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Abstract and Applied Analysis
Volume 2014 (2014), Article ID 858704, 9 pages
On the Paranormed Nörlund Sequence Space of Nonabsolute Type
1Department of Mathematics, Uşak University, 1 Eylül Campus, 64200 Uşak, Turkey
2Department of Mathematics, Fatih University, Hadımköy Campus, Büyükçekmece, 34500 İstanbul, Turkey
Received 3 December 2013; Accepted 2 February 2014; Published 26 March 2014
Academic Editor: M. Mursaleen
Copyright © 2014 Medine Yeşilkayagil and Feyzi Başar. 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.
Maddox defined the space of the sequences such that , in Maddox, 1967. In the present paper, the Nörlund sequence space of nonabsolute type is introduced and proved that the spaces and are linearly isomorphic. Besides this, the alpha-, beta-, and gamma-duals of the space are computed and the basis of the space is constructed. The classes and of infinite matrices are characterized. Finally, some geometric properties of the space are investigated.
We denote the space of all sequences of complex entries by . Any vector subspace of is called a sequence space. We write , , and for the spaces of all bounded, convergent, and null sequences, respectively. Also by , , , and , we denote the spaces of all bounded, convergent, absolutely and -absolutely convergent series, respectively.
A linear topological space over the real field is said to be a paranormed space if there is a subadditive function such that , and scalar multiplication is continuous; that is, and imply for all 's in and all 's in , where is the zero vector in the linear space . Assume here and after that is a bounded sequence of strictly positive real numbers with and . Then, the linear spaces and were defined by Maddox in  (see also [2, 3]) as follows: which are the complete spaces paranormed by respectively, where . For simplicity in notation, here and in what follows, the summation without limits runs from to . We assume throughout that , provided , and denote the collection of all finite subsets of by .
For the sequence spaces and , define the set by With the notation of (3), the alpha-, beta-, and gamma-duals of a sequence space , which are, respectively, denoted by , , and , are defined by
If a sequence space paranormed by 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 then called the expansion of with respect to and written as .
Let , be any two sequence spaces, and let be an infinite matrix of 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 By , we denote the class of all matrices such that . Thus, if and only if the series on the right side of (6) converges for each and every , and we have for all . Also, we write for the sequence in the th row of .
Now, following Peyerimhoff [4, pp. 17–19] and Mears , we give short knowledge on the Nörlund means. Let be a sequence of nonnegative real numbers with and write for all . Then, the Nörlund means with respect to the sequence is defined by the matrix which is given by for all . It is known that the Nörlund matrix is a Toeplitz matrix if and only if , as , and is reduced in the case to the matrix of arithmetic means. Additionally, for for all , the method is reduced to the Cesàro method of order , where Let and define for by The inverse matrix of the matrix is given by Mears in  as follows: for all . Also, one can derive by straightforward calculation for all that
The rest of this paper is organized as follows.
In Section 2, the complete paranormed Nörlund sequence space is introduced and proved that is linearly isomorphic to the space and the basis for the space is determined. Section 3 is devoted to the alpha-, beta-, and gamma-duals of the space . In Section 4, the classes and of infinite matrices are characterized, where denotes any given sequence space. In Section 5, the rotundity of the space is characterized and some results related to this concept are given. In the final section of the paper, the significance of the space is mentioned and further suggestions are recorded.
2. The Nörlund Sequence Space of Nonabsolute Type
In this section, we define the Nörlund sequence space and prove that is linearly isomorphic to the space , where for all . Finally, we give the basis for the space .
Let be any sequence space. Then, the matrix domain of an infinite matrix in is defined by In , Choudhary and Mishra defined the sequence space which consists of all sequences such that -transforms of them are in , where is defined by Başar and Altay  examined the space which was formerly defined by Başar  as the set of all series whose sequences of partial sums are in the space . With the notation of (12), the spaces and can be redefined by
In , Başar and Altay defined the sequence space which consists of all sequences such that -transforms of them are in , where is the matrix of Riesz mean; that is,
In , Wang defined the sequence space consisting of all sequences whose -transforms are in which is a Banach space with the norm
Now, we introduce the Nörlund sequence space defined by It is natural that the space can also be defined with the notation of (12) that .
Define the sequence by the -transform of a sequence ; that is,
Theorem 1. is a complete linear metric space paranormed by defined by
Proof. Since this can be shown by a routine verification, we omit the detail.
