# On Opial-Type Inequalities for Superquadratic Functions and Applications in Fractional Calculus

Authors: Waqas Nazeer; Ghulam Farid; Zabidin Salleh; Ayesha Bibi
Journal: Mathematical Problems in Engineering
 (2021)

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## Abstract

We have studied the Opial-type inequalities for superquadratic functions proved for arbitrary kernels. These are estimated by applying mean value theorems. Furthermore, by analyzing specific functions, the fractional integral and fractional derivative inequalities are obtained.


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## Body

## 1. Introduction and Preliminary Results

Opial’s inequality is very important and useful in the study of differential and difference equations. It was introduced by Opial in 1960. It is stated in the following theorem.Theorem 1.
(see [1]). Let ψ∈C10,c be such that ψ0=0 and ψt>0 for t∈0,c. Then,(1)∫0cψtψ′tdt≤c4∫0hψ′t2dt.
Here,c/4 is the best possible constant.Many authors have been working continuously on Opial’s inequality and succeeded to establish very interesting results. For its numerous generalizations and extensions, we refer the readers to [2–14] and references therein. Opial-type inequalities are useful in the study of difference and differential equations, for example, in the uniqueness of initial value problems, in the existence and uniqueness of boundary value problems, and setting upper bounds of their solutions. For a historical survey of the inequalities given after the publication of inequality (1), one can study the book by Agarwal and Pang; see [2]. Mitrinović and Pečarić, in 1988, proved Opial-type inequalities for convex functions with respect to the power function by defining two classes of functions involving general kernels; see [15, 16]. Motivated by these inequalities, recently, we have studied such inequalities for convex functions by considering a generalized class of functions with arbitrary kernel [17]. Also, for special kernels, fractional Opial-type inequalities are proved for different fractional integral and derivative operators [18, 19].Our aim in this paper is to study Opial-type inequalities for superquadratic functions which are also connected with inequalities that hold for convex functions. The established results estimate the Opial-type inequalities with the help of some suitably defined functions. The definitions and results needful for the presentation of results of this paper are given as follows.Definition 1.
(see [4]). A function ψ:I⟶ℝ is said to be convex if for all x,y∈I and all t∈0,1,(2)ψtx+1−ty≤tψx+1−tψy,holds, where I is an interval in ℝ.Definition 2.
(see [20]). A function ψ:0,∞⟶ℝ is called a superquadratic function provided for all x≥0, there exists a real constant Cx such that(3)ψy≥ψx+Cxy−x+ψy−x,∀y≥0.
The following lemma explores convexity from the superquadratic function.Lemma 1.
(see [20]). Let ψ be a superquadratic function with Cx as in Definition 2. Then, we have(i)
ψ0≤0(ii)
Cx=ψ′x where ψ is differentiable for x>0 and ψ0=ψ′0=0(iii)
ψ is convex and ψ0=ψ′0=0 if ψ≥0
LetU1u,k denote the class of functions w:a,b⟶ℝ having representation(4)wx=∫axkx,tutdt,where u is a continuous function and k is an arbitrary nonnegative kernel such that kx,t=0 for t<x and ux>0 implies wx>0 for every x∈a,b. Let U2u,k denote the class of functions w:a,b⟶ℝ having representation(5)wx=∫xbkx,tutdt,where u is a continuous function and k is an arbitrary nonnegative kernel such that kx,t=0 for t<x and ux>0 implies wx>0 for every x∈a,b.Theorem 2.
(see [15, 16]). Let ψ:0,∞⟶ℝ be a differentiable function such that, for p2>1, the function ψx1/p2 is convex and ψ0=0. Let w∈U1u,k where ∫axkx,tp1dt1/p1≤K1 and 1/p1+1/p2=1. Then,(6)∫abwx1−p2ψ′wxuxp2dx≤p2K1p2ψK1∫abuxp2dx1/p2.The reverse of the above inequality holds whenψx1/p2 is concave.A similar result for the classU2u,k is given as follows.Theorem 3.
