# Operator-valued Schatten spaces and Quantum entropies

Salman Beigi & Milad M. Goodarzi  School of Mathematics, Institute for
Research in Fundamental Sciences (IPM), P.O. Box 19395-5746, Tehran, Iran
<EMAIL_ADDRESS>School of Mathematics, Institute for Research in
Fundamental Sciences (IPM), P.O. Box 19395-5746, Tehran, Iran
<EMAIL_ADDRESS>
###### Abstract.

Operator-valued Schatten spaces were introduced by G. Pisier as a
noncommutative counterpart of vector-valued $\ell_{p}$-spaces. This family of
operator spaces forms an interpolation scale which makes it a powerful and
convenient tool in a variety of applications. In particular, as the norms
coming from this family naturally appear in the definition of certain entropic
quantities in Quantum Information Theory (QIT), one may apply Pisier’s theory
to establish some features of those quantities. Nevertheless, it could be
quite challenging to follow the proofs of the main results of this theory from
the existing literature. In this article, we attempt to fill this gap by
presenting the underlying concepts and ideas of Pisier’s theory in an almost
self-contained way which we hope to be more accessible, especially for the QIT
community at large. Furthermore, we describe some applications of this theory
in QIT. In particular, we prove a new uniform continuity bound for the quantum
conditional Rényi entropy.

###### Contents

  1. 1 Introduction
  2. 2 Interpolation theory
    1. 2.1 Calderón’s complex interpolation method
    2. 2.2 A subharmonicity lemma
  3. 3 Operator spaces
    1. 3.1 Dual operator space
    2. 3.2 Interpolation of operator spaces
    3. 3.3 Schatten class operator spaces
    4. 3.4 Column and row operator spaces
  4. 4 Haagerup tensor product
    1. 4.1 Haagerup tensor product of column and row operator spaces
    2. 4.2 Interpolation and the Haagerup tensor product
  5. 5 Operator-valued Schatten spaces
    1. 5.1 Pisier’s variational formula
    2. 5.2 Reiteration and duality
    3. 5.3 Pisier’s formula for the space $\mathcal{S}_{p}[\mathcal{S}_{q}]$
  6. 6 Applications in quantum information theory
    1. 6.1 Quantum conditional Rényi entropy
    2. 6.2 Rényi coherent information
    3. 6.3 A correlation measure
  7. A Kuratowski–Ryll-Nardzewski selection theorem
  8. B Wiener–Masani factorization theorem
    1. B.1 Hardy spaces
    2. B.2 Vector-valued Hardy spaces
    3. B.3 Invariant subspaces
    4. B.4 Factorization of operator-valued functions

##  1\. Introduction

In recent years, tools from noncommutative analysis have proven fruitful in
understanding the primitives of Quantum Information Theory (QIT), where–mostly
due to the noncommutative nature of the theory–even finding viable quantum
analogs of certain information-theoretic quantities turns out to be
nontrivial, and various challenges arise to be overcome (see, e.g., [11, 10,
40, 5, 2]). Analytical methods have been exploited in classical information
theory just as well, where the quantum phenomena are nonexistent (see, e.g.,
[35, 23, 20, 24]).

Rényi divergence (or Rényi relative entropy) is a useful concept in
information theory [38]. A quantum analog of this entropic measure is tricky
to define; since arbitrary density operators $\rho,\sigma$ do not necessarily
commute, there are multiple candidates for a definition of quantum Rényi
relative entropy. One way is to define it as the so-called _sandwiched Rényi
divergence_ [40, 31] given by

$D_{\alpha}(\rho\|\sigma)=\frac{\alpha}{\alpha-1}\log\|\sigma^{\frac{1-\alpha}{2\alpha}}\rho\sigma^{\frac{1-\alpha}{2\alpha}}\Big{\|}_{\alpha},$

where $\|\cdot\|_{\alpha}$ denotes the $\alpha$-Schatten norm. For
applicability and operational significance [40, 30], this quantity has to
satisfy certain properties (e.g., data-processing inequality) [31, 15, 4].
Since the Schatten norms form an interpolation scale, the above expression of
sandwiched Rényi divergences allows one to utilize the complex interpolation
method in establishing the aforementioned properties [4].

Consider a composite quantum system with bipartite Hilbert space
$\mathcal{H}_{X}\otimes\mathcal{H}_{Y}$. In analogy with the classical case, a
quantum generalization of the conditional Rényi entropy can be defined, for a
density operator $\rho_{XY}$ on $\mathcal{H}_{X}\otimes\mathcal{H}_{Y}$, using
the (sandwiched) Rényi divergence [31]:

$H_{\alpha}(X|Y)_{\rho}:=-\inf D_{\alpha}(\rho_{XY}\|I_{X}\otimes\sigma_{Y}),$

where the infimum runs over all density operators $\sigma_{Y}$ acting on
$\mathcal{H}_{Y}$. Expressing this quantity in terms of the Schatten norms as
above, the norms on the _operator-valued Schatten spaces_ 111These are the
discrete case of non-commutative vector valued $L_{p}$-spaces of [33]. fully
incarnate in the guise of this divergence (see below). These spaces were first
introduced by Pisier [33] as a noncommutative counterpart to the vector-valued
space $\ell_{p}(\mathcal{X})$ of $p$-summable sequences in a Banach space
$\mathcal{X}$ for $1\leq p\leq\infty$. Roughly speaking, Pisier’s theory
interweaves the interpolating structure of Schatten classes $\mathcal{S}_{p}$
with an operator space structure222This operator space structure is essential
for the theory to have all the desired properties, such as duality. on
$\mathcal{X}$ to construct a whole family of operator spaces, denoted by
$\mathcal{S}_{p}[\mathcal{X}]$, whose elements act on a tensor product Hilbert
space, say $\mathcal{H}_{Y}\otimes\mathcal{H}_{X}$.333The precise definition
of $\mathcal{S}_{p}[\mathcal{X}]$ is given in Section 5.

For our purposes in this work, the most relevant norms are obtained when
$\mathcal{X}$ is taken to be the Schatten class operator space
$\mathcal{S}_{q}$, $1\leq q\leq\infty$. In this way we obtain a two-parameter
family of spaces $\mathcal{S}_{p}[\mathcal{S}_{q}]$, $1\leq p,q\leq\infty$,
which form an interpolation scale. Assuming $p\leq q$, the norm of an operator
$M_{XY}$ in $\mathcal{S}_{p}[\mathcal{S}_{q}]$ is quantitatively characterized
by the variational expression [33, 18]

$\displaystyle\|M_{XY}\|_{\mathcal{S}_{p}[\mathcal{S}_{q}]}=\|M_{XY}\|_{(p,q)}:=\inf\|A_{X}\|_{\mathcal{S}_{2r}}\|T_{XY}\|_{\mathcal{S}_{q}}\|B_{Y}\|_{\mathcal{S}_{2r}},$
(1.1)

where the infimum is taken over all factorizations $M_{XY}=(A_{X}\otimes
I_{Y})T_{XY}(B_{X}\otimes I_{Y})$, and $r$ is defined by

$\frac{1}{p}=\frac{1}{q}+\frac{1}{r}.$

The variational formula (1.1) is the subtle bridge that links the norm on
$\mathcal{S}_{p}[\mathcal{S}_{q}]$ with quantum entropies as it allows to
formulate the quantum conditional Rényi entropy in terms of the
$(1,\alpha)$-norm:

$\displaystyle
H_{\alpha}(X|Y)_{\rho}=\frac{\alpha}{1-\alpha}\log\|\rho_{YX}\big{\|}_{(1,\alpha)}.$

This new formulation makes it possible, as we will see, to utilize the power
of methods and techniques from interpolation theory and operator space theory
to investigate characteristics of these entropic measures. From a broader
perspective, this connection suggests potential applications of Pisier’s
theory in other areas of QIT as well.

Based on the above observation, we believe that it is worth the effort to make
the theory of operator-valued Schatten spaces more accessible, particularly to
the QIT community at large; this is our primary goal in this paper. To this
end, we describe all the necessary constituent components of the theory and
provide detailed proofs of the major results in a self-contained way, whereas
the unnecessary parts (with respect to QIT) are dismissed. Although for the
most parts we do not confine ourselves to finite-dimensional spaces, we will
not refrain from doing so when it contributes to simplicity of presentation
and clarity of the main ideas. To showcase advantages of this theory, we also
provide some applications in QIT, specifically in the study of quantum Rényi
quantities. In particular, we establish a new uniform continuity bound for the
quantum conditional Rényi entropy.

We remark that following Pisier [33] we use the theory of operator spaces to
define operator-valued Schatten spaces and prove their properties. An
alternative approach for defining these spaces might be taking the concrete
formula (1.1) as the starting point, without appealing to operator spaces to
avoid complications.444This suggestion was raised in a discussion with Marco
Tomamichel and Alexander McKinlay and was an impetus for this work.
Nevertheless, this approach does not seem to be effective since given (1.1) as
the starting definition, proving that these norms form an interpolation family
is not easy from scratch. This approach appears to be even more disappointing
once we realize that even proving the triangle for (1.1) is not
straightforward. Thus, to utilize the full power of operator-valued Schatten
spaces in applications, we need the theory of operator spaces.

The rest of this paper is organized as follows. In Section 2 we present a
concise description of the complex interpolation theory due to Calderón [9],
and prove a subharmonicity lemma that will be crucial in the proof of a main
theorem. Next, in Section 3 we review the theory of operator spaces and
describe the natural operator space structure of Schatten spaces. We also
introduce the _column_ and _row_ operator spaces that play pivotal roles in
the construction of $\mathcal{S}_{p}[\mathcal{X}]$. Section 4 is devoted to
the Haagerup tensor product of operator spaces, where we show how Schatten
spaces can be identified with the Haagerup tensor product of column and row
operator spaces. In Section 5 we finally give the definition of operator-
valued Schatten spaces based on the Haagerup tensor product. We explain why
they form an interpolation scale and prove the aforementioned variational
characterization for their norms. Section 6 is devoted to applications of
these spaces in QIT. Here, we provide short proofs of some existing results as
well as prove some new ones based on the theory developed in the previous
sections.

One of the key ingredients in the development of Pisier’s theory is the fact
that the Haagerup tensor product “commutes” with interpolation which is
explicitly stated and proven in Theorem 4.2. This theorem is the most
technical result in this paper, for which we give a detailed proof based on
the work of Kouba [21]. The proof relies heavily on a certain factorization of
operator-valued functions due to Wiener and Masani. As we found the
Wiener–Masani theorem to be another cornerstone of the theory, we devote
Appendix B to the proof of this theorem that is borrowed from [16].

### Acknowledgements.

MMG is supported by a fund from MSRT. SB is thankful to Marco Tomamichel and
Alexander McKinlay for discussions on this work.

##  2\. Interpolation theory

Banach spaces in this paper are all considered to be complex. For Banach
spaces $\mathcal{X},\mathcal{Y}$ we denote by $B(\mathcal{X},\mathcal{Y})$ the
Banach space of all bounded linear operators from $\mathcal{X}$ to
$\mathcal{Y}$, equipped with the usual operator norm

$\|T\|_{B(\mathcal{X},\mathcal{Y})}=\sup\big{\\{}\|Tx\|_{\mathcal{Y}}:x\in{X},\|x\|_{\mathcal{X}}\leq
1\big{\\}}.$

When $\mathcal{X}=\mathcal{Y}$, let
$B(\mathcal{X})=B(\mathcal{X},\mathcal{Y})$. We denote the dual of
$\mathcal{X}$ by $\mathcal{X}^{*}=B(\mathcal{X},\mathbb{C})$.

The space of $m\times n$ matrices with entries in $\mathcal{X}$ is denoted by
$M_{m,n}(\mathcal{X})$. For simplicity we denote $M_{m,n}(\mathbb{C})$ by
$M_{m,n}$. The norm of a matrix $A\in M_{m,n}$ is denoted by $\|A\|_{m,n}$.
When $m=n$ we let $M_{n}(\mathcal{X})=M_{n,n}(\mathcal{X})$.

Unless otherwise explicitly stated, the equality $\mathcal{X}=\mathcal{Y}$
means that $\mathcal{X},\mathcal{Y}$ are the same linear spaces with
_equivalent norms_. We denote by $\ell_{2}^{n}$ the Hilbert space of
$n$-tuples of complex numbers, and let $\ell_{2}=\ell_{2}^{\infty}$. For a
Hilbert space $\mathcal{H}$ and $n\in\mathbb{N}$, we let
$\mathcal{H}^{n}=\oplus_{i=1}^{n}\mathcal{H}=\ell_{2}^{n}\otimes_{2}\mathcal{H}$
where $\otimes_{2}$ denotes the _Hilbertian tensor product_ .

An element $T\in B(\mathcal{H})$ is positive (positive semidefinite) if it is
of the form $T=X^{*}X$ for some $X\in B(\mathcal{H})$, where $X^{*}$ is the
adjoint of $X$. We shall write $T\geq 0$ to denote that $T$ is positive.

###  2.1. Calderón’s complex interpolation method

In this section, we present a brief overview of the complex interpolation
method due to Calderón [9]. We refer the reader to [7, 25] for more details on
interpolation theory.

We write $\mathcal{X}\hookrightarrow\mathcal{Y}$ if there is a continuous
embedding from $\mathcal{X}$ into $\mathcal{Y}$. A couple of Banach spaces
$(\mathcal{X},\mathcal{Y})$ is said to be _compatible_ if there is a Hausdorff
topological vector space $\mathcal{V}$ such that
$\mathcal{X},\mathcal{Y}\hookrightarrow\mathcal{V}$. For such a compatible
couple we may consider $\mathcal{X}+\mathcal{Y}$ as a linear subspace of
$\mathcal{V}$, and for any $v\in\mathcal{X}+\mathcal{Y}$ define

$\|v\|_{\mathcal{X}+\mathcal{Y}}:=\inf\\{\|x\|_{\mathcal{X}}+\|y\|_{\mathcal{Y}}:x\in\mathcal{X},y\in\mathcal{X},v=x+y\\}.$

This norm turns $\mathcal{X}+\mathcal{Y}$ into a Banach space.

In order to prepare the setup, consider the strip

$\mathbb{S}:=\\{z\in\mathbb{C}:0\leq\text{Re}(z)\leq 1\\},$

in the complex plane, and let $\mathcal{X}$ be a Banach space. A function
$f:\mathbb{S}\rightarrow\mathcal{X}$ is said to be _holomorphic_ in the
interior of $\mathbb{S}$, if for every $x^{*}\in\mathcal{X}^{*}$ the complex
function $z\mapsto\langle x^{*},f(z)\rangle$ is holomorphic in the interior of
$\mathbb{S}$. It is not hard to verify that the _maximum principle_ holds for
such holomorphic functions. Namely, for a bounded continuous function
$f:\mathbb{S}\rightarrow\mathcal{X}$ that is holomorphic in the interior of
$\mathbb{S}$, we have

$\sup_{z\in
S}\|f(z)\|_{\mathcal{X}}\leq\max\big{\\{}\sup_{t\in\mathbb{R}}\|f(it)\|_{\mathcal{X}},\sup_{t\in\mathbb{R}}\|f(1+it)\|_{\mathcal{X}}\big{\\}}.$
(2.1)

Let $(\mathcal{X},\mathcal{Y})$ be a compatible couple of complex Banach
spaces. We denote by $\mathscr{F}(\mathcal{X},\mathcal{Y})$ the space of all
bounded continuous functions $f:\mathbb{S}\to\mathcal{X}+\mathcal{Y}$ that are
holomorphic in the interior of $\mathbb{S}$ such that

$t\mapsto f(it)\in\mathcal{X},\text{ and }t\mapsto f(1+it)\in\mathcal{Y},$

are continuous functions. We equip $\mathscr{F}(\mathcal{X},\mathcal{Y})$ with
the norm

$\|f\|_{\mathscr{F}(\mathcal{X},\mathcal{Y})}:=\max\big{\\{}\sup_{t\in\mathbb{R}}\|f(it)\|_{\mathcal{X}},\sup_{t\in\mathbb{R}}\|f(1+it)\|_{\mathcal{Y}}\big{\\}}.$

Using the maximum principle mentioned above, it is not difficult to verify
that $\mathscr{F}(\mathcal{X},\mathcal{Y})$ is a Banach space.

We are now ready to define the complex interpolation spaces
$[\mathcal{X},\mathcal{Y}]_{\theta}$, $0\leq\theta\leq 1$.

######  Definition 2.1.

Let $(\mathcal{X},\mathcal{Y})$ be a compatible couple of Banach spaces. For
every $\theta\in[0,1]$, we define $[\mathcal{X},\mathcal{Y}]_{\theta}$ to be
the linear space

$[\mathcal{X},\mathcal{Y}]_{\theta}:=\big{\\{}f(\theta):f\in\mathscr{F}(\mathcal{X},\mathcal{Y})\big{\\}},$

endowed with the norm

$\|v\|_{[\mathcal{X},\mathcal{Y}]_{\theta}}:=\inf\big{\\{}\|f\|_{\mathscr{F}(\mathcal{X},\mathcal{Y})}:f(\theta)=v,f\in\mathscr{F}(\mathcal{X},\mathcal{Y})\big{\\}}.$

It can be verified that $[\mathcal{X},\mathcal{Y}]_{\theta}$ is a Banach
space.

From the definition it is clear that
$[\mathcal{X},\mathcal{X}]_{\theta}=\mathcal{X}$ with identical norms.
Moreover, we have (see, [7, Theorem 4.2.1])

$[\mathcal{X},\mathcal{Y}]_{\theta}=[\mathcal{Y},\mathcal{X}]_{1-\theta}.$

_Reiteration theorem_ is another interesting property of interpolation spaces.
Let $(\mathcal{X}_{0},\mathcal{X}_{1})$ be a compatible couple and let
$\mathcal{X}_{\theta}=[\mathcal{X}_{0},\mathcal{X}_{1}]_{\theta}$,
$0<\theta<1$. Then, a special case of the reiteration theorem ([7, Theorem
4.6.1]) says that if $\mathcal{X}_{0}\subseteq\mathcal{X}_{1}$ as vector
spaces, for $0\leq\theta_{0},\theta_{1}\leq 1$ we have

$[\mathcal{X}_{\theta_{0}},\mathcal{X}_{\theta_{1}}]_{\theta}=\mathcal{X}_{(1-\theta)\theta_{0}+\theta\theta_{1}},\qquad
0<\theta<1.$ (2.2)

Interpolation spaces behave nicely under taking the Banach dual as well.
Suppose that one of the spaces $\mathcal{X}_{0},\mathcal{X}_{1}$ of a
compatible couple $(\mathcal{X}_{0},\mathcal{X}_{1})$ are finite-dimensional
(or more generally, reflexive). Then, letting
$\mathcal{X}_{\theta}=[\mathcal{X}_{0},\mathcal{X}_{1}]_{\theta}$,
$0<\theta<1$, we have

$\mathcal{X}_{\theta}^{*}=[\mathcal{X}^{*}_{0},\mathcal{X}^{*}_{1}]_{\theta},$

where $\mathcal{X}^{*}_{\theta}$ is the Banach dual of $\mathcal{X}_{\theta}$
(see, [7, Corollary 4.5.2]).

######  Remark 2.2.

For every $v\in[\mathcal{X},\mathcal{Y}]_{\theta}$ we have

$\displaystyle\|v\|_{[\mathcal{X},\mathcal{Y}]_{\theta}}:=\inf_{\stackrel{{\scriptstyle
f\in\mathscr{F}(\mathcal{X},\mathcal{Y})}}{{f(\theta)=v}}}\Big{(}\sup_{t}\|f(it)\|_{\mathcal{X}}\Big{)}^{1-\theta}\Big{(}\sup_{t}\|f(1+it)\|_{\mathcal{Y}}\Big{)}^{\theta}.$
(2.3)

To prove this, we note that by definition

$\|v\|_{[\mathcal{X},\mathcal{Y}]_{\theta}}:=\inf_{\stackrel{{\scriptstyle
f\in\mathscr{F}(\mathcal{X},\mathcal{Y})}}{{f(\theta)=v}}}\max\big{\\{}\sup_{t}\|f(it)\|_{\mathcal{X}},\sup_{t}\|f(1+it)\|_{\mathcal{Y}}\big{\\}},$

and

$\Big{(}\sup_{t}\|f(it)\|_{\mathcal{X}}\Big{)}^{1-\theta}\Big{(}\sup_{t}\|f(1+it)\|_{\mathcal{Y}}\Big{)}^{\theta}\leq\max\big{\\{}\sup_{t}\|f(it)\|_{\mathcal{X}},\sup_{t}\|f(1+it)\|_{\mathcal{Y}}\big{\\}}.$

On the other hand, given $f\in\mathscr{F}(\mathcal{X},\mathcal{Y})$ with
$f(\theta)=v$, the map $f_{\alpha}(z)=e^{\alpha(z-\theta)}f(z)$, for arbitrary
$\alpha\in\mathbb{R}$, also belongs to $\mathscr{F}(\mathcal{X},\mathcal{Y})$
and satisfies $f_{\alpha}(\theta)=v$. Moreover, for every $t\in\mathbb{R}$ we
have
$\|f_{\alpha}(it)\|_{\mathcal{X}}=e^{-\alpha\theta}\|f(it)\|_{\mathcal{X}}$
and
$\|f_{\alpha}(1+it)\|_{\mathcal{Y}}=e^{\alpha(1-\theta)}\|f(1+it)\|_{\mathcal{Y}}$.
Then, optimizing over the choice of $\alpha\in\mathbb{R}$, we have

$\inf_{\alpha\in\mathbb{R}}\max\big{\\{}\sup_{t}\|f_{\alpha}(it)\|_{\mathcal{X}},\sup_{t}\|f_{\alpha}(1+it)\|_{\mathcal{Y}}\big{\\}}=\Big{(}\sup_{t}\|f(it)\|_{\mathcal{X}}\Big{)}^{1-\theta}\Big{(}\sup_{t}\|f(1+it)\|_{\mathcal{Y}}\Big{)}^{\theta}.$

Putting these together, (2.3) follows. $\Box$

######  Remark 2.3.

Considering the constant map $f(z)=v$, equation (2.3) shows that

$\displaystyle\|v\|_{[\mathcal{X},\mathcal{Y}]_{\theta}}\leq\|v\|_{\mathcal{X}}^{1-\theta}\|v\|_{\mathcal{Y}}^{\theta}.$
(2.4)

$L_{p}$-spaces are important examples of complex interpolation. Let
$(\Omega,\mu)$ be a $\sigma$-finite measure space, and for $1\leq p<\infty$
let $L_{p}(\Omega)=L_{p}(\Omega,\mathbb{C})$ be the spaces of complex
measurable functions $h$ on $\Omega$ with
$\int|h|^{p}\differential\mu<+\infty$, equipped with norm
$\|h\|_{p}=\big{(}\int|h|^{p}\differential\mu\big{)}^{1/p}$. For $p=+\infty$,
we define $L_{\infty}(\Omega,\mathbb{C})$ with norm
$\|h\|_{\infty}=\operatorname*{ess\,sup}|h|$.555$\operatorname*{ess\,sup}|h|:=\inf\\{a\in\mathbb{R}:g(z)\leq
a\ \text{for almost all}\ z\in\Omega\\}$. Then, $L_{p}(\Omega)$ for any $1\leq
p\leq+\infty$ is a Banach space and for $1\leq p_{0},p_{1}\leq\infty$,
$0<\theta<1$ we have

$\displaystyle[L_{p_{0}}(\Omega),L_{p_{1}}(\Omega)]_{\theta}=L_{p_{\theta}}(\Omega),$
(2.5)

with identical norms, where666As a convention we set $\frac{1}{\infty}=0$.

