A Note Around Operator Bellman Inequality

TL;DR

This paper extends the operator Bellman inequality using the Kantorovich constant, providing a new mathematical estimate in the field of operator inequalities.

Contribution

It introduces an extended form of the operator Bellman inequality incorporating the Kantorovich constant, advancing theoretical understanding.

Findings

Extended operator Bellman inequality derived

Incorporates Kantorovich constant for estimation

Contributes to mathematical operator inequality theory

Abstract

In this paper, we shall give an extension of operator Bellman inequality. This result is estimated via Kantorovich constant.

Equations

$${\left(A^r-\sum _{i=1}^n{a}_i^r\right)}^{\frac{1}{r}}+{\left(B^r-\sum _{i=1}^n{b}_i^r\right)}^{\frac{1}{r}}\le {\left({\left(A+B\right)}^r-\sum _{i=1}^n{\left(a_i+b_i\right)}^r\right)}^{\frac{1}{r}}.$$

$$\Phi \left({\left(I-A\right)}^r{\nabla }_v{\left(I-B\right)}^r\right)\le \Phi {\left(I-A{\nabla }_{v}B\right)}^r$$

$$\Phi {\left(I-A{\nabla }_{v}B\right)}^r\le \Phi \left({\left(I-A\right)}^r{\nabla }_v{\left(I-B\right)}^r\right),r\in \left[-1,0\right]\cup \left[1,2\right].$$

$$\left(\begin{array}{cc}-0.25& -0.25\\ -0.25& 0.25\end{array}\right)=\Phi {\left(I-A{\nabla }_{v}B\right)}^r\nleqq \Phi \left({\left(I-A\right)}^r{\nabla }_v{\left(I-B\right)}^r\right)=\left(\begin{array}{cc}-1.5& -1\\ -1& 0\end{array}\right).$$

$$\Phi {\left(I-A{\nabla }_{v}B\right)}^r$$

$$\le \Phi \left({\left(I-A\right)}^r{♯}_v{\left(I-B\right)}^r\right)$$

$$\le \Phi \left({\left(I-A\right)}^r{\nabla }_v{\left(I-B\right)}^r\right)$$

$$A{♯}_{v}B=A^{\frac{1}{2}}{\left(A^{-\frac{1}{2}}B{A}^{-\frac{1}{2}}\right)}^v{A}^{\frac{1}{2}}v\in \left[0,1\right].$$

$$⟨f\left(A\right)x,x⟩\le f\left(⟨Ax,x⟩\right)$$

$$K\left(m,M,f\right)\Phi \left(f\left(A\right)\right)\le f\left(\Phi \left(A\right)\right)\le \frac{1}{K\left(m,M,f\right)}\Phi \left(f\left(A\right)\right)$$

$$K\left(m,M,f\right)=\max \left\{\frac{1}{f\left(t\right)}\left(\frac{M-t}{M-m}f\left(m\right)+\frac{t-m}{M-m}f\left(M\right)\right):m\le t\le M\right\}.$$

$$\Phi \left(f\left(A\right)\right){\nabla }_v\Phi \left(f\left(B\right)\right)\le \frac{1}{K{\left(m,M,f\right)}^2}f\left(\Phi \left(A{\nabla }_{v}B\right)\right)$$

$$⟨f\left(A\right){\nabla }_{v}f\left(B\right)x,x⟩$$

$$\le f\left(⟨A{\nabla }_{v}Bx,x⟩\right)\text{(since f is concave)}$$

$$\le \frac{1}{K\left(m,M,f\right)}⟨f\left(A{\nabla }_{v}B\right)x,x⟩\text{(by RHS of eqref1)}$$

$$f\left(A\right){\nabla }_{v}f\left(B\right)\le \frac{1}{K\left(m,M,f\right)}f\left(A{\nabla }_{v}B\right)$$

$$\Phi \left(f\left(A\right){\nabla }_{v}f\left(B\right)\right)$$

$$\le \frac{1}{K\left(m,M,f\right)}\Phi \left(f\left(A{\nabla }_{v}B\right)\right)$$

