On the Operator Jensen Inequality for Convex Functions

TL;DR

This paper extends the operator Jensen inequality to a broader class of convex functions, including non-operator convex functions, providing new generalizations and their reverses, with various special cases analyzed.

Contribution

It introduces a novel generalization of the Jensen inequality for convex functions beyond operator convexity, including reverse inequalities.

Findings

Established a new operator Jensen inequality for convex functions.

Derived reverse inequalities related to the new Jensen inequality.

Discussed several special cases of the generalized inequalities.

Abstract

This paper is mainly devoted to studying operator Jensen inequality. More precisely, a new generalization of Jensen inequality and its reverse version for convex (not necessary operator convex) functions have been proved. Several special cases are discussed as well.

Equations

$$f\left(\Phi \left(A\right)\right)\le \Phi \left(f\left(A\right)\right).$$

$$f\left(\sum _{i=1}^n{\Phi }_i\left(A_i\right)\right)\le \sum _{i=1}^n{\Phi }_i\left(f\left(A_i\right)\right).$$

$$f\left(⟨\sum _{i=1}^n{\Phi }_i\left(A_i\right)x,x⟩\right)\le ⟨\sum _{i=1}^n{\Phi }_i\left(f\left(A_i\right)\right)x,x⟩$$

$$\sum _{i=1}^n{\Phi }_i\left(f\left(A_i\right)\right)\le \beta 1_{\mathcal{K}}+f\left(\sum _{i=1}^n{\Phi }_i\left(A_i\right)\right)$$

$$\beta =\max \left\{\frac{f\left(M\right)-f\left(m\right)}{M-m}t+\frac{Mf\left(m\right)-mf\left(M\right)}{M-m}-f\left(t\right):m\le t\le M\right\}.$$

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

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

$$f\left(⟨\sum _{i=1}^n{\Phi }_i\left(A_i\right)x,x⟩\right)\le ⟨f\left(\sum _{i=1}^n{\Phi }_i\left(A_i\right)\right)x,x⟩.$$

$$C_s\left(t-s\right)+f\left(s\right)\le f\left(t\right),\left(t\in J\right).$$

$$\sum _{i=1}^n{\Phi }_i\left(f\left(A_i\right)\right)\le f\left(\sum _{i=1}^n{\Phi }_i\left(A_i\right)\right)+\delta 1_{\mathcal{K}}$$

$$\delta =\sup \left\{⟨\sum _{i=1}^n{\Phi }_i\left(C_{A_i}{A}_i\right)x,x⟩-⟨\sum _{i=1}^n{\Phi }_i\left(A_i\right)x,x⟩⟨\sum _{i=1}^n{\Phi }_i\left(C_{A_i}\right)x,x⟩:x\in \mathcal{K};\Vert x\Vert =1\right\}.$$

$$f\left(A_i\right)\le f\left(t\right)1_{\mathcal{H}}+C_{A_i}{A}_i-t{C}_{A_i}.$$

$$\sum _{i=1}^n{\Phi }_i\left(f\left(A_i\right)\right)\le f\left(t\right)1_{\mathcal{K}}+\sum _{i=1}^n{\Phi }_i\left(C_{A_i}{A}_i\right)-t\sum _{i=1}^n{\Phi }_i\left(C_{A_i}\right).$$

$$⟨\sum _{i=1}^n{\Phi }_i\left(f\left(A_i\right)\right)x,x⟩\le f\left(t\right)+⟨\sum _{i=1}^n{\Phi }_i\left(C_{A_i}{A}_i\right)x,x⟩-t⟨\sum _{i=1}^n{\Phi }_i\left(C_{A_i}\right)x,x⟩.$$

$$⟨\sum _{i=1}^n{\Phi }_i\left(f\left(A_i\right)\right)x,x⟩$$

$$+⟨\sum _{i=1}^n{\Phi }_i\left(C_{A_i}{A}_i\right)x,x⟩-⟨\sum _{i=1}^n{\Phi }_i\left(A_i\right)x,x⟩⟨\sum _{i=1}^n{\Phi }_i\left(C_{A_i}\right)x,x⟩.$$

