Contents - Linköping Universitycourses.mai.liu.se/GU/TATA66/Dokument/wavelets-2015.pdf · AN INTRODUCTION TO WAVELETS BENGT OVE TURESSON Contents 1. Introduction 1 2. Preliminaries

AN INTRODUCTION TO WAVELETS

BENGT OVE TURESSON

Contents

1. Introduction 12. Preliminaries 23. Wavelets 44. Multiresolution Analyses 55. Orthonormal Sequences and Riesz Systems 66. Orthonormal and Riesz Systems Generated by Translates 87. The Scaling Equation and the Structure Constants 138. Generating a Multiresolution Analysis 159. Construction of Wavelets 1810. Regular Wavelets with Compact Support 23

1. Introduction

Consider the Fourier expansion of a function f ∈ L2(0, 1):

f(x) =∑k∈Z

cke2πikx,

where the series converges in L2(0, 1) and the coefficients ck are given by

ck =

∫ 1

0

f(x)e−2πikx dx, k ∈ Z.

The basis functions e2πikx are perfectly localized with respect to frequency and theFourier expansion is therefore a very useful tool for studying problems that concernfrequency. If we interpret f as a signal, we may for instance find out what thedominating frequencies in the signal are by comparing the size of the coefficients.Assuming that the coefficients ck in the expansion decay rapidly, the signal mayfurthermore be compressed effectively by ignoring coefficients corresponding to highfrequencies. We can also find out what frequencies contribute most to the energycontent of the signal from Parseval’s identity:∑

k∈Z

|ck|2 =

∫ 1

0

|f(x)|2 dx.

Here, the right-hand side is the energy of the signal f .The functions in the basis are, however, not at all localized with respect to x,

that is, with respect to space or time, depending on what meaning we give to the

Date: January 28, 2015.

1

2 BENGT OVE TURESSON

variable x. A consequence of this fact is that the Fourier coefficients depend onthe behaviour of f on the entire interval (0, 1); one single discontinuity of f will forinstance affect the size of all coefficients ck and make the Fourier series convergeslowly, thus making compression hard. It is also difficult or impossible to investigatelocal properties of f without summing the series and just look at the sequence ofFourier coefficients.

Wavelet expansions provide an alternative to Fourier series expansions. In suchexpansions, the basis functions are in general well-localized with respect to fre-quency and better localized with respect to space or time than the basis functionsoccurring in Fourier series expansions. A wavelet expansion is in fact a two-scaleexpansion. This feature makes it possible to study local phenomena of functions orsignals with high accuracy.

For simplicity, we will study wavelet expansions of non-periodic functions on R,where our main tool will be the Fourier transform on L2(R). It is also possibleto do wavelet expansions of multidimensional signals, but in this exposition, werestrict ourselves to one-dimensional signals.

2. Preliminaries

The scalar field of any abstract Hilbert space H will be the complex numbers. Thenorm of a vector x ∈ H is denoted by ‖x‖ and the inner-product between twovectors x, y ∈ H is denoted by (x, y). If x1, ... , xn is a finite sequence in H ofpairwise orthogonal vectors, meaning that (xk, xl) = 0 if k 6= l, then Pythagoras’theorem holds true: ∥∥∥∥ n∑

k=1

xk

∥∥∥∥2 =n∑k=1

‖xk‖2. (2.1)

This identity is proved simply by expanding the left-hand side using the propertiesof the inner-product. A sequence (xk)k∈Z is said to be orthonormal if the vectorsare pairwise orthogonal and every vector has norm 1. Suppose that (xk)k∈Z is anorthonormal sequence in H and that (ak)k∈Z is a sequence of complex numbers.Using Pythagoras’ theorem and the completeness of H, is easy to show that

x =∑k∈Z

akxk ∈ H if and only if∑k∈Z

|ak|2 <∞,

in which case

‖x‖2 =∑k∈Z

|ak|2.

We will call the last identity Parseval’s theorem. If (xk)k∈Z orthonormal se-quence in H, then Bessel’s inequality:∑

k∈Z

|(x, xk)|2 ≤ ‖x‖2

holds for every vector x ∈ H. A consequence of Bessel’s inequality is the fact thatthe series

∑k∈Z(x, xk)xk is convergent in H for every vector x ∈ H. When this

series equals x for any vector x ∈ H, then (xk)k∈Z is called an orthonormal basisfor H.

The following are the most important examples of Hilbert spaces.

AN INTRODUCTION TO WAVELETS 3

(i) Let `2(Z) denote the space of all sequences a = (ak)k∈Z of complex numbersthat satisfy

‖a‖2 =

Å∑k∈Z

|ak|2ã1/2

<∞.

This is a Hilbert space with the inner-product defined by

(a, b) =∑k∈Z

akbk, a, b ∈ `2(Z).

(ii) Let L2(T) denote the space of all measurable, 2π-periodic functions f on Rsuch that

‖f‖2 =

Å∫ 2π

0

|f(x)|2 dxã1/2

<∞.


(f, g) =1

2π

∫ 2π

0

f(x)g(x) dx, f, g ∈ L2(T).

It is well-known that if f ∈ L2(T), then

f(x) =∑k∈Z

f(k)eikx,

where the series converges in L2(T) and the Fourier coefficients f(k) aredefined by

f(k) =1

2π

∫ 2π

0

f(x)e−ikx dx, k ∈ Z.

The identity∑k∈Z

f(k)g(k) =1

2π

∫ 2π

0

f(x)g(x) dx, f, g ∈ L2(T),

is a consequence of Parseval’s theorem.(iii) Let L2(R) denote space of all measurable functions f on R such that

‖f‖2 =

Å∫R

|f(x)|2 dxã1/2

<∞.


(f, g) =

∫R

f(x)g(x) dx, f, g ∈ L2(R).

Let furthermore L1(R) denote the Banach spaces of all measurable functions fon R such that

‖f‖1 =

∫R

|f(x)| dx <∞.

The Fourier transform f of a function f ∈ L1(R) is defined by

f(ξ) =

∫R

f(x)e−ixξ dx, ξ ∈ R.


If the Fourier transform f of a function f ∈ L1(R) is known and belongs to L1(R),then f can be reconstructed through the inversion formula:

f(x) =1

2π

∫R

f(ξ)eiξx dξ for a.e. x ∈ R.

We will use the well-known fact that the Fourier transform can be extended to L2(R)

and denote the Fourier transform of f ∈ L2(R) by f . A consequence of the defini-

tion of the Fourier transform on L2(R) is the fact that if f ∈ L2(R), then f alsobelongs to L2(R). One of the most useful results concerning the Fourier transformon L2(R) is Plancherel’s theorem:∫

R

f(x)g(x) dx =1

2π

∫R

f(ξ)g(ξ) dξ, f, g ∈ L2(R).

The inversion formula for L2(R) takes the form

f(x) =1

2πpv

∫R

f(ξ)eiξx dξ for a.e. x ∈ R,

where the principal value integral is defined by

pv

∫R

f(ξ)eiξx dξ = limN→∞

∫|ξ|≤N

f(ξ)eiξx dξ.

Suppose that ψ ∈ L2(R). We shall use the notation

ψj,k(x) = 2j/2ψ(2jx− k), x ∈ R,

where j, k ∈ Z. It is straight-forward to verify that

‖ψj,k‖2 = ‖ψ‖2 (2.2)

and ‘ψj,k(ξ) = 2−j/2e−i2−jkξψ(2−jξ), ξ ∈ R, (2.3)

for all j, k ∈ Z.Let S the denote the Schwartz class, i.e., the collection of all infinitely differen-

tiable functions f on R such that

supx∈R|tjf (k)(x)| <∞ for all integers j, k ≥ 0.

It is a well-known fact and easy to prove that the Fourier transform is a bijectionbetween S and S . In a proof in the last section, the space S ′ appears. This spaceis the dual space of S , consisting of so-called tempered distributions.

We finally let χI denote the characteristic function for an interval I ⊂ R anduse the notation

sincx =sinx

x, x 6= 0.

