We prove some Hardy-type inequalities via an approach that involves
constructing auxiliary sequences.Comment: 10 page

Bennett

Hardy

Knopp

Levin

Peng Gao

Redheffer

English

arXiv

Crossref

Hardy-type inequalities via auxiliary sequences

We prove some Hardy-type inequalities via an approach that involves constructing

auxiliary sequences

Gao, Peng

Victoria University Eprints Repository

HARDY-TYPE INEQUALITIES VIA AUXILIARY SEQUENCESPENG GAOAbstract. We prove some Hardy-type inequalities via an approach that involves constructingauxiliary sequences.1. IntroductionSuppose throughout that p 6= 0, 1p + 1q = 1. Let lp be the Banach space of all complex sequencesa = (an)n≥1 with norm||a|| := (∞∑n=1|an|p)1/p <∞.The celebrated Hardy’s inequality ([5, Theorem 326]) asserts that for p > 1,(1.1)∞∑n=1∣∣ 1nn∑k=1ak∣∣p ≤ ( pp− 1)p∞∑k=1|ak|p.Hardy’s inequality can be regarded as a special case of the following inequality:∞∑j=1∣∣ ∞∑k=1cj,kak∣∣p ≤ U ∞∑k=1|ak|p,in which C = (cj,k) and the parameter p are assumed fixed (p > 1), and the estimate is to hold forall complex sequences a. The lp operator norm of C is then defined as the p-th root of the smallestvalue of the constant U :||C||p,p = U1p .Hardy’s inequality thus asserts that the Cesa´ro matrix operator C, given by cj,k = 1/j, k ≤ jand 0 otherwise, is bounded on lp and has norm ≤ p/(p− 1). (The norm is in fact p/(p− 1).)We say a matrix A is a summability matrix if its entries satisfy: aj,k ≥ 0, aj,k = 0 for k > j and∑jk=1 aj,k = 1. We say a summability matrix A is a weighted mean matrix if its entries satisfy:aj,k = λk/Λj , 1 ≤ k ≤ j; Λj =j∑i=1λi, λi ≥ 0, λ1 > 0.Hardy’s inequality (1.1) now motivates one to determine the lp operator norm of an arbitrarysummability matrix A. For examples, the following two inequalities were claimed to hold by Bennett( [1, p. 40-41]; see also [2, p. 407]):∞∑n=1∣∣∣ 1nαn∑i=1(iα − (i− 1)α)ai∣∣∣p ≤ ( αpαp− 1)p ∞∑n=1|an|p,(1.2)∞∑n=1∣∣∣ 1∑ni=1 iα−1n∑i=1iα−1ai∣∣∣p ≤ ( αpαp− 1)p ∞∑n=1|an|p,(1.3)Date: January 3, 2007.2000 Mathematics Subject Classification. Primary 26D15.Key words and phrases. Hardy’s inequality.12 PENG GAOwhenever α > 0, p > 1, αp > 1.No proofs of the above two inequalities were supplied in [1]-[2] and recently, the author [4] andBennett himself [3] proved inequalities (1.2) for p > 1, α ≥ 1, αp > 1 and (1.3) for p > 1, α ≥ 2 or0 < α ≤ 1, αp > 1 independently.We point out here that Bennett in fact was able to prove (1.2) for p ≥ 1, α > 0, αp > 1 (see [3,Theorem 1] with β = 1 there) which now leaves the case p > 1, 1 < α < 2 of inequality (1.3) theonly case open to us. For this, Bennett expects inequality (1.3) to hold for 1 + 1/p < α < 2 (seepage 830 of [3]) and as a support, Bennett [3, Theorem 18] has shown that inequality (1.3) holdsfor α = 1 + 1/p, p ≥ 1.In this paper, we will study inequality (1.3) using a method of Knopp [6] which involves con-structing auxiliary sequences. We will partially resolve the remaining case p > 1, 1 < α < 2 ofinequality (1.3) by proving in Section 2 the following:Theorem 1.1. Inequality (1.3) holds for p ≥ 2, 1 ≤ α ≤ 1 + 1/p or 1 < p ≤ 4/3, 1 + 1/p ≤ α ≤ 2.We shall leave the explanation of Knopp’s approach in detail in Section 2 by pointing out herethat it can be applied to prove other types of inequalities similar to that of Hardy’s. As an example,we note that Theorem 359 of [5] states:Theorem 1.2. For 0 < p < 1 and an ≥ 0,∞∑n=1( 1n∞∑k=nak)p ≥ pp ∞∑n=1apn.The constant pp in Theorem 1.2 is not best possible and this was fixed by Levin and Stecˇkin [7,Theorem 61] for 0 < p ≤ 1/3 in the followingTheorem 1.3. For 0 < p ≤ 1/3 and an ≥ 0,∞∑n=1( 1n∞∑k=nak)p ≥ ( p1− p)p ∞∑n=1apn.We shall give another proof of this result in Section 3 using Knopp’s approach. We point outhere for each 1/3 < p < 1, Levin and Stecˇkin also gave a better constant than the one pp given inTheorem 1.2. For example, when p = 1/2, they gave√3/2 instead of 1/√2. In Section 4, we shallfurther improve this constant by proving the followingTheorem 1.4. For an ≥ 0,∞∑n=1( 1n∞∑k=nak)1/2 ≥ √7/3 ∞∑n=1a1/2n .In our proofs of Theorems 1.1-1.2, certain auxiliary sequences are constructed and there canbe many ways to construct such sequences. In Section 5, we give an example regarding thesepossibilities by answering a question of Bennett.2. Proof of Theorem 1.1We begin this section by explaining Knopp’s idea [6] on proving Hardy’s inequality (1.1). Infact, we will explain this more generally for the case involving weighted mean matrices. For realnumbers λ1 > 0, λi ≥ 0, i ≥ 2, we write Λn =∑ni=1 λi and we are looking for a positive constant Usuch that(2.1)∞∑n=1∣∣∣ 1Λnn∑k=1λkak∣∣∣p ≤ U ∞∑k=1|ak|pHARDY-TYPE INEQUALITIES VIA AUXILIARY SEQUENCES 3holds for all complex sequences a with p > 1 being fixed. Knopp’s idea is to find an auxiliarysequence w = {wi}∞i=1 of positive terms such that by Ho¨lder’s inequality,( n∑k=1λk|ak|)p=( n∑k=1λk|ak|w− 1p∗k · w1p∗k)p≤( n∑k=1λpk|ak|pw−(p−1)k)( n∑j=1wj)p−1so that∞∑n=1∣∣∣ 1Λnn∑k=1λkak∣∣∣p ≤ ∞∑n=11Λpn( n∑k=1λpk|ak|pw−(p−1)k)( n∑j=1wj)p−1=∞∑k=1w−(p−1)k λpk( ∞∑n=k1Λpn( n∑j=1wj)p−1)|ak|p.Suppose now one can find for each p > 1 a positive constant U , a sequence w of positive termswith wp−1n /λpn decreasing to 0, such that for any integer n ≥ 1,(2.2) (w1 + · · ·+ wn)p−1 < UΛpn(wp−1nλpn− wp−1n+1λpn+1),then it is easy to see that inequality (2.1) follows from this. When λn = 1 for all n, Knopp’s choicefor w is given by wn =(n−1−1/pn−1)and one can show that (2.2) holds in this case with U = (p∗)pand Hardy’s inequality (1.1) follows from this.We now want to apply Knopp’s approach to prove Theorem 1.1. For this, we replace α − 1 byα and rewrite (1.3) as∞∑n=1∣∣∣ 