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Improved Heinz inequality and its application

Journal of Inequalities and Applications20122012:113

https://doi.org/10.1186/1029-242X-2012-113

• Accepted: 23 May 2012
• Published:

Abstract

We obtain an improved Heinz inequality for scalars and we use it to establish an inequality for the Hilbert-Schmidt norm of matrices, which is a refinement of a result due to Kittaneh.

Mathematical Subject Classification 2010: 26D07; 26D15; 15A18.

Keywords

• Heinz inequality
• convex function
• Hilbert-Schmidt norm

1. Introduction

Let M n be the space of n × n complex matrices and ||·|| stand for any unitarily invariant norm on M n . So, ||UAV|| = ||A|| for all A M n and for all unitary matrices U, V M n . If A = [a ij ] M n , then
${∥A∥}_{2}={\left(\sum _{i,j=1}^{n}|{a}_{ij}{|}^{2}\right)}^{1/2}$

is the Hilbert-Schmidt norm of matrix A. It is known that the Hilbert-Schmidt norm is unitarily invariant.

The classical Young's inequality for nonnegative real numbers says that if a, b ≥ 0 and 0 ≤ v ≤ 1, then
${a}^{v}{b}^{1-v}\le va+\left(1-v\right)\phantom{\rule{2.77695pt}{0ex}}b$
(1.1)
with equality if and only if a = b. Young's inequality for scalars is not only interesting in itself but also very useful. If $v=\frac{1}{2}$, by (1.1), we obtain the arithmetic-geometric mean inequality
$2\sqrt{ab}\le a+b.$
(1.2)
Kittaneh and Manasrah  obtained a refinement of Young's inequality as follows:
${a}^{v}{b}^{1-v}+{r}_{0}{\left(\sqrt{a}-\sqrt{b}\right)}^{2}\le va+\left(1-v\right)b,$
(1.3)

where r0 = min {v, 1 − v}.

Let a, b ≥ 0 and 0 ≤ v ≤ 1. The Heinz means are defined as follows:
${H}_{v}\left(a,\phantom{\rule{2.77695pt}{0ex}}b\right)=\frac{{a}^{v}{b}^{1-v}+{a}^{1-v}{b}^{v}}{2}.$
It follows from the inequalities (1.1) and (1.2) that the Heinz means interpolate between the geometric mean and the arithmetic mean:
$\sqrt{ab}\le {H}_{v}\left(a,\phantom{\rule{2.77695pt}{0ex}}b\right)\le \frac{a+b}{2}.$
(1.4)

The second inequality of (1.4) is known as Heinz inequality for nonnegative real numbers.

As a direct consequence of the inequality (1.3), Kittaneh and Manasrah  obtained a refinement of the Heinz inequality as follows:
${H}_{v}\left(a,\phantom{\rule{2.77695pt}{0ex}}b\right)+{r}_{0}{\left(\sqrt{a}-\sqrt{b}\right)}^{2}\le \frac{a+b}{2},$
(1.5)

where r0 = min {v, 1 − v}.

Bhatia and Davis  proved that if A, B, X M n such that A and B are positive semidefinite and if 0 ≤ v ≤ 1, then
$2∥{A}^{1/2}X{B}^{1/2}∥\phantom{\rule{2.77695pt}{0ex}}\le \phantom{\rule{2.77695pt}{0ex}}∥{A}^{v}X{B}^{1-v}+{A}^{1-v}X{B}^{v}∥\phantom{\rule{2.77695pt}{0ex}}\le \phantom{\rule{2.77695pt}{0ex}}∥AX+XB∥.$
(1.6)
This is a matrix version of the inequality (1.4). Kittaneh  proved that if A, B, X M n such that A and B are positive semidefinite and if 0 ≤ v ≤ 1, then
$∥{A}^{v}X{B}^{1-v}+{A}^{1-v}X{B}^{v}∥\phantom{\rule{2.77695pt}{0ex}}\le \phantom{\rule{2.77695pt}{0ex}}4{r}_{0}∥{A}^{1/2}X{B}^{1/2}∥+\phantom{\rule{2.77695pt}{0ex}}\left(1-2{r}_{0}\right)∥AX+XB∥,$
(1.7)

where r0 = min {v, 1 − v}. This is a refinement of the second inequality in (1.6).

