Bounded Domains of Generalized Riesz Methods with the Hahn Property

({0, 1} N ∩ E) β = E β = l 1 then the two last properties coincide.Wewill 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 matrixHp, 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.


Preliminaries and Introduction
We start with some preliminaries.For other notations and preliminary results we refer the reader to [1][2][3].
Let  denote the set of all sequences of 0s and 1s and let () denote the linear hull of  ∩ .
An FK-space is a sequence space endowed with a complete, metrizable, locally convex topology under which all coordinate mappings  = (  ) →   ( ∈ N) are continuous.
A sequence space  is said to have the Hahn property, the separable Hahn property, and the matrix Hahn property, if () ⊂  implies  ⊂  whenever  is any FK-space, a separable FK-space, and a matrix domain   , respectively.Obviously, the Hahn property implies the separable Hahn property, and the latter implies the matrix Hahn property.
If  has the matrix Hahn property then ()  =   (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  to be a solid sequence space containing  (the set of all finite sequences) and satisfying   = ℓ 1 , then ()  =   implies the separable Hahn property of  (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  be a solid sequence space containing  and satisfying ()  =   = ℓ 1 .Then  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  = (  ).In particular, the following sets are double sequence spaces: We denote by  1 the set of double sequences of zeros and ones in C be0 ; that is, The space of all double sequences Ω can be identified with the space of all sequences  using a suitable isomorphism  : Ω →  (see (2) in [5] for a possible definition of ).
Theorem 1.There exists a solid sequence space  containing  and satisfying ()  =   = ℓ 1 which does not have the Hahn property.
Taking the explained situation concerning the sufficiency of ()  =   = ℓ 1 for the Hahn property of  into consideration, it is mathematically interesting to research for classes of sequence spaces  with the property that ()  =   = ℓ 1 implies the Hahn property of  (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.

Bounded Domains of Riesz Methods
Let  = (  ) be a real sequence with The Riesz matrix   = (  ) (associated with ) is defined by The summability method corresponding   is called Riesz method.
In Section 1 we have seen that the relation ()  =   = ℓ 1 does not imply in general the Hahn property of  even for a solid space .This result suggests the following problem due to Boos Proof.In view of  ⊂ ℓ ∞  we have sup  | ∑      | < ∞ for each  ∈ , so in particular (  )  ∈   = ℓ 1 for each  ∈ N.
Assume on contrary that there exists an Aiming to a contradiction we will construct by induction two index sequences (  ) and (  ), and then with the help of them we will define  ∈   \ ℓ 1 .First of all set  1 := 0 and Then choose and set Now suppose that  1 , . . .,   and  1 , . . .,  −1 are already chosen for  > 1.Since  ⊂ ℓ ∞  , then We choose   > We are going to verify  ∈   in contradiction to   = ℓ 1 .
For that end let  ∈  and ,  ∈ N with  >  be fixed.We choose for any  =  1 , . . .,  2 − 1.So for the second term on the righthand side of ( 16) we have the estimate For the first and the third term on the right-hand side of (16) we have the estimates (20) so the series ∑      converges and  ∈   .
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.
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  = (  ) be an infinite matrix, and let  be a subset of  with  ⊂ ⟨⟩ ⊂ ℓ ∞  and   = ℓ 1 .Let  be a subset of ℓ ∞ such that   = ℓ 1 and  ∩  ⊂   .Furthermore, let (,   ) be an F-space with a dense subset ∩.Then  ⊂   .
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   is KG if and only if ℓ ∞ ∩    has the Hahn property (Theorem 3.2.1 in [7]).

Bounded Domains of Generalized Riesz Methods
In the previous section we demonstrated that for  =   the relation (ℓ ∞ ∩   )  = ℓ 1 is equivalent to the Hahn property of ℓ ∞ ∩   .Now we consider two more conditions being equivalent to these properties for  =   (cf.[6, Corollary 3.9 (ii)], [7, Theorem 3.1.1]): (a)  = (  ) has spreading rows; that is, (sup ∈N   ) ∈  0 ; (b) there exists a sequence of partitions of N (called an admissible partition sequence) such that lim sup Recall that So in fact conditions (a) and (b) are equivalent for  =   .Note that for any nonnegative and regular for null sequences matrix  condition (b) is equivalent to the following condition (see [7] for the corresponding definitions).
(  ) Density   defined by the matrix  is strongly nonatomic on the power set (N) of N.
Now we replace a Riesz matrix   with a more general matrix obtained as a row submatrix of   .More precisely, we consider the matrix  , = ( , ), called generalized Riesz matrix, defined by where (  ) is any fixed index sequence and  = (  ) is a sequence of positive reals with  ∉ ℓ 1 .Boos and Leiger showed (cf.Theorem 2.1 in [8]) that in the case of some sufficient conditions on terms of  , the properties (a) and (b) are equivalent for  =  , .These conditions covered all possible cases except lim sup 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.
Theorem 10.Let  = (  ) be a sequence of positive reals with  ∉ ℓ 1 .Then the following conditions are equivalent: (ii)   , is strongly nonatomic.
The following example demonstrates that the assumption of Problem 3 that T ⊂  0 , in the case of (24) is not true in general.
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  be an at most countable set, and let  = {  |  ∈ } be a partition of N. Let (  )  be the sequence of all elements of   arranged in the ascending order ( ∈ ).We introduce the notation For the proof of the main result we need two lemmas.
Lemma 12 (cf.Lemma 3.6 in [6]).Let  ∈ N and  1 , .Proof.To get a proof of the statement just replace everywhere  with   in the proof of Lemma 3.7 [6], the case of   .
In the next theorem we generalize Theorem 3.8 in [6] in the case of Riesz matrices   .The proof requires nontrivial refinements of the methods used in the corresponding part of the proof in [6].
Since  , = (  ) has spreading rows, then Let  0 ∈ N be such that   < 1 for  ≥  0 .For every  ≥  0 we choose the minimal integer   ∈ N such that

Proposition 7 .
Let   be a nonnegative regular Hausdorff matrix.Then ℓ ∞ ∩    has the Hahn property if and only if Problem 2. Does the relation (ℓ ∞ ∩    )  = ℓ 1 imply the Hahn property of ℓ ∞ ∩    ?