Hermite-Hadamard type inequalities for harmonically (α,m) -convex functions via fractional integrals
aa r X i v : . [ m a t h . C A ] M a y HERMITE-HADAMARD TYPE INEQUALITIES FORHARMONICALLY ( α, m ) -CONVEX FUNCTIONS VIAFRACTIONAL INTEGRALS MEHMET KUNT AND ˙IMDAT ˙IS¸CAN
Abstract.
In this paper, some Hermite-Hadamard type inequalities are es-tablished for harmonically ( α, m )-convex functions via fractional integrals andsome Hermite-Hadamard type inequalities are obtained for these classes offunctions. introduction Let f : I ⊂ R → R be a convex function defined on the interval I of real numbersand a, b ∈ I with a < b . The following inequality is well known in the literature asHermite-Hadamard integral inequality for convex functions f (cid:18) a + b (cid:19) ≤ b − a Z ba f ( x ) dx ≤ f ( a ) + f ( b )2 . (1.1)Both inequalities hold in the reversed direction if f is concave. Note that, someof the classical inequalities for means can be obtained from appropriate particularselections of the mapping f . For some results which generalize, improve and extendthe inequalities (1 .
1) we refer the reader to the recent paper [1]-[5] and referencestherein.In [1], ˙I¸scan gave definition of harmonically convex functions and establishedsome Hermite-Hadamard type inequalities for harmonically convex functions asfollows:
Definition 1.
Let I ⊂ R \ { } be a real interval. A function f : I → R is said tobe harmonically convex, if f (cid:18) xytx + (1 − t ) y (cid:19) ≤ tf ( y ) + (1 − t ) f ( x ) (1.2) for all x, y ∈ I and t ∈ [0 , . If the inequality in (1 . is reversed, then f is said tobe harmonically concave. Theorem 1. [1] . Let f : I ⊂ R \ { } → R be a harmonically convex function and a, b ∈ I with a < b . If f ∈ L [ a, b ] then the following inequalities hold: f (cid:18) aba + b (cid:19) ≤ abb − a Z ba f ( x ) x dx ≤ f ( a ) + f ( b )2 . Theorem 2. [1] . Let f : I ⊂ (0 , ∞ ) → R be a differentiable function on I ◦ , a, b ∈ I with a < b, and f ′ ∈ L [ a, b ] . If | f ′ | q is harmonically convex on [ a, b ] for q ≥ , then (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) f ( a ) + f ( b )2 − abb − a Z ba f ( x ) x dx (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) Mathematics Subject Classification.
Primary 26D15; Secondary 26A51.
Key words and phrases.
Hermite-Hadamard type inequalities, harmonically ( α, m )-convexfunctions, fractional integrals. ≤ ab ( b − a )2 λ − /q (cid:2) λ | f ′ ( a ) | q + λ | f ′ ( b ) | q (cid:3) /q , (1.3) where λ = 1 ab − b − a ) ln ( a + b ) ab ! ,λ = − b ( b − a ) + 3 a + b ( b − a ) ln ( a + b ) ab ! ,λ = 1 a ( b − a ) − b + a ( b − a ) ln ( a + b ) ab ! = λ − λ . Theorem 3. [1] . Let f : I ⊂ (0 , ∞ ) → R be a differentiable function on I ◦ , a, b ∈ I with a < b, and f ′ ∈ L [ a, b ] . If | f ′ | q is harmonically convex on [ a, b ] for q > , p + q = 1 , then (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) f ( a ) + f ( b )2 − abb − a Z ba f ( x ) x dx (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≤ ab ( b − a )2 (cid:18) p + 1 (cid:19) /p (cid:2) µ | f ′ ( a ) | q + µ | f ′ ( b ) | q (cid:3) /q , (1.4) where µ = (cid:2) a − q + b − q [( b − a ) (1 − q ) − a ] (cid:3) b − a ) (1 − q ) (1 − q ) ,µ = (cid:2) b − q − a − q [( b − a ) (1 − q ) + b ] (cid:3) b − a ) (1 − q ) (1 − q ) . In [8], Miha¸sen gave definition of ( α, m )-convex functions as follows:
Definition 2.
The function f : [0 , b ] → R , b > , is said to be ( α, m ) -convex where ( α, m ) ∈ [0 , , if we have f ( tx + m (1 − t ) y ) ≤ t α f ( x ) + m (1 − t α ) f ( y ) for all x, y ∈ [0 , b ] and t ∈ [0 , . It can be easily that for ( α, m ) ∈ { (0 , , ( α, , (1 , , (1 , m ) , (1 , , ( α, } oneobtains the following classes of functions: increasing, α -starshaped, starshaped, m -convex, convex, α -convex.For recent results and generalizations concerning ( α, m )-convex functions werefer the reader to paper [8]-[12] and references therein.In [6], ˙I¸scan gave definition of harmonically ( α, m )-convex functions as follows: Definition 3.
