8.5 FOURIER SERIES
8.3
101
THE TRIGONOMETRIC FUNCTIONS
• skipped
8.4
THE ALGEBRAIC COMPLETENESS OF THE COMPLEX FIELD
• skipped
8.5
FOURIER SERIES
8.9 Orthogonal Functions, Orthonormal: Let {
of complex functions on [a, b], such that
Z
Then {
n}
for all n, {
n },
n = 1, 2, 3, . . . , be a sequence
b
n (x) m (x) dx
n 6= m.
= 0,
a
is said to be an orthogonal system of functions on [a, b]. In addition, if
Z
n}
b
a
|
n (x)|
2
dx = 1
is said to be orthonormal.
Example
(a) (2⇡)
1/2
e
inx
, n = 0, ±1, ±2, . . . , is an orthonormal system on [ ⇡, ⇡].
(b) 1, cos x, sin x, cos 2x, sin 2x, . . . form an orthogonal system on [ ⇡, ⇡].
(c) The sequence
1 cos x sin x cos 2x sin 2x
p , p , p , p , p ,...,
⇡
⇡
⇡
⇡
2⇡
form an orthonormal system on [ ⇡, ⇡].
8.10 Fourier Series: Let {
n}
cn =
be orthonormal on [a, b]. Put
Z
b
f (t)
n (t) dt,
n = 1, 2, 3, . . . .
a
We call cn the nth Fourier coefficient of f relative to {
f (x) ⇠
1
X
cn
n }.
We write
n (x)
n=1
and call this series the Fourier series of f (relative to {
n }).
102
8 SOME SPECIAL FUNCTIONS
8.11 Theorem Let {
n}
be orthonormal on [a, b]. Let
sn =
n
X
cm
m (x)
m=1
be the nth partial sum of the Fourier series of f , and suppose
tn (x) =
n
X
m m (x).
m=1
Then
Z
Z
b
a
sn |2 dx 
|f
b
a
tn |2 dx,
|f
and equality holds if and only if
m
Proof Since {
n}
= cm ,
m = 1, . . . , n.
is orthonormal, elementary calculations give
!
Z b
Z b
n
X
f tn dx =
f·
dx
m m
a
a
=
n
X
m=1
b
Z
m
m=1
f
m
a
dx =
n
X
cm
m,
m=1
and
Z
b
a
|tn |2 dx =
=
Z
Z
b
tn tn dx
a
b
a
!
n
X
m m
(f
tn ) · (f
m=1
n
X
·
m m
m=1
!
dx =
n
X
m=1
|
m|
2
.
Hence
Z
b
a
|f
tn |2 dx =
=
=
=
Z
Z
Z
Z
b
a
b
2
|f | dx
a
b
|f |2 dx
a
b
a
|f |2 dx
Z
tn ) dx
b
f tn dx
a
n
X
m=1
n
X
m=1
cm
m
Z
b
f tn dx +
a
n
X
cm
m=1
|cm |2 +
n
X
m=1
|
m
+
Z
b
|tn |2 dx
a
n
X
m=1
m
cm |2 .
|
m|
2
8.5 FOURIER SERIES
In particular, if
m
= cm , we have
Z
These give
Z
Z
b
sn |2 dx =
|f
a
b
a
103
|f
tn |2 dx =
Z
b
|f |2 dx
a
a
sn |2 dx +
f (x) ⇠
n
X
|
m=1
The results follow this inequality.
8.12 Theorem (Bessel Inequality) If {
|cm |2 .
m=1
b
|f
n
X
m
cm |2 .
n}
is orthonormal on [a, b], and if
1
X
cn
n (x),
n=1
then the Bessel inequality holds:
1
X
n=1
|cn |2 
Z
b
|f (x)|2 dx.
a
In particular, lim cn = 0.
n!1
Proof In the proof of Theorem 8.11, we have
0
Z
b
a
|f
sn |2 dx =
Z
n
X
b
a
|f |2 dx
m=1
|cm |2 .
