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Article

# Hermite–Hadamard and Fejér Inequalities for Co-Ordinated (F,G)-Convex Functions on a Rectangle

by
Małgorzata Chudziak
and
Marek Żołdak
*,†
College of Natural Sciences, Institute of Mathematics, University of Rzeszów, Pigonia 1, 35-310 Rzeszów, Poland
*
Author to whom correspondence should be addressed.
These authors contributed equally to this work.
Symmetry 2020, 12(1), 13; https://doi.org/10.3390/sym12010013
Submission received: 14 November 2019 / Revised: 8 December 2019 / Accepted: 15 December 2019 / Published: 19 December 2019
(This article belongs to the Special Issue Symmetry in Functional Equations and Inequalities)

## Abstract

:
We introduce the notion of a co-ordinated $( F , G )$-convex function defined on an interval in $R 2$ and we prove the Hermite–Hadamard and Fejér type inequalities for such functions.
MSC:
26A51; 26B25

## 1. Introduction

The celebrated inequality states that, if $f : [ a , b ] → R$ is a convex function, then
$f a + b 2 ≤ 1 b − a ∫ a b f ( x ) d x ≤ f ( a ) + f ( b ) 2 .$
Furthermore, if $p : [ a , b ] → [ 0 , ∞ )$ is an integrable function symmetric with respect to $a + b 2$, that is
$p ( a + b − x ) = p ( x ) for x ∈ [ a , b ] ,$
then the following weighted generalization of the Hermite–Hadamard inequality is known as the Fejér inequality
$f a + b 2 ≤ ∫ a b f ( x ) p ( x ) d x ∫ a b p ( x ) d x ≤ f ( a ) + f ( b ) 2 .$
Dragomir [1] established a counterpart of the Hermite–Hadamard inequality for co-ordinated convex functions, that is functions $f : [ a , b ] × [ c , d ] → R$ which are convex with respect to each variable separately. It has been proven in [1] that for such functions, the following inequalities hold
$f a + b 2 , c + d 2 ≤ 1 2 1 b − a ∫ a b f x , c + d 2 d x + 1 d − c ∫ c d f a + b 2 , y d y$
$≤ 1 ( b − a ) ( d − c ) ∫ a b ∫ c d f ( x , y ) d y d x$
$≤ 1 4 1 b − a ∫ a b f ( x , c ) d x + 1 b − a ∫ a b f ( x , d ) d x + 1 d − c ∫ c d f ( a , y ) d y + 1 d − c ∫ c d f ( b , y ) d y$
$≤ f ( a , c ) + f ( a , d ) + f ( b , c ) + f ( b , d ) 4 .$
Refinement versions of these inequalities have been presented in [1,2,3].
A counterpart of the Fejér inequality for co-ordinated convex functions has been formulated by Alomari and Darus [4]. They proved that if $p : [ a , b ] × [ c , d ] → [ 0 , ∞ )$ is an integrable function symmetric with respect to the lines $x = a + b 2$ and $y = c + d 2$, i.e.,
$p ( a + b − x , y ) = p ( x , y ) for x ∈ [ a , b ] , y ∈ [ c , d ]$
and
$p ( x , c + d − y ) = p ( x , y ) for x ∈ [ a , b ] , y ∈ [ c , d ] ,$
then for every co-ordinated convex function the following inequalities hold
$f a + b 2 , c + d 2 ≤ ∫ a b ∫ c d f ( x , y ) p ( x , y ) d y d x ∫ a b ∫ c d p ( x , y ) d y d x ≤ f ( a , c ) + f ( a , d ) + f ( b , c ) + f ( b , d ) 4 .$
In recent years, several modifications of the notion of convexity were studied by many authors (see e.g., [5,6,7,8,9]). The following general definition was introduced in [10].
Definition 1.