3. Wavelets

The following definition may be the most important in this text.

Definition 3.1. A function ψ ∈ L2(R) is called a wavelet if the system (ψj,k)j,k∈Zis an orthonormal basis for L2(R). This basis is then called a wavelet basisfor L2(R).


At this stage, it is not at all clear that wavelets exist. Giving examples of waveletsand providing general tools for constructing wavelets with good properties withregards to localization and regularity are in fact the main objectives of this text.

Suppose that (ψj,k)j,k∈Z is a wavelet basis for L2(R). This means that everyfunction f ∈ L2(R) can be written as

f(x) =∑j,k∈Z

cj,kψj,k(x),

where the series converges in L2(R) and the coefficients cj,k are given by

cj,k = (f, ψj,k) =

∫R

f(x)ψj,k(x) dx, j, k ∈ Z.

4. Multiresolution Analyses

We will in this exposition present a general framework for constructing wavelets.In this approach, the concept of a multiresolution analysis is central.

Definition 4.1. A multiresolution analysis is a sequence (Vj)j∈Z of closedsubspaces of L2(R) such that

(i) Vj ⊂ Vj+1 for every j ∈ Z;(ii)

⋃j∈Z Vj is dense in L2(R);

(iii)⋂j∈Z Vj = {0};

(iv) f( · ) ∈ Vj if and only if f(2−j · ) ∈ V0;(v) f( · ) ∈ V0 implies that f( · − k) ∈ V0 for every k ∈ Z;(vi) there exists a function φ ∈ V0 such that the system (φ0,k)k∈Z is an or-

thonormal basis for V0.

Axiom (i) is of course about monotonicity of the subspaces Vj . The third axiom issometimes called the separation axiom. Axiom (iv) and (v) say that the spaces Vjshould behave well under scaling and that the space V0 is translation invariant,respectively. The function φ ∈ V0 in (vi) is called a scaling function. Let usmention that these axioms are not completely independent, but we will not go into this isaue here.

Remark 4.2. Notice that (iv) and (vi) imply that the system (φj,k)k∈Z is an or-thonormal basis for Vj . Indeed, if k 6= l, then a change of variables shows that

(φj,k, φj,l) =

∫ ∞−∞

2jφ(2jx− k)φ(2jx− l) dx =

∫ ∞−∞

φ(x− k)φ(x− l) dx

= (φ0,k, φ0,l) = 0.

Equation (2.2) also shows that ‖φj,k‖2 = ‖φ‖2 = 1 for every k ∈ Z. Moreover, iff ∈ Vj , then f(2−j · ) ∈ V0, and therefore

f(2−jx) =∑k∈Z

ckφ(x− k) in L2(R)

for some sequence (ck)k∈Z ∈ `2(Z), so that

f(x) =∑k∈Z

ckφ(2jx− k) =∑k∈Z

c′kφj,k(x) in L2(R),

where c′k = 2−j/2ck, k ∈ Z.


Example 4.3. Our first example of a multiresolution analysis is the so-called Haarsystem.1 Let Vj be the class of functions f ∈ L2(R) such that each function f isconstant on every interval of the form [2−jn, 2−j(n+ 1)), n ∈ Z. It is not so hardto show that each subspace Vj is a closed subset of L2(R). Out of the axioms inDefinition 4.1, (i), (iv), and (v) are obvious.

Since step functions are dense in L2(R), (ii) holds.To prove (iii), suppose that f ∈

⋂j∈Z Vj . Since f ∈ Vj for every j, f is constant

on every interval [0, 2−j). Letting j → −∞, this shows that f is constant on [0,∞)and therefore that f = 0 on [0,∞) since f ∈ L2(R). In a similar manner, one showsthat f = 0 on (−∞, 0], so f = 0.

Finally, take φ = χ[0,1), which will be our scaling function. Then the se-quence (φ0,k)k∈Z is orthonormal since different functions in the sequence have dis-joint supports. Also, if f ∈ V0, then

f(x) =∑k∈Z

f(k)φ(x− k) for every x ∈ R. (4.1)

Since f is constant on every interval [k, k + 1), we have that

c0,k = (f, φ0,k) =

∫R

f(x)φ(x− k) dx =

∫ k+1

k

f(x)φ(x− k) dx = f(k)

for every k ∈ Z. It therefore follows from Bessel’s inequality that∥∥∥∥f − ∑|k|≤N

c0,kφ0,k

∥∥∥∥22

≤∑

|k|≥N+1

|f(k)|2 −→ 0 as N →∞

because ∑k∈Z

|f(k)|2 =∑k∈Z

∫ k+1

k

|f(x)|2 dx = ‖f‖22 <∞,

which shows that (4.1) holds in L2(R), thus establishing (vi).

Example 4.4. Let Vj be the class of functions f ∈ L2(R) which are continuousand linear on every interval [2−jn, 2−j(n + 1)), n ∈ Z. The properties (i)–(v) inDefinition 4.1 are verified more or less as in the previous example. We will comeback to the choice of the scaling function and the proof of (vi) in the next section.This example relates to approximation with piecewise linear splines.

5. Orthonormal Sequences and Riesz Systems

In the present section, we study orthonormal sequences in Hilbert spaces. We alsoconsider so-called Riesz systems, which are closely related to orthonormal sequences.We begin by showing that a system in a Hilbert space is orthonormal if and only ifa version of Pythagoras theorem holds.

Lemma 5.1. Suppose that H is a Hilbert space. Then a sequence (xk)k∈Z ofvectors in H is orthonormal if and only if∥∥∥∥∑

k∈Z

akxk

∥∥∥∥2 =∑k∈Z

|ak|2 (5.1)

1This system was introduced by the Hungarian mathematician Alfred Haar in 1909 and pro-vides an orthormal basis for L2(0, 1).


for every sequence (ak)k∈Z ∈ `2(Z) such that ak 6= 0 for not more than a finitenumber of elements in the sequence.

Proof. The necessity part follows directly form Pythagoras’ theorem (2.1).To prove the converse, take am = 1 and ak = 0 for k 6= m. Then (5.1) shows

that ‖xm‖ = 1. Then take am = 1, an = −1, and ak = 0 for k 6= m,n in (5.1):

2 = ‖xm − xn‖2 = 2− 2 Re(xm, xn),

so Re(xm, xn) = 0. Replacing xm by ixm, we obtain that Im(xm, xn) = 0. Thisshows that (xm, xn) = 0. �

For a Riesz system, the left- and right-hand sides of (5.1) do not have to beequal, but should be equivalent in the following sense.

Definition 5.2. Suppose that H is a Hilbert space. A sequence (xk)k∈Z of vectorsin H is called a Riesz system if there exist two constants 0 < C ≤ D such that

C∑k∈Z

|ak|2 ≤∥∥∥∥∑k∈Z

akxk

∥∥∥∥2 ≤ D∑k∈Z

|ak|2 (5.2)

for every sequence (ak)k∈Z ∈ `2(Z) such that ak 6= 0 for not more than a finitenumber of elements in the sequence.

Remark 5.3.

(a) According to Lemma 5.1, any orthonormal sequence in a Hilbert space is aRiesz system with C = D = 1.

(b) Notice that every Riesz system is linearly independent. Indeed, supposethat (xk)k∈Z is a Riesz system and that

ak1xk1 + ...+ aknxkn = 0

for some vectors in the sequence and some complex coefficients. Then (5.2)shows that |ak1 |2 + ...+ |akn |2 = 0, so ak1 = ... = akn = 0.

Example 5.4. Take H = L2(R) and let φ be the “tent function” defined by

φ(x) = (1− |x|)χ[−1,1](x), x ∈ R.

If (ak)k∈Z ∈ `2(Z) is a sequence as in Definition 5.2, then∥∥∥∥∑k∈Z

akφ0,k

∥∥∥∥22

=∑n∈Z

∫ n+1

n

|an(n+ 1− x) + an+1(x− n)|2 dx

=1

3

∑n∈Z

(|an|2 + |an+1|2 + Re(anan+1)).