(see [15, 16]). Let ψ:0,∞⟶ℝ be a differentiable function such that, for p2>1, the function ψx1/p2 is convex and ψ0=0. Let w∈U2u,k where ∫xbkx,tp1dt1/p1≤K2 and 1/p1+1/p2=1. Then,(7)∫abwx1−p2ψ′wxuxp2dx≤p2K2p2ψK2∫abuxp2dx1/p2.The reverse of the above inequality holds whenψx1/p2 is concave.Recently, the following Opial-type inequalities for superquadratic functions are established.Theorem 4.
(see [21]). Let ψ:0,∞⟶ℝ be a differentiable function such that ψ′x/x is increasing and ψ0=0. Let w∈U1u,k, kx,t≤K, and M<∫axutdt. Then, the following inequality holds for the superquadratic function ψ:(8)∫abψ′wxuxwxdx≤1K2MψK∫abutdt.Theorem 5.
(see [21]). Under the conditions of Theorem 4, in addition, if ψ is nonnegative, then we have(9)∫abψ′wxuxwxdx≤1K2MψK∫abutdt≤1b−aMK2∫abψKb−autdt.Theorem 6.
(see [21]). Let ψ:0,∞⟶ℝ be a differentiable function such that ψ′x/x is increasing and ψ0=0. Let w∈U1u,k, ∫axkx,tp1dt1/p1≤K, M<∫axutp2dt2/p2−1, and 1/p1+1/p2=1. Then, the following inequality holds for superquadratic function ψ:(10)∫abψ′wxuxp2wxdx≤p2K2MψK∫abutp2dt1/p2.Theorem 7.
(see [21]). Under the conditions of Theorem 6, in addition, if ψ is nonnegative, then we have(11)∫abψ′wxuxp2wxdx≤p2K2MψK∫abutp2dt1/p2≤p2K2Mb−a∫abΨb−a1/p2Kutdt.The following lemmas provide classes of convex as well as superquadratic functions which will be useful to establish results of this paper.Lemma 2.
(see [22]). Let ψ∈C2I, where I⊆0,∞ and m′≤ψ″η≤M′ for all η∈I. Then, the functions ψ1andψ2 defined by(12)ψ1x=M′x22−ψx,ψ2x=ψx−m′x22,are convex.Lemma 3.
(see [23]). Let ψ∈C2I, I⊆0,∞ such that m′≤ηψ″η−ψ′η/η2≤M′, for all η∈I. Consider the functions ψ1andψ2 defined by(13)ψ1x=M′x32−ψx,ψ2x=ψx−m′x32.Then, the functionsψ1′x/x and ψ1′x/x are increasing. Also, if ψj0=0,j=1,2, then the functions ψj,j=1,2, are superquadratic.Lemma 4.
(see [22]). Let I⊆0,∞, ψ∈C2I, hx=xp2,p2>1, and m′≤ηψ″η−p2−1ψ′η/p2η2p2−1≤M′ for all η∈I. Then, the functions ψ1andψ2 defined by(14)ψ1x=M′x2p22−ψx,ψ2x=ψx−m′x2p22,are convex with respect to hx=xp2; that is, ψjx1/p2, j=1,2, are convex.Next, we define the Riemann–Liouville fractional integral and Caputo fractional derivative as follows.Definition 3.
(see [24]). Let −∞<a<b<∞ and h∈L1a,b. Then, the left-sided and right-sided Riemann–Liouville fractional integrals of order γ>0 are defined by(15)Ia+γhx=1Γγ∫axx−tγ−1htdt,x>a,Ib−γhx=1Γγ∫xbt−xγ−1htdt,x<b,where Γ. is the gamma function.Definition 4.
