# Bounded Domains of Generalized Riesz Methods with the Hahn Property

Authors: Maria Zeltser
Journal: Journal of Function Spaces and Applications
 (2013)

---

## Abstract

In 2002  Bennett et al. started the investigation to which extent sequence spaces are determined by the sequences of 0s and 1s that they contain. In this relation they defined three types of Hahn properties for sequence spaces: the Hahn property, separable Hahn property, and matrix Hahn property. In general all these three properties are pairwise distinct. If a sequence spaceE is solid and (0,1ℕ∩E)β=Eβ=ℓ1 then the two last properties coincide. We will show that even on these additional assumptions the separable Hahn property and the Hahn property still do not coincide. However if we assume E to be the bounded summability domain of a regular Riesz matrix Rp or a regular nonnegative Hausdorff matrix Hp, then this assumption alone guarantees that E has the Hahn property. For any (infinite) matrix A the Hahn property of its bounded summability domain is related to the strongly nonatomic property of the density dA defined by A. We will find a simple necessary and sufficient condition for the density dA defined by the generalized Riesz matrix Rp,m to be strongly nonatomic. This condition appears also to be sufficient for the bounded summability domain of Rp,m to have the Hahn property.


---

## Body

## 1. Preliminaries and Introduction

We start with some preliminaries. For other notations and preliminary results we refer the reader to [1–3].Letχ denote the set of all sequences of 0s and 1s and let χ(E) denote the linear hull of χ∩E.An FK-space is a sequence space endowed with a complete, metrizable, locally convex topology under which all coordinate mappingsx=(xj)→xk(k∈ℕ) are continuous.A sequence spaceE is said to have the Hahn property, the separable Hahn property, and the matrix Hahn property, if χ(E)⊂F implies E⊂F whenever F is any FK-space, a separable FK-space, and a matrix domain cB, respectively. Obviously, the Hahn property implies the separable Hahn property, and the latter implies the matrix Hahn property.IfE has the matrix Hahn property then χ(E)β=Eβ (cf. Theorem 5.1 in [1]), but, in general, the inverse implication does not hold even for monotone sequence spaces (see Theorem 1.1 in [4]). Still if we ask E to be a solid sequence space containing φ (the set of all finite sequences) and satisfying Eβ=ℓ1, then χ(E)β=Eβ implies the separable Hahn property of E (see [2, Theorem 6]). This result suggests the following problem due to Boos and Leiger (cf. Problem 3 in [2]). Now, we formulate this question as a problem and give a negative answer to it.Problem 1.
LetE be a solid sequence space containing φ and satisfying χ(E)β=Eβ=ℓ1. Then E has the separable Hahn property. Does it have the Hahn property?To answer this problem we consider some facts from the theory of double sequence spaces.A double sequence space is a linear subspace ofΩ, the space of all real double sequences x=(xkl). In particular, the following sets are double sequence spaces:
(1)ℳu≔{x∈Ω∣∥x∥∞≔supk,l|xk,l|<∞},𝒞be0≔{x∈Ω∣(xkl)k∈ℓ∞(l∈ℕ),(limsupk|xk,l|)l∈c0}.
We denote by ξ1 the set of double sequences of zeros and ones in 𝒞be0; that is,
(2)ξ1≔{0,1}ℕ×ℕ∩𝒞be0={0,1}ℕ×ℕ∩(ℳu∩𝒞be0).The space of all double sequencesΩ can be identified with the space of all sequences ω using a suitable isomorphism T:Ω→ω (see (2) in [5] for a possible definition of T).Theorem 1.
There exists a solid sequence spaceE containing φ and satisfying χ(E)β=Eβ=ℓ1 which does not have the Hahn property.Proof.
