# q,h-Opial-Type Inequalities via Hahn Operators

Authors: Ammara Nosheen; Anum Saba; Khuram Ali Khan; Michael Kikomba Kahungu
Journal: Discrete Dynamics in Nature and Society
 (2022)

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## Abstract

In this paper, the well-known Hölder’s inequality is proved via Hahn differential and integral operators, which is a helping tool to establish some Opial-type inequalities via Hahn’s calculus. The weight functions involved in these Opial-type inequalities are positive and monotone. In search of applications, some new as well as some existing inequalities in the literature are obtained by applying suitable limits.


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## Body

## 1. Introduction

In 1960, first time Opial’s inequality was founded by Opial [1]. He established the following important integral inequality.(1)∫cdvtv′tdt≤d−44∫cdv′t2dt,where v is an absolutely continuous function on c,d and vc=vd=0, the constant d−c/4 is the most suitable. Equality (1) holds, if and only if(2)vt=at−c,c≤v≤d−c2;ad−t,d−c2≤v≤d,,where a is a constant.Furthermore, the proof of Opial’s inequality is simplified by Olech [2], Beescak [3], Levison [4], Pederson [5], Mallows [6]. Levison [7] proved that if v is an absolutely continuous function on 0,d with v0=0, in (1), then,(3)∫0dvtv′tdt≤d2∫0dv′t2dt.For the extension of (1), Beescak [3] demonstrated that if u is an absolutely continuous function on c,x1 with uc=0, then,(4)∫cx1utu′tdt≤12∫cx11ptdt∫cx1ptu′t2dt,where pt is a continuous and positive function with ∫cx1dt/ptdt<∞, and if ud=0, then,(5)∫x1dutu′tdt≤12∫x1d1ptdt∫x1dptu′t2dt.Keng [8] generalized the inequality (3) in the following form: If u is an absolutely continuous function with uc=0, then,(6)∫cdutju′tdt≤d−cjj+1∫cdu′tj+1dt,where j is a positive integer. Extensions of Beescak’s inequalities (4) and (5) are proved by Yang [9]. He assumed that if u is absolutely continuous on c,x1 with uc=0, then,(7)2∫cx1mtutu′tdt≤∫cx11rtdt∫cx1rtmtu′t2dt,where r is continuous and positive function with ∫cx11/rtdt<∞ and m is positive, bounded and nonincreasing function on c,x1. If ud=0, then,(8)2∫x1dmtutu′tdt≤∫x1d1rtdt∫x1drtmtu′t2dt,where m is positive, bounded and nondecreasing function on x1,d. Further he assumed that if u is an absolutely continuous function on c,d with uc=ud=0, then,(9)∫cdmtutu′tdt≤12K∫cdrtmtu′t2dt,where K is defined by K=∫cx1dt/rt=∫x1ddt/rt and mt is a positive bounded function on interval c,d. Further mt is nonincreasing on c,x1 and mt is nondecreasing on x1,d. Also, another extension of Opial’s inequality is given by Yang [9] is generalization of (6). He proved that, if uc=ud=0, then,(10)∫cdutju′tkdt≤kj+kd−c2j∫cdu′tj+kdt,where j,k≥1. The following inequality is presented by Lee [10], which is a generalization of the inequality (7). It demonstrates that if u is an absolutely continuous function on c,x1 with given condition uc=0, then,(11)j+k∫cx1rkk−1xmkxuxju′xkdx≤k∫cx1r−kxdxj∫cx1rkj+k−1xmkxu′xj+kdx,where j≥0,k≥1 and r is positive with ∫cx1dt/rkt<∞ and m is nonincreasing and positive on c,x1. He also proved a generalization of inequality (8) and suppose that if u is an absolutely continuous function on x1,d and ud=0, then,(12)j+k∫x1drkk−1xmkxuxju′xkdx≤k∫x1dr−kxdxj∫x1drkj+k−1xmkxu′xj+kdx,where k≥1 and j≥0 and r⋅ is positive function with ∫x1ddt/rkt<∞ and m⋅ is nondecreasing and positive on x1,d. Lee [10] has combined (11) and (12) with uc=ud=0, to find extension of (9) in the following form:(13)j+k∫cdrkk−1xmkxuxju′xkdx≤kC∫cdrkj+k−1xmkxu′xj+kdx,where C=∫cx1r−kxdxj=∫x1dr−kxdxj. Further discussion on Opial-type inequalities may include the work: some Opial inequalities in q-calculus [11] by Mirković, q-Opial-type inequality by Alp, et al., in [12], refinements of Opial-type inequalities in two variables [13], Opial-type inequalities for conformable fractional integrals [14,15], dynamic Opial inequalities on time scales [16–19], Opial-type inequalities in (p, q)-calculus by Li et al., in [20], interval valued Opial-type inequalities by Zhao et al., in [21].The history of quantum calculus is 300 years old. It is considered the most difficult subject to engage in mathematics by Bernoulli and Euler. The quantum has been derived from “quantus” a Latin word meaning “how much”, commonly quantum deals with the measurement to its smallest unit. Quantum calculus treats the sets of nondifferentiable functions without using limits.Theh-calculus h>0, deals with calculus of finite differences (Boole [22]). For the study of h−calculus, readers are suggested to Milne Thomson [23]. Hahn’s calculus unifies h-calculus and q-calculus, initiated by Hahn [24]. It is utilized to construct families of orthogonal polynomials and to deal with some approximation problems [25].In 2015, Saker et al. ([18], Theorems 3.3 and 3.4) initiated the study of dynamic versions of (11) and (12) on time scales (a time scale is a closed subset of real line). In 2019, Fatma et al., [16] have also studied (11)–(13) with the help of time scales calculus. In this paper, we present q,h−analogues of Opial-type inequalities proved in [10,16,18] with the help of Hahn’s calculus.The paper is arranged as follows: In Section 2, some basic concepts of Hahn’s calculus and useful lemmas are presented. In Section 3, Hölder inequality and some Opial-type inequalities for monotonic functions via Hahn integrals (also called Jackson Nörlund integrals) are established. Section 4 consists of the conclusion of the paper.


