A generalized Hölder-type inequalities for measurable operators

Abstract

We prove a generalized Hölder-type inequality for measurable operators associated with a semi-finite von Neumann algebra which is a generalization of the result shown by Bekjan (Positivity 21:113–126, 2017). This also provides a generalization of the unitarily invariant norm inequalities for matrix due to Bhatia–Kittaneh, Horn–Mathisa, Horn–Zhan and Zou under a cohyponormal condition.

1 Introduction

Let $\mathbb{M}_{n}$ be the space of $n\times n$ complex matrices. A norm $|\!|\!|\cdot |\!|\!|$ on $\mathbb{M}_{n}$ is called unitarily invariant if $|\!|\!|UAV|\!|\!|=|\!|\!|A|\!|\!|$ for all $A\in \mathbb{M}_{n}$ and all unitary matrices $U, V\in \mathbb{M}_{n}$. Let $A, B\in \mathbb{M}_{n}$. In 1990, Bhatia and Kittaneh [6] established an arithmetic–geometric mean inequality for unitarily invariant norms, i.e.,

$$\begin{aligned} \big|\!\big|\!\big|A^{*}B\big|\!\big|\!\big|\leq \frac{1}{2} \big|\!\big|\!\big|AA^{*}+BB^{*}\big|\!\big|\!\big|. \end{aligned}$$

(1.1)

Using tensor algebra techniques, a strengthening inequality of (1.1) was presented by Bhatia and Davis [5]

$$\begin{aligned} \big|\!\big|\!\big|A^{*}XB\big|\!\big|\!\big|\leq \frac{1}{2} \big|\!\big|\!\big|AA^{*}X+XBB^{*}\big|\!\big|\!\big| \end{aligned}$$

(1.2)

for $A, B, X\in \mathbb{M}_{n}$. On the other hand, let $A,B \in M_{n}$ and $r>0$, Horn and Mathisa proved in [15] the following Cauchy–Schwarz inequality for unitarily invariant norms

$$\begin{aligned} \big|\!\big|\!\big|\bigl\vert A^{*}B \bigr\vert ^{r}\big|\!\big|\!\big|^{2}\leq \big|\!\big|\!\big|\bigl(AA^{*} \bigr)^{r}\big|\!\big|\!\big|\big|\!\big|\!\big|\bigl(BB^{*} \bigr)^{r}\big|\!\big|\!\big|. \end{aligned}$$

(1.3)

Let $A, B\in \mathbb{M}_{n}$ and $\frac{1}{p}+\frac{1}{q}=1$, $p, q>1$, $r\geq 0$. With the properties of C-S semi-norms in hand, Horn and Zhan [16] established a stronger version of inequality (1.3) as follows:

$$\begin{aligned} \big|\!\big|\!\big|\bigl\vert A^{*}B \bigr\vert ^{r}\big|\!\big|\!\big|\leq \big|\!\big|\!\big|\bigl(AA^{*} \bigr)^{\frac{rp}{2}}\big|\!\big|\!\big|^{\frac{1}{p}} \big|\!\big|\!\big|\bigl(BB^{*} \bigr)^{ \frac{rq}{2}}\big|\!\big|\!\big|^{\frac{1}{q}}. \end{aligned}$$

(1.4)

which is the Hölder inequality for unitarily invariant norms. In particular, these authors also showed in [16] that

$$\begin{aligned} \big|\!\big|\!\big|\bigl\vert A^{*}XB \bigr\vert ^{r}\big|\!\big|\!\big|\leq \big|\!\big|\!\big|\bigl( \vert A \vert ^{p}X \bigr)^{r}\big|\!\big|\!\big|^{\frac{1}{p}} \big|\!\big|\!\big|\bigl(X \vert B \vert ^{q}\bigr)^{r} \big|\!\big|\!\big|^{\frac{1}{q}}. \end{aligned}$$

(1.5)

Subsequently, a considerable different proofs, equivalent statements, along with some generalizations, refinements, and applications of inequalities (1.1)–(1.4) were discussed by many authors. We refer to [1–3, 5, 15, 20] for more information on this topic and historical references.

Let $A, B\in \mathbb{M}_{n}$ and $\frac{1}{p}+\frac{1}{q}=1$, $p, q>1$, $\alpha \in [0, 1]$, $r\geq 0$ and let $T_{X}(\alpha )=\alpha AA^{*}X+(1-\alpha )XBB^{*}$. In 2015, by majorization techniques, Audenaert [2] prove an inequality that interpolates between the arithmetic–geometric mean and Cauchy–Schwarz matrix norm inequalities

$$\begin{aligned} \big|\!\big|\!\big|\bigl\vert A^{*}B \bigr\vert ^{r}\big|\!\big|\!\big|\leq \big|\!\big|\!\big|\bigl(T_{1}(\alpha ) \bigr)^{\frac{rp}{2}}\big|\!\big|\!\big|^{ \frac{1}{p}} \big|\!\big|\!\big|\bigl(T_{1}(1- \alpha )\bigr)^{\frac{rq}{2}}\big|\!\big|\!\big|^{\frac{1}{q}}. \end{aligned}$$

(1.6)

Recently, Zou [20] presented the inequality for unitarily invariant norms

$$\begin{aligned} \big|\!\big|\!\big|\bigl\vert A^{*}XB \bigr\vert ^{2r}\big|\!\big|\!\big|\leq \big|\!\big|\!\big|\bigl(T_{X}(\alpha ) \bigr)^{rp}\big|\!\big|\!\big|^{\frac{1}{p}} \big|\!\big|\!\big|\bigl(T_{X}(1- \alpha )\bigr)^{rq}\big|\!\big|\!\big|^{\frac{1}{q}}, \end{aligned}$$

(1.7)

which is a unified version of inequalities (1.1) and (1.6).

By the concept of uniform Hardy–Littlewood majorization Bekjan [8] gave a Hölder-type inequality (1.4) for τ-measurable operators associated with a semi-finite von Neumann algebra and for symmetric Banach spaces norm. In this paper, we will give a generalized Hölder-type inequality (1.7) for τ-measurable operators under a cohyponormal condition by adopting a technique similar to the one used by Bekjan and Zou. This is a generalization of Bekjan’s result in [8].

