Unbounded linear operators

Jan Dereziński

Department of Mathematical Methods in Physics

Warsaw University
Hoża 74, 00-682, Warszawa, Poland

Lecture notes, version of Jan. 2007

January 30, 2007

Contents
1 Unbounded operators 2
1.1 Relations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2
1.2 Linear pseudooperators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3
1.3 Closed operators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3
1.4 Closable operators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4
1.5 Perturbations of closed operators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5
1.6 Invertible unbounded operators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6
1.7 Spectrum of unbounded operators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7
1.8 Examples of unbounded operators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8
1.9 Pseudoresolvents . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9

2 One-parameter semigroups in Banach spaces 10

2.1 (M, β)-type semigroups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10
2.2 Generator of a semigroup . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10
2.3 Norm continuous semigroups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12
2.4 Essential domains of generators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12
2.5 Operators of (M, β)-type . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13
2.6 The Hille-Philips-Yosida theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13
2.7 Semigroups of contractions and dissipative operators . . . . . . . . . . . . . . . . . . . . . 15

3 Unbounded operators in Hilbert spaces 16

3.1 Graph scalar product . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16
3.2 The adjoint of an operator . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16
3.3 Inverse of the adjoint operator . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18
3.4 Maximal operators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18
3.5 Dissipative operators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19
3.6 Hermitian operators I . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20
3.7 Self-adjoint operators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21
3.8 Essentially self-adjoint operators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22
3.9 Scale of Hilbert spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22

1
 3.10 Relative operator boundedness . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22
3.11 Relative form boundedness . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24
3.12 Non-maximal operators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24
3.13 Hermitian operators II . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25

4 Sesquilinear forms 26
4.1 Sesquilinear forms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26
4.2 Closed positive forms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27
4.3 Closable positive forms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27
4.4 Operators associated with positive forms . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28
4.5 Polar decomposition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28
4.6 Sectorial forms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29
4.7 Operators associated with sectorial forms . . . . . . . . . . . . . . . . . . . . . . . . . . . 30
4.8 Perturbations of sectorial forms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30
4.9 Friedrichs extensions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31

5 Aronszajn-Donoghue and Friedrichs Hamiltonian and their renormalization 32

5.1 Aronszajn Donoghue Hamiltonians . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32
5.2 Aronszajn-Donoghue Hamiltonians and extensions of Hermitian operators . . . . . . . . . 33
5.3 Aronszajn-Donoghue Hamiltonians and extensions of positive forms . . . . . . . . . . . . . 33
5.4 Friedrichs Hamiltonian . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34

6 Discrete and essential spectrum 36

6.1 Extended discrete and essential spectrum . . . . . . . . . . . . . . . . . . . . . . . . . . . 36
6.2 Operators with a compact resolvent . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36
6.3 Stability of essential spectrum . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37
These lecture notes are a continuation of the notes “Bounded operators”. We develop the theory of
unbounded operators in Banach, and especially Hilbert spaces. We avoid using more advanced tools such
as locally convex topologies and applications of the Baire category theorem.

1 Unbounded operators
1.1 Relations
Let X, Y be sets. R is called a relation iff R ⊂ Y × X. We will also write R : X → Y . (Note the inversion
of the direction). An example of a relation is the identity 1X := {(x, x) : x ∈ X} ⊂ X × X.
Introduce the “projections”

Y × X 3 (y, x) 7→ πY (y, x) := y ∈ Y,

Y × X 3 (y, x) 7→ πX (y, x) := x ∈ X,
and the “flip”
Y × X 3 (y, x) 7→ τ (y, x) := (x, y) ∈ X × Y.
The domain of R is defined as DomR := πX R, its range is Ran R = πY R, the inverse of R is defined as
R−1 := τ R ⊂ X × Y . If S ⊂ Z × Y , then the superposition of S and R is defined as S ◦ R ⊂ Z × X,
S ◦ R := {(z, x) ∈ Z × X : ∨ (z, y) ∈ S, (y, x) ∈ R}.
y∈Y
If X0 ⊂ X, then the restriction of R to X0 is defined as

R := R ∩ Y ×X0 .
X0

2
If, moreover, Y0 ⊂ Y , then
R := R ∩ Y0 ×X0 .
X0 →Y0

We say that a relation R is right-unique, if for any x ∈ X πY (R ∩ Y ×{x}) is one-element. We say

that R has a maximal domain if DomR = X.

Proposition 1.1 a) If R, S are right-unique, then so is S ◦ R.

b) If R, S have a maximal domain, then so does S ◦ R.

A right unique relation will be also called a pseudo-transformation (-operator, etc). Instead of writing
(y, x) ∈ R, we will then write y = R(x) or, in some contexts, y = Rx. We also introduce the graph of R:

Gr R := {(y, x) ∈ Y × X : y = R(x), x ∈ DomR}.

Note that strictly speaking Gr R = R. The difference of Gr R and R lies only in their syntactic role.
Note that a superposition of pseudotransformations is a pseudotransformation.
We say that a pseudotransformation is injective if it is left-unique. The inverse of a pseudotransfor-
mation is a pseudotransformation iff it is injective.
A transformation is a pseudotransformation with a maximal domain. The composition of transfor-
mations is a transformation.
We say that a transformation R is bijective iff it is left-unique and Ran R = Y . The inverse of a
transformation is a transformation iff it is bijective.

Proposition 1.2 Let R ⊂ X ×Y and S ⊂ Y ×X be transformations such that R◦S = 1Y and S◦R = 1X .
Then S and R are bijections and S = R−1 .

1.2 Linear pseudooperators

Let X , Y be vector spaces.

Proposition 1.3 1) A linear subspace V ⊂ Y ⊕ X is a graph of a certain pseudooperator iff (y, 0) ∈ V

implies y = 0.
2) A pseudooperator A is injective iff (0, x) ∈ Gr A implies x = 0.

From now on by an “operator” we will mean a “pseudooperator”. To say that A is a true operator
we will write DomA = X .

1.3 Closed operators

Let X , Y be Banach spaces.

Theorem 1.4 Let A : X → Y be an operator. The following conditions are equivalent:

(1) Gr A is closed in Y × X .
(2) If xn → x, xn ∈ DomA and Axn → y, then x ∈ DomA and y = Ax.
(3) For some p ∈ [1, ∞], DomA with the norm
1
kxkA,p := (kAxkp + kxkp ) p .

is a Banach space.

3
Proof. The equivalence of (1), (2) and (3) is obvious, if we note that

DomA 3 x 7→ (Ax, x) ∈ Gr A

is a bijection. 2

Definition 1.5 An operator satisfying the above conditions is called closed.

We will say that A : X → Y is bounded iff

kAxk ≤ ckxk, x ∈ DomA. (1.1)

B(X , Y) will denote all bounded operators from X to Y with the domain equal to X .

Proposition 1.6 A bounded operator A is closed iff DomA is closed.

Theorem 1.7 Let B ∈ B(X , Y) be invertible and A : X → Y closed. Then BA is closed on DomA and
AB is closed on B −1 DomA.

Proof. We check that

kxkA ≤ max(1, kB −1 k)kxkBA ,

kxkBA ≤ max(1, kBk)kxkA ,

kBxkA ≤ max(1, kBk)kxkAB ,

kxkAB ≤ max(1, kB −1 k)kBxkA .

2

Theorem 1.8 If A is closed and injective, then so is A−1 .

Proof. The flip τ : Y × X → X × Y is continuous. 2

1.4 Closable operators

Theorem 1.9 Let A : X → Y be an operator. The following conditions are equivalent:
(1) There exists a closed operator B such that B ⊃ A.
(2) (Gr A)cl is the graph of an operator.
(3) (y, 0) ∈ (Gr A)cl ⇒ y = 0.
(4) (xn ) ⊂ DomA, xn → 0, Axn → y implies y = 0.

Definition 1.10 An operator A satisfying the conditions of Theorem 1.9 is called closable. If the con-
ditions of Theorem 1.9 hold, then the operator whose graph equals (Gr A)cl is denoted by Acl and called
the closure of A.

Proof of Theorem 1.9 To show (2)⇒(1) it suffices to take as B the operator Acl . Let us show
(1)⇒(2). Let B be a closed operator such that A ⊂ B. Then (Gr A)cl ⊂ (Gr B)cl = Gr B. But
(y, 0) ∈ Gr B ⇒ y = 0, hence (y, 0) ∈ (Gr A)cl ⇒ y = 0. Thus (Gr A)cl is the graph of an
operator. 2
As a by-product of the above proof, we obtain

4
Proposition 1.11 If A is closable, B closed and A ⊂ B, then Acl ⊂ B.

Proposition 1.12 Let A be bounded. Then A is closable, DomAcl = (DomA)cl and Acl x = limn Axn
for xn → x, xn ∈ DomA. Besides, Acl satisfies (1.1).
n→∞

Let A be a closed operator. We say that a linear subspace D is an essential domain for A iff D is
dense in DomA in the graph topology. In other words, D is an essential domain for A, if
 cl
A = A.
D

Theorem 1.13 (1) If A ∈ B(X , Y), then a linear subspace D ⊂ X is an essential domain for A iff it
is dense in X (in the usual topology).
(2) If A is closed, has a dense domain and D is its essential domain, then D is dense in X .

1.5 Perturbations of closed operators

Definition 1.14 Let B, A : X → Y. We say that B is bounded relatively to A iff DomA ⊂ DomB and
there exist constants a, b such that

kBxk ≤ akAxk + bkxk, x ∈ DomA. (1.2)

The infimum of a satisfying (1.2) is called the A-bound of B. In other words: the A-bound of B equals

kBxk
inf sup .
c>0 x∈DomA\{0} kAxk + ckxk

In particular, if B is bounded, then its A-bound equals 0.

If A is unbounded, then its A-bound equals 1.
In the case of Hilbert spaces it is more convenient to use the following condition to define the relative
boundedness:

Theorem 1.15 B is bounded relatively to A with the A-bound a iff DomA ⊂ DomB and
1/2
kBxk2

inf sup < ∞. (1.3)
c>0 x∈DomA\{0} kAxk2 + ckxk2

If this is the case, then (1.3) equals the A-bound of B

Proof. For any  > 0 we have

21
kAxk2 + c2 kxk2
≤ kAxk + ckxk

21
≤ (1 + 2 )kAxk2 + c2 (1 + −2 )kxk2 .
2

Theorem 1.16 Let A be closed and let B be bounded relatively to A with the A-bound less than 1. Then
A + B with the domain DomA is closed. All essential domains of A are essential domains of A + B

5
Proof. We know that
kBxk ≤ akAxk + bkxk
for some a < 1 and b. Hence

k(A + B)xk + kxk ≤ (1 + a)kAxk + (1 + b)kxk

and
(1 − a)kAxk + kxk ≤ kAxk − kBxk + (1 + b)kxk ≤ k(A + B)xk + (1 + b)kxk.
Hence the norms kAxk + kxk and k(A + B)xk + kxk are equivalent on DomA. 2

Theorem 1.17 Suppose that A, C are two operators with the same domain DomA = DomC = D
satisfying
k(A − C)xk ≤ a(kAxk + kCxk) + bkxk
for some a < 1. Then
(1) A is closed on D iff C is closed on D.
(2) A is closable on D iff C is closable on D and then the domains of Acl and C cl coincide.

