# Fourier series and the discrete Fourier transform

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```							Fourier series and the discrete Fourier transform
802647S

Lecture Notes
1st Edition
Second printing

Valeriy Serov
University of Oulu
May 7, 2010

Edited by Markus Harju
Contents
1 Preliminaries                                                        1

2 Formulation of Fourier series                                        7

3 Fourier coeﬃcients and their properties                             11

4 Convolution and Parseval equality                                   16

e
5 Fej´r means of Fourier series. Uniqueness of the Fourier series.    18

6 Riemann-Lebesgue lemma                                              22

7 Fourier series of square-integrable function. Riesz-Fischer theorem. 25

o
8 Besov and H¨lder spaces                                             32

9 Absolute convergence. Bernstein and Peetre theorems.                38

10 Dirichlet kernel. Pointwise and uniform convergence.               43

11 Formulation of discrete Fourier transform and its properties.      58

12 Connection between discrete Fourier transform and Fourier transform. 65

13 Some 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
mπx                  mπx                     mπx
sin        ,          cos       ,           ei       L    ,      m = 1, 2, . . .
L                    L
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
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,                   1 < p < ∞, 1 < p′ < ∞
p p
then their product is integrable on [a, b] and
b                                      b                       1/p            b               1/p′
p                                   p′
|f (x)g(x)|dx ≤                        |f (x)| dx                             |g(x)| dx          .
a                                      a                                      a

o
This inequality is called the H¨lder’s inequality for integrals. The Fubini’s theorem
states that
b       d                                    d       b
F (x, y)dydx =                               F (x, y)dxdy,
a       c                                    c       a

1
where F (x, y) ∈ L1 ((a, b) × (c, d)) is positive.
n
If f1 , f2 , . . . , fn are p-integrable on [a, b] for 1 ≤ p < ∞ then so is                                                       j=1   fj and
n                  p       1/p       n                                              1/p
b                                                            b
p
fj (x) dx                   ≤                          |fj (x)| dx                   .
a       j=1                                   j=1            a

o
This inequality is called Minkowski’s inequality. As a consequence of H¨lder’s inequal-
ity we obtain the generalized Minkowski’s inequality
p       1/p                                                        1/p
b        d                                              d             b
p
F (x, y)dy dx                     ≤                          |F (x, y)| dx                     dy.         (1.2)
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
a+T                            T
f (x)dx =                 f (x)dx                                                 (1.3)
a                              0
for any a ∈ R.
Proof. Let ﬁrst a > 0. Then
a+T                                    a+T                        a
f (x)dx =                             f (x)dx −                f (x)dx
a                                      0                              0
T                          a+T                               a
=                 f (x)dx +                    f (x)dx −                       f (x)dx .
0                          T                                 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
a+T                                   0                      a+T
f (x)dx =                        f (x)dx +                  f (x)dx
a                                     a                     0
0                      T                        T
=                 f (x)dx +              f (x)dx −                     f (x)dx
a                     0                            a+T
T                          0                     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),                 x ∈ D(f ).

2
Lemma 1.2. If f is integrable on any ﬁnite interval and if it is even then
a                                  a
f (x)dx = 2                       f (x)dx
−a                                 0

for any a > 0. Similarly, if f is odd then
a
f (x)dx = 0
−a

for any a > 0.
Proof. Since
a                              a                            0
f (x)dx =                      f (x)dx +                    f (x)dx
−a                             0                            −a

then changing variables in the second integral we obtain
a                              a                            a
f (x)dx =                      f (x)dx +                    f (−x)dx.
−a                          0                            0

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
|f (xj ) − f (xj−1 )| ≤ c0 .
j=1

The number                                                          n
Vab (f )    :=           sup                   |f (xj ) − f (xj−1 )|            (1.5)
x0 ,x1 ,...,xn
j=1

is called the full variation of f on the interval [a, b]. For any x ∈ [a, b] we can also
deﬁne Vax (f ) by (1.5).
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
b
g(x)df (x),                                         (1.6)
a

where f ∈ BV [a, b] and g is an arbitrary continuous function. The integral (1.6) can
be understood as
b                                  n
g(x)df (x) = lim                   g(ξj )(∆j α − ∆j β),
a                      ∆→0
j=1

where f = α − β, ∆j α = α(xj ) − α(xj−1 ), 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             |f (x + h) − f (x)|,         h > 0.               (1.7)
{x∈[a,b]:x+h∈[a,b]}

Deﬁnition 1.6. A function f belongs to H¨lder space C α [a, b], 0 < α ≤ 1, if
o

ωh (f ) ≤ Chα

o
with some constant C > 0. This inequality is called the H¨lder condition with exponent
α.
1
Deﬁnition 1.7. We say that f belongs to Sobolev space Wp (a, b), 1 ≤ p < ∞ if
f ∈ Lp (a, b) and there is g ∈ Lp (a, b) such that
x
f (x) =                 g(t)dt + C                                 (1.8)
a

with some constant C.

4
1
Lemma 1.3. Suppose that f ∈ Wp (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 (1.8) holds with some
constant C. Hence for ﬁxed x ∈ [a, b] with x + h ∈ [a, b] we have
x+h
f (x + h) − f (x) =                                     g(t)dt.
x

It implies that
x+h
|f (x + h) − f (x)| ≤                                 g(t)dt → 0,                  h→0
x

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
n                                  n             xj                         n             xj                       b
|f (xj ) − f (xj−1 )| =                          g(t)dt ≤                                 |g(t)|dt =              |g(t)|dt.
j=1                             j=1           xj−1                          j=1        xj−1                     a

b
Hence, Deﬁnition 1.5 is satisﬁed with constant c0 = a |g(t)|dt.
o
If 1 < p < ∞ then using H¨lder’s inequality for integrals we obtain for h > 0 that
x+h                                     x+h             1/p′        x+h                  1/p
p
|f (x + h) − f (x)| ≤                      |g(t)|dt ≤                               dt                         |g(t)| dt
x                                       x                           x
b                     1/p
1−1/p                               p
≤ h                                 |g(t)| dt               ,
a

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 f ∈ Lp (a, b), 1 < p < ∞ implies that f ∈ L1 (a, b) then we may conclude
1
that any f ∈ Wp (a, b), 1 < p < ∞ is of bounded variation.
1
Remark. Since any f ∈ Wp (a, b), 1 ≤ p < ∞ is continuous then the constant C in (1.8)
is equal to f (a).

Deﬁnition 1.8. Two functions u and v are said to be orthogonal on a ≤ x ≤ b if the
product uv is integrable and
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 ≤ x ≤ b.

5
Lemma 1.4. The functions
mπx                   mπx
1,     sin       ,        cos          ,             m = 1, 2, . . .
L                     L
form a mutually orthogonal set on the interval [−L, L] as well as on the interval [0, 2L].
In fact,
L                                                2L
mπx     nπx                                               mπx     nπx                            0,     m=n
cos     cos     dx =                                cos           cos     dx =                                  ,          (1.9)
−L      L       L                             0                   L       L                             L,     m=n
L                                                    2L
mπx     nπx                                       mπx     nπx
cos           sin     dx =                            cos       sin     dx = 0,                         (1.10)
−L              L       L                         0               L       L
L                                            2L
mπx     nπx                                   mπx     nπx                                0,     m=n
sin       sin     dx =                        sin       sin     dx =                                                (1.11)
−L             L       L                     0               L       L                                 L,     m=n
and
L                                  2L                             L                                     2L
mπx                                 mπx                               mπx                                mπx
sin       dx =                      sin       dx =                 cos          dx =                     cos       dx = 0. (1.12)
−L            L                   0               L                 −L              L                  0               L
Proof. Due to Lemma 1.1 it is enough to prove the equalities (1.9), (1.10), (1.11) and
(1.12) only for integrals over [−L, L]. Let us derive, for example, (1.10). Using the
equality
1
cos α sin β = (sin(α + β) − sin(α − β))
2
we have for m = n that
L                                                      L                                                 L
mπx     nπx      1                                     (m + n)πx      1                                   (m − n)πx
cos       sin     dx =                                 sin             dx −                               sin             dx
−L            L       L       2                       −L                L          2                       −L              L
L                                    L
1         − cos (m+n)πx
L                               1        − cos (m−n)πx
L
=              (m+n)π
−                   (m−n)π
=0
2                                                  2
L                          −L                          L        −L

since cosine is even. If m = n we have
L                                                    L
mπx     nπx      1                                  2mπx
cos        sin     dx =                              sin        dx = 0
−L               L       L       2                      −L           L
since sine is odd. Other identities can be proved in a similar manner and are left to
nπx
Remark. This lemma holds trivially for the functions ei                                              L    , n = 0, ±1, ±2, . . .. Namely,
L                                      2L
nπx         mπx                         nπx         mπx                0,  n=m
ei    L    e−i    L    dx =             ei    L    e−i    L    dx =
−L                                  0                                            2L, n = m.

6
2     Formulation of Fourier series
Let us consider a series of the form
∞
a0              mπx          mπx
+     am cos     + bm sin     .                                      (2.1)
2    m=1
L            L

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              mπx          mπx
f (x) :=      +     am cos     + bm sin     .                                (2.2)
2    m=1
L            L

To determine am and bm we proceed as follows: assuming that the integration can be
legitimately carried out term by term (it will be, for example, if ∞ (|am | + |bm |) <
m=1
∞) we obtain
L                                  L                     ∞        L
nπx      a0                nπx                                   mπx     nπx
f (x) cos     dx =               cos     dx +     am                 cos       cos     dx
−L            L       2          −L      L       m=1               −L          L       L
∞            L
mπx     nπx
+          bm          sin       cos     dx
m=1           −L          L       L

for each ﬁxed n = 1, 2, . . . . It follows from the orthogonality relations (1.9), (1.10) and
(1.12) that the only nonzero term on the right hand side is the one for which m = n
in the ﬁrst summation. Hence
L
1                       nπx
an =               f (x) cos       dx.                           (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
L
1                         nπx
bn =               f (x) sin       dx.                           (2.4)
L        −L                L
Using (1.12) we can easily obtain that
L
1
a0 =                f (x)dx.                                (2.5)
L     −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
∞
a0              mπx          mπx
f (x) ∼    +     am cos     + bm sin     .                                  (2.6)
2    m=1
L            L

7
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               mπx
f (x) ∼    +     am cos                             (2.7)
2    m=1
L
and if f is odd then
∞
mπx
f (x) ∼           bm sin         .                                        (2.8)
m=1
L
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
∞
a0
f (x) ∼        +     (am cos mx + bm sin mx) ,                                          (2.9)
2    m=1

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.

Thus, 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.

Thus, h(x) is odd and its Fourier (sine) series (2.8) represents f on [0, L].

3. Deﬁne a function f (t) on the interval [−π, π] as the superposition
tL L
f (t) = f                +      .
2π   2
If f (0) = f (L) then we may extend f to be periodic with period L. Then
π                       π                                        L
1                         1                 tL L                1 2π
a0 (f ) =               f (t)dt =                f       +         dt =                  f (x) dx
π     −π                  π       −π        2π   2              π L    0
L
2
=               f (x) dx := a0 (f ),
L     0

8
L
2                         2mπx
am (f ) = (−1)m                     f (x) cos        dx = (−1)m am (f ),
L    0                     L
and
L
2                         2mπx
bm (f ) = (−1)m                     f (x) sin        dx = (−1)m bm (f ).
L     0                    L
Hence,
∞
a0
f (t) ∼    +     (am cos mt + bm sin mt)
2    m=1

and, at the same time,
∞
a0                    2mπx          2mπx
f (x) ∼    +     (−1)m am cos      + bm sin                           ,
2    m=1
L             L

tL
where a0 , am and bm are the same and x =                   2π
+ L.
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
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) ∼                cn einx ,                        (2.10)
n=−∞

where the coeﬃcients cn = cn (f ) are equal to

 a2 + bn ,

n
2i
n = 1, 2, . . .
a0
cn (f ) = 2 ,            n=0                                           (2.11)
 a−n b−n

2
− 2i , n = −1, −2, . . . .

The formulas (2.3), (2.4), (2.5) and (2.11) imply that
π
1
cn (f ) =                 f (x)e−inx dx                     (2.12)
2π       −π

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)

Exercise 3. Prove formulas (2.10), (2.11), (2.12) and (2.13).

