# An Alternate Proof of the De Branges Theorem on Canonical Systems

Authors: Keshav Raj Acharya
Journal: ISRN Mathematical Analysis
 (2014)

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## Abstract

The aim of this paper is to show that, in the limit circle case, the defect index of a symmetric relation induced by canonical systems, is constant onℂ. This provides an alternative proof of the De Branges theorem that the canonical systems with trH1 imply the limit point case. To this end, we discuss the spectral theory of a linear relation induced by a canonical system.


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## Body

## 1. Introduction

This paper deals with the canonical systems of the following form:(1)Ju′(x)=zH(x)u(x),z∈ℂ.
Here J=(0-110) and H(x) is a 2×2 positive semidefinite matrix whose entries are locally integrable. For fixed z∈ℂ, a function u(·,z):[0,N]→ℂ2 is called a solution if u is absolutely continuous and satisfies (1). Consider the Hilbert space as follows:
(2)L2(H,ℝ+)={f(x,z)=(f1(x)f2(x)):∥f∥<∞},
provided with an inner product 〈f,g〉=∫0∞‍f(x)*H(x)g(x)dx.The canonical systems (1) on L2(H,ℝ+) have been studied by Hassi et al., Winkler, and Remling in [1–4] in various contexts. The Jacobi and Schrödinger equations can be written into canonical systems with appropriate choice of H(x). In addition, canonical systems are closely connected with the theory of the De Branges spaces and the inverse spectral theory of one-dimensional Schrödinger operators; see [3]. We believe that the extensions of the theories from these equations to the canonical systems are to be of general interest.If the system (1) can be written in the form of
(3)H(x)-1Ju′=zu,
then we may consider this as an eigenvalue equation of an operator on L2(H,R+). But H(x) is not invertible in general. Instead, the system (1) induces a linear relation that may have a multivalued part. Therefore, we consider this as an eigenvalue problem of a linear relation induced by (1) on L2(H,R+).For somez∈ℂ, if the canonical system (1) has all solutions in L2(H,ℝ+), we say that the system is in the limit circle case, and if the system has unique solution in L2(H,ℝ+), we say that the system is in limit point case. The basic results in this paper are the following theorems.Theorem 1.
In the limit circle case, the defect indexβ(ℛ0) of the symmetric relation ℛ0, induced by (1), is constant on ℂ.The immediate consequence of the Theorem1 is the following theorem.Theorem 2 (De Branges).
The canonical systems withtr⁡H≡1 prevail the limit point case.Theorem2 has been proved in [5] by function theoretic approach. However the proof was not easily readable to me and we thought of providing an alternate and simple proof of the theorem.In order to prove the main theorems we use the results from the papers [1, 3, 4] and we use the spectral theory of a linear relation from [6].Letℋ be a Hilbert space over ℂ and denote by ℋ2 the Hilbert space ℋ⊕ℋ. A linear relation ℛ={(f,g):f,g∈ℋ} on ℋ is a subspace of ℋ2. The adjoint of ℛ on ℋ is a closed linear relation defined by
(4)ℛ*={(h,k)∈ℋ2:〈g,h〉=〈f,k〉∀(f,g)∈ℛ}.
A linear relation 𝒮 is called symmetric if 𝒮⊂𝒮* and self-adjoint if 𝒮=𝒮*. The theory of such relations can be found in [5–8]. The regularity domain of ℛ is the following set:
(5)Γ(ℛ)={z∈ℂ:∃C(z)>0such that∥(zf-g)∥hhh≥C(z)∥f∥∀(f,g)∈ℛ}.
The following theorem has been derived from [6].Theorem 3.
Suppose𝒯 is a self-adjoint relation and suppose z∈Γ(𝒯); then
(6)ℋ={zf-g:(f,g)∈𝒯}.The defect indexβ(ℛ,z) is the dimension of defect space:
(7)R(z-ℛ)⊥={zf-g:(f,g)∈ℛ}⊥.
It has been shown in [6] that the defect index β(ℛ,z) is constant on each connected subset of Γ(ℛ). Moreover, if ℛ is symmetric, then the defect index is constant in the upper and lower half-planes. In addition, it is worth mentioning here the following theorem from [6] which provides us with the condition for a symmetric relation on a Hilbert space to have self-adjoint extension.Theorem 4.
(1) A symmetric relationℛ possesses self-adjoint extension if and only if β(ℛ,z) on lower and upper half-planes is equal.
