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

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

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## Abstract

The aim of this paper is to develop the conditions for a symmetric
relation in a Hilbert spaceℋ to have self-adjoint extensions in terms of defect
indices and discuss some spectral theory of such linear relation.


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## Body

## 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].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 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(1)Ju′(x)=zH(x)u(x),z∈ℂ,
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
(2)H(x)-1Ju′=zu
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
(3)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*. From now on, we write relation to mean linear relation.A relationℛ is called isometry if
(4)〈f1,f2〉=〈g1,g2〉,∀(f1,g1),(f2,g2)∈ℛ
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 ℛz=R(z-ℛ).It is clear to see thatN(ℛ*,z¯)=ℛz⊥.In Section2, 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

Letℛ be a relation on a Hilbert space ℋ. The set
(5)Γ(ℛ)={z∈ℂ:thereexistsaC(z)>0suchthat∥(zf-g)∥≥C(z)∥f∥,∀(f,g)∈ℛ}
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)
ifℛ is symmetric, then ℂ-ℝ⊂Γ(ℛ);(iii)
Γ(ℛ) is open.The subspaceℛz⊥ is called the defect space of ℛ and z. The cardinal number β(ℛ,z)=dim⁡ℛz⊥ is called the defect index of ℛ and z.Theorem 1.
The defect indexβ(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.
LetQz denote the orthogonal projection onto ℛ¯z. We first show that ∥Qz-Qz0∥→0 as z→z0, for any z0∈Γ(ℛ). Let z0∈Γ(ℛ); then there is a constant C(z0)>0 such that
(6)∥f∥≤C(z0)∥z0f-g∥,
for all (f,g)∈ℛ. For |z-z0|<1/(2C(z0)) and all (f,g)∈ℛ, we have
(7)∥f∥≤C(z0)∥z0f-g∥≤(∥zf-g∥+|z-z0|∥f∥)≤C(z0)∥zf-g∥+12∥f∥∴∥f∥≤C(z0)∥zf-g∥.
For h∈R(Qz0)⊥=ℛz0⊥,
(8)∥Qzh∥=sup⁡{|〈h,zf-g〉|:zf-g∈ℛz,∥zf-g∥≤1}=sup⁡{|〈h,(z-z0)f〉|:zf-g∈ℛz,∥zf-g∥≤1}.∴∥Qzh∥≤∥h∥|z-z0|C(z0).
Similarly, for h∈R(Qz)⊥=ℛz⊥,
(9)∥Qz0h∥≤∥h∥|z-z0|C(z0).
It follows that
(10)∥Qz-Qz0∥≤2|z-z0|C(z0).
Let Pz denote an orthogonal projection onto ℛ¯z⊥; then
(11)∥pz-pz0∥=∥Qz-Qz0∥≤2|z-z0|C(z0)⟶0asz⟶z0.
Hence, if we choose 0<ϵ<1/(2C(z0)), then ∥pz-pz0∥<1, for |z-z0|<ϵ. Therefore, dim⁡ℛz⊥=dim⁡ℛz0⊥. It follows that
(12)β(ℛ,z)=β(ℛ,z0)for|z-z0|<ϵ.
If ℛ is symmetric, then the upper and lower half-planes are connected subsets of Γ(ℛ); therefore, the defect index is constant there.Lettingℛ 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 relationℛ on ℋ is defined by the relation
(13)𝒱={(g+if,g-if):(f,g)∈ℛ}.
Then clearly D(𝒱)=R(ℛ+i) and R(𝒱)=R(ℛ-i).Theorem 2.
Ifℛ 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 ui=gi+ifi and vi=gi-ifi, for (fi,gi)∈ℛ,i=1,2; then
(14)〈u1,u2〉=〈g1+if1,g2+if2〉=〈g1,g2〉+〈g1,if2〉+〈if1,g2〉+〈if1,if2〉=〈g1,g2〉+i〈g1,f2〉-i〈f1,g2〉-i2〈f1,f2〉=〈g1-if1,g2-if2〉=〈v1,v2〉.
