have to overcome, mainly in the case of non-unimodular groups.

Let us explain a bit more in details the technical complications that we meet. Recall [@PisXu] that to a von Neumann algebra ${\mathcal{M}}$ one can associate non-commutative $L^p$ spaces $L^p({\mathcal{M}})$ for $1 \leq p \leq \infty$. There are several different ways to define $L^p({\mathcal{M}})$, but they all yield (completely) isometric spaces. When ${\mathcal{M}}= B({\mathcal{H}})$, $L^p({\mathcal{M}})$ is most naturally realized as the Schatten class ${\mathcal{S}}_p({\mathcal{H}})$ (which is contained in $B({\mathcal{H}})$). When ${\mathcal{M}}$ carries a normal faithful finite trace $\tau$ (for example ${\mathcal{M}}= {\mathcal{L}}G$ for a discrete group $G$), the natural choice is to see $L^p({\mathcal{M}})$ as the completion of ${\mathcal{M}}$ for the norm $\|x\|_p = \tau(|x|^p)^{1/p}$ ($L^p({\mathcal{M}})$ contains ${\mathcal{M}}$). When $G$ is a locally compact group, there is a natural weight on ${\mathcal{L}}G$, the Plancherel weight, which is a trace if and only if $G$ is unimodular. Although ${\mathcal{L}}G$ is often semi-finite, it is more natural to view the $L^p$-spaces of ${\mathcal{L}}G$ as contructed from this weight. As is well-known, the non-commutative $L^p$-spaces then have the property that $L^p$ and $L^q$ do not naturally intersect for $p \neq q$. It turns out that the most convenient way to define the $L^p$-Fourier multiplier is by realizing $L^p({\mathcal{L}}G)$ in the Connes-Hilsum construction (Subsection \[subsect=Connes-Hilsum\]) associated to the Plancherel weight on ${\mathcal{L}}G'  = {\mathcal{R}}G$ (the von Neumann algebra of $L^2G$ associated to the right translations). Indeed, this realizes $L^p({\mathcal{L}}G)$ as a space of unbounded operators on $B(L^2 G)$, and the naive definition of $T_\phi^p$ would just be the restriction of $T_\phi$ to $L^p({\mathcal{L}}G)$. Since in general (*i.e.* unless $G$ is unimodular) $L^p({\mathcal{L}}G) \cap B(L^2(G)) = \{0\}$ this is a bit to naive, and the precise definition along these lines is given in the Definition-Proposition \[dfprop=def\_\_Tphip\].

An important observation we make is that although $L^p({\mathcal{L}}G)$ might not contain nonzero bounded operators, its corners in fact lie in the Schatten class ${\mathcal{S}}_p$. More precisely, for $F \subset G$ relatively compact we can (carefully) define the corner $P_F x P_F$ of any element $x \in L^p({\mathcal{L}}G)$, and we prove in Theorem \[thm=contraction\_well\_defined\] that $P_F x P_F \in {\mathcal{S}}_p(L^2F)$ with norm at most $|F|^{1/p}\|x\|_{L^p({\mathcal{L}}G)}$. When $G$ is amenable, one then chooses a Følner net $F_n \subset G$ and consider the maps $x \mapsto |F_n|^{-1/p} P_{F_n} x P_{F_n}$.

In all the definitions and statements we give, $L^p({\mathcal{L}}G)$ will be the Connes-Hilsum space, but the theory of interpolation [@TerpII], [@Izumi] of non-commutative $L^p$-spaces associated to a weight plays an important role in our proofs. This is a bit technical. Therefore, to make the paper readable we collect (and prove when necessary) all the results we need from interpolation in Theorem \[thm=interpolation\_BB\], and in the proofs we use this theorem as an abstract black box.

The structure of the paper is as follows. Section \[Sect-Impossible\], which is independent from the rest of the paper, gives a sufficient (and necessary) condition for a discrete group to be amenable in terms of intertwiners of Schur and Fourier multipliers. Section \[Sect-Preliminaries\] recalls the necessary results and fixes the notation. In Section \[Sect=transference\], we prove that the inequality $\geq$ in Question \[Que-One\] holds for any locally compact group. Section \[Sect=amenable\] gives the affirmative answer to Question \[Que-One\] for amenable groups. The last Section \[Sect=interpolation\] on interpolation states and then proves all the results needed in the first sections on interpolation of non-commutative $L^p$-spaces.

Characterizing amenability by intertwining Schur and Fourier multipliers {#Sect-Impossible}
========================================================================

In this section we prove the following characterization of the amenability of a discrete group. It takes away our hope to give a direct positive answer to Question \[Que-One\] for non-amenable discrete groups.

Recall that the Schur multiplier $M_\phi^p$ and the Fourier multiplier $T_\phi^p$ were defined in the introduction for discrete groups. We define them more generally in Section \[Sect-Preliminaries\].

\[Thm-CounterExample\] Let $\Gamma$ be a discrete group, let ${\mathcal{H}}$ be a Hilbert space and let $p\geq 4$ be an even integer. Let ${\mathcal{U}}$ be a non-trivial ultrafilter on some set. Assume that there exists an isometric embedding $$\label{Eq-IntertwinerCounter}
j_p: L^p({\mathcal{L}}\Gamma) \to \prod_{\mathcal U} {\mathcal{S}}_p(\ell^2 \Gamma \otimes
  {\mathcal{H}})$$ such that for every $\phi :\Gamma \to \C$ of finite support, $j_p \circ T_\phi^p = (\prod_{{\mathcal{U}}} M_\phi^{p} \otimes id) \circ j_p$. Then, $\Gamma$ is amenable.

Conversely, if $\Gamma$ is amenable such an embedding exists for all $1 \leq p \leq \infty$ and we can take ${\mathcal{H}}=\C$.

In fact the proof below shows more generally that the same conclusion holds if there is such an embedding of $L^p({\mathcal{L}}(\Gamma))$ into $\prod_{\mathcal U} L^p( B(\ell^2 \Gamma)   \otimes  
{\mathcal{M}}_\alpha)$ with a net $({\mathcal{M}}_\alpha,\tau_\alpha)$ of semi-finite von Neumann algebras.

To keep the notation simple, we consider the case $p=4$ and write $j$ for $j_4$. The same proof works for every $p \in 2\N$.

Fix $\gamma \in \Gamma$, and let $(\Xi_\alpha)$ be a representative of $j(\lambda(\gamma))$. Express that $j \circ T_\phi^4 = (\prod_{{\mathcal{U}}} M_\phi^{p} \otimes id) \circ j$ for $\phi = \delta_\gamma$ a Dirac mass point. This implies that $j(\lambda(\gamma))=((M_{\delta_\gamma}^4 \otimes id)
(\Xi_\alpha))_{\alpha}$, so that we can assume that $(M_{\delta_\gamma} \otimes id) (\Xi_\alpha) = \Xi_\alpha$ for all $\alpha$. In other words, if we see $\Xi_\alpha$ as an infinite matrix with values in ${\mathcal{S}}_p({\mathcal{H}})$, its $(s,t)$ entry is zero unless $st^{-1}=\gamma$. Equivalently, there is an element $\xi \in
\ell^p(\Gamma;{\mathcal{S}}_p({\mathcal{H}}))\subset {\mathcal{S}}_p(\ell^2 \Gamma \otimes {\mathcal{H}})$ such that $\Xi_\alpha = (\lambda(\gamma) \otimes id) \xi$. Let us denote this $\xi$ by $\xi_\gamma^{(\alpha)}$. Using that $1 =
\|\lambda(\gamma)\|_{L^4({\mathcal{L}}(G))} = \lim_{\alpha, \mathcal U}
\|\xi_\gamma^{(\alpha)}\|_{\ell^4(\Gamma; {\mathcal{S}}_4({\mathcal{H}}))}$, we can as well assume that $\|\xi_\gamma^{(\alpha)}\|_{\ell^4(\Gamma; {\mathcal{S}}_4({\mathcal{H}}))}=1$. Since $\gamma \in\Gamma$ was arbitrary, we can choose such a representative for all $\gamma$.

