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Hardy-type inequalities quantum calculus DOCTORAL THESIS Serikbol Shaimardan Mathematics

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Hardy-type inequalities quantum calculus

Department of Engineering Sciences and Mathematics

ISSN 1402-1544ISBN 978-91-7790-240-9 (print)ISBN 978-91-7790-241-6 (pdf)

Luleå University of Technology 2018

DOCTORA L T H E S I S

Serikbol Shaimardan H

ardy-type inequalities quantum calculus

Serikbol Shaimardan

Mathematics

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Hardy-type inequalities quantum calculus

Serikbol Shaimardan

Luleå University of TechnologyDepartment of Engeneering Sciences and Mathematics

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Printed by Luleå University of Technology, Graphic Production 2018

ISSN 1402-1544 ISBN 978-91-7790-240-9 (print)ISBN 978-91-7790-241-6 (pdf)

Luleå 2018

www.ltu.se

2010 Mathematics Subject Classification. 35B27, 76D08

Key words and phrases. Inequalities, Hardy-type inequalities,Riemann-Liouville operator, Integral operator, q-analysis,q-analog, weights, h-calculus, h-integral, discrete fractional

calculus

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Abstract

This PhD thesis deals with fractional Hardy-type inequalitiesand some new Hardy-type inequalities for the Hardy operator andRiemann-Liouville fractional integral operator and fractional Hardy-type inequalities in quantum calculus, which are given in the frameof q-calculus and h-calculus.

The thesis contains five papers (papers A, B, C, D and E) andan introduction , which put these papers into a more general frame.In particular, in this introduction we give a brief history of quantumcalculus and a short description as basis for the rest of the quantumcalculus.

In paper A we study some q-analogs of Hardy-type inequalitiesfor the Riemann-Liouville fractional integral operator of order n ∈N and find necessary and sufficient conditions of the validity ofthese inequalities for all non-negative real functions.

In paper B we define the q-analog of the classical Hardy operatorand we characterize the q-analog of the weighted Hardy inequalitiesfor all possible values of the parameters p and r in q-calculus. Wealso study the corresponding dual results.

In paper C we consider a q-analog of the operator I defined by

If(x) :=

x∫

0

lnx

x− sf(s)

sds,

which is called the fractional integral operator of infinitesimal order.Moreover, we characterize the q-analog of the following Hardy-type

iii

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iv ABSTRACT

integral inequality:

(1)

∞∫

0

ur(x)

x∫

0

tγ−1 lnx

x− tf(t)dt

r

dx

1r

≤ C

∞∫

0

fp(x)dx

1p

, ∀f(·) ≥ 0,

where u(.) is a weight i.e. a measurable function, which is positivea.e. in (0,∞). In fact, we derive necessary and sufficient conditionsfor the validity of the q-analog of the inequality (1) in q-analysis forthe case 1 < p ≤ r <∞ and γ > 1

p. We also consider the problem

to find the best constant in the q-analog of inequality (1).In paper D we consider that the first power weighted version of

Hardy’s inequality which can be rewritten as∞∫

0

xα−1

x∫

0

1

tαf(t)dt

p

dx ≤(

p

p− α− 1

)p ∞∫

0

fp(x)dx,

for f ≥ 0, p ≥ 1 and α < p−1, where the constant C =(

pp−α−1

)pis

sharp. In this paper we prove and discuss some discrete analogues ofHardy-type inequalities in fractional h-discrete calculus. Moreover,we prove that the corresponding constants are sharp.

In paper E we consider the fractional order Hardy-type inequal-ity in the following form:∞∫

0

∞∫

0

|f(x)− f(y)|p

|x− y|1+pαdxdy

p

≤ C

∞∫

0

|f ′(x)|p x(1−α)pdx

p

,

for 0 < α < 1 and 1 < p < ∞ in fractional h-discrete calculus,

where C = 21p α−1

(p−pα)1p

. A discrete analogue of this inequality, namely

a new fractional order Hardy inequality in h-discrete calculus, isproved and discussed. Moreover, we prove that the same constantis sharp also in this case.

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Preface

The main part of this PhD thesis consists of five papers (pa-pers A, B, C, D and E), which deal with a new fractional Hardyinequality and also some new Hardy-type inequalities in quantumcalculus. The thesis also contains an introduction, which put thesepapers to a more general frame.

[A] L. E. Persson and S. Shaimardan, Some new Hardy-typeinequalities for Riemann-Liouville fractional q -integral op-erator, J. Inequal. Appl. (2015), 2015:296, 17pp.

[B] A. O. Baiarystanov, L. E. Persson, S. Shaimardan and A.Temirkhanova, Some new Hardy-type inequalities in q -analysis, J. Math. Inequal. 10(2016), no. 3, 761-781.

[C] S. Shaimardan, A Hardy-type inequality for the fractionalintegral operator in q-analysis, Eurasian Math. J., 7 (2016),no. 1, 5-16.

[D] L. E. Persson, R. Oinarov and S. Shaimardan, Hardy-typeinequalities in fractional h-discrete calculus, J. Inequal.Appl. (2018), 2018:73, 14pp.

[E] S. Shaimardan, Fractional order Hardy-type inequality infractional h-discrete calculus, Research Report, Depart-ment of Mathematics, Lulea University of Technology, 2018(submitted)

Remark. Paper [E] is a slightly improved version of the originalReport 1 from 2018.

v

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Acknowledgment

First of all I want to express my deep gratitude to my scien-tific supervisors Professor Lars-Erik Persson (Department of Engi-neering Sciences and Mathematics, Lulea University of Technology,Sweden) and Professor Ryskul Oinarov (Eurasian National Univer-sity, Kazakhstan) for their valuable remarks and attention to mywork and their constant support.

Moreover, I thank my other supervisor Professor Peter Wall(Department of Engineering Sciences and Mathematics, Lulea Uni-versity of Technology, Sweden), which has given me the possibilityto work in this international PhD program and provided me withnecessary invitations and related support. My sincere thanks alsoto Professor Natasha Samko, who has given me several generousand valuable remarks on my introduction and helped me with myRussian-English translations.

I also thank PhD Ainur Temirkhanova for helping me with manypractical things.

Moreover, I thank Lulea University of Technology and L. N. Gu-milyov Eurasian National University for giving me an opportunityto participate in their partnership program in research and post-graduate education in mathematics. I also thank both universitiesfor financial support which made my studies possible.

Furthermore, I would like to thank everyone at the Departmentof Mathematics at Lulea University of Technology for helping mein different ways and for the warm and friendly atmosphere.

Finally, and most important, I want to express my deepest grat-itude to my wife Aigerim Edenova for all support and understand-ing.

vii

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Introduction

”The calculus was the first achievement of modern mathemat-ics and it is difficult to overestimate its importance. I think it de-fines more unequivocally than anything else the inception of modernmathematics, and the system of mathematical analysis, which is itslogical development, still constitutes the greatest technical advancein exact thinking.” - John von Neumann.

Calculus or infinitesimal calculus has a fascinating history. Inthe 17th century, I. Newton and G. Leibniz independently inventedcalculus based on the concept of limit (but elements of it have al-ready appeared in ancient Greece). The usual meaning of limit im-plies that space and time are continuous, and we have maintainedthat all natural processes happen continuously on smooth curvesand surfaces. However, the atomic theory in physics and chemistryin the 19th century paved that the nature process of dividing itinto ever smaller parts will terminate in an indivisible or an atom,a part which, lacking proper parts itself, cannot be further divided.In a word, continua are divisible without limit or infinitely divisi-ble. This becomes the origin of developing another type of calculusbased on finite difference principle, or calculus without limit whichis quantum calculus (the calculus of finite differences was developedat the same time).

In mathematics, the quantum calculus is equivalent to usualinfinitesimal calculus without the concept of limits or the investi-gation of calculus without limits (quantum is from the Latin word”quantus” and literally it means how much, in Swedish ”Kvant”).It has two major branches, q-calculus and the h-calculus. And bothof them were worked out by P. Cheung and V. Kac [36] in the earlytwentieth century.

1

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2 INTRODUCTION

1. Background and further development

h-calculus: One of the popular quantum calculus is h-calculus.This calculus is the study of the definitions, properties, and ap-plications of related concepts, the fractional calculus and discretefractional calculus. However, the investigation for fractional calcu-lus was studied already by G. Leibniz after that G. L’Hospital in1695 asked him: ” what would be the one-half derivative of x?” (see[74]). In 1772, J.L. Lagrange introduced the differential operatorsof integer order and wrote (see [72]):

dm

dxmdn

dxny =

dn+m

dxn+my.

In 1819, S.F. Lacroix developed a more mathematical generaliz-ing from a case of integer order [73]. Namely he presented the nthderivative in the following form:

Dnxm =dn

dxn(xm) =

m!

(n+m)!xm−n,

with y = xm and m,n ∈ Z such that m ≥ n. Replacing the factorialsymbol by the Gamma function, he developed the formula for thefractional derivative of a power function:

Dαxβ =Γ (β + 1)

Γ (β − α + 1)xβ−α.

where α and β are fractional numbers. Then he gave the examplethat the derivative of order 1

2for y = x is as follows:

d12x

dx12

=2√x√π.

Note that this interesting result of S.F. Lacroix is the same asthe Riemann-Liouville definition of a fractional derivative.

However, this above authors did not define derivatives of arbi-trary order and they gave no applications or examples. The firstapplication was presented by N.H. Abel [2] in 1823. He applied thefractional calculus in the solution of an integral equation. Abel’s so-lution was so elegant that it attracted the attention of J. Liouville.In 1832 he took the first step to solve differential equations involv-ing fractional operators (see [75]). Moreover, he gave his definition

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INTRODUCTION 3

of a fractional derivative:

Dαxa =(−1)αΓ (a+ α)

Γ (a)x−a−α, a > 0,

for α ∈ R. He was successful in applying this definition to prob-lems in potential theory. One of the most useful advances in thedevelopment of fractional calculus was due to a paper written byB. Riemann during his student days. Seeking to generalize a Tay-lor series in 1853, he derived a different definition that involved thedefinite integral in the following form [116]:

Dαf(x) =1

Γ (α)

x∫

c

(x− t)α−1f(t)dt+ ψ(x), a > 0.

Because of the ambiguity in the lower limit of integration c, headded to his definition a complementary function ψ(x).

In the period 1900-1970 a modest amount of published worksappeared on the subject of fractional calculus. Some of the con-tributors were G.H. Hardy and J.E. Littlewood [56], M. Riesz [108],S. Samko [110] and H. Weyl [90].

The theory of discrete fractional calculus is far less developed.It seems as no significant work appeared in this area until someones primarily devoted to applications of the fractional calculusto ordinary and partial differential equations appeared, with themajority of interest shown only within the past thirty years. In1974, J.B. Diaz and T. J. Osler [38] introduced a discrete fractionaldifference operator defined as an infinite series, a generalization ofthe binomial formula for the Nth-order difference operator ∆N inthe following form:

∆αf(x) =∞∑

k=0

(−1)

(N

k

)f(x+ α + k),

More results concerning the process to develop the analogoustheory for fractional finite differences was proved by K.S. Millerand B. Ross [87]. In particulary, they presented more rules forcomposing fractional sums. In 2007, this direction was studied byF.M. Atici and P.W. Eloe, which in [20] discussed some of the

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4 INTRODUCTION

properties of this factorial function in the following form:

t(α) =Γ (t+ 1)

Γ (t+ 1− α),

for α ∈ R, which is generalization of factorial polynomial:

t(n) =n−1∏

j=0

(t− j) =Γ (t+ 1)

Γ (t+ 1− n),

where Γ denotes the special gamma function and if t + 1 − j = 0for some j, then we assume the product to be zero. We shall usethe convention that division at a pole yields zero.

During the last two decades, the h-calculus has been successfullyapplied to several fields within mathematics, see e.g. [3], [4], [37],[71], [83], [96], [104] and [109] the references therein. Finally, wemention that h- calculus is also important in applied fields such aseconomics, engineering and physics (see, e.g. [8], [9], [77], [79], [82],[97]).

q-calculus: The most common language of quantum calculusis based on the q-calculus, but this kind of calculus had alreadybeen represented by L. Euler [44] in the 18th century. The study ofq-calculus started in 1748 when L. Euler [44] considered the infinite

product (q; q)−1∞ =∞∏k=0

11−qk+1 , |q| < 1, as a generating function for

p(n) (the partition function p(n) is the number of ways to writen as a sum of integers). Furthermore, he discovered the first twoq-exponential functions, a prelude to the q-binomial theorem (see[42]). One hundred years later the progress of investigation contin-ued under E. Heine, who in 1846 considered a generalization of thehypergeometric (q-hypergeometric) series (see [46]), given by theformula

(2) 2φ1 =∞∑

n=0

(a; q)n(b; q)n(q; q)n(c; q)n

zn, | z |< 1,

where the q-shifted factorial is defined by

(a; q)n =

1, n = 0,n−1∏m=0

(1− aqm), n ∈ N,

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INTRODUCTION 5

and the series (2) converges absolutely for 0 <| q |< 1. When q → 1we get the Gauss’ series.

The q-analog of the gamma function was obtained by J. Thomae[127] and later by F.H. Jackson [58] in the following form:

Γq(x) =(q; q)∞(qx; q)∞

(1− q)1−x, 0 < q < 1,

for x ∈ R \ 0,−1,−2, · · · .In 1908 F.H. Jackson [59] (see also [36]) reintroduced the Euler-

Jackson q-difference operator

Dqf(x) =f(x)− f(qx)

(1− q)x , x ∈ (0, b), q ∈ C \ 1,

for f : [0, b) −→ R, 0 ≤ b <∞. It is clear that if f(x) is differen-tiable, then lim

q−→1Dqf(x) = f ′(x). The q-derivative is a discretiza-

tion of the ordinary derivative and therefore has immediate appli-cations in numerical analysis. At the same time the investigation ofthe q-difference equations theory considered in intensive works es-pecially by C.R. Adams [1], R.D. Carmichael [35] and F.H. Jackson[61] and others.

Moreover, in 1910 F.H. Jackson [60] gave the more general q-integral definition

(3)

x∫

0

f(t)dqt = (1− q)x∞∑

k=0

qkf(qkx), x ∈ (0,∞),

and the improper q-integral of a function f(x) : [0,∞)→ R, by theformula

(4)

∞∫

0

f(t)dqt = (1− q)∞∑

k=−∞qkf(qk).

for 0 < |q| < 1. Note that the series on the right hand sides of (3)and (4) converge absolutely.

The study the q-fractional calculus started from the works ofR.P. Agarwal [5], [6] , W.A. Al-Salam [11], [12], W.A. Al-Salam andVerma [14]. They introduced several types of fractional q-integraloperators and fractional q-derivatives and here in particular was

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6 INTRODUCTION

defined the fractional q-integral of the Riemann-Liouville type Iq,αdefined by

Iq,αf(x) :=xα−1

Γq(α)

x∫

0

(qs/x; q)α−1f(s)dqs, x > 0,(5)

with q ∈ C \ 1, where (qs/x; q)α = (qs/x;q)∞(qαqs/x;q)∞

, α ∈ R+.

In 1987, the applications of q-calculus in the area of approxi-mation theory was initiated by A. Lupas [76], who first introducedq-Bernstein polynomials and the development of this direction hasbeen remarkable, and the most important things can be found in thebook [19]. We also mention the work of T. Ernst [42], where he pre-sented applications of the q-calculus in many subjects, like umbralcalculus, oscillations in q-calculus, interpolation theory, quantumgroups, quantum algebras, hypergeometric series, complex analysisand particle physics.

Today the interest in this subject has exploded and the q-calculus has in the last twenty years served as a bridge betweenmathematics and physics. The q-calculus has numerous applica-tions in various fields of mathematics e.g dynamical systems, num-ber theory, combinatorics, special functions, fractals and also forscientific problems in some applied areas such as computer sci-ence, quantum mechanics and quantum physics (see e.g. [18], [23],[42], [43], [45]). Most of the additional information can be foundin the work of G. Gasper and M. Rahman [46], where was givensimpler proofs of many results (for example: q-Clausen’s formula,q-orthogonal polynomials, q-analogues of various product formu-las, etc.) and important applications to other fields (for exam-ple: modern algebra, real and complex variables, number theory,etc.). Moreover, for the further development and recent results inq-calculus we refer to the books [18], [19], [36] and [42] and thereferences given therein.

The first results concerning integral inequalities in q-calculuswere proved in 2004 by H. Gauchman [47] (for example: Stef-fensen’s, Hermite-Hadamard’s, Iyengar’s and Chebyshev’s inequal-ities). Later on some further q-analogs of the classical inequali-ties have been proved in the papers [68], [86] and [123]. We also

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INTRODUCTION 7

pronounce the recent book [15] by G.A. Anastassiou, where manyimportant q-inequalities are proved and discussed.

An essential part of this PhD thesis is devoted to obtain nec-essary and sufficient conditions for the validity of Hardy-type in-equalities in q-calculus. So we focus now our interest on the historyand references concerning weighted Hardy-type inequalities.

The theory of Hardy-type inequalities is a wonderful mathemat-ical subject with a proud history and it is very applicable both inmathematics and to problems in many areas outside the mathemat-ical scienes. The investigation of Hardy-type inequalities began in1915 with the work of G.H. Hardy. In 1925 he proved the followingresults (see [52]):

Theorem 0.1. Let p > 1 and an∞n=1 be a sequence of non-

negative real numbers, such that the series∞∑n=1

apn converges. Then

the inequality∞∑

n=1

(1

n

n∑

k=1

ak

)p

≤(

p

p− 1

)p ∞∑

n=1

apn,(6)

holds.

Theorem 0.2. Let p > 1 and f is a non-negative p-integrablefunction on (0, ∞). Then f is integrable over the interval (0, x)for all x > 0 and the inequality

∞∫

0

1

x

x∫

0

f(t)dt

p

dx ≤(

p

p− 1

)p ∞∫

0

fp(x)dx,(7)

holds.

In fact, by applying Theorem 0.2 with step functions we findthat Theorem 0.2 implies Theorem 0.1.

The inequalities (6) and (7) are called the discrete Hardy in-equality and the continuous Hardy inequality, respectively. More-

over, the constant

(p

p− 1

)pin both inequalities (6) and (7) is sharp

in the sense that it can not be replaced by any smaller number.Nowadays, a lot of books and articles have been dedicated to

the investigation and generalization of Hardy inequalities. The first

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8 INTRODUCTION

book on the Hardy inequality was the book of G.H. Hardy, J.E.Littlewood and G. Polya [55] in 1934. The first book which wascompletely devoted to the Hardy inequality, was published in 1990by B. Opic and A. Kufner [95]. We also mention here the recentbook of A. Kufner, L.-E. Persson and N. Samko [69], which is de-voted to give a basic overview of weighted Hardy-type inequalitiesincluding the most recent developments and open questions (seealso [98]). A description of the most important steps in the devel-opment of Hardy-type inequalities has been described by A. Kufner,L. Maligranda and L.-E. Persson [70].

In 1928 G.H. Hardy [53] proved the first weighted form of in-equality (7) as follows:

Theorem 0.3. Let p > 1, α < p− 1. Then the inequality∞∫

o

1

x

x∫

0

f(t)dt

p

xαdx ≤(

p

p− α− 1

)p ∞∫

0

fp(x)xαdx(8)

holds for all measurable non-negative functions f . Moreover, the

constant

(p

p− α− 1

)pis the best possible.

It was later on discovered that Theorem 0.3 is not a genuinegeneralization of Theorem 0.2. In fact, both inequalities (7) and(8) can equivalently be transformed to the same basic Hardy-typeinequality (which in turn follows easily from Jensen’s inequality, see[99]).

During the recent decades the inequalities (6) and (7) have beendeveloped to the following forms:

( ∞∑

n=1

un

∣∣∣∣∣n∑

k=1

ak

∣∣∣∣∣

r) 1r

≤ C

( ∞∑

n=1

|an|pvn) 1

p

,(9)

b∫

a

∣∣∣∣∣∣

x∫

a

f(t)dt

∣∣∣∣∣∣

r

u(x)dx

1r

≤ C

b∫

a

|f(x)|pv(x)dx

1p

,(10)

respectively, which are called weighted Hardy-type inequalities.In 1930 G.H. Hardy and J.E. Littlewood [54] studied inequality

(10) with parameters p and r, 1 < p < r < ∞. More exactly, for

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INTRODUCTION 9

the interval (0, ∞), they considered the weight functions v(x) ≡ 1,u(x) = xs−r with s = r−p

pand obtained inequality (10). Moreover,

G.A. Bliss found the best constant in this case [31]. Since thisresult by Bliss from 1930 it has been an open question to findthe best constant for other power weights than v(x) = 1. Thisproblem was recently solved by L.E. Persson and S. Samko [100].The investigation of (10) started for the case p = q in the papers ofP.R. Beesack [25], [26], J. Kadlec and A. Kufner [63], V.R. Portnov[106], V.N. Sedov [111], F.A. Sysoeva [124] and others.

It should be noted that in 1969 G. Talenti [125] and G.A.Tomaselli [128] obtained that the condition

supt>0

∞∫

t

u(x)dx

1p

t∫

0

v1−p′(y)dy

1p′

<∞,

is necessary and sufficient for the validity of inequality (10) in thecase p = r, where p′ = p

p−1 . Nowadays this condition is called the

Muckenhoupt condition in honour of B. Muckenhoupt, who in 1972presented a very nice proof of this result in [89] even in a moregeneral form with 1 ≤ p = r <∞.

The study of the case with different parameters p and r wasstarted by J. S. Bradley in [32]. He considered inequality (10) on(0, ∞) and proved that the condition

supt>0

∞∫

t

u(x)dx

1r

t∫

0

v1−p′(y)dy

1p′

<∞,

is necessary for (10) to hold for all 1 ≤ p, r ≤ ∞ and that it is alsosufficient for 1 ≤ p ≤ r <∞.

In thelast century weighted Hardy-type inequalities have beenintensively studied by several authors e.g. K.F. Andersen andB. Muckenhoupt [16], G. Bennett [29], R.K. Juberg [62], V.M. Kok-ilashvili [67], V.G. Maz’ya [84], L.-E. Persson and V.D. Stepanov[103], G. Sinnamon [112], [113], G. Sinnamon and V.D. Stepanov[114], V.D. Stepanov [118] and others. Moreover, for more infor-mation we refer to the books [30],[55], [69], [70], [88], [95] and tothe PhD theses of Z. Abdikalikova [13], A. Abylayeva [10], S. Barza

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10 INTRODUCTION

[24], M. Nassyrova [91], C. A. Okpoti [94], O. Popova [105], D.V.Prokhorov [107], A. Temirkhanova [126], E. Ushakova [129] and A.Wedestig [130].

A more general version of the inequality (10) with non-negativekernel K(·, ·) have been studied by many authors e.g. F. Martin-Reyes and E. Sawyer [81], R. Oinarov [92], [93], L.-E. Persson andV.D. Stepanov [103], D.V. Prokhorov [107], V.D. Stepanov [119]and [121]. In the papers [120] and [122] V.D. Stepanov studied theinequality (10) for the Riemann-Liouville integral operator. Theseworks have given an impulse for further development of such Hardy-type inequalities also in q-calculus.

The first result related to inequality (9) belongs to K.F. An-dersen and H.P. Heinig ([17], Theorem 4.1). In 1985 H.P. Heinig[57] obtained a sufficient condition for the validity of inequality(9). In 1987-1991 G. Bennett [27], [28] and [29] gave a full char-acterization of the weighted inequality (9), except for the case0 < r < 1 < p < ∞. The remaining case was characterized byM.S. Braverman and V.D. Stepanov [33] in 1994. C.A. Okpoti [94]in his PhD thesis proved that for the case 1 < p ≤ r <∞ there areeven infinite many conditions characterizing (9).

In 2014, the first Hardy-type inequality in q-calculus was ob-tained by L. Maligranda, R. Oinarov and L.-E. Persson [78]. Theyproved that the q-analog of Theorem 0.3 is the following:

Theorem 0.4. Let α < p−1p

. If either 1 ≤ p <∞ and f ≥ 0 or

p < 0 and f > 0, then the inequality

(11)

∞∫

0

xp(α−1)

x∫

0

t−αf(t)dqt

p

dqx ≤ C

∞∫

0

fp(x)dqx,

holds with constant

C =1

[p−1p− α]pq

,

where [α]q = 1−qα1−q , α ∈ R.

In the case when 0 < p < 1 the inequality (11) for f > 0 holdsin the reverse direction with the same constant C. Moreover, in allthe three cases the constant C is the best possible.

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INTRODUCTION 11

Remark 0.5. The constant in the inequality (11) is smallerthan the one in (8). In fact, if α < 1 − 1/p with p > 1 or p < 0,then

(12)1

[(p− 1)/p− α]q<

1

p− pα− 1,

for α > −1p. Inequality (12) is reversed if α < −1

p. For α = −1

p

both sides in (12) are equal to 1.Estimate (12) means that (1−q)/(1−q(1−p)/p−α) < p/(p−pα−1)

for any 0 < q < 1, which is true since the function h(q) := p(1 −q(p−1)/p−α)/(p−pα−1)+q−1 has the derivative h′(q) = −q−1/p−α+1 < 0 for α > −1/p, and so h(q) > h(1) = 0.

Remark 0.6. From Theorem 0.4 with α = 0 we obtain theq-analog of the inequality (7)

(13)

∞∫

0

1

x

x∫

0

f(t)dqt

p

dqx ≤1

[(p− 1)/p]pq

∞∫

0

fp(x)dqx, f ≥ 0,

if p > 1 or p < 0 and f > 0. Moreover, the constant 1[(p−1)/p]pq is

best possible and 1[(p−1)/p]pq <

(pp−1

)p.

Moreover, the following generalization of the inequality (14)with the operator (5) involved was also obtained in [78]:

Theorem 0.7. Let p > 1 and α > 0. Then the inequality

∞∫

0

1

xαΓ(α)

x∫

0

(x− qt)α−1q f(t)dqt

p

dqx ≤ C

∞∫

0

fp(x)dqx,

holds with best constant

C =

[Γq(1− 1

p)

Γq(α + 1− 1p)

]p,

where

(x− qt)α−1q = xα−1(qt/x; q)α−1,

(b/a; q)α−1 =(qt/x; q)∞

(qα−1b/a; q)∞, α ∈ R.

