Some new Hermite-Hadamard type inequalities for MT-convex functions on differentiable coordinates

P.O. Mohammed

doi:10.1016/j.jksus.2017.07.011

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Original article

30 (

); 258-262

doi:

10.1016/j.jksus.2017.07.011

Some new Hermite-Hadamard type inequalities for MT-convex functions on differentiable coordinates

P.O. Mohammed

Department of Mathematics, College of Education, University of Sulaimani, Sulaimani, Iraq

Received: 2017-5-13, Accepted: 2017-7-29, Published: 2017-04-01

Disclaimer:
This article was originally published by Elsevier and was migrated to Scientific Scholar after the change of Publisher.

Peer review under responsibility of King Saud University.

Abstract

In this paper, we introduce the notion of $MT$ -convex functions on co-ordinates and establish some new integral inequalities of Hermite-Hadamard type for $MT$ -convex functions on co-ordinates on a rectangle $Δ$ in the plane $R^{2}$ .

Keywords

Hermite-Hadamard type inequality

Differentiable co-ordinated convex functions

26D15

26A51

26A33

26A42

1 Introduction

Let us recall some definitions of various convex functions that are known in the literature.

Definition 1.1

Definition 1.1 Guo et al., 2016; Sarikaya et al., 2016

A function $f : I \subseteq R \to R$ is said to be convex on the interval I, if for all $x, y \in I$ and $t \in (0, 1)$ it satisfies the following inequality:

(1.1)

f (tx + (1 - t) y) ⩽ tf (x) + (1 - t) f (y) .

Definition 1.2

Definition 1.2 Tunç et al., 2013; Park, 2015

A function $f : I \subseteq R \to R$ is said to be MT-convex on I, if it is nonnegative and for all $x, y \in I$ and $t \in (0, 1)$ it satisfies the following inequality:

(1.2)

f (tx + (1 - t) y) ⩽ \frac{\sqrt{t}}{2 \sqrt{1 - t}} f (x) + \frac{\sqrt{1 - t}}{2 \sqrt{t}} f (y) .

Example of such functions are:

The functions $f, g : (1, \infty) \to R$ , where $f (x) = x^{p} and g (x) = {(1 + x)}^{p}, p \in (0, \frac{1}{1000})$
The function $h : [1, \frac{3}{2}] \to R$ , where $h (x) = {(1 + x^{2})}^{q}, q \in (0, \frac{1}{1000}) .$

Notice that these functions are not convex.

Definition 1.3

Definition 1.3 Guo et al., 2016

If $(X, A)$ is a measurable space, then $f : X \to R$ is measurable if $f^{- 1} (B) \in A$ for every Borel set $B \in B (R)$ . A function $f : R^{n} \to R$ is Lebesgue measurable if $f^{- 1} (B)$ is a Lebesgue measurable subset of $R^{n}$ for every Borel subset B of $R$ .

Let us now consider a formal definition for co-ordinated convex functions:

Definition 1.4

Definition 1.4 Dragomir et al., 2000; Dragomir, 2001

A function $f : Δ \to R$ is said to be convex on the co-ordinates on $Δ = [a, b] \times [c, d] \subseteq R^{2}$ with $a < b$ and $c < d$ if for all $t, λ \in (0, 1)$ and $(x, y), (z, w) \in Δ$ satisfies the following inequality:

(1.3)

f (tx + (1 - t) z, λ y + (1 - λ) w) ⩽ t λ f (x, y) + t (1 - λ) f (x, w) + (1 - t) λ f (z, y) + (1 - t) (1 - λ) f (z, w) .

Definition 1.5

Definition 1.5 Samko et al., 1993

The incomplete beta function is defined by $B_{x} (a, b) = \int_{0}^{x} z^{a - 1} {(1 - z)}^{b - 1} dz, a, b > 0 .$

For $z = 1$ , the incomplete beta function coincides with the complete beta function.

Throughout this paper we denote by $L_{1} (Δ)$ the set of all Lebesgue integrable functions on $Δ$ as indicated by the authors in Guo et al. (2016). Some integral inequalities of Hermite-Hadamard type for co-ordinated convex functions on the rectangle in the plane $R^{2}$ may be recited as follows:

Theorem 1.1

Theorem 1.1 Dragomir et al., 2000; Dragomir, 2001, Theorem 2.2

Let $f : Δ = [a, b] \times [c, d] \subseteq R^{2} \to R$ be convex on the co-ordinates on $Δ$ with $a < b$ and $c < d$ . Then $\begin{matrix} f (\frac{a + b}{2}, \frac{c + d}{2}) & ⩽ \frac{1}{2} [\frac{1}{b - a} \int_{a}^{b} f (x, \frac{c + d}{2}) dx + \frac{1}{d - c} \int_{c}^{d} f (\frac{a + b}{2}, y) dy] \\ ⩽ \frac{1}{(b - a) (d - c)} \int_{a}^{b} \int_{c}^{d} f (x, y) dy dx \\ ⩽ \frac{1}{4} [\frac{1}{b - a} (\int_{a}^{b} f (x, c) dx + \int_{a}^{b} f (x, d) dx) + \frac{1}{d - c} \int_{c}^{d} (\int_{c}^{d} f (a, y) dx + \int_{c}^{d} f (b, y) dx)] \\ ⩽ \frac{1}{4} [f (a, c) + f (b, c) + f (a, d) + f (b, d)] . \end{matrix}$