Let $\eta_\gamma^{(\alpha)}$ be the unit vector in $\ell^4 \Gamma$ defined by $\eta_\gamma^{(\alpha)} =
(\|\xi_\gamma^{(\alpha)}(s)\|_{{\mathcal{S}}_4({\mathcal{H}})})_{s \in \Gamma}$. We claim that, $$\label{eq=invariant}
\forall \gamma,s \in \Gamma, \lim_{\alpha, \mathcal U} \|\lambda_s 
(\eta_\gamma^{(\alpha)})^2 - (\eta_\gamma^{(\alpha)})^2\|_{\ell^2\Gamma} = 0.$$ 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 $$\begin{aligned}
  \|j(x)\|_{\prod_{\mathcal U} {\mathcal{S}}_4(\ell^2 \Gamma \otimes
  {\mathcal{H}})} ^4 & = & \sum_{s,\gamma,\widetilde \gamma} x_\gamma {x_{\gamma s}} {x_{\widetilde \gamma}} x_{\widetilde \gamma s} \lim_{\alpha, {\mathcal{U}}} {\mbox{Re\,}}\: {\mathrm{Tr}}\left(s^{-1}.(\xi_\gamma^{(\alpha)} \xi_{\widetilde \gamma}^{(\alpha)*}) \xi_{\widetilde \gamma s}^{(\alpha)}\xi_{\gamma s}^{(\alpha)*}\right)\\
& \leq & \sum_{s,\gamma,\widetilde \gamma} x_\gamma {x_{\gamma s}} {x_{\widetilde \gamma}} x_{\widetilde \gamma s} 1= \|x\|_{L^4({\mathcal{L}}G)}^4.\end{aligned}$$ Hence if $\gamma,\widetilde \gamma,\gamma s,\widetilde \gamma s$ belong to the support of $x$, $$\lim_{\alpha,{\mathcal{U}}} {\mbox{Re\,}}\: {\mathrm{Tr}}\left(s^{-1}.(\xi_\gamma^{(\alpha)} \xi_{\widetilde \gamma}^{(\alpha)*}) \xi_{\widetilde \gamma s}^{(\alpha)}\xi_{\gamma s}^{(\alpha)*}\right) = 1,$$ so that by the uniform convexity Lemma \[lemma=uniform\_convexity\] below $$\label{eq=limit_terme_a_terme}
\lim_{\alpha, \mathcal U}\|\lambda_s (\eta_{\widetilde \gamma s}^{(\alpha)})^2 - (\eta_{\gamma}^{(\alpha)})^2\|_{\ell^2(\Gamma)} = 0.$$ Since $x$ was arbitrary, holds for all $s,\gamma,\widetilde \gamma \in \Gamma$. This proves , and hence that $\Gamma$ is amenable.

The converse was proved in [@NeuRic].

\[lemma=uniform\_convexity\] Let $\Gamma$ be a discrete group and let ${\mathcal{H}}$ be a Hilbert space. For all $\epsilon >0$, there exists $\delta>0$ such that the following holds: Let $s \in \Gamma$ and $\xi_i \in \ell^4(\Gamma;{\mathcal{S}}_4({\mathcal{H}}))$, $i=1\dots 4$ be unit vectors, and denote by $\eta_i$ the unit vector $(\|\xi_i(t)\|_{{\mathcal{S}}_4({\mathcal{H}})})_{t \in \Gamma}$ in $\ell^4 \Gamma$. If $${\rm Re}\: {\mathrm{Tr}}\left(s^{-1}.(\xi_1 \xi_2) \xi_3 \xi_4\right) \geq 1-\delta,$$ then $\|\lambda_s \eta_3^2 - \eta_1^2\|_{\ell^2\Gamma} \leq \epsilon$.

In the proof we write $o(1)$ for a quantity that goes to $0$ as $\delta$ goes to $0$. By Hölder’s inequality in ${\mathcal{S}}_4({\mathcal{H}})$, $$\begin{aligned}
1-\delta\leq{\mbox{Re\,}}\: {\mathrm{Tr}}\left(s^{-1}.(\xi_1 \xi_2) \xi_3
  \xi_4\right)&\leq& \langle \lambda_{s^{-1}}(\eta_1 \eta_2), \eta_3
  \eta_4\rangle\\ & \leq & \|\eta_1 \eta_2\|_{\ell^2\Gamma} \|\eta_3
  \eta_4\|_{\ell^2\Gamma} \leq 1.\end{aligned}$$ In particular, $\langle \eta_1^2,\eta_2^2\rangle = \|\eta_1 \eta_2\|_{\ell^2\Gamma}^2 \geq
(1-\delta)^2$, which clearly implies that $\|\eta_1^2 - \eta_2^2\|_{\ell^2\Gamma} =
o(1)$. The elementary inequality $\|a-b\|_{\ell^4\Gamma} \leq \sqrt{\|a^2 - b^2\|_{\ell^2\Gamma}}$ valid for all $a,b:\Gamma \to \R_+$ in turn implies $\|\eta_1 -
\eta_2\|_{\ell^4\Gamma} = o(1)$. Similarly, $\|\eta_3 - \eta_4\|_{\ell^4\Gamma}  =o(1)$. By the first line of the series of inequalities above, we get $\langle
\lambda_{s^{-1}} \eta_1^2, \eta_3^2 \rangle =1+o(1)$, which indeed implies $\|\lambda_s \eta_3^2 - \eta_1^2\|_{\ell^2\Gamma} = o(1)$.

Preliminaries {#Sect-Preliminaries}
=============

We collect the necessary results and fix the notation. In particular, we introduce $L^p$-Fourier multipliers for any locally compact group.

General notation
----------------

For a Hilbert space ${\mathcal{H}}$, ${\mathcal{S}}_p({\mathcal{H}})$ denotes the Schatten class, i.e. the non-commutative $L^p$-space associated with $B({\mathcal{H}})$. We use brackets $[\: ]$ to denote the closure of an operator and $\cdot$ for the strong product of (unbounded) operators. Recall that $a \cdot b$, when it exists, is the closure of the operator defined on $\{x \in {\mathrm{dom}}(b), b(x) \in {\mathrm{dom}}(a)\}$ by $x \mapsto a(b(x))$. For $\varphi$ a normal, semi-finite, faithful weight on a von Neumann algebra ${\mathcal{M}}$, we set ${\mathfrak{n}_\varphi}= \{ x \in {\mathcal{M}}\mid \varphi(x^\ast x ) < \infty \}$ and ${\mathfrak{m}_\varphi}= {\mathfrak{n}_\varphi}^\ast {\mathfrak{n}_\varphi}$.

Integral operators
------------------

Let $(X,d x)$ be a measure space. A (bounded) integral operator $A$ on $L^2(X,d x)$ is a bounded operator for which there is a measurable function $(s,t) \mapsto A_{s,t}$ (called the kernel of $A$) on $X \times X$ such that, for all $\xi \in L^2(X,d x)$,

-   $t \mapsto A_{s,t} \xi(t) \in L^1(X,d x)$ for almost every $s \in X$.

-   $A \xi (s) = \int_X A_{s,t} \xi(t)$ for almost every $s \in X$.

By Fubini the kernel is only defined up to an almost everywhere zero function. The set of integral operators is a self-adjoint subspace of $B(L^2(X))$, and the kernel of the adjoint is given by $(A^*)_{s,t} = \overline{A_{t,s}}$. The Schatten ${\mathcal{S}}_2$-class consists of all integral operators with kernel belonging to $L^2(X \times X)$, and for such operator, $$\|A\|_{{\mathcal{S}}_2(L^2X)} = \left(\iint |A_{s,t}|^2 ds dt\right)^{1/2}.$$ More generally if $A,B \in {\mathcal{S}}^2(L^2X)$, $$\label{eq=formula_for_trace} {\mathrm{Tr}}(AB) =\iint A_{s,t} B_{t,s} dt ds.$$

Schur multipliers {#subsection=Schur}
-----------------

Let $(X, \mu)$ be a measure space and let $\psi: X \times X \rightarrow \mathbb{C}$ be an essentially bounded measurable function. We identify ${\mathcal{S}}_2(L^2 X )$ linearly with $L^2(X \times X)$ by the previous paragraph. Then, we obtain a bounded map $$L^2(X \times X) \rightarrow  L^2(X \times X): (a_{x,y})_{x,y \in X} \mapsto (\psi(x,y) a_{x,y})_{x,y \in X},$$ which canonically determines a bounded map on ${\mathcal{S}}_2(L^2 X)$. If it maps ${\mathcal{S}}_2(L^2 X)\cap {\mathcal{S}}_p(L^2 X)$ to ${\mathcal{S}}_p(L^2 X)$ and it extends boundedly to ${\mathcal{S}}_p(L^2X)$, it will be called a (Schur) multiplier of ${\mathcal{S}}_p(L^2X)$.

We will use the following result, which is a minor modification of [@LafSal Theorem 1.19].

\[thm=mult\_de\_symb\_continu\] Let $\mu$ be a Radon measure on a locally compact space $X$, and $\psi:X \times X \to \C$ a continuous function. Let $1 \leq p
  \leq \infty$ and $K>0$. The following are equivalent:

(i) \[asser=norm\_locCompact\] $\psi$ defines a bounded multiplier on ${\mathcal{S}}_p(L^2(X,\mu))$ with norm less than $K$.

(ii) \[asser=norm\_parties\_semifinies\] For every $\sigma$-finite measurable subset $X_0$ in $X$, $\psi$ restricts to a bounded multiplier on ${\mathcal{S}}_p(L^2( X_0, \mu))$ with norm less than $K$.

(iii) \[asser=norm\_parties\_finies\] For any finite subset $F=\{x_1,\dots,x_N\}$ in $X$ belonging to the support of $\mu$, the multiplier $(\psi(x_i,x_j))$ is bounded on ${\mathcal{S}}_p(\ell^2 F)$ with norm less than $K$.

