Hermite–Hadamard type inequalities for F convex function involving fractional integrals

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R E S E A R C H

Open Access

Hermite–Hadamard type inequalities for

F

-convex function involving fractional

integrals

Pshtiwan Othman Mohammed

1*

_{and Mehmet Zeki Sarikaya}

2

*_{Correspondence:}

pshtiwansangawi@gmail.com 1_{Department of Mathematics,}

College of Education, University of Sulaimani, Sulaimani, Iraq Full list of author information is available at the end of the article

Abstract

In this study, the familyFandF-convex function are given with its properties. In view of this, we establish some new inequalities of Hermite–Hadamard type for

diﬀerentiable function. Moreover, we establish some trapezoid type inequalities for functions whose second derivatives in absolute values areF-convex. We also show that through the notion ofF-convex we can ﬁnd some new Hermite–Hadamard type and trapezoid type inequalities for the Riemann–Liouville fractional integrals and classical integrals.

MSC: 26A51; 26D15; 35A23

Keywords: F-convex;

λ

ϕ-preinvex; Integral inequalities

1 Introduction

A functionf :I⊆R→Ris said to be convex on the intervalI, if for allx,y∈Iandt∈(0, 1) it satisﬁes the following inequality:

ftx+ (1 –t)y≤tf(x) + (1 –t)f(y). (1)

Convex functions play an important role in the ﬁeld of integral inequalities. For convex functions, many equalities and inequalities have been established, but one of the most im-portant ones is the Hermite–Hadamard’ integral inequality, which is deﬁned as follows [1]: Let f :I⊆R→Rbe a convex function with a<banda,b∈I. Then the Hermite– Hadamard inequality is given by

f

a+b 2

≤ 1

b–a

b

a

f(x)dx≤f(a) +f(b)

2 . (2)

In recent years, a number of mathematicians have devoted their efforts to generalizing, refining, counterparting, and extending the Hermite–Hadamard inequality (2) for differ-ent classes of convex functions and mappings. The Hermite–Hadamard inequality (2) is established for the classical integral, fractional integrals, conformable fractional integrals and most recently for generalized fractional integrals; see for details and applications [2–8] and the references therein.

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The concepts of classical convex functions have been extended and generalized in sev-eral directions, such as quasi-convex [9], pseudo-convex [10],MT-convex [11] strongly convex [12],-convex [13],s-convex [14],h-convex [15], andλϕ-preinvex [16]. Recently,

Samet [17] has deﬁned a new concept of convexity that depends on a certain function sat-isfying some axioms, generalizing diﬀerent types of convexity, including-convex func-tions,α-convex functions,h-convex functions, and so on, as stated in the next section.

2 Review of the family ofF

We address the family ofFof mappingsF:R×R×R×[0, 1]→Rsatisfying the following axioms:

(A1) Ifui∈L1(0, 1),i= 1, 2, 3, then, for everyλ∈[0, 1], we have

1

0

Fu1(t),u2(t),u3(t),λ

dt=F

1

0

u1(t)dt,

1

0

u2(t)dt,

1

0

u3(t)dt,λ

.

(A2) For everyu∈L1_{(0, 1)}_,_w_∈_L∞_{(0, 1)}_and_(z

1,z2)∈R2, we have

1

0

Fw(t)u(t),w(t)z1,w(t)z2,t

dt=TF,w

1

0

w(t)u(t)dt,z1,z2

,

whereTF,w:R×R×R→Ris a function that depends on(F,w), and it is nondecreasing with respect to the ﬁrst variable.

(A3) For any(w,u1,u2,u3)∈R4,u4∈[0, 1], we have

wF(u1,u2,u3,u4) =F(wu1,wu2,wu3,u4) +Lw,

whereLw∈Ris a constant that depends only onw.

Deﬁnition 2.1 Letf : [a,b]→R, (a,b)∈R2,a<b, be a given function. We say thatf is a convex function with respect to someF∈F(orF-convex function) iﬀ

Fftx+ (1 –t)y,f(x),f(y),t≤0, (x,y,t)∈[a,b]×[a,b]×[0, 1].

Remark1 Suppose that (a,b)∈R2_with_a_<_b.

(i) Letf : [a,b]→Rbe anε-convex function, that is [18],

ftx+ (1 –t)y≤tf(x) + (1 –t)f(y), (x,y,t)∈[a,b]×[a,b]×[0, 1].

Deﬁne the functionsF:R×R×R×[0, 1]→Rby

F(u1,u2,u3,u4) =u1–u4u2– (1 –u4)u3–ε (3)

andTF,w:R×R×R×[0, 1]→Rby

TF,w(u1,u2,u3) =u1–

1

0

tw(t)dt

u2–

1

0

(1 –t)w(t)dt

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For

Lw= (1 –w)ε, (5)

it will be seen thatF∈Fand

Fftx+ (1 –t)y,f(x),f(y),t=ftx+ (1 –t)y–tf(x) – (1 –t)f(y) –ε≤0,

that is,f is anF-convex function. Particularly, takingε= 0we show that iff is a convex function thenf is anF-convex function with respect toFdeﬁned above. (ii) Letf : [a,b]→Rbeλϕ-preinvex function according toϕand bifunctionη,

0≤ϕ≤π 2,λ∈(0,

1

2], that is [16],

fu+teiϕη(v,u)

≤ √

t

2√1 –tf(v) +

(1 –λ)√1 –t

2λ√t f(u), (u,v,t)∈[a,b]×[a,b]×(0, 1).

