# The Strong Convex Functions and Related Inequalities

Authors: Xue Wang; Absar ul Haq; Muhammad Shoaib Saleem; Sami Ullah Zakir
Journal: Journal of Function Spaces
 (2022)

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## Abstract

The study of convex functions is one of the most researched of the classical fields. Analysis of the geometric characteristics of these functions is a core area of research in this field; however, a paradigm shift in this research is the application of convexity in optimization theory. The Jensen-Mercer type inequalities are studied extensively in recent years. In the present paper, we extend Jensen-Mercer type inequalities for strong convex function. Some improved inequalities in Hölder sense are also derived. The previously established results are generalized and strengthened by our results.


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## Body

## 1. Introduction and Preliminary Results

Convex functions and their consequences are useful in the establishment of different kinds of inequalities; therefore, they are considered the base of theory of inequalities in mathematical analysis. A real valued functionψ:I⟶ℝ is said to be convex on the interval I⊂ℝ, if
(1)ψzx+1−zy≤zψx+1−zψyholds for all x,y∈I and z∈0,1. The function ψ is said to be concave if reverse of inequality (1) holds.Convex functions are also very important in the fields of mathematical analysis, mathematical statistics, and optimization theory. These functions motivate towards a nice theory named convex analysis (see [1–3]). Convex functions have been defined in various ways by using different techniques, for example, by support function, by chords joining two points, and Jensen’s inequality. Inequality (1) represents the convex function analytically and provides encouragement to define further general notions.The study of convex functions [4–14] began with Jensen’s thought-provoking concepts and interesting work over the period from 1905 to 1906. It is used in the analysis as an efficient tool for solving optimization issues. Additionally, inequalities involving convex functions are very stimulating in the development of different sections of mathematics, such as mathematical finance, economics, management sciences, and optimization theory.If the functionψ:I⊂ℝ is convex, then the inequality
(2)ψc+d2≤1d−c∫cdψα1dα1≤ψc+ψd2is called the Hermite-Hadamard inequality [15, 16].Definition 1 (Convex set) (see [17]).
A setI is considered to be convex if the line segment between any two points in I lies in I; i.e., ∀α1,α2∈I,∀∈0,1(3)lα1+1−lα2∈I.Authors [18–20] expanded on the idea of a strongly convex function by replacing the nonnegative term with a real-valued nonnegative function and defined it as follows:Definition 2 (Strongly convex function (see [18])).
A functionψ:I⊂ℝ⟶ℝ is strongly convex, if
(4)ψlα1+1−lα2≤lψα1+11−lψα2−l1−lMα1−α2
holds for allα1,α2∈I and l∈0,1.Definition 3 (Riemann-Liouville fractional integral).
For a functionψ:I⊂ℝ⟶ℝ, the Riemann-Liouville fractional integral operator of order ξ≤0 with c≤0 is defined as
(5)Jcξψα1=1Γξ∫cα1α1−lζ−1ψldl,Jc0ψα1=ψα1.
Many scholars have recently analyzed a variety of inequalities by using the Riemann-Liouville fractional integrals (see [21–24]).Definition 4 (Hadamard fractional integral).
For a functionψ:I⊂ℝ⟶ℝ, the Hadamard fractional integral of order ζ≤ℝ+ for all α1>1 is defined as
(6)JH1,α1−ζψα1=1Γξ∫1α1lnα1lζ−1ψldl,where Γζ=∫0∞e−llζ−1dl.Definition 5 (Conformable fractional integral).
Letζ∈0,1 and 0≤c<d. A function ψ:c,d⟶ℝ is ζ-fractional integrable on c,d if the integral
(7)∫cdψα1dζα1=∫cdψα1α1ζ−1exists and is finite.This paper is aimed at establishing Hermite-Jensen-Mercer type inequalities and some other inequalities including improved Hölder inequality for strong convex function.


## 2. New Hermite-Jensen-Mercer Type Inequalities

Theorem 6.
Letζ,ξ>0 and ψ:ϕ,φ⟶ℝ be a strong convex function. Then, the inequality
(8)ψϕ+φ−α1+α22≤2ζξ/k−1ζξ/kΓkξ+kα2−α1ζξ/kJkξϕ+φ−α1+α2/2+ζψϕ+φ−α2+Jkξϕ+φ−α1+α2/2+ζψϕ+φ−α2−a2Mα2−α1≤ψϕ+ψφ−ψα2+ψα22−a2ψϕ+φ−α1−α22
holds for allα1,α2∈ϕ,φ.Proof.
To prove that the first inequality holds, take(9)ψϕ+φ−α11+α212=ψ2ϕ+2φ−α11−α212.
Sinceψ is a strong convex function, so
(10)ψϕ+φ−α11+α212ψϕ+φ−α11+α212≤12ψϕ+φ−α112+12ψϕ+φ−α212−a1212α21−α1122≤12ψϕ+φ−α112+ψϕ+φ−α212−a2α21−α1122.
Supposeα11=l/2α1+2−l/2α2 and α21=2−l/2α1+l/2α2; then, for α1,α2∈ϕ,φ and l∈0,1, we have
(11)ψϕ+φ−l/2α1−2−l/2α22+ψϕ+φ−2−l/2α1−l/2α22−a2α1−α2221−l22ψϕ+φ−α11+α212≤ψϕ+φ−α112+ψϕ+φ−α212−a2α21−α1122.
Multiplying both sides of Equation (11) with 1−1−lζ/ζξ/k−11−lζ−1, we get
(12)2ψϕ+φ−α11+α2121−1−lζζξ/k−11−lζ−1≤ψϕ+φ−α112+ψϕ+φ−α212−a2α21−α11221−1−lζζξ/k−11−lζ−1.
