Open Access

Some generalizations of operator inequalities for positive linear maps

Journal of Inequalities and Applications20162016:27

https://doi.org/10.1186/s13660-016-0976-5

Received: 9 October 2015

Accepted: 15 January 2016

Published: 27 January 2016

Abstract

In this paper, we generalize some operator inequalities for positive linear maps due to Lin (Stud. Math. 215:187-194, 2013) and Zhang (Banach J. Math. Anal. 9:166-172, 2015).

Keywords

operator inequalities Kantorovich inequalities positive linear maps

MSC

47A63 47A30

1 Introduction

Throughout this paper, let M, \({M}'\), m, \({m}'\) be scalars, I be the identity operator, and \(\mathcal{B}(\mathcal{H})\) be the set of all bounded linear operators on a Hilbert space \((\mathcal{H},\langle\cdot,\cdot\rangle)\). The operator norm is denoted by \(\Vert \cdot \Vert \). We write \(A\ge0\) if the operator A is positive. If \(A-B\ge0\), then we say that \(A\ge B\). For \(A,B>0\), we use the following notation:
  • \(A\mathbin{\nabla}_{\mu}B=(1-\mu)A+\mu B\), \(A\mathbin{\sharp}_{\mu}B=A^{\frac{1}{2}}(A^{-\frac{1}{2}}BA^{-\frac{1}{2}})^{\mu}A^{\frac{1}{2}}\), where \(0\le\mu\le1\).

When \(\mu=\frac{1}{2}\) we write \(A\mathbin{\nabla} B\) and \(A\mathbin{\sharp} B\) for brevity for \(A\mathbin{\nabla}_{\frac{1}{2}} B\) and \(A\mathbin{\sharp}_{\frac{1}{2}} B\), respectively; see Kubo and Ando [3].

A linear map Φ is positive if \(\Phi(A)\ge0\) whenever \(A\ge0\). It is said to be unital if \(\Phi(I)=I\). We say that Φ is 2-positive if whenever the \(2\times2\) operator matrix \(\bigl [{\scriptsize\begin{matrix}{} A & B \cr {B^{\ast}} & C \end{matrix}} \bigr ]\) is positive, then so is \(\bigl [ {\scriptsize\begin{matrix}{} {\Phi(A)} & {\Phi(B)} \cr {\Phi(B^{\ast})} & {\Phi(C)} \end{matrix}}\bigr ]\).

Let \(0< m\le A\), \(B\le M\) and Φ be positive unital linear map. Lin [1], Theorem 2.1, proved the following reversed operator AM-GM inequalities:
$$\begin{aligned}& \Phi^{2}\biggl(\frac{A+B}{2}\biggr)\le K^{2}(h)\Phi^{2}(A\mathbin{\sharp} B), \end{aligned}$$
(1.1)
$$\begin{aligned}& \Phi^{2}\biggl(\frac{A+B}{2}\biggr)\le K^{2}(h) \bigl(\Phi(A) \mathbin{\sharp}\Phi(B)\bigr)^{2}, \end{aligned}$$
(1.2)
where \(K(h)=\frac{(h+1)^{2}}{4h}\) with \(h=\frac{M}{m}\) is the Kantorovich constant.

Can the inequalities (1.1) and (1.2) be improved? Lin [1], Conjecture 4.2, conjectured that the constant \(K(h)\) can be replaced by the Specht ratio \(S(h)=\frac{h^{\frac{1}{h-1}}}{e\log h^{\frac{1}{h-1}}}\) in (1.1) and (1.2), which remains as an open question.

Zhang [2], Theorem 2.6, generalized (1.1) and (1.2) when \(p\ge2\):
$$\begin{aligned}& \Phi^{2p}\biggl(\frac{A+B}{2}\biggr)\le \frac {(K(M^{2}+m^{2}))^{2p}}{16M^{2p}m^{2p}}\Phi ^{2p}(A \mathbin{\sharp} B), \end{aligned}$$
(1.3)
$$\begin{aligned}& \Phi^{2p}\biggl(\frac{A+B}{2}\biggr)\le \frac {(K(M^{2}+m^{2}))^{2p}}{16M^{2p}m^{2p}}\bigl(\Phi (A) \mathbin{\sharp} \Phi(B)\bigr)^{2p}. \end{aligned}$$
(1.4)

We will present some operator inequalities which are generalizations of (1.1), (1.2), (1.3), and (1.4) in the next section.