SetE≔T(ℳu∩𝒞be0). Then E is a solid sequence space containing φ and failing to have the Hahn property (see Example 3.7 in [5]). In view of Theorem 3.2 in [5] we have also χ(E)β=(T(ξ1))β=ℓ1 as well as Eβ=ℓ1.Taking the explained situation concerning the sufficiency ofχ(E)β=Eβ=ℓ1 for the Hahn property of E into consideration, it is mathematically interesting to research for classes of sequence spaces E with the property that χ(E)β=Eβ=ℓ1 implies the Hahn property of E (or another equivalent condition). In that sense we consider on the base of related results in [6, 7] the bounded domains of Riesz methods in Section 2 and, more general, of the generalized Riesz methods in Section 3.


## 2. Bounded Domains of Riesz Methods

Letp=(pn) be a real sequence with
(3)p1>0,pk≥0(k∈ℕ),Pn≔∑k=1n‍pk(n∈ℕ).
The Riesz matrix Rp=(rnk) (associated with p) is defined by
(4)rnk≔{pkPnifk≤n,0otherwise(n,k∈ℕ).
The summability method corresponding Rp is called Riesz method.The Riesz matrixRp is conservative, and it is either regular (being equivalent to p∉ℓ1) or coercive. If p∈ℓ1, then ℓ∞∩cRp has the Hahn property since ℓ∞∩cRp=ℓ∞. If p∉ℓ1, then ℓ∞∩cRp has the Hahn property if and only if (pn/Pn)∈c0 (cf. Corollary 3.9 in [6]).In Section1 we have seen that the relation χ(E)β=Eβ=ℓ1 does not imply in general the Hahn property of E even for a solid space E. This result suggests the following problem due to Boos and Leiger (cf. Problem 3 in [7, Section 4]).Problem 2.
Does the relationχ(ℓ∞∩cRp)β=ℓ1 imply the Hahn property of ℓ∞∩cRp?Aiming to a positive result we first prove the following more general theorem.Theorem 2.
LetB=(bnk) be an infinite matrix such that there exists a subset F of χ with φ⊂〈F〉⊂ℓB∞ and Fβ(=〈F〉β)=ℓ1. Then ℓ∞⊂ℓB∞ (being equivalent to supn∑k‍|bnk|<∞).Proof.
In view ofF⊂ℓB∞ we have supn|∑k‍bnkxk|<∞ for each x∈F, so in particular (bnk)k∈Fβ=ℓ1 for each n∈ℕ.
Assume on contrary that there exists anx∈ℓ∞∖ℓB∞. Hence supn∑k‍|bnk|=∞ since
(5)|∑k‍bnkxk|≤∥x∥∞∑k‍|bnk|(x∈ℓ∞,n∈ℕ).
Aiming to a contradiction we will construct by induction two index sequences(ki) and (ni), and then with the help of them we will define w∈Fβ∖ℓ1. First of all set k1≔0 and A1≔0. Now choose n1∈ℕ such that
(6)∑k‍|bn1k|>12(A1+1)+12.
Then choose k2>k1 such that
(7)∑k=k2+1∞‍|bn1k|<12,
and set
(8)B1≔∑k=k1+1k2‍|bn1k|.
Now suppose that k1,…,ki and n1,…,ni-1 are already chosen for i>1. Since φ⊂ℓB∞, then
(9)Ai≔supn∑k=1ki‍|bnk|<∞.
We chooseni>ni-1 such that
(10)∑k‍|bnik|>i2(Ai+1)+12i
and then we take ki+1>ki such that
(11)∑k=ki+1+1∞‍|bnik|<12i.
So
(12)Bi≔∑k=ki+1ki+1‍|bnik|>i2(Ai+1),AiBi<1i2.
Now we define a w=(wk) by
(13)wk≔bnikBiiforki+1≤k≤ki+1,i∈ℕ
and note that w∉ℓ1 because
(14)∑k‍|wk|=∑i‍1Bii∑k=ki+1ki+1‍|bnik|=∑i1i=∞.
We are going to verify w∈Fβ in contradiction to Fβ=ℓ1. For that end let u∈F and s,t∈ℕ with t>s be fixed. We choose i1,i2∈ℕ such that ki1-1<s≤ki1 and ki2<t≤ki2+1. If ki1-1=ki2 we have
(15)|∑k=st‍wkuk|=|∑k=st‍bni1-1kukBi1-1(i1-1)|≤1Bi1-1(i1-1)∑k=ki1-1+1ki1‍|bni1-1k|=1i1-1⟶0,ass,t⟶∞.