(see [24]). Let γ>0 and γ∉1,2,3,…, n=γ+1, and h∈ACna,b. Then, the left-sided and right-sided Caputo fractional derivatives of order γ are defined by(16)DCa+γhx=1Γn−γ∫axhntx−tγ−n+1dt,x>a,DCb−γhx=−1nΓn−γ∫xbhntt−xγ−n+1dt,x<b.
In [25], Andrić et al. presented the composition identities for the left-sided and right-sided Caputo fractional derivatives; these are stated in the following two lemmas.Lemma 5.
Letμ>γ≥0, l=μ+1, and m=γ+1 for γ,μ∉N0; m=γ and l=μ for γ,μ∈N0. Let h∈ACma,b be such that hja=0 for j=m,m+1,…,l−1. Let DCa+μh,DCa+γh∈L1a,b. Then, we have(17)DCa+γhx=1Γμ−γ∫axx−tμ−γ−1DCa+γhtdt,x∈a,b.Lemma 6.
Letμ>γ≥0, l=μ+1, and m=γ+1 for γ,μ∉N0; m=γ and l=μ for γ,μ∈N0. Let h∈ACma,b be such that hjb=0 for j=m,m+1,…,l−1. Let DCb−μh,DCb−γh∈L1a,b. Then, we have(18)DCb−γhx=1Γμ−γ∫xbt−xμ−γ−1DCb−γhtdt,x∈a,b.In the upcoming section, we prove the mean value theorems for the estimation of the nonnegative differences of the inequalities given in Theorems4–7. In Section 3, we prove the fractional versions of the results of Section 2 for Riemann–Liouville integrals and Caputo fractional derivatives.


## 2. Main Results

First, we define the linear functionalΦjψw,u;K,j=1,2, from nonnegative differences of Opial-type inequalities for superquadratic functions given in Theorems 4 and 5 as follows:(19)Φ1ψw,u;K=ψK∫abutdt−K2M∫abψ′wxuxwxdx,Φ2ψw,u;K=∫abψKb−autdt−b−aMK2∫abψ′wxuxwxdx.Remark 1.
Under the assumptions of Theorems4 and 5, we have Φjψw,u;K≥0 for j=1,2.Theorem 8.
With the same assumptions as given in Theorem4, furthermore, let I⊆0,∞ be a compact interval and ψ∈C2I. Also, let m′≤ψ′x/x≤M′ where infx∈Iψ″x=m′ and supx∈Iψ″x=M′. Then, there exists η1∈I such that the following result holds:(20)Φ1ψw,u;K=η1ψ″η1−ψ′η13η12Φ1x3w,u;K.Proof.
By replacingψ with ψ1 (defined in Lemma 3) in Theorem 4, one can have the following inequality:(21)∫abM′wx2−ψ′wxuxwxdx≤1MK2M′3K∫abutdt3−ψK∫abutdt.
From inequality (21), after simplification, one can obtain(22)Φ1ψw,u;K≤M′3Φ1x3w,u;K.
Similarly, if we takeψ2 from Lemma 3 instead of ψ in Theorem 4, then the following inequality holds:(23)Φ1ψw,u;K≥m′3Φ1x3w,u;K.
The above two inequalities lead to the following inequality:(24)m′≤3Φ1ψw,u;KΦ1x3w,u;K≤M′.
Therefore, there existsη1∈I such that the following equation is obtained:(25)η1ψ″η1−ψ′η1η12=3Φ1ψw,u;KΦ1x3w,u;K,which gives equation (20).Theorem 9.
With the same assumptions onψ1andψ2 as given for ψ in Theorem 4, furthermore, if I⊆0,∞ is a compact interval and ψ1,ψ2∈C2I where Φ1x3w,u;K≠0, then there exists η1∈I such that the following result holds:(26)Φ1ψ1w,u;KΦ1ψ2w,u;K=η1ψ1″η1−ψ1′η1η1ψ2″η1−ψ2′η1,where denominators should not be zero.Proof.
Let us define the functionf by f=λ1ψ1−λ2ψ2, where λ1 and λ2 are given by(27)λ1=ϕ1ψ2w,u;K,λ2=ϕ1ψ1w,u;K.