2 Preliminaries

Let $L_{0}$ be the set of all Lebesgue measurable functions on $(0,\infty )$. A Banach space $E\subseteq L_{0}$ with the norm $\|\cdot \|_{E}$ satisfying the condition that $f\in E$ and $\|f\|_{E}\leq \|g\|_{E}$ whenever $0\leq f\leq g$, $f\in L_{0}$ and $g\in F$, is said to be a Banach function space. A Banach function space $E\subseteq L_{0}$ is called a symmetric Banach function space if it follows from $f\in L_{0}$, $g\in E$ and $f^{*}\leq g^{*}$ that $f\in E$ and $\|f\|_{E}\leq \|g\|_{E}$, where

$$ f^{*}(t)=\inf \bigl\{ s>0: d_{f}(s)=m\bigl\{ r: \bigl\vert f(r) \bigr\vert >s\bigr\} \leq t\bigr\} ,\quad t>0, $$

and m denotes the Lebesgue measure on $(0,\infty )$. The symmetric Banach function space E is called fully if and only if $f\in E$, $g\in L_{0}$ and $\int _{0}^{t}f^{*}(s)\,ds\geq \int _{0}^{t}g^{*}(s)\,ds$ give us that $g\in E$ and $\|f\|_{E}\geq \|g\|_{E}$. We say that E has order continuous norm if for every net $\{f_{i}\}_{i\in \varLambda }\subseteq E$ such that $f_{i}\downarrow 0$ we have $\|f_{i}\|_{E}\downarrow 0$. In particular, a symmetric Banach function space which has order continuous norm is automatically fully symmetric. For $0< r<\infty$, $E^{(r)}$ will denote the quasi-Banach spaces defined by

$$ E^{(r)}:=\bigl\{ g\in L_{0}: \vert g \vert ^{r}\in E\bigr\} \quad \text{and}\quad \Vert g \Vert _{E^{(r)}}= \bigl\Vert \vert g \vert ^{r} \bigr\Vert _{E}^{\frac{1}{r}}. $$

For $r>0$, we know from [17] that if E is a symmetric Banach function space, then $E^{(r)}$ is a symmetric quasi-Banach space, and if E has order continuous norm, then $E^{(r)}$ has order continuous norm.

We suppose that $\mathcal{M}$ is a semi-finite von Neumann algebra, namely a von Neumann algebra equipped with a semi-finite, faithful and normal trace τ. We will denote by 1 the identity in $\mathcal{M}$ and $P(\mathcal{M})$ 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 $\mathcal{M}$. Let $e^{\bot }_{s}(|x|)=e_{(s, \infty )}(|x|)$ be the spectral projection of $|x|$ associated with the interval $(s, \infty )$. If x is affiliated with $\mathcal{M}$, x will be called τ-measurable if and only if $\tau (e^{\bot }_{s}(|x|))<\infty$ for some $s>0$. The set of all τ-measurable operators will be denoted by $L_{0}(\mathcal{M})$.

Definition 2.1

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

$$ \mu _{t}(x)=\inf \bigl\{ \Vert xe \Vert : e \text{ is a projection in } \mathcal{M} \text{ with } \tau \bigl(e^{\bot }\bigr)\leq t\bigr\} . $$

We will denote simply by $\lambda (x)$ and $\mu (x)$ the functions $t\rightarrow \lambda _{t}(x)$ and $t\rightarrow \mu _{t}(x)$, respectively. The generalized singular number function $t\rightarrow \mu _{t}(x)$ is decreasing right-continuous. For $x, y\in L_{0}(\mathcal{M})$ and $u, v\in \mathcal{M}$, we obtain

$$ \mu (x)=\mu \bigl( \vert x \vert \bigr)=\mu \bigl(x^{*}\bigr),\qquad \mu (uxv)\leq \Vert u \Vert \Vert v \Vert \mu (x). $$

(2.1)

Moreover, let f be a continuous increasing function on $[0, \infty )$ with $f(0)=0$. It follows from [11, Lemma 2.5, Lemma 2.6 and Corollary 2.8] that

$$ \mu \bigl(f\bigl( \vert x \vert \bigr)\bigr)=f\bigl(\mu \bigl( \vert x \vert \bigr)\bigr) $$

(2.2)

and

$$ \tau \bigl(f\bigl( \vert x \vert \bigr)\bigr)= \int _{0}^{\tau (1)} f\bigl(\mu _{t}(x) \bigr)\,dt. $$

(2.3)

See [11] for basic properties and detailed information on generalized singular number of x.

Let E be a symmetric Banach function space on $(0,\infty )$. We define

$$ E(\mathcal{M})=\bigl\{ x\in L_{0}(\mathcal{M}): \mu (x)\in E\bigr\} \quad \text{and}\quad \Vert x \Vert _{E(\mathcal{M})}= \bigl\Vert \mu (x) \bigr\Vert _{E}. $$

Then $(E(\mathcal{M}), \|\cdot \|_{E(\mathcal{M})})$ is a noncommutative symmetric Banach function space. If $E=L^{p}$, then $(E(\mathcal{M}), \|\cdot \|_{E(\mathcal{M})})$ is the usual noncommutative $L_{p}$ spaces $(L^{p}(\mathcal{M}), \|\cdot \|_{p})$. For $0< r<\infty$, we define

$$ E(\mathcal{M})^{(r)}=\bigl\{ x\in L_{0}(\mathcal{M}): \vert x \vert ^{r}\in E( \mathcal{M})\bigr\} \quad \text{and}\quad \Vert x \Vert _{E(\mathcal{M})^{(r)}}= \bigl\Vert \vert x \vert ^{r} \bigr\Vert _{E( \mathcal{M})}^{\frac{1}{r}}. $$

As is shown in [10, Proposition 3.1], if E is a symmetric Banach function space, then $E^{(r)}(\mathcal{M})=E(\mathcal{M})^{(r)}$, where

$$ E^{(r)}(\mathcal{M})=\bigl\{ x\in L_{0}(\mathcal{M}): \mu (x)\in E^{(r)}\bigr\} $$

and $\|x\|_{E^{(r)}(\mathcal{M})}=\|\mu (x)\|_{E^{(r)}}$. It is well known that $E(\mathcal{M})^{(r)}$ is also a noncommutative fully symmetric Banach function space when $r\geq 1$ and E is fully (cf. [19]).

