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Journal of Function Spaces and Applications
Volume 2012, Article ID 713534, 19 pages
http://dx.doi.org/10.1155/2012/713534
Research Article

Best Constants between Equivalent Norms in Lorentz Sequence Spaces

1Department of Mathematics, Karlstad University, 65188 Karlstad, Sweden
2Department of Mathematics and Computer Science, Technical University of Civil Engineering Bucharest, 020 396 Bucharest, Romania
3Department of Mathematics, Luleå University of Technology, 97 187 Luleå, Sweden

Received 19 March 2010; Accepted 24 May 2010

Academic Editor: Nicolae Popa

Copyright © 2012 S. Barza et al. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

Abstract

We find the best constants in inequalities relating the standard norm, the dual norm, and the norm 𝑥(𝑝,𝑠)=inf{𝑘𝑥(𝑘)𝑝,𝑠}, where the infimum is taken over all finite representations 𝑥=𝑘𝑥(𝑘) in the classical Lorentz sequence spaces. A crucial point in this analysis is the concept of level sequence, which we introduce and discuss. As an application, we derive the best constant in the triangle inequality for such spaces.

1. Introduction

The study of Lorentz spaces goes back to the work of Lorentz [1, 2] (see also [3, 4] for more recent results concerning functional properties of Lorentz spaces). They play an important role in the theory of Banach function spaces, in particular in the interpolation theory of linear operators.

Let 1<𝑝< and 1𝑠. For a sequence 𝑥=(𝑥𝑛)𝑛, the decreasing rearrangement 𝑥=(𝑥𝑛)𝑛 is obtained by rearranging (|𝑥𝑛|)𝑛 in decreasing order. We recall the definition of Lorentz sequence spaces𝑝,𝑠=𝑥𝑥𝑝,𝑠=𝑛=1𝑥𝑛𝑠𝑛𝑠/𝑝11/𝑠,<(1.1) with the usual modification if 𝑠=. Lorentz proved that 𝑝,𝑠 is a norm if and only if 1𝑠𝑝< and that the space 𝑝,𝑠 is always normable (i.e., there exists a norm which is equivalent to 𝑝,𝑠) when 1<𝑝<𝑠 (for the remaining cases, it is known that 𝑝,𝑠 cannot be endowed with an equivalent norm).

The study of normability for 𝑝<𝑠 was carried out by means of the maximal norm:𝑥𝑝,𝑠=𝑥𝑛𝑛𝑝,𝑠=𝑛=1𝑛𝑠/𝑝11𝑛𝑛𝑖=1𝑥𝑖𝑠1/𝑠(1.2) defined in terms of the discrete Hardy operator. It is well known that 𝑝,𝑠 is always a norm. Moreover, one can prove that 𝑝,𝑠 is equivalent to 𝑝,𝑠 with the following estimates:𝑥𝑝,𝑠𝑥𝑝,𝑠𝑝𝑥𝑝,𝑠,(1.3) see, for example, [5, 6].

As a consequence of the fact that 𝑝,𝑠 is equivalent to a norm, it is easy to see that it is a quasinorm satisfying the triangle inequality uniformly in the numbers of terms; that is, there exists a constant 𝐶𝑝𝑠>0 such that, for every finite collection {𝑥(𝑘)}𝑘=1,2,,𝑁𝑝,𝑠, it yields that𝑁𝑘=1𝑥(𝑘)𝑝,𝑠𝐶𝑁𝑝𝑠𝑘=1𝑥(𝑘)𝑝,𝑠.(1.4)

Also the converse result holds, namely that if (1.4) holds, then 𝑝,𝑠 is equivalent with a norm and an alternative equivalent norm is given by means of the following decomposition norm:𝑥(𝑝,𝑠)=inf𝑁𝑘=1𝑥(𝑘)𝑝,𝑠𝑥=𝑁𝑘=1𝑥(𝑘).(1.5)

One can prove that (𝑝,𝑠) is always a norm and that (𝑝,𝑠) is equivalent with 𝑝,𝑠, 1𝑝,𝑠. If 1𝑠𝑝, then (𝑝,𝑠)=𝑝,𝑠 (see, e.g., [7, 8]). We also remark that the best constant 𝐶𝑝𝑠 in the inequality𝑥𝑝,𝑠𝐶𝑝𝑠𝑥(𝑝,𝑠)(1.6) is the same as the optimal one in (1.4).

It is natural to consider also the following norm defined in terms of Köthe duality, which we will call the dual norm: 𝑥𝑝,𝑠=sup𝑛=1𝑥𝑛𝑦𝑛𝑦𝑝,𝑠.=1(1.7) The dual norm 𝑝,𝑠 is a norm indeed, equivalent to 𝑝,𝑠 and, moreover, if 1𝑠𝑝, then𝑥𝑝,𝑠=𝑥𝑝,𝑠=𝑥(𝑝,𝑠).(1.8)

In the recent papers [9, 10], the authors considered estimates between the dual norm, decomposition norm, and the norm which defines the Lorentz spaces over nonatomic resonant measure spaces. The main reason for these consideration was that the technique based on the norm defined in terms of the maximal function (or Hardy operator) does not give the best constant in the triangle inequality with 𝑛-terms. In [9], the case of the classical Lorenz spaces 𝐿𝑝,𝑠 was considered, while in [10] similar results were proved for the weighted Lorentz space Γ𝑝(𝑤), where 𝑤 is an increasing weight function.

The aim of this paper is to treat similar problems in the context of Lorentz spaces of sequences. In particular, in this paper, we introduce the notion of level sequence, which corresponds to the notion of level function introduced by Halperin in [11] and Lorentz in [12]. We would like to pronounce that our results are not corollaries of the results proved in [9] although some of the techniques are similar (cf. our Example 3.5). One of the main applications of our results is that we obtain the best constant 𝐶𝑝𝑠 in the triangle inequality (1.4). This constant was found by a different approach in [13] where the best constants in 𝑞-convexity and 𝑞-concavity inequalities were found for Lorentz and Marcinkiewicz spaces of functions and sequences.

