# Characterization of Eigenvalues in Spectral Gap for Singular Differential Operators

Authors: Zhaowen Zheng; Wenju Zhang
Journal: Abstract and Applied Analysis
 (2012)

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## Abstract

The spectral properties forn order differential operators are considered. When given a spectral gap (a,b) of the minimal operator T0 with deficiency index r, arbitrary m points βi(i=1,2,…,m) in (a,b), and a positive integer function p such that ∑i=1mp(βi)≤r, T0 has a self-adjoint extension T̃ such that each βi(i=1,2,…,m) is an eigenvalue of T̃ with multiplicity at least p(βi).


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## Body

## 1. Introduction

In this paper, we consider the followingnth-order formal symmetric differential expression:
(1.1)τy=w-1{∑j=0[n/2]‍(-1)j(pjy(j))(j)+∑j=0[n-1/2]‍(-1)j[(qjy(j))(j+1)-(qj*y(j+1))(j)]},
on (α,β), where -∞≤α<β≤+∞, y=y(t) is a complex-valued m-vector function, pj(t),qj(t) and w(t) are measurable and locally integrable m×m matrices, pj(t),w(t) are Hermitian, and w(t)>0.a.e.t∈(α,β).The spectral properties ofnth-order differential expression, particularly the distribution of eigenvalues, have been widely researched in these years (see [1–8] and references cited therein).Let us recall some known results due to  Neumann [1], Stone [2], Friedrichs [3] and Krein [4]. For convenience of the reader in addition to these original sources we will also give text book references [7].An open interval(a,b) with -∞<a<b<∞ is called a spectral gap of a symmetric operator A if
(1.2)∥(A-a+b2)f∥w≥b-a2∥f∥w∀f∈D(A).∥·∥w and 〈·,·〉w are defined in Section 2. It is easy to find that (a+b)/2 is a real regular point of A. If 〈Af,f〉w≥b∥f∥w2, we shall also say that (-∞,b) is a spectral gap of A (the latter definition is not generally used but convenient for our purpose; and it is justified since each interval (a,b) with a<b is a spectral gap of A). Let (a,b) be a spectral gap of A, then there exists a self-adjoint extension A~ of A (for instance the famous Friedrichs extension) such that (a,b)⊂Γ(A~) (here Γ(A~) is the regular-form domain of A~ which will be defined in Definition 2.4), this is the reason why we call (a,b) a spectral gap of A.Suppose in addition that the deficiency index of a symmetric operatorA is equal to n(n∈ℕ). Let A~ be a self-adjoint extension of A. The sum of the multiplicities of the eigenvalues of A~ within the interval (a,b) is at most n, and no point of the continuous spectrum of A~ lies in the interval (a,b) (cf. [5, Theorem 8.19 and Corollary 2 in Section 8.3]). Conversely, given any finite subset ℰ of (a,b) and positive integers p(λ), λ∈ℰ, such that ∑λ∈ℰ‍p(λ)≤n there exists a self-adjoint extension A~ of A such that σ(A~)∩(a,b)=ℰ and the multiplicities of λ as an eigenvalue of A~ are equal p(λ) for each λ∈ℰ (cf. [4]). If one only requires that (a,b) is some interval within the set of real regular points of A, then the corresponding statement is false. For instance, one can give a symmetric operator A^, such that each λ∈ℝ is a regular point of A^, the deficiency index of A^ is equal one and each self-adjoint extension A~ of A^ has a periodic point spectrum with period p independent of A~ (cf. [4]).This paper consists of three sections including the introduction. In Section2, we present some preliminary materials that include definitions and theorems needed for the rest of the paper. In Section 3, we give three main results in the study of self-adjoint extensions of a minimal operator generated by differential expression τ. Firstly, we present a partial self-adjointness of the minimal operator T0. Secondly, if (a,b) is a spectral gap of T0, for all β1,β2,…βm∈(a,b)(m≤r,r is the deficiency index of T0) and positive integer function p satisfying ∑i=1m‍p(βi)=r, T0 has a self-adjoint extension T~ with the following properties:(i)
σp(T~)∩(a,b)={β1,β2,…βm};(ii)
eachβi(i=1,2,…m)isaneigenvalueofT~withmultiplicityequaltop(βi);(iii)
T~haspurepointspectrumwithin(a,b). Finally, given a symmetric operator T0 with the deficiency index being equal to r, (a,b)⊂ℝ, β1,β2,…βm∈(a,b)(m≤r) being real regular points of T0,  (the interval (a,b) need not be a spectral gap of T0) and positive integer function p satisfying ∑i=1m‍p(βi)≤r, T0 has a self-adjoint extension T~ with the properties that each βi(i=1,2,…m) is an eigenvalue of T~ with multiplicity at least βi.


