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.
Similar content being viewed by others
1 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 \(q\).
2 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
and
For \(s \in {\mathbb {R}}\) let \(W(s)\) be the unitary operator on \( L_2({\mathbb {R}},{\mathcal {H}})\) which is defined by
The dual action \({\theta} \) is then defined by
The elements of \({\mathcal {M}}\) are the fixed points under \({\theta} \) when \( {\mathcal {M}}\) is identified with \({\pi ({\mathcal {M}})}\):
The crossed product \({\mathcal {M}} \rtimes _{\sigma ^{\varphi }} {\mathbb {R}}\) has a unique normal faithful semifinite trace \({\tau} \) which satisfies
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
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:
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
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
Especially, when \( \frac{1}{p} + \frac{1}{q} = 1\) we get for \( a \in L_p({\mathcal {M}}) \) and \( b \in L_{q}({\mathcal {M}}) \)
Moreover, the space \( L_{q}({\mathcal {M}}) \) is isometrically isomorphic to the dual space of \( L_p({\mathcal {M}}) \). We denote this duality by
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
By (2.2),
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:
-
(i)
\({ -a \le b \le a} \).
-
(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
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 \({ n \in {\mathbb {N}}} \)
Therefore, for \({ \xi , \eta \in p_n{\mathcal {H}}} \), we have
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:
-
(i)
The operators \({a}\) and \({b}\) are positive.
-
(ii)
\({ \text {supp}(a)\cdot c_1 \cdot \text {supp}(b) = c_1} \) and \( {\text {supp}(a)\cdot c_2 \cdot \text {supp}(b) = c_2} \).
-
(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
and
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
Then there is an operator \( {y \in {\mathcal {M}} \otimes M_n} \) such that
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
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
This implies
We replace \( {\eta} \) in (2.6) by \( {e^{\text {i}t}\eta} \), and choose a suitable value for \( {t \in {\mathbb {R}}} \) to get
Then we replace \( {\xi }\) by \({ \lambda \xi} \), \({ \eta }\) by \( {\frac{1}{\lambda }\eta} \) in (2.7), minimize over \({ \lambda \in {\mathbb {R}}_+} \), and get
So, we can define the sesquilinear form
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
This shows that \( {B} \) is well defined, and by (2.8), we get
Thus, \({ B} \) can be extended to a bounded sesquilinear form
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
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 \( {i \in {\mathbb {N}} }\),
and
Since \({ u }\) commutes with \({ f} \), \({ g }\), supp\({(f)}\), and supp\({(g)}\), we get for all \({ i \in {\mathbb {N}}}\)
and therefore
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
Since
and
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
and
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
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
Then
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
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
\({\square }\)
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. \(\square \)
3 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
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
Theorem 3.3
Let \( 1 \le p \le \infty \text { and } n \in {\mathbb {N}} \). Then the following holds:
-
(i)
If \( {x, y \in L_p({\mathcal {M}}) \otimes M_n}\), then \({ \Vert x+y\Vert _{p,n} \le \Vert x\Vert _{p,n} + \Vert y\Vert _{p,n}} \).
-
(ii)
If \({ x \in L_p({\mathcal {M}}) \otimes M_n}\), then \({ \Vert x^*\Vert _{p,n} = \Vert x\Vert _{p,n}}\).
-
(iii)
If \({ x \in L_p({\mathcal {M}}) \otimes M_n}\) and \({\alpha \in M_n }\), then
$$\begin{aligned} \Vert \alpha x\Vert _{p,n} \le \Vert \alpha \Vert \Vert x\Vert _{p.n} \text { and } \Vert x \alpha \Vert _{p,n} \le \Vert \alpha \Vert \Vert x\Vert _{p,n}. \end{aligned}$$ -
(iv)
If \( {x \in L_p({\mathcal {M}}) \otimes M_n}\) and \({ \lambda \in {\mathbb {C}}} \), then \({ \Vert \lambda x \Vert _{p,n} = |\lambda | \Vert x\Vert _{p,n} }\).