The paper is organized as follows. In Section 2, we state and discuss several technical lemmas, which will be used in the proofs of our theorems. In Section 3, we introduce and discuss one of the key tools used in this paper, the so-called level sequence. In Theorem 3.7 of Section 3, we prove optimal estimates between the quasinorm in 𝑝,𝑠 of 𝑥=(𝑥𝑛)𝑛 and of its level sequence 𝑥=(𝑥𝑛)𝑛. Section 4 is dedicated to the study of the dual norm 𝑥𝑝,𝑠, 1<𝑝<𝑠, in relation to the other norms. For instance, one of the main results in Section 4 is that the dual norm of 𝑥=(𝑥𝑛)𝑛𝑝,𝑠 coincides with the quasinorm of the level sequence 𝑥𝑝,𝑠 (see Theorem 4.2). Moreover, 𝑥𝑝,𝑠𝐶𝑝𝑠𝑥𝑝,𝑠, where 𝐶𝑝𝑠 is the optimal constant which appears in Theorem 4.4. The main result of Section 5, that is, that the dual norm coincides with the decomposition norm is given in Theorem 5.2. Finally, in the last section, we give some complementary results and remarks, as, for example, Theorem 6.1, which gives us the best constant in the “triangle inequality” for the quasinorm 𝑥𝑝,𝑠. Moreover, we point out that our results can be given in a somewhat more general setting (see Remark 6.3 and Theorem 6.4).

Throughout this paper, given two sequences 𝑥=(𝑥𝑛)𝑛, 𝑦=(𝑦𝑛)𝑛, we denote that 𝑥𝑦 if𝑛𝑖=1𝑥𝑖𝑛𝑖=1𝑦𝑖,𝑛1(1.9) and we use the notation 𝑥𝑦 if𝑥𝑛𝑦𝑛,𝑛1.(1.10) The number 𝑝 stands for the conjugate index of 𝑝; that is, 1/𝑝+1/𝑝=1 and , stand for the sets of nonnegative, respectively, positive integers.

2. Some Lemmas

First, we recall the following well-known majorization lemma (see, e.g., [14, page 9]).

Lemma 2.1. Let 𝑢,𝑣,𝑤 be nonnegative sequences and suppose that 𝑤 is decreasing. If 𝑛𝑘=1𝑢𝑘𝑛𝑘=1𝑣𝑘(𝑛=1,2,),(2.1) then 𝑛𝑘=1𝑢𝑘𝑤𝑘𝑛𝑘=1𝑣𝑘𝑤𝑘(𝑛=1,2,).(2.2)

We also need the following auxiliary statement related to the dual norm.

Lemma 2.2. Let 𝑥=(𝑥𝑛)𝑛𝑝,𝑠, 1<𝑝<, 1𝑠. Then, 𝑥𝑝,𝑠=𝑥𝑝,𝑠.(2.3)

A proof can be found, for example, in [15, pages 45–49].

The next lemmas are crucial in the proofs of some of our main results and are also of independent interest.

Lemma 2.3. Let 𝑥=(𝑥𝑛)𝑛𝑝,s, 1<𝑝<,1𝑠. Then, the following statements hold.(a)The equality 𝑥(𝑝,𝑠)=inf𝑘𝑥(𝑘)𝑝,𝑠(2.4) holds, where the infimum is taken over all finite nonnegative sequences |𝑥𝑛|=𝑘𝑥𝑛(𝑘).(b) If 0𝑦𝑥, then 𝑦(𝑝,𝑠)𝑥(𝑝,𝑠).(c) If 0𝑦(𝑘)𝑥 and 𝑦(𝑘)𝑥 as 𝑘, then 𝑦(𝑘)(𝑝,𝑠)𝑥(𝑝,𝑠) as 𝑘.

Proof. The proof is similar to that of Lemma  2.7 in [9], so we omit the details.

3. Level Sequences

The notion of level function was introduced in the early 1950s by Halperin [11] and Lorentz [12] and generalized more recently by Sinnamon in, for example, [1618] and Mastylo and Sinnamon in [19]. Based on the extension given by Lorentz, the optimal constant in the triangle inequality in Lorentz spaces 𝐿𝑝,𝑠(𝑅,𝜇), where (𝑅,𝜇) is a totally 𝜎-finite nonatomic measure space and 1<𝑝<𝑠, was found.

Inspired by these results, we will use in this paper the concept of level sequence with respect to a given sequence and study Lorentz sequence spaces in this frame.

Definition 3.1. Let 𝜑=(𝜑𝑛)𝑛 be a sequence of positive numbers and Φ𝑛=𝑛𝑖=1𝜑𝑖. A sequence 𝑋𝑋=(𝑛)𝑛 is called 𝜑-concave if, for arbitrary 𝑛1,𝑛2, we have that 𝑋𝑛𝐿𝑛, where 𝐿𝑛=𝐶Φ𝑛+𝐷, 𝐶 and 𝐷 are constants, such that 𝐿 “interpolate” 𝑋 in 𝑛1, 𝑛2; that is, 𝐿𝑛1=𝑋𝑛1,𝐿𝑛2=𝑋𝑛2.(3.1)

Proposition 3.2. The following statements are equivalent: 𝑋𝑋=𝑛𝑛𝑋is𝜑-concave,(3.2)𝑛𝑋𝑛1Φ𝑛Φ𝑛1𝑋𝑛2𝑋𝑛1Φ𝑛2Φ𝑛1,1𝑛1<𝑛𝑛2𝑋,(3.3)𝑛𝑋𝑛1Φ𝑛Φ𝑛1𝑋𝑛2𝑋𝑛Φ𝑛2Φ𝑛,1𝑛1𝑛<𝑛2𝑋,(3.4)𝑛1𝑋𝑚1Φ𝑛1Φ𝑚1𝑋𝑛2𝑋𝑚2Φ𝑛2Φ𝑚2,𝑚1𝑚2,𝑛1𝑛2,𝑚1𝑛1,𝑚2𝑛2.(3.5)

Proof. The proof of this proposition is literally the same as the proof of the equivalence of the corresponding statements in the case of functions, so we omit the details (see, e.g., [12]).

