A Beurling-Blecher-Labuschagne theorem for Haagerup noncommutative $L^p$ spaces

TL;DR

This paper extends classical Beurling-Blecher-Labuschagne theorems to Haagerup noncommutative L^p spaces, characterizing invariant subspaces and outer operators within this noncommutative framework.

Contribution

It establishes a Beurling-Blecher-Labuschagne type theorem for invariant subspaces of Haagerup noncommutative L^p spaces and characterizes outer operators in these spaces.

Findings

Proves a Beurling-Blecher-Labuschagne theorem for Haagerup noncommutative L^p spaces.

Provides a characterization of outer operators in Haagerup noncommutative H^p-spaces.

Extends classical invariant subspace theory to noncommutative L^p settings.

Abstract

Let $\mathcal{M}$ be a $\sigma$-finite von Neumann algebra, equipped with a normal faithful state $\varphi$, and let $\mathcal{A}$ be maximal subdiagonal subalgebra of $\mathcal{M}$ and $1\le p<\8$. We prove a Beurling-Blecher-Labuschagne type theorem for $\mathcal{A}$-invariant subspaces of Haagerup noncommutative $L^p(\mathcal{M})$ and give a characterization of outer operators in Haagerup noncommutative $H^{p}$-spaces associated with $\mathcal{A}$.

Equations

$$\mathcal{N}=\mathcal{M}{⋊}_{{\sigma }^{\phi }}\mathbb{R}$$

$$\left\{\pi (x):x\in \mathcal{M}\right\}\cup \left\{\lambda (s):s\in \mathbb{R}\right\},$$

$$(\pi (x)\xi )(t)={\sigma }_{-t}^{\phi }(x)\xi (t),\forall \xi \in L^2(\mathbb{R},\mathcal{H}),\forall t\in \mathbb{R},$$

$$(\lambda (s)\xi )(t)=\xi (t-s),\forall \xi \in L^2(\mathbb{R},\mathcal{H}),\forall t\in \mathbb{R}.$$

$$\lambda (t)\pi (x)\lambda (t{)}^{\ast }=\pi ({\sigma }_t^{\phi }(x)),\forall t\in \mathbb{R},\forall x\in \mathcal{M}.$$

$${\sigma }_t^{\phi }(x)=\lambda (t)x{\lambda }^{\ast }(t),x\in \mathcal{M},t\in \mathbb{R}.$$

$${\hat{\sigma }}_t(x)=W(t)x{W}^{\ast }(t),\forall x\in \mathcal{N},\forall t\in \mathbb{R},$$

$$W(t)(\xi )(s)=e^{-its}\xi (s),\forall \xi \in L^2(\mathbb{R},\mathcal{H}),\forall s,t\in \mathbb{R}.$$

$${\hat{\sigma }}_t(x)=x\text{mbox}and{\hat{\sigma }}_t(\lambda (s))=e^{-ist}\lambda (s),\forall t\in \mathbb{R}.$$

$$\mathcal{M}=\{x\in \mathcal{N}:{\hat{\sigma }}_t(x)=x,\forall t\in \mathbb{R}\}.$$

$$\tau \circ {\hat{\sigma }}_t=e^{-t}\tau ,\forall t\in \mathbb{R}.$$

$$\hat{\phi }(x)=\tau (Dx),x\in {\mathcal{N}}_{+}.$$

$$\lambda (t)=D^{it},\forall t\in \mathbb{R}.$$

$$L^p(\mathcal{M},\phi )=\{x\in L^0(\mathcal{N},\tau ):{\hat{\sigma }}_t(x)=e^{-\frac{t}{p}}x,\forall t\in \mathbb{R}\}.$$

$$L^{\infty }(\mathcal{M},\phi )=\mathcal{M}.$$

$$\hat{\psi }(x)=\tau (D_{\psi }x),x\in {\mathcal{N}}_{+}.$$

$$D_{\psi }\in L^0(\mathcal{N},\tau )$$

$${\hat{\sigma }}_t(D_{\psi })=e^{-t}{D}_{\psi },\forall t\in \mathbb{R}.$$

$$D_{\psi }\in L^1(\mathcal{M},\phi {)}_{+}.$$

$$tr(D_{\psi })=\psi (1),\psi \in {\mathcal{M}}_{\ast }.$$

$$\Vert x{\Vert }_p=\Vert |x{|}^p{\Vert }_1^{\frac{1}{p}},\forall x\in L^p(\mathcal{M},\phi ),$$

$$(L^p(\mathcal{M},\phi ),\Vert \cdot {\Vert }_p)$$

$$\Vert x{\Vert }_p=\Vert x^{\ast }{\Vert }_p=\Vert |x|{\Vert }_p,\forall x\in L^p(\mathcal{M},\phi ).$$

$$L^p(\mathcal{M})\times L^q(\mathcal{M})\to L^r(\mathcal{M}),$$

$$tr(xy)=tr(yx),\forall x\in L^p(\mathcal{M}),\forall y\in L^q(\mathcal{M}).$$

$$L^p(\mathcal{M})=[D^{\frac{1-\eta }{p}}\mathcal{M}{D}^{\frac{\eta }{p}}{]}_p.$$

$${\mathcal{A}}_0=\{x\in \mathcal{A}:\mathcal{E}(x)=0\}$$

$$\mathcal{A}=\{x\in \mathcal{M}:\mathcal{E}(yxz)=\mathcal{E}(yxz)=0,\forall y\in \mathcal{A},\forall z\in {\mathcal{A}}_0\}.$$

$${\sigma }_t^{\phi }(\mathcal{A})=\mathcal{A},\forall t\in \mathbb{R}.$$

$$H^p(\mathcal{A})=[\mathcal{A}{D}^{\frac{1}{p}}{]}_p,H_0^p(\mathcal{A})=[{\mathcal{A}}_0{D}^{\frac{1}{p}}{]}_p.$$

Full text

1 Introduction

Arveson introduced his notion of subdiagonal subalgebras of von Neumann algebras (see A ), in effect, subdiagonal algebras are the noncommutative analogue of weak* Dirichlet algebras (for the definition of weak* Dirichlet algebras see SW ). For the finite and semi-finite case, most results on the classical Hardy spaces on the torus have been established in this noncommutative setting. We refer to A ; BX ; B1 ; B2 ; BL1 ; BL2 ; BL3 ; BL4 ; PX ; Sa (see also BL4 for more historical references). It is natural to consider the case of $\sigma$-finite von Neumann algebras. But, the transition from finite or semifinite to $\sigma$-finite von Neumann algebras is not trival, need some new techniques and some changes. For some results for this case, see BM ; JOS ; J1 ; J2 ; L2 ; X .

Let ${\mathcal{M}}$ be a finite von Neumann and ${\mathcal{A}}$ be its Arveson’s maximal subdiagonal subalgebras. In BL3 , Blecher and Labuschagne extended the classical Beurling’s theorem to describe closed ${\mathcal{A}}$-invariant subspaces in noncommutative space $L^{p}({\mathcal{M}})$ with $1\leq p\leq\infty$. Sager Sa extended the work of Blecher and Labuschagne from a finite von Neumann algebra to semifinite von Neumann algebras, proved a Beurling-Blecher-Labuschagne theorem for ${\mathcal{A}}$-invariant spaces of $L^{p}({\mathcal{M}})$ when $0<p\leq\infty$. The Beurling theorem has been generalized to the setting of unitarily invariant norms on finite and semifinite von Neumann algebras (see B2 , CHS , SL ).

When ${\mathcal{A}}$ is subdiagonal subalgebra of $\sigma$-finite von Neumann ${\mathcal{M}}$, Labu-

schagne L2 showed that a Beurling type theory of invariant subspaces of noncommutative $H^{2}$-spaces holds true. A motivation for this paper is to extend the result in L2 to the setting of the Haagerup noncommutative $L^{p}$-spaces for $1\leq p<\infty$.

Blecher and Labuschagne BL1 studied outer operators of the noncommutative $H^{p}$-spaces associated with Arveson’s subdiagonal subalgebras. They proved inner-outer factorization theorem and characterizations of outer operators for the case $1\leq p<\infty$ (for the case $p<1$, see BX ). In BL2 , they extended their generalized inner-outer factorization theorem in BL1 and established characterizations of outer operators that are valid even in the case of operators with zero determinant. In this paper, we apply Labuschagne’s Beurling type theorem for ${\mathcal{A}}$-invariant subspaces of Haagerup noncommutative $L^{2}$-spaces to prove a Blecher-Labuschagne theorem for outer operators in Haagerup noncommutative $H^{p}$-spaces ($1\leq p<\infty$).

The organization of the paper is as follows. In Section 2, we give some definitions and related results of Haagerup noncommutative $L^{p}$-spaces and $H^{p}$-spaces. A Blecher-Labuschagne-Beurling type theorem for Haagerup noncommutative $L^{p}$-spaces is presented in Section 3. In Section 4, we give characterizations of outer operators in Haagerup noncommutative $H^{p}$-spaces.

