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
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[email protected] 