In the next theorem, we prove the existence of the level sequence and derive its properties. Since, as far as we know, this theorem has not been proved earlier in this context, we give the entire proof here.

Theorem 3.3. Let 𝜑=(𝜑𝑛)𝑛 be a positive sequence and Φ𝑛=𝑛𝑖=1𝜑𝑖. Let 𝑥=(𝑥𝑛)𝑛 be a positive sequence and suppose that lim𝑛𝑛𝑖=1𝑥𝑖𝑛𝑖=1𝜑𝑖=0.(3.6)
Then, there exists a unique nonnegative sequence 𝑥=(𝑥𝑛)𝑛 satisfying the following conditions:(a)(𝑥𝑛/𝜑𝑛)𝑛 is decreasing;(b)𝑥𝑥;(c) the set {𝑛𝑥𝑛𝑥𝑛}=𝑘=1𝐼𝑘, where {𝐼𝑘} are disjoint sets of positive integers, 𝐼𝑘={𝑛𝑘,,𝑛𝑘+𝑚𝑘} such that 𝑖𝐼𝑘𝑥𝑖=𝑖𝐼𝑘𝑥𝑖 and 𝑥𝑖/𝜑𝑖=𝜆𝑘, for all 𝑖𝐼𝑘.

Remark 3.4. The unique sequence 𝑥=(𝑥𝑛)𝑛 in Theorem 3.3 is called the level sequence of 𝑥=(𝑥𝑛)𝑛 with respect to 𝜑=(𝜑𝑛)𝑛.

Proof. (a) Let 𝑋𝑛=𝑛𝑖=1𝑥𝑖, 𝑛. Condition (3.6) implies the existence of a sequence 𝑋𝑋=(𝑛)𝑛, which is a 𝜑-concave majorant of 𝑋. We denote 𝑋𝑛𝑋=inf𝑛 and from (3.5) we get that 𝑋𝑘1𝑋𝑘2Φ𝑘1Φ𝑘21,for𝑘1𝑘2,(3.7) that is, 𝑋=(𝑋𝑘)𝑘 is increasing. Thus, there exists a unique positive sequence (𝑥𝑛)𝑛 such that 𝑋𝑛=𝑛𝑖=1𝑥𝑖, with 𝑥1=𝑋1 and 𝑥𝑛=𝑋𝑛𝑋𝑛1.(3.8)
According to (3.5), we have that (𝑋𝑛𝑋𝑛1)/(Φ𝑛Φ𝑛1)(𝑋𝑛+1𝑋𝑛)/(Φ𝑛+1Φ𝑛), which means that (𝑥𝑛/𝜑𝑛)𝑛 is decreasing and this proves (a).
(b) This statement follows directly from (3.8) and the definition of 𝑋𝑛.
(c) Let us first note that two consecutive terms 𝑋𝑘1 and 𝑋𝑘 are equal to 𝑋𝑘1 and 𝑋𝑘, respectively, if and only if 𝑥𝑘=𝑥𝑘, 𝑘1.
Now, we assume that there exists 𝑗 such that 𝑋𝑗>𝑋𝑗 and we denote 𝑛=max{𝑘<𝑗𝑋𝑘=𝑋𝑘and𝑋𝑘1=𝑋𝑘1}+1 and 𝑚=min{𝑘>𝑗𝑋𝑘=𝑋𝑘and𝑋𝑘+1=𝑋𝑘+1}𝑛+1. Now, we take 𝐿 to “interpolate” 𝑋 in 𝑛1 and 𝑛+𝑚1. Then, 𝐿𝑋 and 𝑋𝐿. It follows that 𝐿𝑗=𝑋𝑗=𝐶Φ𝑗+𝐷 for 𝑗{𝑛1,,𝑛+𝑚1}.
Thus, 𝑥𝑗=𝐶𝜑𝑗, 𝑗{𝑛,,𝑛1+𝑚}, 𝑥𝑛+𝑚=𝑥𝑛+𝑚 and 𝑥𝑛1=𝑥𝑛1. Since 𝑋𝑛1=𝑋𝑛1,𝑋𝑛+𝑚1=𝑋𝑛+𝑚1,(3.9) we obtain that 𝑛+𝑚1𝑖=𝑛𝑥𝑖=𝑛+𝑚1𝑖=𝑛𝑥𝑖(3.10) and the proof is complete.

The following example shows that the results for sequences do not follow from the results for function simply by taking step functions:𝑓(𝑡)=𝑛=0𝑥𝑛𝜒[𝑛,𝑛+1)(𝑡).(3.11)

Example 3.5. For 𝑓(𝑡)=𝜒[0,1](𝑡), the level function is 𝑓(𝑡)=(1𝛼)𝑡𝛼𝜒[0,1](𝑡),(3.12) where 𝛼=1𝑠/𝑝 so that 𝑓𝑝,𝑠=𝑠𝑝1/𝑠.(3.13)
If 𝑥=(𝑥𝑛)𝑛, where 𝑥𝑛=1if𝑛𝑘,0if𝑛>𝑘,(3.14) then 𝑥𝑛=𝑘𝑘𝑖=1𝑖𝛼𝑛𝛼if𝑛𝑘,0,otherwise,𝑥𝑝,𝑠=𝑘𝑘𝑛=1𝑛𝛼1/𝑠.(3.15)

We also need the following technical lemma.

