SOME GENERALIZED INTEGRAL INEQUALITIES VIA ...

SOME GENERALIZED INTEGRAL INEQUALITIES VIA FRACTIONAL INTEGRALS MEHMET ZEKI SARIKAYA, HÜSEYIN BUDAK, AND FUAT USTA Abstract. The main goal of this paper is to introduce a new integral de…nition concerned with fractional calculus. Then we establish generalized Hermite-Hadamard type integral inequalities for convex function using proposed fractional integrals. The results presented in this paper provide extensions of those given in earlier works.

1. Introduction & Preliminaries The inequalities discovered by C. Hermite and J. Hadamard for convex functions are very signi…cant in the literature (see, e.g.,[11, p.137], [5]). These inequalities state that if f : I ! R is a convex function on the interval I of real numbers and a; b 2 I with a < b, then Z b a+b 1 f (a) + f (b) (1.1) f : f (x)dx 2 b a a 2 Both inequalities hold in the reversed direction if f is concave. We note that Hadamard’s inequality may be regarded as a re…nement 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 re…nements and generalizations have been studied (see, for example, [1, 2, 5, 6, 11, 15, 16]). On the other hand a number of mathematicians have studied the fractional integral inequalities and their applications using Riemann–Liouville fractional integrals. For results connected with Hermite–Hadamard type inequalities involving fractional integrals one can see [3, 4, 8, 12, 13, 14, 17, 18, 19]. In the following we present a brief synopsis of all necessary de…nitions and results that will be required. More details, one can consult [7, 9, 10]. De…nition 1. Let f 2 L1 [a; b]: The Riemann-Liouville fractional integrals Ja+ f and Jb f of order > 0 with a 0 are de…ned by Z x 1 1 (x t) f (t)dt; x > a Ja+ f (x) = ( ) a and Z b 1 1 Jb f (x) = (t x) f (t)dt; x < b ( ) x 0 respectively. Here, ( ) is the Gamma function and Ja+ f (x) = Jb0 f (x) = f (x): Key words and phrases. Hermite-Hadamard’s inequalities, Riemann-Liouville fractional integral, integral inequalities. 2010 Mathematics Sub ject Classi…cation. 26D07, 26D10, 26D15, 26A33. 1

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M EHM ET ZEKI SARIKAYA, HÜSEYIN BUDAK, AND FUAT USTA

In [13], Sarikaya et al. proved a variant of Hermite–Hadamard’s inequalities in Riemann-Liouville fractional integral forms as follows: Theorem 1. Let f : [a; b] ! R be a positive function with 0 a < b and f 2 L1 [a; b] : If f is a convex function on [a; b], then the following inequalities for fractional integrals hold: (1.2) with

f

a+b 2

( + 1) Ja+ f (b) + Jb f (a) 2 (b a)

f (a) + f (b) 2

> 0:

Remark 1. For

= 1, inequality (1.2) reduces to inequality (1.1).

Meanwhile, Sarikaya et al.[13] presented the following important integral identity including the …rst-order derivative of f to establish many interesting HermiteHadamard-type inequalities for convex functions via Riemann-Liouville fractional integrals of the order > 0: Lemma 1. Let f : [a; b] ! R be a di¤ erentiable mapping on (a; b) with a < b: If f 0 2 L1 [a; b] ; then the following equality for fractional integrals holds:

=

f (a) + f (b) 2 Z b a 1 [(1 2 0

( + 1) Ja+ f (b) + Jb f (a) 2 (b a) t)

t ] f 0 (ta + (1

t)b) dt:

Using this Lemma, the authors obtained the following fractional integral inequality in [13]: Theorem 2. Let f : [a; b] ! R be a di¤ erentiable mapping on (a; b) with a < b: If jf 0 j is convex on [a; b], then the following inequality for fractional integrals holds: f (a) + f (b) 2

( + 1) Ja+ f (b) + Jb f (a) 2 (b a)

(1.3) b a 2 ( + 1)

1

1 2

[f 0 (a) + f 0 (b)] :

The aim of this paper is presented a new de…nition concerned with fractional integrals and we establish generalized Hermite-Hadamard type integral inequalities for convex function involving this fractional integrals. 2. Main Findings & Cumulative Results In order to obtain our results, let us start with some notations given in [8]. Let f : I ! R be a function such that a; b 2 I and 0 < a < b < 1: Throughout this article, we suppose that F (x) = f (x) + fe(x) and fe(x) = f (a + b x); for x 2 [a; b]: Then it is easy to show that if f is a convex function, then F is also convex function. We are now give a new generalized de…nitions concerned with fractional integrals as follows:

SOM E GENERALIZED INTEGRAL INEQUALITIES VIA FRACTIONAL INTEGRALS

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De…nition 2. Let u : [a; b] ! R be an increasing and positive monotone function on (a; b) and f; u 2 L [a; b] with a < b: The generalized Riemann-Liouville fractional ;k integrals Ja+;u f and Jb ;k;u f of order > 0 with a 0 are de…ned by ;k Ja+;u