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12 INTRODUCTION

Remark 0.8. Up to now there is no sharp discrete analogueof the inequality (8). For examples, the following two inequalitieswere claimed to hold by G. Bennett([27, p. 40-41]; see also [28, p.407]):

∞∑

n=1

[1

n1−α

n∑

k=0

[kα−1 − (k − 1)α−1

]ak

]p≤

[1− α

p− αp− 1

]p ∞∑

n=1

apn, an ≥ 0,

and

∞∑

n=1

1n∑i=1

i−α

n∑

k=1

k−αak

p

≤[

1− αp− αp− 1

]p ∞∑

n=1

apn, an ≥ 0,

whenever α > 0, p > 1, αp > 1. Both the inequalities were provedindependently by P. Gao [49, Corollary 3.1-3.2]( see also [50, Theo-rem 1.1] and [51, Theorem 6.1]) for p ≥ 1 and some special cases ofα (This means that there are still some regions of parameters withno proof of (8)). Moreover, in [78, Theorems 2.1 and 2.3] was gotproved an other sharp discrete analogue of the inequality (8) in thefollowing form:

∞∑

n=−∞

[1

qnλ

n∑

k=0

qkλak

]p≤ 1

(1− qλ)p∞∑

n=−∞apn, an ≥ 0,

and

∞∑

n=1

[1

qnλ

n∑

k=0

qkλak

]p≤ 1

(1− qλ)p∞∑

n=1

apn, an ≥ 0,

for 0 < q < 1, p ≥ 1 and α < 1− 1/p, where λ := 1− 1/p− α.

Remark 0.9. For p ≥ 1 and α < 1 − 1/p we establish theh-analogue of inequality (8) in fractional h-discrete calculus with

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INTRODUCTION 13

sharp constant in the following form (see [101, Theorem 3.1]):

(14)

∞∫

0

x(α−1)h

δh(x)∫

0

f(t)dht

t(α)h

p

dhx

≤(

p

p− αp− 1

)p ∞∫

0

fp(x)dhx, f ≥ 0,

which is an other discrete analogue of the inequality (8). By usingdefinitions of h-integral and factorial function in (14) we find that

∞∑

n=0

(n(α−1)

n∑

k=0

akk(α)

)p

≤(

p

p− αp− 1

)p ∞∑

n=0

apk, ak ≥ 0.

Hence, by limh→0

n(α)h = nα, we have that the discrete analogue of

the inequality (8) which is the more suitable variant other discreteanalogue above.

There have been of great interest recently on difference equa-tions in quantum calculus. It is caused by the development of thetheory of q-calculus and h-calculus and also by its applications,see [1], [7], [19], [20], [22], [71], [104], [109]. Moreover, quantumcalculus play increasingly important roles in the modeling of someengineering and science problems, as shown in [34], [36], [38], [40],[42], [43], [48], [65], [66], [85], [110]. It has been established that, inmany situations, these models in quantum calculus are more suit-able than the analogous models with integer derivatives and limit.See [39], for details. Since the integral inequalities, with explicitestimates of the sharp constants are so important in the study ofproperties of solutions of differential and integral equations, theirfinite difference (or discrete) analogues are also useful in the studyof properties of solutions of finite difference and fractional differenceequations equations in quantum calculus. One of the best knownand widely used inequalities in the study of difference equations isHardy-type inequalities, see e.g [52] and [53]. In this connection werefer to [69] and [70].

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14 INTRODUCTION

2. A short description of the main contributions in thisPhD thesis.

In paper A (see also [102])we study an operator Iq,n of the fol-lowing form:

Iq,nf(x) =1

Γq(n)

∞∫

0

X(0,x](s)Kn−1(x, s)f(s)ds, ,

where n ∈ N and Kn−1(x, s) = (x−qs)n−1q . The conjugate operatorI∗q,n is defined by

I∗q,nf(s) :=1

Γq(n)

∞∫

0

X[s,∞)(x)Kn−1(x, s)f(x)dx.

Let 1 < r, p ≤ ∞. Then the q-analog of the Hardy-type in-equality for the operator Iq,n is of the following form:

(15)

∞∫

0

ur(x) (Iq,nf(x))r dqx

1r

≤ C

∞∫

0

vp(x)fp(x)dqx

1p

.

The dual inequality of the inequality (15) reads:

(16)

∞∫

0

ur(x)(I∗q,nf(x)

)rdqx

1r

≤ C∗

∞∫

0

vp(x)fp(x)dqx

1p

.

For 0 ≤ m ≤ n− 1,m+1, n ∈ N, we use the following notations:

Qn−1m =

∞∫

0

∞∫

0

X(0,z](s)Kp′m(z, s)v−p

′(s)dqs

p(r−1)p−r

×

∞∫

0

X[z,∞)(x)Krn−m−1(x, z)u

r(x)dqx

pp−r

× Dq

∞∫

0

X(0,z](s)Kp′m(z, s)v−p

′(s)dqs

p−rpr

,

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INTRODUCTION 15

Qn−1m =

∞∫

0

∞∫

0

X(0,z](s)Krm(z, s)ur(s)dqs

rp−r

×

∞∫

0

X[z,∞)(x)K−p′

n−m−1(z, x)v−p′(x)dqx

r(p−1)p−r

× Dq

∞∫

0

X(0,z](s)Krm(z, s)ur(x)dqs

p−rpr

,

Hn−1m = sup

z>0

∞∫

0

X[z,∞)(x)Krn−m−1(x, z)ur(x)dqx

1r

∞∫

0

X(0,z](s)Kp′m(z, s)v−p

′(s)dqs

1p′

,

Hn−1m = sup

z>0

∞∫

0

X(0,z](x)Krm(z, x)ur(x)dqx

1r

∞∫

0

X[z,∞)(s)Kp′n−m−1(s, z)v

−p′(s)dqs

1p′

,

A+(z) =

∞∫

0

X[z,∞)(x)ur(x)

∞∫

0

X(0,z](t)Kp′n−1(x, t)v

−p′(t)dqt

rp′

dqx

1r

,

A−(z) =

∞∫

0

X(0,z](t)v−p′(t)

∞∫

0

X[z,∞)(x)Krn−1(x, t)u

r(x)dqx

p′r

dqt

1p′

,

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16 INTRODUCTION

A+(z) =

∞∫

0

X[z,∞)(t)v−p′(t)

∞∫

0

X(0,z](x)Krn−1(t, x)ur(x)dqx

p′r

dqt

1p′

,

A−(z) =

∞∫

0

X(0,z](x)ur(x)

∞∫

0

X[z,∞)(t)Kp′n−1(t, x)v−p

′(t)dqt

rp′

dqx

1r

,

Hn−1 = max0≤k≤n−1

Hn−1k , Hn−1 = max

0≤k≤n−1Hn−1k , A+

q = supz>0

A+(z),

A−q = supz>0

A−(z), A+q = sup

z>0A+(z), A−q = sup

z>0A−(z),

Qn−1 = max0≤k≤n−1

Qn−1k and Qn−1 = max

0≤k≤n−1Qn−1k .

Our main results read:

Theorem 0.10. (i) Let 1 < r < p < ∞. Then the inequality(15) holds if and only if Qn−1 <∞. Moreover, Qn−1 ≈ C, where Cis the best constant in (15).

(ii) Let 1 < p ≤ r < ∞. Then the inequality (15) holds if andonly if at least one of the conditions Hn−1 < ∞ or A+

q < ∞ orA−q < ∞ holds. Moreover, Hn−1 ≈ A+

q ≈ A−q ≈ C, where C is thebest constant in (15).

Theorem 0.11. (i) Let 1 < r < p < ∞. Then the inequality(16) holds if and only if Qn−1 < ∞. Moreover, Qn−1 ≈ C∗, whereC∗ is the best constant in (16) .

(ii) Let 1 < p ≤ r < ∞. Then the inequality (16) holds if andonly if at least one of the conditions Hn−1 < ∞ or A+

q < ∞ orA−q <∞ holds. Moreover, Hn−1 ≈ A+

q ≈ A−q ≈ C, where C is thebest constant in (16).

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INTRODUCTION 17

In paper B (see also [21]) we consider the inequality of the fol-lowing form:

(17)

∞∫

0

u(x)

∞∫

0

X(0,x](t)v(t)f(t)dqt

r

dqx

1r

≤ C

∞∫

0

fp(x)dqx

1p

,

and derive necessary and sufficient conditions (of Muckenhoupt-Bradley type) for the validity of the inequality (17) with all possiblepositive values of the parameters r and p. Here

∞∫

0

X(0,x](t)f(t)dqt = (1− q)∑

qi≤xqif(qi).

One main result reads:

Theorem 0.12. Let 1 < p ≤ r < ∞. Then the inequality (17)holds if and only if

D1 = supz>0

∞∫

0

X[z,∞)(x)ur(x)dqx

1r∞∫

0

X(0,z](t)vp′(t)dqt

1p′

<∞

or

D2 = supz>0

∞∫

0

X(0,z](t)vp′(t)dqt

− 1p

∞∫

0

X(0,z](x)ur(x)

∞∫

0

X(0,z](t)vp′(t)dqt

r

dqx

1r

<∞

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18 INTRODUCTION

or

D3 = supz>0

∞∫

0

X[z,∞)(x)ur(x)dqx

− 1r′

∞∫

0

X[z,∞)(t)vp′(t)

∞∫

0

X[z,∞)(x)ur(x)dqx

p′

dqt

1p′

<∞.

Moreover, for the sharp constant in (17) we have that C ≈ D1 ≈D2 ≈ D3.

We also study the dual inequality of the inequality (17) as fol-lows:(18)∞∫

0

v(t)

∞∫

0

X[x,∞)(x)u(x)g(x)dqx

p′

dqt

1p′

≤ C

∞∫

0

gr′(t)dqt

1r′

,

where

∞∫

0

X[x,∞)(t)f(t)dqt = (1− q)∑

qi≥xqif(qi).

In this case our main result is the following:

Theorem 0.13. Let 1 < p ≤ r < ∞. Then the inequality (18)holds if and only if

D∗1 = supz>0

∞∫

0

X(0,z](x)ur(x)dqx

1r∞∫

0

X[z,∞)(t)vp′(t)dqt

1p′

<∞

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INTRODUCTION 19

or

D∗2 = supz>0

∞∫

0

X[z,∞)(t)vp′(t)dqt

− 1p

∞∫

0

X[z,∞)(x)ur(x)

∞∫

0

X[z,∞)(t)vp′(t)dqt

r

dqx

1r

<∞

or

D∗3 = supz>0

∞∫

0

X(0,z](x)ur(x)dqx

− 1r′

∞∫

0

X(0,z](t)vp′(t)

∞∫

0

X(0,z](x)ur(x)dqx

p′

dqt

1p′

<∞.

Moreover, for the sharp constant in (18) we have that C ≈ D∗1 ≈D∗2 ≈ D∗3.

Concerning other possible parameters of p and r we have thefollowing complement of Theorem 0.10 (Theorem 0.11):

Theorem 0.14. (i). Let 0 < p ≤ 1, p ≤ r < ∞. Then theinequality (17) holds if and only if

D4 = supz>0

∞∫

0

X[z,∞)(x)ur(x)dqx

1r∞∫

0

X(qz,z](t)vp′(t)dqt

1p′

<∞.

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20 INTRODUCTION

(ii). Let 1 < p < ∞, 0 < r < p. Then the inequality (17) holds ifand only if

D5 =

∞∫

0

∞∫

0

X(0,z](t)vp′(t)dqt

r(p−1)p−r

∞∫

0

X[z,∞)(x)ur(x)dqx

rp−r

ur(z)dqz

p−rpr

<∞.

(iii). Let 0 < r < p = 1. Then the inequality (17) is satisfied if andonly if

D6 =

∞∫

0

supy<z

∞∫

0

X(qy,y](t)v(t)

(1− q)tdqt

r1−r

∞∫

0

X[z,∞)(x)ur(x)dqx

r1−r

ur(z)dqz

1−rr

<∞.

In all cases (i)-(iii), for the best constant in (17) it yields thatC ≈ Di, i = 4, 5, 6, respectively.

Theorem 0.15. (i). Let 0 < p ≤ 1, p ≤ r < ∞. Then theinequality (18) holds if and only if

D∗4 = supz>0

∞∫

0

X(0,z](x)ur(x)dqx

1r∞∫

0

X[z,q−1z)(t)vp′(t)dqt

1p′

<∞.

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INTRODUCTION 21

(ii). Let 1 < p < ∞, 0 < r < p. Then the inequality (18) holds ifand only if

D∗5 =

∞∫

0

∞∫

0

X(0,z](x)ur(x)dqx

rp−r

∞∫

0

X[z,∞)(t)vp′(t)dqt

r(p−1)p−r

ur(z)dqz

)

p−rpr <∞.

(iii). Let 0 < r < p = 1. Then the inequality (18) holds if and onlyif

D∗6 =

∞∫

0

supy≥z

∞∫

0

X[y,q−1y)(t)v(t)

(1− q)tdqt

r1−r

∞∫

0

X(0,z](x)ur(x)dqx

r1−r

ur(z)dqz

1−rr

<∞.

In all cases (i)-(iii), for the best constant in (18) it yields thatC ≈ D∗i , i = 4, 5, 6, respectively.

In paper C (see also [115]) we investigate the inequality of thefollowing form:

(19)

∞∫

0

ur(x)

x∫

0

sγ−1 lnqx

x− qsf(s)dqs

r

dqx

1r

≤ C

∞∫

0

fp(s)dqs

1p

, ∀f(·) ≥ 0,

where

lnqx

x− s :=∞∑

j=1

( sx)j

[j]q.

and C is a positive finite constant independent of f .

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22 INTRODUCTION

Our main results for the inequality (19) read as follows:

Theorem 0.16. Let 1 < p ≤ r <∞, γ > 1p. Then the inequal-

ity (19) holds if and only if B1 <∞, where

B1 := supx>0

xγ+ 1

p′

∞∫

0

X[x,∞)(t)ur(t)

trdqt

1r

,

Moreover, B1 ≈ C , where C is best constant in (19).

Theorem 0.17. Let 0 < r < p < ∞, 1 < p and γ > 1p. Then

the inequality (19) holds if and only if B2 <∞, where

B2 :=

∞∫

0

xγ+

1p′

∞∫

0

X[x,∞)(t)ur(t)

trdqt

1r

prp−r

dqx

p−rpr

.

Moreover, B2 ≈ C , where C is best constant in (19).

Remark 0.18. The only q-analogs of Hardy-type inequalities sofar are that in Theorem A and those presented in this PhD thesis.Hence, it remains a great number of open questions.

In paper D (see also [101]) we consider the first power weightedversion of Hardy’s inequality (see (8)) in slightly rewritten form.Our first main result is the following h-integral analogue of theinequality (8) reads for 1 ≤ p <∞:

Theorem 0.19. Let α < p−1p

and 1 ≤ p < ∞. Then the in-

equality

(20)

∞∫

0

x(α−1)h

δh(x)∫

0

f(t)dht

t(α)h

p

dhx

≤[

p

p− αp− 1

]p ∞∫

0

fp(x)dhx,

holds for all f ≥ 0. Moreover, the constant[

pp−αp−1

]pis the best

possible in (20).

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INTRODUCTION 23

Our second main result is the following h-integral analogue ofthe reversed form of (7) for 0 < p < 1.

Theorem 0.20. Let α < p−1p

and 0 < p < 1 . Then the in-

equality

(21)

∞∫

0

fp(x)dhx ≤

[p− pα− 1

p

]p ∞∫

0

x(α−1)h

δh(x)∫

0

f(t)dht

t(α)h

p

dhx,

holds for all f ≥ 0. Moreover, the constant[p−pα−1

p

]pis the best

possible in (21).

Remark 0.21. The only h-analogs of Hardy-type inequalitiesso far are that in Theorem A and those presented in this PhD thesis.Hence, it remains a great number of open questions.

I the paper E (see also [117]) we consider the fractional orderHardy-type inequality in the following form:

∞∫

0

∞∫

0

|f(x)− f(y)|p

|x− y|1+pαdxdy

p

≤ C

∞∫

0

|f ′(x)|p x(1−α)pdx

p

for 0 < α < 1 and 1 < p < ∞ in fractional h-discrete calculus,

where C = 21p α−1

(p−pα)1p

.

Concerning this inequality and similar ones we refer to Chapter5 of the book [69] by A. Kufner, L.-E. Persson and N. Samko. Inthis case our main result read:

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24 INTRODUCTION

Theorem 0.22. Let 1 < p < ∞, 0 < α < 1 and f(x) =DhF (x). Then the following inequality

(22)

∞∫

0

∞∫

0

|F (x)− F (y)|pdhxdhy[(|x− y|+ 3h)

( 1p+α)

h

]p

1p

≤ C

∞∫

0

|f(x)|p dhx[(x+ h)

(α−1)h

]p

1p

,

holds with constant C = 21p α−1

(p−pα)1p

. Moreover, this constant sharp.

Remark 0.23. The inequality (22) is the only fractional orderHardy inequality in h-calculus in the literature so also have theseare many open questions for further research (s.g. Chapter 5 of thebook [69]).

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Paper A

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Persson and Shaimardan Journal of Inequalities and Applications ( 2015) 2015:296 DOI 10.1186/s13660-015-0816-z

RESEARCH Open Access

Some new Hardy-type inequalities forRiemann-Liouville fractional q-integraloperatorLars-Erik Persson1,2* and Serikbol Shaimardan3

*Correspondence: [email protected]å University of Technology,

Luleå, 971 87, Sweden2Narvik University College, P.O. Box

385, Narvik, 8505, Norway

Full list of author information is

available at the end of the article

AbstractWe consider the q-analog of the Riemann-Liouville fractional q-integral operator of

order n ∈ N. Some new Hardy-type inequalities for this operator are proved and

discussed.

MSC: Primary 26D10; 26D15; secondary 33D05; 39A13

Keywords: inequalities; Hardy-type inequalities; Riemann-Liouville operator; integral

operator; q-calculus; q-integral

1 IntroductionIn FH Jackson defined q-derivative and definite q-integral [] (see also []). It was the

starting point of q-analysis. Today the interest in the subject has exploded. The q-analysis

has numerous applications in various fields of mathematics, e.g., dynamical systems, num-

ber theory, combinatorics, special functions, fractals and also for scientific problems in

some applied areas such as computer science, quantum mechanics and quantum physics

(see, e.g., [–]). For further development and recent results in q-analysis, we refer to the

books [, ] and [] and the references given therein. The first results concerning integral

inequalities in q-analysis were proved in by Gauchman []. Later on some further

q-analogs of the classical inequalities have been proved (see [–]). Moreover, in

Maligranda et al. [] derived a q-analog of the classical Hardy inequality. Further devel-

opment of Hardy’s original inequality from (see [] and []) has been enormous.

Some of the most important results and applications have been presented and discussed

in the books [, ] and []. Hence, it seems to be a huge new research area to investigate

which of these so-called Hardy-type inequalities have their q-analogs.

The aim of this paper is to obtain some q-analogs of Hardy-type inequalities for the

Riemann-Liouville fractional integral operator of order n ∈ N and to find necessary and

sufficient conditions of the validity of these inequalities for all non-negative real functions

(see Theorems . and .).

The paper is organized as follows. In order not to disturb our discussions later on, some

preliminaries are presented in Section . Themain results can be found in Section , while

the detailed proofs are given in Section .

© 2015 Persson and Shaimardan. This article is distributed under the terms of the Creative Commons Attribution 4.0 Interna-tional License (http://creativecommons.org/licenses/by/4.0/), which permits unrestricted use, distribution, and reproduction in anymedium, provided you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commonslicense, and indicate if changes were made.

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Persson and Shaimardan Journal of Inequalities and Applications ( 2015) 2015:296 Page 2 of 17

2 PreliminariesFirst we recall some definitions and notations in q-analysis from the recent books [, ]

and [].

Let q ∈ (, ). Then a q-real number [α]q is defined by

[α]q := – qα

– q, α ∈ R,

where limq→–qα

–q= α.

The q-analog of the binomial coefficients is defined by

[n]q! :=

if n = ,

[]q × []q × · · · × [n]q if n ∈N ,

[n

k

]

q

:=[n]q!

[n – k]q![k]q!.

We introduce the q-analog of a polynomial in the following way:

(x – a)nq :=

if n = ,

(x – a)(x – qa) · · · (x – qn–a) if n ∈N ,()

(x – a)n+mq = (x – a)mq(x – qma

)nq, n,m = , , , . . . . ()

The q-gamma function Ŵq is defined by

Ŵq(n + ) := [n]q!, n ∈N .

For f : [,b)→ R, < b≤ ∞, we define the q-derivative as follows:

Dqf (x) :=f (x) – f (qx)

( – q)x, x ∈ [,b).

Clearly, if the function f (x) is differentiable at a point x ∈ (, ), then limq→Dqf (x) = f ′(x).

Let < a ≤ b < ∞. The definite q-integral (also called the q-Jackson integral) of a func-

tion f (x) is defined by the formulas

∫ a

f (x)dqx := ( – q)a

∞∑

k=

qk f(qka

). ()

Moreover, the improper q-integral of a function f (x) is defined by

∫ ∞

f (x)dqx := ( – q)

∞∑

k=–∞

qk f(qk

), ()

provided that the series on the right-hand sides of () and () converge absolutely.

Suppose that f (x) and g(x) are two functions which are defined on (,∞). Then

∫ ∞

f (x)Dq

(g(x)

)=

∞∑

j=

f(qj

)(g(qj

)– g

(qj+

)). ()

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Let be a subset of (,∞) and X(t) denote the characteristic function of . For all z:

< z < ∞, we have that

∫ ∞

X(,z](t)f (t)dqt = ( – q)∑

qi≤z

qif(qi

)()

and∫ ∞

X[z,∞)(t)f (t)dqt = ( – q)∑

qi≥z

qif(qi

). ()

Al-Salam (see [] and also []) introduced the fractional q-integral of the Riemann-

Liouville operator Iq,n of order n ∈N by

Iq,nf (x) :=

Ŵq(n)

∫ x

Kn–(x, s)f (s)dqs,

where Kn–(x, s) = (x – qs)n–q .

Next we will present a lemma (Lemma .) concerning discrete Hardy-type inequalities

which are proved in []. In this paper all authors studied inequalities of the form

(∞∑

j=

urj((Snf )j

)r)

r

≤ C

(∞∑

i=

vpi f

pi

) p

, ∀f ≥ ()

for the n-multiple discrete Hardy operator with weights of the form

(Snf )j =

∞∑

k=j

ω,k

k∑

k=

ω,k

k∑

k=

ω,k · · ·

kn–∑

kn–=

ωn–,kn–

kn–∑

i=

fi =

∞∑

i=j

An–(i, j)fi,

where u = ui∞i=, v = vi

∞i=, ωi = ωi,k

∞k= are positive sequences of real numbers (i.e.,

weight sequences). She also studied inequality () for the operator S∗n defined by

(S∗nf

)i:=

i∑

j=

fiAn–,(i, j),

which is the conjugate to the operator Sn, where An–,(i, j) ≡ for n = and

An–,(i, j) =

i∑

kn–=j

ωn–,kn–

i∑

kn–=kn–

ωn–,kn– · · ·

i∑

k=k

ω,k

for n≥ .

We consider the following Hardy-type inequalities:

(∞∑

j=–∞

urj((Snf )j

)r)

r

≤ C

(∞∑

i=–∞

vpi f

pi

) p

()

and

(∞∑

i=–∞

uri((S∗nf

)i

)r)

r

≤ C∗

(∞∑

i=–∞

vpi f

pi

) p

. ()

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In the sequel, for any p > , the conjugate number p′ is defined by p′ := p/(p – ), and

the considered functions are assumed to be non-negative. Moreover, the symbol M ≪ K

means that there exists α > such thatM ≤ αK , where α is a constant which may depend

only on parameters such as p, q, r. Similarly, the case K ≪ M. If M ≪ K ≪ M, then we

writeM ≈ K .

Lemma .

(i) Let < p≤ r < ∞ and n≥ . Then inequality () holds if and only if

A(n) = max≤m≤n–Am(n) < ∞, where

Am(n) = supk∈Z

(∞∑

j=k

Ap′

m,(j,k)v–p′

j

) p′

(k∑

i=–∞

Arn–,m+(k, i)u

ri

) r

, n ∈N.

Moreover, A(n) ≈ C, where C is the best constant in ().

(ii) Let < p≤ r < ∞ and n≥ . Then inequality () holds if and only if

A∗(n) = max≤m≤n–A

∗m(n) < ∞, where

A∗m(n) = sup

k∈Z

(∞∑

i=k

Arm,(i,k)u

ri

) r(

k∑

j=–∞

Ap′

n–,m+(k, j)v–p′

j

) p′

, n ∈N.

Moreover, A∗(n) ≈ C, where C is the best constant in ().

We also need the corresponding result for the case < r < p < ∞, which was proved in

[].

Lemma .

(i) Let < r < p <∞ and n≥ . Then inequality () holds if and only if

B(n) = max≤m≤n–Bm(n) < ∞, where

Bm(n) =

∞∑

i=–∞

(∞∑

j=i

Ap′

m,(j, i)v–p′

j

) p(r–)p–r

(i∑

k=–∞

Arn–,m+(i,k)u

rk

) pp–r

× +

(∞∑

j=i

Ap′

m,(j, i)v–p′

j

) p–rpr

, +Ei,j = Ei,j – Ei,j+,n ∈N.

Moreover,B(n) ≈ C, where C is the best constant in ().

(ii) Let < r < p <∞ and n≥ . Then inequality () holds if and only if

B∗(n) = max≤m≤n–B

∗m(n) < ∞, where

B∗m(n) =

∞∑

i=–∞

(∞∑

j=i

Arm,(j, i)u

rj

) rp–r

(i∑

k=–∞

Ap′

n–,m+(i,k)v–p′

k

) r(p–)p–r

× +

(∞∑

j=i

Arm,(j, i)u

rj

) p–rpr

, +Ei,j = Ei,j – Ei,j+,∀n ∈N.

Moreover,B∗(n) ≈ C∗, where C∗ is the best constant in ().

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Persson and Shaimardan Journal of Inequalities and Applications ( 2015) 2015:296 Page 5 of 17

Let (a(n)i,j ) be amatrix whose elements are non-negative and non-increasing in the second

index for all i, j: ∞ > i ≥ j > –∞, and the entries of the matrix a(n)i,j satisfy the following (so-

called discrete Oinarov condition):

a(n)i,j ≈

n∑

γ=

a(γ )i,k d

n,γk,j , γ = , , . . . ,n – ,n ∈N ()

for all ∞ > i≥ k ≥ j > –∞.

Remark . Note that the matrices (dγ ,mk,j ), γ = , , . . . ,m, m ≥ , are arbitrary non-

negative matrices which satisfy () (see []).

Moreover, in [] necessary and sufficient conditions for inequalities () and () were

proved for matrix operators with a matrix (a(n)i,j ) which satisfies (). For our purposes we

need such characterization on the following form.

Lemma .

