HERMITE-HADAMARD INEQUALITIES INVOLVING THE GAUSS HYPERGEOMETRIC FUNCTION MEHMET ZEKI SARIKAYA Abstract. The goal of this study is obtained the new generalization Hermite-Hadamard-Fejer inequalities involving the Gauss hypergeometric function. The results presented here would provide some fractional inequalities involving Saigo, Erdelyi-Kober and Riemann-Liouville type fractional operators.
1. Introduction The theory of convex functions is the most important theory in mathematics which touches almost all branches of mathematics such as optimization theory, control theory, operations research, geometry, functional analysis, information theory etc. This theory plays a signi…cant role in other branches of sciences such as economics, …nance, engineering and management sciences. The following inequality is well known in the literature as the Hermite-Hadamard integral inequality (see, [7]):
(1.1)
f
a+b 2
1 b
a
Z
b
f (x)dx
a
f (a) + f (b) 2
where f : I R ! R is a convex function on the interval I of real numbers and a; b 2 I with a < b. The most well-known inequalities related to the integral mean of a convex function f are the Hermite Hadamard inequalities or its weighted versions, the so-called Hermite-Hadamard-Fejer inequalities. For some results which generalize, improve, and extend the inequalities (1.1) and Hermite-HadamardFejer, (see [6], [11], [12], [15]-[22]) which has generated a wide range of directions for extension and a rich mathematical literature. Fractional calculus is a branch of mathematics, which has a long history and has recently gone through a period of rapid development. Many earlier works on the subject of fractional calculus contain interesting accounts of the theories of fractional calculus operators and their applications in diverse research areas. These considerations have led various researchers in the …eld of integral inequalities to explore certain extensions and generalizations by involving fractional calculus operators. One may, for instance, refer to such type of works in the books [2], [8], [9] and the papers [1], [3], [4], [5], [13]-[18], [23]. De…nition 1. A real-valued function f (x) (x > 0) is said to be in the space C ( 2 R), if there exists a real number p > such that f (x) = xp (x); where (x) 2 C(0; 1). De…nition 2. For real numbers I0;x; ; f (x) =
x
> 0; Z
( )
and ; the Saigo hypergeometric fractional integral operator x
(x
1
t)
2 F1
+ ;
; ;1
0
t x
f (t)dt
with the Gauss hpergeometric function 2 F1 (a; b; c; z) in the kernel is de…ned 2 F1
(a; b; c; z) =
1 X (a)n (b)n z n ; (a)n = (c)n n! n=0
(a + n) : (a)
This operator has been initially introduced by Saigo in a series of his papers for studying boundary value problems for partial di¤erential equations, especially for the Euler-Darboux equation, see [13]. Key words and phrases. Convex function, Hermite-Hadamard inequality, Gauss hypergeometric function. 2010 Mathematics Sub ject Classi…cation 26A09;26A33;26D10;26D15;33E20. 1
2
M EHM ET ZEKI SARIKAYA
In [4], Curiel and Gaue introduced the following fractional hypergeometric operator 2 Z x x t 1 (1.2) I0;x; ; ; f (x) = (x t) t 2 F1 + + ; ; ;1 f (t)dt ( ) x 0 where > 0; > 1; , 2 R and f is a real-valued continuous function. In this paper, …stly we de…ne a new de…nition of fractional hypergeometric fractional operators as follows. De…nition 3. Let > 0; > 1; , 2 R, and f : [a; b] ! R be a real-valued continuous function with 0 a < b. Then, the generalized fractional hypergeometric operators are de…ned by Ia+;
(1.3) =
(b
; ;
f (b)
a)
2
Z
( ) Ib ;
=
(b
(t
a) (b
t)
1
2 F1
+
+ ;
; ;
b b
t a
f (t)dt
(b
t) (t
a)
1
2 F1
+
+ ;
; ;
t b
a a
f (t)dt:
a
and (1.4)
b
; ;
f (a)
a)
2
( )
Z
b
a
By de…nition of the Gauss hypergeometric function, it can be seen in [10] Z y (m) (n) (m + n k l) m+n t dt = y (1.5) tm 1 (y t)n 1 2 F1 k; l; n; 1 y (m + n k) (m + n l) 0
1
:
Remark 1. If we take a = 0 in (1.3), the operator (1.3) reduces to the operator (1.2). Remark 2. If we take type operators Ja+;
;
f (b)
= 0 in De…nition 3, the operators (1.3) and (1.4) reduce to Saigo fractional = =
Jb ;
;
f (a)
= =
Remark 3. If we take = 0; Kober fractional operators ;
Ia+; (b
; ;0
f (b) Z
a) ( )
Ib ; (b
; ;0
b
(b
t)
1
2 F1
+ ;
; ;
b b
t a
f (t)dt;
(t
a)
1
2 F1
+ ;
; ;
t b
a a
f (t)dt:
a
f (a) Z b
a) ( )
a
= 0 in De…nition 3, the operators (1.3) and (1.4) reduce to Erdelyi;0; ;0
Ja+ f (b) = Ia+
Jb ; f (a) = Ib ;0;
;0
f (b) = f (a) =
Remark 4. If we take = 0; = Riemann-Liouville fractional operators ;
(b (b
a) ( ) a) ( )
Z
b
(b
t)
1
(t
a) f (t)dt;
(t
a)
1
(b
t) f (t)dt:
a
Z
b
a
in De…nition 3, the operators (1.3) and (1.4) reduce to
(Ja+ f ) (b) = Ia+
; ;0
(Jb f ) (a) = Ib ;
; ;0
f (b) =
f (a) =
1 ( ) 1 ( )
Z
b
(b
t)
1
f (t)dt;
(t
a)
1
f (t)dt:
a
Z
b
a
In this study, we obtain the new generalization Hermite-Hadamard and Hermite-Hadamard-Fejer inequalities involving the Gauss hypergeometric function. New results obtained from the this study would provide some fractional inequalities involving Saigo, Erdelyi-Kober and Riemann-Liouville type fractional operators.
