A note on some inequalities for positive linear maps

TL;DR

This paper improves and generalizes inequalities involving positive linear maps and operators, providing tighter bounds and an enhanced operator Pólya-Szegö inequality for specific operator orderings.

Contribution

It introduces new bounds for operator inequalities under certain orderings and extends the operator Pólya-Szegö inequality with improved constants.

Findings

Derived new inequalities for positive linear maps with explicit bounds.

Generalized the operator Pólya-Szegö inequality with improved constants.

Applicable to operators within specific orderings and bounds.

Abstract

We improve and generalize some operator inequalities for positive linear maps. It is shown, among other inequalities, that if $0<m\le B\le m'<M'\le A\le M$ or $0<m\le A\le m'<M'\le B\le M$, then for each $2\le p<\infty $ and $\nu \in \left[ 0,1 \right]$, \begin{equation*} {{\Phi }^{p}}\left( A{{\nabla }_{\nu }}B \right)\le {{\left( \frac{K\left( h \right)}{{{4}^{\frac{2}{p}-1}}{{K}^{r}}\left( h' \right)} \right)}^{p}}{{\Phi }^{p}}\left( A{{\#}_{\nu }}B \right), \end{equation*} and \begin{equation*} {{\Phi }^{p}}\left( A{{\nabla }_{\nu }}B \right)\le {{\left( \frac{K\left( h \right)}{{{4}^{\frac{2}{p}-1}}{{K}^{r}}\left( h' \right)} \right)}^{p}}{{\left( \Phi \left( A \right){{\#}_{\nu }}\Phi \left( B \right) \right)}^{p}}, \end{equation*} where $r=\min \left\{ \nu ,1-\nu \right\}$, $h=\frac{M}{m}$ and $h'=\frac{M'}{m'}$. We also obtain an improvement of operator P\'olya-Szeg\"o inequality.

Equations

$${\Phi }^p\left(A{\nabla }_{\nu }B\right)\le {\left(\frac{K\left(h\right)}{4^{\frac{2}{p}-1}{K}^r\left(h^{\prime }\right)}\right)}^p{\Phi }^p\left(A{\#}_{\nu }B\right),$$

$${\Phi }^p\left(A{\nabla }_{\nu }B\right)\le {\left(\frac{K\left(h\right)}{4^{\frac{2}{p}-1}{K}^r\left(h^{\prime }\right)}\right)}^p{\left(\Phi \left(A\right){\#}_{\nu }\Phi \left(B\right)\right)}^p,$$

$$A{\nabla }_{\nu }B=\left(1-\nu \right)A+\nu B,A{\#}_{\nu }B=A^{\frac{1}{2}}{\left(A^{-\frac{1}{2}}B{A}^{-\frac{1}{2}}\right)}^{\nu }{A}^{\frac{1}{2}},$$

$${\Phi }^2\left(\frac{A+B}{2}\right)\le K^2\left(h\right){\Phi }^2\left(A\#B\right),$$

$${\Phi }^2\left(\frac{A+B}{2}\right)\le K^2\left(h\right){\left(\Phi \left(A\right)\#\Phi \left(B\right)\right)}^2,$$

$${\Phi }^2\left(\frac{A+B}{2}\right)\le \frac{K^2\left(h\right)}{K\left(h^{\prime }\right)}{\Phi }^2\left(A\#B\right),$$

$${\Phi }^2\left(\frac{A+B}{2}\right)\le \frac{K^2\left(h\right)}{K\left(h^{\prime }\right)}{\left(\Phi \left(A\right)\#\Phi \left(B\right)\right)}^2.$$

$${\Phi }^p\left(\frac{A+B}{2}\right)\le {\left(\frac{{\left(M+m\right)}^2}{4^{\frac{2}{p}}Mm}\right)}^p{\Phi }^p\left(A\#B\right),$$

$${\Phi }^p\left(\frac{A+B}{2}\right)\le {\left(\frac{{\left(M+m\right)}^2}{4^{\frac{2}{p}}Mm}\right)}^p{\left(\Phi \left(A\right)\#\Phi \left(B\right)\right)}^p.$$

$$A\le \alpha B\iff \Vert A^{\frac{1}{2}}{B}^{-\frac{1}{2}}\Vert \le {\alpha }^{\frac{1}{2}}.$$

