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Examensarbete

An investigation concerning the absolute convergence ofFourier series

Axel Tiger NorqvistLiTH - MAT - EX - - 2016/01 - - SE

An investigation concerning the absolute convergence ofFourier series

MAI, Linkopings Universitet

Axel Tiger Norqvist

LiTH - MAT - EX - - 2016/03 - - SE

Examensarbete: 16 hp

Level: G2

Supervisor: Mats Aigner,MAI, Linkopings Universitet

Examiner: Magnus Herberthson,MAI, Linkopings Universitet

Linkoping: june 2016

Abstract

In this Bachelor’s thesis we present a few results about the absolute convergenceof Fourier series, followed by an example of a differentiable function whoseFourier series does not converge absolutely. In the end we provide a suggestionfor future work on generalizing the given example, and we briefly discuss anissue that has not been given much attention in the existing literature on thesubject.

Keywords: Fourier analysis, Absolute convergence, Lipschitz condition, Bern-stein’s theorem

URL for electronic version:

http://urn.kb.se/resolve?urn=urn:nbn:se:liu:diva-129363

Tiger Norqvist, 2016. v

Acknowledgements

First and foremost I would like to thank my supervisor Mats Aigner − notonly for coming up with an idea for the subject of this thesis, but also forproviding me with great support and feedback throughout the process of writingmy thesis. I would also like to thank my examiner Magnus Herberthson for hiscooperative nature and engagement during the course of writing, and for hissuggestions of improvements on the final product. Lastly I would like to thankmy opponent Karl Nygren, who has been very patient and understanding despitethe delayments that have occurred, mostly due to my own shortcomings.

Tiger Norqvist, 2016. vii

Contents

1 Introduction 11.1 Background and overview . . . . . . . . . . . . . . . . . . . . . . 11.2 Basics on Fourier series . . . . . . . . . . . . . . . . . . . . . . . 1

1.2.1 The space C(T) . . . . . . . . . . . . . . . . . . . . . . . . 21.2.2 Fourier coefficients . . . . . . . . . . . . . . . . . . . . . . 2

2 The absolute convergence of Fourier series 52.1 The space A(T) . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52.2 Lipschitz conditions and Bernstein’s theorem . . . . . . . . . . . 6

3 Differentiability and absolute convergence 93.1 A special function . . . . . . . . . . . . . . . . . . . . . . . . . . 93.2 Estimating the Fourier coefficients of f . . . . . . . . . . . . . . . 10

3.2.1 Estimating I2n . . . . . . . . . . . . . . . . . . . . . . . . . 11

3.2.2 Estimating the contribution from I1n . . . . . . . . . . . . 13

3.2.3 Estimating the size of |I3n| . . . . . . . . . . . . . . . . . . 15

3.2.4 Proof of lemmas 3.2.3, 3.2.4, and 3.2.5 . . . . . . . . . . . 18

4 Closing remarks 23

Appendix A Evaluating an integral 25

Bibliography 27

Tiger Norqvist, 2016. ix

x Contents

Chapter 1

Introduction

1.1 Background and overview

With its origin in the work of Joseph Fourier, who used trigonometric series asa way to simplify heat and wave equations in the beginning of the 19th century,Fourier analysis has become a widely used tool in many industries and areas ofresearch dealing with, for instance, periodic functions and differential equations.

In general, the study of trigonometric series has been a driving force for thedevelopment of mathematics during the last two centuries. Basic notions andresults concerning functions (and more generally, distributions) have been heav-ily influenced by convergence problems concerning Fourier series. The problemsin the theory of trigonometric series also have deep connections with methods inLebesgue integration, which have helped mathematicians in representing func-tions by their Fourier series. Nowadays, this area of mathematical research ispart of what is known as harmonic analysis, which is one of the most intensivelystudied branches in mathematics.

In this text, focus will lie on continuous 2π-periodic functions and the abso-lute convergence of their Fourier series. The study of the space of functions withabsolutely convergent Fourier series has given rise to many interesting resultsin the past, and the area still contains many open problems.

We will make a small contribution to the theory of this subject by givingan example of an everywhere differentiable function whose Fourier series doesnot converge absolutely, as there is a lack of such examples in the existinglitterature concerning Fourier analysis. We will also give a brief overview of someelementary results about the space of continuous functions with an absolutelyconvergent Fourier series, with emphasis on the connection this space has withLipschitz conditions.

1.2 Basics on Fourier series

In the rest of this chapter we go over some basic definitions and results, whichwill be useful in the coming chapters. For more details on the fundamentals ofFourier analysis, and proofs of the results presented, the reader is referred to[1], chapter 1.

Tiger Norqvist, 2016. 1

2 Chapter 1. Introduction

1.2.1 The space C(T)Let T > 0. Recall that a function f : R → C is said to be T -periodic iff(t) = f(t+ T ) for all t ∈ R.

As we will only study continuous 2π-periodic functions in this text, we willhenceforth denote the linear space of continuous 2π-periodic functions by C(T).If a function f ∈ C(T) has a continuous derivative of order k, k ∈ N, we saythat f belongs to Ck(T), which is a linear subspace of C(T).

For estimations made later on, we introduce three different norms on C(T):

Definition 1.2.1. Let f be a function in C(T). We define the 1-norm of f inC(T) as

‖f‖1 =1

2π

∫ π

−π|f(t)|dt, (1.1)

and the 2-norm of f in C(T) as

‖f‖2 =

(1

2π

∫ π

−π|f(t)|2dt

)1/2

. (1.2)

The supremum norm in C(T) is defined as

‖f‖∞ = supt∈[0,2π)

|f(t)|. (1.3)

1.2.2 Fourier coefficients

Definition 1.2.2. For a function f in C(T), we define the Fourier series of f tobe the trigonometric series, given by

S[f ] ∼∞∑

n=−∞f(n)eint (1.4)

where

f(n) =1

2π

∫ π

−πf(t)e−intdt, n ∈ Z, (1.5)

is the n:th Fourier coefficient of f .

The convergence of Fourier series

We say that S[f ] converges at t0 if the partial sums

N∑n=−N

f(n)eint (1.6)

converge as N ∈ N tends to infinity, and we call the limit the sum of S[f ] at t0.It is well known that if S[f ] converges at t0, then f(t0) is equal to the sum ofits Fourier series at t0, as we are only dealing with continuous functions in thistext.

It is not a simple task to describe the general properties of functions whoseFourier series converge at a given point, and this is a central issue in the theoryof trigonometric series. One of the most famous results on this subject is dueto Lennart Carleson, who in 1966 proved that every square-integrable periodicfunction has a Fourier series that converges almost everywhere.

