More results on generalized singular number inequalities of τ-measurable operators

Abstract

In this article we give some generalized singular number inequalities for products and sums of τ-measurable operators. Some related arithmetic-geometric mean and Heinz mean inequalities for a generalized singular number of τ-measurable operators are proved.

Introduction

Let $\mathbb{M}_{n}$ be the space of $n\times n$ complex matrices. Given $A\in\mathbb{M}_{n}$, we define $|A|=(A^{*}A)^{\frac{1}{2}}$. The singular values of A, i.e., the eigenvalues of the operator $|A|$, enumerated in decreasing order, will be denoted by $S_{j}(A)$, $j=1, 2, \ldots, n$. The arithmetic-geometric mean inequality for singular values due to Bhatia and Kittaneh [1] says that

$$ 2S_{j} \bigl(AB^{*} \bigr)\leq S_{j} \bigl(A^{*}A+B^{*}B \bigr),\quad j=1, 2, \ldots, n, $$

(1.1)

holds for any $A, B\in\mathbb{M}_{n}$. In 2000, Zhan [2] proved that

$$ 2S_{j}(A-B)\leq S_{j}(A\oplus B), \quad j=1, 2, \ldots, n, $$

(1.2)

for positive semidefinite matrices $A, B\in\mathbb{M}_{n}$. On the other hand, Tao [3] observed that if $A, B, K\in\mathbb{M}_{n}$ with $\bigl ( {\scriptsize\begin{matrix}{} A&K\cr K^{*}&B \end{matrix}} \bigr )\geq0$, then

$$ 2S_{j}(K)\leq S_{j} \left ( \begin{pmatrix} A&K\\ K^{*}&B \end{pmatrix} \right ),\quad j=1, 2, \ldots, n. $$

(1.3)

It was pointed out in [3] that inequalities (1.1), (1.2), and (1.3) are equivalent. According to inequality (1.3), Audenaert [4] (see also [5]) gave a Heinz mean inequality for singular values, that is, if $A, B\in\mathbb{M}_{n}$ are positive semidefinite matrices and $0\leq r\leq1$, then

$$ S_{j} \bigl(A^{r}B^{1-r}+A^{1-r}B^{r} \bigr)\leq S_{j}(A+B),\quad j=1, 2, \ldots, n. $$

(1.4)

Among other things, in 2012, Albadawi [6] showed that if $A_{i}, B_{i}, X_{i}\in B(\mathcal{H})$ ($i=1, 2,\ldots, n$) with $X_{i}\geq0$, then

(1.5)

holds for $j=1, 2, \ldots$ . Inequality (1.5) yields the well-known arithmetic-geometric mean inequality for singular values as special cases.

Using the notion of the generalized singular number studied by Fack and Kosaki [7], we generalize inequalities (1.1)-(1.5) for τ-measurable operators associated with a semifinite von Neumann algebra $\mathcal{M}$.

Preliminaries

Unless stated otherwise, $\mathcal{M}$ will always denote a semifinite von Neumann algebra acting on a Hilbert space $\mathcal{H}$, with a normal faithful semifinite trace τ. We refer to [7, 8] for noncommutative integration. We denote the identity of $\mathcal{M}$ by 1 and let $\mathcal{P}$ denote the projection lattice of $\mathcal{M}$. A closed densely defined linear operator x in $\mathcal{H}$ with domain $D(x)\subseteq\mathcal{H}$ is said to be affiliated with $\mathcal{M}$ if $u^{*}xu=x$ for all unitary operators u which belong to the commutant $\mathcal{M^{\prime}}$ of $\mathcal {M}$. If x is affiliated with $\mathcal{M}$, we define its distribution function by $\lambda_{s}(x)=\tau(e^{\bot}_{s}(|x|))$ and x will be called τ-measurable if and only if $\lambda_{s}(x)<\infty$ for some $s>0$, where $e^{\bot}_{s}(|x|)=e_{(s, \infty)}(|x|)$ is the spectral projection of $|x|$ associated with the interval $(s, \infty)$. The set of all τ-measurable operators will be denoted by $\overline{ \mathcal{M}}$. The set $\overline{\mathcal{M}}$ is a ∗-algebra with sum and product being the respective closures of the algebraic sum and product.

Definition 2.1

Let $x\in\overline{\mathcal{M}}$ and $t>0$. The ‘tth singular number (or generalized singular number) of x’ $\mu _{t}(x)$ is defined by

$$\mu_{t}(x)=\inf \bigl\{ \|xe\|: e \mbox{ is a projection in } \mathcal{M} \mbox{ with } \tau \bigl(e^{\bot}\bigr)\leq t \bigr\} . $$

From Lemma 2.5 in [7] we see that the generalized singular number function $t\rightarrow\mu_{t}(x)$ is decreasing right-continuous and

$$ \mu_{t}(uxv)\leq\|v\|\|u\|\mu_{t}(x),\quad t>0, $$

(2.1)

for all $u, v\in\mathcal{M}$ and $x\in\overline{\mathcal{M}}$. Moreover,

$$ \mu_{t} \bigl(f(x) \bigr)=f \bigl(\mu_{t}(x) \bigr),\quad t>0, $$

(2.2)

whenever $0\leq x\in\overline{\mathcal{M}}$ and f is an increasing continuous function on $[0,\infty)$ satisfying $f(0) =0$. Proposition 2.2 in [7] implies that

$$ \mu_{t}(x)=\inf \bigl\{ s\geq0; \lambda_{s}(x) \leq t \bigr\} =\inf \bigl\{ s\geq0; \tau \bigl(e_{(s, \infty)}\bigl(|x|\bigr) \bigr)\leq t \bigr\} ,\quad t>0, $$

