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The Scientific World Journal
Volume 2013 (2013), Article ID 349346, 15 pages
Some New Generalized Difference Spaces of Nonabsolute Type Derived from the Spaces and
1Department of Mathematics, Faculty of Arts and Sciences, Fatih University, Hadımköy Campus, Büyükçekmece, 34500 İstanbul, Turkey
2Department of Mathematics & Computer Science, Faculty of Science, Necmettin Erbakan University, Meram Campus, Meram, 42090 Konya, Turkey
Received 13 August 2013; Accepted 2 October 2013
Academic Editors: Y.-M. Chu and I. Stamova
Copyright © 2013 Feyzi Başar and Ali Karaisa. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.
We introduce the sequence space of none absolute type which is a -normed space and space in the cases and , respectively, and prove that and are linearly isomorphic for . Furthermore, we give some inclusion relations concerning the space and we construct the basis for the space , where . Furthermore, we determine the alpha-, beta- and gamma-duals of the space for . Finally, we investigate some geometric properties concerning Banach-Saks type and give Gurarii's modulus of convexity for the normed space .
From the summability theory perspective, the role played by the algebraical, geometrical, and topological properties of the new Banach spaces which are defined by the matrix domain of triangle matrices in sequence spaces is well-known.
By , we denote the space of all real or complex valued sequences. Any vector subspace of is called a sequence space.
A sequence space with a linear topology is called a -space provided that each of the maps defined by is continuous for all , where denotes the complex field and . A -space is called an -space provided is a complete linear metric space. An -space whose topology is normable is called a -space (see ) which contains , the set of all finitely nonzero sequences.
We write , , , and for the spaces of all bounded, almost convergent, convergent, and null sequences, respectively, which are -spaces with the usual supnorm defined by Also, by and , we denote the spaces all -absolutely and absolutely convergent series, respectively, which are -spaces with the usual norm defined by Here, and in what follows, the summation without limits runs from to . Further, we write bs and cs for the spaces of all bounded and convergent series, respectively, which are -spaces with their natural norm .
Let and be two sequence spaces and be an infinite matrix of real or complex numbers , where . Then, we say that defines a matrix transformation from into and we denote it by writing , if for every sequence the sequence , the -transform of is in , where
The notation denotes the class of all matrices such that . Thus, if and only if the series on the right hand side of (3) converges for each and every , and we have for all . The matrix domain of an infinite matrix in a sequence space is defined by An infinite matrix is said to be a triangle if for all and for . The study of matrix domains of triangles has a special importance due to the various properties which they have. For example, if is a triangle and is a -space, then is also a -space with the norm given by for all .
Throughout the paper, we denote the collection of all finite subsets of by . Also, we write for the sequence whose only nonzero term is a 1 in the th place for each .
The approach constructing a new sequence space by means of the matrix domain of a particular triangle has recently been employed by several authors in many research papers. For example, they introduced the sequence spaces and in , and in , and in , and in , and in , and in , in , and in , and in , where , , , , , , and denote Nörlund, arithmetic, Riesz, Euler means, matrix, lambda matrix, and generalized difference matrix, respectively, where .
Recently, there has been a lot of interest in investigating geometric properties of sequence spaces besides topological and some other usual properties. In the literature, there are many papers concerning the geometric properties of different sequence spaces. For example, in , Mursaleen et al. studied some geometric properties of a normed Euler sequence space. Recently, Şimşek and Karakaya  have investigated the geometric properties of the sequence space equipped with Luxemburg norm. Later, Demiriz and Çakan  have studied some geometric properties of the sequence space . For further information on geometric properties of sequence spaces the reader can refer to [15, 16].
The main purpose of the present paper is to introduce the difference sequence spaces of nonabsolute type and derive some related results. We also establish some inclusion relations, where . Furthermore, we determine the alpha-, beta- and gamma-duals of those spaces and construct their bases. We characterize some classes of infinite matrices concerning the spaces and for . Finally, we investigate some geometric properties concerning Banach-Saks type and give Gurarii's modulus of convexity for the normed space .
2. The Sequence Spaces and of Nonabsolute Type
This section is devoted to the examination of the basic topological properties of the sets and . Let throughout that be strictly increasing sequence of positive reals tending to ; that is Let us define the lambda matrix by Recently, Mursaleen and Noman [17, 18] have studied the sequence spaces and of nonabsolute type as follows: With the notation of (4), we can redefine the spaces and by and , where .
Let and be non-zero real numbers, and define the generalized difference matrix by for all . The -transform of a sequence is for all . We note that the matrix can be reduced to the difference matrix in the case and . So, the results related to the domain of the matrix are more general and comprehensive than those of the matrix domain of and include them.
Now, we introduce the new sequence spaces and as follows: By the notation of (4), we can redefine the spaces and as follows: where denotes the generalized difference matrix defined by (8). Now, we may define the triangle matrix by Define the sequence as the -transform of a sequence ; that is, Now, we can redefine the spaces and with the notation of (4) as Also, we derive from the equality (12) that Then, since the sequence spaces and are linearly isomorphic; that is, ; it is trivial that the two-sided implication “ if and only if ” holds, where .
