TAMKANG JOURNAL OF MATHEMATICS
Volume 47, Number 3, 289-322, September 2016
doi:10.5556/j.tkjm.47.2016.1958
-
-
--
-
+
+
-
This paper is available online at http://journals.math.tku.edu.tw/index.php/TKJM/pages/view/onlinefirst
HERMITE-HADAMARD TYPE INEQUALITIES FOR
(p 1 , h 1 )-(p 2 , h 2 )-CONVEX FUNCTIONS ON THE CO-ORDINATES
WENGUI YANG
Abstract. In this paper, we establish some Hermite-Hadamard type inequalities for (p 1 , h1 )(p 2 , h2 )-convex function on the co-ordinates. Furthermore, some inequalities of HermiteHadamard type involving product of two convex functions on the co-ordinates are also
considered. The results presented here would provide extensions of those given in earlier
works.
1. Introduction
The following double integral inequality
f
³a +b ´
2
≤
1
b−a
Zb
f (x)d x ≤
a
f (a) + f (b)
,
2
(1.1)
which holds for any convex function f : [a, b] ⊂ R → R, is well known in the literature as the
Hermite-Hadamard inequality, see [1, 2]. Since Hermite-Hadamard’s inequality for convex
functions has been considered the most useful inequality in mathematical analysis, it has
received renewed attention in recent years and a remarkable variety of refinements and generalizations have been found, we would like to refer the reader to the papers [3, 4, 5, 6, 7, 8,
9, 10, 11, 12] and the references cited therein. For example, Dragomir and Fitzpatrick [13]
proved the variant of the Hermite-Hadamard inequality holds for s-convex functions in the
second sense. Sarikaya et al. [14] obtained that variant of the Hadamard inequality holds for
an h-convex function.
Fang and Shi [15] introduced the definition of (p, h)-convex function.
Definition 1.1. Let h : J → R be a nonnegative and non-zero function. We say that f : I → R is
a (p, h)-convex function or that f belongs to the class g hx(h, p, I ), if f is nonnegative and
1
f ([αx p + (1 − α)y p ] p ) ≤ h(α) f (x) + h(1 − α) f (y),
(1.2)
Received June 21, 2015, accepted September 28, 2015.
2010 Mathematics Subject Classification. 26D15; 26A51.
Key words and phrases. Hermite-Hadamard type inequalities; (p 1 , h1 )-(p 2 , h2 )-convex, co-ordinates,
product of functions.
289
290
WENGUI YANG
for all x, y ∈ I and α ∈ (0, 1). Similarly, if the inequality sign in (1.2) is reversed, then f is said
to be a (p, h)-concave function or belong to the class g hv(h, p, I ).
In [15], Fang and Shi obtained the following Hermite-Hadamard type inequalities of (p, h)convex function.
Theorem 1.1. If f ∈ g hx(h, p, I ) ∩ L 1([a, b]) for a, b ∈ I with a < b, then we have
Z1
Zb
³h a p + b p i p1 ´
1
p
p−1
h(t )d t .
≤
x
f
(x)d
x
≤
[
f
(a)
+
f
(b)]
f
2
bp − ap a
2h( 12 )
0
(1.3)
Furthermore, they established the following inequality of Hermite-Hadamard type involving product of two convex functions.
Theorem 1.2. Suppose that f and g are functions such that f ∈ g hx(h, p, I ),∈ g hx(k, p, I ),
f g ∩ L 1 ([a, b]), and hk ∩ L 1([0, 1]), with a, b ∈ I and a < b, then we have
Z1
Z1
Zb
p
p−1
h(t )k(1 − t )d t ,
h(t )k(t )d t + N (a, b)
x
f (x)g (x)d x ≤ M (a, b)
bp − bp a
0
0
where M (a, b) = f (a)g (a) + f (b)g (b) and N (a, b) = f (a)g (b) + f (b)g (a).
Let us consider now a bidimensional interval ∆ = [a, b] × [c, d ] ∈ R2 with a < b and c < d .
In [16], Dragomir introduced co-ordinated convex function in the following way:
Definition 1.2. A mapping f : ∆ → R is said to be convex on the co-ordinates on ∆ if the
inequality
f (t x + (1 − t )y, r u + (1 − r )w ) ≤ t r f (x, u) + t (1 − r ) f (x, w ) + r (1 − t ) f (y, u) + (1 − t )(1 − r ) f (y, w ),
holds for all t , r ∈ [0, 1] and (x, u), (y, w ) ∈ ∆.
For such a mapping Dragomir [16] proved the following Hermite-Hadamard type inequalities:
Theorem 1.3. Suppose that f : ∆ → R is convex on the co-ordinates on ∆. Then one has the
inequalities:
f
Zd ³
³ a + b c + d ´ 1 ³ 1 Zb ³ c + d ´
1
a +b ´ ´
≤
dx +
f x,
f
,
,y dy
2
2
2 b−a a
2
d −c c
2
Zb Zd
1
≤
f (x, y)d xd y
(b − a)(d − c) a c
Z
Zb
b
1
1³ 1
f (x, c)d x +
f (x, d )d x
≤
4 b−a a
b−a a
Zd
Zd
´
1
1
f (a, y)d y +
f (b, y)d y
+
d −c c
d −c c
f (a, c) + f (a, d ) + f (b, c) + f (b, d )
.
≤
4
HERMITE-HADAMARD TYPE INEQUALITIES
291
This idea of Dragomir inspired many researchers to extend various generalizations of
convex functions on co-ordinates. In 2008, Alomari and Darus [17] defined the co-ordinated
s-convexity in the second sense as follows:
Definition 1.3. A mapping f : ∆ → R is said to be s-convex in the second sense on the coordinates on ∆ if the inequality
f (t x + (1 − t )y, r u + (1 − r )w )
≤ t s r s f (x, u) + t s (1 − r )s f (x, w ) + r s (1 − t )s f (y, u) + (1 − t )s (1 − r )s f (y, w ),
holds for all t , r ∈ [0, 1], (x, u), (y, w ) ∈ ∆ and for some fixed s ∈ [0, 1].
For such a mapping they proved the following Hermite-Hadamard type inequalities:
Theorem 1.4. Suppose that f : ∆ → R is co-ordinated s-convex on ∆. Then one has the inequalities:
4
s−1
Zd ³
³ 1 Zb ³ c + d ´
³a +b c +d ´
a +b ´ ´
1
s−2
f x,
f
,
,y dy
≤2
dx +
f
2
2
b−a a
2
d −c c
2
Zb Zd
1
≤
f (x, y)d xd y
(b − a)(d − c) a c
Zb
Zb
1
1 ³ 1
f (x, c)d x +
f (x, d )d x
≤
2(s + 1) b − a a
b−a a
Zd
Zd
´
1
1
f (a, y)d y +
f (b, y)d y
+
d −c c
d −c c
f (a, c) + f (a, d ) + f (b, c) + f (b, d )
≤
.
(s + 1)2
For refinements and counterparts of other type convex functions on the co-ordinates, see
[18, 19, 20, 21, 22, 23, 24, 25].
In [26], Latif and Alomari established Hadamard-type inequalities for product of two convex functions on the co-ordinates as follow:
Theorem 1.5. Suppose that f , g : ∆ → [0, ∞) are convex functions on the co-ordinates on ∆.
Then one has the inequalities:
Zb Zd
1
f (x, y)g (x, y)d xd y
(b − a)(d − c) a c
1
1
1
≤ L(a, b, c, d ) + M (a, b, c, d ) + N (a, b, c, d ),
9
18
36
and
4f
Zb Zd
³a +b c +d ´
1
f (x, y)g (x, y)d xd y
≤
,
2
2
(b − a)(d − c) a c
292
WENGUI YANG
+
7
2
5
L(a, b, c, d ) + M (a, b, c, d ) + N (a, b, c, d ),
36
36
9
where
L(a, b, c, d ) = f (a, c)g (a, c) + f (b, c)g (b, c) + f (a, d )g (a, d ) + f (b, d )g (b, d ),
M (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)
+ f (b, c)g (a, c) + f (b, d )g (a, d ) + f (a, c)g (b, c) + f (a, d )g (b, d ),
N (a, b, c, d ) = f (b, c)g (a, d ) + f (b, d )g (a, c) + f (a, c)g (b, d ) + f (a, d )g (b, c).
In [27], Ödemir and Akdemir established Hermite-Hadamard-type inequalities for product of convex functions and s-convex functions of 2-variables on the co-ordinates as follow:
Theorem 1.6. f : ∆ → [0, ∞) be s 1 -convex function on the co-ordinates and g : ∆ → [0, ∞) be
s 2 -convex function on the co-ordinates with a < b, c < d for some fixed s 1 , s 2 ∈ (0, 1]. Then one
has the inequalities:
Zb Zd
1
f (x, y)g (x, y)d xd y
(b − a)(d − c) a c
1
B (s 1 + 1, s 2 + 2)
≤
M (a, b, c, d ) + [B (s 1 + 1, s 2 + 2)]2 N (a, b, c, d ),
L(a, b, c, d ) +
(s 1 + s 2 + 1)2
s1 + s2 + 1
and let f : ∆ → [0, ∞) be convex function on the co-ordinates and g : ∆ → [0, ∞) be s-convex
function on the co-ordinates with a < b, c < d for some fixed s ∈ (0, 1]. Then one has the inequality:
Zb Zd
³a +b c +d ´
1
f (x, y)g (x, y)d xd y
≤
,
2 f
2
2
(b − a)(d − c) a c
2s + 3
s 2 + 3s + 3
+
L(a,
b,
c,
d
)
+
M (a, b, c, d )
(s + 1)2 (s + 2)2
(s + 1)2 (s + 2)2
s 2 + 4s + 3
N (a, b, c, d ),
+
(s + 1)2 (s + 2)2
2s
where L(a, b, c, d ), M (a, b, c, d ), N (a, b, c, d ) are defined in Theorem 1.5, and B (⋆, ∗) is Beta
function.
Now we first give a definition of (p, h)-convex functions on ∆.
Definition 1.4. Let h : J → R be a nonnegative and non-zero function. A mapping f : ∆ → R is
said to be (p, h)-convex on the co-ordinates on ∆ if the inequality
1¢
1
¡
f [λx p + (1 − λ)y p ] p , [λu p + (1 − λ)w p ] p ≤ h(λ) f (x, u) + h(1 − λ) f (y, w ),
holds for all λ ∈ [0, 1] and (x, u), (y, w ) ∈ ∆.
HERMITE-HADAMARD TYPE INEQUALITIES
293
Furthermore, we introduce the definition of (p 1 , h 1 )-(p 2 , h 2 )-convex functions on the coordinates on ∆.
Definition 1.5. Let h 1 , h 2 : J → R be two nonnegative and non-zero functions. A mapping f :
∆ → R is said to be (p 1 , h 1 )-(p 2 , h 2 )-convex on the co-ordinates on ∆ if 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 (p 1 , h 1 )-convex with
respect to u and (p 2 , h 2 )-convex with respect to v, respectively, for all y ∈ [c, d ] and x ∈ [a, b].
From the above definition it follows that if f is a co-ordinated (p 1 , h 1 )-(p 2 , h 2 )-convex
function, then
1
1 ¢
¡
f [t x p 1 + (1 − t )y p 1 ] p1 , [r u p 2 + (1 − r )w p 2 ] p2
≤ h 1 (t )h 2 (r ) f (x, u) + h 1 (t )h 2 (1 − r ) f (x, w ) + h 1 (1 − t )h 2 (r ) f (y, u)
+h 1 (1 − t )h 2 (1 − r ) f (y, w ).
(1.4)
Similar to the proof of [16, 17], it is easily to see that every (p, h)-convex mapping f : ∆ → R is
(p, h)-(p, h)-convex on the co-ordinates, but converse is not general true.
Motivated by the results mentioned above, the main aim of this paper is to establish
some new Hermite-Hadamard type inequalities for (p 1 , h 1 )-(p 2 , h 2 )-convex function on the
co-ordinates on ∆. Furthermore, some inequalities of Hermite-Hadamard type involving
product of two convex functions on the co-ordinates are also considered. The results presented here would provide extensions of those given in earlier works.
2. Main results
In this section, we will give the Hermite-Hadamard type inequalities by using (p 1 , h 1 )(p 2 , h 2 )-convex functions of two variables on the co-ordinates on ∆.
Theorem 2.1. Suppose that f : ∆ → R is (p 1 , h 1 )-(p 2 , h 2 )-convex function on the co-ordinates
on ∆. Then one has the inequalities:
³h a p 1 + b p 1 i p1 h c p 2 + d p 2 i p1 ´
1
1
2
,
¡1¢ ¡1¢ f
2
2
4h 1 2 h 2 2
Zb Zd
p1p2
≤ p
x p 1 −1 y p 2 −1 f (x, y)d xd y
(b 1 − a p 1 )(d p 2 − c p 2 ) a c
Z1
Z1
h 2 (t )d t .
h 1 (t )d t
≤ [ f (a, c) + f (a, d ) + f (b, c) + f (b, d )]
0
0
(2.1)
Proof. According to (1.4) with x p 1 = sa p 1 + (1 − s)b p 1 , y p 1 = (1 − s)a p 1 + sb p 1 , u p 2 = zc p 2 + (1 −
z)d p 2 , w p 2 = (1 − z)c p 2 + zd p 2 and t = r = 21 , we find that
³h a p 1 + b p 1 i p1 h c p 2 + d p 2 i p1 ´
1
2
,
f
2
2
294
WENGUI YANG
≤ h1
³1´
2
h2
³ 1 ´³
2
1
1
f ([sa p 1 + (1 − s)b p 1 ] p1 , [zc p 2 + (1 − z)d p 2 ] p2 )
1
1
1
1
+ f ([sa p 1 + (1 − s)b p 1 ] p1 , [(1 − z)c p 2 + zd p 2 ] p2 )
+ f ([(1 − s)a p 1 + sb p 1 ] p1 , [zc p 2 + (1 − z)d p 2 ] p2 )
1 ´
1
+ f ([(1 − s)a p 1 + sb p 1 ] p1 , [(1 − z)c p 2 + zd p 2 ] p2 ) .
(2.2)
Integrating (2.2) with respect to (s, z) on [0, 1] × [0, 1], we obtain
f
³h a p 1 + b p 1 i p1 h c p 2 + d p 2 i p1 ´
1
2
,
2
2
³ 1 ´ ³ 1 ´³ Z1 Z1
1
1
f ([sa p 1 + (1 − s)b p 1 ] p1 , [zc p 2 + (1 − z)d p 2 ] p2 )d sd z
h2
≤ h1
2
2
0 0
Z1 Z1
1
1
+
f ([sa p 1 + (1 − s)b p 1 ] p1 , [(1 − z)c p 2 + zd p 2 ] p2 )d sd z
0 0
Z1 Z1
1
1
f ([(1 − s)a p 1 + sb p 1 ] p1 , zc p 2 + (1 − z)d p 2 ] p2 )d sd z
+
0 0
Z1 Z1
´
1
1
f ([(1 − s)a p 1 + sb p 1 ] p1 , (1 − z)c p 2 + zd p 2 ] p2 )d sd z .
+
0
0
(2.3)
Using the change of the variable in (2.3), we get
¡ ¢ ¡ ¢ Z Z
³h a p 1 + b p 1 i p1 h c p 2 + d p 2 i p1 ´
b d
4p 1 p 2 h 1 21 h 2 12
1
2
x p 1 −1 y p 2 −1 f (x, y)d xd y,
f
≤ p
,
2
2
(b 1 − a p 1 )(d p 2 − c p 2 ) a c
which the first inequality is proved. For the proof of the second inequality in (2.1), we first note
that since f is a (p 1 , h 1 )-(p 2 , h 2 )-convex function on the co-ordinates on ∆, then, by using (1.4)
with x = a, y = b, u = c and w = d , it yields that
1 ¢
1
¡
f [t a p 1 + (1 − t )b p 1 ] p1 , [r c p 2 + (1 − r )d p 2 ] p2
≤ h 1 (t )h 2 (r ) f (a, c) + h 1 (t )h 2 (1 − r ) f (a, d ) + h 1 (1 − t )h 2 (r ) f (b, c)
+h 1 (1 − t )h 2 (1 − r ) f (b, d ).
(2.4)
Similarly, we have
1 ¢
1
¡
f [t a p 1 + (1 − t )b p 1 ] p1 , [(1 − r )c p 2 + r d p 2 ] p2
≤ h 1 (t )h 2 (1 − r ) f (a, c) + h 1 (t )h 2 (r ) f (a, d ) + h 1 (1 − t )h 2 (1 − r ) f (b, c)
+h 1 (1 − t )h 2 (r ) f (b, d ),
(2.5)
1
1 ¢
¡
f [(1 − t )a p 1 + t b p 1 ] p1 , [r c p 2 + (1 − r )d p 2 ] p2
≤ h 1 (1 − t )h 2 (r ) f (a, c) + h 1 (1 − t )h 2 (1 − r ) f (a, d ) + h 1 (t )h 2 (r ) f (b, c)
+h 1 (t )h 2 (1 − r ) f (b, d ),
(2.6)
HERMITE-HADAMARD TYPE INEQUALITIES
and
295
1
1 ¢
¡
f [(1 − t )a p 1 + t b p 1 ] p1 , [(1 − r )c p 2 + r d p 2 ] p2
≤ h 1 (1 − t )h 2 (1 − r ) f (a, c) + h 1 (1 − t )h 2 (r ) f (a, d ) + h 1 (t )h 2 (1 − r ) f (b, c)
+h 1 (t )h 2 (r ) f (b, d ).
