An application of the weighted discrete Hardy inequality
aa r X i v : . [ m a t h . A P ] F e b AN APPLICATION OF THE WEIGHTED DISCRETE HARDYINEQUALITY
GIAO BUI, FERNANDO L ´OPEZ-GARC´IA, AND VAN TRAN
Abstract.
In a note published in 1925, G. H. Hardy stated the inequality ∞ X n =1 n n X k =1 a k ! p ≤ (cid:18) pp − (cid:19) p ∞ X n =1 a pn , for any non-negative sequence { a n } n ≥ , and p >
1. This inequality is knownin the literature as the classical discrete Hardy inequality. It has been widelystudied and several applications and new versions have been shown.In this work, we use a characterization of a weighted version of this in-equality to exhibit a sufficient condition for the existence of solutions of thedifferential equation div u = f in weighted Sobolev spaces over a certain planeirregular domain. The solvability of this equation is fundamental for the anal-ysis of the Stokes equations.The proof follows from a local-to-global argument based on a certain de-composition of functions which is also of interest for its applications to otherinequalities or related results in Sobolev spaces, such as the Korn inequality. Introduction
Given p > the discrete Hardy inequality states ∞ X n =1 n n X k =1 a k ! p ≤ (cid:18) pp − (cid:19) p ∞ X n =1 a pn , (1.1)for any non-negative sequence { a n } n ≥ , where the constant in the inequality ( p/ ( p − p is optimal. This inequality has been widely studied and many generalizationshave been shown. In this article, we use a weighted version known as the weighteddiscrete Hardy inequality which says: ∞ X n =1 u n n X k =1 a k ! p ! /p ≤ C ∞ X n =1 v n a pn ! /p . (1.2)The existence of a constant C , that makes inequality (1.2) valid for any non-negative sequence { a n } n ≥ , depends only on p and the sequence weights { u n } n ≥ and { v n } n ≥ . There are several characterizations of the sequence weights in theprevious inequality such as the one published in [3] that states that the constant C Mathematics Subject Classification.
Primary: 26D15; Secondary: 46E35,76D07.
Key words and phrases.
Discrete Hardy inequality, Divergence operator, Decomposition offunctions, Cusps, Weights, Stokes equations.Partially Supported by nsf-dms 1247679 grant PUMP: Preparing Undergraduates throughMentoring towards PhDs.
1N APPLICATION OF THE HARDY INEQUALITY 2 in (1.2) exists if and only if A = sup k ≥ ∞ X i = k u i ! /p k X i =1 v − qi ! /q < ∞ , where q = p/ ( p − u = f for weights ν ( x ) , ν ( x ) : Ω → R > , where Ω is the planardomain Ω := { ( x , x ) ∈ R : 0 < x < < x < x γ } , (1.3)for γ ≥
1. Specifically, we are looking for sufficient conditions on the weights ν ( x ) x = x γ x x Ω and ν ( x ) such that, for any f ∈ L (Ω , ν ( x )) with vanishing mean value, thereexists a solution u of div u = f in the Sobolev spaces H (Ω , ν ( x )) := C ∞ (Ω) with the following estimate Z Ω | D u ( x ) | ν ( x ) d x ≤ C Z Ω | f ( x ) | ν ( x ) d x, (1.4)where D u ( x ) denotes the differential matrix of u . The weights considered heresatisfy that ν ( x ) = x γ − ν ( x ) and ν , ν depend only on the first component of x (i.e. ν ( x ) = ν ( x ) and ν ( x ) = ν ( x )). Notice that if γ >
1, the domain Ω hasa singularity (cusp) at the origin, while the domain is regular (convex) if γ = 1.The factor x γ − in the definition of ν ( x ) is there to deal with the singularityat the origin and disappears when Ω is regular ( γ = 1), in which case we have thesame weights in both sides of the estimate (1.4). The exponent in the factor x γ − is optimal in the following sense: if ν ( x ) = 1 and ν ( x ) = x a , with a < γ − u = f with estimate (1.4) fails in general (we refer to [2] forcounterexamples).The solvability of the divergence equation is fundamental for the variational anal-ysis of the Stokes equations and strongly depends on the geometry of the domain,which has been studied in Lipschitz domains, star-shaped domains with respectto a ball, John domains, H¨older- α domains, among others. We refer to [1] andreferences therein for an extensive description of the solvability of this equation ondomains under several geometric conditions. The domain Ω of our interest and N APPLICATION OF THE HARDY INEQUALITY 3 defined in (1.3) was already considered in [7, 12]. The authors in [7] use the Piolatransform of an explicit solution on a regular domain whose analysis required theuse of the theory of singular integral operators and Muckenhoupt weights. In [12],the author uses a technique similar to the one treated in this article, where thediscrete weighted Hardy inequality (1.2) is replaced by a Hardy-type operator onweighted L p (Ω) spaces. The reason to work with (1.2) instead of the Hardy-typeoperator defined in [12] relies on the simplicity of the discrete inequality and thecharacterization of the weights for which the inequality remains valid.Now, in order to prove our main results, we decompose Ω into a collection ofinfinitely many regular (star-shaped with respect to a ball) subdomains { Ω i } i ≥ where the weights can be assumed to be constant. In that case the solvability ofthe divergence equation has been proved. Then, we extend by zero the solutionsin Ω i to the whole domain and add them up to obtain a solution in Ω. Inequality(1.2) appears when we estimate the norm of the “global solution” in terms of theestimation of the “local solutions”. The decomposition { Ω i } i ≥ of Ω mentionedabove is: Ω i := { ( x , x ) ∈ Ω : 2 − ( i +2) < x < − i } . (1.5)This is the main result of the paper. Theorem 1.1.
