On a Result of Levin and Stečkin

Abstract

The following inequality for 0<𝑝<1 and 𝑎_𝑛≥0 originates from a study of Hardy, Littlewood, and Pólya: ∑^∞𝑛=1((1/𝑛)∑^∞𝑘=𝑛𝑎_𝑘)^𝑝≥𝑐_𝑝∑^∞𝑛=1𝑎^𝑝𝑛. Levin and Stečkin proved the previous inequality with the best constant 𝑐_𝑝=(𝑝/(1−𝑝))^𝑝 for 0<𝑝≤1/3. In this paper, we extend the result of Levin and Stečkin to 0<𝑝≤0.346.

1. Introduction

Let 𝑝>1, and 𝑙^𝑝 be the Banach space of all complex sequences 𝐚=(𝑎_𝑛)_𝑛≥1. The celebrated Hardy's inequality [1, Theorem 326] asserts that for 𝑝>1 and any 𝐚∈𝑙^𝑝^∞_𝑛=1|||||1𝑛^𝑛_𝑘=1𝑎_𝑘|||||^𝑝≤𝑝𝑝−1^𝑝∞_𝑘=1||𝑎_𝑘||^𝑝.(1.1)

As an analogue of Hardy’s inequality, Theorem 345 of [1] asserts that the following inequality holds for 0<𝑝<1 and 𝑎_𝑛≥0 with 𝑐_𝑝=𝑝^𝑝:^∞_𝑛=11𝑛^∞_𝑘=𝑛𝑎_𝑘^𝑝≥𝑐_𝑝∞_𝑛=1𝑎^𝑝𝑛.(1.2) It is noted in [1] that the constant 𝑐_𝑝=𝑝^𝑝 may not be best possible, and a better constant was indeed obtained by Levin and Stečkin [2, Theorem 61]. Their result is more general as they proved, among other things, the following inequality [2, Theorem 62], valid for 0<𝑟≤𝑝≤1/3 or 1/3<𝑝<1,𝑟≤(1−𝑝)²/(1+𝑝) with 𝑎_𝑛≥0:^∞_𝑛=11𝑛^𝑟^∞_𝑘=𝑛𝑎_𝑘^𝑝≥𝑝1−𝑟^𝑝∞_𝑛=1𝑎^𝑝𝑛𝑛^𝑟−𝑝.(1.3) We note here that the constant (𝑝/(1−𝑟))^𝑝 is best possible, as shown in [2] by setting 𝑎_𝑛=𝑛^{−1−(1−𝑟)/𝑝−𝜖} and letting 𝜖→0⁺. This implies inequality (1.2) for 0<𝑝≤1/3 with the best possible constant 𝑐_𝑝=(𝑝/(1−𝑝))^𝑝. On the other hand, it is also easy to see that inequality (1.2) fails to hold with 𝑐_𝑝=(𝑝/(1−𝑝))^𝑝 for 𝑝≥1/2. The point is that in these cases 𝑝/(1−𝑝)≥1 so one can easily construct counterexamples.

A simpler proof of Levin and Stečkin's result (for 0<𝑟=𝑝≤1/3) is given in [3]. It is also pointed out there that, using a different approach, one may be able to extend their result to 𝑝 slightly larger than 1/3; an example is given for 𝑝=0.34. The calculation however is more involved, and therefore it is desirable to have a simpler approach. For this, we let 𝑞 be the number defined by 1/𝑝+1/𝑞=1 and note that by the duality principle (see [4, Lemma 2], but note that our situation is slightly different since we have 0<𝑝<1 with an reversed inequality), the case 0<𝑟<1,0<𝑝<1 of inequality (1.3) is equivalent to the following one for 𝑎_𝑛>0:^∞_𝑛=1𝑛^{𝑛(𝑟−𝑝)/𝑝}_𝑘=1𝑎_𝑘𝑘^𝑟/𝑝^𝑞≤𝑝1−𝑟^𝑞∞_𝑛=1𝑎^𝑞𝑛.(1.4) The above inequality can be regarded as an analogue of a result of Knopp [5, 6], which asserts that Hardy's inequality (1.1) is still valid for 𝑝<0 if we assume 𝑎_𝑛>0. We may also regard inequality (1.4) as an inequality concerning the factorable matrix with entries 𝑛^{(𝑟−𝑝)/𝑝}𝑘^−𝑟/𝑝 for 𝑘≤𝑛 and 0 otherwise. Here we recall that a matrix 𝐴=(𝑎_𝑛𝑘) is factorable if it is a lower triangular matrix with 𝑎_𝑛𝑘=𝑎_𝑛𝑏_𝑘 for 1≤𝑘≤𝑛. We note that the approach in [7] for the 𝑙^𝑝 norms of weighted mean matrices can also be easily adopted to treat the 𝑙^𝑝 norms of factorable matrices, and it is our goal in this paper to use this similar approach to extend the result of Levin and Stečkin. Our main result is the following.

Theorem 1. Inequality (1.2) holds with the best possible constant 𝑐_𝑝=(𝑝/(1−𝑝))^𝑝 for any 1/3<𝑝<1/2 satisfying 2^{𝑝/(1−𝑝)}1−𝑝𝑝^1/(1−𝑝)−1−𝑝𝑝−1+3−21/𝑝^1/(1−𝑝)≥0.(1.5) In particular, inequality (1.2) holds for 0<𝑝≤0.346.

It readily follows from Theorem 1 and our discussions above that we have the following dual version of Theorem 1.

Corollary 1. Inequality (1.4) holds with 𝑟=𝑝 for any 1/3<𝑝<1/2 satisfying (1.5) and the constant is best possible. In particular, inequality (1.4) holds with 𝑟=𝑝 for 0<𝑝≤0.346.

