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Abstract and Applied Analysis
Volume 2013 (2013), Article ID 739319, 11 pages
Research Article

Certain Sequence Spaces over the Non-Newtonian Complex Field

Department of Mathematics, Faculty of Arts and Sciences, Fatih University, The Hadimköy Campus, Büyükçekmece, 34500 Istanbul, Turkey

Received 4 February 2013; Accepted 5 April 2013

Academic Editor: Victor Kovtunenko

Copyright © 2013 Sebiha Tekin 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.


It is known from functional analysis that in classical calculus, the sets , , , and of all bounded, convergent, null and -absolutely summable sequences are Banach spaces with their natural norms and they are complete according to the metric reduced from their norm, where . In this study, our main goal is to construct the spaces , , , and over the non-Newtonian complex field and to obtain the corresponding results for these spaces, where .

1. Preliminaries, Background, and Notations

A complete ordered field is a system consisting of a set , four binary operations , , , for , and an ordering relation for , all of which behave with respect to the set exactly as +, −, , , behave with respect to the set of real numbers. We call the realm of the complete ordered field, [1, page 32]. A complete ordered field is called arithmetic if its realm is a subset of . A bijective function with domain and range a subset of is called a generator. For example, the identity function , exponential function, and the function are generators.

Bashirov et al. [2] have recently emphasized on the multiplicative calculus and gave the results with applications corresponding to the well-known properties of derivative and integral in the classical calculus. Quite recently, Uzer [3] has extended the multiplicative calculus to the complex valued functions and gave the statements of some fundamental theorems and concepts of multiplicative complex calculus, and demonstrated some analogies between the multiplicative complex calculus and classical calculus by theoretical and numerical examples. Bashirov and Rıza [4] have studied the multiplicative differentiation for complex-valued functions and established the multiplicative Cauchy-Riemann conditions. Bashirov et al. [5] have investigated various problems from different fields which can be modelled more efficiently using multiplicative calculus, in place of Newtonian calculus.

Let be a generator with range . An arithmetic with range , and operations and ordering relation defined as follows, is called -arithmetic. Let . Then, we define the operations addition , subtraction , multiplication , division , and ordering () as follows: With the above new operations, is an -arithmetic. In other words, one can easily show that is a complete ordered field. As was seen, -generator generates -arithmetic. For example, the identity function generates classical arithmetic, and exponential function generates geometric arithmetic. Each generator generates exactly one arithmetic and each arithmetic is generated by exactly one generator. We denote -zero by   and -one by which are obtained from and , respectively. and numbers obtained by successive -addition of to with numbers obtained by successive of from are called . Thus, are given as follows: Thus, we have for all that . Let . If then is called -positive and if then is called . The -absolute value of is defined by For any elements and in with , the set of all elements in such that is called an , is denoted by , has -extent of , and has the -interior consisting of all elements in such that . Let be an infinite sequence of the elements in . Then there is at most one element in such that every with in its contains all but finitely many terms of . If there is such a number , then is said to be to , which is called the -limit of . In other words, and be two arbitrarily selected generators and let (“star”)  also be the ordered pair of arithmetics . is a . Definitions given for are also valid for . For example, is defined by means of and their .

In the   is used for arguments and is used for values; in particular, changes in arguments and values are measured by and , respectively. The operators of the are applied only to functions with arguments in and values in . The of a function at an element in is, if it exists, the unique number in such that for every sequence of arguments of distinct from , if is to , then to and is denoted by . That is, A function is at a point in if and only if is an argument of and When and are the identity function , the concepts of and are identical with those of classical limit and classical continuity.

The isomorphism from to is the unique function that possesses the following three properties:(i)is one to one. (ii) is from onto . (iii)For any numbers and in , It turns out that for every in , and that for every integer . Since, for example, , it should be clear that any statement in can readily be transformed into a statement in .

2. Non-Newtonian Complex Field and Some Inequalities

Let and be arbitrarily chosen elements from corresponding arithmetics. Then the ordered pair is called as a -point. The set of all -points is called the set of non-Newtonian complex numbers and is denoted by ; that is, The binary operations addition and multiplication of non-Newtonian complex numbers and are defined, as follows: where and with Then we have the following.

Theorem 1. is a field.

Proof. A straightforward calculation leads to the following statements:(i) is an Abelian group;(ii) is an Abelian group; (iii) the operation is distributive over the operation which conclude that is a field.

Let . Then the number is called the -square of and is denoted by . Let be a non-negative number in . Then, is called the -square root of and is denoted by . The -distance between two arbitrarily elements and of the set is defined by Up to now, we know that is a field and the distance between two points in is computed by the function , defined by (10). Now, we define the -norm and next derive some required inequalities in the sense of non-Newtonian complex calculus.

