Symmetrized p-convexity and Related Some Integral Inequalities

TL;DR

This paper introduces symmetrized p-convex functions and establishes Hermite-Hadamard type inequalities for them, expanding the theoretical framework of convexity and integral inequalities.

Contribution

It proposes the novel concept of symmetrized p-convexity and derives related integral inequalities, extending existing convexity theories.

Findings

Introduction of symmetrized p-convex functions

Derivation of Hermite-Hadamard type inequalities for these functions

Expansion of convexity and inequality theory

Abstract

In this paper, the author introduces the concept of the symmetrized p-convex function, gives Hermite-Hadamard type inequalities for symmetrized p-convex functions.

Equations

$$f(tx+(1-t)y)\le tf(x)+(1-t)f(y)$$

$$f\left(\frac{xy}{tx+(1-t)y}\right)\le tf(y)+(1-t)f(x)$$

$$f\left(\frac{2ab}{a+b}\right)\le \frac{ab}{b-a}\underset{a}{\overset{b}{\int }}\frac{f(x)}{x^2}dx\le \frac{f(a)+f(b)}{2}.$$

$$f\left({\left[t{x}^p+(1-t)y^p\right]}^{1/p}\right)\le tf(x)+(1-t)f(y)$$

$$\left(f\circ g\right)\left(t{x}^p+(1-t)y^p\right)\le t\left(f\circ g\right)(x^p)+(1-t)\left(f\circ g\right)(y^p),g(x)=x^{1/p},$$

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

$$w(x)=w\left({\left[a^p+b^p-x^p\right]}^{1/p}\right)$$

$$f\left({\left[\frac{a^p+b^p}{2}\right]}^{1/p}\right)\underset{a}{\overset{b}{\int }}\frac{w(x)}{x^{1-p}}dx\le \underset{a}{\overset{b}{\int }}\frac{f(x)w(x)}{x^{1-p}}dx$$

$$J_{a+}^{\alpha }f(x)=\frac{1}{\Gamma (\alpha )}\underset{a}{\overset{x}{\int }}{\left(x-t\right)}^{\alpha -1}f(t)dt,a<x<b$$

$$J_{b-}^{\alpha }f(x)=\frac{1}{\Gamma (\alpha )}\underset{x}{\overset{b}{\int }}{\left(t-x\right)}^{\alpha -1}f(t)dt,a<x<b$$

$$f\left({\left[\frac{a^p+b^p}{2}\right]}^{1/p}\right)\left[J_{a^p+}^{\alpha }\left(w\circ g\right)(b^p)+J_{b^p-}^{\alpha }\left(w\circ g\right)(a^p)\right]$$

$$f\left({\left[\frac{a^p+b^p}{2}\right]}^{1/p}\right)\left[J_{b^p+}^{\alpha }\left(w\circ g\right)(a^p)+J_{a^p-}^{\alpha }\left(w\circ g\right)(b^p)\right]$$

$$\stackrel{⌣}{f}(x):=\frac{1}{2}\left[f(x)+f\left(a+b-x\right)\right],x\in \left[a,b\right].$$

$$\stackrel{\sim }{f}(x):=\frac{1}{2}\left[f(x)-f\left(a+b-x\right)\right],x\in \left[a,b\right].$$

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

$$f\left(\frac{a+b}{2}\right)\le \stackrel{⌣}{f}\left(x\right)=\frac{1}{2}\left[f(x)+f(a+b-x)\right]\le \frac{f(a)+f(b)}{2}$$

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

$$\frac{1}{2}\left[f\left(\frac{x+y}{2}\right)+f\left(a+b-\frac{x+y}{2}\right)\right]$$

$$\stackrel{⌣}{\mathrm{Hf}}(x):=\frac{1}{2}\left[f(x)+f\left(\frac{abx}{\left(a+b\right)x-ab}\right)\right],x\in \left[a,b\right].$$

$$f\left(\frac{2ab}{a+b}\right)\le \frac{ab}{b-a}\underset{a}{\overset{b}{\int }}\frac{f(x)}{x^2}dx\le \frac{f(a)+f(b)}{2}.$$

$$f\left(\frac{2ab}{a+b}\right)\le \stackrel{⌣}{\mathrm{Hf}}\left(x\right)=\frac{1}{2}\left[f(x)+f\left(\frac{abx}{\left(a+b\right)x-ab}\right)\right]\le \frac{f(a)+f(b)}{2}$$