An alternative proof of Theorem 1 is given in Section 3, via an approach using the duality principle. In Section 4, we will study some inequalities which can be regarded as generalizations of (1.2). Motivations for considerations for such inequalities come both from their integral analogues as well as from their counterparts in the 𝑙^𝑝 spaces. As an example, we consider the following inequality for 0<𝑝<1,0<𝛼<1/𝑝:^∞_𝑛=1^∞_𝑘=𝑛𝛼𝑘^𝛼−1𝑛^𝛼𝑎_𝑘^𝑝≥𝛼𝑝1−𝛼𝑝^𝑝∞_𝑛=1𝑎^𝑝𝑛.(1.6) As in the case of (1.2), the above inequality does not hold for all 0<𝑝<1,0<𝛼<1/𝑝. In Section 4, we will however prove a result concerning the validity of (1.6) that can be regarded as an analogue to that of Levin and Stečkin's concerning the validity of (1.2).

Inequality (1.6) is motivated partially by integral analogues of (1.2), as we will explain in Section 4. It is also motivated by the following inequality for 𝑝>1,𝛼𝑝>1, 𝑎_𝑛≥0:^∞_𝑛=1^𝑛_𝑘=1𝛼𝑘^𝛼−1𝑛^𝛼𝑎_𝑘^𝑝≤𝛼𝑝𝛼𝑝−1^𝑝∞_𝑛=1𝑎^𝑝𝑛.(1.7) The above inequality is in turn motivated by the following inequality:^∞_𝑛=1^𝑛_𝑘=1𝑘^𝛼−1∑^𝑛𝑖=1𝑖^𝛼−1𝑎_𝑘^𝑝≤𝛼𝑝𝛼𝑝−1^𝑝∞_𝑛=1𝑎^𝑝𝑛.(1.8) Inequality (1.8) was first suggested by Bennett [8, pages 40-41]; see [9] and the references therein for recent progress on this. We point out here that it is easy to see that inequality (1.7) implies (1.8) when 𝛼>1; hence, it is interesting to know that, for which values of𝛼's, inequality (1.7) is valid. We first note that, on setting 𝑎₁=1 and 𝑎_𝑛=0 for 𝑛≥2 in (1.7) that it is impossible for it to hold when 𝛼 is large for fixed 𝑝. On the other hand, when 𝛼=1, inequality (1.7) becomes Hardy's inequality, and hence one may expect it to hold for 𝛼 close to 1, and we will establish such a result in Section 5.

2. Proof of Theorem 1

First we need a lemma.

Lemma 1. The following inequality holds for 0≤𝑦≤1 and 1/2<𝑡<1: 1+2𝑦𝑡^1+𝑡−(1+𝑦)^−𝑡1+(2𝑡−2𝑡1)𝑦^1+𝑡−𝑦𝑡≥0.(2.1)

Proof. We set 𝑥=𝑦/2𝑡 so that 0≤𝑥≤1, and we recast the above inequality as 𝑓(𝑥,𝑡)∶=(1+𝑥)^1+𝑡−(1+2𝑡𝑥)^−𝑡(1+(2𝑡−1)𝑥)^1+𝑡−2𝑥≥0.(2.2) Direct calculation shows that 𝑓(0,𝑡)=(𝜕𝑓/𝜕𝑥)(0,𝑡)=0 and 𝜕²𝑓𝜕𝑥²(𝑥,𝑡)=∶=𝑡(𝑡1(1++𝑡)𝑡()1(1++𝑥𝑥))^𝑡−1_𝑡−11−(1+2𝑡𝑥)^−𝑡−2(1+(2𝑡−1)𝑥)^𝑡−1(1+𝑥)^1−𝑡𝑔(𝑥,𝑡).(2.3) Note that 𝜕𝑔𝜕𝑥(𝑥,𝑡)=(1+2𝑡𝑥)^−𝑡−3(1+(2𝑡−1)𝑥)^𝑡−2(1+𝑥)^−𝑡×2(4𝑡−1)+4𝑡(4𝑡−1)𝑥+2𝑡(2𝑡−1)(𝑡+2)𝑥²≥0.(2.4) As 𝑔(0,𝑡)=0, it follows that 𝑔(𝑥,𝑡)≥0 for 0≤𝑥≤1 which in turn implies the assertion of the lemma.

