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On some Hadamard-type inequalities for ( h 1 , h 2 )-preinvex functions on the co-ordinates

Abstract

We introduce the class of ( h 1 , h 2 )-preinvex functions on the co-ordinates, and we prove some new inequalities of Hermite-Hadamard and Fejér type for such mappings.

MSC:26A15, 26A51, 52A30.

1 Introduction

A function f:IR, IR is an interval, is said to be a convex function on I if

f ( t x + ( 1 t ) y ) tf(x)+(1t)f(y)
(1.1)

holds for all x,yI and t[0,1]. If the reversed inequality in (1.1) holds, then f is concave.

Many important inequalities have been established for the class of convex functions, but the most famous is the Hermite-Hadamard inequality. This double inequality is stated as follows:

f ( a + b 2 ) 1 b a a b f(x)dx f ( a ) + f ( b ) 2 ,
(1.2)

where f:[a,b]R is a convex function. The above inequalities are in reversed order if f is a concave function.

In 1978, Breckner introduced an s-convex function as a generalization of a convex function [1].

Such a function is defined in the following way: a function f:[0,)R is said to be s-convex in the second sense if

f ( t x + ( 1 t ) y ) t s f(x)+ ( 1 t ) s f(y)
(1.3)

holds for all x,y, t[0,1] and for fixed s(0,1].

Of course, s-convexity means just convexity when s=1.

In [2], Dragomir and Fitzpatrick proved the following variant of the Hermite-Hadamard inequality, which holds for s-convex functions in the second sense:

2 s 1 f ( a + b 2 ) 1 b a a b f(x)dx f ( a ) + f ( b ) s + 1 .
(1.4)

In the paper [3] a large class of non-negative functions, the so-called h-convex functions, is considered. This class contains several well-known classes of functions such as non-negative convex functions and s-convex in the second sense functions. This class is defined in the following way: a non-negative function f:IR, IR is an interval, is called h-convex if

f ( t x + ( 1 t ) y ) h(t)f(x)+h(1t)f(y)
(1.5)

holds for all x,yI, t(0,1), where h:JR is a non-negative function, h0 and J is an interval, (0,1)J.

In the further text, functions h and f are considered without assumption of non-negativity.

In [4] Sarikaya, Saglam and Yildirim proved that for an h-convex function the following variant of the Hadamard inequality is fulfilled:

1 2 h ( 1 2 ) f ( a + b 2 ) 1 b a a b f(x)dx [ f ( a ) + f ( b ) ] 0 1 h(t)dt.
(1.6)

In [5] Bombardelli and Varošanec proved that for an h-convex function the following variant of the Hermite-Hadamard-Fejér inequality holds:

a b w ( x ) d x 2 h ( 1 2 ) f ( a + b 2 ) a b f ( x ) w ( x ) d x ( b a ) ( f ( a ) + f ( b ) ) 0 1 h ( t ) w ( t a + ( 1 t ) b ) d t ,
(1.7)

where w:[a,b]R, w0 and symmetric with respect to a + b 2 .

A modification for convex functions, which is also known as co-ordinated convex functions, was introduced by Dragomir [6] as follows.

Let us consider a bidimensional Δ=[a,b]×[c,d] in R 2 with a<b and c<d. A mapping f:ΔR is said to be convex on the co-ordinates on Δ if the partial mappings f y :[a,b]R, f y (u)=f(u,y) and f x :[c,d]R, f x (v)=f(x,v) are convex for all x[a,b] and y[c,d].

In the same article, Dragomir established the following Hadamard-type inequalities for convex functions on the co-ordinates:

f ( a + b 2 , c + d 2 ) 1 ( b a ) ( d c ) a b c d f ( x , y ) d x d y f ( a , c ) + f ( b , c ) + f ( a , d ) + f ( b , d ) 4 .
(1.8)

The concept of s-convex functions on the co-ordinates was introduced by Alomari and Darus [7]. Such a function is defined in following way: the mapping f:ΔR is s-convex in the second sense if the partial mappings f y :[a,b]R and f x :[c,d]R are s-convex in the second sense.