$$\Vert AB\Vert \le \frac{1}{4}{\Vert A+B\Vert }^2.$$

$${\Phi }^{-1}\left(A\right)\le \Phi \left(A^{-1}\right).$$

$$\Phi \left(A{\#}_{\nu }B\right)\le \Phi \left(A\right){\#}_{\nu }\Phi \left(B\right).$$

$$K^r\left(h^{\prime }\right)\left(A^{-1}{\#}_{\nu }{B}^{-1}\right)\le A^{-1}{\nabla }_{\nu }{B}^{-1},$$

$$K^r\left(h\right)a^{1-\nu }{b}^{\nu }\le a{\nabla }_{\nu }b,$$

$$\underset{h^{\prime }\le x\le h}{\min }{K}^r\left(x\right)X^{\nu }\le \left(1-\nu \right)I+\nu X,$$

$$\underset{h^{\prime }\le x\le h}{\min }{K}^r\left(x\right){\left(A^{\frac{1}{2}}{B}^{-1}{A}^{\frac{1}{2}}\right)}^{\nu }\le \left(1-\nu \right)I+\nu A^{\frac{1}{2}}{B}^{-1}{A}^{\frac{1}{2}}.$$

$$K^r\left(h^{\prime }\right){\left(A^{\frac{1}{2}}{B}^{-1}{A}^{\frac{1}{2}}\right)}^{\nu }\le \left(1-\nu \right)I+\nu A^{\frac{1}{2}}{B}^{-1}{A}^{\frac{1}{2}}.$$

$${\Phi }^2\left(A{\nabla }_{\nu }B\right)\le {\left(\frac{K\left(h\right)}{K^r\left(h^{\prime }\right)}\right)}^2{\Phi }^2\left(A{\#}_{\nu }B\right),$$

$${\Phi }^2\left(A{\nabla }_{\nu }B\right)\le {\left(\frac{K\left(h\right)}{K^r\left(h^{\prime }\right)}\right)}^2{\left(\Phi \left(A\right){\#}_{\nu }\Phi \left(B\right)\right)}^2,$$

$$A+Mm{A}^{-1}\le M+m,$$

$$B+Mm{B}^{-1}\le M+m.$$

$$\left(1-\nu \right)A+\left(1-\nu \right)Mm{A}^{-1}\le \left(1-\nu \right)M+\left(1-\nu \right)m,$$

$$\nu B+\nu Mm{B}^{-1}\le \nu M+\nu m.$$

$$A{\nabla }_{\nu }B+Mm\left(A^{-1}{\nabla }_{\nu }{B}^{-1}\right)\le M+m.$$

$$\Phi \left(A{\nabla }_{\nu }B\right)+Mm\Phi \left(A^{-1}{\nabla }_{\nu }{B}^{-1}\right)\le M+m.$$

$$\Vert \Phi \left(A{\nabla }_{\nu }B\right){\Phi }^{-1}\left(A{\#}_{\nu }B\right)\Vert \le \frac{K\left(h\right)}{K^r\left(h^{\prime }\right)}.$$

$$\Vert \Phi \left(A{\nabla }_{\nu }B\right)Mm{K}^r\left(h^{\prime }\right){\Phi }^{-1}\left(A{\#}_{\nu }B\right)\Vert$$

$$\le \frac{1}{4}{\Vert \Phi \left(A{\nabla }_{\nu }B\right)+Mm{K}^r\left(h^{\prime }\right){\Phi }^{-1}\left(A{\#}_{\nu }B\right)\Vert }^2\text{(by Lemma ref6)}$$

$$\le \frac{1}{4}{\Vert \Phi \left(A{\nabla }_{\nu }B\right)+Mm{K}^r\left(h^{\prime }\right)\Phi \left(A^{-1}{\#}_{\nu }{B}^{-1}\right)\Vert }^2\text{(by Lemma ref7)}$$

$$\le \frac{1}{4}{\Vert \Phi \left(A{\nabla }_{\nu }B\right)+Mm\Phi \left(A^{-1}{\nabla }_{\nu }{B}^{-1}\right)\Vert }^2\text{(by Lemma ref1)}$$

$$\le \frac{1}{4}{\left(M+m\right)}^2\text{(by eqref4)},$$

Full text

[TABLE]

and

[TABLE]

where $r=\min\left\{\nu,1-\nu\right\}$, $h=\frac{M}{m}$ and $h^{\prime}=\frac{M^{\prime}}{m^{\prime}}$. We also obtain an improvement of operator Pólya-Szegö inequality.

