# Embeddings of operator ideals into $\mathcal{L}_{p}-$spaces on finite von
Neumann algebras

M. Junge  1409 West Green Street, Urbana, Illinois, 61801,USA
<EMAIL_ADDRESS>,  F. Sukochev  School of Mathematics and Statistics,
UNSW, Kensington, NSW 2052, Australia<EMAIL_ADDRESS>and  D. Zanin
School of Mathematics and Statistics, UNSW, Kensington, NSW 2052, Australia
<EMAIL_ADDRESS>
###### Abstract.

Let $\mathcal{L}(H)$ be the $*$-algebra of all bounded operators on an
infinite dimensional Hilbert space $H$ and let
$(\mathcal{I},\|\cdot\|_{\mathcal{I}})$ be an ideal in $\mathcal{L}(H)$
equipped with a Banach norm which is distinct from the Schatten-von Neumann
ideal $\mathcal{L}_{p}(\mathcal{H})$, $1\leq p<2$. We prove that $\mathcal{I}$
isomorphically embeds into an $L_{p}$-space $\mathcal{L}_{p}(\mathcal{R}),$
$1\leq p<2,$ (here, $\mathcal{R}$ is the hyperfinite II1-factor) if its
commutative core (that is, Calkin space for $\mathcal{I}$) isomorphically
embeds into $L_{p}(0,1).$ Furthermore, we prove that an Orlicz ideal
$\mathcal{L}_{M}(H)\neq\mathcal{L}_{p}(H)$ isomorphically embeds into
$\mathcal{L}_{p}(\mathcal{R}),$ $1\leq p<2,$ if and only if it is an
interpolation space for the Banach couple
$(\mathcal{L}_{p}(H),\mathcal{L}_{2}(H)).$ Finally, we consider isomorphic
embeddings of $(\mathcal{I},\|\cdot\|_{\mathcal{I}})$ into $L_{p}$-spaces
associated with arbitrary finite von Neumann algebras.

###### Key words and phrases:

Noncommutative independence, Banach space embeddings

###### 1991 Mathematics Subject Classification:

Primary 46L53; secondary 81S25

The first author is partially supported by NSF grant DMS-1201886. The second
and third authors are supported by ARC

##  1\. Introduction

Classifying subspaces of Banach spaces is in general a very difficult task,
and so far can only be achieved for very special examples of classical
function spaces. A beautiful example is Dacuhna-Castelle characterization of
subspaces of $L_{1}(0,1)$ with a symmetric basis, as average of Orlicz spaces.
This characterization is genuinely based on the analysis of independent copies
and, hence, is probabilistic in nature. Noncommutative function spaces appear
naturally in operator theory and noncommutative geometry, and hence it is
reasonable to expect a similar striking result for symmetric subspaces of
noncommutative $L_{1}-$spaces or even noncommutative $L_{p}-$spaces in the
interesting range for $1\leq p<2.$ Again M. Kadec discovery of $q-$stable
random variables resulting in embedding $l_{q}$ into $L_{p}(0,1)$ serves as
inspiration for our results. However, the noncommutative notion of
independence is much more ambiguous in the noncommutative setting. Therefore
we prefer to use more functional analytic tools, in particular (several
versions) of the Kruglov operator introduced in [5] (see [47] for an
alternative approach using Poisson processes going back to [32]).

A detailed study of symmetric subspaces of $L_{1}$ was done by J. Bretagnolle
and D. Dacunha-Castelle (see [11, 12, 19]). They proved that, for every given
mean zero $f\in L_{p}$, the sequence $\\{f_{k}\\}_{k=1}^{\infty}$ of its
independent copies is equivalent, in $L_{p}$, to the standard basis of some
Orlicz sequence space $l_{M}$ [19, Theorem 1, p.X.8]. Later some of these
results were independently rediscovered by M. Braverman [14, 15]. Note that
the methods used in [11, 12, 19, 14, 15] depend heavily on the techniques
related to the theory of random processes. In a recent paper [8], a different
approach based on methods and ideas from the interpolation theory of operators
and the usage of so-called Kruglov operator [4, 5, 6, 7] is suggested.

The main topic of this paper is the study of (symmetric) subspaces of the
noncommutative analogue of the space $L_{p}(0,1),$ i.e. the space
$\mathcal{L}_{p}(\mathcal{R},\tau)$ of $\tau$-measurable affiliated with the
hyperfinite II1-factor $\mathcal{R}$ of finite $p$-norm. Indeed, the
hyperfinite and finite von Neumann algebra $\mathcal{R}$ is equipped with a
unique tracial state $\tau$ and the couple $(\mathcal{R},\tau)$ may be
considered as a natural noncommutative analogue of the pair
$(L_{\infty}(0,1),dm)$ (here, $dm$ is the Lebesgue measure). In particular,
the trace $\tau$ is normal and faithful (see next section for details). More
generally, closed $\tau$-measurable operators affiliated to $\mathcal{R}$ form
a $*$-algebra $\mathcal{S}(\mathcal{R},\tau)$ which is analogous to the
algebra $L_{0}(0,1)$ of all (unbounded) Lebesgue measurable functions on
$(0,1)$. The precise definition of the $L_{p}$-spaces associated with
$\mathcal{R}$ mimics the classical one

$\mathcal{L}_{p}(\mathcal{R}):=\\{A\in\mathcal{S}(\mathcal{R},\tau):\
\tau(|A|^{p})<\infty\\}.$

The separable space $\mathcal{L}_{p}(\mathcal{R})$ plays a role in the class
of noncommutative $L_{p}$-spaces associated with semifinite von Neumann
algebras (see e.g. [53]) similar to that of the role of the space $L_{p}(0,1)$
in the class of all $L_{p}$-spaces on $\sigma$-finite measure spaces.

The class of symmetric subspaces of the space $\mathcal{L}_{p}(\mathcal{R})$
is naturally understood as subspaces isomorphic to Banach ideals in
$\mathcal{L}(H),$ the only von Neumann factor of the type $I_{\infty}.$
Indeed, the latter class corresponds to that of symmetric sequence spaces [40,
48]. For example, when symmetric sequence space in question is $l_{q}$, then
the corresponding Banach ideal is the classical Schatten-von Neumann class
$\mathcal{L}_{q}(H)$. Thus, the precise form of the main question studied in
this article is as follows.

######  Question 1.

Which Banach ideals in $\mathcal{L}(H)$ admit an isomorphic embedding into
$\mathcal{L}_{p}(\mathcal{R})?$

The following result taken from [61, Theorem 3.1], [29, Theorem 1.1 and
Corollary 1.2], [34] encapsulates all available information to date.

######  Theorem 2.

Let $\mathcal{R}$ be the hyperfinite II1-factor. We have

  1. (a)

$\mathcal{L}_{2}(H)$ is the only Banach ideal in $\mathcal{L}(H)$
isomorphically embeddable into $\mathcal{L}_{p}(\mathcal{R})$ for $p\geq 2.$

  2. (b)

$\mathcal{L}_{p}(H)$ does not isomorphically embed into
$\mathcal{L}_{p}(\mathcal{R})$ for $1\leq p<2.$

  3. (c)

$\mathcal{L}_{q}(H)$ isomorphically embeds into $\mathcal{L}_{p}(\mathcal{R})$
for $1\leq p<q<2.$

In this paper, we investigate this question in the setting $1\leq p<2$. To
state our main results, we need a notion of commutative core of an ideal,
which is also frequently called a Calkin space (see e.g. [48]). Fix an
orthonormal basis in $H$ (the particular choice of a basis is inessential) and
identify the algebra $l_{\infty}$ of all bounded sequences with the subalgebra
of all diagonal operators on $H$ with respect to this basis. Given an ideal
$\mathcal{I}\in\mathcal{L}(H),$ we define its commutative core by

$I:=\mathcal{I}\cap l_{\infty}$

(that is, the commutative core $I$ corresponding to $\mathcal{I}$ is the
collection of all diagonal operators from $\mathcal{L}(H)$ belonging to
$\mathcal{I}$). If $(\mathcal{I},\|\cdot\|_{\mathcal{I}})$ is a Banach ideal,
then its commutative core $I$ is equipped with the induced norm
$\|x\|_{{I}}:=\|x\|_{\mathcal{I}}$, $x\in{I}.$ It can be shown that
$(I,\|\cdot\|_{I})$ is a symmetric sequence space. We are now able to state
the main results of the paper.

######  Theorem 3.

A separable Banach ideal $\mathcal{I}\neq\mathcal{L}_{p}(H)$ in
$\mathcal{L}(H)$ admits an isomorphic embedding into
$\mathcal{L}_{p}(\mathcal{R}),$ $1\leq p<2,$ provided that its commutative
core $I$ admits an isomorphic embedding into $L_{p}(0,1).$

The separability condition, as well as the condition
$\mathcal{I}\neq\mathcal{L}_{p}(H)$ are obviously essential (see Theorem 2 (b)
above).

The result of Theorem 3 heavily relies on the classification of Orlicz ideals
which admit an isomorphic embedding into $\mathcal{L}_{p}(\mathcal{R})$ and
which is a perfect noncommutative analogue of results from [11, 12, 19, 8].

######  Theorem 4.

Let $\mathcal{R}$ be the hyperfinite II1-factor, let
$\mathcal{L}_{M}(H)\neq\mathcal{L}_{p}(H)$ be an Orlicz ideal and let $1\leq
p<2.$ The following conditions are equivalent.

  1. (i)

The space $\mathcal{L}_{M}(H)$ isomorphically embeds into
$\mathcal{L}_{p}(\mathcal{R}).$

  2. (ii)

The space $\mathcal{L}_{M}(H)$ is $p$-convex and $2$-concave.

  3. (iii)

The Orlicz function $M$ on $(0,\infty)$ is equivalent on $[0,1]$ to some
$p$-convex and $2$-concave Orlicz function.

  4. (iv)

The space $\mathcal{L}_{M}(H)$ is an interpolation space for the couple
$(\mathcal{L}_{p}(H),\mathcal{L}_{2}(H)).$

######  Remark 5.

The condition $\mathcal{L}_{M}(H)\neq\mathcal{L}_{p}(H)$ guarantees that
$M(t)\not\sim t^{p}$ as $t\to 0.$

Our second main result may be viewed as a converse to Theorem 3 under an
additional assumption that the hyperfinite factor $\mathcal{R}$ is replaced
with an arbitrary finite von Neumann algebra $\mathcal{N}.$

######  Theorem 6.

For a separable Banach ideal $\mathcal{I}\neq\mathcal{L}_{p}(H)$ in
$\mathcal{L}(H),$ the following conditions are equivalent

  1. (i)

$\mathcal{I}$ admits an isomorphic embedding into
$\mathcal{L}_{p}(\mathcal{M},\sigma),$ $1\leq p<2,$ for some finite von
Neumann algebra $(\mathcal{M},\sigma).$

  2. (ii)

The commutative core $I$ of the ideal $\mathcal{I}$ admits an isomorphic
embedding into $\mathcal{L}_{p}(\mathcal{N},\tau),$ $1\leq p<2,$ for some
finite von Neumann algebra $(\mathcal{N},\tau).$

######  Remark 7.

We do not claim that the von Neumann algebras in (i) and (ii) are the same.

The following question remains open.

######  Question 8.

Suppose $\mathcal{I}$ admits an isomorphic embedding into
$\mathcal{L}_{p}(\mathcal{M},\sigma),$ $1\leq p<2,$ for some finite von
Neumann algebra $(\mathcal{M},\sigma).$ Is it true that the commutative core
$I$ of the ideal $\mathcal{I}$ admits an isomorphic embedding into
$L_{p}(0,1)?$

Finally, we are compelled to say a few words about our techniques and methods,
which we believe are of wider applicability and of interest in their own
right. Our main technical instrument is the Kruglov operator (used, e.g. in
the proof of the most difficult implication (iii)$\Rightarrow$(i) in Theorem
4) which is a replacement of integration with respect to the Poisson process
(see e.g. [47]). In the special setting of the commutative von Neumann algebra
$(L_{\infty}(0,1),dm)$, such an operator was firstly introduced in [4, 5] (see
also the survey [7]). In the noncommutative setting, this operator is defined
and studied in this paper. In Section 3, we introduce the Kruglov operator
affiliated with finite von Neumann algebras, basically mimicking the approach
from [5]. In Section 4, we introduce such an operator affiliated already with
a semifinite non-finite von Neumann algebra and this is an important extension
from our viewpoint. We emphasize that the germ of our approach to this
definition can be found already in [34], although our present presentation and
approach are quite different from those given in [34]. Theorem (6) makes
crucial, and extends, the work of Kwapien and Schütt [45].

##  2\. Preliminaries

Let $H$ be a complex infinite-dimensional separable Hilbert space equipped
with an inner product $\langle\cdot,\cdot\rangle_{H}$ and $\mathcal{L}(H)$ be
the $C^{*}$-algebra of all bounded linear operators on $H$ equipped with the
operator norm $\|\cdot\|_{\infty}$. We denote by ${\rm Tr}$ the classical
trace on $\mathcal{L}(H)$. For every compact operator $A\in\mathcal{L}(H),$
let $\mu(A)=\\{\mu(k,A)\\}_{k\geq 0}$ be the decreasing sequence of singular
values of the operator $A,$ counted with multiplicities.

For $A\in\mathcal{L}(H)$ we use the following standard notations

$|A|:=\sqrt{A^{*}A},\quad\Re A:=\frac{A+A^{*}}{2},\quad\Im
A:=\frac{A-A^{*}}{2i},$

and for $A=A^{*}\in\mathcal{L}(H)$ we set

$A_{+}:=\frac{|A|+A}{2},\quad A_{-}:=\frac{|A|-A}{2}.$

The notations $\mathcal{M}$ and $\mathcal{N}$ usually stand for finite or
semifinite von Neumann algebras. By $\mathbbm{1}_{\mathcal{M}}$ (resp.
$\mathbbm{1}_{\mathcal{N}}$) we denote the unit element of the algebra
$\mathcal{M}$ (resp. $\mathcal{N}$). For brevity we use simply $\mathbbm{1}$
instead of both $\mathbbm{1}_{\mathcal{M}}$ or $\mathbbm{1}_{\mathcal{N}}.$
This should not lead to any confusion.

For positive real numbers $a,b$ and some constant $C>0$ we write $a{\sim}b$,
if the inequality $Ca\leq b\leq a/C$ holds. When we need to specify the
constant $C$, we will use the notation $a\stackrel{{\scriptstyle
C}}{{\sim}}b.$

###  2.1. Calkin correspondence

Fix an orthonormal basis in $H$ (the particular choice of a basis is
inessential). We identify the algebra $l_{\infty}$ of bounded complex
sequences with the subalgebra of all diagonal operators in $\mathcal{L}(H)$
with respect to the chosen basis. Given an ideal
$\mathcal{I}\in\mathcal{L}(H),$ we define its commutative core
$I=\mathcal{I}\cap l_{\infty}$ (that is, the commutative core is the
collection of all diagonal operators in $\mathcal{I}$). A famous result due to
Calkin [17] asserts that one can recover the original ideal $\mathcal{I}$ by
the formula

$\mathcal{I}=\\{A\in\mathcal{L}(H):\ \mu(A)\in I\\}.$

The crucial property of the commutative core $I$ is that,

$x\in I,\ \ \mu(y)=\mu(x)\Longrightarrow y\in I.$ (1)

In fact, Calkin theorem [17] states that every ideal in $l_{\infty}$ with
property (1) is a commutative core of some ideal in $\mathcal{L}(H).$

Whenever an ideal $\mathcal{I}$ in $\mathcal{L}(H)$ is equipped with a Banach
norm111For definiteness, we always assume that
$\|A\|_{\infty}\leq\|A\|_{\mathcal{I}}$ for every $A\in\mathcal{I}.$
$\|\cdot\|$ such that the multiplication $A\to AB$ is continuous mapping on
$\mathcal{I}$ for every $B\in\mathcal{I},$ we call $(\mathcal{I},\|\cdot\|)$ a
Banach ideal. For every Banach ideal $\mathcal{I}$ in $\mathcal{L}(H),$ there
exists an equivalent unitarily invariant norm on $\mathcal{I}$ given by the
formula

$\|A\|_{\mathcal{I}}=\sup_{\begin{subarray}{c}B,C\in\mathcal{L}(H)\\\
\|B\|_{\infty},\|C\|_{\infty}\leq 1\end{subarray}}\|BAC\|.$

That is, we have
$\|AU\|_{\mathcal{I}}=\|UA\|_{\mathcal{I}}=\|A\|_{\mathcal{I}}$ for all
$A\in\mathcal{I}$ and for all unitary $U\in\mathcal{L}(H).$ In particular, we
have

$A\in\mathcal{I},\ \mu(B)=\mu(A)\Longrightarrow B\in\mathcal{I},\
\|B\|_{\mathcal{I}}=\|A\|_{\mathcal{I}}.$

It follows immediately that

$x\in I,\ \ \mu(y)=\mu(x)\Longrightarrow y\in I,\ \ \|y\|_{I}=\|x\|_{I},$ (2)

where $\|\cdot\|_{I}$ is the norm on $I$ induced by $\|\cdot\|_{\mathcal{I}}.$
An arbitrary Banach ideal $I$ in $l_{\infty}$ equipped with a norm
$\|\cdot\|_{I}$ satisfying (2) is called symmetric sequence space (see [46,
48]).

The question whether Calkin correspondence specializes to symmetric sequence
spaces and Banach ideals in $\mathcal{L}(H)$ was fully resolved in [40] (see
also [48]).

######  Theorem 9.

[40, Theorem 8.11] If $(I,\|\cdot\|_{I})$ is a symmetric sequence space, then
the corresponding ideal $\mathcal{I}$ in $\mathcal{L}(H)$ equipped with a
functional

$\|\cdot\|_{\mathcal{I}}:A\to\|\mu(A)\|_{I},\quad A\in\mathcal{I}$

is a Banach ideal in $\mathcal{L}(H).$

A variant of Calkin correspondence exists for arbitrary semifinite von Neumann
algebra and symmetric function spaces on $(0,\infty)$ (see [47, 48]). Theorem
9 also works in this generality (see [40, 48]). However, in this paper, we
deal only with $\mathcal{L}_{p}-$spaces on semifinite von Neumann algebras,
which can be defined without resorting to the general machinery of [40, 48]
(see Subsection 2.7 below).

Let $m\geq 1$ and let $\alpha\in\mathbb{C}^{m}$. We shall frequently view a
finite sequence $\alpha=(\alpha(0),\dots,\alpha({m-1}))$ as an element of
symmetric space $I$ by identifying $\alpha$ with an element
$(\alpha(0),\dots,\alpha({m-1}),0,\dots)$.

###  2.2. Banach ideals in $\mathcal{L}(H):$ examples

The best known example of a Banach ideal in $\mathcal{L}(H)$ is a Schatten-von
Neumann ideal $\mathcal{L}_{p}(H),$ $1\leq p<\infty,$ defined by setting

$\mathcal{L}_{p}(H)=\\{A\in\mathcal{L}(H):\ \mu(A)\in
l_{p}\\},\quad\|A\|_{p}=\|\mu(A)\|_{p},$

where $l_{p}$ is a Banach space of all $p$-summable sequences and
$\|\cdot\|_{p}$ is the standard norm on $l_{p}.$

Given an Orlicz function $M$ (that is, an even convex function on $\mathbb{R}$
such that $M(0)=0$), we define an Orlicz sequence ideal $l_{M}$ in
$l_{\infty}$ as the space of all bounded sequences $x$ such that

$M\left(\frac{x}{\lambda}\right)\in l_{1}$

for some $\lambda>0.$ An Orlicz sequence ideal equipped with the norm

$\|x\|_{l_{M}}=\inf\left\\{\lambda:\
\left\|M\left(\frac{x}{\lambda}\right)\right\|_{1}\leq 1\right\\}$

is a symmetric sequence space. An Orlicz ideal $\mathcal{L}_{M}(H)$ in
$\mathcal{L}(H)$ is now defined by setting

$\mathcal{L}_{M}(H)=\\{A\in\mathcal{L}(H):\ \mu(A)\in
l_{M}\\},\quad\|A\|_{\mathcal{L}_{M}}=\|\mu(A)\|_{l_{M}}.$

###  2.3. Dilation operator

let us recall that for $x=\\{x(k)\\}_{k\geq 0}\in l_{\infty},$ the discrete
dilation operators is defined as follows

$D_{n}(x):=\\{\underbrace{x(0),\ldots,x(0)}_{\mbox{$n$
times}},\underbrace{x(1),\ldots,x(1)}_{\mbox{$n$ times}},\ldots\\},\quad
D_{1/n}(x):=\Big{\\{}\frac{1}{n}\sum_{l=kn}^{(k+1)n-1}x(l)\Big{\\}}_{k\geq 0}$
(3)

and satisfies

$\|D_{n}\|_{l_{p}\to l_{p}}=n^{1/p},\ \ \|D_{1/n}\|_{l_{p}\to
l_{p}}=n^{-1/p},\ \ \ 1\leq p<\infty.$ (4)

The continuous version of dilation operator is defined below. For every
$u\in(0,\infty),$ the dilation operator $D_{u}$ on $L_{1}(0,\infty)$ is
defined by

$(D_{u}x)(s):=x\Big{(}\frac{s}{u}\Big{)},\quad s\in(0,\infty),\quad x\in
L_{1}(0,\infty).$ (5)

For $u\in(0,1),$ the dilation operator $D_{u}$ on $L_{1}(0,1)$ is defined by

$(D_{u}x)(s):=\left\\{\begin{array}[]{cl}x(\frac{s}{u}),&s\in(0,u)\\\
0,&s\in[u,1)\end{array}\right.,\quad x\in L_{1}(0,1).$

For the general properties of dilation operator see [47].

###  2.4. Singular value function

In this paper, $\mathcal{M}$ always stands for a semifinite von Neumann
algebra equipped with a faithful normal semifinite trace $\tau.$ Denote by
$\mathcal{P(M)}$ the lattice of all projections from $\mathcal{M}$ and by
$\mathcal{P}_{fin}(\mathcal{M})$ the set of all $\tau$-finite projections. An
unbounded densely defined operator is said to be affiliated with $\mathcal{M}$
if it commutes with all elements in the commutant of $\mathcal{M}.$ A closed
densely defined operator $A$ affiliated with $\mathcal{M}$ is called
$\tau$-measurable if, for every $n\geq 0,$ we have $\tau(E_{|A|}(n,\infty))\to
0$ as $n\to\infty$, where by $E_{|A|}(n,\infty)$ we denote the spectral
projection of the operator $|A|.$ If, in particular,
$\tau(\mathbbm{1})<\infty,$ then every closed operator affiliated with
$\mathcal{M}$ is $\tau$-measurable. We denote the ∗-algebra of all
$\tau$-measurable operators by $\mathcal{S}(\mathcal{M},\tau)$ with the strong
sum and the strong product (see [26, 48]). By $t_{\tau}$ we denote the measure
topology on $\mathcal{S}(\mathcal{M},\tau).$ Observe that the algebra
$\mathcal{N}$ is dense in $\mathcal{S}(\mathcal{M},\tau)$ with respect to the
measure topology $t_{\tau}$ (see e.g. [21] and references therein). For
$x=x^{*}\in\mathcal{S}(\mathcal{M},\tau)$ by $\exp(ix)$ we denote the series
$\sum_{j=0}^{\infty}\frac{(ix)^{j}}{j!},$ which converges in the measure
topology.

For every $A\in\mathcal{S}(\mathcal{M},\tau),$ its singular value function
$\mu(A):t\to\mu(t,A),$ $t>0,$ is defined by the formula (see e.g. [26])

$\mu(t,A):=\inf\\{\|A(\mathbbm{1}-p)\|_{\infty}:\ \tau(p)\leq t\\}.$

Equivalently, the function $\mu(A)$ can be defined in terms of the
distribution function $d_{|A|}$ of $|A|.$ For an arbitrary self-adjoint
$B\in\mathcal{S}(\mathcal{M},\tau),$ its distribution function $d_{B}$ is
defined by setting

$d_{B}(s):=\tau(E_{B}(s,\infty)),\quad s\in\mathbb{R},$

where $E_{B}$ denotes the spectral measure of the operator $B.$ One can define
$\mu(A)$ as the right inverse of the function $d_{|A|},$ i.e.

