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New general integral inequalities for quasi-geometrically convex functions via fractional integrals

Abstract

In this paper, the author introduces the concept of the quasi-geometrically convex functions, gives Hermite-Hadamard’s inequalities for GA-convex functions in fractional integral forms and defines a new identity for fractional integrals. By using this identity, the author obtains new estimates on generalization of Hadamard et al. type inequalities for quasi-geometrically convex functions via Hadamard fractional integrals.

MSC:26A33, 26A51, 26D15.

1 Introduction

Let a real function f be defined on some nonempty interval I of a real line . The function f is said to be convex on I if inequality

f ( t x + ( 1 t ) y ) tf(x)+(1t)f(y)
(1)

holds for all x,yI and t[0,1].

We recall that the notion of quasi-convex function generalizes the notion of convex function. More exactly, a function f:[a,b]RR is said to be quasi-convex on [a,b] if

f ( t x + ( 1 t ) y ) max { f ( x ) , f ( y ) }

for all x,y[a,b] and t[0,1]. Clearly, any convex function is a quasi-convex function. Furthermore, there exist quasi-convex functions which are not convex (see [1]).

The following inequalities are well known in the literature as the Hermite-Hadamard inequality, the Ostrowski inequality and the Simpson inequality, respectively.

Theorem 1.1 Let f:I R R be a convex function defined on the interval I of real numbers, and let a,bI with a<b. The following double inequality holds:

f ( a + b 2 ) 1 b a a b f(x)dx f ( a ) + f ( b ) 2 .
(2)

Theorem 1.2 Let f:I R R be a mapping differentiable in I , the interior of I, and let a,b I with a<b. If | f (x)|M, x[a,b], then the following inequality holds:

|f(x) 1 b a a b f(t)dt| M b a [ ( x a ) 2 + ( b x ) 2 2 ]

for all x[a,b].

Theorem 1.3 Let f:[a,b] R be a four times continuously differentiable mapping on (a,b) and f ( 4 ) = sup x ( a , b ) | f ( 4 ) (x)|<. Then the following inequality holds:

| 1 3 [ f ( a ) + f ( b ) 2 + 2 f ( a + b 2 ) ] 1 b a a b f(x)dx| 1 2 , 880 f ( 4 ) ( b a ) 4 .

The following definitions are well known in the literature.

Definition 1.1 ([2, 3])

A function f:I(0,)R is said to be GA-convex (geometric-arithmetically convex) if

f ( x t y 1 t ) tf(x)+(1t)f(y)

for all x,yI and t[0,1].

Definition 1.2 ([2, 3])

A function f:I(0,)(0,) is said to be GG-convex (called in [4] a geometrically convex function) if

f ( x t y 1 t ) f ( x ) t f ( y ) ( 1 t )

for all x,yI and t[0,1].

We will now give definitions of the right-hand side and left-hand side Hadamard fractional integrals which are used throughout this paper.

Definition 1.3 Let fL[a,b]. The right-hand side and left-hand side Hadamard fractional integrals J a + α f and J b α f of order α>0 with b>a0 are defined by

J a + α f(x)= 1 Γ ( α ) a x ( ln x t ) α 1 f(t) d t t ,a<x<b

and

J b α f(x)= 1 Γ ( α ) x b ( ln t x ) α 1 f(t) d t t ,a<x<b,

respectively, where Γ(α) is the Gamma function defined by Γ(α)= 0 e t t α 1 dt (see [5]).

In recent years, many authors have studied error estimations for Hermite-Hadamard, Ostrowski and Simpson inequalities; for refinements, counterparts, generalization see [4, 620].

In this paper, the concept of the quasi-geometrically convex function is introduced, Hermite-Hadamard’s inequalities for GA-convex functions in fractional integral forms are established, and a new identity for Hadamard fractional integrals is defined. By using this identity, author obtains a generalization of Hadamard, Ostrowski and Simpson type inequalities for quasi-geometrically convex functions via Hadamard fractional integrals.