Let n ! 1, we have the Bessel inequality.
8.13 Trigonometric Series: We shall limit our attention to the orthonormal system
{(2⇡) 1/2 einx } on [ ⇡, ⇡]. Consider functions f that have period 2⇡ and that are
Riemann-integrable on [ ⇡, ⇡]. The Fourier series of f is given by
f (x) ⇠
where
cn =
1
2⇡
Z
1
X
cn einx ,
n= 1
⇡
f (x)e
inx
dx.
⇡
Let
sN (x) = sN (f ; x) =
N
X
n= N
be the N th partial sum of the Fourier series of f .
cn einx
104
8 SOME SPECIAL FUNCTIONS
For the convenience of argument, we introduce the Dirichlet kernel
N
X
DN (x) =
einx = e
iN x
=e
iN x
n= N
[1 + eix + · · · + e2iN x ]
·
sin(N + 12 )x
e2iN x+ix
=
.
1 eix
sin( x2 )
1
By the definition,
N
X
sN (f ; x) =
1
2⇡
n= N
1
=
2⇡
1
2⇡
=
Z
Z
Z
⇡
⇡
f (t)
⇡
f (t)e
int
dteinx
⇡
N
X
ein(x
t)
dt
n= N
⇡
f (t)DN (x
t) dt.
⇡
For f being 2⇡-periodic, the last integral is the same as
Z ⇡
1
sN (f ; x) =
f (x t)DN (t) dt.
2⇡
⇡
8.14 Theorem Suppose f is Riemann-integrable on [ ⇡, ⇡] and 2⇡-periodic. If, for
some x, there are constants > 0 and M < 1 such that
|f (x + t)
for all t 2 (
f (x)|  M |t|
, ), then
lim sN (f ; x) = f (x).
N !1
Proof Since
1
2⇡
Z
⇡
DN (x) dx =
⇡
we have
sN (f ; x)
1
f (x) =
2⇡
=
1
2⇡
If we put
g(t) =
Z
Z
1
2⇡
Z
⇡
N
X
⇡
f (x
t)DN (t) dt
⇡
⇡
[f (x
⇡
8
>
< f (x
>
:0,
t)
einx dx = 1,
⇡ n= N
f (x)] ·
t) f (x)
,
sin(t/2)
1
2⇡
Z
⇡
f (x)DN (t) dt
⇡
sin(N + 12 )t
dt.
sin(t/2)
0 < |t|  ⇡,
t = 0,
8.5 FOURIER SERIES
105
by the hypothesis, it is clear that g is continuous at t = 0, so that g is Riemannintegrable. From
Z ⇡
Z ⇡
1
1
sN (f ; x) f (x) =
[g(t) cos(t/2)] sin N t dt +
[g(t) sin(t/2)] cos N t dt,
2⇡
2⇡
⇡
⇡
by Theorem 8.12, we know that both integrals tends to 0 as N ! 1.
8.14’ We can similarly prove a weaker result:
Theorem Suppose f is Riemann-integrable on [ ⇡, ⇡] and 2⇡-periodic. If, for some
x, both f (x ) and f (x+) exist, and there exists a constants > 0 and M < 1 such
that
|f (x + t) f (x+)|  M t,
|f (x t) f (x )|  M t,
for all 0 < t < , then
lim sN (f ; x) =
N !1
f (x+) + f (x )
.
2
• Corollary: If f (x) = 0 for all x in some segment J, then lim sN (f ; x) = 0 for every
x 2 J.
This result show that the behavior of {sN (f ; x)} depends only on the values of f in
the neighborhood of x. This is very di↵erent from power series.