## 2. Preliminaries

In this section, we introduce notations, definitions, and some theorems that are needed in this paper.First, we define the following space:(2.1)L2(α,β;w):={f:∫αβ‍f*(t)w(t)f(t)<+∞},
with the inner product
(2.2)〈f,g〉w=∫αβ‍g*(t)w(t)f(t)dt,
where the weight function w(t) is the same as that in (1.1). Denote ∥f∥w=(〈f,f〉w)1/2 for f∈L2(α,β;w). If f∈L2(α,β;w), then f is called square integrable. Here, we note that if w is singular, L2(α,β;w) is a quotient space in the sense that y=z if and only if ∥y-z∥w=0. In this case, L2(α,β;w) is a Hilbert space.Now we introduce the maximal operatorT and minimal operator T0 generated by the expression τ.Definition 2.1.
The maximal operatorT generated by τ is defined by
(2.3)D(T)={f∈L2(α,β;w):f{0},f{1},…f{n-1}∈ACloc(α,β),τf∈L2(α,β;w)},Tf=τfforf∈D(T),
where ACloc(α,β) denotes the collection of functions on (α,β) which are absolutely continuous locally. Roughly speaking the jth quasi-derivative u{j} will be collection of terms that, if differentiated (n-j) times, is “part" of the differential expression rτu (see [7] for details). We know that T is densely defined and closed.Definition 2.2.
The preminimal operatorT00 generated by τ is defined by
(2.4)D(T00)={f∈D(T):fhascompactsupportin(α,β)},T00f=τf=Tfforf∈D(T00).
Obviously, T00⊂T and T00 is Hermitian. It is easy to know that D(T00) is dense, so T00 is symmetric, and T00 is not closed [8]. The closure of T00 is called the (closed) minimal operator denoted by T0.Definition 2.3.
Given a linear operatorA with domain and range in a Hilbert space ℍ, the resolvent set of A is
(2.5)ρ(A)={λ∈ℂ∣(λ-A)-1∈B(ℍ)},
the spectrum of A is the set σ(A)=ℂ∖ρ(A). Denote σp={λ:(λ-A)-1doesnotexist},σc={λ:(λ-A)-1existsandR(λ-A)¯=ℍ,but(λ-A)-1isnotcontinuous},σr={λ:(λ-A)-1existsandR(λ-A)¯≠ℍ}.Obviously,σ(A)=σp∪σc∪σr (see [8]).Definition 2.4.
LetA be a linear operator in a Hilbert space ℍ. The set of regular points of A, is called the regular-form domain of A, denoted by γ(A):
(2.6)Γ(A)={z∈ℂ:A-ziscontinuouslyinvertible}={z∈ℂ:thereexistsav=v(z)>0suchthat∥(A-z)f∥≥v∥f∥,∀f∈D(A)}
it is an open subset of ℂ containing ℂ∖ℝ.Definition 2.5.
For a closed operatorA in a Hilbert space ℍ, the essential spectrum of A is defined as
(2.7)σe(A)={λ∈σ(A):R(λ-A)≠R(λ-A)¯}.We assume thatT0 has real regular points, that is, Γ(T0)∩ℝ≠∅. In this case the deficiency index
(2.8)n(T0,z)=n(T0):=dimker(T-z)=dimR(T0-z¯)⊥
does not depend on the special choice of the regular point of T0 and consequently T0 has self-adjoint extensions.The kernelN(T-z) is the vector space of those solutions of the differential equation (τ-z)u=0 which are elements of L2(α,β;w). Since the space of all solutions of (τ-z)u=0 (which in general is not contained in L2(α,β;w)) has dimension p:=n×m, this implies
(2.9)0≤n(T0,z)≤p.Definition 2.6.
We say that an operatorA has pure point spectrum within (a,b) if σ(A)∩(a,b)=σp(A)¯∩(a,b).Proposition 2.7.
LetA be a self-adjoint operator in a Hilbert space ℍ. One has ρ(A)=Γ(A).Proof.
From Definitions2.3–2.5, we have ρ(A)⊆Γ(A) and σp, σc do not belong to Γ(A), so ρ(A)∪(σr∩Γ(A))=Γ(A), A is densely defined since it is a self-adjoint operator, so σr=∅, then we get ρ(A)=Γ(A).The following results give the relation between deficiency index and self-adjoint extension.Proposition 2.8.
IfT00, T0, T are the preminimal operator, minimal operator and maximal operator generated by τ, and T~ is one of the self-adjoint extension of T0, then One has:
(2.10)T00⊂T00¯=T0⊂T~=(T~)*⊂T=T0*.Proposition 2.9.
IfA is Hermitian, then ℂ∖ℝ⊂Γ(A) (cf. [5, Proposition 2, page 229]).Proposition 2.10.
The deficiency index is constant on each connected subset ofΓ(A). If A is Hermitian, then the deficiency index is constant in the upper and lower half-planes (cf. [5, Theorem 8.1]).Proposition 2.11.
A closed symmetric operator is self-adjoint if and only if its deficiency index is equal to0.Proposition 2.12.
LetA be a symmetric operator, (a)
A has self-adjoint extension if and only if its deficiency indices are equal,(b)
Γ(A)∩ℝ≠ϕ, then A has self-adjoint extensions,(c)
ifA is semibounded, then A has self-adjoint extensions,We can find the proof from Theorem 8.8 in [7].Proposition 2.13 (see [8]).
LetA be a closed symmetric operator in L2(α,β;w) and A1=A*. Set N0(λ-A1)=N(λ-A1)∩D(A). If there exists a λ∈ℝ such that λ∉σe(A), then A has a self-adjoint extension and
(2.11)dimN(λ-A1)=n(A)+dimN0(λ-A1).
Furthermore, there exists a self-adjoint extension A^ of A such that
(2.12)[N(λ-A1)⊖N0(λ-A1)]∩D(A^)={0}.


