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ON HERMITE-HADAMARD TYPE INEQUALITIES ASSOCIATED WITH THE

GENERALIZED FRACTIONAL INTEGRALS

FATM A ERTU ¼

GRA L, MEHMET ZEKI SARIKAYA, AND HÜSEYIN BUDAK

Abstract. In this paper, we obtain new generalization of Hermite-Hadamard inequalities via gener-

alized fractional integrals de…ned by Sarikaya and Ertu¼

gral in [30]. We establish some midpoint and

trap ezoid type inequalities for functions whose …rst derivatives in absolute value are convex involving

generalized fractional integrals.

1. Introduction

The inequalities discovered by C. Hermite and J. Hadamard for convex functions are considerable

signi…cant in the literature (see, e.g.,[6], [11], [25, p.137]). These inequalities state that if f:I!Ris

a convex function on the interval Iof real numbers and a; b 2Iwith a < b, then

(1.1) fa+b

21

baZb

a

f(x)dx f(a) + f(b)

2:

Both inequalities hold in the reversed direction if fis 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.

The Hermite-Hadamard inequality, which is the …rst fundamental result for convex mappings with

a natural geometrical interpretation and many applications, has drawn attention much interest in

elementary mathematics. A number of mathematicians have devoted their e¤orts to generalise, re…ne,

counterpart and extend it for di¤erent classes of functions such as using convex mappings.

The overall structure of the study takes the form of six sections including introduction. The remain-

der of this work is organized as follows: we …rst mention some works which focus on Hermite-Hadamard

inequality. In Section 2, we summarize the generalized fractional integrals de…ned by Sarikaya and

Ertu¼

gral along with the very …rst results. In section 3 new Hermite-Hadamard type inequalities for

generalized fractional integrals are proved. In Section 4 and Section 5 midpoint and trapezoid type

inequalities for functions whose …rst derivatives in absolute value are convex via generalized fractional

integrals are presented, respectively. Some conclusions and further directions of research are discussed

in Section 6.

In this paper, we obtain the new generalized Hermite-Hadamard type inequality for the generalized

fractional integrals mentioned in next section.

2. New Generalized Fractional Integral Operators

In this section, we summarize the generalized fractional integrals de…ned by Sarikaya and Ertu¼

gral

in [30].

Let’s de…ne a function ': [0;1)![0;1)satis…ying the following conditions :

Z1

0

'(t)

tdt < 1:

We de…ne the following left-sided and right-sided generalized fractional integral operators, respectively,

as follows:

(2.1) a+I'f(x) = Zx

a

'(xt)

xtf(t)dt; x > a;

Key words and phrases. Hermite-Hadamard inequality, midpoint inequality, fractional integral operators, convex

function.

2010 Mat hem atics Sub ject Classi…cation. 26D15, 26B25, 26D10.

1

2 FATM A ERT U ¼

GR AL, M EH MET ZE KI SA RIKAYA , AND HÜ SE YIN B UD AK

(2.2) bI'f(x) = Zb

x

'(tx)

txf(t)dt; x < b:

The most important feature of generalized fractional integrals is that they generalize some types of

fractional integrals such as Riemann-Liouville fractional integral, k-Riemann-Liouville fractional inte-

gral, Katugampola fractional integrals, conformable fractional integral, Hadamard fractional integrals,

etc. These important special cases of the integral operators (2.1) and (2.2) are mentioned below.

i) If we take '(t) = t; the operator (2.1) and (2.2) reduce to the Riemann integral as follows:

Ia+f(x) = Zx

a

f(t)dt; x > a;

Ibf(x) = Zb

x

f(t)dt; x < b:

ii) If we take '(t) = t

();the operator (2.1) and (2.2) reduce to the Riemann-Liouville fractional

integral as follows:

I

a+f(x) = 1

()Zx

a

(xt)1f(t)dt; x > a;

I

bf(x) = 1

()Zb

x

(tx)1f(t)dt; x < b:

iii) If we take '(t) = 1

kk()t

k;the operator (2.1) and (2.2) reduce to the k-Riemann-Liouville

fractional integral as follows:

I

a+;k f(x) = 1

kk()Zx

a

(xt)

k1f(t)dt; x > a;

I

b;kf(x) = 1

kk()Zb

x

(tx)

k1f(t)dt; x < b

where

k() = Z1

0

t1etk

kdt; R()>0

and

k() = k

k1

k;R()>0; k > 0

are given by Mubeen and Habibullah in [23].

iv) If we take

'(t) = 1

()t(xt)sxs+1 ts+11

and

'(t) = 1

()t(tx)sts+1 xs+11;

in the operators (2.1) and (2.2), respectively, then the (2.1) and (2.2) reduce to the Katugampola

fractional operators as follows for > 0and s6=1is a real numbers:

I

a+;s f(x) = (s+ 1)1

()Zx

axs+1 ts+11tsf(t)dt; x > a;

