Lecture Notes Methods of Mathematical Physics MATH...

Lecture NotesMethods of Mathematical Physics

MATH 535Instructor: Ivan Avramidi

New Mexico Institute of Mining and Technology

Socorro, NM 87801

September 19, 2013

Textbooks:

• S. Hassani, Mathematical Physics (Springer, 1999)

• L. Debnath and P. Mikusinski, Introduction to Hilbert Spaces with Applica-tions (Academic Press, 1999)

Author: Ivan Avramidi; File: mathphyshass1.tex; Date: November 20, 2013; Time: 16:51

Contents

1 Preliminaries 31.1 Preliminaries . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3

2 Finite-Dimensional Vector Spaces 52.1 Vectors and Linear Transformations . . . . . . . . . . . . . . . . 5

2.1.1 Vector Spaces . . . . . . . . . . . . . . . . . . . . . . . . 52.1.2 Inner Product and Norm . . . . . . . . . . . . . . . . . . 82.1.3 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . 112.1.4 Linear Transformations. . . . . . . . . . . . . . . . . . . 122.1.5 Algebras . . . . . . . . . . . . . . . . . . . . . . . . . . 14

2.2 Operator Algebra . . . . . . . . . . . . . . . . . . . . . . . . . . 192.2.1 Algebra of Operators on a Vector Space . . . . . . . . . . 192.2.2 Derivatives of Functions of Operators . . . . . . . . . . . 232.2.3 Self-Adjoint and Unitary Operators . . . . . . . . . . . . 272.2.4 Trace and Determinant . . . . . . . . . . . . . . . . . . . 282.2.5 Finite Difference Operators . . . . . . . . . . . . . . . . . 302.2.6 Projection Operators . . . . . . . . . . . . . . . . . . . . 312.2.7 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . 33

2.3 Matrix Representation of Operators . . . . . . . . . . . . . . . . 362.3.1 Matrices . . . . . . . . . . . . . . . . . . . . . . . . . . . 362.3.2 Operation on Matrices . . . . . . . . . . . . . . . . . . . 382.3.3 Inverse Matrix . . . . . . . . . . . . . . . . . . . . . . . 432.3.4 Trace . . . . . . . . . . . . . . . . . . . . . . . . . . . . 442.3.5 Determinant . . . . . . . . . . . . . . . . . . . . . . . . . 442.3.6 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . 47

2.4 Spectral Decomposition . . . . . . . . . . . . . . . . . . . . . . . 492.4.1 Direct Sums . . . . . . . . . . . . . . . . . . . . . . . . . 492.4.2 Invariant Subspaces . . . . . . . . . . . . . . . . . . . . . 50

I

II CONTENTS

2.4.3 Eigenvalues and Eigenvectors . . . . . . . . . . . . . . . 522.4.4 Spectral Decomposition . . . . . . . . . . . . . . . . . . 552.4.5 Functions of Operators . . . . . . . . . . . . . . . . . . . 582.4.6 Polar Decomposition . . . . . . . . . . . . . . . . . . . . 602.4.7 Real Vector Spaces . . . . . . . . . . . . . . . . . . . . . 612.4.8 Heisenberg Algebra, Fock Space and Harmonic Oscillator 632.4.9 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . 69

3 Banach Spaces 713.1 Normed Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . 713.2 Banach Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . 783.3 Linear Mappings . . . . . . . . . . . . . . . . . . . . . . . . . . 833.4 Banach Fixed Point Theorem . . . . . . . . . . . . . . . . . . . . 89

4 Hilbert Spaces 914.1 Inner Product Spaces . . . . . . . . . . . . . . . . . . . . . . . . 91

4.1.1 Norm in an Inner Product Space. . . . . . . . . . . . . . . 944.2 Hilbert Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . 96

4.2.1 Strong and Weak Convergence . . . . . . . . . . . . . . . 984.3 Orthogonal and Orthonormal Systems . . . . . . . . . . . . . . . 100

4.3.1 Properties of Orthonormal Systems . . . . . . . . . . . . 1024.3.2 Orthonormal Complements and Projection Theorem . . . 1054.3.3 Separable Hilbert Spaces . . . . . . . . . . . . . . . . . . 107

4.4 Trigonometric Fourier Series . . . . . . . . . . . . . . . . . . . . 1114.5 Linear Functionals and the Riesz Representation Theorem . . . . 114

4.5.1 Homework . . . . . . . . . . . . . . . . . . . . . . . . . 116

5 Operators on Hilbert Spaces 1175.1 Examples of Operators . . . . . . . . . . . . . . . . . . . . . . . 117

5.1.1 Homework . . . . . . . . . . . . . . . . . . . . . . . . . 1215.2 Bilinear Functionals and Quadratic Forms . . . . . . . . . . . . . 1225.3 Adjoint and Self-Adjoint Operators . . . . . . . . . . . . . . . . . 1245.4 Normal, Isometric and Unitary Operators . . . . . . . . . . . . . 1275.5 Positive Operators . . . . . . . . . . . . . . . . . . . . . . . . . . 1325.6 Projection Operators . . . . . . . . . . . . . . . . . . . . . . . . 1405.7 Compact Operators . . . . . . . . . . . . . . . . . . . . . . . . . 1445.8 Eigenvalues and Eigenvectors . . . . . . . . . . . . . . . . . . . . 1485.9 Spectral Decomposition . . . . . . . . . . . . . . . . . . . . . . . 155

mathphyshass1.tex; November 20, 2013; 16:51; p. 1

CONTENTS 1

5.10 Unbounded Operators . . . . . . . . . . . . . . . . . . . . . . . . 1615.10.1 Homework . . . . . . . . . . . . . . . . . . . . . . . . . 170

5.11 The Fourier Transform . . . . . . . . . . . . . . . . . . . . . . . 1715.11.1 Fourier Transform in L1(R) . . . . . . . . . . . . . . . . . 1715.11.2 Fourier Transform in L2(R) . . . . . . . . . . . . . . . . . 174

Bibliography 180

Answers To Exercises 183

Notation 185


2 CONTENTS


Chapter 1

Preliminaries

1.1 Preliminaries• Sets, subsets, empty set, proper subset, universal set, union, intersection,

power set, complement, Cartesian product

• Equivalence relations, equivalence classes, partitions, quotient set

• Maps, domain, codomain, image, preimage, graph, range, injections, sur-jections, bijections, binary operations

• Example.

• Metric spaces, sequences, convergence, Cauchy sequence, completeness

• Cardinality, countably infinite sets, uncountable sets

• Mathematical induction

3

4 CHAPTER 1. PRELIMINARIES


Chapter 2

Finite-Dimensional Vector Spaces

2.1 Vectors and Linear Transformations

2.1.1 Vector Spaces

• A vector space consists of a set E, whose elements are called vectors, anda field F (such as R or C), whose elements are called scalars. There are twooperations on a vector space:

1. Vector addition, + : E × E → E, that assigns to two vectors u, v ∈ Eanother vector u + v, and

2. Multiplication by scalars, · : R × E → E, that assigns to a vectorv ∈ E and a scalar a ∈ R a new vector av ∈ E.

The vector addition is an associative commutative operation with an addi-tive identity. It satisfies the following conditions:

1. u + v = v + u, ∀u, v, ∈ E

2. (u + v) + w = u + (v + w), ∀u, v,w ∈ E

3. There is a vector 0 ∈ E, called the zero vector, such that for any v ∈ Ethere holds v + 0 = v.

4. For any vector v ∈ E, there is a vector (−v) ∈ E, called the oppositeof v, such that v + (−v) = 0.

The multiplication by scalars satisfies the following conditions:

5

6 CHAPTER 2. FINITE-DIMENSIONAL VECTOR SPACES

1. a(bv) = (ab)v, ∀v ∈ E, ∀a, bR,

2. (a + b)v = av + bv, ∀v ∈ E, ∀a, bR,

3. a(u + v) = au + av, ∀u, v ∈ E, ∀aR,

4. 1 v = v ∀v ∈ E.

• The zero vector is unique.

• For any u, v ∈ E there is a unique vector denoted by w = v − u, called thedifference of v and u, such that u + w = v.

• For any v ∈ E,0v = 0 , and (−1)v = −v .

• Let E be a real vector space and A = e1, . . . , ek be a finite collection ofvectors from E. A linear combination of these vectors is a vector

a1e1 + · · · + akek ,

where a1, . . . , an are scalars.

• A finite collection of vectorsA = e1, . . . , ek is linearly independent if

a1e1 + · · · + akek = 0

implies a1 = · · · = ak = 0.

• A collection A of vectors is linearly dependent if it is not linearly inde-pendent.

• Two non-zero vectors u and v which are linearly dependent are also calledparallel, denoted by u||v.

• A collectionA of vectors is linearly independent if no vector ofA is a linearcombination of a finite number of vectors fromA.

• LetA be a subset of a vector space E. The span ofA, denoted by spanA,is the subset of E consisting of all finite linear combinations of vectors fromA, i.e.

spanA = v ∈ E | v = a1e1 + · · · + akek , ei ∈ A, ai ∈ R .

We say that the subset spanA is spanned byA.


2.1. VECTORS AND LINEAR TRANSFORMATIONS 7

• Theorem 2.1.1 The span of any subset of a vector space is a vector space.

• A vector subspace of a vector space E is a subset S ⊆ E of E which isitself a vector space.

• Theorem 2.1.2 A subset S of E is a vector subspace of E if and only ifspan S = S .

• Span ofA is the smallest subspace of E containingA.

• A collection B of vectors of a vector space E is a basis of E if B is linearlyindependent and spanB = E.

• A vector space E is finite-dimensional if it has a finite basis.

• Theorem 2.1.3 If the vector space E is finite-dimensional, then the numberof vectors in any basis is the same.

• The dimension of a finite-dimensional real vector space E, denoted bydim E, is the number of vectors in a basis.

• Theorem 2.1.4 If e1, . . . , en is a basis in E, then for every vector v ∈ Ethere is a unique set of real numbers (vi) = (v1, . . . , vn) such that

v =

n∑i=1

viei = v1e1 + · · · + vnen .

• The real numbers vi, i = 1, . . . , n, are called the components of the vectorv with respect to the basis ei.

• It is customary to denote the components of vectors by superscripts, whichshould not be confused with powers of real numbers

v2 , (v)2 = vv, . . . , vn , (v)n .

Examples of Vector Subspaces

• Zero subspace 0.

• Line with a tangent vector u:

S 1 = span u = v ∈ E | v = tu, t ∈ R .



• Plane spanned by two nonparallel vectors u1 and u2

S 2 = span u1,u2 = v ∈ E | v = tu1 + su2, t, s ∈ R .

• More generally, a k-plane spanned by a linearly independent collection ofk vectors u1, . . . ,uk

S k = span u1, . . . ,uk = v ∈ E | v = t1u1 + · · · + tkuk, t1, . . . , tk ∈ R .

• An (n − 1)-plane in an n-dimensional vector space is called a hyperplane.

• Examples of vector spaces: P[t], Pn[t], Mm×n, Ck([a, b]), C∞([a, b])

2.1.2 Inner Product and Norm• A complex vector space E is called an inner product space if there is a

function (·, ·) : E × E → R, called the inner product, that assigns to everytwo vectors u and v a complex number (u, v) and satisfies the conditions:∀u, v,w ∈ E, ∀a ∈ C:

1. (v, v) ≥ 0

2. (v, v) = 0 if and only if v = 0

3. (u, v) = (v,u)

4. (u + v,w) = (u,w) + (v,w)

5. (u, av) = a(u, v)

A finite-dimensional real inner product space is called a Euclidean space.

• Examples: On C([a, b])

( f , g) =

∫ b

af (t)g(t)w(t)dt

where w is a positive continuous real-valued function called the weightfunction.

• The Euclidean norm is a function || · || : E → R that assigns to every vectorv ∈ E a real number ||v|| defined by

||v|| =√

(v, v).



• The norm of a vector is also called the length.

• A vector with unit norm is called a unit vector.

• The natural distance function (a metric) is defined by

d(u, v) = ||u − v||

• Example.

• Theorem 2.1.5 For any u, v ∈ E there holds

||u + v||2 = ||u||2 + 2Re(u, v) + ||v||2 .

• If the norm satisfies the parallelogram law

||u + v||2 + ||u − v||2 = 2||u||2 + 2||v||2

then the inner product can be defined by

(u, v) =14

||u + v||2 − ||u − v||2 − i||u + iv||2 + i||u − iv||2

• Theorem 2.1.6 A normed linear space is an inner product space if and only

if the norm satisfies the parallelogram law.

• Theorem 2.1.7 Every finite-dimensional vector space can be turned intoan inner product space.

• Theorem 2.1.8 Cauchy-Schwarz’s Inequality. For any u, v ∈ E thereholds

|(u, v)| ≤ ||u|| ||v|| .

The equality

|(u, v)| = ||u|| ||v||

holds if and only if u and v are parallel.

• Corollary 2.1.1 Triangle Inequality. For any u, v ∈ E there holds

||u + v|| ≤ ||u|| + ||v|| .



• In real vector space the angle between two non-zero vectors u and v isdefined by

cos θ =(u, v)||u|| ||v||

, 0 ≤ θ ≤ π .

Then the inner product can be written in the form

(u, v) = ||u|| ||v|| cos θ .

• Two non-zero vectors u, v ∈ E are orthogonal, denoted by u ⊥ v, if

(u, v) = 0.

• A basis e1, . . . , en is called orthonormal if each vector of the basis is aunit vector and any two distinct vectors are orthogonal to each other, that is,

(ei, e j) =

1, if i = j0, if i , j .

• Theorem 2.1.9 Every Euclidean space has an orthonormal basis.

• Let S ⊂ E be a nonempty subset of E. We say that x ∈ E is orthogonal toS , denoted by x ⊥ S , if x is orthogonal to every vector of S .

• The setS ⊥ = x ∈ E | x ⊥ S

of all vectors orthogonal to S is called the orthogonal complement of S .

• Theorem 2.1.10 The orthogonal complement of any subset of a Euclideanspace is a vector subspace.

• Two subsets A and B of E are orthogonal, denoted by A ⊥ B, if everyvector of A is orthogonal to every vector of B.

• Let S be a subspace of E and S ⊥ be its orthogonal complement. If everyelement of E can be uniquely represented as the sum of an element of S andan element of S ⊥, then E is the direct sum of S and S ⊥, which is denotedby

E = S ⊕ S ⊥ .

• The union of a basis of S and a basis of S ⊥ gives a basis of E.



2.1.3 Exercises1. Show that if λv = 0, then either v = 0 or λ = 0.

2. Prove that the span of a collection of vectors is a vector subspace.

3. Show that the Euclidean norm has the following properties

(a) ||v|| ≥ 0, ∀v ∈ E;

(b) ||v|| = 0 if and only if v = 0;

(c) ||av|| = |a| ||v||, ∀v ∈ E,∀a ∈ R.

4. Parallelogram Law. Show that for any u, v ∈ E

||u + v||2 + ||u − v||2 = 2(||u||2 + ||v||2

)5. Show that any orthogonal system in E is linearly independent.

6. Gram-Schmidt orthonormalization process. Let G = u1, · · · ,uk be a linearlyindependent collection of vectors. Let O = v1, · · · , vk be a new collection ofvectors defined recursively by

v1 = u1,

v j = u j −

j−1∑i=1

vi(vi,u j)||vi||

2 , 2 ≤ j ≤ k,

and the collection B = e1, . . . , ek be defined by

ei =vi

||vi||.

Show that: a) O is an orthogonal system and b) B is an orthonormal system.

7. Pythagorean Theorem. Show that if u ⊥ v, then

||u + v||2 = ||u||2 + ||v||2 .

8. Let B = e1, · · · en be an orthonormal basis in E. Show that for any vector v ∈ E

v =

n∑i=1

ei(ei, v)

and

||v||2 =

n∑i=1

(ei, v)2 .



9. Prove that the orthogonal complement of a subset S of E is a vector subspace of E.

10. Let S be a subspace in E. Prove that

a) E⊥ = 0, b) 0⊥ = E, c) (S ⊥)⊥ = S .

11. Show that the intersection of orthogonal subsets of a Euclidean space is eitherempty or consists of only the zero vector. That is, for two subsets A and B, ifA ⊥ B, then A ∩ B = 0 or ∅.

2.1.4 Linear Transformations.• A linear transformation from a vector space V to a vector space W is a

mapT : V → W

satisfying the condition:

T (αu + βv) = αTu + βTv

for any u, v ∈ V and α, β ∈ C.

• Zero transformation maps all vectors to the zero vector.

• The linear transformation is called an endomorphism (or a linear operator)if V = W.

• The linear transformation is called a linear functional if W = C.

• A linear transformation is uniquely determined by its action on a basis.

• The set of linear transformations from V to W is a vector space denoted byL(V,W).

• The set of endomorphisms (operators) on V is denoted by End (V) or L(V).

• The set of linear functionals on V is called the dual space and is denotedby V∗.

• Example.

• The kernel (null space) (denoted by Ker T ) of a linear transformation T :V → W is the set of vectors in V that are mapped to zero.



• Theorem 2.1.11 The kernel of a linear transformation is a vector space.

• The dimension of a finite-dimensional kernel is called the nullity of thelinear transformation.

null T = dim Ker T

• Theorem 2.1.12 The range of a linear transformation is a vector space.

• The dimension of a finite-dimensional range is called the rank of the lineartransformation.

rank T = dim Im T

• Theorem 2.1.13 Dimension Theorem. Let T : V → W be a linear trans-formation between finite-dimensional vector spaces. Then

dim Ker T + dim Im T = dim V .

• Theorem 2.1.14 A linear transformation is injective if and only if its kernelis zero.

• An endomorphism of a finite-dimensional space is bijective if it is eitherinjective or surjective.

• Two vector spaces are isomorphic if they can be related by a bijective lineartransformation (which is called an isomorphism).

• An isomorphism is called an automorphism if V = W.

• The set of all automorphisms of V is denoted by Aut (V) or GL(V).

• A linear surjection is an isomorphism if and only if its nullity is zero.

• Theorem 2.1.15 An isomorphism maps linearly independent sets onto lin-early independent sets.

• Theorem 2.1.16 Two finite-dimensional vector spaces are isomorphic ifand only if they have the same dimension.

• All n-dimensional complex vector spaces are isomorphic to Cn.

• All n-dimensional real vector spaces are isomorphic to Rn.



• The dual basis fi in the dual space V∗ is defined by

fi(e j) = δi j,

where e j is the basis in V .

• Theorem 2.1.17 The dual space V∗ is isomorphic to V.

• The dual (or the pullback) of a linear transformation T : V → W is thelinear transformation T ∗ : W∗ → V∗ defined for any g ∈ W∗ by

(T ∗g)v = g(Tv), v ∈ V .

• Graph.

• If T is surjective then T ∗ is injective.

• If T is injective then T ∗ is surjective.

• If T is an isomorphism then T ∗ is an isomorphism.

2.1.5 Algebras• An algebra A is a vector space together with a binary operation called mul-

tiplication satisfying the conditions:

u(αv + βw) = αuv + βuw

(αv + βw)u = αvu + βwu

for any u, v,w, α, β ∈ C.

• Examples. Matrices, functions, operators.

• The dimension of the algebra is the dimension of the vector space.

• The algebra isassociative if

u(vw) = (uv)w

and commutative ifuv = vu



• An algebra with identity is an algebra with an identity element 1 satisfying

u1 = 1u = u

for any u ∈ A.

• An element v is a left inverse of u if

vu = 1

and the right inverse ifuv = 1.

• Example. Lie algebras.

• An operator D : A→ A on an algebra A is called a derivation if it satisfies

D(uv) = (Du)v + uDv

• Example. Let A = Mat(n) be the algebra of square matrices of dimensionn with the binary operation being the commutator of matrices.

• It is easy to show that for any matrices A, B,C the following identity (Jacobiidentity) holds

[A, [B,C]] + [B, [C, A]] + [C, [A, B]] = 0

• Let C be a fixed matrix. We define an operator AdC on the algebra by

AdC B = [C, B]

Then this operator is a derivation since for any matrices A, B

AdC[A, B] = [AdCA, B] + [A, AdC B]

• A linear transformation T : A → B from an algebra A to an algebra B iscalled an algebra homomorphism if

T (uv) = T (u)T (v)

for any u, v ∈ A.



• An algebra homomorphism is called an algebra isomorphism if it is bijec-tive.

• Example. The isomorphism of the Lie algebra so(3) and R3 with the crossproduct.

Let Xi, i = 1, 2, 3 be the antisymmetric matrices defined by

(Xi) jk = ε j

ik .

They form an algebra with respect to the commutator

[Xi, X j] = εki jXk .

We define a map T : R3 → so(3) as follows. Let v = viei be a vector in R3.Then

T (v) = viXi .

let R3 be equipped with the cross product. Then

T (v × u) = (Tv)(Tu)

Thus T is an isomorphism (linear bijective algebra homomorphism).

• Any finite dimensional vector space can be converted into an algebra bydefining the multiplication of the basis vectors by

eie j =

n∑k=1

Cki jek

where Cki j are some scalars called the structure constants of the algebra.

• Example. Lie algebra su(2).

Pauli matrices are defined by

σ1 =

(0 11 0

), σ2 =

(0 −ii 0

), σ3 =

(1 00 −1

). (2.1)

They are Hermitian traceless matrices satisfying

σiσ j = δi jI + iεi jkσk . (2.2)



They satisfy the following commutation relations

[σi, σ j] = 2iεi jkσk (2.3)

and the anti-commutation relations

σiσ j + σ jσi = 2δi jI (2.4)

Therefore, Pauli matrices form a representation of Clifford algebra in 2 di-mensions.

The matricesJi = −

i2σi (2.5)

are the generators of the Lie algebra su(2) with the commutation relations

[Ji, J j] = εki jJk (2.6)

Algebra homomorphism Λ : su(2) → so(3) is defined as follows. Letv = viJi ∈ su(2). Then Λ(v) is the matrix defined by

Λ(v) = viXi .

• Example. Quaternions. The algebra of quaternions H is defined by (herei, j, k = 1, 2, 3)

e20 = e0 , e2

i = −e0 , e0ei = eie0 = ei ,

eie j = εki jek i , j

There is an algebra homomorphism ρ : H→ su(2)

ρ(e0) = I, ρ(e j) = −iσ j

• A subspace of an algebra is called a subalgebra if it is closed under algebramultiplication.

• A subset B of an algebra A is called a left ideal if AB ⊂ B, that is, for anyu ∈ A and any v ∈ B, uv ∈ B.

• A subset B of an algebra A is called a right ideal if BA ⊂ B, that is, for anyu ∈ A and any v ∈ B, vu ∈ B.



• A subset B of an algebra A is called a two-sided ideal if it is both left andright ideal, that is, if ABA ⊂ B, or for any u,w ∈ A and any v ∈ B, uvw ∈ B.

• Every ideal is a subalgebra.

• A proper ideal of an algebra with identity cannot contain the identity ele-ment.

• A proper left ideal cannot contain an element that has a left inverse.

• If an ideal does not contain any proper subideals then it is the minimal ideal.

• Examples. Let x be and element of an algebra A. Let Ax be the set definedby

Ax = ux|u ∈ A

Then Ax is a left ideal.

• Similarly xA is a right ideal and AxA is a two-sided ideal.


2.2. OPERATOR ALGEBRA 19

2.2 Operator Algebra

2.2.1 Algebra of Operators on a Vector Space• A linear operator on a vector space E is a mapping L : E → E satisfying

the condition ∀u, v ∈ E, ∀a ∈ R,

L(u + v) = L(u) + L(v) and L(av) = aL(v).

• Identity operator I on E is defined by

I v = v, ∀v ∈ E

• Null operator 0 : E → E is defined by

0v = 0, ∀v ∈ E

• The vector u = L(v) is the image of the vector v.

• If S is a subset of E, then the set

L(S ) = u ∈ E | u = L(v) for some v ∈ S

is the image of the set S and the set

L−1(S ) = v ∈ E | L(v) ∈ S

is the inverse image of the set A.

• The image of the whole space E of a linear operator L is the range (or theimage) of L, denoted by

Im(L) = L(E) = u ∈ E | u = L(v) for some v ∈ E .

• The kernel Ker(L) (or the null space) of an operator L is the set of allvectors in E which are mapped to zero, that is

Ker (L) = L−1(0) = v ∈ E | L(v) = 0 .

• Theorem 2.2.1 For any operator L the sets Im(L) and Ker (L) are vectorsubspaces.



• The dimension of the kernel Ker (L) of an operator L

null (L) = dim Ker (L)

is called the nullity of the operator L.

• The dimension of the range Im(L) of an operator L

rank (L) = dim Ker (L)

is called the rank of the operator L.

• Theorem 2.2.2 For any operator L on an n-dimensional Euclidean spaceE

rank (L) + null (L) = n

• The set L(E) of all linear operators on a vector space E is a vector spacewith the addition of operators and multiplication by scalars defined by

(L1 + L2)(x) = L1(x) + L2(x), and (aL)(x) = aL(x) .

• The product of the operators A and B is the composition of A and B.

• Since the product of operators is defined as a composition of linear map-pings, it is automatically associative, which means that for any operators A,B and C, there holds

(AB)C = A(BC) .

• The integer powers of an operator are defined as the multiple compositionof the operator with itself, i.e.

A0 = I A1 = A, A2 = AA, . . .

• The operator A on E is invertible if there exists an operator A−1 on E, calledthe inverse of A, such that

A−1A = AA−1 = I .

• Theorem 2.2.3 Let A and B be invertible operators. Then:

(A−1)−1 = A , (AB)−1 = B−1A−1 .



• The operators A and B are commuting if

AB = BA

and anti-commuting ifAB = −BA .