1.2. Basics on Fourier series 3

Definition 1.2.3. The Fourier series of f is said to be absolutely convergent if

∞∑n=−∞

|f(n)| <∞. (1.7)

One direct consequence of the definition above is that if we know that afunction has an absolutely convergent Fourier series, then we also know that forevery t ∈ R, S[f ] is an absolutely convergent series. This implies that f is equalto the sum of S[f ] everywhere on R.

Elementary properties of Fourier coefficients

One fundamental result, which is a direct consequence of the definition of Fouriercoefficients, is that for all f ∈ C(T) and all n ∈ Z, we have that

|f(n)| ≤ ‖f‖1 . (1.8)

From the definition of Fourier coefficients we also get that the operationf 7→ f is linear, that is

(α · f + β · g) (n) = α · f(n) + β · g(n), (1.9)

for any f, g ∈ C(T) and α, β ∈ R.The next result provides formulas for calculating Fourier coefficients for

translations as well as for the derivative of f , which may not be as obviousas the previous statement.

Proposition 1.2.1. For f ∈ C(T), let fh be defined by fh(t) = f(t − h) forh ∈ R. Then

fh = e−inhf(n). (1.10)

Assuming that f ∈ C1(T), we can calculate the Fourier coefficients of f ′ bythe formula

f ′(n) = inf(n). (1.11)

Finally, we present Parseval’s identity:

Theorem 1.2.2 (Parseval’s identity). Suppose that f ∈ C(T). Then

∞∑n=−∞

|f(n)|2 = ‖f‖22 . (1.12)

4 Chapter 1. Introduction

Chapter 2

The absolute convergence ofFourier series

In this chapter, we go over a few elementary results for the space of functionsin C(T) with an absolutely convergent Fourier series. What will be covered hereis for the most part centered around Lipschitz conditions, which will be definedlater on.

For more results about the space of functions with an absolutely convergentFourier series, we refer to [2], as it stands as one of the most comprehensivepresentations of the theory of trigonometric series.

2.1 The space A(T)

By A(T) we mean the subspace of functions in C(T) which have an absolutelyconvergent Fourier series. We introduce the norm ‖ · ‖A for functions in A(T)as

‖f‖A =

∞∑n=−∞

|f(n)|. (2.1)

The first theorem presented is one of the most fundamental results aboutabsolutely convergent Fourier series.

Theorem 2.1.1. If f ∈ C1(T), then f ∈ A(T).

Proof. Having that f ′(n) = in · f(n), combined with the fact that f ′ ∈ C(T),Parseval’s identity yields

∞∑n=−∞

n2|f(n)|2 =

∞∑n=−∞

|f ′(n)|2 = ‖f ′‖22 <∞. (2.2)

Using the result from equation (2.2), coupled with the Cauchy-Bunyakovsky-


6 Chapter 2. The absolute convergence of Fourier series

Schwarz inequality, we end up with:

‖f‖A = |f(0)|+∑n 6=0

|f(n)| = |f(0)|+∑n 6=0

|n||f(n)| · 1

|n|

≤ |f(0)|+

(∑n 6=0

n2|f(n)|2)1/2

·

(∑n 6=0

1

n2

)1/2

<∞. (2.3)

Thus f ∈ A(T), as proposed.

Remark: Parseval’s identity is valid for any function in L2(T), where L2(T)is the linear space of 2π-periodic square-integrable functions, that is

f ∈ L2(T)⇐⇒∫ 2π

0

|f(t)|2dt <∞.

It is a simple matter to modify the proof above to be valid for any continuousfunction with a derivative given by a function in L2(T).

As noted in [2], chapter 6, theorem (3.8), we can generalize the above theo-rem even further than this. It turns out that it is enough for f to be absolutelycontinuous, and f ′ to be in Lp(T), p > 1, to ensure that S[f ] converges abso-lutely. This, however, cannot be proven using the same techniques as above.

2.2 Lipschitz conditions and Bernstein’s theo-rem

The space Λα(T), defined below, was originally studied by Lipschitz while de-veloping results about the convergence of Fourier series. We will, however, onlystudy Λα(T) in the light of what it says about the absolute convergence ofFourier series.

Definition 2.2.1. Assume that f ∈ C(T). If f satisfies the condition

supt∈R,h 6=0

|f(t+ h)− f(t)||h|α

<∞

for some constant α ∈ (0, 1), then f is said to fulfill the Lipschitz condition oforder α. We define Λα(T) as the space of functions that fulfill this property.The norm on Λα(T) is defined as

‖f‖Λα = ‖f‖∞ + supt∈R,h 6=0

|f(t+ h)− f(t)||h|α

. (2.4)

We will in a moment show that if f ∈ Λα(T), it follows that f(n) = O(n−α

).

However, before we can do this we need to define the modulus of continuity andthe integral modulus of a function in C(T):

Definition 2.2.2. For a function f ∈ C(T), the modulus of continuity of f isdefined as

ω(f, h) = sup|δ|≤h

‖f(t+ δ)− f(t)‖∞ . (2.5)

2.2. Lipschitz conditions and Bernstein’s theorem 7

The integral modulus of continuity of f is defined as

Ω(f, h) = sup|δ|≤h

‖f(t+ δ)− f(t)‖1 . (2.6)

Worth noting is that for every continuous 2π-periodic function f , we havethat Ω(f, h) ≤ ω(f, h), which follows directly from the definitions above.

The integral modulus of a function can be used to estimate the magnitudeof Fourier coefficients, which the next theorem shows:

Theorem 2.2.1. For n 6= 0, |f(n)| ≤ 12Ω(f, π|n|

).

Proof. We have that

f(n) =1

2π

∫ π

−πf(t)e−intdt =

−1

2π

∫ π

−πf(t)e−in(t+π/n)dt

=−1

2π

∫ π

−πf(t− π

n

)e−intdt.

Through simple manipulations we get

f(n) =1

2

(1

2π

∫ π

−πf(t)e−intdt

)+

1

2

(−1

2π

∫ π

−πf(t− π

n

)e−intdt

)=

1

2

(1

2π

∫ π

−π

(f(t)− f

(t− π

n

))e−intdt

).

Thus

|f(n)| ≤ 1

2

(1

2π

∫ π

−π

∣∣∣f(t)− f(t− π

n

)∣∣∣ dt) ≤ 1

2Ω

(f,

π

|n|

).

Now we can easily prove the previously mentioned result about Λα(T):

Corollary 1. If f ∈ Λα(T), then f(n) = O(n−α

).

Proof. By the definition of Lipschitz conditions, we have that there is a constantC ≥ 0 such that

supt∈R,h 6=0

|f(t+ h)− f(t)||h|α

= C.