I

b;sf(x) = (s+ 1)1

()Zb

xxs+1 ts+11tsf(t)dt; ; x < b

are given by Katugampola in [18].

v) If we take '(t) = t(xt)1, the operator (2.1) reduces to the conformable fractional operators

as follows:

I

af(x) = Zx

a

t1f(t)dt =Zx

a

f(t)dt; x > a; 2(0;1)

is given by Khalil et.al in [19].

vi) If we take

'(t) = 1

()

[(log xlog(xt)]1

xt

ON H ERM IT E- HA DA MAR D TY PE INE QUA LITIES... 3

and

'(t) = 1

()t[(log(tx)log x]1

tx;

in the operators (2.1) and (2.2), respectively, the operator (2.1) and (2.2) reduce to the right-sided and

left-sided Hadamard fractional integrals as follows:

I

a+f(x) = 1

()Zx

a

(log xlog t)1f(t)

tdt; 0< a < x < b;

I

bf(x) = 1

()Zb

x

(log tlog x)1f(t)

tdt; 0< a < x < b:

vii) If we take '(t) = t

exp 1

tin the operators (2.1) and (2.2), respectively, the operator (2.1)

and (2.2) reduce to the right-sided and left-sided fractional integral operators with exponential kernel

for 2(0;1) as follows:

I

a+f(x) = 1

Zx

a

exp 1

(xt)f(t)dt; a < x;

I

bf(x) = 1

Zb

x

exp 1

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

are de…ned by Kirane and Torebek in [21].

In [30], Sar¬kaya and Ertu¼

gral also establish the following Hermite-Hadamard inequality for the

generalized fractional integral operators:

Theorem 1. Let f: [a; b]!Rbe a convex function on [a; b]with a < b, then the following inequalities

for fractional integral operators hold

(2.3) fa+b

21

2(1) [a+I'f(b) +bI'f(a)] f(a) + f(b)

2

where the mapping : [0;1] !Ris de…ned by

(x) =

x

Z

0

'((ba)t)

tdt:

3. Hermite-Hadamard Type Inequalities for Generalized Fractional Integral

Operators

In this section, we will present a new theorem for Hermite-Hadamard type inequalities associated

with the generalized fractional integral operators which is the generalization of previous work.

Theorem 2. Let f: [a; b]!Rbe a function with a<band f2L1[a; b]:If fis a convex function

on [a; b];then we have the following inequalities for generalized fractional integral operators:

(3.1) fa+b

21

2(1) a+I'f(a+b

2) +bI'f(a+b

2)f(a) + f(b)

2

where the mapping : [0;1] !Ris de…ned by

(x) =

x

Z

0

'ba

2t

tdt:

Proof. Since fis a convex function on [a; b];we have for x; y 2[a; b]

fx+y

2f(x) + f(y)

2:

For x=1t

2a+1+t

2band y=1+t

2a+1t

2b; we obtain

(3.2) 2fa+b

2f1t

2a+1 + t

2b+f1 + t

2a+1t

2b:

4 FATM A ERT U ¼

GR AL, M EH MET ZE KI SA RIKAYA , AND HÜ SE YIN B UD AK

Multiplying both sides of (3.2) by '(ba

2t)

t;then integrating the resulting inequality with respect to t

over [0;1] ;we get

2fa+b

21

Z

0

'ba

2t

tdt

1

Z

0

'ba

2t

tf1t

2a+1 + t

2bdt +

1

Z

0

'ba

2t

tf1 + t

2a+1t

2bdt:

For u=1t

2a+1+t

2band v=1+t

2a+1t

2b; we obtain

2fa+b

2(1)dt

b

Z

a+b

2

'ua+b

2

ua+b

2

f(u)du +

a+b

2

Z

a

'a+b

2v

a+b

2vf(v)dv

=a+I'f(a+b

2) +bI'f(a+b

2)

and the …rst inequality is proved.

For the proof of the second inequality (3.1), we …rst note that if fis a convex function, it yields

f1t

2a+1 + t

2b1t

2f(a) + 1 + t

2f(b)

and

f1 + t

2a+1t

2b1 + t

2f(a) + 1t

2f(b):

By adding these inequalities together, one has the following inequality:

(3.3) f1t

2a+1 + t

2b+f1 + t

2a+1t

2bf(a) + f(b):

Then multiplying both sides of (3.3) by '(ba

2t)

tand integrating the resulting inequality with respect

to tover [0;1] ;we obtain

1

Z

0

'ba

2t

tf1t

2a+1 + t

2bdt+

1

Z

0

'ba

2t

tf1 + t

2a+1t

2bdt [f(a) + f(b)]

1

Z

0

'ba

2t

tdt:

That is, a+I'f(a+b

2) +bI'f(a+b

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

Hence, the proof is completed.