8.15 Theorem Suppose f is continuous and 2⇡-periodic. For any ✏ > 0, there is a
trigonometric polynomial P such that
|P (x)
f (x)| < ✏
for all real x, where a trigonometric polynomial is a finite sum of the form
a0 +
N
X
(an cos nx + bn sin nx).
n=1
Proof Consider
✓
◆n
1 1 + cos x
un (x) =
,
n = 1, 2, 3, . . . ,
↵n
2
◆n
Z ⇡ ✓
1 + cos x
=
dx. Then un (x)
0, un (x + 2⇡) = un (x), and
2
⇡
Z
where ↵n
Z ⇡
un (x) dx = 1. We claim that for any fixed
⇡
2 (0, ⇡),
|x|⇡
un (x) dx ! 0
as n ! 1. In fact, since 1 + cos x is decreasing on [0, ⇡], for any x 2 [ , ⇡] and
y 2 [0, /2], we have
1 + cos x
1 + cos

,
1 + cos y
1 + cos( /2)
106
8 SOME SPECIAL FUNCTIONS
so that
1 + cos x < r(1 + cos y),
1 + cos
< 1. Taking powers and dividing by ↵n , we have un (x) 
1 + cos( /2)
n
r un (y). This implies
Z /2
Z /2
un (x) =
un (x) dy  rn
un (y) dy  rn ! 0,
2
0
0
where r =
as
Z n ! 1. Thus un (x) ! 0 uniformly on [ , ⇡] as n ! 1. By Theorem 7.16,
un (x) dx ! 0.
|x|⇡
By Euler’s formula eix = cos x + i sin x, we see that un (x) =
n
X
me
imx
for some
m= n
constants
m.
For f continuous and 2⇡-periodic, if we put
Z ⇡
Pn (x) =
un (y)f (x y) dy,
⇡
then Pn is a trigonometric polynomial for each n, since
Z x ⇡
Pn (x) =
un (x z)f (z)( 1) dz
=
=
=
Z
x+⇡
x+⇡
n
X
me
x ⇡ m= n
Z ⇡ X
n
me
⇡ m= n
n
X
me
m= n
imx
im(x z)
im(x z)
Z
f (z) dz
f (z) dz
⇡
f (z)e
inz
dz.
⇡
Let ✏ > 0 be given. We shall prove that there is N such that n
|Pn (x)
f (x)| < ✏,
for all x 2 [ ⇡, ⇡]. Then the assertion of the theorem follows.
Since f is uniformly continuous on [ 2⇡, 2⇡], there exists
x 2 [ ⇡, ⇡] and y with |y| < ,
|f (x
Put M =
that n
sup
1<y<1
N implies
y)
N implies
> 0 such that for all
f (x)| < ✏/2.
|f (y)| < 1. Then, from the discussion above, there is N such
Z
un (y) dy < ✏/(4M ).
|y|⇡
8.5 FOURIER SERIES
Thus, for n
|Pn (x)
N,
f (x)| =

=
Z
Z
Z
Z
⇡
un (y)f (x
y) dy
⇡
⇡
⇡
un (y) |f (x
|y|<
Z
un (y) |f (x
y)
⇡
un (y)f (x) dy
⇡
f (x)| dy
Z
f (x)| dy +
un (y) |f (x
|y|⇡
Z
un (y) dy + 2M
un (y) dy
y)
✏
·
2 |y|<
✏
✏
< · 1 + 2M ·
= ✏.
2
4M
<
107
y)
f (x)| dy
|y|⇡
• piecewise continuous, piecewise di↵erentiable, continuously di↵erentiable: A function
f is said piecewise continuous on [a, b], if there is a partition P = {x0 , . . . , xn } of
[a, b] such that f is continuous on each (xi 1 , xi ). Similarly, a function f is said
piecewise di↵erentiable on [a, b] if f 0 exists on each (xi 1 , xi ). A function f is said
continuously di↵erentiable on [a, b] if f 0 is continuous on [a, b].
8.15U Theorem Suppose f is continuous, 2⇡-periodic, and piecewise continuously di↵erentiable. Then sN converges to f uniformly.
Proof Without loss of generality, we assume that f 0 has only one discontinuity at
a 2 [ ⇡, ⇡]. In the case of more discontinuities, the proof is similar.