This implies that

ω(f,π

n

)≤ C ·

(πn

)α= O

(n−α

),

which in turn gives

|f(n)| ≤ Ω(f,π

n

)≤ ω

(f,π

n

)= O

(n−α

).

The corollary above gives us a rough estimate for the magnitude of theFourier coefficients for functions in Λα(T). However, this cannot be used todetermine if a function is in A(T) or not; after all, the rate of convergence givenby the terms being O

(n−α

)for α ∈ (0, 1) is not good enough to ensure the

absolute convergence of a series.A more powerful result is needed, namely, Bernstein’s theorem, which states

that it is enough for α to be greater than 1/2 to ensure that f ∈ Λα(T) has anabsolutely convergent Fourier series.

8 Chapter 2. The absolute convergence of Fourier series

Theorem 2.2.2 (Bernstein’s theorem). Assume that f ∈ Λα(T), for someα > 1/2. Then f ∈ A(T), and

‖f‖A ≤ cα ‖f‖Λα ,

where the constant cα only depends on α.

Proof. Let fh(t) = f(t− h). Then we have that (fh − f)(n) = (e−inh − 1)f(n).If we take h = 2π/(3 · 2k) and 2k ≤ |n| ≤ 2k+1, we have that e−inh = e±i2πγ/3

for some γ ∈ [1, 2]. This yields

∣∣e−inh − 1∣∣ ≥ ∣∣∣ei 2π3 − 1

∣∣∣ =

√9

4+

3

4=√

3 ≥ 1,

which in turn gives us, coupled with Parseval’s identity, that∑2k≤|n|<2k+1

|f(n)|2 ≤∑n

∣∣e−inh − 1∣∣2 |f(n)|2 =

∑n

| (fh − f)(n)|2

= ‖fh − f‖22 ≤ ‖fh − f‖2∞ ≤

(2π

3 · 2k

)2α

‖f‖2Λα . (2.7)

Noting that the left hand side in (2.7) is a sum of finitely many terms, theCauchy-Bunyakovsky-Schwarz inequality yields∑

2k≤|n|<2k+1

|f(n)| =∑

2k≤|n|<2k+1

1 · |f(n)| ≤ 2(k+1)/2

(2π

3 · 2k

)α‖f‖Λα

=√

2 · 2k(1/2−α)

(2π

3

)α‖f‖Λα . (2.8)

As α > 1/2, we get that 21/2−α < 1, so using (2.8) for k = 0, 1, 2, ... coupled

with the formula for geometric sums and the fact that |f(0)| ≤ ‖f‖Λα , we get:

‖f‖A ≤(√

2 ·(

2π

3

)α1

1− 21/2−α + 1

)‖f‖Λα . (2.9)

If we set

cα =√

2 ·(

2π

3

)α1

1− 21/2−α + 1,

the theorem follows.

Chapter 3

Differentiability andabsolute convergence

In the previous chapter we saw a few results about the absolute convergenceof Fourier series. From theorem 2.1.1 we can conclude that f ∈ A(T) if f ′ iswell-behaved enough in some sense. However, as Bernstein’s theorem shows,differentiability of f ∈ C(T) is not even a necessary condition for the absoluteconvergence of S[f ]. All we need is that f ∈ Λα(T) for some α > 1/2.

What happens if we know that f ′ is well-defined everywhere? Is differentia-bility of f sufficient to guarantee that S[f ] converges absolutely? The answer tothis question is no, and in this chapter we will give an example of an everywheredifferentiable function whose Fourier series does not converge absolutely.

3.1 A special function

Let η ∈ C1(R) be a function such that η = 1 for t ≤ 1/2, η = 0 for t ≥ 1, and0 ≤ η ≤ 1 for all t.

In the rest of this chapter we will examine the 2π-periodic function f , whichis generated by

f(t) =

0 if − π ≤ t ≤ 0,

t2e−i/t4

η(t/ε) if 0 < t < π,(3.1)

where ε > 0 less than π will be chosen later on.By this definition, it is clear that f is continuous. As f is the product

of continuously differentiable functions on R \ 2πZ, it is also clear that f iscontinuously differentiable on these intervals as well. We also have that f ′ iswell-defined at t = 0, with f ′(0) = 0, as

|f ′(0)| =∣∣∣∣ limh→0

f(h)− f(0)

h

∣∣∣∣ ≤ limh→0

∣∣∣∣h2

h

∣∣∣∣ = 0.

Worth noting, however, is that for t ∈ (0, ε/2)

f ′(t) = 2te−i/t4

+4i

t3e−i/t

4

,

so f ′ is not continuous at 0, as |f ′(t)| → ∞ when t→ 0+.


10 Chapter 3. Differentiability and absolute convergence

3.2 Estimating the Fourier coefficients of f

The n:th Fourier coefficient of f is given by the formula

2πf(n) =

∫ π

−πf(t)e−intdt. (3.2)

However, the primitive function of f(t)e−int cannot be expressed in terms of el-ementary functions, and there is no obvious way to calculate the integral aboveexplicitly. Therefore we will estimate |f(n)| for large positive values of n, to

show that for large enough values of n, |f(n)| > c/n for some constant c > 0.

To make things simpler in the long run, we define the function

gn(t) =1

t4+ nt, t > 0, (3.3)

for all n ∈ Z. With this, we can rewrite (3.2) as

2πf(n) =

∫ π

0

t2e−ign(t)η(t/ε)dt. (3.4)

Calculating the derivative of gn yields

g′n(t) = − 4

t5+ n, (3.5)

and calculating the second and the third derivatives gives us

g′′n(t) =20

t6, g(3)

n (t) = −120

t7. (3.6)

Inspired by the stationary phase method for evaluating integrals, we have con-structed f in such a way that f(t)e−int has only one stationary phase, i.e. onepoint where g′n is zero, for positive n and no stationary phase when n is negativeor zero. Everywhere else on (0, ε), f has been made to oscillate rapidly for largevalues of n.

The idea is that the contribution to f(n) from any interval, except for someneighbourhood around the stationary phase of f(t)e−int, will be small due to therapid oscillations of f when n is large. Therefore it will be of great importanceto determine where g′n is zero.

Noting that g′n(t) is always negative for positive t if n ≤ 0, we solve theequation g′n(t) = 0 for n > 0:

g′n(t) = 0⇐⇒ n =4

t5⇐⇒ t =

(4

n

)1/5

, (3.7)

so we set

tn =

(4

n

)1/5

. (3.8)

It is finally time to begin estimating the Fourier coefficients of f . To do this,we split up the integral in (3.4) into three parts, so that 2πf(n) = I1

n + I2n + I3

n,

3.2. Estimating the Fourier coefficients of f 11

where I1n, I

2n and I3

n are given by

I1n =

∫ tn

0

f(t)e−intdt, (3.9)

I2n =

∫ tn+εn

tn


I3n =

∫ ε

tn+εn


where ε (see (3.1)) and εn will be specified later on.