## 2. Some Essentials of Hahn’s Calculus

The generalization of both quantum calculus andh-calculus is another kind of quantum calculus called Hahn’s quantum calculus. Hahn’s difference operator is defined by Wolfgang Hahn in 1945 [24].Definition 1.
Leth>0and0<q<1. Defineh∘=h/1−qand letIbe any interval inRcontainingh∘. Supposeu:I⟶R. Theq,h−derivativeofuis described by(14)Dq,hut1=uqt1+h−ut1qt1+h−t1,where t1≠h°, if t1=h°, then, Dq,huh°=u′h°. Dq,h is the q,h−derivative of u at t1 and u is q,h−differential over I, if Dq,hut1 exists for all t1∈I. Generally, when q⟶1, we get h-derivative of u⋅ at t1.(15)Δhut1=ut1+h−ut1h.
Whenh=0, we get q-derivative of u⋅ at t1.(16)Dqut1=uqt1−ut1q−1t1.
Ift1≠0 and Dqu0=u′0, on condition that u′0 exists. It is noted that, by making both restrictions q⟶1,h=0 simultaneously we get usual derivative of u⋅ at t1.(17)limq⟶1Dq,0ut1=u′t1.
The arithmetical properties of Hahn-differential are simply concluded in the following theorem, which is given in [26].Theorem 1.
Assumeu,vboth are Hahn-differentiable (q,h-differentiable) att1∈I, then,(18)Dq,hut1=0;where u is a constant.(19)Dq,hu+vt1=Dq,hut1+Dq,hvt1;(20)Dq,huvt1=Dq,hut1vt1+uqt1+hDq,hvt1;(21)uqt1+h=ut1+t1q−1+hDq,hut1;(22)Dq,huvt1=Dq,hut1vt1−ut1Dq,hvt1vt1vqt1+h;
Provided byvt1vqt1+h≠0.Theorem 2.
(Chain Rule involving Hahn-differential operator) (see [27]).
Consideru:I⟶R is Hahn−differentiable and continuous. Let v:R⟶R is continuously differentiable. Then, there must exists c between t1 and qt1+h, such that,(23)Dq,hvout1=v′ucDq,hut1.
The right inverse of Hahn-differential operator ([28], Chapter 6) is as follows:Definition 2.
LetIbe any closed interval of real numbers, which containsh°. Suppose thatg:I⟶Randc,d∈I, such that,c<d. TheHahn−integralofgfromctodis defined by,(24)∫cdgt1dq,ht1=∫h°dgt1dq,ht1−∫h°cgt1dq,ht1,where(25)∫h°xgt1dq,ht1=x1−q−h∑k=0∞qkgxqk+hkq,x∈I,
kq=1−qk/1−q and the series on the right hand side is convergent at x=c, x=d.
The following properties of Hahn’s integral are given in ([28], Lemma 6.2.2).Lemma 1.
Assumeu,v:I⟶RareHahn−integrablefunctions on intervalI,l∈Randc,d,λ∈I, wherelis a constant. Then,(26)∫ccut1dq,ht1=0;(27)∫cdlut1dq,ht1=l∫cdut1dq,ht1;(28)∫cdut1dq,ht1=−∫dcut1dq,ht1;(29)∫cdut1dq,ht1=∫cλut1dq,ht1+∫λdut1dq,ht1;(30)∫cdut1+vt1dq,ht1=∫cdut1dq,ht1+∫cdvt1dq,ht1.
Next result can be found in ([28], Lemma 6.2.8).Lemma 2.
Assumeu,v:I⟶RareHahn−integrableandy∈I. Denotemq=1−qm/1−q. Ifut≤vt∀t,c,d∈yqm+hmqm=0∞, then,(31)∫cdutdq,ht≤∫cdvtdq,ht.
In particular, (31) leads to the following inequality. If u:I⟶R is Hahn−integrable, y∈I, then ∀t,c,d∈yqm+hmqm=0∞, one has that,(32)∫cdutdq,ht≤∫cdutdq,ht.
The followingq,h integration by parts formula can be found in ([28], Lemma 6.2.8).Lemma 3.
Ifu,v:I⟶Rare continuous ath°. Then,(33)∫cdut1Dq,hvt1dq,ht1=ut1vt1|cd−∫cdDq,hut1vqt1+hdq,ht1.


## 3. Opial Inequalities for Monotone Functions

### 3.1. Hölder’s Inequality via Hahn’s Calculus

The Hölder inequality plays a fundamental role in the field of mathematics. Different variants can be found in [29,30]. All through this section, j and i are conjugate to each other’s, as 1/j+1/i=1. In order to extend Opial-type inequalities by using Hahn calculus, we first prove the Hölder’s inequality involving Hahn calculus.Theorem 3.
Ifa,b∈Iandu,v:a,b⟶Rare continuous functions, then,(34)∫abutvtdq,ht≤∫abutjdq,ht1/i∫abvtidq,ht1/i,where j>1 and i=j/j−1.Proof.
For nonnegativeα,β real numbers, the basic inequality holds.(35)α1/jβ1/i≤αj+βi,
Now consider,(36)∫abutjdq,ht∫abvtidq,ht≠0.
In (35), choose,(37)αt=utj∫abuτjdq,hτ,βt=vti∫abvτidq,hτ.
Use (35) and integrating the resultant inequality from a to b to obtain(38)∫abα1/jtβ1/itdq,ht=∫abutj∫abuτjdq,hτ1/ivti∫abvτidq,hτ1/idq,ht≤∫ab1jutj∫abuτjdq,hτ+1ivti∫abvτidq,hτdq,ht=1j∫abutj∫abuτjdq,hτdq,ht+1i∫abvti∫abvτidq,hτdq,ht=1.
Hence, we get (34). The proof is complete.
Next, consider the notationsy˜m=yqm+hmq, y^=y+mh and yˇ=yqm, where y∈a,x1,b.Remark 1.
In form of sums (34) can be written as(39)b1−q−h∑m=0∞qmub˜mvb˜m−a1−q−h∑m=0∞qmua˜mva˜m≤b1−q−h∑m=0∞qmub˜mj−a1−q−h∑m=0∞qmua˜mji/j×b1−q−h∑m=0∞qmvb˜mi−a1−q−h∑m=0∞qmva˜mi1/i.
Whenh=0 in (39), it recaptures the Hölder’s inequality in q-calculus [31].(40)∑m=0∞qmbubˇvbˇ−auaˇvaˇ≤∑m=0∞qmbubˇj−auaˇji/j∑m=0∞qmbvbˇi−avaˇi1/i.
By usingq⟶1 in (39), we get Hölder’s inequality in h-discrete calculus, which is ([32], Theorem 3.1).(41)∑m=0∞ua^va^−ub^vb^≤∑m=0∞ua^j−ub^j1/j∑m=0∞va^i−vb^i1/i.When h=0, q⟶1 in (34), we obtain Hölder’s inequality in classical calculus, which can be found in [33].


