Jim Lambers
MAT 773
Fall Semester 2018-19
Lecture 5 and 6 Notes
These notes correspond to Section 1.3 in the text.
Convergence Theorems for Fourier Series
The Riemann-Lebesgue Lemma
Theorem 21 (Riemann-Lebesgue Lemma) Let f be piecewise continuous on [a, b].
Then
Z b
Z b
f (x) sin kx dx = 0.
f (x) cos kx dx = lim
lim
k→∞ a
k→∞ a
Proof: In the case where f is continuously differentiable on [a, b], we have
Z b
Z
f (b) sin kb − f (a) sin ka 1 b 0
−
f (x) cos kx dx =
f (x) sin kx dx
k
k a
a
which approaches zero as k → ∞. The case of integrating f (x) sin kx is similar. 2
Convergence at a Point of Continuity
Let f have the Fourier series F (x) = a0 +
of f converges to f at x if
lim
N →∞
a0 +
P∞
N
X
k=1 ak
cos kx + bk sin kx. We say that the Fourier series
!
ak cos kx + bk sin kx
= f (x).
k=1
Theorem 22 Let f be continuous and 2π-periodic. If f 0 (x) exists, then the Fourier
series of f at x converges to f (x).
P
To prove this theorem, we let SN (x) = a0 + N
k=1 ak cos kx + bk sin kx be the truncated Fourier
series of f (x) on [−π, π]. Using the identity for the cosine of a difference, and the formulas for the
Fourier coefficients, this series can be rewritten as
SN (x) = a0 +
N
X
ak cos kx + bk sin kx
k=1
Z π
N Z π
π
X
1
1
1
f (t) dt +
f (t) cos kt dt cos kx +
f (t) sin kt dt sin kx
2π −π
π −π
π −π
k=1
!
Z
N
1 π
1 X
f (t)
+
cos kt cos kx + sin kt sin kx dt
π −π
2
k=1
!
Z
N
1 π
1 X
+
cos(k(t − x)) dt.
f (t)
π −π
2
Z
=
=
=
k=1
1
The following lemma will help us to evaluate the summation.
Lemma 23 For u ∈ [−π, π],

 sin((N + 1/2)u)
1
2 sin(u/2)
+ cos u + cos 2u + · · · + cos N u =

2
N + 1/2
u 6= 0,
u = 0.
Proof: The case where u = 0 is straightforward, so we assume u 6= 0. Using the formula for the
partial sum of a geometric series,
N
X
rN +1 − 1
rk =
,
r−1
k=0
we obtain
N
N
k=1
k=0
1 X
1 X
+
cos ku = − +
cos ku
2
2
N
X
1
eiku
= − + Re
2
1
= − + Re
2
k=0
N
X
(eiu )k
k=0
ei(N +1)u
1
−1
= − + Re
2
eiu − 1
ei(N +1)u/2 ei(N +1)u/2 − e−i(N +1)u/2
1
= − + Re
2
eiu/2
eiu/2 − e−iu/2
sin((N + 1)u/2)
1
= − + Re eiN u/2
2
sin(u/2)
1
sin((N + 1)u/2)
= − + cos(N u/2)
.
2
sin(u/2)
Using a product-to-sum identity, we obtain
N
1 X
1
1
(N + 1)u N u
(N + 1)u N u
+
cos ku = − +
sin
+
+ sin
−
2
2 2 sin(u/2)
2
2
2
2
k=1
1
1
[sin((N + 1/2)u) + sin (u/2)]
= − +
2 2 sin(u/2)
sin((N + 1/2)u)
=
.
2 sin(u/2)
2
Substituting the result of Lemma 23 into SN (x) yields
Z π
Z π
1
sin((N + 1/2)(t − x))
SN (x) =
f (t)
dt =
f (t)PN (t − x) dt
2π −π
sin((t − x)/2)
−π
2
where
PN (u) =
1 sin((N + 1/2)u)
.
2π
sin(u/2)
The following result pertaining to PN (u) will be helpful.
Lemma 24
Z
π
PN (u) du = 1.
−π
Proof: Because PN (u) is an even function, we have, by Lemma 23,
Z π
Z π
PN (u) du = 2
PN (u) du
−π
0
Z
1 π sin((N + 1/2)u)
=
du
π 0
sin(u/2)
Z
2 π1
=
+ cos u + cos 2u + · · · + cos N u du
π 0 2
1
1
2 u
+ sin u + sin 2u + · · · +
sin N u
=
π 2
2
N
= 1.
