MATH 321: Real Variables II
Lecture #30
University of British Columbia
Lecture #30:
Instructor:
Scribe:
March 26, 2008
Dr. Joel Feldman
Peter Wong
Theorem. (Pointwise Convergence of Fourier Series)
Assume: f : R → C be 2π-periodic and Riemann integrable on [−π, π]; x ∈ R
|f (x + t) − f (x+ ) ≤ M |t|
for all 0 < x < δ
|f (x + t) − f (x− ) ≤ M |t|
for all − δ < x < 0
f (x+ )
f (x− ) = limt→0
t<0
for some δ > 0 and M < ∞, then limn→∞ Sn = 21 [f (x+ ) + f (x− )]
Proof. Last class we defined

f (x−t)−f (x+ )


 sin(t/2) , if t ∈ [−π, 0),
(x− )
g(t) = f (x−t)−f
, if t ∈ (0, π],
sin(t/2)


0,
if t = 0.
and showed that
Sn (x) − 12 [f (x+ ) − f (x− )]
Z π
i
1
= 4πi
g(t)e 2 t eint dt −
=
−π
1
(nth
Fourier
2i
1
4πi
π
Z
i
g(t)e− 2 t e−int dt
−π
coefficient of g(t)e(i/2)t ) −
1
2i (−nth
Fourier coefficient of g(t)e−(i/2)t )
so the Theorem follows from
Lemma. g(t) is Riemann integrable on [−π, π].
Proof. Let ε > 0, ε < δ. Note that
1
sin(t/2)
(1) On [−π, −ε] ∪ [ε, π], g(t) is Riemann-integrable since f (x − t), f (x± ),
are integrable.
(2) There is a constant M 0 > 0 such that |g(t)| ≤ M 0 for all t ∈ [−ε, ε] because
M(t/2) |g(t)| ≤ 2 sin(t/2)
≤ M0
since limt→0
Z
t
sin(t/2)
π
−π
g(t) dt −
Z
= 2. Thus,
π
g(t) dt =
−π
=
Z
Z
−ε
g(t) dt +
−π
ε
g(t) dt −
ε
−ε
Z
ε
−ε
π
g(t) dt −
Z
−ε
−π
g(t) dt −
Z
ε
g(t) dt −
−ε
Z
π
g(t) dt
ε
ε
g(t) dt
ε
−ε
R
g(t) dt +
g(t) dt < M 0 (2ε),
Z
for all 0 < ε < δ. This means
Z
ε
−ε
−ε
R
R ε
Since −ε g(t) dt < M 0 (2ε) and Z
g(t) dt −
Z
R
= . Hence, g ∈ R.
ε
g(t) dt ≤ 4M 0 ε
−ε
2
MATH 321: Lecture #30
Theorem. (Uniform Convergence of Fourier Series)
If f : R → C is 2π-periodic and is piecewise C 1 , then Sn (f, x) converges uniformly to f on any closed interval
that contains no point of discontinuity of f ,
Definition. f is piecewise C 1 on [−π, π] if there exist n ∈ N and −π = x0 < x1 < x2 < · · · < xn = π such that
(1) for each 1 ≤ j ≤ n, f has at least one continuous derivative on (xj−1 , xj )
(2) limt→0± f (xj + t) and limt→0± f 0 (xj + t) exist. That is,
f (x+
j ),
f (x−
j ),
f 0 (x+
j ),
f 0 (x−
j )
−π = x0
x2
x3
x4 = π
m=−∞ cm e
imx
, then f 0 has
x1
all exist.
Proof of Theorem:
Step 1: f is continuous and piecewise C 1 .
P∞
imx
m=−∞ im cm e
If f has Fourier series expansion
P∞
MATH 321: Real Variables II
Lecture #31
University of British Columbia
Lecture #31:
Instructor:
Scribe:
March 28, 2008
Dr. Joel Feldman
Peter Wong
Theorem. (Uniform Convergence of Fourier Series)
If f : R → C is 2π-periodic and is piecewise C 1 , then the Fourier series of f converges uniformly on any closed
interval that does not contain a point of discontinuity of f .
[
][
][
]
Proof. Step 1: If f is continuous and piecewise C 1 , then the Fourier series of f converges uniformly to f .
Step (1a): If f has Fourier series
P∞
n=−∞
an einx , then f 0 has Fourier series
P∞
n=−∞
in an einx .
Proof. The n-th Fourier coefficient of f 0 is
Z π
1
f 0 (x)e−inx dx
2π −π
n Z
1 X xk 0
f (x)e−inx dx
=
2π
xk−1
k=1
(where f 0 (x) is continuous on [−π, π] except at −π = x0 < x1 < x2 < · · · < xn = π)
Z xk −ε
n
1 X
=
lim
f 0 (x)e−inx dx
ε→0 x
2π
+ε
k−1
k=1
)
(
xk −ε
Z xk −ε
n
X
1
−inx
−inx + in
f (x)e
dx
=
lim f (x)e
ε→0
2π
xk−1
xk−1 +ε
k=1
x−
n
n Z
X
k =xk
1 X xk
1
−inx +(in)
=
f (x)e−inx dx
f (x)e
+
2π
2π
x
x
=x
k−1
k−1
k=1
k=1
k−1
|
{z
}
{z
}
|
the xk with k = 1, . . . , n − 1 contributions cancel;
and the x0 = −π, xn = π terms cancel by periodicity
1 Rπ
f (x)e−inx dx
an = 2π −π
= 0 + in an = in an .
Step (1b): If f is piecewise C 1 and continuous, the Fourier series converges uniformly. We know
∞
X
an einx
n=−∞
converges pointwise to f (x) since |f (x + t) − f (x)| ≤ sup |f 0 ||t| ≤ M |t| and we know
∞
X
n=−∞
in an einx
2
MATH 321: Lecture #31
converges in the mean to f 0 (x). So for each x,
∞
N
N
X
X
X
X
inx inx
inx
inx
an e an e = an e −
an e = f (x) −
n=−∞
n=−N
n=−N
|n|>N
s
s
X
X
X X
1
1 n an ≤
|n an |2
|an | =
≤
n
n2
But
P∞
1
n=−∞ n2
converges and
lim
N →∞
Step 2: Let F (x) =
(
π−x
2π ,
0,
P∞
n=−∞
s X
|n|>N
|n|>N
|n|>N
1
n2
|n an |2 converges to
=0
and
|n|>N
lim
N →∞
1
2π
Rπ
s X
−π
|n|>N
|f 0 (x)|2 dx < ∞. So,
|n an |2
|n|>N
0 < x < 2π
x = 0, 2
y
1
2
b
−3π
b
b
−π
−2π
π
2π
3π
x
and F (x + 2π) = F (x) for all x ∈ R. By Problem Set 9, #3, F (x) has Foureir series
∞
X
1
2πi
1 inx
ne
n=−∞
n6=0
which converges uniformly on [ε, 2π − ε] for any ε > 0.
Step 3: (Proof of theorem) Suppose f has jump discontinuities on [−π, π) at p1 , . . . , pm . Let, for 1 ≤ j ≤ m,
−
let δj = f (p+
j ) − f (pj ). Write
f (x) =
n
X
δj F (x − pj )
+
f (x) −
δj F (x − pj )
j=1
j=1
|
n
X
{z
}
FS conv. uniformly except at pj
|
{z
}
continuous FS converges uniformly