Linnceus University
Sweden
Degree project
Wiener-Levy theorem
S
imple proof of Wiener's lemma and
Wiener-Levy theorem
Author: Jose Eduardo Vasquez
Supervisor: Joachim Toft
Examiner: Andrei Khrennikov
Term: VT21
Subject: Mathematics
Abstract
Wiener-L´evy Theorem
Jos´
e Eduardo V´
asquez
May 2021
Contents
1 Introduction 2
2 Preliminaries 3
2.1 Limits and Continuity . . . 3
2.2 Convergence of Series . . . 4
2.3 Introduction to Complex Analysis . . . 6
2.4 Cauchy–Bunyakovsky–Schwarz Inequality . . . 8
3 Introduction to Fourier Series 9 3.1 Fourier Series . . . 9
3.2 Existence and Conditions for Fourier Series . . . 10
3.3 Parseval’s Theorem . . . 11
3.3.1 Basel Problem . . . 13
4 Wiener’s Lemma and Wiener-L´evy Theorem 14 4.1 Wiener’s Lemma . . . 14
4.2 Wiener-L´evy Theorem . . . 17
A Introduction to Lebesgue Integrals 21 B Some Proofs 23 B.1 Proof of Lemma 2.9 . . . 23
B.2 Proof of Lemma 3.8 . . . 23
B.3 Proof of Theorem 3.9 . . . 25
1
Introduction
Wiener-L´evy Theorem is a theorem in Fourier analysis that studies the convergence of Fourier series. This theorem is named after Norbert Wiener and Paul L´evy. Norbert Wiener was an American mathematician and philosopher from Massachusetts Institute of Technology [2]. In the year 1932 he wrote a book called ”Tauberian Theorems” where he first proved the Wiener’s lemma which states that if a function f has an absolutely convergent Fourier series and if it is not equal to zero at any point, then its reciprocal 1/f has an absolutely convergent Fourier series [1]. Later Paul L´evy, a French mathematician [14], gave a more general result called the Wiener-L´evy theorem which states the following [3].
Theorem 1.1 (Wiener-L´evy theorem). Suppose that the Fourier series of a function f , converges absolutely, and that the values (in general, complex) of f (x) lie on a curve C, and that φ(z) is an analytic (not necessarily single-valued) function of a complex variable regular at every point of C. Then the Fourier series of φ(f ) converges absolutely.
The original purpose for studying Wiener’s lemma was to formulate and prove the fundamental Tauberian theorems concerning averages ([1] p.4). These Tauberian theo-rems have some applications in the analytic theory of numbers. Using these theotheo-rems, it was demonstrated that the Riemann zeta function ξ(σ + τ i) must be non-zero in the line σ = 1. Later it was understood that converting theorems into Tauberian form was too complicated and it appeared to not have a direct relation to the prime number theorem. Some applications of Wiener-L´evy theorem are found in for example stability problems for certain Volterra integral equations [7].
Wiener-L´evy theorem gives us a weaker condition for absolute convergence of Fourier series. An even stronger version of this theorem was found by Marcinkiewicz ([9] p.251), but we shall stick to Wiener-L´evy theorem.
These theorems of convergences are most notably covered by Zygmund and Katznel-son. In contrast to the classical Fourier analysis approach that Zygmund used, Katznelson later covered these theorems using Banach algebras [8]. This thesis will cover the classical approach. In consideration for the reader, the thesis provides detailed steps. The thesis was designed to make it possible for any undergraduate science student to be able to comprehend the Wiener-L´evy theorem.
2
Preliminaries
We introduce the following preliminaries which are necessary in proving Wiener-L´evy theorem. We acknowledge the following citations throughout this section [11] [10] [5].
2.1
Limits and Continuity
We recall the reader with some basic definitions on limits and continuity.
Definition 2.1 (Right-limit and left-limit). A function f has a right-limit or left-limit L when x approaches a, if for every > 0, there exists a δ > 0 such that
|f (x) − L| < ,
whenever a < x < a + δ (for right-limit) or a − δ < x < a (for left-limit). We write the right-limit and the left-limit respectively as
f (a+) = lim
x→a+f (x), f (a−) = limx→a−f (x).
Definition 2.2 (Right differentiable and left differentiable). Suppose the right-limit f (a+) exists. Then f is called right differentiable if the following exists.
f+0 (a) = lim
h→0+
f (a + h) − f (a+)
h .
Similarly suppose the left-limit f (a−) exists. Then f is called left differentiable if the following exists.
f−0 (a) = lim
h→0−
f (a + h) − f (a−)
h .
Definition 2.3 (Continuous function). Let f be a function defined in a neighbourhood of z0 ∈ C. Then f is continuous at z0 if
lim
z→z0
f (z) = f (z0).