$\mu(t;A)=\inf\\{s:d_{|A|}(s)\leq t\\},\quad t\geq 0.$

In addition, if $A_{0},\dots,A_{n-1}\in S(\mathcal{M},\tau)$, then (see e.g.
[48])

$\mu\Big{(}\sum_{i=0}^{n-1}\mu(A_{i})\Big{)}\leq
D_{n}\sum_{i=0}^{n-1}\mu(A_{i}).$ (6)

When $\mathcal{M}=\mathcal{L}(H)$ and $\tau$ is the standard trace ${\rm Tr},$
and when the operator $A$ is compact, the function $\mu(A)$ is a step function
whose values are the singular values of the operator $A.$

We call the operators $A=A^{*},B=B^{*}\in\mathcal{S}(\mathcal{M},\tau)$
equimeasurable if $d_{A_{+}}=d_{B_{+}}$ and $d_{A_{-}}=d_{B_{-}}.$ Observe
that if $\tau(\mathbbm{1})<\infty,$ then operators
$A=A^{*},B=B^{*}\in\mathcal{S}(\mathcal{M},\tau)$ are equimeasurable if and
only if $d_{A}=d_{B}.$ For equimeasurable operators
$A=A^{*},B=B^{*}\in\mathcal{S}(\mathcal{M},\tau),$ we also have
$\mu(A)=\mu(B).$ The notion of equimeasurability for the operators from
different algebras is defined similarly. Finally, we call the operator
$A=A^{*}\in\mathcal{S}(\mathcal{M},\tau)$ symmetrically distributed if it is
equimeasurable with the operator $-A.$

For every $A\in\mathcal{S}(\mathcal{M},\tau),$ the spectral projection
$E_{|A|}(0,\infty)$ is called the support projection of $A$ (written ${\rm
supp}(A)$). If $\tau({\rm supp}(A))<\infty,$ then we call $A$ finitely
supported.

In the special case when $\mathcal{M}=L_{\infty}(0,1)$ and $\tau$ is a
Lebesgue integral, the definition of distribution function (and related
definition of equimeasurability) coincides with the classical one. In this
case the singular valued function $\mu(A)$ defined above coincides with a
classic notion of decreasing rearrangement (see e.g. [46]).

The following notion is a fundamental concept in classical symmetric function
space theory (we refer the interested reader to the book [48, Chapter II] for
detailed discussion of this important notion).

######  Definition 10.

Let $\mathcal{M}$ be a semifinite von Neumann algebra and let
$A,B\in\mathcal{S}(\mathcal{M},\tau).$ The operator $B$ is said to be
submajorized (in the sense of Hardy, Littlewood and Polya) by $A$ and written
$B\prec\prec A$ if

$\int_{0}^{t}\mu(s,B)ds\leq\int_{0}^{t}\mu(s,A)ds,\quad t\geq 0.$

When $\mathcal{M}$ is the algebra $l_{\infty}$ ($L_{\infty}(0,1),$
$L_{\infty}(0,\infty)$) the definition above reduces to the classical notion
of submajorization of sequences (functions).

The following property of submajorization (see e.g. [43, II.6.1]) is used
below

$\sum_{k=1}^{\infty}x_{k}\prec\prec\sum_{k=1}^{\infty}\mu(x_{k}),\quad
x_{k}\in L_{\infty}(0,\infty)\ (x_{k}\in L_{\infty}(0,1)),\ \ k\geq 1.$ (7)

Here, it is assumed that the series on the right hand side converges in
$L_{\infty}(0,\infty)$ (in $L_{\infty}(0,1)$). This guarantees that the series
in the left hand side converges in $L_{\infty}(0,\infty)$ (in
$L_{\infty}(0,1)$).

###  2.5. Properties of the Orlicz norm

For the general theory of Orlicz spaces we refer to [54]. Let us present the
following simple property that is used frequently in the text. If two Orlicz
functions $M_{1}$ and $M_{2}$ satisfies

$M_{1}(t)\leq M_{2}(t),\quad t\in(0,\infty),$

then for any $x\in l_{M_{2}}$, we have that $x\in l_{M_{1}}$ and
$\|x\|_{l_{M_{1}}}\leq\|x\|_{l_{M_{2}}}.$

Two Orlicz functions $M_{1}$ and $M_{2}$ on $(0,\infty)$ are said to be
equivalent on $[0,1]$ ($M_{1}\sim M_{2}$), if there exists a positive finite
constant $C$ such that

$C^{-1}M_{1}(t)\leq M_{2}(t)\leq CM_{1}(t),\ \ \ t\in[0,1].$

An Orlicz function $M$ on $(0,\infty)$ is said to be

  1. (i)

$p$-convex if the function $t\to M(t^{1/p}),$ $t>0,$ is convex.

  2. (ii)

$q$-concave if the function $t\to M(t^{1/q}),$ $t>0,$ is concave.

The proof of lemma below follows immediately from [8, Lemma 5] by substituting
$st=s_{1}$ and $t=s_{2}$ in that lemma.

######  Lemma 11.

Let $1\leq p<q<\infty.$ An Orlicz function $M$ on $(0,\infty)$ is equivalent
on $[0,1]$ to a $p$-convex and $q$-concave Orlicz function if and only if it
satisfies the conditions

$\frac{M(s_{1})}{s_{1}^{p}}\leq{\rm
const}\cdot\frac{M(s_{2})}{s_{2}^{p}},\quad 0<s_{1}\leq s_{2}\leq 1.$ (8)
$\frac{M(s_{1})}{s_{1}^{q}}\geq{\rm
const}\cdot\frac{M(s_{2})}{s_{2}^{q}},\quad 0<s_{1}\leq s_{2}\leq 1.$ (9)

If $M$ itself is $p$-convex and $q$-concave, then the constants in (8) and (9)
are equal to $1.$

######  Remark 12.

Every Orlicz space has the Fatou property (see p.64 in [43]). Every symmetric
sequence space with the Fatou property coincides with its second Köthe dual
(see p.65 in [43]). In particular, the norm in such a space is monotone with
respect to the submajorization.

###  2.6. $p$-concavity, $q$-convexity, upper $p$-estimate and lower
$q$-estimate of Banach ideals

Let $1\leq p,q<\infty$. The Banach ideal $\mathcal{I}\subseteq\mathcal{L}(H)$
is said to be $p$-convex (respectively, $q$-concave) if there exists a
constant $M>0$ such that, for any finite sequence
$\\{x_{k}\\}_{k=0}^{n}\subseteq\mathcal{I}$,

$\Big{\|}\Big{(}\sum_{k=0}^{n}|x_{k}|^{p}\Big{)}^{1/p}\Big{\|}_{\mathcal{I}}\leq
M\Big{(}\sum_{k=0}^{n}\|x_{k}\|_{\mathcal{I}}^{p}\Big{)}^{1/p},$ (10)

respectively,

$\Big{(}\sum_{k=0}^{n}\|x_{k}\|_{\mathcal{I}}^{q}\Big{)}^{1/q}\leq
M\Big{\|}\Big{(}\sum_{k=0}^{n}|x_{k}|^{q}\Big{)}^{1/q}\Big{\|}_{\mathcal{I}}.$
(11)

These definitions are similar to the usual notions of $p$-convexity and
$q$-concavity for Banach lattices as given in [47].

If $x\in\mathcal{L}(H)$, then the right and left support projections of $x$,
denoted by $r(x)$ and $l(x)$ respectively, are the projections onto the
orthogonal complements of the kernels of $x,x^{*}$ respectively. The operators
$x,y\in\mathcal{L}(H)$ are said to be left (respectively, right) disjointly
supported if $r(x)r(y)=0$ (respectively, $l(x)l(y)=0$).

We now make the following definition. Let $1\leq p,q<\infty$. The Banach ideal
$\mathcal{I}\subseteq\mathcal{L}(H)$ is said to satisfy an upper $p$-estimate
(respectively, lower $q$-estimate) if there exists a constant $K>0$ such that,
for every finite sequence $x_{0},\dots,x_{n}$ of pairwise left disjointly
supported elements of $\mathcal{I}$, it follows that

$\Big{\|}\sum_{k=0}^{n}x_{i}\Big{\|}_{\mathcal{I}}\leq
K\Big{(}\sum_{k=0}^{n}\|x_{k}\|_{\mathcal{I}}^{p}\Big{)}^{1/p}$ (12)

respectively,

$K\Big{\|}\sum_{k=0}^{n}x_{i}\Big{\|}_{\mathcal{I}}\geq\Big{(}\sum_{k=0}^{n}\|x_{k}\|_{\mathcal{I}}^{q}\Big{)}^{1/q}.$
(13)

We remark that “left” can be replaced throughout by “right” since $x,y$ are
left disjointly supported if and only if $x^{*},y^{*}$ are right disjointly
supported. For classical (commutative) counterparts of these notions for
Banach lattices we again refer to [47]. For the study of connections between
just defined properties of $\mathcal{I}$ and

corresponding classical properties of its commutative core $I$ we refer to [3]
and [22].

###  2.7. Type and cotype inequalities in the noncommutative
$\mathcal{L}_{p}-$spaces

A noncommutative $L_{p}$-space associated with an algebra $(\mathcal{M},\tau)$
(written $\mathcal{L}_{p}(\mathcal{M})$) is defined by setting

$\mathcal{L}_{p}(\mathcal{M})=\\{A\in\mathcal{S}(\mathcal{M},\tau):\
\tau(|A|^{p})<\infty\\},\quad\|A\|_{p}=\tau(|A|^{p})^{1/p}.$

Observe that $A\in\mathcal{L}_{p}(\mathcal{M})$ if and only if $\mu(A)$
belongs to the Lebesgue space $L_{p}(0,\infty).$ We also have
$\|A\|_{p}=\|\mu(A)\|_{p}$ for every $A\in\mathcal{L}_{p}(\mathcal{M}).$

For reader’s convenience, we show the elementary right hand side estimate of
(15) below. The (substantially harder) proof of the left hand side estimate is
due to Tomczak-Jaegermann and can be found in [62] (see also [25]). For
brevity, we state this result using the language and notations from the theory
of tensor products of von Neumann algebras. Recall that the Rademacher
functions are defined as follows

$r_{k}(t):=\mathrm{sign}\sin(2^{k+1}\pi t),\ \ \ t\in[0,1],\ \ \
k=0,1,2,\ldots$ (14)

(see also [46, 47]).

######  Theorem 13.

Let $n\geq 0,$ let $\mathcal{M}$ be a semifinite von Neumann algebra and let
$A_{k}\in\mathcal{L}_{p}(\mathcal{M}),$ $0\leq k\leq n.$ If $1\leq p<2,$ then

${\rm
const}\cdot\Big{(}\sum_{k=0}^{n}\|A_{k}\|_{p}^{2}\Big{)}^{1/2}\leq\Big{\|}\sum_{k=0}^{n}A_{k}\otimes
r_{k}\Big{\|}_{p}\leq\Big{(}\sum_{k=0}^{n}\|A_{k}\|_{p}^{p}\Big{)}^{1/p}.$
(15)

Here $r_{k},$ $k\geq 0,$ are Rademacher functions and the norm in the middle
is the usual $L_{p}$-norm in the $L_{p}$-space
$\mathcal{L}_{p}(\mathcal{M}\bar{\otimes}L_{\infty}(0,1),\tau\otimes dm).$

###### Proof of the right hand side of (15).

For $p=1,$ it follows from the triangle inequality in
$\mathcal{L}_{1}(\mathcal{M}\bar{\otimes}L_{\infty}(0,1))$ that

$\Big{\|}\sum_{k=0}^{n}A_{k}\otimes
r_{k}\Big{\|}_{1}\leq\sum_{k=0}^{n}\|A_{k}\|_{1},\quad n\geq 0.$

For $p=2,$ it is immediate that

$\Big{\|}\sum_{k=0}^{n}A_{k}\otimes
r_{k}\Big{\|}_{2}=\Big{(}\sum_{k=0}^{n}\|A_{k}\|_{2}^{2}\Big{)}^{1/2},\quad
n\geq 0.$

The right hand side inequality of (15) now follows by applying complex
interpolation method (see e.g. [42]).∎

###  2.8. Independent (noncommutative) random variables

Let $\mathcal{M}$ be a finite von Neumann algebra equipped with a faithful
normal trace such that $\tau(\mathbbm{1})=1,$ where we denote the unit element
of $\mathcal{M}$ by $\mathbbm{1}.$ Self-adjoint operators affiliated with
$\mathcal{M}$ are called (noncommutative) random variables.

Commuting unital subalgebras $\mathcal{A}_{k},$ $1\leq k\leq n,$ of
$\mathcal{M}$ are called 222This definition is taken from [52] (see Definition
5.1 there). independent if

$\tau(A_{1}A_{2}\cdots A_{n})=\tau(A_{1})\tau(A_{2})\cdots\tau(A_{n})$

whenever $A_{k}\in\mathcal{A}_{k},$ $1\leq k\leq n.$ Random variables
$A_{k}\in\mathcal{S}(\mathcal{M},\tau),$ $1\leq k\leq n,$ are called
independent if the von Neumann subalgebras $\mathcal{A}_{k},$ $1\leq k\leq n,$
generated by spectral projections of $A_{k}$ are independent.

###  2.9. Quantization of the algebra $\mathcal{M}$

For a finite (semifinite) von Neumann algebra we use the following notations.

$\mathcal{M}^{\otimes
k}=\underbrace{\mathcal{M}\bar{\otimes}\cdots\bar{\otimes}\mathcal{M}}_{\mbox{$k$
times}},\quad\tau^{\otimes
k}=\underbrace{\tau\otimes\cdots\otimes\tau}_{\mbox{$k$ times}},\quad
k\in\mathbb{N},$

for von Neumann algebra tensor products. It is a standard fact (see e.g.
Proposition 11.2.20 and Proposition 11.2.21 in [39]) that
$\mathcal{M}^{\otimes k}$ is a finite (semifinite) von Neumann algebra and
$\tau^{\otimes k}$ is a faithful normal finite (semifinite) trace on
$\mathcal{M}^{\otimes k}$. For convenience of notations, we set
$\mathcal{M}^{\otimes 0}$ to be $\mathbb{C}$ and $\tau^{\otimes
0}:\mathbb{C}\to\mathbb{C}$ to be identity mapping. Define the positive linear
mapping

$E_{k}:\mathcal{M}^{\otimes k}\to\mathcal{M}^{\otimes k},\quad
E_{k}(x_{1}\otimes\cdots\otimes
x_{k}):=\frac{1}{k!}\sum_{\rho}x_{\rho(1)}\otimes\cdots\otimes
x_{\rho(k)},\quad k\in\mathbb{N},$ (16)

where $\rho$ runs over the set $\mathfrak{S}_{k}$ of all permutations of
$\\{1,\cdots,k\\}.$ It can be directly verified that

$\mathcal{M}_{s}^{\otimes k}:=E_{k}(\mathcal{M}^{\otimes k})$

is a von Neumann subalgebra and that $E_{k}:\mathcal{M}^{\otimes
k}\to\mathcal{M}^{\otimes k}_{s}$ is the conditional expectation. For
convenience of notations, we set $\mathcal{M}^{\otimes 0}_{s}$ to be
$\mathbb{C}.$

Define the (symmetric or bosonic) quantized von Neumann algebra
$M_{s}(\mathcal{M})$ by setting

$M_{s}(\mathcal{M}):=\bigoplus_{k=0}^{\infty}\mathcal{M}^{\otimes k}_{s}.$
(17)

if $\tau(\mathbbm{1})<\infty,$ then the expression

$\sigma:=\exp(-\tau(\mathbbm{1}))\bigoplus_{k=0}^{\infty}\frac{1}{k!}\tau^{\otimes
k}$ (18)

defines a trace on $\bigoplus_{k=0}^{\infty}\mathcal{M}^{\otimes k}$ (and,
obviously on the $*$-subalgebra $M_{s}(\mathcal{M})$). Observe that
$\sigma(\mathbbm{1})=1.$ For a $\tau$-finite projection $e\in\mathcal{M}$ we
will also consider the algebra $M_{s}(e\mathcal{M}e)$ equipped with the trace

$\sigma_{e}:=\exp(-\tau({e}))\bigoplus_{k=0}^{\infty}\frac{1}{k!}\tau^{\otimes
k}.$ (19)

Note that $\sigma_{e}$ is defined on
$\bigoplus_{k=0}^{\infty}(e\mathcal{M}e)^{\otimes k}.$ By $p_{k}$ we denote
natural projections of the algebra $M_{s}(\mathcal{M})$ given by

$p_{k}:=0\oplus\ldots\oplus 0\oplus{\mathbbm{1}^{\otimes k}}\oplus
0\oplus\ldots\in M_{s}(\mathcal{M}),\quad k\geq 0.$ (20)

We identify the algebra $\mathcal{M}^{\otimes k}_{s}$ with a subalgebra of
$M_{s}(\mathcal{M})$ by means of the projection $p_{k}$.

###  2.10. Inductive limits of $C^{*}$-algebras

We refer the reader to the books [59] and [51] for the details of the
inductive limit construction used in Section 4 (note that in [51] the notion
of direct limit is used instead of inductive limit).

######  Lemma 14.

[59, Proposition 1.23.2] Let $\\{\mathcal{A}_{n}\\}_{n\geq 0}$ be
$C^{*}$-algebras. Suppose that, for every $n\leq m,$ there exists a
${}^{*}-$unital isomorphic embedding
$\pi_{n,m}:\mathcal{A}_{n}\to\mathcal{A}_{m}.$ If

$\pi_{n,k}=\pi_{m,k}\circ\pi_{n,m}\quad n\leq m\leq k,$

then there exists a $C^{*}$-algebra $A$ and a net of ∗-isomorphic embeddings
$\pi_{n}:\mathcal{A}_{n}\to A,$ $n\geq 0,$ such that

$\pi_{n}=\pi_{m}\circ\pi_{n,m}\quad 0\leq n\leq m$ (21)

and $\bigcup_{n\geq 1}\pi_{n}(\mathcal{A}_{n})$ is uniformly dense in $A$.

The algebra $A$ from Lemma 14 is recall the inductive limit of
$\\{\mathcal{A}_{n}\\}_{n\geq 0}.$ In parallel, we construct a inductive limit
of states.

######  Lemma 15.

[59, Definition 1.23.10] Let the $C^{*}$-algebra $A$ be the inductive limit of
$C^{*}$-algebras $\mathcal{A}_{n},$ $n\geq 0.$ If states $\sigma_{n}$ on
$\mathcal{A}_{n},$ $n\geq 0,$ are such that

$\sigma_{n}=\sigma_{m}\circ\pi_{n,m},\quad 0\leq n\leq m,$

then there exists a state $\sigma$ on $A$ such that

$\sigma_{n}=\sigma\circ\pi_{n},\quad n\geq 0.$ (22)

If the states $\sigma_{n},$ $n\geq 0,$ are tracial, then so is the limit state
$\sigma$.

Recall that a pair $(\mathcal{A},\sigma)$ is called a noncommutative
probability space, if $\mathcal{A}$ is a von Neumann algebra equipped with
faithful tracial state $\sigma$.

######  Lemma 16.

Let noncommutative probability spaces $(\mathcal{A}_{n},\sigma_{n}),n\geq 0,$
be as in Lemmas 14 and 15, let $A$ and $\sigma$ be the inductive limits of
$\mathcal{A}_{n}$ and $\sigma_{n}$, respectively. Then $\sigma$ induces a
tracial state on von Neumann algebra generated by cyclic representation
$(\theta,H)$ of $A$ with respect to $\sigma$, such that

$\sigma\circ\theta\circ\pi_{n}=\sigma_{n}.$ (23)

###### Proof.

Let $(\theta,H)$ be cyclic representation of $A$ with respect to $\sigma$ and
let $\mathcal{A}:=\overline{\theta(A)}^{wo}$. For every $x\in\mathcal{A}$ we
set $\sigma(x):=\langle x\xi,\xi\rangle_{H}$, where $\xi$ be the cyclic vector
of the representation $(\theta,H)$. One can easily prove that in this way we
can extend the state $\sigma$ defined on $A$ up to a faithful state on
$\mathcal{A}$. By construction of $\sigma$ and (22) we have
$\sigma\circ\theta\circ\pi_{n}=\sigma_{n}.$ It remains to show that this
extension is in fact trace on $\mathcal{A}$. Let firstly $x\in\mathcal{A},y\in
A$, then there exists a sequence $\\{x_{n}\\}_{n\geq 0}\subset A$ such that
$x_{n}\xrightarrow{wo}x$. Since the state $\sigma$ is tracial on $A$ we have
$\sigma(x_{n}y)=\sigma(yx_{n})$. Hence, $wo$-convergence of $x_{n}$ to $x$
implies that $\sigma(xy)=\sigma(yx)$. Further, for arbitrary
$x,y\in\mathcal{A}$ we have $\sigma(xy_{n})=\sigma(y_{n}x)$, where
$y_{n}\xrightarrow{wo}y$, and therefore, $\sigma(xy)=\sigma(yx)$ for all
$x,y\in\mathcal{A}$. ∎

######  Lemma 17.

If $(\mathcal{A}_{n},\sigma_{n}),(\mathcal{A},\sigma),n\geq 0,$ are as in
Lemma 16 and in addition, every $\mathcal{A}_{n}$ is hyperfinite, then
$\mathcal{A}$ is also hyperfinite.