F(u1,u2,u3,u4) =u1– √

u4

2√1 –u4

u3–

(1 –λ)√1 –u4

2λ√u4

u2 (6)

TF,w(u1,u2,u3) =u1–

1

0 √

t

2√1 –tw(t)dt

u3

–1 –λ

λ

1

0 √

1 –t 2√t w(t)dt

u2. (7)

ForLw= 0, it will be seen thatF∈Fand

Ffu+teiϕη(v,u),f(u),f(v),t

=fu+teiϕ_η_(v,_u)_– √

t

2√1 –tf(v) –

(1 –λ)√1 –t

2λ√t f(u) –ε≤0,

that isf is anF-convex function.

(iii) Leth:I→Rbe a given function which is not identical to0, whereIis an interval inRsuch that(0, 1)⊆I. Letf : [a,b]→[0,∞)be anh-convex function, that is,

ftx+ (1 –t)y≤h(t)f(x) +1 –λ

λ h(1 –t)f(y), (x,y,t)∈[a,b]×[a,b]×[0, 1].

F(u1,u2,u3,u4) =u1–h(u4)u3–

1 –λ

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TF,w(u1,u2,u3) =u1–

1

0

h(t)w(t)dt

u3–

1 –λ λ

1

0

h(1 –t)w(t)dt

u2. (9)

ForLw= (1 –w)ε, it will be seen thatF∈Fand

Fftx+ (1 –t)y,f(x),f(y),t

=ftx+ (1 –t)y–h(t)f(x) –1 –λ

λ h(1 –t)f(y) –ε≤0,

that isf is anF-convex function.

Recently Samet [17] established some integral inequalities of Hermite–Hadamard type viaF-convex functions.

Theorem 1([17, Theorem 3.1]) Let f : [a,b]→R, (a,b)∈R2,a<b,be an F-convex func-tion,for some F∈F.Suppose that F∈L1_(a,_b)._Then

F

f

a+b 2

, 1 b–a

b

a

f(x)dx,1 2

≤0,

TF,1

1 b–a

b

a

f(x)dx,f(a),f(b)

≤0.

Theorem 2 ([17, Theorem 3.4]) Let f :Io_⊆_R_→_R_{be a diﬀerentiable mapping on I}o_,

(a,b)∈Io_×_Io_,_a_<_b._{Suppose that}

(i) |f|isF-convex on[a,b],for someF∈F;

(ii) the functiont∈(0, 1)→Lw(t)belongs toL1(a,b),wherew(t) =|1 – 2t|.

Then

TF,w

2 b–a

f(a) +₂f(b)– 1 b–a

b

a

f(x)dx,f(a),f(b)

+

1

0

Lw(t)dt≤0.

Theorem 3 ([17, Theorem 3.5]) Let f :Io_⊆_R_→_R_{be a diﬀerentiable mapping on I}o_,

(a,b)∈Io_×_Io_,_a_<_{b and let p}_{> 1.}_{Suppose that}_|_f_|p/(p–1)_{is F-convex on}_[a,_b],_{for some}

F∈Fand F∈Lp/(p–1)(a,b).Then

TF,1

A(p,f),f(a) p p–1_,_f_(b)

p p–1≤_0,

where

A(p,f) =

2 b–a

p

p–1

(p+ 1)p1–1f(a) +f(b)

2 –

1 b–a

b

a

f(x)dx p p–1

.

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Theorem 4([17, Corollary 4.3]) Let f :Io_⊆_R_→_R_{be a diﬀerentiable mapping on I}o_,

(a,b)∈Io×Io,a<b.Suppose that the function|f|isε-convex on[a,b],ε≥0.Then

f(a) +₂f(b)– 1 b–a

b

a

f(x)dx≤(b–a)

_|

f(a)|+|f(b)|

8 +

ε

4 .

Theorem 5([17, Corollary 4.9]) Let f :Io_⊆_R_→_R_{be a diﬀerentiable mapping on I}o_,

(a,b)∈Io_×_Io_,_a_<_b._{Suppose that the function}_|_f_|_is_α_{-convex on}_[a,_b],_α_∈_{(0, 1].}_Then

f(a) +₂f(b)– 1 b–a

b

a

f(x)dx

≤ (b–a)

2(α+ 1)(α+ 2)

2–α+αf(a)α(α+ 1) + 2(1 – 2

–α₎

2 f

_(b) _.

Theorem 6([17, Corollary 4.14]) Let f :Io⊆R→Rbe a diﬀerentiable mapping on Io, (a,b)∈Io_×_Io_,_a_<_b._{Suppose that the function}_|_f_|_{is h-convex on}_[a,_b]._Then

f(a) +₂f(b)– 1 b–a

b

a

f(x)dx≤(b–a)

1

0

h(t)dt

_|

f(a)|+|f(b)| 2

.

For more recent results on integral inequalities of Hermite–Hadamard type concerning theF-convex functions, we refer the interested reader to [19] and the references therein.

In the sequel, we recall the concepts of the left-sided and right-sided Riemann- Liouville fractional integrals of the orderα> 0.

Deﬁnition 2.2 ([20]) Suppose thatf ∈L([a,b]). The left and right Riemann–Liouville fractional integrals denoted byJα

a+f andJ_bα–f of orderα> 0 are deﬁned by

J_aα+f(x) =

1

Γ(α)

x

a

(x–t)α–1f(t)dt, x>a,

and

J_bα–f(x) =

1

Γ(α)

b

x

(t–x)α–1f(t)dt, x<b,

respectively, where Γ(α) is the gamma function deﬁned by Γ(α) =₀∞e–t_tα–1_dt _and

J_b0–f(x) =J_b0–f(x) =f(x).