The equivalence of (\[asser=norm\_locCompact\]) and (\[asser=norm\_parties\_finies\]) of Theorem \[thm=mult\_de\_symb\_continu\] was proved in [@LafSal Theorem 1.19] under the additional assumption that $\mu$ is $\sigma$-finite. In the current theorem, this implies the equivalence of (\[asser=norm\_parties\_semifinies\]) and (\[asser=norm\_parties\_finies\]). The implication (\[asser=norm\_locCompact\]) implies (\[asser=norm\_parties\_semifinies\]) is trivial. Assume (\[asser=norm\_parties\_semifinies\]). Let $x \in {\mathcal{S}}_p(L^2(X,\mu))$. The supports of $x$ and $x^\ast$ are $\sigma$-finite projections in $B(L^2X)$. So, let $X_0 \subset X$ be $\sigma$-finite such that the support projections of $x$ and $x^\ast$ project onto spaces contained in $L^2(X_0, \mu)$. Then, the multiplier $\psi$ applied to $x$ is equal to the restriction of $\psi$ to $X_0 \times X_0$ applied to $x$. Hence, (\[asser=norm\_locCompact\]) follows.

Let $G$ be a locally compact group. Let $\phi \in {M_{\mathcal{CB}}A}(G)$ and set $\check{\phi} \in L^\infty(G \times G)$ by $\check{\phi}(s,t) = \phi(s t^{-1})$. Then, $\check{\phi}$ is a Schur multiplier acting on ${\mathcal{S}}_p(L^2G)$, see [@LafSal] for details. We will denote this map by $M_\phi^p$.

The von Neumann algebra of $G$. {#subsect=groupvna}
-------------------------------

Let $G$ be a locally compact group equipped with a left Haar measure, and denote by $\Delta:G\to (0,\infty)$ the modular function. Recall (see [@TakII p. 65]) that $\Delta$ is the group morphism satisfying the following equations for all compactly supported continuous function $f:G \to \C$ : $$\label{eq=modular1} \int f(ts) dt = \Delta(s)^{-1} \int f(t) dt,$$ $$\label{eq=modular2} \int f(t^{-1}) dt = \int f(t) \Delta(t)^{-1} dt.$$

Denote by $\lambda$ and $\rho$ the left and right regular representations of $G$ : for $s \in G$, $\lambda_s$ and $\rho_s$ are unitaries on $L^2(G)$ given by $\lambda_s \xi(t) = \xi(s^{-1}t)$ and $\rho_s \xi(t) = \sqrt{\Delta(s)} \xi(ts)$. The (left) von Neumann algebra of $G$ is defined by ${\mathcal{L}}G = \lambda(G)''$. Its commutant is $\rho(G)''$, the right von Neumann algebra of $G$. If $f \in L^1(G)$, the formulas $\langle \lambda(f)\xi,\eta\rangle = \int f(s) \langle \lambda_s \xi,\eta\rangle ds$ and $\langle \rho(f)\xi,\eta\rangle = \int f(s) \langle \rho_s \xi,\eta\rangle ds$ define operators $\lambda(f) \in {\mathcal{L}}G$ and $\rho(f) \in \rho(G)''$ that are integral operators on $L^2(G)$ with kernel $$\label{eq=kernel_lambdaf} (\lambda(f))_{s,t} = \Delta(t^{-1}) f(st^{-1})\ ,\ (\rho(f))_{s,t} = \sqrt{\Delta(s^{-1}t)} f(s^{-1}t).$$ As a consequence, $\lambda(f)^\ast= \lambda(f^\ast)$ and $\rho(f)^\ast=\rho(f^\ast)$ where $f^\ast \in L^1(G)$ is given by $$\label{eq=def_of_xistar}f^\ast(s) = \overline{f(s^{-1})} \Delta(s^{-1}).$$

When $G$ is discrete, ${\mathcal{L}}G$ is finite and carries a natural trace. More generally when $G$ is unimodular ${\mathcal{L}}G$ carries a natural semifinite trace. In general, although ${\mathcal{L}}G$ might be semifinite, it is not equipped with a natural trace, but rather with its natural weight, given by $\varphi (x^*x) = \|f\|_{L^2G}^2$ if there is $f \in
L^2G$ such that $x \xi = f \ast \xi$ for all $\xi \in L^2G$ and $\varphi(x^*x)=\infty$ otherwise. (it is an easy exercice to check that $\varphi$ is a trace if and only if $G$ is unimodular). It is convenient to work with this weight. A natural weight on the commutant $\rho(G)''$ of ${\mathcal{L}}G$ in $B(L^2 G)$ is given by $\psi(x^*x) = \|f\|_{L^2G}^2$ if $x = \rho(f)$ with $f
\in L^2G$ and $\psi(x^*x)=\infty$ otherwise.

The non-commutative $L^p$-space of ${\mathcal{L}}G$. {#subsect=Connes-Hilsum}
----------------------------------------------------

We recall the Connes-Hilsum construction [@Con], [@Hilsum] of $L^p({\mathcal{L}}G)$ in the particular case when ${\mathcal{L}}G$ is the von Neumann algebra of $G$. The space $D(L^2 G,\psi)$ of $\psi$-bounded elements of $L^2G$ is the set of functions $\xi \in L^2G$ such that $f \mapsto \xi \ast f$ is bounded on $L^2G$. We will also denote this operator by $\lambda(\xi)$. Given a weight $\omega$ on ${\mathcal{L}}G$, the spatial derivative $d\omega/d\psi$ is the unique positive self-adjoint operator such that $\omega(\lambda(\xi) \lambda(\xi)^\ast) = \| (d\omega/d\psi)^{1/2} \xi\|_{L^2G}^2$ for all $\psi$-bounded $\xi$. For example, gives that $d\varphi/d\psi = \Delta$. The spatial derivative of an element $\omega$ of the predual is defined by $d \omega/d\psi = u d |\omega|/d\psi$ where $\omega = u |\omega|$ is the polar decomposition of $\omega$. The set $L^1({\mathcal{L}}G)$ defined as $\{ d\omega/d\psi,\omega \in {\mathcal{L}}G_*\}$ is then a linear space (the sum being the closure of the sum), and we denote $\int (d\omega/d\psi) d\psi = \omega(1)$. For an arbitrary $1 \leq p<\infty$, $L^p({\mathcal{L}}G)$ is defined as the set of closed densily defined operators $T$ with polar decomposition $T=u|T|$ satisfying $u \in {\mathcal{L}}G$, $|T|^p \in L^1({\mathcal{L}}G)$, and one denotes $\|T\|_{L^p({\mathcal{L}}G)} = (\int |T|^p d\psi)^{1/p}$. In general the non-zero elements of $L^p({\mathcal{L}}G)$ are not bounded operators. Hilsum [@Hilsum] proved that the sum of two elements of $L^p({\mathcal{L}}G)$ is densely defined, closable, that its closure belongs to $L^p({\mathcal{L}}G)$, and that for this linear structure if $p\geq 1$, $L^p({\mathcal{L}}G)$ is a Banach space for the norm $\| \cdot \|_{L^p({\mathcal{L}}G)}$. Moreover, for $0\leq p,q,r \leq \infty$ with $1/r=1/p+1/q$ and $a \in L^p({\mathcal{L}}G)$ and $b \in L^q({\mathcal{L}}G)$, $ab$ is closable and its closure (still denoted by $ab$) belongs to $L^r({\mathcal{L}}G)$ and this product is associative. Lastly, if $r=1$, $\int a b d\psi=\int b a d\psi$, and if $p\neq \infty$ the pairing $\langle a,b\rangle=\int a b d\psi$ realizes $L^q({\mathcal{L}}G)$ isometrically as the dual of $L^p({\mathcal{L}}G)$.

Apart from what we just recalled, we will use some facts from [@Hilsum Proposition 11] that we collect in the following proposition.

\[prop=reminders\_hilsum\] We have the following properties.

1.  We have $u D(L^2G,\psi)\subset D(L^2G,\psi)$ for every $u \in {\mathcal{L}}G$.

2.  If $x \in L^p({\mathcal{L}}G)$ with $2 \leq p \leq \infty$, then $D(L^2G,\psi) \subset {\mathrm{dom}}(x)$.

3.  Let $1 < p \leq \infty$. For any $\xi,\eta \in D(L^2G,\psi)$ there is a bounded linear map $\omega_{\xi,\eta}^p: L^p({\mathcal{L}}G) \to \C$ satisfying $\omega_{\xi,\eta}^p(x) = \langle |x|^{1/2} \xi,|x|^{1/2} u^* \eta \rangle$ if $x= u |x|$ is the polar decomposition of $x \in L^p({\mathcal{L}}G)$.

As a direct consequence, using the inclusion $L^2 F   \subset D(L^2G,\psi)$ and the closed graph theorem we get the following.

\[prop=def\_of\_PFxPF\] Let $F \subset G$ be a relatively compact Borel subset with positive measure, and $P_F:L^2G  \to L^2 F$ the orthogonal projection.