Definition 2.4 (Piecewise continuous function). Let I be an interval broken into a finite number of subintervals I1, I2, ..., In. Suppose that a function f is continuous on each of
2.2
Convergence of Series
Let us begin by considering the following infinite sum
S =
∞
X
n=0
an,
where an ∈ R. If this sum approaches a finite limit L which is not infinity, L < ∞, we
have convergence of this series. On the other hand, if this series does not have a limit, we have divergence. We shall look at some examples of convergent and divergent series.
We begin with the alternating harmonic series and it converges to the limit ln 2 ([10] p.533). S1 = ∞ X k=1 (−1)k−1 k = 1 1− 1 2+ 1 3− 1 4 + ... = ln 2 (2.1) The following series is called the harmonic series and it diverges to infinity ([10] p.507).
S2 = ∞ X k=1 (−1)k−1 k = 1 1 + 1 2 + 1 3 + 1 4 + ... = ∞
Notice how both of these series are very similar. The only difference is that in the second series S2, we have the sum of the absolute values. By taking the absolute values, we made
a convergent series to diverge.
Let us define an absolutely convergent series.
Definition 2.5 (Absolute convergence of a series). For a real (or complex) infinite series
S =
∞
X
n=0
an
to be absolutely convergent, the sum of the absolute values has to approach a finite limit L ∈ R. In other words S = ∞ X n=0 |an| = L < ∞.
Similarly, for improper integrals, a function f is absolutely convergent if Z ∞
0
|f (x)| dx = L.
We see from our previous example that the alternating harmonic series (2.1), although convergent, is not absolutely convergent by this definition. Hence it is a conditionally convergent series.
We shall begin looking at pointwise convergence and uniform convergence. The latter one is a more strict criterion. Consequently, it preserves continuity and differentiability when dealing with integrals of sequences of functions ([11] p.92).
Definition 2.6 (Pointwise convergence). Let Ω ⊂ R and let {fm}∞m=1 be a sequence of
functions Ω → C such that
where each fm for m = 1, 2, 3, ... is a function from Ω to C.
Then the sequence (2.2) is called pointwise convergent if lim
m→∞fm(x),
exists for every x ∈ Ω. By sampling all these limits for each x ∈ Ω, we get the following limit function f (x) = lim m→∞fm(x), x ∈ Ω, or in other formulation lim m→∞|fm(x) − f (x)| = 0, x ∈ Ω.
Definition 2.7 (Uniform convergence). Let {fm} and Ω be as before. Then the sequence
(2.2) is called uniformly convergent to the function f if lim
m→∞supx∈Ω(|fm(x) − f (x)|) = 0, x ∈ Ω.
There are many tests that determine the convergence of a series such as the comparison test, ratio test, integral test, etc. . It suffices to look at the comparison theorem for integrals in order to understand the proof of Wiener-L´evy theorem.
Lemma 2.8 (Comparison theorem for integrals). Let −∞ ≤ a < b ≤ ∞ and suppose f and g are continuous on the interval (a, b) where they satisfy 0 < f (x) ≤ g(x). If the improper integral of g(x) converges, then so does the improper integral of f (x)
Z b a f (x) dx ≤ Z b a g(x) dx = L.
Similarly, if the improper integral of f (x) diverges then so does the improper integral of g(x).
Proof. The proof of this lemma is straightforward and is available in ([10] p.365).
We will formulate an important lemma which will occur in many occasions throughout the proof of Wiener-L´evy.
Lemma 2.9 (Geometric series). Let |t| < 1, where t ∈ C. Then the sum of the successive powers of t from zero to infinity is convergent. Moreover, it converges to
∞ X k=0 tk= 1 1 − t, (2.3) where k is an integer.
2.3
Introduction to Complex Analysis
We begin by defining sets. Throughout this subsection, the definitions and theorems are mainly taken from the complex analysis book [5]. Let us remind the reader that an interior point z0 ∈ C is inside a set D ∈ C such that there is some circular neighbourhood
of z0that is completely contained in D. Additionally, a point z0of the set D is a boundary
point if every neighbourhood of z0 contains at least one point of D and at least one point
not in D.
Definition 2.10 (Open and closed sets). A set D is called open if every point of this set is an interior point. For example, the set of all points satisfying
|z − z0| < ,
where > 0 is real, is an open set.
A set D is called closed if it contains all of its boundary points. For example, the set of all points satisfying
|z − z0| ≤ ,
is a closed set.
Definition 2.11 (Bounded sets). A set D is called bounded if there exists a positive real number R such that
|z| < R,
for every z in D. In other words, D is bounded if it is contained in some neighbourhood of the origin.
Next, we shall define derivatives of complex valued functions.
Definition 2.12 (Complex differentiation). Let f be a complex valued function defined in a neighbourhood of z0. Then its derivative at the point z0 is given by
f0(z0) := lim ∆z→0
f (z0+ ∆z) − f (z0)
∆z , provided this limit exists.
The problem here is that ∆z is a complex number. Hence it can approach zero in many different ways, in fact in infinitely many ways.
We shall introduce the Cauchy-Riemann equations which consists of partial deriva-tives. The Cauchy-Riemann equations will let us formulate an important theorem which solves the problem mentioned above.
