Characterization of Eigenvalues in Spectral Gap for Singular Differential Operators

Abstract

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

1. Introduction

In this paper, we consider the following 𝑛th-order formal symmetric differential expression: 𝜏𝑦=𝑤^−1[𝑛/2]_𝑗=0(−1)^𝑗𝑝_𝑗𝑦^(𝑗)(𝑗)+^[𝑛−1/2]_𝑗=0(−1)^𝑗𝑞_𝑗𝑦^{(𝑗)(𝑗+1)}−𝑞^∗_𝑗𝑦^{(𝑗+1)(𝑗)}, (1.1) on (𝛼,𝛽), where −∞≤𝛼<𝛽≤+∞, 𝑦=𝑦(𝑡) is a complex-valued 𝑚-vector function, 𝑝_𝑗(𝑡),𝑞_𝑗(𝑡) and 𝑤(𝑡) are measurable and locally integrable 𝑚×𝑚 matrices, 𝑝_𝑗(𝑡),𝑤(𝑡) are Hermitian, and 𝑤(𝑡)>0.𝑎.𝑒.𝑡∈(𝛼,𝛽).

The spectral properties of 𝑛th-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 𝐴−𝑎+𝑏2𝑓_𝑤≥𝑏−𝑎2𝑓_𝑤∀𝑓∈𝐷(𝐴). (1.2)‖⋅‖_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‖²_𝑤, 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 operator 𝐴 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 Section 2, 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,…𝛽_𝑚∈(𝑎,𝑏)(𝑚≤𝑟,𝑟 is the deficiency index of 𝑇₀) and positive integer function 𝑝 satisfying ∑^𝑚_𝑖=1𝑝(𝛽_𝑖)=𝑟, 𝑇₀ has a self-adjoint extension 𝑇 with the following properties:(i)𝜎_𝑝(𝑇)∩(𝑎,𝑏)={𝛽₁,𝛽₂,…𝛽_𝑚}; (ii)each𝛽_𝑖(𝑖=1,2,…𝑚)isaneigenvalueof𝑇withmultiplicityequalto𝑝(𝛽_𝑖); (iii)𝑇haspurepointspectrumwithin(𝑎,𝑏). Finally, given a symmetric operator 𝑇₀ with the deficiency index being equal to 𝑟, (𝑎,𝑏)⊂ℝ, 𝛽₁,𝛽₂,…𝛽_𝑚∈(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 ∑^𝑚_𝑖=1𝑝(𝛽_𝑖)≤𝑟, 𝑇₀ has a self-adjoint extension 𝑇 with the properties that each 𝛽_𝑖(𝑖=1,2,…𝑚) is an eigenvalue of 𝑇 with multiplicity at least 𝛽_𝑖.

2. Preliminaries

In this section, we introduce notations, definitions, and some theorems that are needed in this paper.

First, we define the following space: 𝐿₂𝛼,𝛽;𝑤∶=𝑓∶^𝛽_𝛼𝑓^∗(𝑡)𝑤(𝑡)𝑓(𝑡)<+∞, (2.1) with the inner product 𝑓,𝑔_𝑤=^𝛽_𝛼𝑔^∗(𝑡)𝑤(𝑡)𝑓(𝑡)𝑑𝑡, (2.2) 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 operator 𝑇 and minimal operator 𝑇₀ generated by the expression 𝜏.

