Self-Adjoint Extension and Spectral Theory of a Linear Relation in a Hilbert Space
The aim of this paper is to develop the conditions for a symmetric relation in a Hilbert space ℋ to have self-adjoint extensions in terms of defect indices and discuss some spectral theory of such linear relation.
In this paper, we discuss the theory of linear relations in a Hilbert space. These linear relations were first studied by Arnes, Coddington, Dijksma, de Snoo, and Hassi et al in [1–4]. It has also been studied extensively more recently in . The theory has particular interest because, in some of the application problems, a linear operator can have multivalued part; for example, see [6, 7]. Here, we concentrate on establishing the conditions for symmetric relations to have self-adjoint extensions in terms of defect indices. Moreover, we discuss the spectral theory of such self-adjoint relations. The analogous treatment on operator theory of some of the theorems on this paper can be found in .
Let be a Hilbert space over and denote by the Hilbert space . A linear relation on is a subspace of . The graph of an operator is an example of a linear relation but note that a relation can have multivalued part.
These relations have been used in some of the eigenvalue problems in ordinary differential equations. For example, the canonical systems where and is a positive semidefinite matrix whose entries are locally integrable, induce a multivalued linear relation. For instance, we may think of writing the systems in the form and consider it as an operator on a Hilbert space. But is not invertible in general therefore can not be considered as an eigenvalue equation of an operator. Instead, the system induces a linear relation that may have a multivalued part. This is one of the main motivations for our work in this paper. The boundary value problem of such canonical systems has been studied by using linear relations; see [6, 7, 9].
and are respectively defined as the domain and range of the relation .
denotes the inverse relation. The adjoint of on is a closed linear relation defined by A linear relation is called symmetric if and self-adjoint if . From now on, we write relation to mean linear relation.
A relation is called isometry if and is unitary if it is isometry and .
Let and .
It is clear to see that .
2. Defect Indices and Self-Adjoint Extension
Let be a relation on a Hilbert space . The set is defined as the regularity domain of and is defined as the Spectral Kernel of . Note that(i) if and only if is a bounded linear operator on ;(ii)if is symmetric, then ;(iii) is open.
The subspace is called the defect space of and . The cardinal number is called the defect index of and .
Theorem 1. The defect index is constant on each connected subset of . If is symmetric, then the defect index is constant on the upper and lower half-planes.
Proof. Let denote the orthogonal projection onto . We first show that as , for any . Let ; then there is a constant such that for all . For and all , we have For , Similarly, for , It follows that Let denote an orthogonal projection onto ; then Hence, if we choose , then , for . Therefore, . It follows that If is symmetric, then the upper and lower half-planes are connected subsets of ; therefore, the defect index is constant there.
Letting be a symmetric relation on a Hilbert space, for , the defect index and, for , the defect index are written as a pair and are called the defect indices of .
The Cayley transform of a symmetric relation on is defined by the relation Then clearly and .
Theorem 2. If is a symmetric relation on and is the Cayley transform of , then, (1) is isometry;(2) and ;(3) is multi-valued if and only if .
Proof. Let ; then and , for ; then
It follows from the definition of Cayley transform.
Suppose that is multivalued; then there is such that . It follows by definition of that . Hence, . On the other hand, let ; then ; then, by , . Hence, is multivalued.
Theorem 3. A relation on is the Cayley transform of a symmetric relation if and only if has the following properties. (1) is an isometric relation.(2). The relation is given by .
Proof. If is the Cayley transform of , then, by Theorem 2, satisfies the properties and . Conversely, supposing that has properties and , we show that is a symmetric relation.
Suppose ; then Since is an isometry, for any , this implies that Hence, is symmetric.
Theorem 4. A symmetric relation is self-adjoint if and only if is unitary.
Proof. We show that the is self-adjoint if and only if
Since is symmetric, we always have . Let ; then and imply that there is such that . So , so that . That is
This implies . Hence, is self-adjoint.
