Refinements of Hermite–Hadamard Inequalities for Continuous Convex Functions via (p,q)-Calculus

Abstract

In this paper, we present some new refinements of Hermite–Hadamard inequalities for continuous convex functions by using $(p,q)$-calculus. Moreover, we study some new $(p,q)$-Hermite–Hadamard inequalities for multiple integrals. Many results given in this paper provide extensions of others given in previous research.

1. Introduction

Mathematical inequalities play important roles in the study of mathematics as well as in other areas of mathematics because of their wide applications in mathematics and physics; see [1,2,3] for more details. One of the most significant functions used to study many interesting inequalities is convex functions, which are defined as follows:

Let $I\subset \mathbb{R}$ be a non-empty interval. The function $f:I\to \mathbb{R}$ called as convex, if

$$f(ta+(1-t\left)b\right)\le tf\left(a\right)+(1-t)f\left(b\right)$$

holds for every $a,b\in I$ and $t\in [0,1]$.

In recent years, many researchers have been fascinated in the study of convex functions and, particularly, one of the well-known inequality for convex functions known as the Hermite–Hadamard inequality, which is defined as follows:

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

(1)

Inequality (1) was introduced by C. Hermite [4] and investigated by J. Hadamard [5] in 1893. So far, the Hermite–Hadamard inequality and a variety of refinements of Hermite–Hadamard inequalities have been extensively studied by many researchers; see [6,7,8,9,10,11,12,13,14,15,16,17,18] and the references therein for more details.

The study of calculus with no limits is called quantum calculus (in short, q-calculus). The main objective of studying q-calculus is to obtain the q-analoques of mathematical objects that can be recaptured by taking q tending toward 1. In the past few years, the topic of q-calculus has become an interesting topic for many researchers, and new results of q-calculus can be found in [19,20,21,22,23,24,25,26,27,28,29,30,31,32,33,34,35,36,37,38,39,40,41], and the references cited therein.

The generalization of q-calculus is post quantum calculus or, sometimes, is called $(p,q)$-calculus. $(p,q)$-calculus is known as two-parameter quantum calculus, for which applications plays significant roles in mathematics and physics such as combinatorics, fractals, special functions, number theory, dynamical systems, and mechanics, among others. In $(p,q)$-calculus, we obtain the q-calculus formula for the case $p=1$ and obtain the original of mathematical formula when q tends towards 1.

Recently, Tunç and Göv [42,43,44] studied the concept of $(p,q)$-calculus over the intervals $[a,b]$ and gave some new definitions of $(p,q)$-derivatives and $(p,q)$-integrals. Moreover, they also derived some of its properties and many integral inequalities as in (1), which is called $(p,q)$-Hermite–Hadamard inequality, and some new results on $(p,q)$-calculus of several important integral inequalities. Next, Mehmet Kunt et al. [45] proved the left side of the $(p,q)$-Hermite–Hadamard inequality through $(p,q)$-differentiable convex and quasi-convex functions, and then, they had some new $(p,q)$-Hermite–Hadamard inequalities.

In 2019, Prabseang et al. [46] established some new $(p,q)$-calculus of Hermite–Hadamard inequalities for the double integral and refinements of the Hermite–Hadamard inequality for $(p,q)$-differentiable convex functions. In the last few years, the topic of $(p,q)$-calculus has been investigated extensively by many researchers, and a variety of new results can be found in the literature (see [47,48,49,50,51,52,53,54,55,56,57,58,59,60,61,62,63,64] and the references cited therein).

In 2020, Prabseang et al. [65] established some new refinement of quantum Hermite–Hadamard inequalities, which have been expanded to integration on a finite interval of an n-dimensional. Some new refinements of $(p,q)$-Hermite–Hadamard inequalities for convex functions are given.

In this paper, we aim to propose some new refinements of Hermite–Hadamard inequalities via $(p,q)$-calculus that have been expanded to integration on a finite interval of an n-dimensional. We obtain some new refinements of $(p,q)$-Hermite–Hadamard inequalities for convex functions and the results in special cases for $p=1$ and $q\to 1$.

Before presenting our main results in Section 3, we introduce the definitions and results from $(p,q)$-calculus in Section 2. Finally, Section 4 concludes the paper by summarizing the results.

2. Preliminaries

In this section, the basic definitions used in our study are discussed. Throughout this paper, let $[a,b]\subseteq \mathbb{R}$ be an interval and $0<q<p\le 1$ be constants. The following definitions for the $(p,q)$-derivative and $(p,q)$-integral were given in [42,43].

Definition 1.

If $f:[a,b]\to \mathbb{R}$ is a continuous function, then the $(p,q)$-derivative of function f at x is defined by

$$\begin{array}{cc}\hfill {}_a{D}_{p,q}f\left(x\right)& =\frac{f(px+(1-p\left)a\right)-f(qx+(1-q\left)a\right)}{(p-q)(x-a)},x\ne a\hfill \\ \hfill {}_a{D}_{p,q}f\left(a\right)& =\underset{x\to a}{lim}{}_a{D}_{p,q}f\left(x\right).\hfill \end{array}$$

(2)

If ${}_a{D}_{p,q}f\left(x\right)$ exists for all $x\in [a,b]$, then the function f is called $(p,q)$-differentiable on $[a,b]$.