Definition 2.1. The maximal operator 𝑇 generated by 𝜏 is defined by 𝐷(𝑇)=𝑓∈𝐿₂𝛼,𝛽;𝑤∶𝑓^{0},𝑓^{1},…𝑓^{𝑛−1}∈𝐴𝐶_loc𝛼,𝛽,𝜏𝑓∈𝐿₂,(𝑇),𝛼,𝛽;𝑤𝑇𝑓=𝜏𝑓for𝑓∈𝐷 (2.3) where 𝐴𝐶_loc(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 operator 𝑇₀₀ generated by 𝜏 is defined by 𝐷𝑇₀₀=(𝑇),𝑇𝑓∈𝐷∶𝑓hascompactsupportin𝛼,𝛽₀₀𝑓=𝜏𝑓=𝑇𝑓for𝑓∈𝐷𝑇₀₀. (2.4) 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 operator 𝐴 with domain and range in a Hilbert space ℍ, the resolvent set of 𝐴 is 𝜌(𝐴)=𝜆∈ℂ∣(𝜆−𝐴)⁻¹∈𝐵(ℍ), (2.5) the spectrum of 𝐴 is the set 𝜎(𝐴)=ℂ⧵𝜌(𝐴). Denote   𝜎_𝑝={𝜆∶(𝜆−𝐴)⁻¹doesnotexist},    𝜎_𝑐={𝜆∶(𝜆−𝐴)⁻¹existsand𝑅(𝜆−𝐴)=ℍ,but(𝜆−𝐴)⁻¹isnotcontinuous},  𝜎_𝑟={𝜆∶(𝜆−𝐴)⁻¹existsand𝑅(𝜆−𝐴)≠ℍ}.

Obviously, 𝜎(𝐴)=𝜎_𝑝∪𝜎_𝑐∪𝜎_𝑟 (see [8]).

Definition 2.4. Let 𝐴 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): Γ(𝐴)==(𝑧))𝑓𝑓(𝐴𝑧∈ℂ∶𝐴−𝑧iscontinuouslyinvertible𝑧∈ℂ∶thereexists𝑎𝑣=𝑣>0suchthat(𝐴−𝑧≥𝑣,∀𝑓∈𝐷) (2.6) it is an open subset of ℂ containing ℂ⧵ℝ.

Definition 2.5. For a closed operator 𝐴 in a Hilbert space ℍ, the essential spectrum of 𝐴 is defined as 𝜎_𝑒(𝐴)=𝜆∈𝜎(𝐴)∶𝑅(𝜆−𝐴)≠𝑅(𝜆−𝐴). (2.7)

We assume that 𝑇₀ has real regular points, that is, Γ(𝑇₀)∩ℝ≠∅. In this case the deficiency index 𝑛𝑇₀,𝑧=𝑛𝑇₀∶=dimker(𝑇−𝑧)=dim𝑅𝑇₀−𝑧^⟂ (2.8) does not depend on the special choice of the regular point of 𝑇₀ and consequently 𝑇₀ has self-adjoint extensions.

The kernel 𝑁(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 0≤𝑛𝑇₀,𝑧≤𝑝. (2.9)

Definition 2.6. We say that an operator 𝐴 has pure point spectrum within (𝑎,𝑏) if 𝜎(𝐴)∩(𝑎,𝑏)=𝜎_𝑝(𝐴)∩(𝑎,𝑏).

Proposition 2.7. Let 𝐴 be a self-adjoint operator in a Hilbert space ℍ. One has 𝜌(𝐴)=Γ(𝐴).

Proof. From Definitions 2.3–2.5, we have 𝜌(𝐴)⊆Γ(𝐴) and 𝜎_𝑝, 𝜎_𝑐 do not belong to Γ(𝐴), so 𝜌(𝐴)∪(𝜎_𝑟∩Γ(𝐴))=Γ(𝐴), 𝐴 is densely defined since it is a self-adjoint operator, so 𝜎_𝑟=∅, then we get 𝜌(𝐴)=Γ(𝐴).

The following results give the relation between deficiency index and self-adjoint extension.

Proposition 2.8. If 𝑇₀₀, 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: 𝑇₀₀⊂𝑇₀₀=𝑇₀⊂𝑇=𝑇^∗⊂𝑇=𝑇^∗₀. (2.10)

Proposition 2.9. If 𝐴 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 of Γ(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 to 0.

Proposition 2.12. Let 𝐴 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)if 𝐴 is semibounded, then 𝐴 has self-adjoint extensions,

We can find the proof from Theorem 8.8 in [7].

