#### Abstract

The sequence space was introduced by Maddox (1967). Quite recently, the sequence space of nonabsolute type has been introduced and studied which is the domain of the double sequential band matrix in the sequence space by Nergiz and Başar (2012). The main purpose of this paper is to investigate the geometric properties of the space , like rotundity and Kadec-Klee and the uniform Opial properties. The last section of the paper is devoted to the conclusion.

#### 1. Introduction

By , we denote the space of all real-valued sequences. 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 convergent, and -absolutely convergent series, respectively, where .

Assume here and after that is a bounded sequence of strictly positive real numbers with and . Then, the linear space was defined by Maddox [1] (see also Simons [2] and Nakano [3]) as follows: which is complete paranormed space paranormed by For simplicity in notation, here and in what follows, the summation without limits runs from to .

Quite recently, Nergiz and Başar [4] have introduced the space of nonabsolute type which consists of all sequences whose -transforms are in the space , where is defined by for all , where and are the convergent sequences. We should record that the double sequential band matrices were used for determining its fine spectrum over some sequence spaces by Kumar and Srivastava in [5, 6], Panigrahi and Srivastava in [7], and Akhmedov and El-Shabrawy in [8]. The reader may refer to Nergiz and Başar [4, 9] for relevant terminology and additional references on the space , since the present paper is a natural continuation of them. Here and after, for short we write instead of . In the special case for all , the space is reduced to the space ; that is,

#### 2. The Rotundity of the Space

The rotundity of Banach spaces is one of the most important geometric property in functional analysis. For details, the reader may refer to [10–12]. In this section, we characterize the rotundity of the space and give some results related to this concept.

*Definition 1. *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 2. *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 3. *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 4. *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 .

Proposition 5. *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 on the space *.

*Proof. *Consider the modular on . (i)Let , then . So, we have
(ii)Let . Then, for all . So, we have
(iii)Let . Then, for all . So, we have
(iv)By (ii) and (iii), one can immediately see for that
By passing to limit as in (11), we have . Hence, is right continuous. If , by (i) we have
By letting in (12), we observe that . Hence, is also left continuous, and so, it is continuous.

Proposition 6. *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 *.

* Proof. *Let . (i)Let be such that . By the definition of , there exists an such that and . From Parts (i) and (ii) of Proposition 5, we obtain
Since is arbitrary, we have (i).(ii)If we choose such that , then . By the definition of and Part (i) of Proposition 5, we have
So, for all . This implies that . (iii)Since is continuous, by Theorem 1.4 of [12] we directly have (iii). (iv)This follows from Parts (i) and (iii). (v)This follows from Parts (ii) and (iii).

Now, we consider the space equipped with the Luxemburg norm given by

Theorem 7. * is a Banach space with Luxemburg norm. *

*Proof. *Let and for all . Then, . Therefore, for all .

For , for all . Hence, and .

Let and and be a nonempty subset of . Since , there exists such that . Obviously, . We assume that and . Then, . Since , we get
which contradicts the assumption. Hence, we obtain that if , then . This means that . Thus, we conclude that if and only if .

Now, let and . Then, we have
Therefore, we obtain
That is, and for all . So, .

If we take and instead of and , respectively, then we obtain that
Hence, we get . This also holds when .

To prove the triangle inequality, let and be given. Then, there exist and such that and . Since is convex,
Therefore, . Then, we have . Since was arbitrary, we obtain . Hence, is a norm on .

Now, we need to show that every Cauchy sequence in is convergent according to the Luxemburg norm. Let be a Cauchy sequence in and . Thus, there exists such that for all . By Part (i) of Proposition 6, we have
for all . This implies that
Then, for each fixed and for all ,
Hence, the sequence is a Cauchy sequence in . Since is complete, there is a such that as . Therefore, as by (22), we have
for all .

Now, we have to show that is an element of . Since as , we have
Then, we see by (21) that for all . This implies that as . So, we have . Therefore, the sequence space is complete with respect to Luxemburg norm. This completes the proof.

Theorem 8. *The space is rotund if and only if for all . *

*Proof. *Let be rotund and choose such that for . Consider the following sequences given by
Then, obviously and
By Part (iii) of Proposition 6, 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 6, we have
which gives that , and
Also, we obtain from (29) that
Since , we have
This implies that
for all . Since the function is strictly convex for all , it follows by (32) that for all . Hence, . That is, the sequence space is rotund.

Theorem 9. *Let . Then, the following statements hold: *(i)* and imply . *(ii)* and imply . *

*Proof. *Let . (i)Suppose that with . Then, . By Part (ii) of Proposition 6, implies . That is, . Since , by Part (i) of Proposition 5, we get . Thus, we have . (ii)Let and . Then, . By Part (i) of Proposition 6, implies . That is, . If , then . If , then by Part (ii) of Proposition 5, we have . This means that .

Theorem 10. *Let be a sequence in . Then, the following statements hold: *(i)* implies . *(ii)* implies . *

*Proof. *Let be a sequence in . (i)Let and . Then, there exists such that for all . By Parts (i) and (ii) of Theorem 9, implies and implies for all . This means and for all there exists such that . That is, . (ii)We assume that and . Then, there exists a subsequence of such that for all . By Part (i) of Theorem 9, and imply . Thus, for all . Hence, we obtain that implies .

Theorem 11. *Let and . If as and as for all , then as . *

*Proof. *Let be given. Since , there exists such that
It follows from the fact
that there exists such that for all and for all ,
and for all ,
Therefore, we obtain from (33), (35), and (36) that
This means that as . By Part (ii) of Theorem 10, as implies as . Hence, as .

Theorem 12. *The sequence space has the Kadec-Klee property. *

*Proof. *Let and such that and are given. By Part (ii) of Theorem 10, we have as . Also implies . By Part (iii) of Proposition 6, we obtain . Therefore, we have as .

Since and defined by is continuous, as for all . Therefore, as .

Since any weakly convergent sequence in is convergent, the sequence space has the Kadec-Klee property.

Theorem 13. *For any , the space has the uniform Opial property. *

*Proof. *Let and be given such that . Also let and . There exists such that
Hence, we have
Furthermore, we have
which yields that
For any weakly null sequence , since as for each , there exists such that for all ,
Therefore, for all ,
Moreover,
Then, we have
This means that has the uniform Opial property.

#### 3. Conclusion

The sequence spaces and of nonabsolute type consisting of all sequences such that is in the Maddox' spaces and were introduced by Başar et al. [13], where is a sequence such that for all and the rotundity of the space was examined.

The sequence space of nonabsolute type consisting of all sequences such that was studied by Aydın and Başar [14], and some results related to the rotundity of the space were given.

Quite recently, the sequence space of nonabsolute type consisting of all sequences such that was defined by Aydın and Başar [15], and emphasized the rotundity of the space together with some related results.

Although the sequence spaces and are not comparable, since the double sequential band matrix reduces to the generalized difference matrix in the special case and , the new space is more general than the space . Similarly, the sequence space is also reduced to the space in the case and . So, the results on the space are much more comprehensive than the results on the space . Additionally, the corresponding theorems on the Kadec-Klee property of the space and the uniform Opial property of the space were not given by Başar et al. [13] and Aydın and Başar [15] which make the present paper significant.

#### Acknowledgments

The main results of this paper were presented in part at the conference First International Conference on Analysis and Applied Mathematics (ICAAM 2012) held on October 18–21, 2012 in Gümüşhane, Turkey, at the University of Gümüşhane.