New Refinement of the Operator Kantorovich Inequality

Abstract

We focus on the improvement of operator Kantorovich type inequalities. Among the consequences, we improve the main result of the paper [H.R. Moradi, I.H. Gümüş, Z. Heydarbeygi, A glimpse at the operator Kantorovich inequality, Linear Multilinear Algebra, doi:10.1080/03081087.2018.1441799].

1. Notation and Preliminaries

At the beginning of this paper, we cite the following inequality which is called the operator Kantorovich inequality [1]:

$$\mathrm{\Phi }\left(A^{-1}\right)\le \frac{{\left(M+m\right)}^2}{4Mm}\mathrm{\Phi }{\left(A\right)}^{-1}$$

(1)

where $\mathrm{\Phi }$ is a normalized positive linear map from $\mathcal{B}\left(\mathcal{H}\right)$ to $\mathcal{B}\left(\mathcal{K}\right)$, (we represent $\mathcal{H}$ and $\mathcal{K}$ as complex Hilbert spaces throughout the paper) and A is a positive operator with spectrum contained in $\left[m,M\right]$ with $0<m<M$. This is a non-commutative analogue of the classical inequality [2],

$$⟨Ax,x⟩⟨A^{-1}x,x⟩\le \frac{{\left(M+m\right)}^2}{4Mm}$$

where $x\in \mathcal{H}$ is a unit vector.

In recent years, various attempts have been made by many authors to improve and generalize the operator Kantorovich inequality. One may see the basic references [3,4,5] and the excellent survey [6] on this topic. In [7], it was shown that

$$\mathrm{\Phi }\left(A^{-1}\right)\le \mathrm{\Phi }\left(m^{\frac{A-MI}{M-m}}{M}^{\frac{mI-A}{M-m}}\right)\le \frac{{\left(M+m\right)}^2}{4Mm}\mathrm{\Phi }{\left(A\right)}^{-1}.$$

(2)

The main aim of the present short paper is to improve both inequalities in (2). Actually, we prove that

$$\begin{array}{cc}\hfill \mathrm{\Phi }\left(A^{-1}\right)& \le \mathrm{\Phi }\left({\left(A-{\left(\sqrt{m}-\sqrt{M}\right)}^{2}r\left(A\right)\right)}^{-1}\right)\hfill \\ & \le \mathrm{\Phi }\left({\left(m^{\frac{A-MI}{M-m}}{M}^{\frac{mI-A}{M-m}}\right)}^{-1}\right)\hfill \\ & \le \frac{{\left(M+m\right)}^2}{4Mm}\mathrm{\Phi }{\left(A\right)}^{-1}-\left(\frac{{\left(\sqrt{M}-\sqrt{m}\right)}^2}{Mm}\right)r\left(A\right)\hfill \end{array}$$

where $r\left(A\right)=\min \left\{\frac{MI-A}{M-m},\frac{A-mI}{M-m}\right\}=\frac{1}{2}I-\frac{1}{M-m}\left|A-\frac{M+m}{2}I\right|$.

In what follows, an operator means a bounded linear one acting on a complex Hilbert space $\mathcal{H}$. As customary, we reserve m, M for scalars and I for the identity operator. A self-adjoint operator A is said to be positive if $⟨Ax,x⟩\ge 0$ holds for all $x\in \mathcal{H}$. A linear map $\mathrm{\Phi }$ is positive if $\mathrm{\Phi }\left(A\right)\ge 0$ whenever $A\ge 0$. It is said to be normalized if $\mathrm{\Phi }\left(I\right)=I$. We denote by $\sigma \left(A\right)$ the spectrum of the operator A.

2. Main Results

Before we present the proof of our theorems, we begin with a general observation. We say that a non-negative function f on $\left[0,\infty \right)$ is geometrically convex [8] when

$$f\left(a^{1-v}{b}^v\right)\le f{\left(a\right)}^{1-v}f{\left(b\right)}^v$$

(3)

for all $a,b>0$ and $v\in \left[0,1\right]$. Equivalently, a function f is geometrically convex if and only if the associated function $F\left(y\right)=\log \left(f\left(e^y\right)\right)$ is convex.

Example 1

([9] Example 2.12). Given real numbers $c_i\ge 0$ and $p_i\in \left(-\infty ,0\right]\cup \left[1,\infty \right)$ for $i=1,\cdots ,n$, the function $f\left(t\right)={\sum }_{i=1}^n{c}_i{t}^{p_i}$ is geometrically convex on $\left(0,\infty \right)$.