2 Main results

Let f:I(0,)R be a differentiable function on I , the interior of I, throughout this section we will take

I f ( x , λ , α , a , b ) = ( 1 λ ) [ ln α x a + ln α b x ] f ( x ) + λ [ f ( a ) ln α x a + f ( b ) ln α b x ] Γ ( α + 1 ) [ J x α f ( a ) + J x + α f ( b ) ] ,

where a,bI with a<b, x[a,b], λ[0,1], α>0 and Γ is the Euler Gamma function.

Definition 2.1 A function f:I(0,) R is said to be quasi-geometrically convex on I if

f ( x t y 1 t ) sup { f ( x ) , f ( y ) } ,

for any x,yI and t[0,1].

Remark 2.1 Clearly, any GA-convex and geometrically convex functions are quasi-geometrically convex functions. Furthermore, there exist quasi-geometrically convex functions which are neither GA-convex nor geometrically convex. In that context, we point out an elementary example. The function f:(0,4]R,

f(x)= { 1 , x ( 0 , 1 ] , ( x 2 ) 2 , x [ 1 , 4 ]

is neither GA-convex nor geometrically convex on (0,4], but it is a quasi-geometrically convex function on (0,4].

Proposition 2.1 If f:I(0,) R is convex and nondecreasing, then it is quasi-geometrically convex on I.

Proof This follows from

f ( x t y 1 t ) f ( t x + ( 1 t ) y ) t f ( x ) + ( 1 t ) f ( y ) sup { f ( x ) , f ( y ) } ,

for all x,yI and t[0,1]. □

Proposition 2.2 If f:I(0,) R is quasi-convex and nondecreasing, then it is quasi-geometrically convex on I. If f:I(0,) R is quasi-geometrically convex and nonincreasing, then it is quasi-convex on I.

Proof These conclusions follows from

f ( x t y 1 t ) f ( t x + ( 1 t ) y ) sup { f ( x ) , f ( y ) }

and

f ( t x + ( 1 t ) y ) f ( x t y 1 t ) sup { f ( x ) , f ( y ) }

for all x,yI and t[0,1], respectively. □

Hermite-Hadamard’s inequalities can be represented for GA-convex functions in fractional integral forms as follows.

Theorem 2.1 Let f:I(0,)R be a function such that fL[a,b], where a,bI with a<b. If f is a GA-convex function on [a,b], then the following inequalities for fractional integrals hold:

f( a b ) Γ ( α + 1 ) 2 ( ln b a ) α { J a + α f ( b ) + J b α f ( a ) } f ( a ) + f ( b ) 2
(3)

with α>0.

Proof Since f is a GA-convex function on [a,b], we have for all x,y[a,b] (with t=1/2 in inequality (1)),

f( x y ) f ( x ) + f ( y ) 2 .

Choosing x= a t b 1 t , y= b t a 1 t , we get

f( a b ) f ( a t b 1 t ) + f ( b t a 1 t ) 2 .
(4)

Multiplying both sides of (4) by t α 1 , then integrating the resulting inequality with respect to t over [0,1], we obtain

f ( a b ) α 2 { 0 1 f ( a t b 1 t ) d t + 0 1 f ( b t a 1 t ) d t } = α 2 { a b ( ln b ln u ln b ln a ) α 1 f ( u ) d u u ln b a + a b ( ln u ln a ln b ln a ) α 1 f ( u ) d u u ln b a } = α Γ ( α ) 2 ( ln b a ) α { J a + α f ( b ) + J b α f ( a ) } = Γ ( α + 1 ) 2 ( ln b a ) α { J a + α f ( b ) + J b α f ( a ) } ,

and the first inequality is proved.

For the proof of the second inequality in (3), we first note that if f is a convex function, then for t[0,1], it yields

f ( a t b 1 t ) tf(a)+(1t)f(b)

and

f ( b t a 1 t ) tf(b)+(1t)f(a).