9
Exercise 4. Find the    Fourier series of

−1,
          −π ≤ x < 0
a) sgn(x) = 0,        x=0


1,       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

1 − x, 0 ≤ x ≤ 1
f (x) =
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
ak sin kx.
k=1

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
ak cos kx.
k=1

10
3      Fourier coeﬃcients and their properties
Deﬁnition 3.1. We say that the trigonometric series
∞
cn einx
n=−∞

a) converges pointwise if for each x ∈ [−π, π] the limit

lim            cn einx
N →∞
|n|≤N

exists,
b) converges uniformly in x ∈ [−π, π] if the limit

lim            cn einx
N →∞
|n|≤N

exists uniformly,
c) converges absolutely if the limit

lim               |cn |
N →∞
|n|≤N

exists or, equivalently, if
∞
|cn | < ∞.
n=−∞

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.
π                                 π
1                                 1
|cn (f )| =             f (x)e−inx dx ≤                   |f (x)|dx,   (3.1)
2π   −π                           2π   −π

where the upper bound does not depend on n. Suppose that a sequence {cn }∞
n=−∞ is
such that                         ∞
|cn | < ∞.
n=−∞

Then the series                                ∞
cn einx
n=−∞

converges uniformly in x ∈ [−π, π] and deﬁnes a continuous function
∞
f (x) :=             cn einx                        (3.2)
n=−∞

11
whose Fourier coeﬃcients are the sequence {cn }∞               ∞
n=−∞ = {cn (f )}n=−∞ . More generally,
suppose that
∞
|n|k |cn | < ∞
n=−∞

for some integer k > 0. Then the series (3.2) deﬁnes a function which is k times
diﬀerentiable with                    ∞
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
∞
1
=               rn einx                      (3.4)
1 − reix         n=0

and this series converges absolutely. Using the deﬁnition of the Fourier coeﬃcients we
obtain                            π
n     1       e−inx
r =                   dx, n = 0, 1, 2, . . .
2π −π 1 − reix
and                                     π
1           einx
0=                       dx,           n = 1, 2, . . . .
2π   −π    1 − reix
From the representation (3.4) we may conclude also that
∞                               ∞
1 − r cos x                  1                        n  1 1
= Re                      =     r cos nx = +       r|n| einx        (3.5)
1 − 2r cos x + r2            1 − reix         n=0
2 2 n=−∞

and
∞                              ∞
r sin x                     1                       n       i
= Im                        =     r sin nx = −        r|n| sgn(n)einx . (3.6)
1 − 2r cos x + r2             1 − reix          n=1
2 n=−∞

Exercise 8. Verify the formulas (3.5) and (3.6).
The formulas (3.5) and (3.6) can be rewritten as
∞
1 − r2
r|n| einx =                      := Pr (x)                 (3.7)
n=−∞
1 − 2r cos x + r2

and                     ∞
2r sin x
−i          sgn(n)r|n| einx =                       := Qr (x).            (3.8)
n=−∞
1 − 2r cos x + r2

12
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
π                                                         π
|n|      1                 −inx                          |n|      1
r         =             Pr (x)e       dx,     −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 x2 + x2 < 1, where x1 + ix2 = reix with 0 ≤ r < 1 and x ∈ [−π, π].
1    2

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                                                N
a k ∆k b = a N b N − a M b M −                    bk−1 ∆k a                (3.9)
k=M +1                                            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
∞
cn einx                                          (3.10)
n=0

converges for any x ∈ [−π, π] \ {0}.
n    ikx
Proof. Let bn =            k=0 e    .   Since

1 − eix(n+1)
bn =                ,      x=0
1 − eix
then
2            1
|bn | ≤              =
|1 − eix |   | sin x |
2

13
for x ∈ [−π, π] \ {0}. Applying (3.9) shows that for M < N we have
N                             N                k                  k−1                  N
ikx                                           ilx              ilx
ck e         =                ck               e       −         e        =            c k ∆k b
k=M +1                        k=M +1            l=0                 l=0                k=M +1
N
= c N bN − c M bM −                              bk−1 ∆k c.
k=M +1

Thus
N                                                                   N
ikx
ck e            ≤ cN |bN | + cM |bM | +                             |bk−1 |∆k c
k=M +1                                                             k=M +1
N
1
≤                     cN + cM +                        |ck − ck−1 |
| sin x |
2                                   k=M +1
1                            2cM
=          x (cN + cM + cM − cN ) =           →0
| sin 2 |                      | sin x |
2

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.
2 |x|
Proof. If π ≥ |x| ≥ δ > 0 then sin x ≥
2                        π 2
δ
≥ π.
Theorem 1 implies that, for example, the series
∞                                         ∞
cos nx                                      sin nx
and
n=1
log(2 + n)                            n=1
log(2 + n)

converge for all x ∈ [−π, π] \ {0}.

Modulus of continuity and tail sum
For the trigonometric series (2.10) with
∞
|cn | < ∞
n=−∞

we introduce the tail sum by

En :=                |ck |,        n = 0, 1, 2, . . . .                             (3.11)
|k|>n

14
There is a good connection between (1.7) and (3.11). Indeed, if f (x) denotes this series
and h > 0 we have
∞
|f (x + h) − f (x)| ≤             |cn ||einh − 1| ≤                |cn ||einh − 1| +             |cn ||einh − 1|
n=−∞                           |n|h≤1                         |n|h>1

≤ h                 |n||cn | + 2               |cn | := I1 + I2 ,
|n|≤[1/h]                  |n|>[1/h]

where [x] denotes the entire part of x. If we denote [1/h] by Nh then I2 = 2ENh and
Nh
I1 = h            |n| E|n|−1 − E|n| = −h                    l (El − El−1 ) .
|n|≤Nh                                       l=1

Using (3.9) in the latter sum we obtain
Nh                                                               Nh
I1 = −h Nh ENh − 0 · E0 −               En−1 (n − (n − 1))               = −hNh ENh + h                 En−1 .
n=1                                                           n=1

1
Since hNh = h[1/h] ≤ h ·    h
= 1 then these formulas for I1 and I2 imply that
Nh                                 Nh −1
1
ωh (f ) ≤ 2ENh − hNh ENh + h                  En−1 ≤ 2ENh          +               En .           (3.12)
n=1
Nh      n=0

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 ) =               1
(3.13)
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.

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
H¨lder space C α [−π, π].
o

15
4     Convolution and Parseval equality
Let the trigonometric series (2.10) be such that
∞
|cn | < ∞.
n=−∞

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 integrable and
π                        ∞               π                           ∞
1                     1                                    inx
f (x)g(x)dx =                cn (f )         g(x)e        dx =             cn (f )c−n (g),   (4.1)
2π   −π               2π     n=−∞               −π                       n=−∞

where integration of the series term by term is justiﬁed by the uniform convergence of
Fourier series. Putting g = f in (4.1) yields
π                     ∞                                ∞
1               2
|f (x)| dx =             cn (f )c−n (f ) =                 |cn (f )|2          (4.2)
2π   −π                  n=−∞                           n=−∞

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
π                              ∞
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
π
1
(f1 ∗ f2 )(x) =                f1 (y)f2 (x − y)dy                              (4.3)
2π      −π

and where the integral is convergent for almost every x.

16
Proof. We note ﬁrst that the convolution (4.3) is well deﬁned as an L1 -function by the
Fubini theorem. Indeed,
π        π                                         π                π−y
|f1 (y)| · |f2 (x − y)|dy dx =            |f1 (y)|             |f2 (z)|dz dy
−π    −π                                          −π                −π−y
π                 π
=           |f1 (y)|         |f2 (z)|dz dy
−π                −π

by Lemma 1.1. The Fourier coeﬃcients of the convolution (4.3) are equal to
π                                      π   π
1                   −inx            1
cn (f1 ∗ f2 ) =              (f1 ∗ f2 )(x)e     dx =                    f1 (y)f2 (x − y)dy e−inx dx
2π −π                              (2π)2 −π    −π
π             π
1
=                  f1 (y)        f2 (x − y)e−inx dx dy
(2π)2 −π            −π
π             π−y
1
=                  f1 (y)           f2 (z)e−in(y+z) dz dy
(2π)2 −π            −π−y
π                   π
1
=                  f1 (y)e−iny          f2 (z)e−inz 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) =           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 x2 + x2 = r2 < 1 with x1 + ix2 = reix .
1    2

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→1−0

exists uniformly in x.

17
5        e
Fej´r means of Fourier series. Uniqueness of the
Fourier series.
Let us denote the partial sum of Fourier series of f ∈ L1 (−π, π) by

SN (f ) :=               cn (f )einx
|n|≤N

e
for each N = 0, 1, 2, . . .. The Fej´r means are deﬁned by

S0 (f ) + · · · + SN (f )
σN (f ) :=                             .
N +1
Writing this out in detail we see that
N                                          N
ikx
(N + 1)σN (f ) =                      ck (f )e       =                  ck (f )eikx
n=0 |k|≤n                        |k|≤N n=|k|

=            ck (f )(N + 1 − |k|)eikx
|k|≤N

which gives the useful representation

|k|
σN (f ) =                 1−               ck (f )eikx .                    (5.1)
N +1
|k|≤N

e
The Fej´r kernel is
|k|
KN (x) :=                  1−                  eikx .                     (5.2)
N +1
|k|≤N

The sum (5.2) can be calculated precisely as
2
1                     +1
sin N2 x
KN (x) =                                       .                     (5.3)
N +1                 sin x
2

Exercise 15. Prove the identity (5.3).

Exercise 16. Prove that                       π
1
KN (x)dx = 1.                                     (5.4)
2π     −π

We can rewrite σN (f ) from (5.1) also as
∞
|k|
σN (f ) =          1[−N,N ] (k) 1 −                        ck (f )eikx ,
k=−∞
N +1

18
where
1, |k| ≤ N
1[−N,N ] (k) =
0, |k| > N.
Let us assume now that f is periodic. Then Exercise 12 and Theorem 2 lead to
π                                           ∞
1                                                                                     |k|
f (y)KN (x − y)dy =                         1[−N,N ] (k) 1 −                           ck (f )eikx .            (5.5)
2π     −π                                         k=−∞
N +1
Exercise 17. Prove (5.5).
e
Hence, the Fej´r 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
π                                          1/p
p
lim                    |σN (f )(x) − f (x)| dx                     = 0.                              (5.7)
N →∞             −π

If, in addition, f has right and left limits f (x0 ± 0) at a point x0 ∈ [−π, π] then
1
lim σN (f )(x0 ) = (f (x0 + 0) + f (x0 − 0)) .                 (5.8)
N →∞                2
Proof. Let us ﬁrst prove (5.7). Indeed, (5.4) and (5.6) give
π                                          1/p             π                                                1/p
|σN (f )(x) − f (x)|p dx                      =            |(f ∗ KN )(x) − f (x)|p dx
−π                                                           −π
π            π                                              π                               p         1/p
1                                              1
=                          KN (y)f (x − y)dy −                            KN (y)f (x)dy dx
−π   2π    −π                                       2π   −π
π         π                                                p           1/p
1
=                               KN (y)(f (x − y) − f (x))dy dx
2π           −π        −π
π                                                            p          1/p
1
≤                                 KN (y)(f (x − y) − f (x))dy dx
2π       −π           |y|<δ
π                                                                   p          1/p
1
+                                     KN (y)(f (x − y) − f (x))dy dx                                  := I1 + I2 .
2π             −π       π≥|y|>δ

Using the generalized Minkowski’s inequality we obtain that
π                                           1/p
1
I1 ≤                        KN (y)                   |f (x − y) − f (x)|p dx                      dy
2π      |y|<δ                        −π
π                                         1/p              π
1
≤ sup                     |f (x − y) − f (x)|p dx                                    KN (y)dy
|y|<δ           −π                                               2π     −π
π                                         1/p
p
= sup                     |f (x − y) − f (x)| dx                      →0                                      (5.9)
|y|<δ           −π

19
as δ → 0 since f ∈ Lp (−π, π). Quite similarly,
π                              1/p
1
I2 ≤                        KN (y)                    |f (x − y) − f (x)|p dx         dy
2π     π≥|y|>δ                         −π
π                        1/p
p                   1
≤ 2             |f (x)| dx                                  KN (y)dy                 (5.10)
−π                              2π     π≥|y|>δ

since f is periodic. The next step is to note that
2           2
y             2 |y|              δ
sin2     ≥                          ≥
2             π 2                π
for π ≥ |y| > δ. That’s why (5.3) leads to

1     1                          1  π2
KN (y) ≤            ·                   y   ≤       · 2
N + 1 sin2               2
N +1 δ

and
1                               π 2 2(π − δ)     π2       π2
KN (y)dy ≤           ·        < 2        < √ →0
2π    π≥|y|>δ                   2πδ 2 N + 1    δ (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))
2
π
1                          1
=       KN (y) f (x0 − y) − (f (x0 + 0) + f (x0 − 0)) dy
2π −π                       2
π
1
=       KN (y)g(y)dy,                                                                   (5.11)
2π −π
where
1
g(y) = f (x0 − y) − (f (x0 + 0) + f (x0 − 0)) .
2
Since Fej´r kernel KN (y) is even we can rewrite the right hand side of (5.11) as
e
π
1
KN (y)h(y)dy,
2π     0

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
π
1
lim                   KN (y)h(y)dy = 0.
N →∞ 2π       0

20
We will proceed as in the proof of (5.7) i.e. we split
π                              δ                                                π
1                             1                                               1
KN (y)h(y)dy =                 KN (y)h(y)dy +                                   KN (y)h(y)dy := I1 + I2 .
2π   0                        2π    0                                         2π     δ

The ﬁrst term can be estimated as
π
1                                                        1
|I1 | ≤     sup |h(y)|                            KN (y)dy =        sup |h(y)| → 0
2π 0≤y≤δ                          0                       2 |y|≤δ

as δ → +0 due to (5.12). For I2 we have
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].