1. Introduction

Let us introduce a notation and state a few elementary facts that will be helpful in the ensuing discussion. Throughout this paper, we reserve $M,M^{\prime},m,m^{\prime}$ for real numbers and $I$ for the identity operator. Other capital letters denote general elements of the ${{C}^{*}}$-algebra $\mathcal{B}\left(\mathcal{H}\right)$ of all bounded linear operators on a complex separable Hilbert space $\left(\mathcal{H},\left\langle\cdot,\cdot\right\rangle\right)$. Also, we identify a scalar with the unit multiplied by this scalar. $\left\|\cdot\right\|$ denote the operator norm. An operator $A$ is said to be positive (strictly positive) if $\left\langle Ax,x\right\rangle\geq 0$ for all $x\in\mathcal{H}$ ($\left\langle Ax,x\right\rangle>0$ for all $x\in\mathcal{H}\backslash\left\{0\right\}$) and write $A\geq 0$ ($A>0$). $A\geq B$ ($A>B$) means $A-B\geq 0$ ($A-B>0$). A linear map $\Phi:\mathcal{B}\left(\mathcal{H}\right)\to\mathcal{B}\left(\mathcal{K}\right)$ is called positive if $\Phi\left(A\right)\geq 0$ whenever $A\geq 0$. It is said to be unital if $\Phi\left(I\right)=I$. As a matter of convenience, we use the following notations to define the weighted arithmetic and geometric means for operators:

[TABLE]

where $A,B>0$ and $\nu\in\left[0,1\right]$.

Lin [8] reduced the study of squared operator inequalities to that of some norm inequalities.

According to the celebrated paper by Lin [8, Theorem 2.1], if $0<m\leq A,B\leq M$, then

[TABLE]

and

[TABLE]

where $K\left(h\right)=\frac{{{\left(h+1\right)}^{2}}}{4h}$ and $h=\frac{M}{m}$.

Related to this, Xue and Hu [10, Theorem 2] proved that if $0<m\leq A\leq m^{\prime}<M^{\prime}\leq B\leq M$, then

[TABLE]

and

[TABLE]

where $K\left(h\right)=\frac{{{\left(h+1\right)}^{2}}}{4h},\text{ }K\left(h^{\prime}\right)=\frac{{{\left(h^{\prime}+1\right)}^{2}}}{4h^{\prime}},\text{ }h=\frac{M}{m}$ and $h^{\prime}=\frac{M^{\prime}}{m^{\prime}}$.

We will get a stronger result than (1.3)-(1.4) (see Theorem 2.1).

Lin’s results was further generalized by several authors. Among them, Fu and He [6, Theorem 4] generalized (1.1) and (1.2) to the power of $p$ ($2\leq p<\infty$) as follows:

[TABLE]

and

[TABLE]

It is interesting to ask whether the inequalities (1.5) and (1.6) can be improved. This is an another motivation of the present paper (Theorem 2.2). We close the paper by improving operator Pólya-Szegö inequality (Theorem 2.5).

2. Main Results

We give some Lemmas before we give the main theorems of this paper:

Lemma 2.1**.**

[2, Lemma 2.3]** Let $A,B>0$ and $\alpha>0$, then

[TABLE]

Lemma 2.2**.**

[3, Theorem 1]** Let $A,B>0$. Then the following norm inequality holds:

[TABLE]

We need the following inequality, which is due to Choi. It can be found, for example, in [5, p. 41].

Lemma 2.3**.**

Let $A>0$. Then for every positive unital linear map $\Phi$,

[TABLE]

The following is attributed to Ando [1]:

Lemma 2.4**.**

Let $\Phi$ be any (not necessary unital) positive linear map and $A,B$ be positive operators. Then for every $\nu\in\left[0,1\right]$,

[TABLE]

The following basic lemma is essentially known as in [14, Theorem 7], but our expression is a little bit different from those in [14]. For the sake of convenience, we give it a slim proof.

Lemma 2.5**.**

Let $0<m\leq B\leq m^{\prime}<M^{\prime}\leq A\leq M$ or $0<m\leq A\leq m^{\prime}<M^{\prime}\leq B\leq M$, then

[TABLE]

for each $\nu\in\left[0,1\right]$ and $r=\min\left\{\nu,1-\nu\right\}$, $h=\frac{M}{m}$ and $h^{\prime}=\frac{M^{\prime}}{m^{\prime}}$.

