On Anderson–Taylor type inequalities for τ-measurable operators

Abstract

The main purpose of this article is to establish Anderson–Taylor type inequalities for τ-measurable operators. Some related results for M-operators are also obtained.

1 Introduction

Let $\mathbb{M}_{n}(\mathrm{C})$ be the space of $n\times n$ complex matrices. For two Hermitian matrices $A, B\in\mathbb{M}_{n}(\mathrm{C})$, $A>B$ ($A \geq B$) means $A-B$ is positive (semi) definite. Then $A> 0$ ($A\geq0$) means A is positive (semi) definite. Of course, $B> A$ ($B \geq A$) is not distinguished from $A>B$ ($A \geq B$). And we call the comparison of Hermitian matrices in this way Löewner partial order. Let $A\geq0$, thus it has a unique square root $A^{\frac{1}{2}}\geq0$. Let trA denote the trace of A. In view of the applications in probability theory, Anderson and Taylor [1, Proposition 1] proved a quadratic inequality for real numbers. In 1983, Olkin [9, Proposition] established a stronger matrix version of Anderson–Taylor inequality as well as a related trace inequality. Using the well-known arithmetic-geometric mean inequality for singular values due to Bhatia and Kittaneh [3], Zhan [10] gave a trace inequality for sums of positive semi-definite matrices, which is a generalization of Anderson–Taylor quadratic inequality for real numbers. Recently, Lin [8] provided a complement to Olkin’s generalization of Anderson–Taylor trace inequality and some related results for M-matrices.

In this article we consider τ-measurable operators affiliated with a finite von Neumann algebra equipped with a normal faithful finite trace τ. By virtue of the method of Lin and Zhan [8, 10], based on the notion of generalized singular value studied by Fack and Kosaki [5], we obtain generalizations of results in [8] and [10] with regard to Anderson–Taylor type inequalities for τ-measurable operators case.

2 Preliminaries

Unless stated otherwise, throughout the paper $\mathcal{M}$ will always denote a finite von Neumann algebra acting on the complex separable Hilbert space $\mathcal {H}$, with a normal faithful finite trace τ. We denote the identity in $\mathcal{M}$ by 1 and let $\mathcal{P}$ denote the projection lattice of $\mathcal{M}$. The 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 u which belong to the commutant $\mathcal{M^{\prime}}$ of $\mathcal {M}$. If x is affiliated with $\mathcal{M}$, then x is said to be τ-measurable if for every $\epsilon>0$ there exists a projection $e\in\mathcal{M}$ such that $e(\mathcal{H})\subseteq D(x)$ and $\tau(\mathbf{1}-e)<\epsilon$. The set of all τ-measurable operators will be denoted by $L_{0}(\mathcal{M}, \tau)$, or simply $L_{0}(\mathcal{M})$. The set $L_{0}(\mathcal{M})$ is a ∗-algebra with sum and product being the respective closures of the algebraic sum and product. The space $L_{0}(\mathcal{M})$ is a partially ordered vector space under the ordering $x\geq0$ defined by $(x\xi, \xi)\geq0$, $\xi\in D(x)$. When $x\geq0$ is invertible, we write $x>0$.

Recall that the geometric mean of two positive definite operators x and y, denoted by $x\sharp y$, is the positive definite solution of the operator equation $zy^{-1}z=x$ and it has the explicit expression

$$ x\sharp y=y^{\frac{1}{2}}\bigl(y^{-\frac{1}{2}}xy^{-\frac{1}{2}} \bigr)^{\frac {1}{2}}y^{\frac{1}{2}}. $$

From this, we find that $x\sharp y=y\sharp x$ and the monotonicity property: $x\sharp y\geq x\sharp z$, whenever $y\geq z>0$ and $x>0$. One of the motivations for geometric mean is the following arithmetic mean-geometric mean inequality:

$$ \frac{x+y}{2}\geq x\sharp y. $$

A remarkable property of the geometric mean is a maximal characterization which is a generalization of the result in [4, Theorem 3.7]; see Lemma 3.4 in Sect. 3 for more details.