In the same paper, they proved the following inequality for an s-convex function:

4 s 1 f ( a + b 2 , c + d 2 ) 1 ( b a ) ( d c ) a b c d f ( x , y ) d x d y f ( a , c ) + f ( b , c ) + f ( a , d ) + f ( b , d ) ( s + 1 ) 2 .
(1.9)

For refinements and counterparts of convex and s-convex functions on the co-ordinates, see [610].

The main purpose of this paper is to introduce the class of ( h 1 , h 2 )-preinvex functions on the co-ordinates and establish new inequalities like those given by Dragomir in [6] and Bombardelli and Varošanec in [5].

Throughout this paper, we assume that considered integrals exist.

2 Main results

Let f:XR and η:X×X R n , where X is a nonempty closed set in R n , be continuous functions. First, we recall the following well-known results and concepts; see [1116] and the references therein.

Definition 2.1 Let uX. Then the set X is said to be invex at u with respect to η if

u+tη(v,u)X

for all vX and t[0,1].

X is said to be an invex set with respect to η if X is invex at each uX.

Definition 2.2 The function f on the invex set X is said to be preinvex with respect to η if

f ( u + t η ( v , u ) ) (1t)f(u)+tf(v)

for all u,vX and t[0,1].

We also need the following assumption regarding the function η which is due to Mohan and Neogy [11].

Condition C Let XR be an open invex subset with respect to η. For any x,yX and any t[0,1],

η ( y , y + t η ( x , y ) ) = t η ( x , y ) , η ( x , y + t η ( x , y ) ) = ( 1 t ) η ( x , y ) .

Note that for every x,yX and every t 1 , t 2 [0,1] from Condition C, we have

η ( y + t 2 η ( x , y ) , y + t 1 η ( x , y ) ) =( t 2 t 1 )η(x,y).

In [12], Noor proved the Hermite-Hadamard inequality for preinvex functions

f ( a + 1 2 η ( b , a ) ) 1 η ( b , a ) a a + η ( b , a ) f(x)dx f ( a ) + f ( b ) 2 .
(2.1)

Definition 2.3 Let h:[0,1]R be a non-negative function, h0. The non-negative function f on the invex set X is said to be h-preinvex with respect to η if

f ( u + t η ( v , u ) ) h(1t)f(u)+h(t)f(v)

for each u,vX and t[0,1].

Let us note that:

− if η(v,u)=vu, then we get the definition of an h-convex function introduced by Varošanec in [3];

− if h(t)=t, then our definition reduces to the definition of a preinvex function;

− if η(v,u)=vu and h(t)=t, then we obtain the definition of a convex function.

Now let X 1 and X 2 be nonempty subsets of R n , let η 1 : X 1 × X 1 R n and η 2 : X 2 × X 2 R n .

Definition 2.4 Let (u,v) X 1 × X 2 . We say X 1 × X 2 is invex at (u,v) with respect to η 1 and η 2 if for each (x,y) X 1 × X 2 and t 1 , t 2 [0,1],

( u + t 1 η 1 ( x , u ) , v + t 2 η 2 ( y , v ) ) X 1 × X 2 .

X 1 × X 2 is said to be an invex set with respect to η 1 and η 2 if X 1 × X 2 is invex at each (u,v) X 1 × X 2 .

Definition 2.5 Let h 1 and h 2 be non-negative functions on [0,1], h 1 0, h 2 0. The non-negative function f on the invex set X 1 × X 2 is said to be co-ordinated ( h 1 , h 2 )-preinvex with respect to η 1 and η 2 if the partial mappings f y : X 1 R, f y (x)=f(x,y) and f x : X 2 R, f x (y)=f(x,y) are h 1 -preinvex with respect to η 1 and h 2 -preinvex with respect to η 2 , respectively, for all y X 2 and x X 1 .

If η 1 (x,u)=xu and η 2 (y,v)=yv, then the function f is called ( h 1 , h 2 )-convex on the co-ordinates.

