# 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. (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. (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. (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. (1)

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

2. (2)

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

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

(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

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

□

## 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

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

□

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## Acknowledgements

The authors would like to thank the editor and anonymous referees for their helpful comments and suggestions on the quality improvement of the manuscript. This research is supported by the National Natural Science Foundation of China No. 11401507 and the Natural Science Foundation of Xinjiang University (Starting Fund for Doctors, Grant No. BS150202).

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Correspondence to Yazhou Han.

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The authors declare that there is no conflict of interests regarding the publication of this paper.

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All authors contributed equally to the writing of this paper. All authors read and approved the final manuscript.

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Han, Y., Shao, J. More results on generalized singular number inequalities of τ-measurable operators. J Inequal Appl 2016, 144 (2016). https://doi.org/10.1186/s13660-016-1085-1

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• DOI: https://doi.org/10.1186/s13660-016-1085-1

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### Keywords

• generalized singular number
• von Neumann algebra
• τ-measurable operator