# Traces on symmetrically normed operator ideals

F. Sukochev  School of Mathematics and Statistics, University of New South
Wales, Sydney, 2052, Australia<EMAIL_ADDRESS>and  D. Zanin  School
of Mathematics and Statistics, University of New South Wales, Sydney, 2052,
Australia<EMAIL_ADDRESS>
###### Abstract.

For every symmetrically normed ideal $\mathcal{E}$ of compact operators, we
give a criterion for the existence of a continuous singular trace on
$\mathcal{E}$. We also give a criterion for the existence of a continuous
singular trace on $\mathcal{E}$ which respects Hardy-Littlewood majorization.
We prove that the class of all continuous singular traces on $\mathcal{E}$ is
strictly wider than the class of continuous singular traces which respect
Hardy-Littlewood majorization. We establish a canonical bijection between the
set of all traces on $\mathcal{E}$ and the set of all symmetric functionals on
the corresponding sequence ideal. Similar results are also proved in the
setting of semifinite von Neumann algebras.

###### Key words and phrases:

Symmetric functionals, singular traces

###### 2000 Mathematics Subject Classification:

47L20, 47B10, 46L52

##  1\. Introduction

In his groundbreaking paper [6], J. Dixmier proved the existence of positive
singular traces (that is, linear positive unitarily invariant functionals
which vanish on all finite dimensional operators) on the algebra $B(H)$ of all
bounded linear operators acting on infinite-dimensional separable Hilbert
space $H.$ Namely, if $\psi:\mathbb{R}_{+}\to\mathbb{R}_{+}$ is a concave
increasing function such that

(1) $\lim_{t\to\infty}\frac{\psi(2t)}{\psi(t)}=1,$

then there is a singular trace $\tau_{\omega},$ defined for every positive
compact operator $A\in B(H)$ by setting

(2) $\tau_{\omega}(A)=\omega(\frac{1}{\psi(n)}\sum_{k=1}^{n}s_{k}(A)).$

Here, $\\{s_{k}(A)\\}_{k\in\mathbb{N}}$ is the sequence of singular values of
the compact operator $A\in B(H)$ taken in the descending order and $\omega$ is
an arbitrary dilation invariant generalised limit on the algebra $l_{\infty}$
of all bounded sequences. This trace is finite on $0\leq A\in B(H)$ if and
only if $A$ belongs to the Marcinkiewicz ideal (see e.g. [14],[15],[27])

$\mathcal{M}_{\psi}:=\\{A\in B(H):\
\sup_{n\in\mathbb{N}}\frac{1}{\psi(n)}\sum_{k=1}^{n}s_{k}(A)<\infty\\}.$

In [18], Dixmier’s result was extended to an arbitrary Marcinkiewicz ideal
$\mathcal{M}_{\psi}$ with the following condition on $\psi$

(3) $\liminf_{t\to\infty}\frac{\psi(2t)}{\psi(t)}=1.$

All the traces defined above by formula (2) vanish on the ideal
$\mathcal{L}_{1}$ consisting of all compact operators $A\in B(H)$ such that
$\sum_{k=1}^{\infty}s_{k}(A)<\infty$.

An ideal $\mathcal{E}$ of algebra $B(H)$ is said to be symmetrically normed if
$\\{s_{k}(B)\\}_{k\in\mathbb{N}}\leq\\{s_{k}(A)\\}_{k\in\mathbb{N}}$ and
$A\in\mathcal{E}$ implies that $\|B\|_{\mathcal{E}}\leq\|A\|_{\mathcal{E}}$
(see [14], [15], [29]111 We have to caution the reader that in Theorem 1.16 of
[29] the assertion $(b)$ does not hold for the norm of an arbitrary
symmetrically normed ideal $\mathcal{E}$ (see e.g. corresponding
counterexamples in [19, p. 83])., [28], [20]). Since the ideal
$\mathcal{M}_{\psi}$ is just a special example of symmetrically normed
operator ideal, the following question (suggested in [18], [16], [17], [7])
arises naturally.

######  Question 1.

Which symmetrically normed operator ideals admit a nontrivial singular
trace222 In this paper, we exclusively deal with positive traces?

In analyzing Dixmier’s proof of the linearity of $\tau_{\omega}$ given by (1),
it was observed in [18] (see also [3]) that $\tau_{\omega}$ possesses the
following fundamental property, namely if $0\leq A,B\in\mathcal{M}_{\psi}$ are
such that

(4) $\sum_{k=1}^{n}s_{k}(B)\leq\sum_{k=1}^{n}s_{k}(A),\quad\forall
n\in\mathbb{N},$

then $\tau_{\omega}(B)\leq\tau_{\omega}(A).$ Such a class of traces was termed
“fully symmetric”in [20], [30] (see also earlier papers [8],[25], where the
term “symmetric”was used). It is natural to consider such traces only on fully
symmetrically normed operator ideals $\mathcal{E}$ (that is, on symmetrically
normed operator ideals $\mathcal{E}$ satisfying the condition: if $A,B$
satisfy (4) and $A\in\mathcal{E},$ then $B\in\mathcal{E}$ and
$\|B\|_{\mathcal{E}}\leq\|A\|_{\mathcal{E}}$). In fact, it was established in
[8] that every Marcinkiewicz ideal $\mathcal{M}_{\psi}$ with $\psi$ satisfying
the condition (3) possesses fully symmetric traces. Furthermore, in the recent
paper [18], the following unexpected result was established. If $\psi$
satisfies the condition (3), then every fully symmetric trace on
$\mathcal{M}_{\psi}$ is a Dixmier trace $\tau_{\omega}$ for some $\omega.$

The following question ( also suggested in [18], [7], [16], [17]) arises
naturally.

######  Question 2.

Which fully symmetrically normed operator ideals admit a nontrivial singular
trace which is fully symmetric?

In papers [16],[17] the following two problems (closely related to Question 1
and Question 2) were also suggested.

######  Question 3.

Which fully symmetrically normed operator ideals admit a trace which is not
fully symmetric?

Let us fix an orthonormal basis $\\{e_{n}\\}_{n\in\mathbb{N}}$ in $H.$ An
operator $A\in B(H)$ is called diagonal if $(Ae_{n},e_{m})=0$ for every $n\neq
m.$

######  Question 4.

Let the mapping $\varphi:\mathcal{E}\to\mathbb{C}$ be unitarily invariant.
Suppose that $\varphi$ is linear on the subset of all diagonal operators from
$\mathcal{E}.$ Does it imply that $\varphi$ is a trace on $\mathcal{E}?$

In some very special cases (for principal ideals contained in
$\mathcal{L}_{1},$ which are, strictly speaking, not symmetrically normed
ideals), Question 3 was answered in the affirmative333We are grateful to the
referee for this remark. in [33]. In [20], question 3 was answered in the
affirmative for the special case of Marcinkiewicz ideals under the assumption
(1). It should be pointed out that the method used in [20] cannot be extended
to an arbitrary Marcinkiewicz ideal $\mathcal{M}_{\psi}$ and, furthermore,
cannot be extended to a general symmetrically normed operator ideal. Question
4 was answered in [20] in full generality using deep results from [11, 10]
(see also [9]).

The following theorem is the main result of this paper. It yields answers to
Questions 1–3. In the course of the proof of Theorem 5, we also present a new
(and very simple) proof answering Question 4. Prior to stating Theorem 5, we
make a few preliminary observations, for which we are grateful to the referee.

Any trace $\varphi:\mathcal{E}\to\mathbb{C}$ obeys the condition

$\frac{1}{m}\varphi(A^{\oplus m})=\varphi(A),\quad A\in\mathcal{E},m\geq 1.$

Here, the direct sum $A^{\oplus m}$ is formed with respect to some arbitrary
Hilbert space isomorphism $H^{\oplus m}\simeq H.$ Thus, traces are closely
related to the following convex (see Lemma 11 below) functional on
$\mathcal{E}.$

$\pi:A\to\lim_{m\to\infty}\frac{1}{m}\|A^{\oplus m}\|_{\mathcal{E}},\quad
A\in\mathcal{E}.$

The non-triviality of the functional $\pi:\mathcal{E}\to\mathbb{R}$ is an
obvious necessary condition for the existence of a trace.

######  Theorem 5.

Let $\mathcal{E}$ be a symmetrically normed operator ideal. Consider the
following conditions.

  1. (1)

There exist nontrivial singular traces on $\mathcal{E}.$

  2. (2)

There exist nontrivial singular traces on $\mathcal{E},$ which are fully
symmetric.

  3. (3)

There exist nontrivial singular traces on $\mathcal{E},$ which are not fully
symmetric.

  4. (4)

$\mathcal{E}\neq\mathcal{L}_{1}$ and there exist an operator $A\in\mathcal{E}$
such that

(5) $\lim_{m\to\infty}\frac{1}{m}\|A^{\oplus m}\|_{\mathcal{E}}>0.$

  1. (i)

The conditions $(1)$ and $(4)$ are equivalent for every symmetrically normed
operator ideal $\mathcal{E}.$

  2. (ii)

The conditions $(1),$ $(2)$ and $(4)$ are equivalent for every fully
symmetrically normed operator ideal $\mathcal{E}.$

  3. (iii)

The conditions $(1)-(4)$ are equivalent for every fully symmetrically normed
operator ideal $\mathcal{E}$ equipped with a Fatou norm.

Recall that the norm on a symmetrically normed operator ideal $\mathcal{E}$ is
called a Fatou norm if the unit ball of $\mathcal{E}$ is closed with respect
to strong (or, equivalently, weak) operator convergence. Observe that
classical ideals (such as Schatten-von Neumann ideals $\mathcal{L}_{p},$
Marcinkiewicz, Orlicz and Lorentz ideals [14], [15], [29]) have a Fatou norm.
In fact, in some standard references on the subject (e.g. Simon’s book [29]),
the requirement that symmetrically normed operator ideal has a Fatou norm
appears to be a part of the definition. Similarly, in the book [24], devoted
to the study of symmetric444termed there “rearrangement invariant”. function
spaces (which are a commutative counterpart of symmetrically normed operator
ideals), an assumption that the norm is a Fatou norm is incorporated into the
definition [24, p. 118].

The proof of Theorem 5 is given in Section 7. In fact, in this paper we will
prove a more general result for symmetric spaces associated with semifinite
von Neumann algebras. The precise statements are given in Section 4 (see
Theorems 23, 28, 29), Section 5 (see Theorems 33, 35, 36) and Section 6 (see
Theorems 47, 48). The appendix contains the proof of important technical
results for which we were unable to find a suitable reference. We also present
a new and short proof of the Figiel-Kalton theorem from [13].

Finally, we say a few words about our proof and its relation to the previous
results in the literature. Our strategy is based on the approach from recent
papers [30] and [21], where condition (5) was connected to the geometry of
$\mathcal{E}$ (see also [2]). The condition (5) is easy to verify in concrete
situations. For example, the following corollary of Theorem 5 strengthens the
main result of [20] and complements earlier results of J. Varga [32].

######  Corollary 6.

Every Marcinkiewicz ideal $\mathcal{M}_{\psi}$ with $\psi$ satisfying the
condition (3) admits a trace which is not fully symmetric.

Indeed, it is proved in [1, Proposition 2.3] that the condition $(4)$ of
Theorem 5 is equivalent to the condition (3) for the Marcinkiewicz ideal
$\mathcal{M}_{\psi}.$ Some examples of symmetrically normed operator ideals,
which are not Marcinkiewicz ideals, possessing symmetric traces were presented
in [7]. These results are also an immediate corollary of Theorem 5.

For completeness, we note that the assertion $(ii)$ in Theorem 5 holds for a
wider class of relatively fully symmetrically normed operator ideals. The
latter class is defined as follows: if $A,B\in\mathcal{E}$ are such that (4)
holds, then $\|B\|_{\mathcal{E}}\leq\|A\|_{\mathcal{E}}.$ It coincides with
the class of all symmetrically normed subspaces of a fully symmetric operator
ideal (see [19])

##  2\. Definitions and preliminaries

The theory of singular traces on symmetric operator ideals rests on some
classical analysis which we now review for completeness.

As usual, $L_{\infty}(0,\infty)$ is the set of all bounded Lebesgue measurable
functions on the semi-axis equipped with the uniform norm. Given a function
$x\in L_{\infty}(0,\infty),$ one defines its decreasing rearrangement
$t\to\mu(t,x)$ by the formula (see e.g. [22])

$\mu(t,x)=\inf\\{s\geq 0:\ m(\\{x>s\\})\leq t\\}.$

Let $H$ be a Hilbert space and let $B(H)$ be the algebra of all bounded
operators on $H$ equipped with the uniform norm.

Let $\mathcal{M}\subset B(H)$ be a semi-finite von Neumann algebra equipped
with a fixed faithful and normal semi-finite trace $\tau.$ $\mathcal{M}$ is
said to be atomic (see [31, Definition 5.9]) if every nonzero projection in
$\mathcal{M}$ contains a nonzero minimal projection. $\mathcal{M}$ is said to
be atomless if there is no minimal projections in $\mathcal{M}.$

For every $A\in\mathcal{M},$ the generalised singular value function
$t\to\mu(t,A)$ is defined by the formula (see e.g. [12])

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

If, in particular, $\mathcal{M}=B(H),$ then $\mu(A)$ is a step function and,
therefore, can be identified with the sequence of singular numbers of the
operators $A$ (the singular values are the eigenvalues of the operator
$|A|=(A^{*}A)^{1/2}$ arranged with multiplicity in decreasing order).

