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Article

New Integral Inequalities via the Katugampola Fractional Integrals for Functions Whose Second Derivatives Are Strongly η-Convex

1
Department of Mathematics and Computer Science, Alabama State University, Montgomery, AL 36101, USA
2
Department of Mathematics, Tuskegee University, Tuskegee, AL 36088, USA
*
Author to whom correspondence should be addressed.
Mathematics 2019, 7(2), 183; https://doi.org/10.3390/math7020183
Submission received: 28 December 2018 / Revised: 3 February 2019 / Accepted: 12 February 2019 / Published: 15 February 2019
(This article belongs to the Special Issue Inequalities)

Abstract

:
In this paper, we introduced some new integral inequalities of the Hermite–Hadamard type for functions whose second derivatives in absolute values at certain powers are strongly η -convex functions via the Katugampola fractional integrals.

1. Introduction

Let I be an interval in R . A function f : I R is said to be convex on I if
f ( t x + ( 1 t ) y ) t f ( x ) + ( 1 t ) f ( y )
for all x , y I and t [ 0 , 1 ] . The following inequalities which hold for convex functions is known in the literature as the Hermite–Hadamard type inequality.
Theorem 1
([1]). If f : [ a , b ] R is convex on [ a , b ] with a < b , then
f a + b 2 1 b a a b f ( x ) d x f ( a ) + f ( b ) 2 .
Many authors have studied and generalized the Hermite–Hadamard inequality in several ways via different classes of convex functions. For some recent results related to the Hermite–Hadamard inequality, we refer the interested reader to the papers [2,3,4,5,6,7,8,9,10,11].
In 2016, Gordji et al. [12] introduced the concept of η -convexity as follows:
Definition 1
([12]). A function f : I R is said to be η-convex with respect to the bifunction η : R × R R if
f ( t x + ( 1 t ) y ) f ( y ) + t η ( f ( x ) , f ( y ) )
for all x , y I and t [ 0 , 1 ] .
Remark 1.
If we take η ( x , y ) = x y in Definition 1, then we recover the classical definition of convex functions.
In 2017, Awan et al. [13] extended the class of η -convex functions to the class of strongly η -convex functions as follows:
Definition 2
([13]). A function f : I R is said to be strongly η-convex with respect to the bifunction η : R × R R with modulus μ 0 if
f ( t x + ( 1 t ) y ) f ( y ) + t η ( f ( x ) , f ( y ) ) μ t ( 1 t ) ( x y ) 2
for all x , y I and t [ 0 , 1 ] .
Remark 2.
If η ( x , y ) = x y in Definition 2, then we have the class of strongly convex functions.
For some recent results related to the class of η -convex functions, we refer the interested reader to the papers [8,12,13,14,15,16].
Definition 3
([17]). The left- and right-sided Riemann–Liouville fractional integrals of order α > 0 of f are defined by
J a + α f ( x ) : = 1 Γ ( α ) a x ( x t ) α 1 f ( t ) d t
and
J b α f ( x ) : = 1 Γ ( α ) x b ( t x ) α 1 f ( t ) d t
with a < x < b and Γ ( · ) is the gamma function given by
Γ ( x ) : = 0 t x 1 e t d t , R e ( x ) > 0
with the property that Γ ( x + 1 ) = x Γ ( x ) .
Definition 4
([18]). The left- and right-sided Hadamard fractional integrals of order α > 0 of f are defined by
H a + α f ( x ) : = 1 Γ ( α ) a x ln x t α 1 f ( t ) t d t
and
H b α f ( x ) : = 1 Γ ( α ) x b ln t x α 1 f ( t ) t d t .
Definition 5.
X c p ( a , b ) ( c R , 1 p ) denotes the space of all complex-valued Lebesgue measurable functions f for which f X c p < , where the norm · X c p is defined by