### 3.2. Opial-Type Inequalities ona,x1

Theorem 4.
LetIbe any interval,y∈Ianda,x1∈yqm+hmqm=0∞. Assumel⋅,s⋅are positive continuous functions ona,x1.s⋅is non-increasing function ona,x1and∫ax1l−kzdq,hz<∞. Furthermore, supposeuis a continuous function ona,x1withua=0. Then,∀j≥1,k≥1, and∀t,z∈yqm+hmqm=0∞, we have,(42)j+k∫ax1lkk−1zskzuzjDq,huzkdq,hz≤k∫ax1l−kzdq,hzj∫ax1lkj+k−1zskzDq,huzj+kdq,hz.Proof.
Let us consider the following integral(43)Tz=∫azlkk−1tsk2/j+ktDq,hutkdq,ht.
SinceTa=0, one has that,(44)Dq,hTz=lkk−1zsk2/j+kzDq,huzk.
Equation (32) implies,(45)uz=uz−ua=∫azDq,hutdq,ht≤∫azDq,hutdq,ht.
Use of Hölder inequality (34) with indices k and k/k−1, provides,(46)uz≤∫azl1−ktlk−1tDq,hutdq,ht≤∫azl1−ktk/k−1dq,htk−1/k∫azlk−1tkDq,hutkdq,ht1/k=∫azl−ktdq,htk−1/k∫azlkk−1tDq,hutkdq,ht1/k.
By taking powerj on both sides, we have(47)uzj≤∫azl−ktdq,htjk−1/k∫azlkk−1tDq,hutkdq,htj/k.
Sinces⋅ is nonincreasing and positive on a,x1, one has that,(48)sjk/j+kzuzj≤sjk/j+kz∫azl−ktdq,htjk−1/k∫azlkk−1tDq,hutkdq,htj/k≤∫azl−ktdq,htjk−1/k∫azlkk−1tsk2/j+ktDq,hutkdq,htj/k=∫azl−ktdq,htjk−1/kTj/kz.
Multiply (48) by lkk−1zsk2/j+kzDq,huzk≥0, to get(49)lkk−1zsjk/j+kzsk2/j+kzuzjDq,huzk≤lkk−1z∫azl−ktdq,htjk−1/kTj/kzsk2/j+kzDq,huzk.
Multiply (49) by j+k, integrate it from a to x1 and use monotonicity of ∫azl−ktdq,ht to obtain(50)j+k∫ax1lkk−1zskzuzjDq,huzkdq,hz≤j+k∫ax1l−ktdq,htjk−1/k∫ax1Tj/kzDq,hTzdq,hz.
Now from Chain rule (23), we have(51)Dq,hTj+k/kz=j+kkTj/kcDq,hTz,c∈z,qz+h.
SinceDq,hTz≥0 and z≤c, one gets(52)Dq,hTj+k/kz=j+kkTj/kcDq,hTz≥j+kkTj/kzDq,hTz.From (50) and (52), it is noted that Ta=0 and(53)j+k∫ax1lkk−1zskzuzjDq,huzkdq,hz≤k∫ax1l−kzdq,hzjk−1/k∫ax1Dq,hTj+k/kzdq,hz,
(by using (43)),(54)≤k∫ax1l−kzdq,hzjk−1/kTj+k/kx1=k∫ax1l−kzdq,hzjk−1/k∫ax1lkk−1zsk2/j+kDq,huzkdq,hzj+k/k.
Denote,(55)Lx1≐∫ax1lkk−1zsk2/j+kDq,huzkdq,hzj+k/k.
By applying Hölder inequality (34) with indices j+k/j and j+k/k on L, one gets(56)L=∫ax1l−jk/j+kzlk21−1/j+kzsk2/j+kzDq,huzkdq,hzj+k/k≤∫ax1l−jk/j+kzj+k/jdq,hzj/k∫ax1lk21−1/j+kzsk2/j+kzDq,huzkj+k/kdq,hz=∫ax1l−kzdq,hzj/k∫ax1lkj+k−1zskzDq,huzj+kdq,hz.
By combining (54) and (56), we get(57)j+k∫ax1lkk−1zskzuzjDq,hzkdq,hz≤∫ax1l−kzdq,hzj∫ax1lkj+k−1zskzDq,huzj+kdq,hz.
Hence, (42) is obtained.Corollary 1.
Case 1.Whenq⟶1, then (42) reduces to the following Opial inequality inh−discretecalculus.(58)∑m=0∞lkk−1a^ska^uja^Δhua^k−lkk−1x^1skx^1ujx^1Δhux^1k≤L1∑m=0∞lkj+k−1a^ska^Δhua^j+k−lkj+k−1x^1skx^1Δhux^1j+k,where L1=khj/j+k∑m=0∞l−ka^−l−kx^1j.Case 2.
Whenh=0in (42), it is converted to the following Opial inequality in q-calculus.(59)∑m=0∞qkx1lkk−1xˇ1skxˇ1ujxˇ1Δhuxˇ1k−alkk−1aˇskaˇujaˇΔhuaˇk≤L2∑m=0∞qkx1lkj+k−1xˇ1skxˇ1Δhuxˇ1j+k−alkj+k−1aˇskaˇΔhuaˇj+k,where L2=k1−qj/j+k∑m=0∞qkx1l−kxˇ1−al−kaˇjCase 3.
Whenq⟶1andh=0in (42). We get (11), which is ([10], Theorem 1.1), ([16], Corollary 3.2).(60)j+k∫ax1lkk−1zskzuzju′zkdz≤k∫ax1l−kxdzj∫ax1ljj+k−1zskzu′zj+kdz.Case 4.
Whenq⟶1,h=0andj=k=1. We obtain (7) given by Yang in ([9], Theorem 3), ([16], Remark 3.3).(61)2∫ax1szuzu′zdz≤∫ax1l−1zdz∫ax1lzszu′z2dz.Remark 2.
Ifh=1in (26), then, it recaptures Opial-type discrete inequality given in ([16], Corollary 1).


