# Integral Inequalities Involving Strongly Convex Functions

Authors: Ying-Qing Song; Muhammad Adil Khan; Syed Zaheer Ullah; Yu-Ming Chu
Journal: Journal of Function Spaces
 (2018)

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## Abstract

We study the notions of strongly convex function as well asF-strongly convex function. We present here some new integral inequalities of Jensen’s type for these classes of functions. A refinement of companion inequality to Jensen’s inequality established by Matić and Pečarić is shown to be recaptured as a particular instance. Counterpart of the integral Jensen inequality for strongly convex functions is also presented. Furthermore, we present integral Jensen-Steffensen and Slater’s inequality for strongly convex functions.


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## Body

## 1. Introduction and Preliminaries

The word “convexity” is the most important, natural, and fundamental notations in mathematics. Convex functions were presented by Johan Jensen over100 years ago. Over the past few years, multiple generalizations and extensions have been made for convexity. These extensions and generalizations in the theory of inequalities have made valuable contributions in many areas of mathematics. Some new generalized concepts in this point of view are quasiconvex [1], strongly convex [2], approximately convex [3], logarithmically convex [4], midconvex functions [5], pseudoconvex [6], φ-convex [7], λ-convex [8], h-convex [9], delta-convex [10], Schur convex [11–15], and others [16–19].The main ingredient of our investigation is the strongly convex function [2]. Let Ψ be the real function defined on interval I and c be positive number, then we say that the function Ψ is strongly convex with modulus c on I if(1)Ψηz+1-ηy≤ηΨz+1-ηΨy-cη1-ηz-y2for all y,z∈I and η∈[0,1].Every strongly convex function is convex, but the converse is not true in general. Strongly convex functions have been utilized for proving the convergence of a gradient type algorithm for minimizing a function. They play a significant role in mathematical economics, approximation theory, and optimization theory. Many applications and properties of them can be found in [2, 9, 20]. In 2016, Adamek [21] further generalized the notion of strongly convex function. They replaced the nonnegative term cz-y2 by a nonnegative real valued function F and defined it as follows: function Ψ is said to be F-strongly convex function if(2)Ψηz+1-ηy≤ηΨz+1-ηΨy-η1-ηFz-yfor all y,z∈I and η∈0,1. From [22], we also have(3)Ψy+Ψ+′yz-y+Fz-y≤Ψz,where Ψ is F-strongly convex function.In literature the following inequality is well-known as Jensen inequality.Theorem 1 (see [4]).
Let(Λ,X,μ) be a measure space with 0<μ(Λ)<∞ and Ψ:I→R be convex function. Suppose ϕ:Λ→I is such that ϕ, Ψ(ϕ)∈L1(μ), then one has(4)Ψ1μΛ∫Λϕdμ≤1μΛ∫ΛΨϕdμ.In 1981, Slater proved a companion inequality to the Jensen inequality [23].Theorem 2.
Let(Λ,X,μ) be a measure space with 0<μ(Λ)<∞ and Ψ:I→R be increasing and convex function. Suppose ϕ:Λ→I is such that Ψ(ϕ), Ψ+′(ϕ), and ϕΨ+′(ϕ)∈L1(μ). If ∫Λ‍Ψ+′(ϕ)dμ≠0, then one has(5)1μΛ∫ΛΨϕdμ≤Ψ∫ΛϕΨ+′ϕdμ∫ΛΨ+′ϕdμ.In the case when Ψ is strictly convex, then equality holds in (5) if and only if ϕ is constant almost everywhere on Λ.Remark 3.
Some improvements and reversions of Slater’s inequality are given in [24, 25].The following inequality is the integral analogue of another companion inequality to the Jensen inequality.Theorem 4 (see [26]).
Let(Λ,X,μ) be a measure space with 0<μ(Λ)<∞ and Ψ:I→R be convex function. Suppose ϕ:Λ→I is such that ϕ, Ψ(ϕ), Ψ+′(ϕ), and ϕΨ+′(ϕ)∈L1(μ) and (6)ϕ¯=1μΛ∫ΛϕdμΨ¯=1μΛ∫ΛΨϕdμ.Then the following inequalities hold:(7)0≤Ψ¯-Ψϕ¯≤1μΛ∫Λϕ-ϕ¯Ψ+′ϕdμ.If the function Ψ is strictly convex, then equality holds in (7) if and only if ϕ is constant almost everywhere on Λ.Matić and Pečarić established a general inequality from which one can directly obtain inequalities (5) and (7).Theorem 5 (see [27]).
Let all the assumptions of Theorem4 be fulfilled. If d1,d2∈I, then one has(8)Ψd1+Ψ+′d1ϕ¯-d1≤Ψ¯≤Ψd2+1μΛ∫Λϕ-d2Ψ+′ϕdμ.Also, when Ψ is strictly convex, then equality in the left side in (8) holds if and only if ϕ=d1 almost everywhere on Λ, while equality in the right side in (8) holds if and only if ϕ=d2 almost everywhere on Λ.Remark 6.
Under the assumptions of Theorem4, let ∫Λ‍Ψ+′(ϕ)dμ≠0 and Ψ¯¯=∫Λ‍ϕΨ+′(ϕ)dμ/∫Λ‍Ψ+′(ϕ)dμ∈I, then by setting d2=Ψ¯¯ in (8), we get Slater’s inequality (5), and similarly by setting d2=ϕ¯ in (8), we get (7).Merentes and Nikodem improved the Jensen inequality for strongly convex functions as follows.Theorem 7 (see [28]).
Let(Λ,X,μ) be a probability measure space and Ψ:I→R be strongly convex function with modulus c. Suppose ϕ:Λ→I is a Lebesgue integrable function and θ¯=∫Λ‍ϕdμ. Then the following inequality holds:(9)0≤∫ΛΨϕdμ-Ψθ¯-c∫Λϕ-θ¯2dμ.For more recent results related to strongly convex function and Jensen type inequalities we recommend [22, 29–34].This paper is organized as follows. In Section2, we establish general inequalities for F-strongly convex function as well as strongly convex functions. As a consequence, we obtain integral Jensen inequality and Slater’s inequality for strongly convex functions. Also by the virtue of these general inequalities we deduce converse of Jensen inequality. In Section 3, we give some properties of strongly convex functions. By using these properties of strongly convex functions we prove Jensen-Steffensen and Slater’s type inequalities.


