#### 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 on . This provides an alternative proof of the De Branges theorem that the canonical systems with 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: Here and is a positive semidefinite matrix whose entries are locally integrable. For fixed , a function is called a solution if is absolutely continuous and satisfies (1). Consider the Hilbert space as follows: provided with an inner product .

The canonical systems (1) on 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 . 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 then we may consider this as an eigenvalue equation of an operator on . But 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 .

For some , if the canonical system (1) has all solutions in , we say that the system is in the limit circle case, and if the system has unique solution in , 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 of the symmetric relation , 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 prevail the limit point case.*

Theorem 2 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].

Let be a Hilbert space over and denote by the Hilbert space . A linear relation on is a subspace of . The adjoint of on is a closed linear relation defined by A linear relation is called symmetric if and self-adjoint if . The theory of such relations can be found in [5–8]. The regularity domain of is the following set: The following theorem has been derived from [6].

Theorem 3. * Suppose is a self-adjoint relation and suppose ; then
*

The defect index is the dimension of defect space: It has been shown in [6] that the defect index is constant on each connected subset of . 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 [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 relation possesses self-adjoint extension if and only if on lower and upper half-planes is equal.**(2) A symmetric extension of is self-adjoint if and only if is an -dimensional extension of .*

The resolvent set for a closed relation is the following set: and the spectrum of is .

We call the* spectral kernel* of . The following theorem from [6] shows the relation between the spectral kernel and spectrum of a self-adjoint relation.

Theorem 5. *If is a self-adjoint relation and , then one has the following* * .* * If , then .* * . *

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 and Proof of the Main Theorems

Consider that a relation in the Hilbert space is induced by (1) as and is called the maximal relation. This relation is made up of pairs of equivalence classes , such that there exists a locally absolutely continuous representative of again denoted by and a representative of , again denoted by , such that a.e. on . The adjoint relation is defined by and is called the minimal relation. It has been shown in [1] that is close and symmetric. Moreover, and .

Lemma 6 (see [1]). *For each there exists such that , have compact support and , .*

Lemma 7 (see [1]). *Let . Then the following limit exists:
*

Lemma 8. *The minimal relation is given by
*

*Proof. *By Lemma 7, we get
On the other hand, let . By Lemma 6 for any there exists such that has compact support and . So
This implies that . This would also force the following:

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

*Remark 9. *The defect index of the minimal relation is equal to the number of linearly independent solutions of the system (1) whose class lies in . Therefore, in the limit circle case, the defect indices of are .

Since has equal defect indices, by Theorem 4, it has self-adjoint extensions say . Consider a relation as follows: on a compact interval .

Lemma 10. * is a self-adjoint relation.*

*Proof. *Clearly is a symmetric relation because of the boundary conditions at and . We will show that is a -dimensional extension of . Then by Theorem 4, is a self-adjoint relation. By Lemma 6, for and , there exist and in such that and and the support of and is contained in . Clearly and are linearly independent but and, are not in . This shows that . Because of the boundary condition at and , is -dimensional restriction of . Hence . Therefore, is -dimensional extension of so that is a self-adjoint relation.

Let and be the solution of the canonical system (1) on with the initial values: For there is a unique such that satisfying the following: This is well defined because does not satisfy the boundary condition at ; otherwise, will be an eigenvalue of some self-adjoint relation .

Next, we describe the spectrum of . Let and define It is not hard to see that.

Lemma 11. *Using the notation above one has
*

Lemma 12. *Let ; then is a bounded linear operator and is defined by
**
where
*

*Proof. *Let . We show that solves the inhomogeneous equation as follows:
for . Here
and , . Then on differentiation we get
Then

On the other hand denote as
then by Theorem 3, for some so that . So also satisfies the inhomogeneous problem and ; it satisfies the boundary condition which implies that for some scalar . We have
Now
Since both and satisfy the same boundary condition at , now
So
By uniqueness we must have . Moreover, is a bounded linear operator.

Now define a map by Here is the unique positive semidefinite square root of . Then is an isometry and hence maps unitarily onto the range . Define an integral operator on as The kernel is square integrable since So is a Hilbert-Schmidt operator and thus compact. Notice that . This implies that is self-adjoint.

Lemma 13 (see [3]). *Let , , and then the following statements are equivalent. *(1)*. *(2)*, and the unique with solves .*

*Proof. *For all we have
lies in . Then . Now if (1) holds, then . So for unique and
for . In other words,
Conversely if (11) holds,
And then

Lemma 14. *Let . For any , if , then solves . Conversely, if and solves , then .*

*Proof. *Let ; then . It follows that
This means that solves
Conversely suppose and solves
That is, so that . So there is such that and
Hence

By Lemma 13, we see that there is a one-to-one correspondence of eigenvalues (eigenfunctions) for the operator and . As is compact operator, it has only discrete spectrum consisting of only eigenvalues. Since is unitarily equivalent to ; that is, has only discrete spectrum consisting of only eigenvalues. Then, by Theorem 5, has only discrete spectrum. By Lemma 14, the spectrum of consists only of eigenvalues. Hence we have

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 the defect indices of are . Suppose such that . Such function clearly exists.

Consider the following relation:

Lemma 15. * defines a self-adjoint extension of .*

*Proof. *Let be a solution of the system (1) with some initial values and as above. Define near the neighborhood of ; that is, , and otherwise. Similarly, in the neighborhood of 0 and otherwise. Then clearly and are linearly independent. Clearly . Since is at least -dimensional restriction of ,
Hence is a self-adjoint relation.

We next discuss the spectrum of . Let and be two linearly independent solutions of the system (1) with Let and as above write satisfying . Let and Then as in Lemma 11 we have

Lemma 16. *Let ; then the resolvent operator is given by
**
where
*

*Proof. *Let ; then solves the following inhomogeneous equation:
This is clear by differentiating
On the other hand denote as ; then, by Theorem 3, for some so that , and hence satisfies the inhomogeneous equation. Since ,
We also have and for some scalar . But we also have
Here
Hence . By uniqueness we have .

Now define a map by . is isometry and maps unitarily onto the range .

Define an integral operator on by Then as before the kernel is square integrable. This means that Hence is a Hilbert Schmidt operator and so it is a compact operator. The following two lemmas are extended from the bounded interval to and the proofs are exactly the same as the proofs of Lemmas 13 and 14.

Lemma 17 (see [3]). *Let , , and then the following statements are equivalent. *(1)*. *(2)*, and the unique with solves .*

Lemma 18. *Let . For any , if , then solves . Conversely, if and solves , then .*

Again, by Lemma 17, we have a one-to-one correspondence of eigenvalues (eigenfunctions) for the operator and . As is compact operator, it has only discrete spectrum consisting of only eigenvalues and possibly zero. Since is unitarily equivalent to ; that is, has only discrete spectrum consisting of only eigenvalues. Then, by Theorem 5, has only discrete spectrum. By Lemma 18 the spectrum of consists of only eigenvalues.

With these theories in hand, we are now ready to prove the main theorems.

*Proof of Theorem 1. *Since is a symmetric relation, the defect index is constant on upper and lower half-planes. In the limit circle case, if is in upper or lower half-planes, . Suppose for some . Since is open, , and hence . Since, for each , is self-adjoint extension of , . In the limit circle case, consists of only eigenvalues. Therefore, is an eigenvalue for all boundary conditions at . However, this is impossible unless .

*Proof of Theorem 2. *Suppose it prevails the limit circle case. By Theorem 1, the defect index for all . In other words, for any , all solutions of (1) are in . In particular, the constant solutions and of (1) when are in . But this is not possible because .

#### Conflict of Interests

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