The Hermite-Hadamard inequality for r-convex functions

Abstract

In this paper, we establish the Hermite-Hadamard inequality for r-convex functions. We prove that r-convexity implies s-convexity ($0\le r\le s$). As a result, we obtain a refinement of the Hermite-Hadamard inequality for an r-convex function ($0\le r\le 1$). We also investigate the Hermite-Hadamard inequality for the product of an r-convex function f and an s-convex function g.

MSC: 26D15, 26D10.

1 Introduction

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

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

(1)

is known as the Hermite-Hadamard inequality (see [1] for more information). Since then, some refinements of the Hermite-Hadamard inequality on convex functions have been extensively investigated by a number of authors (e.g., [2, 3] and [4]). In [5], the first author obtained a new refinement of the Hermite-Hadamard inequality for convex functions. The Hermite-Hadamard inequality was generalized in [6] to an r-convex positive function which is defined on an interval $[a,b]$. A positive function f is called r-convex on $[a,b]$, if for each $x,y\in [a,b]$ and $t\in [0,1]$,

$$f(tx+(1-t)y)\le \{\begin{array}{cc}{}^{\frac{1}{r}},\hfill & r\ne 0,\hfill \\ {}^t[f(y)]^{1-t},\hfill & r=0.\hfill \end{array}$$

It is obvious 0-convex functions are simply log-convex functions and 1-convex functions are ordinary convex functions. One should note that if f is r-convex in $[a,b]$, then $f^r$ is a convex function ($r>0$).

Some refinements of the Hadamard inequality for r-convex functions could be found in [7] and [8]. In [9], Bessenyei studied Hermite-Hadamard-type inequalities for generalized 3-convex functions. In [7], the authors showed that if f is r-convex in $[a,b]$ and $0<r\le 1$, then

$$\frac{1}{b-a}{\int }_a^{b}f(x)dx\le \frac{r}{r+1}{[f^r(a)+f^r(b)]}^{\frac{1}{r}}.$$

(2)

In this paper, first we show that if f is r-convex in $[a,b]$ and $r\ge 1$, then

$$\frac{1}{b-a}{\int }_a^{b}f(x)dx\le {\left[\frac{1}{2}(f^r(a)+f^r(b))\right]}^{\frac{1}{r}}.$$

(3)

In Theorem 2.3, we prove the following inequality for r-convex functions:

$$\frac{1}{b-a}{\int }_a^{b}f(x)dx\le \frac{r}{r+1}\cdot \frac{f^{r+1}(b)-f^{r+1}(a)}{f^r(b)-f^r(a)}(r>0).$$

(4)

The inequality (4) is an extension and refinement of (2) and (3). In Theorem 2.4, we show that r-convexity implies s-convexity ($0\le r\le s$). We employ this result in Theorem 2.6 and Corollary 2.7 to refine the Hermite-Hadamard inequality by r-convexity ($0\le r\le 1$). Finally, we generalize some results in [7] without using Minkowski’s inequality. Indeed, we obtain refinements for the product of an r-convex function f and an s-convex function g ($r,s\ge 0$).

2 Main results

Theorem 2.1 Let $f:[a,b]⟶(0,\mathrm{\infty })$ be r-convex and $r\ge 1$. Then the following inequality holds:

$$\frac{1}{b-a}{\int }_a^{b}f(x)dx\le \frac{1}{2}{\left[(f^r(a)+f^r(b))\right]}^{\frac{1}{r}}.$$

Proof Since $r\ge 1$, by Jensen’s inequality, we have

$${(\frac{1}{b-a}{\int }_a^{b}f(x)dx)}^r\le \frac{1}{b-a}{\int }_a^b{f}^r(x)dx.$$

By convexity of $f^r$ and the right side of (1), we obtain

$$\frac{1}{b-a}{\int }_a^b{f}^r(x)dx\le \frac{1}{2}(f^r(a)+f^r(b)).$$

Thus,

$$\frac{1}{b-a}{\int }_a^{b}f(x)dx\le {\left[\frac{1}{2}(f^r(a)+f^r(b))\right]}^r.$$

