Completely order bounded maps on non-commutative \({\varvec{L_p}}\)-spaces

Abstract

We define norms on \(L_p({\mathcal {M}}) \otimes M_n\) where \({\mathcal {M}}\) is a von Neumann algebra and \(M_n\) is the space of complex \(n \times n\) matrices. We show that a linear map \(T: L_p({\mathcal {M}}) \rightarrow L_q({\mathcal {N}})\) is decomposable if \({\mathcal {N}}\) is an injective von Neumann algebra, the maps \(T \otimes Id_{M_n}\) have a common upper bound with respect to our defined norms, and \(p = \infty \) or \(q = 1\). For \(2p< q < \infty \) we give an example of a map \(T\) with uniformly bounded maps \(T \otimes Id_{M_n}\) which is not decomposable.

Introduction

Completely positive maps on von Neumann algebras have been studied extensively and there are some nice results on such maps (see e.g. [12], Ch. IV.3 and [10], Ch. 11). When combining the order structure with the vector space structure, it is natural to investigate the linear span of completely positive maps. These are called decomposable maps in [4], §1. The decomposable maps from a von Neumann algebra \({\mathcal {M}}\) into an injective von Neumann algebra \({\mathcal {N}}\) are the completely bounded maps. This has been shown at about the same time by Haagerup [4], Paulsen [8], and Wittstock [14]. Completely bounded maps use the operator norm on matrices of elements of a von Neumann algebra.

For a von Neumann algebra \({\mathcal {M}}\), the non-commutative \(L_p\)-space \(L_p({\mathcal {M}}),\) \(1 \le p < \infty ,\) can be realized as (in general) unbounded operators on a Hilbert space. Therefore the \(L_p\)-space itself as well as matrices with entries of a \(L_p\)-space have a natural order given by positive operators. This allows us to define completely positive maps and decomposable maps from one \(L_p\)-space into another one. Then there should be a description of decomposable maps using some norm conditions on matrices of \(L_p\)-spaces.

This question has been partially answered by Pisier [9] for linear maps from the Schatten classes \(S_p\) to \(S_p\), and Arhancet and Kriegler [1] for linear maps from \(L_p({\mathcal {M}})\) to \(L_p({\mathcal {N}})\), where \( {\mathcal {M}}\) and \({\mathcal {N}}\) are semifinite, approximately finite dimensional von Neumann algebras. Junge and Ruan [6] give a description of the decomposable norm for finite rank maps from \(L_p({\mathcal {M}})\) to \(L_p({\mathcal {N}})\), where \( {\mathcal {M}}\) and \({\mathcal {N}}\) are arbitrary von Neumann algebras. All cited articles require the same \(p \)-index for domain and range space.

Our idea is to derive a norm on matrices of \(L_p\)-spaces from the order structure. This norm is quite similar to the norm used in [1] and [9] (see [9] equation (1.5)). Using this norm, we can characterize decomposable maps from \({\mathcal {M}}\) to \(L_q({\mathcal {N}})\) where \(1 \le q \le \infty \) and from \(L_p({\mathcal {M}})\) to \(L_1({\mathcal {N}})\) where \(1 \le p \le \infty \). In both cases \({\mathcal {N}}\) must be injective. Then we give an example of von Neumann algebras \({\mathcal {M}}\) and \({\mathcal {N}}\) and a linear map from \(L_p({\mathcal {M}})\) to \(L_q({\mathcal {N}})\) where \( 1 \le p,q < \infty , q > 2p\) which is completely order bounded with respect to our matricial norms but not decomposable. Therefore this norm cannot characterize decomposable maps for all combinations of \(p\) and \(q\).

The structure of the article is as follows: In Sect. 2, we give a short description of non-commutative \(L_p\)-spaces in the Haagerup-Terp construction. We also show some properties of matrices of operators. In Sect. 3, we define our norm on matrices and derive some properties of this norm. In Sect. 4, we define completely order bounded maps and show that they are decomposable for some combinations of \(p\) and \(q\).

Non-commutative \(L_p\)-spaces

Short descriptions of non-commutative \(L_p\)-spaces in the Haagerup-Terp construction can be found in several publications, but the main source is still [13]. Here, we cite some basic facts from [13] we will use. Let \({\mathcal {M}}\) be a von Neumann algebra acting on a Hilbert space \({\mathcal {H}}\), and let \({\varphi} \) be a normal faithful semifinite weight on \({\mathcal {M}}\) with modular automorphism group \({\sigma ^{\varphi }_t}\). Then the crossed product \({\mathcal {M}} \rtimes _{\sigma ^{\varphi }} {\mathbb {R}}\) acts on the Hilbert space \(L_2({\mathbb {R}},{\mathcal {H}})\) and is the von Neumann algebra generated by the operators \({\pi (x)}\) and \({\lambda (s)}\) where

$$\begin{aligned} (\pi (x)\xi )(t) = \sigma ^{\varphi }_{-t}(x)\xi (t), \quad x \in {\mathcal {M}}, t \in {\mathbb {R}}, \xi \in L_2({\mathbb {R}},{\mathcal {H}}) \end{aligned}$$

and

$$\begin{aligned} (\lambda (s)\xi )(t) = \xi (t-s), \quad s,t \in {\mathbb {R}}, \xi \in L_2({\mathbb {R}},{\mathcal {H}}). \end{aligned}$$

For \(s \in {\mathbb {R}}\) let \(W(s)\) be the unitary operator on \( L_2({\mathbb {R}},{\mathcal {H}})\) which is defined by

$$\begin{aligned}\ (W(s)\xi )(t) = e^{-\text {i}st}\xi (t), \quad s,t \in {\mathbb {R}}, \xi \in L_2({\mathbb {R}},{\mathcal {H}}). \end{aligned}$$

The dual action \({\theta} \) is then defined by

$$\begin{aligned}\ \theta _s(x) = W(s)xW(s)^*, \quad x \in {\mathcal {M}} \rtimes _{\sigma ^{\varphi }} {\mathbb {R}}, s \in {\mathbb {R}}. \end{aligned}$$

The elements of \({\mathcal {M}}\) are the fixed points under \({\theta} \) when \( {\mathcal {M}}\) is identified with \({\pi ({\mathcal {M}})}\):

$$\begin{aligned}\ \pi ({\mathcal {M}}) = \{ x \in {\mathcal {M}} \rtimes _{\sigma ^{\varphi }}{\mathbb {R}}| \theta _s(x) = x \text { for all } s \in {\mathbb {R}} \}. \end{aligned}$$

The crossed product \({\mathcal {M}} \rtimes _{\sigma ^{\varphi }} {\mathbb {R}}\) has a unique normal faithful semifinite trace \({\tau} \) which satisfies

$$\begin{aligned} \tau (\theta _s(x)) = e^{-s}\tau (x) \text { for all } x \in ({\mathcal {M}} \rtimes _{\sigma ^{\varphi }} {\mathbb {R}})_+, s \in {\mathbb {R}}. \end{aligned}$$

The existence of the trace \({\tau} \) allows to consider the \({\tau} \)-measurable operators. These are all closed densely defined operators \(a\) affiliated with \({\mathcal {M}} \rtimes _{\sigma ^{\varphi }} {\mathbb {R}}\) which satisfy: For every \({\varepsilon > 0}\) there exists a projection \(p \in {\mathcal {M}} \rtimes _{\sigma ^{\varphi }} {\mathbb {R}}\) such that \( pL_2({\mathbb {R}},{\mathcal {H}}) \subseteq {\mathcal {D}}(a)\) and \({\tau (1-p) \le \varepsilon} \). A subspace \({\mathcal {D}}\) of \(L_2({\mathbb {R}},{\mathcal {H}})\) is called \( \tau \)-dense if for every \( \varepsilon > 0 \) there is a projection \( p \in {\mathcal {M}} \rtimes _{\sigma ^{\varphi }} {\mathbb {R}}\) such that \( pL_2({\mathbb {R}},{\mathcal {H}}) \subseteq {\mathcal {D}} \) and \( \tau (1-p) \le \varepsilon \). Thus the \( \tau \)-measurable operators are those which are affiliated with \({\mathcal {M}} \rtimes _{\sigma ^{\varphi }} {\mathbb {R}}\) and have a \( \tau \)-dense domain. The closure of a \( \tau \)-measurable operator restricted to a \( \tau \)-dense subspace is unique. Therefore, for proving some property of a \( \tau \)-measurable operator it suffices to prove the property on a \( \tau \)-dense subspace which is contained in the domain of the operator. The \( \tau \)-measurable operators form a topological \( * \)-algebra. When two \( \tau \)-measurable operators are added or multiplied, we have to create the closure of the sum or the product which always exist and are unique. More details for \( \tau \)-measurable operators can be found in [13], Chapter I or [12], Chapter IX.2.

The action \( \theta \) can be extended to all \( \tau \)-measurable operators. The space \( L_p({\mathcal {M}}) \), \( 1 \le p \le \infty \), consists of all \( \tau \)-measurable operators \(a\) for which

$$\begin{aligned}\ \theta _s(a) = e^{-\frac{s}{p}}(a) \text { for all } s \in {\mathbb {R}}. \end{aligned}$$

There is a one-to-one correspondence between the elements of the predual \({\mathcal {M}}_{*}\) and the elements of \( L_1({\mathcal {M}})\). This correspondence defines the linear functional \(tr: L_1({\mathcal {M}}) \rightarrow {\mathbb {C}}\) by \(tr(h_\varphi ) = \varphi (1)\), where \(h_\varphi \in L_1({\mathcal {M}})\) is the operator assigned to \({\varphi \in {\mathcal {M}}_{*}}\). This linear functional has the following properties:

$$\begin{aligned} \begin{aligned}&tr(a) \ge 0 \text { if } a \ge 0,\\&tr(a) = 0 \text { and } a \ge 0 \text { implies } a = 0,\\&|tr(a)| \le tr(|a|) \text { for all } a \in L_1({\mathcal {M}}). \end{aligned} \end{aligned}$$
(2.1)

If \( a \in L_p({\mathcal {M}}) \) has the polar decomposition \( a = u|a| \), then \( u \in {\mathcal {M}} \) and \( |a| \in L_p({\mathcal {M}}) \). For \(1 \le p < \infty \), the norm on \( L_p({\mathcal {M}}) \) is given by

$$\begin{aligned}\ \Vert a\Vert _p = tr(|a|^p)^{\frac{1}{p}}, \quad a\in L_p({\mathcal {M}}). \end{aligned}$$

For \(p = \infty \) and \(a \in {\mathcal {M}}\) the norm \({\Vert a\Vert _\infty} \) is the usual operator norm. If \( \frac{1}{p} + \frac{1}{q} = \frac{1}{r}, a \in L_p({\mathcal {M}}) \) and \( b \in L_q({\mathcal {M}}) \) then \( ab \in L_r({\mathcal {M}}) \) and

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

Especially, when \( \frac{1}{p} + \frac{1}{q} = 1\) we get for \( a \in L_p({\mathcal {M}}) \) and \( b \in L_{q}({\mathcal {M}}) \)

$$\begin{aligned} \begin{aligned}&tr(ab) = tr(ba),\\&|tr(ab)| \le \Vert ab\Vert _1 \le \Vert a\Vert _p \Vert b\Vert _{q},\\&tr(ab)\ge 0 \text { for all } b \in L_q({\mathcal {M}})_+ \text { if and only if } a \ge 0. \end{aligned} \end{aligned}$$
(2.2)

Moreover, the space \( L_{q}({\mathcal {M}}) \) is isometrically isomorphic to the dual space of \( L_p({\mathcal {M}}) \). We denote this duality by

$$\begin{aligned}\ \langle a,b \rangle = tr(ab) \text { for } a \in L_p({\mathcal {M}}) \text { and } b \in L_{q}({\mathcal {M}}). \end{aligned}$$

If \({\mathcal {M}}\) has a normal faithful semifinite trace \({\tau} \), we get the \({L_p}\)-spaces \({L_p({\mathcal {M}}, \tau ) = \{a | a \text { is }\tau -\text {measurable}, \tau (|a|^p) < \infty \}}\). It is shown in [13], pp. 62 - 63 that \({L_p({\mathcal {M}})}\) is isomorphic to \(\{a \otimes f_p | a \in L_p({\mathcal {M}}, \tau ), f_p: {\mathbb {R}} \rightarrow {\mathbb {R}}, s \mapsto \exp (s/p)\}\). Therefore we can switch to the easier to handle spaces \(L_p({\mathcal {M}}, \tau )\) for semifinite von Neumann algebras.

For \( n \in {\mathbb {N}} \), we denote the complex \( n \times n \) matrices by \( M_n \) with the usual trace \( Tr \). If \( a = [a_{ij}] \) is an \( n \times n \) matrix of \( \tau \)-measurable operators and each \( a_{ij} \) acts on the Hilbert space \( {\mathcal {K}} = L_2({\mathbb {R}}, {\mathcal {H}})\), then \( a \) acts on the Hilbert space \( {\mathcal {K}}^n \). The operator \( a \) is densely defined, has a unique closure which we denote again by \( a \), and is \( \tau \otimes Tr \)-measurable. Especially, the elements of \( L_p({\mathcal {M}}) \otimes M_n \) are \( \tau \otimes Tr \)-measurable operators. The elements of \( L_p({\mathcal {M}}) \otimes M_n \) have a natural ordering as being densely defined operators on the Hilbert space \({\mathcal {K}}^n\). The positive operators in \( L_p({\mathcal {M}}) \) will be denoted by \( L_p({\mathcal {M}})_+ \), and the positive operators in \( L_p({\mathcal {M}}) \otimes M_n \) will be denoted by \( (L_p({\mathcal {M}}) \otimes M_n)_+ \). Following the description in [7], p. 70, \( L_p({\mathcal {M}}) \otimes M_n \) is linearly isomorphic to \( L_p({\mathcal {M}} \otimes M_n) \) and this isomorphism maps the positive cone \((L_p({\mathcal {M}}) \otimes M_n )_+\) to \(L_p({\mathcal {M}} \otimes M_n)_+\). Furthermore for \( \frac{1}{p} + \frac{1}{p'} = 1 \), this isomorphism defines the duality of \(L_p({\mathcal {M}}) \otimes M_n\) and \(L_{p'}({\mathcal {M}}) \otimes M_n\) by

$$\begin{aligned}\ \langle a,b\rangle = \sum _{i,j = 1}^{n}\langle a_{ij},b_{ji}\rangle , a = [a_{ij}] \in L_p({\mathcal {M}}) \otimes M_n, b = [b_{ij}] \in L_{p'}({\mathcal {M}}) \otimes M_n. \end{aligned}$$

By (2.2),

$$\begin{aligned} a \ge 0 \text { if and only if } \langle a,b\rangle \ge 0 \text { for all } b \in (L_{p'}({\mathcal {M}}) \otimes M_n)_+. \end{aligned}$$
(2.3)

Next, we describe some relationships between matrices and their elements.

Lemma 2.1

Let \( {\mathcal {M}}_1 \) be a von Neumann algebra acting on a Hilbert space \( {\mathcal {H}} \) with a normal faithful semifinite trace \( {\tau} \), and let \({a} \) and \({ b }\) be self-adjoint \({ \tau }\)-measurable operators. Then the following are equivalent:

  1. (i)

    \({ -a \le b \le a} \).

  2. (ii)

    The matrix \( \begin{bmatrix} a &{} b \\ b &{} a \end{bmatrix} \) is positive.

Proof

We show first that (ii) implies (i): Let \( {\mathcal {D}}(a) \) and \( {\mathcal {D}}(b) \) denote the domains of \({ a} \) and \({ b} \). Let \( {\mathcal {D}} = {\mathcal {D}}(a) \cap {\mathcal {D}}(b) \). Then \( {\mathcal {D}} \) is a \({ \tau }\)-dense subspace of \( {\mathcal {H}} \). For \({ \xi \in {\mathcal {D}}} \), we get

$$\begin{aligned} 0 \le \dfrac{1}{2} \left( \begin{bmatrix} a &{} b \\ b &{} a \end{bmatrix} \begin{bmatrix} \xi \\ \xi \end{bmatrix} \Bigg \vert \begin{bmatrix} \xi \\ \xi \end{bmatrix} \right) = \left( \left( a + b\right) \xi | \xi \right) . \end{aligned}$$
(2.4)

Hence \({ -a \le b} \) on \( {\mathcal {D}} \). By replacing the vector \({ \left[ {\begin{matrix} \xi \\ \xi \end{matrix}}\right]} \) in (2.4) with \( {\left[ {\begin{matrix} \xi \\ -\xi \end{matrix}}\right]} \), we get \({ b \le a} \).

For the implication i) \({ \Rightarrow} \) ii), we assume first that \({ a }\) and \({ b }\) are bounded. Then it follows from [3], Proposition 1.3.5 that \( \left[ {\begin{matrix} a &{} b \\ b &{} a \\ \end{matrix}} \right] \) is positive. Since \( {\mathcal {D}}(a) \cap {\mathcal {D}}(b) \) is \({ \tau }\)-dense, there is a sequence of projections \({ \left( p_n\right) ^{\infty }_{n = 1}}\) in \( {\mathcal {M}}_1 \) such that \({ p_n \le p_{n+1}} \) for all \({ n \in {\mathbb {N}} }\), \({ \tau (1-p_n) \rightarrow 0} \) as \({ n \rightarrow \infty} \), and \({ p_n{\mathcal {H}} \subseteq {\mathcal {D}}(a) \cap {\mathcal {D}}(b) }\). Hence \( {p_n a p_n }\) and \({ p_n b p_n} \) are bounded operators and for all \({ n \in {\mathbb {N}}} \)

$$\begin{aligned} -p_nap_n \le p_nbp_n \le pap_n. \end{aligned}$$

Therefore, for \({ \xi , \eta \in p_n{\mathcal {H}}} \), we have

$$\begin{aligned} \left( \begin{bmatrix} a &{} b \\ b &{} a \end{bmatrix} \begin{bmatrix} \xi \\ \eta \end{bmatrix} \bigg {|} \begin{bmatrix} \xi \\ \eta \end{bmatrix} \right)&= \left( \begin{bmatrix} a &{} b \\ b &{} a \end{bmatrix} \begin{bmatrix} p_n\xi \\ p_n\eta \end{bmatrix} \bigg {|} \begin{bmatrix}p_n\xi \\ p_n\eta \end{bmatrix} \right) \\&=\left( \begin{bmatrix} p_nap_n &{} p_nbp_n \\ p_nbp_n &{} p_nap_n \end{bmatrix} \begin{bmatrix} \xi \\ \eta \end{bmatrix} \bigg {|} \begin{bmatrix} \xi \\ \eta \end{bmatrix} \right) \ge 0. \end{aligned}$$

Since the union \({ \bigcup \limits _{n \in {\mathbb {N}}} p_n{\mathcal {H}}}\) is \({ \tau}\)-dense, \({\left[ {\begin{matrix} a &{} b \\ b &{} a \end{matrix}} \right]} \) is positive. \({\square} \)

For \( {n \in {\mathbb {N}}, 1_n} \) denotes the unit matrix in \( {M_n} \). For a \({\tau} \)-measurable operator \({a}\) let supp\({(a)}\) denote the smallest projection which fulfills \({\text {supp}(a)\cdot a = a \cdot \text {supp}(a) = a} \). If \( {a \in L_p({\mathcal {M}})} \) then supp\({ (a) \in {\mathcal {M}}} \) although \({ a }\) is \( {\tau} \)-measurable with respect to a larger algebra (see [13], Proposition II.4).

Lemma 2.2

Let \( {\mathcal {M}}_1 \) be a von Neumann algebra with a normal faithful semifinite trace \( {\tau} \). Let \( {a, b, c_1, c_2}\) be \({ \tau} \)-measurable operators for which holds:

  1. (i)

    The operators \({a}\) and \({b}\) are positive.

  2. (ii)

    \({ \text {supp}(a)\cdot c_1 \cdot \text {supp}(b) = c_1} \) and \( {\text {supp}(a)\cdot c_2 \cdot \text {supp}(b) = c_2} \).

  3. (iii)

    \( {ac_1b = ac_2b} \).

Then \( {c_1 = c_2} \).