Remark 1. Under assumption of Theorem 2 with '(t) = t; then inequalities 3.1 reduce to inequalities

1.1.

Corollary 1. Under assumption of Theorem 2 with '(t) = t

();then, we have the following inequal-

ities

fa+b

221 (+ 1)

(ba)a+J'fa+b

2+bJ'fa+b

2f(a) + f(b)

2:

Corollary 2. Under assumption of Theorem 2 with '(t) = t

k

kk();then, we have the following in-

equalities

fa+b

2k(+k) 2

k1

(ba)

kIa+;kfa+b

2+Ib;';kfa+b

2f(a) + f(b)

2:

ON H ERM IT E- HA DA MAR D TY PE INE QUA LITIES... 5

4. Midpoint Type Inequalities for Differentiable Functions with Generalized

Fractional Integral Operators

In this section, …rstly we need to give a lemma for di¤erentiable functions which will help us to prove

our main theorems. Then, we present some midpoint type inequalities which are the generalization of

those given in earlier works.

Lemma 1. Let f: [a; b]!Rbe di¤erentiable function on (a; b)with a < b: If f02L[a; b];then we

have the following identity for generalized fractional integral operators:

1

2(1) a+I'f(a+b

2) +bI'f(a+b

2)fa+b

2

(4.1)

=ba

4(0) 2

4

1

Z

0

(t)f01t

2a+1 + t

2bdt

1

Z

0

(t)f01 + t

2a+1t

2bdt3

5

where the mapping (t)is de…ned by

(x) =

1

Z

x

'ba

2t

tdt:

Proof. Integrating by parts gives

I1=

1

Z

0

(t)f01t

2a+1 + t

2bdt(4.2)

=2

ba(t)f1t

2a+1 + t

2b

1

0

+2

ba

1

Z

0

'ba

2t

tf1t

2a+1 + t

2bdt

=2

ba(0)fa+b

2+2

babI'fa+b

2

and similarly we get

(4.3) I2=

1

Z

0

(t)f01 + t

2a+1t

2bdt =2

ba(0)fa+b

22

baa+I'fa+b

2:

By subtracting equation (4.3) from (4.2), we have

ba

4(0) (I1I2) = 2 a+I'f(a+b

2) + bI'f(a+b

2)4(0)fa+b

2:

By re-arranging the last equality above, we get the desired result.

Corollary 3. Under assumption of Lemma (1) with '(t) = t; then we have the following inequalities

1

baZb

a

f(x)dxfa+b

2=ba

4Z1

0

tf01t

2a+1 + t

2bdt +Z1

0

tf01 + t

2a+1t

2bdt:

Corollary 4. Under assumption of Lemma (1)with '(t) = t

();then we have the following inequalities

21 (+ 1)

(ba)a+Jfa+b

2+bJa+b

2fa+b

2

=ba

4Z1

0

tf01t

2a+1 + t

2bdt +Z1

0

tf01 + t

2a+1t

2bdt:

6 FATM A ERT U ¼

GR AL, M EH MET ZE KI SA RIKAYA , AND HÜ SE YIN B UD AK

Corollary 5. Under assumption of Lemma (1)with '(t) = t

k

kk();then we have the following inequal-

ities

2

k1k(+k)

(ba)

kI

a+;kfa+b

2+I

b;k a+b

2fa+b

2

=ba

4Z1

0

t

k1f01t

2a+1 + t

2bdt +Z1

0

t

k1f01 + t

2a+1t

2bdt

Theorem 3. Let f: [a; b]!Rbe di¤erentiable function on (a; b)with a < b: If jf0jis convex function,

then we have the following inequality for generalized fractional integral operators:

1

2(1) a+I'f(a+b

2) +bI'f(a+b

2)fa+b

2

ba

4(0) 0

@

1

Z

0

j(t)jdt1

A[jf0(a)j+jf0(b)j]

where the mapping (t)is de…ned as in Theorem 2.

Proof. From Lemma 1, by using the convexity of jf0j, we have

1

2(1) a+I'f(a+b

2) + bI'f(a+b

2)fa+b

2

ba

4(0) 2

4

1

Z

0

j(t)jf01t

2a+1 + t

2bdt +

1

Z

0

j(t)jf01 + t

2a+1t

2bdt3

5(f5)

ba

4(1) 2

4

1

Z

0

j(t)j1t

2jf0(a)j+1 + t

2jf0(b)jdt +

1

Z

0

j(t)j1 + t

2jf0(a)j+1t

2jf0(b)jdt3

5

=ba

4(0) 0

@

1

Z

0

j(t)jdt1

A[jf0(a)j+jf0(b)j]:

This completes the proof.