In Definition 1, if $a=0$, then ${}_0{D}_{p,q}f=D_{p,q}f$, which is defined by

$$\begin{array}{cc}\hfill D_{p,q}f\left(x\right)& =\frac{f\left(px\right)-f\left(qx\right)}{(p-q)x},x\ne 0.\hfill \end{array}$$

(3)

In addition, if $p=1$ in (3), then it reduces to $D_{q}f$, which is the q-derivative of the function f; see [32].

Example 1.

Define function $f:[a,b]\to \mathbb{R}$ by $f\left(x\right)=x^2+x+C$, where $C\in \mathbb{R}$. Then, for $x\ne a$, we have

$$\begin{array}{cc}\hfill {}_a{D}_{p,q}(x^2+x+C)& =\frac{\left[{(px+(1-p)a)}^2+(px+(1-p)a)+C\right]}{(p-q)(x-a)}\hfill \\ \hfill & -\frac{\left[{(qx+(1-q)a)}^2+(qx+(1-q)a)+C\right]}{(p-q)(x-a)}\hfill \\ \hfill & =\frac{(p+q)x^2+2ax[1-(p+q)]+a^2[(p+q)-2]+(x-a)}{(x-a)}\hfill \\ \hfill & =\frac{x(p+q)(x-a)-a(p+q)(x-a)+2a(x-a)+(x-a)}{(x-a)}\hfill \\ \hfill & =(p+q)(x-a)+2a+1.\hfill \end{array}$$

(4)

Definition 2.

Let $f:[a,b]\to \mathbb{R}$ be a continuous function. Then, the $(p,q)$-integral on $[a,b]$ is defined by

$${\int }_a^{x}f\left(t\right){}_a{d}_{p,q}t=(p-q)(x-a)\sum _{n=0}^{\infty }\frac{q^n}{p^{n+1}}f\left(\frac{q^n}{p^{n+1}}x+\left(1-\frac{q^n}{p^{n+1}}\right)a\right),$$

(5)

for $x\in [a,b]$. If $a=0$ and $p=1$ in (5), then we have the classical q-integral; see [32].

Example 2.

Define function $f:[a,b]\to \mathbb{R}$ by $f\left(x\right)=Ax+B$, where $A,B\in \mathbb{R}$. Then, we have

$$\begin{array}{cc}\hfill {\int }_a^{b}f\left(x\right){}_a{d}_{p,q}x& ={\int }_a^b(Ax+B){}_a{d}_{p,q}x\hfill \\ \hfill & =A(p-q)(b-a)\sum _{n=0}^{\infty }\frac{q^n}{p^{n+1}}\left(\frac{q^n}{p^{n+1}}b+\left(1-\frac{q^n}{p^{n+1}}\right)a\right)\hfill \\ \hfill & +B(p-q)(b-a)\sum _{n=0}^{\infty }\frac{q^n}{p^{n+1}}\hfill \\ \hfill & =\frac{A(b-a)(b-a(1-p-q\left)\right)}{p+q}+B(b-a).\hfill \end{array}$$

(6)

In addition, the following definition for the $(p,q)$-integral of the function of two variables can be defined; we referred to [47].

Definition 3.

Let $f:[a,b]\times [c,d]\subset {\mathbb{R}}^2\to \mathbb{R}$ be a continuous function, then the definite $(p,q)$-integral on $[a,b]\times [c,d]$ is defined by

$$\begin{array}{cc}\hfill {\int }_c^t{\int }_a^{s}f{(x,y)}_a{d}_{p,q}x{}_a{d}_{p,q}y=& {(p-q)}^2(s-a)(t-c)\hfill \\ \hfill & \times \sum _{m=0}^{\infty }\sum _{n=0}^{\infty }\frac{q^{m+n}}{p^{m+n+2}}f\left(\frac{q^n}{p^{n+1}}s+\left(1-\frac{q^n}{p^{n+1}}\right)a,\frac{q^m}{p^{m+1}}t+\left(1-\frac{q^m}{p^{m+1}}\right)a\right),\hfill \end{array}$$

(7)

for $(s,t)\in [a,b]\times [c,d]$.

The proofs of the following theorems were given in [42,43].

Theorem 1.

Let $f:[a,b]\to \mathbb{R}$ be a continuous function. Then, we have the following:

(i) 

${\displaystyle {}_a{D}_{p,q}{\int }_a^{x}f\left(t\right){}_a{d}_{p,q}t=f\left(x\right)}$;

(ii) 

${\displaystyle {\int }_c^x{}_a{D}_{p,q}f\left(t\right){}_a{d}_{p,q}t=f\left(x\right)-f\left(c\right)}$ for $c\in (a,x)$.

Theorem 2.

