Some new inequalities of Hermite-Hadamard type for h-convex

arXiv:1402.3079v1 [math.CA] 13 Feb 2014

SOME NEW INEQUALITIES OF HERMITE-HADAMARD TYPE FOR h−CONVEX FUNCTIONS ON THE CO-ORDINATES VIA FRACTIONAL INTEGRALS. ¨ ULM ¨ ¨ ¸♣ ERHAN SETN , M. ZEKI SARIKAYA , AND HATICE OG US Abstract. By making use of the identity obtained by Sarıkaya, some new Hermite-Hadamard type inequalities for h-convex functions on the co-ordinates via fractional integrals are established. Our results have some relationships with the results of Sarıkaya([16])

1. INTRODUCTION Let f : I ⊆ R → R be a convex function defined on the interval I of real numbers and a, b ∈ I with a < b , then Z b 1 f (a) + f (b) a+b )≤ . f (x)dx ≤ f( 2 b−a a 2

holds, the double inequality is well known in the literature as Hermite-Hadamard inequality [6]: Both inequalities hold in the reversed direction if f is concave. For recent results, generalizations and new inequalities related to the Hermite-Hadamard inequality see ( [4], [12],[14], [18]) The classical Hermite- Hadamard inequality provides estimates of the mean value of a continuous convex function f : [a, b] → R Let us now consider a bidemensional interval ∆ := [a, b] × [c, d] in R2 with a < b and c < d. A mapping f : ∆ → R is said to be convex on if the following inequality: f (tx + (1 − t)z, ty + (1 − t)w) ≤ tf (x, y) + (1 − t)f (z, w) holds, for all (x, y), (z, w) ∈ ∆ and t ∈ [0, 1] . If the inequality reversed then f is said to be concave on ∆. A function f : ∆ → R is said to be on the co-ordinates on ∆ if the partial mappings fy : [a, b] → R, fy (u) = f (u, y) and fx : [c, d] → R, fx (v) = f (x, v) are convex where defined for all x ∈ [a, b] and y ∈ [c, d] . (see [5]) A formal definition for co-ordinated convex function may be stated as follows: Definition 1. A function f : ∆ → R will be called co-ordinated convex on ∆ , for all t, k ∈ [0, 1] and (x, u) , (y, w) ∈ ∆ ,if the following inequality holds: f (tx + (1 − t) y, ku + (1 − k) w) ≤

tkf (x, u) + k (1 − t) f (y, u) + t (1 − k) f (x, w) + (1 − t) (1 − k) f (y, w)

2000 Mathematics Subject Classification. 26A33, 26A51, 26D15. Key words and phrases. Riemann-Liouville fractional integrals, Hermite-Hadamard type inequality, h-convex functions on the co-ordinates. 1

2

¨ ULM ¨ ¨ ¸♣ ERHAN SETN , M. ZEKI SARIKAYA , AND HATICE OG US

Clearly, every convex mapping is convex on the co-ordinates, but the converse is not generally true ([5]). Some interesting and important inequalities for convex functions on the co-ordinates can be found in ([9], [10], [17]) In [1], Alomari and Darus established the following definition of s–convex function in the second sense on co–ordinates. 2

Definition 2. Consider the bidimensional interval ∆ := [a, b] × [c, d] in [0, ∞) with a < b and c < d. The mapping f : ∆ → R is s−convex on ∆ if s

