1. No category

CHAPTER 10
Fourier Series
In the late eighteenth century, it was well-known that complicated functions
could sometimes be approximated by a sequence of polynomials. Some of the leading
mathematicians at that time, including such luminaries as Daniel Bernoulli, Euler
and d’Alembert began studying the possibility of using sequences of trigonometric
functions for approximation. In 1807, this idea opened into a huge area of research
when Joseph Fourier used series of sines and cosines to solve several outstanding
partial di↵erential equations of physics.1
In particular, he used series of the form
1
X
an cos nx + bn sin nx
n=0
to approximate his solutions. Series of this form are called trigonometric series,
and the ones derived from Fourier’s methods are called Fourier series. Much of
the mathematical research done in the nineteenth and early twentieth century was
devoted to understanding the convergence of Fourier series. This chapter presents
nothing more than the tip of that huge iceberg.
1. Trigonometric Polynomials
Definition 10.1. A function of the form
(111)
p(x) =
n
X
↵k cos kx +
k
sin kx
k=0
is called a trigonometric polynomial. The largest value of k such that |↵k |+ k | =
6 0 is
the degree of the polynomial. Denote by T the set of all trigonometric polynomials.
Evidently, all functions in T are 2⇡-periodic and T is closed under addition
and multiplication by real numbers. Indeed, it is a real vector space, in the sense of
linear algebra and the set {sin nx : n 2 N} [ {cos nx : n 2 !} is a basis for T .
The following theorem can be proved using integration by parts or trigonometric
identities.
Theorem 10.2. If m, n 2 Z, then
Z ⇡
(112)
sin mx cos nx dx = 0,
⇡
1
Fourier’s methods can be seen in most books on partial di↵erential equations, such as [3]. For
example, see solutions of the heat and wave equations using the method of separation of variables.
10-1
10-2
CHAPTER 10. FOURIER SERIES
8
>
<0,
sin mx sin nx dx = 0,
>
⇡
:
⇡
Z
(113)
⇡
and
Z
(114)
m 6= n
m = 0 or n = 0
m = n 6= 0
8
>
<0,
cos mx cos nx dx = 2⇡
>
⇡
:
⇡
⇡
m 6= n
m=n=0.
m = n 6= 0
If p(x) is as in (111), then Theorem 10.2 shows
Z ⇡
2⇡↵0 =
p(x) dx,
⇡
and for n > 0,
⇡↵n =
Z
⇡
p(x) cos nx dx,
⇡
⇡
n
=
Z
⇡
p(x) cos nx dx.
⇡
Combining these, it follows that if
Z
Z
1 ⇡
1
an =
p(x) cos nx dx and bn =
⇡
⇡
⇡
for n 2 !, then
(115)
p(x) =
⇡
p(x) sin nx dx
⇡
1
a0 X
+
an cos nx + bn sin nx.
2
n=1
(Remember that all but a finite number of the an and bn are 0!)
At this point, the logical question is whether this same method can be used to
represent a more general 2⇡-periodic function. For any function f , integrable on
[ ⇡, ⇡], the coefficients can be defined as above; i.e., for n 2 !,
Z
Z
1 ⇡
1 ⇡
(116)
an =
f (x) cos nx dx and bn =
f (x) sin nx dx.
⇡
⇡
⇡
⇡
The numbers an and bn are called the Fourier coefficients of f . The problem is
whether and in what sense an equation such as (115) might be true. This turns out
to be a very deep and difficult question with no short answer.2 Because we don’t
know whether equality in the sense of (115) is true, the usual practice is to write
(117)
f (x) ⇠
1
a0 X
+
an cos nx + bn sin nx,
2
n=1
indicating that the series on the right is calculated from the function on the left
using (116). The series is called the Fourier series for f .
2Many people, including me, would argue that the study of Fourier series has been the
most important area of mathematical research over the past two centuries. Huge mathematical
disciplines, including set theory, measure theory and harmonic analysis trace their lineage back to
basic questions about Fourier series. Even after centuries of study, research in this area continues
unabated.
