# Best Constants between Equivalent Norms in Lorentz Sequence Spaces

Authors: S. Barza; A. N. Marcoci; L. E. Persson
Journal: Journal of Function Spaces and Applications
 (2012)

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## Abstract

We find the best constants in inequalities relating the standard norm, the dual norm, and the norm∥x∥(p,s):=inf{∑k∥x(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.


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## Body

## 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.Let1<p<∞ and 1≤s≤∞. 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 spaces(1.1)lp,s={x:‖x‖p,s:=(∑n=1∞(xn*)sns/p-1)1/s<∞},
with the usual modification if s=∞. Lorentz proved that ∥·∥p,s is a norm if and only if 1≤s≤p<∞ and that the space ℓp,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 ℓp,s cannot be endowed with an equivalent norm).The study of normability forp<s was carried out by means of the maximal norm:(1.2)‖x‖p,s*=‖(xn)n‖p,s*=(∑n=1∞ns/p-1(1n∑i=1nxi*)s)1/s
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:(1.3)‖x‖p,s≤‖x‖p,s*≤p′‖x‖p,s,
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,…,N⊂ℓp,s, it yields that(1.4)‖∑k=1Nx(k)‖p,s≤Cps∑k=1N‖x(k)‖p,s.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:(1.5)‖x‖(p,s):=inf⁡{∑k=1N‖x(k)‖p,s:x=∑k=1Nx(k)}.One can prove that∥·∥(p,s) is always a norm and that ∥·∥(p,s) is equivalent with ∥·∥p,s, 1≤p,s≤∞. If 1≤s≤p, then ∥·∥(p,s)=∥·∥p,s (see, e.g., [7, 8]). We also remark that the best constant Cps in the inequality(1.6)‖x‖p,s≤Cps‖x‖(p,s)
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 thedual norm:(1.7)‖x‖p,s′=sup⁡{∑n=1∞xnyn:‖y‖p′,s′=1}.
The dual norm ∥·∥p,s′ is a norm indeed, equivalent to ∥·∥p,s and, moreover, if 1≤s≤p, then(1.8)‖x‖p,s′=‖x‖p,s=‖x‖(p,s).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 Section2, 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 ℓp,s of x=(xn)n and of its level sequence x∘=(xn∘)n. Section 4 is dedicated to the study of the dual norm ∥x∥p,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)n∈ℓp,s coincides with the quasinorm of the level sequence ∥x∘∥p,s (see Theorem 4.2). Moreover, ∥x∥p,s≤Cps∥x∥p,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 ∥x∥p,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 sequencesx=(xn)n, y=(yn)n, we denote that x≺y if(1.9)∑i=1nxi≤∑i=1nyi,∀n≥1
and we use the notation x≤y if(1.10)xn≤yn,∀n≥1.
The number p′ stands for the conjugate index of p; that is, 1/p+1/p′=1 and ℕ, ℕ* 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.
Letu,v,w be nonnegative sequences and suppose that w is decreasing. If
(2.1)∑k=1nuk≤∑k=1nvk(n=1,2,…),
then
(2.2)∑k=1nukwk≤∑k=1nvkwk(n=1,2,…).We also need the following auxiliary statement related to the dual norm.Lemma 2.2.
Letx=(xn)n∈ℓp,s, 1<p<∞, 1≤s≤∞. Then,
(2.3)‖x‖p,s′=‖x*‖p,s′.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.
Letx=(xn)n∈ℓp,s, 1<p<∞,1≤s≤∞. Then, the following statements hold.(a)
The equality(2.4)‖x‖(p,s)=inf⁡{∑k‖x(k)‖p,s}
holds, where the infimum is taken over all finite nonnegative sequences |xn|=∑kxn(k).(b)
If0≤y≤x, then ∥y∥(p,s)≤∥x∥(p,s).(c)
If0≤y(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, [16–18] 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.
Letφ=(φn)n be a sequence of positive numbers and Φn=∑i=1nφi. A sequence X̃=(X̃n)n is called φ-concave if, for arbitrary n1,n2∈ℕ, we have that X̃n≥Ln, where Ln=CΦn+D, C and D are constants, such that L “interpolate” X̃ in n1, n2; that is,
(3.1)Ln1=X̃n1,Ln2=X̃n2.Proposition 3.2.