Corollary 2. Any periodic Lp -function, 1 ≤ p < ∞, can be approximated by the
trigonometric polynomials (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
π
lim                |σN (f )(x) − f (x)| dx = 0
N →∞          −π

by (5.7) then in the case when all Fourier coeﬃcients are zero, we have
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
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
π                              π+x                                       π
f (x + z)e−inz dz =                f (y)e−in(y−x) dy = einx             f (y)e−iny dy      (6.2)
−π                            −π+x                                   −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,
π                            π+π/n                       π
−iny                              −iny
2πcn (f ) =         f (y)e      dy =                 f (y)e       dy =        f (t + π/n)e−int e−iπ dt
−π                           −π+π/n                       −π

by Lemma 1.1. Hence
π
−4πcn (f ) =               (f (t + π/n) − f (t)) e−int dt.                     (6.3)
−π

If f is continuous on the interval [−π, π] then

sup |f (t + π/n) − f (t)| → 0,              n → ∞.
t∈[−π,π]

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
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
ε
sup lim |cn (f )| ≤      .
n→∞               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
π
lim            f (x + z)g(z)e−inz dz = 0
n→∞        −π

and
π                                                π
lim           f (x + z)g(z) sin(nz)dz = lim                    f (x + z)g(z) cos(nz)dz = 0
n→∞     −π                                           n→∞   −π

uniformly in x ∈ [−π, π].

Exercise 18. Prove this Corollary-

o
Exercise 19. Show that if f satisﬁes the H¨lder condition with exponent α ∈ (0, 1]
−α
then cn (f ) = O(|n| ) as n → ∞.

o
Exercise 20. Suppose that f satisﬁes the H¨lder 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
π                                 1/p
p
ωp (δ) := sup                 |f (x + h) − f (x)| dx              .
|h|≤δ         −π

π
1
|cn (f )| ≤                  |f (x + π/n) − f (x)| dx
4π   −π
π                                     1/p
(2π)1−1/p                                               p              1
≤                               |f (x + π/n) − f (x)| dx                ≤ (2π)−1/p ωp (π/n),
4π                 −π                                               2

o
where we have used H¨lder’s inequality in the penultimate step.

Exercise 22. Suppose that ωp (δ) ≤ Cδ α for some C > 0 and α > 1. Prove that f is
constant almost everywhere.
Hint. First show that ωp (2δ) ≤ 2ωp (δ), 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) ∼            cn einx ,
n=−∞

where cn are the Fourier coeﬃcients cn (f ). The series in the right hand side is consid-
ered formally in the sense that we know nothing about its convergence. However, the
limit                          π                        π
lim          cn (f )einx dx =     f (x)dx                   (6.4)
N →∞   −π |n|≤N                           −π

exists. Indeed,
π                                     π                                π
cn (f )einx dx = c0 (f )        dx +               cn (f )        einx dx = 2πc0 (f )
−π |n|≤N                              −π          0<|n|≤N               −π
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. Riesz-
Fischer theorem.
The set of square-integrable functions L2 (−π, π) is a linear Euclidean space equipped
with inner product
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
π
1
|f (x) − g(x)|2 dx = (f − g, f − g)L2 (−π,π) .
2π      −π

In particular, if g(x) =              |n|≤N   bn einx is a trigonometric polynomial then
π                                     π                                        π
1                                       1                    1
|f (x) − g(x)|2 dx =                |f (x)|2 dx +                              |g(x)|2 dx
2π    −π                                2π −π                2π                     −π
π
1
−    2Re       f (x)g(x)dx
2π       −π
π                                        π
1                    1
=        |f (x)|2 dx +                                     bn einx            bk e−ikx dx
2π −π                2π                     −π |n|≤N                |k|≤N
π
1
−    2Re                   f (x)           bn e−inx dx
2π                 −π            |n|≤N
π                                              π
1                              1
=                   |f (x)|2 dx +                     |bn |2        dx
2π      −π                     2π                       −π
|n|≤N

− 2Re                  bn cn (f )
|n|≤N
π
1
=                   |f (x)|2 dx +               |bn |2 − 2Re                bn cn (f )
2π      −π                     |n|≤N                         |n|≤N

+              |cn (f )|2 −              |cn (f )|2
|n|≤N                   |n|≤N
π
1
=                   |f (x)|2 dx −               |cn (f )|2 +             |bn − cn (f )|2 .
2π      −π                     |n|≤N                    |n|≤N

This equality has the following consequences:
1. The minimum error is
π                                          π
1                                         1
min                          |f (x)−g(x)|2 dx =                         |f (x)|2 dx−                |cn (f )|2 (7.2)
g(x)=   |n|≤N   bn einx   2π    −π                                  2π     −π                       |n|≤N

and it is attained when bn = cn (f ).

25
2. For N = 1, 2, . . . it is true that
π
21
|cn (f )| ≤                        |f (x)|2 dx
2π                −π
|n|≤N

and, in particular,
∞                                      π
1
|cn (f )|2 ≤                       |f (x)|2 dx.                       (7.3)
n=−∞
2π       −π

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 con-
verges in L2 (−π, π) i.e.
2
π
1
lim                   f (x) −               cn (f )einx dx = 0                              (7.4)
N →∞ 2π       −π                   |n|≤N

and the Parseval equality
π                              ∞
1                     2
|f (x)| dx =                    |cn (f )|2                             (7.5)
2π    −π                             n=−∞

holds.
Proof. By Bessel’s inequality (7.3) we have for any f ∈ L2 (−π, π) that
2
π
1
cn (f )einx −                 cn (f )einx 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
π
1
lim                    F (x) −                cn (f )einx 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
                     
π                                    π
2πcn (F ) =           F (x)e−inx dx =                       F (x) −                ck (f )eikx  e−inx dx
−π                                    −π                   |k|≤N
π
+             ck (f )            ei(k−n)x dx.
|k|≤N                 −π

26
o
If N > |n| then the last sum is equal to 2πcn (f ). Thus, by H¨lder’s inequality,

π
2π|cn (F ) − cn (f )| ≤                     F (x) −            ck (f )eikx dx
−π              |k|≤N
                                             2   1/2
√               π
≤         2π            F (x) −             ck (f )eikx dx            →0
−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 → ∞.

Corollary (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
∞
|n|2 |cn (f )|2 < ∞.                                             (7.6)
n=−∞
1
Then f ∈ W2 (−π, π) with Fourier series for f ′ (x) as
∞
′
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
∞
g(x) ∼               incn (f )einx .
n=−∞

Integrating term by term we obtain
π
g(x)dx = 0.
−π

x
Let F (x) :=    −π
g(t)dt. Then, for n = 0, we have
π          x                                      π                 π
1                                                     1
cn (F ) =                           g(t)dt e−inx dx =                      g(t)              e−inx dx dt
2π        −π      −π                                  2π    −π             t
π                                                         π
1                        e−inπ e−int                     1 1                                  1
=                g(t)            +                    dt =                 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                             x
f (x) =     g(t)dt + constant,
−π
2                 1
where g ∈ L (−π, π). Therefore, f ∈ W2 (−π, π) and
∞
′
f (x) = g(x) ∼               incn (f )einx .
n=−∞

Corollary. Under the conditions of Theorem 7, it is true that
∞
|cn (f )| < ∞.
n=−∞

Proof. Due to (7.6) we have
∞
1                                1          1
|cn (f )| = |c0 (f )| +           |cn (f )| ≤ |c0 (f )| +              |n|2 |cn (f )|2 +                     < ∞,
n=−∞                                    n=0
2   n=0
2   n=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
π
1
cn (f )einx − f (x) dx =                    |cn (f )|2 .                   (7.8)
2π    −π |n|≤N                                            |n|>N

Exercise 23. Prove (7.8).
Using Parseval’s equality again we have
π                                         ∞                                             ∞
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

|cn (f )|2 = O(N −2α ),             N = 1, 2, . . .                             (7.10)
|n|>N

with 0 < α < 1 if and only if
∞
|einh − 1|2 |cn (f )|2 = O(|h|2α )                                        (7.11)
n=−∞

for |h| small enough.

28
Proof. From (7.9) we have for any integer M > 0 that
π
1
|f (x + h) − f (x)|2 dx ≤             n2 h2 |cn (f )|2 + 4               |cn (f )|2       (7.12)
2π       −π                                   |n|≤M                              |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
In :=              |ck (f )|2
|k|≤n

we have
M                                                  M
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 .
1≤|n|≤M                                          n=1                                                n=2

By hypothesis,
I∞ − In = O(n−2α ),               n = 1, 2, . . . .
Thus,
M
2
M IM −                   (2n − 1) I∞ + O((n − 1)−2α )
n=2
M                         M
2                 −2α
=M            I∞ + O(M           ) − I∞         (2n − 1) −                (2n − 1)O((n − 1)−2α )
n=2                          n=2
M                             M
= O(M 2−2α ) − I∞ M 2 −                    (2n − 1) −                     (2n − 1)O((n − 1)−2α )
n=2                            n=2
=1
= O(M 2−2α ) + O(M 2−2α ) = O(M 2−2α ).
Exercise 24. Prove that
M
1.         n=1 (2n       − 1) = M 2
M
2.         n=2   O((2n − 1)(n − 1)−2α ) = O(M 2−2α ) for 0 < α < 1.
Combining these two estimates we may conclude from (7.12) that there is C > 0
such that           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
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
∞
(1 − cos(nh))|cn (f )|2 ≤ Ch2α
n=−∞

with some C > 0. Integrating this inequality with respect to h over the interval
[0, l], l > 0 we have
∞                          l                                      l
2
|cn (f )|                (1 − cos(nh))dh ≤ C                    h2α dh
n=−∞                     0                                      0

or                           ∞
sin(nl)
|cn (f )|2 l −                        ≤ Cl2α+1
n=−∞
n
or                            ∞
sin(nl)
|cn (f )|2 1 −                    ≤ Cl2α .
n=−∞
nl
It follows that
sin(nl)           1
Cl2α ≥            |cn (f )|2 1 −                        ≥                 |cn (f )|2 .   (7.13)
nl             2
|n|l≥2                                                 |n|l≥2

Taking l = 2/N for integer N > 0 in (7.13) yields

|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
π
|f (x + h) − f (x)|2 dx ≤ Ch2
−π

with some C > 0. Prove that
∞
|n|2 |cn (f )|2 < ∞
n=−∞

1
and, therefore, f ∈ W2 (−π, π).