Then,f∈C2I; by applying Theorem 8 for f, it follows that there exists η1 such that we have(28)0=λ1η1ψ1″η1−ψ1′η1−λ2η1ψ2″η1−ψ2′η1Φ1x3w,u;K.
From this, one can get the required equation.Theorem 10.
With the same assumptions as given in Theorem5, furthermore, if I⊆0,∞ is a compact interval and ψ∈C2I, then there exists η2∈I such that the following result holds:(29)Φ2ψw,u;K=ψ″η22Φ2x2w,u;K.Proof.
By replacingψ with ψ1 from Lemma 2 in Theorem 5, we get the following inequality:(30)∫abM′|wx|−ψ′wxuxwxdx≤1b−aMK2×M′2∫abKb−autdt2−∫abψKb−autdt.
Similarly, adopting the method for functionalΦ2ψw,u;K as we did for Φ1ψw,u;K in the proof of Theorem 8, one can see there exists η2∈I such that equation (29) holds.Theorem 11.
With the same assumptions onψ1andψ2 as given for ψ in Theorem 5, furthermore if I⊆0,∞ is a compact interval and ψ1,ψ2∈C2I where Φ2x2w,u;K≠0, then there exists η2∈I such that we have(31)Φ2ψ1w,u;KΦ2ψ2w,u;K=ψ1″η2ψ2″η2,where denominators should not be zero.Proof.
Let us define the functiong by g=λ1ψ1−λ2ψ2, where λ1 and λ2 are given by(32)λ1=Φ2ψ1w,u;K,λ2=Φ2ψ2w,u;K.
Then,g∈C2I; by applying Theorem 10 for g, it follows that there exists η2 such that we have(33)0=λ1ψ1″η2−λ2ψ2″η2Φ2x2w,u;K.
From this, one can get the required equation.
Next, we define functionalΠjψw,u;K,j=1,2, from nonnegative differences of Opial-type inequalities for superquadratic functions given in Theorems 6 and 7 as follows:(34)Π1ψw,u;K=ψK∫abutp2dt1/p2−K2Mp2∫abψ′wxuxp2wxdx,Π2ψw,u;K=∫abψKb−a1/p2utdt−K2Mb−ap2∫abψ′wxuxp2wxdx.Remark 2.
Under the assumptions of Theorems6 and 7, we have Πjψw,u;K≥0 for j=1,2.Theorem 12.
With the same assumptions of Theorem6 on ψ, furthermore, if I⊆0,∞ is a compact interval and ψ∈C2I, then there exists η3∈I such that we have(35)Π1ψw,u;K=η3ψ″η3−ψ′η33η32Π1x3w,u;K.Proof.
By replacingψ with ψ1 from Lemma 3 in Theorem 6, we get the following inequality:(36)∫abM′wx2−ψ′wxuxp2wxdx≤p2MK2M′3K∫abutp2dt1/p23−ψK∫abutp2dt1/p2.
From inequality (36), after simplification, one can obtain(37)Π1ψw,u;K≤M′3Π1x3w,u;K.
Similarly, if we applyψ2 from Lemma 3 instead of ψ in Theorem 6, then the following inequality holds:(38)Π1ψw,u;K≥m′3Π1x3w,u;K.
The above two inequalities lead to the following inequality:(39)m′≤3Π1ψw,u;KΠ1x3w,u;K≤M′.
Therefore, there existsη3∈I such that the following equation is obtained:(40)η3ψ″η3−ψ′η3η32=3Π1ψw,u;KΠ1x3w,u;K,which gives the required equation.Theorem 13.
With the same assumptions onψ1andψ2 as given for ψ in Theorem 6, furthermore, if I⊆0,∞ is a compact interval and ψ1,ψ2∈C2I where Π1x3w,u≠0, then there exists η3∈I such that we have(41)Π1ψ1w,u;KΠ1ψ2w,u;K=η3ψ1″η3−ψ1η3η3ψ2″η3−ψ2η3,where denominators should not be zero.Proof.
