Eotvos Lorand Tudomanyegyetem

Faculty of Science

David Ferenczi

Approximation in Function spaces

BSc Thesis

Supervisor:

Zsigmond Tarcsay

Department of Applied Analysis and Computational Mathematics

Budapest, 2018

Acknowledgments

I would like to express my gratitude to my supervisor, Zsigmond Tarcsay, for his

inspiring thoughts, and contributions to my thesis. I am also grateful to my family, for

always believing me, and helping me reaching my goals. Last but not least I want to thank

my friends for encouraging me during my studies.

2

Contents

1 Elementary Approximation 5

1.1 Weierstrass Theorems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5

1.2 The Stone-Weierstrass Theorem . . . . . . . . . . . . . . . . . . . . . . . . 11

1.3 Korovkin’s First Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . 13

2 Fourier series 18

2.1 Basics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18

2.2 Abstract Fourier Series In Hilbert Spaces . . . . . . . . . . . . . . . . . . 23

2.3 Uniform Convergence Of Fourier Series . . . . . . . . . . . . . . . . . . . . 25

2.4 Fejer-summation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28

3 Solving Partial Differential Equations With Fourier Series 32

3.1 The Wave Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34

3.2 The Heat Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36

Bibliography 39

3

Introduction

Many problems in mathematics can be answered easier by representing a function

with a function sequence. The main aim of my thesis is to give a brief summary on

certain approximation theorems, and give examples on solving problems with help of

approximation.

In the first chapter we analyze the notion of uniform convergence on the Banach space

of continuous functions. We are going to prove the first and second Weierstrass approx-

imation theorems about uniform approximation of continuous functions by polynomials,

resp. trigonometric polinomials. After that we prove that continuous functions are dense

in certain Banach spaces as well, and as an application prove a theorem on the convergence

of quadrature formulas.

The second chapter is mainly about Fourier series. After proving well known theorems,

we will consider different convergence notions, and give a sufficient condition for f to have

its Fourier series converge uniformly. We will prove a theorem on Fourier coefficients,

namely the Riemann-Lebesgue lemma and Fejer’s theorem as well.

In the third chapter we prove another sufficient condition for f to have a convergent

Fourier series, and solve two partial differential equations, namely the wave and heat

equation in one dimension with the help of Fourier series.

4

Chapter 1

Elementary Approximation

Let I ⊆ R be an arbitrary interval, and f : I → R be an arbitrary function, and

the following (Tn)n,a(x) linear operator the following: for every f ∈ C∞, and for every

n ∈ N, a ∈ I:

Tn,a(f)(x) =n∑j=0

f (j)(a) · xj

j!.

According to the Taylor-theorem, if f is analytical in a, limn→∞

Tn,a(f) = f . During this

chapter, we will prove, that if f ∈ C(I), where I is a compact interval, it can be approx-

imated, prove, that periodic functions can be approximated as well, and we will show,

that approximation can be done in compact metric spaces.

1.1 Weierstrass Theorems

Definition 1.1.1. An f : R −→ R function is affine, if there exists α, β ∈ R, such that

f(x) = αx+ β.

Notation 1.1.2. Let I be a compact interval and f : I → R be a bounded function

‖f‖∞ := supt∈I|f(t)|

Two reminders about uniform convergence:

Definition 1.1.3. Let fn, f , be E → R function. We say fn → f uniformly, if

(∀ε ∈ R) (∃N ∈ N+) (∀x ∈ E) (∀n ≥ N) : |fn(x)− f(x)| ≤ ε.

It is a well known fact, that uniform convergence is metrizable, namely,

fn → f uniformly ⇐⇒ ‖fn − f‖∞ → 0.

5

Theorem 1.1.4 (Weierstrass Approximation Theorem). Let f : [a, b] −→ R be a con-

tinuos function. Then there exists a (pn)n∈N polynomial sequence, such that pn −→ f

uniformly on [a, b].

Proof. Let R = b− a and define

q(t) =

R2 − t2 , |t| ≤ R

0 , |t| > R

Lemma 1.1.5. For every δ ∈ R+ : limn→∞

∫|t|≥δ

q(t)ndt∫Rq(t)ndt

= 0

Proof. Let δ ≥ R. In this case

∫|t|≥δ

q(t)ndt = 0, since q(t) = 0 ∀t ∈ {|t| ≥ R}. Let δ ≤ R.

In this case q has the following properties:

1. is continuos

2. even

3. positive on the (0, R) open intervall

4. strictly decreasing in (0, R) and 0 in R \ (−R,R)

All of these properties can be read from the definition of q.∫|t|≥δ

q(t)ndt ≤ (2R− 2δ) · q(δ)n ≤ 2Rq(δ)n (1.1)

∫Rq(t)ndt ≥

∫|t|≤ δ

2

q(t)ndt ≥ δq

(δ

2

)n(1.2)

Equation (1.1) is true, due to monotonicity, and the length of the interval. In equation

(1.2) the integral is less because: {| t |≤ δ2} ⊂ R. Combining these we obtain

0 ≤

∫|t|≥δ

q(t)ndt∫Rq(t)ndt

≤ 2R

δ

(q(δ)

q( δ2)

)n

and since 0 ≤ q(δ) ≤ q

(δ2

), we get:

limn→∞

2R

δ

(q(δ)

q( δ2)

)n= 0

6

Now back to proving the Weierstrass theorem. We shall assume, that:

f(a) = f(b) = 0, (1.3)

and f(t) = 0 ∀t ∈ R \ [a, b]. If not, by adding affine functions we can achieve it. Since

[a, b] is a compact interval and f is continuous, we know that f is uniformly continuous,

due to Heine’s theorem. Let:

ω(f, δ) = sup{|f(x)− f(t)| : |x− t| ≤ δ}. (1.4)

Due to uniform continuity:

limδ→0

ω(f, δ) = 0. (1.5)

Let:

cn =

∫Rq(t)ndt and Qn(t) =

1

cn· q(t)n ∀n ∈ N+, (1.6)

which has the following properties:

i) Qn ≥ 0, since q(t) ≥ 0 for every t ∈ R.

ii) Qn(t) = 0 ∀t ∈ R, if |t| ≥ R

iii)

∫RQn(t)dt =

1

cn

∫Rq(t)ndt =

cncn

= 1

iv) limn→∞

∫|t|≥δ

Qn(t)dt = 0 according to Lemma 1.1.5.

Let pn(x) =

∫Rf(t)Qn(x− t)dt ∀x ∈ R

|f(x)− pn(x) | =∣∣∣∣ ∫

Rf(x)− f(t)Qn(x− t)dt

∣∣∣∣=

∣∣∣∣ ∫Rf(x)Qn(x− t)− f(t)Qn(x− t)dt

∣∣∣∣=

∣∣∣∣ ∫R(f(x)− f(t))Qn(x− t)dt

∣∣∣∣≤∫|x−t|≤δ

|f(x)− f(t)|Qn(x− t)dt+

∫|x−t|≥δ

|f(x)− f(t)|Qn(x− t)dt

≤ ω(f, δ) + 2 ‖f‖∞∫|s≥δ

Qn(s)ds.

1. we should choose δ > 0 such, that ω(f, δ) ≤ ε2.

2. And let N be such, that ∀n ≥ N : 2 ‖f‖∞∫|s|≥δ

Qn(s)ds ≤ ε

2.

7

Then, |f(x)− pn(x)| ≤ ε ∀x ∈ [a, b] ∀n ≥ N , which means pn → f uniformly.

The only thing left, is to show, that (pn) is a polynomial for every n ∈ N. Since f(x) = 0

∀x ∈ R \ [a, b] the following is true:

pn(x) =

∫Rf(t)Qn(x− t)dt =

∫ x+R

x−Rf(t)

1

cn(R2 − (x− t)2)ndt

=

∫ b

a

f(t)1

cn(R2 − (x− t)2)ndt.

We can see, that the integrand is a polynomial of x, and according to the Newton-Leibniz

theorem: it’s integral will also be a polynomial of x.

Corollary 1.1.6. This also proves, that C([a, b]) is separable, since polynomials with

rational coefficienses are:

1. dense in C([a, b])

2. countable.

Now we will show a similar achievement, by Weierstrass for periodic functions.

Definition 1.1.7. Let f : R −→ R. We say, that f ∈ C2π, if f is a continuous, periodic

function with a period length of 2π.

Definition 1.1.8. Let p : R −→ R. We say, that p is a trigonometrical polynomial, if

p(x) =n∑j=0

aj sin(jx) + bj cos(jx) ∀x ∈ R,where aj, bj ∈ R.

Theorem 1.1.9 (Weierstrass second approximation theorem). For every f ∈ C2π there

exists a sequence (pn)n∈N of trigonometric polynomials, such that pn → f uniformly.