Equivalently, $\mu(A)$ can be defined in terms of the distribution function
$d_{A}$ of $A.$ That is, setting

$d_{A}(s)=\tau(E^{|A|}(s,\infty)),\quad s\geq 0,$

we obtain

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

Here, $E^{|A|}$ denotes the spectral measure of the operator $|A|.$

Using the Jordan decomposition, every operator $A\in B(H)$ can be uniquely
written as

$A=(\Re A)_{+}-(\Re A)_{-}+i(\Im A)_{+}-i(\Im A)_{-}.$

Here, $\Re A:=1/2(A+A^{*})$ (respectively, $\Im A:=1/2i(A-A^{*})$) for any
operator $A\in B(H)$ and $B_{+}=BE^{B}(0,\infty)$ ( respectively,
$B_{−}=-BE^{B}(-\infty,0)$) for any self-adjoint operator $B\in B(H).$ Recall
that $\Re A,\Im A\in\mathcal{M}$ for every $A\in\mathcal{M}$ and
$B_{+},B_{-}\in\mathcal{M}$ for every self-adjoint $B\in\mathcal{M}.$

Further, we need to recall the important notion of Hardy–Littlewood
majorization. Let $A,B\in(L_{1}+L_{\infty})(\mathcal{M}).$ The operator $B$ is
said to be majorized by $A$ and written $B\prec\prec A$ if and only if

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

We have (see [12])

$A+B\prec\prec\mu(A)+\mu(B)\prec\prec 2\sigma_{1/2}\mu(A+B)$

for every positive operators $A,B\in(L_{1}+L_{\infty})(\mathcal{M}).$

If $s>0,$ the dilation operator $\sigma_{s}$ is defined by setting

$(\sigma_{s}(x))(t)=x(\frac{t}{s}),\quad t>0$

in the case of the semi-axis. In the case of the interval $(0,1),$ the
operator $\sigma_{s}$ is defined by

$(\sigma_{s}x)(t)=\begin{cases}x(t/s),&t\leq\min\\{1,s\\}\\\ 0,&s<t\leq
1.\end{cases}$

Similarly, in the sequence case, we define an operator $\sigma_{n}$ by setting

$\sigma_{n}(a_{1},a_{2},\cdots)=(\underbrace{a_{1},\cdots,a_{1}}_{\mbox{$n$
times}},\underbrace{a_{2},\cdots,a_{2}}_{\mbox{$n$ times}},\cdots)$

and an operator $\sigma_{1/2}$ by setting

$\sigma_{1/2}:(a_{1},a_{2},a_{3},a_{4},\cdots)\to(\frac{a_{1}+a_{2}}{2},\frac{a_{3}+a_{4}}{2},\cdots).$

######  Definition 7.

The Banach space $E(\mathcal{M},\tau)\subset(L_{1}+L_{\infty})(\mathcal{M})$
is said to be a symmetric operator space if the following conditions hold.

  1. (1)

Given $A\in E(\mathcal{M},\tau)$ and $B\in(L_{1}+L_{\infty})(\mathcal{M})$
with $\mu(B)=\mu(A),$ we have $B\in E(\mathcal{M},\tau)$ and
$\|B\|_{E}=\|A\|_{E}.$

  2. (2)

Given $0\leq A\in E(\mathcal{M},\tau)$ and $0\leq
B\in(L_{1}+L_{\infty})(\mathcal{M})$ with $B\leq A,$ we have $B\in
E(\mathcal{M},\tau)$ and $\|B\|_{E}\leq\|A\|_{E}.$

The space $E(\mathcal{M},\tau)$ is called fully symmetric if for every $A\in
E(\mathcal{M},\tau)$ and every $B\in(L_{1}+L_{\infty})(\mathcal{M})$ with
$B\prec\prec A,$ we have $B\in E(\mathcal{M},\tau)$ and
$\|B\|_{E}\leq\|A\|_{E}.$

The norm on a symmetric space $E(\mathcal{M},\tau)$ is a Fatou norm if the
unit ball of $E(\mathcal{M},\tau)$ is closed with respect to strong (or,
equivalently, weak) operator convergence. Every symmetric space equipped with
a Fatou norm is necessarily fully symmetric.

A linear functional $\varphi:E(\mathcal{M},\tau)\to\mathbb{C}$ is said to be
symmetric if $\varphi(B)=\varphi(A)$ for every positive $A,B\in
E(\mathcal{M},\tau)$ such that $\mu(B)=\mu(A).$ A linear functional
$\varphi:E(\mathcal{M},\tau)\to\mathbb{C}$ is said to be fully symmetric if
$\varphi(B)\leq\varphi(A)$ for every positive $A,B\in E(\mathcal{M},\tau)$
such that $B\prec\prec A.$ Every fully symmetric functional is symmetric and
bounded. The converse fails [20].

A functional $\varphi:E(\mathcal{M},\tau)\to\mathbb{C}$ is called singular if
$\varphi=0$ on $(L_{1}\cap L_{\infty})(\mathcal{M}).$ If
$E(\mathcal{M},\tau)\not\subset L_{1}(\mathcal{M}),$ then every symmetric
functional is singular.

If $E=E(0,\infty)$ and if $\varphi:E\to\mathbb{R}$ is a symmetric functional,
then $s\varphi(x)=\varphi(\sigma_{s}x)$ for every $x\in E.$ If $E=E(0,1)$ and
if $\varphi:E\to\mathbb{R}$ is a singular symmetric functional, then
$s\varphi(x)=\varphi(\sigma_{s}x)$ for every $x=\mu(x)\in E.$

Let $E$ be a fully symmetric Banach space either on the interval $(0,1)$ or on
the semi-axis. We need the notion of an expectation operator (see [2]).

Let $\mathcal{A}=\\{A_{k}\\}$ be a (finite or infinite) sequence of disjoint
sets of finite measure and denote by $\mathfrak{A}$ the collection of all such
sequences. Denote by $A_{\infty}$ the complement of $\cup_{k}A_{k}.$

The expectation operator $\mathbf{E}(\cdot|\mathcal{A}):L_{1}+L_{\infty}\to
L_{1}+L_{\infty}$ is defined by setting

$\mathbf{E}(x|\mathcal{A})=\sum_{k}\frac{1}{m(A_{k})}(\int_{A_{k}}x(s)ds)\chi_{A_{k}}.$

Note that we do not require $A_{\infty}$ to have finite measure.

Every expectation operator is a contraction both in $L_{1}$ and $L_{\infty}.$
Therefore,

$\mathbf{E}(x|\mathcal{A})\prec\prec x,\quad x\in L_{1}+L_{\infty}.$

It follows that $\mathbf{E}(\cdot|\mathcal{A})$ is also contraction in $E.$

It will be convenient to introduce the following notation. If $\mathcal{A}$ is
a discrete subset of the semi-axis (i.e. a subset without limit points inside
$(0,\infty)$), then the elements of $\mathcal{A}\cup\\{0\\}$ partition the
semi-axis. This partition consists of a (finite or infinite) sequence of sets
of finite measure. We identify this partition with the set $\mathcal{A}.$
Elements of $\mathcal{A}$ will be called nodes of the partition $\mathcal{A}.$
The corresponding averaging operator will be denoted by
$\mathbf{E}(\cdot|\mathcal{A}).$

Let $E$ be a symmetric Banach space either on the interval $(0,1)$ or on the
semi-axis. Define the sets

$\mathcal{D}_{E}={\rm Lin}(\\{x\in E:\ x=\mu(x)\\})=\\{\mu(a)-\mu(b),\ a,b\in
E\\},$ $Z_{E}={\rm Lin}(\\{x_{1}-x_{2}:\ 0\leq x_{1},x_{2}\in E,\
\mu(x_{1})=\mu(x_{2})\\}).$

Let $C$ be a Hardy operator defined by setting

$(Cx)(t)=\frac{1}{t}\int_{0}^{t}x(s)ds.$

The following theorem was proved in [13]. For convenience of the reader, we
give a new and simple proof in the appendix.

######  Theorem 8.

Let $E$ be a symmetric space on the semi-axis and let $x\in\mathcal{D}_{E}.$
We have $x\in Z_{E}$ if and only if $Cx\in E.$ A similar assertion is also
valid for the interval $(0,1)$ provided that $\int_{0}^{1}x(s)ds=0.$

The following uniform submajorization was introduced by Kalton and Sukochev in
[19].

Let $x,y\in L_{1}(0,1)$ (or $x,y\in(L_{1}+L_{\infty})(0,\infty)$). We say that
$y\lhd x$ if there exists $m\in\mathbb{N}$ such that

(6) $\int_{ma}^{b}\mu(s,y)ds\leq\int_{a}^{b}\mu(s,x)ds,\quad\forall ma\leq b.$

Let $x,y\in l_{\infty}.$ We say that $y\lhd x$ if there exists
$m\in\mathbb{N}$ such that

(7) $\sum_{k=ma+1}^{b}\mu(k,y)\leq\sum_{k=a+1}^{b}\mu(k,x)\quad\forall
ma+1\leq b.$

The following important theorem was proved in [19] (see Theorem 5.4 and
Theorem 6.3 there).

######  Theorem 9.

Let $x,y\in L_{1}(0,1)$ or $x,y\in(L_{1}+L_{\infty})(0,\infty)$ or $x,y\in
l_{\infty}$ be such that $y\lhd x.$ For every $\varepsilon>0,$ the function
$(1-\varepsilon)y$ belongs to a convex hull of the set $\\{z:\
\mu(z)\leq\mu(x)\\}.$

This theorem led to the following fundamental result (see [19]).

######  Theorem 10.

Let $E=E(0,1)$ (or $E=E(0,\infty)$ or $E=E(\mathbb{N})$) be a symmetric Banach
space either on the interval $(0,1)$ or on the semi-axis or on $\mathbb{N}.$
It follows that the corresponding set $E(\mathcal{M},\tau)$ is a symmetric
Banach space.

Also, the uniform submajorization permits us to prove the convexity of the
functional $\pi:\mathcal{E}\to\mathbb{R}$ defined in Section 1.

######  Lemma 11.

The functional $\pi:\mathcal{E}\to\mathbb{R}$ is convex on every symmetrically
normed operator ideal $\mathcal{E}.$

###### Proof.

Let $E$ be the corresponding symmetrically normed ideal of $l_{\infty}.$ For
every $A,B\in\mathcal{E},$ it follows from Proposition 8.6 of [19] that
$\mu(A+B)\lhd\mu(A)+\mu(B).$ Hence,
$\sigma_{m}\mu(A+B)\lhd\sigma_{m}(\mu(A)+\mu(B)).$ By Theorem 9, we have

$\|\sigma_{m}\mu(A+B)\|_{E}\leq\|\sigma_{m}(\mu(A)+\mu(B))\|_{E}\leq\|\sigma_{m}\mu(A)\|_{E}+\|\sigma_{m}\mu(B)\|_{E}.$

Note that $\|A^{\oplus m}\|_{\mathcal{E}}=\|\sigma_{m}\mu(A)\|_{E}.$ Dividing
by $m$ and letting $m\to\infty,$ we obtain

$\pi(A+B)\leq\pi(A)+\pi(B).$

∎

##  3\. Lifting of symmetric functionals

In this section, we explain a canonical bijection between symmetric
functionals and traces. In what follows, we require that a semifinite von
Neumann algebra $\mathcal{M}$ be either atomless or atomic with traces of all
atoms being $1.$

For an atomless von Neumann algebra $\mathcal{M},$ we have (see e.g. [12])

$\int_{0}^{t}\mu(s,A)ds=\sup\\{\tau(p|A|):\ p\in P(\mathcal{M}),\
\tau(p)=t\\},\quad A\in\mathcal{M}.$

For a atomic von Neumann algebra $\mathcal{M},$ we have (see e.g. [12])

$\sum_{k=1}^{m}\mu(k,A)=\sup\\{\tau(p|A|):\ p\in P(\mathcal{M}),\
\tau(p)=m\\},\quad A\in\mathcal{M}.$

In either case, this implies a remarkable inequality (see e.g. [12])

(8) $\mu(A+B)\prec\prec\mu(A)+\mu(B)\prec\prec 2\sigma_{1/2}\mu(A+B),\quad
0\leq A,B\in(L_{1}+L_{\infty})(\mathcal{M}).$

######  Lemma 12.

Let $E=E(0,1)$ (or $E=E(0,\infty)$ or $E=E(\mathbb{N})$) be a symmetric Banach
space either on the interval $(0,1)$ or on the semi-axis or on $\mathbb{N}.$
If $x,y\in E_{+}$ are such that $y\lhd x,$ then $\varphi(y)\leq\varphi(x)$ for
every positive symmetric functional $\varphi$ on $E.$

###### Proof.

Fix $\varepsilon>0.$ By Theorem 9, there exist $z_{k}\in E,$ $1\leq k\leq n,$
and positive numbers $\lambda_{k},$ $1\leq k\leq n,$ such that
$\mu(z_{k})\leq\mu(x)$ for every $1\leq k\leq n$ and

$(1-\varepsilon)y=\sum_{k=1}^{n}\lambda_{k}z_{k},\quad\sum_{k=1}^{n}\lambda_{k}=1.$

Since $\varphi$ is positive and symmetric, it follows that

$\varphi(z_{k})\leq\varphi(|z_{k}|)=\varphi(\mu(z_{k}))\leq\varphi(\mu(x))=\varphi(x).$

Therefore, $(1-\varepsilon)\varphi(y)\leq\varphi(x).$ Since $\varepsilon>0$ is
arbitrarily small, the assertion follows. ∎

The following assertion is essentially known. However, we provide the full
proof for readers convenience.

######  Lemma 13.

Let $\mathcal{M}$ be a semifinite atomless von Neumann algebra and let
$A,B\in(L_{1}+L_{\infty})(\mathcal{M},\tau)$ be positive operators.

$\int_{2a}^{b}\mu(s,A+B)ds\leq\int_{a}^{b}(\mu(s,A)+\mu(s,B))ds,\quad\forall
2a\leq b,$
$\int_{2a}^{b}(\mu(s,A)+\mu(s,B))ds\leq\int_{2a}^{2b}\mu(s,A+B)ds,\quad\forall
2a\leq b.$

Similar assertion is valid for atomic von Neumann algebra $\mathcal{M}.$

###### Proof.

Applying inequality (8) to the operators $A,B,$ we obtain that

$\int_{0}^{b}\mu(s,A+B)ds\leq\int_{0}^{b}(\mu(s,A)+\mu(s,B))ds$

and

$\int_{0}^{2a}\mu(s,A+B)ds\geq\int_{0}^{a}(\mu(s,A)+\mu(s,B))ds.$

Subtracting this inequalities, we obtain

$\int_{2a}^{b}\mu(s,A+B)ds\leq\int_{a}^{b}(\mu(s,A)+\mu(s,B))ds.$

Proof of the second inequality is identical. ∎

The following theorem answers Question 4 in the affirmative, as also does [20,
Theorem 5.2]. The proof below is very simple and based on a completely
different approach.

######  Theorem 14.

Let $E=E(0,1)$ (or $E=E(0,\infty)$ or $E=E(\mathbb{N})$) be a symmetric Banach
space either on the interval $(0,1)$ or on the semi-axis or on $\mathbb{N}$
and let $E(\mathcal{M},\tau)$ be the corresponding symmetric Banach operator
space.