Definition 2.1

Let $x\in L_{0}(\mathcal{M})$ and $t>0$. The “tth singular number (or generalized s-number) of x” is defined by

$$\mu_{t}(x)=\inf\bigl\{ \Vert xe \Vert : e \in\mathcal{P}, \tau( \mathbf{1}-e)\leq t\bigr\} . $$

We will denote simply by $\mu(x)$ the function $t\rightarrow\mu_{t}(x)$. The generalized singular number function $t\rightarrow\mu_{t}(x)$ is decreasing right-continuous. Furthermore,

$$\begin{aligned} \mu(uxv)\leq \Vert v \Vert \Vert u \Vert \mu(x),\quad u,v\in \mathcal{M}, x\in L_{0}(\mathcal{M}) \end{aligned}$$

(2.1)

and

$$\begin{aligned} \mu\bigl(f(x)\bigr)=f\bigl(\mu(x)\bigr) \end{aligned}$$

(2.2)

whenever $0\leq x\in L_{0}(\mathcal{M})$ and f is an increasing continuous function on $[0,\infty)$ satisfying $f(0)=0$. See [5] for basic properties and detailed information on the generalized s-numbers.

Let $\mathbb{M}_{2}(\mathcal{M})$ denote the linear space of $2\times 2$ matrices

$$ x= \begin{bmatrix} x_{11} & x_{12} \\ x_{21} & x_{22} \end{bmatrix} $$

with entries $x_{jk}\in\mathcal{M}$, $j, k=1,2$. Let $\mathcal{H}^{2}=\mathcal{H}\oplus\mathcal{H}$, then $\mathbb{M}_{2}(\mathcal{M})$ is a von Neumann algebra in the Hilbert space $\mathcal{H}^{2}$. For $x\in\mathbb{M}_{2}(\mathcal{M})$, define $\tau_{2}(x)=\sum_{j=1}^{2}\tau(x_{jj})$. Then $\tau_{2}$ is a normal faithful finite trace on $\mathbb{M}_{2}(\mathcal{M})$. The direct sum of operators $x_{1}, x_{2}\in L_{0}(\mathcal{M})$, denoted by $\bigoplus_{j=1}^{2} x_{j}$, is the block-diagonal operator matrix defined on $\mathcal{H}^{2}$ by

$$ \bigoplus_{j=1}^{2} x_{j}= \begin{bmatrix} x_{1} & 0 \\ 0 & x_{2} \end{bmatrix} . $$

3 Anderson–Taylor type inequalities

To present our main results, we firstly give the following lemma. Since it is easy to obtain in a similar way to [9, Lemma], we omit the proof.

Lemma 3.1

Let $x, y\in\mathcal{M}$ with $x>0$, $y\geq0$, then

$$\begin{aligned} (x+y)^{-1}y(x+y)^{-1}\leq x^{-1}-(x+y)^{-1}. \end{aligned}$$

(3.1)

Our next result provides an operator generalization of a quadratic inequality for a matrix.

Theorem 3.2

Let $z, x_{j}\in\mathcal{M}$ with $z>0$ and $x_{j}\geq0$ ($j=1,2,\ldots,n$), then

$$\begin{aligned} z^{-1}>\sum_{k=1}^{n} \Biggl(z+\sum_{j=1}^{k} x_{j} \Biggr)^{-1}x_{k}\Biggl(z+\sum_{j=1}^{k} x_{j}\Biggr)^{-1} . \end{aligned}$$

(3.2)

Proof

Let $x=z+x_{0}+\sum_{j=1}^{k-1} x_{j}$, $y=x_{k}$, with $x_{0}\equiv0$. By an application of (3.1) we obtain

$$\begin{aligned} &\sum_{k=1}^{n}\Biggl(z+x_{0}+ \sum_{j=1}^{k} x_{j} \Biggr)^{-1} x_{k} \Biggl(z+x_{0}+\sum _{j=1}^{k} x_{j}\Biggr)^{-1} \\ &\quad\leq\sum_{k=1}^{n} \Biggl(\Biggl(z+\sum _{j=0}^{k-1}x_{j} \Biggr)^{-1}-\Biggl(z+\sum_{j=0}^{k} x_{j}\Biggr)^{-1} \Biggr) \\ &\quad=z^{-1}-\Biggl(z+\sum_{j=1}^{n} x_{j}\Biggr)^{-1} \\ &\quad< z^{-1}, \end{aligned}$$