Kittaneh and Manasrah [10] Theorem 2.1 obtained a refinement of the weighted arithmetic-geometric mean inequality as follows:

$$a^{1-v}{b}^v\le \left(1-v\right)a+vb-r{\left(\sqrt{a}-\sqrt{b}\right)}^2$$

(4)

where $r=\min \left\{v,1-v\right\}$.

Now, if f is a decreasing geometrically convex function, then

$$\begin{array}{cc}\hfill f\left(\left(1-v\right)a+vb\right)& \le f\left(\left(\left(1-v\right)a+vb\right)-r{\left(\sqrt{a}-\sqrt{b}\right)}^2\right)\hfill \\ & \le f\left(a^{1-v}{b}^v\right)\hfill \\ & \le f{\left(a\right)}^{1-v}f{\left(b\right)}^v\hfill \\ & \le \left(1-v\right)f\left(a\right)+vf\left(b\right)-r{\left(\sqrt{f\left(a\right)}-\sqrt{f\left(b\right)}\right)}^2\hfill \\ & \le \left(1-v\right)f\left(a\right)+vf\left(b\right)\hfill \end{array}$$

(5)

where the first inequality follows from the inequality $\left(1-v\right)a+vb-r{\left(\sqrt{a}-\sqrt{b}\right)}^2\le \left(1-v\right)a+vb$ and the fact that f is decreasing function, in the second inequality we used (4), the third inequality is obvious by (3), and the fourth inequality again follows from (4) by interchanging a by $f\left(a\right)$ and b by $f\left(b\right)$.

Of course, each decreasing geometrically convex function is also convex. However, the converse does not hold in general.

The inequality (5) applied to $a=m$, $b=M$, $1-v=\frac{M-t}{M-m}$, and $v=\frac{t-m}{M-m}$ gives

$$\begin{array}{cc}\hfill f\left(t\right)& \le f\left(t-{\left(\sqrt{m}-\sqrt{M}\right)}^{2}r\left(t\right)\right)\hfill \\ & \le f\left(m^{\frac{M-t}{M-m}}{M}^{\frac{t-m}{M-m}}\right)\hfill \\ & \le f{\left(m\right)}^{\frac{M-t}{M-m}}f{\left(M\right)}^{\frac{t-m}{M-m}}\hfill \\ & \le \frac{M-t}{M-m}f\left(m\right)+\frac{t-m}{M-m}f\left(M\right)-{\left(\sqrt{f\left(m\right)}-\sqrt{f\left(M\right)}\right)}^{2}r\left(t\right)\hfill \\ & \le \frac{M-t}{M-m}f\left(m\right)+\frac{t-m}{M-m}f\left(M\right)\hfill \end{array}$$

(6)

with $r\left(t\right)=\min \left\{\frac{t-m}{M-m},\frac{M-t}{M-m}\right\}=\frac{1}{2}-\frac{1}{M-m}\left|t-\frac{M+m}{2}\right|$ whenever $t\in \left[m,M\right]$.

In order to establish our promised refinement of the operator Kantorovich inequality, we also use the well-known monotonicity principle for bounded self-adjoint operators on Hilbert space (see, e.g., [6] (p. 3)): If $A\in \mathcal{B}\left(\mathcal{H}\right)$ is a self-adjoint operator, then

$$f\left(t\right)\le g\left(t\right),t\in \sigma \left(A\right)\Rightarrow f\left(A\right)\le g\left(A\right)$$

(7)

provided that f and g are real-valued continuous functions. Under the same assumptions, $h\left(t\right)=\left|t\right|$ implies $h\left(A\right)=\left|A\right|$.

Now, we are in a position to state and prove our main results. We remark that the following theorem can be regarded as an extension of [5] Remark 4.14 to the context of geometrical convex functions.

Theorem 1.

Let $A\in \mathcal{B}\left(\mathcal{H}\right)$ be a self-adjoint operator with $\sigma \left(A\right)\subseteq \left[m,M\right]$ for some scalars m, M with $0<m<M$ and Φ be a normalized positive linear map from $\mathcal{B}\left(\mathcal{H}\right)$ to $\mathcal{B}\left(\mathcal{K}\right)$. If f is strictly positive decreasing geometrically convex function, then

$$\begin{array}{cc}\hfill \mathrm{\Phi }\left(f\left(A-{\left(\sqrt{m}-\sqrt{M}\right)}^{2}r\left(A\right)\right)\right)& \le \mathrm{\Phi }\left(f\left(m^{\frac{MI-A}{M-m}}{M}^{\frac{A-mI}{M-m}}\right)\right)\hfill \\ & \le \mu \left(m,M,f\right)f\left(\mathrm{\Phi }\left(A\right)\right)-{\left(\sqrt{f\left(m\right)}-\sqrt{f\left(M\right)}\right)}^2\mathrm{\Phi }\left(r\left(A\right)\right)\hfill \end{array}$$

where $r\left(A\right)=\min \left\{\frac{A-mI}{M-m},\frac{MI-A}{M-m}\right\}=\frac{1}{2}I-\frac{1}{M-m}\left|A-\frac{M+m}{2}I\right|$ and

$$\mu \left(m,M,f\right)=\max \left\{\frac{1}{f\left(t\right)}\left(\frac{M-t}{M-m}f\left(m\right)+\frac{t-m}{M-m}f\left(M\right)\right):t\in \left[m,M\right]\right\}.$$