Let $f,g:[a,b]\to \mathbb{R}$ be continuous functions and $\alpha \in \mathbb{R}$. Then, we have the following:

(i) 

${\displaystyle {\int }_a^x\left[f\right(t)+g(t\left)\right]{}_a{d}_{p,q}t={\int }_a^{x}f\left(t\right){}_a{d}_{p,q}t+{\int }_a^{x}g\left(t\right){}_a{d}_{p,q}t}$;

(ii) 

${\displaystyle {\int }_a^x{\left(\alpha f\right)\left(t\right)}_a{d}_{p,q}t=\alpha {\int }_a^x{f\left(t\right)}_a{d}_{p,q}t}$;

(iii) 

${\displaystyle {\int }_c^{x}f{(pt+(1-p)a)}_a{D}_{p,q}g\left(t\right){}_a{d}_q{t=\left(fg\right)|}_c^x-{\int }_c^{x}g{(qt+(1-q)a)}_a{D}_{p,q}f\left(t\right){}_a{d}_{p,q}t}$ for $c\in (a,x)$.

3. Main Results

In this section, we present refinements of Hermite–Hadamard inequalities for continuous convex functions via $(p,q)$-calculus on the interval $J:=[a,pb+(1-p\left)a\right]$.

Theorem 3.

Let $f:J\to \mathbb{R}$ be a continuous convex function. Then, we have

$$\begin{array}{cc}\hfill f\left(\frac{qa+pb}{p+q}\right)& \le \frac{1}{p^2{(b-a)}^2}{\int }_a^{pb+(1-p)a}{\int }_a^{pb+(1-p)a}f\left(\frac{x+y}{2}\right){}_a{d}_{p,q}{x}_a{d}_{p,q}y\hfill \\ \hfill & \le \frac{1}{p^2{(b-a)}^2}{\int }_a^{pb+(1-p)a}{\int }_a^{pb+(1-p)a}\frac{1}{2}\left[f\left(\frac{\alpha x+\beta y}{\alpha +\beta }\right)+f\left(\frac{\beta x+\alpha y}{\alpha +\beta }\right)\right]{}_a{d}_{p,q}{x}_a{d}_{p,q}y\hfill \\ \hfill & \le \frac{1}{p(b-a)}{\int }_a^{pb+(1-p)a}f{\left(x\right)}_a{d}_{p,q}x\hfill \end{array}$$

(8)

for all $\alpha ,\beta \ge 0$ with $\alpha +\beta >0$.

Proof. 

Since f is convex on J, for all $x,y\in J$ and $\alpha ,\beta \ge 0$ with $\alpha +\beta >0$, we have

$$\begin{array}{cc}\hfill f\left(\frac{x+y}{2}\right)& \le \frac{1}{2}\left[f\left(\frac{\alpha x+\beta y}{\alpha +\beta }\right)+f\left(\frac{\beta x+\alpha y}{\alpha +\beta }\right)\right]\hfill \\ \hfill & \le \frac{f\left(x\right)+f\left(y\right)}{2}.\hfill \end{array}$$

(9)

Taking double $(p,q)$-integration on both sides of (9) on $J^2$, we obtain the second part of (8).

On the other hand, by using Jensen’s inequality, we have

$$\begin{array}{cc}\hfill & f\left(\frac{1}{p^2{(b-a)}^2}{\int }_a^{pb+(1-p)a}{\int }_a^{pb+(1-p)a}\left(\frac{x+y}{2}\right){}_a{d}_{p,q}{x}_a{d}_{p,q}y\right)\hfill \\ \hfill & \le \frac{1}{p^2{(b-a)}^2}{\int }_a^{pb+(1-p)a}{\int }_a^{pb+(1-p)a}f\left(\frac{x+y}{2}\right){}_a{d}_{p,q}{x}_a{d}_{p,q}y.\hfill \end{array}$$

Since

$$\frac{1}{p^2{(b-a)}^2}{\int }_a^{pb+(1-p)a}{\int }_a^{pb+(1-p)a}\left(\frac{x+y}{2}\right){}_a{d}_{p,q}{x}_a{d}_{p,q}y=\frac{qa+pb}{p+q},$$

this yields the first part of (8). This completes the proof. □

Remark 1.

If $p=1$, then (8) reduces to

$$\begin{array}{cc}\hfill f\left(\frac{qa+b}{1+q}\right)& \le \frac{1}{{(b-a)}^2}{\int }_a^b{\int }_a^{b}f\left(\frac{x+y}{2}\right){}_a{d}_q{x}_a{d}_{q}y\hfill \\ \hfill & \le \frac{1}{{(b-a)}^2}{\int }_a^b{\int }_a^b\frac{1}{2}\left[f\left(\frac{\alpha x+\beta y}{\alpha +\beta }\right)+f\left(\frac{\beta x+\alpha y}{\alpha +\beta }\right)\right]{}_a{d}_q{x}_a{d}_{q}y\hfill \\ \hfill & \le \frac{1}{b-a}{\int }_a^{b}f\left(x\right){}_a{d}_{q}x,\hfill \end{array}$$