In [21], authors established the following Hermite–Hadamard type inequalities for F-convex functions involving a Riemann–Liouville fractional:

Theorem 7 Let I⊆Rbe an interval,f :Io_⊆_R_→_R_{be a diﬀerentiable mapping on I}o_,

a,b∈Io_,_a_<_b._{If f is F-convex on}_[a,_b],_{for some F}_∈_F,_{then we have}

F

f

a+b 2

,Γ(α+ 1) (b–a)α J

α a+f(b),

Γ(α+ 1) (b–a)αJ

α b–f(a),

1 2

+

1

0

Lw(t)dt≤0,

TF,w

Γ(α+ 1) (b–a)α

J_aα+f(b) +J_bα–f(a)

,f(a) +f(b),f(a) +f(b)

+

1

0

Lw(t)dt≤0,

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Theorem 8 Let I⊆Rbe an interval,f :Io_⊆_R_→_R_{be a diﬀerentiable mapping on I}o_,

a,b∈Io,a<b.If f is F-convex on[a,b],for some F∈Fand the function t∈(0, 1)→Lw(t)

belongs to L1(a,b),where w(t) =|(1 –t)α–tα|.Then we have the inequality

TF,w

2 b–a

f(a) +₂f(b)– Γ(α+ 1) 2(b–a)α

J_aα+f(b) +J_bα–f(a),f(a),f(b)

+

1

0

Lw(t)dt≤0.

The following deﬁnitions will be useful for this study [20].

Deﬁnition 2.3 The Euler beta function is deﬁned as follows:

B(a,b) =

1

0

ta–1(1 –t)b–1dt, a,b> 0.

The incomplete beta function is deﬁned by

Bx(a,b) =

x

0

ta–1(1 –t)b–1dt, a,b> 0.

Note that, forx= 1, the incomplete beta function reduces to the Euler beta function. Also, the following three lemmas are important to obtain our main results.

Lemma 1([22, Lemma 4]) Let f : [a,b]→Rbe a once diﬀerentiable mappings on(a,b) with a<b,η(b,a) > 0.If f∈L[a,a+eiϕ_η_(b,_a)],_{then the following equality for the fractional}

integral holds:

f(a) +f(a+eiϕ_η_(b,_a))

2 –

Γ(α+ 1)) 2(eiϕ_η_(b,_a))α

Jα a+f

a+eiϕη(b,a)+Jα

(a+eiϕ_η(b,a))–f(a)

=e

iϕ_η_(b,_a)

2

1

0

(1 –t)α_–_tα_f_a_{+ (1 –}_t)eiϕ_η_(b,_a)_dt.

Lemma 2([16, Lemma 5]) Let f : [a,b]→Rbe a once diﬀerentiable mappings on(a,b) with a<b,η(b,a) > 0.If f∈L[a,a+eiϕ_η_(b,_a)],_{then the following equality for the fractional}

integral holds:

f(a) +f(a+eiϕ_η_(b,_a))

2 –

Γ(α+ 1)) 2(eiϕ_η_(b,_a))α

J_aα+f

a+eiϕη(b,a)+J_(a+eα iϕ_η(b,a))–f(a)

=(e

iϕ_η_(b,_a))2

2(α+ 1)

1

0

1 – (1 –t)α+1–tα+1fa+ (1 –t)eiϕη(b,a)dt.

Lemma 3([22]) For t∈[0, 1],we have

(1 –t)m≤21–m–tm, for m∈[0, 1],

(1 –t)m≥21–m–tm, for m∈[1,∞).

In this study, using theλϕ-preinvexity of the function, we establish new inequalities of

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3 Hermite–Hadamard type inequalities for differentiable functions

In this section, we establish some inequalities of Hermite–Hadamard type forF-convex functions in fractional integral forms.

Theorem 9 Let I⊆Rbe an open invex set with respect to bifunctionη:I×I→R,where

η(b,a) > 0.Let f: [0,b]→Rbe a diﬀerentiable mapping.Suppose that|f|is measurable, decreasing,λϕ-preinvex function on I,and F-convex on[a,b],for some F∈Fand the

func-tion t∈(0, 1)→Lw(t)belongs to L1(0, 1),where w(t) =|(1 –t)α–tα|.Then

TF,w

2 eiϕ_η_(b,_a)

f(a) +f(a+eiϕη(b,a)) 2

– Γ(α+ 1) 2(eiϕ_η_(b,_a))α

J_aα+f

a+eiϕη(b,a)+J_(a+eα iϕ_η(b,a))–f(a),f(a),f(b)

+

1

0

Lw(t)dt≤0. (10)

Proof Since|f|isF-convex, we have

Ffa+ (1 –t)eiϕη(b,a),f(a),f(b),t≤0, t∈[0, 1].

Multiplying this inequality byw(t) =|(1 –t)α_–_tα_|_{and using axiom (A3), we have}

Fw(t)fa+ (1 –t)eiϕη(b,a),w(t)f(a),w(t)f(b),t+Lw(t)≤0, t∈[0, 1].

Integrating over [0, 1] and using axiom (A2), we get

TF,w

1

0

(1 –t)α_–_tα_f_a_{+ (1 –}_t)eiϕ_η_(b,_a)_dt,_f_(a)_,_f_(b)_,_t

+

1

0

Lw(t)dt≤0, t∈[0, 1].

But from Lemma1we have

2 eiϕ_η_(b,_a)

f(a) +f(a+eiϕη(b,a)) 2

– Γ(α+ 1)) 2(eiϕ_η_(b,_a))α

J_aα+f

≤

1

0

(1 –t)α–tαfa+ (1 –t)eiϕη(b,a)dt.