(i) Let < p≤ r < ∞ and the entries of the matrix (a(n)i,j ) satisfy condition (). Then

inequality () for the operator (A–f )j :=∑∞

i=j a(n)i,j fi, j ∈ Z, holds if and only if at least

one of the conditionsB+ < ∞ orB– < ∞ holds, where

B– = sup

k∈Z

(∞∑

i=k

v–p′

i

(k∑

j=–∞

(a(n)i,j

)rurj

) p′

r)

p′

,

B+ = sup

k∈Z

(k∑

j=–∞

urj

(∞∑

i=k

(a(n)i,j

)p′

v–p′

i

) rp′

) r

.

Moreover,B+ ≈B– ≈ C, where C is the best constant in ().

(ii) Let < p≤ r < ∞. Let the entries of the matrix (a(n)i,j ) satisfy condition (). Then

inequality () for the operator (A+f )i :=∑i

j=–∞ a(n)i,j fj, i ∈ Z, holds if and only if at

least one of the conditions A+ <∞ or A– < ∞ holds, where

A– = sup

k∈Z

(∞∑

i=k

uri

(k∑

j=–∞

(a(n)i,j

)p′

v–p′

j

) rp′

) r

,

A+ = sup

k∈Z

(k∑

j=–∞

v–p′

j

(∞∑

i=k

(a(n)i,j

)ruri

) p′

r)

p′

.

Moreover, A+ ≈A– ≈ C, where C is the best constant in ().

3 Themain resultsLet < r,p ≤ ∞. Then the q-analog of the two-weighted inequality for the operator Iq,n

of the form

(∫ ∞

ur(x)(Iq,nf (x)

)rdqx

) r

≤ C

(∫ ∞

vp(x)f p(x)dqx

) p

()

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Persson and Shaimardan Journal of Inequalities and Applications ( 2015) 2015:296 Page 6 of 17

has several applications in various fields of science. In the classical analysis two-weighted

estimates for the Riemann-Liouville fractional operator were derived by Stepanov for the

case with parameters greater than one (see [, ]).

We consider the operator Iq,n of the following form:

Iq,nf (x) =

Ŵq(n)

∫ ∞

X(,x](s)Kn–(x, s)f (s)dqs,

which is defined for all x > . Although it does not coincide with the operator Iq,n (they

coincide at the points x = qk , k ∈ Z), we have the equality

∫ ∞

ur(x)(Iq,nf (x)

)rdqx =

∫ ∞

ur(x)(Iq,nf (x)

)rdqx.

Therefore, inequality () can be rewritten as

(∫ ∞

ur(x)(Iq,nf (x)

)rdqx

) r

≤ C

(∫ ∞

vp(x)f p(x)dqx

) p

. ()

Its conjugate operator I∗q,n can be defined by

I∗q,nf (s) :=

Ŵq(n)

∫ ∞

X[s,∞)(x)Kn–(x, s)f (x)dqx,

with the same kernel. The dual inequality of inequality () reads as follows:

(∫ ∞

ur(x)(I∗q,nf (x)

)rdqx

) r

≤ C∗

(∫ ∞

vp(x)f p(x)dqx

) p

, ()

where C and C∗ are positive constants independent of f and u(·), v(·) are positive real-

valued functions on (,∞), i.e., weight functions. In what follows we investigate inequal-

ities () and ().

Let N =N∪ . Then, for ≤ m ≤ n – ,m,n ∈ N, we use the following notations:

Qn–m =

∫ ∞

(∫ ∞

X(,z](s)Kp′

m (z, s)v–p′

(s)dqs

) p(r–)p–r

×

(∫ ∞

X[z,∞)(x)Krn–m–(x, z)u

r(x)dqx

) pp–r

×Dq

(∫ ∞

X(,z](s)Kp′

m (z, s)v–p′

(s)dqs

) p–rpr

,

Qn–m =

∫ ∞

(∫ ∞

X(,z](s)Krm(z, s)u

r(s)dqs

) rp–r

×

(∫ ∞

X[z,∞)(x)K–p′

n–m–(z,x)v–p′

(x)dqx

) r(p–)p–r

×Dq

(∫ ∞

X(,z](s)Krm(z, s)u

r(x)dqs

) p–rpr

,

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Persson and Shaimardan Journal of Inequalities and Applications ( 2015) 2015:296 Page 7 of 17

Hn–m = sup

z>

(∫ ∞

X[z,∞)(x)Krn–m–(x, z)u

r(x)dqx

) r(∫ ∞

X(,z](s)Kp′

m (z, s)v–p′

(s)dqs

) p′

,

Hn–m = sup

z>

(∫ ∞

X(,z](x)Krm(z,x)u

r(x)dqx

) r(∫ ∞

X[z,∞)(s)Kp′

n–m–(s, z)v–p′

(s)dqs

) p′

,

A+(z) =

(∫ ∞

X[z,∞)(x)ur(x)

(∫ ∞

X(,z](t)Kp′

n–(x, t)v–p′

(t)dqt

) rp′

dqx

) r

,

A–(z) =

(∫ ∞

X(,z](t)v–p′

(t)

(∫ ∞

X[z,∞)(x)Krn–(x, t)u

r(x)dqx

) p′

r

dqt

) p′

.

A+(z) =

(∫ ∞

X[z,∞)(t)v–p′

(t)

(∫ ∞

X(,z](x)Krn–(t,x)u

r(x)dqx

) p′

r

dqt

) p′

,

A–(z) =

(∫ ∞

X(,z](x)ur(x)

(∫ ∞

X[z,∞)(t)Kp′

n–(t,x)v–p′

(t)dqt

) rp′

dqx

) r

,

Hn– = max≤k≤n–

Hn–k , Hn– = max

≤k≤n–Hn–

k ,

A+q = sup

z>A+(z), A–

q = supz>

A–(z), A+q = sup

z>A+(z), A–

q = supz>

A–(z),

Qn– = max≤k≤n–

Qn–k and Qn– = max

≤k≤n–Qn–

k .

Our main results read as follows.

Theorem .

(i) Let < r < p <∞. Then inequality () holds if and only if Qn– < ∞.Moreover,

Qn– ≈ C, where C is the best constant in ().

(ii) Let < p≤ r < ∞. Then inequality () holds if and only if at least one of the

conditions Hn– < ∞ or A+q < ∞ or A–

q <∞ holds.Moreover, Hn– ≈ A+q ≈ A–

q ≈ C,

where C is the best constant in ().

Theorem .

(i) Let < r < p <∞. Then inequality () holds if and only ifQn– <∞.Moreover,

Qn– ≈ C∗, where C∗ is the best constant in ().

(ii) Let < p≤ r < ∞. Then inequality () holds if and only if at least one of the

conditionsHn– < ∞ or A+q <∞, or A–

q < ∞ holds.Moreover,Hn– ≈A+q ≈A–

q ≈ C,

where C is the best constant in ().

For the proofs of these results, we need the following lemmata of independent interest.

Lemma . Let x, t, s: < s ≤ t ≤ x < ∞. Then

max≤m≤n–

Kn–m–(x, t)Km(t, s)≤ Kn–(x, s) ≤

n–∑

m=

[n –

m

]

q

Kn–m–(x, t)Km(t, s) ()

for m: ≤ m ≤ n – , n,m – ∈N and where Kn–(x, s) = (x – qs)n–q .

Lemma . Let f and g be non-negative functions on (,∞), α,β ∈R and

I(z) :=

(∫ ∞

X(,z](t)f (t)dqt

)α(∫ ∞

X[z,∞)(x)g(x)dqx

.

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Then

supz>

I(z) = ( – q)α+β supk∈Z

(∞∑

j=k

qjf(qj

))α(

k∑

i=–∞

qig(qi

))β

, ()

where at least one of α and β is non-negative.

This result was proved in [], but for the readers’ convenience we will include in Sec-

tion a proof which is slightly simpler than that in the Russian version given in [].

Lemma . Let α,β ∈R+, K(·, ·) be a non-negative function and

I+(z) :=

(∫ ∞

X[z,∞)(x)g(x)

(∫ ∞

X(,z](t)K (x, t)f (t)dqt

dqx

,

I–(z) :=

(∫ ∞

X(,z](t)f (t)

(∫ ∞

X[z,∞)(x)K (x, t)g(x)dqx

dqt

.

Then

supz>

I+(z) = supk∈Z

(( – q)

k∑

j=–∞

qjg(qj

)(( – q)

∞∑

i=k

qiK(qj,qi

)f(qi

))α)β

()

and

supz>

I–(z) = supk∈Z

(( – q)

∞∑

j=k

qjf(qj

)(( – q)

k∑

j=–∞

qjK(qj,qi

)g(qj

))α)β

. ()

Lemma . Let Qn–m ,Qn–

m <∞ for <m ≤ n – . Then

Qn–m =

∞∑

i=–∞

(( – q)

∞∑

t=i

qtKp′

m

(qi,qt

)v–p

′(qt

)) p(r–)

p–r

×

(( – q)

i∑

j=–∞

qjK rn–m–

(qj,qi

)ur

(qj

)) p

p–r

× +

(∞∑

n=i

( – q)qnK rm

(qi,qn

)v–p

′(qn

)) p–r

pr

and

Qn–m =

∞∑

i=–∞

(( – q)

∞∑

t=i

qtK rm

(qi,qt

)ur

(qt

)) r

p–r

×

(( – q)

i∑

j=–∞

qjKp′

n–m–

(qj,qi

)v–p

′(qj

)) r(p–)

p–r

× +

(∞∑

n=i

( – q)qnK rm

(qi,qn

)ur

(qn

)) p–r

pr

,

where +En,i = En,i – En,i+.

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4 ProofsProof of Lemma . Let < s ≤ t ≤ x < ∞. First we prove the lower estimate. By using ()

we find that

Kn–m–(x, t)Km(t, s) = (x – qt)n–m–q (t – qs)mq

≤ (x – qs)n–m–q (x – qs)mq

≤ (x – qs)n–m–q

(x – qn–ms

)mq

= (x – qs)n–q = Kn–(x, s)

for < s≤ t ≤ x <∞ and ≤ m ≤ n – ,m – ,n ∈N. Hence,

max≤m≤n–

Kn–m–(x, t)Km(t, s)≤ Kn–(x, s),

and the lower estimate in () is proved.

According to () we get that K(x, t)K(t, s) = K(x, s) ≡ for n = . Moreover, we have

that

K(x, s) = (x – qs)q < (x – qt)q + (t – qs)q

=

m=

[

m

]

q

K–m–(x, t)Km(t, s)

for n = .

This means that the inequality

Kn–(x, s) <

n–∑

m=

[n –

m

]

q

Kn–m–(x, t)Km(t, s) ()

holds for n = .Our aim is now to use induction, andwe assume that () holds for n = l–,

l ≥ , and we will prove that it then holds also for n = l.

We use our induction assumption, make some calculations and obvious estimates and

find that

Kl–(x, s) = Kl–(x, s)(x – ql–s

)

<

(l–∑

m=

[l –

m

]

q

Kl–m–(x, t)Km(t, s)

)(x – ql–s

)

<

l–∑

m=

[l –

m

]

q

Kl–m–(x, t)Km(t, s)(x – ql–m–t + ql–m–t – ql–s

)

=

l–∑

m=

[l –

m

]

q

Kl–m–(x, t)Km(t, s)(x – ql–m–t

)

+

l–∑

m=

[l –

m

]

q

Kl–m–(x, t)Km(t, s)ql–m–

(t – qm+s

)

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=

[l –

]

q

Kl–(x, t)K(t, s) +

l–∑

m=

[l –

m

]

q

Kl–m–(x, t)Km(t, s)

+

l–∑

m=

ql–m–

[l –

m –

]

q

Kl–m–(x, t)Km(t, s) +

[l –

l –

]

q

K(x, t)Kl–(t, s)

=

[l –

]

q

Kl–(x, t)K(t, s)

+

l–∑

m=

(ql–m–

[l –

m –

]

q

+

[l –

m

]

q

)Kl–m–(x, t)Km(t, s)

+

[l –

l –

]

q

K(x, t)Kl(t, s).

Since, for any m ≥ (ql–m–[l –

m –

]q+

[l –

m

]q=

[l –

m

]q), we get that

Kl–(x, s) <

l–∑

m=

[l –

m

]

q

Kl–m–(x, t)Km(t, s).

Hence, () holds also with n = l which, by the induction axiom, means that also the

upper estimate in () is proved. The proof is complete.

Proof of Lemma . From () and () it follows that

I(z) = ( – q)α+β

(∑

qj≤z

qjf(qj

))α(∑

qi≥z

qig(qi

))β

.

If z = qk , then, for k ∈ Z,

I(z) = ( – q)α+β

(∞∑

j=k

qjf(qj

))α(

k∑

i=–∞

qig(qi

))β

.

If qk < z < qk–, then, for k ∈ Z,

I(z) = ( – q)α+β

(∞∑

j=k

qjf(qj

))α(

k–∑

i=–∞

qig(qi

))β

.

Hence, for k ∈ Z and β > , we find that

supqk≤z<qk–

I(z) = ( – q)α+β

(∞∑

j=k

qjf(qj

))α(

k∑

i=–∞

qig(qi

))β

.

Therefore

supz>

I(z) = supk∈Z

supqk≤z<qk–

I(z)

= ( – q)α+β supk∈Z

(∞∑

j=k

qjf(qj

))α(

k∑

i=–∞

qig(qi

))β

. ()

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Persson and Shaimardan Journal of Inequalities and Applications ( 2015) 2015:296 Page 11 of 17

We have proved that () holds wherever β > .

Next we assume that α > . Let qk+ < z < qk , k ∈ Z. Then we get that

I(z) = ( – q)α+β supk∈Z

(∞∑

j=k+

qjf(qj

))α(

k∑

i=–∞

qig(qi

))β

,

and analogously as above we find that

supqk+<z≤qk

I(z) = ( – q)α+β

(∞∑

j=k

qjf(qj

))α(

k∑

i=–∞

qig(qi

))β

,

and () holds also for the case α > . The proof is complete.

Proof of Lemma . Let z = qk , k ∈ Z. By using () and () we have that

I+(z) =

(( – q)

k∑

j=–∞

qjg(qj

)(( – q)

∞∑

i=k

qiK(qj,qi

)f(qi

))α)β

.

For the cases qk+ < z < qk , k ∈ Z and qk < z < qk–, k ∈ Z, we find that

I+(z) =

(( – q)

k∑

j=–∞

qjg(qj

)(( – q)

∞∑

i=k+

qiK(qj,qi

)f(qi

))α)β

and

I+(z) =

(( – q)

k–∑

j=–∞

qjg(qj

)(( – q)

∞∑

i=k

qiK(qj,qi

)f(qi

))α)β

,

respectively.

Hence, we conclude that

supqk+<z<qk–

I+(z) =

(( – q)

k∑

j=–∞

qjg(qj

)(( – q)

∞∑

i=k

qiK(qj,qi

)f(qi

))α)β

.

Since supz> I+(z) = supk∈Z supqk+<z<qk– I

+(z), we find that () holds. The identity ()

can be proved in a similar way as (). The proof is complete.

Proof of Lemma . Without loss of generality we may assume that Qn–m < ∞. By using

(), () and () we can deduce that

Qn–m =

∞∑

i=–∞

(∫ ∞

X(,qi](s)Kp′

m

(qi, s

)v–p

(s)dqs

) p(r–)p–r

×

(∫ ∞

X[qi ,∞)(x)Krn–m–

(x,qi

)ur(x)dqx

) pp–r

×

(∫ ∞

X(,qi](s)Kp′

m

(qi, s

)v–p

(s)dqs

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Persson and Shaimardan Journal of Inequalities and Applications ( 2015) 2015:296 Page 12 of 17

∫ ∞

X(,qi+](s)Kp′

m

(qi+, s

)v–p

(s)dqs

) p–rpr

=

∞∑

i=–∞

(( – q)

∞∑

t=i

qtKp′

m

(qi,qt

)v–p

′(qt

)) p(r–)

p–r

×

(( – q)

i∑

j=–∞

qjK rn–m–

(qj,qi

)ur

(qj

)) p

p–r

× +

(∞∑

n=i

( – q)qnK rm

(qi,qn

)v–p

′(qn

)) p–r

pr

,

and the first equality in Lemma . is proved.

The second inequality can be proved in a similar way, so we leave out the details. The

proof is complete.

Proof of Theorem . By using formulas () and () we find that inequality () can be

rewritten as

(∞∑

j=–∞

( – q)r+qjur(qj

)(

∞∑

i=j

qif(qi

)Kn–

(qj,qi

))r)

r

≤ C

(∞∑

i=–∞

( – q)qif p(qi

)vp

(qi

))

p

. ()

Let

urj = ( – q)r+qjur(qj

), fi = qif

(qi

),

vpi = ( – q)qi(–p)vp

(qi

), W (n)(i, j) = Kn–

(qj,qi

).

()

Then we get that inequality () can be rewritten as the discrete weighted Hardy-type

inequality (see, e.g., [])

(∞∑

j=–∞

urj

(∞∑

i=j

W (n)(i, j)fi

)r) r

≤ C

(∞∑

i=–∞

vpi a

pi

) p

. ()

Hence, inequality () is equivalent to inequality (), where (W (n)(i, j)) is the non-

negative triangular matrix which has entries W (n)(i, j) ≥ for j ≤ i and W (n)(i, j) ≡ for

j > i and is non-decreasing in the first index for all i≥ j > –∞.

First we will prove that, for n ∈N ,

( – q)n–i∑

kn–=j

[n – ]qqkn–

i∑

kn–=kn–

[n – ]qqkn– · · ·

i∑

k=k

[]qqk =W (n)(i, j). ()

We will use induction and first we note that W ()(i, j) = (qj – qi)q ≡ for n = . If n = ,

then

( – q)

i∑

k=j

qk =

i∑

k=j

[]q(qk – qk+

)=

(qj – qi+

)q=W ()(i, j).

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Persson and Shaimardan Journal of Inequalities and Applications ( 2015) 2015:296 Page 13 of 17

Assume now that formula () holds for n – ∈ N , i.e., that

( – q)n–i∑

kn–=j

[n – ]qqkn–

i∑

kn–=kn–

[n – ]qqkn– · · ·

i∑

k=k

[]qqk =W (n–)(i, j).

By using this induction assumption we find that

( – q)n–i∑

kn–=j

[n – ]qqkn–

i∑

kn–=kn–

[n – ]qqkn– · · ·

i∑

k=k

[]qqk

= ( – q)[n – ]q

i∑

kn–=j

qkn–W (n–)(i, j)

=( – qn–

) i∑

kn–=j

qkn–(qkn– – qi+

)n–q

.

Since (qkn– – qi+)n–q – (qkn–+ – qi+)n–q = ( – qn–)qkn– (qkn– – qi+)n–q , we get that ()

holds also for n. Hence, by the induction axiom, we conclude that () holds for each

n ∈N .

Let wm,k = [m]q(qkm – qkm+),m = , , , . . . ,n – . Then, by using (), we have that

W (n)(i, j) =

i∑

kn–=j

wn–,kn–

i∑

kn–=kn–

wn–,kn– · · ·

i∑

k=k

w,k . ()

Therefore, we see that the matrix operator in (), defined by

(Sf )j :=

∞∑

i=j

W (n)(i, j)fi, j ∈ Z,

is an n-multiple discrete Hardy operator with weights (see ()).

Therefore, Lemma . and Lemma . can be used.

(i) Let < r < p < ∞. Then, based on Lemma ., it follows that inequality () holds if

and only if Qn– = max≤m≤n– Qn–m < ∞, where

Qn–m =

∞∑

i=–∞

(∞∑

j=i

(W (m+)(j, i)

)p′

v–p′

j

) p(r–)p–r

×

(i∑

k=–∞

(W (n–m)(i,k)

)rurk

) pp–r

× +

(∞∑

j=i

(W (m+)

)p′

(j, i)v–p′

j

) p–rpr

.

Since inequality () is equivalent to inequality (), we conclude that the condition

Qn– < ∞ is a necessary and sufficient condition for the validity of inequality (). More-

over, Qn– ≈ C.

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By using the definitions () in Qn–m , we get that

Qn–m =

∞∑

i=–∞

(( – q)

∞∑

j=i

qjKp′

m

(qi,qj

)v–p

′(qj

)) p(r–)

p–r

×

(( – q)

i∑

k=–∞

qkK rn–m–

(qk ,qi

)ur

(qk

)) p

p–r

× +

(∞∑

j=i

( – q)qjKp′

m

(qi,qj

)v–p

′(qj

)) p–r

pr

.

By using Lemma ., we find that

Qn–m =

∫ ∞

(∫ ∞

X(;z](s)Kp′

m (z, s)v–p′

(s)dqs

) p(r–)p–r

×

(∫ ∞

X[z,∞)(x)Krn–m–(x, z)u

r(x)dqx

) pp–r

×Dq

(∫ ∞

X(;z](s)Kp′

m (z, s)v–p′

(s)dqs

) p–rpr

,

i.e., that Qn– =Qn–. Then we find that inequality () holds if and only ifQn– < ∞. More-

over, Qn– ≈ C, where C is the best constant in (). Thus the proof of the statement (i) of

Theorem . is complete.

(ii) Let < p ≤ r < ∞. Then from Lemma . it follows that inequality () holds if and

only if Hn– = max≤m≤n– Hn–m < ∞ holds, where

Hn–m = sup

k∈Z

(∞∑

i=k

(W (m+)(i,k)

)p′

v–p′

i

) p′

×

(k∑

j=–∞

(W (n–m)(k, j)

)rurj

) r

, n ∈N.

If x = qj, s = qi, t = qk , for j ≤ k ≤ i, then, by Lemma . and (), we obtain that (recall

thatW (n)(i, j) = Kn–(qj,qi))

W (n)(i, j) ≤

n–∑

m=

[n –

m

]

q

W (n–m)(k, j)W (m+)(i,k). ()

Since, again by Lemma ., max≤m≤n–W(m+)(i,k)W (n–m)(k, j) ≤ W (n)(i, j), it follows

that

W (n)(i, j) ≥ h(n)

n–∑

m=

[n –

m

]

q

W (m+)(i,k)W (n–m)(k, j), ()

where h(n) =(∑n–

m=

[n –

m

]q

)–.

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Persson and Shaimardan Journal of Inequalities and Applications ( 2015) 2015:296 Page 15 of 17

According to () and () we have that

W (n)(i, j) ≈

n–∑

m=

W (m)(i,k)W (n,m)(k, j),

whereW (n,m)(k, j) =[n –

m

]qW (n–m–)(k, j).

Therefore, we have proved that the matrix (W (n)(i, j)) in () satisfies the Oinarov con-

dition () and Lemma . can be used.

Hence, we have the following necessary and sufficient conditions for the validity of in-

equality ():

A+ = supk∈Z

(k∑

j=–∞

urj

(∞∑

i=k

(W (n)(i, j)

)p′

v–p′

i

) rp′

) r

< ∞

or

A– = supk∈Z

(∞∑

i=k

v–p′

i

(k∑

j=–∞

(W (n)(i, j)

)rurj

) p′

r)

p′

< ∞.

Since, again by Lemma ., max≤m≤n–W(m+)(i,k)W (n–m)(k, j) ≤ W (n)(i, j), we get that

Hn– = supk∈Z

(k∑

j=–∞

uj

(∞∑

i=k

(max

≤m≤n–

(W (n–m)(k, j)W (m+)(i,k)

))p′

v–p′

i

) rp′

) r

(k∑

j=–∞

uj

(∞∑

i=k

(W (n)(i, j)

)p′

v–p′

i

) rp′

) r

= A+.

Moreover, by () we have that

A+ ≤ supk∈Z

(k∑

j=–∞

uj

(∞∑

i=k

(n–∑

m=

[n –

m

]

q

W (n–m)(k, j)W (m+)(i,k)

)p′

v–p′

i

) rp′

) r

n–∑

m=

[n –

m

]

q

Hn– ≪ Hn–.

Hence, A+ ≈ Hn–. In a similar way it can be proved that Hn– ≈ A–.

Since inequality () is equivalent to inequality (), we get that inequality () holds if

and only if at least one of the conditions A+ < ∞ or A– < ∞, or Hn– < ∞ holds.

Now, using notations () in Hn–m , we obtain that

Hn–m = ( – q)

r +

p′ sup

k∈Z

(k∑

j=–∞

qjur(qj

)K rn–m–

(qj,qk

))

r

×

(∞∑

i=k

qtv–p′(qi

)Kp′

m

(qk ,qi

))

p′

.

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Persson and Shaimardan Journal of Inequalities and Applications ( 2015) 2015:296 Page 16 of 17

The parameters and functions in Hn–m satisfy all the conditions of Lemma.. Therefore,

we find that

Hn–m = sup

z>

(∫ ∞

X[z,∞)(x)Krn–m–(x, z)u

r(x)dqx

) r

×

(∫ ∞

X(,z](s)Kp′

m (z, s)v–p′

(s)dqs

) p′

,

i.e., that Hn–m =Hn–

m and Hn– =Hn– = max≤m≤n–Hn–m < ∞.

In a similar way as above, by using Lemma . and (), we get thatA+q = A+ andA–

q = A–.

Hence, we obtain that inequality () holds if and only if at least one of the conditions

Hn– < ∞ or A+q < ∞, or A–

q < ∞ holds. Moreover, A+q ≈ A–

q ≈ Hn– ≈ C, where C is the

best constant in (). Also the proof of the statement (ii) of Theorem . is complete.

Proof of Theorem . In a similar way as in the proof of Theorem ., by using (), () and

(), we can prove that we have the following discrete Hardy-type inequality:

(∞∑

i=–∞

uri

(i∑

j=–∞

W(n)n–,(i, j)fi

)r) r

≤ C

(∞∑

i=–∞

vpi a

pi

) p

, ()

which is equivalent to inequality ().

(i) Let < r < p < ∞. By using Lemma . and Lemma ., we can in a similar way as

in the proof of Theorem .(i) derive that inequality () holds if and only if Qn– < ∞

holds. Moreover,Qn– ≈ C∗, where C∗ is the best constant in (). The proof of part (i) is

complete.

(ii) Let < p ≤ r < ∞. By using Lemma ., Lemma ., Lemma . and Lemma .,

we can, analogously as in the proof of the (ii)-part, prove that inequality () holds if and

only if at least one of the conditions Hn– < ∞ or A+q < ∞, or A–

q < ∞ holds. Moreover,

Hn– ≈ A–q ≈ A+

q ≈ C∗, where C∗ is the best constant in (). The proof of part (ii) is

complete.

Competing interests

The authors declare that they have no competing interests.

Authors’ contributions

All authors have on equal level discussed and posed the research questions in this paper. SS is the main author

concerning the proofs of the main results and typing of the manuscript. L-EP has put the results into a more general

frame in the introduction and instructed how to write the paper in this final form. All authors read and approved the final

manuscript.