$\frac{1}{p_{\theta}}=\frac{1-\theta}{p_{0}}+\frac{\theta}{p_{1}}.$

The following theorem states an important bound on the operator norm of
bounded maps between interpolation spaces.

######  Theorem 2.4 ([25]).

Let $(\mathcal{X}_{0},\mathcal{X}_{1})$ and
$(\mathcal{Y}_{0},\mathcal{Y}_{1})$ be compatible couples, and $0<\theta<1$.
If
$T:\mathcal{X}_{0}+\mathcal{X}_{1}\rightarrow\mathcal{Y}_{0}+\mathcal{Y}_{1}$
is a linear operator such that $T|_{\mathcal{X}_{j}}\in
B(\mathcal{X}_{j},\mathcal{Y}_{j})$, $j=0,1$, then
$T|_{[\mathcal{X}_{0},\mathcal{X}_{1}]_{\theta}}\in
B([\mathcal{X}_{0},\mathcal{X}_{1}]_{\theta},[\mathcal{Y}_{0},\mathcal{Y}_{1}]_{\theta})$,
and

$\|T\|_{B([\mathcal{X}_{0},\mathcal{X}_{1}]_{\theta},[\mathcal{Y}_{0},\mathcal{Y}_{1}]_{\theta})}\leq\|T\|_{B(\mathcal{X}_{0},\mathcal{Y}_{0})}^{1-\theta}\|T\|_{B(\mathcal{X}_{1},\mathcal{Y}_{1})}^{\theta}.$
(2.6)

###  2.2. A subharmonicity lemma

Let $\mathbb{D}$ be the unit disk in the complex plane and
$\partial\mathbb{D}$ be its boundary. Then, the _Poisson kernel_ for
$\mathbb{D}$ is given by

$P(z,e^{it}):=\real\left(\frac{e^{it}+z}{e^{it}-z}\right)=\frac{1-|z|^{2}}{|e^{it}-z|^{2}},$

for $z=re^{i\theta}\in\mathbb{D},e^{it}\in\partial\mathbb{D}$. It is well-
known that if $f\in L_{1}(\partial\mathbb{D})$ and

$F(z):=\frac{1}{2\pi}\int_{-\pi}^{\pi}f(e^{it})P(z,e^{it})\differential t,$
(2.7)

then,

$\lim_{r\rightarrow 1}F(re^{i\theta})=f(e^{i\theta}),\qquad\text{a.e. on}\
\partial\mathbb{D}.$

The Poisson kernel for $\mathbb{S}$ may be obtained by carrying $P$ over to
$\mathbb{S}$ using the conformal map $\phi:\mathbb{D}\rightarrow\mathbb{S}$
defined by

$\phi(z)=\frac{1}{i\pi}\log(i\frac{1+z}{1-z}).$ (2.8)

Indeed, for any holomorphic function $f:\mathbb{S}\rightarrow\mathbb{C}$, the
function $f\circ\phi:\mathbb{D}\to\mathbb{C}$ is also holomorphic. Thus, using
(2.7) we have

$f(z)=\frac{1}{2\pi}\int_{-\pi}^{\pi}f\circ\phi(e^{it})\cdot
P\big{(}\phi^{-1}(z),e^{it}\big{)}\differential t.$

We note that $\phi$ maps the boundary of $\mathbb{D}$ onto the boundary of
$\mathbb{S}$. Then, letting $\phi(e^{it})=b+is$ with $b\in\\{0,1\\}$ and
$s\in\mathbb{R}$, and rewriting the integral in terms of $s$, we find that

$f(z)=\sum_{b=0}^{1}\int_{-\infty}^{+\infty}f(b+is)\cdot\frac{P\big{(}\phi^{-1}(z),\phi^{-1}(b+is)\big{)}}{2\cosh(\pi
s)}\differential s.$

Consequently, the Poisson kernel of $\mathbb{S}$ is given by the functions
$P_{b}(x+iy,s)$ for $b=0,1$, where

$\displaystyle P_{b}(x+iy,s)$
$\displaystyle:=\frac{P\big{(}\phi^{-1}(z),\phi^{-1}(b+is)\big{)}}{2\cosh(\pi
s)}$ $\displaystyle=\frac{\sin(\pi
x)}{2\big{(}\cosh(\pi(y-s))-\cos(\pi(x-b))\big{)}}.$

We note that $P_{b}(x+iy,s)\geq 0$ and it can be verified that

$\int_{-\infty}^{+\infty}P_{0}(x+iy,s)\differential
s=1-x,\qquad\quad\int_{-\infty}^{+\infty}P_{1}(x+iy,s)\differential s=x.$

######  Lemma 2.5.

Let $(\mathcal{X}_{0},\mathcal{X}_{1})$ be a compatible couple and let
$\mathcal{X}_{\theta}=[\mathcal{X}_{0},\mathcal{X}_{1}]_{\theta}$ for
$0<\theta<1$.

  * (i)

Let $f(z)\in\mathscr{F}(\mathcal{X}_{0},\mathcal{X}_{1})$. Then, we have

$\log\|f(\theta)\|_{\mathcal{X}_{\theta}}\leq\sum_{b=0}^{1}\int_{-\infty}^{+\infty}P_{b}(\theta,s)\log\|f(b+is)\|_{\mathcal{X}_{b}}\differential
s.$

  * (ii)

If $\mathcal{X}_{0},\mathcal{X}_{1}$ are finite-dimensional (or more
generally, reflexive), then the same inequality holds whenever $f(z)$ is
bounded on $\mathbb{S}$ and holomorphic in its interior (though not
necessarily continuous up to the boundary).

The first part of this lemma is proven in [7, Lemma 4.3.2]. Here, we prove the
second part for the sake of completeness.

###### Proof of (ii).

Let $\mathcal{X}^{*}_{\theta}$ be the Banach dual of $\mathcal{X}_{\theta}$.
Then, there is $\ell\in\mathcal{X}^{*}_{\theta}$ such that

$\|\ell\|_{\mathcal{X}^{*}_{\theta}}=1,\qquad\|f(\theta)\|_{\mathcal{X}_{\theta}}=|\langle\ell,f(\theta)\rangle|.$

Since $\mathcal{X}^{*}_{\theta}=[\mathcal{X}^{*}_{0},\mathcal{X}^{*}_{1}]$,
for any $\epsilon>0$ there is $g(z)$ such that $g(\theta)=\ell$ and

$\max\big{\\{}\sup_{t}\|g(it)\|_{\mathcal{X}^{*}_{0}},\sup_{t}\|g(1+it)\|_{\mathcal{X}^{*}_{1}}\big{\\}}\leq\|\ell\|_{\mathcal{X}^{*}_{\theta}}+\epsilon=1+\epsilon.$

Now define $\psi(z):=\langle g(z),f(z)\rangle$. We note that $\psi(z)$ is
holomorphic and bounded. Then, by the subharmonicity of holomorphic functions
we have

$\displaystyle\|f(\theta)\|_{\mathcal{X}_{\theta}}$
$\displaystyle=|\psi(\theta)|$
$\displaystyle\leq\sum_{b=0}^{1}\int_{-\infty}^{+\infty}P_{b}(\theta,s)\log|\psi(b+is)|\differential
s$
$\displaystyle\leq\sum_{b=0}^{1}\int_{-\infty}^{+\infty}P_{b}(\theta,s)\log\big(\|g(b+is)\|_{\mathcal{X}_{b}^{*}}\cdot\|f(b+is)\|_{\mathcal{X}_{b}}\big{missing})\differential
s$
$\displaystyle\leq\sum_{b=0}^{1}\int_{-\infty}^{+\infty}P_{b}(\theta,s)\log\big((1+\epsilon)\|f(b+is)\|_{\mathcal{X}_{b}}\big{missing})\differential
s$
$\displaystyle\leq\log(1+\epsilon)+\sum_{b=0}^{1}\int_{-\infty}^{+\infty}P_{b}(\theta,s)\log\|f(b+is)\|_{\mathcal{X}_{b}}\differential
s.$

The desired inequality follows once we note that $\epsilon>0$ is arbitrary. ∎

##  3\. Operator spaces

We now give a brief overview of operator space theory and refer to [34] for
more information. Recall that $B(\mathcal{H})$ is the space of bounded
operators acting on the Hilbert space $\mathcal{H}$. Then,
$M_{n}(B(\mathcal{H}))$ is the space of $n\times n$ matrices with entries in
$B(\mathcal{H})$. The spaces $M_{n}(B(\mathcal{H}))$, $n\geq 1$, are equipped
with a whole sequence of norms naturally associated with $B(\mathcal{H})$.
Indeed, there is a natural way to turn $M_{n}(B(\mathcal{H}))$ into a
$C^{*}$-algebra by identifying it with $B(\mathcal{H}^{n})$. To see this, let
us denote a generic element of $M_{n}(B(\mathcal{H}))$ by $[X_{ij}]$ where
$X_{ij}\in B(\mathcal{H})$. Then for $[X_{ij}],[Y_{ij}]$ in
$M_{n}(B(\mathcal{H}))$, we put

$[X_{ij}]\cdot[Y_{ij}]=\left[\sum_{k=1}^{n}X_{ik}Y_{kj}\right]\ \ \
\text{and}\ \ \ [X_{ij}]^{*}=[X_{ji}^{*}].$

Next, to introduce the norm on $M_{n}(B(\mathcal{H}))$, we think of
$X=[X_{ij}]\in M_{n}(B(\mathcal{H}))$ as an element of
$M_{n}(B(\mathcal{H}))\cong B(\mathcal{H}^{n})$ as $C^{*}$-algebras, and
define

$\|X\|_{n}:=\|X\|_{M_{n}(B(\mathcal{H}))}=\sup\left\\{\bigg{(}\,\sum_{i=1}^{n}\Big{\|}\sum_{j=1}^{n}X_{ij}(v_{j})\Big{\|}^{2}\,\bigg{)}^{\frac{1}{2}}:v_{1},\dots,v_{n}\in\mathcal{H},\sum_{j=1}^{n}\|v_{j}\|^{2}\leq
1\right\\}.$

We can similarly define norms on $M_{m,n}(B(\mathcal{H}))$.

The above matrix norms satisfy two crucial properties. For $A,B^{*}\in
M_{m,n}$ and $X\in M_{n}(B(\mathcal{H}))$, by considering matrix
multiplication in the usual way, we can think of $AXB$ as an element of
$M_{m}(B(\mathcal{H}))$.777Equivalently, $AXB$ can be understood as $(A\otimes
I_{\mathcal{H}})X(B\otimes I_{\mathcal{H}})$ where $I_{\mathcal{H}}$ is the
identity map acting on $\mathcal{H}$. Also, for $X\in M_{m}(B(\mathcal{H}))$
and $Y\in M_{n}(B(\mathcal{H}))$, we denote by $X\oplus Y$ the element of
$M_{m+n}(B(\mathcal{H}))$ given by

$X\oplus Y=\begin{bmatrix}X&0\\\ 0&Y\end{bmatrix}.$

Then, it is readily verified that the norms on $M_{n}(\mathcal{X})$, where
$\mathcal{X}=B(\mathcal{H})$, satisfy the following properties:

  * (i)

$\|AXB\|_{n}\leq\|A\|_{m,n}\|X\|_{n}\|B\|_{n,m}$ for all $A,B^{*}\in M_{m,n}$,
$X\in M_{n}(\mathcal{X})$;

  * (ii)

$\left\|X\oplus Y\right\|_{m+n}=\max\\{\|X\|_{n},\|Y\|_{m}\\}$ for all $X\in
M_{n}(\mathcal{X})$, $Y\in M_{m}(\mathcal{X})$.

We note that the above two properties hold not only in $B(\mathcal{H})$, but
also in all of its linear subspaces. Conversely, Ruan’s theorem [34] states
that any Banach space $\mathcal{X}$, with a sequence of norms $\|\cdot\|_{n}$
on $M_{n}(\mathcal{X})$ for $n\geq 1$ that satisfy properties (i) and
(ii),888In view of Ruan’s theorem, these spaces are sometimes termed as
abstract operator spaces as opposed to concrete operator spaces (see
Definition 3.1.) “embeds” into $B(\mathcal{H})$ for some Hilbert space
$\mathcal{H}$.

######  Definition 3.1.

A (concrete) operator space is a closed subspace of $B(\mathcal{H})$ for some
Hilbert space $\mathcal{H}$.

Ruan’s theorem may be precisely formulated using the language of completely
bounded maps. Let $\phi:\mathcal{X}\rightarrow\mathcal{Y}$ be a linear map
between operator spaces $\mathcal{X},\mathcal{Y}$. The $n$-th _amplification_
$\phi_{n}:M_{n}(\mathcal{X})\rightarrow M_{n}(\mathcal{Y})$ of $\phi$ is
defined by $\phi_{n}([X_{ij}])=[\phi(X_{ij})]$ for $[X_{ij}]\in
M_{n}(\mathcal{X})$. Then $\phi$ is said to be _completely bounded_ (CB) if
$\|\phi\|_{cb}:=\sup_{n\geq 1}\|\phi_{n}\|<\infty$. We say that
$\mathcal{X}=\mathcal{Y}$ _completely isometrically_ if there is an invertible
map $\phi$ with $\|\phi\|_{cb}=\|\phi^{-1}\|_{cb}=1$.

As mentioned above, by Ruan’s theorem any Banach space $\mathcal{X}$ with a
sequence of norms $\|\cdot\|_{n}$ on $M_{n}(\mathcal{X})$ for $n\geq 1$
satisfying properties (i) and (ii) above, is completely isometric to an
operator space, i.e., by Definition 3.1 can be identified by a closed subspace
of some $B(\mathcal{H})$ .

###  3.1. Dual operator space

There is a well-behaved notion of dual space in the category of operator
spaces and completely bounded maps. Let $\mathcal{X},\mathcal{Y}$ be two
operator spaces, and let $CB(\mathcal{X},\mathcal{Y})$ be the Banach space of
all completely bounded maps from $\mathcal{X}$ into $\mathcal{Y}$. Then,
$CB(\mathcal{X},\mathcal{Y})$ enjoys a natural operator space structure which
we now describe. To this end, we identify $M_{n}(CB(\mathcal{X},\mathcal{Y}))$
with $CB(\mathcal{X},M_{n}(\mathcal{Y}))$ as follows: for $\Phi=[\phi_{ij}]\in
M_{n}(CB(\mathcal{X},\mathcal{Y}))$, let
$\widetilde{\Phi}:\mathcal{X}\rightarrow M_{n}(\mathcal{Y})$ be given by
$\widetilde{\Phi}(X):=\big{[}\phi_{ij}(X)\big{]}$. Then, we can define a norm
on $M_{n}(CB(\mathcal{X},\mathcal{Y}))$ by letting

$\|\Phi\|_{n}:=\big{\|}\widetilde{\Phi}\big{\|}_{CB(\mathcal{X},M_{n}(\mathcal{Y}))}.$

It is not hard to verify that $CB(\mathcal{X},\mathcal{Y})$ with this sequence
of norms satisfies Ruan’s axioms (properties (i) and (ii) above), and by
Ruan’s theorem is an operator space.

Considering the special case $\mathcal{Y}=\mathbb{C}$, we find that
$CB(\mathcal{X},\mathbb{C})$ is an operator space. On the other hand, one may
observe that $CB(\mathcal{X},\mathbb{C})$ coincides with
$\mathcal{X}^{*}=B(\mathcal{X},\mathbb{C})$ with identical norms as Banach
spaces. Therefore, the Banach dual $\mathcal{X}^{*}$ of any operator space has
a canonical operator space structure.

###  3.2. Interpolation of operator spaces

Let $(\mathcal{X}_{0},\mathcal{X}_{1})$ be a compatible couple of Banach
spaces. If further $\mathcal{X}_{0},\mathcal{X}_{1}$ are operator spaces, then
$M_{n}(\mathcal{X}_{0})$ and $M_{n}(\mathcal{X}_{1})$ with their specific
norms form a compatible couple of Banach spaces. Thus, we may consider the
complex interpolation space
$[M_{n}(\mathcal{X}_{0}),M_{n}(\mathcal{X}_{1})]_{\theta}$, for every $n\geq
1$, and endow
$\mathcal{X}_{\theta}:=[\mathcal{X}_{0},\mathcal{X}_{1}]_{\theta}$ with a
matrix norm structure as follows:

$\big{\|}(x_{ij})\big{\|}_{M_{n}(\mathcal{X}_{\theta})}:=\big{\|}(x_{ij})\big{\|}_{[M_{n}(\mathcal{X}_{0}),M_{n}(\mathcal{X}_{1})]_{\theta}},\
\ \ (x_{ij})\in M_{n}(\mathcal{X}_{\theta}).$

It can be shown that these norms satisfy Ruan’s axioms, so
$\mathcal{X}_{\theta}$ turns into an operator space [32, 34].

The reiteration identity (2.2) holds completely isometrically for a compatible
couple of operator spaces $\mathcal{X}_{0},\mathcal{X}_{1}$. Moreover, Theorem
2.6 extends to operator spaces as follows.

######  Proposition 3.2 ([32]).

Let $(\mathcal{X}_{0},\mathcal{X}_{1})$ and
$(\mathcal{Y}_{0},\mathcal{Y}_{1})$ be a pair of compatible operator spaces,
and define the operator spaces
$\mathcal{X}_{\theta}=[\mathcal{X}_{0},\mathcal{X}_{1}]$ and
$\mathcal{Y}_{\theta}=[\mathcal{Y}_{0},\mathcal{Y}_{1}]$ as above. Let
$T:\mathcal{X}_{0}+\mathcal{X}_{1}\rightarrow\mathcal{Y}_{0}+\mathcal{Y}_{1}$
be a linear map such that $T|_{\mathcal{X}_{j}}\in
CB(\mathcal{X}_{j},\mathcal{Y}_{j})$, $j=0,1$. Then,
$T|_{\mathcal{X}_{\theta}}\in CB(\mathcal{X}_{\theta},\mathcal{Y}_{\theta})$
and

$\|T\|_{CB(\mathcal{X}_{\theta},\mathcal{Y}_{\theta})}\leq\|T\|_{CB(\mathcal{X}_{0},\mathcal{Y}_{0})}^{1-\theta}\|T\|_{CB(\mathcal{X}_{1},\mathcal{Y}_{1})}^{\theta}.$
(3.1)

###  3.3. Schatten class operator spaces

Let $\mathcal{H}$ be a Hilbert space, and let
$\mathcal{S}_{\infty}(\mathcal{H})\subseteq B(\mathcal{H})$ be the space of
compact operators on $\mathcal{H}$ equipped with the operator norm. Note that
$\mathcal{S}_{\infty}(\mathcal{H})$, as a closed subspace of $B(\mathcal{H})$,
is an operator space.

Let $\mathcal{S}_{1}(\mathcal{H})$ be the space of trace class operators
acting on $\mathcal{H}$ equipped with the norm

$\|X\|_{\mathcal{S}_{1}(\mathcal{H})}=\tr|X|,$

where $|X|=\sqrt{X^{*}X}$. It is well-known that
$\mathcal{S}_{1}(\mathcal{H})$ is the (Banach) dual of
$\mathcal{S}_{\infty}(\mathcal{H})$.999We note the dual of
$\mathcal{S}_{1}(\mathcal{H})$ is $B(\mathcal{H})$ which is not equal to
$\mathcal{S}_{\infty}(\mathcal{H})$ when the dimension of $\mathcal{H}$ is
infinite. Therefore, since the dual of any operator space is an operator
space, $\mathcal{S}_{1}(\mathcal{H})$ inherits an operator space structure.

The Schatten class $\mathcal{S}_{p}(\mathcal{H})$ is indeed defined for any
$p\geq 1$; let $\mathcal{S}_{p}(\mathcal{H})$ be the space of operators $X$ in
$B(\mathcal{H})$ with $\tr|X|^{p}<\infty$, equipped with norm

$\|X\|_{\mathcal{S}_{p}(\mathcal{H})}=\Big{(}\tr|X|^{p}\Big{)}^{1/p}.$

When $\mathcal{H}=\ell_{2}$ (resp. $\mathcal{H}=\ell_{2}^{d}$) we denote
$\mathcal{S}_{p}(\mathcal{H})$ by $\mathcal{S}_{p}$ (resp.
$\mathcal{S}_{p}^{d}$). The space $\mathcal{S}_{p}$ is indeed a noncommutative
version of the ordinary $L_{p}$-space (in the discrete case). Moreover, (2.5)
is generalized to Schatten classes with similar ideas [34]. In particular, we
have

$\displaystyle\mathcal{S}_{p}(\mathcal{H})=[\mathcal{S}_{\infty}(\mathcal{H}),\mathcal{S}_{1}(\mathcal{H})]_{1/p},\qquad
1<p<\infty,$ (3.2)

with identical norms.

######  Remark 3.3.

We argued that
$\mathcal{S}_{1}(\mathcal{H}),\mathcal{S}_{\infty}(\mathcal{H})$ are operator
spaces, so if (3.2) is interpreted as an identity between operator spaces,
then $\mathcal{S}_{p}(\mathcal{H})$ for any $1\leq p\leq\infty$ is endowed
with an operator space structure.

###  3.4. Column and row operator spaces

Let $\mathcal{H}_{1},\mathcal{H}_{2}$ be two Hilbert spaces. Then, the map

$\displaystyle X\mapsto\begin{bmatrix}0&0\\\ X&0\end{bmatrix},$ (3.3)

is an _isometric_ embedding of $B(\mathcal{H}_{1},\mathcal{H}_{2})$ into
$B(\mathcal{H}_{1}\oplus\mathcal{H}_{2})$. Therefore,
$B(\mathcal{H}_{1},\mathcal{H}_{2})$ is an operator space. We note that
extending this map as an embedding of
$M_{n}\big{(}B(\mathcal{H}_{1},\mathcal{H}_{2})\big{)}$ in
$M_{n}\big{(}B(\mathcal{H}_{1}\oplus\mathcal{H}_{2})\big{)}$, the induced norm
on $M_{n}\big{(}B(\mathcal{H}_{1},\mathcal{H}_{2})\big{)}$ coincides with the
norm of $B(\mathcal{H}_{1}^{n},\mathcal{H}_{2}^{n})$. This means that to
understand the operator space structure of
$B(\mathcal{H}_{1},\mathcal{H}_{2})$ we can indeed ignore the above embedding
and directly work with $B(\mathcal{H}_{1},\mathcal{H}_{2})$ [34, p. 22].

