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.,

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

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

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

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

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

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:

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

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

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

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 We refer to [13, 5, 15, 20] for more information on this topic and historical references.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

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

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

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

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

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

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

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


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

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

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


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\(x^{*}y\in E(\mathcal{M})^{(r)}\),

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

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



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


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

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

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


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

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

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

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

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


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

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

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


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

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


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

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


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

and hence

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


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


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

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

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


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

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

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

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


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

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


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


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

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

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


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


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

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