Operator Jensen's inequality for operator superquadratic functions

TL;DR

This paper introduces the concept of operator superquadratic functions for positive Hilbert space operators, explores their properties, and establishes a non-commutative Jensen's inequality with generalizations.

Contribution

It defines operator superquadratic functions, provides examples, and proves a non-commutative Jensen's inequality with extensions to positive unital linear maps.

Findings

Defined operator superquadratic functions with key properties

Established a non-commutative Jensen's inequality for these functions

Generalized the inequality to positive unital linear maps

Abstract

In this work, an operator superquadratic function (in operator sense) for positive Hilbert space operators is defined. Several examples with some important properties together with some observations which are related to the operator convexity are pointed out. Equivalent statements of a non-commutative version of Jensen's inequality for operator superquadratic function are established. A generalization of the main result to any positive unital linear map is also provided.

Equations

$$\mathrm{id}\otimes \Phi :{\mathbb{C}}^{k\times k}\otimes \mathcal{B}\left(\mathcal{H}\right)\to {\mathbb{C}}^{k\times k}\otimes \mathcal{B}\left(\mathcal{K}\right),$$

$$\displaystyle\left({\begin{array}[]{*{20}c}{A_{11}}& \cdots&{A_{1k}}\\ \vdots& \ddots& \vdots\\ {A_{k1}}& \cdots&{A_{kk}}\\ \end{array}}\right)\mapsto\left({\begin{array}[]{*{20}c}{\Phi\left({A_{11}}\right)}& \cdots&{\Phi\left({A_{1k}}\right)}\\ \vdots& \ddots& \vdots\\ {\Phi\left({A_{k1}}\right)}& \cdots&{\Phi\left({A_{kk}}\right)}\\ \end{array}}\right).$$

$$f\left(t\alpha +\left(1-t\right)\beta \right)\le tf\left(\alpha \right)+\left(1-t\right)f\left(\beta \right),$$

$$f\left(t\right)\le \frac{f\left(y\right)-f\left(x\right)}{y-x}\left(t-x\right)+f\left(x\right)$$

$$f\left(t\right)\ge f\left(x\right)+\lambda \left(t-x\right)$$

$$f\left(t\right)\ge f\left(x\right)+C_x\left(t-x\right)+f\left(\left|t-x\right|\right)$$

$$f\left(\int \phi d\mu \right)\le \int f\left(\phi \left(s\right)\right)-f\left(\left|\phi \left(s\right)-\int \phi d\mu \right|\right)d\mu \left(s\right)$$

$$\left[x_i\right]f$$

$$\left[x_0,x_1,\dots ,x_k\right]f$$

$$\left[x_0\right]f$$

$$\left[x_0,x_1\right]f$$

$$\left[x_0,x_1,x_2\right]f$$

$$f\left(\lambda A+\left(1-\lambda \right)B\right)\le \lambda f\left(A\right)+\left(1-\lambda \right)f\left(B\right)$$

$$f\left(t\right)={\beta }_0+{\beta }_{1}t+\frac{1}{2}{\beta }_2{\int }_{-1}^1\frac{t^2}{1-\alpha t}d\mu \left(t\right)$$

$$f\left(\sum _{j=1}^n{\lambda }_j{x}_j\right)\le \sum _{j=1}^n{\lambda }_{j}f\left(x_j\right)$$

$$\displaystyle A=\left({\begin{array}[]{*{20}c}{x_{1}}&{}\hfil&0\\ {}\hfil& \ddots&{}\hfil\\ 0&{}\hfil&{x_{n}}\\ \end{array}}\right)\qquad{\rm{and}}\qquad x=\left({\begin{array}[]{*{20}c}{\sqrt{\lambda_{1}}}\\ \vdots\\ {\sqrt{\lambda_{n}}}\\ \end{array}}\right)$$

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

$${\Phi }^2\left(A\right)\le \Phi \left(A^2\right)$$

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

$$\displaystyle A=\left({\begin{array}[]{*{20}c}{x_{1}}&{}\hfil&0\\ {}\hfil& \ddots&{}\hfil\\ 0&{}\hfil&{x_{n}}\\ \end{array}}\right)\qquad{\rm{and}}\qquad V=\left({\begin{array}[]{*{20}c}{\sqrt{\lambda_{1}}}&{\cdots}&0\\ {\vdots}& \ddots&{}\hfil\\ \sqrt{\lambda_{n}}&{}\hfil&{0}\\ \end{array}}\right),$$

$$f\left(V^{\ast }AV\right)\le V^{\ast }f\left(A\right)V.$$

$$\begin{gathered} f\left(\alpha A+\left(1-\alpha \right)B\right) \\ \le \alpha \left[f\left(A\right)-f\left(\left(1-\alpha \right)\left|A-B\right|\right)\right]+\left(1-\alpha \right)\left[f\left(B\right)-f\left(\alpha \left|A-B\right|\right)\right] \end{gathered}$$

