LECTURE 12: BESSEL’S INEQUALITY AND PARSEVAL’S THEOREM
MINGFENG ZHAO
July 24, 2015
The L2 space and Bessel’s inequality
Definition 1. The space L2 ([−L, L]) is the set of all functions f (x) on [α, β] such that
Z L
|f (x)|2 dx < ∞.
−L
For any f ∈ L2 ([−L, L]), define
E(f ) =
1
L
Z
L
|f (x)|2 dx.
−L
Remark 1. Let f (x) be a bounded function on [−L, L], then f (x) ∈ L2 ([−L, L]).
For f ∈ L2 ([−L, L]), we can define the Fourier series of f by:
∞
nπ i
nπ a0 X h
+
x + bn sin
x ,
an cos
2
L
L
n=1
where
(1)
an
=
(2)
bn
=
1
L
Z
L
nπ x dx,
L
−L
Z
nπ 1 L
x dx,
f (x) sin
L −L
L
f (x) cos
∀n ≥ 0,
∀n ≥ 1.
For any N ≥ 1, let
N
SN (x) =
nπ nπ i
a0 X h
an cos
+
x + bn sin
x .
2
L
L
n=1
Theorem 1 (Bessel’s inequality). Let f ∈ L2 ([−L, L]), then
N
a0 X 2
+
[an + b2n ] ≤ E(f ),
2
n=1
∀n ≥ 1.
In particular, we have
∞
a20 X 2
+
[an + b2n ] ≤ E(f ) < ∞.
2
n=1
1
2
MINGFENG ZHAO
Proof. In fact, for any N ≥ 1, we have
E(f − SN )
=
=
L
1
L
Z
1
L
"Z
|f (x) − SN (x)|2 dx
−L
L
−L
|f (x)|2 dx − 2
Z
L
Z
f (x)SN (x) dx +
−L
#
L
|SN (x)|2 dx
−L
≥ 0.
Then
" Z
#
Z L
L
1
2
2
f (x)SN (x) dx −
|SN (x)| dx ≤ E(f ).
L
−L
−L
Notice that
Z
L
f (x)SN (x)
=
−L
=
=
=
Z
#
N
nπ nπ i
a0 X h
+
an cos
x + bn sin
x
dx
f (x)
2
L
L
−L
n=1
Z
L
|SN (x)|2 dx
=
−L
=
=
"
L
Z L
Z
Z L
N
N
nπ nπ X
X
a0 L
bn
an
x dx +
f (x) sin
x dx
f (x) dx +
f (x) cos
2 −L
L
L
−L
−L
n=1
n=1
#
"
N
N
a20 X 2 X 2
+
an +
bn
By (1) and (2)
L
2
n=1
n=1
"
#
N
a20 X 2
2
L
+
[an + bn ]
2
n=1
Z L N h
nπ i2
nπ a0 X
x + bn sin
x dx
an cos
+
2
L
L
−L
n=1
# "
#
Z L"
N
N
N
N
nπ X
nπ mπ X
mπ X
a0
a0 X
+
an cos
x +
bn sin
x ·
+
am cos
x +
bm sin
x
dx
2
L
L
2
L
L
−L
n=1
n=1
m=1
m=1
Z L" 2
N
N
mπ a X
mπ a0
a0 X
0
+
am cos
x +
bm sin
x
4
2 m=1
L
2 m=1
L
−L
+
N
N X
N
nπ mπ nπ X
a0 X
an cos
x +
an am cos
x cos
x
2 n=1
L
L
L
n=1 m=1
N
N X
N
N X
N
nπ X
nπ nπ X
nπ mπ a0 X
+
bn sin
x +
bn am sin
x cos
x +
bn bm sin
x sin
x
2 n=1
L
L
L
L
L
n=1 m=1
n=1 m=1
=
a20
4
Z
L
−L
Z L
Z L
N
N
mπ mπ a0 X
a0 X
dx +
am
cos
x dx +
bm
sin
x dx
2 m=1
L
2 m=1
L
−L
−L
Z L
Z L
N
N X
N
nπ nπ mπ X
a0 X
cos
x dx +
an am
cos
x cos
x dx
an
2 n=1
L
L
L
−L
−L
n=1 m=1
#
dx
LECTURE 12: BESSEL’S INEQUALITY AND PARSEVAL’S THEOREM
3
Z L
Z L
N
N X
N
nπ nπ nπ X
a0 X
+
bn
sin
x dx +
x cos
x dx
bn am
sin
2 n=1
L
L
L
−L
−L
n=1 m=1
+
N X
N
X
n=1 m=1
Z
L
bn bm
sin
−L
mπ nπ x sin
x dx
L
L
N
N
X
X
a20
·L+0+0+0+
a2n · L + 0 + 0 +
b2n · L
2
n=1
n=1
"
#
N
a2 X 2
= L· 0 +
[an + b2n ] .
2
n=1
=
So we get
N
a20 X 2
+
[an + b2n ] ≤ E(f ).
2
n=1
Parseval’s Theorem
Theorem 2 (Parseval’s identity). Let f ∈ L2 ([−L, L]), then
∞
a20 X 2
+
[an + b2n ] = E(f ).
2
n=1
Moreover, we have
Z
L
|f (x) − SN (x)|2 dx = 0.
lim
N →∞
−L
Proof. It’s a little technical to prove the Parseval’s identity, so here we assume it. By the computations in the proof of
Theorem 1, then
1
L
Z
L
2
|f (x) − SN (x)| dx
=
−L
=
=
#
N
a20 X 2
2
|f (x)| dx −
+
[an + bn ]
2
−L
n=1
"
#
∞
N
a20 X 2
a20 X 2
2
2
+
+
[an + bn ] −
[an + bn ]
2
2
n=1
n=1
1
L
Z
"
L
2
∞
X
[a2n + b2n ]
n=N +1
→ 0,
as N → ∞,
By Theorem 1.
4
MINGFENG ZHAO
Theorem 3 (Parseval’s identity for odd functions). Let f ∈ L2 ([0, L]) and f (x) =
∞
X
bn sin
n=1
2
L
∞
X
L
Z
|f (x)|2 dx =
0
nπ x in 0 < x < L, then
L
b2n .
n=1

