Bounded Domains of Generalized Riesz Methods with the Hahn Property

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 space 𝐸 is solid and ({0,1}^ℕ∩𝐸)^𝛽=𝐸^𝛽=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 𝑅_𝑝 or a regular nonnegative Hausdorff matrix 𝐻_𝑝, 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 𝑑_𝐴 defined by 𝐴. We will find a simple necessary and sufficient condition for the density 𝑑_𝐴 defined by the generalized Riesz matrix 𝑅_𝑝,𝑚 to be strongly nonatomic. This condition appears also to be sufficient for the bounded summability domain of 𝑅_𝑝,𝑚 to have the Hahn property.

1. Preliminaries and Introduction

We start with some preliminaries. For other notations and preliminary results we refer the reader to [1–3].

Let 𝜒 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 mappings 𝑥=(𝑥_𝑗)→𝑥_𝑘(𝑘∈ℕ) are continuous.

A sequence space 𝐸 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 𝑐_𝐵, respectively. Obviously, the Hahn property implies the separable Hahn property, and the latter implies the matrix Hahn property.

If 𝐸 has the matrix Hahn property then 𝜒(𝐸)^𝛽=𝐸^𝛽 (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 𝐸^𝛽=ℓ¹, then 𝜒(𝐸)^𝛽=𝐸^𝛽 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. Let 𝐸 be a solid sequence space containing 𝜑 and satisfying 𝜒(𝐸)^𝛽=𝐸^𝛽=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 of Ω, the space of all real double sequences 𝑥=(𝑥_𝑘𝑙). In particular, the following sets are double sequence spaces: ℳ_𝑢∶=𝑥∈Ω∣‖𝑥‖_∞∶=sup_𝑘,𝑙𝑥_𝑘,𝑙,𝒞<∞_be0∶=𝑥∈Ω∣𝑥_𝑘𝑙𝑘∈ℓ^∞(𝑙∈ℕ),limsup_𝑘𝑥_{𝑘,𝑙𝑙}∈𝑐₀. (1) We denote by 𝜉₁ the set of double sequences of zeros and ones in 𝒞_be0; that is, 𝜉₁∶={0,1}^ℕ×ℕ∩𝒞_be0={0,1}^ℕ×ℕ∩ℳ_𝑢∩𝒞_be0. (2)

The space of all double sequences Ω 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 space 𝐸 containing 𝜑 and satisfying 𝜒(𝐸)^𝛽=𝐸^𝛽=ℓ¹ which does not have the Hahn property.

Proof. Set 𝐸∶=𝑇(ℳ_𝑢∩𝒞_be0). 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 𝐸^𝛽=ℓ¹.

Taking the explained situation concerning the sufficiency of 𝜒(𝐸)^𝛽=𝐸^𝛽=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=𝐸^𝛽=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

Let 𝑝=(𝑝_𝑛) be a real sequence with 𝑝₁>0,𝑝_𝑘(),𝑃≥0𝑘∈ℕ_𝑛∶=^𝑛_𝑘=1𝑝_𝑘(𝑛∈ℕ). (3) The Riesz matrix 𝑅_𝑝=(𝑟_𝑛𝑘) (associated with 𝑝) is defined by 𝑟_𝑛𝑘𝑝∶=𝑃^𝑘_𝑛()if𝑘≤𝑛,0otherwise𝑛,𝑘∈ℕ. (4) The summability method corresponding 𝑅_𝑝 is called Riesz method.

The Riesz matrix 𝑅_𝑝 is conservative, and it is either regular (being equivalent to 𝑝∉ℓ¹) or coercive. If 𝑝∈ℓ¹, then ℓ^∞∩𝑐_𝑅𝑝 has the Hahn property since ℓ^∞∩𝑐_𝑅𝑝=ℓ^∞. If 𝑝∉ℓ¹, then ℓ^∞∩𝑐_𝑅𝑝 has the Hahn property if and only if (𝑝_𝑛/𝑃_𝑛)∈𝑐₀ (cf. Corollary 3.9 in [6]).

In Section 1 we have seen that the relation 𝜒(𝐸)^𝛽=𝐸^𝛽=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 relation 𝜒(ℓ^∞∩𝑐_𝑅𝑝)^𝛽=ℓ¹ imply the Hahn property of ℓ^∞∩𝑐_𝑅𝑝?

Aiming to a positive result we first prove the following more general theorem.

Theorem 2. Let 𝐵=(𝑏_𝑛𝑘) be an infinite matrix such that there exists a subset 𝐹 of 𝜒 with 𝜑⊂⟨𝐹⟩⊂ℓ^∞_𝐵 and 𝐹^𝛽(=⟨𝐹⟩^𝛽)=ℓ¹. Then ℓ^∞⊂ℓ^∞_𝐵 (being equivalent to sup_𝑛∑_𝑘|𝑏_𝑛𝑘|<∞).