Integrating the above inequality with respect tol over the range 0,1 and then combining the result with the integral operator yield
(13)2ψϕ+φ−α11+α212∫011−1−lζζξ/k−11−lα−1≤∫01ψϕ+φ−α112+ψϕ+φ−α212−c2α21−α11221−1−lζζξ/k−11−lζ−1dl=∫01ψϕ+φ−α1121−1−lζζξ/k−11−lζ−1dl+∫01ψϕ+φ−α2121−1−lζζξ/k−11−lζ−1dl−∫01a2α21−α11221−1−lζζξ/k−11−lζ−1dl.
Now, by altering the variables, we can obtain(14)∫ϕ+φ−α2ϕ+φ−α1+α2/21−2ϕ+2φ−α1+α2−2l1/α2−α2ζζξ/k−1×2ϕ+2φ−α1+α2−2l1α2−α1ζ−1ψl1α2−α12dl+∫ϕ+φ−α1ϕ+φ−α1+α2/21−2l2−2ϕ+2φ−α1+α2/α2−α2ζζξ/k−1×2l2−2ϕ+2φ−α1+α2α2−α1ζ−1ψl2α2−α12dl+∫α1−α2α1+α21−2l3/α2−α1ζζξ/k−1l3α1−α2aψl3α2−α1dl.
So,(15)∫011−1−lζζξ/k−11−lζ−1dl=1ξ/kζξ/k,Jκξϕ+ζψα2=1κΓκξ∫ϕα2α2−ϕζ−l−ϕζζξ/k−1ψll−ϕ1−ζdl,Jκξφ−ζψα2=1κΓκξ∫ϕφφ−α2ζ−φ−lζζξ/k−1ψlφ−l1−ζdl.
Therefore,(16)2ψϕ+φ−α1+α221ξ/kζξ/k≤2α2−α1ζξ/kΓkξkξJϕ+φ−α1+α2/2+ζSϕ+φ−α2+ΓkξkξJϕ+φ−α1+α2/2−ζψϕ+φ−α1−a2Mα21−α11.
As a result, the first inequality of (8) is proved.
We can prove the second inequality of (8) by using strong convexity of ψ for l over 0,1.(17)ψϕ+φ−α1=ψϕ+φ+ψα1−12aMϕ+φ−α1,(18)ψϕ+φ−α2=ψϕ+φ+ψα1−12aMϕ+φ−α2,(19)ψϕ+φ−l2α1+2−l2α2=ψϕ+φ+l2ψα1+2−l2ψα2−a2Mϕ+φ−l2α1+2−l2α2,(20)ψϕ+φ−2−l2α1+5l2α2=ψϕ+φ+2−l2ψα1+l2ψα2−a2Mϕ+φ−2−l2α1+l2α2.
Adding (17) and (20) leads to
(21)ψϕ+φ−l2α1+2−l2α2+ψϕ+φ−2−l2α1+l2α2≤2ψϕ+φ+ψl2α1+2−l2α2+ψ2−l2α1+l2α2−a22ϕ+2φ−α2−α1.
Multiply 9 with1−1−lζ/ζξ/k−11−lζ−1, and integrating the obtained inequality w.r.t to l over [0,1] gives
(22)∫011−1−lζζξ/k−11−lζ−1ψϕ+φ−l2α1+2−l2α2+ψϕ+φ−2−l2α1+l2α2dl≤2ψϕ+φ+ψl2α1+2−l2α2+ψ2−l2α1+l2α2−a22ϕ+2φ−α2−α1∫011−1−lζζξ/k11−lζ−1−2α2−α1ζξ/k×ΓkξkξJϕ+φ−α1+α2/2+ζψϕ+φ−α2+ΓkξkξJϕ+φ−α1+α2/2−ζψϕ+φ−α1−a2Mα21−α11dl≤1ξ/kζξ/k2ψϕ+φ−ψα1+ψα2−a22ϕ+2φ−α2−α1.
This completes the proof.Remark 7.
It is obvious from Theorem6 that
(1)
Theorem 2.1 of [25] is obtained if we take a=0, α1=x, and α2=y in Theorem 6(2)
Theorem 2.1 of [26] is obtained if we take a=0, k=1, α1=θ, and α2=ϑ in Theorem 6(3)
Theorem 2 of [27] is obtained by taking a=0, α=k=1, α1=θ, and α2=ϑ in Theorem 6Theorem 8.
Letζ,ξ>0 and ψ:ϕ,φ⟶ℝ be a strong convex function. Then, the inequalities
(23)ψϕ+φ−α1+α22≤ψϕ+ψφ−ζξ/kΓkξ+k2α2−α1ζξ/k×Jkξϕ+φ−α1+ζψα2+Jkξϕ+φ−α2−ζψα1−a2Mα2−α1≤ψϕ+ψφ−ψα1+α22−a2α2−α12and
(24)ψϕ+φ−α1+α22≤ζξ/kΓkξ+k2α2−α1ζξ/k×Jkξϕ+φ−α1+ζψα2+Jkξϕ+φ−α2ζ−ψα1−a2Mα2−α1≤ψϕ+ψφ−ψα1+ψα22−a2α2−α12
hold∀α1,α2∈ϕ,φ.Proof.
