††thanks:  The work of the third and fourth authors was supported by the
Australian Research Council.

# Logarithmic submajorization, uniform majorization and Hölder type
inequalities for $\tau$-measurable operators

PG Dodds
College of Science and Engineering, Flinders University
GPO Box 2100, Adelaide 5001, Australia<EMAIL_ADDRESS>TK Dodds
College of Science and Engineering, Flinders University
GPO Box 2100, Adelaide 5001, Australia<EMAIL_ADDRESS>FA
Sukochev
School of Mathematics and Statistics, UNSW
Kensington, NSW 2052, Australia
<EMAIL_ADDRESS>D.Zanin
School of Mathematics and Statistics, UNSW
Kensington, NSW 2052, Australia
<EMAIL_ADDRESS>Dedicated to the memory of W.A.J. Luxemburg

###### Abstract.

We extend the notion of the determinant function $\Lambda$, originally
introduced by T.Fack for $\tau$-compact operators, to a natural algebra of
$\tau$-measurable operators affiliated with a semifinite von Neumann algebra
which coincides with that defined by Haagerup and Schultz in the finite case
and on which the determinant function is shown to be submultiplicative.
Application is given to Hölder type inequalities via general Araki-Lieb-
Thirring inequalities due to Kosaki and Han and to a Weyl-type theorem for
uniform majorization.

###### Key words and phrases:

measurable operators, logarithmic submajorisation, Fuglede-Kadison theorem,
uniform majorisation, semifinite von Neumann algebra

###### 1991 Mathematics Subject Classification:

Primary 46L52; 47A63 Secondary 46E30,47A30

##  1\. Introduction

In their study of Brown measures of unbounded operators affiliated with a
finite von Neumann algebra $({\mathcal{M}},\tau)$, Haagerup and Schultz
extended the classical Fuglede-Kadison determinant

$\Delta(x)=\exp\left(\int_{0}^{\tau({\bf 1})}\log t\ d(\tau
e^{|x|})(s)\right),\quad x\in{\mathcal{M}}$

to a multiplicative functional $\Delta$ on the subalgebra
${\mathcal{M}}^{\Delta}$ consisting of those $\tau$-measurable operators $x$
affiliated with ${\mathcal{M}}$ for which

$\tau(\log_{+}|x|)=\int_{0}^{\infty}\log_{+}td(\tau e^{|x|})(t)<\infty$

Here $e^{|x|}(\cdot)=\chi_{(\cdot)}(|x|)$ is the spectral measure of $|x|$ and
$\tau e^{|x|}(\cdot)=\tau(e^{|x|}(\cdot))$ is the corresponding Borel measure
supported on the spectrum $\sigma(|x|)$ of $|x|$.

A principal aim of this paper is to extend the Haagerup-Schultz approach from
the setting of finite von Neumann algebras to the semifinite case. Our
approach is via the determinant function $\Lambda(\cdot)$ given by setting

$\Lambda(t;x)=\exp\left(\int_{0}^{t}\log\mu(s;x)\,ds\right),\quad t>0,\quad
x\in L_{\log_{+}}(\tau).$

Here $\mu(\cdot;x)$ is the generalised singular value function (see below) and
$L_{\log_{+}}(\tau)$ is the natural extension of the algebra
${\mathcal{M}}^{\Delta}$ of Haagerup-Schults to the general semifinite
setting. The determinant function $\Lambda$ goes back to A.Grothendieck [20]
and was studied by Fack [14],[15] in the case of $\tau$-compact operators and
subsequently by Fack and Kosaki [17] for operators satisfying a ”Lorentz
space”-type condition(see the definition of this condition in the domments
following Proposition 3.1 below).

Basic properties of the class $L_{\log_{+}}(\tau)$ are studied in the third
section where it is noted that this class is an algebra properly containing
the space of $\tau$-compact operators as well as those $\tau$-measurable
operators satisfying a ”Lorentz space”-type condition. One of the principal
results of the paper is given in Theorem 4.2 which establishes the
submultiplicativity of the determinant $\Delta$ on the algebra
$L_{\log_{+}}(\tau)$. A principal ingredient of the proof is the
multiplicativity of the Haagerup-Schultz functional $\Delta$ on the algebra
${\mathcal{M}}^{\Delta}$.

In subsequent sections, attention is directed to Hölder type inequalities for
$\tau$-measurable operators. Principal results in Section 5 include a
submajorization inequality of Hölder type given in Proposition 5.5 and
corresponding symmetric norm inequalities given in Proposition 5.7, which are
established using techniques based on logarithmic submajorization. See, for
example, Hiai [24] in the finite-dimensional setting. While these results
extend inequalities for unitarily invariant norms given in [29] Theorem 3, the
techniques given there do not extend to the more general setting and a crucial
tool is a strengthened version [22] given in Proposition 5.1 of a
submajorization inequality of Araki-Lieb-Thirring type due to Kosaki [30] in
the setting of semi-finite von Neumann algebras and based on properties of the
Fuglede-Kadison determinant. This section then concludes with a convexity
inequality (Proposition 5.11) which is due to Hiai and Zhan [HZ] in the matrix
setting for unitarily invariant norms.

We show in Section 6 (Theorem 6.3) that Weyl’s submajorization inequality for
the singular values of a product ( [17] Theorem 4.2(iii)) continues to hold
for uniform majorization, a notion introduced by Kalton and Sukochev [28]
which is a strengthening of the classical notion of submajorization in the
sense of Hardy, Littlewood and Polya. This permits us to give a simple proof
of generalized Hölder inequalities for general symmetric spaces proved
recently by Bekjan and Ospanov [2] and Albadawi [1].

This paper is dedicated to the memory of Wim Luxemburg. It is with great
affection that the first two authors recall his Caltech lectures on the theory
of integration which introduced them to the paper of Grothendieck, and to the
ideas of decreasing rearrangements of operators, and the determinant function
which plays a central role in this paper.

##  2\. Preliminaries and notation

Throughout this paper $\mathcal{M}\subseteq{\mathcal{B}}({\mathcal{H}})$ will
denote a semifinite von Neumann algebra on some Hilbert space ${\mathcal{H}}$
(here, ${\mathcal{B}}({\mathcal{H}})$ is the algebra of all bounded linear
operators on ${\mathcal{H}}$ equipped with the operator norm denoted by
$\|\cdot\|_{\infty}$ or $\|\cdot\|_{{\mathcal{B}}({\mathcal{H}})}$). Unless
otherwise stated, it will be assumed throughout that $\mathcal{M}$ is equipped
with a fixed faithful normal semifinite trace $\tau$. For standard facts
concerning von Neumann algebras, we refer to [6], [36]. The identity in
$\mathcal{M}$ is denoted by 1 and we denote by
${\mathcal{P}}\left({\mathcal{M}}\right)$ the complete lattice of all (self-
adjoint) projections in $\mathcal{M}$. A linear operator
$x:{\mathcal{D}}(x)\to{\mathcal{H}}$, with domain
${\mathcal{D}}(x)\subseteq{\mathcal{H}}$, is said to be affiliated with
$\mathcal{M}$ if $yx\subseteq xy$ for all $y$ in the commutant
$\mathcal{M}^{\prime}$ of $\mathcal{M}$ (equivalently, $ux=xu$ for all unitary
$u$ in $\mathcal{M}^{\prime}$). . For any self-adjoint operator $x$ on
$\mathcal{H}$, its spectral measure is denoted by $e^{x}$. A self-adjoint
operator $x$ is affiliated with $\mathcal{M}$ if and only if
$e^{x}\left(B\right)\in\mathcal{P}\left(\mathcal{M}\right)$ for any Borel set
$B\subseteq\mathbb{R}$.

The closed and densely defined operator $x$, affiliated with $\mathcal{M}$, is
called $\tau$-measurable if and only if there exists a number $s\geq 0$ such
that

$\tau\left(e^{\left|x\right|}(s,\infty)\right)<\infty.$

The collection of all $\tau$-measurable operators is denoted by $S(\tau)$.
With the sum and product defined as the respective closures of the algebraic
sum and product, it is well known that $S(\tau)$ is a *-algebra. For
$\epsilon,\ \delta>0$, we denote by $V(\epsilon,\delta)$ the set of all $x\in
S(\tau)$ for which there exists an orthogonal projection $p\in P(\mathcal{M})$
such that $p({\mathcal{H}})\subseteq{\mathcal{D}}(x),\
\|xp\|_{{\mathcal{B}}({\mathcal{H}})}\leq\epsilon$ and
$\tau({\textbf{1}}-p)\leq\delta$. The sets $\\{V(\epsilon,\delta):\epsilon,\
\delta>0\\}$ form a base at $0$ for a metrizable Hausdorff topology on
$S(\tau)$, which is called the measure topology. Equipped with this topology,
$S(\tau)$ is a complete topological $*$-algebra. These facts and their proofs
can be found in the papers [32], [37]. See also [12]

The collection of all closed, densely defined operators $x$ in $H$, affiliated
with the von Neumann algebra $\mathcal{M}$ and satisfying
$\tau\left(e^{\left|x\right|}\left(\lambda,\infty\right)\right)<\infty$ for
all $\lambda>0$, will be denoted by $S_{0}\left(\tau\right)$. The elements of
$S_{0}\left(\tau\right)$ are sometimes called $\tau$-compact operators.
Evidently, each $x\in S_{0}\left(\tau\right)$, is $\tau$-measurable, that is,
$S_{0}\left(\tau\right)\subseteq S\left(\tau\right)$.