3.2.1 Estimating I2n

Estimating I2n will be done by approximating gn(t) with its Taylor expansion at

tn, on Lagrange form:

gn(t) = ωn +g′′n(tn)

2(t− tn)2 +

g(3)n (ξ)

3!(t− tn)3 (3.12)

for some ξ between tn and t, where ωn = gn(tn). We approximate e−ign(t) with

e−iωn−ig′′n(tn)(t−tn)2/2, and we get that∣∣∣e−ign(t) − e−iωn−ig

′′n(tn)(t−tn)2/2

∣∣∣=∣∣∣e−iωn−ig′′n(tn)(t−tn)2/2(e−ig

(3)n (ξ)(t−tn)3/3! − 1)

∣∣∣=∣∣∣e−ig(3)n (ξ)(t−tn)3/3! − 1

∣∣∣ =∣∣∣ei(20(t−tn)3/ξ7) − 1

∣∣∣= 2

∣∣∣∣sin(10

ξ7(t− tn)3

)∣∣∣∣ ≤ 20

ξ7(t− tn)3 ≤ 20

(tn)7(t− tn)3, (3.13)

for t greater than tn.By choosing εn = n−α for some α > 1/5, we get that for sufficiently large

values of n, tn + εn ≤ 2tn ≤ ε/2, and this yields∣∣∣∣I2n −

∫ tn+εn

tn

t2e−iωn−ig′′n(tn)(t−tn)2/2dt

∣∣∣∣=

∣∣∣∣∫ tn+εn

tn

t2(e−ign(t) − e−iωn−ig

′′n(tn)(t−tn)2/2

)dt

∣∣∣∣≤∫ tn+εn

tn

t2∣∣∣e−ign(t) − e−iωn−ig

′′n(tn)(t−tn)2/2

∣∣∣ dt≤ 20(tn + εn)2

(tn)7

∫ tn+εn

tn

(t− tn)3dt =5(tn + εn)2

(tn)7· (εn)4

≤ 20(εn)4

(tn)5= 5n1−4α. (3.14)

Thus, with

fn(t) = t2 exp

(−ign(tn)− ig

′′n(tn)

2(t− tn)2

), (3.15)


we get that

I2n =

∫ tn+εn

tn

fn(t)dt+ r(n), (3.16)

where r(n) = O(n1−4α

). If α is chosen to be greater than 1/2, we note that

r(n) = o(n−1

), which simply means that n · r(n)→ 0 as n→∞.

Calculating the integral of fn is fairly straightforward:∫ tn+εn

tn

fn(t)dt =

∫ tn+εn

tn

t2 exp

(−ign(tn)− ig

′′n(tn)

2(t− tn)2

)dt

= e−iωn∫ tn+εn

tn

t2 exp

(−ig

′′n(tn)

2(t− tn)2

)dt

= e−iωn∫ εn

0

(u+ tn)2 exp

(−ig

′′n(tn)

2u2

)du

= e−iωn∫ εn

0

(u2 + 2utn + (tn)2) exp

(−ig

′′n(tn)

2u2

)du (3.17)

Treating the expression in (3.17) termwise gives us

1 :

∣∣∣∣e−iωn ∫ εn

0

u2 exp

(−ig

′′n(tn)

2u2

)du

∣∣∣∣≤∫ εn

0

u2dt =(εn)3

3= O

(n−3α

)(3.18)

2 :

∣∣∣∣e−iωn ∫ εn

0

2utn exp

(−ig

′′n(tn)

2u2

)du

∣∣∣∣≤ tn

∫ εn

0

2udu = tn(εn)2 = O(n−1/5−2α

)(3.19)

3 :

∫ εn

0

(tn)2 exp

(−ig

′′n(tn)

2u2

)du

= (tn)2

∫ εn

0

exp

(−10i

(tn)6u2

)du

=(tn)5

√10

∫ ξn

0

e−iu2

du =4/√

10

n

∫ ξn

0

e−iu2

du (3.20)

where ξn in (3.20) is given by ξn = (√

10/43/5)n3/5−α. Thus, if α is chosen sothat 1/2 < α < 3/5, we have that (see proposition A.0.6 in the appendix formore details)∫ ξn

0

e−it2

dt = e−iπ/4∫ ξn

0

e−t2

dt+O(nα−3/5

)−→

√π

2

(1

2− i

2

)(3.21)

as n tends to infinity. Combining all the above gives us that for sufficiently largevalues of n, we get that

I2n = e−iωn

K

n+ o(n−1

), (3.22)

where the constant K is given by

K = limn→∞

(4 · e−iπ/4√

10

∫ ξn

0

e−t2

dt

)=

√2π

5e−iπ/4. (3.23)


From this we can conclude that

|K| >√

6

5, (3.24)

and also thatArg(K) = −π

4(3.25)

for large values of n, where Arg(z) is the principal argument of the complexnumber z. The argument of K will be used in the following section in order toavoid having to examine the size of I1

n.

3.2.2 Estimating the contribution from I1n

In this section we show that for large enough values of n,∣∣I1n + I2

n

∣∣ ≥ 3

5n, (3.26)

and this will be done through examining the imaginary part of the integral

eiωn · I1n. (3.27)

We note that ifIm(eiωn · I1

n) ≤ 0,

the statement in (3.26) follows, as we then have that for sufficiently large valuesof n, ∣∣I1

n + I2n

∣∣ ≥ ∣∣Im(eiωn(I1n + I2

n))∣∣ =

∣∣Im(eiωn · I1n) + Im(eiωn · I2

n)∣∣

≥ sin(π/4)| Im(eiωn · I2n)| ≥

√3/5

n≥ 3

5n, (3.28)

since Arg(K) = −π/4.Before we can prove that Im(eiωn · I1

n) ≤ 0, we need the following lemma:

Lemma 3.2.1. Let ρ(t) ≥ 0 be a decreasing continuous function, and let φ(t)be a strictly increasing function such that φ′ is continuously increasing on theinterval [a, b], where φ(a) = 2kπ and φ(b) = φ(a) + 2π for k ∈ Z. Then∫ b

a

ρ(t) sin(φ(t))dt ≥ 0. (3.29)

Proof. As φ is strictly increasing, it is clear that φ is bijective on [a, b]. Let

s = φ(t)⇐⇒ t = φ−1(s) (3.30)

for t ∈ [a, b]. Then it is clear that s is strictly increasing, and takes on the values2kπ, (2k+ 1)π and 2(k+ 1)π when t takes on the values a, ξ and b respectively,where ξ ∈ (a, b).