### 3.3. Opial-Type Inequalities onx1,b

Theorem 5.
LetIbe any interval,y∈Iandx1,b∈yqm+hmqm=0∞. Assumel⋅,s⋅are positive continuous functions onx1,b.s⋅is nondecreasing onx1,band∫x1bl−kzdq,hz<∞. Furthermorer, assumeu⋅is a continuous function onx1,bwithub=0, then,∀j≥1,k≥1,z,t∈yqm+hmqm=0∞, we have,(62)j+k∫x1blkk−1zskzuqz+hjDq,huzkdq,hz≤k∫x1bl−kzdq,hzj∫x1blkj+k−1zskzDq,huzj+kdq,hz.Proof.
Let us consider the following integral:(63)Tz≐−∫zblkk−1tsk2/j+ktDq,hutkdq,ht.
By usingTb=0, we get(64)Dq,hTz=lkk−1zsk2/j+kzDq,huzk.
Use of (32) gives(65)uqz+h=−uqz+h=ub−uqz+h=∫qz+hbDq,hutdq,ht≤∫qz+hbDq,hutdq,ht.
Use of (34) with indices k and k/k−1 implies(66)uqz+h≤∫qz+hbl1−ktlk−1tDq,hutdq,ht≤∫qz+hbl1−ktk/k−1dq,htk−1/k∫qz+hblk−1tkDq,hutkdq,ht1/k=∫qz+hbl−ktdq,htk−1/k∫qz+hblkk−1tDq,hutkdq,ht1/k.
By taking powerj on both sides, one gets(67)uqz+hj≤∫qz+hbl−ktdq,htjk−1/k∫qz+hblkk−1tDq,hutkdq,htj/k.
Sinces⋅ is nondecreasing and positive on x1,b, therefore,(68)sjk/j+kzuqz+hj≤sjk/j+kz∫qz+hbl−ktdq,htjk−1/k∫qz+hblkk−1tDq,hutkdq,htj/k≤∫qz+hbl−ktdq,htjk−1/k∫qz+hblkk−1tsk2j+ktDq,hutkdq,htj/k,
((63)implies),(69)=∫qz+hbl−ktdq,htjk−1/k−Tqz+hj/k.
Multiply both sides of (69) by lkk−1zsk2/j+kzDq,huzk≥0, to find,(70)lkk−1zsk2/j+kzsjk/j+kzuqz+hjDq,huzk≤lkk−1z∫qz+hbl−ktdq,htjk−1/k−Tqz+hj/ksk2/j+kzDq,huzk.
(70) can be written in the following subsequent form,(71)lkk−1zskzuqz+hjDq,huzk≤∫qz+hbl−ktdq,htjk−1/k−Tqz+hj/kDq,hTz.
Multiply (71) by j+k, integrate from x1 to b and use monotonicity of ∫qz+hbl−ktdq,ht to find,(72)j+k∫x1blkk−1zskzuqz+hjDq,huzkdq,hz≤j+k∫x1bl−ktdq,htjk−1/k∫x1b−Tqz+hj/kDq,hTzdq,hz.
Now, from Chain rule (23), we have(73)−Dq,h−Tj+k/kz=j+kk−Tj/kcDq,hTz,c∈z,qz+h.
SinceDq,hTz≥0 and c≤qz+h, one gets(74)Dq,h−−Tj+k/kz=j+kk−Tj/kcDq,hTz≥j+kk−Tqz+hDq,hTz.
It is noted thatTb=0. (72) and (74) yields,(75)j+k∫x1blkk−1zskzuqz+hjDq,huzkdq,hz≤k∫x1bl−kxdq,hzjk−1/k∫x1bDq,h−−Tj+k/kzdq,hz≤k∫x1bl−kzdq,hzjk−1/k−Tj+k/kx1,
(by using (70)),(76)=k∫x1bl−kzdq,hzjk−1/k∫x1blkk−1zsk2/j+kzDq,huzkdq,hzj+k/k.
Again use (34) with indices j+k/k and j+k/j to get(77)∫x1blkk−1zsk2/j+kDq,huzkdq,hzj+k/k=∫x1bl−jk/j+kzlk21−1/j+kzsk2/j+kzDq,huzkdq,hzj+k/k≤∫x1bl−jk/j+kzj+k/jdq,hzj/k∫x1blk21−1/j+kzsk2/j+kzDq,huzkj+k/kdq,hz=∫x1bl−kzdq,hzj/k∫x1blkj+k−1zskzDq,huzj+kdq,hz.
By combining (76) and (77), we get(78)j+k∫x1blkk−1zskzuqz+hjDq,huzkdq,hz≤k∫x1bl−kzdq,hzj∫x1blkj+k−1zskzDq,huzj+kdq,hz.
Hence, (62) is proved.Corollary 2.
Case I: Whenq⟶1, then, (62) is converted to the following Opial’s inequality inhdiscrete-calculus.(79)∑m=0∞lkk−1x^1skx^1ujx^1Δhux^1k−lkk−1b^skb^ujb^Δhub^k≤L3∑m=0∞lkj+k−1x^1skx^1Δhux^1j+k−lkj+k−1b^skb^Δhub^j+k,where L3=khj/j+k∑m=0∞l−kx^1−l−kb^j.
Moreover, ifh=1 in (79), it recaptures ([16], Corollary 3.8).
Case II.Whenh=0in (62), it is converted to the following Opial inequality in q-calculus.(80)∑m=0∞qkblkk−1bˇskbˇujbˇΔhubˇk−x1lkk−1xˇ1skxˇ1ujxˇ1Δhuxˇ1k≤L4∑m=0∞qkblkj+k−1bˇskbˇΔhubˇj+k−x1lkj+k−1xˇ1skxˇ1Δhuxˇ1j+k,where L4=k1−qj/j+k∑m=0∞qkbl−kb−x1l−kxˇ1j.
Case III.Whenq⟶1andh=0in (62), we obtain (12) ([10], Theorem 1.2), ([16], Corollary 3.6).(81)j+k∫x1blkk−1skxuzju′zkdz≤k∫x1bl−kzdzj∫x1blkj+k−1zskzu′zj+kdz.
Case IV:Ifq⟶1, h=0 and j=k=1 in (62). We obtain (8) ([9], Theorem 3), ([16], Remark 3.7).(82)2∫x1bszuzu′zdz≤∫x1bl−1zdz∫x1blzszu′z2dz.


