U.U.D.M. Project Report 2013:9

Examensarbete i matematik, 15 hpHandledare och examinator: Wolfgang Staubach

Maj 2013

Department of MathematicsUppsala University

The Hilbert Transform

Axel Husin

THE HILBERT TRANSFORM

AXEL HUSIN

Abstract. We will investigate various aspects of the Hilbert transform, study some of

its properties and also point out how it can be used in connection to the study of the

norm convergence of Fourier series.

Contents

1. Introduction 2

2. Background material 2

2.1. Measure spaces 2

2.2. Definition of the integral 3

2.3. Properties of the integral 4

2.4. Lp spaces 5

2.5. Convolutions 7

2.6. Operator norm 8

3. Introduction to Fourier analysis 8

3.1. Fourier series on Lp 8

3.2. Parsevals formula 9

3.3. The Fourier transform on L1 11

3.4. The inversion theorem 12

3.5. The Fourier transform on L2 14

4. Relation to harmonic functions 16

4.1. Cauchy type integrals 16

4.2. Introduction to harmonic functions 17

4.3. A first glance at the Hilbert transform 18

4.4. Poisson formula for the unit disk 18

4.5. Poisson formula for the upper half plane 20

5. The Hilbert transform on the real line 22

5.1. Definition 22

5.2. Properties of the Hilbert transform 22

5.3. Some Hilbert transforms 23

5.4. The Riesz inequality on L2 24

6. Normconvergence of Fourier series 25

6.1. The Hilbert transform as a multiplier 26

6.2. The Hilbert transform is bounded on L2 28

6.3. A counterexample in L∞(T) 29

References 291

1. Introduction

This text is supposed to introduce the reader to the Hilbert transform, reading this will

make it easier to later understand more advanced texts on the subject. We will study

the Hilbert transform on the real line and on the unit circle. However, by the Riemann

mapping theorem it is possible to do similar constructions on more general domains.

First we will remind the reader of some background material like the Lebesgue integral,

Fourier series and the Fourier transform followed by a short study of harmonic functions

that will motivate the definition of the Hilbert transform.

Then we define the Hilbert transform on the real line and study some of its main proper-

ties. Among the properties we will see how it relates to the Fourier transform and show

that it is an isometry from L2(R) to L2(R).

To close this text we finally study the Hilbert transform on the unit circle and show how

it can be used to prove norm convergence of Fourier series in Lp(T) for 1 < p <∞.

2. Background material

2.1. Measure spaces. We will use the Lebesgue integral in this text. So we remind

the reader about the definition of a measure space and the Lebesgue integral. However,

this is a short introduction and is not intended to give the reader a full understanding of

measure theory, for the full story and the details omitted here, see e.g. [6]. The French

mathematician Henri Lebesgue (1875-1941) introduced his integral in 1904 and also laid

the foundation of measure theory.

We start with the definition of a measure space. Let X be a set and Σ ⊂ P (X) a σ-

algebra over X, that means that Σ is non empty and closed under the complement and

countable unions. The sets in Σ are called measurable sets. Next we need to assign a

measure to all the sets in Σ, the measure of the sets in Σ is denoted by µ(E) (where

E ∈ Σ). Formally we let µ : Σ→ R ∪ ∞ be a function satisfying

µ(E) ≥ 0

µ(∅) = 0

µ

(∞⊔k=1

Ek

)=∞∑k=1

µ(Ek),

(2.1)

where the sets Ek are disjoint. The triple (X,Σ, µ) is then called a measure space.

From the equations (2.1) we can derive some properties for the measure function, for

example the following:

Theorem 2.1. Let (X,Σ, µ) be a measure space and let A ⊂ B be measurable sets, then

µ(B \ A) = µ(B)− µ(A) and µ(A) ≤ µ(B).

Proof. The set B \ A is measurable and A t (B \ A) = B is a disjoint union, so by (2.1)

it yields that µ(A) + µ(B \ A) = µ(B) and hence µ(B \ A) = µ(B) − µ(A). And since

µ(B \ A) ≥ 0 we get µ(A) ≤ µ(B).

A very important example of a measure space which we will use throughout the text is

the measure set (R, σ, λ) with λ denoting the Lebesgue measure.2

First define the outer Lebesgue measure λ∗ by

λ∗(E) = inf

∞∑k=1

(bk − ak), E ⊂∞⋃k=1

[ak, bk]

.

This makes λ∗ a translation invariant measure on R and one has λ([a, b]) = b − a, but

λ∗ will by the axiom of choice not satisfy property 3 in (2.1), this will be shown in a

moment. To remedy this situation we define σ ⊂ P (R) by

σ = E |λ∗(A) = λ∗(A ∩ E) + λ∗(A ∩ Ec) for every A ⊂ R

and we let λ(E) = λ∗(E) for every E ⊂ σ. We skip the proof that σ actually is a

σ-algebra and that λ satisfies the properties in equation (2.1).

In a similar way we can define the measure space ([0, 2π], σ, λ) which also will be important

later on.

Now we answer the question why not every subset of R is measurable. The reason for this

is that the outer Lebesgue measure λ∗ will not satisfy the properties in (2.1). Actually,

it is impossible to find any translation invariant measure function µ satisfying (2.1) and

µ([a, b]) = b − a. Assume on the contrary that there is such a function µ and consider

for example the set R/Q of equivalence classes, where x ∈ R and y ∈ R are in the same

equivalence class if and only if x−y ∈ Q. Each equivalence class E has a representative in

[0, 1], since if we take any representative r ∈ E we can add a rational number q ∈ Q such

that r+ q ∈ [0, 1] and r+ q ∈ E. Now use the axiom of choice to pick one representative

in the interval [0, 1] from every equivalence class, and collect them in a set V . The set V

we constructed is called a Vitali set. We note that

[0, 1] ⊂⊔k∈Q−1≤k≤1

V + k ⊂ [−1, 2]

and by taking measures of the sets we get

1 ≤∑k∈Q−1≤k≤1

µ(V + k) =∑k∈Q−1≤k≤1

µ(V ) ≤ 3,

which is a contradiction since the infinite sum must satisfy∑µ(V ) = 0 or

∑µ(V ) =∞.

This shows that λ∗ does not satisfy property 3 in (2.1).

2.2. Definition of the integral. Let (X,Σ, µ) be a measure space. A function f : X →R is called measurable if f−1(U) is measurable set for every open set U ⊂ R. We state

without proof some simple but important properties of measurable functions.

Theorem 2.2.

(1) The characteristic function χE of a measurable set E is measurable.

(2) Simple functions f =∑n

k=1 ckχEk , where Ek are measurable sets, are measurable.

(3) Piecewise continuous functions are measurable.

(4) If f and g are measurable, then λf, |f |p, f + g, fg, f/g, min(f, g) and max(f, g)

are measurable, where λ ∈ R and p > 0 (the quotient part f/g only holds if g is

nonzero).

(5) If fk is sequence of measurable functions and fk → f pointwise, then f is

measurable.3

We define the integral in the following three steps.

1. Let f be a nonnegative simple function, f =∑n

k=1 ckχEk where ck > 0 and Ek are

measurable. Then we define ∫X

fdµ =n∑k=1

ckλ(Ek).

The third property in equation (2.1) ensures that this is a well defined.

2. Let f : X → R be a nonnegative measurable function, then we define∫X

fdµ = sup

∫X

gdµ, 0 ≤ g ≤ f and g is simple

.

It can be shown that the definitions in step 1 and 2 coincide for the integral of a nonneg-

ative simple function.

3. Let f : X → R be a measurable function where∫X|f |dµ <∞, then we call f integrable

and define ∫X

fdµ =

∫X

f+dµ−∫X

f−dµ,

where f+ = max(f, 0) and f− = −min(f, 0).

Complex-valued functions can be dealt with by separating it into its real and imaginary

part. We say that a function f : X → C is measurable if Re(f) and Im(f) are measurable,

and we say that f is integrable if Re(f) and Im(f) are integrable, and in that case we

define∫Xfdµ =

∫X

Re(f)dµ+ i∫X

Im(f)dµ.

2.3. Properties of the integral. Some of the most important properties of the Lebesgue

integral are given in the following theorem which we state without proof.

Theorem 2.3.

(1) If f and g are integrable, then∫Xλfdµ = λ

∫Xfdµ and

∫X

(f + g)dµ =∫Xfdµ+∫

Xgdµ.

(2) If f is integrable and g is measurable with |g| ≤ f , then |∫Xgdµ| ≤

∫Xfdµ.

(3) If f is integrable, then |∫Xfdµ| ≤

∫X|f |dµ.

(4) If f is integrable, then∫X|f |dµ = 0 if and only if f = 0 almost everywhere.

From now on, the statement f = 0 almost everywhere means that f is nonzero on a set

of measure 0.