(
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 relation𝒱 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.
If𝒱 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
(15)〈i(f1+g1),(f2-g2)〉=-i(〈f1,f2〉-〈f1,g2〉+〈g1,f2〉-〈g1,g2〉).
Since 𝒱 is an isometry, for any (f1,g1),(f2,g2)∈𝒱,〈f1,f2〉=〈g1,g2〉, this implies that
(16)〈i(f1+g1),(f2-g2)〉=-i〈g1,g2〉+i〈f1,g2〉-i〈g1,f2〉+i〈f1,f2〉=-i(〈g1-f1,g2〉+〈g1-f1,f2〉)=〈f1-g1,i(f2+g2)〉.
Hence, ℛ is symmetric.Theorem 4.
A symmetric relationℛ is self-adjoint if and only if 𝒱 is unitary.Proof.
We show that theℛ is self-adjoint if and only if
(17)R(ℛ+i)=R(ℛ-i)=ℋ.
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
(18)(f+h)∈N(ℛ*,i)=R((ℛ+i)⊥)={0}.
This implies f=-h∈D(ℛ). Hence, ℛ is self-adjoint.
Conversely, suppose thatℛ is self-adjoint. Let
(19)h∈R(ℛ-i)⊥=N(ℛ*,-i)=N(ℛ,-i).
So (h,-ih)∈ℛ. But
(20)〈-ih,h〉=〈h,ih〉⟹i〈h,h〉=-i〈h,h〉.
Hence, we must have h=0. So R(ℛ-i)=ℋ. Similarly, R(ℛ+i)=ℋ.Theorem 5.
Letℛ 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
(21)𝒱′={(f+h,g+k):(f,g)∈𝒱,(h,k)∈𝒱~},D(𝒱′)=R(ℛ′+i)=R(ℛ+i)⊕F+,R(𝒱′)=R(ℛ′-i)=R(ℛ-i)⊕F-.
The spaces F+ and F- have the same dimension.(2)
The relation𝒱′ 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
(22)∥g+k∥2=∥g∥2+∥k∥2=∥f∥2+∥h∥2=∥f+h∥2.
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
(23)D(𝒱′)=R(𝒱′)=ℋ.
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
(24)dim⁡(R(ℛ+i)⊥)=dim⁡(R(ℛ-i)⊥).By definition of Cayley transform, it is clear that if𝒱1 and 𝒱2 are the Cayley transforms of any two symmetric relations ℛ1 and ℛ2, then
(25)ℛ1⊂ℛ2ifandonlyif𝒱1⊂𝒱2.Theorem 6.
Letℛ 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
(26)D(ℛ′)=D(ℛ)+{g+𝒱^g:g∈F+}.(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
(27)𝒱′={(f+h,g+k):(f,g)∈𝒱,(h,k)∈𝒱~},D(𝒱′)=R(ℛ′+i)=R(ℛ+i)⊕F+,R(𝒱′)=R(ℛ′-i)=R(ℛ-i)⊕F-.
Then by definition of the Cayley transform, we see that
(28)D(ℛ′)=R(I-𝒱′)=(I-𝒱′)D(𝒱′)=(I-𝒱′)R(i+ℛ′)=(I-𝒱′)(R(i+ℛ)⊕F+)=(I-𝒱′)(D(𝒱)⊕F+)=(I-𝒱)D(𝒱)+(I-𝒱′)F+=D(ℛ)+{g-𝒱~g:g∈F+}.
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 that𝒯 is a self-adjoint relation and suppose that z∈Γ(𝒯); then
(29)ℋ={zf-g:(f,g)∈𝒯}.Proof.
We will show thatR(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
(30)∥zf-g∥≥C(z)∥f∥.
Let vn∈R(z-𝒯) and vn→vinℋ. Suppose that fn∈D(𝒯) such that (fn,gn)∈𝒯 and vn=zfn-gn so that (fn,vn)∈z-𝒯. But from the above relation we have
(31)∥vn-vm∥=∥z(fn-fm)-(gn-gm)∥≥C∥fn-fm∥.
It follows that fn is a Cauchy sequence in ℋ, and it converges to some f in ℋ. Hence, (fn,vn)→(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
(32)ℋ=R(z-𝒯)⊕R(z-𝒯)⊥.
We next show that R(z-𝒯)⊥={0}. Let h∈R(z-𝒯)⊥=N(𝒯,z¯); then (h,z¯h)∈𝒯. But 0=∥z¯h-z¯h∥≥C(z¯)∥h∥ implies h=0 a. e..Let𝒯 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
(33)∥T∥=sup⁡∥zf-g∥=1∥T(zf-g)∥=sup⁡∥zf-g∥=1∥f∥≤1C(z)
and 𝒯 is given by
(34)𝒯={(Tf,zTf-f):f∈ℋ}.