We now describe a general approach towards establishing inequality (1.3) for 0<𝑟<1,0<𝑝<1. A modification from the approach in Section 3 of [3] shows that, in order for (1.3) to hold for any given 𝑝 with 𝑐_𝑝,𝑟(=(𝑝/(1−𝑟))^𝑝), it suffices to find a sequence 𝐰 of positive terms for each 0<𝑟<1 and 0<𝑝<1, such that for any integer 𝑛≥1𝑛^{(𝑝−𝑟)/(1−𝑝)}𝑤₁+⋯+𝑤_𝑛^{−1/(1−𝑝)}≤𝑐^{−1/(1−𝑝)𝑝,𝑟}𝑤𝑛^{𝑛−1/(1−𝑝)𝑟/(1−𝑝)}−𝑤^{−1/(1−𝑝)𝑛+1}(𝑛+1)^{𝑟/(1−𝑝)}.(2.5) We note here that if we study the equivalent inequality (1.4) instead, then we can also obtain the above inequality from inequality (2.2) of [3], on setting Λ_𝑛=𝑛^{−(𝑟−𝑝)/𝑝},𝜆_𝑛=𝑛^−𝑟/𝑝 there. For the moment, we assume that 𝑐_𝑝,𝑟 is an arbitrary fixed positive number, and, on setting 𝑏^𝑛𝑝−1=𝑤_𝑛/𝑤_𝑛+1, we can recast the above inequality as ^𝑛_𝑛𝑘=1_𝑖=𝑘𝑏^𝑖𝑝−1^{−1/(1−𝑝)}≤𝑐^{−1/(1−𝑝)𝑝,𝑟}𝑛^{(𝑟−𝑝)/(1−𝑝)}𝑛𝑏^{𝑛𝑟/(1−𝑝)}−1(𝑛+1)^{𝑟/(1−𝑝)}.(2.6) The choice of 𝑏_𝑛 in Section 3 of [3] suggests that, for optimal choices of the 𝑏_𝑛's, we may have asymptotically 𝑏_𝑛∼1+𝑐/𝑛 as 𝑛→+∞ for some positive constant 𝑐 (depending on 𝑝). This observation implies that 𝑛^1/(1−𝑝) times the right-hand side expression above should be asymptotically a constant. To take the advantage of possible contributions of higher-order terms, we now further recast the above inequality as𝑛+1𝑎^𝑛_𝑛𝑘=1_𝑖=𝑘𝑏^𝑖𝑝−1^{−1/(1−𝑝)}≤𝑐^{−1/(1−𝑝)𝑝,𝑟}𝑛^{(𝑟−𝑝)/(1−𝑝)}(𝑛+𝑎)^1/(1−𝑝)𝑛𝑏^{𝑛𝑟/(1−𝑝)}−1(𝑛+1)^{𝑟/(1−𝑝)},(2.7) where 𝑎 is a constant (may depend on 𝑝) to be chosen later. It will also be clear from our arguments below that the choice of 𝑎 will not affect the asymptotic behavior of 𝑏_𝑛 to the first order of magnitude. We now choose 𝑏_𝑛 so that𝑛^{(𝑟−𝑝)/(1−𝑝)}(𝑛+𝑎)^1/(1−𝑝)𝑛𝑏^{𝑛𝑟/(1−𝑝)}−1(𝑛+1)^{𝑟/(1−𝑝)}=𝑐^{−𝛼/(1−𝑝)𝑝,𝑟},(2.8) where 𝛼 is a parameter to be chosen later. This implies that 𝑏_𝑛=𝑐^{−𝛼/(1−𝑝)𝑝,𝑟}𝑛^{𝑝/(1−𝑝)}(𝑛+𝑎)^1/(1−𝑝)+𝑛^{𝑟/(1−𝑝)}(𝑛+1)^{𝑟/(1−𝑝)}.(2.9) For the so-chosen 𝑏_𝑛's, inequality (2.7) becomes^𝑛_𝑛𝑘=1_𝑖=𝑘𝑏^𝑖𝑝−1≥(𝑛+𝑎)𝑐^{1+𝛼𝑝,𝑟}.(2.10) We first assume that the above inequality holds for 𝑛=1. Then induction shows that it holds for all 𝑛 as long as 𝑏^𝑛1−𝑝≤𝑛+𝑎+𝑐^{−(1+𝛼)𝑝,𝑟}−1𝑛+𝑎.(2.11) Taking into account the value of 𝑏_𝑛, the above becomes (for 0≤𝑦≤1 with 𝑦=1/𝑛)1+𝑎+𝑐^{−(1+𝛼)𝑝,𝑟}−1𝑦^1/(1−𝑝)−(1+𝑦)^{−𝑟/(1−𝑝)}(1+𝑎𝑦)^1/(1−𝑝)−𝑐^{−𝛼/(1−𝑝)𝑝,𝑟}𝑦≥0.(2.12) The first-order term of the Taylor expansion of the left-hand side expression above implies that it is necessary to have 𝑐^{−(1+𝛼)𝑝,𝑟}−(1−𝑝)𝑐^{−𝛼/(1−𝑝)𝑝,𝑟}+𝑟−1≥0.(2.13) For fixed 𝑐_𝑝,𝑟, the left-hand side expression above is maximized when 𝛼=1/𝑝−1 with value 𝑝𝑐^{−1/𝑝𝑝,𝑟}+𝑟−1. This suggests for us to take 𝑐_𝑝,𝑟=(𝑝/(1−𝑟))^𝑝. From now on we fix 𝑐_𝑝,𝑟=(𝑝/(1−𝑟))^𝑝 and note that in this case (2.12) becomes1+𝑎+1−𝑝𝑟−1𝑦^1/(1−𝑝)−(1+𝑦)^{−𝑟/(1−𝑝)}(1+𝑎𝑦)^1/(1−𝑝)−1−𝑟𝑝𝑦≥0.(2.14)

We note that the choice of 𝑎=0 in (2.14) with 𝑟=𝑝 reduces to that considered in Section 3 of [3] (in which case the case 𝑛=1 of (2.10) is also included in (2.14)). Moreover, with 𝑎=0 in the above inequality and following the treatment in Section 3 of [3], one is able to improve some cases of the previously mentioned result of Levin and Stečkin concerning the validity of (1.3). We will postpone the discussion of this to the next section and focus now on the proof of Theorem 1. Since the cases 0<𝑝≤1/3 of the assertion of the theorem are known, we may assume 1/3<𝑝<1/2 from now on. In this case we set 𝑟=𝑝 in (2.14), and Taylor expansion shows that it is necessary to have 𝑎≥(3−1/𝑝)/2 in order for inequality (2.14) to hold. We now take 𝑎=(3−1/𝑝)/2 and write 𝑡=𝑝/(1−𝑝) to see that inequality (2.14) is reduced to (2.1) and Lemma 1 now implies that inequality (2.14) holds in this case. Inequality (1.2) with the best possible constant 𝑐_𝑝=(𝑝/(1−𝑝))^𝑝 thus follows for any 1/3<𝑝<1/2 as long as the case 𝑛=1 of (2.10) is satisfied, which is just inequality (1.5), and this proves the first assertion of Theorem 1.

For the second assertion, we note that inequality (1.5) can be rewritten as2^𝑡𝑡𝑡^−𝑡−1≥(1+𝑎)^1/(1−𝑝),(2.15) where 𝑡 is defined as above. Note that 1/2<𝑡<1 for 1/3<𝑝<1/2 and both 2^𝑡/𝑡 and 𝑡^−𝑡−1 are decreasing functions of 𝑡. It follows that the left-hand side expression of (2.15) is a decreasing function of 𝑝. Note also that for fixed 𝑎, the right-hand side expression of (2.15) is an increasing function of 𝑝<1. As 𝑎=(3−1/𝑝)/2 in our case, it follows that one just needs to check the above inequality for 𝑝=0.346 and the assertion of the theorem now follows easily.