Theorem 1.2

Theorem 1.2 Guo et al., 2015, Theorem 2.1

Let $f : Ω \subseteq R^{2} \to R$ be a twice partial differentiable mapping on $Ω^{o}$ (the interior of $Ω$ ) and let $Δ = [a, b] \times [c, d] \subseteq Ω^{o}$ with $a < b, c < d$ and $\frac{\partial^{2} f}{\partial x \partial y} \in L_{1} (Δ)$ . If ${|\frac{\partial^{2} f}{\partial x \partial y}|}^{q}$ is convex on the co-ordinates on $Δ$ and $q ⩾ 1$ , then the following inequality holds: $|I (f)| ⩽ \frac{1}{4} {(\frac{1}{9})}^{\frac{1}{q}} \{g_{q} (1, 2, 2, 4) + g_{q} (4, 2, 2, 1) + g_{q} (2, 1, 4, 2) + g_{q} (2, 4, 1, 2)\},$ where $I (f) = \frac{16}{(b - a) (d - c)} [f (\frac{a + b}{2}, \frac{c + d}{2}) - \frac{1}{d - c} \int_{c}^{d} f (\frac{a + b}{2}, y) dy - \frac{1}{b - a} \int_{a}^{b} f (x, \frac{c + d}{2}) dx + \frac{1}{(b - a) (d - c)} \int_{c}^{d} \int_{a}^{b} f (x, y) dx dy],$ and $g_{q} (r_{1}, r_{2}, r_{3}, r_{4}) = {[r_{1} {|f_{xy} (a, c)|}^{q} + r_{2} {|f_{xy} (a, d)|}^{q} + r_{3} {|f_{xy} (b, c)|}^{q} + r_{4} {|f_{xy} (b, d)|}^{q}]}^{\frac{1}{q}} .$

For more information on integral inequalities of the Hermite-Hadamard type for various kinds of convex functions, the reader is referred to the recently published papers (Park, 2013; Guo et al., 2016; Meftah and Boukerrioua, 2015; Xi and Qi, 2015; Bai et al., 2016 ), and the closely related references therein.

In this paper, we will establish more integral inequalities of the Hermite-Hadamard type for MT-convex functions on the co-ordinates on a rectangle $Δ$ in the plane $R^{2}$ .

2 A definition and a lemma

Motivated by Definitions 1.1 and 1.3, we introduce the notion of “co-ordinated MT-convex function”.

Definition 2.1

We say that a function $f : Δ \to R$ is MT-convex on the co-ordinates on $Δ = [a, b] \times [c, d] \subseteq R^{2}$ with $a < b$ and $c < d$ , if it is nonnegative and for all $t, λ \in (0, 1)$ and $(x, y), (z, w) \in Δ$ it satisfies the following inequality:

(2.1)

\begin{matrix} f (tx + (1 - t) z, λ y + (1 - λ) w) ⩽ \frac{\sqrt{t λ}}{4 \sqrt{(1 - t) (1 - λ)}} f (x, y) \\ + \frac{\sqrt{t (1 - λ)}}{4 \sqrt{λ (1 - t)}} f (x, w) + \frac{\sqrt{λ (1 - t)}}{4 \sqrt{t (1 - λ)}} f (z, y) + \frac{\sqrt{(1 - t) (1 - λ)}}{4 \sqrt{t λ}} f (z, w) . \end{matrix}

Now, we give an example to show that a function can be MT-convex on the co-ordinates on $Δ$ without being convex on the co-ordinates on $Δ$ . The function $f (x, y) : (1, \infty) \times (1, \infty) \to R$ , where $f (x, y) = x^{c} + y^{c} for c \in (0, \frac{1}{1000})$ is MT-convex on the co-ordinates on $Δ = (1, \infty) \times (1, \infty)$ while this is not convex on the co-ordinates on $Δ$ .

In order to prove our main results, we need the following lemma.

Lemma 2.1

Let $f : Ω \subseteq R^{2} \to R$ be a twice partial differentiable mapping on $Ω^{o}$ and let $Δ = [a, b] \times [c, d] \subseteq Ω^{o}$ with $a < b, c < d$ and $\frac{\partial^{2} f}{\partial x \partial y} \in L_{1} (Δ)$ . Then the following equality holds:

(2.2)

\begin{matrix} I (f) : & = \frac{16}{(b - a) (d - c)} [f (\frac{a + b}{2}, \frac{c + d}{2}) - \frac{1}{d - c} \int_{c}^{d} f (\frac{a + b}{2}, y) dy - \frac{1}{b - a} \int_{a}^{b} f (x, \frac{c + d}{2}) dx + \frac{1}{(b - a) (d - c)} \int_{c}^{d} \int_{a}^{b} f (x, y) dx dy] \\ = \int_{0}^{1} \int_{0}^{1} t λ f_{xy} (\frac{t}{2} a + (1 - \frac{t}{2}) b, \frac{λ}{2} c + (1 - \frac{λ}{2}) d) dt d λ + \int_{0}^{1} \int_{0}^{1} t λ f_{xy} ((1 - \frac{t}{2}) a + \frac{t}{2} b, (1 - \frac{λ}{2}) c + \frac{λ}{2} d) dt d λ - \int_{0}^{1} \int_{0}^{1} t λ f_{xy} (\frac{t}{2} a + (1 - \frac{t}{2}) b, (1 - \frac{λ}{2}) c + \frac{λ}{2} d) dt d λ - \int_{0}^{1} \int_{0}^{1} t λ f_{xy} ((1 - \frac{t}{2}) a + \frac{t}{2} b, \frac{λ}{2} c + (1 - \frac{λ}{2}) d) dt d λ . \end{matrix}

