$\ell^1$-contractive maps on noncommutative $L^p$-spaces

TL;DR

This paper characterizes $ ext{l}^1$-contractive maps on noncommutative $L^p$-spaces, linking them to isometries and positivity conditions, and extends classical factorization results to this setting.

Contribution

It establishes that $ ext{l}^1$-contractivity characterizes certain $L^p$-isometries and shows that positivity or separating properties imply $ ext{l}^1$-contractivity.

Findings

Yeadon's factorization theorem applies to $L^2$-isometries iff they are $ ext{l}^1$-contractive.

Contractive operators that are 2-positive or separating are automatically $ ext{l}^1$-contractive.

The work extends classical isometry characterizations to the noncommutative $L^p$-space setting.

Abstract

Let $T\colon L^p({\mathcal M})\to L^p({\mathcal N})$ be a bounded operator between two noncommutative $L^p$-spaces, $1\leq p<\infty$. We say that $T$ is $\ell^1$-bounded (resp. $\ell^1$-contractive) if $T\otimes I_{\ell^1}$ extends to a bounded (resp. contractive) map from $L^p({\mathcal M};\ell^1)$ into $L^p({\mathcal N};\ell^1)$. We show that Yeadon's factorization theorem for $L^p$-isometries, $1\leq p\not=2 <\infty$, applies to an isometry $T\colon L^2({\mathcal M})\to L^2({\mathcal N})$ if and only if $T$ is $\ell^1$-contractive. We also show that a contractive operator $T\colon L^p({\mathcal M})\to L^p({\mathcal N})$ is automatically $\ell^1$-contractive if it satisfies one of the following two conditions: either $T$ is $2$-positive; or $T$ is separating, that is, for any disjoint $a,b\in L^p({\mathcal M})$ (i.e. $a^*b=ab^*=0)$, the images $T(a),T(b)$ are disjoint as well.

Equations

$$T(a)=wBJ(a),$$

$$T\overline{\otimes}I_{\ell^{1}}\colon L^{p}(\mbox{${\mathcal{M}}$};\ell^{1})\longrightarrow L^{p}(\mbox{${\mathcal{N}}$};\ell^{1}).$$

$$L^{p}(\mathcal{M}):=\bigl{\{}a\in L^{0}(\mathcal{M}):\tau_{\tiny\mbox{${\mathcal{M}}$}}(\lvert a\rvert^{p})<\infty\bigr{\}}.$$

$$\Vert ab{\Vert }_r\le \Vert a{\Vert }_p\Vert b{\Vert }_q.$$

$$\langle a,b\rangle=\tau_{\tiny{\mbox{${\mathcal{M}}$}}}(ab),\qquad a\in L^{p}(\mathcal{M}),\ b\in L^{p^{\prime}}(\mathcal{M}),$$

$$\left(\begin{array}{ccccc}{b}_1& 0& \dots & 0& \dots \\ ⋮& ⋮& & ⋮& \\ b_n& 0& \dots & 0& \dots \\ ⋮& ⋮& & ⋮& \end{array}\right)$$

$$\Bigl{\|}\sum_{n=1}^{\infty}E_{n1}\otimes b_{n}\Bigr{\|}_{L^{p}(\tiny{\mbox{${\mathcal{M}}$}};\ell^{2}_{c})}=\Bigl{\|}\sum_{n=1}^{\infty}b_{n}^{*}b_{n}\Bigr{\|}_{\frac{p}{2}}^{\frac{1}{2}}$$

$$\Bigl{\|}\sum_{n=1}^{\infty}E_{1n}\otimes a_{n}\Bigr{\|}_{L^{p}(\tiny{\mbox{${\mathcal{M}}$}};\ell^{2}_{r})}=\Bigl{\|}\sum_{n=1}^{\infty}a_{n}a_{n}^{*}\Bigr{\|}_{\frac{p}{2}}^{\frac{1}{2}}.$$

$$\Bigl{\|}\sum_{n=1}^{N}E_{n1}\otimes b_{n}\Bigr{\|}_{L^{p}(\tiny{\mbox{${\mathcal{M}}$}};\ell^{2}_{c})}\leq K.$$

$$\Bigl{\|}\sum_{k,n=1}^{N}u_{kn}u_{kn}^{*}\Bigr{\|}_{p}\leq K\qquad\hbox{and}\qquad\Bigl{\|}\sum_{k,n=1}^{N}v^{*}_{kn}v_{kn}\Bigr{\|}_{p}\leq K$$

$$\bigl{\|}(x_{n})_{n\geq 1}\bigr{\|}_{L^{p}(\mathcal{M};\ell^{1})}=\inf\left\{\sup_{N}\Bigl{\|}\sum_{k,n=1}^{N}u_{kn}u_{kn}^{*}\Bigr{\|}_{p}^{\frac{1}{2}}\sup_{N}\Bigl{\|}\sum_{k,n=1}^{N}v^{*}_{kn}v_{kn}\Bigr{\|}_{p}^{\frac{1}{2}}\right\},$$

$$\Bigl{\|}\sum_{n=1}^{\infty}a_{n}a_{n}^{*}\Bigr{\|}_{p}<1\qquad\hbox{and}\qquad\Bigl{\|}\sum_{n=1}^{\infty}b_{n}^{*}b_{n}\Bigr{\|}_{p}<1.$$

$$x_n=\sum _{k=1}^{\infty }{u}_{kn}{v}_{kn},\text{for all}n\ge 1.$$

$$u_n:=(u_{kn}{)}_{k\ge 1}\in L^{2p}(\mathcal{M};{\ell }_r^2)and{v}_n:=(v_{kn}{)}_{k\ge 1}\in L^{2p}(\mathcal{M};{\ell }_c^2)$$

$$\Vert {\phi }_n{\Vert }_{\infty }\le 1,\Vert {\psi }_n{\Vert }_{\infty }\le 1,u_n=|u_n^{\ast }|{\phi }_{n}and{v}_n={\psi }_n|v_n|.$$

$$x_n=u_n{v}_n=|u_n^{\ast }|{\phi }_n{\psi }_n|v_n|=a_n{b}_n$$

$$a_n{a}_n^{\ast }=|u_n^{\ast }{|}^2=u_n{u}_n^{\ast }=\sum _{k=1}^{\infty }{u}_{kn}{u}_{kn}^{\ast },$$

$$b_n^{\ast }{b}_n$$

$$\le |v_n{|}^2=v_n^{\ast }{v}_n=\sum _{k=1}^{\infty }{v}_{kn}^{\ast }{v}_{kn},$$

$$\bigl{\|}(x_{n})_{n\geq 1}\bigr{\|}_{L^{p}(\tiny\mbox{${\mathcal{M}}$};\ell^{1})}\,=\,\Bigl{\|}\sum_{n=1}^{\infty}x_{n}\Bigr{\|}_{p}.$$

$$\Bigl{\|}\sum_{n=1}^{\infty}a_{n}b_{n}\Bigr{\|}_{p}\leq\Bigl{\|}\sum_{n=1}^{\infty}a_{n}a_{n}^{*}\Bigr{\|}_{p}^{\frac{1}{2}}\Bigl{\|}\sum_{n=1}^{\infty}b_{n}^{*}b_{n}\Bigr{\|}_{p}^{\frac{1}{2}}.$$

$$L^p(\mathcal{M})\otimes {\ell }^1\subset L^p(\mathcal{M};{\ell }^1)$$

$$T\overline{\otimes }{I}_{{\ell }^1}:L^p(\mathcal{M};{\ell }^1)⟶L^p(\mathcal{N};{\ell }^1).$$

$$\|\left(T(x),T(y)\right)\|_{L^{p}(\tiny{\mbox{${\mathcal{N}}$}};\ell^{1}_{2})}\leq\|(x,y)\|_{L^{p}\left(\tiny{\mbox{${\mathcal{M}}$}};\ell^{1}_{2}\right)}.$$

$$L^p(M;{\ell }^1)\simeq L^p(M)[\mathrm{Max}({\ell }^1)]isometrically,$$

$$\bigl{\|}T\otimes I_{M_{n}}\colon L^{p}(\mathcal{M})[M_{n}]\longrightarrow L^{p}(\mathcal{N})[M_{n}]\bigr{\|}\,\leq K.$$

$$\bigl{\|}T\overline{\otimes}I_{E}\colon L^{p}(\mathcal{M})[E]\longrightarrow L^{p}(\mathcal{N})[E]\bigr{\|}\,\leq\|T\|_{reg}.$$

$$\bigl{\|}({}^{t}a_{n})_{n}\bigr{\|}_{L^{2p}(\tiny{\mbox{${\mathcal{M}}$}};\ell^{2}_{c})}=\bigl{\|}(a_{n})_{n}\bigr{\|}_{L^{2p}(\tiny{\mbox{${\mathcal{M}}$}};\ell^{2}_{r})}\quad\hbox{and}\quad\bigl{\|}({}^{t}b_{n})_{n}\bigr{\|}_{L^{2p}(\tiny{\mbox{${\mathcal{M}}$}};\ell^{2}_{r})}=\bigl{\|}(b_{n})_{n}\bigr{\|}_{L^{2p}(\tiny{\mbox{${\mathcal{M}}$}};\ell^{2}_{c})}.$$

$$S^{p}[E]\,\simeq\,R\bigl{(}\tfrac{1}{p}\bigr{)}\otimes_{h}E\otimes_{h}R\bigl{(}1-\tfrac{1}{p}\bigr{)}.$$

$$\bigl{(}T\otimes I_{R}\bigr{)}\Bigl{(}\sum_{k=1}^{n}e_{k}\otimes e_{k}\otimes e_{1}\Bigr{)}\,=\,\sum_{k=1}^{n}e_{1}\otimes e_{k}\otimes e_{k},$$

$$\Bigl{\|}\sum_{k=1}^{n}e_{k}\otimes e_{k}\Bigr{\|}_{R\otimes_{h}R\bigl{(}1-\tfrac{1}{p}\bigr{)}}\,\leq K\,\Bigl{\|}\sum_{k=1}^{n}e_{k}\otimes e_{k}\Bigr{\|}_{R\bigl{(}\frac{1}{p}\bigr{)}\otimes_{h}R}.$$

Full text

1. Introduction

Let ${\mathcal{M}}$ and ${\mathcal{N}}$ be two semifinite von Neumann algebras. For any $1\leq p<\infty$, consider the associated noncommutative $L^{p}$-spaces L^{p}(\mbox{{\mathcal{M}}}) and L^{p}(\mbox{{\mathcal{N}}}). A remarkable theorem of Yeadon [26] (see Theorem 3.1 below) asserts that if $p\not=2$ and T\colon L^{p}(\mbox{{\mathcal{M}}})\to L^{p}(\mbox{{\mathcal{N}}}) is a linear isometry, then there exist a normal Jordan homomorphism J\colon\mbox{{\mathcal{M}}}\to\mbox{{\mathcal{N}}}, a positive operator $B$ affiliated with ${\mathcal{N}}$ and a partial isometry w\in\mbox{{\mathcal{N}}} such that $w^{*}wB=B$, $J(a)$ commutes with $B$ for all a\in\mbox{{\mathcal{M}}}, and

[TABLE]

for all a\in\mbox{{\mathcal{M}}}\cap L^{p}(\mbox{{\mathcal{M}}}).

This striking factorization property is the noncommutative version of the celebrated description of isometries on classical (=commutative) $L^{p}$-spaces due to Banach [1] and Lamperti [13]. We refer to the books [3] and [4] for details on these results, complements and historical background.