Considering spacial cases of the above construction, we find that both
$B(\mathbb{C},\mathcal{H})$ and $B(\mathcal{H},\mathbb{C})$ are operator
spaces. Indeed, for any Hilbert space $\mathcal{H}$, the Banach spaces

$\mathcal{H}_{c}:=B(\mathbb{C},\mathcal{H})\quad\text{ and
}\quad\mathcal{H}_{r}:=B(\mathcal{H}^{*},\mathbb{C}),$

take operator space structures. $\mathcal{H}_{c}$ is called the _column
operator space_ and $\mathcal{H}_{r}$ is called the _row operator space_ [33,
p. 25].

We note that elements of $\mathcal{H}_{c}$ and $\mathcal{H}_{r}$ can be
identified with those of $\mathcal{H}$. To this end, for any $v\in\mathcal{H}$
define $v^{c}\in\mathcal{H}_{c}$ and $v^{r}\in\mathcal{H}_{r}$ by

$v^{c}(\alpha)=\alpha v,\qquad v^{r}(w)=\langle
w,v\rangle,\qquad\alpha\in\mathbb{C},w\in\mathcal{H}^{*}.$

Then, $v^{c}\mapsto v^{r}$ provides an isometry between $\mathcal{H}^{c}$ and
$\mathcal{H}^{r}$ as Banach spaces. Nevertheless, the operator space
structures of $\mathcal{H}^{c}$ and $\mathcal{H}^{r}$ do not coincide under
this map. To confirm this, suppose that $n\leq\dim\mathcal{H}$, and define
$v_{ij}\in\mathcal{H}$, $1\leq i,j\leq n$ as follows: let $v_{ij}=0$ if $j\neq
1$ and let $\\{v_{11},\dots,v_{n1}\\}\subset\mathcal{H}$ be orthonormal. Then,
we have

$\displaystyle\big{\|}\big{(}v_{ij}^{c}\big{)}\big{\|}_{M_{n}(\mathcal{H}_{c})}$
$\displaystyle=\sup_{\stackrel{{\scriptstyle\alpha_{1},\dots,\alpha_{n}\in\mathbb{C}}}{{\sum|\alpha_{k}|^{2}=1}}}\Big{\|}\Big{(}\sum_{j}\alpha_{j}v_{ij}\Big{)}\Big{\|}_{\mathcal{H}^{n}}$
$\displaystyle=\sup_{\stackrel{{\scriptstyle\alpha_{1},\dots,\alpha_{n}\in\mathbb{C}}}{{\sum|\alpha_{k}|^{2}=1}}}\big{\|}\big{(}\alpha_{1}v_{i1}\big{)}\big{\|}_{\mathcal{H}^{n}}$
$\displaystyle=\big{\|}\big{(}v_{i1}\big{)}\big{\|}_{\mathcal{H}^{n}}$
$\displaystyle=\Big{(}\sum_{i}\|v_{i1}\|^{2}_{\mathcal{H}}\Big{)}^{1/2}$
$\displaystyle=\sqrt{n}.$

On the other hand, we have

$\displaystyle\big{\|}\big{(}v_{ij}^{r}\big{)}\big{\|}_{M_{n}(\mathcal{H}_{r})}$
$\displaystyle=\sup_{\stackrel{{\scriptstyle
w_{1},\dots,w_{n}\in\mathcal{H}^{*}}}{{\sum\|w_{k}\|^{2}=1}}}\Big{\|}\Big{(}\sum_{j}\langle
w_{j},v_{ij}\rangle\Big{)}\Big{\|}$
$\displaystyle=\sup_{\stackrel{{\scriptstyle
w_{1},\dots,w_{n}\in\mathcal{H}^{*}}}{{\sum\|w_{k}\|^{2}=1}}}\big{\|}\big{(}\langle
w_{1},v_{i1}\rangle\big{)}\big{\|}$
$\displaystyle=\sup_{\stackrel{{\scriptstyle
w_{1},\dots,w_{n}\in\mathcal{H}^{*}}}{{\sum\|w_{k}\|^{2}=1}}}\Big{(}\sum_{i}|\langle
w_{1},v_{i1}\rangle|^{2}\Big{)}^{1/2}$
$\displaystyle\stackrel{{\scriptstyle(a)}}{{=}}\sup_{\stackrel{{\scriptstyle
w_{1},\dots,w_{n}\in\mathcal{H}^{*}}}{{\sum\|w_{k}\|^{2}=1}}}\|w_{1}\|_{\mathcal{H}^{*}}$
$\displaystyle=1,$

where in (a) we use the fact that $\\{v_{11},\dots,v_{n1}\\}$ is orthonormal.

For simplicity of notation and later use, we denote

$\mathcal{C}=(\ell_{2})_{c},\quad\mathcal{R}=(\ell_{2})_{r}.$

In the finite dimension case of $\mathcal{H}=\ell_{2}^{d}$ we denote the
associated spaces by $\mathcal{C}^{d},\mathcal{R}^{d}$. We note that
$\mathcal{C}$ can be identified with the space of column vectors, and
$\mathcal{R}$ can be identified with the space of row vectors. In the
following we denote elements of $\mathcal{C},\mathcal{R}$ with superscripts
$v^{c},w^{r}$ to highlight that they are column and row vectors, respectively.

######  Remark 3.4.

In the following, we will also need $(\ell^{*}_{2})_{c},(\ell_{2}^{*})_{r}$.
Nevertheless, since these spaces can again be identified by column and row
vectors, we abuse the above notations and denote them by
$\mathcal{C},\mathcal{R}$, respectively.

##  4\. Haagerup tensor product

Let $X=[X_{ij}]\in M_{n,m}(\mathcal{X})$ and $Y=[Y_{ij}]\in
M_{m,n}(\mathcal{Y})$, where $\mathcal{X},\mathcal{Y}$ are two operator
spaces. Then, the element $X\odot Y\in M_{n}(\mathcal{X}\otimes\mathcal{Y})$
is defined by

$X\odot Y:=\left[\sum_{k=1}^{m}X_{ik}\otimes Y_{kj}\right].$

Using this notation for $Z\in M_{n}(\mathcal{X}\otimes\mathcal{Y})$, we define
its _Haagerup matrix norm_ by

$\|Z\|_{h,n}:=\inf\big{\\{}\|X\|_{n,m}\|Y\|_{m,n}:Z=X\odot Y\big{\\}},$

where the infimum is taken over all $m$ and all $X,Y$ as above.

We verify that $\|\cdot\|_{h,n}$ satisfies the triangle inequality and is thus
an actual norm [14]. For $Z_{i}\in M_{n}(\mathcal{X}\otimes\mathcal{Y})$,
$i=1,2$, and arbitrary $\epsilon>0$, let $X_{i}\in M_{n,m_{i}}(\mathcal{X})$
and $Y_{i}\in M_{m_{i},n}(\mathcal{Y})$ be such that $Z_{i}=X_{i}\odot Y_{i}$
and

$\displaystyle\|X_{i}\|_{n,m_{i}}\|Y_{i}\|_{m_{i},n}\leq\|Z_{i}\|_{h,n}+\epsilon.$
(4.1)

After scaling $X_{i},Y_{i}$, we can assume that
$\|X_{i}\|_{n,m_{i}}=\|Y_{i}\|_{m_{i},n}$. Next, define

$\widetilde{X}=\big{[}X_{1},X_{2}\big{]}\in
M_{n,m_{1}+m_{2}}(\mathcal{X}),\qquad\widetilde{Y}=\begin{bmatrix}Y_{1}\\\
Y_{2}\end{bmatrix}\in M_{m_{1}+m_{2},n}(\mathcal{Y}).$

Then, we have $Z_{1}+Z_{2}=\widetilde{X}\odot\widetilde{Y}$, and

$\displaystyle\|Z_{1}+Z_{2}\|_{h,n}$
$\displaystyle\leq\big{\|}\widetilde{X}\big{\|}_{n,m_{1}+m_{2}}\big{\|}\widetilde{Y}\big{\|}_{m_{1}+m_{2},n}$
$\displaystyle=\big{\|}\widetilde{X}\widetilde{X}^{*}\big{\|}_{n}^{1/2}\big{\|}\widetilde{Y}^{*}\widetilde{Y}\big{\|}_{n}^{1/2}$
$\displaystyle=\big{\|}X_{1}X_{1}^{*}+X_{2}X_{2}^{*}\big{\|}_{n}^{1/2}\big{\|}Y^{*}_{1}Y_{1}+Y^{*}_{2}Y_{2}\big{\|}_{n}^{1/2}$
$\displaystyle\leq\frac{1}{2}\Big{(}\big{\|}X_{1}X_{1}^{*}+X_{2}X_{2}^{*}\big{\|}_{n}+\big{\|}Y^{*}_{1}Y_{1}+Y^{*}_{2}Y_{2}\big{\|}_{n}\Big{)}$
$\displaystyle\leq\frac{1}{2}\Big{(}\big{\|}X_{1}X_{1}^{*}\big{\|}_{n}+\big{\|}X_{2}X_{2}^{*}\big{\|}_{n}+\big{\|}Y^{*}_{1}Y_{1}\big{\|}_{n}+\big{\|}Y^{*}_{2}Y_{2}\big{\|}_{n}\Big{)}$
$\displaystyle\leq\frac{1}{2}\Big{(}\big{\|}X_{1}\big{\|}^{2}_{n}+\big{\|}X_{2}\big{\|}^{2}_{n}+\big{\|}Y_{1}\big{\|}^{2}_{n}+\big{\|}Y_{2}\big{\|}^{2}_{n}\Big{)}$
$\displaystyle\leq\|Z_{1}\|_{h,n}+\|Z_{2}\|_{h,n}+2\epsilon,$

where in the last line we use (4.1) and
$\|X_{i}\|_{n,m_{i}}=\|Y_{i}\|_{m_{i},n}$. Then,
$\|Z_{1}+Z_{2}\|_{h,n}\leq\|Z_{1}\|_{h,n}+\|Z_{2}\|_{h,n}$ since $\epsilon>0$
is arbitrary.

It can also be verified that this matrix norm structure satisfies Ruan’s
axioms. Thus, after completion in this norm, we obtain an operator space
structure on $\mathcal{X}\otimes\mathcal{Y}$ called the _Haagerup tensor
product_ which we denote by $\mathcal{X}\otimes_{h}\mathcal{Y}$.

The Haagerup tensor product is associative, that is,
$(\mathcal{X}\otimes_{h}\mathcal{Y})\otimes_{h}\mathcal{Z}=\mathcal{X}\otimes_{h}(\mathcal{Y}\otimes_{h}\mathcal{Z})$
completely isometrically. Nonetheless, it is _not_ commutative as in general
$\mathcal{X}\otimes_{h}\mathcal{Y}$ and $\mathcal{Y}\otimes_{h}\mathcal{X}$
carry different norms.

###  4.1. Haagerup tensor product of column and row operator spaces

In the following proposition we collect some properties of Haagerup tensor
product with respect to column and row Hilbert spaces, which will be used in
the next section.

######  Proposition 4.1.

Let $\mathcal{H},\mathcal{K}$ be two Hilbert spaces. Then,

  * (i)

we have the complete isometries

$\displaystyle\mathcal{H}_{c}\otimes_{h}\mathcal{K}_{c}=(\mathcal{H}\otimes_{2}\mathcal{K})_{c}\quad\text{
and
}\quad\mathcal{K}_{r}\otimes_{h}\mathcal{H}_{r}=(\mathcal{H}\otimes_{2}\mathcal{K})_{r},$
(4.2)

  * (ii)

and

$\displaystyle\mathcal{S}_{\infty}(\mathcal{H})=\mathcal{H}_{c}\otimes_{h}(\mathcal{H}^{*})_{r}\quad\text{
and
}\quad\mathcal{S}_{1}(\mathcal{H})=\mathcal{H}_{r}\otimes_{h}(\mathcal{H}^{*})_{c}.$
(4.3)

In particular, $\mathcal{S}_{\infty}=C\otimes_{h}\mathcal{R}$ and
$\mathcal{S}_{1}=\mathcal{R}\otimes_{h}\mathcal{C}$ completely isometrically.

  * (iii)

Moreover,
$M_{1,d}(\mathcal{C}^{d})=M_{d,1}(\mathcal{R}^{d})=\mathcal{S}_{\infty}^{d}$
and $M_{1,d}(\mathcal{R}^{d})=M_{d,1}(\mathcal{C}^{d})=\mathcal{S}_{2}^{d}$
isometrically.

###### Proof.

Part (i) is straightforward. We prove (ii) in the special case of
$\mathcal{H}=\ell_{2}^{d}$; generalization to arbitrary $\mathcal{H}$ is
similar.

We first show that
$\mathcal{S}_{\infty}^{d}=\mathcal{C}^{d}\otimes_{h}\mathcal{R}^{d}$. Recall
that $\mathcal{C}^{d}$ is the space of column vectors, and $\mathcal{R}^{d}$
is the space of row vectors. Then, for $v^{c}\in\mathcal{C}^{d}$ and
$w^{r}\in\mathcal{R}^{d}$, their multiplication $v^{c}w^{r}$ makes sense and
forms a $d\times d$ matrix, i.e., an element of $\mathcal{S}_{\infty}^{d}$.
Define the linear map
$\phi:\mathcal{C}^{d}\otimes_{h}\mathcal{R}^{d}\to\mathcal{S}_{\infty}^{d}$ by

$\phi(v^{c}\otimes w^{r}):=v^{c}w^{r}.$

We claim that $\phi$ is a complete isometry between
$\mathcal{C}^{d}\otimes_{h}\mathcal{R}^{d}$ and $\mathcal{S}_{\infty}^{d}$. To
this end, we prove that $\phi=\phi_{1}$ is an isometry; extending the proof
for $\phi_{n}$, the $n$-th amplification of $\phi$, is immediate.

Let $Z\in\mathcal{C}^{d}\otimes_{h}\mathcal{R}^{d}$. Then

$\displaystyle\|Z\|_{\mathcal{C}^{d}\otimes_{h}\mathcal{R}^{d}}$
$\displaystyle=\inf\big{\\{}\|X\|_{M_{1,m}(\mathcal{C}^{d})}\|Y\|_{M_{m,1}(\mathcal{R}^{d})}:\,X\in
M_{1,m}(\mathcal{C}^{d}),Y\in M_{m,1}(\mathcal{R}^{d}),Z=X\odot Y\big{\\}}.$

Put

$X=[v_{1}^{c}\dots v_{m}^{c}]\quad\text{ and }\quad
Y=\begin{bmatrix}w_{1}^{r}\\\ \vdots\\\ w_{m}^{r}\end{bmatrix}.$

Then,

$\|X\|_{M_{1,m}(\mathcal{C}^{d})}=\big{\|}[v_{1}^{c}\dots
v_{m}^{c}]\big{\|}_{d,m}\quad\text{ and
}\quad\|Y\|_{M_{m,1}(\mathcal{R}^{d})}=\left\|\begin{bmatrix}w_{1}^{r}\\\
\vdots\\\ w_{m}^{r}\end{bmatrix}\right\|_{m,d}.$

On the other hand, $Z=X\odot Y$ corresponds to
$\phi(Z)=\sum_{i}v^{c}_{i}w^{r}_{i}$. Therefore,
$\|Z\|_{\mathcal{C}^{d}\otimes_{h}\mathcal{R}^{d}}$ equals the infimum of

$\big{\|}[v_{1}^{c}\dots
v_{m}^{c}]\big{\|}_{d,m}\left\|\begin{bmatrix}w^{r}_{1}\\\ \vdots\\\
w_{m}^{r}\end{bmatrix}\right\|_{m,d},$

subject to $\phi(Z)=\sum_{i}v^{c}_{i}w^{r}_{i}$. Note that this is nothing but
the operator norm of $\phi(Z)$, i.e.,
$\|\phi(Z)\|_{\mathcal{S}_{\infty}^{d}}$.

Next, we show that
$\mathcal{S}_{1}^{d}=\mathcal{R}^{d}\otimes_{h}\mathcal{C}^{d}$. As before,
thinking of $\mathcal{R}^{d}$ and $\mathcal{C}^{d}$ as row and column vectors,
we can define the linear map
$\psi:\mathcal{R}^{d}\otimes_{h}\mathcal{C}^{d}\to\mathcal{S}_{1}^{d}$ by

$\psi(v^{r}\otimes w^{c}):=w^{c}v^{r}.$

Again we only show that $\psi$ is an isometry. To see this, let
$Z\in\mathcal{R}^{d}\otimes_{h}\mathcal{C}^{d}$. Then

$\|Z\|_{\mathcal{R}^{d}\otimes_{h}\mathcal{C}^{d}}=\inf\\{\|X\|_{M_{1,m}(\mathcal{R}^{d})}\|Y\|_{M_{m,1}(\mathcal{C}^{d})}:\,X\in
M_{1,m}(\mathcal{R}^{d}),Y\in M_{m,1}(\mathcal{C}^{d}),Z=X\odot Y\\}.$

Put

$X=[v_{1}^{c}\dots v_{m}^{c}]\quad\text{ and }\quad
Y=\begin{bmatrix}w_{1}^{c}\\\ \vdots\\\ w_{m}^{c}\end{bmatrix}.$

Then,

$\displaystyle\|X\|_{M_{1,m}(\mathcal{R}^{d})}=\big{\|}[v_{1}^{c}\dots
v_{m}^{c}]\big{\|}_{1,md}=\Big{(}\sum_{i}\|v^{r}_{i}\|^{2}\Big{)}^{1/2},$
(4.4)

and

$\displaystyle\|Y\|_{M_{m,1}(\mathcal{C}^{d})}=\left\|\begin{bmatrix}w^{c}_{1}\\\
\vdots\\\
w^{c}_{m}\end{bmatrix}\right\|_{md,1}=\Big{(}\sum_{i}\|w^{c}_{i}\|^{2}\Big{)}^{1/2}.$
(4.5)

On the other hand, $Z=X\odot Y$ corresponds to
$\psi(Z)=\sum_{i}w^{c}_{i}v^{r}_{i}$. Therefore,
$\|Z\|_{\mathcal{R}^{d}\otimes_{h}\mathcal{C}^{d}}$ equals the infimum of

$\Big{(}\sum_{i}\|v^{r}_{i}\|^{2}\Big{)}^{1/2}\Big{(}\sum_{i}\|w^{c}_{i}\|^{2}\Big{)}^{1/2},$

subject to $\psi(Z)=\sum_{i}w^{c}_{i}v^{r}_{i}$. An application of Hölder’s
inequality shows that this is equal to $\|\psi(Z)\|_{\mathcal{S}_{1}^{d}}$.

Part (iii) follows readily once we note that $M_{1,d}(\mathcal{C}^{d})$ and
$M_{d,1}(\mathcal{R}^{d})$ are naturally identified with the space of $d\times
d$ matrices, and by definition, their norms are the same as the norm of
$\mathcal{S}_{\infty}^{d}$. The isometries
$M_{1,d}(\mathcal{R}^{d})=M_{d,1}(\mathcal{C}^{d})=\mathcal{S}_{2}^{d}$ are
proven essentially in (4.4) and (4.5). ∎

###  4.2. Interpolation and the Haagerup tensor product

Let $(\mathcal{X}_{0},\mathcal{X}_{1})$ and
$(\mathcal{Y}_{0},\mathcal{Y}_{1})$ be two compatible couples of operator
spaces. Then, it can be verified that
$(\mathcal{X}_{0}\otimes_{h}\mathcal{Y}_{0},\mathcal{X}_{1}\otimes_{h}\mathcal{Y}_{1})$
is also a compatible couple of operator spaces. The following theorem
describes interpolation spaces obtained from this couple.

######  Theorem 4.2 ([32, 21]).

Let $(\mathcal{X}_{0},\mathcal{X}_{1})$ and
$(\mathcal{Y}_{0},\mathcal{Y}_{1})$ be two compatible couples of operator
spaces. Then, for every $0<\theta<1$, we have a complete isometry

$[\mathcal{X}_{0}\otimes_{h}\mathcal{Y}_{0},\mathcal{X}_{1}\otimes_{h}\mathcal{Y}_{1}]_{\theta}=[\mathcal{X}_{0},\mathcal{X}_{1}]_{\theta}\otimes_{h}[\mathcal{Y}_{0},\mathcal{Y}_{1}]_{\theta}.$
(4.6)

$($i.e., the Haagerup tensor product and interpolation are commuting
operations.$)$

This theorem plays a pivotal role in the construction and understanding of
operator-valued Schatten spaces. Its proof is based on the _Wiener–Masani
theorem_. Here, assuming the Wiener–Masani theorem, we present the proof in
the finite-dimensional case based on [21]. Also, due to its important, we
devote Appendix B to the proof of the Wiener–Masani theorem.

###### Proof.

To simplify notation, we denote
$\mathcal{X}_{\theta}=[\mathcal{X}_{0},\mathcal{X}_{1}]_{\theta}$,
$\mathcal{Y}_{\theta}=[\mathcal{Y}_{0},\mathcal{Y}_{1}]_{\theta}$ and
$\mathcal{Z}_{\theta}=[\mathcal{Z}_{0},\mathcal{Z}_{1}]_{\theta}$ where
$\mathcal{Z}_{j}=\mathcal{X}_{j}\otimes_{h}\mathcal{Y}_{j}$, $j=0,1$. Then,
(4.6) is tantamount to

$\|W\|_{M_{n}(\mathcal{Z}_{\theta})}=\|W\|_{M_{n}(\mathcal{X}_{\theta}\otimes_{h}\mathcal{Y}_{\theta})},\
\ \ n\geq 1.$

We assume that all the spaces are finite-dimensional,101010For a
generalization to infinite dimensions the reader is referred to [32]. and
prove this identity only for $n=1$, as the generalization to $n>1$ is
straightforward.

Let us first show that
$\|W\|_{\mathcal{Z}_{\theta}}\leq\|W\|_{\mathcal{X}_{\theta}\otimes_{h}\mathcal{Y}_{\theta}}$.
Let $\epsilon>0$ be arbitrary. By definition we have

$\|W\|_{\mathcal{X}_{\theta}\otimes_{h}\mathcal{Y}_{\theta}}=\inf\big{\\{}\|X\|_{M_{1,n}(\mathcal{X}_{\theta})}\cdot\|Y\|_{M_{n,1}(\mathcal{Y}_{\theta})}:W=X\odot
Y\big{\\}},$

and

$\|X\|_{M_{1,n}(\mathcal{X}_{\theta})}=\|X\|_{[M_{1,n}(\mathcal{X}_{0}),M_{1,n}(\mathcal{X}_{1})]_{\theta}}.$

Therefore, by definition of the norm in the interpolation space, there is a
holomorphic function $f(z)$ such that $f(\theta)=X$ and

$\Big{(}\sup_{t}\|f(it)\|_{M_{1,n}(\mathcal{X}_{0})}\Big{)}^{1-\theta}\Big{(}\sup_{t}\|f(1+it)\|_{M_{1,n}(\mathcal{X}_{1})}\Big{)}^{\theta}<(1+\epsilon)\|X\|_{M_{1,n}(\mathcal{X}_{\theta})}.$
(4.7)

Similarly, there is a holomorphic function $g(z)$ such that $g(\theta)=Y$ and

$\Big{(}\sup_{t}\|g(it)\|_{M_{n,1}(\mathcal{Y}_{0})}\Big{)}^{1-\theta}\Big{(}\sup_{t}\|g(1+it)\|_{M_{n,1}(\mathcal{Y}_{1})}\Big{)}^{\theta}<(1+\epsilon)\|Y\|_{M_{n,1}(\mathcal{Y}_{\theta})}.$
(4.8)

Then, the function $h(z):=f(z)\odot g(z)$ is holomorphic, $h(\theta)=W$, and

$\|W\|_{\mathcal{Z}_{\theta}}\leq\Big{(}\sup_{t}\|h(it)\|_{\mathcal{Z}_{0}}\Big{)}^{1-\theta}\Big{(}\sup_{t}\|h(1+it)\|_{\mathcal{Z}_{1}}\Big{)}^{\theta}.$
(4.9)

Since $\mathcal{Z}_{j}=\mathcal{X}_{j}\otimes_{h}\mathcal{Y}_{j}$ ($j=0,1$),
by definition of the Haagerup norm we have

$\|h(j+it)\|_{\mathcal{Z}_{j}}\leq\|f(j+it)\|_{M_{1,n}(\mathcal{X}_{j})}\cdot\|g(j+it)\|_{M_{n,1}(\mathcal{Y}_{j})}.$
(4.10)

Finally, combining (4.7), (4.8), (4.9), (4.10) and letting
$\epsilon\rightarrow 0$ yields

$\|W\|_{\mathcal{Z}_{\theta}}\leq\|X\|_{M_{1,n}(\mathcal{X}_{\theta})}\cdot\|Y\|_{M_{n,1}(\mathcal{Y}_{\theta})},$

which in turn gives us the desired inequality upon minimizing over all $X,Y$
as above.