Proof. In view of 𝐹⊂ℓ^∞_𝐵 we have sup_𝑛|∑_𝑘𝑏_𝑛𝑘𝑥_𝑘|<∞ for each 𝑥∈𝐹, so in particular (𝑏_𝑛𝑘)_𝑘∈𝐹^𝛽=ℓ¹ for each 𝑛∈ℕ.
Assume on contrary that there exists an 𝑥∈ℓ^∞⧵ℓ^∞_𝐵. Hence sup_𝑛∑_𝑘|𝑏_𝑛𝑘|=∞ since _𝑘𝑏_𝑛𝑘𝑥_𝑘≤‖𝑥‖_∞𝑘𝑏_𝑛𝑘𝑥∈ℓ^∞,𝑛∈ℕ. (5)
Aiming to a contradiction we will construct by induction two index sequences (𝑘_𝑖) and (𝑛_𝑖), and then with the help of them we will define 𝑤∈𝐹^𝛽⧵ℓ¹. First of all set 𝑘₁∶=0 and 𝐴₁∶=0. Now choose 𝑛₁∈ℕ such that _𝑘𝑏_𝑛1𝑘>1²𝐴₁+1+12. (6) Then choose 𝑘₂>𝑘₁ such that ^∞_𝑘=𝑘2+1𝑏_𝑛1𝑘<12, (7) and set 𝐵₁∶=^𝑘2_𝑘=𝑘1+1𝑏_𝑛1𝑘. (8) Now suppose that 𝑘₁,…,𝑘_𝑖 and 𝑛₁,…,𝑛_𝑖−1 are already chosen for 𝑖>1. Since 𝜑⊂ℓ^∞_𝐵, then 𝐴_𝑖∶=sup_𝑛^𝑘𝑖_𝑘=1𝑏_𝑛𝑘<∞. (9)
We choose 𝑛_𝑖>𝑛_𝑖−1 such that _𝑘𝑏_𝑛𝑖𝑘>𝑖²𝐴_𝑖+1+12^𝑖 (10) and then we take 𝑘_𝑖+1>𝑘_𝑖 such that ^∞_{𝑘=𝑘𝑖+1+1}𝑏_𝑛𝑖𝑘<12^𝑖. (11) So 𝐵_𝑖∶=^𝑘𝑖+1_{𝑘=𝑘𝑖+1}𝑏_𝑛𝑖𝑘>𝑖²𝐴_𝑖,𝐴+1^𝑖𝐵_𝑖<1𝑖². (12) Now we define a 𝑤=(𝑤_𝑘) by 𝑤_𝑘𝑏∶=^𝑛𝑖𝑘𝐵_𝑖𝑖for𝑘_𝑖+1≤𝑘≤𝑘_𝑖+1,𝑖∈ℕ (13) and note that 𝑤∉ℓ¹ because _𝑘𝑤_𝑘=_𝑖1𝐵_𝑖𝑖^𝑘𝑖+1_{𝑘=𝑘𝑖+1}𝑏_𝑛𝑖𝑘=_𝑖1𝑖=∞. (14) We are going to verify 𝑤∈𝐹^𝛽 in contradiction to 𝐹^𝛽=ℓ¹. For that end let 𝑢∈𝐹 and 𝑠,𝑡∈ℕ with 𝑡>𝑠 be fixed. We choose 𝑖₁,𝑖₂∈ℕ such that 𝑘_𝑖1−1<𝑠≤𝑘_𝑖1 and 𝑘_𝑖2<𝑡≤𝑘_𝑖2+1. If 𝑘_𝑖1−1=𝑘_𝑖2 we have ^𝑡_𝑘=𝑠𝑤_𝑘𝑢_𝑘=^𝑡_𝑘=𝑠𝑏_{𝑛𝑖1−1𝑘}𝑢_𝑘𝐵_𝑖1−1𝑖₁≤1−1𝐵_𝑖1−1𝑖₁−1^𝑘𝑖1_{𝑘=𝑘𝑖1−1+1}𝑏_{𝑛𝑖1−1𝑘}=1𝑖₁−1⟶0,as𝑠,𝑡⟶∞. (15) If 𝑘_𝑖1≤𝑘_𝑖2 we get ^𝑡_𝑘=𝑠𝑤_𝑘𝑢_𝑘=^𝑘𝑖1_𝑘=𝑠𝑏_{𝑛𝑖1−1𝑘}𝑢_𝑘𝐵_𝑖1−1𝑖₁+−1^𝑖2−1_𝑖=𝑖1^𝑘𝑖+1_{𝑘=𝑘𝑖+1}𝑏_𝑛𝑖𝑘𝑢_𝑘𝐵_𝑖𝑖+^𝑡_{𝑘=𝑘𝑖2+1}𝑏_{𝑛𝑖2𝑘}𝑢_𝑘𝐵_𝑖2𝑖₂. (16) Since 𝑢∈𝐹⊂ℓ^∞_𝐵, we have sup_𝑛|[𝐵𝑢]_𝑛|∶=𝑀<∞; hence, ^𝑘𝑖+1_{𝑘=𝑘𝑖+1}𝑏_𝑛𝑖𝑘𝑢_𝑘𝐵_𝑖𝑖≤1𝐵_𝑖𝑖[𝐵𝑢]_𝑛𝑖+^𝑘𝑖_𝑘=1𝑏_𝑛𝑖𝑘+^∞_{𝑘=𝑘𝑖+1+1}𝑏_𝑛𝑖𝑘≤1𝐵_𝑖𝑖𝑀+𝐴_𝑖+12^𝑖<1+𝑀+2^−𝑖𝑖³ (17) for any 𝑖=𝑖₁,…,𝑖₂−1. So for the second term on the right-hand side of (16) we have the estimate ^𝑖2−1_𝑖=𝑖1^𝑘𝑖+1_{𝑘=𝑘𝑖+1}𝑏_𝑛𝑖𝑘𝑢_𝑘𝐵_𝑖𝑖≤^𝑖2−1_𝑖=𝑖11+𝑀+2^−𝑖𝑖³. (18) For the first and the third term on the right-hand side of (16) we have the estimates ^𝑘𝑖1_𝑘=𝑠𝑏_{𝑛𝑖1−1𝑘}𝑢_𝑘𝐵_𝑖1−1𝑖₁≤1−1𝐵_𝑖1−1𝑖₁−1^𝑘𝑖1_{𝑘=𝑘𝑖1−1+1}𝑏_{𝑛𝑖1−1𝑘}=1𝑖₁,−1_{𝑡𝑘=𝑘𝑖2+1}𝑏_{𝑛𝑖2𝑘}𝑢_𝑘𝐵_𝑖2𝑖₂≤1𝑖₂. (19) Hence ^𝑡_𝑘=𝑠𝑤_𝑘𝑢_𝑘≤1𝑖₁+−1^𝑖2−1_𝑖=𝑖11+𝑀+2^−𝑖𝑖³+1𝑖₂⟶0,as𝑠,𝑡⟶∞; (20) so the series ∑_𝑘𝑤_𝑘𝑢_𝑘 converges and 𝑤∈𝐹^𝛽.

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. Let 𝐵 be a matrix satisfying 𝒯⊂𝑐_𝐵, then 𝑐₀⊂𝑐_𝐵. Moreover, if 𝒯∪{𝑒}⊂𝑐_𝐵; then 𝐵 is conservative and 𝐵∈𝐾𝐺 (i.e., each matrix 𝐴 with 𝜒(ℓ^∞∩𝑐_𝐵)⊂𝑐_𝐴 is conservative).

Proof. Taking 𝐹=𝒯 in Theorem 2 in view of 𝒯^𝛽=ℓ¹ we have ℓ^∞⊂ℓ^∞_𝐵, so sup_𝑛∑_𝑘|𝑏_𝑛𝑘|<∞. Since 𝑒^𝑘∈⟨𝒯⟩⊂𝑐_𝐵 it follows that  𝑐₀⊂𝑐_𝐵.
The second part of the corollary follows immediately from the first one.