which completes the proof of (3.2). □

An immediate consequence from (3.2) is as follows:

$$\begin{aligned} &\sum_{k=1}^{n} \tau \Biggl( \Biggl(z+\sum _{j=1}^{k} x_{j} \Biggr)^{-1}x_{k}\Biggl(z+\sum_{j=1}^{k} x_{j}\Biggr)^{-1} \Biggr) \\ &\quad=\sum_{k=1}^{n} \tau \Biggl(x_{k} \Biggl(z+\sum_{j=1}^{k} x_{j} \Biggr)^{-2} \Biggr) \\ &\quad< \tau\bigl(z^{-1}\bigr). \end{aligned}$$

Furthermore, under the same condition as in Theorem 3.2 we observe that

$$\begin{aligned} z^{-1}>\sum_{k=1}^{n} \Biggl(z+\sum_{j=1}^{k} x_{j} \Biggr)^{-1}x_{k}\Biggl(z+\sum_{j=1}^{k} x_{j}\Biggr)^{-1}. \end{aligned}$$

(3.3)

In what follows, we first give an inequality complementary to (3.3). To obtain it, we need several lemmas.

Lemma 3.3

Let $x, z\in\mathcal{M}$ with $x, z>0$ and $x=x^{*}$, $z=z^{*}$, and let x be invertible, $y\in\mathcal{M}$. Then the $2\times2$ operator matrix

$$\begin{aligned} \begin{bmatrix} x & y \\ y^{*} & z \end{bmatrix} \end{aligned}$$

is positive semi-definite if and only if $z\geq y^{*}x^{-1}y$.

Proof

Put $D=z-y^{*}x^{-1}y$, thus

$$\begin{aligned} \begin{bmatrix} x & y \\ y^{*} & z \end{bmatrix} = \begin{bmatrix} x & y \\ y^{*} & y^{*}x^{-1}y \end{bmatrix} + \begin{bmatrix} 0 & 0 \\ 0 & D \end{bmatrix} . \end{aligned}$$

Note that

$$\begin{aligned} \begin{bmatrix} x & y \\ y^{*} & y^{*}x^{-1}y \end{bmatrix} = \begin{bmatrix} x^{\frac{1}{2}} & x^{-\frac{1}{2}}y \\ 0 & 0 \end{bmatrix} ^{*} \cdot \begin{bmatrix} x^{\frac{1}{2}} & x^{-\frac{1}{2}}y \\ 0 & 0 \end{bmatrix} \geq0. \end{aligned}$$

(3.4)

Hence, the fact that D is positive semi-definite is sufficient to ensure that $\left[\begin{array}{cc}x& y\\ y^{\ast }& z\end{array}\right]$ is positive semi-definite.

On the other hand, it is also evident from (3.4) that for any $\nu\in\mathbb{C}^{n}$, the vector $\left[\begin{array}{c}{x}^{-1}y\nu \\ \nu \end{array}\right]$ belongs to the null space of $\left[\begin{array}{cc}x& y\\ y^{\ast }& y^{\ast }{x}^{-1}y\end{array}\right]$, therefore,

$$\begin{aligned} \left\langle \begin{bmatrix} x^{-1}y\nu\\ \nu \end{bmatrix} , \begin{bmatrix} x & y \\ y^{*} & z \end{bmatrix} \cdot \begin{bmatrix} x^{-1}y\nu\\ \nu \end{bmatrix} \right\rangle =\langle\nu,D\nu\rangle, \end{aligned}$$

consequently, the positive semi-definiteness of D is necessary to ensure that $\left[\begin{array}{cc}x& y\\ y^{\ast }& z\end{array}\right]$ is positive semi-definite. □

Lemma 3.3 says that the set of positive Hermitian operators $z\in\mathcal{M}$ such that $\left[\begin{array}{cc}x& y\\ y^{\ast }& z\end{array}\right]$ is positive semi-definite has a minimum, namely $z=y^{*}x^{-1}y$.