Let us define the functionh by h=ω1ψ1−ω2ψ2, where ω1 and ω2 are given by(42)ω1=Π2ψ1w,u;K,ω2=Π2ψ2w,u;K.
Then,h∈C2I; by applying Theorem 12 for h, it follows that there exists η3 such that we have(43)0=ω1η3ψ1″η3−ψ1η33η32−ω2η3ψ2″η3−ψ2η33η32Π2x3w,u;K.
From this, one can get the required equation.Theorem 14.
With the same assumptions of Theorem7 on ψ, furthermore, if I⊆0,∞ is a compact interval and ψ∈C2I, then there exists η4∈I such that the following result holds:(44)Π2ψw,u;K=η4ψ″η4−p2−1ψη42p2η42p2−1Π2x2p2w,u;K.Proof.
By replacingψ with ψ1 from Lemma 3 in Theorem 4, we get the following inequality:(45)∫abM′p2wx2p2−1−ψ′wxuxp2wxdx≤p2b−aMK2×M′2∫abKb−a1/p2utdt2p2−∫abψKb−a1/p2utdt.
Rest of the proof can be followed from the proof of Theorem12.Theorem 15.
With the same assumptions onψ1andψ2 as given for ψ in Theorem 7, furthermore, if I⊆0,∞ is a compact interval and ψ1,ψ2∈C2I where Π1x2p2w,u;K≠0, then there exists η4∈I such that we have(46)Π1ψ1w,u;KΠ2ψ2w,u;K=η4ψ1″η4−p2−1ψ1′η4η4ψ2″η4−p2−1ψ2′η4,where denominators should not be zero.Proof.
The proof can be followed from the proof of Theorem13.


## 3. Fractional Versions of Mean Value Theorems

In this section, we give applications of results proved in Section2 by choosing particular kernels. We get fractional versions of mean value theorems by applying definitions of fractional integral/derivative operators. First, we give results for Riemann–Liouville fractional integrals.Theorem 16.
With same assumptions of Theorem4 on ψ, furthermore, let I⊆0,∞ be a compact interval and ψ∈C2I. Also, let u∈L1a,b be a Riemann–Liouville fractional integral of order γ, where γ≥1; then, there exists η1∈I such that the following result holds:(47)Φ1ψIa+γu,u;b−aγ−1Γγ=η1ψ″η1−ψ′η13η12Φ1x3Ia+γu,u;b−aγ−1Γγ.Proof.
Let us define the kernelkx,t as follows:(48)kx,t=x−tγ−1Γγ,t∈a,x,0,t∈x,b.
Furthermore, we takew as follows:(49)wx=Ia+γux=1Γγ∫axx−tγ−1utdt.
It is clear that, forγ≥1, 0≤kx,t≤x−aγ−1/Γγ≤b−aγ−1/Γγ, t∈a,x, x∈a,b. By choosing K=b−aγ−1/Γγ and applying Theorem 8 for the kernel given in (48), one can get the required result.Theorem 17.
With the same assumptions of Theorem4 on ψ1andψ2, furthermore, let I⊆0,∞ be a compact interval and ψ1,ψ2∈C2I. Also, let u∈L1a,b be a Riemann–Liouville fractional integral of order γ where γ≥1; then, there exists η1 such that we have(50)ϕ1ψ1Ia+γu,u;b−aγ−1/Γγϕ1ψ2Ia+γu,u;b−aγ−1/Γγ=η1ψ1″η1−ψ1′η1η1ψ2″η1−ψ2′η1,where denominators should not be zero.Proof.
It is easy to prove by applying Theorem9 for the kernel defined in (48) and using the function w given by (49).Theorem 18.