Proof.

From [14, Corollary 3], for any $a,b>0$ and $\nu\in\left[0,1\right]$ we have

[TABLE]

where $h=\frac{b}{a}$ and $r=\min\left\{\nu,1-\nu\right\}$. Taking $a=1$ and $b=x$, then utilizing the continuous functional calculus and the fact that $0<h^{\prime}I\leq X\leq hI$, we have

[TABLE]

for any $x>0$. Replacing $X$ with $1<h^{\prime}=\frac{M^{\prime}}{m^{\prime}}\leq{{A}^{\frac{1}{2}}}{{B}^{-1}}{{A}^{\frac{1}{2}}}\leq\frac{M}{m}=h$

[TABLE]

Since $K\left(x\right)$ is an increasing function for $x>1$, then

[TABLE]

Now by multiplying both sides by ${{A}^{-\frac{1}{2}}}$ we deduce the desired inequality (2.1).

For the case of $0<m\leq A\leq m^{\prime}<M^{\prime}\leq B\leq M$, since the function $K\left(x\right)$ is decreasing for $x<1$ and $K\left(x\right)=K\left(\frac{1}{x}\right)$, similarly we obtain inequality (2.1). ∎

Inequalities (1.3) and (1.4) can be generalized by means of weighted parameter $\nu\in\left[0,1\right]$ as follows:

Theorem 2.1**.**

If $0<m\leq B\leq m^{\prime}<M^{\prime}\leq A\leq M$ or $0<m\leq A\leq m^{\prime}<M^{\prime}\leq B\leq M$, then for each $\nu\in\left[0,1\right]$ we have

[TABLE]

and

[TABLE]

where $r=\min\left\{\nu,1-\nu\right\}$, $K\left(h\right)=\frac{{{\left(h+1\right)}^{2}}}{4h},K\left(h^{\prime}\right)=\frac{{{\left(h^{\prime}+1\right)}^{2}}}{4h^{\prime}},h=\frac{M}{m},h^{\prime}=\frac{M^{\prime}}{m^{\prime}}$.

Proof.

According to the hypothesis we have

[TABLE]

and

[TABLE]

Also the following inequalities holds true

[TABLE]

and

[TABLE]

Now summing up (2.5) and (2.6) we obtain

[TABLE]

Applying positive linear map $\Phi$ we can write

[TABLE]

With inequality (2.7) in hand, we are ready to prove (2.3).

By Lemma 2.1, it is enough to prove that

[TABLE]

By computation, we have

[TABLE]

which leads to (2.3). Now we prove (2.4). The operator inequality (2.4) is equivalent to

[TABLE]

Compute

[TABLE]

Thus, we complete the proof. ∎

Remark 2.1**.**

Inequalities (1.3) and (1.4) are two special cases of Theorem 2.1 by taking $\nu=\frac{1}{2}$. However, our inequalities in Theorem 2.1 are tighter than that in (1.1) and (1.2).

To achieve the second result, we state for easy reference the following fact obtaining from [1, Theorem 3] that will be applied below.

Lemma 2.6**.**

Let $A$ and $B$ be positive operators. Then

[TABLE]

for each $1\leq r<\infty$.

Our promised refinement of inequalities (1.5) and (1.6) can be stated as follows.

Theorem 2.2**.**

If $0<m\leq B\leq m^{\prime}<M^{\prime}\leq A\leq M$ or $0<m\leq A\leq m^{\prime}<M^{\prime}\leq B\leq M$, then for each $2\leq p<\infty$ and $\nu\in\left[0,1\right]$ we have

[TABLE]

and

[TABLE]

where $r=\min\left\{\nu,1-\nu\right\}$, $h=\frac{M}{m}$, $h^{\prime}=\frac{M^{\prime}}{m^{\prime}}$.

Proof.

It can be easily seen that the operator inequality (2.9) is equivalent to

[TABLE]

By simple computation

[TABLE]

which leads to (2.9). The desired inequality (2.10) is equivalent to

[TABLE]

The result will follow from

[TABLE]

as required. ∎

Remark 2.2**.**

Notice that the Kantorovich’s constant $K\left(h\right)=\frac{{{\left(h+1\right)}^{2}}}{4h},\text{ }h>0$ is an increasing function on $\left[1,\infty\right)$. Moreover $K\left(h\right)\geq 1$ for any $h>0$. Therefore, Theorem 2.2 is a refinement of the inequalities, (1.5) and (1.6) for $2\leq p<\infty$.