## 2. Jensen’s Type Inequalities

We start this section to give the following general theorem.Theorem 8.
Let(Λ,X,μ) be a measure space with 0<μ(Λ)<∞ and Ψ:I→R be F-strongly convex function. Suppose ϕ:Λ→I is such that , Ψ(ϕ), Ψ+′(ϕ), and ϕΨ+′(ϕ)∈L1(μ) and also ϕ¯=1/μ(Λ)∫Λ‍ϕdμ, Ψ¯=1/μ(Λ)∫Λ‍Ψ(ϕ)dμ. If d1,d2∈I, then one has(10)Ψd1+Ψ+′d1ϕ¯-d1+1μΛ∫ΛFϕ-d1dμ≤Ψ¯≤Ψd2+1μΛ∫Λϕ-d2Ψ+′ϕdμ-1μΛ∫ΛFd2-ϕdμ.Proof.
SinceΨ is strongly convex function, therefore(11)Ψy-Ψ+′yy-z+Fz-y≤Ψz.Letting y→ϕ and z→d2, in (11), we get(12)Ψϕ-Ψ+′ϕϕ-d2+Fd2-ϕ≤Ψd2.Taking integral of (12) and then dividing by μ(Λ), we obtain(13)Ψ¯≤Ψd2+1μΛ∫Λϕ-d2Ψ+′ϕdμ-1μΛ∫ΛFd2-ϕdμ.Similarly, rearranging (11), we get(14)Ψy+Ψ+′yz-y+Fz-y≤Ψz.Letting y→d1 and z→ϕ, in (14), we have(15)Ψd1+Ψ+′d1ϕ-d1+Fϕ-d1≤Ψϕ.Taking integral of (15) and then dividing by μ(Λ), we get(16)Ψd1+Ψ+′d1ϕ¯-d1+1μΛ∫ΛFϕ-d1dμ≤Ψ¯.Combining (13) and (16), we obtain (10).By virtue of Theorem8, we can deduce some new and interesting consequences.Proposition 9.
Suppose that all the assumptions of Theorem8 are satisfied. Then(17)1μΛ∫ΛΨϕdμ≤1μΛ∫ΛϕΨ+′ϕdμ+infz∈IΨz-zμΛ∫ΛΨ+′ϕdμ-infz∈I1μΛ∫ΛFz-ϕdμ.Proof.
If we sety=ϕ in (11) and taking integral over Λ and then dividing by μ(Λ), we have(18)1μΛ∫ΛΨϕdμ-1μΛ∫ΛϕΨ+′ϕdμ+zμΛ∫ΛΨ+′ϕdμ+1μΛ∫ΛFz-ϕdμ≤Ψz,or equivalently(19)1μΛ∫ΛΨϕdμ≤1μΛ∫ΛϕΨ+′ϕdμ+Ψz-zμΛ∫ΛΨ+′ϕdμ-1μΛ∫ΛFz-ϕdμ.Taking the infimum over z∈I, we obtain (17).Proposition 10.
Suppose that all the assumptions of Theorem8 are satisfied and x¯=1/μ(Λ)∫Λ‍Ψ+′(ϕ)dμ, then(20)0≤Ψ¯-Ψϕ¯-1μΛ∫ΛFϕ-ϕ¯dμ≤infz∈IΨz-zx¯+1μΛ∫ΛϕΨ+′ϕdμ-Ψϕ¯-infz∈I1μΛ∫ΛFz-ϕdμ-1μΛ∫ΛFϕ-ϕ¯dμ≤1μΛ∫ΛϕΨ+′ϕdμ-ϕ¯x¯-infz∈I1μΛ∫ΛFz-ϕdμ-1μΛ∫ΛFϕ-ϕ¯dμ.Proof.