$$\frac{1}{2}\left[f\left(\frac{2xy}{x+y}\right)+f\left(\frac{2abxy}{2xy\left(a+b\right)-ab(x+y)}\right)\right]$$

$$P_{\left(f;p\right)}(x):=\frac{1}{2}\left[f(x)+f\left({\left[a^p+b^p-x^p\right]}^{1/p}\right)\right],x\in \left[a,b\right].$$

$$A{P}_{\left(f;p\right)}(x):=\frac{1}{2}\left[f(x)-f\left({\left[a^p+b^p-x^p\right]}^{1/p}\right)\right],x\in \left[a,b\right].$$

$$P_{\left(f;p\right)}({\left[t{x}^p+(1-t)y^p\right]}^{1/p})$$

$$f({\left[t{u}^p+(1-t)v^p\right]}^{1/p})$$

$$PC[a,b]⫋SPC[a,b].$$

$$\stackrel{⌣}{\left(f\circ g\right)}(tx+(1-t)y)$$

$$P_{\left(f;p\right)}({\left[t{u}^p+(1-t)v^p\right]}^{1/p})\le t{P}_{\left(f;p\right)}(u)+(1-t)P_{\left(f;p\right)}(v)$$

$$=$$

Full text

1. Introduction

Let real function $f$ be defined on some nonempty interval $I$ of real line $\mathbb{R}$. The function $f$ is said to be convex on $I$ if inequality

[TABLE]

holds for all $x,y\in I$ and $t\in\left[0,1\right].$

In [3], the author, gave definition Harmonically convex and concave functions as follow.

Definition 1**.**

Let $I\subset\mathbb{R}\backslash\left\{0\right\}$ be a real interval. A function $f:I\rightarrow\mathbb{R}$ is said to be harmonically convex, if

[TABLE]

for all $x,y\in I$ and $t\in[0,1]$. If the inequality in (1.2) is reversed, then $f$ is said to be harmonically concave.

The following result of the Hermite-Hadamard type holds for harmonically convex functions.

Theorem 1** ([3]).**

Let $f:I\subset\mathbb{R}\backslash\left\{0\right\}\rightarrow\mathbb{R}$ be a harmonically convex function and $a,b\in I$ with $a<b.$ If $f\in L[a,b]$ then the following inequalities hold

[TABLE]

The  above inequalities are sharp.

In [4], the author gave the definition of $p$-convex function as follow:

Definition 2** ([4]).**

Let $I\subset\left(0,\infty\right)$ be a real interval and $p\in\mathbb{R}\backslash\left\{0\right\}.$ A function $f:I\rightarrow\mathbb{R}$ is said to be a p-convex function, if

[TABLE]

for all $x,y\in I$ and $t\in[0,1]$. If the inequality in (1.4) is reversed, then $f$ is said to be $p$-concave.

According to Definition 2, It can be easily seen that for $p=1$ and $p=-1$, $p$-convexity reduces to ordinary convexity and harmonically convexity of functions defined on $I\subset\left(0,\infty\right)$, respectively.

Since the condition (1.4) can be written as

[TABLE]

then we observe that $f:I\subseteq\left(0,\infty\right)\rightarrow\mathbb{R}$ is $p$-convex on $I$ if and only if $f\circ g$ is convex on $g^{-1}(I):=\{z^{p},z\in I\}$ .

Example 1**.**

Let $f:\left(0,\infty\right)\rightarrow\mathbb{R},\ f(x)=x^{p},p\neq 0,$ and $g:\left(0,\infty\right)\rightarrow\mathbb{R},\ g(x)=c,~{}\ c\in\mathbb{R},$ then $f$ and $g$ are both $p$-convex and $p$-concave functions.

In [2, Theorem 5], if we take $I\subset\left(0,\infty\right)$, $p\in\mathbb{R}\backslash\left\{0\right\}$ and $h(t)=t$ , then we have the following Theorem.