## 3. Main Results

LetT0, T be the minimal operator and maximal operator generated by τ in (1.1) and let the deficiency index of T0 be equal to r(0<r≤p). In this section we assume that T0 has real regular points. That is, Γ(T0)∩ℝ≠∅.Definition 3.1.
A closed subspaceℳ of L2(α,β;w) is called a reducing subspace of T0 if Pf∈D(T0), and T0Pf=PT0f for all f∈D(T0), where P denotes the orthogonal projection in D(T0) onto ℳ.Obviously along withℳ the orthogonal complement ℳ⊥ of ℳ is also a reducing subspace of T0. It is easy to see that the closed span of an orthogonal system of eigenvectors of T0 is a reducing subspace of T0. If ℳ is a reducing subspace of T0, then the part of T0 in ℳ, that is, the restriction of T0 to D(T0)∩ℳ may be (and in the following will be) regarded as an operator in the Hilbert space ℳ.SinceΓ(T0)∩ℝ≠∅, there exists a real point λ of T0, and the positive and negative deficiency indices of T0 are equal. We choose a one-dimensional subspace ℒλ of ker(T0*-λ) (= ker(T-λ)), set
(3.1)D(T0′):=D(T0)+˙ℒλ,T0′(f+g):=T0f+λg∀f∈D(T0)and∀g∈ℒλ.
Obviously T0′ is a symmetric extension of T0, and n(T0′,z)=n(T0,z)-1 for each regular point z of T0′ and ℒλ the corresponding eigenspace. Since the graph of T0′ is a one-dimensional extension of the closed graph of T0, therefore the operator T0′ is closed [6].Theorem 3.2.
Assume that the deficiency index ofT0 is equal to r, let (a,b) be a spectral gap of T0, and let T0′ be defined as above.(1)
IfT0′ has an eigenvalue μ∈(a,b), then (a,b) is a spectral gap of the restriction of T0′ to the Hilbert space ker(T0′-μ)⊥.(2)
The deficiency index of the restriction ofT0′ to the Hilbert space ker(T0′-μ)⊥ is equal to r-1.Proof.
(1) First we consider the case that-∞<a<b<+∞. Without loss of generality, we may assume that (a,b)=(-v,v). We have to show that
(3.2)∥T0′f∥≥v∥f∥∀f∈D(T0′)∩ker(T0′-μ)⊥.
Assume that ∥T0′f∥<v∥f∥ for some f∈D(T0′)∩ker(T0′-μ)⊥. Let g∈ker(T0′-μ)  and g≠0. Since ker(T0′-μ) reduces T0′,
(3.3)∥T0′(c1f+c2g)∥2=|c1|2∥T0′f∥2+|c2|2∥T0′g∥2=|c1|2∥T0′f∥2+|c2|2μ∥g∥2<|c1|2v2∥f∥2+|c2|2v2∥g∥2=v2∥c1f+c2g∥2,
for all (c1,c2)∈ℂ2∖(0,0). Since f and g span a two-dimensional subspace of D(T0') and dimD(T0′)∖D(T0)=1, it follows that ∥T0′h∥<v∥h∥ for some h∈D(T0), this is a contradiction to the hypothesis that (-v,v) is a spectral gap of T0. This completes the proof in the case that -∞<a<b<+∞.The proof in the semibounded case is similar. One only has to replace expression of the form∥T0f∥ by those of the form 〈T0f,f〉, here we omit the details.(2) Sincen(T0′,z)=n(T0,z)-1, that is, n(T0′,z)=r-1, and the restriction of T0′ to the Hilbert space ker(T0′-μ) is self-adjoint, so the deficiency index of the restriction of T0′ onto Hilbert space ker(T0′-μ)⊥ is equal to r-1.Theorem 3.3.