$$f\left(\frac{A+B}{2}\right)+f\left(\frac{\left|A-B\right|}{2}\right)-\frac{f\left(A\right)+f\left(B\right)}{2}$$

$$=\left[\alpha \frac{A+B}{2}+\beta \right]+\left[\alpha \frac{\left|A-B\right|}{2}+\beta \right]-\frac{\alpha A+\beta +\alpha B+\beta }{2}$$

$$=\alpha \frac{\left|A-B\right|}{2}+\beta \ge 0.$$

$$\displaystyle A=\left({\begin{array}[]{*{20}c}2&1\\ 1&1\\ \end{array}}\right)\qquad\text{and}\qquad B=\left({\begin{array}[]{*{20}c}1&0\\ 0&0\\ \end{array}}\right)$$

$$\frac{A^3+B^3}{2}-{\left(\frac{A+B}{2}\right)}^3-{\left(\frac{\left|A-B\right|}{2}\right)}^3$$

$${\left(\alpha A+\left(1-\alpha \right)B\right)}^2\le \alpha A^2+\left(1-\alpha \right)B^2-\alpha {\left(1-\alpha \right)}^2{\left|A-B\right|}^2-\left(1-\alpha \right){\alpha }^2{\left|A-B\right|}^2$$

$$\iff {\alpha }^2{A}^2+{\left(1-\alpha \right)}^2{B}^2+\alpha \left(1-\alpha \right)\left(AB+BA\right)\le \alpha A^2+\left(1-\alpha \right)B^2$$

$$-\left[\alpha {\left(1-\alpha \right)}^2+\left(1-\alpha \right){\alpha }^2\right]{\left(A-B\right)}^2$$

$$\iff \alpha \left(\alpha -1\right)A^2+\alpha \left(\alpha -1\right)B^2+\alpha \left(\alpha -1\right)\left(AB+BA\right)\le \alpha \left(\alpha -1\right){\left(A-B\right)}^2$$

$$\iff \alpha \left(\alpha -1\right){\left(A+B\right)}^2\le \alpha \left(\alpha -1\right){\left(A-B\right)}^2$$

$$\iff {\left(A+B\right)}^2\ge {\left(A-B\right)}^2\mathrm{for}\alpha \left(\alpha -1\right)<0$$

$$\iff \left|A+B\right|\ge \left|A-B\right|g\left(t\right)=\sqrt{t}\mathrm{is}\mathrm{operator}\mathrm{monotone}$$

$$f\left(\alpha A+\left(1-\alpha \right)B\right)\le \alpha \left[f\left(A\right)-f\left(\left(1-\alpha \right)\left|A-B\right|\right)\right]+\left(1-\alpha \right)\left[f\left(B\right)-f\left(\alpha \left|A-B\right|\right)\right]$$

$$\begin{gathered} f\left(\alpha A+\left(1-\alpha \right)⟨Ax,x⟩1_{\mathcal{H}}\right)\le \alpha \left[f\left(A\right)-f\left(\left(1-\alpha \right)\left|A-⟨Ax,x⟩1_{\mathcal{H}}\right|\right)\right] \\ +\left(1-\alpha \right)\left[f\left(⟨Ax,x⟩\right)-f\left(\alpha \left|A-⟨Ax,x⟩1_{\mathcal{H}}\right|\right)\right]. \end{gathered}$$

Full text

1. Introduction

Let $\mathscr{B}\left(\mathscr{H}\right)$ be the Banach algebra of all bounded linear operators defined on a complex Hilbert space $\left(\mathscr{H};\left\langle\cdot,\cdot\right\rangle\right)$  with the identity operator  $1_{\mathscr{H}}$ in $\mathscr{B}\left(\mathscr{H}\right)$. A bounded linear operator $A$ defined on $\mathscr{H}$ is selfadjoint if and only if $\left\langle{Ax,x}\right\rangle\in\mathbb{R}$ for all $x\in\mathscr{H}$. The spectrum of an operator $A$ is the set of all $\lambda\in\mathbb{C}$  for which the operator $\lambda I-A$ does not have a bounded linear operator inverse, and is denoted by $\operatorname{sp}\left(A\right)$. Consider the real vector space $\mathscr{B}\left(\mathscr{H}\right)_{sa}$ of self-adjoint operators on $\mathscr{H}$ and its positive cone $\mathscr{B}\left(\mathscr{H}\right)^{+}$ of positive operators on $\mathscr{H}$. Also, $\mathscr{B}\left(\mathscr{H}\right)_{sa}^{I}$ denotes the convex set of bounded self-adjoint operators on the Hilbert space $\mathscr{H}$ with spectra in a real interval $I$. A partial order is naturally equipped on $\mathscr{B}\left(\mathscr{H}\right)_{sa}$ by defining $A\leq B$ if and only if $B-A\in \mathscr{B}\left(\mathscr{H}\right)^{+}$.  We write $A>0$ to mean that $A$ is a strictly positive operator, or equivalently, $A\geq 0$ and $A$ is invertible. When $\mathscr{H}=\mathbb{C}^{n}$, we identify $\mathscr{B}\left(\mathscr{H}\right)$ with the algebra $\mathfrak{M}_{n\times n}$ of $n$-by-$n$ complex matrices. Then, $\mathfrak{M}^{+}_{n\times n}$ is just the cone of $n$-by-$n$ positive semidefinite matrices.