If ki1≤ki2 we get
(16)∑k=st‍wkuk=∑k=ski1‍bni1-1kukBi1-1(i1-1)+∑i=i1i2-1‍∑k=ki+1ki+1‍bnikukBii+∑k=ki2+1t‍bni2kukBi2i2.
Since u∈F⊂ℓB∞, we have supn|[Bu]n|≔M<∞; hence,
(17)|∑k=ki+1ki+1‍bnikukBii|≤1Bii(|[Bu]ni|+∑k=1ki‍|bnik|+∑k=ki+1+1∞‍|bnik|)≤1Bii(M+Ai+12i)<1+M+2-ii3
for any i=i1,…,i2-1. So for the second term on the right-hand side of (16) we have the estimate
(18)|∑i=i1i2-1‍∑k=ki+1ki+1‍bnikukBii|≤∑i=i1i2-1‍1+M+2-ii3.
For the first and the third term on the right-hand side of (16) we have the estimates
(19)|∑k=ski1‍bni1-1kukBi1-1(i1-1)|≤1Bi1-1(i1-1)∑k=ki1-1+1ki1‍|bni1-1k|=1i1-1,|∑k=ki2+1t‍bni2kukBi2i2|≤1i2.
Hence
(20)|∑k=st‍wkuk|≤1i1-1+∑i=i1i2-1‍1+M+2-ii3+1i2⟶0,ass,t⟶∞;
so the series ∑k‍wkuk converges and w∈Fβ.The last theorem provides us an alternative way (cf. Theorem 2.3.8 in [3]) to show that every matrix summing all thin sequences is conservative for null sequences. Let 𝒯 denote the set of all thin sequences. Then we have the following.Corollary 3.
LetB be a matrix satisfying 𝒯⊂cB, then c0⊂cB. Moreover, if 𝒯∪{e}⊂cB; then B is conservative and B∈KG (i.e., each matrix A with χ(ℓ∞∩cB)⊂cA is conservative).Proof.
TakingF=𝒯 in Theorem 2 in view of 𝒯β=ℓ1 we have ℓ∞⊂ℓB∞, so supn∑k‍|bnk|<∞. Since ek∈〈𝒯〉⊂cB it follows that  c0⊂cB.
The second part of the corollary follows immediately from the first one.The following two corollaries are basic for the answer to Problem2.Corollary 4.
LetB=(bnk) be an infinite matrix, and let F be a subset of χ with φ⊂〈F〉⊂ℓB∞ and Fβ=ℓ1. Let E be a subset of ℓ∞ such that Eβ=ℓ1 and F∩E⊂cB. Furthermore, let (E,τE) be an F-space with a dense subset F∩E. Then E⊂cB.Proof.
We consider the functionalsfn:E→ℝ with fn(x)=∑k‍bnkxk(n∈ℕ). By Theorem 2 the sequence (fn) is pointwise bounded on E⊂ℓ∞, and by our assumptions it converges pointwise on the dense subset F∩E of E. Hence by generalized version of Banach-Steinhaus theorem (cf. Theorem 6.8.6 in [3]) the sequence (fn) converges pointwise on E. So E⊂cB.Corollary 5.
LetA be a conservative matrix. Then A∈KG if and only if χ(cA∩ℓ∞)β=ℓ1.Proof.
SupposeA∈KG and let u∈χ(cA∩ℓ∞)β. We set bnk≔uk(k,n∈ℕ). Then χ(cA∩ℓ∞)⊂cB; hence, B is conservative. Therefore u∈cβ=ℓ1. Hence χ(cA∩ℓ∞)β⊂ℓ1. On the other hand in view of conservativity of A we have χ(cA∩ℓ∞)β⊃(cA∩ℓ∞)β=ℓ1.