  1. (1)

If $\varphi$ is a positive symmetric functional on $E,$ then there exists a
positive symmetric functional $\mathcal{L}(\varphi)$ on $E(\mathcal{M},\tau)$
such that $\varphi(x)=\mathcal{L}(\varphi)(A)$ for all positive $x\in E$ and
$A\in E(\mathcal{M},\tau)$ such that $\mu(A)=\mu(x).$

  2. (2)

If $\varphi$ is a positive symmetric functional on $E(\mathcal{M},\tau),$ then
there exists a positive symmetric functional $\mathcal{L}^{-1}(\varphi)$ on
$E$ such that $\varphi(A)=\mathcal{L}^{-1}(\varphi)(x)$ for all positive $x\in
E$ and $A\in E(\mathcal{M},\tau)$ such that $\mu(A)=\mu(x).$

###### Proof.

We will only prove (1). Proof of (2) is identical.

Let $A,B\in E_{+}(\mathcal{M},\tau).$ It follows from Lemma 13 that

$\mu(A+B)\lhd\mu(A)+\mu(B)\lhd 2\sigma_{1/2}\mu(A+B).$

It follows from Lemma 12 that

$\varphi(\mu(A+B))\leq\varphi(\mu(A)+\mu(B))\leq\varphi(2\sigma_{1/2}\mu(A+B))=\varphi(\mu(A+B)).$

It follows that $\mathcal{L}(\varphi)$ is additive on
$E_{+}(\mathcal{M},\tau).$ We than extend it to $E(\mathcal{M},\tau)$ by
linearity. ∎

Theorem 14 provides a very natural bijection between the set of all symmetric
functionals on $E$ and that on $E(\mathcal{M},\tau),$ observed first for the
case of fully symmetric functionals in [8]. Next corollary follows
immediately.

######  Corollary 15.

Let $E$ and $E(\mathcal{M},\tau)$ be as in Theorem 14. The functional
$\varphi$ is fully symmetric on $E$ if and only if $\mathcal{L}(\varphi)$ is a
fully symmetric functional on $E(\mathcal{M},\tau).$

We also need a lifting between sequence and function spaces. The following
space was introduced in [21].

Let $\mathcal{A}=\\{[n-1,n]\\}_{n\in\mathbb{N}}$ be a partition of the semi-
axis. Clearly, $\mathbf{E}(\cdot|\mathcal{A})$ maps $L_{1}+L_{\infty}$ into
the set of step functions which can be identified with sequences.

######  Proposition 16.

Let $E$ be a symmetric Banach sequence space and let $F$ be the linear space
of all such functions $x\in L_{\infty}$ for which
$\mathbf{E}(\mu(x)|\mathcal{A})\in E.$ The space $F$ equipped with the norm

$\|x\|_{F}=\|x\|_{\infty}+\|\mathbf{E}(\mu(x)|\mathcal{A})\|_{E}$

is a symmetric Banach function space.

The fact that the space $F$ is a Banach space is non-trivial. Proof of this
fact was missing in both [19] and [21]. We include it in the appendix.

Below, we assume that $E$ is embedded into $F.$

######  Theorem 17.

Let $E=E(\mathbb{N})$ be a symmetric Banach sequence space and let $F$ be the
corresponding function space.

  1. (1)

If $\varphi$ is a positive symmetric functional on $E,$ then there exists a
positive symmetric functional $\mathcal{L}(\varphi)$ on $F$ such that
$\varphi(\mathbf{E}(\mu(x)|\mathcal{A}))=\mathcal{L}(\varphi)(x)$ for all
positive $x\in F.$

  2. (2)

If $\varphi$ is a positive symmetric functional on $F,$ then its restriction
on $E$ is a positive symmetric functional. This restriction is an inverse
operation for the $\mathcal{L}$ in (1).

###### Proof.

Let us prove (1)

$\varphi(\sigma_{1/2}a)=1/2\varphi(a_{1},a_{3},\cdots)+1/2\varphi(a_{2},a_{4},\cdots)=$
$=1/2\varphi(a_{1},0,a_{2},0,\cdots)+1/2\varphi(0,a_{2},0,a_{4},\cdots)=1/2\varphi(a)$

for every $a\in E.$

Let $x,y\in F$ be positive. It follows from Lemma 50 that

$\mathbf{E}(\mu(x+y)|\mathcal{A})\lhd\mathbf{E}(\mu(x)+\mu(y)|\mathcal{A})\lhd
2\sigma_{1/2}\mathbf{E}(\mu(x+y)|\mathcal{A}).$

It follows from Lemma 12 that

$\varphi(\mathbf{E}(\mu(x+y)|\mathcal{A}))=\varphi(\mathbf{E}(\mu(x)+\mu(y)|\mathcal{A}))$

and (1) follows.

The first assertion of (2) is trivial. Clearly,
$\mu(x)-\mathbf{E}(\mu(x)|\mathcal{A})\in(L_{1}\cap L_{\infty})(0,\infty).$ If
$E\neq l_{1},$ then $\varphi(y)=0$ for every $y\in(L_{1}\cap
L_{\infty})(0,\infty)$ and every symmetric functional $\varphi$ on $F.$ If
$E=l_{1},$ then $F=(L_{1}\cap L_{\infty})(0,\infty)$ and the only symmetric
functional on both spaces is an integral. The second assertion of (2) follows.
∎

##  4\. Existence of symmetric functionals

In this section, we present results concerning existence of symmetric
functionals on symmetric function spaces. The main results of this section are
Theorem 23, Theorem 28 and Theorem 29.

We need the following variation of the Hahn-Banach theorem.

######  Lemma 18.

Let $E$ be a partially ordered linear space and let $p:E\to\mathbb{R}$ be
convex and monotone functional. For every $x_{0}\in E,$ there exists a
positive linear functional $\varphi:E\to\mathbb{R}$ such that $\varphi\leq p$
and $\varphi(x_{0})=p(x_{0}).$

###### Proof.

The existence of $\varphi$ follows from the Hahn-Banach theorem. We only have
to prove that $\varphi\geq 0.$ If $z\geq 0,$ then $\varphi(x_{0}-z)\leq
p(x_{0}-z).$ Therefore,

$\varphi(z)\geq\varphi(x_{0})-p(x_{0}-z)=p(x_{0})-p(x_{0}-z)\geq 0$

due to the fact that $z\geq 0$ and $p$ is monotone. ∎

Define operators
$M_{m}:(L_{1}+L_{\infty})(0,\infty)\to(L_{1}+L_{\infty})(0,\infty)$ (or,
$M_{m}:L_{1}(0,1)\to L_{1}(0,1)$) by setting

$(M_{m}x)(t)=\frac{1}{t\log(m)}\int_{t/m}^{t}x(s),\quad m\geq 2.$

######  Lemma 19.

If $0\leq x\in L_{1}+L_{\infty}$ (or, $0\leq x\in L_{1}(0,1)$), then

$\int_{a}^{b/m}x(s)ds\leq\int_{a}^{b}(M_{m}x)(s)ds\leq\int_{a/m}^{b}x(s)ds$

provided that $ma\leq b.$ In particular, $m^{-1}\sigma_{m}x\lhd M_{m}x\lhd x$
provided that $x=\mu(x).$

###### Proof.

Clearly,

$\int_{a}^{b}(M_{m}x)(s)ds=\frac{1}{\log(m)}\int_{a}^{b}\int_{t/m}^{t}x(s)ds\frac{dt}{t}=$
$=\frac{1}{\log(m)}\int_{a/m}^{b}\int_{\max\\{a,s\\}}^{\min\\{ms,b\\}}\frac{dt}{t}x(s)ds=\frac{1}{\log(m)}\int_{a/m}^{b}x(s)\log(\frac{\min\\{ms,b\\}}{\max\\{a,s\\}})ds.$

The integrand does not exceed $x(s)\log(m)$ and the second inequality follows
immediately. The integrand is positive and is equal to $x(s)\log(m)$ for
$s\in(a,b/m).$ The first inequality follows. ∎

######  Corollary 20.

If $E$ is a symmetric Banach function space either on the interval $(0,1)$ or
on the semi-axis, then $M_{m}:E\to E$ is a contraction for $m\in\mathbb{N}.$

###### Proof.

Let $x=\mu(x)\in E.$ It follows from Lemma 19 that $M_{m}x\lhd x.$ It follows
from theorem 9 that, for every $\varepsilon>0,$ the function
$(1-\varepsilon)M_{m}x$ belongs to a convex hull of the set $\\{z:\
\mu(z)\leq\mu(x)\\}.$ Therefore, $M_{m}x\in E$ and
$(1-\varepsilon)\|M_{m}x\|_{E}\leq\|x\|_{E}.$ Since $\varepsilon$ is
arbitrarily small, the assertion follows. ∎

######  Lemma 21.

Let $E$ be a symmetric Banach space either on the interval $(0,1)$ or on the
semi-axis. Let $p:\mathcal{D}_{E}\to\mathbb{R}$ be convex and monotone
functional. If $p=0$ on $Z_{E}\cap\mathcal{D}_{E},$ then $p$ extends to a
convex monotone functional $p:E\to\mathbb{R}$ by setting

$p(x)=p(\mu(x_{+})-\mu(x_{-})).$

Also, $p(x)=0$ for every $x\in Z_{E}.$

###### Proof.

If $x\in\mathcal{D}_{E},$ then $x-\mu(x_{+})+\mu(x_{-})\in
Z_{E}\cap\mathcal{D}_{E}.$ Therefore, $p(x-\mu(x_{+})+\mu(x_{-}))=0$ and, due
to the convexity of $p,$ $p(x)=p(\mu(x_{+})-\mu(x_{-})).$ This proves the
correctness of the definition.

For $x,y\in E,$ we have

$\mu((x+y)_{+})-\mu((x+y)_{-})-\mu(x_{+})+\mu(x_{-})-\mu(y_{+})+\mu(y_{-})\in
Z_{E}\cap\mathcal{D}_{E}.$

It follows that

$p(\mu((x+y)_{+})-\mu((x+y)_{-})-\mu(x_{+})+\mu(x_{-})-\mu(y_{+})+\mu(y_{-}))=0$

and

$p(x+y)=p(\mu((x+y)_{+})-\mu((x+y)_{-}))=$
$=p(\mu(x_{+})-\mu(x_{-})+\mu(y_{+})-\mu(y_{-}))\leq p(x)+p(y).$

Since $p$ is monotone on $\mathcal{D}_{E},$ then $p(y)\leq 0$ for every $0\geq
y\in\mathcal{D}_{E}.$ It follows that $p(y)=p(-\mu(y))\leq 0$ for $0\geq y\in
E.$ Therefore, $p(x+y)\leq p(x)+p(y)\leq p(x)$ for every $0\geq y\in E.$ ∎

######  Lemma 22.

Let $E$ be a symmetric Banach space either on the interval $(0,1)$ or on the
semi-axis. The functional

$p:x\to\limsup_{m\to\infty}\|(M_{m}x)_{+}\|_{E},\quad x\in\mathcal{D}_{E}$

satisfies the assumptions of Lemma 21. Also, for every $x\in\mathcal{D}_{E},$
we have $p(x)\leq\|x\|_{E}.$

###### Proof.

It follows from Corollary 20 that

$\|(M_{m}x)_{+}\|_{E}\leq\|M_{m}x\|_{E}\leq\|x\|_{E},\quad x\in E.$

It follows that

$p(x)=\limsup_{m\to\infty}\|(M_{m}x)_{+}\|_{E}\leq\|x\|_{E},\quad
x\in\mathcal{D}_{E}.$

Clearly, the mappings $x\to(M_{m}x)_{+}$ are convex and monotone. So are the
mappings $x\to\|(M_{m}x)_{+}\|_{E}.$ Therefore,
$p:\mathcal{D}_{E}\to\mathbb{R}$ is a convex and monotone functional.

If $x\in Z_{E}\cap\mathcal{D}_{E},$ then by Theorem 8 $|Cx|\in E.$ Therefore,

$(M_{m}x)(t)\leq\frac{1}{\log(m)}(|\frac{1}{t}\int_{0}^{t/m}x(s)ds|+|\frac{1}{t}\int_{0}^{t}x(s)ds|)\leq$
$\leq\frac{1}{\log(m)}(\frac{1}{m}\sigma_{m}|Cx|+|Cx|)(t).$

Since $\|\sigma_{m}\|_{E\to E}\leq m$ (see [22, Theorem II.4.5]), it follows
that

$\|(M_{m}x)_{+}\|_{E}\leq\frac{2}{\log(m)}\|Cx\|_{E}$

and $p(x)=0.$ ∎

######  Theorem 23.

Let $E=E(0,\infty)$ be a symmetric Banach space on the semi-axis. For a given
$0\leq x\in E,$ there exists a symmetric linear functional
$\varphi:E\to\mathbb{R}$ such that

$\varphi(x)=\lim_{m\to\infty}\frac{1}{m}\|\sigma_{m}(\mu(x))\|_{E}.$

###### Proof.

Without loss of generality, $x=\mu(x).$ Let $p$ be the convex monotone
functional constructed in Lemma 22. It follows from Lemma 18 that there exist
a positive linear functional $\varphi$ on $E$ such that $\varphi\leq p$ and
$\varphi(x)=p(x).$ Since $p(z)=0$ for every $z\in Z_{E},$ it follows that
$\varphi(z)=0$ for every $z\in Z_{E}.$ Therefore, $\varphi$ is a symmetric
functional.

Since $\varphi(z)\leq p(z)\leq\|z\|_{E}$ for every $z=\mu(z)\in E,$ it follows
that $\|\varphi\|_{E^{*}}\leq 1.$ Therefore,

$\varphi(x)=\varphi(\frac{1}{m}\sigma_{m}x)\leq\frac{1}{m}\|\sigma_{m}x\|_{E}.$

Passing $m\to\infty,$ we obtain

$\varphi(x)\leq\lim_{m\to\infty}\frac{1}{m}\|\sigma_{m}\mu(x)\|_{E}.$

On the other hand, It follows from Lemma 19 that $m^{-1}\sigma_{m}x\lhd
M_{m}x.$ Therefore,

$p(x)=\limsup_{m\to\infty}\|M_{m}x\|_{E}\geq\lim_{m\to\infty}\frac{1}{m}\|\sigma_{m}\mu(x)\|_{E}.$

The assertion follows immediately. ∎

Consider the functional $\pi:E\to E$ (identical to the one defined in Section
1).

(9) $\pi(x)=\lim_{m\to\infty}\frac{1}{m}\|x^{\oplus m}\|_{E},\quad x\in E.$

Note that $\pi(-x)=\pi(x)$ for every $x\in E.$ If $p$ is a functionals defined
in Lemma 22, then $p(-x)=0$ for positive $x\in E.$ Therefore, $p\neq\pi.$
However, the assertion below follows from Theorem 23.

