# On Integral Inequalities Involving Generalized Lipschitzian Functions

Authors: Mohaemd Jleli; Bessem Samet
Journal: Journal of Function Spaces
 (2020)

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## Abstract

A new class of mappings that includes the class of Lipschitzian mappings is introduced. For this kind of mappings, new integral inequalities of Hadamard’s type are obtained. Our results are extensions of many previous contributions related to integral inequalities for Lipschitzian mappings.


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## Body

## 1. Introduction

A functionω:I⊂ℝ→ℝ is said to be L-Lipschitzian, where L>0, if(1)ωσ−ωκ≤Lσ−κ,for all σ,κ∈I. In [1, 2], some integral inequalities of Hadamard’s type involving L-Lipschitzian functions were derived. In particular, it was shown that if ω is L-Lipschitzian in I=α,β, α<β, then(2)1β−α∫αβωσdσ−ωα+β2≤Lβ−α4,(3)1β−α∫αβωσdσ−ωα+ωβ2≤Lβ−α4.In [3], some Hadamard-type and Bullen-type inequalities were obtained for the same class of functions. We recall below the main results established in [3]. Let ω be L-Lipschitzian in I=α,β.(i)
LetM,N∈I with M≤N. Let ξ∈0,1, W=1−ξα+ξβ, and(4)WξM,N=M−α2+β−N2+sgnN−WN−W2+sgnW−MW−M2,where sgn denotes the signum function. Then,(5)ξωM+1−ξωN−1β−α∫αβωσdσ≤LWξM,N2β−αNotice that ifξ=1/2 and M=N=α+β/2, then (5) reduces to (2). Moreover, if ξ=1/2 and M,N=α,β, then (5) reduces to (3).(ii)
LetP,Q,R∈I, P≤Q≤R and ξii=13⊂0,1, where ξ1+ξ2+ξ3=1. Let W1=1−ξ1α+ξ1β, W2=ξ3α+ξ1+ξ2β and(6)Wξ1,ξ2,ξ3P,Q,R=P−α2+sgnW1−PW1−P2+sgnQ−W1Q−W12+sgnW2−QW2−Q2+sgnR−W2R−W22+β−R2Then,(7)ξ1ωP+ξ2ωQ+ξ3ωR−1β−α∫αβωσdσ≤LWξ1,ξ2,ξ3P,Q,R2β−α.Notice that ifP,Q,R=α,α+β/2,β and ξ1,ξ2,ξ3=1/4,1/2,1/4, then (7) reduces to the Bullen-type inequality(8)12ωα+β2+ωα+ωβ2−1β−α∫αβωσdσ≤Lβ−α8.Moreover, ifP,Q,R=α,α+β/2,β and ξ1,ξ2,ξ3=1/6,2/3,1/6, then (7) reduces to the Bullen-type inequality (see [4])(9)164ωα+β2+ωα+ωβ−1β−α∫αβωσdσ≤5Lβ−α36.For other works related to integral inequalities for Lipschitzian functions, see, for example, [5–11] and the references therein.Motivated by the above mentioned results, in this paper, we obtain some Hadamard-type integral inequalities for a new class of functions which includes the class of Lipschitzian functions.