Proposition 2.13 (see [8]). Let 𝐴 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∉𝜎_𝑒(𝐴), then 𝐴 has a self-adjoint extension and dim𝑁𝜆−𝐴₁=𝑛(𝐴)+dim𝑁₀𝜆−𝐴₁. (2.11) Furthermore, there exists a self-adjoint extension 𝐴 of 𝐴 such that 𝑁𝜆−𝐴₁⊖𝑁₀𝜆−𝐴₁∩𝐷𝐴={0}. (2.12)

3. Main Results

Let 𝑇₀, 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 subspace ℳ 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 with ℳ 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.

Since Γ(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 ℒ_𝜆 of ker(𝑇^∗₀−𝜆) (= ker(𝑇−𝜆)), set 𝐷𝑇^′₀∶=𝐷𝑇₀+̇ℒ_𝜆,𝑇^′₀𝑓+𝑔∶=𝑇₀𝑓+𝜆𝑔∀𝑓∈𝐷𝑇₀and∀𝑔∈ℒ_𝜆. (3.1) Obviously 𝑇^′₀ is a symmetric extension of 𝑇₀, and 𝑛(𝑇^′₀,𝑧)=𝑛(𝑇₀,𝑧)−1 for each regular point 𝑧 of 𝑇^′₀ and ℒ_𝜆 the corresponding eigenspace. Since the graph of 𝑇^′₀ is a one-dimensional extension of the closed graph of 𝑇₀, therefore the operator 𝑇^′₀ is closed [6].

Theorem 3.2. Assume that the deficiency index of 𝑇₀ is equal to 𝑟, let (𝑎,𝑏) be a spectral gap of 𝑇₀, and let 𝑇^′₀ be defined as above.(1)If 𝑇^′₀ has an eigenvalue 𝜇∈(𝑎,𝑏), then (𝑎,𝑏) is a spectral gap of the restriction of 𝑇^′₀ to the Hilbert space ker(𝑇^′₀−𝜇)^⟂.(2)The deficiency index of the restriction of 𝑇^′₀ to the Hilbert space ker(𝑇^′₀−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 𝑇^′₀𝑓≥𝑣𝑓∀𝑓∈𝐷𝑇^′₀∩ker𝑇^′₀−𝜇^⟂. (3.2) Assume that ‖𝑇^′₀𝑓‖<𝑣‖𝑓‖ for some 𝑓∈𝐷(𝑇^′₀)∩ker(𝑇^′₀−𝜇)^⟂. Let 𝑔∈ker(𝑇^′₀−𝜇) and 𝑔≠0. Since ker(𝑇^′₀−𝜇) reduces 𝑇^′₀, 𝑇^′₀𝑐₁𝑓+𝑐₂𝑔²=𝑐₁²𝑇^′₀𝑓²+𝑐₂²𝑇^′₀𝑔²=𝑐₁²𝑇^′₀𝑓²+𝑐₂²𝜇𝑔²<𝑐₁²𝑣²𝑓²+𝑐₂²𝑣²𝑔²=𝑣²𝑐₁𝑓+𝑐₂𝑔², (3.3) for all (𝑐₁,𝑐₂)∈ℂ²⧵(0,0). Since 𝑓 and 𝑔 span a two-dimensional subspace of 𝐷(𝑇_0′) and dim𝐷(𝑇^′₀)⧵𝐷(𝑇₀)=1, it follows that ‖𝑇^′₀ℎ‖<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) Since 𝑛(𝑇^′₀,𝑧)=𝑛(𝑇₀,𝑧)−1, that is, 𝑛(𝑇^′₀,𝑧)=𝑟−1, and the restriction of 𝑇^′₀ to the Hilbert space ker(𝑇^′₀−𝜇) is self-adjoint, so the deficiency index of the restriction of 𝑇^′₀ onto Hilbert space ker(𝑇^′₀−𝜇)^⟂ is equal to 𝑟−1.