(10)

see also [65]. Additionally, if $q\to 1$ in (10), then (10) reduces to

$$\begin{array}{cc}\hfill f\left(\frac{a+b}{2}\right)& \le \frac{1}{{(b-a)}^2}{\int }_a^b{\int }_a^{b}f\left(\frac{x+y}{2}\right)dxdy\hfill \\ \hfill & \le \frac{1}{{(b-a)}^2}{\int }_a^b{\int }_a^b\frac{1}{2}\left[f\left(\frac{\alpha x+\beta y}{\alpha +\beta }\right)+f\left(\frac{\beta x+\alpha y}{\alpha +\beta }\right)\right]dxdy\hfill \\ \hfill & \le \frac{1}{b-a}{\int }_a^{b}f\left(x\right)dx,\hfill \end{array}$$

which readily appeared in [66].

Theorem 4.

Let $f:J\to \mathbb{R}$ be a continuous convex function. Then, we have

$$\begin{array}{cc}\hfill f\left(\frac{qa+pb}{p+q}\right)& \le \frac{1}{p^n{(b-a)}^n}{\int }_a^{pb+(1-p)a}\cdots {\int }_a^{pb+(1-p)a}f\left(\frac{x_1+\cdots +x_n}{n}\right){}_a{d}_{p,q}{x}_1\cdots {}_a{d}_{p,q}{x}_n\hfill \\ \hfill & \le \frac{1}{p^{n-1}{(b-a)}^{n-1}}{\int }_a^{pb+(1-p)a}\cdots {\int }_a^{pb+(1-p)a}f\left(\frac{x_1+\cdots +x_{n-1}}{n-1}\right){}_a{d}_{p,q}{x}_1\cdots {}_a{d}_{p,q}{x}_{n-1}\hfill \\ \hfill & ⋮\hfill \\ \hfill & \le \frac{1}{p(b-a)}{\int }_a^{pb+(1-p)a}f\left(x\right){}_a{d}_{p,q}x,\hfill \end{array}$$

(11)

for all $n\in \mathbb{N}$ with $n\ge 3$.

Proof. 

Since

$$\frac{x_1+\cdots +x_n}{n}=\frac{1}{n}\left[\left(\frac{x_1+\cdots +x_{n-1}}{n-1}\right)+\left(\frac{x_2+\cdots +x_n}{n-1}\right)+\cdots +\left(\frac{x_n+\cdots +x_{n-2}}{n-1}\right)\right],$$

and by using Jensen’s inequality, we have

$$f\left(\frac{x_1+\cdots +x_n}{n}\right)\le \frac{1}{n}\left[f\left(\frac{x_1+\cdots +x_{n-1}}{n-1}\right)+f\left(\frac{x_2+\cdots +x_n}{n-1}\right)+\cdots +f\left(\frac{x_n+\cdots +x_{n-2}}{n-1}\right)\right].$$

Taking $(p,q)$-integration on both sides of the above inequality on $J^n$, we obtain

$$\begin{array}{cc}\hfill & {\int }_a^{pb+(1-p)a}\cdots {\int }_a^{pb+(1-p)a}f\left(\frac{x_1+\cdots +x_n}{n}\right){}_a{d}_{p,q}{x}_1\cdots {}_a{d}_{p,q}{x}_n\hfill \\ \hfill & \le \frac{1}{n}\left[{\int }_a^{pb+(1-p)a}\cdots {\int }_a^{pb+(1-p)a}f\left(\frac{x_1+\cdots +x_{n-1}}{n-1}\right){}_a{d}_{p,q}{x}_1\cdots {}_a{d}_{p,q}{x}_n+\cdots \right.\hfill \\ \hfill & \left.+{\int }_a^{pb+(1-p)a}\cdots {\int }_a^{pb+(1-p)a}f\left(\frac{x_n+\cdots +x_{n-2}}{n-1}\right){}_a{d}_{p,q}{x}_1\cdots {}_a{d}_{p,q}{x}_n\right].\hfill \end{array}$$

On the other hand, we get

$$\begin{array}{cc}\hfill & {\int }_a^{pb+(1-p)a}\cdots {\int }_a^{pb+(1-p)a}f\left(\frac{x_1+\cdots +x_{n-1}}{n-1}\right){}_a{d}_{p,q}{x}_1\cdots {}_a{d}_{p,q}{x}_n\hfill \\ \hfill & ={\int }_a^{pb+(1-p)a}\cdots {\int }_a^{pb+(1-p)a}f\left(\frac{x_2+\cdots +x_n}{n-1}\right){}_a{d}_{p,q}{x}_1\cdots {}_a{d}_{p,q}{x}_n\hfill \\ \hfill & ⋮\hfill \\ \hfill & ={\int }_a^{pb+(1-p)a}\cdots {\int }_a^{pb+(1-p)a}f\left(\frac{x_n+\cdots +x_{n-2}}{n-1}\right){}_a{d}_{p,q}{x}_1\cdots {}_a{d}_{p,q}{x}_n\hfill \\ \hfill & =p(b-a){\int }_a^{pb+(1-p)a}\cdots {\int }_a^{pb+(1-p)a}f\left(\frac{x_1+\cdots +x_{n-1}}{n-1}\right){}_a{d}_{p,q}{x}_1\cdots {}_a{d}_{p,q}{x}_{n-1}.\hfill \end{array}$$

