RKHS 11.12.07

Introduction to reproducing kernel Hilbertspaces and the continuous wavelet transform

Moti Levy

December 11, 2007

Supervisor: Dr. Vadim GrinsteinThe Open University

i

ii

Contents

0.1 Foreword . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . iv

1 Reproducing kernel Hilbert spaces 11.1 De�nitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.2 Examples of reproducing kernel Hilbert Spaces . . . . . . . . . 2

1.2.1 Finite dimensional Hilbert space . . . . . . . . . . . . . 21.2.2 Sobolev spaces . . . . . . . . . . . . . . . . . . . . . . 3

1.3 General Theory . . . . . . . . . . . . . . . . . . . . . . . . . . 81.4 Characterization of reproducing kernels . . . . . . . . . . . . . 14

2 Instances of RKHS 212.1 Bergman spaces . . . . . . . . . . . . . . . . . . . . . . . . . . 21

2.1.1 The weighted Bergman spaces on the unit disk . . . . . 242.1.2 The weighted Bergman spaces on the upper half plane 25

2.2 Hardy spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . 262.2.1 Hardy space on the unit disk, H2 (D) . . . . . . . . . . 262.2.2 Hardy space on the upper half plane, H2 (�+) . . . . . 32

2.3 Bandlimited signals and Paley-Wiener spaces . . . . . . . . . . 362.3.1 The space of bandlimited functions . . . . . . . . . . . 362.3.2 Paley-Wiener spaces . . . . . . . . . . . . . . . . . . . 362.3.3 Sampling of band limited functions . . . . . . . . . . . 40

2.4 Wavelet theory . . . . . . . . . . . . . . . . . . . . . . . . . . 422.4.1 The continuous wavelet transform . . . . . . . . . . . . 432.4.2 Calderon�s reproducing formula . . . . . . . . . . . . . 59

3 Appendix: The Fourier transform 633.1 Fourier transform on L1 (R) . . . . . . . . . . . . . . . . . . . 633.2 Properties of the Fourier transform . . . . . . . . . . . . . . . 633.3 Fourier transform operator on L2 (R) . . . . . . . . . . . . . . 65

3.3.1 Plancherel theorem . . . . . . . . . . . . . . . . . . . . 663.3.2 Fourier transform in the complex plane . . . . . . . . . 67

3.4 The Paley-Wiener theorems . . . . . . . . . . . . . . . . . . . 69

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iv CONTENTS

0.1 Foreword

The motivation for this document is the second chapter of Daubechies�book[3, Chapter 2.5] on the continuous wavelet transform, where I �rst met theconcept of reproducing kernel Hilbert spaces (RKHS). Soon I have foundout that RKHS appear in many areas such as wavelet theory, di¤erentialequation theory, machine learning and more.Daubechies describes the reproducing kernel Hilbert space underlying the

continuous wavelet transform. It turns out that the continuous wavelet trans-form maps L2 (R) to a RKHS. She also gives an example where a continuouswavelet transform maps a Hardy space to a Hilbert space of analytic func-tions.Chapter 1 gives an introduction to theory of RKHS and brief review of

some of its applications. I relied mainly on a tutorial by V. I. Paulsen [4].I also used the chapter on RKHS in the book of Weiss, Lindenstaruss andPazi (in Hebrew)[6, Chpater 1.9]Chapter 2 gives examples of RKHS (the classical Hardy and Bergman

spaces and the space of band-limited functions). Also, the examples inDaubechies�book are elaborated.To make this document more self contained, Chapter 3 contains an ap-

pendix on the concepts we need from Fourier analysis and the Paley-Wienertheorems.Words of caution: This document does not contain any original ma-

terial and was not written by a professional mathematician. Please readcarefully. If something seems to be new, it is probably an error.

Chapter 1

Reproducing kernel Hilbertspaces

1.1 De�nitions

Before we de�ne reproducing kernel Hilbert space (RKHS) we de�ne theDirac evaluation functional.

De�nition 1 The Dirac evaluation functional is a linear functional ��that evaluates each function in a Hilbert space H of complex-valued functionson a set S; at the point �,

�� (f) := f (�) : (1.1)

The functional �� is bounded if there exists a number M� such that

j�� (f)j �M� kfk (1.2)

for all f 2 H .

De�nition 2 A Hilbert space H of complex-valued functions on a set S en-dowed with a bounded Dirac evaluation functional �� for every � 2 S iscalled a reproducing kernel Hilbert space (RKHS).

There are many Hilbert spaces of functions which are not RKHS. A simpleexample is the space L2 ([0; 1]). The vectors of this space are equivalenceclasses of functions which di¤er on a subset of measure zero, so there is nomeaning to the value of a function at a certain point. Let us take the subspaceC ([0; 1]), the continuous functions on [0; 1], and de�ne the usual L2 norm onthis space, i.e., kfk2 :=

R 10jf j2 dt; then after completion we get the Hilbert

space L2 ([0; 1]) :We can evaluate the functions of this subspace at each point

1

2 CHAPTER 1. REPRODUCING KERNEL HILBERT SPACES

of [0; 1], yet the linear evaluation functional of this subspace is not bounded;Given any point � 2 [0; 1] we can construct a sequence fn 2 C ([0; 1]) ; suchthat kfnk = 1, and limn fn (�) =1.Now let H be a RKHS. The Riesz representation theorem, states that if

' is a bounded linear functional on a Hilbert space, then there is a uniquevector u inH such that ' (f) = hf; uiH for all f 2 H [1, Chapter II, Theorem5.2]. It follows that there exists a unique vector in H, which will be denotedas K (�; �) ; such that

�� (f) = f (�) = hf;K (�; �)i :

De�nition 3 The function k� := K (�; �) is called reproducing element(sometimes called the representer) of the evaluation functional ��.

De�nition 4 The set of all reproducing elements K (�; �) of H is called thereproducing kernel of H.

Obviously, by the Riesz representation theorem, the reproducing kernel,which will be denoted byK (�; �) ; is unique. The name "Reproducing KernelHilbert Space" is justi�ed by the existence of reproducing kernel for a Hilbertspace in which the Dirac evaluation functional �� is bounded.

1.2 Examples of reproducing kernel HilbertSpaces

1.2.1 Finite dimensional Hilbert space

The set S is f1; 2; : : : ; kg and we consider the space of functions

F := ff j f : f1; 2; : : : ; kg ! Cg = Ck:

The inner product is

hf; gi :=kXj=1

f (j) g (j):

F is a Hilbert space (it is �nite dimensional linear space) and

j�j (f)j = jf (j)j � kfk =

vuut kXj=1

jf (j)j2:

In this simple example the representer �� is �n;� and the reproducing kernelis K (�; �) = ��;�:

1.2. EXAMPLES OF REPRODUCING KERNEL HILBERT SPACES 3

1.2.2 Sobolev spaces

The Sobolev1 space2 W 1;2 ([0; 1]), is a subspace of the Lebesgue space L2 ([0; 1]) ;

W 1;2 ([0; 1]) :=nf 2 L2 ([0; 1])

�� f is absolutely continuous on [0; 1] ; f 0 2 L2 ([0; 1])owith inner product de�ned as

hf; gi :=Z[0;1]

�f (t) g (t) + f

0(t) g0 (t)

�dt:

The norm is de�ned as

kfkW 1;2 :=

�Z[0;1]

�jf (t)j2 +

��f 0 (t)��2� dt� 12

:

Since f 2 W 1;2 is absolutely continuous then f is equal almost everywhereto the Lebesgue integral of its derivative, hence,

f (y)� f (x) =Z y

x

f0(t) dt

1Sobolev, Sergei L�vovich (1908- 1989) was a Russian mathematician, working in math-ematical analysis and partial di¤erential equations. Sobolev introduced notions which arenow fundamental in several di¤erent areas of mathematics. Sobolev spaces can be de-�ned by growth conditions on Fourier transforms; they and their embedding theoremsare an important subject in functional analysis. Generalized functions (later known asdistributions), were �rst introduced by Sobolev in 1935 for weak solutions, and furtherdeveloped by Laurent Schwartz; they rede�ned the notion of di¤erentiation. Both thesedevelopments grew directly out of his work on partial di¤erential equations.

2The Sobolev space W k;p ([0; 1]) is de�ned to be the subset of Lp ([0; 1]) such that fand its weak derivatives up to some order k have a �nite Lp norm, for given p � 1.Some care must be taken to de�ne derivatives in the proper sense. In the one-dimensionalproblem it is enough to assume f (k�1) is di¤erentiable almost everywhere and is equalalmost everywhere to the Lebesgue integral of its derivative. With this de�nition, theSobolev spaces admit a natural norm,

kfkk;p =kXi=0

f (i) Lp=

kXi=0

�Z 1

0

��f (i) (t)��p� 1p

W k;p equipped with the norm k�kk;p is a Banach space. It turns out that it is enoughto take only the �rst and last in the sequence, i.e., the norm de�ned by

kfkk;p = kfkLp + f (k)

Lp

is equivalent to the norm above.


To see that the Dirac evaluation functional of W 1;2 is bounded, we get fromthe Cauchy-Schwartz inequality in L2 ([0; 1]) ;

jf (x)� f (y)j =��Z y

x

f0(t) dt

�� Z y

x

��f 0 (t)�� dt�

�Z y

x

1dt

� 12�Z y

x

��f 0 (t)��2 dt� 12

� jx� yj12

�Z y

x

��f 0 (t)��2 dt� 12

It follows that f 2 W 1;2 implies that f 2 Lip 12([0; 1]) ; that is f satis�es

Hölder condition of order 12:

jf (x)j � jf (y)j+ jf (x)� f (y)j

� jf (y)j+ jx� yj12

�Z y

x

��f 0 (t)��2 dt� 12

� jf (y)j+�Z

[0;1]

��f 0 (t)��2 dt� 12

We integrate with respect to y,Z[0;1]

jf (x)j dy �Z[0;1]

jf (y)j dy +Z[0;1]

�Z[0;1]

��f 0 (t)��2 dt� 12

!dy

and we get,

jf (x)j �Z[0;1]

jf (y)j dy +�Z

[0;1]

��f 0 (t)��2 dt� 12

We use again Cauchy-Schwartz inequalityZ[0;1]

jf (y)j dy ��Z

[0;1]

jf (t)j2 dt� 1

2

jf (x)j ��Z

[0;1]

jf (t)j2 dt� 1

2

+

�Z[0;1]

��f 0 (t)��2 dt� 12

(1.3)

�p2

s�Z[0;1]

jf (t)j2 dt�+

�Z[0;1]

jf 0 (t)j2 dt�

=p2 kfkW 1;2


We have proved that evaluation functional is bounded.It remains to show that W 1;2 ([0; 1]) is complete. Suppose ffng is a

Cauchy sequence in W 1;2 ([0; 1]) :

kfn � fmk2W 1;2 :=

Z[0;1]

�jfn (t)� fm (t)j2 +

��f 0n (t)� f 0m (t)��2� dtHence kfn � fmkW 1;2 ! 0 implies thatZ

[0;1]

jfn (t)� fm (t)j2 dt! 0;

i.e., ffng is a Cauchy sequence in L2 ([0; 1]) ; which by completeness ofL2 ([0; 1]) ; converges in norm to f 2 L2 ([0; 1]) ; andZ

[0;1]

��f 0n (t)� f 0m (t)��2 dt! 0;

i.e.,�f0n

is a Cauchy sequence in L2 ([0; 1]) ; which by completeness of

L2 ([0; 1]) ; converges in norm to g 2 L2 ([0; 1]) ; f 0n � g L2! 0:

By inequality (1.3), the sequence ffng must be pointwise Cauchy, i.e.,

jfn (x)� fm (x)j ! 0;

therefore we may de�ne a function by setting f (x) := limn fn (x) : Clearlyf (x) = limn fn (x) is equal in norm to the limit of the Cauchy sequence ffngmentioned above.Now limn

f 0n � g L2 = 0 implies that limn

f0n � g; �[0;x]

�L2= 0; hence

limn

Z x

0

f0

n (t) dt = limn

Df0

n; �[0;x]

EL2=g; �[0;x]

�L2=

Z x

0

g (t) dt (1.4)

But since ffng are absolutely continuous,

limn

Z x

0

f0

n (t) dt = limnfn (x)� lim

nfn (0) = f (x)� f (0) ;

so we have

f (x)� f (0) =Z x

0

g (t) dt;


which implies that f (x) is absolutely continuous and that f0(t) is equal to

g (t) almost everywhere in [0; 1] : Hence f0(t) 2 L2 ([0; 1]) : This complete the

proof that f (x) 2 W 1;2 ([0; 1]) :We conclude that W 1;2 is RKHS.Reproducing kernels of Sobolev spaces are intrinsically tied to solutions of

ordinary or partial di¤erential equations with boundary value conditions. To�nd the kernel function of W 1;2 we �rst formally solve a di¤erential equationand then show that the function we obtain by this formal solution, belongsto W 1;2. To �nd k� (t), we apply integration by parts to see that

f (�) = hf; k�i =Z[0;1]

�f (t) k� (t) + f

0(t) k

0� (t)

�dt

=

Z[0;1]

f (t) k� (t) dt+�f (t) k

0� (t)

��10�Z[0;1]

f (t) k00� (t) dt

Now if we impose the boundary conditions k0� (0) = k

0� (1) = 0 we get

f (�) =

Z[0;1]

f (t) k� (t) dt�Z[0;1]

f (t) k00� (t) dt (1.5)

If we let �� = � (t� �) denote the formal Dirac-delta function, then

f (�) =

Z[0;1]

f (t) ��dt;

and equation (1.5) becomesZ[0;1]

f (t)�� (t)� k� (t) + k

00

� (t)�dt = 0;

(assuming that k� is real).Thus we need to solve the boundary value problem

k00

� � k� = ��k0

� (0) = k0

� (1) = 0 (1.6)

The solution of this problem is called the Green�s function corresponding tothe boundary value problem (1.6).The general solution of the homogenous equation is

u (t) = Aet +Be�t (1.7)


where A;B are constant. To �nd a solution of the non-homogenous equationwe use the method of variation of parameters3. We assume that A;B arefunctions of t and solve the system:

A0(t) et +B

0(t) e�t = 0

A0(t) et �B0

(t) e�t = �� (t� �) (1.8)

The solution of this system is

A0(t) = �1

2e�t� (t� �)

B0(t) =

1

2et� (t� �) (1.9)

Hence

v (t) =1

2

tZ0

��e�u+t� (u� �) + eu�t� (u� �)

�du (1.10)

is a solution of the non-homogenous equation. Now the general solution is

k� (t) = Aet +Be�t +1

2

tZ0

��e�u+t� (u� �) + eu�t� (u� �)

�du

=

�Aet +Be�t if t � �Aet +Be�t + 1

2

��e��+t + e��t

�if t � �

(1.11)

k0

� (0) = 0 = A�B

k0

� (1) = 0 = Ae�Be�1 � 12

�e��+1 + e��1

�A = B =

e

2 (e2 � 1)�e��+1 + e��1

�

k� (t) =

(e

2(e2�1)�e��+1 + e��1

�(et + e�t) if t � �

e2(e2�1)

�e��+1 + e��1

�(et + e�t) + 1

2

��e��+t + e��t

�if t � �

=

(1

2(e2�1)�e2�� + e�

�(et + e�t) if t � �

12(e2�1) (e

2�t + et)�e� + e��

�if t � �

3Invented by Lagrange 1774,1775.


Now we check that

K (�; �) :=

(1

2(e2�1)�e2�� + e�

�(e� + e��) if � � �

12(e2�1) (e

2�� + e�)�e� + e��

�if � � �

(1.12)satis�es the conditions to be the reproducing kernel for W 1;2.

kK (t; �)k2W 1;2 =

Z[0;1]

�jK (t; �)j2 +

��K 0(t; �)

��2� dt=

�e2�� + e�

�24 (e2 � 1)2

Z �

0

��et + e�t

�2+�et � e�t

�2�dt

+

�e� + e��

�24 (e2 � 1)2

Z 1

�

��e2�t + et

�2+�et � e2�t

�2�dt

< 1

hf; k�i =

Z[0;1]

�f (t)K (t; �) + f

0(t)K

0(t; �)

�dt

=

�e2�� + e�

�2 (e2 � 1)

Z �

0

f (t)�et + e�t

�dt+

�e2�� + e�

�2 (e2 � 1)

Z �

0

f0(t)�et � e�t

�dt

+

�e�� + e�

�2 (e2 � 1)

Z 1

�

f (t)�et + e2�t

�dt+

�e�� + e�

�2 (e2 � 1)

Z 1

�

f0(t)�et � e2�t

�dt

=

�e2�� + e�

�2 (e2 � 1)

Z �

0

f (t)�et + e�t

�dt�

�e2�� + e�

�2 (e2 � 1)

Z �

0

f (t)�et + e�t

�dt

+

�e2�� + e�

�2 (e2 � 1) f (t)

�et � e�t

��0

+

�e�� + e�

�2 (e2 � 1)

Z 1

�

f (t)�et + e2�t

�dt�

�e�� + e�

�2 (e2 � 1)

Z 1

�

f (t)�et � e2�t

�dt

+

�e�� + e�

�2 (e2 � 1) f (t)

�et � e2�t

��1�

=

�e2�� + e�

� �e� � e��

�2 (e2 � 1) f (�)�

�e�� + e�

� �e� � e2��

�2 (e2 � 1) f (�)

= f (�)

1.3 General Theory

Let S be a set and let H be a RKHS with kernel K: In this section wewill present some properties of the RKHS and will show that the kernel

1.3. GENERAL THEORY 9

completely determines the space H.