Remark 2. Under assumption of Theorem 3 with '(t) = t; then inequalities 4.1 reduce to inequalities

1.1

1

baZb

a

f(x)dx fa+b

2ba

4jf0(a)j+jf0(b)j

2:

Corollary 6. Under assumption of Theorem 3 with '(t) = t

();then we have the following inequalities

21 (+ 1)

(ba)a+Jfa+b

2+bJa+b

2fa+b

2(ba)

(+ 1) jf0(a)j+jf0(b)j

2

Corollary 7. Under assumption of Theorem 3 with '(t) = t

k

kk()then we have the following inequal-

ities

2

k1k(+k)

(ba)

kI

a+;kfa+b

2+I

b;k a+b

2fa+b

2(ba)1

k

(+ 1) jf0(a)j+jf0(b)j

2

ON H ERM IT E- HA DA MAR D TY PE INE QUA LITIES... 7

Theorem 4. Let f: [a; b]!Rbe di¤ erentiable function on (a; b)with a < b: If jf0jq; q > 1;is convex

function, then we have the following inequality for generalized fractional integral operators:

1

2(1) a+I'f(a+b

2) + bI'f(a+b

2)fa+b

2

(4.4)

ba

4(1) 0

@

1

Z

0

j(t)jpdt1

A

1

p"jf0(a)jq+ 3 jf0(b)jq

41

q

+3jf0(a)jq+jf0(b)jq

41

q#

ba

22

q(1) 0

@

1

Z

0

j(t)jpdt1

A

1

p

[jf0(a)j+jf0(b)j]

where 1

p+1

q= 1 and the mapping is de…ned as in Theorem 2.

Proof. Taking modulus of (4.1) and using the well-known Hölder inequality, we obtain

1

2(1) a+I'f(a+b

2) + bI'f(a+b

2)fa+b

2

(4.5)

ba

4(1) 2

4

1

Z

0

j(t)jf01t

2a+1 + t

2bdt +

1

Z

0

j(t)jf01 + t

2a+1t

2bdt3

5

ba

4(1) 0

@

1

Z

0

j(t)jpdt1

A

1

p

2

6

40

@

1

Z

0f01t

2a+1 + t

2b

q

dt1

A

1

q

+0

@

1

Z

0f01 + t

2a+1t

2b

q

dt1

A

1

q3

7

5:

Since jf0jq; q > 1;is convex, we have

(4.6)

1

Z

0f01t

2a+1 + t

2b

q

dt

1

Z

01t

2jf0(a)jq+1 + t

2jf0(b)jqdt =jf0(a)jq+ 3 jf0(b)jq

4

and similarly

(4.7)

1

Z

0f01 + t

2a+1t

2b

q

dt 3jf0(a)jq+jf0(b)jq

4:

By substituting inequalities (4.6) and (4.7) into (4.5), we obtain the …rst inequalty in (4.4).

For the proof of second inequality, let a1=jf0(a)jq; b1= 3 jf0(b)jq; a2= 3 jf0(a)jqand b2=

jf0(b)jq:Using the fact that

(4.8)

n

X

k=1

(ak+bk)s

n

X

k=1

as

k+

n

X

k=1

bs

k;0s < 1

and 1+31

q4;then the desired result can be obtained straightforwardly.

Corollary 8. Under assumption of Lemma (4)with '(t) = t; then we hve the following inequalities

1

baZb

a

f(x)dx fa+b

2ba

22

p1

p+ 11

p

[jf0(a)j+jf0(b)j]:

8 FATM A ERT U ¼

GR AL, M EH MET ZE KI SA RIKAYA , AND HÜ SE YIN B UD AK

Corollary 9. Under assumption of Lemma (4)with '(t) = t

();then we hve the following inequalities

21 (+ 1)

(ba)a+Jfa+b

2+bJa+b

2fa+b

2

212

q

(ba) (p+ 1) 1

p

[jf0(a)j+jf0(b)j]:

Theorem 5. Let f: [a; b]!Rbe di¤erentiable function on (a; b)with a < b: If jf0jq; q 1;is a

convex function, then we have the following inequality for generalized fractional integral operators:

1

2(1) a+I'f(a+b

2) + bI'f(a+b

2)fa+b

2

ba

22+ 1

q(1) 0

@

1

Z

0

j(t)jdt1

A

11

q

hB1jf0(a)jq+B2jf0(b)jq1

q+B2jf0(a)jq+B1jf0(b)jq1

qi

where the mapping (t)is de…ned as in Theorem 2 and the constants B1and B2are de…ned by

B1=

1

Z

0

j(t)j(1 t)dt and B2=

1

Z

0

j(t)j(1 + t)dt.

Proof. The case of q= 1 is obvious from Theorem 3.