Let ω : Ω → R be an admissible weight in the sense of Definition2.1, for p = 2 , such that the following weighted Hardy inequality is valid for anynon-negative sequence { d n } n ≥ : ∞ X j =1 u i j X i =1 d i ! ≤ C H ∞ X j =1 u j d j , where u i := | Ω i | ω (2 − i ) . Then, there exists a constant C such that for any f in L (Ω , ω − ( x )) , with vanish-ing mean value, there exists a solution u : Ω ⊂ R → R of the equation div u = f in H (Ω , x γ − ω − ( x )) such that Z Ω | D u ( x ) | x γ − ω − ( x ) d x ≤ C Z Ω | f ( x ) | ω − ( x ) d x. Moreover, C ≤ γ γ C ω C H . Remark 1.2.
The strong connection between the solvability of the equation div u = f and the validity of the Korn inequality in the second case is well-known (see[8, 10, 1]). Thus, it is worth observing that in [4] the authors use the weighteddiscrete Hardy inequality (1.2) to prove the validity of the Korn inequality on do-mains with a single singularity on the boundary by using a different local-to-globalargument.The following result considers the case where the weights are power functions. Corollary 1.3 (Power weights) . Let Ω ⊂ R be the domain defined in (1.3) and β > − γ − . Then, there exists a positive constant C such that for any f ∈ N APPLICATION OF THE HARDY INEQUALITY 4 L (Ω , ω ( x ) − ) , with R Ω f = 0 , there exists a solution u ∈ H (cid:16) Ω , x γ − ω ( x ) − ) (cid:17) of div u = f that satisfies Z Ω | D u ( x ) | x γ − ω ( x ) − d x ≤ C Z Ω | f ( x ) | ω ( x ) − d x, (1.6) where ω ( x ) := x β . Moreover, if β ≤ , the constant C in (1.6) satisfies thefollowing estimate: C ≤ M − − ( β + γ +12 ) , where the constant M is independent of β . Notice that the distance from ( x , x ) in Ω to the origin is comparable to x ,thus the weights here can be understood as powers of the distance to the origin orthe cusp if γ >
1. Indeed, x ≤ q x + x ≤ √ x . for all ( x , x ) ∈ Ω.The existence of a solution of the divergence equation in this planar domain Ωwith the estimate (1.6) was first obtained in [7, Theorem 4.1] for β in (cid:0) − γ − , γ − (cid:1) ,and later in [12, Theorem 5.1] for β ≥
0. In this case, we recover both results as acorollary of our main theorem. In addition, an estimate of the constant that boundsits blow-up as β tends to − γ − is exhibited. Finally, notice that if β ≤ − γ − then L (Ω , x − β ) L (Ω) and the vanishing mean value condition in the divergenceproblem is not well-defined. Hence, the condition β > − γ − is optimal for thecurrent setting. For an example of a non-integrable function in L (Ω , x − β ), when β ≤ − γ − , one can consider f ( x ) = (1 − ln( x )) − x − γ − .The following result considers the case where the weights are powers of a loga-rithmic function. Corollary 1.4 (Powers of logarithmic weights) . Let Ω ⊂ R be the domain definedin (1.3) and α ∈ R . Then, there exists a positive constant C such that for any f ∈ L (Ω , ω ( x ) − ) , with R Ω f = 0 , there exists a solution u ∈ H (cid:16) Ω , x γ − ω ( x ) − ) (cid:17) of div u = f that satisfies Z Ω | D u ( x ) | x γ − ω ( x ) − d x ≤ C Z Ω | f ( x ) | ω ( x ) − d x, where ω ( x ) := (1 − ln( x )) α . The article is organized as follows: In Chapter 2, we show that the weighteddiscrete Hardy inequality, with some appropriate weights, implies the validity of acertain decomposition of functions in which our local-to-global argument is based.The main result in this chapter might be of interest for applications to other in-equalities and related results in Sobolev spaces. In this chapter, we consider general1 < p, q < ∞ , with p + q = 1. Then, we use the estimate of the constant in thedivergence equation provided by Costabel and Dauge [6] for p = q = 2 to proveTheorem 1.1. In Chapter 3, we prove the validity of the corollaries stated in the in-troduction that claim the solvability of the divergence equation in weighted spacesfor power weights and powers of logarithmic weights. N APPLICATION OF THE HARDY INEQUALITY 5
The novelty of this work lies in the use of the well-studied weighted discreteHardy inequality to get new sufficient conditions on the weights that imply thesolvability of the divergence equation, recovering the existing results in [7, 12] whenthe weights are powers of the distance to the cusp/origin. The second corollaryusing powers of logarithmic weights is also new.2.
