ON THE MIDPOINT TYPE INEQUALITIES VIA GENERALIZED FRACTIONAL INTEGRAL OPERATOR HATICE YALDIZ AND MEHMET ZEKI SARIKAYA Abstract. In this paper, using a general class of fractional integral operators, we establish some new fractional integral inequalities of midpoint type. The main results are used to derive Hermite-Hadamard type inequalities involving the general class of fractional integral operators.
1. Introduction Let f : I R ! R be a convex mapping de…ned on the interval I of real numbers and a; b 2 I, with a < b: The following double inequality is well known in the literature as the Hermite-Hadamard inequality [5]: Z b a+b 1 f (a) + f (b) (1.1) f f (x) dx : 2 b a a 2 The most well-known inequalities related to the integral mean of a convex function are the Hermite Hadamard inequalities. In [2], Kirmaci proved the following results connected with the left part of (1.1). Theorem 1. Let f : I R ! R be a di¤ erentiable mapping on I , a; b 2 I with a < b. If jf 0 j is convex on [a; b], then we have Z b 1 a+b (b a) (1.2) f (x)dx f (jf 0 (a)j + jf 0 (b)j) : b a a 2 8 In [8], Sarikaya et al. gave the following interesting Riemann-Liouville integral inequalities of Hermite-Hadamard-type: Theorem 2. 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: (1.3) with
f
a+b 2
( + 1) Ja+ f (b) + Jb f (a) 2 (b a)
f (a) + f (b) 2
> 0:
Meanwhile, in [13] Iqbal et al. gave the following results connected with the left part of Riemann-Liouville integral inequalities of Hermite-Hadamard type (1.3). Key words and phrases. Fractional integral operator, convex function and Hermite-Hadamard inequality. 2010 Mathematics Sub ject Classi…cation 26A33;26D10;26D15;33E20. 1
2
HATICE YALDIZ AND M EHM ET ZEKI SARIKAYA
Lemma 1. Let f : [a; b] ! R be a di¤ erentiable function on (a; b): If f 0 2 L1 [a; b] ; then the following identity for Riemann-Liouville fractional integrals holds: (1.4)
4
( + 1) b aX Ja+ f (b) + Jb f (a) = Ik ; : 2 2 (b a)
a+b 2
f
k=1
where 1
Z2
I1 =
I3 =
1
t f 0 (tb + (1
t)a) dt;
I2 =
0
Z1
( t ) f 0 (ta + (1
t)b) dt;
0
1) f 0 (tb + (1
(t
Z2
t)a) dt; I4 =
1 2
Z1
(1
t ) f 0 (ta + (1
t)b) dt:
1 2
Theorem 3. Let f : [a; b] ! R be a di¤ erentiable function on (a; b) with a < b: If jf 0 j is convex on [a; b], then the following inequality for Riemann-Liouville fractional integrals holds for 0 < 1: ( + 1) Ja+ f (b) + Jb f (a) 2 (b a)
a+b 2
f
(1.5) b 2
+1
a [jf 0 (a)j + jf 0 (b)j] : ( + 1)
For some recent results connected with fractional integral inequalities see([8][11]) In [7], Raina de…ned the following results connected with the general class of fractional integral operators. (1.6)
F
;
(x) = F
(0); (1);::: ;
(x) =
1 X
(k) xk ( k+ )
k=0
( ;
> 0; jxj < R) ;
where the coe¢ cents (k) (k 2 N0 = N[ f0g) is a bounded sequence of positive real numbers and R is the set of real numbers. With the help of (1.6), Raina and Agarwal et. al de…ned the following left-sided and right-sided fractional integral operators, respectively, as follows: Z x 1 (1.7) J ; ;a+;! '(x) = (x t) F ; [! (x t) ] '(t)dt; x > a; a