We expand f as a Fourier series:
f (x) ⇠
with sN (f ; x) =
N
X
n= N
cn e
inx
1
X
cn einx ,
n= 1
1
and cn =
2⇡
Z
⇡
f (x)e
inx
dx.
⇡
The proof is divided into two steps: first we prove that the series
1
X
cn einx
n= 1
converges uniformly to a function, say S(x); next we prove f (x) = S(x) for all x.
To prove the
P Fourier series converges uniformly, by Theorem 7.10, we only need to
prove that
|cn | < 1. Since f is continuously di↵erentiable on ( ⇡, a) and (a, ⇡),
we integrate by parts on each of these intervals:
Z a
Z ⇡
2⇡cn =
f (x)e inx dx +
f (x)e inx dx
⇡
a
"
#
Z a
a
e inx
e inx
0
= f (x)
f (x)
dx
in x= ⇡
in
⇡

Z ⇡
⇡
e inx
e inx
+ f (x)
f 0 (x)
dx .
in x=a
in
a
108
8 SOME SPECIAL FUNCTIONS
Put
1
2⇡
=
n
Z
⇡
f 0 (x)e
inx
dx.
⇡
Since f ( ⇡) = f (⇡), the above calculation gives
n
cn =
in
n = ±1, ±2, ±3, . . . .
,
By Bessel’s inequality for f 0 , we have
N
X
n= N
2
n| 
|
1
2⇡
Z
⇡
⇡
|f 0 (x)|2 dx.
By the Schwarz inequality (Theorem 1.35),
X
1|n|N
which implies
say S(x).
|cn | =
P
X
1|n|N
1
|
n
n|
0
@
0
@
|cn | converges. Thus,
X
1|n|N
X
1|n|N
P
11/2 0
1A @ X
n2
1|n|N
11/2
Z
1A
1
·
n2
2⇡
⇡
⇡
|
n|
11/2
2A
|f 0 (x)|2 dx,
cn einx converges uniformly to a function,
Next we prove that f (x) = S(x). In fact, since
1
X
cn einx converges to S(x) uni-
1
formly, by Theorem 7.12, S is continuous. Notice that, by Theorem 8.11, the value
of the integral
Z ⇡
|f (x) sN (f ; x)|2 dx
⇡
decreases as N increases. Hence
Z
lim
N !1
⇡
⇡
|f (x)
sN (f ; x)|2 dx
exists. By the first step, we know that sN converges uniformly to S, thus, for any
N,
Z ⇡
Z ⇡
Z ⇡
2
2
|f (x) S(x)| dx = lim
|f (x) sN (f ; x)| dx 
|f (x) sN (f ; x)|2 dx.
N !1
⇡
⇡
⇡
This implies that, by Theorem 8.11, for any trigonometric polynomial P (x),
Z ⇡
Z ⇡
|f (x) S(x)|2 dx 
|f (x) P (x)|2 dx.
⇡
⇡
8.5 FOURIER SERIES
109
Let ✏ > 0 be given. By Theorem 8.15, there is a trigonometric polynomial P such
that |P (x) f (x)| < ✏ for all x. Thus,
Z ⇡
|f (x) S(x)|2 dx < 2⇡✏2 ,
⇡
which implies the value of the integral is zero. We conclude that f (x) = S(x) for all
x, since f (x) and S(x) are continuous.
8.16 Theorem (Parseval’s Theorem) Suppose f and g are 2⇡-periodic Riemannintegrable function, and
f (x) ⇠
Then
1
X
cn einx ,
g(x) ⇠
n= 1
1
N !1 2⇡
lim
Z
1
X
ne
inx
.
n= 1
⇡
⇡
|f (x)
sN (f ; x)|2 dx = 0.
Moreover, the following equalities hold:
1
2⇡
Z
and
1
2⇡
⇡
Z
f (x)g(x) dx =
⇡
1
X
cn
n,
n= 1
⇡
⇡
|f (x)|2 dx =
1
X
n= 1
|cn |2 .
