• Sonuç bulunamadı

On hermite-hadamard type inequalities via katugampola fractional integrals

N/A
N/A
Protected

Academic year: 2021

Share "On hermite-hadamard type inequalities via katugampola fractional integrals"

Copied!
13
0
0

Yükleniyor.... (view fulltext now)

Tam metin

(1)

ON HERMITE-HADAMARD TYPE INEQUALITIES VIA KATUGAMPOLA FRACTIONAL INTEGRALS

H. YALDIZ, §

Abstract. In this paper, we give new definitons related to Katugampola fractional tegral for two variables functions. We are interested in giving the Hermite–Hadamard in-equality for a rectangle in plane via convex functions on co-ordinates involving Katugam-pola fractional integral.

Keywords: Convex function, co-ordinated convex function, Hermite-Hadamard inequal-ities, Katugampola fractional integral.

AMS Subject Classification: 26A33, 26D10, 26D16

1. Introduction

The most well-known inequalities related to the integral mean of a convex function are the Hermite Hadamard inequalities. Let f : I ⊂ R → R be convex function defined on the interval I of real numbers and a, b ∈ I, with a < b. Then the following double inequality is known in the literature as the Hermite-Hadamard’s inequality for convex functions [8]:

f a + b 2  ≤ 1 b − a b Z a f (x) dx ≤ f (a) + f (b) 2 . (1)

The inequalities (1) have become an important cornerstone in mathematical analysis and optimization and many uses of these inequalities have been discovered in a variety of settings. Many generalizations and extensions of the Hermite-Hadamard inequality exist in the literatures (see [5]).

Let us consider a bidimensional interval ∆ =: [a, b] × [c, d] in R2 with a < b and c < d. A function f : ∆ ⊂ R2 → R is said to be convex on ∆ if for all (x, y), (z, w) ∈ ∆ and t ∈ [0, 1], it satifies the following inequality:

f (tx + (1 − t) z, ty + (1 − t) w) ≤ t f (x, y) + (1 − t) f (z, w).

A modification for convex function on ∆ was defined by Dragomir [4], as follows:

Department of Mathematics, Kamil ¨Ozda˘g Science Faculty, Karamano˘glu Mehmetbey University, Karaman, TURKEY.

e-mail: yaldizhatice@gmail.com; ORCID: http://orcid.org/0000-0002-9606-1371. § Manuscript received: October 7, 2017; accepted: March 28, 2018.

TWMS Journal of Applied and Engineering Mathematics Vol.9, No.4 c I¸sık University, Department of Mathematics, 2019; all rights reserved.

(2)

A function f : ∆ → R is said to be 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 convex

where defined for all x ∈ [a, b] and y ∈ [c, d].

A formal defination for co-ordinated convex function may be stated as follows:

Definition 1.1. A function f : ∆ → R is called co-ordinated convex on ∆, for all (x, u), (y, v) ∈ ∆ and t, s ∈ [0, 1], if it satifies the following inequality:

f (tx + (1 − t) y, su + (1 − s) v) (2)

≤ ts f (x, u) + t(1 − s)f (x, v) + s(1 − t)f (y, u) + (1 − t)(1 − s)f (y, v).

Note that every convex function f : ∆ → R is co-ordinated convex but the converse is not generally true (see, [4]).

For recent developments about Hermite-Hadamard’s inequality for some convex func-tions on the co-ordinates, please refer to ([2],[9],[15],[19]and[20]). Also several inequalities for convex functions on the co-ordinates see the references ( [1],[6],[14],[16],[17]).