 □

Corollary 2.2 Let $f:[a,b]⟶(0,\mathrm{\infty })$ be a 1-convex function. Then

$$\frac{1}{b-a}{\int }_a^{b}f(x)dx\le \frac{1}{2}(f(b)+f(a)).$$

(5)

Theorem 2.3 Let $f:[a,b]⟶(0,\mathrm{\infty })$ be r-convex and $r\ge 0$. Then the following inequalities hold:

$$\frac{1}{b-a}{\int }_a^{b}f(x)dx\le \{\begin{array}{cc}\frac{r}{r+1}(\frac{f^{r+1}(b)-f^{r+1}(a)}{f^r(b)-f^r(a)}),\hfill & r\ne 0,\hfill \\ ln\frac{f(b)}{f(a)},\hfill & r=0.\hfill \end{array}$$

Proof First, let $r>0$. Since f is r-convex, for all $t\in [0,1]$, we have

$$f(ta+(1-t)b)\le {[t{f}^r(a)+(1-t)f^r(b)]}^{\frac{1}{r}}.$$

It is easy to observe that

$$\begin{array}{rcl}\frac{1}{b-a}{\int }_a^{b}f(x)dx& =& {\int }_0^{1}f(ta+(1-t)b)dt\\ \le & {\int }_0^1{[t{f}^r(a)+(1-t)f^r(b)]}^{\frac{1}{r}}dt\\ =& {\int }_0^1{[t(f^r(a)-f^r(b))+f^r(b)]}^{\frac{1}{r}}dt.\end{array}$$

By substitution $t(f^r(a)-f^r(b))+f^r(b)=z$, we obtain

$$\begin{array}{rcl}\frac{1}{b-a}{\int }_a^{b}f(x)dx& \le & \frac{1}{f^r(b)-f^r(a)}{\int }_{f^r(a)}^{f^r(b)}{z}^{\frac{1}{r}}dz\\ =& \frac{1}{f^r(b)-f^r(a)}\cdot \frac{1}{1+\frac{1}{r}}{\left[z^{1+\frac{1}{r}}\right]}_{f^r(a)}^{f^r(b)}\\ =& \frac{r}{r+1}\left(\frac{f^{r+1}(b)-f^{r+1}(a)}{f^r(b)-f^r(a)}\right).\end{array}$$

For $r=0$, we have

$$f(tx+(1-t)y)\le {[f(x)]}^t{[f(y)]}^{1-t}.$$

So,

$$\begin{array}{rcl}\frac{1}{b-a}{\int }_a^{b}f(x)dx& =& {\int }_0^{1}f(ta+(1-t)b)dt\\ \le & {\int }_0^1{[f(a)]}^t{[f(b)]}^{1-t}dt\\ =& f(b){\int }_0^1{\left[\frac{f(a)}{f(b)}\right]}^{t}dt\\ =& f(b){\left[\frac{f(a)}{f(b)}\right]}^{t}ln\frac{f(a)}{f(b)}{|}_0^1\\ =& [f(b)-f(a)]ln\frac{f(b)}{f(a)}.\end{array}$$

The proof is completed. □

With the hypotheses of Theorem 2.3, if $f(a)=f(b)$, its proving process shows that $\frac{1}{b-a}{\int }_a^{b}f(x)dx$ can be dominated by $f(a)$ where $r\ge 0$.

Note that if we put $r=1$ in Theorem 2.3, we can obtain again the inequality (5).

Theorem 2.4 Let $f:[a,b]⟶(0,\mathrm{\infty })$ be r-convex on $[a,b]$ and $0\le r\le s$. Then f is s-convex. In particular, if f is r-convex and $0\le r\le 1$, then f is convex.

In order to prove the above theorem, we need the following lemma.