The Jensen-Mercer inequality dictates that(25)ψϕ+φ−α11+α212≤ψϕ+ψφ−ψα11+ψα212−a2Mα2−α1,
∀α11,α21∈ϕ,φ.Takingα11=lα1+1−lα2 and α21=1−lα1+lα2 for α1,α2∈ϕ,φ and l∈0,1 in (25), we get
(26)ψϕ+φ−α1+α22≤ψϕ+ψφ−ψlα1+1−lα2+ψ1−lα1+lα22−a2Mα2−α1.Multiplying the above inequality by1−1−lζ/ζξ/k1−lζ−1 and integrating the obtained inequality with respect to l over [0,1] give
(27)ψϕ+φ−α1+α22∫011−1−laζζξ/k1−lζ−1dl≤∫011−1−lζζξ/k1−lζ−1×ψϕ+ψφ−ψlα1+1−lα2+ψ1−lα1+lα22−a2Mα2−α1dl,that is,
(28)ψϕ+φ−α1+α22≤ψϕ+ψφ−ζξ/kΓkξ+k2α2−α1ζξ/k×Jkξα1+ζψα2+Jkξα2−ζψα1−a2Mα1+α2.That proves the first inequality of (24).To prove the second inequality of (24), from the definition of strong convexity of ψ, for l∈0,1, we get
(29)ψα1+α22=ψlα1+1−lα2+1−lα1+lα22≤ψlα1+1−lα2+ψ1−lα1+lα22−a2Mα1−α2.Multiplying the above inequality by1−1−lζ/ζξ/k1−lζ−1 and then integrating with respect to l over [0,1], we have
(30)ψα1+α22∫011−1−lζζξ/k1−lζ−1≤∫011−1−lζζξ/k1−lζ−1×ψlα1+1−lα2+ψ1−lα1+lα22−a2Mα1−α2dl,where
(31)ψα1+α22≤Jkξα1+ζψα2+kξJα2−ζψα1−a2Mα1+α2ζξ/kΓkξ+k2α2−α1ζξ/kimplies
(32)−ψα1+α22≥−Jkξα1+ζψα2+kξJα2−ζψα1a2Mα1+α2ζξ/kΓkξ+k2α2−α1ζξ/k.Addingψϕ+ψφ on both side of above inequality,
(33)ψϕ+ψφ−ψα1+α22≥ψϕ+ψφ−Jkξα1+ζψα2+kξJα2−ζψα1ζξ/kΓkξ+k2α2−α1ζξ/k,which gives (23).To prove (24), using strong convexity of ψ, we get
(34)ψϕ+φ−α11+α212=ψϕ+φ−α11+ϕ+φ−α212≤ψϕ+φ−α112+ψϕ+φ−α212−a2Mα2−α1,(35)∀α11,α21∈ϕ,φ.Letα11=lα1+1−lα2 and α21=1−lα1+lα2; then, (34) leads
(36)ψϕ+φ−α1+α22≤ψϕ+φ−lα1+1−lα22+ψϕ+φ−1−lα1+lα22−a2Mα1+α2.Multiplying the above inequality by1−1−lζ/ζξ/k1−lζ−1 and then integrating with respect to l over [0,1], we have
(37)ψϕ+φ−α1+α22∫011−1−lζζξ/k1−lζ−1dl≤∫011−1−lζζξ/k1−lζ−1ψϕ+φ−lα1+1−lα22+ψϕ+φ−1−lα1+lα22−a2Mα1+α2dl,which can be written as
(38)ψϕ+φ−α1+α22≤ζξ/kΓkξ+k2α2−α1ζξ/kJkξα1+ζψα2+kξJα2−ζψα1−a2Mα1+α2.If follows from the definition of strong convexity ofψ that
(39)ψlϕ+φ−α1+1−lϕ+φ−α2≤lψϕ+φ−α1+1−lψϕ+φ−α2−la21−lMα1−α2,(40)ψ1−lϕ+φ−α1+lϕ+φ−α2≤1−lψϕ+φ−α1+lψϕ+φ−α2−al1−lMα1−α2.Adding the above two inequalities and with the help of Jensen-Mercer inequality, we have(41)ψlϕ+φ−α1+1−lϕ+φ−α2+ψ1−lϕ+φ−α1+lϕ+φ−α2≤lψϕ+φ−α1+1−lψϕ+φ−α2−l1−lMα1−α2+1−lψϕ+φ−α1+lψϕ+φ−α2−l1−lMα1−α2,ψlϕ+φ−α1+1−lϕ+φ−α2+ψ1−lϕ+φ−α1+ψϕ+φ−α2≤ψϕ+φ−α1+ψϕ+φ−α2−2al1−lMα2−α1≤2ψϕ+ψφ−ψα1+ψα2−2aMα2−α1.Multiplying the above inequality by1−1−lζ/ζξ/k1−lζ−1 and then integrating with respect to l over [0,1], we have
(42)∫011−1−lζζξ/k1−lζ−1Slϕ+φ−α1+1−lϕ+φ−α2+ψ1−lϕ+φ−α1+lϕ+φ−α2dl≤2ψϕ+ψφ−ψα1+ψα2−a2Mα2−α1∫011−1−lζζξ/k1−lζ−1dl,that is,(43)ζξ/kΓkξ+k2α2−α1ζξ/kJkξα1+ζψα2+kξJα2−ζψα1−a2Mα1+α2≤ψϕ+ψφ−ψα1+ψα22−a2Mα2−α12.Combining (38) and (43) leads to (24).Remark 9.
Leta=0,ζ=ξ=k=1; then, Theorem 8 leads to
(44)ψϕ+φ−α1+α22≤ψϕ+ψφ−∫01ψlα1+1−lα2dl≤ψϕ+ψφ−ψα1+α22,(45)ψϕ+φ−α1+α22≤1α2−α2∫α1α2ψϕ+φldl≤ψϕ+ψφ−ψψα1+ψα22,which was proved in Theorem 2.1 of [28].Lemma 10.
Letζ,ξ>0,ϕ<φ and ψ:ϕ,φ⟶R be a differentiable mapping such that ψ′∈Lϕ,φ. Then, the inequality
(46)2ζξ/k−1ζξ/kΓkξ+kα2−α1ζξ/kJkξϕ+φ−α1+α2/2ψϕ+φ−α1+Jkξϕ+φ−α1+α2/2ψϕ+φ−α2xψϕ+φ−α1+α22=α2−α1ζξ/k4∫011−1−lζζξ/k×ψ′ϕ+φ−2−l2α1+l2α2−ψ′ϕ+φ−l2α1+2−l2α2
holds∀α1,α2∈ϕ,φ.Proof.