With the same assumptions of Theorem5 on ψ, furthermore, let I⊆0,∞ be a compact interval and ψ∈C2I. Also, let u∈L1a,b be a Riemann–Liouville fractional integral of order γ where γ≥1; then, there exists η2∈I such that the following result holds:(51)Φ2ψIa+γu,u;b−aγ−1Γγ=ψ″η22Φ2x2Ia+γu,u;b−aγ−1Γγ.Proof.
The proof is similar to the proof of Theorem16.Theorem 19.
With the same assumptions of Theorem5 on ψ1andψ2, furthermore, let I⊆0,∞ be a compact interval and ψ1,ψ2∈C2I. Also, let u∈L1a,b be a Riemann–Liouville fractional integral of order γ where γ≥1; then, there exists η2 such that we have(52)Φ2ψ1Ia+γu,u;b−aγ−1/ΓγΦ2ψ2Ia+γu,u;b−aγ−1/Γγ=ψ1″η2ψ2″η2,where denominators should not be zero.Proof.
It is easy to prove by applying Theorem11 for the kernel defined in (48) and using the function w given by (49).Theorem 20.
With the same assumptions of Theorem6 on ψ, furthermore, let I⊆0,∞ be a compact interval and ψ∈C2I. Let u∈L1a,b be a Riemann–Liouville fractional integral of order γ where γ≥1/p2; then, there exists η3∈I such that the following result holds:(53)Π1ψIa+γu,u;b−aγ−1/p2Γγp1γ−1+11/p1=η3ψ″η1−ψ′η33η32×Π1x3Ia+γu,u;b−aγ−1/p2Γγp1γ−1+11/p1.Proof.
Let us consider the kernelkx,t as defined in (48) and w given in (49). If we set Qx=∫axkx,tp1dt1/p1, then Qx=x−aγ−1/p2/Γγp1γ−1+11/p1. Furthermore, Q is increasing for γ>1/p2, on a,b; therefore, we have Qx≤b−aγ−1/p2/Γγp1γ−1+11/p1=K. By applying Theorem 12, we get the required result.Theorem 21.
With the same assumptions of Theorem6 on ψ1andψ2, furthermore, let I⊆0,∞ be a compact interval and ψ1,ψ2∈C2I. Furthermore, let u∈L1a,b be a Riemann–Liouville fractional integral of order γ. If γ>1/p2 and Π2x3Ia+γu,u≠0, then there exists η3∈I such that the following equality holds:(54)Π1ψ1Ia+γu,u;b−aγ−1/p2/Γγp1γ−1+11/p1Π1ψ2Ia+γu,u;b−aγ−1/p2/Γγp1γ−1+11/p1=η3ψ1″η3−ψ1η3η3ψ2″η3−ψ2η3,where denominators should not be zero.Proof.
It is easy to prove by applying Theorem13 for the kernel defined in (48) and using the function w given by (49).Theorem 22.
With the same assumptions of Theorem7 on ψ, furthermore, let I⊆0,∞ be a compact interval and ψ∈C2I. Let u∈L1a,b be a Riemann–Liouville fractional integral of order γ. If γ>1/p2, then there exists η4∈I such that the following result holds:(55)Π2ψIa+γu,u;b−aγ−1/p2Γγp1γ−1+11/p1=η4ψ″η4−p2−1ψ′η42p2η42p2−1×Π2x2p2Ia+γu,u;b−aγ−1/p2Γγp1γ−1+11/p1.Proof.
The proof is similar to the proof of Theorem20.Theorem 23.
With the same assumptions of Theorem7 on ψ1andψ2, furthermore, let I⊆0,∞ be a compact interval and ψ1,ψ2∈C2I. Let u∈L1a,b be a Riemann–Liouville fractional integral of order γ. If γ>1/p2 and Π2x2p2Ia+γu,u≠0, then there exists η4∈I such that the following equality holds:(56)Π1ψ1Ia+γu,u;b−aγ−1/p2/Γγp1γ−1+11/p1Π1ψ2Ia+γu,u;b−aγ−1/p2/Γγp1γ−1+11/p1=η4ψ1″η4−p2−1ψ1η4η4ψ2″η4−p2−1ψ2η4,where denominators should not be zero.Proof.
