Self-Adjoint Extension and Spectral Theory of a Linear Relation in a Hilbert Space

Abstract

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.

1. Introduction

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 [5]. 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 [8].

Let ℋ 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 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 𝐽𝑢^′(𝑥)=𝑧𝐻(𝑥)𝑢(𝑥),𝑧∈ℂ, (1) where 𝐽=⁰⁻¹₁₀ and 𝐻(𝑥) is a 2×2 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 𝐻(𝑥)⁻¹𝐽𝑢^′=𝑧𝑢 (2) and consider it as an operator on a Hilbert space. But 𝐻(𝑥) 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].

𝐷(ℛ)={𝑓∈ℋ∶(𝑓,𝑔)∈ℛ} and 𝑅(ℛ)={𝑔∈ℋ∶(𝑓,𝑔)∈ℛ} are respectively defined as the domain and range of the relation ℛ.

ℛ⁻¹={(𝑔,𝑓)∶(𝑓,𝑔)∈ℛ} denotes the inverse relation. The adjoint of ℛ on ℋ is a closed linear relation defined by ℛ^∗=(ℎ,𝑘)∈ℋ²∶𝑔,ℎ=𝑓,𝑘,∀𝑓,𝑔∈ℛ. (3) A linear relation 𝒮 is called symmetric if 𝒮⊂𝒮^∗ and self-adjoint if 𝒮=𝒮^∗. From now on, we write relation to mean linear relation.

A relation ℛ is called isometry if 𝑓₁,𝑓₂=𝑔₁,𝑔₂,∀𝑓₁,𝑔₁,𝑓₂,𝑔₂∈ℛ (4) and ℛ is unitary if it is isometry and 𝐷(ℛ)=𝑅(ℛ)=ℋ.

Let (𝑧−ℛ)={(𝑓,𝑧𝑓−𝑔)∶(𝑓,𝑔)∈ℛ},𝑁(ℛ,𝑧)={𝑓∶(𝑓,𝑧𝑓)∈ℛ} and ℛ_𝑧=𝑅(𝑧−ℛ).

It is clear to see that 𝑁(ℛ^∗,𝑧)=ℛ^⟂_𝑧.

In Section 2, we establish the condition for a symmetric relation to have self-adjoint extensions in terms of defect indices and in Section 3 we discuss spectral theory.

2. Defect Indices and Self-Adjoint Extension

Let ℛ be a relation on a Hilbert space ℋ. The set Γ(ℛ)={(𝑧)(𝑧𝑓𝑧∈ℂ∶thereexistsa𝐶>0suchthat𝑧𝑓−𝑔≥𝐶),∀𝑓,𝑔∈ℛ (5) is defined as the regularity domain of ℛ and 𝑆(ℛ)=ℂ−Γ(ℛ) is defined as the Spectral Kernel of ℛ. Note that(i)𝑧∈Γ(ℛ) if and only if (𝑧−ℛ)⁻¹ is a bounded linear operator on ℋ;(ii)if ℛ is symmetric, then ℂ−ℝ⊂Γ(ℛ);(iii)Γ(ℛ) is open.

The subspace ℛ^⟂_𝑧 is called the defect space of ℛ and 𝑧. The cardinal number 𝛽(ℛ,𝑧)=dimℛ^⟂_𝑧 is called the defect index of ℛ and 𝑧.

Theorem 1. The defect index 𝛽(ℛ,𝑧) is constant on each connected subset of Γ(ℛ). If ℛ is symmetric, then the defect index is constant on the upper and lower half-planes.