(1.8)
J
; ;b
;! '(x) =
Z
b
(t
x)
1
x
F
;
[! (t
x) ] '(t)dt; x < b;
where ; > 0; ! 2 R, and ' (t) is such that the integrals on the right side exists. It is easy to verify that J ; ;a+;! '(x) and J ; ;b ;! '(x) are bounded integral operators on L (a; b), if M :=F
(1.9)
; +1
[! (b
a) ] < 1:
In fact, for ' 2 L (a; b), we have (1.10)
J
; ;a+;! '(x) 1
M (b
a) k'k1
ON THE M IDPOINT TYPE INEQUALITIES
3
and J
(1.11)
; ;b ;! '(x)
where
0
k'kp := @
M (b
1
Zb a
p
a) k'k1 ; 1 p1
j' (t)j dtA :
The importance of these operators stems indeed from their generality. Many useful fractional integral operators can be obtained by specializing the coe¢ cient (k). Here, we just point out that the classical Riemann-Liouville fractional integrals Ia+ and Ib of order de…ned by (see, [3, 4, 6]) Z x 1 1 (1.12) (Ia+ ') (x) := (x t) '(t)dt (x > a; > 0) ( ) a and (1.13)
(Ib ') (x) :=
1 ( )
Z
b
(t
x)
1
'(t)dt (x < b;
> 0)
x
follow easily by setting (1.14)
= ,
(0) = 1, and w = 0
in (1.7) and (1.8), and the boundedness of (1.12) and (1.13) on L (a; b) is also inherited from (1.10) and (1.11), (see, [1]). In [12], Yaldiz and Sarikaya proved the following theorem Hermite-Hadamard type inequalities involving the general class of fractional integral operators. Theorem 4. Let ' : [a; b] ! R be a convex function on [a; b] with a < b, then the following inequalities for fractional integral operators hold: '
a+b 2
(1.15) 1 2 (b
a) F
; +1
[! (b
a) ]
h
J
; ;a+;! '
(b) + J
; ;b ;! '
i (a)
' (a) + ' (b) : 2 with
> 0:
Our aim in this paper is to continue our work on the Hermite-Hadamard type inequalities involving the general class of fractional integral operators. We state and prove some re…nements on the left side of the inequality (1.15). 2. Main Results In this section, using fractional integral operators, we start with stating and proving the fractional integral counterparts Lemma 1, Theorem 3 and Theorem 4. Then some other re…nements will folllow. We begin by the following lemma:
4
HATICE YALDIZ AND M EHM ET ZEKI SARIKAYA
Lemma 2. Let ' : [a; b] ! R be a di¤ erentiable mapping on (a; b) with a < b and ; > 0; ! 2 R: If '0 2 L [a; b] ; then the following equality for fractional integrals holds: a+b 2
'
1 2 (b
a) F
(2.1) =
a) 1 2F ; +1 [! (b
a) ]
1 2F ;1 +1 [! (b
1
(k) :=
(k) 2 k
a) ]
J
;
;a+;! ' (b) + J
Hk
Z1
d F dt
; +1
[! (b
a) t ] ' ((1
Z1
d F dt
; +1
[! (b
a) t ] ' (ta + (1
t) a + tb) dt
1 2
a) ]
t)b) dt:
1 2
and 1
Z2
H1 =
t F
; +1
a) t ] '0 ((1
[! (b
t)a + tb) dt;
0
1
H2 =
H3 =
Z2
t
0 Z1
t
Z1
1
F
; +1
[! (b
a) t ] '0 (ta + (1
t)b) dt;
1 F
; +1
[! (b
a) t ] '0 ((1
F
; +1
[! (b
a) t ] '0 (ta + (1
t)a + tb) dt;
1 2
H4 =
t
t)b) dt:
1 2
Proof. Here, we apply integration by parts, then we have 1
H1
=
Z2
t F
; +1
a) t ] '0 ((1
[! (b
t)a + tb) dt
0
(2.2) =
; ;b
k=1
a) ]
1 2F ;1 +1 [! (b where
4 X
(b
[! (b
1
; +1
h
1 F 2 (b a)
b
; +1
!
1
[! (b
a 2
'
a+b 2
1
1 b
a
Z2 0
t
F
;
a) t ] ' ((1
t)a + tb) dt;
i ' (a) ;!
ON THE M IDPOINT TYPE INEQUALITIES
5
1
H2
Z2
=
F
t
; +1
a) t ] '0 (ta + (1
[! (b
t)b) dt
0
(2.3) 1 F 2 (b a)
=
b
; +1
!