######  Lemma 24.

Let $E=E(0,\infty)$ be a symmetric Banach space on the semi-axis. Let $p$ and
$\pi$ be the convex functionals on $E$ defined in Lemma 22 and (9),
respectively. For every positive $x\in E,$ we have $p(x)=\pi(x).$

###### Proof.

For every $x\in E,$ consider the functional $\varphi$ constructed in Theorem
23. By construction, we have $\varphi(x)=p(x)=\pi(x).$ ∎

If $E\not\subset L_{1}(0,\infty),$ then the functional $\varphi$ constructed
in Theorem 23 is necessarily singular. The case $E\subset L_{1}$ requires more
detailed treatment.

######  Lemma 25.

Let $E$ be a symmetric (respectively, fully symmetric) Banach function space
either on the interval $(0,1)$ or on the semi-axis. Let
$\\{\varphi_{i}\\}_{i\in\mathbb{I}}\in E^{*}$ be a net and let $\varphi\in
E^{*}$ be such that $\varphi_{i}\to\varphi$ $*-$weakly.

  1. (1)

If every $\varphi_{i}$ is symmetric, then $\varphi$ is symmetric.

  2. (2)

If every $\varphi_{i}$ is fully symmetric, then $\varphi$ is fully symmetric.

###### Proof.

Let each $\varphi_{i}$ be symmetric. If $0\leq x_{1},x_{2}\in E$ are such that
$\mu(x_{1})=\mu(x_{2}),$ then

$\varphi(x_{1})=\lim_{i\in\mathbb{I}}\varphi_{i}(x_{1})=\lim_{i\in\mathbb{I}}\varphi_{i}(x_{2})=\varphi(x_{2}).$

Hence, $\varphi$ is symmetric.

Let each $\varphi_{i}$ be fully symmetric. Thus, $\varphi_{i}(x)\leq 0$ for
every $x\in\mathcal{D}_{E}$ such that $Cx\leq 0.$ Therefore,
$\varphi(x)=\lim_{i\in\mathbb{I}}\varphi_{i}(x)\leq 0$ for every
$x\in\mathcal{D}_{E}$ such that $Cx\leq 0.$

Let $x_{1},x_{2}\in E$ be positive elements such that $x_{1}\prec\prec x_{2}.$
Therefore, $z=\mu(x_{1})-\mu(x_{2})\in\mathcal{D}_{E}$ and $Cz\leq 0.$ It
follows from above that $\varphi(z)\leq 0.$ Hence, $\varphi$ is a fully
symmetric functional. ∎

######  Lemma 26.

Let $E$ be a symmetric (respectively, fully symmetric) Banach function space
either on the interval $(0,1)$ or on the semi-axis and let $\varphi$ be a
symmetric (respectively, fully symmetric) functional on $E.$ The formula

$\varphi_{sing}(x)=\lim_{n\to\infty}\varphi(\mu(x)\chi_{(0,1/n)}),\quad 0\leq
x\in E.$

defines a singular symmetric (respectivley, fully symmetric) linear functional
on $E.$

###### Proof.

If $x,y\in E$ are positive functions, then

$\mu(x+y)\chi_{(0,1/n)}\lhd(\mu(x)+\mu(y))\chi_{(0,1/n)}\lhd
2\sigma_{1/2}\mu(x+y)\chi_{(0,1/n)}.$

Taking the limit as $n\to\infty,$ we derive from Lemma 12 that

$\varphi_{sing}(\mu(x+y))=\varphi_{sing}(\mu(x)+\mu(y)).$

Since $\varphi$ is symmetric, it follows that

$\varphi_{sing}(x+y)=\varphi_{sing}(\mu(x+y))=\varphi_{sing}(\mu(x)+\mu(y))=\varphi_{sing}(x)+\varphi_{sing}(y).$

Hence, $\varphi_{sing}$ is an additive functional on $E_{+}.$ Therefore, it
extends to a linear functional on $E.$ Clearly, $\varphi_{sing}$ is symmetric.
Second assertion is trivial. ∎

In fact, the construction in Lemma 26 gives a singular part of the functional
$\varphi$ as defined by Yosida-Hewitt theorem.

######  Lemma 27.

Let $E=E(0,\infty)\subset L_{1}(0,\infty)$ be a symmetric Banach function
space on the semi-axis and let $\varphi$ be a symmetric functional on $E.$ If
$\varphi_{sing}$ is a functional constructed in Lemma 26, then
$\varphi-\varphi_{sing}$ is a normal functional (that is, an integral).

###### Proof.

It is clear that

$0\leq\varphi_{sing}(z)\leq\|z\|_{\infty}\lim_{n\to\infty}\varphi(\chi_{(0,1/n)})=0$

for every positive $z\in(L_{1}\cap L_{\infty})(0,\infty).$ It follows that
$\varphi_{sing}(\mu(x)\chi_{(1/n,\infty)})=0$ for every $x\in E.$ Therefore,

(10)
$(\varphi-\varphi_{sing})(x)=\lim_{n\to\infty}\varphi(\mu(x)\chi_{(1/n,\infty)})=\lim_{n\to\infty}(\varphi-\varphi_{sing})(\mu(x)\chi_{(1/n,\infty)}).$

On the other hand, for every positive $z\in(L_{1}\cap L_{\infty})(0,\infty)$
with $\|z\|_{\infty}=1,$ we have $z\prec\chi_{(0,\|z\|_{1})}.$ It is proved in
[30, Theorem 23] that $z$ belongs to the closure (in the topology of
$L_{1}\cap L_{\infty}$) of the set $\\{u\geq 0:\
\mu(u)=\chi_{(0,\|z\|_{1})}\\}.$ Thus,

$(\varphi-\varphi_{sing})(z)=(\varphi-\varphi_{sing})(\chi_{(0,\|z\|_{1})})=\|z\|_{1}(\varphi-\varphi_{sing})(\chi_{(0,1)}).$

By linearity,

(11)
$(\varphi-\varphi_{sing})(z)=(\varphi-\varphi_{sing})(\chi_{(0,1)})\cdot\int_{0}^{\infty}z(s)ds,\quad\forall
z\in(L_{1}\cap L_{\infty})(0,\infty).$

It follows that from (10) and (11) that

$(\varphi-\varphi_{sing})(x)=\lim_{n\to\infty}\int_{1/n}^{\infty}\mu(s,x)ds\cdot(\varphi-\varphi_{sing})(0,1)=\int_{0}^{1}x(s)ds\cdot(\varphi-\varphi_{sing})(\chi_{(0,1)})$

for every positive function $x\in E.$ The assertion follows immediately. ∎

######  Theorem 28.

Let $E\subset L_{1}(0,\infty)$ be a symmetric Banach space on the semi-axis.
For a given $0\leq x\in E,$ there exists a singular symmetric linear
functional $\varphi_{sing}$ such that

$\varphi_{sing}(x)=\lim_{m\to\infty}\frac{1}{m}\|\sigma_{m}(\mu(x))\chi_{(0,1)}\|_{E}.$

###### Proof.

Apply Theorem 23 to the function $\mu(x)\chi_{(0,1/n)}.$ It follows that there
exists a symmetric linear functional $\varphi_{n}$ such that
$\|\varphi_{n}\|_{E^{*}}\leq 1$ and

$\varphi_{n}(\mu(x)\chi_{(0,1/n)})=\lim_{m\to\infty}\frac{1}{m}\|\sigma_{m}(\mu(x)\chi_{(0,1/n)})\|_{E}\geq\lim_{m\to\infty}\frac{1}{m}\|\sigma_{m}(\mu(x))\chi_{(0,1)}\|_{E}.$

Since the unit ball in $E^{*}$ is $*-$weakly compact (Banach-Alaoglu theorem),
there exists a convergent subnet $\psi_{i}=\varphi_{F(i)},$ $i\in\mathbb{I},$
of the sequence $\varphi_{n},$ $n\in\mathbb{N}.$ Let $\psi_{i}\to\varphi.$ It
follows from Lemma 25 that $\varphi$ is a symmetric functional.

By the definition of a subnet (see [26, Section IV.2]), for every fixed
$n\in\mathbb{N},$ there exists $i_{n}\in\mathbb{I}$ such that $F(i)>n$ for
every $i>i_{n}.$ Thus, for every $i>i_{n},$ we have

$\psi_{i}(\mu(x)\chi_{(0,1/n)})\geq\varphi_{F(i)}(\mu(x)\chi_{(0,1/F(i))})\geq\lim_{m\to\infty}\frac{1}{m}\|\sigma_{m}(\mu(x))\chi_{(0,1)}\|_{E}.$

The subnet $\psi_{i},$ $i_{n}<i\in\mathbb{I}$ converges to the same limit
$\varphi.$ Therefore,

$\varphi(\mu(x)\chi_{(0,1/n)})\geq\lim_{m\to\infty}\frac{1}{m}\|\sigma_{m}(\mu(x))\chi_{(0,1)}\|_{E}.$

Now, taking the limit as $n\to\infty,$ we obtain the inequality

$\varphi_{sing}(x)\geq\lim_{m\to\infty}\frac{1}{m}\|\sigma_{m}(\mu(x))\chi_{(0,1)}\|_{E},$

where $\varphi_{sing}$ is a singular symmetric functional defined in Lemma 26.
The opposite inequality is trivial. ∎

######  Theorem 29.

Let $E$ be a symmetric Banach space on the interval $(0,1).$ For a given
$0\leq x\in E,$ there exists a singular symmetric linear functional
$\varphi_{sing}$ such that

$\varphi_{sing}(x)=\lim_{m\to\infty}\frac{1}{m}\|\sigma_{m}(\mu(x))\|_{E}.$

###### Proof.

Let $F$ be a symmetric Banach space on the semi-axis with a norm given by the
formula

$\|x\|_{F}=\|\mu(x)\chi_{(0,1)}\|_{E}+\|x\|_{1},\quad\forall x\in F.$

Clearly, $F\subset L_{1}(0,\infty).$ Applying Theorem 28, we obtain a
symmetric singular functional $\varphi$ on $F$ such that

$\varphi(x)=\lim_{m\to\infty}\frac{1}{m}\|\sigma_{m}(\mu(x))\chi_{(0,1)}\|_{F}=\lim_{m\to\infty}\frac{1}{m}\|\sigma_{m}(\mu(x))\|_{E}.$

∎

##  5\. Existence of fully symmetric functionals

In this section, we present results concerning existence of fully symmetric
functionals on fully symmetric function spaces. The main results of this
section are Theorem 33, Theorem 35 and Theorem 36.

######  Lemma 30.

Let $E$ be a symmetric Banach function space either on the interval $(0,1)$ or
on the semi-axis. If $x,z\in\mathcal{D}_{E}$ are such that $Cx\leq Cz,$ then
$CM_{m}x\leq CM_{m}z.$

###### Proof.

Let $x=\mu(a)-\mu(b)$ and $z=\mu(c)-\mu(d)$ with $a,b,c,d\in E.$ It follows
from assumption $Cx\leq Cz$ that $C(\mu(a)+\mu(d))\leq C(\mu(b)+\mu(c))$ or,
equivalently, $\mu(a)+\mu(d)\prec\prec\mu(b)+\mu(c).$

Arguing as in Lemma 19, we have

$\int_{0}^{t}(M_{m}z)(s)ds=\int_{0}^{t}z(s)h(s,t)ds$

with

$h(s,t)=\left\\{\begin{aligned} 1,\quad 0\leq s\leq t/m\\\
\frac{\log(t/s)}{\log(m)},\quad t/m\leq s\leq t\end{aligned}\right.$

It is now clear that

$\int_{0}^{t}M_{m}(\mu(a)+\mu(d))(s)ds=\int_{0}^{t}(\mu(s,a)+\mu(s,d))h(s,t)ds,$
$\int_{0}^{t}M_{m}(\mu(b)+\mu(c))(s)ds=\int_{0}^{t}(\mu(s,b)+\mu(s,c))h(s,t)ds.$

Clearly, $h$ is positive and decreasing with respect to $s.$ It follows from
[22, Equality 2.36] that

$M_{m}(\mu(a)+\mu(d))\prec\prec M_{m}(\mu(b)+\mu(c))$

and the assertion follows. ∎

######  Lemma 31.

Let $E$ be a fully symmetric Banach function space either on the interval
$(0,1)$ or on the semi-axis and let $x=\mu(x)\in E.$ If $z\in\mathcal{D}_{E}$
is such that $Cx\leq Cz,$ then $p(x)\leq p(z).$

###### Proof.

Since $M_{m}x$ is decreasing, it follows from Lemma 30 that

$\int_{0}^{t}\mu(s,M_{m}x)ds=\int_{0}^{t}(M_{m}x)(s)ds\leq\int_{0}^{t}(M_{m}z)_{+}(s)ds\leq\int_{0}^{t}\mu(s,(M_{m}z)_{+})ds.$

Therefore, $(M_{m}x)_{+}=M_{m}x\prec\prec(M_{m}z)_{+}.$ The assertion follows
now from the definition of the functional $p.$ ∎

######  Lemma 32.

Let $E$ be a fully symmetric Banach function space either on the interval
$(0,1)$ or on the semi-axis. Let $p$ be the functional constructed in Lemma
22. The functional

$q(x)=\inf\\{p(z):\ z\in\mathcal{D}_{E},\ Cx\leq Cz\\},\quad
x\in\mathcal{D}_{E}$

satisfies the assumptions of Lemma 21.

###### Proof.

It is clear from the definition of $q$ that $q\leq p$ and that $q$ is a
positive functional.

We claim that $q$ is convex on $\mathcal{D}_{E}.$ Let
$x_{1},x_{2}\in\mathcal{D}_{E}.$ Fix $\varepsilon>0$ and select
$z_{1},z_{2}\in\mathcal{D}_{E}$ such that $Cx_{i}\leq Cz_{i}$ and
$p(z_{i})\leq q(x_{i})+\varepsilon$ for $i=1,2.$ Thus, $C(x_{1}+x_{2})\leq
C(z_{1}+z_{2})$ and

$q(x_{1}+x_{2})\leq p(z_{1}+z_{2})\leq p(z_{1})+p(z_{2})\leq
q(x_{1})+q(x_{2})+2\varepsilon.$

Since $\varepsilon$ is arbitrarily small, the claim follows.

