A note on l^pnorms of weighted mean matrices

Abstract

We present some results concerning the l^pnorms of weighted mean matrices. These results can be regarded as analogues to a result of Bennett concerning weighted Carleman's inequalities.

2000 Mathematics Subject Classification: Primary 47A30.

1. Introduction

Let throughout that $p\ne 0,\frac{1}{p}+\frac{1}{q}=1$. For p ≥ 1, let l^pbe the Banach space of all complex sequences a = (a _n )_{n ≥ 1}with norm

$${∥\mathsf{\text{a}}∥}_p:={\left(\sum _{n=1}^{\infty }{\left|a_n\right|}^p\right)}^{1/p}<\infty .$$

The celebrated Hardy's inequality [1], Theorem 326] asserts that for p > 1,

$$\sum _{n=1}^{\infty }{\left|\frac{1}{n}\sum _{k=1}^n{a}_k\right|}^p\le {\left(\frac{p}{p-1}\right)}^p\sum _{n=1}^{\infty }{\left|a_n\right|}^p.$$

(1.1)

Hardy's inequality can be regarded as a special case of the following inequality:

$${∥C\cdot a∥}_p^p=\sum _{n=1}^{\infty }{\left|\sum _{k=1}^{\infty }{c}_{n,k}{a}_k\right|}^p\le U\sum _{n=1}^{\infty }{\left|a_n\right|}^p,$$

(1.2)

in which C = (c_n,k) and the parameter p > 1 are assumed fixed, and the estimate is to hold for all complex sequences a ∈ l^p. The l^poperator norm of C is then defined as

$${∥C∥}_{p,p}=\underset{{∥\mathsf{\text{a}}∥}_p=1}{\text{sup}}{∥C\cdot \mathsf{\text{a}}∥}_p.$$

It follows that inequality (1.2) holds for any a ∈ l^pwhen U^1/p≥ ||C||_p,pand fails to hold for some a ∈ l^pwhen U^1/p< ||C||_p,p. Hardy's inequality thus asserts that the Cesáro matrix operator C, given by c_n,k= 1/n, k ≤ n and 0 otherwise, is bounded on l^pand has norm ≤ p/(p - 1). (The norm is in fact p/(p - 1).)

We say a matrix A = (a_n,k) is a lower triangular matrix if a_n,k= 0 for n < k and a lower triangular matrix A is a summability matrix if a_n,k≥ 0 and ${\sum }_{k=1}^n{a}_{n,k}=1$. We say a summability matrix A is a weighted mean matrix if its entries satisfy:

$$a_{n,k}={\lambda }_k/{\Lambda }_n,1\le k\le n;{\Lambda }_n=\sum _{i=1}^n{\lambda }_i,{\lambda }_i\ge 0,{\lambda }_1>0.$$

(1.3)

Hardy's inequality (1.1) now motivates one to determine the l^poperator norm of an arbitrary summability or weighted mean matrix A. Gao [2] proved the following result:

Theorem 1.1. Let 1 < p < ∞ be fixed. Let A be a weighted mean matrix given by (1.3). If for any integer n ≥ 1, there exists a positive constant 0 < L < p such that

$$\frac{{\Lambda }_{n+1}}{{\lambda }_{n+1}}\le \frac{{\Lambda }_n}{{\lambda }_n}{\left(1-\frac{L{\lambda }_n}{p{\Lambda }_n}\right)}^{1-p}+\frac{L}{p},$$

(1.4)

then ||A||_p,p≤ p/(p-L).

It is easy to see that the above result implies the following well-known result of Cartlidge [3] (see also [4], p. 416, Theorem C]):

Theorem 1.2. Let 1 < p < ∞ be fixed. Let A be a weighted mean matrix given by (1.3). If

$$L=\underset{n}{\text{sup}}\left(\frac{{\Lambda }_{n+1}}{{\lambda }_{n+1}}-\frac{{\Lambda }_n}{{\lambda }_n}\right)<p,$$

(1.5)

then ||A||_p,p≤ p/(p-L).

The above result of Cartlidge are very handy to use when determining l^pnorms of certain weighted mean matrices. We refer the readers to the articles [2, 5–7] for more recent developments in this area.