A linear map  is defined to be   $\Phi:\mathscr{B}\left(\mathscr{H}\right)\to\mathscr{B}\left(\mathscr{K}\right)$ which preserves additivity and homogeneity, i.e.,  $\Phi\left(\lambda_{1}A+\lambda_{2}B\right)=\lambda_{1}\Phi\left(A \right)+\lambda_{2}\Phi\left(B\right)$ for any $\lambda_{1},\lambda_{2}\in\mathbb{C}$  and $A,B\in\mathscr{B}\left(\mathscr{H}\right)$. The linear map is positive $\Phi:\mathscr{B}\left(\mathscr{H}\right)\to\mathscr{B}\left(\mathscr{K}\right)$ if it preserves the operator order, i.e., if $A\in\mathscr{B}^{+}\left(\mathscr{H}\right)$ then $\Phi\left(A\right)\in\mathscr{B}^{+}\left(\mathscr{K}\right)$, and in this case we write $\mathcal{B}[\mathscr{B}\left(\mathscr{H}\right),\mathscr{B}\left(\mathscr{K}\right)]$. Obviously, a positive linear map $\Phi$ preserves the order relation, namely $A\leq B\Longrightarrow\Phi\left(A\right)\leq\Phi\left(B\right)$ and preserves the adjoint operation $\Phi\left(A^{*}\right)=\Phi\left(A\right)^{*}$. Moreover, $\Phi$ is said to be  unital  if it preserves the identity operator,   in this case we write $\mathcal{B}_{n}[\mathscr{B}\left(\mathscr{H}\right),\mathscr{B}\left(\mathscr{K}\right)]$.

A linear map $\Phi:\mathscr{B}\left(\mathscr{H}\right)\to\mathscr{B}\left(\mathscr{K}\right)$  induces  another map

[TABLE]

in a natural way. If $\mathbb{C}^{k\times k} \otimes\mathscr{B}\left(\mathscr{H}\right)$ is identified with the $C^{*}$-algebra $\mathscr{B}^{k\times k}\left(\mathscr{H}\right)$ of $k\times k$–matrices  with entries in $\mathscr{B}\left(\mathscr{H}\right)$ then ${\rm{id}}\otimes\Phi$ act as:

[TABLE]

We say that $\Phi$  is $k$-positive if ${\rm{id}}\otimes\Phi$  is a positive map, and $\Phi$ is called completely positive if $\Phi$  is $k$-positive for all $k$.

1.1. Superquadratic functions

A function $f:J\to\mathbb{R}$ is called convex iff

[TABLE]

for all points $\alpha,\beta\in J$ and all $t\in[0,1]$. If $-f$ is convex then we say that $f$ is concave. Moreover, if $f$ is both convex and concave, then $f$ is said to be affine.

Geometrically, for two point $\left(x,f\left(x\right)\right)$ and $\left(y,f\left(y\right)\right)$  on the graph of $f$ are on or below the chord joining the endpoints  for all $x,y\in I$, $x<y$. In symbols, we write

[TABLE]

for any $x\leq t\leq y$ and $x,y\in J$.

Equivalently, given a function $f:J\to\mathbb{R}$, we say that $f$ admits a support line at $x\in J$ if there exists a $\lambda\in\mathbb{R}$ such that

[TABLE]

for all $t\in J$.

The set of all such $\lambda$ is called the subdifferential of $f$ at $x$, and it’s denoted by $\partial f$. Indeed, the subdifferential gives us the slopes of the supporting lines for the graph of $f$. So that if $f$ is convex then $\partial f(x)\neq\emptyset$ at all interior points of its domain.

From this point of view  Abramovich et al. [2] extend the above idea for what they called superquadratic functions. Namely, a function $f:[0,\infty)\to\mathbb{R}$ is called superquadratic provided that for all $x\geq 0$ there exists a constant $C_{x}\in\mathbb{R}$ such that

[TABLE]

for all $t\geq 0$. We say that $f$ is subquadratic if $-f$ is superquadratic. Thus, for a superquadratic function we require that $f$ lie above its tangent line plus a translation of $f$ itself. If $f$ is differentiable and satisfies $f(0)=f^{\prime}(0)=0$, then one sees easily that the $C_{x}$ appearing in the definition is necessarily $f^{\prime}(x)$, (see [1]).

Prima facie, superquadratic function  looks  to be stronger than the convex function itself but if $f$ takes negative values then it maybe considered as a weaker function. Therefore, if $f$ is superquadratic and non-negative, then $f$ is convex and increasing [2] (see also [4]).

Moreover,   the following result holds for superquadratic function.