Lemma 2.5 If $0\le \alpha \le 1$ and $0\le r\le s$, then the following inequalities hold for every pair of non-negative real numbers x and y:

$$x^{\alpha }{y}^{1-\alpha }\le {(\alpha x^r+(1-\alpha )y^r)}^{\frac{1}{r}}\le {(\alpha x^s+(1-\alpha )y^s)}^{\frac{1}{s}}.$$

(6)

Proof The left side of the inequality is clear by Young’s inequality. The right side is obvious if either x or y equals zero. So, let $x>0$ and $y>0$. Consider $f:[0,\mathrm{\infty })⟶R$ defined by

$$f(t)={(\alpha t^r+1-\alpha )}^s-{(\alpha t^s+1-\alpha )}^r.$$

Then $f^{\prime }(t)=rs\alpha [t^{r-1}{(\alpha t^r+1-\alpha )}^{s-1}-t^{s-1}{(\alpha t^s+1-\alpha )}^{r-1}]$. So, $t=1$ is a critical point of f. By an easy calculation, we see that $f^{″}(1)=rs\alpha (1-\alpha )(r-s)\le 0$. It follows that f attains its maximum at $t=1$. Thus, $f(t)\le f(1)=0$. This shows that

$${(\alpha t^r+1-\alpha )}^s\le {(\alpha t^s+1-\alpha )}^r.$$

Now, if we put $t=\frac{x}{y}$ in the above inequality, we get

$${(\alpha x^r+(1-\alpha )y^r)}^s\le {(\alpha x^s+(1-\alpha )y^s)}^r.$$

Therefore, we can deduce the right side of (6) by taking rs th root. □

Proof of Theorem 2.4 Since f is r-convex, by Lemma 2.5 for all $x,y\in [a,b]$ and $t\in [0,1]$, we have

$$f(tx+(1-t)y)\le \{\begin{array}{cc}t{[f^r(x)+(1-t)f^r(y)]}^{\frac{1}{r}}\le {[t{f}^s(x)+(1-t)f^s(y)]}^{\frac{1}{s}},\hfill & 0<r\le s,\hfill \\ {}^t[f(y)]^{1-t}\le {[t{f}^s(x)+(1-t)f^s(y)]}^{\frac{1}{s}},\hfill & 0=r\le s.\hfill \end{array}$$

Hence, f is s-convex. □

Theorem 2.6 Let $f:[a,b]⟶(0,\mathrm{\infty })$ be r-convex on $[a,b]$ and $0\le r\le s$. Then the following inequalities hold:

$$\frac{1}{b-a}{\int }_a^{b}f(x)dx\le \{\begin{array}{cc}\frac{r}{r+1}\frac{f^{r+1}(b)-f^{r+1}(a)}{f^r(b)-f^r(a)}\le \frac{s}{s+1}(\frac{f^{s+1}(b)-f^{s+1}(a)}{f^s(b)-f^s(a)}),\hfill & 0<r\le s,\hfill \\ ln\frac{f(b)}{f(a)}\le \frac{s}{s+1}(\frac{f^{s+1}(b)-f^{s+1}(a)}{f^s(b)-f^s(a)}),\hfill & 0=r<s.\hfill \end{array}$$

Proof The left side of the inequalities is clear by Theorem 2.3. For the right side, by the inequality in (6), we have

$${[t{f}^r(x)+(1-t)f^r(b)]}^{\frac{1}{r}}\le {[t{f}^s(a)+(1-t)f^s(b)]}^{\frac{1}{s}}.$$

By integrating it on $[0,1]$, we obtain

$${\int }_0^1[t{f}^r(a)+(1-t)f^r(b)]dt\le {\int }_0^1{[t{f}^s(a)+(1-t)f^s(b)]}^{\frac{1}{t}}dt.$$

Thus,

$$\frac{r}{r+1}\left(\frac{f^{r+1}(b)-f^{r+1}(a)}{f(b)-f(a)}\right)\le \frac{s}{s+1}\left(\frac{f^{s+1}(b)-f^{s+1}(a)}{f^s(b)-f^s(a)}\right).$$

Also, another inequality can be deduced by integrating the inequalities in (6) if we replace x and y by $f(x)$ and $f(y)$, respectively. □

Corollary 2.7 Let $f:[a,b]⟶(0,\mathrm{\infty })$ be r-convex and $0\le r\le 1$. Then

$$\begin{array}{rcl}f\left(\frac{a+b}{2}\right)& \le & \frac{1}{b-a}{\int }_a^{b}f(x)dx\le [f(b)-f(a)]ln\frac{f(b)}{f(a)}\\ \le & \frac{r}{r+1}\left(\frac{f^{r+1}(b)-f^{r+1}(a)}{f^r(b)-f^r(a)}\right)\le \frac{f(a)+f(b)}{2}.\end{array}$$

In other words, when f is r-convex and $0\le r\le 1$, we can refine the Hermite-Hadamard inequalities through Theorem  2.6.