Lemma 3.6. Let 1𝑝<𝑠< and 𝐶𝑝𝑠=sup𝑚,𝑛1𝑛𝑚+𝑛1𝑘=𝑚𝑘𝑠/𝑝11/𝑠1𝑛𝑚+𝑛1𝑘=𝑚𝑘𝑠/𝑝11/𝑠.(3.16) Then, 𝐶𝑝𝑠=𝑝𝑠1/𝑠𝑝𝑠1/𝑠.(3.17)

Proof. We prove first that 𝐶𝑝𝑠=sup𝑛1𝑛𝑛𝑘=1𝑘𝑠/𝑝11/𝑠1𝑛𝑛𝑘=1𝑘𝑠/𝑝11/𝑠.(3.18) The inequality 𝐶𝑝𝑠sup𝑛1𝑛𝑛𝑘=1𝑘𝑠/𝑝11/𝑠1𝑛𝑛𝑘=1𝑘𝑠/𝑝11/𝑠(3.19) is trivial. For the converse inequality, it is enough to prove that 𝑛𝑘=1𝑘𝑠/𝑝11/𝑠𝑛𝑘=1𝑘𝑠/𝑝11/𝑠𝑚+𝑛1𝑘=𝑚𝑘𝑠/𝑝11/𝑠𝑚+𝑛1𝑘=𝑚𝑘𝑠/𝑝11/𝑠(3.20) for all 𝑚𝑛.
The above inequality can be obviously written in the form 𝑛𝑘=1𝑘𝑠/𝑝1𝑠/𝑠𝑛𝑘=1𝑘𝑠/𝑝1𝑛𝑘=1(𝑘+𝑚1)𝑠/𝑝1𝑠/𝑠𝑛𝑘=1(𝑘+𝑚1)𝑠/𝑝1.(3.21) Let us fix 𝑚1. Observe that for 𝑚=1 we have equality, so we may suppose that 𝑚>1. If we denote 𝑥𝑘=𝑘𝑠/𝑝1, 𝑧𝑘=(𝑘+𝑚1)𝑠/𝑝1, and 𝑟=𝑠/𝑠<0, we have to prove that 𝑛𝑘=1𝑥𝑘𝑛𝑘=1𝑥𝑟𝑘1/𝑟𝑛𝑘=1𝑧𝑘𝑛𝑘=1𝑧𝑟𝑘1/𝑟.(3.22)
Observe that both (𝑥𝑘)𝑘 and (𝑧𝑘)𝑘 are decreasing sequences. Denote 𝑋𝑛=𝑛𝑘=1𝑥𝑛, 𝑍𝑛=𝑛𝑘=1𝑧𝑛, and 𝐶=𝑋𝑛/𝑍𝑛.
Since (𝑥𝑘/𝑧𝑘)𝑘 is decreasing, so is also (𝑋𝑘/𝑍𝑘)𝑘, and hence 𝑋𝑘/𝑍𝑘𝑋𝑛/𝑍𝑛=𝐶 for 𝑘𝑛1. Applying Karamata’s inequality (see, e.g., [20]) for the decreasing sequences (𝑥𝑘)𝑘 and 𝑦𝑘=𝐶𝑧𝑘 and for the convex function Φ(𝑡)=𝑡𝑟, 𝑟<0, we get that 𝑛𝑘=1𝑥𝑟𝑘𝑛𝑘=1𝐶𝑧𝑘𝑟(3.23) or 𝑛𝑘=1𝑥𝑟𝑘𝐶𝑟𝑛𝑘=1𝑧𝑟𝑘.(3.24) The last inequality implies 𝑛𝑘=1𝑥𝑟𝑘1/𝑟𝑛𝑘=1𝑥𝑘𝑛𝑘=1𝑧𝑘𝑛𝑘=1𝑧𝑟𝑘1/𝑟,since𝑟<0.(3.25) Since 𝑚 was arbitrary, we get the desired inequality and (3.18) is proved. Moreover, it is known that with 𝐶𝑝𝑠 defined by (3.18) it yields that (3.17) holds (see Theorem  15 in [13]). The proof is complete.

The main result of this section contains optimal estimates between the norm of a sequence and of its level sequence with respect to the sequence 𝜑=(𝜑𝑛)𝑛 with 𝜑𝑛=𝑛1+𝑠/𝑝, 𝑛1.

Theorem 3.7. Let 1<𝑝<, 𝑝<𝑠, and let 𝑥=(𝑥𝑛)𝑝,𝑠 be a nonnegative and decreasing sequence, and let 𝑥=(𝑥𝑛)𝑛 be the level sequence with respect to the sequence 𝜑=(𝜑𝑛), where 𝜑𝑛=𝑛𝛼, 𝛼=1𝑠/𝑝. Then, 𝑥𝑝,𝑠𝑥𝑝,𝑠𝑝𝑠1/𝑠𝑝𝑠1/𝑠𝑥𝑝,𝑠.(3.26) The constants in (3.26) are optimal.

