ISRN.MATHEMATICAL.ANALYSIS ISRN Mathematical Analysis 2090-4665 Hindawi Publishing Corporation 704607 10.1155/2014/704607 704607 Research Article An Alternate Proof of the De Branges Theorem on Canonical Systems Acharya Keshav Raj Karakostas G. L. Zhu C. Department of Mathematics Southern Polytechnic State University 1100 South Marietta Pkwy, Marietta, GA 30060 USA spsu.edu 2014 342014 2014 21 12 2013 18 03 2014 3 4 2014 2014 Copyright © 2014 Keshav Raj Acharya. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

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 on . This provides an alternative proof of the De Branges theorem that the canonical systems with tr H1 imply the limit point case. To this end, we discuss the spectral theory of a linear relation induced by a canonical system.

1. Introduction

This paper deals with the canonical systems of the following form: (1)Ju(x)=zH(x)u(x),z. Here J=(0-110) and H(x) is a 2×2 positive semidefinite matrix whose entries are locally integrable. For fixed z, a function u(·,z):[0,N]2 is called a solution if u is absolutely continuous and satisfies (1). Consider the Hilbert space as follows: (2)L2(H,+)={f(x,z)=(f1(x)f2(x)):f<}, provided with an inner product f,g=0f(x)*H(x)g(x)dx.

The canonical systems (1) on L2(H,+) have been studied by Hassi et al., Winkler, and Remling in  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 . 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 some z, if the canonical system (1) has all solutions in L2(H,+), we say that the system is in the limit circle case, and if the system has unique solution in L2(H,+), 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 index β(0) of the symmetric relation 0, induced by (1), is constant on .

The immediate consequence of the Theorem 1 is the following theorem.

Theorem 2 (De Branges).

The canonical systems with trH1 prevail the limit point case.

Theorem 2 has been proved in  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 .

Let be a Hilbert space over and denote by 2 the Hilbert space . A linear relation ={(f,g):f,g} on is a subspace of 2. The adjoint of on is a closed linear relation defined by (4)*={(h,k)2:g,h=f,k(f,g)}. A linear relation 𝒮 is called symmetric if 𝒮𝒮* and self-adjoint if 𝒮=𝒮*. The theory of such relations can be found in . The regularity domain of is the following set: (5)Γ()={z:C(z)>0  such that  (zf-g)hhhC(z)f(f,g)}. The following theorem has been derived from .

Theorem 3.

Suppose 𝒯 is a self-adjoint relation and suppose zΓ(𝒯); then (6)={zf-g:(f,g)𝒯}.

The defect index β(,z) is the dimension of defect space: (7)R(z-)={zf-g:(f,g)}. It has been shown in  that the defect index β(,z) is constant on each connected subset of Γ(). Moreover, if 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  which provides us with the condition for a symmetric relation on a Hilbert space to have self-adjoint extension.

Theorem 4.

(1) A symmetric relation possesses self-adjoint extension if and only if β(,z) on lower and upper half-planes is equal.

(2) A symmetric extension of is self-adjoint if and only if is an β(,z)-dimensional extension of .

The resolvent set for a closed relation is the following set: (8)ρ()={={(Tf,zTf-f):f}}z:TB()such  thathh0h={(Tf,zTf-f):f}}, and the spectrum of is σ()=-ρ().

We call S()=-Γ() the spectral kernel of . The following theorem from  shows the relation between the spectral kernel and spectrum of a self-adjoint relation.

Theorem 5.

If 𝒯 is a self-adjoint relation and T=(𝒯-z)-1, zΓ(𝒯) then one has the following

(1)S(𝒯)=σ(𝒯).

(2) If λS(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 on <inline-formula><mml:math xmlns:mml="http://www.w3.org/1998/Math/MathML" id="M71"><mml:msup><mml:mrow><mml:mi>L</mml:mi></mml:mrow><mml:mrow><mml:mn>2</mml:mn></mml:mrow></mml:msup><mml:mo mathvariant="bold">(</mml:mo><mml:mi>H</mml:mi><mml:mo mathvariant="bold">,</mml:mo><mml:msub><mml:mrow><mml:mi>ℝ</mml:mi></mml:mrow><mml:mrow><mml:mo mathvariant="bold">+</mml:mo></mml:mrow></mml:msub><mml:mo mathvariant="bold">)</mml:mo></mml:math></inline-formula> and Proof of the Main Theorems

Consider that a relation in the Hilbert space L2(H,+) is induced by (1) as (9)={(f,g)(L2(H,+))2:fAC,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 +. The adjoint relation 0=* is defined by (10)0={(f,g)(L2(H,+))2:g,hhh0h=f,k  (h,k)(L2(H,+))2}, and is called the minimal relation. It has been shown in  that 0 is close and symmetric. Moreover, 0 and (0)*=.