Theorem 3.3. Suppose the deficiency index of 𝑇₀ is equal to 𝑟 and 𝜆 is a real regular point of 𝑇₀. Let 𝑓₁,𝑓₂,…𝑓_𝑚∈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 𝑇^𝑗₀(𝑗=1,2,…𝑚) with the following properties: 𝐷𝑇^𝑗₀∶=𝐷𝑇₀+̇span𝑓₁+̇⋯+̇span𝑓_𝑚,𝑇^𝑗₀𝑓+𝑔₁+𝑔₂+⋯+𝑔_𝑚=𝑇₀𝑓+𝜆𝑔₁+𝑔₂+⋯+𝑔_𝑚,∀𝑓∈𝐷𝑇₀and∀𝑔_𝑗∈span𝑓_𝑗𝑗=1,2…𝑚. (3.4)(2)𝑚≤𝑟. (3)If 𝑚=𝑟, then 𝑇₀ has a self-adjoint extension.

Proof. (1) Since span{𝑓₁} is a one-dimensional subspace of ker(𝑇−𝜆), we define 𝐷𝑇¹₀∶=𝐷𝑇₀+̇span𝑓₁,𝑇¹₀𝑓+𝑔∶=𝑇𝑓+𝜆𝑔∀𝑓∈𝐷𝑇₀and∀𝑔∈span𝑓₁. (3.5) Obviously, 𝑇¹₀ is a symmetric extension of 𝑇₀, 𝑛(𝑇¹₀,𝑧)=𝑛(𝑇₀,𝑧)−1=𝑟−1, and 𝜆 is an eigenvalue of 𝑇¹₀, and span {𝑓₁} is the corresponding eigenspace, and 𝑇¹₀ is also closed. By induction, we get 𝑛𝑇^𝑗₀,𝑧=𝑛𝑇₀,𝑧−𝑗=𝑟−𝑗, (3.6) and 𝑇^𝑗₀(𝑗=1,2…𝑚) is a closed symmetric operator.
(2) When 𝑚>𝑟, then 𝑛(𝑇^𝑚₀,𝑧)=𝑛(𝑇₀,𝑧)−𝑚=𝑟−𝑚<0, that is a contradiction, so we have 𝑚≤𝑟.
(3) When 𝑚=𝑟, then 𝑛(𝑇^𝑚₀,𝑧)=𝑛(𝑇₀,𝑧)−𝑚=𝑟−𝑟=0, so from Proposition 2.11, we have that 𝑇^𝑚₀ is a self-adjoint extension of 𝑇₀.

Theorem 3.4. Assume that the deficiency index of 𝑇₀ is equal to 𝑟. If (𝑎,𝑏) is a spectral gap of 𝑇₀, then for all 𝛽₁,𝛽₂,…𝛽_𝑚∈(𝑎,𝑏) (where 𝑚≤𝑟) and positive integer function 𝑝 satisfying ∑^𝑚_𝑖=1𝑝(𝛽_𝑖)=𝑟, the minimal operator 𝑇₀ has a self-adjoint extension 𝑇 with the following properties: (i)𝜎_𝑝(𝑇)∩(𝑎,𝑏)={𝛽₁,𝛽₂,…𝛽_𝑚}, (ii)each 𝛽_𝑗(𝑗=1,2,…𝑚) is aneigenvalue of 𝑇 with multiplicity 𝑝(𝛽_𝑗),(iii)𝑇 has pure point spectrum within (𝑎,𝑏).