In this article, we first present a refinement of the inequality (1.5). After that, we use it to establish a refinement of the inequality (1.7) for the Hilbert-Schmidt norm.

2. A refinement of the inequality (1.5)

In this section, we give a refinement of the inequality (1.5). To do this, we need the following lemma.

Lemma 2.1. [4, 5] Let f(x) be a real valued convex function on an interval [a, b]. For any x1, x2 [a, b], we have
$f\left(x\right)\le \frac{f\left({x}_{2}\right)-f\left({x}_{1}\right)}{{x}_{2}-{x}_{1}}x-\frac{{x}_{1}f\left({x}_{2}\right)-{x}_{2}f\left({x}_{1}\right)}{{x}_{2}-{x}_{1}},\phantom{\rule{2.77695pt}{0ex}}x\in \left({x}_{1},\phantom{\rule{2.77695pt}{0ex}}{x}_{2}\right).$
Theorem 2.1. Let a, b ≥ 0 and 0 ≤ v ≤ 1. If r0 = min {v, 1 − v}, then
$2{H}_{v}\left(a,\phantom{\rule{2.77695pt}{0ex}}b\right)\le \left\{\begin{array}{c}\left(1-4{r}_{0}\right)\left(a+b\right)+4{r}_{0}\left({a}^{1/4}{b}^{3/4}+{a}^{3/4}{b}^{1/4}\right)\phantom{\rule{2.77695pt}{0ex}},\phantom{\rule{1em}{0ex}}v\in \left[0,\phantom{\rule{2.77695pt}{0ex}}\frac{1}{4}\right]\cup \left[\frac{3}{4},1\right],\\ 2\phantom{\rule{2.77695pt}{0ex}}\left(4{r}_{0}-1\right)\sqrt{ab}+2\left(1-2{r}_{0}\right)\left({a}^{1/4}{b}^{3/4}+{a}^{3/4}{b}^{1/4}\right)\phantom{\rule{2.77695pt}{0ex}},\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}v\in \left[\frac{1}{4},\phantom{\rule{2.77695pt}{0ex}}\frac{3}{4}\right].\end{array}\right\$
(2.1)
Proof. It is known that as a function of v, H v (a, b) is convex and attains its minimum at $v=\frac{1}{2}$. Let
$f\left(v\right)=2{H}_{v}\left(a,b\right)={a}^{v}{b}^{1-v}+{a}^{1-v}{b}^{v},\phantom{\rule{1em}{0ex}}0\le v\le 1.$
Obviously, f(v) is convex. For $0\le v\le \frac{1}{4}$, since f(v) is convex on 0, by Lemma 2.1, we have
$f\left(v\right)\le \frac{f\left(\frac{1}{4}\right)-f\left(0\right)}{\frac{1}{4}-0}v-\frac{0f\left(\frac{1}{4}\right)-\frac{1}{4}f\left(0\right)}{\frac{1}{4}-0},$
which is equivalent to
$f\left(v\right)\le 4\left(f\left(\frac{1}{4}\right)-f\left(0\right)\right)v+f\left(0\right).$
That is,
$f\left(v\right)\le \left(\mathsf{\text{1}}-\mathsf{\text{4}}v\right)f\left(0\right)+\mathsf{\text{4}}vf\left(\frac{1}{4}\right).$
So,
${a}^{v}{b}^{1-v}+{a}^{1-v}{b}^{v}\le \left(\mathsf{\text{1}}-\mathsf{\text{4}}{r}_{0}\right)\left(a+b\right)+\mathsf{\text{4}}{r}_{0}\left({a}^{1/4}{b}^{3/4}+{a}^{3/4}{b}^{1/4}\right).$
For $\frac{3}{4}\le v\le 1$, similarly, we have
