Norm inequalities for operators related to the Cauchy-Schwarz and Heinz inequalities

Abstract

We present some refinements of the Cauchy-Schwarz and Heinz inequalities for operators by utilizing a refinement of the Hermite-Hadamard inequality.

1 Introduction

Let $B(\mathcal{H})$ be the $C^{*}$-algebra of all bounded linear operators on a complex separable Hilbert space $\mathcal{H}$. Let $\Vert \cdot \Vert$ denote a unitarily invariant norm defined on a two-sided ideal $\tau_{\Vert \cdot \Vert }$ that is included in $C_{\infty}$ (the set of compact operators); it has the basic property $\Vert UAV\Vert =\Vert A\Vert$ for every $A\in\tau_{\Vert \cdot \Vert }$ and all unitary operators $U, V\in B(\mathcal{H})$. For $A\in B(\mathcal{H})$, $\vert A\vert =(A^{*}A)^{\frac{1}{2}}$, where $A^{*}$ is the conjugate operator of A.

Hiai and Zhan [1] proved that if $A, B, X\in B(\mathcal{H})$ with $A, B$ positive and $X\in\tau_{\Vert \cdot \Vert }$, $r>0$, then the function $f(v)=\Vert \vert A^{v}XB^{1-v}\vert ^{r}\Vert \cdot \Vert \vert A^{1-v}XB^{v}\vert ^{r}\Vert$ is convex on the interval $[0,1]$ and attains its minimum at $v=\frac{1}{2}$ maximum at $v=0$ and $v=1$. Consequently, it is decreasing on $[0,\frac{1}{2}]$ and increasing on $[\frac{1}{2},1]$; moreover, $f(v)=f(1-v)$ for $v\in [0,1]$. Thus for every unitarily invariant norm $\Vert \cdot \Vert$, we have the operator inequality

$$ \bigl\Vert \bigl\vert A^{\frac{1}{2}}XB^{\frac{1}{2}} \bigr\vert ^{r} \bigr\Vert ^{2}\leq \bigl\Vert \bigl\vert A^{v}XB^{1-v} \bigr\vert ^{r} \bigr\Vert \cdot \bigl\Vert \bigl\vert A^{1-v}XB^{v} \bigr\vert ^{r} \bigr\Vert \leq \bigl\Vert \vert AX\vert ^{r} \bigr\Vert \cdot \bigl\Vert \vert XB\vert ^{r} \bigr\Vert $$

(1)

for $A, B, X\in B(\mathcal{H})$ with $A, B$ positive, $X\in\tau_{\Vert \cdot \Vert }$, $r>0$, and $v\in[0,1]$. The inequality $\Vert \vert A^{\frac{1}{2}}XB^{\frac{1}{2}}\vert ^{r}\Vert ^{2} \leq \Vert \vert AX\vert ^{r}\Vert \cdot \Vert \vert XB\vert ^{r}\Vert$ is often called the Cauchy-Schwarz norm inequality due to Bhatia and Davis [2].

The Heinz means for operators are defined by $g(v)=\Vert A^{v}XB^{1-v}+A^{1-v}XB^{v}\Vert$ for $A, B, X\in B(\mathcal{H})$ with $A, B$ positive, $X\in\tau_{\Vert \cdot \Vert }$, and $v\in[0,1]$, where $\Vert \cdot \Vert$ is a unitarily invariant norm. Bhatia and Davis [3] obtained the inequality

$$ 2 \bigl\Vert A^{\frac{1}{2}}XB^{\frac{1}{2}} \bigr\Vert \leq \bigl\Vert A^{v}XB^{1-v}+A^{1-v}XB^{v} \bigr\Vert \leq \Vert AX+XB\Vert $$

(2)

for $A, B, X\in B(\mathcal{H})$ with $A, B$ positive, $X\in\tau_{\Vert \cdot \Vert }$, and $v\in[0,1]$. In fact, they proved that $g(v)=\Vert A^{v}XB^{1-v}+A^{1-v}XB^{v}\Vert$ is a convex function of v on the interval $[0,1]$, attains its minimum at $v=\frac{1}{2}$ and maximum at $v=0$ and $v=1$. Hence, it is decreasing on $[0,\frac{1}{2}]$ and increasing on $[\frac{1}{2},1]$; moreover, $g(v)=g(1-v)$ for $v\in [0,1]$. The second inequality in (2) is one of the most essential inequalities in operator theory, which is often called the Heinz inequality.

Recently, using the convexity of the function $f(v)=\Vert \vert A^{v}XB^{1-v}\vert ^{r}\Vert \cdot \Vert \vert A^{1-v}XB^{v}\vert ^{r}\Vert$ ($v\in[0,1]$), Burqan [4], Theorem 2, obtained a refinement of inequality (1): Let $A, B, X\in B(\mathcal{H})$ with $A, B$ positive and $X\in\tau_{\Vert \cdot \Vert }$. Then, for every unitarily invariant norm $\Vert \cdot \Vert$,

$$\begin{aligned} \bigl\Vert \bigl\vert A^{\frac{1}{2}}XB^{\frac{1}{2}} \bigr\vert ^{r} \bigr\Vert ^{2} \leq& \bigl\Vert \bigl\vert A^{\frac{2v+1}{4}}XB^{\frac{3-2v}{4}} \bigr\vert ^{r} \bigr\Vert \cdot \bigl\Vert \bigl\vert A^{\frac{3-2v}{4}}XB^{\frac{2v+1}{4}} \bigr\vert ^{r} \bigr\Vert \\ \leq&\frac{1}{\vert 1-2v\vert }\biggl\vert \int_{v}^{1-v} \bigl\Vert \bigl\vert A^{\mu}XB^{1-\mu} \bigr\vert ^{r} \bigr\Vert \cdot \bigl\Vert \bigl\vert A^{1-\mu}XB^{\mu} \bigr\vert ^{r} \bigr\Vert \,d\mu\biggr\vert \\ \leq&\frac{1}{2} \bigl[ \bigl\Vert \bigl\vert A^{v}XB^{1-v} \bigr\vert ^{r} \bigr\Vert \cdot \bigl\Vert \bigl\vert A^{1-v}XB^{v} \bigr\vert ^{r} \bigr\Vert + \bigl\Vert \bigl\vert A^{\frac{1}{2}}XB^{\frac{1}{2}} \bigr\vert ^{r} \bigr\Vert ^{2} \bigr] \\ \leq& \bigl\Vert \bigl\vert A^{v}XB^{1-v} \bigr\vert ^{r} \bigr\Vert \cdot \bigl\Vert \bigl\vert A^{1-v}XB^{v} \bigr\vert ^{r} \bigr\Vert \\ \leq& \bigl\Vert \vert AX\vert ^{r} \bigr\Vert \cdot \bigl\Vert \vert XB\vert ^{r} \bigr\Vert , \end{aligned}$$

