Schur multipliers on $\mathcal{B}(L^p,L^q)$

TL;DR

This paper characterizes Schur multipliers on operators between L^p and L^q spaces, extending classical definitions and providing new insights into their structure and inclusion relationships.

Contribution

It introduces a new characterization of Schur multipliers on (L^p,L^q) spaces for 1 < q t p < t, generalizing classical cases.

Findings

Characterization of Schur multipliers via Bochner space representations.

Extension of Schur multiplier definitions to measure spaces.

New inclusion relationships between Schur multiplier spaces.

Abstract

Let $(\Omega_1, \mathcal{F}_1, \mu_1)$ and $(\Omega_2, \mathcal{F}_2, \mu_2)$ be two measure spaces and let $1 \leq p,q \leq +\infty$. We give a definition of Schur multipliers on $\mathcal{B}(L^p(\Omega_1), L^q(\Omega_2))$ which extends the definition of classical Schur multipliers on $\mathcal{B}(\ell_p,\ell_q)$. Our main result is a characterization of Schur multipliers in the case $1\leq q \leq p \leq +\infty$. When $1 < q \leq p < +\infty$, $\phi \in L^{\infty}(\Omega_1 \times \Omega_2)$ is a Schur multiplier on $\mathcal{B}(L^p(\Omega_1), L^q(\Omega_2))$ if and only if there are a measure space (a probability space when $p\neq q$) $(\Omega,\mu)$, $a\in L^{\infty}(\mu_1, L^{p}(\mu))$ and $b\in L^{\infty}(\mu_2, L^{q'}(\mu))$ such that, for almost every $(s,t) \in \Omega_1 \times \Omega_2$, $$\phi(s,t)=\left\langle a(s), b(t) \right\rangle.$$ Here, $L^{\infty}(\mu_1, L^{r}(\mu))$ denotes the Bochner space on $\Omega_1$ valued in $L^r(\mu)$. This result is new, even in the classical case. As a consequence, we give new inclusion relationships between the spaces of Schur multipliers on $\mathcal{B}(\ell_p,\ell_q)$.

Equations

$$\varphi (s,t)=⟨a(s),b(t)⟩.$$

$$\begin{array}[t]{lrcl}&T_{m}:\mathcal{B}(\ell_{p},\ell_{q})&\longrightarrow&\mathcal{B}(\ell_{p},\ell_{q})\\ &[a_{ij}]_{i,j\geq 1}&\longmapsto&[m_{ij}a_{ij}]_{i,j\geq 1}\end{array}$$

$$\forall i,j\in \mathbb{N},c_{ij}=⟨x_j,y_i⟩\text{and}\underset{i}{\sup }\Vert y_i{\Vert }_{p^{\prime }}\underset{j}{\sup }\Vert x_j{\Vert }_p\le C.$$

$$\begin{array}[t]{lccc}X_{J}:&L^{2}(\Omega_{1})&\longrightarrow&L^{2}(\Omega_{2})\\ &f&\longmapsto&\displaystyle\int_{\Omega_{1}}J(t,\cdot)f(t)\,\text{d}\mu_{1}(t)\end{array}$$

$$\begin{array}[t]{lccc}R_{\phi}:&S^{2}(L^{2}(\Omega_{1}),L^{2}(\Omega_{2}))&\longrightarrow&S^{2}(L^{2}(\Omega_{1}),L^{2}(\Omega_{2}))\\ &X_{J}&\longmapsto&X_{\phi J}\end{array}$$

$$R_{\varphi }:\mathcal{K}(L^2({\Omega }_1),L^2({\Omega }_2))⟶\mathcal{K}(L^2({\Omega }_1),L^2({\Omega }_2)),$$

$$\underset{s\in {\Omega }_1}{\mathrm{essup}}\sum _{i=1}^{+\infty }|a_i(s){|}^2<C,\underset{t\in {\Omega }_2}{\mathrm{essup}}\sum _{i=1}^{+\infty }|b_i(t){|}^2<C$$

$$\varphi (s,t)=\sum _{i=1}^{+\infty }{a}_i(s)b_i(t).$$

$$T_{\varphi }:L^{p^{\prime }}({\Omega }_1)\otimes L^q({\Omega }_2)\to L^{p^{\prime }}({\Omega }_1)\stackrel{\vee }{\otimes }{L}^q({\Omega }_2)$$

$$\varphi (s,t)=⟨a(s),b(t)⟩,$$

$$\Vert z{\Vert }_{\wedge }:=\inf \left\{\sum \Vert x_i\Vert \Vert y_i\Vert \right\},$$

$$z=\sum _i{x}_i\otimes y_i.$$

$$(X\stackrel{\wedge }{\otimes }Y{)}^{\ast }=\mathcal{B}(X,Y^{\ast }).$$

$$L^1(\Omega ;Y)=L^1(\Omega )\stackrel{\wedge }{\otimes }Y.$$

$$L^1(\Omega ;Y{)}^{\ast }=\mathcal{B}(L^1(\Omega ),Y^{\ast }).$$

$$L^1(\Omega ,Y{)}^{\ast }=L^{\infty }(\Omega ,Y^{\ast }).$$

$$L^{\infty }(\Omega ,Y^{\ast })=\mathcal{B}(L^1(\Omega ),Y^{\ast }),$$

$$\begin{array}[t]{lrcl}&L^{\infty}(\Omega,Y^{*})&\longrightarrow&\mathcal{B}(L^{1}(\Omega),Y^{*}).\\ &g&\longmapsto&\left[f\in L^{1}(\Omega)\mapsto\displaystyle{\int_{\Omega}f(t)g(t)\text{d}\mu(t)}\right]\end{array}$$

$$L^1(\Omega ,L^1({\Omega }^{\prime }))=L^1(\Omega \times {\Omega }^{\prime }).$$

$$\mathcal{B}(L^1(\Omega ),L^{\infty }({\Omega }^{\prime }))=L^{\infty }(\Omega \times {\Omega }^{\prime }),$$

$$\begin{array}[t]{lrcl}&L^{\infty}(\Omega\times\Omega^{\prime})&\longrightarrow&\mathcal{B}(L^{1}(\Omega),L^{\infty}(\Omega^{\prime})).\\ &\psi&\longmapsto&\left[f\in L^{1}(\Omega)\mapsto\displaystyle{\int_{\Omega}f(t)\psi(t,\cdot)\text{d}\mu(t)}\right]\end{array}$$

$$⟨z,x^{\ast }\otimes y^{\ast }⟩=\sum _i{x}^{\ast }(x_i)y^{\ast }(y_i).$$

$$\Vert z{\Vert }_{\vee }=\underset{\Vert x^{\ast }\Vert \le 1,\Vert y^{\ast }\Vert \le 1}{\sup }|⟨z,x^{\ast }\otimes y^{\ast }⟩|.$$

$$\tilde{u}(x)=\sum _i{x}_i^{\ast }(x)y_i,\forall x\in X.$$

$$X^{\ast }\stackrel{\vee }{\otimes }Y=\mathcal{K}(X,Y)$$

$$L^{p^{\prime }}({\Omega }_1)\stackrel{\vee }{\otimes }{L}^q({\Omega }_2)=\mathcal{K}(L^p({\Omega }_1),L^q({\Omega }_2)).$$

$$(L^{p^{\prime }}({\Omega }_1)\stackrel{\vee }{\otimes }{L}^q({\Omega }_2){)}^{\ast \ast }=(L^p({\Omega }_1)\stackrel{\wedge }{\otimes }{L}^{q^{\prime }}({\Omega }_2){)}^{\ast }=\mathcal{B}(L^p({\Omega }_1),L^q({\Omega }_2)).$$

$$T_{\varphi }:L^{p^{\prime }}({\Omega }_1)\otimes L^q({\Omega }_2)\to \mathcal{B}(L^p({\Omega }_1),L^q({\Omega }_2))$$

$$[T_{\varphi }(f\otimes g)](h)=\left({\int }_{{\Omega }_1}\varphi (s,\cdot )f(s)h(s)\text{d}{\mu }_1(s)\right)g(\cdot )\in L^q({\Omega }_2),$$

$$L^{p^{\prime }}({\Omega }_1)\otimes L^q({\Omega }_2)\subset L^{p^{\prime }}({\Omega }_1,L^q({\Omega }_2))$$

Full text

[TABLE]

Here, $L^{\infty}(\mu_{1},L^{r}(\mu))$ denotes the Bochner space on $\Omega_{1}$ valued in $L^{r}(\mu)$. This result is new, even in the classical case. As a consequence, we give new inclusion relationships between the spaces of Schur multipliers on $\mathcal{B}(\ell_{p},\ell_{q})$.

1. Introduction

If $1\leq r<+\infty$, we denote by $\ell_{r}$ the Banach space of $r-$summable sequences $(x_{i})_{i\geq 1}\subset\mathbb{C}$ (that is, $\sum_{i}|x_{i}|^{r}<+\infty$) endowed with the norm $\|x\|_{\ell_{r}}=\left(\sum_{i}|x_{i}|^{r}\right)^{1/r}$. Let $\ell_{\infty}$ be the Banach space of bounded sequences $(y_{i})_{i\geq 1}\subset\mathbb{C}$ with the norm $\|y\|_{\ell_{\infty}}=\sup_{i}|y_{i}|$. If $n\in\mathbb{N}$, we denote by $\ell_{r}^{n}$ the $n-$dimensional versions of the spaces introduced before.