Thus,

$$\begin{array}{cc}\hfill & \frac{1}{p^n{(b-a)}^n}{\int }_a^{pb+(1-p)a}\cdots {\int }_a^{pb+(1-p)a}f\left(\frac{x_1+\cdots +x_n}{n}\right){}_a{d}_{p,q}{x}_1\cdots {}_a{d}_{p,q}{x}_n\hfill \\ \hfill & \le \frac{1}{p^{n-1}{(b-a)}^{n-1}}{\int }_a^{pb+(1-p)a}\cdots {\int }_a^{pb+(1-p)a}f\left(\frac{x_1+\cdots +x_{n-1}}{n-1}\right){}_a{d}_{p,q}{x}_1\cdots {}_a{d}_{p,q}{x}_{n-1},\hfill \end{array}$$

which shows the middle part of (11).

On the other hand, by Jensen’s inequality, we have

$$\begin{array}{cc}\hfill & f\left(\frac{1}{p^n{(b-a)}^n}{\int }_a^{pb+(1-p)a}\cdots {\int }_a^{pb+(1-p)a}\left(\frac{x_1+\cdots +x_n}{n}\right){}_a{d}_{p,q}{x}_1\cdots {}_a{d}_{p,q}{x}_n\right)\hfill \\ \hfill & \le \frac{1}{p^n{(b-a)}^n}{\int }_a^{pb+(1-p)a}\cdots {\int }_a^{pb+(1-p)a}f\left(\frac{x_1+\cdots +x_n}{n}\right){}_a{d}_{p,q}{x}_1\cdots {}_a{d}_{p,q}{x}_n.\hfill \end{array}$$

Since

$$\frac{1}{p^n{(b-a)}^n}{\int }_a^{pb+(1-p)a}\cdots {\int }_a^{pb+(1-p)a}\left(\frac{x_1+\cdots +x_n}{n}\right){}_a{d}_{p,q}{x}_1\cdots {}_a{d}_{p,q}{x}_n=\frac{qa+pb}{p+q},$$

this yields the first part of (11). This completes the proof. □

Remark 2.

If $p=1$, then (11) reduces to

$$\begin{array}{cc}\hfill f\left(\frac{qa+b}{1+q}\right)& \le \frac{1}{{(b-a)}^n}{\int }_a^b\cdots {\int }_a^{b}f\left(\frac{x_1+\cdots +x_n}{n}\right){}_a{d}_q{x}_1\cdots {}_a{d}_q{x}_n\hfill \\ \hfill & \le \frac{1}{{(b-a)}^{n-1}}{\int }_a^b\cdots {\int }_a^{b}f\left(\frac{x_1+\cdots +x_{n-1}}{n-1}\right){}_a{d}_q{x}_1\cdots {}_a{d}_q{x}_{n-1}\hfill \\ \hfill & ⋮\hfill \\ \hfill & \le \frac{1}{b-a}{\int }_a^{b}f\left(x\right){}_a{d}_{q}x;\hfill \end{array}$$

(12)

see also [65]. In addition, if $q\to 1$ in (12), then (12) reduces to

$$\begin{array}{cc}\hfill f\left(\frac{a+b}{2}\right)& \le \frac{1}{{(b-a)}^n}{\int }_a^b\cdots {\int }_a^{b}f\left(\frac{x_1+\cdots +x_n}{n}\right)d{x}_{1}d\cdots d{x}_n\hfill \\ \hfill & \le \frac{1}{{(b-a)}^n}{\int }_a^b\cdots {\int }_a^{b}f\left(\frac{x_1+\cdots +x_{n-1}}{n-1}\right)d{x}_1\cdots d{x}_{n-1}\hfill \\ \hfill & ⋮\hfill \\ \hfill & \le \frac{1}{b-a}{\int }_a^{b}f\left(x\right)dx,\hfill \end{array}$$

which readily appeared in [66].

Corollary 1.

Let $f:J\to \mathbb{R}$ be a continuous convex function. Then, we have

$$\begin{array}{cc}\hfill f\left(\frac{qa+pb}{p+q}\right)& \le \frac{1}{p^2{(b-a)}^2}{\int }_a^{pb+(1-p)a}{\int }_a^{pb+(1-p)a}f\left(\frac{x_1+x_2}{2}\right){}_a{d}_{p,q}{x}_1{}_a{d}_{p,q}{x}_2\hfill \\ \hfill & \le \frac{1}{p(b-a)}{\int }_a^{pb+(1-p)a}f\left(x\right){}_a{d}_{p,q}x.\hfill \end{array}$$

(13)

Remark 3.