The following statements are equivalent:(3.2)X̃=(X̃n)nisφ-concave,(3.3)X̃n-X̃n1Φn-Φn1≥X̃n2-X̃n1Φn2-Φn1,1≤n1<n≤n2,(3.4)X̃n-X̃n1Φn-Φn1≥X̃n2-X̃nΦn2-Φn,1≤n1≤n<n2,(3.5)X̃n1-X̃m1Φn1-Φm1≥X̃n2-X̃m2Φn2-Φm2,m1≤m2,n1≤n2,m1≠n1,m2≠n2.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.
Letφ=(φn)n be a positive sequence and Φn=∑i=1nφi. Let x=(xn)n be a positive sequence and suppose that
(3.6)lim⁡n→∞∑i=1nxi∑i=1nφi=0.
Then, there exists a unique nonnegative sequencex∘=(xn∘)n satisfying the following conditions:(a)
(xn∘/φn)n is decreasing;(b)
x≺x∘;(c)
the set{n:xn∘≠xn}=∪k=1∞Ik, where {Ik} are disjoint sets of positive integers, Ik={nk,…,nk+mk} such that ∑i∈Ikxi=∑i∈Ikxi∘ and xi∘/φi=λk, for all i∈Ik.Remark 3.4.
The unique sequencex∘=(xn∘)n in Theorem 3.3 is called the level sequence of x=(xn)n with respect to φ=(φn)n.Proof.
(a) LetXn=∑i=1nxi, n∈ℕ*. Condition (3.6) implies the existence of a sequence X̃=(X̃n)n, which is a φ-concave majorant of X. We denote Xn∘:=infX̃n and from (3.5) we get that
(3.7)X̃k1X̃k2≤Φk1Φk2≤1,fork1≤k2,
that is, X∘=(Xk∘)k is increasing. Thus, there exists a unique positive sequence (xn)n such that Xn∘=∑i=1nxi∘, with x1∘=X1∘ and
(3.8)xn∘:=Xn∘-Xn-1∘.
According to (3.5), we have that (Xn∘-Xn-1∘)/(Φn-Φn-1)≥(Xn+1∘-Xn∘)/(Φn+1-Φn), 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 termsXk-1∘ and Xk∘ are equal to Xk-1 and Xk, respectively, if and only if xk=xk∘, k≥1.
Now, we assume that there existsj∈ℕ* such that Xj∘>Xj and we denote n:=max{k<j:Xk∘=XkandXk-1∘=Xk-1}+1 and m:=min{k>j:Xk∘=XkandXk+1∘=Xk+1}-n+1. Now, we take L to “interpolate” X∘ in n-1 and n+m-1. Then, L≥X and X∘≥L. It follows that Lj=Xj∘=CΦj+D for j∈{n-1,…,n+m-1}.
Thus,xj∘=Cφj, j∈{n,…,n-1+m}, xn+m∘=xn+m and xn-1∘=xn-1. Since
(3.9)Xn-1∘=Xn-1,Xn+m-1∘=Xn+m-1,
we obtain that
(3.10)∑i=nn+m-1xi∘=∑i=nn+m-1xi
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:(3.11)f(t)=∑n=0∞xnχ[n,n+1)(t).Example 3.5.
Forf(t)=χ[0,1](t), the level function is
(3.12)f∘(t)=(1-α)t-αχ[0,1](t),
where α=1-s′/p′ so that
(3.13)‖f∘‖p,s=(s′p′)1/s′.
Ifx=(xn)n, where
(3.14)xn={1ifn≤k,0ifn>k,
then
(3.15)xn∘={k∑i=1ki-αn-αifn≤k,0,otherwise,‖x∘‖p,s=k(∑n=1kn-α)-1/s′.We also need the following technical lemma.Lemma 3.6.
Let1≤p<s<∞ and
(3.16)Cps=sup⁡m,n∈N*(1n∑k=mm+n-1ks/p-1)1/s(1n∑k=mm+n-1ks′/p′-1)1/s′.