Using the inequality

2|Re(anan+1)| ≤ 2|an||an+1| ≤ |an|2 + |an+1|2,we obtain that ∥∥∥∥∑

k∈Z

akxk

∥∥∥∥22

≤ 1

2

∑n∈Z

(|an|2 + |an+1|2) =∑n∈Z|an|2

and ∥∥∥∥∑k∈Z

akxk

∥∥∥∥22

≥ 1

6

∑n∈Z

(|an|2 + |an+1|2) =1

3

∑n∈Z|an|2.


This shows that (φ0,k)k∈Z is a Riesz system with C = 13 and D = 1.

6. Orthonormal and Riesz Systems Generated by Translates

We next consider systems in L2(R), generated by all integer translates of a func-tion φ ∈ L2(R), and prove an equivalent condition for such a system to be a Rieszsystem. This condition is given in terms of the Fourier transform of φ.

Theorem 6.1. Suppose that φ ∈ L2(R). Then (φ0,k)k∈Z is a Riesz systemin L2(R) with constants 0 < C ≤ D if and only if

C ≤∑k∈Z

|φ(ξ + 2kπ)|2 ≤ D for a.e. ξ ∈ R. (6.1)

Remark 6.2. Notice that∫ 2π

0

∑k∈Z

|φ(ξ + 2kπ)|2 dξ =∑k∈Z

∫ 2π

0

|φ(ξ + 2kπ)|2 dξ =∑k∈Z

∫ 2(k+1)π

2kπ

|φ(ξ)|2 dξ

=

∫R

|φ(ξ)|2 dξ <∞.

This shows that the series in (6.1) is defined and finite a.e. on (0, 2π) and hence a.e.on R since it is periodic with period 2π. The series is the so-called periodization

of the square of the absolute value of φ, which therefore belongs to L1(T). Underthe assumption (6.1), the periodization belongs to L2(T).

Before proving this theorem, let us mention the following important corollary.

Corollary 6.3. Suppose that φ ∈ L2(R). Then (φ0,k)k∈Z is an orthonormal se-quence in L2(R) if and only if∑

k∈Z

|φ(ξ + 2kπ)|2 = 1 for a.e. ξ ∈ R.

Proof of Theorem 6.1. Suppose that (ak)k∈Z ∈ `2(Z) and ak 6= 0 for not more thana finite number of elements in the sequence. Put

Φ(x) =∑k∈Z

akφ(x− k), x ∈ R.

The Fourier transform of Φ is

Φ(ξ) = m(ξ)φ(ξ), ξ ∈ R,

where m is the trigonometric polynomial given by

m(ξ) =∑k∈Z

ake−ikξ, ξ ∈ R.

Plancherel’s theorem now shows that

2π‖Φ‖22 =

∫R

|Φ(ξ)|2 dξ =

∫R

|m(ξ)|2|φ(ξ)|2 dξ =∑l∈Z

∫ 2(l+1)π

2lπ

|m(ξ)|2|φ(ξ)|2 dξ

=∑l∈Z

∫ 2π

0

|m(ξ)|2|φ(ξ + 2lπ)|2 dξ =

∫ 2π

0

|m(ξ)|2Å∑l∈Z

|φ(ξ + 2lπ)|2ãdξ,


that is ∥∥∥∥∑k∈Z

akφ0,k

∥∥∥∥22

=1

2π

∫ 2π

0

|m(ξ)|2Å∑l∈Z

|φ(ξ + 2lπ)|2ãdξ. (6.2)

If we suppose that (6.1) holds, then since

1

2π

∫ 2π

0

|m(ξ)|2 dξ =∑k∈Z

|ak|2

according to Parseval’s Theorem, (6.2) immediately shows that (φ0,k)k∈Z is a Rieszsystem.

Conversely, assuming that (φ0,k)k∈Z is a Riesz system, then (6.2) implies that

C

∫ 2π

0

|m(ξ)|2 dξ ≤∫ 2π

0

|m(ξ)|2Å∑l∈Z

|φ(ξ + 2lπ)|2ãdξ ≤ D

∫ 2π

0

|m(ξ)|2 dξ

for every trigonometric polynomial m. Replacing m by a sequence of trigonometricpolynomials that converges boundedly to the characteristic function of an inter-val [a, b] ⊂ (0, 2π), we obtain that

C ≤ 1

b− a

∫ b

a

Å∑l∈Z

|φ(ξ + 2lπ)|2ãdξ ≤ D.

If we finally let (a, b) shrink to a point ξ0 ∈ (0, 2π), then Lebesgue’s differentiationtheorem shows that the inequality in (6.1) holds a.e. �

Example 6.4. Consider again the Haar system in Example 4.3 with scaling func-tion φ = χ[0,1). The Fourier transform of φ is

φ(ξ) = e−iξ/2sin ξ/2

ξ/2, ξ 6= 0.

Since (φ0,k)k∈Z is orthonormal in L2(R), Corollary 6.3 shows that

1 =∑k∈Z

|φ(ξ + 2kπ)|2 = 4∑k∈Z

sin2 ξ/2

(ξ + 2kπ)2

for a.e. ξ ∈ R. Since the right-hand side in this identity is uniformly convergent onany closed interval, which does not contain multiples of 2π, and therefore representsa continuous function on such an interval, the identity holds for every ξ ∈ R, whichis not an multiple of 2π. This shows that

1

4 sin2 ξ/2=∑k∈Z

1

(ξ + 2kπ)2, ξ /∈ 2πZ, (6.3)

which is the expansion of the function (4 sin2 ξ/2)−1 into partial fractions.

Example 6.5. Take f(x) = χ(−π,π)(x), x ∈ R. Then

f(ξ) = 2sinπξ

ξ= 2π sincπξ, ξ ∈ R.

The inversion formula therefore shows that if φ(x) = sincπx, x ∈ R, then

φ(ξ) = χ(−π,π)(ξ), ξ ∈ R.


Notice that ∑k∈Z

|φ(ξ + 2kπ)|2 =∑k∈Z

χ(−π,π)(ξ + 2kπ) = 1

for a.e. ξ ∈ R. This implies that (φ0,k)k∈Z is an orthonormal system in L2(R). Wewill call this system the Shannon system since it is connected with Shannon’ssampling theorem.

The following corollary gives a substitute for Parseval’s theorem for Riesz systemsof the form (φ0,k)k∈Z, where φ ∈ L2(R). The equality in Parseval’s theorem is herereplaced by two inequalities.

Corollary 6.6. Suppose that φ ∈ L2(R) and that (φ0,k)k∈Z is a Riesz systemin L2(R) with constants 0 < C ≤ D . Then

C∑k∈Z

|ak|2 ≤∥∥∥∥∑k∈Z

akφ0,k

∥∥∥∥22

≤ D∑k∈Z

|ak|2 (6.4)

for any sequence (ak)k∈Z ∈ `2(Z).

Proof. Let (ak)k∈Z ∈ `2(Z). Since (φ0,k)k∈Z is a Riesz system,

C∑|k|≤N

|ak|2 ≤∥∥∥∥ ∑|k|≤N

akφ0,k

∥∥∥∥22

≤ D∑|k|≤N

|ak|2 (6.5)

for any integer N ≥ 0. As in the proof of Theorem 6.1 (see in particular (6.2)), wehave that ∥∥∥∥ ∑

|k|≤N

akφ0,k

∥∥∥∥22

=1

2π

∫ 2π

0

|mN (ξ)|2Å∑l∈Z

|φ(ξ + 2lπ)|2ãdξ,

where

mN (ξ) =∑|k|≤N

ake−ikξ, ξ ∈ R.

According to Parseval’s theorem, mN tends to the function m ∈ L2(T), given by

m(ξ) =∑k∈Z

ake−ikξ, ξ ∈ R,

in L2(T) as N → ∞. It therefore follows from (6.1) that the middle memberin (6.5) tends to the middle member in (6.4) as N →∞. The left- and right-handside in (6.5) obviously tend to the left- and right-hand side in (6.4), respectively. �

The next lemma describes the closed linear span of a Riesz system. The closedlinear span of a subset of L2(R) is of course the closure in L2(R) of the linear spanof the subset.