Now we present some theorems (without proofs), concerning the integral of the pointwise

limit of a sequence of functions. These important theorems clarify some of the advantages

of the Lebesgue integral as compared to the Riemann integral.

Theorem 2.4 (Fatou’s lemma). Let fk be a sequence of nonnegative integrable func-

tions converging pointwise to f and∫Xfkdµ ≤M for some M <∞. Then f is integrable

and∫Xfdµ ≤M .

Theorem 2.5 (Monotone convergence theorem (MCT)). Let fk be an increasing se-

quence of nonnegative measurable functions converging pointwise to f . Then f is mea-

surable and∫Xfdµ = lim

k→∞

∫Xfkdµ.

4

Theorem 2.6 (Dominated convergence theorem (DCT)). Let fk be a sequence of in-

tegrable functions converging pointwise to f and |fk| ≤ g for some integrable function g.

Then f is integrable and∫Xfdµ = lim

k→∞

∫Xfkdµ.

In what follows we will denote by∫f(x)dx, the integral of the function f with respect

to the Lebesgue measure λ on R . We also state the following theorem, concerning

interchange of order of integration in double integrals, only in the case of the Lebesgue

measure .

Theorem 2.7 (Fubini-Tonelli). Let f(x, y) be function on R2 such that either of the

following integrals, ∫ ∞−∞

(∫ ∞−∞|f(x, y)|dx

)dy

or ∫ ∞−∞

(∫ ∞−∞|f(x, y)|dy

)dx

converge. Then one has∫ ∞−∞

(∫ ∞−∞

f(x, y)dx

)dy =

∫ ∞−∞

(∫ ∞−∞

f(x, y)dy

)dx.

2.4. Lp spaces. Since we will define the Hilbert transform on Lp spaces, this section

covers some basic material concerning these spaces which will be used later on.

Let 1 ≤ p <∞ and (X,Σ, µ) denote a measure space, then we define Lp(X) as the space

of complex-valued measurable functions f : X → C where∫X|f |pdµ < ∞. It can be

shown that Lp(X) is a vector space by noting that if f ∈ Lp(X), then |λf |p = |λ|p|f |p so

λf is also in Lp(X), and if f, g ∈ Lp(X), then f+g ∈ Lp(X) by the Minkowski inequality

2.12. We also equip Lp(X) with the seminorm ‖f‖p = (∫X|f |pdµ)1/p where we can use

the Minkowski inequality 2.12 to see that the triangle inequality holds.

The seminormed vector space L∞(X) is defined in a slightly different way. We define the

space L∞(X) as the set of essentially bounded measurable functions, that is measurable

functions bounded up to a set of measure zero. And we use the seminorm ‖f‖∞ =

infC ≥ 0 : |f(x)| ≤ C for almost every x.If we let N be the subspace of Lp(X) including all functions on X which are zero almost

everywhere, we define the quotient space Lp(X) = Lp(X)/N . This yields that f , g ∈Lp(X) are equal to each other iff they are equal almost everywhere.

It can be shown that for 1 ≤ p ≤ ∞, Lp(X) is Banach space, that is a complete normed

vector space, this fact will be used many times throughout this text.

If X is a space of finite measure, then f ∈ Lp(X) implies f ∈ Lp′(X) for p′ ≤ p, that is

since |f |p′ < (|f | + 1)p′ ≤ (|f | + 1)p and since both f and the constant function 1 are in

Lp(X), then (|f | + 1)p is integrable since Lp(X) is a vector space. However, if X does

not have finite measure then the implication above does not hold. As an example we can

take the function f(x) = 1/(1 + |x|),which is in L2(R) but not in L1(R).

Definition 2.8. One says that the sequence fk ∈ Lp(X) converges in the norm or in the

mean to a function f , if, for every ε > 0, one has for large enough k∫X

|fk − f |pdµ < ε.

5

Since Lp(X) is a Banach space, it follows that f will also belong to Lp(X). In this

connection we also have the following result.

Lemma 2.9. Let fk be a sequence of functions in Lp(X) such that fk → f pointwise

and fk → g in the Lp norm. Then f = g almost everywhere.

Proof. First we note that |fk − g|p → |f − g|p pointwise. Since fk → g in Lp , Fatou’s

lemma yields that ∫X

|f − g|pdµ < ε.

Since this is valid for every ε > 0 we get∫X

|f − g|pdµ = 0.

And by property 4 in Theorem 2.3 we get |f − g|p = 0 almost everywhere and hence

f = g almost everywhere.

Some inequalities that will be important later on are the Holder and the Minkowski

inequalities.

Theorem 2.10 (Holder’s inequality). Let f ∈ Lp(X) and g ∈ Lq(X) where 1/p+1/q = 1

and 1 ≤ p, q ≤ ∞, then ∫X

|fg|dµ ≤ ‖f‖p‖g‖q.

Proof. If p = ∞ and q = 1, then |fg| ≤ ‖f‖∞|g| almost everywhere so∫X|fg|dµ ≤

‖f‖∞∫X|g|dµ = ‖f‖∞‖g‖1, similarily if p = 1 and q = ∞. Now assume 1 < p, q < ∞.

We start with the special case ‖f‖p = ‖g‖q = 1. We can estimate the integrand fg with

Young’s inequality, which will be proved later:

|f(t)g(t)| ≤ |f(t)|p

p+|g(t)|q

q.

Taking the integral of both sides gives∫X

|fg|dµ ≤‖f‖ppp

+‖g‖qqq

= 1 = ‖f‖p‖g‖q.

Now let’s move on to the general case. If ‖f‖p = 0 or ‖g‖q = 0 the result is trivial since

we have fg = 0 almost everywhere (by property 4 in Theorem 2.3) so the integral on

the left hand side will vanish. Assume ‖f‖p > 0 and ‖g‖q > 0 and apply the version of

Holder’s inequality we just proved to the functions f/‖f‖p and g/‖g‖q to get

(2.2)

∫X

∣∣∣∣ f

‖f‖pg

‖g‖q

∣∣∣∣ dµ ≤ 1.

Holder’s inequality follows by multiplying (2.2) with ‖f‖p‖g‖q.

In the proof above we used Young’s inequality, which says that for nonnegative a and b

ab ≤ ap

p+bq

q,

where 1/p+ 1/q = 1 and 1 < p, q <∞.6

One proof goes something like this. If b = 0 the inequality is trivial, so assume b > 0 and

divide both sides by bq to geta

bq−1≤ ap

pbq+

1

q.

Now let t = a/bq−1, that gives 0 < t < ∞ and tp = ap/bp(q−1) = ap/bq by using the

relation 1/p+ 1/q = 1. The inequality can now equivalently be written as

tp

p− t+

1

q≥ 0.

We can verify the inequality by examining the differentiable function h(t) = tp/p−t+1/q.

We have h(0) = 1/q and limt→∞

h(t) = ∞ and h′(t) = tp−1 − 1, so h(1) = 0 is the global

minimum of h.

Remark 2.11. The special case of p = q = 2 in Holder’s inequality is called Cauchy-

Schwarz inequality.

Theorem 2.12 (Minkowski’s inequality). Let f, g ∈ Lp(X) where 1 ≤ p ≤ ∞, then

‖f + g‖p ≤ ‖f‖p + ‖g‖p.

Proof. If p = 1 the result follows directly by taking the triangle inequality under the

integral sign and if p =∞ we can again use the triangle inequality to get |f(x) + g(x)| ≤|f(x)|+ |g(x)| ≤ ‖f‖∞ + ‖g‖∞ almost everywhere, and hence ‖f + g‖∞ ≤ ‖f‖∞ + ‖g‖∞.

Now assume 1 < p <∞ and let q be defined by 1/p+1/q = 1 and use Holder’s inequality

to get

‖f + g‖pp =

∫X

|f + g|pdµ =

∫X

|f + g||f + g|p−1dµ

≤∫X

(|f |+ |g|)|f + g|p−1dµ

=

∫X

|f ||f + g|p−1dµ+

∫X

|g||f + g|p−1dµ

≤ (‖f‖p + ‖g‖p)(∫

X

|f + g|q(p−1)dµ

)1/q

= (‖f‖p + ‖g‖p)‖f + g‖p−1p .

Dividing both sides by ‖f + g‖p−1p gives the Minkowski inequality.

2.5. Convolutions. The convolution is one of the most fundamental operations in math-

ematical analysis.

Definition 2.13. The convolution of the functions f and g is defined by

f ∗ g(x) =

∫ ∞−∞

f(t)g(x− t)dt.

It can easily be seen that the convolution is commutative i.e. f ∗ g(x) = g ∗ f(x). This

can be shown by a simple change of variables in the definition above.

One also has the following special case of the so called Young’s inequality.7

Theorem 2.14. If f ∈ Lp(R), 1 ≤ p ≤ ∞ and g ∈ L1(R), then f ∗ g(x) is in Lp(R) and

‖f ∗ g‖p ≤ ‖f‖p‖g‖1.