In the following, unless stated otherwise, we will keep all previous notations throughout the paper, and we always assume that E is a symmetric Banach function space on $(0,\infty )$ with order continuous norm.

3 Main results

We start this section with several lemmas which will be used in our proof. From [12, Theorem 3.3] and [13, Lemma 3.4] we have the following two results.

Lemma 3.1

Let$x, y\in \mathcal{M}$and$\alpha \in [0, 1]$. Then

$$\begin{aligned} \mu _{s}\bigl( \bigl\vert x^{*}y \bigr\vert \bigr) \leq \mu _{s}\bigl(\alpha \vert x \vert ^{\frac{1}{\alpha }}+(1- \alpha ) \vert y \vert ^{\frac{1}{1-\alpha }}\bigr). \end{aligned}$$

Lemma 3.2

Let$x, y\in \mathcal{M}$such thatxyis a self-adjoint operator. For every$r>0$, we obtain

$$ \int _{0}^{t}\mu _{s}(xy)^{r}\,ds \leq \int _{0}^{t}\mu _{s}(yx)^{r}\,ds,\quad t>0. $$

Remark 3.3

If x, y are normal operators in $L_{0}(\mathcal{M})$, then $\mu _{s}(xy)=\mu _{s}(yx)$, $s>0$. Indeed, we conclude from (2.1) and (2.2) (see also [11, Lemma 2.5]) that

$$\begin{aligned} \mu _{t}(xy)&=\mu _{t}\bigl( \vert xy \vert ^{2}\bigr)^{\frac{1}{2}}=\mu _{t} \bigl(y^{*}x^{*}xy\bigr)^{ \frac{1}{2}}=\mu _{t}\bigl(y^{*}xx^{*}y\bigr)^{\frac{1}{2}} \\ &=\mu _{t}\bigl( \bigl\vert \bigl(y^{*}x \bigr)^{*} \bigr\vert ^{2}\bigr)^{\frac{1}{2}}=\mu _{t}\bigl( \bigl\vert y^{*}x \bigr\vert ^{2}\bigr)^{ \frac{1}{2}} =\mu _{t} \bigl(x^{*}yy^{*}x\bigr)^{\frac{1}{2}} \\ &=\mu _{t}\bigl(x^{*}y^{*}yx \bigr)^{\frac{1}{2}}= \mu _{t}\bigl( \vert yx \vert ^{2}\bigr)^{ \frac{1}{2}}=\mu _{t}(yx). \end{aligned}$$

Recall that an operator $x \in L_{0}(\mathcal{M})$ is said to be hyponormal if $x^{*}x\geq xx^{*}$, cohyponormal if $x^{*}$ is hyponormal.

Lemma 3.4

Let$x, y\in \mathcal{M}$and$r\geq 0$. If$\alpha \in [0, 1]$and$xx^{*}(yy^{*})^{\alpha }$is cohyponormal, then

$$\begin{aligned} \int _{0}^{t}\mu _{s}\bigl( \bigl\vert x^{*}y \bigr\vert ^{r}\bigr)\,ds\leq \int _{0}^{t}\mu _{s}\bigl( \alpha xx^{*}+(1-\alpha )yy^{*}\bigr)^{\frac{r}{2}}\mu _{s}\bigl((1-\alpha ) xx^{*}+ \alpha yy^{*} \bigr)^{\frac{r}{2}}\,ds,\quad t>0. \end{aligned}$$

Proof

By (2.2) and Lemma 3.2 we have

$$\begin{aligned} \int _{0}^{t}\mu _{s}\bigl( \bigl\vert x^{*}y \bigr\vert ^{r}\bigr)\,ds= \int _{0}^{t}\mu _{s} \bigl(y^{*}xx^{*}y\bigr)^{ \frac{r}{2}}\,ds \leq \int _{0}^{t}\mu _{s} \bigl(xx^{*}yy^{*}\bigr)^{\frac{r}{2}}\,ds. \end{aligned}$$

Since $xx^{*}(yy^{*})^{\alpha }$ is cohyponormal, [8, Corollary 4.5] yields

$$ \mu _{s}\bigl(xx^{*}yy^{*}\bigr)=\mu _{s}\bigl(\bigl[xx^{*}\bigl(yy^{*} \bigr)^{\alpha }\bigr]\bigl(yy^{*}\bigr)^{1- \alpha }\bigr) \leq \mu _{s}\bigl(\bigl(yy^{*}\bigr)^{\alpha }xx^{*} \bigl(yy^{*}\bigr)^{1-\alpha }\bigr), $$

and hence, [11, Theorem 4.2(iii)] and Lemma 3.1 tell us that

$$\begin{aligned} \int _{0}^{t}\mu _{s}\bigl( \bigl\vert x^{*}y \bigr\vert ^{r}\bigr)\,ds&\leq \int _{0}^{t}\mu _{s} \bigl(xx^{*}yy^{*} \bigr)^{\frac{r}{2}}\,ds \\ &= \int _{0}^{t}\mu _{s}\bigl( \bigl(yy^{*}\bigr)^{\alpha }xx^{*} \bigl(yy^{*}\bigr)^{1-\alpha }\bigr)^{ \frac{r}{2}}\,ds \\ &\leq \int _{0}^{t}\mu _{s}\bigl( \bigl(yy^{*}\bigr)^{\alpha }\bigl(xx^{*} \bigr)^{1-\alpha }\bigr)^{ \frac{r}{2}}\mu _{s}\bigl( \bigl(xx^{*}\bigr)^{\alpha }\bigl(yy^{*} \bigr)^{1-\alpha }\bigr)^{ \frac{r}{2}}\,ds \\ &\leq \int _{0}^{t}\mu _{s}\bigl((1- \alpha ) xx^{*}+\alpha yy^{*}\bigr)^{ \frac{r}{2}}\mu _{s}\bigl(\alpha xx^{*}+(1-\alpha )yy^{*} \bigr)^{\frac{r}{2}}\,ds. \end{aligned}$$