Lemma 1**.**

[2]** Let $f$ be superquadratic function. Then

1. (1)

$f\left(0\right)\leq 0$** 2. (2)

If $f$ is differentiable and $f(0)=f^{\prime}(0)=0$, then $C_{x}=f^{\prime}(x)$  for all $x\geq 0$. 3. (3)

If $f(x)\geq 0$ for all $x\geq 0$, then $f$ is convex and $f(0)=f^{\prime}(0)=0$.

The next result gives a sufficient condition when convexity (concavity) implies super(sub)quaradicity.

Lemma 2**.**

[2]** If $f^{\prime}$ is convex (concave) and $f(0)=f^{\prime}(0)=0$, then is super(sub)quadratic. The converse of is not true.

Remark 1**.**

In general, non-negative subquadratic functions does not imply concavity. In other words,  there exists a subquadratic function which is convex. For example, $f(x)=x^{p}$, $x\geq 0$ and $1\leq p\leq 2$ is subquadratic and convex.

Among others,  Abramovich et al. [2] proved that the inequality

[TABLE]

holds for all probability measures $\mu$ and all nonnegative, $\mu$-integrable functions $\varphi$ if and only if $f$ is superquadratic.  For more details the reader may refer to [4], [5], and [22].

1.2. Operator convexity and Jensen inequality

Let $f$ be a real-valued function defined on $J$. A $k$-th order divided difference of $f$ at distinct points $x_{0},\cdots,x_{k}$ in $J$ may be defined recursively by

[TABLE]

For instance, the first $3$-divided differences are given as follows:

[TABLE]

A function $f:J\to\mathbb{R}$ is said to be matrix monotone of degree $n$ or $n$-monotone, if for every $A,B\in\mathfrak{M}_{n\times n}$, it is true that $A\leq B$ $\Longleftrightarrow$ $f\left(A\right)\leq\left(B\right)$. Similarly, $f$ is said to be operator monotone If $f$ is $n$-monotone for all $n\in\mathbb{N}$. Also, $f$ is called operator convex if it is matrix convex ($n$-convex for all $n$); i.e., if for every pair of selfadjoint operators $A,B\in\mathfrak{M}_{n\times n}$ we have

[TABLE]

for all $\lambda\in\left[0,1\right]$. If the inequality is reversed then $f$ is called operator concave. In case we have general Hilbert space $\mathscr{H}$, the above definition holds for every pair of bounded selfadjoint operators $A$ and $B$ in $\mathscr{B}\left(\mathscr{H}\right)$, whose spectra ontianed in $J$.  For more details see [23] and the recent survey [14].

In 1955, Bendat and Sherman [13] have shown that $f$ is operator convex on the open interval $(-1,1)$ if and only if it has a (unique) representation

[TABLE]

for $\beta_{2}\geq 0$ and some probability measure  $\mu$ on $[-1,1]$ (it could be Borel measure). In particular, $f$ must be analytic with $f(0)=\beta_{0}$, $f^{\prime}(0)=\beta_{1}$ and $f^{\prime\prime}(0)=\beta_{2}$.

We recall that the celebrated Löwner-Heinz inequality reads that:

Lemma 3**.**

Let $A,B\in\mathscr{B}\left( \mathscr{H}\right)^{+}$ such that $A\geq B$, then $A^{r}\geq B^{r}$ for all $r\in\left[0,1\right]$.

On the other hand the mapping $t\mapsto t^{p}$ $(p>1)$ is not operator monotone, for more details see [10], [17] and [19].

The classical Jensen’s inequality for reals states that

[TABLE]

valid for all real valued convex function $f$ defined on $[m,M]$, for every $x_{1},\cdots,x_{n}\in[m,M]$ and every positive real numbers $\lambda_{j}$ $(1\leq j\leq n)$ such that $\sum_{j=1}^{n}{\lambda_{j}}=1$.

The inequality \reftagform@1.5 would be rephrases under matrix situation by putting

[TABLE]

then the classical Jensen  inequality \reftagform@1.5 is expressed as

[TABLE]

which is one of the operator version of the classical Jensen’s inequality, see [17].

Kadison [21] established his famous non-commutative version  of the previous inequality where he proved that for every selfadjoint matrix $A$ the inequality

[TABLE]

for every positive unital linear map $\Phi:\mathfrak{M}_{n\times n}\left(\mathbb{C}\right)\to\mathfrak{M}_{k\times k}\left(\mathbb{C}\right)$.

This inequality was generalized later  by Davis  in [16], where he obtained that this is true when $f$ is a matrix convex function and $\Phi$  is completely positive; i.e.,

[TABLE]

The latter restriction about complete positivity of $\Phi$  was removed by Choi [15] who proved that \reftagform@1.7  remains valid for all positive unital linear maps   provided $f$ is matrix convex.