Suppose(47)P=P1−P2α2−α14ζξ/k,where
(48)P1=∫011−1−lζζξ/kψ′ϕ+φ−2−l2α1+l2α2dl,P2=∫011−1−lζζξ/kψ′ϕ+φ−l2α1+2−l2α2dl.
Using integration by parts, we get(49)=−2ζξ/kα2−α1ψϕ+φ−α1+α22+2ζξ/kζξ/kα2−α1∫011−1−lζζξ/kψ′ϕ+φ−2−l2α1+l2α2dl=−2ζξ/kα2−α1ψϕ+φ−α1+α22+2ζξ/kζξ/kα2−α1ζξ/k+1∫ϕ+φ−α1−α22ϕ+φ−α1α2−α22ζ−l1−ϕ+φ−α1+α22ζξ/k−1×ψl1α1−ϕ+φ−α1+α2/21−ζdl1=−2ζξ/kα2−α1ψϕ+φ−α1+α22+2y−xζξ/k+1Γkξ+kζξ/k−1kξJϕ+φ−α1+α2/2+ζψϕ+φ−α1.
Similarly, using integration by parts forP2, we get
(50)P2=∫011−1−lζζξ/kψ′ϕ+φ−l2α1+2−l2α2dl=2ζξ/kα2−α1ψϕ+φ−α1+α22−2y−xζξ/k+1Γkξ+kζξ/k−1kξJϕ+φ−α1+α2/2ζ−ψϕ+φ−α2.
Therefore, inequality (43) follows from (47)–(50).Remark 11.
(1)
If we takeK=1, α1=θ, and α2=υ in Lemma 10, then we can get Lemma 3.1 of [26](2)
If we takeζ=K=1, α1=θ, and α2=υ, then Lemma 10 reduces to Lemma 1.1 of [29]Lemma 12.
Letζ,ξ>0,ϕ<φ and ψ:ϕ,φ⟶ℝ be a differentiable mapping such that ψ′∈ψϕ,φ. Then, the identity
(51)ψϕ+φ−α1+ψϕ+φ−α22−ζξ/kΓkξ+k2α2−α1ζξ/k×Jkξϕ+φ−α2ζ+ψϕ+φ−α1+Jkξϕ+φ−α2ζ+ψϕ+φ−α1=α2−α1ζξ/k2∫011−1−lζζξ/k−1−lζζξ/k×ψ′ϕ+φ−lα1+1−lα2dl
holds∀α1,α2∈ϕ,φ.Proof.
Suppose(52)P=α2−α1ζξ/k2∫011−1−lζζξ/k−1−lζζξ/k×ψ′ϕ+φ−lα1+1−lα2dl=α2−α1ζξ/k2P1−P2.
Then, we clearly see that(53)P1=∫011−1−lζζξ/k×ψ′ϕ+φ−lα1+1−lα2dla=1αξ/kψϕ+φ−α1α2−α1−ξ/kα2−α1∫011−1−lζζξ/k−1×1−lζ−1ψϕ+φ−lα1+1−lα2dl=1αξ/kψϕ+φ−α1α2−α1−Γkξ+kα2−α1ζξ/k+1Jkζϕ+φ−α2ζ+ψϕ+φ−α2,
and(54)P2=∫011−lζζξ/k×ψ′ϕ+φ−lα1+1−lα2dl=−1ζξ/kψϕ+φ−α2α2−α1+ξ/kα2−α1∫011−lζζξ/k−1×lζ−1ψϕ+φlα1+1−lydl=−1ζξ/kψϕ+φ−α2α2−α1+Γkξ+kα2−α1ζξ/k+1Jkξϕ+φ−α2ζ−ψϕ+φ−α1.
Therefore, identity (51) follows from (52)–(54).Corollary 13.
If we takeζ=ξ=k=1, then Lemma 12 leads to the equality
(55)ψϕ+φ−α1+ψϕ+φ−α22−1α2−α1∫ϕ+φ−α2ϕ+φ−α1ψladl=α2−α12∫012l−1S′ϕ+φ−lα1+1−lydl.Remark 14.
If we takeα1=ϕ=θ and α2=φ=υ in Corollary 13, then (55) becomes
(56)ψϕ+ψφ2−1φ−ϕ∫01ψldl=ϕ−φ2∫012l−1ψ′1−lϕ+lφdl,which was proved in Lemma 2.1 of [30].Theorem 15.
Letζ,ξ>0,ϕ<φ and ψ:ϕ,φ⟶ℝ be a differentiable mapping such that ψ′∈Lϕ,φ and ψ′q is a convex mapping on ϕ,φ. Then, the inequality
(57)2ζξ/k−1ζξ/kΓkξ+kα2−α1ζξ/kJkξϕ+φ−α2+α1/2ζ+ψϕ+φ−α1+Jkξϕ+φ−α2+α1/2ζ+ψϕ+φ−α2−ψϕ+φ−α2+α12≤α2−α14ζξ/kψ′ϕ+ψ′φ1ζξ/k+1Bξk+1,1ζ−ψ′α112ζξ/k+1Bξk+1,2ζ+Bξk+1,1ζ+ψ′α212ζξ/k+1Bξk+1,2ζ−Bξk+1,1ζ−a4∫011−1−lζζξ/kl2−ldl+ψ′ϕ+ψ′φ1ζξ/k+1Bξk+1,2ζ−ψ′α112ζξ/k+1Bξk+1,1ζ−Bξk+1,2ζ+ψ′α212ζξ/k+1Bξk+1,2ζ+Bξk+1,1ζ−a4∫011−1−lζζξ/kl2−ldlholds for all α2,α1∈ϕ,φ.Proof.