To prove the reversed inequality, we assume without loss of generality that
$\mathcal{X}_{0},\mathcal{X}_{1}$ have the same dimension, say $d$, and
identify them with $M_{d,1}$ as vector spaces. Likewise, we identify
$\mathcal{Y}_{0},\mathcal{Y}_{1}$ with $M_{1,d^{\prime}}$ as vector spaces
with the same dimension $d^{\prime}$. Then, considering the canonical
identifications

$M_{1,n}(\mathcal{X}_{b})\cong M_{1,n}\otimes M_{d,1}\cong M_{d,n},$

and

$M_{n,1}(\mathcal{Y}_{b})\cong M_{n,1}\otimes M_{1,d^{\prime}}\cong
M_{n,d^{\prime}},$

we observe that for $\widetilde{X}\in M_{1,n}(\mathcal{X}_{b})$ and
$\widetilde{Y}\in M_{n,1}(\mathcal{Y}_{b})$, we have
$\widetilde{X}\odot\widetilde{Y}=XY$ where $X\in M_{d,n}$ and $Y\in
M_{n,d^{\prime}}$ correspond, respectively, to $\widetilde{X}$ and
$\widetilde{Y}$ via the above-mentioned identifications. In the rest of the
proof, we assume that $d\leq d^{\prime}$. Also, applying the polar
decomposition we assume with no loss of generalization that $n=d$.

Let $\epsilon>0$ be arbitrary. Then, by definition there is a holomorphic
function $h(z)$ such that $h(\theta)=W$ and

$\max\big{\\{}\sup_{t}\|h(b+it)\|_{\mathcal{Z}_{b}}:b=0,1\big{\\}}<(1+\epsilon)\|W\|_{\mathcal{Z}_{\theta}}.$
(4.11)

Note that

$\|h(b+it)\|_{\mathcal{Z}_{b}}=\inf\big{\\{}\|A\|_{M_{1,d}(\mathcal{X}_{b})}\cdot\|B\|_{M_{d,1}(\mathcal{Y}_{b})}:\,h(b+it)=AB,A\in
M_{d,d},B\in M_{d,d^{\prime}}\big{\\}},$

where as mentioned above we have identified $M_{1,d}(\mathcal{X}_{b})\cong
M_{d,d}$ and $M_{d,1}(\mathcal{Y}_{b})\cong M_{d,d^{\prime}}$. Then, applying
Lemma A.1, we obtain bounded measurable functions
$A^{(b)}(b+it),B^{(b)}(b+it)$ such that for every $t\in\mathbb{R}$ and
$b=0,1$,

$h(b+it)=A^{(b)}(b+it)B^{(b)}(b+it),$ (4.12)
$\|A^{(b)}(b+it)\|_{M_{1,d}(\mathcal{X}_{b})}\leq 1+\frac{\epsilon}{2},$
(4.13)
$\|B^{(b)}(b+it)\|_{M_{d,1}(\mathcal{Y}_{b})}\leq\|h(b+it)\|_{\mathcal{Z}_{b}}.$
(4.14)

Next, using the Wiener–Masani theorem (see Corollary B.10) we can find a
bounded holomorphic function $F(z)$ such that for almost every
$t\in\mathbb{R}$,

$F(b+it)F(b+it)^{*}=A^{(b)}(b+it)A^{(b)}(b+it)^{*}+\frac{\epsilon^{2}}{4\|I\|_{M_{1,d}(\mathcal{X}_{b})}^{2}}I.$
(4.15)

It follows from (4.13), (4.15) and the $C^{*}$-property of the norms that for
almost every $t\in\mathbb{R}$,

$\|F(b+it)\|_{M_{1,d}(\mathcal{X}_{b})}\leq\left(\|A^{(b)}(b+it)\|_{M_{1,d}(\mathcal{X}_{b})}^{2}+\frac{\epsilon^{2}}{4}\right)^{\frac{1}{2}}\leq
1+\epsilon.$ (4.16)

We note that by the Wiener–Masani theorem $F(z)$ is invertible almost
everywhere. Thus, we can define measurable functions $S^{(b)}(\cdot)$,
$b=0,1$, such that

$S^{(b)}(b+it)=F(b+it)^{-1}A^{(b)}(b+it),$

for almost every $t$. Then, $S^{(b)}(b+it)S^{(b)}(b+it)^{*}\leq I$ for almost
every $t$ since $A^{(b)}(b+it)A^{(b)}(b+it)^{*}\leq F(b+it)F(b+it)^{*}$.

Now, define the holomorphic function $G(z)$ on $\mathbb{S}$ by

$G(z):=F(z)^{-1}\cdot h(z).$

By (4.12) and the preceding paragraph we obtain

$\displaystyle G(b+it)$ $\displaystyle=F(z)^{-1}\cdot
A^{(b)}(b+it)B^{(b)}(b+it)$ $\displaystyle=S^{(b)}(b+it)B^{(b)}(b+it).$

This implies that $G(z)$ is bounded, and by (4.14) for almost every $t$ we
have

$\|G(b+it)\|_{M_{d,1}(\mathcal{Y}_{b})}\leq\|B^{(b)}(b+it)\|_{M_{d,1}(\mathcal{Y}_{b})}\leq\|h(b+it)\|_{\mathcal{Z}_{b}}.$
(4.17)

Finally, using (4.11), (4.16), (4.17), and part (ii) of Lemma 2.5 we obtain

$\displaystyle\|W\|_{\mathcal{X}_{\theta}\otimes_{h}\mathcal{Y}_{\theta}}$
$\displaystyle=\|h(\theta)\|_{\mathcal{X}_{\theta}\otimes_{h}\mathcal{Y}_{\theta}}$
$\displaystyle\leq\|F(\theta)\|_{M_{1,d}(\mathcal{X}_{b})}\|G(\theta)\|_{M_{d,1}(\mathcal{Y}_{b})}$
$\displaystyle\leq\exp\sum_{b}\int
P_{b}(\theta,s)\log\Big(\|F(b+is)\|_{M_{1,d}(\mathcal{X}_{b})}\|G(b+is)\|_{M_{d,1}(\mathcal{Y}_{b})}\Big{missing})\differential
s$ $\displaystyle\leq\exp\sum_{b}\int
P_{b}(\theta,s)\log\Big((1+\epsilon)\|h(b+is)\|_{\mathcal{Z}_{b}}\Big{missing})\differential
s$ $\displaystyle\leq\exp\sum_{b}\int
P_{b}(\theta,s)\log\bigg((1+\epsilon)\Big{(}\sup_{t}\|h(b+it)\|_{\mathcal{Z}_{b}}\Big{)}\bigg{missing})\differential
s$
$\displaystyle=(1+\epsilon)\Big{(}\sup_{t}\|h(it)\|_{\mathcal{Z}_{0}}\Big{)}^{1-\theta}\Big{(}\sup_{t}\|h(1+it)\|_{\mathcal{Z}_{1}}\Big{)}^{\theta}$
$\displaystyle\leq(1+\epsilon)\max\big{\\{}\sup_{t}\|h(b+it)\|_{\mathcal{Z}_{b}}:b=0,1\big{\\}}$
$\displaystyle\leq(1+\epsilon)^{2}\|W\|_{\mathcal{Z}_{\theta}}.$

Letting $\epsilon\rightarrow 0$, the desired inequality is obtained. ∎

For a Hilbert space $\mathcal{H}$, the operator spaces
$(\mathcal{H}_{c},\mathcal{H}_{r})$ form a compatible couple by identifying
$v^{c}\in\mathcal{H}_{c}$ and $v^{r}\in\mathcal{H}_{r}$, for
$v\in\mathcal{H}$. Then, we may consider their interpolation spaces. Moreover,
by the above theorem
$(\mathcal{H}_{c}\otimes_{h}(\mathcal{H}^{*})_{r},\mathcal{H}_{r}\otimes_{h}(\mathcal{H}^{*})_{c})$
is a compatible couple and by equations (3.2) and (4.3) we have

$\displaystyle\mathcal{S}_{p}(\mathcal{H})$
$\displaystyle=\big{[}\mathcal{S}_{\infty}(\mathcal{H}),\mathcal{S}_{1}(\mathcal{H})\big{]}_{1/p}$
$\displaystyle=\big{[}\mathcal{H}_{c}\otimes_{h}(\mathcal{H}^{*})_{r},\mathcal{H}_{r}\otimes_{h}(\mathcal{H}^{*})_{c}\big{]}_{1/p}$
$\displaystyle=\big{[}\mathcal{H}_{c},\mathcal{H}_{r}\big{]}_{1/p}\otimes_{h}\big{[}(\mathcal{H}^{*})_{r},(\mathcal{H}^{*})_{c}\big{]}_{1/p},$

completely isometrically. This equation is a generalization of (4.3) for
arbitrary $p$, and motivates defining:

$\mathcal{R}(\theta):=[\mathcal{R},\mathcal{C}]_{\theta}=[\mathcal{C},\mathcal{R}]_{1-\theta},\
\ \ \mathcal{R}(0):=\mathcal{R},\ \ \ \mathcal{R}(1):=\mathcal{C}.$ (4.18)

Then, we have

$\displaystyle\mathcal{S}_{p}=\mathcal{R}\Big{(}1-\frac{1}{p}\Big{)}\otimes_{h}\mathcal{R}\Big{(}\frac{1}{p}\Big{)}.$
(4.19)

##  5\. Operator-valued Schatten spaces

Our goal in this section is to construct and study
$\mathcal{S}_{p}[\mathcal{X}]$, for an operator space $\mathcal{X}$, as the
space of matrices with entries in $\mathcal{X}$ equipped with a norm that is
closely related to the $p$-Schatten norm. Construction of this space for
$p=\infty$ and finite-size matrices is easy; we can let
$\mathcal{S}_{\infty}^{d}[\mathcal{X}]$ be the same space as
$M_{d}(\mathcal{X})$ with the norm given by the operator space structure of
$\mathcal{X}$.111111This is the reason why we need operator spaces to define
vector-valued Schatten spaces. Definition of
$\mathcal{S}_{1}^{d}[\mathcal{X}]$ involves the _projective operator space
tensor product_ that has not been discussed in this paper. Then,
$\mathcal{S}_{p}[\mathcal{X}]$ can be defined using interpolation between
$\mathcal{S}_{\infty}^{d}[\mathcal{X}]$ and
$\mathcal{S}_{1}^{d}[\mathcal{X}]$. This is the approach of [33] for defining
operator-valued Schatten spaces. Here, based on the result of [33] we take a
different path to define these spaces that uses less tools and is easier to
understand.

######  Definition 5.1.

Let $\mathcal{X}$ be an operator space and $1\leq p\leq\infty$ The
$\mathcal{X}$-valued $p$-Schatten space $\mathcal{S}_{p}[\mathcal{X}]$ is
defined to be the operator space

$\mathcal{S}_{p}[\mathcal{X}]:=\mathcal{R}\Big{(}1-\frac{1}{p}\Big{)}\otimes_{h}\mathcal{X}\otimes_{h}\mathcal{R}\Big{(}\frac{1}{p}\Big{)},$

where $\mathcal{R}(\theta)$ is defined in (4.18).
$\mathcal{S}_{p}^{d}[\mathcal{X}]$ for $d\geq 1$ is defined similarly.

###  5.1. Pisier’s variational formula

Note that for $\mathcal{X}=\mathbb{C}$ this definition matches (4.19). Also,
as shown in the following, $\mathcal{S}_{\infty}[\mathcal{X}]$ can be
considered as the space of $\mathcal{X}$-valued matrices with the norm induced
by the operator space structure of $\mathcal{X}$.

######  Proposition 5.2.

Let $\\{e_{i}^{c}:\,i\\}$ and $\\{e^{r}_{j}:\,j\\}$ be computational bases for
$\mathcal{C}^{d}$ and $\mathcal{R}^{d}$, respectively. Define the linear map
$\phi:\mathcal{S}^{d}_{\infty}[\mathcal{X}]\to M_{d}(\mathcal{X})$ by

$\phi(e_{i}^{c}\otimes X\otimes e_{j}^{r})=(Y_{i^{\prime}j^{\prime}}),\qquad
Y_{i^{\prime}j^{\prime}}=\delta_{i,i^{\prime}}\delta_{j,j^{\prime}}X.$

Then, $\phi$ is a complete isometry.

###### Proof.

Here, we only show that $\phi$ is an isometry. By definition for
$T\in\mathcal{S}^{d}_{\infty}[\mathcal{X}]$ we have

$\|T\|_{\mathcal{S}^{d}_{\infty}[\mathcal{X}]}=\inf\Big{\\{}\|(a_{i}^{c})\|_{M_{1,n}(\mathcal{C}^{d})}\|(X_{ij})\|_{M_{n}(\mathcal{X})}\|(b_{j}^{r})\|_{M_{n,1}(\mathcal{R}^{d})}:\,T=\sum
a_{i}^{c}\otimes X_{ij}\otimes b_{j}^{r}\Big{\\}}.$

We note that $(a_{i}^{c})\in M_{1,n}(\mathcal{C}^{d})$ can be thought of as a
matrix $A\in M_{d,n}$ with the same norm (see, part (iii) of Proposition 4.1).
Similarly, $(b_{j}^{r})\in M_{n,1}$ can be thought of as a matrix $B\in
M_{n,d}$ with the same norm. Then, a simple computation shows that the
condition $T=\sum a_{i}^{c}\otimes X_{ij}\otimes b_{j}^{r}$ is equivalent to

$\phi(T)=A(X_{ij})B.$

On the other hand, since $\mathcal{X}$ is an operator space, by Ruan’s axioms
we have

$\|\phi(T)\|_{M_{d}(\mathcal{X})}\leq\|A\|_{d,n}\|(X_{ij})\|_{M_{n}(\mathcal{X})}\|B\|_{n,d}.$

Taking infimum over the representation $\phi(T)=A(X_{ij})B$, we find that
$\|T\|_{\mathcal{S}_{\infty}^{d}[\mathcal{X}]}=\|\phi(T)\|_{M_{d}(\mathcal{X})}$.
∎

Viewing elements of $\mathcal{S}_{p}[\mathcal{X}]$ as $\mathcal{X}$-valued
matrices goes beyond $p=\infty$. To this end, recall that
$\mathcal{R}(\theta)$ is the interpolation space of $\mathcal{C},\mathcal{R}$,
so the bases $\\{e_{i}^{c}:\,i\\}$ and $\\{e^{r}_{j}:\,j\\}$ considered in the
above proposition induce a basis for $\mathcal{R}(\theta)$. Then, let
$\\{e_{i}^{(1-\theta)}:\,i\\}$ and $\\{e_{i}^{\theta}:\,i\\}$ be the bases for
$\mathcal{R}^{d}(1-\theta)$ and $\mathcal{R}^{d}(\theta)$, respectively. Then,
we can define the map $\phi:\mathcal{S}_{p}[\mathcal{X}]\to
M_{d}(\mathcal{X})$

$\phi\big{(}e^{(1-\theta)}_{i}\otimes X\otimes
e^{\theta}_{j}\big{)}=(Y_{i^{\prime}j^{\prime}}),\qquad
Y_{i^{\prime}j^{\prime}}=\delta_{i,i^{\prime}}\delta_{j,j^{\prime}}X,$

and identify elements of $\mathcal{S}^{d}_{p}[\mathcal{X}]$ with those of
$M_{d}(\mathcal{X})$.131313We note that this identification does not preserve
the norm for arbitrary $p$. Similarly, $\mathcal{S}_{p}(\mathcal{X})$ can be
thought of as infinite-size matrices with entries in $\mathcal{X}$.

######  Theorem 5.3 (Pisier’s formula [33]).

Let $1\leq p<\infty$. Then, for every $T\in\mathcal{S}_{p}[\mathcal{X}]$ we
have

$\|T\|_{\mathcal{S}_{p}[\mathcal{X}]}=\inf\|A\|_{\mathcal{S}_{2p}}\|X\|_{\mathcal{S}_{\infty}[\mathcal{X}]}\|B\|_{\mathcal{S}_{2p}},$
(5.1)

where the infimum is taken over all $A,B\in\mathcal{S}_{2p}$ and
$X\in\mathcal{S}_{\infty}[\mathcal{X}]$ such that $T=AXB$. Here, as explained
above, we understand the identity $T=AXB$ by thinking of $T,X$ as
$\mathcal{X}$-valued matrices.

###### Proof.

Considering $\mathcal{S}_{p}^{d}[\mathcal{X}]$ that is easier to work with, we
note that this theorem is a generalization of Proposition 5.2. Indeed,
following the proof of Proposition 5.2 for arbitrary $p$, we note that
$\|(X_{ij})\|_{M_{n}(\mathcal{X})}=\|(X_{ij})\|_{\mathcal{S}^{d}_{\infty}(\mathcal{X})}$.
Then, we need to evaluate $\|A\|_{M_{1,n}(\mathcal{R}^{d}(1-\theta))}$ and
$\|B\|_{M_{n,1}(\mathcal{R}^{d}(\theta))}$, where $\theta=\frac{1}{p}$. We
first pad $A$ with zeros if necessary in order to assume that $n=d$. Next, by
the definition of the interpolation of operator spaces we have

$\displaystyle M_{1,n}(\mathcal{R}^{n}(1-\theta))$
$\displaystyle=\big{[}M_{1,n}(\mathcal{R}^{n}),M_{1,n}(\mathcal{C}^{n})\big{]}_{1-\theta}$
$\displaystyle=\big{[}\mathcal{S}_{2}^{n},\mathcal{S}_{\infty}^{n}\big{]}_{1-\theta}$
$\displaystyle=\mathcal{S}_{2p}^{n},$

where the second line follows from part (iii) of Theorem 4.1, and the third
line follows from (3.2) and the reiteration theorem. Therefore,
$\|A\|_{M_{1,n}(\mathcal{R}^{d}(1-\theta))}=\|A\|_{\mathcal{S}_{2p}^{n}}$.
Similarly, it can shown that $\|B\|_{M_{n,1}(\mathcal{R}^{d}(\theta))}$ equals
the $2p$-norm of $B$. Then, the desired result follows.

∎

######  Lemma 5.4.

Suppose that $T\in\mathcal{S}_{p}[\mathcal{X}]$ is self-adjoint, i.e.,
$T^{*}=T$. Then, in (5.1) we can assume with no loss of generality that
$B=A^{*}$.

###### Proof.

We need to show that for any $\epsilon>0$ there is a factorization
$T=\hat{A}\hat{X}\hat{A}^{*}$ such that

$\|T\|_{\mathcal{S}_{p}[\mathcal{X}]}+\epsilon\geq\|\hat{A}\|_{\mathcal{S}_{2p}}\|\hat{X}\|_{\mathcal{S}_{\infty}[\mathcal{X}]}\|\hat{A}^{*}\|_{\mathcal{S}_{2p}}.$

Following Theorem 5.3 assume that $T=AXB$ is such that
$\|A\|_{\mathcal{S}_{2p}}=\|B\|_{\mathcal{S}_{2p}}$ and

$\|T\|_{\mathcal{S}_{p}[\mathcal{X}]}+\epsilon\geq\|A\|_{\mathcal{S}_{2p}}\|X\|_{\mathcal{S}_{\infty}[\mathcal{X}]}\|B\|_{\mathcal{S}_{2p}}.$

Define, $\hat{A}=(A~{}B^{*})$ and

$\hat{X}=\frac{1}{2}\begin{bmatrix}0&X\\\ X^{*}&0\end{bmatrix}.$

Then, we have $T=\hat{A}\hat{X}\hat{A}^{*}$, and

$\displaystyle\|\hat{A}\|_{\mathcal{S}_{2p}}\|\hat{X}\|_{\mathcal{S}_{\infty}[\mathcal{X}]}\|\hat{A}^{*}\|_{\mathcal{S}_{2p}}$
$\displaystyle=\frac{1}{2}\|AA^{*}+B^{*}B\|_{S_{p}}\|X\|_{\mathcal{S}_{\infty}[\mathcal{X}]}$
$\displaystyle\leq\frac{1}{2}\big{(}\|AA^{*}\|_{\mathcal{S}_{p}}+\|B^{*}B\|_{\mathcal{S}_{p}}\big{)}\|X\|_{\mathcal{S}_{\infty}[\mathcal{X}]}$
$\displaystyle=\|A\|_{\mathcal{S}_{2p}}\|B\|_{\mathcal{S}_{2p}}\|X\|_{\mathcal{S}_{\infty}[\mathcal{X}]}$
$\displaystyle\leq\|T\|_{\mathcal{S}_{p}[\mathcal{X}]}+\epsilon,$

where in the third line we use
$\|A\|_{\mathcal{S}_{2p}}=\|B\|_{\mathcal{S}_{2p}}$.

∎

###  5.2. Reiteration and duality

The following theorem is a generalization of (3.2).