For $x\in S(\tau)$, the (generalised) singular value function
$\mu(x)=\mu(\cdot;x)=\mu(\cdot;|x|)$ is defined by

$\mu\left(t;x\right):=\inf\left\\{s\geq 0:d(s;|x|)\leq t\right\\},\quad t\geq
0,$

where

$d(s;|x|):=\tau\left(e^{\left|x\right|}\left(s,\infty\right)\right),\quad
s\geq 0.$

It follows directly that the singular value function $\mu(x)$ is a decreasing,
right-continuous function on the positive half-line $[0,\infty)$. Moreover,
$\mu(uxv)\leq\|u\|_{\infty}\|v\|_{\infty}\mu(x)$ for all $u,v\in\mathcal{M}$
and $x\in S(\tau)$ and $\mu(f(x))=f(\mu(x))$ whenever $0\leq x\in S(\tau)$ and
$f$ is an increasing continuous function on $[0,\infty)$ which satisfies
$f(0)=0$.

It should be observed that a sequence $\left\\{x_{n}\right\\}_{n=1}^{\infty}$
in $S\left(\tau\right)$ converges to zero for the measure topology if and only
if $\mu\left(t;x_{n}\right)\rightarrow 0$ as $n\rightarrow\infty$ for all
$t>0$.

If $m$ denotes Lebesgue measure on the semiaxis $[0,\infty)$, and if we
consider $L^{\infty}(m)$ as an Abelian von Neumann algebra acting via
multiplication on the Hilbert space ${\mathcal{H}}=L^{2}(m)$, with the trace
given by integration with respect to $m$, then $S(m)$ consists of all
measurable functions on $[0,\infty)$ which are bounded except on a set of
finite measure. In this case for $f\in S(m)$, the generalized singular value
function $\mu(f)$ is precisely the classical decreasing rearrangement of the
function $|f|$, which is usually denoted by $f^{*}$. In this setting,
convergence for the measure topology coincides with the usual notion of
convergence in measure.

If $\mathcal{M}={\mathcal{B}}({\mathcal{H}})$ and $\tau$ is the standard
trace, then $S(\tau)=\mathcal{M}$, the measure topology coincides with the
operator norm topology. If $x\in S(\tau)$, then $x$ is compact if and only if
$\lim_{t\to\infty}\mu(t;x)=0$; in this case,

$\mu_{n}(x)=\mu\left(t;x\right),\quad t\in[n,n+1),\quad n=0,1,2,\dots,$

and the sequence $\\{\mu_{n}(x)\\}_{{}_{n=0}}^{\infty}$ is just the sequence
of eigenvalues of $|x|$ in non-increasing order and counted according to
multiplicity.

The real vector space $S_{h}\left(\tau\right)=\left\\{x\in
S\left(\tau\right):x=x^{\ast}\right\\}$ is a partially ordered vector space
with the ordering defined by setting $x\geq 0$ if and only if $\left\langle
x\xi,\xi\right\rangle\geq 0$ for all $\xi\in\mathcal{D}\left(x\right)$. The
positive cone in $S_{h}\left(\tau\right)$ will be denoted by
$S\left(\tau\right)^{+}$. If $0\leq x_{\alpha}\uparrow_{\alpha}\leq x$ holds
in $S\left(\tau\right)^{+}$, then $\sup_{\alpha}x_{\alpha}$ exists in
$S\left(\tau\right)^{+}$. The trace $\tau$ extends to $S\left(\tau\right)^{+}$
as a non-negative extended real-valued functional which is positively
homogeneous, additive, unitarily invariant and normal. This extension is given
by

$\tau\left(x\right)=\int_{0}^{\infty}\mu\left(t;x\right)dt,\ \ \ x\in
S\left(\tau\right)^{+},$

and satisfies $\tau\left(x^{\ast}x\right)=\tau\left(xx^{\ast}\right)$ for all
$x\in S\left(\tau\right)$. It should be observed that if $f$ is an increasing
continuous function on $\left[0,\infty\right)$ satisfying $f\left(0\right)\geq
0$, then

$f(\mu(x))=\mu(f(|x|))$ (2.1)

and if $f(0)=0$ then

$\tau\left(f\left(\left|x\right|\right)\right)=\int_{0}^{\infty}\mu\left(t;f\left(\left|x\right|\right)\right)dt=\int_{0}^{\infty}f\left(\mu\left(t;x\right)\right)dt$
(2.2)

for all $x\in S\left(\tau\right)$.

If $x,y\in S(\tau)$ then $x$ is said to be submajorised by $y$ (in the sense
of Hardy, Littlewood and Polya) if and only if

$\int_{0}^{t}\mu(s;x)ds\leq\int_{0}^{t}\mu(s;y)ds$

for all $t\geq 0$. We write $x\prec\prec y$, or equivalently,
$\mu(x)\prec\prec\mu(y)$.

A linear subspace $E\subseteq S(\tau)$, equipped with a norm $\|\cdot\|_{E}$
will be called symmetrically normed if $x\in E$, $y\in S\left(\tau\right)$ and
$\mu\left(y\right)\leq\mu\left(x\right)$ imply that $y\in E$ and
$\left\|y\right\|_{E}\leq\left\|x\right\|_{E}$. If a symmetrically normed
space is Banach, then it will be simply called a  symmetric space. If
$\mathcal{M}$ is $L_{\infty}(m)$, then a symmetrically normed space
$E\subseteq S(m)$ will be called, for simplicity, a symmetrically normed space
on $[0,\infty)$. If $E\subseteq S(\tau)$ is a symmetrically normed space, then
the embedding of $E$ into $S(\tau)$ is continuous from the norm topology of
$E$ to the measure topology on $S(\tau)$.

A wide class of non-commutative symmetrically normed spaces may be constructed
as follows. If $E\subseteq S(m)$ is symmetrically normed space on
$[0,\infty)$, set

$E\left(\tau\right)=\left\\{x\in S\left(\tau\right):\mu\left(x\right)\in
E\right\\},\quad\|x\|_{E(\tau)}:=\|\mu(x)\|_{E}$

It may be shown ([28]; see also [9], [11]), that
$(E(\tau),\|\cdot\|_{E(\tau)})$ is symmetrically normed and is a Banach space
if $E$ is a Banach space.

Important special cases of this construction occur when $E$ is taken to be the
familiar Lebesgue space $L_{p},1\leq p\leq\infty$. In this paricular case, the
spaces $E(\tau)=L_{p}(\tau)$ are the familiar non-commutative $L_{p}$-spaces
and in this setting we have that $L_{\infty}(\tau)={\mathcal{M}}$. In the
special case that ${\mathcal{M}}$ is $B({\mathcal{H}})$ the corresponding non-
commutative $L_{p}$ spaces are the familiar Schatten classes
${\mathfrak{S}}_{p}$. As is well-known, the space $L_{1}(\tau)$ may be
identified with the von Neumann algebra predual ${\mathcal{M}}_{*}$ of
${\mathcal{M}}$ with respect to trace duality. We observe that, whenever
$E\subseteq S(m)$ is a symmetric space, then the continuous inclusions

$L_{1}(\tau)\cap{\mathcal{M}}\subseteq E(\tau)\subseteq
L_{1}(\tau)+{\mathcal{M}}$

hold and that, if $x\in S(\tau)$, then $x\in L_{1}(\tau)+{\mathcal{M}}$ if and
only if $\int_{0}^{t}\mu(s;x)ds<\infty$ for at least one $t>0$ or,
equivalently, for all $t>0$.

For further details and proofs, we refer the reader to [9], [7], [28],
[11],[12],[31].

##  3\. The algebra $L_{\log_{+}}(\tau)$

Throughout, we will write $L_{1},\ L_{\infty}$ rather than
$L_{1}(m),L_{\infty}(m)$ for the usual Lebesgue spaces on the semiaxis
$[0,\infty)$ equipped with Lebesgue measure $m$. We set

$L_{\log_{+}}(\tau):=\\{x\in S(\tau):\log_{+}|x|\in
L_{1}(\tau)+\mathcal{M}\\}$ (3.1)

Here,

$\log_{+}t=\max\\{\log t,0\\},\quad t>0.$

Observing that

$\mu(\log_{+}|x|)=\log_{+}\mu(x),\quad x\in S(\tau),$

it follows that, if $x\in S(\tau)$, then

$\displaystyle x\in L_{\log_{+}}(\tau)\quad$
$\displaystyle\iff\quad\int_{0}^{t}\log_{+}\mu(x)dm<\infty,\quad\forall t>0$
(3.2) $\displaystyle\iff\quad\log_{+}\mu(x)\in L_{1}+L_{\infty}$ (3.3)
$\displaystyle\iff\quad\int_{0}^{1}\log_{+}\mu(x)dm<\infty.$ (3.4)

It is clear that if $x\in L_{\log}(\tau)$, then $x^{*},|x|,uxv\in
L_{\log}(\tau)$ for all unitary $u,v\in{\mathcal{M}}$; moreover, if $x\geq 0$
and $\alpha>0$ then the equality $\mu(x^{\alpha})=\mu(x)^{\alpha}$ readily
implies that $x^{\alpha}\in L_{\log}(\tau)$.

If $\tau({\bf 1})<\infty$, then it follows from [12] Chapter III , remarks
preceding Lemma 3.6, that

$\tau(\log_{+}|x|)=\int_{0}^{\tau({\bf
1})}\log_{+}\mu(x)dm=\int_{0}^{\tau({\bf 1})}\log_{+}sd(\tau e^{|x|})(s).$

Consequently, if $\tau({\bf 1})<\infty$, then the class $L_{\log_{+}}(\tau)$
coincides with the class ${\mathcal{M}}^{\Delta}$ introduced by Haagerup and
Schultz [21].

We now observe that the class $L_{\log_{+}}(\tau)$ is an algebra. First recall
that, if $s>0$ then the dilation operator $\sigma$ is defined on $S(m)$ by
setting

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

If $x,y\in S(\tau)$, then

$\mu(x+y)\leq\sigma_{2}(\mu(x))+\sigma_{2}(\mu(y)),\quad\mu(xy)\leq\sigma_{2}(\mu(x))\sigma_{2}(\mu(y))$
(3.5)

See, for example, [17] Lemma 2.5. See also [31].

######  Proposition 3.1.

If $x,y\in L_{\log_{+}}(\tau)$, then $x+y,xy\in L_{\log_{+}}(\tau)$.