$$0$$

$$\le \underset{\Vert x\Vert =1}{\underset{x\in \mathcal{K}}{\sup }}\left\{⟨\sum _{i=1}^n{\Phi }_i\left(C_{A_i}{A}_i\right)x,x⟩-⟨\sum _{i=1}^n{\Phi }_i\left(A_i\right)x,x⟩⟨\sum _{i=1}^n{\Phi }_i\left(C_{A_i}\right)x,x⟩\right\}$$

$$⟨\sum _{i=1}^n{\Phi }_i\left(f\left(A_i\right)\right)x,x⟩$$

$$\le ⟨f\left(\sum _{i=1}^n{\Phi }_i\left(A_i\right)\right)x,x⟩+\delta \text{(by eqref18)}.$$

$$f\left(\sum _{i=1}^n{\Phi }_i\left(A_i\right)\right)\le \sum _{i=1}^n{\Phi }_i\left(f\left(A_i\right)\right)+\zeta 1_{\mathcal{K}}$$

$$\zeta =\sup \left\{⟨C_{{\sum }_{i=1}^n{\Phi }_i\left(A_i\right)}\sum _{i=1}^n{\Phi }_i\left(A_i\right)x,x⟩-⟨\sum _{i=1}^n{\Phi }_i\left(A_i\right)x,x⟩⟨C_{{\sum }_{i=1}^n{\Phi }_i\left(A_i\right)}x,x⟩:x\in \mathcal{K};\Vert x\Vert =1\right\}.$$

$$f\left(\sum _{i=1}^n{\Phi }_i\left(A_i\right)\right)\le f\left(t\right)1_{\mathcal{K}}+C_{{\sum }_{i=1}^n{\Phi }_i\left(A_i\right)}\sum _{i=1}^n{\Phi }_i\left(A_i\right)-t{C}_{{\sum }_{i=1}^n{\Phi }_i\left(A_i\right)}.$$

$$⟨f\left(\sum _{i=1}^n{\Phi }_i\left(A_i\right)\right)x,x⟩\le f\left(t\right)+⟨C_{{\sum }_{i=1}^n{\Phi }_i\left(A_i\right)}\sum _{i=1}^n{\Phi }_i\left(A_i\right)x,x⟩-t⟨C_{{\sum }_{i=1}^n{\Phi }_i\left(A_i\right)}x,x⟩.$$

$$⟨f\left(\sum _{i=1}^n{\Phi }_i\left(A_i\right)\right)x,x⟩$$

$$+⟨C_{{\sum }_{i=1}^n{\Phi }_i\left(A_i\right)}\sum _{i=1}^n{\Phi }_i\left(A_i\right)x,x⟩-⟨\sum _{i=1}^n{\Phi }_i\left(A_i\right)x,x⟩⟨C_{{\sum }_{i=1}^n{\Phi }_i\left(A_i\right)}x,x⟩.$$

$$0$$

$$\le \underset{\Vert x\Vert =1}{\underset{x\in \mathcal{K}}{\sup }}\left\{⟨C_{{\sum }_{i=1}^n{\Phi }_i\left(A_i\right)}\sum _{i=1}^n{\Phi }_i\left(A_i\right)x,x⟩-⟨\sum _{i=1}^n{\Phi }_i\left(A_i\right)x,x⟩⟨C_{{\sum }_{i=1}^n{\Phi }_i\left(A_i\right)}x,x⟩\right\}$$

$$⟨f\left(\sum _{i=1}^n{\Phi }_i\left(A_i\right)\right)x,x⟩$$

$$\le ⟨\sum _{i=1}^n{\Phi }_i\left(f\left(A_i\right)\right)x,x⟩+\zeta \text{(by eqref19)}$$