######  Theorem 5.5.

Let $(\mathcal{X}_{0},\mathcal{X}_{1})$ be a compatible couple of operator
spaces and $1\leq p_{0},p_{1}\leq\infty$. Then, we have completely
isometrically

$\mathcal{S}_{p_{\theta}}[\mathcal{X}_{\theta}]=\big{[}\mathcal{S}_{p_{0}}[\mathcal{X}_{0}],\,\mathcal{S}_{p_{1}}[\mathcal{X}_{1}]\big{]}_{\theta},$
(5.2)

where $p_{\theta}$ is given by
$\frac{1}{p_{\theta}}=\frac{1-\theta}{p_{0}}+\frac{\theta}{p_{1}}$ and
$\mathcal{X}_{\theta}=[\mathcal{X}_{0},\mathcal{X}_{1}]_{\theta}$. In
particular, if $\mathcal{X}$ is an operator space, we have completely
isometrically

$\mathcal{S}_{p_{\theta}}[\mathcal{X}]=\big{[}\mathcal{S}_{p_{0}}[\mathcal{X}],\,\mathcal{S}_{p_{1}}[\mathcal{X}]\big{]}_{\theta}.$
(5.3)

###### Proof.

Using the reiteration theorem (2.2), it suffices to prove the theorem for
$p_{0}=\infty$ and $p_{1}=1$. For this special case we have
$p_{\theta}=1/\theta$ and by Theorem 4.2

$\displaystyle\mathcal{S}_{p_{\theta}}[\mathcal{X}_{\theta}]$
$\displaystyle=\mathcal{R}(1-\theta)\otimes_{h}\Big{(}\mathcal{X}_{\theta}\otimes_{h}\mathcal{R}(\theta)\Big{)}$
$\displaystyle=\mathcal{R}(1-\theta)\otimes_{h}\Big{[}\mathcal{X}_{0}\otimes_{h}\mathcal{R}(0),\,\mathcal{X}_{1}\otimes_{h}\mathcal{R}(1)\Big{]}_{\theta}$
$\displaystyle=\Big{[}\mathcal{R}(1)\otimes_{h}\mathcal{X}_{0}\otimes_{h}\mathcal{R}(0),\,\mathcal{R}(0)\otimes_{h}\mathcal{X}_{1}\otimes_{h}\mathcal{R}(1)\Big{]}_{\theta}$
$\displaystyle=\big{[}\mathcal{C}\otimes_{h}\mathcal{X}_{0}\otimes_{h}\mathcal{R},\,\mathcal{R}\otimes_{h}\mathcal{X}_{1}\otimes_{h}\mathcal{C}\big{]}_{\theta}$
$\displaystyle=\big{[}\mathcal{S}_{\infty}[\mathcal{X}_{0}],\,\mathcal{S}_{1}[\mathcal{X}_{1}]\big{]}_{\theta}.$

∎

The following proposition is the generalization of Hölder’s duality.

######  Proposition 5.6.

Let $1<p\leq\infty$ and $1=\frac{1}{p}+\frac{1}{p^{\prime}}$. Then,

$\mathcal{S}_{p}[\mathcal{X}]^{*}=\mathcal{S}_{p^{\prime}}[\mathcal{X}^{*}],$

completely isometrically.

###### Proof.

Again, we prove the theorem in the finite-dimensional case, i.e.,
$\mathcal{S}_{p}^{d}[\mathcal{X}]^{*}=\mathcal{S}^{d}_{p^{\prime}}[\mathcal{X}^{*}]$.
We know that for operator spaces $\mathcal{Y}_{1},\mathcal{Y}_{2}$ which _one_
of them is finite-dimensional we have
$(\mathcal{Y}_{1}\otimes_{h}\mathcal{Y}_{2})^{*}=\mathcal{Y}_{1}^{*}\otimes_{h}\mathcal{Y}_{2}^{*}$
completely isometrically [34, Corollary 5.8]. Using this result twice we
obtain

$\displaystyle\mathcal{S}^{d}_{p}[\mathcal{X}]^{*}$
$\displaystyle=\Big{(}\mathcal{R}^{d}(1-1/p)\otimes_{h}\mathcal{X}\otimes_{h}\mathcal{R}^{d}(1/p)\Big{)}^{*}$
$\displaystyle=\mathcal{R}^{d}(1-1/p)^{*}\otimes_{h}\Big{(}\mathcal{X}\otimes_{h}\mathcal{R}^{d}(1/p)\Big{)}^{*}$
$\displaystyle=\mathcal{R}^{d}(1-1/p)^{*}\otimes_{h}\mathcal{X}^{*}\otimes_{h}\mathcal{R}^{d}(1/p)^{*},$

completely isometrically. Next, we have

$\mathcal{R}^{d}(1/p)^{*}=[\mathcal{R}^{d},\mathcal{C}^{d}]_{1/p}^{*}=\big{[}\big{(}\mathcal{R}^{d}\big{)}^{*},\big{(}\mathcal{C}^{d}\big{)}^{*}\big{]}_{1/p}=\big{[}\mathcal{C}^{d},\mathcal{R}^{d}\big{]}_{1/p}=\big{[}\mathcal{R}^{d},\mathcal{C}^{d}\big{]}_{1/p^{\prime}}=\mathcal{R}^{d}(1/p^{\prime}),$

completely isometrically. Here, in the second equality we use the fact that
for a compatible couple $(\mathcal{Y}_{1},\mathcal{Y}_{2})$ of finite-
dimensional operator spaces we have
$[\mathcal{Y}_{1},\mathcal{Y}_{2}]_{\theta}^{*}=[\mathcal{Y}_{1}^{*},\mathcal{Y}_{2}^{*}]_{\theta}$
completely isometrically [34, Theorem 2.7.4]. We also for the third equality
use $(\mathcal{C}^{d})^{*}=\mathcal{R}^{d}$ and
$(\mathcal{R}^{d})^{*}=\mathcal{C}^{d}$ that are easy to prove [34, p. 41].
Then, we have

$\displaystyle\mathcal{S}^{d}_{p}[\mathcal{X}]^{*}=\mathcal{R}^{d}(1/p)\otimes_{h}\mathcal{X}^{*}\otimes_{h}\mathcal{R}^{d}(1/p^{\prime})=S^{d}_{p^{\prime}}[\mathcal{X}^{*}],$

completely isometrically. ∎

We can extend the definition of $\mathcal{S}_{p}[\mathcal{X}]$ for an
arbitrary Hilbert space $\mathcal{H}$. We may define

$\mathcal{S}_{\infty}(\mathcal{H})[\mathcal{X}]:=\mathcal{H}_{c}\otimes_{h}\mathcal{X}\otimes_{h}(\mathcal{H}^{*})_{r}$

and

$\mathcal{S}_{1}(\mathcal{H})[\mathcal{X}]:=\mathcal{H}_{r}\otimes_{h}\mathcal{X}\otimes_{h}(\mathcal{H}^{*})_{c}.$

Then $\mathcal{S}_{p}(\mathcal{H})[\mathcal{X}]$ may be defined, via
interpolation, by

$\mathcal{S}_{p}(\mathcal{H})[\mathcal{X}]:=\big{[}\mathcal{S}_{\infty}(\mathcal{H})[\mathcal{X}],\mathcal{S}_{1}(\mathcal{H})[\mathcal{X}]\big{]}_{\frac{1}{p}},$

which by Theorem 4.2 is equivalent to
$\mathcal{S}_{p}(\mathcal{H})[\mathcal{X}]=[\mathcal{H}_{c},\mathcal{H}_{r}]_{1/p}\otimes_{h}\mathcal{X}\otimes_{h}[(\mathcal{H}^{*})_{r},(\mathcal{H}^{*})_{c}]_{1/p}$.
We note that $\mathcal{S}_{p}(\mathcal{H})[\mathcal{X}]$ matches
$\mathcal{S}_{p}[\mathcal{X}]$ for $\mathcal{H}=\ell_{2}$ (see Remark 3.4).
Moreover, the preceding results can be readily extended to this more general
setting.

###  5.3. Pisier’s formula for the space $\mathcal{S}_{p}[\mathcal{S}_{q}]$

We now consider operator-valued Schatten spaces in the spacial case of
$\mathcal{X}=\mathcal{S}_{q}$.

######  Lemma 5.7 ([33]).

Let $\mathcal{X}$ be an operator space and $\mathcal{H},\mathcal{K}$ be
Hilbert spaces. Also, let $1\leq p\leq\infty$. Then, we have

$\mathcal{S}_{p}(\mathcal{H})[\mathcal{S}_{p}(\mathcal{K})[\mathcal{X}]]=\mathcal{S}_{p}(\mathcal{H}\otimes_{2}\mathcal{K})[\mathcal{X}],$
(5.4)

completely isometrically. In particular,

$\mathcal{S}_{p}[\mathcal{S}_{p}]=\mathcal{S}_{p}(\ell_{2}\otimes_{2}\ell_{2}).$
(5.5)

###### Proof.

For $p=\infty$ we have

$\displaystyle\mathcal{S}_{\infty}(\mathcal{H})[\mathcal{S}_{\infty}(\mathcal{K})[\mathcal{X}]]$
$\displaystyle=\mathcal{H}_{c}\otimes_{h}\mathcal{S}_{\infty}(\mathcal{K})[\mathcal{X}]\otimes_{h}(\mathcal{H}^{*})_{r}$
$\displaystyle=(\mathcal{H}_{c}\otimes_{h}\mathcal{K}_{c})\otimes_{h}\mathcal{X}\otimes_{h}\big{(}(\mathcal{K}^{*})_{r}\otimes_{h}(\mathcal{H}^{*})_{r}\big{)}$
$\displaystyle=(\mathcal{H}\otimes_{2}\mathcal{K})_{c}\otimes_{h}\mathcal{X}\otimes_{h}\big{(}(\mathcal{H}\otimes_{2}\mathcal{K})^{*}\big{)}_{r}\qquad\qquad\text{(Proposition~{}\ref{prop:h-tensor-
R-C})}$
$\displaystyle=\mathcal{S}_{\infty}(\mathcal{H}\otimes_{2}\mathcal{K})[\mathcal{X}].$

The case $p=1$ is similar. The general case is now obtained by interpolation:

$\displaystyle\mathcal{S}_{p}(\mathcal{H})[\mathcal{S}_{p}(\mathcal{K})[\mathcal{X}]]$
$\displaystyle=\mathcal{S}_{p}(\mathcal{H})\Big{[}\big{[}\mathcal{S}_{\infty}(\mathcal{K})[\mathcal{X}],\mathcal{S}_{1}(\mathcal{K})[\mathcal{X}]\big{]}_{\frac{1}{p}}\Big{]}$
$\displaystyle=\Big{[}\mathcal{S}_{\infty}(\mathcal{H})[\mathcal{S}_{\infty}(\mathcal{K})[\mathcal{X}]],\mathcal{S}_{1}(\mathcal{H})[\mathcal{S}_{1}(\mathcal{K})[\mathcal{X}]]\Big{]}_{\frac{1}{p}}\
\ \ \ \ \text{(Theorem \ref{interpolation})}$
$\displaystyle=\big{[}\mathcal{S}_{\infty}(\mathcal{H}\otimes_{2}\mathcal{K})[\mathcal{X}],\mathcal{S}_{1}(\mathcal{H}\otimes_{2}\mathcal{K})[\mathcal{X}]\big{]}_{\frac{1}{p}}$
$\displaystyle=\mathcal{S}_{p}(\mathcal{H}\otimes_{2}\mathcal{K})[\mathcal{X}].$

To obtain (5.5), apply (5.4) with $\mathcal{H}=\mathcal{K}=\ell_{2}$ and
$\mathcal{X}=\mathbb{C}$. ∎

In the following theorem we state an equivalent characterization of the norm
in the space $\mathcal{S}_{p}[\mathcal{S}_{q}]$.

######  Theorem 5.8.

[33, 18] let $1\leq p\leq q\leq\infty$ and define $r$ by

$\frac{1}{p}=\frac{1}{q}+\frac{1}{r}.$

Then, the followings hold:

  * (i)

For every $X$ in $\mathcal{S}_{p}[\mathcal{S}_{q}]$ define

$\|X\|_{(p,q)}:=\inf\|A\|_{\mathcal{S}_{2r}}\|Y\|_{\mathcal{S}_{q}(\ell_{2}\otimes_{2}\ell_{2})}\|B\|_{\mathcal{S}_{2r}},$
(5.6)

where the infimum is taken over all $A,B\in\mathcal{S}_{2r}$ and
$Y\in\mathcal{S}_{q}(\ell_{2}\otimes_{2}\ell_{2})$ such that $X=AYB$. Then, we
have $\|X\|_{(p,q)}=\|X\|_{\mathcal{S}_{p}[\mathcal{S}_{q}]}$. We note that
the multiplication $AYB$ is understood as before, and indeed is equal to
$(A\otimes I_{\ell_{2}})Y(B\otimes I_{\ell_{2}})$, where $I_{\ell_{2}}\in
B(\ell_{2})$ is the identity operator.

  * (ii)

For every $X$ in $\mathcal{S}_{q}[\mathcal{S}_{p}]$ define

$\|X\|_{(q,p)}:=\sup\|AXB\|_{\mathcal{S}_{p}(\ell_{2}\otimes_{2}\ell_{2})},$
(5.7)

where the supremum is taken over all $A,B\in\mathcal{S}_{2r}$ such that
$\|A\|_{\mathcal{S}_{2r}},\|B\|_{\mathcal{S}_{2r}}\leq 1$. Then, we have
$\|X\|_{(q,p)}=\|X\|_{\mathcal{S}_{q}[\mathcal{S}_{p}]}$

We note that it is not immediate that $\|\cdot\|_{(q,p)}$ defined by (5.7)
satisfies the triangle inequality. By the above theorem, however,
$\|\cdot\|_{(q,p)}$ is really a norm and satisfies the triangle inequality.

###### Proof.

(i) We first show that
$\|X\|_{\mathcal{S}_{p}[\mathcal{S}_{q}]}\leq\|X\|_{(p,q)}$. Suppose that
$X\in\mathcal{S}_{p}[\mathcal{S}_{q}]$,
$Y\in\mathcal{S}_{q}(\ell_{2}\otimes_{2}\ell_{2})$ and
$A,B\in\mathcal{S}_{2r}$ are such that $X=AYB$. Suppose also that $Y=CZD$ is
an arbitrary factorization of $Y$ with $C,D\in\mathcal{S}_{2q}$ and
$Z\in\mathcal{S}_{\infty}[\mathcal{S}_{q}]$. Then, $X=AC\cdot Z\cdot DB$ and
by Hölder’s inequality we have $AC,DB\in\mathcal{S}_{2p}$. Hence,

$\displaystyle\|X\|_{\mathcal{S}_{p}[\mathcal{S}_{q}]}$
$\displaystyle\leq\|AC\|_{\mathcal{S}_{2p}}\|Z\|_{\mathcal{S}_{\infty}[\mathcal{S}_{q}]}\|DB\|_{\mathcal{S}_{2p}}\quad\qquad\qquad\qquad\text{(Theorem
\ref{Pisier's formula})}$
$\displaystyle\leq\|A\|_{\mathcal{S}_{2r}}\|C\|_{\mathcal{S}_{2q}}\|Z\|_{\mathcal{S}_{\infty}[\mathcal{S}_{q}]}\|D\|_{\mathcal{S}_{2q}}\|B\|_{\mathcal{S}_{2r}}\
\ \ \ \quad\text{(H\"{o}lder's inequality)}.$

Taking infimum over all $C,D,Z$ as above, and using Theorem 5.3 gives

$\displaystyle\|X\|_{\mathcal{S}_{p}[\mathcal{S}_{q}]}$
$\displaystyle\leq\|A\|_{\mathcal{S}_{2r}}\|Y\|_{\mathcal{S}_{q}[\mathcal{S}_{q}]}\|B\|_{\mathcal{S}_{2r}}$
$\displaystyle=\|A\|_{\mathcal{S}_{2r}}\|Y\|_{\mathcal{S}_{q}(\ell_{2}\otimes_{2}\ell_{2})}\|B\|_{\mathcal{S}_{2r}},$

where in the second line we use Lemma 5.7. Finally, taking infimum over the
choices $A,B,Y$ as above, we obtain
$\|X\|_{\mathcal{S}_{p}[\mathcal{S}_{q}]}\leq\|X\|_{(p,q)}$ as desired.

To prove the reverse inequality, let $\epsilon>0$ be arbitrary. Then, there
exist a $Y\in\mathcal{S}_{\infty}[\mathcal{S}_{q}]$ and
$A,B\in\mathcal{S}_{2p}$ such that $X=AYB$ and
$\|A\|_{\mathcal{S}_{2p}}\|Y\|_{\mathcal{S}_{\infty}[\mathcal{S}_{q}]}\|B\|_{\mathcal{S}_{2p}}\leq\|X\|_{\mathcal{S}_{p}[\mathcal{S}_{q}]}+\epsilon$.
We note by the use of polar decomposition and absorbing unitaries into $Y$
without changing its norm, we may assume that $A,B$ are positive. In this
case, $A^{\frac{p}{q}},B^{\frac{p}{q}}\in\mathcal{S}_{2q}$ and
$A^{\frac{p}{r}},B^{\frac{p}{r}}\in\mathcal{S}_{2r}$ are well-defined. Hence,

$\displaystyle\|X\|_{(p,q)}$ $\displaystyle=\|AYB\|_{(p,q)}$
$\displaystyle=\|A^{\frac{p}{r}}\cdot(A^{\frac{p}{q}}\cdot Y\cdot
B^{\frac{p}{q}})\cdot B^{\frac{p}{r}}\|_{(p,q)}$
$\displaystyle\leq\|A^{\frac{p}{r}}\|_{\mathcal{S}_{2r}}\|A^{\frac{p}{q}}\cdot
Y\cdot
B^{\frac{p}{q}}\|_{\mathcal{S}_{q}(\ell_{2}\otimes_{2}\ell_{2})}\|B^{\frac{p}{r}}\|_{\mathcal{S}_{2r}}$
$\displaystyle=\|A^{\frac{p}{r}}\|_{\mathcal{S}_{2r}}\|A^{\frac{p}{q}}\cdot
Y\cdot
B^{\frac{p}{q}}\|_{\mathcal{S}_{q}[\mathcal{S}_{q}]}\|B^{\frac{p}{r}}\|_{\mathcal{S}_{2r}}\
\ \ \ \qquad\quad\quad\text{(Lemma \ref{lem:Fu})}$
$\displaystyle\leq\|A^{\frac{p}{r}}\|_{\mathcal{S}_{2r}}\|A^{\frac{p}{q}}\|_{\mathcal{S}_{2q}}\|Y\|_{\mathcal{S}_{\infty}[\mathcal{S}_{q}]}\|B^{\frac{p}{q}}\|_{\mathcal{S}_{2q}}\|B^{\frac{p}{r}}\|_{\mathcal{S}_{2r}}\
\ \ \ \ \text{(Theorem \ref{Pisier's formula})}$
$\displaystyle=\|A\|_{\mathcal{S}_{2p}}^{\frac{p}{r}}\|A\|_{\mathcal{S}_{2p}}^{\frac{p}{q}}\|Y\|_{\mathcal{S}_{\infty}[\mathcal{S}_{q}]}\|B\|_{\mathcal{S}_{2p}}^{\frac{p}{q}}\|B\|_{\mathcal{S}_{2p}}^{\frac{p}{r}}$
$\displaystyle=\|A\|_{\mathcal{S}_{2p}}\|Y\|_{\mathcal{S}_{\infty}[\mathcal{S}_{q}]}\|B\|_{\mathcal{S}_{2p}}$
$\displaystyle\leq\|X\|_{\mathcal{S}_{p}[\mathcal{S}_{q}]}+\epsilon.$

Since $\epsilon>0$ is arbitrary, it follows that
$\|X\|_{(p,q)}\leq\|X\|_{\mathcal{S}_{p}[\mathcal{S}_{q}]}$, and the proof of
(i) is complete.

(ii) We prove this part by exploiting the duality relation
$\mathcal{S}_{q}[\mathcal{S}_{p}]=\mathcal{S}_{q^{\prime}}[\mathcal{S}_{p^{\prime}}]^{*}$
given in Proposition 5.6. Using this fact along with part (i) for
$q^{\prime}\leq p^{\prime}$, we compute:

$\displaystyle\|X\|_{\mathcal{S}_{q}[\mathcal{S}_{p}]}$
$\displaystyle=\|X\|_{\mathcal{S}_{q^{\prime}}[\mathcal{S}_{p^{\prime}}]^{*}}$
$\displaystyle=\sup_{Y}\frac{|\tr(Y^{*}X)|}{\|Y\|_{\mathcal{S}_{q^{\prime}}[\mathcal{S}_{p^{\prime}}]}}$
$\displaystyle=\sup_{Y}\frac{|\tr(Y^{*}X)|}{\|Y\|_{(q^{\prime},p^{\prime})}}$
$\displaystyle=\sup_{Y}\ \sup_{\stackrel{{\scriptstyle
A,B,Z}}{{Y=AZB}}}\frac{|\tr(Y^{*}X)|}{\|A\|_{\mathcal{S}_{2r}}\|B\|_{\mathcal{S}_{2r}}\|Z\|_{\mathcal{S}_{p^{\prime}}}}$
$\displaystyle=\sup_{A,B,Z}\frac{\big{|}\tr(B^{*}Z^{*}A^{*}X)\big{|}}{\|A\|_{\mathcal{S}_{2r}}\|B\|_{\mathcal{S}_{2r}}\|Z\|_{\mathcal{S}_{p^{\prime}}}}$
$\displaystyle=\sup_{A,B,Z}\frac{\big{|}\tr\big(Z^{*}(A^{*}XB^{*})\big{missing})\big{|}}{\|A\|_{\mathcal{S}_{2r}}\|B\|_{\mathcal{S}_{2r}}\|Z\|_{\mathcal{S}_{p^{\prime}}}}$
$\displaystyle=\sup_{A,B}\
\frac{\|A^{*}XB^{*}\|_{\mathcal{S}_{p}}}{\|A\|_{\mathcal{S}_{2r}}\|B\|_{\mathcal{S}_{2r}}}$
$\displaystyle=\sup_{A,B}\
\frac{\|AXB\|_{\mathcal{S}_{p}}}{\|A\|_{\mathcal{S}_{2r}}\|B\|_{\mathcal{S}_{2r}}}$
$\displaystyle=\|X\|_{(q,p)}.$

∎

######  Lemma 5.9.

When $X$ is positive, the optimizations in (5.6) and (5.7) can be restricted
to the case that $B=A^{*}$ and they are positive.

###### Proof.

Following [11], when $X$ is positive, by Hölder’s inequality we have

$\displaystyle\|AXB\|_{\mathcal{S}_{p}}$
$\displaystyle\leq\|AX^{\frac{1}{2}}\|_{\mathcal{S}_{2p}}\|X^{\frac{1}{2}}B\|_{\mathcal{S}_{2p}}$
$\displaystyle=\|AXA^{*}\|_{\mathcal{S}_{p}}^{\frac{1}{2}}\|B^{*}XB\|_{\mathcal{S}_{p}}^{\frac{1}{2}}$
$\displaystyle\leq\max\big{\\{}\|AXA^{*}\|_{\mathcal{S}_{p}},\,\|B^{*}XB\|_{\mathcal{S}_{p}}\big{\\}}.$

This shows that the supremum in (5.7) can be restricted to the case of
$B=A^{*}$. Moreover, as argued in the proof of Theorem 5.8 we can apply polar
decomposition to assume that $A$ is positive.

