On a new class of convex functions and integral inequalities

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R E S E A R C H

Open Access

On a new class of convex functions and

integral inequalities

Shanhe Wu

1*

_{, Muhammad Uzair Awan}

2

_{, Muhammad Aslam Noor}

3

_{, Khalida Inayat Noor}

3

_and

Sabah Iftikhar

3

*_{Correspondence:}

[email protected] 1_{Department of Mathematics,} Longyan University, Longyan, China Full list of author information is available at the end of the article

Abstract

The aim of this paper is to introduce a new extension of convexity called

σ

-convexity. We show that the class of

σ

-convex functions includes several other classes of convex functions. Some new integral inequalities of Hermite–Hadamard type are established to illustrate the applications of

σ

-convex functions.

MSC: 26D15; 26D10; 26B25

Keywords: Convex functions;

σ

-convex functions; Hermite–Hadamard inequality; Integral inequalities

1 Introduction

A setK⊂Ris said to be convex if∀x,y∈K,t∈[0, 1], we have (1 –t)x+ty∈K.

A functionf :K→Ris said to be convex in the classical sense, if∀x,y∈K,t∈[0, 1], we have

f(1 –t)x+ty≤(1 –t)f(x) +tf(y).

The theory of convexity plays a vital role in different fields of pure and applied sciences. Consequently the classical concepts of convex sets and convex functions have been gen-eralized in different directions. For more information, see [1–3]. Another aspect due to which the convexity theory has attracted many researchers is its close relation with the-ory of inequalities. Many famous inequalities can be obtained using the concept of convex functions. For details, interested readers are referred to [4–14]. Among these inequalities, Hermite–Hadamard’s inequality, which provides us a necessary and sufficient condition for a function to be convex, is one of the most studied results. This result of Hermite and Hadamard reads as follows:

Theorem 1.1 Let f : [a,b]⊂R→Rbe an integrable convex function.Then

f

a+b 2

≤ 1

b–a

b

a

f(x) dx≤f(a) +f(b)

2 .

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The main motivation of this paper is to introduce the notions ofσ-convex sets andσ -convex functions by using the quasi-arithmetic means which can bring together all the power meansMp forp∈R. More speciﬁcally, we deﬁne theσ-convex functions via the

formula

M[σ](x,y) :=σ–1

(1 –t)σ(x) +tσ(y),

which is associated with the strictly monotonic continuous functionσ. As applications of theσ-convex functions, we derive some new Hermite–Hadamard-like inequalities. In the meantime some important special cases will also be discussed in detail.

2

σ

-convex functions

Let us now introduce new classes ofσ-convex sets andσ-convex functions.

Deﬁnition 2.1 A setQ⊂Ris said to beσ-convex set with respect to strictly monotonic continuous functionσif

M[σ](x,y) :=σ–1

(1 –t)σ(x) +tσ(y)∈Q, ∀x,y∈Q,t∈[0, 1].

Deﬁnition 2.2 A functionf :Q→Ris said to be σ-convex function with respect to strictly monotonic continuous functionσif

fM[σ](x,y)

≤(1 –t)f(x) +tf(y), ∀x,y∈Q,t∈[0, 1]. (2.1)

Note that the functionf is called strictlyσ-convex onQif the above inequality is true as a strict inequality for each distinctxandy∈Qand for eacht∈(0, 1).

The functionf :Q→Ris calledσ-concave (strictlyσ-concave) onQ, if –f isσ-convex (strictlyσ-convex) onQ.

If we taket=1₂ in (2.1), then we have

f

σ–1

σ(x) +σ(y) 2

≤f(x) +f(y)

2 , ∀x,y∈Q. (2.2)