Proof. The cases p = 1 and p =∞ are simple, so assume 1 < p <∞.

Let q be defined by 1/p+ 1/q = 1. We get

|f ∗ g(x)| =∣∣∣∣∫ ∞−∞

f(t)g(x− t)dt∣∣∣∣

≤∫ ∞−∞|f(t)||g(x− t)|1/p|g(x− t)|1/qdt.

(2.3)

By using Holder’s inequality on (2.3) we get

(2.4) |f ∗ g(x)| ≤(∫ ∞−∞|f(t)|p|g(x− t)|dt

)1/p(∫ ∞−∞|g(x− t)|dt

)1/q

.

If we take Lp-norm of both sides of (2.4) and use Fubini’s theorem we get

‖f ∗ g‖p ≤ ‖g‖1/q1

(∫ ∞−∞

∫ ∞−∞|f(t)|p|g(x− t)|dtdx

)1/p

= ‖g‖1/q1

(∫ ∞−∞|f(t)|p

∫ ∞−∞|g(x− t)|dxdt

)1/p

= ‖g‖1/q1 ‖g‖

1/p1 ‖f‖p

= ‖g‖1‖f‖p.

2.6. Operator norm. An interesting property to study for linear operators is the op-

erator norm. We will later show that the Hilbert transform is a bounded operator. The

operator norm is defined as follows.

Definition 2.15. Let L : X → Y be a linear operator between two normed vector spaces,

then we define the operator norm of L by

‖L‖op = sup

‖Lx‖‖x‖

: x ∈ X, x 6= 0

.

If ‖L‖op <∞ we say that L is bounded.

3. Introduction to Fourier analysis

3.1. Fourier series on Lp. The Fourier series of a function is defined as follows.

Definition 3.1. Let f ∈ Lp(T). Then the Fourier series of f is defined by

limn→∞

Snf(x) = limn→∞

n∑k=−n

f(k)eikx,

where Snf are called the partial sums of f and the Fourier coefficients f(k) are defined

by

f(k) =1

2π

∫ 2π

0

f(t)e−iktdt.

8

To simplify the notation we will set∑∞−∞ = lim

n→∞

∑n−n.

The Fourier coefficients will always exist since |f(t)e−ikt| = |f(t)|, where |f(t)| is in L1

since it is in Lp and T has finite measure. So the the main question is whether the Fourier

series converges or not. That will be studied in the last section of this paper.

We define a trigonometric polynomial as a finite Fourier series of the formn∑

k=−n

ckeikx.

The remarkable fact about the trigonometric polynomials is that all functions in Lp(T),

for 1 ≤ p <∞, can be approximated by trigonometric polynomials.

Theorem 3.2. For 1 ≤ p <∞, the set of trigonometric polynomials are a dense subset

of Lp(T), and if f is a trigonometric polynomial the Fourierseries of f will converge to

f in Lp(T).

Proof. We can see that the set of trigonometric polynomials are dense in Lp(T) by noting

that the algebra of the trigonometric polynomials satisfy the condition of the Stone-

Weierstrass theorem, so they are uniformly dense in C(T), and C(T) is in turn dense in

Lp(T) for 1 ≤ p <∞ see e.g. [7]. This proves the first part of the theorem.

Now assume that if f is a trigonometric polynomial

f(x) =n∑

k=−n

ckeikx.

If we calculate the Fourier coefficients we get

f(k) =1

2π

∫ 2π

0

(n∑

j=−n

cjeijt

)e−iktdt

=1

2π

∫ 2π

0

(n∑

j=−n

cjei(j−k)t

)dt

=1

2π

n∑j=−n

∫ 2π

0

cjei(j−k)tdt = ck,

where we interpret ck = 0 for |k| > n. The Fourier series of f will clearly converge to f

since all terms in the series with index larger than n will be zero, and therefore Skf = f

for k ≥ n.

3.2. Parsevals formula. Among the Lp(T) spaces the space L2(T) stands to be a Hilbert

space. This means that one has an inner product in this space that defines the norm and

the inner product enables us to measure angles and makes the geometry of the L2(T)

spaces considerably easier than that of the other Lp(T) spaces. As a result, it turns out

that L2(T) is perfectly suited for the study of the Fourier series.

Definition 3.3. The inner product on L2(T) is defined by

(3.1) 〈f, g〉 =

∫ 2π

0

f(x)g(t)dt.

9

One can easily see that this is well defined since 2|f(x)g(x)| ≤ |f(x)|2 + |g(x)|2. It is

straight forward to show that 3.1 satisfies the conditions for being an inner product. Let

en(x) = einx, n ∈ Z. Then with the above inner product we have that 〈en, em〉 (where

n,m ∈ Z) is 0 if n 6= m and is 2π if n = m, in other words the functions en (where n ∈ Z)

is an orthogonal set with respect to the inner product (3.1). Therefore we can consider

the Fourier coefficients f(n) = 〈f, en〉 / 〈en, en〉 as the coefficient f(n) in the projection

f(n)en of f onto the subspace generated by en.

Remark 3.4. Some authors define the inner product with a factor 1/2π, however we define

it without to make the norm in L2(T) we defined earlier, coincide with the induced norm

from the inner product ‖f‖2 =√〈f, f〉.

Lemma 3.5. Assume that f ∈ L2(T), and that S is any trigonometric polynomial of

degree at most n with n ∈ Z. Then 〈f − Snf, S〉 = 0 and ‖Snf − f‖2 ≤ ‖S − f‖2.

Proof. S(t) can be written as S(t) =∑n−n cke

ikt. Then

〈f − Snf, S〉 =

⟨f −

n∑k=−n

〈f, ek〉 ek2π

,

n∑k=−n

ckek

⟩

=n∑

k=−n

ck 〈f, ek〉 −n∑

k=−n

〈f, ek〉 ck 〈ek, ek〉2π

= 0.

(3.2)

This proves the first part.

By using (3.2) we get 〈f − Snf, S − Snf〉 = 0, and hence by Pythagoras theorem we have

‖S − f‖22 = ‖S − Snf‖2

2 + ‖Snf − f‖22 ≥ ‖Snf − f‖2

2.

The second part in the Lemma follows by taking the square root of both sides.

Now we can prove the norm convergence of the Fourier expansion of L2(T) functions and

also the Parseval identity.

Theorem 3.6. Let f ∈ L2(T), then limn→∞

Snf = f in the L2 norm.

Proof. Let ε > 0 be arbitrarly. Choose a trigonometric polynomial P such that ‖f−P‖2 <

ε and letN be the degree of P . If n ≥ N we get by Lemma 3.5, ‖f−Snf‖2 ≤ ‖f−SNf‖2 ≤‖f − P‖2 < ε, and the statement follows.

Theorem 3.7 (Parseval’s identity). Let f ∈ L2(T), then the sequence f(k) is square

summable and

2π∞∑

k=−∞

|f(k)|2 =

∫ 2π

0

|f(t)|2dt.

Proof. By Lemma 3.5 〈f − Snf, Snf〉 = 0, so we can use Pythagoras theorem to get

‖f‖22 − ‖Snf‖2

2 = ‖f − Snf‖22.

According to Theorem 3.6 ‖f − Snf‖22 → 0 so the result follows if we make the identifi-

cations

‖f‖22 =

∫ 2π

0

|f(t)|2dt10

and, by the orthogonal properties,

‖Snf‖22 =

∫ 2π

0

(n∑

k=−n

f(k)eikt

)(n∑

k=−n

f(k)e−ikt

)dt = 2π

n∑k=−n

|f(k)|2.

3.3. The Fourier transform on L1. Another important linear operator which will be

used in the study of the Hilbert transform is the Fourier transform. Let us review some

of the basic properties of the Fourier transform.

Throughout this section, we will use the notation τa for the translation operator defined

by τaf(x) = f(x− a), and R for the reflection operator Rf(x) = f(−x).

We start by defining the Fourier transform for functions in L1(R).

Definition 3.8. Let f ∈ L1(R). The Fourier transform of f is defined by

Ff(ω) = f(ω) =

∫ ∞−∞

f(t)e−iωtdt.

This is well defined for all ω since the integral is absolutely convergent.

Example 3.9. Calculate the Fourier transform of

fa(x) =

12π

(1− |x|

a

), if |x| ≤ a,

0 , if |x| > a.

Solution:

fa(ω) =

∫ ∞−∞

f(t)e−iωtdt

=1

π

∫ a

0

(1− t

a

)cos(ωt)dt

= − 1

π

∫ a

0

−1

aωsin(ωt)dt

=1

πaω2(− cos(aω) + cos(0))

=1

πaω2(1− cos(aω))

=2 sin2(aω/2)

πaω2.

Remark 3.10. Observe that fa in Example 3.9 is in L1(R) and by letting u = aω/2 and

using a trick from the residue calculus we can calculate

(3.3)

∫ ∞−∞

fa(t)dt =1

π

∫ ∞−∞

sin2 u

u2du = 1.