(3)

where $v\in[0,1]$ and $r>0$.

Using the convexity of the function $g(v)=\Vert A^{v}XB^{1-v}+A^{1-v}XB^{v}\Vert$ ($v\in[0,1]$), Feng [5], Theorem 1, obtained a refinement of the first inequality in (2): Let $A, B, X\in B(\mathcal{H})$ with $A, B$ positive and $X\in\tau_{\Vert \cdot \Vert }$. Then, for every unitarily invariant norm $\Vert \cdot \Vert$,

$$\begin{aligned} 2 \bigl\Vert A^{\frac{1}{2}}XB^{\frac{1}{2}} \bigr\Vert \leq& \frac{1}{\vert 1-2v\vert } \biggl\vert \int_{v}^{1-v} \bigl\Vert A^{\mu}XB^{1-\mu}+A^{1-\mu}XB^{\mu} \bigr\Vert \,d\mu \biggr\vert \\ \leq&\frac{1}{2} \bigl[ \bigl\Vert A^{v}XB^{1-v}+A^{1-v}XB^{v} \bigr\Vert +2 \bigl\Vert A^{\frac{1}{2}}XB^{\frac{1}{2}} \bigr\Vert \bigr] \\ \leq& \bigl\Vert A^{v}XB^{1-v}+A^{1-v}XB^{v} \bigr\Vert , \end{aligned}$$

(4)

where $v\in[0,1]$.

In this note, we present some refinements of norm inequalities (1), (2), (3), and (4), also using the convexity of the functions f and g.

2 Main results

In this section, we mainly present some refinements of the Cauchy-Schwarz and Heinz inequalities for operators. To achieve our goal, we need some preparations.

Let $f:I\rightarrow R$ be a real-valued convex function on the interval $I\subseteq R$. Let $a, b\in I$ with $a< b$. The inequality

$$ f \biggl(\frac{a+b}{2} \biggr)\leq\frac{1}{b-a} \int_{a}^{b}f(x)\,dx\leq\frac{f(a)+f(b)}{2} $$

(5)

is well known as the Hermite-Hadamard inequality, which includes a basic property of convex functions.

Now, we present some lemmas.

Lemma 2.1

Let f be a real-valued convex function on the interval $[a,b]$. Then

$$ \frac{1}{4} \biggl(f \biggl(\frac{7a+b}{8} \biggr)+f \biggl( \frac{5a+3b}{8} \biggr) + f \biggl(\frac{3a+5b}{8} \biggr) +f \biggl( \frac{a+7b}{8} \biggr) \biggr) \leq \frac{1}{b-a} \int_{a}^{b}f(x)\,dx. $$

(6)

Proof

Thanks to the Hermite-Hadamard’s inequality (5), we obtain

$$ f \biggl(\frac{7a+b}{8} \biggr)=f \biggl(\frac{a+\frac{3a+b}{4}}{2} \biggr) \leq \frac{4}{b-a} \int_{a}^{\frac{3a+b}{4}}f(x)\,dx $$

(7)

and

$$ f \biggl(\frac{5a+3b}{8} \biggr)=f \biggl(\frac{\frac{3a+b}{4}+\frac{a+b}{2}}{2} \biggr) \leq \frac{4}{b-a} \int_{\frac{3a+b}{4}}^{\frac{a+b}{2}}f(x)\,dx. $$

(8)

Inequalities (7) and (8) give

$$ f \biggl(\frac{7a+b}{8} \biggr)+f \biggl(\frac{5a+3b}{8} \biggr) \leq \frac{4}{b-a} \int_{a}^{\frac{a+b}{2}}f(x)\,dx. $$

(9)

Similarly, due to the convexity of f on the interval $[\frac{a+b}{2},b]$, we have

$$ f \biggl(\frac{3a+5b}{8} \biggr)+f \biggl(\frac{a+7b}{8} \biggr) \leq \frac{4}{b-a} \int_{\frac{a+b}{2}}^{b}f(x)\,dx. $$

(10)

Hence, by inequalities (9) and (10) we deduce

$$\begin{aligned} &{\frac{1}{4} \biggl(f \biggl(\frac{7a+b}{8} \biggr)+f \biggl( \frac{5a+3b}{8} \biggr)+f \biggl(\frac{3a+5b}{8} \biggr)+f \biggl( \frac{a+7b}{8} \biggr) \biggr)} \\ &{\quad \leq\frac{1}{b-a} \int_{a}^{b}f(x)\,dx.} \end{aligned}$$

This completes the proof. □

Remark 2.2

Let f be a real-valued convex function on the interval $[a,b]$. By the convexity of f on the interval $[a,\frac{a+b}{2}]$ we get

$$\begin{aligned} f \biggl(\frac{3a+b}{4} \biggr) =&f \biggl(\frac{1}{2} \biggl( \frac{7a+b}{8} \biggr)+\frac{1}{2} \biggl(\frac{5a+3b}{8} \biggr) \biggr) \\ \leq& \frac{1}{2}f \biggl(\frac{7a+b}{8} \biggr)+\frac{1}{2}f \biggl(\frac{5a+3b}{8} \biggr). \end{aligned}$$

(11)

Similarly, due to the convexity of f on the interval $[\frac{a+b}{2},b]$, we have

$$ f \biggl(\frac{a+3b}{4} \biggr)\leq\frac{1}{2}f \biggl( \frac{3a+5b}{8} \biggr)+\frac{1}{2}f \biggl(\frac{a+7b}{8} \biggr). $$

(12)

Hence, inequality (6) is a refinement of the inequality

$$\frac{1}{2} \biggl(f \biggl(\frac{3a+b}{4} \biggr)+f \biggl( \frac{a+3b}{4} \biggr) \biggr) \leq\frac{1}{b-a} \int_{a}^{b}f(x)\,dx $$

obtained by Burqan [4], Lemma 1.