The importance of these operators stems indeed from their generality. Many useful fractional integral operators can be obtained by specializing the coefficient σ (k). Here, we just point out that the classical Riemann-Liouville fractional integrals Iaα+ and Ibα− of

order α defined by (see, [7, 12, 13])

(Iaα+ϕ) (x) := 1 Γ (α) Z x a (x − t)α−1ϕ(t)dt (x > a; α > 0) (3) and (Ibα−ϕ) (x) := 1 Γ(α) Z b x (t − x)α−1ϕ(t)dt (x < b; α > 0) (4) Later, in [17], Sarikaya presented the following Hermite–Hadamard-type inequalities for Riemann–Liouville fractional integrals by using convex functions of two variables on the co-ordinates:

Theorem 1.1. Let f : ∆ ⊂ R2 → R be a 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:

(5) f a + b 2 , c + d 2  ≤ Γ (α + 1) Γ (β + 1) 4(b − a)α(d − c)β ×hJaα,β+,c+f (b, d) + J α,β a+,d−f (b, c) + J α,β b−,c+f (a, d) + J α,β b−,d−f (a, c) i ≤ f (a, c) + f (a, d) + f (b, c) + f (b, d) 4 .

(3)

Theorem 1.2. Let f : ∆ ⊂ R2 → R be a 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:

(6) f a + b 2 , c + d 2  ≤ Γ (α + 1) 4(b − a)α  Jaα+f  b,c + d 2  + Jbα−f  a,c + d 2  +Γ (β + 1) 4 (d − c)β  Jcβ+f  a + b 2 , d  + Jdβ−f  a + b 2 , c  ≤ Γ (α + 1) Γ (β + 1) 4(b − a)α(d − c)β ×hJaα,β+,c+f (b, d) + J α,β a+,d−f (b, c) + J α,β b−,c+f (a, d) + J α,β b−,d−f (a, c) i ≤ Γ (α + 1) 8(b − a)α [J α a+f (b, c) + Jaα+f (b, d) + Jbα−f (a, c) + Jbα−f (a, d)] +Γ (β + 1) 8 (d − c)β h Jcβ+f (a, d) + J β c+f (b, d) + J β d−f (a, c) + J β d−f (b, c) i ≤ f (a, c) + f (a, d) + f (b, c) + f (b, d) 4 .

Recently, Katugampola introduced a new fractional integral that generalizes the Riemann-Liouville and the Hadamard fractional integrals in to a single form(see [10],[11],[18]). Definition 1.2. ([10]) Let [a, b] ⊂ R be a finite interval. then, the left and right side Katugampola fractional integrals of order α (> 0) of f ∈ Xcp(a, b) are defined by,

ρIα a+f (x) = ρ1−α Γ(α) x R a tρ−1 (xρ−tρ)1−af (t) dt and ρIbα+f (x) = ρ1−α Γ(α) b R x tρ−1 (tρ−xρ)1−af (t) dt

with a < x < b and ρ > 0, if the integrals exist.

When ρ = 0 we arrive at the standard Riemann-Liouville fractional integral.

In [3], Chen and Katugampola proved the following inequality which is Hermite-Hadamard’s inequalities for the Katugampola fractional integrals:

Theorem 1.3. Let α > 0 and ρ > 0. Let f : [aρ, bρ] → R be a positive function with 0 ≤ a < b and f ∈ Xcp(aρ, bρ). If f is also a convex function on [a, b], then the following

inequalities hold: f a ρ+ bρ 2  ≤ ρ αΓ (α + 1) 2(bρ− aρ)α [ ρIα a+f (bρ) +ρIbα−f (aρ)] ≤ f (aρ) + f (bρ) 2 (7)

where the fractional integrals are considered for the function f (xρ) and evaluated at a and

b, respectively.

Now, we establish new definitons related to Katugampola fractional integrals for two variables functions:

(4)

Definition 1.3. Let f ∈ L1([a, b] × [c, d]) . The Katugampola fractional integralsρ,σIaα,β+,c+f (x, y) , ρ,σIα,β a+,d−f (x, y) , ρ,σI α,β b−,c+f (x, y) , and ρ,σI α,β