In the next result, we give a generalization of [4, Theorem 3.7].

Lemma 3.4

For all positive $x, z\in\mathcal{M}$, the set of all $y\in\mathcal {M}$ such that

$$\begin{aligned} \begin{bmatrix} x & y \\ y & z \end{bmatrix} >0 \end{aligned}$$

has a maximal element, which is $M(x,z)$.

Proof

If

$$\begin{aligned} \begin{bmatrix} x & y \\ y & z \end{bmatrix} >0, \end{aligned}$$

then via Lemma 3.3, $z\geq yx^{-1}y$, and hence

$$\begin{aligned} x^{-\frac{1}{2}}zx^{-\frac{1}{2}}\geq x^{-\frac {1}{2}}yx^{-1}yx^{-\frac{1}{2}}= \bigl(x^{-\frac{1}{2}}yx^{-\frac{1}{2}}\bigr)^{2}. \end{aligned}$$

From the operator monotonicity of the square root functions, it follows that

$$\begin{aligned} x^{\frac{1}{2}}\bigl(x^{-\frac{1}{2}}zx^{-\frac{1}{2}} \bigr)^{\frac{1}{2}}x^{\frac {1}{2}}\leq y. \end{aligned}$$

This shows the maximality property of $M(x,z)$, i.e.,

$$\begin{aligned} x\sharp z =\max \left\{y\biggm| \begin{bmatrix} x & y \\ y^{*} & z \end{bmatrix} \geq0, y=y^{*} \right\}. \end{aligned}$$

 □

Applying Lemma 3.4 to the summation of positive semi-definite operator matrices $\left[\begin{array}{cc}{x}_i& x_i\mathrm{♯}{y}_i\\ x_i\mathrm{♯}{y}_i& y_i\end{array}\right]$, $i=1,2,\ldots,n$, we get the following inequality:

$$\begin{aligned} \Biggl(\sum_{i=1}^{n} x_{i}\Biggr)\sharp\Biggl(\sum_{i=1}^{n} y_{i}\Biggr)\succeq\sum_{i=1}^{n} (x_{i}\sharp y_{i}). \end{aligned}$$

(3.5)

The operator geometric mean has the similar properties to that of matrix geometric mean in [2]. As for the next lemma, its proof is similar to that of [8, Lemma 2.2] and we give it for easy reference.

Lemma 3.5

Let $x, y\in\mathcal{M}$ with $x>0$ and y Hermitian. Then

$$\begin{aligned} x\sharp\bigl(yx^{-1}y\bigr)\geq y. \end{aligned}$$

(3.6)

Proof

We may assume that y is invertible and the general case follows from a continuity argument. In fact, via Lemma 3.4, the notion of geometric mean can be extended to cover the case of positive semi-definite operators. Based on the proof of Lemma 3.3, it is easy to check that

$$\begin{aligned} \begin{bmatrix} x & y \\ y & yx^{-1}y \end{bmatrix} \geq0. \end{aligned}$$

Now from Lemma 3.4, the desired inequality follows. □

Remark 3.6

Observe that inequality (3.6) is surely a refinement of the following inequality for $\mathcal{M}\ni x>0$ and Hermitian operator $y\in \mathcal{M}$:

$$\begin{aligned} x+yx^{-1}y\geq2y. \end{aligned}$$

(3.7)

Lemma 3.7

Let $x, y\in\mathcal{M}$ with $x, y>0$. Then $x\sharp y\geq y$ if and only if $x\geq y$.

Now we are ready to state our main result. It is easy to get this theorem in a similar way to [8, Theorem 2.5], for completeness, we include a simple proof.