The function f : (0 , b ∗ ] → R , b ∗ > , is said to be harmonically ( α, m ) -convex, where α ∈ [0 , and m ∈ (0 , , if f (cid:18) mxymty + (1 − t ) x (cid:19) ≤ t α f ( x ) + m (1 − t α ) f ( y ) (1.5) for all x, y ∈ (0 , b ∗ ] and t ∈ [0 , . If the inequality in (1 . is reversed, then f issaid to be harmonically ( α, m ) -concave. Note that ( α, m ) ∈ { (1 , m ) , (1 , , ( α, } one obtains the following classes offunctions: harmonically m -convex, harmonically convex, harmonically α -convex(or harmonically s -convex in the first sense, if we take s instead of α ). We recall the following inequality and special functions which are known as Betaand hypergeometric function respectively β ( x, y ) = Γ ( x ) Γ ( y )Γ ( x + y ) = Z t x − (1 − t ) y − dt, x, y > , F ( a, b ; c ; z ) = 1 β ( b, c − b ) Z t b − (1 − t ) c − b − (1 − zt ) − a dt,c > b > , | z | < . Lemma 1. [14] , [15] . For < θ ≤ and ≤ a < b we have (cid:12)(cid:12) a θ − b θ (cid:12)(cid:12) ≤ ( b − a ) θ . Following definitions and mathematical preliminaries of fractional calculus the-ory are used further in this paper.
Definition 4.
Let f ∈ L [ a, b ] . The Riemann-Liouville integrals J θa + f and J θb − f oforder θ > with a ≥ are defined by J θa + f ( x ) = 1Γ ( θ ) Z xa ( x − t ) θ − f ( t ) dt, x > a and J θb − f ( x ) = 1Γ ( θ ) Z bx ( t − x ) θ − f ( t ) dt, x < b respectively, where Γ is the Euler Gamma function defined by Γ ( θ ) = R ∞ e − t t θ − dt and J a + f ( x ) = J b − f ( x ) = f ( x ) . Let f : I ⊂ (0 , ∞ ) → R be a differentiable function on I ◦ , throughout this paperwe will take I f ( g ; θ, a, b ) = f ( a ) + f ( b )2 − Γ ( θ + 1)2 (cid:18) abb − a (cid:19) θ × n J θ /a − ( f ◦ g ) (1 /b ) + J θ /b + ( f ◦ g ) (1 /a ) o . where a, b ∈ I with a < b , θ > g ( x ) = 1 /x .In [7], the authors represented Hermite-Hadamard’s inequalities for harmonicallyconvex functions in fractional integral forms as follows: Theorem 4.
Let f : I ⊂ (0 , ∞ ) → R be a function such that f ∈ L [ a, b ] , where a, b ∈ I with a < b . If f is a harmonically convex function on [ a, b ] , then thefollowing inequalities for fractional integrals hold: f (cid:18) aba + b (cid:19) ≤ Γ ( θ + 1)2 (cid:18) abb − a (cid:19) θ ( J θ /a − ( f ◦ g ) (1 /b )+ J θ /b + ( f ◦ g ) (1 /a ) ) ≤ f ( a ) + f ( b )2 with θ > . In [7], the authors gave the following identity for differentiable functions.
Lemma 2.
Let f : I ⊂ (0 , ∞ ) → R be a differentiable function on I ◦ such that f ′ ∈ L [ a, b ] , where a, b ∈ I with a < b . Then the following equality for fractionalintegrals holds: I f ( g ; θ, a, b ) = ab ( b − a )2 Z t θ − (1 − t ) θ ( ta + (1 − t ) b ) f ′ (cid:18) abta + (1 − t ) b (cid:19) dt (1.6) MEHMET KUNT AND ˙IMDAT ˙IS¸CAN
Remark 1.
The identity (1 . is equal the following one I f ( g ; θ, a, b ) = ab ( b − a )2 Z (1 − t ) θ − t θ ( tb + (1 − t ) a ) f ′ (cid:18) abtb + (1 − t ) a (cid:19) dt. (1.7)Because of the wide application of Hermite-Hadamard type inequalities and frac-tional integrals, many researchers extend their studies to Hermite-Hadamard typeinequalities involving fractional integrals. Recent results for this area, we refer thereader to paper [7], [15]-[18] and references therein.In this paper, we aimed to establish Hermite-Hadamard’s inequalities for har-monically ( α, m )-convex functions via fractional integrals. These results have somerelations with [1]. 2. main results Theorem 5.