30
For any integrable function f periodic on the interval [−π, π] let us introduce the
mapping
f → {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 de-
notes all integer numbers and the sequence space lp (Z) consists of sequences {bn }∞
n=−∞
for which                             ∞
|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
∞                         π                p′ /p
p′                       p
|cn (f )| ≤ cp           |f (x)| dx           < ∞.
n=−∞                       −π

31
8               o
Besov and H¨lder spaces
In this chapter we will consider 2π-periodic functions f which are deﬁned via trigono-
metric Fourier series in L2 (−π, π) as
∞
f (x) ∼             cn (f )einx ,              (8.1)
n=−∞

where the Fourier coeﬃcients cn (f ) satisfy the Parseval equality
π                       ∞
1                   2
|f (x)| dx =             |cn (f )|2 ,      (8.2)
2π    −π                       n=−∞

that is, (8.1) can be understood in the sense of L2 (−π, π) as
2
π
lim              f (x) −            cn (f )einx dx = 0.        (8.3)
N →∞       −π              |n|≤N

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
1
proved (see Theorem 7 and Exercise 25) that periodic f belongs to W2 (−π, π) if and
only if
∞
|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
α
W2 (−π, π)

for some α ≥ 0 if
∞
|n|2α |cn (f )|2 < ∞.                  (8.4)
n=−∞

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
∞
                   |n|2α |cn (f )|2            < ∞.   (8.5)
j=0       2j ≤|n|<2j+1

32
Deﬁnition 8.3. We say that a 2π-periodic function f belongs to Nikol’skii space
α
H2 (−π, π)

for some α ≥ 0 if
sup                        |n|2α |cn (f )|2 < ∞.       (8.6)
j=0,1,2,...
2j ≤|n|<2j+1

Deﬁnition 8.4 (See also Deﬁnition 1.6).      1. We say that a 2π-periodic function f
α
o
belongs to H¨lder 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 = 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 con-
stant C > 0 such that for each n = 1, 2, . . . we have

|m|k |cm (f )| ≤ Cn−α                   (8.8)
|m|≥n

o
with some integer k ≥ 0 and some 0 < α < 1 then f belongs to H¨lder space
k+α
C     [−π, π].
The deﬁnitions 8.1-8.3 imply the following equalities and imbeddings:
α              α
1. B2,2 (−π, π) = W2 (−π, π), α ≥ 0.
0
2. W2 (−π, π) = L2 (−π, π).
α              α              α
3. B2,1 (−π, π) ⊂ B2,θ (−π, π) ⊂ H2 (−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 < ∞.
0
5. L2 (−π, π) ⊂ H2 (−π, π).

Exercise 26. Prove imbeddings 3, 4 and 5.

More imbeddings are formulated in the following theorems.

Theorem 9. If α ≥ 0 then
α
C α [−π, π] ⊂ W2 (−π, π).

33
Proof. If α = 0 then
0
C[−π, π] ⊂ L2 (−π, π) = W2 (−π, π).

If α > 0 then Deﬁnition 8.4 implies that
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. Using Parseval
equality we obtain
∞
|n|2k |cn (f )|2 |einh − 1|2 ≤ C|h|2(α−k)
n=−∞

or, equivalently,
∞
2            |n|2k |cn (f )|2 (1 − cos(nh)) ≤ C|h|2(α−k) .
n=−∞

It suﬃces to consider h > 0. If we integrate the last inequality with respect to h > 0
from 0 to l then      ∞
sin(nl)
|n|2k |cn (f )|2 1 −          ≤ Cl2(α−k) .
n=−∞
nl

For |n|l ≤ 2 we obtain
|n|2k |cn (f )|2 ≤ Cl2(α−k)
|n|l≤2

or, equivalently,
|n|2k l2(k−α) |cn (f )|2 ≤ C,
|n|l≤2

where constant C > 0 does not depend on l. Since l ≤ 2/|n| and 2(k − α) < 0 then it
follows that
|n|2α |cn (f )|2 ≤ C
|n|≤2/l

for any l > 0. Letting l → 0+ we obtain

|n|2α |cn (f )|2 < ∞
|n|≤∞

α
i.e. f ∈ W2 (−π, π). This ﬁnishes the proof.

Corollary. If α ≥ 0 then
α
C α [−π, π] ⊂ H2 (−π, π).                    (8.9)

34
Theorem 10. Assume that α > 1/2 and that α − 1/2 is not an integer. Then
α
H2 (−π, π) ⊂ C α−1/2 [−π, π].                                             (8.10)
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} = 1/2. We will assume
k
ﬁrst that {α} = 0, i.e. α = k is integer and k ≥ 1. If f ∈ H2 (−π, π) then there is a
constant C > 0 such that
|m|2k |cm (f )|2 ≤ C                                      (8.11)
2j ≤|m|<2j+1

for each j = 0, 1, 2, . . .. Let us estimate the tail (8.8). Indeed, by Cauchy-Schwarz-
Buniakowsky inequality and (8.11),
∞
k−1
|m|      |cm (f )| ≤                               |m|k−1 |cm (f )|
|m|≥n                               j=j0    2j ≤|m|<2j+1
2j0 ∼n
                                  1/2                            1/2
∞
                                                              1 
≤                                    |m|2k |cm (f )|2        
m2
j=j0        2j ≤|m|<2j+1                                2j ≤|m|<2j+1
2j0 ∼n
                   1/2
∞                                                ∞
√                                  1              √
≤          C                                        ≤        C            2−j/2 ≤ C2−j0 /2 .
m2
j=j0      2j ≤|m|<2j+1                           j=j0
2j0 ∼n                                           2j0 ∼n

Here and throughout, 2j0 ∼ n means that 2j0 ≤ n < 2j0 +1 . Therefore, we obtain
|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].
α
If α > 1/2 is not an integer and α − 1/2 is not an integer then for f ∈ H2 (−π, π)
we have instead of (8.11) the estimate
|m|2k+2{α} |cm (f )|2 ≤ C,                                    (8.12)
2j ≤|m|<2j+1

where k = [α], 0 < {α} < 1 and {α} = 1/2. If k = 0 then 1/2 < α = {α} < 1.
Repeating now the above procedure we obtain easily
                        1/2            1/2
∞
|cm (f )| ≤                                |m|2α |cm (f )|2                               |m|−2α 
|m|≥n                   j=j0      2j ≤|m|<2j+1                                   2j ≤|m|<2j+1
2j0 ∼n
                            1/2
∞                                                   ∞
≤ C                                 |m| −2α 
≤C              2−(α−1/2)j ≤ Cn−(α−1/2)
j=j0        2j ≤|m|<2j+1                             j=j0
2j0 ∼n                                                2j0 ∼n

35
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

|m|k−1 |cm (f )|
|m|≥n
                                       1/2                             1/2
∞
≤                            |m|2k+2{α} |cm (f )|2                         |m|−2−2{α} 
j=j0     2j ≤|m|<2j+1                                    2j ≤|m|<2j+1
2j0 ∼n
∞
≤C              2−j−j{α} 2j/2 ≤ Cn−1/2−{α} .
j=j0
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:

|m|k |cm (f )|
|m|≥n
                                      1/2                            1/2
∞
≤                             |m|2k+2{α} |cm (f )|2                       |m|−2{α} 
j=j0       2j ≤|m|<2j+1                                  2j ≤|m|<2j+1
2j0 ∼n
∞
≤C             2−({α}−1/2)j ≤ Cn−{α}+1/2 .
j=j0
2j0 ∼n

It means that f ∈ C k+{α}−1/2 [−π, π] = C α−1/2 [−π, π]. Hence, this theorem is com-
pletely proved.
Corollary 1. Assume that α = k + 1/2 for some integer k ≥ 1. Then
α
H2 (−π, π) ⊂ C β−1/2 [−π, π]                                  (8.13)
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
1
cn (f ) =                     , n = 0,      c0 (f ) = 1
|n| log(1 + |n|)
1/2
deﬁnes a function from Besov space B2,θ (−π, π) for any 1 < θ < ∞ but not for θ = 1.

36
Exercise 29. Prove that the Fourier series (8.1) with coeﬃcients
1
cn (f ) =                         , n = 0,   c0 (f ) = 1
|n|3/2   log(1 + |n|)
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
1
cn (f ) =             β
, n = 0,      c0 (f ) = 1.
|n|2 log (1 + |n|)

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                           ∞
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 H¨lder space
o
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 the-
orems.
α
We prove ﬁrst that if 2π-periodic function f belongs to Nikol’skii space H2 (−π, π) for
some 0 < α < 1 in the sense of Deﬁnition 8.3 then it is equivalent to the L2 -H¨lder
o
condition of order α, i.e.
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 = 0 small enough. Indeed, due to Parseval
equality we have
π                                             ∞
1                                     2
|f (x + h) − f (x)| dx =                           |cn (f )|2 |einh − 1|2
2π     −π                                        n=−∞

≤             |n|2 |h|2 |cn (f )|2 + 4              |cn (f )|2 , (9.2)
|n|≤2j0                               |n|≥2j0

1
where j0 is chosen so that 2j0 ≤                      |h|
≤ 2j0 +1 . The ﬁrst sum on the right-hand side of
(9.2) is estimated from above as
j0
2                2         2                  2
|h|             |n| |cn (f )|        ≤ |h|                                 |n|2 |cn (f )|2
|n|≤2j0                                       j=0 2j ≤|n|<2j+1
j0
2                     2−2α
≤ |h|                    2j+1                           |n|2α |cn (f )|2
j=0                       2j ≤|n|<2j+1
j0                                           j +2
2                     j+1                (22−2α ) 0 − 22−2α
≤ C|h|                    22−2α            = C|h|2
j=0
22−2α − 1
2−2α
2−2α                     1
≤ C|h|2 2j0                        ≤ C|h|2                       = C|h|2α          (9.3)
|h|
if 0 < α < 1. We have used this condition for α since we considered geometric
progression with multiplier 22−2α = 1. The second sum on the right-hand side of (9.2)
is estimated from above as
∞                                                                  ∞
2α   −2α             2
4                         |n| |n|          |cn (f )|         ≤ 4               2−2αj                  |n|2α |cn (f )|2
j=j0 2j ≤|n|<2j+1                                                  j=j0            2j ≤|n|<2j+1
∞
≤ C               2−2αj ≤ C2−2j0 α ≤ C(2|h|)2α ≤ C|h|2α
j=j0

1
since |h| ≤ 2j0 +1 and the criterion of Deﬁnition 8.3 holds. Thus, (9.1) is proved.
Conversely, if the L2 -H¨lder condition (9.1) is fulﬁlled then Theorem 8 implies for each
o

38
N = 1, 2, . . . the inequality
|cn (f )|2 ≤ CN −2α
|n|≥N

with the same α as in (9.1). But this leads to the inequality
N 2α                |cn (f )|2 ≤ C,
N ≤|n|<2N

where the constant C is independent of N . Thus, we obtain for any integer N > 0
that
|n|2α |cn (f )|2 ≤ C.
N ≤|n|<2N
α
Since N is arbitrary we may conclude that f ∈ H2 (−π, π) for 0 < α ≤ 1 in the sense
of Deﬁnition 8.3. Therefore, the L2 -H¨lder condition (9.1) can be considered as an
o
α
equivalent deﬁnition of Nikol’skii space H2 (−π, π) for 0 < α < 1.
α
Exercise 33. Prove that if f belongs to Nikol’skii space H2 (−π, π) for any non-integer
2
α > 0 in the sense of Deﬁnition 8.3 then the following L -H¨lder condition holds:
o
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 -
o
H¨lder condition with 1/2 < α ≤ 1. Then its trigonometric Fourier series converges
absolutely, i.e.
∞
|cn (f )| < ∞.
n=−∞
2                                            α
Proof. Since the L -H¨lder condition (9.1) is equivalent to f ∈ H2 (−π, π) for 0 < α < 1
o
then there is a constant C > 0 such that
|n|2α |cn (f )|2 ≤ C                                       (9.4)
2j ≤|n|<2j+1

for each j = 0, 1, 2, . . .. Hence we have
∞                                ∞
|cn (f )| = |c0 (f )| +                        |n|α |cn (f )||n|−α
n=−∞                              j=0 2j ≤|n|<2j+1
                               1/2                              1/2
∞
≤ |c0 (f )| +                             |n|2α |cn (f )|2                          |n|−2α 
j=0      2j ≤|n|<2j+1                                 2j ≤|n|<2j+1
∞                                         ∞
√                                       √                              j
≤ |c0 (f )| +         C         2−αj 2j/2 = |c0 (f )| +         C         2−(α−1/2)          <∞
j=0                                       j=0

since α > 1/2. Thus, theorem is proved.

39
α                α
Corollary. Theorem 11 holds for spaces C α [−π, π], B2,θ (−π, π) and H2 (−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
                                  1/2
∞
                  |n||cn (f )|2         < ∞.                              (9.5)
j=0       2j ≤|n|<2j+1

Hence we have
∞                                ∞
|cn (f )| = |c0 (f )| +                        |n|1/2 |cn (f )||n|−1/2
n=−∞                             j=0 2j ≤|n|<2j+1
                                1/2                       1/2
∞
≤ |c0 (f )| +                             |n||cn (f )|2                      |n|−1 
j=0        2j ≤|n|<2j+1                         2j ≤|n|<2j+1
                                1/2
∞
1/2
≤ |c0 (f )| +                             |n||cn (f )|2       2−j 2j
j=0        2j ≤|n|<2j+1
                                1/2
∞
= |c0 (f )| +                             |n||cn (f )|2      <∞
j=0        2j ≤|n|<2j+1

due to (9.5). This proves the theorem.
Corollary. It is true that
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
α                        1/2
H2 (−π, π) ⊂ B2,1 (−π, π)
if α > 1/2.
1
Theorem 13. Assume that 2π-periodic function f belongs to Sobolev space Wp (−π, π)
with some 1 < p < ∞. Then its trigonometric Fourier series converges absolutely.