Proof

By integration by parts, we have $\begin{matrix} \int_{0}^{1} \int_{0}^{1} t λ f_{xy} (\frac{t}{2} a + (1 - \frac{t}{2}) b, \frac{λ}{2} c + (1 - \frac{λ}{2}) d) dt d λ \\ = \frac{4}{(b - a) (d - c)} [f (\frac{a + b}{2}, \frac{c + d}{2}) - \int_{0}^{1} f (\frac{a + b}{2}, \frac{λ}{2} c + (1 - \frac{λ}{2}) d) d λ \\ - \int_{0}^{1} f (\frac{t}{2} a + (1 - \frac{t}{2}) b, \frac{c + d}{2}) dt + \int_{0}^{1} \int_{0}^{1} f (\frac{t}{2} a + (1 - \frac{t}{2}) b, \frac{λ}{2} c + (1 - \frac{λ}{2}) d) dt d λ] \\ = \frac{4}{(b - a) (d - c)} [f (\frac{a + b}{2}, \frac{c + d}{2}) - \frac{2}{d - c} \int_{\frac{c + d}{2}}^{d} f (\frac{a + b}{2}, y) dy \\ - \frac{2}{b - a} \int_{\frac{a + b}{2}}^{b} f (x, \frac{c + d}{2}) dx + \frac{4}{(b - a) (d - c)} \int_{\frac{c + d}{2}}^{d} \int_{\frac{a + b}{2}}^{b} f (x, y) dx dy] . \end{matrix}$

Similarly, we find $\begin{matrix} \int_{0}^{1} \int_{0}^{1} t λ f_{xy} ((1 - \frac{t}{2}) a + \frac{t}{2} b, (1 - \frac{λ}{2}) c + \frac{λ}{2} d) dt d λ \\ = \frac{4}{(b - a) (d - c)} [f (\frac{a + b}{2}, \frac{c + d}{2}) - \frac{2}{d - c} \int_{c}^{\frac{c + d}{2}} f (\frac{a + b}{2}, y) dy \\ - \frac{2}{b - a} \int_{a}^{\frac{a + b}{2}} f (x, \frac{c + d}{2}) dx + \frac{4}{(b - a) (d - c)} \int_{c}^{\frac{c + d}{2}} \int_{a}^{\frac{a + b}{2}} f (x, y) dx dy], \end{matrix}$ $\begin{matrix} \int_{0}^{1} \int_{0}^{1} t λ f_{xy} (\frac{t}{2} a + (1 - \frac{t}{2}) b, (1 - \frac{λ}{2}) c + \frac{λ}{2} d) dt d λ \\ = - \frac{4}{(b - a) (d - c)} [f (\frac{a + b}{2}, \frac{c + d}{2}) - \frac{2}{d - c} \int_{c}^{\frac{c + d}{2}} f (\frac{a + b}{2}, y) dy \\ - \frac{2}{b - a} \int_{\frac{a + b}{2}}^{b} f (x, \frac{c + d}{2}) dx + \frac{4}{(b - a) (d - c)} \int_{c}^{\frac{c + d}{2}} \int_{\frac{a + b}{2}}^{b} f (x, y) dx dy], \end{matrix}$ and $\begin{matrix} \int_{0}^{1} \int_{0}^{1} t λ f_{xy} ((1 - \frac{t}{2}) a + \frac{t}{2} b, \frac{λ}{2} c + (1 - \frac{λ}{2}) d) dt d λ \\ = - \frac{4}{(b - a) (d - c)} [f (\frac{a + b}{2}, \frac{c + d}{2}) - \frac{2}{d - c} \int_{\frac{c + d}{2}}^{d} f (\frac{a + b}{2}, y) dy \\ - \frac{2}{b - a} \int_{a}^{\frac{a + b}{2}} f (x, \frac{c + d}{2}) dx + \frac{4}{(b - a) (d - c)} \int_{\frac{c + d}{2}}^{d} \int_{a}^{\frac{a + b}{2}} f (x, y) dx dy] . \end{matrix}$

This ends the proof. □

3 Some integral inequalities of the Hermite-Hadamard type

Now we start off to establish some integral inequalities of the Hermite-Hadamard type for the above-introduced MT-convex functions on the co-ordinates.

Theorem 3.1

Let $f : Ω \subseteq R^{2} \to R$ be a twice partial differentiable mapping on $Ω^{o}$ (the interior of $Ω$ ) and let $Δ = [a, b] \times [c, d] \subseteq Ω^{o}$ with $a < b, c < d$ and $\frac{\partial^{2} f}{\partial x \partial y} \in L_{1} (Δ)$ . If ${|\frac{\partial^{2} f}{\partial x \partial y}|}^{q}$ is MT-convex on the co-ordinates on $Δ$ and $q ⩾ 1$ , then the following inequality holds: $|I (f)| ⩽ {(\frac{1}{4})}^{1 - \frac{2}{q}} \{g_{q} (a_{1}, a_{2}, a_{2}, a_{3}) + g_{q} (a_{3}, a_{2}, a_{2}, a_{1}) + g_{q} (a_{2}, a_{1}, a_{3}, a_{2}) + g_{q} (a_{2}, a_{3}, a_{1}, a_{2})\},$ where $g_{q} (r_{1}, r_{2}, r_{3}, r_{4}) = {[r_{1} {|f_{xy} (a, c)|}^{q} + r_{2} {|f_{xy} (a, d)|}^{q} + r_{3} {|f_{xy} (b, c)|}^{q} + r_{4} {|f_{xy} (b, d)|}^{q}]}^{\frac{1}{q}},$ and $a_{1} = B_{\frac{1}{2}}^{2} (\frac{5}{2}, \frac{1}{2}), a_{2} = B_{\frac{1}{2}} (\frac{5}{2}, \frac{1}{2}) B_{\frac{1}{2}} (\frac{3}{2}, \frac{3}{2}), a_{3} = B_{\frac{1}{2}}^{2} (\frac{3}{2}, \frac{3}{2}) .$