Theorem 2**.**

Let $f:I\subseteq\left(0,\infty\right)\rightarrow\mathbb{R}$ be a $p$-convex function, $p\in\mathbb{R}\backslash\left\{0\right\}$, and $a,b\in I$ with $a<b.$ If $f\in L[a,b]$ then we have

[TABLE]

Definition 3** ([6]).**

Let $p\in\mathbb{R}\backslash\left\{0\right\}.$ A function $w:\left[a,b\right]\subseteq\left(0,\infty\right)\rightarrow\mathbb{R}$ is said to be p-symmetric with respect to $\left[\frac{a^{p}+b^{p}}{2}\right]^{1/p}$ if

[TABLE]

holds for all $x\in\left[a,b\right].$

In [5], Kunt and İşcan gave Hermite-Hadamard-Fejér type inequalities for $p$-convex functions as follow:

Theorem 3**.**

Let $f:I\subseteq\left(0,\infty\right)\rightarrow\mathbb{R}$ be a $p$-convex function, $p\in\mathbb{R}\backslash\left\{0\right\},a,b\in I$ with $a<b.$ If $f\in L[a,b]$ and $w:[a,b]\rightarrow\mathbb{R}$ is nonnegative, integrable and $p$-symmetric with respect to $\left[\frac{a^{p}+b^{p}}{2}\right]^{1/p}$, then the following inequalities hold:

[TABLE]

Definition 4**.**

Let $f\in L\left[a,b\right]$. The left-sided and right-sided Hadamard fractional integrals $J_{a+}^{\alpha}f$ and $J_{b-}^{\alpha}f$ of oder $\alpha>0$ with $b>a\geq 0$ are defined by

[TABLE]

and

[TABLE]

respectively, where $\Gamma(\alpha)$ is the Gamma function defined by $\Gamma(\alpha)=$ $\mathop{\displaystyle\int}\limits_{0}^{\infty}e^{-t}t^{\alpha-1}dt$ (see [7]).

In [6], the authors presented Hermite–Hadamard-Fejer inequalities for $p$-convex functions in fractional integral forms as follows:

Theorem 4**.**

Let $f:I\subseteq\left(0,\infty\right)\rightarrow\mathbb{R}$ be a $p$-convex function, $p\in\mathbb{R}\backslash\left\{0\right\},\alpha>0$ and $a,b\in I$ with $a<b.$ If $f\in L[a,b]$ and $w:[a,b]\rightarrow\mathbb{R}$ is nonnegative, integrable and $p$-symmetric with respect to $\left[\frac{a^{p}+b^{p}}{2}\right]^{1/p}$, then the following inequalities for fractional integrals hold:

i.) If $p>0$

[TABLE]

with $g(x)=x^{1/p},x\in\left[a^{p},b^{p}\right]$.

ii.) If $p>0$

[TABLE]

with $g(x)=x^{1/p},x\in\left[b^{p},a^{p}\right]$.

For a function $f:[a,b]\mathbb{\rightarrow R}$ we consider the symmetrical transform of $f$ on the interval , denoted by $\overset{\smile}{f}_{[a,b]}$ or simply $\overset{\smile}{f}$, when the interval $[a,b]$ is implicit, which

is defined by

[TABLE]

The anti symmetrical transform of $f$ on the interval $[a,b]$ is denoted by $\overset{\sim}{f}_{[a,b]}$or simply $\overset{\sim}{f}$ as defined by

[TABLE]

It is obvious that for any function $f$ we have $\overset{\smile}{f}+\overset{\sim}{f}=f$.

If $f$ is convex on $[a,b]$, then $\overset{\smile}{f}$ is also convex on $[a,b].$ But, when $\overset{\smile}{f}$ is onvex on $[a,b]$, $f$ may not be convex on $[a,b]$ ( [1]).

In [1], Dragomir introduced symmetrized convexity concept as follow:

Definition 5**.**

A function $f:[a,b]\mathbb{\rightarrow R}$ is said to be symmetrized convex (concave)on $[a,b]$ if symmetrical transform $\overset{\smile}{f}$ is convex (concave) on $[a,b]$.

Dragomir extends the Hermite-Hadamard inequality to the class of symmetrized convex functions as follow:

Theorem 5** ([1]).**

Assume that $f:[a,b]\mathbb{\rightarrow R}$ is symmetrized convex on the interval $[a,b]$, then we have the Hermite-Hadamard inequalities

[TABLE]

Theorem 6** ([1]).**

Assume that $f:[a,b]\mathbb{\rightarrow R}$ is symmetrized convex on the interval $[a,b]$. Then for any $x\in[a,b]$ we have the bounds

[TABLE]

Corollary 1**.**

If $f:[a,b]\mathbb{\rightarrow R}$ is symmetrized convex on the interval $[a,b]$ and $w:[a,b]\rightarrow[0,\infty)$is integrable on $[a,b]$, then

[TABLE]

