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ON HERMITE-HADAMARD TYPE INEQUALITIES ASSOCIATED WITH THE GENERALIZED FRACTIONAL INTEGRALS

Authors:

Abstract

In this paper, we obtain new generalization of Hermite-Hadamard inequalities via generalized fractional integrals de…ned by Sarikaya and Ertu¼ gral in [30]. We establish some midpoint and trapezoid type inequalities for functions whose …rst derivatives in absolute value are convex involving generalized fractional integrals.
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 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 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
2f(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
2f(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
2fa+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 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
2fa+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
2fa+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
2fa+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 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
2fa+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 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 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.
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GR AL, M EH MET ZE KI SA RIKAYA , AND SE YIN B UD AK
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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
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