2 Preliminaries

Our references for modular theory are PT ; T3 , for the Haagerup noncommutative $L^{p}$-spaces are H1 ; Te and for the Haagerup noncommutative $H^{p}$-spaces are J1 ; J2 . Let us recall some basic facts about the Haagerup noncommutative $L^{p}$-spaces and the Haagerup noncommutative $H^{p}$-spaces, and fix the relevant notation used throughout this paper. Throughout this paper $\mathcal{M}$ will always denote a $\sigma$-finite von Neumann algebra on a complex Hilbert space $\mathcal{H}$, equipped with a distinguished normal faithful state $\varphi$. Let $\{\sigma_{t}^{\varphi}\}_{t\in{\mathbb{R}}}$ be the one parameter modular automorphism group of $\mathcal{M}$ associated with $\varphi$. We denote by

[TABLE]

the crossed product of $\mathcal{M}$ by $\{\sigma_{t}^{\varphi}\}_{t\in\mathbb{R}}$. It is well known that $\mathcal{N}$ is the semi-finite von Newmann algebra acting on the Hilbert space $L^{2}(\mathbb{R},\mathcal{H}),$ generated by

[TABLE]

where the operator $\pi(x)$ is defined by

[TABLE]

and the operator $\lambda(s)$ is defined by

[TABLE]

We will identify $\mathcal{M}$ and the subalgebra $\pi(\mathcal{M})$ of $\mathcal{N}$. The operators $\pi(x)$ and $\lambda(t)$ satisfy

[TABLE]

Then

[TABLE]

We denote by $\{\hat{\sigma}_{t}\}_{t\in\mathbb{R}}$ the dual action of $\mathbb{R}$ on $\mathcal{N}$, this is a one parameter automorphism group of $\mathbb{R}$ on $\mathcal{N},$ implemented by the unitary representation $\{W_{t}\}_{t\in\mathbb{R}}$ of $\mathbb{R}$ on $L^{2}(\mathbb{R},\mathcal{H}):$

[TABLE]

where

[TABLE]

Note that the dual action $\hat{\sigma}_{t}$ is uniquely determined by the following conditions: for any $x\in{\mathcal{M}}$ and $s\in\mathbb{R}$,

[TABLE]

Hence

[TABLE]

Let $\tau$ be the unique normal semi-finite faithful trace on $\mathcal{N}$ satisfying

[TABLE]

Also recall that the dual weight $\hat{\varphi}$ of our distinguished state $\varphi$ has the Radon -Nikodym derivative $D$ with respect to $\tau$, which is the unique invertible positive selfadjoint operator on $L^{2}(\mathbb{R},\mathcal{H}),$ affiliated with $\mathcal{N}$ such that

[TABLE]

Recall that the regular representation of the above $\lambda(t)$ is given by

[TABLE]

Now, we define Haagerup noncommutative $L^{p}$-spaces. Let $L^{0}(\mathcal{N},\tau)$ denote the topological $\ast$-algebra of all operators on $L^{2}(\mathbb{R},\mathcal{H})$ measurable with respect to $(\mathcal{N},\tau)$. Then the Haagerup noncommutative $L^{p}$-spaces, $0<p\leq\infty$, are defined by

[TABLE]

The spaces $L^{p}(\mathcal{M},\varphi)$ are closed selfadjoint linear subspaces of $L^{0}(\mathcal{N},\tau)$. It is not hard to show that

[TABLE]

Since for any $\psi\in\mathcal{M}_{\ast}^{+},$ the dual weight $\hat{\psi}$ has a Radon-Nikodym derivative with respect to $\tau,$ denoted by $D_{\psi}:$

[TABLE]

Then

[TABLE]

and

[TABLE]

So

[TABLE]

It is well known that the map $\psi\mapsto D_{\psi}$ on $\mathcal{M}_{\ast}^{+}$ extends to a linear homeomorphism from $\mathcal{M}_{\ast}$ onto $L^{1}(\mathcal{M},\varphi)$ (equipped with the vector space topology inherited from $L^{0}({\mathcal{N}},\tau)$). This permits to transfer the norm on $\mathcal{M}_{\ast}$ into a norm on $L^{1}(\mathcal{M},\varphi)$, denoted by $\|\cdot\|$. Moreover, $L^{1}(\mathcal{M},\varphi)$ is equipped with a distinguished contractive positive linear functional tr, defined by

[TABLE]

Therefore, $\|x\|_{1}=tr(|x|)$ for every $x\in L^{1}(\mathcal{M},\varphi)$.

Let $0<p<\infty$ and $x\in L^{0}({\mathcal{N}},\tau)$. If $x=u|x|$ is the polar decomposition of $x$, then $x\in L^{p}({\mathcal{M}},\varphi)\;\Leftrightarrow\;u\in{\mathcal{M}}\;\mbox{and}\;|x|\in L^{p}({\mathcal{M}},\varphi)\;\Leftrightarrow\;u\in{\mathcal{M}}\;\mbox{and}\;|x|^{p}\in L^{1}({\mathcal{M}},\varphi)$. If we define

[TABLE]

then for $1\leq p<\infty$ (resp. $0<p<1$),

[TABLE]

is a Banach space (resp. a quasi-Banach space), and

[TABLE]

It is proved in H1 and Te that $L^{p}(\mathcal{M},\varphi)$ is independent of $\varphi$ up to isometry. Hence, we denote $L^{p}(\mathcal{M},\varphi)$ by $L^{p}(\mathcal{M})$.

The usual Holder inequality also holds for the $L^{p}(\mathcal{M})$ spaces. It means that the product of $L^{0}({\mathcal{N}},\tau),\;(x,y)\mapsto xy$, restricts to a contractive bilinear map

[TABLE]

where $\frac{1}{r}=\frac{1}{p}+\frac{1}{q}$. In particular, if $\frac{1}{p}+\frac{1}{q}=1$, then the bilinear form $(x,y)\mapsto tr(xy)$ defines a duality bracket between $L^{p}(\mathcal{M})$ and $L^{q}(\mathcal{M})$, for which $L^{q}(\mathcal{M})$ coincides (isometrically) with the dual of $L^{p}(\mathcal{M})$ (if $p\neq\infty$). Moreover, the $tr$ have the following property:

[TABLE]

Let $0<p\leq\infty$. For $K\subset L^{p}(\mathcal{M})$, we denote the closed linear span of $K$ in $L^{p}(\mathcal{M})$ by $[K]_{p}$ (relative to the w*-topology for $p=\infty$) and the set $\{x^{\ast}:\;x\in K\}$ by $J(K)$.

For $0<p<\infty,\;0\leq\eta\leq 1$, we have that

[TABLE]

Let $\mathcal{D}$ be a von Neumann subalgebra of $\mathcal{M}$ and $\mathcal{E}$ be a faithful normal conditional expectation from ${\mathcal{M}}$ onto ${\mathcal{D}}$.

Definition 1

A w*-closed subalgebra $\mathcal{A}$ of $\mathcal{M}$ is called a subdiagonal subalgebra of $\mathcal{M}$ with respect to $\mathcal{E}$(or to $\mathcal{D}$) if

1. (i)

$\mathcal{A}+J(\mathcal{A})$ is w*-dense in $\mathcal{M}$, 2. (ii)

$\mathcal{E}(xy)=\mathcal{E}(x)\mathcal{E}(y),\quad\forall\;x,y\in\mathcal{A},$ 3. (iii)

$\mathcal{A}\cap J(\mathcal{A})=\mathcal{D},$

The algebra $\mathcal{D}$ is called the diagonal of $\mathcal{A}$.

In A , subdiagonal subalgebras are not assumed to be w*-weakly closed. Since the weak* closure of an algebra that is subdiagonal with respect to ${\mathcal{E}}$ will also be subdiagonal with respect to ${\mathcal{E}}$ (see Remark 2.1.2 in A ), we may assume that our subdiagonal subalgebras are always w*-weakly closed (the definition as in J1 ; J2 ; X ). Since ${\mathcal{M}}$ is $\sigma$-finite, we may take a faithful normal state $\phi$ on ${\mathcal{M}}$ such that $\phi\circ{\mathcal{E}}=\phi$. It is well known (cf. T3 ) that the existence of a (unique) normal conditional expectation ${\mathcal{E}}:{\mathcal{M}}\rightarrow{\mathcal{D}}$ such that $\varphi\circ{\mathcal{E}}=\varphi$ is equivalent to $\sigma_{t}^{\varphi}({\mathcal{D}})={\mathcal{D}}$ for all $t\in\mathbb{R}$. Hence, in the rest of this paper $\varphi$ always denotes a normal faithful state satisfying $\varphi\circ{\mathcal{E}}=\varphi$.

If $\mathcal{A}$ is not properly contained in any other subalgebra of $\mathcal{M}$ which is a subdiagonal with respect to $\mathcal{E}$, We call $\mathcal{A}$ is a maximal subdiagonal subalgebra of $\mathcal{M}$ with respect to $\mathcal{E}$ (or to $\mathcal{D}$). Let

[TABLE]

Then by (A, , Theorem 2.2.1), $\mathcal{A}$ is maximal if and only if

[TABLE]

It follows from (JOS, , Theorem 2.4) and (X, , Theorem 1.1) (also see (L2, , Theorem 1.1)) that a subdiagonal subalgebra ${\mathcal{A}}$ of $\mathcal{M}$ with respect to $\mathcal{D}$ is maximal if and only if

[TABLE]

In this paper $\mathcal{A}$ always denotes a maximal subdiagonal subalgebra in $\mathcal{M}$ with respect to $\mathcal{E}$.

Definition 2

For $0<p<\infty$, we define the Haagerup noncommutative $H^{p}$-space that

[TABLE]

If $1\leq p<\infty,\;0\leq\eta\leq 1$, then by (J2, , Proposition 2.1), we have that

[TABLE]

By (BM, , Proposition 2.7), we know that

[TABLE]

It is known that

[TABLE]

Therefore, if $1\leq p,q,r<\infty$ and $\frac{1}{q}+\frac{1}{r}=\frac{1}{p}$, then

[TABLE]

For $1\leq p\leq\infty$, the conditional expectation $\mathcal{E}$ extends to a contractive projection from $L^{p}(\mathcal{M})$ onto $L^{p}(\mathcal{D})$. The extension will be denoted still by $\mathcal{E}$ (see (JX, , Proposition 2.3)). Let

[TABLE]

Then

[TABLE]

Let $\mathcal{M}_{a}$ be the family of analytic vectors in $\mathcal{M}.$ Recall that $x\in\mathcal{M}_{a}$ if only if the function $t\mapsto\sigma_{t}(x)$ extends to an analytic function from $\mathbb{C}$ to $\mathcal{M}.$ $\mathcal{M}_{a}$ is a w*-dense $\ast$-subalgebra of $\mathcal{M}$ (cf. PT ).