Let$F : [ 0 , 1 ] × [ a , b ] × [ a , b ] → R$be a continuous function. A function$f : [ a , b ] → R$is said to be convex with respect to F, or briefly F-convex, provided
$f ( t x + ( 1 − t ) y ) ≤ t f ( x ) + ( 1 − t ) f ( y ) + F ( t , x , y ) for x , y ∈ [ a , b ] , t ∈ [ 0 , 1 ] .$
In particular, if F is of the form
$F ( t , x , y ) = C t ( 1 − t ) | x − y | for x , y ∈ [ a , b ] , t ∈ [ 0 , 1 ] ,$
where $C ∈ ( 0 , ∞ )$, then any function $f : [ a , b ] → R$ satisfying (3) is called approximately convex. Furthermore, if $f : [ a , b ] → R$ satisfies (3) with F given by
$F ( t , x , y ) = − C t ( 1 − t ) ( x − y ) 2 for x , y ∈ [ a , b ] , t ∈ [ 0 , 1 ] ,$
where $C ∈ ( 0 , ∞ )$, then it is called strongly convex with modulus C. For some applications of F-convex functions in the optimization theory and in the theory of partial differential equations we refer to [11] and [12], respectively.
It should be noted here that, although a definition of the F-convex function does not require any additional properties of F, it is reasonable to assume that F is symmetric, that is
$F ( 1 − t , y , x ) = F ( t , x , y ) for x , y ∈ [ a , b ] , t ∈ [ 0 , 1 ] .$
In fact, if f is F-convex then there exists a symmetric function $F s$ such that f is $F s$-convex and
$F s ( t , x , y ) ≤ F ( t , x , y ) for x , y ∈ [ a , b ] , t ∈ [ 0 , 1 ] .$
To find this, one could take
$F s ( t , x , y ) : = min { F ( t , x , y ) , F ( 1 − t , y , x ) } for x , y ∈ [ a , b ] , t ∈ [ 0 , 1 ] .$
Note that F given by (4) or (5) is symmetric. Moreover, a symmetry of F is a necessary condition for the existence of an F-affine function, i.e., a function satisfying equation
$f ( t x + ( 1 − t ) y ) = t f ( x ) + ( 1 − t ) f ( y ) + F ( t , x , y ) for x , y ∈ [ a , b ] , t ∈ [ 0 , 1 ] .$
In what follows we deal with the functions of two variables, which are F-convex with respect to each variable.
Definition 2.
Let$F : [ c , d ] × [ 0 , 1 ] × [ a , b ] × [ a , b ] → R$,$G : [ a , b ] × [ 0 , 1 ] × [ c , d ] × [ c , d ] → R$be continuous functions. We call a function$f : [ a , b ] × [ c , d ] → R$co-ordinated$( F , G )$-convex, provided
$f ( t x 1 + ( 1 − t ) x 2 , y ) ≤ t f ( x 1 , y ) + ( 1 − t ) f ( x 2 , y ) + F ( y , t , x 1 , x 2 ) ,$
$f ( x , t y 1 + ( 1 − t ) y 2 ) ≤ t f ( x , y 1 ) + ( 1 − t ) f ( x , y 2 ) + G ( x , t , y 1 , y 2 )$
for $t ∈ [ 0 , 1 ]$, $x 1 , x 2 ∈ [ a , b ]$, $y 1 , y 2 ∈ [ c , d ]$, $x ∈ [ a , b ]$, $y ∈ [ c , d ]$.
Following the remark formulated above, we restrict our attention to the case where $F ( y , · , · , · )$ for $y ∈ [ c , d ]$ and $G ( x , · , · , · )$ for $x ∈ [ a , b ]$ are symmetric functions, i.e.,
$F ( y , 1 − t , x 2 , x 1 ) = F ( y , t , x 1 , x 2 ) for x 1 , x 2 ∈ [ a , b ] , y ∈ [ c , d ] , t ∈ [ 0 , 1 ]$
and
$G ( x , 1 − t , y 2 , y 1 ) = G ( x , t , y 1 , y 2 ) for x ∈ [ a , b ] , y 1 , y 2 ∈ [ c , d ] , t ∈ [ 0 , 1 ] ,$
respectively. This assumption will not be repeated. Our main aim is to present the Hermite–Hadamard and the Fejér type inequalities for co-ordinated $( F , G )$-convex functions.