Suppose the deficiency index ofT0 is equal to r and λ is a real regular point of T0. Let f1,f2,…fm∈L2(α,β;w) be linearly independent solutions of (τ-λ)f=0. One has the following.(1)
T0 has a j  dimensional symmetric extension T0j(j=1,2,…m) with the following properties:
(3.4)D(T0j):=D(T0)+˙span{f1}+˙⋯+˙span{fm},T0j(f+g1+g2+⋯+gm)=T0f+λ(g1+g2+⋯+gm),∀f∈D(T0)and∀gj∈span{fj}(j=1,2…m).(2)
m≤r.(3)
Ifm=r, then T0 has a self-adjoint extension.Proof.
(1) Since span{f1} is a one-dimensional subspace of ker(T-λ), we define
(3.5)D(T01):=D(T0)+˙span{f1},T01(f+g):=Tf+λg∀f∈D(T0)and∀g∈span{f1}.
Obviously, T01 is a symmetric extension of T0, n(T01,z)=n(T0,z)-1=r-1, and λ is an eigenvalue of T01, and span {f1} is the corresponding eigenspace, and T01 is also closed. By induction, we get
(3.6)n(T0j,z)=n(T0,z)-j=r-j,
and T0j(j=1,2…m) is a closed symmetric operator.
(2) Whenm>r, then n(T0m,z)=n(T0,z)-m=r-m<0, that is a contradiction, so we have m≤r.
(3) Whenm=r, then n(T0m,z)=n(T0,z)-m=r-r=0, so from Proposition 2.11, we have that T0m is a self-adjoint extension of T0.Theorem 3.4.
Assume that the deficiency index ofT0 is equal to r. If (a,b) is a spectral gap of T0, then for all β1,β2,…βm∈(a,b) (where m≤r) and positive integer function p satisfying ∑i=1m‍p(βi)=r, the minimal operator T0 has a self-adjoint extension T~ with the following properties: (i)
σp(T~)∩(a,b)={β1,β2,…βm},(ii)
eachβj(j=1,2,…m) is aneigenvalue of T~ with multiplicity p(βj),(iii)
T~ has pure point spectrum within  (a,b).Proof.
We chooseλ1,λ2,…λr in (a,b), such that #{l≤r:λl=βj}=p(βj)  (j=1,…,m) (#M denote the cardinality of the set M), in other words each βj∈(a,b) occurs exactly p(βj) times in the sequence {λ1,λ2,…λr}.
We choose a one-dimensional spaceℒλ1 of ker(T-λ1) and define
(3.7)D(T1′):=D(T0)+˙ℒλ1,T1′(f+g):=T0f+λg∀f∈D(T0)and∀g∈ℒλ1.
Then T1′ is a closed symmetric extension of T0 with deficiency index being equal to r-1 and λ1 is an eigenvalue of T1′ with eigensubspace being ℒλ1.
Setℋ1=ℒλ1⊥ and denote by M1 and T1 the restriction of T1′ to ℒλ1 and ℋ1, respectively. Obviously M1 is a self-adjoint operator in the one-dimensional Hilbert space ℒλ1 with σ(M1)={λ1}. T1 is a closed symmetric operator in Hilbert space ℋ1, with deficiency index being equal to r-1, and T1′=M1+˙T1. Moreover, by Lemma 3.1, (a,b) is a spectral gap of T1. Thus we can replace L2(α,β;W) by ℋ1, T0 by T1, and λ1 by λ2 in the above conclusions.
Proceeding further in this way by induction, we obtain sequences{ℒλ1,…ℒλr} and {ℋ1,…ℋr} of Hilbert spaces and sequence {M1,…Mr} and {T1′,…Tr′} of the operators with the following properties:(i)
Tl′ are closed symmetric extension of T0 which satisfy: T1′⊂T2′⊂⋯⊂Tr′, Tl′ with deficiency indices r-l  (l=1,2,…r),(ii)