1
[! (b
a
'
2
a+b 2
1
1 b
a
Z2
t
F
;
a) t ] ' (ta + (1
t)b) dt;
0
H3
=
Z1
t
1 F
; +1
a) t ] '0 ((1
[! (b
t)a + tb) dt
1 2
(2.4) 1 F 2 (b a)
= +
1 (b
a) 1
b
a
F
Z1
t
Z1
d F dt
t
F
1
F
a a
'
2
[! (b
;
'
2
b
!
; +1
b
!
; +1
a+b 2
a+b 2
a) t ] ' ((1
t)a + tb) dt
1 2
+
1 b
a
; +1
[! (b
a) t ] ' ((1
t) a + tb) dt;
a) t ] '0 (ta + (1
t)b) dt
1 2
H4
=
Z1
1
; +1
[! (b
1 2
(2.5) 1 F 2 (b a)
= +
1 (b
a) 1
b
a
F
!
; +1
Z1
t
Z1
d F dt
1
!
; +1
F
;
b
a
'
2
b
a 2
[! (b
'
a+b 2
a+b 2
a) t ] ' (ta + (1
t)b) dt
1 2
+
1 b
a
1 2
; +1
[! (b
a) t ] ' (ta + (1
t)b) dt:
6
HATICE YALDIZ AND M EHM ET ZEKI SARIKAYA
Adding above equalities (2.2),(2.3),(2.4) and (2.5), we get 4 X
Hk
=
k=1
(2.6)
2 (b
a)
1 b
a
b
!
a
'
a+b 2
F
; +1
Z1
t
1
Z1
t
1
Z1
d F dt
; +1
[! (b
a) t ] ' (ta + (1
Z1
d F dt
; +1
[! (b
a) t ] ' ((1
2
F
;
[! (b
a) t ] ' (ta + (1
F
;
[! (b
a) t ] ' ((1
t)b) dt
0
1 b
a
t)a + tb) dt
0
+
1 b
a
t)b) dt
1 2
+
1 b
a
t) a + tb) dt:
1 2
Now we use the substitution rule last integrals in (2.6), after by using de…nition of left and right-sided fractional integral operator, we obtain proof of this lemma. Remark 1. If in Lemma 2, we get = , (0) = 1, and w = 0; then the inequalities (2.1) become the inequalities (1.4) of Lemma 1. We have the following results: Theorem 5. Let ' : [a; b] ! R be a di¤ erentiable mapping on (a; b) with a < b and ; > 0; ! 2 R: If j'0 j and j'j are convex on [a; b], then the following inequality for fractional integrals holds: 1 2 (b
a) F
1
; +1
[! (b
a) ]
(2.7)
F and
2
1
;
1 [! (b +1
(k) :=
(k)
h
J
;
;a+;! ' (b) + J
j'0 (a)j + j'0 (b)j F 2 a) ] 1 2
k+ +1 (
k+ +1)
+
1 2
2
; +1
[! (b
; ;b
a) ] +
( + 1; k + 1) and
i ' (a) ;!
t2 j
jt1
t2 j ;
a+b 2
j' (a)j + j' (b)j F 2 3
(k) :=
Proof. Using Lemma 2 and the convexity of j'0 j and j'j, and fact that for and 8t1 ; t2 2 [0; 1] ; jt1
'
3
; +1
(k) 1 2 (0; 1]
1 2
k
[! (b
:
a) ] :
ON THE M IDPOINT TYPE INEQUALITIES
7
we …nd that a+b 2
'
(b
2 (b
a) 2F ;1 +1 [! (b 1
Z2 + t
1 X
k=0
0
Z1 + (1
t)
Z1 + (1
t)
1 2
Z1 + 1 2
Z1 + 1 2
1 X
k=0
1 X
k=0
J
;
;a+;! ' (b) + J
a) F ;1 +1 [! (b a) ] 2 1 Z2 1 X (k) 6 t ! k (b 4 ( k + + 1) a) ] 0
1 X 1 X
k=0
a)
k
t
k=0
(k) ! k (b ( k + + 1)
k=0
1 2
h
1
a)
k
t
(k) ! k (b ( k + + 1)
(k) ! k (b ( k + + 1)
(k) ! k (b ( k + + 1)
(k) ! k (b ( k + + 1)
a)
k
a)
k
kt
kt
k
!