Proof. We choose 𝜆₁,𝜆₂,…𝜆_𝑟 in (𝑎,𝑏), such that #{𝑙≤𝑟∶𝜆_𝑙=𝛽_𝑗}=𝑝(𝛽_𝑗)  (𝑗=1,…,𝑚) (#𝑀 denote the cardinality of the set 𝑀), in other words each 𝛽_𝑗∈(𝑎,𝑏) occurs exactly 𝑝(𝛽_𝑗) times in the sequence {𝜆₁,𝜆₂,…𝜆_𝑟}.
We choose a one-dimensional space ℒ_𝜆1 of ker(𝑇−𝜆₁) and define 𝐷𝑇^′₁∶=𝐷𝑇₀+̇ℒ_𝜆1,𝑇^′₁𝑓+𝑔∶=𝑇₀𝑓+𝜆𝑔∀𝑓∈𝐷𝑇₀and∀𝑔∈ℒ_𝜆1. (3.7) Then 𝑇^′₁ is a closed symmetric extension of 𝑇₀ with deficiency index being equal to 𝑟−1 and 𝜆₁ is an eigenvalue of 𝑇^′₁ with eigensubspace being ℒ_𝜆1.
Set ℋ₁=ℒ^⟂_𝜆1 and denote by 𝑀₁ and 𝑇₁ the restriction of 𝑇^′₁ to ℒ_𝜆1 and ℋ₁, respectively. Obviously 𝑀₁ is a self-adjoint operator in the one-dimensional Hilbert space ℒ_𝜆1 with 𝜎(𝑀₁)={𝜆₁}. 𝑇₁ is a closed symmetric operator in Hilbert space ℋ₁, with deficiency index being equal to 𝑟−1, and 𝑇^′₁=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 {ℒ_𝜆1,…ℒ_𝜆𝑟} and {ℋ₁,…ℋ_𝑟} of Hilbert spaces and sequence {𝑀₁,…𝑀_𝑟} and {𝑇^′₁,…𝑇^′_𝑟} of the operators with the following properties:(i)𝑇^′_𝑙 are closed symmetric extension of 𝑇₀ which satisfy: 𝑇^′₁⊂𝑇^′₂⊂⋯⊂𝑇^′_𝑟, 𝑇^′_𝑙 with deficiency indices 𝑟−𝑙  (𝑙=1,2,…𝑟), (ii)𝑀_𝑙 is a self-adjoint operator on a one-dimensional subspace ℒ_𝜆𝑙 of ℋ_𝑙−1(ℋ₀∶=𝐿₂(𝛼,𝛽;𝑤)),ℋ_𝑙=ℋ_𝑙−1⊖ℒ_𝜆𝑙, with 𝜎(𝑀_𝑙)={𝜆_𝑙} for each 𝜆_𝑙(𝑙=1,2,…𝑟),(iii)𝑇_𝑙 is a closed symmetric operator in the Hilbert space ℋ_𝑙 with spectral gap (𝑎,𝑏) for all 𝑙=1,2…𝑟.(iv)𝑇^′_𝑙={⊕^𝑙_𝑗=1𝑀_𝑗}⊕𝑇_𝑙(𝑙=1,…𝑟). Then we have that 𝑇^′_𝑟 is a closed symmetric extension of 𝑇₀ with deficiency index 0. So 𝑇=𝑇^′_𝑟 is a self-adjoint extension of 𝑇₀ with the required properties.

Theorem 3.5. Assume the deficiency index of 𝑇₀ is equal to 𝑟. Let 𝛽₁,𝛽₂,…𝛽_𝑚∈(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 ∑^𝑚_𝑗=1𝑝(𝛽_𝑗)≤𝑟. Moreover, one assumes that (𝜏−𝜆_𝑖)𝑓=0 has at least 𝑝(𝛽_𝑖) square integrable solutions. Then 𝑇₀ has a self-adjoint extension 𝑇 such that each 𝛽_𝑗(𝑗=1,2,…𝑚) is an eigenvalue of 𝑇 with multiplicity at least equal to 𝑝(𝛽_𝑗).