40
Proof. Since
1             1
Wp1 (−π, π) ⊂ Wp2 (−π, π)
1
for 1 ≤ p2 < p1 , then we may assume without loss of generality that f ∈ Wp (−π, π)
with 1 < p ≤ 2. Then there is a function g ∈ Lp (−π, π) with 1 < p ≤ 2 such that
x                                  π
f (x) =         g(t)dt + f (−π),                   g(t)dt = 0.                           (9.6)
−π                                 −π

As we know, (9.6) leads to
1
cn (f ) =      cn (g),       n = 0.
in
Since g ∈ Lp (−π, π) with 1 < p ≤ 2 then the results of Chapter 7 give

∞                    1/p′
p′
|cn (g)|              < ∞,                                        (9.7)
n=−∞

1        1
where   p
+   p′
o
= 1. The facts (9.6), (9.7) and H¨lder’s inequality imply that

∞                                                                                     1/p                     1/p′
1                                         1                        p′
|cn (f )| = |c0 (f )|+             |cn (g)| ≤ |c0 (f )|+                                    |cn (g)|           <∞
n=−∞                            n=0
|n|                                 n=0
|n|p         n=0

since 1 < p ≤ 2. This ﬁnishes the proof.
1
Remark. For Sobolev space W1 (−π, π) this theorem is not valid, i.e. there is a func-
1
tion f from W1 (−π, π) with absolutely divergent trigonometric Fourier series. More
precisely, we will prove in the next chapter that the function
∞
sin nx
f (x) :=                                                                    (9.8)
n=1
n log(1 + n)

1
belongs to Sobolev space W1 (−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).
1
Theorem 14. Suppose that f ∈ W1 (−π, π) ∩ C α [−π, π] with some 0 < α < 1. Then
its trigonometric Fourier series converges absolutely.
1
Proof. Since f ∈ W1 (−π, π) then f is of bounded variation. The periodicity of f
implies that
π                                               π    N
1                                           1
|f (x + h) − f (x)|2 dx =                            |f (x + kh) − f (x + (k − 1)h)|2 dx, (9.9)
2π     −π                                   2πN      −π k=1

41
where integer N is chosen so that N |h| ≤ 1 for h = 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

π   N
1
|f (x + kh) − f (x + (k − 1)h)|2 dx
2πN   −π k=1

π   N
C|h|α
≤                   |f (x + kh) − f (x + (k − 1)h)|dx
2πN     −π k=1

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     C|h|α+1
=         =                    ≤            =         ≤ 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
∞
eikx
f (x) :=              .
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
∞
eikx
f (x) :=               .
k=1
k 3/2

1
1. Show that f belongs to Nikol’skii space H2 (−π, π) but
π
1                                       4h2      π
|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 conver-
gence.
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

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. Using Euler’s formula we have
N                        N                              2N
1 − ei(2N +1)x
DN (x) =                 einx = e−iN x              ei(n+N )x = e−iN x            eikx = e−iN x
n=−N                     n=−N                            k=0
1 − eix
−iN x      i(N +1)x        i(N +1/2)x         −i(N +1/2)x
e        −e                  e              −e                       sin(N + 1/2)x
=                             =                                        =                 .
1 − eix                     eix/2   − e−ix/2                     sin x/2

Thus, the Dirichlet kernel equals

sin(N + 1/2)x
DN (x) =                      ,           x = 0.                          (10.2)
sin x/2

For x = 0 we have
DN (0) = 2N + 1,
so that (10.2) holds for all x ∈ [−π, π].

Exercise 39. Prove that
1   π
1.   2π   −π
DN (x)dx = 1, N = 0, 1, 2, . . .
1        N
2. KN (x) =       N +1      j=0                                  e
Dj (x), where KN (x) is the Fej´r kernel.

Recall that the trigonometric Fourier partial sum is given by

SN f (x) =             cn (f )einx .                                (10.3)
|n|≤N

The Fourier coeﬃcients of DN (x) are equal to
π
1                                          0, |n| > N
cn (DN ) =               e−inx           eikx dx =
2π    −π                                    1, |n| ≤ N.
|k|≤N

43
Hence, if f is periodic and integrable then the partial sum (10.3) can be rewritten as
(see Exercise 12)
∞                                                             π
1
SN f (x) =          cn (DN )cn (f )einx = (f ∗ DN )(x) =                        DN (x − y)f (y)dy
n=−∞
2π   −π
π                                        π
1                                         1                    sin(N + 1/2)y
=             DN (y)f (x + y)dy =                       f (x + y)                 dy. (10.4)
2π   −π                                   2π   −π                   sin y/2

Exercise 40. Let f be the function
1   x
− 2π ,
2
0<x≤π
f (x) =             1    x
− 2 − 2π , −π ≤ x < 0.

Show that
1
1. (SN f )′ (x) =   2π
(DN (x)       − 1)

2. limN →∞ SN f (x) = f (x), x = 0

3. limN →∞ SN f (0) = 0.

Exercise 41. Prove that
π
1                              4 log N
|DN (x)|dx =           + O(1).
2π       −π                        π2

Since the Dirichlet kernel is an even function (see (10.2)) we can rewrite (10.4) as
π
1                                          sin(N + 1/2)y
SN f (x) =                     (f (x + y) + f (x − y))                  dy.           (10.5)
2π       0                                      sin y/2

Using the normalization of the Dirichlet kernel (see Exercise 39) we have for any
function S(x) that
π
1                                                       sin(N + 1/2)y
SN f (x) − S(x) =                   (f (x + y) + f (x − y) − 2S(x))                          dy.   (10.6)
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  π2
−   ≤ .                                       (10.7)
sin z/2 z   24

44
Proof. First we show that
z     |z|3
≤sin z/2 −                                                 (10.8)
2      48
for all z ∈ [−π, π]. In order to prove this inequality it is enough to show that

x3
0 < x − sin x <
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   2 |z/2 − sin(z/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 → (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.
π
1    2
|f (x + y) + f (x − y) − 2S(x)|                  − dy
0                                                 sin y/2 y
π                          π
π2
≤                 |f (x + y)|dy +            |f (x − y)|dy + 2π|S(x)|
24    0                          0
π
π2
≤                      |f (y)|dy + 2π sup |S(x)| .
24    −π                   x∈[−π,π]

Application of Riemann-Lebesgue lemma (Theorem 5) gives us that
π
1                                                                  1     2
lim                   (f (x + y) + f (x − y) − 2S(x))                       −          sin(N + 1/2)y = 0 (10.9)
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
π
sin(N + 1/2)y
lim            (f (x + y) + f (x − y) − 2S(x))                            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 in-
terval 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
δ
1                                                        sin(N + 1/2)y
SN f (x) − S(x) =                      (f (x + y) + f (x − y) − 2S(x))                   dy + oN (1)
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 ques-
tion is: to which values of S(x) in (10.11) the trigonometric Fourier partial sum may
e
converge? The second part of Theorem 3 shows that the Fej´r means converge to
1
lim σN f (x) =            (f (x + 0) + f (x − 0))
N →∞                     2
for any x ∈ [−π, π] pointwise. But since
N
Sj f (x)
j=0
σN f (x) =
N +1
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
δ
|f (x + y) + f (x − y) − 2S(x)|
dy < ∞                      (10.12)
0                       y
pointwise or uniformly in x ∈ [−π, π]. Then
lim SN f (x) = S(x)                               (10.13)
N →∞

pointwise or uniformly in x ∈ [−π, π].

46
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 H¨lder space C α [−π, π] for some 0 < α ≤ 1.
o
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
1
lim SN f (x) =           (f (x + 0) + f (x − 0)) .        (10.14)
N →∞                      2
Proof. Since f is of bounded variation then the limit
1                          1
lim     (f (x + y) + f (x − y)) = (f (x + 0) + f (x − 0)) := S(x)            (10.15)
y→+0   2                          2
exists. For 0 < y < δ we denote
1
F (y) :=     (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
y
sin(N + 1/2)t
GN (y) :=                            dt,          0 < y ≤ δ.   (10.16)
0             t
It is easy to check that
(N +1/2)y
sin ρ
GN (y) =                                dρ,     0 < y ≤ δ.
0                 ρ
This representation implies that
∞
sin ρ     π
lim GN (y) =                            dρ = .        (10.17)
N →∞                       0         ρ       2

47
For ﬁxed x we have from (10.11) and (10.16) that
δ
1
SN f (x) − S(x) =                   F (y)G′N (y)dy + oN (1).
π         0

Here integration by parts gives
δ
1
SN f (x) − S(x) =         F (y)GN (y)|δ −
0                                   GN (y)dF (y) + oN (1)
π                                            0
δ
1
=      F (δ)GN (δ) −                           GN (y)dF (y) + oN (1),             (10.18)
π                                    0

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
δ
1         π π                                                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.
1
Corollary. If f is periodic and belongs to Sobolev space W1 (−π, π) then its trigono-
metric Fourier series converges to f (x) everywhere.
1
Proof. Since f ∈ W1 (−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
1
trigonometric Fourier series of functions from Sobolev space W1 (−π, π). However,
uniform convergence is valid as we will prove later in this chapter.
Exercise 42. Show that
1
f (x) =            1 ,             |x| < 1/2
log |x|
is of bounded variation but this function does not satisfy condition (10.12) at x = 0.
Hint.
δ
1
dy = +∞.
0       y log y
Exercise 43. Show that
1
f (x) = x sin
x
satisﬁes condition (10.12) at x = 0 but this function is not of bounded variation.

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
b                          b
lim               SN f (x)dx =               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                                   x
F (x) :=    (f (t) − c0 (f ))dt.                 (10.19)
−π
1
It is clear that F (x) belongs to Sobolev space W1 (−π, π) with F (−π) = F (π) = 0
(periodicity) and
F ′ (x) = f (x) − c0 (f ).
This implies
cn (F ′ ) = incn (F ) = cn (f ),                  n = 0,          c0 (F ′ ) = 0.
Corollary of Theorem 16 gives us that F (x) has everywhere convergent trigonometric
Fourier series
cn (f ) inx
F (x) = c0 (F ) +            e .                  (10.20)
n=0
in

In particular, for any −π ≤ a < b ≤ π we have from (10.20) that
b
cn (f ) inb
F (b) − F (a) =                       (e − eina ) =                         cn (f )           einx dx
n=0
in                                n=0                a

or equivalently (see (10.19)),
b                                  a                                                b
(f (x) − c0 (f ))dx −              (f (x) − c0 (f ))dx =                            f (x)dx − (b − a)c0 (f )
−π                                 −π                                            a
b
=                 cn (f )               einx dx.
n=0                       a

Thus, we obtain ﬁnally
b                 ∞                  b                                                  b                                      b
f (x)dx =          cn (f )           einx dx = lim                  cn (f )             einx dx = lim                          SN f (x)dx.
a                                    a                       N →∞                       a                         N →∞         a
n=−∞                                            |n|≤N

This proves the theorem.
Exercise 44. Calculate c0 (F ) for F deﬁned by (10.19).