The next result is known. For easy reference, we give its proof (see the proof of Theorem 2.5 in J1 ).

Lemma 1

Let $\mathcal{A}_{a}$ and ${\mathcal{D}}_{a}$ be respectively the families of analytic vectors in $\mathcal{A}$ and ${\mathcal{D}}$. If $1\leq p<\infty$, then:

1. (i)

$\mathcal{A}_{a}$* is a w*-dense in $\mathcal{A}$, $({\mathcal{A}}_{a})_{0}$ is a w*-dense in ${\mathcal{A}}_{0}$ and ${\mathcal{D}}_{a}$ is a w*-dense in ${\mathcal{D}}$, where $({\mathcal{A}}_{a})_{0}=\{x\in{\mathcal{A}}_{a}:\;\mathcal{E}(x)=0\}$;* 2. (ii)

[TABLE] 3. (iii)

$\mathcal{A}_{a}D^{\frac{1}{p}}$* is dense in $H^{p}({\mathcal{A}})$, $(\mathcal{A}_{a})_{0}D^{\frac{1}{p}}$ is dense in $H^{p}_{0}({\mathcal{A}})$ and ${\mathcal{D}}_{a}D^{\frac{1}{p}}$ is dense in $L^{p}({\mathcal{D}})$.*

Proof

(i) Let $x\in\mathcal{A}$. We define

[TABLE]

By (2.2), $x_{n}\in\mathcal{A}$. Moreover by (PT, , p. 58), $x_{n}\in\mathcal{A}_{a}$ and $x_{n}\rightarrow x$ w*-weakly. Since

[TABLE]

(see (JOS, , p. 313)), a similar argument works for ${\mathcal{A}}_{0}$ and ${\mathcal{D}}$.

(ii) We prove only the first equivalence. The proofs of the two others are similar. Let $x\in\mathcal{A}_{a}$. Then

[TABLE]

whence $D^{\pm\frac{1}{p}}x\subseteq\mathcal{A}_{a}D^{\pm\frac{1}{p}}$. The inverse inclusion can be proved in a similar way.

(iii) Let $p^{\prime}$ be the conjugate index of $p$. If $y\in L^{p^{\prime}}({\mathcal{M}})$ such that $tr(aD^{\frac{1}{p}}y)=0,\;\forall a\in\mathcal{A}_{a}$, then by (i),

[TABLE]

since $D^{\frac{1}{p}}y\in L^{1}({\mathcal{M}})$. Hence, by (2.3),

[TABLE]

By the Hahn-Banach theorem, ${\mathcal{A}}_{a}D^{\frac{1}{p}}$ is dense in $H^{p}({\mathcal{A}})$. Similarly, we can prove the two others.

3 ${\mathcal{A}}$-invariant subspaces of $L^{p}({\mathcal{M}})$

We recall that a right (resp. left) ${\mathcal{A}}$-invariant subspace of $L^{p}({\mathcal{M}})$, is a closed subspace $K$ of $L^{p}({\mathcal{M}})$ such that $K\mathcal{A}\subset K$ (resp. $\mathcal{A}K\subset K$).

In the case when von Neumann algebra ${\mathcal{M}}$ is finite, for a right ${\mathcal{A}}$-invariant subspace $K$ of $L^{2}({\mathcal{M}})$, Blecher and Labuschagne BL3 defined the right wandering subspace of $K$ to be the space $W=K\ominus[K{\mathcal{A}}_{0}]_{2}$; and they say that $K$ is type 1 if $W$ generates $K$ as an ${\mathcal{A}}$-module (that is, $K=[W{\mathcal{A}}]_{2}$) and say that $K$ is type 2 if $W=\{0\}$ (also see NW , but the last notation conflicts with that of NW , where this class of subspaces is decomposed into two further subclasses which Nakazi and Watatani call type II and type III). If $p\neq 2$, Blecher and Labuschagne BL3 defined the wandering quotient to be $K/[K{\mathcal{A}}_{0}]_{p}$, and say that $K$ is type 2 if this is trivial. It turns out that the wandering quotient is an $L^{p}({\mathcal{D}})$-module in the sense of Junge and Sherman (see JS ), and it is isometric to a canonically defined subspace of $K$ which can be called the right wandering subspace of $K$. They say that $K$ is type 1 if this subspace generates $K$ as an ${\mathcal{A}}$-module. For the case $1\leq p<2$ (resp. $p>2$), they have shown that $K$ is type 1 iff $K\cap L^{2}({\mathcal{M}})$ (resp. $[K]_{2}$) is type 1 in the sense of the $L^{2}$ case above.

Now, in the case that $\mathcal{M}$ is a $\sigma$-finite von Neumann algebra. Recall that if $K$ is a right $\mathcal{A}$-invariant subspace of $L^{2}({\mathcal{M}})$, then

[TABLE]

is often called the right wandering subspace of $K$. We say that $K$ is type 1 if $W$ generates $K$ as an $\mathcal{A}$-module (that is $K=[WA]_{2})$ and $K$ is type 2 if $W=\{0\}$ (see L2 ).

Proposition 1

Let $1\leq p,q,r<\infty$, and $K$ be a closed subspace of $L^{p}(\mathcal{M})$. Suppose $\frac{1}{p}-\frac{1}{r}=\frac{1}{q}$ and $K_{r}=\{x\in K:\;xD^{-\frac{1}{r}}\in L^{q}({\mathcal{M}})\}$. If $[K_{r}]_{p}=K$, then

[TABLE]

Proof

(1) If $x\in[K_{r}D^{-\frac{1}{r}}]_{q}$, then there is a sequence $(x_{n})\subset K_{r}$ such that $x_{n}D^{-\frac{1}{r}}\rightarrow x$ in norm in $L^{q}({\mathcal{M}})$. Hence, $x_{n}\rightarrow xD^{\frac{1}{r}}$ in norm in $L^{p}({\mathcal{M}})$. It follows that $[K_{r}D^{-\frac{1}{r}}]_{q}D^{\frac{1}{r}}\subset K$, and so $[[K_{r}D^{-\frac{1}{r}}]_{q}D^{\frac{1}{r}}]_{p}\subset K$. On the other hand, since $K_{r}\subset[K_{r}D^{-\frac{1}{r}}]_{q}D^{\frac{1}{r}}$, $K=[K_{r}]_{p}\subset[[K_{r}D^{-\frac{1}{r}}]_{q}D^{\frac{1}{r}}]_{p}$. Therefore, we obtain the desired result.

Lemma 2

Let $1\leq p<\infty$, and let $K$ be an $\mathcal{A}$-invariant subspace of $L^{p}({\mathcal{M}})$.

1. (i)

If $1\leq q,r<\infty$ and $\frac{1}{p}-\frac{1}{r}=\frac{1}{q}$, then $[K_{r}D^{-\frac{1}{r}}]_{q}$ is a right $\mathcal{A}$-invariant subspace of $L^{q}({\mathcal{M}})$, where $K_{r}=\{x\in K:\;xD^{-\frac{1}{r}}\in L^{q}({\mathcal{M}})\}$. 2. (ii)

If $1\leq q,r<\infty$ and $\frac{1}{p}+\frac{1}{r}=\frac{1}{q}$, then $[KD^{\frac{1}{r}}]_{q}$ is a right $\mathcal{A}$-invariant subspace of $L^{q}({\mathcal{M}})$.

Proof

(i) It is clear that $[K_{r}D^{-\frac{1}{r}}]_{q}\subset L^{q}({\mathcal{M}})$. Using (ii) of Lemma 1, we get that

[TABLE]

On the other hand, for any $a\in{\mathcal{A}}_{a}$ and $x\in K_{r}$, we have that $xa\in K$. By (3.1), there is an element $a^{\prime}$ of ${\mathcal{A}}_{a}$ such that $xaD^{-\frac{1}{r}}=xD^{-\frac{1}{r}}a^{\prime}$. It follows that $xaD^{-\frac{1}{r}}\in L^{q}({\mathcal{M}})$, and so $xa\in K_{r}$. Hence, $K_{r}{\mathcal{A}}_{a}\subset K_{r}$. From (3.1) follows that $K_{r}D^{-\frac{1}{r}}\mathcal{A}_{a}\subset K_{r}D^{-\frac{1}{r}}$ and

[TABLE]

Now if $a\in{\mathcal{A}}$, then by (i) in Lemma 1, we have a sequence $(a_{n})$ in ${\mathcal{A}}_{a}$ such that $a_{n}\rightarrow a$ w*-weakly. Hence,

[TABLE]

where $q^{\prime}$ is the conjugate index of $q$. Since the weak closure of $K_{r}D^{-\frac{1}{r}}\mathcal{A}_{a}$ is equal to $[K_{r}D^{-\frac{1}{r}}\mathcal{A}_{a}]_{q}$,

[TABLE]

Using (3.2), we get

[TABLE]

Therefore,

[TABLE]

(ii) can be proved in a similar way.

Using same method as in the proof of Lemma 2, we get the following result.