Lemma 6 (see [<xref ref-type="bibr" rid="B6">1</xref>]).

For each c,d2 there exists (ϕ0,ψ0),(ϕN,ψN) such that ϕ0, ϕN have compact support and ϕ(0+)=c,  ϕ(N-)=d.

Lemma 7 (see [<xref ref-type="bibr" rid="B6">1</xref>]).

Let (f,g),(h,k). Then the following limit exists: (11)limxh(x)Jf(x)=h(0+)Jf(0+)-[f,k-g,h].

Lemma 8.

The minimal relation 0 is given by (12)0={(f,g):f(0+)=0,limxf*(x)Jh(x)hhh=0  (h,k)limxf*(x)}.

Proof.

By Lemma 7, we get (13){(f,g):f(0+)=0,limxf*(x)Jh(x)h=0  (h,k)limxf*}0. On the other hand, let (f,g)R0. By Lemma 6 for any u2 there exists (ϕ,ψ) such that ϕ has compact support and ϕ(0+)=c. So (14)0=f,ψ-g,ϕ=limxf*(x)Jϕ(x)-ϕ(0+)Jf(0+)=uJf(0+). This implies that f(0+)=0. This would also force the following: (15)limxf*(x)Jh(x)=0(h,k).

Note that the dimension of the solution space of the system (1) is two.

Remark 9.

The defect index β(0) of the minimal relation 0 is equal to the number of linearly independent solutions of the system (1) whose class lies in L2(H,+). Therefore, in the limit circle case, the defect indices of 0 are (2,2).

Since 0 has equal defect indices, by Theorem 4, it has self-adjoint extensions say 𝒯. Consider a relation as follows: (16)𝒯α,β={(f,g):f1(0)sinα+f2(0)cosα=0,hhhf1(N)sinβ+f2(N)cosβ=0,α,β(0,π]}, on a compact interval [0,N].

Lemma 10.

𝒯 α , β is a self-adjoint relation.

Proof.

Clearly 𝒯α,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 0. Then by Theorem 4, 𝒯α,β is a self-adjoint relation. By Lemma 6, for c=(-cosαsinα) and w=(-cosβsinβ)2, there exist ϕ0 and ϕN in D() 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(0). This shows that D(0)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(). Hence D(𝒯α,β)=D(0)+L(ϕ0,ϕN). Therefore, 𝒯α,β is 2-dimensional extension of 0 so that 𝒯α,β is a self-adjoint relation.

Let u(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 zC+ 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)cosβ=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 of 𝒯α,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)cosαT(x,z)(cosα-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.

Let zΓ(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)*if  0<txwα(t,z-)f(x,z-0)if  x<tN.

Proof.

Let y(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 denote g(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)=(cosα-sinα)c(z) for some scalar c(z). We have (29)y(0,z)=1sinα+m(z)cosα(cosα-sinα)f(x,z-),h. Now (30)f(·,z-),h=f(·,z-),h-f(·,z-),zg+f(·,z-),zg=f(·,z-),h-zg+zf(·,z-),g=-0Nf(x,z-)*H(zg-h)dx+z0Nf(x,z-)*Hgdx=-0Nf(x,z-)*Jgdx-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)(cosα-sinα)c(z). So (32)y(0,z)=1sinα+m(z)cosα(cosα-sinα)(1,m(z))×(0-110)(cosα-sinα)c(z)=1sinα+m(z)cosα(cosα-sinα)(m(z),-1)×(cosα-sinα)c(z)=(cosα-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 map V: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 on L2(I,[0,N]) as (34)(f)(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)0NL*Ldxdt0NVG*(VG)*dxdt0NG*Gdxdt<. So is a Hilbert-Schmidt operator and thus compact. Notice that L(x,t)=L*(t,x). This implies that is self-adjoint.

Lemma 13 (see [<xref ref-type="bibr" rid="B8">3</xref>]).

Let fL2(I,[0,N]), λ0, and then the following statements are equivalent.

f=λ-1f.

fR(V), and the unique yL2(H,[0,N]) with Vy=f solves (𝒯α,β-z)-1y=λy.

Proof.

For all gL2(I,[0,N]) we have (36)(g)(x)=H1/2(x)w(x)where  w(x)=0NG(x,t)H1/2(t)g(t)dt lies in L2(H,[0,N]). Then R()R(V). Now if (1) holds, then f=λfR(V). So f=V(y) for unique yL2(H,[0,N]) and (37)f(x)=H1/2(x)y(x)=λy(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)-λ0NG(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.

Let z. For any λz, if (f,λf)𝒯α,β, then f solves (𝒯α,β-z)-1y=1/(λ-z)y. Conversely, if yL2(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 yL2(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=yg=y+zλy. Hence (44)(λy,y+zλy)𝒯α,β(y,(z+1λy))𝒯α,β.