By the dominated convergence theorem we get for all δ > 0

(3.4) lima→∞

∫|t|≥δ

fa(t)dt = 0.

We say that the sequence fa is an approximation of identity since it satisfies (3.3) and

(3.4).11

Theorem 3.11. If f ∈ L1(R) is differentiable and f ′ ∈ L1(R), then f ′(ω) = iωf(ω).

Proof. Since f ∈ L1(R) we can choose sequences an and bn such that limn→∞

an = −∞,

limn→∞

bn =∞ and limn→∞

|f(an)| = limn→∞

|f(bn)| = 0. By partial integration we get

f ′(ω) = limn→∞

∫ bn

an

f ′(t)e−iωtdt

= limn→∞

([f(t)e−iωt

]bnan− iω

∫ bn

an

f(t)e−iωtdt

)= iωf(ω).

There is a simple formula for the Fourier transform of a convolution.

Theorem 3.12. Let f ∈ L1(R) and g ∈ L1(R), then

f ∗ g(ω) = f(ω)g(ω).

Proof. We can use Fubini’s theorerm since f ∗ g ∈ L1(R).∫ ∞−∞

(∫ ∞−∞

f(t)g(x− t)dt)e−iωxdx =

∫ ∞−∞

f(t)

(∫ ∞−∞

g(x− t)e−iωxdx)dt

=

∫ ∞−∞

f(t)

(∫ ∞−∞

g(u)e−iω(u+t)du

)dt

=

∫ ∞−∞

f(t)g(ω)e−iωtdt

= f(ω)g(ω).

3.4. The inversion theorem. We start first with the following lemma:

Lemma 3.13 (Hatmoving lemma). Let f, g ∈ L1(R), then∫ ∞−∞

f(x)g(x)dx =

∫ ∞−∞

f(x)g(x)dx.

Proof. ∫ ∞−∞

f(t)g(t)dt =

∫ ∞−∞

[∫ ∞−∞

f(x)e−itxdx

]g(t)dt

=

∫ ∞−∞

f(x)

[∫ ∞−∞

g(t)e−ixtdt

]dx

=

∫ ∞−∞

f(x)g(x)dx.

The changce of integration order is allowed by Fubini’s theorem since∫ ∞−∞

∫ ∞−∞|f(x)g(t)e−itx|dxdt =

∫ ∞−∞|f(x)|dx

∫ ∞−∞|g(t)|dt <∞.

12

The following lemma concerning density and continuity will be quite useful for us.

Lemma 3.14.

(a) Let 1 ≤ p <∞. Then the space C0 of continuous functions with compact support

on R is dense in Lp(R).

(b) Let 1 ≤ p <∞. Then lim|h|→0

‖τhf − f‖p = 0.

Proof. (a) follows from Theorem 1.3 in [8] and (b) follows from Proposition 1.4 in [8].

Remark 3.15. One can in fact show that the space Ck0 , k = 0, 1, 2, . . . ,∞ of k times

continuously differentiable functions with compact support on R is dense in Lp(R), 1 ≤p <∞, see e.g. [8].

Theorem 3.16. Assume f ∈ L1(R) and f ∈ L1(R), then F−1Ff = f almost everywhere.

The inverse Fourier transform F−1 is defined by

F−1f(x) =1

2πRF(x) =

1

2πf(−x) =

1

2π

∫ ∞−∞

f(ω)eixωdω.

Proof. Let

(3.5) ka(x) =

12π

(1− |x|

a

), if |x| ≤ a,

0 , if |x| > a.

Then, by Example 3.9,

(3.6) ka(t) =2 sin2(at/2)

πat2.

Let f ∈ L1(R). By Lemma 3.13 and the fact that f(x− t)(ω) = f(−ω)e−ixω (where the

Fourier transform on the left hand side is taken with respect to the variable t) we get∫ ∞−∞

f(x− t)ka(t)dt =

∫ ∞−∞

f(−t)e−itxka(t)dt =

∫ ∞−∞

f(t)eitxka(t)dt.

By the dominant convergence theorem the right hand side tends as a→∞ pointwise to

F−1Ff(x), and by Fubini’s theorem the left hand side tends to f in L1 since

‖f ∗ ka − f‖1 ≤∫ ∞−∞

∫ ∞−∞|f(x− t)− f(x)|ka(t)dtdx

=

∫ ∞−∞

(∫ ∞−∞|f(x− t)− f(x)|dx

)ka(t)dt

=1

π

∫ ∞−∞‖τ2u/af − f‖1

sin2 u

u2du.

Moreover ‖τ2u/af − f‖1 ≤ 2‖f‖1 and ‖τ2u/af − f‖1 → 0 (Lemma 3.14 part (b)), so we

can use the dominated convergence theorem to see that ‖f ∗ ka− f‖1 → 0. This implies,

by Lemma 2.9 that F−1Ff = f almost everywhere.

As a consequence of this we have the important Plancherel theorem.13

Theorem 3.17. Let f ∈ L1(R) ∩ L2(R), then ‖f‖22 = 2π‖f‖2

2, or in integral notation∫ ∞−∞|f(ω)|2dω = 2π

∫ ∞−∞|f(t)|2dt.

Proof. The proof of this uses a similar technique as the proof of Theorem 3.16 so we start

with defining the functions ka as in equation (3.5). Let g = f ∗ Rf , then by Theorem

3.12, g = f Rf = |f |2. We can also see that g is uniformly continuous since

|g(x)− g(y)| =∫ ∞−∞

f(t)f(t− x)dt−∫ ∞−∞

f(t)f(t− y)dt

≤∫ ∞−∞|f(t)||f(t− x)− f(t− y)|dt

≤ ‖f‖2‖τxf − τyf‖2

= ‖f‖2‖τx−yf − f‖2,

where we have used Cauchy-Schwarz inequality. Now letting |x − y| → 0, Lemma 3.14

part (b) yields that |g(x)− g(y)| → 0.

By Lemma 3.13 we get ∫ ∞−∞

g(t)ka(t)dt =

∫ ∞−∞|f(t)|2ka(t)dt.

The right hand side tends as a → ∞ to 12π‖f‖2

2 by the monotone convergence theorem.

Moreover the left hand side tends to g(0) = ‖f‖22. That is since g is continuous at 0 so

that for every ε > 0 there exists a δ > 0 such that |g(t)− g(0)| < ε/2 for all |t| < δ, and

we get for big enough a that

∣∣∣∣∫ ∞−∞

g(t)ka(t)dt− g(0)

∣∣∣∣ ≤ ∫ ∞−∞|g(t)− g(0)|ka(t)dt

≤∫|t|≤δ|g(t)− g(0)|ka(t)dt+

∫|t|≥δ

2|g|max|t|≥δ

(ka(t))dt

≤ ε

2+ε

2= ε.

This works for all ε > 0 so we must have ‖f‖22 = 2π‖f‖2

2.

3.5. The Fourier transform on L2. Now we define the Fourier transform for functions

in L2(R). Since the integral in definition 3.8 is not absolutely convergent for all functions

in L2(R) we need to make the definition in terms of sequences.

Definition 3.18. Let f ∈ L2(R). Then since L1(R) ∩ L2(R) is dense in L2(R) there is

a sequence of functions fn ∈ L1(R) ∩ L2(R) converging to f in L2-sense. The Fourier

transform of f is defined by

(3.7) f = limn→∞

fn,

where the limit is to be taken in L2.14

Since fn is a Cauchy sequence one can show that fn is also a Cauchy sequence by

‖fn − fm‖22 = ‖ fn − fm‖2

2 = 2π‖fn − fm‖22, so the limit in (3.7) will always exist. Now

we show that the limit (3.7) is unique. Assume both fn and gn are sequences of

L1(R)∩L2(R) functions converging to f in L2 and set f = limn→∞

fn and g = limn→∞

gn, then

‖f − g‖2 ≤ ‖f − fn‖2 + ‖fn − gn‖2 + ‖gn − g‖2

= ‖f − fn‖2 +√

2π‖fn − gn‖2 + ‖gn − g‖2

≤ ‖f − fn‖2 +√

2π‖fn − f‖2 +√

2π‖f − gn‖2 + ‖gn − g‖2,

where all the terms go to zero, so f = g.

Remark 3.19. For functions in L1(R)∩L2(R) this new definition coincides with definition

3.8 since we can get convergence to such functions by a constant sequence.

Now it is a simple task to extend the inversion theorem, Plancherel theorem and the

convolution theorem to L2(R).

Theorem 3.20. Let f ∈ L2(R), then∫ ∞−∞|f(ω)|2dω = 2π

∫ ∞−∞|f(t)|2dt.