(2) A symmetric extensionℛ′ of ℛ is self-adjoint if and only if ℛ′is an β(ℛ,z)-dimensional extension of ℛ.The resolvent set for a closed relationℛ is the following set:
(8)ρ(ℛ)={ℛ={(Tf,zTf-f):f∈ℋ}}z∈ℂ:∃T∈B(ℋ)suchthathh0hℛ={(Tf,zTf-f):f∈ℋ}},
and the spectrum of ℛ is σ(ℛ)=ℂ-ρ(ℛ).We callS(ℛ)=ℂ-Γ(ℛ) thespectral kernel of ℛ. The following theorem from [6] shows the relation between the spectral kernel and spectrum of a self-adjoint relation.Theorem 5.
If𝒯 is a self-adjoint relation and T=(𝒯-z)-1, z∈Γ(𝒯) then one has the following(1) S(𝒯)=σ(𝒯).(2) If λ∈S(𝒯), then 1/(z-λ)∈S(T).(3)S(T)⊂σ(T).In the next section we discuss the linear relation induced by a canonical system and prove our main theorems.


## 2. Relation Induced by a Canonical System onL2(H,ℝ+) and Proof of the Main Theorems

Consider that a relationℛ in the Hilbert space L2(H,ℝ+) is induced by (1) as
(9)ℛ={(f,g)∈(L2(H,ℝ+))2:f∈AC,Jf′=Hg},
and is called the maximal relation. This relation is made up of pairs of equivalence classes (f,g), such that there exists a locally absolutely continuous representative of f again denoted by f and a representative of g, again denoted by g, such that Jf′=Hg a.e. on ℝ+. The adjoint relation ℛ0=ℛ* is defined by
(10)ℛ0={(f,g)∈(L2(H,ℝ+))2:〈g,h〉hh0h=〈f,k〉∀(h,k)∈ℛ(L2(H,ℝ+))2},
and is called the minimal relation. It has been shown in [1] that ℛ0 is close and symmetric. Moreover, ℛ0⊂ℛ and (ℛ0)*=ℛ.Lemma 6 (see [1]).
For eachc,d∈ℂ2 there exists (ϕ0,ψ0),(ϕN,ψN)∈ℛ such that ϕ0, ϕN have compact support and ϕ(0+)=c,  ϕ(N-)=d.Lemma 7 (see [1]).
Let(f,g),(h,k)∈ℛ. Then the following limit exists:
(11)lim⁡x→∞h(x)Jf(x)=h(0+)Jf(0+)-[〈f,k〉-〈g,h〉].Lemma 8.
The minimal relationℛ0 is given by
(12)ℛ0={(f,g)∈ℛ:f(0+)=0,lim⁡x→∞f*(x)Jh(x)hhh=0∀(h,k)∈ℛlim⁡x→∞f*(x)}.Proof.
By Lemma7, we get
(13){(f,g)∈ℛ:f(0+)=0,lim⁡x→∞f*(x)Jh(x)h=0∀(h,k)∈ℛlim⁡x→∞f*}⊂ℛ0.
On the other hand, let (f,g)∈R0. By Lemma 6 for any u∈ℂ2 there exists (ϕ,ψ)∈ℛ such that ϕ has compact support and ϕ(0+)=c. So
(14)0=〈f,ψ〉-〈g,ϕ〉=lim⁡x→∞f*(x)Jϕ(x)-ϕ(0+)Jf(0+)=uJf(0+).
This implies that f(0+)=0. This would also force the following:
(15)lim⁡x→∞f*(x)Jh(x)=0∀(h,k)∈ℛ.Note that the dimension of the solution space of the system (1) is two.Remark 9.
The defect indexβ(ℛ0) of the minimal relation ℛ0 is equal to the number of linearly independent solutions of the system (1) whose class lies in L2(H,ℝ+). Therefore, in the limit circle case, the defect indices of ℛ0 are (2,2).Sinceℛ0 has equal defect indices, by Theorem 4, it has self-adjoint extensions say 𝒯. Consider a relation as follows:
(16)𝒯α,β={(f,g)∈ℛ:f1(0)sinα+f2(0)cos⁡α=0,hhhf1(N)sinβ+f2(N)cos⁡β=0,α,β∈(0,π]},
on a compact interval [0,N].Lemma 10.
𝒯
α
,
β is a self-adjoint relation.Proof.
