New inequalities for operator convex functions

Abstract

The aim of this paper is to present some new inequalities of Hermite-Hadamard type inequalities for operator convex functions. In this paper, we use elementary operations and give some inequalities related to the Hermite-Hadamard type. We conclude that the results given in this work are the generalization of the recent results.

MSC:26D15, 47A63.

1 Introduction

Let $f:[a,b]⟶\mathbb{R}$ be a convex function, then the inequality

$$f\left(\frac{a+b}{2}\right)\le \frac{1}{b-a}{\int }_a^{b}f(x)dx\le \frac{f(a)+f(b)}{2},a,b\in \mathbb{R},$$

(1)

is known in the literature as the Hermite-Hadamard inequality (see [1, 2] for more information).

Let X be a vector space, $x,y\in X$, $x\ne y$ and $[x,y]=\{(1-t)x+ty,t\in [0,1]\}$. We consider the function $f:[x,y]⟶\mathbb{R}$ and the associated function

$$g(x,y):[0,1]⟶\mathbb{R},g(x,y)(t):=f[(1-t)x+ty],t\in [0,1].$$

Note that f is convex on $[x,y]$ if and only if $g(x,y)$ is convex on $[0,1]$.

For any convex function defined on a segment $[x,y]\subset X$, we have the Hermite-Hadamard integral inequality

$$f\left(\frac{x+y}{2}\right)\le {\int }_0^{1}f[(1-t)x+ty]dt\le \frac{f(x)+f(y)}{2},$$

(2)

which can be derived from the classical Hermite-Hadamard inequality (1) for the convex function $g(x,y):[0,1]⟶\mathbb{R}$.

A real-valued continuous function f on an interval I is said to be operator convex (operator concave) if

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

in the operator order for all $\lambda \in [0,1]$ and for every self-adjoint operator A and B on a Hilbert space H whose spectra are contained in I. Notice that a function f is operator concave if −f is operator convex.

In recent years, many authors have been interested in giving some refinements and extensions of the Hermite-Hadamard inequality in (1). For more about convex functions and the Hermite-Hadamard inequality, see [3–6].

The author in [7] shows some new integral inequalities analogous to the well-known Hermite-Hadamard inequality. We give a general form of the second of these inequalities and show that the inequalities therein are satisfied for operator convex functions.

The author in [8] shows some new Hermite-Hadamard inequalities similar to Pachpatte’s results.

Pachpatte (2003) gives some integral inequalities analogous to the well-known Hermite-Hadamard inequality by using a fairly elementary analysis in [7].

Theorem 1 Let f and g be real-valued, nonnegative and convex functions on $[a,b]$. Then

• (i)
  $$\frac{1}{b-a}{\int }_a^{b}f(x)g(x)dx\le \frac{1}{3}M(a,b)+\frac{1}{6}N(a,b),$$

  (3)
• (ii)
  $$2f\left(\frac{a+b}{2}\right)g\left(\frac{a+b}{2}\right)\le \frac{1}{b-a}{\int }_a^{b}f(x)g(x)dx+\frac{1}{6}M(a,b)+\frac{1}{3}N(a,b),$$

  (4)

where $M(a,b)=f(a)g(a)+f(b)g(b)$, $N(a,b)=f(a)g(b)+f(b)g(a)$.

Tunç (2012) gives an inequality for convex functions in [8] as follows.

Theorem 2 Let $f,g:[a,b]⟶\mathbb{R}$ be two convex functions. Then

(5)

where $M(a,b)=f(a)g(a)+f(b)g(b)$, $N(a,b)=f(a)g(b)+f(b)g(a)$.

Tunç (2012) gives another inequality for convex functions in [8], too.