## 2. The Class of Generalized Lipschitzian Functions

Letω:I⊂ℝ→ℝ be a given function and σ0∈I be fixed.(10)∣ωσ−ωκ∣≤L∣σ−σ0∣+∣κ−σ0∣,for all σ,κ∈I.(i)
ω is L,σ0-Lipschitzian(ii)
For allσ∈I,(11)∣ωσ−ωσ0∣≤L∣σ−σ0∣(12)∣ωσ−ωκ∣≤∣ωσ−ωσ0∣+∣ωκ−ωσ0∣≤L∣σ−σ0∣+L∣κ−σ0∣=L∣σ−σ0∣+∣κ−σ0∣,which shows that ω is L,σ0-Lipschitzian.Definition 1.
We say thatωisL,σ0-Lipschitzian, whereL>0, ifProposition 2.
LetL>0. The following statements are equivalent:Proof.
Suppose thatω is L,σ0-Lipschitzian. Taking κ=σ0 in (10), (ii) follows. Suppose now that ω satisfies (11). Then, for all σ,κ∈I,Remark 3.
It can be easily seen that ifωisL-Lipschitzian, then ω is L,σ-Lipschitzian, for all σ∈I.Let us denote byLI the set of functions ω:I→ℝ such that there exists L>0 for which ω is L-Lipschitzian. Moreover, we denote by LI~ the set of functions ω:I→ℝ such that there exist L>0 and σ0∈I for which ω is L,σ0-Lipschitzian. The following example shows that(13)LIÖLI~.(14)ωσ=σif0≤σ≤14,12−σif14≤σ≤12,σ−12if12≤σ≤34,σ−34if34<σ≤1.Example 1.
Letω:I=0,1→ℝbe the function defined byOne observes thatω is not continuous at σ=3/4, which shows that ω∈LI. On the other hand, one has(15)ωσ−ω12=∣ωσ∣=ωσ≤σ−12,for all σ∈I, which shows that ω is 1,1/2-Lipschitzian. Hence, ω∈LI~.


