Fourier series and the discrete Fourier transform
802647S
Lecture Notes
1st Edition
Fourth printing
Valery Serov
University of Oulu
2014
Edited by Markus Harju
Contents
1 Preliminaries
1
2 Formulation of Fourier series
7
3 Fourier coefficients
11
4 Convolution and Parseval equality
16
5 Fejér means and uniqueness of Fourier series
18
6 Riemann-Lebesgue lemma
22
7 Fourier series of square-integrable function
25
8 Besov and Hölder spaces
32
9 Absolute convergence
38
10 Pointwise and uniform convergence
43
11 Discrete Fourier transform
58
12 Discrete and usual Fourier transform
65
13 Applications of discrete Fourier transform
72
Index
77
i
1
Preliminaries
Definition 1.1. A function f (x) of one variable x is said to be periodic with period
T > 0 if the domain D(f ) of f contains x + T whenever it contains x and, if for any
x ∈ D(f ) it holds that
f (x + T ) = f (x).
(1.1)
Remark. If also x − T ∈ D(f ) then
f (x − T ) = f (x).
It follows that if T is a period of f then mT is also a period for any integer m > 0.
The smallest value of T > 0 for which (1.1) holds is called the fundamental period of
f.
For example, the functions
sin
mπx
,
L
mπx
,
L
cos
ei
mπx
L
,
m = 1, 2, . . .
are periodic with fundamental period T = 2L
. Note also that they are periodic with
m
common period 2L.
If some function f is defined on the interval [a, a + T ], T > 0 and f (a) = f (a + T ),
then f can be extended periodically with period T on the whole line as
f (x) := f (x − mT ),
x ∈ [a + mT, a + (m + 1)T ],
m = 0, ±1, ±2, . . . .
Therefore, we may assume from now on that any periodic function is defined on the
whole line.
We say that f is p-integrable, 1 ≤ p < ∞, on the interval [a, b] if
Z b
|f (x)|p dx < ∞.
a
The set of all such functions is denoted by Lp (a, b). When p = 1 we say that f is
integrable. If f is p-integrable and g is p′ -integrable on [a, b], where
1
1
+ ′ = 1,
p p
1 < p < ∞, 1 < p′ < ∞
then their product is integrable on [a, b] and
Z
b
a
|f (x)g(x)|dx ≤
Z
b
p
a
|f (x)| dx
1/p Z
b
a
p′
|g(x)| dx
1/p′
.
This inequality is called the Hölder’s inequality for integrals. The Fubini’s theorem
states that
Z dZ b
Z bZ d
F (x, y)dxdy,
F (x, y)dydx =
a
c
c
1
a
where F (x, y) ∈ L1 ((a, b) × (c, d)) is positive.
P
If f1 , f2 , . . . , fn are p-integrable on [a, b] for 1 ≤ p < ∞ then so is their sum nj=1 fj
and
p !1/p
1/p
Z b X
n
n Z b
X
p
fj (x) dx
≤
.
|fj (x)| dx
a a
j=1
j=1
This inequality is called Minkowski’s inequality. As a consequence of Hölder’s inequality we obtain the generalized Minkowski’s inequality
p !1/p Z d Z b
1/p
Z b Z d
p
F (x, y)dy dx
≤
dy.
(1.2)
|F (x, y)| dx
a
c
c
a
Exercise 1. Prove (1.2).
Lemma 1.1. If f is periodic with period T > 0 and if it is integrable on any finite
interval then
Z T
Z a+T
f (x)dx
(1.3)
f (x)dx =
0
a
for any a ∈ R.
Proof. Let first a > 0. Then
Z a
Z a+T
Z a+T
f (x)dx
f (x)dx −
f (x)dx =
0
0
a
Z a+T
Z T
Z
=
f (x)dx +
f (x)dx −
0
T
a
f (x)dx .
0
The difference in the brackets is equal to zero due to periodicity of f . Thus, (1.3) holds
for a > 0.
If a < 0 then we proceed similarly as
Z a+T
Z 0
Z a+T
f (x)dx
f (x)dx +
f (x)dx =
0
a
a
Z 0
Z T
Z T
=
f (x)dx +
f (x)dx −
f (x)dx
a
0
a+T
Z 0
Z T
Z T
=
f (x)dx +
f (x)dx −
f (x)dx .
0
a
a+T
Again, the periodicity of f implies that the difference in brackets is zero. Thus lemma
is proved.
Definition 1.2. Let us assume that the domain of f is symmetric with respect to {0},
i.e. if x ∈ D(f ) then −x ∈ D(f ). A function f is called even if
f (−x) = f (x),
x ∈ D(f )
and odd if
f (−x) = −f (x),
2
x ∈ D(f ).
Lemma 1.2. If f is integrable on any finite interval and if it is even then
Z a
Z a
f (x)dx = 2
f (x)dx
−a
0
for any a > 0. Similarly, if f is odd then
Z a
f (x)dx = 0
−a
for any a > 0.
Proof. Since
Z
a
f (x)dx =
−a
Z
a
f (x)dx +
0
Z
0
f (x)dx
−a
then changing variables in the second integral we obtain
Z a
Z a
Z a
f (−x)dx.
f (x)dx +
f (x)dx =
0
0
−a
Now the result of this lemma follows from Definition 1.2.
Definition 1.3. The notations f (c ± 0) are used to denote the right and left limits
f (c ± 0) := lim f (x).
x→c±
Definition 1.4. A function f is said to be piecewise continuous on an interval [a, b] if
there are x0 , x1 , . . . , xn such that a = x0 < x1 < · · · < xn = b and
1. f is continuous on each subinterval (xj−1 , xj ), j = 1, 2, . . . , n
2. f (x0 + 0), f (xn − 0) and f (xj ± 0), j = 1, 2, . . . , n − 1 exist.
Definition 1.5. A function f is said to be of bounded variation on an interval [a, b],
denoted by BV [a, b], if there is c0 ≥ 0 such that, for any {x0 , x1 , . . . , xn } with a =
x0 < x1 < · · · < xn = b it holds that
n
X
j=1
|f (xj ) − f (xj−1 )| ≤ c0 .
The number
Vab (f )
:=
sup
x0 ,x1 ,...,xn
n
X
j=1
|f (xj ) − f (xj−1 )|
(1.4)
is called the full variation of f on the interval [a, b]. For any x ∈ [a, b] we can also
define Vax (f ) by (1.4).
Exercise 2. Prove that
3
1. Vax (f ) is monotone increasing in x
2. for any c ∈ (a, b) we have Vab (f ) = Vac (f ) + Vcb (f ).
If f is real-valued then Exercise 2 implies that Vax (f ) − f (x) is monotone increasing
in x. Indeed, for h > 0 we have that
Vax+h (f ) − f (x + h) − (Vax (f ) − f (x)) = Vax+h (f ) − Vax (f ) − (f (x + h) − f (x))
= Vxx+h (f ) − (f (x + h) − f (x))
≥ Vxx+h (f ) − |f (x + h) − f (x)| ≥ 0.
As an immediate consequence we obtain that any real-valued function f ∈ BV [a, b]
can be represented as the difference of two monotone increasing functions as
f (x) = Vax (f ) − (Vax (f ) − f (x)) .
This fact allows us to define the Stieltjes integral
Z b
g(x)df (x),
(1.5)
a
where f ∈ BV [a, b] and g is an arbitrary continuous function. The integral (1.5) is
defined as
Z b
n
X
g(x)df (x) = lim
g(ξj )(f (xj ) − f (xj−1 )),
a
∆→0
j=1
where j = 1, 2, . . . , n, a = x0 < x1 < · · · < xn = b, ξj ∈ [xj−1 , xj ] and ∆ =
max1≤j≤n (xj − xj−1 ).
Let us introduce the modulus of continuity of f by
ωh (f ) :=
sup
{x∈[a,b]:x+h∈[a,b]}
|f (x + h) − f (x)|,
h > 0.
(1.6)
Definition 1.6. A function f belongs to Hölder space C α [a, b], 0 < α ≤ 1, if
ωh (f ) ≤ Chα
with some constant C > 0. This inequality is called the Hölder condition with exponent
α.
Definition 1.7. We say that f belongs to Sobolev space Wp1 (a, b), 1 ≤ p < ∞ if
f ∈ Lp (a, b) and there is g ∈ Lp (a, b) such that
Z x
g(t)dt + C
(1.7)
f (x) =
a
with some constant C.
4
Lemma 1.3. Suppose that f ∈ Wp1 (a, b), 1 ≤ p < ∞. Then f is of bounded variation.
Moreover, if p = 1 then f is also continuous and if 1 < p < ∞ then f ∈ C 1−1/p [a, b].
Proof. Let first p = 1. Then there is an integrable g such that (1.7) holds with some
constant C. Hence for fixed x ∈ [a, b] with x + h ∈ [a, b] we have
Z
f (x + h) − f (x) =
It follows that
Z
|f (x + h) − f (x)| = x+h
x
x+h
g(t)dt.
x
g(t)dt → 0,
h→0
since g is integrable. This proves the continuity of f . At the same time for any
{x0 , x1 , . . . , xn } such that a = x0 < x1 < · · · < xn = b we have
Z b
n
n Z xj
n Z xj
X
X
X
|f (xj ) − f (xj−1 )| =
g(t)dt ≤
|g(t)|dt =
|g(t)|dt.
xj−1
a
xj−1
j=1
j=1
j=1
Rb
Hence, Definition 1.5 is satisfied with constant c0 = a |g(t)|dt and f is of bounded
variation.
If 1 < p < ∞ then using Hölder’s inequality for integrals we obtain for h > 0 that
Z
|f (x + h) − f (x)| ≤
x+h
|g(t)|dt ≤
x
≤ h
1−1/p
Z
b
Z
x+h
dt
x
p
a
|g(t)| dt
1/p
1/p′ Z
x+h
p
x
|g(t)| dt
1/p
,
where 1/p + 1/p′ = 1. By Definition 1.6 it means that f ∈ C 1−1/p [a, b]. Lemma is
proved.
Remark. Since any f ∈ Wp1 (a, b), 1 ≤ p < ∞ is continuous then the constant C in (1.7)
is equal to f (a).
Definition 1.8. Two functions u and v are said to be orthogonal on [a, b] if the product
uv is integrable and
Z
b
u(x)v(x)dx = 0.
a
A set of functions is said to be mutually orthogonal if each distinct pair in the set is
orthogonal on [a, b].
Lemma 1.4. The functions
1,
sin
mπx
,
L
cos
mπx
,
L
5
m = 1, 2, . . .
form a mutually orthogonal set on the interval [−L, L] as well as on the interval [0, 2L].
In fact,
(
Z L
Z 2L
0, m 6= n
mπx
mπx
nπx
nπx
cos
,
(1.8)
cos
cos
dx =
cos
dx =
L
L
L
L
L, m = n
−L
0
Z
L
Z
2L
mπx
nπx
sin
dx = 0,
L
L
0
(
Z L
Z 2L
0, m 6= n
mπx
nπx
mπx
nπx
sin
sin
dx =
sin
sin
dx =
L
L
L
L
L, m = n
−L
0
and
Z L
nπx
mπx
sin
dx =
cos
L
L
−L
mπx
dx =
sin
L
−L
Z
2L
0
mπx
sin
dx =
L
Z
cos
L
mπx
cos
dx =
L
−L
Z
2L
cos
0
(1.9)
(1.10)
mπx
dx = 0. (1.11)
L
Proof. Due to Lemma 1.1 it is enough to prove the equalities (1.8), (1.9), (1.10) and
(1.11) only for integrals over [−L, L]. Let us derive, for example, (1.9). Using the
equality
1
cos α sin β = (sin(α + β) − sin(α − β))
2
we have for m 6= n that
Z
Z
Z L
nπx
1 L
(m + n)πx
1 L
(m − n)πx
mπx
sin
dx =
sin
dx −
sin
dx
cos
L
L
2 −L
L
2 −L
L
−L
!
!
(m+n)πx L
(m−n)πx L
−
cos
1
1 − cos
L
L
=
−
=0
(m+n)π
(m−n)π
2
2
L
−L
L
−L
since cosine is even. If m = n we have
Z
Z L
nπx
1 L
2mπx
mπx
sin
dx =
dx = 0
sin
cos
L
L
2 −L
L
−L
since sine is odd. Other identities can be proved in a similar manner and are left to
the readers. Lemma is proved.
Remark. This lemma holds also for the functions ei
nπx
L
, n = 0, ±1, ±2, . . . in the form
(
Z L
Z 2L
nπx
0,
n 6= m
nπx
mπx
mπx
ei L e−i L dx =
ei L e−i L dx =
2L, n = m.
−L
0
6
2
Formulation of Fourier series
Let us consider a series of the form
∞
a0 X mπx mπx
+
+ bm sin
am cos
.
2
L
L
m=1
(2.1)
This series consists of 2L-periodic functions. Thus, if the series (2.1) converges for all
x, then the function to which it converges will also be 2L-periodic. Let us denote this
limiting function by f (x) i.e.
a0 X mπx mπx
+
+ bm sin
am cos
.
2
L
L
m=1
∞
f (x) :=
(2.2)
To determine am and bm we proceed as follows: assuming that the
P integration can be
legitimately carried out term by term (it will be, for example, if ∞
m=1 (|am | + |bm |) <
∞) we obtain
Z L
Z
Z L
∞
X
nπx
a0 L
nπx
nπx
mπx
f (x) cos
dx =
dx +
cos
dx
cos
am
cos
L
2 −L
L
L
L
−L
−L
m=1
Z L
∞
X
nπx
mπx
+
cos
dx
bm
sin
L
L
−L
m=1
for each fixed n = 1, 2, . . . . It follows from the orthogonality relations (1.8), (1.9) and
(1.11) that the only nonzero term on the right hand side is the one for which m = n
in the first summation. Hence
Z
1 L
nπx
an =
f (x) cos
dx, n = 1, 2, . . . .
(2.3)
L −L
L
A similar expression for bn is obtained by multiplying (2.2) by sin nπx
and integrating
L
termwise from −L to L. The result is
Z
nπx
1 L
dx, n = 1, 2, . . . .
(2.4)
f (x) sin
bn =
L −L
L
Using (1.11) we can easily obtain that
1
a0 =
L
Z
L
f (x)dx.
(2.5)
−L
Definition 2.1. Let f be integrable (not necessarily periodic) on the interval [−L, L].
The Fourier series of f is the trigonometric series (2.1), where the coefficients a0 , am
and bm are given by (2.5), (2.3) and (2.4), respectively. In that case we write
mπx a0 X mπx
+
+ bm sin
f (x) ∼
am cos
.
2
L
L
m=1
∞
7
(2.6)
Remark. This definition does not imply that the series (2.6) converges to f or that f
is periodic.
Definition 2.1 and Lemma 1.2 imply that if f is even on [−L, L] then the Fourier
series of f has the form
∞
a0 X
mπx
f (x) ∼
+
(2.7)
am cos
2
L
m=1
and if f is odd then
f (x) ∼
∞
X
bm sin
m=1
mπx
.
L
(2.8)
The series (2.7) and (2.8) are called the Fourier cosine series and Fouries sine series,
respectively.
If L = π then the Fourier series (2.6) ((2.7) and (2.8)) transforms to
∞
f (x) ∼
a0 X
+
(am cos mx + bm sin mx) ,
2
m=1
(2.9)
where the coefficients a0 , am and bm are given by (2.3), (2.4) and (2.5) with L = π.
There are different approaches if function f is defined on a nonsymmetric interval
[0, L] with an arbitrary L > 0.
1. Even extension. Define a function g(x) on the interval [−L, L] as
(
f (x),
0≤x≤L
g(x) =
f (−x), −L ≤ x < 0.
Then g(x) is even and its Fourier (cosine) series (2.7) represents f on [0, L].
2. Odd extension. Define a function h(x) on the interval [−L, L] as
(
f (x),
0≤x≤L
h(x) =
−f (−x), −L ≤ x < 0.
Then h(x) is odd and its Fourier (sine) series (2.8) represents f on [0, L].
3. Define a function fe(t) on the interval [−π, π] as the superposition
L
tL
+
.
fe(t) = f
2π
2
If f (0) = f (L) then we may extend f to be periodic with period L. Then
Z
Z
Z
1 π e
1 2π L
1 π
tL L
e
a0 ( f ) =
dt =
+
f (t)dt =
f
f (x) dx
π −π
π −π
2π
2
π L 0
Z
2 L
f (x) dx := a0 (f ),
=
L 0
8
and
2
am (fe) = (−1)
L
Hence,
2
bm (fe) = (−1)m
L
m
Z
L
f (x) cos
2mπx
dx = (−1)m am (f ),
L
f (x) sin
2mπx
dx = (−1)m bm (f ).
L
0
Z
L
0
∞
a0 X
fe(t) ∼
+
(am cos mt + bm sin mt)
2
m=1
and, at the same time,
∞
a0 X
2mπx
2mπx
m
f (x) ∼
+
+ bm sin
am cos
,
(−1)
2
L
L
m=1
where a0 , am and bm are the same and x =
tL L
+ .
2π
2
These three alternatives allow us to consider (for simplicity) only the case of symmetric
interval [−π, π] such that the Fourier series will be of the form (2.9) i.e.
∞
a0 X
f (x) ∼
+
(am cos mx + bm sin mx) .
2
m=1
Using Euler’s formula we will rewrite this series in the complex form
f (x) ∼
∞
X
cn einx ,
(2.10)
n=−∞
where the coefficients cn = cn (f ) are equal to

an b n


+ ,
n = 1, 2, . . .


 a2
2i
0
,
n=0
cn (f ) =
2



a
b

 −n − −n , n = −1, −2, . . . .
2
2i
The formulas (2.3), (2.4), (2.5) and (2.11) imply that
Z π
1
f (x)e−inx dx
cn (f ) =
2π −π
(2.11)
(2.12)
for n = 0, ±1, ±2, . . .. We call cn (f ) the nth Fourier coefficient of f . It can be checked
that
cn (f ) = c−n (f ).
(2.13)
9
Exercise 3. Prove formulas (2.10), (2.11), (2.12) and (2.13).
Exercise 4. Find the