Theorem 7** ([1]).**

Assume that $f:[a,b]\subseteq\left(0,\infty\right)\rightarrow\mathbb{R}$ is symmetrized convex on the interval $[a,b].$ Then for any $x,y\in[a,b]$ with $x\neq y$ we have the Hermite-Hadamard inequalities

[TABLE]

For a function $f:[a,b]\subseteq\mathbb{R\diagdown}\left\{0\right\}\mathbb{\rightarrow}\mathbb{\mathbb{C}}$, we consider the symmetrical transform of $f$ on the interval , denoted by $\overset{\smile}{Hf}_{[a,b]}$ or simply $\overset{\smile}{Hf}$, when the interval $[a,b]$ as defined by

[TABLE]

Definition 6** ([8]).**

A function $f:I\subseteq\mathbb{R\diagdown}\left\{0\right\}\mathbb{\rightarrow R}$ is said to be symmetrized harmonic convex (concave)on $[a,b]$ if $\overset{\smile}{Hf}$ is harmonic convex (concave) on $I$.

The similars of above results given for the class of symmetrized convex functions, in [8] it has been obtained by Wu et al. for the class of symmetrized harmonic convex functions as follow:

Theorem 8** ([8]).**

Assume that $f:[a,b]\subseteq\mathbb{R\diagdown}\left\{0\right\}\mathbb{\rightarrow R}$ is symmetrized harmonic convex and integrable on the interval $[a,b]$. Then we have the Hermite-Hadamard type İşcan inequalities

[TABLE]

Theorem 9** ([8]).**

Assume that $f:[a,b]\subseteq\mathbb{R\diagdown}\left\{0\right\}\mathbb{\rightarrow R}$ is symmetrized harmonic convex on the interval $[a,b]$. Then for any $x\in[a,b]$ we have the bounds

[TABLE]

Theorem 10** ([8]).**

Assume that $f:[a,b]\subseteq\mathbb{R\diagdown}\left\{0\right\}\rightarrow\mathbb{R}$ is symmetrized harmonic convex on the interval $[a,b].$ Then for any $x,y\in[a,b]$ with $x\neq y$ we have the Hermite-Hadamard inequalities

[TABLE]

Motivated by the above results, in this paper we introduces the concept of the symmetrized $p$-convex function and establish some Hermite-Hadamard type inequalities. Some examples of interest are provided as well.

2. Symmetrized $p$-Convexity

For a function $f:[a,b]\subseteq\left(0,\infty\right)\mathbb{\rightarrow R}$ we consider the $p$-symmetrical transform of $f$ on the interval , denoted by $P_{\left(f;p\right),[a,b]}$ or simply $P_{\left(f;p\right)}$, when the interval $[a,b]$ is implicit, which

is defined by

[TABLE]

The anti $p$-symmetrical transform of $f$ on the interval $[a,b]$ is denoted by $AP_{\left(f;p\right),[a,b]}$ or simply $AP_{\left(f;p\right)}$ as defined by

[TABLE]

It is obvious that for any function $f$ we have $P_{\left(f;p\right)}+AP_{\left(f;p\right)}=f$. Also it is seen that $P_{\left(f;1\right)}(x)=\frac{1}{2}\left[f(x)+f\left(a+b-x\right)\right]=\overset{\smile}{f}(x)$ and $P_{\left(f;-1\right)}(x)=\frac{1}{2}\left[f(x)+f\left(\frac{abx}{\left(a+b\right)x-ab}\right)\right]=\overset{\smile}{Hf}(x).$

If $f$ is $p$-convex on $[a,b]$, then $P_{\left(f;p\right)}$ is also $p$-convex on $[a,b].$ Indeed, for any $x,y\in[a,b]$ and $t\in\left[0,1\right]$ we have

[TABLE]

Remark 1**.**

If $P_{\left(f;p\right)}$ is $p$-convex on $[a,b]$ for a function $f:[a,b]\subseteq\left(0,\infty\right)\mathbb{\rightarrow R}$, then the function $f$ is nor necessary $p$-convex on $[a,b]$. For example, let $p=-1$, Consider the function $f(x)=-\ln x,x\in\left(0,\infty\right)$. The function $f$ is not $-1$-convex (or harmonically convex), but $P_{\left(f;-1\right)}$ is $-1$-convex [8].