Proof. 

On account of the assumptions, from parts of (6), we have

$$\begin{array}{cc}\hfill f\left(t-{\left(\sqrt{m}-\sqrt{M}\right)}^{2}r\left(t\right)\right)& \le f\left(m^{\frac{M-t}{M-m}}{M}^{\frac{t-m}{M-m}}\right)\hfill \\ & \le L\left(t\right)-{\left(\sqrt{f\left(m\right)}-\sqrt{f\left(M\right)}\right)}^{2}r\left(t\right)\hfill \end{array}$$

(8)

where

$$L\left(t\right)=\frac{M-t}{M-m}f\left(m\right)+\frac{t-m}{M-m}f\left(M\right).$$

Note that inequality (8) holds for all $t\in \left[m,M\right]$. On the other hand, $\sigma \left(A\right)\subseteq \left[m,M\right]$, which, by virtue of monotonicity principle (7) for operator functions, yields the series of inequalities

$$\begin{array}{cc}\hfill f\left(A-{\left(\sqrt{m}-\sqrt{M}\right)}^{2}r\left(A\right)\right)& \le f\left(m^{\frac{MI-A}{M-m}}{M}^{\frac{A-mI}{M-m}}\right)\hfill \\ & \le L\left(A\right)-{\left(\sqrt{f\left(m\right)}-\sqrt{f\left(M\right)}\right)}^{2}r\left(A\right).\hfill \end{array}$$

It follows from the linearity and the positivity of the map $\mathrm{\Phi }$ that

$$\begin{array}{cc}\hfill \mathrm{\Phi }\left(f\left(A-{\left(\sqrt{m}-\sqrt{M}\right)}^{2}r\left(A\right)\right)\right)& \le \mathrm{\Phi }\left(f\left(m^{\frac{MI-A}{M-m}}{M}^{\frac{A-mI}{M-m}}\right)\right)\hfill \\ & \le \mathrm{\Phi }\left(L\left(A\right)\right)-{\left(\sqrt{f\left(m\right)}-\sqrt{f\left(M\right)}\right)}^2\mathrm{\Phi }\left(r\left(A\right)\right).\hfill \end{array}$$

Now, by using [5] Corollary 4.12 we get

$$\begin{array}{cc}\hfill \mathrm{\Phi }\left(f\left(A-{\left(\sqrt{m}-\sqrt{M}\right)}^{2}r\left(A\right)\right)\right)& \le \mathrm{\Phi }\left(f\left(m^{\frac{MI-A}{M-m}}{M}^{\frac{A-mI}{M-m}}\right)\right)\hfill \\ & \le \mathrm{\Phi }\left(L\left(A\right)\right)-{\left(\sqrt{f\left(m\right)}-\sqrt{f\left(M\right)}\right)}^2\mathrm{\Phi }\left(r\left(A\right)\right)\hfill \\ & \le \mu \left(m,M,f\right)f\left(\mathrm{\Phi }\left(A\right)\right)-{\left(\sqrt{f\left(m\right)}-\sqrt{f\left(M\right)}\right)}^2\mathrm{\Phi }\left(r\left(A\right)\right).\hfill \end{array}$$

This completes the proof.  □

As discussed extensively in [6] Cahpter 2, for $f\left(t\right)=t^p$, we have

$$\begin{array}{cc}\hfill \mu \left(m,M,t^p\right)& =\max \left\{\frac{1}{t^p}\left(\frac{M-t}{M-m}{m}^p+\frac{t-m}{M-m}{M}^p\right):t\in \left[m,M\right]\right\}\hfill \\ & =\frac{\left(m{M}^p-M{m}^p\right)}{\left(p-1\right)\left(M-m\right)}{\left(\frac{p-1}{p}\frac{M^p-m^p}{m{M}^p-M{m}^p}\right)}^p.\hfill \end{array}$$

Now, the following fact can be easily deduced from Theorem 1 and Example 1.

Corollary 1.