Lemma 3

Let $1\leq p<\infty$, and let $K\subset L^{p}({\mathcal{M}})$. If $1\leq q,r<\infty$ and $\frac{1}{p}+\frac{1}{r}=\frac{1}{q}$, then

[TABLE]

and

[TABLE]

Lemma 4

Let $1\leq p<\infty$. If $1<q,r<\infty$ and $\frac{1}{p}-\frac{1}{r}=\frac{1}{q}$, then

[TABLE]

Proof

Let $x\in H^{p}({\mathcal{A}})D^{-\frac{1}{r}}\cap L^{q}({\mathcal{M}})$. Then there is an element $y\in H^{p}({\mathcal{A}})$ such that $x=yD^{-\frac{1}{r}}$. If $q^{\prime}$ (resp. $p^{\prime}$) is the conjugate index of $q$ (resp. $p$), then $\frac{1}{q^{\prime}}=\frac{1}{p^{\prime}}+\frac{1}{r}$. Hence,

[TABLE]

Using (2.3), we get $x\bot J(H^{q^{\prime}}_{0}({\mathcal{A}}))$. By (J2, , Corollary 3.4) (or (BM, , (2.13))), $x\in H^{q}({\mathcal{A}})$, and so $H^{p}({\mathcal{A}})D^{-\frac{1}{r}}\cap L^{q}({\mathcal{M}})\subset H^{q}({\mathcal{A}})$. Conversely, from $H^{q}({\mathcal{A}})D^{\frac{1}{r}}\subset H^{p}({\mathcal{A}})$ it follows that $H^{p}({\mathcal{A}})D^{-\frac{1}{r}}\cap L^{p}({\mathcal{M}})\supset H^{q}({\mathcal{A}})$. Thus, we obtain the first result. The second result follows analogously.

Definition 3

Let $1\leq p<\infty$, and let $K$ be a right $\mathcal{A}$-invariant subspace of $L^{p}({\mathcal{M}})$.

1. (i)

If $1\leq p\leq 2,\;\frac{1}{p}-\frac{1}{r}=\frac{1}{2}$ and $W$ is the right wandering subspace of $[KD^{-\frac{1}{r}}\cap L^{2}({\mathcal{M}})]_{2}$, we define the right wandering subspace of K to be the $L^{p}$-closure of $WD^{\frac{1}{r}}$. 2. (ii)

If $2\leq p<\infty,\;\frac{1}{p}+\frac{1}{r}=\frac{1}{2}$ and $W$ is the right wandering subspace of $[KD^{\frac{1}{r}}]_{2}$, we define the right wandering subspace of K to be the $L^{p}$-closure of $W_{r}D^{-\frac{1}{r}}$, where $W_{r}=\{x\in W:\;xD^{-\frac{1}{r}}\in L^{p}({\mathcal{M}})\}$.

If $K$ is a right $\mathcal{A}$-invariant subspace of $L^{p}(\mathcal{M})$, we say that $K$ is type 1 if the right wandering subspace of $K$ generates $K$ as an $\mathcal{A}$-module, and $K$ is type 2 if $1\leq p<2$ (resp. $p>2$) and $K=[K\mathcal{A}_{0}]_{p}$ (resp. $[KD^{\frac{1}{r}}]_{2}=[KD^{\frac{1}{r}}\mathcal{A}_{0}]_{2}$, where $\frac{1}{p}+\frac{1}{r}=\frac{1}{2}$).

To extend the result in L2 to the setting of the Haagerup noncommutative $L^{p}$-spaces ($1\leq p<\infty$), we will use the column $L^{p}$-sum studied by Junge and Sherman JS to investigate this: If $X$ is a subspace of $L^{p}({\mathcal{M}})$, and if $\{X_{i}:i\in I\}$ is a collection of subspaces of X, which together densely span $X$, with the property that $X^{*}_{i}X_{j}=\{0\}$ if $i\neq j$, then we say that $X$ is the internal column $L^{p}$-sum $\oplus_{i}^{col}X_{i}$.

Theorem 3.1

Let $1\leq p<2$ and $K$ be a right $\mathcal{A}$-invariant subspace of $L^{p}(\mathcal{M})$. Suppose $\frac{1}{p}-\frac{1}{r}=\frac{1}{2}$ and $K_{r}=\{x\in K:\;xD^{-\frac{1}{r}}\in L^{2}({\mathcal{M}})\}$. If $[K_{r}]_{p}=K$, then:

1. (i)

$K$* may be written uniquely as an $L^{p}$-column sum $Z\oplus^{col}[Y{\mathcal{A}}]_{p}$, where $Z$ is a type 2 right $\mathcal{A}$-invariant subspace of $L^{p}({\mathcal{M}})$, $Y$ is the right wandering subspace of $K$ such that $Y=[Y{\mathcal{D}}]_{p}$ and $J(Y)Y\subset L^{\frac{p}{2}}({\mathcal{D}})$.* 2. (ii)

If $K\neq\{0\}$ then K is type 1 if and only if $K=\oplus_{i}^{col}u_{i}H^{p}(\mathcal{A})$, for $u_{i}$ partial isometries with mutually orthogonal ranges and $u_{i}^{*}u_{i}\in\mathcal{D}$. 3. (iii)

If $K=K_{1}\oplus^{col}K_{2}$ where $K_{1}$ and $K_{2}$ are types 2 and 1 respectively, then the right wandering subspace for $K$ equals the right wandering subspace for $K_{2}$. 4. (iv)

The wandering quotient $K/[K{\mathcal{A}}_{0}]p$ is isometrically ${\mathcal{D}}$-isomorphic to the right wandering subspace of K. 5. (v)

The wandering subspace $W$ of $K$ is an $L^{p}({\mathcal{D}})$-module in the sense of Junge and Sherman.

Proof

(i) By Lemma 2, $K^{\prime}=[K_{r}D^{-\frac{1}{r}}]_{2}$ is a right $\mathcal{A}$-invariant subspace of $L^{2}({\mathcal{M}})$. Using Theorem 2.3 and 2.8 in L2 , we have that

[TABLE]

where $Z^{\prime}$ is a type 2 right $\mathcal{A}$-invariant subspace of $L^{2}({\mathcal{M}})$ and $Y^{\prime}$ is the right wandering subspace of $K^{\prime}$ with $Y^{\prime}=[Y^{\prime}{\mathcal{D}}]_{2}$ and $J(Y^{\prime})Y^{\prime}\subset L^{1}({\mathcal{D}})$. Let $Z=[Z^{\prime}D^{\frac{1}{r}}]_{p}$ and $Y=[Y^{\prime}D^{\frac{1}{r}}]_{p}$. By Lemma 2 and Definition 3, $Z$ is a right $\mathcal{A}$-invariant subspaces of $L^{p}({\mathcal{M}})$ and $Y$ is the right wandering subspace of $K$. Using Lemma 3, we know that $[[Y^{\prime}{\mathcal{A}}]_{2}D^{\frac{1}{r}}]_{p}=[Y{\mathcal{A}}]_{p}$. For any $x\in Z^{\prime},y\in[Y^{\prime}{\mathcal{A}}]_{2}$, we have that $x^{*}y=0$, and so

[TABLE]

Hence, $J(Z)[Y{\mathcal{A}}]_{p}=\{0\}$. On the other hand, by Proposition 1, $K=[K^{\prime}D^{\frac{1}{r}}]_{p}$. Therefore,

[TABLE]

Since $Z^{\prime}=[Z^{\prime}{\mathcal{A}}_{0}]_{2},\;Y^{\prime}=[Y^{\prime}{\mathcal{D}}]_{2}$, by Lemma 3,

[TABLE]

and

[TABLE]

Since

[TABLE]

it follows that $J(Y)Y\subset L^{\frac{p}{2}}({\mathcal{D}})$.

Now we prove the uniqueness. Suppose that $Z_{1}$ is a type 2 right $\mathcal{A}$-invariant subspace of $L^{p}({\mathcal{M}})$ and $Y_{1}$ is the right wandering subspace of $K$ such that

[TABLE]

Since $Y_{1}$ is the right wandering subspace of $K$, by Definition 3, $Y_{1}=[Y_{1}^{\prime}D^{\frac{1}{r}}]_{p}$, where $Y_{1}^{\prime}$ is the right wandering subspace of $[KD^{-\frac{1}{r}}\cap L^{2}({\mathcal{M}})]_{2}=[K_{r}D^{-\frac{1}{r}}]_{2}$. By by the uniqueness assertion in Theorem 2.3 of L2 , $Y^{\prime}=Y^{\prime}_{1}$. It follows that $Y_{1}=Y$. From $K=Z_{1}\oplus^{col}[Y{\mathcal{A}}]_{p}=Z\oplus^{col}[Y{\mathcal{A}}]_{p}$, we obtain that $Z_{1}=Z$.

(ii) Let $K\neq\{0\}$ and $K$ is type 1. From the proof of (1), we know that $[K_{r}D^{-\frac{1}{r}}]_{2}$ is type 1. So, by (L2, , (ii) of Theorem 2.8), there are partial isometries $u_{i}$ with mutually orthogonal ranges such that $u_{i}^{*}u_{i}\in{\mathcal{D}}$,

[TABLE]

Using Proposition 1 and (2.3), we get

[TABLE]

Conversely, let for $u_{i}$ as above,

[TABLE]

By Lemma 4, $[H^{p}({\mathcal{A}})D^{-\frac{1}{r}}\cap L^{2}({\mathcal{M}})]_{2}=H^{2}(\mathcal{A})$. Hence,

[TABLE]

So

[TABLE]

Hence, the right wandering subspace $W$ of $[K_{r}D^{-\frac{1}{r}}]_{2}$ satisfies

[TABLE]

By Definition 3 and (2.6), $\oplus_{i}^{col}u_{i}L^{p}({\mathcal{D}})$ is the right wandering subspace of $K$. Since

[TABLE]

$K$ is type 1.

