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: 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].
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\(A\mathbin{\nabla}_{\mu}B=(1-\mu)A+\mu B\), \(A\mathbin{\sharp}_{\mu}B=A^{\frac{1}{2}}(A^{-\frac{1}{2}}BA^{-\frac{1}{2}})^{\mu}A^{\frac{1}{2}}\), where \(0\le\mu\le1\).
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 \(\bigl [ {\scriptsize\begin{matrix}{} {\Phi(A)} & {\Phi(B)} \cr {\Phi(B^{\ast})} & {\Phi(C)} \end{matrix}}\bigr ]\).
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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:
where \(K(h)=\frac{(h+1)^{2}}{4h}\) with \(h=\frac{M}{m}\) is the Kantorovich constant.
$$\begin{aligned}& \Phi^{2}\biggl(\frac{A+B}{2}\biggr)\le K^{2}(h)\Phi^{2}(A\mathbin{\sharp} B), \end{aligned}$$
(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)
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.
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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)
\(0< m\le A\le{m}'<{M}'\le B\le M\),
(2)
\(0< m\le B\le{m}'<{M}'\le A\le M\).
Then
for all
\(\mu\in[0,1]\), where
\(r=\min(\mu,1-\mu)\)
and
\({h}'=\frac{{M}'}{{m}'}\).
$$A\mathbin{\nabla}_{\mu}B\ge K^{r}\bigl({h}'\bigr)A \mathbin{\sharp}_{\mu}B $$
Theorem 1
Let
\(0< m\le A\le{m}'<{M}'\le B\le M\). Then
where
\(K({h}')=\frac{({h}'+1)^{2}}{4{h}'}\)
with
\({h}'=\frac {{M}'}{{m}'}\).
$$ \frac{A+B}{2}+MmK^{\frac{1}{2}}\bigl({h}'\bigr) (A\mathbin{\sharp} B)^{-1}\le M+m, $$
(2.4)
Proof
It is easy to see that
then
Similarly,
Summing up the above two inequalities, we get
By \((A\mathbin{\sharp} B)^{-1}=A^{-1}\mathbin{\sharp} B^{-1}\) and Lemma 4, we have
This completes the proof. □
$$\frac{1}{2}(M-A) (m-A)A^{-1}\le0, $$
$$Mm\frac{A^{-1}}{2}+\frac{A}{2}\le\frac{M+m}{2}. $$
$$Mm\frac{B^{-1}}{2}+\frac{B}{2}\le\frac{M+m}{2}. $$
$$\frac{A+B}{2}+Mm\frac{A^{-1}+B^{-1}}{2}\le M+m. $$
$$\begin{aligned} \frac{A+B}{2}+MmK^{\frac{1}{2}}\bigl({h}'\bigr) (A\mathbin{\sharp} B)^{-1} =& \frac{A+B}{2}+MmK^{\frac {1}{2}}\bigl({h}' \bigr) \bigl(A^{-1}\mathbin{\sharp} B^{-1}\bigr) \\ \le& \frac{A+B}{2}+Mm\frac{A^{-1}+B^{-1}}{2} \\ \le& M+m. \end{aligned}$$
Theorem 2
Let
\(0< m\le A\le{m}'<{M}'\le B\le M\). Then for every positive unital linear map Φ,
and
where
\(K(h)=\frac{(h+1)^{2}}{4h}\), \(K({h}')=\frac{({h}'+1)^{2}}{4{h}'}\), \(h=\frac{M}{m}\), and
\({h}'=\frac{{M}'}{{m}'}\).
$$ \Phi^{2}\biggl(\frac{A+B}{2}\biggr)\le \frac{K^{2}(h)}{K({h}')}\Phi^{2}(A\mathbin{\sharp} B) $$
(2.5)
$$ \Phi^{2}\biggl(\frac{A+B}{2}\biggr)\le \frac{K^{2}(h)}{K({h}')}\bigl(\Phi(A)\mathbin{\sharp}\Phi(B)\bigr)^{2}, $$
(2.6)
Proof
The inequality (2.5) is equivalent to
Compute
That is,
Thus, (2.7) holds. The proof of (2.6) is similar, we omit the details.