## 2. Results

In this section, we prove the Hermite–Hadamard type inequalities for $( F , G )$-convex functions. Our proof is based on some methods used in [1,3]. We begin with the result establishing the Hermite–Hadamard type inequalities for F-convex functions. It will be useful in further considerations.
Theorem 1.
Let$F : [ 0 , 1 ] × [ a , b ] × [ a , b ] → R$be a continuous symmetric function (cf. (6)). If$f : [ a , b ] → R$is an integrable F-convex function then
$f a + b 2 ≤ 1 b − a ∫ a b f ( t ) d t + 1 b − a ∫ a b F 1 2 , x , a + b − x d x$
and
$1 b − a ∫ a b f ( t ) d t ≤ f ( a ) + f ( b ) 2 + ∫ 0 1 F ( t , a , b ) d t .$
Proof.
Assume that $f : [ a , b ] → R$ is an integrable F-convex function. In view of (3), we obtain
$1 b − a ∫ a b f ( s ) d s = ∫ 0 1 f ( t a + ( 1 − t ) b ) d t ≤ 1 2 f ( a ) + 1 2 f ( b ) + ∫ 0 1 F ( t , a , b ) d t ,$
which gives (8). Note also that, as f is F-convex, we have
$f x + y 2 ≤ f ( x ) + f ( y ) 2 + F 1 2 , x , y for x , y ∈ [ a , b ] .$
Setting in (9) $x = t a + ( 1 − t ) b$, $y = t b + ( 1 − t ) a$, where $t ∈ [ 0 , 1 ]$, and integrating obtained in this way inequality with respect to t, we obtain (7). □
Now, we are going to formulate and prove the Hermite–Hadamard type inequalities for co-ordinated $( F , G )$-convex functions.
Theorem 2.
Assume that$f : [ a , b ] × [ c , d ] → R$is an integrable co-ordinated$( F , G )$-convex function. Then:
$f a + b 2 , c + d 2 ≤ 1 2 1 b − a ∫ a b f x , c + d 2 d x + 1 d − c ∫ c d f a + b 2 , y d y + R 1 ,$
where
$R 1 = 1 2 1 b − a ∫ a b F c + d 2 , 1 2 , x , a + b − x d x + 1 d − c ∫ c d G a + b 2 , 1 2 , y , c + d − y d y ;$
$1 2 1 b − a ∫ a b f x , c + d 2 d x + 1 d − c ∫ c d f a + b 2 , y d y ≤ 1 ( b − a ) ( d − c ) ∫ a b ∫ c d f ( x , y ) d y d x + R 2 ,$
where
$R 2 = 1 2 ( b − a ) ( d − c ) ∫ a b ∫ c d G x , 1 2 , y , c + d − y d y d x + ∫ a b ∫ c d F y , 1 2 , x , a + b − x d y d x ;$
$1 ( b − a ) ( d − c ) ∫ a b ∫ c d f ( x , y ) d y d x ≤ 1 4 1 b − a ∫ a b f ( x , c ) d x + 1 b − a ∫ a b f ( x , d ) d x + 1 d − c ∫ c d f ( a , y ) d y + 1 d − c ∫ c d f ( b , y ) d y + R 3 ,$
where
$R 3 = 1 2 1 b − a ∫ a b ∫ 0 1 G ( x , t , c , d ) d t d x + 1 d − c ∫ c d ∫ 0 1 F ( y , t , a , b ) d t d y ;$
and
$1 4 1 b − a ∫ a b f ( x , c ) d x + 1 b − a ∫ a b f ( x , d ) d x + 1 d − c ∫ c d f ( a , y ) d y + 1 d − c ∫ c d f ( b , y ) d y ≤ f ( a , c ) + f ( a , d ) + f ( b , c ) + f ( b , d ) 4 + R 4 ,$
where