By settingd1=ϕ¯ and d2=z∈I in (10), we have(21)Ψϕ¯+1μΛ∫ΛFϕ-ϕ¯dμ≤Ψ¯≤Ψz+1μΛ∫Λϕ-zΨ+′ϕdμ-1μΛ∫ΛFz-ϕdμ.Indeed, the following equivalent form of (21) is(22)0≤Ψ¯-Ψϕ¯-1μΛ∫ΛFϕ-ϕ¯dμ≤Ψz+1μΛ∫Λϕ-zΨ+′ϕdμ-Ψϕ¯-1μΛ∫ΛFz-ϕdμ-1μΛ∫ΛFϕ-ϕ¯dμ.Taking the infimum over z∈I, we can easily derive the first and the second inequality in (20). The remaining third inequality in (20) follows because(23)infz∈IΨz-zx¯≤Ψϕ¯-ϕ¯x¯.Corollary 11.
Let(Λ,X,μ) be a measure space with 0<μ(Λ)<∞ and Ψ:I→R be strongly convex function with modulus c. Suppose ϕ:Λ→I is such that ϕ, Ψ(ϕ), Ψ+′(ϕ), and ϕΨ+′(ϕ)∈L1(μ) and also ϕ¯=1/μ(Λ)∫Λ‍ϕdμ, Ψ¯=1/μ(Λ)∫Λ‍Ψ(ϕ)dμ. If d1,d2∈I, then one has(24)Ψd1+Ψ+′d1ϕ¯-d1+cμΛ∫Λϕ-d12dμ≤Ψ¯≤Ψd2+1μΛ∫Λϕ-d2Ψ+′ϕdμ-cμΛ∫Λϕ-d22dμ.Remark 12.
If we putd1=ϕ¯ in (24) and take probability measure space, then we obtain integral Jensen inequality (9) for strongly convex function.In the following corollary, we obtain integral Slater’s inequality for strongly convex function.Corollary 13.
SupposeΨ, ϕ, μ(Λ), ϕ¯, and Ψ¯ are stated as in Corollary 11 and assume that ∫Λ‍Ψ+′(ϕ)dμ≠0; also if ϕ¯¯=∫Λ‍ϕΨ+′(ϕ)dμ/∫Λ‍Ψ+′(ϕ)dμ∈I, then(25)Ψ¯≤Ψϕ¯¯-cμΛ∫Λϕ-ϕ¯¯2dμ.Proof.
By settingd2=ϕ¯¯ in (24), we deduced(26)Ψ¯≤Ψ∫ΛϕΨ+′ϕdμ∫ΛΨ+′ϕdμ+1μΛ∫Λϕ-∫ΛϕΨ+′ϕdμ∫ΛΨ+′ϕdμΨ+′ϕdμ-cμΛ∫Λϕ-∫ΛϕΨ+′ϕdμ∫ΛΨ+′ϕdμ2dμ.Since 1/μ(Λ)∫Λ‍ϕ-∫Λ‍ϕΨ+′(ϕ)dμ/∫Λ‍Ψ+′(ϕ)dμΨ+′(ϕ)dμ=0, therefore (26) is equivalent to (25).In the following corollary, we obtain a converse of the Jensen inequality for strongly convex function.Corollary 14.
SupposeΨ, ϕ, μ(Λ), ϕ¯, and Ψ¯ are stated as in Corollary 11, then one has(27)0≤Ψ¯-Ψϕ¯-cμΛ∫Λϕ-ϕ¯2dμ≤1μΛ∫Λϕ-ϕ¯Ψ+′ϕdμ-2cμΛ∫Λϕ-ϕ¯2dμ.Proof.
By settingd1=d2=ϕ¯ in (24), we obtain (27).