−1,
a) sgn(x) = 0,


1,
Fourier series of
−π ≤ x < 0
x=0
0 < x ≤ π.
b) |x|, −1 ≤ x ≤ 1.
c) x, −1 ≤ x ≤ 1.
(
0, −L ≤ x ≤ 0
d) f (x) =
L, 0 < x ≤ L.
Exercise 5. Suppose that
f (x) =
(
1 − x, 0 ≤ x ≤ 1
0,
1 < x ≤ 2.
Find the Fourier cosine and sine series of f (x).
Exercise 6. Show that if N is odd then sinN x can be written as a finite sum of the
form
N
X
ak sin kx.
k=1
It means that this finite sum is the Fourier series of sinN x and the coefficients ak
(which are real) are the Fourier coefficients of sinN x.
Exercise 7. Show that if N is odd then cosN x can be written as a finite sum of the
form
N
X
ak cos kx.
k=1
10
3
Fourier coefficients and their properties
Definition 3.1. We say that a trigonometric series
∞
X
cn einx
n=−∞
a) converges pointwise if for each x ∈ [−π, π] the limit
X
lim
cn einx
N →∞
|n|≤N
exists,
b) converges uniformly in x ∈ [−π, π] if the limit
X
lim
cn einx
N →∞
|n|≤N
exists uniformly,
c) converges absolutely if the limit
X
lim
N →∞
exists or, equivalently, if
∞
X
n=−∞
|n|≤N
|cn |
|cn | < ∞.
These three different types of convergence appear frequently in the sequel and
they are presented above from the weakest to strongest. In other words, absolute
convergence implies uniform convergence which in turn implies pointwise convergence.
If f is integrable on the interval [−π, π] then the Fourier coefficients cn (f ) are
uniformly bounded with respect to n = 0, ±1, ±2, . . . i.e.
Z
Z π
1 π
1
−inx
|cn (f )| =
|f (x)|dx,
(3.1)
f (x)e
dx ≤
2π −π
2π −π
where the upper bound does not depend on n. Suppose that a sequence {cn }∞
n=−∞ is
such that
∞
X
|cn | < ∞.
n=−∞
Then the series
∞
X
cn einx
n=−∞
11
converges uniformly in x ∈ [−π, π] and defines a continuous function
f (x) :=
∞
X
cn einx
(3.2)
n=−∞
∞
whose Fourier coefficients are {cn }∞
n=−∞ = {cn (f )}n=−∞ . More generally, suppose that
∞
X
n=−∞
|n|k |cn | < ∞
for some integer k > 0. Then the series (3.2) defines a function which is k times
differentiable with
∞
X
f (k) (x) =
(in)k cn einx
(3.3)
n=−∞
being a continuous function. This follows from the fact that the series (3.3) converges
uniformly with respect to x ∈ [−π, π].
Let us consider one more useful example where the Fourier coefficients are applied.
If 0 ≤ r < 1 then the geometric progression series gives
∞
X
1
=
rn einx
1 − reix
n=0
(3.4)
and this series converges absolutely. Using the definition of the Fourier coefficients we
obtain
Z π −inx
e
1
n
dx, n = 0, 1, 2, . . .
r =
2π −π 1 − reix
and
1
0=
2π
Z
π
−π
einx
dx,
1 − reix
n = 1, 2, . . . .
From the representation (3.4) we may conclude also that
1 − r cos x
= Re
1 − 2r cos x + r2
1
1 − reix
∞
1 1 X |n| inx
=
r e
r cos nx = +
2 2 n=−∞
n=0
∞
X
n
(3.5)
and
r sin x
= Im
1 − 2r cos x + r2
1
1 − reix
∞
i X |n|
=
r sgn(n)einx . (3.6)
r sin nx = −
2
n=−∞
n=1
∞
X
n
Exercise 8. Verify the formulas (3.5) and (3.6).
12
The formulas (3.5) and (3.6) can be rewritten as
∞
X
r|n| einx =
n=−∞
and
−i
∞
X
1 − r2
:= Pr (x)
1 − 2r cos x + r2
sgn(n)r|n| einx =
n=−∞
2r sin x
:= Qr (x).
1 − 2r cos x + r2
(3.7)
(3.8)
Definition 3.2. The function Pr (x) is called the Poisson kernel while Qr (x) is called
the conjugate Poisson kernel .
Since the series (3.7) and (3.8) converge absolutely we have
Z π
Z π
1
1
|n|
−inx
|n|
r =
Pr (x)e
dx and − i sgn(n)r =
Qr (x)e−inx dx,
2π −π
2π −π
where n = 0, ±1, ±2, . . ..
Exercise 9. Prove that both Pr (x) and Qr (x) are solutions of the Laplace equation
ux 1 x 1 + ux 2 x 2 = 0
in the disk x21 + x22 < 1, where x1 + ix2 = reix with 0 ≤ r < 1 and x ∈ [−π, π].
Given a sequence an , n = 0, ±1, ±2, . . . define
∆n a = an − an−1 .
Then for any two sequences an and bn and for any integers M < N the formula
N
X
k=M +1
ak ∆ k b = aN b N − aM b M −
N
X
bk−1 ∆k a
(3.9)
k=M +1
holds. The formula (3.9) is called summation by parts.
Exercise 10. Prove (3.9).
Summation by parts allows us to investigate the convergence of special type of
trigonometric series.
Theorem 1. Suppose that cn > 0, n = 0, 1, 2, . . ., cn ≥ cn+1 and limn→∞ cn = 0. Then
the trigonometric series
∞
X
cn einx
(3.10)
n=0
converges for any x ∈ [−π, π] \ {0}.
13
Proof. Let bn =
Pn
k=0
eikx . Since
bn =
1 − eix(n+1)
,
1 − eix
then
|bn | ≤
x 6= 0
2
1
=
ix
|1 − e |
| sin x2 |
for x ∈ [−π, π] \ {0}. Applying (3.9) shows that for M < N we have
!
k
N
k−1
N
N
X
X
X
X
X
ilx
ikx
ilx
e −
ck e
=
e
ck
=
ck ∆ k b
k=M +1
l=0
k=M +1
= c N bN − c M bM −
l=0
N
X
k=M +1
bk−1 ∆k c.
k=M +1
Thus
N
N
X
X
ikx ≤
c
|b
|
+
c
|b
|
+
c
e
|bk−1 |∆k c
N N
M M
k
k=M +1
k=M +1
1
≤
| sin x2 |
=
cN + cM +
N
X
k=M +1
|ck − ck−1 |
!
2cM
1
→0
x (cN + cM + cM − cN ) =
| sin 2 |
| sin x2 |
as N > M → ∞. This proves the theorem.
Corollary. Under the same assumptions as in Theorem 1, the trigonometric series
(3.10) converges uniformly for all π ≥ |x| ≥ δ > 0.
Proof. If π ≥ |x| ≥ δ > 0 then sin x2 ≥ π2 |x|
≥ πδ .
2
Theorem 1 implies that, for example, the series
∞
X
n=1
cos nx
log(2 + n)
and
∞
X
n=1
converge for all x ∈ [−π, π] \ {0}.
Modulus of continuity and tail sum
For the trigonometric series (2.10) with
∞
X
n=−∞
|cn | < ∞
14
sin nx
log(2 + n)
we introduce the tail sum by
En :=
X
|k|>n
|ck |,
n = 0, 1, 2, . . . .
(3.11)
There is a good connection between the modulus of continuity (1.6) and (3.11). Indeed,
if f (x) denotes the series (2.10) and h > 0 we have
|f (x + h) − f (x)| ≤
∞
X
n=−∞
≤ h
|cn ||einh − 1| =
X
|n|≤[1/h]
|n||cn | + 2
X
|n|h≤1
X
|n|>[1/h]
|cn ||einh − 1| +
X
|n|h>1
|cn ||einh − 1|
|cn | := I1 + I2 ,
where [x] denotes the entire part of x. If we denote [1/h] by Nh then I2 = 2ENh and
I1 = h
X
|n|≤Nh
|n| E|n|−1 − E|n| = −h
Nh
X
l=1
l (El − El−1 ) .
Using (3.9) in the latter sum we obtain
I1 = −h Nh ENh − 0 · E0 −
Since hNh = h[1/h] ≤ h ·
1
h
Nh
X
n=1
En−1 (n − (n − 1))
!
= −hNh ENh + h
Nh
X
En−1 .
n=1
= 1 then these formulas for I1 and I2 imply that
ωh (f ) ≤ 2ENh − hNh ENh + h
Nh
X
n=1
En−1 ≤ 2ENh
Nh −1
1 X
+
En .
Nh n=0
(3.12)
Since En → 0 as n → ∞ then the inequality (3.12) implies that ωh (f ) → 0 as h → 0.
Moreover, if En = O(n−α ) for some 0 < α ≤ 1 then
(
O(hα ),
0<α<1
ωh (f ) =
(3.13)
1
O(h log h ), α = 1.
Here and throughout, the notation A = O(B) on a set X means that |A| ≤ C|B| on
the set X with some constant C > 0. Similarly, A = o(B) means that A/B → 0.
Exercise 11. Prove the second relation in (3.13).
We summarize (3.13) as follows: if the tail of the trigonometric series (2.10) behaves
as O(n−α ) for some 0 < α < 1 then the function f to which it converges belongs to
Hölder space C α [−π, π].
15
4
Convolution and Parseval equality
Let the trigonometric series (2.10) be such that
∞
X
n=−∞
|cn | < ∞.
Then the function f to which it converges is continuous and periodic. If g(x) is any
continuous function then the product f g is also continuous and hence integrable on
[−π, π] and
1
2π
Z
Z π
∞
∞
X
1 X
inx
f (x)g(x)dx =
cn (f )
g(x)e dx =
cn (f )c−n (g),
2π n=−∞
−π
−π
n=−∞
π
(4.1)
where integration of the series term by term is justified by the uniform convergence of
Fourier series. Putting g = f in (4.1) yields
1
2π
Z
π
−π
|f (x)|2 dx =
∞
X
cn (f )c−n (f ) =
n=−∞
∞
X
n=−∞
|cn (f )|2
(4.2)
by (2.13).
Definition 4.1. Equality (4.2) is called the Parseval equality for the trigonometric
Fourier series.
The formula (4.1) can be generalized as follows.
Exercise 12. Let periodic f be defined by absolutely convergent Fourier series (2.10)
and let g be integrable and periodic. Prove that
Z π
∞
X
1
f (x)g(y − x)dx =
cn (f )cn (g)einy
2π −π
n=−∞
and that this series converges absolutely.
The generalization of Exercise 12 is given by the following theorem.
Theorem 2. If f1 and f2 are two periodic L1 -functions then
cn (f1 )cn (f2 ) = cn (f1 ∗ f2 ),
where f1 ∗ f2 denotes the convolution
1
(f1 ∗ f2 )(x) =
2π
Z
π
−π
f1 (y)f2 (x − y)dy
and where the integral converges for almost every x.
16
(4.3)
Proof. We note first that the convolution (4.3) is well defined as an L1 -function by the
Fubini theorem. Indeed,
Z π−y
Z π Z π
Z π
|f1 (y)| · |f2 (x − y)|dy dx =
|f1 (y)|
|f2 (z)|dz dy
−π
−π
−π
−π−y
Z π
Z π
|f1 (y)|
=
|f2 (z)|dz dy
−π
−π
by Lemma 1.1. The Fourier coefficients of the convolution (4.3) are equal to
Z π
Z π Z π
1
1
−inx
cn (f1 ∗ f2 ) =
(f1 ∗ f2 )(x)e
dx =
f1 (y)f2 (x − y)dy e−inx dx
2π −π
(2π)2 −π
−π
Z π
Z π
1
−inx
f1 (y)
f2 (x − y)e
dx dy
=
(2π)2 −π
−π
Z π−y
Z π
1
−in(y+z)
f1 (y)
=
f2 (z)e
dz dy
(2π)2 −π
−π−y
Z π
Z π
1
−iny
−inz
=
f1 (y)e
f2 (z)e
dz dy = cn (f1 )cn (f2 )
(2π)2 −π
−π
by Lemma 1.1. Thus, theorem is proved.
Exercise 13. Prove that if f1 and f2 are integrable and periodic then their convolution
is symmetric and periodic.
Exercise 14. Let f be a periodic L1 -function. Prove that
(f ∗ Pr )(x) = (Pr ∗ f )(x) =
∞
X
r|n| cn (f )einx
n=−∞
and that Pr ∗ f satisfies the Laplace equation i.e.
(Pr ∗ f )x1 x1 + (Pr ∗ f )x2 x2 = 0,
where x21 + x22 = r2 < 1 with x1 + ix2 = reix .
Remark. We are going to prove in Chapter 10 that for any periodic and continuous
function f the limit
lim (f ∗ Pr )(x) = f (x)
r→1−
exists uniformly in x.
17
5
Fejér means of Fourier series. Uniqueness of the
Fourier series.
Let us denote the partial sum of Fourier series of f ∈ L1 (−π, π) by
X
SN (f ) :=
cn (f )einx
|n|≤N
for each N = 0, 1, 2, . . .. The Fejér means are defined by
S0 (f ) + · · · + SN (f )
.
N +1
σN (f ) :=
Writing this out in detail we see that
N X
X
(N + 1)σN (f ) =
ck (f )e
ikx
=
n=0 |k|≤n
X
=
|k|≤N
N
X X
ck (f )eikx
|k|≤N n=|k|
(N + 1 − |k|)ck (f )eikx
which gives the useful representation
σN (f ) =
X |k|≤N
The Fejér kernel is
KN (x) :=
|k|
1−
N +1
X |k|≤N
ck (f )eikx .
|k|
1−
N +1
eikx .
(5.1)
(5.2)
The sum (5.2) can be calculated precisely as
1
KN (x) =
N +1
sin N2+1 x
sin x2
!2
.
(5.3)
Exercise 15. Prove the identity (5.3).
Exercise 16. Prove that
1
2π
Z
π
KN (x)dx = 1.
(5.4)
−π
We can rewrite σN (f ) from (5.1) also as
σN (f ) =
∞
X
k=−∞
1[−N,N ] (k) 1 −
18
|k|
N +1
ck (f )eikx ,
where
1[−N,N ] (k) =
(
1, |k| ≤ N
0, |k| > N.
Let us assume now that f is periodic. Then Exercise 12 and Theorem 2 lead to
Z π
∞
X
1
|k|
ck (f )eikx .
(5.5)
f (y)KN (x − y)dy =
1[−N,N ] (k) 1 −
2π −π
N
+
1
k=−∞
Exercise 17. Prove (5.5).
Hence, the Fejér means can be represented as
σN (f )(x) = (f ∗ KN )(x) = (KN ∗ f )(x).
(5.6)
The properties (5.3), (5.4) and (5.6) allow us to prove
Theorem 3. Let f ∈ Lp (−π, π) be periodic with 1 ≤ p < ∞. Then
Z π
1/p
p
lim
|σN (f )(x) − f (x)| dx
= 0.
N →∞
(5.7)
−π
If, in addition, f has right and left limits f (x0 ± 0) at a point x0 ∈ [−π, π] then
1
(5.8)
lim σN (f )(x0 ) = (f (x0 + 0) + f (x0 − 0)) .
N →∞
2
Proof. Let us first prove (5.7). Indeed, (5.4) and (5.6) give
Z π
1/p Z π
1/p
p
p
|σN (f )(x) − f (x)| dx
=
|(f ∗ KN )(x) − f (x)| dx
−π
=
Z
−π
p 1/p
Z π
Z π
1
1
KN (y)f (x − y)dy −
KN (y)f (x)dy dx
2π
2π −π
−π
−π
p 1/p
Z π Z π
1
dx
=
K
(y)(f
(x
−
y)
−
f
(x))dy
N
2π
−π
−π
p 1/p
Z π Z
1
dx
K
(y)(f
(x
−
y)
−
f
(x))dy
≤
N
2π
|y|<δ
−π
p 1/p
Z π Z
1
dx
+
K
(y)(f
(x
−
y)
−
f
(x))dy
:= I1 + I2 .
N
2π
−π
π≥|y|>δ
π
Using the generalized Minkowski’s inequality we obtain that
Z π
1/p
Z
1
p
KN (y)
|f (x − y) − f (x)| dx
dy
I1 ≤
2π |y|<δ
−π
Z π
1/p
Z π
1
p
≤ sup
KN (y)dy
|f (x − y) − f (x)| dx
2π −π
|y|<δ
−π
Z π
1/p
p
= sup
|f (x − y) − f (x)| dx
→0
|y|<δ
−π
19
(5.9)
as δ → 0 since f ∈ Lp (−π, π). Quite similarly,
I2
Z π
1/p
Z
1
p
≤
KN (y)
|f (x − y) − f (x)| dx
dy
2π π≥|y|>δ
−π
Z π
1/p
Z
1
p
KN (y)dy
≤ 2
|f (x)| dx
2π π≥|y|>δ
−π
(5.10)
since f is periodic. The next step is to note that
2 2
δ
2 |y|
2 y
sin
≥
≥
2
π 2
π
for π ≥ |y| > δ. That’s why (5.3) leads to
KN (y) ≤
and
1
2π
Z
π≥|y|>δ
KN (y)dy ≤
1
1
·
N + 1 sin2
y
2
≤
π2
1
· 2
N +1 δ
π2
π2
π 2 2(π − δ)
√
→0
·
<
<
2πδ 2 N + 1
δ 2 (N + 1)
N
as N → ∞ if we choose δ = N −1/4 . From (5.9) and (5.10) we may conclude that (5.7)
is proved. In order to prove (5.8) we use (5.4) to represent the difference
1
σN (f )(x0 ) − (f (x0 + 0) + f (x0 − 0))
Z π 2 1
1
=
KN (y) f (x0 − y) − (f (x0 + 0) + f (x0 − 0)) dy
2π −π
2
Z π
1
=
KN (y)g(y)dy,
2π −π
(5.11)
where
1
(f (x0 + 0) + f (x0 − 0)) .
2
Since Fejér kernel KN (y) is even we can rewrite the right hand side of (5.11) as
Z π
1
KN (y)h(y)dy,
2π 0
g(y) = f (x0 − y) −
where h(y) = g(y) + g(−y). It is clear that h(y) is L1 -function. But we have more,
namely,
lim h(y) = 0.
(5.12)
y→0+
Our task now is to prove that
1
lim
N →∞ 2π
Z
π
KN (y)h(y)dy = 0.
0
20
We will proceed as in the proof of (5.7) i.e. we split
1
2π
Z
π
0
1
KN (y)h(y)dy =
2π
Z
δ
1
KN (y)h(y)dy +
2π
0
Z
π
KN (y)h(y)dy := I1 + I2 .
δ
The first term can be estimated as
1
|I1 | ≤
sup |h(y)|
2π 0≤y≤δ
Z
π
KN (y)dy =
0
1
sup |h(y)| → 0
2 |y|≤δ
as δ → 0+ due to (5.12). For I2 we have
Z π
1 π 2 1
|I2 | ≤
|h(y)|dy → 0
2π δ N + 1 0
as N → ∞ if we choose, for example, δ = N −1/4 . Thus, theorem is completely
proved.
Corollary 1. If f is periodic and continuous on the interval [−π, π] then
lim σN (f )(x) = f (x)
N →∞
uniformly in x ∈ [−π, π].
p
Corollary 2. Any periodic
1 ≤ p < ∞, can be approximated by the
P L -function,
ikx
trigonometric polynomials |k|≤N bk e (which are infinitely many times differentiable
functions).
Theorem 4 (Uniqueness of Fourier series). If periodic f ∈ L1 (−π, π) has Fourier
coefficients identically zero then f = 0 almost everywhere.
Proof. Since
lim
N →∞
Z
π
−π
|σN (f )(x) − f (x)| dx = 0
by (5.7) then in the case when all Fourier coefficients are zero, we have
Z π
|f (x)| dx = 0.
−π
It means that f = 0 almost everywhere. This proves the theorem.
21
6
Riemann-Lebesgue lemma
Theorem 5 (Riemann-Lebesgue lemma). If f is periodic with period 2π and belongs
to L1 (−π, π) then
Z π
lim
f (x + z)e−inz dz = 0
(6.1)
n→∞
−π
uniformly in x ∈ R. In particular, cn (f ) → 0 as n → ∞.
Proof. Since f is periodic with period 2π then
Z π+x
Z π
Z
−in(y−x)
inx
−inz
f (y)e
dy = e
f (x + z)e
dz =
−π+x
−π
π
f (y)e−iny dy
(6.2)
−π
by Lemma 1.1. Formula (6.2) shows that to prove (6.1) it is enough to show that the
Fourier coefficients cn (f ) tend to zero as n → ∞. Indeed,
2πcn (f ) =
Z
π
f (y)e
−iny
dy =
−π
Z
π+π/n
f (y)e
−iny
dy =
−π+π/n
Z
π
f (t + π/n)e−int e−iπ dt
−π
by Lemma 1.1. Hence
−4πcn (f ) =
Z
π
−π
(f (t + π/n) − f (t)) e−int dt.
(6.3)
If f is continuous on the interval [−π, π] then
sup |f (t + π/n) − f (t)| → 0,
t∈[−π,π]
n → ∞.
Hence cn (f ) → 0 as n → ∞. In case f is an arbitrary L1 -function we let ε > 0. Then
we can define a continuous function g (see Corollary 2 of Theorem 3) such that
Z π
|f (x) − g(x)| dx < ε.
−π
Write
cn (f ) = cn (g) + cn (f − g).
The first term tends to zero as n → ∞ since g is continuous, whereas the second term
is less than ε/(2π). It implies that
sup lim |cn (f )| ≤
n→∞
ε
.
2π
Since ε is arbitrary then
lim |cn (f )| = 0.
n→∞
This fact together with (6.2) gives (6.1). Theorem is thus proved.
22
Corollary. Let f be as in Theorem 5. If periodic g is continuous on [−π, π] then
Z π
lim
f (x + z)g(z)e−inz dz = 0
n→∞
−π
and
lim
n→∞
Z
π
f (x + z)g(z) sin(nz)dz = lim
n→∞
−π
Z
π
f (x + z)g(z) cos(nz)dz = 0
−π
uniformly in x ∈ [−π, π].
Exercise 18. Prove this Corollary.
Exercise 19. Show that if f satisfies the Hölder condition with exponent α ∈ (0, 1]
then cn (f ) = O(|n|−α ) as n → ∞.
Exercise 20. Suppose that f satisfies the Hölder condition with exponent α > 1.
Prove that f ≡ constant.
Exercise 21. Let f (x) = |x|α , where −π ≤ x ≤ π and 0 < α < 1. Prove that
cn (f ) ≍ |n|−1−α as n → ∞.
Remark. The notation a ≍ b means that there exist 0 < c1 < c2 such that
c1 |a| < |b| < c2 |a|.
Let us introduce for any 1 ≤ p < ∞ and for any periodic f ∈ Lp (−π, π) the
Lp -modulus of continuity of f by
ωp,δ (f ) := sup
|h|≤δ
Z
π
−π
p
|f (x + h) − f (x)| dx
1/p
.
The equality (6.3) leads to
Z π
1
|f (x + π/n) − f (x)| dx
|cn (f )| ≤
4π −π
Z π
1/p
(2π)1−1/p
1
p
≤
|f (x + π/n) − f (x)| dx
≤ (2π)−1/p ωp,π/n (f ),
4π
2
−π
where we have used Hölder’s inequality in the penultimate step.
Exercise 22. Suppose that ωp,δ (f ) ≤ Cδ α for some C > 0 and α > 1. Prove that f is
constant almost everywhere.
Hint. First show that ωp,2δ (f ) ≤ 2ωp,δ (f ), then iterate this to obtain a contradiction.
23
Suppose that f ∈ L1 (−π, π) but not necessarily periodic. We can consider Fourier
series corresponding to f i.e.
f (x) ∼
∞
X
cn einx ,
n=−∞
where cn are the Fourier coefficients cn (f ). The series in the right hand side is considered formally in the sense that we know nothing about its convergence. However, the
limit
Z π X
Z π
inx
lim
cn (f )e dx =
f (x)dx
(6.4)
N →∞
exists. Indeed,
Z π X
cn (f )e
inx
−π |n|≤N
dx = c0 (f )
−π |n|≤N
=
Z
−π
Z
π
dx +
−π
X
0<|n|≤N
π
cn (f )
Z
π
einx dx = 2πc0 (f )
−π
f (x)dx.
−π
The existence of the limit (6.4) shows us that we can always integrate the Fourier series
of an L1 -function term by term.
24
7
Fourier series of square-integrable function. RieszFischer theorem.
The set of square-integrable functions L2 (−π, π) is a linear Euclidean space equipped
with inner product
Z π
1
(f, g)L2 (−π,π) =
f (x)g(x)dx.
(7.1)
2π −π
We can measure the degree of approximation by the mean square distance
Z π
1
|f (x) − g(x)|2 dx = (f − g, f − g)L2 (−π,π) .
2π −π
P
In particular, if g(x) = |n|≤N bn einx is a trigonometric polynomial then
Z π
Z π
Z π
1
1
1
2
2
|f (x) − g(x)| dx =
|f (x)| dx +
|g(x)|2 dx
2π −π
2π −π
2π −π
Z π
1
2Re
f (x)g(x)dx
−
2π
−π
Z π
Z π X
X
1
1
2
=
bk e−ikx dx
|f (x)| dx +
bn einx
2π −π
2π −π
|n|≤N
|k|≤N
Z π
X
1
f (x)
2Re
bn e−inx dx
−
2π
−π
|n|≤N
Z π
Z π
1 X
1
2
2
|f (x)| dx +
|bn |
dx
=
2π −π
2π
−π
|n|≤N
X
− 2Re
bn cn (f )
|n|≤N
π
Z
X
X
1
bn cn (f )
|f (x)|2 dx +
|bn |2 − 2Re
2π −π
|n|≤N
|n|≤N
X
X
+
|cn (f )|2 −
|cn (f )|2
=
|n|≤N
=
1
2π
Z
|n|≤N
π
−π
|f (x)|2 dx −
X
|n|≤N
|cn (f )|2 +
X
|n|≤N
|bn − cn (f )|2 .
This equality has the following consequences:
1. The minimum error is
Z π
Z π
X
1
1
2
2
|f
(x)−g(x)|
dx
=
|f
(x)|
dx−
|cn (f )|2 (7.2)
min
P
2π −π
g(x)= |n|≤N bn einx 2π −π
|n|≤N
and it is attained when bn = cn (f ).
25
2. For N = 1, 2, . . . it is true that
Z π
X
1
2
|cn (f )| ≤
|f (x)|2 dx
2π −π
|n|≤N
and, in particular,
∞
X
1
|cn (f )| ≤
2π
n=−∞
2
Z
π
−π
|f (x)|2 dx.
(7.3)
This inequality is called Bessel’s inequality.
It turns out that (7.3) holds with equality sign. This is Parseval equality for f ∈
L2 (−π, π) which we state as
Theorem 6. For any periodic f ∈ L2 (−π, π) with period 2π its Fourier series converges in L2 (−π, π) i.e.
2
Z π
X
1
inx
f (x) −
lim
cn (f )e dx = 0
(7.4)
N →∞ 2π −π |n|≤N
and the Parseval equality
1
2π
Z
π
2
−π
|f (x)| dx =
∞
X
n=−∞
|cn (f )|2
(7.5)
holds.
Proof. By Bessel’s inequality (7.3) we have for any f ∈ L2 (−π, π) that
2
Z πX
X
X
1
inx
inx
cn (f )e −
cn (f )e dx =
|cn (f )|2 → 0
2π −π |n|≤N
|n|≤M
M +1≤|n|≤N
as N > M → ∞. Due to completeness of trigonometric polynomials in L2 (−π, π) (see
Corollary 2 of Theorem 3) we may conclude now that there is F ∈ L2 (−π, π) such that
2
Z π
X
1
inx F
(x)
−
c
(f
)e
lim
n
dx = 0.
N →∞ 2π −π |n|≤N
It remains to show that F (x) = f (x) almost everywhere. To do this, we compute the
Fourier coefficients cn (F ) by writing