###### Proof.

Let $(\theta,H)$ be as in Lemma 16. For a given $n\geq 0,$ select finite
dimensional subalgebras $\mathcal{A}_{n,m},$ $m\geq 0,$ in $\mathcal{A}_{n}$
such that their union is weak∗ dense in $\mathcal{A}_{n}.$ Define finite
dimensional subalgebras $\mathcal{B}_{n,m},$ $n,m\geq 0,$ in $\mathcal{A}$ by
setting $\mathcal{B}_{n,m}=(\theta\circ\pi_{n})(\mathcal{A}_{n,m}).$ We have
that $\bigcup_{n}\mathcal{B}_{n,m}$ is weak∗ dense in
$(\theta\circ\pi_{n})(\mathcal{A}_{n})$, since trace preserving
$*$-homomorphism $\theta\circ\pi_{n}$ from $\mathcal{A}_{n}$ onto
$(\theta\circ\pi_{n})(\mathcal{A}_{n})$ is weak$*$-continuous [16, Theorem
2.4.23]. Consequently, $\bigcup_{n,m}\mathcal{B}_{n,m}$ is weak∗ dense in
$\bigcup_{n}(\theta\circ\pi_{n})(\mathcal{A}_{n})$. By Lemma 14 the union
$\bigcup_{n}\pi_{n}(\mathcal{A}_{n})$ is uniformly dense in the algebra $A$,
and therefore $\bigcup_{n}(\theta\circ\pi_{n})(\mathcal{A}_{n})$ is weak∗
dense in $\mathcal{A}$. Thus, $\bigcup_{n}\mathcal{B}_{n,m}$ is weak∗ dense in
$\mathcal{A}$ and, therefore, $\mathcal{A}$ is also hyperfinite. ∎

###  2.11. Ultraproducts of von Neumann algebras and $\mathcal{L}_{p}-$spaces

Let $(\mathcal{N}_{k},\tau_{k}),$ $k\geq 0,$ be noncommutative probability
spaces. Consider the $C^{*}-$algebra

$l_{\infty}(\\{\mathcal{N}_{k}\\}_{k\geq 0})=\Big{\\{}\\{x(k)\\}_{k\geq
0}\in\prod_{k\geq 0}\mathcal{N}_{k}:\ \sup_{k\geq
0}\|x(k)\|_{\infty}<\infty\Big{\\}}.$

We fix a free ultrafilter $\omega$ on $\mathbb{Z}_{+}$ and equip
$l_{\infty}(\\{\mathcal{N}_{k}\\}_{k\geq 0})$ with a tracial state
$\Big{(}\tau_{k}\Big{)}_{k\geq 0}^{\omega}$ defined by the formula

$\Big{(}\tau_{k}\Big{)}_{k\geq 0}^{\omega}(\\{x(k)\\}_{k\geq
0})=\lim_{k\to\omega}\\{\tau_{k}(x(k))\\}_{k\geq 0}.$

Consider the ideal $J$ in $l_{\infty}(\\{\mathcal{N}_{k}\\}_{k\geq 0})$
defined by the formula

$J=\Big{\\{}x\in l_{\infty}(\\{\mathcal{N}_{k}\\}_{k\geq 0}):\
\Big{(}\tau_{k}\Big{)}_{k\geq 0}^{\omega}(|x|^{2})=0\Big{\\}}.$

The quotient $C^{*}-$algebra

$\Big{(}\mathcal{N}_{k}\Big{)}_{k\geq 0}^{\omega}\stackrel{{\scriptstyle
def}}{{=}}l_{\infty}(\\{\mathcal{N}_{k}\\}_{k\geq 0})/J$

is, in fact, a von Neumann algebra (see e.g. [50, 64]). The tracial state
$\Big{(}\tau_{k}\Big{)}_{k\geq 0}^{\omega}$ generates a faithful tracial state
on $\Big{(}\mathcal{N}_{k}\Big{)}_{k\geq 0}^{\omega}$ (which we also denote by
$\Big{(}\tau_{k}\Big{)}_{k\geq 0}^{\omega}$). Thus,
$(\Big{(}\mathcal{N}_{k}\Big{)}_{k\geq
0}^{\omega},\Big{(}\tau_{k}\Big{)}_{k\geq 0}^{\omega})$ becomes a
noncommutative probability space. For a special case
$\mathcal{N}_{k}=\mathcal{N},$ $\tau_{k}=\tau,$ it is common to write
$\mathcal{N}^{\omega}$ and $\tau^{\omega}.$

To define a Banach space
$\Big{(}\mathcal{L}_{p}(\mathcal{N}_{k},\tau_{k})\Big{)}_{k\geq 0}^{\omega},$
we equip the set of all bounded sequences $\\{x(k)\\}_{k\geq 0}\in\prod_{k\geq
0}\mathcal{L}_{p}(\mathcal{N}_{k},\tau_{k})$ with a seminorm

$\|\\{x(k)\\}_{k\geq
0}\|_{p}=\lim\limits_{k\to\omega}\|x(k)\|_{p},\quad\\{x(k)\\}_{k\geq
0}\in\prod_{k\geq 0}\mathcal{L}_{p}(\mathcal{N}_{k},\tau_{k})\mbox{ is
bounded}.$

We define $\Big{(}\mathcal{L}_{p}(\mathcal{N}_{k},\tau_{k})\Big{)}_{k\geq
0}^{\omega}$ to be a quotient space over the linear subspace

$\Big{\\{}\\{x(k)\\}_{k\geq 0}:\
\lim\limits_{k\to\omega}\|x(k)\|_{p}=0\Big{\\}}$

and equip it with a quotient norm.

It is tempting to make an identification

$\Big{(}\mathcal{L}_{p}(\mathcal{N}_{k},\tau_{k})\Big{)}_{k\geq
0}^{\omega}=\mathcal{L}_{p}(\Big{(}\mathcal{N}_{k}\Big{)}_{k\geq
0}^{\omega},\Big{(}\tau_{k}\Big{)}_{k\geq 0}^{\omega}).$ (24)

Unfortunately, such an identification does not hold. In fact, we have (see
Lemma 2.13 in [29] and diagram following it) that

$\Big{(}\mathcal{L}_{p}(\mathcal{N}_{k},\tau_{k})\Big{)}_{k\geq
0}^{\omega}\supset\mathcal{L}_{p}(\Big{(}\mathcal{N}_{k}\Big{)}_{k\geq
0}^{\omega},\Big{(}\tau_{k}\Big{)}_{k\geq 0}^{\omega}).$ (25)

However, the following lemma shows how to use the uniform integrability
condition to deduce that a certain element of
$\Big{(}\mathcal{L}_{p}(\mathcal{N}_{k},\tau_{k})\Big{)}_{k\geq 0}^{\omega}$
belongs to the $\mathcal{L}_{p}-$space
$\mathcal{L}_{p}(\Big{(}\mathcal{N}_{k}\Big{)}_{k\geq
0}^{\omega},\Big{(}\tau_{k}\Big{)}_{k\geq 0}^{\omega}).$

The sequence $\\{y(k)\\}_{k\geq 0}\in\prod_{k\geq
0}\mathcal{L}_{1}(\mathcal{N}_{k},\tau_{k})$ is said to be uniformly
integrable if, for every $\varepsilon>0,$ there exists $K$ such that
$\tau(|y(k)|E_{|y(k)|}(K,\infty))\leq\varepsilon$ for all $k\geq 0.$

######  Lemma 18.

If the sequence $x=\\{x(k)\\}_{k\geq 0}\in\prod_{k\geq
0}\mathcal{L}_{p}(\mathcal{N}_{k},\tau_{k})$ is such that the sequence
$\\{|x(k)|^{p}\\}_{k\geq 0}\in\prod_{k\geq
0}\mathcal{L}_{1}(\mathcal{N}_{k},\tau_{k})$ is uniformly integrable, then

$x\in\mathcal{L}_{p}(\Big{(}\mathcal{N}_{k}\Big{)}_{k\geq
0}^{\omega},\Big{(}\tau_{k}\Big{)}_{k\geq 0}^{\omega}).$

###### Proof.

Without loss of generality, we have $x(k)\geq 0,$ $k\geq 0.$ For a given
$t>0,$ consider a operator $x_{t}=\\{x(k)E_{x(k)}[0,t]\\}_{n\geq
0}\in\Big{(}\mathcal{N}_{k}\Big{)}_{k\geq 0}^{\omega}.$ We have that

$\|x-x_{t}\|_{\Big{(}\mathcal{L}_{p}(\mathcal{N}_{k},\tau_{k})\Big{)}_{k\geq
0}^{\omega}}=\lim\limits_{k\to\omega}\|x(k)E_{x(k)}(t,\infty)\|_{p}\leq(\sup_{k\geq
0}\tau(x(k)^{p}E_{x(k)}(t,\infty)))^{1/p}.$

Since the sequence $\\{x(k)^{p}\\}_{k\geq 0},$ is uniformly integrable, it
follows that the right hand side tends to $0$ as $t\to\infty.$ Thus, $x$
admits an approximation in
$\Big{(}\mathcal{L}_{p}(\mathcal{N}_{k},\tau_{k})\Big{)}_{k\geq 0}^{\omega}$
by bounded operators. By Lemma 2.13 in [29], we have that
$x\in\mathcal{L}_{p}(\Big{(}\mathcal{N}_{k}\Big{)}_{k\geq
0}^{\omega},\Big{(}\tau_{k}\Big{)}_{k\geq 0}^{\omega}).$ ∎

##  3\. Noncommutative Kruglov operator ($\tau$-finite case)

The Kruglov operator was originally introduced in the setting of classical
(commutative) probability theory in [4, 5]. The origins of this construction
go back to Braverman [13] and Kruglov [44]. Note that in commutative setting
the Kruglov operator is originally defined on the space of all measurable
functions (see [5] and [6]). In order to achieve main objective of this paper
we will consider this operator on $L_{1}$-spaces associated with semifinite
von Neumann algebras.

Let $\Omega=\prod_{k=0}^{\infty}(0,1)$ be the (infinite dimensional) hypercube
equipped with the product Lebesgue measure $dm^{\infty}.$ The Kruglov operator
$K$ acts from $L_{1}(0,1)$ to $L_{1}(\Omega)$ by the following formula

$Kx=\sum_{k=1}^{\infty}\sum_{m=1}^{k}\chi_{A_{k}}\otimes\chi_{(0,1)}^{\otimes(m-1)}\otimes
x\otimes\chi_{(0,1)}^{\otimes\infty},\quad x\in L_{1}(0,1),$ (26)

where $A_{k},$ $k\geq 0,$ are pairwise disjoint sets with
$m(A_{k})=\frac{1}{e\cdot k!}$ for all $k\geq 0$ (so that $\cup_{k\geq
0}A_{k}=(0,1)$) and where $\chi_{B}$ is the indicator function of the
measurable set $B\subset(0,1).$

The following assertion is proved in [8]

######  Theorem 19.

The Kruglov operator $K:L_{1}(0,1)\to L_{1}(\Omega)$ defined in (26) satisfies
the following properties.

  1. (i)

$K$ maps positive random variables into positive ones.

  2. (ii)

$K$ maps symmetrically distributed random variables into symmetrically
distributed ones.

  3. (iii)

If $0\leq x,y\in L_{1}(0,1)$ are equimeasurable, then so are $Kx,Ky\in
L_{1}(\Omega).$

  4. (iv)

If $x_{k}\in L_{1}(0,1),$ $1\leq k\leq n,$ are pairwise disjointly supported
functions, then the functions $Kx_{k}\in L_{1}(0,1),$ $1\leq k\leq n,$ are
independent.

The main aim of this section is to construct the noncommutative Kruglov
operator in a $\tau$-finite von Neumann algebra. Recall that if
$x_{k}\in\mathcal{S}(\mathcal{M},\tau),$ $k\geq 0,$ if $e_{k}\in\mathcal{M},$
$k\geq 0,$ are pairwise orthogonal projections, and if $y_{k}\in
e_{k}\mathcal{S}(\mathcal{M},\tau)e_{k}$ are such that
$\mu(y_{k})=\mu(x_{k}),$ $k\geq 0,$ then we write

$\bigoplus_{k=0}^{\infty}x_{k}:=\sum_{k=0}^{\infty}y_{k}.$ (27)

If $x_{0},x_{1},\ldots\in\mathcal{S}(\mathcal{M},\tau)$ then we view
$\bigoplus_{k=0}^{\infty}x_{k}$ as an element of
$\mathcal{S}\big{(}\bigoplus_{k=0}^{\infty}\mathcal{M},\bigoplus_{k=0}^{\infty}\tau\big{)}.$

Let $\mathcal{N}$ be a von Neumann algebra equipped with a faithful normal
finite trace $\tau.$ For every $x\in\mathcal{S}(\mathcal{N},\tau),$ set

$\mathscr{K}x=0\oplus\bigoplus_{k=1}^{\infty}\left(\sum_{m=1}^{k}\mathbbm{1}^{\otimes(m-1)}\otimes
x\otimes\mathbbm{1}^{\otimes(k-m)}\right).$ (28)

Here, $\mathbbm{1}$ denotes the unit element in $\mathcal{N}.$ By definition
$\mathscr{K}x$ belongs to
$\mathcal{S}\Big{(}\bigoplus_{k=0}^{\infty}\mathcal{N}^{\otimes
k},\sigma\Big{)}$ for every $x\in\mathcal{S}(\mathcal{N},\tau),$ where
$\sigma$ is defined by (18). The operator

$\mathscr{K}:\mathcal{S}(\mathcal{N},\tau)\to\mathcal{S}\Big{(}\bigoplus_{k=0}^{\infty}\mathcal{N}^{\otimes
k},\sigma\Big{)}$

defined by (28) is said to be the (noncommutative) Kruglov operator.

The following lemma shows that the values of the Kruglov operator belong in
fact to a smaller algebra.

######  Lemma 20.

For any $x\in\mathcal{S}(\mathcal{N},\tau),$ we have
$\mathscr{K}x\in\mathcal{S}(M_{s}(\mathcal{N}),\sigma),$ where the algebra
$M_{s}(\mathcal{N})$ is defined by (17).

###### Proof.

In order to prove the assertion of the lemma it is enough to show that
$y_{k}(x)\in\mathcal{S}(\mathcal{N}^{\otimes k}_{s},\tau^{\otimes k}),$ for
any $x\in\mathcal{S}(\mathcal{N},\tau),$ where

$y_{k}(x):=\sum_{m=1}^{k}\mathbbm{1}^{\otimes(m-1)}\otimes
x\otimes\mathbbm{1}^{\otimes(k-m)},\quad k\geq 1.$ (29)

Let us first show that

$y_{k}(x)\in\mathcal{N}^{\otimes k}_{s}\quad\text{for every}\quad
x\in\mathcal{N},\quad k\geq 1.$ (30)

Indeed, since $x\in\mathcal{N},$ it follows that
$y_{k}(x)\in\mathcal{N}^{\otimes k}.$ Consider the element

$z_{k}(x):=k(x\otimes\mathbbm{1}^{\otimes(k-1)})\in\mathcal{N}^{\otimes k},\ \
\ k\geq 1.$

Then by the definition of $E_{k}$ (see (16)), we have

$E_{k}(z_{k}(x))=k\cdot\frac{1}{k!}\sum_{\rho}\tilde{x}_{\rho(1)}\otimes\cdots\otimes\tilde{x}_{\rho(k)},\
\ k\geq 1,$ (31)

where $\tilde{x}_{\rho(j)}=x,$ if $\rho(j)=1$ and
$\tilde{x}_{\rho(j)}=\mathbbm{1}$ otherwise, $1\leq j\leq k,$ and $\rho$ runs
over the set of all permutations $\mathfrak{S}_{k}$. Expanding the right hand
side of (31), we obtain

$E_{k}(z_{k}(x))=k\cdot\frac{1}{k!}\left((k-1)!(x\otimes\mathbbm{1}^{\otimes(k-1)})+(k-1)!(\mathbbm{1}\otimes
x\otimes\mathbbm{1}^{\otimes(k-2)})+\ldots\right.\\\
\left.+(k-1)!(\mathbbm{1}^{\otimes(k-1)}\otimes
x)\right)=\sum_{m=1}^{k}\mathbbm{1}^{\otimes(m-1)}\otimes
x\otimes\mathbbm{1}^{\otimes(k-m)}=y_{k}(x),\ \ k\geq 1,$

that completes the proof of (30).

Now let $x\in\mathcal{S}(\mathcal{N},\tau).$ Fix $k\geq 1$ and consider
$x_{n}=xE_{x}([-n,n])$ and $y_{k}(x_{n})$, $n\geq 1.$ Since
$x_{n}\in\mathbb{N},$ by (30) we have that
$y_{k}(x_{n})\in\mathcal{N}^{\otimes k}_{s}$ for every $n\geq 1.$ Now observe
that the sequence $\\{y_{k}(x_{n})\\}_{n=1}^{\infty}$ converges to $y_{k}(x)$
in the measure topology $t_{\tau^{\otimes k}}$. Since
$\mathcal{S}(\mathcal{N}^{\otimes k}_{s},\tau^{\otimes k})$ is complete with
respect to the measure topology, we obtain that
$y_{k}(x)\in\mathcal{S}(\mathcal{N}^{\otimes k}_{s},\tau^{\otimes k}).$ ∎

The following proposition collects some of important properties of the
operator $\mathscr{K}.$ Recall that $[x,y]:=xy-yx,$
$x,y\in\mathcal{S}(\mathcal{N},\tau)$.

######  Proposition 21.

  1. (i)

$\mathscr{K}x\in\mathcal{L}_{1}(M_{s}(\mathcal{N}),\sigma)$ and
$\sigma(\mathscr{K}x)=\tau(x)$ for every
$x\in\mathcal{L}_{1}(\mathcal{N},\tau).$

  2. (ii)

For any $x,y\in\mathcal{S}(\mathcal{N},\tau)$, we have that
$[\mathscr{K}x,\mathscr{K}y]=\mathscr{K}[x,y].$

  3. (iii)

$\mathscr{K}x\geq 0$ whenever $x\geq 0;$

  4. (iv)

$(\mathscr{K}x)^{*}=\mathscr{K}x$ whenever $x=x^{*}.$

###### Proof.

To prove (i), we show that $\sigma(\mathscr{K}x)=\tau(x)$ for every
$x\in\mathcal{L}_{1}(\mathcal{N},\tau).$ Indeed,

$\displaystyle\sigma(\mathscr{K}x)$
$\displaystyle=\sum_{k=1}^{\infty}\frac{1}{\exp{(\tau(\mathbbm{1}))}\cdot k!}\
\tau^{\otimes k}\left(\sum_{m=1}^{k}\mathbbm{1}^{\otimes(m-1)}\otimes
x\otimes\mathbbm{1}^{\otimes(k-m)}\right)$
$\displaystyle=\sum_{k=1}^{\infty}\frac{k\cdot\tau(x)\tau(\mathbbm{1})^{k-1}}{\exp(\tau(\mathbbm{1}))\cdot
k!}=\frac{\tau(x)}{\exp(\tau(\mathbbm{1}))}\sum_{k=1}^{\infty}\frac{\tau(\mathbbm{1})^{k-1}}{(k-1)!}=\tau(x).$

(ii). A straightforward computation yields

$\displaystyle\Big{(}\sum_{m_{1}=1}^{k}1^{\otimes(m_{1}-1)}\otimes x\otimes
1^{\otimes(k-m_{1})}\Big{)}\Big{(}\sum_{m_{2}=1}^{k}1^{\otimes(m_{2}-1)}\otimes
y\otimes 1^{\otimes(k-m_{2})}\Big{)}$
$\displaystyle=\sum_{m_{1}=1}^{k-1}\sum_{m_{2}=m_{1}+1}^{k}1^{\otimes(m_{1}-1)}\otimes
x\otimes 1^{\otimes(m_{2}-m_{1}-1)}\otimes y\otimes 1^{\otimes(k-m_{2})}$
$\displaystyle+\sum_{m_{2}=1}^{k-1}\sum_{m_{1}=m_{2}+1}^{k}1^{\otimes(m_{2}-1)}\otimes
y\otimes 1^{\otimes(m_{1}-m_{2}-1)}\otimes x\otimes 1^{\otimes(k-m_{1})}$
$\displaystyle+\sum_{m=1}^{k}1^{\otimes(m-1)}\otimes xy\otimes
1^{\otimes(k-m)}\ \ $

for every $k\geq 1$. It follows that, for every $k\geq 1,$

$\displaystyle\Big{(}\sum_{m=1}^{k}1^{\otimes(m-1)}\otimes x\otimes
1^{\otimes(k-m)}\Big{)}\Big{(}\sum_{m=1}^{k}1^{\otimes(m-1)}\otimes y\otimes
1^{\otimes(k-m)}\Big{)}$
$\displaystyle-\Big{(}\sum_{m=1}^{k}1^{\otimes(m-1)}\otimes y\otimes
1^{\otimes(k-m)}\Big{)}\Big{(}\sum_{m=1}^{k}1^{\otimes(m-1)}\otimes x\otimes
1^{\otimes(k-m)}\Big{)}$ $\displaystyle=\sum_{m=1}^{k}1^{\otimes(m-1)}\otimes
xy\otimes 1^{\otimes(k-m)}-\sum_{m=1}^{k}1^{\otimes(m-1)}\otimes yx\otimes
1^{\otimes(k-m)}$
$\displaystyle=\sum_{m=1}^{k}1^{\otimes(m-1)}\otimes[x,y]\otimes
1^{\otimes(k-m)}.$

Thus, we have

$[\mathscr{K}x,\mathscr{K}y]=0\oplus\bigoplus_{k=1}^{\infty}\left(\Big{(}\sum_{m=1}^{k}1^{\otimes(m-1)}\otimes
x\otimes 1^{\otimes(k-m)}\Big{)}\Big{(}\sum_{m=1}^{k}1^{\otimes(m-1)}\otimes
y\otimes 1^{\otimes(k-m)}\Big{)}\right.$
$\left.-\Big{(}\sum_{m=1}^{k}1^{\otimes(m-1)}\otimes y\otimes
1^{\otimes(k-m)}\Big{)}\Big{(}\sum_{m=1}^{k}1^{\otimes(m-1)}\otimes x\otimes
1^{\otimes(k-m)}\Big{)}\right)=\mathscr{K}[x,y].$

Property (iii) follows from the definition of the operator $\mathscr{K}$.
Property (iv) is a straightforward subsequence of (iii). ∎

The following result shows the connection between noncommutative and
commutative Kruglov operators.

######  Lemma 22.

Let $\tau(\mathbbm{1})=1.$ Then for every positive
$x\in\mathcal{L}_{1}(\mathcal{N},\tau),$ we have
$\mu(K\mu(x))=\mu(\mathscr{K}x).$

###### Proof.

Fix $k\in\mathbb{N}$. Observe that the operators $y_{k}(x)$ defined by (29)
with respect to the algebra $(\mathcal{N}^{\otimes k},\tau^{\otimes k})$ and

$\sum_{m=1}^{k}\chi_{(0,1)}^{\otimes(m-1)}\otimes\mu(x)\otimes\chi_{(0,1)}^{\otimes(k-m)},$

with respect to $(\Omega,dm^{\infty})$ are equimeasurable (see e.g. [18]).
Therefore, the operator $y_{k}(x)$ with respect to the algebra
$(\mathcal{N}^{\otimes k},\frac{1}{e^{\tau(\mathbbm{1})}\cdot
k!}\cdot\tau^{\otimes k})$ is equimeasurable with the operator

$\chi_{A_{k}}\otimes\sum_{m=1}^{k}\chi_{(0,1)}^{\otimes(m-1)}\otimes\mu(x)\otimes\chi_{(0,1)}^{\otimes(k-m)},$

with respect to $(\Omega,dm^{\infty})$, where $A_{k}$ is the same as in (26).
Thus, the operators $K\mu(x)$ and $\mathscr{K}x$ are equimeasurable as a
direct sum of equimeasurable operators. ∎

It is not difficult to see that there is a much stronger connection than
indicated by Lemma 22 between the commutative and noncommutative Kruglov
operators. The following theorem states this connection in full generality.
Since we do not use this theorem in this paper, we omit the proof.

######  Theorem 23.

Suppose that $\mathcal{N}=L_{\infty}(0,1).$ Then there exists trace preserving
unital $*$-homomorphism $\gamma:M_{s}(\mathcal{N})\to L_{\infty}(\Omega)$ such
that its $L_{1}$-extension $\tilde{\gamma}$ satisfies
$\tilde{\gamma}(\mathscr{K}(x))=Kx$ for any $x\in L_{1}(0,1).$

The following important lemma delivers the expression for the characteristic
function of the random variable $\mathscr{K}x$ in terms of that of the random
variable $x$ (compare with the Kruglov property studied in [13, 7]).