The work presented in this paper was originally motivated by the following question, concerning the case $p=2$: what are the linear isometries T\colon L^{2}(\mbox{{\mathcal{M}}})\to L^{2}(\mbox{{\mathcal{N}}}) which admit a Yeadon type factorization, that is, isometries for which there exist $J,B,w$ as above such that (1) holds true for any a\in\mbox{{\mathcal{M}}}\cap L^{2}(\mbox{{\mathcal{M}}})?

This issue leads us to introduce a new property, called $\ell^{1}$-boundedness, which is defined as follows. Consider the $\ell^{1}$-valued noncommutative $L^{p}$-space L^{p}(\mbox{{\mathcal{M}}};\ell^{1}) introduced by Junge [7] (see also [20] and [9]). Let T\colon L^{p}(\mbox{{\mathcal{M}}})\to L^{p}(\mbox{{\mathcal{N}}}) be a bounded operator. We say that $T$ is $\ell^{1}$-bounded if $T\otimes I_{\ell^{1}}$ extends to a bounded map

[TABLE]

We further say that $T$ is $\ell^{1}$-contractive if the map $T\overline{\otimes}I_{\ell^{1}}$ is a contraction. The main result of this paper (Theorem 4.2 below) is that an isometry T\colon L^{2}(\mbox{{\mathcal{M}}})\to L^{2}(\mbox{{\mathcal{N}}}) is $\ell^{1}$-contractive if and only if it admits a Yeadon type factorization.

To explain the relevance of Theorem 4.2 we note that $\ell^{1}$-boundedness is a noncommutative analogue of regularity for maps acting on commutative $L^{p}$-spaces. (We refer to [18, Chapter 1] for definitions and background on regular maps.) It follows that Theorem 4.2 is a noncommutative extension of the well-known result stating that a linear isometry between commutative $L^{2}$-spaces is a Lamperti operator if and only if it is contratively regular, if and only if it is a subpositive contraction (see e.g [15]).

The proof of Yeadon’s theorem heavily relies on the fact that for $p\not=2$, any linear isometry T\colon L^{p}(\mbox{{\mathcal{M}}})\to L^{p}(\mbox{{\mathcal{N}}}) has the following property: if $a,b\in L^{p}(M)$ are disjoint, that is $a^{*}b=ab^{*}=0$, then $T(a)$ and $T(b)$ are disjoint as well. Such maps are called separating in the present paper. We show that a bounded operator L^{p}(\mbox{{\mathcal{M}}})\to L^{p}(\mbox{{\mathcal{N}}}) is separating if and only if it admits a Yeadon type factorization.

The concept of $\ell^{1}$-boundedness is interesting in its own sake and this paper aims at studying some of its main features. We show in particular that a contractive operator T\colon L^{p}(\mbox{{\mathcal{M}}})\to L^{p}(\mbox{{\mathcal{N}}}) is automatically $\ell^{1}$-contractive either if $T$ is separating (see Theorem 3.15) or if $T$ is $2$-positive (see Proposition 5.1).

2. Notion of $\ell^{1}$-boundedness and background

In this section, we provide some background on noncommutative $L^{p}$-spaces and on the $\ell^{1}$-valued spaces L^{p}(\mbox{{\mathcal{M}}};\ell^{1}). Then we introduce the notions of $\ell^{1}$-boundedness and $\ell^{1}$-contractivity and establish some preliminary results.

Let $\mathcal{M}$ be a semifinite von Neumann algebra equipped with a normal semifinite faithful trace \tau_{\tiny{\mbox{{\mathcal{M}}}}}. We briefly recall the noncommutative $L^{p}$-spaces $L^{p}(\mathcal{M})$, $0<p\leq\infty$, associated with (\mathcal{M},\tau_{\tiny{\mbox{{\mathcal{M}}}}}) and some of their basic properties. The reader is referred to the survey [16] and references therein for details and further properties.

If $\mathcal{M}\subset B(\mathcal{H})$ acts on some Hilbert space $\mathcal{H}$, the elements of $L^{p}(\mathcal{M})$ can be viewed as closed densely defined (possibly unbounded) operators on $\mathcal{H}$. More precisely, let $\mathcal{M}^{\prime}$ denote the commutant of $\mathcal{M}$ in $B(\mathcal{H})$. A closed densely defined operator $a$ is said to be affiliated with $\mathcal{M}$ if $a$ commutes with every unitary of $\mathcal{M}^{\prime}$. An affiliated operator $a$ is called measurable (with respect to (\mathcal{M},\tau_{\tiny{\mbox{{\mathcal{M}}}}})) if there is a positive number $\lambda>0$ such that \tau_{\tiny{\mbox{{\mathcal{M}}}}}(\varepsilon_{\lambda})<\infty, where $\varepsilon_{\lambda}=\chi_{[\lambda,\infty)}(\lvert a\rvert)$ is the projection associated with the indicator function of $[\lambda,\infty)$ in the Borel functional calculus of $\lvert a\rvert$. Then the set $L^{0}(\mathcal{M})$ of all measurable operators forms a $*$-algebra (see e.g. [24, Chapter I] for a proof and also for the definitions of algebraic operations on $L^{0}(\mathcal{M})$). We proceed with defining $L^{p}(\mathcal{M})$ as a subspace of $L^{0}(\mathcal{M})$. First note that for any $a\in L^{0}(\mathcal{M})$ and any $0<p<\infty$, the operator $\lvert a\rvert^{p}=(a^{*}a)^{\frac{p}{2}}$ belongs to $L^{0}(\mathcal{M})$. If $L^{0}(\mathcal{M})^{+}$ denotes the positive cone of $L^{0}(\mathcal{M})$, that is the set of all positive operators in $L^{0}(\mathcal{M})$, the trace \tau_{\tiny{\mbox{{\mathcal{M}}}}} extends to a positive tracial functional on $L^{0}(\mathcal{M})^{+}$, taking values in $[0,\infty]$, also denoted by \tau_{\tiny{\mbox{{\mathcal{M}}}}}. For any $0<p<\infty$, the noncommutative $L^{p}$-space, $L^{p}(\mathcal{M})$, associated with (\mathcal{M},\tau_{\tiny{\mbox{{\mathcal{M}}}}}), is

[TABLE]

For $a\in L^{p}(\mathcal{M})$, let \|a\|_{p}:=\tau_{\tiny\mbox{{\mathcal{M}}}}(\left\lvert a\rvert^{p}\right)^{\frac{1}{p}}. For $1\leq p<\infty$, $\|\cdot\|_{p}$ defines a complete norm, and for $p<1$, a complete $p$-norm. We let $L^{\infty}(\mathcal{M}):=\mathcal{M}$, equipped with its operator norm $\|\,\cdotp\|_{\infty}$.

For any $0<p\leq\infty$ and any $a\in L^{p}(\mathcal{M})$, the adjoint operator $a^{*}$ belongs to $L^{p}(\mathcal{M})$ and $\|a^{*}\|_{p}=\|a\|_{p}$. Furthermore, we have that $a^{*}a\in L^{\frac{p}{2}}(\mathcal{M})$ and $\lvert a\rvert\in L^{p}(\mathcal{M})$, with $\|\lvert a\rvert\|_{p}=\|a\|_{p}$. More generally, for any $0<p,q,r\leq\infty$ with $\frac{1}{p}+\frac{1}{q}=\frac{1}{r}$, we have that ab\in L^{r}(\mbox{{\mathcal{M}}}) if a\in L^{p}(\mbox{{\mathcal{M}}}) and b\in L^{q}(\mbox{{\mathcal{M}}}), with Hölder’s inequality

[TABLE]

For any $1\leq p<\infty$, let $p^{\prime}:=\frac{p}{p-1}$ be the conjugate number of $p$. Then by (2), $ab$ belongs to L^{1}(\mbox{{\mathcal{M}}}) for any $a\in L^{p}(\mathcal{M})$ and $b\in L^{p^{\prime}}(\mathcal{M})$. Further the duality pairing

[TABLE]

yields an isometric isomorphism $L^{p}(\mathcal{M})^{*}=L^{p^{\prime}}(\mathcal{M})$. In particular, we may identify $L^{1}(\mathcal{M})$ with the (unique) predual $\mathcal{M}_{*}$ of $\mathcal{M}$. These duality results will be used without further reference in the paper.

We let $L^{p}(\mathcal{M})^{+}:=L^{0}(\mathcal{M})^{+}\cap L^{p}(\mathcal{M})$ denote the positive cone of $L^{p}(\mathcal{M})$. A bounded operator $T\colon L^{p}(\mathcal{M})\to L^{p}(\mathcal{N})$ between two noncommutative $L^{p}$-spaces is called positive if it maps $L^{p}(\mathcal{M})^{+}$ into L^{p}(\mathcal{\mbox{{\mathcal{N}}}})^{+} .

If $\mathcal{M}=B(\mathcal{H})$, the algebra of all bounded operators on $\mathcal{H}$, and \tau_{\tiny\mbox{{\mathcal{M}}}}=tr, the usual trace on $B(\mathcal{H})$, then the associated noncommutative $L^{p}$-space is the Schatten class $S^{p}(\mathcal{H})$. If \mbox{{\mathcal{M}}}=L^{\infty}(\Omega,\mbox{{\mathcal{F}}},\mu) is the commutative von Neumann algebra associated with a measure space (\Omega,\mbox{{\mathcal{F}}},\mu), then L^{p}(\mbox{{\mathcal{M}}}) coincides with the classical $L^{p}$-space L^{p}(\Omega,\mbox{{\mathcal{F}}},\mu).

Let $tr$ denote the usual trace on $B(\ell^{2})$ and consider the von Neumann algebra tensor product B(\ell^{2})\overline{\otimes}\mbox{{\mathcal{M}}}, equipped with the normal semifinite faithful trace tr\overline{\otimes}\tau_{\tiny\mbox{{\mathcal{M}}}} (see [23, Chapter V, Proposition 2.14]). Any element of L^{p}(B(\ell^{2})\overline{\otimes}\mbox{{\mathcal{M}}}) can be regarded as an infinite matrix $(b_{ij})_{i,j\geq 1}$, with $b_{ij}\in L^{p}(\mathcal{M})$. We let L^{p}(\mbox{{\mathcal{M}}};\ell^{2}_{c}) denote the subspace of L^{p}(B(\ell^{2})\overline{\otimes}\mbox{{\mathcal{M}}}) consisting of all matrices whose entries off the first column are all zero. We regard this space as a sequence space by saying that a sequence $(b_{n})_{n\geq 1}$ of L^{p}(\mbox{{\mathcal{M}}}) belongs to L^{p}(\mbox{{\mathcal{M}}};\ell^{2}_{c}) if the infinite matrix

[TABLE]

represents an element of L^{p}(B(\ell^{2})\overline{\otimes}\mbox{{\mathcal{M}}}). Similarly, we define L^{p}(\mbox{{\mathcal{M}}};\ell^{2}_{r}) as the subspace of L^{p}(B(\ell^{2})\overline{\otimes}\mbox{{\mathcal{M}}}) consisting of all matrices whose entries off the first row are all zero.

We let $E_{ij}$, $i,j\geq 1$, denote the usual matrix units of $B(\ell^{2})$, and regard S^{p}(\ell^{2})\otimes L^{p}(\mbox{{\mathcal{M}}}) as a subspace of L^{p}(B(\ell^{2})\overline{\otimes}\mbox{{\mathcal{M}}}) in the usual way. For any finitely supported sequence $(a_{n})_{n\geq 1}$ and $(b_{n})_{n\geq 1}$ of L^{p}(\mbox{{\mathcal{M}}}), we have

[TABLE]

and

[TABLE]

When $1\leq p<\infty$, elements of L^{p}(\mbox{{\mathcal{M}}};\ell^{2}_{c}) and L^{p}(\mbox{{\mathcal{M}}};\ell^{2}_{r}) can be approximated by finitely supported sequences, thanks to the following (easy) result.