Lemma 6.7. Suppose that φ ∈ L2(R) and (φ0,k)k∈Z is a Riesz system in L2(R).Then a function f ∈ L2(R) belongs to the closed linear span of (φ0,k)k∈Z if andonly if

f(x) =∑k∈Z

akφ0,k(x) (6.6)

in L2(R) for some sequence (ak)k∈Z ∈ `2(Z).


Proof. The sufficiency part is straightforward since any partial sum to the seriesin (6.6) belongs to the linear span of (φ0,k)k∈Z and converges to f in L2(T).

Now suppose that f is in the closed linear span of (φ0,k)k∈Z. Then there existsa sequence (fn)∞n=1, where each function fn belongs to the linear span of (φ0,k)k∈Z,i.e.,

fn(x) =∑k∈Z

a(n)k φ0,k(x), x ∈ R,

and a(n)k 6= 0 for not more than a finite number of k, such that fn → f in L2(R).

Put an = (a(n)k )k∈Z for n = 1, 2, ... . According to (5.2),

C‖am − an‖22 ≤ ‖fm − fn‖22,which shows that an converges to some sequence a = (ak)k∈Z in `2(Z). Put

gK(x) =∑|k|≤K

akφ0,k(x), x ∈ R,

for K = 0, 1, ... . Then, given ε > 0, choose n so large that

‖f − fn‖2 < ε and ‖a− an‖2 < ε.

Now, if K is so large that a(n)k = 0 for |k| ≥ K + 1, then

‖f − gK‖2 ≤ ‖f − fn‖2 + ‖fn − gK‖2 < ε+√D‖an − a‖2 < (1 +

√D)ε.

This proves that (6.6) holds in L2(R). �

Remark 6.8. Suppose that (φ0,k)k∈Z is a Riesz system in L2(R). Then (φj,k)k∈Zis also a Riesz system for every j ∈ Z since∥∥∥∥∑

k∈Z

akφj,k

∥∥∥∥22

=

∫R

∣∣∣∣∑k∈Z

ak2j/2φ(2jx− k)

∣∣∣∣2 dx =

∫R

∣∣∣∣∑k∈Z

akφ(x− k)

∣∣∣∣2 dx=

∥∥∥∥∑k∈Z

akφ0,k

∥∥∥∥22

for any sequence (ak)k∈Z ∈ `2(Z) with not more than a finite number of non-zeroof elements. The lemma therefore shows that the closed linear span of (φj,k)k∈Zconsists of all functions f ∈ `2(Z) of the form

f(x) =∑k∈Z

akφj,k(x),

where (ak)k∈Z ∈ `2(Z) and the series converges in L2(R).

The next theorem gives a way to orthogonalize a Riesz system. Starting with aRiesz system, one obtains an orthonormal system with the same closed linear spanas the original system.

Theorem 6.9. Suppose that φ ∈ L2(R) and (φ0,k)k∈Z is a Riesz system in L2(R).

(a) There exists a sequence (ak)k∈Z ∈ `2(Z) and a function Φ ∈ L2(R) of theform

Φ(x) =∑k∈Z

akφ0,k(x), x ∈ R, (6.7)

such that (Φ0,k)k∈Z is an orthonormal system in L2(R).


(b) The closed linear span of the system (φ0,k)k∈Z equals the closed linear spanof (Φ0,k)k∈Z.

Proof. (a) Put

Φ(ξ) =φ(ξ)(∑

l∈Z |φ(ξ + 2lπ)|2)1/2 , ξ ∈ R. (6.8)

According to Theorem 6.1 (see (6.1)), the denominator in this quotient is bounded

from away from 0 from below and from above, which means that Φ is defined and

that Φ belongs to L2(R). It follows that the inverse Fourier transform of Φ alsobelongs to L2(R). Since the denominator has period 2π, we also have that∑

k∈Z

|Φ(ξ + 2kπ)|2 = 1

for a.e. ξ ∈ R. Corollary 6.3 now shows that (Φ0,k)k∈Z is an orthonormal systemin L2(R). Let m denote the function

m(ξ) =

Å∑l∈Z

|φ(ξ + 2lπ)|2ã−1/2

, ξ ∈ R.

Then m has period 2π and is bounded and therefore belongs to L2(T), so

m(ξ) =∑k∈Z

ake−ikξ

for some sequence (ak)k∈Z ∈ `2(Z). Inverting (6.8), we therefore obtain (6.7).(b) Consider the equation ∑

k∈Z

bkφ0,k =∑k∈Z

ckΦ0,k, (6.9)

where either (bk)k∈Z ∈ `2(Z) is unknown or (ck)k∈Z ∈ `2(Z) is unknown. TakingFourier transforms, we see that this equation is equivalent to

B(ξ)φ(ξ) = C(ξ)Φ(ξ) = C(ξ)m(ξ)φ(ξ), ξ ∈ R,

where

B(ξ) =∑k∈Z

bke−ikξ and C(ξ) =

∑k∈Z

cke−ikξ.

Suppose first that (bk)k∈Z is known. We then get a solution to equation (6.9) bytaking C = B/m ∈ L2(T) and choosing (c−k)k∈Z as the Fourier coefficients of C.This shows that the closed linear span of (Φ0,k)k∈Z is a subset of the closed linearspan of (φ0,k)k∈Z. For the opposite inclusion suppose that (ck)k∈Z is known. Thenchoose B = mC and (b−k)k∈Z as the Fourier coefficients of C. �

Example 6.10. Let us return to Example 5.4, where

φ(x) = (1− |x|)χ[−1,1](x), x ∈ R.

A simple calculation shows that

φ(ξ) = 4sin2 ξ/2

ξ2, ξ 6= 0.


It follows that∑k∈Z

|φ(ξ + 2kπ)|2 = 16 sin4 ξ/2∑k∈Z

1

(ξ + 2kπ)4, ξ /∈ 2πZ.

Differentiating (6.3) twice, we obtain that∑k∈Z

1

(ξ + 2kπ)4=

1 + cos2 ξ/2

3 sin4 ξ/2,

so ∑k∈Z

|φ(ξ + 2kπ)|2 =16

3(1 + cos2 ξ/2).

Applying (6.8), we therefore see that the function Φ therefore is given by

Φ(ξ) =

√3 sin2 ξ/2

ξ2√

1 + cos2 ξ/2=

√3

4√

1 + cos2 ξ/2φ(ξ).

It is at least in principle possible to find Φ by inverting this formula by expandingthe function

m(ξ) =

√3

4√

1 + cos2 ξ/2, ξ ∈ R,

into a Fourier series with period 2π.

7. The Scaling Equation and the Structure Constants

Let φ ∈ L2(R) be the scaling function for a multiresolution analysis (Vj)j∈Z. Then,for every function f ∈ V1, there exists a sequence (ak)k∈Z ∈ `2(Z) such that

f(x) =∑k∈Z

akφ1,k(x) (7.1)

in L2(R). The coefficients ak are of course given by inner products:

ak = (f, φ1,k) = 21/2∫R

f(x)φ(2x− k) dx, k ∈ Z. (7.2)

In particular, since φ ∈ V0 ⊂ V1,

φ(x) =∑k∈Z

ckφ1,k(x) (7.3)

in L2(R). The equation (7.3) is called the scaling equation and the numbers ckare known as the structure constants.

Example 7.1. The scaling function for the Haar system is φ = χ[0,1). Since

φ(x) = φ(2x) + φ(2x− 1) for every x ∈ R,

the scaling equation is

φ(x) = 1√2φ1,0(x) + 1√

2φ1,1(x)

and the structure constants are c0 = c1 = 1√2

and ck = 0 for k 6= 0, 1.

We leave the proof of the following proposition as an exercise to the reader.