######  Lemma 24.

Let $\mathcal{N}$ be a finite von Neumann algebra equipped with a finite trace
$\tau.$ The Kruglov operator
$\mathscr{K}:\mathcal{L}_{1}(\mathcal{N},\tau)\to\mathcal{L}_{1}(M_{s}(\mathcal{N}),\sigma)$
satisfies the condition

$\sigma(\exp(i\mathscr{K}x))=\exp(\tau(\exp(ix)-\mathbbm{1})),\quad
x=x^{*}\in\mathcal{L}_{1}(\mathcal{N},\tau),$

where $\sigma$ is defined by (18) and $\mathbbm{1}$ is the unit element of the
algebra $\mathcal{N}.$

###### Proof.

Let $x=x^{*}\in\mathcal{L}_{1}(\mathcal{N},\tau)$ and let $p_{k}$ ($k\geq 0$)
be a projection of the algebra $M_{s}(\mathcal{N})$ defined in (20). By
definition of $\sigma$ we have that

$\sigma(\exp(i\mathscr{K}x))=\exp(-\tau(\mathbbm{1}))\sum_{k=0}^{\infty}\frac{1}{k!}\tau^{\otimes
k}(p_{k}\exp(i\mathscr{K}x)).$ (32)

Let $y\in\mathcal{S}(M_{s}(\mathcal{N}),\sigma).$ Observing that
$p_{k}y^{m}=(p_{k}y)^{m}$ for all $m\geq 0$ and $k\geq 0,$ we infer that
$p_{k}(\exp(iy))=\exp(ip_{k}y)$. It follows that

$p_{k}\exp(i\mathscr{K}x)=\exp(ip_{k}\mathscr{K}x),\quad k\geq 0.$ (33)

Next we will show that

$\tau^{\otimes k}(\exp(ip_{k}\mathscr{K}x))=\tau^{k}(\exp(ix)),\quad k\geq 0.$
(34)

Indeed, for every $x=x^{*}\in\mathcal{L}_{1}(\mathcal{N},\tau)$ and $k\geq 0$
we have

$\exp\left(i\sum_{m=1}^{k}\mathbbm{1}^{\otimes(m-1)}\otimes
x\otimes\mathbbm{1}^{\otimes(k-m)}\right)=\prod_{m=1}^{k}\exp\left(\mathbbm{1}^{\otimes(m-1)}\otimes
ix\otimes\mathbbm{1}^{\otimes(k-m)}\right).$ (35)

Considering each term of the latter product separately for every $k\geq 0$ and
$1\leq m\leq k,$ we obtain

$\displaystyle\exp(\mathbbm{1}^{\otimes(m-1)}\otimes
ix\otimes\mathbbm{1}^{\otimes(k-m)})$ (36)
$\displaystyle=\sum_{j=0}^{\infty}\frac{1}{j!}\Big{(}\mathbbm{1}^{\otimes(m-1)}\otimes
ix\otimes\mathbbm{1}^{\otimes(k-m)}\Big{)}^{j}=\mathbbm{1}^{\otimes(m-1)}\otimes\left(\sum_{j=0}^{\infty}\frac{(ix)^{j}}{j!}\right)\otimes\mathbbm{1}^{\otimes(k-m)}$
$\displaystyle=\mathbbm{1}^{\otimes(m-1)}\otimes\underbrace{\exp(ix)}_{m}\otimes\mathbbm{1}^{\otimes(k-m)}.$

Combining (35) and (36) we arrive at

$\exp\left(i\sum_{m=1}^{k}\mathbbm{1}^{\otimes(m-1)}\otimes
x\otimes\mathbbm{1}^{\otimes(k-m)}\right)=\exp(ix)^{\otimes k}.$

Hence, for $k\geq 0$ we obtain

$\tau^{\otimes k}(\exp(ip_{k}\mathscr{K}x))=\tau^{\otimes
k}\left(\exp\left(i\sum_{m=1}^{k}\mathbbm{1}^{\otimes(m-1)}\otimes
x\otimes\mathbbm{1}^{\otimes(k-m)}\right)\right)=\tau^{k}(\exp(ix)),$

which completes the proof of (34). Applying subsequently (32), (33) and (34)
we conclude

$\displaystyle\sigma(\exp(i\mathscr{K}x))=\exp(-\tau(\mathbbm{1}))\sum_{k=0}^{\infty}\frac{1}{k!}\tau^{\otimes
k}(p_{k}\exp(i\mathscr{K}x))$
$\displaystyle=\exp(-\tau(\mathbbm{1}))\sum_{k=0}^{\infty}\frac{1}{k!}\tau^{\otimes
k}(\exp(ip_{k}\mathscr{K}x))=\exp(-\tau(\mathbbm{1}))\sum_{k=0}^{\infty}\frac{\tau^{k}(\exp(ix))}{k!}$
$\displaystyle=\exp(\tau(\exp(ix)-\mathbbm{1})).\qed$

The following result is the noncommutative extension of Theorem 19.

######  Theorem 25.

The operator
$\mathscr{K}:\mathcal{L}_{1}(\mathcal{N},\tau)\to\mathcal{L}_{1}(M_{s}(\mathcal{N}),\sigma)$
satisfies the following properties.

  1. (i)

$\mathscr{K}$ maps positive random variables into positive ones.

  2. (ii)

$\mathscr{K}$ maps symmetrically distributed random variables into
symmetrically distributed ones.

  3. (iii)

If random variables $x,y\in\mathcal{L}_{1}(\mathcal{N},\tau)$ are
equimeasurable, then so are
$\mathscr{K}x,\mathscr{K}y\in\mathcal{L}_{1}(M_{s}(\mathcal{N}),\sigma).$

  4. (iv)

If random variables $x_{k}\in\mathcal{L}_{1}(\mathcal{N},\tau),$ $1\leq k\leq
n,$ are such that $x_{j}x_{k}=0$ for $j\neq k,$ then random variables
$\mathscr{K}x_{k}\in\mathcal{L}_{1}(M_{s}(\mathcal{N}),\sigma),$ $1\leq k\leq
n,$ are independent (in the sense of definition given in Subsection 2.8).

###### Proof.

The property (i) is proved in Proposition 21 (iii). The property (ii) is a
straightforward consequence of (iii) and (iv). The property (iii) is also
straightforward. Indeed, since

$\sum_{m=1}^{k}1^{\otimes(m-1)}\otimes x\otimes 1^{\otimes(k-m)}\ \
\mbox{and}\ \ \sum_{m=1}^{k}1^{\otimes(m-1)}\otimes y\otimes
1^{\otimes(k-m)},$

are equimeasurable, whenever $x,y\in\mathcal{S}(\mathcal{N},\tau)$ are, the
equimeasurability of $\mathscr{K}x$ and $\mathscr{K}y$ follows from (28).

Next, we concentrate on (iv). Throughout the proof we fix
$x_{k}=x_{k}^{*}\in\mathcal{L}_{1}(\mathcal{N},\tau),$ $1\leq k\leq n,$ such
that $x_{j}x_{k}=0$ for $j\neq k.$ First we show that

$\exp\Big{(}i\sum_{k=1}^{n}\lambda_{k}x_{k}\Big{)}-\mathbbm{1}=\sum_{k=1}^{n}\Big{(}\exp(i\lambda_{k}x_{k})-\mathbbm{1}\Big{)},\quad\lambda_{k}\in\mathbb{R},$
(37)

where $\mathbbm{1}$ is a unit element of the algebra $\mathcal{N}.$ Indeed,
the condition on $x_{k}\in\mathcal{L}_{1}(\mathcal{N},\tau),$ $1\leq k\leq n,$
implies that there exists a commutative von Neumann subalgebra
$\mathcal{N}_{0}$ such that $x_{k}\in\mathcal{L}_{1}(\mathcal{N}_{0},\tau),$
$1\leq k\leq n.$ By spectral theorem,
$x_{k}\in\mathcal{L}_{1}(\mathcal{N}_{0},\tau),$ $1\leq k\leq n,$ can be
realised as functions (with disjoint support) on some measure space. For these
functions equality (37) is verified pointwise.

Therefore, applying $\tau$ and $\exp$ to both parts in (37), we obtain

$\exp\Big{(}\tau\Big{(}\exp\Big{(}i\sum_{k=1}^{n}\lambda_{k}x_{k}\Big{)}-\mathbbm{1}\Big{)}\Big{)}=\prod_{k=1}^{n}\exp(\tau(\exp(i\lambda_{k}x_{k})-\mathbbm{1})).$
(38)

Applying Lemma 24 to both expressions below, we obtain

$\exp\Big{(}\tau\Big{(}\exp\Big{(}i\sum_{k=1}^{n}\lambda_{k}x_{k}\Big{)}-\mathbbm{1}\Big{)}\Big{)}=\sigma\Big{(}\exp\Big{(}i\sum_{k=1}^{n}\lambda_{k}\mathscr{K}x_{k}\Big{)}\Big{)}$

and

$\prod_{k=1}^{n}\exp(\tau(\exp(i\lambda_{k}x_{k})-\mathbbm{1}))=\prod_{k=1}^{n}\sigma(\exp(i\lambda_{k}\mathscr{K}x_{k})).$

Using (38), we obtain

$\sigma\Big{(}\exp\Big{(}i\sum_{k=1}^{n}\lambda_{k}\mathscr{K}x_{k}\Big{)}\Big{)}=\prod_{k=1}^{n}\sigma(\exp(i\lambda_{k}\mathscr{K}x_{k})).$
(39)

By Lemma 21 (ii), the operators $\mathscr{K}x_{k},$ $1\leq k\leq n,$ commute.
Hence,

$\exp\Big{(}i\sum_{k=1}^{n}\lambda_{k}\mathscr{K}x_{k}\Big{)}=\prod_{k=1}^{n}\exp(i\lambda_{k}\mathscr{K}x_{k}).$
(40)

Combining (39) and (40), we arrive at

$\sigma\Big{(}\prod_{k=1}^{n}\exp(i\lambda_{k}\mathscr{K}x_{k})\Big{)}=\prod_{k=1}^{n}\sigma(\exp(i\lambda_{k}\mathscr{K}x_{k})).$
(41)

Let $f_{k},$ $1\leq k\leq n,$ be Schwartz functions. Denoting by $\mathcal{F}$
and $\mathcal{F}^{-1}$ the Fourier transform and the inverse Fourier transform
respectively, we have

$f_{k}(\mathscr{K}x_{k})=(\mathcal{F}^{-1}\mathcal{F}f_{k})(\mathscr{K}x_{k})=\int_{\mathbb{R}}(\mathcal{F}f_{k})(\lambda_{k})\
\exp(i\lambda_{k}\mathscr{K}x_{k})d\lambda_{k},$ (42)

where the latter integral is understood as $L_{1}$-limit of the series

$2^{-n}\sum_{m=-2^{2n}}^{2^{2n}}(\mathcal{F}f_{k})\Big{(}\frac{m}{2^{n}}\Big{)}\exp\Big{(}i\frac{m}{2^{n}}\mathscr{K}x_{k}\Big{)},\quad\mbox{when}\quad
n\to\infty.$

The fact that this series converges in $\mathcal{L}_{1}(H)$ can be found in
[30, Theorem 6.2]. Therefore, using (41), we obtain

$\displaystyle\begin{split}\sigma\Big{(}\prod_{k=1}^{n}f_{k}(\mathscr{K}x_{k})\Big{)}&=\sigma\Big{(}\prod_{k=1}^{n}\int_{\mathbb{R}}(\mathcal{F}f_{k})(\lambda_{k})\exp(i\lambda_{k}\mathscr{K}x_{k})d\lambda_{k}\Big{)}\\\
&=\sigma\Big{(}\int_{\mathbb{R}^{n}}\prod_{k=1}^{n}\Big{(}(\mathcal{F}f_{k})(\lambda_{k})\exp(i\lambda_{k}\mathscr{K}x_{k})\Big{)}d\lambda_{1}\ldots
d\lambda_{n}\Big{)}\\\
&=\int_{\mathbb{R}^{n}}\prod_{k=1}^{n}(\mathcal{F}f_{k})(\lambda_{k})\cdot\sigma\Big{(}\prod_{k=1}^{n}\exp(i\lambda_{k}\mathscr{K}x_{k})\Big{)}d\lambda_{1}\ldots
d\lambda_{n}\\\
&\stackrel{{\scriptstyle\eqref{xdg2}}}{{=}}\int_{\mathbb{R}^{n}}\prod_{k=1}^{n}(\mathcal{F}f_{k})(\lambda_{k})\cdot\prod_{k=1}^{n}\sigma(\exp(i\lambda_{k}\mathscr{K}x_{k}))\
d\lambda_{1}\ldots d\lambda_{n}\\\
&=\prod_{k=1}^{n}\Big{(}\int_{\mathbb{R}}(\mathcal{F}f_{k})(\lambda_{k})\cdot\sigma(\exp(i\lambda_{k}\mathscr{K}x_{k}))\
d\lambda_{k}\Big{)}\\\
&=\prod_{k=1}^{n}\sigma\Big{(}\int_{\mathbb{R}}(\mathcal{F}f_{k})(\lambda_{k})\
\exp(i\lambda_{k}\mathscr{K}x_{k})\ d\lambda_{k}\Big{)}\\\
&=\prod_{k=1}^{n}\sigma(f_{k}(\mathscr{K}x_{k})).\end{split}$ (43)

Let $\mathcal{A}_{k}$ be the von Neumann subalgebra in $\mathcal{M}$ generated
by the spectral family of the element $\mathscr{K}x_{k},$ $1\leq k\leq n.$.
Since Schwartz functions are weak∗ dense in $L_{\infty}(\mathbb{R}),$ it
follows that the set

$\\{f(\mathscr{K}x_{k}),\ f\mbox{ is Schwartz function}\\}$

is weak∗ dense in $\mathcal{A}_{k}.$ Hence, by (43), we arrive at

$\sigma\Big{(}\prod_{k=1}^{n}u_{k}\Big{)}=\prod_{k=1}^{n}\sigma(u_{k}),\quad
u_{k}\in\mathcal{A}_{k},\quad 1\leq k\leq n.$

By definition given in Subsection 2.8, elements $\mathscr{K}x_{k},$ $1\leq
k\leq n,$ are independent. ∎

##  4\. Noncommutative Kruglov operator ($\tau-$infinite case)

The main objective of this section is to adapt definition (28) given in the
preceding section for $\tau$-finite von Neumann algebras to the setting when
$\mathcal{N}$ is a semifinite von Neumann algebra. This is a much harder task
and its accomplishment is based on the approach from [34].

Fix a semifinite (nonfinite) von Neumann algebra $(\mathcal{N},\tau)$ with a
faithful normal semifinite trace $\tau$. For a fixed $\tau$-finite projection
$e$ from $\mathcal{N}$ and $n\in\mathbb{N}$ we consider the mapping
$\alpha_{n,k}:e\mathcal{N}e\to\mathcal{N}^{\otimes n},0\leq k\leq n,$ defined
by setting

$\alpha_{n,k}(x)=\binom{n}{k}E_{n}(x^{\otimes
k}\otimes(\mathbbm{1}-e)^{\otimes(n-k)}),\quad x\in e\mathcal{N}e,$ (44)

where $E_{n}$ is the conditional expectation from $\mathcal{N}^{\otimes n}$
onto $\mathcal{N}^{\otimes n}_{s}$ defined by (16).

######  Lemma 26.

For every $e\in\mathcal{P}_{fin}(\mathcal{N})$ and $k,n\in\mathbb{N},0\leq
k\leq n$, the mapping $\alpha_{n,k}:e\mathcal{N}e\to\mathcal{N}^{\otimes n}$
defined by (44) preserves multiplication.

###### Proof.

We denote by $2^{n}$ the set of all subsets of $\\{1,\cdots,n\\}.$ Since for
every $x\in e\mathcal{N}e$ the mapping $E_{n}(x^{\otimes
k}\otimes(\mathbbm{1}-e)^{\otimes(n-k)})$ contains $k!(n-k)!$ equal summands
(see (16)), by (44) we infer that

$\alpha_{n,k}(x)=\binom{n}{k}E_{n}(x^{\otimes
k}\otimes(\mathbbm{1}-e)^{\otimes(n-k)})=\sum_{\begin{subarray}{c}A\in
2^{n}\\\ |A|=k\end{subarray}}\mathcal{Z}_{A}^{x,\mathbbm{1}-e},$ (45)

where

$\mathcal{Z}_{A}^{x,y}=z_{1}\otimes\dots\otimes z_{n}\in\mathcal{N}^{\otimes
n},\quad z_{i}=\left\\{\begin{aligned} x,&\quad i\in A\\\
y,&\quad\mathrm{otherwise}\end{aligned}\right.\quad 1\leq i\leq n.$ (46)

Therefore,

$\alpha_{n,k}(x)\alpha_{n,k}(y)=\sum_{\begin{subarray}{c}A_{1},A_{2}\in
2^{n}\\\
|A_{1}|=|A_{2}|=k\end{subarray}}\mathcal{Z}_{A_{1}}^{x,\mathbbm{1}-e}\mathcal{Z}_{A_{2}}^{y,\mathbbm{1}-e}$

If $A_{1}\neq A_{2},$ then, , we may assume that there exists $l\in A_{1}$
such that $l\notin A_{2}$ and, since $x(\mathbbm{1}-e)=0,$ it follows that the
corresponding summand is $0.$ Hence,

$\alpha_{n,k}(x)\alpha_{n,k}(y)=\sum_{\begin{subarray}{c}A\in 2^{n}\\\
|A|=k\end{subarray}}\mathcal{Z}_{A}^{x,\mathbbm{1}-e}\mathcal{Z}_{A}^{y,\mathbbm{1}-e}=\sum_{\begin{subarray}{c}A\in
2^{n}\\\
|A|=k\end{subarray}}\mathcal{Z}_{A}^{xy,\mathbbm{1}-e}=\alpha_{n,k}(xy).$

∎

Now, we construct a $*-$homomorphism from $(e\mathcal{N}e)^{\otimes k}_{s}$
into $\mathcal{N}^{\otimes n}_{s},$ $e\in\mathcal{P}_{fin}(\mathcal{N}),$
$n,k\in\mathbb{N}$. We define the linear mapping
$\pi_{n,k}:(e\mathcal{N}e)^{\otimes k}\to\mathcal{N}_{s}^{\otimes n},0\leq
k\leq n,$ by the following rule:

$\pi_{n,k}^{e,\mathbbm{1}}:z\to\binom{n}{k}E_{n}(z\otimes(\mathbbm{1}-e)^{\otimes(n-k)}),\quad
z\in(e\mathcal{N}e)^{\otimes k}.$ (47)

In what follows, we frequently suppress the dependence of
$\pi_{n,k}^{e,\mathbbm{1}}$ on $e$ and $\mathbbm{1}$ and simply write
$\pi_{n,k}$.

It is clear that for $\alpha_{n,k}$, defined by (44), the equality

$\alpha_{n,k}(x)=\pi_{n,k}(x^{\otimes k}),\,x\in e\mathcal{N}e$ (48)

holds.

We need the following technical lemma, whose proof consists of tedious
algebraic computations and therefore is omitted.

######  Lemma 27.

Let $n\geq k$. We have

  1. (i)

$\pi_{n,k}(x)=\pi_{n,k}\circ E_{k}(x),\,x\in(e\mathcal{N}e)^{\otimes k}$;

  2. (ii)

For $u=u_{1}\otimes\cdots\otimes u_{k}\in(e\mathcal{N}e)^{\otimes k}$ we have

$E_{k}(u)=\frac{1}{k!}\frac{\partial^{k}}{\partial^{k}\lambda}\Big{(}\sum_{i=1}^{k}\lambda_{i}u_{i}\Big{)}^{\otimes
k},\lambda_{i}\in\mathbb{R},$

where we use the notation $\partial^{k}\lambda$ as a shorthand for
$\partial\lambda_{1}\cdots\partial\lambda_{k}$;

  3. (iii)

$\pi_{n,k}\left(\frac{\partial^{k}}{\partial^{k}\lambda}(\sum_{i=1}^{k}\lambda_{i}u_{i})^{\otimes
k}\right)=\frac{\partial^{k}}{\partial^{k}\lambda}\pi_{n,k}((\sum_{i=1}^{k}\lambda_{i}u_{i})^{\otimes
k}).$

######  Lemma 28.

Let $e\in\mathcal{P}_{fin}(\mathcal{N}),\,n,k\in\mathbb{N},0\leq k\leq n$ and
let $\pi_{n,k}$ be defined by (47). Then, the restriction of $\pi_{n,k}$ onto
the algebra $(e\mathcal{N}e)^{\otimes k}_{s}$ is a $*$-homomorphism from
$(e\mathcal{N}e)^{\otimes k}_{s}$ into $\mathcal{N}^{\otimes n}_{s}$.

###### Proof.

For the proof of this lemma we firstly show that for all
$u,v\in(e\mathcal{N}e)^{\otimes k}$ the equality

$\pi_{n,k}(u)\pi_{n,k}(v)=\pi_{n,k}(E_{k}(u)E_{k}(v))$ (49)

holds. Fix $u=u_{1}\otimes\cdots\otimes u_{k}$ and
$v=v_{1}\otimes\cdots\otimes v_{k}$. By Lemma 27 (ii) we have

$E_{k}(u)E_{k}(v)=\frac{1}{k!^{2}}\frac{\partial^{2k}}{\partial^{k}\lambda\partial^{k}\mu}\left(\Big{(}\sum_{i=1}^{k}\lambda_{i}u_{i}\Big{)}\Big{(}\sum_{j=1}^{k}\mu_{j}v_{j}\Big{)}\right)^{\otimes
k}.$ (50)

Therefore, again applying Lemma 27, we obtain

$\displaystyle\begin{split}\pi_{n,k}(u)\pi_{n,k}(v)&\stackrel{{\scriptstyle(i)}}{{=}}\pi_{n,k}(E_{k}(u))\pi_{n,k}(E_{k}(v))\\\
&=\frac{1}{k!}\pi_{n,k}\Big{(}\frac{\partial^{k}}{\partial^{k}\lambda}\Big{(}\sum_{i=1}^{k}\lambda_{i}u_{i}\Big{)}^{\otimes
k}\Big{)}\cdot\frac{1}{k!}\pi_{n,k}\Big{(}\frac{\partial^{k}}{\partial^{k}\mu}\Big{(}\sum_{j=1}^{k}\mu_{j}u_{j}\Big{)}^{\otimes
k}\Big{)}\\\
&\stackrel{{\scriptstyle(iii)}}{{=}}\frac{1}{k!^{2}}\frac{\partial^{2k}}{\partial^{k}\lambda\partial^{k}\mu}\pi_{n,k}\Big{(}\Big{(}\sum_{i=1}^{k}\lambda_{i}u_{i}\Big{)}^{\otimes
k}\Big{)}\pi_{n,k}\Big{(}\Big{(}\sum_{j=1}^{k}\mu_{j}u_{j}\Big{)}^{\otimes
k}\Big{)}\\\
&\stackrel{{\scriptstyle\eqref{alpha_pi}}}{{=}}\frac{1}{k!^{2}}\frac{\partial^{2k}}{\partial^{k}\lambda\partial^{k}\mu}\alpha_{n,k}\Big{(}\sum_{i=1}^{k}\lambda_{i}u_{i}\Big{)}\alpha_{n,k}\Big{(}\sum_{j=1}^{k}\mu_{j}u_{j}\Big{)}.\end{split}$

Since by Lemma 26 the mapping $\alpha_{n,k}$ preserves multiplication, it
follows that

$\begin{split}\pi_{n,k}(u)\pi_{n,k}(v)&=\frac{1}{k!^{2}}\frac{\partial^{2k}}{\partial^{k}\lambda\partial^{k}\mu}\alpha_{n,k}\Big{(}\Big{(}\sum_{i=1}^{k}\lambda_{i}u_{i}\Big{)}\Big{(}\sum_{j=1}^{k}\mu_{j}u_{j}\Big{)}\Big{)}\\\
&\stackrel{{\scriptstyle\eqref{alpha_pi}}}{{=}}\frac{1}{k!^{2}}\frac{\partial^{2k}}{\partial^{k}\lambda\partial^{k}\mu}\pi_{n,k}\Big{(}\Big{(}\Big{(}\sum_{i=1}^{k}\lambda_{i}u_{i}\Big{)}\Big{(}\sum_{j=1}^{k}\mu_{j}u_{j}\Big{)}\Big{)}^{\otimes
k}\Big{)}\\\
&=\pi_{n,k}\Big{(}\frac{1}{k!^{2}}\frac{\partial^{2k}}{\partial^{k}\lambda\partial^{k}\mu}\Big{(}\Big{(}\sum_{i=1}^{k}\lambda_{i}u_{i}\Big{)}\Big{(}\sum_{j=1}^{k}\mu_{j}u_{j}\Big{)}\Big{)}^{\otimes
k}\Big{)}\\\ &\stackrel{{\scriptstyle\eqref{second
linear}}}{{=}}\pi_{n,k}(E_{k}(u)E_{k}(v)).\end{split}$

Since $E_{k}$ is a conditional expectation from $(e\mathcal{N}e)^{\otimes k}$
into $(e\mathcal{N}e)_{s}^{\otimes k}$, for every
$u,v\in(e\mathcal{N}e)_{s}^{\otimes k}$ we have $E_{k}(u)=u,E_{k}(v)=v$, and
therefore by (49) we obtain

$\pi_{n,k}(u)\pi_{n,k}(v)=\pi_{n,k}(E_{k}(u)E_{k}(v))=\pi_{n,k}(uv).$

Since $E_{n}(x^{*})=(E_{n}(x))^{*},x\in\mathcal{N}^{\otimes n}$, one can
easily obtain that $\pi_{n,k}(u^{*})=\pi_{n,k}(u)^{*}$ for all
$u\in(e\mathcal{N}e)_{s}^{\otimes k}$. Thus, $\pi_{n,k}$ is a $*$-homomorphism
from $(e\mathcal{N}e)_{s}^{\otimes k}$ into $\mathcal{N}_{s}^{\otimes n}$.