(2.3)

and

$$ \lambda_{\mu_{t}(x)}(x)\leq t,\quad t>0. $$

(2.4)

The space $\overline{\mathcal{M}}$ is a partially ordered vector space under the ordering $x\geq0$ defined by $(x\xi, \xi)\geq0$, $\xi\in D(x)$. The trace τ on $\mathcal{M}^{+}$ (the positive part of $\mathcal {M}$) extends uniquely to an additive, positively homogeneous, unitarily invariant, and normal functional $\widetilde{\tau}: \overline{\mathcal{M}}\rightarrow[0, \infty]$, which is given by $\widetilde{\tau}(x)=\int_{0}^{\infty}\mu_{t}(x)\,dt$, $x\in\mathcal{M}^{+}$. This extension is also denoted by τ. Further,

$$\tau \bigl(f(x) \bigr)= \int_{0}^{\infty}f \bigl(\mu_{t}(x) \bigr)\,dt $$

whenever $0\leq x \in\overline{\mathcal{M}}$ and f is non-negative Borel function which is bounded on a neighborhood of 0 and satisfies $f(0) = 0$. See [7, 9] for basic properties and detailed information on the generalized singular number. For $0< p<\infty$, $L^{p}(\mathcal{M})$ is defined as the set of all densely defined closed operators x affiliated with $\mathcal{M}$ such that

$$\|x\|_{p}=\tau \bigl(|x|^{p} \bigr)^{\frac{1}{p}}= \biggl( \int_{0}^{\infty}\mu_{t}(x)^{p}\,dt \biggr)^{\frac {1}{p}}< \infty. $$

As usual, we put $L^{\infty}(\mathcal{M};\tau)=\mathcal{M}$ and denote by $\|\cdot\|_{\infty}$ ($=\|\cdot\|$) the usual operator norm. It is well known that $L^{p}(\mathcal{M})$ is a Banach space under $\| \cdot\|_{p}$ ($1\leq p\leq\infty$) (cf. [8]).

Let $\mathbb{M}_{n}(\mathcal{M})$ denote the linear space of $n\times n$ matrices

with entries $x_{ij}\in\mathcal{M}$, $i,j=1,2,\ldots,n$. Let $\mathcal{H}^{n}=\bigoplus_{i=1}^{n}\mathcal{H}$. Then $\mathbb{M}_{n}(\mathcal{M})$ is a von Neumann algebra in the Hilbert space $\mathcal{H}^{n}$. For $x\in\mathbb{M}_{n}(\mathcal {M})$, define $\tau_{n}(x)=\sum_{i=1}^{n}\tau(x_{ii})$, then $\tau_{n}$ is a normal faithful semifinite trace on $\mathbb{M}_{n}(\mathcal{M})$. The direct sum of operators $x_{1}, x_{2},\ldots, x_{n}\in \overline{\mathcal {M}}$, denoted by $\bigoplus_{i=1}^{n}x_{i}$, is the block-diagonal operator matrix defined on $\mathcal{H}^{n}$ by

Arithmetic-geometric mean and Heinz mean inequalities for generalized singular number of τ-measurable operators

Let $x\in \overline{\mathcal{M}}$ and $d_{\mu(x)}(t)$ be the classical distribution function of $s\rightarrow\mu_{s}(x)$. By Proposition 1.2 of [10], we deduce

$$\lambda_{t}(x)=d_{\mu(x)}(t)=m \bigl( \bigl\{ s\in(0, \infty): \mu_{s}(x)>t \bigr\} \bigr),\quad t>0, $$

where m is the Lebesgue measure on $(0, \infty)$. Since $s\rightarrow\mu_{s}(x)$ is non-increasing and continuous from the right (see, Lemma 2.5 of [7]), we have

$$\lambda_{t}(x)=\inf \bigl\{ s>0: \mu_{s}(x)\leq t \bigr\} , \quad t>0. $$

Moreover,

$$ \mu_{\lambda_{s}(x)}(x) \leq s,\quad s>0. $$

(3.1)

The following lemma, which includes a basic property of generalized singular number, plays a central role in our investigation.

Lemma 3.1

Let $x_{i}\in \overline{\mathcal{M}}$, $i=1, 2, \ldots, n$. Then

$$\begin{aligned} \mu_{t} \Biggl(\bigoplus_{i=1}^{n} x_{i} \Biggr)=\inf \Biggl\{ \max \bigl\{ \mu_{s_{1}}(x_{1}), \mu_{s_{2}}(x_{2}), \ldots, \mu_{s_{n}}(x_{n}) \bigr\} : s_{i}\geq0, \sum_{i=1}^{n}s_{i} \leq t \Biggr\} . \end{aligned}$$

Moreover,

$$\begin{aligned} \mu_{t} \Biggl(\bigoplus_{i=1}^{n} x_{i} \Biggr)=\inf \Biggl\{ \max \bigl\{ \mu_{s_{1}}(x_{1}), \mu_{s_{2}}(x_{2}), \ldots, \mu_{s_{n}}(x_{n}) \bigr\} : s_{i}\geq0, \sum_{i=1}^{n}s_{i}=t \Biggr\} . \end{aligned}$$

Proof

Let $s_{i}\geq0$ with $\sum_{i=1}^{n}s_{i}\leq t$. By (2.4), we get

$$\tau \Biggl(\bigoplus_{i=1}^{n}e_{(\mu_{s_{i}}(x_{i}), \infty)} \bigl(|x_{i}| \bigr) \Biggr)\leq\sum_{i=1}^{n}s_{i} \leq t. $$