By adding these inequalities, we have

f ( a t b 1 t ) +f ( b t a 1 t ) f(a)+f(b).
(5)

Then multiplying both sides of (5) by t α 1 , and integrating the resulting inequality with respect to t over [0,1], we obtain

0 1 f ( a t b 1 t ) t α 1 dt+ 0 1 f ( b t a 1 t ) t α 1 dt [ f ( a ) + f ( b ) ] 0 1 t α 1 dt,

i.e.,

Γ ( α + 1 ) ( ln b a ) α { J a + α f ( b ) + J b α f ( a ) } f(a)+f(b).

The proof is completed. □

In order to prove our main results, we need the following identity.

Lemma 2.1 Let f:I(0,)R be a differentiable function on I such that f L[a,b], where a,bI with a<b. Then for all x[a,b], λ[0,1] and α>0, we have:

I f ( x , λ , α , a , b ) = a ( ln x a ) α + 1 0 1 ( t α λ ) ( x a ) t f ( x t a 1 t ) d t b ( ln b x ) α + 1 0 1 ( t α λ ) ( x b ) t f ( x t b 1 t ) d t .
(6)

Proof By integration by parts and twice changing the variable, for xa, we can state that

a ln x a 0 1 ( t α λ ) ( x a ) t f ( x t a 1 t ) d t = 0 1 ( t α λ ) d f ( x t a 1 t ) = ( t α λ ) f ( x t a 1 t ) | 0 1 α ( ln x a ) α a x ( ln u a ) α 1 f ( u ) u d u = ( 1 λ ) f ( x ) + λ f ( a ) Γ ( α + 1 ) ( ln x a ) α J x α f ( a ) ,
(7)

and for xb, similarly, we get

b ln b x 0 1 ( t α λ ) ( x b ) t f ( x t b 1 t ) d t = 0 1 ( t α λ ) d f ( x t b 1 t ) = ( t α λ ) f ( x t b 1 t ) | 0 1 α ( ln b x ) α x b ( ln b u ) α 1 f ( u ) u d u = ( 1 λ ) f ( x ) + λ f ( b ) Γ ( α + 1 ) ( ln b x ) α J x + α f ( b ) .
(8)

Multiplying both sides of (7) and (8) by ( ln x a ) α and ( ln b x ) α , respectively, and adding the resulting identities, we obtain the desired result. For x=a and x=b, the identities

I f (a,λ,α;a,b)=b ( ln b a ) α + 1 0 1 ( t α λ ) ( a b ) t f ( a t b 1 t ) dt,

and

I f (b,λ,α;a,b)=a ( ln b a ) α + 1 0 1 ( t α λ ) ( b a ) t f ( b t a 1 t ) ,

can be proved respectively easily by performing an integration by parts in the integrals from the right-hand side and changing the variable. □

Theorem 2.2 Let f:I(0,)R be a differentiable function on I such that f L[a,b], where a,b I with a<b. If | f | q is quasi-geometrically convex on [a,b] for some fixed q1, x[a,b], λ[0,1] and α>0, then the following inequality for fractional integrals holds:

| I f ( x , λ , α , a , b ) | A 1 1 1 q ( α , λ ) { a ( ln x a ) α + 1 ( sup { | f ( x ) | q , | f ( a ) | q } ) 1 q B 1 1 q ( x , α , λ , q ) + b ( ln b x ) α + 1 ( sup { | f ( x ) | q , | f ( b ) | q } ) 1 q B 2 1 q ( x , α , λ , q ) } ,
(9)

where

A 1 ( α , λ ) = 2 α λ 1 + 1 α + 1 α + 1 λ , B 1 ( x , α , λ , q ) = 0 1 | t α λ | ( x a ) q t d t , B 2 ( x , α , λ , q ) = 0 1 | t α λ | ( x b ) q t d t .