We remark here that, in the proof of Theorem 1, instead of choosing 𝑏_𝑛 to satisfy (2.8) (with 𝑟=𝑝 and 𝑐_𝑝,𝑝=(𝑝/(1−𝑝))^𝑝 there), we can choose 𝑏_𝑛 for 𝑛≥2 so that (𝑛+𝑐)^1/(1−𝑝)𝑛1^{𝑝/(1−𝑝)}−1(𝑛+1)^{𝑝/(1−𝑝)}𝑏_𝑛=(1−𝑝𝑝).(2.16) Moreover, note that we can also rewrite (2.7) for 𝑛≥2 as (with 𝑎 replaced by 𝑐 and 𝑟=𝑝, 𝑐_𝑝,𝑝=(𝑝/(1−𝑝))^𝑝) 𝑛+1𝑐^𝑛−1_{𝑘=1𝑛−1𝑖=𝑘}𝑏^𝑖𝑝−1+1^{−1/(1−𝑝)}≤1−𝑝𝑝^{𝑝/(1−𝑝)}(𝑛+𝑐)^1/(1−𝑝)𝑛1^{𝑝/(1−𝑝)}−1(𝑛+1)^{𝑝/(1−𝑝)}𝑏_𝑛.(2.17) If we further choose 𝑏₁ so that 1=1−𝑝𝑝^{𝑝/(1−𝑝)}1−21^{𝑝/(1−𝑝)}𝑏₁,(2.18) then, repeating the same process as in the proof of Theorem 1, we find that the induction part (with 𝑐=(1/𝑝−1)/2 here) leads back to inequality (2.14) (with 𝑟=𝑝 and 𝑎=(3−1/𝑝)/2 there) while the initial case (corresponding to 𝑛=2 here) is just (2.15), so this approach gives another proof of Theorem 1.

We end this section by pointing out the relation between the treatment in Sections 3 and 4 in [3] on inequality (1.2). We note that it is shown in Section 3 of [3] that, for any 𝑁≥1 and any positive sequence 𝐰, we have^𝑁_𝑛=1𝑎^𝑝𝑛≤^𝑁_𝑛=1𝑤_𝑛⎛⎜⎜⎝^𝑁_𝑘=𝑛^𝑘_𝑖=1𝑤_𝑖^{−1/(1−𝑝)}⎞⎟⎟⎠^1−𝑝^𝑁_𝑘=𝑛𝑎_𝑘^𝑝.(2.19) We now use 𝑤_𝑛=∑^𝑛𝑘=1𝑤_𝑘−∑^{𝑛−1𝑘=1}𝑤_𝑘 and set (with 𝜈_𝑁=0) 𝜈_𝑛=∑^{𝑁𝑘=𝑛+1}∑^𝑘𝑖=1𝑤_𝑖^1/(𝑝−1)∑^𝑛𝑖=1𝑤_𝑖^1/(𝑝−1)(2.20) to see that inequality (2.19) leads to (with 𝜈₀=0) ^𝑁_𝑛=1𝑎^𝑝𝑛≤^𝑁_𝑛=11+𝜈^𝑛^1−𝑝−𝜈^{1−𝑝𝑛−1}^𝑁_𝑘=𝑛𝑎_𝑘^𝑝.(2.21) The above inequality is essentially what is used in Section 4 of [3].

3. An Alternative Proof of Theorem 1

In this section we give an alternative proof of Theorem 1, using the following.

Lemma 2 (see, Lemma 2.4 [10]). Let {𝜆_𝑖}^∞𝑖≥1,{𝑎_𝑖}^∞𝑖≥1 be two sequences of positive real numbers, and let 𝑆_𝑛=∑^𝑛𝑖=1𝜆_𝑖𝑎_𝑖. Let 0≠𝑝<1 be fixed and let {𝜇_i}^∞𝑖≥1,{𝜂_𝑖}^∞𝑖≥1 be two positive sequences of real numbers such that 𝜇_𝑖≤𝜂_𝑖 for 0<𝑝<1 and 𝜇_𝑖≥𝜂_𝑖 for 𝑝<0, then for 𝑛≥2^𝑛−1_𝑖=2𝜇^𝑖−𝜇^𝑞𝑖+1−𝜂^𝑞𝑖+1^1/𝑞𝑆^𝑖1/𝑝+𝜇_𝑛𝑆^𝑛1/𝑝≤𝜇^𝑞2−𝜂^𝑞2^1/𝑞𝜆^11/𝑝𝑎^11/𝑝+^𝑛_𝑖=2𝜂_𝑖𝜆^𝑖1/𝑝𝑎^𝑖1/𝑝.(3.1)

Following the treatment in Section 4 of [3], on first setting 𝜂_𝑖=𝜆^{𝑖−1/𝑝}, then a change of variables: 𝜇_𝑖↦𝜇_𝑖𝜂_𝑖 followed by setting 𝜇^𝑞𝑖−1=𝜈_𝑖 and lastly a further change of variable: 𝑝↦1/𝑝, we can transform inequality (3.1) to the following inequality (with 𝜈₁=0 here): ^𝑛−1_𝑖=11+𝜈^𝑖^1−𝑝𝜆^𝑝𝑖−𝜈^{1−𝑝𝑖+1}𝜆^𝑝𝑖+1𝑆^𝑝𝑖+1+𝜈^𝑛^1−𝑝𝜆^𝑝𝑛𝑆^𝑝𝑛≤^𝑛_𝑖=1𝑎^𝑝𝑖.(3.2) Here the 𝜈_𝑖's are arbitrary nonnegative real numbers for 2≤𝑖≤𝑛. On setting 𝜈_𝑛+1 to be any non-negative real number, we deduce immediately from the above inequality the following:^𝑛_𝑖=11+𝜈^𝑖^1−𝑝𝜆^𝑝𝑖−𝜈^{1−𝑝𝑖+1}𝜆^𝑝𝑖+1𝑆^𝑝𝑖≤^𝑛_𝑖=1𝑎^𝑝𝑖.(3.3)