Proof. Let 𝑄_𝑧 denote the orthogonal projection onto ℛ_𝑧. We first show that ‖𝑄_𝑧−𝑄_𝑧0‖→0 as 𝑧→𝑧₀, for any 𝑧₀∈Γ(ℛ). Let 𝑧₀∈Γ(ℛ); then there is a constant 𝐶(𝑧₀)>0 such that 𝑓≤𝐶𝑧₀𝑧₀𝑓−𝑔, (6) for all (𝑓,𝑔)∈ℛ. For |𝑧−𝑧₀|<1/(2𝐶(𝑧₀)) and all (𝑓,𝑔)∈ℛ, we have 𝑓≤𝐶𝑧₀𝑧₀𝑓−𝑔≤𝑧𝑓−𝑔+𝑧−𝑧₀𝑓≤𝐶𝑧₀+1𝑧𝑓−𝑔2𝑓∴𝑓≤𝐶𝑧₀𝑧𝑓−𝑔. (7) For ℎ∈𝑅(𝑄_𝑧0)^⟂=ℛ^⟂_𝑧0, 𝑄_𝑧ℎ=supℎ,𝑧𝑓−𝑔∶𝑧𝑓−𝑔∈ℛ_𝑧,𝑧𝑓−𝑔≤1=supℎ,𝑧−𝑧₀𝑓∶𝑧𝑓−𝑔∈ℛ_𝑧,.∴𝑄𝑧𝑓−𝑔≤1_𝑧ℎ≤‖ℎ‖𝑧−𝑧₀𝐶𝑧₀. (8) Similarly, for ℎ∈𝑅(𝑄_𝑧)^⟂=ℛ^⟂_𝑧, 𝑄_𝑧0ℎ≤‖ℎ‖𝑧−𝑧₀𝐶𝑧₀. (9) It follows that 𝑄_𝑧−𝑄_𝑧0≤2𝑧−𝑧₀𝐶𝑧₀. (10) Let 𝑃_𝑧 denote an orthogonal projection onto ℛ^⟂_𝑧; then 𝑝_𝑧−𝑝_𝑧0=𝑄_𝑧−𝑄_𝑧0≤2𝑧−𝑧₀𝐶𝑧₀⟶0as𝑧⟶𝑧₀. (11) Hence, if we choose 0<𝜖<1/(2𝐶(𝑧₀)), then ‖𝑝_𝑧−𝑝_𝑧0‖<1, for |𝑧−𝑧₀|<𝜖. Therefore, dimℛ^⟂_𝑧=dimℛ^⟂_𝑧0. It follows that 𝛽(ℛ,𝑧)=𝛽ℛ,𝑧₀for𝑧−𝑧₀<𝜖. (12) If ℛ is symmetric, then the upper and lower half-planes are connected subsets of Γ(ℛ); therefore, the defect index is constant there.

Letting ℛ be a symmetric relation on a Hilbert space, for 𝑧∈ℂ⁺, the defect index 𝑚=𝛽(ℛ,𝑧) and, for 𝑤∈ℂ⁻, the defect index 𝑛=𝛽(ℛ,𝑤) are written as a pair (𝑚,𝑛) and are called the defect indices of ℛ.

The Cayley transform of a symmetric relation ℛ on ℋ is defined by the relation 𝒱=𝑔+𝑖𝑓,𝑔−𝑖𝑓∶𝑓,𝑔∈ℛ. (13) Then clearly 𝐷(𝒱)=𝑅(ℛ+𝑖) and 𝑅(𝒱)=𝑅(ℛ−𝑖).

Theorem 2. If ℛ is a symmetric relation on ℋ and 𝒱 is the Cayley transform of ℛ, then, (1)𝒱 is isometry;(2)𝑅(𝐼−𝒱)=𝐷(ℛ) and ℛ={(𝑓−𝑔,𝑖(𝑓+𝑔))∶(𝑓,𝑔)∈𝒱};(3)ℛ is multi-valued if and only if 𝑁(𝐼−𝒱)≠{0}.