Then,
(3.17)Cps=(ps)1/s(p′s′)1/s′.Proof.
We prove first that(3.18)Cps=sup⁡n∈N*(1n∑k=1nks/p-1)1/s(1n∑k=1nks′/p′-1)1/s′.
The inequality
(3.19)Cps≥sup⁡n∈N*(1n∑k=1nks/p-1)1/s(1n∑k=1nks′/p′-1)1/s′
is trivial. For the converse inequality, it is enough to prove that
(3.20)(∑k=1nks/p-1)1/s(∑k=1nks′/p′-1)1/s′≥(∑k=mm+n-1ks/p-1)1/s(∑k=mm+n-1ks′/p′-1)1/s′
for all m≥n.
The above inequality can be obviously written in the form(3.21)(∑k=1nks/p-1)s′/s(∑k=1nks′/p′-1)≥(∑k=1n(k+m-1)s/p-1)s′/s(∑k=1n(k+m-1)s′/p′-1).
Let us fix m≥1. Observe that for m=1 we have equality, so we may suppose that m>1. If we denote xk=ks′/p′-1, zk=(k+m-1)s′/p′-1, and r=-s/s′<0, we have to prove that
(3.22)∑k=1nxk(∑k=1nxkr)1/r≥∑k=1nzk(∑k=1nzkr)1/r.
Observe that both(xk)k and (zk)k are decreasing sequences. Denote Xn=∑k=1nxn, Zn=∑k=1nzn, and C=Xn/Zn.
Since(xk/zk)k is decreasing, so is also (Xk/Zk)k, and hence Xk/Zk≥Xn/Zn=C for k≤n-1. 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
(3.23)∑k=1nxkr≥∑k=1n(Czk)r
or
(3.24)∑k=1nxkr≥Cr∑k=1nzkr.
The last inequality implies
(3.25)(∑k=1nxkr)1/r≤∑k=1nxk∑k=1nzk(∑k=1nzkr)1/r,sincer<0.
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 sequenceφ=(φn)n with φn=n-1+s′/p′, n≥1.Theorem 3.7.
Let1<p<∞, p<s≤∞, and let x=(xn)∈ℓp,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=n-α, α=1-s′/p′. Then,
(3.26)‖x∘‖p,s≤‖x‖p,s≤(ps)1/s(p′s′)1/s′‖x∘‖p,s.
The constants in (3.26) are optimal.Proof.
Assume first thats<∞ and consider the left-hand side inequality in (3.26). By applying Theorem 3.3, we have that xj∘/j-α=λk, j∈Ik. Since α=(s/p-1)/(s-1), (xj∘)s-1js/p-1=λks-1, applying Hölder’s inequality, we obtain that
(3.27)∑j∈Ik(xj∘)sjs/p-1=λks-1∑j∈Ikxj∘=(∑j∈Ikj-α)1-s(∑j∈Ikxj)s-1(∑j∈Ikxj∘)=(∑j∈Ikj-α)1-s(∑j∈Ikxj)s≤∑j∈Ikxjsjs/p-1.
By the above estimate and Theorem3.3, we get the left-hand side inequality in (3.26). Let us now consider the sequence ψ=(ψn)n, ψn=xns-1ns/p-1,n≥1, and let ψ̃=(ψ̃n)n be the level sequence of ψ=(ψn)n with respect to φ=(φn)n, φn=1, n≥1. By applying Theorem 3.3, Lemma 2.1, and Hölder’s inequality, we obtain that
(3.28)‖x‖p,ss=∑n=1∞xnψn≤∑n=1∞xnψ̃n≤∑n=1∞xn∘ψ̃n≤‖x∘‖p,s‖ψ̃‖p′,s′.
We note that to obtain the right-hand side inequality in (3.26) it is sufficient to prove that
(3.29)‖ψ̃‖p′,s′≤(ps)1/s(p′s′)1/s′‖x‖p,ss-1.
LetE:={n≥1:ψ̃n=ψn}. Then, we have that
(3.30)N*-E=⋃kIk,
where {Ik} are disjoint and such that
(3.31)ψ̃n=1|Ik|+1∑j∈Ikψj,∀n∈Ik.