$f\left(v\right)\le \frac{f\left(1\right)-f\left(\frac{3}{4}\right)}{1-\frac{3}{4}}v-\frac{\frac{3}{4}f\left(1\right)-f\left(\frac{3}{4}\right)}{1-\frac{3}{4}},$
which is equivalent to
$f\left(v\right)\le \mathsf{\text{4}}\left(f\left(\mathsf{\text{1}}\right)-f\left(\frac{3}{4}\right)\right)v-\mathsf{\text{3}}f\left(\mathsf{\text{1}}\right)+\mathsf{\text{4}}f\left(\frac{3}{4}\right).$
That is,
$f\left(v\right)\le \left(\mathsf{\text{4}}v-\mathsf{\text{3}}\right)f\left(\mathsf{\text{1}}\right)+\mathsf{\text{4}}\left(\mathsf{\text{1}}-v\right)f\left(\frac{3}{4}\right).$
So,
${a}^{v}{b}^{1-v}+{a}^{1-v}{b}^{v}\le \left(\mathsf{\text{1}}-\mathsf{\text{4}}{r}_{0}\right)\left(a+b\right)+\mathsf{\text{4}}{r}_{0}\left({a}^{\mathsf{\text{1}}/\mathsf{\text{4}}}{b}^{\mathsf{\text{3}}/\mathsf{\text{4}}}+{a}^{\mathsf{\text{3}}/\mathsf{\text{4}}}{b}^{\mathsf{\text{1}}/\mathsf{\text{4}}}\right).$
If $\frac{1}{4}\le v\le \frac{1}{2}$, then by Lemma 2.1, we have
$f\left(v\right)\le \frac{f\left(\frac{1}{2}\right)-f\left(\frac{1}{4}\right)}{\frac{1}{2}-\frac{1}{4}}v-\frac{\frac{1}{4}f\left(\frac{1}{2}\right)-\frac{1}{2}f\left(\frac{1}{4}\right)}{\frac{1}{2}-\frac{1}{4}},$
and so
$f\left(v\right)\le \left(\mathsf{\text{4}}v-\mathsf{\text{1}}\right)f\left(\frac{1}{2}\right)+2\left(1-2v\right)f\left(\frac{1}{4}\right),$
which is equivalent to
${a}^{v}{b}^{1-v}+{a}^{1-v}{b}^{v}\le 2\left(\mathsf{\text{4}}{r}_{0}-1\right)\sqrt{ab}+2\left(1-2{r}_{0}\right)\left({a}^{\mathsf{\text{1}}/\mathsf{\text{4}}}{b}^{\mathsf{\text{3}}/\mathsf{\text{4}}}+{a}^{3/\mathsf{\text{4}}}{b}^{1/\mathsf{\text{4}}}\right).$
If $\frac{1}{2}\le v\le \frac{3}{4}$, similarly, we have
$f\left(v\right)\le \frac{f\left(\frac{3}{4}\right)-f\left(\frac{1}{2}\right)}{\frac{3}{4}-\frac{1}{2}}v-\frac{\frac{1}{2}f\left(\frac{3}{4}\right)-\frac{3}{4}f\left(\frac{1}{2}\right)}{\frac{3}{4}-\frac{1}{2}},$
and so
$f\left(v\right)\le \left(3-4v\right)f\left(\frac{1}{2}\right)+2\left(2v-1\right)f\left(\frac{3}{4}\right),$
which is equivalent to
$f\left(v\right)\le \left(\mathsf{\text{4}}{r}_{0}-\mathsf{\text{1}}\right)f\left(\frac{1}{2}\right)+2\left(1-2{r}_{0}\right)f\left(\frac{3}{4}\right).$
That is,
${a}^{v}{b}^{1-v}+{a}^{1-v}{b}^{v}\le \mathsf{\text{2}}\left(\mathsf{\text{4}}{r}_{0}-\mathsf{\text{1}}\right)\sqrt{ab}+\mathsf{\text{2}}\left(\mathsf{\text{1}}-\mathsf{\text{2}}{r}_{0}\right)\left({a}^{\mathsf{\text{1}}/\mathsf{\text{4}}}{b}^{\mathsf{\text{3}}/\mathsf{\text{4}}}+{a}^{3/\mathsf{\text{4}}}{b}^{1/\mathsf{\text{4}}}\right).$