Proof. (1) Let (𝑢₁,𝑣₁),(𝑢₂,𝑣₂)∈𝒱; then 𝑢_𝑖=𝑔_𝑖+𝑖𝑓_𝑖 and 𝑣_𝑖=𝑔_𝑖−𝑖𝑓_𝑖, for (𝑓_𝑖,𝑔_𝑖)∈ℛ,𝑖=1,2; then 𝑢₁,𝑢₂=𝑔₁+𝑖𝑓₁,𝑔₂+𝑖𝑓₂=𝑔₁,𝑔₂+𝑔₁,𝑖𝑓₂+𝑖𝑓₁,𝑔₂+𝑖𝑓₁,𝑖𝑓₂=𝑔₁,𝑔₂+𝑖𝑔₁,𝑓₂−𝑖𝑓₁,𝑔₂−𝑖²𝑓₁,𝑓₂=𝑔₁−𝑖𝑓₁,𝑔₂−𝑖𝑓₂=𝑣₁,𝑣₂. (14)
(2) It follows from the definition of Cayley transform.
(3) Suppose that ℛ is multivalued; then there is 𝑔∈ℋ,𝑔≠0 such that (0,𝑔)∈ℛ. It follows by definition of 𝒱 that (𝑔,𝑔)∈𝒱. Hence, 𝑔∈𝑁(𝐼−𝒱). On the other hand, let 𝑔∈𝑁(𝐼−𝒱),𝑔≠0; then (𝑔,𝑔)∈𝒱; then, by (2), (0,2𝑖𝑔)∈ℛ. Hence, ℛ is multivalued.

Theorem 3. A relation 𝒱 on ℋ is the Cayley transform of a symmetric relation ℛ if and only if 𝒱 has the following properties. (1)𝒱 is an isometric relation.(2)𝑅(𝐼−𝒱)=𝐷(ℛ). The relation ℛ is given by ℛ={(𝑓−𝑔,𝑖(𝑓+𝑔))∶(𝑓,𝑔)∈𝒱}.

Proof. If 𝒱 is the Cayley transform of ℛ, then, by Theorem 2, 𝒱 satisfies the properties (1) and (2). Conversely, supposing that 𝒱 has properties (1) and (2), we show that ℛ={(𝑓−𝑔,𝑖(𝑓+𝑔))∶(𝑓,𝑔)∈𝒱} is a symmetric relation.
Suppose (𝑓₁−𝑔₁,𝑖(𝑓₁+𝑔₁)),(𝑓₂−𝑔₂,𝑖(𝑓₂+𝑔₂))∈ℛ; then 𝑖𝑓₁+𝑔₁,𝑓₂−𝑔₂=−𝑖𝑓₁,𝑓₂−𝑓₁,𝑔₂+𝑔₁,𝑓₂−⟨𝑔₁,𝑔₂⟩. (15) Since 𝒱 is an isometry, for any (𝑓₁,𝑔₁),(𝑓₂,𝑔₂)∈𝒱,⟨𝑓₁,𝑓₂⟩=⟨𝑔₁,𝑔₂⟩, this implies that 𝑖𝑓₁+𝑔₁,𝑓₂−𝑔₂=−𝑖𝑔₁,𝑔₂+𝑖𝑓₁,𝑔₂−𝑖𝑔₁,𝑓₂+𝑖𝑓₁,𝑓₂=−𝑖𝑔₁−𝑓₁,𝑔₂+𝑔₁−𝑓₁,𝑓₂=𝑓₁−𝑔₁,𝑖𝑓₂+𝑔₂. (16) Hence, ℛ is symmetric.

Theorem 4. A symmetric relation ℛ is self-adjoint if and only if 𝒱 is unitary.

Proof. We show that the ℛ is self-adjoint if and only if 𝑅(ℛ+𝑖)=𝑅(ℛ−𝑖)=ℋ. (17) Since ℛ is symmetric, we always have ℛ⊂ℛ^∗. Let (𝑓,𝑔)∈ℛ^∗; then 𝑖𝑓−𝑔∈ℋ and 𝑅(ℛ−𝑖)=ℋ imply that there is (ℎ,𝑘)∈ℛ such that 𝑘−𝑖ℎ=𝑖𝑓−𝑔. So 𝑖(𝑓+ℎ)=𝑘+𝑔, so that (𝑓+ℎ,𝑖(𝑓+ℎ))∈ℛ^∗. That is 𝑓+ℎ∈𝑁ℛ^∗,𝑖=𝑅(ℛ+𝑖)^⟂={0}. (18) This implies 𝑓=−ℎ∈𝐷(ℛ). Hence, ℛ is self-adjoint.
Conversely, suppose that ℛ is self-adjoint. Let ℎ∈𝑅(ℛ−𝑖)^⟂=𝑁ℛ^∗,−𝑖=𝑁(ℛ,−𝑖). (19) So (ℎ,−𝑖ℎ)∈ℛ. But ⟨−𝑖ℎ,ℎ⟩=⟨ℎ,𝑖ℎ⟩⟹𝑖⟨ℎ,ℎ⟩=−𝑖⟨ℎ,ℎ⟩. (20) Hence, we must have ℎ=0. So 𝑅(ℛ−𝑖)=ℋ. Similarly, 𝑅(ℛ+𝑖)=ℋ.