∂u ∂x = ∂v ∂y, ∂u ∂y = − ∂v ∂x. (2.4)
Theorem 2.13 (Complex differentiation). Let f (z) = u(x, y) + iv(x, y), where u and v are functions of two variables, be defined in some open set D containing the point z0 ∈ C.
Suppose the first partial derivatives of u and v exist in D, are continuous at z0, and satisfy
the Cauchy-Riemann equations at z0. Then f is differentiable at z0.
Proof. The proof is tedious and will not be presented in this thesis as it is not the main focus of it. However, it can be found in ([5] p.74).
Definition 2.14 (Analytic function). Let D ⊂ C be open and let φ be a complex-valued function on D. Then φ is called analytic (on D) if it has a derivative for every z ∈ D.
We note that if the first partial derivatives of f (z) = u(x, y) + iv(x, y) are continuous and satisfy the Cauchy-Riemann equations at all points of a domain D, then f is analytic. Consequently, this means that an analytic function is differentiable in D ([5] p.74).
Another interesting property is the following.
Theorem 2.15. If φ is analytic in a domain D, then all its derivatives φ0, φ00, ..., φ(n) exist and are analytic in D.
Proof. We refer to [5] p.209 for the proof.
Definition 2.16 (Definition of a single-valued function). A function is single-valued if it is one-to-one or many-to-one.
Definition 2.17 (Regular function). The function φ(z) is called regular on the closed set D, if it is analytic and single-valued in a neighbourhood of D.
Definition 2.18 (Smooth curve). A point set γ in the complex plane is said to be a smooth curve if it is the range of some continuous complex-valued function z = z(t), where a ≤ t ≤ b, that satisfies the following conditions:
1. z(t) has a continuous derivative on [a, b], 2. z0(t) never vanishes on [a, b],
3a. z(t) is one-to-one on [a, b], OR
3b. z(t) is one-to-one on the half-open interval [a, b), but z(b) = z(a) and z0(b) = z0(a). We note that a smooth curve, together with a specific ordering of its points, is called a directed smooth curve.
We continue by defining a contour.
Definition 2.19 (Contour). A contour Γ is either a single point z0 or a finite sequence
of directed smooth curves (γ1, γ2, ..., γn) such that the terminal point of γk coincides with
the initial point of γk+1 for each k = 1, 2, ..., n − 1. In the latter case we can write
Γ = γ1+ γ2+ ... + γn.
Theorem 2.22 (Cauchy’s Integral formula). Let Γ be a simple closed positively oriented contour. Suppose f is analytic in some simply connected domain D containing Γ and z0
is any point inside Γ. Then
f (z0) = 1 2πi Z Γ f (z) z − z0 dx. (2.5)
Proof. The proof of this theorem can be found in ([5] p. 204).
2.4
Cauchy–Bunyakovsky–Schwarz Inequality
Cauchy–Bunyakovsky–Schwarz inequality can be considered as one of the most important inequalities in mathematics as it is useful in many areas [13]. We give a special case proof following the steps from ([13] p.2) in Euclidean space Rnwhere we have the standard dot
product ui· vi = u1v1+ u2v2+ u3v3+ ... + unvn= n X n=i uivi.
Lemma 2.23 (Cauchy–Bunyakovsky–Schwarz inequality). For un, vn ∈ R we have n X i=1 uivi !2 ≤ n X i=1 u2i ! n X i=1 vi2 ! . (2.6)
Proof. Evidently the left-hand side of (2.6) is zero when the right-hand side is zero. Hence we may assume that the sums of the right-hand side of (2.6) are non-zero.
0 ≤ (u1x + v1)2+ (u2x + v2)2+ ... + (unx + vn)2 = n X i=1 u2i ! x2+ 2 n X i=1 uivi ! x + n X i=1 v2i ! .
Since this polynomial is non-negative, it must have at most one real root. Hence the discriminant is 4 n X i=1 uivi !2 − 4 n X i=1 u2i ! n X i=1 vi2 ! ≤ 0,
3
Introduction to Fourier Series
We begin this section by introducing the Fourier series. Then we shall look at some of the conditions for the Fourier series of a function to exist. Finally, we will prove the Parseval’s theorem, which we will use when proving Wiener’s lemma. Throughout this section we use the definitions, theorems and proofs from [12].
3.1
Fourier Series
Before looking at Fourier series, we shall have the following definition.
Definition 3.1 (T-periodic function). A function f on R is called T-periodic if f (x + T ) = f (x),
for every x ∈ R.
Now we shall look at a Fourier polynomial.
Definition 3.2 (Fourier polynomial). A T -periodic Fourier polynomial is a function of the following form
f (x) = N X k=−N ckeikθx, where ck ∈ C and θ = 2π/T .