Proof

By putting \({c = c_1 - c_2 }\), we may assume that \({c_2 = 0}\). First let \({a = b}\). Then ii) can be formulated as \({\text {supp}(a)\cdot c \cdot \text {supp}(a) = c}\) and iii) as \({aca = 0}\). This means that the left support and the right support of \({c}\) are less than or equal to supp\({(a)}\). Thus \({c}\) fulfills the conditions of [11], Lemma 2.2 (c), and therefore \(c = 0.\) For the general case, we put \({a' = \left[ {\begin{matrix} a &{} 0 \\ 0 &{} b \end{matrix}}\right] }\) and \({c' = \left[ {\begin{matrix} 0 &{} c \\ 0 &{} 0 \end{matrix}}\right].}\) Then

$$\begin{aligned} \text {supp}(a') = \begin{bmatrix} \text {supp}(a) &{} 0 \\ 0 &{} \text {supp}(b) \end{bmatrix}, \\ a'c'a' = \begin{bmatrix} a &{} 0 \\ 0 &{} b \end{bmatrix} \begin{bmatrix} 0 &{} c \\ 0 &{} 0 \end{bmatrix} \begin{bmatrix} a &{} 0 \\ 0 &{} b \end{bmatrix} = \begin{bmatrix} 0 &{} acb \\ 0 &{} 0 \end{bmatrix} = 0, \end{aligned}$$

and

$$\begin{aligned} \text {supp}(a')\cdot c' \cdot \text {supp}(a') = \begin{bmatrix} 0 &{} \text {supp}(a)\cdot c \cdot \text {supp}(b) \\ 0 &{} 0 \end{bmatrix} = 0. \end{aligned}$$

Hence, by the first part of the proof, \({c' = 0}\) which implies \({c = 0.}\) \({\square} \)

Theorem 2.3

Let \( {\mathcal {M}} \) be a von Neumann algebra with a normal faithful semifinite weight \({ \varphi} \) and acting on the Hilbert space \( {\mathcal {H}} \), \({ 1 \le p\le \infty} \), \({n \in {\mathbb {N}} }\), \({ f, g \in L_p({\mathcal {M}})_+} \), and \( {x \in L_p({\mathcal {M}}) \otimes M_n} \) such that

$$\begin{aligned} \begin{bmatrix} f \otimes 1_n &{} x \\ x^* &{} g \otimes 1_n \end{bmatrix} \ge 0. \end{aligned}$$

Then there is an operator \( {y \in {\mathcal {M}} \otimes M_n} \) such that

$$\begin{aligned} x = (f^{\frac{1}{2}} \otimes 1_n)y(g^{\frac{1}{2}} \otimes 1_n). \end{aligned}$$
(2.5)

We have \( {\Vert y\Vert _\infty \le 1} \) and \( {y} \) is unique subject to the condition \((\text {supp}(f) \otimes 1_n)\cdot y \cdot (\text {supp}(g)\otimes 1_n) = y\) and to (2.5). If \( {f = g} \) and \({ x} \) is self-adjoint, then \({ y} \) is self-adjoint. If \({ f = g} \) and \({x}\) is positive, then \({y}\) is positive.

Proof

Let \( {\mathcal {M}}_1 = {\mathcal {M}} \rtimes _{\sigma ^{\varphi }} {\mathbb {R}}\), \( {\mathcal {K}} = L_2({\mathbb {R}}, {\mathcal {H}}) \), and \({ \tau} \) be the canonical trace on \( {\mathcal {M}}_1 \). Let \({x = [x_{ij}]}\) and

$$\begin{aligned} {\mathcal {D}} = {\mathcal {D}}(f) \cap {\mathcal {D}}(f^{\frac{1}{2}}) \cap {\mathcal {D}}(g) \cap {\mathcal {D}}(g^{\frac{1}{2}}) \cap \bigcap \limits _{i,j=1}^{n} {\mathcal {D}}(x_{ij}). \end{aligned}$$

Then \( {\mathcal {D}} \) is \( {\tau} \)-dense in \( {\mathcal {K}} \) and consequently, \( {\mathcal {D}}^n \) is \({ \tau \otimes Tr} \)-dense in \( {\mathcal {K}}^n \). For \( {\xi , \eta \in {\mathcal {D}}^n} \), we get

$$\begin{aligned} 0&\le \left( \begin{bmatrix} f \otimes 1_n &{} x \\ x^* &{} g \otimes 1_n \end{bmatrix} \begin{bmatrix} \xi \\ \eta \end{bmatrix} \Bigg \vert \begin{bmatrix} \xi \\ \eta \end{bmatrix} \right) \\&= ((f \otimes 1_n) \xi | \xi ) + (x \eta | \xi ) + (x^* \xi | \eta ) + ((g \otimes 1_n) \eta | \eta ). \end{aligned}$$

This implies

$$\begin{aligned} - 2 \text {Re}((x \eta | \xi )) \le ((f \otimes 1_n) \xi | \xi ) + ((g \otimes 1_n) \eta | \eta ). \end{aligned}$$
(2.6)

We replace \( {\eta} \) in (2.6) by \( {e^{\text {i}t}\eta} \), and choose a suitable value for \( {t \in {\mathbb {R}}} \) to get

$$\begin{aligned} 2 |(x \eta | \xi )| \le ((f \otimes 1_n) \xi | \xi ) + ((g \otimes 1_n) \eta | \eta ). \end{aligned}$$
(2.7)

Then we replace \( {\xi }\) by \({ \lambda \xi} \), \({ \eta }\) by \( {\frac{1}{\lambda }\eta} \) in (2.7), minimize over \({ \lambda \in {\mathbb {R}}_+} \), and get

$$\begin{aligned} |(x \eta | \xi )|^2 \le ((f \otimes 1_n) \xi | \xi ) ((g \otimes 1_n) \eta | \eta ). \end{aligned}$$
(2.8)

So, we can define the sesquilinear form

$$\begin{aligned} \begin{aligned} B:~&(g^{\frac{1}{2}} \otimes 1_n) {\mathcal {D}}^n \times (f^{\frac{1}{2}} \otimes 1_n) {\mathcal {D}}^n \rightarrow {\mathbb {C}} \\&((g^{\frac{1}{2}} \otimes 1_n)\eta , (f^{\frac{1}{2}} \otimes 1_n)\xi ) \mapsto (x\eta |\xi ). \end{aligned} \end{aligned}$$

If \({\xi , \xi ', \eta , \eta ' \in {\mathcal {D}}^n} \) with \( {(f^{\frac{1}{2}} \otimes 1_n)\xi = (f^{\frac{1}{2}} \otimes 1_n )\xi '} \) and \( {(g^{\frac{1}{2}} \otimes 1_n)\eta = (g^{\frac{1}{2}} \otimes 1_n)\eta '} \), we get

$$\begin{aligned} |(x\eta |\xi ) - (x\eta '|\xi ')|&\le |(x\eta |\xi -\xi ')| + |(x(\eta -\eta ')|\xi ')|\\&\le ((f \otimes 1_n)(\xi - \xi ')|\xi - \xi ')^{\frac{1}{2}} ((g \otimes 1_n)\eta |\eta )^{\frac{1}{2}}\\&\quad + ((f \otimes 1_n) \xi '|\xi ')^{\frac{1}{2}} ((g \otimes 1_n)(\eta - \eta ')|\eta - \eta ')^{\frac{1}{2}}\\&= 0. \end{aligned}$$

This shows that \( {B} \) is well defined, and by (2.8), we get

$$\begin{aligned} \begin{aligned}&|B((g^{\frac{1}{2}} \otimes 1_n)\eta , (f^{\frac{1}{2}} \otimes 1_n)\xi )|^2 = |(x\eta |\xi )|^2 \\&\quad \le ((g^{\frac{1}{2}} \otimes 1_n ) \eta | (g^{\frac{1}{2}} \otimes 1_n) \eta ) ((f^{\frac{1}{2}} \otimes 1_n) \xi | (f^{\frac{1}{2}}\otimes 1_n) \xi ). \end{aligned} \end{aligned}$$

Thus, \({ B} \) can be extended to a bounded sesquilinear form

$$\begin{aligned} B:~(\text {supp}(g) \otimes 1_n){\mathcal {K}}^n \times (\text {supp}(f) \otimes 1_n){\mathcal {K}}^n \rightarrow {\mathbb {C}} \end{aligned}$$

with norm \({ \Vert B\Vert \le 1} \). Next we extend \({B}\) to a bounded sesquilinear form on \( {\mathcal {K}}^n \times {\mathcal {K}}^n \) by \({ B(\eta , \xi ) = B((\text {supp}(g) \otimes 1_n)\eta , (\text {supp}(f) \otimes 1_n)\xi ) \text { for } \xi , \eta \in {\mathcal {K}}^n}\). Then \( {B }\) has still norm \({ \Vert B\Vert \le 1} \). Therefore, there is an operator \({ y \in {\mathcal {B}}({\mathcal {K}}^n)} \) with \({ (y\eta | \xi ) = B(\eta ,\xi )}\) and \( {\Vert y\Vert _\infty = \Vert B\Vert \le 1} \). By construction, we have

$$\begin{aligned} (\text {supp}(f) \otimes 1_n) \cdot y \cdot (\text {supp}(g) \otimes 1_n) = y . \end{aligned}$$

We still have to show that \( {y \in {\mathcal {M}} \otimes M_n} \). Let \({ u \in {\mathcal {M}}_1'} \) be a unitary element of the commutant of \( {\mathcal {M}}_1 \), \({ \xi , \eta \in {\mathcal {K}}^n}\). Then there are sequences \({ (\xi _i)_{i=1}^{\infty } }\) and \({ (\eta _i)_{i=1}^{\infty }} \) with \({ \xi _i, \eta _i \in {\mathcal {D}}^n} \) for all \( {i \in {\mathbb {N}} }\),

$$\begin{aligned} (\text {supp}(f) \otimes 1_n)\xi = \lim _{i \rightarrow \infty } (f^{\frac{1}{2}} \otimes 1_n) \xi _i, \end{aligned}$$

and

$$\begin{aligned} (\text {supp}(g) \otimes 1_n)\eta = \lim _{i \rightarrow \infty } (g^{\frac{1}{2}} \otimes 1_n) \eta _i. \end{aligned}$$

Since \({ u }\) commutes with \({ f} \), \({ g }\), supp\({(f)}\), and supp\({(g)}\), we get for all \({ i \in {\mathbb {N}}}\)

$$\begin{aligned} \begin{aligned}&((u^* \otimes 1_n)y(u \otimes 1_n)(g^{\frac{1}{2}} \otimes 1_n ) \eta _i | (f^{\frac{1}{2}} \otimes 1_n) \xi _i) \\&= ((u^* \otimes 1_n)(f^{\frac{1}{2}} \otimes 1_n )y (g^{\frac{1}{2}} \otimes 1_n) (u \otimes 1_n) \eta _i| \xi _i) \\&= ((u^* \otimes 1_n)x(u \otimes 1_n) \eta _i| \xi _i) \\&= (x \eta _i | \xi _i) \\&= ((f^{\frac{1}{2}} \otimes 1_n)y(g^{\frac{1}{2}} \otimes 1_n ) \eta _i| \xi _i) \end{aligned} \end{aligned}$$

and therefore

$$\begin{aligned} \begin{aligned}&((u^* \otimes 1_n)y(u \otimes 1_n)\eta | \xi ) \\&= ((u^* \otimes 1_n) (\text {supp}(f) \otimes 1_n) y (\text {supp}(g) \otimes 1_n)(u \otimes 1_n) \eta | \xi )\\&= ((u^* \otimes 1_n) y (u \otimes 1_n)(\text {supp}(g) \otimes 1_n) \eta | (\text {supp}(f) \otimes 1_n) \xi )\\&= \lim _{i \rightarrow \infty } ((u^* \otimes 1_n)y(u \otimes 1_n)(g^{\frac{1}{2}} \otimes 1_n) \eta _i | (f^{\frac{1}{2}} \otimes 1_n) \xi _i) \\&= \lim _{i \rightarrow \infty } ((f^{\frac{1}{2}} \otimes 1_n )y(g^{\frac{1}{2}} \otimes 1_n) \eta _i| \xi _i) \\&= (y\eta |\xi ). \end{aligned} \end{aligned}$$

Every operator in \( {\mathcal {M}}_1^{'} \) can be written as a finite linear combination of unitaries and \( {({\mathcal {M}}_1 \otimes M_n)^{'} = {\mathcal {M}}_1^{'} \otimes {\mathbb {C}}} \). Hence \({ y \in {\mathcal {M}}_1 \otimes M_n} \). We still have to show that \( {y \in {\mathcal {M}} \otimes M_n }\). Let \( {y = [y_{ij}]} \) with \({y_{ij} \in {\mathcal {M}}_1} \) for \({ i, j \in \{1, \dots n\}}\). Let \({s \in {\mathbb {R}}} \). Then we have

$$\begin{aligned} f^{\frac{1}{2}} y_{ij} g^{\frac{1}{2}}&= x_{ij} = e^{\frac{s}{p}} \theta _s(x_{ij}) \\&= e^{\frac{s}{p}} \theta _s(f^{\frac{1}{2}}) \theta _s(y_{ij})\theta _s(g^{\frac{1}{2}}) = f^{\frac{1}{2}}\theta _s(y_{ij})g^{\frac{1}{2}}. \end{aligned}$$

Since

$$\begin{aligned} y_{ij} = \text {supp}(f) \cdot y_{ij} \cdot \text {supp}(g) \end{aligned}$$

and

$$\begin{aligned} \theta _s(y_{ij}) = \text {supp}(f) \cdot \theta _s(y_{ij}) \cdot \text {supp}(g), \end{aligned}$$

we can apply Lemma 2.2 and get \({ y_{ij} = \theta _s(y_{ij})} \). Hence \({ y_{ij} \in {\mathcal {M}} }\). For the uniqueness of the decomposition, suppose that there is another \({ {\tilde{y}} \in {\mathcal {M}} \otimes M_n} \) such that

$$\begin{aligned} x = (f^{\frac{1}{2}} \otimes 1_n) {\tilde{y}} (g^{\frac{1}{2}} \otimes 1_n) \end{aligned}$$

and

$$\begin{aligned} (\text {supp}(f) \otimes 1_n) \cdot {\tilde{y}} \cdot (\text {supp}(g) \otimes 1_n) = {\tilde{y}}. \end{aligned}$$

Then we can apply Lemma 2.2 and get \({ y = {\tilde{y}}} \). Now assume that \({ x} \) is self-adjoint and \({ f = g} \). Since \({ x = x^*} \), we get

$$\begin{aligned} \begin{aligned} (f^{\frac{1}{2}} \otimes 1_n) y (f^{\frac{1}{2}} \otimes 1_n )&= x = \frac{1}{2}(x + x^*) \\&= (f^{\frac{1}{2}} \otimes 1_n) \frac{1}{2} (y + y^*) (f^{\frac{1}{2}} \otimes 1_n). \end{aligned} \end{aligned}$$

Again, we apply Lemma 2.2, use the uniqueness of \({ y} \), and get \({y = y^*. }\) Finally, assume that \({x}\) is positive. For \({\xi \in {\mathcal {K}}^n}\) there is a sequence \({(\xi _i)_{i=1}^{\infty }} \) with \({\xi _i \in {\mathcal {D}}^n}\) for all \({i \in {\mathbb {N}}}\) and

$$\begin{aligned} (\text {supp}(f) \otimes 1_n)\xi = \lim _{i \rightarrow \infty } (f^{\frac{1}{2}} \otimes 1_n) \xi _i. \end{aligned}$$

Then

$$\begin{aligned} \begin{aligned} (y\xi |\xi )&= (y(\text {supp}(f) \otimes 1_n)\xi |(\text {supp}(f) \otimes 1_n)\xi ) \\&= \lim _{i \rightarrow \infty } (y(f^{\frac{1}{2}} \otimes 1_n ) \xi _i|(f^{\frac{1}{2}} \otimes 1_n) \xi _i| \xi _i) \\&= \lim _{i \rightarrow \infty } (x\xi _i|\xi _i) \ge 0. \end{aligned} \end{aligned}$$

Since \({\xi} \) was arbitrary, \({y}\) is positive. \({\square} \)

Lemma 2.4

Let \( {\mathcal {M}} \) be a von Neumann algebra, \( {1 \le p \le \infty} \), \({n \in {\mathbb {N}}}\), and \( {x \in L_p({\mathcal {M}}) \otimes M_n}\). Then there exists \({ f \in L_p({\mathcal {M}})_+} \) such that

$$\begin{aligned} \begin{bmatrix} f \otimes 1_n &{} x \\ x^* &{} f \otimes 1_n \end{bmatrix} \ge 0. \end{aligned}$$

Proof

Since every \( {x \in L_p({\mathcal {M}}) \otimes M_n}\) is a finite linear combination of elements of the form \({ y \otimes \alpha} \), where \( {y \in L_p({\mathcal {M}})_+ }\) and \( {\alpha \in M_n} \), it suffices to prove the statement for \({ x = y \otimes \alpha} \). Let \({ \Vert \alpha \Vert} \) denote the usual maximum norm of an \( n \times n \) matrix acting on \( {\mathbb {C}}^n \). Then we get

$$\begin{aligned} \begin{aligned}&\begin{bmatrix} \Vert \alpha \Vert y \otimes 1_n &{} y \otimes \alpha \\ y \otimes \alpha ^* &{} \Vert \alpha \Vert y \otimes 1_n \end{bmatrix} \\& = \begin{bmatrix} y^{\frac{1}{2}} \otimes 1_n &{} 0 \\ 0 &{} y^{\frac{1}{2}} \otimes 1_n \end{bmatrix} \begin{bmatrix} \Vert \alpha \Vert 1_n &{} \alpha \\ \alpha ^* &{} \Vert \alpha \Vert 1_n \end{bmatrix} \begin{bmatrix} y^{\frac{1}{2}} \otimes 1_n &{} 0 \\ 0 &{} y^{\frac{1}{2}} \otimes 1_n \end{bmatrix} \ge 0. \end{aligned} \end{aligned}$$

\({\square }\)

Lemma 2.5

Let \( {\mathcal {M}} \) be a von Neumann algebra, \( {1 \le p \le \infty} \), \({ n \in {\mathbb {N}}}\), and \({ x \in (L_p({\mathcal {M}}) \otimes M_n)_+ }\). Then \({ x} \) is a finite sum of matrices of the form \({ [x_i^*x_j]} \) where \( {x_i \in L_{2p}({\mathcal {M}}) }\) for \({ i \in \{1, \dots , n\} }\).

Proof

By Lemmas 2.4 and 2.3, there exists \({ f \in L_p({\mathcal {M}})_+ }\) and \( {y \in ({\mathcal {M}} \otimes M_n)_+} \)such that \({x = (f^{\frac{1}{2}} \otimes 1_n)y(f^{\frac{1}{2}} \otimes 1_n).}\). By applying [12], page 193, Lemma 3.1, to \({y}\) we get the desired result. \(\square \)

Matrix norms on non-commutative \({L_p}\)-spaces

In this section \( {\mathcal {M}} \) always denotes a von Neumann algebra without any further restrictions. For each \( {n \in {\mathbb {N}}} \), we will define a norm on \({ L_p({\mathcal {M}}) \otimes M_n} \) and derive some properties of this norm.

Definition 3.1

Let \( {1 \le p \le \infty} \), \( {n \in {\mathbb {N}} }\), and \({ x \in L_p({\mathcal {M}}) \otimes M_n} \). Then we define

$$\begin{aligned} \Vert x \Vert _{p,n} = \inf \left\{ \frac{1}{2} \left( \Vert f\Vert _p + \Vert g\Vert _p \right) \Big \vert f, g \in L_p({\mathcal {M}})_+, \begin{bmatrix} f \otimes 1_n &{} x \\ x^* &{} g \otimes 1_n \end{bmatrix} \ge 0 \right\} . \end{aligned}$$
(3.1)

Remark 3.2

By Lemma 2.4, the set on the right side of (3.1) is not empty, and therefore, the infimum is well defined.

If \( {p = \infty} \), the norm \( {\Vert x\Vert _{\infty , n}} \) is identical to the usual operator norm of \( {x} \) considered as a bounded operator on a Hilbert space.

The combination of this norm definition with Theorem 2.3 shows that this norm is quite similar to the norm used in [9], Eq. (1.5).

If \( {x \in L_p({\mathcal {M}}) \otimes M_n, x = \sum _{j=1}^{k} x_j \otimes \beta _j} \) and \({ \alpha \in M_n} \), then we have \({ \alpha x = \sum _{j=1}^{k} x_j \otimes \alpha \beta _j} \) and \( {x\alpha = \sum _{j=1}^{k} x_j \otimes \beta _j\alpha} \). For \({n \in {\mathbb {N}}, i, j \in \{1,\ldots ,n\}}\) let

$$\begin{aligned} \varepsilon _{ij} \text { be the } n \times n \text { matrix with 1 at position } (i,j) \text { and 0 else.} \end{aligned}$$
(3.2)

Theorem 3.3

Let \( 1 \le p \le \infty \text { and } n \in {\mathbb {N}} \). Then the following holds:

  1. (i)

    If \( {x, y \in L_p({\mathcal {M}}) \otimes M_n}\), then \({ \Vert x+y\Vert _{p,n} \le \Vert x\Vert _{p,n} + \Vert y\Vert _{p,n}} \).