It is easy to prove by applying Theorem15 for the kernel defined in (48) and using the function w given by (49).
Next, we give the results for Caputo fractional derivatives using their composition identities.Theorem 24.
With the same assumptions of Theorem4 on ψ, furthermore, let I⊆0,∞ be a compact interval and Ψ∈C2I. Also, let l=μ+1 and m=γ+1, for γ,μ∉N0 and u∈ACma,b such that uja=0 for j=m,m+1,…,l−1. Let DCa+μu,DCa+γu∈L1a,b. Then, for γ≤μ−1, the following result holds:(57)Φ1ψDCa+γu,DCa+μu;b−aμ−γ−1Γμ−γ=η1ψ1″η1−ψ1′η12×Φ1x3DCa+γu,DCa+μu;b−aμ−γ−1Γμ−γ.Proof.
Let us consider the kernelkx,t as follows:(58)kx,t=x−tμ−γ−1Γμ−γ,t∈a,x,0,t∈x,b.
Furthermore, we takew as follows:(59)wx=DCa+γux=1Γμ−γ∫axx−tμ−γ−1DCa+μutdt.
It is clear that, forγ≤μ−1, 0≤kx,t≤x−aμ−γ−1/Γμ−γ≤b−aμ−γ−1/Γμ−γ, t∈a,x, x∈a,b. By choosing K=b−aμ−γ−1/Γμ−γ and applying Theorem 8 for the kernel given in (59), one can get the required result.Theorem 25.
With the same assumptions of Theorem4 on ψ1andψ2, furthermore, let I⊆0,∞ be a compact interval and ψ1,ψ2∈C2I. Let l=μ+1 and m=γ+1, for γ,μ∉N0. Also, let DCa+μu,DCa+μu∈L1a,b, γ≤μ−1, and ϕ1x3Da+γu,Da+μu≠0. Then, there exists η1∈I such that we have(60)Φ1ψ1DCa+γu,DCa+μu;b−aμ−γ−1/Γμ−γΦ1ψ2DCa+γu,DCa+μu;b−aμ−γ−1/Γμ−γ=η1ψ1″η1−ψ1′η1η1ψ2″η2−ψ2′η1,where denominators should not be zero.Proof.
It is easy to prove by applying Theorem9 for the kernel defined in (58) and using function w given by (59).Theorem 26.
With the same assumptions of Theorem5 on ψ, furthermore, let I⊆0,∞ be a compact interval and ψ∈C2I. Also, let l=μ+1 and m=γ+1, for γ,μ∉N0, and u∈ACma,b such that uja=0 for j=m,m+1,…,l−1. Let DCa+μu,DCa+γu∈L1a,b. Then, the following result holds for γ≤μ−1:(61)Φ2ψDCa+γu,DCa+μu;b−aμ−γ−1Γμ−γ=ψ″η22Φ2x2DCa+γu,DCa+μu.Proof.
The proof is similar to the proof of Theorem24.Theorem 27.
With the same assumptions of Theorem5 on ψ1andψ2, furthermore, let I⊆0,∞ be a compact interval and ψ1,ψ2∈C2I. Let l=μ+1 and m=γ+1, for γ,μ∉N0. Also, let DCa+μu,DCa+μu∈L1a,b, γ≤μ−1, and ϕ2x2Da+γu,Da+μu≠0. Then, there exists η2∈I such that we have(62)Φ2ψ1DCa+γu,DCa+μu;b−aμ−γ−1/Γμ−γΦ2ψ2DCa+γu,DCa+μu;b−aμ−γ−1/Γμ−γ=ψ1″η2ψ2″η2,where denominators should not be zero.Proof.
It is easy to prove by applying Theorem11 for the kernel defined in (58) and using function w given by (59).Theorem 28.
