Construction and Stability of Riesz Bases

As is well known Riesz basis is not only a base but also a special frame. The research of frame and Riesz basis plays important role in theoretical research of wavelet analysis [1]; because of the redundancy of frame and Riesz basis, they have been extensively applied in signal denoising, feature extraction, robust signal processing, and so on. Therefore, construction of Riesz basis has attractedmuch attention of the researchers due to their wide applications. In 1934, Paley andWiener studied the problem of finding sequences {λn} for which {exp(iλnx)} is a Riesz basis in L2[−π, π] [2]. Since then many results on the Riesz basis have been obtained [3–5]. Also the Riesz basis of the systems of sines and cosines in L2[0, π] and Riesz basis associated with Sturm-Liouville problems have been studied in many papers [6–12]; moreover, on the problems of expansion of eigenfunctions, we refer to [13–18] and references cited therein. Motivated by these works, on the one hand, we construct two groups of Riesz bases {1} ∪ {cos(2nx)} ∪ {sin(2nx)} and {sin((2n − 1)x)} ∪ {cos((2n − 1)x)} and study the stability of them. On the other, we consider the problem of finding a new sequence associated with eigenfunctions of Sturm-Liouville problem −y󸀠󸀠 + qy = λy, on [0, π] ; y (0) = y (π) = 0, (1)


Introduction
As is well known Riesz basis is not only a base but also a special frame. The research of frame and Riesz basis plays important role in theoretical research of wavelet analysis [1]; because of the redundancy of frame and Riesz basis, they have been extensively applied in signal denoising, feature extraction, robust signal processing, and so on. Therefore, construction of Riesz basis has attracted much attention of the researchers due to their wide applications.

Riesz Bases Generated by Sines and Cosines
Let us first recall some basic concepts.
Using the above lemmas, we obtain the following lemmas.
Proof. Let ( ) = cos( ),̃( ) = sin(̃) and ( ) = cos( ),̃( ) = sin(̃). Suppose that { } ∪ {̃} is a Riesz basis in 2 [0, ]. By Lemma 3, we find that the sequences { } and {̃} are separated, respectively. Using (6), we get that the sequences { } and {̃} are also separated, respectively. Therefore, we can assume and there is a positive such that ≥ and̃≥ for all ∈ N. Since we obtain that  (9), we have that is a Hilbert-Schmidt operator. Furthermore, by Lemma 2, it is sufficient to prove that 1 is a regular point of in order to prove that { } ∪ {̃} is a Riesz basis. Assume that 1 is not a regular point of . By the compactness of , − is not one to one; i.e., there exists a sequence { } ∪ { } ∈ ℓ 2 , not identically zero, such that Let ∈ C such that ̸ = ± , ±̃for all ∈ N. Then, the series is convergent uniformly on [0, ]. Similarly, also converges uniformly on [0, ]. Because of ( ) = − 2 ( ) , Journal of Function Spaces 3 we can deduce that When → ∞, the sequence on the right-hand side of (15) converges to − 2 ( ) in 2 [0, ]. This shows that ( ) is twice differentiable and ( ) = − 2 ( ) for all ∈ [0, ]. Due to we obtain that where ( ) = (0) and V( ) = −1 (0). The functions ( ) and V( ) are meromorphic and not identically zero, respectively. Thus it has at most countably many zeros. If ( )V( ) ̸ = 0, by (12) and (17) Thus Combining (18) Similarly, using the method in (i), the desired results can be obtained. The proof is completed.  Proof. (i) By the assumptions (i) of Theorem 6, we have Hence the result follows from Theorem 5 and Lemma 4. The proof of the second part of this theorem follows in a similar manner.
It is well known that (see, for example, [19]) the eigenvalues of problem (30) are  Proof. (i) Using integration by parts we obtain This clearly vanishes for = . If ̸ = , then ̸ = , and we can use to obtain (ii) Again, integration by parts yields where = ( , ) ( = 1, 2). If ̸ = , we may use to obtain If = , then eigenfunction is a multiple of solution 2 ; hence [ 2 , ] = 0, and by the Wronskian identity, we get

Data Availability
No data were used to support this study.

Conflicts of Interest
The authors declare that there are no conflicts of interest.