Proposition 5 Properties of the reproducing kernel K (�; �) :

a) K (�; �) = hK (�; �) ; K (�; �)ib) K (�; �) = K (�; �)

c) K (�; �) = kK (�; �)k2 � 0 for all � 2 Sd) For all f 2 H and � 2 S; jf (�)j �

pK (�; �) kfk

where equality holds only if f = �K (�; �) ; � is a scalar:

e) jK (�; �)j �pK (�; �)

pK (�; �)

Proof. Property a) follows directly from the de�nition of the reproducingkernel.Properties b) and c) are direct consequences of property a).Property d) follows Cauchy-Schwartz inequality and a):

jf (�)j = jhf;K (�; �)ij � kK (�; �)k kfk =pK (�; �) kfk :

Finally, substitute f (�) = K (�; �) in d) to get e).

Proposition 6 Let H be a RKHS on a set S with kernel K: Then the linearspan of the reproducing elements k� := K (�; �) is dense in H.

Proof. The linear span of the functions fk�g�2S is the subspace generatedby the linear combinations of the form

mXi=1

�ik�i (1.13)

where �i 2 C and �i 2 S, m 2 N. Denote by M the closure of the linearspan of fk�g�2S. We will show that M = H.Clearly M � H is a closed subspace of H. A function f 2 H is orthogonalto M if and only if hf;K (�; �)i = 0 for every � 2 S. But hf;K (�; �)i = f (�),hence a function f 2 H is orthogonal to M if and only if f (�) = 0 for every� 2 S; that is if and only if f = 0: Now by [1, Chapter I, Theorem 8.2], givenf 2 H; there exists a unique w 2 M and a unique v orthogonal to M suchthat f = w+ v:We have already shown that v = 0; hence f = w 2M whichimplies that H �M.

Next we show that for RKHS, convergence in the norm implies pointwiseconvergence.


Lemma 7 Let H be a RKHS on S and let ffng be a sequence in H. Iflimn kfn � fk = 0; then limn fn (�) = f (�) for every � 2 S.

Proof.

jfn (�)� f (�)j = jhfn; K (�; �)i � hf;K (�; �)ij = jhfn � f;K (�; �)ij

By Cauchy-Schwartz inequality

jhfn � f;K (�; �)ij � kfn � fk kK (�; �)k :

It follows that if kfn � fk ! 0 then jfn (�)� f (�)j ! 0:

The next proposition shows that the kernel K completely determines thespace H.

Proposition 8 Let H1 and H2 be RKHS�s on S with kernels K1 (�; �) andK2 (�; �) respectively. If K1 (�; �) = K2 (�; �) for all �; � 2 S then H1 = H2and kfk1 = kfk2 for every f:

Proof. LetK (�; �) := K1 (�; �) = K2 (�; �). LetW1 := span fk� 2 H1 j � 2 S gand W2 := span fk� 2 H2 j � 2 S g.By proposition 6, W1 and W2 are dense in H1 and H2 respectively. Forany f 2 W1 we have f (�) =

Pj �jk�j (�) with k�j (�) 2 H1, but k�j (�) are

also in H2, hence f (�) 2 W2 and W1 � W2: By similar argument we haveW2 � W1: It follows that W1 =W2.Also for f 2 W1,

kfk21 =Xi;j

�i�j

Dk�i ; k�j

E1=Xi;j

�i�jK��j; �i

�=

Xi;j

�i�j

Dk�i ; k�j

E2= kfk22 :

(We used above property (5, a)) K (�; �) = hK (�; �) ; K (�; �)i):Thus kfk1 = kfk2, for all f 2 W1 =W2:Finally, if f 2 H1 then there exists a sequence of functions ffng � W1 withkf � fnk1 ! 0. Since ffng is Cauchy sequence in W1 it is also Cauchysequence in W2. So there exists a function g 2 H2 with kg � fnk2 ! 0.But by the lemma above kg � fnk2 ! 0 implies f (�) = limn fn (�) = g (�) :Thus, every function f 2 H1 is also in H2; and by analogous argument everyg 2 H2 is also in H1. Hence, H1 = H2.


The proposition above and the fact that the kernel is unique implies thattwo RKHS�s are equal if and only if their kernels are the same.

Hereafter we con�ne our discussion to separable Hilbert spaces.

Theorem 9 Let H be a separable reproducing kernel Hilbert space with re-producing kernel K (�; �). If fung1n=1 is an orthonormal basis4 for H, thenK (�; �) =

P1n=1 un (�)un (�):

Proof. By [1, Chapter I, Theorem 11.3] k� =P1

n=1 hk�; uniun =P1

n=1 hun; k�iun =P1n=1 un (�)un; where this series converges in the norm. But since in RKHS,

convergence in the norm implies pointwise convergence, we have k� =P1

n=1 un (�)unwith pointwise convergence. It follows that

k� (�) = K (�; �) =1Xn=1

un (�)un (�);

or

K (�; �) =1Xn=1

un (�)un (�):

The theorem above can be generalized to a set of functions which neednot be an orthonormal basis but are a frame.

De�nition 10 A set of vectors fuig1i=1 in a Hilbert spaceH is called a frameif there are two constants A;B such that the following inequality holds for allvectors v 2 H,

A kvk2 �1Xi=1

jhv; uiij2 � B kvk2 : (1.14)

Note that A = B = 1 for an orthonormal basis, thus an orthonormalbasis is a special case of a frame.However, the vectors in a frame need not be linearly independent. A

frame with A = B is called an A-tight frame and a frame with A = B = 1;is called 1-tight frame or Parseval frame.

4A Hilbert space contains an orthonormal basis if and only if it is separable.[1, ChapterI, Theorem 16.1]


De�nition 11 A set of vectors fuig1i=1 in a Hilbert space H is called a Par-seval frame if the following equality holds for all vectors v 2 H,

1Xi=1

jhv; uiij2 = kvk2 : (1.15)

The following result shows one of the most common way that Parsevalframe arises.

Proposition 12 Let H be a Hilbert space, let H0 � H be a closed subspaceand let P0 denote the orthogonal projection of H onto H0. If fung1n=1 is anorthonormal basis for H then fP0 (un)g1n=1 is a Parseval frame for H0:

Proof. Let v 2 H0, then P0 (v) = v. hv; uni = hP0 (v) ; uni = hv; P0 (un)i 5:Thus kvk2 =

P1i=1 jhv; uiij

2 =P1

i=1 jhv; P0 (ui)ij2 and the result follows.

The following result gives equivalent de�nition of Parseval frame.

Proposition 13 Let H be a separable Hilbert space and let fuig1i=1 � Hbe a set of vectors. Then fuig1i=1 is a Parseval frame if and only if v =P1

i=1 hv; uiiui for every v 2 H.

Proof. Let feig1i=1 denote the canonical orthonormal basis of l2:We de�ne afunction A which mapsH into l2, as follows: for v 2 H , Av :=

P1i=1 hv; uii ei.

A is a linear isometry from H to l2 if kAvk = kvk for all v 2 H.Since kAvk =

P1i=1 jhv; uiij

2 ; it follows that A is a linear isometry if andonly if fuig1i=1 is a Parseval frame.Note that hv; A�eki = hAv; eki = hv; uki, hence for all v 2 H, hv; A�ek � uki =0; which implies that A�ek = uk. By [1, Chapter VIII, Theorem 3.1] A is anisometry if and only if A�A = IH. But A�A = IH if and only if

v = A�Av = A�

1Xi=1

hv; uii ei

!=

1Xi=1

hv; uiiA�ei =1Xi=1

hv; uiiui (1.16)

for every v 2 H.Thus, we have that fuig1i=1 is a Parseval frame if and only if A is an isometryif and only if A�A = IH if and only if v =

P1i=1 hv; uiiui.

Theorem 9 allows to express the kernel by an orthonormal basis. Thefollowing theorem gives a more general way to compute the kernel, that is toexpress the kernel by a Parseval frame.

5P is orthogonal projection if and only if P 2 = P and P is self adjoint.


Theorem 14 Let H be a separable RKHS on S with reproducing kernelK (�; �) : Then fuig1i=1 � H is a Parseval frame if and only if K (�; �) =P1

i=1 ui (�)ui (�); where the series converges pointwise.

Proof. Suppose fuig1i=1 is a Parseval frame. Then

k� =1Xi=1

hk�; uiiui =1Xi=1

hui; k�iui =1Xi=1

ui (�)ui

K (�; �) = hK (�; �) ; K (�; �)i = hk�; k�i

=

*k�;

1Xi=1

ui (�)ui

+=

1Xi=1

ui (�) hk�; uii

=1Xi=1

ui (�) hui; k�i =1Xi=1

ui (�)ui (�):

Conversely, suppose K (�; �) =P1

i=1 ui (�)ui (�): Let v =Pm

i=1 �ik�i a �nitelinear combination of reproducing elements. Then

kvk2 =

*mXj=1

�jk�j ;mXi=1

�ik�i

+=

mXj=1

mXi=1

�i�j

Dk�j ; k�i

E(1.17)

=mXj=1

mXi=1

�i�jK��i; �j

�=

mXj=1

mXi=1

�i�j

1Xn=1

un (�i)un��j�

=mXj=1

mXi=1

�i�j

1Xn=1

un; k�i

� Dk�j ; un

E=

1Xn=1

mXj=1

�j

Dk�j ; un

E mXi=1

�iun; k�i

�=

1Xn=1

*mXi=1

�ik�i ; un

+*un;

mXi=1

�ik�i

+=

1Xn=1

hv; uni hun; vi =1Xn=1

jhun; vij2 :

By Proposition 6, the set of �nite linear combinations of reproducing ele-ments is dense inH. For every vector v 2H there is a sequence

�vk =

Pmki=1 �i;kk�i

1k=1

which converges in norm to v; i.e., kv � vkk ! 0 as k !1.By continuity of the norm we have kvkk ! kvk ; and by continuity of theinner product we have jhw; vkij ! jhw; vij :Now it is easily seen that if we take a limit of norm convergent sequence ofvectors on both sides of identity 1.17, then we obtain the identity for thelimit vector, too. Thus the condition to be Parseval frame is met by the setfuig1i=1.


1.4 Characterization of reproducing kernels

In this subsection the necessary and su¢ cient conditions for a functionK (�; �)to be a reproducing kernel for some RKHS will be obtained.

De�nition 15 Let A = (ai;j) be a n�n complex matrix. Then A is positive(written A � 0) if for every �1; : : : �n 2 C,

nXi=1

nXj=1

�i�jai;j � 0:

Remarks:1) We can consider the matrix A as a linear operator on the Hilbert spaceCn. Then A is positive if and only if hAx; xi � 0 for every vector x =(�1; : : : �n) 2 Cn : In fact

hAx; xi =nXi=1

nXj=1

�i�jai;j:

2) Complex matrix A is positive if and only if A = A�; that is the matrixA is Hermitian, and every eigenvalue � of A satis�es � � 0: If A = A� andevery eigenvalue � of A satis�es � > 0 then A is called strictly positive(written A > 0). See [1, Chapter III, Section 9 ].

De�nition 16 Let S be a set and let K : S � S ! C be a function oftwo variables. Then K is called a kernel function provided that for everyn and for every choice of n distinct points, f�1; : : : ; �ng � S; the matrix�K��i; �j

�� 0.

Proposition 171) If K1 and K2 are kernel functions from S � S to C then K1 +K2 is akernel function.

2) If K is a kernel function and f : S ! C is any function, then K3 :=f (�)K (�; �) f (�) is kernel function.

Proof.1) Since K1 and K2 are kernel functions we have

hK1u; ui � 0 and hK2u; ui � 0:

By de�nition of K1 +K2; we have (K1 +K2)u = K1u+K2u; hence

h(K1 +K2)u; ui = hK1u+K2u; ui = hK1u; ui+ hK2u; ui � 0:

1.4. CHARACTERIZATION OF REPRODUCING KERNELS 15

2)

nXi=1

nXj=1

�i�jK3

��i; �j

�=

nXi=1

nXj=1

�i�jf (�i)K��i; �j

�f��j�

=nXi=1

nXj=1

(�if (�i))��jf

��j��K��i; �j

�=

nXi=1

nXj=1

(�if (�i))��jf

��j��K��i; �j

�� 0:

Proposition 18 Let S be a set and let H be a RKHS on S with reproducingkernel K. Then K is a kernel function.

Proof. Let f�1; : : : ; �ng � S and �1; : : : �n 2 C. Then

nXi=1

nXj=1


�=

nXi=1

nXj=1

�i�jK��; �j�; K (�; �i)

�=

nXi=1

nXj=1

�jK

��; �j�; �iK (�; �i)

�=

*nXj=1

�jK��; �j�;

nXi=1

�iK (�; �i)+

=

*nXj=1

�jk�j;

nXi=1

�ik�i

+=

nXj=1

�jk�j

2

� 0:

Thus we have shown that the Gram matrix of the reproducing kernel ispositive.

Although the above proposition is elementary, it has converse which isquite deep.

Theorem 19 (Moore-Aronszajn) Let S be a set and let K : S � S ! C bea kernel function. Then there exists a reproducing kernel Hilbert space H offunctions on S such that K is the reproducing kernel of H.


Proof. We �rst give an outline of the proof.The proof proceeds in three steps.The �rst step is to consider the set of all �nite linear combinations of k� (�) :=

K (�; �) that is, W is the set of linear combinations W :=nPm

j=1 �jk�j (�)o

for any m 2 N, and de�ne a sesquilinear map6, ' : W �W ! C.The second step is to show that ' is indeed an inner product on W .The third step is to complete the pre-Hilbert spaceW to into a Hilbert spaceH of functions and to check that the completionH has the RKHS properties.

Now we begin with the detailed proof.For each � 2 S, set k� (�) := K (�; �) and let W be the space spanned bythe set fk� j � 2 S g ; that is, W is the set of linear combinations W :=nPm

j=1 �jk�j (�)ofor any m 2 N.

We will show that there is well-de�ned sesquilinear map, ' : W �W ! C,de�ned by

'

0@ m0X

j=1

�jk�0j;mXi=1

�ik�i

1A :=m0X

i=1

mXj=1

�j�iK��i; �

0

j

�;

where �j; �i are scalars.To see that ' is well-de�ned onW we must show that if f =

Pmj=1 �jk�j (�) is

identically zero as a function on S then ' (f; w) = ' (w; f) = 0 for anyw 2 W: Since W is spanned by the functions k� it is enough to show that' (f; k�) = ' (k�; f) = 0: But by de�nition,

' (f; k�) = '

mXj=1

�jk�j; k�

!=

mXj=1

�jK��; �j

�=

mXj=1

�jk�j (�) = f (�) = 0:

6A sesquilinear map on a complex vector space V is a map � : V � V ! C, that islinear in one argument and conjugate-linear in the other.Speci�cally a map � : V � V ! C is sesquilinear if:

�(x+ y; u+ v) = � (x; u) + � (x; v) + � (y; u) + � (y; v)� (�x; y) = �� (x; y)� (x; �y) = �� (x; y) :


Similarly,

' (k�; f) = '

k�;

mXj=1

�jk�j

!=

mXj=1

�iK (�i; �) =

mXj=1

�iK (�i; �) =

mXj=1

�iK (�; �i) = f (�) = 0:

Moreover, if ' (f; w) = 0 for every w 2 W then taking w = k�; it follows that' (f; k�) = f (�) = 0:Thus ' (f; w) = 0 for every w 2 W if and only if f is identically zero as afunction on S:Next, we check that ' satis�es all the properties of an inner product on W :

Let x :=Pm

0

j=1 �jk�0j; y :=

Pmi=1 �ik�i and z :=

Pm00

l=1 lk�00l(a)

' (x+ y; z) = '

0@ m0X

j=1

�jk�0j+

mXi=1

�ik�i ;m00X

l=1

lk�00l

1A= '

0@ m0X

j=1

�jk�0j;m00X

l=1

lk�00l

1A+ '

0@ mXi=1

�ik�i ;m00X

l=1

lk�00l

1A= ' (x; z) + ' (y; z) :

(b) Let

� 2 C:' (�x; y) = '

0@� m0X

j=1

�jk�0j;mXi=1

�ik�i

1A = '

0@ m0X

j=1

��jk�0j;

mXi=1

�ik�i

1A

=m0X

i=1

mXj=1

��i�iK��i; �

0

j

�= �

m0X

i=1

mXj=1

�i�iK��i; �

0

j

�= �' (x; y) :

c)

' (x; y) = '

0@ m0Xj=1

�jk�0j;

mXi=1

�ik�i

1A =

m0X

i=1

mXj=1

�i�iK��i; �

0

j

�

=

m0X

i=1

mXj=1

�i�iK��i; �

0j

�=

m0X

i=1

mXj=1

�i�iK��0

j; �i;�= ' (y; x) :


d) Since K is positive then

' (x; x) =mXi=1

mXj=1


�� 0:

So far we have shown that ' de�nes a semi-de�nite inner product. The normis de�ned as:

mXj=1

�jk�j

=vuut'

mXj=1

�jk�j ;

mXi=1

�ik�i

!=

vuut mXi=1

mXj=1


�:

Fix f 2 W and let w 2 W; then by the same proof as for the Cauchy-Schwartz inequality,j' (f; w)j � kfk kwk. If ' (f; f) = 0 then kfk = 0 andit follows that ' (f; w) = 0: Similarly j' (w; f)j � kfk kwk ; so ' (f; f) = 0implies ' (w; f) = 0. Thus one sees that ' (f; w) = 0 and ' (w; f) = 0 forall w 2 W if and only if ' (f; f) = 0: Hence, we see that ' (f; f) = 0 if andonly if f is the function which is identically zero. By establishing the factthat ' (f; f) = 0 if and only if f = 0, we have shown that ' de�nes an innerproduct on W:

Therefore the semi-de�nite norm Pm

j=1 �jk�j

=qPmi=1

Pmj=1 �i�jK

��i; �j

�is actually a norm andW is a pre-Hilbert space, which can be completed intoa Hilbert space H of functions.Next, one has to check that the reproducing property carries over to thecompletion. To this end, let h 2 H and let ffng be a Cauchy sequencewhich converges to h. Clearly, ' (fn; k�) = fn (�) : Hence, jfn (�)� fm (�)j =j' (fn; k�)� ' (fm; k�)j = j' (fn � fm; k�)j :Now by Cauchy-Schwartz inequality,

j' (fn � fm; k�)j � kfn � fmk kk�k = kfn � fmkpK (�; �):

It follows that

jfn (�)� fm (�)j �pK (�; �) kfn � fmk ;

which implies that ffng converges pointwise, so we are allowed to de�neh (�) := limn!1 fn (�) : This value is independent of the particular Cauchysequence chosen.