(iii) Set $K_{1}^{(r)}=\{x\in K_{1}:\;xD^{-\frac{1}{r}}\in L^{2}({\mathcal{M}})\}$ and $K_{2}^{(r)}=\{x\in K_{2}:\;xD^{-\frac{1}{r}}\in L^{2}({\mathcal{M}})\}$. If $x\in K_{r}$, then there exist $z\in K_{1}$ and $y\in K_{2}$ such that $x=z+y$ and $z^{*}y=0$. It follows that $|xD^{-\frac{1}{r}}|^{2}=|zD^{-\frac{1}{r}}|^{2}+|yD^{-\frac{1}{r}}|^{2}$, and so $|xD^{-\frac{1}{r}}|\geq|zD^{-\frac{1}{r}}|$, $|xD^{-\frac{1}{r}}|\geq|yD^{-\frac{1}{r}}|$. Since $xD^{-\frac{1}{r}}\in L^{2}({\mathcal{M}})\subset L^{0}({\mathcal{N}})$, we get $zD^{-\frac{1}{r}},\>yD^{-\frac{1}{r}}\in L^{0}({\mathcal{N}})$. On the other hand,

[TABLE]

Hence,

[TABLE]

so that

[TABLE]

Moreover,

[TABLE]

Thus $zD^{-\frac{1}{r}},\>yD^{-\frac{1}{r}}\in L^{2}(\mathcal{M})$, i.e., $z\in K_{1}^{(r)}$ and $y\in K_{2}^{(r)}$.

Next, we prove that $[K_{1}^{(r)}]_{p}=K_{1}$. To this end let $P:K\rightarrow K_{1}$ be the projection operator. From the above, we know that $P(K_{r})\subset K_{1}^{(r)}$. If $a\in K_{1}$, then $a\in K$. Since $[K_{r}]_{p}=K$, there exists a sequence $(a_{n})\subset K_{r}$ such that $a_{n}\rightarrow a$. Hence $P(a_{n})\rightarrow P(a)=a$. It follows that $a\in[K_{1}^{(r)}]_{p}$, Therefore, $[K_{1}^{(r)}]_{p}=K_{1}$. Similarly, $[K_{2}^{(r)}]_{p}=K_{2}$.

$[K_{r}D^{-\frac{1}{r}}]_{2}$ is a right $\mathcal{A}$-invariant subspace of $L^{2}({\mathcal{M}})$ and

[TABLE]

From the proof of (1), it follows that $[K_{1}^{(r)}D^{-\frac{1}{r}}]_{2}$ and $[K_{2}^{(r)}D^{-\frac{1}{r}}]_{2}$ are types 2 and 1 respectively. By (L2, , Proposition 2.7), the right wandering subspace for $[K_{r}D^{-\frac{1}{r}}]_{2}$ equals the right wandering subspace for $[K_{2}^{(r)}D^{-\frac{1}{r}}]_{2}$. By Definition 3, we obtain the desired result.

(iv) By (i), (ii) and (iii), we get that

[TABLE]

where $Z$ is a type 2, and $u_{i}$ are partial isometries with mutually orthogonal ranges such that $u_{i}^{*}u_{i}\in\mathcal{D}$ and $\oplus_{i}^{col}u_{i}L^{p}({\mathcal{D}})$ is the right wandering subspace of $K$. Using the properties of ${\mathcal{E}}$, similar to the proof (2) of Theorem 4.5 in BL3 , we prove the desired result. We omit the details.

(v) Since $J(W)W\subset L^{\frac{p}{2}}({\mathcal{D}})$, $W$ is a right $L^{p}({\mathcal{D}})$-module with inner product $\langle\xi,\eta\rangle=\xi^{*}\eta$ (see (JS, , Definition 3.3)).

Lemma 5

Let $2<p<\infty,\;\frac{1}{p}+\frac{1}{r}=\frac{1}{2}\;(r>2)$ and $K$ be a right $\mathcal{A}$-invariant subspace of $L^{p}(\mathcal{M})$. If $Y$ is the right wandering subspace of $[KD^{\frac{1}{r}}]_{2}$, then $[Y_{r}]_{2}=Y$, where $Y_{r}=\{x\in Y:\;xD^{-\frac{1}{r}}\in L^{p}({\mathcal{M}})\}$.

Proof

Let $K^{\prime}=[KD^{\frac{1}{r}}]_{2}$. Then $K^{\prime}=[K^{\prime}{\mathcal{A}}_{0}]_{2}\oplus Y$. By (L2, , Theorem 2.3 and 2.8), $Y=\oplus_{i}^{col}u_{i}L^{2}({\mathcal{D}})$ where $u_{i}$ are partial isometries with mutually orthogonal ranges such that $u_{i}^{*}u_{i}\in\mathcal{D}$. Since $\oplus_{i}^{col}u_{i}L^{p}({\mathcal{D}})D^{\frac{1}{r}}\subset Y_{r}$, using (2.6), we get $[Y_{r}]_{2}=Y$.

Similar to Theorem 3.1, we have the following result.

Theorem 3.2

Let $2<p<\infty,\;\frac{1}{p}+\frac{1}{r}=\frac{1}{2}$ and $K$ be a right $\mathcal{A}$-invariant subspace of $L^{p}(\mathcal{M})$. If $K=[[KD^{\frac{1}{r}}]_{2}D^{-\frac{1}{r}}\cap L^{p}({\mathcal{M}})]_{p}$, then:

1. (i)

$K$* may be written uniquely as an $L^{p}$-column sum $Z\oplus^{col}[Y{\mathcal{A}}]_{p}$, where $Z$ is a type 2 right $\mathcal{A}$-invariant subspace of $L^{p}({\mathcal{M}})$, $Y$ is the right wandering subspace of $K$ such that $Y=[Y{\mathcal{D}}]_{p}$ and $J(Y)Y\subset L^{\frac{p}{2}}({\mathcal{D}})$.* 2. (ii)

If $K\neq\{0\}$ then K is type 1 if and only if $K=\oplus_{i}^{col}u_{i}H^{p}(\mathcal{A})$, for $u_{i}$ partial isometries with mutually orthogonal ranges and $u_{i}^{*}u_{i}\in\mathcal{D}$. 3. (iii)

If $K=K_{1}\oplus^{col}K_{2}$ where $K_{1}$ and $K_{2}$ are types 2 and 1 respectively, then the right wandering subspace for $K$ equals the right wandering subspace for $K_{2}$. 4. (iv)

The wandering quotient $K/[K{\mathcal{A}}_{0}]p$ is isometrically ${\mathcal{D}}$-isomorphic to the right wandering subspace of K. 5. (v)

The wandering subspace $W$ of $K$ is an $L^{p}({\mathcal{D}})$-module in the sense of Junge and Sherman.

Proof

(i) By Lemma 2, $K^{\prime}=[KD^{\frac{1}{r}}]_{2}$ is a right $\mathcal{A}$-invariant subspace of $L^{2}({\mathcal{M}})$. Using Theorem 2.3 and 2.8 in L2 , we have that

[TABLE]

where $Z^{\prime}$ is a type 2 right $\mathcal{A}$-invariant subspace of $L^{2}({\mathcal{M}})$ and $Y^{\prime}$ is the right wandering subspace of $K^{\prime}$ with $Y^{\prime}=[Y^{\prime}{\mathcal{D}}]_{2}$ and $J(Y^{\prime})Y^{\prime}\subset L^{1}({\mathcal{D}})$. For simplicity, we set

[TABLE]

Let $Z=[Z_{r}D^{-\frac{1}{r}}]_{p}$ and $Y=[Y_{r}D^{-\frac{1}{r}}]_{p}$. By Lemma 2 and Definition 3, $Z$ is a right $\mathcal{A}$-invariant subspaces of $L^{p}({\mathcal{M}})$ and $Y$ is the right wandering subspace of $K$. We notice that $K=[[KD^{\frac{1}{r}}]_{2}D^{-\frac{1}{r}}\cap L^{p}({\mathcal{M}})]_{p}$ implies that $K=[K_{r}D^{-\frac{1}{r}}]_{p}$.

Since $KD^{\frac{1}{r}}\subset K_{r}$, we get $[K_{r}]_{2}=K^{\prime}$. We use same method as in the proof of (iii) of Theorem 3.1 to obtain that $Z^{\prime}=[Z_{r}]_{2}$, $X^{\prime}=[X_{r}]_{2}$ and

[TABLE]

We have that

[TABLE]

Hence,

[TABLE]

i.e., $Z$ is a type 2 right $\mathcal{A}$-invariant subspace of $L^{p}({\mathcal{M}})$. By Lemma 1, we have that $Y_{r}{\mathcal{D}}_{a}\subset Y_{r}$,

[TABLE]

and $[Y_{r}D^{-\frac{1}{r}}{\mathcal{D}}_{a}]_{p}=[Y_{r}D^{-\frac{1}{r}}{\mathcal{D}}]_{p}$. Therefore, it follows that

[TABLE]

Since

[TABLE]

we deduce that $J(Y)Y\subset L^{\frac{p}{2}}({\mathcal{D}})$.

Now we prove that

[TABLE]

By (L2, , Theorem 2.8), there are partial isometries $u_{i}$ with mutually orthogonal ranges such that $|u_{i}|\in{\mathcal{D}}$,

[TABLE]

Using Lemma 4, we get that

[TABLE]

and

[TABLE]

So, it follows that $[X_{r}D^{-\frac{1}{r}}]_{p}=[Y{\mathcal{A}}]_{p}$.

