Abstract

Most recently, the generalized difference operator of order three was defined and its domain in Hahn sequence space was calculated. In this paper, the spaces and are introduced as the domain of generalized difference operator of order three in the sequence spaces and . Then, some topological properties of and are given, and some inclusion relations are shown. Additionally, algebraic dual, , , and dual spaces of and are computed. In the last section, the classes and of matrix transformations are characterized, where and .

1. Preliminaries and Notations

The set of all complex valued sequences is denoted by and, according to the classification of each subset of , is called a sequence space. In the literature of sequences, the set which is called the set of all bounded sequences, the set which is called the set of all convergent sequences, and the set which is called the set of all null sequences are called classical sequence spaces. If a sequence space is a complete metric space with continuous coordinates, then it is called -space. A normed -spaces is called a -space. Therefore, the classical sequence spaces , , and are -spaces with respect to the norm defined by . Moreover, the spaces of all absolutely summable, convergent series, null series, and bounded series are denoted by , , , and , respectively, where .

The space of absolutely summable sequences which is denoted by and the space of all sequences of bounded variation which is denoted by are defined, respectively, as follows:and it is a -space with its norm :and it is a -space with respect to the norm . On the one hand, the sequence space is defined as the backward difference operator domain on the sequence space , where , for all . On the other hand, we can also represent aswhich is the forward difference operator domain on the sequence space , where , for all . The space and the inclusions are strictly held.

The alpha-dual , beta-dual , and gamma-dual of a sequence space are defined by

The duals of the spaces and are defined by

Let be an infinite matrix and . We writeand then, we say that defines a matrix transformation from into as if , for every . We denote the set of all infinite matrices that map the sequence space into the sequence space by . Thus, if and only if the right side of (6) converges for every , that is, , for all , and we have , for all .

If a normed sequence space contains a sequence with the following property that, for every , there is a unique sequence of scalars such thatthen is called a Schauder basis for . The series which has the sum is then called the expansion of with respect to and written as .

If is an -space, , and is a basis for , then is said to have property, where is a sequence whose only term in place is 1; the others are zero, for each and . If is dense in , then is called -space; thus, implies .

Let be a sequence space and be an infinite matrix. Then, the matrix domain of an infinite matrix in the sequence space is defined by

Wilansky (Theorem 4.4.2, p. 66 of [1]) defined that if is a sequence space, then the continuous dual of the space is given by

It is well known that (see [2, 3]).

2. The New Difference Sequence Spaces and

Now, we define the new difference sequence spaces and as the domain of generalized difference matrix of order three in the sequence space and . Then, we show that and are spaces and they are linearly isomorphic to the sequence spaces and , respectively, and we show that is a linear and bounded operator over the sequence spaces and to prove the inclusion relations among and and and , respectively.

The difference matrix of order one was defined by Kizmaz [4] as , and he studied its domain on classical sequence spaces. The generalized difference operator of order two was defined by Dutta and Baliarsing [5] as , and they studied its spectrum on the sequence space . Moreover, Baliarsing and Dutta [5] defined the generalized difference operator of order two as and studied its spectral subdivisions over the sequence spaces and . Then, again, Dutta and Baliarsing [6] defined generalized difference operator of order three as and studied its spectrum over the sequence spaces and .

The generalized difference matrix of order three was defined by Malkowsky et al. [7] as

The matrix transforms a sequence by

Then, most recently, the generalized difference matrix domain in Hahn sequence space was calculated and studied by Tug et al. [8].

Now, we define the following difference sequence spaces as the set of all sequences whose -transforms are in the sequence spaces and as follows:

We define the sequence by the transform of the sequence asfor all . The generalized difference matrix of order three is a triangle; then, it is invertable and the inverse is unique. Therefore, we obtain by considering the relation between the terms of and (13), and are zero terms thatwhere is defined by

Theorem 1. The sequence spaces and are -spaces with respect to the norm:respectively.

Proof. Since and are -spaces and is a triangle matrix, we obtain from the Theorem 4.3.2 of Wilansky (p. 61 of [1]) that and are also -spaces.

Relation (14) between the terms of the sequences and is given by the following calculation:

Thus, the following equation which was derived from (17) is given byand we calculate roots of equation (18) as one real and two complex as follows:

Then, we have the following simple calculations by random three roots , and of equation (18):

Theorem 2. The sequence space is linearly isomorphic to the sequence spaces , i.e., .