Let $m=(m_{ij})_{i,j\geq 1}$ be a bounded family of complex numbers and let $1\leq p,q\leq+\infty$. We say that $m$ is a Schur multiplier on $\mathcal{B}(\ell_{p},\ell_{q})$ if for any matrix $[a_{ij}]_{i,j\geq 1}$ in $\mathcal{B}(\ell_{p},\ell_{q})$, the matrix $[m_{ij}a_{ij}]_{i,j\geq 1}$ defines an element of $\mathcal{B}(\ell_{p},\ell_{q})$. An application of the Closed Graph theorem shows that $m$ is a Schur multiplier if and only if the mapping

[TABLE]

is bounded. By definition, the norm of the Schur multiplier $m$ is the norm of $T_{m}$.

There is a well-known characterization of Schur multipliers on $\mathcal{B}(\ell_{2})$ (see for instance [11, Theorem 5.1]) which can be extended to the case $\mathcal{B}(\ell_{p})$ as follows.

Theorem 1.1**.**

[11, Theorem 5.10]** Let $\phi=(c_{ij})_{i,j\in\mathbb{N}}\subset\mathbb{C}$, $C\geq 0$ be a constant and let $1\leq p<\infty$. The following are equivalent :

1. (i)

$\phi$* is a Schur multiplier on $\mathcal{B}(\ell_{p})$ with norm $\leq C$.* 2. (ii)

There is a measure space $(\Omega,\mu)$ and elements $(x_{j})_{j\in\mathbb{N}}$ in $L^{p}(\mu)$ and $(y_{i})_{i\in\mathbb{N}}$ in $L^{p^{\prime}}(\mu)$ such that

[TABLE]

Denote by $\mathcal{M}(p,q)$ the space of Schur multipliers on $\mathcal{B}(\ell_{p},\ell_{q})$. In [3], Bennett gives some results about the inclusions between the spaces $\mathcal{M}(p,q)$. In the same paper, he also gives a necessary and sufficient condition for a family $m$ to belong to $\mathcal{M}(p,q)$, using the theory of absolutely summing operators. Theorem 1.1 provides a different type of characterization, which is more explicit and useful.

Let $(\Omega_{1},\mu_{1})$ and $(\Omega_{2},\mu_{2})$ be two $\sigma$-finite measure spaces. The space $L^{2}(\Omega_{1}\times\Omega_{2})$ can be identified with the space $S^{2}(L^{2}(\Omega_{1}),L^{2}(\Omega_{2}))$ of Hilbert-Schmidt operators. If $J\in L^{2}(\Omega_{1}\times\Omega_{2})$, the operator

[TABLE]

is a Hilbert-Schmidt operator and $\|X_{J}\|_{2}=\|J\|_{L^{2}}$. Moreover, any element of $S^{2}(L^{2}(\Omega_{1}),L^{2}(\Omega_{2}))$ has this form.

Let $\phi\in L^{\infty}(\Omega_{1}\times\Omega_{2})$. We may associate the operator

[TABLE]

whose norm is equal to $\|\phi\|_{\infty}$. We say that $\phi$ is a Schur multiplier on $\mathcal{B}(L^{2}(\Omega_{1}),L^{2}(\Omega_{2}))$ if $R_{\psi}$ extends to a (necessarily unique) bounded operator still denoted by

[TABLE]

where $\mathcal{K}(L^{2}(\Omega_{1}),L^{2}(\Omega_{2}))$ denotes the space of compact operators from $L^{2}(\Omega_{1})$ into $L^{2}(\Omega_{2})$. When $\phi$ is a Schur multiplier, the norm of $\phi$ is by definition the norm of $R_{\phi}$ as an operator from $\mathcal{K}(L^{2}(\Omega_{1}),L^{2}(\Omega_{2}))$ into itself.

A characterization similar to the one in Theorem 1.1 holds in this setting. The following result was established by Peller [9].

Theorem 1.2**.**

Let $\phi\in L^{\infty}(\Omega_{1}\times\Omega_{2})$ and $C>0$. The following are equivalent :

1. (i)

$\phi$* is a Schur multiplier and $\|R_{\phi}\|<C$.* 2. (ii)

There exist families $(a_{i})_{i\geq 1}\subset L^{\infty}(\Omega_{1})$ and $(b_{i})_{i\geq 1}\subset L^{\infty}(\Omega_{2})$ such that

[TABLE]

and for almost every $(s,t)\in\Omega_{1}\times\Omega_{2}$,

[TABLE]

See also [12] for another formulation of this theorem and results about Schur multipliers in the measurable case.

In this article, we define more generally Schur multipliers on $\mathcal{B}(L^{p}(\Omega_{1}),L^{q}(\Omega_{2}))$ for some measure spaces $(\Omega_{1},\mu_{1})$ and $(\Omega_{2},\mu_{2})$. To any $\phi\in L^{\infty}(\Omega_{1},\Omega_{2})$, we associate a linear mapping

[TABLE]

and we say that $\phi$ is a Schur multiplier if $T_{\phi}$ is bounded. When $\Omega_{1}=\Omega_{2}=\mathbb{N}$ with the counting measures, $T_{\phi}$ corresponds to $(\ref{MapSchur})$.

In the case $1\leq q\leq p\leq+\infty$, we characterize the elements of $L^{\infty}(\Omega_{1}\times\Omega_{2})$ which are Schur multipliers on $\mathcal{B}(L^{p}(\Omega_{1}),L^{q}(\Omega_{2}))$. We prove that if $1<q\leq p<+\infty$, $\phi$ is a Schur multiplier if and only if there are a measure space (a probability space when $p\neq q$) $(\Omega,\mu)$, $a\in L^{\infty}(\mu_{1},L^{p}(\mu))$ and $b\in L^{\infty}(\mu_{2},L^{q^{\prime}}(\mu))$ such that, for almost every $(s,t)\in\Omega_{1}\times\Omega_{2}$,

[TABLE]

where $L^{\infty}(\mu_{1},L^{r}(\mu))$ is the Bochner space valued in $L^{r}(\mu)$.

This result is new, even in the setting of classical Schur multipliers on $\mathcal{B}(\ell_{p},\ell_{q})$, and is of different nature than the characterization of Bennett. As a consequence, we give in the last section of this article new results of comparisons for the spaces $\mathcal{M}(p,q)$.

1.1. Notations

Let $X$ and $Y$ be Banach spaces.

If $z\in X\otimes Y$, the projective tensor norm of $z$ is defined by

[TABLE]

where the infimum runs over all finite families $(x_{i})_{i}$ in $X$ and $(y_{i})_{i}$ in $Y$ such that

[TABLE]

The completion $X\overset{\wedge}{\otimes}Y$ of $(X\otimes Y,\|.\|_{\wedge})$ is called the projective tensor product of $X$ and $Y$. Note that the projective tensor product is commutative, that is $X\overset{\wedge}{\otimes}Y=Y\overset{\wedge}{\otimes}X$.

The mapping taking any functional $\omega\colon X\otimes Y\to\mathbb{C}$ to the operator $u\colon X\to Y^{*}$ defined by $\langle u(x),y\rangle=\omega(x\otimes y)$ for any $x\in X,y\in Y$, induces an isometric identification

[TABLE]

We refer to [7, Chapter 8, Corollary 2] for this fact.

Let $(\Omega,\mu)$ be a localizable measure space and let $L^{p}(\Omega;Y)$ denote the Bochner space of $p-$integrable functions from $\Omega$ into $Y$. By [7, Chapter 8, Example 10], the natural embedding $L^{1}(\Omega)\otimes Y\subset L^{1}(\Omega;Y)$ extends to an isometric isomorphism

[TABLE]

By (2), this implies

[TABLE]

Assume that $Y^{*}$ has the Radon-Nikodym property (in short, $Y^{*}$ has RNP). In this case,

[TABLE]

The latter implies that

[TABLE]

and the isometric isomorphism is given by

[TABLE]

Assume now that $Y=L^{1}(\Omega^{\prime})$ where $(\Omega^{\prime},\mu^{\prime})$ is a localizable measure space. Then, an application of Fubini Theorem gives

[TABLE]

Using equality $(\ref{L1tensor})$, we deduce that

[TABLE]

and the correspondence is given by

[TABLE]

For $\psi\in L^{\infty}(\Omega\times\Omega^{\prime})$, denote by $u_{\psi}$ the corresponding element of $\mathcal{B}(L^{1}(\Omega),L^{\infty}(\Omega^{\prime}))$.

If $z=\sum_{i}x_{i}\otimes y_{i}\in X\otimes Y$, $x^{*}\in X^{*}$ and $y^{*}\in Y^{*}$, we write

[TABLE]

Then, the injective tensor norm of $z\in X\otimes Y$ is given by

[TABLE]

The completion $X\overset{\vee}{\otimes}Y$ of $(X\otimes Y,\|.\|_{\vee})$ is called the injective tensor product of $X$ and $Y$.

In this paper, we will often identify $X^{*}\otimes Y$ with the finite rank operators from $X$ into $Y$ as follow. If $u=\sum_{i}x_{i}^{*}\otimes y_{i}\in X^{*}\otimes Y$, we define $\tilde{u}:X\rightarrow Y$ by

[TABLE]

Then, it is easy to check that $\|u\|_{\vee}=\|\tilde{u}\|_{\mathcal{B}(X,Y)}$.

