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Some Hermite-Hadamard and Ostrowski type inequalities for fractional integral operators with exponential kernel

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Volume 23, Number 1, June 2019

Available online at http://acutm.math.ut.ee

Some Hermite–Hadamard and Ostrowski type

inequalities for fractional integral operators with

exponential kernel

H¨useyin Budak, Mehmet Zeki Sarikaya, Fuat Usta, and H¨useyin Yildirim

Abstract. We firstly establish Hermite–Hadamard type integral in-equalities for fractional integral operators. Secondly, we give new gener-alizations of fractional Ostrowski type inequalities through convex func-tions via H¨older and power means inequalities. In accordance with this purpose, we use fractional integral operators with exponential kernel.

1. Introduction

The study of various types of integral inequalities has been the focus of great attention for well over a century by a number of scientists, interested both in pure and applied mathematics. On the other hand, the concept of fractional calculus got wide use in numerous physical and other applications, as viscoelastic materials, fluid flow, diffusive transport, electrical networks, electromagnetic theory, probability, and others. We will summarize these two concepts below.

1.1. Hermite–Hadamard inequality. The inequalities discovered by C. Hermite and J. Hadamard for convex functions are of considerable signif-icance in the literature (see, e.g., [8, p. 137], [2]). These inequalities state that if f : I → R is a convex function on an interval I of real numbers and a, b ∈ I with a < b, then f a + b 2  ≤ 1 b − a Z b a f (x)dx ≤ f (a) + f (b) 2 . (1.1) Received January 9, 2018.

2010 Mathematics Subject Classification. 26D15; 26B25; 26D10; 26A51.

Key words and phrases. Hermite–Hadamard inequality; Ostrowski inequality; fractional integral operators; convex function.

http://dx.doi.org/10.12697/ACUTM.2019.23.03 Corresponding author: Fuat Usta

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Both inequalities hold in the reverse direction if f is concave. We note that Hadamard’s inequality may be regarded as a refinement of the concept of convexity, and it follows easily from Jensen’s inequality. Hadamard’s inequality for convex functions has received renewed attention in recent years and a remarkable variety of refinements and generalizations have been found (see, for example, [3], [9], [10], and the references therein).

1.2. Ostrowski inequality. One of the many fundamental mathematical discoveries of A. M. Ostrowski [7] is the following classical integral inequality associated with differentiable mappings.

Theorem 1. Let f : [a, b] → R be a differentiable mapping on (a, b) whose derivative f0: (a, b) → R is bounded on (a, b), i.e., kf0k:= sup

t∈(a,b)

|f0(t)| <

∞. Then, we have the inequality f (x) − 1 b − a b Z a f (t)dt ≤ " 1 4+ x −a+b2 2 (b − a)2 # (b − a) f0

for all x ∈ [a, b]. The constant 14 is the best possible.

Ostrowski’s inequality has applications in quadrature, probability and op-timization theory, stochastic, statistics, information, and integral operator theory. During the past few years, a number of scientists have focused on Ostrowski’s type inequalities for functions of bounded variation (see, for example, [1], [4], [6], [11], and the references therein).

1.3. Fractional integral operators with exponential kernel. Recently, Kirane and Torebek [5] introduced a new class of fractional integrals as fol-lows.

Definition 1. Let f ∈ L1(a, b) . The fractional integrals Ia+α and Ib−α of

order α ∈ (0, 1) are defined by Ia+α f (x) : = 1 α x Z a exp  −1 − α α (x − t)  f (t)dt, x > a, and Ib−α f (x) : = 1 α b Z x exp  −1 − α α (t − x)  f (t)dt, x < b.

The remainder of this work is organized as follows. In Section 2, we will present a new Hermite–Hadamard type integral inequality via fractional cal-culus mentioned above. In Section 3, new Ostrowski type integral inequali-ties are proved via fractional integral operators with exponential kernel.