$$\delta =\sup \left\{⟨\sum _{i=1}^n{\Phi }_i\left(f^{\prime }\left(A_i\right)A_i\right)x,x⟩-⟨\sum _{i=1}^n{\Phi }_i\left(A_i\right)x,x⟩⟨\sum _{i=1}^n{\Phi }_i\left(f^{\prime }\left(A_i\right)\right)x,x⟩:x\in \mathcal{K};\Vert x\Vert =1\right\}$$

$$\zeta$$

$$-⟨\sum _{i=1}^n{\Phi }_i\left(A_i\right)x,x⟩⟨f^{\prime }\left(\sum _{i=1}^n{\Phi }_i\left(A_i\right)\right)x,x⟩:x\in \mathcal{K};\Vert x\Vert =1\}.$$

$$\sum _{i=1}^n{\Phi }_i\left(A_i^p\right)\le {\left(\sum _{i=1}^n{\Phi }_i\left(A_i\right)\right)}^p+p\delta 1_{\mathcal{K}}$$

Full text

1. Introduction

Let $\mathcal{B}\left(\mathcal{H}\right)$ be the $C^{*}$–algebra of all bounded linear operators on a Hilbert space $\mathcal{H}$. As customary, we reserve $m$, $M$ for scalars and ${{\mathbf{1}}_{\mathcal{H}}}$ for the identity operator on $\mathcal{H}$. A self-adjoint operator $A$ is said to be positive (written $A\geq 0$) if $\left\langle Ax,x\right\rangle\geq 0$ holds for all $x\in\mathcal{H}$ also an operator $A$ is said to be strictly positive (denoted by $A>0$) if $A$ is positive and invertible. If $A$ and $B$ are self-adjoint, we write $B\geq A$ in case $B-A\geq 0$. The Gelfand map $f\left(t\right)\mapsto f\left(A\right)$ is an isometrical $*$–isomorphism between the ${{C}^{*}}$–algebra $C\left(sp\left(A\right)\right)$ of continuous functions on the spectrum $sp\left(A\right)$ of a selfadjoint operator $A$ and the ${{C}^{*}}$–algebra generated by $A$ and the identity operator ${{\mathbf{1}}_{\mathcal{H}}}$. If $f,g\in C\left(sp\left(A\right)\right)$, then $f\left(t\right)\geq g\left(t\right)$ ($t\in sp\left(A\right)$) implies that $f\left(A\right)\geq g\left(A\right)$.

For $A,B\in\mathcal{B}\left(\mathcal{H}\right)$, $A\oplus B$ is the operator defined on $\mathcal{B}\left(\mathcal{H}\oplus\mathcal{H}\right)$ by $\left(\begin{matrix}A&0\\ 0&B\\ \end{matrix}\right)$. A linear map $\Phi:\mathcal{B}\left(\mathcal{H}\right)\to\mathcal{B}\left(\mathcal{K}\right)$ is positive if $\Phi\left(A\right)\geq 0$ whenever $A\geq 0$. It’s said to be unital if $\Phi\left({{\mathbf{1}}_{\mathcal{H}}}\right)={{\mathbf{1}}_{\mathcal{K}}}$. A continuous function $f$ defined on the interval $J$ is called an operator convex function if $f\left(\left(1-v\right)A+vB\right)\leq\left(1-v\right)f\left(A\right)+vf\left(B\right)$ for every $0<v<1$ and for every pair of bounded self-adjoint operators $A$ and $B$ whose spectra are both in $J$.

The well known operator Jensen inequality states (sometimes called the Choi–Davis–Jensen inequality):

[TABLE]

It holds for every operator convex $f:J\to\mathbb{R}$, self-adjoint operator $A$ with spectra in $J$, and unital positive linear map $\Phi$ [3, 5].