Author details1Luleå University of Technology, Luleå, 971 87, Sweden. 2Narvik University College, P.O. Box 385, Narvik, 8505, Norway.3L.N. Gumilyov Eurasian National University, Munaytpasov St. 5, Astana, 010008, Kazakhstan.

Acknowledgements

We thank both careful referees and Professor Ryskul Oinarov for generous advice, which have improved the final version

of this paper.

Received: 6 May 2015 Accepted: 7 September 2015

References

1. Jackson, FH: On q-definite integrals. Q. J. Pure Appl. Math. 41, 193-203 (1910)2. Cheung, P, Kac, V: Quantum Calculus. Edwards, Ann Arbor (2000)

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3. Annaby, MH, Mansour, ZS: q-Fractional Calculus and Equations. Springer, Heidelberg (2012)

4. Bangerezako, G: Variational calculus on q-nonuniform lattices. J. Math. Anal. Appl. 306, 161-179 (2005)5. Ernst, T: A Comprehensive Treatment of q-Calculus. Birkhäuser, Basel (2012)

6. Ernst, T: The history of q-calculus and a new method. PhD thesis, Uppsala University (2001)

7. Exton, H: q-Hypergeometric Functions and Applications. Halsted, New York (1983)

8. Gauchman, H: Integral inequalities in q-calculus. Comput. Math. Appl. 47, 281-300 (2004)9. Krasniqi, V, Erratum: Several q-integral inequalities. J. Math. Inequal. 5(3), 451 (2011)10. Miao, Y, Qi, F: Several q-integral inequalities. J. Math. Inequal. 3, 115-121 (2009)11. Stankovic, MS, Rajkovic, PM, Marinkovic, SD: On q-fractional derivatives of Riemann-Liouville and Caputo type (2013).

arXiv:0909.0387

12. Sulaiman, WT: New types of q-integral inequalities. Adv. Pure Math. 1, 77-80 (2011)13. Maligranda, L, Oinarov, R, Persson, L-E: On Hardy q-inequalities. Czechoslov. Math. J. 64, 659-682 (2014)14. Hardy, GH: Note on a theorem of Hilbert. Math. Z. 6, 314-317 (1920)15. Hardy, GH: Notes on some points in the integral calculus, LX. An inequality between integrals. Messenger Math. 54,

150-156 (1925)

16. Kokilashvili, V, Meskhi, A, Persson, L-E: Weighted Norm Inequalities for Integral Transforms with Product Kernels. Nova

Science Publishers, New York (2010)

17. Kufner, A, Maligranda, A, Persson, L-E: The Hardy Inequality - About Its History and Some Related Results. University of

West Bohemia, Plzen (2007)

18. Kufner, A, Persson, L-E: Weighted Inequalities of Hardy Type. World Scientific, River Edge (2003)

19. Al-Salam, WA: Some fractional q-integrals and q-derivatives. Proc. Edinb. Math. Soc. 15, 135-140 (1966/1967)20. Oinarov, R, Temirkhanova, AM: Boundedness and compactness of a class of matrix operators in weighted sequence

spaces. J. Math. Inequal. 2, 555-570 (2008)21. Kalybay, A, Oinarov, R, Temirkhanova, A: Boundedness of n-multiple discrete Hardy operators with weights for

1 < q < p <∞. J. Funct. Spaces Appl. 41, 1-9 (2013)22. Oinarov, R, Taspaganbetova, Z: Criteria of boundedness and compactness of a class of matrix operators. J. Inequal.

Appl. 2012, 53 (2012)23. Stepanov, VD: Two-weighted estimates of Riemann-Liouville integrals. Math. USSR, Izv. 36, 669-681 (1991)24. Stepanov, VD: Weighted inequalities for a class of Volterra convolution operators. J. Lond. Math. Soc. 45, 232-242

(1992)

25. Baiaristanov, AO, Shaimardan, S, Temirkhanova, A: Weighted Hardy inequalities in quantum analysis. Vestin. KarGU,

Math. Ser. 70, 35-45 (2013)26. Persson, L-E, Ragusa, MA, Samko, N, Wall, P: Commutators of Hardy operators in vanishing Morrey spaces. In: 9th

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Conference Proceedings, vol. 1493, pp. 859-866 (2012)

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Paper B

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Journal of

Mathematical

Inequalities

Volume 10, Number 3 (2016), 761–781 doi:10.7153/jmi-10-62

SOME NEW HARDY–TYPE INEQUALITIES IN q–ANALYSIS

A. O. BAIARYSTANOV, L. E. PERSSON, S. SHAIMARDAN AND A. TEMIRKHANOVA

(Communicated by J. Pecaric)

Abstract. We derive necessary and sufficient conditions (of Muckenhoupt-Bradley type) for thevalidity of q -analogs of (r, p) -weighted Hardy-type inequalities for all possible positive valuesof the parameters r and p . We also point out some possibilities to further develop the theory ofHardy-type inequalities in this new direction.

1. Introduction

G. H. Hardy announced in 1920 [17] and finally proved in 1925 [18] (also see [19,p. 240]) his famous inequality

∞∫

0

1

x

x∫

0

f (t)dt

p

dx 6(

pp−1

)p ∞∫

0

f p(x)dx, p > 1, (1.1)

for all non-negative functions f (in the sequel we assume that all functions are non-

negative). The constant(

pp−1

)pin (1.1) is sharp. Since then it has been an enormous

activity to develop and apply what is today known as Hardy-type inequalities, see e.gthe books [21], [23] and [24] and the references there.

One central problem in this development was to characterize the weights u(x) andυ(x) so that the more general Hardy-type inequality

∞∫

0

x∫

0

f (t)dt

r

u(x)dx

1r

6C

∞∫

0

f p(x)υ(x)dx

1p

(1.2)

holds for some constant C and various parameters p and r .To make our introduction clear we just concentrate on the case 1 6 p 6 r < ∞ . In

this case e.g the following result is well-known:

Mathematics subject classification (2010): 26D10, 26D15, 33D05, 39A13.Keywords and phrases: Inequalities, Hardy-type inequalities, Riemann-Liouville operator, integral op-

erator, q -analysis, q -analog, weights.

c© , ZagrebPaper JMI-10-62

761

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762 A. O. BAIARYSTANOV, L. E. PERSSON, S. SHAIMARDAN AND A. TEMIRKHANOVA

PROPOSITION A. Let 1 < p 6 r < ∞ . Then the inequality (1.2) holds if and onlyif

A1 := sup0<x<∞

(U(x))1r (V (x))

1p′ < ∞

or

A2 := sup0<x<∞

x∫

0

u(t)V r(t)dt

1r

V− 1p (x)< ∞

or

A3 := sup0<x<∞

∞∫

x

υ1−p′(t)U p′(t)dt

1p′

U− 1r′ (x)< ∞,

where U(x) =∞∫x

u(t)dt , V (x) =x∫

0υ1−p′(t)dt , p′ = p

p−1 and r′ = rr−1 . Moreover, for

the sharp constant in (1.2) we have that C ≈ A1 ≈ A2 ≈ A3 .

REMARK 1.1. A nice proof of the condition A1 < ∞ was given in 1978 by J. S.Bradley [9]. The case p = r was proved by B. Muckenhoupt [28] already in 1972 . Thecondition A2 < ∞ was proved in 2002 by L. E. Persson and V. D. Stepanov [30], butwas for the case p = r proved by G. A. Tomaselli [34] already in 1969. The conditionA3 < ∞ is just the dual condition of the condition A2 < ∞ .

In the beginning G. H. Hardy was most occupied with the discrete version of (1.1).The discrete version of (1.2) reads:

(∞

∑n=1

(n

∑k=1

fk

)r

un

) 1r

6C

(∞

∑n=1

f pn υn

) 1p

, (1.3)

where u = un and υ = υn are non-negative weight sequences and the questionis to characterize all such weight sequence so that (1.3) holds for an arbitrary non-negative sequence f = fn (in the sequel we assume that the considered sequences arenon-negative).

It is interesting that the similar results as that in Proposition A for the discrete casewas independently proved by G. Bennett [6] in 1987 (see also [2], [8] and [22, Theorem7]). It reads:

PROPOSITION B. Let 1 < p 6 r < ∞ . Then the inequality (1.3) holds if and onlyif

B1 := supn∈N

U1r

n V1p′

n < ∞

or

B2 := supn∈N

(n

∑k=1

ukVrk

) 1r

V− 1

pn < ∞

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SOME NEW HARDY-TYPE INEQUALITIES IN q -ANALYSIS 763

or

B3 := supn∈N

(∞

∑k=n

υ1−p′k U p′

k

) 1p′

U− 1

r′n < ∞,

where Un =∞

∑k=n

uk and Vn =n∑

k=1υ

1−p′k .

Moreover, for the sharp constant C in (1.3) it yields that C ≈ B1 ≈ B2 ≈ B3 .

For our purposes we will consider the inequality (1.3) on the following differentbut equivalent form:

(∞

∑n=1

(un

n

∑k=1

υk fk

)r) 1r

6C

(∞

∑n=1

f pn

) 1p

, (1.4)

with the obvious changes of the conditions Bi < ∞, i = 1,2,3.In 1910, F. H. Jackson defined q -derivative and definite q -integral [20] (see also

[11]). It was the starting point of q -analysis. Today the interest in the subject hasexploded. The q -analysis has numerous applications in various fields of mathematicse.g dynamical systems, number theory, combinatorics, special functions, fractals andalso for scientific problems in some applied areas such as computer science, quantummechanics and quantum physics (see e.g. [3], [5], [12], [13] and [14]). For the fur-ther development and recent results in q -analysis we refer to the books [3], [11] and[12] and the references given therein. The first results concerning integral inequalitiesin q -analysis were proved in 2004 by H. Gauchman [15]. Later on some further q -analogs of the classical inequalities have been proved (see [22], [27], [32] and [33]).We also pronounce the recent book [1] by G.A. Anastassiou, where many importantq-inequalities are proved and discussed. Moreover, in 2014 L. Maligranda, R. Oinarovand L.-E. Persson [26] derived a q -analog of the classical Hardy inequality (1.1) andsome related inequalities. It seems to be a huge new research area to investigate whichof these so called Hardy-type inequalities have their q -analogs.

One main aim in this paper is to prove the q -analog of the results in PropositionsA and B (see our Theorem 3.1). We will also prove the corresponding characterizationfor other possible values of the parameters p and r (see our Theorem 3.3). We alsoprove the corresponding dual results (see Theorem 3.2 and Theorem 3.4).

Our paper is organized as follows: The main results are stated Section 3 and provedin Section 4. In order not to disturb our discussions there some preliminaries are givenin Section 2. In particular, we present some basic facts from q -analysis and also state

Proposition B on a formally more general form namely where∞

∑1

is replaced by∞

∑−∞

(see

Proposition 2.2). We also state this result for other parameters which is important forour proof of the Theorem 3.3 (see Proposition 2.3). Finally, in Section 5 we presentsome remarks and in particular point out the possibility to generalize our results evento modern forms of Propositions A and B, where these three conditions even can bereplaced by four scales of conditions (For the continuous case, see the review article[25] and for the discrete case see [29]).

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764 A. O. BAIARYSTANOV, L. E. PERSSON, S. SHAIMARDAN AND A. TEMIRKHANOVA

2. Preliminaries

2.1. Some basic facts in q -analysis

This subsection gives the definitions and notions of q -analysis [11] (see also [12]).Let the function f defined on (0,b) , 0 < b 6 ∞ and 0 < q < 1. Then

Dq f (x) :=f (x)− f (qx)(1−q)x

, x ∈ (0,b) (2.1)

is called the q -derivative of the function f . This definition was introduced by F. H.Jackson in 1910.

Let x ∈ (0,b) . Then

x∫

0

f (t)dqt := (1−q)x∞

∑k=0

qk f (xqk), (2.2)

is called q -integral or Jackson integral.If b = ∞ the improper q -integral is defined by

∞∫

0

f (t)dqt := (1−q)∞

∑k=−∞

qk f (qk). (2.3)

The integrals (2.2) and (2.3) are meaningful, if the series on the right hand sidesconverge.

Let 0 < a < b 6 ∞ . Then we have that

b∫

a

f (t)dqt :=

b∫

0

f (t)dqt −a∫

0

f (t)dqt. (2.4)

We also need the following fact:

PROPOSITION 2.1. Let k ∈ Z . Then

∞∫

qk+1

f (t)dqt = (1−q)k

∑j=−∞

q j f (q j). (2.5)

Proof of Proposition 2.1. By using (2.2), (2.3) and (2.4) with b = ∞ , a = qk+1 wehave that

∞∫

qk+1

f (t)dqt =

∞∫

0

f (t)dqt −qk+1∫

0

f (t)dqt

= (1−q)∞

∑j=−∞

q j f (q j)− (1−q)∞

∑i=0

qi+k+1 f (qi+k+1)

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SOME NEW HARDY-TYPE INEQUALITIES IN q -ANALYSIS 765

= (1−q)∞

∑j=−∞

q j f (q j)− (1−q)∞

∑i=k+1

qi f (qi)

= (1−q)k

∑j=−∞

q j f (q j),

i.e. (2.5) holds. The proof is complete.

Let Ω be a subset of (0,∞) and XΩ(t) denote the characteristic function of theset Ω . Let z > 0. Then from (2.3) we can deduce that

∞∫

0

X(0,z](t) f (t)dqt = (1−q)∞

∑i=−∞

qiX(0,z](qi) f (qi) = (1−q) ∑

qi6z

qi f (qi), (2.6)

and∞∫

0

X[z,∞)(t) f (t)dqt = (1−q) ∑qi>z

qi f (qi). (2.7)

Moreover,∞∫

0

X(qz,z](t) f (t)dqt = (1−q)qk f (qk), (2.8)

for qk 6 z < qk−1 , k ∈ Z ,

∞∫

0

X[z,q−1z)(t) f (t)dqt = (1−q)qm f (qm), (2.9)

for qm+1 < z 6 qm , m ∈ Z .

2.2. An important variant of Proposition B

We consider the inequality:

(∞

∑n=−∞

(un

n

∑k=−∞

υk fk

)r) 1r

6C

(∞

∑n=−∞

f pn

) 1p

, fn > 0. (2.10)

We need the following formal extension of Proposition B, of independent interest:

PROPOSITION 2.2. Let 1 < p 6 r < ∞ . Then the inequality (2.10) holds if andonly if

C1 = supn∈Z

(∞

∑k=n

urk

) 1r(

n

∑i=−∞

υp′i

) 1p′

< ∞ (2.11)

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766 A. O. BAIARYSTANOV, L. E. PERSSON, S. SHAIMARDAN AND A. TEMIRKHANOVA

or

C2 = supn∈Z

(n

∑i=−∞

υp′i

)− 1p(

n

∑k=−∞

urk

(k

∑i=−∞

υp′i

)r) 1r

< ∞ (2.12)

or

C3 = supn∈Z

(∞

∑k=n

urk

)− 1r′

∑i=n

υp′i

(∞

∑k=i

urk

)p′

1p′

< ∞. (2.13)

Moreover, for the sharp constant C in (2.10) it yields that C ≈C1 ≈C2 ≈C3 .

This proposition is even equivalent to Proposition B, which can be seen from theproof below we give for the reader’s convenience.

Proof of Proposition 2.2. Let Z= Z∪+∞∪−∞, N=N∪+∞ . The func-tion ϕ : Z→N , given by

∀n ∈ Z : ϕ(n) =

+∞ n =+∞,2n n > 0,−2n+3 n 6 0,1 n =−∞,

is a bijection.Therefore, ϕ(n) = m , m = 1,2, · · · and ϕ(k) = j , j = 1,2, · · ·m , so that

(∞

∑n=−∞

(un

n

∑k=−∞

υk fk

)r) 1r

=

(ϕ(∞)

∑ϕ(n)=ϕ(−∞)

(uϕ(n)

ϕ(n)

∑ϕ(k)=ϕ(−∞)

υϕ(k) fϕ(k)

)r) 1r

=

(∞

∑m=1

(um

m

∑j=1

υ j f j

)r) 1r

, (2.14)

and

(∞

∑n=−∞

f pn

) 1p

=

(ϕ(∞)

∑ϕ(n)=ϕ(−∞)

f pϕ(n)

) 1p

=

(∞

∑m=1

f pm

) 1p

, (2.15)

where fm = fϕ(n) , um = uϕ(n) , υ j = υϕ(k) .By (2.14) and (2.15), we obtain that (2.10) holds if and only if the inequality

(∞

∑m=1

(um

m

∑j=1

υ j f j

)r) 1r

6C

(∞

∑m=1

f pm

) 1p

(2.16)

holds.

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SOME NEW HARDY-TYPE INEQUALITIES IN q -ANALYSIS 767

Let 1 < p 6 r < ∞ . By Proposition B we get that the inequality (2.16) holds if andonly if

B1 = supm∈N

(∞

∑j=m

urj

) 1r(

m

∑i=1

υp′i

) 1p′

< ∞

holds. Moreover, since the function ϕ−1 : N→ Z is a bijection, we find that

B1 = supn∈Z

(∞

∑k=n

urk

) 1r(

n

∑i=−∞

υp′i

) 1p′

=C1. (2.17)

Hence, according to (2.14), (2.15) and (2.17), we obtain that the inequality (2.10)holds if and only if C1 < ∞ . Moreover, by Proposition B we find that C ≈C1 , where Cis the sharp constant in (2.10).

The proofs of the facts that also C2 < ∞ and C3 < ∞ are necessary and sufficientconditions for the characterization of (2.10), and also that C ≈C2 ≈C3 , are similar sowe leave out the details. The proof is complete.

We also need the corresponding result for other cases of possible parameters pand r .

PROPOSITION 2.3. (i). Let 0 < p 6 1, p 6 r < ∞ . Then the inequality (2.10)holds if and only if

C4 = supn∈Z

(∞

∑k=n

urk

) 1r

υn < ∞. (2.18)

(ii). Let 1 < p < ∞, 0 < r < p. Then the inequality (2.10) holds if and only if

C5 =

∑n=−∞

(n

∑i=−∞

υp′i

) r(p−1)p−r

(∞

∑k=n

urk

) rp−r

urn

p−rpr

< ∞. (2.19)

(iii). Let 0 < r < p = 1 . Then the inequality (2.10) is satisfied if and only if

C6 =

∑n=−∞

maxi6n

υr

1−ri

(∞

∑k=n

urk

) r1−r

urn

1−rr

< ∞. (2.20)

In all cases (i)–(iii) for the best constant in (2.10) it yields that C ≈ Bi, i = 4,5,6 ,respectively.

Proof of Proposition 2.3. By using well-known characterizations (see [6], [7], [8],

[10], [16] and [21, p. 58]) for the cases (i)–(iii) where∞

∑−∞

is replaced by∞

∑1

, the proof

can be performed exactly as the proof of Proposition 2.2. We leave out the details.

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768 A. O. BAIARYSTANOV, L. E. PERSSON, S. SHAIMARDAN AND A. TEMIRKHANOVA

2.3. Some q -analogs of weighted Hardy-type inequalities

Let 0 < r, p 6 ∞ . Then the q -analog of the discrete Hardy-type inequality of theform (1.4) can be rewritten in the following way:

∞∫

0

u(x)

x∫

0

v(t) f (t)dqt

r

dqx

1r

6C

∞∫

0

f p(x)dqx

1p

. (2.21)

By Proposition 2.1 we find that the inequality (2.21) can be rewritten on the fol-lowing dual form:

∞∫

0

v(x)

∞∫

qx

u(t)g(t)dqt

p′

dqx

1p′

6C

∞∫

0

gr′(x)dqx

1r′

. (2.22)

We see that the (2.22) lacks some symmetry as in classical analysis.

We consider the operator (Hq f ) (x) =∞∫0

X(0,x](t)v(t) f (t)dqt , which is defined for

all x > 0. Although it does not coincide with the operatorx∫

0v(t) f (t)dqt (they coincide

at the points x = qk , k ∈ Z) we have the equality

∞∫

0

u(x)

x∫

0

v(t) f (t)dqt

r

dqx =

∞∫

0

u(x)

∞∫

0

X(0,x](t)v(t) f (t)dqt

r

dqx.

Therefore, the inequality (2.21) can be rewritten as

∞∫

0

u(x)

∞∫

0

X(0,x](t)v(t) f (t)dqt

r

dqx

1r

6C

∞∫

0

f p(x)dqx

1p

, (2.23)

which will be called the q -integral analog of the weighted Hardy-type inequality. Thedual inequality of the inequality (2.23) (equivalent of (2.22)) reads:

∞∫

0

v(t)

∞∫

0

X[x,∞)(x)u(x)g(x)dqx

p′

dqt

1p′

6C

∞∫

0

gr′(t)dqt

1r′

.

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SOME NEW HARDY-TYPE INEQUALITIES IN q -ANALYSIS 769

3. The main results

Our main result reads:

THEOREM 3.1. Let 1 < p 6 r < ∞ . Then the inequality (2.23) holds if and onlyif

D1 = supz>0

∞∫

0

X[z,∞)(x)ur(x)dqx

1r

∞∫

0

X(0,z](t)vp′(t)dqt

1p′

< ∞

or

D2 = supz>0

∞∫

0

X(0,z](t)vp′(t)dqt

− 1p

∞∫

0

X(0,z](x)ur(x)

∞∫

0

X(0,z](t)vp′(t)dqt

r

dqx

1r

< ∞

or

D3 = supz>0

∞∫

0

X[z,∞)(x)ur(x)dqx

− 1r′

∞∫

0

X[z,∞)(t)vp′(t)

∞∫

0

X[z,∞)(x)ur(x)dqx

p′

dqt

1p′

< ∞.

Moreover, for the sharp constant in (2.23) we have that C ≈ D1 ≈ D2 ≈ D3 .

Next, we will consider the corresponding inequality

∞∫

0

u(x)

∞∫

0

X[x,∞)(t)v(t) f (t)dqt

r

dqx

1r

6C

∞∫

0

f p(x)dqx

1p

, (3.1)

for the dual operator of Hq .

THEOREM 3.2. Let 1 < p 6 r < ∞ . Then the inequality (3.1) holds if and only if

D∗1 = sup

z>0

∞∫

0

X(0,z](x)ur(x)dqx

1r

∞∫

0

X[z,∞)(t)vp′(t)dqt

1p′

< ∞

or

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770 A. O. BAIARYSTANOV, L. E. PERSSON, S. SHAIMARDAN AND A. TEMIRKHANOVA

D∗2 = sup

z>0

∞∫

0

X[z,∞)(t)vp′(t)dqt

− 1p

∞∫

0

X[z,∞)(x)ur(x)

∞∫

0

X[z,∞)(t)vp′(t)dqt

r

dqx

1r

< ∞

or

D∗3 = sup

z>0

∞∫

0

X(0,z](x)ur(x)dqx

− 1r′

∞∫

0

X(0,z](t)vp′(t)

∞∫

0

X(0,z](x)ur(x)dqx

p′

dqt

1p′

< ∞.

Moreover, for the sharp constant in (3.1) we have that C ≈ D∗1 ≈ D∗

2 ≈ D∗3 .

Concerning other possible parameters of p and r we have the following comple-ment of Theorem 3.1:

THEOREM 3.3. (i). Let 0 < p 6 1, p 6 r < ∞ . Then the inequality (2.23) holdsif and only if

D4 = supz>0

∞∫

0

X[z,∞)(x)ur(x)dqx

1r

∞∫

0

X(qz,z](t)vp′(t)dqt

1p′

< ∞.

(ii). Let 1 < p < ∞ , 0 < r < p. Then the inequality (2.23) holds if and only if

D5 =

∞∫

0

∞∫

0

X(0,z](t)vp′(t)dqt

r(p−1)p−r

∞∫

0

X[z,∞)(x)ur(x)dqx

rp−r

ur(z)dqz

p−rpr

< ∞.

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SOME NEW HARDY-TYPE INEQUALITIES IN q -ANALYSIS 771

(iii). Let 0 < r < p = 1 . Then the inequality (2.23) is satisfied if and only if

D6 =

∞∫

0

supy<z

∞∫

0

X(qy,y](t)v(t)

(1−q)tdqt

r1−r

∞∫

0

X[z,∞)(x)ur(x)dqx

r1−r

ur(z)dqz

1−rr

< ∞.

In all cases (i)–(iii), for the best constant in (2.23) it yields that C ≈Di , i = 4,5,6 ,respectively.

Finally, the corresponding complement of Theorem 3.2 reads:

THEOREM 3.4. (i). Let 0 < p 6 1, p 6 r < ∞ . Then the inequality (3.1) holds ifand only if

D∗4 = sup

z>0

∞∫

0

X(0,z](x)ur(x)dqx

1r

∞∫

0

X[z,q−1z)(t)vp′(t)dqt

1p′

< ∞.

(ii). Let 1 < p < ∞ , 0 < r < p. Then the inequality (3.1) holds if and only if

D∗5 =

∞∫

0

∞∫

0

X(0,z](x)ur(x)dqx

rp−r

∞∫

0

X[z,∞)(t)vp′(t)dqt

r(p−1)p−r

ur(z)dqz

p−rpr

< ∞.

(iii). Let 0 < r < p = 1 . Then the inequality (3.1) holds if and only if

D∗6 =

∞∫

0

supy>z

∞∫

0

X[y,q−1y)(t)v(t)

(1−q)tdqt

r1−r

∞∫

0

X(0,z](x)ur(x)dqx

r1−r

ur(z)dqz

1−rr

< ∞.

In all cases (i)–(iii), for the best constant in (3.1) it yields that C ≈ D∗i , i = 4,5,6 ,

respectively.

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772 A. O. BAIARYSTANOV, L. E. PERSSON, S. SHAIMARDAN AND A. TEMIRKHANOVA

To prove these theorems, we need some Lemmas of independent interest:

LEMMA 3.5. Let f and g be nonnegative functions and

I(z) :=

∞∫

0

X(0,z](t) f (t)dqt

α

∞∫

0

X[z,∞)(x)g(x)dqx

β

,

for α, β ∈ R, and where at least one of the numbers α, β is positive. Then

supz>0

I(z) = (1−q)α+β supk∈Z

(∞

∑j=k

q j f (q j)

)α( k

∑i=−∞

qig(qi)

. (3.2)

LEMMA 3.6. Let α, β ∈ R+ ,

I+(z) :=

∞∫

0

X(0,z](x) f (x)dqx

α

∞∫

0

X[z,q−1z)(t)g(t)dqt

β

,

and

I−(z) :=

∞∫

0

X[z,∞)(x) f (x)dqx

α

∞∫

0

X(qz,z](t)g(t)dqt

β

.