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2. Hermite–Hadamard type inequalities via fractional integral operators with exponential kernel

In this section, we will present some significant results for Hermite– Hadamard type inequalities with fractional integral operators with exponen-tial kernel. Throughout this section, we denote A : = 1−αα b−a2  for α ∈ (0, 1). Theorem 2. Let f : [a, b] → R be a function with 0 ≤ a < b and f ∈ L1[a, b] . If f is a convex function on [a, b], then we have the

follow-ing inequalities for fractional integral operators with exponential kernel: f a + b 2  ≤ 1 − α 2 [1 − exp {−A}]  Iα (a+b 2 )+ f (b) + Iα (a+b 2 )− f (a)  ≤ f (a) + f (b) 2 . (2.1) Proof. Since f is a convex function on [a, b] , we have

f x + y 2



≤ f (x) + f (y)

2 , x, y ∈ [a, b] . For x = 2ta +2−t2 b and y = 2−t2 a + t2b, we obtain

2f a + b 2  ≤ f t 2a + 2 − t 2 b  + f 2 − t 2 a + t 2b  . (2.2)

Multiplying both sides of (2.2) by exp {−At} and integrating the resulting inequality with respect to t over [0, 1] , we get

2f a + b 2 Z1 0 exp {−At} dt ≤ 1 Z 0 exp {−At} f t 2a + 2 − t 2 b  dt + 1 Z 0 exp {−At} f 2 − t 2 a + t 2b  dt.

For u = 2ta +2−t2 b and v = 2−t2 a +2tb, we obtain 2 Af  a + b 2  [1 − exp {−A}] ≤ 2 b − a b Z a+b 2 exp  −1 − α α (b − u)  f (u) du + 2 b − a a+b 2 Z a exp  −1 − α α (v − a)  f (v) dv = 2α b − a  Iα (a+b 2 )+ f (b) + Iα (a+b 2 )− f (a)  , and the first inequality is proved.

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For the proof of the second inequality in (2.1), we first note that if f is a convex function, then

f t 2a + 2 − t 2 b  ≤ t 2f (a) + 2 − t 2 f (b) and f 2 − t 2 a + t 2b  ≤ 2 − t 2 f (a) + t 2f (b). By adding these inequalities, we have

f t 2a + 2 − t 2 b  + f 2 − t 2 a + t 2b  ≤ f (a) + f (b). (2.3) Then multiplying both sides of (2.3) by exp {−At} and integrating the re-sulting inequality with respect to t over [0, 1] , we obtain

1 Z 0 exp {−At} f t 2a + 2 − t 2 b  dt + 1 Z 0 exp {−At} f 2 − t 2 a + t 2b  dt ≤ [f (a) + f (b)] 1 Z 0 exp {−At} dt = 1

A[1 − exp {−A}] [f (a) + f (b)] . That is, 2α b − a  Iα (a+b 2 )+ f (b) + I(αa+b 2 )− f (a)  ≤ 1

A[1 − exp {−A}] [f (a) + f (b)] .

The proof is complete. 

Remark 1. Since lim

α→1

1−α 2[1−exp{−A}] =

1

b−a, the inequality (2.1) reduces to

the classical Hermite–Hadamard inequality (1.1).

Lemma 1. Let f : [a, b] → R be a differentiable function on (a, b) with a < b. If f0 ∈ L [a, b] , then we have the following identity for generalized fractional integral operators with exponential kernel:

1 − α 2 [1 − exp {−A}]  Iα (a+b 2 )+ f (b) + I(αa+b 2 )− f (a)  − f a + b 2  = 1 2 [1 − exp {−A}]     b Z a+b 2  1 − exp  −1 − α α (b − u)  f0(u) du − a+b 2 Z a  1 − exp  −1 − α α (u − a)  f0(u) du   . (2.4)

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Proof. Integrating by parts gives J1 : = b Z a+b 2  1 − exp  −1 − α α (b − u)  f0(u) du = f (u)  1 − exp  −1 − α α (b − u)  b a+b 2 + 1 − α α b Z a+b 2 exp  −1 − α α (b − u)  f (u) du = − [1 − exp {−A}] f a + b 2  + (1 − α) I(αa+b 2 )+ f (b). (2.5) Similarly, we get J2 : = a+b 2 Z a  1 − exp  −1 − α α (u − a)  f0(u) du = [1 − exp {−A}] f a + b 2  − (1 − α) Iα (a+b 2 )− f (a). (2.6)

By subtracting (2.6) from (2.5), we have J1− J2 = −2 [1 − exp {−A}] f  a + b 2  + (1 − α)  Iα (a+b 2 )+ f (b) + I(αa+b 2 )− f (a)  .