BecauseTF,wis nondecreasing with respect to the ﬁrst variable so that

TF,w

2 eiϕ_η_(b,_a)

f(a) +f(a+₂eiϕη(b,a))

– Γ(α+ 1) 2(eiϕ_η_(b,_a))α

Jα a+f

a+eiϕη(b,a)+Jα

(a+eiϕ_η(b,a))–f(a),f(a),f(b)

+

1

0

Lw(t)dt≤0, t∈[0, 1].

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Remark2 If we chooseη(b,a) =b–aandϕ= 0 in Theorem9, we get

TF,w

2 b–a

f(a) +₂f(b)–Γ(α+ 1)) 2(b–a)α

J_aα+f(b) +J_bα–f(a),f(a),f(b)

+

1

0

Lw(t)dt≤0.

Corollary 1 Under the assumptions of Theorem9,if|f|isε-convex,then we have

f(a) +f(a+₂eiϕη(b,a))– Γ(α+ 1)) 2(eiϕ_η_(b,_a))α

J_aα+f

≤eiϕη(b,a)

2(α+ 1)

1 – 1 2α

f(a)+f(b)+ 2ε.

Proof Using (5) withw(t) =|(1 –t)α_–_tα_|_{, we ﬁnd}

1

0

Lw(t)dt=ε

1

0

1 –(1 –t)α–tαdt

=ε

1

2

0

1 – (1 –t)α₊_tα_dt₊

1

2

1

1 – (1 –t)α₊_tα_dt

=ε

1 – 2

α+ 1

1 –1

α .

From (4) withw(t) =|(1 –t)α_–_tα_|_{, we have}

TF,w(u1,u2,u3)

=u1–

1

0

t(1 –t)α–tαdt

u2–

1

0

(1 –t)(1 –t)α–tαdt

u3–ε

=u1–

1

α+ 1

1 – 1

α

(u2+u3) –ε,

foru1,u2,u3∈R. Hence, by Theorem9, we have

0≥TF,w

2 eiϕ_η_(b,_a)

– Γ(α+ 1) 2(eiϕ_η_(b,_a))α

Jα a+f

a+eiϕ_η_(b,_a)₊_Jα

(a+eiϕ_η(b,a))–f(a),f(a),f(b)

+

1

0

Lw(t)dt

= 2

eiϕ_η_(b,_a)

f(a) +f(a+eiϕη(b,a)) 2

– Γ(α+ 1) 2(eiϕ_η_(b,_a))α

J_aα+f

– 1

α+ 1

1 –1

α

f(a)+f(b)+ 2ε.

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Remark3 In Corollary1, if we choose

(a) η(b,a) =b–aandϕ= 0, we get

f(a) +₂f(b)–Γ(α+ 1)) 2(b–a)α

≤ b–a

2(α+ 1)

1 – 1 2α

f(a)+f(b)+ 2ε.

(b) η(b,a) =b–a,ϕ= 0, andε= 0, we get

f(a) +₂f(b)–Γ(α+ 1)) 2(b–a)α

≤ b–a

2(α+ 1)

1 – 1 2α

f(a)+f(b)

which is given by [18].

Corollary 2 Under the assumptions of Theorem9,if|f|isλϕ-preinvex,then we have

J_aα+f

≤eiϕη(b,a)

4

B1 2

1 2,α+

1 2

–B1 2

α+1 2,

1

2 f

_(a)₊1 –λ

λ f

_(b)_.

Proof Using (7) withw(t) =|(1 –t)α_–_tα_|_{, we have}

TF,w(u1,u2,u3)

=u1–

1

0 √

t

2√1 –t(1 –t)

α

–tαdt

u3–

1 –λ λ

1

0 √

1 –t 2√t (1 –t)

α

–tαdt

u2

=u1–

1 2

B1 2

1 2,α+

1 2

–B1 2

α+1 2,

1

2 u2+

1 –λ λ u3

0≥TF,w

2 eiϕ_η_(b,_a)

– Γ(α+ 1) 2(eiϕ_η_(b,_a))α

J_aα+f

+

1

0

Lw(t)dt

= 2

eiϕ_η_(b,_a)

f(a) +f(a+eiϕη(b,a)) 2

– Γ(α+ 1) 2(eiϕ_η_(b,_a))α

Jα a+f

a+eiϕη(b,a)+Jα

–1 2

B1 2

1 2,α+

1 2

–B1 2

α+1 2,

1

2 f

_(a)₊1 –λ

λ f

(10)

This leads to

f(a) +f(a+₂eiϕη(b,a))– Γ(α+ 1) 2(eiϕ_η_(b,_a))α

J_aα+f

≤eiϕη(b,a)

4

B1 2

1 2,α+

1 2

–B1 2

α+1 2,

1

2 f

_(a)₊1 –λ

λ f

_(b)_.

Thus, the proof is done.

f(a) +₂f(b)–Γ(α+ 1)) 2(b–a)α

≤b–a

4

B1 2

1 2,α+

1 2

–B1 2

α+1 2,

1

2 f

_(a)₊1 –λ

λ f

_(b)_.

(b) η(b,a) =b–a,ϕ= 0, andλ=1₂, we get

f(a) +₂f(b)–Γ(α+ 1)) 2(b–a)α

≤b–a

4

B1 2

1 2,α+

1 2

–B1 2

α+1 2,

1

2 f

_(a)₊_f_(b)_.

Corollary 3 Under the assumptions of Theorem9,if|f|is h-convex,then we have

J_aα+f

≤eiϕη(b,a)

2

1

0

h(t)(1 –t)α–tαdtf(a)+1 –λ

λ f

_(b)_.