Next we shall look at the coefficients ck. Let T = 2π and n be a fixed integer. Then
by multiplying with e−inx we get
f (x)e−inx = N X k=−N ckeikxe−inx ⇒ Z π −π f (x)e−inxdx = N X k=−N ck Z π −π ei(k−n)xdx.
Let m ∈ Z. We have for m 6= 0 Z π −π eimxdx = e imx im π −π = 1 im e imπ− e−imπ = 0, (3.1) and for m = 0 Z π −π e0dx = Z π −π 1 dx = 2π.
Hence, only the term for which we have (k − n) = 0 survives giving
N X k=−N ck Z π −π ei(k−n)xdx = 2πcn
3.2
Existence and Conditions for Fourier Series
Next we shall look at the properties of the function f for which we have a Fourier series. As usual, we begin with a definition.
Definition 3.3 (Trigonometric polynomial). A trigonometric polynomial is a function of the form fN(x) = N X k=−N ck(fN)eikθx,
where θ = 2π/T . The order of this trigonometric polynomial is N and it is T-periodic. We get the coefficients ck(fN) using the same steps as for the coefficients for the
Fourier polynomial 3.2. ck(f ) = 1 T Z T2 −T2 f (x)e−ikθxdx. (3.2) Definition 3.4 (Formal Fourier series). Suppose f : R → C is a T-periodic function. Suppose furthermore that f is piecewise continuous and bounded. Then the formal Fourier series is formulated as follows
S[f ](x) = lim N →∞ N X k=−N ck(f )eikθx, ck(f ) = 1 T Z T2 −T 2 f (x)e−ikθxdx,
where ck are the Fourier coefficients and θ = 2π/T .
Now that we have established the formal Fourier series, we shall remind the reader with the following definitions.
Definition 3.5 (Absolute convergence of Fourier series). A T -periodic function f : R → C has an absolutely convergent Fourier series whenever
lim N →∞ N X k=−N |ck(f )eikθx| < ∞.
In other words, whenever
lim N →∞ N X k=−N |ck(f )| < ∞,
since |eikθx| = 1 for every real x.
Consider the next theorem.
Theorem 3.6. Let f be a T -periodic piecewise continuous bounded function on R, with x ∈ R and T > 0. Suppose further that f (x−), f (x+), f0+(x) and f0−(x) exists. Then the Fourier series S[f ] exists and
S[f ](x) = f (x+) + f (x−)
2 .
Corollary 3.7. Let f be a T -periodic continuos function on R, with x ∈ R and T > 0. Suppose f0+(x) and f0−(x) exists. Then
S[f ](x) = f (x).
3.3
Parseval’s Theorem
Our next goal is to formulate and proof a theorem called the Parseval’s theorem, but we have to make some preparations. We start by giving a well-known inequality called the Bessel’s inequality.
Lemma 3.8 (Bessel’s inequality). Suppose f is T -periodic such that Z T2
−T 2
|f (x)|2dx,
converges and we have the following Fourier coefficients
ck(f ) = 1 T Z T2 −T 2 f (x)e−ikθxdx. Then ∞ X k=−∞ |ck(f )|2 ≤ 1 T Z T2 −T 2 |f (x)|2dx. (3.3)
Proof. The proof of this lemma is found at the appendix section B.2.
Next we shall formulate a theorem which will help us make Bessel’s inequality into an equality.
Theorem 3.9. Let f : R → C be a T -periodic continuous function with piecewise con-tinuous derivatives on R, with x ∈ R, T > 0 and θ = 2π/T . Then S[f ](x) exists and has the following uniformly convergent series
f (x) = S[f ](x) = ∞ X k=1 c−k(f )e−ikθx ! + c0(f ) + ∞ X k=1 ck(f )eikθx ! and ∞ X k=−∞ |ck(f )|, converges.
Proof. The proof of this theorem is found at the appendix section B.3. Lemma 3.10. Let f be as in the previous theorem. Then
ck(f0) = ikθck(f ). (3.4)
Proof. Integration by parts gives us
Since we have θ = 2π/T and f −T2 = f T 2, we get e−ikθT2 = e−ikπ = (−1)k. Similarly eikπ = (−1)k, and by (3.2) ck(f0) = 0 + ikθck(f ),
giving the result.
The preparations are finished, hence let us formulate and proof the Parseval’s theorem. Theorem 3.11 (Parseval’s theorem). Let f (x) : R → C be 2π-periodic and x ∈ R. Suppose f has a Fourier series given by
∞
X
k=−∞
ck(f )eikx,
where ck are the Fourier coefficients of f . Then, by Parseval’s theorem, we have the
following 1 2π Z π −π |f (x)|2dx = ∞ X k=−∞ |ck(f )|2. (3.5)
Proof. We start from the left-hand side of (3.5) giving 1 2π Z π −π |f (x)|2dx = 1 2π Z π −π f (x) · f (x) dx = 1 2π Z π −π ∞ X j=−∞ cj(f )eijx ! · ∞ X k=−∞ ck(f )eikx ! dx.