1∑ni=1 iαn∑i=1iαai∣∣∣p ≤ ( (α+ 1)p(α+ 1)p− 1)p ∞∑n=1|an|p.Note that we are interested in the case 0 ≤ α ≤ 1 here. From our discussions above, we are lookingfor a sequence w of positive terms with wp−1n /λpn decreasing to 0, such that for any integer n ≥ 1,(2.3) (w1 + · · ·+ wn)p−1 <( (α+ 1)p(α+ 1)p− 1)p( n∑i=1iα)p(wp−1nnαp− wp−1n+1(n+ 1)αp).Following Knopp’s choice, we define a sequence w such that(2.4) wn+1 =n+ α− 1/pnwn, n ≥ 1.Note that the above sequence is uniquely determined for any given positive w1 and therefore wemay assume w1 = 1 here. We note further that we need α > −1/p∗ in order for wn > 0 for all nand we also point out that it is easy to show by induction that(2.5)n∑i=1wi =n+ α− 1/p1 + α− 1/pwn.Moreover, one can easily check thatwp−1nnαp= O(n−α−1/p∗),so that wp−1n /λpn decreases to 0 as n approaches infinity as long as α > −1/p∗.Now we need a lemma on sums of powers, which is due to Levin and Stecˇkin [7, Lemma 1, 2,p.18]:4 PENG GAOLemma 2.1. For an integer n ≥ 1,n∑i=1ir ≥ 1r + 1n(n+ 1)r, 0 ≤ r ≤ 1,(2.6)n∑i=1ir ≥ rr + 1nr(n+ 1)r(n+ 1)r − nr , r ≥ 1.(2.7)Inequality (2.7) reverses when −1 < r ≤ 1.We note here only the case r ≥ 0 for (2.7) was proved in [7] but one checks easily that the proofextends to the case r > −1.As we are interested in 0 ≤ α ≤ 1 here, we can now combine (2.4)-(2.6) to deduce that inequality(2.3) will follow from(1 +α− 1/pn)p−1 <n1 + α− 1/p((1 +1n)αp − (1 + α− 1/pn)p−1).We can simplify the above inequality further by recasting it as(2.8)(1 +α+ 1/p∗n)1/p(1 +α− 1/pn)1/p∗<(1 +1n)α.Now we define for fixed n ≥ 1, p > 1,f(x) = x ln(1 + 1/n)− 1pln(1 +x+ 1/p∗n)− 1p∗ln(1 +x− 1/pn).It is easy to see here that inequality (2.8) is equivalent to f(α) > 0. It is also easy to see that f(x)is a convex function of x for 0 ≤ x ≤ 1 and that f(1/p) = 0. It follows from this that if f ′(1/p) ≤ 0then f(x) > 0 for 0 ≤ x < 1/p and if f ′(1/p) ≥ 0 then f(x) > 0 for 1/p < x ≤ 1. We havef ′(1/p) = ln(1 + 1/n)− 1n+1pn(n+ 1).We now use Taylor expansion to conclude for x > 0,(2.9) x− x2/2 < ln(1 + x) < x− x2/2 + x3/3.It follows from this that for p ≥ 2, n ≥ 2,f ′(1/p) < − 12n2+13n3+1pn(n+ 1)≤ − 12n2+13n3+12n(n+ 1)=13n3− 12n2(n+ 1)≤ 0.and for n = 1,f ′(1/p) = ln 2− 1 + 12p≤ ln 2− 1 + 14< 0,It’s also easy to check that for 1 < p ≤ 4/3, n = 1,f ′(1/p) = ln 2− 1 + 12p> 0.For n ≥ 2, 1 < p ≤ 4/3, by using the first inequality of (2.9) we getf ′(1/p) > − 12n2+1pn(n+ 1)≥ 0.This now enables us to conclude the proof of Theorem 1.1.HARDY-TYPE INEQUALITIES VIA AUXILIARY SEQUENCES 53. Another Proof of Theorem 1.3We use the idea of Levin and Stecˇkin in the proof of Theorem 62 in [7] to find an auxiliarysequence w = {wi}∞i=1 of positive terms so that for any finite summation from n = 1 to N withN ≥ 1, we haveN∑n=1apn =N∑n=1apn∑ni=1win∑k=1wk =N∑n=1wnN∑k=napk∑ki=1wi.On letting N →∞, we then