Moreover, if $Y$ has the approximation property (see e.g. [6] for the definition), [6, Theorem 1.4.21] gives the isometric identification

[TABLE]

where $\mathcal{K}(X,Y)$ denotes the space of compact operators from $X$ into $Y$.

Let $(\Omega_{1},\mathcal{F}_{1},\mu_{1})$ and $(\Omega_{2},\mathcal{F}_{2},\mu_{2})$ be two localizable measure spaces. Let $1\leq p<\infty$ and $1\leq q\leq\infty$. Then $L^{q}(\Omega_{2})$ has the approximation property so that we have

[TABLE]

Finally, if we assume that $1<p,q<+\infty$, then by [5, Theorem 2.5] and $(\ref{dualproj})$,

[TABLE]

2. Definition of Schur multipliers on $\mathcal{B}(L^{p},L^{q})$

Let $(\Omega_{1},\mathcal{F}_{1},\mu_{1})$ and $(\Omega_{2},\mathcal{F}_{2},\mu_{2})$ be two localizable measure spaces and let $\phi\in L^{\infty}(\Omega_{1}\times\Omega_{2})$. Let $1\leq p,q\leq\infty$ and denote by $p^{\prime}$ and $q^{\prime}$ their conjugate exponents.

Let

[TABLE]

be defined for any elementary tensor $f\otimes g\in L^{p^{\prime}}(\Omega_{1})\otimes L^{q}(\Omega_{2})$ by

[TABLE]

for all $h\in L^{p}(\Omega_{1})$.

We have an inclusion

[TABLE]

given by $f\otimes g\mapsto[s\in\Omega_{1}\mapsto f(s)g]$. Under this identification, $T_{\phi}$ is the multiplication by $\phi$. Note that $L^{p^{\prime}}(\Omega_{1},L^{q}(\Omega_{2}))$ is invariant by multiplication by an element of $L^{\infty}(\Omega_{1}\times\Omega_{2})$ and that we have a contractive inclusion

[TABLE]

Therefore, $T_{\phi}$ is valued is in $L^{p^{\prime}}(\Omega_{1})\overset{\vee}{\otimes}L^{q}(\Omega_{2})$. Using the identification

[TABLE]

given by $(\ref{tensorop})$, we deduce that the elements of $L^{p^{\prime}}(\Omega_{1})\overset{\vee}{\otimes}L^{q}(\Omega_{2})$ are compact operators as limits of finite rank operators for the operator norm.

Definition 2.1**.**

We say that $\phi$ is a Schur multiplier on $\mathcal{B}(L^{p}(\Omega_{1}),L^{q}(\Omega_{2}))$ if there exists a constant $C\geq 0$ such that for all $u\in L^{p^{\prime}}(\Omega_{1})\otimes L^{q}(\Omega_{2})$,

[TABLE]

that is, if $T_{\phi}$ extends to a bounded operator

[TABLE]

In this case, the norm of $\phi$ is by definition the norm of $T_{\phi}$.

Remark 2.2*.*

By $\mathcal{E}_{1}$ (resp. $\mathcal{E}_{2}$) we denote the space of simple functions on $\Omega_{1}$ (resp. $\Omega_{2}$). By density of $\mathcal{E}_{1}\otimes\mathcal{E}_{2}$ in $L^{p^{\prime}}(\Omega_{1})\overset{\vee}{\otimes}L^{q}(\Omega_{2})$, $T_{\phi}$ extends to a bounded operator from $L^{p^{\prime}}(\Omega_{1})\overset{\vee}{\otimes}L^{q}(\Omega_{2})$ into itself if and only if it is bounded on $\mathcal{E}_{1}\otimes\mathcal{E}_{2}$ equipped with the injective tensor norm.

Assume that $1<p,q<+\infty$. By $(\ref{injcompact})$ we have

[TABLE]

so that $\phi$ is a Schur multiplier on $\mathcal{B}(L^{p}(\Omega_{1}),L^{q}(\Omega_{2}))$ if and only if $T_{\phi}$ extends to a bounded operator

[TABLE]

In this case, considering the bi-adjoint of $T_{\phi}$, we obtain by $(\ref{bidualcomp})$ a $w^{*}-$continuous mapping

[TABLE]

which extends $T_{\phi}$. This explains the terminology ’$\phi$ is a Schur multiplier on $\mathcal{B}(L^{p}(\Omega_{1}),L^{q}(\Omega_{2}))$’.

Classical Schur multipliers : Assume that $\Omega_{1}=\Omega_{2}=\mathbb{N}$ and that $\mu_{1}$ and $\mu_{2}$ are the counting measures. An element $\phi\in L^{\infty}(\mathbb{N}^{2})$ is given by a family $c=(c_{ij})_{i,j\in\mathbb{N}}$ of complex numbers, where $c_{ij}=\phi(j,i)$. In this situation, the mapping $T_{\phi}$ is nothing but the classical Schur multiplier

[TABLE]

When this mapping is bounded from $\mathcal{B}(\ell_{p},\ell_{q})$ into itself, we will denote it by $T_{c}$.

Notations : If $(\Omega,\mathcal{F},\mu)$ is a measure space and $n\in\mathbb{N}^{*}$, we denote by $\mathcal{A}_{n,\Omega}$ the collection of $n-$tuples $(A_{1},\ldots,A_{n})$ of pairwise disjoint elements of $\mathcal{F}$ such that

[TABLE]

If $A=(A_{1},\ldots,A_{n})\in\mathcal{A}_{n,\Omega}$ and $1\leq p\leq+\infty$, denote by $S_{A,p}$ the subspace of $L^{p}(\Omega)$ generated by $\chi_{A_{1}},\ldots,\chi_{A_{n}}$. Then $S_{A,p}$ is $1-$complemented in $L^{p}(\Omega)$, and a norm one projection from $L^{p}(\Omega)$ into $S_{A,p}$ is given by the conditional expectation

[TABLE]

Note that the mapping

[TABLE]

is an isometric isomorphism between $S_{A,p}$ and $\ell_{p}^{n}$.

Proposition 2.3**.**

Let $(\Omega_{1},\mathcal{F}_{1},\mu_{1})$ and $(\Omega_{2},\mathcal{F}_{2},\mu_{2})$ be two measure spaces and let $\phi\in L^{\infty}(\Omega_{1}\times\Omega_{2})$. The following are equivalent :

1. (i)

$\phi$* is a Schur multiplier on $\mathcal{B}(L^{p}(\Omega_{1}),L^{q}(\Omega_{2}))$.* 2. (ii)

For all $n,m\in\mathbb{N}^{*}$, for all $A=(A_{1},\ldots,A_{n})\in\mathcal{A}_{n,\Omega_{1}},B=(B_{1},\ldots,B_{m})\in\mathcal{A}_{m,\Omega_{2}}$, write

[TABLE]

Then the Schur multipliers on $\mathcal{B}(\ell_{p}^{n},\ell_{q}^{m})$ associated with the families $\phi_{A,B}=(\phi_{ij})$ are uniformly bounded with respect to $n,m,A$ and $B$.

In this case, $\|T_{\phi}\|=sup_{n,m,A,B}\|T_{\phi_{A,B}}\|<+\infty$.

Proof.

$(i)\Rightarrow(ii)$. Assume first that $\phi$ is a Schur multiplier on $\mathcal{B}(L^{p}(\Omega_{1}),L^{q}(\Omega_{2}))$ with $\|T_{\phi}\|\leq 1$. Let $n,m\in\mathbb{N}^{*},A=(A_{1},\ldots,A_{n})\in\mathcal{A}_{n,\Omega_{1}}$ and $B=(B_{1},\ldots,B_{m})\in\mathcal{A}_{m,\Omega_{2}}$. Let $c=\sum_{i,j}c(i,j)e_{j}\otimes e_{i}\in\ell_{p^{\prime}}^{n}\otimes\ell_{q}^{m}\simeq\mathcal{B}(\ell_{p}^{n},\ell_{q}^{m})$.

Let $\varphi_{A,p}:S_{A,p}\rightarrow\ell_{p}^{n}$ and $\psi_{B,q}:S_{B,q}\rightarrow\ell_{q}^{m}$ be the isometries defined in $(\ref{isolp})$. Then $\tilde{c}:=\psi_{B,q}^{-1}\circ c\circ\varphi_{A,p}:S_{A,p}\rightarrow S_{B,q}$ satisfies $\|\tilde{c}\|=\|c\|$ and we have

[TABLE]

where $\tilde{c}(i,j)=\dfrac{c(i,j)}{\mu_{1}(A_{j})^{1/p^{\prime}}\mu_{2}(B_{i})^{1/q}}$.

The operator $u:=\psi_{B,q}\circ P_{B,q}\circ T_{\phi}(\tilde{c})_{|S_{A,p}}\circ\varphi_{A,p}^{-1}:\ell_{p}^{n}\rightarrow\ell_{q}^{m}$ satisfies

[TABLE]

and by assumption

[TABLE]

so that

[TABLE]

Let us prove that $u=T_{\phi_{A,B}}(c)$ where $T_{\phi_{A,B}}$ is the Schur multiplier associated with the family $(\phi_{ij})$.