f (λx + (1 − λ) z, λy + (1 − λ) w) ≤ λs f (x, y) + (1 − λ) f (z, w) holds for all (x, y), (z, w) ∈ ∆ with λ ∈ [0, 1] , and for some fixed s ∈ (0, 1] . A function f : ∆ → R is s-convex on ∆ is called co-ordinated s-convex on ∆ if the partial mappings fy : [a, b] → R, fy (u) = f (u, y) and fx : [c, d] → R, fx (v) = f (x, v) are s-convex for all x ∈ [a, b] and y ∈ [c, d] with some fixed s ∈ (0, 1] . In [11], Latif and Alamori give the notion of h-convexity of a function f on a rectangle from the plane R2 and h-convexity on the co-ordinates on a rectangle from the plane R2 as follows Definition 3. Let us consider a bidimensional interval ∆ := [a, b] × [c, d] in R2 with a < b and c < d. Let h : J ⊆ R → R be a positivie function. A mapping f : ∆ := [a, b] × [c, d] → R is said to be h-convex on ∆, if f is non-negative and if the following inequality: f (λx + (1 − λ) z, λy + (1 − λ) w) ≤ h (λ) f (x, y) + h (1 − λ) f (z, w) holds, for all (x, y), (z, w) ∈ ∆ with λ ∈ (0, 1) . Let us denote this class of functions by SX (h, ∆) . The function f is said to be h-concave if the inequality reversed. We denot this class of functions by SV (h, ∆) . A function f : ∆ → R is said to be h-convex on the co-ordinates on ∆ if the partial mappings fy : [a, b] → R, fy (u) = f (u, y) and fx : [c, d] → R, fx (v) = f (x, v) are h-convex where defined for all x ∈ [a, b] and y ∈ [c, d]. A formal definition of h-convex functions may also be stated as follows: Definition 4. [11] A function f : ∆ → R is said to be h-convex on the co-ordinates on ∆ , if the following inequality: (1.1) f (tx + (1 − t) y, ku + (1 − k) w) ≤ h (t) h (k) f (x, u) + h (k) h (1 − t) f (y, u) + h (t) h (1 − k) f (x, w) + h (1 − t) h (1 − k) f (y, w) holds for all t, k ∈ [0, 1] and (x, u) , (x, w) , (y, u) , (y, w) ∈ ∆. Obviously, if h(α) = α, then all the non-negative convex (concave) functions on ∆ belong to the class SX(h, ∆) (SV (h, ∆)) and if h(α) = αs , where s ∈ (0, 1), then the class of s-convex on ∆ belong to the class SX(h, ∆). Similarly we can say that if h(α) = α, then the class of non-negative convex (concave) functions on the co-ordinates on ∆ is contained in the class of h-convex (concave) functions on the co-ordinates on ∆ and if h(α) = αs ,where s ∈ (0, 1), then the class of s-convex functions on the co-ordinates on ∆ is contained in the class of h-convex functions on the co-ordinates on ∆.

HERMITE-HADAMARD TYPE INEQUALITIES FOR H-CONVEX FUNCTION

3

In the following we will give some necessary definitions which are used further in this paper. More details, one can consult [[7], [8], [13]] Definition 5. Let f ∈ L1 [a, b]. The Riemann-Liouville integrals Jaα+ f and Jbα− f order α > 0 and a ≥ 0 are defined by Z x 1 (x − t)α−1 f (t) dt, x > a Jaα+ f (x) = Γ (α) a and Jbα− f (x) =

1 Γ (α)

Z

b

α−1

(t − x)

f (t) dt, x < b

x

respectively. Here Γ (α) is the Gamma function and Ja0+ f (x) = Jb0− f (x) = f (x) . Definition 6. Let f ∈ L1 ([a, b] × [c, d]). the Riemann-Liouville integrals Jaα,β + ,c+ , α,β α,β Jaα,β + ,d− , Jb− ,c+ and Jb− ,d− of order α, β > 0 with a, c ≥ 0 are defined by

Jaα,β + ,c+ f (x, y) =

=

1 Γ (α) Γ (β)

Z

a

xZ y

(x − t)

α−1

β−1

(y − s)

f (t, s) dsdt, x > a, y > c

c

Jaα,β + ,d− f (x, y) Z xZ d 1 α−1 β−1 (x − t) (s − y) f (t, s) dsdt, x > a, y < d Γ (α) Γ (β) a y

Jbα,β − ,c+ f (x, y) Z bZ y 1 α−1 β−1 = (t − x) (y − s) f (t, s) dsdt, x < b, y > c Γ (α) Γ (β) x c Jbα,β − ,d− f (x, y) Z bZ d 1 α−1 β−1 (t − x) (s − y) f (t, s) dsdt, x < b, y < d = Γ (α) Γ (β) x y respectively. Here, Γ is the Gama function, 0,0 0,0 0,0 Ja0,0 + ,c+ f (x, y) = Ja+ ,d− f (x, y) = Jb− ,c+ f (x, y) = Jb− ,d− f (x, y) = f (x, y)

and Ja1,1 + ,c+ f (x, y) =

Z

a

x

Z

y

f (t, s) dsdt.

c

For some recent results connected with fractional integral inequalities see ([2], [3], [15], [19]). In ([16]), Sarıkaya establish the following inequalities of Hadamard’s type for co-ordinated convex mapping on a rectangle from the plane R2 :