Finally, if we let h�; �i := ' (�; �) denote the inner product de�ned on H;then for h as above (since inner product is continuous), we have hh; k�i =limn hfn; k�i = limn ' (fn; k�) = limn fn (�) = h (�) : Thus the Dirac evalua-tion functional �� is de�ned for every � 2 S:


Now we show that �� is bounded. �� (h) = h (�) = hh; k�i : By Cauchy-Schwartz inequality

j�� (h)j = jh (�)j = jhh; k�ij � khk kk�k = khkpK (�; �):

It was shown that any other Hilbert space with the same reproducing kernelhas to be isometrically isomorphic7.We conclude that H is simply the completion of the linear span of the rep-resenters fk� j � 2 S g endowed with the inner product*X

j

�jk�0j;Xi

�ik�i

+:=Xi

Xj

�i�iK��i; �

0

j

�:

7Hilbert spaces H1 and H2 de�ned over the same �eld are isomerically isomorphic ifthere is a bijective linear opeartor T : H1 ! H2 such that for all x; y 2 H1, hTx; Tyi =hx; yi :


Chapter 2

Instances of RKHS

Reproducing kernel Hilbert spaces occur and are useful in many di¤erent con-text and branches of Mathematics. To name a few, RKHS occur in approx-imation and interpolation, statistical (machine) learning, complex analyticfunction theory, system theory and wavelet theory.In this chapter, we give several instances of RKHS. We begin with three

spaces of analytic functions in one complex variable, which turn out to havea reproducing kernel Hilbert space structure. The �rst class of spaces willbe the Bergman spaces1 on open domains in C: The second class of spaces isthe Hardy spaces. The third space will be the Paley-Wiener space P whichis the space of all entire functions of exponential type at most . Then wewill continue with examples of RKHS related to wavelet theory.

2.1 Bergman spaces

Bergman spaces can be considered with weights. Since the treatment ofweighted Bergman spaces causes no extra e¤ort, we will begin with weightedBergman spaces.Let G be a complex domain, i.e. open and connected subset of C, and

1The spaces are named for Stephan Bergman (1895 - 1977). It is probably becauseof S. Bergman�s book �The Kernel Function and Conformal Mapping�(AMS,1950) thatprompted the use of the term �Bergman spaces�, although the study of such spaces hadbegun much earlier.Stefan Bergman entered Berlin University in 1921. He was in�uenced by von Mises at thistime and for the rest of his career. He worked on potential theory applied to electricalengineering. Bergman used the theory of integral equations as developed by Schmidt andHilbert. Bergman is best known for his kernel function which he invented in 1922 whileat Berlin University, now known as the Bergman kernel. He is also known for applicationsof the kernel function to conformal mappings. In fact he spent most of his life developingproperties and applications of his kernel function, as well as those of its associated metric.

21

22 CHAPTER 2. INSTANCES OF RKHS

w : G! R a positive measurable function that is locally bounded below2 bya strictly positive constant. We call such w a regular weight on G:L2w (G) is the Hilbert space of functions which are square integrable with theregular weight w on the domain G; i.e.,

f (z) 2 L2w (G) if

ZZG

jf (z)j2w (z) dxdy <1:

De�nition 20 The weighted Bergman space A2w (G) is the subspace of L2w (G)

of analytic functions on G such that

kfk2A2w(G) :=ZZG

jf (z)j2w (z) dxdy <1: (2.1)

The Bergman space for the trivial weight w = 1 will be denoted by A2 (G) :

We can de�ne an inner product on A2w (G) by

hf; gi :=ZZG

f (z) g (z)w (z) dxdy: (2.2)

We will show later that A2w (G) is indeed a Hilbert space. Now we show thatthe evaluation functional �z : A2w (G)! C is bounded.

Proposition 21 Let �0 2 G; such that the closed diskDr (�0) := f� 2 C j j� � �0j � rgis contained in G; then for every f 2 A2w (G) ;

jf (�0)j �1

cp�rkfkA2w(G) (2.3)

Proof. By Gauss mean value theorem [7, Chapter 5.6], for any r � � > 0;we have

f (�0) =1

2�

Z 2�

0

f��0 + �ei�

�d�: (2.4)

We replace 12�by 1

�r2

�R r0�d��and obtain

f (�0) =1

�r2

Z r

0

Z 2�

0

f��0 + �ei�

��d�d�

=1

�r2

ZZDr(�0)

f (z) dxdy:

2A function f de�ned on some topological space X with real or complex values is calledlocally bounded, if for any x0 in X there exists a neighborhood A of x0 such that f(A) isa bounded set, that is, for some number Mx0 > 0 one has jf(x)j �M for all x in A.A function is locally bounded below by a positive number if for any x0 in X there exists

a neighborhood A of x0 such that f(x) �Mx0 > 0 for all x 2 A.

2.1. BERGMAN SPACES 23

By Cauchy-Schwartz inequality,

jf (�0)j2 =

��1

�r2

ZZDr(�0)

f (z) dxdy

��2

�

0B@ 1

�r2

ZZDr(�0)

jf (z)j2 dxdy

1CA0B@ 1

�r2

ZZDr(�0)

dxdy

1CA=

1

�r2

ZZDr(�0)

jf (z)j2 dxdy:

Since w is a regular weight, we can �nd a constant c2 > 0 such that c2 < w (z)for all z 2 Dr (�0) :

jf (�0)j2 � 1

c2�r2

ZZDr(�0)

jf (z)j2wdxdy

=1

c2�r2kfk2A2w(G) :

This proves the Bergman�s estimate (2.3).

The Bergman�s estimate (2.3) implies that the evaluation functional �z isbounded for every z 2 G:

Now we show that A2w (G) is complete.

Proposition 22 A2w (G) is a Hilbert space.

Proof. Let ffng be a Cauchy sequence in A2w (G) :It follows from the Bergman�s estimate (2.3) that for any compact subset Kof G, we can �nd a constant MK such that

jf (z)j �MK kfkA2w(G) for all z 2 K: (2.5)

By inequality (2.5) we have

jfn (z)� fm (z)j �MK kfn � fmkA2w(G) (2.6)

where z 2 K a compact subset of G:Inequality (2.6) implies that the sequence ffn (z)g converges pointwise to afunction f (z) on G, and moreover, the sequence ffn (z)g converges uniformly


to f (z) on every compact subset of G.By Weierstrass theorem [7, Theorem 10.5], the limit f (z) is analytic functionon G:Since L2w (G) is complete then the Cauchy sequence ffn (z)g converges to afunction g (z) 2 L2w (G) : Therefore, there is a subsequence ffnk (z)g whichconverges3 almost everywhere to g (z) [8]: This implies that f (z) = g (z)almost everywhere on G; i.e., f (z) 2 L2 (G) and since it is analytic, f (z) 2A2w (G) :We conclude that fn (z)! f (z) 2 A2w (G) which implies thatA2w (G)is complete.

It follows from propositions 21 and 22 that A2w (G) is indeed a reproducingkernel Hilbert space.

2.1.1 The weighted Bergman spaces on the unit disk

A special useful case of weighted Bergman space is the space A2� (D) ; wherethe domain G is D, the open unit disk, and the regular weight, wD is de�nedas,

wD (z) := (�+ 1)�1� jzj2

��; �1 < � 2 R:

A function f (z) is in A2� (D) if f (z) is analytic in D and

(�+ 1)

ZZD

jf (z)j2�1� jzj2

��dxdy <1: (2.7)

When � = 0 we get the "unweighted" Bergman space A2 (D) on the unitopen disk.Now we �nd the reproducing kernel of A2� (D).

In view of theorem 9, we �nd �rst an orthonormal basis of A2� (D) :Since the measure

�1� jzj2

��dxdy is rotation invariant, we suspect that

the monomials fzng1n=0 are an orthogonal system.

hzm; zni = (�+ 1)ZDzmzn

�1� jzj2

��dxdy:

By substituting z = rei�; we get

hzm; zni = (�+ 1)

Z 2�

0

Z 1

0

rm+nei(m�n)��1� r2

��rdrd� (2.8)

=

�0 if m 6= n

2� (�+ 1)R 10r2m+1 (1� r2)� dr if m = n

(2.9)

3fn ! f in Lp implies that fn ! f in measure. fn ! f in measure implies that ffnghas a subsequence which converges a.e.

2.1. BERGMAN SPACES 25

Now we substitute u = r2 in order to evaluate hzm; zmi ;

2 (�+ 1)

Z 1

0

r2m+1�1� r2

��dr = (�+ 1)

Z 1

0

um (1� u)� du

= (�+ 1)B (m+ 1; �+ 1) = (�+ 1)� (m+ 1)� (�+ 1)

� (m+ �+ 2)

=� (m+ 1)� (�+ 2)

� (m+ �+ 2)

hzm; zmi = �� (m+ 1)� (�+ 2)

� (m+ �+ 2)= �

�m+ �+ 1

m

��1(2.10)

Thus the monomials�q

(m+�+1m )p�

zm�1n=0

form an orthonormal basis forA2� (D).

We denote um (z) :=

q(m+�+1m )p�

zm; and by Theorem 9 the kernel is given by

K(�)D (�; �) =

1Xn=1

un (�)un (�) =1Xn=0

q�n+�+1

n

�p�

�n

q�n+�+1

n

�p�

�n (2.11)

=1

�

1Xn=0

�n+ �+ 1

n

��n�n =

1

�

�1

1� ��

��+2(2.12)

(for the last inequality, one can check the Taylor expansion4 of (1� z)��).

2.1.2 The weighted Bergman spaces on the upper halfplane

Another useful special case of weighted Bergman space is A2� (�+), where the

domain G is �+, the upper half plane, and the regular weight is de�ned as

w�+ (z) := (Im (z))� � 1 < � 2 R. (2.13)

A function f (z) is in A2� (�+) if f (z) is analytic in �+ andZZ

�+

jf (z)j2 (Im (z))� dxdy <1: (2.14)

4The Taylor expansion of 1(1�z)� is

1 + �z + �(�+1)2 z2 + � � �+ �(�+1)��(�+n�1)

n! zn + � � �


2.2 Hardy spaces

2.2.1 Hardy space on the unit disk, H2 (D)

Hardy space H2 (D) on the unit disk is de�ned as the space of the analyticfunctions f : D! C such that

kfk2H2(D) := sup0<r<1

1

2�

Z �

��

��f �rei��2 d� <1: (2.15)

From the work of Fatou, a student of Lebesgue, and his successors oneknows that each function5 f (z) in H2 (D) has an associated boundary func-tion, de�ned almost everywhere on @D in terms of radial limit6, i.e., lim

r!1�f (rz) :

Because of this, H2 (D) can be identi�ed with a subspace of L2 ([��; �]). Thissubspace consists of the functions in L2 ([��; �]) whose Fourier coe¢ cientswith negative indices vanish. A function in H2 (D) can be reconstructedfrom its boundary function by means of the Poisson integral, or the Cauchyintegral. On our way to show that H2 (D) is indeed a RKHS, we will showthat the space H2 (D) can be described as the space of analytic functionsin D whose Taylor coe¢ cients at the origin are square summable and theorthonormal basis of H2 (D) are the monomials fzmg1m=0.The rest of this subsection is devoted to proving these statements.

Proposition 23 The space H2 (D) together with the norm k�kH2(D) is a re-producing kernel Hilbert space with kernel K (z; w) = 1

1�wz ; more precisely,if f 2 H2 (D) ; then

f (z) = sup0<r<1

1

2�

Z �

��K�z; rei�

�f�rei��d�; for all z 2 D: (2.16)

Proof. Let f be an analytic function from D to C, then it can be expandedas a power series f (z) =

P1n=0 anz

n; and for 0 < jzj < 1; the power series

5Actually this statement is true for Hardy spaces Hp; 0 < p � 1; where the norm is

de�ned as kfkp := limr!1�

n12�

R ��f �rei��po 1

p

: For 1 < p � 1, Hp is Banach space and

for p = 2, H2 is Hilbert space.6Also, in terms of nontangential limit.

2.2. HARDY SPACES 27

converges uniformly. So we have for 0 < r < 1;

M2 (f ; r) : =1

2�

Z �

��

��f �rei��2 d� = 1

2�

Z �

��f�rei��f (rei�)d�

=1

2�

Z �

��

1Xn=0

anrnei�n

! 1Xm=0

amrme�i�m

!d�

=1

2�

Z �

��

1Xm;n=0

anamrn+mei�(n�m)

!d�

=1X

m;n=0

anam1

2�

Z �

��rn+mei�(n�m)d�

=1Xn=0

janj2 r2n: (2.17)

The interchange of the order of summation and integration is justi�ed bythe uniform convergence of the power series and we used the orthogonalityproperty of feintg ;

1

2�

Z �

��ei�(n�m)d� = �n;m: (2.18)

It follows from (2.17) that the function M2 (f ; r) is monotonically increasingfunction of r in [0; 1) ; so we can replace the supremum over (0; 1) with lim

r!1r<1

.

IfP1

n=0 janj2 <1 then

P1n=0 janj

2 r2n <1 for 0 � r < 1 and it follows that12�

R ��

��f �rei��2 d� <1 for 0 � r < 1:

Thus sup0<r<112�

R ��

��f �rei��2 d� < 1 which implies that f 2 H2 (D) :On the other hand, if f 2 H2 (D) then M2 (f ; r) is bounded monotonicallyincreasing function of r: By the monotone convergence principle, the functionM2 (f ; r) has a limit when r ! 1� and this implies that

P1n=0 janj

2 <1: Sowe have established that f 2 H2 (D) if and only if

P1n=0 janj

2 converges.Now we want to check if the space H2 (D) meets the conditions of RKHS.

(i) The functions in H2 (D) are well de�ned on the open disk D, so H2 (D) islinear space of functions on D with pointwise algebraic operations.(ii) We can de�ne a map L : H2 (D)! l2 (Z) by letting L (f) := (a0; a1; : : :).This map is linear and preserves the inner product. Hence we see thatH2 (D)can be identi�ed with the Hilbert space, l2 (Z+) where Z+ := f0; 1; 2; : : :g thenatural numbers, N together with 0, and it follows that H2 (D) is a Hilbertspace.(iii) The set polynomials (in z) is a dense subset of H2 (D) and the set of


monomials pm (z) := zm is an orthonormal system, since

hzm; zni = sup0<r<1

1

2�

Z �

��rmeim�rne�in�d�

= sup0<r<1

rm+n

2�

Z �

��ei(m�n)�d� = sup

0<r<1rm+n�m;n = �m;n

so we conclude that fzmg1m=0 is an orthonormal basis of H2 (D).Motivated by Theorem 9 we set

K (�; �) =1Xn=0

pn (�) pn (�) =1Xn=0

�n�n=

1

1� ��(2.19)

This function is called the Szegö7 kernel on the disk.The representer k� : D! C, k� (z) := K (z; �) is contained in H2 (D) since

kk�k2H2(D) �P1

n=0

��n��2 = 11�j�j2 <1; and we �nd for f 2 H

2 (D)

hf; k�i =* 1Xn=0

anzn;

1Xn=0

�nzn

+=

1Xn=0

an�n=

1Xn=0

an�n = f (�) (2.20)

By Cauchy-Schwartz inequality

jf (�)j � kk�kH2(D) kfk �1q

1� j�j2kfk (2.21)

This implies the continuity of k�; the point evaluation functional. ThusH2 (D) is a RKHS on D. By Theorem 9, the function K (�; �) is equal tothe reproducing kernel of H2 (D).