Proof. Let

q(t) =

1 + cos(t), |t| ≤ π

0, |t| > π

Lemma 1.1.10. For every δ > 0, limn→∞

∫|t|≥δ

q(t)ndt∫Rq(t)ndt

= 0

Proof. It is the same, like in Lemma 1.1.5. Let δ ≥ π. In this case

∫|t|≥δ

q(t)ndt = 0, since

q(t) = 0 ∀t ∈ {|t| ≥ π}.Let 0 < δ ≤ π. In this case q has the following properties:

1. is continuous

8

2. even

3. positive on the (0, π) open interval

4. strictly decreasing in (0, π) and 0 in R \ (−π, π)

All of these properties can be read from the definition of q.∫|t|≥δ

q(t)ndt ≤ (2π − 2δ) · q(δ)n ≤ 2πq(δ)n (1.7)

∫Rq(t)ndt ≥

∫|t|≤ δ

2

q(t)ndt ≥ δq

(δ

2

)n(1.8)

Equation (1.7) is true, due to q’s monotonicity, and the length of the interval. In equation

(1.8) the integral is less because: {| t |≤ δ2} ⊂ R. combining these:

0 ≤

∫|t|≥δ

q(t)ndt∫Rq(t)ndt

≤ 2π

δ

(q(δ)

q( δ2)

)n

and since:0 ≤ q(δ) ≤ q

(δ2

)we get:

limn→∞

2π

δ

(q(δ)

q( δ2)

)n= 0

We shall assume, that:

f(−π) = f(π) = 0, (1.9)

Since [−π, π] is a compact interval, and f is continuos, we know that, f is uniformly

continous, due to Heine’s theorem. Let:

ω(f, δ) = sup{|f(x)− f(t)| : |x− t| ≤ δ}. (1.10)

Due to uniform continuity:

limδ→0

ω(f, δ) = 0. (1.11)

Let:

cn =

∫Rq(t)ndt and Qn(t) =

1

cn· q(t)n ∀n ∈ N+, (1.12)

which has the following properties:

i) Qn ≥ 0, since q(t) ≥ 0 ∀t ∈ R.

ii) Qn(t) = 0 for every t ∈ R, |t| ≥ R

9

iii)

∫RQn(t)dt =

1

cn

∫Rq(t)ndt =

cncn

= 1

iv) limn→∞

∫|t|≥δ

Qn(t)dt = 0 according to the Lemma.

Let pn(x) =

∫Rf(t)Qn(x− t)dt ∀x ∈ R

|f(x)− pn(x) | =∣∣∣∣ ∫

Rf(x)− f(t)Qn(x− t)dt

∣∣∣∣=

∣∣∣∣ ∫Rf(x)Qn(x− t)− f(t)Qn(x− t)dt

∣∣∣∣=

∣∣∣∣ ∫R(f(x)− f(t))Qn(x− t)dt

∣∣∣∣≤∫|x−t|≤δ

|f(x)− f(t)|Qn(x− t)dt+

∫|x−t|≥δ

|f(x)− f(t)|Qn(x− t)dt

≤ ω(f, δ) + 2 ‖f‖∞∫|s≥δ

Qn(s)ds.

1. we should choose δ ≥ 0 such, that ω(f, δ) ≤ ε2.

2. And let N be such, that ∀n ≥ N : 2 ‖f‖∞∫|s|≥δ

Qn(s)ds ≤ ε

2.

Therefore, |f(x)− pn(x)| ≤ ε ∀x ∈ R ∀n ≥ N . The only thing left is to show that (pn)n∈N

is a trigonometrical polynomial.

Lemma 1.1.11. If p, q ∈ C2π are trigonometrical polynomials, then pq ∈ C2π is also a

trigonometrical polynomial.

Proof. Since p(x) =n∑j=0

aj sin(jx) + bj cos(jx), and q(x) =m∑j=0

aj sin(jx) + bj cos(jx), it

is enough to show, that

cos(mt) cos(nt), sin(mt) sin(nt), cos(mt) sin(mt)

are also trigonometric polynomials.

i) cos(nt) cos(mt) = cos((m−n)t)+cos((n+m)t)2

ii) sin(nt) sin(mt) = cos((m−n)t)−cos((n+m)t)2

iii) sin(nt) cos(mt) = cos((m−n)t)−cos((n+m)t)2

due to the additional formulas.

10

Exploiting this, and the definition of Qn we obtain:

pn(x) =

∫Rf(t)Qn(x− t)dt =

1

cn

∫ x+π

x−πf(t)(1 + cos(x− t))n,

=1

cn

∫ π

−πf(t)(1 + cos(x− t))n,

=1

cn

∫ π

−πf(t)(1 + cos(x)cos(t) + sin(x)sin(t))n,

=n∑j=0

ak cos(kx) + bk sin(kx).

1.2 The Stone-Weierstrass Theorem

We have seen that every f ∈ C([a, b]) can be approximated, where [a, b] is a compact

interval. We proceed by proving density of continuous functions in certain metric spaces,

with the help of our previous results.

Definition 1.2.1. Let K be a compact metric space. A linear subset M ⊂ C(K) is a

subalgebra, if for every f, g ∈M , fg ∈M stands.

Definition 1.2.2. Let K be a compact metric space, and M ⊂ C(K) a subspace . If for

every f, g ∈M , implies that f ∧ g, and f ∨ g ∈M , M is a vector lattice, where

f ∧ g(t) = max(f(t), g(t)), and f ∨ g(t) = min(f(t), g(t)).

Remark 1.2.3. Polynomials and trigonometric polynomials form a subalgebra in C[a, b].

Proposition 1.2.4. Let K be a compact metric space, and M ⊂ C(K) be a subspace.

If for every h ∈M , |h| ∈M stands, then M is a vector lattice.

Proof. We need to show, that f ∧ g and f ∨ g is in M , which is true since

f ∧ g(t) = maxK{f(t), g(t)} =

f(t), f(t) ≥ g(t)

g(t), g(t) > f(t)=

f(t), f(t)− g(t) ≥ 0

g(t), g(t)− f(t) > 0,

and

f ∨ g(t) = minK{f(t), g(t)} =

f(t), f(t) ≤ g(t)

g(t), g(t) < f(t)=

f(t), f(t)− g(t) ≤ 0

g(t), g(t)− f(t) < 0,

from which we get, that f ∧ g =f + |f − g|+ g

2, and f ∨ g =

f − |f − g|+ g

2.

11

Theorem 1.2.5 (Stone-Weierstrass Theorem). Let K be a compact, metric space. If

M ⊂ C(K) is a subalgebra, for which stands:

i) every constant function is in M ,

ii) M separates the points of K, which means:

∀x, y ∈ K ∃h ∈M : h(x) 6= h(y),

then M is dense in C(K).

Proof. First we will show, that: fn → f, gn → g =⇒ fngn → fg uniformly, as n→∞.

‖fg − fngn‖∞ ≤ ‖f − fn‖∞ ‖g‖∞ + ‖fn‖∞ ‖g − gn‖∞ → 0. (1.1)

From this we know, that

1. M is a subalgebra

2. The closure: M is also a subalgebra.

Now we will show, that M is a vector lattice. Let h ∈ M be aribtary, and T ∈ R, such

that ‖h‖∞ ≤ T . Since M is a vector space we only have to show, that |h| ∈ M stands.

Let I := [−T, T ], we wish to approximate h on I.

According to the first Weierstrass approximation theorem:

∃(pn)n∈N : limn→∞

pn = |x|. (1.2)

pn ◦ h ∈M stands, since pn(h) =n∑j=0

ajhj, and M is a closed subalgebra.

Let f ∈ C(K), ε ∈ R+ be arbitrary, and h(t), such that h(x) 6= h(y), for every x, y ∈ K,

if x 6= y. We define the function fxy for every x, y ∈ K:

fxy(t) := f(x)h(t)− h(y)

h(x)− h(y)+ f(y)

h(t)− h(x)

h(y)− h(x). (1.3)

Therefore:fxy(x) = f(x) and fxy(y) = f(y). We continue by defining the following set:

Uy := {z ∈ K : fxy(z) > f(z)− ε} ∀y ∈ K. (1.4)

i) K ⊂⋃y∈K

Uy

ii) Uy is an open set for every y ∈ K

iii) K is compact

12

Because of these, we can choose a finite semicover:K ⊂n⋃i=1

Uyi , and let fx := maxi=1,...,n

{fxyi}.From

this we know, that fx ≥ f − ε, and fx(x) = f(x).

Let

Vx := {z ∈ K : fx(z) < f(z) + ε} ∀x ∈ K (1.5)

This is an open set, K is compact, we can choose a finite semicover, Vxi (i = 1..n), and

g := mini=1,...,n

{fxi}. Combining these:

∀t ∈ K : f(t)− ε ≤ g(t) ≤ f(t) + ε, (1.6)

since g is in both U and V . but this also means that:

‖f − g‖∞ ≤ ε =⇒ g → f uniformly.

1.3 Korovkin’s First Theorem

Definition 1.3.1. An operator T ∈ L(C([0, 1]), C([0, 1])) is positive, if for every f ∈C(K), for which f(t) ≥ 0 for every t ∈ K, then Lf(t) ≥ 0 is also true.

Theorem 1.3.2 (Korovkin’s First Theorem). Let (Tn)n∈N ∈ L(C(K), C(K)) be a positive

operator sequence, and set xi(t) := ti(i = 0, 1, 2). If Tn(xi) → xi for i = 0, 1, 2 then

Tn(x)→ x uniformly for every x ∈ C[0, 1].

Proof. Let x ∈ C([0, 1]) be arbitrary. Since [0, 1] is compact, and x is continous, it is also

uniformly continuous. Which means:

(∀ε ≥ 0) (∃δ ≥ 0) : |s− t| ≤ δ =⇒ |x(s)− x(t)| ≤ ε (1.1)

We shall show, that: |x(s)− x(t)| ≤ ε+ α(t− s)2 for a suitable α.