Now we consider establishing inequality (1.3) for 0<𝑟<1,1/3<𝑝<1/2 in general, and, as has been pointed out in Section 1, we know this is equivalent to establishing inequality (1.4). Now, in order to establish inequality (1.4), it suffices to consider the cases of (1.4) with the infinite summations replaced by any finite summations, say from 1 to 𝑁≥1 there. We now set 𝑛=𝑁,𝑝=𝑞,𝜆_𝑖=𝑖^−𝑟/𝑝 in inequality (3.3) to recast it as (with 𝜈₁=0, 𝑆_𝑛=∑^𝑛𝑘=1𝑘^−𝑟/𝑝𝑎_𝑘 here) ^𝑁_𝑛=11+𝜈^𝑛^1/(1−𝑝)𝑛^{𝑟/(1−𝑝)}−𝜈^{1/(1−𝑝)𝑛+1}(𝑛+1)^{𝑟/(1−𝑝)}𝑆^𝑞𝑛≤^𝑁_𝑛=1𝑎^𝑞𝑛.(3.4) Comparing the above inequality with (1.4), we see that inequality (1.4) holds as long as we can find non-negative 𝜈_𝑛's (with 𝜈₁=0) such that1+𝜈_𝑛^1/(1−𝑝)𝑛^{𝑟/(1−𝑝)}−𝜈^{1/(1−𝑝)𝑛+1}(𝑛+1)^{𝑟/(1−𝑝)}≥𝑛^{(𝑝−𝑟)/(1−𝑝)}𝑝1−𝑟^{𝑝/(1−𝑝)}.(3.5) Now, on setting for 𝑛≥2, 𝜈_𝑛=𝑛+𝑎−1(1−𝑟)/𝑝,(3.6) and 𝑦=1/𝑛, we see easily that inequality (3.5) can be transformed into inequality (2.14). In the case of 𝑟=𝑝, we further set 𝑎=(3−1/𝑝)/2 to see that the validity of (2.14) established for this case in Section 2 ensures the validity of (3.5) for 𝑛≥2. Moreover, with the above chosen 𝜈₂ with 𝑟=𝑝 and 𝑎=(3−1/𝑝)/2, the 𝑛=1 case of (3.5) is easily seen to be equivalent to inequality (1.5), and hence this provides an alternative proof of Theorem 1.

4. A Generalization of Theorem 1

Let 0<𝑝<1,𝛼<1/𝑝, and let 𝑓(𝑥) be a non-negative function. We note the following identity:^∞0𝑥1^𝛼^∞𝑥𝑓(𝑡)𝑡^𝛼−1𝑑𝑡^𝑝𝑑𝑥=𝑝1−𝛼𝑝^∞0𝑥1^𝛼^∞𝑥𝑓(𝑡)𝑡^𝛼−1𝑑𝑡^𝑝−1𝑓(𝑥)𝑑𝑥.(4.1) In the above expression, we assume, 𝑓 is taken so that all the integrals converge. The case of 𝛼=1 is given in the proof of Theorem 337 of [1], and the general case is obtained by some changes of variables. As in the proof of Theorem 337 of [1], we then deduce the following inequality (with the same assumptions as above): ^∞0𝑥1^𝛼^∞𝑥𝑓(𝑡)𝑡^𝛼−1𝑑𝑡^𝑝𝑑𝑥≥𝑝1−𝛼𝑝^𝑝^∞0𝑓^𝑝(𝑥)𝑑𝑥.(4.2) The above inequality can also be deduced from Theorem 347 of [1] (see also [11, equation (2.4)]). Following the way how Theorem 338 is deduced from Theorem 337 of [1], we deduce similarly from (4.1) the following inequality for 0<𝑝<1,0<𝛼<1/𝑝, and 𝑎_𝑛≥0: ^∞_𝑛=1𝑛1^𝛼∞_𝑘=𝑛((𝑘+1)^𝛼−𝑘^𝛼)𝑎_𝑘^𝑝≥𝛼𝑝1−𝛼𝑝^∞_𝑛=1𝑛1^𝛼∞_𝑘=𝑛((𝑘+1)^𝛼−𝑘^𝛼)𝑎_𝑘^𝑝−1𝑎_𝑛.(4.3) The dash over the summation on the left-hand side expression above (and in what follows) means that the term corresponding to 𝑛=1 is to be multiplied by 1+1/(1−𝛼𝑝). It's easy to see here the constant is best possible (on taking 𝑎_𝑛=𝑛^{−1/𝑝−𝜖} and letting 𝜖→0⁺). By Hölder's inequality, the above inequality readily implies the following inequality: ^∞_𝑛=1𝑛1^𝛼∞_𝑘=𝑛((𝑘+1)^𝛼−𝑘^𝛼)𝑎_𝑘^𝑝≥𝛼𝑝1−𝛼𝑝^𝑝∞_𝑛=1𝑎^𝑝𝑛.(4.4)