Write $u(i,j):=\psi_{B,q}\circ P_{B,q}\circ T_{\phi}(\chi_{A_{j}}\otimes\chi_{B_{i}})_{|S_{A,p}}\circ\varphi_{A,p}^{-1}$. We have

[TABLE]

Let $1\leq k\leq n$.

[TABLE]

so that $[u(i,j)](e_{k})=0$ if $k\neq j$ and if $k=j$ then

[TABLE]

It follows that

[TABLE]

that is, $u=T_{\phi_{A,B}}(c)$. We conclude thanks to the inequality $(\ref{equa1})$.

$(ii)\Rightarrow(i)$. Assume now that the assertion $(ii)$ is satisfied and show that $\phi$ is a Schur multiplier. By Remark 2.2, we just need to show that $T_{\phi}$ is bounded on $\mathcal{E}_{1}\otimes\mathcal{E}_{2}$. Let $v\in\mathcal{E}_{1}\otimes\mathcal{E}_{2}$ and write $\alpha=sup_{n,m,A,B}\|T_{c}\|$. We will show that $\|T_{\phi}(v)\|\leq\alpha\|v\|$. By density, it is enough to prove that for any $h_{1}\in\mathcal{E}_{1},h_{2}\in\mathcal{E}_{2}$,

[TABLE]

By assumption, there exist $n,m\in\mathbb{N}^{*},A=(A_{1},\ldots,A_{n})\in\mathcal{A}_{n,\Omega_{1}},B=(B_{1},\ldots,B_{m})\in\mathcal{A}_{m,\Omega_{2}}$ and complex numbers $v(i,j),a_{i},b_{j}$ such that

[TABLE]

Equation $(\ref{equa2})$ can be rewritten as

[TABLE]

Consider $\tilde{v}:=\psi_{B,q}\circ v\circ\varphi_{A,p}^{-1}:\ell_{p}^{n}\rightarrow\ell_{q}^{m}$ and $z:=\psi_{B,q}\circ P_{B,q}\circ T_{\phi}(v)_{|S_{A,p}}\circ\phi_{A,p}^{-1}:\ell_{p}^{n}\rightarrow\ell_{q}^{m}$. The computations made in the first part of the proof show that $z=T_{m}(\tilde{v})$ where $m$ is the family $(\phi_{ij})$.

Now, let $x:=\varphi_{A,p}(h_{1})$ and $y:=\psi_{B,q^{\prime}}(h_{2})$. Since $T_{m}$ is bounded with norm smaller than $\alpha$ we have

[TABLE]

An easy computation shows that the left-hand side on this equality is nothing but the left-hand side of the inequality $(\ref{equa3})$. Finally, the right-hand side of the inequalities $(\ref{equa3})$ and $(\ref{equa4})$ are equal, which concludes the proof.

∎

3. $(p,q)-$Factorable operators

Let $X$ and $Y$ be Banach spaces.

3.1. Dual norm.

[4, Chapter 15]. Let $M\subset X$ and $N\subset Y$ be finite dimensional subspaces (in short, f.d.s). If $u=\sum_{i=1}^{n}x_{i}\otimes y_{i}\in M\otimes N$ and $v=\sum_{j=1}^{m}x_{j}^{*}\otimes y_{j}^{*}\in M^{*}\otimes N^{*}$ we set

[TABLE]

Let $\alpha$ be a tensor norm on tensor products of finite dimensional spaces. We define, for $z\in M\otimes N$,

[TABLE]

Now, for $z\in X\otimes Y$, we set

[TABLE]

$\alpha^{\prime}$ defines a tensor norm on $X\otimes Y$, called the dual norm of $\alpha$.

In the sequel, we will write $\alpha^{\prime}(z)$ instead of $\alpha^{\prime}(z,X,Y)$ for the norm of an element $z\in X\otimes Y$ when there is no possible confusion.

3.2. Lapresté norms.

[4, Proposition 12.5]. Let $s\in[1,\infty]$. If $x_{1},x_{2},\ldots,x_{n}\in X$, we define

[TABLE]

Let $p,q\in[1,\infty]$ with $\dfrac{1}{p}+\dfrac{1}{q}\geq 1$ and take $r\in[1,\infty]$ such that

[TABLE]

Denote by $p^{\prime}$ and $q^{\prime}$ the conjugate of $p$ and $q$. For $z\in X\otimes Y$, we define

[TABLE]

Then $\alpha_{p,q}$ is a norm on $X\otimes Y$ and we denote by $X\otimes_{\alpha_{p,q}}Y$ its completion.

3.3. $(p,q)-$Factorable operators.

If $T\in\mathcal{B}(X,Y^{*})$ and $\xi=\sum_{i}x_{i}\otimes y_{i}\in X\otimes Y$, then in accordance with $(\ref{dualproj})$ we set

[TABLE]

Definition 3.1**.**

Let $1\leq p,q\leq\infty$ such that $\dfrac{1}{p}+\dfrac{1}{q}\geq 1$. Let $T\in\mathcal{B}(X,Y^{*})$. We say that $T\in\mathcal{L}_{p,q}(X,Y^{*})$ if there exists a constant $C\geq 0$ such that

[TABLE]

*In this case, we write $L_{p,q}(T)=\inf\left\{C\ |\ C\ \text{satisfying}\ (\ref{defLpq})\right\}.$

Then $(\mathcal{L}_{p,q}(X,Y^{*}),L_{p,q})$ is a Banach space, called the space of $(p,q)-$Factorable operators.*

For a general definition of the spaces $\mathcal{L}_{p,q}(X,Y)$ (including the case when the range is not a dual space), see [4, Chapter 17].

Since $Y^{*}$ is 1-complemented in its bidual, [4, Theorem 18.11] gives the following result.

Theorem 3.2**.**

*Let $1\leq p,q\leq\infty$ such that $\dfrac{1}{p}+\dfrac{1}{q}\geq 1$. Let $T\in\mathcal{B}(X,Y^{*})$. The two following statements are equivalent :

$(i)$ $T\in\mathcal{L}_{p,q}(X,Y^{*})$.

$(ii)$ There are a measure space $(\Omega,\mu)$ (a probability space when $\dfrac{1}{p}+\dfrac{1}{q}>1$), operators $R\in\mathcal{B}(X,L^{q^{\prime}}(\mu))$ and $S\in\mathcal{B}(L^{p}(\mu),Y^{*}))$ such that $T=S\circ I\circ R$*

[TABLE]

*where $I:L^{q^{\prime}}(\mu)\rightarrow L^{p}(\mu)$ is the inclusion mapping (well defined because $q^{\prime}\geq p$).

In this case, $L_{p,q}(T)=\inf\|S\|\|R\|$ over all such factorizations.*

Remark 3.3*.*

Here we consider the case when $\dfrac{1}{p}+\dfrac{1}{q}=1$. Denote by $p^{\prime}$ the conjugate exponent of $p$. We have $T\in\mathcal{L}_{p,p^{\prime}}(X,Y^{*})$ if and only if there are a measure space $(\Omega,\mu)$, operators $R\in\mathcal{B}(X,L^{p}(\mu))$ and $S\in\mathcal{B}(L^{p}(\mu),Y^{*})$ such that $T=SR$

[TABLE]

We usually write $\Gamma_{p}(X,Y^{*})$ instead of $\mathcal{L}_{p,p^{\prime}}(X,Y^{*})$. Such operators are called $p-$factorable.

Remark 3.4*.*

Suppose that $X=L^{1}(\lambda)$ and $Y=L^{1}(\nu)$ for some localizable measure spaces $(\Omega_{1},\lambda)$ and $(\Omega_{2},\nu)$. Consider $T\in\mathcal{B}(L^{1}(\lambda),L^{\infty}(\nu))$. By $(\ref{LinfB})$, there exists $\psi\in L^{\infty}(\lambda\times\nu)$ such that

[TABLE]

(See subsection 1.1 for the notation.)

1. (i)

If $1<q<+\infty$, $L^{q^{\prime}}(\mu)$ has RNP so by (5),

[TABLE]

It means that if $R\in\mathcal{B}(X,L^{q^{\prime}}(\mu))$, there exists $a\in L^{\infty}(\lambda,L^{q^{\prime}}(\mu))$ such that

[TABLE] 2. (ii)

If $1<p<+\infty$, then using $(\ref{dualproj})$, $(\ref{L1tensor})$ and $(\ref{L1tensorcor})$ we obtain

[TABLE]

Thus, if $S\in\mathcal{B}(L^{p}(\mu),L^{\infty}(\nu))$, there exists $b\in L^{\infty}(\nu,L^{p^{\prime}}(\mu))$ such that

[TABLE]

We deduce that if $1<p,q<+\infty$, there exist $a\in L^{\infty}(\lambda,L^{q^{\prime}}(\mu))$ and $b\in L^{\infty}(\nu,L^{p^{\prime}}(\mu))$ such that for almost every $(s,t)\in\Omega_{1}\times\Omega_{2}$,

[TABLE]

If $T$ satisfies Theorem 3.2, the latter implies that for all $f\in L^{1}(\lambda)$,

[TABLE]

Using the same identifications we have for the following cases :

1. (1)

If $q=1$ and $1<p<+\infty$, then there exist $a\in L^{\infty}(\lambda\times\mu)$ and $b\in L^{\infty}(\nu,L^{p^{\prime}}(\mu))$ such that for almost every $(s,t)\in\Omega_{1}\times\Omega_{2}$,

[TABLE] 2. (2)

If $1<q<+\infty$ and $p=+\infty$, then there exist $a\in L^{\infty}(\lambda,L^{q^{\prime}}(\mu))$ and $b\in L^{\infty}(\nu\times\mu)$ such that for almost every $(s,t)\in\Omega_{1}\times\Omega_{2}$,

[TABLE] 3. (3)

If $q=1$ and $p=+\infty$, then there exist $a\in L^{\infty}(\lambda\times\mu)$ and $b\in L^{\infty}(\nu\times\mu)$ such that for almost every $(s,t)\in\Omega_{1}\times\Omega_{2}$,

[TABLE]

3.4. Finite dimensional case.

If $X$ and $Y$ are finite dimensional, it follows from the very definition of the dual norm that

[TABLE]

The next theorem describes the elements of this space.