Proof Since | f | q is quasi-geometrically convex on [a,b], for all t[0,1],

| f ( x t a 1 t ) | q sup { | f ( x ) | q , | f ( a ) | q }

and

| f ( x t b 1 t ) | q sup { | f ( x ) | q , | f ( b ) | q } .

Hence, using Lemma 2.1 and power mean inequality, we get

| I f ( x , λ , α , a , b ) | a ( ln x a ) α + 1 ( 0 1 | t α λ | d t ) 1 1 q ( 0 1 | t α λ | ( x a ) q t sup { | f ( x ) | q , | f ( a ) | q } d t ) 1 q + b ( ln b x ) α + 1 ( 0 1 | t α λ | d t ) 1 1 q ( 0 1 | t α λ | ( x b ) q t sup { | f ( x ) | q , | f ( b ) | q } d t ) 1 q , | I f ( x , λ , α , a , b ) | ( 0 1 | t α λ | d t ) 1 1 q | I f ( x , λ , α , a , b ) | × { a ( ln x a ) α + 1 ( sup { | f ( x ) | q , | f ( a ) | q } ) 1 q ( 0 1 | t α λ | ( x a ) q t d t ) 1 q | I f ( x , λ , α , a , b ) | + b ( ln b x ) α + 1 ( sup { | f ( x ) | q , | f ( b ) | q } ) 1 q ( 0 1 | t α λ | ( x b ) q t d t ) 1 q } | I f ( x , λ , α , a , b ) | A 1 1 1 q ( α , λ ) { a ( ln x a ) α + 1 ( sup { | f ( x ) | q , | f ( a ) | q } ) 1 q B 1 1 q ( x , α , λ , q ) | I f ( x , λ , α , a , b ) | + b ( ln b x ) α + 1 ( sup { | f ( x ) | q , | f ( b ) | q } ) 1 q B 2 1 q ( x , α , λ , q ) } ,

which completes the proof. □

Corollary 2.1 Under the assumptions of Theorem  2.2 with q=1, inequality (9) reduces to the following inequality:

| I f ( x , λ , α , a , b ) | { a ( ln x a ) α + 1 B 1 ( x , α , λ , 1 ) sup { | f ( x ) | , | f ( a ) | } + b ( ln b x ) α + 1 B 2 ( x , α , λ , 1 ) sup { | f ( x ) | , | f ( b ) | } } .

Corollary 2.2 Under the assumptions of Theorem  2.2 with α=1, inequality (9) reduces to the following inequality:

( ln b a ) 1 | I f ( x , λ , α , a , b ) | | ( 1 λ ) f ( x ) + λ [ f ( a ) ln x a + f ( b ) ln b x ln b a ] 1 ln b a a b f ( u ) u d u | ( ln b a ) 1 ( 2 λ 2 2 λ + 1 2 ) 1 1 q { a ( ln x a ) 2 B 1 1 q ( x , 1 , λ , q ) ( sup { | f ( x ) | q , | f ( a ) | q } ) 1 q + b ( ln b x ) 2 B 2 1 q ( x , 1 , λ , q ) ( sup { | f ( x ) | q , | f ( b ) | q } ) 1 q } ,

where

B 1 ( x , 1 , λ , q ) = h λ ( ( x a ) q ) , B 2 ( x , 1 , λ , q ) = h λ ( ( x b ) q ) , h ( u , λ ) = 2 u λ u 1 ( ln u ) 2 + ( 1 λ ) u λ ln u , u ( 0 , ) { 1 } ,
(10)

specially for x= a b , we get

| ( 1 λ ) f ( a b ) + λ ( f ( a ) + f ( b ) 2 ) 1 ln b a a b f ( u ) u d u | ln b a 4 ( 2 λ 2 2 λ + 1 2 ) 1 1 q { a h 1 q ( ( b a ) q 2 , λ ) ( sup { | f ( a b ) | q , | f ( a ) | q } ) 1 q + b h 1 q ( ( a b ) q 2 , λ ) ( sup { | f ( a b ) | q , | f ( b ) | q } ) 1 q } .
(11)

Corollary 2.3 In Theorem  2.2,

  1. 1.