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2. THE RIEMANN LEBESGUE LEMMA
10-3
Figure 1. This shows f (x) = |x|, s1 (x) and s3 (x), where sn (x) is
the nth partial sum of the Fourier series for f .
Example 10.1. Let f (x) = |x|. Since f is an even functions and sin nx is odd,
Z
1 ⇡
bn =
|x| cos nx dx = 0
⇡
⇡
for all n 2 N. On the other hand,
1
an =
⇡
for n 2 Z. Therefore,
Z
⇡ 4 cos x
2
⇡
(See Figure 10.1.)
|x| ⇠
8
<⇡,
|x| cos nx dx = 2(cos n⇡
:
⇡
n2 ⇡
n=0
⇡
4 cos 3x
9⇡
4 cos 5x
25⇡
1)
4 cos 7x
49⇡
,
n2N
4 cos 9x
+ ···
81⇡
There are at least two fundamental questions arising from (117): Does the
Fourier series of f converge to f ? Can f be recovered from its Fourier series, even
if the Fourier series does not converge to f ? These are often called the convergence
and representation questions, respectively. The next few sections will give some
partial answers.
2. The Riemann Lebesgue Lemma
We learned early in our study of series that the first and simplest convergence
test is to check whether the terms go to zero. For Fourier series, this is always the
case.
Rb
Theorem 10.3 (Riemann-Lebesgue Lemma). If f is a function such that a f
exists, then
Z b
Z b
lim
f (t) cos ↵t dt = 0 and lim
f (t) sin ↵t dt = 0.
↵!1
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a
↵!1
a
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10-4
CHAPTER 10. FOURIER SERIES
Proof. Since the two limits have similar proofs, only the first will be proved.
Let " > 0 and P be a generic partition of [a, b] satisfying
Z b
"
0<
f D (f, P ) < .
2
a
For mi = glb {f (x) : xi 1 < x < xi }, define a function g on [a, b] by g(x) = mi
Rb
when xi 1  x < xi and g(b) = mn . Note that a g = D (f, P ), so
Z b
"
(118)
0<
(f g) < .
2
a
Choose
n
(119)
↵>
Since f
4X
|mi |.
" i=1
g,
Z
b
f (t) cos ↵t dt =
a



Z
Z
Z
Z
b
(f (t)
g(t)) cos ↵t dt +
a
Z
b
(f (t)
g(t)) cos ↵t dt +
a
b
g(t) cos ↵t dt
a
Z
b
g(t) cos ↵t dt
a
n
b
(f
g) +
(f
g) +
a
1X
mi (sin(↵xi )
↵ i=1
sin(↵xi
1 ))
n
b
a
2X
|mi |
↵ i=1
Use (118) and (119).
" "
+
2 2
="
<
⇤
Corollary 10.4. If f is integrable on [ ⇡, ⇡] with an and bn the Fourier
coefficients of f , then an ! 0 and bn ! 0.
3. The Dirichlet Kernel
Suppose f is integrable on [ ⇡, ⇡] and 2⇡-periodic on R, so the Fourier series
of f exists. The partial sums of the Fourier series are written as sn (f, x), or more
simply sn (x) when there is only one function in sight. To be more precise,
n
sn (f, x) =
a0 X
+
(ak cos kx + bk sin kx) .
2
k=1
Notice sn is a trigonometric polynomial of degree at most n.
We begin with the following calculation.
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3. THE DIRICHLET KERNEL
10-5
15
12
9
6
3
-p
-
p
p
2
2
p
-3
Figure 2. The Dirichlet kernel Dn (s) for n = 1, 4, 7.
n
a0 X
+
(ak cos kx + bk sin kx)
2
k=1
Z ⇡
Z
n
X
1
1 ⇡
=
f (t) dt +
(f (t) cos kt cos kx + f (t) sin kt sin kx) dt
2⇡
⇡
⇡
⇡
k=1
!