With the same assumptions of Theorem6 on ψ, furthermore, let I⊆0,∞ be a compact interval and ψ∈C2I. Also, let l=μ+1 and m=γ+1, for γ,μ∉N0, and u∈ACma,b such that uja=0 for j=m,m+1,…,l−1. Let DCa+μu,DCa+γu∈L1a,b. Then, for γ<μ−1/p2, the following result holds:(63)Π1ψDCa+γu,DCa+μu;b−aμ−γ−1/p2Γμ−γp1γ−1+11/p1=η3ψ″η3−ψ′η33η32×Π1x3DCa+γu,DCa+μu;b−aμ−γ−1/p2Γμ−γp1γ−1+11/p1.Proof.
Let us consider the kernelkx,t as defined in (58) and w given in (59). If we set Qx=∫axkx,tp1dt1/p1, then Qx=x−aμ−γ−1/p2/Γμ−γp1μ−γ−1+11/p1. Furthermore, Q is increasing for γ<μ−1/p2, on a,b; therefore, we have Qx≤b−aμ−γ−1/p2/Γμ−γp1μ−γ−1+11/p1=K. By applying Theorem 12, we get the required result.Theorem 29.
With the same assumptions of Theorem6 on ψ1andψ2, furthermore, let I⊆0,∞ be a compact interval and ψ1,ψ2∈C2I. Let l=μ+1 and m=γ+1, for γ,μ∉N0. Also, let DCa+μu,DCa+μu∈L1a,b, γ≤μ−1, and Π2x3Da+γu,Da+μu≠0. Then, there exists η3∈I such that, for γ<μ−1/p2, we have(64)Π1ψ1DCa+γu,DCa+μu;b−aμ−γ−1/p2/Γμ−γp1γ−1+11/p1Π1ψ2DCa+γu,DCa+μu;b−aμ−γ−1/p2/Γμ−γp1γ−1+11/p1=η3ψ1″η3−ψ1′η3η3ψ2″η3−ψ2′η3,where denominators should not be zero.Proof.
It is easy to prove by applying Theorem13 for the kernel defined in (58) and using function w given by (59).Theorem 30.
With the same assumptions of Theorem7 on ψ, furthermore, let I⊆0,∞ be a compact interval and ψ∈C2I. Also, let l=μ+1 and m=γ+1, for γ,μ∉N0, and u∈ACma,b such that uja=0 for j=m,m+1,…,l−1. Let DCa+μu,DCa+γu∈L1a,b. Then, for γ<μ−1/p2, the following result holds:(65)Π2ψDCa+γu,DCa+μu;b−aμ−γ−1/p2Γμ−γp1γ−1+11/p1=η4ψ″η4−p2−1ψ1η42p2η42p2−1×Π2x2p2DCa+γu,DCa+μu;b−aμ−γ−1/p2Γμ−γp1γ−1+11/p1.Proof.
The proof is similar to the proof of Theorem28.Theorem 31.
With the same assumptions of Theorem7 on ψ1andψ2, furthermore, let I⊆0,∞ be a compact interval and ψ1,ψ2∈C2I. Let l=μ+1 and m=γ+1, for γ,μ∉N0. Also, let DCa+μu∈Lqa,b and DCa+μ∈L1a,b, γ≤μ−1, and Πjx2p2Da+γu,Da+μu≠0. Then, there exists η4∈I such that, for γ<μ−1/p2, we have(66)Π2ψ1DCa+γu,DCa+μu;b−aμ−γ−1/p2/Γμ−γp1γ−1+11/p1Π2ψ1DCa+γu,DCa+μu;b−aμ−γ−1/p2/Γμ−γp1γ−1+11/p1=η4ψ1″η4−p2−1ψ1′η4η4ψ2″η4−p2−1ψ2′η4,where denominators should not be zero.Proof.
It is easy to prove by applying Theorem15 for the kernel defined in (58) and using function w given by (59).



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*Source: 6379883-2021-06-30.xml*