De�nition 24 L2 (@D) denotes the Hilbert space of square-integrable Lebesgue-measurable complex-valued functions on the unit circle @D with pointwisealgebraic operations and inner product de�ned as

hf; gi :=Z �

��f�ei��g (ei�)d� (2.22)

L2 (@D) is isometrically isomorphic to the space L2 ([��; �]) :7Gabor Szegö (January 20, 1895 �August 7, 1985) was a Jewish Hungarian mathe-

matician.Szegö was born in Kunhegyes, Hungary. His most important work centered on Toeplitz

matrices and orthogonal polynomials. He was a teacher of John von Neumann. His bookon orthogonal polynomials published in 1939 is a classic within its �eld and is still widelyused as a reference in polynomial theory.


In view of this correspondence between L2 (@D) and L2 ([��; �]) and sincen1p2�einxois an orthonormal basis in L2 ([��; �]) ; we can say that,

Proposition 25 The sequence fzng1n=�1 is an orthonormal basis in L2 (@D).

We shall write bf (n) for the n-th Fourier coe¢ cient of f with respect tothe orthonormal basis fzng1n=�1.Thus for f 2 L2 (@D) ;

bf (n) = hf; zni

=

Z �

��f�ei��e�in�d� (2.23)

If f 2 L2 (@D) then

f (z) =1X

n=�1

bf (n) zn (2.24)

where equality is in the sense of the L2 (@D) norm.Now we ask which functions in L2 (@D) might have extensions to analyticfunctions in the open unit disk D. We may take the subspace of function forwhich bf (n) = 0 for n = �1;�2; : : : :De�nition 26 H2 (@D) is the closed subspacen

f 2 L2 (@D)�� bf (n) = 0 for n < 0

o(2.25)

of L2 (@D).

The basic properties of H2 (D) are summarized in the following theorem:

Theorem 27 (a) Let f be an analytic function from D to C, with the powerseries expansion

f (z) =1Xn=0

anzn z 2 D, (2.26)

then f 2 H2 (D) if and only ifP1

n=0 janj2 <1;

in that case kfkH2(D) =�P1

n=0 janj2 1

2 .(b) If f 2 H2 (D) then(i) f has radial limits ef �ei�� := lim

r!1�f�rei��at almost all points of D;

(ii) the function ef �ei�� is in L2 (@D) ;


(iii) the n-th Fourier coe¢ cient of ef is an if n � 0 and 0 if n < 0;(iv)

limr!1�

ef �ei�� f �rei�� L2(@D)

= 0; (2.27)

c) The mapping f ! ef is an isometry of H2 (D) onto H2 (@D).

Proof. We have proved (a) in Proposition 23.Now suppose f 2 H2 (D). For 0 < s < 1, de�ne a function fs on @D by

fs�ei��:= f

�sei��=

1Xn=0

ansnein� (2.28)

Since f 2 H2 (D) thenP1

n=0 janj2 <1: As we mentioned before,

n1p2�ein�o

is an orthonormal system in L2 (@D) ; therefore the Riesz-Fischer theorem8

ensures that there exists a function g 2 L2 (@D) such that bg (n) = an forn � 0 and bg (n) = 0 for n < 0 i.e., g 2 H2 (@D) by our de�nition.The Fourier coe¢ cients of g � fs are (1� sn) an for n � 0: By Parseval�stheorem,

kg � fsk2L2(@D) =1Xn=1

(1� sn)2 janj2 (2.29)

For 0 < s � 1, (1� sn)2 < 1; henceP1

n=1 (1� sn)2 janj2 is uniformly

convergent by Weierstrass�M -test. It follows that

lims!1

1Xn=1

(1� sn)2 janj2 =1Xn=1

lims!1

(1� sn)2 janj2 = 0

We conclude thatlims!1kg � fskL2(@D) = 0: (2.30)

Now we make a small digression and introduce the Poisson kernel Pr (�) ;

Pr (�) :=1� r2

1� 2r cos � + r2; 0 � r < 1 and � 2 R. (2.31)

For any �xed s 2 (0; 1) ; f (sz) is analytic in the disk D�0; 1

s

�: Let z 2 D and

z := rei�:We prove now one of the properties of Poisson integral of harmonicfunction that

f (sz) =1

2�

Z �

��Pr (� � t) fs

�eit�dt (2.32)

8The Riesz-Fiseher Theorem: Let f'1; '2; : : :g be an orthomrmd system in a Hilbertspace H. Suppose (�1; �2; : : :) 2 l2 (i.e.,

P1i=1 j�ij

2< 1). Then �i = hx; 'ii ; i = 1; 2; : : :

for some x 2 H.


Since for any �xed s 2 (0; 1) ; f (sz) is analytic in the disk D�0; 1

s

�; the

Cauchy�s theorem implies equations 2.33 and 2.34:

f (z0) =1

2�i

ZD(0;s)

f (z)

z � z0dz; (2.33)

where z0 = srei�:

1

2�i

ZD(0;s)

f (z)

z � z1dz = 0 (2.34)

where z1 = 1rsei�:

We subtract 2.34 from 2.33 and obtain 2.32:

f (z0)� 0 = f�srei�

�=

1

2�i

ZD(0;s)

f (z)

�1

z � z0� 1

z � z1

�dz

=1

2�

Z �

��f�seit�� 1

seit � srei� �1

seit � 1rsei�

�seitdt

=1

2�

Z �

��f�seit�� 1

eit � rei� �r

reit � ei�

�eitdt

=1

2�

Z �

��f�seit� 1

1� 2r cos (� � t) + r2dt

=1

2�

Z �

��f�seit�Pr (� � t) dt:

Now we evaluate the di¤erence between f (sz) and 12�

R �� Pr (� � t) g (e

it) dt;

and apply Schwartz inequality on 12�

R �� Pr (� � t) (fs (e

it)� g (eit)) dt;�� 12�Z �

��Pr (� � t)

�fs�eit�� g

�eit��dt

��=

��fs �eit�� g �eit� ; Pr (� � t)��L2(@D)� kfs � gk kPr (� � t)k = kfs � gk :

lims!112�

R �� Pr (� � t) (fs (e

it)� g (eit)) dt = 0; which implies that

lims!1

f (sz) = f (z) =1

2�

Z �

��Pr (� � t) g

�eit�dt (2.35)

So f is the Poisson integral of g:We use a property of the Poisson integral about radial limits (without proof):If u 2 L1 (@D) and U

�rei��:= 1

2�

R �� Pr (� � t)u (e

it) dt then

limr!1�

U�rei��= u

�ei��

(2.36)


almost everywhere.By 2.36 and 2.35 we get,

limr!1�

f�rei��= g

�eit�

(2.37)

It follows that f has radial limits at almost all points of @D and they areequal to g (eit) : So we denote g (eit) by ef �ei�� and this proves part (b).Now we prove that the mapping f ! ef is unitary mapping of H2 (D)

onto H2 (@D).From lims!1

ef � fs L2(@D)

= 0 we conclude that ef � lims!1 fs

2L2(@D)

=

0; so that ef

L2(@D)= klims!1 fskL2(@D) : But klims!1 fskL2(@D) = lims!1

kfskL2(@D) = lims!112�

R ��

��f �sei��2 d� = kfkH2(D) : This shows that the

mapping f ! ef is an isometry. To complete the proof we will show now thatthe mapping is onto. Suppose g 2 H2 (@D), thus bg (n) = 0 for all n < 0; andput

f (z) :=1Xn=0

bg (n) zn: (2.38)

Then f 2 H2 (D) by (a) and the proof of (b) shows that ef = g:

2.2.2 Hardy space on the upper half plane, H2 (�+)

Hardy space H2 (�+) on the upper half plane is de�ned as the space of theanalytic functions f : �+! C such that

kfk2H2(�+) := sup0<y

ZRjf (x+ iy)j2 dx <1: (2.39)

Proposition 28 The space H2 (�+) with the norm k�k2H2(�+) is a reproduc-ing kernel Hilbert space with kernel K�+ (z; w) =

14�

2iz�w :

If f 2 H2 (�+) ; then

f (z) = sup0<y

ZRK�+ (z; x+ iy) f (x+ iy) dx; for all z 2 �+: (2.40)

Proof. Let h 2 L2 (R+), i.e., a function in L2 (R) which vanishes on (�1; 0).The complex Fourier transform of h is de�ned as

f (z) :=1p2�

ZR+h (t) eiztdt; z 2 �+: (2.41)


If z 2 �+ then jeiztj =��ei(x+iy)t�� = jeixtj je�ytj = e�yt; which shows that

jf (z)j = 1p2�

��ZR+h (t) eiztdt

�� 1p

2�

Z 1

0

jh (t)j��eizt�� dt = 1p

2�

Z 1

0

jh (t)j e�ytdt

� 1p2�kh (t)kL2(R+)

sZ 1

0

e�2ytdt

=1p2�2y

kh (t)kL2(R+) <1:

so the integral in (2.41) exists as a Lebesgue integral. A straightforwardapplication of Fubini and Morera�s Theorems9 shows that f (z) is analyticfunction on �+ :Z�

f (z) dz =

Z�

�ZR+h (t) eiztdt

�dz =

ZR+h (t)

0@Z�

eiz!dz

1A dt =

ZR+h (t)�0dt = 0;

where � is triangular path in �+:We may rewrite (2.41),

f (x+ iy) :=1p2�

ZR+

�h (t) e�yt

�eixtdt; for all y > 0:

By Plancherel Theorem (Theorem 41),ZRjf (x+ iy)j2 dx =

ZR+

��h (t) e�yt��2 dtZRjf (x+ iy)j2 dx =

ZR+jh (t)j2 e�2ytdt � kh (t)kL2(R+) ;

9Morera�s Theorem: If a (single-valued) function f (z) of a complex variable z in adomain D is continuous and if its integral over any closed recti�able contour � � D isequal to zero, that is, if

(�)Z�

f (z) dz = 0

then f (z) is an analytic function in D:The conditions of Morera�s theorem can be weakened by restricting the requirement on

vanishing integrals (*) to those taken over the boundary of any triangle � = @� that iscompactly contained in D, i.e, such that � � � � D.Morera�s theorem is an (incomplete) converse of the Cauchy integral theorem and is oneof the basic theorems in the theory of analytic functions. See [7] Theorem 5.12


thus f (z) 2 H2 (�+).

kfk2H2(�+) = sup0<y

ZR+jh (t)j2 e�2ytdt = lim

y!0y>0

ZR+jh (t)j2 e�2ytdt

By the Dominated Convergence Theorem,

limy!0y>0

ZR+jh (t)j2 e�2ytdt =

ZR+limy!0y>0

jh (t)j2 e�2ytdt =ZR+jh (t)j2 dt = kh (t)kL2(R+) :

Thus the mapping �;

� : L2�R+�! H2

��+�;

de�ned by

f (z) :=1p2�

ZR+h (t) eiztdt; z 2 �+

is an isometry.The mapping � is onto. This is the consequence of a classical theorem by Pa-ley and Wiener (see the Appendix for a proof of the �rst Paley-Wiener theo-rem), which states that for every function f in the Hardy spaceH2 (�+) thereexists a function h (t) 2 L2 (R+) such that f (z) := 1p

2�

RR+ h (t) e

iztdt; z 2�+:Thus � is unitary mapping of L2 (R+) onto H2 (�+). So we established thatH2 (�+) is indeed a Hilbert space.We still have to prove that H2 (�+) is a reproducing kernel Hilbert space.To this end we set

z (r) :=1p2�e�irz; for z 2 �+: (2.42)

Then for a �xed z 2 �+, the function z (r) is in L2 (R+) sinceZ 1

0

�� 1p2�e�irz

��2 dr = 1

2�

Z 1

0

��e�irz��2 dr = 1

2�

Z 1

0

e�2rdr =1

4�:

(�h) (z) =1p2�

ZR+h (t) eiztdt = hh; ziL2(R+) : (2.43)

Since � is unitary we have

hh; ziL2(R+) = h�h;� ziH2(�+) (2.44)


which implies that � z is the reproducing element of H2 (�+) and that the

evaluation functional is bounded. This proves thatH2 (�+) is indeed a RKHSwith the reproducing kernel

K (z; w) = h� w;� ziH2(�+) = h w; ziL2(R+)

=

Z 1

0

�1p2�e�irw

��1p2�eirz�dr

=1

2�

Z 1

0

e�ir(w�z)dr =1

2�

i

z � w:

We now prove a corollary of the Paley-Wiener theorem, which we willlater use in Example 37 of Wavelet transform.

Corollary 29 Given f 2 H2 (�+) and setting fy (x) := f (x+ iy) ; thenfy 2 L2 (R) for all y > 0 and the limit f0 := limy!0 fy exists in the normof L2 (R) : Moreover, the mapping f 7! f0 de�nes an isometric embeddingH2 (�+) ,! L2 (R).

Proof. Given f 2 H2 (�+) : Let h 2 L2 (R+) be ��1 (f) ; i.e.,

f (z) :=1p2�

ZR+h (t) eiztdt; z 2 �+:

Set f0 (x) := 1p2�

RR+ h (r) e

irxdr (in the sense of the Fourier transform in L2;not to be read pointwise in x).

(fy � f0) (x) =1p2�

ZR+

�e�ry � 1

�h (r) eirxdr:

By Plancherel�s Theorem,

kfy � f0k2L2(R) =ZR+

��e�ry � 1�h (r)��2 dr = ZR+

�1� e�ry

�2 jh (r)j2 drand by the Dominated Convergence Theorem,

limy!0+

kfy � f0k2L2(R) = limy!0+

ZR+

�1� e�ry

�2 jh (r)j2 dr=

ZR+

limy!0+

�1� e�ry

�2 jh (r)j2 dr = 0:It follows that f0 := limy!0 fy exists in the norm of L2 (R).

kfkH2(�+) = khkL2(R+) = kf0kL2(R) (2.45)

hence the mapping f 7! f0 de�nes an isometric embedding of H2 (�+) intoL2 (R).


2.3 Bandlimited signals and Paley-Wiener spaces

2.3.1 The space of bandlimited functions

A function f (t) 2 L2 (R) is called bandlimited if its Fourier transform bf (!) :=F (f (t)) has compact support, i.e., bf (!) = 0 for j!j > , which means thatbf (!) vanishes outside the band [�;]. Let us denote this space of ban-dlimited functions by B:To show that B is a RKHS we have to show �rst that B is a closed subspaceof L2 (R) :The set of functionsnbf 2 L2 (R) ��supp bf � [�;] o

is a closed subspace of L2 (R) ; since it is a subspace of L2 (R) and everyCauchy sequence of elements of this subspace converges to an element in thesubspace.By Plancherel�s theorem10, the Fourier transform F is a unitary oper-

ator11 on L2 (R) and the inverse Fourier transform, F�1; can be obtainedby letting (F�1g) (x) = (Fg) (�x) for all g 2 L2 (R). In other words, everyfunction in L2 (R) can be written as the Fourier transform bf (!) of a functionf (t) 2 L2 (R) and bf (!)

L2(R)= kf (t)kL2(R) :

B is the image of the subspacenbf 2 L2 (R) ��supp bf � [�;] o under the

inverse Fourier operatorF�1: It follows thatB is a closed subspace of L2 (R),since a unitary operator (in this case F�1) carries a closed subspace to aclosed subspace. A closed subspace of a Hilbert space is a Hilbert space;thus B is a Hilbert space.In the next subsection we will use a fundamental theorem in Fourier analysis,the Paley-Wiener theorem, to show that B is a reproducing kernel Hilbertspace.

2.3.2 Paley-Wiener spaces

Paley-Wiener theorem, characterizes those functions in L2 (R) ; which areFourier transforms of functions that vanish outside of a given interval [�;].10See Apendix for discussion on Fourier operator on L2 (R) :11A linear isometry which maps H onto H is called a unitary operator.An operator U , which has the property that kUxk = kxk for all x 2 H is called an

isometry.