Proof. Define B := C − A and F (t) := A + tB with the domain D. For 0 ≤ t ≤ 1, we have

kBxk ≤ a(kAxk + kCxk) + bkxk

= a (k(F (t) − tB)xk + k(F (t) + (1 − t)B)xk) + bkxk

≤ 2akF (t)xk + akBxk + bkxk

Hence
2a b
kBxk ≤ kF (t)xk + kxk.
1−a 1−a
Therefore, if |s| < 1−a
2a and t, t + s ∈ [0, 1], then F (t + s) is closed iff F (t) is closed. 2

1.6 Invertible unbounded operators

Definition 1.18 We say that an operator A is invertible iff A−1 ∈ B(Y, X ).

Theorem 1.19 Let A be closed. Suppose that for some c > 0

kAxk ≥ ckxk, x ∈ DomA. (1.4)

Then Ran A is closed. If Ran A = Y , then A is invertible and

kA−1 k ≤ c−1

Proof. Let yn ∈ Ran A and yn → y. Let Axn = yn . Then xn is a Cauchy sequence. Hence there exists
limn→∞ xn := x. But A is closed, hence Ax = y. Therefore, Ran A is closed. 2

Corollary 1.20 Let A be closed. Suppose that for some c > 0

kAxk ≥ ckxk,

and Ran A is dense in Y. Then A is invertible

6
Theorem 1.21 Let A be closable and for some c (1.4) holds. Then (1.4) holds for Acl as well.

Theorem 1.22 (1) Let A be injective and DomB ⊃ DomA. Then B has the A-bound less than
kBA−1 k.
(2) If, moreover, kBA−1 k < 1, then A + B with the domain DomA is closed, invertible and
∞
X
(A + B)−1 = (−1)j A−1 (BA−1 )j .
j=0

Proof. Let a := kBA−1 k. By the estimate

kBxk ≤ akAxk, x ∈ DomA,

we see that B has the A-bound less than or equal to a. This proves (1).
Assume now that a < 1. Let
X n
Cn := (−1)j A−1 (BA−1 )j .
j=0

Then limn→∞ Cn =: C exists.

Let y ∈ Y. Clearly, limn→∞ Cn y = Cy.

(A + B)Cn y = y + (−1)n (BA−1 )n+1 y → y.

But A + B is closed, hence Cy ∈ Dom(A + B) and (A + B)Cy = y.

If x ∈ Dom(A + B), then

Cn (A + B)x = x + (−1)n A−1 (BA−1 )n Ax → x.

Hence C(A + B)y = y. 2

Proposition 1.23 Let A and B be invertible and DomB ⊃ DomA. Then

B −1 − A−1 = B −1 (A − B)A−1 .

1.7 Spectrum of unbounded operators

Let A be an operator on X . We define the resolvent set of A as

rsA := {z ∈ C : z − A is invertible }.

We define the spectrum of A as spA := C\rsA.

We say that x ∈ X is an eigenvector of A with the eigenvalue λ ∈ C iff x ∈ DomA, x 6= 0 and Ax = λx.
The set of eigenvalues is called the point spectrum of A and denoted spp A. Clearly, spp A ⊂ spA.
Let Ccomp denote the Riemann sphere (the one-point compactification of C). In the case of unbounded
operators it is sometimes convenient to use the “extended spectrum”, which is a subset of Ccomp , instead
of the usual spectrum—a subset of C.
The extended resolvent set is defined as rsext A := rsA ∪ {∞} if A ∈ B(X ) and rsext A := rsA, if A is
unbounded. The extended spectrum is defined as

spext A = Ccomp \rsext A.

If A ∈ B(X ), we set (∞ − A)−1 = 0.

7
Theorem 1.24 (1) If λ, µ ∈ rsA, then

(λ − A)−1 − (µ − A)−1 = (µ − λ)(λ − A)−1 (µ − A)−1 .

(2) If λ ∈ rsA and k(λ − A)−1 k = c, then {z : |z − λ| < c−1 } ⊂ rsA.

−1
(3) k(z − A)−1 k ≥ (dist(z, spA)) .
(4) If rsA is nonempty, then A is closed.
(5) spext A is a compact subset of Ccomp .
(6) (z − A)−1 is analytic on rsext A.
(7) (z − A)−1 cannot be analytically extended to a larger subset of Ccomp than rsext (A).
(8) spext (A) 6= ∅
(9) Ran (z − A)−1 does not depend on z ∈ rsA and equals DomA.
(10) Ker(z − A)−1 = {0}.

Proof. Let us show (4). If λ ∈ rs(A), then λ − A is invertible, hence closed. 2

It is easy to define the functional calculus for unbounded operators with a non-empty resolvent set.
The definition usual definition can be repeated verbatim, replacing spA with spext (A). Theorem ??
remains valid except that the point about convergent power series should be dropped

Proposition 1.25 Suppose that rsA is non-empty and DomA is dense. Then DomA2 is dense.

Proof. Let z ∈ rsA. (z − A)−1 is a bounded operator with a dense range and DomA is dense. Hence
(z − A)−1 DomA is dense. If x ∈ DomA, then A(z − A)−1 x = (z − A)−1 Ax ∈ DomA Hence (z −
A)−1 DomA ⊂ DomA2 . 2

Theorem 1.26 Let A and B be operators on X with A ⊂ B, A 6= B. Then rsA ⊂ spB, and hence
rsB ⊂ spA.

Proof. Let λ ∈ rsA. Let x ∈ DomB\DomA. We have Ran (λ − A) = X , hence there exists y ∈ DomA
6 rsB. 2
such that (λ − A)y = (λ − B)x. Hence (λ − B)y = (λ − B)x. Hence λ ∈

1.8 Examples of unbounded operators

Example 1.27 Let I be an infinite set and (ai )i∈I be an unbounded complex sequence. Let C0 (I) be the
space of sequences with a finite number of non-zero elements. For 1 ≤ p < ∞ we define the operator

Lp (I) ⊃ C0 (I) 3 x 7→ Ax ∈ Lp (I)

by the formula
(Ax)i = ai xi .
p
(We can use C∞ (I) instead of L (I), then p = ∞ in the formulas below). Then the operator A is
unbounded and non-closed. Besides,

spp (A) = {ai : i ∈ I},

spA = C.

8
 The closure of A has the domain

DomAcl := {(xi )i∈I ∈ Lp (I) : |ai xi |p < ∞}

P
i∈I (1.5)

We then have
spp (Acl ) = {ai : i ∈ I},
spAcl = {ai : i ∈ I}cl .
To prove this let D be the rhs of (1.5) and x ∈ D. Then there exists a countable set I1 such that i 6∈ I1
implies xi = 0. We enumerate the elements of I1 : i1 , i2 , . . .. Define xn ∈ C0 (I) setting xnij = xij
for j ≤ n and xni = 0 for the remaining indices. Then limn→∞ xn = x and Axn → Ax. Hence,
{(x, Ax) : x ∈ D} ⊂ (Gr A)cl .
If xn belongs to (1.5) and (xn , Axn ) → (x, y), then xni → xi and ai xni = (Axn )i → yi . Hence
yi = ai xi . Using that y ∈ Lp (I) we see that x belongs to (1.5).

Example 1.28 Let p−1 + q −1 = 1, 1 < p ≤ ∞ and let (wi )i∈I be a sequence that does not belong to
Lq (I). Let C0 (I) be as above. Define
X
Lp (I) ⊃ C0 (I) 3 x 7→ hw|xi := xi wi ∈ C.
i∈I

Then hw| is non-closable.

q
It is sufficient to assume that I = N and define vin := wi (P|wn
i| n
q , i ≤ n, vi = 0, i > n. Then
i=1 |wi | )
Pn 1
hw|v n i = 1 and kv n kp = ( i=1 |wi |q )− q → 0. Hence (0, 1) belongs to the closure of the graph of the
operator.

1.9 Pseudoresolvents
Definition 1.29 Let Ω ⊂ C be open. Then the continuous function

Ω 3 z 7→ R(z) ∈ B(X )

is called a pseudoresolvent if
R(z1 ) − R(z2 ) = (z2 − z1 )R(z1 )R(z2 ). (1.6)

Evidently, every resolvent of a closed operator is a pseudoresolvent.

Proposition 1.30 Let Ω 3 z 7→ Rn (z) ∈ B(X ) be a sequence of pseudoresolvents and R(z) := s− limn→∞ Rn (z).
Then R(z) is a pseudoresolvent.

Theorem 1.31 Let Ω 3 z 7→ R(z) ∈ B(X ) be a pseudoresolvent. Then

(1) R := Ran R(z) does not depend on z ∈ Ω.
(2) N := KerR(z) does not depend on z ∈ Ω.
(3) R(z) is an analytic function and
d
R(z) = −R(z)2 .
dz
(4) R(z) is a resolvent of a certain operator iff N = {0}. The domain of this operator equals R.

Proof. Let us prove (4)⇐. Fix z1 ∈ Ω. If N = {0}, then every element of R can be uniquely represented
as R(z1 )x, x ∈ X . Define HR(z1 )x := −x + z1 R(z1 )x. By formula (1.6) we check that the definition does
not depend on z1 . 2

9
2 One-parameter semigroups in Banach spaces
2.1 (M, β)-type semigroups
Let X be a Banach space.
Definition 2.1 [0, ∞[3 t 7→ W (t) ∈ B(X ) is called a strongly continuous one-parameter semigroup iff
(1) W (0) = 1;
(2) W (t1 )W (t2 ) = W (t1 + t2 ), t1 , t2 ∈ [0, ∞[;
(3) limt↓0 W (t)x = x, x ∈ X ;
(4) for some t0 > 0, kW (t)k < M , 0 ≤ t ≤ t0 .

Remark 2.2 Using the Banach-Steinhaus Theorem one can show that (4) follows from the remaining
points.

Theorem 2.3 If W (t) is a strongly continuous semigroup, then

[0, ∞[×X 3 (t, x) 7→ W (t)x ∈ X

is a continuous function. Besides, there exist constants M , β such that

kW (t)k ≤ M eβt . (2.7)

Proof. By (4), for t ≤ nt0 we have kW (t)k ≤ M n . Hence, kW (t)k ≤ M exp( tt0 log M ). Therefore, (2.7)
is satisfied.
Let tn → t and xn → x. Then

kW (tn )xn − W (t)xk ≤ kW (tn )xn − W (tn )xk + kW (tn )x − W (t)xk

≤ M eβtn kxn − xk + M eβ min(tn ,t) kW (|t − tn |)x − xk.

2
We say that the semigroup W (t) is (M, β)-type, if the condition (2.7) is satisfied.
Clearly, if W (t) is (M, β)-type, then W (t)e−βt is (M, 0)-type. Since W (0) = 1, no semigroups (M, β)
exist for M < 1.

2.2 Generator of a semigroup

If W (t) is a strongly continuous one-parameter semigroup, we define

DomA := {x ∈ X : there exists lim t−1 (W (t)x − x)}.

t→0

The operator A with the domain DomA is defined by the formula

Ax := lim t−1 (W (t)x − x).

t→0

A will be called the generator of W (t).