Z π
Z π
X
F (x) −
ck (f )eikx  e−inx dx
F (x)e−inx dx =
2πcn (F ) =
−π
−π
+
X
|k|≤N
ck (f )
Z
π
ei(k−n)x dx.
−π
26
|k|≤N
If N > |n| then the last sum is equal to 2πcn (f ). Thus, by Hölder’s inequality,
Z π
X
ikx
F (x) −
dx
2π|cn (F ) − cn (f )| ≤
c
(f
)e
k
−π |k|≤N
2 1/2

Z
π
X
√
ikx
F (x) −
ck (f )e dx → 0
≤
2π 
−π |k|≤N
as N → ∞, i.e. cn (F ) = cn (f ) for all n = 0, ±1, ±2, . . .. Theorem 4 (uniqueness of
Fourier series) implies now that F = f almost everywhere. Parseval’s equality follows
from (7.2) if we let N → ∞.
CorollaryP(Riesz-Fischer theorem). Suppose {bn }∞
n=−∞ is a sequence of complex num∞
2
bers with n=−∞ |bn | < ∞. Then there is a unique f ∈ L2 (−π, π) such that bn =
cn (f ).
Proof. Completely the same as the proof of Theorem 6.
Theorem 7. Suppose that f ∈ L2 (−π, π) is periodic with period 2π and that its Fourier
coefficients satisfy
∞
X
|n|2 |cn (f )|2 < ∞.
(7.6)
n=−∞
Then f ∈
W21 (−π, π)
with Fourier series for f ′ (x) as
∞
X
′
f (x) ∼
incn (f )einx .
(7.7)
n=−∞
Proof. Since (7.6) holds then by Riesz-Fischer theorem there is a unique g(x) ∈
L2 (−π, π) such that
∞
X
g(x) ∼
incn (f )einx .
n=−∞
Integrating term by term we obtain
Z
Let F (x) :=
Z
π
g(x)dx = 0.
−π
x
−π
g(t)dt. Then, for n 6= 0, we have
Z π
Z π Z x
Z π
1
1
−inx
−inx
cn (F ) =
g(t)
g(t)dt e
dx =
e
dx dt
2π −π
2π −π
−π
t
−inπ
Z π
Z π
1 1
e−int
1
1
e
dt =
+
g(t)
g(t)e−int dt = cn (g).
=
2π −π
−in
in
in 2π −π
in
27
On the other hand, cn (g) = incn (f ). Thus, by uniqueness of Fourier series, we obtain
that F (x) − f (x) = constant almost everywhere or f ′ (x) = g(x) almost everywhere. It
means that
Z x
f (x) =
g(t)dt + constant,
−π
2
where g ∈ L (−π, π). Therefore, f ∈ W21 (−π, π) and
′
f (x) = g(x) ∼
∞
X
incn (f )einx .
n=−∞
Corollary. Under the conditions of Theorem 7, it is true that
∞
X
n=−∞
|cn (f )| < ∞.
Proof. Due to (7.6) we have
∞
X
n=−∞
|cn (f )| = |c0 (f )| +
X
n6=0
|cn (f )| ≤ |c0 (f )| +
1X 1
1X 2
< ∞,
|n| |cn (f )|2 +
2 n6=0
2 n6=0 |n|2
where we have used the basic inequality 2ab ≤ a2 + b2 for real numbers a and b.
Using Parseval’s equality (7.5) we can obtain for any periodic f ∈ L2 (−π, π) and
for any N = 1, 2, . . . that
2
Z πX
X
1
inx
dx =
c
(f
)e
−
f
(x)
|cn (f )|2 .
(7.8)
n
2π −π |n|≤N
|n|>N
Exercise 23. Prove (7.8).
Using Parseval’s equality again we have
Z π
∞
∞
X
X
1
2
2
|f (x + h) − f (x)| dx =
|cn (f (x + h) − f (x))| =
|eihn − 1|2 |cn (f )|2 .
2π −π
n=−∞
n=−∞
(7.9)
Theorem 8. Suppose f ∈ L2 (−π, π) is periodic with period 2π. Then
X
|cn (f )|2 = O(N −2α ), N = 1, 2, . . .
(7.10)
|n|>N
with 0 < α < 1 if and only if
∞
X
n=−∞
|einh − 1|2 |cn (f )|2 = O(|h|2α )
for |h| small enough.
28
(7.11)
Proof. From (7.9) we have for any integer M > 0 that
Z π
X
X
1
|f (x + h) − f (x)|2 dx ≤
n2 h2 |cn (f )|2 + 4
|cn (f )|2
2π −π
|n|≤M
(7.12)
|n|>M
if M h ≤ 1. If (7.10) holds then the second sum is O(M −2α ). To estimate the first sum
we use summation by parts. Writing
X
In :=
|ck (f )|2
|k|≤n
we have
M
M
X
X
X
2
2
2
2
2
2
n |cn (f )| = M IM −0·I0 −
In−1 (n −(n−1) ) = M IM − (2n−1)In−1 −I0 .
n=1
1≤|n|≤M
n=2
By hypothesis,
I∞ − In = O(n−2α ),
Thus,
2
M IM −
M
X
n=2
=M
2
(2n − 1) I∞ + O((n − 1)−2α )
I∞ + O(M
−2α
= O(M 2−2α ) − I∞
|
) − I∞
M2 −
M
X
n=2
M
X
n=2
n = 1, 2, . . . .
(2n − 1) −
!
(2n − 1) −
{z
}
=1
M
X
n=2
M
X
n=2
(2n − 1)O((n − 1)−2α )
(2n − 1)O((n − 1)−2α )
= O(M 2−2α ) + O(M 2−2α ) = O(M 2−2α ).
Exercise 24. Prove that
PM
2
1.
n=1 (2n − 1) = M
PM
−2α
2.
) = O(M 2−2α ) for 0 < α < 1.
n=2 O((2n − 1)(n − 1)
Combining these two estimates we may conclude from (7.12) that there is C > 0
such that
Z π
1
|f (x + h) − f (x)|2 dx ≤ C h2 M 2−2α + M −2α .
2π −π
Since 0 < α < 1 then choosing M = [1/|h|] we obtain
Z π
1
|f (x + h) − f (x)|2 dx ≤ C h2 [1/|h|]2−2α + [1/|h|]−2α
2π −π
≤ C h2 (1/|h|)2−2α + (1/|h| − {1/|h|})−2α
≤ C |h|2α + (1/|h| − 1)−2α
= C |h|2α + |h|2α /(1 − |h|)2α ≤ C|h|2α
29
if |h| ≤ 1/2. Here {1/|h|} denotes the fractional part of 1/|h| i.e. {1/|h|} = 1/|h| −
[1/|h|] ∈ [0, 1).
Conversely, if (7.11) holds then
∞
X
n=−∞
(1 − cos(nh))|cn (f )|2 ≤ Ch2α
with some C > 0. Integrating this inequality with respect to h over the interval
[0, l], l > 0 we have
∞
X
n=−∞
or
|cn (f )|
∞
X
n=−∞
or
n=−∞
Cl
2α
≥
X
|n|l≥2
Z
l
(1 − cos(nh))dh ≤ C
0
|cn (f )|
∞
X
It follows that
2
2
|cn (f )|
|cn (f )|
2
2
sin(nl)
l−
n
sin(nl)
1−
nl
sin(nl)
1−
nl
l
h2α dh
0
≤ Cl2α+1
Z
≤ Cl2α .
≥
1 X
|cn (f )|2 .
2
|n|l≥2
Taking l = 2/N for integer N > 0 in (7.13) yields
X
|cn (f )|2 ≤ CN −2α .
|n|≥N
This finishes the proof.
Remark. If α = 1 then (7.11) implies (7.10) but not vice versa.
Exercise 25. Suppose that periodic f ∈ L2 (−π, π) satisfies the condition
Z π
|f (x + h) − f (x)|2 dx ≤ Ch2
−π
with some C > 0. Prove that
∞
X
n=−∞
|n|2 |cn (f )|2 < ∞
and, therefore, f ∈ W21 (−π, π).
30
(7.13)
For any integrable function f periodic on the interval [−π, π] let us introduce the
mapping
f 7→ {cn (f )}∞
n=−∞ ,
where cn (f ) are the Fourier coefficients of f . This mapping is a linear transformation.
Formula (3.1) says that this mapping is bounded from L1 (−π, π) to l∞ (Z). Here Z denotes all integer numbers and the sequence space lp (Z) consists of sequences {bn }∞
n=−∞
for which
∞
X
|bn |p < ∞
n=−∞
if 1 ≤ p < ∞ and supn∈Z |bn | < ∞ if p = ∞.
The Parseval equality (7.5) shows that it is also bounded from L2 (−π, π) to l2 (Z).
Due to an interpolation theorem we may conclude that this mapping is bounded from
′
Lp (−π, π) to lp (Z) for any 1 < p < 2, 1/p + 1/p′ = 1 and
∞
X
n=−∞
p′
|cn (f )| ≤ cp
Z
π
p
−π
31
|f (x)| dx
p′ /p
< ∞.
8
Besov and Hölder spaces
In this chapter we will consider 2π-periodic functions f which are defined via trigonometric Fourier series in L2 (−π, π) as
f (x) ∼
∞
X
cn (f )einx ,
(8.1)
n=−∞
where the Fourier coefficients cn (f ) satisfy the Parseval equality
1
2π
Z
π
2
−π
|f (x)| dx =
∞
X
n=−∞
|cn (f )|2 ,
that is, (8.1) can be understood in the sense of L2 (−π, π) as
2
Z π
X
inx lim
f (x) −
cn (f )e dx = 0.
N →∞ −π |n|≤N
(8.2)
(8.3)
We will introduce new spaces of functions (that are actually subspaces of L2 (−π, π))
in terms of Fourier coefficients. The motivation of such approach is the following: we
proved (see Theorem 7 and Exercise 25) that periodic f belongs to W21 (−π, π) if and
only if
∞
X
|n|2 |cn (f )|2 < ∞.
n=−∞
This fact and Parseval equality justify the following definitions.
Definition 8.1. We say that a 2π-periodic function f belongs to Sobolev space
W2α (−π, π)
for some α ≥ 0 if
∞
X
n=−∞
|n|2α |cn (f )|2 < ∞.
(8.4)
Definition 8.2. We say that a 2π-periodic function f belongs to Besov space
α
B2,θ
(−π, π)
for some α ≥ 0 and some 1 ≤ θ < ∞ if