By Theorem 3.9 we have that both of these sums are uniformly convergent. Hence we can change the order of integration which gives
1 2π Z π −π |f (x)|2dx = ∞ X j,k=−∞ cj(f )ck(f ) · 1 2π Z π −π eijxe−ikxdx.
By an earlier result (3.1), we see that only the term for which we have (j −k) = 0 survives. We get 1 2π Z π −π |f (x)|2dx = ∞ X j,k=−∞ cj(f )ck(f ) · 1 2π · 2π dx. = ∞ X k=−∞ |ck(f )|2,
We can make an extension of Parseval’s theorem which will be useful in proving Wiener-L´evy theorem.
Corollary 3.12. Let f (x) : R → C be 2π-periodic with piecewise continuous and bounded derivatives. Then 1 2π Z π −π |f0(x)|2dx = ∞ X k=−∞ (k · |ck(f )|)2. (3.6)
Proof. By Lemma 3.10 and Parseval’s theorem (3.5) 1 2π Z π −π |f0(x)|2dx = ∞ X k=−∞ |ck(f0)|2 = ∞ X k=−∞ |ikck(f )|2 = ∞ X k=−∞ (k · |ck(f )|)2. 3.3.1 Basel Problem
The Basel problem is a very old problem first posed in the year 1650. It is named after the hometown of Leonhard Euler who first solved the following problem [6].
∞ X k=1 1 k2 = π2 6 (3.7)
4
Wiener’s Lemma and Wiener-L´
evy Theorem
We are first going to proof Wiener’s 1/f theorem and then extend it to Wiener-L´evy theorem by making some arguments. The proof might contain some topics that are unfamiliar to the less educated reader, so we provide all the help needed in order to get a solid understanding of this proof. The steps for these proofs are based on [4] and [3].
4.1
Wiener’s Lemma
Sometimes in literature, Wiener’s lemma is referred to as Wiener’s 1/f theorem. Through-out this research we will use Wiener’s 1/f . First, we formulate the Wiener’s 1/f theorem and it goes as follows.
Theorem 4.1 (Wiener’s 1/f theorem). Let f : R → C be 2π-periodic and x ∈ R. If f (x) 6= 0, and if the Fourier series of f converges absolutely, so does the Fourier series of 1/f .
Proof. Let f (x) =P∞
n=−∞aneinx. We work with the norm k · k on Fourier series, defined
by kf k = ∞ X −∞ |an|,
where an∈ R are some coefficients.
For some functions f : R → C and g : R → C both 2π-periodic, we have the triangle inequality
kf + gk ≤ kf k + kgk. (4.1) We shall find an estimate of the supremum of f . We divide it into two parts |a0| and
P
n6=0|an|, respectively. We have for every integer n the following
|an| = 1 2π Z 2π 0 f (x)e−inxdx ≤ 1 2π Z 2π 0 |f (x)| dx ≤ 1 2πsup |f (x)| · 2π =−π≤x≤πsup |f (x)|. Hence we get |an| ≤ sup −π≤x≤π |f (x)|. (4.2)
Here and in the rest of the proof, all suprema are taken over x ∈ [−π, π].
For the remainding terms (which completes both parts of the estimation), we have by Cauchy–Bunyakovsky–Schwarz inequality (2.6) the following
X n6=0 |an| !2 ≤ X n6=0 1 n2 ! X n6=0 n2|an|2 ! .
By identifying that the first sum is actually two times the Basel problem (3.7) we get
By Corollary 3.12, we get X n6=0 |an| !2 = π 2 3 · 1 2π Z π −π |f0(x)|2dx ≤ π 2 3 sup |f 0 (x)|2 ≤ 4 sup |f0(x)|2. We proceed to take the square roots giving
X
n6=0
|an| ≤ 2 sup |f0(x)|.
Hence we obtain the following estimate for our function f
sup |f | ≤ kf k ≤ sup |f | + 2 sup |f0|. (4.3) Now, we suppose that kf k < ∞, namely that f is absolutely convergent, and that f (x) has no zeros. The latter is a necessary condition as we want to prove that 1/f is absolutely convergent as well.
We may assume that |f (x)| ≥ 1 for every x without any loss of generality. This is because for some convergent function f0, we have |f0(x)| ≤ c for some c ∈ R. If c < 1,
then we can consider f1 = 1/f0 for which we would get |f1| ≥ (1/c) ≥ 1.