### 3.4. Opial-Type Inequalities ona,b

Theorem 6.
SupposeIis any interval,y∈Ianda,b∈yqm+hmqm=0∞. Assumel⋅,s⋅are positive continuous functions ona,b.s⋅is nonincreasing ona,x1ands⋅is nondecreasing onx1,b. SupposeC=max∫ax1l−kzdq,hzj,∫x1bl−kzdq,hzj. Furthermorer, chooseu⋅to be a nondecreasing continuous function ona,bwith conditionsua=ub=0. Then,∀j≥1,k≥1, we hav,(83)j+k∫ablkk−1zskzuzjDq,huzkdq,hx≤kC∫ablkj+k−1zskzDq,huzj+kdq,hz.Proof.
Let us consider the following integral:(84)j+k∫ablkk−1zskzuzjDq,huzkdq,hz=j+k∫ax1lkk−1zskzuzjDq,huzkdq,hz+j+k∫x1blkk−1zskzuzjDq,huzkdq,hz.
Sinces is nonincreasing on a,x1, and s is nondecreasing on x1,b, therefore by using (42) and (62) in (84), one gets(85)j+k∫ablkk−1zskzuzjDq,huzkdq,hz≤j+k∫ax1lkk−1zskzuzjDq,huzkdq,hz+j+k∫x1blkk−1zskzuqz+hjDq,huzkdq,hz≤k∫ax1l−kzdq,hzj∫ax1lkj+k−1zskzDq,huzj+kdq,hz+k∫x1bl−kzdq,hzj∫x1blkj+k−1zskzDq,huzj+kdq,hz≤kC∫ax1lkj+k−1zskzDq,huzj+kdq,hz+∫x1blkj+k−1zskzDq,huzj+kdq,hz=kC∫ablkj+k−1zskzDq,huzj+kdq,hz,which is the required inequality (83).Corollary 3.
Case I: Whenq⟶1, then, the inequality (83) becomes the Opial’s inequality inh−calculus.(86)∑m=0∞lkk−1a^ska^uja^Δhua^k−lkk−1b^skb^ujb^Δhub^k≤C1∑m=0∞lkj+k−1a^ska^Δhua^j+k−lkj+k−1b^skb^Δhub^j+k,where C1=maxL1,L3.
Moreover, ifh=1 in (86), it recaptures ([16], Corollary 3.11).
Case II.Whenh=0in (83), it is converted to the following Opial inequality in q-calculus.(87)∑m=0∞qkblkk−1bˇskbˇujbˇΔhubˇk−alkk−1aˇskaˇujaˇΔhuaˇk≤C2∑m=0∞qkblkj+k−1bˇskbˇΔhubˇj+k−alkj+k−1aˇskaˇΔhuaˇj+k,where C2=maxL2,L4.
Case III.Whenq⟶1andh=0in the inequality (83). We get (13) ([10], Theorem 2), ([16], Corollary 3.10).(88)j+k∫ablkk−1zskzuzju′zkdz≤kC∫ablkj+k−1zskzu′zj+kdz.
Case IV.Whenq⟶1, h=0 and j=k=1 in (83). Then, we obtain the Opial’s inequality (9).(89)2∫abszuz‖u′zdz≤C∫ablzszu′z2dz.