θ/2
∞
X
X

|n|2α |cn (f )|2  < ∞.
j=0
2j ≤|n|<2j+1
32
(8.5)
Definition 8.3. We say that a 2π-periodic function f belongs to Nikol’skii space
H2α (−π, π)
for some α ≥ 0 if
sup
j=0,1,2,...
X
2j ≤|n|<2j+1
|n|2α |cn (f )|2 < ∞.
(8.6)
Definition 8.4 (See also Definition 1.6).
1. We say that a 2π-periodic function f
α
belongs to Hölder space C [−π, π] for some non-integer α > 0 if f is continuous
on the interval [−π, π], there is a continuous derivative f (k) of order k = [α] on
the interval [−π, π] and for all h 6= 0 small enough, we have
sup |f (k) (x + h) − f (k) (x)| ≤ C|h|α−k ,
(8.7)
x∈[−π,π]
where the constant C > 0 does not depend on h.
2. By the space C k [−π, π] for integer k > 0 we mean 2π-periodic functions f that
have continuous derivatives f (k) of order k on the interval [−π, π].
Remark. We use later the following sufficient condition (see (3.13)): if there is a constant C > 0 such that for each n = 1, 2, . . . we have
X
|m|k |cm (f )| ≤ Cn−α
(8.8)
|m|≥n
with some integer k ≥ 0 and some 0 < α < 1 then f belongs to Hölder space
C k+α [−π, π].
The definitions 8.1-8.3 imply the following equalities and imbeddings:
α
1. B2,2
(−π, π) = W2α (−π, π), α ≥ 0.
2. W20 (−π, π) = L2 (−π, π).
α
α
3. B2,1
(−π, π) ⊂ B2,θ
(−π, π) ⊂ H2α (−π, π), α ≥ 0, 1 ≤ θ < ∞.
0
0
4. B2,θ
(−π, π) ⊂ L2 (−π, π), 1 ≤ θ ≤ 2 and L2 (−π, π) ⊂ B2,θ
(−π, π), 2 ≤ θ < ∞.
5. L2 (−π, π) ⊂ H20 (−π, π).
Exercise 26. Prove imbeddings 3, 4 and 5.
More imbeddings are formulated in the following theorems.
Theorem 9. If α ≥ 0 then
and
C α [−π, π] ⊂ W2α (−π, π)
C α [−π, π] ⊂ H2α (−π, π).
33
Proof. Let us prove the first claim only for integer α ≥ 0. If α = 0 then
C[−π, π] ⊂ L2 (−π, π) = W20 (−π, π).
If α = k > 0 is an integer then Definition 8.4 implies that
Z π
1
|f (k−1) (x + h) − f (k−1) (x)|2 dx ≤ Ch2 .
2π −π
Using Parseval equality we obtain
∞
X
n=−∞
or
4
∞
X
n=−∞
It follows that
|n|2k−2 |cn (f )|2 |einh − 1|2 ≤ Ch2
|n|2k−2 |cn (f )|2 sin2 (nh/2)) ≤ Ch2 .
X
|nh|≤2
or
|n|2k−2 |cn (f )|2 n2 h2 ≤ Ch2
X
|n|≤2/|h|
Letting h → 0 yields
∞
X
n=−∞
∈ W2k (−π, π).
[α]
|n|2k |cn (f )|2 ≤ C.
|n|2k |cn (f )|2 ≤ C
i.e. f
For non-integer α one needs to use interpolation between the
spaces C [−π, π] and C [α]+1 [−π, π].
Now let us consider the second claim. As above, for f ∈ C α [−π, π], we have
Z π
1
|f (k) (x + h) − f (k) (x)|2 dx ≤ C|h|2(α−k) ,
2π −π
where k = [α] if α is not an integer and k = α − 1 if α is an integer. By Parseval
equality,
∞
X
|n|2k |cn (f )|2 |einh − 1|2 ≤ C|h|2(α−k)
n=−∞
or
2
∞
X
n=−∞
|n|2k |cn (f )|2 (1 − cos(nh)) ≤ C|h|2(α−k) .
It suffices to consider h > 0. If we integrate the last inequality with respect to h > 0
from 0 to l then
∞
X
sin(nl)
2k
2
|n| |cn (f )| 1 −
≤ Cl2(α−k) .
nl
n=−∞
34
For 1 ≤ |n|l ≤ 2 we obtain
X
|n|2k |cn (f )|2 ≤ Cl2(α−k)
X
|n|2k l2(k−α) |cn (f )|2 ≤ C,
1≤|n|l≤2
or, equivalently,
1≤|n|l≤2
where constant C > 0 does not depend on l. Since 2(k − α) < 0 then it follows that
X
|n|2α |cn (f )|2 ≤ C
1/l≤|n|≤2/l
for any l > 0. Choosing l = 2−j we obtain
X
|n|2α |cn (f )|2 ≤ C
2j ≤|n|≤2j+1
i.e. f ∈ H2α (−π, π). This finishes the proof.
Theorem 10. Assume that α > 1/2 and that α − 1/2 is not an integer. Then
H2α (−π, π) ⊂ C α−1/2 [−π, π].
(8.9)
Proof. Let k = [α] so that α = k + {α}, where {α} denotes the fractional part of α.
Note that, in general, 0 ≤ {α} < 1 and in this theorem {α} 6= 1/2. We will assume
first that {α} = 0, i.e. α = k is integer and k ≥ 1. If f ∈ H2k (−π, π) then there is a
constant C > 0 such that
X
|m|2k |cm (f )|2 ≤ C
(8.10)
2j ≤|m|<2j+1
for each j = 0, 1, 2, . . .. Let us estimate the tail (8.8). Indeed, by Cauchy-SchwarzBuniakowsky inequality and (8.10),
X
|m|≥n
|m|k−1 |cm (f )| ≤
∞
X
j=j0
2j0 ∼n
≤
∞
X
j=j0
2j0 ∼n
≤
√
C
X
2j ≤|m|<2j+1


X
|m|k−1 |cm (f )|
2j ≤|m|<2j+1
∞
X
j=j0
2j0 ∼n


1/2 

|m|2k |cm (f )|2 
X
2j ≤|m|<2j+1
35
1/2
1 
m2
≤
X
2j ≤|m|<2j+1
√
C
∞
X
j=j0
2j0 ∼n
1/2
1 
m2
2−j/2 ≤ C2−j0 /2 .
Here and throughout, 2j0 ∼ n means that 2j0 ≤ n < 2j0 +1 . Therefore, we obtain
X
|m|k−1 |cm (f )| ≤ Cn−1/2 ,
|m|≥n
where the constant C > 0 is independent of n. It means that (see (8.8)) f belongs to
C k−1+1/2 [−π, π] = C k−1/2 [−π, π].
If α > 1/2 is not an integer and α − 1/2 is not an integer then for f ∈ H2α (−π, π)
we have instead of (8.10) the estimate
X
|m|2k+2{α} |cm (f )|2 ≤ C,
(8.11)
2j ≤|m|<2j+1
where k = [α], 0 < {α} < 1 and {α} 6= 1/2. If k = 0 then 1/2 < α = {α} < 1.
Repeating now the above procedure we obtain easily

1/2 
1/2
∞
X
X
X
X

|cm (f )| ≤
|m|2α |cm (f )|2  
|m|−2α 
|m|≥n
2j ≤|m|<2j+1
j=j0
2j0 ∼n
≤ C
∞
X
j=j0
2j0 ∼n

X

2j ≤|m|<2j+1
2j ≤|m|<2j+1
|m|
1/2
−2α 
≤C
∞
X
j=j0
2−(α−1/2)j ≤ Cn−(α−1/2)
2j0 ∼n
i.e. we have again that f ∈ C α−1/2 [−π, π].
For the case [α] = k ≥ 1, α is not an integer and α − 1/2 is not an integer either
we consider two cases: 0 < {α} < 1/2 and 1/2 < {α} < 1. In the first case we have
X
|m|≥n
|m|k−1 |cm (f )| ≤

∞
X

X
2j ≤|m|<2j+1
j=j0
2j0 ∼n
1/2 
|m|2k+2{α} |cm (f )|2 
≤C
∞
X
j=j0

X
2j ≤|m|<2j+1
1/2
|m|−2−2{α} 
2−j−j{α} 2j/2 ≤ Cn−1/2−{α} .
2j0 ∼n
This means again that f ∈ C k−1+1/2+{α} [−π, π] = C α−1/2 [−π, π]. In the second case
1/2 < {α} < 1 we proceed as follows:
X
|m|≥n
|m|k |cm (f )| ≤
∞
X
j=j0
2j0 ∼n


X
2j ≤|m|<2j+1
1/2 
|m|2k+2{α} |cm (f )|2 
≤C
∞
X
j=j0
2j0 ∼n
36

X
2j ≤|m|<2j+1
1/2
|m|−2{α} 
2−({α}−1/2)j ≤ Cn−{α}+1/2 .
It means that f ∈ C k+{α}−1/2 [−π, π] = C α−1/2 [−π, π]. Hence, this theorem is completely proved.
Corollary 1. Assume that α = k + 1/2 for some integer k ≥ 1. Then
H2α (−π, π) ⊂ C β−1/2 [−π, π]
(8.12)
for any 1/2 < β < α.
Corollary 2. Assume that α > 1/2 and α − 1/2 is not an integer. Then
α
B2,θ
(−π, π) ⊂ C α−1/2 [−π, π]
for any 1 ≤ θ < ∞.
Exercise 27. Prove Corollaries 1 and 2.
Exercise 28. Prove that the Fourier series (8.1) with coefficients
cn (f ) =
1
, n 6= 0,
|n| log(1 + |n|)
c0 (f ) = 1
1/2
defines a function from Besov space B2,θ (−π, π) for any 1 < θ < ∞ but not for θ = 1.
Exercise 29. Prove that the Fourier series (8.1) with coefficients
cn (f ) =
|n|3/2
1
, n 6= 0,
log(1 + |n|)
c0 (f ) = 1
1
defines a function from Besov space B2,θ
(−π, π) for any 1 < θ < ∞ but not for θ = 1.
Exercise 30. Consider the Fourier series (8.1) with coefficients
cn (f ) =
|n|2
1
, n 6= 0,
log (1 + |n|)
β
c0 (f ) = 1.
Prove that
3/2
1. f ∈ H2 (−π, π) if β ≥ 0
3/2
2. f ∈ W2 (−π, π) if β > 1/2
3. f ∈ C 1 [−π, π] if β > 1 but f ∈
/ C 1 [−π, π] if β ≤ 1.
Exercise 31. Let
∞
X
ak cos(bk x)
k=0
be a trigonometric series, where b = 2, 3, . . . and 0 < a < 1. Prove that the series
defines a function from C 1 [−π, π] if 0 < ab < 1 and a function from Hölder space
C γ [−π, π], γ < 1 if ab = 1.
Exercise 32. Assume that a = 1/b2 in Exercise 31. Is it true that f ∈ C 1 [−π, π]?
37
9
Absolute convergence. Bernstein and Peetre theorems.
We prove first that if 2π-periodic function f belongs to Nikol’skii space H2α (−π, π) for
some 0 < α < 1 in the sense of Definition 8.3 then it is equivalent to the L2 -Hölder
condition of order α, i.e.
Z π
1
|f (x + h) − f (x)|2 dx ≤ K|h|2α
(9.1)
2π −π
with some constant K > 0 and for all h 6= 0 small enough. Indeed, due to Parseval
equality we have
1
2π
Z
∞
X
π
2
−π
|f (x + h) − f (x)| dx =
where j0 is chosen so that 2j0 ≤
(9.2) is estimated from above as
|h|
2
X
|n|≤2j0
2
|n| |cn (f )|
2
n=−∞
1
|h|
≤ |h|
2
≤ |h|
2
|cn (f )|2 |einh − 1|2
≤
X
|n|2 |h|2 |cn (f )|2 + 4
|n|≤2j0
X
|n|≥2j0
|cn (f )|2 , (9.2)
≤ 2j0 +1 . The first sum on the right-hand side of
j0
X
X
j=0 2j ≤|n|<2j+1
j0
X
2j+1
j=0
≤ C|h|
2
j0
X
2−2α
j=0
2−2α
X
2j ≤|n|<2j+1
j+1
|n|2α |cn (f )|2
j +2
(22−2α ) 0 − 22−2α
22−2α − 1
2−2α
1
2
≤ C|h|
= C|h|2α
(9.3)
|h|
22−2α
≤ C|h|2 2j0
|n|2 |cn (f )|2
= C|h|2
if 0 < α < 1. We have used this condition for α since we considered geometric
progression with multiplier 22−2α 6= 1. The second sum on the right-hand side of (9.2)
is estimated from above as
4
∞
X
X
j=j0 2j ≤|n|<2j+1
2α
|n| |n|
−2α
|cn (f )|
2
≤ 4
∞
X
2−2αj
j=j0
≤ C
∞
X
j=j0
X
2j ≤|n|<2j+1
|n|2α |cn (f )|2
2−2αj ≤ C2−2j0 α ≤ C(2|h|)2α ≤ C|h|2α
1
since |h|
≤ 2j0 +1 and the criterion of Definition 8.3 holds. Thus, (9.1) is proved.
Conversely, if the L2 -Hölder condition (9.1) is fulfilled then Theorem 8 implies for each
38
N = 1, 2, . . . the inequality
X
|n|≥N
|cn (f )|2 ≤ CN −2α
with the same α as in (9.1). But this leads to the inequality
X
|cn (f )|2 ≤ C,
N 2α
N ≤|n|<2N
where the constant C is independent of N . Thus, we obtain for any integer N > 0
that
X
|n|2α |cn (f )|2 ≤ C.
N ≤|n|<2N
Since N is arbitrary we may conclude that f ∈ H2α (−π, π) for 0 < α ≤ 1 in the sense
of Definition 8.3. Therefore, the L2 -Hölder condition (9.1) can be considered as an
equivalent definition of Nikol’skii space H2α (−π, π) for 0 < α < 1.
Exercise 33. Prove that if f belongs to Nikol’skii space H2α (−π, π) for any non-integer
α > 0 in the sense of Definition 8.3 then the following L2 -Hölder condition holds:
Z π
1
|f (k) (x + h) − f (k) (x)|2 dx ≤ K|h|2α−2k
2π −π
with some constant K > 0 and k = [α].
Theorem 11 (Bernstein, 1914). Assume that 2π-periodic function f satisfies the L2 Hölder condition with 1/2 < α ≤ 1. Then its trigonometric Fourier series converges
absolutely, i.e.
∞
X
|cn (f )| < ∞.
n=−∞
Proof. Since the L -Hölder condition (9.1) is equivalent to f ∈ H2α (−π, π) for 0 < α < 1
then there is a constant C > 0 such that
X
|n|2α |cn (f )|2 ≤ C
(9.4)
2
2j ≤|n|<2j+1
for each j = 0, 1, 2, . . .. Hence we have
∞
X
n=−∞
|cn (f )| = |c0 (f )| +
∞
X
≤ |c0 (f )| +
∞
X
≤ |c0 (f )| +
X
j=0 2j ≤|n|<2j+1
j=0
√
C


|n|α |cn (f )||n|−α
X
2j ≤|n|<2j+1
∞
X
j=0
39

|n|2α |cn (f )|2 
2−αj 2j/2 = |c0 (f )| +
since α > 1/2. Thus, theorem is proved.
1/2 
√
C
X
2j ≤|n|<2j+1
∞
X
j=0
1/2
|n|−2α 
2−(α−1/2)
j
<∞
α
Corollary. Theorem 11 holds for spaces C α [−π, π], B2,θ
(−π, π) and H2α (−π, π) for any
α > 1/2 and 1 ≤ θ < ∞.
Exercise 34. Prove this Corollary.
Theorem 12 (Peetre, 1967). Assume that 2π-periodic function f belongs to Besov
1/2
space B2,1 (−π, π). Then its trigonometric Fourier series converges absolutely.
1/2
Proof. If f ∈ B2,1 (−π, π) then
∞
X
j=0
Hence we have
∞
X
n=−∞