Let α =2 ‖x‖∞δ

. If |t − s| ≤√δ, than |x(s) − x(t)| ≤ ε is true, due to the uniform

continuity of x. If |t− s| ≥√δ:

|x(s)− x(t)| ≤ |x(s)|+ |x(t)| < ε+ 2 ‖x‖∞ = ε+2 ‖x‖∞ δ

δ< ε+ α(t− s)2. (1.2)

By defining yt(s) := (t− s)2 (1.2) can be rewritten as:

− ε− αyt ≤ x− x(t) ≤ ε+ αyt ∀t ∈ [0, 1] (1.3)

Now we apply the Tn operator sequence, and use it’s positivity:

− εTn(x0)− αTn(yt) ≤ Tn(x)− x(t)Tn(x0) ≤ εTn(x0) + αTn(yt) ∀t ∈ [0, 1], (1.4)

13

which shows:

|Tn(x)− x(t)Tn(x0)| ≤ |εTn(x0) + αTn(yt)| ∀t ∈ [0, 1], (1.5)

First we will prove pointwise convergence, therefore let t be an arbitrary point from [0, 1].

|Tn(x)(t)− x(t)Tn(x0)(t)| ≤ |εTn(x0)(t) + αTn(yt)(t)| = ε,

since Tn(xi)→ xi (i = 0, 1, 2), and yt(t) = 0, which delivers the pointwise convergence for

every continous function in [0, 1]. The uniform converge is true as well, since:

yt = αx0 + βx1 + γx2 therefore: Tnyt − yt = Tn((t− s)2)− (t− s)2

= Tn(αx0 + βx1 + γx2)− αx0 + βx1 + γx2

= α(Tn(x0)− x0) + β(Tn(x1)− x1) + γ(Tn(x2)− x2),

since x0, x1, x2 span the quadratic polynomials. From this we can see that Tn(yt) → yt

uniformly, therefore ∀x ∈ C[0, 1] : Tnx→ x uniformly.

We proceed by proving a theorem on quadrature formulas, for which we mention the

well known Banach-Steinhaus theorem, without a proof, which can be read in [3].

Theorem 1.3.3. Let X be a Banach space, Y be a normed space. An (An)n∈N ∈ L(X, Y )

operator sequence is pointwise bounded if and only if (An)n∈N is uniformly bounded.

With the help of this, we can prove a theorem of Szego, namely:

Theorem 1.3.4 (Szego). Let Qn(x) ∈ L(C([0, 1]),R) be for every x ∈ C([0, 1]) the

following: Qn(x) =n∑i=0

αix(ti), where αi ∈ R, ti ∈ [0, 1], and ti 6= tj, if i 6= j. The

following are equivalent:

i) Qn(x)→∫ 1

0

x(t)dt for every x ∈ C([0, 1]),

ii) Qn(p)→∫ 1

0

p(t)dt for every p polynomial, and supn

n∑i=0

|αi| <∞.

Proof. We start with showing: ‖Qn‖ =n∑i=0

|αi|. Since Qn ∈ L(C([0, 1]),R),

‖Qn‖ = inf{c : |Qn(x)| ≤ c ‖x‖∞ ∀x ∈ C([0, 1])},

the following is true:

|Qn(x)| =∣∣∣∣ n∑i=0

x(ti)αi

∣∣∣∣ ≤ n∑i=0

|x(ti)||αi| ≤ ‖x‖∞n∑i=0

|αi|

14

and |Qn(1)| =n∑i=0

|αi|, therefore ‖Qn‖ =n∑i=0

|αi|. We proceed by showing i) ⇒ ii).

Since every p polynomial is continuous we only have to show, that supn

n∑i=0

|αi| < ∞,

which stands since C([0, 1]) is a Banach space, R is a normed space, and Qn is pointwise

bounded, thus according to the Banach-Steinhaus theorem Qn is uniformly bounded, i.e.

sup ‖an‖ <∞ .

We continue by showing, implication ii)⇒ i). Namely for every polynomial p, and every

continuous function x the following stands:

Qn(p)→∫ 1

0

p(t)dtn∑i=0

|αi| <∞⇒ Qn(x)→∫ 1

0

x(t)dt.

Due the the continuity of x Theorem 1.1.4 can be used: let pn : [0, 1]→ R, such polynomial

that ‖pn − x‖∞ → 0 , as n→∞. Therefore:

|Qn(pn)−Qn(x)| =∣∣∣∣ n∑i=0

αi(pn(ti)− x(ti))

∣∣∣∣ ≤ ‖pn − x‖∞ n∑i=0

|αi| ≤n∑i=0

|αi|ε, (1.6)

thus Qn(pn)→ Qn(x), as n→∞. We proceed by showing:∫ 1

0

pn(t)dt→∫ 1

0

x(t)dt, as n→∞,

which stands due to:∣∣∣∣ ∫ 1

0

pn(t)−∫ 1

0

x(t)dt

∣∣∣∣ ≤ ∫ 1

0

|pn(t)− x(t)|dt ≤ ε. (1.7)

From (1.6), and (1.7) we get i).

Here we mention Korovkin’s second theorem without proving. It will be proved in

Chapter 2.

Theorem 1.3.5 (Korovkin’s Second Theorem). Let (Tn)n∈N ∈ L(C2π, C2π) be a positive

operator sequence, and x0(t) := 1 x1(t) := cos(t) x2(t) := sin(t). Then

Tn(xi)→ xi(i = 0, 1, 2) =⇒ Tnx→ x ∀x ∈ C2π.

We have seen certain approximation methods already, but none of them could be used

in practice. The next theorem will give a new, constructive proof for Theorem 1.1.4.

Definition 1.3.6. Let I be a closed, bounded interval in R,and f ∈ C(I).

(Bnf)(x) :=n∑k=0

(n

k

)f

(k

n

)xk(1− x)n−k ∀x ∈ I ∀n ∈ N+. (1.8)

Bnf(x) is the so called n-th Bernstein polynomial.

15

Theorem 1.3.7 (Bernstein’s Approximation Theorem). For every f ∈ C(I), Bn(f)→ f

uniformly:

Proof. We assume, that I = [0, 1] for easier calculation. Bn is linear, since:

(Bn(cf))(x) :=n∑k=0

(n

k

)cf

(k

n

)xk(1− x)n−k (1.9)

= cn∑k=0

(n

k

)f

(k

n

)xk(1− x)n−k = c(Bnf)c ∈ R (1.10)

(Bn(f + g))(x) :=n∑k=0

(n

k

)(f

(k

n

)+ g

(k

n

))xk(1− x)n−k (1.11)

=n∑k=0

(n

k

)f

(k

n

)xk(1− x)n−k +

n∑k=0

(n

k

)g

(k

n

)xk(1− x)n−k = (Bnf) + (Bng) g ∈ C(I).

(1.12)

Bn is positive, since if f ≥ 0, Bnf(x) ≥ 0 holds. Due to the binomial theorem:

(Bn1)(x) =n∑k=0

(n

k

)xk(1− x)n−k = (x+ 1− x)n = 1

(Bnid)(x) =n∑k=0

(n

k

)k

nxk(1− x)n−k =

n∑k=0

(n− 1)!

(k − 1)!(n− k)!xk(1− x)n−k

=n∑k=1

(n− 1

k − 1

)xk(1− x)n−k = x(x+ 1− x)n−1 = x.

And according to Korovkin’s first theorem, by proving convergence for id2, we will get

the result, that Bn converges on [0, 1]. We start, by showing:

Bn

(id2 − id

n

)(x) =

n∑k=0

(n

k

)k(k − 1)

n2xk(1− x)n−k

=n∑k=0

n!k(k − 1)

k!(n− k)!n2xk(1− x)n−k

=n− 1

n

n∑k=2

(n− 2

k − 2

)xk(1− x)n−k

=n− 1

nx2

From this, and the fact, that Bn is a linear operator we get:

Bn(id2) =n− 1

nid2 +

id

n.

16

and finally: ∥∥id2 −Bn(id2)∥∥∞ =

1

n

∥∥id2 − id∥∥∞ → 0,

therefore Bn(f)→ f uniformly for every f ∈ C([0, 1]).

17

Chapter 2

Fourier series

It is a well known fact from elementary analysis, that not every f : R→ R integrable

function’s Fourier series converge uniformly. In this chapter we will inspect, under what

conditions can uniform convergence be achieved, and inspect the general theory of Fourier

series.

2.1 Basics

Definition 2.1.1. Let f ∈ C2π .The Fourier series of f is the following:

a02

+∞∑k=1

ak cos(kt) + bk sin(kt),

where

ak =1

π

∫ π

−πf(t) cos(kt)dt and bk =

1

π

∫ π

−πf(t) sin(kt)dt.

Definition 2.1.2. ai, bj,where i ∈ N, j ∈ N+, are called the corresponding Fourier

coefficients.

Remark 2.1.3. In order to determine, whether we are talking about a series, or the sum

of the series, we will write∑j∈N

fj, when we mean the series, and write∞∑j=0

fj, when we

mean the sum of the series.