Lemma 2.1**.**

Let $1\leq p<\infty$ and suppose that $(b_{n})_{n\geq 1}$ is a sequence in $L^{p}(\mathcal{M})$. The following are equivalent:

• (i)

$(b_{n})_{n\geq 1}$* belongs to $L^{p}(\mathcal{M};\ell_{c}^{2})$.*

• (ii)

There exists a positive constant $K$ such that for every $N\geq 1$,

[TABLE]

• (iii)

The series $\sum_{n\geq 1}E_{n1}\otimes b_{n}$ converges in $L^{p}(\mathcal{M};\ell^{2}_{c})$.

Moreover, the same result holds with $\ell^{2}_{c}$ replaced by $\ell^{2}_{r}$ and $E_{n1}$ replaced by $E_{1n}$.

Remark 2.2**.**

By (3) and the Cauchy convergence test, we see that a sequence $(b_{n})_{n\geq 1}$ in $L^{p}(\mathcal{M})$ satisfies the assertion (iii) of Lemma 2.1 if and only if the series $\sum_{n}b_{n}^{*}b_{n}$ converges in L^{\frac{p}{2}}(\mbox{{\mathcal{M}}}). In this case, the identity (3) holds true for $(b_{n})_{n\geq 1}$.

Let $1\leq p<\infty$. In [7], Junge defined $L^{p}(\mathcal{M};\ell^{1})$ as the space of all sequences $x=(x_{n})_{n\geq 1}$ in $L^{p}(\mathcal{M})$ for which there exist families $(u_{kn})_{k,n\geq 1}$, $(v_{kn})_{k,n\geq 1}\in L^{2p}(\mathcal{M})$ and a positive constant $K$ such that

[TABLE]

for any $N\geq 1$, and $\sum_{k=0}^{\infty}u_{kn}v_{kn}=x_{n}$, for all $n\geq 1$. (The convergence of the series is ensured by (5) and Lemma 2.1.) He showed that this a Banach space when equipped with the norm

[TABLE]

where the infimum is taken over all families $(u_{kn})_{k,n\geq 1}$ and $(v_{kn})_{k,n\geq 1}$ as above.

The following alternative description is well-known to specialists (and implicit in [17, pp. 537-538]). We give a proof for the sake of completeness.

Lemma 2.3**.**

Suppose that $1\leq p<\infty$ and that $(x_{n})_{n\geq 1}$ is a sequence in $L^{p}(\mathcal{M})$. Then the following are equivalent:

• (i)

$(x_{n})_{n\geq 1}$* belongs to $L^{p}(\mathcal{M};\ell^{1})$ and $\|(x_{n})_{n\geq 1}\|_{L^{p}(\mathcal{M};\ell^{1})}<1$.*

• (ii)

There exist sequences $(a_{n})_{n\geq 1}$ and $(b_{n})_{n\geq 1}$ in $L^{2p}(\mathcal{M})$ such that $a_{n}b_{n}=x_{n}$ for all $n\geq 1$, the series $\sum_{n}a_{n}a_{n}^{*}$ and $\sum_{n}b_{n}^{*}b_{n}$ converge in L^{p}(\mbox{{\mathcal{M}}}), and we have

[TABLE]

Proof.

The assertion “(ii) $\Rightarrow$ (i)” is obvious. Conversely assume (i) and consider $u_{kn},v_{kn}$ in L^{2p}(\mbox{{\mathcal{M}}}) satisfying (5) for some $K<1$, and such that

[TABLE]

We regard L^{p}(\mbox{{\mathcal{M}}}) as a subspace of L^{p}(B(\ell^{2})\overline{\otimes}\mbox{{\mathcal{M}}}) by identifying any b\in L^{p}(\mbox{{\mathcal{M}}}) with $E_{11}\otimes b$. We set

[TABLE]

for all $n\geq 1$, that we regard as elements of L^{2p}(B(\ell^{2})\overline{\otimes}\mbox{{\mathcal{M}}}). Then $x_{n}=u_{n}v_{n}$ for all $n\geq 1$.

By polar decomposition, there exist $\varphi_{n}\in L^{\infty}(\mathcal{M};\ell^{2}_{r})$ and $\psi_{n}\in L^{\infty}(\mathcal{M};\ell^{2}_{c})$ such that

[TABLE]

If we let $a_{n}=\lvert u_{n}^{*}\rvert$ and $b_{n}=\varphi_{n}\psi_{n}\lvert v_{n}\rvert$, then we have

[TABLE]

for all $n\geq 1$. Further $a_{n},b_{n}$ belong to L^{2p}(\mbox{{\mathcal{M}}}). Next we have

[TABLE]

hence for any $N\geq 1$, $\|\sum_{n=1}^{N}a_{n}a_{n}^{*}\|_{p}\leq K$. By Lemma 2.1 and Remark 2.2, this implies the convergence of the series of the $a_{n}a_{n}^{*}$ in L^{p}(\mbox{{\mathcal{M}}}), with \bigl{\|}\sum_{n=1}^{\infty}a_{n}a_{n}^{*}\bigr{\|}_{p}\leq K. Likewise, since $(\varphi_{n}\psi_{n})^{*}(\varphi_{n}\psi_{n})\leq 1$, we have

[TABLE]

from which we deduce that the series of the $b_{n}^{*}b_{n}$ converges in L^{p}(\mbox{{\mathcal{M}}}), with \bigl{\|}\sum_{n=1}^{\infty}b_{n}^{*}b_{n}\bigr{\|}_{p}\leq K. This proves (ii). ∎

When dealing with positive sequences, the study of the L^{p}(\mbox{{\mathcal{M}}},\ell^{1})-norm is simple. We learnt the following result from [25].

Lemma 2.4**.**

Let $1\leq p<\infty$, let $(x_{n})_{n\geq 1}$ be a sequence of L^{p}(\mbox{{\mathcal{M}}}) and assume that $x_{n}\geq 0$ for any $n\geq 1$. The following are equivalent.

• (i)

$(x_{n})_{n\geq 1}$* belongs to L^{p}(\mbox{{\mathcal{M}}};\ell^{1}).*

• (ii)

The series $\sum_{n}x_{n}\,$ converges in L^{p}(\mbox{{\mathcal{M}}}).

Further in this case, we have

[TABLE]

Proof.

It follows from (3) and (4) that for any finitely supported families $(a_{n})_{n\geq 1}$ and $(b_{n})_{n\geq 1}$ in L^{2p}(\mbox{{\mathcal{M}}}), we have

[TABLE]

The assertion “$(i)\Rightarrow(ii)$” and the inequality $\geq$ in (6) follow at once (here we do not need any positivity assumption on the $x_{n}$).

Assume conversely that the series $\sum_{n}x_{n}\,$ converges in L^{p}(\mbox{{\mathcal{M}}}) and set $a_{n}=b_{n}=x_{n}^{\frac{1}{2}}$. Then the convergence of $\sum_{n}a_{n}a_{n}^{*}$ and $\sum_{n}b_{n}^{*}b_{n}$ are trivial and $x_{n}=a_{n}b_{n}$ for all $n\geq 1$. This implies (i), as well as the inequality $\leq$ in (6). ∎

We remark that for any $c=(c_{n})_{n\geq 1}\in\ell^{1}$ and any x\in L^{p}(\mbox{{\mathcal{M}}}), the sequence $(c_{n}x)_{n\geq 1}$ belongs to L^{p}(\mbox{{\mathcal{M}}};\ell^{1}). Further the mapping $x\otimes c\mapsto(c_{n}x)_{n\geq 1}$ extends to an embedding

[TABLE]

and with this convention, $L^{p}(\mathcal{M})\otimes\ell^{1}$ is a dense subspace of $L^{p}(\mathcal{M};\ell^{1})$.

Let $(e_{k})_{k\geq 1}$ denote the canonical basis of $\ell^{1}$. Then we let L^{p}(\mbox{{\mathcal{M}}};\ell^{1}_{2}) be the direct sum L^{p}(\mbox{{\mathcal{M}}})\oplus L^{p}(\mbox{{\mathcal{M}}}) equipped with the norm \|(x,y)\|=\|x\otimes e_{1}+y\otimes e_{2}\|_{L^{p}(\tiny{\mbox{{\mathcal{M}}}};\ell^{1})}, for any x,y\in L^{p}(\mbox{{\mathcal{M}}}).

Throughout the paper we will consider two semifinite von Neumann algebras ${\mathcal{M}}$ and ${\mathcal{N}}$ equipped with normal semifinite faithful traces \tau_{\tiny{\mbox{{\mathcal{M}}}}} and \tau_{\tiny{\mbox{{\mathcal{N}}}}}, respectively, and we will consider various bounded operators L^{p}(\mbox{{\mathcal{M}}})\to L^{p}(\mbox{{\mathcal{N}}}), for $1\leq p<\infty$.

Definition 2.5**.**

We say that a bounded operator $T\colon L^{p}(\mathcal{M})\to L^{p}(\mathcal{N})$ is

1. (i)

$\ell^{1}$-bounded if $T\otimes I_{\ell^{1}}$ extends to a bounded map

[TABLE]

In this case, the norm of $T\overline{\otimes}I_{\ell^{1}}$ is called the $\ell^{1}$-bounded norm of $T$ and is denoted by $\|T\|_{\ell^{1}}$; 2. (ii)

$\ell^{1}$-contractive if it is $\ell^{1}$-bounded and $\|T\|_{\ell^{1}}\leq 1$; 3. (iii)

$\ell^{1}_{2}$-contractive if for every x,y\in L^{p}(\mbox{{\mathcal{M}}}), we have

[TABLE]

Remark 2.6**.**

In the case $p=1$, we note that L^{1}(\mbox{{\mathcal{M}}};\ell^{1})\simeq\ell^{1}(L^{1}(\mbox{{\mathcal{M}}})) isometrically. This implies that any bounded operator T\colon L^{1}(\mbox{{\mathcal{M}}})\to L^{1}(\mbox{{\mathcal{N}}}) is automatically $\ell^{1}$-bounded, with $\|T\|_{\ell^{1}}=\|T\|$.

In the rest of this section, we compare $\ell^{1}$-boundedness with Pisier’s notion of complete regularity. Let us recall that for a hyperfinite von Neumann algebra ${\mathcal{M}}$ and an operator space $E$, Pisier [20, Chapter 3] introduced a vector valued noncommutative $L^{p}$-space L^{p}(\mbox{{\mathcal{M}}})[E]. Let ${\rm Max}(\ell^{1})$ be $\ell^{1}$ equipped with its so-called maximal operator space structure (see e.g. [19, Chapter 3]). It turns out that

[TABLE]

when ${\mathcal{M}}$ is hyperfinite (see [7, 9]).