Bhatia and Davis [4] proved that if \(0< m\le A\le M\) and X and Y are two partial isometries on \(\mathcal{H}\) whose final spaces are orthogonal to each other. Then for every 2-positive unital linear map Φ,
$$ \Phi\bigl(X^{\ast}AY\bigr)\Phi\bigl(Y^{\ast}AY \bigr)^{-1}\Phi\bigl(Y^{\ast}AX\bigr)\le \biggl( \frac{M-m}{M+m}\biggr)^{2}\Phi\bigl(X^{\ast}AX\bigr). $$
(1.5)
Lin [5], Conjecture 3.4, conjectured that the following inequality could be true:
$$\bigl\Vert {\Phi\bigl(X^{\ast}AY\bigr)\Phi\bigl(Y^{\ast}AY \bigr)^{-1}\Phi\bigl(Y^{\ast}AX\bigr)\Phi \bigl(X^{\ast}AX\bigr)^{-1}} \bigr\Vert \le\biggl(\frac{M-m}{M+m} \biggr)^{2}. $$
Recently, Fu and He [6], Theorem 5, proved
$$ \bigl\Vert {\Phi\bigl(X^{\ast}AY\bigr)\Phi \bigl(Y^{\ast}AY\bigr)^{-1}\Phi\bigl(Y^{\ast}AX\bigr) \Phi \bigl(X^{\ast}AX\bigr)^{-1}} \bigr\Vert \le \frac{1}{4}\biggl(\biggl(\frac{M-m}{M+m}\biggr)^{2}M+ \frac{1}{m}\biggr)^{2}. $$
(1.6)

We will get a stronger result than (1.6).

2 Main results

We begin this section with the following lemmas.

Lemma 1

[7]

Let \(A,B>0\). Then the following norm inequality holds:
$$ \Vert {AB} \Vert \le\frac{1}{4}\Vert {A+B} \Vert ^{2}. $$
(2.1)

Lemma 2

[8]

Let \(A>0\). Then for every positive unital linear map Φ,
$$ \Phi\bigl(A^{-1}\bigr)\ge\Phi^{-1}(A). $$
(2.2)

Lemma 3

[9]

Let \(A,B>0\). Then, for \(1\le r<\infty\),
$$ \bigl\Vert {A^{r}+B^{r}} \bigr\Vert \le \bigl\Vert {(A+B)^{r}} \bigr\Vert . $$
(2.3)

Lemma 4

([10], Theorem 7)

Suppose that two operators A, B and positive real numbers m, \({m}'\), M, \({M}'\) satisfy either of the following conditions:
  1. (1)

    \(0< m\le A\le{m}'<{M}'\le B\le M\),

     
  2. (2)

    \(0< m\le B\le{m}'<{M}'\le A\le M\).

     
Then
$$A\mathbin{\nabla}_{\mu}B\ge K^{r}\bigl({h}'\bigr)A \mathbin{\sharp}_{\mu}B $$
for all \(\mu\in[0,1]\), where \(r=\min(\mu,1-\mu)\) and \({h}'=\frac{{M}'}{{m}'}\).

Theorem 1

Let \(0< m\le A\le{m}'<{M}'\le B\le M\). Then
$$ \frac{A+B}{2}+MmK^{\frac{1}{2}}\bigl({h}'\bigr) (A\mathbin{\sharp} B)^{-1}\le M+m, $$
(2.4)
where \(K({h}')=\frac{({h}'+1)^{2}}{4{h}'}\) with \({h}'=\frac {{M}'}{{m}'}\).