∎

The following auxiliary lemma is used in Proposition 30 below.

######  Lemma 29.

Let $1\leq k\leq n.$ For every $x\in\mathcal{N},$ we have

$\sum_{k=1}^{n}k\binom{n}{k}E_{n}(x\otimes
e^{\otimes(k-1)}\otimes(\mathbbm{1}-e)^{\otimes(n-k)})=\sum_{k=1}^{n}\mathbbm{1}^{\otimes(k-1)}\otimes
x\otimes\mathbbm{1}^{\otimes(n-k)}\in\mathcal{N}^{\otimes n}_{s}.$

###### Proof.

It is sufficient to consider only the terms on the LHS having $x$ in the first
position, since all other terms can be treated similarly. Therefore, by
definition of $E_{n}$ (see (16)), we have to prove the equality

$x\otimes\Big{(}\sum_{k=1}^{n}k\binom{n}{k}\frac{1}{n}E_{n-1}(e^{\otimes(k-1)}\otimes(\mathbbm{1}-e)^{\otimes(n-k)})\Big{)}=x\otimes\mathbbm{1}^{n}.$

By (45), we have that

$\binom{n-1}{k-1}E_{n-1}(e^{\otimes(k-1)}\otimes(\mathbbm{1}-e)^{\otimes(n-k)})=\sum_{\begin{subarray}{c}A\in
2^{n-1}\\\ |A|=k-1\end{subarray}}e_{A},$

where $e_{A}$ is defined by (46). Hence,

$LHS=x\otimes\Big{(}\sum_{k=1}^{n}k\binom{n}{k}\frac{1}{n\binom{n-1}{k-1}}\sum_{\begin{subarray}{c}A\in
2^{n-1}\\\ |A|=k-1\end{subarray}}e_{A}\Big{)}=x\otimes\Big{(}\sum_{A\in
2^{n-1}}e_{A}\Big{)}=x\otimes\mathbbm{1}^{\otimes(n-1)}.$

The second assertion follows from the proof of Lemma 20. ∎

Let $(\mathcal{N},\tau)$ be a semifinite (nonfinite) von Neumann algebra with
faithful normal semifinte trace $\tau$, let $e\leq f$ be arbitrary
$\tau$-finite projections from $\mathcal{N}$ and let
$M_{s}(e\mathcal{N}e),M_{s}(f\mathcal{N}f)$ be symmetric quantized algebras
with corresponding finite traces $\sigma_{e},\sigma_{f}$ given by (19). Denote
by $\mathscr{K}_{e}$ (respectively, $\mathscr{K}_{f}$) the corresponding
Kruglov operators on $e\mathcal{N}e$ (respectively, on $f\mathcal{N}f$)
constructed in Section 3 (see (28)).

The following proposition is the key result used in the construction of the
Kruglov operator $\mathcal{K}$ on $\mathcal{L}_{1}(\mathcal{N}).$ It allows us
to pass from Kruglov operator defined on a smaller algebra, to the Kruglov
operator on a larger algebra.

######  Proposition 30.

Let $(\mathcal{N},\tau)$ be a semifinite von Neumann algebra. For all
$\tau$-finite projections $e\leq f\in\mathcal{N},$ there exists a unital
$*-$homomorphism

$\pi_{e,f}:S(M_{s}(e\mathcal{N}e),\sigma_{e})\to
S(M_{s}(f\mathcal{N}f),\sigma_{f})$

such that

  1. (i)

for all $z\in M_{s}(e\mathcal{N}e)$, we have

$\sigma_{e}(z)=\sigma_{\mathbbm{1}}\circ\pi_{e,\mathbbm{1}}(z),$ (51)

in particular
$\pi_{e,f}:\mathcal{L}_{1}(M_{s}(e\mathcal{N}e),\sigma_{e})\to\mathcal{L}_{1}(M_{s}(f\mathcal{N}f),\sigma_{f})$;

  2. (ii)

for all $x\in\mathcal{L}_{1}(e\mathcal{N}e,\tau)$ the equality

$\pi_{e,f}(\mathscr{K}_{e}(x))=\mathscr{K}_{f}(x)$ (52)

holds.

###### Proof.

Without loss of generality, we can assume that $f=\mathbbm{1}.$ Let $p_{k}$ be
the natural projection from $M_{s}(e\mathcal{N}e)=\bigoplus_{k\geq
0}(e\mathcal{N}e)^{\otimes k}_{s}$ into $(e\mathcal{N}e)_{s}^{\otimes k}.$
Define the mapping $\pi_{e,\mathbbm{1}}:M_{s}(e\mathcal{N}e)\to
M_{s}(\mathcal{N})$ by setting

$\pi_{e,\mathbbm{1}}=\bigoplus_{n\geq 0}\sum_{k=0}^{n}\pi_{n,k}\circ p_{k},$
(53)

where $\pi_{n,k}$ is defined by (47).

Let $k\neq l,k,l\leq n$ and $u=u_{1}\otimes\dots\otimes
u_{k}\in(e\mathcal{N}e)^{\otimes k},v=v_{1}\otimes\dots\otimes
v_{l}\in(e\mathcal{N}e)^{\otimes l}$. Each term of the product

$E_{n}(u_{1}\otimes\dots\otimes
u_{k}\otimes(\mathbbm{1}-e)^{\otimes(n-k)})E_{n}(v_{1}\otimes\dots\otimes
v_{l}\otimes(\mathbbm{1}-e)^{\otimes(n-l)})$

contains the product $u_{i}(\mathbbm{1}-e)=0$ or $(\mathbbm{1}-e)v_{i}=0$ on
some $j$-th place of tensor product, and therefore,

$\pi_{n,k}(u)\pi_{n,l}(v)=\binom{n}{k}\binom{n}{l}E_{n}(u\otimes(\mathbbm{1}-e)^{\otimes(n-k)})E_{n}(v\otimes(\mathbbm{1}-e)^{\otimes(n-l)})=0$

for all $u\in(e\mathcal{N}e)^{\otimes k},v\in(e\mathcal{N}e)^{\otimes l}.$
Consequently, since $p_{k}(z_{1})\in(e\mathcal{N}e)_{s}^{\otimes
k},p_{l}(z_{2})\in(e\mathcal{N}e)_{s}^{\otimes l},z_{1},z_{2}\in
M_{s}(e\mathcal{N}e)$ we have

$(\pi_{n,k}\circ p_{k})(z_{1})\cdot(\pi_{n,l}\circ p_{l})(z_{2})=0,\quad
z_{1},z_{2}\in M_{s}(e\mathcal{N}e),k\neq l,k,l\leq n.$ (54)

Since, by Lemma 28, $\pi_{n,k}:(e\mathcal{N}e)^{\otimes
k}_{s}\to\mathcal{N}^{\otimes n}_{s},$ $n\geq k,$ is a $*-$homomorphism, for
every $z_{1},z_{2}\in M_{s}(e\mathcal{N}e),$ we obtain

$\displaystyle\Big{(}\sum_{k=0}^{n}(\pi_{n,k}\circ
p_{k})(z_{1})\Big{)}\Big{(}\sum_{l=0}^{n}(\pi_{n,l}\circ
p_{l})(z_{2})\Big{)}=\sum_{k,l=0}^{n}(\pi_{n,k}\circ
p_{k})(z_{1})(\pi_{n,l}\circ p_{l})(z_{2})$
$\displaystyle\stackrel{{\scriptstyle\eqref{pi_n,k,l}}}{{=}}\sum_{k=0}^{n}(\pi_{n,k}\circ
p_{k})(z_{1})(\pi_{n,k}\circ p_{k})(z_{2})=\sum_{k=0}^{n}(\pi_{n,k}\circ
p_{k})(z_{1}z_{2}),$

and similarly

$\sum_{k=0}^{n}(\pi_{n,k}\circ p_{k})(z^{*})=(\sum_{k=0}^{n}(\pi_{n,k}\circ
p_{k})(z))^{*},z\in M_{s}(e\mathcal{N}e).$

Therefore, the mapping

$\pi_{e,\mathbbm{1}}=\bigoplus_{n\geq 0}\sum_{k=0}^{n}\pi_{n,k}\circ
p_{k}:M_{s}(e\mathcal{N}e)\to M_{s}(\mathcal{N})$

is a $*-$homomorphism as a direct sum of $*$-homomorphisms. In addition, (see
also the end of the proof of Lemma (29)),

$\sum_{k=1}^{n}\pi_{n,k}(\mathbbm{1}^{\otimes
k})\stackrel{{\scriptstyle\eqref{alpha_pi}}}{{=}}\sum_{k=1}^{n}\alpha_{n,k}(\mathbbm{1})\stackrel{{\scriptstyle\eqref{E_n_hatx}}}{{=}}\sum_{k=1}^{n}\sum_{\begin{subarray}{c}A\in
2^{n}\\\ |A|=k\end{subarray}}e_{A}=\sum_{A\in 2^{n}}e_{A}=\mathbbm{1}^{\otimes
n}.$

That is $\pi_{e,\mathbbm{1}}$ is a unital $*$-homomorphism from
$M_{s}(e\mathcal{N}e)$ into $M_{s}(f\mathcal{N}f)$.

Let us show now, that the $*$-homomorphism $\pi_{e,\mathbbm{1}}$ is trace
preserving, i.e.

$\sigma_{e}(z)=\sigma_{\mathbbm{1}}\circ\pi_{e,\mathbbm{1}}(z),\quad z\in
M_{s}(e\mathcal{N}e).$

Clearly, this assertion needs to be verified on each direct summand of
$M_{s}(e\mathcal{N}e)$.

Let $z=(e\mathcal{N}e)_{s}^{\otimes k}$. Since $p_{m}(z)=0$ for all $m\neq k$,
by the definition of $\sigma_{e},$ (see (19)) we have

$\sigma_{e}(z)=\exp(-\tau(e))\frac{\tau^{\otimes k}(z)}{k!}.$

Again, by the definition of $\sigma_{\mathbbm{1}},$ we have

$\displaystyle\begin{split}(\sigma\circ\pi_{e,\mathbbm{1}})(z)&=\sigma\Big{(}\bigoplus_{n\geq
0}\sum_{k=0}^{n}\pi_{n,k}\circ p_{k}(z)\Big{)}\\\
&=\sigma\Big{(}\bigoplus_{n\geq
k}\pi_{n,k}(z)\Big{)}=\exp(-\tau(\mathbbm{1}))\sum_{n\geq
k}\frac{1}{n!}\tau^{\otimes n}(\pi_{n,k}(z))\\\
&=\exp(-\tau(\mathbbm{1}))\sum_{n\geq k}\frac{1}{n!}\binom{n}{k}\tau^{\otimes
k}(z)\tau^{n-k}(e-\mathbbm{1})=\exp(-\tau(e))\frac{\tau^{\otimes
k}(z)}{k!},\end{split}$

that is $\sigma_{e}(z)=(\sigma_{\mathbbm{1}}\circ\pi_{e,\mathbbm{1}})(z).$

Thus the $*$-homomorphism $\pi_{e,\mathbbm{1}}:M_{s}(e\mathcal{N}e)\to
M_{s}(\mathcal{N})$ is trace preserving, and therefore (see e.g. [23])
$\pi_{e,\mathbbm{1}}$ can be extended to a unital $*$-homomorphism from
$S(M_{s}(e\mathcal{N}e),\sigma_{e})$ into $S(M_{s}(\mathcal{N}),\sigma)$ In
particular, $\pi_{e,\mathbbm{1}}$ maps
$\mathcal{L}_{1}(M_{s}(e\mathcal{N}e),\sigma_{e})$ into
$\mathcal{L}_{1}(M_{s}(\mathcal{N}),\sigma)$. We preserve the notation
$\pi_{e,\mathbbm{1}}$ for this extension.

For the proof of equality (52) fix $x\in\mathcal{L}_{1}(e\mathcal{N}e,\tau)$.
By the definition of $\mathcal{K}_{e}$ (see (28)), we have
$p_{0}(\mathscr{K}_{e}x)=0$ and

$p_{k}(\mathscr{K}_{e}x)=\sum_{m=1}^{k}e^{\otimes(m-1)}\otimes x\otimes
e^{\otimes(k-m)},\quad k\geq 1.$

It follows now from Lemma 29 that

$\displaystyle\begin{split}\pi_{e,\mathbbm{1}}(\mathscr{K}_{e}x)&\stackrel{{\scriptstyle\eqref{pi
def}}}{{=}}\bigoplus_{n\geq 0}\sum_{k=1}^{n}\pi_{n,k}\circ
p_{k}(\mathscr{K}_{e}x)\\\
&\stackrel{{\scriptstyle\eqref{pi_n,k}}}{{=}}0\oplus\bigoplus_{n\geq
1}\sum_{k=1}^{n}\binom{n}{k}\sum_{m=1}^{k}E_{n}(e^{\otimes(m-1)}\otimes
x\otimes e^{\otimes(k-m)}\otimes(\mathbbm{1}-e)^{\otimes(n-k)})\\\
&=0\oplus\bigoplus_{n\geq 1}\sum_{k=1}^{n}k\binom{n}{k}E_{n}(x\otimes
e^{\otimes(k-1)}\otimes(\mathbbm{1}-e)^{\otimes(n-k)})\\\
&\stackrel{{\scriptstyle L.\ref{strange
equality}}}{{=}}0\oplus\bigoplus_{n\geq
1}\sum_{k=1}^{n}\mathbbm{1}^{\otimes(k-1)}\otimes
x\otimes\mathbbm{1}^{\otimes(n-k)}=\mathscr{K}_{\mathbbm{1}}x.\end{split}$

∎

Note that, the equality (51) can be viewed as a noncommutative analogue of the
compatibility of measures in Kolmogorov extension theorem [27].

The following lemma shows that $*$-homomorphisms constructed in Proposition 30
are well-compatible, that allows us to construct the inductive limit of
algebras of the form $M_{s}(e\mathcal{N}e),e\in\mathcal{P}_{fin}(\mathcal{N})$
(see the proof of Theorem 32).

######  Lemma 31.

We have $\pi_{e,g}(z)=(\pi_{f,g}\circ\pi_{e,f})(z)$ whenever $e\leq f\leq
g,z\in M_{s}(e\mathcal{N}e).$

###### Proof.

Let $x\in e\mathcal{N}e$ and let $n\geq k\geq m.$ We have (cf. (45) and (48)),

$\pi_{k,m}^{e,f}(x^{\otimes m})=\sum_{\begin{subarray}{c}A_{1}\in 2^{k}\\\
|A_{1}|=m\end{subarray}}\mathcal{Z}_{A_{1}}^{x,f-e},$

where $\mathcal{Z}_{A_{1}}^{x,f-e}$ is defined in (46), and therefore,

$\pi_{n,k}^{f,g}(\pi_{k,m}^{e,f}(x^{\otimes
m}))=\binom{n}{k}E_{n}\Big{(}\Big{(}\sum_{\begin{subarray}{c}A_{1}\in 2^{k}\\\
|A_{1}|=m\end{subarray}}\mathcal{Z}_{A_{1}}^{x,f-e}\Big{)}\otimes(g-f)^{\otimes(n-k)}\Big{)}=$
$=\sum_{\begin{subarray}{c}A_{1}\in 2^{k}\\\
|A_{1}|=m\end{subarray}}\sum_{\begin{subarray}{c}A_{2}\in 2^{n}\\\
A_{2}\supset A_{1}\\\
|A_{2}|=k\end{subarray}}\mathcal{Z}_{A_{1},A_{2}}^{x,f-e,g-f},$

where

$\mathcal{Z}_{A_{1},A_{2}}^{x,f-e,g-f}=z_{1}\otimes\dots\otimes
z_{n}\in\mathcal{N}^{\otimes n},\quad z_{i}=\left\\{\begin{aligned} x,&\quad
i\in A_{1}\\\ f-e,&\quad i\in A_{2}\setminus A_{1}\\\
g-f,&\quad\mathrm{otherwise}\end{aligned}\right.\quad 1\leq i\leq n,$

And hence we obtain,

$\sum_{k=m}^{n}\pi_{n,k}^{f,g}(\pi_{k,m}^{e,f}(x^{\otimes
m}))=\sum_{\begin{subarray}{c}A_{1}\in 2^{n}\\\
|A_{1}|=m\end{subarray}}\sum_{\begin{subarray}{c}A_{2}\in 2^{n}\\\
A_{2}\supset A_{1}\end{subarray}}\mathcal{Z}_{A_{1},A_{2}}^{x,f-e,g-f}.$ (55)

Next we shall prove that for a fixed $A_{1}\in 2^{n}$ such that $|A_{1}|=m,$
we have

$\mathcal{Z}_{A_{1}}^{x,g-e}=\sum_{\begin{subarray}{c}A_{2}\in 2^{n}\\\
A_{2}\supset A_{1}\end{subarray}}\mathcal{Z}_{A_{1},A_{2}}^{x,f-e,g-f}.$ (56)

Indeed, observing that the following formula

$(y+z)^{\otimes(n-m)}=\sum_{A\in 2^{n-m}}\mathcal{Z}_{A}^{y,z},\ \ \mbox{for
all}\ \ y,z\in\mathcal{N}.$ (57)

holds, that can be easily established by induction, plugging $x$ on fixed
$A_{1}$ positions between elements of tensors on the both sides of (57), we
obtain that

$\mathcal{Z}_{A_{1}}^{x,y+z}=\sum_{\begin{subarray}{c}A\in 2^{n}\\\ A\cap
A_{1}=\varnothing\end{subarray}}\mathcal{Z}_{A_{1},A_{1}\cup
A}^{x,y,z}=\sum_{\begin{subarray}{c}A_{2}\in 2^{n}\\\ A_{2}\supset
A_{1}\end{subarray}}\mathcal{Z}_{A_{1},A_{2}}^{x,y,z},\ \ \mbox{for all}\ \
y,z\in\mathcal{N}.$ (58)

Now taking $y=f-e,$ $z=g-f$ in (58), we arrive at (56). Combining (55) and
(56), we obtain that

$\sum_{k=m}^{n}\pi_{n,k}^{f,g}(\pi_{k,m}^{e,f}(x^{\otimes
m}))=\sum_{\begin{subarray}{c}A_{1}\in 2^{n}\\\
|A_{1}|=m\end{subarray}}Z_{A_{1}}^{x,g-e}.$

Consequently, we have that

$\pi_{n,m}^{e,g}(x^{\otimes
m})=\sum_{k=m}^{n}\pi_{n,k}^{f,g}(\pi_{k,m}^{e,f}(x^{\otimes m})),\,x\in
e\mathcal{N}e.$ (59)

By polarization identity every element $z_{1}\otimes\dots\otimes
z_{m},z_{i}\in(e\mathcal{N}e),1\leq i\leq m,$ is a (finite) linear combination
of the elements of the form $x^{\otimes m}$, and hence equality (59) holds for
these elements too. Since elements of the form $z_{1}\otimes\dots\otimes
z_{m},z_{i}\in(e\mathcal{N}e),1\leq i\leq m$ are dense in
$(e\mathcal{N}e)_{s}^{\otimes m}$ (in the weak∗ topology) and by construction
the $*-$homomorphism $\pi_{n,k}$ is continuous in the weak∗ topology (see
(47)), we infer the equality (59) for all $z\in(e\mathcal{N}e)_{s}^{\otimes
m}.$

Since for every $z\in(e\mathcal{N}e)_{s}^{\otimes m},$ we have $p_{k}(z)=0$
for $k\neq m$ and $p_{m}(z)=z,$ it follows from construction of the mapping
$\pi_{e,g}$ that

$\displaystyle\begin{split}\pi_{f,g}(\pi_{e,f}(z))&\stackrel{{\scriptstyle\eqref{pi
def}}}{{=}}\pi_{f,g}(\bigoplus_{k\geq 0}\sum_{m=0}^{k}(\pi_{k,m}^{e,f}\circ
p_{m})(z))=\pi_{f,g}(\bigoplus_{k\geq m}\pi_{k,m}^{e,f}(z))\\\
&=\bigoplus_{n\geq
m}\sum_{k=m}^{n}\pi_{n,k}^{f,g}(\pi_{k,m}^{e,f}(z))\stackrel{{\scriptstyle\eqref{compatpi1}}}{{=}}\bigoplus_{n\geq
m}\pi_{n,m}^{e,g}(z)=\pi_{e,g}(z).\end{split}$

for every $z\in(e\mathcal{N}e)_{s}^{\otimes m}.$ It follows immediately that

$\pi_{e,g}(z)=\pi_{f,g}(\pi_{e,f}(z)),\quad z\in M_{s}(e\mathcal{N}e).$

∎

The following theorem is the main result of this section. It presents the
construction of Kruglov operator for arbitrary semifinite von Neumann algebra.
The idea of the construction used in Theorem 32 is adopted from [34].

######  Theorem 32.

Let $(\mathcal{N},\tau)$ be a semifinite von Neumann algebra. There exists a
finite von Neumann algebra $(\mathcal{M},\sigma)$ and a positive bounded
linear operator
$\mathscr{K}:\mathcal{L}_{1}(\mathcal{N},\tau)\to\mathcal{L}_{1}(\mathcal{M},\sigma)$
such that

$\sigma(\exp(i\mathscr{K}x))=\exp(\tau(\exp(ix)-\mathrm{1})),\quad
x=x^{*}\in\mathcal{L}_{1}(\mathcal{N},\tau).$ (60)

If $\mathcal{N}$ is hyperfinite, then so is $\mathcal{M}.$

###### Proof.

Let $e_{n}\in\mathcal{N}$ be a sequence of $\tau-$finite projections such that
$e_{n}\uparrow\mathbbm{1}$. For all $n\in\mathbb{N}$ consider the finite von
Neumann algebra $(e_{n}\mathcal{N}e_{n},\tau_{e_{n}})$. By Proposition 30 and
Lemma 31 there exist unital $*$-homomorphisms

$\pi_{e_{n},e_{m}}:M_{s}(e_{n}\mathcal{N}e_{n})\to
M_{s}(e_{m}\mathcal{N}e_{m}),n\leq m$

such that $\pi_{e_{n},e_{m}}=\pi_{e_{k},e_{m}}\circ\pi_{e_{n},e_{k}}$ for all
$n\leq k\leq m$. Hence, by Lemma 14 there exists the inductive limit of unital
$C^{*}$-algebras

$M=\lim_{n}M_{s}(e_{n}\mathcal{N}e_{n})$

and canonical $*$-homomorphisms $\pi_{n}:M_{s}(e_{n}\mathcal{N}e_{n})\to M$.
Furthermore, since every algebra $M_{s}(e_{n}\mathcal{N}e_{n})$ is equipped
with finite trace $\sigma_{e_{n}}$ and by Proposition 30 $*$-homomorphisms
$\pi_{e_{n},e_{k}},k\leq n,$ preserve these traces, by Lemma 15 the equality
$\sigma=\lim_{n}\sigma_{e_{n}}$ defines a tracial state on $M$. By Lemma 16
the trace $\sigma$ can be extended on the von Neumann algebra
$\mathcal{M}:=\overline{\theta(M)}^{wo}$, where $\theta$ is cyclic
representation of $M$ for $\sigma$ on a Hilbert space $H$. Thus
$(\mathcal{M},\sigma)$ is a finite von Neumann algebra with faithful normal
finite trace $\sigma$. In addition, if $\mathcal{N}$ is hyperfinite, then
every von Neumann algebra $e_{n}\mathcal{N}e_{n}$ is hyperfinite.
Consequently, by Lemma 17 the algebra $\mathcal{M}$ is also hyperfinite.