Lemma 2.3

Let f be a real-valued convex function on the interval $[a,b]$. Then

$$\begin{aligned} \frac{1}{b-a} \int_{a}^{b}f(x)\,dx \leq&\frac{1}{8} \biggl(f(a)+2f \biggl(\frac{3a+b}{4} \biggr)+2f \biggl(\frac{a+b}{2} \biggr) \\ &{}+2f \biggl(\frac{a+3b}{4} \biggr)+f(b) \biggr). \end{aligned}$$

(13)

Proof

By the Hermite-Hadamard’s inequality (5) we obtain

$$\begin{aligned} \frac{1}{b-a} \int_{a}^{b}f(x)\,dx =&\frac{1}{b-a} \biggl\{ \int_{a}^{\frac{3a+b}{4}}f(x)\,dx+ \int_{\frac{3a+b}{4}}^{\frac{a+b}{2}}f(x)\,dx \\ &{}+ \int_{\frac{a+b}{2}}^{\frac{a+3b}{4}}f(x)\,dx + \int_{\frac{a+3b}{4}}^{b}f(x)\,dx \biggr\} \\ \leq&\frac{1}{b-a} \biggl\{ \frac{f(a)+f (\frac{3a+b}{4} )}{2}\cdot\frac{b-a}{4} + \frac{f (\frac{3a+b}{4} )+f (\frac{a+b}{2} )}{2}\cdot\frac{b-a}{4} \\ &{}+\frac{f (\frac{a+b}{2} )+f (\frac{a+3b}{4} )}{2}\cdot\frac{b-a}{4}+ \frac{f (\frac{a+3b}{4} )+f(b)}{2}\cdot \frac{b-a}{4} \biggr\} \\ =&\frac{1}{8} \biggl(f(a)+2f \biggl(\frac{3a+b}{4} \biggr) +2f \biggl(\frac{a+b}{2} \biggr)+2f \biggl(\frac{a+3b}{4} \biggr)+f(b) \biggr). \end{aligned}$$

This completes the proof. □

Remark 2.4

Let f be a real-valued convex function on the interval $[a,b]$. Thanks to the convexity of f on the interval $[a,\frac{a+b}{2}]$ and $[\frac{a+b}{2},b]$, respectively, we obtain

$$ f \biggl(\frac{3a+b}{4} \biggr)=f \biggl(\frac{a+\frac{a+b}{2}}{2} \biggr) \leq \frac{f(a)+f (\frac{a+b}{2} )}{2} $$

(14)

and

$$ f \biggl(\frac{a+3b}{4} \biggr)=f \biggl(\frac{\frac{a+b}{2}+b}{2} \biggr) \leq \frac{f(b)+f (\frac{a+b}{2} )}{2}. $$

(15)

Hence, combining inequalities (14) and (15), we get

$$\begin{aligned} &{\frac{1}{2} \biggl(f(a)+2f \biggl( \frac{3a+b}{4} \biggr) +2f \biggl( \frac{a+b}{2} \biggr) +2f \biggl( \frac{a+3b}{4} \biggr)+f(b) \biggr)} \\ &{\quad \leq \biggl(f(a)+2f \biggl(\frac{a+b}{2} \biggr)+f(b) \biggr).} \end{aligned}$$

(16)

Hence, inequality (13) is a refinement of the inequality

$$\frac{1}{b-a} \int_{a}^{b}f(x)\,dx\leq\frac{1}{4} \biggl(f(a)+2f \biggl(\frac{a+b}{2} \biggr)+f(b) \biggr) $$

obtained by Feng [5], Lemma 2.

Combining Lemmas 2.1 and 2.3, we obtain the following lemma.

Lemma 2.5

Let f be a real-valued convex function on the interval $[a,b]$. Then

$$\begin{aligned} &{\frac{1}{4} \biggl(f \biggl(\frac{7a+b}{8} \biggr)+f \biggl( \frac{5a+3b}{8} \biggr)+f \biggl(\frac{3a+5b}{8} \biggr)+f \biggl( \frac{a+7b}{8} \biggr) \biggr)} \\ &{\quad \leq\frac{1}{b-a} \int_{a}^{b}f(x)\,dx} \\ &{\quad \leq \frac{1}{8} \biggl(f(a)+2f \biggl(\frac{3a+b}{4} \biggr) +2f \biggl(\frac{a+b}{2} \biggr)} \\ &{\qquad{} +2f \biggl(\frac{a+3b}{4} \biggr)+f(b) \biggr).} \end{aligned}$$

(17)

Remark 2.6

By inequalities (11), (12), and (16) it is easy to see that inequality (17) is a refinement of inequality (5).

Based on Lemma 2.5, we obtain the following theorems.