Let f : I ⊂ (0 , ∞ ) → R be a differentiable function on I ◦ , a, b/m ∈ I ◦ with a < b , m ∈ (0 , and f ′ ∈ L [ a, b ] . If | f ′ | q is harmonically ( α, m ) -convex on [ a, b/m ] for some fixed q ≥ ,with α ∈ [0 , , then | I f ( g ; θ, a, b ) | ≤ ab ( b − a )2 C − /q ( θ ; a, b ) × (cid:2) C ( θ ; α ; a, b ) | f ′ ( a ) | q + mC ( θ ; α ; a, b ) | f ′ ( b/m ) | q (cid:3) /q (2.1) where C ( θ ; a, b ) = b − θ + 1 (cid:20) F (cid:0) , θ + 1; θ + 2; 1 − ab (cid:1) + F (cid:0) , θ + 2; 1 − ab (cid:1) (cid:21) , C ( θ ; α ; a, b ) = " β ( θ +1 ,α +1) b F (cid:0) , θ + 1; θ + α + 2; 1 − ab (cid:1) + b − θ + α +1 2 F (cid:0) , θ + α + 2; 1 − ab (cid:1) , C ( θ ; α ; a, b ) = C ( θ ; a, b ) − C ( θ ; α ; a, b ) .Proof. Let A t = tb + (1 − t ) a , B u = ua + (1 − u ) b . Since | f ′ | q is harmonically( α, m )-convex, using (1 . (cid:12)(cid:12)(cid:12)(cid:12) f ′ (cid:18) abA t (cid:19)(cid:12)(cid:12)(cid:12)(cid:12) q = (cid:12)(cid:12)(cid:12)(cid:12) f ′ (cid:18) abtb + (1 − t ) a (cid:19)(cid:12)(cid:12)(cid:12)(cid:12) q = (cid:12)(cid:12)(cid:12)(cid:12) f ′ (cid:18) ma ( b/m ) mt ( b/m ) + (1 − t ) a (cid:19)(cid:12)(cid:12)(cid:12)(cid:12) q ≤ t α | f ′ ( a ) | q + m (1 − t α ) | f ′ ( b/m ) | q . (2.2) From (1 . . | I f ( g ; θ, a, b ) | ≤ ab ( b − a )2 Z (cid:12)(cid:12)(cid:12) (1 − t ) θ − t θ (cid:12)(cid:12)(cid:12) A t (cid:12)(cid:12)(cid:12)(cid:12) f ′ (cid:18) abA t (cid:19)(cid:12)(cid:12)(cid:12)(cid:12) dt ≤ ab ( b − a )2 Z (1 − t ) θ + t θ A t (cid:12)(cid:12)(cid:12)(cid:12) f ′ (cid:18) abA t (cid:19)(cid:12)(cid:12)(cid:12)(cid:12) dt ≤ ab ( b − a )2 Z (1 − t ) θ + t θ A t dt ! − /q × Z (1 − t ) θ + t θ A t (cid:12)(cid:12)(cid:12)(cid:12) f ′ (cid:18) abA t (cid:19)(cid:12)(cid:12)(cid:12)(cid:12) q dt ! /q ≤ ab ( b − a )2 Z (1 − t ) θ + t θ A t dt ! − /q × (cid:16)R
10 (1 − t ) θ + t θ A t t α dt (cid:17) | f ′ ( a ) | q + m (cid:16)R
10 (1 − t ) θ + t θ A t (1 − t α ) dt (cid:17) | f ′ ( b/m ) | q /q (2.3)calculating following integrals, we have Z (1 − t ) θ + t θ A t dt = Z u θ + (1 − u ) θ B u du = b − θ + 1 (cid:20) F (cid:0) , θ + 1; θ + 2; 1 − ab (cid:1) + F (cid:0) , θ + 2; 1 − ab (cid:1) (cid:21) = C ( θ ; a, b ) (2.4) Z (1 − t ) θ + t θ A t t α dt = Z u θ + (1 − u ) θ B u (1 − u ) α du = " β ( θ +1 ,α +1) b F (cid:0) , θ + 1; θ + α + 2; 1 − ab (cid:1) + b − θ + α +1 2 F (cid:0) , θ + α + 2; 1 − ab (cid:1) = C ( θ ; α ; a, b ) (2.5) Z (1 − t ) θ + t θ A t (1 − t α ) dt = Z (1 − t ) θ + t θ A t dt − Z (1 − t ) θ + t θ A t t α dt = C ( θ ; a, b ) − C ( θ ; α ; a, b )= C ( θ ; α ; a, b ) (2.6)Thus, if we use (2 . .
6) in (2 .
3) we get the inequality of (2 .
1) and this com-pletes the proof. (cid:3)
Corollary 1.
In Theorem 5, (a)
If we take α = 1 , m = 1 we have the following inequality for harmonicallyconvex functions: | I f ( g ; θ, a, b ) | ≤ ab ( b − a )2 C − /q ( θ ; a, b ) × (cid:2) C ( θ ; 1; a, b ) | f ′ ( a ) | q + C ( θ ; 1; a, b ) | f ′ ( b ) | q (cid:3) /q , MEHMET KUNT AND ˙IMDAT ˙IS¸CAN (b)
If we take α = 1 we have the following inequality for harmonically m -convexfunctions: | I f ( g ; θ, a, b ) | ≤ ab ( b − a )2 C − /q ( θ ; a, b ) × (cid:2) C ( θ ; 1; a, b ) | f ′ ( a ) | q + mC ( θ ; 1; a, b ) | f ′ ( b/m ) | q (cid:3) /q , (c) If we take m = 1 we have the following inequality for harmonically α -convexfunctions: | I f ( g ; θ, a, b ) | ≤ ab ( b − a )2 C − /q ( θ ; a, b ) × (cid:2) C ( θ ; α ; a, b ) | f ′ ( a ) | q + C ( θ ; α ; a, b ) | f ′ ( b ) | q (cid:3) /q . When 0 < θ ≤
1, using Lemma 1 we shall give another result for harmonically( α, m )-convex functions as follows:
Theorem 6.