A decomposition of functions and applications
We name a weight ν : Ω → R a positive and Lebesgue-measurable function, anda sequence weight { ν i } i ≥ a sequence of positive real numbers. We will denote by x = ( x , x ) a general point in R . Definition 2.1.
A weight ω : Ω → R is called admissible if ω p ∈ L (Ω) and thereexists a uniform constant C ω such thatess sup x ∈ Ω i ω ( x ) ≤ C ω ess inf x ∈ Ω i ω ( x ) , (2.1)for all i ≥
0. Notice that admissible weights are subordinate to a partition { Ω i } i ≥ of Ω introduced in (1.5), and 1 < p < ∞ . Examples 2.2.
The function ω ( x ) := x β , where β > − γ − p , is an admissible weightwith C ω = 2 | β | , where β := − γ − p . Definition 2.3.
Given g : Ω → R integrable function with vanishing mean value,i.e. R g = 0, we refer by a C -orthogonal decomposition of g subordinate to { Ω i } i ≥ to a collection of integrable functions { g i } i ≥ with the following properties:(1) g ( x ) = P i ≥ g i ( x ) . (2) supp( g i ) ⊂ Ω i , for all i ≥ R Ω i g i = 0, for all i ≥ . The letter C in the previous definition refers to the space of constant functions.Notice that having vanishing mean value could also be understood as being or-thogonal to the functions in C . Other applications of this type of decomposition offunctions require to have orthogonality to other spaces (see [13, 14]). We also referthe reader to [9] for applications to a fractional Poincar´e type inequality.We show the existence of a C -orthogonal decomposition by using a constructiveargument introduced in [12]. Let us describe the idea of this argument assumingthat Ω is the union of the first three subdomains in partition defined in (1.5). Thus,let f ∈ L (Ω) be a function with vanishing mean value. Then, using a partition ofthe unity { φ i } ≤ i ≤ subordinate to { Ω i } ≤ i ≤ we can write g as: g = f + f + f = gφ + gφ + gφ . However, this partition might not be orthogonal to C . In order to get this propertywe make the following arrangements: g = f + (cid:18) f + χ B | B | Z Ω f (cid:19) + (cid:18) f − χ B | B | Z Ω f (cid:19)| {z } f − h , where B := Ω ∩ Ω . Note that the function f − h has its support in Ω and R f − h = 0. Finally, we repeat the process with the first two functions. Thus, if N APPLICATION OF THE HARDY INEQUALITY 6 B := Ω ∩ Ω we have that f = f − h z }| {(cid:18) f + χ B | B | Z Ω ∪ Ω f + f (cid:19) + (cid:18) f + χ B | B | Z Ω f − χ B | B | Z Ω ∪ Ω f + f (cid:19)| {z } f − h + (cid:18) f − χ B | B | Z Ω f (cid:19)| {z } f − h , (2.2)obtaining the claimed decomposition. Observe that we have used the vanishingmean value of f only to prove that f − h integrates zero.Now, let us introduce the following weighted discrete Hardy-type inequalities: ∞ X j =1 | Ω j | ω pj j X i =1 d i ! p ≤ C pH ∞ X j =1 | Ω j | ω pj d pj , (2.3)and ∞ X i =1 | Ω i | − q ω − qi ∞ X j = i b j q ≤ C qH ∞ X i =1 | Ω i | − q ω − qi b qi . (2.4)The first one is inequality (1.2) to the p power where the sequence weight u n = v n = | Ω n | ω pn , and the second one is its dual version. The following lemma followsfrom this duality. Lemma 2.4.