Theorem 2.4 (1) A is a closed densely defined operator;

(2) W (t)DomA ⊂ DomA and W (t)A = AW (t);
(3) If W1 (t) and W2 (t) are two different semigroups, then their generators are different.

10
 By (3), if A is an operator, which is a generator of a semigroup W (t), then such W (t) is unique. We
will write W (t) =: etA .
Proof of Theorem 2.4 (2). Let x ∈ DomA. Then

lim s−1 (W (s) − 1)W (t)x = W (t) lim s−1 (W (s) − 1)x = W (t)Ax. (2.8)
s↓0 s↓0

Hence the limit of the left hand side of (2.8) exists. Hence W (t)x ∈ DomA and AW (t)x = W (t)Ax. 2

Lemma 2.5 For x ∈ X put Z t

Bt x := t−1 W (s)xds.
0
Then
(1) s− limt→0 Bt = 1.
(2) Bt W (s) = W (s)Bt .
(3) If x ∈ X , then Bt x ∈ DomA,
ABt x = t−1 (W (t)x − x), (2.9)

(4) For x ∈ DomA, ABt x = Bt Ax.

Proof. (1) follows by

Z t
Bt x − x = t−1 (W (s)x − x)ds → 0.
0 t↓0

(2) is obvious. (4) is proven as Theorem 2.4 (2). To prove (3) we note that

u−1 (W (u) − 1)Bt x = t−1 (W (t) − 1)Bu x → t−1 (W (t)x − x).

u↓0

2
Proof of Theorem 2.4 (1), (3) The density of DomA follows by Lemma 2.5 (1) and (3).
Let us show that A is closed. Let xn → x and Axn → y. Using the boundedness of Bt A = ABt
n→∞ n→∞
we get
Bt y = lim Bt Axn = Bt Ax.
n→∞

Hence
y = lim Bt y = lim Bt Ax = Ax.
t↓0 t↓0
2

Proposition 2.6 Let W (t) be a semigroup and A its generator. Then, for any x ∈ DomA there exists a
unique solution of
d
[0, t0 ] 3 t 7→ x(t) ∈ DomA, x(t) = Ax(t), (2.10)
dt
(for t = 0 the derivative is right-sided). The solution is given by x(t) = W (t)x.

Proof. Let us show that x(t) := W (t)x solves (2.10). We already know that the right-sided derivative
equals Ax(t). It suffices to prove the same about the left-sided derivative. For 0 ≤ u ≤ t we have

(−u)−1 (W (t − u)x − W (t)x) = W (t − u)u−1 (W (u) − 1)x →u→0 W (t)Ax = AW (t)x.

11
 Let us show now the uniqueness. Let x(t) solve (2.10) . Let y(s) := W (t − s)x(s). Then
d
y(s) = W (t − s)Ax(s) − AW (t − s)x(s) = 0
ds
Hence y(s) does not depend on s. At s = t it equals x(t), and at s = 0 it equals W (t)x. 2
Proof of Theorem 2.4 (3) By Theorem 2.6 (2), W (t) is uniquely determined by A on DomA. But
W (t) is bounded and DomA is dense, hence W (t) is uniquely determined. 2

2.3 Norm continuous semigroups

Theorem 2.7 (1) If A ∈ B(X ), then C 3 z 7→ ezA is a norm continuous group and A is its generator.
(2) If a one-parameter semigroup W (t) is norm continuous, then its generator is bounded.

Proof. (1) follows by the functional calculus.

Let us show (2). W (t) is norm continuous, hence limt→0 Bt = 1. Therefore, for 0 < t < t0

kBt − 1k < 1.

Hence Bt is then invertible.

We know that for x ∈ DomA
t−1 (W (t) − 1)x = Bt Ax.
For 0 ≤ t < t0 we can write this as

Ax = t−1 Bt−1 (W (t) − 1)x.

Hence kAxk ≤ ckxk. 2

2.4 Essential domains of generators

Theorem 2.8 Let W (t) be a strongly continuous one-parameter semigroup and let A be its generator.
Let D ⊂ DomA be dense in X and W (t)D ⊂ D, t > 0. Then D is dense in DomA in the graph
topology—in other words, D is an essential domain of A.

Lemma 2.9 (1) For x ∈ X , kBt xkDomA ≤ (Ct−1 + 1)kxk;

(2) For x ∈ DomA, limt↓0 kBt x − xkDomA = 0;
(3) W (t) is a strongly continuous semi-group on DomA equipped with the graph norm.
(4) If D̃ is a closed subspace in DomA invariant wrt W (t), then it is invariant also wrt Bt .

Proof. (1) follows by Lemma 2.5 (3).

(2) follows by Lemma 2.5 (1) and because B(t) commutes with A.
(3) follows from the fact that W (t) is a strongly continuous semigroup on X , preserves DomA and
commutes with A.
To show (4), note that Bt x is defined using an integral involving W (s)x. W (s)x depends continuously
on s in the topology of DomA, as follows by (3). Hence this integral (as Riemann’s integral) is well defined.
Besides, Bt x belongs to the closure of the space spanned by W (s)x, 0 ≤ s ≤ t. 2
Proof of Theorem 2.8. Let x ∈ DomA, xn ∈ D and xn → x in X . Let D̃ be he closure of D in
n→∞
DomA. Then Bt xn ∈ D̃, by Lemma 2.9 (4). By Lemma 2.9 (1) we have

kBt xn − Bt xkDomA ≤ Ct kxn − xk.

12
Hence Bt x ∈ D̃. By Lemma 2.9 (2)
kBt x − xkDomA → 0.
t↓0

Hence, x ∈ D̃. 2

2.5 Operators of (M, β)-type

Theorem 2.10 Let A be a densely defned operator. Then the following conditions are equivalent:
(1) [β, ∞[⊂ rs(A) and

k(x − A)−m k ≤ M |x − β|−m , m = 1, 2, . . . , x ∈ [β, ∞[

(2) {z ∈ C : Rez > β} ⊂ rs(A) and

k(z − A)−m k ≤ M |Rez − β|−m , m = 1, 2, . . . , z ∈ {z ∈ C : Rez > β}.

Proof. It suffices to prove (1)⇒(2). Let (1) be satisfied. It suffices to assume that β = 0. Let z = x + iy.
Then for t > 0
(z − A)−m = (x + t − A)m (1 + (iy − t)(x + t − A)−1 )−m
∞
 
P −m−j j −m
= (x + t − A) (iy − t) .
j=0 j
 
−m
Using the fact that has an alternating sign we get
j
 
−m
P∞ −m−j j j −m
k(z − A) k ≤ j=0 |x + t| (−1) |iy − t|
j
 −m
= M |x + t|m 1 − |iy−t|
x+t

= M (x + t − |iy − t|)−m → M x−m .

t→∞

Definition 2.11 We say that an operator A is (M, β)-type, iff the conditions of Theorem 2.10 are sat-
isfied.

Obviously, if A is of (M, β)-type, then A − β is of (M, 0)-type.

2.6 The Hille-Philips-Yosida theorem

Theorem 2.12 If W (t) is a semigroup of (M, β)-type, then its generator A is also of (M, β)-type.
Besides, Z ∞
−1
(z − A) = e−tz W (t)dt, Rez > β.
0

13
Proof. Set Z ∞
R(z)x := e−zt W (t)xdt.
0

Let y = R(z)x. Then

u−1 (W (u) − 1)y

Ru R∞
= −u−1 ezu 0 e−zt W (t)xdt + u−1 (ezu − 1) 0 e−zt W (t)xdt → −x + zy.
u↓0

Hence y ∈ DomA and (z − A)R(z)x = x.

d
Suppose now that x ∈ Ker(z − A). Then xt := ezt x ∈ DomA satisfies dt xt = Axt . Hence xt = W (t)x.
Rezt
But kxt k = e kxk, which is impossible.
By the formula
Z ∞ Z ∞
(z − A)−m = ··· e−z(t1 +···+tm ) W (t1 + · · · + tm )dt1 · · · dtm
0 0

we get the estimate

Z ∞ Z ∞
−m
k(z − A) k≤ ··· M e−(z−β)(t1 +···+tm ) dt1 · · · dtm = M |z − β|−m .
0 0

Theorem 2.13 If A is an operator of (M, β)-type, then it is the generator of a semigroup W (t). This
semigroup is of (M, β)-type.
To simplify, let us assume that β = 0 (which does not restrict the generality). Then we have the
formula
 −n
t
W (t) = s− lim 1 − A ,
n→∞ n
−n
t2

t
W (t)x − 1 − A x ≤ M kA2 xk, x ∈ DomA2 .
n 2

Proof. Set  −n

t
Vn (t) := 1− A .
n
Let us first show that
s− lim Vn (t) = 1. (2.11)
t↓0

To prove (2.11) it suffices to prove that

s− lim(1 − sA)−1 = 1. (2.12)

s↓0

We have (1 − sA)−1 − 1 = (s−1 − A)−1 A. Hence for x ∈ DomA

k(1 − sA)−1 x − xk ≤ M s−1 kAxk,

which proves (2.12).

14
 Let us list some other properties of Vn (t): for Ret > 0, Vn (t) is holomorphic, kVn (t)k ≤ M and
 −n−1
d t
Vn (t) = A 1 − A .
dt n

To show that Vn (t)x is a Cauchy sequence for x ∈ Dom(A2 ), we compute

Vn (t)x − Vm (t)x = lims↓0 Vn (t − s)Vm (s)x − lims↑t Vn (t − s)Vm (s)x

R t− d
= lim↓0  ds Vn (t − s)Vm (s)x
R t−  
= lim↓0  − Vn0 (t − s)Vm (s) + Vn (t − s)Vm0 (s) x
R t−  s t−s

−n−1
−m−1 2
= lim↓0  n − m 1 − t−s
n A 1 − ns A A x.

Hence for x ∈ Dom(A2 )

Rt s t−s
kVn (t)x − Vm (t)xk ≤ kA2 xk 0
|m − 2
n |M ds

1 t2
= M 2 ( n1 + m) 2 .

By the Proposition 1.25, Dom(A2 ) is dense in X . Therefore, there exists a limit uniform on [0, t0 ]

s− lim Vn (t) =: W (t),

n→∞

which depends strongly continuously on t.

Finally, let us show that W (t) is a semigroup with the generator A. To this end it suffices to show
that for x ∈ DomA
d
W (t)x = AW (t)x. (2.13)
dt
But x ∈ DomA Z t+u 
s −1
Vn (t + u)x = Vn (t)x + A 1− A Vn (s)xds
t n
Hence passing to the limit we get
Z t+u
W (t + u)x = W (t)x + AW (s)xds.
t

This implies (2.13). 2

2.7 Semigroups of contractions and dissipative operators

Theorem 2.14 Let A be a closed operator on X . Then the following conditions are eqivalent:
(1) A is a generator of a semigroup of contractions, eg. ketA k ≤ 1
(2) The operator A is of (1, 0)-type.
(3) ]0, ∞[⊂ rs(A) and
k(µ − A)−1 k ≤ µ−1 , µ ∈ R, µ > 0,

(4) {z ∈ C : Rez > 0} ⊂ rs(A) and

k(z − A)−1 k ≤ |Rez|−1 , z ∈ C, Rez > 0.