We claim that $q$ is monotone on $\mathcal{D}_{E}.$ Let
$x_{1},x_{2}\in\mathcal{D}_{E}$ be such that $x_{1}\leq x_{2}.$ Fix
$\varepsilon>0$ and select $z\in\mathcal{D}_{E}$ such that $Cx_{2}\leq Cz$ and
$p(z)\leq q(x_{2})+\varepsilon.$ Thus, $Cx_{1}\leq Cx_{2}\leq Cz$ and
$q(x_{1})\leq p(z)\leq q(x_{2})+\varepsilon.$ Since $\varepsilon$ is
arbitrarily small, the claim follows.

For $x\in Z_{E}\cap\mathcal{D}_{E},$ we have $0\leq q(x)\leq p(x)=0$ and,
therefore, $q(x)=0.$ s ∎

The following theorem is the first main result of this section.

######  Theorem 33.

Let $E=E(0,\infty)$ be a fully symmetric Banach space on the semi-axis. For a
given $0\leq x\in E,$ there exists a fully symmetric linear functional
$\varphi:E\to\mathbb{R}$ such that

$\varphi(x)=\lim_{m\to\infty}\frac{1}{m}\|\sigma_{m}(\mu(x))\|_{E}.$

###### Proof.

Without loss of generality, $x=\mu(x).$ Let $q$ be the convex monotone
functional constructed in Lemma 32. It follows from Lemma 18 that there exist
a positive linear functional $\varphi$ on $E$ such that $\varphi\leq q$ and
$\varphi(x)=q(x).$

It is clear that $\varphi\leq q\leq p.$ Since $p(z)=0$ for every $z\in Z_{E},$
it follows that $\varphi(z)=0$ for every $z\in Z_{E}.$ Therefore, $\varphi$ is
a symmetric functional. For every $z\in\mathcal{D}_{E}$ with $Cz\leq 0,$ we
have $\varphi(z)\leq q(z)\leq p(0)=0.$

Let $x_{1},x_{2}\in E$ be positive elements such that $x_{1}\prec\prec x_{2}.$
Therefore, $z=\mu(x_{1})-\mu(x_{2})\in\mathcal{D}_{E}$ and $Cz\leq 0.$ It
follows from above that $\varphi(z)\leq 0.$ Hence, $\varphi$ is a fully
symmetric functional.

Since $\varphi(z)\leq q(z)\leq p(z)\leq\|z\|_{E}$ for every $z=\mu(z)\in E,$
it follows that $\|\varphi\|_{E^{*}}\leq 1.$ Therefore,

$\varphi(x)=\varphi(\frac{1}{m}\sigma_{m}x)\leq\frac{1}{m}\|\sigma_{m}x\|_{E}.$

Passing $m\to\infty,$ we obtain

$\varphi(x)\leq\lim_{m\to\infty}\frac{1}{m}\|\sigma_{m}\mu(x)\|_{E}.$

On the other hand, $q(x)=p(x)$ by Lemma 31. By Lemma 19, we have
$m^{-1}\sigma_{m}x\lhd M_{m}x.$ Therefore,

$\varphi(x)=q(x)=p(x)=\limsup_{m\to\infty}\|M_{m}x\|_{E}\geq\lim_{m\to\infty}\frac{1}{m}\|\sigma_{m}\mu(x)\|_{E}.$

The assertion follows immediately. ∎

If $\pi:E\to E$ is a convex functional defined in (9), then $\pi(-x)=\pi(x)$
for every $x\in E.$ If $q$ is a functional defined in Lemma 32, then $q(-x)=0$
for positive $x\in E.$ Therefore, $q\neq\pi.$ However, the assertion below
follows from Theorem 33.

######  Lemma 34.

Let $E=E(0,\infty)$ be a fully symmetric Banach space on the semi-axis. Let
$q$ and $\pi$ be the convex functionals on $E$ defined in Lemma 32 and (9),
respectively. For every positive $x\in E,$ we have $q(x)=\pi(x).$

###### Proof.

For every $x\in E,$ consider the functional $\varphi$ constructed in Theorem
33. By construction, we have $\varphi(x)=q(x)=p(x)=\pi(x).$ ∎

The proofs of the two following theorems are very similar to that of Theorem
28 (respectively, Theorem 29) and are, therefore, omitted. The only difference
is that the reference to Theorem 23 (respectively, Theorem 28) has to be
replaced with the reference to Theorem 33 (respectively, Theorem 35).

######  Theorem 35.

Let $E\subset L_{1}(0,\infty)$ be a fully symmetric Banach space on the semi-
axis. For a given $0\leq x\in E,$ there exists a singular fully symmetric
linear functional $\varphi_{sing}$ such that

$\varphi_{sing}(x)=\lim_{m\to\infty}\frac{1}{m}\|\sigma_{m}(\mu(x))\chi_{(0,1)}\|_{E}.$

######  Theorem 36.

Let $E\subset L_{1}(0,1)$ be a fully symmetric Banach space on the interval
$(0,1).$ For a given $0\leq x\in E,$ there exists a singular fully symmetric
linear functional $\varphi_{sing}$ such that

$\varphi_{sing}(x)=\lim_{m\to\infty}\frac{1}{m}\|\sigma_{m}(\mu(x))\|_{E}.$

##  6\. The sets of symmetric and fully symmetric functionals are different

In this section, we demonstrate that the sets of symmetric and fully symmetric
functionals on a given fully symmetric space $E$ are distinct (provided that
one of these sets is non-empty). The main results are Theorem 47 and Theorem
48.

Let $x=\mu(x)\in(L_{1}+L_{\infty})(0,\infty)$ (or $x=\mu(x)\in L_{1}(0,1)$)
and let $X(t)=\int_{0}^{t}x(s)ds.$ For every $\theta>0,$ let $a_{n}(\theta)$
be such that

$X(a_{n}(\theta))=(3/2)^{n}\theta$

for every $n\in\mathbb{Z}$ such that $a_{n}(\theta)$ does exist. Given a
sequence
$\kappa=\\{\kappa_{n}\\}_{n\in\mathbb{Z}}\in(\mathbb{N}\cup\\{\infty\\})^{\mathbb{Z}},$
let

$\mathcal{B}_{\kappa,\theta}=\\{\kappa_{n}a_{3n}(\theta),\mbox{ where
}n\in\mathbb{Z}\mbox{ is such that
}\kappa_{n}^{2}a_{3n}(\theta)<a_{3n+1}(\theta)\\}.$

If $\kappa_{n}=m$ for all $n\in\mathbb{N},$ we write $\mathcal{B}_{m,\theta}$
instead of $\mathcal{B}_{\kappa,\theta}.$ Also, set

$\mathcal{A}_{m}=\\{ma_{n}(1):\quad m^{2}a_{n}(1)<a_{n+1}(1),\
n\in\mathbb{Z}\\}.$

######  Lemma 37.

If $x=\mu(x)\in L_{1}+L_{\infty}$ and if $\mathcal{C}_{i},$ $1\leq i\leq k,$
are discrete sets, then

$\mathbf{E}(x|\cup_{i=1}^{k}\mathcal{C}_{i})\prec\prec\sum_{i=1}^{k}\mathbf{E}(x|\mathcal{C}_{i}).$

###### Proof.

It is sufficient to verify

$\int_{0}^{t}\mathbf{E}(x|\cup_{i=1}^{k}\mathcal{C}_{i})(s)ds\leq\sum_{i=1}^{k}\int_{0}^{t}\mathbf{E}(x|\mathcal{C}_{i})(s)ds$

only at the nodes of $\mathbf{E}(x|\cup_{i=1}^{k}\mathcal{C}_{i}),$ that is at
the nodes of $\mathbf{E}(x|\mathcal{C}_{i})$ for every $i.$ However, if
$t\in\mathcal{C}_{i}$ for some $i,$ then

$\int_{0}^{t}\mathbf{E}(x|\cup_{i=1}^{k}\mathcal{C}_{i})(s)ds=X(t)=\int_{0}^{t}\mathbf{E}(x|\mathcal{C}_{i})(s)ds$

and we are done. ∎

We will need the following lemma.

######  Lemma 38.

If $x=\mu(x)\in L_{1}+L_{\infty}$ and if $\kappa\geq\kappa^{\prime}$ (that is
$\kappa_{n}\geq\kappa_{n}^{\prime}$ for every $n$), then

(12)
$\mathbf{E}(x|\mathcal{B}_{\kappa,\theta})\prec\prec\frac{3}{2}\mathbf{E}(x|\mathcal{B}_{\kappa^{\prime},\theta}).$

###### Proof.

Let $n\in\mathbb{Z}$ be such that
$\kappa_{n}^{2}a_{3n}(\theta)<a_{3n+1}(\theta).$ It follows that
$\kappa_{n}^{\prime 2}a_{3n}(\theta)<a_{3n+1}(\theta).$ Therefore,

$\int_{0}^{\kappa_{n}a_{3n}(\theta)}\mathbf{E}(x|\mathcal{B}_{\kappa,\theta})(s)ds\leq\int_{0}^{a_{3n+1}(\theta)}x(s)ds=3/2\int_{0}^{a_{3n}(\theta)}x(s)ds\leq$
$\leq
3/2\int_{0}^{\kappa_{n}^{\prime}a_{3n}(\theta)}x(s)ds=3/2\int_{0}^{\kappa_{n}^{\prime}a_{3n}(\theta)}\mathbf{E}(x|\mathcal{B}_{\kappa^{\prime},\theta})(s)ds.$

Hence, we have

(13) $\int_{0}^{t}\mathbf{E}(x|\mathcal{B}_{\kappa,\theta})(s)ds\leq
3/2\int_{0}^{t}\mathbf{E}(x|\mathcal{B}_{\kappa^{\prime},\theta})(s)ds$

for every $t$ being a node of the partition $\mathcal{B}_{\kappa,\theta}.$
Thus, (13) holds for every $t>0.$ ∎

######  Remark 39.

The inequality (12) holds if $\kappa_{n}\geq\kappa_{n}^{\prime}$ only for such
$n\in\mathbb{Z}$ that satisfy the inequality
$\kappa_{n}^{2}a_{3n}(\theta)<a_{3n+1}(\theta).$

######  Lemma 40.

Let $E$ be a fully symmetric Banach function space either on the interval
$(0,1)$ or on the semi-axis. Let $x=\mu(x)\in E$ and $y=\mu(y)\in E$ be such
that $\varphi(y)\leq\varphi(x)$ for every positive symmetric functional
$\varphi\in E^{*}.$ There exists $0\leq u_{m}\in E$ such that $u_{m}\to 0$ in
$E$ and

$\int_{ma}^{b}y(s)ds\leq\int_{a}^{mb}(x+u_{m})(s)ds,\quad\forall ma\leq b.$

###### Proof.

Let $p$ be a convex positive functional considered in Lemma 22. By Lemma 18,
there exists a positive functional $\varphi\in E^{*}$ such that $\varphi\leq
p$ and $\varphi(y-x)=p(y-x).$ We have $p(z)=0$ for every $z\in Z_{E}$ and,
therefore, $\varphi(z)=0$ for every $z\in Z_{E}.$ Therefore, $\varphi$ is a
positive symmetric linear functional on $E.$

By the assumption, $\varphi(y-x)\leq 0$ and, therefore, $p(y-x)=0.$ Hence, by
the definition of $p,$ we have $u_{m}=(M_{m}(y-x))_{+}\to 0$ in $E.$ Clearly,
$M_{m}y\leq M_{m}x+u_{m}.$ It follows from Lemma 19 that

$\int_{ma}^{b}y(s)ds\leq\int_{ma}^{mb}(M_{m}y)(s)ds\leq\int_{ma}^{mb}(M_{m}x+u_{m})(s)ds\leq\int_{a}^{mb}(x+u_{m})(s)ds.$

∎

For each sequence $\kappa$ and $\lambda>0,$ we define the sequence
$\kappa^{\lambda}$ by setting

$\kappa^{\lambda}_{n}=\begin{cases}\kappa_{n},\qquad\kappa_{n}\geq\lambda\\\
\infty,\qquad\kappa_{n}<\lambda.\end{cases}$

######  Lemma 41.

If $m\in\mathbb{N},$ $x=\mu(x)\in L_{1}+L_{\infty}$ and $0\leq u\in
L_{1}+L_{\infty}$ are such that

$\int_{ma}^{b}\mathbf{E}(x|\mathcal{B}_{\kappa,\theta})(s)ds\leq\int_{a}^{mb}(x+u)(s)ds,\quad\forall
ma\leq b\in\mathbb{R},$

then

(14)
$m^{-1}\sigma_{m}\mathbf{E}(x|\mathcal{B}_{\kappa^{100m},\theta})\prec\prec
30\mu(u).$

###### Proof.

If $\kappa^{100m}_{n}=\infty$ for every $n\in\mathbb{Z},$ then
$\mathbf{E}(x|\mathcal{B}_{\kappa^{100m},\theta})=0$ and the assertion is
trivial.