Now letB be a matrix satisfying χ∩cA⊂cB. Take E=c0 and F=χ(cA∩ℓ∞) in Corollary 4. Then since χ(cA∩ℓ∞)∩c0=φ is dense in (c0,∥∥∞), we have c0⊂cB. Moreover, since e∈χ∩cA⊂cB, it follows that c⊂cB. Hence A∈KG.Now we give a positive answer to Problem2.Proposition 6.
LetRp be regular. Then ℓ∞∩cRp has the Hahn property if and only if χ(ℓ∞∩cRp)β=ℓ1.Proof.
In view of Corollary5 we have χ(ℓ∞∩cRp)β=ℓ1 if and only if Rp∈KG, but the last condition is equivalent to the Hahn property of ℓ∞∩cRp (cf. Corollary 3.9 in [6]).In the same way as for Riesz matrices we obtain a similar result for nonnegative Hausdorff matrices, where we make use that a nonnegative regular Hausdorff matrixHp is KG if and only if ℓ∞∩cHp has the Hahn property (Theorem 3.2.1 in [7]).Proposition 7.
LetHp be a nonnegative regular Hausdorff matrix. Then ℓ∞∩cHp has the Hahn property if and only if χ(ℓ∞∩cHp)β=ℓ1.


## 3. Bounded Domains of Generalized Riesz Methods

In the previous section we demonstrated that forA=Rp the relation χ(ℓ∞∩cA)β=ℓ1 is equivalent to the Hahn property of ℓ∞∩cA. Now we consider two more conditions being equivalent to these properties for A=Rp (cf. [6, Corollary 3.9 (ii)], [7, Theorem 3.1.1]): (a)
A=(ank) has spreading rows; that is, (supk∈ℕank)∈c0;(b)
there exists a sequence(21)(𝒩t)=({Nνt∣ν∈ℕνt}),νt∈ℕ(t∈ℕ)of partitions ofℕ (called an admissible partition sequence) such that
(22)limsupt→∞limsupn→∞sup1≤ν≤νtAnνt=0whereAnνt≔∑k∈Nνt‍ank.Recall that𝒩t={Nνt∣ν∈ℕνt} is called a partition of ℕ if ℕ=⋃ν=1νt‍Nνt and Nνt∩Nμt=∅ if ν≠μ.So in fact conditions (a) and (b) are equivalent forA=Rp. Note that for any nonnegative and regular for null sequences matrix A condition (b) is equivalent to the following condition (see [7] for the corresponding definitions).(b′)
DensitydA defined by the matrix A is strongly nonatomic on the power set P(ℕ) of ℕ.Now we replace a Riesz matrixRp with a more general matrix obtained as a row submatrix of Rp. More precisely, we consider the matrix Rp,m=(an,k), called generalized Riesz matrix, defined by
(23)ank≔{pkPmnif1≤k≤mn,0otherwise,(n,k∈ℕ),
where (mn) is any fixed index sequence and p=(pi) is a sequence of positive reals with p∉ℓ1.Boos and Leiger showed (cf. Theorem 2.1 in [8]) that in the case of some sufficient conditions on terms of Rp,m the properties (a) and (b) are equivalent for A=Rp,m. These conditions covered all possible cases except
(24)limsupnPmnPmn+1=1,liminfnPmnPmn+1=0.
Note that even in this case the implication (b)⇒(a) holds (cf. Theorem 2.1 in [8]). So only the implication (a)⇒(b) in case (24) was under the question.In relation with this in [8], Problem 2.3 Boos and Leiger posed the following.Problem 3.
Complete the distinction of cases in [8, Theorem 2.1]. May be the relation 𝒯⊂c0Rp,m hold?First we answer the first question in Problem3 and then demonstrate that the relation 𝒯⊂c0Rp,m does not hold in general for generalized Riesz matrices Rp,m satisfying (24).For our first aim we apply—in the same way as Boos and Leiger in [8]—the following lemma due to Kuttner and Parameswaran (cf. [9, Lemma 2]).Lemma 8.