(2.7)
Adding the inequalities (2.4)-(2.7), we have
1
1 ¢
1
1 ¢
¡
¡
f [t a p 1 + (1 − t )b p 1 ] p1 , [r c p 2 + (1 − r )d p 2 ] p2 + f [t a p 1 + (1 − t )b p 1 ] p1 , [(1 − r )c p 2 + r d p 2 ] p2
1
1 ¢
1
1 ¢
¡
¡
+ f [(1 − t )a p 1 + t b p 1 ] p1 , [r c p 2 + (1 − r )d p 2 ] p2 + f [(1 − t )a p 1 + t b p 1 ] p1 , [(1 − r )c p 2 + r d p 2 ] p2
≤ [h 1 (t )h 2 (r ) + h 1 (t )h 2 (1 − r ) + h 1 (1 − t )h 2 (r ) + h 1 (1 − t )h 2 (1 − r )][ f (a, c) + f (a, d )
+ f (b, c) + f (b, d )].
(2.8)
Integrating (2.8) with respect to (t , r ) on [0, 1] × [0, 1] and using the change of the variable, we
get
p1 p2
(b p 1 − a p 1 )(d p 2 − c p 2 )
Zb Zd
x p 1 −1 y p 2 −1 f (x, y)d xd y
a c
Z1
Z1
h 2 (t )d t .
h 1 (t )d t
≤ [ f (a, c) + f (a, d ) + f (b, c) + f (b, d )]
0
0
The proof is completed.
By using different method, the following inequalities will be obtained.
Theorem 2.2. Suppose that f : ∆ → R is (p 1 , h 1 )-(p 2 , h 2 )-convex function on the co-ordinates
on ∆. Then one has the inequalities:
³h a p 1 + b p 1 i p1 h c p 2 + d p 2 i p1 ´
1
1
2
,
¡1¢ ¡1¢ f
2
2
4h 1 2 h 2 2
Zb
³ h c p 2 + d p 2 i p1 ´
p1
2
p 1 −1
≤
x
f x,
dx
1
p
p
1
1
2
4h 2 ( 2 )(b − a ) a
Zd
³h a p 1 + b p 1 i p1 ´
p2
1
p 2 −1
y
f
+
,y dx
2
4h 1 ( 21 )(d p 2 − d p 2 ) c
Zb Zd
p1 p2
x p 1 −1 y p 2 −1 f (x, y)d xd y
≤ p
(b 1 − a p 1 )(d p 2 − c p 2 ) a c
Zb
´ Z1
³ Zb
p1
p 1 −1
p 1 −1
≤
h 2 (t )d t
x
f
(x,
d
)d
x
x
f
(x,
c)d
x
+
2(b p 1 − a p 1 ) a
0
a
Zd
´ Z1
³ Zd
p2
p 2 −1
p 2 −1
h 1 (t )d t
y
f (b, y)d y
y
f (a, y)d y +
+
2(d p 2 − c p 2 ) c
0
c
Z1
Z1
h 2 (t )d t .
h 1 (t )d t
≤ [ f (a, c) + f (a, d ) + f (b, c) + f (b, d )]
0
0
(2.9)
296
WENGUI YANG
Proof. Since f : ∆ → R is (p 1 , h 1 )-(p 2 , h 2 )-convex function on the co-ordinates on ∆, it follows
that the mapping f x : [c, d ] → R, f x (y) = f (x, y), is (p 2 , h 2 )-convex on [c, d ] for all x ∈ [a, b].
Then by using inequalities (1.3), we can write
Zd
Z1
³h c p 2 + d p 2 i p1 ´
p2
1
2
p 2 −1
≤ p
fx
y
f x (y)d y ≤ [ f x (c) + f x (d )]
h 2 (t )d t , ∀x ∈ [a, b].
2
d 2 − c p2 c
2h 2 ( 21 )
0
That is,
Zd
³ h c p 2 + d p 2 i p1 ´
p2
2
y p 2 −1 f (x, y)d y
f
x,
≤
2
d p2 − c p2 c
2h 2 ( 21 )
Z1
h 2 (t )d t , ∀x ∈ [a, b].
≤ [ f (x, c) + f (x, d )]
1
0
Multiplying both sides of (2.10) by
p 1 x p 1 −1
b p 1 −a p 1
(2.10)
and integrating with respect to x over [a, b], respec-
tively, we have
³ h c p 2 + d p 2 i p1 ´
2
p 1 −1
x
f
x,
dx
2
2h 2 ( 21 )(b p 1 − a p 1 ) a
Zb Zd
p1 p2
x p 1 −1 y p 2 −1 f (x, y)d xd y
≤ p
(b 1 − a p 1 )(d p 2 − c p 2 ) a c
Zb
´ Z1
³ Zb
p1
p 1 −1
p 1 −1
h 2 (t )d t .
x
f
(x,
d
)d
x
≤ p
x
f
(x,
c)d
x
+
b 1 − a p1 a
0
a
p1
Zb
(2.11)
A similar arguments applied for the mapping f y : [a, b] → R, f y (x) = f (x, y), we get
p2
2h 1 ( 21 )(d p 2 − d p 2 )
Zd
c
y p 2 −1 f
³h a p 1 + b p 1 i p1
1
2
´
,y dx
Zb Zd
p1 p2
x p 1 −1 y p 2 −1 f (x, y)d xd y
(b p 1 − a p 1 )(d p 2 − c p 2 ) a c
Zd
´ Z1
³ Zd
p2
p 2 −1
p 2 −1
h 1 (t )d t .
y
f (b, y)d y
y
f (a, y)d y +
≤ p
d 2 − c p2 c
0
c
≤
(2.12)
Summing the inequalities (2.11) and (2.12), we get the second and the third inequalities in
(2.9).
Now, by using the first inequality in (1.3), we also have
Zb
³h a p 1 + b p 1 i p1 h c p 2 + d p 2 i p1 ´
³ h c p 2 + d p 2 i p1 ´
1
p1
1
2
2
p 1 −1
f
,
≤
x
f
x,
d x, (2.13)
2
2
b p1 − a p1 a
2
2h 1 ( 12 )
and
Zd
³h a p 1 + b p 1 i p1 ´
³h a p 1 + b p 1 i p1 h c p 2 + d p 2 i p1 ´
p2
1
2
1
p 2 −1
y
f
,
≤
, y d y. (2.14)
2
2
d p2 − c p2 c
2
2h 2 ( 12 )
1
f
Multiplying both sides of (2.13) and (2.14) by
1
4h2 ( 12 )
obtained results, we get the first inequality in (2.9).
and
1
,
4h1 ( 12 )
respectively, and adding the
HERMITE-HADAMARD TYPE INEQUALITIES
297
Finally, by using the second inequality in (1.3), we can also state that
Zb
Z1
p1
p 1 −1
x
f
(x,
c)d
x
≤
[
f
(a,
c)
+
f
(b,
c)]
h 1 (t )d t ,
b p1 − a p1 a
0
Z1
Zb
p1
p 1 −1
h 1 (t )d t ,
x
f
(x,
d
)d
x
≤
[
f
(a,
d
)
+
f
(b,
d
)]
b p1 − a p1 a
0
Zd
Z1
p2
p 2 −1
y
f (a, y)d y ≤ [ f (a, c) + f (a, d )]
h 2 (t )d t ,
d p2 − c p2 c
0
and
p2
p
d 2 − c p2
Zd
y p 2 −1 f (b, y)d y ≤ [ f (b, c) + f (b, d )]
c
Z1
0
h 2 (t )d t ,
which give, by addition, the last inequality in (2.9). The proof is completed.
Remark 1. In Theorem 2.2, letting p 1 = p 2 = 1 and h 1 (t ) = h 2 (t ) = t , Theorem 2.2 reduces to
Theorem 1.3. In Theorem 2.2, letting p 1 = p 2 = 1 and h 1 (t ) = h 2 (t ) = t s , Theorem 2.2 reduces
to Theorem 1.4.
Similarly, from the proof of Theorem 2.1, we can obtain the following theorem.
Theorem 2.3. Suppose that f : ∆ → R is (p 1 , h 1 )-(p 2 , h 2 )-convex function on the co-ordinates
on ∆. Then one has the inequalities:
³h a p 1 + b p 1 i p1 h c p 2 + d p 2 i p1 ´ Zb Zd
1
1
2
x p 1 −1 y p 2 −1 w (x, y)d xd y
,
¡1¢ ¡1¢ f
2
2
4h 1 2 h 2 2
a c
Zb Zd
x p 1 −1 y p 2 −1 f (x, y)w (x, y)d xd y
≤
a
c
≤ [ f (a, c)+ f (a, d )+ f (b, c)+ f (b, d )]
ZbZd
a
c
x p 1 −1 y p 2 −1 h 1
³ x p 1 −a p 1 ´ ³ y p 2 −c p 2 ´
h2 p
w (x, y)d xd y.
b p 1 −a p 1
d 2 −c p 2
¡£ p1 p1 ¤ 1 £ c p2 +d p2 ¤ 1 ¢
p1
p2
where w : ∆ → [0, ∞) is a symmetric function with respect to a +b
,
, that is,
2
2
1 ¢
1 £
1 £
¤
¤
¤1¢
¡£
¤
¡£ p
p2
p 2 p2
p1
p 1 p1
p2
p 2 p2
p 1 p1
1
, r c +(1−r )d
= w (1− t )a + t b
, r c +(1−r )d
=
w t a +(1− t )b
¡£ p
¤1 £
¤1¢
¡£
¤1 £
¤1¢
p 1 p1
p2
p 2 p2
p1
p 1 p1
p2
p 2 p2
1
w t a + (1 − t )b
, (1 − r )c + r d
= w (1 − t )a + t b
, (1 − r )c + r d
.
3. Some inequalities involving product of two convex functions
In this section, we will consider some inequalities of Hermite-Hadamard type involving
product of two convex functions on the co-ordinates on ∆.
Theorem 3.1. Suppose that f , g : ∆ → [0, ∞) are (p 1 , h 1 )-(p 2 , h 2 )-convex and (p 1 , k 1 )-(p 2 , k 2 )convex functions on the co-ordinates on ∆, respectively. Then one has the inequality:
p1p2
p
p
(b 1 − a 1 )(d p 2 − c p 2 )
Zb Zd
a
c
x p 1 −1 y p 2 −1 f (x, y)g (x, y)d xd y
298
WENGUI YANG
Z1
h 2 (t )k 2 (t )d t
h 1 (t )k 1 (t )d t
0
0
Z1
Z1
h 2 (t )k 2 (1 − t )d t
h 1 (t )k 1 (t )d t
+M 2 (a, b, c, d )
0
0
Z1
Z1
+M 3 (a, b, c, d )
h 1 (t )k 1 (1 − t )d t
h 2 (t )k 2 (t )d t
0
0
Z1
Z1
h 2 (t )k 2 (1 − t )d t ,
h 1 (t )k 1 (1 − t )d t
+M 4 (a, b, c, d )
≤ M 1 (a, b, c, d )
Z1
(3.1)
0
0
where
M 1 (a, b, c, d ) = f (a, c)g (a, c) + f (b, c)g (b, c) + f (a, d )g (a, d ) + 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 (p 1 , h 1 )-(p 2 , h 2 )-convex on the co-ordinates and g is (p 1 , k 1 )-(p 2 , k 2 )-convex
on the coordinates on ∆, it follows that
1
1 ¢
¡
f [t a p 1 + (1 − t )b p 1 ] p1 , [r c p 2 + (1 − r )d p 2 ] p2
≤ h 1 (t )h 2 (r ) f (a, c) + h 1 (t )h 2 (1−r ) f (a, d ) + h 1 (1−t )h 2 (r ) f (b, c) + h 1 (1−t )h 2 (1−r ) f (b, d ), (3.2)
and
1
1 ¢
¡
g [t a p 1 + (1 − t )b p 1 ] p1 , [r c p 2 + (1 − r )d p 2 ] p2
≤ k 1 (t )k 2 (r )g (a, c) + k 1 (t )k 2 (1−r )g (a, d ) + k 1 (1−t )k 2 (r )g (b, c) + k 1 (1−t )k 2 (1−r )g (b, d ). (3.3)
Multiplying (3.2) and (3.3) and integrating the obtained result with respect to (t , r ) on [0, 1] ×
[0, 1], we obtain our inequality (3.1).
Theorem 3.2. Suppose that f , g : ∆ → [0, ∞) are (p 1 , h 1 )-(p 2 , h 2 )-convex and (p 1 , k 1 )-(p 2 , k 2 )convex functions on the co-ordinates on ∆, respectively. Then one has the inequality:
³h a p 1 + b p 1 i p1 h c p 2 + d p 2 i p1 ´ ³h a p 1 + b p 1 i p1 h c p 2 + d p 2 i p1 ´
1
2
1
2
g
,
,
2
2
2
2
4h 1 ( 12 )h 2 ( 12 )k 1 ( 21 )k 2 ( 21 )
Zb Zd
p1 p2
x p 1 −1 y p 2 −1 f (x, y)d xd y
− p
(b 1 − a p 1 )(d p 2 − c p 2 ) a c
≤ Ω1 (h 1 , h 2 , k 1 , k 2 )M 1 (a, b, c, d ) + Ω2 (h 1 , h 2 , k 1 , k 2 )M 2 (a, b, c, d )
1
f
+Ω3 (h 1 , h 2 , k 1 , k 2 )M 3 (a, b, c, d ) + Ω4 (h 1 , h 2 , k 1 , k 2 )M 4 (a, b, c, d ),
(3.4)
where M 1 (a, b, c, d ), M 2 (a, b, c, d ), M 3 (a, b, c, d ) and M 4 (a, b, c, d ) are defined in Theorem 3.1,
and
Ω1 (h 1 , h 2 , k 1 , k 2 ) =
Z1
0
h 1 (t )h 2 (t )d t
Z1
0
k 1 (t )k 2 (1 − t )d t +
Z1
0
h 1 (t )h 2 (1 − t )d t
Z1
0
k 1 (t )k 2 (t )d t
HERMITE-HADAMARD TYPE INEQUALITIES
299
Z1
Z1
k 1 (t )k 2 (1 − t )d t ,
h 1 (t )h 2 (1 − t )d t
+
0
0
Z1
Z1
Z1
Z1
k 1 (t )k 2 (t )d t
h 1 (t )h 2 (1 − t )d t
k 1 (t )k 2 (t )d t +
h 1 (t )h 2 (t )d t
Ω2 (h 1 , h 2 , k 1 , k 2 ) =
0
0
0
0
Z1
Z1
+
h 1 (t )h 2 (1 − t )d t
k 1 (t )k 2 (1 − t )d t ,
0
0
Z1
Z1
Z1
Z1
k 1 (t )k 2 (1 − t )d t
h 1 (t )h 2 (t )d t
k 1 (t )k 2 (t )d t +
h 1 (t )h 2 (t )d t
Ω3 (h 1 , h 2 , k 1 , k 2 ) =
0
0
0
0
Z1
Z1
+
h 1 (t )h 2 (1 − t )d t
k 1 (t )k 2 (1 − t )d t ,
0
and
Ω4 (h 1 , h 2 , k 1 , k 2 ) =
0
Z1
Z1
Z1
h 1 (t )h 2 (t )d t
k 1 (t )k 2 (t )d t +
h 1 (t )h 2 (t )d t
0
0
Z1
Z1
+
h 1 (t )h 2 (1 − t )d t
k 1 (t )k 2 (t )d t .
0
0
Z1
0
k 1 (t )k 2 (1 − t )d t
0
Proof. Since f is (p 1 , h 1 )-(p 2 , h 2 )-convex on the co-ordinates and g is (p 1 , k 1 )-(p 2 , k 2 )-convex
on the coordinates on ∆, it follows that f y : [a, b] → [0, ∞), f y (x) = f (x, y) and f x : [c, d ] →
[0, ∞), f x (y) = f (x, y) are are (p 1 , h 1 )-convex and (p 2 , h 2 )-convex on [a, b] and [c, d ], respectively, where x ∈ [a, b], y ∈ [c, d ]. Similarly, g y : [a, b] → [0, ∞), g y (x) = g (x, y) and g x : [c, d ] →
[0, ∞), g x (y) = g (x, y) are are (p 1 , k 1 )-convex and (p 2 , k 2 )-convex on [a, b] and [c, d ], respectively, where x ∈ [a, b], y ∈ [c, d ].