Theorem 2.1.4. Let a0 +∑k∈N+

ak cos(kt) + bk sin(kt) be uniformly convergent on R. if

a0 +∞∑k=1

ak cos(kt) + bk sin(kt) = f(t), then f is continuous, and

a0 =1

2π

∫ π

−πf(t)dt,

ak =1

π

∫ π

−πf(t) cos(kt)dt , bk =

1

π

∫ π

−πf(t) sin(kt)dt.

18

We will need two lemmas to prove this.

Lemma 2.1.5. If fn → f uniformly on H ⊂ R, and fn is continuous in a ∈ H =⇒ f is

continuous in a ∈ H.

Proof. We have to show, that:

(∀ε ≥ 0) (∃δ ≥ 0) : |t− a| ≤ δ ⇒ |f(t)− f(a)| ≤ ε.

To this aim fix ε > 0, then by exploiting uniform convergence there is N ∈ N, so that

|fn(t)− f(t)| ≤ ε3

for each n > N and t ∈ H.

By exploiting fn’s continousity, let δ > 0, such that, |fN(t)− fN(a)| ≤ ε

3,

if |t− a| ≤ δ. Combining these:

|f(t)− f(a)| =|f(t)− fn(t) + fn(t)− fn(a) + fn(a)− f(a)|

≤|f(t)− fn(t)|+ |fn(t)− fn(a)|+ |fn(a)− f(a)|

≤ε.

Which proves f ’s continuousity.

Lemma 2.1.6. Let n ∈ N+.

i)

∫ π

−πsin2(nt)dt =

∫ π

−πcos2(nt)dt = π

ii) n 6= m

∫ π

−πsin(nt) cos(mt)dt = 0

iii) n 6= m

∫ π

−πsin(nt) sin(mt)dt =

∫ π

π

cos(nt) cos(mt)dt = 0

Proof. ∫ π

−πsin2(nt)dt =

∫ π

−π

1

2− cos(2nt)

2dt =

[t

2− sin(2nt)

4n

]2π0

= π∫ π

−πcos2(nt)dt =

∫ π

−π1− sin2(nt)dt = 2π − π,

thus (i) is proved.∫ π

−πcos(mt) sin(nt)dt =

[m sin(mt) sin(nt) + n cos(mt) cos(nt)

m2 − n2

]π−π

= 0,

∫ π

−πcos(nt) cos(mt)dt =

[m sin(mt) cos(nt)− n cos(mt) sin(nt)

m2 − n2

]π−π

= 0,∫ π

−πsin(mt) sin(nt)dt =

[n sin(mt) cos(nt)−m cos(mt) sin(nt)

m2 − n2

]π−π

= 0,

since sin and cos ∈ C2π.

19

Now we proceed, by proving Theorem 2.1.4.

Proof. Since a0 +∑k∈N+

ak cos(kt) + bk sin(kt) is continuous, by exploiting Lemma 2.1.5,

f is continuous, and according to the theorem on integrating uniformly convergent se-

quences, both side of the equation: a0 +∞∑k=1

ak cos(kt) + bk sin(kt) = f(t) can be inte-

grated. By doing so we get:∫ π

−πa0 +

∫ π

−π

∞∑k=1

ak cos(kt) + bk sin(kt)dt =

∫ π

−πf(t)dt, (2.1)

which gives us: a0 =

∫ π

−π

f(t)

2πdt. We proceed, by showing, that: for every m ≥ 0

a0 cos(mt) +∞∑k=1

ak cos(kt) cos(mt) + bk sin(kt) cos(mt) (2.2)

converges uniformly on R.

Let ε ≥ 0, due to the uniform convergence of a0 +∑k∈N+

ak cos(kt) + bk sin(kt), there is

N ∈ N, such that for every n ≥ N∣∣∣∣f(t)− a0 +n∑k=1

ak cos(kt) + bk sin(kt)dt

∣∣∣∣ ≤ ε.

Since | cos(mt)| ≤ 1:∣∣∣∣f(t) cos(mt)− a0 cos(mt) +n∑k=1

(ak cos(kt) cos(mt) + bk sin(kt) cos(mt)

∣∣∣∣ ≤ ε,

for every n ∈ N, and every x ∈ R,which means, that,(2.2) converges uniformly, therefore

it can be integrated. This gives us the following:∫ π

−πf(t) cos(mt)dt = πam.

The same can be done with sin(mx), which gives:∫ π

−πf(x) sin(mx)dx = πbm,

which means the theorem has been proved.

This theorem sounds a bit strange, since it states, that if the Fourier series converges

uniformly, it converges to f , where the Fourier-coefficients can be obtained by integration.

However, if we take an arbitrary f ∈ C2π, we can calculate its Fourier coefficients, but

will not know whether the Fourier series will converge uniformly to f or not. Emphasizing

this problem, we will proceed, by showing that, there are f ∈ C2π, such that, it’s Fourier

series will not even converge pointwise.

20

Remark 2.1.7. When calculating the Fourier coefficients, f doesn’t have to be integrated

on [−π, π]. Any 2π long interval will do,since if f is periodical, with a period length of

2π,and integrable on [−π, π] then: :∫ a+2π

a

f(t)dt =

∫ π

−πf(t) ∀a ∈ R

Lemma 2.1.8. Let f ∈ C2π. Let Smf(x) be the m-th partial sum of f ’s Fourier series,

i.e.,

a0 +m∑k=1

ak cos(kx) + bk sin(kx).

Then Smf(x) =1

2π

∫ π

−πDm(x− t)f(t)dt, where Dm ∈ C2π, and

Dm(2s) =

sin((2m+ 1)s)

sin(s)if s 6= 0

2m+ 1 if s = 0

(2.3)

and Dm(s) is the so called Dirichlet-kernel.

Remark 2.1.9. Continuity is granted, since lims→0

sin((2m+ 1)s)

sin(s)= 2m+ 1.

Proof.

Smf(x) =a02

+m∑k=1

ak cos(kx) + bk sin(kx)

=1

2π

∫ π

−π

(1 + 2

m∑k=1

cos(kx) cos(kt) + sin(kx) sin(kt)

)f(t)dt

=1

2π

∫ π

−π

(1 + 2

m∑k=1

cos(k(x− t)))f(t)dt.

The only thing left is to show, that:

1 + 2m∑k=1

cos(2ks) =sin((2m+ 1)s)

sin(s)∀m ∈ N.

We prove by induction. If m = 0 then 1 = sin(s)sin(s)

= 1. Suppose the proposition stands for

m = m: 1 + 2m∑j=1

cos(2js) =sin((2m+ 1)s)

sin(s)

Consider m = m+ 1.

2 sin(s) cos(2(m+ 1)s) = sin((2m+ 3)s)− sin((2m+ 1)s) ∀m ∈ N.

After dividing by sin(s), and exploiting the inductional assumption, the proof has been

completed.

21

We proceed, by defining the following ϕ : (C2π(R), ‖·‖∞)→ R. functional. Let f ∈ C2π

be arbitrary, then ϕm(f) := Smf(0).Since ϕ is a functional, the following stands:

‖ϕ‖ = inf{c ∈ R+ : |ϕ(f)| ≤ c ‖f‖ : ∀f ∈ C2π}.

Lemma 2.1.10. ϕm is continuous, and ‖ϕm‖ → ∞.

Proof. Since |aj|, |bi| ≤ ‖f‖∞ ∀i ∈ N+, ∀j ∈ N, due to Smf ’s definition:

‖Smf‖∞ ≤(

2m+1

2

)2 ‖f‖∞ = (4m+ 1) ‖f‖∞ ⇒ ‖ϕm‖ ≤ 4m+ 1 <∞

Let f(2s) := (sign(sin(s)) sin((2m+ 1)s). It can be easily seen, that f ∈ C2π, and

‖f‖∞ = 1.

ϕm(f) =1

2π

∫ π

−πDm(−t)f(t)dt =

1

π

∫ π2

−π2

Dm(−2s)f(2s)ds (2.4)

=1

π

∫ π2

−π2

sin2((2m+ 1)s)

| sin(s)|ds =

2

π

∫ π2

0

sin2((2m+ 1)s)

sin(s)ds (2.5)

≥ 2

π

∫ π2

0

sin2((2m+ 1)s)

sds =

2

π

∫ (2m+1)π2

0

sin2(s)

sds (2.6)

≥ 2

π

m∑j=1

∫ jπ

(j−1)π

sin2(s)

sds ≥ 2

π

∫ π

0

m∑j=1

sin2(s)

jπds (2.7)

=1

π

m∑j=1

1

j. (2.8)

(2.4) is true due to the definition of the Dirichlet-Kernel, and changing the integration

variable, in (2.5) we used f ’s definition,and that sin2 is an even function. In (2.6) we

exploited the periodicity of sin, in (2.7) we used the integral’s additivity, and the fact,

that the sum and integral can be interchanged. From these we get, that ‖ϕm‖ ≥ ϕm(f) >

1

π

m∑j=1

1

j→∞.

Combining these we get:

Theorem 2.1.11. There are f ∈ C2π, whose Fourier-series doesn’t converge to f point-

wise.

Proof. We prove indirect.

(∀f ∈ C2π) : ϕm(f)→ f(0).

We will use the Banach-Steinhaus theorem, which states the following:

Let X be a Banach space, Y be a normed space. An (An)n∈N ∈ L(X, Y ) operator sequence

is bounded pointwise if and only if (An)n∈N is bounded uniformly.

Using this and the fact, that C2π is a Banach space, and R is a normed space we get, that

‖ϕm‖ <∞, which contradicts the previous lemma.