The functionf is calledσ-Jensen (or mid)-convex function. We now discuss some special cases of Deﬁnition2.2. Case I. If we takeσ(x) =lnx, then condition (2.1) becomes

fx1–tyt≤(1 –t)f(x) +tf(y), ∀x,y∈[a,b]⊂(0,∞),t∈[0, 1],

which is the concept of geometric convexity as considered in [1]. Case II. If we takeσ(x) =1_x, then condition (2.1) becomes

f

xy tx+ (1 –t)y

≤(1 –t)f(x) +tf(y), ∀x,y∈[a,b]⊂(0,∞),t∈[0, 1],

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Case III. If we takeσ(x) =xp₍_p_{> 0), then condition (}_2.1_{) becomes}

f(1 –t)xp+typ1p≤_{(1 –}_t₎_f₍_x_{) +}_tf₍_y_), ∀_x_,_y∈_[_a_,_b_]⊂_(0,∞_),_t∈_{[0, 1],}

which is the concept ofp-convexity as considered in [16]. Case IV. If we takeσ(x) =ex_{, then condition (}_2.1_{) becomes}

fln(1 –t)ex+tey≤(1 –t)f(x) +tf(y), ∀x,y∈[a,b],t∈[0, 1],

which is the concept oflog-exponential convex functions on [a,b].

3 Applications of

σ

-convex functions to integral inequalities

In this section, we show a representative application ofσ-convex functions. We will estab-lish some new integral inequalities of Hermite–Hadamard type viaσ-convex functions.

LetI= [a,b] unless otherwise speciﬁed andσbe a continuous diﬀerentiable and strictly monotonic function in its domain. We denote byR+_{the set of positive real numbers.}

Theorem 3.1 Suppose that f :I→Ris an integrableσ-convex function with respect to the functionσ,then we have the following inequalities:

f

σ–1

σ(a) +σ(b) 2

≤ 1

σ(b) –σ(a)

b

a

f(x)σ(x) dx

≤f(a) +f(b)

2 . (3.1)

Proof Sincef is aσ-convex function, we have

f

σ–1

σ(x) +σ(y) 2

≤f(x) +f(y)

2 .

Substitutingx=σ–1_{((1 –}_t₎_σ₍_a_{) +}_t_σ₍_b_{)) and}_y₌_σ–1₍_t_σ₍_a_{) + (1 –}_t₎_σ₍_b_{)) in the above}

inequality, we have

f

σ–1

σ(a) +σ(b) 2

≤f(σ–1((1 –t)σ(a) +tσ(b))) +f(σ–1(tσ(a) + (1 –t)σ(b)))

2 . (3.2)

Integrating both sides of (3.2) with respect toton [0, 1], we get

f

σ–1

σ(a) +σ(b) 2

≤ 1

σ(b) –σ(a)

b

a

f(x)σ(x) dx. (3.3)

Similarly, in light of the assumption in Theorem3.1thatf is theσ-convex function, we have

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Integrating both sides of the above inequality with respect toton [0, 1], we obtain

1

σ(b) –σ(a)

b

a

f(x)σ(x) dx≤f(a) +f(b)

2 . (3.4)

Combining (3.3) and (3.4) completes the proof of Theorem3.1.

Theorem 3.2 Let f :I→R+_{be an integrable}_σ_{-convex function with respect to the}

func-tionσ.Then

2f(a)

σ(b) –σ(a)

b

a

σ(b) –σ(x)

σ(b) –σ(a)

f(x)σ(x) dx

+ 2f(b)

σ(b) –σ(a)

b

a

σ(x) –σ(a)

σ(b) –σ(a)

f(x)σ(x) dx

≤ 1

σ(b) –σ(a)

b

a

f2(x)σ(x) dx+f

2₍_a_{) +}_f₍_a₎_f₍_b_{) +}_f2₍_b₎

3

≤2[f2(a) +f(a)f(b) +f2(b)]

3 . (3.5)

Proof Using the arithmetic-geometric means inequality gives

2fσ–1(1 –t)σ(a) +tσ(b)(1 –t)f(a) +tf(b)

≤fσ–1(1 –t)σ(a) +tσ(b)2+(1 –t)f(a) +tf(b)2

=fσ–1(1 –t)σ(a) +tσ(b)2+ (1 –t)2f2(a) +t2f2(b) + 2t(1 –t)f(a)f(b).