f X c p = a b | t c f ( t ) | p d t t 1 / p ( 1 p < )
and for p =
f X c = e s s sup a t b | t c f ( t ) | .
In 2011, Katugampola [19] introduced a new fractional integral operator which generalizes the Riemann–Liouville and Hadamard fractional integrals as follows:
Definition 6.
Let [ a , b ] R be a finite interval. Then, the left- and right-sided Katugampola fractional integrals of order α > 0 of f X c p ( a , b ) are defined by
ρ I a + α f ( x ) : = ρ 1 α Γ ( α ) a x t ρ 1 ( x ρ t ρ ) 1 α f ( t ) d t
and
ρ I b α f ( x ) : = ρ 1 α Γ ( α ) x b t ρ 1 ( t ρ x ρ ) 1 α f ( t ) d t
with a < x < b and ρ > 0 , if the integrals exist.
Remark 3.
It is shown in [19] that the Katugampola fractional integral operators are well-defined on X c p ( a , b ) .
Theorem 2
([19]). Let α > 0 and ρ > 0 . Then for x > a
  • lim ρ 1 ρ I a + α f ( x ) = J a + α f ( x ) ,
  • lim ρ 0 + ρ I a + α f ( x ) = H a + α f ( x ) .
Similar results also hold for right-sided operators.
For more information about the Katugampola fractional integrals and related results, we refer the interested reader to the papers [19,20,21]. Recently, Chen and Katugampola [20] introduced several integral inequalities of Hermite–Hadamard type for functions whose first derivatives in absolute value are convex functions via the Katugampola fractional integrals. We present two of their results here for the purpose of our discussion. The first result of importance to us employs the following lemma.
Lemma 1
([20]). Let α > 0 , ρ > 0 and f : [ a ρ , b ρ ] R be a differentiable mapping on ( a ρ , b ρ ) with 0 a < b . Then the following equality holds if the fractional integrals exist:
f ( a ρ ) + f ( b ρ ) α ρ ρ α 1 Γ ( α ) ( b ρ a ρ ) α ρ I a + α f ( b ρ ) + ρ I b α f ( a ρ ) = b ρ a ρ α 0 1 t ρ ( α + 1 ) 1 f ( ( 1 t ρ ) a ρ + t ρ b ρ ) f ( t ρ a ρ + ( 1 t ρ ) b ρ ) d t .
By using Lemma 1, the authors proved the following result.
Theorem 3
([20]). Let f : [ a ρ , b ρ ] R be a differentiable mapping on ( a ρ , b ρ ) with 0 a < b . If | f | is convex on [ a ρ , b ρ ] , then the following inequality holds:
f ( a ρ ) + f ( b ρ ) 2 ρ α Γ ( α + 1 ) 2 ( b ρ a ρ ) α ρ I a + α f ( b ρ ) + ρ I b α f ( a ρ ) b ρ a ρ 2 ( α + 1 ) | f ( a ρ ) | + | f ( b ρ ) | .
The second result of importance to us also uses the following lemma.
Lemma 2
([20]). Let α > 0 , ρ > 0 and f : [ a ρ , b ρ ] R be a differentiable mapping on ( a ρ , b ρ ) with 0 a < b . Then the following equality holds if the fractional integrals exist:
f ( a ρ ) + f ( b ρ ) 2 ρ α Γ ( α + 1 ) 2 ( b ρ a ρ ) α ρ I a + α f ( b ρ ) + ρ I b α f ( a ρ ) = b ρ a ρ 2 0 1 [ ( 1 t ρ ) α t ρ α ] t ρ 1 f ( t ρ a ρ + ( 1 t ρ ) b ρ ) d t .
By using Lemma 2, the authors proved the following result.
Theorem 4
([20]). Let f : [ a ρ , b ρ ] R be a differentiable mapping on ( a ρ , b ρ ) with 0 a < b . If | f | is convex on [ a ρ , b ρ ] , then the following inequality holds:
| f ( a ρ ) + f ( b ρ ) 2 ρ α Γ ( α + 1 ) 2 ( b ρ a ρ ) α [ ρ I a + α f ( b ρ ) + ρ I b α f ( a ρ ) ] | b ρ a ρ 2 ρ ( α + 1 ) 1 1 2 α | f ( a ρ ) | + | f ( b ρ ) | .
Remark 4.
It is important to note that Lemmas 1 and 2 are corrected versions of [20] (Lemma 2.4 and Equation (14)).
Our purpose in this paper is to provide some new estimates for the right hand side of the inequalities in Theorems 3 and 4 for functions whose second derivatives in absolute value at some powers are strongly η -convex.