Let{z1,z2,…,zn} for any n∈ℕ be a set of nonnegative real numbers. Let Zn≔∑k=1n‍zk and suppose that Bn>0 with zk≤Bn(k∈ℕn), and let t∈ℕ be arbitrarily given. Then one can divide the set ℕn into t (pairwise disjoint) subsets N1,N2,…,Nt (some of them may be empty), such that
(25)∑k∈Ns‍zk≤1tZn+Bn(s=1,2,…,t).Proposition 9.
Suppose thatRp,m=(ank) has spreading rows; that is,
(26)limn1Pmnsup1≤k≤mnpk=0,(27)liminfnPmnPmn+1=0;
then there exists an admissible partition sequence of ℕ satisfying (22).Proof.
We setAn≔Pmn. For a given t∈ℕ in view of (27) we can choose the minimal index n1 such that An1/A1>(t+1)/t. Afterwards (27) we choose the minimal index n2>n1 such that An2/An1>(t+1)/t. Continuing in the same way we choose the minimal index ni>ni-1 such that Ani/Ani-1>(t+1)/t. In view of (26) it follows that mni+1-mni→∞. Again by (26) we can choose αt∈ℕ, αt≥n1 such that
(28)pkAn<1t2foreachk∈ℕmn,n≥αt;
also we can find βt∈ℕ, βt>αt such that
(29)AαtAn<1t2forn≥βt.
By Lemma8 for any positive integer s≥αt we can divide the set Is≔(mns,mns+1]∩ℕ into t2 disjoint subsets (some of them may be empty), say Nst(1),Nst(1),…,Nst(t2) such that (cf. (28))
(30)∑k∈Nst(ν)‍pk≤1t2∑k∈Is‍pk+1t2Ans+1≤2t2Ans+1.
Now, setting
(31)Nνt≔⋃s‍Nst(ν),Nt≔{Nνt∣ν∈ℕt2},
we get an admissible partition sequence 𝒩≔(𝒩t)t of ℕ. We verify that this partition satisfies (22). Given n≥βt we find r∈ℕ such that nr≤n<nr+1. By construction of (ni) it follows that Anr≥tAn/(t+1). Therefore
(32)1An∑k∈Nνt,k≤mn‍pk≤AαtAn+1An∑s=αtr-1‍∑k∈Nst(ν)‍pk+An-AnrAn≤1t2+2t21An∑s=1r-1‍Ans+1+1t+1≤1t2+2t2AnrAn∑s=0∞‍(tt+1)s+1t+1≤1t2+2t211-(t/(t+1))+1t+1=2t+3t2+1t+1.
Therefore
(33)limsupt→∞limsupn→∞supν∈ℕt21An∑k∈Nνt,k≤mn‍pk=0.Now combining the obtained result with Theorem 2.1 in [8] we get the following.Theorem 10.
Letp=(pi) be a sequence of positive reals with p∉ℓ1. Then the following conditions are equivalent:(i)
Rp,m has spreading rows; that is, limn(1/Pmn)sup1≤k≤mnpk=0;(ii)
dRp,m is strongly nonatomic.The following example demonstrates that the assumption of Problem3 that 𝒯⊂c0Rp,m in the case of (24) is not true in general.Example 11.
We setm1≔1, pm1≔1, m2≔m1+1, pm2≔2-m2, m3≔m2+22[Pm2], pm3≔1, pm2+2k≔1 for k=1,…,2[Pm2]-1, and pk≔2-k for m2<k<m3, k∉{m2+2k∣k=1,…,2[Pm2]-1}.
Suppose thatm2n-2, m2n-1, and pk for k≤m2n-1 are defined. We set m2n≔m2n-1+1 and m2n+1≔m2n+(n+1)2[Pm2n]. Then we set pm2n≔2-m2n, pm2n+1≔1, pm2n+k(n+1)≔1(k=1,…,[Pm2n](n+1)-1), and pk≔2-k for m2n<k<m2n+1, k∉{m2n+k(n+1)∣k=1,…,[Pm2n](n+1)-1}. Then
(34)Pm2n+1Pm2n≥Pm2n+(n+1)[Pm2n]Pm2n⟶∞,Pm2n-1Pm2n⟶1.