Let us choose a sufficiently large partial sum P such that kP − f k ≤ 1 3. (4.4) Consider 1 f = 1 P − (P − f ) = 1 P · 1 1 − P −fP . By the formula for geometric series (2.3)
1 f = 1 P · ∞ X n=0 P − f P n = 1 P · ∞ X n=1 P − f P n−1 = ∞ X n=1 (P − f )n−1 Pn . (4.5)
We proceed to obtain the estimates from (4.3) in terms of our new series that we have considered previously. From the inequality (4.4) we get
From |f (x)| ≥ 1
|f (x)| − |P (x) − f (x)| ≥ 2 3, and by the triangle inequality (4.1) we get
|P (x) − f (x)| = |f (x) − P (x)| ≥ |f (x)| − |P (x)| ⇒ |P (x)| ≥ |f (x)| − |P (x) − f (x)| ≥ 2 3 ⇒ 1 P (x) ≤ 3 2 ⇒ 1 P (x)n ≤ 3 2 n . Continuing now for the supremum
sup 1 Pn ≤ 3 2 n , (4.6)
which gives the first term of our estimate (4.3). Now, we continue to get the second term to complete this step. We begin by
d dx 1 P (x)n = 1 P (x)n+1 · (−n) · (P 0 (x))
and applying the modulus and using (4.6) we get
sup d dx 1 P (x)n ≤ 3 2 n+1 · (−n) · sup P0(x) = nA 3 2 n+1 , where A = sup |P0(x)|. By equation (4.3) we get 1 Pn ≤ sup 1 Pn + 2 sup d dx 1 P (x)n = 3 2 n + 2nA 3 2 n+1 = 3 2 n 1 + 2nA · 3 2 = (1 + 3nA) 3 2 n . (4.7)
Similarly for the numerator of our considered series (4.5) we get k(P − f )n−1k ≤ kP − f kn−1≤ 1
3 n−1
.
But since 1 f = ∞ X n=1 (P − f )n−1 Pn ≤ ∞ X n=1 (P − f )n−1 Pn ≤ ∞ X n=1 3 + 9An 2n < ∞, (4.8)
we have convergence in norm. This is because the terms of the last infinite sum are bounded since they approach 0 as n → ∞. Consequently, 1/f is absolutely convergent. This completes the proof for Wiener’s 1/f theorem.
4.2
Wiener-L´
evy Theorem
An extension of Wiener’s 1/f theorem was first proposed by Paul L´evy, hence giving the name of Wiener-L´evy theorem. We need some preparations and we start with the following lemma.
Lemma 4.2. Let g : R → C be a two times continuously differentiable 2π-periodic function with expansion
g(x) = ∞ X −∞ akeikx, and let kgk = ∞ X −∞ |ak|, M g = sup 0≤x≤2π |g(x)|. Then kgk ≤ 4(M g00+ M g). (4.9) We observe that kgk in Lemma 4.2 is a norm.
Proof. From now on, all suprema are taken over x ∈ [0, 2π].
Suppose that {gi}∞i=1 is a sequence of 2π periodic functions such that
P∞
i=1kgjk is
convergent. ThenP∞
i=1gi and g1g2 are 2π periodic functions, and
X i gi ≤X i kgik, kg1g2k ≤ kg1kkg2k. (4.10)
In fact, the first inequality is the triangle inequality. The second inequality can be derived considering g1 =Pkckeikx and g2 =Plckeilx. We have
Let j = k − l. By taking the norm and using the triangle inequality we get kg1g2k = X k X l ck−ldl ≤X k X l |ck−l||dl| =X j X l |cj||dj| = X j |cj| ! X l |dl| ! = kg1kkg2k.
In order to estimate kgk, we decompose it as kgk =X
k6=0
|ak| + |a0|.
Since |a0| ≤ sup |g(x)| = M g (from (4.2)), we get
kgk ≤ X k6=0 |k2· a k| · 1 k2 ! + M g. (4.11)
For estimating the latter series, we consider M g00. Since g is two times continuously differentiable g00(x) = d 2 d2x ∞ X −∞ akeikx ! = d dx X k6=0 ik · akeikx ! =X k6=0 (−k2· ak)eikx. Hence (4.2) gives sup k6=0 |k2· a k| ≤ M g00.
Using the previous inequality in (4.11) we get
kgk ≤ X k6=0 |k2· a k| · 1 k2 ! + M g ≤ M g00·X k6=0 1 k2 + M g. Since (3.7) X k6=0 1 k2 = 2π2 6 , we get kgk ≤ M g00· 2π 6 + M g ≤ 4M g 00 + M g ≤ 4(M g00+ M g).
We also need the following triangle inequality for integral expression.
Lemma 4.3. Suppose for some 2π-periodic function g(x, θ) : R → C, where 0 ≤ θ ≤ 2π, we have kg(x, θ)k ≤ C, where C is finite. Then
Proof. The proof is straightforward and is available in books on integral calculus [5]. Now that we have finished with our preparations, we begin to state the Wiener-L´evy theorem.
Theorem 4.4 (Wiener-L´evy). Let f : R → C be 2π-periodic and S[f ] be its Fourier series expansion. Suppose that S[f ] converges absolutely, and that the values (in general, complex) of f (x) lie on a curve C, and that φ(z) is an analytic (not necessarily single-valued) function of a complex variable regular at every point of C. Then S[φ(f )] converges absolutely.