If $p=1$, then (13) reduces to

$$\begin{array}{cc}\hfill f\left(\frac{qa+b}{1+q}\right)& \le \frac{1}{{(b-a)}^2}{\int }_a^b{\int }_a^{b}f\left(\frac{x_1+x_2}{2}\right){}_a{d}_q{x}_1{}_a{d}_q{x}_2\hfill \\ \hfill & \le \frac{1}{b-a}{\int }_a^{b}f\left(x\right){}_a{d}_{q}x;\hfill \end{array}$$

(14)

see also [65]. In addition, if $q\to 1$ in (14), then (14) reduces to

$$\begin{array}{cc}\hfill f\left(\frac{a+b}{2}\right)& \le \frac{1}{{(b-a)}^2}{\int }_a^b{\int }_a^{b}f\left(\frac{x_1+x_2}{2}\right)d{x}_{1}d{x}_2\hfill \\ \hfill & \le \frac{1}{b-a}{\int }_a^{b}f\left(x\right)dx,\hfill \end{array}$$

which readily appeared in [67].

Theorem 5.

Let $f:J\to \mathbb{R}$ be a continuous convex function. Then, we have

$$\begin{array}{cc}\hfill f\left(\frac{qa+pb}{p+q}\right)& \le \frac{1}{p^n{(b-a)}^n}{\int }_a^{pb+(1-p)a}\cdots {\int }_a^{pb+(1-p)a}f\left(\frac{t_1{x}_1+\cdots +t_n{x}_n}{T_n}\right){}_a{d}_{p,q}{x}_1\cdots {}_a{d}_{p,q}{x}_n\hfill \\ \hfill & \le \frac{1}{p(b-a)}{\int }_a^{pb+(1-p)a}f{\left(x\right)}_a{d}_{q}x,\hfill \end{array}$$

(15)

for all $t_i\ge 0$$(i=1,2,\dots ,n)$ with ${\displaystyle \sum _{i=1}^n{t}_i=T_n>0}$ and $n\in \mathbb{N}$.

Proof. 

By Jensen’s inequality, we have

$$f\left(\frac{t_1{x}_1+\cdots +t_n{x}_n}{T_n}\right)\le \frac{1}{T_n}\left[t_{1}f\left(x_1\right)+\cdots +t_{n}f\left(x_n\right)\right]$$

for all $x_i\in J$ and $t_i\ge 0$, where $i=1,2,\dots ,n$. Taking $(p,q)$-integration on both sides of the above inequality on $J^n$, we obtain

$$\begin{array}{cc}\hfill & {\int }_a^{pb+(1-p)a}\cdots {\int }_a^{pb+(1-p)a}f\left(\frac{t_1{x}_1+\cdots +t_n{x}_n}{T_n}\right){}_a{d}_{p,q}{x}_1\cdots {}_a{d}_{p,q}{x}_n\hfill \\ \hfill & \le \frac{1}{T_n}{\int }_a^{pb+(1-p)a}\cdots {\int }_a^{pb+(1-p)a}[t_{1}f\left(x_1\right)+\cdots +t_{n}f\left(x_n\right)]{}_a{d}_{p,q}{x}_1\cdots {}_a{d}_{p,q}{x}_n\hfill \\ \hfill & =p^{n-1}{(b-a)}^{n-1}{\int }_a^{pb+(1-p)a}f\left(x\right){}_a{d}_{p,q}x,\hfill \end{array}$$

which yields the second part of (17).

On the other hand, by Jensen’s inequality, we have

$$\begin{array}{cc}\hfill & f\left(\frac{1}{p^n{(b-a)}^n}{\int }_a^{pb+(1-p)a}\cdots {\int }_a^{pb+(1-p)a}\left(\frac{t_1{x}_1+\cdots +t_n{x}_n}{T_n}\right){}_a{d}_{p,q}{x}_1\cdots {}_a{d}_{p,q}{x}_n\right)\hfill \\ \hfill & \le \frac{1}{p^n{(b-a)}^n}{\int }_a^{pb+(1-p)a}\cdots {\int }_a^{pb+(1-p)a}f\left(\frac{t_1{x}_1+\cdots +t_n{x}_n}{T_n}\right){}_a{d}_{p,q}{x}_1\cdots {}_a{d}_{p,q}{x}_n.\hfill \end{array}$$

Since

$$\frac{1}{p^n{(b-a)}^n}{\int }_a^{pb+(1-p)a}\cdots {\int }_a^{pb+(1-p)a}\left(\frac{t_1{x}_1+\cdots +t_n{x}_n}{T_n}\right){}_a{d}_{p,q}{x}_1\cdots {}_a{d}_{p,q}{x}_n=\frac{qa+pb}{p+q},$$

this yields the first part of (15). This completes the proof. □

Remark 4.