It follows from Lemma10, Jensen-Mercer inequality, power mean inequality, and the convexity of function ψ′q that
(58)2ζξ/k−1ζξ/kΓkξ+kα2−α1ζξ/kJkξϕ+φ−α2+α1/2ζ+ψϕ+φ−α1+Jkξϕ+φ−α2+α1/2ζ+ψϕ+φ−α2−ψϕ+φ−α2+α12≤α2−α14ζξ/k∫011−1−lζζξ/kψ′ϕ+φ−2−l2α1+l2α2dl+∫011−1−lζζξ/kψ′ϕ+φ−l2α1+2−l2α2dl.
Using the definition of strong convexity(59)≤α2−α14ζξ/k∫011−1−lζζξ/kψ′ϕ+ψ′φ−2−l2ψ′α1+l2ψ′α2−al22−l2α2−α12dl+∫011−1−lζζξ/kψ′ϕ+ψ′φ−l2ψ′α1+2−l2ψ′α2−al22−l2α2−α12dl≤α2−α14ζξ/kψ′ϕ+ψ′φ∫011−1−lζζξξ/k/kdl−ψ′α1∫011−1−lζζξ/k2−l2dl+ψ′α2∫011−1−lζζξ/kl2dl−α2−α12∫011−1−lζζξ/kal22−l2dl+ψ′ϕ+ψ′φ∫011−1−lζζξ/kdl−ψ′α1∫011−1−lζζξ/kl2dl+ψ′α2∫011−1−lζζξ/k2−l2dl−α2−α12∫011−1−lζζξ/kal22−l2dl.
Therefore, inequality (57) can be derived after some simple calculation.Remark 16.
From Theorem15, we clearly see that
(1)
If we takea=0, α1=x, and α2=y in Theorem 15, then we get Theorem 3.1 of [25](2)
If we takea=0, ζ=k=1, α1=θ, and α2=ϑ in Theorem 15, then we get Theorem 3.1 of [26](3)
If we takea=0, ζ=k=1, α1=θ, and α2=ϑ in Theorem 15, then we get Theorem 5 of [29] in the case of q=1Theorem 17.
Letq>1,ζ,ξ>0,ϕ<0 and ψ:ϕ,φ⟶ℝ be a differentiable mapping such that ψ′∈Lϕ,φ and ψ′q is a convex mapping on ϕ,φ. Then, the inequality
(60)2ζξ/k−1ζξ/kΓkξ+kα2−α1ζξ/k+Jkξϕ+φ−α2+α1/2ζ+ψϕ+φ−α1+kξJϕ+φ−α2+α1/2ζ+ψϕ+φ−α2−ψϕ+φ−α2+α12≤α2−α14ζξ/k1ζξ/k+1Bξk+1,1ζ1−1/q×ψ′ϕq+ψ′φq1ζξ/k+1Bξk+1,1ζ−ψ′α1q12ζξ/k+1Bξk+1,1ζ+Bξk+1,2ζ+ψ′α2q12ζξ/k+1Bξk+1,2ζ−Bξk+1,1ζ−a4∫011−1−lζζξ/kl2−ldl+ψ′ϕq+ψ′φq1ζξ/k+1×Bξk+1,2ζ−ψ′α1q12ζξ/k+1Bξk+1,2ζ−Bξk+1,1ζ+ψ′α2q12ζξ/k+1Bξk+1,1ζ+Bξk+1,2ζ−a4∫011−1−lζζξ/kl2−ldl1/q
holds for allα2,α1∈ϕ,φ.Proof.
It follows from Lemma10, Jensen-Mercer inequality, power-mean inequality, and the convexity of function ψ′q that
(61)2ζξ/k−1ζξ/kΓkξ+kα2−α1ζξ/kJkξϕ+φ−α2+α1/2ζ+ψϕ+φ−α1+Jkξϕ+φ−α2+α1/2ζ+ψϕ+φ−α2−ψϕ+φ−α1+α22≤α2−α14ζξ/k∫011−1−lζζξ/kdl1−1/q×∫011−1−lζζξ/kψ′ϕ+φ−l2ζ+2−l2yqdl1/q+∫011−1−lζζξ/kdl1−1/q×∫011−1−lζζξ/kψ′ϕ+φ−2−la2ζ+l2yqdl1/q.
Using definition of strong convexity, we have(62)≤α2−α14ζξ/k∫011−1−lζζξ/kψ′ϕq+ψ′φq−2−l2ψ′α1q+l2ψ′α2q−al22−l2α2−α12dl+∫011−1−lζζξ/kψ′ϕq+ψ′φq−l2ψ′α1q+2−l2ψ′α2q−al22−l2α2−α12dl≤α2−α14ζξ/kψ′ϕq+ψ′φq∫011−1−laζζξ/kdl−ψ′α1q∫011−1−lζζξ/k2−l2dl+ψ′α2q∫011−1−lζζξ/kl2dl−α2−α12∫011−1−lζζξ/kal22−l2dl+ψ′ϕq+ψ′φq∫011−1−lζζξ/kdl−ψ′α1q∫011−1−lζζξ/kl2dl+ψ′α2q∫011−1−lζζξ/k2−l2dl−α2−α12∫011−1−lζζξ/kal22−l2dl.
Making simple simplification, we get (60) from (61).Remark 18.
Theorem17 leads to
(1)
If we takea=0, α1=x, and α2=y in Theorem 17, then we get Theorem 2.12 of [25](2)
If we takea=0, k=1, α1=x=ϕ, and α2=y=φ in Theorem 17, then we get Theorem 3.1 of [26](3)
If we takea=0, ζ=k=1, α1=x=ϕ, and α2=y=φ in Theorem 17, then we get Theorem 5 of [29] in the case of q=1Theorem 19.