Clearly𝒯α,β is a symmetric relation because of the boundary conditions at 0 and N. We will show that 𝒯α,β is a 2-dimensional extension of ℛ0. Then by Theorem 4, 𝒯α,β is a self-adjoint relation. By Lemma 6, for c=(-cos⁡αsinα) and w=(-cos⁡βsinβ)∈ℂ2, there exist ϕ0 and ϕN in D(ℛ) such that ϕ0(0+)=c and ϕN(N-)=w and the support of ϕ0 and ϕN is contained in [0,N]. Clearly ϕ0 and ϕN are linearly independent but ϕ0 and, ϕN are not in D(ℛ0). This shows that D(ℛ0)⊂D(R0)+L(ϕ0,ϕN)⊂D(𝒯α,β). Because of the boundary condition at 0 and N, D(𝒯α,β) is 2-dimensional restriction of D(ℛ). Hence D(𝒯α,β)=D(ℛ0)+L(ϕ0,ϕN). Therefore, 𝒯α,β is 2-dimensional extension of ℛ0 so that 𝒯α,β is a self-adjoint relation.Letu(x,z) and v(x,z) be the solution of the canonical system (1) on [0,N] with the initial values:
(17)u(0,z)=(10),v(0,z)=(01).
For z∈C+ there is a unique m(z) such that f(x,z)=u(x,z)+m(z)v(x,z) satisfying the following:
(18)f1(N,z)sinβ+f2(N,z)cos⁡β=0.
This is well defined because u does not satisfy the boundary condition at N; otherwise, z will be an eigenvalue of some self-adjoint relation 𝒯α,β.Next, we describe the spectrum of𝒯α,β. Let
(19)T(x,z)=(u1(x,z)v1(x,z)u2(x,z)v2(x,z)),T(0,z)=(1001),
and define
(20)wα(x,z)=1sinα+m(z)cos⁡αT(x,z)(cos⁡α-sinα).
It is not hard to see that.Lemma 11.
Using the notation above one has(21)f(x,z)wα(x,z-)*-wα(x,z)f(x,z-)*=T(x,z)JT(x,z-)*=J.Lemma 12.
Letz∈Γ(𝒯α,β); then (𝒯α,β-z)-1 is a bounded linear operator and is defined by
(22)(𝒯α,β-z)-1h(x)=∫0N‍G(x,t,z)H(t)h(t)dt,
where
(23)G(x,t,z)={f(x,z)wα(t,z-0)*if0<t≤xwα(t,z-)f(x,z-0)ifx<t≤N.Proof.
Lety(x,z)=∫0N‍G(x,t,z)H(t)h(t)dt. We show that y(x,z) solves the inhomogeneous equation as follows:
(24)Jy′=zHy-Hh,
for a.e.x>0. Here
(25)y(x,z)=∫0x‍f(x,z)wα(t,z-)*H(t)h(t)dt+∫xN‍wα(x,z)f(t,z-)*H(t)h(t)dt,
and Jf′=zHf, Jwα′=zHwα. Then on differentiation we get
(26)y′(x,z)=f(x,z)wα(x,z-)*H(x)h(x)+f′(x,z)∫0x‍wα(t,z-)*H(t)h(t)dt-wα(x,z)f(x,z-)*H(x)h(x)+wα′(x,z)∫xN‍f(t,z-)*H(t)h(t)dt.
Then
(27)Jy′(x,z)=Jf(x,z)wα(x,z-)*H(x)h(x)+Jf′(x,z)∫0xwα(t,z-)*H(t)h(t)dt-Jwα(x,z)f(x,z-)*H(x)h(x)+Jwα′(x,z)∫xNf(t,z-)*H(t)h(t)dt=Jf(x,z)wα(x,z-)*H(x)h(x)+zHf(x,z)∫0xwα(t,z-)*H(t)h(t)dt-Jwα(x,z)f(x,z-)*H(x)h(x)+zHwα(x,z)∫xNf(t,z-)*H(t)h(t)dt=J(f(x,z)wα(x,z-)*-wα(x,z)f(x,z-)*)Hh+zH(∫0xf(x,z)wα(t,z-)*H(t)h(t)dthhhhhh+∫xNwα(x,z)f(t,z-)*H(t)h(t)dt)=JJHh+zHy=zHy-Hh.
On the other hand denoteg(x,z) as
(28)g(x,z)=(𝒯α,β-z)-1h(x);
then by Theorem 3, h(x)=zu-v for some (u,v)∈𝒯α,β so that (g,zg-h)∈𝒯α,β. So g(x,z) also satisfies the inhomogeneous problem and g(x,z)∈D(𝒯α,β); it satisfies the boundary condition which implies that g(0,z)=(cos⁡α-sinα)c(z) for some scalar c(z). We have
(29)y(0,z)=1sinα+m(z)cos⁡α(cos⁡α-sinα)〈f(x,z-),h〉.