Consequently the matrixRp,m satisfies condition (24).
On the other hand forx∈𝒯 having xn≔1 when pn=1 and xn≔0 otherwise (n∈ℕ) we have
(35)[Ax]m2n+1≥1Pm2n+1∑k=1[Pm2n](n+1)‍pm2n+k(n+1)≥(n+1)[Pm2n]Pm2n+(n+1)[Pm2n]+∑i2-i⟶1.
Hencex∉c0Rp,m.As well as for Riesz matrices in the case of a generalized Riesz matrix the assumption that the matrix has spreading rows implies that its bounded summability domain has the Hahn property. To prove it we will adjust the methods developed in [6, Theorem 3.8].LetI be an at most countable set, and let N={Ni∣i∈I} be a partition of ℕ. Let (nik)k be the sequence of all elements of Ni arranged in the ascending order (i∈I). We introduce the notation
(36)bs(N)≔{x∈ω∣∀j∈I:∥x∥bs(N)≔supj∥(xk)k∈Nj∥bs<∞}.For the proof of the main result we need two lemmas.Lemma 12 (cf. Lemma 3.6 in [6]).
Letn∈ℕ and p1,…,pn be a (finite) sequence of numbers. Then there exists a partition M1,…,Mt of {1,…,n} such that t≤2n and (pi)i∈Mj is monotone (j=1,…,t).The second required lemma generalizes Lemma 3.7 in [6] concerning Riesz matrices.Lemma 13.
LetN={Ni∣i∈I} be a partition of ℕ. If (37)1Pmn∑s∈ℕNs∩[1,mn]≠∅(∑j∈ℕns,j+1≤mn|pnsj-pns,j+1|+∑j∈ℕnsj≤mn<ns,j+1pnsj)⟶0(n⟶∞),
thenbs(Ni,I)⊂c0Rp,m⊂cRp,m.Proof.
To get a proof of the statement just replace everywheren with mn in the proof of Lemma 3.7 [6], the case of Rp.In the next theorem we generalize Theorem 3.8 in [6] in the case of Riesz matrices Rp. The proof requires nontrivial refinements of the methods used in the corresponding part of the proof in [6].Theorem 14.
Let(pn) be a positive sequence such that Pn→∞ and Rp,m=(ank) has spreading rows. Then cRp,m∩ℓ∞ has the Hahn property.Proof.
By Lemma13 it is sufficient to verify the existence of a partition N={Ni∣i∈I} of ℕ satisfying the condition in the lemma. Aiming to that we set k~0≔k0≔1 and suppose that k1,…,ki,k~1,…,k~i are already chosen. Since Pn→∞, we can choose
(38)k~i+1≔min{k∈ℕ∣PmkPmki≥2}.
If Pmk~i+1/Pmki≤8 or k~i+1=ki+1 we set ki+1≔k~i+1. Otherwise we set ki+1≔k~i+1-1 and ki+2≔k~i+2≔k~i+1. In the latter case we have Pmki+1/Pmki<2 and ki+2=ki+1+1. Hence in both cases
(39)Pmki+2Pmki≥2,Pmki+1Pmki≤8orki+1=ki+1.
Since Rp,m=(ank) has spreading rows, then
(40)ɛi≔max{pj∣j≤mki}Pmki⟶0.
Let i0∈ℕ be such that ɛi<1 for i≥i0.
For everyi≥i0 we choose the minimal integer ji∈ℕ such that mki1/ji<ɛi-1/2. Since ɛi=max{pk∣k≤mki}/Pmki≥1/mki, then mki≥ɛi-1 (i≥i0). Hence ji>2 (i≥i0). Note that mki1/ji≥ɛi-1/4(i≥i0). (Otherwise mki1/[(ji+2)/2]≤mki2/ji<ɛi-1/2, so ji/2+1≥[ji/2]+1≥ji, which is equivalent to ji≤2 and contradicts ji>2).
Set(41)αi≔Pmki+1mki+1,βi≔max{pk∣k≤mki+1}(i∈ℕ).