b−,d−f (x, y) of order α, β > 0 are defined by

ρ,σIα,β a+,c+f (x, y) : = ρ1−α Γ (α) σ1−β Γ (β) x Z a y Z c tρ−1sσ−1 (xρ− tρ)1−a(yσ − sσ)1−βf (t, s)dsdt, x > a, y > c, ρ,σIα,β a+,d−f (x, y) : = ρ1−α Γ (α) σ1−β Γ (β) x Z a d Z y tρ−1sσ−1 (xρ− tρ)1−a(sσ− yσ)1−βf (t, s)dsdt, x > a, y < d, ρ,σIα,β b−,c+f (x, y) : = ρ1−α Γ (α) σ1−β Γ (β) b Z x y Z c tρ−1sσ−1 (tρ− xρ)1−a(yσ− sσ)1−βf (t, s)dsdt, x < b, y > c, and ρ,σIα,β b−,d−f (x, y) : = ρ1−α Γ (α) σ1−β Γ (β) b Z x d Z y tρ−1sσ−1 (tρ− xρ)1−a(sσ− yσ)1−βf (t, s)dsdt, x < b, y < d. with a < x < b and c < y < d with ρ > 0.

In this paper, we are interested to give the Hermite–Hadamard inequality for a rectangle in plane via convex functions on co-ordinates involving Katugampola fractional integrals. We also study some properties of mappings associated with the Hermite–Hadamard in-equality for convex functions on co-ordinates.

2. Hermite Hadamard Type Inequalities for Katugampola fractional integrals

In this section, we will give Hermite-Hadamard type inequalities for the Katugampola fractional integrals by using co-ordinated convex functions.

Theorem 2.1. Let α, β > 0 and ρ, σ > 0. Let f : ∆ρ,σ ⊂ R2 → R be a co-ordinated

convex on ∆ρ,σ := [aρ, bρ] × [cσ, dσ] in R2 with 0 ≤ a < b, 0 ≤ c < d and f ∈ L1(∆ρ,σ).

Then the following inequalities hold:

(8) f a ρ+ bρ 2 , cσ+ dσ 2  ≤ ρ ασβΓ (α + 1) Γ (β + 1) 4(bρ− aρ)α(dσ− cσ)β ×hρ,σIaα,β+,c+f (b ρ, dσ) +ρ,σIα,β a+,d−f (bρ, cσ) +ρ,σI α,β b−,c+f (a ρ, dσ) +ρ,σIα,β b−,d−f (aρ, cσ) i ≤ f (a ρ, cσ) + f (aρ, dσ) + f (bρ, cσ) + f (bρ, dσ) 4 .

(5)

with a < x < b and c < y < d.

Proof. According to (2) with xρ = tρaρ+ (1 − tρ)bρ, yρ= (1 − tρ)aρ+ tρbρ, uσ = sσcσ + (1 − sσ)dσ, wσ = (1 − sσ)cσ+ sσdσ and t = s = 12, we find that

f a ρ+ bρ 2 , cσ + dσ 2  ≤ 1 4[f (t ρaρ+ (1 − tρ)bρ, sσcσ+ (1 − sσ)dσ) +f (tρaρ+ (1 − tρ)bρ, (1 − sσ)cσ+ sσdσ) +f ((1 − tρ)aρ+ tρbρ, sσcσ+ (1 − sσ)dσ) (9) +f ((1 − tρ)aρ+ tρbρ, (1 − sσ)cσ+ sσdσ)].

Multiplying both sides of (9) by tαρ−1sβσ−1, then integrating with respect to (t, s) on [0, 1] × [0, 1], we obtain 4f a ρ+ bρ 2 , cσ+ dσ 2 Z1 0 1 Z 0 tαρ−1sβσ−1dsdt ≤ 1 Z 0 1 Z 0 tαρ−1sβσ−1f (tρaρ+ (1 − tρ)bρ, sσcσ+ (1 − sσ)dσ)dsdt + 1 Z 0 1 Z 0 tαρ−1sβσ−1f (tρaρ+ (1 − tρ)bρ, (1 − sσ)cσ+ sσdσ)dsdt + 1 Z 0 1 Z 0 tαρ−1sβσ−1f ((1 − tρ)aρ+ tρbρ, sσcσ + (1 − sσ)dσ)dsdt + 1 Z 0 1 Z 0 tαρ−1sβσ−1f ((1 − tρ)aρ+ tρbρ, (1 − sσ)cσ+ sσdσ)dsdt.