Then

supz>0

I+(z) = (1−q)α+β supk∈Z

(∞

∑i=k

qi f (qi)

)α (qkg(qk)

)β, (3.3)

and

supz>0

I−(z) = (1−q)α+β supk∈Z

(k

∑i=−∞

qi f (qi)

)α (qkg(qk)

)β. (3.4)

LEMMA 3.7. Let f , ϕ and g be nonnegative functions. Then

D ≡∞∫

0

∞∫

0

X[z,∞)(t) f (t)dqt

α

∞∫

0

X(0,z](x)g(x)dqx

β

ϕ(z)dqz

= (1−q)α+β∞

∑k=−∞

(

k

∑i=−∞

qi f (qi)

)α(∞

∑j=k

q jg(q j)

qkϕ(qk)

,

for α, β ∈ R .

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SOME NEW HARDY-TYPE INEQUALITIES IN q -ANALYSIS 773

LEMMA 3.8. Let k ∈ Z , α ∈ R and

F(y) :=

∞∫

0

X[y,q−1y)(t) f (t)dqt

α

.

Thensupy>qk

F(y) = (1−q)α supi6k

(qi f (qi)

)α. (3.5)

4. Proofs

Proof of Lemma 3.5. From (2.6) and (2.7) it follows that

I(z) = (1−q)α+β

(∑

q j6z

q j f (q j)

)α(∑

qi>z

qig(qi)

.

If z = qk , then, for k ∈ Z ,

I(z) = I(qk) = (1−q)α+β

(∞

∑j=k

q j f (q j)

)α( k

∑i=−∞

qig(qi)

.

If qk < z < qk−1, then, for k ∈ Z ,

I(z) = (1−q)α+β

(∞

∑j=k

q j f (q j)

)α( k−1

∑i=−∞

qig(qi)

.

Hence, for k ∈ Z and β > 0 we find that

supqk6z<qk−1

I(z) = I(qk) = (1−q)α+β

(∞

∑j=k

q j f (q j)

)α( k

∑i=−∞

qig(qi)

.

Therefore

supz>0

I(z) = supk∈Z

supqk6z<qk−1

I(z)

= (1−q)α+β supk∈Z

(∞

∑j=k

q j f (q j)

)α( k

∑i=−∞

qig(qi)

.

We have proved that (3.1) holds wherever β > 0.Next we assume that α > 0. Let qk+1 < z < qk , k ∈ Z . Then we get that

I(z) = (1−q)α+β supk∈Z

(∞

∑j=k+1

q j f (q j)

)α( k

∑i=−∞

qig(qi)

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774 A. O. BAIARYSTANOV, L. E. PERSSON, S. SHAIMARDAN AND A. TEMIRKHANOVA

and analogously as above we find that

supqk+1<z6qk

I(z) = I(qk) = (1−q)α+β

(∞

∑j=k

q j f (q j)

)α( k

∑i=−∞

qig(qi)

and (3.1) holds also for the case α > 0. The proof is complete.

Proof of Lemma 3.6. According to (2.6) and (2.9) we have that

I+(qk) = (1−q)α+β

(∞

∑i=k

qi f (qi)

)α (qkg(qk)

)β,

for z = qk, k ∈ Z , and

I+(z) = (1−q)α+β

(∞

∑i=k+1

qi f (qi)

)α (qkg(qk)

)β,

for qk+1 < z < qk, k ∈ Z .Therefore,

supqk+1<z6qk

I+(z) = (1−q)α+β

(∞

∑i=k

qi f (qi)

)α (qkg(qk)

)β.

Since supz>0 I+(z) = supk∈Z

supqk+1<z6qk

I+(z) , we conclude that (3.3) holds.

Next, by using (2.7) and (2.8) we find that

I−(qk) = (1−q)α+β

(k

∑i=−∞

qi f (qi)

)α (qkg(qk)

)β, (4.1)

for z = qk, k ∈ Z , and

I−(z) = (1−q)α+β

(k−1

∑i=−∞

qi f (qi)

)α (qkg(qk)

)β,

for qk < z < qk−1, k ∈ Z .Thus,

supqk6z<qk−1

I−(z) = (1−q)α+β

(k

∑i=−∞

qi f (qi)

)α (qkg(qk)

)β.

Since supz>0

I−(z) = supk∈Z

supqk6z<qk−1

I−(z) , we have that (3.4) holds. The proof is com-

plete.

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SOME NEW HARDY-TYPE INEQUALITIES IN q-ANALYSIS 775

Proof of Lemma 3.7. By using (2.3), (2.6) and (2.7), we have that

D = (1−q)∞

∑k=−∞

qk

∞∫

0

X[qk,∞)(t) f (t)dqt

α

∞∫

0

X(0,qk](x)g(x)dqx

β

ϕ(qk)

= (1−q)α+β∞

∑k=−∞

qk

(k

∑i=−∞

qi f (qi)

)α(∞

∑j=k

q jg(q j)

ϕ(qk).

The proof is complete.

Proof of Lemma 3.8. By using (2.9), we get that

F(qk) =

∞∫

0

X[qk,qk−1)(t) f (t)dqt

α

= (1−q)α(

qk f (qk))α

, (4.2)

for y = qk , k ∈ Z , and

supy>qk

F(y) = supi6k

supqi<y6qi−1

F(y)

= (1−q)α supi6k

(qi−1 f (qi−1)

= (1−q)α supi6k−1

(qi f (qi)

)α, (4.3)

for i 6 k and qi < y 6 qi−1 .From (4.2) and (4.3) it follows that

supy>qk

F(y) = maxsupy>qk

F(y),F(qk)= (1−q)α supi6k

(qi f (qi)

)α.

Thus, (3.5) holds so the proof is complete.

Proof of Theorem 3.2. By using (2.3) and (2.7), we have that

∞∫

0

f p(x)dqx

1p

= (1−q)1p

(∞

∑j=−∞

q j f p(q j)

) 1p

, (4.4)

and

∞∫

0

u(x)

∞∫

0

X[x,∞)(t)v(t) f (t)dqt

r

dqx

1r

= (1−q)1r

∑j=−∞

q jur(q j)

∞∫

0

X[q j ,∞)(t)v(t) f (t)dqt

r

1r

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776 A. O. BAIARYSTANOV, L. E. PERSSON, S. SHAIMARDAN AND A. TEMIRKHANOVA

= (1−q)1+ 1r

(∞

∑j=−∞

q jur(q j)

(∑

qi>q j

qiv(qi) f (qi)

)r) 1r

= (1−q)1+ 1r

(∞

∑j=−∞

q jur(q j)

(j

∑i=−∞

qiv(qi) f (qi)

)r) 1r

. (4.5)

By now using (3.1), (4.4) and (4.5) we find that

(1−q)1p′+

1r

(∞

∑j=−∞

q j

(u(q j)

j

∑i=−∞

qiv(qi) f (qi)

)r) 1r

6C

(∞

∑j=−∞

q j f p(q j)

) 1p

.

Let

q j f p(q j) = f pj , v j = q

jp′ v(q j)(1−q)

1p′ , u j = (1−q)

1r q

jr u(q j), j ∈ Z. (4.6)

Then we see that the inequality (3.1) is equivalent to the inequality (2.10). Thebest constants in inequalities (3.1) and (2.10) are the same.

Since the inequality (3.1) is equivalent to the inequality (2.10) we can use Propo-sition 2.2 to conclude that the inequality (3.1) holds if and only if at least one of theconditions C1 < ∞ , C2 < ∞ and C3 < ∞ holds. Moreover, for the best constant C in(3.1) it yields that C ≈C1 ≈C2 ≈C3 .

Hence, according to Lemma 3.5 we have that

C1 = supn∈Z

(∞

∑k=n

urk

) 1r(

n

∑i=−∞

vp′i

) 1p′

= (1−q)1r +

1p′ sup

n∈Z

(∞

∑k=n

qkur(qk)

) 1r(

n

∑i=−∞

qivp(qi)

) 1p′

= supz>0

∞∫

0

X(0,z](x)ur(x)dqx

1r

∞∫

0

X[z,∞](t)vp′(t)dqt

1p′

= D∗1.

In particular, C ≈ D∗1 . Moreover, by arguing as above and using Lemma 3.6 we

obtain that C2 ≈ D∗2 and C3 ≈ D∗

3 . Hence, for the best constant C in (3.1) it yields thatC ≈ D∗

1 ≈ D∗2 ≈ D∗

3 . The proof is complete.

Proof of Theorem 3.4. In a similarly way as in the proof of Theorem 3.2, by using(2.3), (2.7) and (4.6), we find that the inequality (2.10) is equivalent to the inequality(3.1).

Since the inequality (3.1) is equivalent to the inequality (2.10) we can use Propo-sition 2.3 to conclude that the inequality (3.1) holds if and only if the conditions (2.18),(2.19) and (2.20) hold, for considered cases 0 < p < 1, p 6 r; 1 < p < ∞ , 0 < r < pand 0 < r < p = 1 , respectively.

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SOME NEW HARDY-TYPE INEQUALITIES IN q-ANALYSIS 777

Next, we prove that the conditions (2.18), (2.19) and (2.20) are equivalent to theconditions D∗

4 < ∞ , D∗5 < ∞ and D∗

6 < ∞ , respectively.By using Lemma 3.6 from (2.18) and (4.6) we obtain that

C4 = supn∈Z

(∞

∑k=n

urk

) 1r

vn = (1−q)1r +

1p′ sup

n∈Z

(∞

∑k=n

qkur(qk)

) 1r (

qnvp′(qn)) 1

p′

= supz>0

∞∫

0

X(0,z](x)ur(x)dqx

1r

∞∫

0

X[z,−q−1z)(t)vp′(t)dqt

1p′

= D∗4.

Moreover, by Lemma 3.7 we have that

C5 =∞

∑n=−∞

(n

∑i=−∞

vp′i

) r(p−1)p−r

(∞

∑k=n

urk

) rp−r

urn

= (1−q)rp

p−r+1∞∫

n=−∞

(n

∑i=−∞

qivp′(qi)

) r(p−1)p−r

(∞

∑k=n

qkur(qk)

) rp−r

qnur(qn)

=

∞∫

0

∞∫

0

X[z,∞)(t)vp′(t)dqt

r(p−1)p−r

∞∫

0

X(0,z](x)ur(x)dqx

rp−r

ur(z)dqz = D∗5

Now let p = 1 so that p′ = ∞ . Then vi = v(qi) in (4.6). By Lemma 3.8 we findthat

maxi6n

vr

1−ri = (max

i6nv(qi))

r1−r =

((1−q)max

i6n

qiv(qi)

(1−q)qi

) r1−r

=

sup

y>qn

∞∫

0

X[y,q−1y)(t)v(t)

(1−q)tdqt

r1−r

= supy>qn

∞∫

0

X[y,q−1y)(t)v(t)

(1−q)tdqt

r1−r

.

Therefore,

C6 =∞

∑n=−∞

maxi6n

vr

1−ri

(∞

∑k=n

urk

) r1−r

urn

=∞

∑n=−∞

qn maxi6n

vr

1−ri

((1−q)

∑k=n

qkurk(q

k)

) r1−r

ur(qn)

= (1−q)∞

∑n=−∞

qn supy>qn

∞∫

0

X[y,q−1y)(t)v(t)

(1−q)tdqt

r1−r

×

∞∫

0

X(0,qn](x)ur(x)dqx

r1−r

ur(qn)

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778 A. O. BAIARYSTANOV, L. E. PERSSON, S. SHAIMARDAN AND A. TEMIRKHANOVA

=

∞∫

0

supy>z

∞∫

0

X[y,q−1y)(t)v(t)

(1−q)tdqt

r1−r

∞∫

0

X(0,z](x)ur(x)dqx

r1−r

ur(z)dqz

= D∗6.

Thus, in all cases (i)–(iii), for the best constant in (3.1) it yields that C ≈ D∗i ,

i = 4,5,6, respectively. The proof is complete.

Proof of Theorem 3.1. As in the proof of Theorem 3.2 we get that the inequality(2.23) is equivalent to the inequality

(∞

∑j=−∞

(u j

∑i= j

vi fi

)r) 1r

6C

(∞

∑j=−∞

f pj

) 1p

. (4.7)

By using standard dual arguments the characterizations similar to those in Propo-sition 2.2 hold also in this situation (see e.g. [16, p. 59]). Here it is even simpler to justput ui = u−i , vi = v−i , fi = f−i , i ∈ Z , and note that then (4.7) reads

(∞

∑j=−∞

(u j

j

∑i=−∞

vi fi

)r) 1r

6C

(∞

∑j=−∞

f pj

) 1p

. (4.8)

Now use Proposition 2.2, and find that the inequality (4.8) holds if and only if oneof the conditions Ci < ∞ , 1 6 i 6 3 holds. Note that here Ci , 1 6 i 6 3, are defined byjust in the expressions for Ci inserting u j , v j , j ∈ Z . Moreover, for the best constantC in (4.8) it yields that C ≈ C1 ≈ C2 ≈ C3 .

Next, by replacing u j and v j by u j and v j , j ∈ Z , in the expressions Ci , 1 6i 6 3, respectively, we obtain the corresponding characterizations for the validity of theinequality (4.7). In a similar way as in the proof of Theorem 3.2, from the equivalenceof inequalities (2.23) and (4.7) and using Lemma 3.6 we find that the inequality (2.23)holds if and only if D1 < ∞ or D2 < ∞ or D3 < ∞ holds. Moreover, for the bestconstant C in (2.23) it yields that C ≈ D1 ≈ D2 ≈ D3 . The proof is complete.

Proof of Theorem 3.3. The equivalence between (4.7) and (4.8) holds in the casetoo. Hence, by arguing exactly as in proof of Theorem 3.1 but using Proposition 2.3instead of Proposition 2.2 the proof can be done analogously, so we leave out the details.

5. Final remarks

REMARK 5.1. Assume that v(t) = 0, u(t) = 0, f (t) = 0, t > 1 and the integralsin the expressions Di , D∗

i , 1 6 i 6 6 are replaced by the integrals from zero to one andthe sets [z,∞) , [z,q−1z) are replaced by the sets [z,1] , [z,minq−1z,1] , respectively.Then, by using Theorem 3.1, Theorem 3.2, Theorem 3.3 and Theorem 3.4, we obtain

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SOME NEW HARDY-TYPE INEQUALITIES IN q -ANALYSIS 779

that the corresponding characterizations for the validity of the inequalities

1∫

0

u(x)

1∫

0

X(0,x](t)v(t) f (t)dqt

r

dqx

1r

6C

1∫

0

f p(t)dqt

1p

,

and

1∫

0

u(x)

1∫

0

X[x,1](t)v(t) f (t)dqt

r

dqx

1r

6C

1∫

0

f p(t)dqt

1p

,

for all parameters r and p in these theorems.

REMARK 5.2. Note that nowadays it is known that the conditions Bi < ∞ , i =1,2,3, in Proposition B are special cases of more general conditions. More exactlythese conditions can be replaced by infinite many conditions, namely the followingfour scales of conditions (see [29] and also [17, p. 60]):

B1(s) := supn∈N

(n

∑k=1

v1−p′k

) (s−1)p

∑k=n

uk

(k

∑m=1

v1−p′m

) r(p−s)p

1r

< ∞,

for s satisfying 1 < s 6 p ;

B∗1(s) := sup

n∈N

(∞

∑k=n

uk

) (s−1)r′

n

∑k=1

v1−p′k

(∞

∑m=k

um

) p′(r′−s)r′

1r′

< ∞,

for s satisfying 1 < s 6 r′ ;

B2(s) := supn∈N

(n

∑k=1

v1−p′k

)−s

n

∑k=1

uk

(k

∑m=1

v1−p′m

)r( 1p′ +s)

1r

< ∞,

for s satisfying 0 < s 6 1p ;

B∗2(s) := sup

n∈N

(∞

∑k=n

uk

)−s

∑k=n

v1−p′k

(∞

∑m=k

um

)p′( 1r +s)

1p′

< ∞,

for s satisfying 0 < s 6 1r′ . Note that B1(p) = B∗

1(r′) = B1 , B2(

1p ) = B2 and B∗

2(1r′ ) =

B3 .Our results in Theorems 3.1 and 3.2 can be generalized in a corresponding way

namely that the three alternative conditions in these theorems can be replaced by infinitemany equivalent conditions.

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780 A. O. BAIARYSTANOV, L. E. PERSSON, S. SHAIMARDAN AND A. TEMIRKHANOVA

REMARK 5.3. The corresponding alternative conditions for the parameters in Pro-position 2.3 are not known except for the continuous case r < p , p > 1 where even fourscales of such alternative equivalent conditions are known (see [31]). Hence, at the mo-ment only in this case it seems to be possible to generalize Theorems 3.3 and 3.4 in thisdirection.

REMARK 5.4. Some similar results as those in this paper can found in [4] (inRussian). However, the results in this paper are more complete and putted to a moregeneral frame. The proofs are also different and more precise and clear.

Acknowledgements. This research has been done within the agreement betweenLule°a University of Technology, Sweden and L. N. Gumilyev Eurasian National Uni-versity, Kazakhstan. We thank both these universities for financial and other support.We also thank Professors R. Oinarov and V. D. Stepanpv for several generous adviseswhich has improved the final version of this paper. The third author was partially sup-ported by project 5495/GF4 of the Scientific Committee of Ministry of Education andScience of the Republic of Kazakhstan and project RFFI 16-31-50042.

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[30] L.-E. PERSSON AND V. D. STEPANOV, Weighted integral inequalities with the geometric mean op-erator, J. Inequal. Appl., 7 (5): 727–746, 2002.

[31] L.-E. PERSSON, V. D. STEPANOV AND P. WALL, Some scales of equivalent chrarcterizations ofHardy’s inequal: the case q < p , Math. Inequal. Appl. 10 (2007), no. 2, 267–279.

[32] M. S. STANKOVIC, P. M. RAJKOVIC AND S. D. MARINKOVIC, On q-fractional derivatives ofRiemann-Liouville and Caputo type, arXiv:0909.0387.

[33] W. T. SULAIMAN, New types of q -integral inequalities, Adv. Pure Math. 1 (2011), 77–80.[34] G. TOMASELLI, A class of inequalities, Boll. Un. Mat. Ital. 2 (1969), 622–631.

(Received February 17, 2015) A. O. BaiarystanovEurasian National University

Munaytpasov st., 5, 010008 Astana, Kazakhstane-mail: oskar [email protected]

L. E. PerssonDepartment of Engineering Sciences and Mathematics

Lule°a University of TechnologySE-971 87, Lule°a, Sweden

andUiT, The Artic University of Norway

P. O. Box 385, N-8505, Narvik, Norwaye-mail: [email protected]

S. ShaimardanEurasian National University

Munaytpasov st., 5, 010008 Astana, Kazakhstane-mail: [email protected]

A. TemirkhanovaEurasian National University

Munaytpasov st., 5, 010008 Astana, Kazakhstane-mail: [email protected]

Journal of Mathematical [email protected]

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Paper C

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EURASIAN MATHEMATICAL JOURNAL

ISSN 2077-9879Volume 7, Number 1 (2016), 84 – 99

HARDY-TYPE INEQUALITIES FOR THE FRACTIONAL INTEGRAL

OPERATOR IN q-ANALYSIS

S. Shaimardan

Communicated by R. Oinarov

Key words: Hardy-type inequalities, integral operator, q-analysis, q-integral.

AMS Mathematics Subject Classification: 26D10, 26D15, 33D05, 39A13.

Abstract. We obtain necessary and sufficient conditions for the validity of a certianHardy-type inequality involving q-integrals.

1 Introduction

The q-derivative or Jackson’s derivative, is a q-analogue of the ordinary derivative.q-differentiation is the inverse of Jackson’s q-integration. It was introduced by F. H.Jackson [11] (see also [7]). He was the first to develop q-analysis. After that manyq-analogue of classical results and concepts were studied and their applications areinvestigated.

Concerning recent results on q-analysis and its applications we also refer to theresent book by T. Ernst [8]. Some integral inequalities were obtained by H. Gauchman[10]. A Hardy-type inequality in q-analysis was recently obtained by L. Maligranda,R. Oinarov and L-E. Persson [13].

In this paper we prove a new Hardy-type inequality in which the Hardy operatoris replaced by the q-analogue of the infinitesimal fractional operator (see [1] and (1.3)below).

In classical analysis, the hypergeometric function (Gaussian function) is defined for|z| < 1 by the power series [9]:

2F1 (α, β; γ; z) =∞∑

n=0

(α)n(β)n

(γ)n

zn

n!, ∀α, β, γ ∈ C,

where (α)n is the Pochhammer symbol, which is defined by:

(α)0 = 1, (α)n = α(α + 1) · · · (α + n − 1), n > 0.

If B denotes the Beta function, then

2F1 (α − 1, β; γ; z) =1

B(β, γ − β)

1∫

0

xβ−1(1 − x)γ−β−1(1 − zx)2−αdx,

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Some Hardy-type inequalities for the fractional integral operator in q-analysis 85

where Re(γ) > Re(β) > 0. When β = γ we have that

2F1 (α − 1, β; β; z) = (1 − z)α−1.

Let α+β < γ, γ 6= 0,−1,−2, · · · . Then the following generalized fractional integraloperator was introduced in [14]:

Iγ,βα f(x) =

xα−1

Γ(α)

x∫

0

2F1

(α − 1, β; γ;

s

x

)ds, (1.1)

where Γ(·) denotes the Gamma function. If β = γ then the operator

Iαf(x) :=xα−1

Γ(α)

x∫

0

2F1

(α − 1, β; β;

s

x

)ds,

is called the Riemann-Liouville fractional integral operator. When γ = 1, β = 2, wehave that

If(x) := limα−→0

Γ(α)I1,2α f(x) =

x∫

0

lnx

x − s

f(s)

sds, (1.2)

which is called the infinitesimal fractional integral operator [1].The purpose of this paper is to find a q-analogue of operator (1.2) and to prove a

q-analogue of the following Hardy-type integral inequality [1]:

∞∫

0

ur(x)

x∫

0

tγ−1 lnx

x − tf(t)dt

r

dx

1r

≤ C

∞∫

0

fp(x)dx

1p

, ∀f(·) ≥ 0, (1.3)

where C > 0 is independent of f and u is a positive real valued function on (0,∞)briefly a weight function. We derive necessary and sufficient conditions for the validityof a q-analogue of inequality (1.3) in q-analysis for the case 1 < p < ∞, 0 < r < ∞ andγ > 1

p(see Theorem 3.1 and Theorem 3.2). We also consider the problem of finding

the best constant in a q-analogue of inequality (1.3).The paper is organized as follows: We present some preliminaries in Section 2. The

main results and detailed proofs are presented in Section 3.

2 Preliminaries

First we recall definitions and notions of the theory of q-analysis, our main referencesare the books [7], [8] and [9].

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86 S. Shaimardan

Let 0 < q < 1 be fixed.For a real number α ∈ R, the q-real number [α]q is defined by

[α]q =1 − qα

1 − q, α ∈ R.

It is clear that limq−→1

1−qα

1−q= α.

The q-analogue of the power (a − b)k is defined by

(a − b)0q = 1, k ∈ N, (a − b)k

q =k−1∏

i=0

(a − qib), ∀ a, b ∈ R,

and

(1 − b)αq :=

(1 − b)∞q(1 − qαb)∞q

, ∀ b, α ∈ R. (2.1)

and by using well-known relations this can also be written as

(1 − b)αq =

1

(1 − qαb)−αq

, ∀b, α ∈ R. (2.2)

The q-hypergeometric function 2Φ1 is defined by ([9]):

2Φ1

[qα qβ

;q ;xqγ

]:=

∞∑

n=0

(qα; q)nq (qβ; q)n

q

(qγ; q)nq (q; q)n

q

xn, |x| < 1,

where (qα; q)nq =

n−1∏i=0

(1 − qi+α) and γ 6= 0,−1,−2, · · · . Moreover, this series converges

absolutely and limq→1

(qα;q)nq

(1−q)n = (a)n, so

limq→1

2Φ1

[qα qβ

;q ;xqγ

]=2 F1 (α, β; γ; x) .

For f : [0, b) −→ R, 0 ≤ b < ∞, the q-derivative is defined by:

Dqf(x) :=f(x) − f(qx)

(1 − q)x, x ∈ (0, b), (2.3)

and Dqf(0) = f ′(0) provided f ′(0) exists. It is clear that if f(x) is differentiable, thenlim

q−→1Dqf(x) = f ′(x).

Definition 1. The q-Taylor series of f(x) at x = c is defined by

f(x) :=∞∑

j=0

(Dj

q

)(c)

(x − c)jq

[j]q!,

where

[j]q! =

1, if j = 0,[1]q × [2]q × · · · × [j]q, if j ∈ N.

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Some Hardy-type inequalities for the fractional integral operator in q-analysis 87

The definite q-integral or the q-Jackson integral of a function f is defined by theformula

x∫

0

f(t)dqt := (1 − q)x∞∑

k=0

qkf(qkx), x ∈ (0, b), (2.4)

and the improper q-integral of a function f(x) : [0,∞) → R, is defined by the formula

∞∫

0

f(t)dqt := (1 − q)∞∑

k=−∞

qkf(qk). (2.5)

Note that the series in the right hand sides of (2.4) and (2.5) converge absolutely.

Definition 2. The function

Γq(α) :=

∞∫

0

xα−1E−qxq dqx, α > 0,

is called the q-Gamma function, where E−qxq = (1 − (1 − q)x)∞q .

We have that

Γq(α + 1) = [α]qΓq(α),

for any α > 0.

Definition 3. The function

Bq(α, β) :=

1∫

0

tα−1(1 − qt)β−1q dqt, α, β > 0,

is called the q-Beta function. Note that

Bq(α, β) =Γq(α)Γq(β)

Γq(α + β),

for α, β > 0.

Let Ω be a subset of (0,∞) and XΩ(t) denote the characteristic function of Ω. Forall z > 0, we have that(see [5]):

∞∫

0

X(0,z](t)f(t)dqt = (1 − q)∑

qi≤z

qif(qi), (2.6)

∞∫

0

X[z,∞)(t)f(t)dqt = (1 − q)∑

qi≥z

qif(qi). (2.7)

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88 S. Shaimardan

R.P. Agarwal and W.A. Al-Salam (see [2], [3] and [4]) introduced several types offractional q-integral operators and fractional q-derivatives. In particular, they definedthe q-analogue of the fractional integral operator of the Riemann-Liouville type by

Iq,αf(x) =xα−1

Γq(α)

x∫

0

(1 −qs

x)α−1q f(s)dqs, α ∈ R

+.

Using formula (2.2), we can rewrite Iq,α as follows:

Iq,αf(x) =xα−1

Γq(α)

x∫

0

f(s)

(1 − qα sx)1−αq

dqs, α ∈ R+. (2.8)

Out next goal is to define a q-analogue of ln xx−s

, but for this we need the followingresult of independent interest.