By re-arranging the last equality above, we get the desired result.  Theorem 3. Let f : [a, b] → R be a function with 0 ≤ a < b and f ∈ L1[a, b] . If f0 is bounded on [a, b], then we have the inequality

1 − α 2 [1 − exp {−A}]  Iα (a+b 2 )+ f (b) + I(αa+b 2 )− f (a)  − f a + b 2  ≤ (1 − α) (b − a) − 2α [1 − exp {−A}] 2 (1 − α) [1 − exp {−A}] f0 . Proof. Using Lemma 1, we have

M : = 1 − α 2 [1 − exp {−A}]  Iα (a+b 2 )+ f (b) + Iα (a+b 2 )− f (a)  − f a + b 2 

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≤ 1 2 [1 − exp {−A}]     b Z a+b 2  1 − exp  −1 − α α (b − u)  f0(u) du + a+b 2 Z a  1 − exp  −1 − α α (u − a)  f0(u) du   .

Since f0 is bounded on [a, b] , we deduce that

M ≤ kf 0k ∞ 2 [1 − exp {−A}]     b Z a+b 2  1 − exp  −1 − α α (b − u)  du + a+b 2 Z a  1 − exp  −1 − α α (u − a)  du    = (1 − α) (b − a) − 2α [1 − exp {−A}] 2 (1 − α) [1 − exp {−A}] f0 ,

which completes the proof. 

Remark 2. Since lim α→1 1 − α 2 [1 − exp {−A}] = 1 b − a and lim α→1 (1 − α) (b − a) − 2α [1 − exp {−A}] 2 (1 − α) [1 − exp {−A}] = b − a 4 , we have the midpoint inequality

1 b − a b Z a f (x)dx − f a + b 2  ≤ b − a 4 f0 .

3. Ostrowski type inequalities via fractional integral operators with exponential kernel

Throughout this section, we denote θa: = 1−αα (x − a) and θb: = 1−αα (b − x)

for α ∈ (0, 1).

Lemma 2. Let f : [a, b] → R be a differentiable function on (a, b) with a < b. If f0 ∈ L [a, b] , then we have the following identity for generalized

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fractional integral operators with exponential kernel: f (x) − 1 − α 2 − exp {−θa} − exp {−θb} Iα x−f (a) + Ix+α f (b)  = x − a 2 − exp {−θa} − exp {−θb} 1 Z 0 [1 − exp {−θat}] f0(tx + (1 − t)a)dt − b − x 2 − exp {−θa} − exp {−θb} 1 Z 0 [1 − exp {−θbt}] f0(tx + (1 − t)b)dt.

Proof. Integrating by parts, we have J3 : = 1 Z 0 [1 − exp {−θat}] f0(tx + (1 − t)a)dt = 1 x − a[1 − exp {−θat}] f (tx + (1 − t)a) 1 0 − θa x − a 1 Z 0 exp {−θat} f (tx + (1 − t)a)dt = 1 − exp {−θa} x − a f (x) − 1 − α x − aI α x−f (a). Similarly, J4: = 1 Z 0 [1 − exp {−θbt}] f0(tx + (1 − t)b)dt = −1 − exp {−θb} b − x f (x) + 1 − α b − xI α x+f (b). It follows that (x − a)J3− (b − x)J4 2 − exp {−θa} − exp {−θb} = f (x) − 1 − α 2 − exp {−θa} − exp {−θb} Iα x−f (a) + Ix+α f (b) ,

which completes the proof. 