2. Main Results

To prove the main results of this paper, we need the following lemmas which are extensions of Lemmas 1 and 2 for the second derivative case of the function f.
Lemma 3.
Let α > 0 , ρ > 0 and f : [ a ρ , b ρ ] R be a twice differentiable mapping on ( a ρ , b ρ ) with 0 a < b . Then the following equality holds if the fractional integrals exist:
f ( a ρ ) + f ( b ρ ) α ρ ρ α 1 Γ ( α ) ( b ρ a ρ ) α ρ I a + α f ( b ρ ) + ρ I b α f ( a ρ ) = ( b ρ a ρ ) 2 α ( α + 1 ) [ 0 1 1 t ρ ( α + 1 ) t ρ 1 f ( ( 1 t ρ ) a ρ + t ρ b ρ ) d t 0 1 t ρ ( α + 2 ) 1 f ( t ρ a ρ + ( 1 t ρ ) b ρ ) d t ] .
Proof. 
Let
I 1 = 0 1 1 t ρ ( α + 1 ) t ρ 1 f ( ( 1 t ρ ) a ρ + t ρ b ρ ) d t
and
I 2 = 0 1 t ρ ( α + 2 ) 1 f ( t ρ a ρ + ( 1 t ρ ) b ρ ) d t .
By using integration by parts we have that
I 1 = 0 1 1 t ρ ( α + 1 ) t ρ 1 f ( ( 1 t ρ ) a ρ + t ρ b ρ ) d t = 1 ρ ( b ρ a ρ ) [ 1 t ρ ( α + 1 ) ] f ( ( 1 t ρ ) a ρ + t ρ b ρ ) | 0 1 + ρ ( α + 1 ) ρ ( b ρ a ρ ) 0 1 t ρ ( α + 1 ) 1 f ( ( 1 t ρ ) a ρ + t ρ b ρ ) d t = 1 ρ ( b ρ a ρ ) f ( a ρ ) + ( α + 1 ) ( b ρ a ρ ) 0 1 t ρ ( α + 1 ) 1 f ( ( 1 t ρ ) a ρ + t ρ b ρ ) d t .
By a similar argument, one gets:
I 2 = 1 ρ ( b ρ a ρ ) f ( a ρ ) + ( α + 1 ) ( b ρ a ρ ) 0 1 t ρ ( α + 1 ) 1 f ( t ρ a ρ + ( 1 t ρ ) b ρ ) d t .
Using (4) and (5), we have
I 1 I 2 = ( α + 1 ) ( b ρ a ρ ) 0 1 t ρ ( α + 1 ) 1 f ( ( 1 t ρ ) a ρ + t ρ b ρ ) f ( t ρ a ρ + ( 1 t ρ ) b ρ ) d t .
The desired identity in (3) follows from (6) by using (1) and rearranging the terms.  □
Lemma 4.
Let α > 0 , ρ > 0 and f : [ a ρ , b ρ ] R be a twice differentiable mapping on ( a ρ , b ρ ) with 0 a < b . Then the following equality holds if the fractional integrals exist:
f ( a ρ ) + f ( b ρ ) 2 ρ α Γ ( α + 1 ) 2 ( b ρ a ρ ) α ρ I a + α f ( b ρ ) + ρ I b α f ( a ρ ) = ( b ρ a ρ ) 2 2 ( α + 1 ) 0 1 [ 1 ( 1 t ρ ) α + 1 t ρ ( α + 1 ) ] t ρ 1 f ( t ρ a ρ + ( 1 t ρ ) b ρ ) d t .
Proof. 
We start by considering the following computation which is a direct application of integration by parts.
0 1 [ 1 ( 1 t ρ ) α + 1 t ρ ( α + 1 ) ] t ρ 1 f ( t ρ a ρ + ( 1 t ρ ) b ρ ) d t = 1 ρ ( a ρ b ρ ) [ 1 ( 1 t ρ ) α + 1 t ρ ( α + 1 ) ] f ( t ρ a ρ + ( 1 t ρ ) b ρ ) | 0 1 ρ ( α + 1 ) ρ ( a ρ b ρ ) 0 1 [ ( 1 t ρ ) α t ρ α ] t ρ 1 f ( t ρ a ρ + ( 1 t ρ ) b ρ ) d t = ( α + 1 ) ( b ρ a ρ ) 0 1 [ ( 1 t ρ ) α t ρ α ] t ρ 1 f ( t ρ a ρ + ( 1 t ρ ) b ρ ) d t .
The intended identity in (7) follows from (8) by using (2) and rearranging the terms.  □
We are now in a position to prove our main results.
Theorem 5.
Let α > 0 , ρ > 0 and f : [ a ρ , b ρ ] R be a twice differentiable mapping on ( a ρ , b ρ ) with 0 a < b . If | f | q is strongly η-convex with modulus μ 0 for q 1 , then the following inequality holds:
f ( a ρ ) + f ( b ρ ) 2 ρ α Γ ( α + 1 ) 2 ( b ρ a ρ ) α ρ I a + α f ( b ρ ) + ρ I b α f ( a ρ ) ( b ρ a ρ ) 2 2 ( α + 1 ) [ α + 1 α + 2 1 1 q ( α + 1 α + 2 | f ( a ρ ) | q + α + 1 2 ( α + 3 ) η | f ( b ρ ) | q , | f ( a ρ ) | q μ ( b ρ a ρ ) 2 ( α + 1 ) 2 + 5 ( α + 1 ) 6 ( α + 4 ) ( α + 3 ) ) 1 q + 1 α + 2 1 1 q ( 1 α + 2 | f ( b ρ ) | q + 1 α + 3 η | f ( a ρ ) | q , | f ( b ρ ) | q μ ( b ρ a ρ ) 2 ( α + 3 ) ( α + 4 ) ) 1 q ] .
Proof. 
Using Lemma 3, the well-known power mean inequality and the strong η -convexity of | f | q , we obtain
f ( a ρ ) + f ( b ρ ) α ρ ρ α 1 Γ ( α ) ( b ρ a ρ ) α ρ I a + α f ( b ρ ) + ρ I b α f ( a ρ ) ( b ρ a ρ ) 2 α ( α + 1 ) [ 0 1 1 t ρ ( α + 1 ) t ρ 1 f ( ( 1 t ρ ) a ρ + t ρ b ρ ) d t + 0 1 t ρ ( α + 2 ) 1 f ( t ρ a ρ + ( 1 t ρ ) b ρ ) d t ] ( b ρ a ρ ) 2 α ( α + 1 ) [ 0 1 1 t ρ ( α + 1 ) t ρ 1 d t 1 1 q × 0 1 1 t ρ ( α + 1 ) t ρ 1 f ( ( 1 t ρ ) a ρ + t ρ b ρ ) q d t 1 q + 0 1 t ρ ( α + 2 ) 1 d t 1 1 q 0 1 t ρ ( α + 2 ) 1 f ( t ρ a ρ + ( 1 t ρ ) b ρ ) d t 1 q ] ( b ρ a ρ ) 2 α ( α + 1 ) [ 0 1 1 t ρ ( α + 1 ) t ρ 1 d t 1 1 q × ( 0 1 1 t ρ ( α + 1 ) t ρ 1 ( | f ( a ρ ) | q + t ρ η | f ( b ρ ) | q , | f ( a ρ ) | q μ t ρ ( 1 t ρ ) ( b ρ a ρ ) 2 ) d t ) 1 q + 0 1 t ρ ( α + 2 ) 1 d t 1 1 q ( 0 1 t ρ ( α + 2 ) 1 ( | f ( b ρ ) | q + t ρ η | f ( a ρ ) | q , | f ( b ρ ) | q μ t ρ ( 1 t ρ ) ( b ρ a ρ ) 2 ) d t ) 1 q ] = ( b ρ a ρ ) 2 α ( α + 1 ) [ 0 1 1 t ρ ( α + 1 ) t ρ 1 d t 1 1 q ( | f ( a ρ ) | q 0 1 1 t ρ ( α + 1 ) t ρ 1 d t + η | f ( b ρ ) | q , | f ( a ρ ) | q 0 1 1 t ρ ( α + 1 ) t 2 ρ 1 d t μ ( b ρ a ρ ) 2 0 1 1 t ρ ( α + 1 ) t 2 ρ 1 ( 1 t ρ ) d t ) 1 q
+ 0 1 t ρ ( α + 2 ) 1 d t 1 1 q ( | f ( b ρ ) | q 0 1 t ρ ( α + 2 ) 1 d t + η | f ( a ρ ) | q , | f ( b ρ ) | q 0 1 t ρ ( α + 3 ) 1 d t μ ( b ρ a ρ ) 2 0 1 t ρ ( α + 3 ) 1 ( 1 t ρ ) d t ) 1 q ] .
The desired inequality follows from the above estimation and observing that:
0 1 1 t ρ ( α + 1 ) t ρ 1 d t = α + 1 ρ ( α + 2 ) , 0 1 1 t ρ ( α + 1 ) t 2 ρ 1 d t = α + 1 2 ρ ( α + 3 ) ,
0 1 1 t ρ ( α + 1 ) t 2 ρ 1 ( 1 t ρ ) d t = ( α + 1 ) 2 + 5 ( α + 1 ) 6 ρ ( α + 4 ) ( α + 3 ) , 0 1 t ρ ( α + 2 ) 1 d t = 1 ρ ( α + 2 )
0 1 t ρ ( α + 3 ) 1 d t = 1 ρ ( α + 3 ) a n d 0 1 t ρ ( α + 3 ) 1 ( 1 t ρ ) d t = 1 ρ ( α + 3 ) ( α + 4 ) .
This completes the proof of Theorem 5.  □
Corollary 1.
Let α > 0 , ρ > 0 and f : [ a ρ , b ρ ] R be a twice differentiable mapping on ( a ρ , b ρ ) with 0 a < b . If | f | q is convex for q 1 , then the following inequality holds:
f ( a ρ ) + f ( b ρ ) 2 ρ α Γ ( α + 1 ) 2 ( b ρ a ρ ) α ρ I a + α f ( b ρ ) + ρ I b α f ( a ρ ) ( b ρ a ρ ) 2 2 ( α + 1 ) [ α + 1 α + 2 1 1 q ( ( α + 1 ) ( α + 4 ) 2 ( α + 2 ) ( α + 3 ) | f ( a ρ ) | q + α + 1 2 ( α + 3 ) | f ( b ρ ) | q + 1 α + 2 1 1 q 1 ( α + 2 ) ( α + 3 ) | f ( b ρ ) | q + 1 α + 3 | f ( a ρ ) | q 1 q ] .
Proof. 
The result follows directly from Theorem 5 if we take η ( x , y ) = x y and μ = 0 .  □
Theorem 6.
Let α > 0 , ρ > 0 and f : [ a ρ , b ρ ] R be a twice differentiable mapping on ( a ρ , b ρ ) with 0 a < b . If | f | q is strongly η-convex with modulus μ 0 for q > 1 , then the following inequalities hold:
f ( a ρ ) + f ( b ρ ) 2 ρ α Γ ( α + 1 ) 2 ( b ρ a ρ ) α ρ I a + α f ( b ρ ) + ρ I b α f ( a ρ ) ρ ( b ρ a ρ ) 2 2 ( α + 1 ) [ 1 ρ 0 1 1 u α + 1 s d u 1 s ( 1 ρ | f ( a ρ ) | q + 1 2 ρ η | f ( b ρ ) | q , | f ( a ρ ) | q μ ( b ρ a ρ ) 2 6 ρ ) 1 q + 1 s ρ ( α + 2 ) s + 1 1 s ( | f ( b ρ ) | q + 1 ρ + 1 η | f ( a ρ ) | q , | f ( b ρ ) | q μ ρ ( b ρ a ρ ) 2 ( ρ + 1 ) ( 2 ρ + 1 ) ) 1 q ]
ρ ( b ρ a ρ ) 2 2 ( α + 1 ) [ s ( α + 1 ) ρ ( s ( α + 1 ) + 1 ) 1 s ( 1 ρ | f ( a ρ ) | q + 1 2 ρ η | f ( b ρ ) | q , | f ( a ρ ) | q μ ( b ρ a ρ ) 2 6 ρ ) 1 q + 1 s ρ ( α + 2 ) s + 1 1 s ( | f ( b ρ ) | q + 1 ρ + 1 η | f ( a ρ ) | q , | f ( b ρ ) | q μ ρ ( b ρ a ρ ) 2 ( ρ + 1 ) ( 2 ρ + 1 ) ) 1 q ] ,
where 1 s + 1 q = 1 .
Proof. 
Using Lemma 3, the Hölder’s inequality and the strong η -convexity of | f | q , we obtain
f ( a ρ ) + f ( b ρ ) α ρ ρ α 1 Γ ( α ) ( b ρ a ρ ) α ρ I a + α f ( b ρ ) + ρ I b α f ( a ρ ) ( b ρ a ρ ) 2 α ( α + 1 ) [ 0 1 | 1 t ρ ( α + 1 ) | s t ρ 1 d t 1 s 0 1 t ρ 1 | f ( ( 1 t ρ ) a ρ + t ρ b ρ ) | q d t 1 q + 0 1 t s ρ ( α + 2 ) s d t 1 s 0 1 | f ( t ρ a ρ + ( 1 t ρ ) b ρ ) | q d t 1 q ] ( b ρ a ρ ) 2 α ( α + 1 ) [ 0 1 | 1 t ρ ( α + 1 ) | s t ρ 1 d t 1 s ( 0 1 t ρ 1 ( | f ( a ρ ) | q + t ρ η | f ( b ρ ) | q , | f ( a ρ ) | q μ t ρ ( 1 t ρ ) ( b ρ a ρ ) 2 ) d t ) 1 q + 0 1 t s ρ ( α + 2 ) s d t 1 s ( 0 1 ( | f ( b ρ ) | q + t ρ η | f ( a ρ ) | q , | f ( b ρ ) | q μ t ρ ( 1 t ρ ) ( b ρ a ρ ) 2 ) d t ) 1 q ] = ( b ρ a ρ ) 2 α ( α + 1 ) [ 0 1 1 t ρ ( α + 1 ) s t ρ 1 d t 1 s ( | f ( a ρ ) | q 0 1 t ρ 1 d t + η | f ( b ρ ) | q , | f ( a ρ ) | q 0 1 t 2 ρ 1 d t μ ( b ρ a ρ ) 2 0 1 t 2 ρ 1 ( 1 t ρ ) d t ) 1 q + 0 1 t s ρ ( α + 2 ) s d t 1 s ( | f ( b ρ ) | q 0 1 1 d t + η | f ( a ρ ) | q , | f ( b ρ ) | q 0 1 t ρ d t μ ( b ρ a ρ ) 2 0 1 t ρ ( 1 t ρ ) d t ) 1 q ] = ( b ρ a ρ ) 2 α ( α + 1 ) [ 1 ρ 0 1 1 u α + 1 s d u 1 s ( 1 ρ | f ( a ρ ) | q + 1 2 ρ η | f ( b ρ ) | q , | f ( a ρ ) | q μ ( b ρ a ρ ) 2 6 ρ ) 1 q + 1 s ρ ( α + 2 ) s + 1 1 s ( | f ( b ρ ) | q + 1 ρ + 1 η | f ( a ρ ) | q , | f ( b ρ ) | q μ ρ ( b ρ a ρ ) 2 ( ρ + 1 ) ( 2 ρ + 1 ) ) 1 q ] .
This proves the first inequality. To prove the second inequality, we observe that for any A > B 0 and s 1 , we have ( A B ) s A s B s . Thus, it follows that 1 u α + 1 s 1 u s ( α + 1 ) for all u [ 0 , 1 ] . Hence, we have that
0 1 1 u α + 1 s d u 0 1 1 u s ( α + 1 ) d u = s ( α + 1 ) s ( α + 1 ) + 1 .
This completes the proof.  □
Corollary 2.
Let α > 0 , ρ > 0 and f : [ a ρ , b ρ ] R be a twice differentiable mapping on ( a ρ , b ρ ) with 0 a < b . If | f | q is convex for q > 1 , then the following inequalities hold:
f ( a ρ ) + f ( b ρ ) 2 ρ α Γ ( α + 1 ) 2 ( b ρ a ρ ) α ρ I a + α f ( b ρ ) + ρ I b α f ( a ρ ) ρ ( b ρ a ρ ) 2 2 ( α + 1 ) [ 1 ρ 0 1 | 1 u α + 1 | s d u 1 s 1 2 ρ | f ( a ρ ) | q + 1 2 ρ | f ( b ρ ) | q 1 q + 1 s ρ ( α + 2 ) s + 1 1 s ρ ρ + 1 | f ( b ρ ) | q + 1 ρ + 1 | f ( a ρ ) | q 1 q ] ρ ( b ρ a ρ ) 2 2 ( α + 1 ) [ s ( α + 1 ) ρ ( s ( α + 1 ) + 1 ) 1 s 1 2 ρ | f ( a ρ ) | q + 1 2 ρ | f ( b ρ ) | q 1 q + 1 s ρ ( α + 2 ) s + 1 1 s ρ ρ + 1 | f ( b ρ ) | q + 1 ρ + 1 | f ( a ρ ) | q 1 q ] ,
where 1 s + 1 q = 1 .
Proof. 
The result follows directly from Theorem 6 if we take η ( x , y ) = x y and μ = 0 .  □
Theorem 7.
Let α > 0 , ρ > 0 and f : [ a ρ , b ρ ] R be a twice differentiable mapping on ( a ρ , b ρ ) with 0 a < b . If | f | q is a strongly η-convex function on [ a ρ , b ρ ] with modulus μ 0 for q 1 , then the following inequality holds:
f ( a ρ ) + f ( b ρ ) 2 ρ α Γ ( α + 1 ) 2 ( b ρ a ρ ) α ρ I a + α f ( b ρ ) + ρ I b α f ( a ρ ) ( b ρ a ρ ) 2 2 ρ ( α + 1 ) α α + 2 1 1 q [ α α + 2 | f ( b ρ ) | q + α + 1 2 ( α + 3 ) B ( 2 , α + 2 ) η | f ( a ρ ) | q , | f ( b ρ ) | q μ ( b ρ a ρ ) 2 1 6 2 B ( 2 , α + 3 ) ] 1 q ,
where B ( · , · ) denotes the beta function defined by B ( x , y ) = 0 1 t x 1 ( 1 t ) y 1 d t .
Proof. 
Using Lemma 4, the power mean inequality and the strong η -convexity of | f | q , we obtain
f ( a ρ ) + f ( b ρ ) 2 ρ α Γ ( α + 1 ) 2 ( b ρ a ρ ) α ρ I a + α f ( b ρ ) + ρ I b α f ( a ρ ) ( b ρ a ρ ) 2 2 ( α + 1 ) 0 1 | 1 ( 1 t ρ ) α + 1 t ρ ( α + 1 ) | t ρ 1 | f ( t ρ a ρ + ( 1 t ρ ) b ρ ) | d t ( b ρ a ρ ) 2 2 ( α + 1 ) 0 1 1 ( 1 t ρ ) α + 1 t ρ ( α + 1 ) t ρ 1 d t 1 1 q × 0 1 1 ( 1 t ρ ) α + 1 t ρ ( α + 1 ) t ρ 1 | f ( t ρ a ρ + ( 1 t ρ ) b ρ ) | q d t 1 q ( b ρ a ρ ) 2 2 ( α + 1 ) 0 1 1 ( 1 t ρ ) α + 1 t ρ ( α + 1 ) t ρ 1 d t 1 1 q × ( 0 1 1 ( 1 t ρ ) α + 1 t ρ ( α + 1 ) t ρ 1 ( | f ( b ρ ) | q + t ρ η | f ( a ρ ) | q , | f ( b ρ ) | q μ t ρ ( 1 t ρ ) ( b ρ a ρ ) 2 ) d t ) 1 q ( b ρ a ρ ) 2 2 ( α + 1 ) 0 1 1 ( 1 t ρ ) α + 1 t ρ ( α + 1 ) t ρ 1 d t 1 1 q × ( | f ( b ρ ) | q 0 1 1 ( 1 t ρ ) α + 1 t ρ ( α + 1 ) t ρ 1 d t + η | f ( a ρ ) | q , | f ( b ρ ) | q 0 1 1 ( 1 t ρ ) α + 1 t ρ ( α + 1 ) t 2 ρ 1 d t μ ( b ρ a ρ ) 2 0 1 1 ( 1 t ρ ) α + 1 t ρ ( α + 1 ) t 2 ρ 1 ( 1 t ρ ) d t ) 1 q .
The desired result follows from the above inequality and using the following computations:
0 1 1 ( 1 t ρ ) α + 1 t ρ ( α + 1 ) t ρ 1 d t = 1 ρ 0 1 1 ( 1 u ) α + 1 u α + 1 d u = α ρ ( α + 2 ) ,
0 1 1 ( 1 t ρ ) α + 1 t ρ ( α + 1 ) t 2 ρ 1 d t = 1 ρ 0 1 1 ( 1 u ) α + 1 u α + 1 u d u = 1 ρ 1 2 B ( 2 , α + 2 ) 1 α + 3 = 1 ρ α + 1 2 ( α + 3 ) B ( 2 , α + 2 )
and
0 1 1 ( 1 t ρ ) α + 1 t ρ ( α + 1 ) t 2 ρ 1 ( 1 t ρ ) d t = 1 ρ 0 1 1 ( 1 u ) α + 1 u α + 1 u ( 1 u ) d u = 1 ρ 1 6 2 B ( 2 , α + 3 ) .
This completes the proof of the theorem.  □
Corollary 3.
Let α > 0 , ρ > 0 and f : [ a ρ , b ρ ] R be a twice differentiable mapping on ( a ρ , b ρ ) with 0 a < b . If | f | q is convex for q 1 , then the following inequality holds:
f ( a ρ ) + f ( b ρ ) 2 ρ α Γ ( α + 1 ) 2 ( b ρ a ρ ) α ρ I a + α f ( b ρ ) + ρ I b α f ( a ρ ) ( b ρ a ρ ) 2 2 ρ ( α + 1 ) α α + 2 1 1 q [ B ( 2 , α + 2 ) α 2 + 3 α 2 2 ( α + 2 ) ( α + 3 ) | f ( b ρ ) | q + α + 1 2 ( α + 3 ) B ( 2 , α + 2 ) | f ( a ρ ) | q ] 1 q .