  2. (ii)

    If \({ x \in L_p({\mathcal {M}}) \otimes M_n}\), then \({ \Vert x^*\Vert _{p,n} = \Vert x\Vert _{p,n}}\).

  3. (iii)

    If \({ x \in L_p({\mathcal {M}}) \otimes M_n}\) and \({\alpha \in M_n }\), then

    $$\begin{aligned} \Vert \alpha x\Vert _{p,n} \le \Vert \alpha \Vert \Vert x\Vert _{p.n} \text { and } \Vert x \alpha \Vert _{p,n} \le \Vert \alpha \Vert \Vert x\Vert _{p,n}. \end{aligned}$$
  4. (iv)

    If \( {x \in L_p({\mathcal {M}}) \otimes M_n}\) and \({ \lambda \in {\mathbb {C}}} \), then \({ \Vert \lambda x \Vert _{p,n} = |\lambda | \Vert x\Vert _{p,n} }\).

  5. (v)

    If \({ x = [x_{ij}] \in L_p({\mathcal {M}}) \otimes M_n}\), then

    $$\begin{aligned} \max \left\{ \Vert x_{ij}\Vert _{p} \ \big {|} \ 1 \le i, j \le n \right\} \le \Vert x\Vert _{p,n} \le \sum _{i,j=1}^{n} \Vert x_{ij}\Vert _p. \end{aligned}$$
  6. (vi)

    If \({ x \in L_p({\mathcal {M}}) \otimes M_n}\) with \({ \Vert x\Vert _{p,n} = 0} \), then \({ x = 0} \).

Proof

Let \({x,y \in L_p({\mathcal {M}}) \otimes M_n}\), and \({ \varepsilon > 0 }\). Then there exist \( {f_1, f_2, g_1, g_2 \in L_p({\mathcal {M}})_+} \) such that

$$\begin{aligned} \begin{bmatrix} f_1 \otimes 1_n &{} x \\ x^* &{} g_1 \otimes 1_n \end{bmatrix} \ge 0,\quad \frac{1}{2}\left( \Vert f_1\Vert _p + \Vert g_1\Vert _p\right) \le \Vert x\Vert _{p,n} + \varepsilon , \end{aligned}$$

and

$$\begin{aligned} \begin{bmatrix} f_2 \otimes 1_n &{} y \\ y^* &{} g_2 \otimes 1_n \end{bmatrix} \ge 0,\quad \frac{1}{2}\left( \Vert f_2\Vert _p + \Vert g_2\Vert _p\right) \le \Vert y\Vert _{p,n} + \varepsilon . \end{aligned}$$

Then we get

$$\begin{aligned} \begin{bmatrix} \left( f_1 + f_2\right) \otimes 1_n &{} x+y \\ \left( x+y\right) ^* &{} \left( g_1 + g_2\right) \otimes 1_n \end{bmatrix} \ge 0 \end{aligned}$$

and

$$\begin{aligned} \Vert x+y\Vert _{p,n} \le \frac{1}{2}\left( \Vert f_1 + f_2\Vert _p + \Vert g_1 + g_2\Vert _p\right) \le \Vert x\Vert _{p,n} + \Vert y\Vert _{p,n} + 2\varepsilon . \end{aligned}$$

Since \({\varepsilon} \) is arbitrary, (i) is proved.

To prove (ii), let \({x \in L_p({\mathcal {M}}) \otimes M_n}\) and \( {\varepsilon > 0} \). Then there exist \({ f, g \in L_p({\mathcal {M}})_+} \) such that

$$\begin{aligned} \begin{bmatrix} f \otimes 1_n &{} x \\ x^* &{} g \otimes 1_n \end{bmatrix} \ge 0, \quad \frac{1}{2}\left( \Vert f\Vert _p + \Vert g\Vert _p\right) \le \Vert x\Vert _{p,n} + \varepsilon . \end{aligned}$$

Then we get

$$\begin{aligned} 0 \le \begin{bmatrix} 0 &{} 1 \\ 1 &{} 0 \end{bmatrix} \begin{bmatrix} f \otimes 1_n &{} x \\ x^* &{} g \otimes 1_n \end{bmatrix} \begin{bmatrix} 0 &{} 1 \\ 1 &{} 0 \end{bmatrix} = \begin{bmatrix} g \otimes 1_n &{} x^* \\ x &{} f \otimes 1_n \end{bmatrix}. \end{aligned}$$

Hence, we conclude \({ \Vert x^*\Vert _{p,n} \le \frac{1}{2}\left( \Vert f\Vert _p + \Vert g\Vert _p\right) \le \Vert x\Vert _{p,n} + \varepsilon} \). Since \( {\varepsilon} \) is arbitrary, we get \({ \Vert x^*\Vert _{p,n} \le \Vert x\Vert _{p,n} }\). Since \({ \Vert x\Vert _{p,n} = \Vert x^{**}\Vert _{p,n} \le \Vert x^*\Vert _{p,n}}\), ii) is proved.

Next, we prove iii). Let \( {\alpha \in M_n}\). If \({\alpha = 0 }\) then \({ \alpha x = 0} \) and the inequality is true. So let \( {\alpha \ne 0} \). For \({ \varepsilon > 0} \), there exist \({ f,g \in L_p({\mathcal {M}})_+} \) such that

$$\begin{aligned} \begin{bmatrix} f \otimes 1_n &{} x \\ x^* &{} g \otimes 1_n \end{bmatrix} \ge 0 \text { and } \frac{1}{2}(\Vert f\Vert _p+\Vert g\Vert _p) \le \Vert x\Vert _{p,n} + \varepsilon . \end{aligned}$$

Then we have for \({ \lambda > 0 }\)

$$\begin{aligned} 0&\le \begin{bmatrix} \frac{1}{\lambda }\alpha &{} 0 \\ 0 &{} \lambda 1_n \end{bmatrix} \begin{bmatrix} f \otimes 1_n &{} x \\ x^* &{} g \otimes 1_n \end{bmatrix} \begin{bmatrix} \frac{1}{\lambda }\alpha ^* &{} 0 \\ 0 &{} \lambda 1_n \end{bmatrix} \\&= \begin{bmatrix} \frac{1}{\lambda ^2}\alpha \alpha ^*(f \otimes 1_n) &{} \alpha x \\ (\alpha x)^* &{} \lambda ^2 g \otimes 1_n \end{bmatrix} \\&\le \begin{bmatrix} \frac{1}{\lambda ^2}\Vert \alpha \Vert ^2 f \otimes 1_n &{} \alpha x \\ (\alpha x)^* &{} \lambda ^2 g \otimes 1_n \end{bmatrix}. \end{aligned}$$

We put \({ \lambda ^2 = \Vert \alpha \Vert }\) and get

$$\begin{aligned} \Vert \alpha x\Vert _{p,n} \le \frac{1}{2}\left( \Vert \alpha \Vert \Vert f\Vert _p + \Vert \alpha \Vert \Vert g\Vert _p \right) \le \Vert \alpha \Vert \left( \Vert x\Vert _{p,n} + \varepsilon \right) . \end{aligned}$$

Since \({ \varepsilon }\) was arbitrary, we get the desired result. A similar argument proves the second inequality in iii).

To prove (iv), let \({x \in L_p({\mathcal {M}}) \otimes M_n}\) and \({ \lambda \in {\mathbb {C}}}\). If \({ \lambda = 0 }\), we have

$$\begin{aligned} \Vert \lambda x\Vert _{p,n} = 0 = |\lambda | \Vert x\Vert _{p,n}. \end{aligned}$$

For \({ \lambda \ne 0 }\), we put \({ \alpha = \lambda 1_n }\), apply iii), and get

$$\begin{aligned} \Vert \lambda x\Vert _{p,n} = \Vert \lambda 1_n x\Vert _{p,n} \le |\lambda | \Vert x\Vert _{p,n} \text { and} \\ |\lambda | \Vert x\Vert _{p,n} = |\lambda | \left\| \frac{1}{\lambda }\lambda x\right\| _{p,n} \le |\lambda | \frac{1}{|\lambda |} \Vert x\Vert _{p,n} \le \Vert \lambda x\Vert _{p,n}. \end{aligned}$$

For v), let \( {x = [x_{ij}] \in L_p({\mathcal {M}}) \otimes M_n}\). For \({\varepsilon > 0} \), there exist \({ f, g \in L_p({\mathcal {M}})_+ }\) such that

$$\begin{aligned} \begin{bmatrix} f \otimes 1_n &{} x \\ x^* &{} g \otimes 1_n \end{bmatrix} \ge 0, \quad \frac{1}{2}\left( \Vert f\Vert _p + \Vert g\Vert _p\right) \le \Vert x\Vert _{p,n} + \varepsilon . \end{aligned}$$

By Theorem 2.3, there exists \({y \in {\mathcal {M}} \otimes M_n}\), such that \({ \Vert y\Vert _{\infty } \le 1}\) and \({x = (f^{\frac{1}{2}} \otimes 1_n)y(g^{\frac{1}{2}} \otimes 1_n)}\). Then \({x_{ij} = f^{\frac{1}{2}}y_{ij}g^{\frac{1}{2}}}\) and \({\Vert y_{ij}\Vert _{\infty }\le \Vert y\Vert _{\infty } \le 1} \) for all \({i,j \in \{1,\ldots , n\}}\). Hence

$$\begin{aligned} \begin{aligned} \Vert x_{ij}\Vert _p&= \Vert f^{\frac{1}{2}}y_{ij}g^{\frac{1}{2}}\Vert _p \le \Vert f^{\frac{1}{2}}\Vert _{2p}\Vert g^{\frac{1}{2}}\Vert _{2p} = \sqrt{\Vert f\Vert _p\Vert g\Vert _p}\\&\le \frac{1}{2}\left( \Vert f\Vert _p + \Vert g\Vert _p\right) \le \Vert x\Vert _{p,n} + \varepsilon , \quad i,j \in \{1,\ldots ,n\}. \end{aligned} \end{aligned}$$

For the second inequality of v), let \( {y \in L_p({{\mathcal {M}}}) }\) with polar decomposition \({ y = v|y|}\). Then \({|y^*|v = v|y|, v^*|y^*|v = |y|}\), and

$$\begin{aligned} 0 \le \begin{bmatrix} 1 &{} 0 \\ v^* &{} 0 \end{bmatrix} \begin{bmatrix} |y^*| &{} 0 \\ 0 &{} |y^*| \end{bmatrix} \begin{bmatrix} 1 &{} v \\ 0 &{} 0 \end{bmatrix} = \begin{bmatrix} |y^*| &{} y \\ y^* &{} |y| \end{bmatrix}. \end{aligned}$$
(3.3)

Now it follows that

$$\begin{aligned} \begin{bmatrix} |y^*| \otimes 1_n &{} y \otimes 1_n \\ y^* \otimes 1_n &{} |y| \otimes 1_n \end{bmatrix} \ge 0 \end{aligned}$$

and therefore \( {\Vert y \otimes 1_n \Vert _{p,n} \le \Vert y\Vert _{p}}\). For \({ i \in \{1, \dots , n\} }\), let \({ \varepsilon _{ij}} \) be as in (3.2). Then we get for \( {x = [x_{ij}] \in L_p({\mathcal {M}}) \otimes M_n}\)

$$\begin{aligned} \Vert x\Vert _{p,n} = \left\| \sum _{i,j=1}^{n}x_{ij} \otimes \varepsilon _{ij}\right\| _{p,n} \le \sum _{i,j=1}^{n}\Vert x_{ij} \otimes 1_n \Vert _{p,n} \Vert \varepsilon _{ij}\Vert \le \sum _{i,j=1}^{n}\Vert x_{ij}\Vert _p. \end{aligned}$$

To prove (vi), let \({ \Vert x\Vert _{p,n} = 0 }\). From (v), it follows that \({ x_{ij} = 0} \) for all \({ i,j \in \{1, \dots , n\}} \). Hence \({ x = 0} \). \({\square} \)

The next theorem shows that the infimum in Definition 3.1 is actually a minimum.

Theorem 3.4

Let \( {1 \le p \le \infty} \), \({ n \in {\mathbb {N}}} \), and \( {x \in L_p({\mathcal {M}}) \otimes M_n} \). Then there exist \({ f, g \in L_p({\mathcal {M}})_+} \) such that

$$\begin{aligned} \begin{bmatrix} f \otimes 1_n &{} x \\ x^* &{} g \otimes 1_n \end{bmatrix} \ge 0 \text { and } \Vert f\Vert _p = \Vert g\Vert _p = \Vert x\Vert _{p,n}. \end{aligned}$$

Proof

If \({ x = 0} \), we can take \( {f = g = 0} \). So suppose that \({ x \ne 0} \). Let \( {1 \le q \le \infty} \) and \({ \frac{1}{p} + \frac{1}{q} = 1}\). Let \( {L_q({\mathcal {M}})^*} \) be the dual space of \( {L_q({\mathcal {M}})} \). Note that \( {L_q({\mathcal {M}})^* }\) is \( {L_p({\mathcal {M}})} \) when \({ q < \infty} \) and \( {{\mathcal {M}}^*} \) when \( {q = \infty} \). For \({ \varepsilon > 0} \) we define

The symbol \({ \Vert \cdot \Vert} \) denotes the norm of \( {L_q({\mathcal {M}})^*} \). The sets \({ K_\varepsilon \text {, } \varepsilon > 0} \), have the following properties:

Each \({ K_\varepsilon \ne \emptyset : }\) By definition of \({ \Vert x\Vert _{p,n} }\), there exist \({ 0 \ne f,g \in L_p({\mathcal {M}})_+} \) with

$$\begin{aligned} \begin{bmatrix} f \otimes 1_n &{} x \\ x^* &{} g \otimes 1_n \end{bmatrix} \ge 0 \text { and } \frac{1}{2}(\Vert f\Vert _p + \Vert g\Vert _p) < \Vert x\Vert _{p,n} + \varepsilon . \end{aligned}$$

For \({ \lambda > 0} \), we get

$$\begin{aligned} 0 \le \begin{bmatrix} \lambda &{} 0 \\ 0 &{} \frac{1}{\lambda } \end{bmatrix} \begin{bmatrix} f \otimes 1_n &{} x \\ x^* &{} g \otimes 1_n \end{bmatrix} \begin{bmatrix} \lambda &{} 0 \\ 0 &{} \frac{1}{\lambda } \end{bmatrix} = \begin{bmatrix} \lambda ^2f \otimes 1_n &{} x \\ x^* &{} \frac{1}{\lambda ^2}g \otimes 1_n \end{bmatrix}. \end{aligned}$$

We put \({ \lambda ^2 = \sqrt{\frac{\Vert g\Vert _p}{\Vert f\Vert _p}} }\), \({ f' = \lambda ^2 f }\), and \({ g' = \frac{1}{\lambda ^2}g }\). Then

$$\begin{aligned} \Vert f'\Vert _p = \Vert g'\Vert _p = \sqrt{\Vert f\Vert _p\Vert g\Vert _p} \le \frac{1}{2}(\Vert f\Vert _p+ \Vert g\Vert _p) < \Vert x\Vert _{p,n} + \varepsilon \end{aligned}$$

and

$$\begin{aligned} \begin{bmatrix} f' \otimes 1_n &{} x \\ x^* &{} g' \otimes 1_n \end{bmatrix} \ge 0. \end{aligned}$$

This shows that \({ (f',g') \in K_\varepsilon }\). Next we show that \( {K_\varepsilon} \) is weak\({ ^*} \)-closed for every \({ \varepsilon > 0 }\). We fix \({ \varepsilon > 0} \) and \( {(f,g)} \) as an element of the weak\({ ^* }\) closure of \({ K_\varepsilon} \). For \( {a = \left[ {\begin{matrix} a_{11} &{} a_{12} \\ a_{21} &{} a_{22} \end{matrix}} \right] \in (L_q({\mathcal {M}}) \otimes M_{2n})_+} \) there exist sequences \({ (f_m)_{m=1}^\infty }\) and \({ (g_m)_{m=1}^\infty }\) in \( {K_\varepsilon} \) such that

$$\begin{aligned} \langle f \otimes 1_n , a_{11} \rangle = \lim _{m \rightarrow \infty } \langle f_m \otimes 1_n, a_{11}\rangle \end{aligned}$$

and

$$\begin{aligned} \langle g \otimes 1_n , a_{22} \rangle = \lim _{m \rightarrow \infty } \langle g_m \otimes 1_n, a_{22}\rangle , \end{aligned}$$

and \({(f_m,g_m) \in K_\varepsilon} \) for all \({m \in {\mathbb {N}}.}\) Hence, we get

$$\begin{aligned} \left\langle \begin{bmatrix} f \otimes 1_n &{} x \\ x^* &{} g \otimes 1_n \end{bmatrix} \Bigg \vert a \right\rangle = \lim _{m \rightarrow \infty } \left\langle \begin{bmatrix} f_m \otimes 1_n &{} x \\ x^* &{} g_m \otimes 1_n \end{bmatrix} \Bigg \vert a \right\rangle \ge 0. \end{aligned}$$

Since this holds for every \( {a \in (L_q({\mathcal {M}}) \otimes M_{2n})_+} \), we can apply (2.3), and conclude that

$$\begin{aligned} \begin{bmatrix} f \otimes 1_n &{} x \\ x^* &{} g \otimes 1_n \end{bmatrix} \ge 0. \end{aligned}$$

If \({ q = \infty} \), we have to do this in \({\mathcal {M}}^*\) which is isomorphic to \({L_1({\mathcal {M}}^{**})}\). Especially, \({ f }\) and \({ g} \) are positive. Taking \( {a \in L_q({\mathcal {M}})} \) with \({ \Vert a\Vert _q \le 1} \), we can find a sequence \({ (f_m)_{m=1}^\infty }\) in \({ K_\varepsilon} \) such that

$$\begin{aligned} \langle f, a \rangle = \lim \limits _{m \rightarrow \infty }\langle f_m,a \rangle \end{aligned}$$

Hence \({ |\langle f,a\rangle | \le \Vert x\Vert _{p,n} + \varepsilon} \) and

$$\begin{aligned} \Vert f\Vert = \sup \{|\langle f,a\rangle |~ \Vert a\Vert _q \le 1\} \le \Vert x\Vert _{p,n} + \varepsilon . \end{aligned}$$

Similarly, \( {\Vert g\Vert \le \Vert x\Vert _{p,n} + \varepsilon} \) and therefore \({ (f,g) \in K_\varepsilon} \). By the Banach-Alaoglu theorem, the unit ball of \({ L_q({\mathcal {M}})} \) is compact in the weak\({ ^*} \)-topology. Hence all sets \( {K_\varepsilon} \) are compact in the weak\({^*}\)-topology.

The sets \( {K_\varepsilon} \), \({ \varepsilon > 0} \), have the finite intersection property: Given \({ k \in {\mathbb {N}}} \), \({ \varepsilon _1, \ldots , \varepsilon _k > 0} \), we put \( {\varepsilon = \min \{\varepsilon _1, \ldots , \varepsilon _k\}} \) and get

$$\begin{aligned} K_\varepsilon \subseteq \bigcap \limits _{i=1}^{k} K_{\varepsilon _i}. \end{aligned}$$

Combining the finite intersection property and the weak-\({ ^* }\) compactness, we get

$$\begin{aligned} \bigcap \limits _{\varepsilon > 0} K_{\varepsilon } \ne \emptyset . \end{aligned}$$
(3.4)

Let \({ (f,g)} \) be in the set defined in Eq. (3.4). Then we have

$$\begin{aligned} \begin{bmatrix} f \otimes 1_n &{} x \\ x^* &{} g \otimes 1_n \end{bmatrix} \ge 0 \end{aligned}$$
(3.5)

If \({ q < \infty} \), then \({ f,g \in L_p({\mathcal {M}}) }\). If \({ q = \infty }\), there is a central projection in \( {\mathcal {M}}^{**} \) which works as projection from \( {\mathcal {M}}^* \) to \( {L_1({\mathcal {M}})} \). Hence we may assume that \({f,g \in L_1({\mathcal {M}}) }\). By construction, we have \({ \Vert f\Vert _p, \Vert g\Vert _p \le \Vert x\Vert _{p,n}} \). From Eq. (3.5), we get \({ 2\Vert x\Vert _{p,n} \le \Vert f\Vert _p + \Vert g\Vert _p }\). Combining both gives

$$\begin{aligned} \Vert f\Vert _p = \Vert x\Vert _{p,n} = \Vert g\Vert _p. \end{aligned}$$

\({\square} \)

Theorem 3.5

Let \({ 1 \le p \le \infty }\), \( {n \in {\mathbb {N}}} \), and \({ x = x^* \in L_p({\mathcal {M}}) \otimes M_n} \).