Theorem 3 Let $f,g:[a,b]⟶\mathbb{R}$ be two convex functions. Then

$$\begin{array}{r}\frac{1}{b-a}{\int }_a^b(f\left(\frac{a+b}{2}\right)g(x)+g\left(\frac{a+b}{2}\right)f(x))dx\\ \le \frac{1}{2(b-a)}{\int }_a^{b}f(x)g(x)dx+\frac{1}{12}M(a,b)+\frac{1}{6}N(a,b)+f\left(\frac{a+b}{2}\right)g\left(\frac{a+b}{2}\right),\end{array}$$

(6)

where $M(a,b)=f(a)g(a)+f(b)g(b)$, $N(a,b)=f(a)g(b)+f(b)g(a)$.

Ghazanfari (2012) gives an inequality for two operator convex functions in [9] as follows.

Theorem 4 Let $f,g:I⟶\mathbb{R}$ be operator convex functions on the interval I. Then for any self-adjoint operators A and B on a Hilbert space H with spectra in I, the inequality

(7)

holds for any $x\in H$ with $\parallel x\parallel =1$, where

For further inequalities, see [10–12].

2 Main results

In this section, we give some new Hermite-Hadamard type inequalities for operator convex functions and mention the differences related to the results in recent papers. We emphasize the difference by giving an example.

The following theorem is a generalization for the product of two operator convex functions.

Theorem 5 Let $f,g:I⟶\mathbb{R}$ be operator convex, nonnegative functions on the interval I. Then for any self-adjoint operators A and B with spectra in I, we have the inequality

(8)

where

(9)

(10)

and k is the number of steps.

Proof Let $x\in H$, $\parallel x\parallel =1$ and A, B be two self-adjoint operators with spectra in I. Using the convexity of f, g and the change of variable $u=kt$, we have

$$\begin{array}{rl}〈f((1-t)A+tB)x,x〉& =〈f((1-\frac{u}{k})A+\frac{u}{k}B)x,x〉\\ =〈f((1-u)A+u\frac{(k-1)A+B}{k})x,x〉\\ \le (1-u)〈f(A)x,x〉+u〈f\left(\frac{(k-1)A+B}{k}\right)x,x〉\end{array}$$

(11)

and

$$\begin{array}{rl}〈f(tA+(1-t)B)x,x〉& =〈f(\frac{u}{k}A+(1-\frac{u}{k})B)x,x〉\\ =〈f(u\frac{A+(k-1)B}{k}+(1-u)B)x,x〉\\ \le u〈f\left(\frac{A+(k-1)B}{k}\right)x,x〉+(1-u)〈f(B)x,x〉.\end{array}$$

(12)

By the change of variable $u=kt-1$, we have

$$\begin{array}{rcl}〈f((1-t)A+tB)x,x〉& =& 〈f((1-\frac{u+1}{k})A+\frac{u+1}{k}B)x,x〉\\ =& 〈f((1-u)\frac{(k-1)A+B}{k}+u\frac{(k-2)A+2B}{k})x,x〉\\ \le & (1-u)〈f\left(\frac{(k-1)A+B}{k}\right)x,x〉+u〈f\left(\frac{(k-2)A+2B}{k}\right)x,x〉\end{array}$$

and

$$\begin{array}{rcl}〈f(tA+(1-t)B)x,x〉& =& 〈f(\frac{u+1}{k}A+(1-\frac{u+1}{k})B)x,x〉\\ =& 〈f(u\frac{2A+(k-2)B}{k}+(1-u)\frac{A+(k-1)B}{k})x,x〉\\ \le & u〈f\left(\frac{2A+(k-2)B}{k}\right)x,x〉+(1-u)〈f\left(\frac{A+(k-1)B}{k}\right)x,x〉.\end{array}$$

Similarly, by using the change of variables $u=kt-2,u=kt-3,\dots ,u=kt-(k-2)$, we have some inequalities. By the change of variable $u=kt-(k-1)$, we get

$$\begin{array}{rcl}〈f((1-t)A+tB)x,x〉& =& 〈f((1-\frac{u+k-1}{k})A+\frac{u+k-1}{k}B)x,x〉\\ =& 〈f((1-u)\frac{A+(k-1)B}{k}+uB)x,x〉\\ \le & (1-u)〈f\left(\frac{A+(k-1)B}{k}\right)x,x〉+u〈f(B)x,x〉\end{array}$$

and

$$\begin{array}{rcl}〈f(tA+(1-t)B)x,x〉& =& 〈f(\frac{u+k-1}{k}A+(1-\frac{u+k-1}{k})B)x,x〉\\ =& 〈f(uA+(1-u)\frac{(k-1)A+B}{k}x,x)〉\\ \le & u〈f(A)x,x〉+(1-u)〈f\left(\frac{(k-1)A+B}{k}\right)x,x〉.\end{array}$$