HERM ITE-HADAM ARD INEQUALITIES INVOLVING THE GAUSS HYPERGEOM ETRIC FUNCTION
3
2. Main Results In order to prove our main results we start the following theorem: Theorem 1. Let > 0; > 1; , 2 R. If f : [a; b] ! R be a convex function on [a; b] with 0 a < b, then the following inequalities for the generalized fractional hypergeometric operators hold:. a+b 2
f
( ) ( + 1) ( + 1) (1 ) ( + + + 1) h
( )
(2.1)
2 (b
a)
Ia+;
; ;
f (b) + Ib ;
i f (a)
; ;
( ) ( + 1) ( + 1) f (a) + f (b) : (1 ) ( + + + 1) 2 Proof. For t 2 [0; 1]; let x = ta + (1 (2.2)
t)b; y = (1
a+b 2
f
=f
t)a + tb: The convexity of f yields
x+y 2
f (x) + f (y) 2
i.e., (2.3)
a+b 2
2f
f (ta + (1
t)b) + f ((1
t)a + tb) :
Multiplying both sides of (2.3) by t 1 (1 t) 2 F1 ( + + ; ; ; t), then integrating the resulting inequality with respect to t over (0; 1), we obtain Z 1 a+b 2f t 1 (1 t) 2 F1 ( + + ; ; ; t) dt 2 0 Z1
t
1
(1
t)
2 F1
( +
+ ;
; ; t) f (ta + (1
t)b) dt
0
Z1 + t
1
(1
t)
2 F1
( +
+ ;
; ; t) f ((1
t)a + tb) dt:
0
As consequence, by using (1.5) and by changing the variable, we obtain (2.4)
2f
(b
a+b 2 ( ) a)
h
( ) ( + 1) ( + 1) (1 ) ( + + + 1) Ia+;
; ;
f (b) +
Ib ;
; ;
i f (a)
and the …rst inequality is proved. To prove the other half of the inequality in (2.1), since f is convex, for every t 2 [0; 1], we have, (2.5)
f (ta + (1
t)b) + f ((1
t)a + tb)
f (a) + f (b) :
Then multiplying both sides of (2.5) by t 1 (1 t) 2 F1 ( + + ; ; ; t) and integrating the resulting inequality with respect to t over (0; 1), we obtain h i ( ) Ia+; ; ; f (b) + Ib ; ; ; f (a) (b a) ( ) ( + 1) ( + 1) [f (a) + f (b)] (1 ) ( + + + 1) and the second inequality is proved.
4
M EHM ET ZEKI SARIKAYA
Corollary 1. Under the assumption of Theorem 1 with = 0; = 0; then we have a+b ( + + 1) f (a) + f (b) (2.6) f Ja+; f (b) + Jb ; f (a) : 2 2 2 (b a) Remark 5. If we take = 0; = the inequalities (2.6)) reduce to a+b 2
f
in Theorem 1(or
= 0 in Corollary 1), the inequalities (2.1)(or
( + 1) Ja+ f (b) + Jb f (a) 2 (b a)
f (a) + f (b) 2
which is proved by Sarikaya et al. in [15]. Lemma 1. Let > 0; with 0 a < b, then
>
1;
,
2 R. If g : [a; b] ! R is integrable and symmetric to (a + b)=2
i 1h Ia+; ; ; g (b) + Ib ; ; ; g (a) : 2 Proof. Since g is symmetric to (a + b)=2; then for all x 2 [a; b]; it can write g(a + b Therefore, by changing the variable y = a + b t, it follows that Ib ;
Ib ; = = =
(b
; ;
g (a) = Ia+;
; ;
g (a)
a)
2
Z
( ) (b
Z
2
( ) ; ;
b
(y
a)
(b
t)
g (b) =
1
(b
y)
2 F1
+
+ ;
; ;
a
a)
Ia+;
; ;
b
1
(t
a)
2 F1
+
+ ;
; ;
a
y b
b b
a a
x) = g(x).