  1. (i)

    We have \({ \Vert x\Vert _{p,n} = \inf \left\{ \Vert f\Vert _p | f \in L_p({\mathcal {M}})_+, f \otimes 1_n \pm x \ge 0 \right\}} \).

  2. (ii)

    There exists \({ f \in L_p({\mathcal {M}})_+ }\) such that

    $$\begin{aligned} f \otimes 1_n \pm x \ge 0 \text { and } \Vert f\Vert _p = \Vert x\Vert _{p,n}. \end{aligned}$$

Proof

Let \({A = \inf \left\{ \Vert f\Vert _p | f \in L_p({\mathcal {M}})_+, f \otimes 1_n \pm x \ge 0 \right\} }\). By Theorem 3.4, there exist \({ f, g \in L_p({\mathcal {M}})_+ }\) such that

$$\begin{aligned} \Vert f\Vert _p = \Vert g\Vert _p = \Vert x\Vert _{p,n} \text { and } \begin{bmatrix} f \otimes 1_n &{} x \\ x &{} g \otimes 1_n \end{bmatrix} \ge 0. \end{aligned}$$
(3.6)

It follows then

$$\begin{aligned} 0 \le \begin{bmatrix} 0 &{} 1 \\ 1 &{} 0 \end{bmatrix} \begin{bmatrix} f \otimes 1_n &{} x \\ x &{} g \otimes 1_n \end{bmatrix} \begin{bmatrix} 0 &{} 1 \\ 1 &{} 0 \end{bmatrix} = \begin{bmatrix} g \otimes 1_n &{} x \\ x &{} f \otimes 1_n \end{bmatrix} \end{aligned}$$

Hence we conclude

$$\begin{aligned} \begin{bmatrix} \frac{1}{2}(f+g) \otimes 1_n &{} x \\ x &{} \frac{1}{2}(f+g) \otimes 1_n \end{bmatrix} \ge 0. \end{aligned}$$

Then Lemma 2.1 implies that \({ \frac{1}{2}(f+g) \otimes 1_n \pm x \ge 0} \). This shows that

$$\begin{aligned} A \le \Vert \frac{1}{2}(f+g)\Vert _p \le \Vert x\Vert _{p,n}. \end{aligned}$$

For the converse direction, let \({ \varepsilon > 0 }\). Then there exist \({ f \in L_p({\mathcal {M}})_+} \) such that \({ f \otimes 1_n \pm x \ge 0} \) and \({ \Vert f\Vert _p \le A + \varepsilon }\). It follows from Lemma 2.1 that

$$\begin{aligned} \begin{bmatrix} f \otimes 1_n &{} x \\ x &{} f \otimes 1_n \end{bmatrix} \ge 0. \end{aligned}$$

Hence, we get \({ \Vert x\Vert _{p,n} \le \Vert f\Vert _p \le A + \varepsilon} \). Since \({\varepsilon} \) is arbitrary, we get \( {\Vert x\Vert _{p,n} \le A} \). This proves (i). To prove (ii), we take \({ f }\) and \( {g} \) from (3.6). Then we have

$$\begin{aligned} \frac{1}{2}(f+g) \pm x \ge 0 \end{aligned}$$

and

$$\begin{aligned} \Vert x\Vert _{p,n} \le \frac{1}{2}\Vert f+g\Vert _p \le \frac{1}{2}(\Vert f\Vert _p + \Vert g\Vert _p) = \Vert x\Vert _{p,n}. \end{aligned}$$

This shows that \({ \Vert \frac{1}{2}(f+g) \Vert _p = \Vert x\Vert _{p,n}} \). \({\square} \)

Theorem 3.6

Let \({ 1 \le p \le \infty }\) and \( {x \in L_p({\mathcal {M}}) }\) . Then

$$\begin{aligned} \Vert x\Vert _{p,1} = \Vert x\Vert _p. \end{aligned}$$

Proof

Let \({ x = v|x| }\) be the polar decomposition of \({ x} \). By Eq. (3.3), we get \( \left[ {\begin{matrix} |x^*| &{} x \\ x^* &{} |x| \end{matrix}} \right] \ge 0 \). Hence,

$$\begin{aligned} \Vert x\Vert _{p,1} \le \frac{1}{2}(\Vert x\Vert _p + \Vert x^*\Vert _p) = \Vert x\Vert _p. \end{aligned}$$

To prove the converse inequality, we apply Theorems 3.4 and 2.3 and get \({ f,g \in L_p({\mathcal {M}})_+, \Vert f\Vert _p = \Vert g\Vert _p = \Vert x\Vert _{p,1}} \), \({ y \in {\mathcal {M}}, \Vert y\Vert _\infty \le 1 }\) and \({x = f^{\frac{1}{2}} y g^{\frac{1}{2}}.}\) Hence

$$\begin{aligned} \Vert x\Vert _p = \Vert f^{\frac{1}{2}} y g^{\frac{1}{2}}\Vert _p \le \Vert f\Vert _p^{\frac{1}{2}} \Vert y\Vert _\infty \Vert g\Vert _p^{\frac{1}{2}} = \Vert x\Vert _{p,1}. \end{aligned}$$

\({\square} \)

Completely order bounded maps

In this section, we define completely order bounded maps from \({ L_p} \) to \({ L_q }\) and show the decomposition of such maps for \({ p= \infty , q} \) arbitrary and for \({ p }\) arbitrary, \({ q = 1} \). For \({ 2p< q < \infty }\) we give an example of a completely order bounded map which is not decomposable.

Throughout this Sects. \({ {\mathcal {M}} }\) and \({ {\mathcal {N}}} \) are von Neumann algebras with no further restrictions unless stated explicitly. If \({ 1 \le p,q \le \infty , n \in {\mathbb {N}} }\), and \({ T: L_p({\mathcal {M}}) \rightarrow L_q({\mathcal {N}})}\) is a linear map, then

$$\begin{aligned} T_n: L_p({\mathcal {M}}) \otimes M_n \rightarrow L_q({\mathcal {N}}) \otimes M_n, [x_{ij}] \mapsto [T(x_{ij})]. \end{aligned}$$

We need the notion of decomposable maps which were introduced for \({ C^*} \)-algebras in [5] and extended to non-commutative \({ L_p} \)-spaces in [6, 9]. The above map \({T}\) is called decomposable if there exist completely positive maps \( {S_1,S_2: L_p({\mathcal {M}}) \rightarrow L_q({\mathcal {N}})}\) such that the induced map

$$\begin{aligned} \Phi : L_p({\mathcal {M}}) \otimes M_2 \rightarrow L_q({\mathcal {N}}) \otimes M_2, \begin{bmatrix} x_{11} &{} x_{12} \\ x_{21} &{} x_{22} \end{bmatrix} \mapsto \begin{bmatrix} S_1(x_{11}) &{} T(x_{12}) \\ T(x_{21}^*)^* &{} S_2(x_{22}) \end{bmatrix} \end{aligned}$$
(4.1)

is completely positive. The decomposable norm \({ \Vert T\Vert _{dec} }\) is defined by

$$\begin{aligned} \Vert T\Vert _{dec} = \inf \left\{ \max \left\{ \Vert S_1\Vert , \Vert S_2\Vert \right\} \right\} \end{aligned}$$

where the infimum is taken over all completely positive maps \( {S_1} \) and \( {S_2 }\) in (4.1).

Definition 4.1

Let \({ 1 \le p,q \le \infty} \). A linear map \({ T: L_p({\mathcal {M}}) \rightarrow L_q({\mathcal {N}}) }\) is called completely order bounded, if

$$\begin{aligned} \Vert T\Vert _{cob} = \sup \left\{ \Vert T_n(x)\Vert _{q,n} ~|~ x \in L_p({\mathcal {M}}) \otimes M_n, \Vert x\Vert _{p,n} \le 1,~ n\in {\mathbb {N}} \right\} < \infty . \end{aligned}$$

The name completely order bounded will be justified by Theorem 4.5 where we show that a completely order bounded map maps order intervals to order intervals uniformly over all matrix levels.

Proposition 4.2

Let \( {1 \le p,q \le \infty }\) and \( {T: L_p({\mathcal {M}}) \rightarrow L_q({\mathcal {N}})} \) be completely positive. Then \({ T }\) is completely order bounded and

$$\begin{aligned} \Vert T\Vert _{cob} = \Vert T\Vert = \sup \left\{ \Vert T(x)\Vert _q~|~x \in L_p({\mathcal {M}})_+,~ \Vert x\Vert _p \le 1 \right\} . \end{aligned}$$

Here, \({ \Vert T\Vert} \) means the usual operator norm of \( {T} \) as a bounded operator on a normed vector space.

Proof

Let \({ \Lambda = \sup \left\{ \Vert T(x)\Vert _q~|~x \in L_p({\mathcal {M}})_+,~ \Vert x\Vert _p \le 1 \right\}} \). By Theorem 3.6, we have

$$\begin{aligned} \sup \left\{ \Vert T(x)\Vert _q~|~x \in L_p({\mathcal {M}})_+,~ \Vert x\Vert _p \le 1 \right\} \le \Vert T\Vert \le \Vert T\Vert _{cob}. \end{aligned}$$

For the opposite inequality, let \({ n \in {\mathbb {N}}, x \in L_p({\mathcal {M}}) \otimes M_n} \) with \({ \Vert x\Vert _{p,n} \le 1} \). By Theorem 3.4, there exist \( {f,g \in L_p({\mathcal {M}})_+} \) such that

$$\begin{aligned} \Vert f\Vert _p = \Vert g\Vert _p = \Vert x\Vert _{p,n} \text { and } \begin{bmatrix} f \otimes 1_n &{} x \\ x^* &{} g \otimes 1_n \end{bmatrix} \ge 0. \end{aligned}$$

Since \({ T }\) is completely positive, we have \({ T_n(x^*) = T_n(x)^*} \) and

$$\begin{aligned} \begin{bmatrix} T(f) \otimes 1_n &{} T_n(x) \\ T_n(x)^* &{} T(g) \otimes 1_n \end{bmatrix} \ge 0 \end{aligned}$$

Hence,

$$\begin{aligned} \Vert T_n(x)\Vert _{q,n} \le \frac{1}{2}(\Vert T(f)\Vert _q+\Vert T(g)\Vert _q) \le \Lambda \Vert x\Vert _{p,n} \le \Lambda . \end{aligned}$$

\({\square}\)

Theorem 4.3

Let \({1 \le p, q \le \infty , {\mathcal {M}} \text { and } {\mathcal {N}}}\) be von Neumann algebras, and \({T: L_p({\mathcal {M}}) \rightarrow L_q({\mathcal {N}})}\) be a decomposable map. Then \({T} \) is completely order bounded and \({\Vert T\Vert _{cob} \le \Vert T\Vert _{dec}}\).

Proof

For \({\varepsilon > 0}\) there exist completely positive maps \({S_1, S_2: L_p({\mathcal {M}}) \rightarrow L_q({\mathcal {N}})} \) such that

$$\begin{aligned} \Phi : L_p({\mathcal {M}}) \otimes M_2 \rightarrow L_q({\mathcal {N}}) \otimes M_2, \begin{bmatrix} x_{11} &{} x_{12} \\ x_{21} &{} x_{22} \end{bmatrix} \mapsto \begin{bmatrix} S_1(x_{11}) &{} T(x_{12}) \\ T(x_{21}^*)^* &{} S_2(x_{22}) \end{bmatrix} \end{aligned}$$

is completely positive and \({\Vert S_1\Vert , \Vert S_2\Vert \le \Vert T\Vert _{dec} + \varepsilon} \). By Theorem 3.4, for \({n \in {\mathbb {N}}, x = [x_{ij}]\in L_p({\mathcal {M}}\,\otimes\,M_n}\), there exist \({f, g \in L_p({\mathcal {M}})_+, \Vert f\Vert _p = \Vert g\Vert _p = \Vert x\Vert _{p,n}}\) such that \( {a = \left[ {\begin{matrix} f \otimes 1_n &{} x \\ x^* &{} g \otimes 1_n \end{matrix}} \right] \ge 0 }\). We multiply \(a\) from left and right with a suitable permutation matrix to get \({b = [b_{ij}]}\) with \({b_{ij} = \left[ {\begin{matrix} f &{} x_{ij} \\ x_{ji}^* &{} g \end{matrix}} \right]} \) for \({i, j = 1, \ldots , n}\). Then \({b}\) is positive too, so \({\Phi _n(b)} \) is positive and \({\Phi (b_{ij}) = \left[ {\begin{matrix} S_1(f) &{} T(x_{ij}) \\ T(x_{ji})^* &{} S_2(g) \end{matrix}} \right] \text { for }i, j = 1, \ldots , n}\). Again, we multiply \({\Phi _n(b)}\) from left and right with a permutation matrix and get \( \left[ {\begin{matrix} S_1(f) \otimes 1_n &{} T_n(x) \\ T_n(x)^* &{} S_2(g) \otimes 1_n \end{matrix}} \right] \ge 0 \). Thus

$$\begin{aligned} \Vert T_n(x)\Vert _{q,n} \le \frac{1}{2}(\Vert S_1(f)\Vert _q + \Vert S_2(g)\Vert _q ) \le (\Vert T\Vert _{dec} + \varepsilon ) \Vert x\Vert _{p,n}. \end{aligned}$$

Since \({\varepsilon} \) was arbitrary, the proof is finished. \({\square} \)

Next, we show that the composition of complete order bounded maps is completely order bounded.

Theorem 4.4

Let \({1 \le p_1, p_2, p_3 \le \infty , {\mathcal {M}}_1, {\mathcal {M}}_2, {\mathcal {M}}_3} \) be von Neumann algebras, and \( {T_1: L_{p_1}({\mathcal {M}}_1) \rightarrow L_{p_2}({\mathcal {M}}_2), T_2: L_{p_2}({\mathcal {M}}_2) \rightarrow L_{p_3}({\mathcal {M}}_3)}\) be completely order bounded maps. Then the composition \({ T_2 \circ T_1}\) is completely order bounded and

$$\begin{aligned} \Vert T_2 \circ T_1\Vert _{cob} \le \Vert T_2\Vert _{cob} \Vert T_1\Vert _{cob}. \end{aligned}$$

Proof

Let \( {n \in {\mathbb {N}} }\) and \( {x \in L_{p_1}({\mathcal {M}}_1) \otimes M_n} \) such that \({ \Vert x\Vert _{p_1,n} \le 1 }\). Then

$$\begin{aligned} \Vert T_{2,n}(T_{1,n}(x))\Vert _{p_3,n} \le \Vert T_2\Vert _{cob}\Vert T_{1,n}(x))\Vert _{p_2,n} \le \Vert T_2\Vert _{cob}\Vert T_1\Vert _{cob}. \end{aligned}$$

\({\square} \)

The next theorem justifies the name completely order bounded: Completely order bounded maps map order intervals into order intervals uniformly over all matrix levels.

Theorem 4.5

Let \( {1 \le p, q \le \infty }\), \( {T:L_p({\mathcal {M}}) \rightarrow L_q({\mathcal {N}}) }\) be completely order bounded, and \({ f_1, f_2 \in L_p({\mathcal {M}})_+} \). Then there exist \({ g_1, g_2 \in L_q({\mathcal {N}})_+} \) such that \({ \Vert g_1\Vert _q, \Vert g_2\Vert _q \le \frac{1}{2} \Vert T\Vert _{cob}( \Vert f_1\Vert _p + \Vert f_2\Vert _p)}\) and for all \({n \in {\mathbb {N}}, x \in L_p({\mathcal {M}}) \otimes M_n}\)

$$\begin{aligned} \begin{bmatrix} f_1 \otimes 1_n &{} x \\ x^* &{} f_2 \otimes 1_n \end{bmatrix} \ge 0 \text { implies } \begin{bmatrix} g_1 \otimes 1_n &{} T_n(x) \\ T_n(x)^* &{} g_2 \otimes 1_n \end{bmatrix} \ge 0. \end{aligned}$$
(4.2)

Proof

For \({ q > 1, n\in {\mathbb {N}}, x \in L_p({\mathcal {M}}) \otimes M_n} \) with \({ \left[ {\begin{matrix} f_1 \otimes 1_n &{} x \\ x^* &{} f_2 \otimes 1_n \end{matrix}} \right] \ge 0 }\), we put

$$\begin{aligned} K(x) =&\left\{ (g_1, g_2) \in (L_q({\mathcal {N}})_+ ,L_q({\mathcal {N}})_+ ) \big {|} \right. \\&\quad \begin{bmatrix} g_1 \otimes 1_n &{} T_n(x) \\ T_n(x)^* &{} g_2 \otimes 1_n \end{bmatrix} \ge 0, \\&\quad \Vert g_1\Vert _q,\Vert g_2\Vert _q \le \frac{1}{2}\Vert T\Vert _{cob} (\Vert f_1\Vert _p + \Vert f_2\Vert _p) \left. \right\} . \end{aligned}$$

For \( {q = 1} \) we define a similar set, but take pairs \( {(g_1,g_2) \in {\mathcal {N}}^*_+ \times {\mathcal {N}}^*_+} \) instead of \({ L_1({\mathcal {N}})_+ \times L_1({\mathcal {N}})_+} \). By Theorem 3.4, \( {K(x) }\) is not empty. Further, \( {K(x) }\) is weak\({ ^* }\)-closed. This is proved similarly as in the proof of Theorem 3.4. By the Banach-Alaoglu theorem, the unit ball of \({ L_q({\mathcal {N}})} \) is compact in the weak\({^* }\)-topology for \({q > 1}\). The same holds for \({{\mathcal {N}}^*}\). Hence all sets \({ K(x) }\) are compact in the weak \({ ^*} \)-topology. The sets \({ K(x), x \in L_p({\mathcal {N}}) \otimes M_n, n\in {\mathbb {N}}, }\) have the finite intersection property: For \({ k \in {\mathbb {N}}, n_1, \ldots , n_k \in {\mathbb {N}}} \), and \({ x_i \in L_p({\mathcal {M}}) \otimes M_{n_i}} \) with

$$\begin{aligned} \begin{bmatrix} f_1 \otimes 1_{n_i} &{} x_i \\ x_i^* &{} f_2 \otimes 1_{n_i} \end{bmatrix} \ge 0,~ i\in \{1, \ldots , k\}, \end{aligned}$$

we put all \({x_i }\) in the diagonal matrix \({ x = \text {diag}(x_1, \ldots , x_k) }\) and set \({ n = \sum _{i=1}^{k} n_i} \). Then

$$\begin{aligned} \begin{bmatrix} f_1 \otimes 1_{n} &{} x \\ x^* &{} f_2 \otimes 1_{n} \end{bmatrix} \ge 0. \end{aligned}$$

By Theorem 3.4, there exist \( {g_1, g_2 \in L_q({\mathcal {N}})_+ }\) such that

$$\begin{aligned} \begin{bmatrix} g_1 \otimes 1_n &{} T_n(x) \\ T_n(x)^* &{} g_2 \otimes 1_n \end{bmatrix} \ge 0 \text { and } \Vert g_1\Vert , \Vert g_2\Vert \le \frac{1}{2}\Vert T\Vert _{cob} (\Vert f_1\Vert _p+ \Vert f_2\Vert _p). \end{aligned}$$

Hence, \({ (g_1, g_2) \in K(x_i) \text { for all } i \in \{1, \ldots , k\}}\). We conclude that

$$\begin{aligned} \bigcap \limits _{n \in {\mathbb {N}},~ x \in L_p({\mathcal {M}}) \otimes M_n} K(x) \ne \emptyset . \end{aligned}$$

We take a pair \({ (g_1,g_2) }\) of this set. If \({ q > 1} \), this pair fulfills (4.2). If \({ q = 1 }\) there is a central projection \({ z \in {\mathcal {N}}^{**} }\) which maps \( {\mathcal {N}}^* \) to \( {L_1({\mathcal {N}})} \). Then the pair \({ (zg_1,zg_2)} \) is in \({L_1({\mathcal {N}})_+ \times L_1({\mathcal {N}})_+ }\) and fulfills (4.2). \({\square }\)

Theorem 4.6

Let \({1 \le q \le \infty} \) and let \({ T: {\mathcal {M}} \rightarrow L_q({\mathcal {N}}) }\) be completely order bounded. Then there exist \({[ f,g \in L_q({\mathcal {N}})_+} \) and a completely order bounded map \({ S : {\mathcal {M}} \rightarrow {\mathcal {N}}} \) such that

$$\begin{aligned} T(x) = f^{\frac{1}{2}}S(x)g^{\frac{1}{2}} \text { for all } x \in {\mathcal {M}} \end{aligned}$$

and

$$\begin{aligned} \Vert f\Vert _q, \Vert g\Vert _q \le \Vert T\Vert _{cob},~ \Vert S\Vert _{cob} \le 1. \end{aligned}$$

Note that in case of a linear map from \( {\mathcal {M}} \) to \( {\mathcal {N}} \) completely order bounded is identical to completely bounded.