Theorem 2.7

Let $\Vert \cdot \Vert$ be a unitarily invariant norm, and $A, B, X\in B(\mathcal{H})$ with $A, B$ positive and $X\in\tau_{\Vert \cdot \Vert }$. Then

$$\begin{aligned} &{\frac{1}{2} \bigl\{ \bigl\Vert \bigl\vert A^{\frac{6v+1}{8}}XB^{\frac{7-6v}{8}} \bigr\vert ^{r} \bigr\Vert \cdot \bigl\Vert \bigl\vert A^{\frac{7-6v}{8}}XB^{\frac{6v+1}{8}} \bigr\vert ^{r} \bigr\Vert + \bigl\Vert \bigl\vert A^{\frac{2v+3}{8}}XB^{\frac{5-2v}{8}} \bigr\vert ^{r} \bigr\Vert \cdot \bigl\Vert \bigl\vert A^{\frac{5-2v}{8}}XB^{\frac{2v+3}{8}} \bigr\vert ^{r} \bigr\Vert \bigr\} } \\ &{\quad \leq \frac{1}{\vert 1-2v\vert } \biggl\vert \int_{v}^{1-v} \bigl\Vert \bigl\vert A^{x}XB^{1-x} \bigr\vert ^{r} \bigr\Vert \cdot \bigl\Vert \bigl\vert A^{1-x}XB^{x} \bigr\vert ^{r} \bigr\Vert \,dx \biggr\vert } \\ &{\quad \leq\frac{1}{4} \bigl\{ \bigl\Vert \bigl\vert A^{v}XB^{1-v} \bigr\vert ^{r} \bigr\Vert \cdot \bigl\Vert \bigl\vert A^{1-v}XB^{v} \bigr\vert ^{r} \bigr\Vert + \bigl\Vert \bigl\vert A^{\frac{1}{2}}XB^{\frac{1}{2}} \bigr\vert ^{r} \bigr\Vert ^{2}} \\ &{\qquad{} +2 \bigl\Vert \bigl\vert A^{\frac{2v+1}{4}}XB^{\frac{3-2v}{4}} \bigr\vert ^{r} \bigr\Vert \cdot \bigl\Vert \bigl\vert A^{\frac{3-2v}{4}}XB^{\frac{2v+1}{4}} \bigr\vert ^{r} \bigr\Vert \bigr\} ,} \end{aligned}$$

(18)

where $v\in[0,1]\setminus\{\frac{1}{2}\}$ and $r>0$.

Proof

We first consider the case $0\leq v<\frac{1}{2}$. Applying Lemma 2.5 to the function $f(x)=\Vert \vert A^{x}XB^{1-x}\vert ^{r}\Vert \cdot \Vert \vert A^{1-x}XB^{x}\vert ^{r}\Vert$ on the interval $[v, 1-v]$ and $f(x)=f(1-x)$ for $x\in[0,1]$, we obtain

$$\begin{aligned} \frac{1}{2} \biggl\{ f \biggl(\frac{6v+1}{8} \biggr)+f \biggl( \frac{2v+3}{8} \biggr) \biggr\} \leq&\frac{1}{1-2v} \int_{v}^{1-v}f(x)\,dx \\ \leq&\frac{1}{4} \biggl\{ f(v)+2f \biggl(\frac{2v+1}{4} \biggr)+f \biggl(\frac{1}{2} \biggr) \biggr\} . \end{aligned}$$

Thus, we get inequality (18) for $0\leq v<\frac{1}{2}$.

Similarly, when $\frac{1}{2}< v\leq1$, applying Lemma 2.5 to the function f on the interval $[1-v,v]$ and $f(x)=f(1-x)$ for $x\in[0,1]$, we obtain

$$\begin{aligned} \frac{1}{2} \biggl\{ f \biggl(\frac{6v+1}{8} \biggr)+f \biggl( \frac{2v+3}{8} \biggr) \biggr\} \leq&\frac{1}{2v-1} \int_{1-v}^{v}f(x)\,dx \\ \leq&\frac{1}{4} \biggl\{ f(v)+2f \biggl(\frac{2v+1}{4} \biggr)+f \biggl(\frac{1}{2} \biggr) \biggr\} . \end{aligned}$$

Hence, we get inequality (18) for $\frac{1}{2}< v\leq1$.

This completes the proof. □

The next theorem is a refinement of Heinz’s norm inequalities.

Theorem 2.8

Let $\Vert \cdot \Vert$ be a unitarily invariant norm, and $A, B, X\in B(\mathcal{H})$ with $A, B$ positive and $X\in\tau_{\Vert \cdot \Vert }$. Then

$$\begin{aligned} &{\frac{1}{2} \bigl\{ \bigl\Vert A^{\frac{6v+1}{8}}XB^{\frac{7-6v}{8}}+A^{\frac{7-6v}{8}}XB^{\frac{6v+1}{8}} \bigr\Vert + \bigl\Vert A^{\frac{2v+3}{8}}XB^{\frac{5-2v}{8}}+A^{\frac{5-2v}{8}}XB^{\frac{2v+3}{8}} \bigr\Vert \bigr\} } \\ &{\quad \leq \frac{1}{\vert 1-2v\vert } \biggl\vert \int_{v}^{1-v} \bigl\Vert A^{x}XB^{1-x}+A^{1-x}XB^{x} \bigr\Vert \,dx \biggr\vert } \\ &{\quad \leq\frac{1}{4} \bigl\{ \bigl\Vert A^{v}XB^{1-v}+A^{1-v}XB^{v} \bigr\Vert +2 \bigl\Vert A^{\frac{1}{2}}XB^{\frac{1}{2}} \bigr\Vert } \\ &{\qquad{} +2 \bigl\Vert A^{\frac{2v+1}{4}}XB^{\frac{3-2v}{4}}+A^{\frac{3-2v}{4}}XB^{\frac{2v+1}{4}} \bigr\Vert \bigr\} ,} \end{aligned}$$

(19)

where $v\in[0,1]\setminus\{\frac{1}{2}\}$.