Theorem 5. Let ℛ be a closed symmetric relation on a Hilbert space ℋ and let 𝒱 denote its Cayley transform. One has the following.(1)𝒱^′ is the Cayley transform of a closed symmetric extension ℛ^′ of ℛ if and only if the following holds. There exist closed subspaces 𝐹₋ of 𝑅(ℛ−𝑖)^⟂ and 𝐹₊ of 𝑅(ℛ+𝑖)^⟂ and an isometric relation 𝒱 on F₊⊕𝐹₋ for which 𝒱^′=∶()∈𝒱,𝐷𝒱𝑓+ℎ,𝑔+𝑘𝑓,𝑔∈𝒱,ℎ,𝑘_′=𝑅ℛ^′+𝑖=𝑅(ℛ+𝑖)⊕𝐹₊,𝑅𝒱^′=𝑅ℛ^′−𝑖=𝑅(ℛ−𝑖)⊕𝐹₋. (21) The spaces 𝐹₊ and 𝐹₋ have the same dimension.(2)The relation 𝒱^′ in part (1) is unitary if and only if 𝐹₋=𝑅(ℛ−𝑖)^⟂ and 𝐹₊=𝑅(ℛ+𝑖)^⟂.(3)ℛ possess self-adjoint extension if and only if its defect indices are equal.

Proof. (1) Suppose that 𝒱^′ has the given form. Then 𝒱^′ is isometric relation, since for any (𝑓+ℎ,𝑔+𝑘)∈𝒱^′, we have 𝑔+𝑘²=𝑔²+‖𝑘‖²=𝑓²+‖ℎ‖²=𝑓+ℎ². (22) Hence, we can define a symmetric extension ℛ^′ such that 𝒱^′ is its Cayley transform. Conversely, if 𝒱^′ is the Cayley transform of a symmetric extension ℛ^′ of ℛ, then put 𝐹₋=𝑅(ℛ^′−𝑖)⊖𝑅(ℛ−𝑖),𝐹₊=𝑅(ℛ^′+𝑖)⊖𝑅(ℛ+𝑖) and 𝒱=𝒱^′|_{𝐹+⊕𝐹−}. Then we have the desired properties.
(2) Here we have that 𝒱^′ is unitary if and only if 𝐷𝒱^′=𝑅𝒱^′=ℋ. (23) That is, if and only if 𝐹₋=𝑅(ℛ−𝑖)^⟂ and 𝐹₊=𝑅(ℛ+𝑖)^⟂.
(3) By (1) and (2), 𝒱 possess unitary extension if and only if there exists an isometry relation 𝒱 onto 𝑅(ℛ+𝑖)^⟂⊕𝑅(ℛ−𝑖)^⟂. This happens if and only if dim𝑅(ℛ+𝑖)^⟂=dim𝑅(ℛ−𝑖)^⟂. (24)

By definition of Cayley transform, it is clear that if 𝒱₁ and 𝒱₂ are the Cayley transforms of any two symmetric relations ℛ₁ and ℛ₂, then ℛ₁⊂ℛ₂ifandonlyif𝒱₁⊂𝒱₂. (25)

Theorem 6. Let ℛ be a closed symmetric relation on a Hilbert space with defect indices (𝑚,𝑚). One has the following.(1)ℛ^′ is a symmetric extension of ℛ if and only if the following holds. There are closed subspaces 𝐹₊ of 𝑅(𝑖+ℛ)^⟂ and 𝐹₋ of 𝑅(𝑖−ℛ)^⟂ and an isometric mapping 𝒱 of 𝐹₊ onto 𝐹₋ such that 𝐷ℛ^′=𝐷(ℛ)+𝑔+𝒱𝑔∶𝑔∈𝐹₊. (26)(2)ℛ^′ is self-adjoint if and only if ℛ^′ is an 𝑚-dimensional extension of ℛ.