Proof. Let fn be a sequence of functions in L1(R) ∩ L2(R) that converges to f . Then

by the continuity of the L2-norms (see [8])

‖f‖2 = limn→∞

‖fn‖2 =√

2π limn→∞

‖fn‖2 =√

2π‖f‖2.

Theorem 3.21. Assume f ∈ L2(R), then F−1Ff = f almost everywhere.

Proof. Let fn be a sequence of C20(R) functions such that fn → f . The existence of

such a sequence follows from Remark 3.15. Then, by Theorem 3.11, f ′′n(ω) = −ω2fn(ω)

so |fn(ω)| ≤ ‖f ′′n‖∞/ω2 and, hence, fn ∈ L1(R). We get

‖ 1

2πRf − f‖2 ≤ ‖

1

2πRf − 1

2πRfn‖2 + ‖fn − f‖2 = ‖f − fn‖2 + ‖fn − f‖2,

where the right hand side tends to zero.

Remark 3.22. Since the Fourier transform commutes with reflection we also have

FF−1f = F 1

2πRFf =

1

2πRFFf = F−1Ff = f.

Theorem 3.23. Let f ∈ L2(R) and g ∈ L1(R), then

F [f ∗ g] = FfFg.Proof. Let fn be a sequence of L1(R) ∩ L2(R) converging to f in L2 sense. Then

‖F [f ∗ g]−FfFg‖2

≤ ‖F [f ∗ g]−F [fn ∗ g]‖2 + ‖F [fn ∗ g]−FfnFg‖2 + ‖FfnFg −FfFg‖2

=√

2π‖(f − fn) ∗ g‖2 + 0 + ‖F [fn − f ]Fg‖2

≤√

2π‖(f − fn)‖2‖g‖1 +√

2π‖fn − f‖2‖Fg‖∞,where the right hand side tends to zero.

15

4. Relation to harmonic functions

4.1. Cauchy type integrals. In this section, we shall start by reviewing some facts

from complex analysis and the theory of harmonic functions. To this end, we need to

briefly discuss Cauchy type integrals, that is functions of the form

(4.1) Φ(z) =1

2πi

∫γ

φ(ζ)

ζ − zdζ,

where γ is a smooth curve and φ is a complex valued function defined on γ.

Let γ be a closed curve and D the domain inside γ, and let φ be an holomorphic function

on some open set containing D ∪ γ. Then by the Cauchy integral formula and Cauchy

integral theorem Φ(z) = φ(z) if z ∈ D and Φ(z) = 0 if z 6∈ D∪γ. It is interesting to note

that even if γ is not necessarily closed or if φ is just continuous on γ, then the function

Φ behaves nicely by the following theorem.

Theorem 4.1. Let γ be a finite smooth curve and φ be a continuous function on γ. Then

the function Φ in (4.1) is a holomorphic function on C \ γ and

Φ′(z) =1

2πi

∫γ

φ(ζ)

(ζ − z)2dζ.

Proof. Consider the following difference for small enough |h|:∣∣∣∣Φ(z + h)− Φ(z)

h− 1

2πi

∫γ

φ(ζ)

(ζ − z)2dζ

∣∣∣∣=

∣∣∣∣ 1

2πi

∫γ

(φ(ζ)

h(ζ − z − h)− φ(ζ)

h(ζ − z)− φ(ζ)

(ζ − z)2

)dζ

∣∣∣∣=

∣∣∣∣ h2πi∫γ

φ(ζ)

(ζ − z − h)(ζ − z)2dζ

∣∣∣∣≤ |h|

2π

∫γ

∣∣∣∣ φ(ζ)

(ζ − z − h)(ζ − z)2

∣∣∣∣ |dζ|≤ |h|

2π

m

d3l,

where m is the maximum of |φ| on γ and d is half of the distance from z to γ and l is the

length of γ. The expression clearly tends to zero as h→ 0.

It is interesting to investigate what happens with (4.1) if z approaches a point w on γ

which is not an endpoint. It turns out that in some situations there will exist a limit,

but the limit might be different depending on if we approach the curve from the left

or from the right. If we approach the curve from the left we will denote the limit by

Φ+(w), and if the curve is approached from the right we denote the limit by Φ−(w). The

values Φ+(w) and Φ−(w) are given by Sokhotzki-Plemelj jump theorem. Before we state

Sokhotzki-Plemelj jump theorem we fix our notation for a disk of radius R and define the

principal value of an integral.

Definition 4.2. Let x0, y0 ∈ R and define DR(x0, y0) by

DR(x0, y0) = (x, y) : |(x− x0, y − y0)| < R.16

Definition 4.3. Let γ be a smooth curve and φ be a complex-valued function defined

on γ with a singularity at the point z. Then we define the principal value integral by

PV

∫γ

φ(ζ)dζ = limε→0

∫γε

φ(ζ)dζ,

where γε is the part of γ outside of Dε(z).

Theorem 4.4 (Sokhotzki-Plemelj jump theorem). Let γ be a finite smooth curve and φ

be a complex-valued function defined on γ and satisfying the Holder condition

(4.2) |φ(x)− φ(y)| ≤ C|x− y|α,

for some 0 < α ≤ 1 and C > 0. Let

Φ(z) =1

2πi

∫γ

φ(ζ)

ζ − zdζ

and let w be a point on γ which is not an endpoint. Then both sides of the following

equality will exist and be equal.

(4.3) Φ±(w) =1

2πiPV

∫γ

φ(ζ)

ζ − wdζ ± φ(w)

2.

For the proof of Sokhotzki-Plemelj jump theorem we refer the reader to [5].

4.2. Introduction to harmonic functions. A harmonic function f(x, y) is a function

satisfying the Laplace equation fxx + fyy = 0. The Dirichlet problem in an open domain

Ω is the problem of finding a function f(x, y) which is harmonic in Ω and continuous in

Ω (the closure of Ω) and takes specific values on ∂Ω (the boundary of Ω). The following

theorem shows that if the boundary of Ω is a piecewise smooth closed curve and if a

solution to Dirichlet’s problem exists, then the solution is unique.

Theorem 4.5. Let Ω be an open domain where the boundary is a piecewise smooth closed

curve. If f(x, y) and g(x, y) are harmonic in Ω, continuous on Ω and f(x, y) = g(x, y)

on ∂Ω, then f(x, y) = g(x, y) in Ω.

Proof. Let h(x, y) = f(x, y)− g(x, y). Then h is harmonic in Ω and is equal to 0 on ∂Ω.

Let F = (−hhy, hhx). Since a harmonic function is actually smooth, Greens theorem tells

us that∮∂Ω

F · dr =

∫∫Ω

(∂F2

∂x− ∂F1

∂y

)dxdy

=

∫∫Ω

((∂h

∂x

)2

+

(∂h

∂y

)2)dxdy +

∫∫Ω

(h∂2h

∂x2+ h

∂2h

∂y2

)dxdy.

The integral on the left hand side vanishes since F is zero on ∂Ω, and the right most

integral vanishes since h is harmonic. So we can conclude that∫∫Ω

((∂h

∂x

)2

+

(∂h

∂y

)2)dxdy = 0

17

and therefore that hx and hy are constant 0 on Ω which means that h(x, y) = C (constant)

on Ω. But h is continuous on Ω and h(x, y) = 0 on ∂Ω, so the constant C must be zero.

This means that f(x, y) = g(x, y) on Ω.

4.3. A first glance at the Hilbert transform. From a historical point of view, the

Hilbert transform originated in the work of David Hilbert on integral equations and

boundary value problems in 1905.

The Hilbert transform of a continuous function f defined on the boundary of a simply

connected open domain Ω can be thought of as the values on the boundary of Ω, of the

harmonic conjugate to the solution of the Dirichlet problem with boundary value f . By

the existence of solution of the Dirichlet’s problem (see e.g. [1]), the solution will exist.

And since we are working in a simply connected domain the harmonic conjugate to the

solution will always exists, but will only be unique up to a constant. To make the Hilbert

transform unique we have to choose how to set this constant.

However we will not define the Hilbert transform in this way since we would like to be

able to transform functions for which a solution to Dirichlet’s problem doesn’t exist.

In this section we will find formulas for the Hilbert transform on the unit circle and the

real line.

4.4. Poisson formula for the unit disk. First we solve the Dirichlet problem on a disk

of radius R centered at the origin, and to make it easier for us we assume that there exist

a solution which is harmonic in the closed disk with radius R centered at the origin (this

means that we assume the existence of a harmonic function in an open domain slightly

larger then the disk of radius R centered at the origin). In the actual theorem we don’t

assume the existence of the solution and we only require the solution to be continuous

on the closed disk of radius R (not harmonic), but this variant of the theorem is harder

to prove. For the proof of the full theorem the interested reader is refereed to [1].

Theorem 4.6. Let f(x, y) be real and harmonic on the closed disk DR(0, 0). Then the

values of f in DR(0, 0) are given by

(4.4) f(reiθ) =R2 − r2

2π

∫ 2π

0

f(eit)

R2 − 2Rr cos(θ − t) + r2dt.

