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.
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 .
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:
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 .
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.
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) .
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
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].
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 .
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
≤ 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
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).
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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.