Theorem 4. Let f : [a, b] → R be a function with 0 ≤ a < b and f ∈ L1[a, b] . If |f0|q, q > 1, is a convex function on [a, b] , then we have the

following inequality for fractional integral operators with exponential kernel: f (x) − 1 − α 2 − exp {−θa} − exp {−θb} Iα x−f (a) + Ix+α f (b) 

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≤ 1 2 − exp {−θa} − exp {−θb} × " (x − a)A1(α, p)  |f0(x)|q + |f0(a)|q 2 1q + (b − x)A2(α, p)  |f0(x)|q+ |f0(b)|q 2 1q# , where 1p +1q = 1, and A1(α, p) :=   1 Z 0 [1−exp{−θat}]pdt   1 p , A2(α, p) :=   1 Z 0 [1−exp{−θbt}]pdt   1 p .

Proof. Taking the modulus in Lemma 2 and using the fact that ex > 1, x > 0, we obtain f (x) − 1 − α 2 − exp {−θa} − exp {−θb} Iα x−f (a) + Ix+α f (b)  ≤ x − a 2 − exp {−θa} − exp {−θb} J3+ b − x 2 − exp {−θa} − exp {−θb} J4. (3.1)

Using H¨older’s inequality and convexity of |f0|q, we find that

1 Z 0 [1 − exp {−θat}] f0(tx + (1 − t)a) dt ≤   1 Z 0 [1 − exp {−θat}]pdt   1 p  1 Z 0 f0(tx + (1 − t)a) q dt   1 q ≤ A1(α, p)  |f0(x)|q+ |f0(a)|q 2 1q (3.2) and, similarly, 1 Z 0 [1 − exp {−θbt}] f0(tx + (1 − t)b) dt ≤ A2(α, p) |f 0(x)|q + |f0(b)|q 2 1q . (3.3)

By substituting the inequalities (3.2) and (3.3) into (3.1), we have f (x) − 1 − α 2 − exp {−θa} − exp {−θb} Iα x−f (a) + Ix+α f (b) 

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≤ 1 2 − exp {−θa} − exp {−θb} " (x − a)A1(α, p)  |f0(x)|q + |f0(a)|q 2 1q + (b − x)A2(α, p)  |f0(x)|q+ |f0(b)|q 2 1q# .

This completes the proof. 

Theorem 5. Let f : [a, b] → R be a function with 0 ≤ a < b and f ∈ L1[a, b] . If |f0|q, q ≥ 1, is a convex function on [a, b] , then we have the

inequality f (x) − 1 − α 2 − exp {−θa} − exp {−θb} Iα x−f (a) + Ix+α f (b)  ≤ 1 2 − exp {−θa} − exp {−θb} × ( x − a θ1− 1 q a [1 + θa− exp {−θa}]1− 1 q ×  f0(x) q 1 2 − 1 θ2 a [1 − exp {−θa} (θa+ 1)]  + f0(a) q 1 2− 1 θ2 a [θa+ exp {−θa} − 1] 1q +b − x θ1− 1 q b (1+θb−exp {−θb})1− 1 q  f0(x) q 1 2− 1 θ2b [1−exp {−θb} (θb+1)]  + f0(b) q 1 2 − 1 θ2b [θb+ exp {−θb} − 1] 1q) .

Proof. Using power means inequality and convexity of |f0|q in (3.1), we find that f (x) − 1 − α 2 − exp {−θa} − exp {−θb} Iα x−f (a) + Ix+α f (b)  ≤ x − a 2 − exp {−θa} − exp {−θb}   1 Z 0 [1 − exp {−θat}] dt   1−1q ×   1 Z 0 [1 − exp {−θat}] f0(tx + (1 − t)a) q dt   1 q (3.4)

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+ b − x 2 − exp {−θa} − exp {−θb}   1 Z 0 [1 − exp {−θbt}] dt   1−1q ×   1 Z 0 [1 − exp {−θbt}] f0(tx + (1 − t)b) q dt   1 q .