Another noncommutative operator version of the classical Jensen’s inequality under the situation that

[TABLE]

then the classic Jensen’s inequality is expressed as

[TABLE]

The inequality \reftagform@1.9 was proved by Davis in [16] for all $A\in\mathscr{B}\left(\mathscr{H}\right)$ and every isometry $C$. However, a more informative version was extended by Hansen-Pedersen [19] as follows:

Theorem 1**.**

Let $\mathscr{H}$ and $\mathscr{K}$ be Hilbert space. Let $f$ be a real valued continuous function on an interval $I$. Let $A$ and $A_{j}$ be selfadjoint operators on $\mathscr{H}$ with spectra contained in $I$ $(j=1,2,\cdots,k)$. Then the following conditions are mutually

1. (1)

$f$* is operator convex on $I$ and $f(0)\leq 0$.* 2. (2)

$f\left({C^{*}AC}\right)\leq C^{*}f\left(A\right)C$, for every $A\in\mathscr{B}\left(\mathscr{H}\right)$ and contraction $C\in\mathscr{B}\left(\mathscr{H}\right)$; i.e., $C^{*}C\leq 1_{\mathscr{K}}$. 3. (3)

$f\left({\sum\limits_{j=1}^{n}{C_{j}^{*}A_{j}C_{j}}}\right)\leq\sum\limits_{j=1}^{n}{C_{j}^{*}f\left({A_{j}}\right)C_{j}}$, for all $A_{j}\in\mathscr{B}\left(\mathscr{H}\right)$ and $C_{j}\in\mathscr{B}\left(\mathscr{H}\right)$ with $\sum_{j=1}^{n}C_{j}^{*}C_{j}\leq 1_{\mathscr{H}}$, $(j=1,2,\cdots,k)$. 4. (4)

$f\left({PAP}\right)\leq Pf\left(A\right)P$, for every $A\in\mathscr{B}\left(\mathscr{H}\right)$ and  projection $P$.

Here we give some popular examples of operator convex and concave function [14].

1. (1)

For each $p\in\left[0,1\right]$, $t^{p}$    is operator concave on $\left[0,\infty\right)$. 2. (2)

The function $t\log t$  is operator convex on $\left[0,\infty\right)$.

This work is organized as follows: after this introduction; in Section 2, the operator superquadratic functions for positive Hilbert space operators are introduced and elaborated. Several examples with some important properties together with some observations related to operator convexity are pointed out. In Section 3, A Jensen type inequality is proved. Equivalent statements of a non-commutative version in of Jensen’s inequality for operator superquadratic are also established.   Finally, several trace inequalities for superquadratic functions (in ordinary sense) are provided as well.

2. Operator superquadratic function

Definition 1**.**

Let $I=[0,M]\subseteq\left[0,\infty\right)$.      A real valued continuous function $f(t)$ on an interval $I$ is said to be operator superquadratic function if

[TABLE]

holds for all $\alpha\in\left[0,1\right]$ and for every positive   operators $A$ and $B$ on a Hilbert space $\mathscr{H}$ whose spectra are contained in $I\subset[0,\infty)$.  We say that $f$ is operator subquadratic function if $-f$ is operator superquadratic function. Moreover, if the equality holds in \reftagform@2.1, in this case  we say that $f$ is operator quadratic function.

It’s convenient to note that; if $f$ satisfies \reftagform@2.1, then with $A=x$ and $B=y$ (two positive scalars) one can obtain the Jensen inequality for superquadratic functions and if $f$ is continuous (which is necessary to define an operator functions), then \reftagform@1.4 would imply that $f$ is superquadratic function. Thus, we observe that:

Corollary 1**.**

If $f$ is an operator superquadratic function then $f$ is a real superquadratic function.

Let $f(t)=\alpha t+\beta$,  then $f$ is  operator subquadratic on every bounded interval for all $\alpha,\beta\geq 0$. Indeed, we have

[TABLE]

Moreover,  $g(t)=-f(t)$ is operator superquadratic.

One can easily seen that the function $t\mapsto t^{3}$ is not operator superquadratic nor operator subquadratic function. Simply, assume $f(t)=t^{3}$,  $t\in[0,\infty)$ and let

[TABLE]

then,

[TABLE]

However, the map $t\mapsto t^{2}$ is non-negative operator convex  on $(0,\infty)$ and it is also  operator superquadratic  on $(0,\infty)$. Indeed,  by \reftagform@2.1 we have

[TABLE]

which is true  since $A,B>0$, and this proves that $t^{2}$ is operator superquadratic function.

From the definition of operator superquadratic function  we have

[TABLE]

for any arbitrary positive  operators $A,B\in\mathscr{B}\left(\mathscr{H}\right)$ and each $\alpha\in\left[0,1\right]$.

In particular, by   setting $B=\left\langle{Ax,x}\right\rangle 1_{\mathscr{H}}$ in \reftagform@2.1 we have

[TABLE]

for each positive   operator $A\in\mathscr{B}\left(\mathscr{H}\right)$ and all $\alpha\in\left[0,1\right]$.