$$ \biggl\Vert {\Phi\biggl(\frac{A+B}{2}\biggr) \Phi^{-1}(A\mathbin{\sharp} B)} \biggr\Vert \le \frac{K(h)}{K^{\frac{1}{2}}({h}')}. $$
(2.7)
$$\begin{aligned}& \biggl\Vert {\Phi\biggl(\frac{A+B}{2}\biggr)MmK^{\frac{1}{2}} \bigl({h}'\bigr)\Phi^{-1}(A\mathbin{\sharp} B)} \biggr\Vert \\& \quad \le\frac{1}{4}\biggl\Vert {\Phi \biggl(\frac{A+B}{2} \biggr)+MmK^{\frac{1}{2}}\bigl({h}'\bigr)\Phi^{-1}(A\mathbin{\sharp} B)} \biggr\Vert ^{2} \quad \mbox{(by (2.1))} \\& \quad \le\frac{1}{4}\biggl\Vert {\Phi\biggl(\frac{A+B}{2} \biggr)+MmK^{\frac {1}{2}}\bigl({h}'\bigr)\Phi \bigl((A\mathbin{\sharp} B)^{-1}\bigr)} \biggr\Vert ^{2} \quad \mbox{(by (2.2))} \\& \quad =\frac{1}{4}\biggl\Vert {\Phi\biggl(\frac{A+B}{2}+MmK^{\frac {1}{2}} \bigl({h}'\bigr) (A\mathbin{\sharp} B)^{-1}\biggr)} \biggr\Vert ^{2} \\& \quad \le\frac{1}{4}\bigl\Vert {\Phi(M+m)} \bigr\Vert ^{2} \quad \mbox{(by (2.4))} \\& \quad =\frac{1}{4}(M+m)^{2}. \end{aligned}$$
$$\biggl\Vert {\Phi\biggl(\frac{A+B}{2}\biggr)\Phi^{-1}(A\mathbin{\sharp} B)} \biggr\Vert \le \frac{(M+m)^{2}}{4MmK^{\frac{1}{2}}({h}')}=\frac{K(h)}{K^{\frac{1}{2}}({h}')}. $$
This completes the proof. □
Remark 1
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,
and
where
\(K(h)=\frac{(h+1)^{2}}{4h}\), \(K({h}')=\frac{({h}'+1)^{2}}{4{h}'}\), \(h=\frac{M}{m}\), and
\({h}'=\frac{{M}'}{{m}'}\).
$$ \Phi^{2p}\biggl(\frac{A+B}{2}\biggr)\le \frac{1}{16}\biggl(\frac{K^{2}(h)(M^{2}+m^{2})^{2}}{K({h}')M^{2}m^{2}}\biggr)^{p}\Phi ^{2p}(A\mathbin{\sharp} B) $$
(2.8)
$$ \Phi^{2p}\biggl(\frac{A+B}{2}\biggr)\le \frac{1}{16}\biggl(\frac{K^{2}(h)(M^{2}+m^{2})^{2}}{K({h}')M^{2}m^{2}}\biggr)^{p}\bigl(\Phi(A) \mathbin{\sharp} \Phi (B)\bigr)^{2p}, $$
(2.9)
Proof
By the operator reverse monotonicity of inequality (2.5), we have
where \(L=\frac{K(h)}{K^{\frac{1}{2}}({h}')}\).
$$ \Phi^{-2}(A\mathbin{\sharp} B)\le L^{2}\Phi ^{-2}\biggl(\frac{A+B}{2}\biggr), $$
(2.10)
Compute
That is,
Thus, (2.8) holds. By inequality (2.6), the proof of (2.9) is similar, we omit the details.