Using Theorem 7 in [15] and multiplying both sides of the inequalities by
get
1
,
2k 1 ( 12 )k 2 ( 21 )
we
³h a p 1 + b p 1 i p1 h c p 2 + d p 2 i p1 ´ ³h a p 1 + b p 1 i p1 h c p 2 + d p 2 i p1 ´
1
2
1
2
g
,
,
2
2
2
2
4h 1 ( 21 )h 2 ( 12 )k 1 ( 21 )k 2 ( 12 )
Zb
³ h c p 2 + d p 2 i p1 ´ ³ h c p 2 + d p 2 i p1 ´
p1
2
2
p 1 −1
x
f
x,
g x,
dx
−
2
2
2k 1 ( 12 )k 2 ( 21 )(b p 1 − a p 1 ) a
· ³ h p2
1
1
c + d p 2 i p2 ´ ³ h c p 2 + d p 2 i p2 ´
1
f a,
≤
g a,
2
2
2k 1 ( 12 )k 2 ( 21 )
¸
Z
³ h c p 2 + d p 2 i p1 ´ ³ h c p 2 + d p 2 i p1 ´
1
2
2
h 1 (t )h 2 (1 − t )d t
g b,
+ f b,
2
2
0
· ³ h p2
1
1
1
c + d p 2 i p2 ´ ³ h c p 2 + d p 2 i p2 ´
+
g
b,
f
a,
2
2
2k 1 ( 12 )k 2 ( 21 )
¸
Z
1
1
³ h c p2 + d p2 i p ´ ³ h c p2 + d p2 i p ´
1
2
2
h 1 (t )h 2 (t )d t .
(3.5)
+ f b,
g a,
2
2
0
and
³h a p 1 + b p 1 i p1 h c p 2 + d p 2 i p1 ´ ³h a p 1 + b p 1 i p1 h c p 2 + d p 2 i p1 ´
1
1
2
1
2
f
,
g
,
2
2
2
2
4h 1 ( 21 )h 2 ( 12 )k 1 ( 21 )k 2 ( 12 )
1
f
300
WENGUI YANG
´ ³h a p 1 + b p 1 i p1 ´
1
,
y
g
,y dy
2
2
2h 1 ( 21 )h 2 ( 12 )(d p 2 − c p 2 ) c
· ³h p 1
1
1
1
a + b p 1 i p1 ´ ³h a p 1 + b p 1 i p1 ´
≤
,c g
,c
f
2
2
2h 1 ( 21 )h 2 ( 12 )
³h a p 1 + b p 1 i p1 ´ ³h a p 1 + b p 1 i p1 ´¸ Z1
1
1
k 1 (t )k 2 (1 − t )d t
,d g
,d
+f
2
2
0
· ³h p 1
1
1
1
a + b p 1 i p1 ´ ³h a p 1 + b p 1 i p1 ´
+
,
c
g
,d
f
2
2
2h 1 ( 21 )h 2 ( 12 )
³h a p 1 + b p 1 i p1 ´ ³h a p 1 + b p 1 i p1 ´¸ Z1
1
1
,d g
,c
+f
k 1 (t )k 2 (t )d t .
2
2
0
Zd
p2
−
y p 2 −1 f
³h a p 1 + b p 1 i p1
1
(3.6)
Now, by adding (3.5) and (3.6), we obtain
µh p 1
1
1 ¶ µh p
1
1 ¶
a + b p 1 i p1 h c p 2 + d p 2 i p2
a 1 + b p 1 i p1 h c p 2 + d p 2 i p2
,
g
,
2
2
2
2
2h 1 ( 21 )h 2 ( 12 )k 1 ( 21 )k 2 ( 12 )
¶
µ
¶
µ h p2
Zb
1
1
h c p2 + d p2 i p
c + d p 2 i p2
p1
2
p 1 −1
g
x,
dx
x
f
x,
−
1
1
p
p
1
1
2
2
2k 1 ( 2 )k 2 ( 2 )(b − a ) a
Zd
³h a p 1 + b p 1 i p1 ´ ³h a p 1 + b p 1 i p1 ´
p2
1
1
p 2 −1
y
f
−
,y g
,y dy
1
1
p
p
2
2
2
2
2h 1 ( 2 )h 2 ( 2 )(d − c ) c
· ³ h p2
1
1
1
c + d p 2 i p2 ´ ³ h c p 2 + d p 2 i p2 ´
g
a,
≤
f
a,
2
2
2k 1 ( 12 )k 2 ( 21 )
³ h c p 2 + d p 2 i p1 ´ ³ h c p 2 + d p 2 i p1 ´¸ Z1
2
2
h 1 (t )h 2 (1 − t )d t
g b,
·
+ f b,
2
2
0
· ³ h p2
1
1
1
c + d p 2 i p2 ´ ³ h c p 2 + d p 2 i p2 ´
+
g
b,
f
a,
2
2
2k 1 ( 12 )k 2 ( 21 )
³ h c p 2 + d p 2 i p1 ´ ³ h c p 2 + d p 2 i p1 ´¸ Z1
2
2
g a,
+ f b,
h 1 (t )h 2 (t )d t
2
2
0
· ³h p 1
1
1
a + b p 1 i p1 ´ ³h a p 1 + b p 1 i p1 ´
1
,c g
,c
f
+
2
2
2h 1 ( 12 )h 2 ( 21 )
³h a p 1 + b p 1 i p1 ´ ³h a p 1 + b p 1 i p1 ´¸ Z1
1
1
+f
,d g
,d ·
k 1 (t )k 2 (1 − t )d t
2
2
0
· ³h p 1
1
1
1
a + b p 1 i p1 ´ ³h a p 1 + b p 1 i p1 ´
+
,
c
g
,d
f
2
2
2h 1 ( 12 )h 2 ( 21 )
³h a p 1 + b p 1 i p1 ´ ³h a p 1 + b p 1 i p1 ´¸ Z1
1
1
+f
,d g
,c
k 1 (t )k 2 (t )d t .
(3.7)
2
2
0
1
f
Applying Theorem 1.2 to each term of right hand side of the above inequality (3.7), we have
Zd
³ h c p 2 + d p 2 i p1 ´ ³ h c p 2 + d p 2 i p1 ´
p2
2
2
y p 2 −1 f (a, y)g (a, y)d y
f
a,
g
a,
≤
2
2
d p2 − c p2 c
2k 1 ( 21 )k 2 ( 21 )
1
HERMITE-HADAMARD TYPE INEQUALITIES
+[ f (a, c)g (a, c) + f (a, d )g (a, d )]
Z1
k 1 (t )k 2 (1 − t )d t
+[ f (a, c)g (a, d ) + f (a, d )g (a, c)]
Z1
k 1 (t )k 2 (t )d t ,
0
0
301
(3.8)
Zd
³ h c p 2 + d p 2 i p1 ´ ³ h c p 2 + d p 2 i p1 ´
p2
2
2
y p 2 −1 f (b, y)g (b, y)d y
g b,
≤ p
f b,
1
1
2 − c p2
2
2
d
2k 1 ( 2 )k 2 ( 2 )
c
Z1
+[ f (b, c)g (b, c) + f (b, d )g (b, d )]
k 1 (t )k 2 (1 − t )d t + [ f (b, c)g (b, d )
0
Z1
k 1 (t )k 2 (t )d t ,
(3.9)
+ f (b, d )g (b, c)]
1
0
Zd
h c p 2 + d p 2 i p1 ´ ³ h c p 2 + d p 2 i p1 ´
p2
2
2
f
a,
g
b,
≤
y p 2 −1 f (a, y)g (b, y)d y
2
2
d p2 − c p2 c
2k 1 ( 12 )k 2 ( 21 )
Z1
k 1 (t )k 2 (1 − t )d t
+[ f (a, c)g (b, c) + f (a, d )g (b, d )]
0
Z1
+[ f (a, c)g (b, d ) + f (a, d )g (b, c)]
k 1 (t )k 2 (t )d t ,
(3.10)
1
³
0
Zd
³ h c p 2 + d p 2 i p1 ´ ³ h c p 2 + d p 2 i p1 ´
p2
1
2
2
y p 2 −1 f (b, y)g (a, y)d y
f b,
g a,
≤ p
2
2
d 2 − c p2 c
2k 1 ( 12 )k 2 ( 21 )
Z1
+[ f (b, c)g (a, c) + f (b, d )g (a, d )]
k 1 (t )k 2 (1 − t )d t
0
Z1
k 1 (t )k 2 (t )d t ,
(3.11)
+[ f (b, c)g (a, d ) + f (b, d )g (a, c)]
0
³h a p 1 + b p 1 i p1
Zb
´ ³h a p 1 + b p 1 i p1 ´
p1
1
1
x p 1 −1 f (x, c)g (x, c)d x
f
,
c
g
,
c
≤
2
2
b p1 − a p1 a
2h 1 ( 21 )h 2 ( 21 )
Z1
h 1 (t )h 2 (1 − t )d t
+[ f (a, c)g (a, c) + f (b, c)g (b, c)]
0
Z1
+[ f (a, c)g (b, c) + f (b, c)g (a, c)]
h 1 (t )h 2 (t )d t ,
(3.12)
1
0
³h a p 1 + b p 1 i p1
Zb
´ ³h a p 1 + b p 1 i p1 ´
p1
1
x p 1 −1 f (x, d )g (x, d )d x
f
,d g
,d ≤ p
2
2
b 1 − a p1 a
2h 1 ( 21 )h 2 ( 21 )
Z1
+[ f (a, d )g (a, d ) + f (b, d )g (b, d )]
h 1 (t )h 2 (1 − t )d t + [ f (a, d )g (b, d )
0
Z1
h 1 (t )h 2 (t )d t ,
(3.13)
+ f (b, d )g (a, d )]
1
1
0
Zb
´ ³h a p 1 + b p 1 i p1 ´
p1
1
,
c
g
,
c
≤
x p 1 −1 f (x, c)g (x, d )d x
2
2
b p1 − a p1 a
2h 1 ( 21 )h 2 ( 21 )
Z1
h 1 (t )h 2 (1 − t )d t
+[ f (a, c)g (a, d ) + f (b, c)g (b, d )]
0
Z1
+[ f (a, c)g (b, d ) + f (b, c)g (a, d )]
h 1 (t )h 2 (t )d t ,
(3.14)
1
f
³h a p 1 + b p 1 i p1
1
0
302
WENGUI YANG
and
Zb
´ ³h a p 1 + b p 1 i p1 ´
p1
1
x p 1 −1 f (x, d )g (x, c)d x
,
c
g
,
c
≤
2
2
b p1 − a p1 a
2h 1 ( 21 )h 2 ( 21 )
Z1
h 1 (t )h 2 (1 − t )d t
+[ f (a, d )g (a, c) + f (b, d )g (b, c)]
0
Z1
h 1 (t )h 2 (t )d t .
(3.15)
+[ f (a, d )g (b, c) + f (b, d )g (a, c)]
1
f
³h a p 1 + b p 1 i p1
1
0
Using the inequalities (3.8)-(3.15) in (3.7), we get
³h a p 1 + b p 1 i p1 h c p 2 + d p 2 i p1 ´ ³h a p 1 + b p 1 i p1 h c p 2 + d p 2 i p1 ´
1
2
1
2
g
,
,
2
2
2
2
2h 1 ( 21 )h 2 ( 21 )k 1 ( 12 )k 2 ( 21 )
Zb
³ h c p 2 + d p 2 i p1 ´ ³ h c p 2 + d p 2 i p1 ´
p1
2
2
p 1 −1
x
f
x,
g x,
dx
−
2
2
2k 1 ( 21 )k 2 ( 12 )(b p 1 − a p 1 ) a
Zd
³h a p 1 + b p 1 i p1 ´ ³h a p 1 + b p 1 i p1 ´
p2
1
1
p 2 −1
y
f
−
,y g
,y dy
2
2
2h 1 ( 21 )h 2 ( 12 )(d p 2 − c p 2 ) c
Zd
´ Z1
³ Zd
p2
p 2 −1
p 2 −1
≤ p
h 1 (t )h 2 (1 − t )d t
y
f (b, y)g (b, y)d y
y
f (a, y)g (a, y)d y +
d 2 − c p2 c
0
c
Z
Z
Z
³ d
´ 1
d
p2
p 2 −1
p 2 −1
y
f
(a,
y)g
(b,
y)d
y
+
+ p
y
f
(b,
y)g
(a,
y)d
y
h 1 (t )h 2 (t )d t
d 2 − c p2 c
c
0
Zb
´ Z1
³ Zb
p1
p 1 −1
p 1 −1
k 1 (t )k 2 (1 − t )d t
x
f
(x,
d
)g
(x,
d
)d
x
x
f
(x,
c)g
(x,
c)d
x
+
+ p
b 1 − a p1 a
0
a
Zb
³ Zb
´ Z1
p1
p 1 −1
p 1 −1
+ p
x
f
(x,
c)g
(x,
d
)d
x
+
x
f
(x,
d
)g
(x,
c)d
x
k 1 (t )k 2 (t )d t
b 1 − a p1 a
a
0
Z1
Z1
k 1 (t )k 2 (1 − t )d t
h 1 (t )h 2 (1 − t )d t
+2M 1 (a, b, c, d )
0
0
Z1
Z1
k 1 (t )k 2 (t )d t
h 1 (t )h 2 (1 − t )d t
+2M 2 (a, b, c, d )
0
0
Z1
Z1
+2M 3 (a, b, c, d )
h 1 (t )h 2 (t )d t
k 1 (t )k 2 (1 − t )d t
0
0
Z1
Z1
k 2 (t )k 2 (t )d t ,
(3.16)
h 1 (t )h 2 (t )d t
+2M 4 (a, b, c, d )
1
f
0
and
0
³h a p 1 + b p 1 i p1 h c p 2 + d p 2 i p1 ´ ³h a p 1 + b p 1 i p1 h c p 2 + d p 2 i p1 ´
1
2
1
2
f
,
g
,
1
1
1
1
2
2
2
2
2h 1 ( 2 )h 2 ( 2 )k 1 ( 2 )k 2 ( 2 )
Zb
³ h c p 2 + d p 2 i p1 ´ ³ h c p 2 + d p 2 i p1 ´
p1
2
2
p 1 −1
x
f
x,
−
g x,
dx
1
1
p
p
1
1
2
2
2k 1 ( 2 )k 2 ( 2 )(b − a ) a
Zd
³h a p 1 + b p 1 i p1 ´ ³h a p 1 + b p 1 i p1 ´
p2
1
1
p 2 −1
y
f
,y g
,y dy
−
2
2
2h 1 ( 21 )h 2 ( 12 )(d p 2 − c p 2 ) c
Zd
´ Z1
³ Zd
p2
p 2 −1
p 2 −1
h 1 (t )h 2 (1 − t )d t
y
f
(b,
y)g
(b,
y)d
y
y
f
(a,
y)g
(a,
y)d
y
+
≤ p
d 2 − c p2 c
0
c
1
HERMITE-HADAMARD TYPE INEQUALITIES
303
Zd
´ Z1
³ Zd
p2
p 2 −1
p 2 −1
h 1 (t )h 2 (t )d t
y
f
(b,
y)g
(a,
y)d
y
y
f
(a,
y)g
(b,
y)d
y
+
d p2 − c p2 c
0
c
Zb
³ Zb
´ Z1
p1
p 1 −1
p 1 −1
+ p
x
f (x, c)g (x, c)d x +
x
f (x, d )g (x, d )d x
k 1 (t )k 2(1 − t )d t
b 1 − a p1 a
a
0
Z
Z
Z
´ 1
³ b
b
p1
p 1 −1
p 1 −1
k 1 (t )k 2(t )d t
x
f
(x,
d
)g
(x,
c)d
x
x
f
(x,
c)g
(x,
d
)d
x
+
+ p
b 1 − a p1 a
0
a
Z1
Z1
+2M 1 (a, b, c, d )
h 1 (t )h 2 (1 − t )d t
k 1 (t )k 2 (1 − t )d t
0
0
Z1
Z1
k 1 (t )k 2 (t )d t
h 1 (t )h 2 (1 − t )d t
+2M 2 (a, b, c, d )
0
0
Z1
Z1
k 1 (t )k 2 (1 − t )d t
h 1 (t )h 2 (t )d t
+2M 3 (a, b, c, d )
0
0
Z1
Z1
+2M 4 (a, b, c, d )
h 1 (t )h 2 (t )d t
k 2 (t )k 2 (t )d t .