This completes the proof. □

Now, we give a simple comparison between the upper bound for a v b1-v+ a1-vb v in (1.5) and (2.1). If $v\in \left[0,\frac{1}{4}\right]\cup \left[\frac{3}{4},1\right]$, then
$\begin{array}{ll}\hfill a+b-2{r}_{0}{\left(\sqrt{a}-\sqrt{b}\right)}^{2}& -\left(1-4{r}_{0}\right)\left(a+b\right)-4{r}_{0}\left({a}^{1/4}{b}^{3/4}+{a}^{3/4}{b}^{1/4}\right)\phantom{\rule{2em}{0ex}}\\ =2{r}_{0}\left(a+b+2\sqrt{ab}-2\left({a}^{1/4}{b}^{3/4}+{a}^{3/4}{b}^{1/4}\right)\right)\phantom{\rule{2em}{0ex}}\\ \ge 0.\phantom{\rule{2em}{0ex}}\end{array}$
If $v\in \left[\frac{1}{4},\frac{3}{4}\right]$, then
$\begin{array}{ll}\hfill a+b-2{r}_{0}{\left(\sqrt{a}-\sqrt{b}\right)}^{2}& -2\left(4{r}_{0}-1\right)\sqrt{ab}-2\left(1-2{r}_{0}\right)\left({a}^{1/4}{b}^{3/4}+{a}^{3/4}{b}^{1/4}\right)\phantom{\rule{2em}{0ex}}\\ =\left(1-2{r}_{0}\right)\left(a+b+2\sqrt{ab}-2\left({a}^{1/4}{b}^{3/4}+{a}^{3/4}{b}^{1/4}\right)\right)\phantom{\rule{2em}{0ex}}\\ \ge 0.\phantom{\rule{2em}{0ex}}\end{array}$

So, the inequality (2.1) is a refinement of the inequality (1.5).

3. An application

In this section, we give a refinement of the inequality (1.7) for the Hilbert-Schmidt norm based on the inequality (2.1).

Theorem 3.1. Let A, B, X M n such that A and B are positive semidefinite and suppose

that
$\varphi \left(v\right)=||{A}^{v}X{B}^{1-v}+{A}^{1-v}X{B}^{v}|{|}_{\mathsf{\text{2}}},\phantom{\rule{1em}{0ex}}0\le v\le \mathsf{\text{1}}.$
Then
$\varphi \left(v\right)\le \left\{\begin{array}{c}\left(1-4{r}_{0}\right)\varphi \left(0\right)+4{r}_{0}\varphi \left(\frac{1}{4}\right),\phantom{\rule{1em}{0ex}}v\in \left[0,\phantom{\rule{2.77695pt}{0ex}}\frac{1}{4}\right]\cup \left[\frac{3}{4},1\right]\\ \left(4{r}_{0}-1\right)\varphi \left(\frac{1}{2}\right)+2\left(1-2{r}_{0}\right)\varphi \left(\frac{1}{4}\right),\phantom{\rule{1em}{0ex}}v\in \left[\frac{1}{4},\phantom{\rule{2.77695pt}{0ex}}\frac{3}{4}\right]\end{array}\right\,$
(3.1)

where r0 = min {v, 1 − v}.