Proof

It is easy to see that
$$\frac{1}{2}(M-A) (m-A)A^{-1}\le0, $$
then
$$Mm\frac{A^{-1}}{2}+\frac{A}{2}\le\frac{M+m}{2}. $$
Similarly,
$$Mm\frac{B^{-1}}{2}+\frac{B}{2}\le\frac{M+m}{2}. $$
Summing up the above two inequalities, we get
$$\frac{A+B}{2}+Mm\frac{A^{-1}+B^{-1}}{2}\le M+m. $$
By \((A\mathbin{\sharp} B)^{-1}=A^{-1}\mathbin{\sharp} B^{-1}\) and Lemma 4, we have
$$\begin{aligned} \frac{A+B}{2}+MmK^{\frac{1}{2}}\bigl({h}'\bigr) (A\mathbin{\sharp} B)^{-1} =& \frac{A+B}{2}+MmK^{\frac {1}{2}}\bigl({h}' \bigr) \bigl(A^{-1}\mathbin{\sharp} B^{-1}\bigr) \\ \le& \frac{A+B}{2}+Mm\frac{A^{-1}+B^{-1}}{2} \\ \le& M+m. \end{aligned}$$
This completes the proof. □

Theorem 2

Let \(0< m\le A\le{m}'<{M}'\le B\le M\). Then for every positive unital linear map Φ,
$$ \Phi^{2}\biggl(\frac{A+B}{2}\biggr)\le \frac{K^{2}(h)}{K({h}')}\Phi^{2}(A\mathbin{\sharp} B) $$
(2.5)
and
$$ \Phi^{2}\biggl(\frac{A+B}{2}\biggr)\le \frac{K^{2}(h)}{K({h}')}\bigl(\Phi(A)\mathbin{\sharp}\Phi(B)\bigr)^{2}, $$
(2.6)
where \(K(h)=\frac{(h+1)^{2}}{4h}\), \(K({h}')=\frac{({h}'+1)^{2}}{4{h}'}\), \(h=\frac{M}{m}\), and \({h}'=\frac{{M}'}{{m}'}\).

Proof

The inequality (2.5) is equivalent to
$$ \biggl\Vert {\Phi\biggl(\frac{A+B}{2}\biggr) \Phi^{-1}(A\mathbin{\sharp} B)} \biggr\Vert \le \frac{K(h)}{K^{\frac{1}{2}}({h}')}. $$
(2.7)
Compute
$$\begin{aligned}& \biggl\Vert {\Phi\biggl(\frac{A+B}{2}\biggr)MmK^{\frac{1}{2}} \bigl({h}'\bigr)\Phi^{-1}(A\mathbin{\sharp} B)} \biggr\Vert \\& \quad \le\frac{1}{4}\biggl\Vert {\Phi \biggl(\frac{A+B}{2} \biggr)+MmK^{\frac{1}{2}}\bigl({h}'\bigr)\Phi^{-1}(A\mathbin{\sharp} B)} \biggr\Vert ^{2} \quad \mbox{(by (2.1))} \\& \quad \le\frac{1}{4}\biggl\Vert {\Phi\biggl(\frac{A+B}{2} \biggr)+MmK^{\frac {1}{2}}\bigl({h}'\bigr)\Phi \bigl((A\mathbin{\sharp} B)^{-1}\bigr)} \biggr\Vert ^{2} \quad \mbox{(by (2.2))} \\& \quad =\frac{1}{4}\biggl\Vert {\Phi\biggl(\frac{A+B}{2}+MmK^{\frac {1}{2}} \bigl({h}'\bigr) (A\mathbin{\sharp} B)^{-1}\biggr)} \biggr\Vert ^{2} \\& \quad \le\frac{1}{4}\bigl\Vert {\Phi(M+m)} \bigr\Vert ^{2} \quad \mbox{(by (2.4))} \\& \quad =\frac{1}{4}(M+m)^{2}. \end{aligned}$$
That is,
$$\biggl\Vert {\Phi\biggl(\frac{A+B}{2}\biggr)\Phi^{-1}(A\mathbin{\sharp} B)} \biggr\Vert \le \frac{(M+m)^{2}}{4MmK^{\frac{1}{2}}({h}')}=\frac{K(h)}{K^{\frac{1}{2}}({h}')}. $$
Thus, (2.7) holds. The proof of (2.6) is similar, we omit the details.