We have the following result which is essential in the text.
Theorem 1. The following statements hold. (a)If , then is a complete -normed space with the -norm ; that is, (b)If , then is -space with the norm ; that is,
Proof. (a) Let . It is immediate by the fact that is a complete -normed linear space with the -norm .
(b) Since the sets and endowed with the norms and are -spaces (see [2, Example (b), (c)]) and the matrix is triangle, Theorem of Wilansky [19, page 61] gives the fact that the spaces and are -spaces with the norms in (16) and (17), respectively.
One can easily check that the absolute property does not hold on the space ; that is, for at least one sequence in the space , and this tells us that is nonabsolute type, where and .
Now, one may expect the similar result for the space as was observed for the space and ask the natural question: Is not the space a Hilbert space with The answer is positive and is given by the following theorem.
Theorem 2. Except the case , the space is not an inner product space and hence it is not a Hilbert space, where .
Proof. We have to prove that the space is the only Hilbert space among the spaces for . Since the space is a -space with the norm by Part (b) of Theorem 1 and its norm can be obtained from an inner product; that is, holds for every , the space is a Hilbert space. Let us define the sequences and by Then, we have Thus, it can easily be seen that that is, the norm of the space with does not satisfy the parallelogram equality which means that this norm cannot be obtained from an inner product. Hence, the space with is a Banach space which is not a Hilbert space. This completes the proof.
3. The Inclusion Relations
In this section, we give some inclusion relations between the spaces and and the spaces and , where . We essentially prove that holds and characterize the case in which the inclusion holds for .
Theorem 3. Let . Then, the inclusions strictly holds.
Proof. Let and . This implies that . Since , . So, we have .
Let us consider the sequence with the aid of the sequence defined by Then, since which shows that the inclusion is strict.
Theorem 4. The inclusions strictly hold.
Proof. The inclusion strictly holds by Theorem 3.1 of . So, it is enough to show that the inclusions and are strict, where . Assume that . This means that . Since , which implies that . Hence, the inclusion holds for . Now, we show that this inclusion is strict. Let and define the sequence as follows:
Then, we have for every that
which shows that but . Thus, the sequence is in but not in . Hence, is strict.
Since holds, we have the inclusion . Let us define the sequence by Then, one can easily see for every that which shows that . Thus, is in but not in ; that is, . That is to say that the inclusion is strict. This completes the proof.
Theorem 5. The inclusion strictly holds.
Proof. Let . Then, we have which means that . So, the inclusion holds. Furthermore, consider the sequence defined by Clearly, . Then, we obtain by (12) that which shows that . This implies that . Hence, the inclusion is strict and this completes the proof.
Theorem 6. If the inclusion holds, then , where .
Proof. Assume that the inclusion holds and consider sequence . So, by this hypothesis, . Hence, . Therefore, and we obtain that which shows that . This completes the proof.
Lemma 7 (see [17, Lemma 4.11, page, 43]). If , then
Theorem 8. If , the inclusion strictly holds for .
Proof. Let for . Then, by applying Hölder's inequality, we derive from (12) that which gives that By (34) and Lemma 7, we have Therefore, combining the inequality (35) with Minkowski's inequality, we derive that This shows that . So, the inclusion holds. Now, let us consider the sequence defined by with . Then, since , one can immediately observe that is in but not in . That is, . Thus, we have showed that the inclusion is strict. Similarly, the inclusion also strictly holds in the case , so we omit the details. This completes the proof.
Theorem 9. The sequence spaces and do not include each other.
Proof. It is clear by Theorem 8 that the sequence spaces and are not disjointed. Let us consider the sequence defined by (37). Then, is in but not in . Now, let us define the sequence with . Then, since , is in but not in . This completes the proof.
4. The Basis for the Space
In this section, we begin with defining the concept of the Schauder basis for a normed sequence space and then give the basis of the sequence space , where . Now, we define the Schauder basis of a normed space. If a normed sequence space contains a sequence with the property that for every there is a unique sequence of scalars such that then is called a Schauder basis (or briefly basis) for . The series which has the sum is called the expansion of with respect to , and written as .
Theorem 10. The following statements hold. (i)The space has no Schauder basis. (ii)Define the sequence of elements of the space by for all . Then, the sequence is a basis for the space and every has a unique representation of the form
Proof. (i) It is known that the matrix domain of a normed sequence space has a basis if and only if has a basis whenever is a triangle [21, Remark 2.4]. Since the space has no Schauder basis, has no Schauder basis.
(ii) Let . It is clear that , since for all . Furthermore, let be given. For every nonnegative integer , we put Then, by applying to (41), we get that and therefore, we have for all . Now, for any given , there is a nonnegative integer such that Thus, we have for every that for all which proves that is represented as in (40).
Let us show that the uniqueness of representation for is given by (40). Suppose, on the contrary, that there exists a representation . Since the transformation defined from to by is continuous, we have which contradicts the assumption that for all . That is to say that the representation (40) of is unique.