We are thus motivated to consider the above inequality with the dash sign removed, and this can be regarded as an analogue of inequality (1.2) with 𝑐_𝑝=(𝑝/(1−𝑝))^𝑝, which corresponds to the case 𝛼=1 here. As in the case of (1.2), such an inequality does not hold for all 𝛼 and 𝑝 satisfying 0<𝑝<1 and 0<𝛼<1/𝑝. However, on setting 𝑎_𝑛=𝑛^{−1/𝑝−𝜖} and letting 𝜖→0⁺, one sees easily that if such an inequality holds for certain 𝛼 and 𝑝, then the constant is best possible. More generally, we can consider the following inequality:^∞_𝑛=1∑^𝑛𝑖=1𝐿1^𝛽𝛼−1(𝑖,𝑖−1)^∞_𝑘=𝑛𝐿^𝛽𝛼−1(𝑘±1,𝑘)𝑎_𝑘^𝑝≥𝛼𝑝1−𝛼𝑝^𝑝∞_𝑛=1𝑎^𝑝𝑛,(4.5) where the function 𝐿_𝑟(𝑎,𝑏) for 𝑎>0,𝑏>0,𝑎≠𝑏, and 𝑟≠0,1 (the only case we will concern here) is defined as 𝐿^𝑟𝑟−1(𝑎,𝑏)=(𝑎^𝑟−𝑏^𝑟)/(𝑟(𝑎−𝑏)). It is known [12, Lemma 2.1] that the function 𝑟↦𝐿_𝑟(𝑎,𝑏) is strictly increasing on ℝ. Here we restrict our attention to the plus sign in (4.5) for the case 𝛽>0,max(1,𝛽)≤𝛼 and to the minus sign in (4.5) for the case 0<𝛼<1 and 𝛽≥𝛼. Our remark above implies that in either case (note that 𝐿_𝛽(1,0) is meaningful) ^𝑛_𝑖=1𝐿^𝛽𝛼−1(𝑖,𝑖−1)≤^𝑛_𝑖=1𝐿^𝛼𝛼−1(𝑖,𝑖−1)=𝑛^𝛼𝛼.(4.6) As we also have 𝐿^𝛽𝛼−1(𝑘±1,𝑘)≥𝑘^𝛼−1, we see that the validity of (4.5) follows from that of (1.6). We therefore focus on (1.6) from now on, and we proceed as in Section 3 of [3] to see that in order for inequality (1.6) to hold, it suffices to find a sequence 𝐰 of positive terms for each 0<𝑝<1, such that for any integer 𝑛≥1^𝑛_𝑘=1𝑤_𝑘^1/(𝑝−1)≤𝛼𝑝1−𝛼𝑝^{𝑝/(𝑝−1)}𝛼𝑛^𝛼−1^{𝑝/(1−𝑝)}𝑛𝑤^{𝑛1/(𝑝−1)𝛼𝑝/(1−𝑝)}−𝑤^{1/(𝑝−1)𝑛+1}(𝑛+1)^{𝛼𝑝/(1−𝑝)}.(4.7) We now choose 𝐰 inductively by setting 𝑤₁=1, and for 𝑛≥1𝑤_𝑛+1=𝑛+1/𝑝−𝛼−1𝑛𝑤^𝑛.(4.8) The above relation implies that ^𝑛_𝑘=1𝑤_𝑘=𝑛+1/𝑝−𝛼−11/𝑝−𝛼𝑤^𝑛.(4.9) We now assume 0<𝑝<1/2 and note that, for the so-chosen 𝐰, inequality (4.7) follows (with 𝑥=1/𝑛) from 𝑓(𝑥)≥0 for 0≤𝑥≤1, where𝑓(𝑥)=1+1𝑝−𝛼−1𝑥^1/(1−𝑝)−(1+𝑥)^{−𝛼𝑝/(1−𝑝)}−1−𝛼𝑝𝑝𝑥.(4.10) As 𝑓(0)=𝑓^(0)=0, it suffices to show 𝑓^(𝑥)≥0, which is equivalent to showing 𝑔(𝑥)≥0, where 𝑔(𝑥)=(1/𝑝−𝛼−1)²𝛼((𝛼−1)𝑝+1)^{(1−𝑝)/(1−2𝑝)}(1+𝑥)^{(2+(𝛼−2)𝑝)/(1−2𝑝)}−1+1𝑝−𝛼−1𝑥.(4.11) Now𝑔^(𝑥)=(1/𝑝−𝛼−1)²𝛼((𝛼−1)𝑝+1)^{(1−𝑝)/(1−2𝑝)}2+1(𝛼−−2𝑝2)𝑝(1+𝑥)^{(2+(𝛼−2)𝑝)/(1−2𝑝)−1}−1≥(1/𝑝−𝛼−1)₂𝑝−𝛼−1𝛼((𝛼−1)𝑝+1)^{(1−𝑝)/(1−2𝑝)}2+1(𝛼−−2𝑝2)𝑝−1𝑝−𝛼−1∶=ℎ(𝛼,𝑝).(4.12) Suppose now 𝛼≥1, then, when 1/𝑝≥(𝛼+2)(𝛼+1)/2, we have 1/𝑝≥𝛼(𝛼−1)𝑝+2𝛼+1 since 𝑝<1/2 so that both inequalities 1/𝑝−𝛼−1≥1 and 1/𝑝−𝛼−1≥𝛼((𝛼−1)𝑝+1) are satisfied. In this case we have ℎ(𝛼,𝑝)≥(1/𝑝−𝛼−1)²𝛼((𝛼−1)𝑝+1)^{(1−𝑝)/(1−2𝑝)}−1𝑝−𝛼−1≥𝛼((1(/𝛼𝑝−−1𝛼)−𝑝+1)²1)−1𝑝−𝛼−1≥0.(4.13) It follows that 𝑔′(𝑥)≥0 and as 𝑔(0)≥0, and we conclude that 𝑔(𝑥)≥0, and hence 𝑓(𝑥)≥0. Similar discussion leads to the same conclusion for 0<𝛼<1 when 𝑝≤1/(𝛼+2). We now summarize our discussions above in the following.

Theorem 2. Let 0<𝑝<1/2 and 0<𝛼<1/𝑝. Let ℎ(𝛼,𝑝) be defined as in (4.12). Inequality (1.6) holds for 𝛼,𝑝 satisfying ℎ(𝛼,𝑝)≥0. In particular, when 𝛼≥1, inequality (1.6) holds for 0<𝑝≤2/((𝛼+2)(𝛼+1)). When 0<𝛼≤1, inequality (1.6) holds for 0<𝑝≤1/(𝛼+2).