Proof. 
The result follows directly from Theorem 7 if we take η ( x , y ) = x y and μ = 0 .  □
Theorem 8.
Let α > 0 , ρ > 0 and f : [ a ρ , b ρ ] R be a twice differentiable mapping on ( a ρ , b ρ ) with 0 a < b . If | f | q is a strongly η-convex function on [ a ρ , b ρ ] with modulus μ 0 for q > 1 , then the following inequalities hold:
f ( a ρ ) + f ( b ρ ) 2 ρ α Γ ( α + 1 ) 2 ( b ρ a ρ ) α ρ I a + α f ( b ρ ) + ρ I b α f ( a ρ ) ( b ρ a ρ ) 2 2 ρ ( α + 1 ) 0 1 1 ( 1 u ) α + 1 u α + 1 s d u 1 s × | f ( b ρ ) | q + 1 2 η | f ( a ρ ) | q , | f ( b ρ ) | q μ ρ ( b ρ a ρ ) 2 2 ( ρ + 1 ) ( 2 ρ + 1 ) 1 q ( b ρ a ρ ) 2 2 ρ ( α + 1 ) s ( α + 1 ) 1 s ( α + 1 ) + 1 1 s ( | f ( b ρ ) | q + 1 2 η | f ( a ρ ) | q , | f ( b ρ ) | q μ ρ ( b ρ a ρ ) 2 2 ( ρ + 1 ) ( 2 ρ + 1 ) ) 1 q ,
where 1 s + 1 q = 1 .
Proof. 
Using Lemma 4, the Hölder’s inequality and the strong η -convexity of | f | q , we obtain
f ( a ρ ) + f ( b ρ ) 2 ρ α Γ ( α + 1 ) 2 ( b ρ a ρ ) α ρ I a + α f ( b ρ ) + ρ I b α f ( a ρ ) ( b ρ a ρ ) 2 2 ( α + 1 ) 0 1 | 1 ( 1 t ρ ) α + 1 t ρ ( α + 1 ) | t ρ 1 | f ( t ρ a ρ + ( 1 t ρ ) b ρ ) | d t ( b ρ a ρ ) 2 2 ( α + 1 ) 0 1 1 ( 1 t ρ ) α + 1 t ρ ( α + 1 ) s t ρ 1 d t 1 s × 0 1 t ρ 1 | f ( t ρ a ρ + ( 1 t ρ ) b ρ ) | q d t 1 q ( b ρ a ρ ) 2 2 ( α + 1 ) 0 1 1 ( 1 t ρ ) α + 1 t ρ ( α + 1 ) s t ρ 1 d t 1 s × ( 0 1 t ρ 1 ( | f ( b ρ ) | q + t ρ η | f ( a ρ ) | q , | f ( b ρ ) | q μ t ρ ( 1 t ρ ) ( b ρ a ρ ) 2 ) d t ) 1 q
( b ρ a ρ ) 2 2 ( α + 1 ) 0 1 1 ( 1 t ρ ) α + 1 t ρ ( α + 1 ) s t ρ 1 d t 1 s × ( | f ( b ρ ) | q 0 1 t ρ 1 d t + η | f ( a ρ ) | q , | f ( b ρ ) | q 0 1 t 2 ρ 1 d t μ ( b ρ a ρ ) 2 0 1 t 2 ρ 1 ( 1 t ρ ) d t ) 1 q ,
where
0 1 1 ( 1 t ρ ) α + 1 t ρ ( α + 1 ) s t ρ 1 d t = 1 ρ 0 1 1 ( 1 u ) α + 1 u α + 1 s d u ,
0 1 t ρ 1 d t = 1 ρ , 0 1 t 2 ρ 1 d t = 1 2 ρ a n d 0 1 t 2 ρ 1 ( 1 t ρ ) d t = 1 2 ( ρ + 1 ) ( 2 ρ + 1 ) .
This proves the first inequality. Using a similar argument as in the proof of Theorem 6, we obtain
0 1 1 ( 1 u ) α + 1 u α + 1 s d u 0 1 1 ( 1 u ) s ( α + 1 ) u s ( α + 1 ) d u = 1 2 s ( α + 1 ) + 1 = s ( α + 1 ) 1 s ( α + 1 ) + 1 .
This completes the proof of the theorem.  □
Corollary 4.
Let α > 0 , ρ > 0 and f : [ a ρ , b ρ ] R be a twice differentiable mapping on ( a ρ , b ρ ) with 0 a < b . If | f | q is convex for q > 1 , then the following inequalities hold:
f ( a ρ ) + f ( b ρ ) 2 ρ α Γ ( α + 1 ) 2 ( b ρ a ρ ) α ρ I a + α f ( b ρ ) + ρ I b α f ( a ρ ) ( b ρ a ρ ) 2 2 ρ ( α + 1 ) 0 1 1 ( 1 u ) α + 1 u α + 1 s d u 1 s × 1 2 | f ( b ρ ) | q + 1 2 | f ( a ρ ) | q 1 q ( b ρ a ρ ) 2 2 ρ ( α + 1 ) s ( α + 1 ) 1 s ( α + 1 ) + 1 1 s × 1 2 | f ( b ρ ) | q + 1 2 | f ( a ρ ) | q 1 q ,
where 1 s + 1 q = 1 .
Proof. 
The result follows directly from Theorem 8 if we take η ( x , y ) = x y and μ = 0 .  □

