Hermite-Hadamard-Fejer type inequalities

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Hermite–Hadamard–Fejer type inequalities

Hatice Yaldiz & Mehmet Zeki Sarikaya

To cite this article: Hatice Yaldiz & Mehmet Zeki Sarikaya (2018) Hermite–Hadamard–Fejer type inequalities, Journal of Interdisciplinary Mathematics, 21:7-8, 1547-1561, DOI:

10.1080/09720502.2018.1471806

To link to this article: https://doi.org/10.1080/09720502.2018.1471806

Published online: 17 Dec 2018.

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Hermite–Hadamard–Fejer type inequalities

Hatice Yaldiz †

Department of Mathematics Kamil Ozdag Science Faculty Karamanoglu Mehmetbey University Yunus Emre Campus

Karaman 70100 Turkey

Mehmet Zeki Sarikaya * Department of Mathematics Faculty of Science and Arts Düzce University

Düzce Turkey

Abstract

In this paper, we have established the left hand side of the Hermite–Hadamard–Fejer type inequalities for the class of functions whose derivatives in absolute value at certain powers are convex functions by using fractional integrals.

Subject Classification: (2000) 26D07, 26D10, 26D15, 26A33.

Keywords: Convex function, Hermite–Hadamard inequality, Hermite–Hadamard–Fejer inequality, Riemann–Liouville fractional integral.

1. Introduction

The function f : [a, b] Ã R Æ R, is said to be convex if the following inequality holds

f x(λ + −(1 λ) )y ≤λf x( ) (1+ −λ) ( )f y

for all x, y Œ a, b] and l Œ [0, 1]. We say that f is concave if (– f) is convex. †_{E–mail: yaldizhatice@gmail.com}

*E–mail: sarikayamz@gmail.com (Corresponding Author) Vol. 21 (2018), No. 7 & 8, pp. 1547–1561

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The inequalities discovered by C. Hermite and J. Hadamard for convex functions are very important in the literature (see, e.g. [5]). These inequalities state that if f : I Æ R is a convex function on the interval I of real numbers and a, b Œ I with a < b, then

b a f a f b a b f f x dx b a ( ) ( ) 1 _{( )} _. _(1.1) 2 2 +  +  ≤ ≤   ₋  

∫

Both inequalities hold in the reversed direction if f is concave. We note that Hadamard’s inequality may be regarded as a refinement of the concept of convexity and it follows easily from Jensen’s inequality. Hadamard’s inequality for convex functions has received renewed attention in recent years and a remarkable variety of refinements and generalizations have been found (see, for example, [1], [5], [7], [9], [10], [14], [15]) and the references cited therein.

The most well–known inequalities related to the integral mean of a convex function are the Hermite Hadamard inequalities or its weighted versions, the so–called Hermite–Hadamard–Fejér inequalities (see, [7], [12], [14]). In [6], Fejer gave a weighted generalizatinon of the inequalities (1.1) as the following:

Theorem 1. f : [a, b] Æ R, be a convex function, then the inequality b a b b a a a b f w x dx f a f b f x w x dx w x dx b a ( ) 2 ( ) ( ) 1 _{( ) ( )} _{( )} _(1.2) 2  +      + ≤ ≤ −

∫

holds, where w : [a, b] Æ R is nonnegative, integrable, and symmetric about x=a b₂+ (i.e. w(x) = w(a + b – x)).