-
(v)
If \({ x = [x_{ij}] \in L_p({\mathcal {M}}) \otimes M_n}\), then
$$\begin{aligned} \max \left\{ \Vert x_{ij}\Vert _{p} \ \big {|} \ 1 \le i, j \le n \right\} \le \Vert x\Vert _{p,n} \le \sum _{i,j=1}^{n} \Vert x_{ij}\Vert _p. \end{aligned}$$ -
(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
and
Then we get
and
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
Then we get
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
Then we have for \({ \lambda > 0 }\)
We put \({ \lambda ^2 = \Vert \alpha \Vert }\) and get
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
For \({ \lambda \ne 0 }\), we put \({ \alpha = \lambda 1_n }\), apply iii), and get
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
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
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
Now it follows that
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 \( {x = [x_{ij}] \in L_p({\mathcal {M}}) \otimes M_n}\)
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
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
For \({ \lambda > 0} \), we get
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
and
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
and
and \({(f_m,g_m) \in K_\varepsilon} \) for all \({m \in {\mathbb {N}}.}\) Hence, we get
Since this holds for every \( {a \in (L_q({\mathcal {M}}) \otimes M_{2n})_+} \), we can apply (2.3), and conclude that
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
Hence \({ |\langle f,a\rangle | \le \Vert x\Vert _{p,n} + \varepsilon} \) and
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
Combining the finite intersection property and the weak-\({ ^* }\) compactness, we get
Let \({ (f,g)} \) be in the set defined in Eq. (3.4). Then we have
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
\({\square} \)
Theorem 3.5
Let \({ 1 \le p \le \infty }\), \( {n \in {\mathbb {N}}} \), and \({ x = x^* \in L_p({\mathcal {M}}) \otimes M_n} \).
-
(i)
We have \({ \Vert x\Vert _{p,n} = \inf \left\{ \Vert f\Vert _p | f \in L_p({\mathcal {M}})_+, f \otimes 1_n \pm x \ge 0 \right\}} \).
-
(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
It follows then
Hence we conclude
Then Lemma 2.1 implies that \({ \frac{1}{2}(f+g) \otimes 1_n \pm x \ge 0} \). This shows that
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
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
and
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
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,
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
\({\square} \)
4 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
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
is completely positive. The decomposable norm \({ \Vert T\Vert _{dec} }\) is defined by
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
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
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
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
Since \({ T }\) is completely positive, we have \({ T_n(x^*) = T_n(x)^*} \) and
Hence,
\({\square}\)
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
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
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
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
\({\square} \)
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 \({n \in {\mathbb {N}}, x \in L_p({\mathcal {M}}) \otimes M_n}\)
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
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
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
By Theorem 3.4, there exist \( {g_1, g_2 \in L_q({\mathcal {N}})_+ }\) such that
Hence, \({ (g_1, g_2) \in K(x_i) \text { for all } i \in \{1, \ldots , k\}}\). We conclude that
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
and
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
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
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
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
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
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
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
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
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
\(\square \)
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
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
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
We define the linear functional
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
Then
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
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
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
By Proposition 2.1, we have
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
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
We apply \(\omega \) to (4.9) and get
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
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
\(\square \)
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
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
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
Then we get
This implies
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
and hence
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
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
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
Then we put \(x = -c \otimes y^t, a = c \otimes y^t, b = 0, f = g = 0\), apply (4.11), and get
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
We combine (4.13), (4.14), (4.15), and get
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
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
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
Similarly, we put \(x = \sum _{i,j=1}^{n} x_{ij} \otimes y_{ji}^t,a = 0, b = -x, f = g = 0\) and get
The combination of (4.16), (4.17), and (4.18) gives
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
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
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 \(c \in L_p({\mathcal {M}})_+, y \in M_{k+}\)
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
and
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
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
and the map
is completely positive. Next, we show that the map
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
We write \(x\) as
where \(x_{ijlm} \in L_p({\mathcal {M}}) \otimes M_2, i,j \in \{1, \ldots ,n\},~ l,m = 1,2.\) Then
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
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
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
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
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 \(\ i, j = 1,\ldots , n_l, \ l=1,\ldots ,m\)
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
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 \(l=1,\ldots ,m\)
Now the first sum of the last expression in (4.22) is equal to
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
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
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
Then we can find \((S_{1\delta },S_{2\delta }) \in K(X,\varepsilon )\) such that
and therefore
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
and the real part
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
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
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
Then
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
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
Then we get
Hence
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
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
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
\(\square \)
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
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
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
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
Since \(q > 2p \ge 2\), we have \(\Vert a\Vert _q \le \Vert a\Vert _{2p} \le 1\). Hence
Similarly, we get
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
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
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
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
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
If there is some \(k, 2 \le k \le n\) such that \(b_{k,11} \le n^{-1/2q}\), then
If \(b_{k,11} \ge n^{-1/2q}\) for all \(k \ge 2\), we have
Combining (4.27) and (4.28), we get
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
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
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
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
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
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
which gives a contradiction. Thus \(T\) is not decomposable.