By Lemma 13, we see that there is a one-to-one correspondence of eigenvalues (eigenfunctions) for the operator and (𝒯α,β-z)-1. As is compact operator, it has only discrete spectrum consisting of only eigenvalues. Since (𝒯α,β-z)-1 is unitarily equivalent to R(V); that is, (45)V-1R(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:uα1(N,z)sinβ+uα2cosβ=0}.

We would like to extend this idea over the half line +. First note that we are considering the limit circle case of the system (1). That implies that for any z+ the defect indices of 0 are (2,2). Suppose pD()D(0) such that limxp(x)*Jp(x)=0. Such function clearly exists.

Consider the following relation: (47)𝒯α,p={limxf(x)*Jp(x)=0(f,g):f1(0)sinα+f2(0,z)cosα=0  andhhhlimxf(x)*Jp(x)=0}.

Lemma 15.

𝒯 α , p defines a self-adjoint extension of 0.

Proof.

Let u(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, limxu0(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,p0D(0) and are linearly independent. Clearly D(0)+L(u0,p0)D(𝒯α,p). Since D(𝒯α,p) is at least 2-dimensional restriction of D(), (48)D(𝒯α,p)=D(0)+L(u0,p0). Hence 𝒯α,p is a self-adjoint relation.

We next discuss the spectrum of 𝒯α,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+ and as above write f(x,z)=u(x,z)+m(z)v(x,z)L2(H,+) satisfying limxf(x,z)*JP(x)=0. Let T(x,z)=(u1v1u2v2) and (50)wα(x,z)=1sinα+m(z)cosαT(x,z)(cosα-sinα). 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.

Let zρ(𝒯α,p); then the resolvent operator (𝒯α,p-z)-1 is given by (52)(𝒯α,p-z)-1h(x)=0G(x,t,z)H(t)h(t)dt, where (53)G(x,t,z)={f(x,z)wα(t,z-)*if  0<txwα(t,z-)f(x,z-)if  x<t.

Proof.

Let y(x,z)=0G(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+xwα(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 gD(𝒯α,p), (56)g1(0,z)sinα+g2(0,z)cosα=0,limxg*(x,z)Jp(x,z)=0. We also have limxf*(x,z)Jg(x,z)=0 and g(0,z)=(cosα-sinα)c(z) for some scalar c(z). But we also have (57)y(0,z)=1(m(z)cosα+sinα)(cosαsinα)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)-limxf*(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 map V:L2(H,+)L2(I,+) by Vy=H1/2(x)y(x).   V is isometry and maps unitarily onto the range R(V)L2(I,+).

Define an integral operator on L2(I,+) by (59)(g)(x)=0L(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)0L*L<. Hence 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 + and the proofs are exactly the same as the proofs of Lemmas 13 and 14.

Lemma 17 (see [<xref ref-type="bibr" rid="B8">3</xref>]).

Let fL2(I,R+), λ0, and then the following statements are equivalent.

f=λ-1f.

fR(V), and the unique yL2(H,R+) with Vy=f solves (𝒯α,p-z)-1y=λy.

Lemma 18.

Let z. For any λz, if (y,λy)𝒯α,p, then y solves (𝒯α,p-z)-1y=(1/(λ-z))y. Conversely, if yL2(H,R+) and y solves (Tα,p-z)-1y=λy, then (y,(z+1/λ)y)𝒯α,p.

Again, by Lemma 17, we have a one-to-one correspondence of eigenvalues (eigenfunctions) for the operator and (𝒯α,p-z)-1. As is compact operator, it has only discrete spectrum consisting of only eigenvalues and possibly zero. Since (𝒯α,p-z)-1 is unitarily equivalent to R(V); that is, V-1R(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 Theorem <xref ref-type="statement" rid="thm1.1">1</xref>.

Since 0 is a symmetric relation, the defect index β(0,z) is constant on upper and lower half-planes. In the limit circle case, if z is in upper or lower half-planes, β(0,z)=2. Suppose β(0,λ)<2 for some λ. Since Γ(0) is open, λΓ(0), and hence λS(0). Since, for each α(0,π], 𝒯α,p is self-adjoint extension of 0, λS(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(0,λ)=2.

Proof of Theorem <xref ref-type="statement" rid="thm1.2">2</xref>.

Suppose it prevails the limit circle case. By Theorem 1, the defect index β(0,z)=dimN(,z-)=2 for all z. In other words, for any z, all solutions of (1) are in L2(H,+). In particular, the constant solutions u(x)=(10) and v(x)=(01) of (1) when z=0 are in L2(H,+). But this is not possible because 0u(x)*H(x)u(x)dx+0v(x)*H(x)v(x)dx=0trH(x)dx=.

Conflict of Interests

The author declares that there is no conflict of interests regarding the publication of this paper.

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