Therefore, according to the definition of generalized singular number, we obtain

$$\begin{aligned} \mu_{t} \Biggl(\bigoplus_{i=1}^{n} x_{i} \Biggr)\leq\Biggl\| \bigoplus_{i=1}^{n}x_{i}e_{[0,\mu _{s_{i}}(x_{i})]}\bigl(|x_{i}|\bigr) \Biggr\| \leq\max_{i=1, 2, \ldots, n} \bigl\{ \mu_{s_{i}}(x_{i}) \bigr\} . \end{aligned}$$

For the reverse inclusion, from (2.3), we get

$$\mu_{t} \Biggl(\bigoplus_{i=1}^{n} x_{i} \Biggr)=\inf \Biggl\{ s\geq0: \tau \Biggl(e_{(s, \infty)} \Biggl( \Biggl|\bigoplus_{i=1}^{n} x_{i} \Biggr| \Biggr) \Biggr)\leq t \Biggr\} . $$

Since

$$e_{(s, \infty)} \Biggl(\Biggl|\bigoplus_{i=1}^{n} x_{i}\Biggr| \Biggr)=e_{(s, \infty)} \Biggl(\bigoplus _{i=1}^{n} |x_{i}| \Biggr) =\bigoplus _{i=1}^{n} e_{(s, \infty)} \bigl(|x_{i}| \bigr), $$

we have

$$\mu_{t} \Biggl(\bigoplus_{i=1}^{n} x_{i} \Biggr)=\inf \Biggl\{ s\geq0: \sum_{i=1}^{n} \tau \bigl(e_{(s, \infty )} \bigl(|x_{i}| \bigr) \bigr)\leq t \Biggr\} . $$

Let $s_{i}=\tau(e_{(s, \infty)}(|x_{i}|))$. It follows from inequality (3.1) that $\mu_{s_{i}}(x_{i})\leq s$. Hence

$$\max_{i=1, 2, \ldots, n} \bigl\{ \mu_{s_{i}}(x_{i}) \bigr\} \leq\mu_{t} \Biggl(\bigoplus_{i=1}^{n} x_{i} \Biggr). $$

 □

Remark 3.2

• (1) Let $x\in \overline{\mathcal{M}}$. If $x_{i}=x$, $i=1, 2, \ldots, n$, it follows from Lemma 3.1 that $\mu_{t}(\bigoplus_{i=1}^{n} x_{i})=\mu_{\frac{t}{n}}(x)$, $t>0$.

• (2) Let $x\in \overline{\mathcal{M}}$. If $x_{1}=x$ and $x_{i}=0$, $i= 2, 3, \ldots, n$, it follows from Lemma 3.1 that $\mu_{t}(\bigoplus_{i=1}^{n} x_{i})=\mu_{t}(x)$, $t>0$.

• (3) Let $x_{1}, x_{2}, y_{1}, y_{2}\in\overline{\mathcal{M}}$ such that $\mu_{t}(x_{i})\leq\mu_{t}(y_{i})$, $t>0$, $i=1, 2$. From Lemma 3.1, we deduce $\mu_{t}(x_{1}\oplus x_{2})\leq\mu_{t}(y_{1}\oplus y_{2})$, $t>0$.

As an application of Lemma 3.1 we now obtain the desired generalized singular number inequality (1.3) for τ-measurable operators.

Lemma 3.3

Let $x, y, z\in \overline{\mathcal{M}}$. If $\bigl ( {\scriptsize\begin{matrix}{} x&z\cr z^{*}&y \end{matrix}} \bigr )\geq0$, then

$$2\mu_{t}(z)\leq\mu_{t} \left ( \begin{pmatrix} x&z\\ z^{*}&y \end{pmatrix} \right ),\quad t>0. $$

Proof

Let $N=\bigl ( {\scriptsize\begin{matrix}{} 0&z\cr z^{*}&0 \end{matrix}} \bigr )$, $M=\bigl ( {\scriptsize\begin{matrix}{} x&z\cr z^{*}&y \end{matrix}} \bigr )$, and $U=\bigl ( {\scriptsize\begin{matrix}{} 1&0\cr 0&-1 \end{matrix}} \bigr )$. Then

$$\begin{aligned} \begin{pmatrix} 1&0\\ 0&-1 \end{pmatrix} \begin{pmatrix} x&z\\ z^{*}&y \end{pmatrix} \begin{pmatrix} 1&0\\ 0&-1 \end{pmatrix} &= \begin{pmatrix} x&-z\\ -z^{*}&y \end{pmatrix} \\ &= \begin{pmatrix} x&z\\ z^{*}&y \end{pmatrix} -2N. \end{aligned}$$

Hence $N=\frac{1}{2}(M-UMU^{*})$. Let $N=N^{+}-N^{-}$ be the Jordan decomposition of N. It follows from Lemma 6 of [11] that $\mu_{t}(N^{+})\leq\mu_{t}(\frac{1}{2}M)$, $t>0$, and

$$\mu_{t} \bigl(N^{-} \bigr)\leq\mu_{t} \biggl( \frac{1}{2} UMU^{*} \biggr)\leq \|U\|\bigl\| U^{*}\bigr\| \mu_{t} \biggl( \frac{1}{2}M \biggr) \leq\mu_{t} \biggl(\frac{1}{2}M \biggr),\quad t>0. $$

By Theorem 6 of [12], we have

$$\mu_{t}(N)=\mu_{t} \bigl(N^{+}-N^{-} \bigr)\leq \mu_{t} \bigl(N^{+}\oplus N^{-} \bigr), \quad t>0. $$