$R 4 = 1 4 ∫ 0 1 F ( c , t , a , b ) d t + ∫ 0 1 F ( d , t , a , b ) d t + ∫ 0 1 G ( a , t , c , d ) d t + ∫ 0 1 G ( b , t , c , d ) d t .$
Proof.
Note that, for every $x ∈ [ a , b ]$, the function $f ( x , · )$ is $G ( x , · , · , · )$-convex. Thus, applying Theorem 1, we obtain
$f x , c + d 2 ≤ 1 d − c ∫ c d f ( x , y ) d y + 1 d − c ∫ c d G x , 1 2 , y , c + d − y d y$
$≤ f ( x , c ) + f ( x , d ) 2 + ∫ 0 1 G ( x , t , c , d ) d t + 1 d − c ∫ c d G x , 1 2 , y , c + d − y d y .$
Integrating this inequality with respect to x, we find
$1 b − a ∫ a b f x , c + d 2 d x$
$≤ 1 ( b − a ) ( d − c ) ∫ a b ∫ c d f ( x , y ) d y d x + ∫ a b ∫ c d G x , 1 2 , y , c + d − y d y d x$
$≤ 1 2 ( b − a ) ∫ a b f ( x , c ) d x + ∫ a b f ( x , d ) d x$
$+ 1 b − a ∫ a b ∫ 0 1 G ( x , t , c , d ) d t d x + 1 ( b − a ) ( d − c ) ∫ a b ∫ c d G x , 1 2 , y , c + d − y d y d x .$
Moreover, since for every $y ∈ [ c , d ]$, $f ( · , y )$ is $F ( y , · , · , · )$-convex, using the similar arguments, we conclude that
$1 d − c ∫ c d f a + b 2 , y d y$
$≤ 1 ( b − a ) ( d − c ) ∫ c d ∫ a b f ( x , y ) d x d y + ∫ c d ∫ a b F y , 1 2 , x , a + b − x d x d y$
$≤ 1 2 ( d − c ) ∫ c d f ( a , y ) d y + ∫ c d f ( b , y ) d y$
$+ 1 d − c ∫ c d ∫ 0 1 F ( y , t , a , b ) d t d y + 1 ( b − a ) ( d − c ) ∫ c d ∫ a b F y , 1 2 , x , a + b − x d x d y .$
Adding up these inequalities, we obtain (11) and (12).
Since $f ( · , c + d 2 )$ is $F ( c + d 2 , · , · , · )$-convex and $f ( a + b 2 , · )$ is $G ( a + b 2 , · , · , · )$-convex, taking into account the first inequality in Theorem 1, we have
$f a + b 2 , c + d 2 ≤ 1 b − a ∫ a b f x , c + d 2 d x + 1 b − a ∫ a b F c + d 2 , 1 2 , x , a + b − x d x$
and
$f a + b 2 , c + d 2 ≤ 1 d − c ∫ c d f a + b 2 , y d y + 1 d − c ∫ c d G a + b 2 , 1 2 , y , c + d − y d y .$
Adding them up we obtain (10).
Finally, as $f ( · , c )$, $f ( · , d )$, $f ( a , · )$ and $f ( b , · )$ are $F ( c , · , · , · )$-, $F ( d , · , · , · )$-, $G ( a , · , · , · )$- and $G ( b , · , · , · )$-convex, respectively, applying the second inequality in Theorem 1, we find
$1 b − a ∫ a b f ( x , c ) d x ≤ f ( a , c ) + f ( b , c ) 2 + ∫ 0 1 F ( c , t , a , b ) d t ,$
$1 b − a ∫ a b f ( x , d ) d x ≤ f ( a , d ) + f ( b , d ) 2 + ∫ 0 1 F ( d , t , a , b ) d t ,$
$1 d − c ∫ c d f ( a , y ) d y ≤ f ( a , c ) + f ( a , d ) 2 + ∫ 0 1 G ( a , t , c , d ) d t$
and
$1 d − c ∫ c d f ( b , y ) d y ≤ f ( b , c ) + f ( b , d ) 2 + ∫ 0 1 G ( b , t , c , d ) d t .$
Adding up these inequalities, we obtain (13). □