References
Arhancet, C., Kriegler, C.: Projections, multipliers and decomposable maps on noncommutative \( L^p \)-spaces, arXiv:1707.05591v14 (2018)
Blackadar, B.: Operator Algebras: Theory of C*-Algebras and von Neumann Algebras. Springer, New York (2006)
Effros, E., Ruan, Z.-J.: Operator Spaces. Oxford University Press, Oxford (2000)
Haagerup, U.: Decomposition of completely bounded maps on operator algebras, Unpublished manuscript (1980)
Haagerup, U.: Injectivity and decomposition of completely bounded maps, pp. 170-222, In: Operator Algebras and their connection with Topology and Ergodic Theory. Lecture Notes in Mathematics 1132, Springer (1985)
Junge, M., Ruan, Z.-J.: Decomposable maps on non-commutative \( L_p \)-spaces, 355-381, In: Operator algebras, quantization, and noncommutative geometry. Contemporary Mathematics 365, Journal of the American Mathematical Society, Providence, RI (2004)
Junge, M., Ruan, Z.-J., Xu, Q.: Rigid \(\cal{OL}_p\) structures of non-commutative \(L_p\)-spaces associated with hyperfinite von Neumann algebras. Math. Scand. 95, 63–95 (2005)
Paulsen, V.: Completely bounded maps On \( C^* \)-algebras and invariant operator ranges. Proc. Am. Math. Soc. 86, 91–96 (1982)
Pisier, G.: Regular operators between non-commutative \( L_p \)-spaces. Bull. Sci. Math. 119, 95–118 (1995)
Pisier, G.: Introduction to operator space theroy. Cambridge University Press, Cambridge (2003)
Schmitt, L.M.: The Radon–Nikodym theorem for \( L^p \)-spaces of \(W^*\)-algebras. Publ. RIMS Kyoto Univ. 22, 1025–1034 (1986)
Takesaki, M.: Theory of operator algebras I and II. In: Encyclopaedia of Mathematical Sciences, vol. 124 and 125. Springer (2002 and 2003)
Terp, M.: \( L^p \) spaces associated with von Neumann algebras. Copenhagen Univercity, Copenhagen (1981)
Wittstock, G.: Ein operatorwertiger Hahn–Banach Satz. J. Funct. Anal. 40, 127–150 (1981)
Acknowledgements
The author thanks the anonymous reviewers for pointing out simplifications of some proofs and typing errors.
Funding
Open Access funding enabled and organized by Projekt DEAL.
Author information
Authors and Affiliations
Corresponding author
Additional information
Communicated by Uwe Franz.
Rights and permissions
Open Access This article is licensed under a Creative Commons Attribution 4.0 International License, which permits use, sharing, adaptation, distribution and reproduction in any medium or format, as long as you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons licence, and indicate if changes were made. The images or other third party material in this article are included in the article's Creative Commons licence, unless indicated otherwise in a credit line to the material. If material is not included in the article's Creative Commons licence and your intended use is not permitted by statutory regulation or exceeds the permitted use, you will need to obtain permission directly from the copyright holder. To view a copy of this licence, visit http://creativecommons.org/licenses/by/4.0/.
About this article
Cite this article
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
Received:
Accepted:
Published:
DOI: https://doi.org/10.1007/s43036-021-00145-2