For every i∈ℕ and s=2,…,ji+1-2 we use the notations
(42)Mi1≔{mki<k≤mki+1∣pk≤mki+11/ji+1αi},Mis≔{mki<k≤mki+1∣mki+1(s-1)/ji+1αi<pk≤mki+1s/ji+1αi},Mi,ji+1-1≔{mki<k≤mki+1∣mki+1(ji+1-2)/ji+1αi<pk}.
Set νis≔|Mis|(s=1,…,ji+1-1). Note that
(43)νis≤Pmki+1mki+1(s-1)/ji+1αi=mki+1αimki+1(s-1)/ji+1αi=mki+1(ji+1+1-s)/ji+1.
By Lemma12 for every i∈ℕ and s=1,…,ji+1-1 we may find a partition (Sis1,…,Sisris) of Mis with ris≤2νis such that (pν)ν∈Sist is monotone (t=1,…,ris). Let eisk≔|Sisk| and let (ξiskl)l=1eisk be the finite sequence of all elements of Sisk arranged in the ascending order (i∈ℕ,s=1,…,ji+1-1,k=1,…,ris). Then
(44)∑k=1ris‍(pξisk1+∑l=1eisk-1‍|pξiskl-pξi,s,k,l+1|+pξiskeisk)=∑k=1ris‍2max{pξisk1,pξiskeisk}≤2rismki+1s/ji+1αi≤4νismki+1s/ji+1αi≤4mki+1(ji+1+1-s)/2ji+1mki+1s/ji+1αi=4mki+1(ji+1+1+s)/2ji+1αi
for s=1,…,ji+1-2 and
(45)∑k=1ris‍(pξisk1+∑l=1eisk-1‍|pξiskl-pξi,s,k,l+1|+pξiskeisk)=∑k=1ris‍2max{pξisk1,pξiskeisk}≤4νisβi≤4mki+11/ji+1βi
for s=ji+1-1(i∈ℕ). Hence for every i≥i0 we get(46)1Pmki+1∑s=1ji+1-1‍∑k=1ris‍(pξisk1+∑l=1eisk-1‍|pξiskl-pξi,s,k,l+1|+pξiskeisk)≤1Pmki+1(∑s=1ji+1-2‍4mki+1(ji+1+1+s)/2ji+1αi+4mki+11/ji+1βi)=4mki+1(ji+1+1)/2ji+1αimki+1αi(mki+1(ji+1-1)/2ji+1-1mki+11/2ji+1-1-1)+4βimki+11/ji+1Pmki+1≤4mki+1(-ji+1+1)/2ji+1mki+1(ji+1-1)/2ji+14-1mki+11/2ji+1+4ɛi+1ɛi+1-1/2=16mki+1-1/2ji+1+4ɛi+1-1/2≤16ɛi+11/8+4ɛi+11/2.To define Ni(i∈ℕ) we use the notation N|n≔{k∈N∣k≤n} for a subset N⊂ℕ and n∈ℕ.
SetN1|mki0≔{1,…,mki0} and Nj|mi0≔∅ for j>1. Let s10,…,st00 be all indexes such that Mi0st0≠∅(t=1,…,t0). We set
(47)Nj|mki0+1≔{Nj|mki0∪Si0st0kforj=∑τ=1t-1‍ri0sτ0+k(t=1,…,t0,k=1,…,ri0st0),Nj|mki0forj>∑τ=1t0‍ri0sτ0.
Continuing inductively fori>i0 let s1i-i0,…,sti-i0i-i0 be all indexes such that Misti-i0≠∅(t=1,…,ti-i0). We set
(48)Nj|mki+1≔{Nj|mki∪Sisti-i0kforj=∑τ=1t-1‍risτi-i0+k(t=1,…,ti-i0,k=1,…,ri0sti-i0),Nj|mki0forj>∑τ=1ti-i0‍risτi-i0.
LetI be the set of all indexes i∈ℕ such that Ni≠∅. If I is infinite, then by our construction every Ni is infinite. If I is finite, then without loss of generality we may assume that Ni is infinite (i∈I).