4

Theorem 1. Let f : ∆ ⊂ R2 → R be co-ordinated convex on ∆ := [a, b] × [c, d] in R2 with 0 ≤ a < b, 0 ≤ c < d and f ∈ L1 (∆) . Then one has the inequalities: (1.2) f ≤

≤

a+b c+d , 2 2

Γ (α + 1) Γ (β + 1) 4 (b − a)α (d − c)β h i α,β α,β α,β Jaα,β + ,c+ f (b, d) + Ja+ ,d− f (b, c) + Jb− ,c+ f (a, d) + Jb− ,d− f (a, c) f (a, c) + f (a, d) + f (b, c) + f (b, d) . 4

Theorem 2. Let f : ∆ ⊂ R2 → R be a partial differentiable mapping on ∆ := ∂2 f 2 [a, b] × [c, d] in R with 0 ≤ a < b, 0 ≤ c < d. If ∂t∂k is a convex function on the co-ordinates on ∆, then one has the inequalities: (1.3) f (a, c) + f (a, d) + f (b, c) + f (b, d) 4 +

≤

where

Γ (α + 1) Γ (β + 1) α

β

4 (b − a) (d − c) h i α,β α,β α,β × Jaα,β + ,c+ f (b, d) + Ja+ ,d− f (b, c) + Jb− ,c+ f (a, d) + Jb− ,d− f (a, c) − A

(b − a) (d − c) 4 (α + 1) (β + 1) 2 2 2 2 ∂ f ∂ f ∂ f ∂ f (a, c) + (a, d) + (b, c) + (b, d) × ∂k∂t ∂k∂t ∂k∂t ∂k∂t

A =

i Γ (β + 1) h β β β β J f (a, d) + J f (b, d) + J f (a, c) + J f (b, c) + + − − c c d d 4 (d − c)β +

Γ (α + 1) α α α α α [Ja+ f (b, c) + Ja+ f (b, d) + Jb− f (a, c) + Jb− f (a, d)] . 4 (b − a)

Theorem 3. Let f : ∆ ⊂ R2 → R be a partial mapping on ∆ := 2differentiable ∂ f q [a, b] × [c, d] in R2 with 0 ≤ a < b, 0 ≤ c < d. If ∂t∂k , q > 1, is a convex function


5

on the co-ordinates on ∆, then one has the inequalities: (1.4) f (a, c) + f (a, d) + f (b, c) + f (b, d) 4 ( Γ (α + 1) Γ (β + 1) + α β 4 (b − a) (d − c) h io α,β α,β α,β × Jaα,β −A + ,c+ f (b, d) + Ja+ ,d− f (b, c) + Jb− ,c+ f (a, d) + Jb− ,d− f (a, c) 1q 1 1 [(αp + 1) (βp + 1)] p 4 q 2 q 2 q 2 q q1 2 ∂ f ∂ f ∂ f ∂ f (a, c) + (a, d) + (b, c) + (b, d) × ∂k∂t ∂k∂t ∂k∂t ∂k∂t (b − a) (d − c)

≤

where

A =

i Γ (β + 1) h β Jc+ f (a, d) + Jcβ+ f (b, d) + Jdβ− f (a, c) + Jdβ− f (b, c) β 4 (d − c) +

and

1 p

+

1 q

Γ (α + 1) α α α α α [Ja+ f (b, c) + Ja+ f (b, d) + Jb− f (a, c) + Jb− f (a, d)] 4 (b − a)

= 1.

In order to prove our main results we need the following lemma (see [16]). Lemma 1. Let f : ∆ ⊂ R2 → R be a partial differentiable mapping on ∆ := ∂2 f [a, b] × [c, d] in R2 with 0 ≤ a < b, 0 ≤ c < d. If ∂t∂k ∈ L (∆) , then the following equality holds: (1.5) f (a, c) + f (a, d) + f (b, c) + f (b, d) 4 ( Γ (α + 1) Γ (β + 1) + α β 4 (b − a) (d − c) h io α,β α,β α,β × Jaα,β + ,c+ f (b, d) + Ja+ ,d− f (b, c) + Jb− ,c+ f (a, d) + Jb− ,d− f (a, c) −

i Γ (β + 1) h β Jc+ f (a, d) + Jcβ+ f (b, d) + Jdβ− f (a, c) + Jdβ− f (b, c) β 4 (d − c)