(6)

Using the change of variable in the last integrals, we have 4f a ρ+ bρ 2 , cσ+ dσ 2  1 αρ 1 βσ ≤ 1 (bρ− aρ)α(dσ− cσ)β   b Z a d Z c xρ−1 (bρ− xρ)1−α yσ−1 (dσ− yσ)1−βf (x ρ, yσ) dydx + b Z a d Z c xρ−1 (bρ− xρ)1−α yσ−1 (yσ− cσ)1−βf (x ρ, yσ) dydx + b Z a d Z c xρ−1 (xρ− aρ)1−α yσ−1 (dσ− yσ)1−βf (x ρ, yσ) dydx + b Z a d Z c xρ−1 (xρ− aρ)1−α yσ−1 (yσ− cσ)1−βf (x ρ, yσ) dydx  .

which gives the left hand side inequality in (8). Now we prove the right hand side inequality in (8). For this purpose we first note that if f is a co-ordinated convex on ∆, then we can write by using (2) f (tρaρ+ (1 − tρ)bρ, sσcσ+ (1 − sσ)dσ) ≤ tρsσf (aρ, cσ) + (1 − tρ)sσf (bρ, cσ) + tρ(1 − sσ)f (aρ, dσ) + (1 − tρ)(1 − sσ)f (bρ, dσ), f (tρaρ+ (1 − tρ)bρ, (1 − sσ)cσ+ sσdσ) ≤ tρ(1 − sσ)f (aρ, cσ) + (1 − tρ)(1 − sσ)f (bρ, cσ) + tρsσf (aρ, dσ) + (1 − tρ)sσf (bρ, dσ), f ((1 − tρ)aρ+ tρbρ, sσcσ+ (1 − sσ)dσ) ≤ (1 − tρ)sσf (aρ, cσ) + tρsσf (bρ, cσ) + (1 − tρ)(1 − sσ)f (aρ, dσ) + tρ(1 − sσ)f (bρ, dσ), and f ((1 − tρ)aρ+ tρbρ, (1 − sσ)cσ+ sσdσ) ≤ (1 − tρ)(1 − sσ)f (aρ, cσ) + tρ(1 − sσ)f (bρ, cσ) + (1 − tρ)sσf (aρ, dσ) + tρsσf (bρ, dσ). By adding these inequalities, we get

f (tρaρ+ (1 − tρ)bρ, sσcσ+ (1 − sσ)dσ) + f (tρaρ+ (1 − tρ)bρ, (1 − sσ)cσ+ sσdσ) +f ((1 − tρ)aρ+ tρbρ, sσcσ+ (1 − sσ)dσ) + f ((1 − tρ)aρ+ tρbρ, (1 − sσ)cσ+ sσdσ)

(10) ≤ f (aρ, cσ) + f (aρ, dσ) + f (bρ, cσ) + f (bρ, dσ) .

(7)

Multiplying both sides of (10) by tαρ−1sβσ−1, then integrating with respect to (t, s) on [0, 1] × [0, 1] we obtain 1 Z 0 1 Z 0 tαρ−1sβσ−1f (tρaρ+ (1 − tρ)bρ, sσcσ+ (1 − sσ)dσ)dsdt + 1 Z 0 1 Z 0 tαρ−1sβσ−1f (tρaρ+ (1 − tρ)bρ, (1 − sσ)cσ+ sσdσ)dsdt + 1 Z 0 1 Z 0 tαρ−1sβσ−1f ((1 − tρ)aρ+ tρbρ, sσcσ+ (1 − sσ)dσ)dsdt + 1 Z 0 1 Z 0 tαρ−1sβσ−1f ((1 − tρ)aρ+ tρbρ, (1 − sσ)cσ+ sσdσ)dsdt ≤ [f (aρ, cσ) + f (aρ, dσ) + f (bρ, cσ) + f (bρ, dσ)] 1 Z 0 1 Z 0 tαρ−1sβσ−1dsdt.

Then by using the change of variable we have

Γ (α) Γ (β) ρ1−ασ1−β(bρ− aρ)α(dσ− cσ)β ×hρ,σIaα,β+,c+f (bρ, dσ) +ρ,σI α,β a+,d−f (bρ, cσ) +ρ,σI α,β b−,c+f (aρ, dσ) +ρ,σI α,β b−,d−f (aρ, cσ) i ≤ 1 αρ 1 βσ[f (a ρ, cσ) + f (aρ, dσ) + f (bρ, cσ) + f (bρ, dσ)] .