Theorem 3.5**.**

[4*, Theorem 19.2]** Let $E$ and $F$ be Banach spaces. Let $p,q\in[1,\infty]$ with $\dfrac{1}{p}+\dfrac{1}{q}\geq 1$ and $K\subset B_{E^{*}}$ and $L\subset B_{F^{*}}$ weak$-*$-compact norming sets for E and F, respectively. For $\phi:E\otimes F\rightarrow\mathbb{C}$ the following two statements are equivalent:

$(i)\ \phi\in(E\otimes_{\alpha_{p,q}}F)^{*}$.

$(ii)$ There are a constant $A\geq 0$ and normalized Borel-Radon measures $\mu$ on $K$ and $\nu$ on $L$ such that for all $x\in E$ and $y\in F$,*

[TABLE]

*(if the exponent is $\infty$, we replace the integral by the norm).

In this case, $\|\phi\|_{(E\otimes_{\alpha_{p,q}}F)^{*}}=\inf\left\{A\ |\ A\ \text{as in (ii)}\right\}.$

This theorem will allow us to describe the predual of $\mathcal{L}_{p,q}(\ell_{1}^{n},\ell_{\infty}^{m})$, $n,m\in\mathbb{N}$. Let us apply the previous theorem with $E=\ell_{\infty}^{n}$ and $F=\ell_{\infty}^{m}$. Take $T\in\ell_{1}^{n}\otimes_{\alpha_{p,q}^{\prime}}\ell_{1}^{m}=(\ell_{\infty}^{n}\otimes_{\alpha_{p,q}}\ell_{\infty}^{m})^{*}$ and let

[TABLE]

be a representation of $T$. In the previous theorem, we can take $K=\left\{1,2,\ldots,n\right\}$ and $L=\left\{1,2,\ldots,m\right\}$. In this case, a normalized Borel-Radon measure $\mu$ on $K$ is nothing but a sequence $\mu=(\mu_{1},\ldots,\mu_{n})$ where, for all $i$, $\mu_{i}:=\mu(\left\{i\right\})\geq 0$ and $\sum_{i}\mu_{i}=1$. Similarly, $\nu=(\nu_{1},\ldots,\nu_{m})$ where, for all $i$, $\nu_{i}\geq 0$ and $\sum_{i}\nu_{i}=1$. In this case, the inequality $(\ref{dominated})$ means that for all sequences of complex numbers $x=(x_{i})_{i=1}^{n},y=(y_{j})_{i=j}^{m}$,

[TABLE]

Set $\alpha_{k}=x_{k}\mu_{k}^{1/q^{\prime}}$, $\beta_{k}=y_{k}\nu_{k}^{1/p^{\prime}}$ and define, for $1\leq i\leq n,1\leq j\leq m$, $c(i,j)$ such that $T(i,j)=c(i,j)\mu_{i}^{1/q^{\prime}}\nu_{j}^{1/p^{\prime}}$ (we can assume $\mu_{i}>0$ and $\nu_{j}>0$). Then, the previous inequality becomes

[TABLE]

This means that the operator $c:\ell_{q^{\prime}}^{n}\rightarrow\ell_{p}^{m}$ whose matrix is $[c(i,j)]_{1\leq j\leq m,1\leq i\leq n}$ has a norm smaller than $A$. Moreover, if we see $T$ as a mapping from $\ell_{\infty}^{n}$ into $\ell_{1}^{m}$ the relation between $T$ and $c$ means that $T$ admits the following factorization

[TABLE]

where $d_{\mu}$ and $d_{\nu}$ are the operators of multiplication by $\mu=(\mu_{1}^{1/q^{\prime}},\ldots,\mu_{n}^{1/q^{\prime}})$ and $\nu=(\nu_{1}^{1/p^{\prime}},\ldots,\nu_{m}^{1/p^{\prime}})$. Those operators have norm 1.

Therefore, it is easy to check that

[TABLE]

The elements of $(\ell_{\infty}^{n}\otimes_{\alpha_{p,q}}\ell_{\infty}^{m})^{*}$ are called $(q^{\prime},p^{\prime})-$dominated operators. For more informations about this space in the infinite dimensional case (it is the predual of $\mathcal{L}_{p,q}$), see for instance [4, Chapter 19].

By $(\ref{formulepredual})$ and the fact that $\mathcal{L}_{p,q}(\ell_{1}^{n},\ell_{\infty}^{n})=(\ell_{1}^{n}\otimes_{\alpha_{p,q}^{\prime}}\ell_{1}^{m})^{*}$, we get the following result.

Proposition 3.6**.**

Let $v=[v_{ij}]:\ell_{1}^{n}\rightarrow\ell_{\infty}^{m}$. Then

[TABLE]

where the supremum runs over all $u:\ell_{\infty}^{m}\rightarrow\ell_{1}^{n}$ admitting the factorization

[TABLE]

*with $\|d_{\mu}\|\leq 1,\|d_{\nu}\|\leq 1$ and $\|c\|\leq 1$.

Equivalently,*

[TABLE]

4. The main result

4.1. Schur multipliers and factorization

Let $p,q$ be two positive numbers such that $1\leq q\leq p\leq\infty$. This condition is equivalent to $p,q\in[1,\infty]$ with $\dfrac{1}{q}+\dfrac{1}{p^{\prime}}\geq 1$, so that we can consider the space $\mathcal{L}_{q,p^{\prime}}$.

The following results will allow us to give a description of the functions $\phi$ which are Schur multipliers.

Lemma 4.1**.**

Let $X$, $Y$ be Banach spaces and let $E\subset X,F\subset Y$ be $1-$complemented subspaces of $X$ and $Y$. For any $v\in E\otimes F$, denote by $\tilde{\alpha}^{\prime}_{q,p^{\prime}}(v)$ the $\alpha^{\prime}_{q,p^{\prime}}$-norm of $v$ as an element of $E\otimes F$ and by $\alpha^{\prime}_{q,p^{\prime}}(v)$ the $\alpha^{\prime}_{q,p^{\prime}}$-norm of $v$ as an element of $X\otimes Y$. Then

[TABLE]

Proof.

The inequality $\tilde{\alpha}^{\prime}_{q,p^{\prime}}(v)\geq\alpha^{\prime}_{q,p^{\prime}}(v)$ is easy to prove. For the converse inequality, take $v=\sum_{k}e_{k}\otimes f_{k}\in E\otimes F$ such that $\alpha^{\prime}_{q,p^{\prime}}(v)<1$ and show that $\tilde{\alpha}^{\prime}_{q,p^{\prime}}(v)<1$. By assumption, there exists $M\subset X$ and $N\subset Y$ finite dimensional subspaces such that $v\in M\otimes N$ and

[TABLE]

By assumption, there exist two norm one projections $P$ and $Q$ respectively from $X$ onto $E$ and from $Y$ onto $F$. Set $M_{1}=P(M)\subset E$ and $N_{1}=Q(N)\subset F$. $M_{1}$ and $N_{1}$ are finite dimensional. Moreover, since $v\in E\otimes F$, it is easy to check that $(P\otimes Q)(v)=v$, where, for all $c=\sum_{l}a_{l}\otimes b_{l}\in X\otimes Y$,

[TABLE]

Thus, $v\in M_{1}\otimes N_{1}$. We will show that $\alpha^{\prime}_{q,p^{\prime}}(v,M_{1},N_{1})<1$.

Let $z=\sum_{j=1}^{m}x_{j}^{*}\otimes y_{j}^{*}\in M_{1}^{*}\otimes N_{1}^{*}$ be such that $\alpha_{q,p^{\prime}}(z)<1$ and show that $|\left\langle v,z\right\rangle|\leq\alpha^{\prime}_{q,p^{\prime}}(v)$, so that $\alpha^{\prime}_{q,p^{\prime}}(v,M_{1},N_{1})\leq 1$.