2f(a)

1

0

(1 –t)fσ–1(1 –t)σ(a) +tσ(b)dt

+ 2f(b)

1

0

tfσ–1(1 –t)σ(a) +tσ(b)dt

≤

1

0

f2σ–1(1 –t)σ(a) +tσ(b)dt+f2(a)

1

0

(1 –t)2dt+f2(b)

1

0

t2dt

+ 2f(a)f(b)

1

0

t(1 –t) dt. (3.6)

By making the change of variable, inequality (3.6) can be rewritten as

2f(a)· 1

σ(b) –σ(a)

b

a

σ(b) –σ(x)

σ(b) –σ(a)

f(x)σ(x) dx

+ 2f(b)· 1

σ(b) –σ(a)

b

a

σ(x) –σ(a)

σ(b) –σ(a)

f(x)σ(x) dx

≤ 1

σ(b) –σ(a)

b

a

f2(x)σ(x) dx+f

2₍_a_{) +}_f₍_a₎_f₍_b_{) +}_f2₍_b₎

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On the other hand, sincef is aσ-convex function, we have

fσ–1(1 –t)σ(a) +tσ(b)≤(1 –t)f(a) +tf(b), ∀t∈[0, 1], therefore

1

σ(b) –σ(a)

b

a

f2(x)σ(x) dx

=

1

0

f2σ–1(1 –t)σ(a) +tσ(b)dt

≤

1

0

(1 –t)f(a) +tf(b)2dt=f

2₍_a_{) +}_f₍_a₎_f₍_b_{) +}_f2₍_b₎

3 . (3.8)

Combining (3.7) and (3.8) leads to the inequalities described in Theorem3.2.

Theorem 3.3 Let f :I→R+_{be an integrable}_σ_{-convex function with respect to the}

func-tionσ.Then

1

σ(b) –σ(a)

b

a

f(x)σ(x) dx

≤1

2f

σ–1

σ(a) +σ(b) 2

+ 1

4(σ(b) –σ(a))f(σ–1₍σ(a)+σ(b)

2 ))

b

a

f2(x)σ(x) dx

+ 1

24f(σ–1₍σ(a)+σ(b)

2 ))

f2(a) +f2(b) + 4f(a)f(b). (3.9)

Proof Using the arithmetic-geometric means inequality and theσconvexity off, it follows that

f

σ–1

σ(a) +σ(b) 2

fσ–1(1 –t)σ(a) +tσ(b)+fσ–1tσ(a) + (1 –t)σ(b)

≤f2

σ–1

σ(a) +σ(b) 2

+1 4

fσ–1(1 –t)σ(a) +tσ(b)+fσ–1tσ(a) + (1 –t)σ(b)2

=f2

σ–1

σ(a) +σ(b) 2

+1 4

f2σ–1(1 –t)σ(a) +tσ(b)+f2σ–1tσ(a) + (1 –t)σ(b)

+ 2fσ–1(1 –t)σ(a) +tσ(b)fσ–1tσ(a) + (1 –t)σ(b)

≤f2

σ–1

σ(a) +σ(b) 2

+1 4

f2σ–1(1 –t)σ(a) +tσ(b)+f2σ–1tσ(a) + (1 –t)σ(b)

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f

σ–1

σ(a) +σ(b) 2

1

0

fσ–1(1 –t)σ(a) +tσ(b)dt

+

1

0

fσ–1tσ(a) + (1 –t)σ(b)dt

≤f2

σ–1

σ(a) +σ(b) 2

1

0

dt

+1 4

1

0

f2σ–1(1 –t)σ(a) +tσ(b)dt

+

1

0

f2σ–1tσ(a) + (1 –t)σ(b)dt

+ 2f2(a) +f2(b)

1

0

t(1 –t) dt+ 2f(a)f(b)

1

0

t2+ (1 –t)2dt

.

Performing the change of variable, we get

f

σ–1

σ(a) +σ(b) 2

2

σ(b) –σ(a)

b

a

f(x)σ(x) dx

≤f2

σ–1

σ(a) +σ(b) 2

+ 1

2(σ(b) –σ(a))

b

a

f2(x)σ(x) dx

+f

2₍_a_{) +}_f2₍_b_{) + 4}_f₍_a₎_f₍_b₎

12 .

After a simple computation, one can transform the above inequality to the required

inequality of Theorem3.3.