−



6

Z 1 Z 1 ∂2f (b − a) (d − c) (ta + (1 − t) b, kc + (1 − k) d) dkdt tα k β 4 ∂t∂k 0 0 Z 1Z 1 ∂ 2f α − (1 − t) k β (ta + (1 − t) b, kc + (1 − k) d) dkdt ∂t∂k 0 0 Z 1Z 1 ∂2f (ta + (1 − t) b, kc + (1 − k) d) dkdt − tα (1 − k)β ∂t∂k 0 0 Z 1Z 1 2 α β ∂ f − (1 − t) (1 − k) (ta + (1 − t) b, kc + (1 − k) d) dkdt . ∂t∂k 0 0

=

2. MAIN RESULTS Theorem 4. Let f : ∆ ⊂ R2 → R be h-convex function on the co-ordinates on ∆ := [a, b] × [c, d] in R2 and f ∈ L2 (∆) . The one has the inequalities: (2.1) a+b c+d , f 2 2 2 Γ (α + 1) Γ (β + 1) 1 h α β 2 (b − a) (d − c) h i α,β α,β α,β × Jaα,β + ,c+ f (b, d) + Ja+ ,d− f (b, c) + Jb− ,c+ f (a, d) + Jb− ,d− f (a, c)

≤

2 1 αβ [f (a, c) + f (a, d) + f (b, c) + f (b, d)] h 2 Z 1 Z 1 × tα−1 k α−1

≤

0

0

× [h (t) h (k) + h (t) h (1 − k) + h (1 − t) h (k) + h (1 − t) h (1 − k)] dkdt] . Proof. According to (1.1) with x = t1 a + (1 − t1 ) b, y = (1 − t1 ) a + t1 b, u = k1 c + (1 − k1 ) d, w = (1 − k1 ) c + k1 d and t = k = 21 , we find that (2.2)

a+b c+d , 2 2 2 1 ≤ h 2 f

× [f (t1 a + (1 − t1 ) b, k1 c + (1 − k1 ) d) + f (t1 a + (1 − t1 ) b, (1 − k1 ) c + k1 d) +f ((1 − t1 ) a + t1 b, k1 c + (1 − k1 ) d) + f ((1 − t1 ) a + t1 b, (1 − k1 ) c + k1 d)] .


7

Thus, multiplying both sides of (2.2) by tα−1 k1β−1 , then by integrating with 1 respect to (t1 , k1 ) on [0, 1] × [0, 1] , we obtain a+b c+d 1 f , αβ 2 2 2 1 h 2  R1R1

≤

tα−1 k1α−1 1 × [f (t1 a + (1 − t1 ) b, k1 c + (1 − k1 ) d) +f (t1 a + (1 − t1 ) b, (1 − k1 ) c + k1 d) +f ((1 − t1 ) a + t1 b, k1 c + (1 − k1 ) d) +f ((1 − t1 ) a + t1 b, (1 − k1 ) c + k1 d)] dk1 ds1 . 0

  ×  

0

Using the change of the varible, we get a+b c+d f , 2 2 (Z Z 2 b d αβ 1 α−1 β−1 (b − x) (d − y) f (x, y) dydx ≤ h α β 2 (b − a) (d − c) a c +

Z

b

Z

b

Z

b

a

+

d

Z

d

Z

d

(b − x)

α−1

(y − c)

β−1

f (x, y) dydx

α−1

(d − y)

β−1

f (x, y) dydx

c

a

+

Z

a

(x − a)

c α−1

(x − a)

c

(y − c)

β−1

)

f (x, y) dydx

which the first inequality is proved. For the proof of second inequality (2.1), we first note that if f is a h-convex function on ∆, then, by using (1.1) with x = a, y = b, u = c, w = d, it yields

≤

f (ta + (1 − t) b, kc + (1 − k) d) h (t) h (k) f (a, c) + h (t) h (1 − k) f (a, d) +h (1 − t) h (k) f (b, c) + h (1 − t) h (1 − k) f (b, d)

≤

f ((1 − t) a + tb, kc + (1 − k) d) h (1 − t) h (k) f (a, c) + h (1 − t) h (1 − k) f (a, d) +h (t) h (k) f (b, c) + h (t) h (1 − k) f (b, d)

≤

f (ta + (1 − t) b, (1 − k) c + kd) h (t) h (1 − k) f (a, c) + h (t) h (k) f (a, d) +h (1 − t) h (1 − k) f (b, c) + h (1 − t) h (k) f (b, d)


8

f ((1 − t) a + tb, (1 − k) c + kd) ≤

h (1 − t) h (1 − k) f (a, c) + h (1 − t) h (k) f (a, d) +h (t) h (1 − k) f (b, c) + h (t) h (k) f (b, d) .