Definition 7**.**

A function $f:[a,b]\subseteq\left(0,\infty\right)\mathbb{\rightarrow R}$ is said to be symmetrized $p$-convex ($p$-concave)on $[a,b]$ if $p$-symmetrical transform $P_{\left(f;p\right)}$ is $p$-convex ($p$-concave) on $[a,b]$.

Example 2**.**

Let $a,b\in\mathbb{R}$ with $0<a<b$ and $\alpha\geq 2$. Then the function $f:\left[a,b\right]\rightarrow\mathbb{R},~{}f(x)=\left(x^{p}-a^{p}\right)^{\alpha-1},$ $p>0,$ (or $f(x)=\left(a^{p}-x^{p}\right)^{\alpha-1},$ $p<0$) is $p$-convex on $\left[a,b\right]$. Indeed, for any $u,v\in\left[a,b\right]$ and $t\in\left[0,1\right]$ by convexity of the function $g(\varsigma)=\varsigma^{\alpha-1}$,$\zeta\geq 0$, we have

[TABLE]

Thus $P_{\left(f;p\right)}$ is also $p$-convex on $[a,b]$. Therefore $f$ is symmetrized $p$-convex function.

Example 3**.**

Let $\alpha\geq 2$. Then the function $f:\left[a,b\right]\subseteq\left(0,\infty\right)\rightarrow\mathbb{R},~{}f(x)=\left(b^{p}-x^{p}\right)^{\alpha-1},$ $p>0,$ (or $f(x)=\left(x^{p}-b^{p}\right)^{\alpha-1},$ $p<0$) is $p$-convex on $\left[a,b\right].$Therefore $f$ is symmetrized $p$-convex function.

Example 4**.**

Let $\alpha\geq 2$. Then the function $f:\left[a,b\right]\subseteq\left(0,\infty\right)\rightarrow\mathbb{R},~{}f(x)=\left(x^{p}-a^{p}\right)^{\alpha-1}+\left(b^{p}-x^{p}\right)^{\alpha-1},\ p>0,$ (or $f(x)=\left(a^{p}-x^{p}\right)^{\alpha-1}+\left(x^{p}-b^{p}\right)^{\alpha-1},$ $p<0$) is symmetrized $p$-convex function.

Now if $PC[a,b]$ is the class of $p$-convex functions defined on I and $SPC[a,b]$ is the class of symmetrized $p$-convex functions on $[a,b]$ then

[TABLE]

Also, if $[c,d]\subset[a,b]$ and $f\in SPC[a,b]$, then this does not imply in general that $f\in SPC[c,d]$.

Proposition 1**.**

Let $f:[a,b]\subseteq\left(0,\infty\right)\mathbb{\rightarrow R}$ be a function and $g(x)=x^{1/p},x>0,p\neq 0$. $f$ is symmetrized $p$-convex on the interval $[a,b]$ if and only if $f\circ g$ is symmetrized convex on the interval $g^{-1}\left([a,b]\right)=[a^{p},b^{p}]$ (or $[b^{p},a^{p}]$).

Proof.

Let $f$ be a symmetrized $p$-convex function on the interval $[a,b]$. If we take arbitrary $x,y\in g^{-1}\left([a,b]\right)$, then there exist $u,v\in[a,b]$ such that $x=u^{p}$ and $y=g(v)=v^{p}$

[TABLE]

Since $f$ is a symmetrized $p$-convex function on the interval $[a,b],$ we have

[TABLE]

[TABLE]

By (2) and (2.2), we obtain that $f\circ g$ is symmetrized convex on the interval $[a^{p},b^{p}]$ (or $[b^{p},a^{p}]$).

Conversely, if $f\circ g$ is symmetrized convex on the interval $[a^{p},b^{p}]$ (or $[b^{p},a^{p}]$) then it is easily seen that $f$ is symmetrized $p$-convex on the interval $[a,b]$ by a similar procedure. The details are omitted.

Theorem 11**.**

If $f:[a,b]\subseteq\left(0,\infty\right)\mathbb{\rightarrow R}$ is symmetrized $p$-convex on the interval $[a,b]$, then we have the Hermite-Hadamard inequalities

[TABLE]

Proof.

Since $f:[a,b]\subseteq\left(0,\infty\right)\mathbb{\rightarrow R}$ is symmetrized $p$-convex on the interval $[a,b]$, then by writing the Hermite-Hadamard inequality for the function $P_{\left(f;p\right)}(x)$ we have

[TABLE]

where, it is easily seen that

[TABLE]

and

[TABLE]

Then by (2.4) we get required inequalities.