49
Corollary 1. If f ∈ L1 (−π, π) then the series
cn (f )                       cn (f )(−1)n
and                                                        (10.21)
n=0
n                    n=0
n
converge.
Proof. Follows from (10.20).
Corollary 2. The series
∞
sin(nx)
n=1
log(1 + n)
1
is not a Fourier series of an L -function.
Proof. Let us assume on the contrary that there is f ∈ L1 (−π, π) such that
∞                                ∞                            ∞
sin(nx)    1                    einx     1                     e−inx
f (x) ∼                      =                             −
n=1
log(1 + n)   2i         n=1
log(1 + n) 2i            n=1
log(1 + n)
∞           inx                −∞                 inx
1             e        1       sgn(n)e       1                              sgn(n)einx
=                        +                     =
2i   n=1
log(1 + n) 2i n=−1 log(1 + |n|)   2i                      n=0
log(1 + |n|)

i.e. we have
1 sgn(n)
cn (f ) =               , n = 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. But Corollary 1 of Theorem 17 implies that
∞
sgn(n)        1                          1
=
n=0
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
π                            π
f (x)dx = 0,                 |f (x)|dx = +∞.
−π                         −π

Recall that the Poisson kernel is equal to
1 − r2
Pr (x) =                       ,            0≤r<1
1 − 2r cos x + r2
and its trigonometric Fourier series is
∞
Pr (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)                                          (10.22)
r→1−0

uniformly in x ∈ [−π, π] or
∞
lim               r|n| cn (f )einx = f (x)                              (10.23)
r→1−0
n=−∞

uniformly in x ∈ [−π, π] even if f (x) has no convergent trigonometric Fourier series.
Proof. Using the normalization
π
1
Pr (x)dx = 1
2π         −π

we have
π
1
(Pr ∗ f )(x) − f (x) =                      Pr (y)(f (x − y) − f (x))dy
2π           −π
1
=                          Pr (y)(f (x − y) − f (x))dy
2π           |y|≤δ
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
π
1
|I1 | ≤     sup        |f (x − y) − f (x)|                   Pr (y)dy =        sup         |f (x − y) − f (x)| → 0
x∈[−π,π],|y|≤δ                     2π        −π                   x∈[−π,π],|y|≤δ

as δ → 0. At the same time, I2 can be estimated as
1
|I2 | ≤ 2 max |f (x)|                           Pr (y)dy.
|x|≤π                2π   δ≤|y|≤π

For δ ≤ |y| ≤ π the Poisson kernel can be estimated as
1 − r2                  1 − r2             1−r         1−r         π2 1 − r
Pr (y) =                   =                        ≤             ≤            =          .
1 − 2r cos y + r2   4r sin2 y/2 + (1 − r)2   2r sin2 y/2   2rδ 2 /π 2   2 rδ 2
If we choose δ 4 = 1 − r then I2 is estimated as
√
1            π2 1 − r  π             1−r
|I2 | ≤ max |f (x)|    √     = max |f (x)|      →0
π |x|≤π      2 r 1−r   2 |x|≤π        r
as r → 1 − 0. Hence, the estimates for I1 and I2 show that
lim ((Pr ∗ f )(x) − f (x)) = 0
r→1−0

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 }∞ be a sequence of complex numbers such that
k=0

k|ak | ≤ M,           k = 0, 1, 2, . . . ,                       (10.24)

where the constant M is independent of k. If there is
n
j
lim σn := lim                  1−            aj = a                   (10.25)
n→∞            n→∞
j=0
n+1

then
lim (σn − sn ) = 0,
n→∞
n
where sn =     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 uniform.
Proof. For n < m it is true that
m
(m + 1)σm − (n + 1)σn −                        (m + 1 − j)aj = (m − n)sn .
j=n+1

Indeed,
m
(m + 1)σm − (n + 1)σn −                  (m + 1 − j)aj
j=n+1
m                         n                             m
=         (m + 1 − j)aj −            (n + 1 − j)aj −                (m + 1 − j)aj
j=0                        j=0                         j=n+1
n                          n                           n
=           (m + 1 − j)aj −           (n + 1 − j)aj =              (m − n)aj = (m − n)sn .
j=0                       j=0                          j=0

That’s why,
m
(m + 1)σm − (n + 1)σn −                   (m + 1 − j)aj − (m − n)σn = (m − n)sn − (m − n)σn
j=n+1

or, equivalently,
m
m(σm − σn ) + (σm − σn ) −                     (m + 1 − j)aj = (m − n)(sn − σn )
j=n+1

52
i.e.                                         m
m+1              m+1             j
(σm − σn ) −             1−                           aj = sn − 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             j                  m+1             j                       M
1−             aj    ≤             1−
m − n j=n+1    m+1                 m − n j=n+1    m+1                      j
m
1   1
∼ M (1 + 1/δ)                     −
j=n+1
j m+1
m
1 m−n
= M (1 + 1/δ)                −
j=n+1
j   m+1
m
1      m−n
≤ M (1 + 1/δ)       dξ −
n ξ      m+1
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
|cn (f )| ≤     , n=0
|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 Fej´r
e
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 ,       k = 1, 2, . . . .

53
Then                                       n
sn ≡            ak (x) = SN f (x)
k=0

and                                             n                             n
1                         1
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
2M
|ak (x)| ≤          ,      k = 1, 2, . . . .
k
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].
1
Corollary 2. If f ∈ W1 (−π, π) is periodic then

lim SN f (x) = f (x)
N →∞

uniformly in x ∈ [−π, π].
1
Proof. Since f ∈ W1 (−π, π) is periodic then there is g ∈ L1 (−π, π) such that
x                                  π
f (x) =         g(t)dt + f (−π),                   g(t)dt = 0.
−π                                −π

Thus, f ′ = g and
cn (g) = incn (f )
or, equivalently,
cn (g)   M
|cn (f )| =           ≤     ,          n = 0.
in      |n|
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
∞
sin(nx)
f1 (x) =                                                 (10.27)
n=1
n log(1 + n)

54
and                                                    ∞
cos(nx)
f2 (x) =                      .                         (10.28)
n=1
n log(1 + n)
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
1
show that f2 (x) does not belong to W1 (−π, π) but f1 (x) does.
1                             ′
If we assume on the contrary that f2 ∈ W1 (−π, π) then its derivative f2 has the
Fourier series                               ∞
′               sin(nx)
f2 (x) ∼ −                .
n=1
log(1 + n)
But due to Corollary 2 of Theorem 17 this is not a Fourier series of an L1 -function.
1
This contradiction proves that f2 ∈ W1 (−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                               N                          n                n−1
sin(nx)                1
=                                                sin(kx) −         sin(kx)
n=M +1
n log(1 + n) n=M +1 n log(1 + n)                       k=1               k=1
sin(N x)      sin(M x)
=                 −
N log(1 + N ) M log(1 + M )
N          n
1               1
−                       sin(kx)                        −                      . (10.29)
n=M +1        k=1
(n + 1) log(2 + n) n log(1 + n)

Using the calculation from Exercise 15 we have
n
cos x/2 − cos(n + 1/2)x   sin(nx/2) sin((n + 1)x/2)
sin(kx) =                             =                           .                   (10.30)
k=1
2 sin x/2                  sin x/2

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
sin(nx/2) sin((n + 1)x/2)     n log 2+n + log(2 + n)
1+n
−
n=M +1
sin x/2          n(n + 1) log(1 + n) log(2 + n)
N                                           2+n
sin(nx/2) sin((n + 1)x/2)            log 1+n
=−
n=M +1
sin x/2          (n + 1) log(1 + n) log(2 + n)
N
sin(nx/2) sin((n + 1)x/2)          1
−                                                        = I1 + I2 .
n=M +1
sin x/2          n(n + 1) log(1 + n)

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
[ |x| ]
1
[ |x| ]
1
1
π                   n log (1 + 1/(n + 1))       π                         nn
|I1 | ≤                                               <
2        n=M +1
(n + 1) log(1 + n) log(2 + n)   2               n=M +1
n log2 n
∞
π                     1
<                            →0                                                                    (10.31)
2          n=M +1
n log2 n

as M → ∞ uniformly in x. In the ﬁrst case for I2 we have, by integration by parts,
that
[ |x| ]
1
[ |x| ]
1

π               n n+1 |x|
2            π             1      π                                          1/|x|
dt
|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|         1/|x|
π                 t                             dt
=   |x|                           +
4               log t   M +1         M +1     log2 t
1/|x|
π             1/|x|      M +1                                      dt
=   |x|                 −           +
4           log(1/|x|) log(M + 1)                       M +1     log2 t
π       1       |x|(M + 1)      |x|
≤               +            +  2       (1/|x| + M + 1)
4   log(1/|x|) log(M + 1) log (M + 1)
π        1            1           1             1
<                +            +   2       +     2
4   log(M + 1) log(M + 1) log (M + 1) log (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)        1        π
≤            ≤     ≤ πn.
sin x/2            | sin x/2|   |x|

Then
N                                                    N            1
n log (1 + 1/(n + 1))                                nn
|I1 | ≤ π                                                   <π
(n + 1) log(1 + n) log(2 + n)                         n log2 n
n=[ |x| ]
1
n=[ |x| ]
1

∞
1
< π                                   → 0,      M →∞                                             (10.33)
n log2 n
n=[    1
|x|   ]≥M

56
uniformly in x. For I2 we have, by integration by parts,
N                           ∞
1                        1        1               dt
|I2 | ≤                            2 log n
≤π                2 log t
| sin x/2|               n            |x| [ |x| ]≥M t
1
n=[ |x| ]+1
1

∞         ∞
1          1                            dt
= π        −                 +            2 log2 t
|x|      t log t [ 1 ]≥M   [ |x| ]≥M t
1
|x|

∞
1/|x|           1             dt
= π                         +             2 log2 t
[1/|x|] log[1/|x|] |x| [ |x| ]≥M t
1

[1/|x|] + 1       [1/|x|] + 1 ∞ dt
< π                         +                      2
[1/|x|] log[1/|x|]      [1/|x|]   M t log t
1 + 1/M       1 + 1/M
≤ π                 +             → 0, M → ∞                      (10.34)
log M         log M

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                ∞
cos(nx)
n=1
log(1 + n)
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

′              C
f1 (x) ∼ −            .
x log2 x
1
This singularity is integrable at zero. Thus, the function f1 (x) belongs to W1 (−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) =            cm eimt ,      t ∈ [−π, π],                                       (11.1)
m=−∞

where cm are the Fourier coeﬃcients of x(t).
Let now N be an even positive integer and
2πk                 N          N
tk =       ,      k=−        , . . . , − 1.
N                  2          2
Then x(tk ) is a response at tk i.e.
∞
2πkm
x(tk ) =             cm ei     N    .                                        (11.2)
m=−∞

Since ei2πkl = 1 for integers k and l then the series (11.2) can be rewritten as
∞                            ∞
i 2πk (m−lN )                                                   2πk
cm e   N            =                                       cm ei    N
(m−lN )

m=−∞                           l=−∞ −N/2≤m−lN ≤N/2−1
∞       N/2−1
2πk
=                      cn+lN ei      N
n

l=−∞ n=−N/2
N/2−1                ∞                     N/2−1
i 2πk n                                        2πk
=             e   N             cn+lN =                  ei    N
n
Xn ,   (11.3)
n=−N/2               l=−∞                   n=−N/2

where Xn , n = −N/2, . . . , N/2 − 1 is given by
∞
Xn =             cn+lN .                                                 (11.4)
l=−∞

In the calculation (11.3) we have used the fact that the series (11.1) converges abso-
lutely. Combining (11.2) and (11.3) we obtain
N/2−1
2πkn
xk := x(tk ) =                Xn ei      N      .                                (11.5)
n=−N/2

58
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
2πn         0, k − m = 0, ±N, ±2N, . . .
ei(k−m)    N    =                                        (11.6)
n=−N/2
N, k − m = 0, ±N, ±2N, . . .

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
1               2πkn
Xn =            xk e−i N .                           (11.7)
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
1              2πkn
Xn =            Yk e−i N .                            (11.8)
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 in-
N/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
2πkn
Zk =               Xn ei    N     .           (11.9)
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
Lemma 11.1. The following equalities hold:
1) F −1 (F(Y )) = Y ;
2) F(F −1 (X)) = X;
3)
N/2−1                                            N/2−1
1
F(Xn )k F(Yn )k =                                 Xn Yn .
N
k=−N/2                                           n=−N/2

Proof. Using (11.6), (11.8) and (11.9) we have
                           
N/2−1                                         N/2−1           N/2−1
2πkn             1                                   2πln
 ei 2πkn
F −1 (F(Y ))k =               F(Yl )n ei      N      =                                    Yl e−i    N           N
N
n=−N/2                                         n=−N/2       l=−N/2
                                
N/2−1              N/2−1
1                                           2πn(k−l)
 = 1 Yk N = Yk .
=                  Yl                   ei      N
N                                                          N
l=−N/2            n=−N/2

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).
N/2−1                     N/2−1
1
|Xn |2 =                   |F(Xn )k |2 .
N
n=−N/2                   k=−N/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 we have
N/2−1                               N/2−1                               −N/2+1
−i 2πkn                                −i 2πkn                             2πkn
N F(Xrev )k =            XN −n e       N        =                 X−n e       N     =              Xn ei       N

n=−N/2                              n=−N/2                                n=N/2
N/2                           N/2−1
i 2πkn                               2πkn
=                Xn e     N      =                  Xn ei    N     + XN/2 eiπk − X−N/2 e−iπk
n=−N/2+1                        n=−N/2
N/2−1
2πkn
=             Xn ei    N     = F −1 (X)k
n=−N/2

60
due to periodicity condition (11.10).
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 =                   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−l−1                          N/2−1
−i 2πnm                             2πnm
Ym e     N       =             Ym e−i     N    .
m=−N/2−l                             m=−N/2