Proof. We have that φ(z) is regular at z = f (x). Therefore by analytic continuation (see [5] p.292) there is a ρ > 0 such that φ(z) is regular in each disc |z − f (x)| ≤ 2ρ. Since a regular function implies single-valuedness, this function φ(z) has the same value for each point on these discs independent of the path it was reached by analytic continuation.
Let s(x) be a partial sum of S[f ] of large enough order such that M (s − f ) ≤ ks − f k ≤ 1
2ρ. (4.13)
Since s(x) + ρeiθ is inside the circle |z − f (x)| = 2ρ (for clarification see figure 1 below), we have that φ[s(x) + ρeiθ] is regular (analytic and single-valued) and hence has finite
derivatives at every point on this domain. Therefore, by Theorem 2.15, it is two times continuously differentiable in x and θ. Consequently, it obeys the estimate (4.9).
Figure 1: A view of the different circles which occur.
Since kφ[s(x) + ρeiθ]k ≤ 4(M φ00[s(x) + ρeiθ] + M φ[s(x) + ρeiθ]) = A, for every θ in
0 ≤ θ ≤ 2π where A is finite, Lemma 4.3 gives Z 2π
0
||φ[s(x) + ρeiθ]k dθ < ∞. (4.14)
Consider the following
(s + ρeiθ − f )−1 = ρ−1e−iθ
1 − f − s ρeiθ
By (4.13) we get f − s ρeiθ = |s − f | ρ ≤ 1 2, and the formula for geometric series (2.3) gives
1 − f − s ρeiθ −1 = ∞ X n=0 f − s ρeiθ n = 1 + ∞ X n=1 (f − s)nρ−ne−inθ. Therefore
(s + ρeiθ− f )−1 = ρ−1e−iθ 1 +
∞ X n=1 (f − s)nρ−ne−inθ ! .
Applying the inequality (4.13) and then the formula for geometric series (2.3) gives
k(s + ρeiθ − f )−1k ≤ ρ−1 1 + ∞ X n=1 1 2ρ n ρ−n ! = 2ρ−1. (4.15)
Finally, by integrating along the curve z(θ) = s(x)+ρeiθ, when 0 ≤ θ ≤ 2π, and observing
that dz = iρeiθ, Cauchy’s formula (2.22) gives
φ[f (x)] = 1 2πi I |z−s(x)|=ρ φ(z) z − f (x)dz = 1 2πi Z 2π 0 φ[s(x) + ρeiθ] s(x) + ρeiθ− f (x)iρe
iθdθ.
Applying the modulus and using the inequality (4.10), we get
kφ[f (x)]k ≤ 1 2π Z 2π 0 φ[s(x) + ρeiθ] s(x) + ρe iθ− f (x)−1 ρeiθ dθ. By the inequality (4.15) which we considered and since keiθk = 1, we have
kφ[f (x)]k ≤ 1 2π
Z 2π
0
kφ[s(x) + ρeiθ]k · 2 dθ.
But because we had (4.14)
kφ[f (x)]k ≤ 1 π
Z 2π
0
kφ[s(x) + ρeiθ]k dθ < ∞.
A
Introduction to Lebesgue Integrals
Let us start by giving the simplest definition of an integral. Basically in the simplest case of only one variable, an integral of a non-negative function is the area between the function and the x − axis. A Lebesgue integral is an extension of this notion. A Lebesgue integral handles well higher dimensional functions and can define certain domains of integration better than the regular Riemann integration. Let us give an intuitive interpretation of the Lebesgue integration alongside the Riemann integration [15].
Suppose we are calculating the volume of a mountain. You can see illustrations of these two approaches as we are explaining the intuitive interpretation below (Figure 2 and Figure 3).
1. The Riemann approach: We divide the bottom of the mountain into a grid of arbitrarily small squares, for example of area 1m2. Then we check the altitude of each of these squares by measuring when the center of each of these squares is touching the top surface of the mountain. Finally we multiply the total altitude htotal of each square by
the base area that we had, in other words: htotal× 1m2.
2. The Lebesgue approach: We first draw a contour map of our mountain, with adjacent contour lines being for example 1m of height apart. The volume of a single contour is then approximately the area of the contour times the height 1m. For the total volume, we multiply the total area of the contours with the height of each of these contours.
Figure 2: The contour map of a mountain [16].
Using the Riemann integrals we will encounter problems such as: limit processing, dependence of continuity and the difficulty to expand this approach to higher dimensions. By limit processing we mean
lim n→∞ Z b a fn(x) dx = Z b a lim n→∞fn(x) dx,
where x ∈ Rk. We would need the functions f
n to be uniformly convergent for this
equation to be defined. For example, in our previous example of calculating the volume of a mountain, it is difficult to get a grid of squares from the bottom of the mountain. As it can be seen in Figure 2, it is already more difficult to get a reminiscent shape of the figure’s contour map using only squares. As the integer k from the function f : Rk → R grows, it will be exponentially more difficult to define these spaces as we are partitioning the domain space Rk.