    If we take x= a b , λ= 1 3 , then we get the following Simpson-type inequality for fractional integrals:

    | 1 6 [ f ( a ) + 4 f ( a b ) + f ( b ) ] 2 α 1 Γ ( α + 1 ) ( ln b a ) α [ J a b α f ( a ) + J a b + α f ( b ) ] | ln b a 4 A 1 1 1 q ( α , 1 3 ) { a [ sup { | f ( a b ) | q , | f ( a ) | q } ] 1 q B 1 1 q ( a b , α , 1 3 , q ) + b [ sup { | f ( a b ) | q , | f ( b ) | q } ] 1 q B 2 1 q ( a b , α , 1 3 , q ) } ,

specially for α=1, we get

| 1 6 [ f ( a ) + 4 f ( a b ) + f ( b ) ] 1 ln b a a b f ( u ) u d u | ln b a 4 ( 5 18 ) 1 1 q { a [ sup { | f ( a b ) | , | f ( a ) | } ] 1 q h 1 q ( ( b a ) q 2 , 1 3 ) + b [ sup { | f ( a b ) | q , | f ( b ) | q } ] 1 q h 1 q ( ( a b ) q 2 , 1 3 ) } ,

where h is defined as in (10).

Remark 2.2

  1. 1.

    If we take x= a b , λ=0, then we get the following midpoint- type inequality for fractional integrals:

    | f ( a b ) 2 α 1 Γ ( α + 1 ) ( ln b a ) α [ J a b α f ( a ) + J a b + α f ( b ) ] | ln b a 4 ( 1 α + 1 ) 1 1 q { a [ sup { | f ( a b ) | q , | f ( a ) | q } ] 1 q B 1 1 q ( a b , 1 , 0 , q ) + b [ sup { | f ( a b ) | q , | f ( b ) | q } ] 1 q B 2 1 q ( a b , 1 , 0 , q ) } ,

specially for α=1, we get

| f ( a b ) 1 ln b a a b f ( u ) u d u | 2 1 q ln b a 8 { a [ sup { | f ( a b ) | q , | f ( a ) | q } ] 1 q h 1 q ( ( b a ) q 2 , 0 ) + b [ sup { | f ( a b ) | q , | f ( b ) | q } ] 1 q h 1 q ( ( a b ) q 2 , 0 ) } ,

where h is defined as in (10).

  1. 2.

    If we take x= a b , λ=1, then we get the following trapezoid-type inequality for fractional integrals:

    | f ( a ) + f ( b ) 2 2 α 1 Γ ( α + 1 ) ( ln b a ) α [ J a b α f ( a ) + J a b + α f ( b ) ] | ln b a 4 ( 1 α + 1 ) 1 1 q { a [ sup { | f ( a b ) | q , | f ( a ) | q } ] 1 q B 1 1 q ( a b , α , 1 , q ) + b [ sup { | f ( a b ) | q , | f ( b ) | q } ] 1 q B 2 1 q ( a b , α , 1 , q ) } ,

specially for α=1, we get

| f ( a ) + f ( b ) 2 1 ln b a a b f ( u ) u d u | 2 1 q ln b a 8 { a [ sup { | f ( a b ) | q , | f ( a ) | q } ] 1 q h 1 q ( ( b a ) q 2 , 1 ) + b [ sup { | f ( a b ) | q , | f ( b ) | q } ] 1 q h 1 q ( ( a b ) q 2 , 1 ) } ,

where h is defined as in (10).