## 3. Spectral Theory of a Linear Relation

Definition 8.
Letℛ be a closed relation on a Hilbert space ℋ. We define
(35)ρ(ℛ)={ℛ={(Tf,zTf-f):f∈ℋ}}z∈ℂ:∃T∈B(ℋ),ℛ={(Tf,zTf-f):f∈ℋ}}
to be the resolvent set and σ(ℛ)=ℂ-ρ(ℛ) to be the spectrum of ℛ.Remark 9.
When a relationℛ is an operator on ℋ, then
(36)ρ(ℛ)={z∈ℂ:(z-ℛ)-1∈B(ℋ)}.A complex numberz∈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 σp(ℛ).Remark 10.
For any closed relationℛ on a Hilbert space ℋ, σp(ℛ)⊂σ(ℛ).Let𝒵={(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 ℋs=ℋ/Z. We know that this quotient space is also a Hilbert space with the norm defined by
(37)∥[f]∥=inf⁡g∈Z∥f+g∥.
Define a relation ℛs on ℋs⊕ℋs by ℛs={([f],[g]):(f,g)∈ℛ}. We consider the relation ℛs as the restriction of ℛ on ℋs. By natural isomorphism, the space ℋs is identified as ℋ⊖Z and the relation ℛs as ℛ⊖𝒵. Then clearly ℛs is an operator on ℋs with D(ℛs)=D(ℛ).Theorem 11.
If𝒯 is a self-adjoint relation on ℋ, then
(38)S(𝒯)=σ(𝒯)=σ(𝒯s).Proof.
Lettingz∈Γ(𝒯), then there exists a constant C>0 such that
(39)∥zf-g∥≥C∥f∥,∀(f,g)∈𝒯.
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
(40)∥zf-g∥=∥zTh-zTh+h∥=∥h∥≥C∥Th∥=C∥f∥
for some C>0 and hence z∈Γ(𝒯). Hence, S(𝒯)=σ(𝒯).
Next, assuming thatz∈Γ(𝒯s), then for any ([f],[g])∈𝒯s, there exists a constant C>0 such that
(41)∥z[f]-[g]∥≥C∥[f]∥.
For any (f,g)∈𝒯 we have
(42)∥zf-g∥≥∥z[f]-[g]∥≥C∥[f]∥=C∥f∥.
Hence z∈Γ(𝒯). On the other hand, suppose that z∈Γ(𝒯), then there is a constant C>0 such that
(43)∥zf-g∥≥C∥f∥.
For any ([f],[g])∈𝒯s, we have
(44)∥z[f]-[g]∥=inf⁡u∈Z∥zf-g+u∥=inf⁡u∈Z(∥zf-g∥+∥u∥)≥∥zf-g∥≥C∥f∥=C∥[f]∥.
This implies that z∈Γ(𝒯s). Thus, S(𝒯s)=S(𝒯). Hence, S(𝒯)=σ(𝒯)=σ(𝒯s).Remark 12.
If𝒯 is a self-adjoint relation on ℋ, then σ(𝒯)⊂ℝ.Theorem 13.
Letz∈Γ(𝒯) and T=(z-𝒯)-1. One has the following.(1)
Ifλ∈Γ(T), then (z-(1/λ))∈Γ(𝒯).(2)
Ifλ∈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
(45)∥λ(zf-g)-f∥≥C(λ)∥zf-g∥,∀(f,g)∈𝒯.
Note that λ≠0. For any (f,g)∈𝒯, we have
(46)∥(z-1λ)f-g∥=1|λ|∥zλf-f-λg∥=1|λ|∥λ(zf-g)-f∥≥C(λ)|λ|∥zf-g∥≥C(λ)C(z)|λ|∥f∥.
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)∈𝒯(47)∥zf-g∥=∥(λ-T)-1(λ-T)(zf-g)∥≤∥(λ-T)-1∥∥λ(zf-g)-T(zf-g)∥⟹∥λ(zf-g)-T(zf-g)∥≥1∥(λ-T)-1∥∥zf-g∥.⇒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.



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