The following two corollaries are basic for the answer to Problem 2.

Corollary 4. Let 𝐵=(𝑏_𝑛𝑘) be an infinite matrix, and let 𝐹 be a subset of 𝜒 with 𝜑⊂⟨𝐹⟩⊂ℓ^∞_𝐵 and 𝐹^𝛽=ℓ¹. Let 𝐸 be a subset of ℓ^∞ such that 𝐸^𝛽=ℓ¹ and 𝐹∩𝐸⊂𝑐_𝐵. Furthermore, let (𝐸,𝜏_𝐸) be an F-space with a dense subset 𝐹∩𝐸. Then 𝐸⊂𝑐_𝐵.

Proof. We consider the functionals 𝑓_𝑛∶𝐸→ℝ with 𝑓_𝑛(𝑥)=∑_𝑘𝑏_𝑛𝑘𝑥_𝑘(𝑛∈ℕ). By Theorem 2 the sequence (𝑓_𝑛) 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 (𝑓_𝑛) converges pointwise on 𝐸. So 𝐸⊂𝑐_𝐵.

Corollary 5. Let 𝐴 be a conservative matrix. Then 𝐴∈KG if and only if 𝜒(𝑐_𝐴∩ℓ^∞)^𝛽=ℓ¹.

Proof. Suppose 𝐴∈KG and let 𝑢∈𝜒(𝑐_𝐴∩ℓ^∞)^𝛽. We set 𝑏_𝑛𝑘∶=𝑢_𝑘(𝑘,𝑛∈ℕ). Then 𝜒(𝑐_𝐴∩ℓ^∞)⊂𝑐_𝐵; hence, 𝐵 is conservative. Therefore 𝑢∈𝑐^𝛽=ℓ¹. Hence 𝜒(𝑐_𝐴∩ℓ^∞)^𝛽⊂ℓ¹. On the other hand in view of conservativity of 𝐴 we have 𝜒(𝑐_𝐴∩ℓ^∞)^𝛽⊃(𝑐_𝐴∩ℓ^∞)^𝛽=ℓ¹.
Now let 𝐵 be a matrix satisfying 𝜒∩𝑐_𝐴⊂𝑐_𝐵. Take 𝐸=𝑐₀ and 𝐹=𝜒(𝑐_𝐴∩ℓ^∞) in Corollary 4. Then since 𝜒(𝑐_𝐴∩ℓ^∞)∩𝑐₀=𝜑 is dense in (𝑐₀,‖‖_∞), we have 𝑐₀⊂𝑐_𝐵. Moreover, since 𝑒∈𝜒∩𝑐_𝐴⊂𝑐_𝐵, it follows that 𝑐⊂𝑐_𝐵. Hence 𝐴∈KG.

Now we give a positive answer to Problem 2.

Proposition 6. Let 𝑅_𝑝 be regular. Then ℓ^∞∩𝑐_𝑅𝑝 has the Hahn property if and only if 𝜒(ℓ^∞∩𝑐_𝑅𝑝)^𝛽=ℓ¹.

Proof. In view of Corollary 5 we have 𝜒(ℓ^∞∩𝑐_𝑅𝑝)^𝛽=ℓ¹ if and only if 𝑅_𝑝∈KG, but the last condition is equivalent to the Hahn property of ℓ^∞∩𝑐_𝑅𝑝 (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 matrix 𝐻_𝑝 is KG if and only if ℓ^∞∩𝑐_𝐻𝑝 has the Hahn property (Theorem 3.2.1 in [7]).

Proposition 7. Let 𝐻_𝑝 be a nonnegative regular Hausdorff matrix. Then ℓ^∞∩𝑐_𝐻𝑝 has the Hahn property if and only if 𝜒(ℓ^∞∩𝑐_𝐻𝑝)^𝛽=ℓ¹.

3. Bounded Domains of Generalized Riesz Methods

In the previous section we demonstrated that for 𝐴=𝑅_𝑝 the relation 𝜒(ℓ^∞∩𝑐_𝐴)^𝛽=ℓ¹ is equivalent to the Hahn property of ℓ^∞∩𝑐_𝐴. Now we consider two more conditions being equivalent to these properties for 𝐴=𝑅_𝑝 (cf. [6, Corollary 3.9 (ii)], [7, Theorem 3.1.1]): (a)𝐴=(𝑎_𝑛𝑘) has spreading rows; that is, (sup_𝑘∈ℕ𝑎_𝑛𝑘)∈𝑐₀;(b) there exists a sequence 𝒩_𝑡=𝑁_𝜈𝑡∣𝜈∈ℕ_𝜈𝑡,𝜈_𝑡∈ℕ(𝑡∈ℕ) (21) of partitions of ℕ (called an admissible partition sequence) such that limsup_𝑡→∞limsup_𝑛→∞sup_{1≤𝜈≤𝜈𝑡}𝐴_𝑛𝜈𝑡=0where𝐴_𝑛𝜈𝑡∶=_{𝑘∈𝑁𝜈𝑡}𝑎_𝑛𝑘. (22) Recall that 𝒩_𝑡={𝑁_𝜈𝑡∣𝜈∈ℕ_𝜈𝑡} is called a partition of ℕ if ℕ=⋃^𝜈𝑡_𝜈=1𝑁_𝜈𝑡 and 𝑁_𝜈𝑡∩𝑁_𝜇𝑡=∅ if 𝜈≠𝜇.

So in fact conditions (a) and (b) are equivalent for 𝐴=𝑅_𝑝. 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^′) Density 𝑑_𝐴 defined by the matrix 𝐴 is strongly nonatomic on the power set 𝑃(ℕ) of ℕ.

Now we replace a Riesz matrix 𝑅_𝑝 with a more general matrix obtained as a row submatrix of 𝑅_𝑝. More precisely, we consider the matrix 𝑅_𝑝,𝑚=(𝑎_𝑛,𝑘), called generalized Riesz matrix, defined by 𝑎_𝑛𝑘𝑝∶=𝑃^𝑘_𝑚𝑛if1≤𝑘≤𝑚_𝑛,()0otherwise,𝑛,𝑘∈ℕ, (23) where (𝑚_𝑛) is any fixed index sequence and 𝑝=(𝑝_𝑖) 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 𝑅_𝑝,𝑚 the properties (a) and (b) are equivalent for 𝐴=𝑅_𝑝,𝑚. These conditions covered all possible cases except limsup_𝑛𝑃_𝑚𝑛𝑃_𝑚𝑛+1=1,liminf_𝑛𝑃_𝑚𝑛𝑃_𝑚𝑛+1=0. (24) 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⊂𝑐_{0𝑅𝑝,𝑚} hold?