Step 1. Here we define a one-parameter unitary group $\exp(it\mathscr{K}x)$
for every $x=x^{*}\in e_{k}\mathcal{N}e_{k}$. Fix $k\in\mathbb{N}$ and
$x=x^{*}\in e_{k}\mathcal{N}e_{k}$. Since the $*$-homomorphism
$\pi_{e_{k},e_{n}}$ given by (52) is trace preserving, it follows that it is
continuous in the topology of convergence in measure [23, Proposition 3.3],
and therefore, for all $n\geq k$ we have

$\displaystyle\begin{split}\pi_{e_{k},e_{n}}(\exp(it\mathscr{K}_{e_{k}}x))&=\pi_{e_{k},e_{n}}\left(\sum_{j=0}^{\infty}\frac{(it\mathscr{K}_{e_{k}}x)^{j}}{j!}\right)=\sum_{j=0}^{\infty}\frac{\pi_{e_{k},e_{n}}\left((it\mathscr{K}_{e_{k}}x)^{j}\right)}{j!}\\\
&=\sum_{j=0}^{\infty}\frac{(it)^{j}\left(\pi_{e_{k},e_{n}}(\mathscr{K}_{e_{k}}x)\right)^{j}}{j!}\stackrel{{\scriptstyle\eqref{pi_K}}}{{=}}\sum_{j=0}^{\infty}\frac{(it\mathscr{K}_{e_{n}}x)^{j}}{j!}=\exp(it\mathscr{K}_{e_{n}}x).\end{split}$

Therefore,

$(\theta\circ\pi_{k})(\exp(it\mathscr{K}_{e_{k}}x))\stackrel{{\scriptstyle\eqref{eq_pi_n}}}{{=}}(\theta\circ\pi_{n}\circ\pi_{e_{k},e_{n}})(\exp(it\mathscr{K}_{e_{k}}x))=(\theta\circ\pi_{n})(\exp(it\mathscr{K}_{e_{n}}x))\in\mathcal{M}$

for all $n\geq k.$ Consequently, the formula
$t\to(\theta\circ\pi_{k})(\exp(it\mathscr{K}_{e_{k}}x))$ defines a one-
parameter unitary group in $\mathcal{M},$ which does not depend on $k$. By
Stone’s theorem, there exists a self-adjoint operator $\mathscr{K}x$ on $H$
such that

$\exp(it\mathscr{K}x)=(\theta\circ\pi_{k})(\exp(it\mathscr{K}_{e_{k}}x)),\quad
x=x^{*}\in e_{k}\mathcal{N}e_{k}.$ (61)

In addition, the operator $\mathscr{K}x$ is affiliated with $\mathcal{M}$
since $(\theta\circ\pi_{k})(\exp(it\mathscr{K}_{e_{k}}x))\in\mathcal{M}$.
Thus, for every $x=x^{*}\in e_{k}\mathcal{N}e_{k}$ it is correctly defined a
self-adjoint operator $\mathscr{K}x\in S(\mathcal{M},\sigma)$, since the trace
$\sigma$ finite on $\mathcal{M}$.

Step 2. We claim that the mapping $x\mapsto\mathscr{K}x,x\in
e_{k}\mathcal{N}e_{k},$ is linear. It is clear that $\mathscr{K}(\lambda
x)=\lambda\mathscr{K}(x)$ for all $\lambda\in\mathbb{R}$ and $x=x^{*}\in
e_{k}\mathcal{N}e_{k}$. Let $x=x^{*},y=y^{*}\in e_{k}\mathcal{N}e_{k}.$ By
Trotter formula [57, Theorem VIII.31] (which is applicable to the operators
$\mathscr{K}x$ and $\mathscr{K}y$, since the sum $\mathscr{K}x+\mathscr{K}y$
is understood in strong sense), we have

$\exp(it(\mathscr{K}x+\mathscr{K}y))=(so)-\lim_{m\to\infty}(e^{i\frac{t}{m}\mathscr{K}x}e^{i\frac{t}{m}\mathscr{K}y})^{m},$

where the notation $(so)-\lim_{m\rightarrow\infty}$ stands for the limit in
topology of convergence in measure. By [20, Theorem 2, p. 59],
$*$-homomorphisms on von Neumann algebras are continuous with respect to the
strong operator topology. Therefore,

$\displaystyle\begin{split}\exp(it(\mathscr{K}x+\mathscr{K}y))\\\
&\stackrel{{\scriptstyle\eqref{k
def}}}{{=}}(so)-\lim_{m\to\infty}((\theta\circ\pi_{k})(\exp(i\frac{t}{m}\mathscr{K}_{e_{k}}x))(\theta\circ\pi_{k})(\exp(i\frac{t}{m}\mathscr{K}_{e_{k}}y)))^{m}\\\
&=(so)-\lim_{m\to\infty}(\theta\circ\pi_{k})\Big{(}\exp(i\frac{t}{m}\mathscr{K}_{e_{k}}x)\exp(i\frac{t}{m}\mathscr{K}_{e_{k}}y)\Big{)}^{m}\\\
&=(\theta\circ\pi_{k})\Big{(}(so)-\lim_{m\to\infty}(\exp(i\frac{t}{m}\mathscr{K}_{e_{k}}x)\exp(i\frac{t}{m}\mathscr{K}_{e_{k}}y))^{m}\Big{)}\\\
&=(\theta\circ\pi_{k})(\exp(it(\mathscr{K}_{e_{k}}x+\mathscr{K}_{e_{k}}y)))\\\
&=(\theta\circ\pi_{k})(\exp(it\mathscr{K}_{e_{k}}(x+y)))\\\
&\stackrel{{\scriptstyle\eqref{k
def}}}{{=}}\exp(it\mathscr{K}(x+y)).\end{split}$

Consequently,

$\mathscr{K}(x+y)=\mathscr{K}x+\mathscr{K}y,\quad x=x^{*},y=y^{*}\in
e_{k}\mathcal{N}e_{k}.$ (62)

Thus, the mapping $x\mapsto\mathscr{K}x$ is linear on $e_{k}\mathcal{N}e_{k}$.

Step 3. Let us show that $\mathscr{K}$ acts boundedly with respect to
$L_{1}$-norm. Firstly, we show that for $x=x^{*}\in e_{k}\mathcal{N}e_{k}$ the
operators $\mathscr{K}_{e_{k}}x$ and $\mathscr{K_{e_{k}}}x$ are
equimeasurable. For a Schwartz function $f\in S(\mathbb{R})$, we have (see
(42))

$\displaystyle\begin{split}(\theta\circ\pi_{k})(f(\mathscr{K}_{e_{k}}x))&=(\theta\circ\pi_{k})\int_{\mathbb{R}}(\mathcal{F}(f))(\lambda)\exp(i\lambda\mathscr{K}_{e_{k}}x)d\lambda\\\
&=\int_{\mathbb{R}}(\mathcal{F}(f))(\lambda)(\theta\circ\pi_{k})\exp(i\lambda\mathscr{K}_{e_{k}}x)d\lambda\\\
&\stackrel{{\scriptstyle\eqref{k
def}}}{{=}}\int_{\mathbb{R}}(\mathcal{F}(f))(\lambda)\exp(i\lambda\mathscr{K}x)d\lambda=f(\mathscr{K}x),\end{split}$

where the second equality holds since $(\theta\circ\pi_{k})$ is a trace
preserving $*$-homomorphism (see (23)). For an interval
$(s,t),s,t\in\mathbb{R},s<t,$ choose a sequence $f_{n}$ of Schwartz functions,
such that $f_{n}\downarrow\chi_{(s,t)}$. Then we have

$(\theta\circ\pi_{k})(f_{n}(\mathscr{K}_{e_{k}}x))\rightarrow(\theta\circ\pi_{k})(E_{\mathscr{K}_{e_{k}}x}(s,t))$

and

$f_{n}(\mathscr{K}x)\rightarrow E_{\mathscr{K}x}(s,t)$

in the topology of convergence in measure. Consequently,

$(\theta\circ\pi_{k})(E_{\mathscr{K}_{e_{k}}x}(s,t))=(E_{\mathscr{K}x}(s,t)),s<t.$
(63)

In particular, by general properties of spectral projections, we infer that
equality (63) holds for $t=\infty$. Hence, by (23)

$\sigma_{e_{k}}(E_{\mathscr{K}_{e_{k}}x}(s,\infty))=\sigma((\theta\circ\pi_{k})(E_{\mathscr{K}_{e_{k}}x}(s,\infty)))=\sigma(E_{\mathscr{K}x}(s,\infty))$

for all $s\in\mathbb{R}$. In other words,
$d_{\mathscr{K}_{e_{k}}x}(s)=d_{\mathscr{K}x}(s)$ for all $s\in\mathbb{R}$ and
$x=x^{*}\in e_{k}\mathcal{N}e_{k}$. Thus, the operators $\mathscr{K}_{e_{k}}x$
and $\mathscr{K_{e_{k}}}x$ are equimeasurable, in particular,

$\mu(\mathscr{K}_{e_{k}}x)=\mu(\mathscr{K}(x)),\quad x=x^{*}\in
e_{k}\mathcal{N}e_{k}.$ (64)

Furthermore, from equality (63) we also have that
$(\theta\circ\pi_{k})(E_{\mathscr{K}_{e_{k}}x}(-\infty,0))=(E_{\mathscr{K}x}(-\infty,0))$,
and therefore, $\mathscr{K}x\geq 0$ for all $e_{k}\mathcal{N}e_{k}\ni x\geq
0$. Hence, by Proposition 21 and equality (64) we infer that
$\sigma(\mathscr{K}x)=\sigma_{e_{k}}(\mathscr{K}_{e_{k}}x)=\tau(x).$
Consequently, $\|\mathscr{K}x\|_{1}\leq 4\|x\|_{1}$ for every $x\in
e_{k}\mathcal{N}e_{k}.$ Since the union of $e_{k}\mathcal{N}e_{k},$ $k\geq 0,$
is $\|\cdot\|_{1}-$dense in $\mathcal{L}_{1}(\mathcal{N},\tau)$ [18] it
follows that $\mathscr{K}$ admits a bounded extension

$\mathscr{K}:\mathcal{L}_{1}(\mathcal{N},\tau)\to\mathcal{L}_{1}(\mathcal{M},\sigma).$

Thus, we have constructed the finite von Neumann algebra $\mathcal{M}$ and
operator $\mathscr{K}$ acting from $\mathcal{L}_{1}(\mathcal{N},\tau)$ into
$\mathcal{L}_{1}(\mathcal{M},\sigma)$.

Step 4. We now prove equality (60). For this purpose fix $x=x^{*}\in
e_{k}\mathcal{N}e_{k}.$ We have

$\sigma(\exp(i\mathscr{K}x))\stackrel{{\scriptstyle\eqref{k
def}}}{{=}}(\sigma\circ\pi_{k})(\exp(i\mathscr{K}_{e_{k}}(x)))\stackrel{{\scriptstyle\eqref{eq_sigma_n}}}{{=}}\sigma_{e_{k}}(\exp(i\mathscr{K}_{e_{k}}(x))).$

It follows now from Lemma 24 that

$\sigma(\exp(i\mathscr{K}x))=\exp(\tau_{e_{k}}(\exp(ix)-\mathbbm{1})),\quad
x=x^{*}\in e_{k}\mathcal{N}e_{k}.$ (65)

Using again $\|\cdot\|_{1}$-density of union of $e_{k}\mathcal{N}e_{k},$
$k\geq 0,$ in $\mathcal{L}_{1}(\mathcal{N},\tau),$ for a fixed
$x=x^{*}\in\mathcal{L}_{1}(\mathcal{N},\tau)$ choose a sequence
$\\{x_{k}\\}_{n\geq 1},x_{k}\in e_{k}\mathcal{N}e_{k}$ such that
$\|x_{k}-x\|_{1}\rightarrow 0,k\rightarrow\infty$. It follows from Duhamel’s
formula (see e.g. [9, Lemma 5.2]) that
$\|e_{k}(\exp(ix_{k})-\exp(ix))e_{k}\|_{1}\rightarrow 0,$ as well as
$\|\exp(i\mathscr{K}x_{k})-\exp(i\mathscr{K}x)\|_{1}\rightarrow 0,$ for
$k\rightarrow\infty$. Therefore,

$\sigma(\exp(i\mathscr{K}x_{k}))\rightarrow\sigma(\exp(i\mathscr{K}x))$

and

$\tau_{e_{k}}(\exp(ix_{k})-\mathbbm{1})=\tau(e_{k}(\exp(ix_{k})-\mathbbm{1})e_{k})\rightarrow\tau((\exp(ix)-\mathbbm{1})).$

Thus, by (65) we infer equality (60). ∎

The following corollary lists main properties of the Kruglov operator
constructed in Theorem 32.

######  Corollary 33.

Let $(\mathcal{N},\tau)$ be a semifinite atomless von Neumann algebra, let
$(\mathcal{M},\sigma)$ and $\mathscr{K}$ be the finite von Neumann algebra and
the Kruglov operator constructed in Theorem 32, respectively. The following
assertions hold:

  1. (i)

If self-adjoint operators $x,y\in\mathcal{L}_{1}(\mathcal{N},\tau)$ are
equimeasurable, then
$\mathscr{K}x,\mathscr{K}y\in\mathcal{L}_{1}(\mathcal{M},\sigma)$ are
equimeasurable too. In addition, if
$x=x^{*}\in\mathcal{L}_{1}(\mathcal{N},\tau)$ is equimeasurable with $y\in
L_{1}(0,1),$ then the elements $\mathscr{K}x\in\mathcal{L}_{1}(\mathcal{M})$
and $Ky\in L_{1}(0,1)$ are equimeasurable.

  2. (ii)

The operator $\mathscr{K}$ acts boundedly from
$(\mathcal{L}_{1}\cap\mathcal{L}_{2})(\mathcal{N})$, equipped with the norm
$\|\cdot\|_{\mathcal{L}_{1}\cap\mathcal{L}_{2}}=\|\cdot\|_{\mathcal{L}_{1}}+\|\cdot\|_{\mathcal{L}_{2}}$
into $\mathcal{L}_{2}(\mathcal{M})$.

  3. (iii)

For all $x,y\in(\mathcal{L}_{1}\cap\mathcal{L}_{2})(\mathcal{N})$ the equality
$[\mathscr{K}x,\mathscr{K}y]=\mathscr{K}[x,y]$ holds.

  4. (iv)

If $x_{k}=x_{k}^{*}\in\mathcal{L}_{1}(\mathcal{N},\tau),$ $1\leq k\leq n,$ are
such that $x_{j}x_{k}=0$ for $j\neq k,$ then the random variables
$\mathscr{K}x_{k}\in\mathcal{L}_{1}(\mathcal{M},\sigma),$ $1\leq k\leq n,$ are
independent in the sense of Section 2.8.

  5. (v)

The operator $\mathscr{K}$ maps symmetrically distributed operators from
$\mathcal{L}_{1}(\mathcal{N},\tau)$ into symmetrically distributed operators
from $\mathcal{L}_{1}(\mathcal{M},\sigma)$.

###### Proof.

(i). We have that

$\displaystyle\begin{split}\tau(\exp(ix)-\mathbbm{1})&\stackrel{{\scriptstyle\eqref{xdg1}}}{{=}}\tau(\exp(ix_{+})-\mathbbm{1})+\tau(\exp(ix_{-})-\mathbbm{1})\\\
&=\int_{0}^{\infty}(\exp(i\mu(s,x_{+}))-1)ds+\int_{0}^{\infty}(\exp(i\mu(s,x_{-}))-1)ds\\\
&=\int_{0}^{\infty}(\exp(i\mu(s,y_{+}))-1)ds+\int_{0}^{\infty}(\exp(i\mu(s,y_{-}))-1)ds\\\
&=\tau(\exp(iy_{+})-\mathbbm{1})+\tau(\exp(iy_{-})-\mathbbm{1})\stackrel{{\scriptstyle\eqref{xdg1}}}{{=}}\tau(\exp(iy)-\mathbbm{1}),\
\end{split}$

and therefore by equality (60) we obtain that

$\displaystyle\begin{split}\sigma(\exp(i\mathscr{K}x))&=\exp(\tau(\exp(ix)-\mathbbm{1}))=\exp(\tau(\exp(iy)-\mathbbm{1}))=\sigma(\exp(i\mathscr{K}y))\end{split}$

By [28, Theorem 2.12.1] the preceding equality implies that
$d_{\mathscr{K}x}=d_{\mathscr{K}y}$. Since the von Neumann algebra
$\mathcal{M}$ is finite we infer that the operators $\mathscr{K}x$ and
$\mathscr{K}y$ are equimeasurable. If $x$ is equimeasurable with $y\in
L_{1}[0,1]$, then the assertion is proved similarly using Lemma 22.

(ii). Suppose firstly that $0\leq x\in\mathcal{N}$ with finite support. There
exist a finite sequence of projections
$\\{q_{k}\\}_{k=0}^{N-1}\subset\mathcal{N}$ with $\tau(q_{k})=1$ such that
$x\leq\sum_{k=0}^{N-1}\|x\|_{\infty}q_{k}$. Since the operator $\mathscr{K}$
is positive we have that
$\mathscr{K}x\leq\|x\|_{\infty}\sum_{k=0}^{N-1}\mathscr{K}q_{k}$. Hence

$\mu(\mathscr{K}x)\leq\|x\|_{\infty}\mu(\sum_{k=0}^{N-1}\mathscr{K}q_{k})\leq\|x\|_{\infty}ND_{N}\mu(\mathscr{K}q_{0}).$

In addition by (i) the equality
$\mu(\mathscr{K}q_{0})=\mu(K\mathbbm{1})\in\bigcap_{n\geq 1}L_{n}(\Omega)$
holds [5, Theorem 4.4], and therefore we obtain $\mathscr{K}x\in\bigcap_{n\geq
1}\mathcal{L}_{n}(\mathcal{M})$. Furthermore, by standard arguments one can
prove that
$\frac{d^{2}}{dt^{2}}\sigma(\exp(it\mathscr{K}x))\big{|}_{t=0}=-\sigma((\mathscr{K}x)^{2})$
and
$\frac{d^{2}}{dt^{2}}\exp(\tau(\exp(itx)-\mathbbm{1}))\big{|}_{t=0}=-\tau(x^{2})-\tau^{2}(x)$.
Consequently, by (60) we infer that
$\sigma((\mathscr{K}x)^{2})=\tau(x^{2})+\tau^{2}(x)$, and therefore
$\|\mathscr{K}x\|_{2}\leq\|x\|_{1}+\|x\|_{2}$.

Now, for arbitrary $x\geq 0$ from
$(\mathcal{L}_{1}\cap\mathcal{L}_{2})(\mathcal{N})$ we set
$x_{n}=xE_{x}[\frac{1}{n},n]$. We have that $x_{n}\uparrow x$ and
$x_{n}\xrightarrow{\|\cdot\|_{1}}x$ [18], and therefore
$\mathscr{K}x_{n}\uparrow$ and by Theorem 32
$\mathscr{K}x_{n}\xrightarrow{\|\cdot\|_{1}}\mathscr{K}x$. Since in addition
$\|\mathscr{K}x_{n}\|_{2}\leq\|x\|_{1}+\|x_{2}\|$ for all $n\in\mathbb{N}$ we
obtain that $\|\mathscr{K}x\|_{2}\leq\|x\|_{1}+\|x\|_{2}$ due to Fatou
property of $\mathcal{L}_{2}(\mathcal{M})$ (see e.g. [21]).

(iii). Let $x,y\in(\mathcal{L}_{1}\cap\mathcal{L}_{2})(\mathcal{N})$. Set
$x_{n}=e_{n}xe_{n},y_{n}=e_{n}ye_{n},$ where $e_{n}$ is sequence of
$\tau$-finite projections from the proof of Theorem 32. By Proposition 21
(iii) the equality
$[\mathscr{K}x_{n},\mathscr{K}y_{n}]=\mathscr{K}[x_{n},y_{n}]$ holds. Since
$[x_{n},y_{n}]\xrightarrow{\|\cdot\|_{1}}[x,y]$ [18] we have that
$\mathscr{K}[x_{n},y_{n}]\xrightarrow{\|\cdot\|_{1}}\mathscr{K}[x,y].$ On the
other hand, by (ii) we obtain

$\displaystyle\begin{split}\|[\mathscr{K}x_{n},\mathscr{K}y_{n}]&-[\mathscr{K}x,\mathscr{K}y]\|_{1}\leq\|[\mathscr{K}x_{n}-\mathscr{K}x,\mathscr{K}y_{n}]\|_{1}+\|[\mathscr{K}x,\mathscr{K}y_{n}-\mathscr{K}y]\|_{1}\\\
&\leq
2\|\mathscr{K}x_{n}-\mathscr{K}x\|_{2}\|\mathscr{K}y_{n}\|_{2}+2\|\mathscr{K}x\|_{2}\|\mathscr{K}y_{n}-\mathscr{K}y\|_{2}\\\
&\leq
2\|\mathscr{K}\|^{2}_{\mathcal{L}_{1}\cap\mathcal{L}_{2}\to\mathcal{L}_{2}}(\|x_{n}-x\|_{\mathcal{L}_{1}\cap\mathcal{L}_{2}}\|y_{n}\|_{\mathcal{L}_{1}\cap\mathcal{L}_{2}}+\|x\|_{\mathcal{L}_{1}\cap\mathcal{L}_{2}}\|y_{n}-y\|_{\mathcal{L}_{1}\cap\mathcal{L}_{2}}).\end{split}$

And therefore, since
$\|x_{n}-x\|_{\mathcal{L}_{1}\cap\mathcal{L}_{2}}\rightarrow
0,\|y_{n}-y\|_{\mathcal{L}_{1}\cap\mathcal{L}_{2}}\rightarrow 0$ we infer that

$[\mathscr{K}x_{n},\mathscr{K}y_{n}]\xrightarrow{\|\cdot\|_{1}}[\mathscr{K}x,\mathscr{K}y]$

and, therefore, $[\mathscr{K}x,\mathscr{K}y]=\mathscr{K}[x,y].$

The proof of property (iv) is identical to that of Theorem 25 and is,
therefore, omitted. Property (v) is a straightforward consequence of (i) and
(iv).

∎

######  Corollary 34.

Let $\mathcal{R}$ be the hyperfinite $II_{1}$ factor, $H$ be a separable
infinite-dimensional Hilbert space and let
$\mathcal{R}\bar{\otimes}\mathcal{L}(H)$ be the hyperfinite $II_{\infty}$
factor. Then the operator $\mathscr{K}$ defined in Theorem 32 acts from
$\mathscr{K}:\mathcal{L}_{1}(\mathcal{R}\bar{\otimes}\mathcal{L}(H),\tau\otimes{\rm
Tr})$ into $\mathcal{L}_{1}(\mathcal{R},\tau).$

###### Proof.

Set $\mathcal{N}=\mathcal{R}\bar{\otimes}\mathcal{L}(H)$ in Theorem 32. We
have
$\mathscr{K}:\mathcal{L}_{1}(\mathcal{N},\tau)\to\mathcal{L}_{1}(\mathcal{M},\sigma),$
where $\mathcal{M}$ is hyperfinite. Since every hyperfinite II${}_{1}-$type
von Neumann algebra embeds into $\mathcal{R}$ (see e.g. the proof of Theorem
5.2 in [29]), it follows that
$\mathscr{K}:\mathcal{L}_{1}(\mathcal{N},\tau)\to\mathcal{L}_{1}(\mathcal{R},\tau).$
∎

##  5\. Proof of Theorem 4

In this section we prove Theorem 4. Let us first note that the equivalence
(ii)$\Leftrightarrow$(iii) of Theorem 4 follows from the equivalence
(i)$\Leftrightarrow$(v) of Theorem 52.