Z ⇡
n
X
1
=
f (t) 1 +
2(cos kt cos kx + sin kt sin kx) dt
2⇡
⇡
k=1
!
Z ⇡
n
X
1
=
f (t) 1 +
2 cos k(x t) dt
2⇡
⇡
sn (x) =
k=1
Substitute s = x
Z
1
(120)
=
2⇡
t and use the assumption that f is 2⇡-periodic.
!
n
⇡
X
f (x s) 1 + 2
cos ks ds
⇡
k=1
The sequence of trigonometric polynomials from within the integral,
n
X
(121)
Dn (s) = 1 + 2
cos ks,
k=1
is called the Dirichlet kernel. Its properties will prove useful for determining the
pointwise convergence of Fourier series.
Theorem 10.5. The Dirichlet kernel has the following properties.
(a) Dn (s) is an even 2⇡-periodic function for each n 2 N.
(b) Dn (0) = 2n + 1 for each n 2 N.
(c) |DnZ
(s)|  2n + 1 for each n 2 N and all s.
⇡
1
(d)
Dn (s) ds = 1 for each n 2 N.
2⇡
⇡
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10-6
CHAPTER 10. FOURIER SERIES
(e) Dn (s) =
sin(n + 1/2)s
for each n 2 N and s/2 not an integer multiple
sin s/2
of ⇡.
Proof. Properties (a)–(d) follow from the definition of the kernel.
The proof of property (e) uses some trigonometric manipulation. Suppose n 2 N
and s 6= m⇡ for any m 2 Z.
Dn (s) = 1 + 2
n
X
cos ks
k=1
Use the facts that the cosine is even and the sine is odd.
=
n
X
cos ks +
k= n
=
=
1
sin 2s
1
sin 2s
n
cos 2s X
sin ks
sin 2s
k= n
n ⇣
X
k= n
n
X
k= n
sin
⌘
s
s
cos ks + cos sin ks
2
2
1
sin(k + )s
2
This is a telescoping sum.
=
sin(n + 12 )s
sin 2s
⇤
According to (120),
sn (f, x) =
1
2⇡
Z
⇡
f (x
t)Dn (t) dt.
⇡
Figure 3. This graph shows D50 (t) and the envelope y = ±1/ sin(t/2).
As n gets larger, the Dn (t) fills the envelope more completely.
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4. DINI’S TEST FOR POINTWISE CONVERGENCE
10-7
This is similar to a situation we’ve seen before within the proof of the Weierstrass
approximation theorem, Theorem 9.13. The integral given above is a convolution
integral similar to that used in the proof of Theorem 9.13, although the Dirichlet
kernel isn’t a convolution kernel in the sense of Lemma 9.14 because it doesn’t
satisfy conditions (a) and (c) of that lemma. (See Figure 10.3.)
4. Dini’s Test for Pointwise Convergence
Theorem 10.6 (Dini’s Test). Let f : R ! R be a 2⇡-periodic function integrable
on [ ⇡, ⇡] with Fourier series given by (117). If there is a > 0 and s 2 R such
that
Z
f (x + t) + f (x t) 2s
dt < 1,
t
0
then
1
a0 X
+
(ak cos kx + bk cos kx) = s.
2
k=1
Proof. Since Dn is even,
Z ⇡
1
sn (x) =
f (x t)Dn (t) dt
2⇡
⇡
Z 0
Z ⇡
1
1
=
f (x t)Dn (t) dt +
f (x
2⇡
2⇡ 0
⇡
Z ⇡
1
=
(f (x + t) + f (x t)) Dn (t) dt.
2⇡ 0
t)Dn (t) dt
By Theorem 10.5(d) and (e),
Z ⇡
1
sn (x) s =
(f (x + t) + f (x t) 2s) Dn (t) dt
2⇡ 0
Z ⇡
1
f (x + t) + f (x t) 2s
t
1
=
·
· sin(n + )t dt.
