International Scholarly Research Notices

Volume 2014, Article ID 974934, 6 pages

http://dx.doi.org/10.1155/2014/974934

## Hahn Sequence Space of Modals

^{1}Department of Mathematics, Kamaraj College, Tuticorin, Tamilnadu 628003, India^{2}Department of Mathematics, Dr. G. U. Pope College of Engineering, Sawyerpuram, Tuticorin, Tamilnadu 628251, India

Received 6 June 2014; Revised 14 October 2014; Accepted 15 October 2014; Published 9 November 2014

Academic Editor: Mahdi Sanati

Copyright © 2014 T. Balasubramanian and S. Zion Chella Ruth. 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.

#### Abstract

The history of modal intervals goes back to the very first publications on the topic of interval calculus. The modal interval analysis is used in Computer graphics and Computer Aided Design (CAD), namely, the computation of narrow bounds on Bezier and -Spline curves. Since modal intervals are used in many fields, we introduce a new sequence space called the Hahn sequence space of modal intervals. We have given some new definitions and theorems. Some inclusion relation and some topological properties of this space are investigated. Also dual spaces of this space are computed.

#### 1. Introduction

Interval arithmetic was first suggested by Dwyer [1] in 1951. Furthermore, Moore and Yang [2, 3] have developed applications to differential equations. Chiao in 2002 [4] introduced sequence of interval numbers and defined usual convergence of sequences of interval number. Recently, Zararsız and Şengönül [5] introduced null, bounded, and convergent sequence space of modals. Hahn in 1922 [6] defined space and G. Goes and S. Goes [7] in 1970 studied the functional analytic properties of this space. The Hahn sequence space was initiated by Rao in 1990 [8]. The present paper is devoted to the study of Hahn sequence space of modal intervals.

Let us denote the set of all real valued closed interval by , the set of positive integers by , and the set of all real numbers by . Any element of is called interval number and it is denoted by . That is . An interval number is a closed subset of real numbers. Let and be, respectively, first and last points of the interval number . Therefore, when , is not an interval number. But in modal analysis is a valid interval. A modal is defined by a pair of real numbers . Let us denote the set of all modals by . Let us suppose that . Then the algebraic operations between and are defined in the Kaucher arithmetic [9]. For a modal operator is defined as . Thus, if , then , . Let us suppose that ; then is called symmetric modal if or vice versa.

The set of all modals is a metric space with the metric defined by

If and , then the set is reduced ordinary set of interval numbers which is complete metric space with the metric defined in (1) [4]. If we take and , we obtain the usual metric of with , where .

#### 2. Definitions and Preliminaries

Let be a function from to which is defined by , . Then is called sequence of modals. We will denote the set of all sequences of modals by .

For two sequences of modals and , the addition, scalar product, and multiplication are defined as follows: , , , ,, respectively.

The set is a vector space since the vector space rules are clearly provided. The zero element of is the sequence , all terms of which are zero interval. If then inverse of , according to addition, is .

Let . If a sequence space contains a sequence of modals with the property that for every there is a unique sequence of scalars such that then is called a Schauder modal basis for . The series which has the sum is then called the expansion of with respect to , and we write .

Let and be linear space of modals. Then a function is called a linear transformation if and only if, for all and all , .

Proposition 1. *If , , and are sequences of symmetric modal, then the following equality holds:
*

*Definition 2. *A sequence of modals is said to be convergent to the modal if for each there exists a positive integer such that for all and we denote it by writing . Thus, and .

*Definition 3. *A sequence of modals, , is said to be modal fundamental sequence if for every there exists such that whenever .

*Definition 4. *A sequence of modals is said to be solid if whenever for all and .

*Definition 5. *A sequence of generalized intervals is said to be monotone if contains the canonical preimage of all its step spaces.

*Definition 6 (Weierstrass -test). *Let be given. If there exists an such that and the series converges, then the series is uniformly and absolutely convergent in .

*The following spaces are needed for our work:
*

*3. Main Results*

*Define a sequence which will be frequently used as the -transform of a sequence . That is, , .*

*We introduce the sequence spaces and as the set of all sequences such that -transforms of them are in .*

*That is,
is a normed space with the norm .*

*Example 7. *Consider the sequence defined by
Note that which is convergent.

Also .

Hence .

*Theorem 8. and are complete metric spaces with the metrics and defined by
respectively, where and are the elements of the space or .*

*Proof. *Let be any fundamental sequence in the space , where . Then for a given , there exists a positive integer such that
We obtain for each fixed from (8) that
which leads to the fact that is a fundamental sequence in for every fixed .

Since is complete, converges to as .

Consider the sequence ; we have from (9) for each and that
Take any and take limit as first and then let in (8); we obtain
Therefore .

We have to prove .

Since is a fundamental sequence in , we have
Now,
Hence
Also from (10) and (11), . Hence . Since is an arbitrary fundamental sequence, the space is complete.

Similarly, we can prove is complete space.

*Theorem 9. The space is monotone.*

*Proof. *Let ; it follows from
that

The sequence is monotone increasing; it thus follows from that . This completes the proof.

*Theorem 10. The space is not rotund.*

*Proof. *Consider the sequences and defined by
Then
Also,
Thus and are in and .

Note that , but . Therefore is not rotund.

*Theorem 11. The space is not solid.*

*Proof. *Consider the sequence
Since
it immediately follows that .

However, it is trivial that . , which implies . This completes the proof.

*4. Dual Space of *

*Definition 12. *The , , and of are, respectively, defined by
It is trivial that the following inclusions hold: .

*Theorem 13. Let and be the sets of sequences of modals. Then the following statements hold.(i).(ii).(iii)If then .*

The same results hold for dual also.

*Proof. *Let , , and be sequences of modal intervals.

(i) Suppose ; then for at least one . But implies that for all .

This means that is not a member of . Hence . The proof for (ii) follows similarly.

(iii) Suppose ; then for all . Since , . Thus .

Define the sequence which will be frequently used as the -transform of a sequence .

That is,
The Cesaro space of is the set of all sequences such that the -transforms of them are in . That is, .

*Theorem 14. is a complete metric space with the metric where and are the elements of space .*

*Proof. *Let be any fundamental sequence in the space where . Then for a given , there exists a positive integer such that
We obtain for each fixed from (24) that
which leads to the fact that is a fundamental sequence for every fixed . Since is complete, as .

Consider the sequence . We have from (25) for each and that
For any , taking limit first and letting in (24), we obtain
Finally, we proceed to prove . Since is a fundamental sequence in , we have .

Now,
Hence .

Since is an arbitrary fundamental sequence, the space is complete.

*Theorem 15. The and of are .*

*Proof. *Let and . Since , we have . Therefore for given , there exists such that .

Since , . Thus for all and .

Hence there exists an such that for all and .

Now,
converges uniformly by Weierstrass -test.

Thus
Conversely, suppose . Then the series converges for all . This also holds for the sequence of modals defined by for all .

converges uniformly. Thus .

Hence
From (30) and (31), . This completes the proof.

*Theorem 16. Consider(i) .(ii) .*

*Proof. *(i) Let . Then and .

Consider . Therefore, .

And also since , .

Therefore, . Hence .

(ii) For ,

implies

The last series is convergent since . Hence also and therefore .

Hence
Conversely, (32) implies for
Thus .

Therefore,
Hence from (34) and (36), we have shown that .

Similarly, we can prove other equalities.

*Conflict of Interests*

*Conflict of Interests*

*The authors declare that there is no conflict of interests regarding the publication of this paper.*

*References*

*References*

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