Let $n\in\mathbb{Z}$ be such that
$\kappa_{n}^{2}a_{3n}(\theta)<a_{3n+1}(\theta)$ and $\kappa_{n}\geq 100m.$ It
follows that

(15)
$\int_{0}^{m\kappa_{n}a_{3n}(\theta)}u(s)ds\geq\int_{a_{3n}(\theta)}^{m\kappa_{n}a_{3n}(\theta)}(x+u)(s)ds-\int_{a_{3n}(\theta)}^{m\kappa_{n}a_{3n}(\theta)}x(s)ds.$

By the assumption, we have

(16)
$\int_{a_{3n}(\theta)}^{m\kappa_{n}a_{3n}(\theta)}(x+u)(s)ds\geq\int_{ma_{3n}(\theta)}^{\kappa_{n}a_{3n}(\theta)}\mathbf{E}(x|\mathcal{B}_{\kappa,\theta})(s)ds.$

Note that $m\kappa_{n}a_{3n}(\theta)<a_{3n+1}(\theta).$ It follows from (15)
and (16) that

(17)
$\int_{0}^{m\kappa_{n}a_{3n}(\theta)}u(s)ds\geq\int_{ma_{3n}(\theta)}^{\kappa_{n}a_{3n}(\theta)}\mathbf{E}(x|\mathcal{B}_{\kappa,\theta})(s)ds-\int_{a_{3n}(\theta)}^{a_{3n+1}(\theta)}x(s)ds.$

Let $n^{\prime}$ be the maximal integer number such that $n^{\prime}<n$ and
$\kappa_{n^{\prime}}^{2}a_{3n^{\prime}}(\theta)<a_{3n^{\prime}+1}(\theta).$ It
is clear that

$\kappa_{n^{\prime}}^{2}a_{3n^{\prime}}(\theta)<a_{3n^{\prime}+1}(\theta)\leq
a_{3n-2}(\theta)<ma_{3n}(\theta)$

and

(18)
$\mathbf{E}(x|\mathcal{B}_{\kappa,\theta})=\frac{X(\kappa_{n}a_{3n}(\theta))-X(\kappa_{n^{\prime}}a_{3n^{\prime}}(\theta))}{\kappa_{n}a_{3n}(\theta)-\kappa_{n^{\prime}}a_{3n^{\prime}}(\theta)}\geq\frac{X(a_{3n}(\theta))-X(a_{3n-2}(\theta))}{\kappa_{n}a_{3n}(\theta)}$

on the interval $(ma_{3n}(\theta),\kappa_{n}a_{3n}(\theta)).$

If $\kappa_{n^{\prime}}^{2}a_{3n^{\prime}}(\theta)\geq a_{3n+1}(\theta)$ for
every $n^{\prime}<n,$ then

(19)
$\mathbf{E}(x|\mathcal{B}_{\kappa,\theta})=\frac{X(\kappa_{n}a_{3n}(\theta))}{\kappa_{n}a_{3n}(\theta)}\geq\frac{X(a_{3n}(\theta))}{\kappa_{n}a_{3n}(\theta)}$

on the interval $(ma_{3n}(\theta),\kappa_{n}a_{3n}(\theta)).$

It follows from (17) and (18) (or (19)) that

$\int_{0}^{m\kappa_{n}a_{3n}(\theta)}u(s)ds\geq\frac{\kappa_{n}-m}{\kappa_{n}}\cdot(1-\frac{4}{9})X(a_{3n}(\theta))-\frac{1}{2}X(a_{3n}(\theta)).$

Since $\kappa_{n}\geq 100m,$ it follows that

$\int_{0}^{m\kappa_{n}a_{3n}(\theta)}u(s)ds\geq((1-\frac{1}{100})(1-\frac{4}{9})-\frac{1}{2})X(a_{3n}(\theta))=\frac{1}{20}X(a_{3n}(\theta))=$
$=\frac{1}{30}X(a_{3n+1}(\theta))\geq\frac{1}{30}X(\kappa_{n}a_{3n}(\theta))=\frac{1}{30}\int_{0}^{\kappa_{n}a_{3n}(\theta)}\mathbf{E}(x|\mathcal{B}_{\kappa^{100m},\theta})(s)ds.$

It follows immediately that

(20) $\int_{0}^{t}\mathbf{E}(x|\mathcal{B}_{\kappa^{100m},\theta})(s)ds\leq
30\int_{0}^{mt}u(s)ds\leq 30\int_{0}^{mt}\mu(s,u)ds$

for every $t$ being a node of the partition
$\mathcal{B}_{\kappa^{100m},\theta}.$ Therefore,

$\int_{0}^{t}\mathbf{E}(x|\mathcal{B}_{\kappa^{100m},\theta})(s)ds\leq
30\int_{0}^{mt}\mu(s,u)ds,\quad t>0$

or, equivalently,

$\int_{0}^{t/m}\mathbf{E}(x|\mathcal{B}_{\kappa^{100m},\theta})(s)ds\leq
30\int_{0}^{t}\mu(s,u)ds,\quad t>0.$

The assertion follows immediately. ∎

######  Lemma 42.

Let $E$ be a fully symmetric Banach function space either on the interval
$(0,1)$ or on the semi-axis. If $x=\mu(x)\in E$ is such that
$\varphi(y)\leq\varphi(x)$ for every positive symmetric functional $\varphi$
on $E$ and every $0\leq y\prec\prec x,$ then
$\lambda^{-1}\sigma_{\lambda}\mathbf{E}(x|\mathcal{B}_{\kappa^{\lambda},\theta})\to
0$ as $\lambda\to\infty.$

###### Proof.

Since $\mathbf{E}(x|\mathcal{B}_{\kappa,\theta})\prec\prec x,$ it follows from
the assumption and Lemma 40 that there exists $0\leq u_{m}\to 0$ such that

$\int_{ma}^{b}\mathbf{E}(x|\mathcal{B}_{\kappa,\theta})(s)ds\leq\int_{a}^{mb}(x+u_{m})(s)ds,\quad\forall
ma\leq b\in\mathbb{R}.$

For every $\lambda\geq 100m,$ we have $\kappa^{100m}\leq\kappa^{\lambda}.$ It
follows from Lemma 41 that

$\frac{1}{\lambda}\sigma_{\lambda}\mathbf{E}(x|\mathcal{B}_{\kappa^{\lambda},\theta})\prec\prec\frac{1}{m}\sigma_{m}\mathbf{E}(x|\mathcal{B}_{\kappa^{\lambda},\theta})\stackrel{{\scriptstyle
Lemma\ref{majorant
lemma}}}{{\prec\prec}}\frac{3}{2m}\sigma_{m}\mathbf{E}(x|\mathcal{B}_{\kappa^{100m},\theta})\stackrel{{\scriptstyle\eqref{kappa
estimate}}}{{\prec\prec}}45\mu(u_{m}).$

The assertion now follows immediately. ∎

######  Proposition 43.

Let $E$ be a fully symmetric Banach function space either on the interval
$(0,1)$ or on the semi-axis equipped with a Fatou norm. If $x=\mu(x)\in E$ is
such that $\varphi(y)\leq\varphi(x)$ for every positive symmetric functional
$\varphi$ on $E$ and every $0\leq y\prec\prec x,$ then
$m^{-1}\sigma_{m}\mathbf{E}(x|\mathcal{B}_{m,\theta})\to 0$ as $m\to\infty.$

###### Proof.

For every $m,r\in\mathbb{N},$ set

$\kappa^{m,r}_{n}=\begin{cases}m\qquad 0\leq|n|<r\\\ \infty\qquad
r\leq|n|\end{cases}$

and $\kappa^{m,r}=\\{\kappa^{m,r}_{n}\\}_{n\in\mathbb{Z}}.$ Clearly,
$\mathbf{E}(x|\mathcal{B}_{\kappa^{m,r},\theta})\to\mathbf{E}(x|\mathcal{B}_{m,\theta})$
almost everywhere when $r\to\infty.$ It follows from the definition of Fatou
norm that

$\lim_{r\to\infty}\|\sigma_{m}\mathbf{E}(x|\mathcal{B}_{\kappa^{m,r},\theta})\|_{E}=\|\sigma_{m}\mathbf{E}(x|\mathcal{B}_{m,\theta})\|_{E}.$

Select $r_{m}$ so large that

(21)
$\frac{1}{m}\|\sigma_{m}\mathbf{E}(x|\mathcal{B}_{\kappa^{m,r_{m}},\theta})\|_{E}>\frac{1}{2m}\|\sigma_{m}\mathbf{E}(x|\mathcal{B}_{m,\theta})\|_{E}.$

Now define the sequence $\kappa=\\{\kappa_{n}\\}_{n\in\mathbb{Z}}$ by setting

$\kappa_{n}=\inf_{m\geq 1}\kappa^{m,r_{m}}_{n}=\inf_{r_{m}>|n|}m,\quad
n\in\mathbb{Z}.$

Clearly, $r_{\kappa_{n}}\geq|n|$ and, therefore, $\kappa_{n}\rightarrow\infty$
as $|n|\rightarrow\infty.$ In particular, the set $\\{n:\kappa_{n}<\lambda\\}$
is finite for every $\lambda\in\mathbb{N}.$ Set

$M(\lambda)=\max\\{\lambda,\max_{\kappa_{n}<\lambda}(\frac{a_{3n+1}(\theta)}{a_{3n}(\theta)})^{1/2}\\}.$

If $m>M(\lambda),$ then $m^{2}a_{3n}(\theta)\geq a_{3n+1}(\theta)$ whenever
$\kappa_{n}<\lambda.$ Thus, $\kappa_{n}\geq\lambda$ whenever
$m^{2}a_{3n}(\theta)<a_{3n+1}(\theta).$ Hence,
$\kappa_{n}^{\lambda}=\kappa_{n}$ whenever
$(\kappa_{n}^{m,r_{m}})^{2}a_{3n}(\theta)<a_{3n+1}(\theta).$ Therefore,
$\kappa_{n}^{\lambda}\leq\kappa_{n}^{m,r_{m}}$ for every $n\in\mathbb{Z}$ such
that $(\kappa_{n}^{m,r_{m}})^{2}a_{3n}(\theta)<a_{3n+1}(\theta).$ According to
Remark 39, it follows that

$\mathbf{E}(x|\mathcal{B}_{\kappa^{m,r_{m}},\theta})\prec\prec\frac{3}{2}\mathbf{E}(x|\mathcal{B}_{\kappa^{\lambda},\theta}).$

Since $m\geq\lambda,$ it follows that

(22)
$\frac{1}{m}\sigma_{m}\mathbf{E}(x|\mathcal{B}_{\kappa^{m,r_{m}},\theta})\prec\prec\frac{3}{2\lambda}\sigma_{\lambda}\mathbf{E}(x|\mathcal{B}_{\kappa^{\lambda},\theta}).$

By Lemma 42, for every $\varepsilon>0,$ there exists $\lambda$ such that

(23)
$\frac{1}{\lambda}\|\sigma_{\lambda}\mathbf{E}(x|\mathcal{B}_{\kappa^{\lambda},\theta})\|_{E}<\frac{1}{3}\varepsilon.$

It follows that

$\frac{1}{m}\|\sigma_{m}\mathbf{E}(x|\mathcal{B}_{m,\theta})\|_{E}\stackrel{{\scriptstyle\eqref{qss17}}}{{\leq}}\frac{2}{m}\|\sigma_{m}\mathbf{E}(x|\mathcal{B}_{\kappa^{m,r_{m}},\theta})\|_{E}\stackrel{{\scriptstyle\eqref{qss18}}}{{\leq}}\frac{3}{\lambda}\|\sigma_{\lambda}\mathbf{E}(x|\mathcal{B}_{\kappa^{\lambda},\theta})\|_{E}\stackrel{{\scriptstyle\eqref{qss19}}}{{<}}\varepsilon$

for every $m>M(\lambda).$ Since $\varepsilon>0$ is arbitrarily small, the
assertion follows. ∎

######  Lemma 44.

Let $E$ be a fully symmetric Banach space either on the interval $(0,1)$ or on
the semi-axis equipped with a Fatou norm. If $x=\mu(x)\in E$ is such that
$\varphi(y)\leq\varphi(x)$ for every positive symmetric functional $\varphi$
on $E$ and every $0\leq y\prec\prec x,$ then
$m^{-1}\sigma_{m}\mathbf{E}(x|\mathcal{A}_{m})\to 0$ as $m\to\infty.$

###### Proof.

It is clear that $a_{k}(3/2)=a_{k+1}(1)$ and $a_{k}((3/2)^{2})=a_{k+2}(1)$ for
every $k\in\mathbb{N}.$ It follows that

$\mathcal{B}_{m,1}\cup\mathcal{B}_{m,3/2}\cup\mathcal{B}_{m,(3/2)^{2}}=\mathcal{A}_{m}.$

Therefore, by Lemma 37, we have

(24)
$\mathbf{E}(x|\mathcal{A}_{m})\prec\prec\mathbf{E}(x|\mathcal{B}_{m,1})+\mathbf{E}(x|\mathcal{B}_{m,3/2})+\mathbf{E}(x|\mathcal{B}_{m,(3/2)^{2}}).$

The assertion follows now from Proposition 43. ∎

######  Lemma 45.

Let $x=\mu(x)\in L_{1}+L_{\infty}(0,\infty)$ be a function on the semi-axis.
If $x\notin L_{1}(0,\infty),$ then, for every $t>0$ and every
$m\in\mathbb{N},$ we have

(25)
$X(t)\leq\frac{2}{3}X(m^{4}t)+\frac{3}{2}\int_{0}^{m^{4}t}\mathbf{E}(x|\mathcal{A}_{m})(s)ds.$

###### Proof.

For a given $t>0,$ there exists $n\in\mathbb{Z}$ such that
$t\in[a_{n}(1),a_{n+1}(1)].$ If $a_{n+1}(1)>m^{2}a_{n}(1),$ then

$\int_{0}^{m^{4}t}\mathbf{E}(x|\mathcal{A}_{m})(s)ds\geq\int_{0}^{ma_{n}(1)}\mathbf{E}(x|\mathcal{A}_{m})(s)ds=X(ma_{n}(1))\geq\frac{2}{3}X(t).$

If $a_{n+1}(1)\leq m^{2}a_{n}(1)$ and $a_{n+2}(1)>m^{2}a_{n+1}(1),$ then

$\int_{0}^{m^{4}t}\mathbf{E}(x|\mathcal{A}_{m})(s)ds\geq\int_{0}^{ma_{n+1}(1)}\mathbf{E}(x|\mathcal{A}_{m})(s)ds=X(ma_{n+1}(1))\geq
X(t).$

If $a_{n+2}(1)\leq m^{2}a_{n+1}(1)$ and $a_{n+1}(1)\leq m^{2}a_{n}(1),$ then

$X(m^{4}t)\geq X(a_{n+2}(1))=\frac{3}{2}X(a_{n+1}(1))\geq\frac{3}{2}X(t)$

and the assertion follows. ∎

The situation in the case that $x\in L_{1}$ is slightly more complicated.

######  Lemma 46.

If $x=\mu(x)\in L_{1}(0,1)$ or $x\in L_{1}(0,\infty),$ then there exists
constant $C$ such that for every $t>0$

(26)
$X(t)\leq\frac{2}{3}X(m^{4}t)+\frac{3}{2}\int_{0}^{m^{4}t}\mathbf{E}(x|\mathcal{A}_{m})(s)ds+C\int_{0}^{m^{4}t}\chi_{[0,1]}(s)ds.$

###### Proof.

Consider first the case of the semi-axis. Fix $n_{0}$ such that
$X(a_{n_{0}})\leq 4/9X(\infty).$ For a givne $t\in[a,a_{n_{0}}],$ there exists
$n\in\mathbb{Z}$ such that $n<n_{0}$ and $t\in[a_{n},a_{n+1}].$ Then, the
argument in Lemma 45 applies mutatis mutandi. For every $t\geq a_{n_{0}}$ we
have

$X(t)\leq\frac{X(\infty)}{\min\\{a_{n_{0}},1\\}}\min\\{m^{4}t,1\\}=\frac{X(\infty)}{\min\\{a_{n_{0}},1\\}}\int_{0}^{m^{4}t}\chi_{[0,1]}(s)ds.$

Setting $C=X(\infty)/\min\\{a_{n_{0}},1\\},$ we obtain the assertion.