In this way the proof is completed. 

Remark 2.1. If we set ρ, σ = 1 in Theorem 2.1, then the inequalities (8) become the inequalities (5).

Theorem 2.2. Let α, β > 0 and ρ, σ > 0. Let f : ∆ρ,σ ⊂ R2 → R be a co-ordinated

(8)

Then the following inequalities hold: (11) f a ρ+ bρ 2 , cσ+ dσ 2  ≤ ρ αΓ (α + 1) 4(bρ− aρ)α  ρ,σIα a+f  b,c σ+ dσ 2  +ρ,σ Ibα−f  a,c σ+ dσ 2  +σ βΓ (β + 1) 4 (dσ − cσ)β  ρ,σIβ c+f  aρ+ bρ 2 , d  +ρ,σIdβ−f  aρ+ bρ 2 , c  ≤ ρ ασβΓ (α + 1) Γ (β + 1) 4(bρ− aρ)α(dσ− cσ)β ×hρ,σIaα,β+,c+f (b ρ, dσ) +ρ,σIα,β a+,d−f (bρ, cσ) +ρ,σI α,β b−,c+f (a ρ, dσ) +ρ,σIα,β b−,d−f (aρ, cσ) i ≤ ρ αΓ (α + 1) 8(bρ− aρ)α [ ρ,σIα a+f (b, cσ) +ρ,σ Iaα+f (b, dσ) +ρ,σIbα−f (a, cσ) +ρ,σIbα−f (a, dσ)] +σ βΓ (β + 1) 8 (dσ − cσ)β h ρ,σIβ c+f (aρ, d) +ρ,σI β d−f (aρ, c) +ρ,σI β c+f (bρ, d) +ρ,σI β d−f (bρ, c) i ≤ f (a ρ, cσ) + f (aρ, dσ) + f (bρ, cσ) + f (bρ, dσ) 4 . with a < x < b and c < y < d.

Proof. Since f : ∆ρ,σ → R is co-ordinated convex on ∆ρ := [aρ, bρ] × [cσ, dσ] in R2 with

0 ≤ a < b, 0 ≤ c < d, it follows that the mapping gx: [c, d] → R, gx(y) = f (x, y), is convex

on [c, d] for all x ∈ [a, b]. Then by using inequalities ([3]), we can write

gx  cσ+ dσ 2  ≤ σ βΓ (β + 1) 2(dσ− cσ)β h ρ,σIβ c+gx(dσ) +ρ,σ Idβ−gx(cσ) i ≤ gx(c σ) + g x(dσ) 2 , x ∈ [a, b]. That is, f  x,c σ+ dσ 2  (12) ≤ ρβ 2(dσ− cσ)β   d Z c yσ−1 (dσ− yσ)1−βf (x ρ, yσ) dy + d Z c yσ−1 (yσ− cσ)1−βf (x ρ, yσ) dy   ≤ f (x, c σ) + f (x, dσ) 2 , x ∈ [a, b].

(9)

Then multiplying both sides of (12) by 2(bρρα−aρ)α x ρ−1 (bρ−xρ)1−α and ρα 2(bρ−aρ)α x ρ−1 (xρ−aρ)1−α,