15
(5) {z ∈ C : Rez > 0} ⊂ rsA; besides, if x ∈ DomA, v ∈ X # , kvk = 1 and hv|xi = kxk, then

Rehv|Axi ≤ 0.

(6) There exists z ∈ C with Rez > 0 such that Ran (z − A) = X; besides if x ∈ DomA, then there
∨ v ∈ X # such that kvk = 1, hv|xi = kxk, and

Rehv|Axi ≤ 0.

Proof. The equivalence of (1) and (2) is a special case of Theorems 2.12 and 2.13. The implications
(2)⇒(3) and (2)⇒(4) are obvious, the converse implications are easy.
Let us show (1),(3)⇒(5). We have

Rehv|xi = hv|xi = kxk,

Re hv|etA xi ≤ |hv|etA xi| ≤ kxk.

Hence
Rehv|Axi = lim Ret−1 (hv|etA xi − hv|xi) ≤ 0.
t↓0

We know that if Rez > 0, then z ∈ rs(A). Hence Ran (z − A) = X .

The implication (5)⇒(6) is obvious.
Let us prove (6)⇒(3).

k(z − A)xk ≥ |hv|(z − A)xi|

≥ Rehv|(z − A)xi ≥ Rezhv|xi = kxkRez.

Using Ran (z − A)−1 = X , we conclude that (z − A)−1 exists and k(z − A)−1 k ≤ |Rez|−1 . 2

3 Unbounded operators in Hilbert spaces

3.1 Graph scalar product
Let V, W be Hilbert spaces. Let A : V → W be an operator with domain DomA. It is natural to treat
DomA as a space with a scalar product

(v1 |v2 )A := (v1 |v2 ) + (Av1 |Av2 ).

Clearly, DomA is a Hilbert space with this product iff A is closed.

3.2 The adjoint of an operator

Definition 3.1 Let A : V → W have a dense domain. Then w ∈ DomA∗ , iff the functional

DomA 3 v 7→ (w|Av)

is bounded (in the topology of V). Hence there exists a unique y ∈ V such that

(w|Av) = (y|v), v ∈ V.

We set then
A∗ w = y.

16
 If CV , CW are the Riesz antiisomorphisms, then we have the relationship between the (Banach space)-
conjugate A# and the (Hilbert space)-adjoint A∗ :

A∗ = CV−1 A# CW .

Theorem 3.2 Let A : V → W have a dense domain. Then

(1) A∗ is closed.
(2) DomA∗ is dense in W iff A is closable.
(3) (Ran A)⊥ = KerA∗ .
(4) DomA ∩ (Ran A∗ )⊥ ⊃ KerA.

Proof. Let j : V ⊕ W → W ⊕ V, j(v, w) := (−w, v). Note that j is unitary. We have

Gr A∗ = j(Gr A)⊥ .

Hence Gr A∗ is closed. This proves (1).

Let us prove (2).
w ∈ (DomA∗ )⊥ ⇔ (0, w) ∈ (Gr A∗ )⊥

⇔ (w, ) ∈ (Gr A)⊥⊥ = (Gr A)cl .

Proof of (3):
w ∈ KerA∗ ⇔ (A∗ w|v) = 0, v ∈ V

⇔ (A∗ w|v) = 0, v ∈ DomA

⇔ (w|Av) = 0, v ∈ DomA

⇔ w ∈ (Ran A)⊥ .
Proof of (4)
v ∈ KerA ⇔ (w|Av) = 0, w ∈ W

⇒ (w|Av) = 0, w ∈ DomA∗

⇔ (A∗ w|v) = 0, w ∈ DomA∗

⇔ v ∈ (Ran A∗ )⊥ .

Theorem 3.3 Let A : V → W be closable with a dense domain. Then

(1) A∗ is closed with a dense domain.
(2) A∗ = (Acl )∗ .
(3) (A∗ )∗ = Acl
(4) (Ran A)⊥ = KerA∗ . Hence A∗ is injective iff Ran A is dense.
(5) (Ran A∗ )⊥ = KerA. Hence A is injective iff Ran A∗ is dense.

Proof. (1) was proven in Theorem 3.2.

To see (2) note that
Gr A∗ = j(Gr A)⊥ = j((Gr A)cl )⊥ = Gr Acl∗ .
To see (3) we use

⊥
Gr (A∗∗ ) = j −1 j(Gr A)⊥ = (Gr A)⊥⊥ = (Gr A)cl .

17
 (4) is proven in Theorem 3.2.
To prove (5) note that in the second line of the proof of Theorem 3.2 (4) we can use the fact that
DomA∗ is dense in W to replace ⇒ with ⇔. 2

3.3 Inverse of the adjoint operator

Theorem 3.4 Let A be densely defined, closed, injective and with a dense range. Then
(1) A−1 is densely defined, closed, injective and with a dense range.
(2) A∗ is densely defined, closed, injective and with a dense range.
(3) (A∗ )−1 = (A−1 )∗ .

Proof. (1) and (2) sum up previously proven facts.

To prove (3), recall the maps τ, j : V ⊕ W → W ⊕ V. We have

Gr A∗ = j(Gr A)⊥ , Gr A−1 = τ (Gr A).

Hence
Gr A−1∗ = j(τ (Gr A))⊥ = τ −1 (j(Gr A)⊥ ) = Gr A∗−1 .
2

Theorem 3.5 Let A : V → W be densely defined and closed. Then the following conditions are equiva-
lent:
(1) A is invertible.
(2) A∗ is invertible.
(3) For some c > 0, kAvk ≥ ckvk, v ∈ V and kA∗ wk ≥ ckvk, w ∈ W.
Moreover, spext (A) = spext (A∗ ).

Proof. (1)⇒(2). Let A be invertible. Then A−1 ∈ B(W, V). Hence, A−1∗ ∈ B(V, W).
Clearly, the assumptions of Theorem 3.4 are satisfied, and hence A∗−1 = A−1∗ . Therefore, A∗−1 ∈
B(V, W).
(1)⇐(2). A∗ is also densely defined and closed. Hence the same arguments as above apply.
It is obvious that (1) and (2) imply (3). Let us prove that (3)⇒(1). kA∗ vk ≥ ckvk implies that
KerA∗ = {0}. Hence (Ran A)⊥ is dense. This together with kAvk ≥ ckvk implies that Ran A = W, and
consequently, A is invertible. 2

3.4 Maximal operators

The numerical range of the operator T is defined as

NumT = {(v|T v) : v ∈ DomT, kvk = 1}.

Theorem 3.6 (1) k(z − T )vk ≥ dist(z, NumT )kvk, v ∈ DomT .

(2) If T is a closed operator and z ∈ C\(NumT )cl , then z − T has a closed range.
(3) If z ∈ rsT \NumT , then k(z − T )−1 k ≤ |dist(z, NumT )|−1 .
(4) Let ∆ be a connected component of C\(NumT )cl . Then either ∆ ⊂ spT or ∆ ⊂ rsT .

18
Proof. To prove (1), take (z0 6∈ NumT )cl . Recall that NumT is convex. Hence, replacing T wih αT + β
we can assume that z0 = iν and 0 ∈ NumT ⊂ {Imz ≤ 0}. Thus ν = dist(iν, NumT ) and

k(iν − T )vk2 = (T v|T v) − iν(v|T v) + iν(T v|v) + |ν|2 kvk2

= (T v|T v) − 2νIm(v|T v) + |ν|2 kvk2

≥ |ν|2 kvk2 .

(1) implies (2) and (3).

Let z0 ∈ rsT \NumT . By (3), if r = dist(z0 , NumT ), then {|z − z0 | < r} ⊂ rsT . This proves (4). 2

Definition 3.7 An operator T is called maximal, if spT ⊂ (NumT )cl .

Clearly, if T is a maximal operator, and z 6∈ (NumT )cl , then

k(z − T )−1 k ≤ (dist(z, NumT )))−1 .

If T is bounded, then T is maximal.

Theorem 3.8 Suppose that T is an operator and for any connected component ∆i of C\(NumT )cl we
choose λi ∈ ∆i . Then the following conditions are necessary and sufficient for T to be maximal
(1) For all i, λi 6∈ spT ;
(2) T is closable and for all i, Ran (λi − T ) = V.
(3) T is closed and for all i, Ran (λi − T ) is dense in V.
(4) T is closed and for all i, Ker(λi − T ∗ ) = {0}.

3.5 Dissipative operators

We say that an operator A is dissipative iff

Im(v|Av) ≤ 0, v ∈ DomA.

Equivalently, A is dissipative iff NumA ⊂ {Imz ≤ 0}.

A is maximally dissipative iff A is dissipative and spA ⊂ {Imz ≤ 0}.

Theorem 3.9 Let A be a densely defined operator. Then the following conditions are equivalent:
(1) −iA is the generator of a strongly continuous semigroup of contractions.
(2) A is maximally dissipative.

Proof. (1) ⇒(2) We have

Re(v|e−itA v) ≤ |(v|e−itA v)| ≤ kvk2 .
Hence
Im(v|Av) = −Rei(v|Av)

= Re limt%0 t−1 (v|e−itA v) − kvk2 ≤ 0.

Hence A is dissipative.
We know that the generators of contractions satisfy {Rez > 0} ⊂ rs(−iA).

19
 (2)⇒(1) Let Rez > 0. We have
kvkk(z + iA)vk ≥ |(v|(z + iA)v)|

≥ Re(v|(z + iA)v) ≥ Rezkvk2 .

Hence, noting that z ∈ rsA, we obtain k(z + iA)−1 k ≤ Rez −1 . Therefore, A is an operator of the type
(1, 0). 2

Theorem 3.10 Let A be dissipative. Then the following conditions are equivalent:
(1) A is maximally dissipative.
(2) A is closable and there exists z0 with Imz0 > 0 and Ran (z0 − A) = V.
(3) A is closed and there exists z0 with Imz0 > 0 and Ran (z0 − A) dense in V.
(4) A is closed and there exists z0 with Imz0 > 0 and Ker(z 0 − A∗ ) = {0}.

3.6 Hermitian operators I

An operator A : V → V is hermitian iff
(Aw|v) = (w|Av), w, v ∈ DomA.
Clearly, A is hermitian iff NumA ⊂ R.
Remark 3.11 In a part of literature the term “symmetric” is used instead of “hermitian”.
Theorem 3.12 Let A be densely defined and hermitian. Then A is closable. Besides, one of the
following possibilities is true:
(1) spA ⊂ R.
(2) spA = {Imz ≥ 0}.
(3) spA = {Imz ≤ 0}.
(4) spA = C.