Proposition 7.2. Suppose that (ck)k∈Z ∈ `2(Z) are the structure constants for amultiresolution analysis. Then∑

k∈Z

ckc2m+k =

ß1 if m = 0

0 if m 6= 0.

Theorem 7.3. Suppose that φ ∈ L2(R) is the scaling function for a multiresolutionanalysis (Vj)j∈Z. Then, for every f ∈ V1, there exists a function mf ∈ L2(T) suchthat

f(ξ) = mf ( ξ2 )φ( ξ2 ) for a.e. every ξ ∈ R. (7.4)

The function mf is given by

mf (ξ) =√22

∑k∈Z

ake−ikξ, ξ ∈ R, (7.5)

where the coefficients ak are given by (7.2).

The function mf , that appears in (7.4), is called the filter associated with fand (7.4) the filter identity.

Proof. According to (7.1),

f(x) =∑k∈Z

akφ1,k(x)

in L2(R). Taking Fourier transforms of both sides, using the fact that the Fouriertransform of φ1,k is

21/2∫R

φ(2x− k)e−ixξ dx = 2−1/2∫R

φ(y)e−i(y+k)ξ/2 dx

= 2−1/2e−ikξ/2φ( ξ2 ), (7.6)

we see that

f(ξ) =√22

(∑k∈Z

ake−ikξ/2

)φ( ξ2 ).

If we finally let

mf (ξ) =√22

∑k∈Z

ake−ikξ, ξ ∈ R,

then (7.4) holds and mf ∈ L2(T) since (ak)k∈Z ∈ `2(R). �

Example 7.4. As we saw in Example 7.1, the scaling equation for the Haar sys-tem is

φ(x) = 1√2φ1,0(x) + 1√

2φ1,1(x), x ∈ R.

We also have that (see (7.6)‘φ0,1(ξ) = 2−1/2φ( ξ2 ) and ‘φ1,1(ξ) = 2−1/2e−iξ/2φ( ξ2 )

for ξ ∈ R, so the filter identity is

φ(ξ) = mφ( ξ2 )φ( ξ2 ), where mφ(ξ) =1

2(1 + e−iξ), ξ ∈ R.


Example 7.5. For the Shannon system in Example 6.5, the scaling functionis φ(x) = sincπx, x ∈ R, with

φ(ξ) = χ(−π,π)(ξ), ξ ∈ R.

The filter identity therefore takes the form

χ(−π,π)(ξ) = mφ( ξ2 )χ(−π,π)(ξ2 ) = mφ( ξ2 )χ(−2π,2π)(ξ), ξ ∈ R.

This identity holds if we choose mφ = χ(−π/2,π/2). Since the Fourier coefficientsof mφ are ”mφ(k) =

1

2π

∫ π/2

−π/2eikξ dξ =

sin kπ/2

kπfor k 6= 0

1

2for k = 0

,

the structure constants are

ck =

√2 sin kπ/2

kπ, k 6= 0, and c0 =

√2

2.

Example 7.6.

8. Generating a Multiresolution Analysis

Theorem 8.1. Suppose that φ ∈ L2(R) and

(i) (φ0,k)k∈Z is an orthonormal system in L2(R);(ii) φ satisfies a scaling equation:

φ(x) =∑k∈Z

ckφ1,k(x)

in L2(R) for some sequence (ck)k∈Z ∈ `2(Z);

(iii) φ is continuous at 0 with |φ(0)| = 1.

For j ∈ Z, let Vj denote the closed linear span of (φj,k)k∈Z. Then (Vj)j∈Z is amultiresolution analysis of L2(R).

Remark 8.2. According to Remark 6.8, Vj consists of all functions f ∈ L2(R) suchthat

f(x) =∑k∈Z

akφj,k(x)

in L2(R) for some sequence (ak)k∈Z ∈ `2(Z).

Proof of Theorem 8.1. We need to verify all six axioms in Definition 4.1. Supposefirst that f ∈ Vj , i.e.,

f(x) =∑k∈Z

akφj,k(x) =∑k∈Z

(2j/2ak)φ(2jx− k)

in L2(R) for some sequence (ak)k∈Z ∈ `2(Z). The scaling equation now shows that

φ(2jx− k) =∑l∈Z

clφ1,l(2jx− k) =

∑l∈Z

(21/2cl)φ(2j+1x− 2k − l).

Inserting the last identity in the previous, we see that

f(x) =∑k,l∈Z

(2(j+1)/2akcl)φ(2j+1x− 2k − l),


which after renumbering is seen to be an element of Vj+1. Thus, Vj ⊂ Vj+1.We will next check that f( · ) ∈ Vj if and only if f(2−j · ) ∈ V0. If f ∈ Vj , then

f(2−jx) =∑k∈Z

akφj,k(2−jx) =∑k∈Z

(2j/2ak)φ0,k(x),

which shows that f(2−j · ) ∈ V0. Suppose conversely that f(2−j · ) ∈ V0, i.e.,

f(2−jx) =∑k∈Z

akφ0,k(x).

Then

f(x) =∑k∈Z

akφ(2jx− k) =∑k∈Z

(2−j/2ak)φj,k(x),

showing that f ∈ Vj .The space V0 is also invariant under translations, since if

f(x) =∑k∈Z

φ(x− k), then f(x− l) =∑k∈Z

φ(x− (k + l)).

The remaining parts of the proof are consequences of the following two lemmas. �

Let Pjf denote the orthogonal projection of a function f ∈ L2(R) on Vj for j ∈ Z,i.e.,

Pjf =∑k∈Z

(f, φj,k)φj,k.

Lemma 8.3. Suppose that f ∈ L2(R). Then Pjf → 0 in L2(R) as j → −∞.

This lemma implies that⋂j∈Z Vj = {0}. Indeed, if f ∈ Vj for every j ∈ Z,

then Pjf = f , so that f = limj→−∞ Pjf = 0.

Proof of Lemma 8.3. Put fR = fχ[−R,R], R > 0. Since

‖Pjf‖2 ≤ ‖Pj(f − fR)‖2 + ‖PjfR‖2 ≤ ‖f − fR‖2 + ‖PjfR‖2and the first term in the right-hand side of this inequality tends to 0 as R → ∞,we can assume that f has compact support, i.e., supp f ⊂ [−R,R] for some R > 0.Then, according to Parseval’s identity and Holder’s inequality,

‖Pjf‖22 =∑k∈Z

|(f, φj,k)|2 ≤∑k∈Z

Å∫ R

−R|f(x)||φj,k(x)| dx

ã2≤∑k∈Z

∫ R

−R|f(x)|2 dx

∫ R

−R2j |φ(2jx− k)|2 dx

= ‖f‖22∫ 2jR−k

−2jR−k|φ(y)|2 dy.

Notice finally that the last integral tends to 0 as j → −∞ due to the DominatedConvergence Theorem. �

Lemma 8.4. Suppose that f ∈ L2(R). Then Pjf → f in L2(R) as j →∞.

This lemma implies that⋃j∈Z Vj is dense in L2(R).


Proof of Lemma 8.4. Let fn ∈ C∞c (R) such that fn → f in L2(R). Then f ∈ Sand fn → f in L2(R) according to Plancherel’s theorem. Moreover, since

‖f − Pjf‖2 ≤ ‖f − fn‖2 + ‖fn − Pjfn‖2 + ‖Pj(fn − f)‖2≤ 2‖f − fn‖2 + ‖fn − Pjfn‖2,

we see that it suffices to prove the lemma in the case when f is bounded and has

compact support. Suppose therefore that f is bounded and supp f ⊂ [−R,R] forsome R > 0. Because

‖f − Pjf‖22 = ‖f‖22 − ‖Pjf‖22according to Pythagoras’ theorem, we only need to show that ‖Pjf‖2 → ‖f‖2as j →∞. Suppose that j is so large that 2−jR ≤ π. Applying first Parseval’s andthen Plancherel’s theorems, we obtain that

‖Pjf‖22 =∑k∈Z

|(f, φj,k)|2 =1

(2π)2

∑k∈Z

∣∣∣∣∫ R

−Rf(ξ)φj,k(ξ) dξ

∣∣∣∣2.We then use (2.3) for the Fourier transform of φj,k and change of variables 2−jξ = ω:∫ R

−Rf(ξ)φj,k(ξ) dξ = 2−j/2

∫ R

−Rf(ξ)ei2

−jkξφ(2−jξ) dξ

= 2j/2∫ 2−jR

−2−jR

f(2jω)φ(ω)eikω dω

= 2j/2∫ π

−πf(2jω)φ(ω)eikω dω.