Therefore, from Lemma 3.1 we obtain

$$\mu_{2t}(N)\leq\mu_{2t} \bigl(N^{+}\oplus N^{-} \bigr)\leq \mu_{2t} \biggl(\frac{1}{2}M \oplus\frac{1}{2}M \biggr)= \mu_{t} \biggl(\frac{1}{2}M \biggr),\quad t>0, $$

i.e.,

$$2\mu_{2t} \left ( \begin{pmatrix} 0&z\\ z^{*}&0 \end{pmatrix} \right )=2\mu_{2t}(N)\leq \mu_{t} \left ( \begin{pmatrix} x&z\\ z^{*}&y \end{pmatrix} \right ),\quad t>0. $$

It is clear that $\bigl ( {\scriptsize\begin{matrix}{} 0&1\cr 1&0 \end{matrix}} \bigr )\bigl ( {\scriptsize\begin{matrix}{} 0&z\cr z^{*}&0 \end{matrix}} \bigr )=\bigl ( {\scriptsize\begin{matrix}{} z&0\cr 0&z^{*} \end{matrix}} \bigr )^{*}$ and $\|\bigl ( {\scriptsize\begin{matrix}{} 0&1\cr 1&0 \end{matrix}} \bigr )\|=1$. Then Lemma 2.5 of [7] and Lemma 3.1 imply that

$$2\mu_{t}(z) =2\mu_{2t} \left ( \begin{pmatrix} z&0\\ 0&z^{*} \end{pmatrix} \right )\leq2 \mu_{2t}(N)\leq\mu_{t} \left ( \begin{pmatrix} x&z\\ z^{*}&y \end{pmatrix} \right ),\quad t>0. $$

 □

Combing Lemma 3.3 with the following theorem we see that inequalities (1.1), (1.2), and (1.3) hold for τ-measurable operators.

Theorem 3.4

The following statements are equivalent:

• (1) Let $0\leq x, y \in \overline{\mathcal{M}}$. Then $\mu_{t}(x-y)\leq\mu_{t}(x\oplus y)$, $t>0$.

• (2) For any $x, y\in\overline{\mathcal{M}}$, $2\mu_{t}(xy^{*})\leq\mu_{t}(x^{*}x+y^{*}y)$, $t>0$.

• (3) Let $x, y, z\in \overline{\mathcal{M}}$. If $\bigl ( {\scriptsize\begin{matrix}{} x&z\cr z^{*}&y \end{matrix}} \bigr )\geq0$, then

  $$2\mu_{t}(z)\leq\mu_{t} \left ( \begin{pmatrix} x&z\\z^{*}&y \end{pmatrix} \right ),\quad t>0. $$

Proof

(1) ⇒ (2): For any $x, y\in\overline{\mathcal{M}}$, we write $X=\bigl ( {\scriptsize\begin{matrix}{} x&0\cr y&0 \end{matrix}} \bigr )$, $Y=\bigl ( {\scriptsize\begin{matrix}{} x&0\cr -y&0 \end{matrix}} \bigr )$. Then $X^{*}X=\bigl ( {\scriptsize\begin{matrix}{} x^{*}x+y^{*}y&0\cr 0&0 \end{matrix}} \bigr )$ and $Y^{*}Y=\bigl ( {\scriptsize\begin{matrix}{} x^{*}x+y^{*}y&0\cr 0&0 \end{matrix}} \bigr )$. It follows from Lemma 3.1 and (1) that

$$\begin{aligned} 2\mu_{t} \left ( \begin{pmatrix} yx^{*}&0\\0&xy^{*} \end{pmatrix} \right ) &=2\mu_{t} \left ( \begin{pmatrix} 0&xy^{*}\\yx^{*}&0 \end{pmatrix} \right )=\mu_{t} \bigl(XX^{*}-YY^{*} \bigr) \\ &\leq\mu_{t} \bigl(XX^{*}\oplus YY^{*} \bigr) \\ &=\inf \bigl\{ \max \bigl(\mu_{a} \bigl(XX^{*} \bigr), \mu_{b} \bigl(YY^{*} \bigr) \bigr): a, b\geq0, a+b=t \bigr\} \\ &=\inf \bigl\{ \max \bigl(\mu_{a} \bigl(X^{*}X \bigr), \mu_{b} \bigl(Y^{*}Y \bigr) \bigr): a, b\geq0, a+b=t \bigr\} \\ &=\inf_{ a, b\geq0, a+b=t} \bigl\{ \max \bigl(\mu_{a} \bigl(x^{*}x+y^{*}y \bigr), \mu_{b} \bigl(x^{*}x+y^{*}y \bigr) \bigr) \bigr\} \\ &=\mu_{t} \left ( \begin{pmatrix} x^{*}x+y^{*}y&0\\ 0&x^{*}x+y^{*}y \end{pmatrix} \right ),\quad t>0. \end{aligned}$$

Lemma 3.1 ensures that $2\mu_{t}(xy^{*})\leq\mu_{t}(x^{*}x+y^{*}y)$, $t>0$.

(2) ⇒ (1): Let $0\leq x, y \in \overline{\mathcal{M}}$ and let

$$S= \begin{pmatrix} x^{\frac{1}{2}}&-y^{\frac{1}{2}}\\0&0 \end{pmatrix},\qquad T= \begin{pmatrix} x^{\frac{1}{2}}& y^{\frac{1}{2}}\\0&0 \end{pmatrix}. $$

From (2) we have $2\mu_{t}(ST^{*})\leq\mu_{t}(S^{*}S+T^{*}T)$, $t>0$. Then the result follows from Lemma 3.1.