In [9] in order to prove some inequalities related to Hadamard’s inequality K rmac used the following lemma:

Lemma 1. Let f : I◦_{[a, b] Ã R Æ R, be a differentiable mapping on I}◦_{, a, b Œ I (I}◦

is the interior of I) with a < b. If f¢ Œ ([a, b]), then we have

b a 1 ₁ 2 1 0 ₂ 1 _{f x dx f} a b b a 2 b a tf ta 1 t b dt t 1 f ta 1 t b dt ( ) ( ) ( ( ) ) ( ) ( ( ) ) .  +  −   − _ _   = −  ′ + − + − ′ + −   

∫

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Also, in [9], K rmac obtained the following inequality for differeftiable mappings which are connected with Hermite–Hadamard’s inequality:

Theorem 2. Let f : I◦_{Ã R Æ R be a differentiable mapping on I}◦_{, a, b Œ I with}

a < b. If the mapping |f¢| is convex on [a, b], then we have

b a 1 _{f x dx f} a b b a _{f a} _{f b} b a ( ) 2 8 (| ( )| | ( )|). (1.3)  +  − − _ _ ≤ ′ + ′ −

∫

_ _

Theorem 3. Let f : I◦_{Ã R Æ R, be a differentiable mapping on I}◦_{, a, b Œ I}◦_with

a < b and p > 1. If the mapping

p p '

f 1

| |− _{is convex on [a, b], then} b a p 1 1 p p p p p 1 p 1 p 1 p p p p 1 p 1 1 p 1 _{f x dx f} a b b a 2 b a 1 _{f a} _{3 f b} 4 p 1 3 f a f b b a 4 _{f a} _{f b} 4 p 1 ( ) ( ) _{( )} _{( )} ( ) ( ) [| ( )| | ( )|]. (1.4) − − − − − −  +  −   − _ _      −  ≤  ₊ _{ } ′ + ′ _         _ +_ ′ + ′ _        − ≤  ₊  ′ + ′  

∫

Definition 1. Let f ŒL1 [a, b]. The Riemann–Liouville integrals J faα+ and b

J fα

− of order a > 0 with a ≥ 0 are defined by x a _a J f x_{( ) =} 1 ₍x t₎ 1f t dt x a_{( ) , >} ( ) α α α − + _Γ

∫

− and b b _x J f x_{( ) =} 1 ₍t x₎ 1f t dt x b_{( ) , <} ( ) α α α − − _Γ

∫

−

respectively. Here, G(a) is the Gamma function and J f xa0+ ( ) =J f xb0− ( ) = ( ).f x

In the following we will give some necessary definitions and mathematical preliminaries of fractional calculus theory which are used further in this paper. More details, one can consult [2] – [4], [8], [11], [13] – [15].

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Meanwhile, Sarikaya et al. [15] presented the following important integral identity including the first–order derivative of f to establish many interesting Hermite–Hadamard type inequalities for convexity functions via Riemann–Liouville fractional integrals of the order a > 0.

Theorem 4. Let f : [a, b] Æ R be a positive function with 0 £ a < b and f Œ L1 [a, b]. If f is a convex function on [a, b], then the following inequalities for fractional integrals hold:

a a f a f b a b 1 f J f b J f a 2 2 b a – 2 ( ) ( ) ( _{) [ ( )} _{( )]} _(1.5) ( ) α α α α + +  +  _≤ Γ + ₊ _≤   ₋   with a > 0

It is remarkable that Sarikaya et al. [15] first give the following interesting integral inequalities of Hermite–Hadamard type involving Riemann–Liouville fractional integrals.

Lemma 2. Let f : [a, b] Æ R be a differentiable mapping on (a, b) with a < b. If

f¢ ŒL[a, b] then the following equality for fractional integrals holds:

1 a b 0 f a f b 1 _{J f b J f a} b a 2 2 b a 2 1 t t f ta 1 t b dt ( ) ( ) ( _{) [ ( )} _{( )]} ( ) [( ) ] ( ( ) ) . (1.6) α α α α α α + − + ₋ Γ + ₊ ₌ − − − − ′ + −

∫

In [7], some Hermite–Hadamard–Fejer type integral inequalities for fractional integral proved using the following lemma and theorem.

Lemma 3. If g : [a, b] Æ R is integrable and symmetric to (a + b)/2 with a < b,

then a a a a 1 J g b = J g a = g b g a 2 – – ( ) ( ) [ ( ) ( )] α α α α + + + with a > 0

Theorem 5. Let f : [a, b] Æ R be convex function with a < b and f ŒL [a, b].