Assume that the semifinite von Neumann algebras $\mathcal{M},\mathcal{N}$ are both hyperfinite. Let $T\colon L^{p}(\mathcal{M})\to L^{p}(\mathcal{N})$ be a bounded operator. Following Pisier [21], $T$ is called completely regular if there exists a constant $K\geq 0$ such that for any $n\geq 1$,

[TABLE]

In this case the least possible $K$ is denoted by $\|T\|_{reg}$ and is called the completely regular norm of $T$. It is noticed in [21] that if $T$ is completely regular, then for any operator space $E$, $T\otimes I_{E}$ (uniquely) extends to a bounded operator $T\overline{\otimes}I_{E}$ from $L^{p}(\mathcal{M})[E]$ into $L^{p}(\mathcal{N})[E]$, with

[TABLE]

Combining this fact with (7), we obtain that if $T\colon L^{p}(\mathcal{M})\to L^{p}(\mathcal{N})$ is completely regular, then $T$ is $\ell^{1}$-bounded, with $\|T\|_{\ell^{1}}\leq\|T\|_{reg}$.

The next example shows that the converse is wrong.

Example 2.7**.**

We consider the specific case $\mathcal{M}=B(\ell^{2})$, and we let $T\colon S^{p}(\ell^{2})\to S^{p}(\ell^{2})$ be the transposition map. This map is $\ell^{1}$-contractive. This is an easy fact, which is a special case of Theorem 3.15 below. Here is a direct argument.

Let $(x_{n})_{n\geq 1}$ be in L^{p}(\mbox{{\mathcal{M}}});\ell^{1}) and let $(a_{n})_{n\geq 1}$ and $(b_{n})_{n\geq 1}$ be two sequences belonging to L^{2p}(\mbox{{\mathcal{M}}};\ell^{2}_{r}) and L^{2p}(\mbox{{\mathcal{M}}};\ell^{2}_{c}), respectively, such that $x_{n}=a_{n}b_{n}$ for any $n\geq 1$. Then $T(x_{n})={}^{t}b_{n}{}^{t}a_{n}$ for any $n\geq 1$, $({}^{t}a_{n})_{n\geq 1}$ belongs to L^{2p}(\mbox{{\mathcal{M}}};\ell^{2}_{c}), $({}^{t}b_{n})_{n\geq 1}$ belongs to L^{2p}(\mbox{{\mathcal{M}}};\ell^{2}_{r}) and we both have

[TABLE]

The result follows at once.

Let us now prove that $T$ is not completely regular. We need a little operator space technology, in particular we use the Haagerup tensor product $\otimes_{h}$, the operator spaces $R,C$ and the interpolation spaces $R(\theta)=[C,R]_{\theta}$, $0\leq\theta\leq 1$, for which we refer to [19].

Let $(e_{k})_{k\geq 1}$ be the canonical basis of $\ell^{2}$. As it is outlined in [20, Theorem 1.1 and p.20], for any operator space $E$, the mapping $e_{i}\otimes x\otimes e_{j}\mapsto E_{ij}\otimes x$, for $i,j\geq 1$ and $x\in E$, uniquely extends to an isometric isomorphism

[TABLE]

Assume that $T$ is completely regular and let $K=\|T\|_{reg}$. Apply (8) with $E=R$. For any $n\geq 1$, we have

[TABLE]

hence

[TABLE]

It follows from the calculations in [20, Chapter 1] (see also [8]) that

[TABLE]

isometrically, where $(2p)^{\prime}$ is the conjugate number of $2p$. Let $Q_{n}\in B(\ell^{2})$ be the orthogonal projection onto ${\rm Span}\{e_{1},\ldots,e_{n}\}$. Then $\sum_{k=1}^{n}e_{k}\otimes e_{k}=Q_{n}$ in the above identifications. Hence we obtain that

[TABLE]

Since $\|Q_{n}\|_{q}=n^{\frac{1}{q}}$ for any $1<q<\infty$, we obtain that $n^{1-\frac{1}{2p}}\leq Kn^{\frac{1}{2p}}$, equivalently, $n\leq Kn^{\frac{1}{p}}$, for any $n\geq 1$. This yields a contradiction if $p>1$. In the case $p=1$, the fact that $T$ is not completely regular is obtained by applying (8) with $E=C$ instead of $E=R$.

Remark 2.8**.**

Here we consider the commutative case. Let (\Omega,\mbox{{\mathcal{F}}},\mu) be a measure space. For any operator space $E$, $L^{p}(L^{\infty}(\Omega))[E]$ coincides with the Bochner space $L^{p}(\Omega;E)$. Thus if (\Omega^{\prime},\mbox{{\mathcal{F}}}^{\prime}\mu^{\prime}) is another measure space and $T\colon L^{p}(\Omega)\to L^{p}(\Omega^{\prime})$ is any bounded operator, then $T$ is completely regular (in the above sense) if and only if $T$ is regular in the lattice sense (see [18, Chapter 1] for details and background). Moreover in this case, the completely regular norm of $T$ coincides with its regular norm. It follows from [18, Paragraph 1.2] that $T$ is regular if (and only if) $T\otimes I_{\ell^{1}}$ extends to a bounded map $T\overline{\otimes}I_{\ell^{1}}$ from $L^{p}(\Omega;\ell^{1})$ into $L^{p}(\Omega^{\prime};\ell^{1})$ and in this case, we have $\|T\overline{\otimes}I_{\ell^{1}}\|=\|T\|_{reg}$. Consequently, $T\colon L^{p}(\Omega)\to L^{p}(\Omega^{\prime})$ is $\ell^{1}$-bounded if and only if $T$ is regular and in this case, the $\ell^{1}$-bounded norm of $T$ is equal to its regular norm.

3. Disjointness and separating operators

In [12], Kan introduced the concept of Lamperti operators on commutative $L^{p}$-spaces, which include $L^{p}$-isometries, $1\leq p\neq 2<\infty$, and positive $L^{2}$-isometries. He then proved a structural theorem for such operators. In this section we provide a noncommutative version of this result, as well as a connection with $\ell^{1}$-boundedness.

Let us first recall some facts related to Jordan homomorphisms that we require in this section. (We refer to [22], [5] and [11, Exercices 10.5.21-10.5.31] for general information.) A Jordan homomorphism between von Neumann algebras $\mathcal{M}$ and $\mathcal{N}$ is a linear map $J\colon\mathcal{M}\to\mathcal{N}$ that satisfies $J(a^{2})=J(a)^{2}$ and $J(a^{\ast})=J(a)^{\ast}$, for every $a\in\mathcal{M}$. We say that the Jordan homomorphism J\colon\mbox{{\mathcal{M}}}\to\mbox{{\mathcal{N}}} is a Jordan monomorphism when $J$ is one-to-one. Any Jordan homomorphism is a positive contraction and any Jordan monomorphism is an isometry.

Let $J\colon\mathcal{M}\to\mathcal{N}$ be a Jordan homomorphism and let $\mathcal{D}\subset\mathcal{N}$ be the von Neumann algebra generated by $J(\mathcal{M})$. Let $e:=J(1)$. Then $e$ is a projection and $e$ is the unit of $\mathcal{D}$. According to [22, Theorem 3.3] (see also [5, Corollary 7.4.9.]), there exist projections $g$ and $f$ in the center of $\mathcal{D}$ such that

• (i)

$g+f=e$;

• (ii)

$a\mapsto J(a)g$ is a $\ast$-homomorphism;

• (iii)

$a\mapsto J(a)f$ is an anti-$\ast$-homomorphism;

Let $\mathcal{N}_{1}=g\mathcal{N}g$ and $\mathcal{N}_{2}=f\mathcal{N}f$. We let $\pi\colon\mathcal{M}\to\mathcal{N}_{1}$ and $\sigma\colon\mathcal{M}\to\mathcal{N}_{2}$ be defined by $\pi(a)=J(a)g$ and $\sigma(a)=J(a)f$, for all a\in\mbox{{\mathcal{M}}}. Then, $\mathcal{D}=\mathcal{D}_{1}\mathop{\oplus}\limits^{\infty}\mathcal{D}_{2}$ and $J(a)=\pi(a)+\sigma(a)$, for all a\in\mbox{{\mathcal{M}}}. We will use the suggestive notations

[TABLE]

to refer to such a central decomposition. We note that $J$ is normal (i.e. $w^{*}$-continuous) if and only if $\pi$ and $\sigma$ are normal.

Assume that \mbox{{\mathcal{M}}}\subset B(\mbox{{\mathcal{H}}}) acts on some Hilbert space ${\mathcal{H}}$ and let $x$ be a closed densely defined operator on ${\mathcal{H}}$, affiliated with ${\mathcal{M}}$. If $x$ is self-adjoint, with polar decomposition $x=w\lvert x\rvert$, we let $s(x)$ denote the projection $w^{\ast}w\ (=ww^{\ast})$, called the support of $x$.

The following remarkable characterization of $L^{p}$-isometries, $1\leq p\not=2<\infty$, is at the root of our investigations.

Theorem 3.1** (Yeadon [26]).**

For $1\leq p<\infty,\ p\neq 2$, a bounded operator

[TABLE]

is an isometry if and only if there exist a normal Jordan monomorphism $J\colon\mathcal{M}\to\mathcal{N}$, a partial isometry $w\in\mathcal{N}$, and a positive operator $B$ affiliated with $\mathcal{N}$, which verify the following conditions:

• (a)

$T(a)=wBJ(a)$* for all $a\in\mathcal{M}\cap L^{p}(\mathcal{M})$;*

• (b)

$w^{\ast}w=J(1)=s(B)$;

• (c)

every spectral projection of $B$ commutes with $J(a)$, for all $a\in\mathcal{M}$;

• (d)

$\tau_{\mathcal{M}}(a)=\tau_{\mathcal{N}}(B^{p}J(a))$* for all $a\in\mathcal{M}^{+}$.*

This motivates the introduction of the following concept. Since we would like to consider maps T:L^{p}(\mbox{{\mathcal{M}}})\to L^{p}(\mbox{{\mathcal{N}}}) that are not necessarily isometries we drop part $(d)$ of Theorem 3.1 in our definition.

Definition 3.2**.**

We say that a bounded operator $T\colon L^{p}(\mathcal{M})\to L^{p}(\mathcal{N})$, $1\leq p\leq\infty$, has a “Yeadon type factorization” if there exist a normal Jordan homomorphism $J\colon\mathcal{M}\to\mathcal{N}$, a partial isometry $w\in\mathcal{N}$, and a positive operator $B$ affiliated with $\mathcal{N}$, which verify the following conditions:

• (a)

$T(a)=wBJ(a)$ for all $a\in\mathcal{M}\cap L^{p}(\mathcal{M})$;

• (b)

$w^{\ast}w=J(1)=s(B)$;

• (c)

every spectral projection of $B$ commutes with $J(a)$, for all $a\in\mathcal{M}$.

Remark 3.3**.**

The argument in the last paragraph of the proof of [26, Theorem 2] shows that for an operator $T\colon L^{p}(\mathcal{M})\to L^{p}(\mathcal{N})$ with a Yeadon type factorization, $w$, $B$ and $J$ from Definition 3.2 are uniquely determined by $T$. We call $(w,B,J)$ the Yeadon triple of the operator $T$.

The crucial property that allowed Yeadon to describe $L^{p}$-isometries is that they map disjoint elements to disjoint elements. This property is shared by operators other than isometries and as we show in Proposition 3.11, it characterizes operators with a Yeadon type factorization. We introduce the relevant concepts and supply a few preparatory results.

Definition 3.4**.**

Let $a$ and $b$ be elements in $L^{p}(\mathcal{M})$, $1\leq p\leq\infty$. We say $a$ and $b$ are disjoint if $ab^{*}=a^{*}b=0$.

Lemma 3.5**.**

The elements $a$ and $b$ in $L^{p}(\mathcal{M})$ are disjoint if and only if $\lvert a\rvert$ and $\lvert b\rvert$ are disjoint and $\lvert a^{*}\rvert$ and $\lvert b^{*}\rvert$ are disjoint.