Letq>1,ζ,ξ>0,ϕ<0 and ψ:ϕ,φ⟶ℝ be a differentiable mapping such that ψ′∈Lϕ,φ and ψ′q is a convex mapping on ϕ,φ. Then, the inequality
(63)2ζξ/k−1ζξ/kΓkξ+kα2−α1ζξ/kJkξϕ+φ−α2+α1/2ζ+ψϕ+φ−α1+kξJϕ+φ−α2+α1/2ζ+ψϕ+φ−α2−ψϕ+φ−α2+α12≤α2−α14ζξ/k1ζξ/kp+1Bξkp+1,1ζ1/pψ′ϕq+ψ′φq−3ψ′α1q+ψ′α2q4−al22−l2α2−α121/q+ψ′ϕq+ψ′φq−ψ1α1+3ψ′α2−al22−l2α2−α121/q.Proof.
It follows from Lemma10, Jensen-Mercer inequality, power mean inequality, and the convexity of function ψ′q that
(64)2ζξ/k−1ζξ/kΓkξ+kα2−α1ζξ/kJkξϕ+φ−α2+α1/2ζ+ψϕ+φ−α1+Jkξϕ+φ−α2+α1/2ζ+ψϕ+φ−α2−ψϕ+φ−α1+α22≤α2−α14ζξ/k∫011−1−lζζξ/kpdl1/p×∫01ψ′ϕ+φ−2−l2ζ+l2yqdl1/q+∫01ψ′ϕ+φ−l2ζ+2−l2yqdl1/q.
It follows from the strong convexity ofψ′q(65)≤α2−α14ζξ/k∫011−1−lζζξ/kpdl1/p×∫01ψ′ϕq+ψ′φq−2−l2ψ′α1q+l2ψ′α2q−al22−l2α2−α121/qdl+∫01ψ′ϕq+ψ′φq−l2ψ′α1q+2−l2ψ′α2q−al22−l2α2−α121/qdl≤α2−α14ζξ/k1ζξ/kp+1Bξkp+1,1ζ1/pψ′ϕq+ψ′φq−3ψ′α1q+ψ′α2q4−al22−l2α2−α121/q+ψ′ϕq+ψ′φq−ψ′α1+3ψ′α2−al22−l2α2−α121/q,
which completes the proof.Corollary 20.
Leta=0 and ξ=k=1. Then, Theorem 19 leads to
(66)1α2−α1∫ϕ+φ−α2ϕ+φ−α1ψldl−ψϕ+φ−α1+α22≤121/pψ′ϕq+ψ′φq−3ψ′α1q+ψ′α2q4−al22−l2α2−α121/q+ψ′ϕq+ψ′φq−ψ′α1+3ψ′α2−al22−l2α2−α121/q.Theorem 21.
Letζ,ξ>0,p,q>1 with 1/p+1/q=1,ϕ<φ and ψ:ϕ,φ⟶ℝ be a differentiable mapping such that ψ′∈Lϕ,φ and ψ′q is a convex mapping on ϕ,φ. Then, the inequality
(67)2ζξ/k−1ζξ/kΓkξ+kα2−α1ζξ/kJkξϕ+φ−α2+α1/2ζ+ψϕ+φ−α1+Jkξϕ+φ−α2+α1/2ζ+ψϕ+φ−α2−ψϕ+φ−α2+α12≤α2−α1ζξ/k4ψ′ϕq+ψ′φqBξ/k+1,ξ/kξξ/k+1−Bξ/k+1,2/ξ+Bξ/k+1,1/ξ2ζξ/k+1ψ′α1q+Bξ/k+1,1/ξ+Bξ/k+1,2/ξ2ζξ/k+1ψ′α1q−al22−l2α2−α121/q+ψ′ϕq+ψ′φqBξ/k+1,1/ξξξ/k+1−Bξ/k+1,1/ξ−Bξ/k+1,2/ξ2ζξ/k+1S′α1q+Bξ/k+1,2/ξ+Bξ/k+1,1/ξ2ζξ/k+1ψ′α2q−al22−l2α2−α121/q
holds∀α1,α1∈ϕ,φ.Proof.
It follows from Lemma10, Jensen-Mercer inequality, strong convexity of ψ′q, and Holder integral inequality that
(68)2ζξ/k−1ζξ/kΓkξ+kα2−α1ζξ/kJkξϕ+φ−α2+α1/2ζ+ψϕ+φ−α1+Jkξϕ+φ−α2+α1/2ζ+ψϕ+φ−α2−ψϕ+φ−α2+α12≤α2−α1ζξ/k4∫011dl1/p∫011−1−lζζξ/kψ′ϕ+φ−2−l2α1+l2α2qdl1/q+∫011dl1/p×∫011−1−lζζξ/kψ′ϕ+φ−l2α1+2−l2α2qdl1/q≤α2−α1ζξ/k4∫011−1−lζζξ/kψ′ϕq+ψ′φq−2−l2ψ′α1q+l2ψ′α2q−al22−l2α2−α12q1/q+∫011−1−lζζξ/kψ′ϕq+ψ′φq−l2ψ′α1q+2−l2ψ′α2q−al22−l2α2−α12dl1/q.
By making necessary changes, we get (67).Theorem 22.