First we answer the first question in Problem 3 and then demonstrate that the relation 𝒯⊂𝑐_{0𝑅𝑝,𝑚} does not hold in general for generalized Riesz matrices 𝑅_𝑝,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,…,𝑧_𝑛} for any 𝑛∈ℕ be a set of nonnegative real numbers. Let 𝑍_𝑛∶=∑^𝑛_𝑘=1𝑧_𝑘 and suppose that 𝐵_𝑛>0 with 𝑧_𝑘≤𝐵_𝑛(𝑘∈ℕ_𝑛), and let 𝑡∈ℕ be arbitrarily given. Then one can divide the set ℕ_𝑛 into 𝑡 (pairwise disjoint) subsets 𝑁₁,𝑁₂,…,𝑁_𝑡 (some of them may be empty), such that _{𝑘∈𝑁𝑠}𝑧_𝑘≤1𝑡𝑍_𝑛+𝐵_𝑛(𝑠=1,2,…,𝑡). (25)

Proposition 9. Suppose that 𝑅_𝑝,𝑚=(𝑎_𝑛𝑘) has spreading rows; that is, lim_𝑛1𝑃_𝑚𝑛sup_{1≤𝑘≤𝑚𝑛}𝑝_𝑘=0,liminf_𝑛𝑃_𝑚𝑛𝑃_𝑚𝑛+1=0; (26)(27) then there exists an admissible partition sequence of ℕ satisfying (22).

Proof. We set 𝐴_𝑛∶=𝑃_𝑚𝑛. For a given 𝑡∈ℕ in view of (27) we can choose the minimal index 𝑛₁ such that 𝐴_𝑛1/𝐴₁>(𝑡+1)/𝑡. Afterwards (27) we choose the minimal index 𝑛₂>𝑛₁ such that 𝐴_𝑛2/𝐴_𝑛1>(𝑡+1)/𝑡. Continuing in the same way we choose the minimal index 𝑛_𝑖>𝑛_𝑖−1 such that 𝐴_𝑛𝑖/𝐴_𝑛𝑖−1>(𝑡+1)/𝑡. In view of (26) it follows that 𝑚_𝑛𝑖+1−𝑚_𝑛𝑖→∞. Again by (26) we can choose 𝛼_𝑡∈ℕ, 𝛼_𝑡≥𝑛₁ such that 𝑝_𝑘𝐴_𝑛<1𝑡²foreach𝑘∈ℕ_𝑚𝑛,𝑛≥𝛼_𝑡; (28) also we can find 𝛽_𝑡∈ℕ, 𝛽_𝑡>𝛼_𝑡 such that 𝐴_𝛼𝑡𝐴_𝑛<1𝑡²for𝑛≥𝛽_𝑡. (29)
By Lemma 8 for any positive integer 𝑠≥𝛼_𝑡 we can divide the set 𝐼_𝑠∶=(𝑚_𝑛𝑠,𝑚_𝑛𝑠+1]∩ℕ into 𝑡² disjoint subsets (some of them may be empty), say 𝑁_𝑠𝑡(1),𝑁_𝑠𝑡(1),…,𝑁_𝑠𝑡(𝑡²) such that (cf. (28)) _{𝑘∈𝑁𝑠𝑡(𝜈)}𝑝_𝑘≤1𝑡²_{𝑘∈𝐼𝑠}𝑝_𝑘+1𝑡²𝐴_𝑛𝑠+1≤2𝑡²𝐴_𝑛𝑠+1. (30) Now, setting 𝑁_𝜈𝑡∶=_𝑠𝑁_𝑠𝑡(𝜈),𝑁_𝑡∶=𝑁_𝜈𝑡∣𝜈∈ℕ_𝑡2, (31) we get an admissible partition sequence 𝒩∶=(𝒩_𝑡)_𝑡 of ℕ. We verify that this partition satisfies (22). Given 𝑛≥𝛽_𝑡 we find 𝑟∈ℕ such that 𝑛_𝑟≤𝑛<𝑛_𝑟+1. By construction of (𝑛_𝑖) it follows that 𝐴_𝑛𝑟≥𝑡𝐴_𝑛/(𝑡+1). Therefore 1𝐴_{𝑛𝑘∈𝑁𝜈𝑡,𝑘≤𝑚𝑛}𝑝_𝑘≤𝐴_𝛼𝑡𝐴_𝑛+1𝐴_𝑛^𝑟−1_{𝑠=𝛼𝑡𝑘∈𝑁𝑠𝑡(𝜈)}𝑝_𝑘+𝐴_𝑛−𝐴_𝑛𝑟𝐴_𝑛≤1𝑡²+2𝑡²1𝐴_𝑛^𝑟−1_𝑠=1𝐴_𝑛𝑠+1+1≤1𝑡+1𝑡²+2𝑡²𝐴_𝑛𝑟𝐴_𝑛^∞_𝑠=0𝑡𝑡+1^𝑠+1≤1𝑡+1𝑡²+2𝑡²1((+11−𝑡/𝑡+1))=𝑡+12𝑡+3𝑡²+1.𝑡+1 (32) Therefore limsup_𝑡→∞limsup_𝑛→∞sup_{𝜈∈ℕ𝑡2}1𝐴_{𝑛𝑘∈𝑁𝜈𝑡,𝑘≤𝑚𝑛}𝑝_𝑘=0. (33)

Now combining the obtained result with Theorem 2.1 in [8] we get the following.

Theorem 10. Let 𝑝=(𝑝_𝑖) be a sequence of positive reals with 𝑝∉ℓ¹. Then the following conditions are equivalent:(i)𝑅_𝑝,𝑚 has spreading rows; that is, lim_𝑛(1/𝑃_𝑚𝑛)sup_{1≤𝑘≤𝑚𝑛}𝑝_𝑘=0;(ii)𝑑_{𝑅𝑝,𝑚} is strongly nonatomic.

The following example demonstrates that the assumption of Problem 3 that 𝒯⊂𝑐_{0𝑅𝑝,𝑚} in the case of (24) is not true in general.