For q > 1we proceed as follows. Taking modulus of (4.1) and using well-known power mean

inequality, we obtain

1

2(1) a+I'f(a+b

2) + bI'f(a+b

2)fa+b

2

ba

4(1) 2

4

1

Z

0

j(t)jf01t

2a+1 + t

2bdt +

1

Z

0

j(t)jf01 + t

2a+1t

2bdt3

5

ba

4(1) 0

@

1

Z

0

j(t)jdt1

A

11

q2

6

40

@

1

Z

0

j(t)jf01t

2a+1 + t

2b

q

dt1

A

1

q

+0

@

1

Z

0

j(t)jf01 + t

2a+1t

2b

q

dt1

A

1

q3

7

5:

ON H ERM IT E- HA DA MAR D TY PE INE QUA LITIES... 9

Since jf0jqis convex, we have

1

2(1) a+I'f(a+b

2) + bI'f(a+b

2)fa+b

2

ba

4(1) 0

@

1

Z

0

j(t)jdt1

A

11

q2

6

40

@

1

Z

0

j(t)j1t

2jf0(a)jq+1 + t

2jf0(b)jqdt1

A

1

q

+0

@

1

Z

0

j(t)j1 + t

2jf0(a)jq+1t

2jf0(b)jqdt1

A

1

q3

7

5

=ba

22+ 1

q(1) 0

@

1

Z

0

j(t)jdt1

A

11

q

hB1jf0(a)jq+B2jf0(b)jq1

q+B2jf0(a)jq+B1jf0(b)jq1

qi

which completes the proof.

5. Trapezoid Type Inequalities for Differentiable Functions with

Generalized Fractional Integral Operators

In this section, …rstly we need to give a lemma for di¤erentiable functions which will help us to prove

our main theorems. Then, we present some trapezoid type inequalities which are the generalization of

those given in earlier studies.

Lemma 2. Let f: [a; b]!Rbe di¤erentiable function on (a; b)with a < b: If f02L[a; b];then we

have the following identity for generalized fractional integral operators:

f(a) + f(b)

21

2(1) a+I'f(a+b

2) + bI'f(a+b

2)

(5.1)

=ba

4(1) 2

4

1

Z

0

(t)f01t

2a+1 + t

2bdt

1

Z

0

(t)f01 + t

2a+1t

2bdt3

5

where the mapping (t)is de…ned by

(t) =

t

Z

0

'ba

2u

udu;

with (0) = (1):

Proof. Integrating by parts, we have

I3=

1

Z

0

(t)f01t

2a+1 + t

2bdt(5.2)

=2

ba(t)f1t

2a+1 + t

2b

1

0

2

ba

1

Z

0

'ba

2t

tf1t

2a+1 + t

2bdt

=2

ba(1)f(b)2

babI'fa+b

2

10 FATM A ERT U ¼

GR AL, M EH MET ZE KI SA RIKAYA , AND HÜ SE YIN B UD AK

and similarly we get

(5.3) I4=

1

Z

0

(t)f01 + t

2a+1t

2bdt =2

ba(1)f(a) + 2

baa+I'fa+b

2:

Thus, we have

ba

4(1) (I3I4) = f(a) + f(b)

21

2(1) a+I'f(a+b

2) + bI'fa+b

2:

This completes the proof.

Corollary 10. Under assumption of Lemma (2)with '(t) = t; then we hve the following inequalities

f(a) + f(b)

2

1

baZb

a

f(x)dx =ba

4Z1

0

tf01t

2a+1 + t

2bdt +Z1

0

tf01 + t

2a+1t

2bdt:

Corollary 11. Under assumption of Lemma (2)with '(t) = t

();then we hve the following inequalities

f(a) + f(b)

221 (+ 1)

(ba)a+Jfa+b

2+bJa+b

2

=ba

4Z1

0

tf01t

2a+1 + t

2bdt +Z1

0

tf01 + t

2a+1t

2bdt:

Corollary 12. Under assumption of Lemma (2)with '(t) = t

();then we hve the following inequalities

f(a) + f(b)

22

k1k(+k)

(ba)

kI

a+;kfa+b

2+I

b;k a+b

2

=ba

4Z1

0

t

k1f01t

2a+1 + t

2bdt +Z1

0

t

k1f01 + t

2a+1t

2bdt:

Theorem 6. Let f: [a; b]!Rbe di¤erentiable function on (a; b)with a < b: If jf0jis a convex

function, then we have the following inequality for generalized fractional integral operators:

f(a) + f(b)

21

2(1) a+I'fa+b

2+bI'fa+b

2

ba

4(1) 0

@

1

Z

0

j(t)jdt1

A[jf0(a)j+jf0(b)j]:

Proof. From Lemma 2, by the using convexity of jf0j;we have

f(a) + f(b)

21

2(1) a+I'f(a+b

2) + bI'f(a+b

2)

ba

4(1) 2

4

1

Z

0

j(t)jf01t

2a+1 + t

2bdt +

1

Z

0

j(t)jf01 + t

2a+1t

2bdt3

5

ba

4(1) 2

4

1

Z

0

j(t)j1t

2jf0(a)j+1 + t

2jf0(b)jdt +

1

Z

0

j(t)j1 + t

2jf0(a)j+1t

2jf0(b)jdt3

5

=ba

4(1) 0

@

1

Z

0

j(t)jdt1

A[jf0(a)j+jf0(b)j]

which completes the proof.