Remark 1 From the above definition it follows that if f is a co-ordinated ( h 1 , h 2 )-preinvex function, then

f ( x + t 1 η 1 ( b , x ) , y + t 2 η 2 ( d , y ) ) h 1 ( 1 t 1 ) f ( x , y + t 2 η 2 ( d , y ) ) + h 1 ( t 1 ) f ( b , y + t 2 η 2 ( d , y ) ) h 1 ( 1 t 1 ) h 2 ( 1 t 2 ) f ( x , y ) + h 1 ( 1 t 1 ) h 2 ( t 2 ) f ( x , d ) + h 1 ( t 1 ) h 2 ( 1 t 2 ) f ( b , y ) + h 1 ( t 1 ) h 2 ( t 2 ) f ( b , d ) .

Remark 2 Let us note that if η 1 (x,u)=xu, η 2 (y,v)=yv, t 1 = t 2 and h 1 (t)= h 2 (t)=t, then our definition of a co-ordinated ( h 1 , h 2 )-preinvex function reduces to the definition of a convex function on the co-ordinates proposed by Dragomir [6]. Moreover, if h 1 (t)= h 2 (t)= t s , then our definition reduces to the definition of an s-convex function on the co-ordinates proposed by Alomari and Darus [7].

Now, we will prove the Hadamard inequality for the new class functions.

Theorem 2.1 Suppose that f:[a,a+η(b,a)]R is an h-preinvex function, Condition  C for η holds and a<a+η(b,a), h( 1 2 )>0. Then the following inequalities hold:

1 2 h ( 1 2 ) f ( a + 1 2 η ( b , a ) ) 1 η ( b , a ) a a + η ( b , a ) f(x)dx [ f ( a ) + f ( b ) ] 0 1 h(t)dt.
(2.2)

Proof From the definition of an h-preinvex function, we have that

f ( a + t η ( b , a ) ) h(1t)f(a)+h(t)f(b).

Thus, by integrating, we obtain

0 1 f ( a + t η ( b , a ) ) dtf(a) 0 1 h(1t)dt+f(b) 0 1 h(t)dt= [ f ( a ) + f ( b ) ] 0 1 h(t)dt.

But

0 1 f ( a + t η ( b , a ) ) dt= 1 η ( b , a ) a a + η ( b , a ) f(x)dx.

So,

1 η ( b , a ) a a + η ( b , a ) f(x)dx [ f ( a ) + f ( b ) ] 0 1 h(t)dt.

The proof of the second inequality follows by using the definition of an h-preinvex function, Condition C for η and integrating over [0,1].

That is,

f ( a + 1 2 η ( b , a ) ) = f ( a + t η ( b , a ) + 1 2 η ( a + ( 1 t ) η ( b , a ) , a + t η ( b , a ) ) h ( 1 2 ) [ f ( a + t η ( b , a ) ) + f ( a + ( 1 t ) η ( b , a ) ) ] , f ( a + 1 2 η ( b , a ) ) h ( 1 2 ) [ 0 1 f ( a + t η ( b , a ) ) d t + 0 1 f ( a + ( 1 t ) η ( b , a ) ) ] , f ( a + 1 2 η ( b , a ) ) 2 h ( 1 2 ) 1 η ( b , a ) a a + η ( b , a ) f ( x ) d x .

The proof is complete. □

Theorem 2.2 Suppose that f:[a,a+ η 1 (b,a)]×[c,c+ η 2 (d,c)]R is an ( h 1 , h 2 )-preinvex function on the co-ordinates with respect to η 1 and η 2 , Condition  C for η 1 and η 2 is fulfilled, and a<a+ η 1 (b,a), c<c+ η 2 (d,c), and h 1 ( 1 2 )>0, h 2 ( 1 2 )>0. Then one has the following inequalities:

(2.3)

Proof Since f is ( h 1 , h 2 )-preinvex on the co-ordinates, it follows that the mapping f x is h 2 -preinvex and the mapping f y is h 1 -preinvex. Then, by the inequality (2.2), one has