If g : [a, b] Æ R is nonnegative, integrable and symmetric to (a + b)/2, then the following inequalities for fractional integrals hold

a a a a a a a b f J g b J g a J fg b J fg a 2 f a f b J g b J g a 2 – – – [ ( ) ( )] [ ( )( ) ( )( )] ( ) ( ) [ ( ) ( )] α α α α α α + + +  +  ₊ _≤ ₊     + ≤ +

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with a > 0.

Lemma 4. [16] Suppose a £ t £ b, then for ≥ 1 we have

n n n

b t t a b a .

( − ) (+ − ) ≤ −( )

In this paper, we firstly represented Hermite–Hadamard–Fejer inequality in fractional integral forms which is the weighted generalization of Hermite–Hadamard inequality. Secondly, we obtained some new inequalities connected with the left–hand side of Hermite–Hadamard– Fejer type integral inequality for the fractional integrals.

2. Main Results

Throughout this section, let || ||g =∞ supt a,b∈[ ]| ( )|g x , for the continuous

function g : [a, b] Æ R.

Lemma 5. Let f : [a, b] Æ R be a differentiable mapping on (a, b) with a < b and

f¢ ŒL[a, b] If g : [a, b] Æ R is integrable, then the following equality for fractional integrals holds a b a b a b t t 2 1 1 ' a a a b t t 1 1 ' a b b b 2 a b f J g b J g a J fg b J fg a 2 1 _{s a} _{g s ds} _{b s} _{g s ds f t dt} s a g s ds b s g s ds f t dt , [ ( ) ( )] [ ( )( ) ( )( )] ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) (2.1) α α α α α α α α α + − + − + − − − − +  +  + − +         = _Γ   − + −           _ +  − + −        

∫ ∫

∫

∫ ∫

∫

with a < 0

Proof. It suffices to note that

a b t t a a a b t t a b b b I s a g s ds b s g s ds f t dt s a g s ds b s g s ds f t dt I I 2 1 1 1 1 2 1 2 ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) . (2.2) α α α α + − − − − +   =  − + −  ′       +  − + −  ′     = +

∫ ∫

∫

∫ ∫

∫

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By integration by parts, we get a b t t ' a a a I 2 s a 1g s ds b s 1g s ds f t dt 1 ( ) ( ) ( ) ( ) ( ) α α + − −   =  − + −     

∫ ∫

∫

a b a b a a a b a a b s a g s ds b s g s ds f t a b t g t f t dt 2 2 1 1 2 1 1 ( ) ( ) ( ) ( ) 2 (( ) ( ) ) ( ) ( ) (2.3) α α α α + + − − + − −       + = _ − + − _ _ _       − − + −

∫

and similarly, a t t ' a b b b b b a b a b a b a I s a g s ds b s g s ds f t dt a b s a g s ds b s g s ds f t a b t g t f t dt 1 1 2 2 1 1 2 2 2 1 1 ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 2 (( ) ( ) ) ( ) ( ) (2.4) α α α α α α − − + − − + + + − −   =  − + −            + = _ − + − _ _ _       − − + −

∫ ∫

∫

Putting (2.3) and (2.4) in (2.2), it follows that

( )

}

a b a b a b I I I f J g b J g a J fg b J fg a 1 2= ₂ [ ( ) ( )] ( )( ) ( )( ) . α α α α α + − + −   +   = + Γ _{ } _ +       −_ + _

Multiplying the both sides by _{( ( )) ,}_Γ_α −1 _{we obtain (2.1) which}

completes the proof.