TF,w(u1,u2,u3)

=u1–

1

0

h(t)(1 –t)α–tαdt

u3–

1 –λ λ

1

0

h(1 –t)(1 –t)α–tαdt

u2

=u1–

1

0

h(t)(1 –t)α–tαdt

u3–

1 –λ λ

1

0

h(t)(1 –t)α–tαdt

u2

=u1–

1

0

h(t)(1 –t)α–tαdt

u2+

1 –λ λ u3

foru1,u2,u3∈R. So, by Theorem9, we have

0≥TF,w

2 eiϕ_η_(b,_a)

f(a) +f(a+eiϕη(b,a)) 2

– Γ(α+ 1) 2(eiϕ_η_(b,_a))α

Jα a+f

a+eiϕη(b,a)+Jα

(11)

+

1

0

Lw(t)dt

= 2

eiϕ_η_(b,_a)

f(a) +f(a+eiϕη(b,a)) 2

– Γ(α+ 1) 2(eiϕ_η_(b,_a))α

J_aα+f

–

1

0

h(t)(1 –t)α–tαdtf(a)+1 –λ

λ f

_(b)_.

This leads to

J_aα+f

≤eiϕη(b,a)

2

1

0

h(t)(1 –t)α–tαdtf(a)+1 –λ

λ f

_(b)_.

Theorem 10 Let I ⊆Rbe an open invex set with respect to bifunctionη:I×I→R, whereη(b,a) > 0. Let f : [0,b]→Rbe a diﬀerentiable mapping.Suppose that |f|pp–1 _is

measurable,decreasing,λϕ-preinvex function on I,and F-convex on[a,b],for some F∈F

and|f| ∈L p

p–1_(a,_b)._Then

TF,1

G1(f,p),f(a)

p p–1_,_f_(b)

p

p–1≤_0, ₍₁₁₎

where

G1(f,p) =

2 eiϕ_η_(b,_a)

p

p–1 _α_p_{+ 1}

2 – 21–αp

1

p–1

f(a) +f(a+eiϕη(b,a)) 2

– Γ(α+ 1) 2(eiϕ_η_(b,_a))α

Jα a+f

a+eiϕ_η_(b,_a)₊_Jα

p p–1

.

Proof Since|f|pp–1 _is_{F-convex, we have}

Ffa+ (1 –t)eiϕ_η_(b,_a)pp–1_,_f_(a)

p p–1_,_f_(b)

p

p–1_,_t≤_0, _t∈_{[0, 1].}

Withw(t) = 1 in (A2), we have

TF,1

1

0

fa+ (1 –t)eiϕη(b,a) p

p–1_dt,_f_(a)

p p–1_,_f_(b)

p p–1

≤0, t∈[0, 1].

Using Lemma1and the Hölder inequality, we get

J_aα+f

=e

iϕ_η_(b,_a)

2

1

0

(12)

≤eiϕη(b,a)

2

1

0

(1 –t)α–tαdt

1

p 1

0

fa+ (1 –t)eiϕη(b,a) p p–1_dt

p–1

p

=e

iϕ_η_(b,_a)

2

2 – 21–αp

αp+ 1

1

p 1

0

fa+ (1 –t)eiϕη(b,a) p p–1_dt

p–1

p

or, equivalently,

2 eiϕ_η_(b,_a)

p

p–1 _α_p_{+ 1}

2 – 21–αp

1

p–1

– Γ(α+ 1) 2(eiϕ_η_(b,_a))α

Jα a+f

a+eiϕη(b,a)+Jα

p p–1

≤ 1 0

fa+ (1 –t)eiϕη(b,a) p p–1_dt.

BecauseTF,1is nondecreasing with respect to the ﬁrst variable, we get

TF,1

G1(f,p),f(a)

p p–1_,_f_(b)

p p–1≤_0.

Thus, the proof is completed.

TF,1

2 b–a

p

p–1 _α_p_{+ 1}

2 – 21–αp

1

p–1

f(a) +₂f(b)–Γ(α+ 1)) 2(b–a)α

J_aα+f(b) +J_bα–f(a),

_f_(a)_,_f_(b)_≤_0.

Corollary 4 Under the assumptions of Theorem10,if|f| p

p–1 _is_ε_-convex,_{we have}

f(a) +f(a+eiϕη(b,a))

2 –

Γ(α+ 1)) 2(eiϕ_η_(b,_a))α

J_aα+f

≤eiϕη(b,a)

2

2 – 21–αp

αp+ 1

1

p|_f_(a)| p

p–1₊|_f_(b)|

p p–1

2 +ε

p–1

p .

Proof Using (5) withw(t) = 1, we have

1

0

Lw(t)dt=ε

1

0

1 –w(t)dt= 0. (12)

From (4) withw(t) = 1, we have

TF,1(u1,u2,u3)

=u1–

1

0

t dt

u2–

1

0

(1 –t)dt

u3–ε

=u1–

u2+u3

(13)

0≥TF,1

G1(f,p),f(a)

p p–1_,_f_(b)

p p–1

=G1(f,p) – |

f(a)| p

p–1₊|_f_(b)|

p p–1

2 –ε.

This leads to

2 eiϕ_η_(b,_a)

p

p–1 _α_p_{+ 1}

2 – 21–αp

1

p–1

– Γ(α+ 1) 2(eiϕ_η_(b,_a))α

J_aα+f

p p–1

–|f

_(a)_|pp–1₊|_f_(b)|

p p–1

2 ≤0

or, equivalently,

J_aα+f

≤eiϕη(b,a)

2

2 – 21–αp

αp+ 1

1

p|_f_(a)| p

p–1₊|_f_(b)|

p p–1

2 +ε

p–1

p .

