# Generalized Hölder’s and Minkowski’s Inequalities for Jackson’sq-Integral and Some Applications to the Incomplete q-Gamma Function

Authors: Kwara Nantomah
Journal: Abstract and Applied Analysis
 (2017)

---

## Abstract

We establish some generalized Hölder’s and Minkowski’s inequalities for Jackson’sq-integral. As applications, we derive some inequalities involving the incomplete q-Gamma function.


---

## Body

## 1. Introduction

The classical Hölder’s and Minkowski’s inequalities are usually defined as follows.Definition 1.
Letα>1 and 1/α+1/β=1. Then the discrete and integral forms of Hölder’s inequality are given as(1)∑i=1naibi≤∑i=1naiα1/α∑i=1nbiβ1/βfor sequences ai, bi and(2)∫abftgtdt≤∫abftαdt1/α∫abgtβdt1/βfor continuous function f and g on [a,b].Definition 2.
Letu>1. Then the discrete and integral forms of Minkowski’s inequality are given as(3)∑i=1nai+biu1/u≤∑i=1naiu1/u+∑i=1nbiu1/ufor sequences ai, bi and(4)∫abft+gtudt1/u≤∫abftudt1/u+∫abgtudt1/ufor continuous function f and g on [a,b].These fundamental results are well known in the literature and have been studied intensively by several researchers. Their role in mathematics and related disciplines is invaluable. For instance, they play a pivotal role in classical real and complex analysis, probability theory, statistics, numerical analysis, and so on. Over the past years, various refinements, extensions, and applications have appeared in the literature. In the present work, our objective is to provide some generalized Hölder’s and Minkowski’s inequalities for Jackson’sq-integral. As applications, we derive some inequalities involving the incomplete q-Gamma function. Let us begin with the following auxiliary results.


## 2. Auxiliary Results

We begin with the following inequality which is well known as Young’s inequality.Lemma 3.
Forx,y≥0, a>1, and 1/a+1/b=1, the inequality(5)xy≤xaa+ybbis satisfied.Inequality (5) can be generalized as follows.Lemma 4.
Fork=1,2,…,n, let xk≥0 and αk>1 such that ∑k=1n1/αk=1. Then the inequality(6)∏k=1nxk≤∑k=1nxkαkαkis valid.Inequality (6) can be written in the following form which is known as the weighted AM-GM inequality.Lemma 5.
Fork=1,2,…,n, let Qk≥0 and αk>1 such that ∑k=1n1/αk=1. Then the inequality(7)∏k=1nQk1/αk≤∑k=1nQkαkis valid.Lemma 6 (generalized Hölder’s inequality for sums).
Leti=1,2,3,…,n and k=1,2,3,…,m such that the sums exist. Then the inequality(8)∑i=1n∏k=1mQi,k≤∏k=1m∑i=1nQi,kαk1/αkis valid for αk>1 such that ∑k=1m1/αk=1.Proof.
The inequality is obvious if∑i=1nQi,kαk=0 for each k. So we assume ∑i=1nQi,kαk≠0 and let (9)xi,k=Qi,k∑i=1nQi,kαk1/αk.
Then by the generalized Young’s inequality (6), we obtain (10)∏k=1mQi,k∏k=1m∑i=1nQi,kαk1/αk≤∑k=1m1αkQi,kαk∑i=1nQi,kαk.
By adding thesen inequalities, we obtain (11)∑i=1n∏k=1mQi,k∏k=1m∑i=1nQi,kαk1/αk≤∑k=1m1αk∑i=1nQi,kαk∑i=1nQi,kαk=∑k=1m1αk=1which gives inequality (8).Lemma 7 (generalized Minkowski’s inequality for sums).
Leti=1,2,3,…,n and k=1,2,3,…,m such that the sums exist. Then the inequality(12)∑i=1n∑k=1mQi,ku1/u≤∑k=1m∑i=1nQi,ku1/uis valid for u>1.Proof.
We prove this by mathematical induction onm. If m=2, then (12) reduces to the classical Minkowski’s inequality (3). Thus (12) is valid for m=2. Next, assume that (12) holds for some m≥2. Based on this assumption, we want to show that (12) holds for m+1. We proceed as follows: (13)∑i=1n∑k=1m+1Qi,ku1/u=∑i=1n∑k=1mQi,k+Qi,m+1u1/u≤∑i=1n∑k=1mQi,ku1/u+∑i=1nQi,m+1u1/u≤∑k=1m∑i=1nQi,ku1/u+∑i=1nQi,m+1u1/u=∑k=1m+1∑i=1nQi,ku1/u.Thus (12) holds for m+1 and this completes the proof.Remark 8.
Inequalities (8) and (12) are already known in the literature. See, for instance, [1–3]. Here we offer simple proofs of the results.