Since |f0|q is a convex function, we have

1 Z 0 [1 − exp {−θat}] f0(tx + (1 − t)a) q dt ≤ 1 Z 0 [1 − exp {−θat}]t f0(x) q + (1 − t) f0(a) q dt = f0(x) q 1 2 − 1 θ2 a [1 − exp {−θa} (θa+ 1)]  + f0(a) q 1 2− 1 θ2 a [θa+ exp {−θa} − 1]  (3.5) and, similarly, 1 Z 0 [1 − exp {−θbt}] f0(tx + (1 − t)b) q dt ≤ f0(x) q 1 2 − 1 θb2 [1 − exp {−θb} (θb+ 1)]  + f0(b) q 1 2 − 1 θ2b [θb+ exp {−θb} − 1]  . (3.6)

On the other hand, we have

1 Z 0 [1 − exp {−θat}] dt = 1 θa (1 + θa− exp {−θa}) (3.7) and, similarly, 1 Z 0 [1 − exp {−θbt}] dt = 1 θb (1 + θb− exp {−θb}) . (3.8)

Then, substituting the inequalities (3.5)–(3.8) into (3.4), we obtain the

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Corollary 1. Suppose that the assumptions of Theorem 5 hold for q = 1. Then we have the following inequality for fractional integral operators with exponential kernel: f (x) − 1 − α 2 − exp {−θa} − exp {−θb} Iα x−f (a) + Ix+α f (b)  ≤ x − a 2 − exp {−θa} − exp {−θb} × 1 2− 1 θ2 a [1 − exp {−θa} (θa+ 1)]  f0(x) + 1 2 − 1 θ2 a [θa+ exp {−θa} − 1]  f0(a)  + b − x 2 − exp {−θa} − exp {−θb} × 1 2− 1 θb2[1 − exp {−θb} (θb+ 1)]  f0(x) + 1 2 − 1 θ2 b [θb+ exp {−θb} − 1]  f0(b)  . References

[1] S. S. Dragomir, The Ostrowski integral inequality for mappings of bounded variation, Bull. Austral. Math. Soc. 60(3) (1999), 495–508.

[2] S. S. Dragomir and C. E. M. Pearce, Selected Topics on Hermite-Hadamard Inequali-ties and Applications, RGMIA Monographs, Victoria University, 2000.

[3] M. Jleli and B. Samet, On Hermite-Hadamard type inequalities via fractional integrals of a function with respect to another function, J. Nonlinear Sci. Appl. 9 (2016), 1252– 1260.

[4] A. R. Kashif, M. Shoaib, and M. A. Latif, Improved version of perturbed Ostrowski type inequalities for n-times differentiable mappings with three-step kernel and its application, J. Nonlinear Sci. Appl. 9 (2016), 3319–3332.

[5] M. Kirane and B. T. Torebek, Hermite-Hadamard, Hermite-Hadamard-Fejer, Dragomir-Agarwal and Pachpatte type inequalities for convex functions via fractional integrals, arXiv:1701.00092.

[6] W. Liu, New bounds for the companion of Ostrowski’s inequality and applications, Filomat 28 (2014), 167–178.

[7] A. M. Ostrowski, ¨Uber die Absolutabweichung einer differentierbaren Funktion von ihrem Integralmittelwert, Comment. Math. Helv. 10 (1937), 226–227.

[8] J. E. Peˇcari´c, F. Proschan, and Y. L. Tong, Convex Functions, Partial Orderings and Statistical Applications, Academic Press, Boston, 1992.

[9] M. Z. Sarikaya and H. Budak, Generalized Hermite-Hadamard type integral inequali-ties for fractional integrals, Filomat 30 (2016), 1315–1326.

[10] M. Z. Sarikaya and H. Yildirim, On Hermite-Hadamard type inequalities for Riemann-Liouville fractional integrals, Miskolc Math. Notes 17(2) (2016), 1049–1059.

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[11] 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(7) 2012, 1147– 1154.

Department of Mathematics, Faculty of Science and Arts, D¨uzce Univer-sity, D¨uzce, Turkey

E-mail address: hsyn.budak@gmail.com E-mail address: sarikayamz@gmail.com E-mail address: fuatusta@duzce.edu.tr

Department of Mathematics, Faculty of Science and Arts, University of Kahramanmaras¸ S¨utc¸¨u ˙Imam, 46100, Kahramanmaras¸, Turkey

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