22

In the previous theorem we have seen, that not even pointwise convergence of Fourier

series is guaranteed for arbitrary functions. There are two possible ways to procced

1. strengthening f ’s properties, in order to achieve uniform convergence

2. weakening the notion of convergence.

We continue by showing that every Fourier series converges in L2. In order to do so we

generalize Fourier series in arbitrary Hilbert spaces.

2.2 Abstract Fourier Series In Hilbert Spaces

In this chapter, H is always a Hilbert space, and 〈·, ·〉, is the inner product of the

Hilbert space.

Definition 2.2.1. An (en)n∈N ⊂ H vector sequence is a complete system, if

(∀x ∈ H) : 〈x, en〉 = 0 ∀n ∈ N⇒ x = 0

Definition 2.2.2. An (en)n∈N is a total system, if

span((en)n∈N) = H

From the elementary theory of Hilbert spaces the following theorem is well known:

Let M be a subset in H, the following are equivalent

1. M is a complete set, i.e. M⊥ = 0

2. M is a total system.

Definition 2.2.3. An (en)n∈N ⊂ H vector sequence is

1. orthogonal, if 〈ei, ej〉 = 0 ∀i 6= j ∈ N, and 〈ei, ei〉 = c c ∈ R+

2. orthonormal, if (en)n∈N is orthogonal, and 〈ei, ei〉 = 1 ∀i ∈ N

Proposition 2.2.4. Let (en)n∈N ⊂ H, be an orthogonal sequence, then

∞∑j=0

ej converge ⇐⇒∞∑j=0

‖ej‖2 converge

Proof. Let sn :=n∑j=0

ej, and σn :=n∑j=0

‖ej‖2, and n,m ∈ N, n ≥ m. By exploiting

orthogonality of the vector sequences, we get:

‖sn − sm‖2 =

∥∥∥∥∥n∑

j=m+1

ej

∥∥∥∥∥2

=n∑

j=m+1

‖ej‖2 = |σn − σm|. (2.1)

Since bothH, and R are complete, and on the two sides of (2.1) we have Cauchy-sequences,

they converge under the same conditions.

23

Definition 2.2.5. Let (en)n∈N be a complete orthonormal system, and x ∈ H be arbitrary.∑j∈N

〈x, ej〉ej is called x’s Fourier series, and 〈x, ej〉 are x’s Fourier coefficients.

Theorem 2.2.6. Let H be a Hilbert-space, (en) ⊂ H complete orthonormal system. for

every x ∈ H x’s Fourier series converge, and their sum equals x.

Proof. Let xi := 〈x, ei〉ei ∀i ∈ N, which is a complete orthonormal system. Using 2.2.4:

∞∑i=1

xi converge ⇐⇒∞∑i=1

‖xi‖2 converge

Let sn :=n∑i=1

xi, and Hn := span{e1, . . . en} n ∈ N+. In this case sn ⊂ Hn, and ‖sn‖2 ≤

‖x‖2, since

0 ≤ ‖x− sn‖2 = ‖x‖2 + ‖sn‖2 − 2Re〈x, sn〉 = ‖x‖2 − ‖sn‖2 ,

since

〈x, sn〉 =n∑j=1

〈x, ej〉〈x, ej〉 =n∑j=1

|〈x, ej〉|2 =

∥∥∥∥∥n∑j=1

〈x, ej〉ej

∥∥∥∥∥2

= ‖sn‖2 .

From this, we get the convergence:

n∑i=0

‖xi‖2 = ‖sn‖2 ≤ ‖x‖2 ⇒∞∑i=0

‖xi‖2 <∞⇒∞∑i=1

xi converge.

We have to prove, that the sum is x. Let s :=∞∑i=1

xi, we will show, that s = x. Due to

the fact, that (ei)i∈N, is complete, it is enough to show, that 〈s− x, ej〉 = 0

∀j ∈ N. Using 〈·, ·〉’s continuity:

〈s, ej〉 =

⟨ ∞∑i=1

xi, ej

⟩=

⟨ ∞∑i=1

〈x, ei〉ei, ej⟩

=∞∑i=1

〈x, ei〉〈ei, ej〉 = 〈x, ej〉,

therefore 〈x− s, ej〉 = 0 ∀j ∈ N+.

Remark 2.2.7. This theorem doesn’t bring us closer to giving sufficient conditions for

Fourier series to converge uniformly, since ‖·‖∞ is not generated by an inner product.

Remark 2.2.8. Using this theorem, generalized Fourier series can be inspected, using

arbitrary orthogonal systems. A few examples for it:

1. H := L2(0, π) en(t) =√

2π

sin(nt) ∀n ∈ N

2. H := L2(0, π) e0(t) = 1√πen(t) =

√2π

cos(nt) ∀n ∈ N+

24

3. H := L2(0, 2π) en(t) =√

12π

einx ∀n ∈ N

4. Fourier series, defined in Definition 2.1.1

Remark 2.2.9. In order to avoid misunderstandings, from now on, if we write Fourier

series, we are reffering to the traditional ones, defined in Section 2.1. When we need the

generalized notion of Fourier series, we will refer to it as generalized Fourier series.

2.3 Uniform Convergence Of Fourier Series

Definition 2.3.1. Let f : E → R, if the range of f is finite, then we say that f is simple.

Theorem 2.3.2. The set of continuous functions on [a, b] is a dense set in L2([a, b]).

Proof. Let f ∈ L2([a, b]), and we need g ∈ C([a, b]), such that ‖f − g‖2 ≤ ε. Let A ⊆ [a, b]

be a closed subset, and χA the characteristic function of A, namely:

χA(t) =

0, t /∈ A

1, t ∈ A.

Let h(t) := inft∈[a,b]

|t− y|, for every y ∈ A, and gn(t) :=1

1 + nh(t)for every n ∈ N. For gn

the following hold:

i) gn(t) is continuous for every t ∈ [a, b],

ii) gn(t) = 1, if t ∈ A, and gn(t)→ 0 for every t /∈ A as n→∞.

Exploiting this we get that ‖gn − χA‖2 =

(∫[a,b]\A

g2n

) 12

→ 0 as n → ∞, which means

that characteristic functions of closed sets can be approximated with continuous functions.

From this we obtain, that every simple function can be approximated with continuous

functions.

Let f ∈ L2([a, b]), f ≥ 0, and (sn)n∈N non negative, simple,monotone increasing set of

functions, such that sn(t) → f(t). Since |f − sn|2 ≤ f 2, according to the dominated

convergence theorem ‖f − sn‖2 → 0, as n → ∞. To prove for arbitrary f we have to

decompose f into f+ and f−.

Theorem 2.3.3 (Completeness of Fourier series). Let f ∈ C2π be arbitrary. If ai, bi =

0 ∀i ∈ N+, and a0 = 0 then f = 0, where ai, bi are the corresponding Fourier coefficients.

Proof. We have to show, that the trigonometrical polynomials, are a total system, in

L2([−π, π]), which is equivalent, to being a complete orthonormal system, so we proceed,

25

by showing, that the trigonometrical polynomials, are a complete orthonormal system.

Let f ∈ L2[−π, π], ε > 0, we will need a p trigonometrical polinom , such that, ‖f − p‖2 ≤ε We know from Theorem 2.3.1, that C[−π, π] is dense in L2[−π, π].Let g ∈ C(−π, π)

such that, ‖f − g‖2 ≤ ε.

For this g, we can construct a g, such that g(−π) = g(π) = 0, and ‖g − g‖2 ≤ ε. by

noticing the fact: g is periodical, with a period length of 2π, according to the second

Weierstrass theorem, it can be approximated with a p trigonometrical polynom, such that

‖g − p‖∞ ≤ ε2. From this: ∫ π

−π|g − p|2 ≤ 2π ‖g − p‖2∞ ,

which implies

‖g − p‖2 ≤√

2π ‖g − p‖∞ ≤√

2πε,

therefore

‖f − p‖2 ≤ ‖f − g‖+ ‖g − g‖+ ‖g − p‖ ≤ ε(2 +√

2π).

Which means, that the trigonometric polynomials form a complete orthonormal set in

L2([−π, π]). Using this, let f ∈ C2π, be arbitrary, whose Fourier coefficients are 0. Since

f ’s every Fourier coefficients are 0, f is orthogonal to a complete orthonormal system

in L2, which means, f = 0 a.e. But since f is continuous, it also means, that f = 0

everywhere.

Proposition 2.3.4. Let f ∈ C2π, be arbitrary. If f ’s Fourier series converge uniformly

on R, then a0 +∞∑j=1

aj cos(jt) + bj sin(jt) = f(t) ∀t ∈ R

Proof. a0 +∑j∈N+

aj cos(jt) + bj sin(jt) is continuous, and every Fourier coefficients of a0 +∑j∈N+

aj cos(jt) + bj sin(jt), are identical to f ’s Fourier coefficients, therefore

f(t)− a0 +∑j∈N+

aj cos(jt) + bj sin(jt) ∈ C2π, and the Fourier coefficients of

f(t)−a0+∑j∈N+

aj cos(jt)+bj sin(jt) are 0, therefore f(t)−a0+∑j∈N+

aj cos(jt)+bj sin(jt) = 0

which implies f(t) = a0 +∑j∈N+

aj cos(jt) + bj sin(jt) for every t ∈ R.

