Some generalizations of operator inequalities for positive linear maps

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).

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

• (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).