Theorem 2.8 Let $f,g:[a,b]⟶(0,\mathrm{\infty })$ be r-convex and s-convex functions respectively on $[a,b]$ and $r,s>0$. Then the following inequality holds:

$$\begin{array}{rcl}\frac{1}{b-a}{\int }_a^{b}f(x)g(x)dx& \le & \frac{1}{2}\left(\frac{r}{r+2}\right)\left(\frac{f^{r+2}(b)-f^{r+2}(a)}{f^r(b)-f^r(a)}\right)\\ +\frac{1}{2}\left(\frac{s}{s+2}\right)\left(\frac{g^{s+2}(b)-g^{s+2}(a)}{g^s(b)-g^s(a)}\right)\\ (f(a)\ne f(b),g(a)\ne g(b)).\end{array}$$

Proof Since f is r-convex and g is s-convex, for all $t\in [0,1]$, we have

$$\begin{array}{c}f(ta+(1-t)b)\le {[t{f}^r(a)+(1-t)f^r(b)]}^{\frac{1}{r}},\hfill \\ g(ta+(1-t)b)\le {[t{g}^s(a)+(1-t)g^s(b)]}^{\frac{1}{s}}.\hfill \end{array}$$

Thus,

$$\begin{array}{rcl}\frac{1}{b-a}{\int }_a^{b}f(x)g(x)dx& =& {\int }_0^{1}f(ta+(1-t)b)g(ta+(1-t)b)dt\\ \le & {\int }_0^1{[t{f}^r(a)+(1-t)f^r(b)]}^{\frac{1}{r}}{[t{g}^s(a)+(1-t)g^s(b)]}^{\frac{1}{s}}dt\\ =& {\int }_0^1{[t(f^r(a)-f^r(b))+f^r(b)]}^{\frac{1}{r}}{[t(g^s(a)-g^s(b))+g^s(b)]}^{\frac{1}{s}}dt.\end{array}$$

Applying Cauchy’s inequality, we get

(7)

Similar to the proof of Theorem 2.3 and by substitution $t(f^r(a)-f^r(b))+f^r(b)=z$, we obtain

$${\int }_0^1{[t(f^r(a)-f^r(b))+f^r(b)]}^{\frac{2}{r}}dt=\frac{r}{r+2}\left(\frac{f^{r+2}(b)-f^{r+2}(a)}{f^r(b)-f^r(a)}\right).$$

(8)

Similarly,

$${\int }_0^1{[t(g^s(a)-g^s(b))+g^s(b)]}^{\frac{1}{s}}dt=\frac{s}{s+2}\left(\frac{g^{s+2}(b)-g^{s+2}(a)}{g^s(b)-g^s(a)}\right).$$

(9)

Using (7), (8) and (9), we can obtain the desired result. □

Remark 2.9 If the conditions of Theorem 2.8 hold, and $r\le s$, by Theorem 2.4, f is s-convex. Thus, the result of Theorem 2.8 could be as follows:

$$\frac{1}{b-a}{\int }_a^{b}f(x)g(x)dx\le \frac{1}{2}\left(\frac{s}{s+2}\right)[\frac{f^{s+2}(b)-f^{s+2}(a)}{f^s(b)-f^s(a)}+\frac{g^{s+2}(b)-g^{s+2}(a)}{g^s(b)-g^s(a)}].$$

If $f=g$, we have

$$\frac{1}{b-a}{\int }_a^b{f}^2(x)dx\le \frac{s}{s+2}\left(\frac{f^{s+2}(b)-f^{s+2}(a)}{f^s(b)-f^s(a)}\right).$$

Now, if $f=g$ and $r=s=2$ in Theorem 2.8, we have

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

which is the same result as in [[7], Corollary 2.5]. This shows that Theorem 2.8 is a generalization of [[7], Theorem 2.3]. In fact, the condition $r,s\le 2$ is redundant.