This completes the proof. □

Remark 3.5

Let $x, y\in \mathcal{M}$ and $r\geq 0$, $\alpha \in [0, 1]$. (2.1) now yields $\mu _{t}(xx^{*}yy^{*})=\mu _{t}(yy^{*}xx^{*})$ for all $t>0$. If $yy^{*}(xx^{*})^{\alpha }$ is hyponormal, then from Lemma 3.4 we have

$$\begin{aligned} \int _{0}^{t}\mu _{s}\bigl( \bigl\vert x^{*}y \bigr\vert ^{r}\bigr)\,ds\leq \int _{0}^{t}\mu _{s}\bigl( \alpha xx^{*}+(1-\alpha )yy^{*}\bigr)^{\frac{r}{2}}\mu _{s}\bigl((1-\alpha ) xx^{*}+ \alpha yy^{*} \bigr)^{\frac{r}{2}}\,ds, \quad t>0. \end{aligned}$$

Proposition 3.6

Let$\alpha \in [0, 1]$, $r\geq 0$, $1< p, q<\infty$with$\frac{1}{p}+\frac{1}{q}=1$and let$x, y\in E(\mathcal{M})^{(2r)}$. If$xx^{*}(yy^{*})^{\alpha }$is cohyponormal, then$x^{*}y\in E(\mathcal{M})^{(r)}$,

$$\begin{aligned} \bigl\Vert \bigl\vert x^{*}y \bigr\vert ^{r} \bigr\Vert _{E(\mathcal{M})}\leq \bigl\Vert \bigl(T(\alpha ) \bigr)^{\frac{rp}{2}} \bigr\Vert _{E( \mathcal{M})}^{\frac{1}{p}} \bigl\Vert \bigl(T(1-\alpha )\bigr)^{\frac{rq}{2}} \bigr\Vert _{E( \mathcal{M})}^{\frac{1}{q}}, \end{aligned}$$

(3.1)

where$T(\alpha )=\alpha xx^{*}+(1-\alpha )yy^{*}$.

Proof

If

$$ \bigl\Vert \bigl(\alpha xx^{*}+(1-\alpha )yy^{*} \bigr)^{\frac{rp}{2}} \bigr\Vert _{E(\mathcal{M})}^{ \frac{1}{p}}=\infty $$

or

$$ \bigl\Vert \bigl((1-\alpha ) xx^{*}+\alpha yy^{*} \bigr)^{\frac{rq}{2}} \bigr\Vert _{E(\mathcal{M})}^{ \frac{1}{q}}=\infty , $$

then the inequality (3.1) is obvious, and so we always suppose that

$$ \bigl\Vert \bigl(\alpha xx^{*}+(1-\alpha )yy^{*} \bigr)^{\frac{rp}{2}} \bigr\Vert _{E(\mathcal{M})}^{ \frac{1}{p}}< \infty $$

and

$$ \bigl\Vert \bigl((1-\alpha ) xx^{*}+\alpha yy^{*} \bigr)^{\frac{rq}{2}} \bigr\Vert _{E(\mathcal{M})}^{ \frac{1}{q}}< \infty . $$

First we assume that $x, y\in E(\mathcal{M})^{(2r)}\cap \mathcal{M}$. According to [4, Theorem 3] and Lemma 3.4, we have $x^{*}y\in E(\mathcal{M})^{(r)}$ and

$$\begin{aligned} \bigl\Vert \bigl\vert x^{*}y \bigr\vert ^{r} \bigr\Vert _{E(\mathcal{M})}&= \bigl\Vert \mu _{s}\bigl( \bigl\vert x^{*}y \bigr\vert ^{r}\bigr) \bigr\Vert _{E} \\ &\leq \bigl\Vert \mu _{s}\bigl(\alpha xx^{*}+(1- \alpha )yy^{*}\bigr)^{\frac{r}{2}}\mu _{s}\bigl((1- \alpha ) xx^{*}+\alpha yy^{*}\bigr)^{\frac{r}{2}} \bigr\Vert _{E} \\ &\leq \bigl\Vert \mu _{s}\bigl(\alpha xx^{*}+(1- \alpha )yy^{*}\bigr)^{\frac{pr}{2}} \bigr\Vert _{E}^{ \frac{1}{p}} \bigl\Vert \mu _{s}\bigl((1- \alpha ) xx^{*}+\alpha yy^{*}\bigr)^{ \frac{rq}{2}} \bigr\Vert _{E}^{\frac{1}{q}}. \end{aligned}$$

In the general case, for $y, x\in L_{0}(\mathcal{M})$, let $x=u|x|$ and $y=v|y|$ be the polar decomposition of x and y, respectively. We assume also that $|y|=\int _{0}^{\infty }\lambda \,de_{\lambda }(|y|)$ and $|x|=\int _{0}^{\infty }\lambda \,de_{\lambda }(|x|)$ are the spectral decomposition of $|y|$ and $|x|$, respectively. Set $y_{n}=v\int _{0}^{n} \lambda \,de_{\lambda }(|y|)$ and $x_{n}=u\int _{0}^{n} \lambda \,de_{\lambda }(|x|)$. Then

$$\begin{aligned} \mu _{t}(x-x_{n})\leq \mu _{t}\bigl( \vert x \vert \bigr)\chi _{(0, \tau (e_{[n, \infty )}( \vert x \vert )))},\qquad \vert x-x_{n} \vert = \int _{n}^{\infty }\lambda \,de_{\lambda }\bigl( \vert x \vert \bigr). \end{aligned}$$

From [18, Proposition 21 of Chapter I] and [11, Lemma 3.1] we conclude that $\tau (e_{[n, \infty )}(|x|))\rightarrow 0$ and $\mu _{t}(x-x_{n})\downarrow 0$ as $n\rightarrow \infty$. Similarly, $\mu _{t}(y-y_{n})\downarrow 0$ as $n\rightarrow \infty$. Since E has order continuous norm, we see that

$$\begin{aligned} \bigl\Vert \mu _{t}(y_{n}-y)^{2r} \bigr\Vert ^{\frac{1}{2}}_{E}\downarrow 0,\qquad \bigl\Vert \mu _{t}(x_{n}-x)^{2r} \bigr\Vert ^{\frac{1}{2}}_{E}\downarrow 0 \end{aligned}$$