## 3. Generalizedq-Hölder’s and q-Minkowski’s Inequalities

Jackson’sq-integral from 0 to a and that from 0 to ∞ are, respectively, defined as(14)∫0aftdqt=1-qa∑k=0∞faqkqk,(15)∫0∞ftdqt=1-q∑-∞∞fqkqkprovided that the sums in (14) and (15) converge absolutely [4]. In a generic interval [a,b], the q-integral takes the following form: (16)∫abftdqt=∫0bftdqt-∫0aftdqt.Theorem 9 (generalizedq-Hölder’s inequality).
Letf1,f2,…,fn be functions such that the integrals exist. Then the inequality(17)∫0a∏i=1nfitdqt≤∏i=1n∫0afitαidqt1/αiholds for αi>1 such that ∑i=1n1/αi=1.Proof.
Letαi>1 such that ∑i=1n1/αi=1. Then by relation (14) and inequality (8), we obtain (18)∫0a∏i=1nfitdqt=1-qa∑k=0∞∏i=1nfiaqkqk≤1-qa∏i=1n∑k=0∞fiaqkαiqk1/αi=1-qa∑i=1n1/αi∏i=1n∑k=0∞fiaqkαiqk1/αi=∏i=1n1-qa1/αi∏i=1n∑k=0∞fiaqkαiqk1/αi=∏i=1n1-qa∑k=0∞fiaqkαiqk1/αi=∏i=1n∫0afitαidqt1/αi which concludes the proof.Remark 10.
Letn=2, α1=p, α2=t, f1=f, and f2=g in Theorem 9. Then we obtain the result of Lemma  2.1 of [5].Theorem 11 (generalizedq-Minkowski’s inequality).
Letf1,f2,…,fn be functions such that the integrals exist. Then the inequality(19)∫0a∑i=1nfitudqt1/u≤∑i=1n∫0afitudqt1/uholds for u>1.Proof.
Similarly we apply the principle of mathematical induction. Forn=2, inequality (19) reduces to q-Minkowski’s inequality obtained in Remark  4 of [6]. Assume that (19) holds for n≥2. Based on this assumption, we show that (19) holds for n+1. That is, (20)∫0a∑i=1n+1fitudqt1/u=∫0a∑i=1nfit+fn+1tudqt1/u≤∫0a∑i=1nfitudqt1/u+∫0afn+1tudqt1/u≤∑i=1n∫0afitudqt1/u+∫0afn+1tudqt1/u=∑i=1n+1∫0afitudqt1/u.Thus (19) holds for n+1. This completes the proof.