Theorem 2.10 Let $f,g:[a,b]⟶(0,\mathrm{\infty })$ be 0-convex on $[a,b]$. Then the following inequality holds:

$$\frac{1}{b-a}{\int }_a^{b}f(x)g(x)dx\le [f(b)g(b)-f(a)g(a)]ln\frac{f(b)g(b)}{f(a)g(a)}.$$

Proof Since f and g are 0-convex, for all $t\in [0,1]$, we have

$$\begin{array}{c}f(tx+(1-t)y)\le {[f(x)]}^t{[f(y)]}^{1-t},\hfill \\ g(tx+(1-t)y)\le {[g(x)]}^t{[g(y)]}^{1-t}.\hfill \end{array}$$

For all $x,y\in [0,1]$, and thus

$$\begin{array}{rcl}\frac{1}{b-a}{\int }_a^{b}f(x)g(x)dx& =& {\int }_0^1\left[f(ta+(1-t)b)\right]\left[g(tx+(1-t)y)\right]dt\\ \le & {\int }_0^1{[f(a)]}^t{[f(b)]}^{1-t}{[g(a)]}^t{[g(b)]}^{1-t}dt\\ =& f(b)g(b){\int }_0^1{\left[\frac{f(a)g(a)}{f(b)g(b)}\right]}^{t}dt\\ =& [f(b)g(b)-f(a)g(a)]ln\frac{f(b)g(b)}{f(a)g(a)}.\end{array}$$

 □

Corollary 2.11 With the hypotheses of the above theorem and $f=g$, we have

$$\frac{1}{b-a}{\int }_a^b{f}^2(x)dx\le [f^2(b)-f^2(a)]ln\frac{f^2(b)}{f^2(a)}.$$

Theorem 2.12 Let $f,g:[a,b]⟶(0,\mathrm{\infty })$ be r-convex and 0-convex functions respectively on $[a,b]$ and $r>0$. Then the following inequality holds:

Proof Since f is r-convex and g is 0-convex, for all $t\in [0,1]$, we have

$$\begin{array}{c}f(ta+(1-t)b)\le {[t{f}^r(a)+(1-t)f^r(b)]}^{\frac{1}{r}},\hfill \\ g(tx+(1-t)y)\le {[g(x)]}^t{[g(y)]}^{1-t}.\hfill \end{array}$$

Thus,

$$\begin{array}{rcl}\frac{1}{b-a}{\int }_a^{b}f(x)g(x)dx& =& {\int }_0^{1}f(ta+(1-t)b)g(ta+(1-t)b)dt\\ \le & {\int }_0^1{[t{f}^r(a)+(1-t)f^r(b)]}^{\frac{1}{r}}{[g(a)]}^t{[g(b)]}^{1-t}dt\\ =& {\int }_0^1{[t(f^r(a)-f^r(b))+f^r(b)]}^{\frac{1}{r}}{[g(a)]}^t{[g(b)]}^{1-t}dt.\end{array}$$

Again, Cauchy’s inequality shows that

(10)

We have

$${\int }_0^1{[t(f^r(a)-f^r(b))+f^r(b)]}^{\frac{2}{r}}dt=\frac{r}{r+2}\left(\frac{f^{r+2}(b)-f^{r+2}(a)}{f^r(b)-f^r(a)}\right).$$

(11)

Similar to the proof of Theorem 2.10, we can show that

$${\int }_0^1{[g(a)]}^{2t}{[g(b)]}^{2-2t}dt=\frac{1}{2}[g{(b)}^2-g{(a)}^2]ln\frac{g(b)}{g(a)}.$$

(12)

Using (10), (11) and (12), we can obtain the desired result. □