Proof. The claim is trivial for l = 0. If l > 0 then
N/2−l−1                         N/2−1                           −N/2−1                            N/2−1
−i 2πnm                            −i 2πnm                          −i 2πnm                            2πnm
Ym e    N       =              Ym e        N     +              Ym e        N     −              Ym e−i     N

m=−N/2−l                        m=−N/2                          m=−N/2−l                          m=N/2−l
N/2−1                           N/2−1
−i 2πnm                                 2πn
=              Ym e        N     +             Ym−N e−i        N
(m−N )

m=−N/2                          m=N/2−l
N/2−1                           N/2−1
−i 2πnm                           2πnm
−               Ym e       N     =             Ym e−i     N

m=N/2−l                          m=−N/2

due to periodicity condition (11.10). If l < 0 then
N/2−l−1                          N/2−1                         N/2−l−1                          −N/2−l−1
2πnm                            2πnm                           2πnm                              2πnm
Ym e−i     N    =               Ym e−i     N     +             Ym e−i     N       −              Ym e−i     N

m=−N/2−l                         m=−N/2                          m=N/2                           m=−N/2
N/2−1                         N/2−l−1
−i 2πnm                          2πnm
=               Ym e        N    +             Ym e−i     N

m=−N/2                          m=N/2
N/2−l−1                                   N/2−1
−i 2πn (m−N )                               2πnm
−               Ym−N e        N           =              Ym e−i         N

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                N/2−1
Ym =                 Ym .
m=−N/2−l                   m=−N/2

61
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                   k+1−N/2
(X ∗ Y )k =             Xl Yk−l =                    Yj Xk−j
l=−N/2                  j=k+N/2
N/2−1+(k+1)                          N/2−1
=                    Yj Xk−j =                    Yj Xk−j = (Y ∗ X)k
j=−N/2+(k+1)                         j=−N/2

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
N/2−1                                         N/2−1            N/2−1
1                                −i 2πkn     1                                           2πkn
F(X ∗ Y )n     =                (X ∗ Y )k e           N     =                 Xl               Yk−l e−i    N
N                                            N
k=−N/2                                       l=−N/2            k=−N/2
N/2−1           N/2−l−1
1                                          2πn
=                Xl                  Ym e−i     N
(m+l)
N
l=−N/2        m=−N/2−l
N/2−1                      N/2−l−1
1                   −i 2πnl                           2πnm
=                Xl e     N                    Ym e−i     N    .                             (11.12)
N
l=−N/2                  m=−N/2−l

The above Proposition allows us to rewrite (11.12) as
N/2−1                      N/2−1
1                      −i 2πnl                        2πnm
F(X ∗ Y )n   =                 Xl e       N                 Ym e−i     N    = N F(X)n F(Y )n .
N
l=−N/2                    m=−N/2

Part 2) is proved in a similar manner.
Corollary. For each n = −N/2, . . . , N/2 − 1 it is true that
1
F −1 (X · Y )n =         F −1 (X) ∗ F −1 (Y )                  n
,                     (11.13)
N
N/2−1
where X · Y denotes the sequence {Xk · Yk }k=−N/2 .

62
Proof. Lemmas 11.1 and 11.3 imply that

X ∗Y          = F −1 F X ∗ Y                     = N F −1 F(X) · F(Y )                 .
n                                  n                                     n

Denoting X := F −1 (X) and Y := F −1 (Y ) we obtain easily from the latter equality
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
|Xn − cn | =                            cn+lN − cn =                          cn+lN
n=−N/2                         n=−N/2 l=−∞                               n=−N/2 l=0
N/2−1
≤                   |cn+lN | ≤                |cν |.                      (11.14)
n=−N/2 l=0                      |ν|≥N/2

Similarly, we have

N/2−1                             N/2−1
−1                            i 2πkn                                        2πkn
F        (Xn )k −            cn e    N      =                   (Xn − cn )ei     N

n=−N/2                           n=−N/2
N/2−1
≤                 |Xn − cn | ≤                |cν |.   (11.15)
n=−N/2                     |ν|≥N/2

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 .
m
Theorem 20. If x(t) is periodic and belongs to Sobolev space W2 (−π, π) for some
m = 1, 2, . . . then
1
Xn = cn + o     m−1/2
(11.16)
N
and
N/2−1
−1                                2πkn               1
F        (Xn )k =             cn ei    N     +o                                      (11.17)
N m−1/2
n=−N/2

uniformly in n and k from the set {−N/2, . . . , N/2 − 1}.

63
o
Proof. Using H¨lder’s inequality we have
                        1/2                     1/2

|cν | ≤             |ν|2m |cν |2                   |ν|−2m      .
|ν|≥N/2              |ν|≥N/2                         |ν|≥N/2

The ﬁrst sum in the right hand side tends to zero as N → ∞ due to Parseval equality for
m
function from Sobolev space W2 (−π, π). The second sum can be estimated precisely.
Namely, since for any m = 1, 2, . . . we have
                      1/2
∞              1/2
              |ν|−2m         ≍          t−2m dt          ≍ N −m+1/2
|ν|≥N/2                         N/2

then (11.16) and (11.17) follow from the last estimate and (11.14) and (11.15), respec-
tively.
m
Corollary. An unknown periodic function f ∈ W2 (−π, π), m = 1, 2, . . . can be recov-
ered from its IDFT as

2πk                                  1
f              = F −1 (X)k + o
N                             N m−1/2

uniformly in k from the set {−N/2, . . . , N/2 − 1}.

Exercise 48.     1) Show that
2πn
N/2−1               1,                   a = ei N
F     {ak }k=−N/2          =    1        −N/2 −aN/2       2πn
n        N
(−1)n a −i 2πn , a = ei N
1−ae        N

N/2−1
2) If sequence Y = {Yk }k=−N/2 is real then show that

F(Y )n = F(Y )N −n .

64
12      Connection between discrete Fourier transform
and Fourier transform.
If function f (x) is integrable over the whole line i.e.
∞
|f (x)|dx < ∞
−∞

then its Fourier transform is deﬁned as
∞
1
Ff (ξ) = f (ξ) := √                  f (x)e−ixξ dx.                        (12.1)
2π         −∞

Similarly, inverse Fourier transform of an integrable function g(ξ) is deﬁned as
∞
−1            1
F        g(x) := √              g(ξ)eixξ dξ.                            (12.2)
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
∞                                      ∞
1                                1
f (ξ) = − √           f (x)e−ixξ+iπ dx = − √                   f (y + π/ξ)e−iξy dy.
2π     −∞                        2π           −∞

This fact implies that
∞
1
−2f (ξ) = √                 (f (x + π/ξ) − f (x))e−iξx dx.
2π        −∞

Hence                                       ∞
1
2|f (ξ)| ≤ √               |f (x + π/ξ) − f (x)|dx → 0
2π       −∞

as |ξ| → ∞ since f is integrable on the whole line. This is a well-known property of
integrable functions. Continuity of f (ξ) follows from the representation
∞
1
f (ξ + h) − f (ξ) = √                  f (x)e−ixξ (e−ixh − 1)dx
2π         −∞

and its consequence
1                                     2
|f (ξ + h) − f (ξ)| ≤ √              |f (x)||e−ixh − 1|dx + √                        |f (x)|dx := I1 + I2 .
2π    |xh|<δ                          2π            |xh|>δ

65
For the ﬁrst term I1 we have the estimate
∞
1                               δ
I1 ≤ √               |f (x)||xh|dx < √                         |f (x)|dx → 0,       δ → 0.
2π     |xh|<δ                   2π                 −∞

For the second term I2 we have
2
I2 = √                         |f (x)|dx → 0
2π          |x|>δ/|h|

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
m
we say that f belongs to Sobolev space W1 (R).
1
Exercise 49. Prove that if f ∈ W1 (R) then limx→±∞ f (x) = 0.
1
Theorem 22 (Fourier inversion formula). Suppose that f belongs to W1 (R). Then

F −1 (Ff )(x) = f (x)

at any point x ∈ R.
Proof. First we prove that
∞                             ∞
f (x)g(x)dx =                 f (ξ)g(ξ)dξ
−∞                             −∞

for any two integrable functions f and g. Indeed,
∞                                     ∞              ∞
1
f (x)g(x)dx = √                           f (x)         g(ξ)e−ixξ dξdx
−∞                 2π                 −∞              −∞
∞               ∞                             ∞
1
= √                           g(ξ)          f (x)e−ixξ dxdξ =             f (ξ)g(ξ)dξ
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
n                                    n
1              −ixξ         1 e−ixξ                  1            e−ixn   eixn             2 sin(nx)
g(x) = √             e        dξ = √                       =√                   −              =             .
2π     −n                   2π −ix           −n      2π          −ix     −ix              π    x
Thus, we have the equality
∞                                  n
2                   sin(nx)
f (x)           dx =               f (ξ)dξ,
π   −∞                 x                  −n

66
1
where f ∈ W1 (R). Letting n → ∞ we obtain
∞                                     ∞
sin(nx)       2
p.v.            f (ξ)dξ = lim     dx.                f (x)                                  (12.3)
−∞                       −∞         x
n→∞        π
√
We will prove that the limit in (12.3) is actually equal to 2πf (0). Since
∞
sin(nx)
dx = π
−∞       x

then the limit in (12.3) can be rewritten as

2       ∞
sin(nx)      √                                         2    ∞
sin(nx)
lim                    f (x)            dx =   2πf (0) + lim                                     (f (x) − f (0))            dx
n→∞        π   −∞                  x                    n→∞                            π    −∞                         x
√                                         2     ∞
sin(t)
=   2πf (0) + lim                                     (f (t/n) − f (0))           dt.
n→∞         π    −∞                           t

It remains to show that the latter limit is equal to zero. In order to prove this fact we
split
∞
sin(t)
(f (t/n) − f (0))                 dt
−∞                                 t
sin(t)                                           sin(t)
=               (f (t/n) − f (0))            dt +                  (f (t/n) − f (0))          dt := I1 + I2 .
|t|<1                            t                 |t|>1                          t
1
Since f ∈ W1 (R) then it can be proved that f is continuous. Therefore

sin(t)
|I1 | ≤ sup |f (t/n) − f (0)|                                dt → 0,           n → ∞.
|t|<1                            |t|<1       t

For the second term I2 we ﬁrst change variables back as

sin(nz)
I2 =                (f (z) − f (0))                   dz
|z|>1/n                                z
′               1                                              ′
∞                                   z                                −n                             z
sin(ny)                                                         sin(ny)
=          (f (z) − f (0))                             dy        dz +            (f (z) − f (0))                       dy       dz.
1
n
0          y                        −∞                          0          y

67
Integration by parts in these integrals leads to
z                           ∞               ∞                      z
sin(ny)                                                            sin(ny)
I2 = (f (z) − f (0))                                dy                        −          f ′ (z)                       dydz
0           y                    1/n            1/n                 0          y
z                           −1/n             −1/n                          z
sin(ny)                                                ′                   sin(ny)
+ (f (z) − f (0))                               dy                         −              f (z)                            dydz
0           y                    −∞              −∞                         0          y
∞                                                                       1/n
sin(ny)                                                               sin(ny)
= ( lim f (z) − f (0))                                          dy − (f (1/n) − f (0))                                                dy
z→∞                                    0              y                                                        0            y
∞                   nz
sin(t)
−           f ′ (z)                   dtdz
1/n             0           t
−1/n                                                                               −∞
sin(ny)                                                                           sin(ny)
+ (f (−1/n) − f (0))                                              dy − ( lim f (z) − f (0))                                                         dy
0                y          z→−∞                                                   0               y
−1/n                    nz
sin(t)
−              f ′ (z)                     dtdz.
−∞                   0           t
Since limz→±∞ f (z) = 0 (see Exercise 49) and since f is continuous we obtain (when
n → ∞)
∞                      nz
π                                       ′                       sin(t)
I2   → −f (0) − lim                              f (z)                              dtdz
2 n→∞                    1/n                       0               t
−1/n                            nz
π                                            ′                   sin(t)
− f (0) − lim                                f (z)                             dtdz
2 n→∞                   −∞                              0           t
∞                             nz                                           −1/n                       nz
sin(t)                                                                  sin(t)
= −πf (0) − lim                             f ′ (z)                        dtdz − lim                                f ′ (z)                       dtdz
n→∞       1/n                        0           t         n→∞                       −∞                      0           t
∞                                         0
π               π
= −πf (0) −      f ′ (z) dz +    f ′ (z) dz
0         2     −∞        2
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 conver-
gence theorem. Thus, (12.3) transforms to
∞                     √
p.v.                   f (ξ)dξ =             2πf (0)
−∞

or equivalently,
F −1 (Ff )(0) = f (0).
In order to prove Fourier inversion formula for any x ∈ R let us note that
∞
1
fx (y)(ξ) = f (x + y)(ξ) = √                                                    f (x + y)e−iyξ dy
2π                                           −∞
∞
1
= √                                                    f (z)e−izξ eixξ dz = eixξ f (ξ).
2π                                           −∞

68
Since fx (0) = f (x) then

f (x) = fx (0) = F −1 (Ffx )(0)
∞                                     ∞
1                       1
=√           (Ffx )(ξ)dξ = √                         eixξ f (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
∞
1                   sin(nx)
lim             f (x)           dx = f (0)
n→∞ π     −∞                x
1
for any f ∈ W1 (R).
m
Lemma 12.1. If f belongs to Sobolev space W1 (R) for some m = 1, 2, . . . then
1
f (ξ) = o                                                     (12.4)
|ξ|m
when |ξ| → ∞.
m                                          1
Proof. Since f ∈ W1 (R) then f ′ , f ′′ , . . . , f (m−1) ∈ W1 (R). By Exercise 49 we have
lim f (k) (x) = 0
x→±∞

for each k = 0, 1, . . . , m − 1. This fact and integration by parts give us
∞                                    ∞                 ∞                            ∞
−ixξ          e−ixξ                  1                                1
f (x)e      dx =         f (x)        +                f ′ (x)e−ixξ dx =             f ′ (x)e−ixξ dx
−∞                       −iξ           −∞       iξ   −∞                          iξ   −∞
−ixξ         ∞                      ∞
e                          1
= −       2
f ′ (x)       +      2
f ′′ (x)e−ixξ dx
(iξ)               −∞    (iξ) −∞
∞                                            ∞
1                                             1
=                f ′′ (x)e−ixξ 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 (up to the 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:
R
1
f ∗ (x) := √                 Ff (ξ)eixξ dξ.                              (12.5)
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 Ff (ξ) is smooth function for which (12.4) holds.