2.3. BANDLIMITED SIGNALS AND PALEY-WIENER SPACES 37

We are going to show that if bf (!) vanishes outside of the interval [�;] ;then its inverse Fourier transform f (t) is (equal almost everywhere to) therestriction of an entire function f (z) to the real axis.It is easily veri�ed that f (z) is of exponential type12 at most .To see this, let bf (!) 2 L2 [�;] ; then

f (z) :=

Z�

bf (!) eiz!d! (2.46)

is such a function. A straightforward application of Fubini and Morera�sTheorem shows that f (z) is entire,

Z�

f (z) dz =

Z�

0@ Z�

bf (!) eiz!d!1A dz =

Z�

bf (!)0@Z

�

eiz!dz

1A d! =

Z�

bf (!)�0d! = 0;where � is any triangular domain.The estimate

jf (z)j �Z�

�� bf (!)�� eiz!�� d! = Z�

�� bf (!)�� e� Im(z)!d!� ejIm(z)j

Z�

�� bf (!)�� d!shows that f (z) is of exponential type at most ; and Plancherel�s theoremshows that1Z�1

jf (t)j2 dt = 2�Z�

�� bf (!)��2 d! <1.Paley and Wiener showed that the converse is also true! Any entire

function of exponential type at most ; which is square integrable on the realaxis is the complex Fourier transform of a function which vanishes outside ofthe interval [�;].This is the celebrated Paley-Wiener theorem. For a proof see 3.4.

De�nition 30 Paley-Wiener space P is the space of all entire functions ofexponential type at most ; which are square integrable on the real axis.

12An entire function F (z) is said to be of exponential type if = lim supjzj!1

logjF (z)jjzj <1:


Each function in the space B can be extended to an analytic functionin P, by equation (2.46); and conversely, by Paley-Wiener theorem, therestriction of a function in P is a function in B.

We will show that the set P of all such entire functions is a Hilbert spacein the metric of L2 (R). This Hilbert space is called Paley-Wiener space.Clearly, P is a vector space under pointwise addition and scalar multiplica-tion; it is also an inner product space with respect to the inner product

hf (z) ; g (z)i :=Z 1

�1f (x) g (x)dx:

The norm of f (z) 2 P is de�ned as

kf (z)kP :=sZ 1

�1jf (x)j2 dx

By Plancherel�s theorem, and by the Paley-Wiener theorem it follows thatP is a separable Hilbert space, isometrically isomorphic to L2 ([�;]) ;

kf (z)kP = kf (x)kL2(R) = bf (!)

L2(R)= bf (!)

L2([�;]):

The isomorphism between P and L2 ([�;]) is important for many appli-cation, e.g.,in the next subsection, we shall use it to obtain the Shannon-Whittaker sampling theorem.If f belongs to P then it has the representation

f (z) =1

2�

Z

�� (!) eiz!d!

with � in L2 ([�;]) :Now we arrive at a useful estimate of jf (z)j :

f (x+ iy) =1

2�

Z

�� (!) ei(x+iy)!d! =

1

2�

Z

�� (!) e�y!eix!d!

jf (x+ iy)j � 1

2�

Z

�j� (!)j ejyj!dt

� ejyj1

2�

Z

�j� (!)j d! � ejyj

1

2�

sZ

�j� (!)j2 d!

= ejyjr1

2�

s1

2�

Z

�j� (!)j2 d! =

r1

2�ejyj kfk � ejyj kfk :


jf (x+ iy)j � ejyj kfk : (2.47)

Equipped with 2.47 and with Paley-Wiener theorem, we can show thatP is indeed a reproducing kernel Hilbert space.

Theorem 31 The Paley-Wiener space P is a reproducing kernel Hilbertspace of entire functions. Its reproducing kernel is given by

K (z; w) =sin (z � w)� (z � w) ; (2.48)

and the integral representation

f (z) =

Z 1

�1f (t)

sin (t� z)� (t� z) dt (2.49)

is valid for every function f 2 P.

Proof. Inequality 2.47 shows that the Dirac evaluation functionals on Pare bounded linear functionals. Thus P is a RKHS.If f (z) 2 P; then by Paley-Wiener theorem,

f (w) =

Z

�� (t) eiwtdt =

Z

�� (t) eiwtdt =

� (t) ; eiwt

�L2([�;]) ;

for some function � in L2 ([�;]) and all values of w 2 C.But the Fourier transform operatorF is a unitary operator between L2 ([�;])and P, hence

� (t) ; eiwt�L2([�;]) =

F (� (t)) ;F

�eiwt��:

F (� (t)) =1p2�f (z) and

F�eiwt�=

1p2�

Z

�eiwte�iztdt =

i

z � weitwe�itz

��=2 sin (z � w)p2� (z � w)

� (t) ; eiwt

�L2([�;]) =

F (� (t)) ;F

�eiwt��

=

�1p2�f (z) ;

2 sin (z � w)p2� (z � w)

�=

�f (z) ;

sin (z � w)� (z � w)

�


We obtain

f (w) =

�f (z) ;

sin (z � w)� (z � w)

�which implies that K (z; w) ; the kernel of P is sin(z�w)�(z�w) :

The fact, Paley-Wiener space P is a RKHS with reproducing kernelsin(z�w)�(z�w) implies that B is a RKHS with reproducing kernel given by

K (x;w) =sin (x� w)� (x� w) :

2.3.3 Sampling of band limited functions

Let P� be the Paley-Wiener RKHS of all entire functions of exponential typeat most � that are square integrable on the real axis. We have shown beforethat P� is isometrically isomorphic to L2 ([��; �]).If f (z) 2 P� then by Paley-Wiener theorem,it has the representation

f (z) =1

2�

Z �

�� (t) eiztdt;

with � in L2 ([��; �]).It follows from the fact that � 2 L2 ([��; �]) that � (t) has Fourier series

expansion,

� (t) =1X

n=�1ane

int

an =1

2�

Z �

�� (t) e�intdt

with equality in L2 ([��; �]) norm.We see that an = f (�n) so that

� (t) =1X

n=�1f (n) e�int

Plancherel�s theorem shows that

kfk2 =Z 1

�1jf (x)j2 dx = 1

2�

Z �

��j� (t)j2 dt = k�k2


Parseval�s identity applied to � gives

1Xn=�1

jf (n)j2 = k�k2

thus we obtain

kfk2 =1X

n=�1jf (n)j2

The setn

1p2�einto1n=�1

is an orthonormal basis for L2 ([��; �]) : Since theFourier operator is unitary, then taking the Fourier transform of feintg1n=�1gives us the set

nsin�(z�n)�(z�n)

o1n=�1

which is an orthonormal basis for P�.Accordingly, each function in P� has a unique Fourier series expansion of

the form

f (z) :=1X

n=�1cnsin � (z � n)� (z � n)

withP1

n=�1 jcnj2 < 1: The convergence of the series is in the metric of

P�. But we will now deduce that convergence in the metric of P� im-plies uniform convergence in each horizontal strip on the complex plane,say fz = x+ iy j jyj < M g : This is immediate consequence of the estimate2.47:Denote by fk (z) :=

Pkn=�k cn

sin�(z�n)�(z�n) then fk (z) 2 P� and

jf (z)� fk (z)j � e�M kf (z)� fk (z)kP�

Now converges in P� implies that for every " > 0, there is a number N suchthat for k > N ; kf (z)� fk (z)kP < ": It follows that for all z in the stripfz j jyj < M g and for k > N we have jf (z)� fk (z)j < e�M" , which impliesuniform convergence in the strip.�

f (z) ;sin � (z �m)� (z �m)

�=

Z 1

�1

1Xn=�1

cnsin � (x� n)� (x� n)

!sin � (x�m)� (x�m) dx

Since the convergence is uniform, we can interchange integration with sum-mation�

f (z) ;sin � (z �m)� (z �m)

�=

1Xn=�1

cn

Z 1

�1

sin � (x� n)� (x� n)

sin � (x�m)� (x�m) dx = cm


By property of the kernel function,�f (z) ;

sin � (z �m)� (z �m)

�= f (m)

thus we conclude that cm = f (m) and thereby we obtain the cardinal seriesfor f :

f (z) =1X

n=�1f (n)

sin � (z � n)� (z � n)

= sin �z

1Xn=�1

(�1)n f (n)

� (z � n) : (2.50)

A function in P� can be reconstructed from its values at the integers. Thecardinal series for band limited function were �rst introduced by Whittaker[1915].Equation (2.50) is known in the mathematical theory of communication

as the Shannon-Whittaker sampling theorem.The cardinal series for band limited function in P is

f (z) =1X

n=�1f�n�

� sin �z � n �

��z � n �

�2.4 Wavelet theory

Wavelet theory also gives us an instance of RKHS. We will show that a spaceof wavelet transforms is RKHS.I cannot give here a comprehensive introduction to wavelet theory, but I

cannot avoid telling few introductory words. Wavelet analysis can be de�nedas an alternative to the classical windowed Fourier analysis. In many applica-tions, given a signal f (t) ; one is interested in the frequency content locally intime. One way to get this time-frequency analysis is to use widowed Fouriertransform, in which the signal f (t) is multiplied by a "window" functionh (t) which is di¤erent from zero at the time interval one is interested in.In the wavelet case, one is comparing several magni�cations of this signal,with distinct resolutions. The building blocks of a windowed Fourier analysisare sines and cosines (waves) multiplied by a sliding window. They are usu-ally referred to as time-frequency atoms. In a wavelet analysis, the window isalready oscillating and is called a "mother wavelet". This mother wavelet isno longer multiplied by sines or cosines. Instead it is translated and dilated

2.4. WAVELET THEORY 43

by arbitrary translations and dilations. That is the way the mother waveletgenerates the other wavelets which are the building blocks of a wavelet analy-sis. These dilations are precisely the magni�cations we alluded to, and thebuilding blocks are called time-scale atoms.Fourier analysis, windowed Fourier analysis, and wavelet analysis are

based on an identical recipe. In the three cases, the analysis of a func-tion amounts to computing all the correlations between this function andthe time-frequency or time-scale atoms which are being used. The synthesisis obtained exactly as if these building blocks were an orthonormal basis.We treat here only the continuous wavelet transform which is not so

useful in practical application. In practice, one use discrete values of thedilation and translation, thus getting the discrete wavelet transform and itsderivatives which have found applications in signal processing.

2.4.1 The continuous wavelet transform

We begin by introducing the continuous wavelet transform.For 2 L2 (R) ; let the dilated-translated version of be de�ned by

a;b (t) :=1pjaj

�t� ba

�; 0 6= a 2 R; b 2 R: (2.51)

The mapping W , acting on f 2 L2 (R) de�ned by (W f) (a; b) :=f; a;b

�is often referred to as the continuous wavelet transform induced by : (Thefunction is sometimes called �mother wavelet�).

De�nition 32 The continuous wavelet transform of a function f 2 L2 (R)with respect to is de�ned by

(W f) (a; b) :=f; a;b

�=

ZRf (t) a;b (t)dt (2.52)

The purpose of the factor jaj�12 in the de�nition above is to normalize the

functions of the doubly-indexed sequence� a;b

, i.e.

a;b 2L2(R) := ZR

�� a;b (t)��2 dt = ZR

1

jaj

�� t� ba��2 dt

Substitute y = t�ba. If a > 0 thenZ

R

1

jaj

�� t� ba��2 dt = Z 1

�1

a

jaj j (y)j2 dy =

ZRj (y)j2 dy;


If a < 0 thenZR

1

jaj

�� t� ba��2 dt = Z �1

1

a

jaj j (y)j2 dy =

Z 1

�1

�ajaj j (y)j

2 dy =

ZRj (y)j2 dy:

So a;b 2L2(R) := RR �� a;b (t)��2 dt = RR 1jaj

�� t�ba

��2 dt=RR j (y)j

2 dy = k k2L2(R)(2.53)

It is required thatk k2L2(R) = 1; (2.54)

thus� a;b

is a sequence of functions of unit norm.

The Fourier transform of a;b is denoted by b a;b;b a;b (!) := 1p

2�

ZR a;be

�i!tdt (2.55)

Now we impose additional condition on . 2 L2 (R) is a continuum wavelet if it satis�es the admissibility

condition:C := 2�

ZR

��b (!)��2 d!j!j <1: (2.56)

Remark: If, in addition, 2 L1 (R) ; then the integrability condition 2.56implies that Z

R (t) dt = 0: (2.57)

Indeed, if 2 L1 (R) then its Fourier transform is continuous on R andin particular at ! = 0 where b (0) = 1p

2�

RR (t) dt: Suppose

b (0) 6= 0

thenRR

��b (!)��2 d!j!j is divergent, contrary to condition 2.56, hence b (0) =1p2�

RR (t) dt = 0:

Wavelet transform would not have been useful at all unless a function canbe recovered from its wavelet transform. The next proposition shows how itis done.

Proposition 33 For all f; g 2 L2 (R) ;1Z�1

1Z�1

(W f) (a; b) (W g) (a; b)da db

a2= C hf; giL2(R) (2.58)


Proof. By de�nition of the continuous wavelet transform (2.52) and sincethe Fourier transform operator is unitary (Plancherel�s theorem) we obtain

(W f) (a; b) =f; a;b

�=F (f) ;F

� a;b

��=

D bf (!) ; b a;b (!)EBy the properties of Fourier transform,

b a;b = 1pjaj

\

�t� ba

�=pjaje�i!bb (a!)

so (W f) (a; b) can be rewritten as

(W f) (a; b) =D bf (!) ; b a;bE

=D bf (!) ; ae�i!bb (a!)E = 1Z

�1

bf (!)pjaje�i!bb (a!)d!=

1Z�1

bf (!)pjajb (a!)ei!bd!:Similarly for (W g) (a; b),

(W g) (a; b) =

1Z�1

bg (�)pjajb (a�)ei�bd�:

Now

1Z�1

bf (!)pjajb (a!)ei!bd! can be viewed as the inverse Fourier trans-form of

Fa (!) :=p2�pjaj bf (!) b (a!)

and similarly

1Z�1

bg (�)pjajb (a�)ei�bd� can be viewed as the inverse Fouriertransform of

Ga (�) :=p2�pjajbg (�) b (a�)


Now we rewrite

1Z�1

1Z�1

(W f) (a; b) (W g) (a; b)da dba2

using Fa (!) andGa (�) ;

1Z�1

1Z�1


a2

=

1Z�1

1Z�1

0@ 1Z�1

bf (!)pjajb (a!)ei!bd!1A0@ 1Z

�1

bg (�)pjajb (a�) e�i�bd�1A da db

a2

=

1Z�1

0@ 1Z�1

F�1 (Fa)F�1 (Ga) (b) db

1A da

a2

Again, by Plancherel�s theorem

1Z�1

1Z�1


a2

=

1Z�1

0@ 1Z�1

F�1 (Fa)F�1 (Ga) (b) db

1A da

a2

=

1Z�1

0@ 1Z�1

p2�pjaj bf (!) b (a!)p2�pjajbg (!) b (a!)d!

1A da

a2

= 2�

1Z�1

0@ 1Z�1

bf (!) bg (!) ��b (a!)��2 d!1A da

jaj

Now while we are whispering the "magic words" Fubini�s theorem, we change


the order of integration,

1Z�1

1Z�1


a2

= 2�

1Z�1

0@ 1Z�1

bf (!) bg (!) ��b (a!)��2 d!1A da

jaj

=

1Z�1

bf (!) bg (!)0@2� 1Z

�1

��b (a!)��2 dajaj1A d!

=

0@2� 1Z�1

��b (a!)��2 dajaj1A hf; giL2(R)

(we have used Plancherel�s theorem in the previous step).

A change of variable � = a! in 2�

1Z�1

��b (a!)��2 dajaj shows that

2�

1Z�1

��b (a!)��2 dajaj = 2�ZR

��b (!)��2 d!j!j = C :

and the proof is complete.

We will show in the sequel that equation (2.58) will lead us to the inversionformula

f (t) = C�1

1Z�1

1Z�1

(W f) (a; b) a;b (t)da db

a2(2.59)

to be read "in the weak sense". This formula shows how to reconstructa function from its wavelet transform and is also called "resolution of theidentity" formula.Now what does it mean convergence of the integral (2.59) in the weak

sense? Let us take a small digression and discuss weak convergence innormed linear spaces.


De�nition 34 A sequence fxng in a normed space X is said to be weaklyconvergent if there is an x 2 X such that for every f 2 X 0

(the dual space13

of X)limn!1

f (xn) = f (x) :

Here we are dealing with a Hilbert space, then the de�nition of weakconvergence becomes as follows:

De�nition 35 A sequence ffng in a Hilbert space H is said to be weaklyconvergent if there is an f 2 H such that for every g 2 H

limn!1

hfn; gi = hf; gi :

Now to relate this discussion with (2.59), we write

fn (t) :=1

C

nZ�n

nZ�n


a2:

limn!1

hfn; gi = limn!1

*1

C

nZ�n

nZ�n


a2; g

+

= limn!1

1

C

nZ�n

nZ�n

(W f) (a; b) a;b (t) ; g

� da dba2

= limn!1

1

C

nZ�n

nZ�n

(W f) (a; b)g; a;b (t)

�da dba2

= limn!1

1

C

nZ�n

nZ�n


a2

=1

C

1Z�1

1Z�1


a2:

The weak convergence of the integral implies limn!1 hfn; gi = hf; gi ; sowe conclude that the statement about weak convergence of the integral inequation (2.59) is actually restatement of equation (2.58) .