Corollary 2. Let 0<𝑝<1/2 and 0<𝛼<1/𝑝. Let ℎ(𝛼,𝑝) be defined as in (4.12). When 𝛽>0,max(1,𝛽)≤𝛼, inequality (4.5) holds (where one takes the plus sign) for 𝛼,𝑝 satisfying ℎ(𝛼,𝑝)≥0. In particular, inequality (4.5) holds for 0<𝑝≤2/((𝛼+2)(𝛼+1)). When 0<𝛼<1,𝛽≥𝛼, inequality (4.5) holds (where one takes the minus sign) for 𝛼,𝑝 satisfying ℎ(𝛼,𝑝)≥0. In particular, inequality (4.5) holds for 0<𝑝≤1/(𝛼+2).

We note here a special case of the above corollary: the case 0<𝛼<1 and 𝛽→+∞ leads to the following inequality, valid for 0<𝑝≤1/(𝛼+2): ^∞_𝑛=1∑1^𝑛𝑖=1𝑖^∞𝛼−1_𝑘=𝑛𝑘^𝛼−1𝑎_𝑘^𝑝≥𝛼𝑝1−𝛼𝑝^𝑝∞_𝑛=1𝑎^𝑝𝑛.(4.14)

We further note here that if we set 𝑟=𝛼𝑝 and 𝑎=0 in inequality (2.14), then it is reduced to 𝑓(𝑥)≥0 for 𝑓(𝑥) defined as in (4.10). Since the case 0<𝑟<𝑝≤1/3 is known, we need only to be concerned about the case 𝛼≥1 here and we now have the following improvement of the result of Levin and Stečkin [2, Theorem 62].

Corollary 3. Let 0<𝑝<1/2 and 1≤𝛼<1/𝑝. Let ℎ(𝛼,𝑝) be defined as in (4.12). Inequality (1.3) holds for 𝑟=𝛼𝑝 for 𝛼,𝑝 satisfying ℎ(𝛼,𝑝)≥0. In particular, inequality (1.3) holds for 𝑟=𝛼𝑝 for 𝛼,𝑝 satisfying 0<𝑝≤2/((𝛼+2)(𝛼+1)).

Just as Theorem 1 and Corollary 1 are dual versions to each other, our results above can also be stated in terms of their dual versions, and we will leave the formulation of the corresponding ones to the reader.

5. Some Results on 𝑙^𝑝 Norms of Factorable Matrices

In this section we first state some results concerning the 𝑙^𝑝 norms of factorable matrices. In order to compare our result to that of weighted mean matrices, we consider the following type of inequalities:^∞_𝑛=1^𝑛_𝑘=1𝜆_𝑘Λ_𝑛𝑎_𝑘^𝑝≤𝑈_𝑝∞_𝑛=1𝑎^𝑝𝑛,(5.1) where 𝑝>1,𝑈_𝑝 is a constant depending on 𝑝. Here we assume that the two positive sequences (𝜆_𝑛) and (Λ_𝑛) are independent (in particular, unlike in the weighted mean matrices case, we do not have Λ_𝑛=∑^𝑛𝑘=1𝜆_𝑘 in general). We begin with the following result concerning the bound for 𝑈_𝑝.

Theorem 3. Let 1<𝑝<∞ be fixed in (5.1). Let 𝑎 be a constant such that Λ_𝑛+𝑎𝜆_𝑛>0 for all 𝑛≥1. Let 0<𝐿<𝑝 be a positive constant, and let 𝑏_𝑛=𝑝−𝐿𝑝1+𝑎Λ𝜆^𝑛_𝑛^𝑝−1𝜆_𝑛Λ_𝑛+𝜆^𝑛𝜆_𝑛+1.(5.2) If, for any integer 𝑛≥1, one has ^𝑛_𝑘=1𝜆_𝑘𝑛_𝑖=𝑘𝑏^{𝑖1/(𝑝−1)}≤𝑝𝑝−𝐿Λ^𝑛+𝑎𝜆_𝑛,(5.3) then inequality (5.1) holds with 𝑈_𝑝≤(𝑝/(𝑝−𝐿))^𝑝.

We point out that the proof of the above theorem is analogue to that of Theorem 3.1 of [7], except that, instead of choosing 𝑏_𝑛 to satisfy the equation (3.4) in [7], we choose 𝑏_𝑛 so that 𝜆𝑏^𝑛_𝑛−1𝜆_𝑛+1Λ^𝑝𝑛=𝑝−𝐿𝑝Λ^𝑛+𝑎𝜆_𝑛^𝑝−1.(5.4) We will leave the details to the reader, and we point out that, as in the case of weighted mean matrices in [7], we deduce from Theorem 3 the following.

Corollary 4. Let 1<𝑝<∞ be fixed in (5.1). Let 𝑎 be a constant such that Λ_𝑛+𝑎𝜆_𝑛>0 for all 𝑛≥1. Let 0<𝐿<𝑝 be a positive constant such that the following inequality is satisfied for all 𝑛≥1 (with Λ₀=𝜆₀=0): 𝑝−𝑝𝐿1+𝑎Λ𝜆^𝑛_𝑛^𝑝−1+Λ^𝑛𝜆_𝑛+1≤Λ^𝑛𝜆_𝑛1+𝑎Λ𝜆^𝑛_𝑛^𝑝−11−𝐿𝑝Λ𝜆^𝑛_𝑛+Λ^𝑛−1Λ_𝑛+𝑎𝜆^𝑛−1Λ_𝑛^1−𝑝.(5.5) Then inequality (5.1) holds with 𝑈_𝑝≤(𝑝/(𝑝−𝐿))^𝑝.