Corollary 1. Suppose that all the assumptions of Lemma 5 hold. Then, for g(x)

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a b a b 2 a b a b 2 a b f J f b J f a 2 2 b a 1 _{t a} _{b t} _{b a f t dt} 2 b a t a b t b a f t dt ( ) [ ( )( ) ( )( )] ( ) [( ) ( ) ( ) ] ( ) ( ) [( ) ( ) ( ) ] ( ) . α α α α α α α α α α α α + − + +  +  Γ − +   −     = _ − − − + − ′ − _    + − − − + − ′ _  

∫

Remark 1. If we choose a = 1 in Corollary 1, then we heve b a a b b a a b a b f f t dt t a f t dt t b f t dt b a 2 2 1 ( ) ( ) ( ) ( ) ( ) 2 + + + ₋ ₌ ₋ _′ ₊ ₋ _′ −             _ _  

∫

which is proved by K rmac in [9].

Corollary 2. Suppose that all the assumptions of Lemma 5. hold. Then, for a = 1,

the following inequality holds:

b b a a a b t b t 2 a a a b b 2 a b f g t dt f t g t dt 2 g s ds f t dt g s ds f t dt ( ) ( ) ( ) ( ) ( ) ( ) ( ) . + + + − = ′ + ′                       

∫

∫ ∫

Theorem 6. f : [a, b] Æ R be a differentiable mapping (a, b) and f¢ Œ [a, b] with

a < b. If |f¢| is convex on [a, b] and g : [a, b] Æ R is continuous and symmetric to a b+₂ , then the following inequality for fractional integrals holds

a b a a 1 a b f J g b J g b J fg b J fg a 2 g _{b a} 1 1 1 _{f a} _{f b} 1 2 2 1 – [ ( ) ( )] [ ( )( ) ( )( )] (2.5) ( ) [| ( )| | ( )|] ( ) ( ) ( ) α α α α α α α α α α α + − + + ∞  +  ₊ ₋ ₊       || || ≤ −  + −  ′ + ′ Γ _ + + _ with a > 0

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a a a a a b t t a a a b t t a b b b a b f J g b J g a J fg b J fg a s a g s ds b s g s ds f t dt s a g s ds b s g s ds f t dt – – 2 1 1 1 1 2 [ ( ) ( )] [ ( )( ) ( )( )] 2 1 ₍ ₎ _{( )} ₍ ₎ _{( )} _{( )} ( ) ( ) ( ) ( ) ( ) ( ) α α α α α α α α α + + + − − − − +  + _ ₊ _{ }₋ ₊ _  _{ } _{ } _     ≤ _ − + − _ ′ Γ __ __   − + − ′      

∫ ∫

∫

∫ ∫

∫

a b t a a b t a b b s a b s g s ds f t dt s a b s g s ds f t dt K K 2 1 1 1 1 2 1 2 1 ₍₍ ₎ ₍ _{) ) ( )} _{( )} ( ) (( ) ( ) ) ( ) ( ) 1 ( ), (2.6) ( ) α α α α α α + − − − − +   ≤ _ − + − ′ Γ _    + − + − ′    = + Γ

∫ ∫

Since |f¢| is convex on [a, b], we know that for t Œ [a, b]

b t t a b t t a f t f a b f a f b b a b a b a b a ( ) =  − −  − ( ) − ( ) , ′ ′_ + _ ≤ ′ + ′ − − − −  

and since g : [a, b] Æ R is symmetric to a b₂+ we write g _aa b g a b_b

, , , , 2 2 =  +   +  ∞  ∞      || || || || g

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a b t a a a b a b a s a b a a b s K s a b s g s ds f t dt s a b s g s f t dt ds g s a b s b t _{f t} t a _{f t dt ds} b a b a g b a 2 1 1 1 2 2 1 1 2 1 1 2 1 1 = (( ) ( ) ) ( ) ( ) ( ) ( ) ) ( ) ( ) (( ) ( ) ) ( ) ( ) = ( ) 1 2 ( 1) α α α α α α α α _{α α} + − − + + − − + − − ∞ + + ∞ + − + − ′     ≤ − + − _ ′ _     ≤ − + −      − −  ×_ _ ′ + ′ _ _ − −       − × + +