π
0
2
To show that SN (x) converges to f (x), we must show
Z π
[f (x) − f (t)]PN (t − x) dt → 0
−π
as N → ∞, from Lemma 23. We have
Z π
1
f (x) − f (t)
f (x) − SN (x) =
sin((N + 1/2)(t − x)) dt.
2π −π sin((t − x)/2)
Consider the function
g(t) =
f (t) − f (x)
,
sin((t − x)/2)
t ∈ [−π, π].
This function is continuous, except possibly at t = x. However,
lim g(t) = lim 2
t→x
t→x
f (t) − f (x) (t − x)/2
= 2f 0 (x).
t−x
sin((t − x)/2)
Therefore, g(t) can be made continuous for all t by defining g(x) = 2f 0 (x). We conclude that
Z π
1
f (x) − SN (x) = −
g(t) sin((N + 1/2)(t − x)) dt → 0
2π −π
as N → ∞ by the Riemann-Lebesgue Lemma. This completes the proof of Theorem 22.
3
Convergence at a Point of Discontinuity
Definition 25 The right and left limits of f at x are
f (x + 0) =
lim f (x + h),
h→0+
f (x − 0) =
lim f (x − h).
h→0+
We say that f is right differentiable if the limit
f 0 (x + 0) = lim
h→0+
f (x + h) − f (x + 0)
h
exists, and that f is left differentiable if the limit
f 0 (x − 0) = lim
h→0+
f (x − 0) − f (x − h)
h
exists.
Example 26 Let f (x) be the periodic extension of y = x on [−π, π]. Then f (x) is discontinuous
at x = kπ where k is an integer. We also have
f (π + 0) = −π,
f (π − 0) = π,
and
f 0 (π + 0) = 1,
f 0 (π − 0) = 1.
(
x
f (x) =
π−x
if 0 ≤ x ≤ π/2,
if π/2 ≤ x ≤ π.
2
Example 27 Let
Then f is continuous on [−π, π] but not differentiable at x = π/2. We have f 0 (π/2 − 0) = 1 and
f 0 (π/2 + 0) = −1. 2
Theorem 28 Let f be periodic and piecewise continuous, and let f be left and right
differentiable at x. Then the Fourier series of f at x converges to
f (x + 0) + f (x − 0)
.
2
The proof of this theorem is very similar to that of Theorem 22.
Example 29 Let f be the function from Example 26. By Theorem 28, at x = π the Fourier series
of f (x) converges to
f (π + 0) + f (π − 0)
−π + π
=
= 0.
2
2
This is consistent with the Fourier series of f computed earlier,
∞
X
2(−1)k+1
F (x) =
sin kx
k
k=1
which is zero at x = π. 2
4
Uniform Convergence
Theorem 30 Let f be continuous and piecewise smooth on [−π, π]. Then the Fourier
series of f on [−π, π] converges uniformly to f on [−π, π].
Proof: We prove this theorem under the weaker assumption that f is twice continuously differentiable. Assume f has the Fourier series
∞
X
f (x) = a0 +
ak cos kx + bk sin kx.
k=1
As in the proof of Theorem 22, we let
SN (x) = a0 +
N
X
ak cos kx + bk sin kx
k=1
be the truncated Fourier series of f . We then have
∞
X
f (x) − SN (x) =
ak cos kx + bk sin kx.
k=N +1
We also have
00
f (x) =
∞
X
(−k 2 )(ak cos kx + bk sin kx).
k=1
f 00
Because
is continuous, we must have k 2 ak and k 2 bk converge to zero by the Riemann-Lebesgue
Lemma. As a convergent sequence is bounded, it follows that |k 2 ak | and |k 2 bk | are both bounded
by some number M . We then have
∞
X
|ak | + |bk | ≤
k=1
∞
X
2M
k=1
k2
< ∞.
The result follows from Lemma 33 below. 2
Example 31 Consider the sawtooth wave from Example 27. Its periodic extension is piecewise
smooth. Therefore, by Theorem 30, its Fourier series on [0, π] converges uniformly. 2
Example 32 Consider the Fourier series for f (x) = x on [−π, π]. Its periodic extension is piecewise
smooth but discontinuous, so Theorem 30 does not apply. Its Fourier series does not converge
uniformly. 2
Lemma 33 Let
f (x) = a0 +
∞
X
ak cos kx + bk sin kx
k=1
with
∞
X
|ak | + |bk | < ∞.
k=1
Then the Fourier series converges uniformly and absolutely to f (x).