We note here that by a change of variables $a_k\to a_k^{1/p}$ in (1.1) and on letting p → +∞, one obtains the following well-known Carleman's inequality [8], which asserts that for convergent infinite series ∑a _n with non-negative terms, one has

$$\sum _{n=1}^{\infty }{\left(\prod _{k=1}^n{a}_k\right)}^{\frac{1}{n}}\le e\sum _{n=1}^{\infty }{a}_n,$$

with the constant e being best possible.

It is then natural to study the following weighted version of Carleman's inequality:

$$\sum _{n=1}^{\infty }\left(\prod _{k=1}^n{a}_k^{{\lambda }_k/{\Lambda }_n}\right)\le E\sum _{n=1}^{\infty }{a}_n,$$

(1.6)

where the notations are as in (1.3). The task here is to determine the best constant E so that inequality (1.6) holds for any convergent infinite series ∑a _n with non-negative terms. Note that Cartlidge's result (Theorem 1.2) implies that when (1.5) is satisfied, then for any a ∈ l^p, one has

$$\sum _{n=1}^{\infty }{\left|\sum _{k=1}^n\frac{{\lambda }_k{a}_k}{{\Lambda }_n}\right|}^p\le {\left(\frac{p}{p-L}\right)}^p\sum _{n=1}^{\infty }{\left|a_n\right|}^p.$$

(1.7)

Similar to our discussions above, by a change of variables $a_k\to a_k^{1/p}$ in (1.7) and on letting p → +∞, one obtains inequality (1.6) with E = e^Las long as (1.5) is satisfied with p replaced by +∞ there.

Note that (1.5) can be regarded as the case p → 1⁺ of (1.4) while the case p → +∞ of (1.4) suggests the following result:

Corollary 1.1. Suppose that

$$M=\underset{n}{\text{sup}}\frac{{\Lambda }_n}{{\lambda }_n}\text{log}\left(\frac{{\Lambda }_{n+1}/{\lambda }_{n+1}}{{\Lambda }_n/{\lambda }_n}\right)<+\infty ,$$

then inequality (1.6) holds with E = e^M.

In fact, the above corollary is a consequence of the following nice result of Bennett (see the proof of [5, Theorem 13]):

Theorem 1.3. Inequality (1.6) holds with

$$E=\underset{n}{\text{sup}}\frac{{\Lambda }_{n+1}}{{\lambda }_{n+1}}\prod _{k=1}^n{\left(\frac{{\lambda }_k}{{\Lambda }_k}\right)}^{{\lambda }_k/{\Lambda }_n}.$$

It is shown in the proof of Theorem 13 in [5] that Corollary 1.1 follows from the above theorem. It is also easy to see that M ≤ L for L defined by (1.5) so that Corollary 1.1 provides a better result than what one can infer from Cartlidge's result as discussed above.

Note that the bound given in Theorem 1.3 is global in the sense that it involves all the λ _n 's and it implies the local version Corollary 1.1, in which only the terms Λ _n /λ _n and Λ_n+1/λ_n+1are involved. It is then natural to ask whether one can obtain a similar result for the l^pnorms for p > 1 so that it implies the local version Theorem 1.1. It is our goal in this note to present one such result and as our result is motivated by the result of Bennett, we will first study the limiting case p → +∞, namely weighted Carleman's inequalities in the next section before we move on to the l^pcases in Section 3.

2. A discussion on weighted Carleman's inequalities

In this section, we study the weighted Carleman's inequalities. Our goal is to give a different proof of Theorem 1.3 than that given in [5] and discuss some variations of it. It suffices to consider the cases of (1.6) with the infinite summations replaced by any finite summations, say from 1 to N ≥ 1 here. Our starting point is the following result of Pečarić and Stolarsky [[9], (2.4)], which is an outgrowth of Redheffer's approach in [10]:

$$\sum _{n=1}^N{\Lambda }_n\left(b_n-1\right)G_n+{\Lambda }_N{G}_N\le \sum _{n=1}^N{\lambda }_n{a}_n{b}_n^{{\Lambda }_n/{\lambda }_n},$$