By adding these inequalities we have (2.3) f (ta + (1 − t) b, kc + (1 − k) d) + f ((1 − t) a + tb, kc + (1 − k) d) +f (ta + (1 − t) b, (1 − k) c + kd) + f ((1 − t) a + tb, (1 − k) c + kd) ≤ [f (a, c) + f (b, c) + f (a, d) + f (b, d)] × [h (t) h (k) + h (t) h (1 − k) + h (1 − t) h (k) + h (1 − t) h (1 − k)] . Then, multiplying both sides of (2.3) by tα−1 k β−1 and integrating with respect to (t, k) over [0, 1] × [0, 1] , we get Z

1

0

Z

1

tα−1 k β−1 0

× [f (ta + (1 − t) b, kc + (1 − k) d) + f ((1 − t) a + tb, kc + (1 − k) d) +f (ta + (1 − t) b, (1 − k) c + kd) + f ((1 − t) a + tb, (1 − k) c + kd)] dkdt ≤ [f (a, c) + f (b, c) + f (a, d) + f (b, d)] ×

Z

0

1

Z

1

tα−1 k β−1

0

× [h (t) h (k) + h (t) h (1 − k) + h (1 − t) h (k) + h (1 − t) h (1 − k)] dkdt} . Here, using the change of the variable we have 2 Γ (α + 1) Γ (β + 1) 1 h α β 2 (b − a) (d − c) h i α,β α,β α,β × Jaα,β + ,c+ f (b, d) + Ja+ ,d− f (b, c) + Jb− ,c+ f (a, d) + Jb− ,d− f (a, c) ≤

2 1 αβ [f (a, c) + f (a, d) + f (b, c) + f (b, d)] h 2 Z 1 Z 1 × tα−1 k α−1 0

0

× [h (t) h (k) + h (t) h (1 − k) + h (1 − t) h (k) + h (1 − t) h (1 − k)] dkdt] . The proof is completed.


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Remark 1. If we take h (α) = α in Theorem 4, then the inequality (2.1) becomes the inequality (1.2) of Theorem 1. Corollary 1. If we take h (α) = αs in Theorem 1, we have the following inequality: a+b c+d , f 2 2 2s 1 Γ (α + 1) Γ (β + 1) ≤ α β 2 (b − a) (d − c) h i α,β α,β α,β × Jaα,β + ,c+ f (b, d) + Ja+ ,d− f (b, c) + Jb− ,c+ f (a, d) + Jb− ,d− f (a, c) 2s 1 αβ [f (a, c) + f (a, d) + f (b, c) + f (b, d)] ≤ 2 1 1 + B (α, s + 1) + B (β, s + 1) × α+s β+s

where B is the Beta function,

B (x, y) =

Z

1

tx−1 (1 − t)

y−1

dt.

0

Theorem 5. Let f : ∆ ⊂ R2 → R be a partial differentiable mapping on ∆ := ∂2 f 2 [a, b] × [c, d] in R with 0 ≤ a < b, 0 ≤ c < d. If ∂t∂k is a h-convex function on the co-ordinates on ∆, then one has the inequalities: (2.4) f (a, c) + f (a, d) + f (b, c) + f (b, d) 4 +

≤

(

Γ (α + 1) Γ (β + 1) 4 (b − a)α (d − c)β

h io α,β α,β α,β × Jaα,β − A + ,c+ f (b, d) + Ja+ ,d− f (b, c) + Jb− ,c+ f (a, d) + Jb− ,d− f (a, c)

(b − a) (d − c) 4 Z 1 Z 1 2 ∂ f α β (a, c) (tα + (1 − t) ) k β + (1 − k) h (t) h (k) dkdt × ∂t∂k 0 0 2 Z 1Z 1 ∂ f α β + (b, c) (tα + (1 − t) ) k β + (1 − k) h (1 − t) h (k) dkdt ∂t∂k 0 0 Z 1 Z 1 2 ∂ f α β (a, d) (tα + (1 − t) ) k β + (1 − k) h (t) h (1 − k) dkdt + ∂t∂k 0 0 Z 1 Z 1 2 ∂ f α β α β (b, d) (t + (1 − t) ) k + (1 − k) h (1 − t) h (1 − k) dkdt + ∂t∂k 0 0