Proof. (1) Let 𝒱 and 𝒱^′ be the Cayley transforms of the closed symmetric relation ℛ and its symmetric extension ℛ^′, respectively. By Theorem 5, there exist closed subspaces 𝐹₋ of 𝑅(ℛ−𝑖)^⟂ and 𝐹₊ of 𝑅(ℛ+𝑖)^⟂ and an isometric relation 𝒱 on 𝐹₊⊕𝐹₋ for which 𝒱^′=∶()∈𝒱,𝐷𝒱𝑓+ℎ,𝑔+𝑘𝑓,𝑔∈𝒱,ℎ,𝑘_′=𝑅ℛ^′+𝑖=𝑅(ℛ+𝑖)⊕𝐹₊,𝑅𝒱^′=𝑅ℛ^′−𝑖=𝑅(ℛ−𝑖)⊕𝐹₋. (27) Then by definition of the Cayley transform, we see that 𝐷ℛ^′=𝑅𝐼−𝒱^′=𝐼−𝒱^′𝐷𝒱^′=𝐼−𝒱^′𝑅𝑖+ℛ^′=𝐼−𝒱^′𝑅(𝑖+ℛ)⊕𝐹₊=𝐼−𝒱^′𝐷(𝒱)⊕𝐹₊=(𝐼−𝒱)𝐷(𝒱)+𝐼−𝒱^′𝐹₊=𝐷(ℛ)+𝑔−𝒱𝑔∶𝑔∈𝐹₊. (28) The converse is similar.
(2) By Theorem 5, ℛ^′ is self-adjoint if and only if 𝒱^′ is unitary. This happens if and only if 𝐹₊=𝑅(ℛ+𝑖)^⟂. So, by (1), ℛ^′ is self-adjoint if and only if it is an 𝑚-dimensional extension of ℛ.

Theorem 7. Suppose that 𝒯 is a self-adjoint relation and suppose that 𝑧∈Γ(𝒯); then ℋ=𝑧𝑓−𝑔∶𝑓,𝑔∈𝒯. (29)

Proof. We will show that 𝑅(𝑧−𝒯)={𝑧𝑓−𝑔∶(𝑓,𝑔)∈𝒯} is a closed subspace of ℋ. Since 𝑧∈Γ(𝒯), there is a constant 𝐶(𝑧) such that 𝑧𝑓−𝑔≥𝐶(𝑧)𝑓. (30) Let 𝑣_𝑛∈𝑅(𝑧−𝒯) and 𝑣_𝑛→𝑣inℋ. Suppose that 𝑓_𝑛∈𝐷(𝒯) such that (𝑓_𝑛,𝑔_𝑛)∈𝒯 and 𝑣_𝑛=𝑧𝑓_𝑛−𝑔_𝑛 so that (𝑓_𝑛,𝑣_𝑛)∈𝑧−𝒯. But from the above relation we have 𝑣_𝑛−𝑣_𝑚=𝑧𝑓_𝑛−𝑓_𝑚−𝑔_𝑛−𝑔_𝑚≥𝐶𝑓_𝑛−𝑓_𝑚. (31) It follows that 𝑓_𝑛 is a Cauchy sequence in ℋ, and it converges to some 𝑓 in ℋ. Hence, (𝑓_𝑛,𝑣_𝑛)→(𝑓,𝑣). Since 𝒯 is closed, 𝑓∈𝐷(𝒯) and (𝑓,𝑣)∈𝑧−𝒯 and 𝑣∈𝑅(𝑧−𝒯). Hence, 𝑅(𝑧−𝒯) is closed. So we have ℋ=𝑅(𝑧−𝒯)⊕𝑅(𝑧−𝒯)^⟂. (32) We next show that 𝑅(𝑧−𝒯)^⟂={0}. Let ℎ∈𝑅(𝑧−𝒯)^⟂=𝑁(𝒯,𝑧); then (ℎ,𝑧ℎ)∈𝒯. But 0=‖𝑧ℎ−𝑧ℎ‖≥𝐶(𝑧)‖ℎ‖ implies ℎ=0 a. e..