Notice that the assumption on f implies that the integrand in the last integralbelongs to L2(−π, π). Parseval’s theorem therefore shows that

‖Pjf‖22 =2j

(2π)2

∑k∈Z

∣∣∣∣∫ π

−πf(2jω)φ(ω)eikω dω

∣∣∣∣2 =2j

2π

∫ π

−π|f(2jω)|2|φ(ω)|2 dω

=1

2π

∫ R

−R|f(ξ)|2|φ(2−jξ)|2 dξ.

Using the assumption that φ is continuous at 0, it is easy to show that |φ(2−jξ)| → 1uniformly on [−R,R] as j →∞. This implies that

1

2π

∫ R

−R|f(ξ)|2|φ(2−jξ)|2 dξ −→ 1

2π

∫ R

−R|f(ξ)|2 dξ =

∫R

|f(x)|2 dx

as j →∞. �

Example 8.5. With aid of Theorem 8.1, we will now show how the multiresolutionanalysis connected to the Meyer wavelets is constructed.

Suppose that the function θ is defined on [0, 2π], is decreasing, and satisfies

0 ≤ θ(ξ) ≤ 1 for 0 ≤ ξ ≤ 2π, (8.1)

θ(ξ) = 1 for 0 ≤ ξ ≤ 2π

3, (8.2)

θ(ξ) + θ(2π − ξ) = 1 for 0 ≤ ξ ≤ 2π. (8.3)


Notice that (8.2) and (8.3) imply that θ(ξ) = 0 for 4π3 ≤ ξ ≤ 2π. Then extend

to R by first making an even extension on [−2π, 2π] and then letting θ(ξ) = 0for |ξ| > 2π. Let the scaling function φ be defined by

φ(x) =1

2π

∫ 2π

−2π

»θ(ξ)eixξ dξ, x ∈ R,

so that φ =√θ.

Every ξ ∈ R has a unique representation modulo 2π as ξ = 2mπ+ ξ0, where thequotient m is an integer and the residue ξ0 satisfies 0 ≤ ξ0 < 2π. It follows that∑

k∈Z

|φ(ξ + 2kπ)|2 =∑k∈Z

θ(ξ0 + 2(k +m)π) = θ(ξ0) + θ(ξ0 − 2π)

= θ(ξ0) + θ(2π − ξ0) = 1

because of the facts that θ is even and satisfies (8.3). According to Corollary 6.3,this shows that the sequence (φ0,k)k∈Z is orthonormal.

We next define the filter mφ by first letting

mφ(ξ) =»θ(2ξ), |ξ| ≤ π,

and then extending mφ to R with period 2π. Then mφ ∈ L2(T). Since

θ(ξ) = θ(ξ)θ( ξ2 ) for every ξ ∈ R,

the filter identity

φ(ξ) = mφ( ξ2 )φ( ξ2 )

holds for every ξ ∈ R. Thus, φ satisfies a scaling equation.

Finally, since θ(ξ) = 1 for 0 ≤ ξ ≤ 2π3 , φ is continuous at 0 and φ(0) = 1.

Theorem 8.1 now shows that φ is the scaling function for a multiresolution analysis.

Notice that φ ∈ C∞(R) since φ has compact support. Also, if√θ ∈ C∞(R),

then φ ∈ S .

Example 8.6. The Shannon system in Example 6.5 is in fact a special case of themore general Meyer systems. Indeed, if we choose θ so that θ(ξ) = 1 for |ξ| < π,

i.e., θ = χ(−π,π), then φ = χ(−π,π) and therefore

φ(x) = sincπx, x ∈ R.

9. Construction of Wavelets

Definition 9.1. Given a a multiresolution analysis (Vj)j∈Z of L2(R), let

Wj = Vj+1 Vjdenote the orthogonal complement of Vj in Vj+1 for j ∈ Z. The spaces Wj arecalled detail spaces.

We leave the proof of the following proposition to the reader, which shows that thedetail spaces Wj have the same scaling properties as the spaces Vj .

Proposition 9.2. Suppose that (Vj)j∈Z is a multiresolution analysis of L2(R).Then, for every j ∈ Z, f( · ) ∈Wj if and only if f(2−j · ) ∈W0.


Theorem 9.3. Suppose that (Vj)j∈Z is a multiresolution analysis of L2(R). Then

L2(R) =⊕j∈Z

Wj . (9.1)

The right-hand side in (9.1) is the subspace of L2(R) that consists of all series

f =∑j∈Z

fj

with fj ∈Wj for j ∈ Z, where the convergence is in L2(R).

Proof of Theorem 9.3. We will first show that

L2(R) = V0 ⊕∞⊕j=0

Wj .

Denote the right-hand side in this identity by M . If a function f ∈ L2(R) belongsto M⊥, then f ⊥ V0 and f ⊥ Wj for every integer j ≥ 0. By the definition of Wj

and the monotonicity of the multiresolution analysis, this implies that f ⊥ Vj for

every integer j ∈ Z and therefore that f ⊥⋃j∈Z Vj . Since

⋃j∈Z Vj = L2(R), it

follows that f = 0.We will finally show that

V0 =∞⊕j=1

W−j .

Denote the right-hand side in this identity by N . If a function f ∈ V0 belongsto N⊥, then f ⊥ W−j for every integer j ≥ 1. This then implies that f ∈ V−j forevery integer j ≥ 0 and therefore that f ∈

⋂∞j=0 V−j . Since

⋂∞j=0 V−j = {0}, it

follows that f = 0. �

Theorem 9.3 immediately gives the following important corollary.

Corollary 9.4. Suppose that (Vj)j∈Z is a multiresolution analysis of L2(R). Ifthere exists a function ψ ∈ L2(R) such that (ψ0,k)k∈Z is an orthonormal basisfor W0, then (ψj,k)j,k∈Z is a wavelet basis for L2(R).

Let φ ∈ V0 be the scaling function for a multiresolution analysis (Vj)j∈Z of L2(R).For ψ ∈W0, let mψ ∈ L2(T) be the filter in Theorem 7.3, so that

ψ(ξ) = mψ( ξ2 )φ( ξ2 ) for a.e. every ξ ∈ R. (9.2)

We then define the matrix M by

M(ξ) =

Åmφ(ξ) mψ(ξ)

mφ(ξ + π) mψ(ξ + π)

ã, ξ ∈ R. (9.3)

Theorem 9.5. Suppose that (Vj)j∈Z is a multiresolution analysis of L2(R) withscaling function φ.

(a) If (ψ0,k)k∈Z is an orthonormal system W0, then the matrix M in (9.3) isunitary a.e.

(b) Conversely, if M is unitary a.e. for some function mψ ∈ L2(T) and thefunction ψ ∈ L2(R) is defined by (9.2), then (ψ0,k)k∈Z is an orthonormalsequence in W0.