Let f : I ⊂ (0 , ∞ ) → R be a differentiable function on I ◦ , a, b/m ∈ I ◦ with a < b , m ∈ (0 , and f ′ ∈ L [ a, b ] . If | f ′ | q is harmonically ( α, m ) -convex on [ a, b/m ] for some fixed q ≥ ,with α ∈ [0 , , then | I f ( g ; θ, a, b ) | ≤ ab ( b − a )2 C − /q ( θ ; a, b ) × (cid:2) C ( θ ; α ; a, b ) | f ′ ( a ) | q + mC ( θ ; α ; a, b ) | f ′ ( b/m ) | q (cid:3) /q (2.7) where < θ ≤ and C ( θ ; a, b ) = b − θ +1 2 F (cid:0) , θ + 2; 1 − ab (cid:1) − b − θ +1 2 F (cid:0) , θ + 1; θ + 2; 1 − ab (cid:1) + (cid:0) a + b (cid:1) − θ +1 2 F (cid:16) , θ + 1; θ + 2; b − ab + a (cid:17) , C ( θ ; α ; a, b ) = b − θ + α +1 2 F (cid:0) , θ + α + 2; 1 − ab (cid:1) − β ( θ +1 ,α +1) b F (cid:0) , θ + 1; θ + α + 2; 1 − ab (cid:1) + β ( θ +1 ,α +1)( a + b ) α − F (cid:16) , θ + 1; θ + α + 2; b − ab + a (cid:17) , C ( θ ; α ; a, b ) = C ( θ ; a, b ) − C ( θ ; α ; a, b ) . Proof.
Let A t = tb + (1 − t ) a , B u = ua + (1 − u ) b . From (1 . . | I f ( g ; θ, a, b ) | ≤ ab ( b − a )2 Z (cid:12)(cid:12)(cid:12) (1 − t ) θ − t θ (cid:12)(cid:12)(cid:12) A t (cid:12)(cid:12)(cid:12)(cid:12) f ′ (cid:18) abA t (cid:19)(cid:12)(cid:12)(cid:12)(cid:12) dt ≤ ab ( b − a )2 Z (cid:12)(cid:12)(cid:12) (1 − t ) θ − t θ (cid:12)(cid:12)(cid:12) A t dt − /q × Z (cid:12)(cid:12)(cid:12) (1 − t ) θ − t θ (cid:12)(cid:12)(cid:12) A t (cid:12)(cid:12)(cid:12)(cid:12) f ′ (cid:18) abA t (cid:19)(cid:12)(cid:12)(cid:12)(cid:12) q dt /q ≤ ab ( b − a )2 Z (cid:12)(cid:12)(cid:12) (1 − t ) θ − t θ (cid:12)(cid:12)(cid:12) A t dt − /q × (cid:18)R | (1 − t ) θ − t θ | A t t α dt (cid:19) | f ′ ( a ) | q + m (cid:18)R | (1 − t ) θ − t θ | A t (1 − t α ) dt (cid:19) | f ′ ( b/m ) | q /q (2.8)calculating following integrals by Lemma 1, we have Z (cid:12)(cid:12)(cid:12) (1 − t ) θ − t θ (cid:12)(cid:12)(cid:12) A t dt = Z / (1 − t ) θ − t θ A t dt + Z / t θ − (1 − t ) θ A t dt = Z t θ − (1 − t ) θ A t dt + 2 Z / (1 − t ) θ − t θ A t dt ≤ Z t θ A t dt − Z (1 − t ) θ A t dt + 2 Z / (1 − t ) θ A t dt = Z (1 − u ) θ B u du − Z u θ B u du + Z (1 − u ) θ (cid:0) u b + (cid:0) − u (cid:1) a (cid:1) du = Z (1 − u ) θ B u du − Z u θ B u du + Z v θ (cid:18) a + b (cid:19) − (cid:18) − v (cid:18) b − ab + a (cid:19)(cid:19) − dv = b − θ +1 2 F (cid:0) , θ + 2; 1 − ab (cid:1) − b − θ +1 2 F (cid:0) , θ + 1; θ + 2; 1 − ab (cid:1) + (cid:0) a + b (cid:1) − θ +1 2 F (cid:16) , θ + 1; θ + 2; b − ab + a (cid:17) = C ( θ ; a, b ) (2.9) MEHMET KUNT AND ˙IMDAT ˙IS¸CAN and similarly we get Z (cid:12)(cid:12)(cid:12) (1 − t ) θ − t θ (cid:12)(cid:12)(cid:12) A t t α dt ≤ Z t θ + α A t dt − Z (1 − t ) θ t α A t dt + 2 Z / (1 − t ) θ t α A t dt = Z (1 − u ) θ + α B u du − Z u θ (1 − u ) α B u du + Z (1 − u ) θ (cid:0) u (cid:1) α (cid:0) u b + (cid:0) − u (cid:1) a (cid:1) du = Z (1 − u ) θ + α B u du − Z u θ (1 − u ) α B u du + (cid:0) a + b (cid:1) − α Z v θ (1 − v ) α (cid:18) − v (cid:18) b − ab + a (cid:19)(cid:19) − dv = b − θ + α +1 2 F (cid:0) , θ + α + 2; 1 − ab (cid:1) − β ( θ +1 ,α +1) b F (cid:0) , θ + 1; θ + α + 2; 1 − ab (cid:1) + β ( θ +1 ,α +1)( a + b ) α − F (cid:16) , θ + 1; θ + α + 2; b − ab + a (cid:17) = C ( θ ; α ; a, b ) (2.10) Z (cid:12)(cid:12)(cid:12) (1 − t ) θ − t θ (cid:12)(cid:12)(cid:12) A t (1 − t α ) dt = Z (cid:12)(cid:12)(cid:12) (1 − t ) θ − t θ (cid:12)(cid:12)(cid:12) A t dt − Z (cid:12)(cid:12)(cid:12) (1 − t ) θ − t θ (cid:12)(cid:12)(cid:12) A t t α dt = C ( θ ; a, b ) − C ( θ ; α ; a, b )= C ( θ ; α ; a, b ) (2.11)Thus, if we use (2 . .