5
P∞
k=N +1 |ak |
Proof: Let > 0. Then, there exists N0 > 0 such that for N ≥ N0 ,
SN (x) = a0 +
N
X
+ |bk | < . Let
ak cos kx + bk sin kx.
k=1
Then, we also have, for N ≥ N0 ,
|f (x) − SN (x)| =
∞
X
ak cos kx + bk sin kx ≤
k=N +1
∞
X
|ak | + |bk | < .
k=1
That is, SN (x) converges to f (x). Because the choice of N0 does not depend on x, convergence is
uniform. 2
Convergence in the Mean
Lemma 34 Let V = L2 ([−π, π]) and let f ∈ V . Let
VN = span{1, cos kx, sin kx, . . . , cos N x, sin N x}
and
fN (x) = a0 +
N
X
ak cos kx + bk sin kx
k=1
where ak and bk are the Fourier coefficients of f . Then fN is the orthogonal projection
of f onto VN .
Theorem 35 Let f ∈ L2 ([−π, π]), and let
fN (x) = a0 +
N
X
ak cos kx + bk sin kx
k=1
where ak and bk are the Fourier coefficients of f . Then fN converges to f in L2 ([−π, π]).
Proof: Let > 0 and let g be a smooth 2π-periodic function that approximates f to within /2;
that is, kf − gkL2 < /2. Such a g exists by Lemma 38 below. Let fN and gN be truncated Fourier
series for f and g, respectively. Because g is smooth, gN converges uniformly to g on [−π,
√ π] by
Theorem 30, so there exists an N0 > 0 such that for N ≥ N0 , |g(x) − gN (x)| < /(2 2π) for
x ∈ [−π, π], and therefore kg − gN kL2 ≤ /2. Because fN is the closest function from VN to f in the
L2 norm, where VN = span{1, cos x, sin x, . . .}, we must have kf − fN kL2 ≤ kf − gN kL2 . Putting
all of these results together, we have
kf − fN kL2 ≤ kf − gN kL2 ≤ kf − gkL2 + kg − gN kL2 ≤ .
2
6
Theorem 36 Let f ∈ L2 ([−π, π]), and let
N
X
fN (x) =
αk eikx
k=−N
where
1
αk =
2π
Z
π
f (t)e−ikt dt,
k∈Z
−π
are the complex Fourier coefficients of f . Then fN converges to f in L2 ([−π, π]).
Example 37 All Fourier series presented in the examples in these notes converge in the mean, as
they arise from functions in L2 over an appropriate interval. 2
Lemma 38 Let f ∈ L2 ([−π, π]). Then f can be approximated arbitrarily closely by a
smooth 2π-periodic function.
Theorem 39 (Parseval’s Identity, Real Version) Let f ∈ L2 ([−π, π]) have the
Fourier series
∞
X
f (x) = a0 +
ak cos kx + bk sin kx.
k=1
Then
∞
X
1
kf k2 = 2|a0 |2 +
|ak |2 + |bk |2 .
π
k=1
Theorem 40 (Parseval’s Identity, Complex Version) Let f ∈ L2 ([−π, π]) have the
Fourier series
∞
X
f (x) =
αk eikx .
k=−∞
Then
∞
X
1
kf k2 =
|αk |2 .
2π
k=−∞
Furthermore,
∞
X
1
hf, gi =
αk β k ,
2π
k=−∞
where the βk , k ∈ Z, are the complex Fourier coefficients of g.
Proof: Let
fN (x) =
N
X
αk eikx ,
gN (x) =
k=−N
N
X
k=−N
7
βk eikx .
Then, by the orthogonality of the functions {eikx }k∈Z , we have
Z π
fN (x)gN (x) dx
hfN , gN i =
−π
Z
N
X
π
=
−π
=
αk eikx
k=−N
N
N
X
X
N
X
!
!
β` ei`x
dx
`=−N
αk βk heikx , ei`x
k=−N `=−N
N
X
= 2π
αk βk .
k=−N
We then have
|hf, gi − hfN , gN i| = |hf, gi − hf, gN i + hf, gN i − hfN , gN i
≤ |hf, g − gN i| + |hf − fN , gN i|
≤ kf kkg − gN k + kf − fN kkgN k.
By Theorem 36, hfN , gN i converges to hf, gi, from which the Theorem follows. 2
One consequence of Parseval’s Identity is Bessel’s Inequality
N
X
|αk |2 ≤
k=−N
1
kf k2 .
2π
Example 41 Let f (x) = x on [−π, π]. From Example 29, f has the Fourier series
∞
X
2(−1)k+1
F (x) =
k
k=1
sin kx.
By the real version of Parseval’s identity,
∞ 2
X
2
k=1
k
We conclude that
1
=
π
Z
π
x2 dx =
−π
∞
X
1
π2
=
.
k2
6
k=1
2
Exercises
Chapter 1: Exercises 32, 33, 36, 37, 38, 39
8
2π 2
.
3