Example 11. We set 𝑚₁∶=1, 𝑝_𝑚1∶=1, 𝑚₂∶=𝑚₁+1, 𝑝_𝑚2∶=2^−𝑚2, 𝑚₃∶=𝑚₂+2²[𝑃_𝑚2], 𝑝_𝑚3∶=1, 𝑝_𝑚2+2𝑘∶=1 for 𝑘=1,…,2[𝑃_𝑚2]−1, and 𝑝_𝑘∶=2^−𝑘 for 𝑚₂<𝑘<𝑚₃, 𝑘∉{𝑚₂+2𝑘∣𝑘=1,…,2[𝑃_𝑚2]−1}.
Suppose that 𝑚_2𝑛−2, 𝑚_2𝑛−1, and 𝑝_𝑘 for 𝑘≤𝑚_2𝑛−1 are defined. We set 𝑚_2𝑛∶=𝑚_2𝑛−1+1 and 𝑚_2𝑛+1∶=𝑚_2𝑛+(𝑛+1)²[𝑃_𝑚2𝑛]. Then we set 𝑝_𝑚2𝑛∶=2^−𝑚2𝑛, 𝑝_𝑚2𝑛+1∶=1, 𝑝_{𝑚2𝑛+𝑘(𝑛+1)}∶=1(𝑘=1,…,[𝑃_𝑚2𝑛](𝑛+1)−1), and 𝑝_𝑘∶=2^−𝑘 for 𝑚_2𝑛<𝑘<𝑚_2𝑛+1, 𝑘∉{𝑚_2𝑛+𝑘(𝑛+1)∣𝑘=1,…,[𝑃_𝑚2𝑛](𝑛+1)−1}. Then 𝑃_𝑚2𝑛+1𝑃_𝑚2𝑛≥𝑃_𝑚2𝑛+(𝑛+1)𝑃_𝑚2𝑛𝑃_𝑚2𝑛𝑃⟶∞,^{𝑚2𝑛−1}𝑃_𝑚2𝑛⟶1. (34)
Consequently the matrix 𝑅_𝑝,𝑚 satisfies condition (24).
On the other hand for 𝑥∈𝒯 having 𝑥_𝑛∶=1 when 𝑝_𝑛=1 and 𝑥_𝑛∶=0 otherwise (𝑛∈ℕ) we have [𝐴𝑥]_𝑚2𝑛+1≥1𝑃_𝑚2𝑛+1^{[𝑃𝑚2𝑛](𝑛+1)}_𝑘=1𝑝_{𝑚2𝑛+𝑘(𝑛+1)}≥(𝑛+1)𝑃_𝑚2𝑛𝑃_𝑚2𝑛+(𝑛+1)𝑃_𝑚2𝑛+∑_𝑖2^−𝑖⟶1. (35)
Hence 𝑥∉𝑐_{0𝑅𝑝,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].

Let 𝐼 be an at most countable set, and let 𝑁={𝑁_𝑖∣𝑖∈𝐼} be a partition of ℕ. Let (𝑛_𝑖𝑘)_𝑘 be the sequence of all elements of 𝑁_𝑖 arranged in the ascending order (𝑖∈𝐼). We introduce the notation (𝑁)‖𝑥‖𝑏𝑠∶=𝑥∈𝜔∣∀𝑗∈𝐼∶_{𝑏𝑠(𝑁)}∶=sup_𝑗(𝑥_𝑘)_{𝑘∈𝑁𝑗𝑏𝑠}<∞. (36)

For the proof of the main result we need two lemmas.

Lemma 12 (cf. Lemma 3.6 in [6]). Let 𝑛∈ℕ and 𝑝₁,…,𝑝_𝑛 be a (finite) sequence of numbers. Then there exists a partition 𝑀₁,…,𝑀_𝑡 of {1,…,𝑛} such that 𝑡≤2√𝑛 and (𝑝_𝑖)_{𝑖∈𝑀𝑗} is monotone (𝑗=1,…,𝑡).

The second required lemma generalizes Lemma 3.7 in [6] concerning Riesz matrices.

Lemma 13. Let 𝑁={𝑁_𝑖∣𝑖∈𝐼} be a partition of ℕ. If 1𝑃_{𝑚𝑛𝑁𝑠∈ℕ𝑠∩1,𝑚𝑛≠∅𝑛𝑗∈ℕ𝑠,𝑗+1≤𝑚𝑛}𝑝_𝑛𝑠𝑗−𝑝_{𝑛𝑠,𝑗+1}+_{𝑛𝑗∈ℕ𝑠𝑗≤𝑚𝑛<𝑛𝑠,𝑗+1}𝑝_𝑛𝑠𝑗(),⟶0𝑛⟶∞ (37)
then 𝑏𝑠(𝑁_𝑖,𝐼)⊂𝑐_{0𝑅𝑝,𝑚}⊂𝑐_{𝑅𝑝,𝑚}.

Proof. To get a proof of the statement just replace everywhere 𝑛 with 𝑚_𝑛 in the proof of Lemma 3.7 [6], the case of 𝑅_𝑝.

In the next theorem we generalize Theorem 3.8 in [6] in the case of Riesz matrices 𝑅_𝑝. The proof requires nontrivial refinements of the methods used in the corresponding part of the proof in [6].

Theorem 14. Let (𝑝_𝑛) be a positive sequence such that 𝑃_𝑛→∞ and 𝑅_𝑝,𝑚=(𝑎_𝑛𝑘) has spreading rows. Then 𝑐_{𝑅𝑝,𝑚}∩ℓ^∞ has the Hahn property.