Conversely, suppose that is self-adjoint. Let So . But Hence, we must have . So . Similarly, .
Theorem 5. Let be a closed symmetric relation on a Hilbert space and let denote its Cayley transform. One has the following.(1) is the Cayley transform of a closed symmetric extension of if and only if the following holds. There exist closed subspaces of and of and an isometric relation on for which The spaces and have the same dimension.(2)The relation in part is unitary if and only if and .(3) possess self-adjoint extension if and only if its defect indices are equal.
Proof. Suppose that has the given form. Then is isometric relation, since for any , we have
Hence, we can define a symmetric extension such that is its Cayley transform. Conversely, if is the Cayley transform of a symmetric extension of , then put and . Then we have the desired properties.
Here we have that is unitary if and only if That is, if and only if and .
By and , possess unitary extension if and only if there exists an isometry relation onto . This happens if and only if
By definition of Cayley transform, it is clear that if and are the Cayley transforms of any two symmetric relations and , then
Theorem 6. Let be a closed symmetric relation on a Hilbert space with defect indices . One has the following.(1) is a symmetric extension of if and only if the following holds. There are closed subspaces of and of and an isometric mapping of onto such that (2) is self-adjoint if and only if is an -dimensional extension of .
Proof. Let and be the Cayley transforms of the closed symmetric relation and its symmetric extension , respectively. By Theorem 5, there exist closed subspaces of and of and an isometric relation on for which
Then by definition of the Cayley transform, we see that
The converse is similar.
By Theorem 5, is self-adjoint if and only if is unitary. This happens if and only if . So, by , is self-adjoint if and only if it is an -dimensional extension of .
Theorem 7. Suppose that is a self-adjoint relation and suppose that ; then
Proof. We will show that is a closed subspace of . Since , there is a constant such that Let and . Suppose that such that and so that . But from the above relation we have It follows that is a Cauchy sequence in , and it converges to some in . Hence, . Since is closed, and and . Hence, is closed. So we have We next show that . Let ; then . But implies a. e..
Let be a self-adjoint relation on and . Define by . That is . Then is a bounded linear operator since and is given by
3. Spectral Theory of a Linear Relation
Definition 8. Let be a closed relation on a Hilbert space . We define to be the resolvent set and to be the spectrum of .
Remark 9. When a relation is an operator on , then
A complex number is called an eigenvalue of a relation if there exists such that . The set of all eigenvalues of is called the point spectrum of and is denoted by .
Remark 10. For any closed relation on a Hilbert space , .
Let and be the multivalued part of a relation . Clearly is a closed subspace of . Note that is not dense if is multivalued. Now define the quotient space . We know that this quotient space is also a Hilbert space with the norm defined by Define a relation on by . We consider the relation as the restriction of on . By natural isomorphism, the space is identified as and the relation as . Then clearly is an operator on with .
Theorem 11. If is a self-adjoint relation on , then
Proof. Letting , then there exists a constant such that
For such , we can define as a bounded linear operator on such that . So . On the other hand, let ; then there exists such that . For any , there is such that and . So
for some and hence . Hence, .
Next, assuming that , then for any , there exists a constant such that For any we have Hence . On the other hand, suppose that , then there is a constant such that For any , we have This implies that . Thus, . Hence, .
Remark 12. If is a self-adjoint relation on , then .
Theorem 13. Let and . One has the following.(1)If , then .(2)If , then .(3).
Proof. Let ; then by definition there exists such that
Note that . For any , we have
Let and suppose that . Then . But by . This implies that which is a contradiction.
Let ; then is bounded and is defined on all of ; then for any . This shows that .
Note. We may think of developing further the spectral theory of linear relation analogous to that of operator theory. Any extension would be useful in many of the application problems which induce linear relation instead of linear operator.
Conflict of Interests
The author declares that there is no conflict of interests regarding the publication of this paper.
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