1 2 h 2 ( 1 2 ) f ( x , c + 1 2 η 2 ( d , c ) ) 1 η 2 ( d , c ) c c + η 2 ( d , c ) f ( x , y ) d y [ f ( x , c ) + f ( x , d ) ] 0 1 h 2 ( t ) d t

and

1 2 h 1 ( 1 2 ) f ( a + 1 2 η 1 ( b , a ) , y ) 1 η 1 ( b , a ) a a + η 1 ( b , a ) f ( x , y ) d x [ f ( a , y ) + f ( b , y ) ] 0 1 h 1 ( t ) d t .

Dividing the above inequalities for η 1 (b,a) and η 2 (d,c) and then integrating the resulting inequalities on [a,a+ η 1 (b,a)] and [c,c+ η 2 (d,c)], respectively, we have

1 η 1 ( b , a ) 2 h 2 ( 1 2 ) a a + η 1 ( b , a ) f ( x , c + 1 2 η 2 ( d , c ) ) d x 1 η 1 ( b , a ) η 2 ( d , c ) a a + η 1 ( b , a ) c c + η 2 ( d , c ) f ( x , y ) d x d y 1 η 1 ( b , a ) 0 1 h 2 ( t ) d t [ a a + η 1 ( b , a ) f ( x , c ) d x + a a + η 1 ( b , a ) f ( x , d ) d x ]

and

1 η 2 ( b , a ) 2 h 1 ( 1 2 ) c c + η 2 ( d , c ) f ( a + 1 2 η 1 ( b , a ) , y ) d y 1 η 1 ( b , a ) η 2 ( d , c ) a a + η 1 ( b , a ) c c + η 2 ( d , c ) f ( x , y ) d x d y 1 η 2 ( d , c ) 0 1 h 1 ( t ) d t [ c c + η 2 ( c , d ) f ( a , y ) d y + c c + η 2 ( c , d ) f ( b , y ) d y ] .

Summing the above inequalities, we get the second and the third inequalities in (2.3).

By the inequality (2.2), we also have

1 2 h 2 ( 1 2 ) f ( a + 1 2 η 1 ( b , a ) , c + 1 2 η 2 ( d , c ) ) 1 η 2 ( d , c ) c c + η 2 ( d , c ) f ( a + 1 2 η 1 ( b , a ) , y ) dy

and

1 2 h 1 ( 1 2 ) f ( a + 1 2 η 1 ( b , a ) , c + 1 2 η 2 ( d , c ) ) 1 η 1 ( b , a ) a a + η 1 ( b , a ) f ( x , c + 1 2 η 2 ( d , c ) ) dx,

which give, by addition, the first inequality in (2.3).

Finally, by the same inequality (2.2), we ca also state

1 η 2 ( d , c ) c c + η 2 ( d , c ) f ( a , y ) d y [ f ( a , c ) + f ( a , d ) ] 0 1 h 2 ( t ) d t , 1 η 2 ( d , c ) c c + η 2 ( d , c ) f ( b , y ) d y [ f ( b , c ) + f ( b , d ) ] 0 1 h 2 ( t ) d t , 1 η 1 ( b , a ) a a + η 1 ( b , a ) f ( x , c ) d x [ f ( a , c ) + f ( b , c ) ] 0 1 h 1 ( t ) d t , 1 η 1 ( b , a ) a a + η 1 ( b , a ) f ( x , d ) d x [ f ( a , d ) + f ( b , d ) ] 0 1 h 1 ( t ) d t ,

which give, by addition, the last inequality in (2.3). □

Remark 3 In particular, for η 1 (b,a)=ba, η 2 (d,c)=dc, h 1 ( t 1 )= h 2 ( t 2 )=t, we get the inequalities obtained by Dragomir [6] for functions convex on the co-ordinates on the rectangle from the plane R 2 .