Proof.

Let $a,\ b\in L^{p}(\mathcal{M})$. First we note that $a$ and $b$ are disjoint if and only if

[TABLE]

This implies that $\lvert a\rvert$ and $\lvert b\rvert$ are disjoint if and only if $\text{Ker}(a)^{\perp}=\text{Ker}(\lvert a\rvert)^{\perp}\subseteq\text{Ker}(\lvert b\rvert)=\text{Ker}(b)$. Then $\lvert a^{*}\rvert$ and $\lvert b^{*}\rvert$ are disjoint if and only if $\text{Ker}(a^{*})^{\perp}\subset\text{Ker}(b^{*})$, which is itself equivalent to $\text{Im}(b)\subseteq\text{Im}(a)^{\perp}$. Therefore, $a$ and $b$ are disjoint if and only if $\lvert a\rvert$ and $\lvert b\rvert$ are disjoint and $\lvert a^{*}\rvert$ and $\lvert b^{*}\rvert$ are disjoint. ∎

Remark 3.6**.**

Consider the special case $p=2$ and let $a,b$ be two positive elements in L^{2}(\mbox{{\mathcal{M}}}). Then $a$ and $b$ are disjoint if (and only if) \tau_{\tiny\mbox{{\mathcal{M}}}}(ab)=0, that is, if and only if $a$ and $b$ are orthogonal in the Hilbertian sense. Indeed assume that \tau_{\tiny\mbox{{\mathcal{M}}}}(ab)=0. Then 0=\tau_{\tiny\mbox{{\mathcal{M}}}}(ab)=\tau_{\tiny\mbox{{\mathcal{M}}}}((a^{\frac{1}{2}}b^{\frac{1}{2}})(b^{\frac{1}{2}}a^{\frac{1}{2}})) and $a^{\frac{1}{2}}b^{\frac{1}{2}}$ is the adjoint of $b^{\frac{1}{2}}a^{\frac{1}{2}}$. Since the trace \tau_{\tiny\mbox{{\mathcal{M}}}} is faithful, this implies that $a^{\frac{1}{2}}b^{\frac{1}{2}}=0$. Therefore, $ab=a^{\frac{1}{2}}(a^{\frac{1}{2}}b^{\frac{1}{2}})b^{\frac{1}{2}}=0$. Hence $a$ and $b$ are disjoint.

Definition 3.7**.**

We say that a bounded operator $T\colon L^{p}(\mathcal{M})\to L^{p}(\mathcal{N})$, $1\leq p\leq\infty$, is separating if whenever a,b\in L^{p}(\mbox{{\mathcal{M}}}) are disjoint, then $T(a)$ and $T(b)$ are disjoint.

Lemma 3.8**.**

Any Jordan homomorphism $J\colon\mathcal{M}\to\mathcal{N}$ is separating.

Proof.

Let $J\colon\mathcal{M}\to\mathcal{N}$ be a Jordan homomorphism and consider a decomposition $J=\begin{pmatrix}\pi&0\\ 0&\sigma\end{pmatrix}$ as in (9).

Suppose that $a$ and $b$ are disjoint elements of ${\mathcal{M}}$. Then we have

[TABLE]

Similarly, we can show that $J(a)J(b)^{*}=0$, and therefore $J$ is separating. ∎

Lemma 3.9**.**

Suppose that a bounded operator $T\colon L^{p}(\mathcal{M})\to L^{p}(\mathcal{N})$, $1\leq p<\infty$, is separating on $\mathcal{M}\cap L^{p}(\mathcal{M})$, that is, $T(a)$ and $T(b)$ are disjoint for any disjoint $a$ and $b$ in \mbox{{\mathcal{M}}}\cap L^{p}(\mbox{{\mathcal{M}}}). Then $T$ is separating.

Proof.

Let $a,b\in L^{p}(\mathcal{M})$ with $a^{\ast}b=ab^{\ast}=0$. We let $a=v\lvert a\rvert$ and $b=w\lvert b\rvert$ be the polar decompositions of $a$ and $b$, respectively. By Lemma 3.5, we have $\lvert a\rvert\lvert b\rvert=0$.

For any $n\geq 1$, let $p_{n}:=\chi_{[-n,n]}(\lvert a\rvert)$ be the projection associated with the indicator function of $[-n,n]$ in the Borel functional calculus of $\lvert a\rvert$, and similarly let $q_{n}:=\chi_{[-n,n]}(\lvert b\rvert)$. Let $a_{n}:=ap_{n}$ and $b_{n}:=bq_{n}$. We have

[TABLE]

in $L^{p}(\mathcal{M})$. This implies that $a_{n}\to a$ and $b_{n}\to b$ in $L^{p}(\mathcal{M})$.

Note that for any $n\geq 1$, $a_{n}$ and $b_{n}$ belong to $\mathcal{M}\cap L^{p}(\mathcal{M})$. Further we have

[TABLE]

and

[TABLE]

Thus $a_{n}$ and $b_{n}$ are disjoint.

By assumption this implies that $T(a_{n})^{\ast}T(b_{n})=0$ and $T(a_{n})T(b_{n})^{\ast}=0$. Passing to the limit, we deduce that $T(a)^{\ast}T(b)=0$ and $T(a)T(b)^{\ast}=0$. ∎

From now on, we consider

[TABLE]

For any x\in\mbox{{\mathcal{M}}}, $exe\to x$ in the $w^{*}$-topology of ${\mathcal{M}}$, when $e\to 1$. Further for any $1\leq p<\infty$ and any x\in L^{p}(\mbox{{\mathcal{M}}}), $exe\to x$ in L^{p}(\mbox{{\mathcal{M}}}). Thus ${\mathcal{A}}$ is a $w^{*}$-dense subspace of ${\mathcal{M}}$ and a norm dense subspace of L^{p}(\mbox{{\mathcal{M}}}), for any $1\leq p<\infty$. Lemma 3.10 below is a $w^{*}$-extension result of independent interest, which will be used in the proof of Proposition 3.11.

Given any w\in\mbox{{\mathcal{M}}}^{\ast} and a,b\in\mbox{{\mathcal{M}}}, we let awb\in\mbox{{\mathcal{M}}}^{\ast} be defined by

[TABLE]

Recall e.g. from [23, pp 126-128] the decomposition

[TABLE]

where \mbox{{\mathcal{M}}}^{\ast}_{s} denotes the space of singular functionals on ${\mathcal{M}}$, and \mbox{{\mathcal{M}}}_{*} is the predual of ${\mathcal{M}}$, which coincides with the space of normal functionals on ${\mathcal{M}}$. It is well-known that if $w=w_{n}+w_{s}$ is the aforementioned decomposition of $w$, then $aw_{n}b$ and $aw_{s}b$ are the normal part and the singular part of $awb$, respectively.

Lemma 3.10**.**

Let $Y$ be a dual Banach space. For any bounded operator u\colon\mbox{{\mathcal{A}}}\to Y, the following are equivalent:

• (i)

For every e\in\mbox{{\mathcal{E}}}, the restriction u|_{e\tiny{\mbox{{\mathcal{M}}}}e}\colon e\mbox{{\mathcal{M}}}e\to Y is $w^{*}$-continuous.

• (ii)

There exists a $w^{\ast}$-continuons extension \widehat{u}\colon\mbox{{\mathcal{M}}}\to Y of $u$.

Proof.

The implication “$(ii)\Rightarrow(i)$” is trivial. For the converse, we assume (i).

We first consider the case when $Y=\mathbb{C}$. Suppose that \alpha\in\mbox{{\mathcal{A}}}^{\ast} is such that \alpha|_{e\tiny{\mbox{{\mathcal{M}}}}e} is $w^{*}$-continuous for each e\in\mbox{{\mathcal{E}}}. Using Hahn-Banach, we let w\colon\mbox{{\mathcal{M}}}\to\mathbb{C} be a bounded extension of $\alpha$ to ${\mathcal{M}}$ and we consider its decomposition $w=w_{n}+w_{s}$ according to (10).

For every e\in\mbox{{\mathcal{E}}}, $ewe$ is $w^{*}$-continuous. We noticed that $ewe=ew_{n}e+ew_{s}e$ is the decomposition of $ewe$. Since $ewe$ and $ew_{n}e$ are $w^{*}$-continuous, the singular part $ew_{s}e$ of $ewe$ must be zero, and consequently, $ewe=ew_{n}e$, for every e\in\mbox{{\mathcal{E}}}. This implies that the restriction of $w_{n}$ to ${\mathcal{A}}$ coincides with $\alpha$. Thus $\widehat{\alpha}=w_{n}$ is a $w^{\ast}$-continuous extension of $\alpha$.

For the general case, let v=u^{\ast}|_{Y_{\ast}}\colon Y_{\ast}\to\mbox{{\mathcal{A}}}^{\ast} be the restriction of the adjoint of $u$ to the predual of $Y$. Let

[TABLE]

denote the restriction map taking any \nu\in\mbox{{\mathcal{M}}}_{*} to \nu|_{\mbox{{\mathcal{A}}}}. This is an isometry (by Kaplansky’s theorem, say), whose range coincides with the space of all functionals \mbox{{\mathcal{A}}}\to\mathbb{C} which admit a $w^{*}$-continuous extension to ${\mathcal{M}}$.

For each $\eta\in Y_{\ast}$, \eta\circ u|_{e\tiny{\mbox{{\mathcal{M}}}}e} is $w^{*}$-continuous, for every e\in\mbox{{\mathcal{E}}}. By our argument for the case $Y=\mathbb{C}$, this implies that $\eta\circ u\in\text{Im}(\kappa)$. This means that $v$ is valued in $\text{Im}(\kappa)$. We can therefore consider w=\kappa^{-1}\circ v\colon Y_{\ast}\to\mbox{{\mathcal{M}}}_{\ast} and define \widehat{u}=w^{\ast}\colon\mbox{{\mathcal{M}}}\to Y. By construction, $\widehat{u}$ is $w^{*}$-continuous.

We now claim that $\widehat{u}$ is an extension of $u$. To see this, recall that for any $\eta\in Y_{*}$, the functional $\kappa^{-1}\circ u^{\ast}(\eta)$ is an extension to ${\mathcal{M}}$ of u^{\ast}(\eta)\colon\mbox{{\mathcal{A}}}\to\mathbb{C}. Consequently,

[TABLE]

for any x\in\mbox{{\mathcal{M}}}, any e\in\mbox{{\mathcal{E}}} and any $\eta\in Y_{*}$. This proves the claim. ∎

Proposition 3.11**.**

For $1\leq p<\infty$, a bounded operator $T\colon L^{p}(\mathcal{M})\to L^{p}(\mathcal{N})$ is separating if and only if it has a Yeadon type factorization.

Proof.

Assume that $T$ is separating. We adapt Yeadon’s argument from [26], taking into account that our operators are no longer necessarily isometries.

For any e\in\mbox{{\mathcal{E}}}, let $B_{e}=|T(e)|$ and let $T(e)=w_{e}B_{e}$ be the polar decomposition of $T(e)$. We have

[TABLE]

Set $J(e):=w_{e}^{\ast}w_{e}=s(B_{e})$. If $e$ and $f$ are in ${\mathcal{E}}$ and $ef=0$, then since $T$ is separating we have

[TABLE]

Using (12), the argument in the proof of [26, Theorem 2] shows that J\colon\mbox{{\mathcal{E}}}\to\mbox{{\mathcal{N}}} extends to a linear map J\colon\mbox{{\mathcal{A}}}\to\mbox{{\mathcal{N}}} such that

[TABLE]

for all a\in\mbox{{\mathcal{A}}},

[TABLE]

for all e\in\mbox{{\mathcal{E}}} and x\in\mbox{{\mathcal{M}}}, and

[TABLE]

Note that by (13), the restriction of $J$ to e\mbox{{\mathcal{M}}}e is a Jordan homomorphism for any e\in\mbox{{\mathcal{E}}}. Consequently,

[TABLE]

for any e\in\mbox{{\mathcal{E}}} and any x\in\mbox{{\mathcal{A}}}.