ON H ERM IT E- HA DA MAR D TY PE INE QUA LITIES... 11

Theorem 7. Let f: [a; b]!Rbe di¤erentiable function on (a; b)with a < b: If jf0jq; q > 1;is a

convex function, then we have the following inequality for generalized fractional integral operators:

f(a) + f(b)

21

2(1) a+I'f(a+b

2) +bI'f(a+b

2)

(5.4)

ba

4(1) 0

@

1

Z

0

j(t)jpdt1

A

1

p"jf0(a)jq+ 3 jf0(b)jq

41

q

+3jf0(a)jq+jf0(b)jq

41

q#

ba

22

q(1) 0

@

1

Z

0

j(t)jpdt1

A

1

p

[jf0(a)j+jf0(b)j]

where 1

p+1

q= 1 and the mappingsand are de…ned as above.

Proof. Similar to proof of Theorem 4, by using the well-known Hölder inequality and convexity of jf0jq,

we obtain

f(a) + f(b)

21

2(1) a+I'fa+b

2+bI'fa+b

2

ba

4(1) 2

4

1

Z

0

j(t)jf0t

2a+2t

2bdt +

1

Z

0

j(t)jf02t

2a+t

2bdt3

5

ba

4(1) 0

@

1

Z

0

j(t)jpdt1

A

1

p

"jf0(a)jq+ 3 jf0(b)jq

41

q

+3jf0(a)jq+jf0(b)jq

41

q#:

This completes the proof of …rst inequality in (5.4)

The proof of second inequality in (5.4) is obvious from the inequality (4.8).

Theorem 8. Let f: [a; b]!Rbe di¤ erentiable function on (a; b)with a < b: If jf0jq; q 1;is convex

function, then we have the following inequality for generalized fractional integral operators:

f(a) + f(b)

21

2(1) a+I'f(a+b

2) + bI'f(a+b

2)

ba

22+ 1

q(1) 0

@

1

Z

0

j(t)jdt1

A

11

q

hB5jf0(a)jq+B6jf0(b)jq1

q+B6jf0(a)jq+B5jf0(b)j1

qi

where the mappings as above and the constants B5and B6are de…ned by

B5=

1

Z

0

j(t)j(1 t)dt and B6=

1

Z

0

j(t)j(1 + t)dt.

Proof. The case of the q= 1 is obvious from the Theorem 6.

12 FATM A ERT U ¼

GR AL, M EH MET ZE KI SA RIKAYA , AND HÜ SE YIN B UD AK

For q > 1;using well-known power mean inequality in Lemma 2, we obtain

f(a) + f(b)

21

2(1) a+I'fa+b

2+bI'fa+b

2

ba

4(1) 2

4

1

Z

0

j(t)jf01t

2a+1 + t

2bdt +

1

Z

0

j(t)jf01 + t

2a+1t

2bdt3

5

ba

4(1) 0

@

1

Z

0

j(t)jqdt1

A

11

q

2

6

40

@

1

Z

0

j(t)jf01t

2a+1 + t

2b

q

dt1

A

1

q

+0

@

1

Z

0

j(t)jf01 + t

2a+1t

2b

q

dt1

A

1

q3

7

5:

By the using convexity of jf0jq, we have

f(a) + f(b)

21

2(1) a+I'fa+b

2+bI'fa+b

2

ba

4(1) 0

@

1

Z

0

j(t)jqdt1

A

11

q

2

6

40

@

1

Z

0

j(t)j1t

2jf0(a)jq+1 + t

2jf0(b)jqdt1

A

1

q

+0

@

1

Z

0

j(t)j1 + t

2jf0(a)jq+jf0(b)jq1t

2dt1

A

1

q3

7

5

=ba

22+ 1

q(1) 0

@

1

Z

0

j(t)jqdt1

A

11

qhB5jf0(a)jq+B6jf0(b)jq1

q+B6jf0(a)jq+B5jf0(b)j1

qi:

The proof is completely completed.

6. Concluding Remarks

In this study, we consider the Hermite-Hadamard for convex function involving generalized fractional

integrals de…ned by Sarikaya and Ertu¼

gral in [30]. We also focus on midpoint and trapezoid type

inequalities for functions whose …rst derivatives in absolute value are convex via generalized fractional

integrals. The results presented in this study would provide generalizations of those given in earlier

works.

References

[1] M. Alomari, M. Darus, U. S. K irmaci, Re…nem ents of Hadamard-type inequalities for quasi-convex functions with

applications to trapezoidal formula and to special means, Comput. Math. Appl., 59 (2010), 225–232. 1

[2] G. A. Anastassiou, General fractional Hermite–Hadamard inequalities using m-convexity and (s; m)-convexity,

Frontiers in Time Scales and Inequalities. 2016. 237-255.