From Lemma 3.3 we have (1) ⇒ (3).

(3) ⇒ (1): For any $0\leq x, y\in\overline{\mathcal{M}}$, we have the following unitary similarity transform:

$$\frac{1}{\sqrt{2}} \begin{pmatrix} 1&1\\-1&1 \end{pmatrix} \begin{pmatrix} \frac{x+y}{2}&\frac{x-y}{2}\\ \frac{x-y}{2} &\frac{x+y}{2} \end{pmatrix} \frac{1}{\sqrt{2}} \begin{pmatrix} 1&-1\\1&1 \end{pmatrix} = \begin{pmatrix} x&0\\ 0 &y \end{pmatrix} \geq0. $$

According to (3), we obtain

$$\mu_{t}(x-y)\leq\mu_{t} \left ( \begin{pmatrix} \frac{x+y}{2}&\frac{x-y}{2}\\ \frac{x-y}{2} &\frac{x+y}{2} \end{pmatrix} \right )\leq \mu_{t} \begin{pmatrix} x&0\\0 &y \end{pmatrix},\quad t>0. $$

 □

Lemma 3.5

Let $0\leq x, y\in \overline{\mathcal{M}}$ and $0\leq r\leq1$. Then

$$2\mu_{t} \bigl(x^{1+r}+y^{1+r} \bigr)\geq \mu_{t} \bigl((x+y)^{\frac {1}{2}} \bigl(x^{r}+y^{r} \bigr) (x+y)^{\frac{1}{2}} \bigr),\quad t>0. $$

Proof

Let $0\leq x, y\in \overline{\mathcal{M}}$ and $0\leq r\leq1$. Since $1\leq1+r\leq2$, the function $t\rightarrow t^{1+r}$ is operator convex. Hence

$$\frac{x^{1+r}+y^{1+r}}{2}\geq \biggl(\frac{x+y}{2} \biggr)^{1+r} = \frac{1}{2}(x+y)^{\frac{1}{2}} \biggl(\frac{x+y}{2} \biggr)^{r}(x+y)^{\frac{1}{2}}. $$

Note that $t\rightarrow t^{r}$ ($0\leq r\leq1$) is operator concave, we obtain $\frac{x^{r}+y^{r}}{2}\leq(\frac{x+y}{2})^{r}$. Therefore,

$$x^{1+r}+y^{1+r}\geq\frac{1}{2}(x+y)^{\frac{1}{2}} \bigl(x^{r}+y^{r} \bigr) (x+y)^{\frac{1}{2}}. $$

This completes the proof. □

Based on Lemma 3.5 we now obtain the desired generalized singular number inequality (1.4) for τ-measurable operators.

Theorem 3.6

Let $0\leq r\leq1$ and $0\leq x, y\in L^{1}(\mathcal{M})$. Then

$$ \mu_{t} \bigl(x^{r}y^{1-r}+x^{1-r}y^{r} \bigr)\leq\mu_{t}(x+y),\quad t>0. $$

(3.2)

Proof

Let $0\leq v\leq1$. If we replace x, y by $x^{\frac{1}{1+v}}$, $y^{\frac{1}{1+v}}$, respectively, in Lemma 3.5, we deduce

$$2\mu_{t}(x+y)\geq\mu_{t} \bigl( \bigl(x^{\frac{1}{1+v}}+y^{\frac{1}{1+v}} \bigr)^{\frac{1}{2}} \bigl(x^{\frac{v}{1+v}}+y^{\frac{v}{1+v}} \bigr) \bigl(x^{\frac{1}{1+v}}+y^{\frac{1}{1+v}} \bigr)^{\frac{1}{2}} \bigr). $$

It follows from Lemma 2 of [13] and the fact $x, y\in L^{1}(\mathcal{M})$ that

$$ 2\mu_{t}(x+y)\geq\mu_{t} \bigl( \bigl(x^{\frac{1}{1+v}}+y^{\frac{1}{1+v}} \bigr) \bigl(x^{\frac{v}{1+v}}+y^{\frac{v}{1+v}} \bigr) \bigr). $$

(3.3)

Note that

$$\begin{aligned} &\mu_{t} \left ( \begin{pmatrix} (x^{\frac{1}{1+v}}+y^{\frac{1}{1+v}}) (x^{\frac{v}{1+v}}+y^{\frac{v}{1+v}})&0\\ 0&0 \end{pmatrix} \right ) \\ &\quad=\mu_{t} \left ( \begin{pmatrix} x^{\frac{1}{2+2v}}&y^{\frac{1}{2+2v}}\\ 0 &0 \end{pmatrix} \begin{pmatrix} x^{\frac{1}{2+2v}}&0\\ y^{\frac{1}{2+2v}} &0 \end{pmatrix} \begin{pmatrix} x^{\frac{v}{1+v}}+y^{\frac{v}{1+v}}&0\\ 0 &0 \end{pmatrix} \right ) \\ &\quad=\mu_{t} \left ( \begin{pmatrix} x^{\frac{1}{2+2v}}&0\\y^{\frac{1}{2+2v}} &0 \end{pmatrix} \begin{pmatrix} x^{\frac{v}{1+v}}+y^{\frac{v}{1+v}}&0\\ 0 &0 \end{pmatrix} \begin{pmatrix} x^{\frac{1}{2+2v}}&y^{\frac{1}{2+2v}}\\ 0 &0 \end{pmatrix} \right ) \\ &\quad=\mu_{t} \left ( \begin{pmatrix} x^{\frac{1}{2+2v}}(x^{\frac{v}{1+v}}+y^{\frac{v}{1+v}}) x^{\frac{1}{2+2v}}&x^{\frac{1}{2+2v}}(x^{\frac{v}{1+v}}+y^{\frac{v}{1+v}}) y^{\frac{1}{2+2v}}\\y^{\frac{1}{2+2v}}(x^{\frac{v}{1+v}}+y^{\frac{v}{1+v}}) x^{\frac{1}{2+2v}} &y^{\frac{1}{2+2v}}(x^{\frac{v}{1+v}}+y^{\frac{v}{1+v}}) y^{\frac{1}{2+2v}} \end{pmatrix} \right ). \end{aligned}$$