Notice that the matrix M is unitary at ξ ∈ R if

|mφ(ξ)|2 + |mφ(ξ + π)|2 = 1, (9.4)

|mψ(ξ)|2 + |mψ(ξ + π)|2 = 1, (9.5)

mφ(ξ)mψ(ξ) +mφ(ξ + π)mψ(ξ + π) = 0. (9.6)

Proof of Theorem 9.5. We first prove (a). So let us assume that (ψ0,k)k∈Z is anorthonormal system in W0 and deduce (9.4)–(9.6). Since (φ0,k)k∈Z is orthonormal,we know according to Corollary 6.3 that∑

k∈Z

|φ(2ξ + 2kπ)|2 = 1

for a.e. ξ ∈ R. We then apply the filter identity (7.4) for φ and split the series intosums over even and odd indices, using the fact that mφ has period 2π:∑

k∈Z

|φ(2ξ + 2kπ)|2 =∑l∈Z

|mφ(ξ + 2lπ)|2|φ(ξ + 2lπ)|2

+∑l∈Z

|mφ(ξ + (2l + 1)π)|2|φ(ξ + (2l + 1)π)|2

= |mφ(ξ)|2∑l∈Z

|φ(ξ + 2lπ)|2

+ |mφ(ξ + π)|2∑l∈Z

|φ(ξ + π + 2lπ)|2

= |mφ(ξ)|2 + |mφ(ξ + π)|2.

This establishes (9.4). The proof of (9.5) is identical. Since ψ0,k ∈ W0, ψ0,k

is orthogonal to every function φ0,l. Using this fact together with Plancherel’stheorem, we obtain that∫ ∞

−∞ψ(x− k)φ(x− l) dx =

1

2π

∫ ∞−∞

ψ(ξ)φ(ξ)e−i(k−l)ξ dξ = 0

for all k, l ∈ Z. Put n = k − l and make a periodization of the last integral:∫ ∞−∞

ψ(ξ)φ(ξ)e−inξ dξ =∑k∈Z

∫ 2(k+1)π

2kπ

ψ(ξ)φ(ξ)e−inξ dξ

=∑k∈Z

∫ 2π

0

ψ(ξ + 2kπ)φ(ξ + 2kπ)e−inξ dξ

=

∫ 2π

0

Å∑k∈Z

ψ(ξ + 2kπ)φ(ξ + 2kπ)

ãe−inξ dξ.

This shows that all Fourier coefficients of the 2π-periodic function inside the brack-ets are all zero, and hence that∑

k∈Z

ψ(ξ + 2kπ)φ(ξ + 2kπ) = 0


for a.e. ξ ∈ R. We then replace ξ by 2ξ in the last identity and continue as in thefirst part of the proof:∑

k∈Z

ψ(2ξ + 2kπ)φ(2ξ + 2kπ) =∑l∈Z

mφ(ξ + 2lπ)mψ(ξ + 2lπ)|φ(ξ + 2lπ)|2

+∑l∈Z

mφ(ξ + (2l + 1)π)mψ(ξ + (2l + 1)π)|φ(ξ + (2l + 1)π)|2

= mφ(ξ)mψ(ξ) +mφ(ξ + π)mψ(ξ + π) = 0

for a.e. ξ ∈ R, which proves (9.6).We now turn our attention to the proof of (b). Notice that all steps in the calcu-

lations above can be reversed. This means that if (9.4)–(9.6) hold, then (ψ0,k)k∈Zis an orthonormal system in W0. �

We will now show how the function ψ can be constructed with the aid of Theo-rem 9.5. The orthogonality condition (9.6) gives thatÅ

mψ(ξ)mψ(ξ + π)

ã= α(ξ)

Çmφ(ξ + π)

−mφ(ξ)

å(9.7)

for some function α on R. Since the vectors in the left- and right-hand side bothhave length 1, we see that

|α(ξ)| = 1 for a.e. ξ ∈ R. (9.8)

Moreover, replacing ξ by ξ + π in (9.7), using the fact that both mφ and mψ haveperiod 2π, we have thatÅ

mψ(ξ + π)mψ(ξ)

ã= α(ξ + π)

Çmφ(ξ)

−mφ(ξ + π)

å.

This identity in combination with (9.7) shows that

α(ξ + π) = −α(π) for a.e. ξ ∈ R. (9.9)

One function, that satisfies (9.8) and (9.9), is

α(ξ) = −e−iξ, ξ ∈ R.

With this choice, we obtain that

mψ(ξ) = α(ξ)mφ(ξ + π) = −√22 e−iξ∑k∈Z

cke−ik(ξ+π)

= −√22

∑k∈Z

(−1)kcke−i(1−k)ξ

=√22

∑l∈Z

(−1)lc1−le−ilξ,

and therefore that

ψ(x) =∑k∈Z

(−1)kc1−kφ1,k(x), x ∈ R.

It remains to be shown that ψ generates an orthonormal basis for W0.


Theorem 9.6. Suppose that (Vj)j∈Z is a multiresolution analysis of L2(R) withscaling function φ. If ψ ∈ L2(R) is defined by

ψ(x) =∑k∈Z

(−1)kc1−kφ1,k(x), (9.10)

where (ck)k∈Z are the structure constants in (7.3), or equivalently by

ψ(ξ) = e−i(ξ/2+π)mφ( ξ2 + π)φ( ξ2 ),

then (ψ0,k)k∈Z is an orthonormal basis for W0.

Proof. Suppose that f is a function in W0 with filter identity

f(ξ) = mf ( ξ2 )φ( ξ2 );

see Theorem 7.3. As in the discussion above, the fact that f is orthogonal to V0means that

mf (ξ) = β(ξ)mφ(ξ + π),

where β ∈ L2(T) satisfies β(ξ + π) = −β(ξ) for a.e. ξ ∈ R. Recall that

ψ(ξ) = e−i(ξ/2+π)mφ( ξ2 + π)φ( ξ2 ).

It therefore follows that

f(ξ) = ei(ξ/2+π)β( ξ2 )ψ(ξ).

Let us define

α(ξ) = ei(ξ/2+π)β( ξ2 ), ξ ∈ R.

Then α has period 2π since

α(ξ + 2π) = ei(ξ/2+2π)β( ξ2 + π) = −eiξ/2β( ξ2 ) = ei(ξ/2+π)β( ξ2 ) = α(ξ)

for a.e. ξ ∈ R. Also,∫ 2π

0

|α(ξ)|2 dξ =

∫ 2π

0

|β( ξ2 )|2 dξ

= 2

∫ π

0

|β(ξ)|2(|mφ(ξ)|2 + |mφ(ξ + φ)|2) dξ

= 2

∫ 2π

0

|β(ξ)|2|mφ(ξ + φ)|2 dξ

= 2

∫ 2π

0

|mf (ξ)|2 dξ <∞.

This shows that α ∈ L2(T). Thus,

f(ξ) = α(ξ)ψ(ξ)

or equivalently that

f(x) =∑k∈Z

akψ(x− k) =∑k∈Z

akψ0,k(x)

in L2(R). Hence, (ψ0,k)k∈Z is a basis for W0. �

Example 9.7.


10. Regular Wavelets with Compact Support

Suppose that φ is the scaling function for a multiresolution analysis and moreover

that suppφ ⊂ [−R,R] for some R > 0 and φ(0) 6= 0. As we have seen, φ thensatisfies the scaling equation

φ(x) =∑k∈Z

ckφ1,k(x) in L2(R),

which in turn is equivalent to the identity

φ(ξ) = m( ξ2 )φ( ξ2 ) a.e.,

where the filter m is given by

m(ξ) =√22

∑k∈Z

cke−ikξ, ξ ∈ R.

Notice that

ck =√

2

∫ R

−Rφ(x)φ(2x− k) dx = 0 for |k| ≥ 3R,

which shows that m is a trigonometric polynomial. A direct consequence of thisobservation is the fact that the wavelet ψ, given by

ψ(x) =∑k∈Z

(−1)kc1−kφ1,k(x), x ∈ R,

has compact support.Iterating in the filter identity, we see that

φ(ξ) = m( ξ2 )φ( ξ2 ) = m( ξ2 )m( ξ4 )φ( ξ4 ) = ... = φ( ξ2n )

n∏j=1

m( ξ2j ) a.e.

for n = 1, 2, ... . Since φ has compact support, φ belongs to L1(R) and therefore

is φ continuous on R. Using this together with the assumption that φ(0) 6= 0, wesee that we can let n tend to infinity in the last identity and obtain that

φ(ξ) = φ(0)∞∏j=1

m( ξ2j ) a.e.