1.  If $p \geq 2$, $L^2F \subset {\mathrm{dom}}(x)$ for every $x \in L^p({\mathcal{L}}G)$, and $x \in L^p({\mathcal{L}}G) \to x P_F \in B(L^2 G)$ is a linear bounded map.

2.  If $1 \leq p \leq \infty$ there is a bounded linear map $L^p({\mathcal{L}}G) \to B(L^2 F)$, that maps an element $x =u |x| \in L^p({\mathcal{L}}G)$ to $ (|x|^{1/2}u^*P_F)^* |x|^{1/2} P_F$. We abusively denote this map by $x \mapsto P_F x P_F$.

Completely bounded maps between von Neumann algebras
----------------------------------------------------

Let ${\mathcal{M}}\subset B({\mathcal{H}})$ a von Neumann algebra, $\psi$ a weight on ${\mathcal{M}}'$ and $L^p({\mathcal{M}},\psi)$ the spatial non-commutative $L^p$-space. The commutant of $M_n \otimes {\mathcal{M}}$ in $M_n \otimes B({\mathcal{H}})$ is $1 \otimes {\mathcal{M}}'$, and is therefore equipped with the weight $1 \otimes \psi$. This naturally identifies ${\mathcal{S}}_p^n \otimes L^p({\mathcal{M}},\psi)$ with $L^p(M_n \otimes {\mathcal{M}}, 1 \otimes \psi)$. We use this identification to define a Banach space structure on ${\mathcal{S}}_p^n \otimes L^p({\mathcal{M}},\psi)$.

If ${\mathcal{M}}\subset B({\mathcal{H}}),{\mathcal{N}}\subset B({\mathcal{H}}')$ are von Neumann algebras and $\psi,\psi'$ are weights on ${\mathcal{M}}'$, ${\mathcal{N}}'$, a bounded map $u:L^p({\mathcal{M}},\psi) \to L^p({\mathcal{N}},\psi')$ between the non-commutative $L^p$-spaces (Connes-Hilsum construction) is called completely bounded if $\|u\|_{cb}:=\sup_n \|u_n\|<\infty$. Here $u_n = id \otimes u : {\mathcal{S}}_p^n \otimes L^p({\mathcal{M}},\psi)\to {\mathcal{S}}_p^n \otimes L^p({\mathcal{N}},\psi')$. When ${\mathcal{M}}$ and ${\mathcal{N}}$ are semifinte, one can check that this definition agrees with the natural operator space structure on non-commutative $L^p$-spaces given in [@MR1648908].

Definition of $L^p$ Fourier multipliers {#subsection=Fourier}
---------------------------------------

Recall that the Fourier algebra $A(G) = \left\{ \varphi\colon s\mapsto \langle \lambda_s \xi,\eta\rangle, \xi,\eta \in L^2 G\right\}$ coincides with the predual of ${\mathcal{L}}G$ through the pairing $\langle \varphi,\lambda(f) \rangle = \int \varphi(s) f(s) ds$ for every $f \in L^1(G)$. We record here the explicit isomorphism between the Fourier algebra $A(G)$ and $L^1({\mathcal{L}}G)$.

\[lemma=isom\_AG\_L1\] An element $\varphi \in A(G)$ corresponds (isometrically) to $x_\varphi \in L^1({\mathcal{L}}G)$ that satisfies $P_F x_\varphi P_F = (\varphi(ts^{-1}))_{s,t \in F}$.

Start by considering 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 $\phi \in {M_{\mathcal{CB}}A}(G)$. 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 satisfies $P_F T_\phi^p(x) P_F = M_\phi (P_F x P_F)$ by and again.

The fact that $T_\phi^p$ is completely bounded follows from the argument in [@CanHaa Theorem 1.6] (see the proof of Theorem \[thm=embedding\_in\_ultraproduct\] for details): apply the preceding to the function $\phi(s,k) = \phi(s)$ on $G \times SU(2)$ and use that ${\mathcal{L}}SU(2) \simeq \oplus_{n \geq 1} M_n$.

Transference {#Sect=transference}
============

This section is devoted to the equality $\geq$ of Question \[Que-One\]. The proof relies on transference techniques [@NeuRic Lemma 2.4], which we adapt to non-discrete groups.

\[lem=recover\_phi\_from\_Iphi\] Let $G$ be a locally compact group and let $1 < p < \infty$ be such that the conjugate exponent $q$ satisfies $p/q \in \Q$. Then there is a net $x_\alpha \in L^p({\mathcal{L}}G)$ and $y_\alpha \in L^q({\mathcal{L}}G)$ such that $\|x_\alpha\|_p = \|y_\alpha\|_{q} = 1$ and for all $\phi \in {M_{\mathcal{CB}}A}(G)$, $$\lim_\alpha \langle T_\phi^p x_\alpha,y_\alpha^*\rangle = \phi(e).$$

The proof relies on Theorem \[thm=interpolation\_BB\]. Since $p/q \in \Q$ there exists $r<\infty$ and $k,k' \in \N$ such that $\frac 1 p = \frac k r$ and $\frac 1{q} = \frac{k'}{r}$.

Take a net of $\R^+$-valued fuctions $h_\alpha\in A_c(G)$ such that the support $S_\alpha$ of $h_\alpha$ converge to $\{e\}$. This means that every neighbourhood of $e$ in $G$ 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 $1 \leq p \leq \infty$. 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 $$A_\alpha = (a_{i,j} \lambda(s_i) x_\alpha \lambda(s_j^{-1}))_{i,j \leq k}, B_\alpha = (b_{i,j} \lambda(s_i) y_\alpha \lambda(s_j^{-1}))_{i,j \leq k}.$$ $A_\alpha$ is obtained from $a \otimes x_\alpha$ by conjugating by the unitary $\sum_i e_{i,i} \otimes \lambda(s_i)$, so that $\|A_\alpha\|_{p} = \|a \otimes x_\alpha\|_p = \|a\|_{{\mathcal{S}}_p^k}$. Similarly $\|B_\alpha\|_q = \|b\|_{{\mathcal{S}}_q^k}$. Notice now that $$\lambda(s_i^{-1}) T_\phi^p(\lambda(s_i) x_\alpha \lambda(s_j)^{-1}) \lambda(s_j) = T_{\phi_{i,j}}^p(x_\alpha)$$ where $\phi_{i,j}(s) = \phi(s_i s s_j^{-1})$. In particular $\phi_{i,j} \in {M_{\mathcal{CB}}A}(G)$ and $\phi_{i,j}(e) = \phi(s_i s_j^{-1})$. Lemma \[lem=recover\_phi\_from\_Iphi\] gives $$\begin{aligned}
 \sum_{i,j} \phi(s_i s_j^{-1}) a_{i,j}  b_{j,i} & = & \lim_\alpha \sum_{i,j} a_{i,j}  b_{j,i} \langle T_{\phi_{i,j}}^p(x_\alpha),y_\alpha \rangle \\ & = & \lim_\alpha 
\langle (id \otimes T_\phi^p)(A_\alpha),B_\alpha\rangle.\end{aligned}$$ The claim follows from the inequality $$\begin{aligned}
|\langle (id \otimes T_\phi^p)(A_\alpha),B_\alpha\rangle| &\leq& \Vert T_{\phi}^p \Vert_{{\mathcal{CB}}(L^p({\mathcal{L}}G))} \|A_\alpha\|_{p} \|B_\alpha\|_{q} \\&=& \Vert T_{\phi}^p \Vert_{{\mathcal{CB}}(L^p({\mathcal{L}}G))} \|a\|_{{\mathcal{S}}_p^k} \|b\|_{{\mathcal{S}}_q^k},\end{aligned}$$ valid for every $n$.

Amenable groups {#Sect=amenable}
===============

We prove that the answer to Question \[Que-One\] is affirmative for amenable locally compact groups. This generalizes beyond (amenable) discrete groups the result by Neuwirth and Ricard [@NeuRic].

\[thm=contraction\_well\_defined\] Let $F \subset G$ be a relatively compact Borel subset with positive measure, and $P_F:L_2(G) \to L_2(F)$ the orthogonal projection. Let $1 \leq p \leq \infty$. For any $x \in L^p({\mathcal{L}}G)$, $P_F x P_F \in {\mathcal{S}}_p(L_2 G)$ and $\| P_F x P_F\|_{{\mathcal{S}}_p(L^2F)} \leq |F|^{1/p} \|x\|_{L^p({\mathcal{L}}G)}$.

When $p=\infty$ this is obvious.

