Lecture #25
MATH 321: Real Variables II
University of British Columbia
Lecture #25:
Instructor:
Scribe:
March 10, 2008
Dr. Joel Feldman
Peter Wong
Theorem. If K is a compact metric space, C = { f : K → C | f is continuous } with uniform metric and A ⊂ C
obeys
(1) A is a complex algebra, (i.e., if f ∈ A and c ∈ C, then cf ∈ A)
(2) A vanishes nowhere,
(3) A separates points,
(4) A is self-adjoint, (i.e., f ∈ A =⇒ f ∗ ∈ A)
then A = C.
Application of Stone-Weierstraß Theorem
Corollary. If f : R → C is continuous and 2π-periodic, then f is the uniform limit of a sequence of trigonometric
polynomials. That is,
M
X
∀ε > 0, ∃ trigonometric polynomial
cn e
n=−M
inθ
M
X
inθ such that f (θ) −
cn e < ε for all θ.
n=−M
Reminder.
eiθ = cos θ + i sin θ has period 2π,
Proof. Let
cos θ =
eiθ + e−iθ
,
2
sin θ =
eiθ − e−iθ
,
2i
K = { z ∈ C | |z| = 1 }
CK = { φ : K → C | φ continuous }
A={
M
X
cn z n | M ∈ { 0, 1, . . . } , cn ∈ C, ∀ − M < n < M }
n=−M
Cp = { f : R → C | f continuous, f has period 2π }
with the sup norm. The map φ ∈ CK 7→ f (θ) = φ(eiθ ) ∈ Cp is a bijection that preserves the metric (an isometric
PM
PM
isometry ). In particular, φ(z) = n=−M cn z n 7→ f (θ) = n=−M cn einθ . It suffices to show that A = CK . Since
(1) A is a complex algebra.
(2) A vanishes nowhere (φ(z) = 1 is in A)
(3) A separates points (φ(z) = z is in A)
(4) A is self-adjoint, (z̄ =
1
z
so A = CK by Stone-Weierstraß.
1 z̄z
= |z|2 = (1)2 = 1.
on the unit circle1 . )
2
MATH 321: Lecture #25
Fourier Series (Rudin pp.185–192)
Motivation
Consider the problem of a vibrating string
y(x, t)
x=π
x=0
Let y(x, t) be the amplitude at position x and time t. We are told that
(1) y(0, t) = 0 for all t (left end is tied to a nail)
(2) y(π, t) = 0 for all t (right end is also tied to a nail)
(3) y(x, 0) = f (x) (initial position)
(4)
∂y
∂t (x, 0)
2
= g(x) (initial velocity)
2
∂ y
(5) ρ ∂∂t2y = T ∂x
2 (Assumed Newton’s law for small amplitude and only transverse vibration)
Solving by the method of separation of variables with ρ = T = 1, we get the solution
y(x, t) =
∞
X
An sin(nx) sin(nt) +
∞
X
Bn sin(nx) cos(nt),
n=1
n=1
which obeys (1), (2), and (5) for all An and Bn . Thus, the problem is solved if we can find the An ’s and Bn ’s such
that
∞
X
Bn sin(nx) = f (x)
(3)
nAn sin(nx) = g(x)
(4)
n=1
∞
X
n=1
This leads to the following questions:
1. Given a function f (θ), can it be written in the form f (θ) =
2. If so, what type of convergence do we have?
3. And what are the cn ’s?
P∞
n=−∞ cn e
inθ
?
Lecture #26
MATH 321: Real Variables II
University of British Columbia
Lecture #26:
Instructor:
Scribe:
March 12, 2008
Dr. Joel Feldman
Peter Wong
Fourier Series
Last time, we started asking the following questions:
P∞
(1) Which functions f : R → C can represented as f (x) =
n=−∞ cn e
inx
?
(2) If so, in what ways does the series converge?
(3) What are the coefficients cn ’s?