(3.2)

as $n\rightarrow \infty$. Thus, [4, Theorem 3] gives

$$\begin{aligned} & \bigl\Vert \bigl\vert x_{n}^{*}y_{n}-x^{*}y \bigr\vert ^{r} \bigr\Vert _{E(\mathcal{M})} \\ &\quad = \bigl\Vert x_{n}^{*}y_{n}-x_{n}^{*}y+x_{n}^{*}y-x^{*}y \bigr\Vert ^{r}_{E(\mathcal{M})^{(r)}} \\ &\quad \leq C\bigl\{ \bigl\Vert x_{n}^{*}y_{n}-x_{n}^{*}y \bigr\Vert ^{r}_{E(\mathcal{M})^{(r)}}+ \bigl\Vert x_{n}^{*}y-x^{*}y \bigr\Vert ^{r}_{E(\mathcal{M})^{(r)}}\bigr\} \\ &\quad \leq C\bigl\{ \bigl\Vert x_{n}^{*} \bigr\Vert ^{r}_{E(\mathcal{M})^{(2r)}} \Vert y_{n}-y \Vert ^{r}_{E( \mathcal{M})^{(2r)}} + \bigl\Vert x_{n}^{*}-x^{*} \bigr\Vert ^{r}_{E(\mathcal{M})^{(r)}} \Vert y \Vert ^{r}_{E(\mathcal{M})^{(r)}}\bigr\} \\ &\quad =C\bigl\{ \bigl\Vert \mu _{t}\bigl(x_{n}^{*} \bigr) \bigr\Vert ^{r}_{E^{(2r)}} \Vert \mu _{(}y_{n}-y) \Vert ^{r}_{E^{(2r)}} + \bigl\Vert \mu _{t}\bigl(x_{n}^{*}-x^{*} \bigr) \bigr\Vert ^{r}_{E^{(r)}} \bigl\Vert \mu _{t}(y) \bigr\Vert ^{r}_{E^{(r)}} \bigr\} \\ &\quad =C\bigl\{ \bigl\Vert \mu _{t}(x_{n})^{2r} \bigr\Vert ^{\frac{1}{2}}_{E} \bigl\Vert \mu _{(}y_{n}-y)^{2r} \bigr\Vert ^{\frac{1}{2}}_{E} + \bigl\Vert \mu _{t}(x_{n}-x)^{2r} \bigr\Vert ^{\frac{1}{2}}_{E} \bigl\Vert \mu _{t}(y)^{2r} \bigr\Vert ^{\frac{1}{2}}_{E} \bigr\} , \end{aligned}$$

where the constant C from the triangle inequality in $E(\mathcal{M})^{(r)}$. Therefore, the fact $\|\mu _{t}(x_{n})^{2r}\|^{\frac{1}{2}}_{E}\leq \|\mu _{t}(x)^{2r}\|^{ \frac{1}{2}}_{E}$ and (3.2) imply that $\Vert \vert x_{n}^{*}y_{n}-x^{*}y \vert ^{r} \Vert _{E(\mathcal{M})}\rightarrow 0$ as $n\rightarrow \infty$. Moreover, $\Vert \vert x_{n}^{*}y_{n} \vert ^{r} \Vert _{E(\mathcal{M})}\rightarrow \||x^{*}y|^{r} \|_{E(\mathcal{M})}$ as $n\rightarrow \infty$. In the same manner we can see that

$$ \bigl\Vert \bigl(\alpha x_{n}x^{*}_{n}+(1- \alpha )y_{n}y^{*}_{n}\bigr)^{\frac{rp}{2}} \bigr\Vert _{E( \mathcal{M})}^{\frac{1}{p}} \rightarrow \bigl\Vert \bigl( \alpha xx^{*}+(1-\alpha )yy^{*}\bigr)^{ \frac{rp}{2}} \bigr\Vert _{E(\mathcal{M})}^{\frac{1}{p}} $$

and

$$ \bigl\Vert \bigl((1-\alpha ) x_{n}x_{n}^{*}+ \alpha y_{n}y_{n}^{*}\bigr)^{\frac{rq}{2}} \bigr\Vert _{E(\mathcal{M})}^{\frac{1}{q}} \rightarrow \bigl\Vert \bigl((1- \alpha ) xx^{*}+ \alpha yy^{*}\bigr)^{\frac{rq}{2}} \bigr\Vert _{E(\mathcal{M})}^{\frac{1}{q}}. $$

This completes the proof. □

Remark 3.7

Let $1< p, q<\infty$ with $\frac{1}{p}+\frac{1}{q}=1$. If $\alpha =0$, then $xx^{*}(yy^{*})^{\alpha }=xx^{*}$ is cohyponormal. Therefore, Proposition 3.6 yields $x^{*}y\in E(\mathcal{M})^{(r)}$ and

$$ \bigl\Vert \bigl\vert x^{*}y \bigr\vert ^{r} \bigr\Vert _{E(\mathcal{M})}\leq \bigl\Vert \bigl\vert yy^{*} \bigr\vert ^{\frac{rp}{2}} \bigr\Vert _{E( \mathcal{M})}^{\frac{1}{p}} \bigl\Vert \bigl\vert xx^{*} \bigr\vert ^{\frac{rq}{2}} \bigr\Vert _{E( \mathcal{M})}^{\frac{1}{q}}, $$

which is a main result of [4].