Theorem 3.4 Suppose that f,h:I→R+ _{are two similarly ordered integrable}_σ_-convex

functions with respect to the functionσ,then hf is also theσ-convex function with respect to the functionσ.

Proof Sincef,h:I→R+ _{are two similarly ordered integrable}_σ_{-convex functions, for}

∀x,y∈I,t∈[0, 1], we have

fσ–1(1 –t)σ(x) +tσ(y)hσ–1(1 –t)σ(x) +tσ(y)

≤(1 –t)f(x) +tf(y)(1 –t)h(x) +th(y)

= (1 –t)2f(x)h(x) +t(1 –t)f(x)h(y) +f(y)h(x)+t2f(y)h(y)

= (1 –t)f(x)h(x) +tf(y)h(y) + (1 –t)2f(x)h(x)

+t(1 –t)f(x)h(y) +f(y)h(x)+t2f(y)h(y) – (1 –t)f(x)h(x) –tf(y)h(y)

= (1 –t)f(x)h(x) +tf(y)h(y) –t(1 –t)f(x) –f(y)h(x) –h(y)

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Thus

fσ–1(1 –t)σ(x) +tσ(y)hσ–1(1 –t)σ(x) +tσ(y)

≤(1 –t)f(x)h(x) +tf(y)h(y). (3.10)

Using the deﬁnition ofσ-convex function (Deﬁnition2.2), we conclude thathfis theσ -convex function with respect to the functionσ. The proof of Theorem3.4is complete.

Theorem 3.5 Let f,h:I→R+_{be integrable}_σ_{-convex functions with respect to the}

func-tionσ.Then

1

σ(b) –σ(a)

b

a

f(x)h(x)σ(x) dx

≤1

3M2(a,b) + 1 6N2(a,b)

≤1

6

M1(a,b)

2

+N1(a,b)

2

–f(a)f(b) +h(a)h(b),

where

M1(a,b) =f(a) +f(b), N1(a,b) =h(a) +h(b),

M2(a,b) =f(a)h(a) +f(b)h(b), N2(a,b) =f(a)h(b) +f(b)h(a).

Proof Sincef,h:I→R+_{are integrable}_σ_{-convex functions, we have}

1

σ(b) –σ(a)

b

a

f(x)h(x)σ(x) dx

=

1

0

fσ–1(1 –t)σ(a) +tσ(b)hσ–1(1 –t)σ(a) +tσ(b)dt

≤

1

0

(1 –t)f(a) +tf(b)(1 –t)h(a) +th(b)dt

=f(a)h(a)

1

0

(1 –t)2dt+f(a)h(b) +f(b)h(a)

1

0

t(1 –t) dt+f(b)h(b)

1

0

t2dt

=1 3

f(a)h(a) +f(b)h(b)+1 6

f(a)h(b) +f(b)h(a)

=1

3M2(a,b) + 1

6N2(a,b).

On the other hand, we have

1

σ(b) –σ(a)

b

a

f(x)h(x)σ(x) dx

≤

1

0

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≤

1

0

[(1 –t)f(a) +tf(b)]2_{+ [(1 –}_t₎_h₍_a_{) +}_th₍_b_)]2

2

dt

=1 2

1

0

(1 –t)2f2(a) +h2(a)+t2f2(b) +h2(b)

+ 2t(1 –t)f(a)f(b) +h(a)h(b)dt

=1 6

f(a) +f(b)2+h(a) +h(b)2–f(a)f(b) +h(a)h(b)

=1 6

M1(a,b)

2

+N1(a,b)

2

–f(a)f(b) +h(a)h(b).

This completes the proof of Theorem3.5.

Theorem 3.6 Let f,h:I→R+ be two similarly ordered integrableσ-convex functions with respect to the functionσ.Then

1

σ(b) –σ(a)

b

a

f(x)h(x)σ(x) dx≤1

2M2(a,b), (3.11)

whereM2(a,b) =f(a)h(a) +f(b)h(b).

Proof Using inequality (3.10) gives

fσ–1(1 –t)σ(a) +tσ(b)hσ–1(1 –t)σ(a) +tσ(b)

≤(1 –t)f(a)h(a) +tf(b)h(b).