Given a sequence weight { ω i } i ≥ , inequality (2.4) is valid for anynon-negative sequence { b i } i ≥ if and only if inequality (2.3) is valid for any non-negative sequence { d j } j ≥ , with the same constant C H .Proof. By using the duality between l p and l q , and defining ˜ d j := | Ω j | /p ω j d j and˜ b i := | Ω i | − /p ω − i b i , it follows that inequality (2.3) and (2.4) can be written assup k ˜ d k lp =1 sup k ˜ b k lq =1 ∞ X j =1 ˜ b j | Ω j | /p ω j j X i =1 | Ω i | − /p ω − i ˜ d i ≤ C H (2.5)and sup k ˜ b k lq =1 sup k ˜ d k lp =1 ∞ X i =1 ˜ d i | Ω i | − /p ω − i ∞ X j = i | Ω j | /p ω j ˜ b j ≤ C H (2.6)Finally, one can obtain (2.6) from (2.5), and viceversa, by changing the order ofthe summations. (cid:3) Theorem 2.5.
Let ω : Ω → R be an admissible weight that satisfies (2.3) for thesequence weight ω i := ω (2 − i ) . Then, given g ∈ L (Ω) , with R Ω g = 0 , there exists { g t } t ∈ Γ , a C -decomposition of g subordinate to { Ω i } i ≥ (see Definition 2.3), suchthat ∞ X i =0 Z Ω i | g i ( x ) | q ω − q ( x ) d x ≤ C qd Z Ω | g ( x ) | q ω − q ( x ) d x. (2.7) Moreover, we have the following estimate for the optimal constant C d : C d ≤ /q C ω C H . (2.8) N APPLICATION OF THE HARDY INEQUALITY 7
Proof.
The decomposition treated here follows the example with three subdomainsin page 5. Indeed, let { φ i } i ≥ be a partition of unity subordinate to the collection { Ω i } i ≥ . Namely, a collection of smooth functions such that P i ≥ φ i = 1, 0 ≤ φ i ≤ supp ( φ i ) ⊂ Ω i . Thus, g can be cut-off into g = P i ≥ f i by taking f i = gφ i .This decomposition satisfies (1) and (2) in Definition 2.3 but not necessarily (3).Thus, we make the following modifications to { f i } i ≥ to obtain a collection offunctions that also satisfies (3). Indeed, for any i ≥ g i ( x ) := f i ( x ) + h i +1 ( x ) − h i ( x ) , (2.9)where h i ( x ) := χ i ( x ) | B i | Z W i X k ≥ i f k ,B i := Ω i ∩ Ω i − , (2.10) W i := [ k ≥ i Ω k . We denote by χ i the characteristic function of B i . Notice that the auxiliaryfunction h i is not defined for i = 0, thus g follows in this other way g ( x ) = f ( x ) + h ( x ) . This decomposition was introduced in [12] in a more general way where thenatural numbers in the subindex set is replaced by a set with a partial order givenby a structure of tree (i.e. connected graph without cycles). We also use in thisarticle inequality (2.3) instead of another Hardy type inequality on trees introducedin [12]. Thus, it only remains to show estimate (2.7). Notice that h i and h i +1 havedisjoint supports thus | h i +1 ( x ) − h i ( x ) | q = | h i +1 ( x ) | q + | h i ( x ) | q . Next, using that | a + b | q ≤ q − ( | a | q + | b | q ) for all a, b ∈ R , we have ∞ X i =0 Z Ω i | g i ( x ) | q ω − q ( x ) d x ≤ q − ∞ X i =0 Z Ω i | f i ( x ) | q ω − q ( x ) d x + 2 ∞ X i =1 Z Ω i | h i ( x ) | q ω − q ( x ) d x ! ≤ q Z Ω | g ( x ) | q ω − q ( x ) d x + ∞ X i =1 Z Ω i | h i ( x ) | q ω − q ( x ) d x ! . (2.11)Let us work over the sum on the right hand side in the previous inequality byusing the weighted discrete Hardy inequality. Notice that from the definition of theauxiliary functions in (2.10) and inequality (2.1) in Definition 2.1 it follows that | h i ( x ) | ≤ χ i ( x ) | B i | ∞ X k = i Z Ω i | g | and Z Ω i χ i ( x ) | B i | q ω − q ( x ) d x ≤ C qω ω − qi | B i | − q . N APPLICATION OF THE HARDY INEQUALITY 8
Therefore, since | Ω i | < | B i | for any i ≥
1, the sum in inequality (2.11) isbounded by ∞ X i =1 Z Ω i | h i ( x ) | q ω − q ( x ) d x ≤ C qω ∞ X i =1 | B i | − q ω − qi ∞ X k = i Z Ω k | g | ! q ≤ q − C qω ∞ X i =1 | Ω i | − q ω − qi ∞ X k = i Z Ω k | g | ! q . Next, by using Lemma 2.4 with b i = R Ω i | g | , i ≥
1, and H¨older inequality, we canconclude that ∞ X i =1 Z Ω i | h i ( x ) | q ω − q ( x ) d x ≤ q − C qω C qH ∞ X i =1 | Ω i | − q ω − qi (cid:18)Z Ω i | g | (cid:19) q ≤ q − C qω C qH ∞ X i =1 ω − qi (cid:18)Z Ω i | g | q (cid:19) ≤ q − C qω C qH ∞ X i =1 Z Ω i | g ( x ) | q ω − q ( x ) d x ≤ q C qω C qH Z Ω | g ( x ) | q ω − q ( x ) d x. Finally, from inequality (2.11) it follows (2.7). (cid:3)
In order to prove the solvability of the divergence equation on the subdomainsΩ i we use the following result proved by M. Costabel and M. Dauge in [6] for star-shaped domains. Let us recall the definition of this class of domains. A domain U is star-shaped with respect to a ball B if and only if any segment with an end-pointin U and the other one in B is contained in U . Theorem 2.6.