Ml is a self-adjoint operator on a one-dimensional subspace ℒλl of ℋl-1(ℋ0:=L2(α,β;w)),ℋl=ℋl-1⊖ℒλl, with σ(Ml)={λl} for each λl(l=1,2,…r),(iii)
Tl is a closed symmetric operator in the Hilbert space ℋl with spectral gap (a,b) for all l=1,2…r.(iv)
Tl′={⊕j=1lMj}⊕Tl(l=1,…r).Then we have that Tr′ is a closed symmetric extension of T0 with deficiency index 0. So T~=Tr′ is a self-adjoint extension of T0 with the required properties.Theorem 3.5.
Assume the deficiency index ofT0 is equal to r. Let β1,β2,…βm∈(a,b) (where m≤r, (a,b) need not be a spectral gap of T0) be real regular points of T0 and  let positive integer function p satisfy ∑j=1m‍p(βj)≤r. Moreover, one assumes that (τ-λi)f=0 has at least p(βi) square integrable solutions. Then T0 has a self-adjoint extension T~ such that each βj(j=1,2,…m) is an eigenvalue of T~ with multiplicity at least equal to p(βj).Proof.
Choose a sequenceλ1,…λr in (a,b) such that each βi occurs at least p(βi) times in this sequence (the times ki of the λi appears in the sequence satisfies p(βi)≤k≤k~i, k~i is the number of square integrable solutions of (τ-λi)f=0; the λi which is not in {β1,…,βm} will be selected as a regular point of T satisfying ker(T-λi)≠∅). Since the deficiency index of T0 is equal to r, the negative and positive deficiency indices are equal to r, so we select a normalized orthogonal basis f1+⋯fr+∈R(T0+i)⊥ and f1-⋯fr-∈R(T0-i)⊥. Now, we define by induction, orthonormal sequences g1,…gr∈D(T) as follows:
fork=1, we choose normalized element g1∈ker(T-λ1) with
(3.8)〈f1+,g1〉=0=〈f1-,g1〉.
Suppose for given k<r-1, the g1,…,gk (gj∈ker(T-λj)) are well defined with the properties
(3.9)〈fi+,gj〉=0=〈fi-,gj〉∀1≤i,j≤k,
(those gi are eigenfunctions of T relating to λj, so they are orthogonal if λi≠λj, and if λi=λj, we select different orthogonal eigenfunctions of T relating to λj), we choose a normalized element gk+1∈ker(T-λk+1) with
(3.10)〈gk+1,fj+〉=0=〈gk+1,fj-〉∀1≤j≤k+1.
When k=r-1, we stop the induction process. So {g1,…gr} is an orthonormal sequence with the following properties:(i)
gk∈ker(T-λk) for all k=1,…r,(ii)
fk+∈R(T0+i)⊥ and fk-∈R(T0-i)⊥ for all k=1,…r,(iii)
〈gj,fk+〉=〈gj,fk-〉=0 for all j,k∈{1,…r}.Set
(3.11)D(T~):=D(T0)+span{g1,…gr},T~h:=Th∀h∈D(T~).
Obviously T~ is a symmetric extension of T0 and T~gj=Tgj=λjgj for j=1,…m, so each βj is an eigenvalue of T~ with multiplicity at least equal to p(βj). For every k∈{1,…,r}, we have
(3.12)〈fk+,(T~+i)f〉=〈fk+,(T0+i)f〉=0∀f∈D(T0),〈fk+,(T~+i)gj〉=〈fk+,(λj+i)gj〉=0∀j∈{1,…r}.
It follows that fk+∈R(T~+i)⊥ for all k=1,…,r. Thus we have
(3.13)dimR(T~+i)⊥=dimR(T0+i)⊥-r=0.
Analogously, it can be shown that
(3.14)dimR(T~-i)⊥=dimR(T0-i)⊥-r=0.
Thus dimR(T~+i)⊥=dimR(T~-i)⊥=0 and consequently T~ is a self-adjoint extensions which has the properties mentioned above.



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