Remark 4 If η 1 (b,a)=ba, η 2 (d,c)=dc, and h 1 ( t 1 )= h 2 ( t 2 )= t s , then we get the inequalities obtained by Alomari and Darus in [7] for s-convex functions on the co-ordinates on the rectangle from the plane R 2 .

Theorem 2.3 Let f,g:[a,a+ η 1 (b,a)]×[c,c+ η 2 (d,c)]R with a<a+ η 1 (b,a), c<c+ η 2 (d,c). If f is ( h 1 , h 2 )-preinvex on the co-ordinates and g is ( k 1 , k 2 )-preinvex on the co-ordinates with respect to η 1 and η 2 , then

1 η 1 ( b , a ) η 2 ( d , c ) a a + η 1 ( b , a ) c c + η 2 ( d , c ) f ( x , y ) g ( x , y ) d x d y M 1 ( a , b , c , d ) 0 1 0 1 h 1 ( t 1 ) h 2 ( t 2 ) k 1 ( t 1 ) k 2 ( t 2 ) d t 1 d t 2 + M 2 ( a , b , c , d ) 0 1 0 1 h 1 ( t 1 ) h 2 ( t 2 ) k 1 ( t 1 ) k 2 ( 1 t 2 ) d t 1 d t 2 + M 3 ( a , b , c , d ) 0 1 0 1 h 1 ( t 1 ) h 2 ( t 2 ) k 1 ( 1 t 1 ) k 2 ( t 2 ) d t 1 d t 2 + M 4 ( a , b , c , d ) 0 1 0 1 h 1 ( t 1 ) h 2 ( t 2 ) k 1 ( 1 t 1 ) k 2 ( 1 t 2 ) d t 1 d t 2 ,

where

M 1 ( a , b , c , d ) = f ( a , c ) g ( a , c ) + f ( a , d ) g ( a , d ) + f ( b , c ) g ( b , c ) + f ( b , d ) g ( b , d ) , M 2 ( a , b , c , d ) = f ( a , c ) g ( a , d ) + f ( a , d ) g ( a , c ) + f ( b , c ) g ( b , d ) + f ( b , d ) g ( b , c ) , M 3 ( a , b , c , d ) = f ( a , c ) g ( b , c ) + f ( a , d ) g ( b , d ) + f ( b , c ) g ( a , c ) + f ( b , d ) g ( a , d ) , M 4 ( a , b , c , d ) = f ( a , c ) g ( b , d ) + f ( a , d ) g ( b , c ) + f ( b , c ) g ( a , d ) + f ( b , d ) g ( a , c ) .

Proof Since f is ( h 1 , h 2 )-preinvex on the co-ordinates and g is ( k 1 , k 2 )-preinvex on the co-ordinates with respect to η 1 and η 2 , it follows that

f ( a + t 1 η 1 ( b , a ) , c + t 2 η 2 ( d , c ) ) h 1 ( 1 t 1 ) h 2 ( 1 t 2 ) f ( a , c ) + h 1 ( 1 t 1 ) h 2 ( t 2 ) f ( a , d ) + h 1 ( t 1 ) h 2 ( 1 t 2 ) f ( b , c ) + h 1 ( t 1 ) h 2 ( t 2 ) f ( b , d )

and

g ( a + t 1 η 1 ( b , a ) , c + t 2 η 2 ( d , c ) ) k 1 ( 1 t 1 ) k 2 ( 1 t 2 ) g ( a , c ) + k 1 ( 1 t 1 ) k 2 ( t 2 ) g ( a , d ) + k 1 ( t 1 ) k 2 ( 1 t 2 ) g ( b , c ) + k 1 ( t 1 ) k 2 ( t 2 ) g ( b , d ) .

Multiplying the above inequalities and integrating over [ 0 , 1 ] 2 and using the fact that

0 1 0 1 f ( a + t 1 η 1 ( b , a ) , c + t 2 η 2 ( d , c ) ) g ( a + t 1 η 1 ( b , a ) , c + t 2 η 2 ( d , c ) ) d t 1 d t 2 = 1 η 1 ( b , a ) η 2 ( d , c ) a a + η 1 ( b , a ) c c + η 2 ( d , c ) f ( x , y ) g ( x , y ) d x d y ,

we obtain our inequality. □

In the next two theorems, we will prove the so-called Hermite-Hadamard-Fejér inequalities for an ( h 1 , h 2 )-preinvex function.