Proof

By using Lemma 2.1 and by changing the variables $u = t / 2$ and $v = λ / 2$ , we have $\begin{matrix} | I_{f} | & ⩽ \int_{0}^{1} \int_{0}^{1} t λ |f_{xy} (\frac{t}{2} a + (1 - \frac{t}{2}) b, \frac{λ}{2} c + (1 - \frac{λ}{2}) d)| dt d λ \\ + \int_{0}^{1} \int_{0}^{1} t λ |f_{xy} ((1 - \frac{t}{2}) a + \frac{t}{2} b, (1 - \frac{λ}{2}) c + \frac{λ}{2} d)| dt d λ \\ + \int_{0}^{1} \int_{0}^{1} t λ |f_{xy} (\frac{t}{2} a + (1 - \frac{t}{2}) b, (1 - \frac{λ}{2}) c + \frac{λ}{2} d)| dt d λ \\ + \int_{0}^{1} \int_{0}^{1} t λ |f_{xy} ((1 - \frac{t}{2}) a + \frac{t}{2} b, \frac{λ}{2} c + (1 - \frac{λ}{2}) d)| dt d λ \\ = 16 \{\int_{0}^{\frac{1}{2}} \int_{0}^{\frac{1}{2}} uv |f_{xy} (ua + (1 - u) b, vc + (1 - v) d)| du dv \\ + \int_{0}^{\frac{1}{2}} \int_{0}^{\frac{1}{2}} uv |f_{xy} ((1 - u) a + ub, (1 - v) c + vd)| du dv \\ + \int_{0}^{\frac{1}{2}} \int_{0}^{\frac{1}{2}} uv |f_{xy} (ua + (1 - u) b, (1 - v) c + vd)| du dv \\ + \int_{0}^{\frac{1}{2}} \int_{0}^{\frac{1}{2}} uv |f_{xy} ((1 - u) a + ub, vc + (1 - v) d)| du dv\} . \end{matrix}$

Using the MT-convexity of ${|\frac{\partial^{2} f}{\partial x \partial y}|}^{q}$ on the co-ordinates on $Δ$ and the power-mean integral inequality, we have $\begin{matrix} | I_{f} | ⩽ 16 {(\int_{0}^{\frac{1}{2}} \int_{0}^{\frac{1}{2}} uv du dv)}^{1 - \frac{1}{q}} \{[\int_{0}^{\frac{1}{2}} \int_{0}^{\frac{1}{2}} uv (\frac{\sqrt{uv}}{4 \sqrt{(1 - u) (1 - v)}} {|f_{xy} (a, c)|}^{q} \\ {+ \frac{\sqrt{u (1 - v)}}{4 \sqrt{v (1 - u)}} {|f_{xy} (a, d)|}^{q} + \frac{\sqrt{v (1 - u)}}{4 \sqrt{u (1 - v)}} {|f_{xy} (b, c)|}^{q} + \frac{\sqrt{(1 - u) (1 - v)}}{4 \sqrt{uv}} {|f_{xy} (b, d)|}^{q}) du dv]}^{\frac{1}{q}} \\ + [\int_{0}^{\frac{1}{2}} \int_{0}^{\frac{1}{2}} uv (\frac{\sqrt{(1 - u) (1 - v)}}{4 \sqrt{uv}} {|f_{xy} (a, c)|}^{q} + \frac{\sqrt{v (1 - u)}}{4 \sqrt{u (1 - v)}} {|f_{xy} (a, d)|}^{q} \\ {+ \frac{\sqrt{v (1 - u)}}{4 \sqrt{u (1 - v)}} {|f_{xy} (b, c)|}^{q} + \frac{\sqrt{(1 - u) (1 - v)}}{4 \sqrt{uv}} {|f_{xy} (b, d)|}^{q}) du dv]}^{\frac{1}{q}} \\ + [\int_{0}^{\frac{1}{2}} \int_{0}^{\frac{1}{2}} uv (\frac{\sqrt{u (1 - v)}}{4 \sqrt{v (1 - u)}} {|f_{xy} (a, c)|}^{q} + \frac{\sqrt{uv}}{4 \sqrt{(1 - u) (1 - v)}} {|f_{xy} (a, d)|}^{q} \\ {+ \frac{\sqrt{(1 - u) (1 - v)}}{4 \sqrt{uv}} {|f_{xy} (b, c)|}^{q} + \frac{\sqrt{u (1 - v)}}{4 \sqrt{v (1 - u)}} {|f_{xy} (b, d)|}^{q}) du dv]}^{\frac{1}{q}} \\ + [\int_{0}^{\frac{1}{2}} \int_{0}^{\frac{1}{2}} uv (\frac{\sqrt{v (1 - u)}}{4 \sqrt{u (1 - v)}} {|f_{xy} (a, c)|}^{q} + \frac{\sqrt{(1 - u) (1 - v)}}{4 \sqrt{uv}} {|f_{xy} (a, d)|}^{q} \\ {+ \frac{\sqrt{uv}}{4 \sqrt{(1 - u) (1 - v)}} {|f_{xy} (b, c)|}^{q} + \frac{\sqrt{u (1 - v)}}{4 \sqrt{v (1 - u)}} {|f_{xy} (b, d)|}^{q}) du dv]}^{\frac{1}{q}}\} \end{matrix}$ $⩽ {(\frac{1}{4})}^{1 - \frac{2}{q}} \{g_{q} (a_{1}, a_{2}, a_{2}, a_{3}) + g_{q} (a_{3}, a_{2}, a_{2}, a_{1}) + g_{q} (a_{2}, a_{1}, a_{3}, a_{2}) + g_{q} (a_{2}, a_{3}, a_{1}, a_{2})\} .$