Using the convexity of f, g, we have

$$\begin{array}{rl}〈f\left(\frac{A+B}{2}\right)x,x〉& =〈f(\frac{tA+(1-t)B}{2}+\frac{(1-t)A+tB}{2})x,x〉\\ \le \frac{〈f(tA+(1-t)B)x,x〉+〈f((1-t)A+tB)x,x〉}{2}\end{array}$$

(13)

and

$$\begin{array}{rl}〈g\left(\frac{A+B}{2}\right)x,x〉& =〈g(\frac{tA+(1-t)B}{2}+\frac{(1-t)A+tB}{2})x,x〉\\ \le \frac{〈g(tA+(1-t)B)x,x〉+〈g((1-t)A+tB)x,x〉}{2}.\end{array}$$

(14)

Firstly, if we write the values obtained from the change of variable $u=kt$ in (13) and (14), we get

$$\begin{array}{rcl}〈f\left(\frac{A+B}{2}\right)x,x〉& \le & \frac{〈f(tA+(1-t)B)x,x〉+〈f((1-t)A+tB)x,x〉}{2}\\ =& \frac{〈f(u\frac{A+(k-1)B}{k}+(1-u)B)x,x〉+〈f((1-u)A+u\frac{(k-1)A+B}{k})x,x〉}{2}\end{array}$$

(15)

and

$$\begin{array}{r}〈g\left(\frac{A+B}{2}\right)x,x〉\\ \le \frac{〈g(tA+(1-t)B)x,x〉+〈g((1-t)A+tB)x,x〉}{2}\\ =\frac{〈g(u\frac{A+(k-1)B}{k}+(1-u)B)x,x〉+〈g((1-u)A+u\frac{(k-1)A+B}{k})x,x〉}{2}.\end{array}$$

(16)

If we multiply (15) and (16) and suppose $(1-u)A+u\frac{(k-1)A+B}{k}=X_1$ and $u\frac{A+(k-1)B}{k}+(1-u)B=Y_1$, we get

(17)

If we integrate both sides of inequality (17) over $[0,1]$, we reach

$$\begin{array}{r}〈f\left(\frac{A+B}{2}\right)x,x〉〈g\left(\frac{A+B}{2}\right)x,x〉\\ \le \frac{k}{4}\left[{\int }_0^{1/k}〈f(tA+(1-t)B)x,x〉〈g(tA+(1-t)B)x,x〉dt\right]\\ +\frac{k}{4}\left[{\int }_0^{1/k}〈f((1-t)A+tB)x,x〉〈g((1-t)A+tB)x,x〉dt\right]\\ +\frac{1}{24}[〈f\left(\frac{A+(k-1)B}{k}\right)x,x〉〈g(A)x,x〉+〈f(B)x,x〉〈g\left(\frac{(k-1)A+B}{k}\right)x,x〉]\\ +\frac{1}{12}[〈f\left(\frac{A+(k-1)B}{k}\right)x,x〉〈g\left(\frac{(k-1)A+B}{k}\right)x,x〉+〈f(B)x,x〉〈g(A)x,x〉]\\ +\frac{1}{24}[〈f(A)x,x〉〈g\left(\frac{A+(k-1)B}{k}\right)x,x〉+〈f\left(\frac{(k-1)A+B}{k}\right)x,x〉〈g(B)x,x〉]\\ +\frac{1}{12}[〈f(A)x,x〉〈g(B)x,x〉+〈f\left(\frac{(k-1)A+B}{k}\right)x,x〉〈g\left(\frac{A+(k-1)B}{k}\right)x,x〉].\end{array}$$

If we continue the same operations as above until the change of variable $u=kt-(k-1)$, we have some inequalities. And then, if we sum these obtained inequalities, we get the desired inequality. □

Remark 6 In inequality (8), if we take $k=1$, we get the inequality in (7).