Theorem 2.4 Let f:[a,a+ η 1 (b,a)]×[c,c+ η 2 (d,c)]R be ( h 1 , h 2 )-preinvex on the co-ordinates with respect to η 1 and η 2 , a<a+ η 1 (b,a), c<c+ η 2 (d,c), and w:[a,a+ η 1 (b,a)]×[c,c+ η 2 (d,c)]R, w0, symmetric with respect to

( a + 1 2 η 1 ( b , a ) , c + 1 2 η 2 ( d , c ) ) .

Then

1 η 1 ( b , a ) η 2 ( d , c ) a a + η 1 ( b , a ) c c + η 2 ( d , c ) f ( x , y ) w ( x , y ) d x d y [ f ( a , c ) + f ( a , d ) + f ( b , c ) + f ( b , d ) ] 0 1 0 1 h 1 ( t 1 ) h 2 ( t 2 ) w ( a + t 1 η 1 ( b , a ) , c + t 2 η 2 ( d , c ) ) d t 1 d t 2 .
(2.4)

Proof From the definition of ( h 1 , h 2 )-preinvex on the co-ordinates with respect to η 1 and η 2 , we have

  1. (a)
    f ( a + t 1 η 1 ( b , a ) , c + t 2 η 2 ( d , c ) ) h 1 ( 1 t 1 ) h 2 ( 1 t 2 ) f ( a , c ) + h 1 ( 1 t 1 ) h 2 ( t 2 ) f ( a , d ) + h 1 ( t 1 ) h 2 ( 1 t 2 ) f ( b , c ) + h 1 ( t 1 ) h 2 ( t 2 ) f ( b , d ) ,
  2. (b)
    f ( a + ( 1 t 1 ) η 1 ( b , a ) , c + ( 1 t 2 ) η 2 ( d , c ) ) h 1 ( t 1 ) h 2 ( t 2 ) f ( a , c ) + h 1 ( t 1 ) h 2 ( 1 t 2 ) f ( a , d ) + h 1 ( 1 t 1 ) h 2 ( t 2 ) f ( b , c ) + h 1 ( 1 t 1 ) h 2 ( 1 t 2 ) f ( b , d ) ,
  3. (c)
    f ( a + t 1 η 1 ( b , a ) , c + ( 1 t 2 ) η 2 ( d , c ) ) h 1 ( 1 t 1 ) h 2 ( t 2 ) f ( a , c ) + h 1 ( 1 t 1 ) h 2 ( 1 t 2 ) f ( a , d ) + h 1 ( t 1 ) h 2 ( t 2 ) f ( b , c ) + h 1 ( t 1 ) h 2 ( 1 t 2 ) f ( b , d ) ,
  4. (d)
    f ( a + ( 1 t 1 ) η 1 ( b , a ) , c + t 2 η 2 ( d , c ) ) h 1 ( t 1 ) h 2 ( 1 t 2 ) f ( a , c ) + h 1 ( t 1 ) h 2 ( t 2 ) f ( a , d ) + h 1 ( 1 t 1 ) h 2 ( 1 t 2 ) f ( b , c ) + h 1 ( 1 t 1 ) h 2 ( t 2 ) f ( b , d ) .