inte-grating with respect to x over [a, b], respectively, we get

(13) ρα 2(bρ− aρ)α b Z a xρ−1 (bρ− xρ)1−αf  x,c σ+ dσ 2  dx ≤ ρα 2(bρ− aρ)α σβ 2(dσ− cσ)β   b Z a d Z c xρ−1 (bρ− xρ)1−α yσ−1 (dσ− yσ)1−βf (x ρ, yσ) dydx + b Z a d Z c xρ−1 (bρ− xρ)1−α yσ−1 (yσ− cσ)1−βf (x ρ, yσ) dydx   ≤ ρα 2(bρ− aρ)α   b Z a xρ−1 (bρ− xρ)1−αf (x, c σ)dx + b Z a xρ−1 (bρ− xρ)1−αf (x, d σ)dx  , and (14) ρα 2(bρ− aρ)α b Z a xρ−1 (xρ− aρ)1−αf  x,c σ+ dσ 2  dx ≤ ρα 2(bρ− aρ)α σβ 2(dσ− cσ)β   b Z a d Z c xρ−1 (xρ− aρ)1−α yσ−1 (dσ− yσ)1−βf (x ρ, yσ) dydx + b Z a d Z c xρ−1 (xρ− aρ)1−α yσ−1 (yσ− cσ)1−βf (x ρ, yσ) dydx   ≤ ρα 2(bρ− aρ)α   b Z a xρ−1 (xρ− aρ)1−αf (x, c σ)dx + b Z a xρ−1 (xρ− aρ)1−αf (x, d σ)dx  .

(10)

By similar argument applied for the mapping gy : [a, b] → R, gy(x) = f (x, y), we have σβ 2(dσ− cσ)β d Z c yσ−1 (yσ− cσ)1−βf  aρ+ bρ 2 , y  dy (15) ≤ ρα 2(bρ− aρ)α σβ 2(dρ− cρ)β   b Z a d Z c xρ−1 (xρ− aρ)1−α yσ−1 (yσ− cσ)1−βf (x ρ, yσ) dydx + b Z a d Z c xρ−1 (bρ− xρ)1−α yσ−1 (yσ− cσ)1−βf (x ρ, yσ) dydx   ≤ σβ 2(dσ− cσ)β   d Z c yσ−1 (yσ − cσ)1−βf (a ρ, y)dy + d Z c yσ−1 (yσ− cσ)1−βf (b ρ, y)dy  , and σβ 2(dσ− cσ)β d Z c yσ−1 (dσ− yσ)1−βf  aρ+ bρ 2 , y  dy (16) ≤ ρα 2(bρ− aρ)α σβ 2(dσ− cσ)β   b Z a d Z c xρ−1 (xρ− aρ)1−α yσ−1 (dσ− yσ)1−βf (x ρ, yσ) dydx + b Z a d Z c xρ−1 (bρ− xρ)1−α yσ−1 (dσ− yσ)1−βf (x ρ, yσ) dydx   ≤ σβ 2(dσ− cσ)β   d Z c yσ−1 (dσ− yσ)1−βf (a ρ, y)dy + d Z c yσ−1 (dσ− yσ)1−βf (b ρ, y)dy  .

Adding the inequalities (13)-(16), we obtain

ραΓ (α + 1) 4(bρ− aρ)α  ρ,σIα a+f  b,c σ+ dσ 2  +ρ,σIbα−f  a,c σ+ dσ 2  +σ βΓ (β + 1) 4 (dσ− cσ)β  ρ,σIβ c+f  aρ+ bρ 2 , d  +ρ,σIdβ−f  aρ+ bρ 2 , c  ≤ ρ ασβΓ (α + 1) Γ (β + 1) 4(bρ− aρ)α(dσ− cσ)β ×hρ,σIα,β a+,c+f (bρ, dσ) +ρ,σI α,β a+,d−f (bρ, cσ) +ρ,σI α,β b−,c+f (aρ, dσ) +ρ,σI α,β b−,d−f (aρ, cσ) i

(11)

≤ ρ αΓ (α + 1) 8(bρ− aρ)α [ ρ,σIα a+f (b, cσ) +ρ,σIaα+f (b, dσ) +ρ,σIbα−f (a, cσ) +ρ,σ Ibα−f (a, dσ)] +σ βΓ (β + 1) 8 (dσ− cσ)β h ρ,σIβ c+f (aρ, d) +ρ,σI β c+f (bρ, d) +ρ,σ I β d−f (aρ, c) +ρ,σI β d−f (bρ, c) i .