Proof Let ✏ > 0 be given. We show that for 2⇡-periodic Riemann-integrable function
f , there is a function h that is continuous, 2⇡-periodic and piecewise di↵erentiable
such that
Z ⇡
1
|f (x) h(x)|2 dx < ✏.
2⇡
⇡
In fact, we only prove this assertion for real f . If f is complex, we just need to apply
the conclusion for real and imaginary parts separately.
Since f is Riemann-integrable, by Theorem 6.6, there is a partition P = {x0 , . . . , xn }
of [ ⇡, ⇡], such that
U (P, f )
L(P, f ) =
n
X
(Mi
mi ) X i <
n=1
where Mi =
sup
xi
1 xxi
f (x), mi =
xi
inf
1 xxi
f (x), and M =
xi x
f (xi
xi xi 1
1)
sup
⇡x⇡
define a piecewise linear (continuous) function on [xi
h(x) =
⇡✏
,
M
+
x
xi
1 , xi ]
xi
xi
1
1
by
f (xi ).
|f (x)|. Now we
110
8 SOME SPECIAL FUNCTIONS
The facts that h(xi 1 ) = f (xi 1 ), h(xi ) = f (xi ), and f is 2⇡-periodic imply that h
can be continuously extended to the whole real line. We still denote the extended
function as h. Hence we have a continuous, 2⇡-periodic, and piecewise di↵erentiable
function h. To see the estimate, we notice that on [xi 1 , xi ], mi  h(x)  Mi ,
1  i  n. Hence,
Z ⇡
X
X
|f (x) h(x)|2 dx 
(Mi mi )2 xi  2M
(Mi mi ) xi < 2⇡✏.
⇡
To prove the limit in the theorem, for the function h, we know that, by Theorem
8.15U, there exists N such that n N implies
p
|h(x) sN (h; x)| < ✏
for all x 2 [ ⇡, ⇡]. By the Bessel inequality, we know that
Z ⇡
Z ⇡
Z
1
1
1
2
2
|sN (f ; x) sN (h; x)| dx =
|sN (f h; x)| dx 
2⇡
2⇡
2⇡
⇡
⇡
⇡
⇡
|f (x) h(x)|2 dx < ✏.
Thus, if n N ,
Z ⇡
1
|f (x) sN (f ; x)|2 dx
2⇡
⇡
Z ⇡
1
=
|f (x) h(x) + h(x) sN (h; x) + sN (h; x) sN (f ; x)|2 dx
2⇡
⇡
Z ⇡
Z ⇡
Z ⇡
3

|f (x) h(x)|2 dx +
|h(x) sN (h; x)|2 dx +
|sN (h; x)
2⇡
⇡
⇡
⇡
sN (f ; x)|2 dx
< 3(✏ + ✏ + ✏) = 9✏.
That is,
1
N !1 2⇡
lim
Z
⇡
⇡
|f (x)
sN (f ; x)|2 dx = 0.
Re-writing the integral in the limit gives
Z ⇡
1
|f (x) sN (f ; x)|2 dx
2⇡
⇡
Z ⇡
1
=
[f (x) sN (f ; x)] · [f (x) sN (f ; x)] dx
2⇡
⇡
Z ⇡
Z ⇡
N
X
1
1
=
|f (x)|2 dx
f (x) ·
cn e
2⇡
2⇡
⇡
⇡
n= N
!
Z ⇡
Z ⇡
N
X
1
1
inx
cn e
· f (x) dx +
2⇡
2⇡
⇡
⇡
n= N
=
1
2⇡
Z
⇡
⇡
|f (x)|2 dx
N
X
n= N
|cn |2 ,
inx
!
N
X
dx
n= N
cn e
inx
!
·
N
X
n= N
cn e
inx
!
dx
8.5 FOURIER SERIES
so that
1
2⇡
Z
⇡
⇡
|f (x)|2 dx =
In general, we can similarly prove
Z ⇡
1
lim
[f (x)
N !1 2⇡
⇡
1
X
n= 1
|cn |2 .
sN (f ; x)]g(x) dx = 0,
which leads to the general Parseval’s identity.
111