Assume $p=1$, and take $x \in L_1({\mathcal{L}}G)$. Let $\phi \in A(G)$ corresponding to $x$, and write $\phi(s) =\langle \lambda(s) \xi,\eta\rangle$ for $\xi,\eta \in L^2(G)$ with $\|\xi\| \|\eta\| = \|x\|_{L^1({\mathcal{L}}G)}$. By Lemma \[lemma=isom\_AG\_L1\], $P_F x P_F = ( \langle \lambda(s^{-1})\xi,\lambda(t^{-1}) \eta\rangle)_{s,t \in F} = A B^*$ where $A = (\xi(s t))_{s \in F,t\in G}$ and $B = (\eta(st))_{s \in F, t \in G}$. But $\|A\|_{{\mathcal{S}}_2(L^2G)}^2 = \iint |\xi(st)|^2 1_{s \in F} dt ds = |F| \|\xi\|^2$, and similarly $\|B\|_{{\mathcal{S}}_2(L^2G)} = |F|\|\eta\|^2$, so that $\| P_F x P_F\|_{{\mathcal{S}}_1(L^2G)} \leq \|A\|_{{\mathcal{S}}_2(L^2G)} \|B\|_{{\mathcal{S}}_2(L^2G)} =|F| \|x\|_{L^1({\mathcal{L}}G)}$. This proves the case $p=1$.

The general case follows by the interpolation Theorem \[thm=interpolation\_BB\]: the cases $p=1$ and $p=\infty$ above say that $u_F$ maps $A_0 \to A_{0,F}$ with norm $1$ and $A_1 \to A_{1,F}$ with norm $|F|$. Hence it maps $A_{1/p} \to A_{1/p,F}$ with norm less than $|F|^{1/p}$. By this is exactly saying that $P_F x P_F \in {\mathcal{S}}_p(L^2F)$ and $\|P_F x P_F \|_{{\mathcal{S}}_p(L^2F)} \leq |F|^{1/p} \|x\|_{L_p({\mathcal{L}}G)}$.

\[thm=embedding\_in\_ultraproduct\] Let $G$ be an amenable locally compact group and $1< p <\infty$. Then there is an ultrafilter ${\mathcal{U}}$ on some set and a completely isometric embedding $i:L^p({\mathcal{L}}G) \to \prod_{{\mathcal{U}}} {\mathcal{S}}_p(L^2G)$ that intertwines Fourier and Schur multipliers.

By amenability there exists a Følner net $(F_\alpha)_\alpha$ : $F_\alpha$ are compact subsets with positive measure such that $\lim_\alpha |F_\alpha \cap g F_\alpha|/|F_\alpha| =1$ for all $g \in G$. Choose an ultrafilter ${\mathcal{U}}$ refining the net $\alpha$, denote by $P_\alpha$ the orthogonal projection on $L^2(F_\alpha)$ and consider the map $i_p:L^p({\mathcal{L}}G) \to \prod_{{\mathcal{U}}} {\mathcal{S}}_p(L^2 F_\alpha)$ defined by $i_p(x) = (P_\alpha x P_\alpha/|F_\alpha|^{1/p})_\alpha$. By Theorem \[thm=contraction\_well\_defined\], $i_p$ is a well defined contraction, and it intertwines Fourier and Schur multipliers by definition of Fourier multipliers. We use a duality argument to prove that it is isometric. To do this take $q$ the conjugate exponent of $p$ and consider $x \in L^p({\mathcal{L}}G)$ and $y \in L^q({\mathcal{L}}G)$. We claim that $$\label{eq=ipiq} \langle i_p(x),i_q(y)^*\rangle = \langle x,y^*\rangle.$$ where on the left-side, the pairing is given by $\langle (x_\alpha)_\alpha,(y_\alpha) \rangle =\lim_{{\mathcal{U}}} {\mathrm{Tr}}(x_\alpha y_\alpha)$ and on the right-side the pairing is given by $\langle x,y\rangle = \int xy d\psi$. Since $L^q$ is isometrically the dual of $L^p$ for this pairing, this clearly implies that $i_p$ (and $i_q$) are isometries.

Let us first prove in the particular case that $p=\infty,q=1$, $x = \lambda(f)$ with $f \in L^1(G)$ and $y^*$ corresponds to $\varphi \in  A(G)$. We compute $\langle  i_\infty(x), i_1(y)^* \rangle$, using Lemma \[lemma=isom\_AG\_L1\], and then , $$\begin{split}
&  \langle  i_\infty(x), i_1(y)^* \rangle\\
 = &
\lim_{\alpha,{\mathcal{U}}} \frac{1}{\vert F_\alpha \vert} {\mathrm{Tr}}\left( P_\alpha \lambda(f) P_\alpha y^* P_\alpha \right)\\
= &  \lim_{\alpha,{\mathcal{U}}} \frac{1}{\vert F_\alpha \vert} {\mathrm{Tr}}\left( (f(st^{-1}) \Delta(t^{-1})_{s,t \in F_\alpha} (\varphi(ts^{-1}))_{s,t \in F_\alpha} \right)\\
= &  \lim_{\alpha,{\mathcal{U}}} \frac{1}{\vert F_\alpha \vert} \int_{F_\alpha} \int_{F_\alpha} \Delta(t^{-1}) f(st^{-1}) \varphi(st^{-1}) ds dt \\
 = & \lim_{\alpha,{\mathcal{U}}} \frac{1}{\vert F_\alpha \vert} \int_{G} f(u) \varphi(u)(\int_{F_\alpha} \chi_{F_\alpha}(u^{-1}s) ds) du \\
= & \lim_{\alpha,{\mathcal{U}}}\int_{G} f(u) \varphi(u) \frac{\vert F_\alpha \cap uF_\alpha\vert}{\vert F_\alpha \vert} du \\
= & \int_G f(u) \varphi(u) du =\langle x,y^*\rangle.
\end{split}$$ On the fifth line we made the change of variable $u=st^{-1}$, and the last line is justified by the dominated convergence theorem.

We now use the interpolation Theorem \[thm=interpolation\_BB\] and its notation. We claim that $\langle i_p(j_p(a)) ,i_q(j_q(b))^*\rangle = \langle j_p(a),j_q(b)^*\rangle$ holds for all $p,q$ with $1/p+1/q=1$ and all $a,b \in A_0 \cap A_1$. This will conclude the proof of since by general interpolation theory, $j_p(A_0 \cap A_1)$ is dense in $L^p({\mathcal{L}}G)$ for every $1<p<\infty$. By the interpolation equation and in Theorem \[thm=interpolation\_BB\] neither $\langle i_p(j_p(a)) ,i_q(j_q(b))^*\rangle$ nor $\langle j_p(a),j_q(b)^*\rangle$ depend on $p$, and we know (case $p=\infty$) that they are equal when $a \in E$. These quantities are therefore also (case $p=1$) equal when $b \in E$, and by the norm density of $j_1(E)$ in $L^1({\mathcal{L}}G)$, they are equal for every $b$. This proves .

To prove that $i_p$ is completely isometric, the same proof can be applied. Otherwise we can use a classical argument [@CanHaa Theorem 1.6] and consider $K=SU(2)$. This is very convenient because as is well-known ${\mathcal{L}}K = \oplus_{n \geq 1} M_n$, but all we need is that $K$ is a compact group with irreducible unitary representations of arbitrarily large dimension. Remark that $G \times K$ is amenable and that $F_\alpha \times K$ is a Følner net. By what we just proved, $i_p \otimes id: L^p( {\mathcal{L}}(G \times K)) \to \prod_{{\mathcal{U}}} {\mathcal{S}}_p(L^2(G \times K))$ is isometric. But since ${\mathcal{L}}(G \times K) = {\mathcal{L}}G   \otimes {\mathcal{L}}K \simeq \oplus_n M_n({\mathcal{L}}G)$, this means that $$id \otimes i_p: \ell^p\{ {\mathcal{S}}_p^n \otimes L^p({\mathcal{L}}G), n \geq 1\} \to \prod_{{\mathcal{U}}} \ell^p\{ {\mathcal{S}}_p^n \otimes {\mathcal{S}}_p(L^2 G), n\geq 1\}$$ is isometric, which implies that $i_p$ was completely isometric.

As a straightforward consequence.

\[Cor-MainEquality\] Let $G$ be an amenable group and let $1 \leq p \leq \infty$. Let $\phi \in {M_{\mathcal{CB}}A}(G)$. Then, $$\Vert T_\phi^p \Vert_{B(L^p({\mathcal{L}}G))} \leq \Vert M_\phi^p \Vert_{B({\mathcal{S}}_p(L^2 G))},$$ $$\Vert T_\phi^p \Vert_{{\mathcal{CB}}(L^p({\mathcal{L}}G))} \leq \Vert M_\phi^p \Vert_{{\mathcal{CB}}({\mathcal{S}}_p(L^2 G))}.$$

Interpolation {#Sect=interpolation}
=============

In this section, we collect the necessary results from [@TerpII] and [@Hilsum] in order to interpret non-commutative $L^p$-spaces as interpolation spaces. This is done in Theorem \[thm=interpolation\_BB\]. Using this theorem, we are able to prove our results in a self-contained way. The theorem is technical in nature and its conceptual consequences are stated in the earlier sections.