This completes the proof.

f(a) +₂f(b)–Γ(α+ 1)) 2(b–a)α

Jα

a+f(b) +J_bα–f(a)

≤b–a

2

2 – 21–αp

αp+ 1

1

p|_f_(a)| p

p–1₊|_f_(b)|

p p–1

2 +ε

p–1

p .

f(a) +₂f(b)–Γ(α+ 1)) 2(b–a)α

Jα

≤b–a

2

2 – 21–αp

αp+ 1

1

p|_f_(a)| p

p–1₊|_f_(b)|

p p–1

2

p–1

p .

Corollary 5 Under the assumptions of Theorem10.If|f|p–1p _is_λ

ϕ-preinvex,we have

Jα a+f

a+eiϕη(b,a)+Jα

≤eiϕη(b,a)

2

2 – 21–αp

αp+ 1

1

p_π 4

f(a) p–1

p ₊1 –λ

λ f

_(b)p–1p p–1

(14)

Proof Using (7) withw(t) = 1, we have

TF,1(u1,u2,u3)

=u1–

1

0 √

t 2√1 –tdt

u3–

1 –λ λ

1

0 √

1 –t 2√t dt

u2

=u1–

1 2β

1 2,

3 2

u2+

1 –λ λ u3

=u1–

π

4

u2+

1 –λ λ u3

(14)

0≥TF,1

G1(f,p),f(a)

p–1

p _,_f_(b) p–1

p

=

2 eiϕ_η_(b,_a)

p

p–1 _α_p_{+ 1}

2 – 21–αp

1

p–1

– Γ(α+ 1) 2(eiϕ_η_(b,_a))α

J_aα+f

p p–1

–π 4

f(a) p–1

p ₊1 –λ

λ f

_(b)p–1p

.

This leads to

2 –

Γ(α+ 1) 2(eiϕ_η_(b,_a))α

Jα a+f

a+eiϕη(b,a)+Jα

≤eiϕη(b,a)

2

2 – 21–αp

αp+ 1

1

p_π 4

f(a) p–1

p ₊1 –λ

λ f

_(b)p–1p p–1

p .

f(a) +₂f(b)–Γ(α+ 1)) 2(b–a)α

≤b–a

2

2 – 21–αp

αp+ 1

1

p_π 4

f(a) p–1

p ₊1 –λ

λ f

_(b)p–1p p–1

p .

f(a) +₂f(b)–Γ(α+ 1)) 2(b–a)α

Jα

≤b–a

2

2 – 21–αp

αp+ 1

1

p_π 4f

_(a)p–1p ₊_f_(b) p–1

p p–1

(15)

Corollary 6 Under the assumptions of Theorem10.If|f| p

p–1 _{is h-convex,}_{we have}

J_aα+f

≤eiϕη(b,a)

2

2 – 21–αp

αp+ 1

1

p 1

0

h(t)

p–1

p

_f_(a)p–1p ₊1 –λ

λ f

_(b)p–1p

.

Proof From (9) withw(t) = 1, we have

TF,1(u1,u2,u3)

=u1–

1

0

h(t)dt

u3–

1 –λ λ

1

0

h(1 –t)dt

u2

=u1–

1

0

h(t)dt

u3–

1 –λ λ

1

0

h(t)dt

u2

=u1–

1

0

h(t)dt

u2+

1 –λ λ u3

0≥TF,1

G1(f,p),f(a)

p–1

p _,_f_(b) p–1

p

=

2 eiϕ_η_(b,_a)

p

p–1 _α_p_{+ 1}

2 – 21–αp

1

p–1

×f(a) +f(a+eiϕη(b,a))

2 –

Γ(α+ 1) 2(eiϕ_η_(b,_a))α

Jα a+f

a+eiϕη(b,a)+Jα

–

1

0

h(t)dtf(a) p–1

p ₊1 –λ

λ f

_(b)p–1p

,

that is,

2 –

Γ(α+ 1) 2(eiϕ_η_(b,_a))α

J_aα+f

≤eiϕη(b,a)

2

2 – 21–αp

αp+ 1

1

p

× 1 0

h(t)(1 –t)α–tαdtf(a) p–1

p ₊1 –λ

λ f

_(b)p–1p

.

Theorem 11 Let I ⊆Rbe an open invex set with respect to bifunctionη:I×I→R, whereη(b,a) > 0. Let f : [0,b]→Rbe a diﬀerentiable mapping.Suppose that |f|

p p–1 _is

measurable,decreasing,λϕ-preinvex function on I,and F-convex on[a,b],for some F∈F

and|f| ∈Lpp–1_(a,_b)._Then

TF,w

G2(f,p),f(a)

p p–1_,_f_(b)

p p–1₊

1

0

(16)

where

G2(f,p) =

2 eiϕ_η_(b,_a)

p

p–1 _α_{+ 1}

2 – 21–α

1

p–1

f(a) +f(a+eiϕη(b,a)) 2

– Γ(α+ 1) 2(eiϕ_η_(b,_a))α

J_aα+f

p p–1

for w(t) =|(1 –t)α_–_tα_|_.

Proof Since|f| p

p–1 _is_{F-convex, we have}

Ffa+ (1 –t)eiϕη(b,a) p p–1_,_f_(a)

p p–1_,_f_(b)

p

p–1_,_t≤_0, _t∈_{[0, 1].}

Using (A3) withw(t) =|(1 –t)α_–_tα_|_{, we obtain}

Fw(t)fa+ (1 –t)eiϕη(b,a) p

p–1_,_w(t)_f_(a)

p

p–1_,_w(t)_f_(b)

p p–1_,_t₊_L

w(t)≤0,

t∈[0, 1].