Let 𝒯 be a self-adjoint relation on ℋ and 𝑧∈Γ(𝒯). Define 𝑇∶ℋ→ℋ by 𝑇(𝑧𝑓−𝑔)=𝑓. That is 𝑇=(𝑧−𝒯)⁻¹={(𝑧𝑓−𝑔,𝑓)∶(𝑓,𝑔)∈𝒯}. Then 𝑇 is a bounded linear operator since ‖𝑇‖=sup_{𝑧𝑓−𝑔=1}𝑇𝑧𝑓−𝑔=sup_{𝑧𝑓−𝑔=1}𝑓≤1𝐶(𝑧) (33) and 𝒯 is given by 𝒯=𝑇𝑓,𝑧𝑇𝑓−𝑓∶𝑓∈ℋ. (34)

3. Spectral Theory of a Linear Relation

Definition 8. Let ℛ be a closed relation on a Hilbert space ℋ. We define 𝜌(ℛ)=𝑧∈ℂ∶∃𝑇∈𝐵(ℋ),ℛ=𝑇𝑓,𝑧𝑇𝑓−𝑓∶𝑓∈ℋ (35) to be the resolvent set and 𝜎(ℛ)=ℂ−𝜌(ℛ) to be the spectrum of ℛ.

Remark 9. When a relation ℛ is an operator on ℋ, then 𝜌(ℛ)=𝑧∈ℂ∶(𝑧−ℛ)⁻¹∈𝐵(ℋ). (36)

A complex number 𝑧∈ℂ is called an eigenvalue of a relation ℛ if there exists 𝑓∈ℋ,𝑓≠0 such that (𝑓,𝑧𝑓)∈ℛ. The set of all eigenvalues of ℛ is called the point spectrum of ℛ and is denoted by 𝜎_𝑝(ℛ).

Remark 10. For any closed relation ℛ on a Hilbert space ℋ, 𝜎_𝑝(ℛ)⊂𝜎(ℛ).

Let 𝒵={(0,𝑔)∈ℛ} and 𝑍={𝑔∶(0,𝑔)∈ℛ} be the multivalued part of a relation ℛ. Clearly 𝑍 is a closed subspace of ℋ. Note that 𝐷(ℛ) is not dense if ℛ is multivalued. Now define the quotient space ℋ_𝑠=ℋ/𝑍. We know that this quotient space is also a Hilbert space with the norm defined by 𝑓=inf_𝑔∈𝑍𝑓+𝑔. (37) Define a relation ℛ_𝑠 on ℋ_𝑠⊕ℋ_𝑠 by ℛ_𝑠={([𝑓],[𝑔])∶(𝑓,𝑔)∈ℛ}. We consider the relation ℛ_𝑠 as the restriction of ℛ on ℋ_𝑠. By natural isomorphism, the space ℋ_𝑠 is identified as ℋ⊖𝑍 and the relation ℛ_𝑠 as ℛ⊖𝒵. Then clearly ℛ_𝑠 is an operator on ℋ_𝑠 with 𝐷(ℛ_𝑠)=𝐷(ℛ).

Theorem 11. If 𝒯 is a self-adjoint relation on ℋ, then 𝑆(𝒯)=𝜎(𝒯)=𝜎𝒯_𝑠. (38)