###  5.1. Proof of Theorem 4: (i)$\Rightarrow$(ii)

###### Proof of Theorem 4: (i)$\Rightarrow$(ii).

We first prove that if a Banach ideal $\mathcal{I}$ isomorphically embeds into
$\mathcal{L}_{p}(\mathcal{R})$ ($1\leq p<2$), then $\mathcal{I}$ satisfies an
upper $p$-estimate and a lower $2$-estimate.

Let $T$ be an isomorphic embedding of $\mathcal{I}$ into
$\mathcal{L}_{p}(\mathcal{R})$ ($1\leq p<2$). Fix a constant $C>0$ such that

$C^{-1}\|A\|_{\mathcal{I}}\leq\|T(A)\|_{p}\leq C\|A\|_{\mathcal{I}}\ \ \
\mbox{for every}\ \ A\in\mathcal{I}.$ (66)

Fix $n\geq 0$ and operators $A_{k}\in\mathcal{I},$ $0\leq k\leq n,$ such that
$l(A_{j})l(A_{k})=0$ for $j\neq k.$ Observe that $l(A_{j})l(A_{k})=0$ implies
$A_{j}^{*}A_{k}=0$ for $j\neq k.$ It follows that for almost all $t\in(0,1),$
we have

$\Big{\|}\sum_{k=0}^{n}r_{k}(t)A_{k}\Big{\|}_{\mathcal{I}}=\Big{\|}\sum_{k=0}^{n}A_{k}\Big{\|}_{\mathcal{I}},$
(67)

where $r_{k},$ $0\leq k\leq n,$ are the Rademacher functions. Applying
subsequently (67), (66) and identifying the Bochner space
$L_{p}((0,1),\mathcal{L}_{p}(\mathcal{R}))$ with
$\mathcal{L}_{p}(\mathcal{R}\bar{\otimes}L_{\infty}(0,1),\tau\otimes dm)$ (see
e.g [10, Lemma 6.2]), we obtain

$\Big{\|}\sum_{k=0}^{n}A_{k}\Big{\|}_{\mathcal{I}}=\Big{(}\int_{0}^{1}\Big{\|}\sum_{k=0}^{n}r_{k}(t)A_{k}\Big{\|}_{\mathcal{I}}^{p}dt\Big{)}^{1/p}\leq
C\Big{(}\int_{0}^{1}\Big{\|}\sum_{k=0}^{n}r_{k}(t)T(A_{k})\Big{\|}_{p}^{p}dt\Big{)}^{1/p}\\\
=C\Big{\|}\sum_{k=0}^{n}r_{k}T(A_{k})\Big{\|}_{L_{p}((0,1),\mathcal{L}_{p}(\mathcal{R}))}=C\Big{\|}\sum_{k=0}^{n}T(A_{k})\otimes
r_{k}\Big{\|}_{\mathcal{L}_{p}(\mathcal{R}\bar{\otimes}L_{\infty}(0,1))}.$
(68)

Similarly, we obtain

$\Big{\|}\sum_{k=0}^{n}A_{k}\Big{\|}_{\mathcal{I}}\geq
C^{-1}\Big{\|}\sum_{k=0}^{n}T(A_{k})\otimes r_{k}\Big{\|}_{p},$ (69)

where for brevity we use $\|\cdot\|_{p}$ instead of
$\|\cdot\|_{\mathcal{L}_{p}(\mathcal{R}\bar{\otimes}L_{\infty}(0,1))}$. By
Theorem 13, we have

${\rm
const}\cdot\Big{(}\sum_{k=0}^{n}\|T(A_{k})\|_{p}^{2}\Big{)}^{1/2}\leq\Big{\|}\sum_{k=0}^{n}T(A_{k})\otimes
r_{k}\Big{\|}_{p}\leq\Big{(}\sum_{k=0}^{n}\|T(A_{k})\|_{p}^{p}\Big{)}^{1/p}.$
(70)

Combining (68) and (69) with (70) and (66), we conclude

${\rm
const}\cdot\Big{(}\sum_{k=0}^{n}\|A_{k}\|_{\mathcal{I}}^{2}\Big{)}^{1/2}\leq\Big{\|}\sum_{k=0}^{n}A_{k}\Big{\|}_{\mathcal{I}}\leq{\rm
const}\cdot\Big{(}\sum_{k=0}^{n}\|A_{k}\|_{\mathcal{I}}^{p}\Big{)}^{1/p}.$

Now let $\mathcal{I}=\mathcal{L}_{M}(H)$ isomorphically embeds into
$\mathcal{L}_{p}(\mathcal{R})$ ($1\leq p<2$). Then, as we proved above,
$\mathcal{L}_{M}(H)$ satisfies an upper $p$-estimate and a lower $2$-estimate.
The implication follows from Theorem 52. ∎

###  5.2. Proof of Theorem 4: (iii)$\Leftrightarrow$(iv)

###### Proof of Theorem 4: (iii)$\Rightarrow$(iv).

Let $M$ satisfies (iii). We may assume that $M$ is $p$-convex and $2$-concave.
Define an Orlicz function $M_{1}$ by setting

$M_{1}(t):=M(t^{1/p}),\ \ \ t>0.$

According to Remark 12, for every $0\leq x\in l_{M_{1}}$ and for every $0\leq
y\in l_{\infty},$ we have

$y\prec\prec x\Longrightarrow\|y\|_{l_{M_{1}}}\leq\|x\|_{l_{M_{1}}},$ (71)

for the notion of submajorization (”$\prec\prec$”) see Definition 10. Now if
$0\leq x\in l_{M},$ then $x^{p}\in l_{M_{1}}$ and $y^{p}\in l_{\infty}.$
Applying (71) to $x^{p}$ and $y^{p},$ we obtain that the implication

$y^{p}\prec\prec x^{p}\Longrightarrow\|y\|_{l_{M}}\leq\|x\|_{l_{M}}$

holds. By [49, Theorem 2], we have that $l_{M}$ is an interpolation space for
the couple $(l_{p},l_{\infty}).$

Since by the assumption $M$ is $2$-concave, it similarly follows from [49,
Theorem 3] that $l_{M}$ is an interpolation space for the couple
$(l_{1},l_{2}).$ Applying [2, Corollary], we obtain that $l_{M}$ is an
interpolation space for the couple $(l_{p},l_{2}).$ By [1, Theorem 2.7],
$\mathcal{L}_{M}(H)$ is an interpolation space for the couple
$(\mathcal{L}_{p}(H),\mathcal{L}_{2}(H)).$ ∎

###### Proof of Theorem 4: (iv)$\Rightarrow$(iii).

Suppose that the Orlicz space $\mathcal{L}_{M}(H)$ is an interpolation space
for the couple $(\mathcal{L}_{p}(H),\mathcal{L}_{2}(H)).$ Then again by [1,
Theorem 2.7], the commutative core $l_{M}$ is an interpolation space for the
couple $(l_{p},l_{2}).$ For every $n\geq 1$ and $x=\\{x(k)\\}_{k\geq 0}\in
l_{\infty}$. Using (4), we obtain

$\|D_{n}\|_{l_{M}\to l_{M}}\leq{\rm const}\cdot\max\\{n^{1/2},n^{1/p}\\}={\rm
const}\cdot n^{1/p},\\\ \|D_{1/n}\|_{l_{M}\to l_{M}}\leq{\rm
const}\cdot\max\\{n^{-1/2},n^{-1/p}\\}={\rm const}\cdot n^{-1/2},\quad n\geq
1,$

where $D_{n}$ and $D_{1/n}$ are defined by (3). Therefore, for $m\geq 1$, we
have

$\Big{\|}\sum_{j=0}^{nm-1}e_{j}\Big{\|}_{l_{M}}\leq\|D_{n}\|_{l_{M}\to
l_{M}}\Big{\|}\sum_{j=0}^{m-1}e_{j}\Big{\|}_{l_{M}}\leq{\rm const}\cdot
n^{1/p}\Big{\|}\sum_{j=0}^{m-1}e_{j}\Big{\|}_{l_{M}}$

and

$\Big{\|}\sum_{j=0}^{m-1}e_{j}\Big{\|}_{l_{M}}\leq\|D_{1/n}\|_{l_{M}\to
l_{M}}\Big{\|}\sum_{j=0}^{nm-1}e_{j}\Big{\|}_{l_{M}}\leq{\rm const}\cdot
n^{-1/2}\Big{\|}\sum_{j=0}^{nm-1}e_{j}\Big{\|}_{l_{M}}.$

Thus, we arrive at (118). Arguing as in the proof of Theorem 52
(iv)$\Rightarrow$(v) (see (118) and subsequent arguments), we obtain that $M$
is equivalent to a $p$-convex and $2$-concave Orlicz function. ∎

###  5.3. Proof of Theorem 4: (iii)$\Rightarrow$(i)

The proof of this implication is much more difficult and is in fact completely
different from previous ones. Here we use the concept of Kruglov operators
developed in Sections 3 and 4. Before we proceed with the proof of the
implication, let us present several technical estimates, which are important
for the subsequent exposition.

The following lemma yields a submajorization estimate for the element $Kx,$
$x\in L_{1}(0,1),$ where $K$ is the Kruglov operator defined by (26). Recall
that the notion of submajorization (”$\prec\prec$”) is given in Definition 10
and the dilation operator $D$ is defined by (5).

######  Lemma 35.

For every $x\in L_{1}(0,1),$ we have

$Kx\prec\prec\sum_{n=1}^{\infty}nD_{\frac{1}{e\cdot n!}}\mu(x).$

###### Proof.

Recall (see (26)) that

$Kx=\sum_{n=1}^{\infty}\chi_{A_{n}}\otimes\Big{(}\sum_{m=1}^{n}1^{\otimes(m-1)}\otimes
x\otimes 1^{\otimes\infty}\Big{)},$ (72)

where $A_{n},$ $n\geq 0,$ are disjoint sets with $m(A_{n})=1/(e\cdot n!)$ for
all $n\geq 0.$

It follows from the inequality (7) that

$\sum_{m=1}^{n}1^{\otimes(m-1)}\otimes x\otimes 1^{\otimes\infty}\prec\prec
n\mu(x),\quad n\geq 1$

and, therefore,

$\chi_{A_{n}}\otimes\Big{(}\sum_{m=1}^{n}1^{\otimes(m-1)}\otimes x\otimes
1^{\otimes\infty}\Big{)}\prec\prec nD_{m(A_{n})}\mu(x)=nD_{\frac{1}{e\cdot
n!}}\mu(x),\quad n\geq 1.$

Again applying the inequality (7), we obtain

$Kx\prec\prec\sum_{n=1}^{\infty}nD_{\frac{1}{e\cdot n!}}\mu(x).$

∎

######  Lemma 36.

If $1\leq p\leq 2$ and $x_{k}\in L_{p}(0,1),$ $1\leq k\leq n,$ then

$\Big{\|}\bigoplus_{k=0}^{\infty}Kx_{k}\Big{\|}_{L_{p}+L_{2}}\sim\Big{\|}\bigoplus_{k=0}^{\infty}x_{k}\Big{\|}_{L_{p}+L_{2}}.$

###### Proof.

It follows from Lemma 35 that

$\bigoplus_{k=1}^{\infty}Kx_{k}\prec\prec\bigoplus_{k=1}^{\infty}\Big{(}\sum_{n=1}^{\infty}nD_{\frac{1}{e\cdot
n!}}\mu(x_{k})\Big{)}=\sum_{n=1}^{\infty}n\Big{(}\bigoplus_{k=1}^{\infty}D_{\frac{1}{e\cdot
n!}}\mu(x_{k})\Big{)}.$ (73)

Since the norm $\|\cdot\|_{L_{p}+L_{2}}$ respects submajorization (see e.g.
[43, Theorem 4.10]), using (73) and the triangle inequality in
$(L_{p}+L_{2})(0,\infty),$ we infer that

$\displaystyle\begin{split}\Big{\|}\bigoplus_{k=0}^{\infty}Kx_{k}\Big{\|}_{L_{p}+L_{2}}&\leq\sum_{n=1}^{\infty}n\Big{\|}\bigoplus_{k=1}^{\infty}D_{\frac{1}{e\cdot
n!}}\mu(x_{k})\Big{\|}_{L_{p}+L_{2}}\\\
&=\sum_{n=1}^{\infty}n\Big{\|}D_{\frac{1}{e\cdot
n!}}\Big{(}\bigoplus_{k=1}^{\infty}\mu(x_{k})\Big{)}\Big{\|}_{L_{p}+L_{2}}\\\
&\leq\Big{(}\sum_{n=1}^{\infty}n\big{\|}D_{\frac{1}{e\cdot
n!}}\big{\|}_{L_{p}+L_{2}\to
L_{p}+L_{2}}\Big{)}\Big{\|}\bigoplus_{k=0}^{\infty}x_{k}\Big{\|}_{L_{p}+L_{2}}.\end{split}$
(74)

Taking into account that

$\big{\|}D_{u}\big{\|}_{L_{p}+L_{2}\to L_{p}+L_{2}}\leq u^{1/p},\quad 0<u\leq
1,$

we infer that

$\sum_{n=1}^{\infty}n\big{\|}D_{\frac{1}{e\cdot n!}}\big{\|}_{L_{p}+L_{2}\to
L_{p}+L_{2}}\leq\sum_{n=1}^{\infty}\frac{n}{(e\cdot n!)^{1/p}}\leq{\rm
const}<\infty.$ (75)

Combining (74) and (75), we obtain

$\Big{\|}\bigoplus_{k=0}^{\infty}Kx_{k}\Big{\|}_{L_{p}+L_{2}}\leq{\rm const}\
\Big{\|}\bigoplus_{k=0}^{\infty}x_{k}\Big{\|}_{L_{p}+L_{2}}.$

On the other hand, taking only the first term in (72), we obtain that

$\mu(Kx_{k})\geq\mu(A_{1}\otimes
x_{k}\otimes\mathbbm{1}^{\infty})=D_{m(A_{1})}\mu(x_{k})=D_{1/e}\ \mu(x_{k}),\
\ 1\leq k\leq n.$ (76)

Observing that

$\big{\|}D_{u}x\big{\|}_{L_{p}+L_{2}}\geq u^{1/p}\|x\|_{L_{p}+L_{2}},\ \
\mbox{for all}\ \ x\in(L_{p}+L_{2})(0,\infty),\ \ 0<u\leq 1,$

and applying (76), we conclude that

$\displaystyle\Big{\|}\bigoplus_{k=0}^{\infty}Kx_{k}\Big{\|}_{L_{p}+L_{2}}$
$\displaystyle\geq\Big{\|}\bigoplus_{k=0}^{\infty}D_{1/e}\
\mu(x_{k})\Big{\|}_{L_{p}+L_{2}}$
$\displaystyle=\Big{\|}D_{1/e}\Big{(}\bigoplus_{k=0}^{\infty}\mu(x_{k})\Big{)}\Big{\|}_{L_{p}+L_{2}}\geq{\rm
const}\Big{\|}\bigoplus_{k=0}^{\infty}x_{k}\Big{\|}_{L_{p}+L_{2}}.$

∎

The following lemma is the (noncommutative analogue of) special case of the
inequality due to Johnson and Schechtman (see [33, Theorem 1]). This
“noncommutative Rosenthal inequality”(see [37]) has become popular in
compressed sensing (see e.g. [55]).

######  Lemma 37.

Let $1\leq p\leq 2$ and let $A_{k}\in\mathcal{L}_{p}(\mathcal{R}),$ $1\leq
k\leq n,$ be independent symmetrically distributed random variables. Then

$\Big{\|}\sum_{k=1}^{n}A_{k}\Big{\|}_{p}\sim\Big{\|}\bigoplus_{k=1}^{n}A_{k}\Big{\|}_{\mathcal{L}_{p}+\mathcal{L}_{2}}\
\ \ n\geq 1.$

###### Proof.

Recall that the independence of random variables
$A_{k}\in\mathcal{L}_{p}(\mathcal{R}),$ $1\leq k\leq n,$ implies that
$A_{k}A_{l}=A_{l}A_{k},$ $1\leq k,l\leq n$ (see definition of independence in
Subsection 2.8). Hence, there exists a commutative von Neumann subalgebra
$\mathcal{M}$ in $\mathcal{R}$ such that
$A_{k}\in\mathcal{L}_{p}(\mathcal{M}),$ $1\leq k\leq n.$ Without loss of
generality, we may assume that $\mathcal{M}$ is ∗-isomorphic to
$L_{\infty}(0,1).$ Hence, the assertion can be viewed as the statement about
functions. The latter is proved in [33, Theorem 1]. ∎

The next lemma describes an isomorphic embedding of the Banach space
$(\mathcal{L}_{p}+\mathcal{L}_{2})(\mathcal{R}\overline{\otimes}\mathcal{L}(H))$
into the Banach space $\mathcal{L}_{p}(\mathcal{R}).$ Recall that the von
Neumann algebra $\mathcal{R}\overline{\otimes}\mathcal{L}(H)$ is a hyperfinite
$II_{\infty}$ factor. By $\|\cdot\|_{\mathcal{L}_{p}+\mathcal{L}_{2}}$ we
denote the standard norm on the space
$(\mathcal{L}_{p}+\mathcal{L}_{2})(\mathcal{R}\overline{\otimes}\mathcal{L}(H)).$
For brevity we set $r=r_{1},$ where $r_{1}$ is a Rademacher function defined
in (14).

######  Lemma 38.

Let $1\leq p\leq 2$ and let
$A\in\mathcal{L}_{p}(\mathcal{R}\overline{\otimes}\mathcal{L}(H))$ be finitely
supported. Then

$\|\mathscr{K}(A\otimes r)\|_{p}\sim\|A\|_{\mathcal{L}_{p}+\mathcal{L}_{2}}.$

Moreover, the mapping $\mathscr{R}:A\to\mathscr{K}(A\otimes r)$ extends to an
isomorphic embedding of
$(\mathcal{L}_{p}+\mathcal{L}_{2})(\mathcal{R}\overline{\otimes}\mathcal{L}(H))$
into $\mathcal{L}_{p}(\mathcal{R}).$

###### Proof.

Observe that for every finitely supported
$A\in\mathcal{L}_{p}(\mathcal{R}\overline{\otimes}\mathcal{L}(H)),$ the
element $A\otimes r$ belongs to
$\mathcal{L}_{1}(\mathcal{R}\bar{\otimes}\mathcal{L}(H)).$ In particular,
$\mathscr{K}(A\otimes r)$ is well defined.

Suppose first that $A=A^{*}.$ Since $A$ is finitely supported, it follows that
one can select
$A_{k}=A_{k}^{*}\in\mathcal{L}_{p}(\mathcal{R}\overline{\otimes}\mathcal{L}(H)),$
$1\leq k\leq n,$ such that $A_{k}A_{j}=0$ for $k\neq j,$
$A=\sum_{k=1}^{n}A_{k}$ and $(\tau\otimes{\rm Tr})({\rm supp}(A_{k}))\leq 1.$
Since the random varaibles $A_{k}\otimes r,$ $1\leq k\leq n,$ are
symmetrically distributed, it follows from Corollary 33 (v) and (iv) that the
random variables $\mathscr{K}(A_{k}\otimes r),$ $1\leq k\leq n,$ are
independent and symmetrically distributed. Applying Lemma 37, we infer that

$\|\mathscr{K}(A\otimes
r)\|_{p}=\Big{\|}\sum_{k=1}^{n}\mathscr{K}(A_{k}\otimes
r)\Big{\|}_{p}\sim\Big{\|}\bigoplus_{k=1}^{n}\mathscr{K}(A_{k}\otimes
r)\Big{\|}_{\mathcal{L}_{p}+\mathcal{L}_{2}}.$ (77)

For every $1\leq k\leq n,$ select a function $x_{k}\in L_{p}(0,1)$ which is
equimeasurable with $A_{k}\otimes r.$ By Corollary 33 (i), we have that
$\mathscr{K}(A_{k}\otimes r)$ is equimeasurable with $Kx_{k}.$ Therefore, by
Lemma 36, we have

$\displaystyle\Big{\|}\bigoplus_{k=1}^{n}\mathscr{K}(A_{k}\otimes
r)\Big{\|}_{\mathcal{L}_{p}+\mathcal{L}_{2}}=\Big{\|}\bigoplus_{k=1}^{n}Kx_{k}\Big{\|}_{L_{p}+L_{2}}\stackrel{{\scriptstyle
L.\ref{modified
ms}}}{{\sim}}\Big{\|}\bigoplus_{k=1}^{n}x_{k}\Big{\|}_{L_{p}+L_{2}}=\Big{\|}\bigoplus_{k=1}^{n}A_{k}\Big{\|}_{\mathcal{L}_{p}+\mathcal{L}_{2}}.$
(78)

Combining (77) and (78), we obtain

$\|\mathscr{K}(A\otimes
r)\|_{p}\sim\Big{\|}\bigoplus_{k=1}^{n}A_{k}\Big{\|}_{\mathcal{L}_{p}+\mathcal{L}_{2}}=\|A\|_{\mathcal{L}_{p}+\mathcal{L}_{2}}.$

For a not necessarily self-adjoint finitely supported operator
$A\in\mathcal{L}_{p}(\mathcal{R}\overline{\otimes}\mathcal{L}(H)),$ we have

$\displaystyle\|A\|_{\mathcal{L}_{p}+\mathcal{L}_{2}}$ $\displaystyle\sim\|\Re
A\|_{\mathcal{L}_{p}+\mathcal{L}_{2}}+\|\Im
A\|_{\mathcal{L}_{p}+\mathcal{L}_{2}}$ $\displaystyle\sim\|\mathscr{K}(\Re
A\otimes r)\|_{p}+\|\mathscr{K}(\Im A\otimes
r)\|_{p}\sim\|\mathscr{K}(A\otimes r)\|_{p}.$

The final assertion follows from the fact that the linear subspace of all
elements $A\in\mathcal{L}_{p}(\mathcal{R}\overline{\otimes}\mathcal{L}(H))$
with finite support is dense in
$(\mathcal{L}_{p}+\mathcal{L}_{2})(\mathcal{R}\overline{\otimes}\mathcal{L}(H))$.
∎

By $\mathcal{O}_{p,q}$ we denote the class of $p$-convex and $q$-concave
Orlicz functions $M$ on $\mathbb{R}$ such that $M(1)=1.$ For $1\leq p\leq 2$
we define an Orlicz function $M_{p}\in\mathcal{O}_{p,2}$ by setting

$M_{p}(t):=\begin{cases}t^{2},\quad 0\leq t\leq 1\\\
1+\frac{2}{p}(t^{p}-1),\quad t\geq 1.\end{cases}$ (79)

One can show that $(L_{p}+L_{2})(0,\infty)=L_{M_{p}}(0,\infty)$ and,
therefore,

$(\mathcal{L}_{p}+\mathcal{L}_{2})(\mathcal{R}\bar{\otimes}\mathcal{L}(H))=\mathcal{L}_{M_{p}}(\mathcal{R}\bar{\otimes}\mathcal{L}(H)).$

In what follows, we use the Orlicz norm
$\|\cdot\|_{\mathcal{L}_{p}+\mathcal{L}_{2}}:=\|\cdot\|_{\mathcal{L}_{M_{p}}}$
in the space
$(\mathcal{L}_{p}+\mathcal{L}_{2})(\mathcal{R}\bar{\otimes}\mathcal{L}(H)).$

The following lemma shows how some Orlicz spaces can be embedded into
$(\mathcal{L}_{p}+\mathcal{L}_{2})(\mathcal{R}\overline{\otimes}\mathcal{L}(H)).$

######  Lemma 39.

Let $A\in\mathcal{L}(H)$ be an operator of finite rank and let
$A_{0}\in\mathcal{L}_{p}(\mathcal{R})$ for $1\leq p\leq 2$. Then

$\|A_{0}\otimes
A\|_{\mathcal{L}_{p}+\mathcal{L}_{2}}=\|A\|_{\mathcal{L}_{M}},$

where $\mathcal{L}_{M}$ is an Orlicz ideal associated to the Orlicz function
$M$ defined by setting

$M(t):=\tau(M_{p}(|tA_{0}|)),\ \ \ t>0.$

###### Proof.

Employing standard properties of singular valued function $\mu,$ by definition
of the function $M$, we have

$M(t)=\int_{0}^{1}\mu(s,M_{p}(|tA_{0}|))ds=\int_{0}^{1}M_{p}(t\mu(s,A_{0}))ds,\quad
t>0.$ (80)

Denote, for brevity, $\|A\|_{\mathcal{L}_{M}}=\lambda.$ Observe that there
exists measure preserving mapping
$\mathcal{G}:\mathbb{Z}^{+}\times(0,1)\to(0,\infty),$ $\mathcal{G}(k,s)=k+s$
such that $\mathcal{G}$ extends to the measure preserving ∗-homomorphism from
$L_{\infty}(0,1)\otimes l_{\infty}$ into $L_{\infty}(0,\infty).$ Observe also
that the function $f_{A,A_{0}}:k+s\mapsto\mu(s,A_{0})\mu(k,A)$ is
equimeasurable with $\mu(A_{0}\otimes A).$ Hence, using (80) and the facts
listed above, we obtain

$\displaystyle 1$ $\displaystyle={\rm
Tr}\Big{(}M\Big{(}\frac{|A|}{\lambda}\Big{)}\Big{)}=\sum_{k\geq
0}\mu\Big{(}k,M\Big{(}\frac{|A|}{\lambda}\Big{)}\Big{)}$ (81)
$\displaystyle=\sum_{k\geq
0}M\Big{(}\frac{\mu(k,A)}{\lambda}\Big{)}=\sum_{k\geq
0}\int_{0}^{1}M_{p}\Big{(}\frac{\mu(s,A_{0})\mu(k,A)}{\lambda}\Big{)}ds$
$\displaystyle=\int_{0}^{\infty}M_{p}\Big{(}\frac{\mu(u,A_{0}\otimes
A)}{\lambda}\Big{)}du=({\rm
Tr}\otimes\tau)\Big{(}M_{p}\Big{(}\frac{A_{0}\otimes
A}{\lambda}\Big{)}\Big{)}.$

Now (81) and well-known property of Orlicz norms (see e.g. [54, I, 1.2,
Proposition 11]) implies $\|A_{0}\otimes
A\|_{\mathcal{L}_{M_{p}}}=\|A\|_{\mathcal{L}_{M}},$ that completes the proof
of the lemma. ∎

Recall (see [43]) that the function $\varphi:(0,\infty)\to(0,\infty)$ is
called quasi-concave if

$\varphi(s)\leq\varphi(t),\quad\frac{\varphi(s)}{s}\geq\frac{\varphi(t)}{t},\quad
0\leq s\leq t.$

The following lemma is somewhat similar to the one in [5]. However, the proof
presented here is simpler.