11) in (2 .
8) we get the inequality of (2 .
7) and this completesthe proof. (cid:3)
Remark 2.
If we take θ = 1 , α = 1 , m = 1 in Theorem 6, then inequality (2 . becomes inequality (1 . of Theorem 2. Corollary 2.
In Theorem 6, (a)
If we take α = 1 , m = 1 we have the following inequality for harmonicallyconvex functions: | I f ( g ; θ, a, b ) | ≤ ab ( b − a )2 C − /q ( θ ; a, b ) × (cid:2) C ( θ ; 1; a, b ) | f ′ ( a ) | q + C ( θ ; 1; a, b ) | f ′ ( b ) | q (cid:3) /q , (b) If we take α = 1 we have the following inequality for harmonically m -convexfunctions: | I f ( g ; θ, a, b ) | ≤ ab ( b − a )2 C − /q ( θ ; a, b ) × (cid:2) C ( θ ; 1; a, b ) | f ′ ( a ) | q + mC ( θ ; 1; a, b ) | f ′ ( b/m ) | q (cid:3) /q , (c) If we take m = 1 we have the following inequality for harmonically α -convexfunctions: | I f ( g ; θ, a, b ) | ≤ ab ( b − a )2 C − /q ( θ ; a, b ) × (cid:2) C ( θ ; α ; a, b ) | f ′ ( a ) | q + C ( θ ; α ; a, b ) | f ′ ( b ) | q (cid:3) /q . Theorem 7.
Let f : I ⊂ (0 , ∞ ) → R be a differentiable function on I ◦ , a, b/m ∈ I ◦ with a < b , m ∈ (0 , and f ′ ∈ L [ a, b ] . If | f ′ | q is harmonically ( α, m ) -convex on [ a, b/m ] for some fixed q > ,with α ∈ [0 , , then | I f ( g ; θ, a, b ) | ≤ a ( b − a )2 b (cid:18) θp + 1 (cid:19) /p (cid:18) | f ′ ( a ) | q + mα | f ′ ( b/m ) | q α + 1 (cid:19) /q × " F /p (cid:0) p, θp + 1; θp + 2; 1 − ab (cid:1) + F /p (cid:0) p, θp + 2; 1 − ab (cid:1) (2.12) where p + q = 1 .Proof. Let A t = tb + (1 − t ) a , B u = ua + (1 − u ) b . From (1 . . | I f ( g ; θ, a, b ) | ≤ ab ( b − a )2 "Z (1 − t ) θ A t (cid:12)(cid:12)(cid:12)(cid:12) f ′ (cid:18) abA t (cid:19)(cid:12)(cid:12)(cid:12)(cid:12) dt + Z t θ A t (cid:12)(cid:12)(cid:12)(cid:12) f ′ (cid:18) abA t (cid:19)(cid:12)(cid:12)(cid:12)(cid:12) dt ≤ ab ( b − a )2 (cid:16)R
10 (1 − t ) θp A pt dt (cid:17) /p (cid:16)R (cid:12)(cid:12)(cid:12) f ′ (cid:16) abA t (cid:17)(cid:12)(cid:12)(cid:12) q dt (cid:17) /q + (cid:16)R t θp A pt dt (cid:17) /p (cid:16)R (cid:12)(cid:12)(cid:12) f ′ (cid:16) abA t (cid:17)(cid:12)(cid:12)(cid:12) q dt (cid:17) /q ≤ ab ( b − a )2 (cid:18)Z u θp B pu du (cid:19) /p + Z (1 − u ) θp B pu du ! /p × (cid:18)Z t α | f ′ ( a ) | q + m (1 − t α ) | f ′ ( b/m ) | q dt (cid:19) /q = ab ( b − a )2 (cid:16) K /p + K /p (cid:17) (cid:18) | f ′ ( a ) | q + mα | f ′ ( b/m ) | q α + 1 (cid:19) /q . (2.13)calculating K and K we have K = Z u θp B pu du = b − p θp + 1 F (cid:0) p, θp + 1; θp + 2; 1 − ab (cid:1) (2.14) K = Z (1 − u ) θp B pu du = b − p θp + 1 F (cid:0) p, θp + 2; 1 − ab (cid:1) (2.15)Thus, if we use (2 . .