Remembering that

1 =d

dt(t) =

d

dt

(φ−1(s(t))

)= (φ−1)′(s) · s′(t)⇐⇒ (φ−1)′(s) =

1

φ′(t)(3.31)


for t ∈ (a, b) and s ∈ (0, 2π), it is clear that (φ−1)′ is positive and decreasing on(0, 2π). Let µ = ρ φ−1. It is clear from the definition of ρ that µ is decreasingon (0, 2π).

Through substitution, we get that∫ b

a

ρ(t) sin(φ(t))dt =

∫ 2π

0

µ(s) sin(s)(φ−1)′(s)ds

=

∫ π

0

µ(s) sin(s)(φ−1)′(s)ds+

∫ 2π

π

µ(s) sin(s)(φ−1)′(s)ds

=

∫ π

0

µ(s) sin(s)(φ−1)′(s)ds−∫ π

0

µ(s+ π) sin(s)(φ−1)′(s+ π)ds

≥∫ π

0

µ(π) sin(s)(φ−1)′(π)ds−∫ π

0

µ(π) sin(s)(φ−1)′(π)ds = 0,

and the proof is complete.

Fix n and let

u(t) = − Im(ei(ωn−n(tn−t))f(tn − t)

)= ρ(t) sin(φ(t)), (3.32)

for t ∈ [0, tn), where ρ and φ are given by

ρ(t) = (tn − t)2, (3.33)

φ(t) =1

(tn − t)4+ n(tn − t)− ωn. (3.34)

It is clear that ρ is positive and decreasing on [0, tn), and we have that

φ′(t) =4

(tn − t)5− n ≥ 0, φ′′(t) =

20

(tn − t)6> 0, (3.35)

on [0, tn), so φ is strictly increasing with an increasing derivative.One more thing to note is that by the definition of u, we have that

−∫ tn

0

u(t)dt = Im(eiωn · I1n), (3.36)

and also that u is continuous on [0, tn), with u(t) → 0 as t → t−n . Withthis, it is clear that the integral above converges. Let (bk)∞k=1 be the sequencecharacterized by

bk =

∫ γk

0

u(t)dt, (3.37)

where φ(γk) = 2kπ, and γk → tn as k →∞. As we know that the integral of uconverges on [0, tn), it is also clear that (bk) converges, and

bk →∫ tn

0

u(t)dt, k →∞. (3.38)

Applying lemma 3.2.1 repeatedly on bk for k = 1, 2, ..., it is clear that for everyvalue of k, we have that bk ≥ 0. Thus, as the sequence converges and is boundedbelow by zero, we have that the integral of u on [0, tn) is bounded below by zero,and from this we get that (3.26) follows, as proposed.


3.2.3 Estimating the size of |I3n|To make the estimation of |I3

n| easier, we begin by generating a useful resultabout the estimation of integrals of a function similar to f(t)e−int.

Lemma 3.2.2. Let A be an open interval of positive length, and let ρ, φ ∈C∞(A) such that ρ ≥ 0 and φ is strictly increasing and strictly convex on A.Further, let there be an interval [a, b] ⊂ A such that φ(b) = φ(a) + 2π. Then∣∣∣∣∣

∫ b

a

ρ(x)e−iφ(x)dx

∣∣∣∣∣≤ (b− a)3 max

x,y∈[a,b]

(|ρ(x)|φ

′′(y)

2

)+ (b− a)(ρmax − ρmin)

≤ (b− a)3 maxx,y∈[a,b]

(|ρ(x)|φ

′′(y)

2

)+ (b− a)2 max

x∈[a,b]|ρ′(x)|.

Proof. Let s be the secant line going through the points (a, φ(a)) and (b, φ(b)),or more explicitly,

s(t) = φ(a) +2π

b− a(t− a). (3.39)

We will study the integral I, which is given by

I =

∫ b

a

ρ(x)e−iφ(x)dx

=

∫ b

a

ρ(x)(e−iφ(x) − e−is(x)

)dx+

∫ b

a

ρ(x)e−is(x)dx

=

∫ b

a


)dx+

∫ b

a

(ρ(x)− ρmin) e−is(x)dx, (3.40)

where ρmin in (3.40) is the minimum value of ρ on the interval [a, b].

By the triangle inequality, we get that

|I| ≤

∣∣∣∣∣∫ b

a


)dx

∣∣∣∣∣+

∣∣∣∣∣∫ b

a

(ρ(x)− ρmin) e−is(x)dx

∣∣∣∣∣ . (3.41)

The size of the first term in the right-hand side of the expression in (3.41) canbe estimated in the following way:∣∣∣∣∣

∫ b

a


)dx

∣∣∣∣∣ ≤∫ b

a

|ρ(x)|∣∣∣e−iφ(x) − e−is(x)

∣∣∣ dx≤∫ b

a

|ρ(x)| |φ(x)− s(x)| dx

≤ (b− a) maxx∈[a,b]

|ρ(x)| maxx∈[a,b]

|φ(x)− s(x)| . (3.42)

Let ψ(x) = s(x) − φ(x), x ∈ A. Then ψ(a) = ψ(b) = 0, and as φ is strictlyconvex, we have that ψ is strictly concave, and therefore, ψ > 0 on the interval(a, b). Furthermore, as ψ is strictly concave, and also ψ ∈ C∞(A), we have


that ψ′′ < 0. Thus ψ′ is strictly decreasing on (a, b). Combining these factswith Rolle’s theorem, we get that ψ has a unique maximum on [a, b] at a pointc ∈ (a, b).

Using a Taylor expansion of ψ at c, with the remainder expressed in Lagrangeform, we get that

ψ(x) = ψ(c) +ψ′′(ξ)

2(x− c)2 (3.43)

on the interval [a, b], where ξ is somewhere between c and x.From (3.43) we get that

0 = ψ(a) = ψ(c) +ψ′′(ξa)

2(a− c)2 ⇐⇒ ψ(c) = −ψ

′′(ξa)

2(a− c)2. (3.44)

As ψ′′ = −φ′′, coupled with (3.44) and the fact that (a−c)2 is less than (b−a)2,we get:

ψ(c) ≤ maxx∈[a,b]

φ′′(x)

2(b− a)2. (3.45)

Combining (3.42) and (3.45) gives us:∣∣∣∣∣∫ b

a


)dx

∣∣∣∣∣ ≤ (b− a)3 maxx∈[a,b]

|ρ(x)| maxx∈[a,b]

φ′′(x)

2. (3.46)

Estimating the second term of the right-hand side of the expression in (3.41)gives us ∣∣∣∣∣

∫ b

a

(ρ(x)− ρmin) e−is(x)dx

∣∣∣∣∣ ≤∫ b

a

|ρ(x)− ρmin| dx

≤ (b− a)(ρmax − ρmin) (3.47)

= (b− a)

∫ d

c

ρ′(x)dx

≤ (b− a)2 maxx∈[a,b]

|ρ′(x)|, (3.48)

where ρmax = ρ(d), for some d ∈ [a, b], is the maximum value of r, and also,ρmin = ρ(c) for some c ∈ [a, b]. Thus the proof is complete.