From this point of view \reftagform@2.3, Kian early in [24] and then jointly  with Dragomir in [25]  proved   a finite dimensional operator version of Jensen’s inequality for superquadratic functions (in ordinary sense) under the interpretation that for A=\left({\begin{array}[]{*{20}c}a&0\\ 0&b\\ \end{array}}\right) and x=\left({\begin{array}[]{*{20}c}{\sqrt{\lambda}}\\ {\sqrt{1-\lambda}}\\ \end{array}}\right), then we have $\left\langle{Ax,x}\right\rangle =\lambda a+\left({1-\lambda}\right)b$ if follows that

[TABLE]

Therefore, as a matrix Jensen inequality for a superquadratic function $f:\left[0,\infty\right)\to\mathbb{R}$ we have

[TABLE]

This result was generalized for positive unital linear maps, as follows:

Theorem 2**.**

([25], [7])* Let $A\in\mathscr{B}\left(\mathscr{H}\right)$ be a positive  operator and  $\Phi:\mathscr{B}\left(\mathscr{H}\right)\to\mathscr{B}\left(\mathscr{K}\right)$  be a positive unital linear map. If   $f:\left[0,\infty\right)\to\mathbb{R}$ is super(sub)quadratic function, then we have*

[TABLE]

for every $x\in\mathscr{K}$ with $\|x\|=1$.

The above inequality and other consequences were proved later by the first author of this paper in [7] where different approach is used.

Proposition 1**.**

Let $f$ be an operator superquadratic function on $I$. Then

1. (1)

$f\left(0\right)\leq 0$. 2. (2)

If $f$ is non-negative, then $f$ is operator convex   and $f(0)=0$.

Proof.

1. (1)

Setting $A=B=0$ in \reftagform@2.3 we ge that $f\left(0\right)\leq 0$. 2. (2)

Since $f$ is continuous and non-negative, then from \reftagform@2.3 we have

[TABLE]

which means that $f$ is operator convex. To show that  $f(0)=0$, we have by part (1) $f(0)\leq 0$ and by assumption $f(x)$ is non-negative i.e., $f(x)\geq 0$ for all $x\in I$. In particular, $f(0)\geq 0$. Thus, $f(0)=0$.

∎

Example 1**.**

Let $f(t)=t^{-1}$, then $f$ is non-negative operator convex  on $(0,\infty)$. However, $f$ is not operator superquadratic function on $(0,\infty)$. For instance, let

[TABLE]

Applying  \reftagform@LABEL:eq2.4 for $f(t)=t^{-1}$, we get

[TABLE]

Proposition 2**.**

Let $f$ be a real valued continuous function defined on an interval $\left[0,\infty\right)$. If $f$ is operator  convex  and non-positive then $f$ is operator superquadratic function.

Proof.

Since $f$ is operator convex, then

[TABLE]

But also $f$ is non-positive, so that

[TABLE]

which means that $f$ is operator superquadratic function. ∎

Example 2**.**

Let $f\left(t\right)=t\log\left(t\right)$, $t\in\left[0,\infty\right)$ it well known that $f$ operator convex. Clearly, $f$ is negative for all $t\in\left(0,1\right)\subseteq\left[0,\infty\right)$. Therefore,    $f\left(t\right)=t\log\left(t\right)$ is  operator superquadratic function for all $t\in\left(0,1\right)$.

Proposition 3**.**

Let $f$ be a real valued continuous function defined on an interval $\left[0,\infty\right)$. If $f$ is operator  concave  and non-negative then $f$ is operator subquadratic.

Proof.

Since $f$ is operator concave, then

[TABLE]

But also $f$ is non-negative, so that

[TABLE]

which means that $f$ is operator subquadratic. ∎

Example 3**.**

Let $f:\left(0,\infty\right)\to\left(0,\infty\right)$, given by $f(t)=t^{r}$, $r\in\left[0,1\right]$. Then $f$ is   operator subquadratic  on $(0,\infty)$. But $f$ is also operator concave, so that

[TABLE]

which means $f$ is operator subquadratic  on $(0,\infty)$.

3. Operator Jensen’s inequality

In order to prove our results we need the following Lemmas:

Lemma 4**.**

([17])* If $A\in\mathscr{B}\left(\mathscr{H}\right)$ is selfadjoint and $U$ is unitary, i.e. $U^{*}U=UU^{*}=1_{\mathscr{H}}$, then $f(U^{*}AU)=U^{*}f(A)U$ for every $f$ continuous on the $\operatorname{sp}(A)$.*

Lemma 5**.**

([18])* Define a unitary matrix $E_{n}=\operatorname{diag}\left(\xi,\xi^{2},\cdots,\xi^{n-1},1\right)$ in $\mathfrak{M}_{n}\left(\mathbb{C}\right)\subset\mathscr{B}\left(\mathscr{H}^{n}\right)$, where $\xi=\exp\left(\frac{2\pi i}{n}\right)$. Then for each element $A=\left(a_{ij}\right)\in \mathscr{B}\left(\mathscr{H}^{n}\right)$ we have*

[TABLE]

Lemma 6**.**

([18])* Let $P$ denote the projection in $\mathfrak{M}_{n}$ given by $P_{ij}=n^{-1}$ for all $i$ and $j$, so that $P$ is the projection of rank one on the subspace spanned by the vector $\xi+\xi^{2}+\cdots+\xi^{n}$ in $\mathbb{C}^{n}$, where $\xi,\xi^{2},\cdots,\xi^{n}$ are the standard basis vectors. Then with $E$ as in Lemma 5 we obtain the pairwise orthogonal projections $P_{k}=E^{-k}PE^{k}$, for $1\leq k\leq n$, with $\sum_{k=1}^{n}{P_{k}}=1$.*

To establish our main first result we need the following primary result.