Remark 2. One can easily see that the absolute property does not hold on the space ; that is, for at least one sequence in the space , and this says that is a sequence space of nonabsolute type, where .
Theorem 3. The Nörlund sequence space of nonabsolute type is linearly isomorphic to the space , where for all .
Proof. To prove the theorem, we should show the existence of a linear bijection between the spaces and for . Consider the transformation defined, with the notation of (18), from to by . The linearity of is clear. Further, it is trivial that whenever and hence is injective.
Let us take any and define the sequence by Therefore, we see from (19) that This means that . Consequently, is surjective and is paranorm preserving. Hence, is linear bijection and this says us that the spaces and are linearly isomorphic. Therefore, the proof is completed.
We determine the basis for the paranormed space .
Theorem 4. Define the sequence of the elements of the space for every fixed by Then, the sequence is a basis for the space and any has a unique representation of the form where for all and .
Proof. It is clear that , since
where is the sequence whose only nonzero term is a in the th place for each and .
Let be given. For every nonnegative integer , we put Then, we obtain by applying to (25) with (24) that where . Given , then there is an integer such that for all . Hence, for all which proves that is represented as in (23).
Let us show the uniqueness of the representation for given by (23). Suppose, on the contrary, that there exists a representation . Since the linear transformation , from to , used in the proof of Theorem 3 is continuous, we have at this stage that for all which contradicts the fact that for all . Hence, the representation (23) of is unique. This completes the proof.
3. The Alpha-, Beta-, and Gamma-Duals of the Space
In this section, we determine the alpha-, beta-, and gamma-duals of the space . We will quote some lemmas which are needed in proving our theorems.
Lemma 5 (see , Theorem 5.1.0). The following statements hold.(i)Let for every . Then, if and only if there exists an integer such that (ii)Let for every . Then, if and only if
Lemma 6 (see , Theorem 1). The following statements hold.(i)Let for every . Then, if and only if there exists an integer such that (ii)Let for every . Then, if and only if
Theorem 8. Let for every . Define the sets , , and as follows: Then, the following statements hold:(i);(ii);(iii).
Proof. (i) Let us take . We easily derive with (20) that
where is defined by
for all . Thus, we observe by combining (35) with Part (i) of Lemma 5 that whenever if and only if whenever . This gives the result that .
(ii) Consider the equality where is defined by for all . Thus, we deduce from Part (i) of Lemma 6 with (37) that whenever if and only if whenever . Therefore, we obtain from Part (i) of Lemma 6 that .
(iii) We see from Lemma 7 that whenever if and only if whenever . Therefore, we derive from Lemma 7 that .
Therefore, the proof is completed.
Theorem 9. Let for every . Define the sets and by Then, the following statements hold:(i);(ii);(iii).
4. Some Matrix Transformations Related to the Sequence Space
In the present section, we characterize the matrix transformations from the space into any given sequence space and from a given sequence space into the space . Since for any triangle and any sequence space , it is trivial that the equivalence “ if and only if ” holds.
Now, we can give the following theorem.
Theorem 10. Suppose that the elements of the infinite matrices and are connected with the relation for all and is any given sequence space. Then, if and only if for all and .
Proof. Let be any given sequence space. Suppose that (40) holds between the elements of the matrices and , and take into account that the spaces and are linearly isomorphic.
Let and take any . Then That is, exists and which yields that for each . Hence, exists and thus for all . So, we have that , which leads us to the consequence .
Conversely, let for each and , and take . Then, exists. Therefore, we obtain from the equality that and this shows that . This completes the proof.
By changing the roles of the spaces with in Theorem 10, we have the following.
Theorem 11. Suppose that is any given sequence space and the elements of the infinite matrices and are connected with the relation for all . Then, if and only if .
Proof. Let and consider the following equality: Then, by letting in (44), we have for all . Since , . This completes the proof.
5. The Rotundity of the Space
In functional analysis, the rotundity of Banach spaces is one of the most important geometric properties. For details, the reader may refer to [13–15]. In this section, we give the necessary and sufficient condition in order to the space be rotund and present some results related to this concept.
Definition 12. Let be the unit sphere of a Banach space . Then, a point is called an extreme point if implies for every . A Banach space is said to be rotund (strictly convex) if every point of is an extreme point.
Definition 13. A Banach space is said to have Kadec-Klee property (or property ()) if every weakly convergent sequence on the unit sphere is convergent in norm.