To prove the lemma for (5.6) we follow similar steps as in the proof of
Theorem 5.8. First, by Lemma 5.4, for any $\epsilon>0$ there exists a
factorization $X=AYA^{*}$ such that

$\|A\|_{\mathcal{S}_{2p}}^{2}\|Y\|_{\mathcal{S}_{\infty}[\mathcal{S}_{q}]}\leq\|X\|_{\mathcal{S}_{p}[\mathcal{S}_{q}]}+\epsilon.$

Moreover, once again applying the polar decomposition we can assume that
$A>0$. Let $\hat{A}=A^{\frac{p}{r}}$ and
$\hat{Y}=A^{\frac{p}{q}}YA^{\frac{p}{q}}$. Then, we have
$X=\hat{A}\hat{Y}\hat{A}^{*}$ and
$\|\hat{A}\|_{\mathcal{S}_{2r}}=\|\hat{A}^{*}\|_{\mathcal{S}_{2r}}=\|A\|_{\mathcal{S}_{2p}}^{\frac{p}{r}}$.
Moreover, by (5.1) we have

$\|Y\|_{\mathcal{S}_{\infty}[\mathcal{S}_{q}]}\geq\frac{\|A^{\frac{p}{q}}YA^{\frac{p}{q}}\|_{\mathcal{S}_{q}}}{\|A^{\frac{p}{q}}\|^{2}_{\mathcal{S}_{2q}}}=\frac{\|\hat{Y}\|_{\mathcal{S}_{q}}}{\|A\|_{\mathcal{S}_{2p}}^{\frac{2p}{q}}}.$

Therefore,

$\displaystyle\|\hat{A}\|_{\mathcal{S}_{2r}}\|\hat{Y}\|_{\mathcal{S}_{q}}\|\hat{A}^{*}\|_{\mathcal{S}_{2r}}$
$\displaystyle\leq\|A\|_{\mathcal{S}_{2p}}^{\frac{2p}{r}}\cdot\|A\|_{\mathcal{S}_{2p}}^{\frac{2p}{q}}\cdot\|Y\|_{\mathcal{S}_{\infty}[\mathcal{S}_{q}]}$
$\displaystyle=\|A\|_{\mathcal{S}_{2p}}^{2}\cdot\|Y\|_{\mathcal{S}_{\infty}[\mathcal{S}_{q}]}$
$\displaystyle\leq\|X\|_{\mathcal{S}_{p}[\mathcal{S}_{q}]}+\epsilon.$

As a result, since $\epsilon>0$ is arbitrary, in (5.6) we can again restrict
to the case of $B=A^{*}>0$.

∎

##  6\. Applications in quantum information theory

In this section we present some applications of the above tools in quantum
information theory. Here, we denoted Hilbert spaces (mostly finite-
dimensional) by $\mathcal{H}_{X},\mathcal{H}_{Y}$ where subscripts $X,Y$
indicate quantum subsystems. We add a subscript to an operator $T_{X}\in
B(\mathcal{H}_{X})$ to emphasize that it acts on subsystem $X$, and $I_{X}\in
B(\mathcal{H}_{X})$ is the identity operator. A density matrix (quantum state)
is a positive operator with normalized trace. For a density matrix
$\rho_{XY}\in B(\mathcal{H}_{X}\otimes\mathcal{H}_{Y})$, its marginal states
$\rho_{X}\in B(\mathcal{H}_{X})$ and $\rho_{Y}\in B(\mathcal{H}_{Y})$ are
defined via partial trace. Quantum operations (super-operators) are completely
positive trace-preserving (cptp) maps $\Phi:B(\mathcal{H}_{X})\to
B(\mathcal{H}_{Y})$. We denote the identity super-operator acting on subsystem
$X$ by $\mathcal{I}_{X}$.

Recall that for a distribution $p_{X}$ over discrete set $\mathcal{X}$, its
_(Shannon)_ entropy is defined by

$H(X)=-\sum_{x}p(x)\log p(x).$

In the quantum case, for a density matrix $\rho_{X}$ acting on the finite-
dimensional Hilbert space $\mathcal{H}_{X}$, its _(von Neumann)_ entropy is
defined by

$H(X)=-\tr(\rho_{X}\log\rho_{X}).$

For certain applications it is advantageous to define other forms of entropies
that somehow approximate Shannon or von Neumann entropies. A well-known such
quantity is the Rényi entropy that for a parameter $\alpha$ is given by

$H_{\alpha}(X)=\frac{1}{1-\alpha}\log\tr(\rho_{X}^{\alpha}),$

which in the classical case (when $\rho_{X}$ is diagonal with values
$p(x),x\in\mathcal{X}$, on the diagonal) turns into
$H_{\alpha}(X)=\frac{1}{1-\alpha}\log\big(\sum_{x}p(x)^{\alpha}\big{missing})$.
Rényi entropy is an approximation of the usual entropy since

$\lim_{\alpha\to 1}H_{\alpha}(X)=H(X).$

Now, $L_{p}$-spaces naturally appear in the study of entropic quantities since
Rényi entropy can be stated in terms of the norm in these spaces:

$H_{\alpha}(X)=\frac{\alpha}{1-\alpha}\log\|\rho_{x}\|_{\alpha}.$

Therefore, the properties of $L_{p}$-spaces are applicable in the
understanding of these entropic quantities. We emphasize that although Rényi
quantities can also be defined for $\alpha<1$, hereafter we assume that
$\alpha\geq 1$ so that we can use the fact that $\|\cdot\|_{\alpha}$ is really
a norm.

###  6.1. Quantum conditional Rényi entropy

Given a bipartite distribution $p_{XY}$, the condition entropy $H(X|Y)$ is
defined by $H(X|Y)=\sum_{y}p(y)H(X|Y=b)$ as the average entropy of the
conditional distribution $p_{X|Y=y}$. Generalization of this definition to the
quantum case is tricky since conditional states do not have a well-defined
interpretation. Nevertheless, it can be verified that

$\displaystyle H(X|Y)=H(XY)-H(Y),$ (6.1)

and this formula is readily generalized to the quantum case as the definition
of quantum conditional entropy. The definition of conditional Rényi entropy in
the quantum case is even more tricky since (6.1), called the _chain rule_ , no
longer holds for $\alpha\neq 1$. Thus, we should take a different approach to
define the quantum conditional Rényi entropies.

Divergence is another important quantity in information theory which can be
considered as a parent quantity for entropy. For a density matrix $\rho_{X}$,
and positive operator $\sigma_{X}$, that is not necessarily normalized, we
define their _(Umegaki)_ divergence by

$D(\rho_{X}\|\sigma_{X})=\tr(\rho_{X}\log\rho_{x})-\tr(\rho_{X}\log\sigma_{X}).$

Considering the case where $\rho_{X},\sigma_{x}$ are diagonal, divergence in
the classical case is also derived. Moreover, letting $\sigma_{X}=I_{X}$ to be
the identity operator, we find that $D(\rho_{X}\|I_{X})=-H(\rho_{X})$. The
conditional entropy can also be stated in terms of divergence:

$H(X|Y)=-D(\rho_{XY}\|I_{X}\otimes\rho_{Y}).$

Even more, we have

$\displaystyle H(X|Y)=-\inf_{\sigma_{Y}}D(\rho_{XY}\|I_{X}\otimes\sigma_{Y}),$
(6.2)

where the infimum is taken over all density matrices $\sigma_{Y}$. This
identity can be proven using
$D(\rho_{XY}\|I_{X}\otimes\sigma_{Y})=D(\rho_{XY}\|I_{X}\otimes\rho_{Y})+D(\rho_{Y}\|\sigma_{Y})$
that is easy to verify, and the fact that $D(\rho_{Y}\|\sigma_{Y})\geq 0$ for
any two density matrices $\rho_{Y},\sigma_{Y}$. Equation (6.2) suggests that
having a good definition of quantum Rényi divergence, we can define a
conditional version of quantum Rényi entropy.

Rényi divergence in the classical case is given by

$\displaystyle D_{\alpha}(p_{x}\|q_{X})$
$\displaystyle=\frac{1}{\alpha-1}\log\sum_{x}p(x)^{\alpha}q(x)^{1-\alpha}$
$\displaystyle=\frac{\alpha}{\alpha-1}\log\|q_{X}^{\frac{1-\alpha}{\alpha}}p_{X}\Big{\|}_{\alpha},$

and we have $H_{\alpha}(p_{X})=-D_{\alpha}(p_{X}\|I_{X})$ as expected. Next,
inspired by (6.2) we may define the conditional Rényi entropy by

$\displaystyle H_{\alpha}(X|Y)=-\inf_{q_{Y}}D_{\alpha}(p_{XY}\|I_{A}\otimes
q_{Y}).$ (6.3)

Trying to generalize these definitions to the quantum case a complication
arises: since two density matrices $\rho_{X},\sigma_{X}$ do not necessarily
commute, $\sigma_{X}^{\frac{1-\alpha}{\alpha}}$ and $\rho_{X}$ can be
“multiplied” in different orders. This amounts to different notions of quantum
Rényi divergence, yet here we limit ourselves to the so called _sandwiched_
Rényi divergence [40, 31] defined by

$D_{\alpha}(\rho_{X}\|\sigma_{X})=\frac{\alpha}{\alpha-1}\log\|\sigma_{X}^{\frac{1-\alpha}{2\alpha}}\rho_{X}\sigma_{X}^{\frac{1-\alpha}{2\alpha}}\Big{\|}_{\alpha},$

Based on this and following (6.3) we define the quantum conditional Rényi
entropy by

$\displaystyle H_{\alpha}(X|Y)$
$\displaystyle=-\inf_{\sigma_{Y}}D_{\alpha}(\rho_{XY}\|I_{X}\otimes\sigma_{Y})$
$\displaystyle=-\inf_{\sigma_{Y}}\frac{\alpha}{\alpha-1}\log\|\Big{(}I_{X}\otimes\sigma_{Y}^{\frac{1-\alpha}{2\alpha}}\Big{)}\rho_{XY}\Big{(}I_{X}\otimes\sigma_{Y}^{\frac{1-\alpha}{2\alpha}}\Big{)}\Big{\|}_{\alpha}.$

For $\alpha\geq 1$

$\alpha^{\prime}=\frac{\alpha}{\alpha-1},$

is the Hölder conjugate of $\alpha$, and we have

$\displaystyle H_{\alpha}(X|Y)$
$\displaystyle=-\alpha^{\prime}\log\Big(\inf_{\sigma_{Y}}\Big{\|}\Big{(}I_{X}\otimes\sigma_{Y}^{-\frac{1}{2\alpha^{\prime}}}\Big{)}\rho_{XY}\Big{(}I_{X}\otimes\sigma_{Y}^{-\frac{1}{2\alpha^{\prime}}}\Big{)}\Big{\|}_{\alpha}\Big{missing})$
$\displaystyle=-\alpha^{\prime}\log\Bigg(\inf_{\stackrel{{\scriptstyle
T_{Y}\geq 0}}{{\|T_{Y}\|_{\alpha^{\prime}}=1}}}\Big{\|}\Big{(}I_{X}\otimes
T_{Y}^{-\frac{1}{2}}\Big{)}\rho_{XY}\Big{(}I_{X}\otimes
T_{Y}^{-\frac{1}{2}}\Big{)}\Big{\|}_{\alpha}\Bigg{missing}).$

Then, comparing this equation with (5.6) and using Lemma 5.9 we find that

$\displaystyle
H_{\alpha}(X|Y)=-\alpha^{\prime}\log\|\rho_{YX}\big{\|}_{(1,\alpha)}.$ (6.4)

This is a key identity that connects operator-valued Schatten spaces to
information theoretic quantities.

######  Proposition 6.1.

The followings hold for any bipartite density matrix $\rho_{XY}$:

  1. (i)

$\lim_{\alpha\rightarrow 1^{+}}H_{\alpha}(X|Y)=H(X|Y)$.

  2. (ii)

$\alpha\mapsto H_{\alpha}(X|Y)$ is monotonic non-increasing for $\alpha\geq
1$.

  3. (iii)

$H_{\alpha}(X|Y)$ satisfies data processing inequality. More precisely, for
any positive trace preserving (cptp) map $\Phi:B(\mathcal{H}_{Y})\to
B(\mathcal{H}_{Y^{\prime}})$ we have $H_{\alpha}(X|Y)_{\rho}\leq
H_{\alpha}(X|Y^{\prime})_{\sigma}$ where
$\sigma_{XY^{\prime}}=\mathcal{I}_{A}\otimes\Phi(\rho_{XY})$.

  4. (iv)

$|H_{\alpha}(X|Y)|\leq\log d_{X}$ where $d_{X}=\dim\mathcal{H}_{X}$.

###### Proof.

(i) A straightforward computation shows that it suffices to prove that

$\frac{\differential}{\differential\alpha}\|\rho_{YX}\|_{(1,\alpha)}^{\alpha}\bigg{|}_{\alpha=1}=H(Y)-H(XY),$

which itself is proven in [11, Theorem 17]. Interestingly, this latter fact
was known even before the definition of quantum conditional Rényi entropy.

(ii) If $\alpha\geq\beta\geq 1$ there is $\theta\in[0,1]$ such that
$1/\beta=(1-\theta)+\theta/\alpha$. Therefore, by (2.4) we have

$\|\rho_{YX}\|_{(1,\beta)}\leq\|\rho_{YX}\|_{(1,1)}^{1-\theta}\cdot\|\rho_{YX}\|_{(1,\alpha)}^{\theta}=\|\rho_{YX}\|_{(1,\alpha)}^{\theta},$

where we used $\|\rho_{YX}\|_{(1,1)}=\|\rho_{YX}\|_{1}=1$ (see, Lemma 5.7).
Then, taking the logarithm of both sides and multiplying by $-\beta^{\prime}$
(where $\beta^{\prime}$ is the Hölder conjugate of $\beta$), we obtain
$H_{\beta}(X|Y)\leq H_{\alpha}(X|Y)$.

(iii) Using (6.4) it suffices to show that

$\|\Phi\otimes\mathcal{I}_{X}\|_{(1,\alpha)\rightarrow(1,\alpha)}\leq 1.$

This holds since using [11, Lemma 5] we have
$\|\Phi\otimes\mathcal{I}_{X}\|_{(1,\alpha)\rightarrow(1,\alpha)}=\|\Phi\|_{1\rightarrow
1}$ which equals $1$ because $\Phi$ is positive and trace preserving.

(iv) Let $\mathcal{H}_{X^{\prime}}$ be a Hilbert space isomorphic to
$\mathcal{H}_{X}$, and let $\sigma_{XX^{\prime}}$ be a _purification_ of
$\rho_{XY}$. Then, there is a cptp map $\Phi:B(\mathcal{H}_{X^{\prime}})\to
B(\mathcal{H}_{Y})$ (composition of an isometry and a partial trace) such that
$\mathcal{I}_{X}\otimes\Phi(\sigma_{XX^{\prime}})=\rho_{XY}$. Then, by parts
(ii) and (iii) we have

$H_{\infty}(X|X^{\prime})\leq H_{\alpha}(X|X^{\prime})\leq H_{\alpha}(X|Y).$

Next, by definition we have

$\displaystyle H_{\infty}(X|X^{\prime})$
$\displaystyle=-\log\Bigg(\inf_{\stackrel{{\scriptstyle T_{X^{\prime}}\geq
0}}{{\|T_{X^{\prime}}\|_{1}=1}}}\Big{\|}\Big{(}I_{X}\otimes
T_{X^{\prime}}^{-\frac{1}{2}}\Big{)}\sigma_{XX^{\prime}}\Big{(}I_{X}\otimes
T_{X^{\prime}}^{-\frac{1}{2}}\Big{)}\Big{\|}_{\infty}\Bigg{missing})$
$\displaystyle=-\log\|\Big{(}I_{X}\otimes\sigma_{X^{\prime}}^{-\frac{1}{2}}\Big{)}\sigma_{XX^{\prime}}\Big{(}I_{X}\otimes\sigma_{X^{\prime}}^{-\frac{1}{2}}\Big{)}\Big{\|}_{\infty}$
$\displaystyle\geq-\log\tr\Big(\Big{(}I_{X}\otimes\sigma_{X^{\prime}}^{-\frac{1}{2}}\Big{)}\sigma_{XX^{\prime}}\Big{(}I_{X}\otimes\sigma_{X^{\prime}}^{-\frac{1}{2}}\Big{)}\Big{missing})$
$\displaystyle=-\log d_{X}.$

Therefore, $H_{\alpha}(X|Y)\geq-\log d_{X}$.

We note that for any unitary $U_{X}$ we have
$\|\rho_{YX}\|_{(1,\alpha)}=\|(I_{Y}\otimes U_{X})\rho_{YX}(I_{Y}\otimes
U_{X}^{*})\|_{(1,\alpha)}$. Therefore, choosing $U_{X}$ randomly according to
the Haar measure, and using the convexity of the $(1,\alpha)$-norm, we find
that

$\displaystyle\|\rho_{YX}\|_{(1,\alpha)}$
$\displaystyle=\mathbb{E}\|(I_{Y}\otimes U_{X})\rho_{YX}(I_{B}\otimes
U_{x}^{*})\|_{(1,\alpha)}$ $\displaystyle\geq\|\mathbb{E}(I_{Y}\otimes
U_{X})\rho_{YX}(I_{Y}\otimes U_{X}^{*})\|_{(1,\alpha)}$
$\displaystyle=\Big{\|}\rho_{Y}\otimes\frac{I_{X}}{d_{X}}\Big{\|}_{(1,\alpha)}$
$\displaystyle=\|\rho_{Y}\|_{1}\cdot\Big{\|}\frac{I_{X}}{d_{X}}\Big{\|}_{\alpha}$
$\displaystyle=d_{X}^{-\frac{1}{\alpha^{\prime}}}.$

Using this in the definition of $H_{\alpha}(X|Y)$, we arrive at
$H_{\alpha}(X|Y)\leq\log d_{X}$.

∎

It is well-known that the conditional von Neumann entropy satisfies a
_uniform_ continuity bound [1, 41], i.e., for two quantum states
$\rho_{XY},\sigma_{XY}$ satisfying
$\frac{1}{2}\|\rho_{XY}-\sigma_{XY}\|_{1}\leq\epsilon$ we have

$\big{|}H(X|Y)_{\rho}-H(X|Y)_{\sigma}\big{|}\leq 2\epsilon\log
d_{X}+(1+\epsilon)h\Big{(}\frac{\epsilon}{1+\epsilon}\Big{)},$

where $h(x)=-x\log x-(1-x)\log(1-x)$ is the binary entropy function. The
crucial point in this continuity bound is that it is independent of the
dimension of $\mathcal{H}_{Y}$. Such a continuity bound for the quantum
conditional Rényi entropy is first found in [27] (see also [17, 22] for
related results). Here, we prove a stronger inequality than the result of [27]
with a much simpler proof.

######  Theorem 6.2.

Let $\rho_{XY},\sigma_{XY}$ be two quantum states satisfying
$\frac{1}{2}\|\rho_{XY}-\sigma_{XY}\|_{1}\leq\epsilon$. Then, we have

$\big{|}H_{\alpha}(X|Y)_{\rho}-H_{\alpha}(X|Y)_{\sigma}\big{|}\leq\alpha^{\prime}\log(1+2\epsilon
d_{X}^{2/\alpha^{\prime}}).$

Our bound in this theorem for $\alpha=1$ (in the limit of $\alpha\to 1$)
equals $\frac{2\epsilon\log d_{X}}{1+\epsilon}$ which is of order of
$O(\epsilon\log d_{X})$ for small $\epsilon>0$. On the other hand, the bound
of [27] for $\alpha=1$ is of order $\Omega(\sqrt{\epsilon}\log d_{X})$. This
means that, for $\alpha\approx 1$ and small $\epsilon>0$ our bound is
stronger. Numerical plots also suggest that our bound is tighter even for
larger values of $\alpha$.

###### Proof.

Let us assume with no loss of generality that
$\|\rho_{XY}-\sigma_{XY}\|_{1}=2\epsilon$. Then, there are density matrices
$\mu_{XY},\nu_{XY}$ such that

$\rho_{XY}-\sigma_{XY}=\epsilon(\mu_{XY}-\nu_{XY}).$

Therefore,

$\displaystyle\Big{|}\|\rho_{YX}\|_{(1,\alpha)}-\|\sigma_{YX}\|_{(1,\alpha)}\Big{|}$
$\displaystyle\leq\|\rho_{YX}-\sigma_{YX}\|_{(1,\alpha)}$
$\displaystyle=\epsilon\|\mu_{XY}-\nu_{XY}\|_{(1,\alpha)}$
$\displaystyle\leq\epsilon\big{(}\|\mu_{XY}\|_{(1,\alpha)}+\|\nu_{XY}\|_{(1,\alpha)}\big{)}.$

On the other hand, as a consequence of part (iv) of Proposition 6.1 we have

$\|\mu_{XY}\|_{(1,\alpha)},\|\nu_{XY}\|_{(1,\alpha)}\leq
d_{X}^{1/\alpha^{\prime}}.$

Hence,

$\Big{|}\|\rho_{YX}\|_{(1,\alpha)}-\|\sigma_{YX}\|_{(1,\alpha)}\Big{|}\leq
2\epsilon d_{X}^{1/\alpha^{\prime}}.$

As a result,

$\displaystyle\frac{\|\rho_{YX}\|_{(1,\alpha)}}{\|\sigma_{YX}\|_{(1,\alpha)}}$
$\displaystyle\leq\frac{\|\sigma_{YX}\|_{(1,\alpha)}+2\epsilon
d_{X}^{1/\alpha^{\prime}}}{\|\sigma_{YX}\|_{(1,\alpha)}}$
$\displaystyle=1+\frac{2\epsilon
d_{X}^{1/\alpha^{\prime}}}{\|\sigma_{YX}\|_{(1,\alpha)}}$ $\displaystyle\leq
1+2\epsilon d_{X}^{2/\alpha^{\prime}},$

where in the last inequality we once again use part (iv) of Proposition 6.1.
Taking the logarithm of both sides and multiplying by $\alpha^{\prime}$ yields

$H_{\alpha}(X|Y)_{\sigma}-H_{\alpha}(X|Y)_{\rho}\leq\alpha^{\prime}\log(1+2\epsilon
d_{X}^{2/\alpha^{\prime}}).$

∎

The following theorem is a generalization of the chain rule for quantum von
Neumann entropy.

######  Theorem 6.3.

Let $\alpha,\beta,\gamma\geq 1$ such that

$\frac{\alpha}{\alpha-1}=\frac{\beta}{\beta-1}+\frac{\gamma}{\gamma-1}.$

Then, for any tripartite density matrix $\rho_{XYZ}$ we have

$\displaystyle H_{\alpha}(XY|Z)\geq H_{\beta}(X|YZ)+H_{\gamma}(Y|Z).$ (6.5)

This theorem is first proven in [12]. The following proof, that is much
simpler than the original proof in [12], is found by the first author of this
paper and independently by McKinlay and Tomamichel (private communications).
See also [28] for other results in this direction.