###### Proof.

If $x,y\in L_{\log_{+}}(\tau)$ then, using (3.5),

$\mu(x+y)\leq\sigma_{2}(\mu(x))+\sigma_{2}(\mu(y))\leq\max\\{2\sigma_{2}(\mu(x)),2\sigma_{2}(\mu(y))\\}.$

Consequently,

$\displaystyle\log_{+}(\mu(x+y))$
$\displaystyle\leq\log_{+}\left(\max\\{2\sigma_{2}(\mu(x)),2\sigma_{2}(\mu(y))\\}\right)$
$\displaystyle=\max\\{\log_{+}(2\sigma_{2}(\mu(x)),\log_{+}(2\sigma_{2}(\mu(y))\\}$
$\displaystyle\leq\log_{+}(2\sigma_{2}(\mu(x)))+\log_{+}(2\sigma_{2}(\mu(y)))$
$\displaystyle=\sigma_{2}(\log_{+}(2\mu(x))+\log_{+}(2\mu(y)))$
$\displaystyle\leq\sigma_{2}(\log_{+}\mu(x)+\log_{+}\mu(y)+2\log 2)$

Since

$\log_{+}\mu(x),\log_{+}\mu(y)\in L_{1}+L_{\infty},$

as follows from (3.2), it follows that

$\sigma_{2}(\log_{+}\mu(x)+\log_{+}\mu(y)+2\log 2)\in L_{1}+L_{\infty}.$

This implies that

$\log_{+}\mu(x+y)\in L_{1}+L_{\infty},$

and accordingly, $x+y\in L_{\log_{+}}(\tau)$.

To see that $xy\in L_{\log_{+}}(\tau)$, observe first, again using (3.5), that

$\mu(xy)\leq\sigma_{2}(\mu(x)\mu(y))=\sigma_{2}(\mu(x))\sigma_{2}(\mu(y)).$

Consequently,

$\displaystyle\log_{+}\mu(xy)$
$\displaystyle\leq\log_{+}\left(\sigma_{2}(\mu(x))\sigma_{2}(\mu(y))\right)=\sigma_{2}(\log_{+}\mu(x)+\log_{+}\mu(y))$
$\displaystyle\in L_{1}+L_{\infty},$

and this implies that $xy\in L_{\log_{+}}(\tau)$.

Finally, since $\mu(x)=\mu(x^{*})$, the final assertion of the proposition is
trivial.

∎

It should be noted that if $x\in S(\tau)$ satisfies a (so-called) ”Lorentz
space”-type condition (see [17])

$x\in{\mathcal{M}}\quad{\rm or}\quad\mu(s;x)\leq
Cs^{-\alpha}\quad(C,\alpha>0),\ s>0$ (3.6)

then $x\in L_{\log_{+}}(\tau)$. Indeed, this assertion is clear if
$x\in{\mathcal{M}}$. Assume then that $x\in S(\tau)$ and that $\mu(s;x)\leq
Cs^{-\alpha}$ for some $C,\alpha>0$ and all $s>0$ and for simplicity, suppose
that $C=1$. Observe that

$\int_{0}^{1}\log_{+}\mu(s;x)ds\leq\int_{0}^{1}\log_{+}s^{-\alpha}ds=-\alpha\int_{0}^{1}\log
sds<\infty$

and this suffices to show that $x\in L_{\log_{+}}(\tau)$. In particular, it
follows that if $x\in L^{p}(\tau)$ for some $0<p<\infty$, then $x\in
L_{\log_{+}}(\tau)$. Indeed, as noted in [17], in this case

$\mu(s;x)^{p}\leq s^{-1}\int_{0}^{s}\mu(x)^{p}dm\leq s^{-1}\|x\|_{p}^{p}$

for all $s>0$.

On the other hand, if $x\in L_{\log_{+}}(\tau)$, then it need not be the case
that $x$ satisfies any Lorentz space condition. By way of example, if
$f(t)=\exp(t^{-\frac{1}{2}}),t>0$, then it is evident that $f\in
L_{\log_{+}}$. However, since

$\lim_{t\to 0}t^{-\alpha}\exp(-t^{-{\frac{1}{2}}})=0,\quad\forall\alpha>0,$

it follows that $f$ fails to satisfy any condition of the type (3.6).

One immediate consequence of the preceding is that

$L_{1}(\tau)+{\mathcal{M}}\subseteq L_{\log_{+}}(\tau)\subseteq S(\tau).$

Note that the inclusion on the left hand side is, in general, proper by
observing that, if $f(t)=1/t,t>0$, then $f\in L_{\log_{+}}$ but $\mu(f)\not\in
L_{1}+L_{\infty}$.

The proposition which follows is due to L.G.Brown [5] Proposition 1.11. It is
an extension of the Weyl majorization theorem given in [14] Corollary 4.2
which goes back to the paper of Weyl[33].

######  Proposition 3.2.

Suppose that $f,g:(0,\infty)\to[0,\infty)$ be non-increasing. If $f,g\in
L_{log_{+}}$ then the following statements are equivalent:

$\displaystyle(i)\quad\int_{0}^{t}\log f(s)ds$
$\displaystyle\leq\int_{0}^{t}\log g(s)ds,\quad\forall t>0.$
$\displaystyle(ii)\quad\int_{0}^{\infty}\log_{+}(rf(s))ds$
$\displaystyle\leq\int_{0}^{\infty}\log_{+}(rg(s))ds,\quad\forall
r>0,\quad\forall t>0.$ $\displaystyle(iii)\quad\int_{0}^{t}\varphi(f(s))ds$
$\displaystyle\leq\int_{0}^{t}\varphi(g(s))ds,\quad\forall t>0,$

for all continuous increasing functions $\varphi:[0,\infty)\to[0,\infty)$ such
that $\varphi(0)=0$ and $\varphi\circ\exp$ is convex.

######  Remark 3.3.

It should be observed that a simple consequence of Proposition 3.2 preceding
is that if $x,y\in L_{\log_{+}}(\tau)$ and if $x$ is logarithmically
submajorised by $y$ in the sense that

$\int_{0}^{t}\log\mu(s;x)ds\leq\int_{0}^{t}\log\mu(s;y)ds,\quad\forall t>0,$

then $x$ is submajorised by $y$ in the usual sense, that is,

$\int_{0}^{t}\mu(s;x)ds\leq\int_{0}^{t}\mu(s;y)ds\quad\forall t>0.$

Indeed, one need only take $\varphi(t)=t,t\geq 0$.

##  4\. The determinant function

If $x\in L_{\log_{+}}(\tau)$ then the determinant function $\Lambda(x)$ is
defined by setting, for all $t>0$,

$\Lambda(t;x)=\exp\left(\int_{0}^{t}\log\mu(s;x)\,ds\right).$

Since $x\in L_{\log_{+}}(\tau)$, it follows that the integral is well-defined
in the sense that it is either finite or takes the value $-\infty$. Indeed,

$\Lambda(t;x)=0\Leftrightarrow\int_{0}^{t}\log\mu(s;x)ds=-\infty$

In the case that $x$ is $\tau$-compact, this function, which goes back to
A.Grothendieck [20], was studied by T.Fack [14], [15] and extended further in
[17] to the class of operators satisfying a ”Lorentz space”-type condition.

We remark that, if $\tau({\bf 1})<\infty$, then $\Delta(x):=\Lambda(\tau({\bf
1});x)$ is precisely the extension of the classical Fuglede-Kadison
determinant [16] to $S(\tau)$ given by Haagerup and Schultz [21] who proved
the following fundamental result.

######  Proposition 4.1.

If $\tau({\bf 1})<\infty$ and if $x,y\in L_{\log_{+}}(\tau)$, then

$\Delta(xy)=\Delta(x)\Delta(y).$

Let us note that if $\mathcal{M}=B(H)$ with standard trace and if
$x\in\mathcal{M}$ is compact then

$\Lambda(n+1;x)=\prod_{k=0}^{n}\mu(k;x).$

We note first some elementary properties of the determinant function $\Lambda$
which follow directly from the definition and properties of the singular value
function (see [17] Lemma 2.5) together and the definition (3.1).

  1. (1)

If $0\leq x\in L_{\log_{+}}(\tau)$ and $\alpha>0$, then $x^{\alpha}\in
L_{\log_{+}}(\tau)$ and $\Lambda(x^{\alpha})=\Lambda(x)^{\alpha}.$

  2. (2)

If $x\in L_{\log_{+}}(\tau)$, then $x^{*},|x|,x^{*}x\in L_{\log_{+}}(\tau)$
and

$\Lambda(x)=\Lambda(|x|)=\Lambda(x^{*})=\Lambda(x^{*}x)^{1/2}.$

  3. (3)

If $x\in L_{\log_{+}}(\tau)$, then

$\Lambda(x^{*}x)=\Lambda(xx^{*}).$

  4. (4)

If $x,y\in L_{\log_{+}}(\tau)$ then

$\Lambda(xy)=\Lambda(|x||y^{*}|).$

  5. (5)

If $x\in L_{\log_{+}}(\tau)$, then $uxv\in L_{\log_{+}}(\tau)$ for all unitary
$u,v\in{\mathcal{M}}$, in which case

$\Lambda(uxv)=\Lambda(x).$

For example, to indicate the proof of 4. above, it suffices to show that

$\mu(xy)=\mu(|x||y^{*}|).$

To this end, let $x=u|x|$ be the polar decomposition, and observe that

$\displaystyle\mu(xy)$ $\displaystyle=\mu(u|x|y)\leq\mu(|x|y)$
$\displaystyle=\mu(y^{*}|x|)\leq\mu(|y^{*}||x|)=\mu(|x||y^{*}|)$
$\displaystyle=\mu(u^{*}x|y^{*}|)\leq\mu(x|y^{*}|)$
$\displaystyle=\mu(|y^{*}|x^{*})=\mu(y^{*}x^{*})=\mu(xy),$

using repeatedly [17] Lemma 2.5 (vi).