Letϕ<φ and ψ:ϕ,φ⟶ℝ be a differentiable mapping such that ψ′∈Lϕ,φ and ψ′ is a convex mapping onϕ,φ. Then, one has
(69)ϕ+φ−α1+ψϕ+φ−α22−ζξ/kΓkξ+k2α2−α1ζξ/k×Jkξϕ+φ−α2ζ+ψϕ+φ−α1+Jkξϕ+φ−α1ζ+ψϕ+φ−α2≤α2−α1ζξ/k2ψ′ϕ+ψ′φB1/ξ,ξ/k+1ζξ/k+1−ψ′α1ζξ/k+1B1/2ζ1ζ,ξk+1+B2ζ,ξk+1−B1ζ,ξk+1−ψ′α2ζξ/k+1B1/2ζ1ζ,ξk+1−B2ζ,ξk+1−al1−lα2−α12+ψ′ϕ+ψ′φB1/ξ,ξ/k+1ζξ/k+1−ψ′α1ζξ/k+1B1/2ζ1ζ,ξk+1−B2ζ,ξk+1−ψ′α2ζξ/k+1B1/2ζ1ζ,ξk+1+B2ζ,ξk+1−B1ζ,ξk+1−al1−lα2−α12,∀α2,α1∈ϕ,φ.Proof.
By using Lemma12 and similar arguments as in the proofs of previous theorem, we have
(70)ψϕ+φ−α1+ψϕ+φ−α22−ζξ/kΓkξ+k2α2−α1ζξ/k×Jkξϕ+φ−α2ζ+ψϕ+φ−α1+kξJϕ+φ−α1ζψϕ+φ−α2≤α2−α1ζξ/k2∫011−1−laζζξ/k−1−lζζξ/k×ψ′ϕ+φ−lα1+1−lα2dl≤α2−α1ζξ/k2∫011−1−laζζξ/k−1−lζζξ/k×ψ′ϕ+ψ′φ−lψ′α1+1−lα2−al1−lα2−α12dl≤α2−α1ζξ/k2∫01/21−1−lζζξ/k−1−lζζξ/k×ψ′ϕ+ψ′φ−lψ′α1+1−lα2−al1−lα2−α12dl+∫1/211−1−lζζξ/k−1−lζζξ/k×ψ′ϕ+ψ′φ−lψ′α1+1−lα2−al1−lα2−α12dl,
which completes the proof.


## 3. New Inequalities by Improved Hölder Inequality

Theorem 23.
Letξ,ζ>0,p,q>1 with 1/p+1/q=1,ϕ<φ and ψ:ϕ,φ⟶ℝ be a differentiable mapping such that ψ′∈lϕ,φ and ψ′q is a strong convex mapping on ϕ,φ. Then, one has
(71)2ζξ/k−1ζξ/kΓkξ+kα2−α1ζξ/kJkξϕ+φ−α2+α1/2ζ+ψϕ+φ−α1+Jkξϕ+φ−α2+α1/2ζψϕ+φ−α2−ψϕ+φ−α2+α12≤α2−α1ζξ/k2B2/ζ,ξ/kp+1ζξ/kp+11/pψ′ϕq+ψ′φq2−512ψ′α1q+112ψ′α2q−a5144α2−α12+B1/ζ,ξ/kp+1−B2/ζ,ξ/kp+1ζξ/kp+11/pψ′ϕq+ψ′φq2−13ψ′α1q+16ψ′α2q−a118α2−α12+B2/ζ,ξ/kp+1ζξ/kp+11/pψ′ϕq+ψ′φq2−112ψ′α1q+512ψ′α2q−a5144α2−α12+B1/ζ,ξ/kp+1−B2/ζ,ξ/kp+1ζξ/kp+11/p×ψ′ϕq+ψ′φq2−16ψ′α1q+13ψ′α2q−a118α2−α12.Proof.
It follows from Lemma10, Jensen-Mercer inequality, the convexity of ψ′q, and Holder-Iscan integral inequality given in Theorem 1.4 of [31] that
(72)2ζξ/k−1ζξ/kΓkξ+kα2−α1ζξ/kJkξϕ+φ−α2+α1/2ζ+ψϕ+φ−α1+kξJϕ+φ−α2+α1/2ζ+ψϕ+φ−α2−ψϕ+φ−α2+α12≤α2−α1ζξ/k4∫011−l1−1−lζζξ/kPdl1/P×∫011−lψ′ϕ+φ−2−l2α1+l2qdl1/q∫01l1−1−lζζξ/kPdl1/P∫01lψ′ϕ+φ−2−l2α1+l2qdl1/q+∫011−l1−1−lζζξ/kPdl1/P∫011−lψ′ϕ+φ−l2α1+2−l2qdl1/q∫01l1−1−lζζξ/kPdl1/P×∫01lψ′ϕ+φ−l2α1+2−l2qdl1/q.
Applying definition of strong convexity,(73)≤α2−α1ζξ/k4∫011−l1−1−lζζξ/kPdl1/P∫011−l×ψ′ϕq+ψ′φq−2−l2ψ′α1q+ψ′l2q−al22−l2α2−α12dl1/q+∫01l1−1−lζζξ/kPdl1/P∫01lψ′ϕq+ψ′φq−2−l2ψ′α1q+ψ′l2q−al22−l2α2−α12qdl1/q+∫011−l1−1−lζζξ/kPdl1/P∫011−lψ′ϕq+ψ′φq−l2ψ′α1q+2−l2ψ′α2q−al22−l2α2−α12qdl1/q×∫01l1−1−lζζξ/kPdl1/P∫01lψ′ϕq+ψ′φq−l2ψ′α1q+2−l2ψ′α2q−al22−l2α2−α12qdl1/q.
By some computations, one can get the required result.Theorem 24.
