# Construction and Stability of Riesz Bases

Authors: Yulin Bai; Wanyi Wang; Guixia Wang; Suqin Ge
Journal: Journal of Function Spaces
 (2018)

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## Abstract

We construct some new Riesz bases and consider the stability of them. The investigation is based on the stability of Riesz bases of cosines and sines in the Hilbert spaceL2[0,π].


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## Body

## 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.In 1934, Paley and Wiener 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-1x}∪{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(1)-y′′+qy=λy,on0,π;y0=yπ=0,such that it forms a Riesz basis.


## 2. Riesz Bases Generated by Sines and Cosines

Let us first recall some basic concepts. Let{fn}, n∈N, be a sequence in a Hilbert space H, where N is the set of positive integers. The sequence is calledcomplete if its closed span equals H [5, P. 154]. We say that {fn} is aBessel sequence if ∑n=1∞f,fn2<∞ for every element f∈H and that the sequence {fn} is aRiesz-Fischer sequence if the moment problem f,fn=cn(n=1,2,3,…) admits at least one solution f∈H whenever {cn}∈l2 [5, P. 154].A basis{fn} of Hilbert space is called aRiesz basis if it is obtained from an orthonormal basis by means of a bounded linear invertible operator. Two sequences of elements {fn} and {gn} from Hilbert space H are calledquadratically close if ∑n=1∞fn-gn2<∞ [5, P. 45]. A sequence {λn} of real or complex numbers is said to beseparated if, for some positive number ϵ, λn-λm≥ϵ whenever n≠m [5, P. 98]. A sequence {fn} is called ω-linearly independent if the equality ∑n=1∞cnfn=0 is possible only for cn=0(n≥1) [5, P. 40].Next we need the following lemmas to get our main results.Lemma 1 ([5, P. 155]).
(i)
The sequence{fn} is a Bessel sequence with bound M if and only if the inequality(2)∑ncnfn2≤M∑ncn2holds for every finite systems{cn} of complex numbers.(i)
The sequence{fn} is a Riesz-Fischer sequence with bound m if and only if the inequality(3)m∑ncn2≤∑ncnfn2holds for every finite systems{cn} of complex numbers.Lemma 2 ([6, P. 95]).
Let two sequences{fn} and {gn} be quadratically close and let {fn} be an Riesz basis in H. (i)
If the sequence{gn} is ω-linearly independent, then {gn} is a Riesz basis in H.(ii)
If the sequence{gn} is complete in H, then {gn} is ω-linearly independent.Using the above lemmas, we obtain the following lemmas.Lemma 3.
If{cos⁡λnx}∪{sin⁡(λ~nx)} is a Riesz-Fischer sequence in L2[0,π] with real λn and λ~n, then the sequences {λn} and {λ~n} are separated, respectively.Proof.
Letm be a lower bound of {cos(λnx)}∪{sin(λ~nx)}. With cm=1, ck=-1 and cn=0, dn=0, it follows from (3) that (4)2m≤cos⁡λmx-cos⁡λkx. On the other hand, (5)cos⁡λmx-cos⁡λkx2=∫0πcos⁡λmx-cos⁡λkx2dx≤∫0πλm-λk2x2dx=π33λm-λk2. Thus {λn} is separated by definition.