Letn∈ℕ with ki<n≤ki+1 and i≥i0 be fixed. Then (49)Bn≔1Pmn×∑Ns∩[1,mn]≠∅s∈ℕ‍(∑ns,j+1≤mnj∈ℕ‍|pnsj-pns,j+1|+∑nsj≤mn<ns,j+1j∈ℕ‍pnsj)≤1Pmn×(2∑k=1mki0‍pk+∑τ=i0i‍∑s=1jτ+1-1‍∑k=1rτs‍(pξisk1+∑l=1eτsk-1‍|pξτskl-pξτ,s,k,l+1|+pξiskeisk))≤1Pmn(2Pmki0+∑τ=i0i‍(16ɛτ+11/8+4ɛτ+11/2)Pmτ+1).If ki+1=ki+1 then n=ki+1, so Pn=Pki+1. If ki+1>ki+1 then Pn≥Pki≥8-1Pki+1. Hence
(50)Bn≤8Pki+1(2Pmi0+∑τ=i0i‍(16ɛτ+11/8+4ɛτ+11/2)Pmτ+1)≤162[(i-i0+1)/2]+32∑τ=i0i‍12[(i-τ)/2](4ɛτ+11/8+ɛτ+11/2).
Consider the matrix A=(aiτ) with aiτ≔2[(τ-i)/2] for τ≤i and aiτ≔0 otherwise. Evidently, A is regular for null sequences, so it sums the null sequence (4ɛτ+i01/8+ɛτ+i01/2)τ to zero. Therefore Bn→0(n→∞). So in view of Lemma 13 the inclusion bs(N)⊂cRp,m holds.Now having in mind Proposition 2.8 in [7] we can prove the following theorem.Theorem 15.
Letp=(pi) be a sequence of positive reals with p∉ℓ1. Then one considers the following conditions:(a)
ℓ∞∩cRp,m has the matrix Hahn property,(b)
Rp,m has spreading rows; that is, limn(1/Pmn)sup1≤k≤mnpk=0,(c)
dRp,m is strongly nonatomic,(d)
Rp,m∈KG,(e)
ℓ∞∩cRp,m has the separable Hahn property,(f)
ℓ∞∩cRp,m has the Hahn property,(g)
χ(cRp,m∩ℓ∞)β=ℓ1.
Then the implications (c)⇔(b)⇒(f)⇒(e)⇒(a)⇒(d)⇔(g) hold.Proof.
The implication chain (f)⇒(e)⇒(a)⇒(d) is obvious. (c)⇔(b) is Theorem 10, (b)⇒(f) is Theorem 14, and (d)⇔(g) follows from Corollary 5.The author did not succeed to show that the condition(51)limn1Pmnsup1≤k≤mnpk=0
is necessary for Rp,m∈KG. The following example gives a generalized Riesz matrix Rp,m not having spreading rows and failing to have Rp,m∈KG.Example 16.
Supposep∉ℓ1 and
(52)liminfn(1Pmn)sup1≤k≤mnpk>12.
Then there exists δ>1/2 and n0∈ℕ such that (1/Pmn)sup1≤k≤mnpk≥δ for n≥n0.
Letkn≤mn be an integer such that pkn=sup1≤k≤mnpk. Then
(53)1Pmn∑k≠knk=1mn‍pk≤1-δ<12.
Since Pn→∞ we can choose an index sequence (ni) such that (kni) is increasing. Now if x∈χ∩cRp,m then there exists i0=i0(x) such that xkni=0 for all i≥i0 or xkni=1 for all i≥i0. We set ykni≔(-1)i/i for i∈ℕ and yk≔0 for k∉{kni∣i∈ℕ}. Then the series
(54)∑k=kni0∞‍xkyk=∑i=i0∞‍(-1)iixkni
converges, so y∈χ(cRp,m∩ℓ∞)β∖ℓ1.Problem 4.
In Theorem15 we have shown that the fact that Rp,m has spreading rows implies Rp,m∈KG. Does the inverse implication hold?



---

*Source: 908682-2013-10-21.xml*