15) in (2 .
13) we get the inequality of (2 .
12) and thiscompletes the proof. (cid:3)
Corollary 3.
In Theorem 7, (a)
If we take α = 1 , m = 1 we have the following inequality for harmonicallyconvex functions: | I f ( g ; θ, a, b ) | ≤ a ( b − a )2 b (cid:18) θp + 1 (cid:19) /p (cid:18) | f ′ ( a ) | q + | f ′ ( b ) | q (cid:19) /q × " F /p (cid:0) p, θp + 1; θp + 2; 1 − ab (cid:1) + F /p (cid:0) p, θp + 2; 1 − ab (cid:1) , (b) If we take α = 1 we have the following inequality for harmonically m -convexfunctions: | I f ( g ; θ, a, b ) | ≤ a ( b − a )2 b (cid:18) θp + 1 (cid:19) /p (cid:18) | f ′ ( a ) | q + m | f ′ ( b/m ) | q (cid:19) /q × " F /p (cid:0) p, θp + 1; θp + 2; 1 − ab (cid:1) + F /p (cid:0) p, θp + 2; 1 − ab (cid:1) , (c) If we take m = 1 we have the following inequality for harmonically α -convexfunctions: | I f ( g ; θ, a, b ) | ≤ a ( b − a )2 b (cid:18) θp + 1 (cid:19) /p (cid:18) | f ′ ( a ) | q + α | f ′ ( b ) | q α + 1 (cid:19) /q × " F /p (cid:0) p, θp + 1; θp + 2; 1 − ab (cid:1) + F /p (cid:0) p, θp + 2; 1 − ab (cid:1) . Theorem 8.
Let f : I ⊂ (0 , ∞ ) → R be a differentiable function on I ◦ , a, b/m ∈ I ◦ with a < b , m ∈ (0 , and f ′ ∈ L [ a, b ] . If | f ′ | q is harmonically ( α, m ) -convex on [ a, b/m ] for some fixed q > ,with α ∈ [0 , , then | I f ( g ; θ, a, b ) | ≤ a ( b − a )2 b (cid:18) θp + 1 (cid:19) /p (cid:18) α + 1 (cid:19) /q × F (cid:0) q, α + 2; 1 − ab (cid:1) | f ′ ( a ) | q + m (cid:20) ( α + 1) F (cid:0) q,
1; 2; 1 − ab (cid:1) − F (cid:0) q, α + 2; 1 − ab (cid:1) (cid:21) | f ′ ( b/m ) | q /q (2.16) where p + q = 1 .Proof. Let A t = tb + (1 − t ) a , B u = ua + (1 − u ) b . From (1 . . | I f ( g ; θ, a, b ) | ≤ ab ( b − a )2 Z (cid:12)(cid:12)(cid:12) (1 − t ) θ − t θ (cid:12)(cid:12)(cid:12) A t (cid:12)(cid:12)(cid:12)(cid:12) f ′ (cid:18) abA t (cid:19)(cid:12)(cid:12)(cid:12)(cid:12) dt ≤ ab ( b − a )2 (cid:18)Z (cid:12)(cid:12)(cid:12) (1 − t ) θ − t θ (cid:12)(cid:12)(cid:12) p dt (cid:19) /p × (cid:18)Z A qt (cid:12)(cid:12)(cid:12)(cid:12) f ′ (cid:18) abA t (cid:19)(cid:12)(cid:12)(cid:12)(cid:12) q dt (cid:19) /q ≤ ab ( b − a )2 (cid:18)Z | − t | θp dt (cid:19) /p × (cid:18)Z A qt (cid:2) t α | f ′ ( a ) | q + m (1 − t α ) | f ′ ( b/m ) | q (cid:3) dt (cid:19) /q (2.17)calculating following integrals, we have Z | − t | θp dt = 1 θp + 1 (2.18) Z t α A qt dt = Z (1 − u ) α B qu dt = b − q α + 1 F (cid:0) q, α + 2; 1 − ab (cid:1) (2.19) Z − t α A qt dt = Z − (1 − u ) α B qu dt = b − q F (cid:0) q,
1; 2; 1 − ab (cid:1) − b − q α + 1 F (cid:0) q, α + 2; 1 − ab (cid:1) (2.20)Thus, if we use (2 . .
20) in (2 .
17) we get the inequality of (2 .
16) and thiscompletes the proof. (cid:3)
Remark 3.
If we take θ = 1 , α = 1 , m = 1 in Theorem 8, then inequality (2 . becomes inequality (1 . of Theorem 3. Corollary 4.