We summarize these observations in the following theorem.

Theorem 10.1. Suppose that the φ scaling function for a multiresolution analysis

has compact support and φ(0) 6= 0. Then the filter m is a trigonometric polynomial,the corresponding wavelet ψ has compact support, and

φ(ξ) = φ(0)∞∏j=1

m( ξ2j ) a.e.

We will now show that this argument can essentially be reversed. To this end,let m be a trigonometric polynomial of the form

m(ξ) =

√2

2

l∑k=−l

cke−ikξ, ξ ∈ R (10.1)


such that

|m(ξ)|2 + |m(ξ + π)|2 = 1 for every ξ ∈ R, (10.2)

m(0) = 1, (10.3)

m(ξ) 6= 0 for ξ ∈ [−π2 ,π2 ]. (10.4)

Proposition 10.2. Suppose that m is a trigonometric polynomial of the form (10.1)that satisfies conditions (10.2)–(10.4). Then the product

φ(ξ) =∞∏j=1

m( ξ2j ), ξ ∈ R, (10.5)

converges locally uniformly on R and satisfies φ(0) = 1. Moreover, φ is a continu-ous function.

Proof. Let Pn denote the n-th partial product of the product (10.5), i.e.,

Pn(ξ) =n∏j=1

m( ξ2j ), ξ ∈ R, n = 1, 2, ... .

Then

|m(ξ)− 1| = |m(ξ)−m(0)| ≤√

2

2

l∑k=−l

|ck||e−ikξ − 1| ≤Å√

2

2

l∑k=−l

|k||ck|ã|ξ|

for every ξ ∈ R. Using this observation, we obtain that

|Pk+1(ξ)− Pk(ξ)| = |m( ξ2k+1 )− 1||Pk(ξ)| ≤ C2−(k+1)|ξ|.

It follows that if m > n, then

|Pm(ξ)− Pn(ξ)| ≤ C|ξ|Å

1

2m+ ...+

1

2n+1

ã≤ C2−n|ξ|.

This shows that the product (10.5) converges uniformly on every compact subset

of R and therefore represents a continuous function. Finally, φ(0) = 1 accordingto (10.3). �

We will omit the proof of the following theorem. A consequence of this theoremis the fact that the function φ, implicitly defined by (10.5), belongs to L2(R).

Theorem 10.3. Suppose that m is a trigonometric polynomial that satisfies con-

ditions (10.2)–(10.4). Then the function φ in (10.5) belongs to L2(R).

The next step is to show that φ satisfies a filter equation. This means that φalso satisfies a scaling equation.

Proposition 10.4. Suppose that m is a trigonometric polynomial that satisfiesconditions (10.2)–(10.4). Then

φ(ξ) = m( ξ2 )φ( ξ2 ) a.e.

Proof. By the definition of φ,

φ(ξ) =∞∏j=1

m( ξ2j ) = m( ξ2 )∞∏j=2

m( ξ2j ) = m( ξ2 )φ( ξ2 ) a.e. �


The construction also implies that φ has compact support. A short proof can beobtained by using some distributiton theory.

Theorem 10.5. Suppose that m is a trigonometric polynomial of the form (10.1)that satisfies conditions (10.2)–(10.4). Then suppφ ⊂ [−l, l].

Proof. Put

uj =

√2

2

l∑k=−l

ckδ2−jk, j = 1, 2, ... ,

where δ2−jk is the Dirac delta at 2−jk, considered as an element of S ′. It is notso hard to verify that “uj(ξ) = m( ξ2j ) for ξ ∈ R.

Also,

u1 ∗ ... ∗ un =(√2

2

)n ∑|kj |≤l

ck1 · ... · cknδ2−1k1+...+2−nkn .

Notice that the support of the convolution is a subset of the interval [−l, l]. More-over, ¤�u1 ∗ ... ∗ un =

n∏j=1

m( ξ2j ) for n = 1, 2, ... .

We know that the right-hand side of this identity converges to φ(ξ) locally uni-formly. This together with the fact that the absolut value of each partial productis uniformly bounded by 1 implies that the convergence actually holds in S ′. Itfollows that u1 ∗ ... ∗ un → φ in S ′ and therefore that suppφ ⊂ [−l, l]. �

We will finally show that (φ0,k)k∈Z is an orthonormal system in L2(R) and beginwith a lemma.

Lemma 10.6. Suppose that m is a trigonometric polynomial of the form (10.1)that satisfies conditions (10.2)–(10.4). Then the function

Φ(ξ) =∑k∈Z

|φ(ξ + 2kπ)|2, ξ ∈ R,

is a trigonometric polynomial.

Proof. Plancherel’s formula and a periodization argument shows that∫R

φ(x)φ(x− n) dx =1

2π

∫R

|φ(ξ)|2einξ dξ =1

2π

∫ 2π

0

Φ(ξ)einξ dξ.

The left-hand side is nonzero for not more than a finite number of integers n, whichimplies that all but a finite number of Fourier coefficients of Φ are 0 and thereforethat Φ is a trigonometric polynomial. �

Theorem 10.7. Suppose that m is a trigonometric polynomial of the form (10.1)that satisfies conditions (10.2)–(10.4). Then (φ0,k)k∈Z is an orthonormal systemin L2(R).


Proof. The same argument that was used in the proof of Theorem 9.5 shows that

Φ(2ξ) = |m(ξ)|2Φ(ξ) + |m(ξ + π)|2Φ(ξ + π) for every ξ ∈ R.

Let a = Φ(ξ0) be the minimal value of Φ on [−π, π]. Then

a = Φ(ξ0) = |m( ξ02 )|2Φ( ξ02 ) + |m( ξ02 + π)|2Φ( ξ02 + π) ≥ a.

Due to (10.2) and (10.4), this shows that Φ( ξ02 ) = a. Iterating this argument, we

obtain that Φ( ξ02j ) = a for j = 0, 1, ... and therefore that Φ(0) = a. Any nonzerointeger k can be written in the form k = 2rs, where r is a nonnegative integer and sis an odd integer. Using this together with the fact that m(sπ) = 0, we obtain that

φ(2kπ) =∞∏j=1

m(2r+1−jsπ) = 0,

and therefore thatΦ(0) =

∑k∈Z

|φ(2kπ)|2 = |φ(0)|2 = 1.

This shows that a = 1.A similar argument shows that the maximal value of Φ on [−π, π] is 1 and hence

that Φ is identically 1 on [−π, π]. Due to periodicity of Φ, this means that Φ isidentically 1 on R. �

Corollary 10.8. Suppose that m is a trigonometric polynomial of the form (10.1)that satisfies conditions (10.2)–(10.4). Then φ generates a multiresolution analysisof L2(R) such that both φ and ψ have compact support.

Our final result shows that it is possible to construct compactly supportedwavelets that are as smooth as we want.

Theorem 10.9. For every nonnegative integer r, there exists a multiresolutionanalysis of L2(R) such that both the scaling function φ and the wavelet ψ belongto Crc (R).

Sketch of proof. Choose a trigonometric polynomial m satisfies (10.2)–(10.4) and∞∏j=1

|m( ξ2j )| ≤ C

(1 + |ξ|)r+2for every ξ ∈ R.

Then the scaling function φ, given by

φ(x) =1

2π

∫R

∞∏j=1

m( ξ2j )eiξx dξ, x ∈ R,

belongs to Crc (R). It follows from the formula (9.10) that the wavelet ψ also belongsto Crc (R). �

Remark 10.10. One can show that there does not exist a multiresolution analysisof L2(R) such that both φ and ψ belong to C∞c (R).

Department of Mathematics, Linkoping University, SE-581 83 Sweden

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Contents - Linköping Universitycourses.mai.liu.se/GU/TATA66/Dokument/wavelets-2015.pdf · AN INTRODUCTION TO WAVELETS BENGT OVE TURESSON Contents 1. Introduction 1 2. Preliminaries