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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
Hardy-type inequalities quantum calculus
Serikbol Shaimardan
Luleå University of TechnologyDepartment of Engeneering Sciences and Mathematics
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
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
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.
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
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
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
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
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
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,
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
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
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
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
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
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.
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.
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
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].
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
,
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′
,
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).
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
<∞
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′
<∞
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′
<∞.
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′
<∞.
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 .
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).
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:
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
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: larserik@ltu.se1Luleå 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.
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+
)). ()
Persson and Shaimardan Journal of Inequalities and Applications ( 2015) 2015:296 Page 3 of 17
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
. ()
Persson and Shaimardan Journal of Inequalities and Applications ( 2015) 2015:296 Page 4 of 17
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 ().
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
()
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
,
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
)β
.
Persson and Shaimardan Journal of Inequalities and Applications ( 2015) 2015:296 Page 8 of 17
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
)α
dqx
)β
,
I–(z) :=
(∫ ∞
X(,z](t)f (t)
(∫ ∞
X[z,∞)(x)K (x, t)g(x)dqx
)α
dqt
)β
.
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+.
Persson and Shaimardan Journal of Inequalities and Applications ( 2015) 2015:296 Page 9 of 17
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
)
Persson and Shaimardan Journal of Inequalities and Applications ( 2015) 2015:296 Page 10 of 17
=
[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
. ()
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
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).
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.
Persson and Shaimardan Journal of Inequalities and Applications ( 2015) 2015:296 Page 14 of 17
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
)–.
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′
.
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)
Persson and Shaimardan Journal of Inequalities and Applications ( 2015) 2015:296 Page 17 of 17
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
International Conference on Mathematical Problems in Engineering, Aerospace and Sciences (ICNPAA 2012). AIP
Conference Proceedings, vol. 1493, pp. 859-866 (2012)
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Paper B
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
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 < ∞
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]).
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)
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)
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.
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.
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′
.
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
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
< ∞.
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.
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)
)β
. (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)
)β
qkϕ(qk)
,
for α, β ∈ R .
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)
)β
.
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)
)β
.
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)
)β
.
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)
)β
.
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)
)β
.
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)
)β
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)
)β
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.
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)
)β
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)
)α
= (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
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.
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)
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
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.
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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[3] M. H. ANNABY AND Z. S. MANSOUR, q-fractional calculus and equations, Springer, Heidelberg,2012.
[4] A. O. BAIARYSTANOV, S. SHAIMARDAN AND A. TEMIRKHANOVA, Weighted Hardy inequalities inquantum analysis, Vestinik KarGU, Mathematics series, 2013, no. 2 (70), 35–45, (in Russian).
[5] G. BANGEREZAKO, Variational calculus on q-nonuniform lattices, J. Math. Anal. Appl. 306 (2005),no. 1, 161–179.
[6] G. BENNETT, Some elementary inequalities, Quart. J. Math. Oxford Ser. (2) 38 (1987), no. 152, 401–425.
[7] G. BENNETT, Some elementary inequalities III, Quart. J. Math. Oxford Ser. (2) 42 (1991), no. 166,149–174.
[8] G. BENNETT, Inequalities complimentary to Hardy, Quart. J. Math. Oxford Ser. (2) 49, (1998), 395–432.
[9] J. S. BRADLEY, Hardy inequalities with mixed norms, Canad. Math. Bull. 21 (1978), 405–408.[10] M. S. BRAVERMAN AND V. D. STEPANOV, On the discrete Hardy inequality, Bull. London Math.
Soc. 26 (1994), no. 3, 283–287.[11] P. CHEUNG AND V. KAC, Quantum calculus, Edwards Brothers, Inc., Ann Arbor, MI, USA, 2000.[12] T. ERNST, A comprehensive treatment of q -calculus, Birkhauser/Springer Basel AG, Basel, 2012.[13] T. ERNST, The history of q -calculus and a new method, Licentiate thesis, Uppsala university, 2001.[14] H. EXTON, q-Hypergeometric Functions and Applications, Halstead Press, New York, 1983.[15] H. GAUCHMAN, Integral inequalities in q-calculus, Comput. Math. Appl. 47 (2004), no. 2–3, 281–
300.[16] K.-G. GROSSE-ERDMANN,The Blocking Technique, Weighted Mean Operators and Hardy’s Inequal-
ity, Lecture Notes in Mathematics, Springer Verlag, Berlin, Germany, 1998.[17] G. H. HARDY, Note on a theorem of Hilbert, Math. Z. 6 (1920), 314–317.[18] G. H. HARDY, Notes on some points in the integral calculus. LX. An inequality between integrals,
Messenger of Math. 54 (1925), 150–156.