Proof. A ⊂ A∗ and A∗ is closed. Hence A is closable. 2

Theorem 3.13 Let A be a densely defined operator. Then the following conditions are equivalent:
(1) −iA is the generator of a strongly continuous semigroup of isometries.
(2) A is hermitian and spA ⊂ {Imz ≤ 0}.
Proof. (1)⇒(2) For v ∈ DomA,

0 = ∂t (e−itA v|e−itA v) = −i(Av|v) + i(v|Av).

t=0

(2)⇒(1) We know that e−itA is the generator of a strongly continuous contractive semigroup. For
v ∈ DomA,
0 = ∂t (e−itA v|e−itA v)
Hence, for v ∈ DomA, ke−itA vk2 = kvk2 . 2

Theorem 3.14 Let A be hermitian. Then the following conditions are equivalent:
(1) spA ⊂ {Imz ≤ 0}.
(2) There exists z0 with Imz0 > 0 and Ran (z0 − A) = V.
(3) A is closed and there esxists z0 with Imz0 > 0 and Ran (z0 − A) dense in V.
(4) A is closed and there exists z0 with Imz0 > 0 and Ker(z 0 − A∗ ) = {0}.

20
3.7 Self-adjoint operators
Let T be a densely defined operator on V. T is self-adjoint iff T ∗ = T , that means if for w ∈ W there
exists y ∈ V such that
(y|v) = (w|T v), v ∈ DomT,
then w ∈ DomT and T w = y.

Theorem 3.15 Every self-adjoint operator is hermitian and closed. If A is bounded, then it is self-
adjoint iff it is hermitian.

Theorem 3.16 Fix z± with ±Imz± > 0. Let A be hermitian. Then the following conditions are
necessary and sufficient for A to be self-adjoint:
(1) spA ⊂ R.
(2) z± 6∈ spA.
(3) Ran (z± − A) = V.
(4) A is closed and Ran (z± − A) is dense in V.
(5) A is closed and Ker(z ± − A∗ ) = {0}.

Theorem 3.17 Let λ0 ∈ R. Let A be hermitian. Then the following conditions are sufficient for A to
be self-adjoint:
(1) λ0 6∈ spA.
(2) Ran (λ0 − A) = V.
(3) A is closed and Ran (λ0 − A) is dense in V.
(4) A is closed and Ker(λ0 − A∗ ) = {0}.

Theorem 3.18 Let A be self-adjoint. Then U := (A + i)(A − i)−1 is a unitary operator with spU =
(spext A + i)(spext A − i)−1 .

If f ∈ C(spext A), we can define

f (A) := f (i(U + i)(U − 1)−1 ).

Of course, we can also apply the functional calculus for measurable functions. In particular, the function
spA 3 x 7→ id(x) := x is a measurable function on spA. We have idA = A.

Theorem 3.19 (Stone Theorem) Let A be an operator. Then the following conditions are equiva-
lent:
(1) iA is the generator of a strongly continuous group of unitary operators.
(2) A is self-adjoint.

Proof. To prove (1)⇒(2), suppose that R 7→ U (t) is a strongly continuous unitary group. Let −iA be
its generator. Then [0, ∞[3 U (t), U (−t) are semigroups of contractions with the generators iA and −iA.
By Theorem 3.19, A is hermitian and spA ⊂ R. Hence A is self-adjoint.
(2)⇒(1) follows by the spectral theorem. 2

21
3.8 Essentially self-adjoint operators
Definition 3.20 An operator A : V → V is essentially self-adjoint iff Acl is self-adjoint.

Theorem 3.21 Every essentially self-adjoint operator is hermitian and closable.

Theorem 3.22 Fix z± with ±Imz± > 0 Let A be hermitian. Then the following conditions are necessary
and sufficient for A to be essentially self-adjoint:
(1) A∗ is self-adjoint.
(2) Ran (z± − A) is dense in V.
(3) Ker(z ± − A∗ ) = {0}.

Theorem 3.23 Let λ0 ∈ R. Let A be hermitian. Then the following conditions are sufficient for A to
be self-adjoint:
(1) Ran (λ0 − A) is dense in V.
(2) Ker(λ0 − A∗ ) = {0}.

3.9 Scale of Hilbert spaces

Let B be a positive operator on V with B ≥ 1. We define the family of Hilbert spaces Vα , α ∈ R as
follows. For α ≥ 0, we set Vα := Ran B −α/2 with the scalar product

(v|w)α := (v|B α w),

and V−α := Vα∗ , (where Vα∗ denotes the space of bounded antilinear functionals on Vα ). Note that we
have the identification V = V ∗ , hence both definitions give V0 = V.
It is clear, that for 0 ≤ α ≤ β, V ⊃ Vα ⊃ Vβ . Hence V−α = Vα∗ can be identified with a subspace of
V−β = Vβ∗ . Thus we obtain Vα ⊃ Vβ for any α ≤ β.
Note that for α > 0 V is embedded in V−α and for v, w ∈ V
 
(v|w)−α = B −α/2 v|B −α/2 w .

Moreover, V is dense in V−α .

It is easy to see that, for α ≥ 0, B0−α := B −α is a unitary operator from V0 to V2α . Moreover, the
operator B α , defined by the functional calculus (or as the inverse of the bounded operator B −α , with the
domain V2α extends to a unitary operator from V0 to V−α , which we will denote by B0α . For any α, β,
setting Bαα−β := B0−β (B0−α )−1 we obtain a unitary operator from V2α to V2β . These operators satisfy
the chain rule:
−γ
Bα+2β Bα−β = Bα−γ−β .
Sometimes we will use a different notation: B −α V = V2α . If A is a self-adjoint operator, then we will
use the notation hAi := (1 + A2 )1/2 .

Lemma 3.24 Let 0 ≤ α ≤ 1. Then DomB = {v ∈ Vα : Bα v ∈ V}.

3.10 Relative operator boundedness

Theorem 3.25 Let A be a closed operator and B an operator with DomB ⊃ DomA. Then the following
statements are equivalent:

22
(1) B has the A-bound equal to a1 , that is
 21
kBvk2

inf sup = a1 .
ν>0 v6=0, v∈DomA kAvk2 + ν 2 kvk

(2)

inf kB(A∗ A + ν 2 )−1/2 k = a1 .

ν>0

If, moreover, A is self-adjoint, then the above statements are equivalent to

(3) inf kB(iν − A)−1 k = a1
ν>0

Proof. The equivalence of (1) and (2) is evident.

The equivalence of (2) and (3) for a self-adjoint A is the consequence of the unitarity of

(A2 + ν 2 )−1/2 (iν − A).

Theorem 3.26 (Kato-Rellich) Let A be self-adjoint, B hermitian. Let B be A-bounded with the
A−bound < 1. Then
(1) A + B is self-adjoint on DomA.
(2) If A is essentally self-adjoint on D, then A + B is essentially self-adjoint on D.

Proof. Clearly, A + B is hermitian on DomA. Moreover, for some ν, kB(iν − A)−1 k < 1 and (which
is equivalent by the unitarity of (A − iν)(A + iν)−1 ), kB(−iν − A)−1 k < 1. Hence, iν − A − B and
−iν − A − B are invertible. 2
Let us note an improved version of the notion of the operator boundedness:

Theorem 3.27 Let A be a closed operator and B an operator with DomB ⊃ DomA. Then the following
statements are equivalent:
(1)
 12
kBvk2

a2 = inf sup .
µ,ν>0 v6=0 k(A − µ)vk2 + ν 2 kvk

(2)

inf kB((A − µ)∗ (A − µ) + ν 2 )−1/2 k = a2 .

µ,ν>0

If, moreover, A is self-adjoint, then the above statements are equivalent to

(3) inf kB(µ + iν − A)−1 k = a2
µ,ν>0

Note hat the analog of Theorem 3.26 is true with a1 replaced with a2 .

23
3.11 Relative form boundedness
Theorem 3.28 Let A be a self-adjoint operator. Let B be a bounded operator from (1 + |A|)−1/2 H to
(1 + |A|)1/2 H. Then the following statements are equivalent:
(1)
1 1
inf k(A − µ)2 + ν 2 )− 4 B((A − µ)2 + ν 2 )− 4 k = a3 .
µ,ν

(2)
1 1
inf k(µ + iν − A)− 2 v, (µ + iν − A)− 2 k = a3 .
µ,ν>0

If the conditions of the above theorem are satified, then we say that the A-form-bound of B equals
a3 .

Theorem 3.29 Let A be a self-adjoint operator. Let B have the A-form-bound less than 1. Then there
exists a open subsets in the upper and lower complex half-plane such that the series
∞
X
R(z) := (z − A)−1 (B(z − A)−1 )j
j=0

is convergent. Moreover, R(z) is a resolvent of a self-adjoint operator, which will be called the form sum
1 1
of A and B. If A is bounded from below, then so is A + B and Dom|A + B| 2 = Dom|A| 2 .

3.12 Non-maximal operators

Theorem 3.30 dim Ran (z − A)⊥ = dim Ker(z − A∗ ) is a constant function on connected components
of C\(NumA)cl .

Proof. Let us show that if |α| < λ, then

Ran (iλ − A) ∩ Ran (iλ + α − A)⊥ = {0}. (3.14)

Let w ∈ Ran (iλ − A). Then there exists v ∈ DomA such that

w = (iλ − A)v

and kvk ≤ λ−1 kwk. If moreover, w ∈ Ran (iλ + α − A)⊥ = Ker(−iλ − α − A∗ ), then

0 = ((−iλ + α − A∗ )w|v)
= (w|(iλ − A)v) + α(w|v)
= kwk2 + α(w|v).

But
|kwk2 + α(w|v)| ≥ (1 − |α|/|λ|)kwk2 > 0,
which is a contradiction and completes the proof of (3.14).
Now (3.14) implies that dim Ran (iλ − A) ≤ dim Ran (iλ + α − A). 2

24
3.13 Hermitian operators II
Let A be closed hermitian.
Theorem 3.31 The so-called defect indices

n± := dim Ker(z − A∗ ), z ∈ C±

do not depend on λ. One of the following possibilities is true:

1) spA ⊂ R, n± = 0, A is self-adjoint;
2) spA = {Imz ≥ 0}, n+ 6= 0, n− = 0, A is not self-adjoint;
3) spA = {Imz ≤ 0}, n+ = 0, n− 6= 0, A is not self-adjoint;
4) spA = C n+ 6= 0, n− 6= 0, A is not self-adjoint.

Proof. This is a special case of Theorem 3.30. 2

Definition 3.32 Let A be hermitian and closed. Define on DomA∗ the following scalar product:

(v|w)A∗ := (v|w) + (A∗ v|A∗ w)

and the following antihermitian form:

[v|w]A∗ := (A∗ v|w) − (v|A∗ w).

The A∗ −completeness and the A∗ −orthogonality is defined using the scalar product (·|·)A∗ . A space is
A∗ −hermitian iff [·|·]A∗ vanishes on this subspace.

Theorem 3.33 (1) Every closed extension of A is a restriction of A∗ to an A∗ −closed subspace in

DomA∗ containing DomA.
(2)

DomA∗ = DomA ⊕ Ker(A∗ + i) ⊕ Ker(A∗ − i)

and the components in the above direct sum are A∗ -closed, A∗ −orthogonal and

(w ⊕ w+ ⊕ w− |v ⊕ v+ ⊕ v− )A∗ = (w|v) + (Aw|Av) + 2(w+ |v+ ) + 2(w− |v− ),

[w ⊕ w+ ⊕ w− |v ⊕ v+ ⊕ v− ]A∗ = 2i(w+ |v+ ) − 2i(w− |v− ).