Now, we show the comparison between Theorems 4 and 5 utilizing self-adjoint operators (Hermitian matrices) as follows.

Example 7 Let $A=\left[\begin{array}{cc}1& 0\\ 0& 2\end{array}\right]$, $B=\left[\begin{array}{cc}-0.4& 1\\ 1& 1\end{array}\right]$. Let our operator convex functions be $f(X)=X^2$ and $g(X)=X$. Since $x\in H$ and $\parallel x\parallel =1$, then we can choose x as $x=\left[\begin{array}{c}1\\ 0\end{array}\right]$. From the information given above, for $k=3$, Theorem 5 gives

$$\begin{array}{r}\left(x^{\ast }f\left(\frac{A+B}{2}\right)x\right)\left(x^{\ast }g\left(\frac{A+B}{2}\right)x\right)\\ \le \frac{1}{2}{\int }_0^1〈f(tA+(1-t)B)x,x〉〈g(tA+(1-t)B)x,x〉dt\\ +\frac{1}{24}[(x^{\ast }f(A)x)\left(x^{\ast }g\left(\frac{A+B}{2}\right)x\right)+(x^{\ast }f(B)x)\left(x^{\ast }g\left(\frac{A+B}{2}\right)x\right)\\ +\left(x^{\ast }f\left(\frac{A+B}{2}\right)x\right)(x^{\ast }f(A)x)+\left(x^{\ast }f\left(\frac{A+B}{2}\right)x\right)(x^{\ast }g(B)x)]\\ +\frac{1}{12}[(x^{\ast }f(A)x)(x^{\ast }g(B)x)+2\left(x^{\ast }f\left(\frac{A+B}{2}\right)x\right)\left(x^{\ast }g\left(\frac{A+B}{2}\right)x\right)\\ +(x^{\ast }f(B)x)(x^{\ast }g(A)x)].\end{array}$$

Putting the values of the functions in the above inequality, we get

$$\begin{array}{r}0,102\le \frac{1}{2}{\int }_0^1〈f(tA+(1-t)B)x,x〉〈g(tA+(1-t)B)x,x〉dt+0.1158\\ ⟹{\int }_0^1〈f(tA+(1-t)B)x,x〉〈g(tA+(1-t)B)x,x〉dt\ge -0.0276.\end{array}$$

Theorem 4 gives

$$\begin{array}{rcl}\left(x^{\ast }f\left(\frac{A+B}{2}\right)x\right)\left(x^{\ast }g\left(\frac{A+B}{2}\right)x\right)& \le & \frac{1}{2}{\int }_0^1〈f(tA+(1-t)B)x,x〉〈g(tA+(1-t)B)x,x〉dt\\ +\frac{1}{12}[(x^{\ast }f(A)x)(x^{\ast }g(A)x)+(x^{\ast }f(B)x)(x^{\ast }g(B)x)]\\ +\frac{1}{6}[(x^{\ast }f(A)x)(x^{\ast }g(B)x)+(x^{\ast }f(B)x)(x^{\ast }g(A)x)].\end{array}$$

Putting the values of the functions in the above inequality, we obtain

$$\begin{array}{r}0,102\le \frac{1}{2}{\int }_0^1〈f(tA+(1-t)B)x,x〉〈g(tA+(1-t)B)x,x〉dt+0.1713\\ ⟹{\int }_0^1〈f(tA+(1-t)B)x,x〉〈g(tA+(1-t)B)x,x〉dt\ge -0.1386.\end{array}$$

So, we can conclude that our result, Theorem 5, is more strict than Theorem 4 in this case.

The following theorem is a lower bound for the product of two operator convex functions.