#### 2.2. Fejér Type Inequalities

In order to prove the Fejér type inequalities for co-ordinated $( F , G )$-convex functions we need the following auxiliary result.
Lemma 1.
Assume that$f : [ a , b ] × [ c , d ] → R$is a co-ordinated$( F , G )$-convex function.
(i)
If$[ x 1 , x 2 ] ⊂ [ x 1 ′ , x 2 ′ ] ⊂ [ a , b ]$and$x 1 + x 2 = x 1 ′ + x 2 ′$then
$f ( x 1 , y ) + f ( x 2 , y ) ≤ f ( x 1 ′ , y ) + f ( x 2 ′ , y ) + F y , x 2 ′ − x 1 x 2 ′ − x 1 ′ , x 1 ′ , x 2 ′ + F y , x 2 ′ − x 2 x 2 ′ − x 1 ′ , x 1 ′ , x 2 ′$
for$y ∈ [ c , d ]$.
(ii)
If$[ y 1 , y 2 ] ⊂ [ y 1 ′ , y 2 ′ ] ⊂ [ c , d ]$and$y 1 + y 2 = y 1 ′ + y 2 ′$then
$f ( x , y 1 ) + f ( x , y 2 ) ≤ f ( x , y 1 ′ ) + f ( x , y 2 ′ ) + G x , y 2 ′ − y 1 y 2 ′ − y 1 ′ , y 1 ′ , y 2 ′ + G x , y 2 ′ − y 2 y 2 ′ − y 1 ′ , y 1 ′ , y 2 ′$
for$x ∈ [ a , b ]$.
Proof.
We prove only the first part of the lemma since the proof of the second part is similar. Assume that $[ x 1 , x 2 ] ⊂ [ x 1 ′ , x 2 ′ ] ⊂ [ a , b ]$ and $x 1 + x 2 = x 1 ′ + x 2 ′$. Since
$x 1 = x 2 ′ − x 1 x 2 ′ − x 1 ′ x 1 ′ + x 1 − x 1 ′ x 2 ′ − x 1 ′ x 2 ′$
and
$x 2 = x 2 ′ − x 2 x 2 ′ − x 1 ′ x 1 ′ + x 2 − x 1 ′ x 2 ′ − x 1 ′ x 2 ′ ,$
for every $y ∈ [ c , d ]$, we obtain
$f ( x 1 , y ) + f ( x 2 , y ) ≤ x 2 ′ − x 1 x 2 ′ − x 1 ′ f ( x 1 ′ , y ) + x 1 − x 1 ′ x 2 ′ − x 1 ′ f ( x 2 ′ , y ) + F y , x 2 ′ − x 1 x 2 ′ − x 1 ′ , x 1 ′ , x 2 ′$
$+ x 2 ′ − x 2 x 2 ′ − x 1 ′ f ( x 1 ′ , y ) + x 2 − x 1 ′ x 2 ′ − x 1 ′ f ( x 2 ′ , y ) + F y , x 2 ′ − x 2 x 2 ′ − x 1 ′ , x 1 ′ , x 2 ′$
$= 2 x 2 ′ − ( x 1 + x 2 ) x 2 ′ − x 1 ′ f ( x 1 ′ , y ) + x 1 + x 2 − 2 x 1 ′ x 2 ′ − x 1 ′ f ( x 2 ′ , y ) + F y , x 2 ′ − x 1 x 2 ′ − x 1 ′ , x 1 ′ , x 2 ′ + F y , x 2 ′ − x 2 x 2 ′ − x 1 ′ , x 1 ′ , x 2 ′$
$= f ( x 1 ′ , y ) + f ( x 2 ′ , y ) + F y , x 2 ′ − x 1 x 2 ′ − x 1 ′ , x 1 ′ , x 2 ′ + F y , x 2 ′ − x 2 x 2 ′ − x 1 ′ , x 1 ′ , x 2 ′ .$
In the next theorem we establish the Fejér type inequalities for $( F , G )$-convex functions.
Theorem 3.
Assume that$p : [ a , b ] × [ c , d ] → R$is a positive integrable function symmetric with respect to the lines$x = a + b 2$and$y = c + d 2$(cf. (1) and (2)). If$f : [ a , b ] × [ c , d ] → R$is a continuous co-ordinated$( F , G )$-convex function such that$f p$is integrable on$[ a , b ] × [ c , d ]$then