Easy answers:
(1) (Small part)
P
n∈Z cn e
inx
is 2π-periodic. If f (x) is not 2π-periodic, then it cannot be written as
(3) Since
π
Z
e
imx
dx =
(
2π,
Z
π
−π
We have
Z
π
∞
X
f (x)eimx dx =
−π
cn
eimx π
im −π
if m = 0,
= 0, if m ∈ Z \ { 0 }
−π
n=−∞
∞
X
einx e−imx dx =
n=−∞
cn
Z
π
ei(n−m)x dx
−π
assuming adequate convergence (for moving the integral inside the sum.) But
(
Z π
2π, if n = m,
i(n−m)x
e
dx =
0,
otherwise
−π
So
Rπ
−π
f (x)e−imx dx = 2πcm =⇒ cm =
1
2π
Rπ
−π
cm =
f (x)e−imx dx. If f ∈ R on [−π, π], then
Z
π
f (x)e−imx dx
−π
exists, called the mth -Fourier coefficient of f (x), so
∞
X
cn einx
with each cn =
1
2π
n=−∞
Z
π
f (x)einx dx
−π
is called the Fourier series of f , whether or not it converges.
Remark.
Z
π
ei(n−m)x dx =
−π
is not magic. It is in fact just linear algebra. Recall that
• Cn = { ~v = (v1 , . . . , vn ) | v1 , . . . , vn ∈ C },
P
• (~v , w)
~ = nj=1 vj wj , 1
1 Mathematical
physicists write
Pn
j=1 vj wj
(
2π, if n = m
0,
if n 6= m
P
cn einx .
2
MATH 321: Lecture #26
• kvk2 = (~v , ~v ) =
Pn
j=1
|vj |2
• If the matrix A = [Aij ]1≤i≤n , then A maps ~v ∈ Cn to (A~v )i =
Pn
j=1
Aij vj .
• ~v is called an eigenvector of A with eigenvalue λ. If ~v 6= ~0 and A~v = λ~v .
• The matrix A is called self-adjoint 2 (or Hermitian 3 ) if Aij = Aji for all 1 ≤ i, j ≤ n, or equivalently,
(A~v , w)
~ = (~v , Aw)
~ for all ~v , w
~ ∈ Cn .
Theorem. If A is self-adjoint, then
(a) all eigenvalues of A are real (i.e., A~v = λ~v , ~v 6= 0 =⇒ λ ∈ R)
(b) if A~v = λ~v and Aw
~ = µw
~ for ~v 6= 0, w
~ 6= 0 and λ 6= µ, then ~v ⊥ w
~ (i.e., (~v , w)
~ ≡ 0)
(c) there is an orthonormal4 basis for Cn consisting of eigenvalues of A.
Proof. See the web notes “Normal Matrices”
Connection between Linear Algebra and Fourier series
Cn −→ { f : R → C | f ∈ R on [−π, π],
Z π
(~v , w)
~ −→ (f, g) =
f (x)g(x) dx
f is 2π-periodic }
−π
d
d
d
is self-adjoint since (i dx
f, g) = (f, i dx
g)
dx
Eigenvectors −→ Eigenfunctions en (x) = einx , n ∈ Z
A −→ Differential operator − i
We see that
−i
d inx
e
= −neinx
dx
=⇒
π
Z
ei(n−m)x dx = (einx , eimx ) = 0 for n 6= m
−π
Theorem. (Best Approximation Theorem) Let f : [−π, π] → R be a Riemann integrable function. Let n ∈ N and
let dm ∈ C for each −n ≤ m ≤ n. Write
n
X
t(x) =
dm eimx dx
and
m=−n
with cm =
1
2π
Rπ
−π
π
Z
|f (x) − t(x)|2 dx =
π
|f (x) − s(x)|2 dx +
−π
n
X
m=−n
2
|cm − dm |
(
≥ 0,
= 0,
in Mathematics: For the lack of a better name . . .
if this makes you feel better . . .