Proof. Choose a sequence 𝜆₁,…𝜆_𝑟 in (𝑎,𝑏) such that each 𝛽_𝑖 occurs at least 𝑝(𝛽_𝑖) times in this sequence (the times 𝑘_𝑖 of the 𝜆_𝑖 appears in the sequence satisfies 𝑝(𝛽_𝑖)≤𝑘≤𝑘_𝑖, 𝑘_𝑖 is the number of square integrable solutions of (𝜏−𝜆_𝑖)𝑓=0; the 𝜆_𝑖 which is not in {𝛽₁,…,𝛽_𝑚} will be selected as a regular point of 𝑇 satisfying ker(𝑇−𝜆_𝑖)≠∅). 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 𝑓⁺₁⋯𝑓⁺_𝑟∈𝑅(𝑇₀+𝑖)^⟂ and 𝑓⁻₁⋯𝑓⁻_𝑟∈𝑅(𝑇₀−𝑖)^⟂. Now, we define by induction, orthonormal sequences 𝑔₁,…𝑔_𝑟∈𝐷(𝑇) as follows:
for 𝑘=1, we choose normalized element 𝑔₁∈ker(𝑇−𝜆₁) with 𝑓⁺₁,𝑔₁=0=𝑓⁻₁,𝑔₁. (3.8) Suppose for given 𝑘<𝑟−1, the 𝑔₁,…,𝑔_𝑘 (𝑔_𝑗∈ker(𝑇−𝜆_𝑗)) are well defined with the properties 𝑓⁺_𝑖,𝑔_𝑗=0=𝑓⁻_𝑖,𝑔_𝑗∀1≤𝑖,𝑗≤𝑘, (3.9) (those 𝑔_𝑖 are eigenfunctions of 𝑇 relating to 𝜆_𝑗, so they are orthogonal if 𝜆_𝑖≠𝜆_𝑗, and if 𝜆_𝑖=𝜆_𝑗, we select different orthogonal eigenfunctions of 𝑇 relating to 𝜆_𝑗), we choose a normalized element 𝑔_𝑘+1∈ker(𝑇−𝜆_𝑘+1) with 𝑔_𝑘+1,𝑓⁺_𝑗=0=𝑔_𝑘+1,𝑓⁻_𝑗∀1≤𝑗≤𝑘+1. (3.10) When 𝑘=𝑟−1, we stop the induction process. So {𝑔₁,…𝑔_𝑟} is an orthonormal sequence with the following properties:(i)𝑔_𝑘∈ker(𝑇−𝜆_𝑘) for all 𝑘=1,…𝑟,(ii)𝑓⁺_𝑘∈𝑅(𝑇₀+𝑖)^⟂ and 𝑓⁻_𝑘∈𝑅(𝑇₀−𝑖)^⟂ for all 𝑘=1,…𝑟,(iii)⟨𝑔_𝑗,𝑓⁺_𝑘⟩=⟨𝑔_𝑗,𝑓⁻_𝑘⟩=0 for all 𝑗,𝑘∈{1,…𝑟}.Set 𝐷𝑇∶=𝐷𝑇₀+span𝑔₁,…𝑔_𝑟,𝑇ℎ∶=𝑇ℎ∀ℎ∈𝐷𝑇. (3.11) Obviously 𝑇 is a symmetric extension of 𝑇₀ and 𝑇𝑔_𝑗=𝑇𝑔_𝑗=𝜆_𝑗𝑔_𝑗 for 𝑗=1,…𝑚, so each 𝛽_𝑗 is an eigenvalue of 𝑇 with multiplicity at least equal to 𝑝(𝛽_𝑗). For every 𝑘∈{1,…,𝑟}, we have 𝑓⁺_𝑘,𝑇+𝑖𝑓=𝑓⁺_𝑘,𝑇₀+𝑖𝑓=0∀𝑓∈𝐷𝑇₀,𝑓⁺_𝑘,𝑇+𝑖𝑔_𝑗=𝑓⁺_𝑘,𝜆_𝑗+𝑖𝑔_𝑗=0∀𝑗∈{1,…𝑟}. (3.12) It follows that 𝑓⁺_𝑘∈𝑅(𝑇+𝑖)^⟂ for all 𝑘=1,…,𝑟. Thus we have dim𝑅𝑇+𝑖^⟂=dim𝑅𝑇₀+𝑖^⟂−𝑟=0. (3.13) Analogously, it can be shown that dim𝑅𝑇−𝑖^⟂=dim𝑅𝑇₀−𝑖^⟂−𝑟=0. (3.14) Thus dim𝑅(𝑇+𝑖)^⟂=dim𝑅(𝑇−𝑖)^⟂=0 and consequently 𝑇 is a self-adjoint extensions which has the properties mentioned above.

Acknowledgments

The authors sincerely thank the referee for the constructive suggestions and corrections. This paper was partially supported by the NNSF of China (Grant 11171178, 11271225) and NSF of Shandong (Grant ZR2009AQ010).