## 3.1. Hölder’s Inequality via Hahn’s Calculus

The Hölder inequality plays a fundamental role in the field of mathematics. Different variants can be found in [29,30]. All through this section, j and i are conjugate to each other’s, as 1/j+1/i=1. In order to extend Opial-type inequalities by using Hahn calculus, we first prove the Hölder’s inequality involving Hahn calculus.Theorem 3.
Ifa,b∈Iandu,v:a,b⟶Rare continuous functions, then,(34)∫abutvtdq,ht≤∫abutjdq,ht1/i∫abvtidq,ht1/i,where j>1 and i=j/j−1.Proof.
For nonnegativeα,β real numbers, the basic inequality holds.(35)α1/jβ1/i≤αj+βi,
Now consider,(36)∫abutjdq,ht∫abvtidq,ht≠0.
In (35), choose,(37)αt=utj∫abuτjdq,hτ,βt=vti∫abvτidq,hτ.
Use (35) and integrating the resultant inequality from a to b to obtain(38)∫abα1/jtβ1/itdq,ht=∫abutj∫abuτjdq,hτ1/ivti∫abvτidq,hτ1/idq,ht≤∫ab1jutj∫abuτjdq,hτ+1ivti∫abvτidq,hτdq,ht=1j∫abutj∫abuτjdq,hτdq,ht+1i∫abvti∫abvτidq,hτdq,ht=1.
Hence, we get (34). The proof is complete.
Next, consider the notationsy˜m=yqm+hmq, y^=y+mh and yˇ=yqm, where y∈a,x1,b.Remark 1.
In form of sums (34) can be written as(39)b1−q−h∑m=0∞qmub˜mvb˜m−a1−q−h∑m=0∞qmua˜mva˜m≤b1−q−h∑m=0∞qmub˜mj−a1−q−h∑m=0∞qmua˜mji/j×b1−q−h∑m=0∞qmvb˜mi−a1−q−h∑m=0∞qmva˜mi1/i.
Whenh=0 in (39), it recaptures the Hölder’s inequality in q-calculus [31].(40)∑m=0∞qmbubˇvbˇ−auaˇvaˇ≤∑m=0∞qmbubˇj−auaˇji/j∑m=0∞qmbvbˇi−avaˇi1/i.
By usingq⟶1 in (39), we get Hölder’s inequality in h-discrete calculus, which is ([32], Theorem 3.1).(41)∑m=0∞ua^va^−ub^vb^≤∑m=0∞ua^j−ub^j1/j∑m=0∞va^i−vb^i1/i.When h=0, q⟶1 in (34), we obtain Hölder’s inequality in classical calculus, which can be found in [33].


## 3.2. Opial-Type Inequalities ona,x1

Theorem 4.
LetIbe any interval,y∈Ianda,x1∈yqm+hmqm=0∞. Assumel⋅,s⋅are positive continuous functions ona,x1.s⋅is non-increasing function ona,x1and∫ax1l−kzdq,hz<∞. Furthermore, supposeuis a continuous function ona,x1withua=0. Then,∀j≥1,k≥1, and∀t,z∈yqm+hmqm=0∞, we have,(42)j+k∫ax1lkk−1zskzuzjDq,huzkdq,hz≤k∫ax1l−kzdq,hzj∫ax1lkj+k−1zskzDq,huzj+kdq,hz.Proof.
Let us consider the following integral(43)Tz=∫azlkk−1tsk2/j+ktDq,hutkdq,ht.
SinceTa=0, one has that,(44)Dq,hTz=lkk−1zsk2/j+kzDq,huzk.
Equation (32) implies,(45)uz=uz−ua=∫azDq,hutdq,ht≤∫azDq,hutdq,ht.
Use of Hölder inequality (34) with indices k and k/k−1, provides,(46)uz≤∫azl1−ktlk−1tDq,hutdq,ht≤∫azl1−ktk/k−1dq,htk−1/k∫azlk−1tkDq,hutkdq,ht1/k=∫azl−ktdq,htk−1/k∫azlkk−1tDq,hutkdq,ht1/k.
By taking powerj on both sides, we have(47)uzj≤∫azl−ktdq,htjk−1/k∫azlkk−1tDq,hutkdq,htj/k.
Sinces⋅ is nonincreasing and positive on a,x1, one has that,(48)sjk/j+kzuzj≤sjk/j+kz∫azl−ktdq,htjk−1/k∫azlkk−1tDq,hutkdq,htj/k≤∫azl−ktdq,htjk−1/k∫azlkk−1tsk2/j+ktDq,hutkdq,htj/k=∫azl−ktdq,htjk−1/kTj/kz.
Multiply (48) by lkk−1zsk2/j+kzDq,huzk≥0, to get(49)lkk−1zsjk/j+kzsk2/j+kzuzjDq,huzk≤lkk−1z∫azl−ktdq,htjk−1/kTj/kzsk2/j+kzDq,huzk.
Multiply (49) by j+k, integrate it from a to x1 and use monotonicity of ∫azl−ktdq,ht to obtain(50)j+k∫ax1lkk−1zskzuzjDq,huzkdq,hz≤j+k∫ax1l−ktdq,htjk−1/k∫ax1Tj/kzDq,hTzdq,hz.
Now from Chain rule (23), we have(51)Dq,hTj+k/kz=j+kkTj/kcDq,hTz,c∈z,qz+h.
SinceDq,hTz≥0 and z≤c, one gets(52)Dq,hTj+k/kz=j+kkTj/kcDq,hTz≥j+kkTj/kzDq,hTz.From (50) and (52), it is noted that Ta=0 and(53)j+k∫ax1lkk−1zskzuzjDq,huzkdq,hz≤k∫ax1l−kzdq,hzjk−1/k∫ax1Dq,hTj+k/kzdq,hz,
(by using (43)),(54)≤k∫ax1l−kzdq,hzjk−1/kTj+k/kx1=k∫ax1l−kzdq,hzjk−1/k∫ax1lkk−1zsk2/j+kDq,huzkdq,hzj+k/k.
Denote,(55)Lx1≐∫ax1lkk−1zsk2/j+kDq,huzkdq,hzj+k/k.
By applying Hölder inequality (34) with indices j+k/j and j+k/k on L, one gets(56)L=∫ax1l−jk/j+kzlk21−1/j+kzsk2/j+kzDq,huzkdq,hzj+k/k≤∫ax1l−jk/j+kzj+k/jdq,hzj/k∫ax1lk21−1/j+kzsk2/j+kzDq,huzkj+k/kdq,hz=∫ax1l−kzdq,hzj/k∫ax1lkj+k−1zskzDq,huzj+kdq,hz.
By combining (54) and (56), we get(57)j+k∫ax1lkk−1zskzuzjDq,hzkdq,hz≤∫ax1l−kzdq,hzj∫ax1lkj+k−1zskzDq,huzj+kdq,hz.
Hence, (42) is obtained.Corollary 1.
Case 1.Whenq⟶1, then (42) reduces to the following Opial inequality inh−discretecalculus.(58)∑m=0∞lkk−1a^ska^uja^Δhua^k−lkk−1x^1skx^1ujx^1Δhux^1k≤L1∑m=0∞lkj+k−1a^ska^Δhua^j+k−lkj+k−1x^1skx^1Δhux^1j+k,where L1=khj/j+k∑m=0∞l−ka^−l−kx^1j.Case 2.
Whenh=0in (42), it is converted to the following Opial inequality in q-calculus.(59)∑m=0∞qkx1lkk−1xˇ1skxˇ1ujxˇ1Δhuxˇ1k−alkk−1aˇskaˇujaˇΔhuaˇk≤L2∑m=0∞qkx1lkj+k−1xˇ1skxˇ1Δhuxˇ1j+k−alkj+k−1aˇskaˇΔhuaˇj+k,where L2=k1−qj/j+k∑m=0∞qkx1l−kxˇ1−al−kaˇjCase 3.
Whenq⟶1andh=0in (42). We get (11), which is ([10], Theorem 1.1), ([16], Corollary 3.2).(60)j+k∫ax1lkk−1zskzuzju′zkdz≤k∫ax1l−kxdzj∫ax1ljj+k−1zskzu′zj+kdz.Case 4.
Whenq⟶1,h=0andj=k=1. We obtain (7) given by Yang in ([9], Theorem 3), ([16], Remark 3.3).(61)2∫ax1szuzu′zdz≤∫ax1l−1zdz∫ax1lzszu′z2dz.Remark 2.
Ifh=1in (26), then, it recaptures Opial-type discrete inequality given in ([16], Corollary 1).