(3.17)
+
0
0
By applying Theorem 7 in [15] to
p 1 x p 1 −1
b p 1 −a p 1
1
f
2k 1 ( 12 )k 2 ( 21 )
³ h p p i p1 ´ ³ h p p i p1 ´
2
2
2
2
2
2
x, c +d
g x, c +d
, multiplying both
2
2
and integrating over [a, b], we have
Zb
³ h c p 2 + d p 2 i p1 ´ ³ h c p 2 + d p 2 i p1 ´
p1
2
2
p 1 −1
x
f
x,
g x,
dx
2
2
2k 1 ( 21 )k 2 ( 12 )(b p 1 − a p 1 ) a
Zb Zd
p1p2
x p 1 −1 y p 2 −1 f (x, y)d xd y
− p
(b 1 − a p 1 )(d p 2 − c p 2 ) a c
Zb
´ Z1
³ Zb
p1
p 1 −1
p 1 −1
k 1 (t )k 2 (1 − t )d t
x
f
(x,
d
)g
(x,
d
)d
x
≤ p
x
f
(x,
c)g
(x,
c)d
x
+
b 1 − a p1 a
0
a
Zb
´ Z1
³ Zb
p1
p 1 −1
p 1 −1
k 1 (t )k 2 (t )d t . (3.18)
x
f (x, d )g (x, c)d x
x
f (x, c)g (x, d )d x +
+ p
b 1 − a p1 a
0
a
³h p p i p1 ´ ³h p p i p1 ´
1
1
1
1
a 1 +b 1
1
Similarly by applying Theorem 7 in [15] to
, y g a +b
, y , mul1 f
1
2
2
sides by
tiplying both sides by
p 2 y p 2 −1
d p 2 −c p 2
p2
2h 1 ( 21 )h 2 ( 21 )(d p 2 − c p 2 )
−
(b p 1
p1p2
p
− a 1 )(d p 2
Zd
2h1 ( 2 )h2 ( 2 )
and integrating over [c, d ], we have
y p 2 −1 f
c
− c p2 )
Zb Zd
a
³h a p 1 + b p 1 i p1
1
2
´ ³h a p 1 + b p 1 i p1 ´
1
,y g
,y dy
2
x p 1 −1 y p 2 −1 f (x, y)d xd y
c
Zd
´ Z1
³ Zd
p2
p 2 −1
p 2 −1
h 1 (t )h 2 (1 − t )d t
y
f (b, y)g (b, y)d y
y
f (a, y)g (a, y)d y +
≤ p
d 2 − c p2 c
0
c
Z
Z
Z
³ d
´ 1
d
p2
p 2 −1
p 2 −1
+ p
y
f
(a,
y)g
(b,
y)d
y
+
y
f
(b,
y)g
(a,
y)d
y
h 1 (t )h 2 (t )d t . (3.19)
d 2 − c p2 c
c
0
Adding (3.16)-(3.19), we get
³h a p 1 + b p 1 i p1 h c p 2 + d p 2 i p1 ´ ³h a p 1 + b p 1 i p1 h c p 2 + d p 2 i p1 ´
1
1
2
1
2
f
,
g
,
2
2
2
2
4h 1 ( 12 )h 2 ( 12 )k 1 ( 21 )k 2 ( 12 )
304
WENGUI YANG
Zb Zd
p1p2
x p 1 −1 y p 2 −1 f (x, y)d xd y
(b p 1 − a p 1 )(d p 2 − c p 2 ) a c
Zd
³ Zd
´ Z1
p2
p 2 −1
p 2 −1
≤ p
y
f (a, y)g (a, y)d y +
y
f (b, y)g (b, y)d y
h 1 (t )h 2 (1 − t )d t
d 2 − c p2 c
c
0
Z
Z
Z
´ 1
³ d
d
p2
p 2 −1
p 2 −1
h 1 (t )h 2 (t )d t
y
f
(b,
y)g
(a,
y)d
y
y
f
(a,
y)g
(b,
y)d
y
+
+ p
d 2 − c p2 c
0
c
Zb
´ Z1
³ Zb
p1
p 1 −1
p 1 −1
k 1 (t )k 2 (1 − t )d t
x
f
(x,
d
)g
(x,
d
)d
x
x
f
(x,
c)g
(x,
c)d
x
+
+ p
b 1 − a p1 a
0
a
Zb
´ Z1
³ Zb
p1
p 1 −1
p 1 −1
+ p
k 1 (t )k 2 (t )d t
x
f
(x,
d
)g
(x,
c)d
x
x
f
(x,
c)g
(x,
d
)d
x
+
b 1 − a p1 a
0
a
Z1
Z1
+M 1 (a, b, c, d )
h 1 (t )h 2 (1 − t )d t
k 1 (t )k 2 (1 − t )d t
0
0
Z1
Z1
k 1 (t )k 2 (t )d t
h 1 (t )h 2 (1 − t )d t
+M 2 (a, b, c, d )
0
0
Z1
Z1
+M 3 (a, b, c, d )
h 1 (t )h 2 (t )d t
k 1 (t )k 2 (1 − t )d t
0
0
Z1
Z1
+M 4 (a, b, c, d )
h 1 (t )h 2 (t )d t
k 2 (t )k 2 (t )d t .
(3.20)
−
0
0
Applying Theorem 1.2 to each term of right hand side of the above inequality (3.19), we
have
p2
d p2 − c p2
Zd
y p 2 −1 f (a, y)g (a, y)d y
c
Z1
k 1 (t )k 2 (t )d t
≤ [ f (a, c)g (a, c) + f (a, d )g (a, d )]
0
Z1
k 1 (t )k 2 (1 − t )d t ,
+[ f (a, c)g (a, d ) + f (a, d )g (a, c)]
0
p2
p
2
d − c p2
Zd
y p 2 −1 f (b, y)g (b, y)d y
Z1
k 1 (t )k 2 (t )d t
≤ [ f (b, c)g (b, c) + f (b, d )g (b, d )]
0
Z1
k 1 (t )k 2 (1 − t )d t ,
+[ f (b, c)g (b, d ) + f (b, d )g (b, c)]
c
0
p2
d p2 − c p2
(3.22)
Zd
y p 2 −1 f (a, y)g (b, y)d y
Z1
≤ [ f (a, c)g (b, c) + f (a, d )g (b, d )]
k 1 (t )k 2(t )d t
0
Z1
+[ f (a, c)g (b, d ) + f (a, d )g (b, c)]
k 1 (t )k 2 (1 − t )d t ,
c
0
p2
p
d 2 − c p2
(3.21)
Zd
c
y p 2 −1 f (b, y)g (a, y)d y
(3.23)
HERMITE-HADAMARD TYPE INEQUALITIES
Z1
k 1 (t )k 2 (t )d t
Z1
k 1 (t )k 2 (1 − t )d t ,
+[ f (b, c)g (a, d ) + f (b, d )g (a, c)]
≤ [ f (b, c)g (a, c) + f (b, d )g (a, d )]
0
0
p1
b p1 − a p1
0
a
0
(3.26)
Zb
x p 1 −1 f (x, c)g (x, d )d x
Z1
≤ [ f (a, c)g (a, d ) + f (b, c)g (b, d )]
h 1 (t )h 2 (t )d t
0
Z1
h 1 (t )h 2 (1 − t )d t ,
+[ f (a, c)g (b, d ) + f (b, c)g (a, d )]
a
0
p1
p
b 1 − a p1
(3.25)
Zb
x p 1 −1 f (x, d )g (x, d )d x
Z1
h 1 (t )h 2 (t )d t
≤ [ f (a, d )g (a, d ) + f (b, d )g (b, d )]
0
Z1
+[ f (a, d )g (b, d ) + f (b, d )g (a, d )]
h 1 (t )h 2 (1 − t )d t ,
p1
p
1
b − a p1
(3.24)
Zb
x p 1 −1 f (x, c)g (x, c)d x
a
Z1
≤ [ f (a, c)g (a, c) + f (b, c)g (b, c)]
h 1 (t )h 2 (t )d t
0
Z1
+[ f (a, c)g (b, c) + f (b, c)g (a, c)]
h 1 (t )h 2 (1 − t )d t ,
p1
p
1
b − a p1
305
(3.27)
Zb
x p 1 −1 f (x, d )g (x, c)d x
a
Z1
h 1 (t )h 2 (t )d t
≤ [ f (a, d )g (a, c) + f (b, d )g (b, c)]
0
Z1
+[ f (a, d )g (b, c) + f (b, d )g (a, c)]
h 1 (t )h 2 (1 − t )d t .
0
(3.28)
Using the inequalities (3.20)-(3.28) in (3.19), we get
³h a p 1 + b p 1 i p1 h c p 2 + d p 2 i p1 ´ ³h a p 1 + b p 1 i p1 h c p 2 + d p 2 i p1 ´
1
2
1
2
g
,
,
2
2
2
2
4h 1 ( 21 )h 2 ( 12 )k 1 ( 21 )k 2 ( 12 )
Zb Zd
p1p2
− p
x p 1 −1 y p 2 −1 f (x, y)d xd y
(b 1 − a p 1 )(d p 2 − c p 2 ) a c
≤ Ω1 (h 1 , h 2 , k 1 , k 2 )M 1 (a, b, c, d ) + Ω2 (h 1 , h 2 , k 1 , k 2 )M 2 (a, b, c, d )
1
f
+Ω3 (h 1 , h 2 , k 1 , k 2 )M 3 (a, b, c, d ) + Ω4 (h 1 , h 2 , k 1 , k 2 )M 4 (a, b, c, d ),
which give the desired result (3.4). This completes the proof.
Remark 2. In Theorems 3.1 and 3.2, letting p 1 = p 2 = 1 and h 1 (t ) = h 2 (t ) = k 1 (t ) = k 2 (t ) = t ,
Theorems 3.1 and 3.2 reduce to Theorem 1.5. In Theorems 3.1 and 3.2, letting p 1 = p 2 = 1,
306
WENGUI YANG
and h 1 (t ) = h 2 (t ) = t s1 , k 1 (t ) = k 2 (t ) = t s2 or h 1 (t ) = h 2 (t ) = k 1 (t ) = k 2 (t ) = t s , Theorems 3.1
and 3.2 reduces to Theorem 1.6.
Theorem 3.3. Suppose that f , g : ∆ → [0, ∞) are (p 1 , h 1 )-(p 2 , h 2 )-convex and (p 1 , k 1 )-(p 2 , k 2 )convex functions on the co-ordinates on ∆, respectively. Then one has the inequality:
Zb Zd
³ b p1 − x p1 ´ ³ d p2 − y p2 ´
³
p1 p2
p 1 −1 p 2 −1
x
y
k
g
(a,
c)
k2 p
f (x, y)d xd y
1
(b p 1 − a p 1 )(d p 2 − c p 2 )
b p1 − a p1
d 2 − c p2
a c
Zb Zd
³ b p1 − x p1 ´ ³ y p2 − c p2 ´
k2 p
f (x, y)d xd y
+g (a, d )
x p 1 −1 y p 2 −1 k 1 p
b 1 − a p1
d 2 − c p2
a c
Zb Zd
³ x p1 − a p1 ´ ³ d p2 − y p2 ´
x p 1 −1 y p 2 −1 k 1 p
+g (b, c)
k2 p
f (x, y)d xd y
b 1 − a p1
d 2 − c p2
a c
Zb Zd
´
³ x p1 − a p1 ´ ³ y p2 − c p2 ´
k
f
(x,
y)d
xd
y
+g (b, d )
x p 1 −1 y p 2 −1 k 1 p
2
b 1 − a p1
d p2 − c p2
a c
Zb Zd
³ b p1 − x p1 ´ ³ d p2 − y p2 ´
³
p1 p2
p 1 −1 p 2 −1
x
y
h
f
(a,
c)
h2 p
g (x, y)d xd y
+ p
1
(b 1 − a p 1 )(d p 2 − c p 2 )
b p1 − a p1
d 2 − c p2
a c
Zb Zd
³ b p1 − x p1 ´ ³ y p2 − c p2 ´
+ f (a, d )
x p 1 −1 y p 2 −1 h 1 p
h2 p
g (x, y)d xd y
b 1 − a p1
d 2 − c p2
a c
Zb Zd
³ x p1 − a p1 ´ ³ d p2 − y p2 ´
h2 p
g (x, y)d xd y
x p 1 −1 y p 2 −1 h 1 p
+ f (b, c)
b 1 − a p1
d 2 − c p2
a c
Zb Zd
´
³ x p1 − a p1 ´ ³ y p2 − c p2 ´
h
g
(x,
y)d
xd
y
+ f (b, d )
x p 1 −1 y p 2 −1 h 1 p
2
b 1 − a p1
d p2 − c p2
a c
Zb Zd
p1p2
≤ p
x p 1 −1 y p 2 −1 f (x, y)d xd y
p
(b 1 − a 1 )(d p 2 − c p 2 ) a c
Z1
Z1
+M 1 (a, b, c, d )
h 1 (t )k 1 (t )d t
h 2 (t )k 2 (t )d t
0
0
Z1
Z1
h 2 (t )k 2 (1 − t )d t
h 1 (t )k 1 (t )d t
+M 2 (a, b, c, d )
0
0
Z1
Z1
h 2 (t )k 2(t )d t
h 1 (t )k 1 (1 − t )d t
+M 3 (a, b, c, d )
0
0
Z1
Z1
+M 4 (a, b, c, d )
h 1 (t )k 1 (1 − t )d t
h 2 (t )k 2(1 − t )d t ,
0
0
where M 1 (a, b, c, d ), M 2 (a, b, c, d ), M 3 (a, b, c, d ) and M 4 (a, b, c, d ) are defined in Theorem 3.1.
Proof. Since f is (p 1 , h 1 )-(p 2 , h 2 )-convex on the co-ordinates and g is (p 1 , k 1 )-(p 2 , k 2 )-convex
on the coordinates on ∆, it follows that
1 ¢
1
¡
f [t a p 1 + (1 − t )b p 1 ] p1 , [r c p 2 + (1 − r )d p 2 ] p2 ≤ h 1 (t )h 2 (r ) f (a, c) + h 1 (t )h 2 (1 − r ) f (a, d )
+ h 1 (1 − t )h 2 (r ) f (b, c) + h 1 (1 − t )h 2 (1 − r ) f (b, d ), (3.29)
and
HERMITE-HADAMARD TYPE INEQUALITIES
1
1 ¢
¡
g [t a p 1 + (1 − t )b p 1 ] p1 , [r c p 2 + (1 − r )d p 2 ] p2 ≤ k 1 (t )k 2 (r )g (a, c) + k 1 (t )k 2 (1 − r )g (a, d )
+ k 1 (1 − t )k 2 (r )g (b, c) + k 1 (1 − t )k 2 (1 − r )g (b, d ).
307
(3.30)
By (3.29)-(3.30) and using the elementary inequality, if e ≤ f and p ≤ r , then er + f p ≤ e p + f r
for all e, f , p, r ∈ R, we get the following ienquality
1 ¢³
1
¡
f [t a p 1 + (1 − t )b p 1 ] p1 , [r c p 2 + (1 − r )d p 2 ] p2 k 1 (t )k 2 (r )g (a, c) + k 1 (t )k 2 (1 − r )g (a, d )
´
+k 1 (1 − t )k 2 (r )g (b, c) + k 1 (1 − t )k 2 (1 − r )g (b, d )
1
1 ¢³
¡
+g [t a p 1 + (1 − t )d p 1 ] p1 , [r c p 2 + (1 − r )d p 2 ] p2 h 1 (t )h 2 (r ) f (a, c)
´
+h 1 (t )h 2 (1 − r ) f (a, d ) + h 1 (1 − t )h 2 (r ) f (b, c) + h 1 (1 − t )h 2 (1 − r ) f (b, d )
1 ¢ ¡
1
1 ¢
1
¡
≤ f [t a p 1 + (1 − t )b p 1 ] p1 , [r c p 2 + (1 − r )d p 2 ] p2 g [t a p 1 + (1 − t )b p 1 ] p1 , [r c p 2 + (1 − r )d p 2 ] p2
³
´
+ h 1 (t )h 2 (r ) f (a, c) + h 1 (t )h 2 (1 − r ) f (a, d ) + h 1 (1 − t )h 2 (r ) f (b, c) + h 1 (1 − t )h 2 (1 − r ) f (b, d )
³
× k 1 (t )k 2 (r )g (a, c) + k 1 (t )k 2 (1 − r )g (a, d ) + k 1 (1 − t )k 2 (r )g (b, c)
´
+k 1 (1 − t )k 2 (1 − r )g (b, d ) .
(3.31)
By integrating the above inequality (3.31) on [0, 1]×[0, 1] with respect to t , r and by taking into
account the change of variables t a p 1 + (1 − t )b p 1 = x p 1 and r c p 2 + (1 − r )d p 2 = y p 2 , we obtain
the desired result.
Remark 3. In Theorem 3.3, letting p 1 = p 2 = 1 and h 1 (t ) = h 2 (t ) = k 1 (t ) = k 2 (t ) = t , Theorem
3.3 reduces to Theorem 10 obtained by Ödemir and Akdemir [27].