Remark 3.8

It is necessary for us to remark here that, it can be observed in [7, Lemma 2] without a proof that $\mu (ab)=\mu (ba)$ when $ab, ba\in L^{1}(\mathcal{M})$. However, we are not able to give it a proof at this moment. On the other hand, the authors were informed by an anonymous referee that $\mu (ab)=\mu (ba)$ does not hold even in the matrix case. On account of this, there could be a gap in the proof of [13, Theorem 3.6] and we give a corresponding illustration as follows: Set $r\geq 1$, $\alpha \in [0, 1]$ and let $xx^{*}(yy^{*})^{\alpha }$ be cohyponormal. Using Proposition 3.6 to the case $E=L_{1}$ and $p=q=2$, we have

$$\begin{aligned} \bigl\Vert \bigl\vert x^{*}y \bigr\vert ^{r} \bigr\Vert _{L^{1}(\mathcal{M})}\leq \bigl\Vert \bigl\vert \alpha xx^{*}+(1-\alpha )yy^{*} \bigr\vert ^{r} \bigr\Vert _{L^{1}(\mathcal{M})}^{\frac{1}{2}} \bigl\Vert \bigl\vert (1- \alpha ) xx^{*}+\alpha yy^{*} \bigr\vert ^{r} \bigr\Vert _{L^{1}(\mathcal{M})}^{\frac{1}{2}}, \end{aligned}$$

i.e.,

$$\begin{aligned} \bigl\Vert x^{*}y \bigr\Vert _{L^{r}(\mathcal{M})}^{2} \leq \bigl\Vert \alpha xx^{*}+(1-\alpha )yy^{*} \bigr\Vert _{L^{r}(\mathcal{M})} \bigl\Vert (1-\alpha ) xx^{*}+\alpha yy^{*} \bigr\Vert _{L^{r}( \mathcal{M})}, \end{aligned}$$

which is the result of [14, Theorem 3.6] under a cohyponormal condition.

Theorem 3.9

Let$\alpha \in [0, 1]$and$1< p, q<\infty$with$\frac{1}{p}+\frac{1}{q}=1$. Assume also that$r\geq \max \{\frac{2}{p}, \frac{2}{q}\}$, $x, y\in E(\mathcal{M})^{(2r)}$and$z\in P(\mathcal{M})$. If$zxx^{*}z(zyy^{*}z)^{\alpha }$is cohyponormal, then$x^{*}\mathit{zy}\in E(\mathcal{M})^{(r)}$,

$$\begin{aligned} \bigl\Vert \bigl\vert x^{*}\mathit{zy} \bigr\vert ^{r} \bigr\Vert _{E(\mathcal{M})}\leq \bigl\Vert \big|T_{z}(\alpha ) \big\vert ^{ \frac{rp}{2}} \bigr\Vert _{E(\mathcal{M})}^{\frac{1}{p}} \bigl\Vert \bigl\vert T_{z}(1- \alpha ) \bigr\vert ^{ \frac{rq}{2}} \bigr\Vert _{E(\mathcal{M})}^{\frac{1}{q}}, \end{aligned}$$

where$T_{z}(\alpha )= \alpha xx^{*}z+(1-\alpha )zyy^{*}$.

Proof

Let $T(\alpha )= \alpha xx^{*}+(1-\alpha )yy^{*}$. Then $z\in P(\mathcal{M})$ and Proposition 3.6 force that

$$\begin{aligned} \bigl\Vert \bigl\vert x^{*}zy \bigr\vert ^{r} \bigr\Vert _{E(\mathcal{M})}&= \bigl\Vert \bigl\vert x^{*}z z y \bigr\vert ^{r} \bigr\Vert _{E( \mathcal{M})} \\ &\leq \bigl\Vert \bigl(z T(\alpha )z \bigr)^{\frac{rp}{2}} \bigr\Vert _{E(\mathcal{M})}^{ \frac{1}{p}} \bigl\Vert \bigl(z T(1-\alpha )z \bigr)^{\frac{rq}{2}} \bigr\Vert _{E(\mathcal{M})}^{ \frac{1}{q}} \end{aligned}$$

and

$$ 2\mu _{t}\bigl(z T(\alpha )z\bigr)=\mu \bigl(z \bigl(T(\alpha )z+ zT(\alpha )\bigr)z\bigr)\leq \mu \bigl( T( \alpha )z+ zT(\alpha )\bigr), $$

(3.3)

and hence

$$\begin{aligned} \bigl\Vert \bigl(zT(\alpha )z\bigr)^{\frac{rp}{2}} \bigr\Vert _{E(\mathcal{M})}^{\frac{1}{p}} \leq \biggl\Vert \biggl(\frac{T(\alpha )z+ zT(\alpha )}{2} \biggr)^{\frac{rp}{2}} \biggr\Vert _{E( \mathcal{M})}^{\frac{1}{p}}. \end{aligned}$$

Similarly,

$$ \bigl\Vert \bigl(z T(1-\alpha )z \bigr)^{\frac{rq}{2}} \bigr\Vert _{E(\mathcal{M})}^{\frac{1}{q}} \leq \biggl\Vert \biggl\vert \frac{zT(1-\alpha )+T(1-\alpha )z}{2} \biggr\vert ^{\frac{rq}{2}} \biggr\Vert _{E( \mathcal{M})}^{\frac{1}{q}}. $$

Therefore,

$$ \begin{aligned}[b] & \bigl\Vert \bigl\vert x^{*}zy \bigr\vert ^{r} \bigr\Vert _{E(\mathcal{M})} \\ &\quad \leq \biggl\Vert \biggl\vert \frac{zT(\alpha )+T(\alpha )z}{2} \biggr\vert ^{\frac{rp}{2}} \biggr\Vert _{E( \mathcal{M})}^{\frac{1}{p}} \biggl\Vert \biggl\vert \frac{zT(1-\alpha )+T(1-\alpha )z}{2} \biggr\vert ^{ \frac{rq}{2}} \biggr\Vert _{E(\mathcal{M})}^{\frac{1}{q}}. \end{aligned} $$

(3.4)

A simple computation shows

$$ \frac{T(\alpha )z+zT(\alpha )}{2}=\frac{1}{2} \bigl\{ \alpha xx^{*}z+(1- \alpha )zyy^{*}+\bigl(\alpha xx^{*}z+(1-\alpha )zyy^{*}\bigr)^{*}\bigr\} . $$