Notation 2.3.5. f ∈ Ck2π, if f ∈ C2π ∩ Ck(R)

Lemma 2.3.6. If f ∈ Ck2π, then exists such M ∈ R+:

(∀n ∈ N+) : |an| ≤M

nk, |bn| ≤

M

nk.

26

Proof. We proceed using induction. If k = 0 f ∈ C2π, therefore f is bounded. Let

K := supt∈[−π,π]

{|f(t)|}. From the defintion of the Fourier coefficients, it can be seen, that:

|ai|, |bj| ≤ 2K ∀i ∈ N ∀j ∈ N+. Thus, the property holds if k = 0.

Assume, it is true for k = k, and consider k = k+ 1: f ∈ Ck+12π ⇒ f ′ ∈ Ck

2π. Which means,

that the property holds, for the Fourier coefficients of f ′. Let n ∈ N+.

an =

∫ π

−πf(t) cos(nt)dt =

[f(x)

sin(nt)

n

]π−π−∫ π

−π

f ′(t) sin(nt)

ndt

= 0− 1

n

∫ π

−πf ′(t) sin(nt)dt =⇒

∣∣∣∣ ∫ π

−πf(t) cos(nt)dt

∣∣∣∣ ≤ M

nk+1.

bn =

∫ π

−πf(t) sin(nt)dt =

[f(x)

cos(nt)

n

]π−π−∫ π

−π

f ′(t) cos(nt)

ndt

= 0− 1

n

∫ π

−πf ′(t) cos(nt)dt =⇒

∣∣∣∣ ∫ π

−πf(t) sin(nt)dt

∣∣∣∣ ≤ M

nk+1.

Theorem 2.3.7. Let f ∈ C22π, then f ’s Fourier series converges uniformly to f .

Proof. According to Lemma 2.3.6,

|aj| ≤M

j2, |bj| ≤

M

j2, ∀j ∈ N,

therefore:∣∣∣∣a0 +∑j∈N+

ak cos(jt) + bk sin(jt)

∣∣∣∣ ≤ |a0|+ ∞∑j=1

M(| cos(jt)|+ | sin(jt)|)j2

≤ |a0|+∞∑j=1

2M

j2.

This means, that a0 +∑j∈N+

aj cos(jt) + bj sin(jt) fulfills the Weierstrass-criteria, therefore

a0 +∑j∈N+

aj cos(jt) + bj sin(jt)→ f uniformly.

Our next theorem is on the order of the Fourier coefficients.

Theorem 2.3.8 (Riemann-Lebesgue Lemma). If f ∈ C2π, then limi→∞

ai, bi = 0, where

ai, bi are the corresponding Fourier coefficients.

Proof. We want to show, that

i) bk = limk→∞

∫ π

−πf(t) sin(kt)dt = 0,

27

ii) ak = limk→∞

∫ π

−πf(t) cos(kt)dt = 0,

We start by proving i).

Let ε > 0 be arbitrary. Since f is continuous and periodical with a period length of 2π, we

can use the Second Weierstrass Approximation theorem, therefore let N ∈ N, such that:

‖f − pN‖∞ ≤ ε, where pN is a trigonometrical polynomial. Let k ∈ N, and k > N , then

bk =1

π

∫ π

−πf(t) sin(kt)dt =

1

π

∫ π

−πf(t) sin(kt)− pN(t) sin(kt)dt,

according to Lemma 2.1.6. Exploiting this we get:

|bk| =∣∣∣∣ 1π∫ π

−πf(t) sin(kt)dt

∣∣∣∣ ≤ 1

π

∫ π

−π|(f(t)− pN(t)) sin(kt)|dt

≤ ‖f − pN‖∞1

π2π ≤ 2ε,

thus bk → 0, as k → ∞. Proving ii) starts the same as for i), the only difference arise

during the calculation of the ak coefficients, namely:

ak =1

π

∫ π

−πf(t) cos(kt)dt =

1

π

∫ π

−πf(t) cos(kt)− pN(t) cos(kt)dt,

according to Lemma 2.1.6. Exploiting this we get:

|ak| =∣∣∣∣ 1π∫ π

−πf(t) cos(kt)dt

∣∣∣∣ ≤ 1

π

∫ π

−π|(f(t)− pN(t)) cos(kt)|dt

≤ ‖f − pN‖∞1

π2π ≤ 2ε.

Definition 2.3.9. Let α ∈ (0, 1], f : I → R. We say that f is Holder continuous, if for

every x, y ∈ I exists such C ∈ R, that |f(x) − f(y)| ≤ C|x − y|α. The set of Holder

continuous functions with α ∈ R constant is referred as Lipα(I).

Remark 2.3.10. If α = 1, f is Lipschitz continuous.

Without proving we mention the following:

Remark 2.3.11. Let α ∈ (0, 1], and f ∈ Lipα([−π, π]), where f(−π) = f(π). Then

a0 +∑j∈N+

aj cos(jt) + bj sin(jt)→ f uniformly.

2.4 Fejer-summation

Up to this point we have seen, that for every f ∈ C22π f ’s Fourier series will converge

uniformly to f . We have seen that convergence can be achieved in arbitrary Hilbert spaces,

28

and exploited the fact, that L2([−π, π]) is a Hilbert space.We proceed by modifying Fourier

series, in order to show that for every continuous function f its Fourier series converge

to f uniformly. The motivation for this, is the well known fact, from elementary analysis,

that:

an → a⇒ 1

n

n∑i=1

ai → a.

Definition 2.4.1. Let σn(f) :=1

n+ 1

n∑m=0

Smf , where σn is the so called n-th Fejer-sum.

Lemma 2.4.2. σnf(x) =1

2π

∫ π

−πFn(x− t)f(t)dt, where Fn(2s) =

1

n+ 1

sin2((n+ 1)s)

sin2(s).

Fn is called the Fejer-kernel.

Proof. Since

σn(f) =1

n+ 1

n∑m=0

Smf =1

n+ 1

n∑m=0

Smf =n∑

m=0

1

2π

∫ π

−πDm(x− t)f(t)dt

n+ 1

=1

2π(n+ 1)

∫ π

−π

n∑m=0

Dm(x− t)f(t)dt.

Which means: Fn =

n∑j=0

Dj

n+ 1, therefore it is enough to show:

sin2((n+ 1)s)

sin2(s)=

n∑m=0

sin((2m+ 1)s)

sin(s).

By direct calculation:

n∑m=0

sin(s) sin((2m+ 1)s) =1

2

n∑m=0

cos(2ms)− cos((2m+ 2)s)

=1− cos((2n+ 2)s)

2= sin2((n+ 1)s).

After dividing by sin2(s), we get:

sin2((n+ 1)s)

sin2(s)=

n∑m=0

sin((2m+ 1)s)

sin(s).

Proposition 2.4.3. σn is a positive operator.

Proof. Let f ≥ 0 arbitrary. Since Fnf ≥ 0⇒∫ π

−πFn(t)f(t)dt ≥ 0.

29

Theorem 2.4.4 (Freud). Let K be a compact metric space, h1, h2, . . . hm ∈ C(K), such

functions, that separates K’s points, and h0 := 1. If

‖hi − Lnhi‖∞ → 0 ∀h = (0, 1, 2, . . .m), and

∥∥∥∥∥m∑i=1

h2i − Ln( m∑

i=1

h2i

)∥∥∥∥∥∞

→ 0, (2.1)

then:

‖f − Lnf‖∞ → 0 ∀f ∈ C(K).

Proof. Let f ∈ C(K) be arbitrary, and ε ≥ 0, M := {hi : 1 ≤ i ≤ m}. Since M separates

K’s points, for every x, y ∈ K, we can choose Nx,y, such that:

|f(x)− f(y)| ≤ ε+Nx,y

m∑j=1

|hj(x)− hj(y)|2. (2.2)

Let Gx,y be the following:

Gx,y := {(x′, y′) ∈ K ×K : |f(x′)− f(y′)| ≤ ε+Nx,y

m∑j=1

|hj(x)− hj(y)|2}.

By (2.2) we have K ⊂⋃

x,y∈K

Gx,y, and due to the fact, that K is a compact set, we

can choose a finite subcover. Let⋃i∈I

G ′xi,yi ⊂⋃x,y

Gx,y be a finite subcover, and N be the

maximum of Nxi,yi ’s. Therefore:

|f(x)− f(y)| ≤ ε+N

m∑j=1

|hj(x)− hj(y)|2} ∀x, y ∈ K.

Using this, for a fixed x ∈ K, and arbitrary y ∈ K:

|f(x)(Ln1)(y)− (Lnf)(y)|

≤ ε(Ln1)(y) +Nm∑j=1

h2j(x)(Ln1)(y)− 2Nm∑j=1

hj(x)(Lnhj)(y) +NLn

( m∑j=1

h2j

)(y).

From this, by choosing y = x, we get:

|f − Lnf | = |f − Lnf + fLn1− fLn1| ≤ |f ||1− Ln1|+ |Lnf − fLn1| ≤

≤ |f ||1− Ln1|+Nm∑j=1

h2j(Ln1)− 2Nm∑j=1

hj(Lnhj) +NLn

( m∑j=1

h2j

)→ 0,

because of (2.1).

We give another proof to Korovkin’s first theorem.