Let $A\in \mathcal{B}\left(\mathcal{H}\right)$ be a positive operator with $\sigma \left(A\right)\subseteq \left[m,M\right]$ for some scalars m, M with $0<m<M$ and Φ be a normalized positive linear map from $\mathcal{B}\left(\mathcal{H}\right)$ to $\mathcal{B}\left(\mathcal{K}\right)$. Then for any $p<0$,

$$\begin{array}{cc}\hfill \mathrm{\Phi }\left(A^p\right)& \le \mathrm{\Phi }\left({\left(A-{\left(\sqrt{m}-\sqrt{M}\right)}^{2}r\left(A\right)\right)}^p\right)\hfill \\ & \le \mathrm{\Phi }\left({\left(m^{\frac{A-MI}{M-m}}{M}^{\frac{mI-A}{M-m}}\right)}^p\right)\hfill \\ & \le K\left(m,M,p\right)\mathrm{\Phi }{\left(A\right)}^p-{\left(m^{p/2}-M^{p/2}\right)}^2\mathrm{\Phi }\left(r\left(A\right)\right)\hfill \end{array}$$

where

$$K\left(m,M,p\right)=\frac{(m{M}^p-M{m}^p)}{\left(p-1\right)\left(M-m\right)}{\left(\frac{p-1}{p}\frac{M^p-m^p}{m{M}^p-M{m}^p}\right)}^p.$$

In particular,

$$\begin{array}{cc}\hfill \mathrm{\Phi }\left(A^{-1}\right)& \le \mathrm{\Phi }\left({\left(A-{\left(\sqrt{m}-\sqrt{M}\right)}^{2}r\left(A\right)\right)}^{-1}\right)\hfill \\ & \le \mathrm{\Phi }\left({\left(m^{\frac{A-MI}{M-m}}{M}^{\frac{mI-A}{M-m}}\right)}^{-1}\right)\hfill \\ & \le \frac{{\left(M+m\right)}^2}{4Mm}\mathrm{\Phi }{\left(A\right)}^{-1}-\left(\frac{{\left(\sqrt{M}-\sqrt{m}\right)}^2}{Mm}\right)\mathrm{\Phi }\left(r\left(A\right)\right).\hfill \end{array}$$

We note that $K\left(m,M,-1\right)=\frac{{\left(M+m\right)}^2}{4Mm}$ is the original Kantorovich constant.

Theorem 2.

Let all the assumptions of Theorem 1 hold. Then

$$\begin{array}{cc}\hfill f\left(\mathrm{\Phi }\left(A\right)-{\left(\sqrt{m}-\sqrt{M}\right)}^{2}r\left(\mathrm{\Phi }\left(A\right)\right)\right)& \le f\left(m^{\frac{MI-\mathrm{\Phi }\left(A\right)}{M-m}}{M}^{\frac{\mathrm{\Phi }\left(A\right)-mI}{M-m}}\right)\hfill \\ & \le \mu \left(m,M,f\right)\mathrm{\Phi }\left(f\left(A\right)\right)-{\left(\sqrt{f\left(m\right)}-\sqrt{f\left(M\right)}\right)}^{2}r\left(\mathrm{\Phi }\left(A\right)\right).\hfill \end{array}$$

Proof. 

By applying a standard functional calculus for the operator $\mathrm{\Phi }\left(A\right)$ such that $mI\le \mathrm{\Phi }\left(A\right)\le MI$, we get from (8)

$$\begin{array}{cc}\hfill f\left(\mathrm{\Phi }\left(A\right)-{\left(\sqrt{m}-\sqrt{M}\right)}^{2}r\left(\mathrm{\Phi }\left(A\right)\right)\right)& \le f\left(m^{\frac{MI-\mathrm{\Phi }\left(A\right)}{M-m}}{M}^{\frac{\mathrm{\Phi }\left(A\right)-mI}{M-m}}\right)\hfill \\ & \le \mathrm{\Phi }\left(L\left(A\right)\right)-{\left(\sqrt{f\left(m\right)}-\sqrt{f\left(M\right)}\right)}^{2}r\left(\mathrm{\Phi }\left(A\right)\right).\hfill \end{array}$$