It is proved in [13, Theorem 2.6] that for $4\leq p<\infty$,

[TABLE]

and

[TABLE]

These inequalities can be improved:

Theorem 2.3**.**

If $0<m\leq B\leq m^{\prime}<M^{\prime}\leq A\leq M$ or $0<m\leq A\leq m^{\prime}<M^{\prime}\leq B\leq M$, then for each $4\leq p<\infty$ and $\nu\in\left[0,1\right]$ we have

[TABLE]

and

[TABLE]

where $r=\min\left\{\nu,1-\nu\right\}$, $K\left(h\right)=\frac{{{\left(h+1\right)}^{2}}}{4h},K\left(h^{\prime}\right)=\frac{{{\left(h^{\prime}+1\right)}^{2}}}{4h^{\prime}},h=\frac{M}{m},h^{\prime}=\frac{M^{\prime}}{m^{\prime}}$.

Proof.

It is easily verified that if $0<m\leq T\leq M$, then

[TABLE]

According to the assumption we have

[TABLE]

we can also write

[TABLE]

Using the substitution $T=\Phi\left(A{{\nabla}_{\nu}}B\right)$ in (2.13) we get

[TABLE]

On the other hand, from (2.3) we obtain

[TABLE]

From this one can see that

[TABLE]

Show that

[TABLE]

The validity of this inequality is just inequality (2.11).

Similarly, (2.12) holds by the inequality (2.4). ∎

Remark 2.3**.**

For $\nu=\frac{1}{2}$ such result can be found in [10, Theorem 3].

In [9, Theorem 2.1] the authors gave operator Pólya-Szegö inequality as follows:

Theorem 2.4**.**

Let $\Phi$ be a positive linear map. If $0<m_{1}^{2}\leq A\leq M_{1}^{2}$ and $0<m_{2}^{2}\leq B\leq M_{2}^{2}$ for some positive real numbers ${{m}_{1}}\leq{{M}_{1}}$ and ${{m}_{2}}\leq{{M}_{2}}$, then

[TABLE]

where $m=\frac{{{m}_{2}}}{{{M}_{1}}}$ and $M=\frac{{{M}_{2}}}{{{m}_{1}}}$.

It is worth noting that the inequality (2.16) was first proved in [7, Theorem 4] for matrices under the sandwich assumption $mA\leq B\leq MA$ (see also [4, Theorem 3]). Zhao et al. [12, Theorem 3.2] by using the same strategies of [14] obtained that:

Lemma 2.7**.**

If $A,B\geq 0,\text{ }1<h\leq{{A}^{-\frac{1}{2}}}B{{A}^{-\frac{1}{2}}}\leq h^{\prime}$ or $0<h^{\prime}\leq{{A}^{-\frac{1}{2}}}B{{A}^{-\frac{1}{2}}}\leq h<1$, then

[TABLE]

for all $\nu\in\left[0,1\right]$, where $r=\min\left\{\nu,1-\nu\right\}$.

Now, we try to obtain a new refinement of Theorem 2.4 by using Lemma 2.7.

Theorem 2.5**.**

Let $A,B$ be two positive operators such that $m_{1}^{2}\leq A\leq M_{1}^{2}$, $m_{2}^{2}\leq B\leq M_{2}^{2}$, $m=\frac{{{m}_{2}}}{{{M}_{1}}}$ and $M=\frac{{{M}_{2}}}{{{m}_{1}}}$. If ${{M}_{1}}<{{m}_{2}}$, then

[TABLE]

where

[TABLE]

and $h=\frac{m_{2}^{2}}{M_{1}^{2}}$.

Moreover the inequality (2.17) holds true for ${{M}_{2}}<{{m}_{1}}$ and $h=\frac{M_{2}^{2}}{m_{1}^{2}}$.

Proof.