10

where

A =

i Γ (β + 1) h β β β β J f (a, d) + J f (b, d) + J f (a, c) + J f (b, c) + + − − c c d d β 4 (d − c) +

Γ (α + 1) α α α α α [Ja+ f (b, c) + Ja+ f (b, d) + Jb− f (a, c) + Jb− f (a, d)] . 4 (b − a)

Proof. From Lemma 1, we have

≤

f (a, c) + f (a, d) + f (b, c) + f (b, d) 4 ( Γ (α + 1) Γ (β + 1) + 4 (b − a)α (d − c)β h io α,β α,β α,β × Jaα,β − A + ,c+ f (b, d) + Ja+ ,d− f (b, c) + Jb− ,c+ f (a, d) + Jb− ,d− f (a, c)

2 Z 1 Z 1 (b − a) (d − c) α β ∂ f (ta + (1 − t) b, kc + (1 − k) d) dkdt t k 4 ∂t∂k 0 0 2 Z 1Z 1 ∂ f (ta + (1 − t) b, kc + (1 − k) d) dkdt + (1 − t)α k β ∂t∂k 0 0 2 Z 1Z 1 β ∂ f α + t (1 − k) (ta + (1 − t) b, kc + (1 − k) d) dkdt ∂t∂k 0 0 2 Z 1Z 1 α β ∂ f (ta + (1 − t) b, kc + (1 − k) d) dkdt . − (1 − t) (1 − k) ∂t∂k 0 0

2 ∂ f Since ∂t∂k is h-convex function on the co-ordinates on ∆, then one has: f (a, c) + f (a, d) + f (b, c) + f (b, d) 4 ( Γ (α + 1) Γ (β + 1) + α β 4 (b − a) (d − c) h io α,β α,β α,β × Jaα,β f (b, d) + J f (b, c) + J f (a, d) + J f (a, c) − A + ,c+ a+ ,d− b− ,c+ b− ,d−


11

(b − a) (d − c) 4 Z 1 Z 1 2 ∂ f α β (a, c) (tα + (1 − t) ) k β + (1 − k) h (t) h (k) dkdt × ∂t∂k 0 0 Z 1Z 1 2 ∂ f α β (b, c) (tα + (1 − t) ) k β + (1 − k) h (1 − t) h (k) dkdt + ∂t∂k 0 0 2 Z 1 Z 1 ∂ f α β + (a, d) (tα + (1 − t) ) k β + (1 − k) h (t) h (1 − k) dkdt ∂t∂k 0 0 2 Z 1 Z 1 ∂ f α β α β + (b, d) (t + (1 − t) ) k + (1 − k) h (1 − t) h (1 − k) dkdt . ∂t∂k 0 0

≤

The proof is completed.

Remark 2. If we take h (α) = α in Theorem 5, then the inequality (2.4) becomes the inequality (1.3) of Theorem 2. Corollary 2. If we take h (α) = αs in Theorem 5, we have f (a, c) + f (a, d) + f (b, c) + f (b, d) 4 +

≤

where

(

Γ (α + 1) Γ (β + 1) α

β

4 (b − a) (d − c)

h io α,β α,β α,β Jaα,β − A + ,c+ f (b, d) + Ja+ ,d− f (b, c) + Jb− ,c+ f (a, d) + Jb− ,d− f (a, c)

(b − a) (d − c) 4 2 2 2 2 ∂ f ∂ f ∂ f ∂ f × (a, c) + (b, c) + (a, d) + (b, d) ∂t∂k ∂t∂k ∂t∂k ∂t∂k 1 1 + B (s + 1, α + 1) + B (s + 1, β + 1) × α+s+1 β+s+1

A =

i Γ (β + 1) h β Jc+ f (a, d) + Jcβ+ f (b, d) + Jdβ− f (a, c) + Jdβ− f (b, c) β 4 (d − c) +


and B is the Beta function, B (x, y) =

Z

1

tx−1 (1 − t)

y−1

dt.