We thus have

$$\begin{array}{cc}\hfill f\left(\mathrm{\Phi }\left(A\right)-{\left(\sqrt{m}-\sqrt{M}\right)}^{2}r\left(\mathrm{\Phi }\left(A\right)\right)\right)& \le f\left(m^{\frac{MI-\mathrm{\Phi }\left(A\right)}{M-m}}{M}^{\frac{\mathrm{\Phi }\left(A\right)-mI}{M-m}}\right)\hfill \\ & \le L\left(\mathrm{\Phi }\left(A\right)\right)-{\left(\sqrt{f\left(m\right)}-\sqrt{f\left(M\right)}\right)}^{2}r\left(\mathrm{\Phi }\left(A\right)\right)\hfill \\ & =\mathrm{\Phi }\left(L\left(A\right)\right)-{\left(\sqrt{f\left(m\right)}-\sqrt{f\left(M\right)}\right)}^{2}r\left(\mathrm{\Phi }\left(A\right)\right)\hfill \\ & \le \mu \left(m,M,f\right)\mathrm{\Phi }\left(f\left(A\right)\right)-{\left(\sqrt{f\left(m\right)}-\sqrt{f\left(M\right)}\right)}^{2}r\left(\mathrm{\Phi }\left(A\right)\right)\hfill \end{array}$$

where at the last step we used the basic inequality [5] Corollary 4.12.

Hence, the proof is complete.  □

As a corollary of Theorem 2 we have:

Corollary 2.

Let all the assumptions of Corollary 1 hold. Then for any $p<0$

$$\begin{array}{cc}\hfill \mathrm{\Phi }{\left(A\right)}^p& \le {\left(\mathrm{\Phi }\left(A\right)-{\left(\sqrt{m}-\sqrt{M}\right)}^{2}r\left(\mathrm{\Phi }\left(A\right)\right)\right)}^p\hfill \\ & \le {\left(m^{\frac{MI-\mathrm{\Phi }\left(A\right)}{M-m}}{M}^{\frac{\mathrm{\Phi }\left(A\right)-mI}{M-m}}\right)}^p\hfill \\ & \le K\left(m,M,p\right)\mathrm{\Phi }\left(A^p\right)-{\left(\sqrt{m^p}-\sqrt{M^p}\right)}^{2}r\left(\mathrm{\Phi }\left(A\right)\right).\hfill \end{array}$$

Remark 1.

Notice that the inequalities in Corollary 2 are stronger than the inequalities obtained in [11] Corollary 2.1.

Recall that if f is operator convex, the solidarities [12] or the perspective [13] of f is defined by

$${\mathcal{P}}_f\left(A\mid B\right)=A^{\frac{1}{2}}f\left(A^{-\frac{1}{2}}B{A}^{-\frac{1}{2}}\right)A^{\frac{1}{2}}.$$

Using a series of inequalities (6) we have the upper bounds of the perspective for non-negative decreasing geometrically convex function (not necessary operator convex f). We use the same symbol ${\mathcal{P}}_f\left(A\mid B\right)$ for a simplicity.

Proposition 1.

Let $A,B>0$ with $mA\le B\le MA$ for some scalars $0<m<M$. For a non-negative decreasing geometrically convex function f, we have

$$\begin{array}{cc}{\mathcal{P}}_f\left(A\mid B\right)& \le A^{1/2}f\left(A^{-1/2}B{A}^{-1/2}-{\left(\sqrt{m}-\sqrt{M}\right)}^{2}r(A,B)\right)A^{1/2}\hfill \\ & \le A^{1/2}f\left(m^{\frac{MI-A^{-1/2}B{A}^{-1/2}}{M-m}}{M}^{\frac{A^{-1/2}B{A}^{-1/2}-mI}{M-m}}\right)A^{1/2}\hfill \\ & \le A^{1/2}f{\left(m\right)}^{\frac{MI-A^{-1/2}B{A}^{-1/2}}{M-m}}f{\left(M\right)}^{\frac{A^{-1/2}B{A}^{-1/2}-mI}{M-m}}{A}^{1/2}\hfill \\ & \le \frac{Mf\left(m\right)-mf\left(M\right)}{M-m}A+\frac{f\left(M\right)-f\left(m\right)}{M-m}B-{\left(\sqrt{f\left(m\right)}-\sqrt{f\left(M\right)}\right)}^2{A}^{1/2}r(A,B)A^{1/2}\hfill \\ & \le \frac{Mf\left(m\right)-mf\left(M\right)}{M-m}A+\frac{f\left(M\right)-f\left(m\right)}{M-m}B,\hfill \end{array}$$

where

$$\begin{array}{cc}\hfill r\left(A,B\right)& =\min \left\{\frac{A^{-1/2}B{A}^{-1/2}-mI}{M-m},\frac{MI-A^{-1/2}B{A}^{-1/2}}{M-m}\right\}\hfill \\ & =\frac{1}{2}I-\frac{1}{M-m}\left|A^{-1/2}B{A}^{-1/2}-\frac{M+m}{2}I\right|.\hfill \end{array}$$