(f ) (x) =

1 ( )

Z

1 ( )

Z

and Jb ;k;u (f ) (x) =

x

(x

t)

1

(u(x)

u(t)) f (t)dt; x > a

k

(t

x)

1

(u(t)

u(x)) f (t)dt; x < b

a

b

k

x

provided that the integrals exist, respectively, k 2 N [ f0g. Example 1. If we choose u(t) = t and f (t) = 1; it follows that +k

;k Ja+;t (1) (x) =

(2.1)

(x a) ( + k) ( )

and Jb ;k;t (1) (x) =

(2.2)

(b

+k

x) : ( + k) ( )

Firstly, we present a new Hermite-Hadamard type inequalities for new generalized fractional integrals in the following theorem: Theorem 3. Let f : [a; b] ! R be a convex function on [a; b] and u : [a; b] ! R be an increasing and positive monotone function on (a; b) and f; u 2 L [a; b] with a < b: Then F is also integrable and the following inequalities for fractional integral operators hold (2.3)

f

a+b 2

h

i ;k Ja+;u (1) (b) + Jb ;k;u (1) (a)

i 1 h ;k Ja+;u (F ) (b) + Jb ;k;u (F ) (a) 2

with

h

i f (a) + f (b) ;k Ja+;u (1) (b) + Jb ;k;u (1) (a) 2

> 0 and k 2 N [ f0g:

Proof. Since f is an convex mapping on [a; b] ; we have (2.4)

f

x+y 2

f (x) + f (y) 2

for x; y 2 [a; b] : Now, for t 2 [0; 1] ; let x = ta + (1 we …nd that (2.5)

2f

a+b 2

f (ta + (1

t)b and y = (1

t)b) + f ((1

t)a + tb) :

t)a + tb: Then

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M EHM ET ZEKI SARIKAYA, HÜSEYIN BUDAK, AND FUAT USTA k

Then multiplying both sides of (2.5) by t 1 (u(b) u (ta + (1 t)b)) and integrating the resulting inequality with respect to t over [0; 1] ; we deduce that a+b 2

2f

Z1

t

1

(u(b)

u (ta + (1

k

t)b)) dt

0

Z1

t

+

Z1

1

(u(b)

k

u (ta + (1

t)b)) f (ta + (1

t)b) dt

0

1

t

(u(b)

k

u (ta + (1

t)b)) f ((1

t)a + tb) dt:

0

Using the change of variable y = ta + (1 a+b 2

2f

Ja+;k fe (b) + Ja+;k (f ) (b);

Ja+;k (1) (b)

i.e. (2.6)

a+b 2

2f

t)b; we have

Ja+;k (1) (b)

Ja+;k (F ) (b): k

Similarly, multiplying both sides of (2.5) by t 1 (u ((1 t)a + tb) u(a)) and integrating the resulting inequality with respect to t over [0; 1] ; we obtain a+b 2

2f

Z1

t

1

(u ((1

t)a + tb)

k

u(a)) dt

0

Z1

t

+

Z1

1

(u ((1

k

t)a + tb)

u(a)) f (ta + (1

t)b) dt

k

t)a + tb) dt:

0

t

1

(u ((1

t)a + tb)

u(a)) f ((1

0

Using the change of variable y = (1 (2.7)

2f

a+b 2

t)a + tb; we have

Jb ;k;u (1) (a)

Jb ;k;u (F ) (a):

Summing the inequalities (2.6) and (2.7), we get f

a+b 2

h

i ;k Ja+;u (1) (b) + Jb ;k;u (1) (a)

i 1 h ;k Ja+;u (F ) (b) + Jb ;k;u (F ) (a) : 2

This completes the proof of …rst inequality in (2.3). For the proof of the second inequality in (2.3), since f is convex , we have (2.8)

f (ta + (1

t)b) + f ((1

t)a + tb)

[f (a) + f (b)] :

SOM E GENERALIZED INTEGRAL INEQUALITIES VIA FRACTIONAL INTEGRALS

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k

Multiplying both sides of (2.8) by t 1 (u(b) u (ta + (1 t)b)) and integrating the resulting inequality with respect to t over [0; 1] ; we have Z1

t

+

Z1

1

(u(b)

u (ta + (1

k

t)b)) f (ta + (1

t)b) dt

0

t

1

(u(b)

u (ta + (1

k

t)b)) f ((1

t)a + tb) dt

0

[f (a) + f (b)]

Z1

t

1

(u(b)

u (ta + (1

k

t)b)) dt:

0

Then, we get (2.9)

;k Ja+;u (F ) (b)