 f (x),
if 0 < x < L,
Proof. Consider the odd extension of f : fo (x) =
, since f ∈ L2 ([0, L]), then f0 ∈
 −f (−x), if −L < x < 0.
L2 ([−L, L]), and
1
L
Z
L
|fo (x)|2 dx
2
L
=
−L
fo (x)
=
L
Z
|f (x)|2 dx
0
∞
X
bn (x) sin
n=1
nπ x ,
L
x ∈ (−L, 0)
[
(0, L).
By Theorem 2, then
∞
X
L
1
L
Z
2
L
Z
|f0 (x)|2 dx =
−L
b2n .
n=1
Therefore, we have
L
|f (x)|2 dx =
0
∞
X
b2n .
n=1
Example 1. For f (x) = x in (0, 2), then the Fourier sine series of f is:
∞
nπ 4 X (−1)n+1
x=
sin
x ,
π n=1
n
2
∀x ∈ (0, 2),
that is,
bn =
4 (−1)n+1
·
,
π
n
∀n ≥ 1.
Notice that
2
2
Z
2
2
Z
|f (x)| dx =
0
2
x2 dx =
0
8
.
3
By Parsavel’s identity, then
∞
X
n=1
b2n =
Z
∞
16 X 1
8
2 2
·
=
=
|f (x)|2 dx.
π 2 n=1 n2
3
2 0
So we get
∞
X
π2
1
=
.
n2
6
n=1
LECTURE 12: BESSEL’S INEQUALITY AND PARSEVAL’S THEOREM
5
Example 2. For f (x) = x2 in (−π, π), then the Fourier series of f is:
x=
∞
X
π2
(−1)n
cos(nx),
+4
3
n2
n=1
∀x ∈ (−π, π),
that is,
 2
π


,
if n = 0,

3
an =
n


 4(−1) , if n ≥ 1.
n2
Notice that
1
π
Z
π
1
|f (x)| dx =
π
−π
2
Z
π
2
x dx =
π
−π
4
Z
π
0
x4 dx =
2 π5
2π 4
·
=
.
π 5
5
By Parsavel’s identity, then
∞
∞
X
a20 X 2
π4
1
2π 4
+
an =
+ 16
=
.
4
2
9
n
5
n=1
n=1
So we get
∞
X
1
π4
=
.
4
n
90
n=1
Department of Mathematics, The University of British Columbia, Room 121, 1984 Mathematics Road, Vancouver, B.C.
Canada V6T 1Z2
E-mail address: mingfeng@math.ubc.ca