Theorem 3.8

Let $x_{i}\in\mathcal{M}$ with $x_{i}>0$ for $i=1,2,\ldots,n$. Then

$$\begin{aligned} \sum_{j=1}^{n} \Biggl(\sum _{i=1}^{j} x_{i} \Biggr)x_{j}^{-1}\Biggl(\sum_{i=1}^{j} x_{i}\Biggr)>\frac {1}{2}\sum_{k=1}^{n} \sum_{j=1}^{k}\sum _{i=1}^{j}x_{i}. \end{aligned}$$

(3.8)

Moreover, the constant $1/2$ is best possible.

Proof

Interchange the order of summation and we deduce that

$$\begin{aligned} \sum_{k=1}^{n}\sum _{j=1}^{k}\sum_{i=1}^{j}x_{i} &=\sum_{j=1}^{n} \sum _{k=j}^{n}\Biggl(\sum_{i=1}^{j} x_{i}\Biggr) \\ &=\sum_{j=1}^{n}(n-j+1)\sum _{i=1}^{j} x_{i} \\ &=\sum_{i=1}^{n} x_{i} \sum _{j=1}^{n}(n-j+1) \\ &=\sum_{i=1}^{n} {2 \choose n-i+2} x_{i} \\ &>\frac{1}{2}\sum_{i=1}^{n} (n-i+1)^{2} x_{i}, \end{aligned}$$

i.e.,

$$\begin{aligned} 2\sum_{k=1}^{n}\sum _{j=1}^{k}\sum_{i=1}^{j}x_{i} >\sum_{i=1}^{n} (n-i+1)^{2} x_{i}. \end{aligned}$$

(3.9)

On the other hand, combining (3.5) and (3.9), by an application of Lemma 3.4 we obtain

$$\begin{aligned} \sum_{k=1}^{n}\sum _{j=1}^{k}\sum_{i=1}^{j}x_{i} &=\sum_{j=1}^{n} (n-j+1) \Biggl(\sum _{i=1}^{j} x_{i}\Biggr) \\ &< \sum_{j=1}^{n} \bigl((n-j+1)^{2}x_{j} \bigr)\sharp \Biggl\{ \Biggl(\sum_{i=1}^{j} x_{i}\Biggr)x_{j}^{-1}\Biggl(\sum _{i=1}^{j} x_{i}\Biggr) \Biggr\} \\ &< \Biggl\{ \sum_{j=1}^{n} \bigl((n-j+1)^{2}x_{j}\bigr) \Biggr\} \sharp \Biggl\{ \sum _{j=1}^{n}\Biggl(\sum _{i=1}^{j} x_{i}\Biggr)x_{j}^{-1} \Biggl(\sum_{i=1}^{j} x_{i}\Biggr) \Biggr\} \\ &< 2 \Biggl\{ \sum_{k=1}^{n}\sum _{j=1}^{k}\sum_{i=1}^{j}x_{i} \Biggr\} \sharp \Biggl\{ \sum_{j=1}^{n}\Biggl( \sum_{i=1}^{j} x_{i} \Biggr)x_{j}^{-1}\Biggl(\sum_{i=1}^{j} x_{i}\Biggr) \Biggr\} \\ &= \Biggl\{ \sum_{k=1}^{n}\sum _{j=1}^{k}\sum_{i=1}^{j}x_{i} \Biggr\} \sharp \Biggl\{ 2\sum_{j=1}^{n} \Biggl(\sum_{i=1}^{j} x_{i} \Biggr)x_{j}^{-1}\Biggl(\sum_{i=1}^{j} x_{i}\Biggr) \Biggr\} . \end{aligned}$$

Hence, the assertion follows from Lemma 3.7. □

Regarding the proof that the constant $1/2$ in (3.8) is best possible, it could be organized by using the method applied in [8, Appendix], thus we omit it.

Remark 3.9

Under the same condition as in Theorem 3.8, we have the following inequality:

$$\begin{aligned} \tau \Biggl( \sum_{j=1}^{n} x_{j}^{-1}\Biggl(\sum_{i=1}^{j} x_{i}\Biggr)^{2} \Biggr) >\frac{1}{2}\sum _{k=1}^{n}\sum_{j=1}^{k} \sum_{i=1}^{j}\tau(x_{i}). \end{aligned}$$

(3.10)

4 M-Operators analog

In this section, we extend some results for M-matrix established in [8] to M-operators. From the definition of M-matrix, namely, [7, Definition 2.4.3], we could define the M-operator as follows.