We now show that $J$ admits a normal extension (still denoted by) J\colon\mbox{{\mathcal{M}}}\to\mbox{{\mathcal{N}}}. (Note that Yeadon’s argument in the isometric case does not apply to our general case.) According to Lemma 3.10 it suffices to show that the restriction of $J$ to e\mbox{{\mathcal{M}}}e is normal for any e\in\mbox{{\mathcal{E}}}. To see this, we fix such an $e$ and we let $(ex_{i}e)_{i}$ be a bounded net of $e\mathcal{M}e$ converging to $exe$ in the $w^{*}$-topology of ${\mathcal{M}}$. Then $ex_{i}e\to exe$ in the weak topology of L^{p}(\mbox{{\mathcal{M}}}), hence $w_{e}^{*}T(ex_{i}e)\to w_{e}^{*}T(exe)$ in the weak topology of L^{p}(\mbox{{\mathcal{N}}}). By (14) and (11), this implies that \tau_{\tiny\mbox{{\mathcal{N}}}}(AB_{e}J(ex_{i}e))\to\tau_{\tiny\mbox{{\mathcal{N}}}}(AB_{e}J(exe)) for any A\in L^{p^{\prime}}(\mbox{{\mathcal{N}}}), where $p^{\prime}$ is the conjugate number of $p$. Note that by (16), the restriction of $J$ to e\mbox{{\mathcal{M}}}e is valued in J(e)\mbox{{\mathcal{N}}}J(e). To deduce from the above convergence property that $J(ex_{i}e)\to J(exe)$ in the $w^{*}$-topology of ${\mathcal{N}}$, it therefore suffices to check that

[TABLE]

This is indeed the case, since B_{e}=|T(e)|\in L^{p}(\mbox{{\mathcal{N}}}) and $s(B_{e})=J(e)$.

We note that since $J$ is $w^{*}$-continuous and ${\mathcal{A}}$ is $w^{*}$-dense in ${\mathcal{M}}$, (16) holds true for any e\in\mbox{{\mathcal{E}}} and any x\in\mbox{{\mathcal{M}}}.

We now use the increasing net ${\mathcal{E}}$ and we recall that $e\to 1$ in the $w^{*}$-topology of ${\mathcal{M}}$. Since $J$ is normal, $J(1)$ is the $w^{*}$-limit of $J(e)$. Then using (11) and (15), the same argument as in [10, Theorem 3.1] can be implemented to obtain extensions $B$ (as supremum of the $B_{e}$) and $w$ (as strong limit of the $w_{e}$) which satisfy properties $(b)$ and $(c)$ of Definition 3.2. By (15) we further have

[TABLE]

for any e\in\mbox{{\mathcal{E}}}.

We now aim at showing property $(a)$ of Definition 3.2. For any y\in\mbox{{\mathcal{M}}}^{+}\cap L^{1}(\mbox{{\mathcal{M}}}), using spectral projections, we find a sequence $(e_{n})_{n\geq 1}$ in ${\mathcal{E}}$ such that $e_{n}ye_{n}$ is increasing to $y$ when $n\to\infty$. Since $J$ is normal, this implies that $B^{p}J(e_{n}ye_{n})$ is increasing to $B^{p}J(y)$ when $n\to\infty$. Consequently, using the normality of \tau_{\tiny\mbox{{\mathcal{N}}}}, we obtain that

[TABLE]

Let x\in\mbox{{\mathcal{M}}}\cap L^{p}(\mbox{{\mathcal{M}}}). Consider a decomposition $J=\begin{pmatrix}\pi&0\\ 0&\sigma\end{pmatrix}$ as in (9). Then we have

[TABLE]

Then by a well-known argument (see the proof of the easy implication of [26, Theorem 2]), this implies that

[TABLE]

Using (18) with $y=|x|^{p}$, we deduce that for some sequence $(e_{n})_{n\geq 1}$ in ${\mathcal{E}}$, we have

[TABLE]

Fix e\in\mbox{{\mathcal{E}}}. Combining (14) and (16), we have $T(exe)=w_{e}B_{e}J(x)J(e)$ from which we deduce as in (20) that

[TABLE]

Using (16) again, and (17), we have

[TABLE]

Consequently, using (19), we have

[TABLE]

Applying this with $e=e_{n}$ and passing to the limit, we deduce that wBJ(x)\in L^{p}(\mbox{{\mathcal{N}}}) for any x\in\mbox{{\mathcal{M}}}\cap L^{p}(\mbox{{\mathcal{M}}}), with $\|wBJ(x)\|_{p}\leq 2\|T\|\|x\|_{p}$. This shows the existence of a (necessarily unique) bounded operator T^{\prime}\colon L^{p}(\mbox{{\mathcal{M}}})\to L^{p}(\mbox{{\mathcal{N}}}) such that $T^{\prime}(x)=wBJ(x)$ for any x\in\mbox{{\mathcal{M}}}\cap L^{p}(\mbox{{\mathcal{M}}}). By (14) and (17), $T$ and $T^{\prime}$ coincide on ${\mathcal{A}}$. Since the latter is dense in L^{p}(\mbox{{\mathcal{M}}}), we obtain that $T=T^{\prime}$, hence property $(a)$ of Definition 3.2 is satisfied.

For the converse suppose that $T\colon L^{p}(\mathcal{M})\to L^{p}(\mathcal{N})$ has a Yeadon type factorization, with Yeadon triple $(w,B,J)$, and let us show that $T$ is separating. By Lemma 3.9, it is enough to show that $T$ is separating on \mbox{{\mathcal{M}}}\cap L^{p}(\mbox{{\mathcal{M}}}). Let $a$ and $b$ be disjoint elements in \mbox{{\mathcal{M}}}\cap L^{p}(\mbox{{\mathcal{M}}}), then

[TABLE]

Similarly we can show that $T(a)T(b)^{\ast}=0$, and hence $T(a)$ and $T(b)$ are disjoint. ∎

After a first version of this paper was circulated, we were informed that the ‘only if’ part of Proposition 3.11 was proved independently in [6].

We now give a series of remarks on this statement.

Remark 3.12**.**

(a) In Proposition 3.11, the proof that a separating map $T$ admits a Yeadon type factorization only uses the separation property on positive elements (even on projections with finite trace). Hence a bounded operator $T\colon L^{p}(\mathcal{M})\to L^{p}(\mathcal{N})$ is separating if and only if for any a,b\in L^{p}(\mbox{{\mathcal{M}}})^{+}, $ab=0$ implies that $T(a)^{*}T(b)=T(a)T(b)^{*}=0$.

(b) Let us say that a separating map T\colon L^{p}(\mbox{{\mathcal{M}}})\to L^{p}(\mbox{{\mathcal{N}}}) is $2$-separating if the tensor extension I_{S^{p}_{2}}\otimes T\colon L^{p}(M_{2}(\mbox{{\mathcal{M}}}))\to L^{p}(M_{2}(\mbox{{\mathcal{N}}})) is separating. Combining Proposition 3.11 and the argument in the proof of [10, Proposition 3.2], we obtain that T\colon L^{p}(\mbox{{\mathcal{M}}})\to L^{p}(\mbox{{\mathcal{N}}}) is $2$-separating if and only if the Jordan homomorphism in the Yeadon factorisation of $T$ is multiplicative. This fact was also observed in [6, Theorem 3.4].

Remark 3.13**.**

Here we discuss the commutative case. Let (\Omega,\mbox{{\mathcal{F}}},\mu) and (\Omega^{\prime},\mbox{{\mathcal{F}}}^{\prime},\mu^{\prime}) be two $\sigma$-finite measure spaces. Let \widetilde{\mbox{{\mathcal{F}}}} be the set of classes of ${\mathcal{F}}$ modulo the null sets (i.e. E\in\mbox{{\mathcal{F}}} such that $\mu(E)=0$). We identify any element of ${\mathcal{F}}$ with its class in \widetilde{\mbox{{\mathcal{F}}}}. We define \widetilde{\mbox{{\mathcal{F}}}^{\prime}} similarly.

Recall that a regular set morphism (RSM) from (\Omega,\mbox{{\mathcal{F}}},\mu) into (\Omega^{\prime},\mbox{{\mathcal{F}}}^{\prime},\mu^{\prime}) is a map \varphi\colon\widetilde{\mbox{{\mathcal{F}}}}\to\widetilde{\mbox{{\mathcal{F}}}^{\prime}} satisfying the following two properties:

• (i)

For any E_{1},E_{2}\in\mbox{{\mathcal{F}}}, $E_{1}\bigcap E_{2}=\emptyset\,\Rightarrow\,\varphi(E_{1})\bigcap\varphi(E_{2})=\emptyset$.

• (ii)

For any sequence $(E_{n})_{n\geq 1}$ of pairwise disjoint sets in ${\mathcal{F}}$, \varphi\bigl{(}\bigcup_{n\geq 1}E_{n}\bigr{)}=\bigcup_{n\geq 1}\varphi(E_{n}).

Following [13], Kan [12, Theorem 4.1] showed that a bounded operator $T\colon L^{p}(\Omega)\to L^{p}(\Omega^{\prime})$ is separating if and only if there exist a measurable function $h$ and a regular set morphism $\varphi$ from (\Omega,\mbox{{\mathcal{F}}},\mu) into (\Omega^{\prime},\mbox{{\mathcal{F}}}^{\prime},\mu^{\prime}) such that for every set of finite measure $E$, we have

[TABLE]

(See [12] and [3, Chapter 3] for more on this factorization property.)

There is a well-known correspondence between RSM and normal $*$-homomorphisms on $L^{\infty}$-spaces. Namely, let \pi\colon L^{\infty}(\Omega,\mbox{{\mathcal{F}}},\mu)\to L^{\infty}(\Omega^{\prime},\mbox{{\mathcal{F}}}^{\prime},\mu^{\prime}) be a normal $\ast$-homomorphism. Then for any E\in\mbox{{\mathcal{F}}}, $\pi(\chi_{E})$ is a projection, hence an indicator function. We may therefore define \varphi\colon\widetilde{\mbox{{\mathcal{F}}}}\to\widetilde{\mbox{{\mathcal{F}}}^{\prime}} by $\pi(\chi_{E})=\chi_{\varphi(E)}$. It is easy to check that $\varphi$ is a RSM. It turns out that any regular set morphism is of this form. Indeed let \varphi\colon\widetilde{\mbox{{\mathcal{F}}}}\to\widetilde{\mbox{{\mathcal{F}}}^{\prime}} be a RSM. For any g\in L^{1}(\Omega^{\prime},\mbox{{\mathcal{F}}}^{\prime},\mu^{\prime}), define \nu_{g}\colon\mbox{{\mathcal{F}}}\to\mbox{{\mathbb{C}}} by

[TABLE]

By (ii) and Lebegue’s theorem, $\nu_{g}$ is a complex measure, whose total variation is less than or equal to $\|g\|_{1}$. By (i), $\varphi(\emptyset)=\emptyset$ hence $\nu$ is absolutely continuous with respect to $\mu$. Hence by the Radon-Nikodym theorem, there exists a necessarily unique h_{g}\in L^{1}(\Omega,\mbox{{\mathcal{F}}},\mu) such that

[TABLE]

Moreover $\|h_{g}\|_{1}\leq\|g\|_{1}$. It is clear that the mapping u\colon L^{1}(\Omega^{\prime},\mbox{{\mathcal{F}}}^{\prime},\mu^{\prime})\to L^{1}(\Omega,\mbox{{\mathcal{F}}},\mu) defined by $u(g)=h_{g}$ is linear. The above estimate shows that $u$ is contractive. Set

[TABLE]

By construction, $\pi$ is $w^{*}$-continuous. Further it is easy to check that $\pi$ is a $*$-homomorphism and that $\pi(\chi_{E})=\chi_{\varphi(E)}$ for any E\in\mbox{{\mathcal{F}}}.