Hansen et al. [8] gave a general formulation of (1.1). The discrete version of their result reads as follows: If $f:J\to\mathbb{R}$ is an operator convex function, ${{A}_{1}},\ldots,{{A}_{n}}\in\mathcal{B}\left(\mathcal{H}\right)$ are self-adjoint operators with the spectra in $J$, and ${{\Phi}_{1}},\ldots,{{\Phi}_{n}}:\mathcal{B}\left(\mathcal{H}\right)\to\mathcal{B}\left(\mathcal{K}\right)$ are positive linear mappings such that $\sum\nolimits_{i=1}^{n}{{{\Phi}_{i}}\left({{\mathbf{1}}_{\mathcal{H}}}\right)}={{\mathbf{1}}_{\mathcal{K}}}$, then

[TABLE]

Though in the case of convex function the inequality (1.2) does not hold in general (see [3, Remark 2.6]), we have the following estimate [6, Lemma 2.1]:

[TABLE]

for any unit vector $x\in\mathcal{K}$. For recent results treating the Jensen operator inequality, we refer the reader to [9, 10, 11].

As a converse of (1.2), in [8] (see also [12]), it has been shown that if $f:\left[m,M\right]\to\mathbb{R}$ is a convex function and ${{A}_{1}},\ldots,{{A}_{n}}$ are self-adjoint operators with the spectra in $\left[m,M\right]$, then

[TABLE]

where

[TABLE]

A monograph on the reverse of Jensen inequality and its consequences is given by Furuta et al. in [7].

In this paper, we prove an inequality of type (1.2) without operator convexity assumption. Furthermore, as we can see in (1.4), the constant $\beta$ is dependent on $m$ and $M$. In this paper, we establish another reverse of operator Jensen inequality by dropping this restriction.

2. Operator Jensen-type inequalities without operator convexity

Let $f:J\to\mathbb{R}$ be a convex function, $A\in\mathcal{B}\left(\mathcal{H}\right)$ self-adjoint operator with the spectra in $J$, and let $x\in\mathcal{H}$ be a unit vector. Then from [13],

[TABLE]

Replace $A$ with $\Phi\left(A\right)$, where $\Phi:\mathcal{B}\left(\mathcal{H}\right)\to\mathcal{B}\left(\mathcal{K}\right)$ is a unital positive linear map, we get

[TABLE]

for any unit vector $x\in\mathcal{K}$. Assume that ${{A}_{1}},\ldots,{{A}_{n}}$ are self-adjoint operators on $\mathcal{H}$ with spectra in $J$ and ${{\Phi}_{1}},\ldots,{{\Phi}_{n}}:\mathcal{B}\left(\mathcal{H}\right)\to\mathcal{B}\left(\mathcal{K}\right)$ are positive linear maps with $\sum\nolimits_{i=1}^{n}{{{\Phi}_{i}}\left({{\mathbf{1}}_{\mathcal{H}}}\right)}={{\mathbf{1}}_{\mathcal{K}}}$. Now apply inequality (2.1) to the self-adjoint operator $A$ on the Hilbert space $\mathcal{H}\oplus\cdots\oplus\mathcal{H}$ defined by $A={{A}_{1}}\oplus\cdots\oplus{{A}_{n}}$ and the positive linear map $\Phi$ defined on $\mathcal{B}\left(\mathcal{H}\oplus\cdots\oplus\mathcal{H}\right)$ by $\Phi\left(A\right)={{\Phi}_{1}}\left({{A}_{1}}\right)\oplus\cdots\oplus{{\Phi}_{n}}\left({{A}_{n}}\right)$. Thus,

[TABLE]

Let us also recall that if $f$ is a convex function on an interval $J$, then for each point $\left(s,f\left(s\right)\right)$, there exists a real number ${{C}_{s}}$ such that

[TABLE]

Inequality (2.2), together with (2.3) yield the following theorem.