However, using the Lebesgue integrals, we do not have these problems and we can define better our integrals. This is because we are partitioning the range space R which can be done much easier than partitioning the domain space Rk. In essence, we are
B
Some Proofs
B.1
Proof of Lemma 2.9
Proof of Lemma 2.9. We just have to show that for |t| < 1, we have the equality in (2.3), as this would imply convergence
1
1 − t < ∞.
Now, let us write the first terms of the series and assume that the series approaches some limit R. t0+ t1+ t2+ t3+ ... = R Then ⇒ 1 t + t 0+ t1+ t2+ ... = R t ⇒ t0+ t1+ t2+ t3+ ... = R t − 1 t. But the left-hand side is again the limit R.
⇒ R = R t − 1 t ⇒ R = 1 1 − t But R is exactly the sum we began with.
B.2
Proof of Lemma 3.8
Proof of Lemma 3.8. This proof follows the steps from [12]. Consider the following trigonometric polynomial from Definition 3.3
Since (3.1) we get 1 T Z T2 −T 2 |fN(x)|2dx = N X k=−N ck(fN)ck(fN) = N X k=−N |ck(fN)|2. (B.1)
Now consider the following
0 ≤ Z T2 −T 2 |f (x) − fN(x)|2dx = Z T2 −T 2 (f (x) − fN(x)) · (f (x) − fN(x)) dx = Z T2 −T 2 f (x)f (x) − f (x)fN(x) − fN(x)f (x) + fN(x)fN(x) dx = I1− I2− I3+ I4.
Let us look at these integrals separately
I1 = Z T2 −T 2 |f (x)|2dx I2 = Z T2 −T 2 f (x) · fN(x) dx I3 = Z T2 −T 2 fN(x) · f (x) dx = I2 I4 = Z T2 −T 2 |fN(x)|2dx = T · N X k=−N |ck(fN)|2,
where the last equality follows from (B.1). Continuing we have
⇒ N X k=−N |ck(fN)|2 ≤ 1 T Z T2 −T2 |f (x)|2dx. By letting N → ∞, we get ∞ X k=−∞ |ck(f )|2 ≤ 1 T Z T2 −T 2 |f (x)|2dx.
B.3
Proof of Theorem 3.9
Proof of Theorem 3.9. This proof follows the steps from [12]. By Lemma 3.10 and by Bessel’s inequality (3.3) we have
∞ X k=−∞ |ck(f0)|2 ≤ 1 T Z T2 −T 2 |f0(x)|2dx. ⇒ ∞ X −∞ |kθck(f )|2 ≤ 1 T Z T2 −T 2 |f0(x)|2dx.
We get that the following converges
⇒ ∞ X −∞ k2|ck(f )|2 ≤ 1 θ2T Z T2 −T 2 |f0(x)|2dx < ∞. (B.2)
Now let us consider the following for k 6= 0. We remind the reader the arithmetic and geometric inequality 4xy ≤ (x + y)2.
|ck(f )| = |ck(f ) · k| · 1 |k| ≤ 1 2 |ck(f )k|2 + 1 |k|2 = 1 2 |ck(f )|2k2+ 1 k2
By taking the whole sum and by identifying that one of the sums is two times the Basel problem (3.7) ∞ X k=−∞ |ck(f )| = |c0(f )| + X k6=0 |ck(f )| ≤ |c0(f )| + 1 2 X k6=0 |ck(f )|2k2+ X k6=0 1 k2 ! ⇒ ∞ X k=−∞ |ck(f )| ≤ 1 2 X k6=0 k2|ck(f )|2+ π2 6 . Since we had (B.2), the sum
∞
X
k=−∞
|ck(f )| (B.3)
converges as well.
Consider
uk(x) = ck(f )eikθx, Mk= |ck(f )|,
where by (B.3), the sum
∞
X
k=−∞
Mk
is convergent. Since |eikθx| = 1, we get
|uk(x)| = |ck(f )eikθx| ≤ |ck(f )| · |eikθx| = |ck(f )| = Mk.
Now by Weierstrass M-test ([5] p.263) it follows that
∞ X k=1 c−k(f )e−ikθx, ∞ X k=1 ck(f )eikθx,
are uniformly convergent.
B.4
Proof of Basel problem
Proof of the Basel problem (3.7). Consider a 2π-periodic function f (x) = x. By Parse-val’s theorem (3.5) we have
∞ X k=−∞ |ck|2 = 1 2π Z π −π x2dx. Since ck = 1 2π Z π −π xe−ikxdx,
integration by parts gives us
ck = 1 2π x · e −ikx −ik π −π − Z π −π e−ikx −ik dx = 1 2π π −ik(−1) k− −π −ik (−1)k + 1 ik e−ikx −ik π π = (−1) k −ik + 1 2πk2 · 0 = (−1)k k · i, for k 6= 0. For k = 0 we have
c0 = 1 2π Z π −π 0 · e−ik·0dx = 0.
Hence for k 6= 0 and k = 0 we get |ck|2 =
1
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