Similarly, settingdm=1, dk=-1 and cn=0, dn=0 in (3), we also have that {λ~n} is separated.Lemma 4.
Let{λn}∪{λ~n} and {μn}∪{μ~n}, n∈N, be two sequences of nonnegative real numbers such that λm≠λk, λ~m≠λ~k, μm≠μk, and μ~m≠μ~k for all m≠k and(6)∑n=1∞λn-μn2+∑n=1∞λ~n-μ~n2<∞.Then {cos(λnx)}∪{sin(λ~nx)} is a Riesz basis in L2[0,π] if and only if {cos(μnx)}∪{sin(μ~nx)} is a Riesz basis in L2[0,π].Proof.
Letfn(x)=cos(λnx), f~n(x)=sin(λ~nx) and gn(x)=cos(μnx), g~n(x)=sin(μ~nx). Suppose that {fn}∪{f~n} is a Riesz basis in L2[0,π]. By Lemma 3, we find that the sequences {λn} and {λ~n} are separated, respectively. Using (6), we get that the sequences {μn} and {μ~n} are also separated, respectively. Therefore, we can assume (7)0≤μ1<μ2<μ3⋯,0≤μ~1<μ~2<μ~3⋯ and there is a positive ϵ such that μn≥nϵ and μ~n≥nϵ for all n∈N. Since (8)fn-gn≤πλn-μn,f~n-g~n≤πλ~n-μ~n,we obtain that(9)∑n=1∞fn-gn2+∑n=1∞f~n-g~n2≤π3∑n=1∞λn-μn2+∑n=1∞λ~n-μ~n2<∞;thus two sequences {fn}∪{f~n} and {gn}∪{g~n} are quadratically close. In particular, {fn-gn}∪{f~n-g~n} is a Bessel sequence.
We can define a bounded linear operator(10)T∑n=1∞cnfn+∑n=1∞dnf~n=∑n=1∞cnfn-gn+∑n=1∞dnf~n-g~non L2[0,π], as {fn}∪{f~n} is a Riesz basis. From (9), we have that T is a Hilbert-Schmidt operator. Furthermore, by Lemma 2, it is sufficient to prove that 1 is a regular point of T in order to prove that {gn}∪{g~n} is a Riesz basis.
Assume that1 is not a regular point of T. By the compactness of T, I-T is not one to one; i.e., there exists a sequence {cn}∪{dn}∈l2, not identically zero, such that (11)∑n=1∞cngn+∑n=1∞dng~n=0.Let λ∈C such that λ≠±μn,±μ~n for all n∈N. Then, the series(12)gx=∑n=1∞cnμn2-λ2gnx+∑n=1∞dnμ~n2-λ2g~nxis convergent uniformly on [0,π]. Similarly, (13)g′x=∑n=1∞cnμn2-λ2gn′x+∑n=1∞dnμ~n2-λ2g~n′x=-∑n=1∞cnμnμn2-λ2sin⁡μnx+∑n=1∞dnμ~nμ~n2-λ2cos⁡μ~nx also converges uniformly on [0,π]. Because of (14)gn′′x=-μn2gnx,g~n′′x=-μ~n2g~nx, we can deduce that(15)∑n=1mcnμn2-λ2gn′′x+∑n=1mdnμ~n2-λ2g~n′′x=-∑n=1mcnμn2μn2-λ2gnx-∑n=1mcnμ~n2μ~n2-λ2g~nx=-∑n=1mcngnx+∑n=1mdng~nx-λ2∑n=1mcnμn2-λ2gnx+∑n=1mdnμ~n2-λ2g~nx.When m→∞, the sequence on the right-hand side of (15) converges to -λ2g(x) in L2[0,π]. This shows that g(x) is twice differentiable and g′′(x)=-λ2g(x) for all x∈[0,π]. Due to (16)g0=∑n=1∞cnμn2-λ2,g′0=∑n=1∞dnμ~nμ~n2-λ2,we obtain that(17)gx=uλcos⁡λx+vλsin⁡λx,where u(λ)=g(0) and v(λ)=λ-1g′(0). The functions u(λ) and v(λ) are meromorphic and not identically zero, respectively. Thus it has at most countably many zeros. If u(λ)v(λ)≠0, by (12) and (17), we have that {cos(λx)}∪{sin(λx)} is in the closed linear span of {cos(μnx)}∪{sin(μ~nx)}. Owing to {cos(λx)}∪{sin(λx)} which is continuous about (x,λ), we get that {cos(λx)}∪{sin(λx)} is in the closed linear span of {cos(μnx)}∪{sin(μ~nx)} for all λ∈C. It follows that {sin(nx)}, n∈N, is in the closed linear span of {gn(x)}∪{g~n(x)}, so {gn(x)}∪{g~n(x)} is complete in L2[0,π]. Hence the R(I-T) is dense in L2[0,π]. Using the fact that T is compact, we have that R(I-T)=L2[0,π] and I-T is one to one; this contradicts the assumption.