$f a + b 2 , c + d 2 ≤ ∫ a b ∫ c d f ( x , y ) p ( x , y ) d y d x + K ∫ a b ∫ c d p ( x , y ) d y d x ,$
where
$K = 2 ∫ a a + b 2 ∫ c c + d 2 G x , 1 2 , y , c + d − y p ( x , y ) d y d x$
$+ 2 ∫ a a + b 2 ∫ c c + d 2 G a + b − x , 1 2 , y , c + d − y p ( x , y ) d y d x$
$+ 4 ∫ a a + b 2 ∫ c c + d 2 F c + d 2 , 1 2 , x , a + b − x p ( x , y ) d y d x$
and
$∫ a b ∫ c d f ( x , y ) p ( x , y ) d y d x − L ∫ a b ∫ c d p ( x , y ) d y d x ≤ f ( a , c ) + f ( a , d ) + f ( b , c ) + f ( b , d ) 4 ,$
where
$L = L 1 + L 2 + L 3$
$: = ∫ a a + b 2 ∫ c c + d 2 F y , b − x b − a , a , b + F y , x − a b − a , a , b + F c + d − y , b − x b − a , a , b + F c + d − y , x − a b − a , a , b p ( x , y ) d y d x$
$+ ∫ a a + b 2 ∫ c c + d 2 G a , d − y d − c , c , d + G a , y − c d − c , c , d p ( x , y ) d y d x$
$+ ∫ a a + b 2 ∫ c c + d 2 G b , d − y d − c , c , d + G b , y − c d − c , c , d p ( x , y ) d y d x .$
Proof.
Assume that $f : [ a , b ] × [ c , d ] → R$ is an integrable co-ordinated $( F , G )$-convex function such that $f p$ is integrable. Then, for every $x ∈ [ a , b ]$ and $y ∈ [ c , d ]$, we have
$f a + b 2 , c + d 2 ≤ 1 2 f x , c + d 2 + 1 2 f a + b − x , c + d 2 + F c + d 2 , 1 2 , x , a + b − x$
$≤ 1 4 f ( x , y ) + 1 4 f ( x , c + d − y ) + 1 4 f ( a + b − x , y ) + 1 4 f ( a + b − x , c + d − y )$
$+ 1 2 G x , 1 2 , y , c + d − y + 1 2 G a + b − x , 1 2 , y , c + d − y + F c + d 2 , 1 2 , x , a + b − x .$
Therefore, as p is symmetric with respect to the lines $x = a + b 2$ and $y = c + d 2$, we obtain
$f a + b 2 , c + d 2 ∫ a b ∫ c d p ( x , y ) d y d x = 4 ∫ a a + b 2 ∫ c c + d 2 f a + b 2 , c + d 2 p ( x , y ) d y d x$
$≤ ∫ a a + b 2 ∫ c c + d 2 [ f ( x , y ) + f ( a + b − x , c + d − y ) ] p ( x , y ) d y d x$
$+ ∫ a a + b 2 ∫ c c + d 2 [ f ( x , c + d − y ) + f ( a + b − x , y ) ] p ( x , y ) d y d x + K$
$= ∫ a a + b 2 ∫ c c + d 2 [ f ( x , y ) + f ( a + b − x , c + d − y ) ] p ( x , y ) d y d x$
$+ ∫ a + b 2 b ∫ c c + d 2 [ f ( a + b − x , c + d − y ) + f ( x , y ) ] p ( a + b − x , y ) d y d x + K$