## 3. Jensen-Steffensen Inequality for Riemann-Stieltjes Integrals

To prove the main results of this section, first we prove the following lemma which will play a key role in the proof of main results.Lemma 15.
LetΨ:(γ,β)→R be strongly convex function with modulus c. Suppose y is fixed element from interval (γ,β), then (a)
the functionΔy(ζ):(γ,β)→R defined by(28)Δyζ=Ψζ-Ψy-Ψ+′yζ-y-cζ-y2(b)
and the functionΔ¯y(ζ):(γ,β)→R defined by(29)Δ¯yζ=Ψy-Ψζ-Ψ+′ζy-ζ-cy-ζ2are nonnegative on (γ,β), decreasing on (γ,y] and increasing on [y,β).Proof.
By definition of strongly convexity we have(30)Δyζ=Ψζ-Ψy-Ψ+′yζ-y-cζ-y2≥0,∀ζ,y∈γ,βIt means that Δy(ζ) is nonnegative on (γ,β).
Letγ<ζ1<ζ2≤y. Since Ψ is strongly convex with modulus c, therefore Ψ+′(ζ)-2cζ is increasing and hence(31)Ψ+′ζ2-2cζ2≤Ψ+′y-2cy;it follows(32)Ψ+′yζ1-ζ2≤Ψ+′ζ2ζ1-ζ2-2cζ2ζ1-ζ2+2cyζ1-ζ2.Setting ζ=ζ1 and ζ=ζ2 in (28) and then taking the difference, we get(33)Δyζ1-Δyζ2=Ψζ1-Ψζ2-Ψ+′yζ1-ζ2-cζ1-y2+cζ2-y2≥Ψζ1-Ψζ2-Ψ+′ζ2ζ1-ζ2+2cζ2ζ1-ζ2-2cyζ1-ζ2-cζ1-y2+cζ2-y2by  using32=Ψζ1-Ψζ2-Ψ+′ζ2ζ1-ζ2-cζ1-ζ22≥0.Hence, Δy(ζ) is decreasing on (γ,y].
Also, ify≤ζ1<ζ2<β, then similarly as above by strongly convexity we have(34)Ψ+′y-2cy≤Ψ+′ζ1-2cζ1,from which it follows that(35)Ψ+′yζ1-ζ2≥Ψ+′ζ1ζ1-ζ2-2cζ1ζ1-ζ2+2cyζ1-ζ2.Setting ζ=ζ1 and ζ=ζ2 in (28), then taking the difference we get(36)Δyζ1-Δyζ2=Ψζ1-Ψζ2-Ψ+′yζ1-ζ2-cζ1-y2+cζ2-y2≤Ψζ1-Ψζ2-Ψ+′ζ1ζ1-ζ2+2cζ1ζ1-ζ2-2cyζ1-ζ2-cζ1-y2+cζ2-y2by  using35=Ψζ1-Ψζ2-Ψ+′ζ1ζ1-ζ2-cζ1-ζ22≤0.Hence, Δy(ζ) is increasing on [y,β).
In the same manner as above, we also obtainΔ¯y(ζ) is nonnegative on (γ,β).
Letγ<ζ1<ζ2≤y, then by strongly convexity we have(37)Ψ+′ζ1-2cζ1≤Ψ+′ζ2-2cζ2,(38)thatis,Ψ+′ζ2y-ζ2≥Ψ+′ζ1y-ζ2-2cζ1y-ζ2+2cζ2y-ζ2.Setting ζ=ζ1 and ζ=ζ2 in (29) and then taking the difference we get(39)Δ¯yζ1-Δ¯yζ2=Ψζ2-Ψζ1-Ψ+′ζ1y-ζ1+Ψ+′ζ2y-ζ2-cy-ζ12+cy-ζ22≥Ψζ2-Ψζ1-Ψ+′ζ1y-ζ1+Ψ+′ζ1y-ζ2-2cζ1y-ζ2+2cζ2y-ζ2-cy-ζ12+cy-ζ22by38=Ψζ2-Ψζ1-Ψ+′ζ1ζ2-ζ1-cζ2-ζ12≥0.Hence, Δ¯y(ζ) is decreasing on (γ,y].