Proof. (1) is obvious. In (2) the A∗ −orthogonality and the A∗ −closedness are easy.
Let w ∈ DomA∗ and
w ⊥ DomA ⊕ Ker(A∗ + i) ⊕ Ker(A∗ − i)
in the sense of the product (·|·)A∗ . In particular, for v ∈ DomA we have

0 = (A∗ w|A∗ v) + (w|v) = (A∗ w|Av) + (w|v).

Hence A∗ w ∈ DomA∗ and

A∗ A∗ w = −w.
Therefore,
(A∗ + i)(A∗ − i)w = 0.

25
Thus
(A∗ − i)w ∈ Ker(A∗ + i). (3.15)
If y ∈ Ker(A∗ + i), then

i(y|(A∗ − i)w) = (A∗ η|A∗ w) + (η|w) = (η|w)A∗ = 0

In particular, setting y = (A∗ − i)w and using (3.15), we get

w ∈ Ker(A∗ − i).

But w ⊥ Ker(A∗ − i), hence w = 0. 2

Every A∗ −closed subspace containing DomA is of the form DomA ⊕ Z, where

Z ⊂ Ker(A∗ + i) ⊕ Ker(A∗ − i).

If
A ⊂ B ⊂ A∗ ,
then the subspace Z corresponding to B will be denoted by ZB . We will write

Z per := {v ∈ Ker(A∗ − i) ⊕ Ker(A∗ − i) : [z, v]A∗ = 0, z ∈ Z}.

The subspace Z is A∗ −hermitian iff

Z per ⊃ Z.
Theorem 3.34 We have
ZB ∗ = (ZB )per .
In particular, B is hermitian iff Z is A∗ −hermitian. Every A∗ −hermitian subspace corresponds to a
partial isometry U : Ker(A∗ + i) → Ker(A∗ − i). Then

Z := {z ⊕ U z : z ∈ Ran U ∗ U }.

B is self-adjoint iff U is unitary.

4 Sesquilinear forms
4.1 Sesquilinear forms
Let V, W be complex spaces. We say that t is a sesquilinear quasiform on W × V iff there exist subspaces
Doml t ⊂ W and Domr t ⊂ V such that

Doml t × Domr t 3 (w, v) 7→ t(w, v) ∈ C

is a sesquilinear map. From now on by a sesquilinear form we will mean a sesquilinear quasiform.
We define a form t∗ with the domains Doml t∗ := Domr t, Domr t∗ := Doml t, by the formula t∗ (v, w) :=
t(w, v). If t1 are t2 forms, then we define t1 + t2 with the domain Doml t1 + t2 := Doml t1 ∩ Doml t1 ,
Domr t1 + t2 := Domr t1 ∩ Domr t1 by (t1 + t2 )(w, v) := t1 (w, v) + t2 (w, v). We write t1 ⊂ t2 if Doml t1 ⊂
Doml t2 , Domr t1 ⊂ Domr t2 , and t1 (w, v) = t2 (w, v) w ∈ Doml t1 , v ∈ Domr t1 .
t is bounded iff
|t(w, v)| ≤ ckwkkvk, w ∈ Doml t, v ∈ Domr t.
From now on, we will usually assume that W = V and Doml t = Domr t and the latter subspace will
be simply denoted by Domt.

26
 Recall that the numerical range of the form t is defined as
Numt := {t(v) : v ∈ Domt, kvk = 1}.
We proved that Numt is a convex set.
The form t is bounded iff Numt is bounded. Equivalently, |t(v)| ≤ ckvk2 .
t is bounded from below, if there exists c such that
Numt ⊂ {z : Rez > c}.
t is hermitian iff Numt ⊂ R. The equivalent condition: t(w, v) = t(v, w).
If T is an operator on V, then t(w, v) := (w, T v) with the domain DomT is a form called the form
associated with the operator T . Clearly, Numt = NumT .

4.2 Closed positive forms

Let s be a positive form.
Definition 4.1 We say that s is a closed form iff Doms with the scalar product
(w, v)s := (s + 1)(w, v), w, v ∈ Doms,
is a Hilbert space.
Theorem 4.2 The form s is closed iff for any sequence (xn ) in Doms, if xn → x and s(xn − xm ) → 0,
then x ∈ Doms and s(xn − x) → 0.
Example 4.3 Let A be an operator. Then (Aw|Av) with the domain DomA is a closed form iff A is
closed.

4.3 Closable positive forms

Let s be a positive form.
Definition 4.4 We say that s is a closable form iff there exists a closed form s1 such that s ⊂ s1 .
Theorem 4.5 (1) The form t is closable ⇔ for any sequence (xn ) ⊂ Doms, if xn → 0 and s(xn −
xm ) → 0, then s(xn ) → 0.
(2) If s is closable, then there exists the smallest closed form s1 such that s ⊂ s1 . We will denote it by
scl .
(3) Nums is dense in Numscl
Proof. (1) ⇒ follows immediately from Theorem 4.2.
To prove (1) ⇐, define s1 as follows: v ∈ Doms1 , iff there exists a sequence (vn ) ⊂ Doms such that
vn → v and s(vn − vm ) → 0. We set then s1 (v) := limn→∞ s(vn ) (the limit exists, because s(vn ) is a
Cauchy sequence).
To show that the definition is correct, suppose that wn ∈ Doms, wn → v and s(wn − wm ) → 0.
Then vn − wn − (vm − wm ) → 0 and vn − wn → 0. By the hypothesis we get s(vn − wn ) → 0. Hence,
limn→∞ s(vn ) = limn→∞ s(wn ). Thus the definition of s1 does not depend on the choice of the sequence
vn . It is clear that s1 is a closed form containing s. Hence s is closable.
To prove (2) note that the form s1 constructed above is the smallest closed form containg s. 2

Example 4.6 Let A be an operator. Then (Aw|Av) with the domain DomA is closable iff A is a closable
operator. Then (Acl w|Acl v) with the domain DomAcl is its closure.

27
4.4 Operators associated with positive forms
Let S be a positive self-adjoint operator. We define the form s as follows: Doms := DomS 1/2 and
s(v, w) := (S 1/2 v|S 1/2 w). We will say that s is the form associated with S.

Theorem 4.7 (1) s is a closed form.

(2) DomS is an essential domain for s.
(3) NumS is dense in Nums.

The next theorem describes the converse construction.

Theorem 4.8 Let s be a densely defined closed positive form. Then there exists a unique positive self-
adjoint operator S such that Doms = DomS 1/2 and s(v, w) := (S 1/2 v|S 1/2 w). We will say that S is the
operator associated with the form s.

Proof. For w ∈ V, v ∈ Doms, we have

|(v|w)| ≤ kvkkwk ≤ kvks kwk.

By the Riesz lemma, there exists A : V → Doms such that

(v|w) = (v|Aw)s , (4.16)

kAwk ≤ kAwks ≤ kwk.

KerA = {0}, because Aw = 0 implies (v|w) = 0 for v ∈ Doms, and Doms s dense in V. Besides, A is
self-adjoint. Putting
S := A−1 − 1
we define a positive self-adjoint operator.

s(v, y) = (v|Sy), v ∈ Doms, y ∈ DomS = Ran A.

Let us show that Doms is an essential domain for s. Let v ∈ Doms is s-orthogonal to Ran A = DomS.
Then v is orthogonal to Doms—see (4.16). Hence v = 0.
1 1
Define s1 by Doms1 = DomS 2 and s1 (w, v) = (S 2 w|S 1/2 v). The form s and s1 coincide on DomS ⊂
1
DomS 2 ∩ Doms. We proved above that DomS is an essential domain for s. The form s1 is obviously
closed and Doms1 is an essential domain for s1 . Hence, s1 = s. 2

4.5 Polar decomposition

Theorem 4.9 Let A be a densely defined closed operator. Let B be the operator associated to the form
(Aw|Av). Let Vα be the scale of spaces (B + 1)−α V, so that V1 = DomA and V−1 = V1∗ . Then
(1) A, treated as an operator from V1 to V0 , and denoted A1 , is bounded.
(2) A∗ extends by density to a bounded operator, denoted A∗0 from V0 to V−1 . We have A∗0 = (A1 )∗ .
(3) DomA∗ = {v ∈ V : A∗0 v ∈ V}.
(4) DomB = {v ∈ DomA : Av ∈ DomA∗ } and for v ∈ DomB, Bv = A∗ Av.

28
Proof. (1) is obvious.
To see (2), note that for v ∈ DomA∗ , w ∈ DomA we have

|(A∗ v|w)| = (v|Aw)| ≤ kvkkwk1 .

Hence kA∗ vk−1 ≤ kvk, and so A∗ : V → V−1 is bounded. DomA∗ is dense in V. Hence A∗ extends to a
bounded operator A∗0 : V → V−1 .
To prove (3), let v ∈ V, A∗0 v = w ∈ V. Then there exists (vn ) ⊂ DomA∗ such that vn → v in the
norm of V and A∗ vn → w in the norm of V−1 . Hence for x ∈ DomA,

(vn |Ax) = (A∗ vn |x) → (w|x).

Therefore, (w|x) = (v|Ax). Hence, v ∈ DomA∗ and A∗ v = w.

To prove (4), denote B1 as the extension of B to the operator from V1 to V−1 . Note that B1 is
bounded.
We have B1 = A∗0 A1 . In fact, for v, w ∈ V1

(w|B1 v) = (B 1/2 w|B 1/2 v) = (Aw|Av) = (w|A∗0 A1 v).

Now
DomB = {v ∈ V1 : B1 v ∈ V}

= {v ∈ V1 : A∗0 A1 v ∈ V}

= {v1 : Av ∈ DomA∗ }.
2
Motivated by the above theorem we will write A∗ A for B.

Theorem 4.10 Let A be closed. Then there exist a unique positive operator |A| and a unique partial
isometry U such that KerU = KerA and A = U |A|. We have then Ran U = Ran Acl .

Proof. The operator A∗ A is positive. By the spectral theorem, we can then define
√
|A| := A∗ A.

On Ran |A| the operator U is defined by

U |A|v := Av.
It is isometric, because
k|A|vk2 = (v||A|2 v) = (v|A∗ Av) = kAvk2 ,
and correctly defined. We can extend it to (Ran |A|)cl by continuity. On Ker|A| = (Ran |A|)cl , we extend
it by putting U v = 0. 2

4.6 Sectorial forms

A subset of C of the form Sec(a, θ) := a + {z : | arg z| < θ} with θ < π2 will be called a sector. a will be
called its tip and θ its angle. We say that a form t is sectorial iff there exists a ∈ C and θ < π2 such that
Numt ⊂ Sec(a, θ).

Lemma 4.11 Let t be a sectorial form with the sector given by a, θ. Then
1 1
|(t − a)(w, v)| ≤ (1 + tan θ)Re(t − a)(w) 2 Re(t − a)(v) 2 .