have∞∑n=1apn =∞∑n=1wn∞∑k=napk∑ki=1wi.By Ho¨lder’s inequality, we have∞∑k=napk∑ki=1wi≤( ∞∑k=n( k∑i=1wi)−1/(1−p))1−p( ∞∑k=nak)p.Suppose now one can find a sequence w of positive terms with w−1/(1−p)n n−p/(1−p) decreasing to 0for each 0 < p ≤ 1/3, such that for any integer n ≥ 1,(3.1) (w1 + · · ·+ wn)−1/(1−p) ≤(1− pp)p/(1−p)(w−1/(1−p)nnp/(1−p)− w−1/(1−p)n+1(n+ 1)p/(1−p)),then it is easy to see that Theorem 1.3 follows from this.We now define our sequence w to be(3.2) wn+1 =n+ 1/p− 2nwn, n ≥ 1.Note that the above sequence is uniquely determined for any given positive w1 and therefore wemay assume w1 = 1 here. We note further that wn > 0 for all n as 0 < p ≤ 1/3 and it is easy toshow by induction that(3.3)n∑i=1wi =n+ 1/p− 21/p− 1 wn.Moreover, one can easily check thatw−1/(1−p)nnp/(1−p)= O(n−(1−p)/p),so that w−1/(p−1)n n−p/(1−p) decreases to 0 as n approaches infinity.We now combine (3.2)-(3.3) to recast inequality (3.1) as(n+ 1/p− 2)−1/(1−p) ≤ p1− p(n−p/(1−p) − (n+ 1)−p/(1−p)n1/(1−p)(n+ 1/p− 2)−1/(1−p)).We further rewrite the above inequality as1− pp≤ n−p/(1−p)(n+ 1/p− 2)1/(1−p) − (n+ 1)−p/(1−p)n1/(1−p)= n((1 +1/p− 2n)1/(1−p) − (1 + 1n)−p/(1−p)).It is easy to see that the above inequality follows from f(1/n) ≥ 0 where we define for x ≥ 0,f(x) =(1 + (1/p− 2)x)1/(1−p) − (1 + x)−p/(1−p) − 1− ppx.6 PENG GAOWe now prove that f(x) ≥ 0 for x ≥ 0 for 0 < p ≤ 1/3 and this will conclude the proof of Theorem1.3. We note thatf ′(x) =1/p− 21− p(1 + (1/p− 2)x)p/(1−p)+p1− p(1 + x)−p/(1−p)−1 − 1− pp,f ′′(x) =p(1/p− 2)2(1− p)2(1 + (1/p− 2)x)p/(1−p)−1 − p(1− p)2(1 + x)−p/(1−p)−2.We now define for x ≥ 0,g(x) = (1/p− 2)2(1−p)/(1−2p)(1 + x)(2−p)/(1−2p) − (1 + (1/p− 2)x).It is easy to see that g(x) ≥ 0 implies f ′′(x) ≥ 0. Note that (2− p)/(1− 2p) ≥ 1 so thatg′(x) = (1/p− 2)2(1−p)/(1−2p)(2− p)/(1− 2p)(1 + x)(2−p)/(1−2p)−1 − (1/p− 2)≥ (1/p− 2)2(1−p)/(1−2p) − (1/p− 2) ≥ 0,where the last inequality above follows from 2(1 − p)/(1 − 2p) ≥ 1 and 0 < p ≤ 1/3 so that1/p − 2 ≥ 1. It follows from this that f ′′(x) ≥ 0 and as one checks easily that f ′(0) = 0, whichimplies f ′(x) ≥ 0 so that f(x) ≥ f(0) = 0 which is just what we want to prove.4. Proof of Theorem 1.4We follow our strategy in the previous section to look for a sequence w of positive terms withw−2n n−1 decreasing to 0, such that for any integer n ≥ 1,(4.1) (w1 + · · ·+ wn)−2 ≤ C(w−2nn− w−2n+1(n+ 1)),then it is easy to see that Theorem 1.4 with√7/3 replaced by C−1/2 follows from this.We now define our sequence w to be(4.2) wn+1 =n+ αnwn, n ≥ 1.Here α > −1/2 is our parameter so that we hope to optimize the constant C in (4.1) later bychoosing a suitable α. Similar to our treatment in the previous section, we let w1 = 1 here andnote that wn > 0 for all n and that(4.3)n∑i=1wi =n+ α1 + αwn.Moreover, one can easily check thatw−2nn= O(n−1−2α),so that w−2n n−1 decreases