We finally note that all Jordan homomorphisms on $L^{\infty}$-spaces are $*$-homomorphisms. Thereby, through the aforementioned correspondence, Proposition 3.11 reduces to Kan’s theorem in the case when ${\mathcal{M}}$ and ${\mathcal{N}}$ are commutative.

Remark 3.14**.**

(a) In general, separating operators may not be one-to-one (contrary to isometries). We observe however that if a bounded operator T\colon L^{p}(\mbox{{\mathcal{M}}})\to L^{p}(\mbox{{\mathcal{N}}}) is separating, with Yeadon triple $(w,B,J)$, then $T$ is one-to-one if and only if $J$ is a Jordan monomorphism.

Indeed if $T$ is one-to-one, then J(e)=s\bigl{(}|T(e)|\bigr{)}\not=0 for any e\in\mbox{{\mathcal{E}}}\setminus\{0\}. This implies that for any pairwise disjoint non zero $e_{1},\ldots,e_{m}$ in ${\mathcal{E}}$ and any $\alpha_{1},\ldots,\alpha_{m}$ in ${\mathbb{C}}$, we have

[TABLE]

Hence the restriction of $J$ to ${\mathcal{E}}$ is an isometry. This readily implies that $J$ is an isometry, and hence $J$ is one-to-one.

Assume conversely that $J$ is one-to-one and let $x\in L^{p}(M)$ such that $T(x)=0$. Let $x=u|x|$ be the polar decomposition of $x$ and for any integer $n\geq 1$, let $p_{n}=\chi_{[-n,n]}(|x|)$. Then consdider $x^{\prime}_{n}=xp_{n}$ and $x^{\prime\prime}_{n}=x(1-p_{n})$. We have $x=x^{\prime}_{n}+x^{\prime\prime}_{n}$ hence $T(x^{\prime}_{n})+T(x^{\prime\prime}_{n})=0$. Further we have

[TABLE]

whereas $x^{\prime}_{n}{x^{\prime\prime}_{n}}^{*}=xp_{n}(1-p_{n})x^{*}=0$. Hence $x^{\prime}_{n}$ and $x^{\prime\prime}_{n}$ are disjoint. Since $T$ is separating this implies that $T(x^{\prime}_{n})$ and $T(x^{\prime\prime}_{n})$ are disjoint. Since these elements are opposite to each other, this implies that $T(x^{\prime}_{n})=0$. For any $n\geq 1$, $x^{\prime}_{n}\in L^{p}(M)\cap M$ hence $T(x^{\prime}_{n})=wBJ(x^{\prime}_{n})$. Since $w^{*}w=s(B)=J(1)$, this implies that $J(x^{\prime}_{n})=0$, hence $x^{\prime}_{n}=0$ by our assumption. Since $x^{\prime}_{n}\to x$ in $L^{p}(M)$, we deduce that $x=0$. This shows that $T$ is one-to-one.

(b) We also observe that a separating operator T\colon L^{p}(\mbox{{\mathcal{M}}})\to L^{p}(\mbox{{\mathcal{N}}}) with Yeadon triple $(w,B,J)$ is positive if and only $w$ is positive (if and only if $w$ is a projection). The verification is left to the reader.

The following theorem shows that separating bounded operators are $\ell^{1}$-bounded. The converse does not hold true, this can be easily seen on commutative $L^{p}$-spaces (see Remark 2.8).

Theorem 3.15**.**

Suppose that $T\colon L^{p}(\mathcal{M})\to L^{p}(\mathcal{N})$ is a separating bounded operator, with $1\leq p<\infty$. Then $T$ is $\ell^{1}$-bounded and $\|T\|_{\ell^{1}}=\|T\|$.

Proof.

We apply Proposition 3.11. We let $(w,B,J)$ be the Yeadon triple of the operator $T$. Next according to (9), we let $J=\begin{pmatrix}\pi&0\\ 0&\sigma\end{pmatrix}$ be a central decomposition of $J$ and we let $e,f$ be the central projections such that $\pi=J(\,\cdotp)e$ and $\sigma=J(\,\cdotp)f$.

Let T_{1}\colon L^{p}(\mbox{{\mathcal{M}}})\to L^{p}(\mbox{{\mathcal{N}}}) and T_{2}\colon L^{p}(\mbox{{\mathcal{M}}})\to L^{p}(\mbox{{\mathcal{N}}}) be defined by

[TABLE]

for any x\in L^{p}(\mbox{{\mathcal{M}}}). Then, $T=T_{1}+T_{2}$ and for all u,v\in L^{p}(\mbox{{\mathcal{M}}}), we have

[TABLE]

Let $(x_{n})_{n\geq 1}\in L^{p}(\mathcal{M};\ell^{1})$, with $\|(x_{n})\|_{L^{p}(\mathcal{M};\ell^{1})}<1$. By Lemma 2.3, there exist factorizations $x_{n}=a_{n}b_{n}$, $n\geq 1$, with a_{n},b_{n}\in L^{2p}(\mbox{{\mathcal{M}}}), such that

[TABLE]

The identity (20) and its proof show that for any a\in\mbox{{\mathcal{M}}}\cap L^{2p}(\mbox{{\mathcal{M}}}), we have

[TABLE]

Similarly, $\|B^{\frac{1}{2}}\sigma(a)\|_{2p}^{2p}\leq\|T(\lvert a\rvert^{2})\|_{p}^{p}$. Hence there exist two bounded operators

[TABLE]

such that

[TABLE]

for all $a$ in $\mathcal{M}\cap L^{2p}(\mathcal{M})$. It is clear from above that for any $a$ and $b$ in $L^{2p}(\mathcal{M})$ we have

[TABLE]

We will use this repeatidly in the rest of the proof.

For any $n\geq 1$, we have

[TABLE]

Hence $T(x_{n})=c_{n}d_{n}$, with

[TABLE]

With a similar computation, we obtain

[TABLE]

Let $N\geq 1$ and set $u_{N}=\sum_{n=1}^{N}a_{n}a_{n}^{*}$ and $v_{N}=\sum_{n=1}^{N}b_{n}^{*}b_{n}$. Summing up we obtain

[TABLE]

Appealing to (21), we deduce that

[TABLE]

Similarly,

[TABLE]

Consequently,

[TABLE]

Let $\alpha=\|T_{1}(u_{N})\|_{p}^{p}$ and $\beta=\|T_{1}(v_{N})\|_{p}^{p}$. Since,

[TABLE]

by (21), and $\|u_{N}\|_{p}<1$, we have $\|T_{2}(u)\|_{p}^{p}\leq\|T\|^{p}-\alpha$. Similarly, $\|T_{2}(v)\|_{p}^{p}\leq\|T\|^{p}-\beta.$ Therefore,

[TABLE]

and hence

[TABLE]

This implies that $(T(x_{n}))_{n\geq 1}$ belongs to L^{p}(\mbox{{\mathcal{N}}};\ell^{1}) and that its norm in L^{p}(\mbox{{\mathcal{N}}};\ell^{1}) is less than or equal to $\|T\|$. This yields the boundedness of $T\otimes I_{\ell^{1}}$, as well as the equality $\|T\|_{\ell^{1}}=\|T\|$. ∎

4. Isometries on $L^{2}$-spaces with a Yeadon type factorization

As it is outlined in Section 3, the crucial property that allowed Yeadon to describe isometries between noncommutative $L^{p}$-spaces, $p\not=2$, is that they are separating. To show that every isometry is indeed separating he relied on the property that when $1\leq p\neq 2<\infty$, the equality condition in Clarkson’s inequality, $\displaystyle{\|a+b\|_{p}+\|a-b\|_{p}=2(\|a\|_{p}+\|b\|_{p})}$, holds true if and only if $a$ and $b$ are disjoint. However, this equality always holds true when $p=2$ and this is why the study of disjointness on $L^{2}$-spaces requires a different approach. This is the purpose of Lemma 4.1 below and as a consequence, we will characterize isometries between noncommutative $L^{2}$-spaces which admit a Yeadon type factorization.

Lemma 4.1**.**

Suppose that $a,b\in L^{2}(\mathcal{M})$. Then, $a$ and $b$ are disjoint if and only if we have

[TABLE]

Proof.

First suppose that for disjoint elements $a,b\in L^{2}(\mathcal{M})$ the polar decompositions are given by $a=u\lvert a\rvert$ and $b=v\rvert b\lvert$. Define

[TABLE]

These elements belong to L^{4}(\mbox{{\mathcal{M}}}) and we have $a=a_{1}a_{2}$ and $b=b_{1}b_{2}$. Further we have

[TABLE]

Consequently,

[TABLE]

Now, since $a$ and $b$ are disjoint we have that $\lvert a\rvert\lvert b\rvert=0$, by Lemma 3.5, and so

[TABLE]

Similarly, since $a$ and $b$ are disjoint, we have $ua^{*}bv^{*}=0$, and so

[TABLE]

This implies that \|(a,b)\|_{L^{2}(\tiny\mbox{{\mathcal{M}}};\ell_{2}^{1})}\leq\left(\|a\|^{2}_{2}+\|b\|^{2}_{2}\right)^{1/2}.

Conversely, suppose that $a,b\in L^{2}(\mathcal{M})$ satisfy $\|(a,b)\|_{L^{2}(\mathcal{M};\ell^{1}_{2})}\leq\left(\|a\|_{2}^{2}+\|b\|_{2}^{2}\right)^{1/2}.$ Let $(\varepsilon_{k})_{k}$ be a sequence of positive real numbers with $\lim_{k}\varepsilon_{k}=0$. By Lemma 2.3,

[TABLE]

where the infimum is taken over all factorizations $a=uv$ and $b=wz$, with u,v,w,z\in L^{4}(\mbox{{\mathcal{M}}}). Thus for any $k\geq 1$, we can find u_{k},v_{k},w_{k},z_{k}\in L^{4}(\mbox{{\mathcal{M}}}) such that $a=u_{k}v_{k}$, $b=w_{k}z_{k}$,

[TABLE]

and

[TABLE]

This implies that

[TABLE]

and

[TABLE]

We claim that

[TABLE]

Indeed

[TABLE]

and similarly, \|w_{k}z_{k}\|_{2}^{2}=\langle w_{k}^{*}w_{k},z_{k}z_{k}^{*}\bigr{\rangle}_{L^{2}(\tiny\mbox{{\mathcal{M}}})}. Hence (26) follows by applying the Cauchy-Schwarz inequality in the Hilbertian direct sum L^{2}(\mbox{{\mathcal{M}}})\mathop{\oplus}\limits^{2}L^{2}(\mbox{{\mathcal{M}}}).