Integrating both sides of the above inequality with respect toton [0, 1] yields the

in-equality asserted by Theorem3.6.

Theorem 3.7 Let f,h:I→R+_{be two integrable}_σ_{-convex functions with respect to the}

functionσ.Then

2f

σ–1

σ(a) +σ(b) 2

h

σ–1

σ(a) +σ(b) 2

– 1

σ(b) –σ(a)

b

a

f(x)h(x)σ(x) dx≤1

6M2(a,b) + 1

3N2(a,b), (3.12)

whereM2(a,b) =f(a)h(a) +f(b)h(b),N2(a,b) =f(a)h(b) +f(b)h(a).

Proof Sincef andhare both integrableσ-convex functions, by the same way as in the proof of Theorem3.1, we deduce that

f

σ–1

σ(a) +σ(b) 2

2 ,

h

σ–1

σ(a) +σ(b) 2

≤h(σ–1((1 –t)σ(a) +tσ(b))) +h(σ–1(tσ(a) + (1 –t)σ(b)))

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Hence, we have

f

σ–1

σ(a) +σ(b) 2

h

σ–1

σ(a) +σ(b) 2

≤1

4

×hσ–1(1 –t)σ(a) +tσ(b)+hσ–1tσ(a) + (1 –t)σ(b) =1

4

+fσ–1tσ(a) + (1 –t)σ(b)hσ–1tσ(a) + (1 –t)σ(b)

+fσ–1(1 –t)σ(a) +tσ(b)hσ–1tσ(a) + (1 –t)σ(b)

+fσ–1tσ(a) + (1 –t)σ(b)hσ–1(1 –t)σ(a) +tσ(b)

≤1

4

+(1 –t)f(a) +tf(b)th(a) + (1 –t)h(b)

+tf(a) + (1 –t)f(b)(1 –t)h(a) +th(b).

Integrating both sides of the above inequality with respect toton [0, 1] gives

f

σ–1

σ(a) +σ(b) 2

h

σ–1

σ(a) +σ(b) 2

≤1

4

1

0

+

1

0

fσ–1tσ(a) + (1 –t)σ(b)hσ–1tσ(a) + (1 –t)σ(b)dt

+ 2f(a)h(a) +f(b)h(b)

1

0

t(1 –t) dt

+f(a)h(b) +f(b)h(a)

1

0

t2+ (1 –t)2dt

=1 4

1

0

+

1

0

+1 3

f(a)h(a) +f(b)h(b)+2 3

f(a)h(b) +f(b)h(a)

=1 2

1

σ(b) –σ(a)

b

a

f(x)h(x)σ(x) dx+1

6M2(a,b) + 1 3N2(a,b)

.

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functionσ.Then

f

σ–1

σ(a) +σ(b) 2

h

σ–1

σ(a) +σ(b) 2

≤M2(a,b) +N2(a,b)

4 , (3.13)

Proof From Theorem3.6and Theorem3.7, one can obtain the required result.

strictly increasing functionσ.Then

b

a

b

a

1

0

fσ–1(1 –t)σ(x) +tσ(y)hσ–1(1 –t)h(x) +th(y)σ(x)σ(y) dtdydx

≤2(σ(b) –σ(a))

3

b

a

f(x)h(x)σ(x) dx+(σ(b) –σ(a))

2

12

M2(a,b) +N2(a,b)

, (3.14)

Proof Sincef andhareσ-convex functions, then for∀x,y∈I,t∈[0, 1] we have fσ–1(1 –t)σ(x) +tσ(y)hσ–1(1 –t)σ(x) +tσ(y)

≤(1 –t)f(x) +tf(y)(1 –t)h(x) +th(y)

= (1 –t)2_f₍_x₎_h₍_x_{) +}_t_{(1 –}_t₎_f₍_x₎_h₍_y_{) +}_h₍_x₎_f₍_y₎₊_t2_f₍_y₎_h₍_y_).