Remark 2**.**

In Theorem 11,

i.) if we choose $p=1$, then the inequalities (2.5) reduces to the inequalities (1.8) in Theorem (5).

ii.) if we choose $p=-1$, then the inequalities (2.5) reduces to the inequalities (1.11) in Theorem (8).

Remark 3**.**

By helping Theorem 5 and Proposition 1, the proof of Theorem 11 can also be given as follows :

Since $f:[a,b]\subseteq\left(0,\infty\right)\mathbb{\rightarrow R}$ is symmetrized $p$-convex on the interval $[a,b]$, $f\circ g$ is symmetrized convex on the interval $[a^{p},b^{p}]$ (or $[b^{p},a^{p}]$) with $g(x)=x^{1/p},x>0,p\neq 0.$ So, by Theorem 5 we have

[TABLE]

i.e.

[TABLE]

Theorem 12**.**

If $f:[a,b]\subseteq\left(0,\infty\right)\mathbb{\rightarrow R}$ is symmetrized $p$-convex on the interval $[a,b]$. Then for any $x\in[a,b]$ we have the bounds

[TABLE]

Proof.

Since$P_{\left(f;p\right)}$ is $p$-convex on $[a,b]$ then for any $x\in[a,b]$ we have

[TABLE]

This give us the first inequality in (2.5).

Also, for any $x\in[a,b]$ there exist a number $t_{0}\in[0,1]$ such that $x=\left[t_{0}a^{p}+(1-t_{0})b^{p}\right]^{1/p}$. By the $p$-convexity of $P_{\left(f;p\right)}$ we have

[TABLE]

which gives the second inequality in (2.5).

Remark 4**.**

In Theorem 12,

i.) if we choose $p=1$, then the inequalities (2.5) reduces to the inequalities (1.9) in Theorem (6).

ii.) if we choose $p=-1$, then the inequalities (2.5) reduces to the inequalities (1.12) in Theorem (9).

Remark 5**.**

By helping Theorem 6 and Proposition 1, the proof of Theorem 12 can also be given as follows :

Since $f:[a,b]\subseteq\left(0,\infty\right)\mathbb{\rightarrow R}$ is symmetrized $p$-convex on the interval $[a,b]$, $f\circ g$ is symmetrized convex on the interval $[a^{p},b^{p}]$ with $g(x)=x^{1/p},x>0,p\neq 0$. So, by Theorem 6 we have

[TABLE]

i.e.

[TABLE]

for any $x\in[a,b]$.

Remark 6**.**

If $f:[a,b]\subseteq\left(0,\infty\right)\mathbb{\rightarrow R}$ is symmetrized $p$-convex on the interval $[a,b]$, then we have the bounds

[TABLE]

and

[TABLE]

Corollary 2**.**

If $f:[a,b]\subseteq\left(0,\infty\right)\mathbb{\rightarrow R}$ is symmetrized $p$-convex on the interval $[a,b]$ and $w:[a,b]\rightarrow[0,\infty)$is integrable on $[a,b]$, then

[TABLE]

Moreover, if $w$ is $p$-symmetric with respect to $\left[\frac{a^{p}+b^{p}}{2}\right]^{1/p}$on $[a,b]$, i.e. $w(x)=w(\left[a^{p}+b^{p}-x^{p}\right]^{1/p})$ for all $x\in[a,b]$, then

[TABLE]

Proof.

The inequality (2.6) follows by (2.5) multiplying by $w(x)/x^{1-p}\geq 0$ and integrating over $x$ on $[a,b]$.

By changing the variable, we have

[TABLE]

Since $w$ is $p$-symmetric with respect to $\left[\frac{a^{p}+b^{p}}{2}\right]^{1/p}$, then

[TABLE]

Thus

[TABLE]

and by (2.6) we get (2.7).

Remark 7**.**

The inequality (2.7) is known as weighted generalization of Hermite-Hadamard inequality for $p$-convex functions (it is also given in Theorem 3). It has been shown now that this inequality remains valid for the larger class of symmetrized $p$-convex functions $f$ on the interval $[a,b]$.

Remark 8**.**

We note that by helping Corollary 1 and Proposition 1, the proof of Corollary 2 can also be given. The details is omitted.