Let $1\leq r\leq\infty$ such that

[TABLE]

The condition $\alpha_{q,p^{\prime}}(z)<1$ in $M_{1}^{*}\otimes N_{1}^{*}$ implies that $z$ admits a representation $z=\sum_{j=1}^{m}\lambda_{j}m_{j}^{*}\otimes n_{j}^{*}$ where $m_{j}^{*}\in M_{1}^{*},n_{j}^{*}\in N_{1}^{*}$ and

[TABLE]

Set $\tilde{z}:=\sum_{j=1}^{m}\lambda_{j}P^{*}(m_{j}^{*})\otimes Q^{*}(n_{j}^{*})$ in $M^{*}\otimes N^{*}$. It is easy to check that

[TABLE]

Therefore, $\alpha_{q,p^{\prime}}(\tilde{z},M^{*},N^{*})<1$. Then, the condition $\alpha^{\prime}_{q,p^{\prime}}(v,M,N)<1$ implies that

[TABLE]

Finally, we have

[TABLE]

and therefore

[TABLE]

This proves that $\tilde{\alpha}^{\prime}_{q,p^{\prime}}(v)<1$. ∎

We recall that if $\phi\in L^{\infty}(\Omega_{1}\times\Omega_{2})$, we denote by $u_{\phi}$ the mapping

[TABLE]

Theorem 4.2**.**

Let $(\Omega_{1},\mu_{1})$ and $(\Omega_{2},\mu_{2})$ be two localizable measure spaces and let $\phi\in L^{\infty}(\Omega_{1}\times\Omega_{2})$. Let $1\leq q\leq p\leq\infty$. Then $\phi$ is a Schur multiplier on $\mathcal{B}(L^{p}(\Omega_{1}),L^{q}(\Omega_{2}))$ if and only if the operator $u_{\phi}$ belongs to $\mathcal{L}_{q,p^{\prime}}(L^{1}(\Omega_{1}),L^{\infty}(\Omega_{2}))$. Moreover,

[TABLE]

Proof.

Assume first that $T_{\phi}$ extends to a bounded operator

[TABLE]

with norm $\leq 1$. To prove that $u_{\phi}\in\mathcal{L}_{q,p^{\prime}}(L^{1}(\Omega_{1}),L^{\infty}(\Omega_{2}))$ with $L_{q,p^{\prime}}(u_{\phi})\leq 1$, we have to show that for any $v=\sum_{k}f_{k}\otimes g_{k}\in L^{1}(\Omega_{1})\otimes L^{1}(\Omega_{2})$ with $\alpha^{\prime}_{q,p^{\prime}}(v)<1$ we have

[TABLE]

By density, we can assume that $f_{k},g_{k}$ are simple functions. Hence, with the notations introduced in Section 2 there exist $n,m\in\mathbb{N}^{*},A=(A_{1},\ldots,A_{n})\in\mathcal{A}_{n,\Omega_{1}}$ and $B=(B_{1},\ldots,B_{m})\in\mathcal{A}_{m,\Omega_{2}}$ such that, for all $k$, $f_{k}\in S_{A,1}$ and $g_{k}\in S_{B,1}$.

By Lemma 4.1, the $\alpha^{\prime}_{q,p^{\prime}}$-norm of $v$ as an element of $S_{A,1}\otimes S_{B,1}$ is less than 1.

Let $\varphi_{A,1}:S_{A,1}\rightarrow\ell_{1}^{n}$ and $\psi_{B,1}:S_{B,1}\rightarrow\ell_{1}^{m}$ the isomorphisms defined in $(\ref{isolp})$. Set $v^{\prime}=\sum_{k}\varphi_{A,1}(f_{k})\otimes\psi_{B,1}(g_{k})\in\ell_{1}^{n}\otimes\ell_{1}^{m}$. Since $\varphi_{A,1}$ and $\psi_{B,1}$ are isometries, we have $\alpha^{\prime}_{q,p^{\prime}}(v^{\prime})<1$. Using the identification $(\ref{tensorop})$, we obtain by $(\ref{formulepredual})$ that $v^{\prime}$ admits a factorization

[TABLE]

where $\delta=(\delta_{1},\ldots,\delta_{n})$, $\gamma=(\gamma_{1},\ldots,\gamma_{m})$, $d_{\delta}$ and $d_{\gamma}$ are the operators of multiplication and

[TABLE]

This factorization means that

[TABLE]

Therefore, we have

[TABLE]

We compute

[TABLE]

Define

[TABLE]

where $\tilde{c}(i,j)=c_{i,j}\mu_{1}(A_{j})^{-1/p^{\prime}}\mu_{2}(B_{i})^{-1/q}$.

Using the identification $(\ref{tensorop})$, it is easy to check that we have

[TABLE]

Therefore,

[TABLE]

We have

[TABLE]

where

[TABLE]

Since $\|T_{\phi}\|\leq 1$, we deduce that

[TABLE]

Conversely, assume that $u_{\phi}\in\mathcal{L}_{q,p^{\prime}}(L^{1}(\Omega_{1}),L^{\infty}(\Omega_{2}))$ with $L_{q,p^{\prime}}(u_{\phi})\leq 1$. To prove that $\phi$ is a Schur multiplier, we will use Proposition 2.3. Let $n,m\in\mathbb{N}^{*}$, $A=(A_{1},\ldots,A_{n})\in\mathcal{A}_{n,\Omega_{1}}$ and $B=(B_{1},\ldots,B_{m})\in\mathcal{A}_{m,\Omega_{2}}$. Set

[TABLE]

We want to show that the Schur multiplier on $\mathcal{B}(\ell_{p}^{n},\ell_{q}^{m})$ associated to the family $m=(\phi_{ij})_{i,j}$ has a norm less than $1$. To prove that, let $c=\sum_{i,j}c(i,j)e_{j}\otimes e_{i}\in\mathcal{B}(\ell_{p}^{n},\ell_{q}^{m}),x=(x_{j})_{j=1}^{n},y=(y_{i})_{i=1}^{m}$ in $\mathbb{C}$ be such that $\|c\|\leq 1,\|x\|_{\ell_{p}^{n}}=1,\|y\|_{\ell_{q^{\prime}}}=1$. We have to show that

[TABLE]

This inequality can be rewritten as

[TABLE]

Let $v=\sum_{i,j}x_{j}c(i,j)y_{i}e_{j}\otimes e_{i}$. According to $(\ref{formulepredual})$, $\alpha^{\prime}_{q,p^{\prime}}(v)\leq 1$. Now, let $\tilde{v}=\sum_{i,j}x_{j}c(i,j)y_{i}\varphi_{A,1}^{-1}(e_{j})\otimes\psi_{B,1}^{-1}(e_{i})$. We have

[TABLE]

and

[TABLE]

By assumption, $L_{q,p^{\prime}}(u_{\phi})\leq 1$, which implies that

[TABLE]

and this is precisely the inequality $(\ref{proof1})$. ∎

Theorem 3.2 and Remark 3.4 allow us to reformulate the previous theorem. The following two corollaries are generalizations of Theorem 1.1.

Corollary 4.3**.**

Let $(\Omega_{1},\mu_{1})$ and $(\Omega_{2},\mu_{2})$ be two localizable measure spaces and let $\phi\in L^{\infty}(\Omega_{1}\times\Omega_{2})$. Let $1\leq q\leq p\leq\infty$. The following statements are equivalent :

1. (i)

$\phi$* is a Schur multiplier on $\mathcal{B}(L^{p}(\Omega_{1}),L^{q}(\Omega_{2}))$.* 2. (ii)

There are a measure space (a probability space when $p\neq q$) $(\Omega,\mu)$, operators $R\in\mathcal{B}(L^{1}(\Omega_{1}),L^{p}(\mu))$ and $S\in\mathcal{B}(L^{q}(\mu),L^{\infty}(\Omega_{2}))$ such that $u_{\phi}=S\circ I\circ R$

[TABLE]

where $I$ is the inclusion mapping.

*In the following cases, $(i)$ and $(ii)$ are equivalent to :

If $1<q\leq p<+\infty$ :*

1. (iii)

There are a measure space (a probability space when $p\neq q$) $(\Omega,\mu)$, $a\in L^{\infty}(\mu_{1},L^{p}(\mu))$ and $b\in L^{\infty}(\mu_{2},L^{q^{\prime}}(\mu))$ such that, for almost every $(s,t)\in\Omega_{1}\times\Omega_{2}$,

[TABLE]

If $1=q<p<+\infty$ :

1. (iii)

There are a probability space $(\Omega,\mu)$, $a\in L^{\infty}(\mu_{1}\times\mu)$ and $b\in L^{\infty}(\mu_{2},L^{q^{\prime}}(\mu))$ such that for almost every $(s,t)\in\Omega_{1}\times\Omega_{2}$,

[TABLE]

If $1<q<+\infty$ and $p=+\infty$ :

1. (iii)

There are a probability space $(\Omega,\mu)$, $a\in L^{\infty}(\mu_{1},L^{p}(\mu))$ and $b\in L^{\infty}(\mu_{2}\times\mu)$ such that for almost every $(s,t)\in\Omega_{1}\times\Omega_{2}$,

[TABLE]

If $q=1$ and $p=+\infty$ :

1. (iii)

There are a probability space $(\Omega,\mu)$, $a\in L^{\infty}(\mu_{1}\times\mu)$ and $b\in L^{\infty}(\mu_{2}\times\mu)$ such that for almost every $(s,t)\in\Omega_{1}\times\Omega_{2}$,

[TABLE]

In this case, $\|T_{\phi}\|=\inf\|R\|\|I\|\|S\|=\inf\|a\|\|b\|$.

Remark 4.4*.*

In the previous corollary, the condition $(ii)$ implies that every $\phi\in L^{\infty}(\Omega_{1}\times\Omega_{2})$ is a Schur multiplier on $\mathcal{B}(L^{1}(\Omega_{1}),L^{1}(\Omega_{2}))$ and on $\mathcal{B}(L^{\infty}(\Omega_{1}),L^{\infty}(\Omega_{2}))$.