The same argument applies in the case of the interval $(0,1)$ by replacing
$X(\infty)$ by $X(1).$ ∎

The following two theorems are crucial for the proof of the implication
$(3)\Leftrightarrow(4)$ in Theorem 5.

######  Theorem 47.

Let $E$ be a fully symmetric Banach space either on the interval $(0,1)$ or on
the semi-axis and let $x\in E.$ Suppose that the norm on $E$ is a Fatou norm.
If $\varphi(y)\leq\varphi(x)$ for every positive symmetric functional on $E$
and every $0\leq y\prec\prec x,$ then

(27) $\lim_{m\to\infty}\frac{1}{m}\|\sigma_{m}(\mu(x))\|_{E}=0$

provided that one of the following conditions is satisfied

  1. (1)

$E=E(0,1)$ is a space on the interval $(0,1).$

  2. (2)

$E=E(0,\infty)$ is a space on the semi-axis and $E(0,\infty)\not\subset
L_{1}(0,\infty).$

###### Proof.

Without loss of generality, $x=\mu(x).$ If $x\notin L_{1},$ then by Lemma 45,

$\int_{0}^{t/m^{4}}x(s)ds\leq\frac{2}{3}\int_{0}^{t}x(s)ds+\frac{3}{2}\int_{0}^{t}\mathbf{E}(x|\mathcal{A}_{m})(s)ds,\quad\forall
t>0$

or, equivalently,

$\frac{1}{m^{4}}\sigma_{m^{4}}x\prec\prec\frac{2}{3}x+\frac{3}{2}\mathbf{E}(x|\mathcal{A}_{m}).$

Applying $m^{-1}\sigma_{m}$ to the both parts, we obtain

$\frac{1}{m^{5}}\sigma_{m^{5}}x\prec\prec\frac{2}{3}\frac{1}{m}\sigma_{m}x+\frac{3}{2}\frac{1}{m}\sigma_{m}\mathbf{E}(x|\mathcal{A}_{m}).$

Take norms and let $m\to\infty.$ It follows from Lemma 44 that

$\lim_{m\to\infty}\frac{1}{m}\|\sigma_{m}x\|_{E}\leq\frac{2}{3}\lim_{m\to\infty}\frac{1}{m}\|\sigma_{m}x\|_{E}.$

This proves (27).

If $x\in L_{1}$ and $C$ are as in Lemma 46, then it follows from Lemma 46 that

$\int_{0}^{t/m^{4}}x(s)ds\leq\frac{2}{3}\int_{0}^{t}x(s)ds+\frac{3}{2}\int_{0}^{t}\mathbf{E}(x|\mathcal{A}_{m})(s)ds+C\int_{0}^{t}\chi_{[0,1]}(s)ds,\quad\forall
t>0$

or, equivalently,

$\frac{1}{m^{4}}\sigma_{m^{4}}x\prec\prec\frac{2}{3}x+\frac{3}{2}\mathbf{E}(x|\mathcal{A}_{m})+C\chi_{(0,1)}.$

Applying $m^{-1}\sigma_{m}$ to the both parts, we obtain

$\frac{1}{m^{5}}\sigma_{m^{5}}x\prec\prec\frac{2}{3}\frac{1}{m}\sigma_{m}x+\frac{3}{2}\frac{1}{m}\sigma_{m}\mathbf{E}(x|\mathcal{A}_{m})+C\frac{1}{m}\sigma_{m}\chi_{(0,1)}.$

Take norms and let $m\to\infty.$ For every symmetric space $E$ on the interval
$(0,1)$ and for every symmetric space $E$ on the semi-axis such that
$E\not\subset L_{1}(0,\infty)$ we have
$m^{-1}\|\sigma_{m}\chi_{(0,1)}\|_{E}\to 0.$ It follows from Lemma 44 that

$\lim_{m\to\infty}\frac{1}{m}\|\sigma_{m}x\|_{E}\leq\frac{2}{3}\lim_{m\to\infty}\frac{1}{m}\|\sigma_{m}x\|_{E}$

and again (27) follows. ∎

######  Theorem 48.

Let $E=E(0,\infty)$ be a fully symmetric Banach space on the semi-axis
equipped with a Fatou norm such that $E(0,\infty)\subset L_{1}(0,\infty).$ If
$\varphi(y)\leq\varphi(x)$ for every positive symmetric functional on $E$ and
every $0\leq y\prec\prec x,$ then

(28) $\lim_{m\to\infty}\frac{1}{m}\|\sigma_{m}(\mu(x))\chi_{(0,1)}\|_{E}=0.$

###### Proof.

Fully symmetric Banach space $F$ on the interval $(0,1)$ consists of those
$z\in E$ supported on the interval $(0,1).$ Let $x_{1}=\mu(x)\chi_{(0,1)}\in
F.$ Suppose that

$\lim_{m\to\infty}\frac{1}{m}\|\sigma_{m}(\mu(x))\chi_{(0,1)}\|_{E}>0.$

It clearly follows that

$\lim_{m\to\infty}\frac{1}{m}\|\sigma_{m}(\mu(x_{1}))\|_{F}>0.$

By Theorem 47, there exists $0\leq y_{1}\prec\prec x_{1}$ and a positive
symmetric functional $\varphi\in F^{*}$ such that
$\varphi(y_{1})>\varphi(x_{1}).$ Let $\varphi_{sing}$ be a singular part of
the functional $\varphi$ constructed in Lemma 26. It follows from Lemma 26
that $\varphi_{sing}$ is symmetric. By Lemma 27, the difference
$\varphi-\varphi_{sing}$ is a symmetric normal functional on $F$ (that is, an
integral). Therefore, $\varphi_{sing}(y_{1})>\varphi_{sing}(x_{1}).$

Now we show that the functional $\varphi_{sing}$ can be extended from $F$ to
$E$ by setting

$\varphi_{sing}(z)=\lim_{n\to\infty}\varphi_{sing}(\mu(z)\chi_{(0,1/n)}),\quad
0\leq z\in E.$

Repeating the argument in Lemma 26, we prove that the extension above is
additive on $E_{+}.$ Thus, the functional $\varphi_{sing}\in E^{*}$ is
positive and symmetric. Since $y_{1}\prec\prec x$ and
$\varphi_{sing}(y_{1})>\varphi_{sing}(x_{1})=\varphi_{sing}(x),$ the assertion
follows. ∎

##  7\. Proof of Theorem 5

In this section, we prove an assertion more general then that of Theorem 5.
The assertion of Theorem 5 follows from that of Theorem 49 by setting
$\mathcal{M}=B(H).$

In what follows, the semifinite von Neumann algebra $\mathcal{M}$ is either
atomless or atomic so that the trace of every atom is $1.$

######  Theorem 49.

Let $E(\mathcal{M},\tau)$ be a symmetric operator space. Consider the
following conditions.

  1. (1)

There exist nontrivial positive singular symmetric functionals on
$E(\mathcal{M},\tau).$

  2. (2)

There exist nontrivial singular fully symmetric functionals on
$E(\mathcal{M},\tau).$

  3. (3)

There exist positive symmetric symmetric functional on $E(\mathcal{M},\tau)$
which are not fully symmetric.

  4. (4)

If $E(\mathcal{M},\tau)\not\subset L_{1}(\mathcal{M},\tau),$ then there exists
an operator $A\in E(\mathcal{M},\tau)$ such that

(29) $\lim_{m\to\infty}\frac{1}{m}\|\sigma_{m}\mu(A)\|_{E}>0.$

If $E(\mathcal{M},\tau)\subset L_{1}(\mathcal{M},\tau),$ then there exists an
operator $A\in E(\mathcal{M},\tau)$ such that

(30) $\lim_{m\to\infty}\frac{1}{m}\|(\sigma_{m}\mu(A))\chi_{(0,1)}\|_{E}>0.$

  1. (i)

The conditions (1) and (4) are equivalent for every symmetric operator space
$E(\mathcal{M},\tau).$

  2. (ii)

The conditions (1), (2) and (4) are equivalent for every fully symmetric
operator space $E(\mathcal{M},\tau).$

  3. (iii)

The conditions (1)-(4) are equivalent for every fully symmetric operator space
$E(\mathcal{M},\tau)$ equipped with a Fatou norm.

###### Proof.

Implications $\eqref{second main condition}\Rightarrow\eqref{first main
condition}$ and $\eqref{third main condition}\Rightarrow\eqref{first main
condition}$ are trivial.

$\eqref{first main condition}\Rightarrow\eqref{fourth main condition}$ Let
$E(\mathcal{M},\tau)$ be a symmetric operator space with a singular symmetric
functional $\varphi.$ Let $A\in E(\mathcal{M},\tau)$ be an operator such that
$\varphi(A)\neq 0.$ Without loss of generality, $A\geq 0.$

If $E(\mathcal{M},\tau)\not\subset L_{1}(\mathcal{M},\tau),$ then

$|\varphi(A)|=\frac{1}{m}|\varphi(\underbrace{A\oplus\cdots\oplus
A}_{\mbox{$m$
times}})|\leq\|\varphi\|_{E^{*}(\mathcal{M},\tau)}\cdot\frac{1}{m}\|\sigma_{m}\mu(A)\|_{E}.$

Passing $m\to\infty,$ we obtain the required inequality (29).

Let now $E(\mathcal{M},\tau)\subset L_{1}(\mathcal{M},\tau).$ If $\mathcal{M}$
is atomic, then $E(\mathcal{M},\tau)=L_{1}(\mathcal{M},\tau)$ and the
assertion is trivial. Let $\mathcal{M}$ be atomless. Since $\varphi$ is a
singular functional and

$A-AE_{A}(\mu(\frac{1}{m},A),\infty)\in(L_{1}\cap
L_{\infty})(\mathcal{M},\tau),\quad\forall m\in\mathbb{N},$

we infer that

$|\varphi(A)|=|\varphi(AE_{A}(\mu(\frac{1}{m},A),\infty))|=$
$=\frac{1}{m}|\varphi(\underbrace{AE_{A}(\mu(\frac{1}{m},A),\infty)\oplus\cdots\oplus
AE_{A}(\mu(\frac{1}{m},A),\infty)}_{\mbox{$m$ times}})|\leq$
$\leq\|\varphi\|_{E^{*}(\mathcal{M},\tau)}\cdot\frac{1}{m}\|\sigma_{m}\mu(AE_{A}(\mu(\frac{1}{m},A),\infty))\|_{E}\leq\|\varphi\|_{E^{*}(\mathcal{M},\tau)}\cdot\frac{1}{m}\|(\sigma_{m}\mu(A))\chi_{(0,1)}\|_{E}.$

Passing $m\to\infty,$ we obtain the required inequality (30).

$\eqref{fourth main condition}\Rightarrow\eqref{first main condition}$
Firstly, we assume that the algebra $\mathcal{M}$ is finite. Without loss of
generality, $\tau(1)=1.$ Let $E(\mathcal{M},\tau)$ be a symmetric operator
space and let $E(0,1)$ be the corresponding symmetric function space. By the
assumption, there exists an element $x=\mu(A)\in E(0,1)$ such that
$m^{-1}\sigma_{m}x\not\to 0$ in $E(0,1).$ By Theorem 29, there exists a
positive singular symmetric functional $0\neq\varphi\in E(0,1)^{*}.$ Let
$\mathcal{L}(\varphi)$ be a functional on $E(\mathcal{M},\tau)$ defined in
Theorem 14. Clearly, $\mathcal{L}(\varphi)$ is a nontrivial positive symmetric
functional on $E(\mathcal{M},\tau).$

The case when $\mathcal{M}$ is an infinite atomless von Neumann algebra can be
treated in a similar manner. The only difference is that the reference to
Theorem 29 has to be replaced with the reference to either Theorem 28 or
Theorem 23.

Let $E(\mathcal{M},\tau)$ be a symmetric operator space on a atomic von
Neumann algebra $\mathcal{M}$ and let $E(\mathbb{N})$ be the corresponding
symmetric sequence space. It follows from the assumption that
$E(\mathcal{M},\tau)\neq L_{1}(\mathcal{M},\tau)$ or, equivalently,
$E(\mathbb{N})\neq l_{1}.$ By the assumption, there exists an element
$x=\mu(A)\in E$ such that $m^{-1}\sigma_{m}x\not\to 0$ in $E.$ Let
$F(0,\infty)$ be a symmetric function space constructed in Proposition 16.
Since $E(\mathbb{N})\neq l_{1},$ it follows that $F(0,\infty)\not\subset
L_{1}(0,\infty).$ Recall that the space $E(\mathbb{N})$ is naturally embedded
into the space $F(0,\infty)$ and that the norms $\|\cdot\|_{E}$ and
$\|\cdot\|_{F}$ are equivalent on $E(\mathbb{N}).$ We have $x\in F$ and
$m^{-1}\sigma_{m}x\not\to 0$ in $F(0,\infty).$ By Theorem 23, there exists a
positive symmetric functional $0\leq\varphi\in F(0,\infty)^{*}.$ The
restriction of the functional $\varphi$ to $E(\mathbb{N})$ is a nontrivial
positive symmetric functional on $E(\mathbb{N}).$ Let $\mathcal{L}(\varphi)$
be a functional on $E(\mathcal{M},\tau)$ defined in Theorem 14. Clearly,
$\mathcal{L}(\varphi)$ is a nontrivial positive symmetric functional on
$E(\mathcal{M},\tau).$

$\eqref{fourth main condition}\Rightarrow\eqref{second main condition}$ The
proof is very similar to that of the implication $\eqref{fourth main
condition}\Rightarrow\eqref{first main condition}$ and is, therefore, omitted.
The only difference is that references to Theorem 29, Theorem 28 or Theorem 23
have to be replaced with references to Theorem 36, Theorem 35 or Theorem 33,
respectively.

$\eqref{fourth main condition}\Rightarrow\eqref{third main condition}$
Firstly, we assume that the algebra $\mathcal{M}$ is finite. Without loss of
generality, $\tau(1)=1.$ Let $E(\mathcal{M},\tau)$ be a symmetric operator
space and let $E(0,1)$ be the corresponding symmetric function space. By the
assumption, there exists an element $x=\mu(A)\in E(0,1)$ such that
$m^{-1}\sigma_{m}x\not\to 0$ in $E(0,1).$ By Theorem 47, there exists a
positive symmetric but not fully symmetric functional $\varphi\in E(0,1)^{*}.$
Let $\mathcal{L}(\varphi)$ be a functional on $E(\mathcal{M},\tau)$ defined in
Theorem 14. Clearly, $\mathcal{L}(\varphi)$ is a symmetric but not fully
symmetric functional on $E(\mathcal{M},\tau).$

The case when $\mathcal{M}$ is an infinite atomless von Neumann algebra can be
treated in a similar manner. The only difference is that the reference to
Theorem 47 has to be replaced with the reference to either Theorem 47 or
Theorem 48.