Using lemma 3.2.2, it is possible to arrive at the following results, which willbe proven in section 3.2.4:

Lemma 3.2.3. Let 1/5 < β < α < 3/5. Then, for large values of n,∫ tn+n−β

tn+n−αt2e−ign(t)dt = O

(n3α−2β−8/5

).

Lemma 3.2.4. Let 0 < β < 1/5 < α < 3/5. Then, for large values of n,∫ n−β

tn+n−αt2e−ign(t)dt = O

(n3α−3β−7/5

).


Lemma 3.2.5. Let 0 < β < 1/6. Then, for n ≥ (40πε2)1/(1−6β),∣∣∣∣∫ ε

n−βf(t)e−intdt

∣∣∣∣ ≤ Cη · εn

,

for a constant Cη = 8π(maxt∈[0,1] |η′(t)|+ 5

), which depends only on the choice

of η in connection with the definition of f .

From now on, we will assume that n is at least greater than or equal to(ε/2)−1/β , as well as fulfilling the condition in lemma 3.2.5, for β = 2/15. This isto ensure that lemmas 3.2.3, 3.2.4, and 3.2.5 can be applied freely to f(t)e−ign(t)

in the discussion below. Otherwise we risk having n−β ≥ ε/2, preventing the useof the results above to infer anything about the size of I3

n, due to the definitionof f .

If α and β in lemma 3.2.3 are chosen so that they fulfill the condition

3α− 2β − 8/5 < −1⇐⇒ β >3

2α− 3

10, (3.49)

the integral in lemma 3.2.3 will be o(n−1

).

Let A = a0, a1, a2, a3, a4, where the elements in A are chosen so that(3.49) will be fulfilled when ai−1 = α and ai = β for i = 1, 2, 3, 4. We alsochoose a0 so that 1/2 < a0 < 3/5, in accordance with restrictions set upon εnwhen estimating I2

n, and we set εn = n−a0 . Below we have chosen the elementsof A in the following way:

a0 =61

120, a1 =

19

40, a2 =

17

40, a3 =

7

20, a4 =

1

4. (3.50)

Every element ak fulfills 1/5 < ak < 3/5 for k = 0, 1, 2, 3, 4, and also,ai−1 < ai for i = 1, 2, 3, 4. Let tk = tn + n−ak for k = 0, 1, 2, 3, 4. If we letti−1 take the role of tα, and ti take the role of tβ in lemma 3.2.3 above fori = 1, 2, 3, 4, it is not difficult to verify that∫ t4

t0

t2e−ign(t)dt =

∫ t4

tn+εn

t2e−ign(t)dt = o(n−1

). (3.51)

Next, we take b = 2/15. Take a4 above as α and b as β in lemma 3.2.4.This, together with (3.51), means that∫ n−2/15

tn+εn

f(t)e−intdt = o(n−1

). (3.52)

As 2/15 < 1/6, we can use lemma 3.2.5 to conclude that∣∣∣∣∫ ε

n−2/15

f(t)e−intdt

∣∣∣∣ ≤ Cη · εn

. (3.53)

Set ε = 1/2Cη. Then, combining this with earlier results about the size of|I1n + I2

n| for large enough values of n, we get that when n is sufficiently large,

the size of |f(n)| is bounded below by

2π|f(n)| ≥ 3

5n− Cη · ε

n=

1

10n. (3.54)

Thus, we can conclude that f does not have an absolutely convergent Fourierseries.


3.2.4 Proof of lemmas 3.2.3, 3.2.4, and 3.2.5

In this section we will prove lemmas 3.2.3, 3.2.4, and 3.2.5. The proofs will bevery similar in nature, as the method used in all of them is essentially the same.Therefore, we will only prove lemma 3.2.3 in great detail, and omit some stepsin the proofs of lemmas 3.2.4 and 3.2.5.

We will use the functions Pn and Ln in every proof in this section, whichare defined as

Pn(I) : The number of whole periods exp(−ign(t)) completes

on the interval I; (3.55)

Ln(I) : The maximum length of an interval (a, b) ⊂ I such

that gn(b)− gn(a) = 2π. (3.56)

Proof of lemma 3.2.3: Recall that, in the formulation of lemma 3.2.3, welet the limits of integration be tα = tn + 1/nα for some α ∈ (1/5, 3/5), andtβ = tn + 1/nβ , where 1/5 < β < α.

It is clear that tβ > tα. We shall now try to overestimate the number ofwhole periods the function exp(−ign(t)) completes on the interval I0 = [tα, tβ ].We recall that

g′′n(t) =20

t6> 0 (3.57)

for t > 0, so gn is strictly convex on (tn, π). Furthermore, for t ∈ (tn, π), wehave

g′n(t) = − 4

t5+ n > − 4

(tn)5+ n = 0,

so gn is also strictly increasing on (tn, π). From (3.57) we can also deducethat the function g′n is strictly increasing on (tn, π). This gives us a way tooverestimate the number of whole periods on I0. By simple geometric reasoning,we have that

Pn(I0) ≤ tβ − tα2π/maxt∈I0 g

′n(t)

≤ g′n (tβ)

2π· 1

nβ, (3.58)

and

Ln(I0) ≤ 2π

mint∈I0 g′n(t)

=2π

g′n (tα). (3.59)

A Taylor expansion of g′n around tn, expressed in Lagrange form, yields

g′n(t) = g′′n(tn)(t− tn) +g

(3)n (ξ)

2(t− tn)2, (3.60)

for some ξ between tn and t.

By this, we get that for sufficiently large values of n,

g′n(tα) =20

nα

(1

(tn)6− 3

ξ70 · nα

)≥ 20

nα

(1

(tn)6− 3

(tn)7 · nα

)=

5

41/5n6/5−α

(1− 3

41/5 · nα−1/5

)≥ n6/5−α, (3.61)


and also that

g′n(tβ) =20

nβ

(1

(tn)6− 3

ξ71 · nβ

)≤ 20

(tn)6 · nβ

=5

41/5n6/5−β . (3.62)

Combining (3.58) and (3.59) with (3.61) and (3.62) gives us the estimations

Pn(I0) ≤ n6/5−2β (3.63)

Ln(I0) ≤ 2π · nα−6/5 (3.64)

Let B = b0, b1, ..., bkn be a sequence fulfilling the following conditions:

• b0 = tα,

• bj > bj−1 for j = 1, 2, ..., kn,

• gn(bj) = gn(bj−1) + 2π for j = 1, 2, ..., kn,

• bkn ≤ tβ ,

• gn(tβ)− gn(bkn) < 2π.

