- Research Article
- Open Access
On Generalized Paranormed Statistically Convergent Sequence Spaces Defined by Orlicz Function
© M. Başarir and S. Altundağ. 2009
- Received: 8 May 2009
- Accepted: 26 August 2009
- Published: 27 September 2009
We define generalized paranormed sequence spaces , , , and defined over a seminormed sequence space . We establish some inclusion relations between these spaces under some conditions.
- Banach Space
- Positive Integer
- Real Number
- Natural Number
- Sequence Space
where is the characteristic function of .
In this case, we write or stat lim
Let be a mapping of the set of positive integers into itself. A continuous linear functional on , the space of real bounded sequences, is said to be an invariant mean or -mean if and only if
(1) when the sequence has for all ,
(2) , where
(3) for all
The mappings are one to one and such that for all positive integers and where denotes the th iterate of the mapping at . Thus extends the limit functional on , the space of convergent sequences, in the sense that for all . In that case is translation mapping , a -mean is often called a Banach limit, and , the set of bounded sequences all of whose invariant means are equal, is the set of almost convergent sequences .
If , set It can be shown  that
An Orlicz function is a function , which is continuous, nondecreasing, and convex with for and as . If the convexity of an Orlicz function is replaced by
Lindenstrauss and Tzafriri  used the idea of Orlicz function to construct the sequence space
which is called an Orlicz sequence space.
The space is closely related to the space which is an Orlicz sequence space with for . Orlicz sequence spaces were introduced and studied by Parashar and Choudhary , Bhardwaj and Singh , and many others.
It is well known that since is a convex function and then for all with .
An Orlicz funtion is said to satisfy -condition for all values of , if there exists constant , such that . The -condition is equivalent to the inequality for all values of and for being satisfied .
The notion of paranormed space was introduced by Nakano  and Simons . Later on it was investigated by Maddox , Lascarides , Rath and Tripathy , Tripathy and Sen , Tripathy , and many others.
A sequence space is said to be solid (or normal) if , whenever and for all sequences of scalars with for all .
A sequence space is said to be symmetric if implies , where is a permutation of .
A sequence space is said to be monotone if it contains the canonical preimages of its step spaces.
Throughout the paper will represent a sequence of positive real numbers and a seminormed space over the field of complex numbers with the seminorm . We define the following sequence spaces:
If , for each and then these spaces reduce to the spaces
defined by Tripathy and Sen .
Firstly, we give some results; those will help in establishing the results of this paper.
Lemma 2.1 ().
For two sequences and one has if and only if , where such that
Lemma 2.2 ().
Let and , then the followings are equivalent:
Lemma 2.3 ().
Then is uncountable.
Lemma 2.4 ().
If a sequence space is solid then is monotone.
Let , be scalars and let . Then by (1.7) we have
Hence is a linear space.
The rest of the cases will follow similarly.
Let . Then by the convexity of , we have
Hence from above inequality, we have
For the continuity of scalar multiplication let be any complex number. Then by the definition of we have
Since we have Then
and therefore converges to zero when converges to zero or converges to zero.
Hence the spaces and are paranormed by .
Let be complete seminormed space, then the spaces and are complete.
Using definition of paranorm we get
Hence is a Cauchy sequence in . Therefore for each there exists a positive integer such that
Using continuity of , we find that
Taking infimum of such s we get
for all and . Since and is continuous, it follows that . This completes the proof of the theorem.
Let and be two Orlicz functions satisfying -condition. Then
where , , and .
If we take then implies that . Hence we have by convexity of
Hence by (3.15) it follows that for a given , there exists such that
We prove this part for the case and the other cases will follow similarly.
This completes the proof.
where , , and .
The proof follows from the fact that the zero sequence belongs to each of the classes the sequence spaces involved in the intersection.
The proof of the following result is easy, so omitted.
Let be an Orlicz function which satisfies condition, and let and be two seminorms on . Then
(iii) where , , , and ,
(iv)if is stronger than , then
where , , and .
The spaces are not solid, where and .
To show that the spaces are not solid in general, consider the following example. Let , for all , , where and for all . Then we have for all . Consider the sequence , where is defined by and for each fixed . Hence for and . Let if is odd and , otherwise. Then for and . Thus is not solid for and .
The proof of the following result is obvious in view of Lemma 2.4.
The space is solid as well as monotone for and .
The spaces are not symmetric, where , , and .
To show that the spaces are not symmetric, consider the following examples. Let , for all , , where and for all . Then we have for all . We consider the sequence defined by if and otherwise. Then for and . Let be a rearrangement of , which is defined as if is odd and , otherwise. Then for and .
To show for and , let for all odd and for all even. Let and , where . Let and for all . Then we have for all . We consider
Then for and . Thus the spaces are not symmetric in general, where , , and .
For two sequences and one has if and only if , where such that .
The proof is obvious in view of Lemma 2.1.
The following result is a consequence of the above result.
For two sequences and one has if and only if and , where such that .
The following result is obvious in view of Lemma 2.2.
Let and , then the followings are equivalent:
where the vertical bar indicates the number of elements in the enclosed set.
From the above inequality it follows that .
Conversely let . Let such that
For a given , let .
Since , so , uniformly in , as . There exits a positive integer such that for all . Then for all , we have
This completes the proof of the theorem.
The following result is a consequence of the above theorem.
Since the inclusion relations and are strict, we have the following result.
The spaces and are nowhere dense subsets of
The following result is obvious in view of Lemma 2.3.
The spaces and are not separable.
The authors would like to express their gratitude to the reviewers for their careful reading and valuable suggestions which improved the presentation of the paper.
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