Multiplying both sides of the above inequalities by w(a+ t 1 η 1 (b,a),c+ t 2 η 2 (d,c)), w(a+(1 t 1 ) η 1 (b,a),c+(1 t 2 ) η 2 (d,c)), w(a+ t 1 η 1 (b,a),c+(1 t 2 ) η 2 (d,c)), w(a+(1 t 1 ) η 1 (b,a),c+ t 2 η 2 (d,c)), respectively, adding and integrating over [ 0 , 1 ] 2 , we obtain

4 η 1 ( b , a ) η 2 ( d , c ) a a + η 1 ( b , a ) c c + η 2 ( d , c ) f ( x , y ) w ( x , y ) d x d y [ f ( a , c ) + f ( a , d ) + f ( b , c ) + f ( b , d ) ] 4 0 1 0 1 h 1 ( t 1 ) h 2 ( t 2 ) w ( a + t 1 η 1 ( b , a ) , c + t 2 η 2 ( d , c ) ) d t 1 d t 2 ,

where we use the symmetricity of the w with respect to (a+ 1 2 η 1 (b,a),c+ 1 2 η 2 (d,c)), which completes the proof. □

Theorem 2.5 Let f:[a,a+ η 1 (b,a)]×[c,c+ η 2 (d,c)]R be ( h 1 , h 2 )-preinvex on the co-ordinates with respect to η 1 and η 2 , and a<a+ η 1 (b,a), c<c+ η 2 (d,c), w:[a,a+ η 1 (b,a)]×[c,c+ η 2 (d,c)]R, w0, symmetric with respect to (a+ 1 2 η 1 (b,a),c+ 1 2 η 2 (d,c)). Then, if Condition  C for η 1 and η 2 is fulfilled, we have

f ( a + 1 2 η 1 ( b , a ) , c + 1 2 η 2 ( d , c ) ) a a + η 1 ( b , a ) c c + η 2 ( d , c ) w ( x , y ) d x d y 4 h 1 ( 1 2 ) h 2 ( 1 2 ) a a + η 1 ( b , a ) c c + η 2 ( d , c ) f ( x , y ) w ( x , y ) d x d y .
(2.5)

Proof Using the definition of an ( h 1 , h 2 )-preinvex function on the co-ordinates and Condition C for η 1 and η 2 , we obtain

f ( a + 1 2 η 1 ( b , a ) , c + 1 2 η 2 ( d , c ) ) h 1 ( 1 2 ) h 2 ( 1 2 ) [ f ( a + t 1 η 1 ( b , a ) , c + t 2 η 2 ( d , c ) ) + f ( a + t 1 η 1 ( b , a ) , c + ( 1 t 2 ) η 2 ( d , c ) ) + f ( a + ( 1 t 1 ) η 1 ( b , a ) , c + t 2 η 2 ( d , c ) ) + f ( a + ( 1 t 1 ) η 1 ( b , a ) , c + ( 1 t 2 ) η 2 ( d , c ) ) ] .

Now, we multiply it by w(a+ t 1 η 1 (b,a),c+ t 2 η 2 (d,c)) = w(a+ t 1 η 1 (b,c),c+(1 t 2 ) η 2 (d,c)) = w(a+(1 t 1 ) η 1 (b,a),c+ t 2 η 2 (d,c)) = w(a+(1 t 1 ) η 1 (b,a),c+(1 t 2 ) η 2 (d,c)) and integrate over [ 0 , 1 ] 2 to obtain the inequality

f ( a + 1 2 η 1 ( b , a ) , c + 1 2 η 2 ( d , c ) ) 0 1 0 1 w ( a + t 1 η 1 ( b , a ) , c + t 2 η 2 ( d , c ) ) d t 1 d t 2 = f ( a + 1 2 η 1 ( b , a ) , c + 1 2 η 2 ( d , c ) ) 1 η 1 ( b , a ) η 2 ( d , c ) a a + η 1 ( b , a ) c c + η 2 ( d , c ) w ( x , y ) d x d y 4 h 1 ( 1 2 ) h 2 ( 1 2 ) 1 η 1 ( b , a ) η 2 ( d , c ) a a + η 1 ( b , a ) c c + η 2 ( d , c ) f ( x , y ) w ( x , y ) d x d y ,

which completes the proof. □

Now, for a mapping f:[a,b]×[c,d]R, let us define a mapping H: [ 0 , 1 ] 2 R in the following way:

H(t,r)= 1 ( b a ) ( d c ) a b c d f ( t x + ( 1 t ) a + b 2 , r y + ( 1 r ) c + d 2 ) dxdy.
(2.6)

Some properties of this mapping for a convex on the co-ordinates function and an s-convex on the co-ordinates function are given in [6, 7], respectively. Here we investigate which of these properties can be generalized for ( h 1 , h 2 )-convex on the co-ordinates functions.