Corollary 2.4 Let the assumptions of Theorem  2.2 hold. If | f (x)|M for all x[a,b] and λ=0, then we get the following Ostrowski-type inequality for fractional integrals from inequality (9):

| [ ( ln x a ) α + ( ln b x ) α ] f ( x ) Γ ( α + 1 ) [ J a b α f ( a ) + J a b + α f ( b ) ] | M ( α + 1 ) 1 1 q [ a ( ln x a ) α + 1 B 1 1 q ( x , α , 0 , q ) + b ( ln b x ) α + 1 B 2 1 q ( x , α , 0 , q ) ] .

Theorem 2.3 Let f:I(0,)R be a differentiable function on I such that f L[a,b], where a,b I with a<b. If | f | q is quasi-geometrically convex on [a,b] for some fixed q>1, x[a,b], λ[0,1] and α>0, then the following inequality for fractional integrals holds:

| I f ( x , λ , α , a , b ) | A 2 1 p ( α , λ , p ) { a ( ln x a ) α + 1 p ( sup { | f ( x ) | q , | f ( a ) | q } ) 1 q ( x q a q q ) 1 q + b ( ln b x ) α + 1 p ( sup { | f ( x ) | q , | f ( b ) | q } ) 1 q ( b q x q q ) 1 q } ,
(12)

where

A 2 ( α , λ , p ) = { 1 α p + 1 , λ = 0 , λ α p + 1 α α { β ( 1 α , p + 1 ) + ( 1 λ ) p + 1 p + 1 × 2 F 1 ( 1 α + p + 1 , p + 1 , p + 2 ; 1 λ ) } , 0 < λ < 1 , 1 α β ( p + 1 , 1 α ) , λ = 1 ,

F 1 2 is hypergeometric function defined by

F 1 2 (a,b;c;z)= 1 β ( b , c b ) 0 1 t b 1 ( 1 t ) c b 1 ( 1 z t ) a dt,c>b>0,|z|<1(see [21]),

and 1 p + 1 q =1.

Proof Using Lemma 2.1, the Hölder inequality and quasi-geometrical convexity of | f | q , we get

| I f ( x , λ , α , a , b ) | a ( ln x a ) α + 1 ( 0 1 | t α λ | p d t ) 1 p ( 0 1 ( x a ) q t sup { | f ( x ) | q , | f ( a ) | q } d t ) 1 q + b ( ln b x ) α + 1 ( 0 1 | t α λ | p d t ) 1 p ( 0 1 ( x b ) q t sup { | f ( x ) | q , | f ( b ) | q } d t ) 1 q , | I f ( x , λ , α , a , b ) | ( 0 1 | t α λ | p d t ) 1 p | I f ( x , λ , α , a , b ) | × { a ( ln x a ) α + 1 ( sup { | f ( x ) | q , | f ( a ) | q } ) 1 q ( 0 1 ( x a ) q t d t ) 1 q | I f ( x , λ , α , a , b ) | + b ( ln b x ) α + 1 ( sup { | f ( x ) | q , | f ( b ) | q } ) 1 q ( 0 1 ( x b ) q t d t ) 1 q } | I f ( x , λ , α , a , b ) | A 2 1 p ( α , λ , p ) { a ( ln x a ) α + 1 1 q ( sup { | f ( x ) | q , | f ( a ) | q } ) 1 q ( x q a q q ) 1 q | I f ( x , λ , α , a , b ) | + b ( ln b x ) α + 1 1 q ( sup { | f ( x ) | q , | f ( b ) | q } ) 1 q ( b q x q q ) 1 q } ,

here, it is seen by a simple computation that

A 2 ( α , λ , p ) = 0 1 | t α λ | p d t = { 1 α p + 1 , λ = 0 , λ α p + 1 α α { β ( 1 α , p + 1 ) + ( 1 λ ) p + 1 p + 1 × 2 F 1 ( 1 α + p + 1 , p + 1 , 2 + p ; 1 λ ) } , 0 < λ < 1 , 1 α β ( p + 1 , 1 α ) , λ = 1 .