We claim that $K_{r}D^{-\frac{1}{r}}$ is closed. Indeed, if $x\in[K_{r}D^{-\frac{1}{r}}]_{p}$, then there is a sequence $(y_{n})$ in $K_{r}$ such that $y_{n}D^{-\frac{1}{r}}\rightarrow x$ in norm in $L^{p}({\mathcal{M}})$. It follows that $y_{n}\rightarrow xD^{\frac{1}{r}}$ in norm in $L^{2}({\mathcal{M}})$. Set $y=xD^{\frac{1}{r}}$. It is clear that $y\in K_{r}$. Hence, $x=yD^{-\frac{1}{r}}\in K_{r}D^{-\frac{1}{r}}$, i.e., $K_{r}D^{-\frac{1}{r}}$ is closed. Similarly, we can prove that $Z_{r}D^{-\frac{1}{r}}$ and $X_{r}D^{-\frac{1}{r}}$ are closed. Thus

[TABLE]

Applying (3.3), we obtain that $K=Z\oplus^{col}[Y{\mathcal{A}}]_{p}$. The remainder of the proof can be done the same way as in the proof of Theorem 3.1.

Remark 1

Let $1\leq p<\infty$ and $K$ be a right $\mathcal{A}$-invariant subspace of $L^{p}(\mathcal{M})$. In general, if $1\leq p<2$ and $\frac{1}{p}-\frac{1}{r}=\frac{1}{2}$, then $[K_{r}]_{p}\subset K$; if $2<p<\infty$ and $\frac{1}{p}+\frac{1}{r}=\frac{1}{2}$, then $K\subset[[KD^{\frac{1}{r}}]_{2}D^{-\frac{1}{r}}\cap L^{p}({\mathcal{M}})]_{p}$. It is unknown at the time of this writing whether for the general case, the results in Theorem 3.1 and 3.2 are hold.

We use same method as in the proof of (L2, , Proposition 2.4) to obtain the following result, we give its proof.

Proposition 2

Let $K$ is a right $\mathcal{A}$-invariant subspace of $L^{2}(\mathcal{M})$, and let $W$ be the right wandering subspace of $K$. If $W$ has a cyclic and separating vector for the ${\mathcal{D}}$-action, then there is an isometry $u\in\mathcal{M}$ such that $W=uL^{2}({\mathcal{D}})$.

Proof

By an adaption of an argument from JS (see p.13) there exists an isometric ${\mathcal{D}}$-module isomorphism $\psi:L^{2}({\mathcal{D}})\rightarrow W$. Let $h=\psi(D^{\frac{1}{2}})\in W$. Then

[TABLE]

By (L2, , (5) of Theorem 2.3), $h^{*}h\in L^{1}({\mathcal{D}})$, and so $h^{*}h=D$. Hence there exists an isometry $u$ with initial projection $1$ such that $h=uD^{\frac{1}{2}}$. Since $\psi$ is ${\mathcal{D}}$-module map, we have that

[TABLE]

Since $L^{2}({\mathcal{D}})=[D^{\frac{1}{2}}{\mathcal{D}}]$, it follows that $\psi(L^{2}({\mathcal{D}}))=uL^{2}(D)$. Thus $W=uL^{2}({\mathcal{D}})$ and $u^{\ast}u=1$.

Similar to Proposition 2, we have the following result.

Proposition 3

Let $K$ is a left $\mathcal{A}$-invariant subspace of $L^{2}(\mathcal{M})$, and let $W$ be the left wandering subspace of $K$. If $W$ has a cyclic and separating vector for the ${\mathcal{D}}$-action, then there is a partial isometry $v\in\mathcal{M}$ such that $vv^{*}=1$ and $W=L^{2}({\mathcal{D}})v$.

4 Outer operators of $H^{p}({\mathcal{A}})$

In the case when von Neumann algebra ${\mathcal{M}}$ is finite, from the Beurling-Blecher-Labuschagne theorem follows a generalized ‘inner-outer’ factorization. Let $x\in L^{p}({\mathcal{M}})\;(1\leq p\leq\infty)$ and $K=[x{\mathcal{A}}]_{p}$. If the right-wandering subspace of $K$ (respectively right-wandering quotient of $K$) has a nonzero separating and cyclic vector for the right action of ${\mathcal{D}}$, then $x$ is of the form $x=uh$ for some some outer operator $h\in H^{p}({\mathcal{A}})$ and a unitary $u\in{\mathcal{M}}$ (see the lines before the Closing remark of BL3 ). For more details on outer operators we refer to BX ; BL1 ; BL2 .

In this section, we consider outer operators in the case that $\mathcal{M}$ is a $\sigma$-finite von Neumann algebra. Similar to the finite case, we define the outer operators as following.

Definition 4

Let $0<p\leq\infty$. An operator $h\in H^{p}({\mathcal{A}})$ is called a left outer operator, a right outer operator or a bilaterally outer operator according to $[h{\mathcal{A}}]_{p}=H^{p}({\mathcal{A}})$, $[{\mathcal{A}}h]_{p}=H^{p}({\mathcal{A}})$ or $[{\mathcal{A}}h{\mathcal{A}}]_{p}=H^{p}({\mathcal{A}})$.

Proposition 4

Let $1\leq p<\infty$, and let $h\in H^{p}({\mathcal{A}})$. The following are equivalent:

1. (i)

$h$* is a bilaterally outer operator;* 2. (ii)

$\mathcal{E}(h)$* is a bilaterally outer operator in $L^{p}(\mathcal{D})$ and $[{\mathcal{A}}h{\mathcal{A}}_{0}]_{p}=[{\mathcal{A}}_{0}h{\mathcal{A}}]_{p}=H^{p}_{0}(\mathcal{A})$;* 3. (iii)

$\mathcal{E}(h)$* is a bilaterally outer operator in $L^{p}(\mathcal{D})$ and $\mathcal{E}(h)-h\in[{\mathcal{A}}h{\mathcal{A}}_{0}]_{p}=[{\mathcal{A}}_{0}h{\mathcal{A}}]_{p}$.*

Proof

(i) $\Rightarrow$ (ii). If $h$ is a bilaterally outer operator, then for $D^{\frac{1}{p}}$ there exist two sequence $(a_{n}),\;(b_{n})\subset{\mathcal{A}}$ such that

[TABLE]

By continuity of ${\mathcal{E}}$, we get

[TABLE]

Hence, by (2.5), we have that

[TABLE]

So, $\mathcal{E}(h)$ is a bilaterally outer operator in $L^{p}(\mathcal{D})$. Using (2.3) and (4.1), we deduce that

[TABLE]

(ii) $\Rightarrow$ (iii) is trivial.

(iii) $\Rightarrow$ (i). It is clear that

[TABLE]

and $h\in[{\mathcal{A}}h{\mathcal{A}}]_{p}$. Hence, ${\mathcal{E}}(h)=({\mathcal{E}}(h)-h)+h\in[{\mathcal{A}}h{\mathcal{A}}]_{p}$. It follows that $D^{\frac{1}{p}}\in[{\mathcal{A}}h{\mathcal{A}}]_{p}$. By (2.3), we obtain that $H^{p}({\mathcal{A}})=[{\mathcal{A}}h{\mathcal{A}}]_{p}$.

Similar to Proposition 4, we have the following result.

Proposition 5

Let $1\leq p<\infty$, and let $h\in H^{p}({\mathcal{A}})$. The following are equivalent:

1. (i)

$h$* is a left outer operator (resp. a right outer operator);* 2. (ii)

$\mathcal{E}(h)$* is a left outer operator (resp. a right outer operator) in $L^{p}(\mathcal{D})$ and $[h{\mathcal{A}}_{0}]_{p}=H^{p}_{0}(\mathcal{A})$ (resp. $[{\mathcal{A}}_{0}h]_{p}=H^{p}_{0}(\mathcal{A})$);* 3. (iii)

$\mathcal{E}(h)$* is a left outer operator (resp., a right outer operator) in $L^{p}(\mathcal{D})$ and $\mathcal{E}(h)-h\in[h{\mathcal{A}}_{0}]_{p}$ (resp. $\mathcal{E}(h)-h\in[{\mathcal{A}}_{0}h]_{p}$).*

Proposition 6

Let $1\leq p<\infty$. If $h\in H^{p}({\mathcal{A}})$ is a left outer operator (resp. a right outer operator), then $\mathcal{E}(h)$ and $h$ are left outer operator (resp. a right outer operator) in $L^{p}({\mathcal{M}})$.

Proof

Let $h\in H^{p}({\mathcal{A}})$ be a left outer operator. From Proposition 5 it follows that $[{\mathcal{E}}(h){\mathcal{D}}]_{p}=L^{p}({\mathcal{D}})$. Since $D^{\frac{1}{p}}\in L^{p}({\mathcal{D}})=[{\mathcal{E}}(h){\mathcal{D}}]_{p}$, there is a sequence $(d_{n})$ in ${\mathcal{D}}$ such that ${\mathcal{E}}(h)d_{n}\rightarrow D^{\frac{1}{p}}$ in norm in $L^{p}({\mathcal{M}})$. Therefore, $[{\mathcal{E}}(h){\mathcal{M}}]_{p}=L^{p}({\mathcal{M}})$.

Notice that ${\mathcal{E}}(h)\in H^{p}({\mathcal{A}})=[h{\mathcal{A}}]_{p}$. It follows that there is a sequence $(a_{n})$ in ${\mathcal{A}}$ such that $ha_{n}\rightarrow{\mathcal{E}}(h)$, and so $[h{\mathcal{M}}]_{p}=L^{p}({\mathcal{M}})$. The alternative claim follows analogously.

We will keep all previous notations throughout this section. If $h$ is a left outer operator and it is also a right outer operator, then we call $h$ is an outer operator.

Lemma 6

Let $0<p<\infty$.