Proposition 2.1. Let 0 < s ≤ x < ∞. Then

2Φ1

[q1−α qβ

;q ;qα sx

]=

1

Bq(β, γ)

1∫

0

tβ−1(1 − qt)γ−β−1q

(1 − qαt sx)1−αq

dqt, (2.9)

for β, γ > 0, and

2Φ1

[q1−α qβ

;q ;qα sx

]=

1

(1 − qα sx)1−αq

, (2.10)

for β = γ.

Proof. First we consider equality (2.10). From (2.1) and (2.3), we get that

D1q.s

(1

(1 − qα sx)1−αq

)= D1

q,s

((1 − q s

x)∞q

(1 − qα sx)∞q

)

=

[(1 − q2 s

x)∞q

(1 − qα+1 sx)∞q

−(1 − q s

x)∞q

(1 − qα sx)∞q

]1

(q − 1)s

=(1 − q2 s

x)∞q

(1 − qα sx)∞q

[(1 − qα s

x) − (1 − q s

x)

s(q − 1)

]

=(1 − q2 s

x)∞q

(1 − qα sx)∞q

[qα(q1−α − 1)

x(q − 1)

]

=qα

x[1 − α]q

(1 − qα sx)2−αq

.

Using this relation and induction, one can easily see that

Djq,s

(1

(1 − qα sx)1−αq

)∣∣∣∣s=0

=qjα

xj[1 − α]q[2 − α]q · · · [j − α]q,

for any j ≥ 1. Therefore, we have the q-Taylor expansion (see Definition 1)

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Some Hardy-type inequalities for the fractional integral operator in q-analysis 89

1

(1 − qα sx)1−αq

=∞∑

j=0

[1 − α]q[2 − α]q · · · [j − α]q[j]q!

(qαs

x

)j

=∞∑

j=0

(1 − q1−α)jq

(1 − q)jq

(qαs

x

)j

= 2Φ1

[q1−α qβ

;q ;qα sx

], (2.11)

and (2.10) is proved.By using the same arguments as above we see that

1

(1 − qαt sx)1−αq

=∞∑

n=0

(1 − q1−α)nq

(1 − q)nq

(tqαs

x

)n

,

for x ≥ s, 0 < t ≤ 1. Therefore

1∫

0

tβ−1(1 − qt)γ−β−1q

(1 − qαt sx)1−αq

dqt =∞∑

n=0

(1 − q1−α)nq

(1 − q)nq

(qαs

x

)n1∫

0

tβ+n−1(1 − qt)γ−β−1q dqt

=∞∑

n=0

(1 − q1−α)nq

(1 − q)nq

(qαs

x

)nΓq(β + n)Γq(γ − β)

Γq(γ + n)

=Γq(β)Γq(γ − β)

Γq(γ)

∞∑

n=0

(1 − q1−α)nq (1 − qβ)n

q

(1 − q)nq (1 − qγ)n

q

(qαs

x

)n

= Bq(β, γ)2Φ1

[q1−α qβ

;q ;qα sx

].

and also (2.9) is proved.

By Proposition 2.1, the integral (2.8) can be rewritten as

Iq,αf(x) =xα−1

Γq(α)

x∫

0

2Φ1

[q1−α qβ

;q ;qα sx

]f(s)dqs, α ∈ R

+, β ∈ R.

More generally, we consider the q-analogue of Iγ,βα (see (1.1))

Iγ,βq,α f(x) =

xα−1

Γq(α)

x∫

0

2Φ1

[q1−α qβ

;q ;qα sx

]f(s)dqs, α, β, γ ∈ R

+.

Due to uniform convergence of the series 2Φ1

[q1−α q

;q ;qα sx

q2

]for 0 < α < 1, we get

that

limα→0+

2Φ1

[q1−α q

;q ;qα sx

q2

]s

x= 2Φ1

[q q

;q ; sx

q2

]s

x

=∞∑

j=0

1 − q

1 − qj+1

sj+1

xj+1=

∞∑

j=0

(sx

)j+1

[j + 1]q=

∞∑

j=1

( sx)j

[j]q,

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90 S. Shaimardan

which is the q-analogue of the Taylor series of the function ln xx−s

with s < x.

Definition 4. We define the q-analogue of the function ln xx−s

, 0 < s < x < ∞, asfollows:

lnq

x

x − s:=

∞∑

j=1

( sx)j

[j]q.

Remark 5. We define the q-analog of (1.2) as follows:

Iqf(x) :=

qx∫

0

lnq

x

x − s

f(s)

sdqs, (2.12)

which is called the infinitesimal q-fractional integral operator.Observe that:

limq→1

Iqf(x) =

x∫

0

lnx

x − s

f(s)

sds

.

Hence, from (2.12) we obtain the q-analogue of (1.3) in the following form:

∞∫

0

ur(x)(Iqf(x)

)r

dqx

1r

≤ C

∞∫

0

fp(t)dqt

1p

, ∀f(·) ≥ 0, (2.13)

where C > 0 independent of f .In the q-integral we are allowed to change variables in the form x = tξ for 0 < ξ < ∞

(see [7]). So by making the substitution t = qs, and dqt = qdqs inequality (2.13)becomes

∞∫

0

ur(x)

x∫

0

sγ−1 lnq

x

x − qsf(s)dqs

r

dqx

1r

≤ C

∞∫

0

fp(s)dqs

1p

, ∀f(·) ≥ 0. (2.14)

where f(s) = f(qs), C = qγ− 1p C.

Since inequality (2.13) holds if and only if inequality (2.14) holds, from now on wewill investigate necessary and sufficient conditions the validity of inequality (2.14).Notation. In the sequel, for any p > 1 the conjugate number p′ is defined by p′ :=p/(p − 1). Moreover, the symbol M ≪ K means that there exists α > 0 such thatM ≤ αK, where α is a constant which depend only on the numerical parameters suchas p, q, r. If M ≪ K ≪ M , then we write M ≈ K.

For the proof of our main theorems we will need the following well-known discreteweighted Hardy inequality proved by G. Bennett [6] (see also [12], p.58):

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Some Hardy-type inequalities for the fractional integral operator in q-analysis 91

Theorem A. Let ui∞i=1 and vj

∞j=1 be non-negative sequences of real numbers and

1 < p ≤ r < ∞. Then the inequality

(∞∑

j=−∞

(∞∑

i=j

fi

)r

urj

) 1r

≤ C

(∞∑

i=−∞

vpi f

pi

) 1p

, f ≥ 0, i ∈ Z, (2.15)

with C > 0 independent of fi, i ∈ Z holds if and only if

B1 := supn∈Z

(n∑

j=−∞

urj

) 1r(

∞∑

i=n

v−p′

i

) 1p′

< ∞, p′ =p

p − 1.

Moreover,B1 ≈ C, where C is the best constant in (2.15).Theorem B. Let 0 < r < p < ∞ and 1 < p. Then inequality (2.15) holds if and onlyif B2 < ∞, where

B2 :=

∞∑

k=−∞

v−p′

k

(k∑

i=−∞

uri

) pp−r(

∞∑

i=k

v−p′

i

) p(r−1)p−r

p−rpr

.

Moreover, B2 ≈ C, where C is the best constant in (2.15).Also we need the following lemma ([5]):

Lemma A. Let f , ϕ and g be nonnegative functions. Then

∞∫

0

∞∫

0

X[z,∞)(t)f(t)dqt

α

∞∫

0

X(0,z](x)g(x)dqx

β

ϕ(z)dqz

= (1 − q)α+β

∞∑

k=−∞

(

k∑

i=−∞

qif(qi)

)α(∞∑

j=k

qjg(qj)

qkϕ(qk)

,

for α, β ∈ R.

3 Main results

Our main result reads:

Theorem 3.1. Let 1 < p ≤ r < ∞, γ > 1p. Then the inequality

∞∫

0

ur(x)

x∫

0

sγ−1 lnq

x

x − qsf(s)dqs

r

dqx

1r

≤ C

∞∫

0

fp(s)dqs

1p

, ∀f(·) ≥ 0, (3.1)

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92 S. Shaimardan

with C > 0 independent of f holds if and only if B1 < ∞, where

B1 := supx>0

xγ+ 1

p′

( ∞∫

0

X[x,∞)(t)ur(t)

trdqt

) 1r

,

Moreover, B1 ≈ C , where C is the best constant in (3.1).

Theorem 3.2. Let 0 < r < p < ∞, 1 < p and γ > 1p. Then the inequality (3.1) holds

if and only if B2 < ∞, where

B2 :=

∞∫

0

x

γ+ 1p′

∞∫

0

X[x,∞)(t)ur(t)

trdqt

1r

prp−r

dqx

p−rpr

.

Moreover, B2 ≈ C , where C is the best constant in (3.1).

Remark 6. By using formulas (2.4) and (2.5) in (3.1) we get that

(∞∑

j=−∞

(1 − q)qjur(qj)

(∞∑

i=j

(1 − q)qiγf(qi) lnq

1

1 − qi−j+1

)r) 1r

≤ C

(∞∑

i=−∞

(1 − q)qifp(qi)

) 1P

.

Let

urj = (1 − q)

1+ rp′ qjur(qj), fi = (1 − q)

1p q

ip f(qi), ai,j = lnq

1

1 − qi−j+1. (3.2)

Then we get that inequality (3.1) is equivalent to the discrete weighted Hardy-typeinequality (

∞∑

j=−∞

urj

(∞∑

i=j

qi(γ− 1p)fiai,j

)r) 1r

≤ C

(∞∑

i=−∞

fpi

) 1p

. (3.3)

Note that inequality (3.1) holds if and only if inequality (3.3) holds, so we will obtainthe desired necessary and sufficient conditions for the validity of inequality (3.3).

Our next Lemmas give a characterization of the discrete Hardy-type inequality(3.3).

Lemma 3.1. Let 1 < p ≤ r < ∞, γ > 1p. Then the inequality (3.3) holds if and only

if B1 < ∞, where

B1 := supk∈Z

( ∞∑

i=k

qi(p′γ+1)

) 1p′( k∑

j=−∞

q−jrurj

) 1r

. (3.4)

Moreover, B1 ≈ C, where C is the best constant in (3.3).

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Some Hardy-type inequalities for the fractional integral operator in q-analysis 93

Proof. Necessity. Let us assume that (3.3) holds with some C > 0. From (3.2) andDefinition 4 we get that qi+1/qj ≤ ai,j for j ≤ i. Then

∞∑

j=−∞

urj

(∞∑

i=j

qi(γ− 1p)fiai,j

)r

≥ q

∞∑

j=−∞

q−jrurj

(∞∑

i=j

qi(γ+ 1

p′)fi

)r

.

Moreover,

q

(∞∑

j=−∞

q−jrurj

(∞∑

i=j

qi(γ+ 1

p′)fi

)r) 1r

≤ C

(∞∑

j=−∞

fpi

) 1p

.

Hence, by Theorem A we obtain that

B1 ≪ C. (3.5)

The proof of the necessity is complete.Sufficiency. Let B < ∞ and f ≥ 0 be arbitrary. We will show that inequality (3.3)

holds.We consider two cases separately: 0 < q ≤ 1

2and 1

2< q < 1.

1) Let 0 < q ≤ 12. Let j ≤ k ≤ i. Then from (3.2) and Definition 4 it follows that

aj,j = lnq1

1−q≤ lnq 2 (we note that lnq 2 :=

∞∑n=1

2−n

[n]q), and

q−kak,j − q−iai,j = q−k

∞∑

n=1

(qqk/qj)n

[n]q− q−i

∞∑

n=1

(qqi/qj)n

[n]q

=∞∑

n=1

(q/qj)n

[n]q

(qk(n−1) − qi(n−1)

)≥ 0,

i.e.

q−iai,j ≤ q−kak,j, (3.6)

for j ≤ k ≤ i.Thus by (3.6) we have that

∞∑

j=−∞

urj

(∞∑

i=j

qi(γ− 1p)fiai,j

)r

=∞∑

j=−∞

urj

(∞∑

i=j

qi(γ+ 1

p′)q−iai,jfi

)r

∞∑

j=−∞

q−jrurja

rj,j

(∞∑

i=j

qi(γ+ 1

p′)fi

)r

≤ (lnq 2)r∞∑

j=−∞

q−jrurj

(∞∑

i=j

qi(γ+ 1

p′)fi

)r

∞∑

j=−∞

q−jrurj

(∞∑

i=j

qi(γ+ 1

p′)fi

)r

.

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94 S. Shaimardan

Hence, by Theorem A we obtain that

(∞∑

j=−∞

q−jrurj

(∞∑

i=j

qi(γ+ 1

p′)fi

)r) 1r

≤ B1

(∞∑

j=−∞

fpi

) 1p

,

which means that inequality (3.3) is valid and that C ≪ B1, where C is the bestconstant for which (3.3) holds.

2) Let 12

< q < 1. Then ∃i0 ∈ N such that i0 > 1 and qi0 ≤ 12

< qi0−1. We assumethat Z =

⋃k∈Z

[tk + 1, tk+1] and tk+1 − tk = t0. Then the left hand side of (3.3) can be

written as

∞∑

j=−∞

urj

(∞∑

i=j

qi(γ− 1p)fiai,j

)r

=∑

k

tk+1∑

j=tk+1

urj

(∞∑

i=j

qi(γ− 1p)fiai,j

)r

≈∑

k

tk+1∑

j=tk+1

urj

(tk+2−1∑

i=j

qi(γ− 1p)fiai,j

)r

+∑

k

tk+1∑

j=tk+1

urj

∞∑

i=tk+2

qi(γ− 1p)fiai,j

r

= I1 + I2. (3.7)

To estimate I1 we use Holder’s inequality. We find that

I1 ≤∑

k

tk+1∑

j=tk+1

urj

(tk+2−1∑

i=j

qip′(γ− 1p)ap′

i,j

) rp′(

tk+2−1∑

i=j

fpi

) rp

≤∑

k

tk+1∑

j=tk+1

urj

(∞∑

i=j

qip′(γ− 1p)ap′

i,j

) rp′(

tk+2∑

i=tk+1

fpi

) rp

=∑

k

tk+1∑

j=tk+1

urjq

jr(γ− 1p)

(∞∑

i=0

qip′(γ− 1p)ap′

i,0

) rp′(

tk+2∑

i=tk+1

fpi

) rp

= Crp′

0

k

tk+1∑

j=tk+1

urjq

−jrqjr(γ+ 1

p′)

(tk+2∑

i=tk+1

fpi

) rp

, (3.8)

where C0 :=∞∑i=0

qip′(γ− 1p)ap′

i,0.

Since

M := (1 − q)C0 =

1∫

0

xp′(γ− 1p)

(lnq

1

1 − qx

)p′

dqx < ∞

and

qjr(γ+ 1

p′)≤ q

(tk+1)(γ+ 1p′

)r= q

−(i0−1)(γ+ 1p′

)q

tk+1r(γ+ 1p′

)≤ 2

r(γ+ 1p′

)q

tk+1(p′γ+1) rp′ , (3.9)

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Some Hardy-type inequalities for the fractional integral operator in q-analysis 95

for tk + 1 ≤ j, we get that

I1 ≤ 2r(γ+ 1

p′)M

rp′ [p′γ + 1]

rp′

q

k

(tk+2∑

i=tk+1

fpi

) rp (

qtk+1(p′γ+1)

1 − qp′γ+1

) rp′

tk+1∑

j=−∞

urjq

−jr

≪∑

k

(tk+2∑

i=tk+1

fpi

) rp (

qtk+1(p′γ+1)

1 − qp′γ+1

) rp′

tk+1∑

j=−∞

urjq

−jr

=∑

k

(tk+2∑

i=tk+1

fpi

) rp

∞∑

i=tk+1

qir(p′γ+1)

1p′ ( tk+1∑

j=−∞

urjq

−jr

) 1r

r

≪ Br1

(∞∑

i=−∞

fpi

) rp

. (3.10)

Let j ≤ k ≤ i. Then from (3.2) and Definition 4 it follows that

qkai,k − qjai,j ≥ qj (ai,k − ai,j) = qk

∞∑

n=1

(qi+1/qk

)n

[n]q− qj

∞∑

n=1

(qi+1/qj)n

[n]q

=∞∑

n=1

q(i+1)n

[n]q

(qk(1−n) − qj(1−n)

)≥ 0,

i.e.

qjai,j ≤ qkai,k, (3.11)

for j ≤ k ≤ i.Using (3.6) and (3.11) we find that

1

qi−jai,j ≤

1

qtk+1−tk+2lnq

1

1 − qtk+1−tk=

1

qi0lnq

1

1 − qi0≤ 2 lnq 2,

for j ≤ tk+1 and tk+2 ≤ i.Therefore,

I2 =∑

k

tk+1∑

j=tk+1

urj

∞∑

i=tk+2

qi(γ− 1p)fiai,j

r

=∑

k

tk+1∑

j=tk+1

q−jrurj

∞∑

i=tk+2

qi(γ+ 1

p′) 1

qi−jai,jfi

r

≤ (2 lnq 2)rp′

k

tk+1∑

j=tk+1

q−jrurj

∞∑

i=tk+2

qi(γ+ 1

p′)fi

r

∞∑

j=−∞

q−jrurj

(∞∑

i=j

qi(γ+ 1

p′)fi

)r

.

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96 S. Shaimardan

By using Theorem A we have that

I2 ≪ Br1

(∞∑

i=−∞

fpi

) rp

, (3.12)

Thus, from (3.7), (3.10) and (3.12) it follows that inequality (3.3) is valid and wesee that the best constant C in (3.3) is such that C ≪ B1, which together with (3.5)gives that C ≈ B1.

Lemma 3.2. Let 0 < r < p < ∞ and 1 < p. Then inequality (3.3) holds if and onlyif B2 < ∞, where

B2 :=

∞∑

k=−∞

qk(p′γ+1)

(k∑

i=−∞

uri q

−ir

) pp−r(

∞∑

i=k

qi(p′γ+1)

) p(r−1)p−r

p−rpr

.

Moreover, B2 ≈ C, where C is the best constant in (3.3).

Proof. In a similar way as in the proof of Lemma 3.1. by Theorem B we obtain thatinequality (3.3) is valid and that C ≈ B2. where C is the best constant for which (3.3)holds for 0 < q ≤ 1

2.

In case 12

< q < 1 the necessary part is due to Theorem B. Therefore,

B2 ≪ C. (3.13)

To prove sufficiency we proceed as follows. Applying to (3.8) Holder’s inequalitywith the exponents p

p−rand p

rwe obtain that

I1 ≪∑

k

(tk+2∑

i=tk+1

fpi

) rp

qtk+1(p′γ+1) r

p′

tk+1∑

j=−∞

urjq

−jr

k

qtk+1(p′γ+1)r(p−1)

p−r

(tk+1∑

j=−∞

urjq

−jr

) pp−r

p−rp (∑

k

tk+2∑

i=tk+1

fpi

) rp

≪ Bp−r

p

(∞∑

i=−∞

fpi

) rp

.

Since

B :=∞∑

i=−∞

qi(p′γ+1)r(p−1)

p−r

(i∑

j=−∞

urjq

−jr

) pp−r

∞∑

i=−∞

qi(p′γ+1)

(qi(p′γ+1)

1 − qp′γ+1

) p(r−1)p−r

(i∑

j=−∞

urjq

−jr

) pp−r

= Bpr

p−r

2 ,

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Some Hardy-type inequalities for the fractional integral operator in q-analysis 97

we have that

I1 ≪ Br2

(∞∑

i=−∞

fpi

) rp

(3.14)

From (3.12) and Theorem B it follows that

I2 ≪

∞∑

j=−∞

q−jrurj

(∞∑

i=j

qi(γ+ 1

p′)fi

)r

∞∑

j=−∞

q−jrurj

(∞∑

i=j

qi(γ+ 1

p′)fi

)r

≤ Br2

(∞∑

i=−∞

fpi

) rp

. (3.15)

Thus, from (3.14) and (3.15) it follows that C ≪ B2 which means that the inequality(3.3) is valid, which together with (3.13) gives B2 ≈ C.

Lemma 3.3. Let γ > 1p, and B1 < ∞. Then

αB1 = supk∈Z

(∞∑

i=k

qi(p′γ+1)

) 1p′(

k∑

j=−∞

qj(1−r)ur(qj)

) 1r

, (3.16)

for r > 0, where α = [p′γ + 1]− 1

p′

q (1 − q)− 1

r− 1

p′ .

Proof. Let γ > 1p. By using (2.7) we obtain that

I(x) = xγ+ 1

p′

∞∫

0

X[x,∞)(t)t−rur(t)dqt

1r

= (1 − q)1r x

γ+ 1p′

qj≥x

q(1−r)jur(qj)

1r

,

for ∀r > 0. Then

I(x) = (1 − q)1p′

+ 1r [p′γ + 1]

1p′

q

(∞∑

i=k

qi(p′γ+1)

) 1p′(

k∑

j=−∞

q(1−r)iur(qi)

) 1r

,

for x = qk,∀k ∈ Z. Moreover,

I(x) = (1 − q)1p′

+ 1r [p′γ + 1]

1p′

q

(∞∑

i=k

qi(p′γ+1)

) 1p′(

k−1∑

j=−∞

q(1−r)iur(qi)

) 1r

,

for qk < x < qk−1. Hence

supqk<x≤qk−1

I(x) = (1 − q)1p′

+ 1r [p′γ + 1]

1p′

q

(∞∑

i=k

qi(p′γ+1)

) 1p′(

k∑

j=−∞

q(1−r)iur(qi)

) 1r

,

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98 S. Shaimardan

and

αB1 = α supk∈Z

supqk<x≤qk−1

I(x) = supk∈Z

(∞∑

i=k

qi(p′γ+1)

) 1p′(

k∑

j=−∞

q(1−r)iur(qi)

) 1r

.

We have proved that (3.16) holds.

Next, we prove Theorem 3.1.Proof of Theorem 3.1. First we note that inequality (3.3) is equivalent to inequality(3.1). Moreover, by Lemma 3.1 inequality (3.1) holds if and only if B1 < ∞. From (3.2)and Lemma 3.3 we have that B1 = αB1. which means that B1 ≈ C and inequality(3.1) holds if and only if B1 < ∞.

Proof of Theorem 3.2. In a similar way as in the proof of Theorem 3.1, by Lemma 3.2we have that inequality (3.1) holds if and only if B2 < ∞. From (3.2) and Lemma Awe have that

Bpr

p−r

2 = (1 − q)∞∑

k=−∞

qk(p′γ+1)

((1 − q)

k∑

i=−∞

ur(qi)q−ir

) pp−r

×

((1 − q)

∞∑

i=k

qi(p′γ+1)

) p(r−1)p−r

=

∞∫

0

xp′γ+1

∞∫

0

X[x,∞)(t)ur(t)

trdqt

pp−r

∞∫

0

X(0,x](t)sp′γdqs

p(r−1)p−r

dqx

= [p′γ + 1]−

prp−r

q

∞∫

0

x

γ+ 1p′

∞∫

0

X[x,∞)(t)ur(t)

trdqt

1r

prp−r

dqx

≪ B2,

which means that B2 ≈ C and inequality (3.1) holds if and only if B2 < ∞. The proofis complete.

Acknowledgments

The author thank Professor Ryskul Oinarov (L.N. Gumilyev Eurasian National Uni-versity, Kazakhstan) and Lars-Erik Persson (Department of Engineering Sciences andMathematics, Lulea University of Technology, Sweden) for good advices which haveimproved the final version of this paper.This work was supported by Scientific Committee of Ministry of Education and Sci-ence of the Republic of Kazakhstan, grant no. 5495/GF4. It was also supported bythe Russian Sientific Foundation (project RFFI 16-31-50042).

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Some Hardy-type inequalities for the fractional integral operator in q-analysis 99

References

[1] A.M. Abylaeva, M.Zh. Omirbek, A weighted estimate for an integral operator with a logarithmic

singularity, (Russian) Izv. Nats. Akad. Nauk Resp. Kaz. Ser. Fiz.-Mat. No 1 (2005), 38-47.

[2] R.P. Agarwal, Certain fractional q-integrals and q-derivatives, Proc. Camb. Phil. Soc. 66 (1969),365-370.

[3] W.A. Al-Salam, Some fractional q-integrals and q-derivatives, Proc. Edinb. Math. Soc. 2(1966/1967), 135-140.

[4] M. H. Annaby, Z.S. Mansour, q-fractional calculus and equations, Springer, Heidelberg, 2012.

[5] A.O. Baiaristanov, S. Shaimardan, A. Temirkhanova, Weighted Hardy inequalities in quantum

analysis, Vestinik KarGU. Mathematics series. 2 (2013), no. 70, 35-45 (in Russian).

[6] G. Bennett, Some elementary inequalities, Quart. J. Math. Oxford Ser. 38 (1987), no. 152,401-425.

[7] P. Cheung, V. Kac, Quantum calculus, - Edwards Brothers, Inc., Ann Arbor, MI, USA, 2000.

[8] T. Ernst, A comprehensive treatment of q-calculus, Birkhauser/Springer Basel AG, Basel, 2012.

[9] G. Gasper, M. Rahman, Basic hypergeometric series, Cambridge 1990.

[10] H. Gauchman, Integral inequalities in q-calculus, Comput. Math. Appl. 47 (2004), no. 2-3, 281-300.

[11] F.H. Jackson, On q-definite integrals, Quart. J. Pure Appl. Math. 41 (1910), 193-203.

[12] A. Kufner, L. Maligranda, L-E. Persson, The Hardy inequality. About its history and some related

results, Vydavatelsky Servis, Plzen, 2007.

[13] L. Maligranda, R. Oinarov, L-E. Persson, On Hardy q-inequalities, Czechoslovak Math. J. 64(2014), 659-682.

[14] A.M. Nakhushev, Equations of mathematical biology, M.: Vysshaya shkola, 1995 (in Russian).

Serikbol ShaimardanFaculty of Mechanics and MathematicsL.N. Gumilyov Eurasian National University2 Satpayev St,010000 Astana, KazakhstanE-mail: [email protected]

Received: 20.11.2015

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Paper D

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Persson et al. Journal of Inequalities and Applications ( 2018) 2018:73 https://doi.org/10.1186/s13660-018-1662-6

RESEARCH Open Access

Hardy-type inequalities in fractionalh-discrete calculusLars-Erik Persson1,2*, Ryskul Oinarov3 and Serikbol Shaimardan1,3

*Correspondence: [email protected]å University of Technology,

Luleå, Sweden2UiT The Artic University of Norway,

Narvik, Norway

Full list of author information is

available at the end of the article

AbstractThe first power weighted version of Hardy’s inequality can be rewritten as

∫ ∞

0

(

xα–1∫ x

0

1

tαf (t)dt

)p

dx ≤

[ p

p – α – 1

]p∫ ∞

0

f p(x)dx, f ≥ 0,p ≥ 1,α < p – 1,

where the constant C = [ p

p–α–1]p is sharp. This inequality holds in the reversed

direction when 0 ≤ p < 1. In this paper we prove and discuss some discrete

analogues of Hardy-type inequalities in fractional h-discrete calculus. Moreover, we

prove that the corresponding constants are sharp.