The Theorem which follows is the principal result of this section and is a
refinement of [17] Theorem 4.2 (ii). See also [20], Théoréme 1, where it is
stated for the case that $a,b$ are compact operators in some Hilbert space.

######  Theorem 4.2.

(Weyl inequality) If $x,y\in L_{\log_{+}}(\tau)$, then

$\Lambda(t;xy)\leq\Lambda(t;x)\Lambda(t;y),\quad\forall t>0.$ (4.1)

We shall need the following geometric characterisation of the singular value
function: If $x\in S(\tau)$ and if $t>0$, then

$\mu(t;x)=\inf\\{\|x-z\|_{\infty}:\tau(s(|z|))\leq t\\}.$ (4.2)

See, for example, [17] Proposition 2.4.

We prove first the following result.

######  Lemma 4.3.

If $a,b\in L_{\log_{+}}(\tau)$ are such that $a,b\geq{\bf 1}$, if
$d(1;a),d(1;b)<\infty$ and if $r=e^{a}(1,\infty)\vee e^{b}(1,\infty)$, then

$\Lambda(t;ab)=\Lambda(t;a)\Lambda(t;b),\quad\forall t\geq\tau(r).$

###### Proof.

We set

$p=\chi_{(1,\infty)}(a)=e^{a}(1,\infty),\quad
q=\chi_{(1,\infty)}(b)=e^{b}(1,\infty),\quad r=p\vee q,$

and note that

$\tau(r)\leq\tau(p)+\tau(q)=d(1;a)+d(1;b)<\infty.$

Since $a,b\geq 1$, it follows that

${\bf 1}-p=\chi_{[0,1]}(a)=\chi_{\\{1\\}}(a),\quad{\bf 1}-q=\chi_{\\{1\\}}(b)$

and so

${\bf 1}-r=({\bf 1}-p)\wedge({\bf
1}-q)=\chi_{\\{1\\}}(a)\wedge\chi_{\\{1\\}}(b).$

Consequently,

$a({\bf 1}-r)=a\chi_{\\{1\\}}(a)({\bf 1}-r)=\chi_{\\{1\\}}(a)({\bf 1}-r)=({\bf
1}-r)$ (4.3)

and

$({\bf 1}-r)a=(a({\bf 1}-r))^{*}=({\bf 1}-r).$ (4.4)

Similar reasoning applies with $a$ replaced by $b$ so that also

$({\bf 1}-r)b=(b({\bf 1}-r))^{*}=({\bf 1}-r)$ (4.5)

and it follows, in particular, that $a,b$ commute with $r$.

Using (4.3), (4.5), note further that

$ab=rab+({\bf 1}-r)ab=rab+({\bf 1}-r).$ (4.6)

It will now be shown that

$\mu(t;a)=\mu(t;rar),\quad\mu(t;b)=\mu(t;rbr),\quad t\in(0,\tau(r)).$ (4.7)

Without loss of generality, it may be assumed that $\tau(r)=1$. By [9] Lemma
2.4 (i), it follows that

$\displaystyle\mu(t;a)$ $\displaystyle=\mu(t;pap)=\mu(t;p\cdot rar\cdot p)$
$\displaystyle\leq\mu(t;rar)\leq\mu(t;a),\quad t\in(0,\tau(p)),$

so that

$\mu(t;rar)=\mu(t;a),\quad t\in(0,\tau(p)).$

Similarly,

$\mu(t;rbr)=\mu(t;b),\quad t\in(0,\tau(q)).$

If $\tau(p)=1$, then the first equality in (4.7) follows immediately, and if
$\tau(q)=1$, then the second equality in (4.7) holds. Consequently, it may be
assumed that $\tau(p)<1$ and $\tau(q)<1$.

Suppose now that $\tau(p)\leq t<1$. Observing that $\tau(s(pa))\leq\tau(p)\leq
t$, it follows from (4.2) that

$\mu(t;a)\leq\|a-pa\|_{\infty}=\|a\chi_{[0,1]}(a)\|_{\infty}\leq 1.$

Since $a\geq{\bf 1}$, it follows that $\mu(t;a)\geq 1$ for all $t>0$ and it
now follows that

$\mu(t;a)=1\quad\forall t\geq\tau(p).$ (4.8)

The same argument shows that

$\mu(t;b)=1\quad\forall t\geq\tau(q).$ (4.9)

In particular, $\mu(t;a)=1$ for all $\tau(p)\leq t<1$ and $\mu(t;b)=1$ for all
$\tau(q)\leq t<1$.

At the same time, since $a$ commutes with $r$, it follows that $a\geq rar\geq
r$ since $a\geq 1$. This implies that

$\mu(a)\geq\mu(rar)\geq\mu(r)=\chi_{[0,\tau(r))}.$

Consequently,

$1=\mu(t;a)\geq\mu(t;rar)\geq 1,\quad\tau(p)\leq t<\tau(r)=1$

so that

$\mu(t;rar)=1=\mu(t;a),\quad\tau(p)\leq t<1.$

Similarly,

$\mu(t;rbr)=1=\mu(t;a),\quad\tau(q)\leq t<1.$

This suffices to complete the proof of the equalities in (4.7)

It will now be shown that

$\mu(t;ab)=\mu(t;rabr)=\mu(t;rar\cdot rbr),\quad 0<t<1.$ (4.10)

Note first that

$\displaystyle\begin{split}\mu(rabr)=&\mu(rar\cdot
rbr)=\mu^{\frac{1}{2}}(rbr(rar)^{2}rbr)\\\ &\geq\mu^{\frac{1}{2}}(rbr\cdot
r\cdot rbr)=\mu(rbr)\geq\mu(r)=\chi_{[0,1)}\end{split}$ (4.11)

Observe next that $r$ commutes with $|rabr|$, since it is clear that $r$
commutes with $|rabr|^{2}=rb^{*}a^{*}rabr=rba^{2}br$. From this it follows
that

$r|rabr|r=|rabr|.$ (4.12)

Indeed,

$\displaystyle(r|rabr|r)^{2}$ $\displaystyle=r|rabr|r|rabr|r=r|rabr|^{2}r$
$\displaystyle=rb^{*}a^{*}rabr=rb^{*}a^{*}abr=|rabr|^{2}$

and the equality (4.12) follows by passing to square roots. We may now show
that

$|ab|=r|rabr|r+({\bf 1}-r)=|rabr|+({\bf 1}-r).$ (4.13)

In fact, using (4.6) and (4.12), observe that

$\displaystyle|ab|^{2}$ $\displaystyle=(ab)^{*}(ab)=rb^{*}a^{*}rabr+({\bf
1}-r)$ $\displaystyle=|rabr|^{2}+({\bf 1}-r)=(r|rabr|r)^{2}+({\bf 1}-r)$
$\displaystyle=(r|rabr|r+({\bf 1}-r))^{2}.$

Passing to square roots and again using (4.12) now yields

$\displaystyle|ab|=r|rabr|r+({\bf 1}-r)=|rabr|+({\bf 1}-r).$

Observe now, that if $s>0$, then, using (4.13)

$\displaystyle d(s;ab)$ $\displaystyle=d(s;|rabr|+({\bf 1}-r))$
$\displaystyle=\begin{cases}d(s;|rabr|),&\ s\geq 1,\\\ d(s;|rabr|)+d(s;({\bf
1}-r)),&\ 0<s<1.\end{cases}$ $\displaystyle=\begin{cases}d(s;|rabr|),&\ s\geq
1,\\\ \infty,&\ 0<s<1.\end{cases}$

It follows readily that

$\displaystyle\mu(t;ab)$ $\displaystyle=\inf\\{s\geq 0:d(s;ab)\leq t\\}$
$\displaystyle=\mu(rabr)\chi_{[0,1)}+\chi_{[1,\infty)},$

and this yields (4.10). In particular, it follows also that

$\mu(t;ab)=1,\quad t\geq 1.$ (4.14)

Finally, if $t\geq\tau(r)=1$, then

$\displaystyle\Lambda(t;ab)$
$\displaystyle=\exp\int_{0}^{t}\log\mu(s;ab)ds\overset{\ref{eqnFKk}}{=}\exp\int_{0}^{1}\log\mu(s;ab)ds$
$\displaystyle=\overset{(\ref{eqnFKf})}{=}\exp\int_{0}^{1}\log\mu(a;rabr)ds$
$\displaystyle\overset{Prop\ref{propHS}}{=}\left(\exp\int_{0}^{1}\log\mu(s;rar)ds\right)\left(\exp\int_{0}^{1}\log\mu(s;rbr)ds\right)$
$\displaystyle\overset{(\ref{eqnFKcc})}{=}\left(\exp\int_{0}^{1}\log\mu(s;a)dt\right)\left(\exp\int_{0}^{1}\log\mu(s;b)ds\right)$
$\displaystyle\overset{(\ref{eqnFKkka}),(\ref{eqnFKkkb})}{=}\left(\exp\int_{0}^{t}\log\mu(s;a)dt\right)\left(\exp\int_{0}^{t}\log\mu(s;b)ds\right)$

and this suffices to complete the proof of the Lemma.

∎

We may now prove the following special case of Theorem 4.2.