• The operators A and B are said to be orthogonal to each other if

AB = BA = 0 .

• An operator A is involutive if

A2 = I

idempotent ifA2 = A ,

and nilpotent if for some integer k

Ak = 0 .

• Two operators A and B are equal if for any u ∈ V

Au = Bu

• If Aei = Bei for all basis vectors in V then A = B.

• Operators are uniquely determined by their action on a basis.

• Theorem 2.2.4 An operator A is equal to zero if and only if for any u, v ∈ V

(u, Av) = 0

• Theorem 2.2.5 An operator A is equal to zero if and only if for any u

(u, Au) = 0

Proof: Use (w, Aw) = 0 for w = au + bv with a = 1, b = i and a = i, b = 1.

• Theorem 2.2.6 1. The inverse of an automorphism is unique.



2. The product of two automorphisms is an automorphism.

3. A linear transformation is an automorphism if and only if it maps abasis to another basis.

• Polynomials of Operators.

Pn(T ) = anT n + · · · a1T + a0I,

where I is the identity operator.

• Commutator of two operators A and B is an operator [A, B] defined by

[A, B] = AB − BA

• Theorem 2.2.7 Properties of commutators.

Anti-symmetry[A, B] = −[B, A]

linearity[aA, bB] = ab[A, B]

[A, B + C] = [A, B] + [A,C]

[A + C, B] = [A, B] + [C, B]

right derivation[AB,C] = A[B,C] + [A,C]B

left derivation[A, BC] = [A, B]C + B[A,C]

Jacobi identity

[A, [B,C]] + [B, [C, A]] + [C, [A, B]] = 0

• Consequences[A, Am] = 0

[A, A−1] = 0

[A, f (A)] = 0



• Functions of Operators.

Negative powersT m = T · · · T︸︷︷︸

m

T−m = (T−1)m

T mT n = T m+n

(T m)n = T mn

Let f be an analytic function given by

f (x) =

∞∑k=0

f (k)(x0)k!

(x − x0)k

Then for an operator T

f (T ) =

∞∑k=0

f (k)(x0)k!

(T − x0I)k

• Exponential

exp(T ) =

∞∑k=0

1k!

T k

• Example.

2.2.2 Derivatives of Functions of Operators• A time-dependent operator is a map

H : R→ End (V)

Note that[H(t),H(t′)] , 0

• Example.A : R2 → R2

A(x, y) = (−y, x)



Note thatA2 = −I

soA2n = (−1)nI, A2n+1 = (−1)nA

Thereforeexp(tA) = cos tI + sin tA

which is a rotation by the angle t

exp(tA)(x, y) = (cos t x − sin t y, cos t y + sin t x)

So A is a generator of rotation.

• Derivative of a time-dependent operator is an operator defined by

dHdt

= limh→0

H(t + h) − H(t)h

• Rules of the differentiation

ddt

(AB) =dAdt

B + AdBdt

• Example.ddt

exp(tA) = A exp(tA)

• Exponential of the adjoint. Let

X(t) = etABe−tA

It satisfies the equationddt

X = [A, B]

with initial conditionX(0) = I

Let AdA be defined byAdAB = [A, B]

Then

X(t) = exp(tAdA)B =

∞∑k=0

tk

k![A, · · · [A, B]︸︷︷︸

k



• Duhamel’s Formula.

ddt

exp[H(t)] = exp[H(t)]∫ 1

0exp[−sH(t)]

dH(t)dt

exp[sH(t)]ds

Proof. Let

Y(s) = e−sH ddt

esH

ThenY(0) = 0

andddt

exp[H] = exp[H]Y(1)

We computedYds

= −[H,Y] + H′

So,(∂s + AdH)Y = H′

Therefore

Y(1) = exp(−AdH)∫ 1

0exp(sAdH)H′ds

which can be written in the form

Y(1) =

∫ 1

0e−(1−s)HH′e(1−s)Hds

By changing the variable s→ (1 − s) we get the desired result.

• Particular case. If H commutes with H′ then

∂teH(t) = eH∂tH

• Campbell-Hausdorff Formula.

exp A exp B = exp[C(A, B)]

ConsiderU(s) = eAesB = eC(s)



Of courseC(0) = A

We havedds

U = UB

Also,

∂sU = U∫ 1

0exp[−τC]∂sC exp[τC] dτ

Let

F(z) =

∫ 1

0e−τzdτ =

1 − e−z

zThen

∂sU = UF(AdC)∂sC

ThereforeF(AdC)∂sC = B

Now, let

Ψ(z) =1

F(log z)=

z log zz − 1

Note thateAdC = AdeC = AdeA · AdesB = eAdAesAdB

ThenΨ(eAdAesAdB)F(AdC) = I

Therefore, we get a differential equation

∂sC = Ψ(eAdAesAdB)B

with initial conditionC(0) = A

Therefore,

C(1) = A +

∫ 1

0Ψ(eAdAesAdB)Bds

This gives a power series in AdA, AdB.

• Particular case. If [A, B] commutes with both A and B then

eAeB = eA+B exp(12

[A, B])



2.2.3 Self-Adjoint and Unitary Operators• The adjoint A∗ of an operator A is defined by

(Au, v) = (u, A∗v), ∀u, v ∈ E.

• Theorem 2.2.8 For any two operators A and B

(A∗)∗ = A , (AB)∗ = B∗A∗ .

(A + B)∗ = A∗ + B∗

(aA)∗ = aA∗

• An operator A is self-adjoint (or Hermitian) if

A∗ = A

and anti-selfadjoint if

A∗ = −A

• Every operator A can be decomposed as the sum

A = AS + AA

of its selfadjoint part AS and its anti-selfadjoint part AA

AS =12

(A + A∗) , AA =12

(A − A∗) .

• Theorem 2.2.9 An operator H is Hermitian if and only if (u,Hu) is realfor any u.

• An operator A on E is called positive, denoted by A ≥ 0, if it is selfdadjointand ∀v ∈ E

(Av, v) ≥ 0.

• An operator H is positive definite (H > 0) if it is positive and

(u,Hu) = 0

only for u = 0.



• Example.H = A∗A ≥ 0

• An operator A is called unitary if

AA∗ = A∗A = I .

• An operator U is isometric if for any v ∈ E

||Uv|| = ||v||

• Example.U = exp(A), A∗ = −A

• Unitary operators preserve the inner product.

• Theorem 2.2.10 Let U be a unitary operator on a real vector space E.Then there exists an anti-selfadjoint operator A such that

U = exp A .

• Recall that the operators U and A satisfy the equations

U∗ = U−1 and A∗ = −A.

2.2.4 Trace and Determinant

• The trace of an operator A is defined by

tr A =

n∑i=1

(ei, Aei)

• The determinant of a positive operator on a finite-dimensional space isdefined by

det A = exp(tr log A)



• Propertiestr AB = tr BA

det AB = det A det B

tr (RAR−1) = tr A

det(RAR−1) = det A

• Theorem.ddt

det(I + tA)∣∣∣∣t=0

= tr A

det(I + tA) = I + ttr A + O(t2)

ddt

det A = det A tr(A−1 dA

dt

)• Note that

tr I = n , det I = 1 .

• Theorem 2.2.11 Let A be a self-adjoint operator. Then

det exp A = etr A .

• Let A be a positive definite operator, A > 0. The zeta-function of theoperator A is defined by

ζ(s) = tr A−s =1

Γ(s)

∫ ∞

0dt ts−1tr e−tA .

• Theorem 2.2.12 The zeta-functions has the properties

ζ(0) = n ,

andζ′(0) = − log det A .



2.2.5 Finite Difference Operators

• Let ei be an orthonormal basis. The shift operator E is defined by

Ee1 = 0, Eei = ei−1, i = 1, . . . , n,

that is,

E f =

n−1∑i=1

fi+1ei

or

(E f )i = fi+1

Let

∆ = E − I

∇ = I − E−1

Next, define an operator D by

E = exp(hD)

that is,

D =1h

log E =1h

log(I + ∆) = −1h

log(I − ∇)

Also, define an operator J by

J = ∆D−1

Then

∆ fi = fi+1 − fi

∇ fi = fi − fi−1

• Problem. Compute U(t) = exp[tD2].



2.2.6 Projection Operators• A Hermitian operator P is a projection if

P2 = P

• Two projections P1, P2 are orthogonal if

P1P2 = P2P1 = 0 .

• Let S be a subspace of E and E = S ⊕ S ⊥. Then for any u ∈ E there existunique v ∈ S and w ∈ S ⊥ such that

u = v + w .

The vector v is called the projection of u onto S .

• The operator P on E defined by

Pu = v

is called the projection operator onto S .

• The operator P⊥ defined byP⊥u = w

is the projection operator onto S ⊥.

• The operators P and P⊥ are called complementary projections. They havethe properties:

P∗ = P, (P⊥)∗ = P⊥ ,

P + P⊥ = I ,

P2 = P , (P⊥)2 = P⊥ ,

PP⊥ = P⊥P = 0 .

• More generally, a collection of projections P1, . . . , Pk is a complete or-thogonal system of complimentary projections if

PiPk = 0 if i , k

andk∑

i=1

Pi = P1 + · · · + Pk = I .



• The trace of a projection P onto a vector subspace S is equal to its rank, orthe dimension of the vector subspace S ,

tr P = rank P = dim S .

• Theorem 2.2.13 An operator P is a projection if and only if P is idempotentand self-adjoint.

• Theorem 2.2.14 The sum of projections is a projection if and only if theyare orthogonal.

• The projection onto a unit vector |e〉 has the form

P = |e〉〈e|

• Let |ei〉mi=1 be an orthonormal set and S = span 1≤i≤m|ei〉. Then the opera-

tor

P =

m∑i=1

|ei〉〈ei|

is the projection onto S .

• If |ei〉 is an orthonormal basis then the projections

Pi = |ei〉〈ei|

for a complete orthogonal set.

Examples

• Let u be a unit vector and Pu be the projection onto the one-dimensionalsubspace (line) S u spanned by u defined by

Puv = u(u, v) .

The orthogonal complement S ⊥u is the hyperplane with the normal u. Theoperator Ju defined by

Ju = I − 2Pu

is called the reflection operator with respect to the hyperplane S ⊥u . Thereflection operator is a self-adjoint involution, that is, it has the followingproperties

J∗u = Ju , J2u = I .



The reflection operator has the eigenvalue −1 with multiplicity 1 and theeigenspace S u, and the eigenvalue +1 with multiplicity (n − 1) and witheigenspace S ⊥u .

• Let u1 and u2 be an orthonormal system of two vectors and Pu1,u2 be theprojection operator onto the two-dimensional space (plane) S u1,u2 spannedby u1 and u2

Pu1,u2v = u1(u1, v) + u2(u2, v) .

Let Nu1,u2 be an operator defined by

Nu1,u2v = u1(u2, v) − u2(u1, v) .

ThenNu1,u2 Pu1,u2 = Pu1,u2 Nu1,u2 = Nu1,u2

andN2

u1,u2= −Pu1,u2 .

A rotation operator Ru1,u2(θ) with the angle θ in the plane S u1,u2 is definedby

Ru1,u2(θ) = I − Pu1,u2 + cos θ Pu1,u2 + sin θ Nu1,u2 .

The rotation operator is unitary, that is, it satisfies the equation

R∗u1,u2Ru1,u2 = I .

2.2.7 Exercises1. Prove that the range and the kernel of any operator are vector spaces.

2. Show that

(aA + bB)∗ = aA∗ + bB∗ ∀a, b ∈ R ,

(A∗)∗ = A

(AB)∗ = B∗A∗

3. Show that for any operator A the operators AA∗ and A + A∗ are selfadjoint.

4. Show that the product of two selfadjoint operators is selfadjoint if and only if theycommute.

5. Show that a polynomial p(A) of a selfadjoint operator A is a selfadjoint operator.



6. Prove that the inverse of an invertible operator is unique.

7. Prove that an operator A is invertible if and only if Ker A = 0, that is, Av = 0implies v = 0.

8. Prove that for an invertible operator A, Im(A) = E, that is, for any vector v ∈ Ethere is a vector u ∈ E such that v = Au.

9. Show that if an operator A is invertible, then

(A−1)−1 = A .

10. Show that the product AB of two invertible operators A and B is invertible and

(AB)−1 = B−1A−1

11. Prove that the adjoint A∗ of any invertible operator A is invertible and

(A∗)−1 = (A−1)∗ .

12. Prove that the inverse A−1 of a selfadjoint invertible operator is selfadjoint.

13. An operator A on E is called isometric if ∀v ∈ E,

||Av|| = ||v|| .

Prove that an operator is unitary if and only if it is isometric.

14. Prove that unitary operators preserves inner product. That is, show that if A is aunitary operator, then ∀u, v ∈ E

(Au, Av) = (u, v) .

15. Show that for every unitary operator A both A−1 and A∗ are unitary.

16. Show that for any operator A the operators AA∗ and A∗A are positive.

17. What subspaces do the null operator 0 and the identity operator I project onto?

18. Show that for any two projection operators P and Q, PQ = 0 if and only if QP = 0.

19. Prove the following properties of orthogonal projections

P∗ = P, (P⊥)∗ = P⊥, P⊥ + P = I, PP⊥ = P⊥P = 0 .



20. Prove that an operator is projection if and only if it is idempotent and selfadjoint.

21. Give an example of an idempotent operator in R2 which is not a projection.

22. Show that any projection operator P is positive. Moreover, show that ∀v ∈ E

(Pv, v) = ||Pv||2 .

23. Prove that the sum P = P1+P2 of two projections P1 and P2 is a projection operatorif and only if P1 and P2 are orthogonal.

24. Prove that the product P = P1P2 of two projections P1 and P2 is a projectionoperator if and only if P1 and P2 commute.



2.3 Matrix Representation of Operators

2.3.1 Matrices• Cn is the set of all ordered n-tuples of complex numbers, which can be

assembled as columns or as rows.

• Let v be a vector in a n-dimensional vector space V with a basis ei. Then

v =

n∑i=1

viei

where v1, . . . , vn are complex numbers called the components of the vectorv. The column-vector is an ordered n-tuple of the form

v1

v2...

vn

.We say that the column vector represents the vector v in the basis ei.

• Let < v| = (|v >)∗ be a linear functional dual to the vector |v >. Let < ei| bethe dual basis. Then

< v| =n∑

i=1

vi < ei|

The row-vector (also called a covector) is an ordered n-tuple of the form

(v1, v2, . . . , vn) .

It represents the dual vector < v| in the same basis.

• A set of nm complex numbers Ai j, i = 1, . . . , n; j = 1, . . . ,m, arranged in anarray that has m columns and n rows

A =

A11 A12 · · · A1m

A21 A22 · · · A2m...

.... . .

...An1 An2 · · · Anm

is called a rectangular n × m complex matrix.


2.3. MATRIX REPRESENTATION OF OPERATORS 37

• The set of all complex n × m matrices is denoted by Mat(n,m;C).

• The number Ai j (also called an entry of the matrix) appears in the i-th rowand the j-th column of the matrix A

A =

A11 A12 · · · A1 j · · · A1m

A21 A22 · · · A2 j · · · A2m...

.... . .

......

...

Ai1 Ai2 · · · Ai j · · · Aim

......

......

. . ....

An1 An2 · · · An j · · · Anm

• Remark. Notice that the first index indicates the row and the second index

indicates the column of the matrix.

• The matrix whose all entries are equal to zero is called the zero matrix.

• Finally, we define the multiplication of column-vectors by matrices fromthe left and the multiplication of row-vectors by matrices from the right asfollows.

• Each matrix defines a natural left action on a column-vector and a rightaction on a row-vector.

• For each column-vector v and a matrix A = (Ai j) the column-vector u = Avis given by

u1

u2...ui...

un

=

A11 A12 · · · A1n

A21 A22 · · · A2n...

.... . .

...Ai1 Ai2 · · · Ain...

......

...An1 An2 · · · Ann

v1

v2...vi...

vn

=

A11v1 + A12v2 + · · · + A1nvn

A21v1 + A22v2 + · · · + A2nvn...

Ai1v1 + Ai2v2 + · · · + Ainvn...

An1v1 + An2v2 + · · · + Annvn

• The components of the vector u are

ui =

n∑j=1

Ai jv j = Ai1v1 + Ai2v2 + · · · + Ainvn .



• Similarly, for a row vector vT the components of the row-vector uT = vT Aare defined by

ui =

n∑j=1

v jA ji = v1A1i + v2A2i + · · · + vnAni .

• Let W be an m-dimensional vector space with aa basis fi and A : V → W bea linear transformation. Such an operator defines a n × m matrix (Ai j) by

Aei =

m∑j=1

A jif j

orA ji = (f j, Aei)

Thus the linear transformation A is represented by the matrix Ai j.

The components of a vector v are obtained by acting on the colum vector(vi) from the left by the matrix (A ji), that is,

(Av)i =

n∑j=1

Ai jv j

• Proposition. The vector space L(V,W) of linear transformations A : V →W is isomorphic to the space M(m × n,C) of m × n matrices.

• Proposition. The rank of a linear transformation is equal to the rank of itsmatrix.

2.3.2 Operation on Matrices

• The addition of matrices is defined by

A + B =

A11 + B11 A12 + B12 · · · A1m + B1m

A21 + B21 A22 + B22 · · · A2m + B2m...

.... . .

...An1 + Bn1 An2 + Bn2 · · · Anm + Bnm



and the multiplication by scalars by

cA =

cA11 cA12 · · · cA1m

cA21 cA22 · · · cA2m...

.... . .

...cAn1 cAn2 · · · cAnm

• A n × m matrix is called a square matrix if n = m.

• The numbers Aii are called the diagonal entries. Of course, there are ndiagonal entries. The set of diagonal entries is called the diagonal of thematrix A.

• The numbers Ai j with i , j are called off-diagonal entries; there are n(n−1)off-diagonal entries.

• The numbers Ai j with i < j are called the upper triangular entries. Theset of upper triangular entries is called the upper triangular part of thematrix A.

• The numbers Ai j with i > j are called the lower triangular entries. The setof lower triangular entries is called the lower triangular part of the matrixA.

• The number of upper-triangular entries and the lower-triangular entries isthe same and is equal to n(n − 1)/2.

• A matrix whose only non-zero entries are on the diagonal is called a diago-nal matrix. For a diagonal matrix

Ai j = 0 if i , j .

• The diagonal matrix

A =

λ1 0 · · · 00 λ2 · · · 0...

.... . .

...0 0 · · · λn

is also denoted by

A = diag (λ1, λ2, . . . , λn)



• A diagonal matrix whose all diagonal entries are equal to 1

I =

1 0 · · · 00 1 · · · 0...

.... . .

...0 0 · · · 1

is called the identity matrix. The elements of the identity matrix are

δi j =

1, if i = j

0, if i , j .

• A matrix A of the form

A =

∗ ∗ · · · ∗

0 ∗ · · · ∗...

.... . .

...0 0 · · · ∗

where ∗ represents nonzero entries is called an upper triangular matrix.Its lower triangular part is zero, that is,

Ai j = 0 if i < j .

• A matrix A of the form

A =

∗ 0 · · · 0∗ ∗ · · · 0...

.... . .

...∗ ∗ · · · ∗

whose upper triangular part is zero, that is,

Ai j = 0 if i > j ,

is called a lower triangular matrix.



• The transpose of a matrix A whose i j-th entry is Ai j is the matrix AT whosei j-th entry is A ji. That is, AT obtained from A by switching the roles of rowsand columns of A:

AT =

A11 A21 · · · A j1 · · · An1

A12 A22 · · · A j2 · · · An2...

.... . .

......

...

A1i A2i · · · A ji · · · Ani

......

......

. . ....

A1m A2m · · · A jm · · · Anm

or

(AT )i j = A ji .

• The Hermitian conjugate of a matrix A = (Ai j) is a matrix A∗ = (A∗i j)defined by

(A∗)i j = A ji

• A matrix A is called symmetric if

AT = A

and anti-symmetric ifAT = −A .

• A matrix A is called Hermitian if

A∗ = A

and anti-Hermitian ifA∗ = −A .

• An anti-Hermitian matrix has the form

A = iH

where H is Hermitian.

• A Hermitian matrix has the form

H = A + iB

where A is real symmetric and B is real anti-symmetric matrix.



• The number of independent entries of an anti-symmetric matrix is n(n−1)/2.

• The number of independent entries of a symmetric matrix is n(n + 1)/2.

• The diagonal entries of a Hermitian matrix are real.

• The number of independent real parameters of a Hermitian matrix is n2.

• Every square matrix A can be uniquely decomposed as the sum of its di-agonal part AD, the lower triangular part AL and the upper triangular partAU

A = AD + AL + AU .

• For an anti-symmetric matrix

ATU = −AL and AD = 0 .

• For a symmetric matrixAT

U = AL .

• Every square matrix A can be uniquely decomposed as the sum of its sym-metric part AS and its anti-symmetric part AA

A = AS + AA ,

whereAS =

12

(A + AT ) , AA =12

(A − AT ) .

• The product of square matrices is defined as follows. The i j-th entry ofthe product C = AB of two matrices A and B is

Ci j =

n∑k=1

AikBk j = Ai1B1 j + Ai2B2 j + · · · + AinBn j .

This is again a multiplication of the “i-th row of the matrix A by the j-thcolumn of the matrix B”.

• Theorem 2.3.1 The product of matrices is associative, that is, for any ma-trices A, B, C

(AB)C = A(BC) .

• Theorem 2.3.2 For any two matrices A and B

(AB)T = BT AT , (AB)∗ = B∗A∗ .



2.3.3 Inverse Matrix• A matrix A is called invertible if there is another matrix A−1 such that

AA−1 = A−1A = I .

The matrix A−1 is called the inverse of A.

• Theorem 2.3.3 For any two invertible matrices A and B

(AB)−1 = B−1A−1 ,

and(A−1)T = (AT )−1 .

• A matrix A is called orthogonal if

AT A = AAT = I ,

which means AT = A−1.

•

• A matrix A is called unitary if

A∗A = AA∗ = I ,

which means A∗ = A−1.

• Every unitary matrix has the form

U = exp(iH)

where H is Hermitian.

• A similarity transformation of a matric A is a map

A 7→ UAU−1

where U is a given invertible matrix.

• The similarity transformation of a function of a matrix is equal to the func-tion of the similar matrix

U f (A)U−1 = f (UAU−1) .



2.3.4 Trace

• The trace is a map tr : Mat(n,C) that assigns to each matrix A = (Ai j) acomplex number tr A equal to the sum of the diagonal elements of a matrix

tr A =

n∑k=1

Akk .

• Theorem 2.3.4 The trace has the properties

tr (AB) = tr (BA) ,

andtr AT = tr A , tr A∗ = tr A

• Obviously, the trace of an anti-symmetric matrix is equal to zero.

• The trace is invariant under a similarity transformation.

• A natural inner product on the space of matrices is defined by

(A, B) = tr (A∗B)

2.3.5 Determinant

• Consider the set Zn = 1, 2, . . . , n of the first n integers. A permutation ϕof the set 1, 2, . . . , n is an ordered n-tuple (ϕ(1), . . . , ϕ(n)) of these num-bers.

• That is, a permutation is a bijective (one-to-one and onto) function

ϕ : Zn → Zn

that assigns to each number i from the set Zn = 1, . . . , n another numberϕ(i) from this set.

• An elementary permutation is a permutation that exchanges the order ofonly two numbers.



• Every permutation can be realized as a product (or a composition) of ele-mentary permutations. A permutation that can be realized by an even num-ber of elementary permutations is called an even permutation. A permu-tation that can be realized by an odd number of elementary permutations iscalled an odd permutation.

• Proposition 2.3.1 The parity of a permutation does not depend on therepresentation of a permutation by a product of the elementary ones.

• That is, each representation of an even permutation has even number ofelementary permutations, and similarly for odd permutations.

• The sign of a permutation ϕ, denoted by sign(ϕ) (or simply (−1)ϕ), isdefined by

sign(ϕ) = (−1)ϕ =

+1, if ϕ is even,−1, if ϕ is odd

• The set of all permutations of n numbers is denoted by S n.

• Theorem 2.3.5 The cardinality of this set, that is, the number of differentpermutations, is

|S n| = n! .

• The determinant is a map det : Mat(n,C) → C that assigns to each matrixA = (Ai j) a complex number det A defined by

det A =∑ϕ∈S n

sign (ϕ)A1ϕ(1) · · · Anϕ(n) ,

where the summation goes over all n! permutations.

• The most important properties of the determinant are listed below:

Theorem 2.3.6 1. The determinant of the product of matrices is equal tothe product of the determinants:

det(AB) = det A det B .

2. The determinants of a matrix A and of its transpose AT are equal:

det AT = det A .



3. The determinant of the conjugate matrix is

det A∗ = det A .

4. The determinant of the inverse A−1 of an invertible matrix A is equalto the inverse of the determinant of A:

det A−1 = (det A)−1

5. A matrix is invertible if and only if its determinant is non-zero.

• The determinant is invariant under the similarity transformation.

• The set of complex invertible matrices (with non-zero determinant) is de-noted by GL(n,C).

• A matrix with unit determinant is called unimodular.

• The set of complex matrices with unit determinant is denoted by S L(n,C).

• The set of complex unitary matrices is denoted by U(n).

• The set of complex unitary matrices with unit determinant is denoted byS U(n).

• The set of real orthogonal matrices is denoted by O(n).

• An orthogonal matrix with unit determinant (a unimodular orthogonal ma-trix) is called a proper orthogonal matrix or just a rotation.

• The set of real orthogonal matrices with unit determinant is denoted byS O(n).

• Theorem 2.3.7 The determinant of an orthogonal matrix is equal to either1 or −1.