Multiplying inequalities (24) and (25) and using the fact that \tau_{\tiny{}_{\mbox{{\mathcal{M}}}}}\left(v_{k}^{*}v_{k}z_{k}^{*}z_{k}\right)\geq 0, we obtain that

[TABLE]

is less than or equal to $(1+\varepsilon_{k})\left(\|u_{k}v_{k}\|_{2}^{2}+\|w_{k}z_{k}\|_{2}^{2}\right)^{2}$. Now using (26) we deduce that (27) is less than or equal to

[TABLE]

Now we observe that $\|u_{k}u_{k}^{*}\|_{2}^{2}=\|u_{k}\|^{4}_{4}=\|u_{k}^{*}u_{k}\|_{2}^{2}$ and similarly for $w_{k}$, $v_{k}$ and $z_{k}$. Hence the above inequality reads

[TABLE]

This yields

[TABLE]

It follows from (22) that $(\|u_{k}\|_{4})_{k}$ and $(\|w_{k}\|_{4})_{k}$ are bounded sequences. Hence we have that \tau_{\tiny{}_{\mbox{{\mathcal{M}}}}}(u_{k}u_{k}^{*}w_{k}w_{k}^{*})\to 0 as $k\to\infty$. Writing

[TABLE]

we deduce that $\|u_{k}^{*}w_{k}\|_{2}\to 0$ as $k\to\infty$.

We have $a^{*}b=v_{k}^{*}u_{k}^{*}w_{k}z_{k}$, hence

[TABLE]

By (23), $(\|v_{k}\|_{4})_{k}$ and $(\|z_{k}\|_{4})_{k}$ are bounded sequences, hence the right hand side in the above inequality tends to [math] as $k\to\infty$. We deduce that $a^{*}b=0$.

Finally using \tau_{\tiny{}_{\mbox{{\mathcal{M}}}}}\left(v_{k}^{*}v_{k}z_{k}^{*}z_{k}\right) instead of \tau_{\tiny{}_{\mbox{{\mathcal{M}}}}}\left(u_{k}u_{k}^{*}w_{k}w_{k}^{*}\right), we show as well that $ab^{*}=0$ and therefore, $a$ and $b$ are disjoint. ∎

Theorem 4.2**.**

For a linear isometry $T\colon L^{2}(\mathcal{M})\to L^{2}(\mathcal{N})$, the following statements are equivalent:

1. (i)

$T$* has a Yeadon type factorization.* 2. (ii)

$T$* is $\ell^{1}_{2}$-contractive.* 3. (iii)

$T$* is $\ell^{1}$-contractive.*

Proof.

In the light of Proposition 3.11 and Theorem 3.15, we only need to establish that if (ii) holds true, then $T$ is separating.

Suppose that $T$ is $\ell^{1}_{2}$-contractive. Let $a,b\in L^{2}(\mathcal{M})$ be disjoint elements. By Lemma 4.1,

[TABLE]

Since $T$ is an isometry we have $\|T(a)\|=\|a\|$ and $\|T(b)\|=\|b\|$ and hence

[TABLE]

By Lemma 4.1 again, this implies that $Ta$ and $Tb$ are disjoint. Hence $T$ is separating. ∎

Remark 4.3**.**

(a) As mentioned in Remark 2.8, when $\mathcal{M}=L^{\infty}(\Omega)$ and $\mathcal{N}=L^{\infty}(\Omega^{\prime})$ are commutative, a bounded operator $T\colon L^{2}(\Omega)\to L^{2}(\Omega^{\prime})$ is $\ell^{1}$-contractive if and only if $T$ is regular, with $\|T\|_{reg}\leq 1$. Hence, Theorem 4.2 implies that in the commutative case, an isometry $T\colon L^{2}(\Omega)\to L^{2}(\Omega^{\prime})$ is separating if and only if $\|T\|_{reg}\leq 1$. This result is implicit in [15].

(b)  Let T\colon L^{p}(\mbox{{\mathcal{M}}})\to L^{p}(\mbox{{\mathcal{N}}}), $1\leq p<\infty$, be a separating isometry, with the Yeadon triple $(w,B,J)$. We show in [14] that when $p=2$ and ${\mathcal{M}}$ and ${\mathcal{N}}$ are hyperfinite, then $J$ is multiplicative (equivalently, $J$ is a $*$-homomorphism) if and only if $T$ is completely regular with $\|T\|_{\text{reg}}=1$. This is an $L^{2}$-analog of [10, Theorem 3.1] which says that for $1\leq p\neq 2<\infty$, $J$ is multiplicative if and only if $T$ is a complete isometry.

In [2], Broise showed that every bijective positive isometry between noncommutative $L^{2}$-spaces associated with semifinite factors admits a Yeadon type factorization. Using Proposition 3.11, one can actually obtain the following more general statement.

Corollary 4.4**.**

Suppose that $T\colon L^{2}(\mathcal{M})\to L^{2}(\mathcal{N})$ is a positive isometry. Then $T$ admits a Yeadon type factorization.

Proof.

Let a,b\in L^{2}(\mbox{{\mathcal{M}}}) be positive elements, with $ab=0$. They are orthogonal and isometries preserve orthogonality, hence $T(a)$ and $T(b)$ are orthogonal. Since $T(a)$ and $T(b)$ are positive, Remark 3.6 ensures that $T(a)$ and $T(b)$ are disjoint.

By Remark 3.12 (a) and Proposition 3.11, the above shows that $T$ admits a Yeadon type factorization. ∎

5. positivity and $\ell^{1}$-contractivity

For any $n\geq 2$, we let $S^{p}_{n}=S^{p}(\ell_{2}^{n})$ and we let S^{p}_{n}[L^{p}(\mbox{{\mathcal{M}}})] be the space S^{p}_{n}\otimes L^{p}(\mbox{{\mathcal{M}}}) equipped with the norm and the partial order coming from its identification with the space L^{p}(M_{n}(\mbox{{\mathcal{M}}})), see Section 2.

We say that a bounded operator $T\colon L^{p}(\mathcal{M})\to L^{p}(\mathcal{N})$, $1\leq p<\infty$, is $n$-positive if

[TABLE]

is positive. We say that $T$ is completely positive if it is $n$-positive for all $n\geq 1$.

Proposition 5.1**.**

Suppose that T\colon L^{p}(\mbox{{\mathcal{M}}})\to L^{p}(\mbox{{\mathcal{N}}}) is a $2$-positive contraction, then $T$ is $\ell^{1}$-contractive.

Proof.

Let T\colon L^{p}(\mbox{{\mathcal{M}}})\to L^{p}(\mbox{{\mathcal{N}}}) be a $2$-positive contraction and let $(x_{n})_{n\geq 1}$ be a sequence in L^{p}(\mbox{{\mathcal{M}}}) such that \|(x_{n})_{n\geq 1}\|_{L^{p}(\tiny\mbox{{\mathcal{M}}};\ell^{1})}<1. According to Lemma 2.3, we may choose sequences $(a_{n})_{n\geq 1}$ and $(b_{n})_{n\geq 1}$ in L^{2p}(\mbox{{\mathcal{M}}}) such that $x_{n}=a_{n}b_{n}$ for any $n\geq 1$,

[TABLE]

For any $n\geq 1$, let

[TABLE]

in S^{p}_{2}[L^{p}(\mbox{{\mathcal{M}}})]. Then $z_{n}=\begin{pmatrix}a_{n}&0\\ b_{n}^{*}&0\end{pmatrix}\begin{pmatrix}a_{n}^{*}&b_{n}\\ 0&0\end{pmatrix},$ hence $z_{n}\geq 0$. Therefore by the $2$-positivity of $T$,

[TABLE]

Consider the positive square root

[TABLE]

Then $\alpha_{n},\beta_{n},\delta_{n}$ belong to L^{2p}(\mbox{{\mathcal{N}}}), we have $\alpha_{n}\geq 0$, $\delta_{n}\geq 0$, and

[TABLE]

Using the third equation above and Junge’s definition of L^{p}(\mbox{{\mathcal{N}}};\ell^{1}), we get that

[TABLE]

(The convergence of the series are justified by the next lines.)

We can now apply the first two equations and (28) to deduce that

[TABLE]

This shows that $T$ is $\ell^{1}$-contractive. ∎

Remark 5.2**.**

(a) An obvious consequence of Proposition 5.1 is that if $T$ is a completely positive contraction then it is $\ell^{1}$-contractive.

(b) Let \mbox{{\mathcal{N}}}^{op} be the opposite von Neumann algebra of ${\mathcal{N}}$ and let I_{op}\colon L^{p}(\mbox{{\mathcal{N}}})\to L^{p}(\mbox{{\mathcal{N}}}^{op}) denote the identity map. We say that T\colon L^{p}(\mbox{{\mathcal{M}}})\to L^{p}(\mbox{{\mathcal{N}}}) is $2$-copositive if the operator I_{op}\circ T\colon L^{p}(\mbox{{\mathcal{M}}})\to L^{p}(\mbox{{\mathcal{N}}}^{op}) is $2$-positive. It is easy to check that

[TABLE]

Therefore, Proposition 5.1 implies that any contractive $2$-copositive map L^{p}(\mbox{{\mathcal{M}}})\to L^{p}(\mbox{{\mathcal{N}}}) is $\ell^{1}$-contractive. It therefore follows that if a positive map L^{p}(\mbox{{\mathcal{M}}})\to L^{p}(\mbox{{\mathcal{N}}}) can be written as a convex combination of a contractive $2$-positive map and a contractive $2$-copositive map, then it is $\ell^{1}$-contractive.

We do not know if any positive contraction is $\ell^{1}$-contractive, however we show below that positive operators are $\ell^{1}$-bounded.

Proposition 5.3**.**

Let T\colon L^{p}(\mbox{{\mathcal{M}}})\to L^{p}(\mbox{{\mathcal{N}}}) be a bounded operator. If $T$ is positive, then $T$ is $\ell^{1}$-bounded, with $\|T\|_{\ell^{1}}\leq 4\|T\|$.

Proof.

As in the proof of Proposition 5.1, let $(x_{n})_{n\geq 1}$ be a sequence in L^{p}(\mbox{{\mathcal{M}}}) such that \|(x_{n})_{n\geq 1}\|_{L^{p}(\tiny\mbox{{\mathcal{M}}};\ell^{1})}<1, and let $(a_{n})_{n\geq 1}$ and $(b_{n})_{n\geq 1}$ in L^{2p}(\mbox{{\mathcal{M}}}) such that $x_{n}=a_{n}b_{n}$ for any $n\geq 1$ and (28) holds.

For any $n\geq 1$, we use the polarization identity,

[TABLE]

For $0\leq k\leq 3$ and $n\geq 1$, let $y_{n}^{k}:=(a_{n}^{*}+i^{k}b_{n})^{*}(a_{n}^{*}+i^{k}b_{n})$. Then $y_{n}^{k}\geq 0$ hence $T(y_{n}^{k})\geq 0$. This implies, by Lemma 2.4, that for any $k$,

[TABLE]

(The convergence of the series are justified by the next lines.)

Moreover

[TABLE]

by (3) and (28).

Since

[TABLE]

we deduce that

[TABLE]

This shows that $\|T\|_{\ell^{1}}\leq 4\|T\|$. ∎

Acknowledgements. The first named author is supported by the French “Investissements d’Avenir” program, project ISITE-BFC (contract ANR-15-IDEX-03). This project was carried out during a visit of the second named author at the “Laboratoire de Mathéma-tiques de Besançon” (LmB). She wishes to thank the LmB for hospitality and support. Finally the authors thank the referee for his/her suggestions and careful reading.