[3] A.G. Azpeitia, Convex fun ctions and the Hada mard in equality, Rev. Colombiana Math., 28 (1994), 7-12.

[4] J. de la Cal, J. Carcamob, L. Escauriaza, A general m ultidimensional Hermite-Hadam ard type inequality, J. Math.

Anal. Appl., 356 (2009), 659–663.

[5] H. Chen and U.N. Katugamp ola, Hermite–Hadamard and Hermite–Hadamard–Fejér type inequalities for ge neralized

fractional integrals, J. Math. Anal. Appl. 446 (2017) 1274–1291

[6] S.S. Dragomir and C.E.M. Pearce, Selected Topics on Hermite-Hadamard Inequalities and Applications, RGMIA

Monographs, Victoria University, 2000. Online:[http://rgmia.org/papers/monographs/Master2.pdf].

[7] S.S. Dragomir, R.P. Agarwal, Tw o inequalities for di¤erentiable mappings and applications to special means of real

num bers and to trapezoidal formula, Appl. Math. lett. 11 (5) (1998) 91–95.

[8] G. Farid, A. ur Rehman and M . Zahra, On Hadamard type inequalities for k-fraction al integrals, Konurap J. Math.

2016, 4(2), 79–86.

ON H ERM IT E- HA DA MAR D TY PE INE QUA LITIES... 13

[9] G. Farid, A. Rehman and M. Zahra, On Hadamard inequalities for k-fractio nal in tegrals, Nonlinear Functional

Analysis and Applications Vol. 21, No. 3 (2016), pp. 463-478.

[10] R. G oren‡o, F. Mainardi, Fractional calculus: integral and di¤erential equations of fractional order, Springer Verlag,

Wien (1997), 223-276.

[11] J. Hadamard, E tude sur les p roprie tes des fonctions entieres en particulier d’une fonction consideree par Riem ann ,

J. Math. Pures Appl. 58 (1893), 171-215.

[12] R. H ussain, A. Ali, G. Gulshan, A. Latif and M. Muddassar, Generalized co-ordin ated integral inequalities for

conve x functions by way of k-fraction al derivatives, M iskolc Mathematical Notesa Publications of the university

of Miskolc. (Submitted)

[13] R. H ussain, A. Ali, A. Latif, G. Gulshan, Some k–fractional associates of H ermite–Hadamard’s inequality for

quasi–convex functions and app lications to special means, Fractional Di¤erential Calculus, Volum e 7, Number 2

(2017), 301–309.

[14] M. Iqbal, S. Qaisar and M. Muddassar, A short no te on integral inequality of type Herm ite-Hadamard through

conve xity, J. Computational analaysis and applications, 21(5), 2016, pp.946-953.

[15] ·

I. ·

I¸scan and S. Wu, Hermite-Hadamard type inequalities for harmo nical ly convex functions via fractional integrals,

Appl. Math. Compt., 238 (2014), 237-244.

[16] ·

I. ·

I¸scan, O n generaliza tion of di¤ erent type integral inequalities for s-convex fu nctions via fractio nal integrals,

Math. Sci. Appl., 2 (2014), 55–67. 1

[17] M. Jleli and B. Samet On Hermite-Hadamard type inequalities via fractional integrals of a function with respect to

another function, Journal of Nonlinear Sciences and Applications. 2016, 9(3), 1252-1260.

[18] U.N. Katugampola, New approach to a gen eralized fractional integral, Appl. Math. Comput. 218(3) (2011) 860–865.

[19] R. K halil, M. Al Horani, A. Yousef and M. Sababheh, A n ew d e…nition of fract iona l de rivat iv e, J. Comput. Appl.

Math. 264. pp. 6570, 2014.

[20] A. A. Kilbas, H. M. Srivastava and J. J. Trujillo, Theory and App licatio ns of Fractional Di¤ erential Equations,

North-Holland Mathematics Studies, 204, Elsevier Sci. B.V., Amsterdam, 2006.

[21] M. Kirane, B. T. Torebek, Hermite-Hadamard, Hermite-Hadamard-Fejer, Dragomir-Agarw al and Pachpatte type

inequalities for convex functions via fractional integrals, arXiv:1701.00092.

[22] U. S. Kirmaci, Inequalities for di¤ erentiable mappings and applications to special m eans of real numbers to midpoint

form ula, Appl. Math. Comput., vol. 147, no. 5, pp. 137–146, 2004, doi: 10.1016/S0096-3003(02)00657-4.

[23] S. M ubeen and G. M Habibullah, k-Fractional integrals a nd application, Int. J. Contemp. Math. Sciences, Vol. 7,

2012, no. 2, 89 - 94.