Also, ify≤ζ1<ζ2<β, then by strongly convexity we have(40)Ψ+′ζ1-2cζ1≤Ψ+′ζ2-2cζ2,(41)i.e.,Ψ+′ζ1y-ζ1≥Ψ+′ζ2y-ζ1-2cζ2y-ζ1+2cζ1y-ζ1.Setting ζ=ζ1 and ζ=ζ2 in (29) and then taking the difference we get(42)Δ¯yζ1-Δ¯yζ2=Ψζ2-Ψζ1-Ψ+′ζ1y-ζ1+Ψ+′ζ2y-ζ2-cy-ζ12+cy-ζ22≤Ψζ2-Ψζ1-Ψ+′ζ2y-ζ1+Ψ+′ζ2y-ζ2-2cζ1y-ζ1+2cζ2y-ζ1-cy-ζ12+cy-ζ22by41=Ψζ2-Ψζ1-Ψ+′ζ2ζ1-ζ2+cζ1-ζ22≤0.Hence, Δ¯y(ζ) is increasing on [y,β). This completes the proof.The following lemma is given in [35].Lemma 16.
Letϕ:[γ,β]→R be a nonnegative function and suppose ξ:[γ,β]→R is either a bounded variation or continuous. Also assume that the functions ϕ and ξ have no common discontinuity points.(a)
Ifϕ is increasing on [γ,β], then(43)ϕβinfγ≤τ≤β∫τβdξx≤∫γβϕxdξx≤ϕβsupγ≤τ≤β∫τβdξx.(b)
Ifϕ is decreasing on [γ,β], then(44)ϕγinfγ≤τ≤β∫γτdξx≤∫γβϕxdξx≤ϕγsupγ≤τ≤β∫γτdξx.In the next result, we prove some general integral inequalities for strongly convex functions.Theorem 17.
Supposeϕ:[γ,β]→(κ,ν) is monotonic and continuous function and let Ψ:(κ,ν)→R be a strongly convex function with modulus c. If ξ:[γ,β]→R is either a bounded variation or continuous and satisfying ξ(γ)≤ξ(x)≤ξ(β) for all x∈[γ,β], ξ(β)-ξ(γ)>0, then ϕ¯ and Ψ¯ given by (45)ϕ¯=1ξβ-ξγ∫γβϕxdξx,Ψ¯=1ξβ-ξγ∫γβΨϕxdξxare well defined and ϕ¯∈(κ,ν). Also, if Ψ+′ϕ(x) and ξ have no common discontinuity points, then for d1,d2∈(κ,ν), one has(46)Ψd1+Ψ+′d1ϕ¯-d1+cξβ-ξγ∫γβϕx-d12dξx≤Ψ¯≤Ψd2+1ξβ-ξγ∫γβϕx-d2Ψ+′ϕxdξx-cξβ-ξγ∫γβϕx-d22dξx.Proof.
Under the given conditions in [35], it has been shown that (47)ϕγ,β=ϕγ,ϕβ⊆κ,ν.We define the function λ:[γ,β]→R by λ(x)=Δd1ϕ(x), where Δd1 is defined as in Lemma 15(a), i.e., (48)λx=Δd1ϕx=Ψϕx-Ψd1-Ψ+′d1ϕx-d1-cϕx-d12.From Lemma 15(a) it follows that λ is nonnegative and since ϕ and Ψ(ϕ) are continuous therefore the integral ∫γβ‍λ(x)dξ(x) exists. Thus we discuss the following three cases:(i)
Ifϕ(β)≤d1, since ϕ is increasing on [γ,β] and by Lemma 15(a) Δd1 is decreasing on (κ,d1], therefore λ=Δd1(ϕ(x)) is decreasing on [γ,β]. So using Lemma 16 (b), we obtain (49)∫γβλxdξx≥λγinfγ≤τ≤βξτ-ξγ=0.(ii)
Ifd1≤ϕ(γ), since Δd1 is increasing on [d1,ν) by Lemma 15(a), therefore λ=Δd1(ϕ(x)) is increasing on [γ,β]. So using Lemma 16 (a), we have (50)∫γβλxdξx≥λβinfγ≤τ≤βξβ-ξτ=0.(iii)