Theorem 8 Let $f,g:I⟶\mathbb{R}$ be operator convex, nonnegative functions on the interval I. Then for any self-adjoint operators A and B with spectra in I, we have the inequality

(18)

where

$$\begin{array}{r}M(A,B)=〈f(A)x,x〉〈g(A)x,x〉+〈f(B)x,x〉〈g(B)x,x〉,\\ N(A,B)=〈f(A)x,x〉〈g(B)x,x〉+〈f(B)x,x〉〈g(A)x,x〉.\end{array}$$

Proof Let $x\in H$, $\parallel x\parallel =1$ and A, B be two self-adjoint operators with spectra in I. Define the real-valued functions ${\phi }_{x,A,B}:[0,1]⟶\mathbb{R}$ given by ${\phi }_{x,A,B}(t)=〈f((1-t)A+tB)x,x〉$ and ${\psi }_{x,A,B}:[0,1]⟶\mathbb{R}$ given by ${\psi }_{x,A,B}(t)=〈g((1-t)A+tB)x,x〉$. Since f and g are operator convex functions, then for every $t\in [0,1]$, we have

(19)

(20)

If $a\le b$ and $c\le d$ for $a,b,c,d\in \mathbb{R}$, we have $ad+bc\le ac+bd$. Using this inequality analogous to (19) and (20), we get

(21)

Since ${\phi }_{x,A,B}(t)$ and ${\psi }_{x,A,B}(t)$ are operator convex on $[0,1]$, they are integrable on $[0,1]$ and consequently ${\phi }_{x,A,B}(t){\psi }_{x,A,B}(t)$ is also integrable on $[0,1]$. Integrating both sides of inequality (21) over $[0,1]$, we get

(22)

It can be easily controlled that

$${\int }_0^1{(1-t)}^{2}dt={\int }_0^1{t}^{2}dt=\frac{1}{3},{\int }_0^{1}t(1-t)dt=\frac{1}{6}.$$

When the above equalities are taken into account, the proof is complete. □

Remark 9 In inequality (18), if we take $x=(1-t)A+tB$, $a=0$ and $b=1$, we get the inequality in (5). Our result is more general than (5).

In Theorem 8, we give a lower bound. But now we give both lower and upper bounds for the product of two operator convex functions.

Theorem 10 Let $f,g:I⟶\mathbb{R}$ be operator convex, nonnegative functions on the interval I. Then for any self-adjoint operators A and B with spectra in I, we have the inequality

(23)

where $Z_1$ and $T_2$ are defined in (9) and (10) and k is the number of steps.

Proof Let $x\in H$, $\parallel x\parallel =1$ and A, B be two self-adjoint operators with spectra in I. Using the convexity of f, g and the change of variable $u=kt$, we have (11) and (12). Using the analogous condition that, if $a\le b$ and $c\le d$ for $a,b,c,d\in \mathbb{R}$, we have $ad+bc\le ac+bd$, we obtain

(24)

If we continue the same operations as above until the change of variable $u=kt-(k-1)$, we have some inequalities. And then, if we integrate the multiplication inequalities, we get k inequalities. These inequalities are defined on $[0,\frac{1}{k}),(\frac{1}{k},\frac{2}{k}),\dots ,(\frac{k-1}{k},1]$, respectively. The sum of the integration parts of these k inequalities yields ${\int }_0^1〈f((1-t)A+tB)x,x〉〈g((1-t)A+tB)x,x〉dt$. Thus, the proof is complete. □

Remark 11 Inequality (23) is a general form of inequality (18). When $k=1$ in inequality (23), we get inequality (18).

Theorem 12 Let $f,g:I⟶\mathbb{R}$ be operator convex, nonnegative functions on the interval I. Then for any self-adjoint operators A and B with spectra in I, we have the inequality

(25)

where $Z_1$, $Z_2$, $T_1$ and $T_2$ are defined in (9) and (10) and k is the number of steps.

Proof The proof is obvious from the proofs of Theorem 3 and Theorem 5. □

Remark 13 In Theorem 12, if we take $k=1$, we get (6). Theorem 12 is a generalization of Theorem 3. If we take k as the largest number we can take in Theorem 12, we near the exact solution.