Let U ⊂ R be a domain contained in a ball of radius R , star-shaped with respect to a concentric ball of radius r . Then, for any g ∈ L ( U ) withvanishing mean value there exists a solution u ∈ H ( U ) of the equation div u = g satisfying the estimate (cid:18)Z U | D u ( x ) | d x (cid:19) / ≤ Rr (cid:18)Z U | g ( x ) | d x (cid:19) / . Proof of Theorem 1.1.
Let f be a function in L (Ω , ω − ( x )) with vanishing meanvalue. Notice that, since ω is an admissible weight for p = 2, L (Ω , ω − ( x )) ⊂ L (Ω) and the mean value of f is well-defined. Then, from Theorem 2.5, thereexists a C -decomposition { f i } i ≥ of f subordinate to { Ω i } i ≥ satisfying (2.7). Now,let us assume, to be shown later in this proof, that Ω i is included in a ball withradius R i = 2 − i +1 and star-shaped with respect to a concentric ball A i with radius r i = 2 − γ ( i +2) − /γ . Then, from Theorem 2.6, there exists a solution of div v i = f i in Ω i that satisfies Z Ω i | D v i ( x ) | d x ≤ γ γ γ − i Z Ω i | f i ( x ) | d x. N APPLICATION OF THE HARDY INEQUALITY 9
Hence, by extending v i by zero, the vector field u ( x ) := P i ≥ v i ( x ) satisfies thatdiv u ( x ) = div X i ≥ v i ( x ) = X i ≥ f i ( x ) = f ( x ) , and Z Ω | D u ( x ) | x γ − ω − ( x ) d x ≤ X i ≥ Z Ω i | D v i ( x ) | x γ − ω − ( x ) d x ≤ C ω X i ≥ − i ( γ − ω − (2 − i ) Z Ω i | D v i ( x ) | d x ≤ γ γ C ω X i ≥ ω − (2 − i ) Z Ω i | f i ( x ) | d x ≤ γ γ C ω X i ≥ Z Ω i | f i ( x ) | ω ( x ) − d x ≤ γ γ C ω C H Z Ω | f ( x ) | d x. Finally, let us show that Ω i is included in a ball with radius R i = 2 − i +1 andstar-shaped with respect to a concentric ball A i . Notice that Ω i is included inthe square [0 , − i ] with diameter 2 − i +1 / . Thus, any ball with center at a pointin Ω i and radius R i = 2 − i +1 contains Ω i . We define A i as the ball with radius r i := ρ i / γ , and center c i := (2 − i − r i , r i ), where ρ i = 2 − γ ( i +2) , as shown in Picture1. x x − i − ( i +2) Ω i A i Figure 1. Ω i is a star-shaped domain.Now, given y ∈ Ω i and x ∈ A i , we have to show that the segment xy withend-points at y and x is included in Ω i .Now, the open rectangle D t with sides parallel to the axis and vertices (2 − i , t, t γ ), for 2 − i − ≤ t ≤ − i − r i , is convex, contains B i and is included in Ω i . N APPLICATION OF THE HARDY INEQUALITY 10
Thus, the segment xy is included in Ω i if y belongs to D t , for any t in the interval[2 − i − , − i − r i ].Hence, it is sufficient to prove the case where y = ( y , y ) belongs to the regionabove (or over) the dashed line in Picture 1: 2 − i − r i < y < − i and (2 − i − r i ) γ ≤ y < y γ . Moreover, observe that if the segment xy is not included in Ω i then itsslope must be equal to γt γ − , for some 2 − i − r i < t < − i . Hence, it is sufficientto show that the slope of xy is larger than γ i.e. | y − x || y − x | ≥ γ. Now, it follows from some straightforward estimations that | y − x | ≥ − iγ − | − iγ − y | − | x | , where, | x | ≤ ρ i and | − iγ − y | ≤ | (2 − i ) γ − (2 − i − r i ) γ | < γ r i = ρ i . Then, | y − x || y − x | ≥ γ ρ i − ρ i ρ i /γ ≥ γ. (cid:3) The weighted discrete Hardy inequality