• Theorem. The determinant of a unitary matrix is a complex number ofmodulus 1.

• A set G of invertible matrices forms a group if it is closed under takinginverse and matrix multiplication, that is, if the inverse A−1 of any matrix Ain G belongs to the set G and the product AB of any two matrices A and Bin G belongs to G.



2.3.6 Exercises1. Show that

tr [A, B] = 0

2. Show that the product of invertible matrices is an invertible matrix.

3. Show that the product of matrices with positive determinant is a matrix with posi-tive determinant.

4. Show that the inverse of a matrix with positive determinant is a matrix with positivedeterminant.

5. Show that GL(n,R) forms a group (called the general linear group).

6. Show that GL+(n,R) is a group (called the proper general linear group).

7. Show that the inverse of a matrix with negative determinant is a matrix with nega-tive determinant.

8. Show that: a) the product of an even number of matrices with negative determinantis a matrix with positive determinant, b) the product of odd matrices with negativedeterminant is a matrix with negative determinant.

9. Show that the product of matrices with unit determinant is a matrix with unit deter-minant.

10. Show that the inverse of a matrix with unit determinant is a matrix with unit deter-minant.

11. Show that S L(n,R) forms a group (called the special linear group or the unimod-ular group).

12. Show that the product of orthogonal matrices is an orthogonal matrix.

13. Show that the inverse of an orthogonal matrix is an orthogonal matrix.

14. Show that O(n) forms a group (called the orthogonal group).

15. Show that orthogonal matrices have determinant equal to either +1 or −1.

16. Show that the product of orthogonal matrices with unit determinant is an orthogonalmatrix with unit determinant.

17. Show that the inverse of an orthogonal matrix with unit determinant is an orthogo-nal matrix with unit determinant.



18. Show that S O(n) forms a group (called the proper orthogonal group or the rota-tion group).


2.4. SPECTRAL DECOMPOSITION 49

2.4 Spectral Decomposition

2.4.1 Direct Sums• Let U and W be vector subspaces of a vector space V . Then the sum of the

vector spaces U and W is the space of all sums of the vectors from U andW, that is,

U + W = v ∈ V | v = u + w, u ∈ U,w ∈ W

• If the only vector common to both U and W is the zero vector then the sumU + W is called the direct sum and denoted by U ⊕W.

• Theorem 2.4.1 Let U and W be subspaces of a vector space V. Then V =

U ⊕W if and only if every vector v ∈ V in V can be written uniquely as thesum

v = u + wwith u ∈ U,w ∈ W.

Proof.

• Theorem 2.4.2 Let V = U ⊕W. Then

dim V = dim U + dim W

Proof.

• The direct sum can be naturally generalized for several subspaces so that

V =

r⊕i=1

Ui

• To such a decomposition one naturally associates orthogonal complemen-tary projections Pi on each subspace Ui such that

P2i = I, PiP j = 0 if i , j,

r∑i=1

Pi = I

• A complete orthogonal system of projections defines the orthogonal decom-position of the vector space

V = U1 ⊕ · · · ⊕ Ur ,

where Ui is the subspace the projection Pi projects onto.



• Theorem 2.4.3 1. The dimension of the subspaces Ui are equal to theranks of the projections Pi

dim Ui = rank Pi .

2. The sum of dimensions of the vector subspaces Ui equals the dimen-sion of the vector space V

r∑i=1

dim Ui = dim U1 + · · · + dim Ur = dim V .

• Let M be a subspace of a vector space V . Then the orthogonal complementof M is the vector space M⊥ of all vector in V orthogonal to all vectors inM

M⊥ = v ∈ V | (v,u) = 0 ∀u ∈ M

• Show that M⊥ is a vector space.

• Theorem 2.4.4 Every vector subspace M of V defines the orthogonal de-composition

V = M ⊕ M⊥

such that the corresponding projection operators P and P⊥ are Hermitian.

• Remark. The projections Pi are Hermitian only in inner product spaceswhen the subspaces Ei are mutually orthogonal.

2.4.2 Invariant Subspaces• Let V be a finite-dimensional vector space, M be its subspace and P be the

projection onto the subspace M.

• The subspace M is an invariant subspace of an operator A if it is closedunder the action of this operator, that is, A(M) ⊆ M.

• An invariant subspace M is called proper invariant subspace if M , V .

• Theorem 2.4.5 Let v be a vector in an n-dimensional vector space V andA be an operator on V. Then

M = span v, Av, . . . , An−1v



is an invariant space of A.

Proof.

• Theorem 2.4.6 The subspace M is invariant under an operator A if andonly if M⊥ is invariant under its adjoint A∗.

Proof.

• The vector subspace M reduces the operator A if both M and its orthogonalcomplement M⊥ are invariant subspaces of A.

• If a subspace M reduces an operator A then we write

A = A1 ⊕ A2

where A1 acts on M and A2 acts on M⊥.

• If the subspace M reduces the operator A then, in a natural basis, the matrixrepresentation of the operator A has a block-diagonal form

A =

(A1 00 A2

)• An operator whose matrix can be brought to this form by choosing a basis

is called reducible; otherwise, it is irreducible.

• Theorem 2.4.7 A subspace M reduces an operator A if and only if it isinvariant under both A and A∗.

Proof: follows from above.

• Theorem 2.4.8 A self-adjoint operator is reducible if and only if it has aproper invariant subspace.

• Theorem 2.4.9 The subspace M is invariant under an operator A if andonly if

AP = PA = PAP.

Proof.

• Theorem 2.4.10 The subspace M reduces the operator A if and only if Aand P commute.

Proof.



2.4.3 Eigenvalues and Eigenvectors• Let A be an operator on a vector space V . A scalar λ is an eigenvalue of A

if there is a nonzero vector v in V such that

Av = λv, or (A − λI)v = 0

Such a vector is called an eigenvector corresponding to the eigenvalue λ.

• Theorem 2.4.11 1. The eigenvalues of a Hermitian operator are real.

2. A Hermitian operator is positive if and only if all of its eigenvalues arepositive.

3. The eigenvalues of a unitary operator are complex numbers of unitmodulus.

4. The eigenvalues of a projection operator can be only 0 and 1.

5. The eigenvalues of a self-adjoint involution can be only 1 and −1.

6. The eigenvalues of an anti-symmetric operator can be either 0 or bepurely imaginary, which appear in complex conjugated pairs.

• The eigenspace of A corresponding to the eigenvalue λ is the vector space

Mλ = Ker (A − λI)

• The eigenspace Mλ is the span of all eigenvectors corresponding to theeigenvalue λ.

• The dimension of the eigenspace of the eigenvalue λ is called the multiplic-ity (also called the geometric multiplicity) of λ,

dλ = dim Mλ

• An eigenvalue of multiplicity 1 is called simple (or non-degenerate).

• An eigenvalue of multiplicity greater than 1 is called multiple (or degener-ate).

• The norm of an operator A is defined by

||A|| = supv∈V

||Av||||v||



• An operator A is bounded if it has a final norm.

• In finite dimensions all operators are bounded.

• In finite dimensions the norm of the operator is equal to the absolute valueof its largest eigenvalue

||A|| = max1≤i≤n|λi|

• For an invertible operator A

||A−1|| =1||A||

= max1≤i≤n

1|λi|

• An operator is invertible if it does not have zero eigenvalues.

• The resolvent of an operator A is the operator

R(λ) = (A − λI)−1

The resolvent is defined for all complex numbers λ for which the operatorA − λI is invertible.

• The norm of the resolvent is

||R(λ)|| = max1≤i≤n

1|λi − λ|

• The resolvent set of an operator A is the set of all complex numbers λ ∈ Csuch that the operator (A − λI) is invertible,

ρ(A) = λ ∈ C | (A − λI) is invertible

• The spectrum of an operator A is the complement of the resolvent set,

σ(A) = C − ρ(A)

• In finite dimensions the spectrum of an operator A is equal to the set of alleigenvalues of A,

σ(A) = λ ∈ C | λ is an eigenvalue of A



• The characteristic polynomial of A is defined by

χ(λ) = det(A − λI)

• The eigenvalues of an operator A are the roots of its characteristic polyno-mial

χ(λ) = 0

Theorem 2.4.12 Every operator on a n-dimensional complex vector spacehas exactly n eigenvalues.

• If there are p distinct roots λi then

χ(λ) = (λ1 − λ)m1 · · · (λp − λ)mp

• Here m j is called the algebraic multiplicity of λ j.

• The geometric multiplicity di of an eigenvalue λi is less or equal to thealgebraic multiplicity mi,

di ≤ mi .

• Example. Let A : R2 → R2 be defined by

A(x, y) = (x + y, y)

Then it has one eigenvalue λ = 1 with geometric multiplicity 1 and algebraicmultiplicity 2. The eigenvectors are of the form (a, 0).

Let the operator B be defined by

B(x, y) = (y, 0)

ThenA = I + B

Since B is nilpotentB2 = 0

thenAn = I + nB

andexp(tA) = et(1 + tB)



• An operator A on a vector space V is diagonalizable if there is a basis in Vconsisting of eigenvectors of A.

• In such basis the operator A is represented by a diagonal matrix.

• Theorem 2.4.13 Let A be a diagonalizable operator, λ j, j = 1, . . . , p, beits distinct eigenvalues, Mi be the corresponding eigenspaces and Pi be theprojections onto Mi. Then:

1.

I =

p∑j=1

P j, PiP j = 0 if i , j,

2.

V =

p⊕i=1

Mi

3.

A =

p∑j=1

λiPi .

• In other words, for any

v =

n∑i=1

ei(ei, v) ,

we have

Av =

n∑i=1

λiei(ei, v) .

2.4.4 Spectral Decomposition• An operator is normal if it commutes with its adjoint.

• Both Hermitian and unitary operators are normal.

• Theorem 2.4.14 An operator A is normal if and only if for any v ∈ V

||Av|| = ||A∗v||

Proof.



• Theorem 2.4.15 Let A be a normal operator. Then λ is an eigenvalue of Awith an eigenvector v if and only if λ is an eigenvalue of A∗ with the sameeigenvector v.

Proof.

• Theorem 2.4.16 Let A be a normal operator. Then:

1. every eigenspace Mλ reduces A,

2. the eigenspaces Mλ and Mµ corresponding to distinct eigenvalues λ ,µ are orthogonal.

Proof.

• The projections to the eigenspaces of a normal operator are Hermitian.

• Theorem 2.4.17 Spectral Decomposition Theorem. Let A be a normaloperator on a complex vector space V. Let λ j, j = 1, . . . , p, be the dis-tinct eigenvalues of A, M j be the corresponding eigenspaces and P j be theprojections on M j. Then:

1.

V =

p⊕j=1

M j,

2.p∑

j=1

dim M j = dim V ,

3. the projections are Hermitian, orthogonal and complete

p∑j=1

P j = I

PiP j = 0 if i , j,P∗i = Pi

4. there is the spectral decomposition of the operator

A =

p⊕j=1

A j =

p∑j=1

λ jP j



Proof.

• Theorem 2.4.18 Let A be a normal operator on a complex vector space V.Then:

1. there is an orthonormal basis consisting of eigenvectors of A,

2. the operator A is diagonalizable.

Proof.

• Theorem 2.4.19 A Hermitian operator is diagonalizable by a unitary op-erator, that is, for every Hermitian operator H there is a diagonal operatorD and a unitary operator U such that

H = UDU−1 .

• Two operators A and B are simultaneously diagonalizable if there is thereis a complete system of Hermitian orthogonal projections Pi, i = 1, . . . , pand numbers λi, µi such that

A =

p∑i=1

λiPi,

B =

p∑i=1

µiPi.

• Theorem 2.4.20 Let A a normal operator and Pi be its projections to theeigenspaces. Then an operator B commutes with A if and only if B com-mutes with all projections Pi.

Proof.

• Theorem 2.4.21 Two normal operators are simultaneously diagonalizableif and only if they commute.

• Let A be a self-adjoint operator with distinct eigenvalues λ1, . . . , λp withmultiplicities m j. Then the trace of the operator and the determinant ofthe operator A are defined by

tr A =

p∑i=1

m jλi ,



det A =

p∏j=1

λm j

j .

• The zeta-function of a positive operator A is defined by

ζ(s) =

p∑i=1

m j1λs

i.

• There holdsζ′(0) = − log det A

• The trace of a projection P onto a vector subspace S is equal to its rank, orthe dimension of the vector subspace S ,

tr P = rank P = dim S .

2.4.5 Functions of Operators

• Let A be a normal operator on a vector space V given by its spectral decom-position

A =

p∑i=1

λiPi ,

where Pi are the projections to the eigenspaces..

• Let f : C→ C be a complex function analytic at 0.

• Then one can define the function of the operator A by

f (A) =

p∑i=1

f (λi)Pi .

• The exponential of A is defined by

exp A =

∞∑k=1

1k!

Ak =

p∑i=1

eλi Pi



• The trace of a function of a self-adjoint operator A is then

tr f (A) =

p∑i=1

m j f (λi) .

where m j is the multiplicity of the eigenvalue.

• Let A be a positive definite operator, A > 0. The zeta-function of theoperator A is defined by

ζ(s) = tr A−s =

p∑i=1

m j1λs

i.

• Theorem 2.4.22 For every unitary operator U there is an Hermitian oper-ator H with real eigenvalues λ j and the corresponding projections P j suchthat

U = exp(iH) =

p∑j=1

eiλ j P j

• The positive square root of a positive operator A is defined by

√A =

p∑j=1

√λ jP j

• Theorem 2.4.23 Let A be a normal operator and P j, j = 1, . . . , p be theprojections to eigenspaces. Then

P j =

p∏k, j;k=1

A − λkIλ j − λk

Proof: Let p j(z) be polynomials of the form

p j(z) =

p∏k, j;k=1

z − λk

λ j − λk

• The polynomial p j(z) has (p − 1) roots λk, k , j, that is,

p j(λk) = 0



• Moreover, the polynomial p j(z) satisfies the equation

p j(λ j) = 1.

• That is,p j(λk) = δ jk

and, therefore, we have

p j(A) =

p∑k=1

p j(λk)Pk = P j

• Dimension-independent definition of the determinant of a positive oper-ator:

1.det A = exp tr log A

2.det A = exp[−ζ′(0)]

3.det A−1/2 =

∫Rn

dx exp[−π(x, Ax)]

where dx = dx1 · · · dxn

2.4.6 Polar Decomposition• Theorem 2.4.24 Polar Decomposition Theorem Let A be an operator on

a complex vector space. Then there exist a unique positive operator R anda unitary operator U such that

A = UR.

If the operator A is invertible then the operator U is also unique.

• Proof. Suppose A is invertible. Let

R =√

A∗A

andU = AR−1.



2.4.7 Real Vector Spaces• Theorem 2.4.25 Let A be a symmetric operator on a real vector space V.

Let λ j, j = 1, . . . , p, be the distinct eigenvalues of A, M j be the correspond-ing eigenspaces and P j be the projections on M j. Then:

1.

V =

p⊕j=1

M j,

2.p∑

j=1

dim M j = dim V ,

3. the projections are symmetric, orthogonal and complete

p∑j=1

P j = I

PiP j = 0 if i , j,PT

i = Pi

4. there is the spectral decomposition of the operator

A =

p⊕j=1

A j =

p∑j=1

λ jP j

• Theorem 2.4.26 The only eigenvalues of an orthogonal operator (on a realvector space) are +1 and −1.

Proof.

• Theorem 2.4.27 Let O be an orthogonal operator in a real vector space V.Then there exists an anti-symmetric operator A such that

O = exp A.

• The (complex) diagonal form of an anti-symmetric operator A is

A = diag (0, . . . , 0, iθ1,−iθ1, . . . , iθk,−iθk)



where θi are real.

Therefore

O = exp A = diag(1, . . . , 1, eiθ1 , e−iθ1 , . . . , eiθk , e−iθk

)• Some of the θ′s may be equal to ±π. By separating them we get

O = exp A = diag(1, . . . , 1,−1, . . . ,−1, eiθ1 , e−iθ1 , . . . , eiθk , e−iθp

)where θi , ±π.

• The real block diagonal form of an anti-symmetric operator A is

A = diag (0, . . . , 0, θ1ε, . . . , θkε)

where

ε =

(0 1−1 0

)• Note that

ε2 = −I

andε2n = (−1)nI, ε2n+1 = (−1)nε

• Thereforeexp(θε) = I cos θ + ε sin θ

• Theorem 2.4.28 Spectral Decomposition of Orthogonal Operators onReal Vector Spaces. Let O be an orthogonal operator on a real vectorspace V. Then the only eigenvalues of O are +1 and −1 (possibly multiple)and there exists an orthogonal decomposition

V = V+ ⊕ V− ⊕ V1 ⊕ · · · ⊕ Vp ,

where V+ and V− are the eigenspaces corresponding to the eigenvalues 1and −1, and V1, . . . ,Vp are mutually orthogonal two-dimensional subspacessuch that

dim V = dim V+ + dim V− + 2p .



Let P+, P−, P1, . . . , Pp be the corresponding orthogonal complimentary sys-tem of projections, that is,

P+ + P− +

p∑i=1

Pi = I .

Then there exists a corresponding system of operators N1, . . . ,Np satisfyingthe equations

N2i = −Pi , NiPi = PiNi = Ni ,

NiP j = P jNi = 0 , if i , j

and the angles θ1, . . . θk such that −π < θi < π and

O = P+ − P− +

p∑i=1

Ri(θi)

whereRi(θi) = cos θi Pi + sin θi Ni .

are the two-dimensional rotation operators in the planes corresponding toPi.

• Theorem 2.4.29 Every invertible operator A on a real vector space can bewritten in a unique way as a product

A = OR

of an orthogonal operator O and a symmetric positive operator R.

2.4.8 Heisenberg Algebra, Fock Space and Harmonic Oscilla-tor

• Heisenberg Algebra. The Heisenberg algebra is a 3-dimensional Lie alge-bra with generators X,Y,Z satisfying the commutation relations

[X,Y] = Z, [X,Z] = 0, [X,Z] = 0.

• A representation of the Lie algebra A is a homomorphism ρ : A → L(V)from the Lie algebra to the space of operators on a vector space V such that

ρ([S ,T ]) = [ρ(S ), ρ(T )].



• The Heisenberg algebra can be represented by matrices

X =

0 1 00 0 00 0 0

, Y =

0 0 00 0 10 0 0

, Z =

0 0 10 0 00 0 0

or by the differential operators C∞(R3)→ C∞(R3) defined by

X = ∂x −12

y∂z, Y = ∂y +12

x∂z, Z = ∂z.

• Properties of the Heisenberg algebra

[X,Yn] = nZYn−1

[X, exp(bY)] = bZ exp(bY)

X exp (bY) = exp (bY) (X + bZ)

exp(−bY)X exp(bY) = X + bZ

X exp(−

12

Y2)

= exp(−

12

Y2)

(X − ZY)

[Y, Xn] = nZXn−1

[Y, exp(aX)] = −aZ exp(aX)

exp(aX)Y = (Y + aZ) exp(aX)

exp(aX)Y exp(−aX) = Y + aZ

exp(aX) exp(bY) = exp(abZ) exp(bY) exp(aX)

exp(−bY) exp(aX) exp(bY) = exp(abZ) exp(aX)

exp(aX) exp(bY) exp(−aX) = exp(abZ) exp(bY)

• Campbell-Hausdorff formula

exp(aX + bY) = exp(−

ab2

Z)

exp(aX) exp(bY)

= exp(ab2

Z)

exp(bY) exp(aX)



• Heisenberg group. The Heisenberg group is a 3-dimensional Lie groupwith the generators X,Y,Z.

• An arbitrary element of the Heisenberg group is parametrized by canonicalcoordinates (a, b, c) as

g(a, b, c) = exp(aX + bY + cZ)

Obviously,g(0, 0, 0) = I

and the inverse is defined by

[g(a, b, c)]−1 = g (−a,−b,−c)

• The group multiplication law in the Heisenberg group takes the form

g(a, b, c)g(a′, b′, c′) = g(a + a′, b + b′, c + c′ +

12

(ab′ − a′b))

• Notice that

g(a, b, c) = g(0, 0, c +

ab2

)g(0, b, 0)g(a, 0, 0)

• A representation of a Lie group G is a homomorphism ρ : G → Aut (V)from the group G to the space of invertible operators on a vector space Vsuch that for any g, h ∈ G

ρ(gh) = ρ(g)ρ(h)

andρ(g−1) = [ρ(g)]−1, ρ(e) = I

• Representations of the Heisenberg group.

• The elements of the Heisenberg group could be represented by the upper-triangular matrices. Notice that

X2 = Y2 = Z2 = XZ = YZ = 0



andXY = YX = Z

or 0 a c0 0 b0 0 0

2

=

0 0 ab0 0 00 0 0

, 1 a c

0 1 b0 0 1

3

= 0.

Therefore,

g(a, b, c) =

1 a c + ab0 1 b0 0 1

,• Another representation is defined by the action on functions in R3. Notice

thatexp(aX) f (x, y, z) = f

(x + a, y, z −

a2

y)

exp(bY) f (x, y, z) = f(a, y + b, z +

a2

x)

exp(cZ) f (x, y, z) = f (x, y, z + c)

• Therefore,

g(a, b, c) f (x, y, z) = f(x + a, y + b, z + c +

b2

x −a2

y)

• Fock space.

• Let us define the operatorN = YX

• It is easy to see that

[N,Y] = ZY, [N, X] = −ZX.

• Suppose that there exists a unit vector v0 called the vacuum state such that

Xv0 = 0

• Let us define a sequence of vectors

vn =1√

n!Ynv0



• By using the properties of the Heisenberg algebra it is easy to show that

Yvn =√

n + 1 vn+1, n ≥ 0,Xvn =

√n Zvn−1, n ≥ 1

ThereforeNvn = n Zvn, n ≥ 0

• Let us define vectors

w(b) = exp(bY)v0 =

∞∑n=0

bn

√n!

vn

called the coherent states.

• Then by using the properties of the Heisenberg algebra we get

Xw(b) = bZw(b)

• Now, suppose thatY = X∗

andZ = I

• Then, the vectors vn are orthonormal

(vn, vm) = δnm

and are the eigenvectors of the self-adjoint operator

N = X∗X

with integer eigenvalues n ≥ 0.

• Then the spacespan vn | n ≥ 0

is called the Fock space and the operators X and X∗ are called the annihi-lation and creation operators and the operator N is called the operator ofthe number of particles.



• Note, also that the coherent states are not orthonormal

(w(a),w(b)) = eab

• Finally, we compute the trace of the heat semigroup operator

Tr exp(−tX∗X) =

∞∑n=0

e−tn =1

1 − e−t

• Harmonic oscillator. Let D be an anti-self-adjoint operator and Q be aself-adjoint operator satisfying the commutation relations

[D,Q] = I

• The harmonic oscillator is a quantum system with the (self-adjoint posi-tive) Hamiltonian

H = −12

D2 +12

Q2

• Then the operators

X =1√

2(D + Q), X∗ =

1√

2(−D + Q).

are the creation and annihilation operators.

• The operator of the number of particles is

N = X∗X = −12

D2 +12

Q2 −12

and, therefore, the Hamiltonian is

H = N +12

• The eigenvalues of the Hamiltonian are

λn = n +12

with the eigenvectors vn



• It is clear that the vectors

ψn(t) = e−itλnvn = exp[−it

(n +

12

)]1

2n/2√

n!(−D + Q)nv0

satisfy the equation(i∂t − H)ψn = 0

which is called the Schrodinger equation.

• The vacuum state is determined from the equation

(D + Q)v0 = 0

and has the form

v0 = exp(−

12

Q2)ψ0

with ψ0 satisfyingDψ0 = 0 .

• The heat trace (also called the partition function) for the harmonic oscilla-tor is

Tr exp(−tH) =1

2 sinh(t/2)

2.4.9 Exercises1. Find the eigenvalues of a projection operator.

2. Prove that the span of all eigenvectors corresponding to the eigenvalue λ ofan operator A is a vector space.

3. LetE(λ) = Ker (A − λI) .

Show that: a) if λ is not an eigenvalue of A, then E(λ) = ∅, and b) if λ is aneigenvalue of A, then E(λ) is the eigenspace corresponding to the eigenvalueλ.

4. Show that the operator A−λI is invertible if and only if λ is not an eigenvalueof the operator A.



5. Let T be a unitary operator. Then the operators A and

A = T AT−1

are called similar. Show that the eigenvalues of similar operators are thesame.

6. Show that an operator similar to a selfadjoint operator is selfadjoint and anoperator similar to an anti-selfadjoint operator is anti-selfadjoint.

7. Show that all eigenvalues of a positive operator A are non-negative.

8. Show that the eigenvectors corresponding to distinct eigenvalues of a uni-tary operator are orthogonal to each other.

9. Show that the eigenvectors corresponding to distinct eigenvalues of a self-adjoint operator are orthogonal to each other.

10. Show that all eigenvalues of a unitary operator A have absolute value equalto 1.

11. Show that if A is a projection, then it can only have two eigenvalues: 1 and0.


Chapter 3

Banach Spaces

3.1 Normed Spaces• Function Spaces

• The vector space F(X, E) of all functions f : X → E from a set X into avector space E.