Definition 4.1

Let $x\in\mathcal{M}$ be positive and invertible operator. x is called an M-operator if $x=sI-x_{1}$, where $x_{1}\geq0$ and $s>r(x_{1})$ with $r(x_{1})$ the spectral radius of $x_{1}$.

Next we give the following lemma without the proof, as it is immediate from [6, p. 117].

Lemma 4.2

Let $x, x+y\in\mathcal{M}$ be two M-operators with $y\geq0$. Then

$$\begin{aligned} x^{-1}-(x+y)^{-1}\geq(x+y)^{-1}y(x+y)^{-1}. \end{aligned}$$

(4.1)

Observe that the following result can also be given according to Theorem 3.2 and here we give another proof.

Proposition 4.3

Let $x_{1}, x_{1}+\sum_{i=2}^{n} x_{i} \in\mathcal{M}$ be M-operators with $x_{i}\geq0$ for $i=2,\ldots,n$. Then

$$\begin{aligned} 2x_{1}^{-1}>\sum_{j=1}^{n} \Biggl(\sum_{i=1}^{j} x_{i} \Biggr)^{-1} x_{j} \Biggl(\sum_{i=1}^{j} x_{i}\Biggr)^{-1}. \end{aligned}$$

(4.2)

Proof

Observe that $x_{1}+\sum_{i=2}^{j} x_{i} \in\mathcal{M}$ is an M-operator for $j=2,\ldots,n$. Moreover, we note that (4.2) is the same as

$$\begin{aligned} x_{1}^{-1}>\sum_{j=2}^{n} \Biggl(\sum_{i=1}^{j} x_{i} \Biggr)^{-1} x_{j} \Biggl(\sum_{i=1}^{j} x_{i}\Biggr)^{-1}. \end{aligned}$$

(4.3)

In fact, let $x=\sum_{i=1}^{j-1}x_{i}$, $y=x_{j}$ ($2\leq j\leq n$). By Lemma 4.2 we have

$$\begin{aligned} \Biggl(\sum_{i=1}^{j-1} x_{i} \Biggr)^{-1}-\Biggl(\sum_{i=1}^{j-1} x_{i}+x_{j}\Biggr)^{-1}>\Biggl(\sum _{i=1}^{j-1} x_{i}+x_{j} \Biggr)^{-1}x_{j}\Biggl(\sum_{i=1}^{j-1} x_{i}+x_{j}\Biggr)^{-1}, \end{aligned}$$

i.e.,

$$\begin{aligned} \Biggl(\sum_{i=1}^{j-1} x_{i}\Biggr)^{-1}-\Biggl(\sum_{i=1}^{j} x_{i}\Biggr)^{-1}>\Biggl(\sum_{i=1}^{j} x_{i}\Biggr)^{-1}x_{j}\Biggl(\sum _{i=1}^{j} x_{i}\Biggr)^{-1}. \end{aligned}$$

(4.4)

Summing up from 2 to n on both sides of inequality (4.4), we deduce that

$$\begin{aligned} \sum_{j=2}^{n} \Biggl(\Biggl(\sum _{i=1}^{j-1} x_{i}\Biggr)^{-1}- \Biggl(\sum_{i=1}^{j} x_{i} \Biggr)^{-1} \Biggr)>\sum_{j=2}^{n} \Biggl(\Biggl(\sum_{i=1}^{j} x_{i} \Biggr)^{-1}x_{j}\Biggl(\sum_{i=1}^{j} x_{i}\Biggr)^{-1} \Biggr), \end{aligned}$$

thus we get the desired result. □

Remark 4.4

Under the assumption of Proposition 4.3, take the trace in (4.2) and we immediately derive that

$$\begin{aligned} 2\tau\bigl(x_{1}^{-1}\bigr)>\tau\Biggl(\sum _{j=1}^{n} x_{j} \Biggl(\sum _{i=1}^{j} x_{i} \Biggr)^{-2}\Biggr). \end{aligned}$$

(4.5)