In this chapter, we prove the two corollaries stated in the Introduction about thesolvability of the divergence equation in weighted Sobolev spaces for the weights ω ( x ) = x β and ω ( x ) = (1 − ln( x )) α . Notice that Theorem 1.1 requires p = 2,however, we analyze the general case 1 < p < ∞ since Theorem 2.5, which doesnot have the constraint p = 2, can be used to obtain other inequalities (such as theweighted fractional Poincar´e inequality [9]) in our cuspidal domain Ω.Let us recall the characterization of the weighted discrete Hardy inequalityproved by K. F. Andersen and H. P. Heinig. We also refer to [11, page 56] and[15] for more details. Theorem 3.1.
Let { u i } i ≥ and { v i } i ≥ be sequence weights, and the conjugateexponents < p, q < ∞ , i.e. p + q = 1 , then inequality (1.2) is valid if and only if A = sup k ≥ ∞ X i = k u i ! p k X i =1 v − qi ! q < ∞ . In addition, if C H represents the optimal constant in (1.2) , then A ≤ C H ≤ A Thus, we use the characterization for the validity of the weighted discrete Hardyinequality to determine the exponents β and α for which the previos weights satisfythe sufficient condition in Theorem 1.1: ∞ X j =1 u i j X i =1 d i ! ≤ C H ∞ X j =1 u i d j , where u i := | Ω i | ω (2 − i ) . N APPLICATION OF THE HARDY INEQUALITY 11
Let us start by calculating the measure of the subdomains Ω i : | Ω i | = Z − i − ( i +2) Z x γ d x d x = Z − i − ( i +2) x γ d x = x γ +11 γ + 1 (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) − i − ( i +2) = 1 γ + 1 (cid:16) − i ( γ +1) − − ( i +2)( γ +1) (cid:17) = 1 − − γ +1) γ + 1 2 − i ( γ +1) = C γ − ( γ +1) i , where C γ = 1 − − γ +1) γ + 1 . (3.12)For simplicity, we include some basic calculations on geometric sums which willbe used in the following proofs: ∞ X i = k r i = r k − r , for 0 < r < , k X i =1 r i (1 − q ) = ( r − q ) k +1 − r − q r − q − , for r > , and q > . The following lemma considers the power weights ω ( x ) = x β . Lemma 3.2.
Let Ω ⊂ R be the domain defined in (1.3) and < p, q < ∞ , with p + q = 1 . Then, the weight ω : Ω → R defined by ω ( x ) = x β , with β > − γ − p , isan admissible weight in the sense of Definition 2.1, and satisfies weighted discreteHardy inequality (2.3) where ω i := ω (2 − i ) .Moreover, C H < (cid:18) r (1 − r ) (cid:19) /p (cid:18) r − q − (cid:19) /q , where r := 2 − pβ − γ − . Proof.