13The dual space X0is the space of all bounded linear functionals on X:


As a matter of fact, the inversion formula (2.59) was already known toCalderón [2].

If both f and are "analytic signals"14, i.e., if support of bf and supportof b are contained in the interval [0;1) ; then (W f) (a; b) = 0 if a < 0:To see this, bf (!) = 0 for ! < 0 implies that

(W f) (a; b) =D bf (!) ; b a;bE = 1Z

�1

bf (!)pjajb (a!)ei!bd!:=

1Z0

bf (!)pjajb (a!)ei!bd!; (2.60)

and Supp�b (!)� � [0;1) implies that b (a!) = 0 for a < 0 and ! > 0.

Hence, the inversion formula becomes in this case

f (t) = C�1

1Z0

1Z�1

(W f) (a; b) a;b (t) dbda

a2: (2.61)

14An analytic signal s(t) is a complex signal created by taking a signal x(t) and thenadding in quadrature its Hilbert transform HT (x (t)) ; i.e., x (t) + iHT (x (t)) ; where theHilbert transform is de�ned as

HT (x (t)) :=1

�t� x (t) =

Z 1

�1

x (t� �)��

d�

It is also called the pre-envelope of the real signal and is de�ned as follows

s (t) = x (t) + iHT (x (t))

= a (t) ei'(t)

where its modulus a(t) and phase derivative can serve as estimates for the amplitudeenvelope and instantaneous frequency of x(t).

By taking Fourier transform of both sides of bs (t) := bx (t) + i \HT (x (t)) = bx (!) +i (�i � sgn (!) bx (!)) :By rewriting, the well-known relation between spectra of a real signal and the associated

analytic signal results

bs (!) = � 2bx (!) if ! > 00 if ! < 0

Therefore, the signal spectrum of analytic signal should be restricted to positive fre-quencies.


As a special case of (2.58) we put g = f 2 L2 (R) and obtain

C�1

1Z�1

1Z�1

j(W f) (a; b)j2da db

a2=

ZRjf (x)j2 dx: (2.62)

An equivalent formulation of this fact is that the wavelet transform W

is an isometry from L2 (R) into L2�R2; 1

C

da dba2

�.

The Hilbert space L2�R2; 1

C

da dba2

�is the space of complex-valued functions

f : R� R! C which satisfyR1�1R1�1 jf (x; y)j

2 dxdyx2

<1:The norm of f in L2

�R2; 1

C

da dba2

�is de�ned as C�1

R1�1R1�1 jf (x; y)j

2 dxdyx2:

kW fkL2�R2; 1

C

da dba2

� = kfkL2(R) : (2.63)

The image W L2 (R) of L2 (R) under the wavelet transform W is a

closed subspace15, strictly contained in L2�R2; 1

C

da dba2

�: Thus W L

2 (R)is a Hilbert space which we will denote by H .Now we show that H is indeed a RKHS.

By Cauchy-Schwartz inequality,

j(W f) (a; b)j � kfk a;b = kfkL2(R) : (2.64)

Hence (and by 2.63), we have

j(W f) (a; b)j � kfkL2(R) = kW fkL2�R2; 1

C

da dba2

� (2.65)

which implies that the evaluation functional of H is bounded. It followsthat H is a reproducing kernel Hilbert space.To derive the kernel K

�a; b; a

0; b

0�of H ; we use the fact that W is an

isometry from L2 (R) onto H ;

(W f) (a; b) =f; a;b

�L2(R)

=W f;W a;b

�H

(2.66)

It follows that W a;b is the reproducing element (or the representer of theevaluation functional �(a;b)), hence the kernel is

K�a; b; a

0; b

0�=

W a0 ;b0 ;W a;b

�H

= a0 ;b0 ; a;b

�L2(R)

(2.67)

15An isometry T maps closed subspace to a closed subspace.


Example 36

Let (x) := 2p3��

14 (1� x2) e�x2

2 ; the "Mexican hat" function.

a;b (x) =2p3��

14

�1�

�xa� b�2�

e�(xa�b)

2

2 (2.68)

The kernel of the space L2�R2; 1

C

da dba2

�with "Mexican hat" analyzing

wavelet is,

K�a; b; a

0; b

0�=4

3��

12

Z 1

�1

�1�

�xa� b�2��

1�� xa0� b0

�2�e�

(xa�b)2+

�x

a0 �b

0�22 dx

Example 37

We elaborate a particular case mentioned in section 2.5 of Daubechiesbook [3]. In this particular example, the continuous wavelet transform car-ries a subspace of L2 (R) onto a reproducing kernel Hilbert space of analyticfunctions.We will show that the continuous wavelet transform with a Paul wavelet car-ries the space H2 of functions whose Fourier transform vanishes for negativefrequencies onto the Bergman space A21 (�

+) :

This example involves analyzing wavelets which are used in signal process-ing disciplines, namely the Paul wavelets.The Paul wavelets (analyzing wavelets) are de�ned as follows:

(m) (t) :=2mimm!p� (2m)!

1

(1� it)m+1=2mi2m+1m!p� (2m)!

1

(t+ i)m+1(2.69)

For m = 1 the Paul wavelet is

(t) =�2ip2�

1

(t+ i)2:

Let F denote the Fourier transform

(Ff) (!) := 1p2�

Z 1

�1f (t) e�it!dt:

In order to �nd the Fourier transform of the Paul wavelet, we need to evaluate


the integralR1�1

1(t+z)�+1

e�it!dt.By complex integration method (residue theorem) we obtainZ 1

�1

1

(t+ z)�+1e�it!dt = (�i)�+1 2�

�!!�ei!z; for ! Im (z) > 0: (2.70)

The Fourier transform b (!) of the analyzing wavelet (t) isb (!) � (F ) (!) := 1p

2�

Z 1

�1

�2ip2�

1

(t+ i)2e�it!dt

=�2i2�

Z 1

�1

1

(t+ i)2e�it!dt =

�2i2�

(�2�)!e�!

=

�2i!e�! if ! > 00 if ! � 0 : (2.71)

Clearly satis�es the admissibility condition (2.56), with

C := 2�

ZR

��b (!)��2 d!j!j = 2�Z 1

0

4!e�2!d! = 2�: (2.72)

a;�b (t) : =1pjaj

�t+ b

a

�=

1pa

�2ip2�

1�t+ba+ i�2 = �2ip

2�

a32

(t+ b+ ia)2�[ a;�b

�(!) =

�2ia

32!e�a!ei!b if ! > 00 if ! � 0 : (2.73)

Now we want to �nd the image W H2, where H2 is the space of functions f

such that bf (!) = 0 for ! � 0:Recall from subsection 2.2.2 that the space H2 is a closed subspace of L2 (R)and by Corollary 29, it is an isometric embedding of the Hardy spaceH2 (�+) intoL2 (R).More precisely, we will show that the image of H2 under the mapping

Ber f (t) :=1p2�a�

32 (W f) (a;�b) (2.74)

is a reproducing kernel Hilbert space (actually a weighted Bergman space)and that in this case, Ber is a unitary map between H2 and the Bergmanspace A21 (�

+).The mappingBer is called theBergman transform or the analytic wavelettransform.The wavelet transform W f of a function f 2 H2; is by de�nition


(W f) (a;�b) =f; a;�b

�By Plancherel�s formula,f; a;�b

�=

D bf;[ a;�bE=

D bf; 2ia 32!e�a!ei!bE = �2ia 32 Z 1

0

bf (!)!e�a!e�i!bd!= �2ia 32

Z 1

0

bf (!)!e�i(b�ia)!ed! = �2ia 32 Z 1

0

bf (!)!e�iz!ed!;where z := b+ ia:

Ber f (t) : =1p2�a�

32 (W f) (a;�b) =

1p2�a�

32

��2ia 32

Z 1

0

bf (!)!e�iz!ed!�=�2ip2�

Z 1

0

bf (!)!e�iz!ed! := G (z) : (2.75)

We may rewrite the expression for (W f) (a;�b) ;

(W f) (a;�b) =p2�a

32G (z) : (2.76)

A straightforward application of Fubini and Morera�s Theorem shows thatG (z) analytic in the upper half plane �+ (Im z > 0).To see that G (z) is indeed in A21 (�

+) we will show that1Z0

1Z�1

a jG (b+ ia)j2 da db <1:

Equation (2.62) for analytic signals becomes

C�1

1Z0

1Z�1

j(W f) (a; b)j2da db

a2=

ZRjf (x)j2 dx: (2.77)

We substitute equations (2.75) in (2.77) and obtain

1 >

ZRjf (x)j2 dx = C�1

1Z0

1Z�1

j(W f) (a;�b)j2dbda

a2

=1

2�

1Z0

1Z�1

��p2�a 32G (b+ ia)��2 dbda

a2

=

1Z0

1Z�1

a jG (b+ ia)j2 dbda: (2.78)


The last equation implies that the mappingBer f (t) := 1p2�a�

32 (W f) (a;�b)

can be interpreted as an isometry from H2 into the weighted Bergman spaceA21 (�

+) of all analytic functions on the upper half plane square integrablewith respect to the measure a da db = Im zd (Im z) d (Re z) :

Nowwe wish to show that the isometryBer f (t) := 1p2�a�

32 (W f) (a;�b)

is onto, and therefore is a unitary map.We will prove that any function in this Bergman space is the image of a func-tion in H2 under the mappingW , the wavelet transform with this particular .To this end, we �nd an orthonormal basis fSng of H2 and then show thatthe image of fSng under the Bergman transform is an orthonormal basis ofA21 (�

+) : Now if Ber (H2) ; the image ofH2, contains an orthonormal basis ofA21 (�

+) ; then A21 (�+) � Ber (H2) which implies that Ber is onto A21 (�

+) :Our plan is summarized here:

L2 (R+) F ! H2 (�+)lima!0+�! H2 Ber�! A21 (�

+)T ! A21 (D)

1p(n+1)(n+2)

ln (x)F ! Sn (z)

limy!0+�! Sn (t)Ber�! �2i

p(n+1)(n+2)p2�

n (z) !p(n+1)(n+2)p2�

wn:

(2.79)The generalized Laguerre polynomials are de�ned by the Rodrigues formula

L(k)n (x) :=exx�k

n!

dn

dxn�e�xxk+n

�(2.80)

and this implies

L(k)n (x) =nXj=0

�n+ k

n� j

�(�x)j

j!: (2.81)

The Laguerre functions are de�ned as

l(k)n (x) =

�e�

x2x

k2L

(k)n (x) for x � 0

0 for x < 0(2.82)

It is well known that the generalized Laguerre functionsnl(k)n

o1n=0

are or-

thogonal basis for L2 (R+) :By the Paley-Wiener theorem 43, the Fourier transform is a unitary mappingbetween H2 (�+) and L2 (R+) : Therefore, the analytic Fourier transform of


the functionsnl(k)n

oconstitute an orthogonal basis for the space H2 (�+) :

Let us denote the analytic Fourier transform of l(k)n by

S(k)n (z) := F l(k)n :=1p2�

Z 1

�1l(k)n (x) e�ixzdx: (2.83)

From now on, we will continue this discussion with k = 2.We use the fact Z 1

0

L(2)n (x)xe�xdx = (n+ 1) (n+ 2) ; (2.84)

to normalize ln in order to obtain obtain an orthonormal basis and write

Sn (z) :=1p

(n+ 1) (n+ 2)F�l(2)n�=

1p2� (n+ 1) (n+ 2)

Z 1

�1l(2)n (x) e�ixzdx:

(2.85)

BerSn (z) =�2ip2�

Z 1

0

cSn (!)!e�iz!ed!=�2ip2�

Z 1

0

1p(n+ 1) (n+ 2)

e�!2 !L(2)n (!)!e�iz!d!

=�2ip

2�p(n+ 1) (n+ 2)

Z 1

0

e�!2 !e!!�2

n!

�dn

d!n�e�!!n+2

��!e�iz!d!

=�2ip

2�p(n+ 1) (n+ 2)n!

Z 1

0

e�(iz�12)! dn

d!n��e�!!n+2

��d!

=�2ip

2�p(n+ 1) (n+ 2)n!

L�dn

d!n��e�!!n+2

��iz � 1

2

�Laplace transform 16 of n-th derivative is given by (see Appendix, equations(3.8)),L�dnfdtn

�= snL (f)� sn�1f (0+)� sn�2f 0 (0+)� � � � � f (n�1) (0+) :

so we get,

BerSn (z) =�2ip

2�p(n+ 1) (n+ 2)n!

�iz � 1

2

�nL�e�!!n+2

��iz � 1

2

�;

By the property of the Laplace transform L (eatf (t)) = L (f) (s� a) we goon and write,


p(n+ 1) (n+ 2)

�iz � 1

2

iz + 12

�n�iz +

1

2

��316 L

�dn

dtn

�e�ttn+2

��= snL

�e�ttn+2

�= snL

�tn+2

�(s+ 1) = (n+ 2)!

�ss+1

�n(s+ 1)

�3:


We de�ne fn (z)g,

n (z) :=

�iz � 1

2

iz + 12

�n�iz +

1

2

��3We will show later that fn (z)g1n=0 constitute an orthogonal basis of theweighted Bergman space A21 (�

+) :We �nally express BerSn (z) as follows,


p(n+ 1) (n+ 2)n (z) :

Now we want to �nd a Hilbert space isomorphism

T : A21��+�! A21 (D) : (2.86)

The biholomorphic map ' : D!�+ is de�ned as

z = x+ iy = ' (w) :=i

2

1 + w

1� w

y =1

2

1� u2 � v2

(1� u)2 + v2

'0(w) =

i

(1� w)2

The next step is to check that the mapping T : A21 (�+)! A21 (D) de�ned as

Tf (z) :=i

2

1

(1� w)3f

�i

2

w + 1

w � 1

�; (2.87)

is a unitary mapping from A21 (�+) onto A21 (D). It is clear from de�nition

of T that it is a linear mapping. To see that T is an isometry we verify nowthat kTf (z)kA21(D) = kfkA21(�+) : We note that when z = ' (w) is viewed asR- linear map from R2 to R2 then the Jacobian17 of the transformation of17The Jacobian is @(x;y)

@(u;v) =@x@u

@y@v �

@y@u

@x@v

After substituting Cauchy-Riemann equations @x@u =@y@v ;

@x@v = �

@y@u , we obtain

@(x;y)@(u;v) =�

@x@u

�2+�@x@v

�2: The complex derivative of z = ' (w) is d'

dw = @x@u + i

@y@u . Once again by

Cauchy-Riemann equations, d'dw =@x@u + i

@y@u =

@x@u � i

@x@v�� d'dw

��2 = ��@x@u � i@x@v��2 = �@x@u

�2+

�@x

@v

�2=@ (x; y)

@ (u; v)(2.88)


x and y with respect to u and v; is equal to��'0��2. Thus the transformation

formula shows thatZZG2

F (z) dxdy =

ZZG1

(F � ') (w) @ (x; y)@ (u; v)

dudv

=

ZZG1

(F � ') (w)��'0(w)��2 dudv; (2.89)

for all F 2 L1 (G2; dxdy) :On one hand,

kTf (z)kA21(D) : = 2

ZZD

jTf (z)j2�1� jwj2

�dudv

2

ZZD

�� i2 1

(1� w)3f

�i

2

w + 1

w � 1

��2 �1� jwj2� dudv=

ZZD

��f � i2w + 1w � 1

��2 1

2 j1� wj6�1� jwj2

�dudv

=

ZZD

��f � i2w + 1w � 1

��2 1� u2 � v2

2�(1� u)2 + v

�3dudvand on the other end,

kfkA21(�+) =

ZZ�+

jf (z)j2 ydxdy =ZZD

(f � ') (w)��'0(w)��2 dudv

=

ZZD

��f � i2w + 1w � 1

��2 12 1� u2 � v2(1� u)2 + v2

�� i

(1� w)2

��2 dudv=

ZZD

��f � i2w + 1w � 1

��2 1� u2 � v2

2�(1� u)2 + v

�3dudvso we conclude that kTf (z)kA21(D) = kfkA21(�+). To show that T is onto wechoose g 2 A21 (D) and de�ne a mapping R : A21 (D)! A21 (�

+) by setting

w = '�1 (z) =2z � i2z + i�

'�1�0(z) =

4i

(2z + i)2


so that

Rg (w) :=2

i

�1� 2z � i

2z + i

�3g

�2z � i2z + i

�Clearly, TRg = g; so if we show that Rg 2 A21 (�+) then we are done.To this end, we restate that g 2 A21 (D) ; which implies that

2

ZZD

jg (w)j2�1� jwj2

�dudv <1

but

2

ZZD

jg (w)j2�1� jwj2

�dudv

= 2

ZZ�+

��g �'�1 (z)��2 1� ��2z � i2z + i

��2!��'�1�0 (z)��2 dxdy

= 2

ZZ�+

��g�2z � i2z + i

��2 1�

��2z � i2z + i

��2!