Similarly, assume that{gn(x)}∪{g~n(x)} is a Riesz basis in L2[0,π], then {fn(x)}∪{f~n(x)} is also a Riesz basis in L2[0,π].Now we shall introduce our main results.Theorem 5.
(i) The sequence{1}∪{sin(2nx)}∪{cos(2nx)} is a orthonormal basis and Riesz basis in L2[0,π].
(ii) The sequence{sin((2n-1)x)}∪{cos((2n-1)x)} is a orthonormal basis and Riesz basis in L2[0,π].Proof.
(i) Suppose thatf(x)∈L2[0,π] satisfies(18)∫0πfx·1dx=0,∫0πfxcos⁡2nxdx=0,∫0πfxsin⁡2nxdx=0.Let F(x)=∫0xf(t)dt; integration by parts yields that (19)∫0πfxcos⁡2nxdx=Fxcos⁡2nx0π+2n∫0πFxsin⁡2nxdx=0. Thus (20)∫0πFxsin⁡2nxdx=0. Setting t=2x-π, we obtain(21)∫0πFxsin⁡2nxdx=-1n2∫-ππFt+π2sin⁡ntdt=0,(22)∫0πfxsin⁡2nxdx=-1n2∫-ππF′t+π2sin⁡ntdt=0.Combining (18), (21), and (22), we obtain f(x)≡0. Therefore, {1}∪{cos(2nx)}∪{sin(2nx)}, n∈N, is complete in L2[0,π]. The orthogonality of {1}∪{cos(2nx)}∪{sin(2nx)}, n∈N, will be proved by establishing that {cos(2nx)} and {sin(2mx)} are orthogonal for all m,n∈N, using the fact that {1}∪{cos(2nx)} and {1}∪{sin(2nx)}, n∈N, are the orthogonal sequences in L2[0,π], respectively.
It follows from(23)cos⁡2nx,sin⁡2mx=∫0πcos⁡2nxsin⁡2mxdx=0that cos(2nx) and sin(2mx) are orthogonal for all m,n∈N. Clearly, it is also a Riesz basis in L2[0,π]. This completes the proof of (i).
(ii) Supposef(x)∈L2[0,π], such that (24)∫0πfxsin⁡2n-1xdx=0,∫0πfxcos⁡2n-1xdx=0. Let F(x)=∫0xf(t)dt. By partial integration, (25)∫0πfxsin2n-1xdx=-2n-1∫0πFxcos2n-1xdx=0. Hence (26)∫0πFxcos⁡2n-1xdx=0.Setting t=2x-π, we obtain (27)∫0πFxcos⁡2n-1xdx=-1n2∫-ππFt+π2sin⁡n-12tdt=0,∫0πfxcos⁡2n-1xdx=-1n2∫-ππF′t+π2sin⁡n-12tdt=0.Similarly, using the method in (i), the desired results can be obtained. The proof is completed.Theorem 6.
Letδn∈l2, n∈N. (i)
Ifλn=2n±δn, λ~n=2n∓δn and λm≠λk, λ~m≠λ~k, where m≠k, then the sequences {1}∪{cos(λnx)}∪{sin(λnx)} and {1}∪{cos(λnx)}∪{sin(λ~nx)} are the Riesz basis in L2[0,π], respectively.(ii)
Ifλˇn=(2n-1)±δn, λ^n=(2n-1)∓δn and λˇm≠λˇk, λ^m≠λ^k, where m≠k, then the sequences {sin(λˇnx)}∪{cos(λˇnx)} and {sin(λˇnx)}∪{cos(λ^nx)} are the Riesz basis in L2[0,π], respectively.Proof.
(i) By the assumptions (i) of Theorem6, we have (28)∑n=1∞λn-2n2=∑n=1∞δn2<∞,∑n=1∞λ~n-2n2=∑n=1∞δn2<∞.Therefore (29)∑n=1∞λn-2n2+∑n=1∞λ~n-2n2<∞.Hence the result follows from Theorem 5 and Lemma 4.
The proof of the second part of this theorem follows in a similar manner.