######  Lemma 4.4.

If $a,b\in L_{\log_{+}}(\tau)$ satisfy $\mu(s;a)>0,\mu(s;b)>0$ for every $s>0$
then

$\Lambda(t;ab)\leq\Lambda(t;a)\Lambda(t;b),\quad\forall t>0.$

###### Proof.

Without loss of generality, it may be assumed that $a,b\geq 0$. Let $t>0$ be
given. It will be shown that

$\int_{0}^{t}\log\mu(s;ab)ds\leq\int_{0}^{t}\log\mu(s;a)ds+\int_{0}^{t}\log\mu(s;b)ds$
(4.15)

Since $\mu(t;a)>0$ and $\mu(t;b)>0$, it may be assumed further, by
homogeneity, that $\mu(t;a)=1=\mu(t;b)$. Now set

$x=a\vee{\bf 1},\quad y=b\vee{\bf 1}$ (4.16)

and observe that

$\mu(s;x)=\mu(s;a)\vee 1=\begin{cases}\mu(s;a),&\ s\in(0,t),\\\ 1,&\
s\in(t,\infty).\end{cases}$ (4.17)

and

$\mu(s;y)=\mu(s;b)\vee 1=\begin{cases}\mu(s;b),&\ s\in(0,t),\\\ 1,&\
s\in(t,\infty).\end{cases}$ (4.18)

Now observe that

$\mu(ab)\leq\mu(xy)\quad{\rm and}\quad\mu(xy)\geq 1.$ (4.19)

To see the first inequality, using the fact that $a\geq 0,b\geq 0$, observe
that

$\displaystyle\mu(ab)$
$\displaystyle=\mu(ba^{2}b)^{\frac{1}{2}}\leq\mu(b(a\vee{\bf
1})^{2}b)^{\frac{1}{2}}$ $\displaystyle=\mu((a\vee{\bf 1})b)=\mu(((a\vee{\bf
1})b)^{*})=\mu(b(a\vee{\bf 1}))$ $\displaystyle\leq\mu((b\vee{\bf
1})(a\vee{\bf 1}))=\mu(((b\vee{\bf 1})(a\vee{\bf 1}))^{*})=\mu((a\vee{\bf
1})(b\vee{\bf 1})).$

The second inequality follows by observing that $x^{2}\geq{\bf 1},\
y^{2}\geq{\bf 1}$ and

$\displaystyle\mu(xy)$
$\displaystyle=\mu(yx^{2}y)^{\frac{1}{2}}\geq\mu(y^{2})^{\frac{1}{2}}\geq 1.$

If $t\leq w=\tau(e^{x}(1,\infty)\vee e^{y}(1,\infty))$, observe that

$\displaystyle\int_{0}^{t}\log\mu(s;ab)ds$
$\displaystyle\overset{(\ref{eqnKaab})}{\leq}\int_{0}^{t}\log\mu(s;xy)ds\overset{}{\leq}\int_{0}^{w}\log\mu(s;xy)ds$
$\displaystyle\overset{Lemma\ref{lemFK1}}{=}\int_{0}^{w}\log\mu(s;x)ds+\int_{0}^{w}\log\mu(s;y)ds$
$\displaystyle\overset{(\ref{eqnFKac}),(\ref{eqnFKad})}{=}\int_{0}^{t}\log\mu(s;x)ds+\int_{0}^{t}\log\mu(s;y)ds$
$\displaystyle=\int_{0}^{t}\log\mu(s;a)ds+\int_{0}^{t}\log\mu(s;a)ds.$

∎

Similarly, if $t\geq\tau(e^{x}(1,\infty)\vee e^{y}(1,\infty))$,

$\displaystyle\int_{0}^{t}\log\mu(s;ab)ds$
$\displaystyle\overset{(\ref{eqnKaab})}{\leq}\int_{0}^{t}\log\mu(s;xy)ds$
$\displaystyle\overset{Lemma\ref{lemFK1}}{=}\int_{0}^{t}\log\mu(s;x)ds+\int_{0}^{t}\log\mu(s;y)ds$
$\displaystyle=\int_{0}^{t}\log\mu(s;a)ds+\int_{0}^{t}\log\mu(s;a)ds.$

Proof of Theorem 4.2: Assume that $x,y\in L_{\log_{+}}(\tau)$. Without loss of
generality, it may be assumed that $x,y\geq 0$. Let $t>0$. We need to
establish the equality

$\int_{0}^{t}\log\mu(s;xy)ds\leq\int_{0}^{t}\log\mu(s;x)ds+\int_{0}^{t}\log\mu(s;y)ds.$
(4.20)

For each $n=1,2,\dots$, set $x_{n}=x+\frac{1}{n}{\bf 1},\
y_{n}=y+\frac{1}{n}{\bf 1}$ and note that

$\displaystyle\mu(xy)$
$\displaystyle=\mu(yx^{2}y)^{\frac{1}{2}}\leq\mu(yx_{n}^{2}y)^{\frac{1}{2}}$
$\displaystyle=\mu(x_{n}y)=\mu((x_{n}y)^{*})=\mu(yx_{n})=\mu(x_{n}y^{2}x_{n})^{\frac{1}{2}}$
$\displaystyle\leq\mu(x_{n}y_{n}^{2}x_{n})^{\frac{1}{2}}=\mu(y_{n}x_{n})=\mu(x_{n}y_{n}),\quad
n\geq 1.$

By Lemma 4.4, we now have that

$\int_{0}^{t}\log\mu(s;xy)ds\leq\int_{0}^{t}\log\mu(s;x_{n}y_{n})ds\leq\int_{0}^{t}\log\mu(s;x_{n})ds+\int_{0}^{t}\log\mu(s;y_{n})ds.$

By the monotone convergence theorem,

$\int_{0}^{t}\log\mu(s;x_{n})ds\downarrow_{n}\int_{0}^{t}\log\mu(s;x)ds,\quad\int_{0}^{t}\log\mu(s;y_{n})ds\downarrow_{n}\int_{0}^{t}\log\mu(s;y)ds$

and the inequality (4.20) follows directly.

##  5\.  Hölder type inequalities via logarithmic submajorization

The principal result of this section (Proposition 5.7) is a very general
Hölder inequality which goes back to Kosaki [29] in the special case of trace
ideals. Our approach is based on logarithmic submajorization and while we
follow the ideas of [29], the methods given there fail to apply in the present
setting.

We shall need the following submajorization inequalities [30], [22].

######  Proposition 5.1.

Let $\varphi$ be a continuous increasing function on $[0,\infty)$ such that
$\varphi(0)=0$ and has the property that $\varphi\circ\exp(\cdot)$ is convex.
If $0\leq a,b\in S(\tau)$, then

$\int_{0}^{t}\varphi(\mu(|ab|^{r})dm\leq\int_{0}^{t}\varphi(\mu(a^{r}b^{r}))dm,\quad
r\geq 1\quad t>0.$ (5.1)

and

$\int_{0}^{t}\varphi(\mu(|ab|^{r})dm\geq\int_{0}^{t}\varphi(\mu(a^{r}b^{r}))dm,\quad
0<r\leq 1\quad t>0.$ (5.2)

The preceding Araki-Lieb-Thirring type inequalities were first proved by
Kosaki( [30] Theorem 2) for $\tau$-measurable operators $0\leq a,b$ which are
$\tau$-compact, i.e., the operators $a,b$ satisfy the additional assumption
that $\lim_{t\to\infty}\mu(t;a)=0=\lim_{t\to\infty}\mu(t;b)$. It has been
noted recently by Han [22] Proposition 2.4 that Kosaki’s arguments may be
extended to the general case.

We note the following consequence which shows that the function
$t\to|a^{t}b^{t}|^{\frac{1}{t}}$ is increasing with respect to
submajorization, and is the counterpart for singular-value submajorization of
a result of Wang and Gong [38] for positive semi-definite matrices. See also
[4] Theorem 3.

######  Corollary 5.2.

If $0\leq a,b\in S(\tau)$ and if $0<t\leq u<\infty$, then

$|a^{t}b^{t}|^{\frac{1}{t}}\prec\prec|a^{u}b^{u}|^{\frac{1}{u}}.$ (5.3)

###### Proof.

In equation (5.2), set $r=t/u,\ \varphi(\cdot)=(\cdot)^{\frac{1}{t}}$ to
obtain

$\mu(|a^{\frac{t}{u}}b^{\frac{t}{u}}|^{\frac{1}{t}})=\left[\mu(a^{\frac{t}{u}}b^{\frac{t}{u}})\right]^{\frac{1}{t}}\prec\prec\left[\mu(|ab|^{\frac{t}{u}})\right]^{\frac{1}{t}}=\mu(|ab|^{\frac{1}{u}}).$

The submajorisation (5.3) now follows by replacing $a,b$ by $a^{u},b^{u}$
respectively.

∎

Before proceeding, it is desirable to introduce some additional notation. If
$x,y\in L_{\log_{+}}(\tau)$, then we shall say that $x$ is logarithmically
submajorized by $y$, written $x\prec\prec_{\log}y$, if and only if

$\int_{0}^{t}\log\mu(s;x)ds\leq\int_{0}^{t}\log\mu(s;y)ds$

Observe that $x\prec\prec_{\log}y$ if and only if $\Lambda(x)\leq\Lambda(y)$.
With this notation, it should also be noted, for ease of reference, that the
Weyl inequality

$\Lambda(xy)\leq\Lambda(x)\Lambda(y)$

given in Theorem 4.2(ii) may be reformulated in terms of logarithmic
submajorisation by the inequality

$xy\prec\prec_{\log}\mu(x)\mu(y).$

Indeed, one need only observe that, for all $t>0$,

$\displaystyle\int_{0}^{t}\log\mu(s;xy)ds$ $\displaystyle=\log\Lambda(t;xy)$
$\displaystyle\leq\log\Lambda(t;x)+\log\Lambda(t;y)$
$\displaystyle=\int_{0}^{t}\log\mu(s;x)ds+\int_{0}^{t}\log\mu(s;y)ds$
$\displaystyle=\int_{0}^{t}\log(\mu(s;x)\mu(s;y))ds.$

######  Corollary 5.3.

If $0\leq a,b\in L_{\log_{+}}(\tau)$ then

$\Lambda(|ab|^{r})\leq\Lambda(a^{r}b^{r}),\quad r\geq 1;$ (5.4)

and

$\Lambda(|ab|^{r})\geq\Lambda(a^{r}b^{r}),\quad 0<r\leq 1.$ (5.5)

The Corollary follows immediately from Proposition 5.1 via the implication
(iii)$\Longrightarrow$ (i) of Proposition 3.2.