Proof. By Lemma 13 it is sufficient to verify the existence of a partition 𝑁={𝑁_𝑖∣𝑖∈𝐼} of ℕ satisfying the condition in the lemma. Aiming to that we set 𝑘₀∶=𝑘₀∶=1 and suppose that 𝑘₁,…,𝑘_𝑖,𝑘₁,…,𝑘_𝑖 are already chosen. Since 𝑃_𝑛→∞, we can choose 𝑘_𝑖+1𝑃∶=min𝑘∈ℕ∣^𝑚𝑘𝑃_𝑚𝑘𝑖≥2. (38) If 𝑃_𝑚𝑘_𝑖+1/𝑃_𝑚𝑘𝑖≤8 or 𝑘_𝑖+1=𝑘_𝑖+1 we set 𝑘_𝑖+1∶=𝑘_𝑖+1. Otherwise we set 𝑘_𝑖+1∶=𝑘_𝑖+1−1 and 𝑘_𝑖+2∶=𝑘_𝑖+2∶=𝑘_𝑖+1. In the latter case we have 𝑃_{𝑚𝑘𝑖+1}/𝑃_𝑚𝑘𝑖<2 and 𝑘_𝑖+2=𝑘_𝑖+1+1. Hence in both cases 𝑃_{𝑚𝑘𝑖+2}𝑃_𝑚𝑘𝑖𝑃≥2,^{𝑚𝑘𝑖+1}𝑃_𝑚𝑘𝑖≤8or𝑘_𝑖+1=𝑘_𝑖+1. (39) Since 𝑅_𝑝,𝑚=(𝑎_𝑛𝑘) has spreading rows, then 𝜀_𝑖𝑝∶=max_𝑗∣𝑗≤𝑚_𝑘𝑖𝑃_𝑚𝑘𝑖⟶0. (40) Let 𝑖₀∈ℕ be such that 𝜀_𝑖<1 for 𝑖≥𝑖₀.
For every 𝑖≥𝑖₀ we choose the minimal integer 𝑗_𝑖∈ℕ such that 𝑚^1/𝑗𝑖_𝑘𝑖<𝜀_𝑖^−1/2. Since 𝜀_𝑖=max{𝑝_𝑘∣𝑘≤𝑚_𝑘𝑖}/𝑃_𝑚𝑘𝑖≥1/𝑚_𝑘𝑖, then 𝑚_𝑘𝑖≥𝜀_𝑖⁻¹ (𝑖≥𝑖₀). Hence 𝑗_𝑖>2 (𝑖≥𝑖₀). Note that 𝑚^1/𝑗𝑖_𝑘𝑖≥𝜀_𝑖^−1/4(𝑖≥𝑖₀). (Otherwise 𝑚^1/[(𝑗𝑖_𝑘^+2)/2]_𝑖≤𝑚^2/𝑗𝑖_𝑘𝑖<𝜀_𝑖^−1/2, so 𝑗_𝑖/2+1≥[𝑗_𝑖/2]+1≥𝑗_𝑖, which is equivalent to 𝑗_𝑖≤2 and contradicts 𝑗_𝑖>2).
Set 𝛼_𝑖𝑃∶=^{𝑚𝑘𝑖+1}𝑚_𝑘𝑖+1,𝛽_𝑖∶=max𝑝_𝑘∣𝑘≤𝑚_𝑘𝑖+1(𝑖∈ℕ). (41) For every 𝑖∈ℕ and 𝑠=2,…,𝑗_𝑖+1−2 we use the notations 𝑀_𝑖1∶=𝑚_𝑘𝑖<𝑘≤𝑚_𝑘𝑖+1∣𝑝_𝑘≤𝑚^1/𝑗𝑖+1_𝑘𝑖+1𝛼_𝑖,𝑀_𝑖𝑠∶=𝑚_𝑘𝑖<𝑘≤𝑚_𝑘𝑖+1∣𝑚^{(𝑠−1)/𝑗𝑖+1}_𝑘𝑖+1𝛼_𝑖<𝑝_𝑘≤𝑚^{𝑠/𝑗𝑖+1}_𝑘𝑖+1𝛼_𝑖,𝑀_{𝑖,𝑗𝑖+1−1}∶=𝑚_𝑘𝑖<𝑘≤𝑚_𝑘𝑖+1∣𝑚^{(𝑗𝑖+1−2)/𝑗𝑖+1}_𝑘𝑖+1𝛼_𝑖<𝑝_𝑘. (42) Set 𝜈_𝑖𝑠∶=|𝑀_𝑖𝑠|(𝑠=1,…,𝑗_𝑖+1−1). Note that 𝜈_𝑖𝑠≤𝑃_{𝑚𝑘𝑖+1}𝑚^{(𝑠−1)/𝑗𝑖+1}_𝑘𝑖+1𝛼_𝑖=𝑚_𝑘𝑖+1𝛼_𝑖𝑚^{(𝑠−1)/𝑗𝑖+1}_𝑘𝑖+1𝛼_𝑖=𝑚^{(𝑗𝑖+1+1−𝑠)/𝑗𝑖+1}_𝑘𝑖+1. (43)
By Lemma 12 for every 𝑖∈ℕ and 𝑠=1,…,𝑗_𝑖+1−1 we may find a partition (𝑆_𝑖𝑠1,…,𝑆_{𝑖𝑠𝑟𝑖𝑠}) of 𝑀_𝑖𝑠 with 𝑟_𝑖𝑠≤2√𝜈_𝑖𝑠 such that (𝑝_𝜈)_{𝜈∈𝑆𝑖𝑠𝑡} is monotone (𝑡=1,…,𝑟_𝑖𝑠). Let 𝑒_𝑖𝑠𝑘∶=|𝑆_𝑖𝑠𝑘| and let (𝜉_{𝑖𝑠𝑘𝑙})^{𝑒𝑖𝑠𝑘}_𝑙=1 be the finite sequence of all elements of 𝑆_𝑖𝑠𝑘 arranged in the ascending order (𝑖∈ℕ,𝑠=1,…,𝑗_𝑖+1−1,𝑘=1,…,𝑟_𝑖𝑠). Then ^𝑟𝑖𝑠_𝑘=1𝑝_{𝜉𝑖𝑠𝑘1}+^{𝑒𝑖𝑠𝑘−1}_𝑙=1𝑝_{𝜉𝑖𝑠𝑘𝑙}−𝑝_{𝜉𝑖,𝑠,𝑘,𝑙+1}+𝑝_{𝜉𝑖𝑠𝑘𝑒𝑖𝑠𝑘}=^𝑟𝑖𝑠_𝑘=12max𝑝_{𝜉𝑖𝑠𝑘1},𝑝_{𝜉𝑖𝑠𝑘𝑒𝑖𝑠𝑘}≤2𝑟_𝑖𝑠𝑚^{𝑠/𝑗𝑖+1}_𝑘𝑖+1𝛼_𝑖≤4√𝜈_𝑖𝑠𝑚^{𝑠/𝑗𝑖+1}_𝑘𝑖+1𝛼_𝑖≤4𝑚^{(𝑗𝑖+1+1−𝑠)/2𝑗𝑖+1}_𝑘𝑖+1𝑚^{𝑠/𝑗𝑖+1}_k𝑖+1𝛼_𝑖=4𝑚^{(𝑗i+1+1+𝑠)/2𝑗𝑖+1}_𝑘𝑖+1𝛼_𝑖 (44) for 𝑠=1,…,𝑗_𝑖+1−2 and ^𝑟𝑖𝑠_𝑘=1𝑝_{𝜉𝑖𝑠𝑘1}+^{𝑒𝑖𝑠𝑘−1}_𝑙=1𝑝_{𝜉𝑖𝑠𝑘𝑙}−𝑝_{𝜉𝑖,𝑠,𝑘,𝑙+1}+𝑝_{𝜉𝑖𝑠𝑘𝑒𝑖𝑠𝑘}=^𝑟𝑖𝑠_𝑘=12max𝑝_{𝜉𝑖𝑠𝑘1},𝑝_{𝜉𝑖𝑠𝑘𝑒𝑖𝑠𝑘}≤4√𝜈_𝑖𝑠𝛽_𝑖≤4𝑚^1/𝑗𝑖+1_𝑘𝑖+1𝛽_𝑖 (45) for 𝑠=𝑗_𝑖+1−1(𝑖∈ℕ). Hence for every 𝑖≥𝑖₀ we get1𝑃_{𝑚𝑘𝑖+1}^{𝑗𝑖+1−1𝑟}_𝑠=1^𝑖𝑠_𝑘=1𝑝_{𝜉𝑖𝑠𝑘1}+^{𝑒𝑖𝑠𝑘−1}_𝑙=1𝑝_{𝜉𝑖𝑠𝑘𝑙}−𝑝_{𝜉𝑖,𝑠,𝑘,𝑙+1}+𝑝_{𝜉𝑖𝑠𝑘𝑒𝑖𝑠𝑘}≤1𝑃_{𝑚𝑘𝑖+1}^{𝑗𝑖+1−2}_𝑠=14𝑚^{(𝑗𝑖+1+1+𝑠)/2𝑗𝑖+1}_𝑘𝑖+1𝛼_𝑖+4𝑚^1/𝑗𝑖+1_𝑘𝑖+1𝛽_𝑖=4𝑚^{(𝑗𝑖+1+1)/2𝑗𝑖+1}_𝑘𝑖+1𝛼_𝑖𝑚_𝑘𝑖+1𝛼_𝑖𝑚^{(𝑗𝑖+1−1)/2𝑗𝑖+1}_𝑘𝑖+1−1𝑚^{1/2𝑗𝑖+1}_𝑘𝑖+1+−1−14𝛽_𝑖𝑚^1/𝑗𝑖+1_𝑘𝑖+1𝑃_{𝑚𝑘𝑖+1}≤4𝑚^{(−𝑗𝑖+1+1)/2𝑗𝑖+1}_𝑘𝑖+1𝑚^{(𝑗𝑖+1−1)/2𝑗𝑖+1}_𝑘𝑖+14⁻¹𝑚^{1/2𝑗𝑖+1}_𝑘𝑖+1+4𝜀_𝑖+1𝜀^−1/2_𝑖+1=16𝑚^{−1/2𝑗𝑖+1}_𝑘𝑖+1+4𝜀^−1/2_𝑖+1≤16𝜀^1/8_𝑖+1+4𝜀^1/2_𝑖+1. (46)To define 𝑁_𝑖(𝑖∈ℕ) we use the notation 𝑁|_𝑛∶={𝑘∈𝑁∣𝑘≤𝑛} for a subset 𝑁⊂ℕ and 𝑛∈ℕ.
Set 𝑁₁|_{𝑚𝑘𝑖0}∶={1,…,𝑚_𝑘𝑖0} and 𝑁_𝑗|_𝑚𝑖0∶=∅ for 𝑗>1. Let 𝑠⁰₁,…,𝑠⁰_𝑡0 be all indexes such that 𝑀_𝑖0𝑠⁰_𝑡≠∅(𝑡=1,…,𝑡₀). We set 𝑁_{𝑗𝑚𝑘𝑖0+1}𝑁∶=^{𝑗𝑚𝑘𝑖0}∪𝑆_𝑖0𝑠⁰_𝑡𝑘for𝑗=^𝑡−1_𝜏=1𝑟_{𝑖0𝑠0𝜏}+𝑘𝑡=1,…,𝑡₀,𝑘=1,…,𝑟_𝑖0𝑠⁰_𝑡,𝑁_{𝑗𝑚𝑘𝑖0}for𝑗>^𝑡0_𝜏=1𝑟_{𝑖0𝑠0𝜏}. (47)
Continuing inductively for 𝑖>𝑖₀ let 𝑠^𝑖−𝑖0₁,…,𝑠^𝑖−𝑖0_{𝑡0𝑖−𝑖} be all indexes such that 𝑀_{𝑖𝑠0𝑡}^𝑖−𝑖≠∅(𝑡=1,…,𝑡_𝑖−𝑖0). We set 𝑁_{𝑗𝑚𝑘𝑖+1}𝑁∶=^{𝑗𝑚𝑘𝑖}∪𝑆_{𝑖𝑠0𝑡}^𝑖−𝑖_𝑘for𝑗=^𝑡−1_𝜏=1𝑟_{𝑖𝑠0𝜏}^𝑖−𝑖+𝑘𝑡=1,…,𝑡_𝑖−𝑖0,𝑘=1,…,𝑟_{𝑖0𝑠0𝑡}^𝑖−𝑖,𝑁_{𝑗𝑚𝑘𝑖0}for𝑗>^{𝑡0𝑖−𝑖}_𝜏=1𝑟_{𝑖𝑠0𝜏}^𝑖−𝑖. (48)
Let 𝐼 be the set of all indexes 𝑖∈ℕ such that 𝑁_𝑖≠∅. If 𝐼 is infinite, then by our construction every 𝑁_𝑖 is infinite. If 𝐼 is finite, then without loss of generality we may assume that 𝑁_𝑖 is infinite (𝑖∈𝐼).
Let 𝑛∈ℕ with 𝑘_𝑖<𝑛≤𝑘_𝑖+1 and 𝑖≥𝑖₀ be fixed. Then 𝐵_𝑛1∶=𝑃_𝑚𝑛×_{𝑁𝑠∩𝑛𝑠∈ℕ1,𝑚≠∅𝑛𝑛𝑗∈ℕ𝑠,𝑗+1≤𝑚}𝑝_𝑛𝑠𝑗−𝑝_{𝑛𝑠,𝑗+1}+_{𝑛𝑛𝑗∈ℕ𝑠𝑗≤𝑚<𝑛𝑠,𝑗+1}𝑝_𝑛𝑠𝑗≤1𝑃_𝑚𝑛×2^{𝑚𝑘𝑖0}_𝑘=1𝑝_𝑘+^𝑖_𝜏=𝑖0^{𝑗𝜏+1−1𝑟}_𝑠=1^𝜏𝑠_𝑘=1𝑝_{𝜉𝑖𝑠𝑘1}+^{𝑒𝜏𝑠𝑘−1}_𝑙=1𝑝_{𝜉𝜏𝑠𝑘𝑙}−𝑝_{𝜉𝜏,𝑠,𝑘,𝑙+1}+𝑝_{𝜉𝑖𝑠𝑘𝑒𝑖𝑠𝑘}≤1𝑃_𝑚𝑛2𝑃_{𝑚𝑘𝑖0}+^𝑖_𝜏=𝑖016𝜀^1/8_𝜏+1+4𝜀^1/2_𝜏+1𝑃_𝑚𝜏+1. (49)If 𝑘_𝑖+1=𝑘_𝑖+1 then 𝑛=𝑘_𝑖+1, so 𝑃_𝑛=𝑃_𝑘𝑖+1. If 𝑘_𝑖+1>𝑘_𝑖+1 then 𝑃_𝑛≥𝑃_𝑘𝑖≥8⁻¹𝑃_𝑘𝑖+1. Hence 𝐵_𝑛≤8𝑃_𝑘𝑖+12𝑃_𝑚𝑖0+^𝑖_𝜏=𝑖016𝜀^1/8_𝜏+1+4𝜀^1/2_𝜏+1𝑃_𝑚𝜏+1≤162^{[(𝑖−𝑖0+1)/2]}+32^𝑖_𝜏=𝑖012^{[(𝑖−𝜏)/2]}4𝜀^1/8_𝜏+1+𝜀^1/2_𝜏+1. (50) Consider the matrix 𝐴=(𝑎_𝑖𝜏) with 𝑎_𝑖𝜏∶=2^{[(𝜏−𝑖)/2]} for 𝜏≤𝑖 and 𝑎_𝑖𝜏∶=0 otherwise. Evidently, 𝐴 is regular for null sequences, so it sums the null sequence (4𝜀^1/8_𝜏+𝑖0+𝜀^1/2_𝜏+𝑖0)_𝜏 to zero. Therefore 𝐵_𝑛→0(𝑛→∞). So in view of Lemma 13 the inclusion 𝑏𝑠(𝑁)⊂𝑐_{𝑅𝑝,𝑚} holds.