X
2j ≤|n|<2j+1
|cn (f )| = |c0 (f )| +
∞
X
≤ |c0 (f )| +
∞
X
≤ |c0 (f )| +
= |c0 (f )| +
1/2
|n||cn (f )|2 
X
j=0 2j ≤|n|<2j+1
j=0
∞
X
j=0
∞
X
j=0



X
X
2j ≤|n|<2j+1


X
2j ≤|n|<2j+1
due to (9.5). This proves the theorem.
Corollary. It is true that
(9.5)
|n|1/2 |cn (f )||n|−1/2
2j ≤|n|<2j+1

< ∞.
1/2 
|n||cn (f )|2 
1/2
|n||cn (f )|2 
1/2
|n||cn (f )|2 

X
2j ≤|n|<2j+1
2−j 2j
1/2
1/2
|n|−1 
<∞
1/2
B2,1 (−π, π) ⊂ C[−π, π].
Exercise 35. Prove that the imbedding
1/2
B2,θ (−π, π) ⊂ C[−π, π]
does not hold for 1 < θ < ∞ by considering the function from Exercise 28.
Exercise 36. Prove that
if α > 1/2.
1/2
H2α (−π, π) ⊂ B2,1 (−π, π)
Theorem 13. Assume that 2π-periodic function f belongs to Sobolev space Wp1 (−π, π)
with some 1 < p < ∞. Then its trigonometric Fourier series converges absolutely.
40
Proof. Since
Wp11 (−π, π) ⊂ Wp12 (−π, π)
for 1 ≤ p2 < p1 , then we may assume without loss of generality that f ∈ Wp1 (−π, π)
with 1 < p ≤ 2. Then there is a function g ∈ Lp (−π, π) with 1 < p ≤ 2 such that
Z x
Z π
f (x) =
g(t)dt + f (−π),
g(t)dt = 0.
(9.6)
−π
−π
As we know from the proof of Theorem 7, (9.6) leads to
1
cn (g),
in
cn (f ) =
n 6= 0.
Since g ∈ Lp (−π, π) with 1 < p ≤ 2 then the results of Chapter 7 give
∞
X
n=−∞
where
1
p
+
1
p′
|cn (g)|
p′
!1/p′
< ∞,
(9.7)
= 1. The facts (9.6), (9.7) and Hölder’s inequality imply that
∞
X
X 1
X 1
|cn (f )| = |c0 (f )|+
|cn (g)| ≤ |c0 (f )|+
|n|
|n|p
n=−∞
n6=0
n6=0
!1/p
X
n6=0
|cn (g)|
p′
!1/p′
<∞
since 1 < p ≤ 2. This finishes the proof.
Remark. For Sobolev space W11 (−π, π) this theorem is not valid, i.e. there is a function f from W11 (−π, π) with absolutely divergent trigonometric Fourier series. More
precisely, we will prove in the next chapter that the function
f (x) :=
∞
X
n=1
sin nx
n log(1 + n)
(9.8)
belongs to Sobolev space W11 (−π, π), is continuous on the interval [−π, π] but its
trigonometric Fourier series (9.8) diverges absolutely.
The next theorem is due to Zigmund (1958–1959).
Theorem 14. Suppose that f ∈ W11 (−π, π) ∩ C α [−π, π] with some 0 < α < 1. Then
its trigonometric Fourier series converges absolutely.
Proof. Since f ∈ W11 (−π, π) then f is of bounded variation. The periodicity of f
implies that
1
2π
Z
π
1
|f (x + h) − f (x)| dx =
2πN
−π
2
Z
π
N
X
−π k=1
41
|f (x + kh) − f (x + (k − 1)h)|2 dx, (9.9)
where integer N is chosen so that N |h| ≤ 1 for h 6= 0 small enough. We choose
N = [1/|h|]. Since f ∈ C α [−π, π] the right-hand side of (9.9) can be estimted as
1
2πN
Z
π
N
X
−π k=1
|f (x + kh) − f (x + (k − 1)h)|2 dx
C|h|α
≤
2πN
≤
Z
π
N
X
−π k=1
|f (x + kh) − f (x + (k − 1)h)|dx
C|h|α
C|h|α
−π
π
V−π−1
(f ) + V−π
(f ) + Vππ+1 (f ) 2π ≤
2πN
N
C|h|α
C|h|α
C|h|α+1
C|h|α
=
≤
=
≤ C|h|α+1
=
[1/|h|]
1/|h| − {1/|h|}
1/|h| − 1
1 − |h|
α+1
if |h| ≤ 1/2. This inequality means (see (9.9)) that f ∈ H2 2 (−π, π) with α+1
> 1/2
2
for α > 0. Using Bernstein’s theorem we may conclude that this theorem is proved.
Exercise 37. Let (periodic) f be defined by
f (x) :=
∞
X
eikx
k=1
k
.
1/2
Prove that f belongs to Nikol’skii space H2 (−π, π) but its trigonometric Fourier
series is not absolutely convergent.
Exercise 38. Let (periodic) f be defined by the absolutely convergent Fourier series
∞
X
eikx
f (x) :=
.
3/2
k
k=1
1. Show that f belongs to Nikol’skii space H21 (−π, π) but
Z π
π
4h2
1
|f (x + h) − f (x)|2 dx ≥ 2 log
2π −π
π
|h|
for 0 < |h| < 1, that is, (9.1) does not hold for α = 1.
1
2. Show that f does not belong to Besov space B2,θ
(−π, π) for any 1 ≤ θ < ∞.
42
10
Dirichlet kernel. Pointwise and uniform convergence.
The material of this chapter makes a central part of the theory of trigonometric Fourier
series. In this chapter we will give an answer to the question: to which value the
trigonometric Fourier series converges?
The Dirichlet kernel DN (x) which is defined by symmetric finite trigonometric sum
X
DN (x) :=
einx
(10.1)
|n|≤N
plays the key role in this chapter. If x ∈ [−π, π] \ {0} then DN (x) from (10.1) can be
recalculated as follows. Using Euler’s formula we have
DN (x) =
N
X
n=−N
=
e
N
X
einx = e−iN x
−iN x
ei(n+N )x = e−iN x
n=−N
i(N +1)x
−e
1 − eix
=
e
i(N +1/2)x
eix/2
2N
X
eikx = e−iN x
k=0
−i(N +1/2)x
−e
− e−ix/2
=
1 − ei(2N +1)x
1 − eix
sin(N + 1/2)x
.
sin x/2
Thus, the Dirichlet kernel equals
DN (x) =
sin(N + 1/2)x
,
sin x/2
x 6= 0.
(10.2)
For x = 0 we have
DN (0) = 2N + 1 = lim DN (x),
x→0
so that (10.2) holds for all x ∈ [−π, π].
Exercise 39. Prove that
Z π
1
DN (x)dx = 1, N = 0, 1, 2, . . .
1.
2π −π
P
2. KN (x) = N1+1 N
j=0 Dj (x), where KN (x) is the Fejér kernel (5.2).
Recall that the trigonometric Fourier partial sum is given by
X
SN f (x) =
cn (f )einx .
|n|≤N
The Fourier coefficients of DN (x) are equal to
1
cn (DN ) =
2π
Z
π
e−inx
−π
X
|k|≤N
43
eikx dx =
(
0, |n| > N
1, |n| ≤ N.
(10.3)
Hence, if f is periodic and integrable then the partial sum (10.3) can be rewritten as
(see Exercise 12)
Z π
1
SN f (x) =
cn (DN )cn (f )e = (f ∗ DN )(x) =
DN (x − y)f (y)dy
2π −π
n=−∞
Z π
Z π
1
sin(N + 1/2)y
1
DN (y)f (x + y)dy =
f (x + y)
dy. (10.4)
=
2π −π
2π −π
sin y/2
∞
X
inx
Exercise 40. Let f be the function
(
f (x) =
1
2
x
− 2π
,
0<x≤π
x
− 21 − 2π
, −π ≤ x < 0.
Show that
1. (SN f )′ (x) =
1
(DN (x)
2π
− 1)
2. limN →∞ SN f (x) = f (x), x 6= 0
3. limN →∞ SN f (0) = 0.
Exercise 41. Prove that
1
2π
Z
π
−π
|DN (x)|dx =
4 log N
+ O(1).
π2
Since the Dirichlet kernel is an even function (see (10.2)) we can rewrite (10.4) as
Z π
1
sin(N + 1/2)y
SN f (x) =
dy.
(10.5)
(f (x + y) + f (x − y))
2π 0
sin y/2
Using the normalization of the Dirichlet kernel (see Exercise 39) we have for any
function S(x) that
Z π
1
sin(N + 1/2)y
SN f (x) − S(x) =
dy. (10.6)
(f (x + y) + f (x − y) − 2S(x))
2π 0
sin y/2
Our task is to define S(x) so that the limit of the right-hand side of (10.6) is equal to
zero. We will simplify the problem into two steps. The first simplification is connected
with
Lemma 10.1. For all z ∈ [−π, π] it is true that
1
2 π 2
sin z/2 − z ≤ 24 .
44
(10.7)
Proof. First we show that
z |z|3
sin z/2 − ≤
2
48
for all z ∈ [−π, π]. In order to prove this inequality it is enough to show that
0 < x − sin x <
(10.8)
x3
6
for all 0 < x < π/2. The left inequality is well-known. To prove the right inequality
we introduce h(x) as
h(x) = x − sin x − x3 /6.
Then its derivative satisfies
h′ (x) = 1 − cos x − x2 /2 = 2(sin2 x/2 − x2 /4) < 0
for all 0 < x < π/2. Thus, h(x) is monotone decreasing on the interval [0, π/2]. It
implies that
0 = h(0) > h(x) = x − sin x − x3 /6
for all 0 < x < π/2. This proves (10.8) which in turn yields
1
2 |z/2 − sin(z/2)|
2
|z|3 /24
|z|3 /24
π|z|
π2
=
−
≤
≤
≤
≤
sin z/2 z |z|| sin(z/2)|
|z|| sin(z/2)|
|z||z|/π
24
24
since | sin z/2| ≥ |z|/π for all z ∈ [−π, π]. This finishes the proof.
As an immediate corollary of Lemma 10.1 we obtain for any periodic and integrable
function f that the function
1
2
y 7→ (f (x + y) + f (x − y) − 2S(x))
−
sin y/2 y
is integrable on the interval [0, π] uniformly in x ∈ [−π, π] if S(x) is bounded, i.e.
Z
π
0
1
2
− dy
|f (x + y) + f (x − y) − 2S(x)| sin y/2 y
Z π
Z π
2
π
|f (x + y)|dy +
|f (x − y)|dy + 2π|S(x)|
≤
24
0
0
!
Z π
π2
|f (y)|dy + 2π sup |S(x)| .
≤
24
x∈[−π,π]
−π
Application of Riemann-Lebesgue lemma (Theorem 5) gives us that
Z π
2
1
1
(f (x + y) + f (x − y) − 2S(x))
−
sin(N + 1/2)y = 0 (10.9)
lim
N →∞ 2π 0
sin y/2 y
45
pointwise in x ∈ [−π, π] and even uniformly in x ∈ [−π, π] if S(x) is bounded on the
interval [−π, π].
Thus, we have reduced the question of pointwise or uniform convergence in (10.6)
to proving that
Z π
sin(N + 1/2)y
(f (x + y) + f (x − y) − 2S(x))
lim
dy = 0
(10.10)
N →∞ 0
y
pointwise or uniformly in x ∈ [−π, π].
For the second simplification we consider the contribution to (10.10) from the interval 0 < δ ≤ y ≤ π. Note that the function
f (x + y) + f (x − y) − 2S(x)
y
is integrable in y on the interval [δ, π] uniformly in x ∈ [−π, π] if S(x) is bounded.
Hence, by the Riemann-Lebesgue lemma this contribution tends to zero when N → ∞.
We summarize these two simplifications as
Z
1 δ
sin(N + 1/2)y
SN f (x) − S(x) =
dy + oN (1)
(f (x + y) + f (x − y) − 2S(x))
π 0
y
(10.11)
pointwise or uniformly in x ∈ [−π, π].
Let us assume now that f is a piecewise continuous periodic function. The question is: to which values of S(x) in (10.11) the trigonometric Fourier partial sum may
converge? The second part of Theorem 3 shows that the Fejér means converge to
lim σN f (x) =
N →∞
1
(f (x + 0) + f (x − 0))
2
for any x ∈ [−π, π] pointwise. But since
σN f (x) =
PN
Sj f (x)
N +1
j=0
then SN f (x) may converge (if it converges) only to the value
1
(f (x + 0) + f (x − 0)) .
2
We can obtain some sufficient conditions when the limit in (10.11) exists.
Theorem 15. Suppose that S(x) is chosen so that
Z δ
|f (x + y) + f (x − y) − 2S(x)|
dy < ∞
y
0
(10.12)
pointwise or uniformly in x ∈ [−π, π]. Then
lim SN f (x) = S(x)
N →∞
pointwise or uniformly in x ∈ [−π, π].
46
(10.13)
Proof. The claim follows immediately from (10.11) and Riemann-Lebesgue lemma.
Remark. If in (10.13) we have uniform convergence then S(x) must necessarily be
periodic (S(−π) = S(π)) and continuous on the interval [−π, π].
Corollary. Suppose periodic f belongs to Hölder space C α [−π, π] for some 0 < α ≤ 1.
Then
lim SN f (x) = f (x)
N →∞
uniformly in x ∈ [−π, π].
Proof. Since f ∈ C α [−π, π] then
|f (x + y) + f (x − y) − 2f (x)| ≤ |f (x + y) − f (x)| + |f (x − y) − f (x)| ≤ Cy α
for 0 < y < δ. It means that the condition (10.12) holds with S(x) = f (x) uniformly
in x ∈ [−π, π]. Thus, this corollary follows.
Theorem 16 (Dirichlet, Jordan). Suppose that periodic f is of bounded variation on
the interval [x − δ, x + δ] for some δ > 0 and some fixed x. Then
lim SN f (x) =
N →∞
1
(f (x + 0) + f (x − 0)) .
2
(10.14)
Proof. Since f is of bounded variation then the limit
lim
y→0+
1
1
(f (x + y) + f (x − y)) = (f (x + 0) + f (x − 0)) := S(x)
2
2
(10.15)
exists. For 0 < y < δ we denote
F (y) :=
1
(f (x + y) + f (x − y) − 2S(x)) ,
2
where S(x) is defined by (10.15). Note that F (0) = 0. Let us also denote
Z y
sin(N + 1/2)t
dt, 0 < y ≤ δ.
GN (y) :=
t
0
(10.16)
It is easy to check that
GN (y) =
Z
(N +1/2)y
0
sin ρ
dρ,
ρ
0 < y ≤ δ.
This representation implies that
lim GN (y) =
N →∞
Z
∞
0
47
π
sin ρ
dρ = .
ρ
2
(10.17)
For fixed x we have from (10.11) and (10.16) that
Z
1 δ
F (y)G′N (y)dy + oN (1).
SN f (x) − S(x) =
π 0
Here integration by parts gives
Z δ
1
δ
SN f (x) − S(x) =
GN (y)dF (y) + oN (1)
F (y)GN (y)|0 −
π
0
Z δ
1
GN (y)dF (y) + oN (1),
=
F (δ)GN (δ) −
π
0
(10.18)
where the last integral is well-defined as the Stieltjes integral with respect to function
F (y) of bounded variation and continuous function GN (y). Since the limit (10.17)
holds and GN (y) is continuous, we can consider the limit in (10.18) when N → ∞.
Hence, we obtain
Z
π π δ
1
1
lim (SN f (x) − S(x)) =
F (δ) −
dF (y) = (F (δ) − F (δ) + F (0)) = 0.
N →∞
π
2
2 0
2
This finishes the proof.
Corollary. If f is periodic and belongs to Sobolev space W11 (−π, π) then its trigonometric Fourier series converges pointwise to f (x) everywhere.
Proof. Since f ∈ W11 (−π, π) then it is of bounded variation and continuous on the
interval [−π, π]. In this case the value S(x) from (10.15) equals f (x) at any point
x ∈ [−π, π]. Thus,
lim SN f (x) = f (x)
N →∞
pointwise in x ∈ [−π, π].
Remark. The above proof does not allow us to conclude uniform convergence of the
trigonometric Fourier series of functions from Sobolev space W11 (−π, π). However,
uniform convergence is valid as we will prove later in this chapter.
Exercise 42. Show that
f (x) =
1
1 ,
log |x|
|x| < 1/2
is of bounded variation but this function does not satisfy condition (10.12) at x = 0.
Hint.
Exercise 43. Show that
Z
δ
0
1
dy = +∞.
y log y
1
x
satisfies condition (10.12) at x = 0 but this function is not of bounded variation.
f (x) = x sin
48
We can return now to the question of term by term integration of the trigonometric
Fourier series.
Theorem 17. Suppose f belongs to L1 (−π, π). Then
Z b
Z b
lim
f (x)dx
SN f (x)dx =
N →∞
a
a
for any interval (a, b) ⊂ [−π, π].
Proof. For a given L1 -function f (not necessarily periodic) introduce a new function
F as
Z x
(f (t) − c0 (f ))dt.
(10.19)
F (x) :=
−π
It is clear that F (x) belongs to Sobolev space W11 (−π, π) with F (−π) = F (π) = 0
(periodicity) and
F ′ (x) = f (x) − c0 (f ).
This implies
cn (F ′ ) = incn (F ) = cn (f ),
c0 (F ′ ) = 0.
n 6= 0,
Corollary of Theorem 16 gives us that F (x) has everywhere convergent trigonometric
Fourier series
X cn (f )
einx .
(10.20)
F (x) = c0 (F ) +
in
n6=0
In particular, for any −π ≤ a < b ≤ π we have from (10.20) that
Z b
X
X cn (f )
inb
ina
(e − e ) =
cn (f )
einx dx
F (b) − F (a) =
in
a
n6=0
n6=0
or equivalently (see (10.19)),
Z
Z b
(f (x) − c0 (f ))dx −
−π
a
−π
(f (x) − c0 (f ))dx =
=
Z
b
a
X
n6=0
f (x)dx − (b − a)c0 (f )
cn (f )
Z
b
einx dx.
a
Thus, we obtain finally
Z b
Z b
Z b
Z b
∞
X
X
inx
inx
f (x)dx =
e dx = lim
cn (f )
e dx = lim
cn (f )
SN f (x)dx.
a
n=−∞
a
N →∞
|n|≤N
a
This proves the theorem.