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[19] G. H. HARDY AND J. E. LITTLEWOOD AND G. POLYA, Inequalities. Cambridge MathematicalLibrary, Cambridge University Press, Cambridge, 1934.
[20] F. H. JACKSON, On q-definite integrals, Quart. J. Pure Appl. Math. 41 (1910), 193–203.[21] V. KOKILASHVILI, A. MESKHI AND L.-E. PERSSON, Weighted Norm Inequalities for Integral Trans-
forms with Product Kernels, Nova Science Publishers, New York, 2010.[22] V. KRASNIQI, Erratum, Several q -integral inequalities, J. Math. Inequal. 5 (2011), no. 3, 451 (Erra-
tum to a paper in J. Math. Inequal. 3 (2009), no. 1, 115–212).[23] A. KUFNER, L. MALIGRANDA AND L.-E. PERSSON, The Hardy Inequality. About its History and
Some Related Results, Vydavatelsky Servis, Plzen, 2007.[24] A. KUFNER AND L.-E. PERSSON, Weighted inequalities of Hardy Type, World Scientific Publishing
Co., Inc., River Edge, NJ, 2003.[25] A. KUFNER, L.-E. PERSSON AND N. SAMKO, Some new scales of weight characterizations of
Hardy-type inequalities, Operator theory, pseudo-differential equations and mathematical physics,261–274, Oper. Theory Adv. Appl., 228, Birkhauser/Springer, Basel AG, Basel, 2013.
[26] L. MALIGRANDA, R. OINAROV AND L.-E. PERSSON, On Hardy q-inequalities, CzechoslovakMath. J., 64 (2014), no. 3, 659–682.
[27] Y. MIAO AND F. QI, Several q -integral inequalities, J. Math. Inequal. 3 (2009), no. 1, 115–121.[28] B. MUCKENHOUPT, Hardy’s inequality with weights, Studia Math. 44 (1972), 31–38.[29] C. A. OKPOTI, L.-E. PERSSON AND A. WEDESTIG, Scales of weight characterizations for the dis-
crete Hardy and Carleman type inequalities, In: Proc. Conf. “Function spaces, Differential operatorsand Nonlinear Analysis”, FSDONA 2004 (Milovy, May 28-Jun 2, 2004), Math. Inst. Acad. Sci. ChechRepublic, Prague 2005, 236–258.
[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 62@mail.ru
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: larserik@ltu.se
S. ShaimardanEurasian National University
Munaytpasov st., 5, 010008 Astana, Kazakhstane-mail: serikbol-87@yandex.kz
A. TemirkhanovaEurasian National University
Munaytpasov st., 5, 010008 Astana, Kazakhstane-mail: ainura-t@yandex.kz
Journal of Mathematical Inequalitieswww.ele-math.comjmi@ele-math.com
Paper C
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,
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].
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.
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)
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
qγ
]=
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
qβ
]=
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)
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
qβ
], (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
qγ
].
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
qβ
]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
qγ
]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,
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):
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)
)β
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)
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).
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
.
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)
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
.
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 ,
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
,
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).
Some Hardy-type inequalities for the fractional integral operator in q-analysis 99
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Serikbol ShaimardanFaculty of Mechanics and MathematicsL.N. Gumilyov Eurasian National University2 Satpayev St,010000 Astana, KazakhstanE-mail: serikbol-87@yandex.kz
Received: 20.11.2015
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Paper D
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: larserik@ltu.se1Luleå 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.
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)
Persson et al. Journal of Inequalities and Applications ( 2018) 2018:73 Page 3 of 14
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)
Persson et al. Journal of Inequalities and Applications ( 2018) 2018:73 Page 4 of 14
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))
Persson et al. Journal of Inequalities and Applications ( 2018) 2018:73 Page 5 of 14
≥ 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.
Persson et al. Journal of Inequalities and Applications ( 2018) 2018:73 Page 6 of 14
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.
Persson et al. Journal of Inequalities and Applications ( 2018) 2018:73 Page 7 of 14
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
Persson et al. Journal of Inequalities and Applications ( 2018) 2018:73 Page 8 of 14
×
[(
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.
Persson et al. Journal of Inequalities and Applications ( 2018) 2018:73 Page 9 of 14
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)
Persson et al. Journal of Inequalities and Applications ( 2018) 2018:73 Page 10 of 14
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
Persson et al. Journal of Inequalities and Applications ( 2018) 2018:73 Page 11 of 14
=
[(
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
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.
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
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
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
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
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
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
[Γ
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
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
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
≤ 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
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
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
≤ 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
=
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
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
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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