We need the following generalization of the polar decomposition given in [19]
Theorem 2.7

######  Lemma 5.4.

If $x\in S(\tau)$ has polar decomposition $x=u|x|$ and if $\phi,\psi$ are
Borel functions on ${\mathbb{R}}$ such that
$\phi(\lambda)\psi(\lambda)=\lambda,\ \lambda\in{\mathbb{R}}$ then

$x=\phi(|x^{*}|)u\psi(|x|).$ (5.6)

In particular, if $1/p+1/q=1$, then

$x=|x^{*}|^{\frac{1}{p}}u|x|^{\frac{1}{q}}.$

######  Proposition 5.5.

If $a,b,x\in L_{\log_{+}}(\tau)$ with $a,b\geq 0$, if $1/p+1/q=1$ and if $r>0$
then

$\Lambda(|axb|^{r})\leq\Lambda^{\frac{r}{p}}(a^{p}x)\Lambda^{\frac{r}{q}}(xb^{q}),$

or, equivalently,

$|axb|^{r}\prec\prec_{\log}\mu(|a^{p}x|^{\frac{r}{p}})\mu(|xb^{q}|^{\frac{r}{q}}).$
(5.7)

###### Proof.

If $x=u|x|$ is the polar decomposition, then it follows from Lemma 5.4 that

$axb=a|x^{*}|^{\frac{1}{p}}u|x|^{\frac{1}{q}}b.$

It follows from the inequality (4.1) that

$\Lambda(axb)\leq\Lambda(a|x^{*}|^{\frac{1}{p}})\Lambda(u|x|^{\frac{1}{q}}b)\leq\Lambda(a|x^{*}|^{\frac{1}{p}})\Lambda(|x|^{\frac{1}{q}}b).$

Using Corollary 5.3, observe that

$\displaystyle\Lambda(a|x^{*}|^{\frac{1}{p}})$
$\displaystyle\leq\Lambda^{\frac{1}{p}}(a^{p}|x^{*}|)=\Lambda^{\frac{1}{p}}(|x^{*}|a^{p})=\Lambda^{\frac{1}{p}}(x^{*}a^{p})=\Lambda^{\frac{1}{p}}(a^{p}x)$
$\displaystyle\Lambda(|x|^{\frac{1}{q}}b)$
$\displaystyle\leq\Lambda^{\frac{1}{q}}(|x|b^{q})=\Lambda^{\frac{1}{q}}(xb^{q})$

Consequently,

$\Lambda(axb)\leq\Lambda^{\frac{1}{p}}(a^{p}x)\Lambda^{\frac{1}{q}}(xb^{q}),$

and the assertion of the Proposition now follows by observing that

$\Lambda(axb)^{r}=\Lambda(|axb|^{r}).$

∎

The following generalised Hölder inequality for normed order ideals is well-
known. The simple proof is included for the sake of completeness.

######  Proposition 5.6.

Suppose that $E\subseteq S(m)$ is a a normed order ideal. Suppose that
$r,p,q>0$ and that $1/r=1/p+1/q$. If $|x|^{p}\in E,|y|^{q}\in E$, then
$|xy|^{r}\in E$ and

$\||xy|^{r}\|_{E}^{\frac{1}{r}}\leq\||x|^{p}\|_{E}^{\frac{1}{p}}\||y|^{q}\|_{E}^{\frac{1}{q}}.$

###### Proof.

It suffices to assume that $\|x\|_{E^{p}}=1=\|y\|_{E^{q}}$. Using the well-
known numerical Young’s inequality, it follows that

$\frac{|x|^{r}|y|^{r}}{r}\leq\frac{|x|^{p}}{p}+\frac{|y|^{q}}{q}.$

Since $|x|^{p},|y|^{q}\in E$, it follows from the linearity of $E$ and the
fact that $E$ is an order ideal, that $|x|^{r}|y|^{r}\in E$. Consequently,
using the fact that $E$ is a normed ideal,

$\|\ |xy|^{r}\ \|_{E}\leq r\left(\frac{\|\ |x|^{p}\ \|_{E}}{p}+\frac{\|\
|y|^{q}\ \|_{E}}{q}\right)\leq 1.$

This suffices to complete the proof. ∎

If $E\subseteq S(\tau)$ is a symmetric (Banach) space, then $E\subseteq
L_{1}(\tau)+{\mathcal{M}}$ so that $E\subseteq L_{\log_{+}}(\tau)$. The norm
on the symmetric space $E$ is said to be monotone with respect to logarithmic
submajorization if whenever $x\in E,y\in L_{\log_{+}}(\tau)$ satisfy
$y\prec\prec_{\log}x$, it follows that $y\in E$ and $\|y\|_{E}\leq\|x\|_{E}$.

The principal result of this section now follows. In the setting of trace
ideals, this result may be found in [29] Theorem 3; however, the proof given
in [29] uses a standard trick with anti-symmetric tensors which is not
available in the present more general setting.

######  Proposition 5.7.

Suppose that $E$ is a symmetric space on $[0,\infty)$ whose norm is monotone
with respect to logarithmic submajorisation. Suppose that $x\in S(\tau)$, that
$0\leq a,b\in S(\tau)$ and that $1<p,p^{\prime}<\infty$ with conjugate
exponents $q=p/(p-1),\ q^{\prime}=p^{\prime}/(p^{\prime}-1)$. If $r>0$ and if
$|a^{p}x|^{\frac{rp^{\prime}}{p}},\ |xb^{q}|^{\frac{rq^{\prime}}{q}}\in
E(\tau)$ then $|axb|^{r}\in E(\tau)$ and

$\|\ |axb|^{r}\|_{E(\tau)}\leq\|\
|a^{p}x|^{\frac{rp^{\prime}}{p}}\|_{E(\tau)}^{\frac{1}{p^{\prime}}}\|\
|xb^{q}|^{\frac{rq^{\prime}}{q}}\|_{E(\tau)}^{\frac{1}{q^{\prime}}}.$ (5.8)

###### Proof.

Using the submajorization

$|axb|^{r}\prec\prec_{\log}\mu(|a^{p}x|^{\frac{r}{p}})\mu(|xb^{q}|^{\frac{r}{q}})$

given by (5.7), and observing that

$\frac{1}{(p^{\prime}r)}+\frac{1}{(q^{\prime}r)}=\frac{1}{r},$

it then follows from the first assertion of Proposition 5.6 and the
assumptions

$\displaystyle\mu(|a^{p}x|^{\frac{rp^{\prime}}{p}})$
$\displaystyle=\mu(|a^{p}x|^{\frac{1}{p}})^{rp^{\prime}}\in E$
$\displaystyle\mu(|xb^{q}|^{\frac{rq^{\prime}}{q}})$
$\displaystyle=\mu(|xb^{q}|^{\frac{1}{q}})^{rq^{\prime}}\in E$

that

$\left(\mu(|a^{p}x|^{\frac{1}{p}})\mu(|xb^{q}|^{\frac{1}{q}})\right)^{r}=\mu(|a^{p}x|^{\frac{r}{p}})\mu(|xb^{q}|^{\frac{r}{q}})\in
E.$

Since the norm on $E$ is monotone with respect to logarithmic submajorisation,
it follows further that $|axb|^{r}\in E(\tau)$ and

$\|\ |axb|^{r}\|_{E(\tau)}\leq\
\|\mu(|a^{p}x|^{\frac{r}{p}})\mu(|xb^{q}|^{\frac{r}{q}})\|_{E(\tau)}.$

Now using the second assertion of Proposition 5.6, it follows that

$\displaystyle\|\mu(|a^{p}x|^{\frac{r}{p}})\mu(|xb^{q}|^{\frac{r}{q}})\|_{E(\tau)}$
$\displaystyle\leq\||a^{p}x|^{\frac{rp^{\prime}}{p}}\|_{E(\tau)}^{\frac{1}{p^{\prime}}}\|\
|xb^{q}|^{\frac{rq^{\prime}}{q}}\|_{E(\tau)}^{\frac{1}{q^{\prime}}},$

and this suffices to complete the proof of the inequality (5.8). ∎

Some special cases are worth noting explicitly.

######  Corollary 5.8.

(Hiai [24] p.174 ; [1], Theorem 7)) Let $E$ be a symmetric space on
$[0,\infty)$ whose norm is monotone with respect to logarithmic
submajorisation and suppose that $a,b,x\in S(\tau)$, that $r>0$ and that
$p,q>1$ satisfy $1/p+1/q=1$. If $|a^{*}ax|^{\frac{rp}{2}},\
|xbb^{*}|^{\frac{rq}{2}}\in E(\tau)$ then $|axb|^{r}\in E(\tau)$ and

$\|\ |axb|^{r}\|_{E(\tau)}\leq\|\
|a^{*}ax|^{\frac{rp}{2}}\|_{E(\tau)}^{\frac{1}{p}}\|\
|xbb^{*}|^{\frac{rq}{2}}\|_{E(\tau)}^{\frac{1}{q}}.$ (5.9)

The Corollary follows immediately from Proposition 5.7 by taking $p=q=2$, then
replacing $p^{\prime},q^{\prime}$ by $p,q$ respectively and noting that

$\mu(|axb|^{r})=\mu^{r}(axb)=\mu^{r}(|a|x|b^{*}|)=\mu(|\ |a|x|b^{*}|\ |^{r}).$

As noted in [24], two further special cases of interest are the inequalities
obtained by taking obtained $x={\bf 1}$, and $p=2$ respectively:

######  Corollary 5.9.

Let $E$ be a symmetric space on $[0,\infty)$ whose norm is monotone with
respect to logarithmic submajorization. Suppose that $a,b,x\in S(\tau)$ and
that $r>0$.