Proof. Letting 𝑧∈Γ(𝒯), then there exists a constant 𝐶>0 such that 𝑧𝑓−𝑔≥𝐶𝑓,∀𝑓,𝑔∈𝒯. (39) For such 𝑧, we can define 𝑇=(𝑧−𝒯)⁻¹ as a bounded linear operator on ℋ such that 𝒯={(𝑇ℎ,𝑧𝑇ℎ−ℎ)∶ℎ∈ℋ}. So 𝑧∈𝜌(𝒯). On the other hand, let 𝑧∈𝜌(𝒯); then there exists 𝑇∈𝐵(ℋ) such that 𝒯={(𝑇ℎ,𝑧𝑇ℎ−ℎ)∶ℎ∈ℋ}. For any (𝑓,𝑔)∈𝒯, there is ℎ∈ℋ such that 𝑇ℎ=𝑓 and 𝑧𝑇ℎ−ℎ=𝑔. So 𝑧𝑓−𝑔=‖𝑧𝑇ℎ−𝑧𝑇ℎ+ℎ‖=‖ℎ‖≥𝐶‖𝑇ℎ‖=𝐶𝑓 (40) for some 𝐶>0 and hence 𝑧∈Γ(𝒯). Hence, 𝑆(𝒯)=𝜎(𝒯).
Next, assuming that 𝑧∈Γ(𝒯_𝑠), then for any ([𝑓],[𝑔])∈𝒯_𝑠, there exists a constant 𝐶>0 such that 𝑧𝑓−𝑔≥𝐶𝑓. (41) For any (𝑓,𝑔)∈𝒯 we have 𝑧𝑓−𝑔≥𝑧𝑓−𝑔≥𝐶𝑓=𝐶𝑓. (42) Hence 𝑧∈Γ(𝒯). On the other hand, suppose that 𝑧∈Γ(𝒯), then there is a constant 𝐶>0 such that 𝑧𝑓−𝑔≥𝐶𝑓. (43) For any ([𝑓],[𝑔])∈𝒯_𝑠, we have 𝑧𝑓−𝑔=inf_𝑢∈𝑍𝑧𝑓−𝑔+𝑢=inf_𝑢∈𝑍+‖𝑢‖≥𝑓𝑓.𝑧𝑓−𝑔𝑧𝑓−𝑔≥𝐶=𝐶 (44) This implies that 𝑧∈Γ(𝒯_𝑠). Thus, 𝑆(𝒯_𝑠)=𝑆(𝒯). Hence, 𝑆(𝒯)=𝜎(𝒯)=𝜎(𝒯_𝑠).

Remark 12. If 𝒯 is a self-adjoint relation on ℋ, then 𝜎(𝒯)⊂ℝ.

Theorem 13. Let 𝑧∈Γ(𝒯) and 𝑇=(𝑧−𝒯)⁻¹. One has the following.(1)If 𝜆∈Γ(𝑇), then (𝑧−(1/𝜆))∈Γ(𝒯).(2)If 𝜆∈𝑆(𝒯), then (1/(𝑧−𝜆))∈𝑆(𝑇).(3)𝑆(𝑇)⊂𝜎(𝑇).

Proof. (1) Let 𝜆∈Γ(𝑇); then by definition there exists 𝐶(𝜆)>0 such that 𝜆𝑧𝑓−𝑔−𝑓≥𝐶(𝜆)𝑧𝑓−𝑔,∀𝑓,𝑔∈𝒯. (45) Note that 𝜆≠0. For any (𝑓,𝑔)∈𝒯, we have 1𝑧−𝜆=1𝑓−𝑔|𝜆|=1𝑧𝜆𝑓−𝑓−𝜆𝑔|𝜆|𝜆≥𝐶(𝜆)𝑧𝑓−𝑔−𝑓|𝜆|≥𝐶(𝜆)𝐶(𝑧)𝑧𝑓−𝑔|𝜆|𝑓. (46) So (𝑧−(1/𝜆))∈Γ(𝒯).
(2) Let 𝜆∈𝑆(𝒯) and suppose that (1/(𝑧−𝜆))∉𝑆(𝑇). Then (1/(𝑧−𝜆))∈Γ(𝑇). But by (1),(𝑧−1/(1/(𝑧−𝜆)))∈Γ(𝒯). This implies that 𝜆∈Γ(𝒯) which is a contradiction.
(3) Let 𝜆∈𝜌(𝑇); then (𝜆−𝑇)⁻¹ is bounded and is defined on all of ℋ; then for any (𝑓,𝑔)∈𝒯𝑧𝑓−𝑔=(𝜆−𝑇)⁻¹()≤)𝜆−𝑇𝑧𝑓−𝑔(𝜆−𝑇⁻¹𝜆⟹𝜆≥1𝑧𝑓−𝑔−𝑇𝑧𝑓−𝑔𝑧𝑓−𝑔−𝑇𝑧𝑓−𝑔(𝜆−𝑇)⁻¹𝑧𝑓−𝑔. (47)⇒𝜆∈Γ(𝑇). This shows that 𝑆(𝑇)⊂𝜎(𝑇).

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.