0

Theorem 6. Let f : ∆ ⊂ R2 → R be a partial differentiable mapping on ∆ := 2 q ∂ f [a, b] × [c, d] in R2 with 0 ≤ a < b, 0 ≤ c < d. If ∂t∂k , q > 1, is a h-convex


12

function on the co-ordinates on ∆, then one has the inequalities: (2.5)

≤

f (a, c) + f (a, d) + f (b, c) + f (b, d) 4 ( Γ (α + 1) Γ (β + 1) + α β 4 (b − a) (d − c) h io α,β α,β α,β × Jaα,β − A + ,c+ f (b, d) + Ja+ ,d− f (b, c) + Jb− ,c+ f (a, d) + Jb− ,d− f (a, c) (b − a) (d − c)

1

[(αp + 1) (βp + 1)] p q Z 1 Z 1 2 ∂ f (a, c) × h (t) h (k) dkdt ∂k∂t 0 0 q Z 1 Z 1 2 ∂ f (a, d) h (t) h (1 − k) dkdt + ∂k∂t 0 0 q Z 1 Z 1 2 ∂ f (b, c) h (1 − t) h (k) dkdt + ∂k∂t 0 0 2 q Z ∂ f + (b, d) ∂k∂t

0

where

A =

1 p

+

1 q

Z

1

h (1 − t) h (1 − k) dkdt

0

q1

i Γ (β + 1) h β β β β J f (a, d) + J f (b, d) + J f (a, c) + J f (b, c) + + − − c c d d 4 (d − c)β +

and

1

Γ (α + 1) α [J + f (b, c) + Jaα+ f (b, d) + Jbα− f (a, c) + Jbα− f (a, d)] 4 (b − a)α a

= 1.

Proof. From Lemma 1, we have f (a, c) + f (a, d) + f (b, c) + f (b, d) 4 ( Γ (α + 1) Γ (β + 1) + α β 4 (b − a) (d − c) h io α,β α,β α,β × Jaα,β f (b, d) + J f (b, c) + J f (a, d) + J f (a, c) − A + ,c+ a+ ,d− b− ,c+ b− ,d−


≤

13

2 Z 1 Z 1 ∂ f (b − a) (d − c) (ta + (1 − t) b, kc + (1 − k) d) dkdt tα k β 4 ∂t∂k 0 0 Z 1Z 1 2 α β ∂ f + (1 − t) k (ta + (1 − t) b, kc + (1 − k) d) dkdt ∂t∂k 0 0 2 Z 1Z 1 β ∂ f α (ta + (1 − t) b, kc + (1 − k) d) dkdt + t (1 − k) ∂t∂k 0 0 2 Z 1Z 1 α β ∂ f − (1 − t) (1 − k) (ta + (1 − t) b, kc + (1 − k) d) dkdt . ∂t∂k 0 0

By using the well known H¨ older’s inequality for double integrals, we get

f (a, c) + f (a, d) + f (b, c) + f (b, d) 4 ( Γ (α + 1) Γ (β + 1) + α β 4 (b − a) (d − c) h io α,β α,β α,β × Jaα,β − A + ,c+ f (b, d) + Ja+ ,d− f (b, c) + Jb− ,c+ f (a, d) + Jb− ,d− f (a, c) ≤

(b − a) (d − c) 4

Z

1

Z

1

0

Z

1

Z

1

t

0

pα

(Z

1

0

pβ

(1 − k)

Z

1

pα pβ

t k dkdt

0

dkdt

p1

+

Z

0

p1

+

1

1

Z

Z

1

0

Z

1

pα

(1 − t)

pβ

k dkdt

0

pα

(1 − t)

pβ

(1 − k)

0

dkdt

p1

p1 )

1

0

0

q 2 q ∂ f ∂t∂k (ta + (1 − t) b, kc + (1 − k) d) dkdt

2 q ∂ f Since ∂t∂k is h-convex function on the co-ordinates on ∆, then one has: f (a, c) + f (a, d) + f (b, c) + f (b, d) 4 ( Γ (α + 1) Γ (β + 1) + α β 4 (b − a) (d − c) h io α,β α,β α,β × Jaα,β f (b, d) + J f (b, c) + J f (a, d) + J f (a, c) − A + ,c+ a+ ,d− b− ,c+ b− ,d−


14

≤

(b − a) (d − c) 1

[(αp + 1) (βp + 1)] p q Z 1 Z 1 2 ∂ f × (a, c) h (t) h (k) dkdt ∂k∂t 0 0 q Z 1 Z 1 2 ∂ f (a, d) h (t) h (1 − k) dkdt + ∂k∂t 0 0 2 q Z 1 Z 1 ∂ f + (b, c) h (1 − t) h (k) dkdt ∂k∂t 0 0 2 q Z ∂ f + (b, d) ∂k∂t

0

and the proof is completed.