Lemma 7**.**

Let $w_{1},\ldots,w_{n}$ be positive real numbers  such that $W_{n}=\sum_{k=1}^{n}{w_{k}}$ and let $A_{1},\cdots,A_{n}$ be positive  operators of a Hilbert space $\mathscr{B}\left(\mathscr{H}\right)$ with spectra contained in a real interval $I$. If $f$ is operator superquadratic function  on $I$, then

[TABLE]

In particular useful case,  for $w_{k}=1$ for all $1\leq k\leq n$, we have

[TABLE]

Proof.

Assume $f$ is operator superquadratic. If $n=2$, then the inequality \reftagform@3.1 reduces to \reftagform@2.1 with $\alpha=\frac{w_{1}}{W_{2}}$ and $1-\alpha=\frac{w_{2}}{W_{2}}$. Let us suppose that inequality \reftagform@3.1 holds for $n-1$. Then for $n$-tuples $\left(A_{1},\cdots,A_{n}\right)$ and $\left(w_{1},\cdots,w_{n}\right)$, we have

[TABLE]

and this is exactly equivalent to write, for any $1\leq m\leq n$

[TABLE]

which proves the desired result in \reftagform@3.1. The particular case follows by setting $w_{k}=1$ for all $k=1,\cdots,n$ so that $W_{n}=n$. ∎

Remark 2**.**

The result in Lemma 7 was proved by Mond & Pečarić in [28] for all operator convex functions and all bounded selfdjoint operators whose spectra contained in $I$. Therefore, in case $f$ is positive the inequality \reftagform@3.1 might be considered as a respective extension and new refinement of that result proved in [28].

Theorem 3**.**

Let $f:I\to\mathbb{R}$ be a real-valued continuous function. Let $\left(A_{1},\cdots,A_{n}\right)$ be an $n$-tuple of positive   of a Hilbert space $\mathscr{H}$ with spectra contained in $I$. Then the following conditions are equivalent:

1. (1)

$f$* is operator superquadratic function.* 2. (2)

The inequality

[TABLE]

*holds for every $n$-tuple $\left(C_{1},\cdots,C_{n}\right)$ of operators on     $\mathscr{H}$ that satisfy the condition $\sum\limits_{k=1}^{n}{C_{k}^{*}C_{k}} =1$. * 3. (3)

The inequality

[TABLE]

holds   for every $n$-tuple  $\left(P_{1},\cdots,P_{n}\right)$ of projections on $\mathscr{H}$ with $\sum\limits_{k=1}^{n}{P_{k}} =1$.

Proof.

$(1)\Rightarrow(2)$.   We say that $C=\left(C_{1},\cdots,C_{n}\right)$ is a unitary column if there is a unitary $n\times n$ operator matrix $U=(u_{ij})$, one of whose columns is $(C_{1},\cdots,C_{n})$. Thus, $u_{ij}=C_{i}$ for some $j$ and all $i$.  Assume that we are given a unitary $n$-column $(C_{1},\cdots,C_{n})$, and choose a unitary $U_{n}=(u_{ij})$ in $\mathscr{B}\left(\mathscr{H}^{n}\right)$ such that $u_{kn}=C_{k}$. Let $E=\operatorname{diag}(\xi,\xi^{2},\cdots,\xi^{n-1},1)$ as in Lemma 4 and put $X=\operatorname{diag}\left(A_{1},\cdots,A_{n}\right)$, both regarded as elements in $\mathscr{B}\left(\mathscr{H}^{n}\right)$. Thus, using   the spectral decomposition theorem,  we have

[TABLE]

We note that since  $f\left(\operatorname{diag}\left(y_{1},\cdots,y_{n}\right)\right)=\operatorname{diag}\left(f\left(y_{1}\right),\cdots,f\left(y_{n}\right)\right)$, then

[TABLE]

Using the above facts taking into account Lemmas 4–7 together with the inequality \reftagform@3.2, thus  the operator superquadraticity of $f$, implies that

[TABLE]

It remains to mention that,   when  the column is just unital, we extend it to the unitary  $\left(n+1\right)$-column $\left(C_{1},\cdots,C_{n},0\right)$ and choose $A_{n+1}$ arbitrarily, but with spectrum in $I$, (see [8]). By the first part of the proof we therefore have

[TABLE]

and thus the proof of the statement (2) is completely established.