In the discrete case, the previous corollary can be reformulated as follow.

Corollary 4.5**.**

Let $\phi=(c_{ij})_{i,j\in\mathbb{N}}\subset\mathbb{C}$, $C\geq 0$ be a constant and let $1\leq q\leq p\leq+\infty$. The following are equivalent :

1. (i)

$\phi$* is a Schur multiplier on $\mathcal{B}(\ell_{p},\ell_{q})$ with norm $\leq C$.* 2. (ii)

There exist a measure space (a probability space when $p\neq q$) $(\Omega,\mu)$ and two bounded sequences $(x_{j})_{j}$ in $L^{p}(\mu)$ and $(y_{i})_{i}$ in $L^{q^{\prime}}(\mu)$ such that

[TABLE]

4.2. An application : the main triangle projection

Let $m_{ij}=1$ if $i\leq j$ and $m_{ij}=0$ otherwise. Let $T_{m}$ be the Schur multiplier associated with the family $m=(m_{ij})$. For any infinite matrix $A=[a_{ij}]$, $T_{m}(A)$ is the matrix $[b_{ij}]$ with $b_{ij}=a_{ij}$ if $i\leq j$ and $b_{ij}=0$ otherwise. For that reason, $T_{m}$ is called the main triangle projection. Similary, we define the n-th main triangle projection as the Schur multiplier on $\mathcal{M}_{n}(\mathbb{C})$ associated with the family $m_{n}=(m_{ij}^{n})_{1\leq i,j\leq n}$ where $m_{ij}^{n}=1$ if $i\leq j$ and $m_{ij}^{n}=0$ otherwise. In [8], Kwapień and Pelczyński proved that if $1\leq q\leq p\leq+\infty,p\neq 1,q\neq+\infty$, there exists a constant $K>0$ such that for all $n$,

[TABLE]

and this order of growth is obtained for the Hilbert matrices. Those estimates imply that $T_{m}$ is not bounded on $\mathcal{B}(\ell_{p},\ell_{q})$. Bennett proved in [2] that when $1<p<q<\infty$, $T_{m}$ is bounded from $\mathcal{B}(\ell_{p},\ell_{q})$ into itself.

The results obtained in subsection 4.1 allow us to give a very short proof of the unbounded case.

Proposition 4.6**.**

Let $1\leq q\leq p\leq+\infty,p\neq 1,q\neq+\infty$. Then $T_{m}$ is not bounded on $\mathcal{B}(\ell_{p},\ell_{q})$.

Proof.

Assume that $T_{m}$ is bounded on $\mathcal{B}(\ell_{p},\ell_{q})$. By Corollary 4.3, there exist a measure space $(\Omega,\mu)$, $(a_{n})_{n}\in L^{p}(\mu)$ and $(b_{n})_{n}\in L^{q^{\prime}}(\mu)$ two bounded sequences such that, for all $i,j\in\mathbb{N}$,

[TABLE]

By boundedness, $(a_{n})_{n}$ and $(b_{n})_{n}$ admit an accumluation point $a\in L^{p}(\mu)$ and $b\in L^{q^{\prime}}(\mu)$ respectively for the weak-* topology. Fix $i\in\mathbb{N}$. For all $j\geq i$, we have

[TABLE]

so that we get

[TABLE]

This equality holds for any $i$ hence

[TABLE]

Now fix $j\in\mathbb{N}$. For all $i>j$ we have

[TABLE]

From this, we deduce as above that

[TABLE]

We obtained a contradiction so $T_{m}$ cannot be bounded. ∎

As a consequence, we have, by Proposition 2.3 :

Corollary 4.7**.**

Let $1\leq q\leq p\leq+\infty,p\neq 1,q\neq+\infty$. Let $\Omega_{1}=\Omega_{2}=\mathbb{R}$ with the Lebesgue measure. Then $\phi\in L^{\infty}(\mathbb{R}^{2})$ defined by

[TABLE]

is not a Schur multiplier on $\mathcal{B}(L^{p}(\mathbb{R}),L^{q}(\mathbb{R}))$.

Remark 4.8*.*

One could wonder whether the results of subsection 4.1 can be extended to the case $1\leq p<q\leq+\infty$, that is, if the boundedness of $T_{\phi}$ on $\mathcal{B}(L^{p},L^{q})$ implies that $u_{\phi}$ has a certain factorization. The fact that if $p<q$ the main triangle projection is bounded tells us that $m$ is a Schur multiplier on $\mathcal{B}(\ell_{p},\ell_{q})$. Nevertheless, the argument used in the previous proof shows that $m$ cannot have a factorization like in $(\ref{factotriangle})$. Therefore, the case $p<q$ is more tricky. For the discrete case, one can find in [3, Theorem 4.3] a necessary and sufficient condition for a family $(m_{i,j})\subset\mathbb{C}$ to be a Schur multiplier, for all values of $p$ and $q$, using the theory of $q-$absolutely summing operators.

5. Inclusion theorems

In this section, we denote by $\mathcal{M}(p,q)$ the space of Schur multipliers on $\mathcal{B}(\ell_{p},\ell_{q})$.

First, we recall the inclusion relationships between the spaces $\mathcal{M}(p,q)$. Then we will establish new results as applications of those obtained in Section 4.1.

Theorem 5.1**.**

[3, Theorem 6.1]** Let $p_{1}\geq p_{2}$ and $q_{1}\leq q_{2}$ be given. Then $\mathcal{M}(p_{1},q_{1})\subset\mathcal{M}(p_{2},q_{2})$ with equality in the following cases:

1. (i)

$p_{1}=p_{2}=1$, 2. (ii)

$q_{1}=q_{2}=\infty$, 3. (iii)

$q_{2}\leq 2\leq p_{2}$, 4. (iv)

$q_{2}<p_{1}=p_{2}<2$, 5. (v)

$2<q_{1}=q_{2}<p_{2}$.

Let $(\Omega_{1},\mu_{1})$ and $(\Omega_{2},\mu_{2})$ be two measure spaces. If $\mathcal{M}(p_{1},q_{1})\subset\mathcal{M}(p_{2},q_{2})$, then using Proposition 2.3 we have that any Schur multiplier on $\mathcal{B}(L^{p_{1}}(\Omega_{1}),L^{q_{1}}(\Omega_{2}))$ is a Schur multiplier on $\mathcal{B}(L^{p_{2}}(\Omega_{1}),L^{q_{2}}(\Omega_{2}))$. Hence, the results in the previous theorem hold true for all the Schur multipliers on $\mathcal{B}(L^{p},L^{q})$.

In the sequel, we will need the notion of type for a Banach space $X$, for which we refer e.g. to [1]. Let $(\mathcal{E}_{i})_{i\in\mathbb{N}}$ be a sequence of independent Rademacher random variables. We have the following definition.

Definition 5.2**.**

A Banach space X is said to have Rademacher type p (in short, type p) for some $1\leq p\leq 2$ if there is a constant $C$ such that for every finite set of vectors $(x_{i})_{i=n}^{n}$ in $X$,

[TABLE]

The smallest constant $C$ for which $(\ref{type})$ holds is called the type-p constant of $X$.

We will use the fact that for $1\leq p\leq 2$, $L^{p}$-spaces have type $p$ and if $2<p<+\infty$, $L^{p}$-spaces have type 2 and that those are the best types for infinite dimensional $L^{p}$-spaces (see for instance [1, Theorem 6.2.14]). We will also use the fact that the type is stable by passing to quotients. Namely, if $X$ has type $p$ and $E\subset X$ is a closed subspace, then $X/E$ has type $p$.

Proposition 5.3**.**

$(i)$* If  $1\leq q<p\leq 2$, then*

[TABLE]

Consequently, for any $1\leq r\leq q$,

[TABLE]

$(ii)$* If  $2\leq p<q\leq r$, then*

[TABLE]

$(iii)$* If $1<q<2<p<+\infty$ or $1<p<2<q<+\infty$, then*

[TABLE]

To prove this proposition, we will need the following definitions and lemma.

Definition 5.4**.**

Let $X$ and $Y$ be Banach spaces. A map $s:X\rightarrow Y$ is a quotient map if $s$ is surjective and for all $y\in Y$ with $\|y\|<1$, there exists $x\in X$ such that $\|x\|<1$ and $s(x)=y$. This is equivalent to the fact that the injective map $\hat{s}:X/\ker(s)\rightarrow Y$ induced by $s$ is a surjective isometry.

Definition 5.5**.**

*Let $X$ and $Y$ be Banach spaces, $u\in\mathcal{B}(X,Y)$ and $1\leq p\leq\infty$. We say that $u\in SQ_{p}(X,Y)$ if there exists a closed subspace $Z$ of a quotient of a $L^{p}$-space and two operators $A\in\mathcal{B}(X,Z)$ and $B\in\mathcal{B}(Z,Y)$ such that $u=BA$.

Then $\|u\|_{SQ_{p}}=\inf\|A\|\|B\|$ defines a norm on $SQ_{p}(X,Y)$ and $(SQ_{p}(X,Y),\|.\|_{SQ_{p})}$ is a Banach space.*

Lemma 5.6**.**

Let $W,X,Y,Z$ be Banach spaces and let $u\in\mathcal{B}(X,Y),s\in\mathcal{B}(W,X),v\in\mathcal{B}(Y,Z)$ such that $s$ is a quotient map, $v$ is a linear isometry and $vus\in\Gamma_{p}(W,Z)$. Then $u\in SQ_{p}(X,Y)$.