Hence, the proof is completed. □

Corollary 2.5 Under the assumptions of Theorem  2.3 with α=1, inequality (12) reduces to the following inequality:

| ( 1 λ ) f ( x ) + λ [ f ( a ) ln x a + f ( b ) ln b x ln b a ] 1 ln b a a b f ( u ) u d u | ( ln b a ) 1 ( λ p + 1 + ( 1 λ ) p + 1 p + 1 ) 1 p × { a ( ln x a ) 1 + 1 p ( sup { | f ( x ) | q , | f ( a ) | q } ) 1 q ( x q a q q ) 1 q + b ( ln b x ) 1 + 1 p ( sup { | f ( x ) | q , | f ( b ) | q } ) 1 q ( b q x q q ) 1 q } ,

specially for x= a b , we get

| ( 1 λ ) f ( a b ) + λ ( f ( a ) + f ( b ) 2 ) 1 ln b a a b f ( u ) u d u | 1 2 ( ln b a ( λ p + 1 + ( 1 λ ) p + 1 ) 2 ( p + 1 ) ) 1 p { a ( sup { | f ( a b ) | q , | f ( a ) | q } ) 1 q ( a b q a q q ) 1 q + b ( sup { | f ( a b ) | q , | f ( b ) | q } ) 1 q ( b q a b q q ) 1 q } .
(13)

Corollary 2.6 In Theorem  2.3,

  1. 1.

    If we take x= a b , λ= 1 3 , then we get the following Simpson-type inequality for fractional integrals:

    | 1 6 [ f ( a ) + 4 f ( a b ) + f ( b ) ] 2 α 1 Γ ( α + 1 ) ( ln b a ) α [ J a b α f ( a ) + J a b + α f ( b ) ] | 1 2 ( ln b a ( 1 + 2 p + 1 ) 3 p + 1 ( p + 1 ) 2 ) 1 p { a [ sup { | f ( a b ) | q , | f ( a ) | q } ] 1 q ( a b q a q q ) 1 q + b [ sup { | f ( a b ) | q , | f ( b ) | q } ] 1 q ( b q a b q q ) 1 q } ,

specially for α=1, we get

| 1 6 [ f ( a ) + 4 f ( a b ) + f ( b ) ] 1 ln b a a b f ( u ) u d u | 1 2 ( ln b a ( 1 + 2 p + 1 ) 3 p + 1 ( p + 1 ) 2 ) 1 p × { a [ sup { | f ( a b ) | q , | f ( a ) | q } ] 1 q ( a b q a q q ) 1 q + b [ sup { | f ( a b ) | q , | f ( b ) | q } ] 1 q ( b q a b q q ) 1 q } .

Remark 2.3

  1. 1.

    If we take x= a b , λ=0, then we get the following midpoint- type inequality for fractional integrals:

    | f ( a b ) 2 α 1 Γ ( α + 1 ) ( ln b a ) α [ J a b α f ( a ) + J a b + α f ( b ) ] | 1 2 ( ln b a 2 ( α p + 1 ) ) 1 p { a [ sup { | f ( a b ) | q , | f ( a ) | q } ] 1 q ( a b q a q q ) 1 q + b [ sup { | f ( a b ) | q , | f ( b ) | q } ] 1 q ( b q a b q q ) 1 q } ,

specially for α=1, we get

| f ( a b ) 1 ln b a a b f ( u ) u d u | 1 2 ( ln b a 2 ( p + 1 ) ) 1 p { a [ sup { | f ( a b ) | q , | f ( a ) | q } ] 1 q ( a b q a q q ) 1 q + b [ sup { | f ( a b ) | q , | f ( b ) | q } ] 1 q ( b q a b q q ) 1 q } .
  1. 2.