Now
(30)〈f(·,z-),h〉=〈f(·,z-),h〉-〈f(·,z-),zg〉+〈f(·,z-),zg〉=〈f(·,z-),h-zg〉+z〈f(·,z-),g〉=-∫0N‍f(x,z-)*H(zg-h)dx+z∫0N‍f(x,z-)*Hgdx=-∫0N‍f(x,z-)*Jg′dx-∫0N‍f′(x,z-)*Jgdx=f(0,z-)*Jg(0,z)-f(N,z-)*Jg(N,z).
Since both f(x,z) and g(x,z) satisfy the same boundary condition at f(N,z-)*Jg(N,z)=0, now
(31)f(0,z-)*Jg(0,z)=(1,m(z))(0-110)(cos⁡α-sinα)c(z).
So
(32)y(0,z)=1sinα+m(z)cos⁡α(cos⁡α-sinα)(1,m(z))×(0-110)(cos⁡α-sinα)c(z)=1sinα+m(z)cos⁡α(cos⁡α-sinα)(m(z),-1)×(cos⁡α-sinα)c(z)=(cos⁡α-sinα)c(z)=g(0,z).
By uniqueness we must have g(x,z)=y(x,z). Moreover, (𝒯α,β-z)-1 is a bounded linear operator.Now define a mapV:L2(H,[0,N])→L2(I,[0,N]) by
(33)Vy=H1/2(x)y(x).
Here H1/2(x) is the unique positive semidefinite square root of H(x). Then V is an isometry and hence maps L2(H,[0,N]) unitarily onto the range R(V)⊂L2(I,[0,N]). Define an integral operator ℒ on L2(I,[0,N]) as
(34)(ℒf)(x)=∫0N‍L(x,t)f(t)dt,L(x,t)=H1/2(x)G(x,t)H1/2(t).
The kernel L is square integrable since
(35)∬0N∥L*L∥dxdt≤∬0N∥VG*∥∥(VG)*∥dxdt≤∬0N∥G*∥∥G∥dxdt<∞.
So ℒ is a Hilbert-Schmidt operator and thus compact. Notice that L(x,t)=L*(t,x). This implies that ℒ is self-adjoint.Lemma 13 (see [3]).
Letf∈L2(I,[0,N]), λ≠0, and then the following statements are equivalent. (1)
ℒf=λ-1f.(2)
f∈R(V), and the unique y∈L2(H,[0,N]) with Vy=f solves (𝒯α,β-z)-1y=λy.Proof.
For allg∈L2(I,[0,N]) we have
(36)(ℒg)(x)=H1/2(x)w(x)wherew(x)=∫0N‍G(x,t)H1/2(t)g(t)dt
lies in L2(H,[0,N]). Then R(ℒ)⊂R(V). Now if (1) holds, then f=λℒf∈R(V). So f=V(y) for unique y∈L2(H,[0,N]) and
(37)f(x)=H1/2(x)y(x)=λℒy(x)=λH12(x)∫0N‍G(x,t)H(t)y(t)dt,
for a.e.x∈[0,N]. In other words,
(38)H1/2(x)(y(x)-λ∫0N‍G(x,t)H(t)y(t)dt)=0.
Conversely if (11) holds,
(39)λy=∫0N‍G(x,t)H(t)y(t)dt,
And then H1/2(x)y=(1/λ)∫0NH1/2(x)G(x,t)H(t)y(t)dt.Lemma 14.
Letz∈ℂ. For any λ≠z, if (f,λf)∈𝒯α,β, then f solves (𝒯α,β-z)-1y=1/(λ-z)y. Conversely, if y∈L2(H,[0,N]) and y solves (Tα,β-z)-1y=λy, then (y,(z+1/λ)y)∈𝒯α,β.Proof.
Let(f,λf)∈𝒯α,β; then (f,λf-zf)∈(𝒯α,β-z). It follows that
(40)((λ-z)f,f)∈(𝒯α,β-z)-1⟹(f,1(λ-z))∈(𝒯α,β-z)-1.
This means that f solves
(41)(𝒯α,β-z)-1y=1λ-zy.
Conversely suppose y∈L2(H,[0,N]) and y solves
(42)(𝒯α,β-z)-1y=λy.