Now having in mind Proposition 2.8 in [7] we can prove the following theorem.

Theorem 15. Let 𝑝=(𝑝_𝑖) be a sequence of positive reals with 𝑝∉ℓ¹. Then one considers the following conditions:(a)ℓ^∞∩𝑐_{𝑅𝑝,𝑚} has the matrix Hahn property,(b)𝑅_𝑝,𝑚 has spreading rows; that is, lim_𝑛(1/𝑃_𝑚𝑛)sup_{1≤𝑘≤𝑚𝑛}𝑝_𝑘=0, (c)𝑑_{𝑅𝑝,𝑚} is strongly nonatomic,(d)𝑅_𝑝,𝑚∈KG,(e)ℓ^∞∩𝑐_{𝑅𝑝,𝑚} has the separable Hahn property,(f)ℓ^∞∩𝑐_{𝑅𝑝,𝑚} has the Hahn property,(g)𝜒(𝑐_{𝑅𝑝,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 lim_𝑛1𝑃_𝑚𝑛sup_{1≤𝑘≤𝑚𝑛}𝑝_𝑘=0 (51) is necessary for 𝑅_𝑝,𝑚∈KG. The following example gives a generalized Riesz matrix 𝑅_𝑝,𝑚 not having spreading rows and failing to have 𝑅_𝑝,𝑚∈KG.

Example 16. Suppose 𝑝∉ℓ¹ and liminf_𝑛1𝑃_𝑚𝑛sup_{1≤𝑘≤𝑚𝑛}𝑝_𝑘>12. (52) Then there exists 𝛿>1/2 and 𝑛₀∈ℕ such that (1/𝑃_𝑚𝑛)sup_{1≤𝑘≤𝑚𝑛}𝑝_𝑘≥𝛿 for 𝑛≥𝑛₀.
Let 𝑘_𝑛≤𝑚_𝑛 be an integer such that 𝑝_𝑘𝑛=sup_{1≤𝑘≤𝑚𝑛}𝑝_𝑘. Then 1𝑃_𝑚𝑛^𝑚𝑛_{𝑛𝑘=1𝑘≠𝑘}𝑝_𝑘1≤1−𝛿<2. (53) Since 𝑃_𝑛→∞ we can choose an index sequence (𝑛_𝑖) such that (𝑘_𝑛𝑖) is increasing. Now if 𝑥∈𝜒∩𝑐_{𝑅𝑝,𝑚} then there exists 𝑖₀=𝑖₀(𝑥) such that 𝑥_𝑘𝑛𝑖=0 for all 𝑖≥𝑖₀ or 𝑥_𝑘𝑛𝑖=1 for all 𝑖≥𝑖₀. We set 𝑦_𝑘𝑛𝑖∶=(−1)^𝑖/𝑖 for 𝑖∈ℕ and 𝑦_𝑘∶=0 for 𝑘∉{𝑘_𝑛𝑖∣𝑖∈ℕ}. Then the series ^∞_{𝑘=𝑘𝑛𝑖0}𝑥_𝑘𝑦_𝑘=^∞_𝑖=𝑖0(−1)^𝑖𝑖𝑥_𝑘𝑛𝑖 (54) converges, so 𝑦∈𝜒(𝑐_{𝑅𝑝,𝑚}∩ℓ^∞)^𝛽⧵ℓ¹.

Problem 4. In Theorem 15 we have shown that the fact that 𝑅_𝑝,𝑚 has spreading rows implies 𝑅_𝑝,𝑚∈KG. Does the inverse implication hold?

Acknowledgments

The author is much obliged to Professor J. Boos (FernUniversität in Hagen) for advice and criticism with this paper. The author thanks the reviewer for his/her thorough review and highly appreciates the comments and suggestions, which significantly contributed to improving the quality of the publication. This research was supported by the Estonian Science Foundation (Grant no. 8627), European Regional Development Fund (Centre of Excellence “Mesosystems: Theory and Applications,” TK114), and Estonian Ministry of Education and Research (Project SF0130010s12).