This ends the proof. □

Remark 3.1

Under the assumptions of Theorem 3.1, when $q = 1$ , we have $|I (f)| ⩽ 4 \{g_{q} (a_{1}, a_{2}, a_{2}, a_{3}) + g_{q} (a_{3}, a_{2}, a_{2}, a_{1}) + g_{q} (a_{2}, a_{1}, a_{3}, a_{2}) + g_{q} (a_{2}, a_{3}, a_{1}, a_{2})\} .$

Theorem 3.2

Let $f : Ω \subseteq R^{2} \to R$ be a twice partial differentiable mapping on $Ω^{o}$ and let $Δ = [a, b] \times [c, d] \subseteq Ω^{o}$ with $a < b, c < d$ . If ${|\frac{\partial^{2} f}{\partial x \partial y}|}^{q}$ is MT- convex on the co-ordinates on $Δ$ and $q > 1$ , then the following inequality holds: $|I (f)| ⩽ {(\frac{q - 1}{2 q - 1})}^{2 (1 - \frac{1}{q})} \{g_{q} (b_{1}, b_{2}, b_{2}, b_{3}) + g_{q} (b_{3}, b_{2}, b_{2}, b_{1}) + g_{q} (b_{2}, b_{1}, b_{3}, b_{2}) + g_{q} (b_{2}, b_{3}, b_{1}, b_{2})\},$ where $g_{q} (r_{1}, r_{2}, r_{3}, r_{4})$ is defined in Theorem 3.1 and $b_{1} = B_{\frac{1}{2}}^{2} (\frac{3}{2}, \frac{1}{2}), b_{2} = B_{\frac{1}{2}} (\frac{3}{2}, \frac{1}{2}) B_{\frac{1}{2}} (\frac{1}{2}, \frac{3}{2}), b_{3} = B_{\frac{1}{2}}^{2} (\frac{1}{2}, \frac{3}{2}) .$

Proof

By using Lemma 2.1 and by changing the variables $u = t / 2$ and $v = λ / 2$ , we have $\begin{matrix} | I_{f} | & ⩽ 16 \{\int_{0}^{\frac{1}{2}} \int_{0}^{\frac{1}{2}} uv |f_{xy} (ua + (1 - u) b, vc + (1 - v) d)| du dv \\ + \int_{0}^{\frac{1}{2}} \int_{0}^{\frac{1}{2}} uv |f_{xy} ((1 - u) a + ub, (1 - v) c + vd)| du dv \\ + \int_{0}^{\frac{1}{2}} \int_{0}^{\frac{1}{2}} uv |f_{xy} (ua + (1 - u) b, (1 - v) c + vd)| du dv \\ + \int_{0}^{\frac{1}{2}} \int_{0}^{\frac{1}{2}} uv |f_{xy} ((1 - u) a + ub, vc + (1 - v) d)| du dv\} . \end{matrix}$

Now, by using the MT-convexity of ${|\frac{\partial^{2} f}{\partial x \partial y}|}^{q}$ on the co-ordinates on $Δ$ and Hölder’s inequality, we have $\begin{matrix} | I_{f} | & ⩽ 16 {(\int_{0}^{\frac{1}{2}} \int_{0}^{\frac{1}{2}} {(uv)}^{\frac{q}{q - 1}} du dv)}^{1 - \frac{1}{q}} \{[\int_{0}^{\frac{1}{2}} \int_{0}^{\frac{1}{2}} (\frac{\sqrt{uv}}{4 \sqrt{(1 - u) (1 - v)}} {|f_{xy} (a, c)|}^{q} \\ + {\frac{\sqrt{u (1 - v)}}{4 \sqrt{v (1 - u)}} {|f_{xy} (a, d)|}^{q} + \frac{\sqrt{v (1 - u)}}{4 \sqrt{u (1 - v)}} {|f_{xy} (b, c)|}^{q} + \frac{\sqrt{(1 - u) (1 - v)}}{4 \sqrt{uv}} {|f_{xy} (b, d)|}^{q}) du dv]}^{\frac{1}{q}} \\ + [\int_{0}^{\frac{1}{2}} \int_{0}^{\frac{1}{2}} (\frac{\sqrt{(1 - u) (1 - v)}}{4 \sqrt{uv}} {|f_{xy} (a, c)|}^{q} + \frac{\sqrt{v (1 - u)}}{4 \sqrt{u (1 - v)}} {|f_{xy} (a, d)|}^{q} \\ {+ \frac{\sqrt{v (1 - u)}}{4 \sqrt{u (1 - v)}} {|f_{xy} (b, c)|}^{q} + \frac{\sqrt{(1 - u) (1 - v)}}{4 \sqrt{uv}} {|f_{xy} (b, d)|}^{q}) du dv]}^{\frac{1}{q}} \\ + [\int_{0}^{\frac{1}{2}} \int_{0}^{\frac{1}{2}} (\frac{\sqrt{u (1 - v)}}{4 \sqrt{v (1 - u)}} {|f_{xy} (a, c)|}^{q} + \frac{\sqrt{uv}}{4 \sqrt{(1 - u) (1 - v)}} {|f_{xy} (a, d)|}^{q} \\ + {\frac{\sqrt{(1 - u) (1 - v)}}{4 \sqrt{uv}} {|f_{xy} (b, c)|}^{q} + \frac{\sqrt{u (1 - v)}}{4 \sqrt{v (1 - u)}} {|f_{xy} (b, d)|}^{q}) du dv]}^{\frac{1}{q}} \\ + [\int_{0}^{\frac{1}{2}} \int_{0}^{\frac{1}{2}} (\frac{\sqrt{v (1 - u)}}{4 \sqrt{u (1 - v)}} {|f_{xy} (a, c)|}^{q} + \frac{\sqrt{(1 - u) (1 - v)}}{4 \sqrt{uv}} {|f_{xy} (a, d)|}^{q} \\ {+ \frac{\sqrt{uv}}{4 \sqrt{(1 - u) (1 - v)}} {|f_{xy} (b, c)|}^{q} + \frac{\sqrt{u (1 - v)}}{4 \sqrt{v (1 - u)}} {|f_{xy} (b, d)|}^{q}) du dv]}^{\frac{1}{q}}\} \end{matrix}$ $⩽ {(\frac{q - 1}{2 q - 1})}^{2 (1 - \frac{1}{q})} \{g_{q} (b_{1}, b_{2}, b_{2}, b_{3}) + g_{q} (b_{3}, b_{2}, b_{2}, b_{1}) + g_{q} (b_{2}, b_{1}, b_{3}, b_{2}) + g_{q} (b_{2}, b_{3}, b_{1}, b_{2})\} .$