That is, (y,λy)∈(𝒯α,β-z)-1 so that (λy,y)∈(𝒯α,β-z). So there is (f,g)∈𝒯α,β such that λy=f and
(43)g-zf=y⟹g=y+zλy.
Hence
(44)(λy,y+zλy)∈𝒯α,β⟹(y,(z+1λy))∈𝒯α,β.By Lemma13, we see that there is a one-to-one correspondence of eigenvalues (eigenfunctions) for the operator ℒ and (𝒯α,β-z)-1. As ℒ is compact operator, it has only discrete spectrum consisting of only eigenvalues. Since (𝒯α,β-z)-1 is unitarily equivalent to ℒ⇂R(V); that is,
(45)V-1ℒ⇂R(V)V=(𝒯α,β-z)-1,(𝒯α,β-z)-1 has only discrete spectrum consisting of only eigenvalues. Then, by Theorem 5, 𝒯α,β has only discrete spectrum. By Lemma 14, the spectrum of 𝒯α,β consists only of eigenvalues. Hence we have
(46)σ(𝒯α,β)={z∈ℂ:uα1(N,z)sinβ+uα2cos⁡β=0}.We would like to extend this idea over the half lineℝ+. First note that we are considering the limit circle case of the system (1). That implies that for any z∈ℂ+ the defect indices of ℛ0 are (2,2). Suppose p∈D(ℛ)∖D(ℛ0) such that lim⁡x→∞p(x)*Jp(x)=0. Such function clearly exists.Consider the following relation:(47)𝒯α,p={lim⁡x→∞f(x)*Jp(x)=0(f,g)∈ℛ:f1(0)sin⁡α+f2(0,z)cos⁡⁡α=0andhhhlim⁡x→∞f(x)*Jp(x)=0}.Lemma 15.
𝒯
α
,
p defines a self-adjoint extension of ℛ0.Proof.
Letu(x) be a solution of the system (1) with some initial values and p(x) as above. Define u0(x)=0 near the neighborhood of ∞; that is, lim⁡x→∞u0(x)=0, and u0(x)=u(x) otherwise. Similarly, p0(x)=0 in the neighborhood of 0 and p0(x)=p(x) otherwise. Then clearly u0,p0∉D(ℛ0) and are linearly independent. Clearly D(ℛ0)+L(u0,p0)⊂D(𝒯α,p). Since D(𝒯α,p) is at least 2-dimensional restriction of D(ℛ),
(48)D(𝒯α,p)=D(ℛ0)+L(u0,p0).
Hence 𝒯α,p is a self-adjoint relation.We next discuss the spectrum of𝒯α,p. Let u(x,z) and v(x,z) be two linearly independent solutions of the system (1) with
(49)u(0,z)=(10),v(0,z)=(01).
Let z∈ℂ+ and as above write f(x,z)=u(x,z)+m(z)v(x,z)∈L2(H,ℝ+) satisfying lim⁡x→∞f(x,z)*JP(x)=0. Let T(x,z)=(u1v1u2v2) and
(50)wα(x,z)=1sin⁡α+m(z)cos⁡⁡αT(x,z)(cos⁡⁡α-sin⁡α).
Then as in Lemma 11 we have
(51)f(x,z)wα(x,z-)*-wα(x,z)f(x,z-)*=T(x,z)JT(x,z-)*f(x,z)wα(x,z-)*-wα(x,z)f(x,z-)*=J.Lemma 16.
Letz∈ρ(𝒯α,p); then the resolvent operator (𝒯α,p-z)-1 is given by
(52)(𝒯α,p-z)-1h(x)=∫0∞‍G(x,t,z)H(t)h(t)dt,
where
(53)G(x,t,z)={f(x,z)wα(t,z-)*if0<t≤xwα(t,z-)f(x,z-)ifx<t≤∞.Proof.
Lety(x,z)=∫0∞‍G(x,t,z)H(t)h(t)dt; then y solves the following inhomogeneous equation:
(54)Jy′=zHy-Hh.
This is clear by differentiating
(55)y(x,z)=∫0xf(x,z)wα(t,z-)*H(t)h(t)dt+∫x∞wα(x,z)f(t,z-)*H(t)h(t)dt.
On the other hand denote g(x,z) as g(x,z)=(𝒯α,p-z)-1h(x); then, by Theorem 3, h(x)=zu-v for some (u,v)∈𝒯α,p so that (g,zg-h)∈𝒯α,p, and hence g satisfies the inhomogeneous equation. Since g∈D(𝒯α,p),
(56)g1(0,z)sin⁡α+g2(0,z)cos⁡⁡α=0,lim⁡x→∞g*(x,z)Jp(x,z)=0.