(2.1)

where b is any positive sequence and

$$G_n=\prod _{k=1}^n{a}_k^{{\lambda }_k/{\Lambda }_n}.$$

We now discard the last term on the left-hand side of (2.1) and make a change of variables ${\lambda }_n{a}_n{b}_n^{{\Lambda }_n/{\lambda }_n}\mapsto a_n$ to recast inequality (2.1) as

$$\sum _{n=1}^N{\Lambda }_n\left(b_n-1\right)\left(\prod _{k=1}^n{\lambda }_k^{-{\lambda }_k/{\Lambda }_n}\right)\left(\prod _{k=1}^n{b}_k^{-{\Lambda }_k/{\Lambda }_n}\right)G_n\le \sum _{n=1}^N{a}_n.$$

Now, a further change of variables b _n ↦ λ_n+1b _n /λ _n allows us to recast the above inequality as

$$\sum _{n=1}^N{\Lambda }_n\left(\frac{b_n}{{\lambda }_n}-\frac{1}{{\lambda }_{n+1}}\right)\prod _{k=1}^n{b}_k^{-{\Lambda }_k/{\Lambda }_n}{G}_n\le \sum _{n=1}^N{a}_n.$$

(2.2)

If one now chooses b _n = (Λ_n+1/λ_n+1)/(Λ _n /λ _n ) such that Λ _n (b _n /λ _n - 1/λ_n+1) = 1, then one gets immediately the following:

$$\sum _{n=1}^N\frac{{\lambda }_{n+1}}{{\Lambda }_{n+1}}\prod _{k=1}^n{\left(\frac{{\Lambda }_k}{{\lambda }_k}\right)}^{{\lambda }_k/{\Lambda }_n}{G}_n\le \sum _{n=1}^N{a}_n,$$

from which the assertion of Theorem 1.3 can be readily deduced.

Another natural choice for the values of b _n 's is to set ${\prod }_{k=1}^n{b}_k^{-{\Lambda }_k/{\Lambda }_n}=e^{-M}$. From this we see that $b_n=e^{M{\lambda }_n/{\Lambda }_n}$ and substituting these values for b _n 's we obtain via (2.2):

$$\sum _{n=1}^N{\Lambda }_n\left(\frac{e^{M{\lambda }_n/{\Lambda }_n}}{{\lambda }_n}-\frac{1}{{\lambda }_{n+1}}\right)G_n\le e^M\sum _{n=1}^N{a}_n.$$

(2.3)

One checks easily that the above inequality implies Corollary 1.1.

We now consider a third choice for the b _n 's by setting $b_n=e^{\left({\Lambda }_{n+1}/{\lambda }_{n+1}-{\Lambda }_n/{\lambda }_n\right)/\left({\Lambda }_n/{\lambda }_n\right)}$ and it follows from this and (2.2) that

$$\sum _{n=1}^N{\Lambda }_n\left(\frac{e^{\left({\Lambda }_{n+1}/{\lambda }_{n+1}-{\Lambda }_n/{\lambda }_n\right)/\left({\Lambda }_n/{\lambda }_n\right)}}{{\lambda }_n}-\frac{1}{{\lambda }_{n+1}}\right)e^{-{\sum }_{k=1}^n\frac{{\lambda }_k}{{\Lambda }_n}\left(\frac{{\Lambda }_{k+1}}{{\lambda }_{k+1}}-\frac{{\Lambda }_k}{{\lambda }_k}\right)}{G}_n\le \sum _{n=1}^N{a}_n,$$

from which we deduce the following

Corollary 2.1. Suppose that

$$M=\underset{n}{\text{sup}}\sum _{k=1}^n\frac{{\lambda }_k}{{\Lambda }_n}\left(\frac{{\Lambda }_{k+1}}{{\lambda }_{k+1}}-\frac{{\Lambda }_k}{{\lambda }_k}\right)<+\infty ,$$

then inequality (1.6) holds with E = e^M.

We point out here that the above corollary also provides a better result than what one can infer from Cartlidge's result and it also follows from Theorem 1.3.