Proof. Since f is harmonic on the simply connected closed disk DR(0, 0), f is the real

part of an holomorphic function h(z) = f(z) + ig(z), with g(0) = 0, this makes h unique.

By the Cauchy integral formula and Cauchy integral theorem we get

(4.5) h(z) =1

2πi

∮C

h(ζ)dζ

ζ − z− 1

2πi

∮C

h(ζ)dζ

ζ − z∗,

where C is the positively oriented circle |ζ| = R and z∗ = R2/z is the reflection of z in

C. If we set z = reiθ in (4.5) we get18

h(reiθ) =1

2πi

∮C

h(ζ)dζ

ζ − z− 1

2πi

∮C

h(ζ)dζ

ζ − z∗

=1

2πi

∮C

(1

ζ − z− 1

ζ − z∗

)h(ζ)dζ

=1

2πi

∮C

(1

ζ − z+

z

ζ(ζ − z

))h(ζ)dζ

=1

2πi

∮C

(|ζ|2 − |z|2

ζ|ζ − z|2

)h(ζ)dζ

=1

2π

∫ 2π

0

(R2 − r2

|Reit − reiθ|2

)h(Reit)dt

=1

2π

∫ 2π

0

(R2 − r2

R2 − 2Rr cos(θ − t) + r2

)h(Reit)dt.

(4.6)

Taking the real part of (4.6) gives (4.4).

Now we continue our studies on the unit circle (letting R = 1) to see what happens with

the harmonic conjugate to f(x, y) when we let (x, y) approach the unit circle. To do this

we first write (4.4) in complex form. Observe that if ζ = eit and z = reiθ, then we have

Re

(ζ − zζ + z

)= Re

(e−it(eit + reiθ)

e−it(eit − reiθ)

)= Re

(1 + r cos(θ − t) + ir sin(θ − t)1− r cos(θ − t)− ir sin(θ − t)

)=

1− r2

1− 2r cos(θ − t) + r2.

Now we can write (4.4) with R = 1 as

Re1

2πi

∮C

ζ + z

ζ − zf(ζ)

ζdζ =

1

2π

∫ 2π

0

1− r2

1− 2r cos(θ − t) + r2f(eit)dt = f(z),

where C is the positively oriented circle |ζ| = 1. Let us define h as

(4.7) h(z) =1

2πi

∮C

ζ + z

ζ − zf(ζ)

ζdζ =

1

2πi

∮C

f(ζ)

ζ − zdζ +

z

2πi

∮C

f(ζ)

ζ(ζ − z)dζ.

Then h is an holomorphic function on |z| < 1 (by Theorem 4.1) with real part Re h = f

and imaginary part Im h = g (since Im h(0) = g(0) = 0), so we get h(z) = h(z) for all

|z| < 1. If we let z approach the point w ∈ C on the unit circle in (4.7), then we get from

Sokhotzki-Plemelj jump Theorem 4.3 (observe that in order to apply Sokhotzki-Plemelj

we need that f satisfyes the Holder condition (4.2) on C, so from now on we assume19

that).

h(w) = f(w) + ig(w) =f(w)

2+f(w)

2+

1

2πiPV

∮C

ζ + w

ζ − wf(ζ)

ζdζ

= f(w) +1

2πPV

∫ 2π

0

f(eit)ei(−t/2−θ/2)(eiθ + eit)

ei(−t/2−θ/2)(−eiθ + eit)dt

= f(w) +1

2πPV

∫ 2π

0

f(eit)ei(θ−t)/2 + e−i(θ−t)/2

−ei(θ−t)/2 + e−i(θ−t)/2dt

= f(w) +i

2πPV

∫ 2π

0

f(eit) cot

(θ − t

2

)dt.

Since the real part of h is f , we get

g(w) = Imh(w) =1

2πPV

∫ 2π

0

f(eit) cot

(θ − t

2

)dt.

This formula is what we call the Hilbert transform on the unit circle.

Let us formulate this as a theorem.

Theorem 4.7. Let f(x, y) be real and harmonic on the closed unit disk and let f satisfy

the Holder condition (4.2) on the unit circle. Then the values on the unit circle of the

harmonic conjugate g to f with g(0) = 0 are given by

g(eiθ) =1

2πPV

∫ 2π

0

f(eit) cot

(θ − t

2

)dt.

4.5. Poisson formula for the upper half plane. Now we will develop similar formulas

for the solution of the Dirichlets problem in the upper half plane. In this case the solution

is not unique, we can for example always add the function f(x, y) = y to the solution to

get a new one. However, if we require that the solution is bounded then we get a unique

solution, see [1].

We start by proving a formula for the bounded solutions of Dirichlet’s problem in the

upper half plane. The technique used here is very similar to that in Section 4.4. Again

we assumes the existence of a solution in the closed upper half plane.

Theorem 4.8. Let f(x, y) be the real part of a bounded holomorphic function on the

closed upper half plane y ≥ 0. Then the values of f where y > 0 are given by

(4.8) f(x, y) =y

π

∫ ∞−∞

f(t, 0)

(t− x)2 + y2dt.

Proof. Let h(z) = f(z) + ig(z) be a bounded holomorphic function with real part f . By

the Cauchy integral formula and Cauchy integral theorem we get, for large enough R,

h(z) =1

2πi

∫ R

−R

(1

t− z− 1

t− z

)h(t, 0)dt+

1

2πi

∫CR

(1

ζ − z− 1

ζ − z

)h(ζ)dζ,

where CR is the positively oriented half circle |ζ| = R, y ≥ 0. If we set z = x + iy and

simplify we get

(4.9) h(x, y) =y

π

∫ R

−R

h(t, 0)

(t− x)2 + y2dt+

y

π

∫CR

h(ζ)

(ζ − z)(ζ − z)dζ.

20

If now ‖h‖∞ ≤M , then one has∣∣∣∣∫CR

h(ζ)

(ζ − z)(ζ − z)dζ

∣∣∣∣ ≤ ∫CR

∣∣∣∣ h(ζ)

(ζ − z)(ζ − z)

∣∣∣∣ |dζ|≤∫CR

M

(|ζ| − |z|)2|dζ|

=MRπ

R2(1− zR

)2,

which tends to 0 as R→∞. Hence, as R→∞ it follows from (4.9) that

(4.10) h(x, y) =y

π

∫ ∞−∞

h(t, 0)

(t− x)2 + y2dt.

The proof completes by taking the real part of (4.10).

To see what happens to the harmonic conjugate as y → 0, we write (4.8) in complex form

by noting that if ζ is real we have

Im

(1

ζ − z

)= Im

(1

ζ − x− iy

)=

y

(ζ − x)2 + y2.

And that gives us

Re1

πi

∫ ∞−∞

f(t, 0)

t− zdt =

1

π

∫ ∞−∞

y

(t− x)2 + y2f(t, 0)dt = f(x, y).

We define

(4.11) h(z) =1

πi

∫ ∞−∞

f(t, 0)

t− zdt.

Then h is a holomorohic function in y > 0 (by Theorem 4.1) with Re h = f and

limy→∞

h(x, y) = 0, so if we choose g to have the property limy→∞

g(x, y) = 0 we get h(x, y) =

h(x, y) for all y > 0. If we let z approach the point s ∈ R on the real line in (4.11),

assuming f satisfies the Holder condition (4.2) on R, a modification of the argument

above yields

(4.12) h(s) = f(s) +1

πiPV

∫ ∞−∞

f(t, 0)

t− sdt.

For the details see [4].

Since the real part of (4.12) is f(s), we can prove the following theorem by taking the

imaginary part of (4.12).

Theorem 4.9. Let f(x, y) be the real part of a bounded holomorphic function on the

closed upper half plane and let f satisfy the Holder condition (4.2) on the real line. Then

there exists a harmonic conjugate g to f with limy→∞

g(x, y) = 0 and the values of g on the

real line are given by

g(s, 0) =1

πPV

∫ ∞−∞

f(t, 0)

s− tdt.

21

5. The Hilbert transform on the real line

5.1. Definition. The work we done so far with harmonic functions makes the following

definitions of the Hilbert transform on the real line natural.

Definition 5.1. The Hilbert transform on the real line is defined pointwise by

Hf(x) =1

πPV

∫ ∞−∞

f(t)

x− tdt.

Similarly we can also define the Hilbert transform on the unit circle.

Definition 5.2. The Hilbert transform on T = [0, 2π] is defined pointwise by

Hf(x) =1

2πPV

∫ 2π

0

f(x) cot

(x− t

2

)dt.

Remark 5.3. In the definitions, we are defining the Hilbert transform as a pointwise limit

as ε → 0 in the principal value. But in many applications where f ∈ Lp for 1 < p < ∞it is useful to take the limit in Lp sense. It can be shown that these definitions are

equivalent, see e.g. [3].