MSC: Primary 39A12; secondary 49J05; 49K05

Keywords: Inequality; Integral operator; h-calculus; h-integral; Discrete Fractional

Calculus

1 IntroductionThe theory of fractional h-discrete calculus is a rapidly developing area of great interest

both from a theoretical and applied point of view. Especially we refer to [1–8] and the

references therein. Concerning applications in various fields of mathematics we refer to

[9–16] and the references therein. Finally, we mention that h-discrete fractional calculus

is also important in applied fields such as economics, engineering and physics (see, e.g.

[17–22]).

Integral inequalities have always been of great importance for the development of many

branches of mathematics and its applications. One typical such example is Hardy-type

inequalities, which from the first discoveries of Hardy in the twentieth century now have

been developed and applied in an almost unbelievable way, see, e.g., monographs [23] and

[24] and the references therein. Let us just mention that in 1928 Hardy [25] proved the

following inequality:

∫ ∞

0

(

xα–1

∫ x

0

1

tαf (t)dt

)p

dx≤

(

p

p – α – 1

)p ∫ ∞

0

f p(x)dx, f ≥ 0, (1.1)

for 1 ≤ p < ∞ and α < p – 1 and where the constant [ pp–α–1

]p is best possible. Inequality

(1.1) is just a reformulation of the first power weighted generalization of Hardy’s original

inequality, which is just (1.1)with α = 0 (so that p > 1) (see [26] and [27]). Up to now there is

© The Author(s) 2018. This article is distributed under the terms of the Creative Commons Attribution 4.0 International License(http://creativecommons.org/licenses/by/4.0/), which permits unrestricted use, distribution, and reproduction in anymedium, pro-vided you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons license, andindicate if changes were made.

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Persson et al. Journal of Inequalities and Applications ( 2018) 2018:73 Page 2 of 14

no sharp discrete analogue of inequality (1.1). For example, the following two inequalities

were claimed to hold by Bennett([28, p. 40–41]; see also [29, p. 407]):

∞∑

n=1

[

1

n1–α

n∑

k=0

[

kα–1 – (k – 1)α–1]

ak

]p

[

1 – α

p – αp – 1

]p ∞∑

n=1

apn, an ≥ 0,

and

∞∑

n=1

[

1∑n

k=11

k–α

n∑

k=1

k–αak

]p

[

1 – α

p – αp – 1

]p ∞∑

n=1

apn, an ≥ 0,

whenever α > 0, p > 1, αp > 1. Both inequalities were proved independently by Gao [30,

Corollary 3.1–3.2] (see also [31, Theorem 1.1] and [32, Theorem 6.1]) for p≥ 1 and some

special cases of α (this means that there are still some regions of parameters with no proof

of (1.1)). Moreover, in [33, Theorems 2.1 and 2.3] proved another sharp discrete analogue

of inequality (1.1) in the following form:

∞∑

n=–∞

[

1

qnλ

n∑

k=0

qkλak

]p

≤1

(1 – qλ)p

∞∑

n=–∞

apn, an ≥ 0,

and

∞∑

n=1

[

1

qnλ

n∑

k=0

qkλak

]p

≤1

(1 – qλ)p

∞∑

n=1

apn, an ≥ 0,

for 0 < q < 1, p ≥ 1 and α < 1 – 1/p, where λ := 1 – 1/p – α.

The main aim of this paper is to establish the h-analogue of the classical Hardy-type

inequality (1.1) in fractional h-discrete calculus with sharp constants which is another

discrete analogue of inequality (1.1).

The paper is organized as follows: In order not to disturb our discussions later on some

preliminaries are presented in Sect. 2. The main results (see Theorem 3.1 and Theo-

rem 3.2) with the detailed proofs can be found in Sect. 3.

2 PreliminariesWe state the some preliminary results of the h-discrete fractional calculus which will be

used throughout this paper.

Let h > 0 and Ta := a,a + h,a + 2h, . . ., ∀a ∈ R.

Definition 2.1 (see [34]) Let f : Ta → R. Then the h-derivative of the function f = f (t) is

defined by

Dhf (t) :=f (δh(t)) – f (t)

h, t ∈ Ta, (2.1)

where δh(t) := t + h.

Let fg : Ta →R. Then the product rule for h-differentiation reads (see [34])

Dh

(

f (x)g(x))

:= f (x)Dhg(x) + g(x + h)Dhf (x). (2.2)

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The chain rule formula that we will use in this paper is

Dh

[

xγ (t)]

:= γ

∫ 1

0

[

zx(

δh(t))

+ (1 – z)x(t)]γ–1

dzDhx(t), γ ∈R, (2.3)

which is a simple consequence of Keller’s chain rule [35, Theorem 1.90]. The integration

by parts formula is given by (see [34]) the following.

Definition 2.2 Let f : Ta →R. Then the h-integral (h-difference sum) is given by

∫ b

a

f (x)dhx :=

b/h–1∑

k=a/h

f (kh)h =

b–ah

–1∑

k=0

f (a + kh)h,

for a,b ∈ Ta,b > a.

Definition 2.3 We say that a function g : Ta −→ R, is nonincreasing (respectively, non-

decreasing) on Ta if and only if Dhg(t) ≤ 0 (respectively, Dhg(t) ≥ 0) whenever x ∈ Ta.

Let DhF(x) = f (x). Then F(x) is called a h-antiderivative of f (x) and is denoted by∫

f (x)dhx. If F(x) is a h-antiderivative of f (x), for a,b ∈ Ta,b > a we have (see [36])

∫ b

a

f (x)dhx := F(b) – F(a). (2.4)

Definition 2.4 (see [34]) Let t,α ∈R. Then the h-fractional function t(α)h is defined by

t(α)h := hα

Ŵ( th+ 1)

Ŵ( th+ 1 – α)

,

where Ŵ is Euler gamma function, th/∈ –1,–2,–3, . . . and we use the convention that

division at a pole yields zero. Note that

limh→0

t(α)h = tα .

Hence, by (2.1) we find that

t(α–1)h =

1

αDh

[

t(α)h

]

. (2.5)

Definition 2.5 The function f : (0,∞) → R is said to be log-convex if f (ux + (1 – u)y) ≤

f u(x)f 1–u(y) holds for all x, y ∈ (0,∞) and 0 < u < 1.

Next, we will derive some properties of the h-fractional function, which we need for the

proofs of the main results, but which are also of independent interest.

Proposition 2.6 Let t ∈ T0. Then, for α,β ∈R,

t(α+β)h = t

(α)h (t – αh)

(β)h , (2.6)

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t(pα)h ≤

[

t(α)h

]p≤

(

t + α(p – 1)h)(pα)

h, (2.7)

for 1≤ p <∞, and

[

t(α)h

]p≤ t

(pα)h , (2.8)

for 0 < p < 1.

Proof By using Definition 2.4 we get

t(α+β)h = hα+β

Ŵ( th+ 1)

Ŵ( th+ 1 – α – β)

= hαŴ( t

h+ 1)

Ŵ( th+ 1 – α)

hβŴ( t

h+ 1 – α)

Ŵ( th+ 1 – α – β)

= t(α)h (t – αh)

(β)h ,

i.e. (2.6) holds for α,β ∈ R.

It is well known that the gamma function is log-convex (see, e.g., [37], p. 21). Hence,

[

t(α)h

]p= hpα

[

Ŵ( th+ 1)

Ŵ( th+ 1 – α)

]p

= hpα[

Ŵ( 1p( th+ 1 + α(p – 1)) + (1 – 1

p)( t

h+ 1 – α))

Ŵ( th+ 1 – α)

]p

≤ hpα[

Ŵ1p ( 1

h+ 1 + α(p – 1))Ŵ

1– 1p ( t

h+ 1 – α)

Ŵ( th+ 1 – α)

]p

= hpαŴ( t

h+ 1 + α(p – 1))

Ŵ( th+ 1 – α)

=(

t + α(p – 1)h)(pα)

h

and

[

t(α)h

]p= hpα

[

Ŵ( th+ 1)

Ŵ( th+ 1 – α)

]p

= hpα[

Ŵ( th+ 1)

Ŵ((1 – 1p)( t

h+ 1) + 1

p( th+ 1 – pα))

]p

≥ hpα[

Ŵ( th+ 1)

Ŵ1– 1

p ( th+ 1)Ŵ

1p ( t

h+ 1 – pα)

]p

= hpαŴ( t

h+ 1)

Ŵ( th+ 1 – pα)

= t(pα)h ,

so we have proved that (2.7) holds wherever 1 ≤ p < ∞. Moreover, for 0 < p < 1,

t(pα)h = hpα

Ŵ( th+ 1)

Ŵ( th+ 1 – pα)

= hpαŴ( t

h+ 1)

Ŵ((1 – p)( th+ 1) + p( t

h+ 1 – α))

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≥ hpαŴ( t

h+ 1)

Ŵ(1–p)( th+ 1)Ŵp( t

h+ 1 – α)

=

[

hαŴ( t

h+ 1)

Ŵ( th+ 1 – α)

]p

=[

t(α)h

]p,

so we conclude that (2.8) holds for 0 < p < 1. The proof is complete.

3 Main resultsOur h-integral analogue of inequality (1.1) reads as follows.

Theorem 3.1 Let α < p–1p

and 1≤ p < ∞. Then the inequality

∫ ∞

0

(

x(α–1)h

∫ δh(x)

0

f (t)dht

t(α)h

)p

dhx ≤

(

p

p – αp – 1

)p ∫ ∞

0

f p(x)dhx, f ≥ 0, (3.1)

holds.Moreover, the constant [ pp–αp–1

]p is the best possible in (3.1).

Our secondmain result is the following h-integral analogue of the reversed form of (1.1)

for 0 < p < 1.

Theorem 3.2 Let α < p–1p

and 0 < p < 1. Then the inequality

∫ ∞

0

f p(x)dhx ≤

(

p – pα – 1

p

)p ∫ ∞

0

(

x(α–1)h

∫ δh(x)

0

f (t)dht

t(α)h

)p

dhx, f ≥ 0, (3.2)

holds.Moreover, the constant [ p–pα–1p

]p is the best possible in (3.2).

To prove Theorem 3.1 we need the following lemma, which is of independent interest.

Lemma 3.3 Let α < p–1p, p > 1 and 1

p+ 1

p′ = 1. Then the function

φ(x) :=

[(

x –

(

α +1

p

)

h

)(– 1p )

h

]1p′

[(

x –

(

α –1

p′

)

h

)( 1p′)

h

]1p

, x ∈ T0,

is nonincreasing on T0.

Proof Let α < p–1p

and 1 ≤ p < ∞. Since Ŵ(x) > 0 for x > 0, and using Definition 2.4, we

have

(

x –

(

α +1

p

)

h

)(– 1p )

h

= h– 1p

Ŵ( xh+ 1

p′ – α)

Ŵ( xh+ 1

p+ 1

p′ – α)> 0

and

(

x –

(

α –1

p′

)

h

)( 1p′)

h

= h1p′

Ŵ( xh+ 1 + 1

p′ – α)

Ŵ( xh+ 1 – α)

> 0.

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Denote ξ (x) := (x – (α + 1p)h)

(– 1p )

h and η(x) := (x – (α – 1p′ )h)

( 1p′)

h . Then by using (2.5) we

find that

Dhη(x) =(x – (α – 1

p′ )h)(– 1

p )

h

p′≥ 0 (3.3)

and

Dhξ (x) = –(x – (α + 1

p)h)

(– 1p–1)

h

p≤ 0, (3.4)

From (2.3), (2.6), (3.3) and (3.4) it follows that

Dh

[

ξ (x)]

1p′ =

1

p′

∫ 1

0

[

zξ (x + h) + (1 – z)ξ (x)]– 1

p dzDhξ (x)

≤ –[

ξ (x)]– 1

p(x – (α + 1

p)h)

(– 1p–1)

h

pp′

≤ –[

ξ (x)]

1p′(x – αh)

(–1)h

pp′(3.5)

and

Dh

[

η(x)]1p =

1

p

∫ 1

0

[

zη(x + h) + zη(x)]– 1

p′ dzDhη(x)

≤[

η(x)]– 1

p′(x – (α – 1

p′ )h)(– 1

p )

h

pp′. (3.6)

By using the fact that (x + h – αh)(1)h (x – αh)

(–1)h = 1, η(x + h) ≥ η(x),

η(x)

[(

x –

(

α –1

p′

)

h

)(– 1p )

h

]–1

= (x + h – αh)(1)h ,

for x ∈ T0 and (2.2), (3.3), (3.4), (3.5) and (3.6) we obtain

Dh

(

φ(x))

=[

ξ (x)]

1p′ Dh

[

η(x)]1p +

[

η(x + h)]1– 1

p′ Dh

[

ξ (x)]

1p′

≤[ξ (x)]

1p′ [η(x)]

– 1p′

pp′

[(

x –

(

α –1

p′

)

h

)(– 1p )

h

– η(x)(x – αh)(–1)h

]

=[ξ (x)]

1p′ [η(x)]

– 1p′

pp′

(

x –

(

α –1

p′

)

h

)(– 1p )

h

[

1 – (x + h – αh)(1)h (x – αh)

(–1)h

]

≤ 0.

Hence, we have proved that the function φ(x) is nonincreasing onT0 (see Definition 2.4)

so the proof is complete.

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Proof of Theorem 3.1 Let p > 1. By using Lemma 3.3 and (2.6) in Proposition 2.6 we have

x(α–1)h =

[

x(α–1)h

]1p′

[

x(α–1)h

]1p

=

[

x(α– 1

p′)

h

(

x –

(

α –1

p′

)

h

)(– 1p )

h

]1p′

[

x(α– 1

p′–1)

h

(

x –

(

α –1

p′– 1

)

h

)( 1p′)

h

]1p

=[

x(α– 1

p′)

h

]1p′

[

x(α– 1

p′–1)

h

]1p

×

[(

x + h –

(

α +1

p

)

h

)(– 1p )

h

]1p′

[(

x + h –

(

α –1

p′

)

h

)( 1p′)

h

]1p

=[

x(α– 1

p′)

h

]1p′

[

x(α– 1

p′–1)

h

]1p φ(x + h)

≤[

x(α– 1

p′)

h

]1p′

[

x(α– 1

p′–1)

h

]1p φ(t), (3.7)

for t,x ∈ T0 : t ≤ x. Moreover,

φ(t)

t(α)h

=[

(t – αh)(–α)h

]1p′

[

(t – αh)(–α)h

]1p φ(t)

=

[

(t – αh)(–α)h

(

t –

(

α +1

p

)

h

)(– 1p )

h

]1p′

[

(t – αh)(–α)h

(

t –

(

α –1

p′

)

h

)( 1p′)

h

]1p

=

[(

t –

(

α +1

p

)

h

)(–α– 1p )

h

]1p′

[(

t –

(

α –1

p′

)

h

)( 1p′–α)

h

]1p

. (3.8)

According to (3.7) and (3.8) we have

L(f ) :=

∫ ∞

0

(

x(α–1)h

∫ δh(x)

0

1

t(α)h

f (t)dht

)p

dhx

∫ ∞

0

(

[

x(α– 1

p′)

h

]1p′

[

x(α– 1

p′–1)

h

]1p

∫ δh(x)

0

[(

t –

(

α +1

p

)

h

)(–α– 1p )

h

]1p′

×

[(

t –

(

α –1

p′

)

h

)( 1p′–α)

h

]1p

f (t)dht

)p

dhx

=

∞∑

i=0

h1+p

(

[

(ih)(α– 1

p′)

h

]1p′

[

(ih)(α– 1

p′–1)

h

]1p ×

×

i∑

k=0

[(

kh –

(

α +1

p

)

h

)(–α– 1p )

h

]1p′

[(

kh –

(

α –1

p′

)

h

)( 1p′–α)

h

]1p

f (kh)

)p

= Ip(f ).

Let N0 =N ∪ 0, g = gk∞k=1 ∈ lp′ (N0), g ≥ 0, and ‖g‖lp′ = 1. Moreover, let θ (z) be Heav-

iside’s unit step function (θ (z) = 1 for z ≥ 0 and θ (z) = 0 for z < 0). Then, based on the

duality principle in lp(N0) and the Hölder inequality, we find that

I(f ) = sup‖g‖lp′

=1

i,k

h1+ 1

p giθ (i – k)[

(ih)(α– 1

p′)

h

]1p′

[

(ih)(α– 1

p′–1)

h

]1p

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×

[(

kh –

(

α +1

p

)

h

)(–α– 1p )

h

]1p′

[(

kh –

(

α –1

p′

)

h

)( 1p′–α)

h

]1p

f (kh)

≤ sup‖g‖lp′

=1

(

i,k

hgp′

i θ (i – k)(ih)(α– 1

p′)

h

(

kh –

(

α +1

p

)

h

)(–α– 1p )

h

)1p′

×

(

i,k

h2θ (i – k)(ih)(α– 1

p′–1)

h

(

kh –

(

α –1

p′

)

h

)( 1p′–α)

h

f p(kh)

)1p

= sup‖g‖lp′

=1

Ip′

1 (g)Iq2 (f ). (3.9)

By using Definition 2.3 and combining (2.4), (2.5) and (2.6) we can conclude that

I1(g) =

∞∑

i=0

gp′

i (ih)(α– 1

p′)

h

i∑

k=0

h

(

kh –

(

α +1

p

)

h

)(–α– 1p )

h

=

∞∑

i=0

gp′

i (ih)(α– 1

p′)

h

∫ δh(ih)

0

(

x –

(

α +1

p

)

h

)(– 1p–α)

h

dhx

=1

1p′ – α

∞∑

i=1

gp′

i (ih)(α– 1

p′)

h

∫ δh(ih)

0

Dh

[(

x –

(

α +1

p

)

h

)( 1p′–α)

h

]

dhx

≤1

1p′ – α

∞∑

i=1

gp′

i (ih)(α– 1

p′)

h

(

ih –

(

α –1

p′

)

h

)( 1p′–α)

h

=1

1p′ – α

‖g‖p′

lp′=

11p′ – α

. (3.10)

Furthermore,

I2(f ) =

∞∑

i=0

h(ih)(α– 1

p′–1)

h

i∑

k=0

hf p(kh)

(

kh –

(

α –1

p′

)

h

)( 1p′–α)

h

=

∞∑

k=0

hf p(kh)

(

kh –

(

α –1

p′

)

h

)( 1p′–α)

h

∞∑

i=k

h(ih)(α– 1

p′–1)

h

=1

α – 1p′

∫ ∞

0

f p(x)

(

x –

(

α –1

p′

)

h

)( 1p′–α)

h

∫ ∞

x

Dh

[

t(α– 1

p′)

h

]

dht dhx

=1

1p′ – α

∫ ∞

0

f p(x)

(

x –

(

α –1

p′

)

h

)( 1p′–α)

h

x(α– 1

p′)

h dhx

=1

1p′ – α

∫ ∞

0

f p(x)dhx. (3.11)

By combining (3.9), (3.10) and (3.11) we obtain

L(f )≤

(

p

p – pα – 1

)p ∫ ∞

0

f p(x)dhx, (3.12)

i.e. (3.1) holds.

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Finally, we will prove that the constant [ pp–αp–1

]p is the best possible in inequality (3.1).

Let x,a ∈ T0 be such that a < x, and consider the test function fβ (t) = t(β)h χ[a,∞)(t), t > 0, for

β = – 1p– ε.

Then from (2.4), (2.5) and (2.7) it follows that

∫ ∞

0

fpβ (t)dht =

∫ ∞

a

[

t(β)h

]pdht ≤

∫ ∞

a

(

t + β(p – 1)h)(βp)

hdht

=1

pβ + 1

∫ ∞

a

Dh

[(

t + β(p – 1)h)(βp+1)

h

]

dht

=(a + β(p – 1)h)

(pβ+1)h

|pβ + 1|< ∞.

Since

(∫ δh(x)

0

(t – hα)(–α)h fβ (t)dht

)p

=

(∫ δh(x)

a

(t – hα)(–α+β)h dht

)p

=

(

1

1 – α + β

∫ δh(x)

a

Dh

[

(t – hα)(1–α+β)h

]

dht

)p

=

(

(x + h – hα)(1–α+β)h

1 – α + β

[

1 –(a – hα)

(1–α+β)h

(x + h – hα)(1–α+β)h

])p

(

(x + h – hα)(1–α+β)h

1 – α + β

)p[

1 – p(a – hα)

(1–α+β)h

(x + h – hα)(1–α+β)h

]

,

we have

L(fβ ) ≥

(

1

1 – α + β

)p[∫ ∞

a

[

x(α–1)h (x + h – hα)

(1–α+β)h

]pdhx

– p(a – hα)(1–α+β)h

∫ ∞

a

[x(α–1)h (x + h – hα)

(1–α+β)h ]p

(x + h – hα)(1–α+β)h

dhx

]

=

(

1

1 – α + β

)p[∫ ∞

0

fpβ (x)dhx – p

∫ ∞

a

(a – hα)(1–α+β)h [x

(β)h ]p

(x + h – hα)(1–α+β)h

dhx

]

. (3.13)

By using (2.4), (2.5), (2.6) and (2.7) we obtain

∫ ∞

a

[x(β)h ]p dhx

(x + h – hα)(1–α+β)h

∫ ∞

a

(x + β(p – 1)h)(pβ)h dhx

(x + h – hα)(1–α+β)h

=

∫ ∞

a

(

x + β(p – 1)h)(β(p–1)+α–1)

hdhx

=

∫ ∞

aDh((x + β(p – 1)h)

(β(p–1)+α)h )dhx

β(p – 1) + α

=1

|β(p – 1) + α|

(

a + β(p – 1)h)(β(p–1)+α)

h(3.14)

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and

(a – hα)(1–α+β)h = (a – hα)

(–α)h

(

a – h(pβ + 1))(β(1–p)

ha(pβ+1)h

= (a – hα)(–α)h

(

a – h(pβ + 1))(β(1–p)

h

∫ ∞

a

Dh

[

t(pβ+1)h

]

dht

≤ (a – hα)(–α)h

(

a – h(pβ + 1))(β(1–p)

h|βp + 1|

∫ ∞

a

[

t(β)h

]pdht. (3.15)

According to (2.6), (3.13), (3.14) and (3.15) we can deduce that

L(fβ ) ≥

(

1

1 – α + β

)p[∫ ∞

0

fpβ (x)dhx – θβ (a)

∫ ∞

0

fpβ (x)dhx

]

,

where θβ (a) :=p|βp+1|

|β(p–1)+α|(a + β(p – 1)h)

(β(p–1))h (a – h(pβ + 1))

(β(1–p))h → 0, ε → 0.

Therefore, limε→0L(fβ )

∫ ∞0 f

pβ(x)dhx

≥ limε→0(1

1–α+β)p = ( p

p–pα–1)p, which implies that the con-

stant [ pp–αp–1

]p in (3.1) in sharp.

Let p = 1. By using Definition 2.3 and (2.5) we get

∫ ∞

0

x(α–1)h

∫ δh(x)

0

1

t(α)h

f (t)dht dhx =

∞∑

i=0

h(ih)(α–1)h

i∑

k=0

h(kh – αh)(–α)h f (kh)

=

∞∑

k=0

h(kh – αh)(–α)h f (kh)

∞∑

i=k

h(ih)(α–1)h

=

∫ ∞

0

(t – αh)(–α)h f (t)

∫ ∞

t

x(α–1)h dhxdht

=1

α

∫ ∞

0

(t – αh)(–α)h f (t)

∫ ∞

t

Dh

(

x(α)h

)

dhxdht

= –1

α

∫ ∞

0

f (t)(t – αh)(–α)h t

(α)h dht = –

1

α

∫ ∞

0

f (t)dht,

which means that (3.1) holds even with equality in this case. The proof is complete.

Proof of Theorem 3.2 Let 0 < p < 1. By using (2.4), (2.5) and (2.7) we get

[

x(α–1)h

]p=

[

x(α–1)h

]p–1x(α–1)h

=

[

x(α– 1

p′)

h

(

x –

(

α –1

p′

)

h

)(– 1p )

h

]p–1

x(α– 1

p′–1)

h

(

x + h –

(

α –1

p′

)

h

)( 1p′)

h

≥[

x(α– 1

p′)

h

]p–1x(α– 1

p′–1)

h

(x – (α – 1p′ )h)

( 1p′)

h

[(x – (α – 1p′ )h)

(– 1p )

h ]1–p

≥[

x(α– 1

p′)

h

]p–1x(α– 1

p′–1)

h

(x – (α – 1p′ )h)

( 1p′)

h

(x – (α – 1p′ )h)

(–1–pp )

h

=[

x(α– 1

p′)

h

]p–1x(α– 1

p′–1)

h

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=

[(

x –

(

α –1

p′

)

h

)( 1p′–α)

h

]1–p

x(α– 1

p′–1)

h

[

11p′ – α

]p–1[∫ δh(x)

0

(

t –

(

α +1

p

)

h

)(–α– 1p )

h

dht

]1–p

x(α– 1

p′–1)

h (3.16)

and

[

1

t(α)h

]p

=[

(t – αh)(–α)h

]p–1 1

t(α)h

=

[(

t –

(

α +1

p

)

h

)(–α– 1p )

h

(t – αh)( 1p )

h

]p–1 1

t(α– 1

p′)

h (t – (α – 1p′ )h)

( 1p′)

h

=

[(

t –

(

α +1

p

)

h

)(–α– 1p )

h

]p–1 1

t(α– 1

p′)

h

(t – αh)(– 1

p′)

h

[(t – αh)( 1p )

h ]1–p

[(

t –

(

α +1

p

)

h

)(–α– 1p )

h

]p–1 1

t(α– 1

p′)

h

(t – αh)(– 1

p′)

h

(t – αh)(1–pp )

h

=

[(

t –

(

α +1

p

)

h

)(–α– 1p )

h

]p–1 1

t(α– 1

p′)

h

. (3.17)

Moreover, by using Definition 2.3, (3.16) and (3.17), and applying the Hölder inequality

with powers 1/p and 1/(1 – p), we obtain

L(f )

[ 11p′–α]p–1

∫ ∞

0

x(α– 1

p′–1)

h

[∫ δh(x)

0

(

t –

(

α +1

p

)

h

)(–α– 1p )

h

dht

]1–p

×

[∫ δh(x)

0

1

t(α)h

f (t)dht

]p

dhx

=

∞∑

k=0

h(kh)(α– 1

p′–1)

h

[

k∑

i=0

h

(

ih –

(

α +1

p

)

h

)(–α– 1p )

h

]1–p[ k∑

i=0

hf (ih)

(ih)(α)h

]p

∞∑

k=0

h(kh)(α– 1

p′–1)

h

k∑

i=0

h

[(

ih –

(

α +1

p

)

h

)(–α– 1p )

h

]1–p[f (ih)

(ih)(α)h

]p

=

∞∑

i=0

hf p(ih)

[(

ih –

(

α +1

p

)

h

)(–α– 1p )

h

]1–p[ 1

(ih)(α)h

]p ∞∑

k=i

h(kh)(α– 1

p′–1)

h

=

∫ ∞

0

f p(t)

[(

t –

(

α +1

p

)

h

)(–α– 1p )

h

]1–p[ 1

t(α)h

]p ∫ ∞

t

x(α– 1

p′–1)

h dhxdht

≥1

1p′ – α

∫ ∞

0

f p(t)

[(

t –

(

α +1

p

)

h

)(–α– 1p )

h

]1–p[(

t –

(

α +1

p

)

h

)(–α– 1p )

h

]p–1

×1

t(α– 1

p′)

h

∫ ∞

t

Dh

[

x(α– 1

p′)

h

]

dhxdht

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Persson et al. Journal of Inequalities and Applications ( 2018) 2018:73 Page 12 of 14

=1

1p′ – α

∫ ∞

0

f p(t)ddt,

i.e.