Proof. Assume first that 𝑠< and consider the left-hand side inequality in (3.26). By applying Theorem 3.3, we have that 𝑥𝑗/𝑗𝛼=𝜆𝑘, 𝑗𝐼𝑘. Since 𝛼=(𝑠/𝑝1)/(𝑠1), (𝑥𝑗)𝑠1𝑗𝑠/𝑝1=𝜆𝑘𝑠1, applying Hölder’s inequality, we obtain that 𝑗𝐼𝑘𝑥𝑗𝑠𝑗𝑠/𝑝1=𝜆𝑘𝑠1𝑗𝐼𝑘𝑥𝑗=𝑗𝐼𝑘𝑗𝛼1𝑠𝑗𝐼𝑘𝑥𝑗𝑠1𝑗𝐼𝑘𝑥𝑗=𝑗𝐼𝑘𝑗𝛼1𝑠𝑗𝐼𝑘𝑥𝑗𝑠𝑗𝐼𝑘𝑥𝑠𝑗𝑗𝑠/𝑝1.(3.27)
By the above estimate and Theorem 3.3, we get the left-hand side inequality in (3.26). Let us now consider the sequence 𝜓=(𝜓𝑛)𝑛, 𝜓𝑛=𝑥𝑛𝑠1𝑛𝑠/𝑝1,𝑛1, and let 𝜓=(𝜓𝑛)𝑛 be the level sequence of 𝜓=(𝜓𝑛)𝑛 with respect to 𝜑=(𝜑𝑛)𝑛, 𝜑𝑛=1, 𝑛1. By applying Theorem 3.3, Lemma 2.1, and Hölder’s inequality, we obtain that 𝑥𝑠𝑝,𝑠=𝑛=1𝑥𝑛𝜓𝑛𝑛=1𝑥𝑛𝜓𝑛𝑛=1𝑥𝑛𝜓𝑛𝑥𝑝,𝑠𝜓𝑝,𝑠.(3.28)
We note that to obtain the right-hand side inequality in (3.26) it is sufficient to prove that 𝜓𝑝,s𝑝𝑠1/𝑠𝑝𝑠1/𝑠𝑥𝑠1𝑝,𝑠.(3.29)
Let 𝐸={𝑛1𝜓𝑛=𝜓𝑛}. Then, we have that 𝐸=𝑘𝐼𝑘,(3.30) where {𝐼𝑘} are disjoint and such that 𝜓𝑛=1||𝐼𝑘||+1𝑗𝐼𝑘𝜓𝑗,𝑛𝐼𝑘.(3.31) By Hölder’s inequality, we obtain that 𝑗𝐼𝑘𝜓𝑗𝑗𝐼𝑘𝑗𝑠/𝑝11/𝑠𝑗𝐼𝑘𝑥𝑠𝑗𝑗𝑠/𝑝11/𝑠.(3.32)
We also have that 𝑛𝐸𝜓𝑛𝑠𝑛𝑠/𝑝1=𝑛𝐸𝜓𝑛𝑠𝑛𝑠/𝑝1=𝑛𝐸𝑥𝑠𝑛𝑛𝑠/𝑝1.(3.33)
Hence, by (3.31) and (3.32), it yields that 𝑛𝐼𝑘𝜓𝑛𝑠𝑛𝑠/𝑝11||𝐼𝑘||+1𝑠𝑚𝐼𝑘𝑚𝑠/𝑝1𝑠/𝑠𝑚𝐼𝑘𝑥𝑠𝑚𝑚𝑠/𝑝1,𝑛𝐼𝑘𝑛𝑠/𝑝1=𝐶𝑠𝑝𝑠,𝑘𝑚𝐼𝑘𝑥𝑠𝑚𝑚𝑠/𝑝1,(3.34) where 𝐶𝑝𝑠,𝑘=1||𝐼𝑘||+1𝑛𝐼𝑘𝑛𝑠/𝑝11/𝑠𝑛𝐼𝑘𝑛𝑠/𝑝11/𝑠.(3.35)
Hence, according to Lemma 3.6, it follows that (3.29) holds, which means that the right-hand side inequality in (3.26) is proved. It only remains to prove the sharpness of the obtained inequalities.
We note that the left-hand side inequality in (3.26) becomes equality if, for a fixed 𝑘0, we take 𝑥=(𝑥𝑛)𝑛, where 𝑥𝑛=𝑘0𝑘0𝑖=1𝑖𝛼𝑛𝛼if𝑛𝑘0,0,otherwise.(3.36)
The right-hand side inequality (3.26) becomes equality for 𝑥=(𝑥𝑛)𝑛, where 𝑥𝑛=1if𝑛𝑘0,0,otherwise.(3.37) We have that 𝑥𝑝,𝑠=(𝑘0𝑛=1𝑛𝑠/𝑝1)1/𝑠. It is easy to verify that if 𝑥=(𝑥𝑛)𝑛, where 𝑥𝑛=𝑘0𝑘0𝑖=1𝑖𝛼𝑛𝛼if𝑛𝑘0,0,otherwise,(3.38) then 𝑥𝑝,𝑠=𝑘0𝑘0𝑛=1𝑛𝛼1/𝑠.(3.39) Since 𝑘0 is arbitrary, we get that also the constant on the right-hand side inequality (3.26) is optimal.
Let us now consider the case 𝑠= and, hence, 𝛼=11/𝑝=1/𝑝. By using Hölder’s inequality, we find that 𝑛𝐼𝑘𝜆𝑘𝑛1/𝑝=𝑛𝐼𝑘𝑥𝑛=𝑛𝐼𝑘𝑥𝑘𝑥𝑝,𝑛𝐼𝑘𝑛1/𝑝.(3.40)
Then, 𝜆𝑘𝑥𝑝, for every 𝑘, which implies that 𝑥𝑝,𝑥𝑝,.
On the other hand, for any 𝑛𝐼𝑘, and using again Hölder’s inequality, we obtain that 𝑥𝑛𝑛1/𝑝𝑛𝑖=1𝑥𝑖𝑛1/𝑝1𝑛𝑖=1𝑥𝑖𝑛1/𝑝1𝑛1/𝑝𝑛𝑖=11𝑛𝑖1/𝑝sup1𝑖𝑛𝑖1/𝑝𝑥𝑖.(3.41) It follows that sup𝑛1𝑛1/𝑝𝑥𝑛sup𝑛1𝑛1/𝑝𝑛𝑖=1𝑖1/𝑝𝑛sup1𝑖𝑛𝑖1/𝑝𝑥𝑖=𝑥𝑝,𝐶𝑝,(3.42) where 𝐶𝑝=sup𝑛1sup𝑘=1,𝑛𝑘1/𝑝𝑛𝑘=1𝑘1/𝑝𝑛=sup𝑛1𝑛1/𝑝𝑛𝑘=1𝑘1/𝑝𝑛=𝑝.(3.43) The constants are optimal also in this case. This can easily be proved if we consider the same sequences as in the case 𝑠<. The proof is complete.