3. Conclusions

Four main results related to the Hermite–Hadamard inequality via the Katugampola fractional integrals involving strongly η -convex functions have been introduced. Similar results via the Riemann–Liouville and Hadamard fractional integrals could be derived as particular cases by taking ρ 1 and ρ 0 + , respectively. Several other interesting results can be obtained by considering different bifunctions η and/or the modulus μ as well as different values for the parameters α and ρ .

Author Contributions

S.K., E.R.N and A.M.T contributed equally to this work.

Funding

This research received no external funding.

Acknowledgments

We kindly appreciate the efforts of the anonymous referees for their valuable comments and suggestions.

Conflicts of Interest

The authors declare no conflict of interest.

References

  1. Hadamard, J. Etude sur les properties des fonctions entries et an particular d’une fonction considree par, Riemann. J. Math. Pures. Appl. 1893, 58, 171–215. [Google Scholar]
  2. Alomari, M.; Darus, M.; Dragomir, S.S. New inequalities of Hermite–Hadamard’s type for functions whose second derivatives absolute values are quasiconvex. Tamkang J. Math. 2010, 41, 353–359. [Google Scholar]
  3. Chun, L.; Qi, F. Integral inequalities of Hermite–Hadamard type for functions whose third derivatives are convex. J. Inequal. Appl. 2013, 2013, 451. [Google Scholar] [CrossRef]
  4. Chun, L.; Qi, F. Integral inequalities of Hermite–Hadamard type for functions whose 3rd derivatives are s-convex. Appl. Math. 2012, 3, 1680–1685. [Google Scholar] [CrossRef]
  5. Dragomir, S.S. Two mappings in connection to Hadamard’s inequalities. J. Math. Anal. Appl. 1992, 167, 49–56. [Google Scholar] [CrossRef]
  6. Dragomir, S.S.; Agarwal, R.P. Two inequalities for differentiable mappings and their applications to special means for real numbers and to trapezoidal formula. Appl. Math. Lett. 1998, 11, 91–95. [Google Scholar] [CrossRef]
  7. Farid, G.; Rehman, A.U.; Zahra, M. On Hadamard-type inequalities for k-fractional integrals. Konulrap J. Math. 2016, 4, 79–86. [Google Scholar]
  8. Khan, M.A.; Khurshid, Y.; Ali, T. Hermite–Hadamard inequality for fractional integrals via η-convex functions. Acta Math. Univ. Comen. 2017, LXXXVI (1), 153–164. [Google Scholar]
  9. Nwaeze, E.R. Inequalities of the Hermite–Hadamard type for Quasi-convex functions via the (k,s)-Riemann–Liouville fractional integrals. Fract. Differ. Calc. 2018, 8, 327–336. [Google Scholar] [CrossRef]
  10. Nwaeze, E.R.; Kermausuor, S.; Tameru, A.M. Some new k-Riemann–Liouville Fractional integral inequalities associated with the strongly η-quasiconvex functions with modulus μ≥0. J. Inequal. Appl. 2018, 2018, 139. [Google Scholar] [CrossRef] [PubMed]
  11. Sarikaya, M.Z.; Set, E.; Yaldiz, H.; Basak, N. Hermite–Hadamard’s inequalities for fractional integrals and related fractional inequalities. Math. Comput. Model. 2013, 57, 2403–2407. [Google Scholar] [CrossRef]
  12. Gordji, M.E.; Delavar, M.R.; de la Sen, M. On φ-convex functions. J. Math. Inequal. 2016, 10, 173–183. [Google Scholar] [CrossRef]
  13. Awan, M.U.; Noor, M.A.; Noor, K.I.; Safdar, F. On strongly generalized convex functions. Filomat 2017, 31, 5783–5790. [Google Scholar] [CrossRef]
  14. Gordji, M.E.; Delavar, M.R.; Dragomir, S.S. Some inequalities related to η-convex functions. Preprint RGMIA Res. Rep. Coll. 2015, 18, 1–14. [Google Scholar]
  15. Gordji, M.E.; Dragomir, S.S.; Delavar, M.R. An inequality related to η-convex functions (II). Int. J. Nonlinear Anal. Appl. 2015, 6, 27–33. [Google Scholar]
  16. Nwaeze, E.R.; Torres, D.F.M. Novel results on the Hermite–Hadamard kind inequality for h-convex funxtions by means of the (k, r)-fractional integral operators. In Advances in Mathematical Inequalities and Applications (AMIA); Trends in Mathematics; Dragomir, S.S., Agarwal, P., Jleli, M., Samet, B., Eds.; Birkhäuser: Singapore, 2018; pp. 311–321. [Google Scholar]
  17. Podlubny, I. Fractional Differential Equations: Mathematics in Science and Engineering; Academic Press: San Diego, CA, USA, 1999. [Google Scholar]
  18. Samko, S.G.; Kilbas, A.A.; Marichev, O.I. Fractional Integrals and Derivatives: Theory and Applications; Gordon and Breach: Amsterdam, The Netherlands, 1993. [Google Scholar]
  19. Katugampola, U.N. New approach to a generalized fractional integral. Appl. Math. Comput. 2011, 218, 860–865. [Google Scholar] [CrossRef] [Green Version]
  20. Chen, H.; Katugampola, U.N. Hermite–Hadamard and Hermite–Hadamard–Fejér type inequalities for generalized fractional integrals. J. Math. Anal. Appl. 2017, 446, 1274–1291. [Google Scholar] [CrossRef] [Green Version]
  21. Katugampola, U.N. A new approach to generalized fractional derivatives. Bull. Math. Anal. Appl. 2014, 6, 1–15. [Google Scholar]

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Kermausuor, S.; Nwaeze, E.R.; Tameru, A.M. New Integral Inequalities via the Katugampola Fractional Integrals for Functions Whose Second Derivatives Are Strongly η-Convex. Mathematics 2019, 7, 183. https://doi.org/10.3390/math7020183

AMA Style

Kermausuor S, Nwaeze ER, Tameru AM. New Integral Inequalities via the Katugampola Fractional Integrals for Functions Whose Second Derivatives Are Strongly η-Convex. Mathematics. 2019; 7(2):183. https://doi.org/10.3390/math7020183

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Kermausuor, Seth, Eze R. Nwaeze, and Ana M. Tameru. 2019. "New Integral Inequalities via the Katugampola Fractional Integrals for Functions Whose Second Derivatives Are Strongly η-Convex" Mathematics 7, no. 2: 183. https://doi.org/10.3390/math7020183

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