g (y) dy
t a
g (a + b
t) dt
g (b)
which is completes the proof. Let > 0; > 1; , 2 R. If f : [a; b] ! R be a convex function on [a; b] with 0 a < b: If g : [a; b] ! R is nonnegative, integrable and symmetric to (a + b)=2; then the following inequalities for the generalized fractional hypergeometric operators hold:. i a+b h (2.7) f Ia+; ; ; g (b) + Ib ; ; ; g (a) 2 h
( ) (b
a)
Ia+;
f (a) + f (b) h Ia+; 2 Proof. For t 2 [0; 1]; let x = ta + (1 (2.8)
2f
; ;
(f g) (b) +
Ib ;
; ;
g (b) + Ib ;
; ;
t)b; y = (1
a+b 2
f (ta + (1
; ;
i (f g) (a)
i g (a) :
t)a + tb: Since f is an convex function, then we have t)b) + f ((1
t)a + tb) :
Multiplying both sides of (2.8) by t 1 (1 t) 2 F1 ( + + ; ; ; t) g ((1 t)a + tb), then integrating the resulting inequality with respect to t over (0; 1), we obtain Z 1 a+b 2f t 1 (1 t) 2 F1 ( + + ; ; ; t) g ((1 t)a + tb) dt 2 0 Z1
t
1
(1
t)
2 F1
( +
+ ;
; ; t) f (ta + (1
t)b) g ((1
t)a + tb) dt
0
Z1 + t 0
1
(1
t)
2 F1
( +
+ ;
; ; t) f ((1
t)a + tb) g ((1
t)a + tb) dt:
HERM ITE-HADAM ARD INEQUALITIES INVOLVING THE GAUSS HYPERGEOM ETRIC FUNCTION
As consequence, by changing the variable x = ta + (1 t)b; t = bb xa , and y = (1 t)a + tb; t = yb we obtain Z b y a a+b 1 1 (y a) (b y) 2 F1 + + ; ; ; 2f g (y) dy + 2 b a (b a) a Z b b x 1 1 (b x) (x a) 2 F1 + + ; ; ; f (x) g (a + b x) dx + b a (b a) a Z b 1 y a 1 + (y a) (b y) 2 F1 + + ; ; ; f (y) g (y) dy: + b a (b a) a a+b 2
Ib ;
; ;
a a;
a+b 2 ;
By using symetry of function g about 2f
5
it follows that h ( ) Ia+; (b a)
g (a)
; ;
f g (b) +
Ib ;
; ;
i f g (a)
and the …rst inequality is proved. To prove the other half of the inequality in (2.7), since f is convex, for every t 2 [0; 1], we have, (2.9)
f (ta + (1
t)b) + f ((1
t)a + tb)
f (a) + f (b) :
Then multiplying both sides of (2.9) by t 1 (1 t) 2 F1 ( + + ; ; ; t) g ((1 integrating the resulting inequality with respect to t over (0; 1), we obtain Z1
t
1
(1
t)
2 F1
( +
+ ;
; ; t) g ((1
t)a + tb) f (ta + (1
t)a + tb) and
t)b) dt
0
Z1 + t
1
(1
2 F1
t)
( +
+ ;
; ; t) g ((1
t)a + tb) f ((1
t)a + tb) dt
0
[f (a) + f (b)]
Z1
t
1
(1
t)
2 F1
( +
+ ;
; ; t) g ((1
t)a + tb) dt:
0
By changing the variable x = ta + (1 ( ) (b
a)
h
Ia+;
; ;
t)b; t = Ib ;
f g (b) +
b x b a, ; ;
and y = (1
i f g (a)
t)a + tb; t =
[f (a) + f (b)] Ib ;
y a b a; ; ;
it can be seen
g (a)
and the second inequality is proved. Corollary 2. Under the assumption of Theorem 2 with (2.10)
a+b 2
f
f (a) + f (b) 2
f
a+b 2
Ja+ g(b) + Jb g(a)
which is proved by · Is¸can in [11].