  1. (1)

Let $p,q>1$ satisfy $\frac{1}{p}+\frac{1}{q}=1$. If $|a|^{rp}\in E(\tau)$ and
$|b|^{rq}\in E(\tau)$, then $|ab|^{r}\in E(\tau)$ and

$\|\ |ab|^{r}\|_{E(\tau)}\leq\|\ |a|^{rp}\|_{E(\tau)}^{\frac{1}{p}}\|\
|b|^{rq}\|_{E(\tau)}^{\frac{1}{q}},$ (5.10)

  2. (2)

If $|a^{*}ax|^{r}\in E(\tau)$ and $|xbb^{*}|^{r}\in E(\tau)$ then

$\|\ |axb|^{r}\|_{E(\tau)}^{2}\leq\|\ |a^{*}ax|^{r}\|_{E(\tau)}\|\
|xbb^{*}|^{r}\|_{E(\tau)}.$ (5.11)

The first inequality uses the identity $\mu(bb^{*})=\mu(b^{*}b)$. The second
inequality in the case of trace ideals is due to Bhatia and Davis [4] Theorem
1.

######  Corollary 5.10.

Let $E$ be a symmetric space on $[0,\infty)$ whose norm is monotone with
respect to logarithmic submajorization and suppose that $0\leq a,b\in
S(\tau)$, that $x\in S(\tau)$ and $r>0$. If $0\leq\nu\leq 1$, and if
$|ax|^{r},\ |xb|^{r}\in E(\tau)$, then $|a^{\nu}xb^{1-\nu}|^{r}\in E(\tau)$
and

$\||a^{\nu}xb^{1-\nu}|^{r}\|_{E(\tau)}\leq\|\ |ax|^{r}\|_{E(\tau)}^{\nu}\||\
xb|^{r}\|_{E(\tau)}^{1-\nu}$ (5.12)
$\||a^{\nu}xb^{1-\nu}|^{r}\|_{E(\tau)}\||a^{1-\nu}xb^{\nu}|^{r}\|_{E(\tau)}\leq\|\
|ax|^{r}\|_{E(\tau)}\||\ xb|^{r}\|_{E(\tau)}$ (5.13)

The inequality (5.12) follows directly from Proposition 5.7 by replacing $a,b$
by $a^{\nu},b^{1-\nu}$ respectively, and setting $p,p^{\prime}=1/\nu$. The
inequality 5.13 then follows immediately.

For $n\times n$ matrices, the proposition which now follows is due to Hiai and
Zhan [25]. See also [23] in the case of non-commutative $L_{p}$-spaces.

######  Proposition 5.11.

Let $E$ be a symmetric space on $[0,\infty)$ whose norm is monotone with
respect to logarithmic submajorization and suppose that $x\in S(\tau)$, that
$0\leq a,b\in S(\tau)$ and that $r>0$. If $0\leq\nu\leq 1$, and if $|ax|^{r},\
|xb|^{r}\in E(\tau)$, then the function

$f(\nu)=\||a^{\nu}xb^{1-\nu}|^{r}\|_{E(\tau)}\||a^{1-\nu}xb^{\nu}|^{r}\|_{E(\tau)},\quad\nu\in[0,1],$

is convex on the interval $[0,1]$ and attains its minimum at $\nu=1/2$ and its
maximum at the points $\nu=0,1$. In particular, for all $\nu\in[0,1]$,

$\|\
|a^{\frac{1}{2}}xb^{\frac{1}{2}}|^{r}\|_{E(\tau)}^{2}\leq\||a^{\nu}xb^{1-\nu}|^{r}\|_{E(\tau)}\||a^{1-\nu}xb^{\nu}|^{r}\|_{E(\tau)}\leq\|\
|ax|^{r}\|_{E(\tau)}\|\ |xb|^{r}\|_{E(\tau)}.$ (5.14)

In addition, if $a^{\frac{1}{2}}xb^{\frac{1}{2}}\neq 0$, then $f$ is
logarithmically convex on $[0,1]$.

###### Proof.

We observe first that it follows from (5.13) that $f$ is bounded on $[0,1]$.
Consequently, to show that $f$ is convex on $[0,1]$, it suffices to show that
$f$ is mid-point convex. See, for example, [39], Chapter 5, 25.3. To this end,
suppose that $t\pm s\in[0,1]$ and using (5.11), observe that

$\displaystyle\|\ |a^{t}xb^{1-t}|^{r}\|_{E(\tau)}^{2}$ $\displaystyle=\|\
|a^{s}(a^{t-s}xb^{1-t-s})b^{s}|^{r}\|_{E(\tau)}^{2}$ $\displaystyle\leq\|\
|a^{t+s}xb^{1-(t+s)}|^{r}\|_{E(\tau)}\cdot\|\
|a^{t-s}xb^{1-(t-s)}|^{r}\|_{E(\tau)}$

and

$\displaystyle\|\ |a^{1-t}xb^{t}|^{r}\|_{E(\tau)}^{2}$ $\displaystyle=\|\
|a^{s}(a^{1-t-s}xb^{t-s})b^{s}|^{r}\|_{E(\tau)}^{2}$ $\displaystyle\leq\|\
|a^{1-(t-s)}xb^{t-s}|^{r}\|_{E(\tau)}\cdot\|\
|a^{1-(t+s)}xb^{t+s}|^{r}\|_{E(\tau)}.$

Consequently,

$2f(t)\leq 2\sqrt{f(t+s)f(t-s)}\leq f(t+s)+f(t-s).$ (5.15)

It now follows that $f((\cdot);a,b,x)$ is convex on $[0,1]$ whenever
$|ax|^{r},\ |xb|^{r}\in E(\tau)$. Since $f$ is clearly symmetric about the
point $1/2$, it follows that $f$ attains its minimal value over [0,1] at $1/2$
and its maximal value at the points $0,1$, and this establishes the first part
of the proposition. Finally, if $a^{\frac{1}{2}}xb^{\frac{1}{2}}\neq 0$, then
it follows from the equation (5.14) that the function $\log f$ is bounded on
$[0,1]$. It suffices to show that $\log f$ is midpoint convex. This, however,
is immediate from the first inequality in (5.15). ∎

######  Remark 5.12.

We remark finally that there is a very substantial class of symmetric spaces
$E\subseteq S(\tau)$ whose norm is monotone with respect to submajorization.
Indeed, if $E$ is fully symmetric in the sense that, if $x\in E,y\in S(\tau)$
satisfy $y\prec\prec x$ then $y\in E$ and $\|y\|_{E}\leq\|x\|_{E}$, then it is
easily seen via remark 3.3 that the norm on $E$ is monotone with respect to
logarithmic submajorization. Such spaces are precisely the exact interpolation
spaces for the pair $(L_{1}(\tau),{\mathcal{M}})$. See [8], [7]. If
$E\subseteq S(m)$ is a fully symmetric space on $[0,\infty)$, then it is
follows directly that $E(\tau)\subset S(\tau)$ is fully symmetric. Important
special cases occur if $E\subseteq S(m)$ is one of the familiar
$L_{p}$-spaces, Orlicz spaces or Lorentz spaces. In the case that
${\mathcal{M}}=B({\mathcal{H}})$, then all trace ideals in the sense of
Gohberg and Krein [18] are fully symmetric. While it is well-known [35] that
there are symmetric spaces on the $[0,\infty)$ which are not closed subspaces
of any fully symmetric space, we do not have an explicit example of a
symmetric space on $[0,\infty)$ whose norm is monotone with respect to
logarithmic submajorization but which is not fully symmetric. This seems to be
a difficult problem.

##  6\. A Weyl-type inequality for uniform majorisation and Hölder
inequalities

The following direct consequence of the Weyl inequality given in Theorem 4.2
and the implication (i)$\Longrightarrow$ (iii) of Proposition 3.2 extends [17,
Theorem 4.2(iii)].

######  Theorem 6.1.

If $x,y\in L_{\log_{+}}(\tau)$, and if $f:(0,\infty)\to(0,\infty)$ is any
continuous increasing function $f:[0,\infty)\to[0,\infty)$ for which $f(0)=0$
and $f\circ\exp$ is convex, then

$f(|xy|)\prec\prec f(\mu(x)\mu(y)).$

We strengthen this as follows, using the notion of uniform Hardy-Littlewood
majorization introduced in [28];see also [31] Chapter 3.4. If $x,y\in S(\tau)$
then $x$ is said to be strongly majorized by  $y$, written $x\lhd y$, if there
exists $\lambda>0$ such that

$\int_{\lambda a}^{b}\mu(x)dm\leq\int_{a}^{b}\mu(y)dm,\quad 0<\lambda a\leq
b.$

It is the case that $x\prec\prec y$ if $x\lhd y$, but the converse is not
valid. We shall need the following fundamental result, which is obtained by
combining [28] Proposition 4.3 with [28]Theorem 6.3.

######  Theorem 6.2.

Let $E\subseteq S(m)$ be a symmetric Banach space on $[0,\infty)$. If $y\in
E(\tau)$ and if $x\in S(\tau)$ satisfies $x\lhd y$ then $x\in E(\tau)$ and
$\|x\|_{E(\tau)}\leq\|y\|_{E(\tau)}$

Our proof of the following theorem uses the approach from the proof of [34]
Theorem 7.