## 4. Some Applications to the Incompleteq-Gamma Function

In this section, we derive some inequalities involving the incompleteq-Gamma function. We shall use the notations N0=0,1,2,3,… and ℵ=0,2,4,6,… subsequently.El-Shahed and Salem [7] defined the incomplete q-Gamma function for q∈(0,1) as(21)γqx,z=∫0ztx-1Eq-qtdqt,x>0,z>0and the complementary incomplete q-Gamma function as (22)Γqx,z=∫z1/1-qtx-1Eq-qtdqt,x>0,z≥0,where Eqt=∑n=0∞qnn-1/2tn/[n]q is a q-analogue of the classical exponential function, uq=1-qu/1-q, and nq!=nqn-1q⋯[2]q[1]q. One can easily see that (23)Γqx,0=Γqx,γqx,z+Γqx,z=Γqx, where Γq(x) is the q-Gamma function. Also, the following identities are satisfied: (24)Γq1,z=Eq-z,Γqx+1,z=xqΓqx,z+zxEq-z,γq1,z=1-Eq-z,γqx+1,z=xqγqx,z-zxEq-z. Moreover, (25)Γq0,z=∫z1/1-qt-1Eq-qtdqt=E1z,q,where E1(z,q) is the q-exponential integral [8].Remark 12.
The functionsγq(x,z) and Γq(x,z) can be viewed as both functions of x (for fixed z) and functions of z (for fixed x). For the purpose of this paper, we shall concentrate on γq(x,z) as functions of x.By differentiating (21) m times, we obtain(26)γqmx,z=∫0ztx-1ln⁡tmEq-qtdqt,m∈N0,where γq(0)(x,z)=γq(x,z).Theorem 13.
Fori=1,2,…,n, let αi>1, ∑i=1n1/αi=1, mi∈ℵ, and ∑i=1nmi/αi∈ℵ. Then the inequality(27)γq∑i=1nmi/αi∑i=1nxiαi,z≤∏i=1nγqmixi,z1/αiholds for xi>0.Proof.
By (26) and the generalized q-Hölder’s inequality (17), we obtain (28)γq∑i=1nmi/αi∑i=1nxiαi,z=∫0zt∑i=1nxi/αi-1ln⁡t∑i=1nmi/αiEq-qtdqt=∫0zt∑i=1nxi-1/αiln⁡t∑i=1nmi/αiEq-qt∑i=1n1/αidqt=∫0z∏i=1ntxi-1/αiln⁡tmi/αiEq-qt1/αidqt≤∏i=1n∫0ztxi-1ln⁡tmiEq-qt1/αidqt=∏i=1nγqmixi,z1/αiwhich completes the proof.Remark 14.
Letn=2, α1=a, α2=b, x1=x, and x2=y in Theorem 13. Then, we obtain(29)γqm1/a+m2/bxa+yb,z≤γqm1x,z1/aγqm2y,z1/b.Remark 15.
Letm1=m2=m in (29). Then we obtain(30)γqmxa+yb,z≤γqmx,z1/aγqmy,z1/bwhich implies that the function γq(m)(x,z) is logarithmically convex. Also, since γq(0)(x,z)=γq(x,z), then it follows that γq(x,z) is also logarithmically convex.Remark 16.
Leta=b=2 and x=y in (29). Then we obtain the Turan-type inequality(31)γqm1x,zγqm2x,z≥γqm1+m2/2x,z2.Corollary 17.
Letm∈ℵ. Then the inequality(32)γqmx+y2,z≤γqmx,z+γqmy,z2holds for x,y>0.Proof.
Leta=b=2 in (30). Then, by the AM-GM inequality, we have(33)γqmx+y2,z≤γqmx,zγqmy,z≤γqmx,z+γqmy,z2.Theorem 18.
Fori=1,2,…,n, let mi∈ℵ. Then the inequality(34)∑i=1nγqmixi,z1/u≤∑i=1nγqmixi,z1/uholds for xi>0 and u≥1.Proof.
Note that∑i=1naiu≤∑i=1naiu for ai≥0 and u≥1. Then by the generalized q-Minkowski’s inequality (19), we obtain(35)∑i=1nγqmixi,z1/u=∑i=1n∫0ztxi-1ln⁡tmiEq-qtdqt1/u=∫0z∑i=1ntxi-1/uln⁡tmi/uEq-qt/uudqt1/u≤∫0z∑i=1ntxi-1/uln⁡tmi/uEq-qt/uudqt1/u≤∑i=1n∫0ztxi-1/uln⁡tmi/uEq-qt/uudqt1/u=∑i=1n∫0ztxi-1ln⁡tmiEq-qtdqt1/u=∑i=1nγqmixi,z1/u which completes the proof.Remark 19.
In particular, letn=2, m1=m, m2=n, x1=x, and x2=y in Theorem 18. Then we obtain (36)γqmx,z+γqny,z1/u≤γqmx,z1/u+γqny,z1/u.Theorem 20.
Fori=1,2,…,n, let αi>1, ∑i=1n1/αi=1, mi∈ℵ, and ∑i=1nmi/αi∈ℵ. Then the inequality(37)exp⁡γq∑i=1nmi/αi∑i=1nxiαi,z≤∏i=1nexp⁡γqmixi,z1/αiis satisfied for xi>0.Proof.
By (26) we obtain (38)γq∑i=1nmi/αi∑i=1nxiαi,z-∑i=1nγqmixi,zαi=∫0zt∑i=1nxi/αi-1ln⁡t∑i=1nmi/αiEq-qtdqt-∑i=1n1αi∫0ztxi-1ln⁡tmiEq-qtdqt=∫0z∏i=1ntxi-1/αiln⁡tmi/αiEq-qtdqt-∑i=1n1αi∫0ztxi-1ln⁡tmiEq-qtdqt=∫0z∏i=1ntxi-1ln⁡tmi1/αi-∑i=1n1αitxi-1ln⁡tmiEq-qtdqt≤0 which results from the weighted AM-GM inequality (7). Hence(39)γq∑i=1nmi/αi∑i=1nxiαi,z≤∑i=1nγqmixi,zαi.Then, by exponentiating (39), we obtain the required result (37).Remark 21.
Results of types (27), (34), and (37) which deal with the (q·k)-Gamma function can also be found in [9].


## 5. Conclusion

In this study, we provided simple proofs of the discrete forms of some generalized Hölder’s and Minkowski’s inequalities. Based on these results, we established some generalized Hölder’s and Minkowski’s inequalities for Jackson’sq-integral. Furthermore, by using the established results, we derived some new inequalities involving the incomplete q-Gamma function. We anticipate that the present results will find some applications in q-Calculus as well as other related disciplines.



---

*Source: 9796873-2017-07-16.xml*