Theorem 3.4. Suppose that f , g : ∆ → [0, ∞) are (p 1 , h 1 )-(p 2 , h 2 )-convex and (p 1 , k 1 )-(p 2 , k 2 )convex functions on the co-ordinates on ∆, respectively. Then one has the inequality:
Zb Zb Zd Zd Z1 Z1
p 12 p 22
x p 1 −1 y p 1 −1 u p 2 −1 w p 2 −1
4(b p 1 − a p 1 )2 (d p 2 − c p 2 )2 a a c c 0 0
1
1 ¢
¡
× f [t x p 1 + (1 − t )y p 1 ] p1 , [r u p 2 + (1 − r )w p 2 ] p2
1
1 ¢
¡
×g [t x p 1 + (1 − t )y p 1 ] p1 , [r u p 2 + (1 − r )w p 2 ] p2 d t d r d xd yd ud w
Z1
Z1
Zb Zd
p1 p2
p 1 −1 p 2 −1
≤ p
x
y
f (x, y)g (x, y)d xd y
h 1 (t )k 1 (t )d t
h 2 (t )k 2 (t )d t
(b 1 − a p 1 )(d p 2 − c p 2 ) a c
0
0
+(Θ1 + Θ3 + Θ5 )M 1 (a, b, c, d ) + (Θ1 + Θ4 + Θ5 )M 2 (a, b, c, d )
+(Θ2 + Θ3 + Θ5 )M 3 (a, b, c, d ) + (Θ2 + Θ4 + Θ5 )M 4 (a, b, c, d ),
(3.32)
where M 1 (a, b, c, d ), M 2 (a, b, c, d ), M 3 (a, b, c, d ) and M 4 (a, b, c, d ) are defined in Theorem 3.1,
and
Θ1 =
Z1
0
h 2 (t )d t
Z1
0
k 2 (t )d t
³ Z1
0
h 1 (t )k 1 (t )d t
´2 Z1
0
h 2 (t )k 2 (1 − t )d t ,
308
WENGUI YANG
Z1
h 2 (t )k 2 (1 − t )d t ,
h 1 (t )k 1(1 − t )d t
0
0
0
0
0
Z1
Z1
´2 Z1
³ Z1
h 1 (t )k 1 (1 − t )d t ,
h 2 (t )k 2(t )d t
k 1 (t )d t
h 1 (t )d t
Θ3 =
0
0
0
0
Z1
Z1
Z1
Z1
Z1
Θ4 =
h 1 (t )d t
k 1 (t )d t
h 2 (t )k 2 (t )d r
h 2 (t )k 2 (1 − t )d t
h 1 (t )k 1 (1 − t )d t ,
0
0
0
0
0
Z1
Z1
Z1
Z1
Z1
Z1
h 2 (t )k 2 (1 − t )d t .
h 1 (t )k 1 (1 − t )d t
k 2 (t )d t
k 1 (t )d t
h 2 (t )d t
h 1 (t )d t
Θ5 =
Θ2 =
Z1
h 2 (t )d t
Z1
k 2 (t )d t
h 1 (t )k 1 (t )d t
0
0
0
0
Z1
Z1
0
0
Proof. Since f is (p 1 , h 1 )-(p 2 , h 2 )-convex on the co-ordinates and g is (p 1 , k 1 )-(p 2 , k 2 )-convex
on the coordinates on ∆, it follows that
1
1 ¢
¡
f [t x p 1 + (1 − t )y p 1 ] p1 , [r u p 2 + (1 − r )w p 2 ] p2
≤ h 1 (t )h 2 (r ) f (x, u) + h 1 (t )h 2 (1 − r ) f (x, w ) + h 1(1 − t )h 2 (r ) f (y, u)
+h 1 (1 − t )h 2 (1 − r ) f (y, w ),
and
(3.33)
1 ¢
1
¡
g [t x p 1 + (1 − t )y p 1 ] p1 , [r u p 2 + (1 − r )w p 2 ] p2
≤ k 1 (t )k 2 (r )g (x, u) + k 1 (t )k 2 (1 − r )g (x, w ) + k 1 (1 − t )k 2 (r )g (y, u)
+k 1 (1 − t )k 2 (1 − r )g (y, w ),
(3.34)
for t , r ∈ [0, 1], x, y ∈ [a, b] and u, w ∈ [c, d ]. Because f and g are nonnegative, from (3.33) and
(3.34), we get the inequality
1 ¢ ¡
1
1 ¢
1
¡
f [t x p 1 + (1 − t )y p 1 ] p1 , [r u p 2 + (1 − r )w p 2 ] p2 g [t x p 1 + (1 − t )y p 1 ] p1 , [r u p 2 + (1 − r )w p 2 ] p2
³
´
≤ h 1 (t )h 2 (r ) f (x, u) + h 1 (t )h 2 (1−r ) f (x, w ) + h 1 (1−t )h 2 (r ) f (y, u) + h 1(1−t )h 2 (1−r ) f (y, w )
³
´
× k 1 (t )k 2 (r )g (x, u) + k 1 (t )k 2 (1−r )g (x, w ) + k 1 (1−t )k 2 (r )g (y, u) + k 1 (1−t )k 2 (1−r )g (y, w )
= h 1 (t )h 2 (r )k 1 (t )k 2 (r ) f (x, u)g (x, u) + h 1 (t )h 2 (1 − r )k 1 (t )k 2 (1 − r ) f (x, w )g (x, w )
+h 1 (1−t )h 2 (r )k 1 (1−t )k 2 (r ) f (y, u)g (y, u) + h 1 (1−t )h 2 (1−r )k 1 (1−t )k 2 (1−r ) f (y, w )g (y, w )
+h 1 (t )h 2 (1 − r )k 1 (t )k 2 (r ) f (x, w )g (x, u) + h 1 (t )h 2 (r )k 1 (t )k 2 (1 − r ) f (x, u)g (x, w )
+h 1 (1−t )h 2 (1−r )k 1 (1−t )k 2 (r ) f (y, w )g (y, u) + h 1(1−t )h 2 (r )k 1 (1−t )k 2 (1−r ) f (y, u)g (y, w
+h 1 (t )h 2 (r )k 1 (1 − t )k 2 (r ) f (x, u)g (y, u) + h 1 (1 − t )h 2 (r )k 1 (t )k 2 (r ) f (y, u)g (x, u)
+h 1 (1−t )h 2 (1−r )k 1 (t )k 2 (1−r ) f (y, w )g (x, w ) + h 1(t )h 2 (1−r )k 1 (1−t )k 2 (1−r ) f (x, w )g (y, w )
+h 1 (1 − t )h 2 (1 − r )k 1 (t )k 2 (r ) f (y, w )g (x, u) + h 1(1 − t )h 2 (r )k 1 (t )k 2 (1 − r ) f (y, u)g (x, w )
+h 1 (t )h 2 (1−r )k 1 (1−t )k 2 (r ) f (x, w )g (y, u) + h 1(t )h 2 (r )k 1 (1−t )k 2 (1−r ) f (x, u)g (y, w ). (3.35)
Multiplying both sides of (3.35) by
[0, 1]2 , we have
p 12 p 22
(b p 1 − a p 1 )2 (d p 2 − c p 2 )2
p 12 p 22 x p 1 −1 y p 1 −1 u p 2 −1 w p 2 −1
(b p 1 −a p 1 )2 (d p 2 −c p 2 )2
Zb Zb Zd Zd Z1 Z1
a
a
c
c
0
0
and integrating over [a, b]2 ×[c, d ]2 ×
x p 1 −1 y p 1 −1 u p 2 −1 w p 2 −1
HERMITE-HADAMARD TYPE INEQUALITIES
309
1
1 ¢
¡
× f [t x p 1 + (1 − t )y p 1 ] p1 , [r u p 2 + (1 − r )w p 2 ] p2
1
1 ¢
¡
×g [t x p 1 + (1 − t )y p 1 ] p1 , [r u p 2 + (1 − r )w p 2 ] p2 d t d r d xd yd ud w
Zb Zd
Z1
Z1
4p 1 p 2
p 1 −1 p 2 −1
x
y
f
(x,
y)g
(x,
y)d
xd
y
h
(t
)k
(t
)d
t
≤ p
h 2 (t )k 2 (t )d t
1
1
(b 1 − a p 1 )(d p 2 − c p 2 ) a c
0
0
Zb Zd Zd
4p 1 p 22
x p 1 −1 u p 2 −1 w p 2 −1 f (x, w )g (x, u)d xd ud w
+ p
(b 1 − a p 1 )(d p 2 − c p 2 )2 a c c
Z1
Z1
h 1 (t )k 1 (t )d t
h 2 (t )k 2 (1 − t )d t
0
+
0
4p 12 p 2
Zb Zb Zd
x p 1 −1 y p 1 −1 u p 2 −1 f (x, u)g (y, u)d xd yd u
(b p 1 − a p 1 )2 (d p 2 − c p 2 ) a a c
Z1
Z1
×
h 1 (t )k 1 (1 − t )d t
h 2 (t )k 2(t )d t
0
+
0
4p 12 p 22
Zb Zd
x p 1 −1 w p 2 −1 f (x, w )d xd w
(b p 1 − a p 1 )2 (d p 2 − c p 2 )2 a c
Z1
Z1
×
h 1 (t )k 1 (1 − t )d t
h 2 (t )k 2(1 − t )d t .
0
Zb Zd
a
y p 1 −1 u p 2 −1 g (y, u)d yd u
c
(3.36)
0
Applying Theorems 1.1 and 1.2 to second and third term of right hand side of the inequality (3.36), we have
p 1 p 22
Zb Zd Zd
x p 1 −1 u p 2 −1 w p 2 −1 f (x, w )g (x, u)d xd ud w
(b p 1 − a p 1 )(d p 2 − c p 2 )2 a c c
Zd Zd
Zb
³
´
p 22
p1
p 2 −1 p 2 −1
p 1 −1
= p
u
w
x
f
(x,
w
)g
(x,
u)d
x
d ud w
(d 2 − c p 2 )2 c c
b p1 − a p1 a
Z1
Zd Zd
³
p2
h 1 (t )k 1 (t )d t
u p 2 −1 w p 2 −1 [ f (a, w )g (a, u) + f (b, w )g (b, u)]
≤ p 2p 2
(d 2 − c 2 ) c c
0
Z1
´
+[ f (a, w )g (b, u) + f (b, w )g (a, u)]
h 1 (t )k 1 (1 − t )d t d ud w
0
≤
p 22
³h Zd
w p 2 −1 f (a, w )d w
Zd
u p 2 −1 g (a, u)d u
(d p 2 − c p 2 )2
c
c
Zd
Zd
i Z1
h 1 (t )k 1 (t )d t
u p 2 −1 g (b, u)d u
w p 2 −1 f (b, w )d w
+
0
c
c
Zd
h Zd
p 2 −1
+
w
f (a, w )d w
u p 2 −1 g (b, u)d u
c
c
Zd
Zd
i Z1
´
h 1 (t )k 1 (1 − t )d t
g (a, u)d u
0
c
c
Z1
Z1
Z1
≤ [ f (a, c) + f (a, d )][g (a, c) + g (a, d )]
h 2 (t )d t
k 2 (t )d t
h 1 (t )k 1 (t )d t
0
0
0
Z1
Z1
Z1
h 1 (t )k 1 (t )d t
k 2 (t )d t
h 2 (t )d t
+[ f (b, c) + f (b, d )][g (b, c) + g (b, d )]
+
w
p 2 −1
f (b, w )d w
u
p 2 −1
0
0
0
310
WENGUI YANG
+[ f (a, c) + f (a, d )][g (b, c) + g (b, d )]
Z1
0
h 2 (t )d t
Z1
0
k 2 (t )d t
Z1
0
h 1 (t )k 1 (1 − t )d t
Z1
Z1
Z1
h 1 (t )k 1 (1 − t )d t
k 2 (t )d t
h 2 (t )d t
+[ f (b, c) + f (b, d )][g (a, c) + g (a, d )]
0
0
0
Z1
Z1
Z1
= [M 1(a, b, c, d ) + M 2 (a, b, c, d )]
h 2 (t )d t
k 2 (t )d t
h 1 (t )k 1(t )d t
0
0
0
Z1
Z1
Z1
h 1 (t )k 1 (1 − t )d t ,
k 2 (t )d t
h 2 (t )d t
+[M 3 (a, b, c, d ) + M 4(a, b, c, d )]
and
p 12 p 2
0
0
0
(3.37)
Zb Zb Zd
x p 1 −1 y p 1 −1 u p 2 −1 f (x, u)g (y, u)d xd yd u
(b p 1 − a p 1 )2 (d p 2 − c p 2 ) a a c
Zb Zb
Zd
³
´
p 12
p2
p 2 −1
p 1 −1 p 1 −1
= p
u
f
(x,
u)g
(y,
u)d
u
d xd y
x
y
(b 1 − a p 1 )2 a a
d p2 − c p2 c
Zb Zb
Z1
³
p2
≤ p 1 p 2
x p 1 −1 y p 1 −1 [ f (x, c)g (y, c) + f (x, d )g (y, d )]
h 2 (t )k 2 (t )d t
(b 1 − a 1 ) a a
0
Z1
´
h 2 (t )k 2 (1 − t )d t d xd y
+[ f (x, c)g (y, d ) + f (x, d )g (y, c)]
0
≤
p 12
³h Zb
Zb
p 1 −1
x
f (x, c)d x
y p 1 −1 g (y, c)d y
(b p 1 − a p 1 )2
a
a
Zb
Zb
i Z1
+
x p 1 −1 f (x, d )d x
y p 1 −1 g (y, d )d y
h 2 (t )k 2 (t )d t
a
0
a
Zb
h Zb
p 1 −1
x
f (x, c)d x
+
y p 1 −1 g (y, d )d y
a
a
Zb
Zb
i Z1
´
g (y, c)d t
h 2 (t )k 2 (1 − t )d t
a
a
0
Z1
Z1
Z1
h 2 (t )k 2 (t )d t
k 1 (t )d t
h 1 (t )d t
≤ [ f (a, c) + f (b, c)][g (a, c) + g (b, c)]
0
0
0
Z1
Z1
Z1
+[ f (a, d ) + f (b, d )][g (a, d ) + g (b, d )]
h 1 (t )d t
k 1 (t )d t
h 2 (t )k 2 (t )d t
0
0
0
Z1
Z1
Z1
h 2 (t )k 2 (1 − t )d t
k 1 (t )d t
h 1 (t )d t
+[ f (a, c) + f (b, c)][g (a, d ) + g (b, d )]
0
0
0
Z1
Z1
Z1
h 2 (t )k 2 (1 − t )d t
k 1 (t )d t
h 1 (t )d t
+[ f (a, d ) + f (b, d )][g (a, c) + g (b, c)]
0
0
0
Z1
Z1
Z1
= [M 1(a, b, c, d ) + M 3 (a, b, c, d )]
h 1 (t )d t
k 1 (t )d t
h 2 (t )k 2(t )d t
0
0
0
Z1
Z1
Z1
h 2 (t )k 2 (1 − t )d t .
k 1 (t )d t
h 1 (t )d t
+[M 2 (a, b, c, d ) + M 4(a, b, c, d )]
+
x
p 1 −1
f (x, d )d x
y
p 1 −1
0
0
0
(3.38)
Applying Theorem 2.1 to fourth term of right hand side of the inequality (3.36), we have
p 12 p 22
(b p 1 − a p 1 )2 (d p 2 − c p 2 )2
Zb Zd
a
c
x p 1 −1 w p 2 −1 f (x, w )d xd w
Zb Zd
a
c
y p 1 −1 u p 2 −1 g (y, u)d yd u
HERMITE-HADAMARD TYPE INEQUALITIES
311
Z1
Z1
´
³
h 2 (t )d t
h 1 (t )d t
≤ [ f (a, c) + f (a, d ) + f (b, c) + f (b, d )]
0
0
Z1
Z1
´
³
k 2 (t )d t
k 1 (t )d t
· [g (a, c) + g (a, d ) + g (b, c) + g (b, d )]
0
0
= [M 1 (a, b, c, d ) + M 2(a, b, c, d ) + M 3 (a, b, c, d ) + M 4(a, b, c, d )]
Z1
Z1
Z1
Z1
k 2 (t )d t .
k 1 (t )d t
h 2 (t )d t
h 1 (t )d t
×
0
0
0
0
Using the inequalities (3.37)-(3.39) in (3.36), we get the desired result (3.32).
(3.39)
Remark 4. Suppose that f , g : ∆ → [0, ∞) are convex functions on the co-ordinates on ∆. Then
one has the inequality:
Zb Zb Zd Zd Z1 Z1
9
f (t x + (1 − t )y, r u + (1 − r )w )g (t x + (1 − t )y,
4(b − a)2 (d − c)2 a a c c 0 0
r u + (1 − r )w )d t d r d xd yd ud w
Zb Zd
19
5
11
1
f (x, y)g (x, y)d xd y +
L(a, b, c, d )+ M (a, b, c, d )+
N (a, b, c, d ),
≤
(b − a)(d − c) a c
192
64
192
where L(a, b, c, d ), M (a, b, c, d ) and N (a, b, c, d ) are defined in Theorem 1.5.
Remark 5. Suppose that f , g : ∆ → [0, ∞) are s-convex functions on the co-ordinates on ∆.
Then one has the inequality:
Zb Zb Zd Zd Z1 Z1
(1 + 2s)2
f (t x + (1 − t )y, r u + (1 − r )w )
4(b − a)2 (d − c)2 a a c c 0 0
g (t x + (1 − t )y, r u + (1 − r )w )d t d r d xd yd ud w
Zb Zd
1
≤
f (x, y)g (x, y)d xd y
(b − a)(d − c) a c
³ (1 + 2s)2 [Γ(1 + s)]2 + 2(1 + s)2 Γ(2 + 2s)
+[Γ(1 + s)]2
L(a, b, c, d )
(1 + s)4 [Γ(2 + 2s)]2
(2 + 8s + 9s 2 + 2s 3 )[Γ(1 + s)]2 + (1 + s)2 Γ(2 + 2s)
M (a, b, c, d )
+
(1 + s)4 [Γ(2 + 2s)]2
´
(3 + 6s + 2s 2 )[Γ(1 + s)]2
+
N
(a,
b,
c,
d
)
,
(1 + s)4 [Γ(2 + 2s)]2
where L(a, b, c, d ), M (a, b, c, d ) and N (a, b, c, d ) are defined in Theorem 1.5.