2⇡ 0
t
2
sin 2t
Since t/ sin 2t is bounded on (0, ⇡), Theorem 10.3 shows sn (x)
Corollary 8.11 to finish the proof.
s ! 0. Now use
⇤
Example 10.2. Suppose f (x) = x for ⇡ < x  ⇡ and is 2⇡-periodic on R.
Since f is odd, an = 0 for all n. Integration by parts gives bn = ( 1)n+1 2/n for
n 2 N. Therefore,
1
X
2
f⇠
( 1)n+1 sin nx.
n
n=1
For x 2 ( ⇡, ⇡), let 0 < < min{⇡ x, ⇡ + x}. (This is just the distance from x to
closest endpoint of ( ⇡, ⇡).) Using Dini’s test, we see
Z
Z
f (x + t) + f (x t) 2x
x + t + x t 2x
dt =
dt = 0 < 1,
t
t
0
0
so
1
X
2
(122)
( 1)n+1 sin nx = x for
⇡ < x < ⇡.
n
n=1
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10-8
CHAPTER 10. FOURIER SERIES
p
p
2
-p
-
p
p
2
2
-
p
2p
p
2
-p
Figure 4. This plot shows the function of Example 10.2 and s8 (x)
for that function.
In particular, when x = ⇡/2, (122) gives another way to derive (104). When x = ⇡,
the series converges to 0, which is the middle of the “jump” for f .
This behavior of converging to the middle of a jump discontinuity is typical. To
see this, denote the one-sided limits of f at x by
f (x ) = lim f (t) and f (x+) = lim f (t),
t"x
t#x
and suppose f has a jump discontinuity at x with
f (x ) + f (x+)
s=
.
2
Guided by Dini’s test, consider
Z
f (x + t) + f (x t) 2s
dt
t
0
Z
f (x + t) + f (x t) f (x ) f (x+)
=
dt
t
0
Z
Z
f (x + t) f (x+)
f (x t) f (x )

dt +
dt
t
t
0
0
If both of the integrals on the right are finite, then the integral on the left is also
finite. This amounts to a proof of the following corollary.
Corollary 10.7. Suppose f : R ! R is 2⇡-periodic and integrable on [ ⇡, ⇡].
If both one-sided limits exist at x and there is a > 0 such that both
Z
Z
f (x + t) f (x+)
f (x t) f (x )
dt < 1 and
dt < 1,
t
t
0
0
then the Fourier series of f converges to
f (x ) + f (x+)
.
2
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5. GIBBS PHENOMENON
10-9
Figure 5. This is a plot of s19 (f, x), where f is defined by (123).
The Dini test given above provides a powerful condition sufficient to ensure the
pointwise convergence of a Fourier series. There is a plethora of ever more abstruse
conditions that can be proved in a similar fashion to show pointwise convergence.
The problem is complicated by the fact that there are continuous functions
with Fourier series divergent at a point and integrable functions with Fourier series
diverging everywhere [13]. In Section 6, a continuous function whose Fourier series
diverges is constructed.
5. Gibbs Phenomenon
For x 2 [ ⇡, ⇡) define
(123)
f (x) =
(
|x|
x ,
0,
0<x<⇡
x = 0, ⇡
and extend f 2⇡-periodically to all of R. This function is often called a square wave.
A straightforward calculation gives
f⇠
1
4 X sin(2k 1)x
.
⇡
2k 1
k=1
Corollary 10.7 shows sn (x) ! f (x) everywhere. This convergence cannot be uniform
because all the partial sums are continuous and f is discontinuous at every integer
multiple of ⇡. A plot of s19 (x) is shown in Figure 5. Notice the higher peaks in
the oscillation of sn (x) just before and after the jump discontinuities of f . This
behavior is not unique to f , as it can also be seen in Figure 4. If a function is
discontinuous at some point, the partial sums of its Fourier series will always have
such higher peaks near that point. This behavior is called Gibbs phenomenon.3
Instead of doing a general analysis of Gibbs phenomenon, we’ll only analyze
the simple case shown in the square wave f . It’s basically a calculus exercise.
