/ / Article

Research Article | Open Access

Volume 2013 |Article ID 908682 | 8 pages | https://doi.org/10.1155/2013/908682

# Bounded Domains of Generalized Riesz Methods with the Hahn Property

Revised28 Aug 2013
Accepted04 Sep 2013
Published21 Oct 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 space is solid and 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 to be the bounded summability domain of a regular Riesz matrix or a regular nonnegative Hausdorff matrix , then this assumption alone guarantees that has the Hahn property. For any (infinite) matrix the Hahn property of its bounded summability domain is related to the strongly nonatomic property of the density defined by . We will find a simple necessary and sufficient condition for the density defined by the generalized Riesz matrix to be strongly nonatomic. This condition appears also to be sufficient for the bounded summability domain of 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 .

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 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 ), but, in general, the inverse implication does not hold even for monotone sequence spaces (see Theorem 1.1 in ). Still if we ask to be a solid sequence space containing (the set of all finite sequences) and satisfying , 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 ). 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 . 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 , the space of all real double sequences . In particular, the following sets are double sequence spaces: We denote by the set of double sequences of zeros and ones in ; that is,

The space of all double sequences can be identified with the space of all sequences using a suitable isomorphism (see in  for a possible definition of ).

Theorem 1. There exists a solid sequence space containing and satisfying which does not have the Hahn property.

Proof. Set . Then is a solid sequence space containing and failing to have the Hahn property (see Example 3.7 in ). In view of Theorem 3.2 in  we have also as well as .

Taking the explained situation concerning the sufficiency of for the Hahn property of into consideration, it is mathematically interesting to research for classes of sequence spaces with the property that 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.

#### 2. 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.

The Riesz matrix is conservative, and it is either regular (being equivalent to ) or coercive. If , then has the Hahn property since . If , then has the Hahn property if and only if (cf. Corollary 3.9 in ).

In Section 1 we have seen that the relation does not imply in general the Hahn property of even for a solid space . This result suggests the following problem due to Boos and Leiger (cf. Problem 3 in [7, Section 4]).

Problem 2. Does the relation imply the Hahn property of ?

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

Theorem 2. Let be an infinite matrix such that there exists a subset of with and . Then (being equivalent to ).

Proof. In view of we have for each , so in particular for each .
Assume on contrary that there exists an . Hence since
Aiming to a contradiction we will construct by induction two index sequences and , and then with the help of them we will define . First of all set and . Now choose such that Then choose such that and set Now suppose that and are already chosen for . Since , then
We choose such that and then we take such that So Now we define a by and note that because We are going to verify in contradiction to . For that end let and with be fixed. We choose such that and . If we have If we get Since , we have ; hence, for any . So for the second term on the right-hand 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 Hence so the series converges and .

The last theorem provides us an alternative way (cf. Theorem 2.3.8 in ) to show that every matrix summing all thin sequences is conservative for null sequences. Let denote the set of all thin sequences. Then we have the following.

Corollary 3. Let be a matrix satisfying , then . Moreover, if ; then is conservative and (i.e., each matrix with is conservative).

Proof. Taking in Theorem 2 in view of we have , so . Since it follows that  .
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 and . Let be a subset of such that and . Furthermore, let be an F-space with a dense subset . Then .

Proof. We consider the functionals with . By Theorem 2 the sequence is pointwise bounded on , and by our assumptions it converges pointwise on the dense subset of . Hence by generalized version of Banach-Steinhaus theorem (cf. Theorem 6.8.6 in ) the sequence converges pointwise on . So .

Corollary 5. Let be a conservative matrix. Then if and only if .

Proof. Suppose and let . We set . Then ; hence, is conservative. Therefore . Hence . On the other hand in view of conservativity of we have .
Now let be a matrix satisfying . Take and in Corollary 4. Then since is dense in , we have . Moreover, since , it follows that . Hence .

Now we give a positive answer to Problem 2.

Proposition 6. Let be regular. Then has the Hahn property if and only if .

Proof. In view of Corollary 5 we have if and only if , but the last condition is equivalent to the Hahn property of (cf. Corollary 3.9 in ).

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 has the Hahn property (Theorem 3.2.1 in ).

Proposition 7. Let be a nonnegative regular Hausdorff matrix. Then has the Hahn property if and only if .