According to the assumptions we have

[TABLE]

i.e.,

[TABLE]

Therefore (2.18) implies that

[TABLE]

Simplifying we find that

[TABLE]

Multiplying both sides by ${{A}^{\frac{1}{2}}}$ to get

[TABLE]

By applying positive linear map $\Phi$ in (2.19) we infer

[TABLE]

Utilizing Lemma 2.7 for the case $\nu=\frac{1}{2},\text{ }A=Mm\Phi\left(A\right)$ and $B=\Phi\left(B\right)$, and by taking into account that $m_{1}^{2}\frac{{{M}_{2}}{{m}_{2}}}{{{M}_{1}}{{m}_{1}}}\leq\frac{{{M}_{2}}{{m}_{2}}}{{{M}_{1}}{{m}_{1}}}\Phi\left(A\right)\leq M_{1}^{2}\frac{{{M}_{2}}{{m}_{2}}}{{{M}_{1}}{{m}_{1}}}$ implies $m_{1}^{2}\leq\Phi\left(A\right)\leq M_{1}^{2}$, and $1<\frac{m_{2}^{2}}{M_{1}^{2}}\leq{{\Phi}^{-\frac{1}{2}}}\left(A\right)\Phi\left(B\right){{\Phi}^{-\frac{1}{2}}}\left(A\right)\leq\frac{M_{2}^{2}}{m_{1}^{2}}$ we infer

[TABLE]

where $h=\frac{m_{2}^{2}}{M_{1}^{2}}$.

Combining (2.20) and (2.21), we deduce the desired result (2.17). The case ${{M}_{2}}<{{m}_{1}}$ is similar, we omit the details.

This completes the proof. ∎

The authors would like to pose the following question that is interesting on its own right.

Question 2.1**.**

Is the constant $\gamma$ in Theorem 2.5 sharp?

Example 2.1**.**

Taking $m_{1}^{2}=1.21,\text{ }M_{1}^{2}=16,\text{ }m_{2}^{2}=20.25,\text{ }M_{2}^{2}=25,\text{ }A=\left(\begin{matrix}2&-2\\ -2&7\\ \end{matrix}\right),\text{ }B=\left(\begin{matrix}21&0.5\\ 0.5&21\\ \end{matrix}\right)$ and $\Phi\left(X\right)=\frac{1}{2}tr\left(X\right)\left(X\in{{\mathcal{M}}_{2}}\right)$, by an easy computation we find that

[TABLE]

[TABLE]

and

[TABLE]

which shows that if ${{M}_{1}}<{{m}_{2}}$, then inequality (2.17) is really an improvement of (2.16).

Example 2.2**.**

Assume that $m_{1}^{2}=4,\text{ }M_{1}^{2}=9,\text{ }m_{2}^{2}=0.5,\text{ }M_{2}^{2}=2,\text{ }A=\left(\begin{matrix}6&-1\\ -1&5\\ \end{matrix}\right),\text{ }B=\left(\begin{matrix}1.5&0.5\\ 0.5&1.2\\ \end{matrix}\right)$ and $\Phi\left(X\right)=\frac{1}{2}tr\left(X\right)\left(X\in{{\mathcal{M}}_{2}}\right)$, by an easy computation we find that

[TABLE]

[TABLE]

and

[TABLE]

which shows that if ${{M}_{2}}<{{m}_{1}}$, then inequality (2.17) is really an improvement of (2.16).

The inequality (2.17) can be squared by a similar method as in [11, Theorem 2.3]:

Theorem 2.6**.**

Suppose all the assumptions of Theorem 2.5 be satisfied. Then

[TABLE]

where

[TABLE]

$\alpha={{m}_{1}}{{m}_{2}}$, and $\beta={{M}_{1}}{{M}_{2}}$.

Proof.

According to the assumption one can see that

[TABLE]

and

[TABLE]

where

[TABLE]

Inequality (2.23) implies

[TABLE]

and (2.24) give us

[TABLE]

Hence

[TABLE]

Consider the real function $f\left(t\right)$ on $\left(0,\infty\right)$ defined as

[TABLE]

As a matter of fact, the inequality (2.26) implies that

[TABLE]

One can see that the function $f\left(t\right)$ is decreasing on $\left[\alpha,\beta\right]$. By an easy computation we have

[TABLE]

This function has an maximum point on

[TABLE]

with the maximum value

[TABLE]

Whence

[TABLE]

Notice that

[TABLE]

It is striking that we can get the same inequality (2.22) under the condition ${{M}_{2}}<{{m}_{1}}$.

Hence the proof of Theorem 2.6 is complete. ∎

Acknowledgment. The authors are deeply indebted to Professor Jean Christophe Bourin for calling our attention to the work of Lee [7]. The authors also thank the referee for his useful comments which improved the current paper.