$= ∫ a b ∫ c c + d 2 [ f ( x , y ) + f ( a + b − x , c + d − y ) ] p ( x , y ) d y d x + K$
$= ∫ a b ∫ c c + d 2 f ( x , y ) p ( x , y ) d y d x + ∫ a b ∫ c c + d 2 f ( a + b − x , c + d − y ) p ( x , y ) d y d x + K$
$= ∫ a b ∫ c c + d 2 f ( x , y ) p ( x , y ) d y d x + ∫ a b ∫ c + d 2 d f ( x , y ) p ( a + b − x , c + d − y ) d y d x + K$
$= ∫ a b ∫ c d f ( x , y ) p ( x , y ) d y d x + K .$
Thus, (14) holds.
Furthermore, using again the symmetry of p and applying Lemma 1 to $[ y , c + d − y ] ⊂ [ c , d ]$ and $[ x , a + b − x ] ⊂ [ a , b ]$, where $x ∈ [ a , a + b 2 ]$, $y ∈ [ c , c + d 2 ]$, we have
$f ( a , c ) + f ( a , d ) + f ( b , c ) + f ( b , d ) 4 ∫ a b ∫ c d p ( x , y ) d y d x$
$= ∫ a a + b 2 ∫ c c + d 2 [ f ( a , c ) + f ( a , d ) + f ( b , c ) + f ( b , d ) ] p ( x , y ) d y d x$
$≥ ∫ a a + b 2 ∫ c c + d 2 f ( a , y ) + f ( a , c + d − y ) − G a , d − y d − c , c , d − G a , y − c d − c , c , d$
$+ f ( b , y ) + f ( b , c + d − y ) − G b , d − y d − c , c , d − G b , y − c d − c , c , d p ( x , y ) d y d x$
$≥ ∫ a a + b 2 ∫ c c + d 2 f ( x , y ) + f ( a + b − x , y ) − F y , b − x b − a , a , b − F y , x − a b − a , a , b$
$+ f ( x , c + d − y ) + f ( a + b − x , c + d − y )$
$− F c + d − y , b − x b − a , a , b − F c + d − y , x − a b − a , a , b p ( x , y ) d y d x − ( L 2 + L 3 )$
$= ∫ a a + b 2 ∫ c c + d 2 [ f ( x , y ) + f ( a + b − x , c + d − y ) ] p ( x , y ) d y d x$
$+ ∫ a a + b 2 ∫ c c + d 2 [ f ( a + b − x , y ) + f ( x , c + d − y ) ] p ( x , y ) d y d x − ( L 1 + L 2 + L 3 )$
$= ∫ a a + b 2 ∫ c c + d 2 [ f ( x , y ) + f ( a + b − x , c + d − y ) ] p ( x , y ) d y d x$
$+ ∫ a + b 2 b ∫ c c + d 2 [ f ( a + b − x , c + d − y ) + f ( x , y ) ] p ( a + b − x , y ) d y d x − L$
$= ∫ a b ∫ c c + d 2 [ f ( x , y ) + f ( a + b − x , c + d − y ) ] p ( x , y ) d y d x − L$
$= ∫ a b ∫ c c + d 2 f ( x , y ) p ( x , y ) d y d x + ∫ a b ∫ c c + d 2 f ( a + b − x , c + d − y ) p ( x , y ) d y d x − L$
$= ∫ a b ∫ c c + d 2 f ( x , y ) p ( x , y ) d y d x + ∫ a b ∫ c + d 2 c f ( x , y ) p ( a + b − x , c + d − y ) d y d x − L$
$= ∫ a b ∫ c d f ( x , y ) p ( x , y ) d y d x − L ,$
which gives (15). □