Thus, we proved the second and the third inequalities in (11). Now, using the left side inequality in (7), we also have

f a ρ+ bρ 2 , cσ+ dσ 2  ≤ ρα 2(bρ− aρ)α ×   b Z a xρ−1 (xρ− aρ)1−αf  x,c σ+ dσ 2  dx + b Z a xρ−1 (bρ− xρ)1−αf  x,c σ+ dσ 2  dx   and f a ρ+ bρ 2 , cσ+ dσ 2  ≤ σβ 2(dσ− cσ)β ×   d Z c yσ−1 (yσ− cσ)1−βf  aρ+ bρ 2 , y  dy + d Z c yσ−1 (dσ− yσ)1−βf  aρ+ bρ 2 , y  dy  .

By adding these inequalities, we get

f a ρ+ bρ 2 , cσ+ dσ 2  ≤ ρ αΓ (α + 1) 4(bρ− aρ)α  ρ,σIα a+f  b,c σ+ dσ 2  +ρ,σIbα−f  a,c σ+ dσ 2  +σ βΓ (β + 1) 4 (dσ− cσ)β  ρ,σIβ c+f  aρ+ bρ 2 , d  +ρ,σIdβ−f  aρ+ bρ 2 , c 

which gives the first inequality in (11).

Finally, using the right-hand side inequality in (7), we can state

(17) ρα 2(bρ− aρ)α   b Z a xρ−1 (bρ− xρ)1−αf (x, c σ) dx + b Z a xρ−1 (xρ− aρ)1−αf (x, c σ) dx  ≤ f (aρ, cσ) + f (bρ, cσ) 2 , (18) ρα 2(bρ− aρ)α   b Z a xρ−1 (bρ− xρ)1−αf (x, d σ) dx + b Z a xρ−1 (xρ− aρ)1−αf (x, d σ) dx  ≤ f (aρ, dσ) + f (bρ, dσ) 2 , (19) σβ 2(dσ− cσ)β   d Z c yσ−1 (dσ− yσ)1−βf (a ρ, y)dy + d Z c yσ−1 (yσ− cσ)1−βf (a ρ, y)dy  ≤ f (aρ, cσ) + f (aρ, dσ) 2

(12)

and (20) σβ 2(dσ− cσ)β   d Z c yσ−1 (dσ− yσ)1−βf (b ρ, y)dy + d Z c yσ−1 (yσ− cσ)1−βf (b ρ, y)dy  ≤ f (bρ, cσ) + f (bρ, dσ) 2 .

which give, by addition (17)-(20), the last inequality in (11). 

Remark 2.2. If we set ρ, σ = 1 in Theorem 2.2, then the inequalities (11) become the inequalities (6).

References

[1] Bakula M. K. and Pecaric J., (2006), On the Jensen’s inequality for convex functions on the co-ordinates in a rectangle from the plane, Taiwanese Journal of Mathematics, 10(5), 1271-1292. [2] Chen F., (2014),A note on the Hermite-Hadamard inequality for convex functions on the co-ordinates,

J.of Math. Inequalities, 8(4), 915-923.

[3] Chen, H. and Katugampola U.N., (2017), Hermite-Hadamard and Hermite-Hadamard-Fej˘er type in-equalities for generalized fractional integrals, J. Math. Anal. Appl., 446 , 1274-1291.

[4] Dragomir S. S., (2001), On Hadamard’s inequality for convex functions on the co-ordinates in a rectangle from the plane, Taiwanese Journal of Mathematics, 4, 775-788.

[5] Dragomir S.S. and Pearce C.E.M., (2000), Selected topics on Hermite–Hadamard inequalities and applications, RGMIA Monographs, Victoria University.

[6] Ekinci A., Akdemir A. O. and ¨Ozdemir M. E., (2017), On Hadamard-type inequalities for co-ordinated r-convex functions, AIP Conference Proceedings 1833.

[7] Gorenflo R. and Mainardi F., (1997), Fractional calculus: integral and differential equations of frac-tional order, Springer Verlag, Wien, 223-276.

[8] Hadamard J., (1893), Etude sur les proprietes des fonctions entieres et en particulier d’une fonction considree par, Riemann, J. Math. Pures. et Appl. 58, 171–215.