Remark 3.8. Let 1<𝑝<𝑠. Let 𝑥=(𝑥𝑛)𝑛𝑝,𝑠 be a nonnegative and nonincreasing sequence, and let 𝑥=(𝑥𝑛)𝑛 be the level sequence of 𝑥 with respect to the sequence 𝜑𝑛=𝑛𝛼𝑠,𝛼=1𝑝,𝑛1.(3.44) Then, the equality 𝑥𝑝,𝑠=𝑥𝑝,𝑠(3.45) holds if and only if 𝑥𝑛=𝑥𝑛 for all 𝑛1.
Indeed, inequality (3.27) becomes equality if and only if (𝑥𝑛𝑛𝛼)𝑛 is constant on 𝐼𝑘. Therefore, (3.45) holds if and only if 𝑥𝑛=𝑥𝑛, for all 𝑛1, hence if and only if (𝑥𝑛𝑛𝛼)𝑛 is decreasing.

4. Level Sequences and the Dual Norm

This section is devoted to investigate the dual norm of a sequence defined in (1.7).

In the next proposition, we summarize some well-known properties of the dual norm of a sequence 𝑥=(𝑥𝑛)𝑛𝑝,𝑠.

Proposition 4.1. Let 1<𝑝<, 1𝑠, and 𝑥=(𝑥𝑛)𝑛𝑝,𝑠. Then, the following statements hold.(a) One has 𝑥𝑝,𝑠𝑥𝑝,𝑠.(4.1)(b) If 𝑠𝑝, then 𝑥𝑝,𝑠=𝑥𝑝,𝑠.(4.2)(c) If 𝑝<𝑠 and if the sequence (𝑥𝑛𝑛1𝑠/𝑝)𝑛 is nonincreasing, then (4.2)holds.(d) If 𝑥=(𝑥𝑛)𝑛𝑝,𝑠, then one has that 𝑥𝑝,𝑠inf𝑥𝑧𝑧𝑝,𝑠.(4.3)

Proof. Indeed, (a) follows from Hölder’s inequality.
(b) Take 𝜓=(𝜓𝑛)𝑛 with 𝜓𝑛=(𝑥𝑛)𝑠1𝑛𝑠/𝑝1, 𝑛1. The sequence 𝜓=(𝜓𝑛)𝑛 is nonincreasing and we have that 𝜓𝑠𝑝,𝑠=𝑛=1𝜓𝑠𝑛𝑛𝑠/𝑝1=𝑥𝑠𝑝,𝑠,𝑥𝑠𝑝,𝑠=𝑛=1𝑥𝑛𝜓𝑛.(4.4) The last two equalities imply that 𝑥𝑝,𝑠𝑥𝑝,𝑠 and, by (a), we get equality (4.2).
(c) The proof is similar to that of (b), so we omit the details.
(d) In view of Lemma 2.1, we have that 𝑛=1𝑥𝑛𝑦𝑛inf𝑥𝑧𝑧𝑝,𝑠𝑦𝑝,𝑠,(4.5) where the infimum is taken over all nonnegative sequences 𝑧=(𝑧𝑛)𝑛𝑝,𝑠 such that 𝑥𝑧 and where 𝑦=(𝑦𝑛)𝑛𝑝,𝑠 is a nonnegative and nonincreasing sequence. This implies that (d) holds.

It was proved by Halperin in [11] (see also [12, Theorem  3.6.5]) that we have equality in (4.3) in the case of real functions defined on (0,1) and that the infimum is attained. For 𝑝<𝑠, a complete proof for Lorentz spaces over 𝜎-finite nonatomic measure spaces was given in the recent paper [9]. We remark here that the proofs given in [9, 11, 12] do not cover the result in the case when (𝑅,𝜇) is a totally 𝜎-finite measure space, completely atomic, with all atoms having the same measure. For completeness, we prove this result in the following theorem.

Theorem 4.2. Let 1<𝑝<𝑠, and let 𝑥=(𝑥𝑛)𝑛𝑝,𝑠 be a nonnegative and nonincreasing sequence. Let 𝛼=1𝑠/𝑝 and 𝜑(𝛼)=(𝜑𝑛(𝛼))𝑛, where 𝜑𝑛(𝛼)=𝑛𝛼. Then, one has that 𝑥𝑝,𝑠=inf𝑥𝑧𝑧𝑝,𝑠=𝑥𝑝,𝑠,(4.6) where 𝑥=(𝑥𝑛)𝑛 is the level sequence of 𝑥=(𝑥𝑛)𝑛 with respect to the sequence 𝜑(𝛼)=(𝜑𝑛(𝛼))𝑛.