We now apply the previous corollary to the special case of (5.1) with 𝜆_𝑛=𝛼𝑛^𝛼−1,Λ_𝑛=𝑛^𝛼 for some 𝛼>1. On taking 𝐿=1/𝛼 and 𝑎=0 in Corollary 4 and setting 𝑦=1/𝑛, we see that inequality (1.7) holds as long as we can show for 0≤𝑦≤11−𝑝1𝛼𝛼𝑦+(1−𝑦)^𝛼^𝑝−11−𝑝1𝛼𝛼𝑦+(1+𝑦)^1−𝛼≤1.(5.6) We note first that, as (1−1/𝑝𝛼)𝛼𝑦+(1−𝑦)^𝛼≤(1−1/𝑝𝛼)𝛼𝑦+(1+𝑦)^1−𝛼, we need to have (1−1/𝑝𝛼)𝛼𝑦+(1−𝑦)^𝛼≤1 in order for the above inequality to hold. Taking 𝑦=1 shows that it is necessary to have 𝛼≤1+1/𝑝. In particular, we may assume 1<𝛼≤2 from now on, and it then follows from Taylor expansion that, in order for (5.6) to hold, it suffices to show that1−1𝑝𝑦+𝛼(𝛼2−1)𝑦²^𝑝−11+1−1𝑝𝑦+𝛼(𝛼2−1)𝑦²≤1.(5.7) We first assume 1<𝑝≤2, and in this case we use 1−1𝑝𝑦+𝛼(𝛼2−1)𝑦²^𝑝−1≤1+(𝑝−1)−1𝑝𝑦+𝛼(𝛼2−1)𝑦²(5.8) to see that (5.7) follows from ℎ_1,𝛼,𝑝(𝑦)∶=𝛼(𝛼−1)𝑝2−1−1𝑝²+𝛼(𝛼−1)(𝑝−1)2𝑝(𝑝−2)𝑦+𝛼²(𝛼−1)²4(𝑝−1)𝑦²≤0.(5.9) We now denote 𝛼₁(𝑝)>1 as the unique number satisfying ℎ_1,𝛼1,𝑝(0)=0 and 𝛼₂(𝑝)>1 the unique number satisfying ℎ_1,𝛼2,𝑝(1)=0 and let 𝛼₀(𝑝)=min(𝛼₁(𝑝),𝛼₂(𝑝)). It is easy to see that both 𝛼₁(𝑝) and 𝛼₂(𝑝) are ≤1+1/𝑝 and that, for 1<𝛼≤𝛼₀, we have ℎ_1,𝛼,𝑝(𝑦)≤0 for 0≤𝑦≤1.

Now suppose that 𝑝>2, then we recast (5.7) as1+1−1𝑝𝑦+𝛼(𝛼2−1)𝑦²≤1−1𝑝𝑦+𝛼(𝛼2−1)𝑦²^1−𝑝.(5.10) In order for the above inequality to hold for all 0≤𝑦≤1, we must have 𝛼(𝛼−1)𝑦²/2≤𝑦/𝑝. Therefore, we may from now on assume 𝛼(𝛼−1)≤2/𝑝. Applying Taylor expansion again, we see that (5.10) follows from the following inequality: 1+1−1𝑝𝑦+𝛼(𝛼2−1)𝑦²≤1+(1−𝑝)−1+𝑝(𝑝−1)−(1𝑝/𝑝𝑦)+𝑦+𝛼((𝛼𝛼2(−𝛼1−)𝑦1²)/2)𝑦₂²2.(5.11) We can recast the above inequality as ℎ_2,𝛼,𝑝(𝑦)∶=𝛼(𝛼−1)𝑝2−1−21/𝑝+(𝑝−1)𝛼2(𝛼−1)𝑦−𝑝(𝑝−1)𝛼²(𝛼−1)²𝑦²8≤0.(5.12) We now denote 𝛼₀(𝑝)>1 as the unique number satisfying 𝛼(𝛼−1)≤2/𝑝 and ℎ_2,𝛼0,𝑝(1)=0. It is easy to see that, for 1<𝛼≤𝛼₀, we have ℎ_2,𝛼,𝑝(𝑦)≤0 for 0≤𝑦≤1. We now summarize our result in the following.

Theorem 4. Let 𝑝>1 be fixed, and let 𝛼₀(𝑝) be defined as above, then inequality (1.7) holds for 1<𝛼≤𝛼₀(𝑝).

As we have explained in Section 1, the study of (1.7) is motivated by (1.8). As (1.7) implies (1.8) and the constant (𝛼𝑝/(𝛼𝑝−1))^𝑝 there is best possible (see [9]), we see that the constant (𝛼𝑝/(𝛼𝑝−1))^𝑝 in (1.7) is also best possible. More generally, we note that inequality (4.7) in [9] proposes to determine the best possible constant 𝑈_𝑝(𝛼,𝛽) in the following inequality (𝐚∈𝑙^𝑝,𝑝>1,𝛽≥𝛼≥1):^∞_𝑛=1|||||1∑^𝑛𝑘=1𝐿^𝛽𝛼−1(𝑘,𝑘−1)^𝑛_𝑖=1𝐿^𝛽𝛼−1(𝑖,𝑖−1)𝑎_𝑖|||||^𝑝≤𝑈_𝑝(𝛼,𝛽)^∞_𝑛=1||𝑎_𝑛||^𝑝.(5.13) We easily deduce from Theorem 4 the following.

Corollary 5. Keep the notations in the statement of Theorem 4. For fixed 𝑝>1 and 1<𝛼≤𝛼₀(𝑝), inequality (5.13) holds with 𝑈_𝑝(𝛼,𝛽)=(𝛼𝑝/(𝛼𝑝−1))^𝑝 for any 𝛽≥𝛼.

Acknowledgments

During this work, the author was supported by postdoctoral research fellowships at Nanyang Technological University (NTU). The author is also grateful to the referee for his/her helpful comments and suggestions.