This ends the proof. □

Remark 3.2

Under the assumptions of Theorem 3.2, when $q = 1$ , we have $|I (f)| {lesg}_{q} (b_{1}, b_{2}, b_{2}, b_{3}) + g_{q} (b_{3}, b_{2}, b_{2}, b_{1}) + g_{q} (b_{2}, b_{1}, b_{3}, b_{2}) + g_{q} (b_{2}, b_{3}, b_{1}, b_{2}) .$

Theorem 3.3

Let $f : Ω \subseteq R^{2} \to R$ be a twice partial differentiable mapping on $Ω^{o}$ and let $Δ = [a, b] \times [c, d] \subseteq Ω^{o}$ with $a < b, c < d$ . If ${|\frac{\partial^{2} f}{\partial x \partial y}|}^{q}$ is MT-convex on the co-ordinates on $Δ$ and $q > 1$ , then the following inequality holds: $|I (f)| ⩽ \frac{q - 1}{2 (2 q - 1)} {(\frac{4 (2 q - 1)}{q - 1})}^{\frac{1}{q}} \times \{g_{q} (c_{1}, c_{2}, c_{3}, c_{4}) + g_{q} (c_{4}, c_{3}, c_{2}, c_{1}) + g_{q} (c_{2}, c_{1}, c_{4}, c_{3}) + g_{q} (c_{3}, c_{4}, c_{1}, c_{2})\},$ where $g_{q} (r_{1}, r_{2}, r_{3}, r_{4})$ is defined in Theorem 3.1 and $\begin{matrix} c_{1} = & B_{\frac{1}{2}} (\frac{5}{2}, \frac{1}{2}) B_{\frac{1}{2}} (\frac{3}{2}, \frac{1}{2}), c_{2} = B_{\frac{1}{2}} (\frac{5}{2}, \frac{1}{2}) B_{\frac{1}{2}} (\frac{1}{2}, \frac{3}{2}), \\ c_{3} = & B_{\frac{1}{2}} (\frac{3}{2}, \frac{3}{2}) B_{\frac{1}{2}} (\frac{3}{2}, \frac{1}{2}), c_{4} = B_{\frac{3}{2}} (\frac{3}{2}, \frac{3}{2}) B_{\frac{1}{2}} (\frac{1}{2}, \frac{3}{2}) . \end{matrix}$

Proof

By using Lemma 2.1, the MT-convexity of ${|\frac{\partial^{2} f}{\partial x \partial y}|}^{q}$ on the co-ordinates on $Δ$ , Hölder’s inequality and changing the variables $u = t / 2$ and $v = λ / 2$ , we have $\begin{matrix} | I_{f} | & ⩽ 16 {(\int_{0}^{\frac{1}{2}} \int_{0}^{\frac{1}{2}} u v^{\frac{q}{q - 1}} du dv)}^{1 - \frac{1}{q}} \{[\int_{0}^{\frac{1}{2}} \int_{0}^{\frac{1}{2}} u (\frac{\sqrt{uv}}{4 \sqrt{(1 - u) (1 - v)}} {|f_{xy} (a, c)|}^{q} \\ {+ \frac{\sqrt{u (1 - v)}}{4 \sqrt{v (1 - u)}} {|f_{xy} (a, d)|}^{q} + \frac{\sqrt{v (1 - u)}}{4 \sqrt{u (1 - v)}} {|f_{xy} (b, c)|}^{q} + \frac{\sqrt{(1 - u) (1 - v)}}{4 \sqrt{uv}} {|f_{xy} (b, d)|}^{q}) du dv]}^{\frac{1}{q}} \\ + [\int_{0}^{\frac{1}{2}} \int_{0}^{\frac{1}{2}} u (\frac{\sqrt{(1 - u) (1 - v)}}{4 \sqrt{uv}} {|f_{xy} (a, c)|}^{q} + \frac{\sqrt{v (1 - u)}}{4 \sqrt{u (1 - v)}} {|f_{xy} (a, d)|}^{q} \\ {+ \frac{\sqrt{v (1 - u)}}{4 \sqrt{u (1 - v)}} {|f_{xy} (b, c)|}^{q} + \frac{\sqrt{(1 - u) (1 - v)}}{4 \sqrt{uv}} {|f_{xy} (b, d)|}^{q}) du dv]}^{\frac{1}{q}} \\ + [\int_{0}^{\frac{1}{2}} \int_{0}^{\frac{1}{2}} u (\frac{\sqrt{u (1 - v)}}{4 \sqrt{v (1 - u)}} {|f_{xy} (a, c)|}^{q} + \frac{\sqrt{uv}}{4 \sqrt{(1 - u) (1 - v)}} {|f_{xy} (a, d)|}^{q} \\ {+ \frac{\sqrt{(1 - u) (1 - v)}}{4 \sqrt{uv}} {|f_{xy} (b, c)|}^{q} + \frac{\sqrt{u (1 - v)}}{4 \sqrt{v (1 - u)}} {|f_{xy} (b, d)|}^{q}) du dv]}^{\frac{1}{q}} \\ + [\int_{0}^{\frac{1}{2}} \int_{0}^{\frac{1}{2}} u (\frac{\sqrt{v (1 - u)}}{4 \sqrt{u (1 - v)}} {|f_{xy} (a, c)|}^{q} + \frac{\sqrt{(1 - u) (1 - v)}}{4 \sqrt{uv}} {|f_{xy} (a, d)|}^{q} \\ {+ \frac{\sqrt{uv}}{4 \sqrt{(1 - u) (1 - v)}} {|f_{xy} (b, c)|}^{q} + \frac{\sqrt{u (1 - v)}}{4 \sqrt{v (1 - u)}} {|f_{xy} (b, d)|}^{q}) du dv]}^{\frac{1}{q}}\} \end{matrix}$ $⩽ \frac{q - 1}{2 (2 q - 1)} {(\frac{4 (2 q - 1)}{q - 1})}^{\frac{1}{q}} \times \{g_{q} (c_{1}, c_{2}, c_{3}, c_{4}) + g_{q} (c_{4}, c_{3}, c_{2}, c_{1}) + g_{q} (c_{2}, c_{1}, c_{4}, c_{3}) + g_{q} (c_{3}, c_{4}, c_{1}, c_{2})\} .$