16

j2z + ij4dxdy

= 2

ZZ�+

��2i�1� 2z � i

2z + i

�3g

�2z � i2z + i

��21

4

��1� 2z � i2z + i

��6 1�

��2z � i2z + i

��2!

16

j2z + ij4dxdy

=

ZZ�+

��2i�1� 2z � i

2z + i

�3g

�2z � i2z + i

��21

2

��1� 2z � i2z + i

��6 1�

��2z � i2z + i

��2!

16

j2z + ij4dxdy

=1

2

ZZ�+

��g �'�1 (z)��2 ydxdy:We conclude that ZZ

�+

��g �'�1 (z)��2 ydxdy <1;which implies that Rg 2 A21 (�

+) and this completes the proof that T isunitary.We return to our main proof and �nd the image of (BerSn) (a;�b) under the


mapping T :

T (BerSn) (a;�b) = T

�2ip2�

p(n+ 1) (n+ 2)

�iz � 1

2

iz + 12

�n�iz +

1

2

��3!=

i

2

1

(1� w)3�2ip2�

p(n+ 1) (n+ 2)wn (1� w)3

=

p(n+ 1) (n+ 2)p

2�wn: (2.90)

We recognize immediately (see subsection 2.1.1 on weighted Bergman spaceson the unit disk) that

fT (BerSn) (a;�b)g =(p

(n+ 1) (n+ 2)p2�

wn

)1n=0

constitute an orthonormal basis of the weighted Bergman space A21 (D). Sincethe mapping T is a unitary mapping from A21 (�

+) to A21 (D), we concludethat (

�2ip(n+ 1) (n+ 2)p

2�n (z)

)1n=0

constitute an orthonormal basis of A21 (�+). This fact implies that the range

of the Bergman transform Ber (H2) contains a basis of A21 (�+). It follows

thatA21��+�� Ber

�H2�

and we are done.

2.4.2 Calderon�s reproducing formula

Remark: This section is not related to RKHS and can be skipped.

The inversion formula (2.61)

f (t) = C�1

1Z�1

1Z�1

(W f) (a; b) a;b (t) dbda

a2

is sometimes called in the literature, the "resolution of the identity".

We may consider the integral C�1

1Z�1

1Z�1

(W (�)) (a; b) a;b (t) dbdaa2 as an

identity operator acting on L2 (R) mapping a function f 2 L2 (R) to itself.


To get Calderon�s reproducing formula (as appears in several textbooks)we introduce the functions 'a and e'a :

'a (y) : =1

a

�1

ay

�(2.91)

e'a (y) : =1

a

��1ay

�(2.92)

then the wavelet transform with respect to of f (t) becomes a convolutionof the function f (t) with

pjaje'a;

(W f) (a; b) =

Z 1

�1f (t) e'a (b� t)pjajdt =pjajf � e'a (2.93)

Convolution of two functions f and g which are measurable in R is de�nedby

(f � g) (x) :=Z 1

�1f (u) g (x� u) du x 2 R: (2.94)

This notation of convolution is now used to present the inversion formula(2.59) in a form which is standard in the literature:

f (t) = C�1

Z 1

�1

Z 1

�1(W f) (a; b) a;b

da db

a2

= C�1

Z 1

�1

Z 1

�1

�pjajf � e'a� jaj� 1

2

�t� ba

�da db

a2

= C�1

Z 1

�1

�Z 1

�1(f � e'a)'a (t� b) db� daa

ActuallyR1�1 (f � e'a)'a (t� b) db is the convolution of f � e'a with 'a; so it

follows that

f (t) = C�1

Z 1

�1f � e'a � 'adaa (2.95)

Equation (2.95) is sometimes called the resolution of the identity [3, Chapter2, 2.4.2].

Many authors of papers and books on wavelet theory consider Calderónas one of the forefathers of the theory and they refer to his paper [2]They say that what is called � Calderón�s resolution of the identity�, or

alternatively called � Calderón�s reproducing formula, the continuous ver-sion�is presented in [2]. This paper seems to me very di¢ cult and deep. Its


main concern is interpolation theory in Banach spaces.I searched Calderón�s paper in to �nd a reminiscent of the resolution of theidentity in the context of continuous wavelet transform.

I tried to follow Calderón�s development, in Section 14, pages 125,126 and127 in his paper, with respect to a simple example.Let B be L2 (R) ; n = 1; � yu := u (x� y) ; u (x) 2 L2 (R) :

F (t) = Tu := t�nZR(� yu)'

�1

ty

�dy (2.96)

=

ZRu (x� y) 1

t'

�1

ty

�dy (2.97)

Let ' (y) be in�nitely di¤erentiable, symmetric function in R with momentsof order less than k equal to zero.

F (t) : = Tu :=

ZRu (x� y) 1

t'

�1

ty

�dy (2.98)

= u ��1

t'

�1

ty

��(2.99)

Now I (Tu; u) in Calderón�s paper section 14.2 becomes (for n = 1);

I (Tu; u) = u � 2 +Z 1

1

�ZR� yF

�1

t

� 1 (ty) dy

�dt (2.100)

= u � 2 +Z 1

1

�ZR� y (u � (t' (ty))) 1 (ty) dy

�dt(2.101)

If we de�ne�t (y) := t' (ty)

and choose 1 = '

then we get

I (Tu; u) = u � 2 +Z 1

1

u � �t � �tdt

t: (2.102)

Calderón claims in his paper that there are functions 2 and 1 such that

u = I (Tu; u) (2.103)

i.e.,

u = u � 2 +Z 1

1

u � �t � �tdt

t(2.104)


The last formula has some resemblance to "Calderon�s reproducing formula"as it appears in the wavelet literature18.

18I am still loking for a book or a paper which explains in detail how Calderón�s 42 yearsold paper relates to the continuous wavelet transform.

Chapter 3

Appendix: The Fouriertransform

3.1 Fourier transform on L1 (R)De�nition 38 If f 2 L1 (R), then the Fourier transform of f , denoted bybf; is a complex-valued function de�ned on R and is given by

F (f (t)) := bf (!) := 1p2�

ZRf (t) e�i!tdt: (3.1)

3.2 Properties of the Fourier transform

Proposition 39 Let f 2 L1 (R) : Then(i) Continuity. ! 7�! bf (!) is a continuous function.(ii) Contraction.

bf 1� kfkL1(R) :

(iii) Linearity. \�1f1 + �2f2 = �1 bf1+�2 bf2 for any complex constants �1; �2.(iv) Translation and phase factor. \f (t� a) = e�i!a bf; deibtf = bf (! � b) :(v) Convolution. bf1 bf2 is the Fourier transform of (f1 � f2) (t) := RR f1 (t� u) f2 (u) du:(vi) Dilation. \f (�t) = 1

�bf �!

�

�Proof.(i) Let !n ! !; then fn (t) := f (t) e�i!nt is a sequence of measurable func-tions on R which converges pointwise to f (t) e�i!t on R. It is clear thatjfn (t)j � jf (t) e�i!tj = jf (t)j : Since f 2 L1 (R) then jf (t)j is a non-negativeintegrable function which dominates the sequence ffn (t)g : By the Domi-

63

64 CHAPTER 3. APPENDIX: THE FOURIER TRANSFORM

nated Convergence Theorem [9, p. 42],

limn!1

ZRf (t) e�i!ntdt = lim

n!1

ZRlimn!1

f (t) e�i!ntdt =

ZRf (t) e�i!tdt:

For every sequence f!ng ; !n ! ! implies bf (!n) ! bf (!), hence bf (!) is acontinuous function on R.

(ii)�� bf (!)�� RR jf (t) e�i!tj dt = RR jf (t)j dt := kfkL1(R) :

The Fourier transform is a continuous map from L1 (R) to the bounded con-tinuous functions on R.

(iii) \�1f1 + �2f2 :=1p2�

RR (�1f1 (t) + �2f2 (t)) e

�i!tdt =

= 1p2�

RR �1f1 (t) e

�i!tdt+ 1p2�

RR �2f2 (t) e

�i!tdt =

= �11p2�

RR f1 (t) e

�i!tdt�1 bf1 + �21p2�

RR f2 (t) e

�i!tdt = �1 bf1 + �2 bf2.(iv) \f (t� a) = 1p

2�

RR f (t� a) e

�i!tdt = 1p2�

RR f (u) e

�i!(u+a)dt

= e�i!a 1p2�

RR f (u) e

�i!udt = e�i!a bf:(v) First we show that the convolution of two L1 (R) exists. Let f1; f2 2L1 (R) ; then

RR2 jf1 (u) f2 (v)j dudv =

�RR jf1 (u)j dv

� �RR jf2 (v)j dv

�< 1:

By Fubini theorem1, (we substituted v = x� u)

1 >

ZR2jf1 (u) f2 (v)j dudv =

ZR2jf1 (u) f2 (x� u)j dxdu

=

ZRjf1 (u)j

�ZRjf2 (x� u)j dx

�du =

ZR

�ZRjf1 (u) f2 (x� u)j du

�dx:

SinceRR

�RR jf1 (u) f2 (x� u)j du

�dx �

RR

��RR f1 (u) f2 (x� u) du

�� dx Thuswe have shown that (f1 � f2) (t) 2 L1 (R) and since

RR

�RR jf1 (u) f2 (x� u)j du

�dx �R

R

��RR f1 (u) f2 (x� u) du

�� dx it follows that kf1 � f2kL1(R) � kf1kL1(R) kf2kL1(R).1Fubini�s theorem states that if

RA�B jf (x; y)j dxdy <1; the integral being taken with

respect to a product measure on the space over A�B where A and B are �-�nite measurespaces, thenRA

�RBf (x; y) dy

�dx =

RB

�RAf (x; y) dx

�dy =

RA�B f (x; y) dxdy

the �rst two integrals being iterated integrals, and the third being an integral withrespect to a product measure.

3.3. FOURIER TRANSFORM OPERATOR ON L2 (R) 65

Now we can compute

bf1 (!) bf2 (!) =1

2�

ZRf (u) e�i!udu

ZRf (v) e�i!vdv

=1

2�

ZR2f (u) f (v) e�i!(u+v)dudv

=1

2�

ZR2f (x� v) f (v) e�i!xdxdv

=1p2�

ZR

�1p2�

ZRf (x� v) f (v) dv

�e�i!xdx

= \(f1 � f2). (3.2)

(vi)

\f (�t) =1p2�

ZRf (�t) e�i!tdt =

1p2�

ZRf (u) e�i

!�udu

�

=1

�bf �!

�

�:

3.3 Fourier transform operator on L2 (R)

The fact that the Lebesgue measure of R is in�nite complicates the subjectof Fourier transform on L2 (R). The system fei!tg!2R is not orthonormalbasis in L2 (R), actually ei!t =2 L2 (R) : We cannot decompose functions inL2 (R) by functions which are not in L2 (R) ; so we cannot use Hilbert spacetechniques. If we want to have Fourier transforms for functions in L2 (R) ;we have to �nd a new way to de�ne them.We start with kind of Parseval�s theorem for continuous functions with

compact support (for a proof, see [12, Chapter VI, Lemma 3.1]:

Lemma 40 Let f be a continuous function with compact support on R, thenZR

�� bf (!)��2 d! = ZRjf (t)j2 dt


3.3.1 Plancherel theorem

In its simplest form, Plancherel theorem states that if a function f is inboth L1(R) and L2(R), then its Fourier transform is in L2(R); moreover theFourier transform map is isometric. This implies that the Fourier transformmap, restricted to L1(R)\L2(R); has a unique extension to a linear isometricmap L2(R)! L2(R). This isometry is actually a unitary map.

Theorem 41 (Plancherel). There exists a unique operator F from L2(R)onto L2(R) having the properties:

Ff =1p2�

ZRf (t) e�i!tdt for f 2 L1(R) \ L2(R); (3.3)

kFfkL2(R) = kfkL2(R) : (3.4)

Proof. First we mention that L1(R) \ L2(R) is dense in L2(R) and conse-quently any continuous operator de�ned on L2(R) is determined by its valueon L1(R) \ L2(R).To see this, let f 2 L2(R); we can �nd a sequence f'ng

1n=1 of functions in

L1(R) \ L2(R) such that k'n � fkL2(R) ! 0. Suppose that there are twobounded operators T1 and T2 that agree on their values on L1(R) \ L2(R);i.e., T1'n = T2'n.Then

kT1f � T2fk = k(T1f � T1'n)� (T2f � T2'n)k� kT1 (f � 'n)k+ kT2 (f � 'n)k � (kT1k+ kT2k) kf � 'nk :

It follows that kT1f � T2fk = 0 for all f 2 L2(R), hence T1 = T2:We conclude that there exists at most one operator satisfying 3.3 and 3.4.By the lemma 40, 3.4 is satis�ed for continuous functions with compactsupport, and since continuous functions with compact support are dense inL1(R) \ L2(R) (with respect to the norm k�kL1(R) + k�kL2(R)) then 3.4 holdsfor all f 2 L1(R) \ L2(R):To see this, let f 2 L2(R); we can �nd a sequence f ng

1n=1 of functions in

the set of continuous functions with compact support such that

k n � fkL2(R) ! 0:

By the lemma 40 we have

kF nkL2(R) = k nkL2(R) :

3.3. FOURIER TRANSFORM OPERATOR ON L2 (R) 67

Since F is bounded on L1(R) \ L2(R); then k n � fkL2(R) ! 0 implies thatkF n �FfkL2(R) ! 0 or kF nkL2(R) ! kFfk, hence

k nkL2(R) ! kFfkL2(R) :

Since k nkL2(R) ! kfkL2(R) we get that

kFfkL2(R) = kfkL2(R) :

Now we want to extend the mapping F : L1(R) \ L2(R) ! L2(R) to F :L2(R) ! L2(R). Let f 2 L2(R): We pick again a sequence f'ng

1n=1 of

functions in L1(R) \ L2(R) such that k'n � fkL2(R) ! 0: Clearly f'ng1n=1

is a Cauchy sequence in L1(R) \ L2(R) and fF'ng1n=1 is a Cauchy sequence

in L2(R): Since L2(R) is complete, there is a function in L2(R) which thelimit of F'n. We de�ne Ff := limn!1F'n: It can easily be shown that thede�nition does not depend on the choice of the sequence f'ng

1n=1 :

What is left to show is that F is onto L2(R): The set of twice di¤erentiablecompactly supported functions on R is dense in L2(R): Moreover every twicedi¤erentiable compactly supported function is the Fourier transform of abounded integrable function on R: The set of bounded integrable functionson R is also dense in L2(R). So it follows that the image of F containsa dense subset of L2(R); which implies that the image of F coincides withL2(R).

3.3.2 Fourier transform in the complex plane

In the previous section, the Fourier transform of a function f (t) on R wasde�ned to be a function bf (!) on R. Frequently the function bf (!) can beextended to a function which is analytic in a certain region of the complexplane C. For instance, if f (t) = e�jtj then bf (!) = 1p

2�

R1�1 e�jtjei!tdt =

2p2�

R10

e�t cos 4tdt = 2p2�

11+!2

; which can be extended to the rational func-tion 2p

2�1

1+z2; z 2 C.

If we replace the kernel ei!t with ei!z where z 2 C, then ei!z is entirefunction for all values of real !. So it is of no surprise that the complexFourier transform of functions in L2 (R) ; de�ned as:

f (z) :=1p2�

Z 1

�1F (t) eiztdt; z 2 D (3.5)

may be analytic function for certain domains in C and conditions on F (t) :Now we describe two such classes of analytic Fourier transforms.


The �rst class was introduced already in subsection 2.2.2 which deals withHardy space on the upper half plane. This class of functions are analytic inD = �+ and the condition on of F (t) is F (t) 2 L2 (R+).So let F (t) 2 L2 (R+) and de�ne the complex Fourier transform of F is as

f (z) :=1p2�

ZR+F (t) eiztdt; z 2 �+:

If z 2 �+ then jeiztj =��ei(x+iy)t�� = jeixtj je�ytj = e�yt; which shows that

jf (z)j = 1p2�

��ZR+F (t) eiztdt

�� 1p

2�

Z 1

0

jF (t)j��eizt�� dt = 1p

2�

Z 1

0

jF (t)j e�ytdt

� 1p2�kF (t)kL2(R+)

sZ 1

0

e�2ytdt

=1p2�2y

kF (t)kL2(R+) <1

so the integral in exists as a Lebesgue integral. A straightforward applicationof Fubini and Morera�s shows that f (z) is analytic in �+.Let us rewrite f (z) = 1p

2�

RR+ F (t) e

iztdt in the form

f (x+ iy) =1p2�

ZR+F (t) ei(x+iy)tdt

=1p2�

Z 1

0

e�tyF (t) eixtdt;

regard y as �xed and apply Plancherel theorem. We obtain

1

2�

Z 1

�1jf (x+ iy)j2 dx =

Z 1

0

��e�tyF (t)��2 dt=

Z 1

0

jF (t)j2 e�2tydt

and since e�2ty � 1 for t � 0; y > 0; then1

2�

Z 1

�1jf (x+ iy)j2 dx �

Z 1

0

jF (t)j2 dt (3.6)

The second class was already introduced in subsection 2.3.2. It is shownthere that a band-limited function can be extended to an entire function.Let f(t) be a band limited function, i.e., supp bf � [�;] then the complexFourier transform of f(t); f (z) := 1p

2�

RR+ F (t) e

iztdt is entire function.