Proof. By Proposition 4.1 and Theorem 3.3(b), it is sufficient to prove that 𝑥𝑝,𝑠𝑥𝑝,𝑠.(4.7)
We denote 𝐸={𝑛1𝑥𝑛=𝑥𝑛}. According to Theorem 3.3 it yields that 𝐸=𝑘𝐼𝑘, where 𝐼𝑘 are disjoint finite subsets of positive integers such that 𝑛𝐼𝑘𝑥𝑛=𝑛𝐼𝑘𝑥𝑛.(4.8) Assume first that 𝑠< and let 𝜓=(𝜓𝑛)𝑛, where 𝜓𝑛=(𝑥𝑛)𝑠1𝑛𝑠/𝑝1, 𝑛1. As before, we have that 𝜓𝑠𝑝,𝑠,=𝑛=1𝜓𝑠𝑛𝑛𝑠/𝑝1=𝑥𝑝,𝑠.(4.9) Choose 𝑦=(𝑦𝑛)𝑛, where 𝑦𝑛=𝜓𝑛/𝑥𝑠1𝑝,𝑠. Then, 𝑦𝑝,𝑠=1.
From Theorem 3.3, it follows that, for each 𝑘, 𝑥𝑛=𝜆𝑘𝑛𝛼, 𝑛𝐼𝑘. Thus, 𝜓𝑛=𝜆𝑘𝑠1 and 𝑥𝑠1𝑝,𝑠𝑛𝐼𝑘𝑥𝑛𝑦𝑛=𝜆𝑘𝑠1𝑛𝐼𝑘𝑥𝑛=𝜆𝑘𝑠1𝑛𝐼𝑘𝑥𝑛=𝑛𝐼𝑘𝑥𝑛𝑠𝑛𝑠/𝑝1.(4.10) Moreover, we have that 𝑥𝑠1𝑝,𝑠𝑛𝐸𝑥𝑛𝑦𝑛=𝑛𝐸𝑥𝑛𝑠𝑛𝑠/𝑝1.(4.11) Thus, we obtain that 𝑛=1𝑥𝑛𝑦𝑛=𝑥𝑝,𝑠. This implies (4.7).
Let now 𝑠=. We consider only the nontrivial case 𝐼1={1,2,,𝑚}. Let 𝑦=(𝑦𝑛)𝑛, where 𝑦𝑛=1𝑚𝑖=1𝑖1/𝑝if𝑛𝑚,0,otherwise(4.12) and 𝑦𝑝,1=1. Then, 𝑛=1𝑥𝑛𝑦𝑛=1𝑚𝑖=1𝑖𝑚1/𝑝𝑛=1𝑥𝑛=1𝑚𝑖=1𝑖𝑚1/𝑝𝑛=1𝑥𝑛=𝜆1=𝑥𝑝,.(4.13) This implies (4.7), and the proof is complete.

Remark 4.3. Let 1<𝑝<𝑠 and let 𝑥=(𝑥𝑛)𝑛𝑝,𝑠 be a nonnegative and nonincreasing sequence. Then, by Remark 3.8, the equality (see (4.2)) 𝑥𝑝,𝑠=𝑥𝑝,𝑠(4.14) holds if and only if (𝑥𝑛𝑛𝛼)𝑛 is decreasing. Note also that (4.2) holds in general if 1𝑠<𝑝 (see Proposition 4.1(b)).
The final result in this section gives the sharp estimate of the standard norm via the dual norm.

Theorem 4.4. Let 1<𝑝< and 𝑝<𝑠. Then, for any sequence 𝑥=(𝑥𝑛)𝑛𝑝,𝑠, it yields that 𝑥𝑝,𝑠𝐶𝑝𝑠𝑥𝑝,𝑠,(4.15) where 𝐶𝑝𝑠 is defined by (3.17). The constant is optimal.

Proof. The proof follows immediately from Theorems 4.2 and 3.7.

5. The Decomposition Norm

In this section, we prove that the dual norm and the decomposition norm coincide. The following lemma plays an important role in the proof of the main result, and it was proved in the recent paper [9].

Lemma 5.1. Let 𝛼1𝛼2𝛼𝜈 be positive numbers, and let {𝜂𝑗𝑘} be a (𝑁×𝜈)-matrix of positive numbers, 1𝑗𝑁,1𝑘𝜈. Set 𝛽𝑘=𝑁𝑗=1𝜂𝑗𝑘,𝑘=1,,𝜈.(5.1) Assume that 𝛽1++𝛽𝑘𝛼1++𝛼𝑘(5.2) for any 𝑘=1,,𝜈. Let 𝜂=max𝜂𝑗𝑘. Then, for any 𝑗=1,,𝑁, there exists a permutation {̃𝜂𝑗𝑘}𝜈𝑘=1 of the 𝜈-tuple {𝜂𝑗𝑘}𝜈𝑘=1 such that 𝛼𝑘̃𝛽𝑘̃𝛽+𝜂,𝑘=𝑁𝑗=1̃𝜂𝑗𝑘,(5.3) for any 𝑘=1,,𝜈.

Our main result in this section reads the following.

Theorem 5.2. Let 1<𝑝< and 1𝑠. Then, for any sequence 𝑥=(𝑥𝑛)𝑛𝑝,𝑠, 𝑥𝑝,𝑠=𝑥(𝑝,𝑠).(5.4)