29
 Clearly, if t is sectorial and a is the tip of a sector containing Numt, then Ret + Rea is a positive form.
We say that a sectorial form t is closed iff Ret is closed. We say that a sectorial form t is closable iff Ret
is closable.
It is easy to see, using Lemma 4.11, that Theorems 4.2 and 4.5 remain true if the form s is assumed
to be sectorial.

4.7 Operators associated with sectorial forms

Definition 4.12 An operator T is called sectorial iff the associated form is sectorial, that means, if
NumT ⊂ Sec(a, θ). A maximal sectorial operator is called shortly m-sectorial.

Theorem 4.13 (1) Let t be a sectorial form. Then there exists a unique m-sectorial operator T such
that DomT ⊂ Domt and

t(w, v) = (w|T v), v ∈ DomT, w ∈ Domt.

T is called the operator associated with the form t and denoted by Tt .

(2) DomT is an essential domain for t.
(3) NumT is dense in Numt.
(4) If t is bounded, then so is T .

We will assume that the sector of t has the tip at 0. We will write s := Ret.

Lemma 4.14 There exists an invertible operator B ∈ B(Doms) such hat

(t + 1)(w, v) = s(w, Bv) + (w|Bv).

Proof. By Lemma 4.11, the form t + 1 is bounded in the Hilbert space Doms. Hence there exists
B ∈ B(Doms) such that
(t + 1)(w, v) = (w|Bv)s = (w|Bv) + s(w, Bv).
We have
kvk2s = Re(t + 1)(v) = Re(v|Bv)s ≤ kBvks kvks .
Hence kvks ≤ kBvks . Therefore, Ran B is closed.
If w is orthogonal in Doms to Ran B, then

kwk2s = Re(w|Bw)s = 0.

Hence, w = 0. Therefore, B is invertible. 2

Proof of Theorem 4.13. Let S denote the operator associated with s. Then T := S(B + 1) − 1 with
the domain DomT = DomS satisfies the conditions of the theorem. 2

4.8 Perturbations of sectorial forms

Theorem 4.15 Let t1 and t2 be sectorial forms.
(1) t1 + t2 is also a sectorial form.
(2) If t1 and t2 are closed, then t1 + t2 is closed as well.
(3) If t1 and t2 are closable, then t1 + t2 is closable as well and (t1 + t2 )cl ⊂ tcl cl
1 + t2 .

30
Definition 4.16 Let p, s be forms and let s be positive. We say that p is s-bounded iff Doms ⊂ Domp
and
|p(v)|
b := inf sup < ∞.
c>0 v∈Doms s(v) + ckvk2

The number b is called the s-bound of p.

Theorem 4.17 Let t be sectorial and let p be Ret-bounded with the Ret-bound < 1. Then
(1) The form t + p (with the domain Domt) is sectorial as well.
(2) t is closed ⇔ t + p is closed.
(3) t is closable ⇔ t + p is closable, and then Dom(t + p)cl = Domtcl .

Proof. Let us prove (1). For some b < 1, we have

|p(v)| ≤ bRet(v) + ckvk2 .

Hence
Im(t + p)(v) ≤ |Imt(v)| + |Imp(v)| ≤ (tan θ + b)Ret(v) + ckvk2 ,
Re(t + p)(v) ≥ Ret(v) − |Imp(v)| ≥ (1 − b)Ret(v) − ckvk2 . (4.17)
Hence,
|Im(t + p)(v)| ≤ (1 − b)−1 (tan θ + b) Re(t + p)(v) + ckvk2 + ckvk2 .

This proves that t + p is sectorial.

To see (2) and (3), note that (4.17) and

(1 + b)Ret(v) + ckvk2 ≥ Re(t + p)(v)

prove that the norms k · kt and k · kt+p are equivalent. 2

4.9 Friedrichs extensions

Theorem 4.18 Let T be a sectorial operator. Then the form t(w, v) := (w|T v) is closable.

Proof. It suffices to assume that the tip of the sector of t is 0. Suppose that wn ∈ DomT = Domt,
wn → 0, limn,m→∞ t(wn − wm ) = 0. Then

|t(wn )| ≤ |t(wn − wm , wn )| + |t(wm , wn )|

1 1
≤ (1 + tan θ)(Ret(wn )) 2 (Ret(wn − wm )) 2 + (wm |T wn ).

For any  > 0 there exists N such that for n, m > N we have Ret(wn −wm ) ≤ 2 . Besides, limm→∞ (wm |T wn ) =
0. Therefore,
|t(wn )| ≤ (1 + tan θ)|t(wn )|1/2 .
Hence t(wn ) → 0. 2
Thus there exists a unique m-sectorial operator TFr associated with the form tcl . The operator TFr is
called the Friedrichs extension of T .

31
5 Aronszajn-Donoghue and Friedrichs Hamiltonian and their
renormalization
5.1 Aronszajn Donoghue Hamiltonians
Let H0 be a self-adjoint operator on H, h ∈ H and λ ∈ R.

H := H0 + λ|h)(h|, (5.18)

is a rank one perturbation of H0 . We will call (5.18) the Aronszajn Donoghue Hamiltonian.
We would like to describe how to define the Aronszajn-Donoghue Hamiltonian if h is not necessarily
a bounded functional on H. It will turn out that it is natural to consider 3 types of h:

I. h ∈ H, II. h ∈ H−1 \H, III. h ∈ H−2 \H−1 , (5.19)

where by H−n we denoted the usual scale of spaces associated to the operator H0 , that is H−n :=
hH0 in/2 H, where hH0 i := (1 + H02 )1/2 .
Clearly, in the case I H is self-adjoint on DomH0 . We will see that in the case II one can easily
define H as a self-adjoint operator, but its domain is no longer equal to DomH0 . In the case III, strictly
speaking, the formula (5.18) does not make sense. Nevertheless, it is possible to define a renormalized
Aronszajn-Donoghue Hamiltonian. To do this one needs to renormalize the parameter λ. This procedure
resembles the renormalization of the charge in quantum field theory.
Consider first the case I. We can compute its resolvent. In fact, for z 6∈ spH0 we define an analytic
function
g(z) := −λ−1 + (h|(z − H0 )−1 h). (5.20)
Then for z ∈ Θ := {z ∈ C\spH0 : g(z) 6= 0} and λ 6= 0, the resolvent of the operator H is given by
Krein’s formula
R(z) = (z − H0 )−1 − g(z)−1 (z − H0 )−1 |h)(h|(z − H0 )−1 . (5.21)
For λ = 0, we set Θ = C\spH0 and clearly

R(z) = (z − H0 )−1 . (5.22)

The following theorem will describe how to define the Aronszajn-Donoghue Hamiltonian in all the
cases I, II and III:
Theorem 5.1 Assume that:
(A) h ∈ H−1 , λ ∈ R ∪ {∞} and let R(z) be given by (5.22) or (5.21) with g(z) given by (5.20),
or
(B) h ∈ H−2 , γ ∈ R and let R(z) be given by (5.21) with g(z) given by

g(z) := γ + h|((z − H0 )−1 + H0 (1 + H02 )−1 )h .

Then, for all z ∈ Θ,

(1) R(z) is a bounded operator which fulfills the first resolvent formula;
(2) KerR(z) = {0}, unless h ∈ H and λ = ∞;
(3) Ran R(z) is dense in H, unless h ∈ H and λ = ∞;
(4) R(z)∗ = R(z).
Hence, except for the case h ∈ H, λ = ∞, there exists a unique densely defined self-adjoint operator H
such that R(z) is the resolvent of H.

32
 Another way to define H for the case h ∈ H−2 is the cut-off method. For all k ∈ N we define hk as
in (5.35) and fix the running coupling constant by
−λ−1 2 −1
k := γ + (hk |H0 (1 + H0 ) hk )
and set the cut-off Hamiltonian to be
Hk := H0 + λk |hk )(hk |. (5.23)
Then the resolvent for Hk is given by
Rk (z) = (z − H0 )−1 + gk (z)−1 (z − H0 )−1 |hk )(hk |(z − H0 )−1 , (5.24)
where
gk (z) := −λ−1 −1


k + hk |(z − H0 ) hk . (5.25)
1
Note that λk is chosen in such a way that the renormalization condition (gk (i) + gk (−i)) = γ. holds.
2
It is easy to see that if H0 is bounded from below, then limk→∞ λk = 0. Again, the cut-off Hamiltonian
converges to the renormalized Hamiltonian:
Theorem 5.2 Assume that h ∈ H−2 . Then lim Rk (z) = R(z).
k→∞

Let us assume that h is cyclic. Then the support of the spectral measure of h wrt H0 is spH0 . If
g(β) = 0 and β 6∈ spH0 , then H has an eigenvalue at β and the corresponding eigenprojection equals
1{β} (H) = (h|(β − H0 )−2 h)−1 (β − H0 )−1 |h)(h|(β − H0 )−1 .

5.2 Aronszajn-Donoghue Hamiltonians and extensions of Hermitian opera-

tors
Let H0 be as above and h ∈ H−2 \H. Define Hmin to be the restriction of H0 to
Dom(Hmin ) := {v ∈ Dom(H0 ) = H2 : (h|v) = 0}.
∗
Then Hmin is a closed Hermitian operator. Set Hmax := Hmin . Then
Dom(Hmax ) = Span{DomHmin , (i − H0 )−1 h, (−i − H0 )−1 h, }.
Note that Ker(Hmax ± i) is spanned by
v± := (±i − H0 )−1 h.
Thus the indices of defect of Hmin are (1, 1).
The operators Hγ discussed in the previous subsection are self-adjoint extensions of Hmin . To obtain
Hγ it suffices to increase the domain of Hmin by adding the vector
θγ (i − H0 )−1 h − θγ (i + H0 )−1 h,
γ+(h|H0 (1+H02 )−1h)
where θγ := γ−i(h|(1+H02 )−1 h)
.

5.3 Aronszajn-Donoghue Hamiltonians and extensions of positive forms

Assume now in addition that H0 is positive.
Consider the positive form hmin associated with Hmin . Thus h(v, v) = (v|Hmin v) = (v|H0 v) with the
domain Dom(hmin ) := DomHmin .
Assume first that h ∈ H−1 . The Friedrichs extension of Hmin is Hλ with λ = ∞. The closure of the
form hmin has the domain {v ∈ H1 : (h|v) = 0}.
Assume now that h ∈ H−2 \H−1 . Then the Friedrichs extension of Hmin equals H0 .

33
5.4 Friedrichs Hamiltonian
Let H0 be again a self-adjoint operator on the Hilbert space H. Let  ∈ R and h ∈ H. The following
operator on the Hilbert space C ⊕ H is often called the Friedrichs Hamiltonian:
" #
 (h|
G := . (5.26)
|h) H0

Recall that expression the operators (h| and |h) are defined by

H 3 v 7→ (h|v := (h|v) ∈ C,
(5.27)
C 3 α 7→ |h)α := αh ∈ H.