4 Orthogonal and of unit length.
3 And
cm eimx dx
m=−n
−π
2 Tragedy
n
X
f (x)e−imx dx. Then
Z
where
s(x) =
n
X
|cm − dm |2
m=−n
in the usual sense,
cm = dm for all m ∈ Z
Lecture #27
MATH 321: Real Variables II
University of British Columbia
Lecture #27:
Instructor:
Scribe:
March 14, 2008
Dr. Joel Feldman
Peter Wong
Theorem. (Best Approximation Theorem) Given
f : [−π, π] → R, which is a Riemann integrable on [−π, π]
Z π
1
f (x)e−inx dx
cn = 2π
−π
n
X
sn (x) =
cm eimx ,
tn (x) =
m=−n
kgn k2 =
π
Z
n
X
dm eimx
with some dm ∈ C, −n ≤ m ≤ n
m=−n
1/2
|gn (x)|2 dx
−π
Use the notation (g, h) =
Rπ
−π
g(x)h(x) dx. Note that
kgk22 =
cn =
(e
inx
,e
imx
)=
Z
π
g(x)g(x) dx = (g, g)
−π
inx
1
)
2π (f, e
Z π
i(n−m)x
e
dx =
−π
kf −
tn k22
(
0,
2π,
n 6= m,
n=m
= (f − tn , f − tn ) = (f − sn + sn − tn , f − sn + sn − tn )
= (f − sn , f − sn ) + (f − sn , sn − tn ) + (sn − tn , f − sn ) + (sn − tn , sn − tn )
1st term: (f − sn , f − sn ) = kf − sn k22
n
n
X
X
4th term: (sn − tn , sn − tn ) =
(cm − dm )(c` − d` ) (eimx − ei`x ) =
2π|cm − dm |2
|
{z
}
m=−n
`,m=−n
=
2nd term: (f − sn , sn − tn ) =
=
n
X

 0,

 2π,
m 6= `,
m=`
(cm − dm )(f − sn , eimx )
m=−n
n
X
n
X
c` (eimx − ei`x ) = 0
(cm − dm ) (f, eimx ) −
|
| {z }
{z
}
m=−n
2πcm
`=−n
=
|

 0,

 2π,
{z
2πcm
m 6= `,
m=`
}
3rd term: (sn − tn , f − sn ) = (f − sn , sn − tn ) = 0 = 0
Conclusion: kf − tn k22 = kf − sn k22 + 2π
Pn
m=−n
|cm − dm |2 =⇒ kf − tn k22 ≥ kf − sn k22 with equality if and only
if cm = dm for all −n ≤ m ≤ n.
Corollary. (Bessel’s inequality) kf k22 ≤ 2π
the mean to f .
P∞
−∞
|cn |2 with equatlity if and only if
Pn
m=−n cm e
imx
converges in
2
MATH 321: Lecture #27
Proof. Choose dm = 0 for all m. Then
kf k22
= kf −
sn k22
+ 2π
n
X
|cm |2
m=−n
=⇒ 2π
n
X
=⇒ 2π
m=−n
∞
X
|cm |2 ≤ kf k22
|cm |2 converges and is no more than kf k22 , with equality if and only if lim kf − sn k22 = 0.
n→∞
m=−∞
Corollary. limn→∞ |cn | = 0.
2
n
X
2
Proof. By the nth term test: |cn | + |c−n | =
2
|cm |
m=−n
|
{z
n−1
X
−
|cm |2
m=−(n−1)
}
P
2
→ ∞
m=−∞ |cm | as m→∞
|
{z
}
P
2
→ ∞
m=−∞ |cm | as m→∞
Theorem. (Parseval’s Theorem – Convergence of Fourier Series in the mean) Let f, g : [−π, π] → C be Riemann
integrable on [−π, π]. Let
cn =
1
2π
Z
π
f (x)e−inx dx,
dn =
−π
then
Rπ
(a) limn→∞ −π |f (x) − sn (f, x)|2 dx = 0
Rπ
P
1
2
(b) 2π
|f (x)|2 dx = ∞
n=−∞ |cn |
−π
Rπ
P
1
f (x)g(x) dx = ∞
(c) 2π
n=−∞ cn dn
−π
1
2π
Z
π
−π
g(x)e−inx dx,
and
sn (f, x) =
n
X
m=−n
cm eimx ,