3It is named after the American mathematical physicist, J. W. Gibbs, who pointed it out
in 1899. He was not the first to notice the phenomenon, as the British mathematician Henry
Wilbraham had published a little-noticed paper on it it 1848. Gibbs’ interest in the phenomenon
was sparked by investigations of the American experimental physicist A. A. Michelson who wrote
a letter to Nature disputing the possibility that a Fourier series could converge to a discontinuous
function. The ensuing imbroglio is recounted in a marvelous book by Paul J. Nahin [16].
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10-10
CHAPTER 10. FOURIER SERIES
Figure 6. This is a plot of s9 (f, x), where f is defined by (123).
To locate the peaks in the graph, di↵erentiate the partial sums.
n
n
d 4 X sin(2k 1)x
4X
s02n 1 (x) =
=
cos(2k 1)x
dx ⇡
2k 1
⇡
k=1
k=1
It is left as Exercise 10.10 to show this has a closed form.
2 sin 2nx
s02n 1 (x) =
⇡ sin x
Looking at the numerator, we see s02n 1 (x) has critical numbers at x = k⇡/2n
for k 2 Z. In the interval (0, ⇡), s2n 1 (k⇡/2n) is a relative maximum for odd k and
a relative minimum for even k. (See Figure 6.) The value s2n 1 (⇡/2n) is the height
of the left-most peak. What is the behavior of these maxima?
From Figure 7 it appears they have an asymptotic limit near 1.179. To prove
this, consider the following calculation.
n
⇡
⇣ ⇡ ⌘
4 X sin (2k 1) 2n
s2n 1 f,
=
2n
⇡
2k 1
k=1
⇣
⌘
(2k 1)⇡
n
2n
2 X sin
⇡
=
(2k
1)⇡
⇡
n
k=1
2n
Figure 7. This is a plot of sn (f, ⇡/2n) for n = 1, 2, · · · , 100. The dots
come in pairs because s2n
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= s2n (f, ⇡/2n).
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6. A DIVERGENT FOURIER SERIES
10-11
The last sum is a midpoint Riemann sum for the function sinx x on the interval [0, ⇡]
using a regular partition of n subintervals. Example 9.16 shows
Z
2 ⇡ sin x
dx ⇡ 1.17898.
⇡ 0
x
Since f (0+) f (0 ) = 2, this is an overshoot of a bit less than 9%. There is a
similar undershoot on the other side. It turns out this is typical behavior at points
of discontinuity [21].
6. A Divergent Fourier Series
This is an advanced section that can be omitted.
Eighteenth century mathematicians, including Fourier, Cauchy, Euler, Weierstrass, Lagrange and Riemann, had computed the Fourier series for many functions.
They believed from these examples that the Fourier series of a continuous function
must converge to that function. Fourier went far beyond this claim in his important
1822 book Th´eorie Analytique de la Chaleur. Cauchy even published a flawed proof
of this “fact.”
Finally, in 1873, Paul du Bois-Reymond, settled the question by giving the
construction of a continuous function whose Fourier series diverges at a point [17].
It was finally shown in 1966 by Lennart Carleson [7] that the Fourier series of a
continuous function converges to that function everywhere with the exception of a
set of measure zero.
The problems around the convergence of Fourier series motivated a huge amount
of research that is still going on today. In this section, we look at the tip of
that iceberg by presenting a continuous function F (t) whose Fourier series fails to
converge when t = 0.
6.1. The Conjugate Dirichlet Kernel.
Lemma 10.8. If m, n 2 ! and 0 < |t| < ⇡ for k 2 Z, then
n
X
sin kt =
cos(m
1
2 )t
cos(m + 12 )t
2 sin 2t
k=m
.