Proof. Since every positive semidefinite matrix is unitarily diagonalizable, it follows that there exist unitary matrices U, V M n such that A = U Λ1U* and B = V Λ2V*, where Λ1 = diag (λ1,..., λ n ), Λ2 = diag(µ1,..., µ n ) and λ i , µ i 0, i = 1,..., n. Let
$Y={U}^{*}XV=\left[{y}_{ij}\right].$
If $v\in \left[0,\frac{1}{4}\right]\cup \left[\frac{3}{4},1\right]$, then by (2.1) and the Cauchy-Schwarz inequality, we have
$\begin{array}{ll}\hfill ||{A}^{v}X{B}^{1-v}+{A}^{1-v}X{B}^{v}|{|}_{2}^{2}& =\sum _{i,j=1}^{n}{\left({\lambda }_{i}^{v}{\mu }_{j}^{1-v}+{\lambda }_{i}^{1-v}{\mu }_{j}^{v}\right)}^{2}|{y}_{ij}{|}^{2}\phantom{\rule{2em}{0ex}}\\ \le \sum _{i,j=1}^{n}{\left(\left(1-4{r}_{0}\right)\left({\lambda }_{i}+{\mu }_{j}\right)+4{r}_{0}\left({\lambda }_{i}^{1/4}{\mu }_{j}^{3/4}+{\lambda }_{i}^{3/4}{\mu }_{j}^{1/4}\right)\right)}^{2}|{y}_{ij}{|}^{2}\phantom{\rule{2em}{0ex}}\\ ={\left(1-4{r}_{0}\right)}^{2}\sum _{i,j=1}^{n}{\left({\lambda }_{i}+{\mu }_{j}\right)}^{2}|{y}_{ij}{|}^{2}\phantom{\rule{2em}{0ex}}\\ +16{r}_{0}^{2}\sum _{i,j=1}^{n}{\left({\lambda }_{i}^{1/4}{\mu }_{j}^{3/4}+{\lambda }_{i}^{3/4}{\mu }_{j}^{1/4}\right)}^{2}|{y}_{ij}{|}^{2}\phantom{\rule{2em}{0ex}}\\ +8{r}_{0}\left(1-4{r}_{0}\right)\sum _{i,j=1}^{n}\left({\lambda }_{i}+{\mu }_{j}\right)\left({\lambda }_{i}^{1/4}{\mu }_{j}^{3/4}+{\lambda }_{i}^{3/4}{\mu }_{j}^{1/4}\right)|{y}_{ij}{|}^{2}\phantom{\rule{2em}{0ex}}\\ \le \phantom{\rule{2.77695pt}{0ex}}{\left(1-4{r}_{0}\right)}^{2}{\varphi }^{2}\left(0\right)+16{r}_{0}^{2}{\varphi }^{2}\left(\frac{1}{4}\right)+8{r}_{0}\left(1-4{r}_{0}\right)\varphi \left(0\right)\varphi \left(\frac{1}{4}\right)\phantom{\rule{2em}{0ex}}\\ ={\left(\left(1-4{r}_{0}\right)\varphi \left(0\right)+4{r}_{0}\varphi \left(\frac{1}{4}\right)\right)}^{2}.\phantom{\rule{2em}{0ex}}\end{array}$

If $v\in \left[\frac{1}{4},\frac{3}{4}\right]$, the result follows from the inequality (2.1) and the same method above. This completes the proof. □

Remark. For the Hilbert-Schmidt norm, by the inequality (1.7), we have
$\varphi \left(v\right)\le 2{r}_{0}\varphi \left(\frac{1}{2}\right)+\left(1-2{r}_{0}\right)\varphi \left(0\right).$
So, for $v\in \left[0,\frac{1}{4}\right]\cup \left[\frac{3}{4},1\right]$, we have
$\begin{array}{ll}\hfill 2{r}_{0}\varphi \left(\frac{1}{2}\right)& +\left(1-2{r}_{0}\right)\varphi \left(0\right)-\left(1-4{r}_{0}\right)\varphi \left(0\right)-4{r}_{0}\varphi \left(\frac{1}{4}\right)\phantom{\rule{2em}{0ex}}\\ =2{r}_{0}\left(\varphi \left(\frac{1}{2}\right)+\varphi \left(0\right)-2\varphi \left(\frac{1}{4}\right)\right)\ge 0.\phantom{\rule{2em}{0ex}}\end{array}$
If $v\in \left[\frac{1}{4},\frac{3}{4}\right]$, then
$\begin{array}{ll}\hfill 2{r}_{0}\varphi \left(\frac{1}{2}\right)& +\left(1-2{r}_{0}\right)\varphi \left(0\right)-\left(4{r}_{0}-1\right)\varphi \left(\frac{1}{2}\right)-2\left(1-2{r}_{0}\right)\varphi \left(\frac{1}{4}\right)\phantom{\rule{2em}{0ex}}\\ =\left(1-2{r}_{0}\right)\left(\varphi \left(\frac{1}{2}\right)+\varphi \left(0\right)-2\varphi \left(\frac{1}{4}\right)\right)\ge 0.\phantom{\rule{2em}{0ex}}\end{array}$

So, the inequality (3.1) is a refinement of the inequality (1.7) for the Hilbert-Schmidt norm.

Declarations

Acknowledgements

The authors wish to express their heartfelt thanks to the referees and Professor Gnana Bhaskar Tenali for their detailed and helpful suggestions for revising the manuscript. This research was supported by the Scientific Research Project of Chongqing Three Gorges University (No. 11QN-21).

Authors’ Affiliations

(1)
School of Mathematics and Statistics, Chongqing Three Gorges University, Chongqing, 404100, People's Republic of China

References 