$$\le \frac{1}{K{\left(m,M,f\right)}^2}f\left(\Phi \left(A{\nabla }_{v}B\right)\right)\text{(by LHS of eqref1)}$$

$$\frac{1}{K{\left(m,M,f\right)}^2}f\left(\Phi \left(A{\nabla }_{v}B\right)\right)\le \Phi \left(f\left(A\right)\right){\nabla }_v\Phi \left(f\left(B\right)\right).$$

$${\left(I-\Phi \left(A{\nabla }_{v}B\right)\right)}^r\le K{\left(m,M,{\left(1-t\right)}^r\right)}^2\Phi \left({\left(I-A\right)}^r{\nabla }_v{\left(I-B\right)}^r\right).$$

$$h^{\prime }\left(t\right)=-\frac{2\left(\mu t+\lambda \right)\left(\left(r-1\right)\mu x+\lambda r+\mu \right)}{{\left(1-t\right)}^{2r}\left(t-1\right)},$$

$$h^{\prime }\left(t\right)=0\Rightarrow t_0=-\frac{\lambda }{\mu }\&t_1=-\frac{\lambda r+\mu }{\mu \left(r-1\right)}.$$

$${\left(I-A{\nabla }_{v}B\right)}^r$$

$$=\left(\begin{array}{cc}\left(1-{\sum }_{k=1}^n{a}_k^{\frac{1}{r}}\right){\nabla }_v{\left(1-{\sum }_{k=1}^n{b}_k^{\frac{1}{r}}\right)}^r& 0\\ 0& 0\end{array}\right),$$

$${\left(I-A\right)}^r{\nabla }_v{\left(I-B\right)}^r$$

$$=\left(\begin{array}{cc}{\left(1-{\sum }_{k=1}^n{a}_k^{\frac{1}{r}}\right)}^r{\nabla }_v{\left(1-{\sum }_{k=1}^n{b}_k^{\frac{1}{r}}\right)}^r& 0\\ 0& 0\end{array}\right).$$

$${\left(1-\sum _{k=1}^n{\left(a_k{\nabla }_v{b}_k\right)}^{\frac{1}{r}}\right)}^r$$

$$={\left(1-\left(\sum _{k=1}^n{a}_k^{\frac{1}{r}}{\nabla }_v\sum _{k=1}^n{b}_k^{\frac{1}{r}}\right)\right)}^r$$

$$={\left(\left(1-\sum _{k=1}^n{a}_k^{\frac{1}{r}}\right){\nabla }_v\left(1-\sum _{k=1}^n{b}_k^{\frac{1}{r}}\right)\right)}^r$$

$$\le \xi \left({\left(1-\sum _{k=1}^n{a}_k^{\frac{1}{r}}\right)}^r{\nabla }_v{\left(1-\sum _{k=1}^n{b}_k^{\frac{1}{r}}\right)}^r\right)\text{(by Corollary ref11)}.$$

$${\left(1-\sum _{k=1}^n{\left(\left(1-\frac{M_2}{M_1+M_2}\right)\left(\frac{a_k}{M_1}\right)+\left(\frac{M_2}{M_1+M_2}\right)\left(\frac{b_k}{M_2}\right)\right)}^{\frac{1}{r}}\right)}^r$$

$$\le \xi \left(\left(1-\frac{M_2}{M_1+M_2}\right){\left(1-\sum _{k=1}^n{\left(\frac{a_k}{M_1}\right)}^{\frac{1}{r}}\right)}^r+\frac{M_2}{M_1+M_2}{\left(1-\sum _{k=1}^n{\left(\frac{b_k}{M_2}\right)}^{\frac{1}{r}}\right)}^r\right).$$

$$\frac{1}{M_1+M_2}{\left({\left(M_1+M_2\right)}^{\frac{1}{r}}-\sum _{k=1}^n{\left(a_k+b_k\right)}^{\frac{1}{r}}\right)}^r$$

