In this section, we define the Fibonacci band matrix and introduce the sequence spaces and , where . Also, we present some inclusion theorems and construct the Schauder basis of the space for.
Let be the n th Fibonacci number for every . Then we define the infinite matrix by
Now, we introduce the Fibonacci difference sequence spaces and as the set of all sequences such that their -transforms are in the space and , respectively, i.e.,
With the notation of (1.2), the sequence spaces and may be redefined by
Define the sequence , which will be frequently used, by the -transform of a sequence , i.e.,
Now, we may begin with the following theorem which is essential in the text.
Theorem 3.1 Let . Then is a BK-space with the norm , that is,
Proof Since (3.1) holds, and are BK-spaces with respect to their natural norms and the matrix is a triangle; Theorem 4.3.12 of Wilansky [, p.63] gives the fact that the spaces and are BK-spaces with the given norms, where . This completes the proof. □
Remark 3.2 One can easily check that the absolute property does not hold on the spaces and , that is, and for at least one sequence in the spaces and , and this shows that and are the sequence spaces of non-absolute type, where and .
Theorem 3.3 The Fibonacci difference sequence space of non-absolute type is linearly isomorphic to the space , that is, for .
Proof To prove this, we should show the existence of a linear bijection between the spaces and for . Consider the transformation T defined, with the notation of (3.2), from to by . Then for every . Also, the linearity of T is clear. Further, it is trivial that whenever and hence T is injective.
Furthermore, let for and define the sequence by
Then, in the cases and , we get
respectively. Thus, we have (). Hence, T is surjective and norm preserving. Consequently, T is a linear bijection which shows that the spaces and are linearly isomorphic for . This concludes the proof. □
Now, we give some inclusion relations concerning the space .
Theorem 3.4 The inclusion strictly holds for .
Proof To prove the validity of the inclusion for , it suffices to show the existence of a number such that for every .
Let and . Since the inequalities and hold for every , we obtain with the notation of (3.2),
which together yield, as expected,
for . Further, since the sequence is in , the inclusion is strict for . Similarly, one can easily prove that inequality (3.4) also holds in the case , and so we omit the details. This completes the proof. □
Theorem 3.5 Neither of the spaces and includes the other one, where .
Proof Let and . Then, since and , we conclude that x is in but not in . Now, consider the equation
Then whenever k is odd, which implies that the series is not convergent, where . Thus, is not in for . Additionally, since , the sequence e is in . Hence, the sequence spaces and overlap but neither contains the other, as asserted. □
Theorem 3.6 If , then .
Proof Let and . Then we obtain from Theorem 3.1 that , where y is the sequence given by (3.2). Thus, the well-known inclusion yields . This means that and hence, the inclusion holds. This completes the proof. □
Now, we give a sequence of the points of the space which forms a basis for the space ().
and define the sequence
for every fixed
Then the sequence is a basis for the space , and every has a unique representation of the form
Proof Let . Then it is obvious by (3.5) that () and hence for all .
Further, let be given. For every non-negative integer m, we put
Then we have that
Now, for any given , there is a non-negative integer such that
Therefore, we have for every that
which shows that and hence x is represented as in (3.6).
Finally, let us show the uniqueness of the representation (3.6) of . For this, suppose that . Since the linear transformation T defined from to in the proof of Theorem 3.3 is continuous, we have
Hence, the representation (3.6) of is unique. This concludes the proof. □