######  Lemma 40.

Let $\varphi:[0,1]\to[0,1]$ be a quasi-concave function such that
$\varphi(0)=0$ and $\varphi(1)=1.$ For every $d>1,$ there exists $0\leq x\in
L_{1}(0,1)$ with $\|x\|_{1}\leq 1$ such that

$\frac{1}{4}\varphi(t)\leq\int_{0}^{1}\min\\{x(s),tx^{d}(s)\\}ds\leq\frac{5}{2}\varphi(t),\quad
t\in[0,1].$

###### Proof.

Suppose first that $\varphi^{\prime}(0)=\infty.$ It follows from the
equalities

$\lim_{t\to 0}\varphi(t)=0,\quad\lim_{t\to 0}\frac{\varphi(t)}{t}=\infty$

that one can select sequences $t_{k}\downarrow 0$ and $s_{k}\downarrow 0$ such
that

$\varphi(t_{k})=2^{-k}\varphi(1),\quad\frac{s_{k}}{\varphi(s_{k})}=2^{-k}\frac{1}{\varphi(1)},\quad
k\geq 0.$

Define a sequence $u_{k}\downarrow 0$ by setting $u_{0}=t_{0}=1,$
$u_{1}=s_{1}$ and

$u_{2k}=\max\\{t_{l}:\ t_{l}<u_{2k-1}\\},\quad u_{2k+1}=\max\\{s_{l}:\
s_{l}\leq u_{2k}\\},\quad k\geq 1.$

Step 1. Observe that

$\varphi(u_{2k})\leq
2^{l-k}\varphi(u_{2l}),\quad\frac{\varphi(u_{2k+1})}{u_{2k+1}}\leq
2^{k-l}\frac{\varphi(u_{2l+1})}{u_{2l+1}},\ \ \ l\leq k.$ (82)

Indeed, by definition of the sequence $\\{u_{2k}\\}_{k\geq 1}$ we have that

$u_{2k}=t_{m},\ \ \ u_{2k-2}=t_{n},\ \ \mbox{for some}\ \ n<m.$

The computation

$\varphi(u_{2k})=\varphi(t_{m})=2^{-m}=2^{-m+n}\varphi(t_{n})=2^{-m+n}\varphi(u_{2k-2})\leq
2^{-1}\varphi(u_{2k-2}).$

implies the first inequality in (82). The second one follows similarly.

Step 2. Fix $d>1.$ Since $\varphi(u_{2k})\leq\varphi(t_{k})=2^{-k}$ for all
$k\geq 0,$ it follows that

$\sum_{k\geq
0}\frac{1}{2}u_{2k}^{1/(d-1)}\varphi(u_{2k})\leq\frac{1}{2}\sum_{k\geq
0}\varphi(u_{2k})\leq\frac{1}{2}\sum_{k\geq 0}2^{-k}=1.$ (83)

Equality (83) guarantees that

$\sum_{k\geq
1}m\Big{(}(0,\frac{1}{2}u_{2k}^{1/(d-1)}\varphi(u_{2k}))\Big{)}\leq 1,$

where $m$ is the Lebesgue measure on $(0,1).$ Therefore, choosing the
decreasing order of the summands below, we construct a positive decreasing
function (see (27))

$x:=\bigoplus_{k\geq
0}u_{2k}^{-1/(d-1)}\chi_{(0,\frac{1}{2}u_{2k}^{1/(d-1)}\varphi(u_{2k}))},$
(84)

that belongs to $L_{1}(0,1).$ Equality (83) implies that $\|x\|_{1}\leq 1$.
For the function $x\in L_{1}(0,1)$ constructed above, for $t\in[0,1]$, we have

$\displaystyle 2\int_{0}^{1}\min\\{x(s),tx^{d}(s)\\}ds=\sum_{k\geq
0}u_{2k}^{1/(d-1)}\varphi(u_{2k})\cdot\min\\{u_{2k}^{-1/(d-1)},t\cdot
u_{2k}^{-d/(d-1)}\\}$ (85) $\displaystyle=\sum_{u_{2k}\leq
t}\varphi(u_{2k})+\sum_{u_{2k}>t}t\cdot\frac{\varphi(u_{2k})}{u_{2k}}.$

Thus, to prove the lemma, it is enough to show that the right hand side of
(85) satisfies

$\frac{1}{2}\varphi(t)\leq\sum_{u_{2k}\leq
t}\varphi(u_{2k})+t\cdot\sum_{u_{2k}>t}\frac{\varphi(u_{2k})}{u_{2k}}\leq
5\varphi(t),\ \ \ t\in[0,1].$ (86)

We first show the right hand side of (86). Let $l\in\mathbb{N}$ be such that
$t\in[u_{2l},u_{2l-2}).$

Step 3. Applying (82), we have

$\sum_{u_{2k}\leq
t}\varphi(u_{2k})=\sum_{k=l}^{\infty}\varphi(u_{2k})\leq\sum_{k=l}^{\infty}2^{l-k}\varphi(u_{2l})=2\varphi(u_{2l})\leq
2\varphi(t).$

Similarly,

$\displaystyle\sum_{u_{2k}>t}\frac{\varphi(u_{2k})}{u_{2k}}$
$\displaystyle=\frac{\varphi(u_{2l-2})}{u_{2l-2}}+\sum_{k=0}^{l-2}\frac{\varphi(u_{2k})}{u_{2k}}$
(87)
$\displaystyle\leq\frac{\varphi(u_{2l-2})}{u_{2l-2}}+\sum_{k=0}^{l-2}\frac{\varphi(u_{2k+1})}{u_{2k+1}}\leq\frac{\varphi(u_{2l-2})}{u_{2l-2}}+\sum_{k=0}^{l-2}2^{k-l+2}\frac{\varphi(u_{2l-3})}{u_{2l-3}}$
$\displaystyle\leq\frac{\varphi(u_{2l-2})}{u_{2l-2}}+2\frac{\varphi(u_{2l-3})}{u_{2l-3}}\leq
3\frac{\varphi(u_{2l-2})}{u_{2l-2}}\leq\frac{3\varphi(t)}{t}.$

Thus, we obtain the right hand side of (85).

Step 4. Next we prove the left hand side of (85). We consider two cases
$t\in[u_{2l},u_{2l-1})$ and $t\in[u_{2l-1},u_{2l-2})$ separately. Let
$t\in[u_{2l},u_{2l-1}).$ Since $u_{2l}=t_{m}$ for some $m$ and $t_{m-1}\geq
u_{2l-1},$ it follows from the definition of ${u_{2k}}$ that

$\sum_{u_{2k}\leq
t}\varphi(u_{2k})+\sum_{u_{2k}>t}t\cdot\frac{\varphi(u_{2k})}{u_{2k}}\geq\varphi(u_{2l})=\frac{1}{2}\varphi(t_{m-1})\geq\frac{1}{2}\varphi(u_{2l-1})\geq\frac{1}{2}\varphi(t).$

Let $t\in[u_{2l-1},u_{2l-2}).$ Since $u_{2l-1}=s_{m}$ for some $m$ and
$s_{m-1}\geq u_{2l-2},$ by the definition of ${u_{2k+1}}$, we have that

$\sum_{u_{2k}\leq
t}\varphi(u_{2k})+\sum_{u_{2k}>t}t\cdot\frac{\varphi(u_{2k})}{u_{2k}}\geq$
$\geq t\cdot\frac{\varphi(u_{2l-2})}{u_{2l-2}}\geq
t\cdot\frac{\varphi(s_{m-1})}{s_{m-1}}=\frac{t}{2}\cdot\frac{\varphi(s_{m})}{s_{m}}=\frac{t}{2}\cdot\frac{\varphi(u_{2l-1})}{u_{2l-1}}\geq\frac{t}{2}\cdot\frac{\varphi(t)}{t}=\frac{1}{2}\varphi(t).$

The two steps above prove the assertion in the case
$\varphi^{\prime}(0)=\infty.$

Step 5. Let $\varphi^{\prime}(0)<\infty.$ Then the sequence $\\{s_{k}\\}$ is
finite and, therefore, the sequence $\\{u_{2k}\\}_{k\geq 0}$ is also finite.
Let $m+1$ and $n+1$ be the lengths of the sequences $\\{s_{k}\\}$ and
$\\{u_{2k}\\}_{k\geq 0}$, respectively. If $t\in[u_{2l},u_{2l-2}),$
$l=0,\ldots,n$ then the assertion can be shown similarly (see Steps 1 and 2
above). Suppose that $t\in[0,u_{2n}).$ The right hand side of (86) can be
estimated similarly to (87), as follows

$\sum_{u_{2k}\leq
t}\varphi(u_{2k})+t\cdot\sum_{u_{2k}>t}\frac{\varphi(u_{2k})}{u_{2k}}=t\cdot\sum_{k=0}^{n}\frac{\varphi(u_{2k})}{u_{2k}}\leq
3\varphi(t).$ (88)

Now we estimate the left hand side of (86). Since the sequence $\\{t_{k}\\}$
is infinite, it follows that the recursive procedure stops at $u_{2n}.$
Indeed, if $u_{2n+1}$ were defined, then by definition of $\\{u_{2k}\\},$ we
would have that $u_{2n+2}$ is defined, which contradicts with the choice of
$n$. Since $u_{2n+1}$ is not defined, it follows that there is no $s_{l}$ such
that $s_{l}\leq u_{2n}.$ Therefore, $s_{k}>u_{2n}$ for all $k=1,...,m$ and

$\varphi^{\prime}(0)={\lim_{s\to\,0}}\
\frac{\varphi(s)}{s}<2\frac{\varphi(s_{m})}{s_{m}}\leq
2\frac{\varphi(u_{2n})}{u_{2n}}.$

Thus, we have

$t\cdot\sum_{k=0}^{n}\frac{\varphi(u_{2k})}{u_{2k}}\geq
t\cdot\frac{\varphi(u_{2n})}{u_{2n}}\geq\frac{t}{2}\
\varphi^{\prime}(0)\geq\frac{t}{2}\cdot\frac{\varphi(t)}{t}=\frac{\varphi(t)}{2},$

which together with (88) completes the proof.

∎

Now, we are fully prepared to present the proof the implication.

###### Proof of Theorem 4: (iii)$\Rightarrow$(i).

Let $M$ be equivalent to a $p$-convex and $2$-concave Orlicz function. Without
loss of generality, we may assume that $M\in\mathcal{O}_{p,2}$. Define the
function $\varphi$ by setting

$\varphi(t):=\frac{M\big{(}t^{\frac{1}{2-p}}\big{)}}{t^{\frac{p}{2-p}}},\ \
\mbox{so that}\ \ M(t)=t^{p}\varphi(t^{2-p}),\ \ t\in(0,1),$

Since $M$ is $p$-convex and $2$ concave, application of Lemma 11 implies that
$\varphi$ is quasi-concave. It follows from Lemma 11 and $p$-convexity of $M$
that there exists a limit

$\lim_{t\to 0}\frac{M(t)}{t^{p}}=\lim_{t\to 0}\varphi(t^{2-p}).$

Since $M(t)\not\sim t^{p},$ it follows that the limit above is $0$ and,
therefore, $\varphi(0)=0.$ Thus, the function $\varphi$ satisfies the
conditions of Lemma 40.

Set $d=2/p,$ select $x\in L_{1}(0,1)$ as in Lemma 40 and select a random
variable $A_{0}\in\mathcal{L}_{p}(\mathcal{R})$ such that
$\mu(A_{0})=x^{1/p}.$ Recall that $M_{p}(t)\sim\min\\{t^{p},t^{2}\\}$ for all
$t>0,$ where $M_{p}$ is the Orlicz function defined by (79). we infer from
Lemma 40 that

$M(t)=t^{p}\ \varphi(t^{2-p})\sim
t^{p}\tau(\min\\{|A_{0}|^{p},t^{2-p}|A_{0}|^{2}\\})\sim$
$\sim\tau(\min\\{(t|A_{0}|)^{p},(t|A_{0}|)^{2}\\})\sim\tau(M_{p}(t|A_{0}|)),\quad
t\in(0,1).$

By Lemma 39 and Lemma 38, we obtain

$\|A\|_{\mathcal{L}_{M}}=\|A\otimes
A_{0}\|_{\mathcal{L}_{p}+\mathcal{L}_{2}}\sim\|\mathscr{R}(A\otimes
A_{0})\|_{p}$

for every finite rank operator $A\in\mathcal{L}(H).$ The set of all finite
rank operators is dense in every separable ideal
$\mathcal{I},\|\cdot\|_{\mathcal{I}})$, in particular in the ideal
$\mathcal{L}_{M}(H)$. Thus, the ideal $\mathcal{L}_{M}(H)$ isomorphically
embeds into $\mathcal{L}_{p}(\mathcal{R}).$ ∎

##  6\. Raynaud-Schütt theorem for $1\leq p<2$

In this section, we extend the main result from [56, Theorem 1.1] established
there for $L_{1}$-spaces to the case of $L_{p}$-spaces, $1\leq p<2.$ For
convenience of the reader, we present a complete version of our argument.

The assertion below extends [45, Lemma 2.9]. In this and subsequent sections,
we denote the set of all $q-$concave (respectively, that of $p$-convex and
$2$-concave) Orlicz functions $M$ such that $M(1)=1$ by $\mathcal{O}_{1,q}$
(respectively, by $\mathcal{O}_{p,2}$).

######  Lemma 41.

Let $1\leq q\leq 2$ be fixed. For every $n\geq 1,$ there exists
$M_{n}\in\mathcal{O}_{1,q}$ such that

$\frac{1}{4}\|x\|_{l_{M_{n}}}\leq\sum_{k=0}^{n-1}\mu(k,x)+n^{1-1/q}\Big{(}\sum_{k=n}^{\infty}\mu^{q}(k,x)\Big{)}^{1/q}\leq
4\|x\|_{l_{M_{n}}}$ (89)

for every finitely supported $x\in l_{\infty}.$

###### Proof.

Let $x=\\{x(k)\\}_{k\geq 0}\in l_{\infty}$ be finitely supported and let
$n\geq 1$ be fixed. Without loss of generality we may assume that $x\geq 0.$
By $N_{n}$ we denote a $q$-concave Orlicz function defined by

$N_{n}(t):=\begin{cases}n^{q-1}t^{q},\quad 0\leq t\leq 1/n\\\ qt-(q-1)/n,\quad
t\geq 1/n.\end{cases}$

Denoting

$S(x):=\sum_{k=0}^{n-1}\mu(k,x)+n^{1-1/q}\Big{(}\sum_{k=n}^{\infty}\mu^{q}(k,x)\Big{)}^{1/q},$

we shall show that the Orlicz function $N_{n}$ satisfies

$\frac{1}{4}\|x\|_{l_{N_{n}}}\leq S(x)\leq 2\|x\|_{l_{N_{n}}}.$ (90)

Step 1: We first show the left hand side of (90). Suppose that $S(x)\leq 1.$
Setting $y=\\{y(k)\\}_{k\geq 0}$ and $z=\\{z(k)\\}_{k\geq 0},$ where

$y(k)=\left\\{\begin{array}[]{cl}x(k),&x(k)>1/n\\\
0,&\mbox{otherwise}\end{array}\right.\ \ \mbox{and}\ \
z(k)=\left\\{\begin{array}[]{cl}x(k),&x(k)\leq 1/n\\\
0,&\mbox{otherwise}\end{array}\right.\ \ \ k\geq 0,$

we have $\mu(y),\mu(z)\leq\mu(x).$ Since $S(x)\leq 1,$ it obviously follows
that

$\sum_{k=0}^{n-1}\mu(k,x)\leq 1.$ (91)

Thus, we obtain

$\mu(n-1,y)\leq\frac{1}{n}\sum_{k=0}^{n-1}\mu(k,y)\leq\frac{1}{n}\sum_{k=0}^{n-1}\mu(k,x)\leq\frac{1}{n}.$

According to definition of $y,$ we have that $\mu(n-1,y)=0$ and, therefore,
$\mu(k,y)=0$ for $k\geq n-1.$ Hence, employing again (91), we have

$\sum_{k\geq
0}N_{n}(y(k))=\sum_{y(k)>1/n}\Big{(}qy(k)-\frac{q-1}{n}\Big{)}\leq\sum_{y(k)>1/n}qy(k)=q\sum_{k=0}^{n-1}\mu(k,y)\leq
q,$

which implies $\|y\|_{l_{N_{n}}}\leq q.$

On the other hand, by the assumption, we have

$n^{1-1/q}\Big{(}\sum_{k=n}^{\infty}\mu^{q}(k,z)\Big{)}^{1/q}\leq S(x)\leq 1$

and, therefore,

$\sum_{k=n}^{\infty}\mu^{q}(k,z)\leq n^{1-q}.$ (92)

Since by (92) and by definition of $z$

$\sum_{k\geq
0}z(k)^{q}=\sum_{k=n}^{\infty}\mu^{q}(k,z)+\sum_{k=0}^{n-1}\mu^{q}(k,z)\leq
n^{1-q}+n\cdot n^{-q}=2n^{1-q},$

it follows that

$\sum_{k\geq 0}N_{n}(z(k))=\sum_{z_{k}\leq 1/n}n^{q-1}z(k)^{q}\leq 2.$

Hence, $\|z\|_{l_{N_{n}}}\leq 2.$

Combining the estimates above, we infer that

$\|x\|_{l_{N_{n}}}\leq\|y\|_{l_{N_{n}}}+\|z\|_{l_{N_{n}}}\leq q+2\leq 4,$

which completes the proof of the left hand side of (90).

Step 2: Now we prove the right hand side of (90). Suppose that
$\|x\|_{l_{N_{n}}}\leq 1.$ Setting

$y:=\Big{(}\underbrace{\frac{1}{n}\sum_{k=0}^{n-1}\mu(k,x)}_{\mbox{\small{$n$
times}}},\mu(n,x),\mu(n+1,x),\ldots\Big{)},$

we have that $y\prec\prec x$ and, therefore (see Remark 12),
$\|y\|_{l_{N_{n}}}\leq 1.$ Hence,

$\sum_{y(k)>1/n}y(k)\leq\sum_{y(k)>1/n}\Big{(}qy(k)-\frac{q-1}{n}\Big{)}\leq\sum_{k\geq
0}N_{n}(y(k))\leq 1.$ (93)

Suppose that $y(0)>1/n.$ Then $y(k)>1/n$ for $0\leq k\leq n-1$ and, therefore,
$\sum_{y(k)>1/n}y(k)>1,$ which contradicts with (93). Thus, we conclude that
$y(0)\leq 1/n$ and, therefore, $y(k)\leq 1/n$ for all $k\geq 0.$ Combining the
latter and the fact that $\|y\|_{l_{N_{n}}}\leq 1$, we obtain

$n^{q-1}\sum_{k\geq 0}y(k)^{q}=\sum_{k\geq 0}N_{n}(y(k))\leq 1.$

Hence, by definition of $y$,

$S(x)=\sum_{k=0}^{n-1}\mu(k,x)+n^{1-1/q}\Big{(}\sum_{k=n}^{\infty}\mu^{q}(k,x)\Big{)}^{1/q}\leq
ny(0)+n^{1-1/q}\Big{(}\sum_{k\geq 0}y(k)^{q}\Big{)}^{1/q}\leq$ $\leq
n\cdot\frac{1}{n}+n^{1-1/q}\cdot(n^{1-q})^{1/q}\leq 2,$

which completes the proof of the right hand side of (90).

Step 3: Next we set

$M_{n}(t):=\frac{N_{n}(t)}{N_{n}(1)},\ \ \ t\in[0,\infty).$

Observe that $M_{n}\in\mathcal{O}_{1,q}.$ Moreover, since $1\leq N_{n}(1)\leq
q\leq 2,$ for all $n\geq 1,$ we have that

$M_{n}(t)\leq N_{n}(t)\leq 2\ M_{n}(t),\ \ \ t\in[0,\infty),\ \ n\geq 1.$

Using the latter fact, one can infer that

$\|x\|_{l_{M_{n}}}\leq\|x\|_{l_{N_{n}}}\leq 2\|x\|_{l_{M_{n}}}$ (94)

Combination of (90) and (94) completes the proof of the lemma. ∎

The assertion below extends [45, Lemma 2.10]. Recall that $\mathfrak{S}_{n}$
denotes the group of all permutations of $n$ elements.

######  Lemma 42.

Let $1\leq p\leq 2$ and $n\in\mathbb{N}.$ For every $y\in\mathbb{R}^{n},$
there exists $M_{n,y}\in\mathcal{O}_{p,2}$ such that

$\frac{1}{80n}\|y\|_{p}^{p}\cdot\|x\|_{l_{M_{n,y}}}^{p}\leq\frac{1}{n!}\sum_{\rho\in\mathfrak{S}_{n}}\Big{(}\sum_{k=0}^{n-1}x^{2}(\rho(k))y^{2}(k)\Big{)}^{\frac{p}{2}}\leq\frac{16}{n}\|y\|_{p}^{p}\cdot\|x\|_{l_{M_{n,y}}}^{p}$
(95)

for every $x\in\mathbb{R}^{n}.$

###### Proof.

Let $1\leq p\leq 2,$ $q=2/p,$ $n\in\mathbb{N}$ and $y\in\mathbb{R}^{n}$ be
fixed. For every $x\in\mathbb{R}^{n}$ we obviously have