###### Proof.

We note that the condition on $\alpha,\beta,\gamma$ can be written as
${\alpha^{\prime}}={\beta^{\prime}}+{\gamma^{\prime}}$. Then, for
$\theta=\gamma^{\prime}/\alpha^{\prime}=1-\beta^{\prime}/\alpha^{\prime}$ we
have $\theta\in[0,1]$ and

$\frac{1}{\alpha}=\frac{1-\theta}{1}+\frac{\theta}{\gamma},\qquad\frac{1}{\alpha}=\frac{1-\theta}{\beta}+\frac{\theta}{1}.$

Then, we have

$\Big{[}\mathcal{S}_{1}(\mathcal{H}_{Z})\big{[}\mathcal{S}_{1}(\mathcal{H}_{Y})[S_{\beta}(\mathcal{H}_{X}]\big{]},\,\mathcal{S}_{1}(\mathcal{H}_{Z})\big{[}S_{\gamma}(\mathcal{H}_{Y})[\mathcal{S}_{1}(\mathcal{H}_{X})]\big{]}\Big{]}_{\theta}=\mathcal{S}_{1}(\mathcal{H}_{Z})\big{[}S_{\alpha}(\mathcal{H}_{Y})[S_{\alpha}(X)]\big{]}.$

That is, the $(1,\alpha,\alpha)$-norm is obtained via interpolation of the
$(1,1,\beta)$-norm and the $(1,\gamma,1)$-norm. Therefore, by (2.4) we have

$\|\rho_{ZYX}\|_{(1,\alpha,\alpha)}\leq\|\rho_{ZYX}\|_{(1,1,\beta)}^{1-\theta}\cdot\|\rho_{ZYX}\|_{(1,\gamma,1)}^{\theta}.$

On the other hand, by Lemma 5.7 we have

$\|\rho_{ZYX}\|_{(1,\alpha,\alpha)}=\|\rho_{Z,YX}\|_{(1,\alpha)},$

and

$\|\rho_{ZYX}\|_{(1,1,\beta)}=\|\rho_{ZY,X}\|_{(1,\beta)}.$

Furthermore, as shown in [11] we have

$\|\rho_{ZYX}\|_{(1,\gamma,1)}=\|\rho_{ZY}\|_{(1,\gamma)}.$

Putting all these together we arrive at

$\|\rho_{Z,YX}\|_{(1,\alpha)}\leq\|\rho_{ZY,X}\|_{(1,\beta)}^{1-\theta}\cdot\|\rho_{ZY}\|_{(1,\gamma)}^{\theta}.$

Now taking the logarithm of both sides and multiplying by $-\alpha^{\prime}$,
the desired inequality is obtained. ∎

###  6.2. Rényi coherent information

The coherent information $I(X\rangle Y)$ of a state $\rho_{XY}$ is defined by

$I(X\rangle Y)=H(Y)-H(XY)=-H(X|Y).$

Then, given a cptp map $\Phi:B(\mathcal{H}_{X})\to B(\mathcal{H}_{Y})$, its
coherent information is defined by

$\displaystyle I^{\rm{coh}}(\Phi)=\max_{\rho_{RX}}I(R\rangle Y)_{\sigma},$
(6.6)

where $I(R\rangle Y)$ is computed with respect to

$\sigma_{RY}=\mathcal{I}_{R}\otimes\Phi(\rho_{RX}),$

and $\mathcal{H}_{R}$ is an auxiliary reference Hilbert space. It is well-
known that $Q(\Phi)$, the quantum capacity of $\Phi$, i.e., its optimal rate
for sending quantum information with asymptotically vanishing error is equal
to [39]

$Q(\Phi)=\lim_{n\rightarrow\infty}\frac{1}{n}I^{\rm{coh}}(\Phi^{\otimes n}),$

where $\Phi^{\otimes n}$ is the $n$-fold tensor product of $\Phi$.

Since coherent information is expressed in terms of conditional entropy, we
may define Rényi coherent information based on the Rényi conditional entropy.
Indeed, based on (6.6) we may define the $\alpha$-Rényi coherent information
of a cptp map by

$\displaystyle I^{\rm{coh}}_{\alpha}(\Phi)$
$\displaystyle=\max_{\rho_{RX}}I_{\alpha}(R\rangle Y)_{\sigma}$
$\displaystyle=\alpha^{\prime}\log\Big(\max_{\rho_{RX}}\|\Phi\otimes\mathcal{I}_{R}(\rho_{XR})\|_{(1,\alpha)}\Big{missing}).$

######  Proposition 6.4.

$I^{{\rm
coh}}_{\alpha}(\Phi)=\alpha^{\prime}\log\|\Phi\otimes\mathcal{I}_{R}\|_{(1,1)\rightarrow(1,\alpha)}$.

###### Proof.

We need to show that

$\|\Phi\otimes\mathcal{I}_{R}\|_{(1,1)\rightarrow(1,\alpha)}=\max_{\rho_{RA}}\|\Phi\otimes\mathcal{I}_{R}(\rho_{XR})\|_{(1,\alpha)},$

where the maximum is taken over density matrices $\rho_{XR}$. Since
$\|\rho_{XR}\|_{(1,1)}=\|\rho_{XR}\|_{1}=1$ for any density matrix
$\rho_{XR}$, it suffices to show that the maximum in

$\max_{F_{XR}:\|F\|_{1}=1}\|\Phi\otimes\mathcal{I}_{R}(F_{XR})\|_{(1,\alpha)},$

is achieved at a positive $F$. To prove this we follow similar steps as in the
proof of [11, Theorem 12].

Let $F_{XR}$ be such that $\|F\|_{1}=1$. Then, by the polar decomposition
there exists a unitary $U_{XR}$ and density matrix $\rho_{XR}$ such that
$F=U\rho$. Now, the block matrix

$\begin{bmatrix}U\rho^{1/2}\\\ \\\
\rho^{1/2}\end{bmatrix}\begin{bmatrix}\rho^{1/2}U^{*}&\rho^{1/2}\end{bmatrix}=\begin{bmatrix}U\rho
U^{*}&U\rho\\\ \rho U^{*}&\rho\end{bmatrix},$

is positive. Then, since $\Phi$ is completely positive,

$\begin{bmatrix}\Phi\otimes\mathcal{I}_{R}(U\rho
U^{*})&\Phi\otimes\mathcal{I}_{R}(U\rho)\\\ \Phi\otimes\mathcal{I}_{R}(\rho
U^{*})&\Phi\otimes\mathcal{I}_{R}(\rho)\end{bmatrix},$

is positive. Therefore, using [8, Proposition 1.3.2] there exists a
contraction $K_{YR}$ such that

$\Phi\otimes\mathcal{I}_{R}(U\rho)=\big{(}\Phi\otimes\mathcal{I}_{R}(U\rho
U^{*})\big{)}^{1/2}K_{YR}\big{(}\Phi\otimes\mathcal{I}_{R}(\rho)\big{)}^{1/2}.$

Hence, according to [11, Lemma 9] we we have

$\displaystyle\|\Phi\otimes\mathcal{I}_{R}(F)\|_{(1,\alpha)}$
$\displaystyle\leq\|\Phi\otimes\mathcal{I}_{R}(U\rho
U^{*})\|_{(1,\alpha)}^{1/2}\cdot\|\Phi\otimes\mathcal{I}_{R}(\rho)\|_{(1,\alpha)}^{1/2}$
$\displaystyle\leq\max\Big{\\{}\|\Phi\otimes\mathcal{I}_{R}(U\rho
U^{*})\|_{(1,\alpha)},\|\Phi\otimes\mathcal{I}_{R}(\rho)\|_{(1,\alpha)}\Big{\\}}.$

The proof concludes once we note that both $\rho$ and $U\rho U^{*}$ are
density matrices. ∎

Now that we represent the $\alpha$-Rényi coherent information of a channel in
terms of an operator norm, we may use all the tools for studying operators
norms, such as convexity, interpolation techniques and log-Sobolev
inequalities to investigate the problem of quantum channel capacity. In
particular, it is tempting to use the $\alpha$-Rényi coherent information to
prove strong converse bounds on quantum channel capacity. A natural open
question in this direction is whether $I_{\alpha}^{\rm{coh}}(\cdot)$ is
_additive_ for certain (e.g., degradable) cptp maps or not, i.e., does
$I_{\alpha}^{\rm{coh}}(\Phi_{1}\otimes\Phi_{2})$ equal to
$I_{\alpha}^{\rm{coh}}(\Phi_{1})+I_{\alpha}^{\rm{coh}}(\Phi_{2})$?

We finish this subsection by mentioning some related results to the above
additivity problem. Motivated by the problem of the additivity of the _minimum
output entropy_ of cptp maps, it is shown in [11] that the completely bounded
$(p\to q)$-norm of cptp maps is multiplicative. The proof of this result is
heavily based on the theory of operator-valued Schatten norms. It is also
shown that the _entanglement-assisted_ capacity of cptp maps can be expressed
in terms of the so called completely $p$-summing norms [19] that is again
related to the theory of operator-valued Schatten norms. Inspired by [11], the
theory of _logarithmic-Sobolev_ inequalities of completely bounded norms is
developed in [6]. Also, extending [6], estimations of the decoherence rate of
_non-primitive_ quantum Markov semigroups is derived in [3]. See also [2] for
some related results.

###  6.3. A correlation measure

In this subsection we define yet another information theoretic invariant based
on operator-valued Schatten spaces. Here, we briefly explain the main
properties of this invariant that is first introduced in [29] and refer to the
original paper for more details.

For a bipartite density matrix $\rho_{XY}$ define

$W_{\alpha}(X|Y)=\Big{\|}\rho_{YX}-\rho_{Y}\otimes\frac{I_{X}}{d_{X}}\Big{\|}_{(1,\alpha)},$

where $d_{X}=\dim\mathcal{H}_{X}$ is the dimension of subsystem $X$. Similar
to $H_{\alpha}(X|Y)$, the definition of $W_{\alpha}(X|Y)$ is also in terms of
the $(1,\alpha)$-norm, so it is expected that these two quantities be related.
Indeed, it is shown in [29, Proposition 9] that

$\Big{|}W_{\alpha}(X|Y)-e^{-\frac{1}{\alpha^{\prime}}H_{\alpha}(X|Y)}\Big{|}\leq
d_{X}^{-\frac{1}{\alpha^{\prime}}}.$

We note that if $\rho_{XY}=\rho_{X}\otimes\rho_{Y}$ is a product density
matrix and $\rho_{X}=I_{X}/d_{X}$, then $W_{\alpha}(X|Y)=0$. Thus, assuming
that marginal density matrix $\rho_{X}$ is maximally mixed, then
$W_{\alpha}(X|Y)$ can be thought of as a correlation measure. As a correlation
measure it is expected that $W_{\alpha}(X|Y)$ satisfies the data processing
inequality: if $\Phi:B(\mathcal{H}_{Y})\to B(\mathcal{H}_{Y}^{\prime})$ is a
cptp map, then

$W_{\alpha}(X|Y^{\prime})\leq W_{\alpha}(X|Y).$

This inequality is proven in [29, Theorem 8] based on similar ideas as in the
proof of part (iii) of Proposition 6.1.

######  Theorem 6.5.

[29, Corollary 14] Let $\rho_{XY}$ be a bipartite density matrix and let
$P:\mathcal{H}_{X}\to\mathcal{H}_{X_{0}}$ be an orthogonal projection where
$\mathcal{H}_{X_{0}}\subseteq\mathcal{H}_{X}$ is a subspace. Then, for any
$1\leq\alpha\leq 2$ we have

$\mathbb{E}_{U}\Big{[}\Big{\|}\frac{d_{X}}{d_{X_{0}}}(PU_{X}\otimes
I_{Y})\rho_{XY}(U^{*}_{X}P\otimes
I_{Y})-\frac{I_{X_{0}}}{d_{X_{0}}}\otimes\rho_{Y}\Big{\|}_{1}\Big{]}\leq
2^{\frac{2}{\alpha}-1}d_{X_{0}}^{\frac{1}{\alpha^{\prime}}}W_{\alpha}(X|Y)$

Proof of this theorem for $\alpha=1$ is easy. For $\alpha=2$, it is proven
essentially based on ideas from [13]. Then, the key observation is that, by
Theorem 2.6, from the boundary cases $\alpha\in\\{1,2\\}$ we can infer the
theorem for all values of $1\leq\alpha\leq 2$. See [29] for more details.

Theorem 6.5 is called a decoupling theorem as it says that by applying a
random unitary and a projection on a bipartite density matrix, the two parts
become decoupled. Such decoupling results have several applications in
information theory, e.g., in channel coding and privacy amplification [13, 29,
37].

##  Appendix A Kuratowski–Ryll-Nardzewski selection theorem

Suppose that $\alpha$, $\beta$ and $\gamma$ are norms, respectively, on
$M_{d,n}$, $M_{n,d^{\prime}}$ and $M_{d,d^{\prime}}$ such that

$\gamma(X)=\inf\big{\\{}\alpha(A)\cdot\beta(B):X=AB,A\in M_{d,n},B\in
M_{n,d^{\prime}}\big{\\}}.$

Our goal here is to show that _almost optimizer_ pairs $(A,B)$ can be chosen
to be measurable functions.

######  Lemma A.1.

Let $X(s)$ be a continuous function on $\mathbb{R}$ with values in
$M_{d,d^{\prime}}$. Then, for every $\epsilon>0$ there exist measurable
functions $A(s),B(s)$ such that for every $s\in\mathbb{R}$,

$\displaystyle X(s)=A(s)B(s),$ $\displaystyle\alpha(A(s))\leq 1+\epsilon,$
$\displaystyle\beta(B(s))\leq\gamma(X(s)).$

To prove this lemma we make use of a special case of the Kuratowski–Ryll-
Nardzewski selection theorem [36]:

######  Theorem A.2.

(Kuratowski–Ryll-Nardzewski selection theorem) Let $\emph{cl}(\mathbb{R}^{n})$
be the set of all nonempty closed subsets of $\mathbb{R}^{n}$. Let
$\psi:\mathbb{R}\rightarrow\emph{cl}(\mathbb{R}^{n})$ be a multi-function such
that the set $\\{s\in\mathbb{R}:\psi(s)\cap U\neq\emptyset\\}$ is measurable
for every closed subset $U$ of $\mathbb{R}^{n}$. Then, there is a measurable
function $f:\mathbb{R}\rightarrow\mathbb{R}^{n}$ such that $f(s)\in\psi(s)$
for all $s\in\mathbb{R}$.

###### Proof of Lemma A.1.

Based on the above selection theorem we define $\psi(s)$ to be the set of all
almost optimizer pairs $(A,B)$ satisfying

$\displaystyle X(s)=AB,$ $\displaystyle\alpha(A)\leq 1+\epsilon,$
$\displaystyle\beta(B)\leq\gamma(X(s)).$

We need to show that $\psi(\cdot)$ satisfies the hypotheses of the above
selection theorem. It is straightforward to show that $\psi(s)$ is non-empty
and closed for every $s$. Thus, we need to verify that
$\\{s\in\mathbb{R}:\psi(s)\cap U\neq\emptyset\\}$ is measurable. To this end,
consider the map

$\Psi:(s,A,B)\mapsto\Big{(}\gamma(X(s)-AB),\,\alpha(A),\,\gamma(X(s))-\beta(B)\Big{)}.$

Since $\Psi$ is continuous, the set

$\mathcal{E}:=\Psi^{-1}\Big{(}\\{0\\}\crossproduct[0,1+\epsilon]\crossproduct[0,+\infty)\Big{)},$

is closed. Now, if $U$ is a closed set of pairs $(A,B)$, then
$\mathcal{E}\cap(\mathbb{R}\crossproduct U)$ is closed. Next, observe that

$\\{s\in\mathbb{R}:\psi(s)\cap
U\neq\emptyset\\}=\pi_{1}(\mathcal{E}\cap(\mathbb{R}\crossproduct U)),$

where $\pi_{1}$ denotes the projection map onto the first component.
Therefore, to see the measurability of $\\{s\in\mathbb{R}:\psi(s)\cap
U\neq\emptyset\\}$, it suffices to prove that
$\pi_{1}(\mathcal{E}\cap(\mathbb{R}\crossproduct U))$ is measurable.

Note that any Euclidean space can be written as a countable union of compact
sets. Then, since the intersection of a closed and a compact set is compact,
the closed set $\mathcal{E}\cap(\mathbb{R}\crossproduct U)$ can be written as

$\mathcal{E}\cap(\mathbb{R}\crossproduct U)=\bigcup_{i=1}^{\infty}K_{i}$

where $K_{i}$’s are compact. On the other hand, since a continuous image of a
compact set is compact, $\pi(K_{i})$’s are compact too. This implies that

$\pi_{1}(\mathcal{E}\cap(\mathbb{R}\crossproduct
U))=\bigcup_{i=1}^{\infty}\pi_{1}(K_{i}),$

is a countable union of compact sets, and is therefore measurable. ∎

##  Appendix B Wiener–Masani factorization theorem

This section is dedicated to the Wiener–Masani theorem on factorization of
operator-valued functions. This theorem is used in the proof of Theorem 4.2,
and is one of the main technical tools in the development of the theory of
operator-valued Schatten spaces. Thus, we devote this section to the statement
and proof of this theorem. We follow the approach of [16] for the proof, that
is based on characterization of _invariant subspaces_ of the shift operator.
To this end, we first need a number of ingredients.

###  B.1. Hardy spaces

Following the notation of Section 2, let $\mathbb{D}$ be the unit disk in the
complex plane and let $\partial\mathbb{D}$ be its boundary, equipped with the
normalized Lebesgue measure. Then, $L_{p}=L_{p}(\partial\mathbb{D})$ denotes
the Banach space of all measurable functions
$f:\partial{\mathbb{D}}\rightarrow\mathbb{C}$ such that the norm

$\|f\|_{p}:=\begin{cases}\left(\frac{1}{2\pi}\int_{-\pi}^{\pi}|f(e^{it})|^{p}\differential
t\right)^{\frac{1}{p}},~{}~{}~{}~{}~{}\qquad\qquad 1\leq
p<\infty,\\\\[7.22743pt]
\operatorname*{ess\,sup}|f|,~{}~{}~{}~{}~{}\qquad\qquad\qquad\qquad\ \ \
p=\infty,\end{cases}$

is finite. We note that by Hölder’s inequality, we have $L_{q}\subseteq L_{p}$
for $p<q$.

######  Definition B.1.

The _Hardy space_ $H_{p}$ is the closed subspace of $L_{p}$ consisting of all
functions $f\in L_{p}$ for which there is a function $F$ holomorphic in the
interior of $\mathbb{D}$ whose _radial limit_ coincides with $f$ almost
everywhere.141414The radial limit of $F$ at $e^{it}\in\partial\mathbb{D}$
equals $\lim_{r\to 1^{-}}F(re^{it})$.

There is another description of $H_{p}$ in terms Fourier coefficients and
Fourier series that goes as follows. For each function $f\in L_{1}$, let

$a_{n}=\frac{1}{2\pi}\int_{-\pi}^{\pi}f(e^{it})e^{-nit}\differential t,$

and

$f(e^{it})\sim\sum_{n=-\infty}^{\infty}a_{n}e^{nit},$

be the Fourier coefficients and the Fourier series of $f$, respectively. Then,
$H_{p}$ is the subset of $L_{p}$ consisting of all functions $f$ whose Fourier
coefficients $a_{n}$ vanish for $n<0$.

Define the operator $\mathfrak{S}$ by
$(\mathfrak{S}f)(e^{it})=e^{it}f(e^{it})$. Consequently, for $n\geq 1$ we have
$(\mathfrak{S}^{n}f)(e^{it})=e^{nit}f(e^{it})$. The operator $\chi$ is called
the _shift operator_ since it shifts the Fourier coefficients.

######  Definition B.2.

A function $f\in H_{2}$ is said to be _outer_ if the linear span of the set
$\\{\mathfrak{S}^{n}f:n\geq 0\\}$ is dense in $H_{2}$.

It is well-known that every outer function $f$ can be written in the form

$f(z)=\gamma\exp\left(\frac{1}{2\pi}\int_{-\pi}^{\pi}\frac{e^{it}+z}{e^{it}-z}k(t)\differential
t\right),$

where $\gamma$ is a complex number of modulus $1$, and $k(t)=\log|f(e^{it})|$
almost everywhere.

###  B.2. Vector-valued Hardy spaces

Let $\mathcal{H}$ be a separable (usually finite-dimensional) Hilbert space.
An $\mathcal{H}$-valued function $F:\partial{\mathbb{D}}\to\mathcal{H}$ is
said to be measurable if for every $v\in\mathcal{H}$ the complex function
$e^{it}\mapsto\langle v,F(e^{it})\rangle$ is measurable. Abusing the notation,
let $L_{p}(\mathcal{H})=L_{p}(\partial\mathbb{D},\mathcal{H})$ be the Banach
space of all measurable functions
$F:\partial{\mathbb{D}}\rightarrow\mathcal{H}$ such that the norm

$\|F\|_{p}:=\begin{cases}\left(\frac{1}{2\pi}\int_{-\pi}^{\pi}\|F(e^{it})\|^{p}\differential
t\right)^{\frac{1}{p}},~{}~{}~{}~{}~{}\qquad\qquad 1\leq
p<\infty,\\\\[7.22743pt]
\operatorname*{ess\,sup}\|F(e^{it})\|,~{}~{}~{}~{}~{}\qquad\qquad\qquad\qquad\
\ \ p=\infty\end{cases}$

is finite.151515$\|F(e^{it})\|$ is the norm of vector
$F(e^{it})\in\mathcal{H}$. It is worth noting that the above integral makes
sense since the function $e^{it}\mapsto\|F(e^{it})\|$, as the supremum of
measurable functions, is itself measurable. The case $p=2$ is of particular
interest since $L_{2}(\mathcal{H})$ is a Hilbert space under the inner product

$\langle
F,G\rangle=\frac{1}{2\pi}\int_{-\pi}^{\pi}\big{\langle}F(e^{it}),G(e^{it})\big{\rangle}\differential
t.$

Let $\\{e_{n}:n=1,2,\cdots\\}$ is an orthonormal basis for $\mathcal{H}$, and
for $F\in L_{2}(\mathcal{H})$ define the _coordinate functions_ by

$f_{n}(e^{it}):=\big{\langle}e_{n},F(e^{it})\big{\rangle}.$ (B.1)

Then, we have

$\|F\|_{2}^{2}=\sum_{n}\|f_{n}\|_{2}^{2},$

where $\|f_{n}\|_{2}$ is the norm of $f_{n}$ in
$L_{2}=L_{2}(\partial\mathbb{D},\mathbb{C})$.

For any $F\in L_{p}(\mathcal{H})$, one may define the _Fourier expansion_ of
$F$ as

$F(e^{it})\sim\sum_{n=-\infty}^{\infty}\hat{a}_{n}\cdot e^{nit},$ (B.2)

where $\hat{a}_{n}\in\mathcal{H}$ is given by

$\hat{a}_{n}=\frac{1}{2\pi}\int_{-\pi}^{\pi}F(e^{it})e^{-nit}\differential t.$

We note that letting $\mathcal{H}=\mathbb{C}$, we obtain the usual
$L_{p}=L_{p}(\partial\mathbb{D},\mathbb{C})$ space with the usual Fourier
transform.

######  Definition B.3.

Let $\mathcal{H}$ be a (separable) Hilbert space and $1\leq p\leq\infty$. The
_vector-valud Hardy space_ $H_{p}(\mathcal{H})$ is defined to be the closed
subspace of $L_{p}(\mathcal{H})$ consisting of all functions $F$ for which
$\hat{a}_{n}=0$, for all $n<0$.