Proof.
n
X
k=m
n
1 X
1
sin sin kt
t
2
sin 2 k=m
✓
✓
◆
n
1 X
1
=
cos
k
t
2
sin 2t
kt =
k=m
=
cos m
1
2
t cos n +
2 sin 2t
1
2
✓
◆ ◆
1
cos k +
t
2
t
⇤
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10-12
CHAPTER 10. FOURIER SERIES
Definition 10.9. The conjugate Dirichlet kernel 4 is
n
X
˜ n (t) =
D
(124)
n 2 N.
sin kt,
k=1
We’ll not have much use for the conjugate Dirichlet kernel, except as a convenient
way to refer to sums of the form (124).
Lemma 10.8 immediately gives the following bound.
Corollary 10.10. If 0 < |t| < ⇡, then
1
.
sin 2t
˜ n (t) 
D
6.2. A Sawtooth Wave. If the function f (x) = (⇡ x)/2 on [0, 2⇡) is extended 2⇡-periodically to R, then the graph of the resulting function is often referred
to as a “sawtooth wave”. It has a particularly nice Fourier series:
⇡
x
2
⇠
1
X
sin kx
k
k=1
.
According to Corollary 10.7
1
X
sin kx
k=1
k
=
(
x = 2n⇡, n 2 Z
.
otherwise
0,
f (x),
We’re interested in various partial sums of this series.
Lemma 10.11. If m, n 2 ! with m  n and 0 < |t| < 2⇡, then
n
X
sin kt
1

t .
k
m
sin
2
k=m
Proof.
n
n ⇣
X
X
sin kt
˜ k (t)
=
D
k
k=m
˜k
D
1
(t)
k=m
⌘1
k
Use summation by parts.
=

n
X
k=m
n
X
k=m
˜ k (t)
D
˜ k (t)
D
✓
✓
1
k
1
k+1
1
k
1
k+1
◆
◆
+
+
˜ n (t)
D
n+1
˜n
D
1
(t)
m
˜ n (t)
˜ n 1 (t)
D
D
+
n+1
m
4In this case, the word “conjugate” does not refer to the complex conjugate, but to the
˜ n (t) = 1 + 2 Pn ekit .
harmonic conjugate. They are related by Dn (t) + iD
k=1
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6. A DIVERGENT FOURIER SERIES
10-13
Apply Corollary 10.10.
n ✓
X
1
k
1

2 sin 2t
k=m
✓
1
1
=
t
2 sin 2 m
1
=
.
m sin 2t
1
k+1
◆
1
1
+
n+1 m
◆
1
1
1
+
+
n+1 n+1 m
!
⇤
Proposition 10.12. If n 2 N and 0 < |t| < ⇡, then
n
X
sin kt
k=1
k
 1 + ⇡.
Proof.
n
X
sin kt
k=1
k
=
X sin kt
X sin kt
+
k
k
1
1
1k t

t <kn
X sin kt
+
k
1
1k t
X kt

+
k
1
1k t

X
1k 1t
t+
X sin kt
k
1
t <kn
1
1
t
t sin 2
1
1 2 t
t ⇡2
1+⇡
⇤
6.3. A Continuous Function with a Divergent Fourier Series. The goal
in this subsection is to construct a continuous function F such that lim sup sn (F, 0) =
1. This implies sn (F, 0) does not converge.
Example 10.3. The first step in the construction is to define a sequence of
trigonometric polynomials that form the building blocks for F .
fn (t) =
1
cos t
cos 2t
cos(n 1)t
+
+
+ ···
n n 1 n 2
1
cos(n + 1)t
1
cos(n + 2)t
2
···
cos 2nt
n
Note that,
(125)
May 8, 2015
(
sm (fn , 0) = fn (0) = 0,
sm (fn , 0) > 0,
m 2n
.
1  m < 2n
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10-14
CHAPTER 10. FOURIER SERIES
Rearrange the sum in the definition of fn to see
fn (t) =
n
X
cos(n
k)t
k
k=1
= 2 sin nt
n
X
k=1
cos(n + k)t
sin kt
.
k
This closed form for fn combined with Proposition 10.12 implies the sequence of
functions fn is uniformly bounded:
(126)
|fn (t)| = 2 sin nt
n
X
sin kt
k
k=1
 2 + 2⇡.