Proof

Replacing $f(x)=\Vert \vert A^{x}XB^{1-x}\vert ^{r}\Vert \cdot \Vert \vert A^{1-x}XB^{x}\vert ^{r}\Vert$ by $g(x)=\Vert A^{x}XB^{1-x}+A^{1-x}XB^{x}\Vert$ in the proof of Theorem 2.7, we get the desired result. □

Remark 2.9

Putting $f(x)=\Vert \vert A^{x}XB^{1-x}\vert ^{r}\Vert \cdot \Vert \vert A^{1-x}XB^{x}\vert ^{r}\Vert$ ($x\in[0,1]$), where $\Vert \cdot \Vert$ is a unitarily invariant norm, $A, B, X\in B(\mathcal{H})$ with $A, B$ positive, $X\in\tau_{\Vert \cdot \Vert }$, and $r>0$. By the convexity of f we have $f (\frac{2v+1}{4} )\leq\frac{1}{2} \{f (\frac{6v+1}{8} )+ f (\frac{2v+3}{8} ) \}$ and $\frac{1}{4} \{f(v)+2f (\frac{2v+1}{4} )+f (\frac{1}{2} ) \} \leq\frac{1}{2}\{f(v)+f(\frac{1}{2})\}$ for $v\in[0,1]\setminus\{\frac{1}{2}\}$. So, norm inequality (18) is a refinement of inequality (3) obtained by Burqan [4], Theorem 2.

Remark 2.10

Putting $g(x)=\Vert A^{x}XB^{1-x}+A^{1-x}XB^{x}\Vert$ $(x\in[0,1])$, where $\Vert \cdot \Vert$ is a unitarily invariant norm, and $A, B, X\in B(\mathcal{H})$ with $A, B$ positive and $X\in\tau_{\Vert \cdot \Vert }$. By the convexity of g we have $g (\frac{2v+1}{4} )\leq\frac{1}{2} \{g (\frac{6v+1}{8} )+ g (\frac{2v+3}{8} ) \}$ and $\frac{1}{4} \{g(v)+2g (\frac{2v+1}{4} )+g (\frac{1}{2} ) \} \leq\frac{1}{2}\{g(v)+g(\frac{1}{2})\}$ for $v\in[0,1]\setminus\{\frac{1}{2}\}$. Hence, norm inequality (19) is a refinement of inequality (4) obtained by Feng [5], Theorem 1.

For $0< v<1$, applying Lemma 2.5 to the convex functions $f(x)=\Vert \vert A^{x}XB^{1-x}\vert ^{r}\Vert \cdot \Vert \vert A^{1-x}XB^{x}\vert ^{r}\Vert$ and $g(x)=\Vert A^{x}XB^{1-x}+A^{1-x}XB^{x}\Vert$ on the interval $[0,v]$ when $0< v\leq\frac{1}{2}$ and on the interval $[v,1]$ when $\frac{1}{2}\leq v<1$, respectively, we obtain the following theorems, where $\Vert \cdot \Vert$ is a unitarily invariant norm, $A, B, X\in B(\mathcal{H})$ with $A, B$ positive and $X\in\tau_{\Vert \cdot \Vert }$, and $r>0$.

Theorem 2.11

Let $\Vert \cdot \Vert$ be a unitarily invariant norm, and $A, B, X\in B(\mathcal{H})$ with $A, B$ positive and $X\in\tau_{\Vert \cdot \Vert }$. Then

(i) for $0< v\leq\frac{1}{2}$,

$$\begin{aligned} &{\frac{1}{4} \bigl\{ \bigl\Vert \bigl\vert A^{\frac{v}{8}}XB^{\frac{8-v}{8}} \bigr\vert ^{r} \bigr\Vert \cdot \bigl\Vert \bigl\vert A^{\frac{8-v}{8}}XB^{\frac{v}{8}} \bigr\vert ^{r} \bigr\Vert + \bigl\Vert \bigl\vert A^{\frac{3v}{8}}XB^{\frac{8-3v}{8}} \bigr\vert ^{r} \bigr\Vert } \\ &{\qquad{} \cdot \bigl\Vert \bigl\vert A^{\frac{8-3v}{8}}XB^{\frac{3v}{8}} \bigr\vert ^{r} \bigr\Vert + \bigl\Vert \bigl\vert A^{\frac{5v}{8}}XB^{\frac{8-5v}{8}} \bigr\vert ^{r} \bigr\Vert \cdot \bigl\Vert \bigl\vert A^{\frac{8-5v}{8}}XB^{\frac{5v}{8}} \bigr\vert ^{r} \bigr\Vert } \\ &{\qquad{} + \bigl\Vert \bigl\vert A^{\frac{7v}{8}}XB^{\frac{8-7v}{8}} \bigr\vert ^{r} \bigr\Vert \cdot \bigl\Vert \bigl\vert A^{\frac{8-7v}{8}}XB^{\frac{7v}{8}} \bigr\vert ^{r} \bigr\Vert \bigr\} } \\ &{\quad \leq \frac{1}{v} \int_{0}^{v} \bigl\Vert \bigl\vert A^{x}XB^{1-x} \bigr\vert ^{r} \bigr\Vert \cdot \bigl\Vert \bigl\vert A^{1-x}XB^{x} \bigr\vert ^{r} \bigr\Vert \,dx} \\ &{\quad \leq \frac{1}{8} \bigl\{ \bigl\Vert \vert AX\vert ^{r} \bigr\Vert \cdot \bigl\Vert \vert XB\vert ^{r} \bigr\Vert +2 \bigl\Vert \bigl\vert A^{\frac{v}{4}}XB^{\frac{4-v}{4}} \bigr\vert ^{r} \bigr\Vert \cdot \bigl\Vert \bigl\vert A^{\frac{4-v}{4}}XB^{\frac{v}{4}} \bigr\vert ^{r} \bigr\Vert } \\ &{\qquad{} +2 \bigl\Vert \bigl\vert A^{\frac{v}{2}}XB^{\frac{2-v}{2}} \bigr\vert ^{r} \bigr\Vert \cdot \bigl\Vert \bigl\vert A^{\frac{2-v}{2}}XB^{\frac{v}{2}} \bigr\vert ^{r} \bigr\Vert +2 \bigl\Vert \bigl\vert A^{\frac{3v}{4}}XB^{\frac{4-3v}{4}} \bigr\vert ^{r} \bigr\Vert } \\ &{\qquad{} \cdot \bigl\Vert \bigl\vert A^{\frac{4-3v}{4}}XB^{\frac{3v}{4}} \bigr\vert ^{r} \bigr\Vert + \bigl\Vert \bigl\vert A^{v}XB^{1-v} \bigr\vert ^{r} \bigr\Vert \cdot \bigl\Vert \bigl\vert A^{1-v}XB^{v} \bigr\vert ^{r} \bigr\Vert \bigr\} ;} \end{aligned}$$