Letξ,ζ>0,p,q>1 with 1/p+1q=1,ϕ<φ and ψ:ϕ,φ⟶ℝ be a differentiable mapping such that ψ′∈lϕ,φ and ψ′q is a strong convex mapping on ϕ,φ. Then, one has
(74)2ζξ/k−1ζξ/kΓkξ+kα2−α1ζξ/kJkξϕ+φ−α2+α1/2ζψϕ+φ−α1+Jkξϕ+φ−α2+α1/2ζψϕ+φ−α2−ψϕ+φ−α2+α12≤α2−α1ζξ/k4B2/ζ,ξ/k+1ζξ/k+11−1/qψ′ϕq+ψ′φqB2/ζ,ξ/k+1ζξ/k+1−12ζξ/k+1ψ′α1qB2ζ,ξk+1+B3ζ,ξk+1+12ζξ/k+1ψ′α2qB2ζ,ξk+1−B3ζ,ξk+1−al4ζξ/k+12−l2α2−α121/q+B1/ζ,ξ/k+1−B2/ζ,ξ/k+1ζξ/k+11−1/q×ψ′ϕq+ψ′φqB1/ζ,ξ/k+1−B2/ζ,ξ/k+1ζξ/k+1−12ζξ/k+1ψ′α1qB1ζ,ξk+1−B3ζ,ξk+1+12ζξ/k+1ψ′α2qB1ζ,ξk+1−2B2ζ,ξk+1+B3ζ,ξk+1−al4ζξ/k+12−l2α2−α121/q+B2/ζ,ξ/k+1ζξ/k+11−1/q×ψ′ϕq+ψ′φqB2/ζ,ξ/k+1ζξ/k+1−12ζξ/k+1ψ′α1qB2ζ,ξk+1−B3ζ,ξk+1+12ζξ/k+1ψ′α2qB2ζ,ξk+1−B3ζ,ξk+1−al4ζξ/k+12−l2α2−α121/q+B1/ζ,ξ/k+1−B2/ζ,ξ/k+1ζξ/k+11−1/q×ψ′ϕq+ψ′φqB1/ζ,ξ/k+1−B2/ζ,ξ/k+1ζξ/k+1−12ζξ/k+1ψ′α1qB1ζ,ξk+1−2B2ζ,ξk+1+B3ζ,ξk+1+12ζξ/k+1ψ′α1qB1ζ,ξk+1−B3ζ,ξk+1−al4ζξ/k+12−l2α2−α121/q.Proof.
It follows from Lemma10, Jensen-Mercer inequality, the convexity of ψ′q, and the improved power-mean integral inequality given in Theorem 1.5 of [31] that
(75)2ζξ/k−1ζξ/kΓkξ+kα2−α1ζξ/kJkξϕ+φ−α2+α1/2ζ+ψϕ+φ−α1+kξJϕ+φ−α2+α1/2ζ+ψϕ+φ−α2−ψϕ+φ−α2+α12≤α2−α14ζξ/k∫011−1−lζζξ/kψ′ϕ+φ−2−l2α1+l2α2dl+∫011−1−lζζξ/kψ′ϕ+φ−l2α1+2−l2α2dl.
Using definition of strong convexity,(76)≤α2−α14ζξ/k∫011−1−lζζξ/kψ′ϕ+ψ′φ−2−l2ψ′α1+l2ψ′α2−al22−l2α2−α12dl+∫011−1−lζζξ/kψ′ϕ+ψ′φ−l2ψ′α1+2−l2ψ′α2−al22−l2α2−α12dl≤α2−α14ζξ/kψ′ϕ+ψ′φ∫011−1−lζζξ/kdl−ψ′α1∫011−1−lζζξ/k2−l2dl+ψ′α2∫011−1−lζζξ/kl2dl−α2−α12∫011−1−lζζξ/kal22−l2dl+ψ′ϕ+ψ′φ∫011−1−lζζξ/kdl−ψ′α1×∫011−1−lζζξ/kl2dl+ψ′α2∫011−1−lζζξ/k2−l2dl−α2−α12∫011−1−lζζξ/kal22−l2dl.


## 4. Applications to Special Means

Means are important in applied and pure mathematics, especially they are used frequently in numerical approximation. In literature, they are order in the following way:(77)H≤G≤L≤I≤A.The arithmetic mean of two numbersa, b such that a a≠b is defined as
(78)A=Aa,b=a+b2,a,b∈ℝ.The generalized logarithmic mean is defined as follows:(79)L=Lrra,b=br+1−ar+1r+1b−a,r∈ℝ−1,0,a,b∈ℝ,a≠b,Lpa,b=bp+1−ap+1p+1b−ap,p∈ℝ−1,0Proposition 25.
Assumea,b>0 and a<b; then,
(80)Mpa,b≤L1−pp−1a,b.Lppa,b≤Aholds for p∈−∞,1/0 where
(81)Mpa,b=ap+bp21/p.Proof.
From Theorem23, we have
(82)ap+bp21/p≤pBααbp−apIaCFαψk+IaCFαψk−21−αBαψk≤ϕa+ϕb2,with ψx=ϕx/x1−p holds. Setting fx=x,α=1 and Bα=B1=1 in the above theorem, we obtain
(83)ap+bp21/p≤pb−abp−apbp+1−ap+1p+1b−a≤a+b2.
Now use the following:(84)Lpa,b=bp+1−ap+1p+1b−a1/p.
Forp=p−1, we have
(85)Lp−1a,b=bp−appb−a1/p−1.
This implies that(86)Lp−1p−1a,b=bp−appb−a.
By using these means, we get(87)Mpa,b≤Lp−11−pa,b.Lppa,b≤A.
By using the results of Section3, we get some application to special means.Proposition 26.
Leta,b∈ℝ+,a<b; then,
(88)Aa2,b2−pLp+11+pap,bp≤bp−appaC1a,b+bC2a,b.Proof.
We obtain the result immediately from Theorem23.Proposition 27.
Leta,b∈ℝ+,a<b, then
(89)Aan,bn−pLn−p+1n−1+pap,bp≤bp−ap2pan−1C1a,b+bn−1C2a,b.Proof.
We obtain the result immediately from Theorem24.



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