[24] M. A. Noor and M. U. Awan, Some integral inequ alities for two kinds of convexities via fractional integrals, TJMM,

5(2), 2013, pp. 129-136.

[25] J.E. Peµ

cari´

c, F. Proschan and Y.L. Tong, Conv ex Functions, Partial Orderings and Statistical A pplications, Acad-

emic Press, Boston, 1992.

[26] M. E. Özdemir, M. Avc¬-Ard¬ç and H. Kavurmac¬-Önalan, Hermite-Hadamard type inequalities for s-con vex and

s-concave function s via fraction al integrals, Turkish J.Science,1(1), 28-40, 2016.

[27] M. E. Ödemir, M. Avci, and E. Set, On some inequalities of Herm ite–Hadamard-type via m-convexity, Appl. Math.

Lett. 23 (2010), pp. 1065–1070.

[28] M. E. Ödemir, M . Avci, and H. Kavurmaci, Hermite–Hadamard-type inequalities via (; m)-convexity, Comput.

Math. Appl. 61 (2011), pp. 2614–2620.

[29] A. Saglam, M. Z. Sarikaya ve H. Yildirim, Some new inequalities of H ermite-H adamard’s type, Kyungpook Math-

ematical Journal, 50(2010), 399-410.

[30] M.Z. Sarikaya and F. Ertu¼

gral, On the generalized Hermite -Hadamard inequalities, Annals of the University of

Craiova - Mathematics and Computer Science Series, in press 2019.

[31] M.Z. Sarikaya and H. Yildirim, On Hermite-Hadamard type inequalitie s for Riemann-Liouville fractional integrals,

Miskolc Mathematical Notes, 7(2) (2016), pp. 1049–1059.

[32] M.Z. Sarikaya, E. Set, H. Yaldiz and N., Basak, Hermite -Hadamard’s inequalities for fractional integrals an d

related fractional inequalities, Mathematical and Computer Modelling, DOI:10.1016/j.mcm.2011.12.048, 57 (2013)

2403–2407.

[33] M.Z. Sarikaya and H. Budak, Generalized Hermite-Hadamard type integral inequalities for fractional integrals,

Filomat 30:5 (2016), 1315–1326.

[34] M.Z. Sarikaya, A. Akkurt , H. Budak, M. E. Yildirim and H Yildirim, Hermite-hadamard’s inequalities for con-

form able fractio nal in tegrals. RGMIA Research Report Collection, 2016;19(83).

[35] E. Set, M. E. Ozdemir and M. Z. Sarikaya, New inequalities o f Ostrowski’s type for s-conve x functions in the second

sense with applications, Facta Universitatis, Ser. Math. Inform. Vol. 27, No 1 (2012), 67-82.

[36] E. Set, M. Z. Sarikaya, M. E. Ozdemir and H. Yildirim, The Hermite-Had ama rd’s inequality for som e convex

functions via fractional integrals and related results, Journal of Applied Mathem atics, Statistics and Informatics

(JAMSI), 10(2), 2014.

[37] F. Usta, H. Budak, M.Z. Sar¬kaya and H y¬ld¬r¬m, Some Hermite-Hadam ard and O s-

trowski type in equalities for fractional integral operators with exponential kernel, ResearchGate:

https://www.researchgate.net/publication/314471240.

[38] J. Wang, X. Li, M. Feµ

ckan, Y. Zhou, Hermite–Hadam ard-type inequalitie s for Riemann–L iouville fractional integrals

via tw o kind s of conve xity, Appl. Anal. 92 (11) (2012) 2241–2253.

14 FATM A ERT U ¼

GR AL, M EH MET ZE KI SA RIKAYA , AND HÜ SE YIN B UD AK

[39] J. Wang, X. Li, C. Zhu, Re…nements of Herm ite-Hadamard type inequalitie s involving fractional integrals Bull.

Belg. Math. Soc. Simon Stevin, 20 (2013), 655–666.

[40] Y. Zhang and J. Wang, On some new Hermite-Hadamard inequalities involv ing Rieman nLiouville fractional inte-

grals. J. Inequal. Appl. 2013, 220 (2013).

[41] P. O. Mohammed and M. Z. Sar¬kaya, O n generalized fractional integral inequalities for twice di¤ erentiable convex

functions, Rev. R. Acad. Cienc. Exactas Fís. Nat. Ser. A Mat. RACSAM, in press, May 2019.

Department of Mathematics, Faculty of Science and Arts, Düzce University, Düzce, Turkey

E-m ail ad dress :fatmaertugral14@gmail.com

Department of Mathematics, Faculty of Science and Arts, Düzce University, Düzce-Turkey

E-m ail ad dress :sarikayamz@gmail.com

Department of Mathematics, Faculty of Science and Arts, Düzce University, Düzce, Turkey

E-m ail ad dress :hsyn.budak@gmail.com