If $p=1$, then (15) reduces to

$$\begin{array}{cc}\hfill f\left(\frac{qa+b}{1+q}\right)& \le \frac{1}{{(b-a)}^n}{\int }_a^b\cdots {\int }_a^{b}f\left(\frac{t_1{x}_1+\cdots +t_n{x}_n}{T_n}\right){}_a{d}_q{x}_1\cdots {}_a{d}_q{x}_n\hfill \\ \hfill & \le \frac{1}{b-a}{\int }_a^{b}f{\left(x\right)}_a{d}_{q}x;\hfill \end{array}$$

(16)

see also [65]. In addition, if $q\to 1$ in (16), then (16) reduces to

$$\begin{array}{cc}\hfill f\left(\frac{a+b}{2}\right)& \le \frac{1}{{(b-a)}^n}{\int }_a^b\cdots {\int }_a^{b}f\left(\frac{t_1{x}_1+\cdots +t_n{x}_n}{T_n}\right)d{x}_1\cdots d{x}_n\hfill \\ \hfill & \le \frac{1}{b-a}{\int }_a^{b}f\left(x\right)dx,\hfill \end{array}$$

which readily appeared in [66].

Corollary 2.

Let $f:J\to \mathbb{R}$ be a continuous convex function. Then, we have

$$\begin{array}{cc}\hfill f\left(\frac{qa+pb}{p+q}\right)& \le \frac{1}{p^2{(b-a)}^2}{\int }_a^{pb+(1-p)a}{\int }_a^{pb+(1-p)a}f\left(t_1{x}_1+t_2{x}_2\right){}_a{d}_q{x}_1{}_a{d}_{p,q}{x}_2\hfill \\ \hfill & \le \frac{1}{p(b-a)}{\int }_a^{pb+(1-p)a}f\left(x\right){}_a{d}_{p,q}x.\hfill \end{array}$$

(17)

Remark 5.

If $p=1$, then (17) reduces to

$$\begin{array}{cc}\hfill f\left(\frac{qa+b}{1+q}\right)& \le \frac{1}{{(b-a)}^2}{\int }_a^b{\int }_a^{b}f\left(t_1{x}_1+t_2{x}_2\right){}_a{d}_q{x}_1{}_a{d}_q{x}_2\hfill \\ \hfill & \le \frac{1}{b-a}{\int }_a^{b}f\left(x\right){}_a{d}_{q}x;\hfill \end{array}$$

(18)

see also [65]. In addition, if $q\to 1$ in (18), then (18) reduces to

$$\begin{array}{cc}\hfill f\left(\frac{a+b}{2}\right)& \le \frac{1}{{(b-a)}^2}{\int }_a^b{\int }_a^{b}f\left(t_1{x}_1+t_2{x}_2\right)d{x}_{1}d{x}_2\hfill \\ \hfill & \le \frac{1}{b-a}{\int }_a^{b}f\left(x\right)dx,\hfill \end{array}$$

which readily appeared in [67].

Theorem 6.

Let $f:J\to \mathbb{R}$ be a continuous convex function. Then, the following inequalities

$$\begin{array}{cc}\hfill f\left(\frac{qa+pb}{p+q}\right)& \le \frac{1}{p^n{(b-a)}^n}{\int }_a^{pb+(1-p)a}\cdots {\int }_a^{pb+(1-p)a}f\left(\frac{x_1+\cdots +x_n}{n}\right){}_a{d}_{p,q}{x}_1\cdots {}_a{d}_{p,q}{x}_n\hfill \\ \hfill & \le \frac{1}{p^n{(b-a)}^n}{\int }_a^{pb+(1-p)a}\cdots {\int }_a^{pb+(1-p)a}f\left(\frac{t_1{x}_1+\cdots +t_n{x}_n}{T_n}\right){}_a{d}_{p,q}{x}_1\cdots {}_a{d}_{p,q}{x}_n\hfill \\ \hfill & \le \frac{1}{p(b-a)}{\int }_a^{pb+(1-p)a}f{\left(x\right)}_a{d}_{p,q}x\hfill \end{array}$$

(19)

are valid for all $t_i\ge 0$$(i=1,2,\dots ,n)$ with ${\displaystyle \sum _{i=1}^n{t}_i=T_n>0}$ and $n\in \mathbb{N}$.

Proof. 

Since

$$\frac{x_1+\cdots +x_n}{n}=\frac{1}{n}\left[\left(\frac{t_1{x}_1+\cdots +t_n{x}_n}{T_n}\right)+\cdots +\left(\frac{t_2{x}_1+\cdots +t_1{x}_n}{T_n}\right)\right],$$

we have

$$f\left(\frac{x_1+\cdots +x_n}{n}\right)\le \frac{1}{n}\left[f\left(\frac{t_1{x}_1+\cdots +t_n{x}_n}{T_n}\right)+\cdots +f\left(\frac{t_2{x}_1+\cdots +t_1{x}_n}{T_n}\right)\right],$$

by using Jensen’s inequality. Taking the $(p,q)$-integration on both sides of the above inequality on $J^n$, we obtain