5.2. Properties of the Hilbert transform. In this section we will take a close look at

some of the basic properties of the Hilbert transform on the real line. It follows directly

from the definition of the Hilbert transform that the associated operator is linear.

Another slightly less obvious property is that the Hilbert transform commutes with trans-

lations and positive dilations. Let τa be the translation operator defined by τaf(x) =

f(x − a), and let Sa for a > 0 be the dilation operator Saf(x) = f(ax). We can get

Hτaf = τaHf and HSaf = SaHf by a simple change of variables:

Hτaf(x) =1

πPV

∫ ∞−∞

f(t− a)

x− tdt =

1

πPV

∫ ∞−∞

f(u)

x− a− udu = τaHf(x),

HSaf(x) =1

πPV

∫ ∞−∞

f(at)

x− tdt =

1

πPV

∫ ∞−∞

f(u)

ax− udu = SaHf(x).

If we now let R be the reflection operator Rf(x) = f(−x), the we can by another chance

of variables get HRf = −RHf .

HRf(x) =1

πPV

∫ ∞−∞

f(−t)x− t

dt = − 1

πPV

∫ ∞−∞

f(u)

−x− udu = −RHf(x).

There is no simple formula for the Hilbert transform of a product of two functions.

However, we will discuss the special cases of the Hilbert transform of xf(x) and f(x)/x.

If we want to consider the Hilbert transform of xnf(x) for n ∈ Z we can iterate the

following formulas.

Theorem 5.4. Let f be integrable. Then

(5.1) H(xf(x)) = xHf(x)− 1

π

∫ ∞−∞

f(t)dt.

If we instead require that f/x is integrable, then

(5.2) H

(f(x)

x

)=Hf(x)−Hf(0)

x.

22

Proof.

H(xf(x)) =1

πPV

∫ ∞−∞

tf(t)

x− tdt

=x

πPV

∫ ∞−∞

f(t)

x− tdt− 1

πPV

∫ ∞−∞

(x− t)f(t)

x− tdt

= xHf(x)− 1

π

∫ ∞−∞

f(t)dt

and by using (5.1) on the function f(x)/x we get

Hf(x) = xH

(f(x)

x

)− 1

π

∫ ∞−∞

f(t)

tdt

= xH

(f(x)

x

)+Hf(0).

(5.2) follows by moving Hf(0) to the other side and dividing by x.

The Hilbert transform of a function f can be seen as the convolution of f with k(x) =

(πx)−1, where one has to remember to calculate the integral as a principal value. To

make it formal we let

kε(x) =

(πx)−1 if |x| ≥ ε,

0 if |x| < ε,

and get

Hf(x) = limε→0

Hεf(x) = limε→0

f ∗ kε(x),

where Hεf = f ∗ kε is called the truncated Hilbert transform.

5.3. Some Hilbert transforms. Let us calculate the Hilbert transform of some basic

functions.

The Hilbert transform of a constant c is given by

H(c) =1

πPV

∫ ∞−∞

c

x− tdt

=c

πPV

∫ ∞−∞

1

udu = 0.

We can calculate the Hilbert transform of sinx by

H(sinx) =1

πPV

∫ ∞−∞

sin t

x− tdt

= − 1

πPV

∫ ∞−∞

sin(u+ x)

udu

= −cos(x)

πPV

∫ ∞−∞

sin(u)

udu− sin(x)

πPV

∫ ∞−∞

cos(u)

udu

= − cosx.

By using that the Hilbert transform commutes with translation we get

H(cosx) = H(sin(x+ π/2)) = − cos(x+ π/2) = sin x.23

5.4. The Riesz inequality on L2. Riesz’s inequality is that the Hilbert transform is a

bounded linear operator from Lp(R) to Lp(R) for 1 < p < ∞. The proof of this is hard

and requires interpolation so we will prove it only in the special case where p = 2. The

full proof can be read in [3].

Theorem 5.5. Let f ∈ L2(R). Then ‖Hf‖2 = ‖f‖2 and

FHf(x) = −isgn(ω)Ff.

Proof. Let Hε,η be the double truncated Hilbert transform defined by Hε,ηf = f ∗ kε,η,where

kε,η(x) =

(πx)−1 if ε ≤ |x| ≤ η,

0 elsewhere.

Let f ∈ L2(R). Then Hε,ηf ∈ L2(R) by Theorem 2.14 since f ∈ L2(R) and kε,η ∈ L1(R).

If we take the Fourier transform of Hε,ηf and use Theorem 3.23 we get FHε,ηf = FfFkε,η,where Fkε,η can be calculated by

Fkε,η(ω) =

∫ε≤|t|≤η

e−iωt

πtdt

=

∫ −ε−η

e−iωt

πtdt+

∫ η

ε

e−iωt

πtdt

= −∫ η

ε

e−iωt

πtdt+

∫ η

ε

e−iωt

πtdt

=−2isgn(ω)

π

∫ η|ω|

ε|ω|

sin t

tdt.

We see that as ε→ 0 and η →∞, Fkε,η(ω)→ −isgn(ω) for every ω ∈ R. From that we

can draw the conclusion that there exists a constant C independent of ε, η and ω such

that |Fkε,η(ω)| ≤ C. By Remark 5.3 we have that Hf = limε→0

limη→∞

Hε,ηf , where the limits

are to be taken in L2. We can now show that

(5.3) Hf = limε→0

limη→∞

Hε,ηf = F−1(−isgn(ω)Ff(ω))

by the following observation:

limε→0

limη→∞

∥∥Hε,ηf −F−1(−isgn(ω)Ff(ω))∥∥

2

=1√2π

limε→0

limη→∞‖Fkε,η(ω)Ff(ω)− (−isgn(ω)Ff(ω))‖2

=1√2π

limε→0

limη→∞‖(Fkε,η(ω)− (−isgn(ω)))Ff(ω)‖2 = 0,

where we have used Plancherel’s theorem and the dominated convergence theorem.

If we take the Fourier transform of both sides of equation (5.3) we get

FHf(x) = −isgn(ω)Ff(ω).

By taking the norm on both sides we get

‖FHf‖2 = ‖ − isgn(ω)Ff(ω)‖2 = ‖Ff‖2

24

and if we use Plancherel’s theorem we get

‖Hf‖2 = ‖f‖2.

Some direct application of the last theorem include the following:

Theorem 5.6. Let f ∈ L2(R) be a differentiable function with f ′ ∈ L2(R) and let Hf be

differentiable with (Hf)′ ∈ L2(R) then H[f ′] = (Hf)′.

Proof. The technique of this proof is to compare the Fourier transform of H[f ′] and of

(Hf)′, if they are equal we can take the inverse Fourier transform to get H[f ′] = (Hf)′.

Hf′(ω) = −isgn(ω)f ′(ω) = −isgn(ω)(iω)f(ω) = sgn(ω)f(ω)

and

(Hf)′(ω) = iωHf = iω(−isgn(ω))f(ω) = sgn(ω)f(ω).

Theorem 5.7. If f ∈ L2(R), then HHf = −f .

Proof. Let T be the operator multiplying function with −isgn(x), Tf(x) = −isgn(x)f(x).

Then

HHf = F−1TFF−1TFf = F−1TTFf = −f.

One reason why the Hilbert transform is so important in mathematics is because it is

almost the only translation and dilation invariant bounded linear operator from Lp(R)

to Lp(R). Let us state it as a theorem without proof. The proof can be found in [3].

Theorem 5.8. Let T : Lp(R) → Lp(R) be a translation and dilation invariant bounded

linear operator, then

T = aH + bI,

where I is the identity operator.

6. Normconvergence of Fourier series

In this section we will see that the Hilbert transform can be used to study the convergence

of Fourier series. We will see that the Fourier series of f converges to f in Lp-norm. In

formulas that can be written as

limn→∞

(∫ 2π

0

|f(t)− Snf(t)|pdt)1/p

= 0.

One of the key ideas is that if can show that the operators Sn are uniformly bounded in

Lp(T), then limn→∞

Snf exists and limn→∞

Snf = f .

Theorem 6.1. Let 1 ≤ p <∞ and let f ∈ Lp(T) and assume that the partial summation

operators Sn are uniformly bounded. Then limn→∞

Snf exists and limn→∞

Snf = f .

25

Proof. Assume ‖Sn‖op ≤ C for some 0 < C < ∞. Let ε > 0 and choose a trigonometric

polynomial g such that ‖f − g‖p < ε/(2 + 2C) (this is possible by Theorem 3.2), and let

n be greater then the degree of g. Then

‖f − Snf‖p ≤ ‖f − g‖p + ‖g − Sng‖p + ‖Sng − Snf‖p

≤ ε

2 + 2C+ 0 + C

ε

2 + 2C

<ε

2+ε

2= ε.

Since this statement holds for arbitrary ε > 0 the statement follows.