Combining Lemma 3.1, Lemma 3.3, and inequality (3.3) we deduce

$$\begin{aligned} \mu_{t}(x+y)&\geq\mu_{t} \bigl(x^{\frac{1}{2+2v}} \bigl(x^{\frac{v}{1+v}}+y^{\frac{v}{1+v}} \bigr) y^{\frac{1}{2+2v}} \bigr) \\ &=\mu_{t} \bigl(x^{\frac{2v+1}{2+2v}}y^{\frac{1}{2+2v}}+x^{\frac {1}{2+2v}}y^{\frac{2v+1}{2+2v}} \bigr),\quad 0\leq v\leq1. \end{aligned}$$

Therefore,

$$\mu_{t} \bigl(x^{r}y^{1-r}+x^{1-r}y^{r} \bigr)\leq\mu_{t}(x+y),\quad \frac{1}{2}\leq r\leq \frac{3}{4}. $$

On the one hand, we have

$$\begin{aligned} &\mu_{t} \left ( \begin{pmatrix} (x^{\frac{1}{1+v}}+y^{\frac{1}{1+v}}) (x^{\frac{v}{1+v}}+y^{\frac{v}{1+v}})&0\\ 0 &0 \end{pmatrix} \right ) \\ &\quad=\mu_{t} \left ( \begin{pmatrix} x^{\frac{v}{2+2v}}&0\\y^{\frac{v}{2+2v}} &0 \end{pmatrix} \begin{pmatrix} x^{\frac{1}{1+v}}+y^{\frac{1}{1+v}}&0\\ 0 &0 \end{pmatrix} \begin{pmatrix} x^{\frac{v}{2+2v}}&y^{\frac{v}{2+2v}}\\ 0 &0 \end{pmatrix} \right ). \end{aligned}$$

Repeating the arguments above we get

$$\mu_{t} \bigl(x^{r}y^{1-r}+x^{1-r}y^{r} \bigr)\leq\mu_{t}(x+y),\quad \frac{3}{4}\leq r\leq1. $$

By the symmetry property of inequality (3.2) with respect to $r=\frac{1}{2}$, we see that inequality (3.2) holds for all $0\leq r\leq1$. □

Let $0\leq x, y\in \overline{\mathcal{M}}$. Then Lemma 3.1 and Theorem 3.4 imply that

$$\mu_{t} \bigl((x-y)\oplus0 \bigr)\leq\mu_{t}(x\oplus y), \quad t>0. $$

If $x, y\in \overline{\mathcal{M}}$ with $\mu_{t}(x)\leq\mu_{t}(y)$, $t>0$, Lemma 3.1 gives us that

$$\mu_{t}(x)=\mu_{t}(x\oplus0)\leq\mu_{t}(y\oplus y),\quad t>0. $$

Some examples of such inequalities related to ones discussed above are presented below.

Lemma 3.7

Let $x, y\in\overline{\mathcal{M}}^{sa}: =\{z\in\overline{\mathcal{M}}; z=z^{*}\}$ such that $\pm y\leq x$. If $x\geq0$, then

$$\mu_{t}(y)\leq\mu_{t}(x\oplus x) $$

and

$$\int_{0}^{t}\mu_{t}(y)\,ds\leq \int_{0}^{t}\mu_{s}(x)\,ds,\quad t>0. $$

Proof

Since $\pm y\leq x$, we have $-x\leq y\leq x$. Then Theorem 1 of [14] indicates that $2|y|\leq x+uxu^{*}$ for some unitary $u\in\overline{\mathcal{M}}^{sa}$. From Theorem 4.4 and Lemma 2.5 of [7], we deduce

$$2\mu_{t}(y)\leq\mu_{t} \bigl(x+uxu^{*} \bigr)\leq \mu_{\frac{t}{2}} \bigl(uxu^{*} \bigr)+\mu_{\frac {t}{2}}(x)\leq2 \mu_{\frac{t}{2}}(x) =2\mu_{t}(x\oplus x),\quad t>0, $$

and

$$2 \int_{0}^{t}\mu_{s}(y)\,ds\leq \int_{0}^{t}\mu_{s} \bigl(x+uxu^{*} \bigr)\,ds \leq2 \int_{0}^{t}\mu_{s}(x)\,ds,\quad t>0. $$

 □

We conclude this section with a series of inequalities which are related to the Heinz mean inequality for a generalized singular number of τ-measurable operators.