(ii) for $\frac{1}{2}< v<1$,

$$\begin{aligned} &{\frac{1}{4} \bigl\{ \bigl\Vert \bigl\vert A^{\frac{7v+1}{8}}XB^{\frac{7-7v}{8}} \bigr\vert ^{r} \bigr\Vert \cdot \bigl\Vert \bigl\vert A^{\frac{7-7v}{8}}XB^{\frac{7v+1}{8}} \bigr\vert ^{r} \bigr\Vert + \bigl\Vert \bigl\vert A^{\frac{5v+3}{8}}XB^{\frac{5-5v}{8}} \bigr\vert ^{r} \bigr\Vert } \\ &{\qquad{} \cdot \bigl\Vert \bigl\vert A^{\frac{5-5v}{8}}XB^{\frac{5v+3}{8}} \bigr\vert ^{r} \bigr\Vert + \bigl\Vert \bigl\vert A^{\frac{3v+5}{8}}XB^{\frac{3-3v}{8}} \bigr\vert ^{r} \bigr\Vert \cdot \bigl\Vert \bigl\vert A^{\frac{3-3v}{8}}XB^{\frac{3v+5}{8}} \bigr\vert ^{r} \bigr\Vert } \\ &{\qquad{}+ \bigl\Vert \bigl\vert A^{\frac{v+7}{8}}XB^{\frac{1-v}{8}} \bigr\vert ^{r} \bigr\Vert \cdot \bigl\Vert \bigl\vert A^{\frac{1-v}{8}}XB^{\frac{v+7}{8}} \bigr\vert ^{r} \bigr\Vert \bigr\} } \\ &{\quad \leq\frac{1}{1-v} \int_{v}^{1} \bigl\Vert \bigl\vert A^{x}XB^{1-x} \bigr\vert ^{r} \bigr\Vert \cdot \bigl\Vert \bigl\vert A^{1-x}XB^{x} \bigr\vert ^{r} \bigr\Vert \,dx} \\ &{\quad \leq\frac{1}{8} \bigl\{ \bigl\Vert \vert AX\vert ^{r} \bigr\Vert \cdot \bigl\Vert \vert XB\vert ^{r} \bigr\Vert +2 \bigl\Vert \bigl\vert A^{\frac{3v+1}{4}}XB^{\frac{3-3v}{4}} \bigr\vert ^{r} \bigr\Vert \cdot \bigl\Vert \bigl\vert A^{\frac{3-3v}{4}}XB^{\frac{3v+1}{4}} \bigr\vert ^{r} \bigr\Vert } \\ &{\qquad{} +2 \bigl\Vert \bigl\vert A^{\frac{v+1}{2}}XB^{\frac{1-v}{2}} \bigr\vert ^{r} \bigr\Vert \cdot \bigl\Vert \bigl\vert A^{\frac{1-v}{2}}XB^{\frac{v+1}{2}} \bigr\vert ^{r} \bigr\Vert +2 \bigl\Vert \bigl\vert A^{\frac{v+3}{4}}XB^{\frac{1-v}{4}} \bigr\vert ^{r} \bigr\Vert } \\ &{\qquad{} \cdot \bigl\Vert \bigl\vert A^{\frac{1-v}{4}}XB^{\frac{v+3}{4}} \bigr\vert ^{r} \bigr\Vert + \bigl\Vert \bigl\vert A^{v}XB^{1-v} \bigr\vert ^{r} \bigr\Vert \cdot \bigl\Vert \bigl\vert A^{1-v}XB^{v} \bigr\vert ^{r} \bigr\Vert \bigr\} ,} \end{aligned}$$

where $r>0$.

Theorem 2.12

Let $\Vert \cdot \Vert$ be a unitarily invariant norm, and $A, B, X\in B(\mathcal{H})$ with $A, B$ positive and $X\in\tau_{\Vert \cdot \Vert }$. Then

(i) for $0< v\leq\frac{1}{2}$,

$$\begin{aligned} &{\frac{1}{4} \bigl\{ \bigl\Vert A^{\frac{v}{8}}XB^{\frac{8-v}{8}}+A^{\frac{8-v}{8}}XB^{\frac{v}{8}} \bigr\Vert + \bigl\Vert A^{\frac{3v}{8}}XB^{\frac{8-3v}{8}}+A^{\frac{8-3v}{8}}XB^{\frac{3v}{8}} \bigr\Vert } \\ &{\qquad{} + \bigl\Vert A^{\frac{5v}{8}}XB^{\frac{8-5v}{8}}+A^{\frac{8-5v}{8}}XB^{\frac{5v}{8}} \bigr\Vert + \bigl\Vert A^{\frac{7v}{8}}XB^{\frac{8-7v}{8}}+A^{\frac{8-7v}{8}}XB^{\frac{7v}{8}} \bigr\Vert \bigr\} } \\ &{\quad \leq\frac{1}{v} \int_{0}^{v} \bigl\Vert A^{x}XB^{1-x}+A^{1-x}XB^{x} \bigr\Vert \,dx} \\ &{\quad \leq\frac{1}{8} \bigl\{ \Vert AX+XB\Vert +2 \bigl\Vert A^{\frac{v}{4}}XB^{\frac{4-v}{4}}+A^{\frac{4-v}{4}}XB^{\frac{v}{4}} \bigr\Vert } \\ &{\qquad{} +2 \bigl\Vert A^{\frac{v}{2}}XB^{\frac{2-v}{2}}+A^{\frac{2-v}{2}}XB^{\frac{v}{2}} \bigr\Vert +2 \bigl\Vert A^{\frac{3v}{4}}XB^{\frac{4-3v}{4}}+A^{\frac{4-3v}{4}}XB^{\frac{3v}{4}} \bigr\Vert } \\ &{\qquad{} + \bigl\Vert A^{v}XB^{1-v}+A^{1-v}XB^{v} \bigr\Vert \bigr\} ;} \end{aligned}$$