#### 3. Bounded Domains of Generalized Riesz Methods

In the previous section we demonstrated that for the relation is equivalent to the Hahn property of . 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, ;(b) there exists a sequence of partitions of (called an admissible partition sequence) such that Recall that is called a partition of if and if .

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  for the corresponding definitions). Density defined by the matrix is strongly nonatomic on the power set of .

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 .

Boos and Leiger showed (cf. Theorem 2.1 in ) 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 Note that even in this case the implication (b)(a) holds (cf. Theorem 2.1 in ). So only the implication (a)(b) in case (24) was under the question.

In relation with this in , 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 hold?

First we answer the first question in Problem 3 and then demonstrate that the relation does not hold in general for generalized Riesz matrices satisfying (24).

For our first aim we apply—in the same way as Boos and Leiger in —the following lemma due to Kuttner and Parameswaran (cf. [9, Lemma 2]).

Lemma 8. Let for any be a set of nonnegative real numbers. Let and suppose that with , and let be arbitrarily given. Then one can divide the set into (pairwise disjoint) subsets (some of them may be empty), such that

Proposition 9. Suppose that has spreading rows; that is, then there exists an admissible partition sequence of satisfying (22).

Proof. We set . For a given in view of (27) we can choose the minimal index such that . Afterwards (27) we choose the minimal index such that . Continuing in the same way we choose the minimal index such that . In view of (26) it follows that . Again by (26) we can choose , such that also we can find , such that
By Lemma 8 for any positive integer we can divide the set into disjoint subsets (some of them may be empty), say such that (cf. (28)) Now, setting we get an admissible partition sequence of . We verify that this partition satisfies (22). Given we find such that . By construction of it follows that . Therefore Therefore

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

Theorem 10. Let be a sequence of positive reals with . Then the following conditions are equivalent:(i) has spreading rows; that is, ;(ii) is strongly nonatomic.

The following example demonstrates that the assumption of Problem 3 that in the case of (24) is not true in general.

Example 11. We set , , , , , , for , and for , .
Suppose that , , and for are defined. We set and . Then we set , , , and for , . Then
Consequently the matrix satisfies condition (24).
On the other hand for having when and otherwise we have
Hence .

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 . 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 ). Let and be a (finite) sequence of numbers. Then there exists a partition of such that and is monotone .

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

Lemma 13. Let be a partition of . If
then .

Proof. To get a proof of the statement just replace everywhere with in the proof of Lemma 3.7 , the case of .

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

Theorem 14. Let be a positive sequence such that and has spreading rows. Then has the Hahn property.

Proof. By Lemma 13 it is sufficient to verify the existence of a partition of satisfying the condition in the lemma. Aiming to that we set and suppose that are already chosen. Since , we can choose If or we set . Otherwise we set and . In the latter case we have and . Hence in both cases Since has spreading rows, then Let be such that for .
For every we choose the minimal integer such that . Since , then (). Hence (). Note that . (Otherwise , so , which is equivalent to and contradicts ).
Set For every and we use the notations Set . Note that
By Lemma 12 for every and we may find a partition of with such that is monotone . Let and let be the finite sequence of all elements of arranged in the ascending order . Then for and for . Hence for every we getTo define we use the notation for a subset and .
Set and for . Let be all indexes such that . We set
Continuing inductively for let be all indexes such that . We set
Let be the set of all indexes such that . If is infinite, then by our construction every is infinite. If is finite, then without loss of generality we may assume that is infinite ().
Let with and be fixed. Then If then , so . If then . Hence Consider the matrix with for and otherwise. Evidently, is regular for null sequences, so it sums the null sequence to zero. Therefore . So in view of Lemma 13 the inclusion holds.

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

Theorem 15. Let be a sequence of positive reals with . Then one considers the following conditions:(a) has the matrix Hahn property,(b) has spreading rows; that is, , (c) is strongly nonatomic,(d),(e) has the separable Hahn property,(f) has the Hahn property,(g).
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 is necessary for . The following example gives a generalized Riesz matrix not having spreading rows and failing to have .

Example 16. Suppose and Then there exists and such that for .
Let be an integer such that . Then Since we can choose an index sequence such that is increasing. Now if then there exists such that for all or for all . We set for and for . Then the series converges, so .

Problem 4. In Theorem 15 we have shown that the fact that has spreading rows implies . 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).

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