Theorem 3.3 is thus proved. □

Theorem 3.4

Let $f : Ω \subseteq R^{2} \to R$ be a twice partial differentiable mapping on $Ω^{o}$ and let $Δ = [a, b] \times [c, d] \subseteq Ω^{o}$ with $a < b, c < d$ . If ${|\frac{\partial^{2} f}{\partial x \partial y}|}^{q}$ is an MT- convex on the co-ordinates on $Δ, q ⩾ 1$ and $q ⩾ r, s > 0$ , then the following inequality holds: $|I (f)| ⩽ {(\frac{1}{2})}^{\frac{- (r + s)}{q}} {(\frac{{(q - 1)}^{2}}{(2 q - r - 1) (2 q - s - 1)})}^{1 - \frac{1}{q}} \times \{g_{q} (d_{1}, d_{2}, d_{3}, c_{4}) + g_{q} (d_{4}, d_{3}, d_{2}, d_{1}) + g_{q} (d_{2}, d_{1}, d_{4}, d_{3}) + g_{q} (d_{3}, d_{4}, d_{1}, d_{2})\},$ where $g_{q} (r_{1}, r_{2}, r_{3}, r_{4})$ is as defined in Theorem 3.1 and $\begin{matrix} d_{1} = & B_{\frac{1}{2}} (r + \frac{1}{2}, \frac{1}{2}) B_{\frac{1}{2}} (s + \frac{3}{2}, \frac{1}{2}), d_{2} = B_{\frac{1}{2}} (r + \frac{3}{2}, \frac{1}{2}) B_{\frac{1}{2}} (s + \frac{1}{2}, \frac{3}{2}), \\ d_{3} = & B_{\frac{1}{2}} (r + \frac{1}{2}, \frac{3}{2}) B_{\frac{1}{2}} (s + \frac{3}{2}, \frac{1}{2}), d_{4} = B_{\frac{3}{2}} (r + \frac{1}{2}, \frac{3}{2}) B_{\frac{1}{2}} (s + \frac{1}{2}, \frac{3}{2}) . \end{matrix}$