$$\le \xi \left(\frac{1}{M_1+M_2}{\left(M_1^{\frac{1}{r}}-\sum _{k=1}^n{a}_k^{\frac{1}{r}}\right)}^r+\frac{1}{M_1+M_2}{\left(M_2^{\frac{1}{r}}-\sum _{k=1}^n{b}_k^{\frac{1}{r}}\right)}^r\right).$$

$$\exp \left(\Phi \left(A{\nabla }_{v}B\right)\right)\le K{\left(m,M,\exp \left(t\right)\right)}^2\Phi \left(\exp \left(A\right){\nabla }_v\exp \left(B\right)\right).$$

$$h^{\prime }\left(t\right)=-\frac{2\left(\mu t+\lambda \right)\left(\mu t+\lambda -\mu \right)}{\exp 2x}.$$

Full text

1. Introduction

Let $\mathcal{B}(\mathcal{H})$ be the $C^{*}$ algebra of bounded linear operators on a complex Hilbert space $\mathcal{H}.$ If $A\in\mathcal{B}(\mathcal{H})$ is positive, we write denote $A\geq 0.$ For two self-ajoint operators $A,B\in\mathcal{B}(\mathcal{H})$, we write $A\leq B$ if $B-A\geq 0$. For a real-valued function $f$ of a real variable and a self-adjoint operator $A\in\mathcal{B}\left(\mathcal{H}\right)$, the value $f\left(A\right)$ is understood by means of the functional calculus.

Let $J$ be a real interval of any type. A continuous function $f:J\to\mathbb{R}$ is said to be operator concave if $f\left(A{{\nabla}_{v}}B\right)\geq f\left(A\right){{\nabla}_{v}}f\left(B\right)$ holds for each $v\in\left[0,1\right]$ and every pair of self-adjoint operators $A,B\in\mathcal{B}\left(\mathcal{H}\right)$, with spectra in $J$.

A linear map $\Phi:\mathcal{B}\left(\mathcal{H}\right)\to\mathcal{B}\left(\mathcal{K}\right)$ is said to be positive if $\Phi\left(A\right)\geq 0$ when $A\geq 0$. If, in addition, $\Phi\left(I\right)=I$, it is said to be normalized.

Bellman [1] showed that if $n,r$ are positive integers and $A,B,{{a}_{i}},{{b}_{i}}\left(1\leq i\leq n\right)$ are positive real numbers such that $\sum\nolimits_{i=1}^{n}{a_{i}^{r}}\leq{{A}^{r}}$ and $\sum\nolimits_{i=1}^{n}{b_{i}^{r}}\leq{{B}^{r}}$, then

[TABLE]

The operator Bellman inequality [4, Corollary 2.2] asserts that: If $\Phi$ is a normalized positive linear map on $\mathcal{B}\left(\mathcal{H}\right)$, $A,B$ are contractions (in the sense that $\left\|A\right\|,\left\|B\right\|\leq 1$), then

[TABLE]

where $r,v\in\left[0,1\right]$.

By applying similar method presented in [4], we infer that

[TABLE]

Of course, the above inequality does not hold in general when $r\notin\left[-1,0\right]\cup\left[1,2\right]$. For instance, if we take $\Phi\left(X\right)=X$, $A=\left(\begin{matrix}2&1\\ 1&1\\ \end{matrix}\right)$, $B=\left(\begin{matrix}1&0\\ 0&0\\ \end{matrix}\right)$, $v=\frac{1}{2}$, and $f\left(t\right)={{t}^{3}}$. By a simple calculation, we have

[TABLE]

Very recently, as an extension of inequality (1.1) to the negative parameter, the authors in [7, Theorem 3.2] proved that for any contraction operators $A,B\in\mathcal{B}\left(\mathcal{H}\right)$, $v\in\left[0,1\right]$, and $r\in\left[-1,0\right]$

[TABLE]

holds, where the notation ${{\sharp}_{v}}$ used for the weighted geometric mean between two positive operators, and is defined as follows

[TABLE]

In this paper, we aim to provide a generalization of the inequality (1.1) to $r\notin\left[-1,0\right]\cup\left[1,2\right]$. Additionally, we show a reverse Bellman type inequality of additive type, by using some ideas from [4, 7].