According to [11, Theorem 4.4(ii)] and (2.1), we have

$$\begin{aligned} &\int _{0}^{t}\mu _{s}\biggl( \frac{T(\alpha )z+zT(\alpha )}{2}\biggr)\,ds \\ &\quad \leq \int _{0}^{t} \mu _{s}\biggl( \frac{1}{2}\bigl(\alpha xx^{*}z+(1-\alpha )zyy^{*}\bigr)\biggr)\,ds \\ &\qquad {}+ \int _{0}^{t}\mu _{s}\biggl( \frac{1}{2}\bigl(\alpha xx^{*}z+(1-\alpha )zyy^{*}\bigr)^{*}\biggr)\,ds \\ &\quad = \int _{0}^{t}\mu _{s}\bigl(\alpha xx^{*}z+(1-\alpha )zyy^{*}\bigr)\,ds. \end{aligned}$$

Since $\frac{rp}{2}\geq 1$, from [9, Theorem 2.1] and (2.2) we can assert that

$$\begin{aligned} \int _{0}^{t}\mu _{s}\biggl( \biggl\vert \frac{T(\alpha )z+zT(\alpha )}{2} \biggr\vert ^{ \frac{rp}{2}}\biggr)\,ds&= \int _{0}^{t}\mu _{s}\biggl( \frac{T(\alpha )z+zT(\alpha )}{2}\biggr)^{\frac{rp}{2}}\,ds \\ &\leq \int _{0}^{t}\mu _{s}\bigl(\alpha xx^{*}z+(1-\alpha )zyy^{*}\bigr)^{ \frac{rp}{2}}\,ds \\ &= \int _{0}^{t}\mu _{s}\bigl( \bigl\vert \alpha xx^{*}z+(1-\alpha )zyy^{*} \bigr\vert ^{ \frac{rp}{2}}\bigr)\,ds. \end{aligned}$$

Consequently,

$$\begin{aligned} \biggl\Vert \biggl\vert \frac{zT(\alpha )+T(\alpha )z}{2} \biggr\vert ^{\frac{rp}{2}} \biggr\Vert _{E(\mathcal{M})} \leq \bigl\Vert \bigl\vert \alpha xx^{*}z+(1-\alpha )zyy^{*} \bigr\vert ^{\frac{rp}{2}} \bigr\Vert _{E( \mathcal{M})}. \end{aligned}$$

(3.5)

In the same way as used above, we can also prove that

$$\begin{aligned} \biggl\Vert \biggl\vert \frac{zT(1-\alpha )+T(1-\alpha )z}{2} \biggr\vert ^{\frac{rq}{2}} \biggr\Vert _{E( \mathcal{M})} \leq \bigl\Vert \bigl\vert (1-\alpha ) xx^{*}z+\alpha zyy^{*} \bigr\vert ^{ \frac{rq}{2}} \bigr\Vert _{E(\mathcal{M})}. \end{aligned}$$

(3.6)

Therefore, inequalities (3.4), (3.5) and (3.6) give

$$ \begin{aligned} \bigl\Vert \bigl\vert x^{*}zy \bigr\vert ^{r} \bigr\Vert _{E(\mathcal{M})} \leq \bigl\Vert \bigl\vert \alpha xx^{*}z+(1- \alpha )zyy^{*} \bigr\vert ^{\frac{rp}{2}} \bigr\Vert _{E(\mathcal{M})}^{\frac{1}{p}} \bigl\Vert \bigl\vert (1- \alpha ) xx^{*}z+\alpha zyy^{*} \bigr\vert ^{\frac{rq}{2}} \bigr\Vert _{E(\mathcal{M})}^{ \frac{1}{q}}. \end{aligned} $$

 □

Remark 3.10

Let $\alpha \in [0, 1]$ and $1< p, q<\infty$ with $\frac{1}{p}+\frac{1}{q}=1$. Assume also that $r\geq \max \{\frac{2}{p}, \frac{2}{q}\}$, $x, y\in E(\mathcal{M})^{(2r)}$ and $z\in \mathcal{M}$. We write $T_{z}(\alpha )= \alpha xx^{*}z+(1-\alpha )zyy^{*}$ and we wish to prove

$$\begin{aligned} \bigl\Vert \bigl\vert x^{*}\mathit{zy} \bigr\vert ^{r} \bigr\Vert _{E(\mathcal{M})}\leq \bigl\Vert \bigl\vert T_{z}(\alpha ) \bigr\vert ^{ \frac{rp}{2}} \bigr\Vert _{E(\mathcal{M})}^{\frac{1}{p}} \bigl\Vert \bigl\vert T_{z}(1- \alpha ) \bigr\vert ^{ \frac{rq}{2}} \bigr\Vert _{E(\mathcal{M})}^{\frac{1}{q}}. \end{aligned}$$

However, we do not succeed in proving it at this moment.

Theorem 3.11

Let$r>0$and$x, y \in E(\mathcal{M})^{(2r)}$, $0\leq z\in \mathcal{M}$. Assume also that$\alpha \in [0, 1]$and$1< p, q<\infty$with$\frac{1}{p}+\frac{1}{q}=1$. If$z^{\frac{1}{2}}xx^{*}z^{\frac{1}{2}}(z^{\frac{1}{2}}yy^{*}z^{ \frac{1}{2}})^{\alpha }$is cohyponormal, then$x^{*}\mathit{zy}\in E(\mathcal{M})^{(r)}$and

$$\begin{aligned} \bigl\Vert \bigl\vert x^{*}\mathit{zy} \bigr\vert ^{r} \bigr\Vert _{E(\mathcal{M})}\leq \bigl\Vert \bigl\vert T( \alpha )z \bigr\vert ^{\frac{rp}{2}} \bigr\Vert _{E(\mathcal{M})}^{\frac{1}{p}} \bigl\Vert \bigl\vert T(1-\alpha ) z \bigr\vert ^{\frac{rq}{2}} \bigr\Vert _{E(\mathcal{M})}^{\frac{1}{q}}, \end{aligned}$$

where$T(\alpha )= \alpha xx^{*}+(1-\alpha )yy^{*}$.