Theorem 2.6 Suppose that f:[a,b]×[c,d] is ( h 1 , h 2 )-convex on the co-ordinates. Then:

  1. (i)

    The mapping H is ( h 1 , h 2 )-convex on the co-ordinates on [ 0 , 1 ] 2 ,

  2. (ii)

    4 h 1 ( 1 2 ) h 2 ( 1 2 )H(t,r)H(0,0) for any (t,r) [ 0 , 1 ] 2 .

Proof (i) The ( h 1 , h 2 )-convexity on the co-ordinates of the mapping H is a consequence of the ( h 1 , h 2 )-convexity on the co-ordinates of the function f. Namely, for r[0,1] and for all α,β0 with α+β=1 and t 1 , t 2 [0,1], we have:

H ( α t 1 + β t 2 , r ) = 1 ( b a ) ( d c ) a b c d f ( ( α t 1 + β t 2 , r ) x + ( 1 ( α t 1 + β t 2 ) ) a + b 2 , r y + ( 1 r ) c + d 2 ) d x d y = 1 ( b a ) ( d c ) a b c d f ( α ( t 1 x + ( 1 t 1 ) a + b 2 ) + β ( t 2 x + ( 1 t 2 ) a + b 2 ) , r y + ( 1 r ) c + d 2 ) d x d y h 1 ( α ) 1 ( b a ) ( d c ) a b c d f ( t 1 x + ( 1 t 1 ) a + b 2 , r y + ( 1 r ) c + d 2 ) d x d y + h 1 ( β ) 1 ( b a ) ( d c ) a b c d f ( t 2 x + ( 1 t 2 ) a + b 2 , r y + ( 1 r ) c + d 2 ) d x d y = h 1 ( α ) H ( t 1 , r ) + h 1 ( β ) H ( t 2 , r ) .

Similarly, if t[0,1] is fixed, then for all r 1 , r 2 [0,1] and α,β0 with α+β=1, we also have

H(t,α r 1 +β r 2 ) h 2 (α)H(t, r 1 )+ h 2 (β)H(t, r 2 ),

which means that H is ( h 1 , h 2 )-convex on the co-ordinates.

  1. (ii)

    After changing the variables u=tx+(1t) a + b 2 and v=ry+(1r) c + d 2 , we have

    H ( t , r ) = 1 ( b a ) ( d c ) a b c d f ( t x + ( 1 t ) a + b 2 , r y + ( 1 r ) c + d 2 ) d x d y = 1 ( b a ) ( d c ) u L u U v L v U f ( u , v ) b a u U u L d c v U v L d u d v = 1 ( u U u L ) ( v U v L ) u L u U v L v U f ( u , v ) d u d v 1 4 h 1 ( 1 2 ) h 2 ( 1 2 ) f ( a + b 2 , c + d 2 ) ,

where u L =ta+(1t) a + b 2 , u U =tb+(1t) a + b 2 , v L =rc+(1r) c + d 2 and v U =rd+(1r) c + d 2 , which completes the proof. □

Remark 5 If f is convex on the co-ordinates, then we get H(t,r)H(0,0). If f is s-convex on the co-ordinates in the second sense, then we have the inequality H(t,r) 4 s 1 H(0,0).

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Matłoka, M. On some Hadamard-type inequalities for ( h 1 , h 2 )-preinvex functions on the co-ordinates. J Inequal Appl 2013, 227 (2013). https://doi.org/10.1186/1029-242X-2013-227

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Keywords

  • ( h 1 , h 2 )-preinvex function on the co-ordinates
  • Hadamard inequalities
  • Hermite-Hadamard-Fejér inequalities