At last, the main function can be defined.
F (t) =
1
X
f2n3 (t)
n2
n=1
The Weierstrass M-Test along with (126) implies F is uniformly convergent and
therefore continuous on R. Using (125), consider
Z ⇡
1
3
s2m (F, 0) =
F (t)D2m3 (t) dt
2⇡
⇡
Z ⇡ X
1
f2n3 (t)
1
=
D2m3 (t) dt
2⇡
n2
⇡ n=1
The uniform convergence allows the sum and integration to be reordered.
Z
1
X
1 1
=
n2 2⇡
n=1
=
⇡
⇡
f2n3 (t)D2m3 (t) dt
1
X
1
s m3 f2n3 , 0
n2 2
n=1
Use (125).
1
X
1
=
s 3 f 2n3 , 0
2 2m
n
n=m
>
1
s m3 f2m3 , 0
m2 2
m3
2
1 X1
= 2
m
k
k=1
3
1
> 2 ln 2m
m
= m ln 2
This implies, lim sup sn (F, 0)
May 8, 2015
limm!1 m ln 2 = 1, so sn (F, 0) does not converge.
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´ KERNEL
7. THE FEJER
10-15
7. The Fej´
er Kernel
Since pointwise convergence of the partial sums seems complicated, why not
change the rules of the game? Instead of looking at the sequence of partial sums,
consider a rolling average instead:
n
n (f, x) =
1 X
sn (f, x).
n+1
k=0
The trigonometric polynomials n (f, x) are called the Ces`aro means of the partial
sums. If limn!1 n (f, x) exists, then the Fourier series for f is said to be (C, 1)
summable at x. The idea is that this averaging will “smooth out” the partial sums,
making them more nicely behaved. It is not hard to show that if sn (f, x) converges
at some x, then n (f, x) will converge to the same thing. But there are sequences
for which n (f, x) converges and sn (f, x) does not. (See Exercises 3.19 and 4.26.)
As with sn (x), we’ll simply write n (x) instead of n (f, x), when it is clear
which function is being considered.
We start with a calculation.
n
n (x)
=
=
(*)
=
=
=
1 X
sk (x)
n+1
k=0
Z ⇡
n
1 X 1
f (x t)Dk (t) dt
n+1
2⇡
⇡
k=0
Z ⇡
n
1
1 X
f (x t)
Dk (t) dt
2⇡
n+1
⇡
k=0
Z ⇡
n
1
1 X sin(k + 1/2)t
f (x t)
dt
2⇡
n+1
sin t/2
⇡
k=0
Z ⇡
n
X
1
1
f (x t)
sin t/2 sin(k + 1/2)t dt
2⇡
(n + 1) sin2 t/2 k=0
⇡
Use the identity 2 sin A sin B = cos(A
1
=
2⇡
Z
The sum telescopes.
Z
1
=
2⇡
⇡
f (x
t)
f (x
t)
⇡
⇡
⇡
Use the identity 2 sin2 A = 1
(**)
May 8, 2015
1
=
2⇡
Z
⇡
f (x
⇡
B)
cos(A + B).
n
X
1/2
(cos kt
(n + 1) sin2 t/2 k=0
1/2
(1
(n + 1) sin2 t/2
cos(k + 1)t) dt
cos(n + 1)t) dt
cos 2A.
1
t)
(n + 1)
✓
sin n+1
2 t
sin 2t
◆2
dt
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10-16
CHAPTER 10. FOURIER SERIES
10
8
6
4
2
-p
-
p
2
p
2
p
Figure 8. A plot of K5 (t), K8 (t) and K10 (t).
The Fej´er kernel is the sequence of functions highlighted above; i.e.,
✓
◆2
sin n+1
1
2 t
(127)
Kn (t) =
, n 2 N.