Proof

Since the norm \(\Vert \cdot \Vert _{\infty ,n}\) is the usual operator norm on \({\mathcal {M}} \otimes M_n\), we get \(\big [{\begin{matrix} {\mathbf{1}} \otimes 1_{n} &{}&{} x \\ x^{*} &{}&{} {\mathbf{1}} \otimes 1_n \end{matrix}}\big ] \ge 0\) for all \({n \in {\mathbb {N}}, x \in {\mathcal {M}} \otimes M_n, \Vert x\Vert _{\infty ,n} \le 1}\) where \({\mathbf{1}} \) is the unit in \({\mathcal {M}}\). Hence we can apply Theorem 4.5 with \({f_1 = f_2 = \mathbf{1}} \) and get \({ f, g \in L_q({\mathcal {N}})}\), such that \( {\Vert f\Vert _q, \Vert g\Vert _q \le \Vert T\Vert _{cob}}\) and

$$\begin{aligned} \left[ \begin{array}{cc} f \otimes 1_n &{} T_n(x) \\ T_n(x)^* &{} g \otimes 1_n \end{array} \right] \ge 0 \text { for all } n\in {\mathbb {N}}, x \in {\mathcal {M}} \otimes M_n, \Vert x\Vert _{\infty ,n} \le 1. \end{aligned}$$

Let \({x \in {\mathcal {M}}}\). We apply Theorem 2.3 with \({n= 1}\) to \({x}\), and get a unique \({y \in {\mathcal {N}}, \Vert y\Vert _{\infty } \le 1}\) such that \({T(x) = f^{\frac{1}{2}}yg^{\frac{1}{2}}}\) and \({\text {supp}(f)\cdot y \cdot \text {supp}(g) = y}\). We put \({S(x) = y}\). Then \({S}\) is a map from \({\mathcal {M}}\) to \({\mathcal {N}}\). If \({x, x_1, x_2 \in {\mathcal {M}}, x = x_1 + x_2,}\) then there are according to Theorem 2.3\(y, y_1, y_2 \in {\mathcal {N}}\) such that \(T(x) = f^{\frac{1}{2}}yg^{\frac{1}{2}}, \text {supp}(f) \cdot y \cdot \text {supp}(g) = y\) and \(T(x_i) = f^{\frac{1}{2}}y_ig^{\frac{1}{2}}, \text {supp}(f) \cdot y_i \cdot \text {supp}(g) = y_i, i=1,2\). Then we get

$$\begin{aligned} f^{\frac{1}{2}}yg^{\frac{1}{2}} = T(x) = T(x_1) + T(x_2) = f^{\frac{1}{2}}(y_1+y_2)(g^{\frac{1}{2}}), \\ \text {supp}(f) \cdot (y_1+y_2) \cdot \text {supp}(g) = y_1 + y_2. \end{aligned}$$

Hence we conclude from Lemma 2.2 that \(y = y_1 + y_2\) which means that \(S\) is additive. Similarly, we show that \(S(\lambda x) = \lambda S(x) \text { for } \lambda \in {\mathbb {C}}, x \in {\mathcal {M}}\). Now for \( n \in {\mathbb {N}}, \) let \( x \in {\mathcal {M}} \otimes M_n\) with \(\Vert x\Vert _{\infty ,n} \le 1\). According to Theorem 2.3, there is \(y \in {\mathcal {N}} \otimes M_n\) such that \((\text {supp}(f) \otimes 1_n) \cdot y \cdot (\text {supp}(g) \otimes 1_n) = y\) and \(T_n(x) = (f^{\frac{1}{2}} \otimes 1_n)y(g^{\frac{1}{2}} \otimes 1_n) \). Then, we have

$$\begin{aligned} (f^{\frac{1}{2}} \otimes 1_n)S_n(x)(g^{\frac{1}{2}} \otimes 1_n) = T_n(x) = (f^{\frac{1}{2}} \otimes 1_n)y(g^{\frac{1}{2}} \otimes 1_n). \end{aligned}$$

Hence, we conclude that \(S_n(x) = y\) which shows that \(\Vert S_n(x)\Vert _{\infty ,n} \le 1\). So \(\Vert S\Vert _{cob} \le 1\). \(\square \)

Theorem 4.7

Let \({\mathcal {N}}\) be injective, \( 1 \le q \le \infty \) and let \( T: {\mathcal {M}} \rightarrow L_q({\mathcal {N}}) \) be completely order bounded. Then there exist linear maps \(T_i: {\mathcal {M}} \rightarrow L_q({\mathcal {N}})\) such that the map

$$\begin{aligned} \begin{aligned} \Phi :&{\mathcal {M}} \otimes M_2 \rightarrow L_q({\mathcal {N}}) \otimes M_2 \\&\begin{bmatrix} x_{11} &{} x_{12} \\ x_{21} &{} x_{22} \end{bmatrix} \mapsto \begin{bmatrix} T_1(x_{11}) &{} T(x_{12}) \\ T(x_{21}^*)^* &{} T_2(x_{22}) \end{bmatrix} \end{aligned} \end{aligned}$$

is complete positive and \(\Vert T_1\Vert , \Vert T_2\Vert \le \Vert T\Vert _{cob} \). Thus \(T\) is decomposable and \(\Vert T\Vert _{dec} = \Vert T\Vert _{cob}\).

Proof

By Theorem 4.6, there exist \(f, g \in L_q({\mathcal {N}})_+, \Vert f\Vert _q, \Vert g\Vert _q \le \Vert T\Vert _{cob}\) and \(S: {\mathcal {M}} \rightarrow {\mathcal {N}}, \Vert S\Vert _{cob} \le 1\) such that

$$\begin{aligned} T(x) = f^{\frac{1}{2}}S(x)g^{\frac{1}{2}}, \quad x \in {\mathcal {M}}. \end{aligned}$$

Since for a linear map from \({\mathcal {M}}\) to \({\mathcal {N}}\) completely bounded is the same as completely order bounded, we can apply [14], Theorem 4.5, and get linear maps \(S_1, S_2: {\mathcal {M}} \rightarrow {\mathcal {N}}\) such that \(\Vert S_i\Vert \le 1, i = 1,2\), and the map

$$\begin{aligned} \left[ \begin{array}{cc} x_{11} &{} x_{12} \\ x_{21} &{} x_{22} \end{array} \right] \mapsto \left[ \begin{array}{cc} S_1(x_{11}) &{} S(x_{12}) \\ S(x_{21}^*)^* &{} S_2(x_{22}) \end{array} \right] \end{aligned}$$

is completely positive. The linear maps \(T_i: {\mathcal {M}} \rightarrow L_q({\mathcal {N}})\) where \(T_1(x) = f^{\frac{1}{2}}S_1(x)f^{\frac{1}{2}}, T_2(x) = g^{\frac{1}{2}}S_2(x)g^{\frac{1}{2}}, x \in {\mathcal {M}},\) fulfill \(\Vert T_1\Vert \le \Vert f\Vert _q \Vert S_1\Vert \le \Vert T\Vert _{cob}\), \(\Vert T_2\Vert \le \Vert g\Vert _q \Vert S_2\Vert \le \Vert T\Vert _{cob}\), and

$$\begin{aligned} {\mathcal {M}} \otimes M_2 \ni \left[ \begin{array}{cc} x_{11} &{} x_{12} \\ x_{21} &{} x_{22} \end{array} \right] \mapsto \left[ \begin{array}{cc} T_1(x_{11}) &{} T(x_{12}) \\ T(x_{21}^*)^* &{} T_2(x_{22}) \end{array} \right] \in L_q({\mathcal {N}}) \otimes M_2 \end{aligned}$$

is completely positive. Together with Theorem 4.3 we get \(\Vert T\Vert _{dec} = \Vert T\Vert _{cob}\) \(\square \)

Remark 4.8

Theorem 4.5 in [14] states the decomposition for self-adjoint maps. The proof of Proposition 1.3 in [5] shows that this is also true for maps which are not self-adjoint.

The next goal is to prove that completely order bounded maps from \(L_p({\mathcal {M}})\) to \(L_1({\mathcal {N}}).\) are decomposable. This proof is divided into several steps.

Lemma 4.9

Let \( k \in {\mathbb {N}}\) and \(a \in L_p({\mathcal {M}}) \otimes M_2 \otimes M_{2k}\) be self-adjoint. Then \(a\) can be written in the form

$$\begin{aligned} a = \sum _{j=1}^{n} a_j \otimes \alpha _j \end{aligned}$$

where all \(a_j \in L_p({\mathcal {M}}) \otimes M_2\) and all \(\alpha _j \in M_{2k}\) are self-adjoint.

Proof

We mimic the proof of [12], Lemma IV.4.4. If \(a = \sum _{j=1}^{n} b_j \otimes \beta _j\) with \(b_j \in {\mathcal {M}} \otimes M_2\) and \(\beta _j \in M_{24}\) for \(j = 1, \ldots , n\) and \(a\) is self-adjoint then

$$\begin{aligned} a = \frac{1}{4}\sum _{j=1}^{n}\left( (b_j + b_j^*) \otimes (\beta _j + \beta _j^*) - i(b_j - b_j^*) \otimes i(\beta _j - \beta _j^*) \right) \end{aligned}$$

\(\square \)

Let \(Tr\) be the usual trace on \(M_{2k}\). We define the duality between \(L_1(M_k) \otimes M_2\) and \(M_{2k}\) by \(\langle a,b\rangle = Tr(ab), a \in L_1(M_k) \otimes M_2, b \in M_{2k}.\) If \(n \in {\mathbb {N}}, a = [a_{ij}], b = [b_{ij}],\) where \(a_{ij} \in L_1(M_k) \otimes M_2\) and \( b_{ij} \in M_k \otimes M_2 \) then \(\langle [a_{ij}],[b_{ij}]\rangle = \sum _{i,j=1}^{2k} \langle a_{ij}, b_{ji}\rangle .\) We define the linear functional

$$\begin{aligned} \begin{aligned} \omega :&L_1(M_k) \otimes M_2 \otimes M_{2k} \rightarrow {\mathbb {C}} \\&\sum _{i=1}^{n} a_i \otimes b_i \mapsto \sum _{i=1}^{n} \langle a_i,b_i^t\rangle \end{aligned} \end{aligned}$$
(4.3)

where \(b_i^t\) denotes the transposed matrix of \(b_i.\)

Lemma 4.10

Let \(a \in (L_1(M_k) \otimes M_2 \otimes M_{2k})_+.\) Then \(\omega (a) \ge 0\).

Proof

Let \( a^\frac{1}{2} = \sum _{i=1}^{n} a_i \otimes \alpha _i.\) Then \(a^\frac{1}{2}\) is self-adjoint, \(a = \sum _{i,j=1}^{n} a_i^*a_j \otimes \alpha _i^* \alpha _j,\) and

$$\begin{aligned} \begin{aligned} \omega (a)&= \sum _{i,j=1}^{n} \omega (a_i^*a_j \otimes \alpha _i^* \alpha _j) = \sum _{i,j=1}^{n} \left\langle a_i^*a_j,\alpha _j^t\alpha _i^{*t}\right\rangle \\&= \left\langle \left[ a_i^*a_j\right] , \left[ \alpha _i^t \alpha _j^{*t} \right] \right\rangle \end{aligned} \end{aligned}$$

Since \(\left[ a_i^*a_j\right] \) and \(\left[ \alpha _i^t \alpha _j^{*t} \right] \) are positive matrices, the last expression is positive by (2.3). \(\square \)

For a linear map \(T: L_p({\mathcal {M}}) \rightarrow L_1(M_k)\) we define

$$\begin{aligned} \begin{aligned} {\tilde{T}}:&L_p({\mathcal {M}}) \otimes M_2 \rightarrow L_1(M_k) \otimes M_2\\&\begin{bmatrix} x_{11} &{} x_{12} \\ x_{21} &{} x_{22} \end{bmatrix} \mapsto \begin{bmatrix} 0 &{} T(x_{12}) \\ T(x_{21}^*)^* &{} 0 \end{bmatrix}. \end{aligned}\end{aligned}$$
(4.4)

We define the linear functional

$$\begin{aligned} \begin{aligned} \varphi _T:&L_p({\mathcal {M}}) \otimes M_2 \otimes M_{2k} \rightarrow {\mathbb {C}}\\&x \mapsto \omega \circ {\tilde{T}}_{2k}(x). \end{aligned} \end{aligned}$$
(4.5)

Lemma 4.11

Let \(T: L_p({\mathcal {M}}) \rightarrow L_1(M_k)\) be completely order bounded and \(\varphi _T\) as in (4.5). Let \(a, b \in (L_p({\mathcal {M}}) \otimes M_2 \otimes M_{2k})_+, f, g \in L_p({\mathcal {M}})_+\) such that

$$\begin{aligned} 0 \le a + b \le f \otimes \varepsilon _{11} \otimes 1_{2k} + g \otimes \varepsilon _{22} \otimes 1_{2k}. \end{aligned}$$
(4.6)

Then

$$\begin{aligned} \Vert T\Vert _{cob}( \Vert f\Vert _p + \Vert g\Vert _p ) + \varphi _T ( a - b) \ge 0. \end{aligned}$$

Here \(\varepsilon _{ij}, i,j = 1,2\) are the \(2 \times 2\) matrices defined in (3.2).

Proof

Since \(a-b\) is self-adjoint, by Lemma 4.9, we can write

$$\begin{aligned} a-b = \sum _{i=1}^{n} a_i \otimes \alpha _i \end{aligned}$$

where all \(a_i \in L_p({\mathcal {M}}) \otimes M_2\) and all \(\alpha _i \in M_{2k}\) are self-adjoint. For \(i \in \{1, \ldots ,n\}\) let

$$\begin{aligned} a_i = \begin{bmatrix} a_{i,11} &{} a_{i,12} \\ a_{i,21} &{} a_{i,22} \end{bmatrix}, \alpha _i = \begin{bmatrix} \alpha _{i,11} &{} \alpha _{i,12} \\ \alpha _{i,21} &{} \alpha _{i,22} \end{bmatrix}, \end{aligned}$$

where \(a_{i,st} \in L_p({\mathcal {M}})\) and \(\alpha _{i,st} \in M_k\) for \(s,t = 1,2\). Since \(a_i\) and \(\alpha _i\) are self-adjoint for all \(i\), we have \(a_{i,21}^* = a_{i,12}\), \(\alpha _{i,21}^* = \alpha _{i,12}\), and \(\langle T(a_{i,21}^*)^*,\alpha _{i,21}^t\rangle = \langle T(a_{i,12})^*,\alpha _{i,12}^{*t}\rangle = \overline{\langle T(a_{i,12}),\alpha _{i,12}^t\rangle }\) for all \(i \in \{1, \ldots ,n\}\). Therefore

$$\begin{aligned} \begin{aligned} \varphi _T(a-b)&= \sum _{i=1}^{n} \varphi _T( a_i \otimes \alpha _i) \\&= \sum _{i=1}^{n} \left\langle \begin{bmatrix} 0 &{} T(a_{i,12}) \\ T(a_{i,21}^*)^* &{} 0 \end{bmatrix}, \begin{bmatrix} \alpha _{i,11}^t &{} \alpha _{i,21}^t \\ \alpha _{i,12}^t &{} \alpha _{i,22}^t \end{bmatrix} \right\rangle \\&= \sum _{i=1}^{n} \left( \langle T(a_{i,12}),\alpha _{i,12}^t\rangle + \langle T(a_{i,12})^*, \alpha _{i,12}^{*t}\rangle \right) . \end{aligned} \end{aligned}$$

By Proposition 2.1, we have

$$\begin{aligned} \begin{bmatrix} (f \otimes \varepsilon _{11} + g \otimes \varepsilon _{22}) \otimes 1_{2k} &{} a-b \\ a-b &{} (f \otimes \varepsilon _{11} + g \otimes \varepsilon _{22}) \otimes 1_{2k} \end{bmatrix} \ge 0. \end{aligned}$$
(4.7)

We multiply the matrix in (4.7) from left with the matrix \(\gamma \) and from right with the transposed matrix \(\gamma ^t\), where \(\gamma \) is the \(8k \times 4k\) matrix \(\left[ {\begin{matrix} \varepsilon _{11} \otimes 1_{2k} \\ \varepsilon _{22} \otimes 1_{2k} \end{matrix}}\right] \) and get

$$\begin{aligned} \begin{aligned} 0&\le \varepsilon _{11}(f \otimes \varepsilon _{11} + g \otimes \varepsilon _{22}) \varepsilon _{11} \otimes 1_{2k} + (\varepsilon _{11} \otimes 1_{2k})(a-b)(\varepsilon _{22} \otimes 1_{2k}) \\&\quad + (\varepsilon _{22} \otimes 1_{2k})(a-b)(\varepsilon _{11} \otimes 1_{2k}) + \varepsilon _{22}(f \otimes \varepsilon _{11} + g \otimes \varepsilon _{22}) \varepsilon _{22} \otimes 1_{2k} \\&= f \otimes \varepsilon _{11} \otimes 1_{2k} + \sum _{i=1}^{n} a_{i,12} \otimes \varepsilon _{12} \otimes \alpha _i \\&\quad +\sum _{i=1}^{n} a_{i,21} \otimes \varepsilon _{21} \otimes \alpha _i + g \otimes \varepsilon _{22} \otimes 1_{2k}. \end{aligned} \end{aligned}$$
(4.8)

This inequality can be written as \(\Vert \sum _{i=1}^{n} a_{i,12} \otimes \alpha _i\Vert _{p,2k} \le \frac{1}{2}(\Vert f\Vert _p + \Vert g\Vert _p). \) Hence \(\Vert T(\sum _{i=1}^{n} a_{i,12} \otimes \alpha _i)\Vert _{1,2k} \le \frac{1}{2} \Vert T\Vert _{cob}(\Vert f\Vert _p + \Vert g\Vert _p)\). By Theorem 3.4, there exist \(f_1, g_1 \in L_1(M_k)_+\) such that \( \Vert f_1\Vert _1 = \Vert g_1\Vert _1 \le \frac{1}{2} \Vert T\Vert _{cob}(\Vert f\Vert _p + \Vert g\Vert _p)\) and

$$\begin{aligned} \begin{bmatrix} f_1 \otimes 1_{2k} &{} \sum _{i=1}^{n} T(a_{i,12}) \otimes \alpha _i \\ \sum _{i=1}^{n} T(a_{i,12})^* \otimes \alpha _i &{} g_1 \otimes 1_{2k} \end{bmatrix} \ge 0. \end{aligned}$$
(4.9)

We apply \(\omega \) to (4.9) and get

$$\begin{aligned} \begin{aligned} 0&\le \langle f_1 \otimes \varepsilon _{11}, 1_{2k}\rangle + \langle g_1 \otimes \varepsilon _{22}, 1_{2k}\rangle \\&\quad + \sum _{i=1}^{n} \left( \langle T(a_{i,12}) \otimes \varepsilon _{12}, \alpha _i^t\rangle + \langle T(a_{i,12})^* \otimes \varepsilon _{21},\alpha _i^t\rangle \right) \\&= \Vert f_1\Vert _1 + \Vert g_1\Vert _1 + \sum _{i=1}^{n} (\langle T(a_{i,12}), \alpha _{i,12}^t\rangle + \langle T(a_{i,12})^*, \alpha _{i,12}^{*t}\rangle ) \\&= \Vert f_1\Vert _1 + \Vert g_1\Vert _1 + \varphi _T(a-b). \end{aligned} \end{aligned}$$

Since \(\Vert f_1\Vert _1 + \Vert g_1\Vert _1 \le \Vert T\Vert _{cob}(\Vert f\Vert _p + \Vert g\Vert _p),\) the proof is finished. \(\square \)