Integrating both sides of the above inequality with respect toton [0, 1], we have

1

0

fσ–1(1 –t)σ(x) +tσ(y)hσ–1(1 –t)σ(x) +tσ(y)dt

≤f(x)h(x)

1

0

(1 –t)2dt+f(y)h(y)

1

0

t2dt

+f(x)h(y) +h(x)f(y)

1

0

t(1 –t) dt

=1 3

f(x)h(x) +f(y)h(y)+1 6

f(x)h(y) +h(x)f(y).

Again, integrating both sides of the above inequality over the plane domain{(x,y) :x∈ [a,b],y∈[a,b]}and then using the right-hand side of the Hermite–Hadamard inequality (3.1), we deduce that

b

a

b

a

1

0

fσ–1(1 –t)σ(x) +tσ(y)hσ–1(1 –t)σ(x) +tσ(y)σ(x)σ(y) dtdydx

≤2(σ(b) –σ(a))

3

b

a

f(x)h(x)σ(x) dx+1 6

b

a

h(y)σ(y) dy

b

a

f(x)σ(x) dx

+1 6

b

a

f(y)σ(y) dy

b

a

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≤2(σ(b) –σ(a))

3

b

a

f(x)h(x)σ(x) dx

+(σ(b) –σ(a))

2

6

h(a) +h(b) 2

f(a) +f(b) 2

+

f(a) +f(b) 2

h(a) +h(b) 2

=2(σ(b) –σ(a)) 3

b

a

2

12

f(a) +f(b)h(a) +h(b)

=2(σ(b) –σ(a)) 3

b

a

2

12

M2(a,b) +N2(a,b)

,

which is the required result. The proof of Theorem3.9is complete.

Theorem 3.10 Let f,h:I→Rbe two integrableσ-convex functions with respect to the functionσ.Then

1

σ(b) –σ(a)

b

a

σ(b) –σ(x)

σ(b) –σ(a)

f(a)h(x) +h(a)f(x)σ(x) dx

+ 1

σ(b) –σ(a)

b

a

σ(x) –σ(a)

σ(b) –σ(a)

f(b)h(x) +h(b)f(x)σ(x) dx

≤ 1

σ(b) –σ(a)

b

a

f(x)h(x)σ(x) dx+M2(a,b)

3 +

N2(a,b)

6 , (3.15)

Proof Sincef andhareσ-convex functions, then for∀t∈[0, 1] we have fσ–1(1 –t)σ(a) +tσ(b)≤(1 –t)f(a) +tf(b),

hσ–1(1 –t)σ(a) +tσ(b)≤(1 –t)h(a) +th(b). Utilizing the rearrangement inequality, we obtain

fσ–1(1 –t)σ(a) +tσ(b)(1 –t)h(a) +th(b)

+hσ–1(1 –t)σ(a) +tσ(b)(1 –t)f(a) +tf(b)

≤(1 –t)f(a) +tf(b)(1 –t)h(a) +th(b)

+fσ–1(1 –t)σ(a) +tσ(b)hσ–1(1 –t)σ(a) +tσ(b).

Integrating both sides of the above inequality over the interval [0, 1], we ﬁnd

h(a)

1

0

(1 –t)fσ–1(1 –t)σ(a) +tσ(b)dt+h(b)

1

0

tfσ–1(1 –t)σ(a) +tσ(b)dt

+f(a)

1

0

(1 –t)hσ–1(1 –t)σ(a) +tσ(b)dt

+f(b)

1

0

(12)

≤f(a)h(a)

1

0

(1 –t)2_d_t₊_f₍_b₎_h₍_b₎

1

0

t2dt+f(a)h(b) +f(b)h(a)

1

0

t(1 –t) dt

+

1

0

fσ–1(1 –t)σ(a) +tσ(b)hσ–1(1 –t)σ(a) +tσ(b)dt.