Theorem 2.1**.**

Let $f:J\to\mathbb{R}$ be a monotone convex function, ${{A}_{1}},\ldots,{{A}_{n}}\in\mathcal{B}\left(\mathcal{H}\right)$ self-adjoint operators with the spectra in $J$, and let ${{\Phi}_{1}},\ldots,{{\Phi}_{n}}:\mathcal{B}\left(\mathcal{H}\right)\to\mathcal{B}\left(\mathcal{K}\right)$ be positive linear mappings such that $\sum\nolimits_{i=1}^{n}{{{\Phi}_{i}}\left({{\mathbf{1}}_{\mathcal{H}}}\right)}={{\mathbf{1}}_{\mathcal{K}}}$. Then

[TABLE]

where

[TABLE]

Proof.

Fix $t\in J$. Since $J$ contains the spectra of the ${{A}_{i}}$ for $i=1,\ldots,n$, we may replace $s$ in the inequality (2.3) by ${{A}_{i}}$, via a functional calculus to get

[TABLE]

Applying the positive linear mappings ${{\Phi}_{i}}$ and summing on $i$ from 1 to $n$, this implies

[TABLE]

The inequality (2.5) easily implies, for any $x\in\mathcal{K}$ with $\left\|x\right\|=1$,

[TABLE]

Since $\sum\nolimits_{i=1}^{n}{{{\Phi}_{i}}\left({{\mathbf{1}}_{\mathcal{H}}}\right)}={{\mathbf{1}}_{\mathcal{K}}}$ we have $\left\langle\sum\nolimits_{i=1}^{n}{{{\Phi}_{i}}\left({{A}_{i}}\right)}x,x\right\rangle\in J$ where $x\in\mathcal{K}$ with $\left\|x\right\|=1$. Therefore, we may replace $t$ by $\left\langle\sum\nolimits_{i=1}^{n}{{{\Phi}_{i}}\left({{A}_{i}}\right)}x,x\right\rangle$ in (2.6). This yields

[TABLE]

On ther other hand,

[TABLE]

thanks to (1.3). Therefore,

[TABLE]

This completes the proof. ∎

Remark 2.1**.**

Inequality (2.7) provides the reverse of the inequality (1.3).

In the next theorem, we aim to present operator Jensen-type inequality without operator convexity assumption.

Theorem 2.2**.**

Let all the assumptions of Theorem 2.1 hold, then

[TABLE]

where

[TABLE]

Proof.

Fix $t\in J$. Since $J$ contains the spectra of the ${{A}_{i}}$ for $i=1,\ldots,n$ and $\sum\nolimits_{i=1}^{n}{{{\Phi}_{i}}\left({{\mathbf{1}}_{\mathcal{H}}}\right)}={{\mathbf{1}}_{\mathcal{K}}}$, so the spectra of $\sum\nolimits_{i=1}^{n}{{{\Phi}_{i}}\left({{A}_{i}}\right)}$ is also contained in $J$. Then we may replace $s$ in the inequality (2.3) by $\sum\nolimits_{i=1}^{n}{{{\Phi}_{i}}\left({{A}_{i}}\right)}$, via a functional calculus to get

[TABLE]

This inequality implies, for any $x\in\mathcal{K}$ with $\left\|x\right\|=1$,

[TABLE]

Substituting $t$ with $\left\langle\sum\nolimits_{i=1}^{n}{{{\Phi}_{i}}\left({{A}_{i}}\right)}x,x\right\rangle$ in (2.9). Thus,

[TABLE]

On the other hand,

[TABLE]

thanks to (2.2). Consequently,

[TABLE]

and the proof is complete. ∎

Remark 2.2**.**

Notice that inequality (2.10) can be considered as a converse of inequality (2.2).

3. Some Applications

In this section, we collect some consequences of Theorems 2.1 and 2.2.