• Let Ω ⊂ Rn be an open subset in Rn (it could be the whole space Rn). Thespace of all functions from Ω into C is a vector space. The following aresubspaces of this vector space:

1. P(Ω) (polynomials)2. C(Ω) (continuous functions f : Ω→ C)3. C0(Ω) (continuous functions with support in the interior of Ω, that is,

functions that are equal to zero in a finite neighborhood of the bound-ary of Ω)

4. Ck(Ω) (functions with continuous partial derivatives of order k)5. C∞(Ω) (smooth functions)6. C∞(Ω,V) (vector-valued functions ϕ : Ω → V taking values in a vec-

tor space V)7. Lp(Ω, µ,V), p ≥ 1 (p-integrable vector-valued functions)

||ϕ||p =

∫Ω

dx µ(x) || ϕ(x)||pV

1/p

< ∞

71

72 CHAPTER 3. BANACH SPACES

• Show thatC(Ω) = L∞(Ω)

• Sequence Spaces

• Let N be the set of positive integers. The space F(N,F) of all functionsfrom N into F is the vector space of sequences of scalars. The following aresubspaces of this vector space:

1. l0 (sequences of complex numbers with zero tails, that is, sequencescontaining only finitely many non-zero elements),

2. convergent sequences of complex numbers,

3. bounded sequences of complex numbers,

4. lp, p ≥ 1 (sequences with finite p-norm)

‖ x ‖p=

∞∑n=1

|xn|p

1p

< ∞

5. lp(V), p ≥ 1 (sequences of vectors with finite p-norm)

‖ x ‖p=

∞∑n=1

||xn||pV

1p

< ∞

• Proof: Use Minkowski inequality.

• Theorem 3.1.1 Minkowski’s Inequality Let p ≥ 1. Then for any two se-quences of complex numbers

‖ x + y ‖p≤‖ x ‖p + ‖ y ‖p

Without proof.

• Cartesian Product of Vector Spaces. Let E jnj=1 be a collection of vector

spaces over a field F. The Cartesian product (or product) of vector spacesE j is the space

E = E1 × · · · × En

= (x1, . . . , xn) | x j ∈ E j, 1 ≤ j ≤ n .


3.1. NORMED SPACES 73

• Normed Space. A vector space with a norm is called a normed space.

• One can define different norms on the same vector space.

• A normed space is a pair (E, ‖ · ‖), where E is a vector space and ‖ · ‖ is anorm on E.

• Some vector spaces have standard norms.

• A vector subspace of a normed space is a normed space with the same norm.

• Norm of Uniform Convergence. Let Ω ⊂ Rn be a closed bounded subsetof Rn and C(Ω) be the space of continuous functions on Ω. Norm in C(Ω)

‖ f ‖∞= maxx∈Ω| f (x)|

• For the lp-norm as p→ ∞

‖ x ‖p→ ||x||∞ = supn∈Z+

|xn|

• The norm can be used to define convergence.

• Convergence in a Normed Space. Let (E, ‖ · ‖) be a normed space and (xn)be a sequence of vectors in E. The sequence (xn) converges to x ∈ E if forevery ε > 0 there exists a positive integer M ∈ N such that for every n ≥ Mwe have

‖ xn − x ‖< ε.

Then we write x = limn→∞ xn or xn → x.

• xn → x simply means that ‖ xn − x ‖→ 0 in R.

• Properties of convergence in normed space.

• A convergent sequence has a unique limit.

• If xn → x and λn → λ, then λnxn → λx.

• If xn → x and yn → y, then xn + yn → x + y.

• Not every convergence in a vector space can be defined by a norm.



• Uniform Convergence. Let C(Ω) be the space of all continuous functionson a closed bounded set Ω ⊂ Rn and let ( fn) ∈ C(Ω) be a sequence offunctions in C(Ω). The sequence ( fn) converges uniformly to f if for everyε > 0 there exists a positive integer M = M(ε) ∈ N such that for all x ∈ Ω

and for all n ≥ M we have

| f (x) − fn(x)| < ε.

The norm of uniform convergence defines the uniform convergence, i.e. thesequence ( fn) converges uniformly to f if and only if

‖ fn − f ‖∞= maxx∈Ω| fn(x) − f (x)| → 0.

• Pointwise Convergence. Let C([0, 1]) be the space of continuous functionson the interval [0, 1] and let ( fn) be a sequence of functions in C([0, 1]). Thesequence ( fn) converges pointwise to f if for all x ∈ [0, 1] and for everyε > 0 there exists a positive integer M = M(ε, x) ∈ N such that for all n ≥ Mwe have

| f (x) − fn(x)| < ε.

The pointwise convergence simply means that for every x ∈ [0, 1] the se-quence ( fn(x)) converges to f (x), i.e.

fn(x)→ f (x) or | fn(x) − f (x)| → 0.

• There is no norm on C([0, 1]) which defines the pointwise convergence.

Proof: (by contradiction). Construct a sequence ( fn) of functions such that

1. ‖ fn ‖= 1 for all n ∈ N and

2. fn(x)→ 0 as n→ ∞ ∀x ∈ [0, 1].

• Equivalence of Norms. Two norms on the same vector space E are equiv-alent if they define the same convergence.

That is, the norms ‖ · ‖1 and ‖ · ‖2 are equivalent if for any sequence (xn) inE and x ∈ E,

‖ xn − x ‖1→ 0 if and only if ‖ xn − x ‖2→ 0 .



• Example. R2.

• Theorem 3.1.2 Two norms ‖ · ‖1 and ‖ · ‖2 in a vector space E are equiva-lent if and only if there exist positive real numbers α, β ∈ R+ such that

α ‖ x ‖1≤‖ x ‖2≤ β ‖ x ‖1 for all x ∈ E.

Proof: No proof.

• Every normed space (E, ‖ · ‖) is a metric space (E, d) with the metric

d(x, y) =‖ x − y ‖ .

• A metric space (E, d) is a set E with a metric d. A metric d on a set E is afunction d : E × E → R satisfying the following axioms: ∀x, y, z ∈ E

1. d(x, y) ≥ 0,

2. d(x, y) = 0 if and only if x = y,

3. d(x, z) ≤ d(x, y) + d(y, z) .

• The convergence defined by the norm ‖ · ‖ is the same as the convergencedefined by the metric d(x, y) =‖ x − y ‖.

• The metric defines a topology in E (open and closed sets).

• The basic topological notions can be defined without a metric.

• A topological space (E,T ) is a set E with a topology T . A topology T ona set E is a collection T of subsets of E (called open sets) that contains Eand ∅ and is closed under union and finite intersection.

• Open Balls, Closed Balls, Spheres. Let E be a normed space, x ∈ E andr ∈ R+ a positive real number. We define the following sets:

Open ballB(x, r) = y ∈ E | ‖ x − y ‖< r

Closed ballB(x, r) = y ∈ E | ‖ x − y ‖≤ r

SphereS (x, r) = y ∈ E | ‖ x − y ‖= r

Here x is the center and r is the radius.



• Examples. R2, C([0, 1]), ‖ · ‖∞.

• Open and Closed Sets. A subset S ⊆ E of a normed space E is open if forevery x ∈ S there exist ε > 0 such that B(x, ε) ⊆ S .

A subset S ⊆ E of a normed space E is closed if its complement E \ S isopen.

• Example. Let Ω be a closed bounded set in Rn and C(Ω) be the space ofcontinuous functions on Ω with the norm of uniform convergence ‖ · ‖∞.Let x0 ∈ Ω. The set

g ∈ C(Ω) | g(x) > 0, ∀x ∈ Ω

is open C(Ω), and the sets

g ∈ C(Ω) | g(x) ≥ 0, ∀x ∈ Ω

andg ∈ C(Ω) | g(x0) = 0

are closed in C(Ω).

• Theorem 3.1.3 1. The union of any number of open sets is open.

2. The intersection of a finite number of open sets is open.

3. The union of a finite number of closed sets is closed.

4. The intersection of any number of closed sets is closed.

5. The empty set and the whole space are both open and closed.

Proof: Exercise.

• Theorem 3.1.4 A subset S of a normed space E is closed if and only ifevery sequence of elements of S convergent in E has its limit in S .

Proof: No proof.

• Closure. Let S be a subset of a normed space E. The closure of S (denotedby S or cl S ) is the intersection of all closed sets containing S .

• The closure of a set is a closed set.

• The closure of a set is the smallest closed set which contains S .



• Theorem 3.1.5 Let S be a subset of a normed space E. The closure of S isthe set of limits of all convergent sequences of elements of S .

Proof: Exercise.

• Weierstrass Approximation Theorem. The closure of the set of all poly-nomials on [a, b] is the whole space C([a, b]).

• Dense Subsets. A subset S of a normed space E is dense in E if cl S = E.

• Examples.

1. The set of all polynomials on [a, b] is dense in C([a, b]).

2. The set of all sequences of complex numbers which have only a finitenumber of nonzero terms is dense in lp for any p ≥ 1.

• Theorem 3.1.6 Let S be a subset of a normed space E. The followingconditions are equivalent:

1. S is dense in E.

2. Every element of E is a limit of a convergent sequence in S .

3. Every nonempty open subset of E contains an element of S .

Proof: Exercise.

• Compact Sets. A subset S of a normed space E is compact in E if everysequence in S contains a convergent subsequence whose limit belongs to S .

• Examples. Rn, Cn

• Theorem 3.1.7 Compact sets are closed and bounded.

Proof: No proof.

• Noncompact Closed and Bounded Set. Let C([0, 1]) be the space of con-tinuous functions on [0, 1]. The closed unit ball B(0, 1) is a closed andbounded set. Let xn(t) = tn ∈ B(0, 1) be the sequence of functions of unitnorm. Then (xn) does not have a convergent subsequence. So, the closedunit ball B(0, 1) is not compact.



3.2 Banach Spaces• Cauchy Sequence. A sequence of vectors (xn) in a normed space is a

Cauchy sequence if for every ε > 0 there exists M ∈ N such that forall n,m ≥ M,

‖ xm − xn ‖< ε.

• Theorem 3.2.1 A sequence (xn) is a Cauchy sequence if and only if for anytwo subsequences (xpn) and (xqn)

‖ xpn − xqn ‖→ 0 as n→ ∞.

• Every convergent sequence is Cauchy.

• Not every Cauchy sequence in a normed space E converges to a vector inE.

• Example. Incompleteness.

• Lemma 3.2.1 Let (xn) be a Cauchy sequence of vectors in a normed space.Then the sequence (‖ xn ‖) or real numbers converges.

Proof:

1. We have | ‖ xm ‖ − ‖ xn ‖ | ≤‖ xm − xn ‖. Thus (‖ xn ‖) is Cauchy.

• Every Cauchy sequence is bounded.

Proof: Exercise.

• Banach Space. A normed space E is complete (or Banach space) if everyCauchy sequence in E converges to an element in E.

• Examples. Rn and Cn (with any norm) are complete.

• Theorem 3.2.2 Completeness of lp, p ≥ 1. The space lp of complex se-quences with the norm ‖ · ‖p is complete.

Proof: No proof.


3.2. BANACH SPACES 79

• Theorem 3.2.3 Completeness of C(Ω). The space of complex-valued con-tinuous functions on a closed bounded set Ω ⊂ Rn with the norm

‖ f ‖∞= supx∈Ω| f (x)|

is complete.

Proof: No proof.

• Characteristic function of a set Ω ⊂ Rn is a mapping χΩ : Rn → R definedby

χΩ(x) =

1, if x ∈ Ω

0, if x < Ω(3.1)

• A set Ω ⊂ Rn is called a null set (or a set of measure zero) if its character-istic function is a null function.

• Theorem 3.2.4 a) Every countable set is a null set.

b) A countable union of null sets is a null set.

c) Every subset of a null set is a null set.

• The closure of the set of all points x ∈ Rn for which a function f : Rn → Cis not zero is called the support of f , i.e.

supp f = x ∈ Rn| f (x) , 0 . (3.2)

• An integrable function f ∈ L1(Rn) is called a null function if

|| f ||1 =

∫Rn| f | = 0.

• Two integrable functions are said to be equal almost everywhere if thesupport of their difference is a null set.

• Theorem 3.2.5 Two functions are equal almost everywhere if and only iftheir difference is a null function.

• Strictly speaking, to make L1(Rn) a normed space one has to consider in-stead of functions the equivalence classes of functions.

• Two functions are said to be equivalent if their difference is a null function.



• The equivalence class of an integrable function f ∈ L1(Ω) is the set of allfunctions g ∈ L1(Ω) equivalent to f ,

[ f ] = g ∈ L1(Ω) | || f − g||1 = 0.

• Theorem 3.2.6 The space Lp(Ω), p ≥ 1 is complete.

• Convergent and Absolutely Convergent Series. A series∑∞

n=1 xn con-verges in a normed space E if the sequence of partial sums sn =

∑nk=1 xk

converges in E.

• If∑∞

n=1 ‖ xn ‖< ∞, then the series converges absolutely.

• An absolutely convergent series does not need to converge.

• Theorem 3.2.7 A normed space is complete if and only if every absolutelyconvergent series converges.

Proof: in functional analysis books.

• Theorem 3.2.8 A closed vector subspace of a Banach space is a Banachspace.

Proof: no proof.

• Completion of Normed Spaces

• Let (E, ‖ · ‖) be a normed space. A normed space (E, ‖ · ‖1) is a completionof (E, ‖ · ‖) if there exists a linear injection Φ : E → E, such that

1. for every x ∈ E

‖ x ‖=‖ Φ(x) ‖1,

2. Φ(E) is dense in E,

3. E is complete.

• The space E is defined as the set of equivalence classes of Cauchy sequencesin E.

• Recall that if (xn) is Cauchy then the sequence ||xn|| converges.

• We define an equivalence relation of Cauchy sequences as follows.


3.2. BANACH SPACES 81

• Two Cauchy sequences (xn) and (yn) in E are equivalent, (xn) ∼ (yn), if

lim ‖ xn − yn ‖= 0.

• The set of Cauchy sequences (yn) equivalent to a Cauchy sequence (xn) isthe equivalence class of (xn)

[(xn)] = (yn) ∈ E | (yn) ∼ (xn)

• Obviously we can identify the space E with the set of constant Cauchysequences.

• Then the set of all equivalent classes is

E = E/ ∼ = [(xn)] | (xn) is Cauchy sequence in E

• The addition and multiplication by scalars in E are defined by

[(xn)] + [(yn)] = [(xn + yn)], λ[(xn)] = [(λxn)]

• The norm in E is defined by the limit

‖ [(xn)] ‖1= limn→∞‖ xn ‖,

which exists for every Cauchy sequence.

• This definition is consistent since for any two equivalent Cauchy sequences(xn) and (yn)

‖ [(xn)] ‖1=‖ [(yn)] ‖1

Proof: Exercise.

• The linear injection Φ : E → E is defined by a constant sequence

Φ(x) = [(xn)] such that xn = x, ∀n ∈ N.

• Then Φ is one-to-one.

Proof: Exercise.

• Obviously, ∀x ∈ E, ‖ x ‖=‖ Φ(x) ‖1.



• Claim: Φ(E) is dense in E.

Proof: Since every element [(xn)] of E is the limit of a sequence (Φ(xn)).

• Claim: E is complete.

Proof:

Let (xn) be a Cauchy sequence in E.

Then ∃(xn) in E such that

‖ Φ(xn) − xn ‖1<1n.

• Claim: (xn) is Cauchy sequence in E.

Proof:

‖ xn − xm ‖ = ‖ Φ(xn) − Φ(xm) ‖1

≤ ‖ Φ(xn) − xn ‖1 + ‖ xn − xm ‖1 + ‖ xm − Φ(xm) ‖1

≤ ‖ xn − xm ‖1 +1n

+1m.

• Next, let x = [(xn)].

• Claim:‖ xn − x ‖1→ 0.

Proof:

‖ xn − x ‖1 ≤ ‖ xn − Φ(xn) ‖1 + ‖ Φ(xn) − x ‖1

<1n

+ ‖ Φ(xn) − x ‖1→ 0.

• Homeomorphism. Two topological spaces E1 and E2 are homeomorphicif there exists a bijection Ψ : E1 → E2 from E1 onto E2 such that both Ψ

and Ψ−1 are continuous.

• Isomorphism of Normed Spaces. Two normed spaces (E1, ‖ · ‖1) and(E2, ‖ · ‖2) are isomorphic if there exists a linear homeomorphism Ψ :E1 → E2 from E1 onto E2.

• Theorem 3.2.9 Any two completions of a normed space are isomorphic.

Proof: Read elsewhere.


3.3. LINEAR MAPPINGS 83

3.3 Linear Mappings• Let L : E1 → E2 be a mapping from a vector space E1 into a vector space

E2.

• If x ∈ E1, then L(x) is the image of the vector x.

• If A ⊂ E1 is a subset of E1, then the set

L(A) = y ∈ E2 | y = L(x) for some x ∈ A

is the image of the set A.

• If B ⊂ E2 is a subset of E2, then the set

L−1(B) = x ∈ E1 | L(x) ∈ B

is the inverse image of the set B.

• A mapping L : D(L) → E2 may be defined on a proper subset (called thedomain) D(L) ⊂ E1 of the vector space E1.

• The image of the domain, L(D(L)), of a mapping L is the range of L. Thatis the range of L is

R(L) = y ∈ E2 | y = L(x) for some x ∈ D(L) .

• The null space N(L) (or the kernel Ker(L)) of a mapping L is the set of allvectors in the domain D(L) which are mapped to zero, that is

N(L) = x ∈ D(L) | L(x) = 0 .

• The graph Γ(L) of a mapping L is the set of ordered pairs (x, L(x)), that is

Γ(L) = (x, y) ⊂ E1 × E2 | x ∈ D(L) and y = L(x) .

• Continuous Mappings. A mapping f : E1 → E2 from a normed space E1

into a normed space E2 is continuous at x0 ∈ E1 if any sequence (xn) inE1 converging to x0 is mapped to a sequence f (xn) in E2 that converges tof (x0).



• That is f is continuous at x0 if

‖ xn − x0 ‖→ 0 implies ‖ f (xn) − f (x0) ‖→ 0.

• A mapping f : E1 → E2 is continuous if it is continuous at every x ∈ E1.

• Theorem 3.3.1 The norm ‖ · ‖: E → R in a normed space E is a continuousmapping from E into R.

Proof: If ‖ xn − x ‖→ 0, then

| ‖ xn ‖ − ‖ x ‖ | ≤‖ xn − x ‖→ 0

• Theorem 3.3.2 Let f : E1 → E2 be a mapping from a normed space E1

into a normed space E2. The following conditions are equivalent:

1. f is continuous.

2. The inverse image of any open set of E2 is open in E1.

3. The inverse image of any closed set of E2 is closed in E1.

Proof: Exercise.

• Linear Mappings. Let S ⊂ E1 be a subset of a vector space E1. A mappingL : S → E2 is linear if ∀x, y ∈ S and ∀α, β ∈ F such that αx + βy ∈ S ,

L(αx + βy) = αL(x) + βL(y).

• Theorem 3.3.3 If S is not a vector subspace of E1, then there is a uniqueextension of L : S → E2 to a linear mapping L : span S → E2 from thevector subspace span S to E2.

Proof: The extension L is defined by linearity.

• Thus, one can always assume that the domain of a linear mapping is a vectorspace.

• Theorem 3.3.4 The range, the null space and the graph of a linear map-ping are vector spaces.

Proof: Exercise.



• For any linear mapping L, L(0) = 0. Thus, 0 ∈ N(L) and the null spaceN(L) is always nonempty.

• Theorem 3.3.5 A linear mapping L : E1 → E2 from a normed space E1

into a normed space E2 is continuous if and only if it is continuous at apoint.

Proof:

1. Assume L is continuous at x0 ∈ E1.

2. Let x ∈ E1 and (xn)→ x.

3. Then (xn − x + x0)→ x0.

4. Thus‖ L(xn) − L(x) ‖=‖ L(xn − x + x0) − L(x0) ‖→ 0

• Bounded Linear Mappings. A linear mapping L : E1 → E2 from a normedspace E1 into a normed space E2 is bounded if there is a real number K ∈ Rsuch that for all x ∈ E1,

‖ L(x) ‖≤ K ‖ x ‖ .

• Theorem 3.3.6 A linear mapping L : E1 → E2 from a normed space E1

into a normed space E2 is continuous if and only if it is bounded.

Proof:

1. (I). Assume that L is bounded.

2. Claim: L is continuous at 0.

3. Indeed, xn → 0 implies

‖ L(xn) ‖≤ K ‖ xn ‖→ 0

4. Hence, L is continuous.

5. (II). Assume that L is continuous.

6. By contradiction, assume that L is unbounded.

7. Then, there is a sequence (xn) in E1 such that

‖ L(xn) ‖> n ‖ xn ‖



8. Letyn =

xn

n ‖ xn ‖, n ∈ N

9. Then

‖ yn ‖=1n, and ‖ L(yn) ‖> 1

10. Then yn → 0 but L(yn) 6→ 0.

11. Thus, L is not continuous at zero.

• Remark. For linear mappings, continuity and uniform continuity are equiv-alent.

• The set L(E1, E2) of all linear mappings from a vector space E1 into a vectorspace E2 is a vector space with the addition and multiplication by scalarsdefined by

(L1 + L2)(x) = L1(x) + L2(x), and (αL)(x) = αL(x) .

• The set B(E1, E2) of all bounded linear mappings from a normed space E1

into a normed space E2 is a vector subspace of the space L(E1, E2).

• Theorem 3.3.7 The space B(E1, E2) of all bounded linear mappings L :E1 → E2 from a normed space E1 into a normed space E2 is a normedspace with norm defined by

‖ L ‖= supx∈E1,x,0

‖ L(x) ‖‖ x ‖

= supx∈E1,‖x‖=1

‖ L(x) ‖ .

Proof:

1. Obviously, ‖ L ‖≥ 0.

2. ‖ L ‖=0 if and only if L = 0.

3. Claim: ‖ L ‖ satisfies triangle inequality.

4. Let L1, L2 ∈ B(E1, E2).



5. Then

‖ L1 + L2 ‖ = sup‖x‖=1‖ L1(x) + L2(x) ‖

≤ sup‖x‖=1‖ L1(x) ‖ + sup

‖x‖=1‖ L2(x) ‖

= ‖ L1 ‖ + ‖ L2 ‖

• For any bounded linear mapping L : E1 → E2

‖ L(x) ‖≤‖ L ‖ ‖ x ‖, ∀x ∈ E1 .

• ‖ L ‖ is the least real number K such that

‖ L(x) ‖≤ K ‖ x ‖ for all x ∈ E1.

• The norm defined by ‖ L ‖= supx∈E1,‖x‖=1 ‖ L(x) ‖ is called the operatornorm.

• Convergence with respect to the operator norm is called the uniform con-vergence of operators.

• Strong convergence. A sequence of bounded linear mappings Ln ∈ B(E1, E2)converges strongly to L ∈ B(E1, E2) if for every x ∈ E1 we have

‖ Ln(x) − L(x) ‖→ 0 as n→ ∞.

• Theorem 3.3.8 Uniform convergence implies strong convergence.

Proof: Follows from

‖ Ln(x) − L(x) ‖≤‖ Ln − L ‖ ‖ x ‖

• Converse is not true.

• Theorem 3.3.9 Let E1 be a normed space and E2 be a Banach space. ThenB(E1, E2) is a Banach space.

Proof: see functional analysis books.



• Theorem 3.3.10 Let E1 be a normed space and E2 be a Banach space. LetS ⊂ E1 be a subspace of E1 and L : S → E2 be a continuous linear mappingfrom S into E2. Then:

1. L has a unique extension to a continuous linear mapping L : S → E2

defined on the closure of the domain of the mapping L.

2. If S is dense in E1, then L has a unique extension to a continuouslinear mapping L : E1 → E2.

Proof: see functional analysis.

• Theorem 3.3.11 Let E1 and E2 be normed spaces, S ⊂ E1 be a subspace ofE1 and L : S → E2 be a continuous linear mapping from S into E2. Then:

1. the null space N(L) is a closed subspace of E1.

2. If the domain S is a closed subspace of E1, then the graph Γ(L) of L isa closed subspace of E1 × E2.

Proof: Exercise.

• A bounded linear mapping L : E → F from a normed space E into the scalarfield F is called a functional.

• The space B(E,F) of functionals is called the dual space and denoted by E′

or E∗.

• The dual space is always a Banach space.

Proof: since F is complete.


3.4. BANACH FIXED POINT THEOREM 89

3.4 Banach Fixed Point Theorem• Contraction Mapping. Let E be a normed space and A ⊂ E be a subset of

E. A mapping f : A→ E from A into E is a contraction mapping if thereexists a real number α, such that 0 < α < 1 and ∀x, y ∈ A

‖ f (x) − f (y) ‖≤ α ‖ x − y ‖ .

• A contraction mapping is continuous.

Proof: Exercise.

• If ∀x, y ∈ A‖ f (x) − f (y) ‖≤‖ x − y ‖,

then it is not necessarily a contraction since the contraction constant α maynot exist.

• Theorem 3.4.1 Banach Fixed Point Theorem. Let E be a Banach spaceand A ⊂ E be a closed subset of E. Let f : A→ A be a contraction mappingfrom A into A. Then there exists a unique z ∈ A such that f (z) = z.

• Examples.

Homework

• Exercises: DM[9,10,11,12,14,21,23,26,31,34,36,37,38,39,42,44]




Chapter 4

Hilbert Spaces

4.1 Inner Product Spaces• Inner Product Space. A complex vector space E with an inner product

is called an inner product space (or a pre-Hilbert space, or a unitaryspace).

• Orthogonal Vectors. Let E be an inner product space. Two vectors x, y ∈ Eare orthogonal if (x, y) = 0.

• Notation. The orthogonality of the vectors x and y is denoted by

x ⊥ y

• The relation ⊥ is symmetric, that is, if x ⊥ y then y ⊥ x.

• Examples.