From the conditions set upon B, it is clear that kn = Pn(I0) ≤ n6/5−2β . Wecan now rewrite I0 as

I0 =

kn⋃j=1

[bj−1, bj ]

∪ [bkn , tβ ] (3.65)

Thus,∣∣∣∣∫I0

t2e−ign(t)dt

∣∣∣∣ =

∣∣∣∣∣kn∑1

(∫ bj

bj−1

t2e−ign(t)dt

)+

∫ tβ

bkn

t2e−ign(t)dt

∣∣∣∣∣≤

kn∑1

∣∣∣∣∣∫ bj

bj−1

t2e−ign(t)dt

∣∣∣∣∣+

∣∣∣∣∣∫ tβ

bkn

t2e−ign(t)dt

∣∣∣∣∣ . (3.66)

Splitting the calculations into two parts gives us:

kn∑1

∣∣∣∣∣∫ bj

bj−1

t2e−ign(t)dt

∣∣∣∣∣3.2.2≤

kn∑1

((bj − bj−1)3(bj)

2 g′′n(bj−1)

2+ (bj − bj−1)((bj)

2 − (bj−1)2)

)

≤kn∑1

(Ln(I0))3(bj)2 g′′n(bj−1)

2+ Ln(I0)((bj)

2 − (bj−1)2)︸︷︷︸telescopic terms

=

kn∑1

(Ln(I0))3(bj)2 g′′n(bj−1)

2+ Ln(I0)((bkn)2 − (b0)2). (3.67)


We estimate the individual terms in (3.67) independently from one another.

kn∑1

(Ln(I0))3(bj)2 g′′n(bj−1)

2≤ Pn(I0)(Ln(I0))3(tβ)2 g

′′n(tα)

2. (3.68)

Asg′′n(tα)

2=

10

(tα)6=

10

(41/5 + n1/5−α)6n6/5 ≤ 10n6/5,

and

(tβ)2 = n−2/5

41/5 + n1/5−β︸︷︷︸≤5

2

≤ 25n−2/5,

we get, combined with (3.68), that

kn∑1

(Ln(I0))3(bi)2 g′′n(bi−1)

2≤ C · n3α−2β−8/5, (3.69)

where C = 2000π3.The second term is estimated by

Ln(I0)((bkn)2 − (b0)2) ≤ Ln(I0)((tβ)2 − (tα)2) (3.70)

As

(tβ)2 − (tα)2 =2tnnβ

+1

n2β− 2tnnα− 1

n2α

= n−1/5−β (2 · 41/5(1− nβ−α) + n1/5−β − n1/5+β−2α)︸︷︷︸<10 for large n

≤ 10n−1/5−β ,

we get thatLn(I0)((bkn)2 − (b0)2) ≤ 20πnα−β−7/5. (3.71)

The fact that3α− 2β − 8/5 > α− β − 7/5,

combined with (3.69) and (3.71), gives us

kn∑1

∣∣∣∣∣∫ bi

bi−1

t2 exp(−ign(t))dt

∣∣∣∣∣ = O(n3α−2β−8/5

). (3.72)

Tracing back to (3.66), estimating the final term in that expression, we get∣∣∣∣∣∫ tβ

bkn

t2e−ign(t)dt

∣∣∣∣∣ ≤ (tβ − bkn) · (tβ)2

≤ Ln(I0) · 25n−2/5 ≤ 50πnα−8/5

= O(nα−8/5

)= O

(n3α−2β−8/5

). (3.73)

Now we obtain, through combining (3.66), (3.72), and (3.73), that∫I0

t2e−ign(t)dt = O(n3α−2β−8/5

), (3.74)

and the proof is complete.


Proof of lemma 3.2.4: The method used here will be very similar to theproof above, and some details will be omitted and left for the reader to fill in.

The only significant difference from the eariler proof is that a Taylor expan-sion of g′n centered at tn on the interval [tα, n

−β ] gives a bad estimation of g′nwhen 0 < β < 1/5. However, if t = 1/n1/5−δ for some δ > 0, we can use that

g′n(t) = − 4

t5+ n = −4n1−5δ + n = n(1− 4n−5δ). (3.75)

As 1− 4n−5δ → 1 as n→∞, (3.75) implies that for large values of n, we havethat n/2 ≤ g′n(t) ≤ n if t = 1/n1/5−δ for some δ > 0.

Let I1 = [tα, n−β ], for some positive β < 1/5. Using arguments similar to

those in (3.58) and (3.59), we get that

Pn(I1) ≤ g′n(n−β)

2πn−β = O

(n1−β), (3.76)

Ln(I1) ≤ 2π

g′n(tα)= O

(nα−6/5

). (3.77)

Using constructions and procedures analogous to those in (3.65), (3.66), and(3.67), we immediately get that∣∣∣∣∫

I1

t2e−ign(t)dt

∣∣∣∣≤ Pn(I1)(Ln(I1))3(n−β)2 g

′′n(tα)

2

+ Ln(I1)(n−2β − t2α) + Ln(I1)(n−β)2

≤ Pn(I1)(Ln(I1))3(n−β)2 g′′n(tα)

2+ 2Ln(I1)(n−β)2

= O(n1−β)O(n3α−18/5

)n−2βO

(n6/5

)+O

(nα−6/5

)n−2β

= O(n3α−3β−7/5

)+O

(nα−2β−6/5

). (3.78)

Noting that

3α− 3β − 7/5 > α− 2β − 6/5, β < 1/5 < α, (3.79)

the lemma follows, as we then have that

O(n3α−3β−7/5

)+O

(nα−2β−6/5

)= O

(n3α−3β−7/5

).

Proof of lemma 3.2.5: Recall that on the interval I2 = [n−β , ε], f is givenby

f(t) = t2η(t/ε)e−i/t4

, (3.80)

where η ∈ C1(R) is a cut-off function such that η = 1 for t ≤ 1/2, η = 0 fort ≥ 1, and 0 ≤ η ≤ 1 for all t.