1. (i)

If $h\in H^{p}({\mathcal{A}})$ is an outer operator in $H^{p}(\mathcal{A})$ and $h=u|h|$ is the polar decomposition of $h$, then $u$ is a unitary. 2. (ii)

If $d\in L^{p}({\mathcal{D}})$ is an outer operator in $L^{p}({\mathcal{D}})$ and $d=v|d|$ is the polar decomposition of $d$, then $v$ is a unitary in ${\mathcal{D}}$.

Proof

(i) Since $h$ is a left outer operator, there exists a sequence $(a_{n})\subset{\mathcal{A}}$ such that $ha_{n}\rightarrow D^{\frac{1}{p}}$ in norm in $L^{p}({\mathcal{M}})$. Let $l(h)$ be the left support projection of $h$. Then $l(h)^{\bot}ha_{n}\rightarrow l(h)^{\bot}D^{\frac{1}{p}}$ in norm in $L^{p}({\mathcal{M}})$. On the other hand, $l(h)^{\bot}ha_{n}=0$ for all $n$, and so $l(h)^{\bot}D^{\frac{1}{p}}=0$. Since $D^{\frac{1}{p}}$ is invertible, $l(h)^{\bot}=0$. Hence, $h$ must have dense range, i.e., $uu^{*}=l(h)=1$. Similarly, from the fact that $h$ is a right outer operator, we obtain that $u^{*}u=r(h)=1$, where $r(h)$ is the right support projection of $h$. Thus $u$ is a unitary.

(ii) The proof is similar to the proof of (i).

Theorem 4.1

Let $1\leq p<\infty$, and let $d\in L^{p}({\mathcal{D}})$. The following are equivalent:

1. (i)

$d$* is an outer operator in $L^{p}({\mathcal{D}})$;* 2. (ii)

$d$* is an outer operator in $H^{p}({\mathcal{A}})$;* 3. (iii)

The left and right support projections of $d$ are 1; 4. (iv)

$d$* is an outer operator in $L^{p}({\mathcal{M}})$.*

Proof

(i) $\Rightarrow$ (ii) Since $D^{\frac{1}{p}}\in L^{p}({\mathcal{D}})=[d{\mathcal{D}}]_{p}=[{\mathcal{D}}d]_{p}$, there are sequence $(a_{n})$ and $(b_{n})$ in ${\mathcal{D}}\subset{\mathcal{A}}$ such that $da_{n}\rightarrow D^{\frac{1}{p}}$ and $b_{n}d\rightarrow D^{\frac{1}{p}}$. Hence, $[d{\mathcal{A}}]_{p}=[{\mathcal{A}}d]_{p}=H^{p}({\mathcal{A}})$.

(ii) $\Rightarrow$ (iii) is follows from the proof of Lemma 6.

(iii) $\Rightarrow$ (iv). First we prove $d$ is a left outer operator in $L^{p}({\mathcal{M}})$. Let $p^{\prime}$ be the conjugate index of $p$. If $x\in L^{p^{\prime}}({\mathcal{M}})$ such that $tr(xdz)=$ for all $z\in{\mathcal{M}}$, then $xd=0$. Hence, $x=xdd^{-1}=0$, and so $[d{\mathcal{M}}]_{p}=L^{p}({\mathcal{M}})$. Using the same method, we can prove that $d$ is a right outer operator in $L^{p}({\mathcal{M}})$.

(iv) $\Rightarrow$ (i). Since $D^{\frac{1}{p}}\in[d{\mathcal{M}}]_{p}=[{\mathcal{M}}d]_{p}$, there are sequences $(a_{n})$ and $(b_{n})$ in ${\mathcal{M}}$ such that $da_{n}\rightarrow D^{\frac{1}{p}}$ and $b_{n}d\rightarrow D^{\frac{1}{p}}$ in norm in $L^{p}({\mathcal{M}})$. Using the continuity of ${\mathcal{E}}$, we obtain that $d{\mathcal{E}}(a_{n})\rightarrow D^{\frac{1}{p}}$ and ${\mathcal{E}}(b_{n})d\rightarrow D^{\frac{1}{p}}$ in norm in $L^{p}({\mathcal{D}})$. Hence, we get the desired result.

Corollary 1

Let $1\leq p<\infty$ and $0<r<\infty$. If $d\in L^{p}({\mathcal{D}})$ is an outer operator and $rp\geq 1$, then $|d|^{\frac{1}{r}}\in L^{pr}({\mathcal{D}})$ is an outer operator.

Proof

It is clear that $|d|\in L^{p}({\mathcal{D}})$ is an outer operator. Hence, by Theorem 4.1, $|d|^{\frac{1}{r}}$ is an outer operator.

Corollary 2

Let $1\leq p<\infty$ and $d\in L^{1}({\mathcal{D}})^{+}$ be an outer operator. If $0\leq\eta\leq 1$, then

[TABLE]

and $L^{p}({\mathcal{M}})=[d^{\frac{1-\eta}{p}}{\mathcal{M}}d^{\frac{\eta}{p}}]_{p}$.

Lemma 7

Let $1\leq p<\infty$, $1\leq q,r<\infty$ and $\frac{1}{p}-\frac{1}{r}=\frac{1}{q}$. If $d\in L^{p}({\mathcal{D}})$ is outer and $dD^{-\frac{1}{r}},\;D^{-\frac{1}{r}}d\in L^{q}({\mathcal{M}})$, then $dD^{-\frac{1}{r}},\;D^{-\frac{1}{r}}d\in L^{q}({\mathcal{D}})$ are outer operators.

Proof

Since $dD^{-\frac{1}{r}}\in H^{p}({\mathcal{A}})D^{-\frac{1}{r}}\cap L^{q}({\mathcal{M}})$ and $dD^{-\frac{1}{r}}\in J(H^{p}({\mathcal{A}}))D^{-\frac{1}{r}}\cap L^{q}({\mathcal{M}})$, by Lemma 4, we get $dD^{-\frac{1}{r}}\in H^{q}({\mathcal{A}})\cap J(H^{q}({\mathcal{M}})=L^{q}({\mathcal{D}})$. Similarly, $D^{-\frac{1}{r}}d\in L^{q}({\mathcal{D}})$. Using Theorem 4.1, we obtain the desired result.

Lemma 8

Let $1\leq p<\infty$, $1\leq q,r<\infty$ and $\frac{1}{p}+\frac{1}{r}=\frac{1}{q}$.

1. (i)

If $h\in H^{p}(\mathcal{A})$ is an outer operator, then $hD^{\frac{1}{r}}$ and $D^{\frac{1}{r}}h\in H^{q}({\mathcal{A}})$ are outer operators. 2. (ii)

If $d\in L^{p}({\mathcal{D}})$ is an outer operator, then $dD^{\frac{1}{r}},\;D^{\frac{1}{r}}d\in L^{q}({\mathcal{D}})$ are outer operators.

Proof

(i) We only prove $hD^{\frac{1}{r}}$ is an outer operator. A similar argument works for $D^{\frac{1}{r}}h$. By (2.6), $[H^{p}({\mathcal{A}})D^{\frac{1}{r}}]_{q}=H^{q}({\mathcal{A}})$. We use same method as in the proof of (3) of Lemma 1 to obtain that $[h{\mathcal{A}}_{a}]_{p}=[{\mathcal{A}}_{a}h]_{p}=H^{p}({\mathcal{A}})$. Hence, $[[h{\mathcal{A}}_{a}]_{p}D^{\frac{1}{r}}]_{q}=H^{q}({\mathcal{A}})$. Using Lemma 1, we get

[TABLE]

Thus $hD^{\frac{1}{r}}$ is a left outer operator. Similarly we can show $hD^{\frac{1}{r}}$ is a right outer operator.

(ii) follows analogously.

Proposition 7

Let $1\leq p<\infty$ and $h\in H^{p}({\mathcal{A}})$. Suppose that ${\mathcal{E}}(h)$ is an outer operator in $L^{p}({\mathcal{D}})$ and one of the the following conditions holds.

1. (i)

$\frac{1}{p}-\frac{1}{r}=\frac{1}{2}\;(r>2)$* and $hD^{-\frac{1}{r}},\;D^{-\frac{1}{r}}h\in L^{2}({\mathcal{M}})$;* 2. (ii)

$\frac{1}{p}+\frac{1}{r}=\frac{1}{2}\;(r>2)$.

Then there is a left outer operator $g\in H^{p}({\mathcal{A}})$ and an isometry $u\in{\mathcal{A}}$ such that $h=ug$ (resp. there is a right outer operator $g^{\prime}\in H^{p}({\mathcal{A}})$ and $v\in{\mathcal{A}}$ such that $vv^{*}=1$ and $h=g^{\prime}v$).

Proof

First assume that condition (i) holds. By Lemma 4, we get $hD^{-\frac{1}{r}}\in H^{2}({\mathcal{A}})$. Let $p^{\prime}$ be the conjugate index of $p$. Then for any $d\in{\mathcal{D}}$, we have that

[TABLE]

By (2.5), we get

[TABLE]

Hence, ${\mathcal{E}}(hD^{-\frac{1}{r}})={\mathcal{E}}(h)D^{-\frac{1}{r}}$. On the other hand, by Lemma 7, ${\mathcal{E}}(h)D^{-\frac{1}{r}}$ is an outer operator in $L^{2}({\mathcal{D}})$.