Exercise 44. Calculate c0 (F ) for F defined by (10.19).
49
N →∞
a
Corollary 1. If f ∈ L1 (−π, π) then the series
X cn (f )
X cn (f )(−1)n
and
n
n
n6=0
n6=0
(10.21)
converge.
Proof. Follows from (10.20).
Corollary 2. The series
∞
X
sin(nx)
log(1 + n)
n=1
is not a Fourier series of an L1 -function.
Proof. Let us assume on the contrary that there is f ∈ L1 (−π, π) such that
f (x) ∼
=
∞
∞
∞
X
sin(nx)
einx
1 X
1 X e−inx
=
−
log(1 + n)
2i n=1 log(1 + n) 2i n=1 log(1 + n)
n=1
∞
−∞
1 X
1 X sgn(n)einx
1 X sgn(n)einx
einx
+
=
2i n=1 log(1 + n) 2i n=−1 log(1 + |n|)
2i n6=0 log(1 + |n|)
i.e. we have
1 sgn(n)
, n 6= 0, c0 (f ) = 0.
2i log(1 + |n|)
Since cn (f ) = −c−n (f ) then this trigonometric Fourier series can be interpreted as the
Fourier series of some odd L1 -function. Then Corollary 1 of Theorem 17 implies that
cn (f ) =
X
n6=0
∞
1X
sgn(n)
1
=
2in log(1 + |n|)
i n=1 n log(1 + n)
must be convergent. But this is not true. This contradiction proves this corollary.
Remark. If we define the function f by the series in Corollary 2 then it turns out that
Z π
Z π
f (x)dx = 0,
|f (x)|dx = +∞.
−π
−π
Recall that the Poisson kernel is equal to
Pr (x) =
1 − r2
,
1 − 2r cos x + r2
0≤r<1
and its trigonometric Fourier series is
Pr (x) =
∞
X
r|n| einx .
n=−∞
The Corollary of Theorem 15 shows us that this series converges to Pr (x) uniformly in
x ∈ [−π, π].
50
Theorem 18. Suppose that f ∈ C[−π, π] is periodic. Then
lim (Pr ∗ f )(x) = f (x)
r→1−
uniformly in x ∈ [−π, π] or
lim
r→1−
∞
X
r|n| cn (f )einx = f (x)
(10.22)
(10.23)
n=−∞
uniformly in x ∈ [−π, π] even if f (x) has no convergent trigonometric Fourier series.
Proof. Using the normalization
1
2π
Z
π
Pr (x)dx = 1
−π
we have
Z π
1
(Pr ∗ f )(x) − f (x) =
Pr (y)(f (x − y) − f (x))dy
2π −π
Z
1
Pr (y)(f (x − y) − f (x))dy
=
2π |y|≤δ
Z
1
Pr (y)(f (x − y) − f (x))dy := I1 + I2 .
+
2π δ≤|y|≤π
Since f is continuous on [−π, π] then I1 can be estimated as
Z π
1
|I1 | ≤
sup
|f (x − y) − f (x)|
Pr (y)dy =
sup
|f (x − y) − f (x)| → 0
2π −π
x∈[−π,π],|y|≤δ
x∈[−π,π],|y|≤δ
as δ → 0. At the same time, I2 can be estimated as
Z
1
Pr (y)dy.
|I2 | ≤ 2 max |f (x)|
|x|≤π
2π δ≤|y|≤π
For δ ≤ |y| ≤ π the Poisson kernel can be estimated as
1−r
1−r
1 − r2
1 − r2
π2 1 − r
≤
≤
Pr (y) =
=
=
.
1 − 2r cos y + r2
2rδ 2 /π 2
2 rδ 2
4r sin2 y/2 + (1 − r)2
2r sin2 y/2
If we choose δ 4 = 1 − r then I2 is estimated as
√
1−r
π
π2 1 − r
1
√
= max |f (x)|
→0
|I2 | ≤ max |f (x)|
π |x|≤π
2 r 1−r
2 |x|≤π
r
as r → 1−. Hence, the estimates for I1 and I2 show that
lim ((Pr ∗ f )(x) − f (x)) = 0
r→1−
uniformly in x ∈ [−π, π]. The equality (10.23) follows from this fact and Exercise 12.
This completes the proof.
51
We will prove now the well-known Hardy’s theorem and then apply it to the uniform
convergence of the trigonometric sum SN f (x).
Theorem 19 (Hardy, 1949). Let {ak }∞
k=0 be a sequence of complex numbers such that
k|ak | ≤ M,
k = 0, 1, 2, . . . ,
(10.24)
where the constant M is independent of k. If the limit
n X
j
1−
lim σn := lim
aj = a
n→∞
n→∞
n
+
1
j=0
(10.25)
exists then
where sn =
lim (σn − sn ) = 0,
n→∞
Pn
j=0
aj , i.e. also
lim sn = a.
(10.26)
n→∞
If ak depends on x and (10.24) holds uniformly in x and convergence in (10.25) is
uniform then convergence in (10.26) is also uniform.
Proof. For n < m it is true that
(m + 1)σm − (n + 1)σn −
m
X
(m + 1 − j)aj = (m − n)sn .
j=n+1
Indeed,
(m + 1)σm − (n + 1)σn −
=
m
X
m
X
(m + 1 − j)aj
j=n+1
(m + 1 − j)aj −
j=0
n
X
=
j=0
n
X
(n + 1 − j)aj −
j=0
n
X
(m + 1 − j)aj −
j=0
m
X
(m + 1 − j)aj
j=n+1
n
X
(n + 1 − j)aj =
j=0
(m − n)aj = (m − n)sn .
That’s why,
(m + 1)σm − (n + 1)σn −
m
X
(m + 1 − j)aj − (m − n)σn = (m − n)sn − (m − n)σn
j=n+1
or, equivalently,
m(σm − σn ) + (σm − σn ) −
m
X
(m + 1 − j)aj = (m − n)(sn − σn )
j=n+1
52
i.e.
m m+1
j
m+1 X
aj = s n − σ n .
1−
(σm − σn ) −
m−n
m − n j=n+1
m+1
Let m > n → ∞ such that m/n ∼ 1 + δ (i.e. limm,n→∞ m/n = 1 + δ) with some
positive δ to be chosen. Since σm is a Cauchy sequence by (10.25) then
m+1
(σm − σn ) → 0,
m−n
m > n → ∞.
At the same time, the condition (10.24) implies that
m m m + 1 X
M
j
j
m+1 X
aj ≤
1−
1−
m − n
m+1
m − n j=n+1
m+1 j
j=n+1
m X
1
1
−
∼ M (1 + 1/δ)
j m+1
j=n+1
!
m
X
1 m−n
= M (1 + 1/δ)
−
j
m+1
j=n+1
Z m
1
m−n
≤ M (1 + 1/δ)
dξ −
m+1
n ξ
m m−n
= M (1 + 1/δ) log −
n
m+1
δ
∼ M (1 + 1/δ) log(1 + δ) −
1 + δ + 1/n
2
∼ M (1 + 1/δ) δ − δ /2 + o(δ 2 ) − δ(1 − δ + O(δ 2 ))
= M (1 + 1/δ) δ 2 /2 + o(δ 2 ) ≤ 2M δ
if δ is chosen small enough. Since δ is arbitrary we may conclude that also the second
term converges to zero. This proves the theorem.
Corollary 1. Suppose that f ∈ C[−π, π] is periodic and that its Fourier coefficients
satisfy
M
, n 6= 0
|cn (f )| ≤
|n|
with positive constant M which does not depend on n. Then the trigonometric Fourier
series of f converges to f uniformly in x ∈ [−π, π].
Proof. Since f ∈ C[−π, π] is periodic then Theorem 3 gives the convergence of Fejér
means
lim σN f (x) = f (x)
N →∞
uniformly in x ∈ [−π, π]. Let us denote a0 = c0 (f ) and
ak (x) = ck (f )eikx + c−k (f )e−ikx ,
53
k = 1, 2, . . . .
Then
sn ≡
and
n
X
ak (x) = SN f (x)
k=0
n
n
1 X
1 X
σn f (x) =
Sk f (x) =
sk .
n + 1 k=0
n + 1 k=0
Thus, we are in the frame of Hardy’s theorem because
|ak (x)| ≤
2M
,
k
k = 1, 2, . . . .
Since this inequality is uniform in x ∈ [−π, π] applying Hardy’s theorem we obtain
that
lim sN ≡ lim SN f (x) = f (x)
N →∞
N →∞
uniformly in x ∈ [−π, π].
Corollary 2. If f ∈ W11 (−π, π) is periodic then
lim SN f (x) = f (x)
N →∞
uniformly in x ∈ [−π, π].
Proof. Since f ∈ W11 (−π, π) is periodic then there is g ∈ L1 (−π, π) such that
Z π
Z x
g(t)dt = 0.
g(t)dt + f (−π),
f (x) =
−π
−π
Thus, f ′ = g and
cn (g) = incn (f )
or, equivalently,
cn (g) M
≤
,
|cn (f )| = in |n|
n 6= 0.
Due to imbedding (see Lemma 1.3) the function f is continuous on the interval [−π, π].
Using again Hardy’s theorem we obtain
lim SN f (x) = f (x)
N →∞
uniformly in x ∈ [−π, π]. This finishes the proof.
Let us return to some special trigonometric Fourier series. Namely, we consider
functions f1 (x) and f2 (x) which are defined by the Fourier series
f1 (x) =
∞
X
n=1
sin(nx)
n log(1 + n)
54
(10.27)
and
f2 (x) =
∞
X
n=1
cos(nx)
.
n log(1 + n)
(10.28)
These functions are well-defined for all x ∈ [−π, π] \ {0}, see Theorem 1. In addition,
f1 (0) = 0 whereas f2 (0) is not defined since the series (10.28) diverges at zero. We will
show that f2 (x) does not belong to W11 (−π, π) but f1 (x) does.
If we assume on the contrary that f2 ∈ W11 (−π, π) then its derivative f2′ has the
Fourier series
∞
X
sin(nx)
f2′ (x) ∼ −
.
log(1
+
n)
n=1
But due to Corollary 2 of Theorem 17 this is not a Fourier series of an L1 -function.
This contradiction proves that f2 6∈ W11 (−π, π).
Concerning the series (10.27) let us prove first that it converges uniformly in x ∈
[−π, π] i.e. f1 (x) is (at least) continuous on the interval [−π, π]. Indeed, by summation
by parts we obtain for 0 ≤ M < N that
N
X
N
X
sin(nx)
1
=
n log(1 + n) n=M +1 n log(1 + n)
n=M +1
n
X
k=1
sin(kx) −
n−1
X
sin(kx)
k=1
!
sin(N x)
sin(M x)
−
N log(1 + N ) M log(1 + M )
!
n
X
1
1
−
. (10.29)
sin(kx)
(n
+
1)
log(2
+
n)
n
log(1
+
n)
k=1
=
−
N
X
n=M +1
Using the calculation from Exercise 15 we have
n
X
k=1
sin(kx) =
cos x/2 − cos(n + 1/2)x
sin(nx/2) sin((n + 1)x/2)
=
.
2 sin x/2
sin x/2
(10.30)
The first two terms on the right-hand side of (10.29) converge to zero as N > M → ∞
uniformly in x ∈ [−π, π]. The sum on the right-hand side of (10.29) becomes, using
(10.30),
N
X
n log 2+n
+ log(2 + n)
sin(nx/2) sin((n + 1)x/2)
1+n
−
sin x/2
n(n + 1) log(1 + n) log(2 + n)
n=M +1
N
2+n
X
log 1+n
sin(nx/2) sin((n + 1)x/2)
=−
sin x/2
(n + 1) log(1 + n) log(2 + n)
n=M +1
N
X
1
sin(nx/2) sin((n + 1)x/2)
= I1 + I2 .
−
sin x/2
n(n + 1) log(1 + n)
n=M +1
55
Let us consider two cases: n|x| < 1 and n|x| > 1. In the first case,
sin(nx/2) sin((n + 1)x/2) n|x|/2 · 1
πn
≤
=
.
sin x/2
|x|/π
2
Then
1
1
[ |x|
[ |x|
]
]
n n1
π X
π X
n log (1 + 1/(n + 1))
|I1 | ≤
<
2 n=M +1 (n + 1) log(1 + n) log(2 + n)
2 n=M +1 n log2 n
∞
π X
1
→0
<
2 n=M +1 n log2 n
(10.31)
as M → ∞ uniformly in x. In the first case for I2 we have, by integration by parts,
that
1
1
[ |x|
[ |x|
]
]
Z 1/|x|
X
|x|
n n+1
π X
π
π
dt
1
2
|I2 | ≤
= |x|
< |x|
2 n=M +1 n(n + 1) log(1 + n)
4 n=M +1 log(1 + n)
4
M +1 log t
!
1/|x| Z 1/|x|
t dt
π
+
|x|
=
2
4
log t M +1
M +1 log t
!
Z 1/|x|
M +1
1/|x|
dt
π
|x|
−
+
=
2
4
log(1/|x|) log(M + 1)
M +1 log t
|x|(M + 1)
|x|
1
π
+
+
≤
(1/|x| + M + 1)
4 log(1/|x|) log(M + 1) log2 (M + 1)
1
1
1
1
π
+
+
+
<
4 log(M + 1) log(M + 1) log2 (M + 1) log2 (M + 1)
π
→0
(10.32)
≤
log(M + 1)
uniformly in x as M → ∞. In the second case,
sin(nx/2) sin((n + 1)x/2) π
1
≤
≤
≤ πn.
| sin x/2|
sin x/2
|x|
Then
N
X
N
X
n n1
n log (1 + 1/(n + 1))
<π
(n + 1) log(1 + n) log(2 + n)
n log2 n
1
1
n=[ |x|
n=
]
[ |x| ]
∞
X
1
→ 0, M → ∞
< π
n log2 n
1
n=[ |x| ]≥M
|I1 | ≤ π
56
(10.33)
uniformly in x. For I2 we have, by integration by parts,
Z ∞
1
1
dt
≤π
2
2
1
n log n
|x| [ |x|
]≥M t log t
1
n=[ |x|
]+1
!
∞
Z ∞
1
1 dt
π
−
+
2
2
1
|x|
t log t [ 1 ]≥M
[ |x|
]≥M t log t
|x|
!
Z ∞
1
dt
1/|x|
+
π
2
2
1
[1/|x|] log[1/|x|] |x| [ |x|
]≥M t log t
Z
[1/|x|] + 1 ∞ dt
[1/|x|] + 1
+
π
2
[1/|x|] log[1/|x|]
[1/|x|]
M t log t
1 + 1/M
1 + 1/M
→ 0, M → ∞
+
π
log M
log M
1
|I2 | ≤
| sin x/2|
=
=
<
≤
N
X
(10.34)
uniformly in x. Finally, we may conclude that the trigonometric Fourier series (10.27)
converges uniformly on the interval [−π, π] and therefore it defines continuous function
f1 (x). This series, as well as (10.28), can be differentiated term by term for π ≥ |x| ≥
δ > 0 because the series
∞
X
cos(nx)
log(1 + n)
n=1
converges uniformly (see Corollary of Theorem 1) for π ≥ |x| ≥ δ > 0. It means that
for this interval f1 (x) belongs to C 1 . Thus, it remains to investigate the behaviour
of the series (10.27) when x → 0+. But the estimates (10.31)-(10.34) show us that
(if we choose M ≍ 1/x, x → 0+) the function f1 (x) from (10.27) has the asymptotic
behaviour
C
f1 (x) ∼
.
(10.35)
log x
It is possible to prove (see Zigmund: Trigonometric series, Vol. I, Chapter V, formula
(2.19)) that the asymptotic (10.35) can be differentiated and we obtain
f1′ (x) ∼ −
C
.
x log2 x
This singularity is integrable at zero. Thus, the function f1 (x) belongs to W11 (−π, π).
Exercise 45. Prove that the series (10.27) and (10.28) do not converge absolutely.
57
11
Formulation of discrete Fourier transform and
its properties.
Let x(t) be 2π-periodic continuous signal. Assume that x(t) can be represented by an
absolutely convergent trigonometric Fourier series
x(t) =
∞
X
cm eimt ,
t ∈ [−π, π],
m=−∞
(11.1)
where cm are the Fourier coefficients of x(t).
Let now N be an even positive integer and
tk =
2πk
,
N
k=−
N
N
, . . . , − 1.
2
2
Then x(tk ) is a response at tk i.e.
x(tk ) =
∞
X
cm e i
2πkm
N
.
(11.2)
m=−∞
Since ei2πkl = 1 for integers k and l then the series (11.2) can be rewritten as
∞
X
cm e
(m−lN )
i 2πk
N
=
m=−∞
∞
X
X
cm e i
2πk
(m−lN )
N
l=−∞ −N/2≤m−lN ≤N/2−1
=
∞
X
N/2−1
X
cn+lN ei
2πk
n
N
l=−∞ n=−N/2
N/2−1
=
X
e
n
i 2πk
N
∞
X
N/2−1
cn+lN =
l=−∞
n=−N/2
X
ei
2πk
n
N
Xn ,
(11.3)
n=−N/2
where Xn , n = −N/2, . . . , N/2 − 1 is given by
Xn =
∞
X
cn+lN .
(11.4)
l=−∞
In the calculation (11.3) we have used the fact that the series (11.1) converges absolutely. Combining (11.2) and (11.3) we obtain
N/2−1
xk := x(tk ) =
X
n=−N/2
58
Xn e i
2πkn
N
.
(11.5)
The formula (11.5) can be viewed as an inverse discrete Fourier transform and it
appeared quite naturally in the discretization of a continuous periodic signal. Moreover,
the formula (11.4) becomes the main property of this approach. Since
N/2−1
X
n=−N/2
e
i(k−m) 2πn
N
=
(
0, k − m 6= 0, ±N, ±2N, . . .
N, k − m = 0, ±N, ±2N, . . .
(11.6)
then we can solve the linear system (11.5) with respect to Xn for n = −N/2, . . . , N/2−1
and obtain
N/2−1
2πkn
1 X
xk e−i N .
(11.7)
Xn =
N
k=−N/2
Exercise 46. Prove (11.6) and (11.7).
Actually, the formulas (11.5) and (11.7) give us the inverse and direct discrete
Fourier transforms, respectively.
N/2−1
Definition 11.1. The sequence {Xn }n=−N/2 of complex numbers is called the discrete
N/2−1
Fourier transform (DFT) of the sequence {Yk }k=−N/2 if for each n = −N/2, . . . , N/2−1
we have
N/2−1
2πkn
1 X
(11.8)
Xn =
Yk e−i N .
N
k=−N/2
We use the symbol F for DFT and write
Xn = F(Yk )n
or simply X = F(Y ).
N/2−1
Definition 11.2. The sequence {Zk }k=−N/2 of complex numbers is called the inN/2−1
verse discrete Fourier transform (IDFT) of the sequence {Xn }n=−N/2 if for each k =
−N/2, . . . , N/2 − 1 we have
N/2−1
Zk =
X
Xn e i
2πkn
N
.
n=−N/2
We use the symbol F −1 for IDFT and write
Zk = F −1 (Xn )k
or simply Z = F −1 (X).
The properties of DFT and IDFT are collected in the following lemmas.
59
(11.9)
Lemma 11.1. The following equalities hold:
1) F −1 (F(Y )) = Y ;
2) F(F −1 (X)) = X;
3)
N/2−1
X
k=−N/2
1
F(Xn )k F(Yn )k =
N
N/2−1
X
n=−N/2
Proof. Using (11.6), (11.8) and (11.9) we have
N/2−1
F −1 (F(Y ))k =
=
X
n=−N/2
F(Yl )n ei
2πkn
N
=
1
N
N/2−1
X
n=−N/2


Xn Y n .