Lemma 4.12

Let \(x \in (L_p({\mathcal {M}}) \otimes M_2 \otimes M_n)_+\) and \(y \in (M_k \otimes M_2 \otimes M_n)_+\) where \(x= [x_{ij}]\) with \(x_{ij} \in L_p({\mathcal {M}} \otimes M_2)\) and \(y = [y_{ij}]\) with \(y_{ij} \in M_k \otimes M_2\) for \(i,j \in \{1,\ldots ,n\}\) Then

$$\begin{aligned} \sum _{i,j = 1}^{n} x_{ij} \otimes y_{ji}^t \ge 0. \end{aligned}$$

Proof

By Lemma 2.3, there is \(f \in L_p({\mathcal {M}})_+\) and \(a \in ({\mathcal {M}} \otimes M_2 \otimes M_n)_+\) such that \(x = (f^{\frac{1}{2}} \otimes 1_{2n})a(f^{\frac{1}{2}} \otimes 1_{2n})\). By [12], Lemma IV.3.1, \(a\) can be written as a finite sum of matrices of the form \([a_i^*a_j]\) where \(a_1, \ldots , a_n \in {\mathcal {M}} \otimes M_2\). Similarly, \(y\) can be written as a finite sum of matrices of the form \([y_i^*y_j]\) where \(y_1, \ldots , y_n \in M_k \otimes M_2\). Thus \(\sum _{i,j = 1}^{n} x_{ij} \otimes y_{ij}^t\) is a finite sum of elements of the form \(\sum _{i,j = 1}^{n} (f^{\frac{1}{2}} \otimes 1_2) a_i^*a_j (f^{\frac{1}{2}} \otimes 1_2) \otimes (y_i^*y_j)^t\). Then we have

$$\begin{aligned} \begin{aligned}&\sum _{i,j = 1}^{n} (f^{\frac{1}{2}} \otimes 1_2) a_i^*a_j (f^{\frac{1}{2}} \otimes 1_2) \otimes (y_j^*y_i)^t \\ =&\sum _{i,j=1}^{n} (f^{\frac{1}{2}} \otimes 1_2)a_i^*a_j (f^{\frac{1}{2}} \otimes 1_2) \otimes y_i^ty_j^{*t} \\ =&\left( \sum _{i=1}^{n}a_i(f^{\frac{1}{2}} \otimes 1_2) \otimes y_i^{*t}\right) ^*\left( \sum _{j=1}^{n}a_j(f^{\frac{1}{2}} \otimes 1_2) \otimes y_j^{*t}\right) \ge 0 \end{aligned} \end{aligned}$$

\(\square \)

Proposition 4.13

Let \(T: L_p({\mathcal {M}}) \rightarrow L_1(M_k)\) be completely order bounded and let \({\tilde{T}}\) be as in (4.4). Then there is a linear map \(S: L_p({\mathcal {M}}) \otimes M_2 \rightarrow L_1(M_k) \otimes M_2\) such that \(S \pm {\tilde{T}}\) are completely positive and

$$\begin{aligned} \begin{aligned}&0 \le \left\langle S \left( \begin{bmatrix} c &{} 0 \\ 0 &{} 0 \end{bmatrix} \right) , \begin{bmatrix} y &{} 0 \\ 0 &{} 0 \end{bmatrix} \right\rangle \le \Vert T\Vert _{cob} \Vert c\Vert _p \Vert y\Vert _{\infty } \\&0 \le \left\langle S \left( \begin{bmatrix} 0 &{} 0 \\ 0 &{} c \end{bmatrix} \right) , \begin{bmatrix} 0 &{} 0 \\ 0 &{} y \end{bmatrix} \right\rangle \le \Vert T\Vert _{cob} \Vert c\Vert _p \Vert y\Vert _{\infty } \end{aligned} \end{aligned}$$
(4.10)

for all \(c \in L_p({\mathcal {M}})_+\) and all \(y \in M_{k+}\).

Proof

Let \((L_p({\mathcal {M}}) \otimes M_2 \otimes M_{2k})_h\) denote the self-adjoint part of \( L_p({\mathcal {M}}) \otimes M_2 \otimes M_{2k}\) and let \(\varphi _T\) be the linear functional defined by (4.5). For \(x \in (L_p({\mathcal {M}}) \otimes M_2 \otimes M_{2k})_h\) we define

$$\begin{aligned} \begin{aligned} \theta (x) = \inf \bigl \{&\Vert T\Vert _{cob} (\Vert f\Vert _p + \Vert g\Vert _p) + \varphi _T(a-b) ~\big | \\&a, b \in (L_p({\mathcal {M}}) \otimes M_2 \otimes M_{2k})_+, f, g \in L_p({\mathcal {M}})_+, \\&x + a + b \le f \otimes \varepsilon _{11} \otimes 1_{2k}+ g \otimes \varepsilon _{22} \otimes 1_{2k} \bigr \}. \end{aligned} \end{aligned}$$
(4.11)

By Lemma 2.4, the set on the right side of (4.11) is not empty, so \(\theta \) is well defined. We will show that \(\theta \) is sublinear. To do this, let \(x_1, x_2 \in (L_p({\mathcal {M}}) \otimes M_2 \otimes M_{2k})_h\) and \(\varepsilon > 0\). Then there exist \(a_1, b_1, a_2, b_2 \in (L_p({\mathcal {M}}) \otimes M_2 \otimes M_{2k})_+\) and \(f_1, g_1, f_2, g_2 \in L_p({\mathcal {M}})_+\) such that

$$\begin{aligned} \begin{aligned}&x_1 + a_1 + b_1 \le f_1 \otimes \varepsilon _{11} \otimes 1_{2k} + g_1 \otimes \varepsilon _{22} \otimes 1_{2k}, \\&x_2 + a_2 + b_2 \le f_2 \otimes \varepsilon _{11} \otimes 1_{2k} + g_2 \otimes \varepsilon _{22} \otimes 1_{2k}, \\&\theta (x_1) \ge \Vert T\Vert _{cob}(\Vert f_1\Vert _p + \Vert g_1\Vert _p) + \varphi _T(a_1 - b_1) - \varepsilon , \\&\theta (x_2) \ge \Vert T\Vert _{cob}(\Vert f_2\Vert _p + \Vert g_2\Vert _p) + \varphi _T(a_2 - b_2) - \varepsilon . \end{aligned} \end{aligned}$$

Then we get

$$\begin{aligned} x_1 + x_2 + a_1 + a_2 + b_1 + b_2 \le (f_1 + f_2) \otimes \varepsilon _{11} \otimes 1_{2k} + (g_1 + g_2) \otimes \varepsilon _{22} \otimes 1_{2k}. \end{aligned}$$

This implies

$$\begin{aligned} \theta (x_1 + x_2) \le \theta (x_1) + \theta (x_2) - 2\varepsilon . \end{aligned}$$

Since \(\varepsilon \) was arbitrary, \(\theta \) is sub-additive. Similarly, we show for \( 0 < \lambda \in {\mathbb {R}}\) and \(x \in (L_p({\mathcal {M}}) \otimes M_2 \otimes M_{2k})_h\) that

$$\begin{aligned} \theta (\lambda x) \le \lambda \theta (x) \end{aligned}$$

and hence

$$\begin{aligned} \lambda \theta (x) = \lambda \theta ( \frac{1}{\lambda } \lambda x) \le \lambda \frac{1}{\lambda } \theta (\lambda x) = \theta (\lambda x). \end{aligned}$$

It remains to show that \( \theta (0) = 0\). If we put \( x = a = b = 0, f = g = 0\) in (4.11), we get \(\theta (0) \le 0\). Lemma 4.11 states that every element in the set on the right side of (4.11) is not negative for \(x = 0\). Hence \(\theta (0) = 0\). By the Hahn-Banach theorem there is a real-linear functional

$$\begin{aligned} \psi : (L_p({\mathcal {M}}) \otimes M_2 \otimes M_{2k})_h \rightarrow {\mathbb {R}} \end{aligned}$$
(4.12)

such that \(\psi (x) \le \theta (x)\) for all \(x \in (L_p({\mathcal {M}}) \otimes M_2 \otimes M_{2k})_h\). Now we can extend \(\psi \) to a complex linear functional on \(L_p({\mathcal {M}}) \otimes M_2 \otimes M_{2k}\) by putting \(\psi (x) = \frac{1}{2}\psi (x + x^*) + \frac{1}{2} \text {i} \psi (\text {i} x^* - \text {i} x)\) for \(x \in L_p({\mathcal {M}}) \otimes M_2 \otimes M_{2k} \). For \(c \in (L_p({\mathcal {M}}) \otimes M_2)_+, \Vert c\Vert _{p,2} \le 1\), by Theorem 3.4, there exists \(f \in L_p({\mathcal {M}})_+\) such that

$$\begin{aligned} f \otimes 1_2 - c \ge 0 \text { and } \Vert f\Vert _p \le 1. \end{aligned}$$

For \(y \in (M_{2k})_+, 0 \le y \le 1_{2k}\), we put \(x = c \otimes y^t, a = b = 0\), apply (4.11), and get

$$\begin{aligned} \psi (c \otimes y^t) \le \theta (c \otimes y^t) \le 2\Vert T\Vert _{cob}\Vert f\Vert _p \le 2 \Vert T\Vert _{cob}. \end{aligned}$$
(4.13)

Then we put \(x = -c \otimes y^t, a = c \otimes y^t, b = 0, f = g = 0\), apply (4.11), and get

$$\begin{aligned} \psi (-c \otimes y^t) \le \theta (-c \otimes y^t) \le \varphi _T(c \otimes y^t) \end{aligned}$$
(4.14)

For \(x = -c \otimes y^t, a = 0, b = c \otimes y^t, f = g = 0\), we apply (4.11) and (4.12) and get

$$\begin{aligned} \psi (-c \otimes y^t) \le \theta (-c \otimes y^t) \le - \varphi _T(c \otimes y^t) \end{aligned}$$
(4.15)

We combine (4.13), (4.14), (4.15), and get

$$\begin{aligned} 0 \le \psi (c \otimes y^t) \le 2\Vert T\Vert _{cob}. \end{aligned}$$

By Theorem 2.3 and [2], Proposition II.3.1.2, we can write \(c \in L_p({\mathcal {M}}) \otimes M_2, \Vert c\Vert _{p,2} \le 1\) as sum \(c = c_1 - c_2 + \text {i}(c_3 - c_4)\) where \(c_i \ge 0\) and \(\Vert c_i\Vert _{p,2} \le 1\) for \(i = 1, \ldots ,4\). A similar decomposition holds for \(y \in M_{2k}\). Thus the bilinear map

$$\begin{aligned} B: L_p({\mathcal {M}}) \otimes M_2 \times M_{2k} \rightarrow {\mathbb {C}}, B(c,y) = \psi (c \otimes y^t) \end{aligned}$$

is bounded. Hence there is a linear map \(S: L_p({\mathcal {M}}) \otimes M_2 \rightarrow L_1(M_k) \otimes M_2\) such that \(\langle S(c),y\rangle = \psi (c \otimes y^t)\). Next we show that \( S \pm {\tilde{T}}\) are completely positive. By (2.3), it suffices to show that for \(n \in {\mathbb {N}}\), \( x \in (L_p({\mathcal {M}}) \otimes M_2 \otimes M_n)_+, y \in (M_k \otimes M_2 \otimes M_n)_+\), where \(x = [x_{ij}]\) with \(x_{ij} \in L_p({\mathcal {M}}) \otimes M_2\) and \( y = [y_{ij}] \in M_k \otimes M_2\) the expression \(\langle S_n(x) \pm {\tilde{T}}_n(x),y\rangle \) is positive. Now we have

$$\begin{aligned} \begin{aligned} \langle S_n(x) \pm {\tilde{T}}_n(x),y\rangle&= \sum _{i,j = 1}^{n} \langle S(x_{ij}) \pm {\tilde{T}}(x_{ij}), y_{ji}\rangle \\&= \psi \left(\sum _{i,j=1}^{n} x_{ij} \otimes y_{ji}^t \right) \pm \varphi _T\left(\sum _{i,j=1}^{n} x_{ij} \otimes y_{ji}^t\right). \end{aligned} \end{aligned}$$
(4.16)

By Lemma 4.12, \(\sum _{i,j=1}^{n} x_{ij} \otimes y_{ji}^t\) is positive. We put \(x = - \sum _{i,j=1}^{n} x_{ij} \otimes y_{ji}^t,\) \(a = -x, b = 0, f = g = 0\) in (4.11) and get

$$\begin{aligned} - \psi \left(\sum _{i,j=1}^{n} x_{ij} \otimes y_{ji}^t \right) \le \varphi _T \left(\sum _{i,j=1}^{n} x_{ij} \otimes y_{ji}^t\right). \end{aligned}$$
(4.17)

Similarly, we put \(x = \sum _{i,j=1}^{n} x_{ij} \otimes y_{ji}^t,a = 0, b = -x, f = g = 0\) and get

$$\begin{aligned} - \psi \left(\sum _{i,j=1}^{n} x_{ij} \otimes y_{ji}^t \right) \le - \varphi _T \left(\sum _{i,j=1}^{n} x_{ij} \otimes y_{ji}^t \right). \end{aligned}$$
(4.18)

The combination of (4.16), (4.17), and (4.18) gives

$$\begin{aligned} \langle S_n(x) \pm {\tilde{T}}_n(x), y\rangle \ge 0. \end{aligned}$$

Now let \(c \in L_p({\mathcal {M}})_+\) and \(y \in M_{k+}\). We put \(x = c \otimes \varepsilon _{11} \otimes y^t \otimes \varepsilon _{11}, a = b = 0, f = \Vert y\Vert _{\infty }\cdot c, g = 0\). Then (4.11) gives

$$\begin{aligned} 0 \le \left\langle S \left( \begin{bmatrix} c &{} 0 \\ 0 &{} 0 \end{bmatrix} \right) , \begin{bmatrix} y &{} 0 \\ 0 &{} 0 \end{bmatrix} \right\rangle = \psi (x) \le \theta (x) \le \Vert T\Vert _{cob}\Vert c\Vert _p\Vert y\Vert _{\infty }. \end{aligned}$$

The second part of (4.10) is shown similarly. \(\square \)

Proposition 4.14

Let \( k \in {\mathbb {N}}\) and \(T: L_p({\mathcal {M}}) \rightarrow L_1(M_k)\) be completely order bounded. Then there are linear maps \(S_1, S_2: L_p({\mathcal {M}}) \rightarrow L_1(M_k)\) such that the map

$$\begin{aligned} \begin{aligned} \Phi :~&L_p({\mathcal {M}}) \otimes M_2 \rightarrow L_1(M_k) \otimes M_2 \\&\left[ \begin{array}{cc} x_{11} &{} x_{12} \\ x_{21} &{} x_{22} \end{array} \right] \mapsto \left[ \begin{array}{cc} S_1(x_{11}) &{} T(x_{12}) \\ T(x_{21}^*)^* &{} S_2(x_{22}) \end{array} \right] \end{aligned} \end{aligned}$$

is completely positive and \(\Vert S_1\Vert , \Vert S_2\Vert \le \Vert T\Vert _{cob}\).

Proof

Let \({\tilde{T}}\) be as in (4.4). By Proposition 4.13, there is a linear map \(S:~ L_p({\mathcal {M}}) \otimes M_2 \rightarrow L_1(M_k) \otimes M_2\) such that \(S \pm {\tilde{T}}\) are completely positive and for \(c \in L_p({\mathcal {M}})_+, y \in M_{k+}\)

$$\begin{aligned} \begin{aligned}&0 \le \left\langle S \left( \begin{bmatrix} c &{} 0 \\ 0 &{} 0 \end{bmatrix} \right) , \begin{bmatrix}{} y &{} 0 \\ 0 &{} 0 \end{bmatrix} \right\rangle \le \Vert T\Vert _{cob} \Vert c\Vert _p \Vert y\Vert _{\infty } \\&0 \le \left\langle S \left( \begin{bmatrix} 0 &{} 0 \\ 0 &{} c \end{bmatrix} \right) , \begin{bmatrix} 0 &{} 0 \\ 0 &{} y \end{bmatrix} \right\rangle \le \Vert T\Vert _{cob} \Vert c\Vert _p \Vert y\Vert _{\infty }. \end{aligned} \end{aligned}$$

The next steps are quite similar to the proof of [1], Proposition 3.18. Let \(\alpha \) be the scalar \(1\times 2\) matrix \([1~0]\) and \(\beta \) be the scalar \(1\times 2\) matrix \([0~ 1]\), and let \(\alpha ^*, \beta ^*\) be the adjoined matrices. We put

$$\begin{aligned} S_1: L_p({\mathcal {M}}) \rightarrow L_1(M_k), a \mapsto \alpha S \left( \begin{bmatrix} a &{} 0 \\ 0 &{} 0 \end{bmatrix} \right) \alpha ^* \end{aligned}$$

and

$$\begin{aligned} S_2: L_p({\mathcal {M}}) \rightarrow L_1(M_k), a \mapsto \beta S \left( \begin{bmatrix} 0 &{} 0 \\ 0 &{} a \end{bmatrix} \right) \beta ^* \end{aligned}$$

Since symmetric multiplication with a matrix and its adjoint is completely positive, the maps \(S_1\) and \(S_2\) are completely positive. For \(a \in L_p({\mathcal {M}})_+\), we have

$$\begin{aligned} \begin{aligned} \Vert S_1(a)\Vert _1&= \langle S_1(a),1_k\rangle = \left\langle \alpha S \left( \begin{bmatrix} a &{} 0 \\ 0 &{} 0 \end{bmatrix} \right) \alpha ^*,1_k\right\rangle \\&= \left\langle S \left( \begin{bmatrix} a &{} 0 \\ 0 &{} 0 \end{bmatrix} \right) ,\begin{bmatrix} 1 &{} 0 \\ 0 &{} 0 \end{bmatrix} \right\rangle \le \Vert T\Vert _{cob}\Vert a\Vert _p. \end{aligned} \end{aligned}$$

Hence \(\Vert S_1\Vert \le \Vert T\Vert _{cob}\). Similarly, \(\Vert S_2\Vert \le \Vert T\Vert _{cob}\). Let \(\gamma _1\) be the scalar \(2 \times 4\) matrix \(\left[ {\begin{matrix} 1 &{} 0 &{} 0 &{} 0\\ 0 &{} 0 &{} 0 &{} 1 \end{matrix}} \right] .\) For \(a = [a_{ij}] \in L_p({\mathcal {M}}) \otimes M_2\) we have

$$\begin{aligned} \gamma _1^* \begin{bmatrix} a_{11} &{} a_{12} \\ a_{21} &{} a_{22} \end{bmatrix} \gamma _1 = \begin{bmatrix} a_{11} &{} 0 &{} 0 &{} a_{12} \\ 0 &{} 0 &{} 0 &{} 0 \\ 0 &{} 0 &{} 0 &{} 0 \\ a_{21} &{} 0 &{} 0 &{} a_{22} \end{bmatrix} \end{aligned}$$

and the map

$$\begin{aligned} \Phi _1: L_p({\mathcal {M}}) \otimes M_2 \rightarrow L_p({\mathcal {M}}) \otimes M_4, a \mapsto \gamma _1^* a \gamma _1 \end{aligned}$$

is completely positive. Next, we show that the map

$$\begin{aligned} \begin{aligned} \Phi _2:~&L_p({\mathcal {M}}) \otimes M_4 \rightarrow L_1(M_k) \otimes M_4 \\&\begin{bmatrix} x_{11} &{} x_{12} \\ x_{21} &{} x_{22} \end{bmatrix} \mapsto \begin{bmatrix} S(x_{11}) &{} {\tilde{T}}(x_{12}) \\ {\tilde{T}}(x_{21}) &{} S(x_{22}) \end{bmatrix} \text { where } x_{ij} \in L_p({\mathcal {M}}) \otimes M_2 \end{aligned} \end{aligned}$$

is completely positive. So let \(n \in {\mathbb {N}}\) and \(x = [x_{ij}] \in (L_p({\mathcal {M}})\otimes M_4 \otimes M_n )_+\) where \( x_{ij} \in L_p({\mathcal {M}}) \otimes M_4 \) . Then \(S_{2n}(x) \pm {\tilde{T}}_{2n}(x) \ge 0\), and, by Lemma 2.1

$$\begin{aligned} \begin{bmatrix} S_{2n}(x) &{} {\tilde{T}}_{2n}(x) \\ {\tilde{T}}_{2n}(x) &{} S_{2n}(x) \end{bmatrix} \ge 0. \end{aligned}$$