Similar to the usual complex-valued Hardy space, $H_{p}(\mathcal{H})$ can also
be characterized in terms of holomorphic functions. We say that a vector-
valued function $\phi:\mathbb{D}\rightarrow\mathcal{H}$ is holomorphic if for
every $v\in\mathcal{H}$, the function $z\mapsto\langle v,\phi(z)\rangle$ is
holomorphic. Then, $H_{p}(\mathcal{H})$ is the space of functions $F\in
L_{p}(\mathcal{H})$ for which there is a holomorphic function
$\phi:\mathbb{D}\rightarrow\mathcal{H}$ whose _radial limit_ coincides with
$F$ almost everywhere, i.e.,

$F(e^{it})=\lim_{r\to 1^{+}}\phi(re^{it}),\qquad\text{a.e.}$

###  B.3. Invariant subspaces

The shift operator $\mathfrak{S}$ can also be defined on the space of vector-
valued functions: $(\mathfrak{S}F)(e^{it})=e^{it}F(e^{it})$.

######  Definition B.4.

A closed subspace $\mathcal{M}$ of $L_{2}(\mathcal{H})$ is said to be
_invariant_ if $\mathfrak{S}(\mathcal{M})\subseteq\mathcal{M}$, where
$\mathfrak{S}(\mathcal{M}):=\\{\mathfrak{S}F:\,F\in\mathcal{M}\\}$. We say
that $\mathcal{M}$ is _doubly invariant_ if
$\mathfrak{S}(\mathcal{M})\subseteq\mathcal{M}$ and
$\mathfrak{S}^{-1}(\mathcal{M})\subseteq\mathcal{M}$. Also, $\mathcal{M}$ is
said to be _simply invariant_ if it is invariant but not doubly invariant.

We may also consider operator-valued (matrix-valued) functions. We say that
such a function $U(e^{it})$ is measurable if the function
$e^{it}\mapsto\langle v,U(e^{it})w\rangle$ is measurable for all vectors
$v,w$.

######  Lemma B.5.

Let $\mathcal{M}$ be a simply invariant subspace of $L_{2}(\mathcal{H})$ that
contains no nontrivial doubly invariant subspace. Then, there exist an
auxiliary Hilbert space $\mathcal{K}$ and a measurable operator-valued
function $U$ such that

  * (i)

$U(e^{it}):\mathcal{K}\rightarrow\mathcal{H}$ is an isometry for any
$e^{it}\in\partial{\mathbb{D}}$.

  * (ii)

$\mathcal{M}$ consists of functions $e^{it}\mapsto U(e^{it})F(e^{it})$ for
$F\in H_{2}(\mathcal{K})$.

We note that vector-valued Hardy spaces are simply invariant subspaces and do
not contain any nontrivial doubly invariant subspace. The above lemma says
that all such subspaces are essentially vector-valued Hardy spaces.

###### Proof.

For each $n$, define $\mathcal{M}_{n}:=\mathfrak{S}^{n}(\mathcal{M})$. As
$\mathcal{M}$ is invariant, we have

$\mathcal{M}=\mathcal{M}_{0}\supseteq\mathcal{M}_{1}\supseteq\mathcal{M}_{2}\supseteq\cdots.$
(B.3)

We note that

$\mathcal{M}_{\infty}=\bigcap\limits_{n\geq 0}\mathcal{M}_{n},$

is doubly invariant, so by assumption $\mathcal{M}_{\infty}=0$.

For any $n\geq 0$, let $\mathcal{N}_{n}$ be the orthogonal complement of
$\mathcal{M}_{n+1}$ in $\mathcal{M}_{n}$ (with respect to the inner product of
$L_{2}(\mathcal{H})$). Thus, we have
$\mathcal{M}_{n}=\mathcal{N}_{n}\oplus\mathcal{M}_{n+1}$. Moreover, since
$\mathcal{M}_{\infty}=0$, we have

$\mathcal{M}=\mathcal{N}_{0}\oplus\mathcal{N}_{1}\oplus\mathcal{N}_{2}\oplus\cdots.$
(B.4)

Let $\mathcal{B}_{0}=\\{E_{1},E_{2},\cdots\\}$ be an orthonormal basis for
$\mathcal{N}_{0}$, and note that
$\mathcal{B}_{n}:=\mathfrak{S}^{n}(\mathcal{B}_{0})=\\{\mathfrak{S}^{n}E_{k}:k\geq
1\\}$, for $n\geq 1$, is a subset of
$\mathcal{M}_{n}\subseteq\mathcal{M}_{1}$. Thus, $\mathcal{B}_{0}$ is
orthogonal to $\mathcal{B}_{n}$ for all $n\geq 1$. This means that for every
$j,k$ and $n\geq 1$ we have

$\langle
E_{j},\mathfrak{S}^{n}E_{k}\rangle=\frac{1}{2\pi}\int_{-\pi}^{\pi}e^{nit}\big{\langle}E_{j}(e^{it}),E_{k}(e^{it})\big{\rangle}\differential
t=0.$

Moreover, as $\mathcal{B}_{0}$ is orthonormal, the same equation holds when
$n=0$ and $j\neq k$. More generally, we observe that

$\langle\mathfrak{S}^{n}E_{j},\mathfrak{S}^{m}E_{k}\rangle=\frac{1}{2\pi}\int_{-\pi}^{\pi}e^{(m-n)it}\big{\langle}E_{j}(e^{it}),E_{k}(e^{it})\big{\rangle}dt=\delta_{jk}\cdot\delta_{mn}.$

This equation shows that all the nontrivial Fourier coefficients of the
functions $e^{it}\mapsto\big{\langle}E_{j}(e^{it}),E_{k}(e^{it})\big{\rangle}$
are zero, and hence it is almost everywhere identical to the constant function
$\delta_{jk}$.

We claim that for each $n$, the orthonormal set $\mathcal{B}_{n}$ is a basis
for $\mathcal{N}_{n}$. To this end, we need to show that
$\mathcal{B}_{n}\subseteq\mathcal{N}_{n}=\operatorname*{span}\mathcal{B}_{n}$.
Letting $F\in\mathcal{M}_{n+1}$, by definition there is $G\in\mathcal{M}$ such
that $F=\mathfrak{S}^{n+1}G$. On the other hand, for
$\mathfrak{S}^{n}E_{j}\in\mathcal{B}_{n}$ we have

$\langle\mathfrak{S}^{n}E_{j},F\rangle=\langle\mathfrak{S}^{n}E_{j},\chi^{n+1}G\rangle=\langle
E_{j},\mathfrak{S}G\rangle=0,$

where we used the fact that $\mathfrak{S}G\in\mathcal{M}_{1}$ and that
$E_{j}\in\mathcal{N}_{0}$. Thus, $\mathfrak{S}^{n}E_{j}\in\mathcal{M}_{n}$ and
is orthogonal to $\mathcal{M}_{n+1}$, which means that
$\mathfrak{S}^{n}E_{j}\in\mathcal{N}_{n}$. Therefore,
$\mathcal{B}_{n}\subseteq\mathcal{N}_{n}$.

Next, let $F\in\mathcal{M}_{n}$. Then, there is $G\in\mathcal{M}$ such that
$F=\mathfrak{S}^{n}G$. Since
$G\in\mathcal{M}=\mathcal{N}_{0}\oplus\mathcal{M}_{1}$, there are
$G_{0}\in\mathcal{N}_{0}$ and $G_{1}\in\mathcal{M}$ such that
$G=G_{0}+\mathfrak{S}G_{1}$. Therefore

$F=\mathfrak{S}^{n}G_{0}+\mathfrak{S}^{n+1}G_{1}.$

By definition we have $\mathfrak{S}^{n+1}G_{1}\in\mathcal{M}_{n+1}$. Also,
since $G_{0}\in\mathcal{N}_{0}=\operatorname*{span}\mathcal{B}_{0}$, we have
$\mathfrak{S}^{n}G_{0}\in\operatorname*{span}\mathcal{B}_{n}$. As a result,
$F\in\operatorname*{span}\mathcal{B}_{n}\oplus\mathcal{M}_{n+1}$ for any
$F\in\mathcal{M}_{n}$. Thus, using $\mathcal{B}_{n}\subseteq\mathcal{N}_{n}$
proven above, and the definition of $\mathcal{N}_{n}$, we find that
$\mathcal{N}_{n}=\operatorname*{span}\mathcal{B}_{n}$.

Putting (B.4) and $\mathcal{N}_{n}=\operatorname*{span}\mathcal{B}_{n}$
together, we find that every $F\in\mathcal{M}$ can be expanded as

$F=\sum_{j}\sum_{n=0}^{\infty}a_{nj}\mathfrak{S}^{n}E_{j},\ \ \
\quad\sum_{n,j}|a_{nj}|^{2}<\infty.$ (B.5)

Now, let $\mathcal{K}$ be a Hilbert space isomorphic to $\mathcal{N}_{0}$ with
orthonormal basis $\\{e_{1},e_{2},\cdots\\}$. Every function in
$L_{2}(\mathcal{K})$ can be written in the form $\sum f_{j}e_{j}$ with
$f_{j}\in L_{2}$, $j=1,2,\dots$, being coordinate functions (see, (B.1)).
Define $\hat{U}:L_{2}(\mathcal{K})\rightarrow L_{2}(\mathcal{H})$ by

$\hat{U}\Big{(}\sum f_{j}e_{j}\Big{)}=\sum f_{j}E_{j},$

meaning that $\hat{U}\Big{(}\sum f_{j}e_{j}\Big{)}(e^{it})=\sum
f_{j}(e^{it})E_{j}(e^{it})$. Then, $\hat{U}$ is a well-defined linear
operator. Moreover, since $\langle
E_{j}(e^{it}),E_{k}(e^{it})\rangle=\delta_{jk}$, almost everywhere, $\hat{U}$
is an isometry.

For every $e^{it}$ we define a linear map
$U(e^{it}):\mathcal{K}\rightarrow\mathcal{H}$ as follows. For every
$v\in\mathcal{K}$ let $\hat{v}\in L_{2}(\mathcal{K})$ be the constant function
$\hat{v}(e^{it})=v$. Then, define

$U(e^{it})v:=(\hat{U}\hat{v})(e^{it}).$

It is easy to verify that $e^{it}\mapsto U(e^{it})$ is measurable. We claim
that $U(e^{it})$ is an isometry for almost every $t$. To show this, fix
$v,w\in\mathcal{K}$. Using the fact that $\hat{U}$ is unitary, for every $n$
we obtain

$\displaystyle\big{\langle}\hat{U}\hat{v},\hat{U}(\mathfrak{S}^{n}\hat{w})\big{\rangle}$
$\displaystyle=\langle\hat{v},\mathfrak{S}^{n}\hat{w}\rangle$
$\displaystyle=\frac{1}{2\pi}\int_{-\pi}^{\pi}e^{nit}\langle
v,w\rangle\differential t$ $\displaystyle=\frac{1}{2\pi}\langle
v,w\rangle\int_{-\pi}^{\pi}e^{nit}\differential t$
$\displaystyle=\delta_{n0}\langle v,w\rangle.$ (B.6)

On the other hand, the definition of $U$ implies that
$\hat{U}(\mathfrak{S}^{n}\hat{w})=\mathfrak{S}^{n}\hat{U}(\hat{w})$.
Therefore,

$\displaystyle\langle\hat{U}\hat{v},\hat{U}(\mathfrak{S}^{n}\hat{w})\rangle$
$\displaystyle=\langle\hat{U}\hat{v},\mathfrak{S}^{n}\hat{U}(\hat{w})\rangle$
$\displaystyle=\frac{1}{2\pi}\int_{-\pi}^{\pi}e^{nit}\langle\hat{U}\hat{v}(e^{it}),\hat{U}\hat{w}(e^{it})\rangle\differential
t.$

Consequently, comparing to the previous equation, all the nontrivial Fourier
coefficients of the function
$e^{it}\mapsto\langle\hat{U}\hat{v}(e^{it}),\hat{U}\hat{w}(e^{it})\rangle=\langle
U(e^{it})v,U(e^{it})w\rangle$ vanish. This means that this function is
constant almost everywhere. Thus, we have, almost everywhere,

$\displaystyle\langle U(e^{it})v,U(e^{it})w\rangle$
$\displaystyle=\frac{1}{2\pi}\int_{-\pi}^{\pi}\langle
U(e^{is})v,U(e^{is})w\rangle\differential s$
$\displaystyle=\frac{1}{2\pi}\int_{-\pi}^{\pi}\langle\hat{U}\hat{v}(e^{is}),\hat{U}\hat{w}(e^{is})\rangle\differential
s$ $\displaystyle=\langle\hat{U}\hat{v},\hat{U}\hat{w}\rangle$
$\displaystyle=\langle v,w\rangle,$

where in the last line we use (B.6). This means that $U(e^{it})$ is an
isometry almost everywhere.

We note that $H_{2}(\mathcal{K})=\big{\\{}\sum f_{j}e_{j}:\,f_{j}\in
H_{2},\|f_{j}\|_{2}^{2}<\infty\big{\\}}$, and every $f_{j}\in H_{2}$ can be
written as $f_{j}(e^{it})=\sum_{n\geq 0}a_{nj}e^{nit}$. Therefore, (B.5)
yeilds

$\displaystyle\hat{U}H_{2}(\mathcal{K})$
$\displaystyle=\Big{\\{}\sum_{j}f_{j}E_{j}:f_{j}\in
H_{2}(\mathbb{C}),\sum\|f_{j}\|_{2}^{2}<\infty\Big{\\}}$
$\displaystyle=\Big{\\{}\sum_{j}\sum_{n\geq
0}a_{nj}\mathfrak{S}^{n}E_{j}:\sum|a_{nj}|^{2}<\infty\Big{\\}}$
$\displaystyle=\mathcal{M}.$

Finally, from the definition of $U$ it follows that
$U(e^{it})e_{j}=(\hat{U}e_{j})(e^{it})=E_{j}(e^{it})$. Hence, for every
$F=\sum_{j}f_{j}e_{j}\in H_{2}(\mathcal{K})$ we obtain

$\displaystyle(\hat{U}F)(e^{it})$ $\displaystyle=\Big{(}\sum
f_{j}E_{j}\Big{)}(e^{it})$ $\displaystyle=\sum f_{j}(e^{it})E_{j}(e^{it})$
$\displaystyle=\sum f_{j}(e^{it})U(e^{it})e_{j}$
$\displaystyle=U(e^{it})\Big{(}\sum f_{j}(e^{it})e_{j}\Big{)}$
$\displaystyle=U(e^{it})F(e^{it}).$

That is, $\mathcal{M}$ consists of functions of the form $e^{it}\mapsto
U(e^{it})F(e^{it})$ for $F\in H_{2}(\mathcal{K})$. ∎

######  Remark B.6.

In the above lemma, if $\mathcal{H}$ is finite-dimensional, then so is
$\mathcal{K}$ since $U(e^{it}):\mathcal{K}\rightarrow\mathcal{H}$ is an
isometry almost everywhere.

###  B.4. Factorization of operator-valued functions

In the following we assume that $\mathcal{H}$ is finite-dimensional with
orthonormal basis $\\{e_{1},\dots,e_{d}\\}$. For every
$e^{it}\in\partial\mathbb{D}$ we assume that
$T(e^{it}):\mathcal{H}\rightarrow\mathcal{H}$ is a linear map such that
$e^{it}\mapsto T(e^{it})$ is an operator-valued measurable function. We
further assume that entries of $T$ (in the basis $\\{e_{1},\dots,e_{d}\\}$)
are integrable, i.e., belong to $L_{1}$.

Suppose that $T(e^{it})$ is positive, and uniformly bounded almost everywhere,
i.e., there exists a constant $c>0$ such that

$0\leq T(e^{it})\leq cI,\qquad\text{ almost everywhere}.$

Let $Q(e^{it})=\sqrt{T(e^{it})}$. We note that with the above assumptions $Q$
can be written as the limit of measurable functions (consider the truncated
Taylor expansion of $x\mapsto\sqrt{x}$ at point $c$), and is measurable
itself. We also note that $Q$ can be considered as an element of
$B(L_{2}(\mathcal{H}))$. Indeed, for every $F\in L_{2}(\mathcal{H})$ we have
$(QF)(e^{it})=Q(e^{it})F(e^{it})$ and

$\displaystyle\|QF\|_{2}^{2}$
$\displaystyle=\frac{1}{2\pi}\int_{-\pi}^{\pi}\big{\langle}Q(e^{it})F(e^{it}),Q(e^{it})F(e^{it})\big{\rangle}\differential
t$ $\displaystyle=\frac{1}{2\pi}\int_{-\pi}^{\pi}\langle
F(e^{it}),T(e^{it})F(e^{it})\rangle\differential t$
$\displaystyle\leq\frac{1}{2\pi}\int_{-\pi}^{\pi}c\langle
F(e^{it}),F(e^{it})\rangle\differential t$ $\displaystyle=c\|F\|_{2}^{2}.$

Define

$\mathcal{M}(T)=\overline{\operatorname*{span}\\{\mathfrak{S}^{n}(Qe_{j}):n\geq
0,\,j=1,\dots,d\\}},$

where $Qe_{j}\in L_{2}(\mathcal{H})$ is given by
$(Qe_{j})(e^{it})=Q(e^{it})e_{j}$. Observe that by definition,
$\mathcal{M}(T)$ is a closed and invariant subspace of $L_{2}(\mathcal{H})$.
In fact, $\mathcal{M}(T)$ is the smallest invariant subspace of
$L_{2}(\mathcal{H})$ generated by columns of $Q$.

######  Lemma B.7.

Let $T(e^{it})$ be a measurable operator-valued function as above, and
$\lambda(e^{it})$ be a positive measurable function such that

$\log\lambda(e^{it})\in L_{1},$ (B.7)

and

$0<\lambda(e^{it})I\leq T(e^{it})\leq cI,\qquad\text{a.e. on
$\partial{\mathbb{D}}$},$

for some $c>0$. Then $\mathcal{M}(T)$ does not contain any nontrivial doubly
invariant subspace.

###### Proof.

Observe that by (B.7) there is $\psi\in H_{2}$ such that
$|\psi(e^{it})|=\sqrt{\lambda(e^{it})}$. To construct such a function we use
the Poisson kernel of $\mathbb{D}$ to define

$g(z):=\frac{1}{2\pi}\int_{-\pi}^{\pi}\real\frac{e^{it}+z}{e^{it}-z}\log\sqrt{\lambda(e^{it})}\differential
t.$

We note that $g(e^{it})=\log\sqrt{\lambda(e^{it})}$ almost everywhere, and
that $g(z)$ is harmonic. Then, considering its harmonic conjugate, there is a
holomorphic function $\hat{g}(z)$ with $\real\hat{g}(z)=g(z)$. Now, letting
$\psi(z)=e^{\hat{g}(z)}$, we have almost everywhere

$|\psi(e^{it})|=|e^{\hat{g}(it)}|=|e^{\real\hat{g}(it)}|=|e^{g(it)}|=\sqrt{\lambda(e^{it})}.$

We say that $P\in H_{2}(\mathcal{H})$ is a _polynomial_ if its Fourier
expansion (B.2) consists of finitely many terms, i.e., $P$ is a polynomial in
variable $e^{it}$ with coefficients in $\mathcal{H}$. For such a polynomial
$P\in H_{2}(\mathcal{H})$ we have $\psi P\in H_{2}(\mathcal{H})$ where $(\psi
P)(z)=\psi(z)P(z)$. Define

$\mathcal{N}=\overline{\operatorname*{span}\\{\psi P:\text{$P$
polynomial}\\}}.$ (B.8)

Observe that $\mathcal{N}\subseteq H_{2}(\mathcal{H})$ is an invariant
subspace.

Next, consider the subspace $\mathcal{W}=\operatorname*{span}\\{QP:\,\text{$P$
polynomial}\\}\subseteq L_{2}(\mathcal{H})$ where as before $Q=\sqrt{T}$. We
note that by definition $\overline{\mathcal{W}}=\mathcal{M}(T)$. Define $U$
from $\mathcal{W}$ to $\mathcal{N}$ by $U(QP)=\psi P$. Since
$\sqrt{\lambda(e^{it})}I\leq Q(e^{it})$ almost everywhere, we have

$\displaystyle\|\psi P\|^{2}_{2}$
$\displaystyle=\frac{1}{2\pi}\int_{-\pi}^{\pi}\big{\|}\psi(e^{it})P(e^{it})\big{\|}^{2}$
$\displaystyle=\frac{1}{2\pi}\int_{-\pi}^{\pi}\lambda(e^{it})\big{\|}P(e^{it})\big{\|}^{2}$
$\displaystyle\leq\frac{1}{2\pi}\int_{-\pi}^{\pi}\big{\|}(QP)(e^{it})\big{\|}^{2}$
$\displaystyle=\|QP\|^{2}_{2}.$

This means that $\|U\|\leq 1$, and $U$ can be extended to
$U:\mathcal{M}(T)\rightarrow\mathcal{N}$ while $\|U\|\leq 1$.

We claim that $U$ is one-to-one. To prove this, suppose that
$F\in\mathcal{M}(T)$ and $UF=0$. Since $F\in\mathcal{M}(T)$, there is a
sequence of polynomials $\\{P_{n}:\,n\geq 1\\}\subset H_{2}(\mathcal{H})$ such
that

$\lim_{n\rightarrow\infty}QP_{n}=F\ \ \ \qquad\text{(in
$L^{2}(\mathcal{H})$)},$

and since $QF=0$, we have

$\lim_{n\rightarrow\infty}\psi P_{n}=0\ \ \ \qquad\text{(in
$L^{2}(\mathcal{H})$)}.$

Define $R_{m}=\\{e^{it}:\lambda(e^{it})\geq 1/m\\}$, and let $J_{m}$ be the
characteristic function of $R_{m}$. We compute

$\displaystyle 0$
$\displaystyle\leq\frac{1}{2\pi}\int_{-\pi}^{\pi}\frac{1}{m}J_{m}(e^{it})\big{\|}P_{n}(e^{it})\big{\|}^{2}\,\differential
t$
$\displaystyle\leq\frac{1}{2\pi}\int_{-\pi}^{\pi}\lambda(e^{it})J_{m}(e^{it})\big{\|}P_{n}(e^{it})\big{\|}^{2}\,\differential
t$
$\displaystyle\leq\frac{1}{2\pi}\int_{-\pi}^{\pi}\lambda(e^{it})\big{\|}P_{n}(e^{it})\big{\|}^{2}\,\differential
t$ $\displaystyle=\|\psi P_{n}\|_{2}^{2}.$

Therefore

$\lim_{n\rightarrow\infty}\frac{1}{2\pi}\int_{-\pi}^{\pi}J_{m}(e^{it})\big{\|}P_{n}(e^{it})\big{\|}_{2}^{2}\,\differential
t=0,\ \ \ \quad\forall m.$ (B.9)

On the other hand, we have

$\displaystyle\big{\|}F-QP_{n}\big{\|}_{2}^{2}$
$\displaystyle=\frac{1}{2\pi}\int_{-\pi}^{\pi}\big{\|}F(e^{it})-Q(e^{it})P_{n}(e^{it})\big{\|}^{2}\,\differential
t$
$\displaystyle\geq\frac{1}{2\pi}\int_{-\pi}^{\pi}J_{m}(e^{it})\big{\|}F(e^{it})-Q(e^{it})P_{n}(e^{it})\big{\|}^{2}\,\differential
t$