We also have lim⁡x→∞f*(x,z)Jg(x,z)=0 and g(0,z)=(cos⁡⁡α-sin⁡α)c(z) for some scalar c(z). But we also have
(57)y(0,z)=1(m(z)cos⁡⁡α+sin⁡α)(cos⁡⁡αsin⁡α)〈f(z-),h〉.
Here
(58)〈f(z-),h〉=〈f(z-),h〉-〈f(z-),zg〉-〈z-f(z-),g〉=〈f(z-),h+zg〉-〈z-f(z-),g〉=f*(0,z-)Jg(0,z)-lim⁡x→∞f*(x,z)Jg(x,z)=f*(0,z-)Jg(0,z).
Hence y(0,z)=g(0,z). By uniqueness we have y(x,z)=g(x,z).Now define a mapV:L2(H,ℝ+)→L2(I,ℝ+) by Vy=H1/2(x)y(x).   V is isometry and maps unitarily onto the range R(V)⊂L2(I,ℝ+).Define an integral operatorℒ on L2(I,ℝ+) by
(59)(ℒg)(x)=∫0∞L(x,t)g(t)dt,L(x,t)=H1/2(x)G(x,t,z)H1/2(t).
Then as before the kernel L is square integrable. This means that
(60)∬0∞∥L*L∥<∞.
Hence ℒ is a Hilbert Schmidt operator and so it is a compact operator. The following two lemmas are extended from the bounded interval [0,N] to ℝ+ and the proofs are exactly the same as the proofs of Lemmas 13 and 14.Lemma 17 (see [3]).
Letf∈L2(I,R+), λ≠0, and then the following statements are equivalent. (1)
ℒf=λ-1f.(2)
f∈R(V), and the unique y∈L2(H,R+) with Vy=f solves (𝒯α,p-z)-1y=λy.Lemma 18.
Letz∈ℂ. For any λ≠z, if (y,λy)∈𝒯α,p, then y solves (𝒯α,p-z)-1y=(1/(λ-z))y. Conversely, if y∈L2(H,R+) and y solves (Tα,p-z)-1y=λy, then (y,(z+1/λ)y)∈𝒯α,p.Again, by Lemma17, we have a one-to-one correspondence of eigenvalues (eigenfunctions) for the operator ℒ and (𝒯α,p-z)-1. As ℒ is compact operator, it has only discrete spectrum consisting of only eigenvalues and possibly zero. Since (𝒯α,p-z)-1 is unitarily equivalent to ℒ⇂R(V); that is, V-1ℒ⇂R(V)V=(𝒯α,p-z)-1,(𝒯α,p-z)-1 has only discrete spectrum consisting of only eigenvalues. Then, by Theorem 5, 𝒯α,p has only discrete spectrum. By Lemma 18 the spectrum of 𝒯α,p consists of only eigenvalues.With these theories in hand, we are now ready to prove the main theorems.Proof of Theorem1.
Sinceℛ0 is a symmetric relation, the defect index β(ℛ0,z) is constant on upper and lower half-planes. In the limit circle case, if z is in upper or lower half-planes, β(ℛ0,z)=2. Suppose β(ℛ0,λ)<2 for some λ∈ℝ. Since Γ(ℛ0) is open, λ∉Γ(ℛ0), and hence λ∈S(ℛ0). Since, for each α∈(0,π], 𝒯α,p is self-adjoint extension of ℛ0, λ∈S(𝒯α,p)=σ(𝒯α,p). In the limit circle case, σ(𝒯α,p) consists of only eigenvalues. Therefore, λ is an eigenvalue for all boundary conditions α at 0. However, this is impossible unless β(ℛ0,λ)=2.Proof of Theorem2.
Suppose it prevails the limit circle case. By Theorem1, the defect index β(ℛ0,z)=dim⁡N(ℛ,z-)=2 for all z∈ℂ. In other words, for any z∈ℂ, all solutions of (1) are in L2(H,ℝ+). In particular, the constant solutions u(x)=(10) and v(x)=(01) of (1) when z=0 are in L2(H,ℝ+). But this is not possible because ∫0∞‍u(x)*H(x)u(x)dx+∫0∞‍v(x)*H(x)v(x)dx=∫0∞‍tr⁡H(x)dx=∞.



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*Source: 704607-2014-04-03.xml*