$(2)\Rightarrow(3)$.  Hold.

$(3)\Rightarrow(1)$. Let $A$ and $B$ be positive and bounded linear operators with spectra in $I$ and $0\leq\lambda\leq 1$.

Consider

[TABLE]

[TABLE]

Then $C$ and $D$ are unitary operators on $H\oplus H$. We have

[TABLE]

[TABLE]

[TABLE]

Thus, we have

[TABLE]

Hence, $f$ is operator superquadratic  on $I$ by seeing the $(1,1)$-components.

∎

Remark 3**.**

An operator convex version of Theorem  3 were proved by Hansen & Pedersen in [18]. Therefore, in case $f$ is positive the inequality \reftagform@3.3 could be considered as a new refinement of that result proved in [18].

A refinement of the classical Jensen’s inequality \reftagform@1.9 could be elaborated as follows:

Corollary 2**.**

Let $f:I\to\mathbb{R}$ be a real-valued continuous function. Let $A$ be a positive operator of a Hilbert space $\mathscr{H}$ with spectra contained in $I$. If $f$ is an operator superquadratic function, then the inequality

[TABLE]

holds for every operator $C$ on $\mathscr{H}$ that satisfy the condition $C^{*}C =1$.

Proof.

Follows from Theorem 3 by setting $n=1$. ∎

Remark 4**.**

Let $f:I\to\mathbb{R}$ be a real-valued continuous function. Let $A$ be a positive operator of a Hilbert space $\mathscr{H}$ with spectra contained in $I$. If $f$ is an operator subquadratic function, then the inequality

[TABLE]

holds for every operator $C$ on $\mathscr{H}$ that satisfy the condition $C^{*}C =1$.  Furthermore, by applying the subquadratic function $f(t)=t^{r}$, $t>0$ $(r\in[0,1])$, then we have

[TABLE]

for all $r\in\left[0,1\right]$.

A generalization of \reftagform@3.5 (also, \reftagform@1.7 and \reftagform@1.8) for any positive unital linear map  between two Hilbert spaces having the same dimension is embodied in the following result.

Theorem 4**.**

Let  $\mathscr{H},\mathscr{K}$   be two Hilbert spaces such that ${\rm{dim}}\mathscr{H}={\rm{dim}}\mathscr{K}$.  Let $f:J\to\mathbb{R}$ be a real-valued continuous function. Let $A$ be a positive  of a Hilbert space $\mathscr{H}$ with spectra contained in $J$, and consider  $\Phi:\mathscr{B}\left(\mathscr{H}\right)\to\mathscr{B}\left(\mathscr{K}\right)$ be a positive unital linear map. If $f$ is operator superquadratic function, then the inequality

[TABLE]

holds. If  $f$ is operator subquadratic, then the inequality \reftagform@3.6 is reversed. Thus, the following refinement of \reftagform@1.7

[TABLE]

is valid.

Proof.

Let $A\in\mathscr{B}\left(\mathscr{H}\right)$ be positive. Assume that $\mathcal{A}$ is the $C^{*}$-subalgebra of $\mathscr{B}\left(\mathscr{H}\right)$ generated by $A$ and $1_{\mathscr{H}}$. Without loss of generality, we may assume that $\Phi$ is defined on $\mathcal{A}$. Since every unital positive linear map on a commutative $C^{*}$-algebra is completely positive. It follows   that $\Phi$ is completely positive. So there exists  (by Stinespring’s theorem [29]), some isometry $V:\mathscr{H}\to\mathscr{K}$; and a unital $*$-homomorphism $\rho$ from $\mathcal{A}$ into the $C^{*}$-algebra $\mathscr{B}\left(\mathscr{H}\right)$ such that $\Phi(A)=V^{*}\rho(A)V$. Clearly, $f(\rho(A))=\rho(f(A))$, for all continuous function $f$. Thus,

[TABLE]

which proves the required inequality. The last inequality holds by applying \reftagform@3.6 to the superquaratic function $f(t)=t^{2}$, $\forall t>0$.

∎

The inequality \reftagform@3.6 can be embodied in multiple versions as stated in the following result.

Corollary 3**.**

Let    $\mathscr{H},\mathscr{K}$   be two Hilbert spaces such that ${\rm{dim}}\mathscr{H}={\rm{dim}}\mathscr{K}$.  Let $f:\left[0,\infty\right)\to\mathbb{R}$ be a real-valued continuous function. and consider  $\Phi_{k}:\mathscr{B}\left(\mathscr{H}\right)\to\mathscr{B}\left(\mathscr{K}\right)$  $(k=1,\cdots,n)$ be a positive linear mappings with $\sum\limits_{k=1}^{n}{\Phi_{k}\left(1_{\mathscr{H}}\right)} =1_{\mathscr{K}}$. Then,  $f$ is operator superquadratic function if and only if

[TABLE]

for all positive operators $A_{1}\cdots,A_{n}$ in $\mathscr{B}\left(\mathscr{H}\right)$.

Proof.

The proof is obvious. ∎