Let ${\mathcal{M}}$ be a von Neumann algebra, and consider $L_p({\mathcal{M}})$ ($1 \leq p \leq \infty$) the associated non-commutative $L_p$-spaces together with a bilinear duality bracket $\langle \cdot,\cdot\rangle_{L_p,L_q}$ for $1/p+1/q=1$. For simplicity we will only use this notion in two particular explicit cases: the first is when ${\mathcal{M}}= B({\mathcal{H}})$ for a Hilbert space ${\mathcal{H}}$, and in this case $L_p({\mathcal{M}})$ is ${\mathcal{S}}_p({\mathcal{H}})$ and the duality $\langle A,B\rangle_{L^p, L^q} = {\mathrm{Tr}}(AB)$. The second case is when ${\mathcal{M}}'$ is equipped with a normal faithful weight $\psi$, $L^p({\mathcal{M}})$ is the associated Connes-Hilsum space, and the duality is $\langle a,b\rangle = \int ab d\psi = \int ba d\psi$. The special case ${\mathcal{M}}= {\mathcal{L}}G$ is recalled in Section \[subsect=Connes-Hilsum\]. The special case ${\mathcal{M}}= B({\mathcal{H}})$ and $\psi$ is the canonical trace on $B({\mathcal{H}})'\simeq \C$ will then yield $L^p({\mathcal{M}}) = {\mathcal{S}}_p({\mathcal{H}})$.

Let ${\mathcal{M}}$, $L^p({\mathcal{M}})$ be as above. An *interpolation scale for $\{(L_p({\mathcal{M}}), 1\leq p \leq \infty\}$* is a compatible pair $\overline A= (A_0,A_1)$ and a family of isometric isomorphisms $j_p:A_{1/p} \to L_p({\mathcal{M}})$ such that for every $a ,b \in A_0 \cap A_1$ and every $1 \leq p,q \leq \infty$ satisfying $1/p+1/q=1$, $$\label{eq=duality_compatibility} \langle j_p(a),j_q(b)^*\rangle_{L_p,L_q} = \langle j_1(a),j_\infty(b)^*\rangle_{L_1,L_\infty}.$$

In this definition $A_\theta \subset A_0 + A_1$ stands for the complex interpolation space between $A_0$ and $A_1$ with parameter $\theta$. We write $\Sigma(\overline{A}) =  A_0 + A_1$.

\[thm=interpolation\_BB\] Let $F \subset G$ be a relatively compact Borel subset. There exist interpolation scales $\{\overline A= (A_0,A_1), j_p:A_{1/p} \to L_p({\mathcal{L}}G)\}$, and $\{\overline A_F= (A_{0,F},A_{1,F}), j_{p,F} :A_{1/p,F} \to {\mathcal{S}}_p(L^2(F))\}$ with $A_{1,F} \subset A_{0,F}$ and a bounded map $u_F: \Sigma(\overline A) \to \Sigma(\overline A_F) = A_{0,F}$ such that

(i) \[item=jpF\]for all $1 \leq p \leq \infty$, $j_{p,F}$ extends to a continuous isomorphism $j_{p,F}:A_{0,F} \to B(L^2 F)$, and $j_{p,F} \circ j_{q,F}^{-1}$ commutes with Schur multipliers for every $p,q$.

(ii) \[item=diagram\]The following diagram commutes. $$\label{Diag-BlackBox}
     \xymatrix{
    L_p({\mathcal{L}}G) \ar[d]^{x \mapsto P_F x P_F} & A_{1/p} \ar[l]^{j_p} \ar[d]^{u_F}\\
    B(L^2 F)  & A_{0,F}\ar[l]^{j_{p,F}} }$$

(iii) \[item=intersection\_big\] The space $E = \textrm{span}\{ j_\infty^{-1}(\lambda(f)^\ast \lambda(g)), f,g \in C_c(G)\}$ is contained in $A_0 \cap A_1$. Furthermore, $\{j_1(a),a\in E\}$ is norm dense in $L^1({\mathcal{L}}G)$.

(iv) \[item=semi\_injective\] Let $a \in \Sigma(\overline{A})$. If $u_F(a) = 0$ for every $F \subseteq G$ relatively compact, then $a = 0$.

(v) \[item=intersection\_big2\] For every $a = j_\infty^{-1}(\lambda(f_a)),b =j_\infty^{-1}(\lambda(f_b))\in E$, every $1\leq p,q,r \leq \infty$ such that $1/p+1/q=1/r$, there exist $c=j_\infty^{-1}(\lambda(f_c)),d = j_\infty^{-1}(\lambda(f_d))\in E$ such that $j_p(a)^* = j_p(c)$ and $j_p(a) j_q(b) = j_r(d)$. Moreover, $support(f_c) = support(f_a)^{-1}$, $support(f_d) \subset support(f_a) support(f_b)$, and if $f_a$ and $f_b$ are non-negative so are $f_c$ and $f_d$. If $f_a$ is non-negative, $j_1(a) \in L^1({\mathcal{L}}G)$ corresponds to a non-negative (i.e. point-wise non-negative) element in $A(G)$.

The proof will use Terp’s construction [@TerpII] of non-commutative $L^p$-spaces as interpolation spaces, that we now recall.

The construction starts from a triple $({\mathcal{M}}, \varphi, \psi)$. Here, ${\mathcal{M}}$ is a von Neumann algebra. $\varphi$ is a normal, semi-finite, faithful weight on ${\mathcal{M}}$ and $\psi$ is a normal, semi-finite, faithful weight on the commutant ${\mathcal{M}}'$. Associated with $\varphi$, we have a GNS-triple $({\mathcal{H}}, \pi, \Lambda)$, a modular conjugation $J$, modular operator $\Delta$ and modular automorphism group $\sigma$ (see [@TakII p 92]). We identify ${\mathcal{M}}$ with $\pi({\mathcal{M}})$ and omit $\pi$ in the notation.

We turn $({\mathcal{M}}, {\mathcal{M}}_\ast)$ into a compatible couple of Banach spaces as follows. Let ${\mathcal{M}}\cap {\mathcal{M}}_\ast$ be the set of $x \in {\mathcal{M}}$, for which there exists a $\varphi_x \in {\mathcal{M}}_\ast$ such that $$\forall y,z \in {\mathfrak{n}_\varphi}: \langle \varphi_x, z^\ast y \rangle = \langle J x^\ast J \Lambda(y), \Lambda(z) \rangle.$$ For $x \in {\mathcal{M}}\cap {\mathcal{M}}_\ast$, we set the norm $\Vert x \Vert_{{\mathcal{M}}\cap {\mathcal{M}}_\ast} = \max \{ \Vert x \Vert_{{\mathcal{M}}}, \Vert \varphi_x \Vert_{{\mathcal{M}}_\ast} \}$, in which it becomes a Banach space. Naturally, we find two embeddings $i_\infty: {\mathcal{M}}\cap {\mathcal{M}}_\ast \rightarrow {\mathcal{M}}: x \mapsto x$ and $i_1: {\mathcal{M}}\cap {\mathcal{M}}_\ast: x \mapsto \varphi_x$. Dualizing the embeddings (and restricting to ${\mathcal{M}}_\ast$) we obtain a commutative diagram: $$\xymatrix{
 & {\mathcal{M}}_\ast\ar@{^{(}->}[dr]^{i_\infty^{\ast}}  & \\
   {\mathcal{M}}\cap {\mathcal{M}}_\ast  \ar@{^{(}->}[ur]^{i_1}\ar@{^{(}->}[dr]_{i_\infty}  & &( {\mathcal{M}}\cap {\mathcal{M}}_\ast )^\ast, \\
& {\mathcal{M}}\ar@{^{(}->}[ur]^{i_1^\ast} &}    \label{EqnLpIzu}$$ which turns $({\mathcal{M}}, {\mathcal{M}}_\ast)$ into a compatible couple of Banach spaces. Moreover, $i_\infty^* i_1({\mathcal{M}}\cap {\mathcal{M}}_\ast)$ is the intersection of $i_\infty^\ast({\mathcal{M}}_\ast)$ and $i_1^\ast({\mathcal{M}})$, so that our notation is justified. We have ${\mathfrak{m}_\varphi}\subseteq {\mathcal{M}}\cap {\mathcal{M}}_\ast$.

We write $L^p({\mathcal{M}})$ for the Connes-Hilsum $L^p$-space [@Hilsum]. We let $d = d\varphi/d\psi$ be the spatial derivative. For any $x \in {\mathfrak{n}_\varphi}$, we have $$[xd^{\frac{1}{2p}}] \in L^{2p}({\mathcal{M}}) \textrm{ and } d^{\frac{1}{2p}} x^* \in L^{2p}({\mathcal{M}})$$ c.f. [@TerpII Theorem 26]. The main result of [@TerpII] states that for $1 \leq p \leq \infty$, there exists a unique isometry $j_p: ({\mathcal{M}}, {\mathcal{M}}_\ast)_{1/p} \rightarrow L^p({\mathcal{M}})$ such that for $x,y \in {\mathfrak{n}_\varphi}$, we have $$\label{Eqn-LpIso}
j_p(x^\ast y ) =  d^{\frac{1}{2p}} x^\ast  \cdot [y d^{\frac{1}{2p}}] \in L^{2p}({\mathcal{M}}) \cdot L^{2p}({\mathcal{M}}) \subseteq L^p({\mathcal{M}}).$$ Moreover, for $1 \leq p < \infty$, $j_p$ is an isomorphism ($j_\infty$ is just the identity map from ${\mathcal{M}}$ to ${\mathcal{M}}$). The following proposition is exactly [@TerpII Eqn. (56)].

\[Prop-InterpolationScale\] Let $({\mathcal{M}}, \varphi, \psi)$ be a triple as above. $\{  ({\mathcal{M}}, {\mathcal{M}}_\ast), j_p: ({\mathcal{M}}, {\mathcal{M}}_\ast)_{1/p} \rightarrow L^p({\mathcal{M}}) \}$ is an interpolation scale.