Definition 14. A Banach space is said to have(i)the Opial property if every sequence weakly convergent to satisfies for every with ;(ii)the uniform Opial property if for each , there exists an such that for each with and each sequence in such that and .
Definition 15. Let be a real vector space. A functional is called a modular if(i) if and only if ;(ii) for all scalars with ;(iii) for all and with ;(iv)the modular is called convex if for all and with . A modular on is called(a)right continuous if for all ;(b)left continuous if for all ;(c)continuous if it is both right and left continuous, where
We define on by . If for all , by the convexity of the function for each , is a convex modular on . We consider equipped with Luxemburg norm given by is a Banach space with this norm. This can be shown by the similar way used in the proof of Theorem 7 in .
We establish some basic properties for the modular .
Proposition 16. The modular on satisfies the following properties with for all .(i)If , then and .(ii)If , then .(iii)If , then .(iv)The modular is continuous.
Proof. (i) Let . Then for all . So, we have
(ii) Let . Then for all . So, we have
(iii) Let . Then for all . Therefore, one can easily see that
(iv) If , then we have that is, By passing to limit as in (53), we have . Hence, is right continuous.
If , we have that is, By letting in (55), we have . Hence, is left continuous. Since is both right and left continuous, it is continuous.
Now, we give some relationships between the modular and the Luxemburg norm on .
Proposition 17. For any , the following statements hold.(i)If , then .(ii)If , then .(iii) if and only if .(iv) if and only if .(v) if and only if .(vi)If and , then .(vii)If and , then .
Proof. Let .(i)Let such that . By the definition of in (48), there exists an such that and . So, we have Since is arbitrary, we have from (56).(ii) If we choose such that , then . By the definition of in (48) and Part (iii) of Proposition 16, we have So, for all . This implies that .(iii)Since is continuous, by Theorem 1.4 of  we directly have (iii).(iv)This follows from Parts (i) and (iii).(v)This follows from Parts (ii) and (iii).(vi)This follows from Part (ii) and Part (i) of Proposition 16.(vii)This follows from Part (i) and Part (ii) of Proposition 16.
Theorem 18. The space is rotund if and only if for all .
Proof. Let be rotund and choose such that for all . Consider the following sequences given by
Then, obviously and
By Part (iii) of Proposition 17, which leads us to the contradiction that the sequence space is not rotund. Hence, for all .
Conversely, let and with . By convexity of and Part (iii) of Proposition 17, we have which gives that Also, since and from (61), we obtain that This implies that for all . Since the function is strictly convex for all , it follows by (63) that for all . Hence, . That is, is rotund.
Theorem 19. Let be a sequence in . Then, the following statements hold:(i) implies ;(ii) implies .
Proof. The proof is similar to that of Theorem 10 in .
Theorem 20. Let and . If as and as for all , then as .
Proof. Let be given. Since and , . So, there exists an such that Also, since as , we have Therefore, we obtain from (64) and (65) that . This means that as . This result implies as from Part (ii) of Theorem 19. Hence, as .
Theorem 21. The sequence space has the Kadec-Klee property.
Proof. Let and such that and are given. By Part (i) of Theorem 19, we have as . Also, implies . By Part (iii) of Proposition 17, we obtain . Therefore, we have as .
Since and (or ) defined by is continuous, as . Therefore, as . This completes the proof.
Theorem 22. For any , the space has the uniform Opial property.
Proof. Since the proof can be given by the similar way used in proving Theorem 13 of Nergiz and Başar , we omit the detail.
Wang introduced the sequence space , in . Although the domain of several triangle matrices in the classical sequence spaces , , , and and in the Maddox spaces , , , and was investigated by researchers, the domain of Nörlund mean neither in a normed sequence space nor in a paranormed sequence space was not studied and is still as an open problem. So, we have worked on the domain of Nörlund mean in the Maddox space . Additionally, we emphasize on some geometric properties of the new space . It is obvious that the matrix is not comparable with the matrices , , or . So, the present results are new.
It is clear that by depending on the choice of the sequence space , the characterization of several classes of matrix transformations from the space and into the space can be obtained from Theorems 10 and 11, respectively. As a natural continuation of this paper, we will study the domain of the Nörlund mean in Maddox's spaces , , and .
Conflict of Interests
The authors declare that there is no conflict of interests regarding the publication of this paper.
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