2. Main Results

In order to prove our theorem, we need the following lemmas.

Lemma 1**.**

[5, Theorem 1.2]** If $f:J\to\mathbb{R}$ is a concave function, then

[TABLE]

for any self-adjoint operator $A$ with spectra contained in $J$ and any unit vector $x\in\mathcal{H}$.

Lemma 2**.**

[3, Corollary 4.12]**) Let $A\in\mathcal{B}\left(\mathcal{H}\right)$ such that $m\leq A\leq M$ and $\Phi$ be a normalized positive linear map. If $f:\left[m,M\right]\to\mathbb{R}$ is a concave function, then

[TABLE]

where

[TABLE]

We can now state our first main result.

Proposition 1**.**

Let $A,B\in\mathcal{B}\left(\mathcal{H}\right)$ be two self-adjoint operators such that $m\leq A,B\leq M$ and $\Phi$ be normalized positive linear map. If $f:\left[m,M\right]\to\mathbb{R}$ is a concave function, then for any $v\in\left[0,1\right]$

[TABLE]

where $K\left(m,M,f\right)$ is defined as (2.2).

Proof.

The assumption $m\leq A,B\leq M$ implies $m\leq\left\langle A{{\nabla}_{v}}Bx,x\right\rangle\leq M$ for any $x\in\mathcal{H}$ with $\left\|x\right\|=1$. So

[TABLE]

i.e.,

[TABLE]

which is an inequality of interest in itself. The hypothesis on $\Phi$ ensures that

[TABLE]

which is the desired inequality (2.3). ∎

Proposition 2**.**

Let all the assumptions of Proposition 1 hold except the condition concavity which is changed to convexity. Then

[TABLE]

Letting $f\left(t\right)={{\left(1-t\right)}^{r}}\left(t<1,r\notin[-1,0]\cap[1,2]\right)$ in Proposition 2, we infer the following result:

Theorem 1**.**

Let $A,B\in\mathcal{B}\left(\mathcal{H}\right)$ be two contractions operators such that $m\leq A,B\leq M$ and $\Phi$ be normalized positive linear map. Then for any $v\in[0,1]$ and $r\notin[-1,0]\cap[1,2]$

[TABLE]

Remark 1**.**

In the following we aim to find the value of $K{{\left(m,M,{{\left(1-t\right)}^{r}}\right)}^{2}}$ in Theorem 1. To do this end, it is enough we find the maximum of the function $h\left(t\right)\equiv{{\left(\frac{\mu t+\lambda}{{{\left(1-t\right)}^{r}}}\right)}^{2}}$ for $0<m\leq t\leq M<1$ and $r\notin\left[1,2\right]\cup\left[-1,0\right]$ with $\mu=\frac{{{\left(1-M\right)}^{r}}-{{\left(1-m\right)}^{r}}}{M-m}$ and $\lambda=\frac{M{{\left(1-m\right)}^{r}}-m{{\left(1-M\right)}^{r}}}{M-m}$. Notice that \left\{\begin{array}[]{rr}\mu<0,\lambda>0&\text{ for }r>2\\ \mu>0,\lambda<0&\text{ for }r<-1\\ \end{array}\right.. By an easy computation we infer

[TABLE]

and

[TABLE]

It is not hard to check that $h(t)$ attains its maximum at $t_{1}$ provided that $m\leq t_{1}\leq M$. To see the proof of inequalities ${{t}_{1}}-m\geq 0$ and $M-{{t}_{1}}\geq 0$ we refer the reader to [6].