∫ ∫

∫

f t f t 1 (2.7) 1 1 _{( )} 2( 2) 8 1 1 1 3 _{( )} 2( 2) 1 8 2α ( 1) α α α α α α α +    ₋ _′  ₊         +_ + + − _ _{′ } + + + _    and similarly

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a t a b b a b b a b s b a b a b s K s a b s g s ds f t dt s a b s g s f t dt ds g s a b s b t _{f t} t a _{f t dt ds} b a b a g b a 1 1 2 2 2 1 1 2 1 1 2 2 1 1 = (( ) ( ) ) ( ) ( ) ( ) ( ) ) ( ) ( ) (( ) ( ) ) ( ) ( ) = ( ) 1 1 2 ( 1) α α α α α α α α _{α α} − − + + − − + − − ∞ + + + ∞ + − + − ′     ≤ − + − _ ′ _     ≤ − + −      − −  ′ + ′    ₋ ₋        − × + +

∫ ∫

∫

f t f t 1 (2.8) 1 1 – ( ) 1 2( 2) 8 1 1 3 _{( )} 2( 2) 8 2α ( 1) α α α α α α α +    ₋ _′  ₊ ₊         +_ + − _ _{′ } + + _   

Adding (2.7) and (2.8), we obtain (2.5) which this completes the proof.

Corollary 3. In Theorem 6, if we take a = 1, then inequality (2.6), we have

b b 2 a a a b b a f g x dx f x g x dx 2 8 f a f b g ( ) ( ) ( ) ( ) [| ( )| | ( )|] _∞ (2.9)  +  ₋ _≤ −     ′ + ′

∫

Remark 2. In Corollary 3, if we take g(x) = 1 then inequality (2.9) becomes

inequality (1.3) of Theorem 2.

Theorem 7. Suppose that all the assumptions of Theorem 6 hold. Then, the

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a b a b 1 a b f J g b J g a J g fg b J g fg a 2 f a f b b a _{g .} 8 [ ( ) ( )] [ ( )( ) ( )( )] (2.10) ( ) ( ) ( ) ( ) α α α α α α + − + − + ∞  +  ₊ ₋ ₊      ′ + ′  − _ _ ≤ Γ _ _ with a > 0

Proof. Using Lemma 5, (2.1), Lemma 4 and the convexity of |f¢|, it follows

that

( )

a a a a a b t a a b b a b t a b t a a a b f J g b J g a J fg b J fg a s a b s g s ds f t dt s a b s g s ds f t dt b a g ds f t d – – 2 1 1 1 1 2 1 ₂ [ ( ) ( )] [ ( )( ) ( )( )] 2 1 ₍₍ ₎ ₍ _{) ) ( )} _{( )} (( ) ( ) ) ( ) ( ) ( ) ( ) ( ) α α α α α α α α α α α + + + − − − − + + − ∞  +  + − +         ≤ _ − + − ′ _ Γ _ _        +_ − + − ′ _     − ≤ ′ Γ

∫ ∫

b b a b t a b b a a b t ds f t dt b a g t a f t dt b t f t dt b a g J J 2 1 2 2 1 1 2 (2.11) ( ) ( ) = ( ) ( ) ( ) ( ) ( ) ( ) = { }. ( ) α α α α + + − ∞ + − ∞         + ′                 −  ₋ _′ ₊ ₋ _′    Γ _ _   − + Γ

∫ ∫

∫

By simple computation, we have

a b a b t t a J t a f a f b dt b a b a f a f b b a 2 1 2 ( ) ( ) ( ) ( ) ( ) ( ) , (2.12) 12 24 +      − −  = _ − _ ′ + ′ _ _ − −        ′ ′    = − +    

∫

and

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b a b b t t a J t a f a f b dt b a b a f a f b b a 2 2 2 ( ) ( ) ( ) ( ) ( ) ( ) . (2.13) 24 12 +      − −  = _ − _ ′ + ′ _ _ − −        ′ ′    = − +    

∫

Adding (2.12) and (2.13) in (2.11), we obtain (2.10) which this completes the proof.