5. The Alpha-, Beta- and Gamma-Duals of the Space and
In this section, we give some theorems determining the alpha-, beta- and gamma-duals of the spaces and . We start with the definition of the alpha-, beta- and gamma-duals of a sequence space.
If and are sequences and and are subsets of , then we write , and for the multiplier space of and . One can easily observe for a sequence space with and that the inclusions and hold. The alpha-, beta- and gamma-duals of a sequence space, which are, respectively, denoted by , , and are defined by It is obvious that . Also, it can be easily seen that the inclusions , , and hold, whenever . Now, we may begin with quoting the following lemmas  which are needed in proving Theorems 14–16.
Lemma 11. if and only if (i)For (ii)For
Theorem 14. Define the sets and by Then, and for , where the matrix is defined via the sequence by for all .
Proof. Let and . Then, by using (12), we immediately derive for every that Thus, we observe by (57) that whenever if and only if whenever . This means that if and only if . Therefore, we get by Lemma 11 with instead of that if and only if which leads us to the consequence that , for . Similarly, we get from (57) that if and only if which is equivalent to (50) of Lemma 11 that
Theorem 15. Define the sets , , , , and as follows: where Then,
Proof. Let us consider the equality where the matrix is defined for all by Then, we deduce from (63) with Lemma 12 that whenever if and only if whenever . This means that if and only if , where . Therefore, we derive from (51) and (52) that which shows that for . Since beta-dual of the space for the cases and can be similarly computed, we omit the details. This completes the proof.
Theorem 16. Let . Then,
6. Certain Matrix Mapping Related to the Spaces and
In this section, we characterize the matrix classes , , , , , , where . Also, by means of a given basic lemma, we derive the characterizations of certain other classes. Since the characterization of matrix mapping on the space can be proved in a similar way, we omit the proof for the cases and and consider only the case in the proofs of theorems given in this section.
For an infinite matrix , we write for brevity that Now, we may begin with quoting the following lemmas (see ) which are needed for proving our main results.
Lemma 17. Let be an infinite matrix. Then, the following statements hold. (i) if and only if and (ii)Let . Then, if and only if (67) holds and . (iii) if and only if (67) holds and is uniformly convergent.
Lemma 18. Let . Then, if and only if .
Lemma 19. Let . Then, if and only if .
Proof. (i) Assume that the conditions (68)–(71) hold and take any , where . Then, we have by Theorem 16 that for all and this implies the existence of . Also, it is clear that the associated sequence is in the space .
Let us now consider the following equality derived by using relation (12) from th partial sum of the series as follows: Therefore, by using (68)–(70), we obtain from (75) as that Furthermore, since the matrix is in the class by Lemma 13, we have . Now, by passing to supremum over in (76), we derive by applying Hölder's inequality that which shows that and hence .
Conversely, assume that , where . Then, since for all by the hypothesis, the necessity of (71) is obvious. Since , (76) holds for all sequences and which are connected by relation (12). Let us now consider the continuous linear functionals on by Then, since and are norm isomorphic, it should follow with (76) that for all . This shows that the functionals defined by the rows of on are pointwise bounded. Thus, we deduce by Banach-Steinhaus theorem that these functionals are uniformly bounded, which yields that there exists a constant such that for all . This shows the necessity of the condition (70) which completes the proof of part (i).
Theorem 21. Let be an infinite matrix. Then, the following statements hold. (i) if and only if (68) and (69) hold and (ii)Let . Then, if and only if (68)–(71) hold and (80) also holds. (iii) if and only if (68), (69), and (74) hold, and
Proof. We consider only part (ii). Assume that satisfies the conditions (68)–(71) and (80), and , where . Then, exists and by using (80), we have for every that as which leads us with (70) to the following inequality:
which holds for every . This shows that . Since , we have . Therefore, we derive by applying Hölder's inequality that for each .
Now, for any given , choose a fixed such that Then, it follows (80) that there is such that for every . Thus, by using (76), we get that for all sufficiently large . Hence, as which means that ; that is, .
Conversely, suppose that , where . Then, since , . Thus, the necessity of (68)–(71) is immediately obtained from Theorem 20 which together imply that (76) holds for all sequences . Since by our assumption, we derive by (76) that which means that . Thus the necessity of (80) is immediate by (51) of Lemma 12. This completes the proof of part (ii).
Now, we can mention the sequence space of almost convergent sequences. The shift operator is defined on by for all . A Banach limit is defined on such that (i) for , where for all , (ii), (iii), where . A sequence is said to be almost convergent to the generalized limit if all Banach limits of is , and denoted by . Let be the composition of with itself for times and define for a sequence by Lorentz  proved that if and only if , uniformly in . It is well-known that a convergent sequence is almost convergent such that its ordinary and generalized limits are equal. By and , we denote the space of all almost convergent sequences and series, respectively, that is,
Theorem 22. Let be an infinite matrix. Then, the following statements hold.(i) if and only if (68) and (69) hold, and (ii)Let . Then, if and only if (68)–(71) and (88) hold.(iii) if and only if (68) and (69) hold, and