We note here that the optimal choice for the b _n 's will be to choose them to satisfy

$${\Lambda }_n\left(\frac{b_n}{{\lambda }_n}-\frac{1}{{\lambda }_{n+1}}\right)\prod _{k=1}^n{b}_k^{-{\Lambda }_k/{\Lambda }_n}=e^{-L}.$$

In general, it is difficult to solve for the b _n 's from the above equations. But we can solve b₁ to get b₁ = e^Lλ₁/((e^L- 1)λ₂) and if we set b _n = (Λ_n+1/λ_n+1)/(Λ _n /λ _n ) for n ≥ 2, we can then deduce from (2.2) that

$$e^{-L}{G}_1+\sum _{n=2}^N{\left(\frac{{\Lambda }_2\left(e^L-1\right)}{{\lambda }_1{e}^L}\right)}^{{\lambda }_1/{\Lambda }_N}\frac{{\lambda }_{n+1}}{{\Lambda }_{n+1}}\prod _{k=1}^n{\left(\frac{{\Lambda }_k}{{\lambda }_k}\right)}^{{\lambda }_k/{\Lambda }_n}{G}_n\le \sum _{n=1}^N{a}_n.$$

Note that this gives an improvement upon Theorem 1.3 as long as λ₂/Λ₂ > e^-L. Similarly, one obtains

$$G_1+\sum _{n=2}^N{\left(\frac{{\lambda }_2\left(e^L-1\right)}{{\lambda }_1}\right)}^{{\lambda }_1/{\Lambda }_N}{\Lambda }_n\left(\frac{e^{M{\lambda }_n/{\Lambda }_n}}{{\lambda }_n}-\frac{1}{{\lambda }_{n+1}}\right)G_n\le e^M\sum _{n=1}^N{a}_n.$$

3. The l^pcases

We now return to the discussions on the general l^pcases. Again it suffices to consider the cases of (1.7) with the infinite summations replaced by any finite summations, say from 1 to N ≥ 1 here. We may also assume that a _n ≥ 0 for all n. With the discussions of the previous section in mind, here we seek for an l^pversion of (2.2). Fortunately, this is available by noting that it follows from inequality (4.3) of [7] that

$$\sum _{n=1}^N{\left(\sum _{k=1}^n{w}_k\right)}^{-\left(p-1\right)}\left(\frac{w_n^{p-1}}{{\lambda }_n^p}-\frac{w_{n+1}^{p-1}}{{\lambda }_{n+1}^p}\right){\Lambda }_n^p{A}_n^p\le \sum _{n=1}^N{a}_n^p,$$

(3.1)

where w _n 's are positive parameters and

$$A_n=\frac{{\sum }_{k=1}^n{\lambda }_k{a}_k}{{\Lambda }_n}.$$

By a change of variables $w_n\mapsto {\lambda }_n{w}_n^{1/\left(p-1\right)}$, we can recast inequality (3.1) as

$$\sum _{n=1}^N{\left(\frac{{\sum }_{k=1}^n{\lambda }_k{w}_k^{1/\left(p-1\right)}}{{\Lambda }_n}\right)}^{-\left(p-1\right)}\left(\frac{w_n}{{\lambda }_n}-\frac{w_{n+1}}{{\lambda }_{n+1}}\right){\Lambda }_n{A}_n^p\le \sum _{n=1}^N{a}_n^p,$$

With another change of variables, w _n /w_n+1↦ b _n , we can further recast the above inequality as

$$\sum _{n=1}^N{\left(\frac{{\sum }_{k=1}^n{\lambda }_k{\prod }_{i=k}^n{b}_i^{1/\left(p-1\right)}}{{\Lambda }_n}\right)}^{-\left(p-1\right)}\left(\frac{b_n}{{\lambda }_n}-\frac{1}{{\lambda }_{n+1}}\right){\Lambda }_n{A}_n^p\le \sum _{n=1}^N{a}_n^p.$$

(3.2)

Note that if one makes a change of variables $a_n^p\mapsto a_n$, then inequality (2.2) follows from the above inequality upon letting p → +∞.