$$\begin{array}{cc}\hfill & {\int }_a^{pb+(1-p)a}\cdots {\int }_a^{pb+(1-p)a}f\left(\frac{x_1+\cdots +x_n}{n}\right){}_a{d}_{p,q}{x}_1\cdots {}_a{d}_{p,q}{x}_n\hfill \\ \hfill & \le \frac{1}{n}\left[{\int }_a^{pb+(1-p)a}\cdots {\int }_a^{pb+(1-p)a}f\left(\frac{t_1{x}_1+\cdots +t_n{x}_n}{T_n}\right){}_a{d}_{p,q}{x}_1\cdots {}_a{d}_{p,q}{x}_n+\cdots \right.\hfill \\ \hfill & \left.+{\int }_a^{pb+(1-p)a}\cdots {\int }_a^{pb+(1-p)a}f\left(\frac{t_2{x}_1+\cdots +t_1{x}_n}{T_n}\right){}_a{d}_{p,q}{x}_1\cdots {}_a{d}_{p,q}{x}_n\right].\hfill \end{array}$$

Since

$$\begin{array}{cc}\hfill & {\int }_a^{pb+(1-p)a}\cdots {\int }_a^{pb+(1-p)a}f\left(\frac{t_1{x}_1+\cdots +t_n{x}_n}{T_n}\right){}_a{d}_{p,q}{x}_1\cdots {}_a{d}_{p,q}{x}_n\hfill \\ \hfill & ={\int }_a^{pb+(1-p)a}\cdots {\int }_a^{pb+(1-p)a}f\left(\frac{t_1{x}_1+\cdots +t_n{x}_n}{T_n}\right){}_a{d}_{p,q}{x}_1\cdots {}_a{d}_{p,q}{x}_n\hfill \\ \hfill & ⋮\hfill \\ \hfill & ={\int }_a^{pb+(1-p)a}\cdots {\int }_a^{pb+(1-p)a}f\left(\frac{t_2{x}_1+\cdots +t_1{x}_n}{T_n}\right){}_a{d}_{p,q}{x}_1\cdots {}_a{d}_{p,q}{x}_n.\hfill \end{array}$$

Thus,

$$\begin{array}{cc}\hfill & \frac{1}{p^n{(b-a)}^n}{\int }_a^{pb+(1-p)a}\cdots {\int }_a^{pb+(1-p)a}f\left(\frac{x_1+\cdots +x_n}{n}\right){}_a{d}_{p,q}{x}_1\cdots {}_a{d}_{p,q}{x}_n\hfill \\ \hfill & \le \frac{1}{p^n{(b-a)}^n}{\int }_a^{pb+(1-p)a}\cdots {\int }_a^{pb+(1-p)a}f\left(\frac{t_1{x}_1+\cdots +t_n{x}_n}{T_n}\right){}_a{d}_{p,q}{x}_1\cdots {}_a{d}_{p,q}{x}_n.\hfill \end{array}$$

Using Theorems 4 and 5, we can obtain the desired result. □

Remark 6.

If $p=1$, then (19) reduces to

$$\begin{array}{cc}\hfill f\left(\frac{qa+b}{1+q}\right)& \le \frac{1}{{(b-a)}^n}{\int }_a^b\cdots {\int }_a^{b}f\left(\frac{x_1+\cdots +x_n}{n}\right){}_a{d}_q{x}_1\cdots {}_a{d}_q{x}_n\hfill \\ \hfill & \le \frac{1}{{(b-a)}^n}{\int }_a^b\cdots {\int }_a^{b}f\left(\frac{t_1{x}_1+\cdots +t_n{x}_n}{T_n}\right){}_a{d}_q{x}_1\cdots {}_a{d}_q{x}_n\hfill \\ \hfill & \le \frac{1}{b-a}{\int }_a^{b}f{\left(x\right)}_a{d}_{q}x,\hfill \end{array}$$

(20)

see also [65]. In addition, if $q\to 1$ in (20), then (20) reduces to

$$\begin{array}{cc}\hfill f\left(\frac{a+b}{2}\right)& \le \frac{1}{{(b-a)}^n}{\int }_a^b\cdots {\int }_a^{b}f\left(\frac{x_1+\cdots +x_n}{n}\right)d{x}_1\cdots d{x}_n\hfill \\ \hfill & \le \frac{1}{{(b-a)}^n}{\int }_a^b\cdots {\int }_a^{b}f\left(\frac{t_1{x}_1+\cdots +t_n{x}_n}{T_n}\right)d{x}_1\cdots d{x}_n\hfill \\ \hfill & \le \frac{1}{b-a}{\int }_a^{b}f\left(x\right)dx,\hfill \end{array}$$

which readily appeared in [68].

4. Conclusions

In the present paper, we used $(p,q)$-calculus to establish some new refinements of $(p,q)$-Hermite–Hadamard inequalities, which have been expanded to integration on an n-dimensional finite interval. Many existing results in the literature are deduced as special cases of our results for $p=1$ and $q\to 1.$ The results of this paper are new and significantly contribute to the existing literature on the topic.