Integrating over [0, 1] and using axiom (A2), we obtain

TF,w

1

0

w(t)fa+ (1 –t)eiϕη(b,a) p

p–1_dt,_f_(a)

p p–1_,_f_(b)

p p–1

+

1

0

Lw(t)dt≤0,

t∈[0, 1].

Using Lemma1and the power mean inequality, we get

2 –

Γ(α+ 1) 2(eiϕ_η_(b,_a))α

J_aα+f

=e

iϕ_η_(b,_a)

2

1

0

(1 –t)α_–_tα_f_a_{+ (1 –}_t)eiϕ_η_(b,_a)_dt

≤eiϕη(b,a)

2

1

0

(1 –t)α_–_tα_dt

1

p 1

0

w(t)fa+ (1 –t)eiϕη(b,a) p p–1_dt

p–1

p

=e

iϕ_η_(b,_a)

2

2 – 21–α

α+ 1

1

p 1

0

w(t)fa+ (1 –t)eiϕη(b,a) p p–1_dt

p–1

p

or, equivalently,

2 eiϕ_η_(b,_a)

p

p–1 _α_{+ 1}

2 – 21–α

1

p–1

f(a) +f(a+eiϕη(b,a)) 2

– Γ(α+ 1) 2(eiϕ_η_(b,_a))α

J_aα+f

p p–1

≤

1

0

(17)

BecauseTF,wis nondecreasing with respect to the ﬁrst variable, we ﬁnd

TF,w

G2(f,p),f(a)

p p–1_,_f_(b)

p p–1₊

1

0

Lw(t)dt≤0.

TF,w

2 b–a

p

p–1 _α_{+ 1}

2 – 21–α

1

p–1

f(a) +₂f(b)–Γ(α+ 1)) 2(b–a)α

J_aα+f(b) +J_bα–f(a),

f(a),f(b)

+

1

0

Lw(t)dt≤0.

Corollary 7 Under the assumptions of Theorem11,if|f| p

p–1 _is_ε_-convex,_{we have}

J_aα+f

≤eiϕη(b,a)

2

2 – 21–α

α+ 1

1

p ₂α_{– 1}

2α₍_α_{+ 1)}f

_(a)p–1p ₊_f_(b)

p p–1_{+ 2}_ε

p–1

p .

Proof Using (5) withw(t) =|(1 –t)α_–_tα_|_{, we get}

1

0

Lw(t)dt=ε

1 – 2 2

α_{– 1}

2α₍_α_{+ 1)}

.

From (4) withw(t) =|(1 –t)α_–_tα_|_{, we get}

TF,w(u1,u2,u3)

=u1–

1

0

(1 –t)α_–_tα_{t dt}

u2–

1

0

(1 –t)α_–_tα_{(1 –}_t)_dt

u3–ε

=u1–

2α_{– 1}

2α₍_α_{+ 1)}(u2+u3) –ε,

0≥TF,w

G2(f,p),f(a)

p p–1_,_f_(b)

p p–1₊

1

0

Lw(t)dt

=G2(f,p) –

2α_{– 1}

2α₍_α_{+ 1)}f

_(a)p–1p ₊_f_(b)

p

p–1_–_ε₊_ε

1 – 2 2

α_{– 1}

2α₍_α_{+ 1)}

.

This implies that

J_aα+f

≤eiϕη(b,a)

2

2 – 21–α

α+ 1

1

p ₂α_{– 1}

2α₍_α_{+ 1)}f

_(a)p–1p ₊_f_(b)

p p–1_{+ 2}_ε

p–1

p .

(18)

f(a) +₂f(b)–Γ(α+ 1)) 2(b–a)α

Jα

≤b–a

2

2 – 21–α

α+ 1

1

p ₂α_{– 1}

2α₍_α_{+ 1)}f

_(a)p–1p ₊_f_(b)

p p–1_{+ 2}_ε

p–1

p .

f(a) +₂f(b)–Γ(α+ 1)) 2(b–a)α

≤b–a

2

2 – 21–α

α+ 1

1

p ₂α_{– 1}

2α₍_α_{+ 1)}f

_(a)p–1p ₊_f_(b)

p p–1

p–1

p .

Corollary 8 Under the assumptions of Theorem11.If|f|p–1p _is_λ

ϕ-preinvex,we have

2 –

Γ(α+ 1)) 2(eiϕ_η_(b,_a))α

J_aα+f

≤eiϕη(b,a)

2

2 – 21–α

α+ 1

1 p × _B₁ 2( 1 2,α+

1

2) –B12(α+

1 2, 1 2) 2

f(a) p

p–1₊1 –λ

λ f

_(b)pp–1

p–1

p .

TF,w(u1,u2,u3)

=u1–

1

0 √

t

2√1 –t(1 –t)

α_–_tα_dt

u3–

1 –λ λ

1

0 √

1 –t 2√t (1 –t)

α_–_tα_dt

u2

=u1–

1 2 B1 2 1 2,α+

1 2

–B1 2

α+1 2,

1 2

u2+

1 –λ λ u3

foru1,u2,u3∈R. Now, by Theorem11, we have

0≥TF,w

G2(f,p),f(a)

p–1

p _,_f_(b) p–1

p

=

2 eiϕ_η_(b,_a)

p

p–1 _α_{+ 1}

2 – 21–α

1

p–1

2 –

Γ(α+ 1) 2(eiϕ_η_(b,_a))α

Jα a+f

a+eiϕη(b,a)+Jα

–1 2 B1 2 1 2,α+

1 2

–B1 2

α+1 2,

1 2

f(a) p–1

p ₊1 –λ

λ f

_(b)p–1p

.