Remark 9**.**

Let $a,b,\alpha\in\mathbb{R}$ with $0<a<b$ and $\alpha\geq 2$. Then the function $f:\left[a,b\right]\rightarrow\mathbb{R},~{}f(x)=\left(x^{p}-a^{p}\right)^{\alpha-1},$ $p>0,$ is symmetrized $p$-convex on $\left[a,b\right]$

i.) If we consider the function

[TABLE]

which is symmetrized $p$-convex on $\left[a,b\right]$ in the inequality (2.6), then we have

[TABLE]

for any $w:[a,b]\rightarrow[0,\infty)$ is integrable on $[a,b]$ with $g(x)=x^{1/p},x\in\left[a^{p},b^{p}\right].$

ii.) If we consider the function

[TABLE]

which is $p$-symmetric with respect to $\left[\frac{a^{p}+b^{p}}{2}\right]^{1/p}$ in the inequality (2.7), then we have the following inequalities

[TABLE]

where $g(x)=x^{1/p},x\in\left[a^{p},b^{p}\right].$

iii.) Let $\varphi$ be $p$-symmetric with respect to $\left[\frac{a^{p}+b^{p}}{2}\right]^{1/p}.$ If we consider the function

[TABLE]

which is $p$-symmetric with respect to $\left[\frac{a^{p}+b^{p}}{2}\right]^{1/p}$ in the inequality (2.7), then we have the following inequalities

[TABLE]

which are the same of inequalities in (4). Where $g(x)=x^{1/p},x\in\left[a^{p},b^{p}\right].$

Theorem 13**.**

Assume that $f:[a,b]\subseteq\left(0,\infty\right)\rightarrow\mathbb{R}$ is symmetrized $p$-convex on the interval $[a,b]$ with $p\in\mathbb{R}\diagdown\left\{0\right\}.$ Then for any $x,y\in[a,b]$ with $x\neq y$ we have the Hermite-Hadamard inequalities

[TABLE]

Proof.

Since $P_{\left(f;p\right),[a,b]}$ is $p$-convex on $[a,b]$, then $P_{\left(f;p\right),[a,b]}$ is also $p$-convex on any subinterval $[x,y]$ (or $[y,x]$) where $x,y\in[a,b].$

By Hermite-Hadamard inequalities for convex functions we have

[TABLE]

for any $x,y\in[a,b]$ with $x\neq y$.

By definition of $P_{\left(f;p\right)}$, we have

[TABLE]

[TABLE]

and

[TABLE]

Thus by (2.9) we obtain the desired result (13).

Remark 10**.**

We note that by helping Theorem 7 and Proposition 1, the proof of Theorem 13 can also be given. The details is omitted.

Remark 11**.**

If we take $x=a$ and $y=b$ in (13), then we get (2.3). If, for a given $x\in[a,b]$, we take $y=\left[a^{p}+b^{p}-x^{p}\right]^{1/p}$, then from (13) we get

[TABLE]

where $x\neq\left[\frac{a^{p}+b^{p}}{2}\right]^{1/p}$, provided that $f:[a,b]\subseteq\left(0,\infty\right)\rightarrow\mathbb{R}$ is symmetrized $p$-convex on the interval $[a,b]$.

Multiplying the inequalities (2.10) by $\frac{1}{x^{1-p}}$, then integrating the resulting inequality over x we get the following refinement of the first part of (2.3)

[TABLE]

provided that $f:[a,b]\subseteq\left(0,\infty\right)\rightarrow\mathbb{R}$ is symmetrized $p$-convex on the interval $[a,b]$.

When the function is $p$-convex, we have the following inequalities as well:

Remark 12**.**

If $f:[a,b]\subseteq\left(0,\infty\right)\rightarrow\mathbb{R}$ is $p$-convex, then from (13) we have the inequalities

[TABLE]

for any $x,y\in[a,b]$ with $x\neq y.$

If we multiply the inequalities(12) by $\frac{1}{\left(xy\right)^{1-p}}$ and integrate (12) over $(x,y)$ on the square $[a,b]^{2}$ and divide by $\frac{p^{2}}{\left(b^{p}-a^{p}\right)^{2}}$ , then we get the following refinement of the first Hermite-Hadamard inequality for $p$-convex functions

[TABLE]

Remark 13**.**

In Theorem 13,

i.) if we choose $p=1$, then the inequalities (13) reduces to the inequalities (7) in Theorem (7).

ii.) if we choose $p=-1$, then the inequalities (13) reduces to the inequalities (10) in Theorem (10).