Recall from the previous proof that on I2, we have that n/2 ≤ g′n(t) ≤ n forlarge values of n. Using arguments similar to those in (3.58) and (3.59), we getthat

Pn(I2) ≤ g′n(ε)

2π· ε ≤ n · ε

2π, (3.81)

Ln(I2) ≤ 2π

g′n(n−β)≤ 4π

n. (3.82)


Let r(t) = t2η(t/ε). Using constructions and procedures analogous to thosein (3.65), (3.66), and (3.67), using the more rough estimation in lemma 3.2.2,we get that∣∣∣∣∫

I2

r(t)e−ign(t)dt

∣∣∣∣≤ (Pn(I2))(Ln(I2))3(ε)2 g

′′n(n−β)

2

+ Pn(I2)(Ln(I2))2 maxt∈I2|r′(t)|+ Ln(I2)(ε)2

≤ 320π2ε3n6β−2 +8π(ε2/2 + maxt∈I2 |r′(t)|

)n

. (3.83)

As r(t) = t2η(t/ε), we have that

|r′(t)| =∣∣∣∣ t2ε η′(t/ε) + 2tη(t/ε)

∣∣∣∣ =

∣∣∣∣t( tεη′(t/ε) + 2η(t/ε)

)∣∣∣∣≤ ε

(maxt∈I2|η′(t)|+ 2

), (3.84)

using the fact that t ≤ ε in the last inequality.Remembering that 6β < 1 and combining (3.83) with (3.84), yields that for

n ≥ (40πε2)1/(1−6β),∣∣∣∣∫I2

r(t)e−ign(t)dt

∣∣∣∣≤ 320π2ε3n6β−2 +

8π(ε2/2 + ε (maxt∈I2 |η′(t/ε)|+ 2)

)n

≤ 8πε

n+

8πε(maxt∈[0,1] |η′(t)|+ 4

)n

=8πε

(maxt∈[0,1] |η′(t)|+ 5

)n

. (3.85)

As [0, 1] is a compact interval, and η ∈ C1(R), we know that maxt∈[0,1] |η′(t)|is a finite number. We set Cη = 8π

(maxt∈[0,1] |η′(t)|+ 5

), and the lemma

follows.

Chapter 4

Closing remarks

In this thesis we have managed to give an example of a differentiable functionwhich has a Fourier series that does not converge absolutely. What may beimmediately interesting is how the function f ∈ C(T) given by

f(t) =

0 if − π ≤ t ≤ 0,

tαe−i/tβ

η(t/ε) if 0 < t < π,

behaves for different positive values of α and β. How does the rate of convergencefor f(n) depend on the choice of α and β? This question should be possible toanswer using the stationary phase method combined with techniques similar tothose used to calculate I2

n in the previous chapter, and one would most likelycome to the conclusion that there is a complex constant C 6= 0 such that

f(n) ∼ C · n−(2α+β+2)/(2β+2).

The special case where α = 2 and β = 4 was studied in detail in the previouschapter. This choice of α and β provided us with an everywhere differentiablefunction f , where the rate of convergence for f(n) is similar to that of n−1.One thing to note is that although the derivative of f in this case is well-definedeverywhere, it has the following property:

lima→0+

∫ π

a

|f ′(t)| dt = +∞.

A 2π-periodic function g is said to be in Lp(T), p ≥ 1, if

‖g‖p =

(1

2π

∫ 2π

0

|g(t)|pdt)1/p

<∞.

From this, it is clear that f ′ /∈ L1(T). In fact, if the hypothesis above about the

rate of convergence for f(n) is correct for different choices of α and β, it maynot be possible to find values of α and β such that f ′ ∈ L1(T) while at the sametime f /∈ A(T).

From the remark to theorem 2.1.1 in Chapter 2, we note that it is enoughto have that f ′ ∈ Lp(T), p > 1, to ensure that f ∈ A(T). However, this saysnothing about the case when p = 1. Is it possible to find a function without an


24 Chapter 4. Closing remarks

absolutely convergent Fourier series, whose derivative is in L1(T)? This is nota question with an immediately obvious answer, and it should not be ignored,which it seems to have been in the existing literature dealing with the absoluteconvergence of Fourier series.

Appendix A

Evaluating an integral

We use techniques from complex analysis to calculate the value of the integral

IR =

∫ R

0

e−ix2

dx (A.1)

for R > 0.

Proposition A.0.6. For R > 1, the integral IR can be expressed as

IR = e−iπ/4∫ R

0

e−t2

dt+O(R−1

).

In particular, ∫ ∞0

e−ix2

dx = e−iπ/4∫ ∞

0

e−t2

dt =

√π

2e−iπ/4.

Proof. Let f : C→ C be defined by the equation

f(z) = e−iz2

for every z ∈ C. It is easily verifiable that f is entire, and thus,∫Γ

f(z)dz = 0 (A.2)

for any closed path Γ in the complex plane. Assume that R > 0 is given, andlet the paths LR1 , L

R2 and CR be given by

LR1 : z = t t : 0→ R, (A.3)

LR2 : z = te−iπ/4 t : R→ 0, (A.4)

CR : z = Reit t : 0→ −π/4. (A.5)

Let ΓR = LR1 + CR + LR2 . Then it is clear, by the definitions above, that ΓR isa closed path, and we get that∫

ΓR

f(z)dz = 0⇐⇒∫LR1

f(z)dz =

∫−LR2 −CR

f(z)dz. (A.6)


26 Appendix A. Evaluating an integral

Integrating along LR2 , we get∫LR2

f(z)dz = e−iπ/4∫ 0

R

e−it2 exp(−iπ/2)dt = −e−iπ/4

∫ R

0

e−t2

dt, (A.7)

and integration along CR yields∫CR

f(z)dz =

∫CR

e−iz2

dz =

∫ −π/40

e−iR2ei2t · iReitdt

=i

2R

∫ −π/40

(e−iR

2ei2t2R2ei2t)e−itdt

=i

2R

([e−i(R

2ei2t+t)]−π/4

0+ i

∫ −π/40

e−iR2ei2te−itdt

). (A.8)

Note that ∣∣∣∣∣i∫ −π/4

0

e−iR2ei2te−itdt

∣∣∣∣∣ ≤∫ 0

−π/4

∣∣∣e−iR2ei2te−it∣∣∣ dt

=

∫ 0

−π/4eR

2 sin(2t)dt ≤ π

4. (A.9)

This, together with (A.8), yields that∫CR

f(z)dz = O(R−1

). (A.10)

As ∫L1R

f(z)dz = IR, (A.11)

the proposition follows.

Bibliography

[1] Yitzak Katznelson, (2004), An introduction to Harmonic Analysis, thirdedition, Cambridge University Press, USA.

[2] Antoni Zygmund, (1968), Trigonometric series Volume 1, 2nd edition,Cambridge University Press, USA.


28 Bibliography

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