Remark 2**.**

Assume $n$ is a positive integers, $r>2$, $v\in\left[0,1\right]$ and ${{a}_{k}},{{b}_{k}}\left(1\leq k\leq n\right)$ are positive real numbers such that $0<m\leq\sum\nolimits_{k=1}^{n}{a_{k}^{\frac{1}{r}}}\leq M<1$ and $0<m\leq\sum\nolimits_{k=1}^{n}{b_{k}^{\frac{1}{r}}}\leq M<1$. Take $A=\left(\begin{matrix}\sum\nolimits_{k=1}^{n}{a_{k}^{\frac{1}{r}}}&0\\ 0&1\\ \end{matrix}\right)\in{{M}_{2}}\left(\mathbb{C}\right)$, $B=\left(\begin{matrix}\sum\nolimits_{k=1}^{n}{b_{k}^{\frac{1}{r}}}&0\\ 0&1\\ \end{matrix}\right)\in{{M}_{2}}\left(\mathbb{C}\right)$. Consequently,

[TABLE]

and

[TABLE]

Now,

[TABLE]

Take $v=\frac{{{M}_{2}}}{{{M}_{1}}+{{M}_{2}}}$ and substitute ${{a}_{k}}$ and ${{b}_{k}}$ by $\frac{{{a}_{k}}}{{{M}_{1}}}$ and $\frac{{{b}_{k}}}{{{M}_{2}}}$, respectively. We infer that

[TABLE]

Equivalently,

[TABLE]

By choosing $f\left(t\right)=\exp\left(t\right)$ in Proposition 2 we have:

Corollary 1**.**

Let $A,B\in\mathcal{B}\left(\mathcal{H}\right)$ be two self-adjoint operators such that $m\leq A,B\leq M$ and $\Phi$ be normalized positive linear map. Then

[TABLE]

Remark 3**.**

To find the value of $K{{\left(m,M,\exp\left(t\right)\right)}^{2}}$ we set $h\left(t\right)\equiv{{\left(\frac{\mu t+\lambda}{\exp t}\right)}^{2}}$ for $0<m\leq t\leq M<1$ with $\mu=\frac{\exp M-\exp m}{M-m}$ and $\lambda=\frac{M\exp m-m\exp M}{M-m}$. It is easy to see that $\lambda,\mu>0$, and

[TABLE]

Solving $h^{\prime}(t)=0$, we obtain

[TABLE]

After simple computations, we find that $h(t)$ attains its maximum at $t_{1}$ provided that $m\leq t_{1}\leq M$. For details we refer to [5].

In the following, we aim to prove a complementary inequality for the inequality (2.3). The following Lemmas will play a role later.

Lemma 3**.**

[2, Lemma 3.2]** Let $A,B\in\mathcal{B}\left(\mathcal{H}\right)$ be two positive operators satisfying $0<m\leq A,B\leq M$ for some scalars. If $f:\left[m,M\right]\to\mathbb{R}$ is a continuous concave function, then for each $v\in\left[0,1\right]$

[TABLE]

where

[TABLE]

Lemma 4**.**

[5]** Let $A\in\mathcal{B}\left(\mathcal{H}\right)$ such that $m\leq A\leq M$ and $\Phi$ be a normalized positive linear map. If $f:\left[m,M\right]\to\mathbb{R}$ is a concave function, then

[TABLE]

where

[TABLE]

Theorem 2**.**

Let $A,B\in\mathcal{B}\left(\mathcal{H}\right)$ be two self-adjoint operators such that $m\leq A,B\leq M$ and $\Phi$ be normalized positive linear map. If $f:\left[m,M\right]\to\mathbb{R}$ is a concave function, then

[TABLE]

where $\widetilde{\beta}\left(m,M,f\right)$ is defined as (2.6).

Proof.

We have

[TABLE]

The proof is completed. ∎

Corollary 2**.**

Let $A,B\in\mathcal{B}\left(\mathcal{H}\right)$ be two contractions operators such that $m\leq A,B\leq M$ and $\Phi$ be normalized positive linear map. Then for any $v\in[0,1]$ and $r\notin[-1,0]\cup[1,2]$,

[TABLE]