## 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 We refer to [13, 5, 15, 20] for more information on this topic and historical references.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}
(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 $$T(\alpha )= \alpha xx^{*}+(1-\alpha )yy^{*}$$.