In Theorem 8, (a)
If we take α = 1 , m = 1 we have the following inequality for harmonicallyconvex functions: | I f ( g ; θ, a, b ) | ≤ a ( b − a )2 /q b (cid:18) θp + 1 (cid:19) /p × F (cid:0) q,
1; 3; 1 − ab (cid:1) | f ′ ( a ) | q + (cid:20) F (cid:0) q,
1; 2; 1 − ab (cid:1) − F (cid:0) q,
1; 3; 1 − ab (cid:1) (cid:21) | f ′ ( b ) | q /q , (b) If we take α = 1 we have the following inequality for harmonically m -convexfunctions: | I f ( g ; θ, a, b ) | ≤ a ( b − a )2 /q b (cid:18) θp + 1 (cid:19) /p × F (cid:0) q,
1; 3; 1 − ab (cid:1) | f ′ ( a ) | q + m (cid:20) F (cid:0) q,
1; 2; 1 − ab (cid:1) − F (cid:0) q,
1; 3; 1 − ab (cid:1) (cid:21) | f ′ ( b/m ) | q /q , (c) If we take m = 1 we have the following inequality for harmonically α -convexfunctions: | I f ( g ; θ, a, b ) | ≤ a ( b − a )2 b (cid:18) θp + 1 (cid:19) /p (cid:18) α + 1 (cid:19) /q × F (cid:0) q, α + 2; 1 − ab (cid:1) | f ′ ( a ) | q + (cid:20) ( α + 1) F (cid:0) q,
1; 2; 1 − ab (cid:1) − F (cid:0) q, α + 2; 1 − ab (cid:1) (cid:21) | f ′ ( b ) | q /q . Theorem 9.
Let f : I ⊂ (0 , ∞ ) → R be a differentiable function on I ◦ , a, b/m ∈ I ◦ with a < b , m ∈ (0 , and f ′ ∈ L [ a, b ] . If | f ′ | q is harmonically ( α, m ) -convex on [ a, b/m ] for some fixed q > ,with α ∈ [0 , , then | I f ( g ; θ, a, b ) | ≤ ab ( b − a )2 (1 /p ) − ( a + b ) (cid:18) θp + 1 (cid:19) /p (cid:18) | f ′ ( a ) | q + mα | f ′ ( b/m ) | q α + 1 (cid:19) /q × F (cid:16) p, θp + 1; θp + 2; b − ab + a (cid:17) + F (cid:16) p, θp + 1; θp + 2; a − bb + a (cid:17) /p (2.21) where p + q = 1 . Proof.
Let A t = tb + (1 − t ) a . From (1 . . | I f ( g ; θ, a, b ) | ≤ ab ( b − a )2 Z (cid:12)(cid:12)(cid:12) (1 − t ) θ − t θ (cid:12)(cid:12)(cid:12) A t (cid:12)(cid:12)(cid:12)(cid:12) f ′ (cid:18) abA t (cid:19)(cid:12)(cid:12)(cid:12)(cid:12) dt ≤ ab ( b − a )2 Z | − t | θ A t (cid:12)(cid:12)(cid:12)(cid:12) f ′ (cid:18) abA t (cid:19)(cid:12)(cid:12)(cid:12)(cid:12) dt ≤ ab ( b − a )2 Z | − t | θp A pt dt ! /p (cid:18)Z (cid:12)(cid:12)(cid:12)(cid:12) f ′ (cid:18) abA t (cid:19)(cid:12)(cid:12)(cid:12)(cid:12) q dt (cid:19) /q ≤ ab ( b − a )2 Z | − t | θp A pt dt ! /p × (cid:18)Z t α | f ′ ( a ) | q + m (1 − t α ) | f ′ ( b/m ) | q dt (cid:19) /q = ab ( b − a )2 "Z / (1 − t ) θp A pt dt + Z / (2 t − θp A pt dt /p × (cid:18) | f ′ ( a ) | q + mα | f ′ ( b/m ) | q α + 1 (cid:19) /q (2.22)calculating following integrals, we have Z / (1 − t ) θp A pt dt = 12 Z (1 − u ) θp (cid:0) u b + (cid:0) − u (cid:1) a (cid:1) p du = ( a + b ) − p − p Z v θp (cid:18) − v (cid:18) b − ab + a (cid:19)(cid:19) − p dv = ( a + b ) − p − p ( θp + 1) F (cid:16) p, θp + 1; θp + 2; b − ab + a (cid:17) (2.23) Z / (2 t − θp A pt dt = 12 Z ( u − θp (cid:0) u b + (cid:0) − u (cid:1) a (cid:1) p du = ( a + b ) − p − p Z v θp (cid:18) − v (cid:18) a − bb + a (cid:19)(cid:19) − p dv = ( a + b ) − p − p ( θp + 1) F (cid:16) p, θp + 1; θp + 2; a − bb + a (cid:17) (2.24)Thus, if we use (2 . .
24) in (2 .
22) we get the inequality of (2 .
21) and thiscompletesF the proof. (cid:3)
Corollary 5.