$$\begin{aligned}& \biggl\Vert {\Phi^{p}\biggl(\frac{A+B}{2}\biggr)M^{p}m^{p} \Phi^{-p}(A\mathbin{\sharp} B)} \biggr\Vert \\& \quad \le\frac{1}{4}\biggl\Vert {L^{\frac{p}{2}}\Phi ^{p} \biggl(\frac{A+B}{2}\biggr)+\biggl(\frac{M^{2}m^{2}}{L}\biggr)^{\frac{p}{2}} \Phi^{-p}(A\mathbin{\sharp} B)} \biggr\Vert ^{2}\quad \mbox{(by (2.1))} \\& \quad \le\frac{1}{4}\biggl\Vert {L\Phi^{2}\biggl( \frac{A+B}{2}\biggr)+\frac {M^{2}m^{2}}{L}\Phi ^{-2}(A\mathbin{\sharp} B)} \biggr\Vert ^{p}\quad \mbox{(by (2.3))} \\& \quad \le\frac{1}{4}\biggl\Vert {L\Phi^{2}\biggl( \frac{A+B}{2}\biggr)+LM^{2}m^{2}\Phi ^{-2} \biggl(\frac{A+B}{2}\biggr)} \biggr\Vert ^{p} \quad \mbox{(by (2.10))} \\& \quad \le\frac{1}{4}\bigl(L\bigl(M^{2}+m^{2}\bigr) \bigr)^{p}\quad \mbox{(by [1], (4.7))}. \end{aligned}$$
$$\biggl\Vert {\Phi^{p}\biggl(\frac{A+B}{2}\biggr) \Phi^{-p}(A\mathbin{\sharp} B)} \biggr\Vert \le \frac{1}{4}\biggl( \frac{L(M^{2}+m^{2})}{Mm}\biggr)^{p}=\frac{1}{4}\biggl(\frac {K^{2}(h)(M^{2}+m^{2})^{2}}{K({h}')M^{2}m^{2}} \biggr)^{\frac{p}{2}}. $$
This completes the proof. □
Remark 2
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
Hence,
This completes the proof. □
$$\begin{aligned}& \biggl(\frac{M-m}{M+m}\biggr)^{2}Mm\bigl\Vert {\Phi \bigl(X^{\ast}AY\bigr)\Phi\bigl(Y^{\ast}AY\bigr)^{-1} \Phi \bigl(Y^{\ast}AX\bigr)\Phi\bigl(X^{\ast}AX \bigr)^{-1}} \bigr\Vert \\& \quad \le\frac{1}{4}\biggl\Vert {\Phi\bigl(X^{\ast}AY\bigr)\Phi \bigl(Y^{\ast}AY\bigr)^{-1}\Phi\bigl(Y^{\ast}AX\bigr)+ \biggl(\frac{M-m}{M+m}\biggr)^{2}Mm\Phi\bigl(X^{\ast}AX \bigr)^{-1}} \biggr\Vert ^{2} \quad \mbox{(by (2.1))} \\& \quad \le\frac{1}{4}\biggl\Vert {\biggl(\frac{M-m}{M+m} \biggr)^{2}\Phi\bigl(X^{\ast}AX\bigr)+\biggl(\frac{M-m}{M+m} \biggr)^{2}Mm\Phi\bigl(X^{\ast}AX\bigr)^{-1}} \biggr\Vert ^{2} \quad \mbox{(by (1.5))} \\& \quad \le\frac{1}{4}\biggl(\frac{M-m}{M+m}\biggr)^{4}(M+m)^{2}. \end{aligned}$$
$$\bigl\Vert {\Phi\bigl(X^{\ast}AY\bigr)\Phi\bigl(Y^{\ast}AY \bigr)^{-1}\Phi\bigl(Y^{\ast}AX\bigr)\Phi \bigl(X^{\ast}AX\bigr)^{-1}} \bigr\Vert \le\frac{(M-m)^{2}}{4Mm}. $$
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.
Competing interests
The authors declare that they have no competing interests.
Authors’ contributions
All authors contributed equally to the writing of this paper. All authors read and approved the final manuscript.