In [@Kosaki; @Izumi] it is shown that depending on a [*complex interpolation parameter*]{} $z \in \mathbb{C}$, the triple $({\mathcal{M}}, \varphi, \psi)$ gives a compatible couple $({\mathcal{M}}, {\mathcal{M}}_\ast)$ by taking different embeddings of (subsets of) ${\mathcal{M}}$ into ${\mathcal{M}}_\ast$. These all give rise to an interpolation scale, the special case treated here being $z=0$. Since the Schur and Fourier multipliers in this paper commute with $\sigma$, there is in fact no preference for the choice of the interpolation parameter.

In the next proof we use the following standard fact [@TakII Chapter $\textrm{VII}.3$] on the Plancherel weight $\varphi$ on ${\mathcal{L}}G$. For $\xi \in L^1 G \cap L^2 G$ we have $\lambda(\xi) \in {\mathfrak{n}_\varphi}$ and the mapping $\lambda(L^1 G \cap L^2 G) \rightarrow L^2 G: \lambda(\xi) \mapsto \xi$ is a $\sigma$-weak/norm core for the GNS-map of $\varphi$.

Consider the triple $({\mathcal{L}}G, \varphi, \psi)$, see Sections \[subsect=groupvna\] and \[subsect=Connes-Hilsum\] for notation. We will describe precisely the GNS-triple and and modular conjugation $J$ in the proof of (\[item=semi\_injective\]). Terp’s construction gives a compatible couple $\overline{A} = (A_0,A_1)$ and an interpolation scale $\{ \overline{A}, j_p: A_{1/p} \rightarrow L^p({\mathcal{L}}G)\}$ satisfying .

Consider also $(B(L^2F), {\mathrm{Tr}}_{\Delta}, {\mathrm{Tr}}')$. Here, $${\mathrm{Tr}}_{\Delta}(x) = {\mathrm{Tr}}(\Delta_F^{\frac{1}{2}}  x \Delta_F^{\frac{1}{2}} ), \qquad x \in B(L^2 F)^+,$$ where $\Delta_F$ is the bounded operator given by $\Delta$ restricted to $L^2 F$. Furthermore, ${\mathrm{Tr}}'$ is unique state on $B(L^2F)' = \C$. In this case, $L^p(B(L^2F)) = {\mathcal{S}}_p(L^2 F)$ and we have spatial derivative $d {\mathrm{Tr}}_\Delta / d{\mathrm{Tr}}' = \Delta_F$. We [*define*]{} the map, $$j_{p,F}: B(L^2 F) \rightarrow B(L^2 F): x \mapsto \Delta_F^{\frac{1}{2p}}  x \Delta_F^{\frac{1}{2p}} .$$ Then, the restriction of $j_{p,F}$ to $A_{1/p,F}$ is an isometric isomorphism onto ${\mathcal{S}}_p( L^2 F )$ (by Terp’s construction). In particular, we find an interpolation scale $\{\overline A_F= (A_{0,F},A_{1,F}), j_{p,F} :A_{1/p,F} \to {\mathcal{S}}_p(L^2(F))\}$. The map $j_{p,F} \circ j_{q,F}^{-1}$, being a Schur multiplier, commutes with Schur multipliers. In all, we conclude (\[item=jpF\]).

(\[item=diagram\]) Take $x,y \in {\mathfrak{n}_\varphi}$. Then, for all $1 \leq p \leq \infty$ we have from $$\label{eq=uF_compatible}
j_{p,F}^{-1} (P_F j_p(j_\infty^{-1}(x^\ast y)) P_F) =  
\Delta_F^{-\frac{1}{2p}}  (P_F \Delta^{\frac{1}{2p}} x^\ast \cdot [ y \Delta_F^{\frac{1}{2p}}] P_F) \Delta_F^{-\frac{1}{2p}}   = P_F x^\ast y P_F.$$ In the case $p=1$, [@TerpII Theorem 8] implies that the equation $j_{1,F}^{-1} (P_F j_1(z) P_F) = P_F j_\infty(z) P_F$ holds for all $z \in A_0 \cap A_1$. We can therefore define a bounded map $u_F: A_0 + A_1 \to A_{0,F} = B(L^2 F)$ by $u_F(x_0+x_1) = P_F j_\infty(x_0) P_F + j_{1,F}^{-1} (P_F j_1(x_1) P_F)$ if $x_0\ \in A_0, x_1 \in A_1$. By the definition of $u_F$, then commutes for $p=1$ or $p=\infty$.

For $1< p < \infty$, proves that $j_{p,F}^{-1} (P_F j_p(z) P_F) = u_F(z)$ for all $z \in j_\infty^{-1}({\mathfrak{m}_\varphi})$. Recall that $j_p(j_\infty^{-1}({\mathfrak{m}_\varphi}))$ is dense in $L^p({\mathcal{L}}G)$ (by [@TerpII Theorem 26], Eqn. (\[Eqn-LpIso\]) and Hölder’s inequality). This concludes (\[item=diagram\]).

(\[item=intersection\_big\]) For $\xi \in L^1G \cap L^2G$ we have $\lambda(\xi) \in {\mathfrak{n}_\varphi}$. Hence $\lambda(\xi)^\ast \lambda(\eta) \in {\mathfrak{m}_\varphi}\subseteq A_0 \cap A_1$ for $\xi,\eta \in C_c(G)$. Let $x \in {\mathcal{L}}G$ be such that $\langle j_1 \circ j_\infty^{-1}(\lambda(\xi) \lambda(\eta)), x \rangle = 0$ for all $\xi, \eta \in C_c(G)$. Then, using the notation and [@TerpII Eqn. (38)] in the fourth equality, $$\begin{split}
0 =& \langle j_1 \circ j_\infty^{-1}(\lambda(\xi) \lambda(\eta)), x \rangle 
=  \langle \Delta^{\frac{1}{2}} \lambda(\xi) \cdot [\lambda(\eta) \Delta^{\frac{1}{2}} ], x \rangle \\
= & \int [\lambda(\xi \Delta^{\frac{1}{2}} )\Delta^{\frac{1}{2}}]  \cdot \Delta^{\frac{1}{2}} \lambda(\eta \Delta^{-\frac{1}{2}}) \cdot x d\psi 
=  \varphi(\lambda(\eta \Delta^{-\frac{1}{2}})x\lambda(\xi \Delta^{\frac{1}{2}})) \\
  = & \langle x \Delta^{\frac{1}{2}} \xi ,   \Delta^{\frac{1}{2}}  \eta^\ast \rangle 
\end{split}$$ This implies that $x = 0$, since $C_c(G)$ is dense in $L^2G$ . Hence $j_1(E)$ is dense in $L^1({\mathcal{L}}G)$.