Applying the change of variable to the integrations, we obtain the required result of

The-orem3.10.

functionσ.Then

1

σ(b) –σ(a)

b

a

f

σ–1

σ(a) +σ(b) 2

h(x) +h

σ–1

σ(a) +σ(b) 2

f(x)

σ(x) dx

≤ 1

2(σ(b) –σ(a))

b

a

f(x)h(x)σ(x) dx

+f

σ–1

σ(a) +σ(b) 2

h

σ–1

σ(a) +σ(b) 2

+ 1

12M2(a,b) + 1

6N2(a,b),

Proof Note thatfandhare integrableσ-convex functions. Takingt=1

2in (2.1) and letting

x=σ–1_{((1 –}_t₎_σ₍_a_{) +}_t_σ₍_b_)),_y₌_σ–1₍_t_σ₍_a_{) + (1 –}_t₎_σ₍_b_{)), we have}

f

σ–1

σ(a) +σ(b) 2

2 ,

h

σ–1

σ(a) +σ(b) 2

≤h(σ–1((1 –t)σ(a) +tσ(b))) +h(σ–1(tσ(a) + (1 –t)σ(b)))

2 .

Utilizing the rearrangement inequality, we obtain

1 2f

σ–1

σ(a) +σ(b) 2

hσ–1(1 –t)σ(a) +tσ(b)+hσ–1tσ(a) + (1 –t)σ(b)

+1 2h

σ–1

σ(a) +σ(b) 2

fσ–1(1 –t)σ(a) +tσ(b)

+fσ–1tσ(a) + (1 –t)σ(b)

≤1

4

×hσ–1(1 –t)σ(a) +tσ(b)+hσ–1tσ(a) + (1 –t)σ(b) +f

σ–1

σ(a) +σ(b) 2

h

σ–1

σ(a) +σ(b) 2

(13)

=1 4

+fσ–1(1 –t)σ(a) +tσ(b)hσ–1tσ(a) + (1 –t)σ(b)

+fσ–1tσ(a) + (1 –t)σ(b)hσ–1(1 –t)σ(a) +tσ(b)

+f

σ–1

σ(a) +σ(b) 2

h

σ–1

σ(a) +σ(b) 2

≤1

4

+(1 –t)f(a) +tf(b)th(a) + (1 –t)h(b)

+tf(a) + (1 –t)f(b)(1 –t)h(a) +th(b)

+f

σ–1

σ(a) +σ(b) 2

h

σ–1

σ(a) +σ(b) 2

.

Integrating the ﬁrst and last expressions among the above inequalities over the interval [0, 1], we have

1 2f

σ–1

σ(a) +σ(b) 2

1

0

hσ–1(1 –t)σ(a) +tσ(b)

+hσ–1tσ(a) + (1 –t)σ(b)dt

+1 2h

σ–1

σ(a) +σ(b) 2

1

0

fσ–1(1 –t)σ(a) +tσ(b)

+fσ–1tσ(a) + (1 –t)σ(b)dt

≤1

4

1

0

+

1

0

+1 3

f(a)h(a) +f(b)h(b)+2 3

f(a)h(b) +f(b)h(a)

+f

σ–1

σ(a) +σ(b) 2

h

σ–1

σ(a) +σ(b) 2

.

From the above inequality, by simple computation and arrangement, we obtain the

in-equality asserted by Theorem3.11.

4 Conclusion

(14)

Acknowledgements

The authors would like to thank the anonymous reviewers for their careful reading of our manuscript and their valuable comments and suggestions.

Funding

The research was supported by the National Natural Sciences Foundation of China (Grant No. 11601214) and the Teaching Reform Project of Fujian Provincial Education Department (Grant No. FBJG20180120).

Availability of data and materials

The datasets used or analysed during the current study are available from the corresponding author on reasonable request.

Competing interests

The authors declare that they have no competing interests.

Authors’ contributions

SW and MUA ﬁnished the proofs of the main results and the writing work. MAN, KIN, and SI gave lots of advice on the proofs of the main results and the writing work. All authors read and approved the ﬁnal manuscript.

Author details

1_{Department of Mathematics, Longyan University, Longyan, China.}2_{Department of Mathematics, Government College}

University, Faisalabad, Pakistan.3_{Department of Mathematics, COMSATS University Islamabad, Faisalabad, Pakistan.}

Publisher’s Note

Springer Nature remains neutral with regard to jurisdictional claims in published maps and institutional aﬃliations.

Received: 2 January 2019 Accepted: 16 April 2019 References

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