######  Theorem 6.3.

If $x,y\in L_{\log_{+}}(\tau)$ and $f:[0,\infty)\to[0,\infty)$ is an
increasing and continuous function for which $f(0)=0$ and $f\circ\exp$ is
convex, then

$f(|xy|)\lhd f(\mu(x)\mu(y)).$

###### Proof.

Let $a,b$ be positive scalars with $0\leq 2a<b$. It will suffice to show that

$\int_{2a}^{b}f(\mu(xy)dm\leq\int_{a}^{b}f(\mu(x)\mu(y))dm.$

Without loss of generality, it may be assumed that ${\mathcal{M}}$ is non-
atomic. By [9] Lemma 2.4), there exist projections $p,q\in P({\mathcal{M}})$
with $\tau(p)=\tau(q)=a$ such that

$e^{|x|}\left((\mu(a;x),\infty)\right)\leq p\leq
e^{|x|}\left([\mu(a;x),\infty)\right),$
$e^{|y|}\left((\mu(a;y),\infty)\right)\leq q\leq
e^{|y|}\left([\mu(a;y),\infty)\right),$
$\mu(|x|p)=\mu(x)\chi_{[0,\tau(p))}=\mu(x)\chi_{[0,a)},\quad\mu(|y|q)=\mu(x)\chi_{[0,\tau(q))}=\mu(y)\chi_{[0,a)},$

and

$\mu(|x|({\bf 1}-p))=\mu(\cdot+a,x),\quad\mu(|y|({\bf 1}-q))=\mu(\cdot+a,y)$

Observe that, for all $t>2a$,

$\mu(t;xy)\leq\mu(a;xpy)+\mu(a;x({\bf 1}-p)yq)+\mu(t-2a;x({\bf 1}-p)y({\bf
1}-q)).$

Using [17] Lemma 2.5 (vii), observe that, for all $\epsilon>0$,

$\mu(a+\epsilon;xpy)\leq\mu(a+\epsilon;|x|py)\leq\mu(a;|x|p)\mu(\epsilon;y)=0$

and

$\mu(a+\epsilon;x({\bf 1}-p)yq)\leq\mu(\epsilon;x({\bf 1}-p))\mu(a;|y|q)=0.$

By right-continuity of the singular value function, it follows that

$\mu(a;xpy)=\mu(a;x({\bf 1}-p)yq)=0$

and so

$\mu(t;xy)\leq\mu(t-2a;x({\bf 1}-p)y({\bf 1}-q))\quad t>2a.$

Consequently,

$\displaystyle\int_{2a}^{b}f(\mu(xy))dm$
$\displaystyle\leq\int_{2a}^{b}f(\mu(\cdot-2a;x({\bf 1}-p)y({\bf
1}-q)dm\leq\int_{0}^{b-2a}f(\mu(x({\bf 1}-p)y({\bf 1}-q))dm$
$\displaystyle\leq\int_{0}^{b-2a}f(\mu(x({\bf 1}-p))\mu(y({\bf
1}-q))dm\quad\quad{\rm using\ Theorem~{}\ref{WIp1}}$
$\displaystyle\leq\int_{0}^{b-2a}f(\mu(\cdot+a;x)\mu(\cdot+a;y))dm$
$\displaystyle=\int_{a}^{b-a}f(\mu(x)\mu(y))dm\leq\int_{a}^{b}f(\mu(x)\mu(y))dm$

This concludes the proof. ∎

######  Corollary 6.4.

(cf. [2], Lemma 4) If $x,y\in L_{\log_{+}}(\tau)$ and $r>0$ then

$|xy|^{r}\lhd\mu(x)^{r}\mu(y)^{r}.$

###### Proof.

The inequality follows immediately from Theorem 6.3 by taking $f(t)=t^{r},\
t>0$.

∎

A general Hölder inequality for arbitrary symmetric spaces now follows.

######  Corollary 6.5.

Suppose that $E\subseteq S(m)$ is a symmetric space, that $r,p,q>0$ and that
$1/r=1/p+1/q$. If $|x|^{p},\ |y|^{q}\in E(\tau)$, then $|xy|^{r}\in E(\tau)$
and

$\||xy|^{r}\|_{E(\tau)}^{\frac{1}{r}}\leq\||x|^{p}\|_{E(\tau)}^{\frac{1}{p}}\|\
|y|^{q}\|_{E(\tau)}^{\frac{1}{q}}$

###### Proof.

By assumption, $\mu(x)^{p}\in E,\mu(y)^{q}\in E$ and so by Proposition 5.6
$\mu(x)^{r}\mu(y)^{r}\in E$. The assumptions $|x|^{p},\ |y|^{q}\in
E(\tau)\subseteq L^{1}(\tau)+{\mathcal{M}}$ imply further that
$|x|^{p},|y|^{p}\in L_{\log_{+}}(\tau)$ and so also $x,y\in
L_{\log_{+}}(\tau)$.

By Corollary 6.4, the uniform submajorization

$|xy|^{r}\lhd\mu(x)^{r}\mu(y)^{r}$

together with Theorem 6.2 yields that $|xy|^{r}\in E$ and

$\||xy|^{r}\|_{E(\tau)}\leq\|\mu(x)^{r}\mu(y)^{r}\|_{E}$

The assertion of the corollary now follows from Proposition 5.6.

∎

The preceding Corollary 6.5 is given in [2] Corollary 1 and in [34] Theorem 1
in the case that $r=1$. The present approach is much more direct than that
given in [2].

Let $E\subseteq S(m)$ be a symmetric space and let $x\in S(\tau)$. For ease of
presentation, it will now be convenient to write $\|x\|_{E(\tau)}<\infty$ if
$x\in E(\tau)$ and $\|x\|_{E(\tau)}=+\infty$ otherwise.

######  Proposition 6.6.

Suppose that $E\subseteq S(m)$ is a symmetric space. Let $r>0$ and suppose
that $1<p,q<\infty$ satisfy $1/p+1/q=1$. Let
$\phi,\psi:(0,\infty)\to(0,\infty)$ be Borel functions such that
$\phi(\lambda)\psi(\lambda)=\lambda,\ \lambda>0$. If $a,b,x\in S(\tau)$, then

$\||a^{*}xb|^{r}\|_{E(\tau)}\leq\|\left(a^{*}\phi^{2}(|x^{*}|)a\right)^{\frac{pr}{2}}\|_{E(\tau)}^{\frac{1}{p}}\|\left(b^{*}\psi^{2}(|x|)b\right)^{\frac{qr}{2}}\|_{E(\tau)}^{\frac{1}{q}},$
(6.1)

provided the right hand side is finite.

###### Proof.

Let $a,x,b\in S(\tau)$ and note that it follows from Lemma 5.4 that

$a^{*}xb=a^{*}\phi(|x^{*}|)u\psi(|x|)b,$

where $x=u|x|$ is the polar decomposition. Theorem 6.3 with
$f(\cdot)=(\cdot)^{r}$ implies that

$\displaystyle|a^{*}xb|^{r}$
$\displaystyle\lhd\mu^{r}(a^{*}\phi(|x^{*}|)u)\mu^{r}(\psi(|x|)b)$
$\displaystyle\leq\mu^{r}(\phi(|x^{*}|)a)\mu^{r}(\psi(|x|)b)=\mu((a^{*}\phi^{2}(|x^{*}|)a)^{\frac{r}{2}})\mu((b^{*}\psi^{2}(|x|)b)^{\frac{r}{2}}).$

It now follows that

$\displaystyle\||a^{*}xb|^{r}\|$
$\displaystyle\leq\|\mu((a^{*}\phi^{2}(|x^{*}|)a)^{\frac{r}{2}})\mu((b^{*}\psi^{2}(|x|)b)^{\frac{r}{2}})\|_{E}$
$\displaystyle\leq\|\mu^{p}((a^{*}\phi^{2}(|x^{*}|)a)^{\frac{r}{2}})\|_{E}^{\frac{1}{p}}\|\mu^{q}((b^{*}\psi^{2}(|x|)b)^{\frac{r}{2}})\|_{E}^{\frac{1}{q}}$
$\displaystyle=\|\left(a^{*}\phi^{2}(|x^{*}|)a\right)^{\frac{pr}{2}}\|_{E(\tau)}^{\frac{1}{p}}\|\left(b^{*}\psi^{2}(|x|)b\right)^{\frac{qr}{2}}\|_{E(\tau)}^{\frac{1}{q}},$

and this completes the proof of the proposition. ∎

For the special case that $\phi(\lambda)=\lambda^{1/2}=\psi(\lambda),\
\lambda>0$ , we obtain the following inequality. See [2] Corollary 2.

######  Corollary 6.7.

Suppose that $E\subseteq S(m)$ is a symmetric space. Let $r>0$ and suppose
that $1<p,q<\infty$ satisfy $1/p+1/q=1$. If $a,b,x\in S(\tau)$,then

$\||a^{*}xb|^{r}\|_{E(\tau)}\leq\|\left(a^{*}|x^{*}|a\right)^{\frac{pr}{2}}\|_{E(\tau)}^{\frac{1}{p}}\|\left(b^{*}|x|b\right)^{\frac{qr}{2}}\|_{E(\tau)}^{\frac{1}{q}},$
(6.2)

whenever the right hand side is finite.

A standard argument (see, for example [2] Theorem 4, or [1] Theorem 21) now
yields the following consequence.

######  Corollary 6.8.

Suppose that $E\subseteq S(m)$ is symmetrically normed. Let $r>0$ and suppose
that $1<p,q<\infty$ satisfy $1/p+1/q=1$. Let
$\phi,\psi:(0,\infty)\to(0,\infty)$ be Borel functions such that
$\phi(\lambda)\psi(\lambda)=\lambda,\ \lambda>0$. If $a_{i},x_{i},b_{i}\in
S(\tau),\ 1\leq i\leq n$, then

$\||\sum_{i=1}^{n}a_{i}^{*}x_{i}b_{i}|^{r}\|_{E(\tau)}\leq\left\|\left(\sum_{i=1}^{n}a_{i}^{*}\phi^{2}(|x_{i}^{*}|)a_{i}\right)^{\frac{pr}{2}}\right\|_{E(\tau)}^{\frac{1}{p}}\
\left\|\left(\sum_{i=1}^{n}b_{i}^{*}\psi^{2}(|x_{i}|)b_{i}\right)^{\frac{qr}{2}}\right\|^{\frac{1}{q}},$
(6.3)

whenever the right hand side is finite.

We remark that Proposition 6.6 and Corollary 6.8 may be found in [1], Theorems
19 and 21 respectively, for the special case that ${\mathcal{M}}=B(H)$. See
also [26]. In the present setting of $\tau$-measurable operators, the special
case of Proposition 6.6 and Corollary 6.8 obtained by setting
$\phi(\lambda)=\lambda^{1/2}=\psi(\lambda),\ \lambda>0$ is proved in [2].
However, the present approach via Theorem 6.3 is more direct and transparent.

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