Theorem 8. Suppose that all the assumptions of Theorem 6. hold. Additionally,

we assume that |f¢|q_{q > 1 is convex on [a, b] Then, the following inequality holds}

a a a b 1 1 p 1 1 q q q q q q a b f J g b J g a J fg b J fg b 2 b a 1 g 4 p 1 3 f a f b f a 3 f b 4 4 – [ ( ) ( )] [ ( )( ) ( )( )] ( ) ( ) ( ) ( ) ( ) , (2.14) α α α α α + + − + ∞  +  ₊ ₋ ₊       − ≤ _ _ +         _ ′ + ′ _ _ ′ + ′ _    × +               where 1_p+_q1= 1.

Proof. Using Lemma 5, Hölder’s inequality and Lemma 4, it follows that

( )

a a a a a b _p p a b q t q a a a p p b b b a b t a b a b f J g b J g a J fg b J fg a s a b s g s ds dt f t dt s a b s g s ds dt f – – 1 1 2 2 1 1 1 1 1 2 2 [ ( ) ( )] [ ( )( ) ( )( )] 2 1 ₍₍ ₎ ₍ _{) ) ( )} _{( )} (( ) ( ) ) ( ) α α α α α α α α α + + + + − − − − + +  +  ₊ ₋ ₊                ≤ _ − + − _{ } ′ _ Γ _ _{ } _          +_ − + − _ ′    

∫ ∫

∫

∫ ∫

∫

q q t dt M M 1 1 2 ( ) 1 { }. (2.15) ( )α   _{ }           _{ } = + Γ

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By simple computation, using the convexity of |f¢|q_{we get} a b _p p a b q t q a a a a b p a b q q p a a a b p _p _q a p M s a b s g s ds dt f t dt g b a t a dt f t dt b a b t t a g f a f b p b a b a 1 1 2 2 1 1 1 1 1 2 2 1 1 1 2 1 1 = (( ) ( ) ) ( ) ( ) ( ) ( ) ( ) ( ) 1 _{( )} _{( )} 1 2 α α α α + + − − + + − ∞ + + ∞ ₊      ₋ _{+ −}   _′                      ≤ − _ − _{ } ′ _           − − − ≤   ′ + ′ + − −  

∫ ∫

∫

q q q q q p p dt f a f b b a g p 1 1 1 1 1 1 3 ( ) ( ) ( ) 1 = (2.16) 1 8 2 α+ ∞ ₊                  _′ _{+ ′}    − _ _  ₊ _{ } _  _{ } _ and similarly p q p b b b q a b t a b q q q p p M s a b s g s ds dt f t dt f a f b b a g p 1 1 1 1 2 2 2 1 1 1 1 1 = (( ) ( ) ) ( ) ( ) (2.17) ( ) 3 ( ) ( ) 1 _. 1 8 2 α α α − − + + + ∞ ₊      ₋ _{+ −}   _′               _′ ₊ _′    − _ _ ≤  ₊ _{ } _  _{ } _

∫ ∫

∫

Adding (2.16) and (2.17) in (2.15), we obtain (2.14) which this completes the proof.

Corollary 4. In Theorem 8, if we take a = 1 then inequality (2.14), we have 1 b b 2 p a a 1 1 q q q q q q a b b a 1 f g x dx f x g x dx 2 4 p 1 3 f a f b f a 3 f b g 4 4 ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) . (2.18) ∞    +  ₋ _≤ −     ₊           _ ′ + ′ _ _ ′ + ′ _   +               

∫

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Remark 3. In Corollary 4, if we take g(x) = 1 then inequality (2.18) becomes

inequality (1.4) of Theorem 3.

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Received June, 2017 Revised December, 2017