## 3. Inequalities of Hadamard’s Type

We first fix some notations that will be used throughout this section. Letω:I⊂ℝ→ℝ be L,σ0-Lipschitzian, where I=α,β, α<β, L>0, and σ0∈I. We denote by μσ0 the quantity defined by(16)μσ0=σ0−α+β22+β−α22.(17)∑i=1nξiωMi−1β−α∫αβωσdσ≤Lμσ0β−α+∑i=1nξi∣Mi−σ0∣.(18)∑i=1nξiωMi−1β−α∫αβωσdσ=1β−α∑i=1nξi∫αβωMi−ωσdσ≤1β−α∑i=1nξi∫αβ∣ωMi−ωσ∣dσ.Theorem 4.
Letn∈ℕ(n is a positive integer), Mii=1n⊂I, and ξii=1n⊂0,1 be such that ∑i=1nξi=1. Then,Proof.
Since∑i=1nξi=1, one hasHence, by the assumption onω, one obtains(19)∑i=1nξiωMi−1β−α∫αβωσdσ≤Lβ−α∑i=1nξi∫αβ∣Mi−σ0∣+∣σ−σ0∣dσ=Lβ−α∑i=1nξi∣Mi−σ0∣β−α+∫αβ∣σ−σ0∣dσ,which yields(20)∑i=1nξiωMi−1β−α∫αβωσdσ≤L1β−α∫αβ∣σ−σ0∣dσ+∑i=1nξi∣Mi−σ0∣.On the other hand, an elementary calculation shows that(21)∫αβ∣σ−σ0∣dσ=μσ0Therefore, combining (20) with (21), (17) follows.We investigate below some particular cases of Theorem4.(22)ξωM1+1−ξωM2−1β−α∫αβωσdσ≤Lμσ0β−α+ξ∣M1−σ0∣+1−ξ∣M2−σ0∣.(23)ξωρα+1−ρβ+1−ξωρα+1−ρβ−1β−α∫αβωσdσ≤Lμσ0β−α+ξ∣ρα+1−ρβ−σ0∣+1−ξ∣1−ρα+ρβ−σ0∣.(24)ωM−1β−α∫αβωσdσ≤Lμσ0β−α+∣M−σ0∣.(25)ωρα+1−ρβ−1β−α∫αβωσdσ≤Lμσ0β−α+∣ρα+1−ρβ−σ0∣.Corollary 5.
Letξ∈0,1 and M1,M2∈I. Then,Proof.
In Theorem4, let n=2, ξ1=ξ and ξ2=1−ξ. Then, by (17), one obtains (22).Corollary 6.
Letξ,ρ∈0,1. Then,Proof.
In Corollary5, let M1=ρα+1−ρβ and M2=ρα+1−ρβ. Then, by (22), one obtains (23).Corollary 7.
LetM∈I. Then,Proof.
In Corollary5, let M1=M2=M. Then, by (22), one obtains (24).Corollary 8.
Letρ∈0,1. Then,Proof.
In Corollary7, let M=ρα+1−ρβ. Then, by (24), one obtains (25).Corollary 9.
We have(26)ωα+β2−1β−α∫αβωσdσ≤Lμσ0β−α+α+β2−σ0.Proof.
Takingρ=1/2 in (25), one obtains (26).Corollary 10.
We have(27)ωσ0−1β−α∫αβωσdσ≤Lμσ0β−α.Proof.
In Corollary7, let M=σ0. Then, by (24), one obtains (27).Corollary 11.
Letσ0=α+β/2, that is, ω is L,α+β/2-Lipschitzian. Then,(28)ωα+β2−1β−α∫αβωσdσ≤Lβ−α4.Proof.
By (16), for σ0=α+β/2, one has
μσ0=β−α/22.(14)(52)
Then, by (27), one obtains (28).Remark 12.
Corollary11 shows that the inequality (2) which was obtained in [1, 2] for the class of L-Lipschitzian functions still holds for L,α+β/2-Lipschitzian functions.Corollary 13.
Letξ∈0,1. Then,(29)ξωα+1−ξωβ−1β−α∫αβωσdσ≤Lμσ0β−α+2ξ−1σ0−ξα+β+β.Proof.
In Corollary5, taking M1=α and M2=β, (29) follows.Corollary 14.
We have(30)ωα+ωβ2−1β−α∫αβωσdσ≤Lμσ0β−α+β−α2.Proof.
In (29), taking ξ=1/2, (30) follows.Corollary 15.
Letσ0=α+β/2, that is,ωisL,α+β/2-Lipschitzian. Then,(31)ωα+ωβ2−1β−α∫αβωσdσ≤3Lβ−α4.Proof.
In (30), taking σ0=α+β/2, (31) follows.Corollary 16.
LetM1,M2,M3∈I, and ξ1,ξ2,ξ2∈0,1 be such that ξ1+ξ2+ξ3=1. Then,(32)ξ1ωM1+ξ2ωM2+ξ3ωM3−1β−α∫αβωσdσ≤Lμσ0β−α+ξ1∣M1−σ0∣+ξ2∣M2−σ0∣+ξ3∣M3−σ0∣.Proof.