• Spaces of sequences

• The spaces of sequences l0 (sequences with zero tails) and l2 (sequenceswith finite || · ||2 norm) are 0inner product spaces with the inner product

(x, y) =

∞∑j=1

x jy j

• Spaces of functions

91

92 CHAPTER 4. HILBERT SPACES

• The space C([a, b]) is an inner product space with the inner product

( f , g) =

∫ b

af g

• The space C0(R) (continuous functions with compact support) is an innerproduct space with the inner product

( f , g) =

∫R

f g

• Spaces of square integrable functions

• The space L2([a, b]) is inner product space with the inner product

( f , g) =

∫ b

af g

• The space L2(R) is and inner product space with the inner product

( f , g) =

∫R

f g

• L2([a, b], µ) (space of square integrable functions with the measure µ) withthe inner product

( f , g) =

∫ b

aµ f g ,

where µ > 0 almost everywhere.

• L2(R, µ) (space of square integrable functions with the measure µ) with theinner product

( f , g) =

∫R

µ f g ,

where µ > 0 almost everywhere.

• Let Ω be an open set in Rn (in particular, Ω can be the whole Rn). The spaceL2(Ω) is the set of complex valued functions such that∫

Ω

| f |2 < ∞ .


4.1. INNER PRODUCT SPACES 93

It is an inner product space with the inner product

( f , g) =

∫Ω

f g

• L2(Rn) with the inner product

( f , g) =

∫Rn

f g

• Let Ω be an open set in Rn (in particular, Ω can be the whole Rn) and Vbe a finite-dimensional vector space. Let 〈 , 〉 be the inner product on thevector space V . The space L2(Ω,V, µ) is the set of vector valued functionsf = ( f 1, . . . , f N) on Ω such that∫

Ω

µ 〈 f , f 〉 =

∫Ω

µ

N∑i=1

| f i|2 < ∞ .

It is an inner product space with the inner product

( f , g) =

∫Ω

µ 〈 f , g〉 =

∫Ω

µ

N∑i=1

figi

• Real Inner Product Spaces. The inner product in a real inner productspace is symmetric.

• Direct Sum of Inner Product Spaces. Let E1 and E2 be inner productspaces. The direct sum E = E1 ⊕ E2 of E1 and E2 is an inner product spaceof ordered pairs z = (x, y) with x ∈ E1 and y ∈ E2 with the inner productdefined by

(z1, z2)E = (x1, x2)E1 + (y1, y2)E2 .

• Tensor Products of Inner Product Spaces. Let E1 and E2 be inner productspaces. For each ϕ1 ∈ E1 and ϕ2 ∈ E2 let ϕ1 ⊗ ϕ2 denote the conjugatebilinear form on E1 × E2 defined by

(ϕ1 ⊗ ϕ2)(ψ1, ψ2) = (ψ1, ϕ1)E1(ψ2, ϕ2)E2

where ψ1 ∈ E1 and ψ2 ∈ E2.



• Let E = E1 ⊗ E2 be the set of finite linear combinations of such bilinearforms. An inner product on E can be defined by

(ϕ ⊗ ψ, η ⊗ µ)E = (ϕ, η)E1(ψ, µ)E2

(with ϕ, η ∈ E1 and ψ, µ ∈ E2) and extending by linearity on E.

• Fock Space. Let E be an inner product space. The space

F(E) = C ⊕∞n=1 E ⊗ · · · ⊗ E︸︷︷︸n

is called the Fock space over E.

4.1.1 Norm in an Inner Product Space.

Let E be an inner product space. The norm in E is a functional ‖ · ‖: E → Rdefined by

‖ x ‖=√

(x, x).

• Theorem 4.1.1 Every inner product space is a normed space with the norm‖ x ‖=

√(x, x) and a metric space with the metric d(x, y) =

√(x − y, x − y).

• Theorem 4.1.2 Schwarz’s Inequality. Let E be an inner product space.Then for any x, y ∈ E we have

|(x, y)| ≤‖ x ‖ ‖ y ‖ .

The equality |(x, y)| =‖ x ‖ ‖ y ‖ holds if and only if x and y are linearlydependent.

Proof:

1.

• Corollary 4.1.1 Triangle Inequality. Let E be an inner product space.Then for any x, y ∈ E we have

‖ x + y ‖≤‖ x ‖ + ‖ y ‖ .

Proof:


4.1. INNER PRODUCT SPACES 95

1.

• Theorem 4.1.3 Parallelogram Law. Let E be an inner product space.Then for any x, y ∈ E we have

‖ x + y ‖2 + ‖ x − y ‖2= 2(‖ x ‖2 + ‖ y ‖2

)Proof:

1.

•

Theorem 4.1.4 Pythagorean Theorem. Let E be an inner productspace. If two vectors x, y ∈ E are orthogonal then

‖ x + y ‖2=‖ x ‖2 + ‖ y ‖2 .

Proof:

1.

• Examples.



4.2 Hilbert Spaces• Hilbert Space. An inner product space is called a Hilbert space if it is

complete as a normed space.

• Examples.

• Spaces of sequences

• The space l2 of square summable sequences is complete. Proved before.

• The space l0 of sequences with vanishing tails is not complete. Counterex-ample.

• Spaces of continuous functions

• C([a, b]) is not complete. Counterexample.

• C0(R) (space of continuous functions with compact support) is not com-plete. Counterexample.

• Spaces of square integrable functions.

• Theorem 4.2.1 The spaces L2([a, b]) and L2(R) are complete.

Proof: Functional analysis books.

Take a Cauchy sequence fn in L2 so that

|| fn − fm||2 → 0

as n,m→ ∞.

By Schwarz inequality

|| fn − fm||1 ≤√

(b − a)|| fn − fm||2 .

So, fn is Cauchy in L1. Since L1 is complete it converges to some f ∈ L1.

Clearly, the subsequence fpn is Cauchy in L2.

Then there is a subsequence fpn convergent to f (almost everywhere).

Therefore, f ∈ L2.


4.2. HILBERT SPACES 97

• Theorem 4.2.2 The space L2(R) is complete.

Proof. Construct a subsequence of a Cauchy sequence in L2(R) that con-verges almost everywhere by a diagonal argument.

• Theorem 4.2.3 The space L2([a, b], µ) is complete.

Proof. Rescale functions by√µ.

• More generally,

• Theorem 4.2.4 The space L2(Ω, µ,V) is complete.

• Sobolev Spaces

• Let Ω be an open set in Rn (in particular, Ω can be the whole Rn) and V afinite-dimensional complex vector space.

• Let Cm(Ω,V) be the space of complex vector valued functions that havecontinuous partial derivatives of all orders less or equal to m.

• Letα = (α1, . . . , αn),

α j ∈ N, be a multiindex of nonnegative integers, αi ≥ 0, and let

|α| = α1 + · · · + αn.

• Define

Dα f =∂|α|

∂xα11 · · · ∂xαn

nf .

• Then f ∈ Cm(V,Ω) iff ∀α, |α| ≤ m, ∀i = 1, . . . ,N, ∀x ∈ Ω we have

|Dα fi(x)| < ∞ .

• The space Hm(Ω,V) is the space of complex vector valued functions suchthat ∀α, |α| ≤ m,

Dα f ∈ L2(Ω,V)

i.e. such that ∀α, |α| ≤ m,∫Ω

dx 〈Dα f ,Dα f 〉 =

∫Ω

N∑i=1

|Dα fi(x)|2dx < ∞ .



• It is an inner product space with the inner product

( f , g) =∑

α, |α|≤m

∫Ω

⟨Dα f ,Dαg

⟩=

∑α, |α|≤m

∫Ω

N∑i=1

Dα fiDαgi

• The Sobolev space Hm(Ω,V) is the completion of the space Hm(Ω,V) de-fined above.

4.2.1 Strong and Weak Convergence

• Strong Convergence. A sequence (xn) of vectors in an inner product spaceE is strongly convergent to a vector x ∈ E if

limn→∞‖ xn − x ‖= 0

• Notation. xn → x.

• Theorem 4.2.5 Let E be an inner product space. Then for every y ∈ E thelinear functional ϕy : E → C defined by

ϕy(x) = (y, x) ∀x ∈ E

is continuous (and, therefore, bounded).

Proof: Use Schwarz inequality.

• Weak Convergence. A sequence (xn) of vectors in an inner product spaceE is weakly convergent to a vector x ∈ E if for any y ∈ E

limn→∞

(xn − x, y) = 0

• Notation. xnw→ x

• Theorem 4.2.6 A strongly convergent sequence is weakly convergent to thesame limit.

Proof: Use Schwarz inequality.

• Converse is not true. Counterexample later.


4.2. HILBERT SPACES 99

• Theorem 4.2.7 Let (xn) be a sequence in an inner product space E andx ∈ E. Suppose that

1. xnw→ x and

2. ‖ xn ‖→‖ x ‖.

Then xn → x.

Proof: Easy.

• Theorem 4.2.8 Let S be a subset of an inner product space E such thatspan S is dense in E, (xn) be a bounded sequence in E and x ∈ E. Supposethat for any y ∈ S ,

limn→∞

(xn − x, y) = 0.

Then xnw→ x.

Proof: Let z ∈ E.

Then there is yn ∈ span S such that yn → z.

Then(xn − x, z) = (xn − x, ym) + (xn − x, z − ym)

converges to zero as n,m→ ∞.

• It is possible that xnw→ x but ||xn|| does not converge to ||x||. But at least the

seqence ||xn|| has to be bounded.

• Theorem 4.2.9 Weakly convergent sequences in a Hilbert space are bounded.

Proof: In functional analysis books.

Let fn be a functional defined by

fn(x) = (xn, x).

Then|| fn|| = ||xn||

On another hand such sequence of bounded functionals must be bounded (atheorem).



4.3 Orthogonal and Orthonormal Systems• Orthogonal and Orthonormal Systems. Let E be an inner product space.

A set S of vectors in E is called an orthogonal system if any pair of distinctvectors in S is orthogonal to each other.

• An orthogonal system of unit vectors is an orthonormal system.

• Every orthogonal system can be made orthonormal.

• Let S be a set of vectors in E. We say that x ⊥ S if x ⊥ y for any y ∈ S .

• If x ⊥ S then x ⊥ span S .

• The orthogonal complement of S is the space

S ⊥ = x ∈ E | x ⊥ S

• Theorem 4.3.1 Orthogonal systems are linearly independent.

Proof: Easy.

• Orthonormal Sequence. A sequence of vectors which is an orthonormalsystem is an orthonormal sequence; then

(xn, xm) = δnm

• Examples.

1. Canonical basis in l2. (en)i = (δin)

2. Fourier basis. The functions

fn(x) =1√

2πeinx

are orthonormal in L2([−π, π]).3. Legendre polynomials.

Pn(x) =1

2nn!∂n

x(x2 − 1)n

Show that the functions

fn =

√n +

12

Pn

are orthonormal in L2([−1, 1]).


4.3. ORTHOGONAL AND ORTHONORMAL SYSTEMS 101

4. Hermite polynomials.

Hn = (−1)nex2∂n

xe−x2

Show that the functions

fn(x) =1

√2nn!π1/4

exp(−

x2

2

)Hn(x)

are orthonormal in L2(R).

• Any orthogonal sequence can be always made orthonormal.

• Gram-Schmidt orthonormalization process. Any sequence of linearlyindependent vectors can be made orthonormal.

1. Let (yn) be a linearly independent sequence.

2. Letzn =

yn

||yn||

3. Recall that for any orthonormal sequence en the operator Pn definedby

Pn = I −n−1∑k=1

|ek〉〈ek|

or

Pnx = x −n−1∑k=1

(ek, x)ek

is the projections to the orthogonal complement of span e1, . . . , en−1.

4. Then the sequence (en) defined by

e1 = z1, en = Pnzn = zn −

n−1∑k=1

(ek, zn)ek

is orthonormal.



4.3.1 Properties of Orthonormal Systems• Theorem 4.3.2 Pythagorean Formula. Let E be an inner product space

and xnNn=1 be an orthonormal set in E. Then∣∣∣∣∣∣∣

∣∣∣∣∣∣∣N∑

n=1

en

∣∣∣∣∣∣∣∣∣∣∣∣∣∣2

=

N∑n=1

‖ en ‖2

Proof: By induction.

• Theorem 4.3.3 Bessel’s Equality and Inequality. Let E be an inner prod-uct space and en

Nn=1 be an orthonormal set in E. Then ∀x ∈ E

‖ x ‖2=N∑

n=1

|xn|2 +

∣∣∣∣∣∣∣∣∣∣∣∣∣∣x −

N∑n=1

xnen

∣∣∣∣∣∣∣∣∣∣∣∣∣∣2

andN∑

n=1

|xn|2 ≤‖ x ‖2

where xn = (en, x).

Proof:

• Remarks. Let (en) be an orthonormal sequence in an inner product spaceE.

• The complex numbers xn = (en, x) are called the generalized Fourier coef-ficients of x with respect to the orthonormal sequence (en).

• An orthonormal sequence (en) in E induces a mapping ϕ : E → l2 definedby

ϕ(x) = x = (xn)

• The sequence of Fourier coefficients x = (xn) is square summable, that isx ∈ l2 since for any x ∈ E

∞∑n=1

|xn|2 ≤‖ x ‖2

and, therefore, this series converges.



• The expansion

x ∼∞∑

n=1

xnen

is called the generalized Fourier series of x.

• The question is whether the mapping ϕ is bijective and whether the Fourierseries converges.

• Theorem 4.3.4 Let (en) be an orthonormal sequence in a Hilbert spaceH and (xn) be a sequence of complex numbers. Then the series

∑∞n=1 xnen

converges if and only if (xn) ∈ l2, that is, the series∑∞

n=1 |xn|2 converges.

In this case ∣∣∣∣∣∣∣∣∣∣∣∣∣∣∞∑

n=1

xnen

∣∣∣∣∣∣∣∣∣∣∣∣∣∣2

=

∞∑n=1

|xn|2

Proof: By Pythagorean theorem∣∣∣∣∣∣∣∣∣∣∣∣∣∣

m∑k=n

xkek

∣∣∣∣∣∣∣∣∣∣∣∣∣∣2

=

m∑k=n

|xk|2

• Fourier series of any x ∈ H in a Hilbert space H converges.

• Fourier series of x may converge to a vector different from x!

• Example. Let (en) be an orthonormal sequence. Then fn = e2n is alsoorthonormal. Let x = e1. Then xn = ( fn, x) = 0 and 0 =

∑∞n=1 xn fn , x.

• Let (en) be an orthonormal sequence in an inner product space E. The se-quence of Fourier coefficients xn = (en, x) is square summable, and, there-fore,

limn→∞

(en, x) = 0 ∀x ∈ E

• Thus, every orthonormal sequence weakly converges to zero.

• Orthonormal sequences are not strongly convergent since ‖ en ‖= 1 ∀n ∈ N.



• Complete Orthonormal Sequence. Let E be an inner product space. Anorthonormal sequence (en) in E is complete if ∀x ∈ E the Fourier series ofx converges to x, that is,

x =

∞∑n=1

xnen ,

more explicitly,

limn→∞

∣∣∣∣∣∣∣∣∣∣∣∣∣∣x −

n∑k=1

xkek

∣∣∣∣∣∣∣∣∣∣∣∣∣∣ = 0

• Example. L2([−π, π]) does not imply pointwise convergence.

• Orthonormal Basis. Let E be an inner product space. An orthonormal sys-tem B in E is an orthonormal basis if for any x ∈ E there exists a uniqueorthonormal sequence (en) in B and a unique sequence (xn) of nonzero com-plex numbers such that

x =

∞∑n=1

xnen .

• Remarks.

• A complete orthonormal sequence in an inner product space is an orthonor-mal basis.

• Let E be an inner product space and (en) be a complete orthonormal se-quence. Then the set

S = span en | n ∈ N

is dense in E.

• Theorem 4.3.5 Let H be a Hilbert space. An orthonormal sequence (en) inH is complete if and only if the only vector orthogonal to this sequence isthe zero vector, that is,

span en|n ∈ Z+⊥ = 0.

Proof: Easy. To prove converse, let y =∑

n(en, x)en. Then (x − y, en) = 0.So, x = y.



• Theorem 4.3.6 Parseval’s Formula. Let H be a Hilbert space. An or-thonormal sequence (en) in H is complete if and only if ∀x ∈ H

‖ x ‖2=∞∑

n=1

|xn|2

where xn = (x, en).

Proof: Use Bessel equality.

• Theorem 4.3.7 Let H1 and H2 be Hilbert spaces. If ϕk and ψl are or-thonormal bases for H1 and H2 respectively, then ϕk⊗ψl is an orthonormalbasis for the tensor product H1 ⊗ H2.

• Examples. The orthonormal sequence

fn(x) =1√

2πeinx

is complete in L2([−π, π]). Proof later.

4.3.2 Orthonormal Complements and Projection Theorem• A subspace of a Hilbert space is an inner product space.

• A closed subspace of a Hilbert space is a Hilbert space.

• Orthogonal Complement. Let H be a Hilbert space and S ⊂ H be anonempty subset of H. We say that x ∈ H is orthogonal to S , denotedby x ⊥ S , if ∀y ∈ S , (x, y) = 0.

The setS ⊥ = x ∈ H | x ⊥ S

of all vectors orthogonal to S is called the orthogonal complement of S .

Two subsets A and B of H are orthogonal, denoted by A ⊥ B, if everyvector of A is orthogonal to every vector of B.

• If x ⊥ H, then x = 0, that is

H⊥ = 0, 0⊥ = H .



• If A ⊥ B, then A ∩ B = 0 or ∅.

• Theorem 4.3.8 The orthogonal complement of any subset of a Hilbert spaceis a Hilbert subspace.

Proof:

1. Let H be a Hilbert space and S ⊂ H.

2. Check directly that S ⊥ is a vector subspace.

3. Claim: S ⊥ is closed.

4. Let (xn) be a sequence in S ⊥ such that xn → x ∈ H.

5. Then ∀y ∈ S(x, y) = lim

n→∞(xn, y) = 0.

6. Thus x ∈ S ⊥.

• Remarks.

• S does not have to be a vector subspace.

• Theorem 4.3.9 Orthogonal Projection. Let H be a Hilbert space and Sbe a closed subspace of H. Then ∀x ∈ H there exist unique y ∈ S andz ∈ S ⊥ such that

x = y + z .

Proof: No proof.

• Orthogonal decomposition. If every element of H can be uniquely repre-sented as the sum of an element of S and an element of S ⊥, then H is thedirect sum of S and S ⊥, which is denoted by

H = S ⊕ S ⊥

• The union of a basis of S and a basis of S ⊥ gives a basis of H.

• Orthogonal projection. An orthogonal decomposition H = S ⊕S ⊥ inducesa projection map P : H → S defined by

P(y + z) = y,

where y ∈ S and z ∈ S ⊥.



• Examples.

• Theorem 4.3.10 Let H be a Hilbert space and S be a closed subspace ofH. Then

(S ⊥)⊥ = S .

Proof:

1. Let x ∈ S .

2. Then x ⊥ S ⊥, or x ∈ S ⊥⊥.

3. So, S ⊆ S ⊥⊥.

4. Let x ∈ S ⊥.

5. Since S is closed, there exist y ∈ S and z ∈ S ⊥ such that x = y + z.

6. Then y ∈ S ⊥⊥.

7. Since S ⊥⊥ is a vector space, z = x − y ∈ S ⊥⊥.

8. Since z ∈ S ⊥⊥ and z ∈ S ⊥.

9. Thus, z = 0, and x = y ∈ S .

10. Therefore, S ⊥⊥ ⊆ S .

4.3.3 Separable Hilbert Spaces• Separable Spaces. An infinite-dimensional Hilbert space is separable if it

contains a complete orthonormal sequence.

• Examples.

• L2([−π, π])

• l2

• Example (Non-separable Hilbert Space). Let H be the space of all com-plex valued functions f : R → C on R such that they vanish everywhereexcept a countable number of points in R and∑

f (x),0

| f (x)|2 < ∞.



Define the inner product by

(g, f ) =∑

f (x)g(x),0

g(x) f (x).

Let fn be an orthonormal sequence in H. Then there are non-zero functionsf such that ( f , fn) = 0 for all n ∈ N. Therefore, fn cannot be complete andH is not separable.

• Theorem 4.3.11 Let H be a separable Hilbert space. Then H contains acountable dense subset.

Proof:

1. Let (en) be a complete orthonormal sequence in H.

2. Define the set

S =

n∑k=1

(αk + iβk)ek | αk, βk ∈ Q, n ∈ N

3. Then S is countable.

4. Also, ∀x ∈ H,

limn→∞‖

n∑k=1

(ek, x)ek − x ‖= 0.

5. Therefore, S is dense in H.

• Theorem 4.3.12 Let H be a separable Hilbert space. Then every orthogo-nal set S in H is countable.

Proof:

1. Let S be an orthogonal set in H.

2. Let

S 1 =

x‖ x ‖

| x ∈ S.

3. Then ∀x, y ∈ S 1, x , y,

‖ x − y ‖2= 2.



4. Consider the collection of balls B2−1/2(x) for every x ∈ S 1.

5. Then, for any x, y ∈ S 1, if x , y, then the corresponding balls aredisjoint,

B2−1/2(x) ∩ B2−1/2(y) = ∅.

6. Since H is separable, it has a countable dense subset A.

7. Since A is dense in H it must have at least one point in every ballB2−1/2(x).

8. Therefore, S 1 must be countable.

9. Thus S is countable.

• Unitary Linear Transformations. Let H1 and H2 be Hilbert spaces. Alinear map T : H1 → H2 is unitary if ∀x, y ∈ H1

(T (x),T (y))H2 = (x, y)H1 .

• Hilbert Space Isomorphism. Let H1 and H2 be Hilbert spaces. Then H1

is isomorphic to H2 if there exists a linear unitary bijection T : H1 → H2

(called a Hilbert space isomorphism).

• Remark. Every Hilbert space isomorphism has unit norm

‖ T ‖= 1.

• Theorem 4.3.13 Every infinite-dimensional separable Hilbert space is iso-morphic to l2.

Proof:

1. Let (en) be a complete orthonormal sequence in H.

2. Let x ∈ H.

3. Let xn = (en, x).

4. This defines a linear bijection T : H → l2 by

T (x) = (xn) .

5. Let x, y ∈ H.



6. Then

(T (x),T (y))l2 =

∞∑n=1

xnyn

7. On another hand

(x, y)H = (∞∑

n=1

xnen, y) =

∞∑n=1

xn(en, y) =

∞∑n=1

xnyn

8. Therefore T is unitary, and is, therefore, an isomorphism from H ontol2.

• Remarks.

• Isomorphism of Hilbert spaces is an equivalence relation.

• All separable infinite-dimensional Hilbert spaces are isomorphic.


4.4. TRIGONOMETRIC FOURIER SERIES 111

4.4 Trigonometric Fourier Series• Consider the Hilbert space L2([−π, π]).

• The sequence

ϕn(x) =1√

2πeinx, n ∈ Z

is an orthonormal sequence in L2([−π, π]).

• Note that L2 ⊂ L1. Consider the space L1([−π, π]).

• Identify the elements of L1([−π, π]) with 2π periodic functions on R.

• Then for any f ∈ L1([−π, π]),∫ π

−π

dt f (t) =

∫ π

−π

dt f (t − x)

• For any k ∈ Z+ let Pn be the projections

Pk = |ϕk〉〈ϕk|

such that(Pk f )(x) = (ϕk, f )ϕk(x) =

∫ π

−π

dt Pk(x − t) f (t)

wherePk(x − t) =

12π

eik(x−t)

is the integral kernel of Pk.

• Let Gk be the projections (not orthogonal)

G j =

j∑k=− j

Pk

with the integral kernel

G j(x − t) =1

2π

j∑k=− j

eik(x−t)



• Also define the sequence of operators

Kn =1

n + 1

n∑j=0

G j =1

n + 1

n∑k=−n

(n + 1 − |k|)Pk

with the integral kernel

Kn(x − t) =1

2π(n + 1)

n∑k=−n

(n + 1 − |k|) eikx

called the Fejer’s kernel.

• Lemma 4.4.1 We have

Kn(x) =1

2π(n + 1)

sin2[(n + 1) x

2

]sin2

(x2

)Proof: Direct calculation.

• A sequence Kn of 2π-periodic continuous functions is a summability ker-nel if it satisfies the conditions:

1. ∫ π

−π

dt Kn(t) = 1, ∀n ∈ Z+

2. There is M ∈ R such that ∀n ∈ N∫ π

−π

dt |Kn(t)| ≤ M,

3. For any δ ∈ (0, π)

limn→∞

2π−δ∫δ

dt |Kn(t)| = 0

• A summability kernel converges formally to

limn→∞

Kn(x) =

∞∑k=−∞

δ(x − 2πk)


4.4. TRIGONOMETRIC FOURIER SERIES 113

• Theorem 4.4.1 Let (Kn) be a summability kernel. Then for any f ∈ L1([−π, π])the sequence Fn = Kn f strongly converges to f in L1 norm, that is,

limn→∞||Fn − f || = lim

n→∞||(Kn − I) f || = 0

Proof: Use the properties of the summability kernel. Basically becauseKn → I (delta function).

• Lemma 4.4.2 The Fejer’s kernel is a summability kernel.

Proof: Direct calculation.

• Theorem 4.4.2 Let f ∈ L1([−π, π]). If all Fourier coefficients fn = (ϕn, f )vanish, then f = 0 almost everywhere.

Proof: Let Fn = Kn f . Suppose fn = 0. Then Fn = 0. Since Fn → f , thenf = 0 a.e.

• Theorem 4.4.3 The sequence

ϕn(x) =1√

2πeinx, n ∈ Z

is a complete orthonormal sequence, (an orthonormal basis), in L2([−π, π]).