[

1

p′– α

]p

L(f ) ≥

∫ ∞

0

f p(t)ddt.

Therefore, we deduce that inequality (3.2) holds for all functions f ≥ 0 and the left hand

side of (3.2) is finite.

Finally, we prove that the constant [ p–1p

– α]p in inequality (3.2) is sharp. Let x,a ∈ T0,

be such that a < x, and fβ (t) = t(β)h χ[a,∞)(t), where α – 1 < β < – 1

p. By using (2.4), (2.5) and

(2.8) we find that

∫ ∞

0

fβ (t)dht =

∫ ∞

a

(

t(β)h

)pdht ≤

∫ ∞

a

t(βp)h dht

=1

pβ + 1

∫ ∞

a

Dh

[

t(βp+1)h

]

dht

=a(pβ+1)h

|pβ + 1|< ∞

and

L(fβ ) =

∞∑

i=0

h

[

(ih)(α–1)h

i∑

k=0

h(kh – αh)(–α)h fβ (kh)

]p

=

ah–1

i=0

h

[

(ih)(α–1)h

i∑

k=0

h(kh – αh)(–α)h fβ (kh)

]p

+

∞∑

i= ah

h

[

(ih)(α–1)h

i∑

k=0

h(kh – αh)(–α)h fβ (kh)

]p

=

∫ ∞

a

[

x(α–1)h

∫ δh(x)

0

(t – αh)(–α+β)h dht

]p

dhx

=

[

1

1 – α + β

]p ∫ ∞

a

[

x(α–1)h

∫ δh(x)

0

Dh

[

(t – αh)(1–α+β)h

]

dht

]p

dhx

[

1

1 – α + β

]p ∫ ∞

a

[

x(α–1)h (x + h – αh)

(1–α+β)h

]pdhx

=

[

1

1 – α + β

]p ∫ ∞

a

[

x(β)h

]pdhx =

(

1

1 – α + β

)p ∫ ∞

0

fpβ (x)dhx. (3.18)

From (3.18) its follows that

supα–1≥β≥– 1

p

∫ ∞

0fβ (t)dht

L(fβ )= sup

α–1<β<– 1p

[1 – α + β]p =

[

1

p′– α

]p

,

and this shows that the constant [ p–1p

– α]p in inequality (3.2) is sharp. The proof is com-

plete.

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Persson et al. Journal of Inequalities and Applications ( 2018) 2018:73 Page 13 of 14

Now, let us comment which discrete analogue of Hardy inequality we are getting from

the Hardy h-inequality. Directly from the proof of Theorems 3.1 and 3.2 we obtain the

following discrete inequality, which is of independent interest.

Remark 3.4 On the basis of Definitions 2.4–2.5 we get

∞∑

n=0

[

Ŵ( nhh+ 1)

Ŵ( nhh+ 2 – α)

n∑

k=0

Ŵ( khh+ 1 – α)

Ŵ( nhh+ 1)

ak

]p

(

p

p – αp – 1

)p ∞∑

n=0

apk , ak ≥ 0,

for p≥ 1 and α < 1 – 1/p.

Acknowledgements

This work was supported by Scientific Committee of Ministry of Education and Science of the Republic of Kazakhstan,

grant no. AP05130975.

Competing interests

The authors declare that they have no competing interests.

Authors’ contributions

All authors have on equal level discussed, posed the research questions and proved the results in this paper. Moreover, all

authors have read and approved the final version of this manuscript.

Author details1Luleå University of Technology, Luleå, Sweden. 2UiT The Artic University of Norway, Narvik, Norway. 3L. N. Gumilyev

Eurasian National University, Astana, Kazakhstan.

Publisher’s NoteSpringer Nature remains neutral with regard to jurisdictional claims in published maps and institutional affiliations.

Received: 13 December 2017 Accepted: 17 March 2018

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Paper E

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Fractional order Hardy-type inequality infractional h-discrete calculus

S. Shaimardan

Eurasian National UniversitySatpayev str., 2

010008 Astana, Kazakhstanand

Lulea University of Technology,97187 Lulea, Sweden

1

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S. Shaimardan. Fractional order Hardy-type inequality in fractional h-discrete calculus, Lulea University of Technology, Department of Mathemat-ical sciences, Research Report (2018).

Abstract: We investigate the power weights fractional order Hardy-typeinequality in the following form:

∞∫

0

∞∫

0

|f(x)− f(y)|p

|x− y|1+pα dxdy

p

≤ C

∞∫

0

|f ′(x)|p x(1−α)pdx

p

for 0 < α < 1 and 1 < p < ∞ in fractional h-discrete calculus, where

C = 21p α−1

(p−pα)1p

. For h-fractional function we prove a discrete analogue of above

inequality in fractional h-discrete calculus, is proved and discussed. More-over, we prove that the same constant is sharp also in this case.

AMS subject classification (MSC 2010): Primary. 39A12. Secondary49J05, 49K05.

Keywords and Phrases: Fractional Hardy type inequality, h-derivative,integral operator, h-calculus, h-integral, discrete fractional calculus, sharpconstant.

Note: This report will be submitted for publication elsewhere.

ISSN: 1400-4003

Lulea University of TechnologyDepartment of Mathematical sciencesSE-971 87 Lulea, SWEDEN

2

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1 Introduction

Fractional h-discrete calculus has generated interest in recent years. It is amathematical subject that has proved to be very useful in applied fields suchas such as economics, engineering and physics (see, e.g. [4], [5], [23], [24],[30]). Concerning applications in various fields of mathematics we refer to[2], [3], [6], [7], [12], [13], [14], [16], [17], [18], [22], [25], [26], [27], [29], [31]and the references therein.

It is well known that integral inequalities play important roles in theresearch of qualitative as well as quantitative properties of solutions of dif-ferential equations, difference equations and dynamic equations. One of theexamples is fractional Hardy-type inequalities. In [15], [32], [33], [34] and[35] a series of fractional order Hardy- type inequalities have been presented.We pronounce especially that even Chapter 5 in the new book [20] by A.Kufner, L.-E. Persson and N. Samko is completely devoted to this subject.In particular, it is proved here (see Theorem 5.3) that

∞∫

0

∞∫

0

|f(x)− f(y)|p

|x− y|1+pα dxdy

p

≤ C

∞∫

0

|f ′(x)|p x(1−α)p

p

, (1.1)

for 0 < α < 1, 1 < p < ∞, where C = 21p α−1

(p−pα)1p

is the sharp constant.

Moreover, in [1], [9], [10], [21], [28] some discrete Hardy-type inequalitieshave been established, which can be used as a handy tool in the research ofsolutions of difference equations. Up to now the discrete analogues of thefractional Hardy-type inequalities are not studied. The main aim of thispaper is to establish the h-analogue of the fractional Hardy-type inequality(1.1) in fractional h-discrete calculus with sharp constants which is a discreteanalogue of the inequality (1.1).

The paper is organized as follows: In order not to disturb our discussionslater on some preliminaries are presented in Section 2. The main result (seeTheorem 3.1) with the detailed proof can be found in Section 3.

2 Preliminaries

First we state some preliminary results of the h-discrete fractional calculus,which will be used throughout this paper.

3

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Let h > 0 and Ta = a, a+ h, a+ 2h, · · · , ∀a ∈ R.

Definition 1. Let f : Ta → R. Then the h-derivative of the function f =f(x) is defined by

Dhf(t) :=f(δh(t))− f(t)

h, t ∈ Ta, (2.1)

where δh(t) = t+ h.

See e.g. [8]. The chain rule formula that we will use in this paper is

Dh [xγ(t)] := γ

1∫

0

[zx(δh(t)) + (1− z)x(t)]γ−1 dzDhx(t), γ ∈ R, (2.2)

which is a simple consequence of Keller’s chain rule ([[11], Theorem 1.90]).

Definition 2. Let f : Ta → R. Then the h-integral (h-difference sum) isgiven by

b∫

a

f(x)dhx :=

b/h−1∑

k=a/h

f(kh)h =

b−ah−1∑

k=0

f(a+ kh)h, (2.3)

for a, b ∈ Ta, b > a.

Definition 3. We say that a function g : Ta −→ R, is nonincreasing (re-spectively, nondecreasing) on Ta if and only if Dhg(t) ≤ 0 (respectively,Dhg(t) ≥ 0) whenever t ∈ Ta.

Let DhF (x) = f(x). Then F (x) is called a h-antiderivative of f(x) and isdenoted by

∫f(x)dhx. If F (x) is a h-antiderivative of f(x), for a, b ∈ Ta, b >

a, then we have that (see [18]):

b∫

a

f(x)dhx = F (b)− F (a). (2.4)

Definition 4. Let t, α ∈ R. Then the h-fractional function t(α)h is defined by

t(α)h := hα

Γ( th

+ 1)

Γ( th

+ 1− α),

4

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where Γ is Euler gamma function, th/∈ −1,−2,−3, · · · and we use the

convention that division at a pole yields zero. Note that

limh→0

t(α)h = tα.

Hence, by (2.1) we find that

t(α−1)h =

1

αDh

[t(α)h

]. (2.5)

(a− t− h)(α−1)h = − 1

αDh

[(a− t)(α)

h

]. (2.6)

1

(t+ h)(α+1)h

= − 1

αDh

[1

t(α)h

]. (2.7)

1

(a− t)(α+1)h

=1

αDh

[1

(a− t)(α)h

]. (2.8)

Definition 5. The function f : (0,∞)→ R is said to be log-convex if f(ux+(1− u)y) ≤ fu(x)f 1−u(y), holds for all x, y ∈ (0,∞) and 0 < u < 1.

Next, we will derive some properties of the h-fractional function, whichwe need for the proofs of the main results but which are also of independentinterest.

Proposition 2.1. Let t ∈ T0. Then, for α, β ∈ R,

t(α+β)h = t

(α)h (t− αh)

(β)h , (2.9)

t(pα)h ≤

[t(α)h

]p≤ (t+ α(p− 1)h)

(pα)h , (2.10)

for 1 ≤ p <∞.

5

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Proof. By using Definition 4 we get that

t(α+β)h = hα+β Γ( t

h+ 1)

Γ( th

+ 1− α− β)

= hαΓ( t

h+ 1)

Γ( th

+ 1− α)hβ

Γ( th

+ 1− α)

Γ( th

+ 1− α− β)= t

(α)h (t− αh)

(β)h .

Therefore, (2.9) holds for α, β ∈ R.It’s well known that the gamma function is log-convex (see e.g [19], p 21).

Hence,

[t(α)h

]p= hpα

[Γ( t

h+ 1)

Γ( th

+ 1− α)

]p

= hpα

[Γ(1

p( th

+ 1 + α(p− 1)) + (1− 1p)( th

+ 1− α))

Γ( th

+ 1− α)

]p

≤ hpα

1p ( 1

h+ 1 + α(p− 1))Γ1− 1

p ( th

+ 1− α)

Γ( th

+ 1− α)

]p

= hpαΓ( t

h+ 1 + α(p− 1))

Γ( th

+ 1− α)

= (t+ α(p− 1)h)(pα)h ,

and

[t(α)h

]p= hpα

[Γ( t

h+ 1)

Γ( th

+ 1− α)

]p

= hpα

[Γ( t

h+ 1)

Γ((1− 1p)( th

+ 1) + 1p( th

+ 1− pα))

]p

≥ hpα

[Γ( t

h+ 1)

Γ1− 1p ( t

h+ 1)Γ

1p ( t

h+ 1− pα)

]p

= hpαΓ( t

h+ 1)

Γ( th

+ 1− pα)= t

(pα)h ,

so we have proved that (2.10) holds whenever 1 ≤ p <∞. Let 1 ≤ p ≤ q < ∞ and 1

p+ 1

p′ = 1. Let f = fi∞i=0 be an arbitrary

sequence of real numbers. Moreover, suppose that ui∞0=1, and vi∞i=0 are

6

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weight sequences, i.e., non-negative sequences. To prove our main resultwe use the following result for a standard weighted Hardy inequality, when1 ≤ p ≤ q <∞ (see [1], Theorem 4.1 and e.g. also [20]):

Theorem B. Let 1 ≤ p ≤ q <∞. Then the inequality

( ∞∑

i=1

(i∑

j=1

fj

)q

uqi

) 1q

≤ C

( ∞∑

i=1

(fivi)p

) 1p

,

holds for all sequences f = fi∞i=0, fi ≥ 0, i ≥ 1, with the best constant

C > 0 if and only if B = supk≥1

( ∞∑i=k

uqi

) 1q

(k∑j=1

v−p′

j

) 1p′

< ∞. Moreover,

B ≤ C ≤ p′q1qB.

3 The main result

Our main result reads:

Theorem 3.1. Let 1 < p < ∞, 0 < α < 1 and f(x) = DhF (x). Then thefollowing inequality

∞∫

0

∞∫

0

|F (x)− F (y)|pdhxdhy[(|x− y|+ 3h)

( 1p

+α)

h

]p

1p

≤ C

∞∫

0

|f(x)|p dhx[(x+ h)

(α−1)h

]p

1p

, (3.1)

holds with constant C = 21p α−1

(p−pα)1p

. Moreover, this constant sharp.

The next lemma permits to shorten the proof of our main result:

Lemma 3.1. Let 0 < α < 1, 1 < p <∞ and 1p

+ 1p′ = 1. Then

B := supz∈T0

∞∫

z

dhx[(x+ 3h)

( 1p

+α)

h

]p

1p

δ(z)∫

0

t(α−1)h dht

[δ(t)

(α)h

]− p′p

1p′

<1

α. (3.2)

7

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Proof. Let 0 < α < 1, 1 ≤ p < ∞ and 1p

+ 1p′ = 1. Exploiting (2.1) we

find that Dh

[x

(α)h

]= αx

(α−1)h ≥ 0 for x ∈ T0. Moreover, in view of Definition

3 we see that x(α)h ≤ (x′)(α)

h for x, x′ ∈ T0 such that x ≤ x′. Then, accordingto (2.1), (2.2), (2.7), (2.8) and (2.10), we obtain that

Dh

[1

(x+ 2h)(α)h

]p= p

1∫

0

[z

(x+ 3h)(α)h

+1− z

(x+ 2h)(α)h

]p−1

dzDh

[1

(x+ 2h)(α)h

]

= −pα 1

(x+ 3h)(α+1)h

1∫

0

[z

(x+ 3h)(α)h

+1− z

(x+ 2h)(α)h

]p−1

dz

≤ −pα 1

(x+ 3h)(α+1)h

[1

(x+ 3h)(α)h

]p−1

and[(x+ 3h)

( 1p

+α)

h

]p=

[(x+ 3h)

(α)h

]p [(x+ 3h− αh)

( 1p

)

h

]p

≥[(x+ 3h)

(α)h

]p−1

(x+ 3h)(α)h (x+ 3h− αh)

(1)h

=[(x+ 3h)

(α)h

]p−1

(x+ 3h)(α+1)h

i.e. 1

(x+ 3h)( 1p

+α)

h

p

≤ − 1

pαDh

[1

(x+ 2h)(α)h

]p. (3.3)

Next we note that, by Definition 3 and (2.5), t(α)h ≤ (z + h)

(α)h , for t, z ∈ T0

such that t≤z + h and then, by applying (2.4), (2.5) and (3.3), we get that

Bp ≤ − 1

pαsupz∈T0

(z + 2h)(α)h

∞∫

z

Dh

[1

(x+ 2h)(α)h

]pdhx

δ(z)∫

0

t(α−1)h dht

pp′

8

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≤ 1

pαsupz∈T0

(z + 2h)(α)h

∞∫

z

Dh

[1

(x+ 2h)(α)h

]pdhx

1

α

δ(z)∫

0

Dh

[t(α)h

]dht

pp′

≤ 1

αpsupz∈T0

(z + 2h)(α)h

[(z + 2h)

(α)h

] pp′

[(z + 2h)

(α)h

]p =1

αp,

i.e. (3.2) holds so the proof is complete.

Proof of Theorem 3.1. By using (2.3) we get that

L(F ) :=

∞∫

0

∞∫

0

|F (x)− F (y)|pdhxdhy[(|x− y|+ 3h)

( 1p

+α)

h

]p

=∞∑

k=0

∞∑

i=0

h2 |F (ih)− F (kh)|p[(|ih− kh|+ 3h)

( 1p

+α)

h

]p

≤∞∑

k=0

k∑

i=0

h2 |F (ih)− F (kh)|p[(|ih− kh|+ 3h)

( 1p

+α)

h

]p

+∞∑

k=0

∞∑

i=k

h2 |F (ih)− F (kh)|p[(|ih− kh|+ 3h)

( 1p

+α)

h

]p

= 2∞∑

k=0

∞∑

i=k

h2 |F (ih)− F (kh)|p[(|ih− kh|+ 3h)

( 1p

+α)

h

]p . (3.4)

Let

fm = h |f(mh)| , ui =h

1p

(|ih− kh|+ 3h)( 1p

+α)

h

,

vm =h− 1p′

[[(mh− kh)

(α−1)h

]p−1

(δ(mh)− kh)(α)h

] 1p

9

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and f(x) = DhF (x). Then, from (2.3) and (3.4) it follows that

L(F ) ≤ 2∞∑

k=0

h∞∑

i=k

h[(|ih− kh|+ 3h)

( 1p

+α)

h

]p∣∣∣∣∣i−1∑

m=k

hf(mh)

∣∣∣∣∣

p

≤ 2∞∑

k=0

h

[ ∞∑

i=k

upi

(i∑

m=k

fm

)p]. (3.5)

Moreover, based on Theorem B we obtain that

∞∑

i=k

upi

(i∑

m=k

fm

)p

≤ Bpk

∞∑

m=k

fpmv−p′m , (3.6)

where

Bpk := sup

n≥k

( ∞∑

i=n

uqi

) 1q(

n∑

j=k

v−p′

j

) 1p′

= supn≥k

∞∫

nh

dhx[(x− kh+ 3h)

( 1p

+α)

h

]p

1p

δ(z)∫

nh

(t− kh)(α−1)h dht

[(δ(t)− kh)

(α)h

]− p′p

1p′

≤ Bp.

By combining (3.5) and (3.6) we have that

L(F ) ≤ 2∞∑

k=0

hBp

∞∑

m=k

h |f(mh)|p[(mh− kh)

(α−1)h

]p−1

(δ(mh)− kh)(α)h

. (3.7)

Moreover, by using Definition 3 and (2.6) we obtain that

(δ(mh)− t)(α−1)h ≤ (mh− t)(α−1)

h ,

for t ∈ T0.Hence, in view of (2.1), (2.2) and (2.8) we get that

10

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Dh,t

[1

(δ(mh)− t)(α−1)h

]p= p

1∫

0

[z

(mh− t)(α−1)h

+(1− z)

(δ(mh)− t)(α−1)h

]p−1

dz

× Dh,t

[1

(δ(mh)− t)(α−1)h

]

=p(α− 1)

(δ(mh)− t)(α)h

1∫

0

[z

(mh− t)(α−1)h

+(1− z)

(δ(mh)− t)(α−1)h

]p−1

dz

≤ p(α− 1)

(δ(mh)− t)(α)h

[1

(mh− t)(α−1)h

]p−1

.

Consequently,

1

(δ(mh)− t)(α)h

[1

(mh− t)(α−1)h

]p−1

≤ 1

p(α− 1)Dh,t

[1

(δ(mh)− t)(α−1)h

]p. (3.8)

Thus, by now using Lemma 3.1 and (3.7) and (3.8), we obtain that

L(F ) ≤ 2Bp

∞∑

k=0

h∞∑

m=k

h |f(mh)|p[(mh− kh)

(α−1)h

]p−1

(δ(mh)− kh)(α)h

≤ 2Bp

∞∑

m=0

h |f(mh)|pm∑

k=0

h[(mh− kh)

(α−1)h

]p−1

(δ(mh)− kh)(α)h

≤ 2α−p

p(α− 1)

∞∑

m=0

h |f(mh)|pδ(mh)∫

0

Dh,t

[1

(δ(mh)− t)(α−1)h

]pdht

11

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≤ 2α−p

p(1− α)

∞∑

m=0

h|f(mh)|p[

(mh+ h)(α−1)h

]p

≤ 2α−p

p(1− α)

∞∫

0

|f(x)|p dhx[(x+ h)

(α−1)h

]p ,

which means that inequality (3.1) holds.

Finally, we will show that the constant 21p α−1

(pα−p)1p

in (3.1) sharp. Let x, y, a ∈T0 such that y ≤ a ≤ x− 4h. By Definition 3 we obtain that

(x− y + 2h− αh+

1

p′h

)(1)

h

≤ (x− y + 3h− αh)(1)h ,

(x− y + 2h− (α− 1)h)(1)h ≤ (x+ 4h− αh)

(1)h .

Then, by using (2.2), (2.8), (2.9) and (2.10) we find that

[(|x− y|+ 3h)

( 1p

+α)

h

]p=

[(x− y + 3h)

(α−1)h

]p [(x− y + 2h− αh)

( 1p

)

h

]p

×[(x− y + 3h− (α− 1)h)

(1)h

]p

≤[(x− y + 3h)

(α−1)h

]p−1

× (x− y + 2h)(α−1)h (x− y + 2h− αh+ 1/p′h)

(1)h

×[(x− y + 3h− (α− 1)h)

(1)h

]p

≤[(x− y + 3h)

(α−1)h

]p−1

(x− y + 3h)(α)h

×[(2x− a+ 4h− αh)

(1)h

]p

and

Dh,y

[1

(x− y + 3h)(α−1)h

]p= p

1∫

0

[z

(x− y + 2h)(α−1)h

+1− z

(x− y + 3h)(α−1)h

]p−1

dz

× Dh,y

[1

(x− y + 3h)(α−1)h

]

12

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=

1∫

0

[z

(x− y + 2h)(α−1)h

+1− z

(x− y + 3h)(α−1)h

]p−1

dz

× p(α− 1)

(x− y + 3h)(α)h

≥ −[

1

(x− y + 3h)(α−1)h

]p−1p(1− α)

(x− y + 3h)(α)h

×

[(2x− a+ 4h− αh)

(1)h

]p[(x− a)

(1)h

]p .

Therefore,

1

(x− y + 3h)( 1p

+α)

h

p

≥ −

[(x− a)

(1)h

]−p

p(1− α)Dh

[1

(x− y + 3h)(α−1)h

]p. (3.9)

Assume now on the contrary that there exists a constant C < 21p α−1

(pα−p)1p

such that (3.1) holds for all measurable functions where the right hand sideis finite. We now consider the test function

f0 := χ[a,a′](t) (t− a− h+ αh)(α−1)h ,

for a′ ∈ T0 such that x ≤ a′. Then, by using (2.4), (2.5) and (2.9) we candeduce that

|F (x)− F (y)|p =

∣∣∣∣∣∣

x∫

a

(t− a− h+ αh)(α−1)h dht

∣∣∣∣∣∣

p

=1

αp

∣∣∣∣∣∣

x∫

a

Dh

[(t− a− h+ αh)

(α)h

]dht

∣∣∣∣∣∣

p

=1

αp

[(x− a− h+ αh)

(α)h

]p

=1

αp

[(x− a− h+ αh)

(α−1)h

]p [(t− a)

(1)h

]p, (3.10)

13

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where (−h+ αh)(α)h = hα Γ(α)

Γ(0)= 0 and

∞∫

0

fp0 (x)dhx[(x+ h)

(α−1)h

]p ≤a′∫

a

[(x− a− h+ αh)

(α−1)h

]pdhx

[(x+ h)

(1−α)h

]p <∞.

By combining (2.3) and (3.4) we obtain that

L(F ) :=

∞∫

0

x∫

0

|F (x)− F (y)|pdhxdhy[(|x− y|+ 3h)

( 1p

+α)

h

]p

+

∞∫

0

∞∫

x

|F (x)− F (y)|pdhxdhy[(|x− y|+ 3h)

( 1p

+α)

h

]p

:= I1 + I2. (3.11)

From (3.9) and (3.10) it follows that

I1 ≥a′∫

a

x∫

0

|F (x)− F (y)|pdhxdhy[(|x− y|+ 3h)

( 1p

+α)

h

]p

≥ − α−p

p(1− α)

a′∫

a

[(x− a− h+ αh)

(α−1)h

]p

×a+4h∫

0

Dh,y

[1

(x− y + 3h)(α−1)h

]pdhy

dhx

≥ α−p

p(1− α)

∞∫

0

fp0 (x)dhx[(x+ h)

(1−α)h

]p , (3.12)

where 1

(−h)(α−1)h

= Γ(α−1)Γ(0)

= 0.

In the same way we can deduce that

I2 ≥α−p

p(1− α)

∞∫

0

fp0 (x)dhx[(x+ h)

(1−α)h

]p . (3.13)

14

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By now using (3.11), (3.12) and (3.13) we obtain that

C1p ≥ L(F )

∞∫0

fp0 (x)dhx[(x+h)

(α−1)h

]p

=2α−p

p(1− α),

which contradicts our assumption so we conclude that the constant 21p α−1

(p−pα)1p

in (3.1) sharp. The proof is complete.

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Page 141: DOCTORAL THESIS - DiVA portalltu.diva-portal.org/smash/get/diva2:1258185/FULLTEXT02.pdf · In particular, in this introduction we give a brief history of quantum calculus and a short
Page 142: DOCTORAL THESIS - DiVA portalltu.diva-portal.org/smash/get/diva2:1258185/FULLTEXT02.pdf · In particular, in this introduction we give a brief history of quantum calculus and a short