(\[item=semi\_injective\]) Let $a =a_0+a_1 \in A_0 + A_1$ be such that $u_F(a) = 0$ for all $F \subseteq G$ relatively compact. Write $j_\infty(a_0)=x \in {\mathcal{L}}G$ and $j_1(a_1)=\omega_\phi \in {\mathcal{L}}G_\ast$ corresponding to $\phi \in A(G)$ and $x_\phi \in L^1({\mathcal{L}}G)$. Firstly, let $y = \lambda (\xi), z = \lambda(\eta)$, with $\xi, \eta \in C_c(G)$. We will use the notation . We have $\Lambda(\lambda(\xi)) = \xi \in L^2 G$ and $J \Lambda(\lambda(\xi)) = \xi^\ast \Delta^{\frac{1}{2}} \in L^2 G$. Furthermore, $y^\ast z \in {\mathfrak{m}_\varphi}\subseteq j_\infty(A_0 \cap A_1)$. By Terp’s construction, $\Sigma(\overline{A}) \subseteq (A_0 \cap A_1)^\ast$. We first claim that $$\langle a, j_\infty^{-1}(y^\ast z) \rangle_{(A_0 \cap A_1)^\ast, A_0 \cap A_1}  = 0.$$ Indeed, taking $F$ that contains the supports of $\eta^\ast, \xi^\ast$ and $\xi^\ast \ast \eta$, $$\label{Eqn=earlycomputation}
\begin{split}
& a, j_\infty^{-1}(y^\ast z) \rangle_{(A_0 \cap A_1)^\ast, A_0 \cap A_1} \\
= &
\langle J x^\ast J \Lambda(z), \Lambda(y) \rangle + \langle \phi, y^\ast z \rangle_{A(G), {\mathcal{L}}G} \\
= &
\langle x J \Lambda(y), J \Lambda(z) \rangle + \int_F \phi(s) (\xi^\ast \ast \eta)(s) ds \\
= &
\langle x \Delta^{\frac{1}{2}} \xi^\ast,  \Delta^{\frac{1}{2}} \eta^\ast \rangle + \int_F \phi(t s^{-1}) \overline{\xi(t^{-1})} \Delta(t^{-1}) \eta(s^{-1}) \Delta(s^{-1})  ds.
\end{split}$$ We continue the equation using Lemma \[lemma=isom\_AG\_L1\], $$\begin{split}
& \langle a, j_\infty^{-1}(y^\ast z) \rangle_{(A_0 \cap A_1)^\ast, A_0 \cap A_1} \\
= & \langle \Delta^{\frac{1}{2}} P_F x P_F \Delta^{\frac{1}{2}}  \xi^\ast  ,  \eta^\ast  \rangle + \langle P_F x_\phi P_F   \xi^\ast, \eta^\ast \rangle \\
= & \langle (P_F x P_F + \Delta^{-\frac{1}{2}}  P_F x_\phi P_F \Delta^{-\frac{1}{2}} )\Delta^{\frac{1}{2}}  \xi^\ast  ,  \Delta^{\frac{1}{2}}  \eta^\ast  \rangle \\
  = & \langle u_F(a) \Delta^{\frac{1}{2}}  \xi^\ast  ,  \Delta^{\frac{1}{2}}  \eta^\ast  \rangle =0.
\end{split}$$ We claim that it follows that in fact $$\langle a, w \rangle_{(A_0 \cap A_1)^\ast, A_0 \cap A_1} = 0,$$ for every $w \in A_0 \cap A_1$. By [@TerpII Theorem 8] it is enough to show this when $w \in {\mathfrak{m}_\varphi}$. By linearity we may assume that $w = y^\ast z$ with $y,z \in {\mathfrak{n}_\varphi}$. We thus need to show that $$\label{Eqn=finalstatement}
\langle J x^\ast J \Lambda(z), \Lambda(y) \rangle + \langle \phi, y^\ast z \rangle_{A(G), {\mathcal{L}}G} = 0,$$ c.f. the first three lines of . Since $\lambda(L^1 G \cap L^2 G)$ is a $\sigma$-weak/norm core for $\Lambda$, there are nets $\xi_i, \eta_i$ in $L^1 G \cap L^2 G$ such that $\lambda(\xi_i) \rightarrow y$ $\sigma$-weakly and $\Vert \xi_i - \Lambda(y) \Vert_{L^2 G} \rightarrow 0$ and similarly, $\lambda(\eta_i) \rightarrow z$ $\sigma$-weakly and $\Vert \eta_i - \Lambda(z) \Vert_{L^2 G} \rightarrow 0$. In fact, we may take $\xi_i, \eta_i \in C_c(G)$. Then, clearly follows from what we have proved.

(\[item=intersection\_big2\]) We claim that the required functions are given by: $$f_c(s) = \overline{f_a(s^{-1})} \Delta(s^{-1}), \qquad f_d(s) = (f_a \Delta^{-\frac{1}{2q}} ) \ast (f_b \Delta^{-\frac{1}{2p}}),$$ so that $c=a^\ast$ and $d = \sigma_{\frac{i}{2q}}(a)\sigma_{\frac{i}{2p}}(b)$. For notational convenience, suppose that $a = u^\ast v$, with $u,v \in {\mathfrak{n}_\varphi}$. In the general case one considers a linear combination of such a decomposition. Recall that for any two unbounded operators $x$ and $y$ we have $(xy)^\ast \supseteq x^\ast y^\ast$. Then, using and [@Hilsum Theorem 4.(1)] in the third equation, $$j_p( c ) = j_p(a^\ast)   = \Delta^{\frac{1}{2p}} v^\ast  \cdot [ u  \Delta^{\frac{1}{2p}} ]
=  \left(  \Delta^{\frac{1}{2p}} u^\ast  \cdot [ v  \Delta^{\frac{1}{2p}} ] \right)^\ast = j_p(a)^\ast.$$ Furthermore, write still $a = u^\ast v$ and also $b = x^\ast y$ with $u,v, x,y \in {\mathfrak{n}_\varphi}$. We claim that we have the following equalities, $$\begin{split}
& j_r(d) = j_r(  \sigma_{\frac{i}{2q}}(a)  \sigma_{\frac{i}{2p}}(b)   ) =   \Delta^{\frac{1}{2r}} \sigma_{\frac{i}{2q}}(a)   \cdot   [\sigma_{\frac{i}{2p}}(b)  \Delta^{\frac{1}{2r}} ] \\
 =\!\!\!\!\!{^?} \:\:\: &     
  \Delta^{\frac{1}{2p}} u^\ast \cdot  [v \Delta^{\frac{1}{2p}}] \cdot   \Delta^{\frac{1}{2q}} x^\ast \cdot  [y \Delta^{\frac{1}{2q}}]  
=    j_p(a) j_q(b).
\end{split}$$ Only $=\!\!\!\!\!{^?} \:\:\:$ needs justification, the other equalities are immediate from . Firstly, using [@TerpII Lemma 22], $$\Delta^{\frac{1}{2r}} \sigma_{\frac{i}{2q}} (a) \supseteq \Delta^{\frac{1}{2p}} a \Delta^{\frac{1}{2q}} = \Delta^{\frac{1}{2p}} u^\ast v \Delta^{\frac{1}{2q}}.$$ Using that the left hand side is closed and taking the closure on the right hand side, $\Delta^{\frac{1}{2r}} \sigma_{\frac{i}{2q}} (a) \supseteq \Delta^{\frac{1}{2p}} u^\ast\cdot[ v \Delta^{\frac{1}{2q}} ] $. Since (use for the right-hand side) both sides are in $L^{2r}({\mathcal{L}}G)$ we in fact have equality by [@Hilsum Theorem 4.(1)]. Also, using the same argument in the second equality, $$[ \sigma_{\frac{i}{2p}}(b) \Delta^{\frac{1}{2r}} ]^\ast = \Delta^{\frac{1}{2r}} \sigma_{\frac{i}{2p}} (b) = \Delta^{\frac{1}{2q}} y^\ast \cdot [x \Delta^{\frac{1}{2p}}] = (\Delta^{\frac{1}{2p}} x^\ast \cdot [y \Delta^{\frac{1}{2q}}] )^\ast.$$ So, $[ \sigma_{\frac{i}{2p}}(b) \Delta^{\frac{1}{2r}} ] = \Delta^{\frac{1}{2p}} x^\ast \cdot [y \Delta^{\frac{1}{2q}}] $. Furthermore, $$[v \Delta^{\frac{1}{2q}}] \cdot \Delta^{\frac{1}{2p}} x^\ast \supseteq v \cdot \Delta^{\frac{1}{2r}} x^\ast \subseteq [v \Delta^{\frac{1}{2q}}] \cdot \Delta^{\frac{1}{2p}} x^\ast.$$ Since each instance of this line is in $L^{2}({\mathcal{L}}G)$, the inclusions are equalities by [@Hilsum Theorem 4.(1)]. In all, we have proved that $$\begin{split}
 \Delta^{\frac{1}{2r}} \sigma_{\frac{i}{2q}}(a) \cdot   [\sigma_{\frac{i}{2p}}(b)  \Delta^{\frac{1}{2r}} ] = &
 \Delta^{\frac{1}{2p}} u^\ast\cdot[ v \Delta^{\frac{1}{2q}} ] \cdot \Delta^{\frac{1}{2p}} x^\ast \cdot [y \Delta^{\frac{1}{2q}}] \\ = &
\Delta^{\frac{1}{2p}} u^\ast \cdot  [v \Delta^{\frac{1}{2p}}] \cdot  \Delta^{\frac{1}{2q}} x^\ast \cdot  [y \Delta^{\frac{1}{2q}}], 
\end{split}$$ so that $f_c$ and $f_d$ have the right properties.

The claims about the supports now follow automatically. For the non-negativity, write $a = \lambda(f) \lambda(g), f,g \in C_c(G)$. Let $F \subseteq G$ be compact. We have $P_F j_1(a)P_F =  ( \lambda(f)^* \Delta^{\frac{1}{2}} P_F)^\ast  ( \lambda(g) \Delta^{\frac{1}{2}} P_F) = (\phi(t^{-1}s) )_{s,t \in F}$ for some $\phi \in A(G)$, see Lemma \[lemma=isom\_AG\_L1\]. But $( \lambda(f)^\ast \Delta^{\frac{1}{2}} P_F)^\ast  ( \lambda(g) \Delta^{\frac{1}{2}} P_F)_{s,t}  =  \Delta^{\frac{1}{2}}(st^{-1})   f_a(s t^{-1})$ is nonnegative.

Acknowledgement {#acknowledgement .unnumbered}
---------------

The authors wish to thank Éric Ricard for useful discussions on this paper, and the referee for useful comments.

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