    If we take x= a b , λ=1, then we get the following trapezoid-type inequality for fractional integrals 1 α β(p+1, 1 α ):

    | f ( a ) + f ( b ) 2 2 α 1 Γ ( α + 1 ) ( ln b a ) α [ J a b α f ( a ) + J a b + α f ( b ) ] | 1 2 ( ln b a β ( p + 1 , 1 α ) 2 α ) 1 p × { a [ sup { | f ( a b ) | q , | f ( a ) | q } ] 1 q ( a b q a q q ) 1 q + b [ sup { | f ( a b ) | q , | f ( b ) | q } ] 1 q ( b q a b q q ) 1 q } ,

specially for α=1, we get

| f ( a ) + f ( b ) 2 1 ln b a a b f ( u ) u d u | 1 2 ( ln b a 2 ( p + 1 ) ) 1 p { a [ sup { | f ( a b ) | q , | f ( a ) | q } ] 1 q ( a b q a q q ) 1 q + b [ sup { | f ( a b ) | q , | f ( b ) | q } ] 1 q ( b q a b q q ) 1 q } .

Corollary 2.7 Let the assumptions of Theorem  2.3 hold. If | f (x)|M for all x[a,b] and λ=0, then we get the following Ostrowski-type inequality for fractional integrals from inequality (12):

| [ ( ln x a ) α + ( ln b x ) α ] f ( x ) Γ ( α + 1 ) [ J a b α f ( a ) + J a b + α f ( b ) ] | M ( α p + 1 ) 1 p [ a ( ln x a ) α + 1 p ( x q a q q ) 1 q + b ( ln b x ) α + 1 p ( b q x q q ) 1 q ] .

3 Application to special means

Let us recall the following special means of two nonnegative numbers a, b with b>a:

  1. 1.

    The arithmetic mean

    A=A(a,b):= a + b 2 .
  2. 2.

    The geometric mean

    G=G(a,b):= a b .
  3. 3.

    The logarithmic mean

    L=L(a,b):= b a ln b ln a .
  4. 4.

    The p-logarithmic mean

    L p = L p (a,b):= ( b p + 1 a p + 1 ( p + 1 ) ( b a ) ) 1 p ,pR{1,0}.

Proposition 3.1 For b>a>0, n>0 and q1, we have

| ( 1 λ ) G n + 1 ( a , b ) + λ A ( a n + 1 , b n + 1 ) ( n + 1 ) L ( a , b ) L n n ( a , b ) | ( n + 1 ) ln b a 4 ( 2 λ 2 2 λ + 1 2 ) 1 1 q { a G n ( a , b ) h 1 q ( ( b a ) q 2 , λ ) + b n + 1 h 1 q ( ( a b ) q 2 , λ ) } ,

where h is defined as in (10).

Proof Let f(x)= x n + 1 n + 1 , x>0, n>0 and q1. Then the function | f (x) | q = x n q is quasi-geometrically convex on (0,). Thus, by inequality (11), Proposition 3.1 is proved. □

Proposition 3.2 For b>a>0, n>0 and q>1, we have

| ( 1 λ ) G n + 1 ( a , b ) + λ A ( a n + 1 , b n + 1 ) ( n + 1 ) L ( a , b ) L n n ( a , b ) | n + 1 2 ( ln b a ( λ p + 1 + ( 1 λ ) p + 1 ) 2 ( p + 1 ) ) 1 p { a G n ( a , b ) ( G q ( a , b ) a q q ) 1 q + b n + 1 ( b q G q ( a , b ) q ) 1 q } .

Proof Let f(x)= x n + 1 n + 1 , x>0, n>0 and q>1. Then the function | f (x) | q = x n q is quasi-geometrically convex on (0,). Thus, by inequality (13), Proposition 3.2 is proved. □

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Correspondence to İmdat İşcan.

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Keywords

  • quasi-geometrically convex function
  • Hermite-Hadamard-type inequalities
  • Hadamard fractional integrals