(I) Suppose, in addition to the assumptions in Theorem 2.1, $f$ is differentiable on $J$ whose derivative $f^{\prime}$ is continuous on $J$, then (2.4) and (2.8) hold with

[TABLE]

and

[TABLE]

(II) By setting $f\left(t\right)={{t}^{p}}\left(p\geq 1\right)$ in Theorems 2.1 and 2.2 we find that:

[TABLE]

where

[TABLE]

and

[TABLE]

where

[TABLE]

whenever ${{A}_{1}},\ldots,{{A}_{n}}\in\mathcal{B}\left(\mathcal{H}\right)$ are positive operators and ${{\Phi}_{1}},\ldots,{{\Phi}_{n}}:\mathcal{B}\left(\mathcal{H}\right)\to\mathcal{B}\left(\mathcal{K}\right)$ positive linear mappings such that $\sum\nolimits_{i=1}^{n}{{{\Phi}_{i}}\left({{\mathbf{1}}_{\mathcal{H}}}\right)}={{\mathbf{1}}_{\mathcal{K}}}$.

If the operators ${{A}_{1}},\ldots,{{A}_{n}}$ are strictly positive, then (3.1) and (3.2) are also true for $p<0$.

(III) Assume that ${{w}_{1}},\ldots,{{w}_{n}}$ are positive scalars such that $\sum\nolimits_{i=1}^{n}{{{w}_{i}}}=1$. If we apply Theorems 2.1 and 2.2 for positive linear mappings ${{\Phi}_{i}}:\mathcal{B}\left(\mathcal{H}\right)\to\mathcal{B}\left(\mathcal{H}\right)$ determined by ${{\Phi}_{i}}:T\mapsto{{w}_{i}}T\left(i=1,\ldots,n\right)$, we get

[TABLE]

where

[TABLE]

and

[TABLE]

where

[TABLE]

Choi’s inequality [4, Proposition 4.3] says that

[TABLE]

whenever $B$ is self-adjoint and $A$ is positive invertible. We shall show the following complementary inequality of (3.3):

Proposition 3.1**.**

Let $A,B\in\mathcal{B}\left(\mathcal{H}\right)$ such that $B$ is self-adjoint and $A$ is positive invertible, and let $\Phi:\mathcal{B}\left(\mathcal{H}\right)\to\mathcal{B}\left(\mathcal{K}\right)$ be a unital positive linear mapping. Then

[TABLE]

where

[TABLE]

Proof.

It follows from Theorem 2.1 that

[TABLE]

where

[TABLE]

To a fixed positive $A\in\mathcal{B}\left(\mathcal{H}\right)$ we set

[TABLE]

and notice that $\Psi:\mathcal{B}\left(\mathcal{H}\right)\to\mathcal{B}\left(\mathcal{K}\right)$ is a unital linear map. Now, if $T={{A}^{-\frac{1}{2}}}B{{A}^{-\frac{1}{2}}}$, we infer from (3.5) that

[TABLE]

where

[TABLE]

By multiplying from the left and from the right with $\Phi{{\left(A\right)}^{\frac{1}{2}}}$ we obtain (3.4). ∎

The parallel sum of two positive operators $A$, $B$ is defined as the operator

[TABLE]

A simple calculation shows that (see, e.g., [2, (4.6) and (4.7)])

[TABLE]

If $\Phi$ is any positive linear map, then (see [2, Theorem 4.1.5])

[TABLE]

The following result gives a reverse of inequality (3.7).

Proposition 3.2**.**

Let $A,B\in\mathcal{B}\left(\mathcal{H}\right)$ positive invertible operators and let $\Phi:\mathcal{B}\left(\mathcal{H}\right)\to\mathcal{B}\left(\mathcal{K}\right)$ be unital positive linear mapping. Then

[TABLE]

where

[TABLE]

Proof.

Proposition 3.1 easily implies

[TABLE]

where

[TABLE]

Then we have

[TABLE]

Hence the conclusions follow. ∎

Remark 3.1**.**

A function $f:\left[0,\infty\right)\to\mathbb{R}$ is called superquadratic (see [1, Definition 1]) if for each $s\geq 0$, there exists a real constant ${{C}_{s}}$ such that

[TABLE]

for all $t\geq 0$.

By applying the same arguments as in Theorems 2.1 and 2.2 for definition (3.9), one can obtain stronger estimates than (2.4) and (2.8).

We leave the elaboration of this idea to the interested reader.