Remark 6. Suppose that f , g : ∆ → [0, ∞) are s 1 -convex and s 2 -convex functions on the coordinates on ∆, respectively. Then one has the inequality:
(1 + s 1 + s 2 )2
4(b − a)2 (d − c)2
Zb Zb Zd Zd Z1 Z1
a
a
c
c
0
0
f (t x + (1−t )y, r u + (1−r )w )g (t x + (1−t )y, r u + (1−r )w )
d t d r d xd yd ud w
≤
1
(b − a)(d − c)
Zb Zd
a
c
f (x, y)g (x, y)d xd y + Ξ1L(a, b, c, d ) + Ξ2 M (a, b, c, d ) + Ξ3 N (a, b, c, d ),
312
WENGUI YANG
where L(a, b, c, d ), M (a, b, c, d ) and N (a, b, c, d ) are defined in Theorem 1.5, and
¡
¢
Γ(1 + s 1 )Γ(1 + s 2 ) (1 + s 1 + s 2 )2 Γ(1 + s 1 )Γ(1 + s 2 ) + 2(1 + s 1 )(1 + s 2 )Γ(2 + s 1 + s 2 )
Ξ1 =
,
(1 + s 1 )2 (1 + s 2 )2 [Γ(2 + s 1 + s 2 )]2
(2(1 + s 2 )2 + s 12 (2 + s 2 ) + s 1 (4 + 5s 2 + s 22 ))[Γ(1 + s 1 )]2 [Γ(1 + s 2 )]2
Ξ2 =
(1 + s 1 )2 (1 + s 2 )2 [Γ(2 + s 1 + s 2 )]2
(1 + s 1 )(1 + s 2 )Γ(1 + s 1 )Γ(1 + s 2 )Γ(2 + s 1 + s 2 )
,
+
(1 + s 1 )2 (1 + s 2 )2 [Γ(2 + s 1 + s 2 )]2
(1 + s 1 + s 2 )(3 + 3s 1 + 3s 2 + 2s 1 s 2 )[Γ(1 + s 1 )]2 [Γ(1 + s 2 )]2
Ξ3 =
.
(1 + s 1 )2 (1 + s 2 )2 [Γ(2 + s 1 + s 2 )]2
Theorem 3.5. Suppose that f , g : ∆ → [0, ∞) are (p 1 , h 1 )-(p 2 , h 2 )-convex and (p 1 , k 1 )-(p 2 , k 2 )convex functions on the co-ordinates on ∆, respectively. Then one has the inequality:
p1 p2
p
p
(b 1 − a 1 )(d p 2 − c p 2 )
Zb Zd Z1 Z1
a
c
0
0
x p 1 −1 u p 2 −1 f
³£
t x p 1 + (1 − t )
a p 1 + b p 1 ¤ p1
1 ,
2
c p 2 + d p 2 ¤ p1 ´
2
2
³£
a p 1 + b p 1 ¤ p1 £ p 2
c p 2 + d p 2 ¤ p1 ´
1 , ru
2 d t d r d xd u
×g t x p 1 + (1 − t )
+ (1 − r )
2
2
Zb Zd
Z1
Z1
p1p2
p 1 −1 p 2 −1
x
u
f (x, u)g (x, u)d xd u
≤ p
h 1 (t )k 1 (t )d t
h 2 (t )k 2 (t )d t
(b 1 − a p 1 )(d p 2 − c p 2 ) a c
0
0
+(Λ1 + Λ3 + Λ5 )M 1 (a, b, c, d ) + (Λ1 + Λ4 + Λ5 )M 2 (a, b, c, d )
£
r u p 2 + (1 − r )
+(Λ2 + Λ3 + Λ5 )M 3 (a, b, c, d ) + (Λ2 + Λ4 + Λ5 )M 4 (a, b, c, d ),
(3.40)
where M 1 (a, b, c, d ), M 2 (a, b, c, d ), M 3 (a, b, c, d ) and M 4 (a, b, c, d ) are defined in Theorem 3.1,
and
Z
Z
³
´
¡
1
1 1
1 ¢ 1
1
Λ1 = 2 2h 2 ( )k 2 ( )
h 2 (t )k 2(1 − t )d t
h 2 (t )k 2 (t )d t + h 2 ( ) + k 2 ( )
2
2 0
2
2 0
Z1
Z1
´2
³ Z1
h 1 (t )k 1 (t )d t ,
k 2 (t )d t d t
h 2 (t )
0
0
0
Z
Z
³
´ Z1
¡
1
1 1
1
1 ¢ 1
Λ2 = 2 2h 2 ( )k 2 ( )
h 2 (t )d t
h 2 (t )k 2(1 − t )d t
h 2 (t )k 2 (t )d t + h 2 ( ) + k 2 ( )
2
2 0
2
2 0
0
Z1
Z1
Z1
k 2 (t )d t
h 1 (t )k 1 (t )d t
h 1 (t )k 1 (1 − t )d t ,
0
0
0
Z
Z
´
³
¡
1
1 1
1 ¢ 1
1
h 1 (t )k 1(1 − t )d t
h 1 (t )k 1 (t )d t + h 1 ( ) + k 1 ( )
Λ3 = 2 2h 1 ( )k 1 ( )
2
2 0
2
2 0
Z1
Z1
´2
³ Z1
h 2 (t )k 2 (t )d t ,
h 1 (t )d t
k 1 (t )d t
0
0
0
Z
Z
³
´
¡
1
1 1
1 ¢ 1
1
Λ4 = 2 2h 1 ( )k 1 ( )
h 1 (t )k 1(1 − t )d t
h 1 (t )k 1 (t )d t + h 1 ( ) + k 1 ( )
2
2 0
2
2 0
Z1
Z1
Z1
Z1
h 2 (t )k 2 (1 − t )d t ,
h 2 (t )k 2 (t )d t
k 1 (t )d t
h 1 (t )d t
0
0
0
0
HERMITE-HADAMARD TYPE INEQUALITIES
313
Z1
Z1
³
1
1
1
1
h 2 (t )k 2 (t )d t
Λ5 = 4 4h 1 ( )h 2 ( )k 1 ( )k 2 ( )
h 1 (t )k 1 (t )d t
2
2
2
2 0
0
Z1
Z
1
1 ¡
1
1 ¢ 1
+2h 1 ( )k 1 ( ) h 2 ( ) + k 2 ( )
h 2 (t )k 2 (1 − t )d t
h 1 (t )k 1 (t )d t
2
2
2
2 0
0
Z
Z1
1 ¡
1
1 ¢ 1
1
h 1 (t )k 1 (1 − t )d t
+2h 2 ( )k 2 ( ) h 1 ( ) + k 1 ( )
h 2 (t )k 2 (t )d t
2
2
2
2 0
0
Z
¡
1
1
1
1
1
1
1 ¢ 1
1
h 1 (t )k 1 (1 − t )d t
+ h 1 ( )k 2 ( ) + h 2 ( )k 1 ( ) + h 1 ( )h 2 ( ) + k 1 ( )k 2 ( )
2
2
2
2
2
2
2
2 0
Z1
Z
Z
Z
Z
´ 1
1
1
1
h 2 (t )k 2 (1 − t )d t
h 1 (t )d t
h 2 (t )d t
k 1 (t )d t
k 2 (t )d t .
0
0
0
0
0
Proof. Since f is (p 1 , h 1 )-(p 2 , h 2 )-convex on the co-ordinates and g is (p 1 , k 1 )-(p 2 , k 2 )-convex
on the coordinates on ∆, it follows that
³£
c p 2 + d p 2 ¤ p1 ´
a p 1 + b p 1 ¤ p1 £ p 2
1 , ru
2
+ (1 − r )
f t x p 1 + (1 − t )
2
2
¡ £ c p2 + d p2 ¤ 1 ¢
p2
≤ h 1 (t )h 2 (r ) f (x, u) + h 1 (t )h 2 (1 − r ) f x,
2
¡£ a p 1 +b p 1 ¤ 1 £ c p 2 +d p 2 ¤ 1 ¢
¡£ a p 1 + b p 1 ¤ 1 ¢
p1
p1
p2
, u + h 1 (1−t )h 2 (1−r ) f
,
,(3.41)
+h 1 (1 − t )h 2 (r ) f
2
2
2
and
³£
a p 1 + b p 1 ¤ p1 £ p 2
c p 2 + d p 2 ¤ p1 ´
1 , ru
2
g t x p 1 + (1 − t )
+ (1 − r )
2
2
¡ £ c p2 + d p2 ¤ 1 ¢
p2
≤ k 1 (t )k 2 (r ) f (x, u) + k 1 (t )k 2 (1 − r ) f x,
2
¡£ a p 1 + b p 1 ¤ 1 ¢
¡£ a p 1 +b p 1 ¤ 1 £ c p 2 +d p 2 ¤ 1 ¢
p1
p1
p2
+k 1 (1 − t )k 2 (r ) f
, u + k 1 (1−t )k 2 (1−r ) f
,
, (3.42)
2
2
2
for t , r ∈ [0, 1], x ∈ [a, b] and u ∈ [c, d ]. Because f and g are nonnegative, from (3.41) and
(3.42), we get the inequality
³£
c p 2 + d p 2 ¤ p1 ´
a p 1 + b p 1 ¤ p1 £ p 2
1 , ru
2
+ (1 − r )
f t x p 1 + (1 − t )
2
2
³£
a p 1 + b p 1 ¤ p1 £ p 2
c p 2 + d p 2 ¤ p1 ´
1 , ru
2
×g t x p 1 + (1 − t )
+ (1 − r )
2
2
³
¡ £ c p2 + d p2 ¤ 1 ¢
¡£ a p 1 + b p 1 ¤ 1 ¢
p2
p1
≤ h 1 (t )h 2 (r ) f (x, u) + h 1 (t )h 2 (1 − r ) f x,
+ h 1 (1 − t )h 2 (r ) f
,u
2
2
¡£ a p 1 + b p 1 ¤ 1 £ c p 2 + d p 2 ¤ 1 ¢´
p1
p2
,
+h 1 (1 − t )h 2 (1 − r ) f
2
2
³
¡ £ c p2 + d p2 ¤ 1 ¢
p2
k 1 (t )k 2 (r ) f (x, u) + k 1 (t )k 2 (1 − r ) f x,
2
¡£ a p 1 + b p 1 ¤ 1 ¢
¡£ a p 1 + b p 1 ¤ 1 £ c p 2 + d p 2 ¤ 1 ¢´
p1
p1
p2
+k 1 (1 − t )k 2 (r ) f
, u + k 1 (1 − t )k 2 (1 − r ) f
,
2
2
2
£ c p2 + d p2 ¤ 1
p2
)
= h 1 (t )h 2 (r )k 1 (t )k 2 (r ) f (x, u)g (x, u) + h 1 (t )h 2 (1 − r )k 1 (t )k 2 (1 − r ) f (x,
2
£ a p1 + b p1 ¤ 1
£ a p1 + b p1 ¤ 1
£ c p2 + d p2 ¤ 1
p2
p1
p1
) + h 1 (1 − t )h 2 (r )k 1 (1 − t )k 2 (r ) f (
, u)g (
, u)
g (x,
2
2
2
314
WENGUI YANG
£ a p1 + b p1 ¤ 1 £ c p2 + d p2 ¤ 1
p1
p2
+h 1 (1 − t )h 2 (1 − r )k 1 (1 − t )k 2 (1 − r ) f (
,
)
2
2
£ a p1 + b p1 ¤ 1 £ c p2 + d p2 ¤ 1
£ c p2 + d p2 ¤ 1
p1
p2
p2
g(
,
) + h 1 (t )h 2 (1 − r )k 1 (t )k 2 (r ) f (x,
)g (x, u)
2
2
2
p
p
£c 2 +d 2 ¤ 1
p2
)
+h 1 (t )h 2 (r )k 1 (t )k 2 (1 − r ) f (x, u)g (x,
2
£ a p1 + b p1 ¤ 1 £ c p2 + d p2 ¤ 1
£ a p1 + b p1 ¤ 1
p1
p2
p1
+h 1 (1 − t )h 2 (1 − r )k 1 (1 − t )k 2 (r ) f (
,
)g (
, u)
2
2
2
£ a p1 + b p1 ¤ 1
£ a p1 + b p1 ¤ 1 £ c p2 + d p2 ¤ 1
p1
p1
p2
+h 1 (1 − t )h 2 (r )k 1 (1 − t )k 2 (1 − r ) f (
, u)g (
,
)
2
2
2
£ a p1 + b p1 ¤ 1
p1
, u)
+h 1 (t )h 2 (r )k 1 (1 − t )k 2 (r ) f (x, u)g (
2
£ a p1 + b p1 ¤ 1
p1
+h 1 (1 − t )h 2 (r )k 1 (t )k 2(r ) f (
, u)g (x, u)
2
£ a p1 + b p1 ¤ 1 £ c p2 + d p2 ¤ 1
£ c p2 + d p2 ¤ 1
p1
p2
p2
+h 1 (1 − t )h 2 (1 − r )k 1 (t )k 2 (1 − r ) f (
,
)g (x,
)
2
2
2
£ a p1 + b p1 ¤ 1 £ c p2 + d p2 ¤ 1
£ c p2 + d p2 ¤ 1
p2
p1
p2
)g (
,
)
+h 1 (t )h 2 (1 − r )k 1 (1 − t )k 2 (1 − r ) f (x,
2
2
2
£ a p1 + b p1 ¤ 1 £ c p2 + d p2 ¤ 1
p1
p2
+h 1 (1 − t )h 2 (1 − r )k 1 (t )k 2 (r ) f (
,
)g (x, u)
2
2
p
p
p
£a 1 +b 1 ¤ 1
£ c 2 + d p2 ¤ 1
p1
p2
+h 1 (1 − t )h 2 (r )k 1 (t )k 2(1 − r ) f (
, u)g (x,
)
2
2
£ c p2 + d p2 ¤ 1
£ a p1 + b p1 ¤ 1
p2
p1
+h 1 (t )h 2 (1 − r )k 1 (1 − t )k 2 (r ) f (x,
)g (
, u)
2
2
£ a p1 + b p1 ¤ 1 £ c p2 + d p2 ¤ 1
p1
p2
,
).
(3.43)
+h 1 (t )h 2 (r )k 1 (1 − t )k 2 (1 − r ) f (x, u)g (
2
2
Multiplying both sides of (3.43) by
p 1 p 2 x p 1 −1 u p 2 −1
(b p 1 −a p 1 )(d p 2 −c p 2 )
and integrating over [a, b] × [c, d ] × [0, 1]2,
we have
Zb Zd Z1 Z1
a p 1 + b p 1 ¤ p1
1 ,
2
0 0
a c
£ p
c p 2 + d p 2 ¤ p1 ´
2
r u 2 + (1 − r )
2
³£
a p 1 + b p 1 ¤ p1 £ p 2
c p 2 + d p 2 ¤ p1 ´
1 , ru
2 d t d r d xd u
×g t x p 1 + (1 − t )
+ (1 − r )
2
2
Zb Zd
Z1
p1 p2
p 1 −1 p 2 −1
x
u
f
(x,
u)g
(x,
u)d
xd
u
h 1 (t )k 1 (t )d t
≤ p
(b 1 − a p 1 )(d p 2 − c p 2 ) a c
0
Z1
Zb
³
³ £ c p2 + d p2 ¤ 1 ´ ³ £ c p2 + d p2 ¤ 1 ´
p1
p 1 −1
p2
p2
h 2 (t )k 2 (t )d t + p
x
f
x,
g x,
dx
b 1 − a p1 a
2
2
0
Zd
³£ a p 1 + b p 1 ¤ 1
³£ a p 1 + b p 1 ¤ 1 ´ ³£ a p 1 + b p 1 ¤ 1 ´
p2
p 2 −1
p1
p1
p1
,
u
g
,
u
d
u
+
f
,
u
f
+ p
d 2 − c p2 c
2
2
2
Z1
Z
£ c p 2 + d p 2 ¤ 1 ´ ³£ a p 1 + b p 1 ¤ 1 £ c p 2 + d p 2 ¤ 1 ´´ 1
p2
p1
p2
h 2 (t )k 2 (t )d t
h 1 (t )k 1 (t )d t
g
,
2
2
2
0
0
p1p2
p
p
1
(b − a 1 )(d p 2 − c p 2 )
x p 1 −1 u p 2 −1 f
³£
t x p 1 + (1 − t )
HERMITE-HADAMARD TYPE INEQUALITIES
315
Zb Zd
³ £ c p2 + d p2 ¤ 1 ´
p1 p2
p 1 −1 p 2 −1
p2
x
u
f
x,
g (x, u)d xd u
(b p 1 − a p 1 )(d p 2 − c p 2 ) a c
2
Zb Zd
³ £ c p2 + d p2 ¤ 1 ´
p1p2
p 1 −1 p 2 −1
p2
d xd u
x
u
f
(x,
u)g
x,
+ p
(b 1 − a p 1 )(d p 2 − c p 2 ) a c
2
Zd
³£ a p 1 + b p 1 ¤ 1 £ c p 2 + d p 2 ¤ 1 ´
³£ a p 1 + b p 1 ¤ 1 ´
p2
p 2 −1
p1
p1
p2
,
u
d
u
f
,
u
g
+ p
d 2 − c p2 c
2
2
2
Zd
³£ a p 1 + b p 1 ¤ 1 ´
³£ a p 1 + b p 1 ¤ 1 £ c p 2 + d p 2 ¤ 1 ´´
p2
p 2 −1
p1
p1
p2
u
f
+ p
,
u
d
u
g
,
d 2 − c p2 c
2
2
2
Z1
Z1
h 2 (t )k 2 (1 − t )d t
h 1 (t )k 1 (t )d t
³
+
0
0
Zb Zd
³£ a p 1 + b p 1 ¤ 1 ´
p1 p2
p 1 −1 p 2 −1
p1
x
u
f
(x,
u)g
, u d xd u
(b p 1 − a p 1 )(d p 2 − c p 2 ) a c
2
Zb Zd
³£ a p 1 + b p 1 ¤ 1 ´
p1p2
p 1 −1 p 2 −1
p1
, u g (x, u)d xd u
+ p
x
u
f
(b 1 − a p 1 )(d p 2 − c p 2 ) a c
2
Zb
³£ a p 1 + b p 1 ¤ 1 £ c p 2 + d p 2 ¤ 1 ´
³ £ c p2 + d p2 ¤ 1 ´
p1
p 1 −1
p2
p1
p2
+ p
d
x
f
x
g
x,
,
b 1 − a p1 a
2
2
2
Zb
³ £ c p2 + d p2 ¤ 1 ´
p1
p 1 −1
p2
dx
x
f
x,
+ p
b 1 − a p1 a
2
Z1
³£ a p 1 + b p 1 ¤ 1 £ c p 2 + d p 2 ¤ 1 ´´ Z1
p1
p2
h 1 (t )k 1 (1 − t )d t
h 2 (t )k 2 (t )d t
,
g
2
2
0
0
Z
Z
³ b d
³£ a p 1 + b p 1 ¤ 1 £ c p 2 + d p 2 ¤ 1 ´
p1p2
p 1 −1 p 2 −1
p1
p2
+ p
,
x
u
g
(x,
u)d
xd
u
f
(b 1 − a p 1 )(d p 2 − c p 2 ) a c
2
2
Zb
³£ a p 1 + b p 1 ¤ 1 ´
³ £ c p 2 + d p 2 ¤ 1 ´ Zd
p2
p1
u p 2 −1 f
dx
,u du
x p 1 −1 g x,
+
2
2
c
a
Zb
³£ a p 1 + b p 1 ¤ 1 ´
³ £ c p 2 + d p 2 ¤ 1 ´ Zd
p 1 −1
p2
p1
dx
,u du
u p 2 −1 g
+
x
f x,
2
2
c
a
Zb Zd
³£ a p 1 + b p 1 ¤ 1 £ c p 2 + d p 2 ¤ 1 ´´
p1
p2
x p 1 −1 u p 2 −1 f (x, u)d xd u g
+
,
2
2
a c
Z1
Z1
h 1 (t )k 1 (1 − t )d t
h 2 (t )k 2 (1 − t )d t .