We write \(x\) as

$$\begin{aligned} x = \begin{bmatrix} x_{ij11} &{} x_{ij12} \\ x_{ij21} &{} x_{ij22} \end{bmatrix} \end{aligned}$$

where \(x_{ijlm} \in L_p({\mathcal {M}}) \otimes M_2, i,j \in \{1, \ldots ,n\},~ l,m = 1,2.\) Then

$$\begin{aligned} \begin{bmatrix} S_{2n}(x) &{} {\tilde{T}}_{2n}(x) \\ {\tilde{T}}_{2n}(x) &{} S_{2n}(x) \end{bmatrix} = \begin{bmatrix} S_n(x_{ij11}) &{} S_n(x_{ij12}) &{} {\tilde{T}}_n(x_{ij11}) &{} {\tilde{T}}_n(x_{ij12})\\ S_n(x_{ij21}) &{} S_n(x_{ij22}) &{} {\tilde{T}}_n(x_{ij21}) &{} {\tilde{T}}_n(x_{ij22})\\ {\tilde{T}}_n(x_{ij11}) &{} {\tilde{T}}_n(x_{ij12}) &{} S_n(x_{ij11}) &{} S_n(x_{ij12})\\ {\tilde{T}}_n(x_{ij21}) &{} {\tilde{T}}_n(x_{ij22}) &{} S_n(x_{ij21}) &{} S_n(x_{ij22})\\ \end{bmatrix}. \end{aligned}$$

We multiply this matrix from left by the scalar matrix \(\gamma _2 = \begin{bmatrix} 1 &{} 0 &{} 0 &{} 0 \\ 0 &{} 0 &{} 0 &{} 1 \end{bmatrix}\) and from right the by its adjoined matrix \(\gamma _2^*\), and get

$$\begin{aligned} 0 \le \begin{bmatrix} S_n([x_{ij11}]) &{} {\tilde{T}}_n([x_{ij12}])\\ {\tilde{T}}_n([x_{ij21}]) &{} S_n([x_{ij22}])\\ \end{bmatrix}. \end{aligned}$$
(4.19)

We multiply the matrix in (4.19) from right by the scalar \(2n \times 2n\) matrix \(\gamma _3\) and from left by \(\gamma _3^*\), where \(\gamma _3\) has 1 at position \((l,m)\) when \((l,m) = (2i-1,i)\) or \((l,m) = (2i,n+i)\) for \(i \in \lbrace 1,\ldots ,n \rbrace \) and 0 else. Then the resulting matrix is \(\Phi _{2,n}(x)\). This shows that \(\Phi _2\) is completely positive. Next we define the linear map

$$\begin{aligned} \Phi _3:~ L_1(M_k) \otimes M_4 \rightarrow L_1(M_k) \otimes M_2, b \mapsto \gamma _1 b \gamma _1^* \end{aligned}$$

where \(\gamma _1\) is the \(2 \times 4\) matrix used in the beginning of the proof. Then \(\Phi _3\) is completely positive. Since \( \Phi = \Phi _3 \circ \Phi _2 \circ \Phi _1\), \(\Phi \) is completely positive. This finishes the proof. \(\square \)

For linear maps \(S_1, S_2, T: L_p({\mathcal {M}}) \rightarrow L_1({\mathcal {N}})\) we define

$$\begin{aligned} \begin{aligned} \begin{bmatrix} S_1 &{} T \\ T^* &{} S_2 \end{bmatrix} :~&L_p({\mathcal {M}}) \otimes M_2 \rightarrow L_1({\mathcal {N}}) \otimes M_2 \\&\begin{bmatrix} x_{11} &{} x_{12} \\ x_{21} &{} x_{22} \end{bmatrix} \mapsto \begin{bmatrix} S_1(x_{11}) &{} T(x_{12}) \\ T(x_{21}^*)^* &{} S_2(x_{22}) \end{bmatrix} \end{aligned} \end{aligned}$$
(4.20)

Lemma 4.15

Let \({\mathcal {N}}\) be injective, \(T: L_p({\mathcal {M}}) \rightarrow L_1({\mathcal {N}})\) be completely order bounded, \(m, n_1,\ldots , n_m \in {\mathbb {N}}\), \(x_l \in (L_p({\mathcal {M}}) \otimes M_2 \otimes M_{n_l})_+, y_l \in ({\mathcal {N}} \otimes M_2 \otimes M_{n_l})_+, \text { for } l= 1, \ldots ,m\) and \(\varepsilon > 0\). Then there exist completely positive maps \(S_1, S_2: L_p({\mathcal {M}}) \rightarrow L_1({\mathcal {N}})\) such that \(\Vert S_1\Vert , \Vert S_2\Vert \le \Vert T\Vert _{cob}\) and for the map \(\left[ {\begin{matrix} S_1 &{} T \\ T^* &{} S_2 \end{matrix}} \right] \) defined in (4.20) holds

$$\begin{aligned} \left\langle \begin{bmatrix} S_1 &{} T \\ T^* &{} S_2 \end{bmatrix}_{n_l}(x_l),y_l\right\rangle \ge - \varepsilon , \quad l = 1,\ldots m. \end{aligned}$$

Proof

Let \(x_l = [x_{l,ij}]\), where \(x_{l,ij} \in L_p({\mathcal {M}}) \otimes M_2\), and \(y_l = [y_{l,ij}]\), where \(y_{l,ij} \in {\mathcal {N}} \otimes M_2\), and for \(\ i, j = 1,\ldots , n_l, \ l=1,\ldots ,m\)

$$\begin{aligned} x_{l,ij} = \begin{bmatrix} x_{l,ij11} &{} x_{l,ij12} \\ x_{l,ij21} &{} x_{l,ij22} \end{bmatrix}, y_{l,ij} = \begin{bmatrix} y_{l,ij11} &{} y_{l,ij12} \\ y_{l,ij21} &{} y_{l,ij22} \end{bmatrix}. \end{aligned}$$

Since \({\mathcal {N}}\) is injective, there exist by [2], Theorem IV.2.4.4, \(k \in {\mathbb {N}}\) and completely positive contractions \(\sigma _1: {\mathcal {N}} \rightarrow M_k\) and \(\sigma _2: M_k \rightarrow {\mathcal {N}}\) such that \(\sigma _1\) is continuous in the \(\sigma \)-weak topology and

$$\begin{aligned} \begin{aligned} \Bigg | \sum _{i,j=1}^{n_l}&( \langle T(x_{l,ij21}^*)^*,y_{l,ji12} - \sigma _2 \circ \sigma _1 (y_{l,ji12})\rangle \\&+ \langle T(x_{l,ij12}),y_{l,ji21} - \sigma _2 \circ \sigma _1 (y_{l,ji21}) \rangle ) \Bigg | < \varepsilon , \ l = 1,\ldots ,m. \end{aligned} \end{aligned}$$
(4.21)

Since \(x_l\) and \(y_l\) are positive, we have \(x_{l,ij21}^* = x_{l,ji12}\) and \(y_{l,ij12}^* = y_{l,ji21}\) for all \(l=1,\ldots ,m\). Thus, \(\langle T(x_{l,ij21}^*)^*,y_{l,ji12}\rangle \) and \(\langle T(x_{l,ji12}),y_{l,ij21}\rangle \) are conjugate complex numbers for all \(i,j = 1, \ldots ,n_l, \ l = 1,\ldots ,m\), and the sum on the left side of this inequality is a real number. Hence the sum is greater than \(-\varepsilon \).

Since \(\sigma _1\) is continuous in the \(\sigma \)-weak topology, its adjoint map \(\sigma _1^t\) maps \(L_1(M_k)\) to \(L_1({\mathcal {N}})\) and is a completely positive contraction. Similarly, the adjoint map \(\sigma _2^t\) maps \(L_1({\mathcal {N}})\) to \(L_1(M_k)\) and is a completely positive contraction, and \(\sigma _2^t \circ T\) is completely order bounded with \(\Vert \sigma _2^t \circ T\Vert _{cob} \le \Vert T\Vert _{cob}\). Now, we apply Proposition 4.14 and get linear maps \(S_1' , S_2': L_p({\mathcal {M}}) \rightarrow L_1(M_k)\) such that \(\Vert S_1'\Vert , \Vert S_2'\Vert \le \Vert T\Vert _{cob}\) and \(\left[ {\begin{matrix} S_1' &{} \sigma _2^t \circ T \\ \sigma _2^t \circ T^* &{} S_2' \end{matrix}} \right] \) is completely positive. We put \(S_1 = \sigma _1^t \circ S_1'\) and \(S_2 = \sigma _1^t \circ S_2'\). Then \(\Vert S_1\Vert , \Vert S_2\Vert \le \Vert T\Vert _{cob}\), \(S_1\) and \(S_2\) are completely positive, and for all \(l=1,\ldots ,m\)

$$\begin{aligned} \begin{aligned} \left\langle \begin{bmatrix} S_1 &{} T \\ T^* &{} S_2 \end{bmatrix}_{n_l} (x_l), y_l \right\rangle&= \sum _{i,j=1}^{n_l} \big (\langle S_1(x_{l,ij11}),y_{l,ji11}\rangle \\& \qquad + \langle T(x_{l,ij12}), y_{l,ji21}\rangle \\& \qquad + \langle T(x_{l,ij21}^*)^*, y_{l,ji12}\rangle \\& \qquad + \langle S_2(x_{l,ij22}), y_{l,ij22}\rangle \big ) \\&= \sum _{i,j=1}^{n_l} \big ( \langle \sigma _1^t \circ S_1'(x_{l,ij11}),y_{l,ji11} \rangle \\& \qquad + \langle T(x_{l,ij12}), \sigma _2 \circ \sigma _1(y_{l,ji21})\rangle \\& \qquad + \langle T(x_{l,ij21}^*)^*, \sigma _2 \circ \sigma _1 (y_{l,ji12}) \rangle \\& \qquad + \langle \sigma _1^t \circ S_2'(x_{l,ij22}), y_{l,ij22}\rangle \big ) \\&\quad + \sum _{i,j=1}^{n_l} \big ( \langle T(x_{l,ij21}^*)^*,y_{l,ji12} - \sigma _2 \circ \sigma _1 (y_{l,ji12})\rangle \\& \qquad + \langle T(x_{l,ij12}),y_{l,ji21} - \sigma _2 \circ \sigma _1 (y_{l,ji21}) \rangle \big ). \end{aligned} \end{aligned}$$
(4.22)

Now the first sum of the last expression in (4.22) is equal to

$$\begin{aligned} \left\langle (\sigma _1^t)_{2n_l} \circ \begin{bmatrix} S_1' &{} \sigma _2^t \circ T \\ (\sigma _2^t \circ T)^* &{} S_2' \end{bmatrix}_{n_l} (x_l), y_l \right\rangle , \end{aligned}$$

which is greater than or equal to 0, because it is a composition of two completely positive maps applied to a positive element. The second sum is greater than \(-\varepsilon \) by (4.21). This finishes the proof. \(\square \)

Theorem 4.16

Let \(1 \le p \le \infty , {\mathcal {M}}\) and \({\mathcal {N}}\) be von Neumann algebras, \({\mathcal {N}}\) injective, and \(T: L_p({\mathcal {M}}) \rightarrow L_1({\mathcal {N}})\) be a completely order bounded map. Then there exist linear maps \(S_1, S_2: L_p({\mathcal {M}}) \rightarrow L_1({\mathcal {N}}), \Vert S_1\Vert , \Vert S_2\Vert \le \Vert T\Vert _{cob}\) such that the map

$$\begin{aligned} \begin{aligned} \Phi :&~L_p({\mathcal {M}}) \otimes M_2 \rightarrow L_1({\mathcal {N}}) \otimes M_2 \\&\begin{bmatrix} x_{11} &{} x_{12} \\ x_{21} &{} x_{22} \end{bmatrix} \mapsto \begin{bmatrix} S_1(x_{11}) &{} T(x_{12}) \\ T(x_{21}^*)^* &{} S_2(x_{22}) \end{bmatrix} \end{aligned} \end{aligned}$$

is completely positive.

Proof

Let \({\mathcal {N}}^*\) be the dual space of \({\mathcal {N}}\) and \({\mathcal {B}}(L_p({\mathcal {M}}), {\mathcal {N}}^*)\) be the vector space of all bounded linear maps from \(L_p({\mathcal {M}})\) to \({\mathcal {N}}^*\). By [12], Theorem IV.2.3, \({\mathcal {B}}(L_p({\mathcal {M}}), {\mathcal {N}}^*)\) is isomorphic to the dual space of \(L_p({\mathcal {M}}) \otimes _{\gamma } {\mathcal {N}}\), where \(\Vert \cdot \Vert _{\gamma }\) denotes the projective tensor norm. The weak\(^*\)-topology on \({\mathcal {B}}(L_p({\mathcal {M}}), {\mathcal {N}}^*)\) is given by the seminorms \(|\langle S(x),y\rangle |\), where \(x \in L_p({\mathcal {M}}), y \in {\mathcal {N}},\) and \(S \in {\mathcal {B}}(L_p({\mathcal {M}}), {\mathcal {N}}^*)\). Let \({\mathcal {U}} = \lbrace S \in {\mathcal {B}}(L_p({\mathcal {M}}), {\mathcal {N}}^*) \ | \ \Vert S\Vert \le \Vert T\Vert _{cob} \rbrace \). By the Banach-Alaoglu theorem \({\mathcal {U}}\) is compact in the weak\(^*\)-topology. Hence \({\mathcal {U}} \times {\mathcal {U}}\) is compact in the product weak\(^*\)-topology. For \(\varepsilon > 0, m \in {\mathbb {N}}, n_1,\ldots ,n_m \in {\mathbb {N}}, l = 1, \ldots ,m, X = \lbrace (x_l,y_l) | \ x_l \in (L_p({\mathcal {M}}) \otimes M_2 \otimes M_{n_l})_+, y_l \in ({\mathcal {N}} \otimes M_2 \otimes M_{n_l})_+ \rbrace \), let

$$\begin{aligned} \begin{aligned} K(X,\varepsilon ) = \Bigg \{&(S_1,S_2) \in {\mathcal {U}} \times {\mathcal {U}} \ | \ S_1, S_2 \text { completely positive, } \\&\left\langle \begin{bmatrix} S_1 &{} T \\ T^* &{} S_2 \end{bmatrix}_{n_l} (x_l) , y_l \right\rangle \ge -\varepsilon \text { for all } (x_l,y_l) \in X \Bigg \}. \end{aligned} \end{aligned}$$

We will show that \(K(X,\varepsilon )\) is closed in the weak\(^*\)-topology and hence compact. So let \((S_1,S_2)\) be in the closure of \(K(X,\varepsilon )\). Let \(\delta > 0\) and \((x_l,y_l) \in X\), where

$$\begin{aligned} \begin{aligned}&x_l = [x_{l,ij}], \ x_{l,ij} = \begin{bmatrix} x_{l,ij11} &{} x_{l,ij12} \\ x_{l,ij21} &{} x_{l,ij22} \end{bmatrix} \in L_p({\mathcal {M}}) \otimes M_2, \\&y_l = [y_{l,ij}], \ y_{l,ij} = \begin{bmatrix} y_{l,ij11} &{} y_{l,ij12} \\ y_{l,ij21} &{} y_{l,ij22} \end{bmatrix} \in {\mathcal {N}} \otimes M_2. \end{aligned} \end{aligned}$$

Then we can find \((S_{1\delta },S_{2\delta }) \in K(X,\varepsilon )\) such that

$$\begin{aligned} \begin{aligned}&\sum _{i,j=1}^{n_l}|\langle S_1(x_{l,ij11})-S_{1\delta }(x_{l,ij11}), y_{l,ji11}\rangle |< \delta \\&\sum _{i,j=1}^{n_l}|\langle S_2(x_{l,ij22})-S_{2\delta }(x_{l,ij22}), y_{l,ji22}\rangle | < \delta , \end{aligned} \end{aligned}$$

and therefore

$$\begin{aligned} \begin{aligned}&\left| \left\langle \begin{bmatrix} S_1 &{} T \\ T^* &{} S_2 \end{bmatrix}_{n_l} (x_l) , y_l \right\rangle - \left\langle \begin{bmatrix} S_{1\delta } &{} T \\ T^* &{} S_{2\delta } \end{bmatrix}_{n_l} (x_l) , y_l \right\rangle \right| \\ =&\left| \sum _{i,j=1}^{n_l}\langle S_1(x_{l,ij11}) -S_{1\delta }(x_{l,ij11}), y_{l,ji11}\rangle \right. \\&\left. + \sum _{i,j=1}^{n_l}\langle S_2(x_{l,ij22}) -S_{2\delta }(x_{l,ij22}), y_{l,ji22}\rangle \right| < 2\delta \end{aligned} \end{aligned}$$

Now \(\left\langle \left[ {\begin{matrix} S_{1\delta } &{} T \\ T^* &{} S_{2\delta } \end{matrix}}\right] _{n_l} (x_l) , y_l \right\rangle \) is a positive number. Hence the imaginary part

$$\begin{aligned} \left| Im \left\langle \begin{bmatrix} S_1 &{} T \\ T^* &{} S_2 \end{bmatrix}_{n_l} (x_l) , y_l \right\rangle \right| < 2\delta \end{aligned}$$

and the real part

$$\begin{aligned} Re \left\langle \begin{bmatrix} S_1 &{} T \\ T^* &{} S_2 \end{bmatrix}_{n_l} (x_l) , y_l \right\rangle > \left\langle \begin{bmatrix} S_{1\delta } &{} T \\ T^* &{} S_{2\delta } \end{bmatrix}_{n_l} (x_l) , y_l \right\rangle - 2\delta \ge \varepsilon - 2\delta . \end{aligned}$$

Since \(\delta \) is arbitrary, \(\left\langle \left[ {\begin{matrix} S_1 &{} T \\ T^* &{} S_2 \end{matrix}}\right] _{n_l} (x_l) , y_l \right\rangle \) is a real number which is greater than or equal to \(-\varepsilon \). Similarly, we show that \(\Vert S_1\Vert , \Vert S_2\Vert \le \Vert T\Vert _{cob}\) and \(S_1, S_2\) are completely positive. So \((S_1,S_2) \in K(X,\varepsilon )\).

The sets \(K(X,\varepsilon ), m \in {\mathbb {N}}, X= \{(x_1,y_1),\ldots (x_m,y_m)\}, \varepsilon > 0\) have the finite intersection property. If we have sets \(K(X_l,\varepsilon _l), \ l=1,\ldots ,m\), we put \(X = \bigcup _{l=1}^m X_l\) and \(\varepsilon = \min \lbrace \varepsilon _1,\ldots ,\varepsilon _m \rbrace \). Then \(K(x,\varepsilon ) \subseteq \cap _{l=1}^m K(X_l,\varepsilon _l)\). Combining the finite intersection property and the compactness in the weak*-topology, we get

$$\begin{aligned} \bigcap _{X,\varepsilon } K(X,\varepsilon ) \ne \emptyset . \end{aligned}$$

Any pair \((S_1,S_2)\) in this intersection has the property that \(\left[ {\begin{matrix} S_1 &{} T \\ T^* &{} S_2 \end{matrix}}\right] \) is completely positive and \(\Vert S_1\Vert , \Vert S_2\Vert \le \Vert T\Vert _{cob}\). Since there is a projection \(z \in {\mathcal {N}}^{**}\) which works as a projection from \({\mathcal {N}}^*\) to \(L_1({\mathcal {N}})\), the maps \(S_1' = zS_1\) and \(S_2' = zS_2\) have the desired properties. \(\square \)

We close this section with an example of a completely order bounded map which is not decomposable when \(q > 2p\). The boundaries for \(p\) and \(q\) are not sharp. The goal is to show that there are many combinations of \(p\) and \(q\) for which completely order bounded does not imply decomposable. Since we are working with semifinite von Neumann algebras in the example, we use the \(L_p\)-spaces associated with the trace, because they are easier to handle than the Haagerup-Terp construction. For \(m \in {\mathbb {N}}\), let \(l_q^m\) be the vector space \({\mathbb {C}}^m\) equipped with norm \(\Vert c\Vert _q = \left( \sum _{i=1}^{m}|c_i|^q\right) ^{1/q}\) where \(c = (c_1,\ldots ,c_m)\). Let

$$\begin{aligned} e_1,\ldots ,e_m \text { be the standard base of } l_q^m. \end{aligned}$$
(4.23)

Before we show our example, we need some formulas to estimate the completely order bounded norm and the decomposable norm for linear maps from \(L_p({\mathcal {M}})\) to \(l_q^m\).