Proof: Let f ∈ L2. Then f ∈ L1. Suppose (ϕn, f ) = 0. Then f = 0 a.e. Thatis, f = 0 in L2.

• Let f ∈ L2([−π, π]). The series

f (x) =

∞∑n=−∞

fnϕn(x),

wherefn =

∫ π

−π

dt ϕn(t) f (t),

is the Fourier series. The scalars fn are the Fourier coefficients.

• Fourier series does not converge pointwise!

• Fourier series of a function f ∈ L2([−π, π]) converges almost everywhere.



4.5 Linear Functionals and the Riesz Representa-tion Theorem

• Examples.

• L2([a, b])

• 〈g| is a linear bounded functional defined by

ϕg( f ) = (g, f )

with the norm||ϕg|| = ||g||

• Let x0 ∈ (a, b). Then ϕ( f ) = f (x0) is a linear but unbounded functional.

• Remarks. The set H′ of all bounded linear functionals on a Hilbert spaceis a Banach space, called the dual space.

• The dual space H′ of a Hilbert space H is isomorphic to H.

• If f = 0, then the null space N( f ) = E, and, therefore, (N( f ))⊥ = 0 anddim(N( f ))⊥ = 0.

• Lemma 4.5.1 Let E be an inner product space and f : E → C be a non-zero bounded linear functional on E. Then the orthogonal complement ofthe null space

dim(N( f ))⊥ = 1.

Proof:

1. Since f is bounded and linear it is continuous.

2. Therefore, N( f ) is a closed subspace of E.

3. Thus, (N( f ))⊥ is not empty.

4. Let x, y ∈ (N( f ))⊥ be two nonzero vectors.

5. Then f (x) , 0 and f (y) , 0.

6. Therefore, there exists α , 0 ∈ C such that f (x + αy) = 0.

7. Hence, x + αy ∈ N( f ).


4.5. LINEAR FUNCTIONALS AND THE RIESZ REPRESENTATION THEOREM115

8. Since x, y ∈ (N( f ))⊥, we also have x + αy ∈ (N( f ))⊥.

9. Thus x + αy = 0.

10. Therefore, x and y are linearly dependent, and, therefore,

dim(N( f ))⊥ = 1.

• There holdsE = N( f ) ⊕ (N( f ))⊥

• Theorem 4.5.1 Riesz Representation Theorem. Let H be a Hilbert spaceand f : H → C be a bounded linear functional on H. There exists a uniquex0 ∈ H such that

f (x) = (x0, x)

for all x ∈ H. Moreover,‖ f ‖=‖ x0 ‖ .

Proof:

1. (I). Existence.If f = 0, then x0 = 0.

2. Suppose f , 0.

3. Then dim(N( f ))⊥ = 1.

4. Let u ∈ (N( f ))⊥.

5. Then ∀x ∈ H,x = y + z

where y = x − (x, u)u ∈ N( f ) and z = (x, u)u ∈ (N( f ))⊥.

6. Therefore, f (y) = 0.

7. Further,f (x) = f (z) = (x, u) f (u) = (x, x0),

wherex0 = ( f (u))∗u.

8. (II). Uniqueness. Suppose there exists x0 and x1 such that ∀x ∈ H

f (x) = (x, x0) = (x, x1).



9. Then ∀x ∈ H(x, x0 − x1) = 0.

10. Thus, (x0 − x1) ∈ H⊥ = 0.

11. So, x0 = x1.

12. Finally, we have

‖ f ‖= supx,0

| f (x)|‖ x ‖

= supx,0

|(x, x0)|‖ x ‖

≤‖ x0 ‖ .

13. On another hand

| f (x0)|‖ x0 ‖

=|(x0, x0)|‖ x0 ‖

=‖ x0 ‖ .

14. Thus, ‖ f ‖=‖ x0 ‖.

4.5.1 Homework• Exercises: 3.12[4,5,9,10,11, 12,13,16,17,18, 20,24,25,32,33, 37,38,39,40,41,

45,46,47,48,49, 50,51,52,53,56, 58,60,61]


Chapter 5

Operators on Hilbert Spaces

5.1 Examples of Operators

• Let E be an inner product space. An linear operator is a linear map A : E →E.

• Only linear operators will be considered.

• An operator A : E → E is bounded if ∃K ∈ R such that ∀x ∈ E,

‖ Ax ‖≤ K ‖ x ‖ .

• The norm of an operator A is

‖ A ‖= supx∈E,x,0

‖ Ax ‖‖ x ‖

• A linear operator is bounded if and only if it is continuous.

• Identity Operator I : E → E is defined by

Id x = x, ∀x ∈ E

Obviously,‖ Id ‖= 1 .

117

118 CHAPTER 5. OPERATORS ON HILBERT SPACES

• Null Operator 0 : E → E is defined by

0x = 0, ∀x ∈ E

Obviously,‖ 0 ‖= 0 .

• Operators on finite-dimensional Hilbert spaces.Let ek, (k = 1, . . . , n), be the canonical orthonormal basis in Cn. Let A :E → E be an operator and define

(Ae j, ek) = Ak j .

Then

Ae j =

n∑k=1

ekAk j

and for any x ∈ E

x =

n∑j=1

e jx j, x j = (x, e j)

we have

Ax =

n∑j,k=1

ekAk jx j

There is a one-to-one correspondence between the operators on Cn and then × n complex matrices.

• The trace of the operator A is defined by

tr A =

n∑k=1

(Aek, ek) =

n∑k=1

Akk .

• The adjoint of the operator A is defined by

(A†e j, ek) = (e j, Aek) .

• The matrix (A†) jk of the adjoint operator A† is Hermitian conjugate of thematrix A jk, that is

(A†) jk = (Ak j)∗ .


5.1. EXAMPLES OF OPERATORS 119

• The Hilbert-Schmidt norm of the operator A is defined by

‖ A ‖2HS = tr AA† =

n∑j,k=1

|A jk|2 .

• Show that||A|| ≤ ||A||HS

• Every operator on a finite-dimensional Hilbert space is bounded.

• Differential Operator D : C∞([a, b])→ C∞([a, b]) on the space of smoothfunctions is defined by

(D f )(x) =d fdx

.

The differential operator is unbounded.

• Integral Operator K : L2([a, b])→ L2([a, b]) is defined by

(K f )(x) =

∫ b

ady K(x, y) f (y) .

The function K(x, y) is the kernel of the operator K.

The trace of the operator K is

Tr K =

∫ b

adx K(x, x)

when the integral exists.

The adjoint K† of the operator K is defined by

(K† f )(x) =

∫ b

ady K∗(y, x) f (y) .

The Hilbert-Schmidt norm of the operator K is defined by

‖ K ‖2HS = Tr KK† =

∫ b

a

∫ b

adx dy |K(x, y)|2 .

when it exists.



• An operator K is called Hilbert-Schmidt if ||K||HS < ∞.

• The operator K : L2([a, b])→ L2([a, b]) is bounded if its norm is finite.

Proof: By Schwarz inequality.

• Multiplication Operator µ f : L2([a, b]) → L2([a, b]), where f ∈ C([a, b])is a continuous function called the multiplier, is defined by

(µ f g)(x) = f (x)g(x) .

The operator µ f is bounded and

‖ µ f ‖= maxx∈[a,b]

| f (x)| .

• Two operator A and B on a vector space E are equal if A − B is a nulloperator.

• The set of all operators on a vector space E is a vector space with the addi-tion and multiplication by scalars defined by

(A + B)(x) = A(x) + B(x) , (αA)(x) = αA(x) .

• The product AB of the operators A and B is the composition of A and B.

• The integer powers of an operator are defined as the multiple compositionof the operator with itself, i.e.

A0 = Id A1 = A, A2 = AA, . . .

• In general, AB , BA.

• The operators A and B are commuting operators if AB = BA.

• Operators form an algebra.

• Noncommuting Operators. The differential operator D and the operatorof multiplication by a nonconstant function µ f do not commute.

•

Theorem 5.1.1 The product of bounded operators is bounded and

‖ AB ‖≤‖ A ‖‖ B ‖

Proof:


5.1. EXAMPLES OF OPERATORS 121

1.‖ ABx ‖≤‖ A ‖‖ Bx ‖≤‖ A ‖‖ B ‖‖ x ‖ .

•Theorem 5.1.2 A bounded operator on a separable infinite-dimensional Hilbert space can be represented by an infinite matrix.

Proof:

1. Let Ai j = (Ae j, ei).

2. Then

Ax =

∞∑k=1

(x, ek)Aek

and

(Ax, ek) =

∞∑j=1

Ak j(x, e j) .

5.1.1 Homework• Exercises: 4.13[1,2,3]



5.2 Bilinear Functionals and Quadratic Forms• Bilinear Functional. Let E be a complex vector space. A bilinear func-

tional on E is a mapping φ : E×E → C which is linear in the first argumentand anti-lenar in the second.

• Examples.

• Let E be a normed space and φ be a bilinear functional on E. Then

1. φ is symmetric if ∀x, y ∈ E,

φ(x, y) = φ(y, x)∗.

2. φ is positive if ∀x ∈ E,φ(x, x) ≥ 0.

3. φ is strictly positive if ∀x , 0 ∈ E,

φ(x, x) > 0.

4. φ is bounded if there exists a constant K > 0 such that ∀x, y ∈ E,

|φ(x, y)| ≤ K ‖ x ‖ ‖ y ‖ .

5. The norm of a bounded bilinear functional is

‖ φ ‖= supx,y,0

|φ(x, y)|‖ x ‖ ‖ y ‖

.

so that ∀x, y ∈ E|φ(x, y)| ≤‖ φ ‖ ‖ x ‖ ‖ y ‖ .

• Quadratic Form. Let E be a normed space and φ be a bilinear functionalon E. The quadratic form associated with the bilinear functional φ isthe functional Φ : E → C defined ∀x ∈ E by

Φ(x) = φ(x, x).

A quadratic form on E is bounded if there exists a constant K > 0 such that∀x ∈ E

|Φ(x)| ≤ K ‖ x ‖2 .


5.2. BILINEAR FUNCTIONALS AND QUADRATIC FORMS 123

The norm of a bounded quadratic form is defined by

‖ Φ ‖= supx,0

|Φ(x)|‖ x ‖2

so that ∀x ∈ E|Φ(x)| ≤‖ Φ ‖ ‖ x ‖2 .

• Coercive (Elliptic) Functional. Let E be a normed space, φ be a bilinearfunctional on E and Φ be the associated quadratic form on E. Then φ iselliptic if there exists a constant K > 0 such that ∀x ∈ E

Φ(x) = φ(x, x) ≥ K ‖ x ‖2 .

• Example.

• Theorem 5.2.1 If A, B are operators on E such that ∀x ∈ E,

(Ax, x) = (Bx, x),

then A = B.

Proof: Easy.

• Theorem 5.2.2 Let H be a Hilbert space and φ be a bounded bilinear func-tional on H. There exists a unique bounded operator A on H such that∀x, y ∈ H

φ(x, y) = (x, Ay).

Proof:

1. By Riesz theorem there exists a unique z = Ay ∈ H such that andφ(x, y) = (x, Ay).

2. Claim: the mapping y 7→ z = Ay is bounded linear operator on E.3. We also have

|(x, Ay)| = |φ(x, y)| ≤‖ φ ‖ ‖ x ‖ ‖ y ‖

4. Thus, for x = Ay , 0 we have

‖ Ay ‖2≤‖ φ ‖ ‖ Ay ‖ ‖ y ‖

and‖ Ay ‖≤‖ φ ‖ ‖ y ‖

5. Uniqueness is trivial.



5.3 Adjoint and Self-Adjoint Operators• Adjoint Operator. Let H be a Hilbert space and A a bounded operator on

H. The adjoint operator A∗ of A is defined by

(Ax, y) = (x, A∗y), ∀x, y ∈ H.

• The adjoint operation ∗ : L(H) → L(H) is an operator on the space of allbounded operators, which has the properties

(αA + βB)∗ = α∗A∗ + β∗B∗

(A∗)∗ = A(AB)∗ = B∗A∗

I∗ = I

where α, β ∈ C and I is the identity operator.

• Theorem 5.3.1 Let H be a Hilbert space and A a bounded operator on H.Then the adjoint operator A∗ is bounded and

‖ A ‖=‖ A∗ ‖, and ‖ AA∗ ‖=‖ A ‖2 .

Proof:

1.

• Self-adjoint (Hermitian) Operator. Let H be a Hilbert space and A abounded operator on H. The operator A is self-adjoint if A = A∗.

• Anti-selfadjoint (Anti-Hermitian) Operator. Let H be a Hilbert spaceand A be a bounded operator on H. Then A is anti-self-adjoint if A = −A∗.

• Example.

• Theorem 5.3.2 Let H be a Hilbert space and A a bounded operator on H.Then the operators AA∗ and A + A∗ are self-adjoint.

Proof: Easy.


5.3. ADJOINT AND SELF-ADJOINT OPERATORS 125

• Theorem 5.3.3 Let H be a Hilbert space and A and B be bounded self-adjoint operators on H. The operator AB is self-adjoint if and only if AB =

BA.

Proof: Easy.

• Corollary 5.3.1 Let H be a Hilbert space, A be bounded operator on H andα0, α1, . . . , αn be real constants. Then the operator α0I + α1A + · · · + αnAn

is self-adjoint.

Proof: Exercise.

• Let H be a Hilbert space and D(A) and D(B) be subspaces of H, and letA : D(A) → H and B : D(B) → H be operators. Then B is an adjointoperator of A if

(Ax, y) = (x, By), ∀x ∈ D(A), y ∈ D(B).

• Remarks.

• In general, an adjoint is not unique!

• If D(A) is dense in H then the adjoint is unique.

• Examples.

• Theorem 5.3.4 Let H be a Hilbert space and A be a bounded operator onH. Then there exist unique self-adjoint operators B and C on H such thatA = B + iC and A∗ = B − iC.

Proof:

1.

• Theorem 5.3.5 Let H be a Hilbert space and A be a bounded self-adjointoperator on H. Then

‖ A ‖= supx,0

|(x, Ax)|‖ x ‖2

Proof.



1. LetM = sup

x,0

|(x, Ax)|‖ x ‖2

2. Let ||x|| = 1. Then|(x, Ax)| ≤ ||Ax|| ≤ ||A||

3. So,M ≤ ||A||

4. Now, Let ||x|| = 1 and ||Ax|| , 0.

5. Lety =

Ax||Ax||

6. Then

(x + y, A(x + y)) − (x − y, A(x − y)) = 2(x, Ay) + 2(y, Ax) = 4||Ax||

7. We have for any x|(x, Ax)| ≤ M ||x||2

8. Therefore,

||Ax|| ≤14

M(||x + y||2 + ||x − y||2

)=

12

M(||x||2 + ||y||2

)= M

9. So,||A|| ≤ M

10. Thus||A|| = M


5.4. NORMAL, ISOMETRIC AND UNITARY OPERATORS 127

5.4 Normal, Isometric and Unitary Operators• Inverse Operator. Let E be a vector space E and A be an operator on

E with a domain D(A) and a range R(A). The operator A is invertible ifthere exists an operator A−1 : R(A) → E, called the inverse of A, such that∀x ∈ D(A) and y ∈ R(A),

A−1Ax = x, and AA−1y = y.

• The inverse of an invertible operator is unique.

• Domains and ranges

D(A−1) = R(A), R(A−1) = D(A).

• Kernel of an Operator. Let E be a vector space E and A be an operator onE with a domain D(A) and a range R(A). The kernel of the operator A isthe set of all vectors in E mapped to zero, that is

Ker A = x ∈ E | Ax = 0

• Theorem 5.4.1 Let E be a vector space E and A and B be linear operatorson E. Then:

1. A−1 is a linear operator.

2. A is invertible if and only if Ax = 0 implies x = 0, or Ker A = 0.

3. If A is invertible and x jnj=1 is a collection of linearly independent

vectors, then Ax jnj=1 is a collection of linearly independent vectors.

4. If A and B are invertible, then AB is invertible and

(AB)−1 = B−1A−1

Proof: Easy.

• Corollary 5.4.1 An invertible operator A : E → E on a finite-dimensionalvector space E is surjective, that is R(A) = E.

• This is not true in infinite dimensions.



• Example.

• One-sided shift operator.

• Left and right shifts.

• Two-sided shift operator.

• Inverse of a bounded operator is not necessarily bounded.

A(xn) =

( xn

n

)A is bounded invertible but the inverse A−1 is not bounded.

• Example. l2

• The inverse of an invertible operator on a finite-dimensional vector space isbounded.

• Theorem 5.4.2 Let H be a Hilbert space and A be a bounded invertibleoperator on H such that R(A) = H and A−1 is bounded. Then A∗ is invertibleand

(A∗)−1 = (A−1)∗.

Proof:

1. Show that ∀x ∈ H

(A−1)∗A∗x = A∗(A−1)∗x = x

• Corollary 5.4.2 Let H be a Hilbert space and A be a bounded invertibleself-adjoint operator on H such that R(A) = H and A−1 is bounded. ThenA−1 is self-adjoint.

Proof: Easy.

• Normal Operator. Let H be a Hilbert space and A be a bounded operatoron H. Then A is normal if

AA∗ = A∗A.



• Every self-adjoint operator is normal.

• Theorem 5.4.3 Let H be a Hilbert space and A be a bounded operator onH. Then A is normal if and only if ∀x ∈ H,

‖ Ax ‖=‖ A∗x ‖ .

Proof:

1. If A is normal, then(A∗Ax, x) =‖ A∗x ‖2

2. So, ‖ Ax ‖=‖ A∗x ‖.

3. If ‖ Ax ‖=‖ A∗x ‖, then

(A∗Ax, x) = (A∗Ax, x) .

4. Therefore,AA∗ = A∗A

5. Thus A is normal.

• The condition ‖ Ax ‖=‖ A∗x ‖ is stronger than ‖ A ‖=‖ A∗ ‖.

• Examples. Non-self-adjoint normal operator.

• Theorem 5.4.4 Let H be a Hilbert space, A be a bounded normal operatoron H and α ∈ C. Then αI − A is normal.

Proof: Easy.

• Theorem 5.4.5 Let H be a Hilbert space, A be a bounded operator on Hand B and C be self-adjoint operators such that A = B + iC. Then A isnormal if and only if A and B commute.

Proof: Easy.



• Isometric Operator. Let H be a Hilbert space and A be a bounded operatoron H. Then A is isometric if ∀x ∈ H,

‖ Ax ‖=‖ x ‖ .

• Examples.

• Unitary Operator. Let H be a Hilbert space and A be a bounded operatoron H such that D(A) = R(A) = H. Then A is unitary if

AA∗ = A∗A = I on H.

• Theorem 5.4.6 Let H be a Hilbert space and A be a bounded operator onH. Then A is isometric if and only if A is unitary.

Proof:

1. If ‖ Ax ‖2=‖ x ‖2, then (A∗Ax, x) = (AA∗x, x) = (x, x) for any x ∈ H.

2. So, A is unitary.

3. Similarly, if A is unitary, then A is isometric.

• Isometric operators preserve the inner product.

• Theorem 5.4.7 Let H be a Hilbert space and A be a bounded operator onH. The A is unitary if and only if A is invertible and

A−1 = A∗.

Proof: Easy.

• Theorem 5.4.8 Let H be a Hilbert space and A be a bounded unitary op-erator on H. Then

1. A is isometric.

2. A is normal.

3. A−1 and A∗ are unitary, and, therefore, isometric and normal.



Proof: Easy.

1.

• A normal operator is not necessarily unitary.

• Examples.

• l2. Let (xn)n∈Z be a sequence. The operator Axn = xn−1 is unitary.

• L2([0, 1]). The operator A f (t) = f (1 − t) is unitary.

Homework

• Exercises: 4.13[15,16,17,18,19,20,21,22,23,25,26,27,28]



5.5 Positive Operators• Positive Operator. Let H be a Hilbert space. An operator A is positive if it

is self-adjoint and ∀x ∈ H(Ax, x) ≥ 0.

• Strictly Positive Operator. Let H be a Hilbert space and A be a self-adjointoperator on H. Then A is strictly positive (or positive definite) if ∀x ∈ H,x , 0,

(Ax, x) > 0.

• Examples. L2([0, 1])

• Theorem 5.5.1 Let H be a Hilbert space and A be a bounded operator onH. Then the operators AA∗ and A∗A are positive.

Proof: Easy.

• Theorem 5.5.2 Let H be a Hilbert space and A be a invertible positiveoperator on H. Then the inverse operator A−1 is positive.

Proof:

1. Let y ∈ H.

2. There is an x ∈ H such that Ax = y.

3. We have(A−1y, y) = (x, Ax) ≥ 0 .

• Remarks.

• If A is positive, then we writeA ≥ 0

• If A and B are two self-adjoint operators such that A−B is positive, then wewrite

A − B ≥ 0, or A ≥ B.


5.5. POSITIVE OPERATORS 133

• Proposition 5.5.1 Let H be a Hilbert space, A, B,C,D be self-adjoint op-erators on H and α ∈ R, α ≥ 0. Then

1. If A ≥ B and C ≥ D, then A + C ≥ B + D.

2. If A ≥ 0, then αA ≥ 0.

3. If A ≥ B and B ≥ C, then A ≥ C.

Proof: Exercise.

• Theorem 5.5.3 Let H be a Hilbert space and A be a bounded self-adjointoperator on H such that ‖ A ‖≤ 1. Then A ≤ I.

Proof:

1. We have ∀x ∈ H

((A − I)x, x) = (‖ A ‖ −1) ‖ x ‖2≤ 0 .

2. Thus A ≤ I.

• Corollary 5.5.1 Let H be a Hilbert space and A be a positive operator onH. Then there exists α ∈ R, α > 0, such that I − αA ≥ 0.

Proof: Exercise.

• The product of positive operators is not necessarily positive.

• Examples. R2

• Theorem 5.5.4 Let H be a Hilbert space and A and B be commuting posi-tive operators on H. Then the product AB is a positive operator on H, i.e.AB ≥ 0.

Proof:

1. Let A , 0.

2. Let Pn∞n=1 be a sequence of operators defined by

P1 =A‖ A ‖

, Pn+1 = Pn − P2n = Pn(I − Pn) .



3. The operators Pn are polynomials in A. Therefore, they are self-adjointand commute with A ∀n ∈ N.

4. Claim:

A =‖ A ‖∞∑

n=1

P2n

5. First, we show (by induction)

0 ≤ Pn ≤ I.

6. Then we show that ∀x ∈ H,

∞∑n=1

‖ Pnx ‖2< ∞.

7. Therefore,‖ Pnx ‖→ 0.

8. This leads to the needed representation of A as a series.

9. Now, we compute ∀x ∈ H

(ABx, x) =‖ A ‖∞∑

n=1

(BPnx, Pnx) ≥ 0 .

10. Therefore, AB ≥ 0.

• Remark. This theorem can be proved much easier by using the square roots√A and

√B of the operators A and B as follows: ∀x ∈ H

(ABx, x) = (√

A√

Bx,√

A√

Bx) ≥ 0 .

• Corollary 5.5.2 Let H be a Hilbert space and A and B be self-adjoint op-erators on H such that A ≤ B. Let C be a positive operator on H thatcommutes with both A and B. Then AC ≤ BC.

Proof: Exercise.



• Theorem 5.5.5 Let H be a Hilbert space, α, β ∈ R be positive real numberssuch that 0 < α < β, and A be a positive operator on H such that

αI ≤ A ≤ βI.

Then

1. A is injective.

2. A is surjective, that is, R(A) = H, and, therefore, bijective.

3.1β

I ≤ A−1 ≤1α

I.

Proof:

1. (1). We have α ‖ x ‖2≤ (Ax, x) ≤ β ‖ x ‖2 .

2. Thus, if Ax = 0, then x = 0.

3. Thus, Ker A = 0, and A is injective.

4. (2). Claim: R(A) is closed.

5. Let (yn) be a sequence in R(A) that converges to some y ∈ H.

6. Then there is a sequence (xn) in H such that yn = Axn.

7. We haveα||xn − xm||

2 ≤ ||yn − ym|| ||xn − xm||,

α ‖ xn − xm ‖≤‖ yn − ym ‖ .

8. Therefore, (xn) is Cauchy and converges to some x ∈ H.

9. By continuity of A, we obtain y = Ax.

10. Thus, y ∈ R(A), and, hence, R(A) is closed.

11. Claim: (R(A))⊥ = 0 and R(A) = H.

12. Let y ∈ R(A)⊥.

13. Then ∀x ∈ H, (Ax, y) = 0.

14. Therefore, (Ay, y) = 0.

15. This means that y = 0.



16. Therefore, (R(A))⊥ = 0.

17. Then D(A−1) = R(A) = 0⊥ = H.

18. Finally, we also have

αA−1 ≤ I ≤ βA−1 ,

which leads to the last assertion of the theorem.

• Theorem 5.5.6 Let H be a Hilbert space, B be a self-adjoint operator onH, and Ai

∞i=1 be a sequence of self-adjoint operators on H such that:

1. all operators An, n ∈ N, commute with each other as well as withoperator B, and

2.A1 ≤ A2 ≤ · · · ≤ An ≤ An+1 ≤ · · · ≤ B,

Then there exists a self-adjoint operator A on H such that

limn→∞

Anx = Ax, ∀x ∈ H

andAn ≤ A ≤ B, ∀n ∈ N.

Proof:

1. Let Cn = B − An.

2. ThenC1 ≥ C2 ≥ · · · ≥ 0 .

3. ThenC2

n+1 ≤ CnCn+1 ≤ C2n .

4. Let x ∈ H and an = (C2n x, x) =‖ Cnx ‖2.

5. Then (an) is an nonincreasing sequence of nonnegative real numbers.

6. Therefore, the sequence (an) converges to some α ≥ 0.

7. Hence, as m, n→ ∞ we also have

limm,n→∞

(Cmx,Cnx) = α .