69
◦
m                 m
/
Deﬁnition 12.1. We say that f ∈ W1 (−R, R) if f ∈ W1 (R) and f ≡ 0 for x ∈
(−R, R).
◦
m
Theorem 23. Suppose that f ∈ W1 (−R, R) is supported in a ﬁxed interval [a, b] ⊂
(−R, R) with R > 0 large enough and some m = 2, 3, . . .. Then
N/2−1
2     1                              2n + 1         i
x(2n+1)               1
f (x) =                              Ff                          e   N m/(m+2)   +O
π N m/(m+2)                         N m/(m+2)                            N (2m−2)/(m+2)
n=−N/2
(12.6)
2/(m+2)
uniformly in x ∈ (−R, R) for R = N                            and even N .
/
Proof. Since f (x) = 0 for x ∈ (−R, R) then using Fourier inversion formula (see
Theorem 22) we have
∞                                      R
∗      1                        ixξ        1                              1
f (x)−f (x) = √               Ff (ξ)e         dξ− √              Ff (ξ)eixξ dξ = √                  Ff (ξ)eixξ dξ.
2π     −∞                           2π        −R                   2π       |ξ|>R

Lemma 12.1 implies then that
1
f (x) − f ∗ (x) = o                      .                           (12.7)
Rm−1

Let us 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
2Rn                              2R
ξn =        ,           ξn+1 − ξn =         .
N                               N
Set also
∗       ξn + ξn+1  R
ξn =               = (2n + 1).
2      N
Then we obtain
R
1
f ∗ (x) = √             Ff (ξ)eixξ dξ
2π     −R
N/2−1
1                                    ∗   2R
= √                        ∗
Ff (ξn )eixξn
2π                                      N
n=−N/2
N/2−1          ξn+1
1                                                               ∗
+ √                                                   ∗
Ff (ξ)eixξ − Ff (ξn )eixξn dξ
2π      n=−N/2      ξn

N/2−1
2R                                                                   R3
=                       Ff (R(2n + 1)/N )eixR(2n+1)/N + O                          .   (12.8)
πN                                                                   N2
n=−N/2

70
Exercise 50. Prove that

N/2−1                                                       O   R3
1               ξn+1
ixξ                   ∗               N2
, supp f = [a, b]
√                       Ff (ξ)e         −        ∗
Ff (ξn )eixξn   dξ =
2π                                                               O   R5
n=−N/2   ξn
N2
, supp f ⊂ (−R, R)

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 = −N/2, . . . , N/2 − 1,
N m/(m+2)
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
◦
m
or m > 2. It means that f must belong to Sobolev space W1 (−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

lim             f (x + 2πn)
N →∞
|n|≤N

exists pointwise in x ∈ R. Then
∞
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
π
1                            1
fp (x)e−imx dx = √ Ff (m).                                   (13.2)
2π       −π                   2π
∞
|Ff (m)| < ∞
m=−∞

then                     ∞                                      ∞
1
f (x + 2πn) = √                          Ff (m)eimx .                     (13.3)
n=−∞
2π                m=−∞

In particular, fp (x) is continuous and we have the Poisson identity
∞                                 ∞
1
f (2πn) = √                        Ff (m).                          (13.4)
n=−∞
2π              m=−∞

Proof. Since f ∈ L1 (R) then
π                       π       ∞                              ∞      π
|fp (x)|dx ≤                    |f (x + 2πn)|dx =                     |f (x + 2πn)|dx
−π                   −π n=−∞                                   n=−∞   −π
∞     π+2πn                            ∞
=                           |f (t)|dt =          |f (t)|dt < ∞.
n=−∞         −π+2πn                    −∞

72
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
π                           ∞             π
1                    −imx                1
cm (fp ) =                fp (x)e       dx =                       f (x + 2πn)e−imx dx
2π        −π                        n=−∞
2π       −π
∞              π+2πn                                         ∞          π+2πn
1                         −im(t−2πn)            1              1
=                            f (t)e               dt = √              √                f (t)e−imt dt
n=−∞
2π           −π+2πn                               2π      n=−∞
2π    −π+2πn
∞
1  1                                  1
= √ √                   f (t)e−imt dt = √ Ff (m).
2π 2π             −∞                  2π
Now, if the series
∞
|Ff (m)|
m=−∞

converges then it is equivalent to the fact that
∞
|cm (fp )| < ∞.
m=−∞

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) =             cm (fp )eimx .
m=−∞

It means that (see (13.1))
∞                                  ∞
1
f (x + 2πn) = √                         Ff (m)eimx .
n=−∞
2π           m=−∞

Finally, set x = 0 to obtain the Poisson identity (13.4).
Example. If
1 − x2
e 4t , x ∈ R,
f (x) = √
4πt
where t > 0 is a parameter then it is very well-known that
1    2
Ff (ξ) = √ e−tξ .
2π
Formula (13.3) transforms in this case to
∞                             ∞
1          (x+2πn)2    1                           2
√          e− 4t      =                        e−tm eimx
4πt n=−∞              2π              m=−∞

73
and the Poisson identity transforms to
∞                            ∞
π                2 n2 /t                       2
e−π             =           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
2πλ
1
f (t) = √                       F (ξ)eitξ dξ,                             (13.5)
2π          −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
∞
sin π(t − n)
f (t) =            f (n)                  ,                               (13.6)
n=−∞
π(t − n)

where the fraction is equal to 1 when t = n. If λ > 1/2 we have
∞
sin π(t − n)   1                   2
f (t) −          f (n)                ≤√                      +1                     |F (ξ)|dξ.       (13.7)
n=−∞
π(t − n)      2π                 π          π<|ξ|<2πλ

Proof. Let
∞
Fp (ξ) =                  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                                        imξ     1
F (ξ + 2πn) = √                         f (−m)e               =√                 f (m)e−imξ .
n=−∞
2π           m=−∞
2π       m=−∞

74
Since any trigonometric Fourier series can be integrated term by term we obtain
π                            π     ∞                                              ∞              π
itξ                                           1  itξ
Fp (ξ)e dξ =                     F (ξ + 2πn)e dξ = √                                f (m)         e−i(m−t)ξ dξ
−π                            −π n=−∞                     2π                    m=−∞              −π
∞                            π
1                             ei(t−m)ξ
= √                     f (m)                      ,    m=t
2π         m=−∞
i(t − m)      −π
∞
1                            ei(t−m)π − e−i(t−m)π
= √                     f (m)
2π         m=−∞
i(t − m)
∞
2π                            sin π(t − m)
= √                     f (m)                .
2π         m=−∞
π(t − m)

Now, if λ ≤ 1/2 then F (ξ) for |ξ| ≤ π is equal to its periodization Fp (ξ) (see Deﬁnition
13.1) and
π                √
Fp (ξ)eitξ dξ = 2πf (t).
−π
These equalities imply immediately that
∞
sin π(t − n)
f (t) =           f (n)
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
π
1                                1
f (t) = √                 F (ξ)eitξ dξ + √                           F (ξ)eitξ dξ
2π           −π                  2π            π<|ξ|<2πλ
∞
sin π(t − n)   1
=              f (n)                +√                             F (ξ)eitξ dξ,
n=−∞
π(t − n)      2π              π<|ξ|<2πλ

where                                                          π
1
f (n) = √                   F (ξ)einξ dξ.
2π         −π
Therefore,
∞
sin π(t − n)
f (t) =           f (n)
n=−∞
π(t − n)
∞
sin π(t − n)   1
+            f (n) − f (n)                   +√                               F (ξ)eitξ dξ.
n=−∞
π(t − n)      2π               π<|ξ|<2πλ

Here the middle series is equal to
∞
1                                                      sin π(t − n)
−√                                  F (ξ)einξ dξ
2π     n=−∞          π<|ξ|<2πλ                          π(t − n)

75
or
∞
1                                      sin π(t − n) inξ
−√                          F (ξ)                    e             dξ.
2π           π<|ξ|<2πλ           n=−∞
π(t − n)
Exercise 51. Prove that
∞
sin π(t − n) inξ 2
e = − eitξ .
n=−∞
π(t − n)       π

Hint. Show that
2                            sin π(t − n)
cn − eitξ                   =                 .
π                              π(t − n)
Using Exercise 51 we have
∞
sin π(t − n)           2  1
f (t) =             f (n)                +        √ +√                                F (ξ)eitξ dξ.
n=−∞
π(t − n)           π 2π  2π                 π<|ξ|<2πλ

Thus
∞
sin π(t − n)   1                2
f (t) −             f (n)                ≤√                   +1                    |F (ξ)|dξ.
n=−∞
π(t − n)      2π              π           π<|ξ|<2πλ

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
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
2π           i
πn(2k+1)   1        2π i πn            2πkn     1
f (n) =            Fk e N + O         2
=      e N         Fk ei N + O      ,
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
f (n) ≈       e N F (Fk )n
N
i.e. for even integer N large enough
∞    √
2π i πn −1     sin π(t − n)
f (t) ≈            e N F (Fk )n              .
n=−∞
N                 π(t − n)

76
Index
L2 -H¨lder condition, 38
o                                   Minkowski’s inequality, 2
p
L -modulus of continuity, 23              modulus of continuity, 4
p-integrable function, 1                  mutually orthogonal functions, 5
e
Fej´r kernel, 18
Fej´r means, 18
e                                     Nikol’skii space, 33

absolute convergence, 11                  odd function, 2
orthogonal functions, 5
band-limited signal, 74
Besov space, 32                           Parseval equality, 16, 26, 60
Bessel’s inequality, 26                   partial sum of Fourier series, 18
periodic function, 1
conjugate Poisson kernel, 13              periodization, 72
convolution, 16, 61                       piecewise continuous function, 3
pointwise convergence, 11
Dirichlet kernel, 43                      Poisson identity, 72
discrete Fourier transform, 59            Poisson kernel, 13
even function, 2                          Poisson summation formula, 72

Fourier coeﬃcient, 9                      Riemann-Lebesgue lemma, 65
Fourier cosine series, 8                  Riesz-Fischer theorem, 27
Fourier inversion formula, 66             sequence space, 31
Fourier series, 7                         Sobolev space, 4, 32, 66
Fourier transform, 65                     square-integrable functions, 25
Fouries sine series, 8                    Stieltjes integral, 4
Fubini’s theorem, 1                       summation by parts formula, 13
full variation, 3
function of bounded variation, 3          tail sum, 14
fundamental period, 1                     trigonometric series, 7

generalized Minkowski’s inequality, 2     uniform convergence, 11

o
H¨lder condition with exponent α, 4
o
H¨lder space, 4, 33
o
H¨lder’s inequality, 1

integrable function, 1
inverse discrete Fourier transform, 59
inverse Fourier transform, 65

Laplace equation, 13

mean square distance, 25

77

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