In Theorem 9, (a)
If we take α = 1 , m = 1 we have the following inequality for harmonicallyconvex functions: | I f ( g ; θ, a, b ) | ≤ ab ( b − a )2 (1 /p ) − ( a + b ) (cid:18) θp + 1 (cid:19) /p (cid:18) | f ′ ( a ) | q + | f ′ ( b ) | q (cid:19) /q × F (cid:16) p, θp + 1; θp + 2; b − ab + a (cid:17) + F (cid:16) p, θp + 1; θp + 2; a − bb + a (cid:17) /p , (b) If we take α = 1 we have the following inequality for harmonically m -convexfunctions: | I f ( g ; θ, a, b ) | ≤ ab ( b − a )2 (1 /p ) − ( a + b ) (cid:18) θp + 1 (cid:19) /p (cid:18) | f ′ ( a ) | q + m | f ′ ( b/m ) | q (cid:19) /q × F (cid:16) p, θp + 1; θp + 2; b − ab + a (cid:17) + F (cid:16) p, θp + 1; θp + 2; a − bb + a (cid:17) /p , (c) If we take m = 1 we have the following inequality for harmonically α -convexfunctions: | I f ( g ; θ, a, b ) | ≤ ab ( b − a )2 (1 /p ) − ( a + b ) (cid:18) θp + 1 (cid:19) /p (cid:18) | f ′ ( a ) | q + α | f ′ ( b ) | q α + 1 (cid:19) /q × F (cid:16) p, θp + 1; θp + 2; b − ab + a (cid:17) + F (cid:16) p, θp + 1; θp + 2; a − bb + a (cid:17) /p . References [1] ˙I. ˙I¸scan, Hermite-Hadamard type inequalities for harmonically convex functions, Hacet. J.Math. Stat., 43 (6) (2014), 935-942.[2] ˙I. ˙I¸scan, New estimates on generalization of some integral inequalities for ( α, m )-convex func-tions, Contemp. Anal. Appl. Math., 1 (2) (2013) 253-264.[3] ˙I. ˙I¸scan, New estimates on generalization of some integral inequalities for s -convex functionsand their applications, Int. J. Pure Appl. Math., 86 (4) (2013) 727-746.[4] H. Kavurmacı, M. E. ¨Ozdemir, M. Avcı, New Ostrowski type inequalities for m -convex func-tions and applications, Hacet. J. Math. Stat., 40 (2) (2011) 135-145.[5] E. Set, M. E. ¨Ozdemir, S. S. Dragomir, On Hadamard-type inequalities involving severalkinds of convexity, J. Inequal. Appl. 2010 (2010) 12, http://dx.doi.org/10.1155/2010/286845(Article ID 286245).[6] ˙I. ˙I¸scan, Hermite-Hadamard type inequalities for harmonically ( α, m )-convex functions (inpress).[7] ˙I. ˙I¸scan, S. Wu, Hermite-Hadamard type inequalities for harmonically convex functions viafractional integrals, Appl. Math. Comput., 238 (2014) 237-244.[8] V. G. Miha¸sen, A generalization of the convexity, Seminar on Functional Equations, Approx-imation and Convexity, Cluj-Napoca, Romania, 1993.[9] M. K. Bakula, M. E. ¨Ozdemir, J. Pecaric, Hadamard type inequalities for m -convex and( α, m )-convex functions, J. Inequal. Pure Appl. Math., 9 (4), Article 96, p. 12, 2008.[10] ˙I. ˙I¸scan, A new generalization of some integral inequalities for ( α, m )-convex functions, Math-ematical Sciences, 7 (22) (2013) 1-8.[11] ˙I. ˙I¸scan, Hermite-Hadamard type inequalities for functions whose derivatives are ( α, m )-convex, Int. J. Eng. Appl. Sci., 2 (3) (2013) 69-78.[12] M. E. ¨Ozdemir, H. Kavurmacı, E. Set, Ostorowski’s type inequalities for ( α, m )-convex func-tions, Kyungpook Math. J., 50 (2010) 371-378.[13] A. A. Kilbas, H. M. Srivastava, J. J. Trujillo, Theory and Applications of Fractional Differ-ential Equations, Elsevier, Amsterdam, 2006.[14] A. P. Prudnikov, Y. A. Brychkov, O. J. Marichev, Integral and series, Elementary Functions,vol. 1, Nauka, Moscow, 1981.[15] J. Wang, C. Zho, Y. Zhou, New generalized Hermite-Hadamard type inequalities and appli-cations to special means, J. Inequal. Appl., 2013 (325) (2013) 15pp.[16] J. Wang, X. Li, M. Feˇckan, Y. Zhou, Hermite-Hadamard-type inequalities for Riemann-Liouville fractional integrals via two kinds of convexity, Appl. Anal. 92 (11) (2012) 2241-2253.[17] ˙I. ˙I¸scan, Generalization of different type integral inequalities for s -convex functions via frac-tional integrals, Appl. Anal., 93 (9) (2014) 1846-1862.[18] ˙I. ˙I¸scan, New general integral inequalities for quasi-geometrically convex functions via frac-tional integrals, J. Inequal. Appl., 2013 (491) (2013) 15pp. Department of Mathematics, Faculty of Sciences, Karadeniz Technical University,Trabzon, Turkey
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