Lemma 4.17

Let \(1 \le p< \infty , 1 \le q < \infty , \frac{1}{p} + \frac{1}{p'} = 1, m \in {\mathbb {N}}, g_1,\ldots ,g_m \in L_{p'}({\mathcal {M}})\) and

$$\begin{aligned} T: L_p({\mathcal {M}}) \rightarrow l_q^m, \quad f \mapsto \sum _{k=1}^{m}\langle f,g_k\rangle e_k. \end{aligned}$$

Then

$$\begin{aligned} \Vert T\Vert _{cob} \le \sup \left\{ \left( \sum _{k=1}^{m} \Vert ag_kb\Vert _1^q \right) ^{1/q} \mid a,b \in L_{2p}({\mathcal {M}}), \Vert a\Vert _{2p}, \Vert b\Vert _{2p} \le 1 \right\} . \end{aligned}$$
(4.24)

Proof

Let \(n \in {\mathbb {N}} \) and \( x \in L_p({\mathcal {M}}) \otimes M_n, \Vert x\Vert _{p,n} \le 1\). By Theorem 3.4 and Theorem 2.3, there exist \(a, b \in L_{2p}({\mathcal {M}}), \Vert a\Vert _{2p}, \Vert b\Vert _{2p} \le 1, y \in {\mathcal {M}}\otimes M_n, \Vert y\Vert _{\infty }\le 1\) such that \(x = (b\otimes 1_n)y(a\otimes 1_n)\). Let \(\varepsilon _{ij}\) be as in (3.2) and \(y = \sum _{i,j=1}^{n} y_{ij} \otimes \varepsilon _{ij}.\) Then we have

$$\begin{aligned} T_n(x) = \sum _{k=1}^{m}\sum _{i,j=1}^{n}\langle ag_kb,y_{ij}\rangle e_k \otimes \varepsilon _{ij}. \end{aligned}$$

For \(k = 1, \ldots ,m\) let \(\varphi _k: {\mathcal {M}} \rightarrow {\mathbb {C}}, f \mapsto \langle ag_kb,f\rangle .\) Then \(\varphi _k \) is completely order bounded and \(\Vert \varphi _k\Vert _{cob} = \Vert \varphi _k\Vert = \Vert ag_kb\Vert _1 \). Since \(\Vert y\Vert _{\infty } \le 1\), we have

$$\begin{aligned} 0 \le \begin{bmatrix} \Vert \varphi _k\Vert 1_n &{} (\varphi _k)_n(y) \\ (\varphi _k)_n(y)^* &{} \Vert \varphi _k\Vert 1_n \end{bmatrix}. \end{aligned}$$

Then we get

$$\begin{aligned} \begin{aligned} 0&\le \sum _{k=1}^{m} e_k \otimes \begin{bmatrix} \Vert \varphi _k\Vert 1_n &{} (\varphi _k)_n(y) \\ (\varphi _k)_n(y)^* &{} \Vert \varphi _k\Vert 1_n \end{bmatrix} \\&= \begin{bmatrix} \sum _{k=1}^{m}\Vert \varphi _k\Vert e_k \otimes 1_n &{} T_n(x) \\ T_n(x)^* &{} \sum _{k=1}^{m}\Vert \varphi _k\Vert e_k \otimes 1_n \end{bmatrix} \end{aligned} \end{aligned}$$

Hence

$$\begin{aligned} \Vert T_n(x)\Vert _{q,n} \le \Vert \sum _{k=1}^{m}\Vert ag_kb\Vert _1e_k\Vert _q = (\sum _{k=1}^{m} \Vert ag_kb\Vert _1^q)^\frac{1}{q} \end{aligned}$$

which proves (4.24). \(\square \)

Lemma 4.18

Let \(2 \le q < \infty \), \(\alpha \in M_n\), and \(\varepsilon _{ii}\) be as in (3.2) for \(i=1,\ldots ,n\). Then

$$\begin{aligned} \sum _{i=1}^{n} \Vert \alpha \varepsilon _{ii}\Vert _q^q \le \Vert \alpha \Vert _q^q. \end{aligned}$$

Proof

Let \(\alpha ^*\alpha = \beta = [\beta _{ij}]_{i,j=1}^n \). Let \( |\alpha \varepsilon _{ii}|\) be the positive matrix of the polar decomposition of \(\alpha \varepsilon _{ii} \). Then \(|\alpha \varepsilon _{ii}|^2 = \varepsilon _{ii} \beta \varepsilon _{ii} = \beta _{ii}\varepsilon _{ii} \) which implies \(|\alpha \varepsilon _{ii}|^q = \beta _{ii}^{q/2}\varepsilon _{ii} \) and

$$\begin{aligned} \sum _{i=1}^{n} \Vert \alpha \varepsilon _{ii}\Vert _q^q = \sum _{i=1}^{n} Tr(\beta _{ii}^{q/2}\varepsilon _{ii}) = \sum _{i=1}^{n} \beta _{ii}^{q/2}. \end{aligned}$$

Since \(\beta \) is positive, it has eigenvalues \(\lambda _1, \ldots , \lambda _n \ge 0\), and there is a unitary matrix \(u=[u_{ij}]\in M_n\) such that \(\beta = u^*diag(\lambda _1,\ldots ,\lambda _n)u \). Especially, we have \(\beta _{ii} = \sum _{l=1}^{n}|u_{li}|^2\lambda _l\). From \(u\) being unitary, it follows that \(\sum _{i=1}^{n}|u_{li}|^2 = 1 \) for \(l=1,\ldots ,n\). Since \(q \ge 2\), the function \(\phi :[0,\infty )\rightarrow {\mathbb {R}}, t\mapsto t^{q/2}\) is convex. Hence we get

$$\begin{aligned} \begin{aligned} \sum _{i=1}^{n} \beta _{ii}^{q/2}&= \sum _{i=1}^{n}\phi \Big (\sum _{l=1}^{n}\lambda _l|u_{li}|^2\Big ) \le \sum _{i=1}^{n} \sum _{l=1}^{n}|u_{li}|^2\phi (\lambda _l) \\&= \sum _{l=1}^{n} \lambda _l^{q/2} \sum _{i=1}^{n}|u_{li}|^2 = \sum _{l=1}^{n} \lambda _l^{q/2} = Tr(\beta ^{q/2}) = \Vert \alpha \Vert _q^q. \end{aligned} \end{aligned}$$

\(\square \)

For \(n \in {\mathbb {N}}\) let \(\varepsilon _{ij} \) be as in (3.2) and \(e_i\) as in (4.23). Then we define the linear map

$$\begin{aligned} T: L_p(M_n) \rightarrow l_n^q, \quad T(x) = \sum _{i=1}^{n}Tr((\varepsilon _{1i}+\varepsilon _{i1})x)e_i. \end{aligned}$$
(4.25)

Proposition 4.19

Let \(1 \le p< \infty , 2p< q < \infty \), and \(T\) as in (4.25). Then \(\Vert T\Vert _{cob} \le 2\).

Proof

Let \(\varepsilon > 0 \). By Lemma 4.17, there exist \(a,b \in L_{2p}(M_n), \Vert a\Vert _{2p}, \Vert b\Vert _{2p} \le 1\) such that

$$\begin{aligned} \Vert T\Vert _{cob} \le \Bigg (\sum _{i=1}^{n}\Vert a(\varepsilon _{i1} + \varepsilon _{1i})b\Vert _1^q\Bigg )^{1/q} + \varepsilon . \end{aligned}$$

Since for all \(i=1,\ldots ,n\) we have \(\Vert a(\varepsilon _{i1} + \varepsilon _{1i})b\Vert _1 \le \Vert a\varepsilon _{i1}b\Vert _1 + \Vert a\varepsilon _{1i}b\Vert _1\) we get

$$\begin{aligned} \Bigg (\sum _{i=1}^{n}\Vert a(\varepsilon _{i1} + \varepsilon _{1i})b\Vert _1^q\Bigg )^{1/q} \le \Bigg (\sum _{i=1}^{n}\Vert a\varepsilon _{i1}b\Vert _1^q\Bigg )^{1/q} + \Bigg (\sum _{i=1}^{n}\Vert a\varepsilon _{1i}b\Vert _1^q\Bigg )^{1/q} \end{aligned}$$

Let \(\frac{1}{q}+\frac{1}{q'} = 1\). Then \(\Vert a\varepsilon _{i1}b\Vert _1 = \Vert a\varepsilon _{ii} \varepsilon _{i1}b\Vert _1 \le \Vert a\varepsilon _{ii}\Vert _q \cdot \Vert \varepsilon _{i1}b\Vert _{q'}\). Since \(q > 2p \ge 2\) we have \(q' < 2 \le 2p\). Hence there exists a real number \(s > 1\) such that \(\frac{1}{q'} = \frac{1}{2p} + \frac{1}{s}\). By the generalized Hölder’s inequality, we have\(\Vert \varepsilon _{i1}b\Vert _{q'} \le \Vert \varepsilon _{i1}\Vert _s\Vert b\Vert _{2p} \le 1\). We apply Lemma 4.18 and get

$$\begin{aligned} \sum _{i=1}^{n}\Vert a\varepsilon _{i1}b\Vert _1^q \le \sum _{i=1}^{n}\Vert a\varepsilon _{ii}\Vert _1^q \le \Vert a\Vert _q^q. \end{aligned}$$

Since \(q > 2p \ge 2\), we have \(\Vert a\Vert _q \le \Vert a\Vert _{2p} \le 1\). Hence

$$\begin{aligned} \Bigg (\sum _{i=1}^{n}\Vert a\varepsilon _{i1}b\Vert _1^q\Bigg )^{1/q} \le 1. \end{aligned}$$

Similarly, we get

$$\begin{aligned} \Bigg (\sum _{i=1}^{n}\Vert a\varepsilon _{1i}b\Vert _1^q\Bigg )^{1/q} \le 1. \end{aligned}$$

Since \(\varepsilon \) was arbitrary, this finishes the proof. \(\square \)

Proposition 4.20

Let \(1 \le p< \infty ,~ 1 \le q < \infty \), and \(T\) as in (4.25). Then \(\Vert T\Vert _{dec} \ge n^{1/2q}\).

Proof

Since \(T\) is self-adjoint, we can apply [1], Lemma 2.18 and 2.19, and get

$$\begin{aligned} \Vert T\Vert _{dec} = \inf \{ \Vert S\Vert \mid S: L_p(M_n) \rightarrow l_q^n, S \pm T \text { is completely positive} \}. \end{aligned}$$

In [1], the case \(p = q\) is considered only. But the proof also works for \( p \ne q\). Let \(S:L_p(M_n) \rightarrow l_q^n\) be a linear map such that \(S \pm T\) is completely positive. There exist \(b_1,\ldots , b_n \in M_{n+} \) such that \(S(f) = \sum _{k=1}^{n}\langle b_k,f\rangle e_k\) for all \(f \in L_p(M_n)\). Since \(S \pm T\) are positive, we have

$$\begin{aligned} \sum _{k=1}^{n}\langle b_k \pm (\varepsilon _{1k}+ \varepsilon _{k1}),f\rangle e_k \ge 0 \text { for all } f\in L_p(M_n)_+. \end{aligned}$$

This means that \(b_k \pm (\varepsilon _{1k}+ \varepsilon _{k1}) \ge 0\) for \(k=1,\ldots ,n\). Let \(b_k = [b_{k,ij}]\). For \(k=1\) we have \(b_1-2\varepsilon _{11} \ge 0\) which implies \(b_{1,11} \ge 2\). For \(k > 1 \) we have

$$\begin{aligned} \begin{aligned} 0&\le \begin{bmatrix}e_k^t &{} 0 \\ 0 &{} e_1^t\end{bmatrix} \begin{bmatrix} b_k &{} \varepsilon _{1k} + \varepsilon _{k1} \\ \varepsilon _{1k} + \varepsilon _{k1} &{} b_k \end{bmatrix} \begin{bmatrix}e_k &{} 0 \\ 0 &{} e_1\end{bmatrix} \\&= \begin{bmatrix} e_k^tb_ke_k &{} e_k^t\varepsilon _{k1}e_1 \\ e_1^t\varepsilon _{1k}e_k &{} e_1^tb_ke_1 \end{bmatrix} = \begin{bmatrix} b_{k,kk} &{} 1 \\ 1 &{} b_{k,11} \end{bmatrix} \end{aligned} \end{aligned}$$
(4.26)

We compute the determinant of the last matrix in (4.26) and get \(b_{k,kk} b_{k,11} \ge 1\). Since the diagonal elements of \(b_k\) are positive, we get \(b_{k,kk} \ge \frac{1}{b_{k,11}}\) for all \(k \ge 2\). For \(k=1,\ldots ,n\) we have

$$\begin{aligned} \Vert S(\varepsilon _{kk})\Vert _q^q = \Big \Vert \sum _{j=1}^{n}\langle b_j,\varepsilon _{kk}\rangle e_j\Big \Vert _q^q = \Big \Vert \sum _{j=1}^{n}b_{j,kk} e_j\Big \Vert _q^q = \sum _{j=1}^{n}(b_{j,kk})^q. \end{aligned}$$

For \(k = 1\) we have \(\Vert S(\varepsilon _{11})\Vert _q^q = \sum _{j=1}^{n}(b_{j,11})^q\). For \(k > 1\) we have \(\Vert S(\varepsilon _{kk})\Vert _q^q \ge (b_{k,kk})^q \ge \frac{1}{(b_{k,11})^q}\). Hence

$$\begin{aligned} \Vert S\Vert ^q \ge \max \left\{ \sum _{j=1}^{n} (b_{j,11})^q, \frac{1}{(b_{k,11})^q}, k=2,\ldots ,n\right\} . \end{aligned}$$

If there is some \(k, 2 \le k \le n\) such that \(b_{k,11} \le n^{-1/2q}\), then

$$\begin{aligned} \Vert S\Vert ^q \ge \frac{1}{(b_{k,11})^q} \ge \sqrt{n}. \end{aligned}$$
(4.27)

If \(b_{k,11} \ge n^{-1/2q}\) for all \(k \ge 2\), we have

$$\begin{aligned} \begin{aligned} \Vert S\Vert ^q&\ge \sum _{k=1}^{n}(b_{k,11})^q \ge 2^q + (n-1)n^{-1/2} \\&= \frac{2^q\sqrt{n} + n -1}{\sqrt{n}} \\&\ge \sqrt{n} \end{aligned} \end{aligned}$$
(4.28)

Combining (4.27) and (4.28), we get

$$\begin{aligned} \Vert S\Vert \ge n^{\frac{1}{2q}}. \end{aligned}$$

Since \(S\) was arbitrary with \(S\pm T\) completely positive, this finishes the proof. \(\square \)

Now we can show our counterexample. Let \({\mathcal {M}} = \oplus _{k=1}^{\infty }M_k\). On \({\mathcal {M}}\) we have the semifinite, normal, faithful trace \(\tau _1(x) = \sum _{k=1}^{\infty }Tr_k(x_k)\) where \(x = \oplus _{k=1}^{\infty } x_k\) and \(Tr_k\) is the usual trace on \(M_k\). For any projection \(e \in {\mathcal {M}},\) we have \(\tau _1(e)\ge 1\) and therefore every \(\tau _1\)-measurable operator affiliated with \({\mathcal {M}}\) is a bounded operator. Hence for all \(1 \le p < \infty ,\) we can write \(L_p({\mathcal {M}})= \lbrace x = \oplus _{k=1}^\infty x_k \mid \sum _{k=1}^{\infty }\Vert x_k\Vert _p^p < \infty \rbrace \) with norm \(\Vert x\Vert _p = (\sum _{k=1}^{\infty }\Vert x_k\Vert _p^p)^{1/p}\). Let \({\mathcal {N}} = \oplus _{k=1}^\infty l_{\infty }^k\). Then \(L_q({\mathcal {N}}) = \lbrace f = \oplus _{k=1}^\infty f_k \mid \sum _{k=1}^{\infty } \Vert f_k\Vert _q^q < \infty \rbrace \) with norm \(\Vert f\Vert _q = (\sum _{k=1}^{\infty }\Vert f_k\Vert _q^q)^{1/q}\). For \(1 \le p,q < \infty ,~ q > 2p \) let

$$\begin{aligned} T: L_p({\mathcal {M}}) \rightarrow L_q({\mathcal {N}}), \quad T(\oplus _{k=1}^{\infty }x_k) = \oplus _{k=1}^{\infty } T_k(x_k) \end{aligned}$$

where \(T_k\) is defined as in (4.25). We claim that \(T\) is completely order bounded and \(\Vert T\Vert _{cob} \le 2\). To show this, let \(n \in {\mathbb {N}}\) and \(x \in L_p({\mathcal {M}} \otimes M_n), \Vert x\Vert _{p,n} \le 1 \). According to Theorem 3.4, there exist \(f, g \in L_P({\mathcal {M}})_+\) such that

$$\begin{aligned} \Vert f\Vert _p = \Vert g\Vert _p = \Vert x\Vert _{p,n} \text { and } \begin{bmatrix}f \otimes 1_n &{} x \\ x^* &{} g \otimes 1_n\end{bmatrix} \ge 0. \end{aligned}$$

Then we have \(x = \oplus _{k=1}^{\infty }x_k, x_k \in L_p(M_k) \otimes M_n, f = \oplus _{k=1}^{\infty } f_k, f_k \in L_p(M_k)_+\) and \(g = \oplus _{k=1}^{\infty } g_k, g_k \in L_p(M_k)_+\) with

$$\begin{aligned} \begin{bmatrix}f_k \otimes 1_n &{} x_k \\ x_k^* &{} g_k \otimes 1_n \end{bmatrix} \ge 0 \text { for all } k \in {\mathbb {N}}. \end{aligned}$$

By Proposition 4.19, we have \(\Vert T_k\Vert _{cob}\le 2\) for all \(k \in {\mathbb {N}}\). By definition of completely order boundedness, for all \(k \in {\mathbb {N}}\), there exist \(h_{1,k}, h_{2,k} \in (l_k^q)_+,\) such that

$$\begin{aligned} \begin{bmatrix}h_{1,k} \otimes 1_n &{} T_{k,n}(x_k) \\ T_{k,n}(x_k)^* &{} h_{2,k} \otimes 1_n\end{bmatrix} \ge 0 \text { and } \Vert h_{1,k}\Vert _q, \Vert h_{2,k}\Vert _q \le \frac{1}{2}\Vert T_k\Vert _{cob}(\Vert f_k\Vert _p + \Vert g_k\Vert _p). \end{aligned}$$

We put \(h_1 = \oplus _{k=1}^{\infty }h_{1,k}\) and \(h_2 = \oplus _{k=1}^{\infty }h_{2,k}\). Then \(\Vert h_1\Vert _q, \Vert h_2\Vert _q \le 2\) and

$$\begin{aligned} \begin{bmatrix}h_1 \otimes 1_n &{} T_n(x) \\ T_n(x)^* &{} h_2 \otimes 1_n\end{bmatrix} \ge 0 \end{aligned}$$

which shows that \(\Vert T\Vert _{cob}\le 2\).

Next, suppose that \(T\) is decomposable. Then there exists a completely positive map \(S: L_p({\mathcal {M}}) \rightarrow L_q({\mathcal {N}}),\) such that \(S \pm T\) are completely positive. For every \(j \in {\mathbb {N}}\), the embedding \(I_j: L_p(M_j) \rightarrow L_p({\mathcal {M}}), x \mapsto (\ldots ,0,x,0,\ldots )\) and the projection \(P_j:L_q({\mathcal {N}}) \rightarrow l_q^j,~ \oplus _{k=1}^{\infty } y_k \mapsto y_j\) are completely positive and have norm less than or equal to 1. We have \(T_j = P_j \circ T \circ I_j \) and put \(S_j = P_j \circ S \circ I_j\). Then \(S_j \pm T_j\) are completely positive. We apply Proposition 4.20 and get

$$\begin{aligned} \infty > \Vert S\Vert \ge \Vert S_j\Vert \ge \Vert T_j\Vert _{dec} \ge j^{1/2q} \text { for all } j \in {\mathbb {N}} \end{aligned}$$

which gives a contradiction. Thus \(T\) is not decomposable.

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Acknowledgements

The author thanks the anonymous reviewers for pointing out simplifications of some proofs and typing errors.

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Communicated by Uwe Franz.

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Neuhardt, E. Completely order bounded maps on non-commutative \({\varvec{L_p}}\)-spaces. Adv. Oper. Theory 6, 52 (2021). https://doi.org/10.1007/s43036-021-00145-2

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Keywords

  • Non-commutative \(L_p\)-space
  • Matrix norm
  • Completely positive map
  • Decomposable map

Mathematics Subject Classification

  • 46L07
  • 47L25