This completes the proof. □

Remark 1

Because of \(\frac{K^{2}(h)}{K({h}')}< K^{2}(h)\), inequalities (2.5) and (2.6) are refinements of (1.1) and (1.2), respectively.

Theorem 3

Let \(0< m\le A\le{m}'<{M}'\le B\le M\) and \(2\le p<\infty \). Then for every positive unital linear map Φ,
$$ \Phi^{2p}\biggl(\frac{A+B}{2}\biggr)\le \frac{1}{16}\biggl(\frac{K^{2}(h)(M^{2}+m^{2})^{2}}{K({h}')M^{2}m^{2}}\biggr)^{p}\Phi ^{2p}(A\mathbin{\sharp} B) $$
(2.8)
and
$$ \Phi^{2p}\biggl(\frac{A+B}{2}\biggr)\le \frac{1}{16}\biggl(\frac{K^{2}(h)(M^{2}+m^{2})^{2}}{K({h}')M^{2}m^{2}}\biggr)^{p}\bigl(\Phi(A) \mathbin{\sharp} \Phi (B)\bigr)^{2p}, $$
(2.9)
where \(K(h)=\frac{(h+1)^{2}}{4h}\), \(K({h}')=\frac{({h}'+1)^{2}}{4{h}'}\), \(h=\frac{M}{m}\), and \({h}'=\frac{{M}'}{{m}'}\).

Proof

By the operator reverse monotonicity of inequality (2.5), we have
$$ \Phi^{-2}(A\mathbin{\sharp} B)\le L^{2}\Phi ^{-2}\biggl(\frac{A+B}{2}\biggr), $$
(2.10)
where \(L=\frac{K(h)}{K^{\frac{1}{2}}({h}')}\).
Compute
$$\begin{aligned}& \biggl\Vert {\Phi^{p}\biggl(\frac{A+B}{2}\biggr)M^{p}m^{p} \Phi^{-p}(A\mathbin{\sharp} B)} \biggr\Vert \\& \quad \le\frac{1}{4}\biggl\Vert {L^{\frac{p}{2}}\Phi ^{p} \biggl(\frac{A+B}{2}\biggr)+\biggl(\frac{M^{2}m^{2}}{L}\biggr)^{\frac{p}{2}} \Phi^{-p}(A\mathbin{\sharp} B)} \biggr\Vert ^{2}\quad \mbox{(by (2.1))} \\& \quad \le\frac{1}{4}\biggl\Vert {L\Phi^{2}\biggl( \frac{A+B}{2}\biggr)+\frac {M^{2}m^{2}}{L}\Phi ^{-2}(A\mathbin{\sharp} B)} \biggr\Vert ^{p}\quad \mbox{(by (2.3))} \\& \quad \le\frac{1}{4}\biggl\Vert {L\Phi^{2}\biggl( \frac{A+B}{2}\biggr)+LM^{2}m^{2}\Phi ^{-2} \biggl(\frac{A+B}{2}\biggr)} \biggr\Vert ^{p} \quad \mbox{(by (2.10))} \\& \quad \le\frac{1}{4}\bigl(L\bigl(M^{2}+m^{2}\bigr) \bigr)^{p}\quad \mbox{(by [1], (4.7))}. \end{aligned}$$
That is,
$$\biggl\Vert {\Phi^{p}\biggl(\frac{A+B}{2}\biggr) \Phi^{-p}(A\mathbin{\sharp} B)} \biggr\Vert \le \frac{1}{4}\biggl( \frac{L(M^{2}+m^{2})}{Mm}\biggr)^{p}=\frac{1}{4}\biggl(\frac {K^{2}(h)(M^{2}+m^{2})^{2}}{K({h}')M^{2}m^{2}} \biggr)^{\frac{p}{2}}. $$
Thus, (2.8) holds. By inequality (2.6), the proof of (2.9) is similar, we omit the details.