Proposition 3.8

Let $x, y\in\overline{\mathcal{M}}$. Then

$$ \mu_{t} \bigl(x^{*}y+y^{*}x \bigr)\leq\mu_{t} \bigl( \bigl(x^{*}x+y^{*}y \bigr)\oplus \bigl(x^{*}x+y^{*}y \bigr) \bigr),\quad t>0, $$

(3.4)

and

$$ \mu_{t} \bigl(yx^{*}+xy^{*} \bigr)\leq\mu_{t} \bigl( \bigl(x^{*}x+y^{*}y \bigr)\oplus \bigl(x^{*}x+y^{*}y \bigr) \bigr),\quad t>0. $$

(3.5)

Proof

Since $(x\pm y)^{*}(x\pm y)\geq0$, we have $\pm(x^{*}y+y^{*}x)\leq x^{*}x+y^{*}y$. Thus inequality (3.4) follows from Lemma 3.7. Inequality (3.5) follows from Theorem 6 of [12] and Theorem 3.4(2). □

Corollary 3.9

Let $x, y\in\overline{\mathcal{M}}$ and $0< r\leq\infty$. Then

$$\int_{0}^{t}\mu_{s} \bigl(x^{*}y+y^{*}x \bigr)\,ds\leq \int_{0}^{t}\mu_{s} \bigl(x^{*}x+y^{*}y \bigr)\,ds,\quad t>0, $$

(3.6)

and

$$ \int_{0}^{t}\mu_{s} \bigl(yx^{*}+xy^{*} \bigr)\,ds\leq \int_{0}^{t}\mu_{s} \bigl(x^{*}x+y^{*}y \bigr)\,ds,\quad t>0. $$

(3.7)

Proof

It follows from Lemma 3.7 and the proof of Proposition 3.8. □

Proposition 3.10

Let $x, y\in\overline{\mathcal{M}}$. Then

$$ \mu_{t}(x+y)\leq\mu_{t} \bigl( \bigl(|x|+|y| \bigr) \oplus \bigl( \bigl|x^{*} \bigr|+ \bigl|y^{*} \bigr| \bigr) \bigr), \quad t>0. $$

(3.8)

Proof

Let $x\in\overline{\mathcal{M}}$. Note that $\bigl ( {\scriptsize\begin{matrix}{} |x|&\pm x^{*}\cr \pm x &|x^{*}| \end{matrix}} \bigr )\geq0$. Then

$$\begin{aligned} \begin{pmatrix} |x|+|y|& \pm(x+y)^{*}\\ \pm(x+y) &|x^{*}|+|y^{*}| \end{pmatrix}\geq0. \end{aligned}$$

Thus

$$\begin{aligned} \pm \begin{pmatrix} 0& (x+y)^{*}\\x+y &0 \end{pmatrix}\leq \begin{pmatrix} |x|+|y|&0\\0 &|x^{*}|+|y^{*}| \end{pmatrix}. \end{aligned}$$

By Lemma 3.7, we obtain

$$\begin{aligned} \mu_{t} \bigl((x+y)\oplus(x+y)^{*} \bigr)={}&\mu_{t} \left ( \begin{pmatrix} 0& (x+y)^{*}\\x+y &0 \end{pmatrix} \right ) \\ \leq{}&\mu_{t} \left ( \begin{pmatrix} |x|+|y|&0\\0 &|x^{*}|+|y^{*}| \end{pmatrix} \right . \\ &{}\left .\oplus \begin{pmatrix} |x|+|y|&0\\0 &|x^{*}|+|y^{*}| \end{pmatrix} \right ) \\ ={}&\mu_{\frac{t}{2}} \left ( \begin{pmatrix} |x|+|y|&0\\0 &|x^{*}|+|y^{*}| \end{pmatrix} \right ),\quad t>0. \end{aligned}$$

According to Lemma 2.5 of [7] and Lemma 3.1, we get

$$\mu_{t} \bigl((x+y)\oplus(x+y)^{*} \bigr)=\mu_{\frac{t}{2}}(x+y),\quad t>0. $$

This implies that

$$\mu_{t}(x+y)\leq\mu_{t} \left ( \begin{pmatrix} |x|+|y|&0\\0 &|x^{*}|+|y^{*}| \end{pmatrix} \right ),\quad t>0. $$

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Generalized singular number inequalities for products and sums of τ-measurable operators

In this section, we establish a generalized singular number inequality for τ-measurable operators which yields the well-known arithmetic-geometric mean inequalities as special cases.

The following proposition is a refinement of the inequality in Theorem 3.4(2).

Proposition 4.1

Let $x, y\in \overline{\mathcal{M}}$ and $0\leq z\in\mathcal{M}$. Then

$$\mu_{t} \bigl(xzy^{*} \bigr)\leq\frac{1}{2}\|z\|\mu_{t} \bigl(x^{*}x+y^{*}y \bigr),\quad t>0. $$

Proof

According to Proposition 2.5(vi) of [7] and Theorem 3.2(2), we have

$$\begin{aligned} 2\mu_{t} \bigl(xzy^{*} \bigr)&=2\mu_{t} \bigl(xz^{\frac{1}{2}}z^{\frac{1}{2}}y^{*} \bigr) \leq\mu_{t} \bigl( \bigl|xz^{\frac{1}{2}} \bigr|^{2}+ \bigl|yz^{\frac{1}{2}} \bigr|^{2} \bigr) \\ &= \mu_{t} \bigl(z^{\frac{1}{2}} \bigl(x^{*}x+y^{*}y \bigr)z^{\frac{1}{2}} \bigr) \leq\|z\|\mu_{t} \bigl(x^{*}x+y^{*}y \bigr). \end{aligned}$$

 □

From Proposition 4.1 we now obtain the promised generalized singular number inequality (1.5) for τ-measurable operators.

Proposition 4.2

Let $x_{i}, y_{i}\in \overline{\mathcal{M}}$ and $0\leq z_{i}\in\mathcal{M}$ ($i=1, 2, \ldots, n$). Then

Proof

Let

Then

From Proposition 4.1, we have

$$2\mu_{t} \bigl(AKB^{*} \bigr)\leq\|K\|\mu_{t} \bigl(A^{*}A+B^{*}B \bigr)= \|K\|\mu_{t} \bigl(|T|^{2} \bigr)=\|K\| \mu_{t}(T)^{2}, \quad t>0. $$

Then the result follows from Lemma 3.1. □

Proposition 4.2 includes several generalized singular number inequalities as special cases.