Proof

First it follows from [4, Theorem 3] that $x^{*}\mathit{zy}\in E(\mathcal{M})^{(r)}$. Since z is positive, Proposition 3.6 gives

$$\begin{aligned} \bigl\Vert \bigl\vert x^{*}zy \bigr\vert ^{r} \bigr\Vert _{E(\mathcal{M})}&= \bigl\Vert \bigl\vert x^{*}z^{\frac{1}{2}}z^{ \frac{1}{2}}y \bigr\vert ^{r} \bigr\Vert _{E(\mathcal{M})} \\ &\leq \bigl\Vert \bigl(z^{\frac{1}{2}}T(\alpha )z^{\frac{1}{2}} \bigr)^{\frac{rp}{2}} \bigr\Vert _{E( \mathcal{M})}^{\frac{1}{p}} \bigl\Vert \bigl(z^{\frac{1}{2}}T(1-\alpha )z^{ \frac{1}{2}}\bigr)^{\frac{rq}{2}} \bigr\Vert _{E(\mathcal{M})}^{\frac{1}{q}}, \end{aligned}$$

and hence Lemma 3.2 leads to

$$\begin{aligned} \bigl\Vert \bigl(z^{\frac{1}{2}}T(\alpha )z^{\frac{1}{2}} \bigr)^{\frac{rp}{2}} \bigr\Vert _{E( \mathcal{M})}^{\frac{1}{p}} & = \bigl\Vert z^{\frac{1}{2}}T(\alpha )z^{ \frac{1}{2}} \bigr\Vert _{E(\mathcal{M})^{(\frac{rp}{2})}}^{\frac{r}{2}} \\ &\leq \bigl\Vert T(\alpha )z \bigr\Vert _{E(\mathcal{M})^{(\frac{rp}{2})}}^{ \frac{r}{2}} = \bigl\Vert \bigl\vert T(\alpha )z \bigr\vert ^{\frac{rp}{2}} \bigr\Vert _{E(\mathcal{M})}^{ \frac{1}{p}}. \end{aligned}$$

Similarly,

$$ \bigl\Vert \bigl(z^{\frac{1}{2}}T(1-\alpha )z^{\frac{1}{2}} \bigr)^{\frac{rq}{2}} \bigr\Vert _{E( \mathcal{M})}^{\frac{1}{q}}\leq \| |T(1-\alpha )z|^{\frac{rq}{2}}\|_{E( \mathcal{M})}^{\frac{1}{q}}. $$

Therefore,

$$ \begin{aligned} \bigl\Vert \bigl\vert x^{*}zy \bigr\vert ^{r} \bigr\Vert _{E(\mathcal{M})} \leq \bigl\Vert \bigl\vert T(\alpha )z \bigr\vert ^{ \frac{rp}{2}} \bigr\Vert _{E(\mathcal{M})}^{\frac{1}{p}} \bigl\Vert \bigl\vert T(1-\alpha )z \bigr\vert ^{ \frac{rq}{2}} \bigr\Vert _{E(\mathcal{M})}^{\frac{1}{q}}. \end{aligned} $$

This completes the proof. □

Remark 3.12

(1) Let $\alpha \in [0, 1]$ and $1< p, q<\infty$ with $\frac{1}{p}+\frac{1}{q}=1$ and let $r\geq \max \{\frac{2}{p}, \frac{2}{q}\}$. For $x, y\in E(\mathcal{M})^{(2r)}$ and $z\in P(\mathcal{M})$, write $T_{z}(\alpha )= \alpha xx^{*}z+(1-\alpha )zyy^{*}$ and $T(\alpha )=\alpha xx^{*}+(1-\alpha )yy^{*}$. Assume also that $zxx^{*}z(zyy^{*}z)^{\alpha }$ is cohyponormal. Combining Theorem 3.11 with Theorem 3.9 we have

$$\begin{aligned} \bigl\Vert \bigl\vert x^{*}\mathit{zy} \bigr\vert ^{r} \bigr\Vert _{E(\mathcal{M})}\leq \min \{a, b\}, \end{aligned}$$

where

$$ a= \bigl\Vert \bigl\vert T_{z}(\alpha ) \bigr\vert ^{\frac{rp}{2}} \bigr\Vert _{E(\mathcal{M})}^{\frac{1}{p}} \bigl\Vert \bigl\vert T_{z}(1-\alpha ) \bigr\vert ^{\frac{rq}{2}} \bigr\Vert _{E(\mathcal{M})}^{\frac{1}{q}} $$

and

$$ b= \bigl\Vert \bigl\vert T(\alpha )z \bigr\vert ^{\frac{rp}{2}} \bigr\Vert _{E(\mathcal{M})}^{\frac{1}{p}} \bigl\Vert \bigl\vert T(1- \alpha ) z \bigr\vert ^{\frac{rq}{2}} \bigr\Vert _{E(\mathcal{M})}^{\frac{1}{q}}. $$

(2) Let $r>0$, $x, y\in E(\mathcal{M})^{(2r)}$, $0\leq z\in \mathcal{M}$ and $\alpha \in [0, 1]$, $1< p, q<\infty$ with $\frac{1}{p}+\frac{1}{q}=1$. If $z^{\frac{1}{2}}yy^{*}z^{\frac{1}{2}}(z^{\frac{1}{2}}xx^{*}z^{ \frac{1}{2}})^{\alpha }$ is cohyponormal, then $x^{*}\mathit{zy}\in E(\mathcal{M})^{(r)}$. Moreover, the fact $\mu _{t}(|x^{*}\mathit{zy}|^{r})=\mu _{t}(x^{*}\mathit{zy})^{r}=\mu _{t}(y^{*}zx)^{r}= \mu _{t}(|y^{*}zx|^{r})$ and Theorem 3.11 yields

$$\begin{aligned} \bigl\Vert \bigl\vert y^{*}\mathit{zx} \bigr\vert ^{r} \bigr\Vert _{E(\mathcal{M})}\leq \bigl\Vert \bigl\vert T( \alpha )z \bigr\vert ^{\frac{rp}{2}} \bigr\Vert _{E(\mathcal{M})}^{\frac{1}{p}} \bigl\Vert \bigl\vert T(1-\alpha ) z \bigr\vert ^{\frac{rq}{2}} \bigr\Vert _{E(\mathcal{M})}^{\frac{1}{q}}, \end{aligned}$$

where $T(\alpha )= \alpha xx^{*}+(1-\alpha )yy^{*}$.