We consider the orthogonal projection

[TABLE]

Then $P={\mathcal{E}}|_{[hD^{-\frac{1}{r}}{\mathcal{A}}]_{2}}$ and $[{\mathcal{E}}(hD^{-\frac{1}{r}}){\mathcal{D}}]=[hD^{-\frac{1}{r}}{\mathcal{A}}]_{2}\ominus[hD^{-\frac{1}{r}}{\mathcal{A}}_{0}]_{2}$. It follows that ${\mathcal{E}}(hD^{-\frac{1}{r}})$ is a cyclic separating vector for the wandering subspace $[{\mathcal{E}}(hD^{-\frac{1}{r}}){\mathcal{D}}]_{2}$ of $[hD^{-\frac{1}{r}}{\mathcal{A}}]_{2}$. By Proposition 2, there exists an isometry $u\in{\mathcal{M}}$ such that

[TABLE]

We may write $hD^{-\frac{1}{r}}=uf$, for $f\in H^{2}({\mathcal{A}})$. Then

[TABLE]

i.e., $f$ is a left outer operator. On the other hand,

[TABLE]

Since $f$ is a left outer operator, by Proposition 5, $[f{\mathcal{A}}_{0}]_{2}=H^{2}_{0}({\mathcal{A}})$. Hence, using (2.3) we obtain that $[f{\mathcal{A}}_{0}D^{\frac{1}{2}}{\mathcal{A}}]_{1}=H_{0}^{1}({\mathcal{A}})$. It follows that $0=tr(ua)$ for any $a\in H_{0}^{1}({\mathcal{A}})$. By (2.4), $u\in A$. Let $g=fD^{\frac{1}{r}}$. From the proof of Lemma 8, we know that $g$ is a left outer operator. This gives the desired result. Similarly, we prove the alternative claim.

If condition (ii) holds. The proof is similar to the above.

Lemma 9

If $x\in L^{2}({\mathcal{M}})$ and $u\in{\mathcal{M}}$ is a contraction such that $\|ux\|_{2}=\|x\|_{2}$, then $x=u^{*}ux$.

Proof

We have that $x^{*}u^{*}ux\leq x^{*}x$ and $tr(x^{*}u^{*}ux)=\|ux\|_{2}^{2}=\|x\|_{2}^{2}=tr(x^{*}x)$. Hence,

[TABLE]

so that $x^{*}x=x^{*}u^{*}ux$. Thus $\|(1-u^{*}u)^{\frac{1}{2}}x\|_{2}^{2}=\|x^{*}(1-u^{*}u)x\|_{1}=0$, therefore $(1-u^{*}u)x=(1-u^{*}u)^{\frac{1}{2}}[(1-u^{*}u)^{\frac{1}{2}}x]=0$, and $x=u^{*}ux$.

In the finite case, $h\in H^{2}({\mathcal{A}})$ is a right outer operator if and only if there is a cyclic separating vector for the right action ${\mathcal{D}}$ on the wandering subspace of $[h{\mathcal{A}}]_{2}$ and $\|{\mathcal{E}}(h)\|_{2}=\|P(h)\|_{2}$, where $P$ is the orthogonal projection from $[h{\mathcal{A}}]_{2}$ to $[h{\mathcal{A}}]_{2}\ominus[h{\mathcal{A}}_{0}]_{2}$ (see (BL1, , Proposition 4.8) or (BL2, , The remark after Theorem 4.4)). This result was extend to the case $1\leq p<\infty$ (see (BL2, , Theorem 4.4)).

The following result extends (BL2, , Theorem 4.4) to the Haagerup noncommutative $H^{p}$-space case.

Theorem 4.2

Let $1\leq p,r<\infty$, and let $h\in H^{p}({\mathcal{A}})$.

If $\frac{1}{p}+\frac{1}{r}=\frac{1}{2}$, then $h$ is an outer operator if and only if $\mathcal{E}(h)$ is an outer operator in $L^{p}({\mathcal{D}})$ and $\|{\mathcal{E}}(hD^{\frac{1}{r}})\|_{2}=\|P(hD^{\frac{1}{r}})\|=\|P^{\prime}(hD^{\frac{1}{r}})\|$, where $P$ is the orthogonal projection from $[hD^{\frac{1}{r}}{\mathcal{A}}]_{2}$ to $[hD^{\frac{1}{r}}{\mathcal{A}}]_{2}\ominus[hD^{\frac{1}{r}}{\mathcal{A}}_{0}]_{2}$ and $P^{\prime}$ is the orthogonal projection from $[{\mathcal{A}}hD^{\frac{1}{r}}]_{2}$ to $[{\mathcal{A}}hD^{\frac{1}{r}}]_{2}\ominus[{\mathcal{A}}_{0}hD^{\frac{1}{r}}]_{2}$. 2. 2.

Suppose that $\frac{1}{p}-\frac{1}{r}=\frac{1}{2}$ and $hD^{-\frac{1}{r}}\in L^{2}({\mathcal{M}})$. If $\mathcal{E}(h)$ is an outer operator in $L^{p}({\mathcal{D}})$ and $\|{\mathcal{E}}(hD^{-\frac{1}{r}})\|_{2}=\|P(hD^{-\frac{1}{r}})\|=\|P^{\prime}(hD^{-\frac{1}{r}})\|$, where $P$ is the orthogonal projection from $[hD^{-\frac{1}{r}}{\mathcal{A}}]_{2}$ to $[hD^{-\frac{1}{r}}{\mathcal{A}}]_{2}\ominus[hD^{-\frac{1}{r}}{\mathcal{A}}_{0}]_{2}$ and $P^{\prime}$ is the orthogonal projection from $[{\mathcal{A}}hD^{-\frac{1}{r}}]_{2}$ to $[{\mathcal{A}}hD^{-\frac{1}{r}}]_{2}\ominus[{\mathcal{A}}_{0}hD^{-\frac{1}{r}}]_{2}$, then $h$ is an outer operator.

Proof

(i) $``\Rightarrow"$. Using Proposition 5, we obtain that $\mathcal{E}(h)$ is an outer operator in $L^{p}(\mathcal{D})$. Since ${\mathcal{E}}$ is a contractive projection from $H^{2}({\mathcal{A}})$ onto $L^{2}(\mathcal{D})$ with kernel $H^{2}_{0}({\mathcal{A}})$, we deduce that

[TABLE]

On the other hand, by Lemma 8, $hD^{\frac{1}{r}}$ is outer operator in $H^{2}({\mathcal{A}})$. Using Proposition 5, we obtain that

[TABLE]

Similarly, we can prove $\|{\mathcal{E}}(hD^{\frac{1}{r}})\|_{2}=\|P^{\prime}(hD^{\frac{1}{r}})\|$.

$``\Leftarrow"$. By Proposition 7, $h=ug$, where $g\in H^{p}({\mathcal{A}})$ is a left outer operator and $u\in{\mathcal{A}}$ is an isometry. On the other hand, it is clear that $gD^{\frac{1}{r}}$ is a left outer operator in $H^{2}({\mathcal{A}})$, Hence,

[TABLE]

This gives $\|{\mathcal{E}}(u){\mathcal{E}}(gD^{\frac{1}{r}})\|_{2}=\|{\mathcal{E}}(gD^{\frac{1}{r}})\|_{2}$. Using Proposition 5, we get ${\mathcal{E}}(gD^{\frac{1}{r}})$ is a left outer operator in in $L^{2}({\mathcal{D}})$, and so the left support of ${\mathcal{E}}(gD^{\frac{1}{r}})$ is 1. Applying Lemma 9, we obtain that ${\mathcal{E}}(u)$ is an isometry. On the other hand, we have that ${\mathcal{D}}{\mathcal{E}}(hD^{\frac{1}{r}})={\mathcal{D}}{\mathcal{E}}(u){\mathcal{E}}(gD^{\frac{1}{r}})\subset{\mathcal{D}}{\mathcal{E}}(gD^{\frac{1}{r}})$. Hence,

[TABLE]

i.e., ${\mathcal{E}}(gD^{\frac{1}{r}})$ is a right outer operator. So, ${\mathcal{E}}(gD^{\frac{1}{r}})$ is an outer operator. From ${\mathcal{E}}(hD^{\frac{1}{r}})={\mathcal{E}}(u){\mathcal{E}}(gD^{\frac{1}{r}})$ follows that

[TABLE]

Hence, ${\mathcal{E}}(u){\mathcal{E}}(u^{*})=1$, and so ${\mathcal{E}}(u)$ is a unitary. Therefore, ${\mathcal{E}}((u-{\mathcal{E}}(u))^{*}(u-{\mathcal{E}}(u)))=0$. So $u={\mathcal{E}}(u)\in{\mathcal{D}}$ and $h$ is a left outer operator.

Using the alternative claim of Proposition 7 and the above method, we deduce that $h$ is a right outer operator.

(ii) From the proof of Proposition 7, we know that ${\mathcal{E}}(hD^{-\frac{1}{r}})$ is an outer operator. Using same method as in the proof of (i), we obtain that $hD^{-\frac{1}{r}}$ is an outer operator in $H^{2}({\mathcal{A}})$. Hence, $h$ is an outer operator in $H^{p}({\mathcal{A}})$.

Let $d$ be a positive outer operator in $L^{1}({\mathcal{D}})$ with $\|d\|_{1}=1$. By Theorem 4.1, $d$ is an invertible positive selfadjoint operator. Set

[TABLE]

It is clear that $\phi$ is a normal faithful state on ${\mathcal{M}}$. Since $tr({\mathcal{E}}(x))=tr(x)$ for $x\in L^{1}({\mathcal{M}})$ (see (JX, , (2.4))), we get that

[TABLE]

We denote the dual weight of $\phi$ by $\hat{\phi}$. Then $d$ is the Radon-Nikodym derivative of $\hat{\phi}$ with respect to $\tau$ and

[TABLE]

Hence, the role of $d$ is similar to that of $D$. It follows that if we replace $D$ by $d$ in Section 3 and 4, then the related results still hold.

Acknowledgment

We thank the referees for very useful comments.