This leads to

Jα a+f

a+eiϕη(b,a)+Jα

(19)

≤eiϕη(b,a)

2

2 – 21–α

α+ 1

1

p

×

_B₁ 2(

1 2,α+

1

2) –B12(α+

1 2,

1 2)

2

f(a) p

p–1₊1 –λ

λ f

_(b)pp–1

p–1

p .

f(a) +₂f(b)–Γ(α+ 1)) 2(b–a)α

≤b–a

2

2 – 21–α

α+ 1

1

p

×

_B₁ 2(

1 2,α+

1

2) –B12(α+

1 2,

1 2)

2

f(a) p

p–1₊1 –λ

λ f

_(b)pp–1

p–1

p .

f(a) +₂f(b)–Γ(α+ 1)) 2(b–a)α

≤b–a

2

2 – 21–α

α+ 1

1

p

×

_B₁ 2(

1 2,α+

1

2) –B12(α+

1 2,

1 2)

2 f

_(a)p–1p ₊_f_(b)

p p–1

p–1

p .

Corollary 9 Under the assumptions of Theorem11.If|f|pp–1 _{is h-convex,}_{we have}

J_aα+f

≤eiϕη(b,a)

2

2 – 21–α

α+ 1

1

p

× 1 0

h(t)(1 –t)α–tαdt

p–1

p

f(a) p

p–1₊1 –λ

λ f

_(b)p–1p p–1

p .

Proof From (9) withw(t) =|(1 –t)α_–_tα_|_{, we have}

TF,w(u1,u2,u3)

=u1–

1

0

h(t)(1 –t)α_–

tαdt

u3–

1 –λ λ

1

0

h(1 –t)(1 –t)α_–

tαdt

u2

=u1–

1

0

h(t)(1 –t)α_–_tα_dt

u3–

1 –λ λ

1

0

h(t)(1 –t)α_–_tα_dt

u2

=u1–

1

0

h(t)(1 –t)α_–_tα_dt

u2+

1 –λ λ u3

(20)

0≥TF,w

G2(f,p),f(a)

p–1

p _,_f_(b) p–1

p

=

2 eiϕ_η_(b,_a)

p

p–1 _α_{+ 1}

2 – 21–α

1

p–1

2 –

Γ(α+ 1) 2(eiϕ_η_(b,_a))α

J_aα+f

–

1

0

h(t)dtf(a) p–1

p ₊1 –λ

λ f

_(b)p–1p

,

that is,

Jα a+f

a+eiϕη(b,a)+Jα

≤eiϕη(b,a)

2

2 – 21–α

α+ 1

1

p

×

1

0

h(t)(1 –t)α–tαdt

p–1

p

f(a) p

p–1₊1 –λ

λ f

_(b)p–1p p–1

p .

4 Trapezoid type inequalities for twice differentiable functions

In this section, we establish some trapezoid type inequalities for functions whose second derivatives absolutely values are

Theorem 12 Let f : [0,b]→Rbe a diﬀerentiable mapping and|f|is measurable, de-creasing,λϕ-preinvex function on[0,b]for0≤a<b,η(b,a) > 0andα> 0.Suppose that

F-convex on[0,b],for some F∈F and the function t∈(0, 1)→Lw(t) belongs to L1(0, 1),

where w(t) = 1 – (1 –t)α+1_–_tα+1_._Then

TF,w

2(α+ 1) (eiϕ_η_(b,_a))2

– Γ(α+ 1) 2(eiϕ_η_(b,_a))α

J_aα+f

+

1

0

Lw(t)dt≤0. (16)

Proof Since|f|isF-convex, we can see that

Ffa+ (1 –t)eiϕη(b,a),f(a),f(b),t≤0, t∈[0, 1].

Multiplying this inequality byw(t) = 1 – (1 –t)α+1_–_tα+1_{and using axiom (A3), we have}

(21)

Integrating over [0, 1] and using axiom (A2), we get

TF,w

1

0

w(t)fa+ (1 –t)eiϕη(b,a)dt,f(a),f(b),t

+

1

0

Lw(t)dt≤0,

t∈[0, 1].

Using Lemma2, we have

2(α+ 1) (eiϕ_η_(b,_a))2

– Γ(α+ 1)) 2(eiϕ_η_(b,_a))α

J_aα+f

≤

1

0

1 – (1 –t)α+1–tα+1fa+ (1 –t)eiϕη(b,a)dt.

BecauseTF,wis nondecreasing with respect to the ﬁrst variable so that

TF,w

2(α+ 1) (eiϕ_η_(b,_a))2

– Γ(α+ 1) 2(eiϕ_η_(b,_a))α

J_aα+f

+

1

0

Lw(t)dt≤0, t∈[0, 1].

Remark11 By takingη(b,a) =b–aandϕ= 0 in Theorem12, we obtain

TF,w

2(α+ 1) (b–a)2

f(a) +₂f(b)–Γ(α+ 1)) 2(b–a)α

Jα

a+f(b) +J_bα–f(a),f(a),f(b)

+

1

0

Lw(t)dt≤0.

Corollary 10 Under the assumptions of Theorem12,if|f|isε-convex,then

J_aα+f

≤ α(eiϕη(b,a))2

4(α+ 1)(α+ 2)f

_(a)₊_f_(b)_{+ 2}_ε_.

Proof Using (5) withw(t) = 1 – (1 –t)α+1_–_tα+1_{, we ﬁnd}

1

0

Lw(t)dt=ε

1

0

(1 –t)α+1+tα+1dt= 2ε