Proof.

By assumption, there exist a $L^{p}$-space $U$ and two operators $a\in\mathcal{B}(W,U)$ and $b\in\mathcal{B}(U,Z)$ such that the following diagram commutes

[TABLE]

Since $v$ is an isometry, $V:=v(Y)\subset Z$ is isometrically isomorphic to $Y$. Let $\psi:Y\rightarrow V$ be the isometric isomorphism induced by $v$.

Set $F:=\left\{x\in U\ \text{such that}\ b(x)\in V\right\}.$ Since $vus=ba$, we have, for all $w\in W,v(us(w))=b(a(w))$, so that $a(w)\in F$. This implies that $a(W)\subset F$. We still denote by $a$ the mapping $a:W\rightarrow F$ and by $b$ the restriction of $b$ to $F$. Denote by $\hat{b}$ the mapping $\hat{b}=\psi^{-1}\circ b:F\rightarrow Y$. Then we have the following commutative diagram

[TABLE]

Now, set $E:=\overline{a(\ker(s))}$ and let $Q:F\rightarrow F/E$ be the canonical mapping. Clearly, $Q\circ a:W\rightarrow F/E$ vanishes on $\ker(s)$, so that we have a mapping

[TABLE]

induced by $Q\circ a$.

Since $s$ is a quotient map, we denote by $\widehat{s}$ the isometric isomorphism

[TABLE]

Define

[TABLE]

$\hat{b}$ vanishes on $E$ so that we have a mapping

[TABLE]

Finally, it is easy to check that $u=BA$, that is, we have the following commutative diagram

[TABLE]

which concludes the proof. ∎

Remark 5.7*.*

To prove Lemma 5.6, one can use a result of Kwapień characterizing elements of $SQ_{p}$, as follows : a Banach space $X$ is isomorphic to an $SQ_{q}$-space if and only if there exists a constant $K\geq 1$ such that for any $n\geq 1$, for any $n\times n$ matrix $[a_{ij}]$ and for any $x_{1},\ldots,x_{n}$ in $X$,

[TABLE]

However, the proof presented in this paper also works if we replace in the statement of the lemma $\Gamma_{p}$ (respectively $SQ_{p}$) by the space of operators that can be factorized by some Banach space $L$ (respectively by a subspace of a quotient of $L$).

Proof of Proposition 5.3.

$(i)$. Let $\Omega:=[0,1]$ and $\lambda$ be the Lebesgue measure on $\Omega$. Let $I_{q}:L^{q}(\lambda)\rightarrow L^{1}(\lambda)$ be the inclusion mapping. By the classical Banach space theory (see [1, Theorem 2.3.1] and [1, Theorem 2.5.7]) there exist a quotient map $\sigma:\ell_{1}\twoheadrightarrow L^{q}(\lambda)$ and an isometry $J:L^{1}(\lambda)\hookrightarrow\ell_{\infty}$. Let $\phi\in\ell_{\infty}(\mathbb{N}^{2})$ be such that

[TABLE]

(by $(\ref{LinfB})$ any continuous linear map $\ell_{1}\rightarrow\ell_{\infty}$ is a certain $u_{\phi}$ for $\phi\in L^{\infty}(\mathbb{N}\times\mathbb{N})$). We have the following factorization

[TABLE]

According to Theorem 4.3, $\phi\in\mathcal{M}(q,1)$.

Assume that $\phi\in\mathcal{M}(p,p)$. Then, again by Theorem 4.3, we have $u_{\phi}\in\Gamma_{p}(\ell_{1},\ell_{\infty})$ and therefore, by Lemma 5.6, there exist an $SQ_{p}$-space $X$ and two operators $\alpha\in\mathcal{B}(L^{q}(\lambda),X)$ and $\beta\in\mathcal{B}(X,L^{1}(\lambda))$ such that $I_{q}=\beta\alpha$.

Let $(\mathcal{E}_{i})_{i\in\mathbb{N}}$ be a sequence of independant Rademacher random variables. Let $n\in\mathbb{N}^{*}$ and $f_{1},\ldots,f_{n}\in L^{q}(\lambda)$.

[TABLE]

But $X$ has type $p$ so there exists a constant $C_{1}>0$ such that

[TABLE]

By Khintchine inequality, there exists $C_{2}>0$ such that

[TABLE]

Thus, setting $K:=C_{1}C_{2}\|\alpha\|\|\beta\|$, we obtained the inequality

[TABLE]

Let $E_{1},\ldots,E_{n}$ be disjoint measurable subsets of $[0,1]$ such that for all $1\leq j\leq n,\lambda(E_{j})=\dfrac{1}{n}$. Set $f_{j}:=\chi_{E_{j}}$. Then

[TABLE]

Hence, applying the previous inequality to the $f_{j}$’s, we obtain

[TABLE]

Since $q<p$, this inequality can’t hold for all $n$, so we obtained a contradiction.

Finally, notice that if $1\leq r\leq q$, then by Theorem 5.1, $\mathcal{M}(q,1)\subset\mathcal{M}(q,r)$. Thus, $\mathcal{M}(q,r)\nsubseteq\mathcal{M}(p,p).$

$(ii)$. By Proposition 2.3 and using duality, it is easy to prove that for all $s,t\in[1,\infty],\phi$ is a Schur multiplier on $\mathcal{B}(\ell_{s},\ell_{t})$ if and only if $\tilde{\phi}$ is a Schur multiplier on $\mathcal{B}(\ell_{t^{\prime}},\ell_{s^{\prime}})$, where $\tilde{\phi}$ is defined for all $i,j\in\mathbb{N}$ by $\tilde{\phi}(i,j)=\phi(j,i)$.

Let $2\leq p<q\leq r$. Then $1\leq r^{\prime}\leq q^{\prime}<p^{\prime}\leq 2$. If we assume that $\mathcal{M}(r,q)\subset\mathcal{M}(p,p)$ then the latter implies $\mathcal{M}(q^{\prime},r^{\prime})\subset\mathcal{M}(p^{\prime},p^{\prime})$, which is, by $(i)$, a contradiction. This proves $(ii)$.

$(iii)$. By duality, it is enough to consider the case $1<q<2<p<+\infty$. Assume that $\mathcal{M}(q,q)\subset\mathcal{M}(p,p)$. Using the notations introduced in the proof of $(i)$, let $\sigma:\ell_{1}\rightarrow\ell_{q}$ be a quotient map and $J:\ell_{q}\rightarrow\ell_{\infty}$ be an isometry. Let $\phi\in L^{\infty}(\mathbb{N}\times\mathbb{N})$ be such that

[TABLE]

where $I_{\ell_{q}}:\ell_{q}\rightarrow\ell_{q}$ is the identity map. Then $\phi\in\mathcal{M}(q,q)$. By assumption, $\phi\in\mathcal{M}(p,p)$. By Lemma 5.6, this implies that $I_{\ell_{q}}\in SQ_{p}(\ell_{q},\ell_{q})$. Clearly, this implies that $\ell_{q}$ is isomorphic to an $SQ_{p}$-space. But $\ell_{q}$ does not have type $2$ and any $SQ_{p}$ has type $2$. This is a contradiction, so $\mathcal{M}(q,q)\nsubseteq\mathcal{M}(p,p)$. ∎

Theorem 5.8**.**

We have $\mathcal{M}(q,q)\subset\mathcal{M}(p,p)$ if and only if $1\leq p\leq q\leq 2$ or $2\leq q\leq p\leq+\infty$.

Proof.

By Proposition 5.3 and duality, we only have to show that when $1\leq p\leq q\leq 2$, $\mathcal{M}(q,q)\subset\mathcal{M}(p,p)$.

We saw in the proof Proposition of 5.3 $(iii)$ that if $\mathcal{M}(q,q)\subset\mathcal{M}(p,p)$ then $\ell_{q}$ is isomorphic to an $SQ_{p}$-space. The converse holds true. Indeed, assume that $\ell_{q}$ is isomorphic to an $SQ_{p}$-space. Then by approximation, any $L^{q}$-space is isomorphic to an $SQ_{p}$-space. Hence any element of $\Gamma_{q}(\ell_{1},\ell_{\infty})$ factors through an $SQ_{p}$-space. By the lifting property of $\ell_{1}$ and the extension property of $\ell_{\infty}$, this implies that any element of $\Gamma_{q}(\ell_{1},\ell_{\infty})$ factors through an $L^{p}$-space, that is $\Gamma_{q}(\ell_{1},\ell_{\infty})\subset\Gamma_{p}(\ell_{1},\ell_{\infty})$. By Corollary 4.5, this implies that $\mathcal{M}(q,q)\subset\mathcal{M}(p,p)$.

Assume that $1\leq p\leq q\leq 2$. By [1, Theorem 6.4.19], there exists an isometry from $\ell_{q}$ into an $L^{p}$-space, obtained by using $q-$stable processes. Hence, $\ell_{q}$ is an $SQ_{p}$-space. This concludes the proof.

∎

Problem 5.9*.*

Compare the other spaces of Schur multipliers. For example, if $1<p\leq 2$, do we have

[TABLE]

Acknowledgements. The author was supported by the French “Investissements d’Avenir” program, project ISITE-BFC (contract ANR-15-IDEX-03).