30

Theorem 2.4.5 (Korovkin’s first theorem). If Ln ∈ L(C(I), C(I)) is a positive operator,

and

Lnf → f uniformly, if f = 1, t, t2,

then Lnf → f uniformly, ∀f ∈ C(I).

Proof. We can apply Theorem 2.4.4, with K = I, and m = 1.

Theorem 2.4.6 (Korovkin’s second theorem). If Ln ∈ L(C2π, C2π) is a positive operator,

and

Lnf → f uniformly, if f = 1, cos, sin,

then Lnf → f uniformly, ∀f ∈ C2π.

Proof. We will prove, exploiting Theorem 2.4.4. Let K ⊂ Rm be a compact subset, and

for every x ∈ Rm, xi is the i-th coordinate of x, and hj(x) := xj for every 1 ≤ j ≤ m. The

hi functions separate K’s points, thus it fulfills hi’s demanded properties.

In the following let m = 2, h0 = 1, h1 = x1, h2 = x2, and K = ∂B0(1). Since x21 +x22 = 1,

if Lnf → f uniformly, for f = 1, x1, x2, then Lnf → f for every f ∈ C(K), according

to 2.4.4.

Let T (s) = (cos(s), sin(s)). Using the fact, that f 7→ f ◦ T , is an isometric isomorphism

between C(K), and C2π, and if f ≥ 0 ⇒ T (f) ≥ 0. Since hi ∈ C(K), i = 0, 1, 2, and

h(T (s)) = 1, cos(s), sin(s), we get, that the results for C(K), is equivalent to the results

for C2π, thus the proof has ended.

Theorem 2.4.7 (Fejer). Let f ∈ C2π be arbitrary. The sequence of the Fejer-sums con-

verges uniformly to f:

∀f ∈ C2π ⇒ σn(f)→ f uniformly

where σn(f) =1

n+ 1

n∑m=0

Smf, ∀n ∈ N.

Proof. Using the definition, we get the following:

σn1 =1

n+ 1Sm1 = 1,

σn cos =1

n+ 1Sm cos =

n

n+ 1cos,

σn sin =1

n+ 1Sm sin =

n

n+ 1sin,

and since σn is a positive operator, by using Theorem 2.4.6, we get the theorem.

31

Chapter 3

Solving Partial Differential

Equations With Fourier Series

In this chapter we wish to proceed with our inspections to a rather more applied field.

Our goal is to solve certain partial differential equations using Fourier series, such as the

wave and heat equation.

Theorem 3.0.1 (Dirichlet). Let f ∈ C12π. Then the Fourier series a0 +

∑k∈N+

ak cos(kx) +

bk sin(kx) converges uniformly, to f .

Proof. We will use the properties of the Dirichlet-kernel, namely:

Smf(x) =1

2π

∫ π

−πDm(x− t)f(t)dt, where Dm ∈ C2π, and

Dm(2s) =

sin((2m+ 1)s)

sin(s)if s 6= 0

2m+ 1 if s = 0,

(3.1)

where Smfx = a0 +m∑j=1

aj cos(jx) + bj sin(jx). Let 0 ≤ h ≤ π:

|Smf(t)− t| =∣∣∣∣ 1π∫ π

−π(f(s+ t)− f(t))Dm(s)ds

∣∣∣∣≤ 1

π

(∣∣∣∣ ∫ −h−π

(f(s+ t)− f(t))Dm(s)ds

∣∣∣∣+

∣∣∣∣ ∫ h

−h(f(s+ t)− f(t))Dm(s)ds

∣∣∣∣+

∣∣∣∣ ∫ π

h

(f(s+ t)− f(t))Dm(s)ds

∣∣∣∣) =:1

π(A+B + C).

Let ε > 0 be arbitrary, and 0 < h <ε

π ‖f ′‖∞. Then according to the mean value theorem

32

of integration, f(s+ t)− f(t) = f ′(u)s for some u ∈ [−h, h], hence∣∣∣∣ ∫ h

−h(f(s+ t)− f(t))Dm(s)ds

∣∣∣∣ ≤ ∫ h

−h

f(s+ t)− f(t)

2 sin( |s|2

)|sin

((m+

1

2

)s

)ds

≤∫ h

−h

‖f ′‖∞ |s|2 sin( |s|

2)ds ≤ 2h ‖f ′‖∞

π

2,

since2u

π≤ sinu for every 0 ≤ u ≤ π

2. From this B ≤ ε.

We continue, by estimating C. Let ft(s) =f(s+ t)− f(t)

2 sin( s2)

, and n = m+ 12. It can be easily

seen, that ft ∈ C1[h, π], and supt∈R‖f ′t‖∞ < ∞ and sup

t∈R‖ft‖∞ < ∞. By partial integration

we get: ∣∣∣∣ ∫ π

h

ft(s) sin(ns)ds

∣∣∣∣ =

∣∣∣∣ft(s)−1

n

∣∣∣∣s=πs=h

−∣∣∣∣ ∫ π

h

f ′t(s)−1

nds

∣∣∣∣ ≤ c

n,

for a suitable constant c, therefore there isN ∈ N, such that for every n > N C ≤ ε. We get

A ≤ ε in the same way. Combining these: we get Smf(t)→ f(t) for every t ∈ [−π, π].

Theorem 3.0.2 (Term by Term Differentiation). Let f be a C22π function. If f ′ is in C1

2π,

then the Fourier series a0 +∑k∈N+

ak cos(kx) + bk sin(kx) = f(x) can be differentiated term

by term, and the series so obtained converges uniformly to f ′.

Proof. Since f ′ is in C12π we can use the Theorem 3.0.1.

Lemma 3.0.3. Let f ∈ C2π be an odd function, assuming, that the corresponding Fourier

series converge. Then the corresponding Fourier series is∑k∈N+

bk sin(kt).

Proof. Since the Fourier series converge, we only have to determine the Fourier coefficients,

and show that ai = 0 for every n ∈ N. Which is true, since f is an odd function a0 = 0,

and ak =1

π

∫ π

−πf(t) cos(kt)dt = 0, since f(t) cos(kt) is also an odd function.

Lemma 3.0.4. Let f ∈ C2π be an even function, assuming, that the corresponding Fourier

series converge. Then the corresponding Fourier series is a0 +∑k∈N+

ak cos(kt).

Proof. Since the Fourier series converge, we only have to determine the Fourier coefficients,

and show that bi = 0 for every n ∈ N+. Which is true, since bk =1

π

∫ π

−πf(t) sin(kt)dt = 0,

since f(t) sin(kt) is an odd function.

Definition 3.0.5. Let f ∈ C2π the Fourier series of f is a sine (resp., cosine) series, if it

only consists of sine (resp., cosine) terms.

With the help of this, we can create a sine (resp., cosine) series of any continuous f ∈C([0, π]), by expanding it onto [−π, π], such that f(−x) = −f(x) (resp., f(−x) = f(x)),

if the expanded function’s Fourier series converges.

33

3.1 The Wave Equation

Consider a string, bound at both endpoints, with a length of L. We wish to inspect the

reaction of the string to physical interference. First we need to construct a mathematical

model. Let u(t, x) : R+ × [0, L] → R, u(t, x) ∈ C2 be a function, which shows, where

position x of the string after t time is. We also assume, that u(t, 0) = u(t, L) = 0, since

the string is bound at both end points, and is unable to move. We also consider the

location, and the velocity of the string to be known, at the beggining of the experiment.

Which means u(0, x) = f(x), ∂tu(t, x) = g(x). Since u ∈ C2 both f, g ∈ C2 stands. We

rely on the physical fact, that the partial differential equation, modeling the string is the

following: ∂2t u(t, x) = c2∂2xu(t, x). Combining these, and assuming that π = L (if not, we

transform the system) we get, the following problem:

∂2t u(t, x) =c2∂2xu(t, x) ∀t ∈ R+ ∀x ∈ [0, π] (3.1)

u(0, x) =f(x) ∀x ∈ [0, π] (3.2)

∂tu(t, x) =g(x) ∀x ∈ [0, π] (3.3)

u(t, 0) =u(t, π) = 0 ∀t ∈ R+, (3.4)

where c is a non zero constant, defined by properties, such the thickness, or the tension

of the string.

We look for the solution in a form, such that u(t, x) = T (t)X(x). This reduces our

system to the following: T ′′(t)X(x) = c2T (t)X ′′(x), and after dividing both sides, we

get:T ′′(t)

c2T (t)=X ′′(x)

X(x). It can be easily seen, that the right side of the equation only depends

upon t, while the right side only upon x. From this we know, that both sides have to be a

constant. LetT ′′(t)

c2T (t)=X ′′(x)

X(x)= η, from which we get two separable ordinary differential

equations:

T ′′(t)− c2ηT (x) = 0 (3.5)

X ′′(x)− ηX(x) = 0. (3.6)

This has a solution for every η arbitrary constant.

Before we proceed, we shall inspect the possible values of η. In the case, where η > 0,

we get the solutions in the form e√ηt, which means, as time goes by, the oscillation of the

string grows, which contradicts to the physical experiments. Which gives us, that η < 0.

We rewrite (3.5),(3.6), using η = −η:

T ′′(t) + c2ηT (x) = 0 (3.7)

X ′′(x) + ηX(x) = 0. (3.8)

34

Solving this will give us:

T (t) =a1 cos(√ηtc) + a2 sin(

√ηtc),

X(x) =b1 cos(√ηx) + b2 sin(

√ηx).