One can then deduce Theorem 1.1 by choosing b _n = (1-L λ _n /(p Λ _n ))^-(p-1)in (3.2) (see [2]). It is easy to see that this gives back inequality (2.3) upon letting p → +∞ by a change of variables $a_n^p\mapsto a_n$ and setting L = M. We note here that the b _n 's are so chosen so that the following relations are satisfied:

$$\frac{{\sum }_{k=1}^n{\lambda }_k{\prod }_{i=k}^n{b}_i^{1/\left(p-1\right)}}{{\Lambda }_n}=\frac{p}{p-L}.$$

(3.3)

Now, to get the l^panalogues of Theorem 1.3, we just need to note that in the p → +∞ case, one obtains Corollary 1.1 by setting ${\prod }_{k=1}^n{b}_k^{-{\Lambda }_k/{\Lambda }_n}=e^{-M}$ and the conclusion of Corollary 1.1 follows by requiring that Λ _n (b _n /λ _n - 1/λ_n+1) ≥ 1 for the so chosen b _n 's. If one instead chooses the b _n 's so that the conditions Λ _n (b _n /λ _n - 1/λ_n+1) = 1 are satisfied, then Theorem 1.3 will follow. Now, in the l^pcases, the choice of the b _n 's so that the conditions (3.3) are satisfied implies Theorem 1.1 as (1.4) implies that for the so chosen b _n 's,

$$\left(\frac{b_n}{{\lambda }_n}-\frac{1}{{\lambda }_{n+1}}\right){\Lambda }_n\ge 1-\frac{L}{p}.$$

Thus, in order to obtain an result analogue to Theorem 1.3 for the l^pcases, we are then motivated to take the b _n 's so that the following conditions are satisfied:

$$\left(\frac{b_n}{{\lambda }_n}-\frac{1}{{\lambda }_{n+1}}\right){\Lambda }_n=1-\frac{L}{p}.$$

(3.4)

We then easily deduce the following l^panalogue of Theorem 1.3:

Theorem 3.1. Let 1 < p < ∞ be fixed. Let A be a weighted mean matrix given by (1.3). If for any integer n ≥ 1, there exists a positive constant 0 < L < p such that

$$\sum _{k=1}^n\frac{{\lambda }_k}{{\Lambda }_n}\prod _{i=k}^n{\left(\frac{{\Lambda }_{i+1}/{\lambda }_{i+1}-L/p}{{\Lambda }_i/{\lambda }_i}\right)}^{1/\left(p-1\right)}\le \frac{p}{p-L},$$

then ||A||_p,p≤ p/(p-L).

It is easy to see by induction that Theorem 3.1 implies Theorem 1.1. Of course one should really choose the b _n 's so that the following relations are satisfied:

$${\left(\frac{{\sum }_{k=1}^n{\lambda }_k{\prod }_{i=k}^n{b}_i^{1/\left(p-1\right)}}{{\Lambda }_n}\right)}^{-\left(p-1\right)}\left(\frac{b_n}{{\lambda }_n}-\frac{1}{{\lambda }_{n+1}}\right){\Lambda }_n={\left(\frac{p}{p-L}\right)}^{-p}.$$

In general it is difficult to determine the b _n 's this way but one can certainly solve for b₁ and by choosing other b _n 's so that (3.4) are satisfied, one can obtain a slightly better result than Theorem 3.1, we shall leave the details to the reader.

We note here that the choice of the b _n 's satisfying (3.4) corresponds to the following choice for the a _n 's in Section 4 of [2] (these a _n 's are not to be confused with the a _n 's used in the rest of the article):

$$a_n={\left(\frac{{\Lambda }_{n+1}/{\lambda }_{n+1}-L/p}{{\Lambda }_n/{\lambda }_n}\right)}^{1/\left(p-1\right)}{a}_{n+1},a_1=1.$$

We also note here that in the case of λ _n = L = 1, on choosing b _n 's to satisfy (3.4), we obtain via (3.2) that

$$\sum _{n=1}^{\infty }{\left(\frac{{\sum }_{k=1}^n{\prod }_{i=k}^n{\left(1+\left(1-1/p\right)/i\right)}^{1/\left(p-1\right)}}{n}\right)}^{-\left(p-1\right)}{\left(\frac{1}{n}\sum _{k=1}^n{a}_k\right)}^p\le \left(\frac{p}{p-1}\right)\sum _{n=1}^{\infty }{a}_n^p.$$

This gives an improvement of Hardy's inequality (1.1).