By Hölder’s inequality, we obtain that
(3.32)∑j∈Ikψj≤(∑j∈Ikjs/p-1)1/s(∑j∈Ikxjsjs/p-1)1/s′.
We also have that(3.33)∑n∈E(ψ̃n)s′ns′/p′-1=∑n∈E(ψn)s′ns′/p′-1=∑n∈Exnsns/p-1.
Hence, by (3.31) and (3.32), it yields that
(3.34)∑n∈Ik(ψ̃n)s′ns′/p′-1≤1(|Ik|+1)s′(∑m∈Ikms/p-1)s′/s(∑m∈Ikxmsms/p-1),(∑n∈Ikns′/p′-1)=Cps,ks′∑m∈Ikxmsms/p-1,
where
(3.35)Cps,k=1(|Ik|+1)(∑n∈Ikns/p-1)1/s(∑n∈Ikns′/p′-1)1/s′.
Hence, according to Lemma3.6, 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
(3.36)xn={k0∑i=1k0i-αn-αifn≤k0,0,otherwise.
The right-hand side inequality (3.26) becomes equality for x=(xn)n, where
(3.37)xn={1ifn≤k0,0,otherwise.
We have that ∥x∥p,s=(∑n=1k0ns/p-1)1/s. It is easy to verify that if x∘=(xn∘)n, where
(3.38)xn∘={k0∑i=1k0i-αn-αifn≤k0,0,otherwise,
then
(3.39)‖x∘‖p,s=k0(∑n=1k0n-α)-1/s′.
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 cases=∞ and, hence, α=1-1/p′=1/p. By using Hölder’s inequality, we find that
(3.40)∑n∈Ikλkn-1/p=∑n∈Ikxn∘=∑n∈Ikxk≤‖x‖p,∞∑n∈Ikn-1/p.
Then,λk≤∥x∥p,∞ for every k, which implies that ∥x∘∥p,∞≤∥x∥p,∞.
On the other hand, for anyn∈Ik, and using again Hölder’s inequality, we obtain that
(3.41)xnn1/p≤(∑i=1nxi)n1/p-1≤(∑i=1nxi∘)n1/p-1≤n1/p(∑i=1n1ni-1/p)sup⁡1≤i≤n{i1/pxi∘}.
It follows that
(3.42)sup⁡n≥1{n1/pxn}≤sup⁡n≥1{n1/p∑i=1ni-1/pnsup⁡1≤i≤n(i1/pxi∘)}=‖x∘‖p,∞Cp,
where
(3.43)Cp=sup⁡n≥1(sup⁡k=1,n¯⁡k1/p)(∑k=1nk-1/p)n=sup⁡n≥1n1/p(∑k=1nk-1/p)n=p′.
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.
Let1<p<s≤∞. Let x=(xn)n∈ℓp,s be a nonnegative and nonincreasing sequence, and let x∘=(xn∘)n be the level sequence of x with respect to the sequence
(3.44)φn=n-α,α=1-s′p′,n≥1.
Then, the equality
(3.45)‖x∘‖p,s=‖x‖p,s
holds if and only if xn∘=xn for all n≥1.
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 n≥1, 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 sequencex=(xn)n∈ℓp,s.Proposition 4.1.
Let1<p<∞, 1≤s≤∞, and x=(xn)n∈ℓp,s. Then, the following statements hold.(a)
One has(4.1)‖x‖p,s′≤‖x‖p,s.(b)
Ifs≤p, then
(4.2)‖x‖p,s′=‖x‖p,s.(c)
Ifp<s≤∞ and if the sequence (xn*n1-s′/p′)n is nonincreasing, then (4.2)holds.(d)
Ifx=(xn)n∈ℓp,s, then one has that
(4.3)‖x‖p,s′≤inf⁡x≺z‖z‖p,s.Proof.
Indeed, (a) follows from Hölder’s inequality.
(b) Takeψ=(ψn)n with ψn=(xn*)s-1ns/p-1, n≥1. The sequence ψ=(ψn)n is nonincreasing and we have that
(4.4)‖ψ‖p′,s′s′=∑n=1∞ψns′ns′/p′-1=‖x‖p,ss,‖x‖p,ss=∑n=1∞xn*ψn.
