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
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
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.
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.
2. Definitions and Notations
defined by Tripathy and Sen .
Firstly, we give some results; those will help in establishing the results of this paper.
Lemma 2.1 ().
Lemma 2.2 ().
Lemma 2.3 ().
Lemma 2.4 ().
3. Main Results
The rest of the cases will follow similarly.
Hence from above inequality, we have
Using definition of paranorm we get
This completes the proof.
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.
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.
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 .
The proof is obvious in view of Lemma 2.1.
The following result is a consequence of the above result.
The following result is obvious in view of Lemma 2.2.
where the vertical bar indicates the number of elements in the enclosed set.
This completes the proof of the theorem.
The following result is a consequence of the above theorem.
The following result is obvious in view of Lemma 2.3.
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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