Since u(t, x) = T (t)X(x), (3.1) stands. We proceed with checking the other conditions.

We start with (3.4). Since u(t, 0) = u(t, π) for every t ∈ R+, we get, we only need the sine

terms from X(x).

Only (3.2), and (3.3) is left to be shown. However apart from particular f and g functions,

these conditions can not be satisfied by such T (t)X(x) functions. Therefore we exploit

the fact, that if we have two solutions, their sum will also be a solution to (3.1).

Let η = k2, and uk(t, x) = sin(kx)(a1(k) cos(ckt) + a2(k)(sin(ckt)) for every k, where

a1(k), a2(k) are suitable constants. Let

u(t, x) =∞∑k=1

uk(t, x)

Nevertheless, with this definition a number of problems arise:

i) We do not know the constants a1(k), a2(k).

ii) We do not know, whether∑k∈N+

uk(t, x) converges or not.

iii) Even if∑k∈N+

uk(t, x) converge, we do not know anything about the convergence of its

derivatives.

We start by determining constants a1(k), a2(k), with the help of the initial conditions.

∞∑k=1

uk(0, x) =∞∑k=1

sin(kx)(a1(k) cos(0) + a2 sin(0)) =∞∑k=1

sin(kx)a1(k) = f(x),

and since f ∈ C2[0, π], f can be represented with sine expansion, the Fourier sine series

of f will converge, and a1(k) are the corresponding Fourier coefficients.

∂t

∞∑k=1

uk(0, x) =∞∑k=1

sin(kx)(a1(k)(−ck) sin(0)+a2(ck) cos(0)) =∞∑k=1

ck sin(kx)a2(k) = g(x),

and since g ∈ C2[0, π] and g can be written in a sine expansion, the Fourier sine series of

g will converge, and cka2(k) are the corresponding Fourier sine coefficients.

35

We proceed by showing, that∞∑k=1

uk(t, x) converges uniformly.

∣∣∣∣ ∞∑k=1

uk(t, x)

∣∣∣∣ =

∣∣∣∣ n∑k=1

sin(kx)(a1(k) cos(ckt) + a2(k)(sin(ckt))

∣∣∣∣≤

∞∑k=1

|a1(k) cos(ckt) + a2(k) sin(ckt)|

≤∞∑k=1

|a1(k)|+ |a2(k)| ≤∞∑k=1

M

k2+M

k2<∞,

with suitableM, M constants. Thus∞∑k=1

uk(t, x) = u(t, x) stands for every t ∈ R+, and x ∈

[0, π]. We only have to show, that the derivatives of u(t, x) also converge, namely:

∂t

∞∑k=1

uk(t, x), ∂x

∞∑k=1

uk(t, x), ∂2t

∞∑k=1

uk(t, x), ∂2x

∞∑k=1

uk(t, x).

However the conditions for Theorem 3.0.2 are fulfilled, thus we can differentiate term by

term.

3.2 The Heat Equation

We wish to inspect the model of heat condensation, in a long thin rod. We assume,

both ends of the rod is kept at steady temperature, the length of the rod is L and the

temperature of the rod, in a positon x after t time, is u(t, x) t ∈ R+, x ∈ [0, L], and u is

twice differentiable in x, and once in t. The physical model is the following:

∂tu(t, x) = c∂2xu(t, x)

where c is a constant, representing the conductivity of the rod. We also assume, that

u(t, 0) = T1 and u(t, L) = T2, which expresses, that the temperature does not change at

the ends of the rod. We consider the initial temperature to be known: u(0, x) = f(x), and

f ∈ C2([0, L]), since u(t, x) is twice differentiable in x. This gives us the following system:

∂tu(t, x) = c∂2xu(t, x) ∀t ∈ R+ ∀x ∈ [0, L] (3.1)

u(t, 0) = T1 ∀t ∈ R+ (3.2)

u(t, L) = T2 ∀t ∈ R+ (3.3)

u(0, x) = f(x) ∀x ∈ [0, L]. (3.4)

We start with a simple observation. Let s(x) =T2 − T1L

x+T1, which would be a solution,

if we dismissed (3.4). This solution, does not change through time, thus it is called the

36

steady state solution. Later on we will see, that every system, without external stimulation

tends to this state, as time passes by. We solve this, by considering an alternate system,

such that:

∂tu(t, x) = c∂2xu(t, x) ∀t ∈ R+ ∀x ∈ [0, L] (3.5)

u(t, 0) = 0 ∀t ∈ R+ (3.6)

u(t, L) = 0 ∀t ∈ R+ (3.7)

u(0, x) = f(x)− s(x) = h(x) ∀x ∈ [0, L]. (3.8)

We proceed just like in the case of the wave equation. We look for u(t, x) in such

form: u(t, x) = T (t)X(x). According this, we get (3.5) to be the following: T ′(t)X(x) =

cT (t)X ′′(x), and ofter a division, we get:T ′(t)

cT (t)=

X ′′(x)

X(x). Since the right side de-

pends only upon t, and the right side only upon x, we get that both side is a constant:T ′(t)

cT (t)=X ′′(x)

X(x)= η. From this, we get the following system:

T ′(t)− cηT (t) =0 (3.9)

X ′′(x)− ηX(x) =0. (3.10)

We proceed by inspecting the possible values of η. A solution of (3.9) is T (t) = a exp(cηt),

where a ∈ R is an arbitrary constant. If η > 0, limt→∞

T (t) =∞, thus we assume η ≤ 0. In

the case η = 0, T (t) is constant, therefore we consider η < 0.Exploiting this, we change

our system to the the following:

T ′(t) + cη1T (t) =0 (3.11)

X ′′(x) + η1X(x) =0, (3.12)

where η1 = −η. A solution to (3.11) is T (t) = exp(−cη1t), and to (3.12), is X(x) =

a1 sin(√η1t) + a2 cos(

√η2t). Combining this, we get the following:

u(t, x) = T (t)X(x) = exp(−cη1t)(a1 sin(√η1x) + a2 cos(

√η2x)).

Up to this point, we have a solution to (3.5), for every η1 > 0. We want to have a rich

enough set of solutions, among which the initial conditions can be satisfied with a series

expansion. Therefore let√η1 =

kπ

L. According to this, we have set of solutions in the

following form:

uk(t, x) = Tk(t)Xk(x) = exp

(− ck

2π2t

L2

)(a1 sin

(kπx

L

)+ a2 cos

(kπx

L

)).

Assuming, the rod satisfies conditions (3.6),(3.7),(3.8), we have f(0) = T1, f(L) = T2,

which is equivalent to h(0) = h(L) = 0. We wish to create a solution uinit(t, x), which

37

fulfilles condition (3.8). To do so, we look for the solution represented by its sine expansion:

uinit(t, x) =∞∑k=1

a(k) exp

(− ck

2π2t

L2

)sin

(kπx

L

). (3.13)

The problems are the same, as during the wave equation.

i) We do not know constants a(k).

ii) We do not know, whether∞∑k=1

uk(t, x) converges or not.

iii) Even if∞∑k=1

uk(t, x) converge, we do not know anything about the convergence of its

derivatives.

We start by determining constants a(k), with the help of the sine expansion: ak =2

L

∫ L

0

h(x) sin

(kπx

L

), since h ∈ C2([−π, π]), the Fourier series converge uniformly.

The convergence of∞∑k=1

uk(t, x) now stands, since:

∣∣∣∣ ∞∑k=1

uk(t, x)

∣∣∣∣ =

∣∣∣∣ ∞∑k=1

a(k) exp

(− ck

2π2t

L2

)sin

(kπx

L

)∣∣∣∣ ≤ ∞∑k=1

ak ≤M

k2<∞,

since h ∈ C2([0, π]).

The only thing left to show, is that the derivatives converge, namely:

∂t

∞∑k=1

uk(t, x), ∂x

∞∑k=1

uk(t, x), ∂2x

∞∑k=1

uk(t, x).

However the conditions of Theorem 3.0.2 are fulfilled, thus we can derivate term by term.

To get a solution to the original system, let u(t, x) = s(x) + uinit(t, x), which is a solution

indeed.

38

Bibliography

[1] Rajendra Bhatia. Fourier Series. The Mathematical Association Of America, 2005.

[2] Kunszenti-Kovacs David. Analızis eloadasok III. 2016-2017.

[3] Karatson Janos. Numerikus Funkcionalanalızis. Typotex Kiado, 2014.

[4] Kristof Janos. Analızis eloadasok I-II. 2015-2016.

[5] Robert C. Rogers Michael Renardy. An Introduction to Partial Differential Equations

Second Edition. Springer-Verlag, 2004.

[6] Tim Olson. Applied Fourier Analysis From Signal Processing to Medical Imaging.

Birkhauser, 2010.

[7] Walter Rudin. A Matematikai Analızis Alapjai reprint kiadas. Typotex Kiado, 2010.

[8] Laczkovich Miklos-T. Sos Vera. Analızis I.-II. Nemzeti Tankonyvkiado, 2006.

[9] Komornik Vilmos. Valos analızis eloadasok I.-II. Typotex Kiado, 2003.

[10] Dirk Werner. Funktionalanalysis 7., korrigierte und erweiterte Auflage. Springer-Verlag,

2011.

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