(3.44)
³
+
0
0
Applying Theorems 1.1 and 1.2 to 2nd-16th term of right hand side of the inequality
(3.44), we have
Zb
£ c p2 + d p2 ¤ 1
£ c p2 + d p2 ¤ 1
p1
p 1 −1
p2
p2
)g
(x,
)d x
x
f
(x,
b p1 − a p1 a
2
2
³
£ c p 2 +d p 2 ¤ 1
£ c p 2 +d p 2 ¤ 1
£ c p 2 +d p 2 ¤ 1
£ c p 2 +d p 2 ¤ 1 ´
p2
p2
p2
p2
≤ f (a,
)g (a,
)+ f (b,
)g (b,
)
2
2
2
2
Z1
³
£ c p2 + d p2 ¤ 1
£ c p2 + d p2 ¤ 1
p2
p2
)g (b,
)
× h 1 (t )k 1 (t )d t + f (a,
2
2
0
Z
£ c p2 + d p2 ¤ 1
£ c p2 + d p2 ¤ 1 ´ 1
p2
p2
h 1 (t )k 1 (1 − t )d t
+ f (b,
)g (a,
)
2
2
0
316
WENGUI YANG
³
´
1
1
≤ [ f (a, c) + f (a, d )][g (a, c) + g (a, d )] + [ f (b, c) + f (b, d )][g (b, c) + g (b, d )] 4h 2 ( )k 2 ( )
2
2
Z1
Z1
Z1
h 1 (t )k 1(t )d t
k 2 (t )d t
h 2 (t )d t
0
0
0
³
´
+ [ f (a, c) + f (a, d )][g (b, c) + g (b, d )] + [ f (a, c) + f (a, d )][g (b, c) + g (b, d )]
Z1
Z1
Z1
1
1
×4h 2 ( )k 2 ( )
h 1 (t )k 1 (1 − t )d t
k 2 (t )d t
h 2 (t )
2
2 0
0
0
Z
Z1
Z1
1 1
1
= (M 1 (a, b, c, d )+M 2 (a, b, c, d ))4h 2 ( )k 2 ( ) h 2 (t ) k 2 (t )d t d t
h 1 (t )k 1 (t )d t
2
2 0
0
0
Z1
Z1
Z1
1
1
h 1 (t )k 1 (1 − t )d t ,
+(M 3 (a, b, c, d ) + M 4 (a, b, c, d ))4h 2 ( )k 2 ( )
k 2 (t )d t d t
h 2 (t )
2
2 0
0
0
(3.45)
Zd
£ a p1 + b p1 ¤ 1
£ a p1 + b p1 ¤ 1
p2
p1
p1
, u)(
, u)d u
u p 2 −1 f (
p
p
d 2 −c 2 c
2
2
³ £ a p 1 +b p 1 ¤ 1
³£ a p 1 +b p 1 ¤ 1 ´ ³£ a p 1 +b p 1 ¤ 1 ´ ³£ a p 1 +b p 1 ¤ 1 ´´
p1
p1
p1
p1
≤ f(
, c)g
,c +f
,d g
,d
2
2
2
2
Z1
³ £ a p1 + b p1 ¤ 1
£ a p1 + b p1 ¤ 1
p1
p1
, c)g (
,d)
× h 2 (t )k 2 (t )d t + f (
2
2
0
´ Z1
£ a p1 + b p1 ¤ 1
£ a p1 + b p1 ¤ 1
p1
p1
+f (
h 2 (t )k 2 (1 − t )d t
, d )g (
, c)
2
2
0
³
´
1
1
≤ [ f (a, c) + f (b, c)][g (a, c) + g (b, c)] + [ f (a, d ) + f (b, d )][g (a, d ) + g (b, d )] 4h 1 ( )k 1 ( )
2
2
Z1
Z1
Z1
h 1 (t )
k 1 (t )d t
h 2 (t )k 2 (t )d t
0
0
0
³
´
+ f (a, c) + f (b, c)][g (a, d ) + g (b, d )] + [ f (a, d ) + f (b, d )][g (a, c) + g (b, c)]
Z1
Z1
Z1
1
1
×4h 1 ( )k 1 ( )
h 1 (t )
k 1 (t )d t
h 2 (t )k 2 (1 − t )d t
2
2 0
0
0
Z1
Z1
Z1
1
1
h 1 (t )
k 1 (t )d t
h 2 (t )k 2 (t )d t
= (M 1 (a, b, c, d ) + M 3 (a, b, c, d ))4h 1( )k 1 ( )
2
2 0
0
0
Z1
Z1
Z
1 1
1
+(M 2 (a, b, c, d )+M 4 (a, b, c, d ))4h 1 ( )k 1 ( ) h 1 (t ) k 1 (t )d t h 2 (t )k 2 (1−t )d t , (3.46)
2
2 0
0
0
£ a p1 + b p1 ¤ 1 £ c p2 + d p2 ¤ 1
£ a p1 + b p1 ¤ 1 £ c p2 + d p2 ¤ 1
p1
p2
p1
p2
f(
,
)g (
,
)
2
2
2
2
Z1
Z1
1
1
≤ [ f (a, c) + f (a, d ) + f (b, c) + f (b, d )]4h 1 ( )h 2 ( )
h 1 (t )d t
h 2 (t )d t
2
2 0
0
Z1
Z1
1
1
k 2 (t )d t
k 1 (t )d t
×[g (a, c) + g (a, d ) + g (b, c) + g (b, d )]4k 1 ( )k 2 ( )
2
2 0
0
= [M 1 (a, b, c, d ) + M 2 (a, b, c, d ) + M 3(a, b, c, d ) + M 4 (a, b, c, d )]
Z1
Z1
Z1
Z1
1
1
1
1
×16h 1 ( )h 2 ( )k 1 ( )k 2 ( )
k 2 (t )d t ,
k 1 (t )d t
h 2 (t )d t
h 1 (t )d t
2
2
2
2 0
0
0
0
(3.47)
HERMITE-HADAMARD TYPE INEQUALITIES
317
Zb Zd
£ c p2 + d p2 ¤ 1
p1 p2
p 1 −1 p 2 −1
p2
x
u
f
(x,
)g (x, u)d xd u
(b p 1 − a p 1 )(d p 2 − c p 2 ) a c
2
Zd
³
£ c p2 + d p2 ¤ 1
p2
p 2 −1
p2
)g (a, u)
f
(a,
u
≤ p
d 2 − c p2 c
2
Z
´
1
£ c p2 + d p2 ¤ 1
p2
h 1 (t )k 1 (t )d t
)g (b, u) d u
+ f (b,
2
0
Zd
³ ³ £ c p2 + d p2 ¤ 1 ´
p2
p 2 −1
p2
f a,
u
+ p
g (b, u)
d 2 − c p2 c
2
´ Z1
³ £ c p2 + d p2 ¤ 1 ´
p2
h 1 (t )k 1 (1 − t )d t
g (a, u) d u
+ f b,
2
0
Z1
Z1
³
1
k 2 (t )d t
h 2 (t )d t [g (a, c) + g (a, d )]
≤ [ f (a, c) + f (a, d )]2h 2( )
2 0
0
Z1
Z1
´ Z1
1
h 2 (t )d t [g (b, c) + g (b, d )]
k 2 (t )d t d u
+[ f (b, c) + f (b, d )]2h 2 ( )
h 1 (t )k 1 (t )d t
2 0
0
0
Z1
Z1
³
1
+ [ f (a, c) + f (a, d )]2h 2 ( )
k 2 (t )d t
h 2 (t )d t [g (b, c) + g (b, d )]
2 0
0
Z1
Z1
´ Z1
1
h 1 (t )k 1 (1 − t )d t
k 2 (t )d t
h 2 (t )d t [g (a, c) + g (a, d )]
+[ f (b, c) + f (b, d )]2h 2 ( )
2 0
0
0
Z1
Z1
Z1
1
h 2 (t )d t
k 2 (t )d t
= (M 1 (a, b, c, d ) + M 2 (a, b, c, d ))2h 2 ( )
h 1 (t )k 1 (t )d t
2 0
0
0
Z1
Z1
Z1
1
h 1 (t )k 1 (1 − t )d t , (3.48)
+(M 3 (a, b, c, d ) + M 4(a, b, c, d ))2h 2 ( )
k 2 (t )d t
h 2 (t )d t
2 0
0
0
Zb Zd
£ c p2 + d p2 ¤ 1
p1 p2
p 1 −1 p 2 −1
p2
x
u
f
(x,
u)g
(x,
)d xd u
(b p 1 − a p 1 )(d p 2 − c p 2 ) a c
2
Zd
³
£ c p2 + d p2 ¤ 1
p2
p 2 −1
p2
)
f
(a,
u)g
(a,
u
≤ p
d 2 − c p2 c
2
³ £ c p 2 + d p 2 ¤ 1 ´´ Z1
p2
+ f (b, u)g b,
du
h 1 (t )k 1(t )d t
2
0
Zd
³
£ c p2 + d p2 ¤ 1
p2
p 2 −1
p2
f
(a,
u)g
(b,
u
)
+ p
d 2 − c p2 c
2
Z
³ £ c p 2 + d p 2 ¤ 1 ´´
1
p2
+ f (b, u)g a,
h 1 (t )k 1 (1 − t )d t
du
2
0
Z1
Z
³
1
1
k 2 (t )d t
h 2 (t )d t [g (a, c) + g (a, d )]2k 2( )
≤ [ f (a, c) + f (a, d )]
2 0
0
Z1
Z1
´ Z1
1
k 2 (t )d t d u
h 1 (t )k 1(t )d t
+[ f (b, c) + f (b, d )]
h 2 (t )d t [g (b, c) + g (b, d )]2k 2( )
2 0
0
0
Z1
Z1
³
1
k 2 (t )d t
h 2 (t )d t [g (b, c) + g (b, d )]2k 2( )
+ [ f (a, c) + f (a, d )]
2 0
0
Z
Z1
´ Z1
1 1
h 1 (t )k 1 (1 − t )d t
k 2 (t )d t
h 2 (t )d t [g (a, c) + g (a, d )]2k 2( )
+[ f (b, c) + f (b, d )]
2 0
0
0
318
WENGUI YANG
Z1
Z1
Z1
1
h 1 (t )k 1 (t )d t
k 2 (t )d t
= (M 1 (a, b, c, d ) + M 2 (a, b, c, d ))2k 2( )
h 2 (t )d t
2 0
0
0
Z1
Z1
Z1
1
+(M 3 (a, b, c, d ) + M 4 (a, b, c, d ))2k 2( )
h 1 (t )k 1 (1 − t )d t ,
(3.49)
k 2 (t )d t
h 2 (t )d t
2 0
0
0
Zd
³£ a p 1 + b p 1 ¤ 1 £ c p 2 + d p 2 ¤ 1 ´
£ a p1 + b p1 ¤ 1
p2
p 2 −1
p1
p1
p2
,
u)d
u
f
,
u
g
(
d p2 − c p2 c
2
2
2
³ ³£ a p 1 + b p 1 ¤ 1 ´
³£ a p 1 + b p 1 ¤ 1 £ c p 2 + d p 2 ¤ 1 ´
´ Z1
£ a p1 + b p1 ¤ 1
p1
p1
p1
p2
≤ g
k 2 (t )d t f
,c + g(
,d)
,
2
2
2
2
0
Z1
Z1
1
k 2 (t )d t
k 1 (t )d t
≤ [g (a, c) + g (b, c) + g (a, d ) + g (b, d )]2k 1 ( )
2 0
0
Z1
Z1
1
1
h 1 (t )d t
×[ f (a, c) + f (b, c) + f (a, d ) + f (b, d )]4h 1 ( )h 2 ( )
h 2 (t )d t
2
2 0
0
= [M 1 (a, b, c, d ) + M 2 (a, b, c, d ) + M 3(a, b, c, d ) + M 4 (a, b, c, d )]
Z1
Z1
Z1
Z1
1
1
1
h 1 (t )d t
h 2 (t )d t
k 1 (t )d t
k 2 (t )d t ,
(3.50)
×8h 1 ( )h 2 ( )k 1 ( )
2
2
2 0
0
0
0
Zb
£ a p1 + b p1 ¤ 1 £ c p2 + d p2 ¤ 1
£ a p1 + b p1 ¤ 1
p2
p 1 −1
p1
p1
p2
,
u)d
xg
(
,
)
x
f
(
d p2 − c p2 a
2
2
2
≤ [M 1 (a, b, c, d ) + M 2 (a, b, c, d ) + M 3(a, b, c, d ) + M 4 (a, b, c, d )]
Z1
Z1
Z1
Z1
1
1
1
h 1 (t )d t
h 2 (t )d t
k 1 (t )d t
k 2 (t )d t ,
(3.51)
×8h 1 ( )k 1 ( )k 2 ( )
2
2
2 0
0
0
0
Zb Zd
£ a p1 + b p1 ¤ 1
p1p2
p 1 −1 p 2 −1
p1
x
u
f
(x,
u)g
(
, u)d xd u
(b p 1 − a p 1 )(d p 2 − c p 2 ) a c
2
Zb
³
£ a p1 + b p1 ¤ 1
p1
p 1 −1
p1
, c)
x
f
(x,
c)g
(
≤ p
b 1 − a p1 a
2
´ Z1
£ a p1 + b p1 ¤ 1
p1
,d) dx
h 2 (t )k 2 (t )d t
+ f (x, d )g (
2
0
Zb
³
£ a p1 + b p1 ¤ 1
p1
p 1 −1
p1
+ p
,d)
f
(x,
c)g
(
x
b 1 − a p1 a
2
Z
´
1
£ a p1 + b p1 ¤ 1
p1
h 2 (t )k 2 (1 − t )d t
, c) d x
+ f (x, d )g (
2
0
Z1
Z1
³
1
k 1 (t )d t
≤ [ f (a, c) + f (b, c)]
h 1 (t )d t [g (a, c) + g (b, c)]2k 1( )
2 0
0
Z1
Z1
´ Z1
1
h 1 (t )d t [g (a, d ) + g (b, d )]2k 1( )
+[ f (a, d ) + f (b, d )]
h 2 (t )k 2 (t )d t
k 1 (t )d t d x
2 0
0
0
Z1
Z1
³
1
k 1 (t )d t
h 1 (t )d t [g (a, d ) + g (b, d )]2k 1( )
+ [ f (a, c) + f (b, c)]
2 0
0
Z1
Z1
´ Z1
1
k 1 (t )d t d x
+[ f (a, d ) + f (b, d )]
h 2 (t )k 2 (1 − t )d t
h 1 (t )d t [g (a, c) + g (b, c)]2k 1( )
2 0
0
0
Z1
Z1
Z1
1
h 2 (t )k 2 (t )d t
k 1 (t )d t
h 1 (t )d t
= (M 1 (a, b, c, d ) + M 3 (a, b, c, d ))2k 1( )
2 0
0
0
HERMITE-HADAMARD TYPE INEQUALITIES
319
Z1
Z1
Z
1 1
h 2 (t )k 2 (1 − t )d t ,
(3.52)
k 1 (t )d t
+(M 2 (a, b, c, d ) + M 4(a, b, c, d )2k 1 ( )
h 1 (t )d t
2 0
0
0
Zb Zd
£ a p1 + b p1 ¤ 1
p1 p2
p 1 −1 p 2 −1
p1
x
u
f
(
, u)g (x, u)d xd u
(b p 1 − a p 1 )(d p 2 − c p 2 ) a c
2
Z1
Z1
Z1
1
h 1 (t )d t
≤ (M 1 (a, b, c, d ) + M 3 (a, b, c, d ))2h 1 ( )
k 1 (t )d t
h 2 (t )k 2 (t )d t
2 0
0
0
Z1
Z1
Z1
1
h 2 (t )k 2(1 − t )d t ,
(3.53)
k 1 (t )d t
h 1 (t )d t
+(M 2 (a, b, c, d ) + M 4(a, b, c, d )2h 1 ( )
2 0
0
0
Zb
£ c p2 + d p2 ¤ 1
£ a p1 + b p1 ¤ 1 £ c p2 + d p2 ¤ 1
p1
p 1 −1
p2
p1
p2
x
g
(x,
)d
x
f
(
,
)
b p1 − a p1 a
2
2
2
Z
³
£ c p2 + d p2 ¤ 1 ´ 1
£ a p1 + b p1 ¤ 1 £ c p2 + d p2 ¤ 1
£ c p2 + d p2 ¤ 1
p2
p2
p1
p2
) + g (b,
)
,
)
k 1 (t )d t f (
≤ g (a,
2
2
2
2
0
Z1
Z1
1
k 2 (t )d t
k 1 (t )d t
≤ [g (a, c) + g (b, c) + g (a, d ) + g (b, d )]2k 2 ( )
2 0
0
Z1
Z1
1
1
×[ f (a, c) + f (b, c) + f (a, d ) + f (b, d )]4h 1 ( )h 2 ( )
h 2 (t )d t
h 1 (t )d t
2
2 0
0
= [M 1 (a, b, c, d ) + M 2(a, b, c, d ) + M 3 (a, b, c, d ) + M 4 (a, b, c, d )]
Z1
Z1
Z1
Z1
1
1
1
k 2 (t )d t ,
(3.54)
k 1 (t )d t
h 2 (t )d t
h 1 (t )d t
×8h 1 ( )h 2 ( )k 2 ( )
2
2
2 0
0
0
0
Zb
£ a p1 + b p1 ¤ 1 £ c p2 + d p2 ¤ 1
£ c p2 + d p2 ¤ 1
p1
p 1 −1
p2
p1
p2
)d
xg
(
,
)
x
f
(x,
b p1 − a p1 a
2
2
2
≤ [M 1 (a, b, c, d ) + M 2(a, b, c, d ) + M 3 (a, b, c, d ) + M 4 (a, b, c, d )]
Z1
Z1
Z1
Z1
1
1
1
h 1 (t )d t
h 2 (t )d t
k 1 (t )d t
×8h 2 ( )k 1 ( )k 2 ( )
k 2 (t )d t ,
(3.55)
2
2
2 0
0
0
0
Zd
Zb
£ a p1 + b p1 ¤ 1
£ c p2 + d p2 ¤ 1
p1 p2
p 2 −1
p 1 −1
p2
p1
u
f
(
)d
x
, u)d u
x
g
(x,
(b p 1 − a p 1 )(d p 2 − c p 2 ) a