The last two equalities imply that ∥x∥p,s′≥∥x∥p,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 Lemma2.1, we have that
(4.5)∑n=1∞xnyn≤inf⁡x≺z‖z‖p,s‖y‖p′,s′,
where the infimum is taken over all nonnegative sequences z=(zn)n∈ℓp,s such that x≺z and where y=(yn)n∈ℓp′,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.
Let1<p<s≤∞, and let x=(xn)n∈ℓp,s be a nonnegative and nonincreasing sequence. Let α=1-s′/p′ and φ(α)=(φn(α))n, where φn(α)=n-α. Then, one has that
(4.6)‖x‖p,s′=inf⁡x≺z‖z‖p,s=‖x∘‖p,s,
where x∘=(xn∘)n is the level sequence of x=(xn)n with respect to the sequence φ(α)=(φn(α))n.Proof.
By Proposition4.1 and Theorem 3.3(b), it is sufficient to prove that
(4.7)‖x‖p,s′≥‖x∘‖p,s.
We denoteE={n≥1:xn=xn∘}. According to Theorem 3.3 it yields that ℕ*-E=∪kIk, where Ik are disjoint finite subsets of positive integers such that
(4.8)∑n∈Ikxn=∑n∈Ikxn∘.
Assume first that s<∞ and let ψ=(ψn)n, where ψn=(xn∘)s-1ns/p-1, n≥1. As before, we have that
(4.9)‖ψ‖p′,s,s′=∑n=1∞ψns′ns′/p′-1=‖x∘‖p,s.
Choose y=(yn)n, where yn=ψn/∥x∘∥p,ss-1. Then, ∥y∥p′,s′=1.
From Theorem3.3, it follows that, for each k, xn∘=λkn-α, n∈Ik. Thus, ψn=λks-1 and
(4.10)‖x∘‖p,ss-1∑n∈Ikxnyn=λks-1∑n∈Ikxn=λks-1∑n∈Ikxn∘=∑n∈Ik(xn∘)sns/p-1.
Moreover, we have that
(4.11)‖x∘‖p,ss-1∑n∈Exnyn=∑n∈E(xn∘)sns/p-1.
Thus, we obtain that ∑n=1∞xnyn=∥x∘∥p,s. This implies (4.7).
Let nows=∞. We consider only the nontrivial case I1={1,2,…,m}. Let y=(yn)n, where
(4.12)yn={1∑i=1mi-1/pifn≤m,0,otherwise
and ∥y∥p′,1=1. Then,
(4.13)∑n=1∞xnyn=1∑i=1mi-1/p∑n=1mxn=1∑i=1mi-1/p∑n=1mxn∘=λ1=‖x∘‖p,∞.
This implies (4.7), and the proof is complete.Remark 4.3.
Let1<p<s≤∞ and let x=(xn)n∈ℓp,s be a nonnegative and nonincreasing sequence. Then, by Remark 3.8, the equality (see (4.2))
(4.14)‖x‖p,s′=‖x‖p,s
holds if and only if (xnnα)n is decreasing. Note also that (4.2) holds in general if 1≤s<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.
Let1<p<∞ and p<s≤∞. Then, for any sequence x=(xn)n∈ℓp,s, it yields that
(4.15)‖x‖p,s≤Cps‖x‖p,s′,
where Cps is defined by (3.17). The constant is optimal.Proof.
The proof follows immediately from Theorems4.2 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.
Letα1≥α2≥⋯αν be positive numbers, and let {ηjk} be a (N×ν)-matrix of positive numbers, 1≤j≤N,1≤k≤ν. Set
(5.1)βk=∑j=1Nηjk,k=1,…,ν.
Assume that
(5.2)β1+⋯+βk≥α1+⋯+αk
for any k=1,…,ν. Let η=maxηjk. Then, for any j=1,…,N, there exists a permutation {η̃jk}k=1ν of the ν-tuple {ηjk}k=1ν such that
(5.3)αk≤β̃k+η,β̃k=∑j=1Nη̃jk,
for any k=1,…,ν.Our main result in this section reads the following.Theorem 5.2.