[9] Hwang D. Y., Tseng K. L. and Yang G. S., (2007), Some Hadamard’s inequalities for co-ordinated convex functions in a rectangle from the plane, Taiwanese Journal of Mathematics, 11, 63-73. [10] Katugampola U.N., (2011), New approach to a generalized fractional integrals, Appl. Math. Comput.,

218 (4) , 860-865.

[11] Katugampola U.N., (2014), New approach to generalized fractional derivatives, Bull. Math. Anal. Appl., 6 (4) , 1-15.

[12] Kilbas A. A., Srivastava H. M. and Trujillo J. J., (2006), Theory and applications of fractional differ-ential equations, North-Holland Mathematics Studies, 204, Elsevier Sci. B.V., Amsterdam.

[13] S. Miller and B. Ross, An introduction to the fractional calculus and fractional differential equations, John Wiley & Sons, USA, 1993, p.2.

[14] ¨Ozdemir M. E., Set E. and Sarıkaya M. Z., (2011), Some new Hadamard’s type inequalities for coordinated m-convex and (α, m)-convex functions, Hacettepe J. of Math. and Statistics, 40, 219-229. [15] Sarıkaya M. Z., Set E., Yaldiz H. and Basak N., ( 2013), Hermite–Hadamard’s inequalities for fractional

integrals and related fractional inequalities. ,Math Comput Model., 57(9–10):2403–2407.

[16] Sarıkaya M. Z., Set E., Ozdemir M.E. and Dragomir S. S. , (2011), New some Hadamard’s type inequalities for co-ordinated convex functions, Tamsui Oxford J of Information and Math. Sciences , 28(2), 137-152.

[17] Sarıkaya M. Z., (2014), On the Hermite-Hadamard-type inequalities for co-ordinated convex function via fractional integrals, Integral Transforms and Special Functions, Vol. 25, No. 2, 134-147.

[18] Set E., Dahmani Z. and Mumcu ´I., (2018), New extensions of Chebyshev type inequalities using generalized Katugampola integrals via P˘olya-Szeg¨o inequality, Vol. 8, No. 2, 137-144.

[19] Sarıkaya M. Z. and Yaldız H., (2013), On the Hadamard’s type inequalities for L-Lipschitzian mapping, Konuralp Journal of Mathematics, Volume 1, No. 2, pp. 33-40.

[20] Yaldız H. , Sarıkaya M. Z. and Dahmani Z., (2017), On the Hermite-Hadamard-Fejer-type inequalities for co-ordinated convex functions via fractional integrals, An International Journal of Optimization and Control: Theories & Applications, Vol.7, No.2, pp.205-215.

(13)

Hatice YALDIZ is an Assistant Professor in the Department of Mathematics at Karamano˘glu Mehmetbey University, Karaman/TURKEY. She was in Western Ken-tucky University/USA as a visiting researcher in 2015. Her major research interests include Inequalities, Fractional theory, Discrete analysis and Time scales.

Referanslar

Benzer Belgeler

Etlerde YBU’sı ilk olarak Macfarlane (1973) tarafından gerçekleştirilmiştir ve bu çalışmayı takiben yüksek basınç uygulamasının et ve et ürünlerinin

believe, can be better understood if we see Women in Love in the light of the theories of language and the novel advanced by the Russian formalist Mikhail Bakhtin, whose

‘&#34;M eulders, D. and Plasman, R., Women in Atypical Employment.. comparison with women’s labor market participation among the other countries o f Southern Europe. The situation

Düzce İli fındık bahçelerinde Mayıs böceği popülasyon yoğunluğu ekonomik zarar eşiği açısından incelendiğinde; İl genelinde incelenen 32 bahçenin 3’ünde,

Analysis of variance (ANOVA) results of total color change (ΔE*) values of samples applied with acetic acid, ammonia, hydrogen peroxide and sodium silicate at different

Of the mechanical properties; experiments of compression strength parallel to grain were conducted in accordance with TS 2595 (1977), bending strength in accordance with TS

The comparison results of the Duncan test on the factor levels of moisture content, type of varnish, thermal processing temperature, and thermal processing time,

The aim of this study was to investigate the effect of the Tinuvin derivatives widely used as UV stabilizers in the plastics industry on EPDM rubber.. The EPDM rubber plates