= 0; then we have
Ja+; g (b) + Jb ; g (a)
( + + 1) (b a)
Remark 6. If we take = 0; = (or the inequalities (2.10)) reduce to
= 0;
Ja+; (f g) (b) + Jb ; (f g) (a) Ja+; g (b) + Jb ; g (a) :
in Theorem 2 (or
= 0 in Corollary 2), the inequalities (2.7)
( + 1) Ja+ (f g) (b) + Jb (f g) (a) (b a)
f (a) + f (b) Ja+ g(b) + Jb g(a) 2
6
M EHM ET ZEKI SARIKAYA
References [1] T. Ali, M. A. Khan and Y. Khurshidi, Hermite-Hadamard inequality for fractional integrals via eta-convex functions, Acta Mathematica Universitatis Comenianae, 86(1), (2017), 153-164. [2] G. A. Anastassiou, Advances on fractional inequalities, Springer Briefs in Mathematics, Springer, New York, NY, USA, 2011. [3] D. Baleanu, S. D. Purohit and P. Agarwal, On Fractional Integral Inequalities Involving Hypergeometric Operators, Chinese Journal of Mathematics, Volume 2014, Article ID 609476, 5 pages. [4] L. Curiel and L. Galue, A generalization of the integral operators involving the Gauss hypergeometric function, Revista Tecnica de la Facultad de Ingenieria Univ. del Zulia, 19(2), 17-22, 1996. [5] Z. Dahmani, On Minkowski and Hermite-Hadamard integral inequalities via fractional integration, Ann. Funct. Anal. 1(1) (2010), 51-58. [6] S. S. Dragomir and R.P. Agarwal, Two inequalities for di¤ erentiable mappings and applications to special means of real numbers and to trapezoidal formula, Appl. Math. lett., 11(5) (1998), 91-95. [7] S. S. Dragomir and C. E. M. Pearce, Selected Ttopics on Hermite-Hadamard inequalities and applications, RGMIA Monographs, Victoria University, 2000. [8] 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. [9] V. Kiryakova, Generalized fractional calculus and applications, Longman Sci. & Technical, Harlow, UK (1994). [10] A.M.Mathai, R. K. Saxena, and H. J. Haubold, The H-Function: Theory and Applications, Springer, Dordrecht,The Netherlands, 2010. [11] I·. I·¸scan, Hermite-Hadamard-Fejér type integral inequalities for convex functions via fractional integrals, Stud. Univ. Babes-Bolyai Math. 60(2015), No. 3, 355–366. [12] M.E. Ozdemir, S.S. Dragomir and C. Yildiz, The Hadamard’s inequality for convex function via fractional integrals, Acta Mathematica Scientia 2013,33B(5):1293–1299. [13] M. Saigo, A remark on integral operators involving the Gauss hypergeometric functions, Math. Rep. Kyushu Univ. 11 (1978), 135-143. [14] M.Z. Sar¬kaya and H. Ogunmez, On new inequalities via Riemann-Liouville fractional integration, Abstract an Applied Analysis, 2012 (2012) 10 pages, Article ID 428983. doi:10.1155/2012/428983. [15] M.Z. Sar¬kaya, E. Set, H. Yald¬z and N. Basak, Hermite-Hadamard’s inequalities for fractional integrals and related fractional inequalities, Mathematical and Computer Modelling, 57(9) (2013), 2403-2407. [16] M. Z. Sarikaya and S. Erden, On the Hermite- Hadamard-Fejer type integral inequality for convex fFunction, Turkish Journal of Analysis and Number Theory, 2014, Vol. 2, No. 3, 85-89. [17] M. Z. Sarikaya and H. Yaldiz, On generalization integral inequalities for fractional integrals, Nihonkai Math. J., Vol.25(2014), 93–104. [18] J. Park, Inequalities of Hermite-Hadamard-Fejér type for convex functions via fractional Integrals, International Journal of Mathematical Analysis,Vol. 8, 2014, no. 59, pp. 2927-2937. [19] K. L. Tseng, S. R. Hwang, S. S. Dragomir, Fejér-type inequalities (I), J. Inequal. Appl., 2010 (2010), Art ID: 531976, 7 pages. [20] C.-L. Wang, X.-H. Wang, On an extension of Hadamard inequality for convex functions, Chin. Ann. Math. 3 (1982) 567–570. [21] S. Wasowicz and A. Witkonski, On some inequality of Hermite-Hadamard type, Opuscula Math. 32(2), (2012), pp:591-600. [22] S.-H. Wu, On the weighted generalization of the Hermite-Hadamard inequality and its applications, The Rocky Mountain J. of Math., vol. 39, no. 5, pp. 1741–1749, 2009. [23] G. Wang, P. Agarwal and M. Chand, Certain Grüss type inequalities involving the generalized fractional integral operator, Journal of Inequalities and Applications 2014, 2014:147. Department of Mathematics, Faculty of Science and Arts, Düzce University, Düzce-Turkey E-mail address :
[email protected]