First, let us show that x pβ ∈ L (Ω): Z Z x γ x βp d x d x = Z x βp + γ dx , which is finite if and only if βp + γ > −
1, equivalently, β > − γ − p . Moreover, it iseasy to prove that Condition (2.1) is valid with C ω = 2 | β | , where β := − γ − p .Now, we have to show that the weighted discrete Hardy inequality (2.3) is sat-isfied for the sequence weight ω i := 2 − iβ , with β > − γ − p . Thus, by Theorem 3.1, N APPLICATION OF THE HARDY INEQUALITY 12 it is necessary and sufficient to show that A = sup k ≥ ∞ X i = k | Ω i | − iβp ! /p k X i =1 ( | Ω i | − iβp ) − q ! /q < ∞ . Hence, let us denote | Ω i | − iβp = C γ (cid:16) − ( γ +1) − pβ (cid:17) i =: C γ r i , where C γ was introduced in (3.12). Notice that r ∈ (0 , A = sup k ≥ ∞ X i = k C γ r i ! /p k X i =1 ( C γ r i ) (1 − q ) ! /q = C /p +(1 − q ) /qγ sup k ≥ ∞ X i = k r i ! /p k X i =1 r i (1 − q ) ! /q = sup k ≥ (cid:18) r k − r (cid:19) /p (cid:18) ( r − q ) k +1 − r − q r − q − (cid:19) /q = (cid:18) − r (cid:19) /p (cid:18) r − q r − q − (cid:19) /q sup k ≥ r k/p (cid:16) r k (1 − q )) − (cid:17) /q < (cid:18) − r (cid:19) /p (cid:18) r − q r − q − (cid:19) /q sup k ≥ r k/p r k (1 − q ) /q = (cid:18) − r (cid:19) /p (cid:18) r − q r − q − (cid:19) /q < ∞ . Moreover, using again Theorem 3.1, it follows that C H ≤ A < (cid:18) r (1 − r ) (cid:19) /p (cid:18) r − q − (cid:19) /q , where r := 2 − pβ − γ − . (cid:3) Proof of Corollary 1.3.
It follows from Theorem 1.1 and Lemma 3.2. (cid:3)
Lemma 3.3.
Let Ω ⊂ R be the domain defined in (1.3) and < p, q < ∞ , with p + q = 1 . Then, the weight ω : Ω → R defined by ω ( x ) = (1 − ln( x )) α , with α ∈ R , is an admissible weight in the sense of Definition 2.1, and satisfies theweighted discrete Hardy inequality (2.3) for ω i := ω (2 − i ) .Proof. If α is zero, then ω ( x ) = 1. This weight was studied in Lemma 3.2, for β = 0, which is admissible and satisfies the discrete Hardy inequality (2.3) with C H < (cid:18) r (1 − r ) (cid:19) /p (cid:18) r − q − (cid:19) /q , for r := 2 − ( γ +1) . Thus, we have to consider the case when α is different from 0. N APPLICATION OF THE HARDY INEQUALITY 13
First, let us show that ω p ( x ) = (1 − ln( x )) pα ∈ L (Ω): Z Z x γ (1 − ln( x )) pα d x d x = Z x γ (1 − ln( x )) pα d x . If α is positive, then the function f ( x ) = x γ ((1 − ln( x )) pα tends to 0 as x tends to 0 from the right, then the integral of this continuous function is finite:lim x → + x γ (1 − ln( x )) pα = lim x → + − ln( x ) x − γ/pα ! pα = 0 , since lim x → + − ln( x ) x − γ/pα = lim x → + − x − ( − γ/pα ) x − γ/pα − = lim x → + pαγ x γ/pα = 0 . If α is negative then 0 < x γ (1 − ln( x )) pα <
1, thus ω p ( x ) ∈ L (Ω).Now, let us estimate the constant C ω in inequality (2.1):sup x ∈ Ω i ω ( x ) ≤ C ω inf x ∈ Ω i ω ( x ) . If α is positive, then ω ( x ) is decreasing with respect to x , thensup x ∈ Ω i ω ( x ) = ω (2 − i − ) = (1 + ( i + 2) ln(2)) α inf x ∈ Ω i ω ( x ) = ω (2 − i ) = (1 + i ln(2)) α , hence, ω (2 − i − ) ω (2 − i ) = (cid:18) i ln(2) (cid:19) α ≤ (1 + 2 ln(2)) α . If α is negative, then ω ( x ) is increasing with respect to x , thensup x ∈ Ω i ω ( x ) = ω (2 − i ) = (1 + i ln(2)) α inf x ∈ Ω i ω ( x ) = ω (2 − i − ) = (1 + ( i + 2) ln(2)) α , hence, ω (2 − i ) ω (2 − i − ) = (cid:18) i ln(2) (cid:19) − α ≤ (1 + 2 ln(2)) − α . Thus, C ω := (1 + 2 ln(2)) | α | satisfies estimate (2.1).Third, let us study the weighted discrete Hardy inequality for this weight. We usethe characterization stated in Theorem 3.1 thus we have to estimate the following N APPLICATION OF THE HARDY INEQUALITY 14 supremum A = sup k ≥ ∞ X i = k u i ! p k X i =1 u i − q ! q = sup k ≥ ∞ X i = k − ( γ +1) i (1 + i ln(2)) pα ! p k X i =1 (cid:16) − ( γ +1) i (1 + i ln(2)) pα (cid:17) − q ! q . (3.13)If α is negative, then pα < i ln(2)) pα ≤ (1 + k ln(2)) pα for all i ≥ k .Similarly, pα (1 − q ) > i ln(2)) pα (1 − q ) ≤ (1 + k ln(2)) pα (1 − q ) for all i ≤ k .Thus, A ≤ sup k ≥ (1 + k ln(2)) α + pα (1 − q ) /q ∞ X i = k r i ! p k X i =1 r i (1 − q ) ! q = sup k ≥ ∞ X i = k r i ! p k X i =1 r i (1 − q ) ! q ≤ (cid:18) − r (cid:19) /p (cid:18) r − q r − q − (cid:19) /q < ∞ , where r = 2 − ( γ +1) .For α positive, we define a = pα > f ( t ) = r t (1 + t ln(2)) a . By a straight-forward calculation, it can be seen that f is positive and decreasing for t sufficientlylarge. Thus, there exists k ∈ N such ∞ X i = k − ( γ +1) i (1 + i ln(2)) pα ≤ Z ∞ k − r t (1 + t ln(2)) a d t. for k ≥ k . Next, by using integration by parts, we obtain I := Z ∞ k − r t (1 + t ln(2)) a d t ≤ − r k r ln( r ) (1 + k ln(2)) a + Z ∞ k − r t (1 + t ln(2)) a (cid:20) a ln(2) − ln( r )(1 + t ln(2)) (cid:21) d t. Now, we assume that k is sufficiently large such that the function between bracketsin the previous line is less than 1 /
2. Thus, I ≤ − r k r ln( r ) (1 + k ln(2)) a + 12 I, and 12 I ≤ − r k r ln( r ) (1 + k ln(2)) a . N APPLICATION OF THE HARDY INEQUALITY 15
Thus, it follows that ∞ X i = k r i (1 + i ln(2)) a ≤ Z ∞ k − r t (1 + t ln(2)) a d t = I ≤ − r k r ln( r ) (1 + k ln(2)) a , (3.14)for k ≥ k .Let us study the second sum in the estimation of A in (3.13). Thus, we define˜ r := 2 − ( γ +1)(1 − q ) = r − q = r − qp > , (3.15)and ˜ a := a (1 − q ) = − aq/p < . (3.16)Notice that the function g ( t ) := ˜ r t (1 + t ln(2)) ˜ a is positive and increasing for t sufficiently large. Thus, there exists a constant C > k X i =1 ˜ r i (1 + i ln(2)) ˜ a ≤ C Z k +11 ˜ r t (1 + t ln(2)) ˜ a d t, (3.17)for all k ≥
1. Finally, notice that to show that A in (3.13) is finite it is sufficient toconsider the case where the supremum runs over k ≥ k and estimate its power q .Thus, from (3.14) and (3.17), we havesup k ≥ k ∞ X i = k r i (1 + i ln(2)) a ! qp k X i =1 ˜ r i (1 + i ln(2)) ˜ a ! ≤ C R k +11 ˜ r t (1 + t ln(2)) ˜ a dtr − kqp (1 + k ln(2)) − aqp , for another constant C , which is independent of k , denoted with the same letterfor simplicity.Finally, we calculate the limit of the above quotient as k goes to infinity, under-standing k as a continuous variable. We use for this analysis definitions (3.15) and(3.16), and L’Hospital rule:lim k →∞ R k +11 ˜ r t (1 + t ln(2)) ˜ a dtr − kqp (1 + k ln(2)) − aqp = lim k →∞ R k +11 ˜ r t (1 + t ln(2)) ˜ a dt ˜ r k (1 + k ln(2)) ˜ a = lim k →∞ ˜ r k +1 (1 + ( k + 1) ln(2)) ˜ a ln(˜ r )˜ r k (1 + k ln(2)) ˜ a + ˜ a ln(2)˜ r k (1 + k ln(2)) ˜ a − = lim k →∞ ˜ r (cid:18) k + 1) ln(2)1 + k ln(2) (cid:19) ˜ a r ) + ˜ a ln(2)1+ k ln(2) = ˜ r ln(˜ r ) . Therefore, the sequence is convergent and bounded, which implies that A isfinite. (cid:3) Proof of Corollary 1.4.
It follows immediately from Theorem 1.1 and Lemma 3.3. (cid:3)
N APPLICATION OF THE HARDY INEQUALITY 16
Acknowledgements
We gratefully acknowledge Professor M. Helena Noronha from Cal State Northridgefor creating the PUMP program
DMS-1247679 depending on NSF, which pro-vided support to several students in the California State Universities located inSouthern California to conduct research in Mathematics. We extend our acknowl-edgement to Professor John Rock from Cal Poly Pomona for introducing the authorsto Professor M. Helena Noronha and this outstanding program.
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