Ifϕ(γ)<d1<ϕ(β), since ϕ is continuous on [γ,β], there exists at least one point x¯∈(γ,β) such that ϕ(x¯)=d1. Also by Lemma 15(a), λ is decreasing on [γ,x¯] and λ is increasing on [x¯,β]. Using Lemma 16, we have (51)∫γβλxdξx=∫γx¯λxdξx+∫x¯βλxdξx≥λγinfγ≤τ≤x¯ξx¯-ξγ+λβinfx¯≤τ≤βξβ-ξx¯=0.From the above three subcases we conclude that(52)∫γβλxdξx=∫γβΨϕx-Ψd1-Ψ+′d1ϕx-d1-cϕx-d12dξx≥0.(53)i.e.,Ψd1ξβ-ξγ+Ψ+′d1∫γβϕxdξx-d1ξβ-ξγ+c∫γβϕx-d12dξx≤∫γβΨϕxdξx.Dividing (53) by ξ(β)-ξ(γ)>0, we obtain the left side of the inequality (46). Similarly, if ϕ is decreasing we consider the cases ϕ(γ)≤d1 (λ is increasing on [γ,β]), d1≤ϕ(β) (λ is decreasing on [γ,β]), and ϕ(γ)<d1<ϕ(β) (λ is decreasing on [γ,x¯] and increasing on [x¯,β]). In all three cases we obtain ∫γβ‍λ(x)dξ(x)≥0 from the first inequality in (46) which directly follows.
Similarly, we can prove the right side of the inequality (46). We define the function λ¯:[γ,β]→R by λ¯(x)=Δ¯d2(ϕ(x)), where Δ¯d2 is defined as in Lemma 15(b), i.e., (54)λ¯x=Δ¯d2ϕx=Ψd2-Ψϕx-Ψ+′ϕxd2-ϕx-cd2-ϕx2.Since ϕ is monotonic and continuous, Ψ(ϕ) is continuous and Ψ+′(ϕ) is monotonic which have no common discontinuity point with ξ. Therefore the integral ∫γβ‍λ(x)dξ(x) exists. Using the same process as above we have ∫γβ‍λ¯(x)dξ(x)≥0, which means that (55)∫γβΨd2-Ψϕx-Ψ+′ϕxd2-ϕx-cd2-ϕx2dξx≥0.i.e.,ξβ-ξγΨd2-∫γβΨϕxdξx+∫γβϕx-d2Ψ+′ϕxdξx-c∫γβϕx-d22dξx≥0.If we divide by ξ(β)-ξ(γ)>0, we obtain the second inequality of (46). So, both inequalities of (46) are proved.Now, we are in a situation to obtain the following result.Corollary 18.
Suppose all the assumptions of Theorem17 are satisfied, then one has(56)0≤Ψ¯-Ψϕ¯-cξβ-ξγ∫γβϕx-ϕ¯2dξx≤1ξβ-ξγ∫γβϕx-ϕ¯Ψ+′ϕxdξx-2cξβ-ξγ∫γβϕx-ϕ¯2dξx.Proof.
By settingd1=d2=ϕ¯ in (46), we obtain (56).Remark 19.
If we setc=0 in (56), we obtain Theorem 3.2 in [35].In the following corollary, we obtain integral Slater’s inequality for strongly convex functions.Corollary 20.
Suppose all the assumptions of Theorem17 are satisfied and assume ∫γβ‍Ψ+′(ϕ(x))dξ(x)≠0; also if ϕ¯¯=∫γβ‍ϕxΨ+′(ϕ(x))dξ(x)/∫γβ‍Ψ+′(ϕ(x))dξ(x)∈I, then(57)Ψ¯≤Ψϕ¯¯-cξβ-ξγ∫γβϕx-ϕ¯¯2dξx.Proof.
Similar to the proof of Corollary13, setting d2=ϕ¯¯ in the right hand side of (46), we get (57).Remark 21.
If we setc=0 in (57), we obtain Slater’s inequality for convex functions given in [35].



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