Proof. If 𝑠𝑝, then it is well known that (see, e.g., [7, 15]) 𝑥𝑝,𝑠=𝑥𝑝,𝑠=𝑥(𝑝,𝑠).(5.5) Let 1<𝑝<𝑠. We prove first that 𝑥𝑝,𝑠𝑥(𝑝,𝑠).(5.6) Let 𝑦=(𝑦𝑛)𝑛𝑝,𝑠, and let 𝑥=𝑘𝑥(𝑘).(5.7) Then, by Hölder’s inequality, we have that 𝑛=1||𝑥𝑛𝑦𝑛||𝑘𝑛=1||𝑥𝑛(𝑘)𝑦𝑛||𝑘𝑥(𝑘)𝑝,𝑠𝑦𝑝,𝑠.(5.8) By taking the infimum over all representation (5.7), we obtain (5.6). It remains only to prove that 𝑥(𝑝,𝑠)𝑥𝑝,𝑠.(5.9) In view of Lemma 2.3, it is sufficient to prove (5.9) in the case of positive nonincreasing sequences. We can assume that there exists 𝑛0 such that 𝑥𝑖=0, for 𝑖>𝑛0, 𝑛0. From Theorem 3.3, we have also that 𝑥𝑖=0 for 𝑖>𝑛1, 𝑛1. Denote 𝑘0=max{𝑛0,𝑛1}. According to Theorem 4.2, it yields that 𝑥𝑝,𝑠=𝑥𝑝,𝑠,(5.10) where 𝑥=(𝑥𝑛)𝑛 is the level sequence of 𝑥=(𝑥𝑛)𝑛 with respect to the sequence 𝜑=(𝜑𝑛)𝑛, 𝜑𝑛=𝑛𝛼, 𝛼=1𝑠/𝑝.
Let 𝜀>0. Choose 𝛿=𝛿(𝜀,𝑘0)>0 such that 𝛿<𝜀(1/𝑧𝑝,𝑠), with 𝑧=(𝑧𝑛)𝑛, where 𝑧𝑛=1if𝑛𝑘0,0,otherwise.(5.11)
Choose 𝑁 such that max𝑖𝑥𝑖<𝑁𝛿.(5.12)
By using Lemma 5.1 with 𝛼𝑘=𝑥𝑘, 𝛽𝑘=𝑥𝑘, 𝜂𝑗𝑘=𝑥𝑘/𝑁, 1𝑘𝑘0, 1𝑗𝑁, we have that there exists a permutation of {𝜂𝑗𝑘}, {̃𝜂𝑗𝑘} such that 𝑥𝑘𝑁𝑗=1̃𝜂𝑗𝑘+𝛿,𝑘𝑘0.(5.13)
We define now 𝑡(𝑗)=(𝑡𝑘(𝑗))𝑘, where 𝑡𝑘(𝑗)=̃𝜂𝑗𝑘. Since 𝑡(𝑗) is equimeasurable with 𝑥/𝑁, we have that 𝑡(𝑗)𝑝,𝑠=𝑥𝑝,𝑠𝑁,𝑗𝑁,(5.14) which, in its turn, implies that 𝑥(𝑝,𝑠)𝑁𝑗=1𝑡(𝑗)𝑝,𝑠+𝛿𝑧𝑝,𝑠𝑥𝑝,𝑠+𝜀.(5.15) Since 𝜀 was arbitrary, it follows that (5.9) holds. The proof is complete.

Corollary 5.3. Let 𝑥=(𝑥𝑛)𝑛𝑝,𝑠, 1𝑝<, 1𝑠. Then, 𝑥(𝑝,𝑠)=𝑥(𝑝,𝑠).(5.16)

Proof. Equality (5.16) follows immediately from (5.4) and Lemma 2.2.

6. Further Results and Remarks

From Theorems 4.4 and 5.2, we can obtain the following sharp version of the “triangle inequality.”

Theorem 6.1. Let 1<𝑝<𝑠, and suppose that 𝑥(𝑘)=(𝑥𝑛(𝑘))𝑛𝑝,𝑠, 𝑘=1,,𝑁. Then, one has that 𝑁𝑘=1𝑥(𝑘)𝑝,𝑠𝐶𝑁𝑝𝑠𝑘=1𝑥(𝑘)𝑝,𝑠,(6.1) where 𝐶𝑝𝑠 is given by (3.17) and the constant is optimal.

Remark 6.2. This result can also be derived from a recent result of Kamińska and Parrish [13]. They solved the problem in a completely different way.

Proof. We note that (6.1) is equivalent to the inequality 𝑥𝑝,𝑠𝐶𝑝𝑠𝑥(𝑝,𝑠),(6.2) where 𝑥=(𝑥𝑛)𝑛 is any sequence from 𝑝,𝑠. Inequality (6.2) follows directly from Theorems 4.4 and 5.2. Moreover, for 𝑥=(𝑥𝑛)𝑛 with 𝑥𝑛=1if𝑛𝑘0,0,otherwise,(6.3) for a fixed 𝑘0, we obtain that 𝑥𝑝,𝑠=𝑘0𝑛=1𝑛𝑠/𝑝11/𝑠.(6.4) From Theorems 4.2 and 5.2, we have that 𝑥(𝑝,𝑠)=𝑘0𝑘0𝑛=1𝑛𝑠/𝑝11/𝑠(6.5) and therefore we obtain equality in (6.2), and the proof is complete.

Remark 6.3. We want to pronounce that in this paper we have formulated our results only for 𝑝,𝑠 spaces, but our proofs show that some results are true for much more general spaces. For example, Theorems 3.7, 4.4, 5.2, and 6.1 are true in the more general case of resonant measure space (𝑅,𝜇). For example, Theorem 5.2 can be generalized as follows.

Theorem 6.4. Let 1<𝑝< and 1𝑠. Then, for any function 𝑓𝐿𝑝,𝑠(𝑅,𝜇), where (𝑅,𝜇) is a totally 𝜎-finite resonant measure space (see, e.g., [15, Definition  2.3, page 45]), one has that 𝑓𝑝,𝑠=𝑓(𝑝,𝑠).(6.6)

Proof. Indeed, according to Theorem 5.2, we conclude that equality (6.6) holds for 𝑓𝐿𝑝,𝑠(𝑅,𝜇), where (𝑅,𝜇) is a totally 𝜎-finite measure space, completely atomic, with all atoms having the same measure. Hence, together with Theorem  5.2 in [9] and by applying [15, Theorem  2.7, page 51], we obtain the desired result.

Acknowledgments

The authors would like to thank Professor Nicolae Popa for careful reading of the manuscript and for valuable comments, to Professor A. Kamińska who made us aware of the information given in Remark 6.2 and to Professor G. Sinnamon for giving us some information regarding level functions and for his remarks about some of our Theorems. Moreover the second named author wants to thank the Department of Mathematics at Luleå University of Technology for hospitality and for financial support. A. N. Marcoci was partially supported by CNCSIS-UEFISCSU, project number 538/2009 PNII-IDEI code 1905/2008.

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