Proof

From Lemma 2.1, the MT-convexity of ${|\frac{\partial^{2} f}{\partial x \partial y}|}^{q}$ on the co-ordinates on $Δ$ , Hölder’s inequality and changing the variables $u = t / 2$ and $v = λ / 2$ , we have $\begin{matrix} | I_{f} | & ⩽ 16 {(\int_{0}^{\frac{1}{2}} \int_{0}^{\frac{1}{2}} u^{\frac{q - r}{q - 1}} v^{\frac{q - s}{q - 1}} du dv)}^{1 - \frac{1}{q}} \{[\int_{0}^{\frac{1}{2}} \int_{0}^{\frac{1}{2}} u^{r} v^{s} (\frac{\sqrt{uv}}{4 \sqrt{(1 - u) (1 - v)}} {|f_{xy} (a, c)|}^{q} \\ + {\frac{\sqrt{u (1 - v)}}{4 \sqrt{v (1 - u)}} {|f_{xy} (a, d)|}^{q} + \frac{\sqrt{v (1 - u)}}{4 \sqrt{u (1 - v)}} {|f_{xy} (b, c)|}^{q} + \frac{\sqrt{(1 - u) (1 - v)}}{4 \sqrt{uv}} {|f_{xy} (b, d)|}^{q}) du dv]}^{\frac{1}{q}} \\ + [\int_{0}^{\frac{1}{2}} \int_{0}^{\frac{1}{2}} u^{r} v^{s} (\frac{\sqrt{(1 - u) (1 - v)}}{4 \sqrt{uv}} {|f_{xy} (a, c)|}^{q} + \frac{\sqrt{v (1 - u)}}{4 \sqrt{u (1 - v)}} {|f_{xy} (a, d)|}^{q} \\ {+ \frac{\sqrt{v (1 - u)}}{4 \sqrt{u (1 - v)}} {|f_{xy} (b, c)|}^{q} + \frac{\sqrt{(1 - u) (1 - v)}}{4 \sqrt{uv}} {|f_{xy} (b, d)|}^{q}) du dv]}^{\frac{1}{q}} \\ + [\int_{0}^{\frac{1}{2}} \int_{0}^{\frac{1}{2}} u^{r} v^{s} (\frac{\sqrt{u (1 - v)}}{4 \sqrt{v (1 - u)}} {|f_{xy} (a, c)|}^{q} + \frac{\sqrt{uv}}{4 \sqrt{(1 - u) (1 - v)}} {|f_{xy} (a, d)|}^{q} \\ {+ \frac{\sqrt{(1 - u) (1 - v)}}{4 \sqrt{uv}} {|f_{xy} (b, c)|}^{q} + \frac{\sqrt{u (1 - v)}}{4 \sqrt{v (1 - u)}} {|f_{xy} (b, d)|}^{q}) du dv]}^{\frac{1}{q}} \\ + [\int_{0}^{\frac{1}{2}} \int_{0}^{\frac{1}{2}} u^{r} v^{s} (\frac{\sqrt{v (1 - u)}}{4 \sqrt{u (1 - v)}} {|f_{xy} (a, c)|}^{q} + \frac{\sqrt{(1 - u) (1 - v)}}{4 \sqrt{uv}} {|f_{xy} (a, d)|}^{q} \\ {+ \frac{\sqrt{uv}}{4 \sqrt{(1 - u) (1 - v)}} {|f_{xy} (b, c)|}^{q} + \frac{\sqrt{u (1 - v)}}{4 \sqrt{v (1 - u)}} {|f_{xy} (b, d)|}^{q}) du dv]}^{\frac{1}{q}}\} \end{matrix}$ $⩽ {(\frac{1}{2})}^{\frac{- (r + s)}{q}} {(\frac{{(q - 1)}^{2}}{(2 q - r - 1) (2 q - s - 1)})}^{1 - \frac{1}{q}} \times \{g_{q} (d_{1}, d_{2}, d_{3}, c_{4}) + g_{q} (d_{4}, d_{3}, d_{2}, d_{1}) + g_{q} (d_{2}, d_{1}, d_{4}, d_{3}) + g_{q} (d_{3}, d_{4}, d_{1}, d_{2})\} .$

Theorem 3.4 is thus proved. □

Remark 3.3

Under the assumptions of Theorem 3.4, when $r = s = q$ , we have $|I (f)| ⩽ 4 \{g_{q} (d_{1}, d_{2}, d_{3}, c_{4}) + g_{q} (d_{4}, d_{3}, d_{2}, d_{1}) + g_{q} (d_{2}, d_{1}, d_{4}, d_{3}) + g_{q} (d_{3}, d_{4}, d_{1}, d_{2})\} .$

References

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Introduction

A definition and a lemma

Some integral inequalities of the Hermite-Hadamard type

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[1] Bai S.-P., Qi F., Wang S.-H., . Some new integral inequalities of Hermite-Hadamard type for $(α, m; P)$ -convex functions on co-ordinates. J. Appl. Anal. Comput.. 2016;6(1):171-178.
[Google Scholar]

[2] Dragomir S.S., . On Hadamard inequality for convex functions on the co-ordinates in a rectangle from the plane. Taiwanese J. Math.. 2001;5:775-788.
[Google Scholar]

[3] Dragomir, S.S., Pearce, C.E.M. 2000. Selected topics on Hermite-Hadamard inequalities and applications (RGMIA Monographs), Victoria University. Retrieved fromhttp://rgmia.org/monographs/hermitehadamard.html.

[4] Guo X.-Y., Qi F., Xi B.-Y., . Some new Hermite-Hadamard type inequalities for differentiable co-ordinated convex functions. Cogent Math.. 2015;2:1092195.
[Google Scholar]

[5] Guo X.-Y., Qi F., Xi B.-Y., . Some new inequalities of Hermite-Hadamard type for geometrically mean convex functions on the co-ordinates. J. Comput. Anal. Appl.. 2016;21(1):144-155.
[Google Scholar]

[6] Meftah B., Boukerrioua K., . On some Cebysev type inequalities for functions whose second derivatives are $(h_{1}, h_{2})$ -convex on the co-ordinates. Konuralp J. Math.. 2015;3(2):77-88.
[Google Scholar]

[7] Park J., . Generalized inequalities for convex mappings on the co-ordinates. Int. J. Pure Appl. Math.. 2013;82(5):829-842.
[Google Scholar]

[8] Park J., . Some Hermite-Hadamard type inequalities for MT-convex functions via classical and Riemann-Liouville fractional integrals. Appl. Math. Sci.. 2015;9(101):5011-5026.
[Google Scholar]

[9] Samko S.G., Kilbas A.A., Marichev O.I., . Fractional Integrals and Derivatives, Theory and Applications. Yverdon, Switzerland: Gordon and Breach; 1993.

[10] Sarikaya M.Z., Yaldiz H., Budak H., . Some integral inequalities for convex stochastic processes. Acta Math. Univ. Comenianae. 2016;LXXXV(1):155-164.
[Google Scholar]

[11] Tunç M., Subas Y., Karabayir I., . On some Hadamard type inequalities for MT-convex functions. Int. J. Open Probl. Comput. Sci. Math.. 2013;6:102-113.
[Google Scholar]

[12] Xi B.-Y., Qi F., . Some new integral inequalities of Hermite-Hadamard type for $(\log, (α, m))$ -convex functions on co-ordinates. Stud. Univ. Babeş-Bolyai Math.. 2015;60(4):509-525.
[Google Scholar]