(ii) for $\frac{1}{2}< v\leq 1$,

$$\begin{aligned} &{\frac{1}{4} \bigl\{ \bigl\Vert A^{\frac{7v+1}{8}}XB^{\frac{7-7v}{8}}+A^{\frac{7-7v}{8}}XB^{\frac{7v+1}{8}} \bigr\Vert + \bigl\Vert A^{\frac{5v+3}{8}}XB^{\frac{5-5v}{8}}+A^{\frac{5-5v}{8}}XB^{\frac{5v+3}{8}} \bigr\Vert } \\ &{\qquad{} + \bigl\Vert A^{\frac{3v+5}{8}}XB^{\frac{3-3v}{8}}+A^{\frac{3-3v}{8}}XB^{\frac{3v+5}{8}} \bigr\Vert + \bigl\Vert A^{\frac{v+7}{8}}XB^{\frac{1-v}{8}}+A^{\frac{1-v}{8}}XB^{\frac{v+7}{8}} \bigr\Vert \bigr\} } \\ &{\quad \leq \frac{1}{1-v} \int_{v}^{1} \bigl\Vert A^{x}XB^{1-x}+A^{1-x}XB^{x} \bigr\Vert \,dx} \\ &{\quad \leq\frac{1}{8} \bigl\{ \bigl\Vert A^{v}XB^{1-v}+A^{1-v}XB^{v} \bigr\Vert + 2 \bigl\Vert A^{\frac{3v+1}{4}}XB^{\frac{3-3v}{4}}+A^{\frac{3-3v}{4}}XB^{\frac{3v+1}{4}} \bigr\Vert } \\ &{\qquad{} +2 \bigl\Vert A^{\frac{1+v}{2}}XB^{\frac{1-v}{2}}+A^{\frac{1-v}{2}}XB^{\frac{1+v}{2}} \bigr\Vert + 2 \bigl\Vert A^{\frac{v+3}{4}}XB^{\frac{1-v}{4}}+A^{\frac{1-v}{4}}XB^{\frac{v+3}{4}} \bigr\Vert } \\ &{\qquad{} +\Vert AX+XB\Vert \bigr\} .} \end{aligned}$$

Remark 2.13

Putting $f(x)=\Vert \vert A^{x}XB^{1-x}\vert ^{r}\Vert \cdot \Vert \vert A^{1-x}XB^{x}\vert ^{r}\Vert$, where $\Vert \cdot \Vert$ is a unitarily invariant norm, $A, B, X\in B(\mathcal{H})$ with $A, B$ positive and $X\in\tau_{\Vert \cdot \Vert }$, and $r>0$. We first consider the case $0< v\leq\frac{1}{2}$. Since f is decreasing on the interval $[0,\frac{1}{2}]$, we have

$$ f(v)\leq f \biggl(\frac{v}{2} \biggr). $$

(20)

Using inequalities (12) and (16) for f on the interval $[0,v]$, respectively, we get

$$\begin{aligned} &{\frac{1}{2} \biggl\{ f \biggl(\frac{v}{4} \biggr)+ f \biggl( \frac{3v}{4} \biggr) \biggr\} } \\ &{\quad \leq\frac{1}{4} \biggl\{ f \biggl(\frac{v}{8} \biggr)+ f \biggl( \frac{3v}{8} \biggr)+ f \biggl(\frac{5v}{8} \biggr)+ f \biggl( \frac{7v}{8} \biggr) \biggr\} } \end{aligned}$$

(21)

and

$$\begin{aligned} &{\frac{1}{8} \biggl\{ f(0)+ 2f \biggl( \frac{v}{4} \biggr)+ 2f \biggl(\frac{v}{2} \biggr) + 2f \biggl( \frac{3v}{4} \biggr)+ f(v) \biggr\} } \\ &{\quad \leq\frac{1}{4} \biggl\{ f(0)+ 2f \biggl(\frac{v}{2} \biggr)+ f(v) \biggr\} .} \end{aligned}$$

(22)

By the convexity of f on the interval $[0,v]$ we have

$$ f \biggl(\frac{v}{2} \biggr)\leq \frac{f(\frac{v}{4})+f(\frac{3v}{4})}{2},\qquad f \biggl( \frac{v}{2} \biggr)\leq\frac{f(0)+f(v)}{2}. $$

(23)

Noting that

$$ \frac{f(0)+f(v)}{2}=\frac{f(1)+f(v)}{2}\leq f(1)= \bigl\Vert \vert AX\vert ^{r}\bigr\Vert \cdot\bigl\Vert \vert XB\vert ^{r} \bigr\Vert , $$

(24)

by (i) of Theorem 2.11 we have

$$\begin{aligned} &{\frac{1}{4} \biggl\{ f \biggl(\frac{v}{8} \biggr)+ f \biggl( \frac{3v}{8} \biggr)+ f \biggl(\frac{5v}{8} \biggr)+ f \biggl( \frac{7v}{8} \biggr) \biggr\} } \\ &{\quad \leq\frac{1}{v} \int_{0}^{v}f(x)\,dx} \\ &{\quad\leq\frac{1}{8} \biggl\{ f(0)+ 2f \biggl(\frac{v}{4} \biggr)+ 2f \biggl(\frac{v}{2} \biggr)+ 2f \biggl(\frac{3v}{4} \biggr)+ f(v) \biggr\} .} \end{aligned}$$

(25)