## 3. Riesz Bases Associated with the Eigenfunctions of Strum-Liouville Problems

We consider the Strum-Liouville problem(30)-y′′+qxy=λ2y,x∈0,π,y0=yπ=0,where λ∈C and q(x)∈L2([0,π],R).It is well known that (see, for example, [19]) the eigenvalues of problem (30) are (31)λn=n+O1n and corresponding normalized eigenfunctions are (32)ynx=2πsin⁡nx+O1nTheorem 7.
Letun(x,q)=g1(x,λn)g2(x,λn), n∈N, where gi(x,λn), i=1,2, are the solutions of (30) satisfying the initial conditions (33)g10,λ,q=g2′0,λ,q=1;g1′0,λ,q=g20,λ,q=0.Then, for m,n∈N, we have (i)
ym2,(d/dx)yn2=0;(ii)
um,(d/dx)yn2=(π/2)δmn;(iii)
um,(d/dx)un=0.Proof.
(i) Using integration by parts we obtain(34)ym2,ddxyn2=12∫0πym2yn2′-ym2′yn2dx=∫0πymynym,yndx. This clearly vanishes for m=n. If m≠n, then λn≠λm, and we can use (35)ym,yn′=λm-λnymynto obtain (36)ym2,ddxyn2=12λm-λnym,yn20π=0.
(ii) Again, integration by parts yields(37)um,ddxyn2=12∫0πumyn2′-um′yn2dx=12∫0π2g1g2ynyn′-g1′g2yn2-g1g2′yn2dx=12∫0πg2yng1,yn+g1yng2,yndx, where gi=gi(x,λm)(i=1,2). If m≠n, we may use (38)gi,yn′=λm-λngiyn,i=1,2 to obtain (39)um,ddxyn2=12λm-λng1,yng2,yn0π=0.If m=n, then eigenfunction ym is a multiple of solution g2; hence [g2,ym]=0, and by the Wronskian identity, we get (40)um,ddxym2=12∫0πumym2′-um′ym2dx=12∫0πg2ymg1,ymdx=12∫0πym2g1,g2dx=12π.
(iii) Ifm=n, then the conclusion holds clearly. If m≠n, using the same procedure in (i) we have (41)um,ddxun=12∫0πumun′-um′undx=12λm-λng1x,λm,g1x,λng2x,λm,g2x,λn0π=0.Theorem 8.
For everyq(x)∈L2([0,π],R), the sequence {1}∪{1-πyn2}∪{(π/2n)(d/dx)yn2}, n∈N, is a Riesz basis in L2[0,π].Proof.
It is clear that the element1-πyn2 is not in the closed linear span of {1,1-πym2}, n≠m, as (42)1-πyn2,ddxun=1,ddxun-πyn2,ddxun=π2,but (43)1,ddxun=∫0πdun=g1g20π=0,1-πym2,ddxun=0,m≠n,by Theorem 7. Hence {1}∪{1-πyn2} is ω-linearly independent. Similarly, the sequence {(π/2n)(d/dx)yn2} is ω-linearly independent. It follows from Theorem 7 that, for all m,n∈N, (44)1,ddxyn2=0,1-πym2,ddxyn2=0.Therefore, the two sequences {1}∪{1-πyn2} and {(π/2n)(d/dx)yn2} are mutually perpendicular. Hence, the sequence {1}∪{1-πyn2}∪{(π/2n)(d/dx)yn2} is ω-linearly independent.
By the expression ofyn(x), we have (45)1=1,1-πyn2=cos⁡2nx+O1n,π2nddxyn2=sin⁡2nx+O1n.Thus the sequence {1}∪{1-πyn2}∪{(π/2n)(d/dx)yn2} is quadratically close with the Riesz basis {1}∪{cos(2nx)}∪{sin(2nx)}. The statement follows directly from Lemma 2.


## 4. Conclusion

Riesz bases have been extensively applied in signal denoising, feature extraction, robust signal processing, and also the corresponding inverse problems. This paper gives that{1}∪{cos⁡2nx}∪{sin⁡(2nx)} and {sin⁡((2n-1)x)}∪{cos⁡((2n-1)x)} form a Riesz basis in L2[0,π], respectively. Based on this result, we find that a new sequence associated with eigenfunctions of Sturm-Liouville problem forms a Riesz basis in L2[0,π].



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