This completes the proof. □

Remark 2

Since \(K({h}')>1\), inequalities (2.8) and (2.9) are sharper than (1.3) and (1.4), respectively.

Theorem 4

Let \(0< m\le A\le M\) and let X, Y be two isometries on \(\mathcal{H}\) whose final spaces are orthogonal to each other. Then for every 2-positive unital linear map Φ,
$$ \bigl\Vert {\Phi\bigl(X^{\ast}AY\bigr)\Phi \bigl(Y^{\ast}AY\bigr)^{-1}\Phi\bigl(Y^{\ast}AX\bigr) \Phi \bigl(X^{\ast}AX\bigr)^{-1}} \bigr\Vert \le \frac{(M-m)^{2}}{4Mm}. $$
(2.11)

Proof

Since X is isometric and \(0< m\le A\le M\), \(m\le\Phi(X^{\ast}AX)\le M\) and \(\frac{1}{M}\le \Phi(X^{\ast}AX)^{-1}\le\frac{1}{m}\).

Compute
$$\begin{aligned}& \biggl(\frac{M-m}{M+m}\biggr)^{2}Mm\bigl\Vert {\Phi \bigl(X^{\ast}AY\bigr)\Phi\bigl(Y^{\ast}AY\bigr)^{-1} \Phi \bigl(Y^{\ast}AX\bigr)\Phi\bigl(X^{\ast}AX \bigr)^{-1}} \bigr\Vert \\& \quad \le\frac{1}{4}\biggl\Vert {\Phi\bigl(X^{\ast}AY\bigr)\Phi \bigl(Y^{\ast}AY\bigr)^{-1}\Phi\bigl(Y^{\ast}AX\bigr)+ \biggl(\frac{M-m}{M+m}\biggr)^{2}Mm\Phi\bigl(X^{\ast}AX \bigr)^{-1}} \biggr\Vert ^{2} \quad \mbox{(by (2.1))} \\& \quad \le\frac{1}{4}\biggl\Vert {\biggl(\frac{M-m}{M+m} \biggr)^{2}\Phi\bigl(X^{\ast}AX\bigr)+\biggl(\frac{M-m}{M+m} \biggr)^{2}Mm\Phi\bigl(X^{\ast}AX\bigr)^{-1}} \biggr\Vert ^{2} \quad \mbox{(by (1.5))} \\& \quad \le\frac{1}{4}\biggl(\frac{M-m}{M+m}\biggr)^{4}(M+m)^{2}. \end{aligned}$$
Hence,
$$\bigl\Vert {\Phi\bigl(X^{\ast}AY\bigr)\Phi\bigl(Y^{\ast}AY \bigr)^{-1}\Phi\bigl(Y^{\ast}AX\bigr)\Phi \bigl(X^{\ast}AX\bigr)^{-1}} \bigr\Vert \le\frac{(M-m)^{2}}{4Mm}. $$
This completes the proof. □

Remark 3

Since \(0< m\le M\),
$$\frac{1}{4}\biggl(\biggl(\frac{M-m}{M+m}\biggr)^{2}M+ \frac{1}{m}\biggr)^{2}\ge \biggl(\frac{M-m}{M+m} \biggr)^{2}\frac{M}{m}\ge\biggl(\frac{M-m}{M+m} \biggr)^{2}\frac {(M+m)^{2}}{4Mm}=\frac{(M-m)^{2}}{4Mm}. $$
Thus, (2.11) is tighter than (1.6).

Declarations

Acknowledgements

The authors wish to express their heartfelt thanks to the referees for their detailed and helpful suggestions for revising the manuscript. This research was supported by Scientific Research Fund of Yunnan Provincial Education Department (No. 2014C206Y).

Open Access This article is distributed under the terms of the Creative Commons Attribution 4.0 International License (http://creativecommons.org/licenses/by/4.0/), which permits unrestricted use, distribution, and reproduction in any medium, provided you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons license, and indicate if changes were made.

Authors’ Affiliations

(1)
Oxbridge College, Kunming University of Science and Technology
(2)
Faculty of Science, Kunming University of Science and Technology

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© Xue and Hu 2016