8. Therefore, as m, n→ ∞, we have

‖ Cmx −Cnx ‖2=‖ Cmx ‖2 + ‖ Cnx ‖2 −2(Cmx,Cnx)→ 0 .

9. Therefore, Cnx is Cauchy, and, hence, converges.

10. Thus, Anx also converges for any x ∈ H.

11. Finally, we define the operator A by

Ax = limn→∞

Anx .

12. Then A is self-adjoint, and ∀n ∈ N,

An ≤ A ≤ B .

• Square Root. Let H be a Hilbert space and A be a positive operator on H.A square root of A is a self-adjoint operator B on H satisfying B2 = A.

• Theorem 5.5.7 Let H be a Hilbert space and A be a positive operator onH. Then A has a unique positive square root (denoted by

√A).

The square root√

A commutes with every operator commuting with A.

Proof:

1. Let D = A‖A‖ .

2. Then D ≤ I.

3. Define the sequence (Tn)n∈N by

T0 = 0, Tn+1 = Tn +12

(D − T 2n ) .

4. Claim:0 ≤ T1 ≤ · · · ≤ Tn ≤ · · · ≤ I .

5. We haveTn+1 − Tn =

12

(2 − Tn − Tn−1)(Tn − Tn−1)

So,Tn ≤ Tn+1.



6. Let Cn = I − Tn.

7. Then, we have,

Cn+1 =12

C2n +

12

(I − D) ≥ 0 .

8. AlsoCn+1 −Cn =

12

(Cn + Cn−1)(Cn −Cn−1) .

9. We have C0 = I and C1 = I − 12 D. Therefore,

C1 ≤ C0 .

10. Therefore, by inductionCn+1 ≤ Cn .

11. The sequence Cn is a decreasing sequence of self-adjoint operatorssqueezed between 0 and I, and, therefore, converges.

12. Therefore, the sequence Tn converges to a positive operator T .

13. As n→ ∞ we obtainT 2 = D .

14. Let B =√‖ A ‖T . Then

B2 = A .

15. The operator B is positive and commutes with every operator that com-mutes with A.

16. Uniqueness. Suppose there are two positive operators B1 and B2 suchthat

B21 = B2

2 = A.

17. ThenB1B2 = B2B1 = A

So,B1(B − 1 − B2) = B2(B1 − B2) = 0

Then(B1 − B2)2 = B1(B1 − B2) − B2(B2 − B − 2) = 0

So, B1 = B2.

18. Let x ∈ H and y = (B1 − B2)x.



19. Then

(B1y, y) + (B2y, y) = ((B1 + B2)y, y) = ((B21 − B2

2)x, y) = 0 .

20. Since both B1 and B2 are positive, then

(B1y, y) = (B2y, y) = 0 .

21. Now let C1 be a square root of B1 and C2 be a square root of B2.

22. Then

0 = (B1y, y) = (C1y,C1y), and 0 = (B2y, y) = (C2y,C2y) .

23. This means that

C1y = B1y = C2y = B2y = 0 .

24. Finally, for any x ∈ H,

‖ (B1 − B2)x ‖2= ((B1 − B2)2x, x) = ((B1 − B2)y, x) = 0 .

25. Thus B1 = B2.

Homework

• Exercises: 4.13[29,30,32,33,35,36,37,38]



5.6 Projection Operators• Orthogonal Projection Operator. Let H be a Hilbert space and S be a

closed subspace of H. Then H = S ⊕ S ⊥ and for any x ∈ H we have

x = y + z,

where y ∈ S and z ∈ S ⊥.

• The vector y is called the projection of x onto S .

• The orthogonal projection operator onto S is an operator P on H definedby

Px = y .

That isP|S = I and PS⊥ = 0 .

• Remarks.

• Projection is a linear operator.

• Projection is bounded and‖ P ‖≤ 1.

• Zero operator is the projection onto the zero subspace 0 and ‖ 0 ‖= 0.

• The identity operator is the projection onto the whole space H.

• A nonzero projection operator has the unit norm

‖ P ‖= 1.

•

P|S⊥ = 0, P|S = I .

• Orthogonality of Projection Operators. Let H be a Hilbert space and Pand Q be two projections operators. Then P and Q are orthogonal if

PQ = QP = 0 .

• For any two projection operators P and Q, PQ = 0 if and only if QP = 0.


5.6. PROJECTION OPERATORS 141

• The operator P⊥ = I − P is the projection onto S ⊥. It is called the comple-mentary projection.

• For orthogonal projections we have

P∗ = P, (P⊥)∗ = P⊥, P⊥ + P = I, PP⊥ = P⊥P = 0 .

• Examples. l2, L2([−π, π]).

• Idempotent Operator. An operator A is idempotent if

A2 = A .

• Projection operators are idempotent.

• Not every idempotent operator is a projection (if it is not self-adjoint).

• Example.

• Theorem 5.6.1 Let H be a Hilbert space and P be a bounded operator onH. Then P is a projection if and only if P is idempotent and self-adjoint.

Proof:

1. (I). Let P be a projection onto a closed subspace S .

2. Then P is idempotent.

3. Let x, y ∈ H. Then

(Px, y) = (Px, Py) = (x, Py)

4. Thus P is self-adjoint.

5. (II). Let P be a self-adjoint idempotent operator.

6. Let S be a subspace of H defined by

S = x ∈ H | Px = x .

7. Then S is closed (since P is bounded).

8. The idempotency leads then to P|S = I.

9. Similarly, P|S⊥ = 0.



10. Thus P is the projection onto S .

• Corollary 5.6.1 Let H be a Hilbert space, S be a closed subspace of H andP be the projection onto S . Then ∀x ∈ H

(Px, x) =‖ Px ‖2 .

Proof: Easy.

• The sum of two projections is not necessarily a projection.

• Example.

• Theorem 5.6.2 Let H be a Hilbert space, R and S be closed subspaces ofH and PR and PS be the projections onto R and S respectively. Then PR andPS are orthogonal if and only if R ⊥ S .

Proof:

1. (I). Let PRPS = 0.

2. Then R ⊥ S since for any x ∈ R and y ∈ S

(x, y) = (PRx, PS y) = 0 .

3. (II). If R ⊥ S , then PRPS = 0.

• Theorem 5.6.3 Let H be a Hilbert space, R and S be closed subspaces ofH and PR and PS be the projections onto R and S respectively. Then the sumP = PR+PS is a projection operator if and only if PR and PS are orthogonal.The sum of the orthogonal projections PR and PS is the projection onto thedirect sum R ⊕ S ,

PR + PS = PR⊕S .

Proof:

1. (I). Let P be a projection.


5.6. PROJECTION OPERATORS 143

2. Then P2 = P andPRPS = 0 .

3. (II). Let PRPS = 0.

4. Then P is idempotent and self-adjoint.

5. Thus P is a projection.

6. Finally, P|R⊕S = I and P|(R⊕S )⊥ = 0.

7. Thus P is the projection onto R ⊕ S .

• Theorem 5.6.4 Let H be a Hilbert space, R and S be closed subspaces of Hand PR and PS be the projections onto R and S respectively. Let P = PRPS .Then P is a projection operator if and only if PR and PS commute. In thiscase P is the projection onto R ∩ S ,

PRPS = PR∩S .

Proof: Do not prove.

• Theorem 5.6.5 Let H be a Hilbert space, R and S be closed subspaces ofH and PR and PS be the projections onto R and S respectively. Then thefollowing conditions are equivalent:

1. R ⊂ S .

2. PS PR = PR.

3. PRPS = PR.

4. ‖ PRx ‖≤‖ PS x ‖ ∀x ∈ H.

Proof: Do not prove.

Homework

• Exercises: 4.13[39,40,41,42,43,44,45,46]



5.7 Compact Operators• Compact Operator. An operator on a Hilbert space is compact (or com-

pletely continuous) if the image of every bounded sequence in H containsa convergent subsequence.

• Remark. Every operator on a finite-dimensional space is compact.

• Example.

• Theorem 5.7.1 Every compact operator is bounded.

Proof:

1. If A : H → H is not bounded, then there is a sequence (xn) in H suchthat ‖ xn ‖= 1, n ∈ N, and ‖ Axn ‖→ ∞ as n→ ∞.

2. Then (Axn) does not contain a convergent subsequence.

• Remark. Not every bounded operator is compact.

• The identity operator on an infinite-dimensional Hilbert space is not com-pact.

• Projection operator on finite-dimensional subspaces are compact.

• Examples.

• Theorem 5.7.2 Integral operators in L2([a, b]) with continuous kernels arecompact.

Proof: No proof. Read the textbook.

• Theorem 5.7.3 The set of all compact operators on a Hilbert space is avector space.

Proof: Exercise.

• Theorem 5.7.4 A product of a compact operator and a bounded operatoron a Hilbert space is a compact operator.

Proof:


5.7. COMPACT OPERATORS 145

1. Let A be a compact operator and B be a bounded operator on a Hilbertspace H.

2. (I). Let (xn) be a bounded sequence in H.

3. Then (Bxn) is bounded and (ABxn) contains a convergent subsequence.

4. Thus, AB is compact.

5. (II). We have (Axn) contains a convergent subsequence (Axnk).

6. Therefore, the sequence BAxnk converges.

7. Thus BA is compact.

• Finite Dimensional Operator. An operator is finite-dimensional if it hasa finite-dimensional range.

• Theorem 5.7.5 Finite-dimensional bounded operators are compact.

Proof:

1. Let A be a finite-dimensional bounded operator and R(A) be its range.

2. Let P be projection onto R(A).

3. Then A = PA.

4. Since A is bounded and P compact, the product A = PA is compact aswell.

• Theorem 5.7.6 The limit of a convergent sequence of compact operators iscompact.

Proof: No proof. Read the book.

• Corollary 5.7.1 The limit of a convergent sequence of finite-dimensionaloperators on a Hilbert space is a compact operator.

Proof: Obvious.

• Theorem 5.7.7 The adjoint of a compact operator on a Hilbert space iscompact.

Proof:



1. Let A be a compact operator on a Hilbert space H.

2. Let (xn) be a bounded sequence in H such that for any n ∈ N, ‖ xn ‖≤

M.

3. Let yn = A∗xn, n ∈ N.

4. Then (yn) is bounded.

5. Since A is compact, there exists a subsequence (ypn) such that Aypn

converges in H.

6. Claim: ∀n,m ∈ N,

‖ ypm − ypn ‖2≤ 2M ‖ Aypn − Aypm ‖→ 0 .

7. Thus, (ypn) is Cauchy in H and, hence, converges.

8. Thus, A∗ is compact.

• Theorem 5.7.8 An operator on a Hilbert space is compact if and only ifit maps every weakly convergent sequence into a strongly convergent se-quence.

Proof: No proof. Read the book.

• Corollary 5.7.2 Let A be a compact operator on a Hilbert space H and (en)be an orthonormal sequence in H. Then Ten → 0 as n→ ∞.

Proof: Obvious.

• Remarks.

• An operator A is compact if and only if for any two weakly convergentsequences xn

w→ x and yn

w→ y there holds

(xn, Ayn)→ (x, A, y)

• The inverse of a compact invertible operator on an infinite-dimensionalHilbert space is unbounded.

• Compactness of operators is a stronger version of continuity.

• That is why compact operators are also called completely continuous oper-ators.


5.7. COMPACT OPERATORS 147

Homework

• Exercises: 4.13[48,49,50]



5.8 Eigenvalues and Eigenvectors• Eigenvalue. Let A be an operator on a complex vector space E. A complex

number λ is called an eigenvalue of the operator A if there is a non-zerovector u ∈ E such that

Au = λu .

The vector u is called the eigenvector corresponding to the eigenvalue λ.

• Example. Projection operator.

• Remarks.

• There are infinitely many eigenvectors corresponding to an eigenvalue.

• Theorem 5.8.1 Let A be an operator on a complex vector space E andλ be an eigenvalue of the operator A. The collection of all eigenvectorscorresponding to the eigenvalue λ is a vector space.

Proof: Exercise.

• The set of all eigenvectors corresponding to the eigenvalue λ is called theeigenvalue space (or eigenspace) of λ.

• The dimension of the eigenspace of the eigenvalue λ is called the multiplic-ity of λ.

• An eigenvalue of multiplicity one is called simple (or non-degenerate)

• An eigenvalue of multiplicity greater than one is called multiple (or degen-erate). The multiplicity is then called the degree of degeneracy.

• The problem of finding the eigenvalues and the eigenvectors is the eigen-value problem.

• Example. Let H = L2([0, 2π]), f (x) = sin x, g(x) = cos x and A : H → Hbe defined by

Au = (u, f )g + (u, g) f .

Show that A has exactly one non-zero eigenvalue λ = π of multiplicity 2and the eigenvalue λ = 0 of infinite multiplicity. Find the eigenvectors. check this!!!

• Remarks.


5.8. EIGENVALUES AND EIGENVECTORS 149

• In finite dimensions the operator A − λI is invertible if and only if λ is notan eigenvalue of the operator A.

• In finite dimensions the operator (A − λI)−1 is bounded if and only if λ isnot an eigenvalue.

• Resolvent and Spectrum. The operator

R(λ) = (A − λI)−1

is called the resolvent of the operator A.

• The values of λ ∈ C for which the resolvent Rλ is well defined and boundedare called regular points of A. The set of regular points ρ(A) is called theresolvent set.

• The set σ(A) of values of λ ∈ Cwhich are not regular is called the spectrumof the operator A, that is,

σ(A) = C − ρ(A).

• The spectral radius of A is defined by

r(A) = sup|λ| | λ ∈ σ(A).

• Remarks.

• Every eigenvalue belongs to the spectrum.

• Not all points in the spectrum are eigenvalues.

• Example. Let E = C([a, b]), u ∈ E and A be an operator on E defined by

(A f )(t) = u(t) f (t) .

Show that:

1. the spectrum of A is exactly the range of u,

2. if u(t) = c is constant, then λ = c is an eigenvalue of A,

3. if u is strictly increasing, then A has no eigenvalues.



• Remarks.

• Theorem 5.8.2 Let A be a bounded operator in H and λ ∈ C be a complexnumber such that ||A|| < |λ|. Then the resolvent R(λ) is a bounded operatorand

1.

R(λ) = −

∞∑n=0

1λn+1 An

and

2.

||R(λ)|| ≤1

|λ| − ||A||.

Proof. Use ||A/λ|| < 1.

• The series

R(λ) = −

∞∑n=0

1λn+1 An

is called the Neumann series.

• If T is an invertible operator on H, then for any operator A the operatorsT AT−1 and A are called similar.

• Theorem 5.8.3 1. The eigenvalues of similar operators are the same.

2. The eigenvalues of a self-adjoint operator are real.

3. The eigenvalues of a positive operator are non-negative.

4. The eigenvalues of a unitary operator have modulus equal to 1.

5. The eigenvalues of a projection are equal to 1 and 0.

Proof: Easy.

• Theorem 5.8.4 Every bounded operator has a finite spectral radius, more-over,

r(A) ≤‖ A ‖ .

• Remark.



• All eigenvalues of a bounded operator lie inside the circle of radius ‖ A ‖ inthe complex plane.

• Theorem 5.8.5 The spectral radius of a bounded self-adjoint operator isequal to its norm.

Proof. Use a sequence xn of unit vectors such that |(xn, Axn)| → ||A||.

Show that there exists λ ∈ σ(A) such that |λ| = ||A||.

• Theorem 5.8.6 The eigenvectors corresponding to distinct eigenvalues ofa self-adjoint or unitary operator are orthogonal.

Proof. Easy.

• Theorem 5.8.7 A compact self-adjoint operator A has an eigenvalue equalto either ‖ A ‖ or − ‖ A ‖.

Proof:

1. Let λ be such that |λ| = ||A||. Assume λ , 0.

2. There is a sequence of unit vectors xn such that (A − λ)xn → 0.

3. There is a subsequence xkn such that Axkn converges.

4. Then xn → x converges to a unit vector x such that Ax = λx.

• Corollary 5.8.1 Let A be a compact self-adjoint operator on a Hilbert spaceH. Then there is a unit vector u ∈ H such that ‖ u ‖= 1 and

|(Au, u)| = sup‖x‖≤1|(Ax, x)| .

Proof:

1. Let u, ‖ u ‖= 1, be the eigenvector corresponding to the eigenvalue‖ A ‖ or − ‖ A ‖.

• Theorem 5.8.8 the eigenspaces corresponding to nonzero eigenvalues of acompact self-adjoint operator are finite dimensional.

Proof. By contradiction. Use an orthonormal sequence in the eigenspace.



• Theorem 5.8.9 Let A be a self-adjoint compact operator on a Hilbert spaceH. Then the set of distinct non-zero eigenvalues of A is either finite or formsa sequence (λn) that converges to 0

limn→∞

λn = 0 .

Proof:

1. Suppose A has infinitely (countably) many distinct eigenvalues (λn)with the corresponding unit eigenvectors (un).

2. Then (un) is an orthonormal sequence.

3. Hence (un) weakly converges to 0.

4. Thus, since A is compactAun → 0

andlimn→∞

λ2n = lim

n→∞‖ Aun ‖

2= 0 .

• Example. Let H = L2([0, 2π]), k be a locally square integrable periodicfunction with period 2π, kt be a function defined by kt(x) = k(t − x) and Abe an operator on H defined by

(A f )(t) = (kt, f ) .

Show that:

1. A is self-adjoint if k(−x) = k(x), and

2. the eigenvalues and the eigenfunctions of A are

λn = (k, un) un(x) = einx, n ∈ Z .

• Theorem 5.8.10 Let H be a Hilbert space and (Pn) be a sequence of pair-wise orthogonal projections on H. Let (λn) be a sequence of complex num-bers converging to 0. Then:

1. The series

A =

∞∑n=1

λnPn

converges in B(H,H) and defines a bounded operator on H.



2. Each λn is an eigenvalue of the operator A. The only other possibleeigenvalue of A is 0.

3. If all λn are real, then A is self-adjoint.

4. If all Pn are finite-dimensional, then A is compact.

Proof:

1. (I). Since B(H,H) is complete we only need to show that sN =∑N

n=1 λnPn

is a Cauchy sequence.

2. Let ε > 0. Then there is n0 ∈ N such that

|λn| < ε for any n > n0 .

3. For any x ∈ H and any k,m ∈ N such that m > k > n0 we obtain

‖

m∑n=k

λnPnx ‖2≤ ε2 ‖

m∑n=k

Pn ‖2‖ x ‖2≤ ε2 ‖ x ‖2 ,

4. So,

‖ sm − sk ‖=‖

m∑n=k

λnPn ‖≤ ε ,

and, hence, sn is a Cauchy sequence.

5. (II). Let k ∈ H and u ∈ Pk(H).

6. ThenAu = λku

and, hence, λk is the eigenvalue of A.

7. Suppose λ is a complex number such that λ , 0, λn for any n ∈ N andu is a vector such that

Au = λu .

8. Let R(A) be the range of A, P be the projection on R(A) and P⊥ be theprojection on the orthogonal complement (R(A))⊥.

9. Then∞∑

n=1

Pn + P⊥ = I .



10. Therefore,

(A − λI)u =

∞∑n=1

(λn − λ)Pnu − λP⊥u = 0 .

11. Since λ , λn for any n and λ , 0, then Pnu = 0 for all n ∈ N andP⊥u = 0.

12. Thus, u = 0 and λ is not an eigenvalue.

13. (III). Suppose λn are real.

14. Then for any x, y ∈ H

(Ax, y) = (x, Ay) .

15. (IV). If all Pn are finite-dimensional, then A is compact since the limitof a convergent sequence of finite-dimensional operators is a compactoperator.

• Approximate Eigenvalue. A complex number λ is called an approximateeigenvalue of an operator A if there exists a sequence of unit vectors (xn)such that ‖ xn ‖= 1, ∀n ∈ N and

limn→∞‖ (A − λI)xn ‖= 0 .

• Every eigenvalue is an approximate eigenvalue.

• Example. Let (en) be an orthonormal basis in H and Pn be the correspond-ing projections onto en. Let λ be a real number and (λn) be a sequence ofreal numbers such that λn , λ and limn→∞ λn = λ. Let A be an operator onH defined by

A =

∞∑n=1

λnPn .

Show that λn is an eigenvalue of A and λ is an approximate eiegenvalue ofA but not an eigenvalue.

Homework

• Exercises: 4.13[51,52,53,54,55,56,57]



5.9 Spectral Decomposition• Theorem 5.9.1 Hilbert-Schmidt Theorem. Let H be an infinite-dimensional

Hilbert space and A be a self-adjoint compact operator on H. Then:

1. there exists an orthonormal system of eigenvectors (un) correspondingto non-zero eigenvalues (λn) such that every vector x ∈ H has a uniquerepresentation

x =

∞∑n=1

αnun + v ,

where (αn) is a sequence in C, and v ∈ Ker A is a vector such that

Av = 0 .

2. If A has infinitely many distinct eigenvalues (λn), then

λn → 0 .

3. That is, there is a direct sum decomposition of the Hilbert space

H =

∞⊕n=1

En ⊕ E0 ,

where En are the eigenspaces corresponding to the eigenvalues λn andE0 = Ker A is the kernel of the operator A..

Proof:

1. Since A is self-adjoint and compact there is an eigenvalue λ1 of A suchthat

|λ1| = supx∈H,‖x‖≤1

|(Ax, x)| .

2. Let E1 ⊂ H be the eigenspace corresponding to λ1.

3. Then E⊥1 is a closed invariant subspace of H.

4. Thus, there exists an eigenvalue λ2 such that

|λ2| = supx∈E⊥1 ,‖x‖≤1

|(Ax, x)| .



5. By induction, at the n-th step, we get the eigenvalues (λ1, . . . , λn) andtheir eigenspaces (E1, . . . , En) and we choose an eigenvalue λn+1 suchthat

|λn+1| = supx∈E⊥n ,‖x‖≤1

|(Ax, x)| .

6. We have|λn| ≥ |λn+1|

andEn ⊥ Ek , for n , k .

7. If at the step k we get an eigenspace Ek such that (Ax, x) = 0 forany x ∈ E⊥k , then this process terminates. The space E0 = E⊥k is theeigenspace of the zero eigenvalue.

8. In this case we have

H = E1 ⊕ · · · ⊕ Ek ⊕ E0 ,

and every vector x ∈ H has a unique representation

x =

k∑j=1

α ju j + v ,

where v ∈ E0 is a zero eigenvector.

9. Suppose that the process does not terminate. Then we get a sequenceof eigenvalues (λn) with noninreasing moduli and the correspondingeigenspaces (En).

10. Then a sequence (un) of unit eigenvectors from En converges weaklyto 0.

11. Since A is compact, the sequence (Aun) converges strongly to 0.

12. Thus|λn| =‖ Aun ‖→ 0 .

13. Now, let

S =

∞⊕n=1

En = Spanun | un ∈ En, n ∈ N .



14. ThenH = S ⊕ E0 ,

and for any x ∈ H there is a unique decomposition

x =

∞∑j=1

α ju j + v ,

where v ∈ E0.

15. Claim: E0 = S ⊥ is the zero eigenspace.

16. Let w ∈ E0 be a unit vector.

17. Then for any n ∈ N

|(Aw,w)| ≤ supx∈E⊥n ‖x‖≤1

|(Ax, x)| = |λn+1| .

18. Since λn → 0, we have (Aw,w) = 0, and therefore, Aw = 0.

• Theorem 5.9.2 Spectral Theorem for Self-Adjoint Compact Operators.Let H be an infinite-dimensional Hilbert space and A be a self-adjoint com-pact operator on H. Then:

1. there exists a complete orthonormal system (vn) (an orthonormal ba-sis) in H consisting of eigenvectors of A corresponding to the eigen-values (λn).

2. Then for every x ∈ H

Ax =

∞∑n=1

λn(x, vn)vn .

Proof:

1. The sequence (vn) we add to the eigenvectors (un) corresponding tonon-zero eigenvalues an orthonormal basis in the zero eigenspace E0.



• Theorem 5.9.3 Let H be a Hilbert space and A and B be two commutingself-adjoint compact operators on H. Then there exists an orthonormalbasis in H consisting of the common eigenvectors of the operators A and B.

Proof:

1. Let λ be an eigenvalue of A and Eλ be the corresponding eigenspace.

2. Then Eλ is an invariant subspace of B and has an orthonromals basisconsisting of the eigenvectors of B. These vectors are also eigenvec-tors of A.

• Theorem 5.9.4 Let H be a Hilbert space and A be a self-adjoint compactoperator on H. Let (λn) be the eigenvalues of A and (vn) be the corre-sponding orthonormal system of eigenvectors. Let (Pn) be the projectionoperators onto the one-dimensional spaces spanned by (vn). Then:

1. for all x ∈ H

x =

∞∑n=1

Pnx,

2. and

A =

∞∑n=1

λnPn .

3. More generally, if p is a polynomial, such that p(0) = 0, then

p(A) =

∞∑n=1

p(λn)Pn .

Proof:

1. The projections Pn are defined by

Pkx = (x, vk)vk ,

so that for any x ∈ H we have

x =

∞∑n=1

Pnx .



or∞∑

n=1

Pn = I .

2. Therefore,

Ax =

∞∑n=1

λnPnx .

3. We immediately obtain

Akx =

∞∑n=1

λknPnx ,

which proves the theorem.

• Function of an Operator. Let f : R → R be a real-valued function on Rsuch that

f (0) = limλ→0

f (λ) = 0 .

Let A be a self-adjoint compact operator on a Hilbert space H given by itsspectral decomposition

A =

∞∑n=1

λnPn .