Corollary 4.3

Let $x_{i}, y_{i}\in \overline{\mathcal{M}}$ and $0\leq z_{i}\in\mathcal {M}$ ($i=1, 2$). Then

$$2\mu_{t} \bigl(x_{1}z_{1}y_{1}^{*}+ x_{2}z_{2}y_{2}^{*} \bigr)\leq \Bigl(\max _{i=1, 2}\|z_{i}\| \Bigr) \mu_{t} \left ( \begin{pmatrix} x_{1}&x_{2} \\ y_{1}&y_{2} \end{pmatrix} \right )^{2},\quad t>0. $$

In particular,

$$2\mu_{t} \bigl(xzy^{*}+ yzx^{*} \bigr)\leq \|z\| \mu_{t} \left ( \begin{pmatrix} x&y \\ y&x \end{pmatrix} \right )^{2},\quad t>0. $$

Proof

The result follows from Proposition 4.2. □

The following inequality is an application of Corollary 4.3.

Corollary 4.4

Let $0\leq x, y\in\overline{\mathcal{M}}$ and $0\leq z\in\mathcal {M}$. Then, for $t>0$,

$$\begin{aligned} \mu_{t} \bigl(x^{\frac{1}{2}}zx^{\frac{1}{2}}+y^{\frac{1}{2}}zy^{\frac{1}{2}} \bigr) \leq\|z\|\mu_{t} \bigl( \bigl(x+ \bigl|y^{\frac{1}{2}}x^{\frac{1}{2}} \bigr| \bigr)\oplus \bigl(y+ \bigl|x^{\frac {1}{2}}y^{\frac{1}{2}} \bigr| \bigr) \bigr). \end{aligned}$$

In particular,

$$\begin{aligned} \mu_{t}(x+y) \leq\mu_{t} \bigl( \bigl(x+ \bigl|y^{\frac{1}{2}}x^{\frac{1}{2}} \bigr| \bigr)\oplus \bigl(y+ \bigl|x^{\frac {1}{2}}y^{\frac{1}{2}} \bigr| \bigr) \bigr) \quad\textit{for all } t>0. \end{aligned}$$

Proof

Let $x_{1}=y_{1}=x^{\frac{1}{2}}$, $x_{2}=y_{2}=y^{\frac{1}{2}}$, and $z_{1}=z_{2}=z$ in Corollary 4.3. Then for all $t>0$

$$\begin{aligned} 2\mu_{t} \bigl(x^{\frac{1}{2}}zx^{\frac{1}{2}}+y^{\frac{1}{2}}zy^{\frac {1}{2}} \bigr)&\leq\|z\| \mu_{t} \left ( \begin{pmatrix} x^{\frac{1}{2}}&y^{\frac{1}{2}} \\ x^{\frac{1}{2}}&y^{\frac{1}{2}} \end{pmatrix} \right )^{2} \\ &=2\|z\| \mu_{t} \left ( \begin{pmatrix} x & x^{\frac{1}{2}}y^{\frac{1}{2}} \\ y^{\frac{1}{2}}x^{\frac{1}{2}}& y \end{pmatrix} \right ) \\ &=2\|z\| \mu_{t}(T_{1}+T_{2}), \end{aligned}$$

where $T_{1}=\bigl ( {\scriptsize\begin{matrix}{} x & 0\cr 0& y \end{matrix}} \bigr )$ and $T_{2}=\bigl ( {\scriptsize\begin{matrix}{} 0& x^{\frac{1}{2}}y^{\frac{1}{2}} \cr y^{\frac{1}{2}}x^{\frac{1}{2}}&0 \end{matrix}} \bigr )$. It follows from the facts that $T_{2}\leq|T_{2}|=\bigl ( {\scriptsize\begin{matrix}{} |y^{\frac{1}{2}}x^{\frac{1}{2}}|&0 \cr 0&|x^{\frac{1}{2}}y^{\frac{1}{2}}| \end{matrix}} \bigr )$ and $T_{1}+|T_{2}|\geq0$ that

$$\begin{aligned} \mu_{t} \bigl(x^{\frac{1}{2}}zx^{\frac{1}{2}}+y^{\frac{1}{2}}zy^{\frac{1}{2}} \bigr) &\leq\|z\| \mu_{t} \bigl(T_{1}+|T_{2}| \bigr),\quad t>0. \end{aligned}$$

This gives the desired inequality. □

The following inequality contains a generalization of the inequality in Theorem 3.4(1).

Corollary 4.5

Let $x, y\in\overline{\mathcal{M}}$ and $0\leq z\in\mathcal{M}$. Then

$$\begin{aligned} \mu_{t} \bigl(xzx^{*}-yzy^{*} \bigr) \leq\|z\|\mu_{t} \bigl(x^{*}x \oplus y^{*}y \bigr) \quad\textit{for all } t>0. \end{aligned}$$

Proof

If we replace $x_{1}$, $x_{2}$, $y_{1}$, $y_{2}$ by x, y, x, −y, respectively, in Corollary 4.3, we deduce

$$2\mu_{t} \bigl(xzx^{*}-yzy^{*} \bigr)\leq\|z\|\mu_{t} \left ( \begin{pmatrix} 2x^{*}x& 0\\ 0&2y^{*}y \end{pmatrix} \right ) \quad\mbox{for all } t>0. $$

 □