to 0 as n approaches infinity.We now combine (4.2)-(4.3) to recast inequality (4.1) as(1 + α)2C≤ (n+ α)2n− n2n+ 1.Or equivalently,(4.4) (1 + 2α)n2 + (2α+ α2)n+ α2 ≥ (1 + α)2Cn(n+ 1).We point out here that the choice α = 0 will lead to C ≥ 2 in order for inequality (4.4) to hold forall n ≥ 1, which corresponds Theorem 1.2 with √7/3 replaced by 2−1/2. The choice α = 1 will leadto C ≥ 4/3 in order for inequality (4.4) to hold for all n ≥ 1, which corresponds Theorem 1.2 withHARDY-TYPE INEQUALITIES VIA AUXILIARY SEQUENCES 7√7/3 replaced by√3/2. We now choose α = 1/2 here with C = 9/7 and one checks readily thatinequality (4.4) holds for such choices and this leads to the desired constant√7/3 in Theorem 1.2.5. Another look at Inequality (1.3)In this section we return to the consideration of inequality (1.3) via our approach in Section 2,which boils down to a construction of a sequence w of positive terms with wp−1n /λpn decreasing to0, such that for any integer n ≥ 1, inequality (2.3) is satisfied. Certainly here the choice for w maynot be unique and in fact in the case α = 0, Bennett asked in [1] (see the paragraph below Lemma4.11) for other sequences, not multiples of Knopp’s, that satisfy (2.3). He also mentioned that theobvious choice, wn = n−1/p, does not work.We point out here even though the choice wn = n−1/p does not satisfy (2.3) when α = 0 for allp > 1, as one can see by considering inequality (2.3) for the case n = 1 with p→ 1+, it neverthelessworks for p ≥ 3, which we now show by first rewriting (2.3) in our case as(5.1)( n∑i=1i−1/p)p−1<( pp− 1)pnp(n−(p−1)/p − (n+ 1)−(p−1)/p).We note that the case n = 1 of (5.1) follows from the case α = 0 of the following inequality,(5.2) 1− 2−(p−1)/p−α >(1− 1(α+ 1)p)p, 0 ≤ α ≤ 1/p.To show (5.2), we see by Taylor expansion, that for p ≥ 2, x < 0,(1 + x)p < 1 + px+p(p− 1)x22.Apply the above inequality with x = −1/(αp+ p), we obtain for p ≥ 3,(1− 1(α+ 1)p)p< 1− 1(α+ 1)+(p− 1)2(α+ 1)2p.Hence inequality (5.2) will follow from1− p− 12(α+ 1)p− 2−(p−1)/p (α+ 1)2α> 0.It is easy to see that when p ≥ 3, the function α 7→ (1 + α)2−α is an increasing function of α for0 ≤ α ≤ 1/p. It follows from this that for 0 ≤ α ≤ 1/p,1− p− 12(α+ 1)p− 2−(p−1)/p (α+ 1)2α> 1− p− 12p− 2−(p−1)/p (1/p+ 1)21/p= 0,and from which inequality (5.2) follows.Now, to show (5.1) holds for all n ≥ 2, p ≥ 3, we first note that for p > 1,n∑i=1i−1/p < 1 +∫ n1x−1/pdx =pp− 1n1−1/p − 1p− 1 .On the other hand, by Hadamard’s inequality, which asserts that for a continuous convex functionf(x) on [a, b],f(a+ b2) ≤ 1b− a∫ baf(x)dx ≤ f(a) + f(b)2,we have for p > 1,n−(p−1)/p − (n+ 1)−(p−1)/p = p− 1p∫ n+1nx−1−1/p∗dx ≥ p− 1p(n+ 1/2)−1−1/p∗.8 PENG GAOHence inequality (5.1) will follow from the following inequality for n ≥ 2,pp− 1n1−1/p − 1p− 1 ≤ p∗n1/p∗(1 +12n)−(1+p∗)/p.It is easy to see that for