Let1<p<∞ and 1≤s≤∞. Then, for any sequence x=(xn)n∈ℓp,s,
(5.4)‖x‖p,s′=‖x‖(p,s).Proof.
Ifs≤p, then it is well known that (see, e.g., [7, 15])
(5.5)‖x‖p,s′=‖x‖p,s=‖x‖(p,s).
Let 1<p<s≤∞. We prove first that
(5.6)‖x‖p,s′≤‖x‖(p,s).
Let y=(yn)n∈ℓp′,s′, and let
(5.7)x=∑kx(k).
Then, by Hölder’s inequality, we have that
(5.8)∑n=1∞|xnyn|≤∑k∑n=1∞|xn(k)yn|≤∑k‖x(k)‖p,s‖y‖p′,s′.
By taking the infimum over all representation (5.7), we obtain (5.6). It remains only to prove that
(5.9)‖x‖(p,s)≤‖x‖p,s′.
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∈ℕ*. From Theorem 3.3, we have also that xi∘=0 for i>n1, n1∈ℕ*. Denote k0=max{n0,n1}. According to Theorem 4.2, it yields that
(5.10)‖x‖p,s′=‖x∘‖p,s,
where x∘=(xn∘)n is the level sequence of x=(xn)n with respect to the sequence φ=(φn)n, φn=n-α, α=1-s′/p′.
Letɛ>0. Choose δ=δ(ɛ,k0)>0 such that δ<ɛ(1/∥z∥p,s), with z=(zn)n, where
(5.11)zn={1ifn≤k0,0,otherwise.
ChooseN∈ℕ* such that
(5.12)max⁡i{xi∘}<N⋅δ.
By using Lemma5.1 with αk:=xk, βk:=xk∘, ηjk:=xk∘/N, 1≤k≤k0, 1≤j≤N, we have that there exists a permutation of {ηjk}, {η̃jk} such that
(5.13)xk≤∑j=1Nη̃jk+δ,∀k≤k0.
We define nowt(j)=(tk(j))k, where tk(j):=η̃jk. Since t(j) is equimeasurable with x∘/N, we have that
(5.14)‖t(j)‖p,s=‖x∘‖p,sN,j≤N,
which, in its turn, implies that
(5.15)‖x‖(p,s)≤∑j=1N‖t(j)‖p,s+δ⋅‖z‖p,s≤‖x∘‖p,s+ɛ.
Since ɛ was arbitrary, it follows that (5.9) holds. The proof is complete.Corollary 5.3.
Letx=(xn)n∈ℓp,s, 1≤p<∞, 1≤s≤∞. Then,
(5.16)‖x‖(p,s)=‖x*‖(p,s).Proof.
Equality (5.16) follows immediately from (5.4) and Lemma 2.2.


## 6. Further Results and Remarks

From Theorems4.4 and 5.2, we can obtain the following sharp version of the “triangle inequality.”Theorem 6.1.
Let1<p<s≤∞, and suppose that x(k)=(xn(k))n∈ℓp,s, k=1,…,N. Then, one has that
(6.1)‖∑k=1Nx(k)‖p,s≤Cps∑k=1N‖x(k)‖p,s,
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
(6.2)‖x‖p,s≤Cps‖x‖(p,s),
where x=(xn)n is any sequence from ℓp,s. Inequality (6.2) follows directly from Theorems 4.4 and 5.2. Moreover, for x=(xn)n with
(6.3)xn={1ifn≤k0,0,otherwise,
for a fixed k0∈ℕ*, we obtain that
(6.4)‖x‖p,s=(∑n=1k0ns/p-1)1/s.
From Theorems 4.2 and 5.2, we have that
(6.5)‖x‖(p,s)=k0(∑n=1k0ns′/p′-1)1/s′
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 forℓp,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.
Let1<p<∞ and 1≤s≤∞. Then, for any function f∈Lp,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
(6.6)‖f‖p,s′=‖f‖(p,s).Proof.
Indeed, according to Theorem5.2, we conclude that equality (6.6) holds for f∈Lp,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.



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*Source: 713534-2012-02-15.xml*