2
2
c
≤ [M 1 (a, b, c, d ) + M 2(a, b, c, d ) + M 3 (a, b, c, d ) + M 4 (a, b, c, d )]
Z1
Z1
Z1
Z1
1
1
×4h 1 ( )k 2 ( )
k 2 (t )d t ,
(3.56)
k 1 (t )d t
h 2 (t )d t
h 1 (t )d t
2
2 0
0
0
0
Zd
Zb
£ a p1 + b p1 ¤ 1
£ c p2 + d p2 ¤ 1
p1 p2
p 2 −1
p 1 −1
p2
p1
)d
x
, u)d u
u
g
(
x
f
(x,
(b p 1 − a p 1 )(d p 2 − c p 2 ) a
2
2
c
≤ [M 1 (a, b, c, d ) + M 2(a, b, c, d ) + M 3 (a, b, c, d ) + M 4 (a, b, c, d )]
Z1
Z1
Z1
Z1
1
1
×4h 2 ( )k 1 ( )
h 1 (t )d t
h 2 (t )d t
k 1 (t )d t
k 2 (t )d t ,
(3.57)
2
2 0
0
0
0
Zb Zd
£ a p1 + b p1 ¤ 1 £ c p2 + d p2 ¤ 1
p1 p2
p1
p2
x p 1 −1 u p 2 −1 g (x, u)d xd u f (
,
)
p
p
p
p
(b 1 − a 1 )(d 2 − c 2 ) a c
2
2
≤ [M 1 (a, b, c, d ) + M 2(a, b, c, d ) + M 3 (a, b, c, d ) + M 4 (a, b, c, d )]
Z1
Z1
Z1
Z1
1
1
k 2 (t )d t ,
(3.58)
k 1 (t )d t
h 2 (t )d t
h 1 (t )d t
×4h 1 ( )h 2 ( )
2
2 0
0
0
0
320
WENGUI YANG
£ a p1 + b p1 ¤ 1 £ c p2 + d p2 ¤ 1
p1
p2
x p 1 −1 u p 2 −1 f (x, u)d xd u g (
,
)
2
2
a c
Z1
1
1
h 1 (t )d t
≤ [M 1 (a, b, c, d ) + M 2 (a, b, c, d ) + M 3(a, b, c, d ) + M 4 (a, b, c, d )]4k 1( )k 2 ( )
2
2 0
Z1
Z1
Z1
k 2 (t )d t .
(3.59)
k 1 (t )d t
h 2 (t )d t
Zb Zd
p1p2
p
p
(b 1 − a 1 )(d p 2 − c p 2 )
0
0
0
Using the inequalities (3.45)-(3.59) in (3.44), we get the desired result (3.40).
Remark 7. Suppose that f , g : ∆ → [0, ∞) are convex functions on the co-ordinates on ∆. Then
one has the inequality:
9
(b − a)(d − c)
Zb Zd Z1 Z1
a
c
0
0
f (t x + (1 − t )
c +d
a +b
, r u + (1 − r )
)
2
2
a +b
c +d
×g (t x + (1 − t )
, r u + (1 − r )
)d t d r d xd u
2
2
Zb Zd
1
37
11
29
≤
f (x, y)g (x, y)d xd y + L(a, b, c, d ) + M (a, b, c, d ) + N (a, b, c, d ),
(b − a)(d − c) a c
48
16
48
where L(a, b, c, d ), M (a, b, c, d ) and N (a, b, c, d ) are defined in Theorem 1.5.
Remark 8. Suppose that f , g : ∆ → [0, ∞) are s-convex functions on the co-ordinates on ∆.
Then one has the inequality:
(1 + 2s)2
(b − a)(d − c)
Zb Zd Z1 Z1
a
c
0
0
f (t x + (1 − t )
a +b
c +d
, r u + (1 − r )
)
2
2
a +b
c +d
×g (t x + (1 − t )
, r u + (1 − r )
)d t d r d xd u
2
2
Zb Zd
1
≤
f (x, y)g (x, y)d xd y + Π1L(a, b, c, d ) + Π2 M (a, b, c, d ) + Π3 N (a, b, c, d ),
(b − a)(d − c) a c
where L(a, b, c, d ), M (a, b, c, d ) and N (a, b, c, d ) are defined in Theorem 1.5, and
23−2s (2s [Γ(1 + s)]2 + Γ(1 + 2s)) 24−4s (21+s [Γ(1 + s)]2 + (1 + 4s )Γ(1 + 2s))
+
,
(1 + s)2 (1 + 2s)Γ(1 + 2s)
(1 + s)4 Γ(1 + 2s)
22−2s (1 + [Γ(1 + s)]2 )(2s [Γ(1 + s)]2 + Γ(1 + 2s)) 24−4s (21+s [Γ(1 + s)]2 + (1 + 4s )Γ(1 + 2s))
+
,
Π2 =
(1 + s)2 (1 + 2s)Γ(1 + 2s)
(1 + s)4 Γ(1 + 2s)
23−2s [Γ(1 + s)]2 (2s [Γ(1 + s)]2 + Γ(1 + 2s)) 24−4s (21+s [Γ(1 + s)]2 + (1 + 4s )Γ(1 + 2s))
+
.
Π3 =
(1 + s)2 (1 + 2s)[Γ(1 + 2s)]2
(1 + s)4 Γ(1 + 2s)
Π1 =
Remark 9. Suppose that f , g : ∆ → [0, ∞) are s 1-convex and s 2 -convex functions on the coordinates on ∆, respectively. Then one has the inequality:
(1 + s 1 + s 2 )2
(b − a)(d − c)
Zb Zd Z1 Z1
a
c
0
0
f (t x + (1 − t )
a +b
c +d
, r u + (1 − r )
)
2
2
a +b
c +d
×g (t x + (1 − t )
, r u + (1 − r )
)d t d r d xd u
2
2
HERMITE-HADAMARD TYPE INEQUALITIES
≤
1
(b − a)(d − c)
Zb Zd
a
c
321
f (x, y)g (x, y)d xd y + Σ1L(a, b, c, d ) + Σ2 M (a, b, c, d ) + Σ3 N (a, b, c, d ),
where L(a, b, c, d ), M (a, b, c, d ) and N (a, b, c, d ) are defined in Theorem 1.5, and
23−s1 −s2 Γ(1 + s 1 + s 2 ) + (22−s1 + 22−s2 )Γ(1 + s 1 )Γ(1 + s 1 )
(1 + s 1 )(1 + s 2 )(1 + s 1 + s 2 )Γ(1 + s 1 + s 2 )
(24−2s1 −2s2 + 22−2s1 + 23−s1 −s2 + 22−2s2 )Γ(1 + s 1 + s 2 ) + 23−s1 −s2 (2−s1 + 2−s2 )Γ(1 + s 1 )Γ(1 + s 1 )
,
+
(1 + s 1 )2 (1 + s 2 )2 Γ(1 + s 1 + s 2 )
¡
¢
(1 + Γ(1 + s 1 )Γ(1 + s 2 )) 22−s1 −s2 Γ(1 + s 1 + s 2 ) + (21−s1 + 21−s2 )Γ(1 + s 1 )Γ(1 + s 1 )
Σ2 =
(1 + s 1 )(1 + s 2 )(1 + s 1 + s 2 )Γ(1 + s 1 + s 2 )
(24−2s1 −2s2 + 22−2s1 + 23−s1 −s2 + 22−2s2 )Γ(1 + s 1 + s 2 ) + 23−s1 −s2 (2−s1 + 2−s2 )Γ(1 + s 1 )Γ(1 + s 1 )
+
,
(1 + s 1 )2 (1 + s 2 )2 Γ(1 + s 1 + s 2 )
¡
¢
Γ(1 + s 1 )Γ(1 + s 2 ) 23−s1 −s2 Γ(1 + s 1 + s 2 ) + (22−s1 + 22−s2 )Γ(1 + s 1 )Γ(1 + s 1 )
Σ3 =
(1 + s 1 )(1 + s 2 )(1 + s 1 + s 2 )Γ(1 + s 1 + s 2 )
4−2s 1 −2s 2
2−2s 1
3−s 1 −s 2
(2
+2
+2
+ 22−2s2 )Γ(1 + s 1 + s 2 ) + 23−s1 −s2 (2−s1 + 2−s2 )Γ(1 + s 1 )Γ(1 + s 1 )
+
.
(1 + s 1 )2 (1 + s 2 )2 Γ(1 + s 1 + s 2 )
Σ1 =
References
[1] J. Pečarić, F. Proschan and Y. L. Tong, Convex functions, partial ordering and statistical applications, Academic Press, New York, 1991.
[2] S. S. Dragomir and C. E. M. Pearce, Selected Topics on Hermite-Hadamard Inequalities and Applications,
RGMIA Monographs, Victoria University, 2000.
[3] S. S. Dragomir, Some new Hermite-Hadamard’s type fractional integral inequalities, J. Comput. Anal. Appl.,
18(2015), 655–661.
[4] M. A. Noor, G. Cristescu and M. U. Awan, Generalized fractional Hermite-Hadamard inequalities for twice
differentiable s-convex functions, Filomat 29(2015), 507–585.
[5] I. Iscan and S. Wu, Hermite-Hadamard type inequalities for harmonically convex functions via fractional
integrals, Appl. Math. Comput., 238 (2014), 237–244.
[6] F. Chen and S. Wu, Some Hermite-Hadamard type inequalities for harmonically s-convex functions, Sci. World
J., 2014 (2014), 279158.
[7] W. Li and F. Qi, Some Hermite-Hadamard type inequalities for functions whose n-th derivatives are (α,m)convex, Filomat, 27 (2013), 1575–1582.
[8] L. Chun and F. Qi, Integral inequalities of Hermite-Hadamard type for functions whose third derivatives are
convex, J. Inequal. Appl., 2013 (2013), 451.
[9] J. Wang, J. Deng and M. Fekan, Hermite-Hadamard-type inequalities for r -convex functions based on the use
of Riemann-Liouville fractional integrals, Ukrain. Math. J., 65 (2013), 193–211.
[10] J. Wang et al., Hermite-Hadamard-type inequalities for Riemann-Liouville fractional integrals via two kinds
of convexity, Appl. Anal., 92 (2013), 2241–2253.
[11] M. Z. Sarikaya et al., Hermite-Hadamard’s inequalities for fractional integrals and related fractional inequalities, Math. Comput. Modelling, 57 (2013), 2403–2407.
[12] E. Set et al., On new inequalities of Hermite-Hadamard-Fejer type for convex functions via fractional integrals,
Appl. Math. Comput., 259 (2015), 875–881.
[13] S. S. Dragomir and S. Fitzpatrick, The Hadamard’s inequality for s-convex functions in the second sense,
Demonstr. Math., 32(1999), 687–696.
322
WENGUI YANG
[14] M. Z. Sarikaya and A. Saglam and H. Yildirim, On some Hadamard-type inequalities for h-convex functions,
J. Math. Inequal., 2(2008), 335–341.
[15] Z. Fang and R. Shi, On the (p,h)-convex function and some integral inequalities, J. Inequal. Appl., 2014 (2014),
45.
[16] S. S. Dragomir, On the Hadamard’s inequality for convex functions on the co-ordinates in a rectangle from the
plane, Taiwan. J. Math., 5 (2001), 775–788.
[17] M. Alomari and M. Darus, The Hadamard’s inequality for s-convex function of 2-variables on the co-ordinates,
Int. J. Math. Anal., 2 (2008), 629–638.
[18] S. Bai and F. Qi, Some inequalities for (s 1 ,m 1 )-(s 2 ,m 2 )-convex functions on the co-ordinates, Glob. J. Math.
Anal., 1 (2013), 22–28.
[19] M. E. Ödemir, C. Yildiz and A. O. Akademir, On the co-ordinated convex functions, Appl. Math. Infor. Sci., 8
(2014), 1085-1091.
[20] D. Y. Hwang, K. L. Tseng and G. S. Yang, Some Hadamard’s inequalities for co-ordinated convex functions in a
rectangle from the plane, Taiwan. J. Math., 11 (2007), 63–73.
[21] B. Xi, J. Hua and F. Qi, Hermite-Hadamard type inequalities for extended s-convex functions on the coordinates in a rectangle, J. Appl. Anal., 20 (2014), 29–39.
[22] M. A. Latif and S. S. Dragomir, On some new inequalities for differentiable co-ordinated convex functions, J.
Inequal. Appl., 2012 (2012), 28.
[23] M. E. Ödemir, M. A. Latif and A. O. Akademir, On some Hadamard-type inequalities for product of two sconvex functions on the co-ordinates, J. Inequal. Appl., 2012 (2012), 21.
[24] M. A. Noor, K. I. Noor and M. U. Awan, Integral inequalities for coordinated Harmonically convex functions,
Complex Var. Elliptic Equ. (2014), http://dx.doi.org/10.1080/17476933.2014.976814.
[25] M. Matloka, On some Hadamard-type inequalities for (h 1 ,h 2 )-preinvex functions on the co-ordinates, J. Inequal. Appl., 2013 (2013), 227.
[26] M. A. Latif and M. Alomari, Hadamard-type inequalities for product two convex functions on the co-ordinates,
Int. Math. Forum, 4 (2009), 2327–2338.
[27] M. E. Ödemir and A. O. Akademir, On the hadamard type inequalities involving product of two convex functions on the co-ordinates, Tamkang J. Math., 46 (2015), 129–142.
Ministry of Public Education, Sanmenxia Polytechnic, Sanmenxia, Henan 472000, China.
E-mail: [email protected]