We would like to describe how to define the Friedrichs Hamiltonian if h is not necessarily a bounded
functional on H. It will turn out that it is natural to consider 3 types of h:

I. h ∈ H, II. h ∈ H−1 \H, III. h ∈ H−2 \H−1 , (5.28)

Clearly, in the case I G is self-adjoint on C ⊕ DomH0 . We will see that in the case II one can easily
define G as a self-adjoint operator, but its domain is no longer equal to C⊕DomH0 . In the case III, strictly
speaking, the formula (5.26) does not make sense. Nevertheless, it is possible to define a renormalized
Friedrichs Hamiltonian. To do this one needs to renormalize the parameter . This procedure resembles
the renormalization of mass in quantum field theory. Let us first consider the case h ∈ H. As we said
earlier, the operator G with DomG = C ⊕ DomH0 is self-adjoint. It is well known that the resolvent of
G can be computed exactly. In fact, for z 6∈ spH0 define the analytic function

g(z) :=  + (h|(z − H0 )−1 h). (5.29)

Then for z ∈ Ω := {z ∈ C\spH0 : g(z) − z 6= 0} the resolvent Q(z) := (z − G)−1 is given by

" #
0 0
Q(z) = (5.30)
0 (z − H0 )−1
(h|(z − H0 )−1
" #
 −1 1
+ z − g(z) .
(z − H0 )−1 |h) (z − H0 )−1 |h)(h|(z − H0 )−1

Theorem 5.3 Assume that:

(A) h ∈ H−1 ,  ∈ R and let Q(z) be given by (5.30) with g(z) defined by (5.29),
or
(B) h ∈ H−2 , γ ∈ R and let Q(z) be given by (5.30) with g(z) defined by

g(z) := γ + h|((z − H0 )−1 + H0 (1 + H02 )−1 )h

  (5.31)
= γ + h|( 2(z−Hi−z0 )(i−H0 )
− i+z
2(z−H0 )(−i−H0 ) )h

Then for all z ∈ Ω :

(1) Q(z) is a bounded operator which fulfills the first resolvent formula (in the terminology of [Ka], Q(z)
is a pseudoresolvent);
(2) KerQ(z) = {0};
(3) Ran Q(z) is dense in C ⊕ H;

34
 (4) Q(z)∗ = Q(z).
Therefore, by [Ka], there exists a unique densely defined self-adjoint operator G such that Q(z) = (z −
G)−1 . More precisely, for any z0 ∈ Ω, DomG = Ran Q(z0 ), and if ϕ ∈ Ran Q(z0 ) and Q(z0 )ψ = ϕ, then

Gϕ := −ψ + z0 Q(z0 )ψ,

Proof. Let z ∈ Ω. It is obvious that Q(z) is bounded and satisfies (4). We easily see that both in the
case (A) and (B) the function g(z) satisfies

g(z1 ) − g(z2 ) = −(z1 − z2 )(h|(z1 − H0 )−1 (z2 − H0 )−1 |h). (5.32)

Direct computations using (5.32) show the first resolvent formula.

Let (α, f ) ∈ C ⊕ H be such that (α, f ) ∈ KerQ(z). Then
 
0 = (z − g(z))−1 α + (h|(z − H0 )−1 f ) , (5.33)
 
0 = (z − H0 )−1 f + (z − H0 )−1 h(z − g(z))−1 α + (h|(z − H0 )−1 f ) . (5.34)

Inserting (5.33) into (5.34) we get 0 = (z − H0 )−1 f and hence f = 0. Now (5.33) implies α = 0, so
KerQ(z) = {0}.
Using (2) and (4) we get (Ran Q(z))⊥ = KerQ(z)∗ = KerQ(z) = {0}. Hence 3) holds. 2
Let h ∈ H−2 and γ ∈ R. Let us impose a cut-off on h. For k ∈ N we define

hk := 1[−k,k] (H0 ) h, (5.35)

where 1[−k,k] (H0 ) is the spectral projection for H0 associated with the interval [−k, k] ⊂ R. Note that
hk ∈ H and hence both (hk | and |hk ) are well defined bounded operators. Set

k := γ + (hk |H0 (1 + H02 )−1 hk ).

For all k ∈ N, the cut-off Friedrichs Hamiltonian

" #
k (hk |
Gk :=
|hk ) H0

is well defined and we can compute its resolvent, Qk (z) := (z − Gk )−1 :

" #
0 0
Qk (z) = (5.36)
0 (z − H0 )−1
(hk |(z − H0 )−1
" #
 −1 1
+ z − gk (z) .
(z − H0 )−1 |hk ) (z − H0 )−1 |hk )(hk |(z − H0 )−1

where
gk (z) := k + (hk |(z − H0 )−1 hk ). (5.37)
1
Note that k is chosen such a way that the following renormalization condition is satisfied: 2 (gk (i) + gk (−i)) =
γ. Let us also mention that if H0 is bounded from below, then limk→∞ k = ∞.

Theorem 5.4 Assume that h ∈ H−2 . Then lim Qk (z) = Q(z), where Q(z) is given by (5.30) and g(z)
k→∞
is given by (5.31).

35
Proof. The proof is obvious if we note that lim k(z−H0 )−1 h−(z−H0 )−1 hk k = 0 and lim gk (z) = g(z).
k→∞ k→∞
2
Thus the cut-off Friedrichs Hamiltonian is norm resolvent convergent to the renormalized Friedrichs
Hamiltonian.
Let us assume that h is cyclic. Then the support of the spectral measure of h wrt H0 is spH0 . If
β = g(β) = 0 and β 6∈ spH0 , then G has an eigenvalue at β. The corresponding projection equals
(h|(β − H0 )−1
 
−2 −1 1
1β (G) = (1 + (h|(β − H0 ) |h)) .
(β − A)−1 |h) (β − H0 )−1 |h)(h|(β − H0 )−1

6 Discrete and essential spectrum

6.1 Extended discrete and essential spectrum
Let X be a Banach space and A a closed operator on X . Recall that we defined the discrete and essential
spectra of A denoted by spd A and spess A.
We say that ∞ belongs to the extended discrete spectrum iff there is a decomposition of X = X0 ⊕ X1
into the direct sum of two closed subspaces such that X1 is nonzero finite dimensional, DomA = X0 and
A maps X0 into itself and A restricted to X0 is bounded. Equivalently, ∞ is a discrete point in spext A
and 1{∞} (A) is finite dimensional. The discrete extended spectrum is denoted by spext
d (A). The essential
extended spectrum is defined as
spext
ess A := sp
ext
A\spext
d A.

Theorem 6.1 Let z0 ∈ rsA. Then

−1 −1 −1 −1
spext
ess (z0 − A) = (z0 − spext
ess A) , spext
d (z0 − A) = (z0 − spext
d A) .

6.2 Operators with a compact resolvent

Theorem 6.2 Let A be an operator with a non-empty resolvent set. Then the following conditions are
equivalent:
(1) (z0 − A)−1 is compact for some z0 ∈ rsA.
(2) (z − A)−1 is compact for all z ∈ rsA.

Proof. We use the resolvent equation

(z − A)−1 = (z0 − A)−1 1 − (z − z0 )(z − A)−1 .

2
When the conditions of Theorem 6.2 are satisfied, then we say that the operator A has a compact
resolvent.
Theorem 6.3 (1) Let A be normal. Then A has a compact resolvent iff spA = spd A.
(2) Let A be bounded from below and self-adjoint. Then A has a compact resolvent iff µn (A) → ∞.
Theorem 6.4 Let f, g ∈ L∞ d
loc (R ), lim|x|→∞ f (x) = ∞ and lim|x|→∞ g(x) = ∞. Then

H := f (x) + g(D)
has a compact resolvent.

36
Proof. Clearly, the functions f, g are bounded from below. Fix  > 0. For r > 0,let

Hr := f (x) + min(g(D), r).

Then for z ∈ C\R the operator (min(g(D), r) − r)(z − f (x) − r)−1 is compact. Hence also the operator

(z − Hr )−1 − (z − f (x) − r)−1 = (z − Hr )−1 (min(g(D), r) − r)(z − f (x) − r)−1

is compact. Thus
spess (Hr ) = spess (f (x) + r) ⊂ [r + inf f, ∞[.
Therefore, there exists N such that for n > N

µn (Hr ) ≥ r −  + inf f.

But Hr ≤ H. Hence
µn (Hr ) ≤ µn (H).
Therefore, for n > N , we have µn (H) ≥ R −  + inf f . Thus µn (H) → ∞. 2

6.3 Stability of essential spectrum

Theorem 6.5 Let V be a Hilbert space and A ∈ B(V) be normal. Then λ ∈ spess A iff there exits a
sequence of vectors vn such that

kvn k = 1, w− lim vn = 0, lim (A − λ)vn = 0. (6.38)

n→∞ n→∞

Proof. ⇒ We know that for any n dim 1B(λ, n1 ) (A) = ∞. Therefore, we can find an orthonormal system
v1 , v2 , . . . such that vn ∈ Ran 1B(λ, n1 ) (A). The sequence v1 , v2 , . . . satisfies (6.38).
⇐ Let  > 0. We have

(1 − 1B(λ,) (A))vn = (1 − 1B(λ,) (A))(A − λ)−1 (A − λ)vn → 0.

Hence
cn := k1B(λ,) (A)vn k → 1.
Let
1
ṽn := 1B(λ,) (A)vn .
cn
Then kṽn k = 1, w− limn→∞ ṽn = 0 and ṽn ∈ Ran 1B(λ,) (A). Hence Span{ṽ1 , ṽ2 , . . .} is infinite dimen-
sional. Thus, 1B(λ,) (A) is infinite dimensional. 2

Definition 6.6 A sequence of vectors vn satisfying the conditions of the above theorem will be called a
Weyl sequence for λ and the operator A.

Theorem 6.7 (Weyl) Let A, B ∈ B(V) be normal and let B − A be compact. Then spess A = spess B.

Proof. Assume that λ ∈ spess A. Then there exists a Weyl sequence v1 , v2 , . . . for λ and the operator A.
We have limn→∞ (B − A)vn = 0. Hence v1 , v2 , . . . is a Weyl sequence for λ and the operator B. 2

37
Theorem 6.8 Assume that A, B are normal operators such that for some z0 6∈ spA ∪ spB the operator
(z0 − A)−1 − (z0 − B)−1 is compact. Then

spess A = spess B.

Proof. By the Weyl theorem spess (z0 − A)−1 = spess (z0 − B)−1 . Then we use Theorem 6.1 to normal
operators A and B. 2

Theorem 6.9 Let f ∈ L∞ d ∞ d

loc (R ), lim|x|→∞ f (x) = ∞ and g ∈ L (R ), lim|x|→∞ g(x) = 0. Then

spess (f (D) + g(x)) = spess (f (D)).

Proof. The operator g(x)(z0 − f (D))−1 is compact. Hence, the operator

(z0 − f (D) − g(x))−1 − (z0 − f (D))−1 = (z0 − f (D) − g(x))−1 g(x)(z0 − f (D))−1

is compact as well. We can thus use Theorem 6.8. 2

References
[Da] Davies, E. B.: One parameter semigroups, Academic Press 1980
[Ka] Kato, T.: Perturbation theory for linear operators, Springer 1966
[RS1] Reed, M., Simon, B.: Methods of Modern Mathematics, vol. 1, Academic Press 1972
[RS2] Reed, M., Simon, B.: Methods of Modern Mathematics, vol. 4, Academic Press 1978

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