Combining inequalities (20), (21), (22), (23), (24), and (25), we have

$$\begin{aligned} f(v) \leq& f \biggl(\frac{v}{2} \biggr) \\ \leq&\frac{1}{2} \biggl\{ f \biggl(\frac{v}{4} \biggr)+ f \biggl( \frac{3v}{4} \biggr) \biggr\} \\ \leq&\frac{1}{4} \biggl\{ f \biggl(\frac{v}{8} \biggr)+ f \biggl( \frac{3v}{8} \biggr)+ f \biggl(\frac{5v}{8} \biggr)+ f \biggl( \frac{7v}{8} \biggr) \biggr\} \\ \leq&\frac{1}{v} \int_{0}^{v}f(x)\,dx \\ \leq&\frac{1}{8} \biggl\{ f(0)+ 2f \biggl(\frac{v}{4} \biggr)+ 2f \biggl(\frac{v}{2} \biggr)+ 2f \biggl(\frac{3v}{4} \biggr)+ f(v) \biggr\} \\ \leq&\frac{1}{4} \biggl\{ f(0)+ 2f \biggl(\frac{v}{2} \biggr)+ f(v) \biggr\} \\ \leq&\frac{f(0)+f(v)}{2} \\ \leq& f(0). \end{aligned}$$

(26)

Similarly, when $\frac{1}{2}< v<1$, since $f(v)$ is increasing on the interval $[\frac{1}{2},1]$, we have $f(v)\leq f(\frac{1+v}{2})$. By the same process as before, we also have

$$\begin{aligned} f(v) \leq& f \biggl(\frac{1+v}{2} \biggr) \\ \leq&\frac{1}{2} \biggl\{ f \biggl(\frac{3v+1}{4} \biggr)+ f \biggl( \frac{v+3}{4} \biggr) \biggr\} \\ \leq&\frac{1}{4} \biggl\{ f \biggl(\frac{7v+1}{8} \biggr)+ f \biggl( \frac{5v+3}{8} \biggr)+ f \biggl(\frac{3v+5}{8} \biggr)+f \biggl( \frac{v+7}{8} \biggr) \biggr\} \\ \leq&\frac{1}{1-v} \int_{v}^{1}f(x)\,dx \\ \leq&\frac{1}{8} \biggl\{ f(v)+ 2f \biggl(\frac{3v+1}{4} \biggr)+ 2f \biggl(\frac{1+v}{2} \biggr)+ 2f \biggl(\frac{v+3}{4} \biggr)+ f(1) \biggr\} \\ \leq&\frac{1}{4} \biggl\{ f(v)+ 2f \biggl(\frac{v+1}{2} \biggr)+ f(1) \biggr\} \\ \leq&\frac{f(1)+f(v)}{2} \\ \leq& f(1). \end{aligned}$$

(27)

Thus, inequalities (26) and (27) are refinements of the second inequality in (1) and also a refinement of Theorem 4 of Burqan [4].

Remark 2.14

Putting $g(x)=\Vert A^{x}XB^{1-x}+A^{1-x}XB^{x}\Vert$, where $\Vert \cdot \Vert$ is a unitarily invariant norm, and $A, B, X\in B(\mathcal{H})$ with $A, B$ positive and $X\in\tau_{\Vert \cdot \Vert }$. Replacing f by g in Remark 2.13, we obtain

$$\begin{aligned} g(v) \leq& g \biggl(\frac{v}{2} \biggr) \\ \leq&\frac{1}{2} \biggl\{ g \biggl(\frac{v}{4} \biggr)+ g \biggl( \frac{3v}{4} \biggr) \biggr\} \\ \leq&\frac{1}{4} \biggl\{ g \biggl(\frac{v}{8} \biggr)+ g \biggl( \frac{3v}{8} \biggr)+ g \biggl(\frac{5v}{8} \biggr)+ g \biggl( \frac{7v}{8} \biggr) \biggr\} \\ \leq&\frac{1}{v} \int_{0}^{v}g(x)\,dx \\ \leq&\frac{1}{8} \biggl\{ g(0)+ 2g \biggl(\frac{v}{4} \biggr)+ 2g \biggl(\frac{v}{2} \biggr)+ 2g \biggl(\frac{3v}{4} \biggr)+ g(v) \biggr\} \\ \leq&\frac{1}{4} \biggl\{ g(0)+ 2g \biggl(\frac{v}{2} \biggr)+ g(v) \biggr\} \\ \leq&\frac{g(0)+g(v)}{2} \\ \leq& g(0), \end{aligned}$$

(28)

where $0< v\leq\frac{1}{2}$.

Similarly, when $\frac{1}{2}< v<1$, we have

$$\begin{aligned} g(v) \leq& g \biggl(\frac{1+v}{2} \biggr) \\ \leq&\frac{1}{2} \biggl\{ g \biggl(\frac{3v+1}{4} \biggr)+ g \biggl( \frac{v+3}{4} \biggr) \biggr\} \\ \leq&\frac{1}{4} \biggl\{ g \biggl(\frac{7v+1}{8} \biggr)+ g \biggl( \frac{5v+3}{8} \biggr)+ g \biggl(\frac{3v+5}{8} \biggr)+g \biggl( \frac{v+7}{8} \biggr) \biggr\} \\ \leq&\frac{1}{1-v} \int_{v}^{1}g(x)\,dx \\ \leq&\frac{1}{8} \biggl\{ g(v)+ 2g \biggl(\frac{3v+1}{4} \biggr)+ 2g \biggl(\frac{1+v}{2} \biggr)+ 2g \biggl(\frac{v+3}{4} \biggr)+ g(1) \biggr\} \\ \leq&\frac{1}{4} \biggl\{ g(v)+ 2g \biggl(\frac{v+1}{2} \biggr)+ g(1) \biggr\} \\ \leq&\frac{g(1)+g(v)}{2} \\ \leq& g(1). \end{aligned}$$

(29)

Therefore, inequalities (28) and (29) are refinements of the second inequality in (2) and also a refinement of Theorem 3 of Feng [5].