1

Z

1

h (1 − t) h (1 − k) dkdt

0

q1

Remark 3. If we take h (α) = α in Theorem 6, then the inequality (2.5) becomes the inequality (1.4) of Theorem 3. Corollary 3. If we take h (α) = αs in Theorem 5, we have f (a, c) + f (a, d) + f (b, c) + f (b, d) 4 +

(

Γ (α + 1) Γ (β + 1) 4 (b − a)α (d − c)β

h io α,β α,β α,β × Jaα,β − A + ,c+ f (b, d) + Ja+ ,d− f (b, c) + Jb− ,c+ f (a, d) + Jb− ,d− f (a, c) (b − a) (d − c)

≤

1

[(αp + 1) (βp + 1)] p ×

where

q 2 q 2 q 2 q ! 1q 2 ∂ f ∂ f ∂ f ∂ f 2 ∂k∂t (a, c) + ∂k∂t (a, d) + ∂k∂t (b, c) + ∂k∂t (b, d) (s + 1)

A =

1

i Γ (β + 1) h β β β β J f (a, d) + J f (b, d) + J f (a, c) + J f (b, c) + + − − c c d d β 4 (d − c)

+

Γ (α + 1) α [J + f (b, c) + Jaα+ f (b, d) + Jbα− f (a, c) + Jbα− f (a, d)] . 4 (b − a)α a References

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15

[5] S.S. Dragomir, On Hadamard’s inequality for convex functions on the co-ordinates in a rectangle from the plane, Taiwanese J. Math. 4 (2001), 775–788. [6] S.S. Dragomir, C.E.M. Pearce, Selected topics on Hermite–Hadamard inequalities and applications, RGMIA Monographs, Victoria University, 2000. [7] R. Gorenflo, F. Mainardi, Fractional calculus: integral and differential equations of fractional order, Springer Verlag, Wien (1997), 223-276. [8] S. Miller and B. Ross, An introduction to the fractional calculus and fractional differential equations, John Wiley& Sons, USA, 1993, p.2. [9] M.A. Latif , M. Alomari, Hadamard-type inequalities for product two convex functions on the co-ordinetes. Int Math Forum. 4(47), 2327–2338 (2009). [10] M.A. Latif and S. Hussain, New inequalities of Ostrowski type for co-ordinated convex functions via fractional integrals, Journal of Fractional Calculus and Appl. 2(9) (2012), 1-15. [11] M.A. Latif, On Hadamard-type inequalities for h-convex functions on the co-ordinates, Int. Journal of Math. Analysis, 3 (33) (2009), 1645 – 1656. [12] C.E.M. Pearce, J. Peˇ cari´ c, Inequalities for differentiable mappings with application to special means and quadrature formula, Appl. Math. Lett. 13 (2000), 51—55. [13] I. Podlubni, Fractional differential equations, Academic Press, San Diego, 1999. [14] M. Z Sarikaya., A. Saglam and H. Yıldırım , On some Hadamard type inequalities for hconvex functions, Journal Math. Ineq. 2(3) (2008), 335-341. [15] M.Z. Sarikaya and H. Yaldiz, On weighted Montogomery identities for Riemann- Liouville fractional integrals, Konuralp Journal of Math., 1(1) (2013), 48-53. [16] M.Z. Sarikaya, On the Hermite-Hadamard-type inequalities for co-ordinated convex function via fractional integrals (submitted). ¨ [17] M.Z. Sarikaya, E. Set, M.E. Ozdemir, S.S. Dragomir, New some Hadamard’s type inequalities for co-ordinated convex functions.Tamsui Oxford Journal of Information and Mathematical Sciences, 28(2) (2012), 137-152. ¨ [18] E. Set, M.E. Ozdemir, S.S. Dragomir, On the Hermite–Hadamard inequality and other integral inequalities involving two functions, J. Inequal. Appl. (2010), p. 9 Article ID 148102. [19] E. Set, New inequalities of Ostrowski type for mappings whose derivatives are s-convex in the second sense via fractional integrals, Comput. Math. Appl., 63 (2012), 1147-1154. N Department

of Mathematics, Faculty of Science and Arts, Ordu University, OrduTURKEY E-mail address: [email protected] Department of Mathematics, Faculty of Science and Arts, Du ¨ zce University, Du ¨ zceTURKEY E-mail address: [email protected] ♣ Department of Mathematics, ¨ zce University, Faculty of Science and Arts, Du ¨ zce-TURKEY Du E-mail address: hatı[email protected]