p > 1, (1 +12n)−(1+p∗)/p ≥ 1− 1 + p∗p12n.Hence it suffices to showpp− 1n1−1/p − 1p− 1 ≤ p∗n1/p∗(1− 1 + p∗p12n),or equivalently, (1 +12p− 2)p ≤ n.It’s easy to check that the right-hand expression above is a decreasing function of p ≥ 3 and isequal to 53/43 < 2 when p = 3. Hence it follows that (5.1) holds for all n ≥ 2, p ≥ 3.We consider lastly inequality (2.3) for other values of α and we take wn = nα−1/p for n ≥ 1 sothat we can rewrite (2.3) as(5.3)( n∑i=1iα−1/p)p−1<( (α+ 1)p(α+ 1)p− 1)p( n∑i=1iα)p(n−(p−1)/p−α − (n+ 1)−(p−1)/p−α).We end our discussion here by considering the case 1 ≤ α ≤ 1+1/p and we apply Lemma 2.1 toobtainn∑i=1iα−1/p ≤ α− 1/pα− 1/p+ 1nα−1/p(n+ 1)α−1/p(n+ 1)α−1/p − nα−1/p =1α− 1/p+ 1(∫ n+1nx−α+1/p−1dx)−1,n∑i=1iα ≥ αα+ 1nα(n+ 1)α(n+ 1)α − nα =1α+ 1(∫ n+1nx−α−1dx)−1We further writen−(p−1)/p−α − (n+ 1)−(p−1)/p−α = (α− 1/p+ 1)∫ n+1nx−α+1/p−2dx,so that inequality (5.3) will follow from∫ n+1nx−α−1dx <(∫ n+1nx−α+1/p−1dx)1−1/p(∫ n+1nx−α+1/p−2dx)1/p.One can easily see that the above inequality holds by Ho¨lder’s inequality and it follows that in-equality (5.3) holds for p > 1, 1 ≤ α ≤ 1 + 1/p. This provides another proof of inequality (1.3) forp > 1, 1 ≤ α ≤ 1 + 1/p.References[1] G. Bennett, Factorizing the classical inequalities, Mem. Amer. Math. Soc., 120 (1996), 1–130.[2] G. Bennett, Inequalities complimentary to Hardy, Quart. J. Math. Oxford Ser. (2), 49 (1998), 395–432.[3] G. Bennett, Sums of powers and the meaning of lp, Houston J. Math., 32 (2006), 801-831.[4] P. Gao, A note on Hardy-type inequalities, Proc. Amer. Math. Soc., 133 (2005), 1977-1984.[5] G. H. Hardy, J. E. Littlewood and G. Po´lya, Inequalities, Cambridge Univ. Press, 1952.[6] K. Knopp, U¨ber Reihen mit positiven Gliedern, J. London Math. Soc., 3 1928, 205-211 and 5 1930, 13-21.[7] V. I. Levin and S.B. Stecˇkin, Inequalities, Amer. Math. Soc. Transl. (2), 14 (1960), 1–29.Department of Computer and Mathematical Sciences, University of Toronto at Scarborough, 1265Military Trail, Toronto Ontario, Canada M1C 1A4E-mail address: penggao@utsc.utoronto.ca

A note on Hardy-type inequalities,

Factorizing the classical inequalities,

Inequalities complimentary to Hardy,

Inequalities,

Sums of powers and the meaning of l p,

Uber Reihen mit positiven

Hardy-Type Inequalities via Auxiliary Sequences

We prove some Hardy-type inequalities via an approach that involves constructing\ud
auxiliary sequences

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AbstractWe prove some Hardy-type inequalities via an approach that involves constructing auxiliary sequences

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Hardy-type inequalities via auxiliary sequences 

Hardy-type Inequalities Via Auxiliary Sequences

Hardy-type Inequalities Via Auxiliary Sequences

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