[t j'0 (a)j + (1
a)
k
a)
k
t
k
!
t
k
!
k 1
!
k 1
!
[(1
[(1
k
!
; ;b
[(1
i ' (a) ;!
t) j'0 (a)j + t j'0 (b)j] dt
t) j'0 (b)j] dt
t) j'0 (a)j + t j'0 (b)j] dt
[t j'0 (a)j + (1
t) j'0 (b)j] dt
t) j' (a)j + t j' (b)j] dt
[t j' (a)j + (1
t) j' (b)j] dt:
Now we use the substitution rule above integrals, after by using de…nition of left and right-sided fractional integral operator. This completes the proof. Remark 2. If in Theorem 5, we get = , (0) = 1, and w = 0; then the inequalities (2.7) become the inequalities (1.5) of Theorem 3. Remark 3. If in Theorem 5, we get = 1, (0) = 1, and w = 0; then, the inequalities (2.7) become the inequalities (1.2) of Theorem 1. References [1] R. P. Agarwal, M.-J. Luo and R. K. Raina, On Ostrowski type inequalities, Fasciculi Mathematici, 204, De Gruyter, doi:10.1515/fascmath-2016-0001, 2016. [2] U. S. Kirmaci, Inequalities for di¤ erentiable mappings and applications to special means of real numbers and to midpoint formula, Applied Mathematics and Computation 147 (2004) 137–146 [3] A. A. Kilbas, H. M. Srivastava and J. J. Trujillo, Theory and applications of fractional di¤ erential equations, North-Holland Mathematics Studies, 204, Elsevier Sci. B.V., Amsterdam, 2006. [4] R. Goren‡o and F. Mainardi, Fractional calculus: integral and di¤ erential equations of fractional order, Springer Verlag, Wien (1997), 223-276. [5] J. Hadamard, Etude sur les proprietes des fonctions entieres et en particulier d’une fonction considree par, Riemann, J. Math. Pures. et Appl. 58 (1893), 171–215.
8
HATICE YALDIZ AND M EHM ET ZEKI SARIKAYA
[6] S. Miller and B. Ross, An introduction to the fractional calculus and fractional di¤ erential equations, John Wiley & Sons, USA, 1993, p.2. [7] R.K. Raina, On generalized Wright’s hypergeometric functions and fractional calculus operators, East Asian Math. J., 21(2) (2005), 191-203. [8] M. Z. Sarikaya, E. Set, H. Yaldiz and N. Basak, Hermite-Hadamard 0 s inequalities for fractional integrals and related fractional inequalities, Mathematical and Computer Modelling, 57 (2013) 2403–2407. [9] M. Z. Sarikaya and H. Yaldiz, On generalization integral inequalities for fractional integrals, Nihonkai Math. J., Vol.25(2014), 93–104. [10] M. Z. Sarikaya, H. Yaldiz and N. Basak, New fractional inequalities of Ostrowski-Grüss type, Le Matematiche, Vol. LXIX (2014)-Fasc. I, pp. 227-235 . [11] M. Z. Sarikaya and H. Yaldiz, On Hermite-Hadamard Type Inequalities for '-convex Functions via Fractional Integrals, Malaysian Journal of Mathematical Sciences, 9(2): (2015) 243-258. [12] H. Yaldiz and M. Z. Sarikaya, On the Hermite-Hadamard type inequalities for fractional integral operator, ( 2016) submitted. [13] M. Iqbal, M. I. Bhatti, and K. Nazeer, Generalization of Inequalities Analogous to Hermite– Hadamard Inequality via Fractional Integrals, Bull. Korean Math. Soc, 52 (2015), No. 3, 707–716. Department of Mathematics, Faculty of Science and Arts, Düzce University, Düzce, Turkey E-mail address :
[email protected] Department of Mathematics, Faculty of Science and Arts, Düzce University, Düzce, Turkey E-mail address :
[email protected]