Then the operator f (A) on H defined by

f (A) =

∞∑n=1

f (λn)Pn

is self-adjoint and compact.

• Example. If all eigenvalues of a self-adjoint compact operator A are non-negative, that is λn ≥ 0, then for any α > 0 we define

Aα =

∞∑n=1

λαn Pn .



• If the function f is not zero at zero but is just bounded at zero, then we canstill define

f (A) =

∞∑n=1

f (λn)Pn .

In this case the operator f (A) is not compact.

• Theorem 5.9.5 Let H be a Hilbert space and A be a self-adjoint operatoron H such that all eigenvalues (λn) of the operator A are non-negative (orpositive) and the eigenvectors (un) of A form an orthonormal basis on H.Then the operator A is positive (or strictly positive).

Proof:

1. Let x be a non-zero in H.

2. Then

(Ax, x) =

∞∑n=1

λn|(x, un)|2 ≥ 0 .


5.10. UNBOUNDED OPERATORS 161

5.10 Unbounded Operators• Let H be a Hilbert space and A be an operator in H.

• We say that A is defined in a Hilbert space if the range of A is a propersubset of H, that is A is not surjective.

• A is unbounded if it is not bounded.

• To show that A is unbounded, one has to find a bounded sequence (xn) in Hsuch that

‖ Axn ‖→ ∞ .

• Unboundedness is equivalent to discontinuity at every point.

• To show that A is unbounded, one has to find a sequence (xn) converging to0 such that the sequence (Axn) does not converge to 0.

• Convention. If the domain D(A) of the operator A is the whole space Hthen we say that

A is an operator on H .

If the domain D(A) of the operator A is a proper subset of H then we saythat

A is an operator in H .

• If A is a bounded operator in a Hilbert space H, then A has a unique exten-sion to a bounded operator defined on the closure of D(A).

• There exists a bounded operator B defined on the closure D(A) such that

Ax = Bx for every x ∈ D(A) .

The operator B is defined by continuity. That is, for any x ∈ D(A) let (xn)be a sequence in D(A) such that xn → x. Then

Bx = limn→∞

Axn .

In this case,‖ A ‖=‖ B ‖ .



• Then the operator B can be extended to a bounded operator C defined onthe whole space H by

C = BPD(B) .

Then ‖ C ‖=‖ B ‖.

• Thus, a bounded operator can always be extended to the whole space H, soone can always assume that the domain of a bounded operator is the wholespace H.

• An unbounded operator does not have a natural unique extension onto theclosure of its domain.

• Extension of Operators. Let E be a vector space and A be an operator inE. An operator B in E is called an extension of the operator A (denoted byA ⊂ B) if

D(A) ⊂ D(B) ,

andAx = Bx for all x ∈ D(A) .

• We have

D(A + B) = D(A) ∩ D(B) and D(AB) = x ∈ D(B) | Bx ∈ D(A)

• We have[(A + B)C] = (AC + BC)

but only(AB + AC) ⊂ [A(B + C)] .

• Densely Defined Operator. Let E be a vector space and A be an operator inE. We say that A is densely defined if D(A) is dense in E, that is D(A) = E .

• Adjoint of a Densely Defined Operator. Let H be a Hilbert space and Abe a densely defined operator in H. Let y ∈ H and ϕy : D(A) → C be alinear functional on D(A) defined for x ∈ D(A) by

ϕy(x) = (Ax, y) .

The adjoint A∗ of A is an operator with the domain



D(A∗) = y ∈ H | ϕy is continuous on D(A)

and such that

(Ax, y) = (x, A∗y) for all x ∈ D(A) and y ∈ D(A∗) .

• If A is not densely defined, then A∗ is not uniquely defined.

• Theorem 5.10.1 Let H be a Hilbert space and A and B be densely definedoperators in H.

1. If A ⊂ B, then B∗ ⊂ A∗.

(The adjoint of a densely defined operator A is the extension of theadjoint of the extension of A).

2. If D(B∗) is dense in H, then B ⊂ B∗∗.

(If the adjoint of a densely defined operator B is densely defined, thenthe adjoint of the adjoint is the extension of the operator B.)

Proof:

1. (I). Let A ⊂ B. Then for all x ∈ D(A) and y ∈ D(B∗)

(Ax, y) = (Bx, y) = (x, B∗y) .

2. We also have for all x ∈ D(A) and y ∈ D(A∗)

(Ax, y) = (x, A∗y) .

3. Thus D(B∗) ⊂ D(A∗) and for all y ∈ D(B∗)

A∗y = B∗y .

4. Therefore,B∗ ⊂ A∗ .

5. (II). Let D(B∗) be dense in H. Then since for all x ∈ D(B) and y ∈D(B∗)

(B∗y, x) = (y, Bx) ,

B∗∗ exists and for all x ∈ D(B∗∗) and y ∈ D(B∗)

(B∗y, x) = (y, B∗∗x) .



6. Therefore, D(B) ⊂ D(B∗∗) and for any x ∈ D(B)

Bx = B∗∗x .

• Theorem 5.10.2 Let H be a Hilbert space and A be a densely defined in-jective operator in H such that the inverse A−1 is densely defined. Then theadjoint A∗ is also injective and

(A∗)−1 = (A−1)∗ .

Proof:

1. Let y ∈ D(A∗). Then for any x ∈ D(A−1

(A−1x, A∗y) = (AA−1x, y) = (x, y) .

2. Thus A∗y ∈ D((A−1)∗) and

(A−1)∗A∗y = y .

3. Then for any y ∈ D((A−1)∗) and x ∈ D(A)

(Ax, (A−1)∗y) = (x, y) .

4. Thus (A−1)∗ ∈ D(A∗) and

A∗(A−1)∗y = y .

• Theorem 5.10.3 Let H be a Hilbert space and A, B and AB be denselydefined operators in H. Then

B∗A∗ ⊂ (AB)∗ .

Proof:

1. Let x ∈ D(AB) and y ∈ D(B∗A∗).



2. Then(ABx, y) = (Bx, A∗y) = (x, B∗A∗y) .

3. Therefore y ∈ D((AB)∗) and

B∗A∗y = (AB)∗y .

• Self-Adjoint Operator. A densely defined operator A in a Hilbert space His self-adjoint if

A∗ = A ,

in particular, D(A∗) = D(A) and

(Ax, y) = (x, Ay) for all x, y ∈ D(A).

• If A is a bounded densely defined operator in H, then A has a unique exten-sion to a bounded operator on H. Then

D(A) = D(A∗) = H.

• For unbounded operators, it is possible that

Ax = A∗x for any x ∈ D(A) ∩ D(A∗), but D(A) , D(A∗).

Then A is not self-adjoint.

• Symmetric Operator. A densely defined operator A in a Hilbert space His symmetric if

(Ax, y) = (x, Ay) for all x, y ∈ D(A) .

• Every self-adjoint operator is symmetric.

• Example. Let H = l2 and A be a self-adjoint injective operator on H definedby

A(xn) =

( xn

n

).



The domain of the inverse operator D(A−1) = R(A) is dense in H. Theinverse operator is defined by

A−1(xn) = (nxn) .

Then A−1 is an unbounded operator. We also have

(A−1)∗ = (A∗)−1 = A−1 .

Thus, A−1 is self-adjoint unbounded operator.

• Example. Let H = L2([0, 1]) and

A = iddt

be the differential operator in H with the domain

D(A) = f ∈ H | f ′ is continuous and f (0) = f (1) = 0 .

Then for any f , g ∈ D(A)

(A f , g) = ( f , Ag) .

Thus A is symmetric.

Let g ∈ H and ϕg be a functional on D(A) defined by

ϕg( f ) = (A f , g) .

Then ϕg is continuous on D(A) for any continuously differentiable functiong ∈ H, not necessarily satisfying g(0) = g(1) = 0. Therefore,

D(A) ⊂ D(A∗)

and A is not self-adjoint.

• Theorem 5.10.4 Let H be a Hilbert space and A be a densely defined op-erator in H. Then A is symmetric if and only if A ⊂ A∗.

Proof:

1. (I). Let A ⊂ A∗. Then for all x, y ∈ D(A)

(Ax, y) = (x, Ay) .



2. Thus A is symmetric.

3. (II). Suppose that A is symmetric. Then for all x, y ∈ D(A)

(Ax, y) = (x, Ay)

and for all x ∈ D(A), y ∈ D(A∗)

(Ax, y) = (x, A∗y) .

4. Thus A ⊂ A∗.

• Let E1 and E2 be vector spaces, D(A) ⊂ E1 and R(A) ⊂ E2. The graph G(A)of an operator A : D(A)→ R(A) is a subset of E1 × E2 defined by

G(A) = (x, Ax) | x ∈ D(A) .

• If A ⊂ B (B is an extension of A), then G(A) ⊂ G(B).

• Closed Operator. Let E1 and E2 be normed spaces and A : E1 → E2 bean operator from E1 into E2. The operator A is closed if its graph G(A) is aclosed subspace of E1 × E2, that is

if xn ∈ D(A), xn → x, and Axn → y, then x ∈ D(A) and Ax = y .

• The domain D(A) of a closed operator does not have to be closed.

• Theorem 5.10.5 Closed Graph Theorem. Every closed operator from aBanach space into a Banach space is bounded.

• The domain of an unbounded operator in a Hilbert space cannot be closed.It cannot be the whole space.

• Theorem 5.10.6 The inverse of a closed operator is closed.

Proof: Obvious.

• Theorem 5.10.7 Let H be a Hilbert space and A be a densely defined op-erator in H. Then the adjoint A∗ is a closed operator.

Proof:



1. Let yn ∈ D(A∗) be a sequence such that

yn → y and A∗yn → z .

2. Then, for any x ∈ D(A),

(Ax, y) = limn→∞

(Ax, yn) = limn→∞

(x, A∗yn) = (x, z) .

3. Therefore, y ∈ D(A∗) and A∗y = z.

• A symmetric operator can be always extended to a closed operator.

• Theorem 5.10.8 Let H be a Hilbert space and A be a densely defined sym-metric operator in H. Then there exist a closed symmetric operator B in Hsuch that A ⊂ B.

A densely defined symmetric operator has a closed symmetric extension.

Proof:

1. Let D(B) be the set of all x ∈ H for which there is a sequence (xn) inD(A) and y ∈ H such that

xn → x and Axn → y .

2. Then D(B) is a vector space and D(A) ⊂ D(B).

3. Let x ∈ D(B) and xn be such that xn → x and the limit y = limn→∞ Axn

exists. Then B is defined by

Bx = limn→∞

Axn .

4. Claim: The value Bx does not depend on the representing sequencexn.

5. Let xn → x and zn → x. Let Axn → y and Azn → w.

6. Then for any u ∈ D(A)

(u, Axn − Azn) = (Au, xn − zn) .

7. As n→ ∞(u, (y − w)) = 0 .



8. Thus, y − w ∈ D(A)⊥.

9. Since D(A) is dense in H, y − w = 0.

10. So, B is well defined.

11. Claim: B is an extension of A.

12. For any x ∈ D(A) we can just take a constant sequence xn = x. ThusBx = Ax.

13. Claim: B is symmetric.

14. Let x, y ∈ D(B). Then there are sequences xn and yn in D(A) such thatxn → x, yn → y and Axn → Bx and Ayn = By.

15. Then(Axn, yn) = (xn, Ayn) ,

and as n→ ∞(Bx, y) = (x, By) .

16. Claim: B is closed.

17. Let x ∈ D(B) and xn be a sequence in D(B) such that xn → x andBxn → y.

18. Claim: x ∈ D(B) and Bx = y.

19. For any m ∈ N there is ym ∈ D(B) such that

‖ xn − ym ‖<1m, and ‖ Bxm − Aym ‖<

1m.

20. Therefore,ym → x, and Aym → y .

21. This means x ∈ D(B) andBx = y .

• Theorem 5.10.9 Let H be a Hilbert space and A be a densely definedclosed operator in H.

1. For any u, v ∈ H there exists unique x ∈ D(A) and y ∈ D(A∗) such that

Ax + y = u and x − A∗y = v .



2. For any v ∈ H, there exists a unique x ∈ D(A∗) such that

A∗Ax + x = v .

Proof:

1. (I). Let H1 = H × H.2. Then G(A) is a closed subspace of H1.3. Thus,

H1 = G(A) ⊕G(A)⊥ .

4. We have (z, y) ∈ G(A)⊥ if and only if for all x ∈ D(A)

(x, z) + (Ax, y) = 0 , or (Ax, y) = (x,−z) ;

equivalently y ∈ D(A∗) and z = −A∗y.5. Thus, if (v, u) ∈ H × H, then there is unique x ∈ D(A) and y ∈ D(A∗)

such that(v, u) = (x, Ax) + (−A∗y, y) .

6. (II). If u = 0 above, then there are unique x ∈ D(A) and y ∈ D(A∗)such that

Ax + y = 0 , and x − A∗y = v .

7. Thus,x + A∗Ax = v .

• Remark. Let H be a Hilbert space and A be a closed operator in H. It ispossible to redefine the inner product on D(A) by

(x, y)1 = (x, y) + (Ax, Ay) .

Then D(A) is complete with respect to the norm

‖ x ‖21=‖ x ‖2 + ‖ Ax ‖2 .

D(A) is a Hilbert space with the inner product (, )1. The operator A is abounded operator on D(A) in this norm.

5.10.1 Homework• Exercises: 4.13[66,67,68,69,70,71]


5.11. THE FOURIER TRANSFORM 171

5.11 The Fourier Transform

5.11.1 Fourier Transform in L1(R)

• The Fourier transform on L2(R) is defined as an extension of the Fouriertransform in L1(R) ∩ L2(R).

•

Definition 5.11.1 Fourier Transform in L1(R). Let L1(R) be thespace of integrable functions on R. The Fourier transform

F : L1(R)→ L1(R)

is a linear operator on L1(R) defined by

(F f )(ω) = (2π)−1/2∫ ∞

−∞

dx e−iωx f (x) .

• Example.

•

Theorem 5.11.1 The Fourier transform of an integrable function is acontinuous function. That is,

F (L1(R)) ⊂ C(R) .

Proof:

1. Let f ∈ L1(R) and h ∈ R.

2. We get an estimate

| f (ω + h) − f (ω)| ≤ (2π)−1/2∫ ∞

−∞

dx |e−ihx − 1|| f (x)| .

Sincelimh→0

∫ ∞

−∞

dx |e−ihx − 1|| f (x)| = 0

f is continuous.

• L1(R) is the normed space with the norm

‖ f ‖1=∫ ∞

−∞

dx | f (x)| .



• Let C0(R) be the space of all continuous functions on R vanishing at infinity,that is

lim|x|→∞

| f (x)| = 0 .

It is a normed space with the uniform convergence norm (or the sup norm)

‖ f ‖∞= supx∈R| f (x)| .

•

Theorem 5.11.2 Let ( fn) be a sequence in L1(R) such that

‖ fn − f ‖1→ 0 .

Then‖ F fn − F f ‖∞→ 0 ,

that is, the sequence (F fn) converges to F f uniformly on R.

Proof:

1. We notice that ∀ω ∈ R

(F f )(ω)| ≤ (2π)−1/2 ‖ f ‖1 .

2. Thus‖ F ( fn − f ) ‖∞≤ (2π)−1/2 ‖ fn − f ‖1 .

•

Theorem 5.11.3 Riemann-Lebesgue Lemma. The Fourier transformof an integrable function is a continuous function vanishing at infinity.That is,

F (L1(R)) ⊂ C0(R) .

Proof:

1. We estimate

f (ω) =12

(2π)−1/2∫ ∞

−∞

dx e−iωx∣∣∣∣∣ f (x) − f

(x −

π

ω

)∣∣∣∣∣=

12

(2π)−1/2∫ ∞

−∞

dx∣∣∣∣∣ f (x) − f

(x −

π

ω

)∣∣∣∣∣ .



• Thus, the Fourier transform is a continuous linear map

F : L1(R)→ C0(R) .

•

Theorem 5.11.4 Let f ∈ L1(R). Then

(F eiαx f (x))(ω) = (F ( f )](ω − α) ,

(F f (x − x0))(ω) = (F ( f ))(ω)e−iωx0 ,

(F f (αx))(ω) =1α

[F ( f )](ω

α

), α > 0 ,

F f (x) = F f (−x) .

Proof: Easy.

• Example (Modulated Gaussian Function).

F

(exp

(iω0x −

x2

2

))(ω) = exp

(−

(ω − ω0)2

2

).

•

Theorem 5.11.5 Let f ∈ L1(R) ∩C0(R) such that f is piecewise differ-entiable and f ′ ∈ L1(R). Then

(F f ′)(ω) = iω(F f )(ω) .

Proof:

1. Integration by parts.

•

Corollary 5.11.1 Let n ∈ N. Let f ∈ L1(R) ∩ C0(R) be a continuousintegrable function such that: 1) f is n times piecewise differentiableand 2) f (k) ∈ L1(R) ∩C0(R) for k = 0, 1, . . . , n and f (n) ∈ L1(R). Then

(F f (k))(ω) = (iω)k(F f )(ω) .

Proof:

1. Induction.



• The convolution of two functions f , g ∈ L1(R) is defined by

( f ∗ g)(x) = (2π)−1/2∫ ∞

−∞

dt f (x − t)g(t) .

•

Theorem 5.11.6 Convolution Theorem. Let f , g ∈ L1(R). Then

F ( f ∗ g) = (F f )(F g) .

Proof:

1. Computation.

5.11.2 Fourier Transform in L2(R)

• L2(R) is the normed space of all square integrable functions with the norm

‖ f ‖22=∫ ∞

−∞

dx | f (x)|2 .

• Let Cc(R) be the space of all continuous functions on R with compact sup-port, that is vanishing outside a bounded interval.

• The space Cc(R) is dense in L2(R).

• Let C be the operator of complex conjugation and Id be the identity operatoron L2(R).

•

Theorem 5.11.7 Fourier transform is a continuous linear mapping

F : Cc(R)→ L2(R)

preserving the L2-norm.

That isF (Cc(R)) ⊂ L2(R)

and for any f ∈ Cc(R)

‖ F f ‖2=‖ f ‖2 .



Proof:

1. Suppose f has a compact support within the interval [−π, π].

2. Then, we compute

‖ f ‖22=‖ F f ‖22 .

3. If f has a compact support but it is not withing the interval [−π, π, thenthere is a λ > 0 such that

g(x) = f (λx)

has a compact support within [−π, π].

4. Then

(F g)(x) =1λ

(F f )( xλ

).

5. Therefore,

‖ f ‖22= λ ‖ g ‖22= λ ‖ F g ‖22=‖ F f ‖22 .

• The Fourier transform is a densely defined operator in L2(R) and has aunique extension to the whole space L2(R).

•

Definition 5.11.2 Fourier Transform in L2(R). Let f ∈ L2(R) and(ϕn) be a sequence in Cc(R) converging to f in L2(R), that is such that

‖ ϕn − f ‖2→ 0 .

The Fourier transform of f is defined by

F f = limn→∞F ϕn ,

where the limit is with respect to the L2-norm.



•

Theorem 5.11.8 Parseval’s Relation. For any f ∈ L2(R)

‖ F f ‖2=‖ f ‖2 .

That is, the Fourier transform is an isometry on L2(R).

More generally, for any f , g ∈ L2(R)

(F f ,F g) = ( f , g) .

Proof: Easy. Every isometry preserves inner product (by polarization iden-tity).

• We define a sequence of operators Fn in L2(R) by

(Fn f )(ω) = (2π)−1/2∫ n

−ndx e−iωx f (x) .

•

Theorem 5.11.9 Let f ∈ L2(R). Then

‖ (Fn − F ) f ‖2→ 0 .

That isFn f → F f

with respect to the L2-norm.

Proof:

1. Let χn(x) be the characteristic function of the interval [−n, n]. Definefor n ∈ N

fn(x) = χn(x) f (x) .

2. Then ‖ f − fn ‖2→ 0 and ‖ F f − F fn ‖2→ 0.

•

Theorem 5.11.10 Let f , g ∈ L2(R). Then

( f ,CFCg) = (F f , g) .

Proof:



1. Letfn = χn f and gn = χng .

2. Then we compute(F fm, gn) = ( fm,F gn) .

3. As n,m→ ∞, we get

(F f , g) = ( f ,F g) .

•Lemma 5.11.1 In L2(R)

CFCF = Id .

Proof:

1. By computation show that

‖ f − F (F f ) ‖22= 0 .

We define a sequence of operators F −1n in L2(R) by

(F −1n f )(ω) = (Fn f )(−ω)

= (2π)−1/2∫ n

−ndx eiωx f (x) .

•

Theorem 5.11.11 Inversion of the Fourier Transform in L2(R). Letf ∈ L2(R). Then

F −1n F f → f ,

where the convergence is with respect to the L2-norm.

Proof:

1. By computation.



• The inverse Fourier transform F −1 is defined by

(F −1 f )(x) = (F f )(−x) = (2π)−1/2∫ ∞

−∞

dω eiωx f (ω) .

•

Corollary 5.11.2 Let f ∈ L1(R) ∩ L2(R). Then almost everywhere in Rit holds

f = F −1(F f ) .

Proof: Follows from previous theorem.

•

Corollary 5.11.3 Duality. Let f ∈ L1(R) ∩ L2(R). Then almost every-where in R it holds

(F 2 f )(x) = f (−x) .

Proof: Easy. Obvious.



•

Theorem 5.11.12 Plancherel’s Theorem. There is a linear boundedoperator

F : L2(R)→ L2(R)

such that for any f , g ∈ L2(R):

1. if f ∈ L1(R), then

(F f )(ω) = (2π)−1/2∫ ∞

−∞

dx e−iωx f (x) ,

2.‖ (F − Fn) f ‖2

n→∞−→ 0 ,

3.‖ (Id − F −1

n F ) f ‖2n→∞−→ 0 ,

4.‖ F f ‖2=‖ f ‖2 ,

5.F ∗ = CFC ,

6.CFCF = FCFC = Id ,

7.F 4 = Id ,

8. F is a unitary operator on L2(R)

F ∗F = Id ,

9. F is a Hilbert space isomorphism of L2(R) onto itself.

Proof:

1. The surjectivity of F follows from the fact that for any f ∈ L2(R),

f = FCFC f = F g ,

where g = CFC f .


180 Bibliography

• Examples.


Bibliography

[1] S. Hassani, Mathematical Physics: A Modern Introduction to its Founda-tions, Springer, New York, 1999

[2] L. Debnath and P. Mikusinski, Introduction to Hilbert Spaces with Appli-cations, 2nd Ed., Academic Press, 1999

[3] E. Kreyszig, Introductory Functional Analysis with Applications, Wiley,New York, 1978

[4] M. Reed and B. Simon, Methods of Modern Mathematical Physics, vol. 1,Functional Analysis, Academic Press, New York, 1972

[5] C. DeVito, Functional Analysis and Linear Operator Theory, Addison-Wesley, 1990

[6] E. Zeidler, Applied Functional Analysis, Springer-Verlag, 1995

[7] R. Courant and D. Hilbert, Methods of Mathematical Physics, vol. 1, Inter-science, 1962

[8] R. Richtmeyer, Principles of Advanced Mathematical Physics, Springer-Verlag, 1978

181

182 BIBLIOGRAPHY


Answers to Exercises

183

184 Answers To Exercises


Notation

NotationLogic

A =⇒ B A implies BA⇐= B A is implied by Biff if and only ifA⇐⇒ B A implies B and is implied by B∀x ∈ X for all x in X∃x ∈ X there exists an x in X such that

Sets and Functions (Mappings)

x ∈ X x is an element of the set Xx < X x is not in Xx ∈ X | P(x) the set of elements x of the set X obeying the property P(x)A ⊂ X A is a subset of XX \ A complement of A in XA closure of set AX × Y Cartesian product of X and Yf : X → Y mapping (function) from X to Yf (X) range of fχA characteristic function of the set A∅ empty setN set of natural numbers (positive integers)Z set of integer numbersQ set of rational numbersR set of real numbers

185

186 Notation

R+ set of positive real numbersC set of complex numbers

Vector Spaces

H ⊕G direct sum of H and GH∗ dual spaceRn vector space of n-tuples of real numbersCn vector space of n-tuples of complex numbersl2 space of square summable sequenceslp space of sequences summable with p-th power

Normed Linear Spaces

||x|| norm of xxn −→ x (strong) convergencexn

w−→ x weak convergence

Function Spaces

supp f support of fH ⊗G tensor product of H and GC0(Rn) space of continuous functions with bounded support in Rn

C(Ω) space of continuous functions on Ω

Ck(Ω) space of k-times differentiable functions on Ω

C∞(Ω) space of smooth (infinitely diffrentiable) functions on Ω

D(Rn) space of test functions (Schwartz class)L1(Ω) space of integrable functions on Ω

L2(Ω) space of square integrable functions on Ω

Lp(Ω) space of functions integrable with p-th power on Ω

Hm(Ω) Sobolev spacesC0(V,Rn) space of continuous vector valued functions with bounded support in Rn

Ck(V,Ω) space of k-times differentiable vector valued functions on Ω


Notation 187

C∞(V,Ω) space of smooth vector valued functions on Ω

D(V,Rn) space of vector valued test functions (Schwartz class)L1(V,Ω) space of integrable vector valued functions functions on Ω

L2(V,Ω) space of square integrable vector valued functions functions on Ω

Lp(V,Ω) space of vector valued functions functions integrable with p-th power on Ω

Hm(V,Ω) Sobolev spaces of vector valued functions

Linear Operators

Dα differential operatorL(H,G) space of bounded linear transformations from H to GH∗ = L(H,C) space of bounded linear functionals (dual space)


188 Notation


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