6.1. The Hilbert transform as a multiplier. We can get one variant of the Fourier

series if we in front of every coefficient in the Fourier series add a weight from the sequence

Λ = λk∞−∞. The linear multiplier operator MΛ is formally defined in the following way.

Definition 6.2. Let Λ = λk∞−∞. Then the multiplier operator MΛ is defined by

MΛf(x) =∞∑

k=−∞

λkf(k)ek,

where ek(x) = eikx. Observe one also has MΛf(k) = λkf(k). This follows from the

uniqueness of the Fourier expansion.

Of course there is no guarantee that all multiplier operators will converge for all functions,

so we have to be a little careful which multiplier operator we work with. One quick

observation is that the operators Sn for the partial sums is a very easy multiplier operator

coming from the sequence Λ = λk, where λk = 1 if |k| ≤ n and λk = 0 if |k| > n.

Let us define the Hilbert transform as a multiplier operator.

Definition 6.3. Let Λ = λk, where λk = −isgn(k) and sgn(k) is defined by

sgn(k) =

1 if k < 0,

0 if k = 0,

−1 if k > 0.

Then we define the Hilbert transform H by Hf = MΛf .

Remark 6.4. By an argument similar to the one in the proof of Theorem 5.5, it is possible

to show that this definition of the Hilbert transform coincides with Definition 5.2.

The Hilbert transform is closely related to partial summation in the following way.

Theorem 6.5. Let f ∈ Lp(T) and let en(x) = einx for n ∈ Z. Then

Snf =1

2

(ie−nH[enf ]− ienH[e−nf ] + f(−n)e−n + f(n)en

).

Proof. First note that eRf(k +R) = f(k) since

eRf(k +R) =

∫ 2π

0

eiRtf(t)e−i(k+R)tdt =

∫ 2π

0

f(t)e−iktdt = f(k).

26

Now we get

1

2

(ie−nH[enf ]− ienH[e−nf ] + f(−n)e−n + f(n)en

)=

1

2

(ie−n

∞∑k=−∞

−isgn(k)enf(k)ek − ien∞∑

k=−∞−isgn(k)e−nf(k)ek + f(−n)e−n + f(n)en

)

=1

2

( ∞∑k=−∞

sgn(k + n)enf(k + n)ek −∞∑

k=−∞sgn(k − n)e−nf(k − n)ek + f(−n)e−n + f(n)en

)

=1

2

( ∞∑k=−∞

sgn(k + n)f(k)ek −∞∑

k=−∞sgn(k − n)f(k)ek +

1

2(f(−n)e−n + f(n)en

)

=

∞∑k=−∞

1

2

((sgn(k + n)− sgn(k − n) + χ−n(k) + χn(k)) f(k)ek

)=

∞∑k=−∞

λkf(k)ek,

where χR(k) is 0 if k 6= R and 1 if k = R. This is the same multiplier operator as the one

for partial summation since we get from an easy calculation that for k < −n or k > n it

yields λk = 0 and for −n < k < n we have that λk = 1 and for k = −n or k = n we find

that λk = 1.

The key point in showing the norm convergence is to show that Sn is uniformly bounded,

since if Sn is uniformly bounded we can then use Theorem 6.1 to get convergence. Now

when we have showed that Sn can be written in terms of the Hilbert transform we wish to

estimate ‖Sn‖op in terms of the ‖H‖op. This type of estimate is possible by the following

theorem.

Theorem 6.6. It yields that ‖Sn‖op ≤ ‖H‖op + 1, with the operator norm taken in Lp.

In particular Sn is uniformly bounded if H is bounded.

Proof. First note that the operator Lf = enf is a isometry since

‖Lf‖p =

(∫T|enf |pdλ

)1/p

=

(∫T|f |pdλ

)1/p

= ‖f‖p,

which specifically means that ‖L‖op = 1, also note that for the projection operators

Lf = f(n)en we have ‖L‖op = 1 since

‖f(n)en‖p =1

2π

∣∣∣∣∫Tfe−ndλ

∣∣∣∣ ‖en‖p≤ 1

2π‖f‖p‖e−n‖q

(∫T|en|pdλ

)1/p

=1

2π‖f‖p(2π)1/q(2π)1/p

= (2π)1/q+1/p−1‖f‖p= ‖f‖p

by Holder’s inequality where 1/p+ 1/q = 1.27

Now we can estimate ‖Sn‖ by

‖Snf‖p =1

2‖ie−nH[enf ]− ienH[e−nf ] + f(−n)e−n + f(n)en‖p

≤ 1

2(‖Hf‖p + ‖Hf‖p + 1 + 1) = ‖Hf‖p + 1.

Thus if we could show that the Hilbert transform is bounded on Lp(T), then we will get

that the partial sums of the Fourier series are uniformly bounded and therefore we can

use Theorem 6.1 to show that the Fourier series of f will converge to f for all functions

f in Lp(T). We formulate this as a theorem.

Theorem 6.7. (Marcel Riesz) If the Hilbert transform is a bounded operator from Lp(T)

to Lp(T), then norm convergence is valid in Lp(T).

6.2. The Hilbert transform is bounded on L2. In this section we will show that

the Hilbert transform is bounded on L2(T). However, in [3] there is a proof that for

1 < p <∞, the Hilbert transform is bounded on Lp(T).

In L2(T) we get as a consequence of Parseval’s identity that we have an explicit formula

for the operator norm in L2(T) of any multiplier operator.

Theorem 6.8. Let Λ = λk be a sequence and assume that the corresponding multiplier

operator MΛ : L2(T)→ L2(T) is well defined. Then ‖MΛ‖ = sup |λk|. In particular, MΛ

is bounded if and only if Λ is bounded.

Proof. It yields that

‖MΛf‖22 = 2π

∞∑k=−∞

|MΛf(k)|2

= 2π∞∑

k=−∞

|λkf(k)|2

≤ 2π sup |λk|2∞∑

k=−∞

|f(k)|2

= sup |λk|2‖f‖22.

Here we have used the Parseval’s identity 3.7 at the first and last equality. On the other

hand, let nk be a sequence of integers such that limk→∞|λnk | = sup |λn|. Then ‖MΛenk‖ =

‖λnkenk‖ = |λnk |‖enk‖ so ‖MΛenk‖/‖enk‖ → sup |λn| and that shows equality.

Using Theorem 6.8, we note that the Hilbert transform is a multiplier operator with

sup |λn| = 1 and is therefore bounded on L2(T). This means by Theorem 6.7 that norm

convergence is valid in L2(T).

As we mentioned earlier, it can be shown that the Hilbert transform is bounded on Lp(T)

for 1 < p <∞. Taking this fact for granted, Theorem 6.7 yields that norm convergence

of the Fourier series is valid in Lp(T). We formulate this as a theorem.28

Theorem 6.9. Let f ∈ Lp(T), where 1 < p <∞. Then

limn→∞

(∫ 2π

0

|f(t)− Snf(t)|pdt)1/p

= 0.

6.3. A counterexample in L∞(T). The endpoints p = 1 and p = ∞ are excluded in

the previous theorem. The reason is that the result of the theorem turns out to be false

in these two cases. We provide a counterexample in the case of L∞(T).

Let f be defined by

f(x) =

1 , if 0 < x < π,

0 , if π < x < 2π.

Then

f(0) =1

2π

∫ 2π

0

f(t)dt = 1/2

and

f(n) + f(−n) =1

2π

∫ 2π

0

f(t)(e−int + eint)dt

=1

π

∫ 2π

0

f(t) cos(nt)dt

=1

π

∫ π

0

cos(nt)dt = 0.

This means that

Snf(0) =n∑

k=−n

f(k) = 1/2.

Since the trigonometric polynomials are continuous, there is for every ε > 0 an interval

[0, δ] of nonzero measure where |f(x)− Snf(x)| ≥ 1/2− ε, and hence ‖f − Snf‖∞ ≥ 1/2

so Snf 6→ f in L∞(T).

References

[1] C.L. Evans. Partial differential equations. Second edition. Graduate Studies in Mathematics, 19.

American Mathematical Society, Providence, RI, 2010.

[2] S. G. Krantz. A panorama of harmonic analysis, Washington University in St. Louis, 1999.

[3] F. W. King. Hilbert Transforms, University of Wisconsin-Eau Claire, 2009.

[4] B. L. Moiseiwitsch. Integral Equations, The Queen’s University of Belfast, 2005.

[5] F. D. Gakhov. Boundary Value Problems, Pergamon Press Ltd, 1966.

[6] W. Rudin. Real and complex analysis. Third edition. McGraw-Hill Book Co., New York, 1987.

[7] A. Torchinsky. Real-variable methods in harmonic analysis. Reprint of the 1986 original. Dover

Publications, Inc., Mineola, NY, 2004.

[8] A. Torchinsky. Real variables. Addison-Wesley Publishing Company, Advanced Book Program, Red-

wood City, CA, 1988.

29