## 3. Discussion

In this paper the Hermite–Hadamard and Fejér type inequalities for co-ordinated $( F , G )$-convex functions are proved. Since every co-ordinated convex function is co-ordinated $( F , G )$-convex (with F and G being identically 0), from our results, one can easily deduce the results by Dragomir [1] and Alomari and Darus [4]. Furthermore, applying Theorems 2 and 3, one can obtain the Hermite–Hadamard and Fejér type inequalities for co-ordinated $( C , D )$-approximately convex functions and co-ordinated $( C , D )$-strongly convex functions defined by
$f ( t x 1 + ( 1 − t ) x 2 , y ) ≤ t f ( x 1 , y ) + ( 1 − t ) f ( x 2 , y ) + D ( y ) t ( 1 − t ) | x 1 − x 2 | ,$
$f ( x , t y 1 + ( 1 − t ) y 2 ) ≤ t f ( x , y 1 ) + ( 1 − t ) f ( x , y 2 ) + C ( x ) t ( 1 − t ) | y 1 − y 2 |$
for $t ∈ [ 0 , 1 ]$, $x 1 , x 2 ∈ [ a , b ]$, $y 1 , y 2 ∈ [ c , d ]$, $x ∈ [ a , b ]$, $y ∈ [ c , d ]$; and
$f ( t x 1 + ( 1 − t ) x 2 , y ) ≤ t f ( x 1 , y ) + ( 1 − t ) f ( x 2 , y ) − D ( y ) t ( 1 − t ) ( x 1 − x 2 ) 2 ,$
$f ( x , t y 1 + ( 1 − t ) y 2 ) ≤ t f ( x , y 1 ) + ( 1 − t ) f ( x , y 2 ) − C ( x ) t ( 1 − t ) ( y 1 − y 2 ) 2$
for $t ∈ [ 0 , 1 ]$, $x 1 , x 2 ∈ [ a , b ]$, $y 1 , y 2 ∈ [ c , d ]$, $x ∈ [ a , b ]$, $y ∈ [ c , d ]$, respectively, where $C : [ a , b ] → ( 0 , ∞ )$ and $D : [ c , d ] → ( 0 , ∞ )$ are given functions.
Note also that from Theorem 1 the Hermite–Hadamard inequalities for approximately convex functions and strongly convex functions can be derived. Finally, applying Theorem 1, with $F ≡ 0$, we obtain the classical Hermite–Hadamard inequality.

## Author Contributions

M.C. and M.Ż. have contributed equally to this paper. All authors have read and agree to the published version of the manuscript.

## Funding

This research received no external funding.

## Conflicts of Interest

The authors declare no conflict of interest.

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MDPI and ACS Style

Chudziak, M.; Żołdak, M. Hermite–Hadamard and Fejér Inequalities for Co-Ordinated (F,G)-Convex Functions on a Rectangle. Symmetry 2020, 12, 13. https://doi.org/10.3390/sym12010013

AMA Style

Chudziak M, Żołdak M. Hermite–Hadamard and Fejér Inequalities for Co-Ordinated (F,G)-Convex Functions on a Rectangle. Symmetry. 2020; 12(1):13. https://doi.org/10.3390/sym12010013

Chicago/Turabian Style

Chudziak, Małgorzata, and Marek Żołdak. 2020. "Hermite–Hadamard and Fejér Inequalities for Co-Ordinated (F,G)-Convex Functions on a Rectangle" Symmetry 12, no. 1: 13. https://doi.org/10.3390/sym12010013

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