## 3.3. Opial-Type Inequalities onx1,b

Theorem 5.
LetIbe any interval,y∈Iandx1,b∈yqm+hmqm=0∞. Assumel⋅,s⋅are positive continuous functions onx1,b.s⋅is nondecreasing onx1,band∫x1bl−kzdq,hz<∞. Furthermorer, assumeu⋅is a continuous function onx1,bwithub=0, then,∀j≥1,k≥1,z,t∈yqm+hmqm=0∞, we have,(62)j+k∫x1blkk−1zskzuqz+hjDq,huzkdq,hz≤k∫x1bl−kzdq,hzj∫x1blkj+k−1zskzDq,huzj+kdq,hz.Proof.
Let us consider the following integral:(63)Tz≐−∫zblkk−1tsk2/j+ktDq,hutkdq,ht.
By usingTb=0, we get(64)Dq,hTz=lkk−1zsk2/j+kzDq,huzk.
Use of (32) gives(65)uqz+h=−uqz+h=ub−uqz+h=∫qz+hbDq,hutdq,ht≤∫qz+hbDq,hutdq,ht.
Use of (34) with indices k and k/k−1 implies(66)uqz+h≤∫qz+hbl1−ktlk−1tDq,hutdq,ht≤∫qz+hbl1−ktk/k−1dq,htk−1/k∫qz+hblk−1tkDq,hutkdq,ht1/k=∫qz+hbl−ktdq,htk−1/k∫qz+hblkk−1tDq,hutkdq,ht1/k.
By taking powerj on both sides, one gets(67)uqz+hj≤∫qz+hbl−ktdq,htjk−1/k∫qz+hblkk−1tDq,hutkdq,htj/k.
Sinces⋅ is nondecreasing and positive on x1,b, therefore,(68)sjk/j+kzuqz+hj≤sjk/j+kz∫qz+hbl−ktdq,htjk−1/k∫qz+hblkk−1tDq,hutkdq,htj/k≤∫qz+hbl−ktdq,htjk−1/k∫qz+hblkk−1tsk2j+ktDq,hutkdq,htj/k,
((63)implies),(69)=∫qz+hbl−ktdq,htjk−1/k−Tqz+hj/k.
Multiply both sides of (69) by lkk−1zsk2/j+kzDq,huzk≥0, to find,(70)lkk−1zsk2/j+kzsjk/j+kzuqz+hjDq,huzk≤lkk−1z∫qz+hbl−ktdq,htjk−1/k−Tqz+hj/ksk2/j+kzDq,huzk.
(70) can be written in the following subsequent form,(71)lkk−1zskzuqz+hjDq,huzk≤∫qz+hbl−ktdq,htjk−1/k−Tqz+hj/kDq,hTz.
Multiply (71) by j+k, integrate from x1 to b and use monotonicity of ∫qz+hbl−ktdq,ht to find,(72)j+k∫x1blkk−1zskzuqz+hjDq,huzkdq,hz≤j+k∫x1bl−ktdq,htjk−1/k∫x1b−Tqz+hj/kDq,hTzdq,hz.
Now, from Chain rule (23), we have(73)−Dq,h−Tj+k/kz=j+kk−Tj/kcDq,hTz,c∈z,qz+h.
SinceDq,hTz≥0 and c≤qz+h, one gets(74)Dq,h−−Tj+k/kz=j+kk−Tj/kcDq,hTz≥j+kk−Tqz+hDq,hTz.
It is noted thatTb=0. (72) and (74) yields,(75)j+k∫x1blkk−1zskzuqz+hjDq,huzkdq,hz≤k∫x1bl−kxdq,hzjk−1/k∫x1bDq,h−−Tj+k/kzdq,hz≤k∫x1bl−kzdq,hzjk−1/k−Tj+k/kx1,
(by using (70)),(76)=k∫x1bl−kzdq,hzjk−1/k∫x1blkk−1zsk2/j+kzDq,huzkdq,hzj+k/k.
Again use (34) with indices j+k/k and j+k/j to get(77)∫x1blkk−1zsk2/j+kDq,huzkdq,hzj+k/k=∫x1bl−jk/j+kzlk21−1/j+kzsk2/j+kzDq,huzkdq,hzj+k/k≤∫x1bl−jk/j+kzj+k/jdq,hzj/k∫x1blk21−1/j+kzsk2/j+kzDq,huzkj+k/kdq,hz=∫x1bl−kzdq,hzj/k∫x1blkj+k−1zskzDq,huzj+kdq,hz.
By combining (76) and (77), we get(78)j+k∫x1blkk−1zskzuqz+hjDq,huzkdq,hz≤k∫x1bl−kzdq,hzj∫x1blkj+k−1zskzDq,huzj+kdq,hz.
Hence, (62) is proved.Corollary 2.
Case I: Whenq⟶1, then, (62) is converted to the following Opial’s inequality inhdiscrete-calculus.(79)∑m=0∞lkk−1x^1skx^1ujx^1Δhux^1k−lkk−1b^skb^ujb^Δhub^k≤L3∑m=0∞lkj+k−1x^1skx^1Δhux^1j+k−lkj+k−1b^skb^Δhub^j+k,where L3=khj/j+k∑m=0∞l−kx^1−l−kb^j.
Moreover, ifh=1 in (79), it recaptures ([16], Corollary 3.8).
Case II.Whenh=0in (62), it is converted to the following Opial inequality in q-calculus.(80)∑m=0∞qkblkk−1bˇskbˇujbˇΔhubˇk−x1lkk−1xˇ1skxˇ1ujxˇ1Δhuxˇ1k≤L4∑m=0∞qkblkj+k−1bˇskbˇΔhubˇj+k−x1lkj+k−1xˇ1skxˇ1Δhuxˇ1j+k,where L4=k1−qj/j+k∑m=0∞qkbl−kb−x1l−kxˇ1j.
Case III.Whenq⟶1andh=0in (62), we obtain (12) ([10], Theorem 1.2), ([16], Corollary 3.6).(81)j+k∫x1blkk−1skxuzju′zkdz≤k∫x1bl−kzdzj∫x1blkj+k−1zskzu′zj+kdz.
Case IV:Ifq⟶1, h=0 and j=k=1 in (62). We obtain (8) ([9], Theorem 3), ([16], Remark 3.7).(82)2∫x1bszuzu′zdz≤∫x1bl−1zdz∫x1blzszu′z2dz.