In Theorem4, let n=3. Then, by (17), one obtains (32).Corollary 17.
Letδ∈0,1. Then,(33)δωα+ωβ2+1−δωα+β2−1β−α∫αβωσdσ≤Lμσ0β−α+δ∣α−σ0∣+∣β−σ0∣2+1−δα+β2−σ0.Proof.
In Corollary16, let M1,M2,M3=α,α+β/2,β and ξ1,ξ2,ξ3=δ/2,1−δ,δ/2. Then, using (32), (33) follows.Corollary 18.
We have(34)14ωα+ωβ+2ωα+β2−1β−α∫αβωσdσ≤Lμσ0β−α+14∣α−σ0∣+∣β−σ0∣+2α+β2−σ0.Proof.
In (33), taking δ=1/2, (34) follows.Corollary 19.
Letσ0=α+β/2, that is,ωisL,α+β/2-Lipschitzian. Then,(35)14ωα+ωβ+2ωα+β2−1β−α∫αβωσdσ≤Lβ−α2.Proof.
In (34), taking σ0=α+β/2, (35) follows.Corollary 20.
We have(36)16ωα+ωβ+4ωα+β2−1β−α∫αβωσdσ≤Lμσ0β−α+∣α−σ0∣+∣β−σ0∣6+23α+β2−σ0.Proof.
In (33), taking δ=1/3, (36) follows.Corollary 21.
Letσ0=α+β/2, that is,ωisL,α+β/2-Lipschitzian. Then,(37)16ωα+ωβ+4ωα+β2−1β−α∫αβωσdσ≤5Lβ−α12.For the next result, letV be the set of functions ϑ:I→ℝ satisfying the following conditions:(P) ϑ∈C1I.(P) ϑ′σ>0, for all σ∈I.Notice that under the above conditions, the functionϑ:I=α,β→ϑα,ϑβ is invertible. We denote by ϑ−1 its inverse.Theorem 22.
Letϑ∈V. Then, for allσ∈I,(38)ωσ−1ϑβ−ϑα∫αβωκϑ′κdκ≤L∣σ−σ0∣+1ϑβ−ϑα∫ϑαϑβ∣ϑ−1τ−σ0∣dτ.Proof.
Takingτ=ϑκ, one obtains(39)∫αβωκϑ′κdκ=∫ϑαϑβωϑ−1τdτ.On the other hand, forσ∈I, one has(40)ωσ=1ϑβ−ϑα∫ϑαϑβωσdτ.Combining (39) with (40), it holds that(41)ωσ−1ϑβ−ϑα∫αβωκϑ′κdκ≤1ϑβ−ϑα∫ϑαϑβωσ−ωϑ−1τdτ.Next, using the assumption onω, one obtains(42)ωσ−ωϑ−1τ≤L∣σ−σ0∣+∣ϑ−1τ−σ0∣,for all τ∈ϑα,ϑβ, which yields(43)∫ϑαϑβωσ−ωϑ−1τdτ≤Lϑβ−ϑα∣σ−σ0∣+∫ϑαϑβ∣ϑ−1τ−σ0∣dτ.Finally, combining (41) and (43), (38) follows.We investigate below some particular cases of Theorem22.Corollary 23.
Letϑ∈V. Then,(44)ωσ0−1ϑβ−ϑα∫αβωκϑ′κdκ≤Lϑβ−ϑα∫ϑαϑβ∣ϑ−1τ−σ0∣dτ.Proof.
Takingσ=σ0 in (38), (44) follows.Corollary 24.
Letα>0. Then, for allσ∈I,(45)ωσ−αββ−α∫αβωκκ2dκ≤L∣σ−σ0∣+αββ−αlnαβσ02+α+βαβ−2σ0σ0.Proof.
Let(46)ϑκ=−1κ,κ∈I.It can be easily seen thatϑ∈V and(47)ϑ−1τ=−1τ,τ∈−1α,−1β.Hence, by Theorem22, one has(48)ωσ−αββ−α∫αβωκκ2dκ≤Lσ−σ0+αββ−α∫−1/α−1/β1τ+σ0dτ.On the other hand,(49)∫−1/α−1/β1τ+σ0dτ=∫1/β1/ασ0−1τdτ=∫1/β1/σ01τ−σ0dτ+∫1/σ01/ασ0−1τdτ=−lnσ0+lnβ−σ01σ0−1β+σ01α−1σ0+lnα−lnσ0=lnαβσ02+α+βαβ−2σ0σ0.Finally, combining (48) with (49), (45) follows.Corollary 25.
Letα>0. Then,(50)ωσ0−αββ−α∫αβωκκ2dκ≤αβLβ−αlnαβσ02+α+βαβ−2σ0σ0.Proof.
Takingσ=σ0 in (45), (50) follows.Corollary 26.
Letα>0. Letσ0=2αβ/α+β, that is, ωisL,2αβ/α+β-Lipschitzian. Then, for all σ∈I,(51)ωσ−αββ−α∫αβωκκ2dκ≤Lσ−2αβα+β+αββ−αlnα+β24αβ.Proof.
Takingσ0=2αβ/α+β in (45), (51) follows.Corollary 27.
Letα>0. IfωisL,2αβ/α+β-Lipschitzian, then,(52)ω2αβα+β−αββ−α∫αβωκκ2dκ≤αβLβ−αlnα+β24αβ.Proof.
Takingσ=2αβ/α+β in (51), (52) follows.



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