N/2−1
X
Yl e−i
2πln
N
l=−N/2
N/2−1
N/2−1
X
2πn(k−l)
1
1 X
Yl 
e i N  = Yk N = Yk .
N
N
l=−N/2
n=−N/2

 ei 2πkn
N
This proves part 1). Part 2) can be proved by the same manner.
Exercise 47. Prove part 3) of Lemma 11.1.
Corollary 1 (Parseval equality).
1
N
N/2−1
X
N/2−1
X
2
n=−N/2
|Xn | =
k=−N/2
|F(Xn )k |2 .
Remark. Due to periodicity of the complex exponential we may extend the values of
Xm , m = −N/2, . . . , N/2 − 1 periodically to any integer by
Xm+lN = Xm ,
l = 0, ±1, ±2, . . . .
(11.10)
N/2−1
Corollary 2. For sequence X = {Xn }n=−N/2 we define
N/2−1
Xrev = {XN −n }n=−N/2 .
Then
F −1 (X) = N F(Xrev ).
Proof. By Definition 11.2 and periodicity condition (11.10) we have
N/2−1
N/2−1
N F(Xrev )k =
=
X
XN −n e
−i 2πkn
N
Xn e
n=−N/2
Xn e i
=
X
Xn e i
n=−N/2
N/2−1
X
X−n e
2πkn
N
= F −1 (X)k .
60
=
X
Xn e i
2πkn
N
n=N/2
N/2−1
i 2πkn
N
n=−N/2+1
=
−N/2+1
−i 2πkn
N
n=−N/2
n=−N/2
N/2
X
=
X
2πkn
N
+ XN/2 eiπk − X−N/2 e−iπk
N/2−1
N/2−1
Definition 11.3. The convolution of sequences X = {Xn }n=−N/2 and Y = {Yn }n=−N/2
is defined as the sequence whose elements are given by
N/2−1
(X ∗ Y )k =
X
Xl Yk−l ,
(11.11)
l=−N/2
where Xn and Yn satisfy the periodicity condition (11.10).
Proposition. For any integer l it is true that
N/2−1
N/2−l−1
X
Ym e
−i 2πnm
N
X
=
Ym e−i
2πnm
N
.
m=−N/2
m=−N/2−l
Proof. The claim is trivial for l = 0. If l > 0 then
N/2−l−1
X
N/2−1
Ym e
−i 2πnm
N
=
m=−N/2−l
X
−N/2−1
Ym e
−i 2πnm
N
+
m=−N/2
=
−
N/2−1
Ym e
−i 2πnm
N
+
X
Ym−N e−i
X
Ym e−i
X
Ym e−i
2πnm
N
m=N/2−l
2πn
(m−N )
N
m=N/2−l
m=−N/2
N/2−1
−
Ym e
−i 2πnm
N
m=−N/2−l
N/2−1
X
N/2−1
X
N/2−1
X
Ym e
−i 2πnm
N
=
m=N/2−l
2πnm
N
m=−N/2
due to periodicity condition (11.10). If l < 0 then
N/2−1
N/2−l−1
X
Ym e
−i 2πnm
N
=
m=−N/2−l
=
N/2−l−1
X
Ym e
X
Ym e−i
−i 2πnm
N
+
X
Ym e−i
m=N/2
N/2−1
N/2−l−1
+
m=−N/2
X
−i 2πnm
N
−
X
m=−N/2
2πnm
N
m=N/2
N/2−l−1
−
Ym e
m=−N/2
2πnm
N
−N/2−l−1
X
Ym−N e
(m−N )
−i 2πn
N
N/2−1
=
X
Ym e−i
m=−N/2
m=N/2
due to periodicity condition (11.10). This proves the Proposition.
Corollary. For any integer l it is true that
N/2−l−1
X
N/2−1
Ym =
m=−N/2−l
X
m=−N/2
61
Ym .
2πnm
N
Ym e−i
2πnm
N
Lemma 11.2. The convolution (11.11) is symmetric i.e.
(X ∗ Y )k = (Y ∗ X)k
for any k = −N/2, . . . , N/2 − 1.
Proof. We have
N/2−1
(X ∗ Y )k =
X
k+1−N/2
X
Xl Yk−l =
l=−N/2
j=k+N/2
N/2−1+(k+1)
=
Yj Xk−j
X
N/2−1
X
Yj Xk−j =
j=−N/2+(k+1)
j=−N/2
Yj Xk−j = (Y ∗ X)k
by the above Corollary.
Lemma 11.3. For each n = −N/2, . . . , N/2 − 1 it is true that
1) F(X ∗ Y )n = N F(X)n F(Y )n ;
2) F −1 (X ∗ Y )n = F −1 (X)n F −1 (Y )n .
Proof. Using (11.11) we have
F(X ∗ Y )n
1
=
N
N/2−1
X
(X ∗ Y )k e
k=−N/2
N/2−1
=
−i 2πkn
N
1 X
Xl
N
l=−N/2
N/2−1
l=−N/2
N/2−l−1
X
N/2−1
N/2−1
X
2πkn
1 X
Xl
Yk−l e−i N
=
N
Ym e−i
k=−N/2
2πn
(m+l)
N
m=−N/2−l
2πnl
1 X
=
Xl e−i N
N
l=−N/2
N/2−l−1
X
Ym e−i
2πnm
N
.
(11.12)
m=−N/2−l
The above Proposition allows us to rewrite (11.12) as
F(X ∗ Y )n
N/2−1
2πnl
1 X
=
Xl e−i N
N
l=−N/2
N/2−1
X
Ym e−i
2πnm
N
m=−N/2
= N F(X)n F(Y )n .
Part 2) is proved in a similar manner.
Corollary. For each n = −N/2, . . . , N/2 − 1 it is true that
F −1 (X · Y )n =
1
F −1 (X) ∗ F −1 (Y ) n ,
N
N/2−1
where X · Y denotes the sequence {Xk · Yk }k=−N/2 .
62
(11.13)
Proof. Lemmas 11.1 and 11.3 imply that
e ∗ Ye = F −1 F X
e ∗ Ye
e · F(Ye ) .
X
= N F −1 F(X)
n
n
n
e := F −1 (X) and Ye := F −1 (Y ) we obtain easily from the latter equality
Denoting X
that
N F −1 (X · Y )n = F −1 (X) ∗ F −1 (Y ) n .
This finishes the proof.
Let us return to the continuous signal x(t), t ∈ [−π, π], which is represented by an
absolutely convergent trigonometric Fourier series (11.1). Formula (11.4) allows us to
obtain
N/2−1
N/2−1 ∞
N/2−1 X
X X
X
X
|Xn − cn | =
cn+lN − cn =
cn+lN n=−N/2 l=−∞
n=−N/2
n=−N/2 l6=0
N/2−1
≤
X X
n=−N/2 l6=0
|cn+lN | ≤
X
|ν|≥N/2
|cν |.
(11.14)
Similarly, we have
N/2−1
N/2−1
X
X
−1
2πkn 2πkn i
i
F (Xn )k −
(Xn − cn )e N cn e N = n=−N/2
n=−N/2
N/2−1
≤
X
n=−N/2
|Xn − cn | ≤
X
|ν|≥N/2
|cν |.
(11.15)
In the formulas (11.14) and (11.15) the numbers cn are the Fourier coefficients of the
N/2−1
N/2−1
signal x(t) and {Xn }n=−N/2 is the DFT of {x(tk )}k=−N/2 with tk = 2πk/N .
Theorem 20. If x(t) is periodic and belongs to Sobolev space W2m (−π, π) for some
m = 1, 2, . . . then
1
Xn = c n + o
(11.16)
N m−1/2
and
N/2−1
F
−1
(Xn )k =
X
cn e
n=−N/2
i 2πkn
N
+o
1
N m−1/2
uniformly in n and k from the set {−N/2, . . . , N/2 − 1}.
63
(11.17)
Proof. Using Hölder’s inequality we have
X
|ν|≥N/2

|cν | ≤ 
X
|ν|≥N/2
1/2 
|ν|2m |cν |2 

X
|ν|≥N/2
1/2
|ν|−2m 
.
The first sum in the right hand side tends to zero as N → ∞ due to Parseval equality for
function from Sobolev space W2m (−π, π). The second sum can be estimated precisely.
Namely, since for any m = 1, 2, . . . we have