Let be an arbitrary element. is called -norm of and is denoted by . In other words, where and . Moreover, since for all we have , is the induced metric from the norm .

Lemma 2 (-Triangle inequality). Let . Then,

Proof. Let . Then, a straightforward calculation gives that Hence, the inequality (12) holds.

Lemma 3. for all .

Proof. Let . In this case, one can observe by a routine verification that as required.

Lemma 4. Let . Then, the following inequality holds:

Proof. Let . Then, one can see that This means that the inequality (15) holds.

Lemma 5 (-Minkowski inequality). Let and for all . Then,

Proof. Let and for all . Then, Let us take . Then, since the equality holds for every fixed , we obtain by (18) and Minkowski inequality in the complex field leads us to
On the other hand, let us prove Indeed, Substituting the relation (24) in (22) we obtain, By using this equality in (21), we get as desired.

Theorem 6. is a complete metric space, where is defined by (10).

Proof. First, we show that , defined by (10), is a metric on .
It is immediate for that (i)Now we show that if and only if for . Indeed, (ii)One can easily establish for all that (iii)We show that the inequality holds for all . In fact,
Therefore, is a metric over .
Now, we can show that the metric space is complete. Let be an arbitrary Cauchy sequence in . In this case, for all there exists an such that for all . Let and . Then, Thus we obtain that On the other hand, since the following inequalities: hold we therefore have by (32) that and . This means that and are Cauchy sequences with real terms. Since is complete, it is clear that for every there exists an such that for all and for every there exists an such that for all .
Define, by . Then, we have Hence, is a complete metric space.

Since is a complete metric space with the metric defined by (10) induced by the norm , as a direct consequence of Theorem 6, we have the following.

Corollary 7. is a Banach space with the norm defined by

3. Sequence Spaces over Non-Newtonian Complex Field

In this section, we define the sets , , , , and of all, bounded, convergent, null, and absolutely sequences over the non-Newtonian complex field which correspond to the sets , , , , and over the complex field , respectively. That is to say that For simplicity in notation, here and in what follows, the summation without limits runs from to . One can easily see that the set forms a vector space over with respect to the algebraic operations addition and scalar multiplication defined on , as follows:

Theorem 8. Define the function by where such that is convergent with for all . Then, is a metric space.

Proof. We show that satisfies the metric axioms on the space of all non-Newtonian complex valued sequences.
(i) First we show that for all .
Because is a metric space, we have ; and . Hence, we obtain that . Moreover, since for all , we have This means that
(ii) We show that iff . In this situation, one can see that
(iii) We show that for . First, we know that is a metric over . Thus,
(iv) We show that holds for . Again, using the fact that is a metric space, it is easy to see by Lemma 4 that as required.

Theorem 9. The set is a sequence space.

Proof. It is trivial that the inclusion holds.
(i) We show that for . Indeed, combining the hypothesis with the fact obtained from Lemma 2, we can easily derive that
Hence, .
(ii) We show that for any and for .
Since by Lemma 3 and , it is immediate that which means that .
Therefore, we have proved that is a subspace of the space .

Theorem 10. Define the relation by
Then, is a complete metric space.

Proof. One can easily show by a routine verification that satisfies the metric axioms on the space . So, we omit the details.
Now, we prove the second part of the theorem. Let be a Cauchy sequence in , where . Then, there exists a positive integer such that for all with . For any fixed , if then In this case for any fixed , is a Cauchy sequence of non-Newtonian complex numbers and since is complete, it converges to a . Define with infinitely many limits . Let us show and , as . Indeed, by (48), by letting , for we obtain that On the other hand, since , there exists such that for all . Hence, by triangle inequality (12), the inequality holds for all which is independent of . Hence, . By (49), since , we obtain . Therefore, the sequence converges to which means that is complete.

Since it is known by Theorem 10 that is a complete metric space with the metric induced by the norm , defined by we have the following.

Corollary 11. is a Banach space with the norm defined by (51).

Now, we give the following lemma required in proving the fact that is a sequence space in the case .

Lemma 12. Let . Then, the inequality holds for all .

Proof. Let and . Then, one can easily see that
as desired.

Theorem 13. The sets and are sequence spaces, where .

Proof. It is not hard to establish by the similar way that and are the sequence spaces. So, to avoid the repetition of the similar statements, we consider only the set .
It is obvious that the inclusion strictly holds.
(i) Let . Then, there exist such that and . Thus, there exist such that
Thus if we set , by (53) we obtain for all that which means that
Therefore, .
(ii) Let and . Since there exists an such that , we have
Thus, for , we have which implies that Hence, .
That is to say that is a subspace of .

Theorem 14. , , and are complete metric spaces, where is defined as follows:

Proof. We consider only the space with .
(i) For , we establish that the two sided implication holds. In fact,