3.4. THE PALEY-WIENER THEOREMS 69

The Laplace transform

The Laplace transform is de�ned by

F (s) := L (f) (s) :=Z 1

0

f (t) e�stdt (3.7)

where t is real and s := x + iy is complex. This is the one-sided Laplacetransform. There is also two-sided Laplace transform obtained by setting thelower integration limit from 0 to �1. Since we have dealt in Example 37only with functions that are supported on (0;1) then we will discuss onlythe one-sided Laplace transform.We observe that the Laplace transform is obtained from the complex

Fourier transform de�ned above by setting iz = �s. Now, if f (t) 2 L2 (R+)then by the previous subsection, the Laplace transform F (s) exists as aLebesgue integral when s 2 RHP (the Right Half of the complex Plane,fs = x+ iy j x > 0 g and F (s) is analytic in the RHP .Moreover, it can be proved that (See [13, Chapter 6, Theorem 1]).

Theorem 42 Let the real-valued function f (t) be sectionally continuos ineach �nite interval and of exponential order B (i.e. there exists constantsB and M such that jf (t)j < Me�Bt when t � 0), then the Laplace transformL (f) (s) is a analytic function of s := x + iy in the half plane x > B: TheLaplace transform is absolutely convergent in the half plane x1 > B and isuniformly convergent with respect to x and y in any half plane x � x1 wherex1 > B:

Now we state (without proof) several properties of the Laplace transformwe have used in Example 37:

L�dnf

dtn

�= snL (f)� sn�1f (0+)� sn�2f 0 (0+)� � � � � f (n�1) (0+) :(3.8)

L�eatf (t)

�= L (f) (s� a) :

L (tn) =n!

sn+1:

3.4 The Paley-Wiener theorems

We will present in this section two theorems of Paley and Wiener.The �rst Paley-Wiener theorem relates H2 (�+) ; the Hardy space on the

upper half plane to the space L2 (R+) : The second Paley-Wiener theorem


relates the space of band-limited functions to P; the Paley-Wiener space ofall entire functions of exponential type which are square integrable on thereal axis.

Theorem 43 (Paley-Wiener) Suppose f (z) 2 H2 (�+) and

sup0<y<1

1

2�

ZRjf (x+ iy)j2 dx = c <1: (3.9)

Then there exists a function F (t) 2 L2 (R+) such that

f (z) =

Z 1

0

F (t) eitzdt z 2 �+ (3.10)

and Z 1

0

jF (t)j2 dt = c: (3.11)

Proof. The proof is after Rudin�s book [11]We begin the proof with a rationale. The function F (t) we are looking formust have the property f (x+ iy) =

R10(F (t) e�yt) eitxdt; so when we regard

y as a positive parameter, we see that f (x+ iy) is the Fourier transform ofF (t) e�yt.By taking formally the inverse Fourier transform of f (x+ iy), we get

F (t) e�yt =1

2�

Z 1

�1f (x+ iy) e�itxdx

or

F (t) =1

2�

Zf (z) e�itzdz: (3.12)

The integral is over an horizontal line in �+; the upper half plane. If ourargument is true, we expect that the integral will not depend on the particularline we happened to choose. This suggests to integrate around a closedrectangular contour with two sides parallel to the x� axis and two sidesparallel to the y� axis. When we let the length of the sides which areparallel to the real axis goes to in�nity, we expect the contribution of thevertical sides to the integral to diminish.

Fix y; 0; y <1: For each � > 0, let �� be a rectangular contour with verticesat �� + i and �� + iy. Since f (z) e�itz is analytic in �+; then by Cauchytheorem Z

��

f (z) e�itzdz = 0:


Now suppose y > 1 (the proof is actually the same for 0 < y < 1). Let usevaluate the contribution of the vertical sides to the integral. Let � (�) bethe integral of f (z) e�itz over the vertical interval from � + i to � + iy.

j� (�)j2 =

��Z �+iy

�+i

f (z) e�itzdz

��2 = ��Z y

1

f (� + iu) e�it(�+iu)du

��2�

�Z y

1

jf (� + iu)j2 du��Z y

1

e2tudu

�De�ne

� (�) :=

Z y

1

jf (� + iu)j2 du:

1

2�

Z 1

�1� (�) d� =

1

2�

Z 1

�1

�Z y

1

jf (� + iu)j2 du�d�

By Fubini�s theorem

1

2�

Z 1

�1� (�) d� =

Z y

1

1

2�

Z 1

�1

�jf (� + iu)j2 d�

�du

Now since sup0<y<112�

RR jf (x+ iy)j2 dx = c, then

1

2�

Z 1

�1� (�) d� �

Z y

1

cdu = c (y � 1) :

The convergence of the improper integral above implies that there is a se-quence of real numbers f�jg such that �j !1 and

� (�j) + � (��j)! 0 as j !1:

j� (�j)j2 + j� (��j)j2 = (� (�j) + � (��j))�Z y

1

e2tudu

�;

hencej� (�j)j2 + j� (��j)j2 ! 0 as j !1;

which implies that

� (�j)! 0 and � (��j)! 0 as j !1: (3.13)

This hold for every t and the sequence f�jg does not depend on t.Let us de�ne the contribution of the horizontal side of the rectangular contourto the integral by

gj (y; t) :=1

2�

Z �j

��jf (x+ iy) e�itxdx:


ThenR��f (z) e�itzdz = 0 implies that

gj (y; t)� gj (1; t) + � (�j)� � (��j) = 0:

We deduce from (3.13) that

limj!1

(gj (y; t)� gj (1; t)) = 0 for �1 < t <1: (3.14)

We write fy (x) := f (x+ iy) : Then sup0<y<112�

RR jf (x+ iy)j2 dx = c <1

implies that fy 2 L2 (R) : By Plancherel�s theorem we know that the sequenceof functions

ngj (y; t) :=

12�

R �j��j f (x+ iy) e�itxdx

o1j=1

converges in L2 (R)

norm to bfy.limj!1

Z 1

�1

�� bfy � gj (y; t)��2 dt = 0; (3.15)

where bfy is the Fourier transform of fy. Convergence in the norm impliesthat there is a subsequence of fgj (y; t)g1b=1 which converges pointwise almosteverywhere to bfy (t).Now we de�ne

F (t) := et bf1 (t) : (3.16)

It follows from (3.14) that

F (t) = ety bfy (t) : (3.17)

Now we apply Plancherel theorem to (3.17),Z 1

�1

�� bfy (t)��2 dt = Z 1

�1

��F (t) e�ty��2 dt = Z 1

�1jfy (t)j2 dt

But by hypothesis,R1�1 jfy (t)j

2 dt =R1�1 jf (t+ iy)j2 dt � c; henceZ 1

�1e�2ty jF (t)j2 dt � c (3.18)

If we let y ! 1 then (3.18) shows that F (t) = 0 almost everywhere in(�1; 0) :If we let y ! 0 then (3.18) shows that

R10jF (t)j2 dt � c:R1

0jF (t)j2 dt � c and bfy (t) = e�tyF (t) implies that bfy (t) 2 L1 (R+)

for y > 0:It is a corollary of Plancherel theorem that if a function f 2 L2 (R) has an


integrable Fourier transform bf 2 L1 (R) then f (t) = 1p2�

R1�1

bf (!) ei!td!.Therefore

fy (t) =1p2�

Z 1

�1bfy (!) ei!td!

or

f (z) =1p2�

Z 1

�1e�!yF (!) ei!xd! =

1p2�

Z 1

�1F (!) ei!zd! z 2 �+:

This proves (3.10), and (3.11) follows from (3.18) and (3.6).

Theorem 44 (Paley-Wiener) Let f (z) be an entire function such that

jf (z)j � Cfejzj

for positive constants C and and all values of z, and

1Z�1

jf (x)j2 dx <1:

Then there exists a function � in L2 [�;] such that

f (z) =

Z�

� (t) eiztdt:

Proof. The proof is taken from [10] and is after Boas2.Let � (t) be the Fourier transform of f (x) ; i.e.,

� (t) :=1

2�

1Z�1

f (x) e�ixtdx;

2Ralph Philip Boas, Jr (August 8, 1913 - July 25, 1992) was a mathematician, teacher,and journal editor. He wrote over 200 papers, mainly in the �elds of real and complexanalysis.He was born in Walla Walla, Washington and got his B.A. degree and Ph.D. at Har-

vard University (Ph.D., 1937; advisor, David Widder). In 1950 he became Professor ofMathematics at Northwestern University, where he stayed until his retirement in 1980.He continued mathematical work after retiring, for instance as co-editor (with GeorgeLeitmann) of the Journal of Mathematical Analysis and Applications from 1985 to 1991.


where the integral is to be interpreted as a limit in the mean in the L2 sense.Then � (t) 2 L2 (R) and by the Fourier inversion formula

f (x) :=

1Z�1

� (t) eixtdt:

So we need only to show that � (t) vanishes almost everywhere outsidethe interval [�;].Let T be positive number. We shall consider the contour integral

I :=

Z�

f (z) e�itzdz;

where t is �xed and � consists of three sides of the rectangle [�T; T ]� [0; T ].Since both f (z) and e�itz are entire functions then their product is an entirefunction of z for each value of t: By Cauchy integral formula applied to thefour sides of the rectangle contour, we have,Z

�[[�T;T ]f (z) e�itzdz = 0

So

I = �Z T

�Tf (x) e�itxdx:

We are going to show that I ! 0 as T !1 whenever jtj > :Since Z T

�Tf (x) e�itxdx! � (t) in the mean (as T !1);

it will follow that � (t) = 0 almost everywhere outside [�;] 3:Suppose �rst that t < �: Integrating over each of the three sides of �; we

3Let fn (t)! 0 for all t 2 A; � (A) <1 andRAjfn � �j2 dt! 0:Z

A

(fn � �)2 dt =ZA

(fn)2dt� 2

ZA

fn�dt+

ZA

(�)2dt �

ZA

jfn � �j2 dt

It follows thatRA(fn)

2dt� 2

RAfn�dt+

RA(�)

2dt! 0: Clearly

RA(fn)

2dt! 0:�Z

A

fn�dt

�2��Z

A

(fn)2dt

��ZA

(�)2dt

�It follows that

RAfn�dt! 0:

HenceRA(�)

2dt = 0 which implies that � = 0 a.e. in A:


obtain

I =

Z T

0

f (T + iy) e�it(T+iy)idy�Z T

�Tf (x+ iT ) e�it(x+iT )dx�

Z T

0

f (�T + iy) e�it(�T+iy)idy

jIj �Z T

0

jf (T + iy)j etydy + etTZ T

�Tjf (x+ iT )j dx+

Z T

0

jf (�T + iy)j etydy

: = I1 + I2 + I3:

We shall estimate the size of each of theses three integrals.To achieve this goal we need a results from complex analysis, mainly on

boundedness of an analytic functions inside an in�nite region. So let us makea long digression and present the following theorems:

Theorem 45 (Phragmén- Lindelöf). Let f (z) be continuous on a closedsector of opening �

�and analytic in the open sector. Suppose that on the

bounding rays of the sector,

jf (z)j �M;

and that for some � < �,jf (z)j � er

�

whenever z lies inside the sector and jzj := r is su¢ ciently large.Then jf (z)j �M throughout the sector.

We can now prove that an entire function of exponential type that isbounded on a line must be bounded on every parallel line. There is no lossof generality in supposing that the given line is the real axis.

Theorem 46 Let f (z) be an entire function such that

jf (z)j � AeBjzj

for all values of z 2 C andjf (x)j �M

for all real values of x: Then

jf (x+ iy)j �MeBjyj:


Proof. Suppose �rst that y > 0. Let " > 0 be an arbitrary positive numberand put

g (z) := ei(B+")zf (z)

thenjg (x)j = jf (x)j �M

for all real x, while jg (iy)j = e�(B+")y jf (iy)j � e�(B+")yAeBy � Ae�"y; sothat

g (iy)! 0 as y !1:Thus g (z) is bounded on the non-negative imaginary axis by say, N:The Phragmén- Lindelöf theorem applied separately to the �rst and secondquadrants, shows that

jg (z)j � max fN;Mgthroughout the upper half-plane. A simple application of the maximummodulus principle4 then shows that N �M; and hence

jg (z)j �M whenever Im z > 0:

Therefore,

f (z) = e�i(B+")zg (z) = e�i(B+")xe(B+")yg (z)

jf (z)j = e(B+")y jg (z)j �Me(B+")y

throughout the upper half-plane. The result follows by letting " ! 0. thecase in which y < 0 can be reduced to the �rst case by considering f (�z).

Theorem 47 If f (z) is an entire function of exponential type and if

f (x)! 0 as jxj ! 1;

thenf (x+ iy)! 0 as jxj ! 1;

uniformly in every horizontal strip.

Proof. By Theorem 46, the function f (z) is uniformly bounded in everyhorizontal strip. Now we use Montel�s Theorem5, (Which I will not prove

4Maximum modulus theorem: (basic version) Let G � C be a connected open set andf : G ! C analytic. If there is any a 2 G with jf (a)j � f (z) for all z 2 G, then f isconstant.So we apply this principle with G equal to the upper-plane.5


here). The result is immediate consequence of Montel�s theorem when weapply the theorem to the two halves of the horizontal strip.

We have �nished our digression and go back to the proof of Paley-Wienertheorem.

Proof. (continuation of Paley-Wiener theorem proof)We �rst deal with I2: We assume �rst that jf (x)j has a �nite upper boundM on the real axis. By Theorem 46,

jf (x+ iT )j �MeT

for all real values of x; and hence

I2 := etTZ T

�Tjf (x+ iT )j dx � 2TMe(t+)T :

Since t < � then the exponent is negative and we conclude that I2 ! 0 asT !1.Now we deal with I1. We write

I1 :=

Z T

0

jf (T + iy)j etydy =Z R

0

jf (T + iy)j etydy +Z T

R

jf (T + iy)j etydy:

Theorem 47 shows that for each �xed R;

f (T + iy)! 0 as T !1;

uniformly in y; 0 � y � R: Consequently,Z R

0

jf (T + iy)j etydy ! 0 as T !1:

It remains only to show thatR TRjf (T + iy)j etydy can be made arbitrarily

small by choosing R and T su¢ ciently large. Again by using Theorem 46,we have jf (T + iy)j �Mey and thereforeZ T

R

jf (T + iy)j etydy �M

Z T

R

e(t+)ydy =M

t+

�e(t+)T � e(t+)R

�Theorem 48 Montel�s Theorem. Let f (z) be an analytic function in the half strip Sde�ned by a < x < b; y > 0: If f (z) is bounded in S and tends to a limit l, as y ! 1;for a certain �xed value � of x between a and b; then f (z) tends to this limit l on everyline x = x0 in S; and indeed f (z)! l uniformly for a+ � � x0 � b� �.


Since t + is negative, Mt+

�e(t+)T � e(t+)R

�approaches zero as R and T

approach in�nity.We have thus shown that I ! 0 as T !1 whenever t < �:The case t > is treated similarly with � in the lower half-plane.

In the general case, that is without assuming that f (x) is bounded on thereal axis, we consider f' (z) de�ned as convolution of f (z) with arbitrarycontinuous function with compact support,

f' (z) :=

Z 1

�1f (z � u)' (u) du:

Now the function f' (z) is an entire function such that jf (z)j � Cf'ejzj

for positive constants Cf' and 6and all values of z, and7

1Z�1

jf' (x)j2 dx <

1: Thus f' (z) satis�es the conditions of Paley-Wiener theorem and it isbounded8 on the real axis.By the convolution property of Fourier transform operator

bf' (t) = bf (t) b' (t)and by what we have proved already above,

bf' (t) = 0 if jtj >

Since ' is arbitrary function, then for every real t; jtj > we can pick afunction '; such that b' (t) 6= 0 and this implies that bf (t) = 0 if jtj > and the proof is complete.

6 jf' (z)j �Rjf (z � u)j j' (u)j du � Cejzj

Rejuj j' (u)j du := Cf'ejzj:

7kf � 'kL2 � kfkL2 k'kL1 � A kfkL2 <18f' (x) =

R1�1 f (x� u)' (u) du = hgx; 'i where gx (u) := f (x� u) :

jf' (x)j � kgxkL2 k'kL2 = kfkL2 k'kL2

Bibliography

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[3] Daubechies Ingrid, "Ten Lectures on Wavelets", SIAM 1992.

[4] Paulsen Vern I., "An Introduction to the Theory of Reproducing KernelHilbert Spaces", 2006;in http://www.math.uh.edu/~vern/rkhs.pdf

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[7] "Complex functions", The Open University of Israel, 1987.

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[9] "Measure Theory",The Open University of Israel, Volume 4.

[10] Young Robert M., "An introduction to nonharmonic Fourier series",Academic Press, 1980.

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