Proof. Suppose that the transformation defined from the space onto as by . The linearity of is clear. Moreover, since gives , for all .
Let us take and consider the sequence with respect to relation (14) asfor every . By considering equations (23)–(26), we have the following transform of the sequence asfor all . Thus, , for all ; then, is surjective. Moreover, for every , we haveHence, is an element of , and clearly, is surjective and preserves the norm. Therefore, . It completes the proof.

Theorem 3. The sequence space is linearly isomorphic to the sequence space , i.e., .

Proof. Suppose that the transformation is defined from the space onto as by . The linearity of is clear. Moreover, since gives , for all , the rest of the proof can be followed by equation (28) in the proof of Theorem 2, and thus, , for all ; then, is surjective. Moreover, for every , we haveHence, is an element of , and clearly, is surjective and preserves the norm. Therefore, . It completes the proof.

Theorem 4. The inclusion strictly holds.

Proof. Suppose that , then . Since , then and it says and it shows the inclusion ) holds. To show that the inclusion is strict, let us define . Then, clearly, , but , since as . This completes the proof.

Lemma 1. The matrix is a bounded linear operator, , from to itself, if and only if the supremum of norms of the columns of is bounded, i.e.,

Theorem 5. is a bounded linear operator.

Proof. The linearity is clear. We should show that which means that satisfies the conditions of Lemma 1 with instead of , that is,This completes the proof.

Lemma 2. The matrix is a bounded linear operator, , from to itself, if and only if

Theorem 6. is a bounded linear operator.

Proof. The linearity is clear. We should show that which means that satisfies the conditions of Lemma 2 with instead of , that is,This completes the proof.

Theorem 7. .

Proof. Since the operator is a bounded and linear operator on the sequence space by Theorem 5, if and only if condition (31) is satisfied. Moreover, , for every . This shows that the inclusion holds.
Moreover, the matrix , the inverse matrix of , which can be reduced from the inverse matrix in the Theorem 2 of [2] by only choosing . Then, we write the following to calculate . Therefore, the operator has the following equation and the following calculations of its roots:where is defined byThe notation gives us the following equation asThen, one real and two complex roots of equation (37) areThen, we have the following simple calculations by random three roots , and of equation (37):for all . We can show that , and if , then and . Thus, we can say that . This completes the proof.

Theorem 8. .

Proof. Since the proof can be done similarly as in the proof of Theorem 7, we omit it.

Theorem 9. Let , for all and . Define the sequence in the sequence space as follows:for every fixed . Then, is a basis for , and there is a unique representation of as

Proof. Since the proof can be done similarly for both sequence space, we consider to prove for the sequence space . First, we need to show that , and it is enough to show , for all . To show thisfor every . Then, clearly, one can see that and then .
Let us take a sequence . Then, we obtain the following representation for every nonnegative integer asThen, the following holds by taking the transform of (43) thatand from (44), we haveTherefore, for every given , there exists an integer such thatfor all . Hence,for all . This proves that .

3. Dual Spaces of the Sequence Spaces and

We begin this section by calculating the algebraic dual space of and , respectively.

Theorem 10. The algebraic dual of the space is the sequence space .

Proof. Let us define with which is a surjective linear map, and is injective since is a basis for . Let , and since , we can writeThen, we haveThus, .
Moreover, since , for every , then . Therefore, .

Theorem 11. The algebraic dual of the space is the sequence space .

Proof. Since the proof can be followed similarly as in Theorem 10, we omit the repetition.

Now, we start with the following lemma which is needed in proving our theorems. Here and after, we denote the collection of all finite subsets of by .

Lemma 3 (see [9]). Let be an infinite matrix over the complex field. Then, the following statements hold:(i) if and only if(i) if and only if(iii) if and only if (50) holds, and(iv) if and only if

Lemma 4 (see [9]). Let be an infinite matrix over the complex field. Then, the following statements hold:(i) if and only if(ii) if and only if(iii) if and only if (55) holds, and(iv) if and only if(v) if and only if(vi) if and only if (57) and (59) hold and

Theorem 12. Let the set be defined asThen, .

Proof. Let the sequence be in . By relation (27), we have the following equality:where the matrix is defined via the sequence as