Let $E(\mathcal{M},\tau)$ be a symmetric operator space on a atomic von
Neumann algebra $\mathcal{M}$ and let $E(\mathbb{N})$ be the corresponding
symmetric sequence space. It follows from the assumption that
$E(\mathcal{M},\tau)\neq L_{1}(\mathcal{M},\tau)$ or, equivalently,
$E(\mathbb{N})\neq l_{1}.$ By the assumption, there exists an element
$x=\mu(A)\in E$ such that $m^{-1}\sigma_{m}x\not\to 0$ in $E.$ Let
$F(0,\infty)$ be a symmetric function space constructed in Proposition 16.
Since $E(\mathbb{N})\neq l_{1},$ it follows that $F(0,\infty)\not\subset
L_{1}(0,\infty).$ Recall that the space $E(\mathbb{N})$ is naturally embedded
into the space $F(0,\infty)$ and that the norms $\|\cdot\|_{E}$ and
$\|\cdot\|_{F}$ are equivalent on $E(\mathbb{N}).$ We have $x\in F$ and
$m^{-1}\sigma_{m}x\not\to 0$ in $F(0,\infty).$ By Theorem 47, there exists a
positive symmetric functional $\varphi\in F(0,\infty)^{*}$ and a function
$0\leq y\prec\prec x$ such that $\varphi(y)>\varphi(x).$ Set
$z=\mathbf{E}(\mu(y)|\\{(n-1,n)\\}_{n\in\mathbb{N}}).$ Clearly, $z\in
E(\mathbb{N})$ and $\varphi(z)=\varphi(y)>\varphi(x).$ Hence, the restriction
of the functional $\varphi$ to $E(\mathbb{N})$ is a positive symmetric but not
fully symmetric functional on $E(\mathbb{N}).$ Let $\mathcal{L}(\varphi)$ be a
functional on $E(\mathcal{M},\tau)$ defined in Theorem 14. Clearly,
$\mathcal{L}(\varphi)$ is a positive symmetric but not fully symmetric
functional on $E(\mathcal{M},\tau).$ ∎

##  8\. Appendix

In this appendix, we set $\mathcal{A}=\\{(n-1,n)\\}_{n\in\mathbb{N}}.$

######  Lemma 50.

If $x,y\in(L_{1}+L_{\infty})(0,\infty)$ are positive functions, then

$\mathbf{E}(\mu(x+y)|\mathcal{A})\lhd\mathbf{E}(\mu(x)|\mathcal{A})+\mathbf{E}(\mu(y)|\mathcal{A})\lhd
2\sigma_{1/2}\mathbf{E}(\mu(x+y)|\mathcal{A}).$

###### Proof.

Recall that

$\mu(x+y)\prec\prec\mu(x)+\mu(y)\prec\prec 2\sigma_{1/2}\mu(x+y).$

It follows that

$\int_{0}^{b}\mu(s,x+y)ds\leq\int_{0}^{b}(\mu(s,x)+\mu(s,y))ds,$
$\int_{0}^{2a}\mu(s,x+y)ds\geq\int_{0}^{a}(\mu(s,x)+\mu(s,y))ds.$

Let now $a,b$ be positive integers. Subtracting the above inequalities, we
obtain

$\int_{2a}^{b}\mathbf{E}(\mu(x+y)|\mathcal{A})(s)ds=\int_{2a}^{b}\mu(s,x+y)ds\leq$
$\leq\int_{a}^{b}(\mu(s,x)+\mu(s,y))ds=\int_{a}^{b}\mathbf{E}(\mu(x)+\mu(y)|\mathcal{A})(s)ds.$

Similarly, we have

$\int_{2a}^{b}\mathbf{E}(\mu(x)+\mu(y)|\mathcal{A})(s)ds\leq\int_{2a}^{2b}\mathbf{E}(\mu(x+y)|\mathcal{A})(s)ds.$

∎

######  Corollary 51.

The quasi-norm in Construction 16 is a norm.

###### Proof.

It follows from Lemma 50 that

$\mathbf{E}(\mu(x+y)|\mathcal{A})\lhd\mathbf{E}(\mu(x)+\mu(y)|\mathcal{A})$

provided that $x,y$ are positive functions. By Theorem 9,

$\|\mathbf{E}(\mu(x+y)|\mathcal{A})\|_{E}\leq\|\mathbf{E}(\mu(x)|\mathcal{A})\|_{E}+\|\mathbf{E}(\mu(y)|\mathcal{A})\|_{E}.$

∎

######  Lemma 52.

Let $y=\mu(y)\in(L_{1}+L_{\infty})(0,\infty).$ It follows that

$\int_{2^{-k}\lambda
a}^{b}y(s)ds\leq\frac{\lambda}{\lambda-1}\int_{a}^{b}(\sigma_{2^{k}}y)(s)ds$

provided that $b\geq\lambda a.$

###### Proof.

Let $\alpha$ be the average value of $y$ on the interval $[2^{-k}\lambda
a,2^{-k}b].$ Clearly, $y\leq\alpha$ on the interval $[2^{-k}\lambda a,b]$ and
$y\geq\alpha$ on the interval $[2^{-k}a,2^{-k}b].$ Thus,
$\sigma_{2^{k}}y\geq\alpha$ on the interval $[a,b].$ Therefore,

$\int_{2^{-k}\lambda a}^{b}y(s)ds\leq(b-2^{-k}\lambda
a)\alpha\leq\frac{\lambda}{\lambda-1}(b-a)\alpha\leq\frac{\lambda}{\lambda-1}\int_{a}^{b}(\sigma_{2^{k}}y)(s)ds.$

∎

######  Theorem 53.

If $\\{x_{n}\\}_{n\in\mathbb{N}}$ be a Cauchy sequence in $F,$ then there
exists $x\in F$ such that $x_{n}\to x$ in $F.$

###### Proof.

For every $k>0,$ there exists $m_{k}$ such that $\|x_{m}-x_{m_{k}}\|_{F}\leq
4^{-k}$ for $m\geq m_{k}.$ Set $y_{k}=x_{m_{k+1}}-x_{m_{k}}.$ Clearly,
$\|y_{k}\|_{F}\leq 4^{-k}$ for every $k\in\mathbb{N}.$ In particular, the
series $\sum_{k=1}^{\infty}y_{k}$ converges in $L_{\infty}(0,\infty).$

Set $z_{n}=\sum_{k=n}^{\infty}\sigma_{2^{k}}\mu(y_{k}).$ We claim that
$z_{n}\in F$ and $z_{n}\to 0$ in $F.$ Indeed,

$\mu(y_{k})\leq\|y_{k}\|_{\infty}\chi_{(0,1)}+T\mathbf{E}(\mu(y_{k})|\mathcal{A}).$

Here, $T$ is a shift to the right. It follows that

$\mathbf{E}(\mu(z_{n})|\mathcal{A})\leq\sum_{k=n}^{\infty}\sigma_{2^{k}}(\|y_{k}\|_{\infty}\chi_{(0,1)}+T\mathbf{E}(\mu(y_{k})|\mathcal{A})).$

Therefore,

$\|z_{n}\|_{F}\leq\|z_{n}\|_{\infty}+\sum_{k=n}^{\infty}2^{k}\|\|y_{k}\|_{\infty}\chi_{(0,1)}+T\mathbf{E}(\mu(y_{k})|\mathcal{A})\|_{E}\leq$
$\leq\|z_{n}\|_{\infty}+\sum_{k=n}^{\infty}2^{k+1}\|y_{k}\|_{F}\leq\frac{1}{3}\cdot
4^{1-n}+2^{2-n}=o(1).$

It follows from Lemma 8.5 of [19] that

$\int_{\lambda
a}^{b}\mu(s,\sum_{k=n}^{\infty}y_{k})ds\leq\sum_{k=n}^{\infty}\int_{2^{-k}\lambda
a}^{b}\mu(s,y_{k})ds.$

It follows from Lemma 52 that

$\int_{2^{-k}\lambda
a}^{b}\mu(s,y_{k})ds\leq\frac{\lambda}{\lambda-1}\int_{a}^{b}(\sigma_{2^{k}}\mu(y_{k}))(s)ds.$

Therefore,

$\int_{\lambda
a}^{b}\mu(s,\sum_{k=n}^{\infty}y_{k})ds\leq\frac{\lambda}{\lambda-1}\int_{a}^{b}z_{n}(s)ds.$

Hence,

$\sum_{k=n}^{\infty}y_{k}\lhd\frac{\lambda}{\lambda-1}z_{n}.$

Since $\lambda>1$ is arbitrarily large, it follows from Theorem 9 that

$\|\sum_{k=n}^{\infty}y_{k}\|_{F}\leq\|z_{n}\|_{F}\to 0.$

Thus, the series $\sum_{k=1}^{\infty}y_{k}$ does converge in $F.$ The
assertion follows immediately.

∎

##  9\. Proof of Figiel-Kalton theorem

The proof of Theorem 8 follows from the combinations of Lemmas below.

######  Lemma 54.

Let $E$ be a symmetric Banach space either on the interval $(0,1)$ or on the
semi-axis. If $x\in Z_{E},$ then $C(\mu(x_{+})-\mu(x_{-}))\in E.$

###### Proof.

Let $x=\sum_{k=1}^{n}(x_{k}-y_{k})$ with $x_{k},y_{k}\in E_{+}$ and
$\mu(x_{k})=\mu(y_{k}),$ $1\leq k\leq n.$ Set

$z=x_{+}+\sum_{k=1}^{n}y_{k}=x_{-}+\sum_{k=1}^{n}x_{k}.$

It follows from the definition of $C$ and (8) that

$C\mu(z)\leq
C(x_{+})+\sum_{k=1}^{n}C\mu(y_{k})=C(\mu(x_{+})-\mu(x_{-}))+C\mu(x_{-})+\sum_{k=1}^{n}C\mu(x_{k}).$

Using the second inequality in (8), we obtain

$\int_{0}^{t}(\mu(s,x_{-})+\sum_{k=1}^{n}\mu(s,x_{k}))ds\leq\int_{0}^{(n+1)t}\mu(s,z)ds\leq\int_{0}^{t}\mu(s,z)ds+nt\mu(t,z).$

Therefore,

$C\mu(z)\leq C\mu(z)+C(\mu(x_{+})-\mu(x_{-}))+n\mu(z).$

It follows that $C(\mu(x_{-})-\mu(x_{+}))\leq n\mu(z).$ Similarly,
$C(\mu(x_{+})-\mu(x_{-}))\leq n\mu(z)$ and the assertion follows. ∎

######  Lemma 55.

Let $E$ be a symmetric Banach space either on the interval $(0,1)$ or on the
semi-axis. If $x\in\mathcal{D}_{E},$ then $C(\mu(x_{+})-\mu(x_{-}))\in Cx+E.$

###### Proof.

Since $x\in\mathcal{D}_{E},$ it follows that $x=\mu(a)-\mu(b)$ with $a,b\in
E.$ Set $u=\mu(a)-x_{+}\geq 0.$ Clearly, $\mu(a)=u+x_{+}$ and
$\mu(b)=u+x_{-}.$ It follows from the definition of $C$ and (8) that

$C\mu(a)\leq
C\mu(u)+C\mu(x_{+})=C(\mu(x_{+})-\mu(x_{-}))+C\mu(u)+C\mu(x_{-}).$

Using the second inequality in (8), we obtain

$C\mu(x_{-})+C\mu(u)\leq C\mu(b)+\mu(b).$

It follows that

$Cx\leq C(\mu(x_{+})-\mu(x_{-}))+\mu(b).$

Similarly,

$Cx\geq C(\mu(x_{+})-\mu(x_{-}))-\mu(a)$

and the assertion follows. ∎

######  Lemma 56.

Let $E=E(0,\infty)$ be a symmetric space on the semi-axis. If
$x\in\mathcal{D}_{E}$ is such that $Cx\in E,$ then $x\in Z_{E}.$

###### Proof.

Define a partition $\mathcal{A}=\\{(2^{n},2^{n+1})\\}_{n\in\mathbb{Z}}$ and
set $x_{1}=\mathbf{E}(x|\mathcal{A}).$ If $x=\mu(a)-\mu(b)$ with $a,b\in E,$
then $x_{1}=\mathbf{E}(\mu(a)|\mathcal{A})-\mathbf{E}(\mu(b)|\mathcal{A}).$
Clearly,

$\mathbf{E}(\mu(a)|\mathcal{A})\leq\sigma_{2}\mu(a)\in
E,\quad\mathbf{E}(\mu(b)|\mathcal{A})\leq\sigma_{2}\mu(b)\in E$

are decreasing functions. It follows that $x_{1}\in\mathcal{D}_{E}.$ It is
easy to see that

$|Cx_{1}-Cx|\leq 2\sigma_{2}(\mu(a)+\mu(b)).$

Therefore, $Cx_{1}\in E.$ Define a function $z\in E$ by setting

$z(t)=(Cx_{1})(2^{n+1}),\quad t\in(2^{n},2^{n+1}).$

Clearly, $x_{1}=2z-\sigma_{2}z\in Z_{E}.$

Consider the function $x-x_{1}$ on the interval $(2^{n},2^{n+1}).$ By Kwapien
theorem [23], there exist positive equimeasurable functions $y_{1n},y_{2n}$
supported on $(2^{n},2^{n+1})$ such that

$\mu(y_{1n})=\mu(y_{2n}),\quad\|y_{1n}\|_{\infty},\|y_{2n}\|_{\infty}\leq
6\|(x-x_{1})\chi_{(2^{n},2^{n+1})}\|_{\infty}.$

Set $y_{1}=\sum_{n\in\mathbb{N}}y_{1n}$ and
$y_{2n}=\sum_{n\in\mathbb{N}}y_{2n}.$ It follows that $y_{1},y_{2}\in E_{+}.$
Since $x-x_{1}=y_{1}-y_{2}$ and $\mu(y_{1})=\mu(y_{2}),$ it follows that
$x-x_{1}\in Z_{E}.$ The assertion follows immediately. ∎

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