## 3.4. Opial-Type Inequalities ona,b

Theorem 6.
SupposeIis any interval,y∈Ianda,b∈yqm+hmqm=0∞. Assumel⋅,s⋅are positive continuous functions ona,b.s⋅is nonincreasing ona,x1ands⋅is nondecreasing onx1,b. SupposeC=max∫ax1l−kzdq,hzj,∫x1bl−kzdq,hzj. Furthermorer, chooseu⋅to be a nondecreasing continuous function ona,bwith conditionsua=ub=0. Then,∀j≥1,k≥1, we hav,(83)j+k∫ablkk−1zskzuzjDq,huzkdq,hx≤kC∫ablkj+k−1zskzDq,huzj+kdq,hz.Proof.
Let us consider the following integral:(84)j+k∫ablkk−1zskzuzjDq,huzkdq,hz=j+k∫ax1lkk−1zskzuzjDq,huzkdq,hz+j+k∫x1blkk−1zskzuzjDq,huzkdq,hz.
Sinces is nonincreasing on a,x1, and s is nondecreasing on x1,b, therefore by using (42) and (62) in (84), one gets(85)j+k∫ablkk−1zskzuzjDq,huzkdq,hz≤j+k∫ax1lkk−1zskzuzjDq,huzkdq,hz+j+k∫x1blkk−1zskzuqz+hjDq,huzkdq,hz≤k∫ax1l−kzdq,hzj∫ax1lkj+k−1zskzDq,huzj+kdq,hz+k∫x1bl−kzdq,hzj∫x1blkj+k−1zskzDq,huzj+kdq,hz≤kC∫ax1lkj+k−1zskzDq,huzj+kdq,hz+∫x1blkj+k−1zskzDq,huzj+kdq,hz=kC∫ablkj+k−1zskzDq,huzj+kdq,hz,which is the required inequality (83).Corollary 3.
Case I: Whenq⟶1, then, the inequality (83) becomes the Opial’s inequality inh−calculus.(86)∑m=0∞lkk−1a^ska^uja^Δhua^k−lkk−1b^skb^ujb^Δhub^k≤C1∑m=0∞lkj+k−1a^ska^Δhua^j+k−lkj+k−1b^skb^Δhub^j+k,where C1=maxL1,L3.
Moreover, ifh=1 in (86), it recaptures ([16], Corollary 3.11).
Case II.Whenh=0in (83), it is converted to the following Opial inequality in q-calculus.(87)∑m=0∞qkblkk−1bˇskbˇujbˇΔhubˇk−alkk−1aˇskaˇujaˇΔhuaˇk≤C2∑m=0∞qkblkj+k−1bˇskbˇΔhubˇj+k−alkj+k−1aˇskaˇΔhuaˇj+k,where C2=maxL2,L4.
Case III.Whenq⟶1andh=0in the inequality (83). We get (13) ([10], Theorem 2), ([16], Corollary 3.10).(88)j+k∫ablkk−1zskzuzju′zkdz≤kC∫ablkj+k−1zskzu′zj+kdz.
Case IV.Whenq⟶1, h=0 and j=k=1 in (83). Then, we obtain the Opial’s inequality (9).(89)2∫abszuz‖u′zdz≤C∫ablzszu′z2dz.


## 4. Conclusions

In this article, Hölder’s inequality is proved with the help of Hahn calculus whose special cases contain Hölder inq-calculus [31], in h-discrete calculus [32], and in usual calculus [33]. Furthermore, the obtained Hölder’s inequality is utilized to find some Opial-type inequalities. Limiting cases of newly proved Opial-type inequalities coincide with some Opial-type inequalities of [9,10,16]. Some extensions of Opial-type inequalities in q-calculus q>1 and in h-discrete calculus h>0 as well [28].



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