;k [f (a) + f (b)] Ja+;u (1) (b): k

Similarly, multiplying both sides of (2.8) by t 1 (u ((1 t)a + tb) u(a)) and integrating the resulting inequality with respect to t over [0; 1] ; we get (2.10)

Jb ;k;u (F ) (a)

[f (a) + f (b)] Jb ;k;u (1) (a):

By adding the inequalities (2.9) and (2.10), we have i h i f (a) + f (b) 1 h ;k ;k Ja+;u (F ) (b) + Jb ;k;u (F ) (a) Ja+;u (1) (b) + Jb ;k;u (1) (a) ; 2 2 which completes the proof. Remark 2. If we choose k = 0 in Theorem 3, then the inequality (2.3) reduces to inequality (1.2). Corollary 1. If we choose u(t) = t in Theorem 3, then we have the inequality for Riemann-Liouville fractional integrals # " +k +k a+b (x a) + (b x) f 2 ( + k) ( ) ( + k) Ja++k (F ) (b) + Jb +k (F ) (a) 2 ( ) " # +k +k (x a) + (b x) f (a) + f (b) : ( + k) ( ) 2 Proof. From De…nition 2 with u(t) = t, we have Z b 1 1 k ;k (2.11) Ja+;t (F ) (b) = (b t) (x t) F (t)dt = ( ) a

( + k) +k Ja+ (F ) (b) ( )

and similarly, (2.12)

Jb ;k;t (F ) (a) =

( + k) +k Jb (F ) (b): ( )

Using the equalities (2.1), (2.2), (2.11) and (2.12) we obtain the desired result.

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M EHM ET ZEKI SARIKAYA, HÜSEYIN BUDAK, AND FUAT USTA

Now, we give an important identity for new generalized fractional integrals in the following theorem: Lemma 2. Let f : [a; b] ! R be a di¤ erentiable function on (a; b) and u : [a; b] ! R be an increasing and positive monotone function on (a; b) with a < b: If f 0 ; u 2 L [a; b] ; then F is also di¤ erentiable and F 2 L [a; b] ; and the following equality holds i h i f (a) + f (b) 1 h ;k ;k (2.13) Ja+;u Ja+;u (F ) (b) + Jb ;k;u (F ) (a) (1) (b) + Jb ;k;u (1) (a) 2 2 =

(b a) 2 ( )

Zb

G(y)F 0 (y)dy

a

where F 0 (y) = f 0 (y)

(2.14)

f 0 (a + b 2

6 = 4

G(y)

y

y) and

1

s

(u(b)

y b

Z1

(1

s)

1

(u(sa + (1

k

(b

a)

;k +1 Ja+

(1) (b)f (b)

( )

(b

a)

Similarly, using the integration by parts, 2 Z1 Zt 4 s 1 (u(b) u (sa + (1 I2 = 0

=

I3

(b

a)

2 Z1 Z1 4 (1 = 0

=

s)

1

( )

(1) (b)f (a) +

(u(sa + (1

a)

;k +1 Jb ;u

(1) (a)f (a)

(b

fe (b):

3

a)

;k +1 Ja+;u

k

t)b) dt

(f ) (b);

3

u (a)) ds5 f 0 (at + (1

s)b)

( ) (b

t)a + tb) dt

s)b)) ds5 f 0 (ta + (1

t

( )

(b

;k +1 Ja+;u

;k +1 Ja+

k

0

( )

3

s)b)) ds5 f 0 ((1

0

( )

3

7 k u (a)) ds7 5:

s)b)

a a

Proof. Integrating by parts, we have 2 Z1 Zt 4 s 1 (u(b) u (sa + (1 I1 = =

7 k s)b)) ds5

u (sa + (1

0

2

6 +6 4

0

3

a a

Zb

a)

;k +1 Jb ;u

(f ) (a):

t)b) dt

SOM E GENERALIZED INTEGRAL INEQUALITIES VIA FRACTIONAL INTEGRALS

and I4

2 Z1 Z1 4 (1 =

1

s)

(u((sa + (1

t

0

( )

=

(b a) Then it follows that (2.15)

;k +1 Jb ;u

I1

(1) (a)f (b) +

I2 + I3 ( )

=

(b

+1

a)

(b

a)

u (a)) ds5 f 0 ((1

( ) (b

3

a)

;k +1 Jb ;u

h

fe (a);

+1

i ;k Ja+;u (1) (b) + Jb ;k;u (1) (a) [f (a) + f (b)] h

i ;k (F ) (b) + Jb ;k;u (F ) (a) : Ja+;u

If we multiply both sides of (2.15) by

(b a) +1 2 ( ) ;

then we obtain the following result.

(2.16)

h

i f (a) + f (b) 1 h i ;k ;k Ja+;u (1) (b) + Jb ;k;u (1) (a) Ja+;u (F ) (b) + Jb ;k;u (F ) (a) 2 2 +1 (b a) 2 ( ) 8 12 t 3