X
|ν|≥N/2
|ν|
1/2
−2m 
≍
Z
∞
t
−2m
N/2
dt
1/2
≍ N −m+1/2
then (11.16) and (11.17) follow from the last estimate and (11.14) and (11.15), respectively.
Corollary. An unknown periodic function f ∈ W2m (−π, π), m = 1, 2, . . . can be recovered from its IDFT as
2πk
1
−1
f
= F (X)k + o
N
N m−1/2
uniformly in k from the set {−N/2, . . . , N/2 − 1}.
Exercise 48.
1) Show that
F
N/2−1
{ak }k=−N/2
n
=
(
2πn
a = ei N
2πn
−N/2 −aN/2
1
a 6= ei N
(−1)n a −i
2πn ,
N
1,
1−ae
N/2−1
2) If sequence Y = {Yk }k=−N/2 is real then show that
F(Y )n = F(Y )N −n .
64
N
12
Connection between discrete Fourier transform
and Fourier transform.
If function f (x) is integrable over the whole line i.e.
Z ∞
|f (x)|dx < ∞
−∞
then its Fourier transform is defined as
1
Ff (ξ) = fb(ξ) := √
2π
Z
∞
f (x)e−ixξ dx.
(12.1)
−∞
Similarly, inverse Fourier transform of an integrable function g(ξ) is defined as
Z ∞
1
−1
g(ξ)eixξ dξ.
(12.2)
F g(x) := √
2π −∞
Theorem 21 (Riemann-Lebesgue lemma). For any integrable function f (x) its Fourier
transform Ff (ξ) is continuous and
lim Ff (ξ) = 0.
ξ→±∞
Proof. Since eiπ = −1 then we have
Z ∞
Z ∞
1
1
−ixξ+iπ
fb(ξ) = − √
f (x)e
dx = − √
f (y + π/ξ)e−iξy dy.
2π −∞
2π −∞
This fact implies that
Hence
1
−2fb(ξ) = √
2π
1
2|fb(ξ)| ≤ √
2π
Z
∞
−∞
Z
(f (x + π/ξ) − f (x))e−iξx dx.
∞
−∞
|f (x + π/ξ) − f (x)|dx → 0
as |ξ| → ∞ since f is integrable on the whole line. This is a well-known property of
integrable functions. Continuity of fb(ξ) follows from the representation
Z ∞
1
b
b
f (x)e−ixξ (e−ixh − 1)dx
f (ξ + h) − f (ξ) = √
2π −∞
and its consequence
1
|fb(ξ + h) − fb(ξ)| ≤ √
2π
Z
|xh|<δ
|f (x)||e
−ixh
65
2
− 1|dx + √
2π
Z
|xh|>δ
|f (x)|dx := I1 + I2 .
For the first term I1 we have the estimate
Z
Z ∞
1
δ
I1 ≤ √
|f (x)|dx → 0,
|f (x)||xh|dx < √
2π |xh|<δ
2π −∞
For the second term I2 we have
2
I2 = √
2π
Z
|x|>δ/|h|
δ → 0.
|f (x)|dx → 0
as |h| → 0. If we choose δ = |h|1/2 then both I1 and I2 tend to zero as |h| → 0. This
finishes the proof.
If function f (x) has integrable derivatives f (k) (x) of order k = 0, 1, 2, . . . , m then
we say that f belongs to Sobolev space W1m (R).
Exercise 49. Prove that if f ∈ W11 (R) then limx→±∞ f (x) = 0.
Theorem 22 (Fourier inversion formula). Suppose that f belongs to W11 (R). Then
F −1 (Ff )(x) = f (x)
at any point x ∈ R.
Proof. First we prove that
Z
∞
f (x)b
g (x)dx =
−∞
Z
∞
−∞
fb(ξ)g(ξ)dξ
for any two integrable functions f and g. Indeed,
Z ∞
Z ∞
Z ∞
1
f (x)b
g (x)dx = √
f (x)
g(ξ)e−ixξ dξdx
2π
−∞
Z−∞
Z −∞
Z ∞
∞
∞
1
−ixξ
fb(ξ)g(ξ)dξ
g(ξ)
f (x)e
dxdξ =
= √
2π −∞
−∞
−∞
by Fubini’s theorem. Suppose now that g(ξ) is given by
(
1, |ξ| < n
g(ξ) =
0, |ξ| > n.
Its Fourier transform is equal to
−ixn
r
Z n
−ixξ n
ixn
e
e
1
e
1
1
2 sin(nx)
=√
e−ixξ dξ = √
−
.
=
gb(x) = √
−ix
π
x
2π −n
2π −ix −n
2π −ix
Thus, we have the equality
r Z ∞
Z n
2
sin(nx)
fb(ξ)dξ,
dx =
f (x)
π −∞
x
−n
66
where f ∈ W11 (R). Letting n → ∞ we obtain
r Z ∞
Z ∞
sin(nx)
2
fb(ξ)dξ = lim
dx.
f (x)
p.v.
n→∞
π −∞
x
−∞
√
We will prove that the limit in (12.3) is actually equal to 2πf (0). Since
Z ∞
sin(nx)
dx = π
x
−∞
(12.3)
then the limit in (12.3) can be rewritten as
r Z ∞
r Z ∞
√
2
2
sin(nx)
sin(nx)
2πf (0) + lim
dx =
dx
lim
f (x)
(f (x) − f (0))
n→∞
n→∞
π −∞
x
π −∞
x
r Z ∞
√
2
sin(t)
2πf (0) + lim
dt.
=
(f (t/n) − f (0))
n→∞
π −∞
t
It remains to show that the latter limit is equal to zero. In order to prove this fact we
split
Z
∞
sin(t)
dt
(f (t/n) − f (0))
t
−∞
Z
Z
sin(t)
sin(t)
=
(f (t/n) − f (0))
dt +
(f (t/n) − f (0))
dt := I1 + I2 .
t
t
|t|<1
|t|>1
Since f ∈ W11 (R) then it can be proved that
Z
|I1 | ≤ sup |f (t/n) − f (0)|
f is continuous. Therefore
sin(t) t dt → 0, n → ∞.
|t|<1
|t|<1
For the second term I2 we first change variables back as
I2 =
=
Z
Z
|z|>1/n
∞
1
n
sin(nz)
dz
z
′
′
Z z
Z −1
n
sin(ny)
sin(ny)
dy dz +
dy dz.
(f (z) − f (0))
y
y
−∞
0
(f (z) − f (0))
(f (z) − f (0))
Z
z
0
67
Integration by parts in these integrals leads to
∞
Z z
Z z
Z ∞
sin(ny) sin(ny)
′
I2 = (f (z) − f (0))
f (z)
dy −
dydz
y
y
0
0
1/n
1/n
−1/n Z −1/n
Z z
Z z
sin(ny) sin(ny)
′
dy dydz
+ (f (z) − f (0))
f (z)
−
y
y
0
0
−∞
−∞
Z ∞
Z 1/n
sin(ny)
sin(ny)
= ( lim f (z) − f (0))
dy − (f (1/n) − f (0))
dy
z→∞
y
y
0
0
Z nz
Z ∞
sin(t)
dtdz
f ′ (z)
−
t
0
1/n
Z −1/n
Z −∞
sin(ny)
sin(ny)
+ (f (−1/n) − f (0))
dy − ( lim f (z) − f (0))
dy
z→−∞
y
y
0
0
Z nz
Z −1/n
sin(t)
′
dtdz.
f (z)
−
t
0
−∞
Since limz→±∞ f (z) = 0 (see Exercise 49) and since f is continuous we obtain (when
n → ∞)
Z ∞
Z nz
π
sin(t)
′
I2 → −f (0) − lim
dtdz
f (z)
n→∞
2
t
1/n
0
Z −1/n
Z nz
sin(t)
π
′
dtdz
f (z)
− f (0) − lim
n→∞
2
t
−∞
0
Z ∞
Z nz
Z −1/n
Z nz
sin(t)
sin(t)
′
′
= −πf (0) − lim
f (z)
dtdz − lim
dtdz
f (z)
n→∞ 1/n
n→∞ −∞
t
t
0
0
Z ∞
Z 0
π
π
′
f (z) dz +
= −πf (0) −
f ′ (z) dz
2
2
0
−∞
π
π
= −πf (0) + f (0) + f (0) = 0.
2
2
Here we used again the fact that limz→±∞ f (z) = 0 and Lebesgue’s dominated convergence theorem. Thus, (12.3) transforms to
Z ∞
√
fb(ξ)dξ = 2πf (0)
p.v.
−∞
or equivalently,
F −1 (Ff )(0) = f (0).
In order to prove Fourier inversion formula for any x ∈ R let us note that
Z ∞
1
[
\
fx (y)(ξ) = f (x + y)(ξ) = √
f (x + y)e−iyξ dy
2π −∞
Z ∞
1
= √
f (z)e−izξ eixξ dz = eixξ fb(ξ).
2π −∞
68
Since fx (0) = f (x) then
f (x) = fx (0) = F −1 (Ffx )(0)
Z ∞
Z ∞
1
1
=√
(Ffx )(ξ)dξ = √
eixξ fb(ξ)dξ = F −1 (Ff )(x).
2π −∞
2π −∞
This finishes the proof.
Remark. As a by-product of the above proof we record the limit
Z
1 ∞
sin(nx)
lim
dx = f (0)
f (x)
n→∞ π −∞
x
for any f ∈ W11 (R).
Lemma 12.1. If f belongs to Sobolev space W1m (R) for some m = 1, 2, . . . then
1
(12.4)
fb(ξ) = o
|ξ|m
when |ξ| → ∞.
Proof. Since f ∈ W1m (R) then f ′ , f ′′ , . . . , f (m−1) ∈ W11 (R). By Exercise 49 we have
lim f (k) (x) = 0
x→±∞
for each k = 0, 1, . . . , m − 1. This fact and repeated integration by parts give us
∞
Z
Z
Z ∞
1 ∞ ′
1 ∞ ′
e−ixξ
−ixξ
−ixξ
f (x)
f (x)e
dx =
f (x)e−ixξ dx
+
f (x)e
dx =
−iξ
iξ
iξ
−∞
−∞
−∞
−∞
∞
Z ∞
−ixξ
e
1
= −
f ′ (x)
+
f ′′ (x)e−ixξ dx
2
2
(iξ)
(iξ)
−∞
−∞
Z ∞
Z ∞
1
1
′′
−ixξ
=
f (x)e
dx = · · · =
f (m) (x)e−ixξ dx
(iξ)2 −∞
(iξ)m −∞
1
= o
|ξ|m
due to Riemann-Lebesgue lemma.
The equality (12.4) allows us to consider (with respect to accuracy of calculations)
the Fourier transform only from the interval (−R, R) with R > 0 large enough i.e. we
may neglect the values of Ff (ξ) for |ξ| > R. This simplification justifies the following
approximation of the inverse Fourier transform:
Z R
1
∗
Ff (ξ)eixξ dξ.
(12.5)
f (x) := √
2π −R
At the same time and without loss of generality we may assume that the function f (x)
has compact support. In that case it can be proved that Ff (ξ) is smooth function for
which (12.4) holds.
69
◦
Definition 12.1. We say that f ∈ W1m (−R, R) if f ∈ W1m (R) and f ≡ 0 for x ∈
/
(−R, R).
◦
Theorem 23. Suppose that f ∈ W1m (−R, R) is supported in a fixed interval [a, b] ⊂
(−R, R) with R > 0 large enough and some m = 2, 3, . . .. Then
f (x) =
r
N/2−1
X
x(2n+1)
2
1
1
2n + 1
i m/(m+2)
+O
eN
Ff
π N m/(m+2)
N m/(m+2)
N (2m−2)/(m+2)
n=−N/2
(12.6)
uniformly in x ∈ (−R, R) for R = N
2/(m+2)
and even N .
Proof. Since f (x) = 0 for x ∈
/ (−R, R) then using Fourier inversion formula (see
Theorem 22) we have
Z ∞
Z R
Z
1
1
1
ixξ
ixξ
∗
Ff (ξ)e dξ− √
Ff (ξ)e dξ = √
Ff (ξ)eixξ dξ.
f (x)−f (x) = √
2π −∞
2π −R
2π |ξ|>R
Lemma 12.1 implies then that
∗
f (x) − f (x) = o
1
Rm−1
.
(12.7)
We divide the interval [−R, R] into N +1 subintervals [ξn , ξn+1 ], n = −N/2, . . . , N/2−1
such that
−R = ξ−N/2 < ξ−N/2+1 < · · · < ξN/2 = R,
where
2R
.
N
ξn =
2Rn
,
N
ξn∗ =
ξn + ξn+1
R
= (2n + 1).
2
N
Set also
ξn+1 − ξn =
Then we obtain
1
f (x) = √
2π
∗
Z
R
−R
Ff (ξ)eixξ dξ
N/2−1
X
1
∗ 2R
= √
Ff (ξn∗ )eixξn
N
2π n=−N/2
N/2−1 Z ξn+1
X
1
∗
+ √
Ff (ξ)eixξ − Ff (ξn∗ )eixξn dξ
2π n=−N/2 ξn
r
3
N/2−1
2R X
R
ixR(2n+1)/N
.
Ff (R(2n + 1)/N )e
+O
=
πN
N2
n=−N/2
70
(12.8)
Exercise 50. Prove that
1
√
2π
X Z
N/2−1
n=−N/2
ξn+1
ξn
 O R32 , supp f = [a, b]
∗
ixξ
∗ ixξn
N5 dξ =
Ff (ξ)e − Ff (ξn )e
O R 2 , supp f ⊂ (−R, R)
N
uniformly in x ∈ [−R, R].
Hint. Use the Taylor expansion for the smooth function Ff (ξ)eixξ at the point ξn∗ .
If we combine (12.8) with (12.7) and choose R = N 2/(m+2) then we obtain (12.6).
This finishes the proof.
Remark. The main part of (12.6) represents some kind of DFT. In order to reconstruct
f at any point x ∈ [−R, R] we need to know only the Fourier transform of this unknown
function at the points
2n + 1
,
N m/(m+2)
n = −N/2, . . . , N/2 − 1,
where m is the smoothness index of f . What is more, the formula (12.6) shows us that
it is effective if
2m − 2
m
>
m+2
m+2
◦
or m > 2. It means that f must belong to Sobolev space W1m (−R, R) with some
m ≥ 3.
71
13
Some applications of discrete Fourier transform.
At first we prove Poisson summation formula.
Definition 13.1. Let f be a function such that
X
lim
f (x + 2πn)
N →∞
|n|≤N
exists pointwise in x ∈ R. Then
∞
X
fp (x) :=
f (x + 2πn)
(13.1)
n=−∞
is called the periodization of f .
Remark. It is clear that fp (x) is periodic with period 2π. Hence we will consider it
only on the interval [−π, π].
Theorem 24 (Poisson summation formula). Suppose that f ∈ L1 (R). Then fp (x)
from (13.1) is finite almost everywhere, satisfies fp (x + 2π) = fp (x) almost everywhere
and is integrable on the interval [−π, π]. The Fourier coefficients of fp (x) are given by
Z π
1
1
(13.2)
fp (x)e−imx dx = √ Ff (m).
2π −π
2π
If in addition
∞
X
m=−∞
then
|Ff (m)| < ∞
∞
1 X
Ff (m)eimx .
f (x + 2πn) = √
2π m=−∞
n=−∞
∞
X
(13.3)
In particular, fp (x) is continuous and we have the Poisson identity
∞
X
∞
1 X
f (2πn) = √
Ff (m).
2π
n=−∞
m=−∞
Proof. Since f ∈ L1 (R) then
Z
π
−π
|fp (x)|dx ≤
=
Z
π
∞
X
|f (x + 2πn)|dx =
−π n=−∞
∞ Z π+2πn
X
n=−∞
−π+2πn
|f (t)|dt =
72
Z
∞ Z
X
n=−∞
∞
−∞
(13.4)
π
−π
|f (x + 2πn)|dx
|f (t)|dt < ∞.
This shows that fp is finite almost everywhere and integrable on [−π, π]. Applying the
same calculation to fp (x)e−imx allows us to integrate term by term to obtain
Z π
∞
X
1
fp (x)e
dx =
f (x + 2πn)e−imx dx
2π
−π
−π
n=−∞
Z
Z π+2πn
∞
∞
π+2πn
X 1
1 X 1
−im(t−2πn)
√
=
f (t)e
dt = √
f (t)e−imt dt
2π
2π
2π
−π+2πn
−π+2πn
n=−∞
n=−∞
Z ∞
1
1
1
f (t)e−imt dt = √ Ff (m).
= √ √
2π 2π −∞
2π
1
cm (fp ) =
2π
Z
π
−imx
Now, if the series
∞
X
m=−∞
|Ff (m)|
converges then it is equivalent to the fact that
∞
X
m=−∞
|cm (fp )| < ∞.
Thus, fp (x) can be represented by its Fourier series at least almost everywhere (and
we can redefine fp (x) so that this representation holds pointwise) i.e.
fp (x) =
∞
X
cm (fp )eimx .
m=−∞
It means that (see (13.1))
∞
X
∞
1 X
f (x + 2πn) = √
Ff (m)eimx .
2π m=−∞
n=−∞
Finally, set x = 0 to obtain the Poisson identity (13.4).
Example. If
1 − x2
e 4t , x ∈ R,
4πt
where t > 0 is a parameter then it is very well-known that
f (x) = √
1
2
Ff (ξ) = √ e−tξ .
2π
Formula (13.3) transforms in this case to
∞
∞
X
(x+2πn)2
1 X −tm2 imx
1
−
4t
√
e
e
e
=
2π m=−∞
4πt n=−∞
73
and the Poisson identity transforms to
r
∞
∞
X
π X −π2 n2 /t
2
e
=
e−tm .
t n=−∞
m=−∞
As an application of the Poisson summation formula we consider the problem of
reconstructing a band-limited signal from its values on the integers.
Definition 13.2. A signal f (t) is called band-limited if it has a representation
1
f (t) = √
2π
Z
2πλ
F (ξ)eitξ dξ,
(13.5)
−2πλ
where λ is a positive parameter and F is some integrable function.
Remark. If we set F (ξ) = 0 for |ξ| > 2πλ then (13.5) is the inverse Fourier transform
of F ∈ L1 (R). In that case f is bounded and continuous.
Theorem 25 (Whittaker, Shannon, Boas). Suppose that F ∈ L1 (R) and F (ξ) = 0 for
|ξ| > 2πλ. If λ ≤ 1/2 then for any t ∈ R we have
f (t) =
∞
X
f (n)
n=−∞
sin π(t − n)
,
π(t − n)
where the fraction is equal to 1 when t = n. If λ > 1/2 we have
Z
∞
X
sin
π(t
−
n)
2
1
f (n)
+1
|F (ξ)|dξ.
f (t) −
≤ √
π(t
−
n)
π
2π
π<|ξ|<2πλ
n=−∞
Proof. Let
Fp (ξ) =
∞
X
(13.6)
(13.7)
F (ξ + 2πn)
n=−∞
be the periodization of F . The formula (13.5) shows that
F(F )(t) = f (−t),
where F(F ) denotes the Fourier transform of F (ξ). Hence, by the Poisson summation
formula, we have (see (13.3))
∞
∞
1 X
1 X
imξ
f (−m)e
=√
f (m)e−imξ .
F (ξ + 2πn) = √
2π
2π
m=−∞
m=−∞
n=−∞
∞
X
74
Since any trigonometric Fourier series can be integrated term by term we obtain
Z π
Z π X
Z π
∞
∞
1 X
itξ
itξ
e−i(m−t)ξ dξ
f (m)
Fp (ξ)e dξ =
F (ξ + 2πn)e dξ = √
2π m=−∞
−π
−π n=−∞
−π
π
∞
1 X
ei(t−m)ξ = √
f (m)
, m 6= t
i(t − m) −π
2π m=−∞
∞
ei(t−m)π − e−i(t−m)π
1 X
f (m)
= √
i(t − m)
2π m=−∞
∞
sin π(t − m)
2π X
f (m)
.
= √
π(t − m)
2π m=−∞
Now, if λ ≤ 1/2 then F (ξ) for |ξ| ≤ π is equal to its periodization Fp (ξ) (see Definition
13.1) and
Z π
√
Fp (ξ)eitξ dξ = 2πf (t).
−π
These equalities imply immediately that
f (t) =
∞
X
f (n)
n=−∞
sin π(t − n)
π(t − n)
so that (13.6) is proved. If λ > 1/2 then we cannot expect that F (ξ) = Fp (ξ) for
|ξ| > π but using Definition 13.1 we have
Z π
Z
1
1
itξ
f (t) = √
F (ξ)e dξ + √
F (ξ)eitξ dξ
2π −π
2π π<|ξ|<2πλ
Z
∞
X
1
sin π(t − n)
e
+√
F (ξ)eitξ dξ,
=
f (n)
π(t
−
n)
2π π<|ξ|<2πλ
n=−∞
where
1
fe(n) = √
2π
Therefore,
f (t) =
∞
X
n=−∞
f (n)
Z
π
F (ξ)einξ dξ.
−π
sin π(t − n)
π(t − n)
Z
∞ sin π(t − n)
X
1
e
+√
F (ξ)eitξ dξ.
+
f (n) − f (n)
π(t
−
n)
2π
π<|ξ|<2πλ
n=−∞
Here the middle series is equal to
∞ Z
1 X
sin π(t − n)
inξ
F (ξ)e dξ
−√
π(t − n)
2π n=−∞ π<|ξ|<2πλ
75
or
1
−√
2π
Z
Exercise 51. Prove that
∞
X
sin π(t − n) inξ
e
F (ξ)
π(t
−
n)
π<|ξ|<2πλ
n=−∞
!
dξ.
∞
X
2
sin π(t − n) inξ
e = − eitξ .
π(t − n)
π
n=−∞
Hint. Show that
2
− eitξ
π
=
sin π(t − n)
+
f (t) =
f (n)
π(t
−
n)
n=−∞
2
1
√ +√
π 2π
2π
cn
Using Exercise 51 we have
∞
X
Thus
sin π(t − n)
.
π(t − n)
Z
F (ξ)eitξ dξ.
π<|ξ|<2πλ
Z
∞
X
1
sin
π(t
−
n)
2
f (n)
+1
|F (ξ)|dξ.
≤ √
f (t) −
π(t
−
n)
π
2π
π<|ξ|<2πλ
n=−∞
This proves the theorem.
Theorem 25 shows that in order to reconstruct a band-limited signal f (t) it is
enough to know the values of this signal at integers f (n). In turn, to evaluate f (n)
it is enough to use IDFT of F (ξ), see (13.5). Indeed, let us assume without loss of
generality that λ = 1/2. Then
Z π
1
f (n) = √
F (ξ)einξ dξ.
2π −π
If F is smooth enough (say F ∈ C 2 [−π, π]) then formula (12.8) gives (R = π and
N >> 1)
√
√
N/2−1
N/2−1
πn(2k+1)
2π X
2π i πn X
1
1
i
i 2πkn
N
N
N
f (n) =
=
,
e
Fk e
+O
+O
Fk e
2
N
N
N
N2
k=−N/2
k=−N/2
where Fk denotes the value of F (ξ) at the point π(2k + 1)/N . Therefore, up to the
accuracy of calculations,
√
2π i πn −1
e N F (Fk )n
f (n) ≈
N
i.e. for even integer N large enough
√
∞
X
sin π(t − n)
2π i πn −1
e N F (Fk )n
.
f (t) ≈
N
π(t − n)
n=−∞
76
Index
L2 -Hölder condition, 38
Lp -modulus of continuity, 23
p-integrable function, 1
absolute convergence, 11
band-limited signal, 74
Besov space, 32
Bessel’s inequality, 26
conjugate Poisson kernel, 13
convolution, 16, 61
Dirichlet kernel, 43
discrete Fourier transform, 59
even extension, 8
even function, 2
Fejér kernel, 18
Fejér means, 18
Fourier coefficient, 9
Fourier cosine series, 8
Fourier inversion formula, 66
Fourier series, 7
Fourier transform, 65
Fouries sine series, 8
Fubini’s theorem, 1
full variation, 3
function of bounded variation, 3
fundamental period, 1
generalized Minkowski’s inequality, 2
Hölder condition with exponent α, 4
Hölder space, 4, 33
Hölder’s inequality, 1
mean square distance, 25
Minkowski’s inequality, 2
modulus of continuity, 4
mutually orthogonal functions, 5
Nikol’skii space, 33
odd extension, 8
odd function, 2
orthogonal functions, 5
Parseval equality, 16, 26, 60
partial sum of Fourier series, 18
periodic extension, 1
periodic function, 1
periodization, 72
piecewise continuous function, 3
pointwise convergence, 11
Poisson identity, 72
Poisson kernel, 13
Poisson summation formula, 72
Riemann-Lebesgue lemma, 65
Riesz-Fischer theorem, 27
right and left limit, 3
sequence space, 31
Sobolev space, 4, 32, 66
square-integrable functions, 25
Stieltjes integral, 4
summation by parts formula, 13
tail sum, 15
trigonometric series, 7
uniform convergence, 11
integrable function, 1
inverse discrete Fourier transform, 59
inverse Fourier transform, 65
Laplace equation, 13
77