(n + 1)
sin 2t
Comparing the lines labeled (*) and (**) in the previous calculation, we see another
form for the Fej´er kernel is
n
(128)
Kn (t) =
1 X
Dk (t).
n+1
k=0
Once again, we’re confronted with a convolution integral containing a kernel:
Z ⇡
1
(x)
=
f (x t)Kn (t) dt.
n
2⇡
⇡
Theorem 10.13. The Fej´er kernel has the following properties.5
(a)
(b)
(c)
(d)
(e)
(f)
Kn (t) is an even 2⇡-periodic function for each n 2 N.
Kn (0) = n + 1 for each n 2 !.
Kn (t)
R ⇡ 0 for each n 2 N.
1
Kn (t) dt = 1 for each n 2 !.
2⇡
⇡
If 0 < < ⇡, then Kn ◆ 0 on [ ⇡, ] [ [ , ⇡].
R
R⇡
If 0 < < ⇡, then ⇡ Kn (t) dt ! 0 and
Kn (t) dt ! 0.
Proof. Theorem 10.5 and (128) imply (a), (b) and (d). Equation (127) implies
(c).
5Compare this theorem with Lemma 9.14.
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´ KERNEL
7. THE FEJER
10-17
Let be as in (e). In light of (a), it suffices to prove (e) for the interval [ , ⇡].
Noting that sin t/2 is decreasing on [ , ⇡], it follows that for  t  ⇡,
✓
◆2
sin n+1
1
2 t
Kn (t) =
(n + 1)
sin 2t
✓
◆2
1
1

(n + 1) sin 2t
1
1

!0
(n + 1) sin2 2
It follows that Kn ◆ 0 on [ , ⇡] and (e) has been proved.
Theorem 9.16 and (e) imply (f).
⇤
Theorem 10.14 (Fej´er). If f : R ! R is 2⇡-periodic, integrable on [ ⇡, ⇡] and
continuous at x, then n (x) ! f (x).
R⇡
R⇡ R
Proof. Since f is 2⇡-periodic and ⇡ f (t) dt exists, so does ⇡ (f (x t)
f (x)) dt. Theorem 8.3 gives an M > 0 so |f (x t) f (x)| < M for all t.
Let " > 0 and choose > 0 such that |f (x) f (y)| < "/3 whenever |x y| < .
By Theorem 10.13(f), there is an N 2 N so that whenever n N ,
Z
Z ⇡
1
"
1
"
Kn (t) dt <
and
Kn (t) dt <
.
2⇡
3M
2⇡
3M
⇡
We start calculating.
|
n (x)
=
Z ⇡
1
f (x t)Kn (t) dt
f (x)Kn (t) dt
2⇡
⇡
⇡
Z ⇡
1
=
(f (x t) f (x))Kn (t) dt
2⇡
⇡
Z
(f (x t) f (x))Kn (t) dt +
(f (x t) f (x))Kn (t) dt
f (x)| =
1
2⇡
Z
⇡
1
2⇡
Z
⇡
+

1
2⇡
Z
(f (x
t)
M
2⇡
This shows
Z
Kn (t) dt +
⇡
n (x)
! f (x).
⇡
(f (x
t)
f (x))Kn (t) dt
f (x))Kn (t) dt +
⇡
+
<
Z
1
2⇡
1
2⇡
Z
Z
⇡
(f (x
|f (x
t)
t)
1
2⇡
Z
(f (x
t)
f (x))Kn (t) dt
f (x))Kn (t) dt
f (x)|Kn (t) dt +
"
" 1
<M
+
3M
3 2⇡
Z
M
2⇡
Z
⇡
Kn (t) dt
Kn (t) dt + M
"
<"
3M
⇤
Theorem 10.14 gives a partial solution to the representation problem.
Corollary 10.15. Suppose f and g are continuous and 2⇡-periodic on R. If f
and g have the same Fourier coefficients, then they are equal.
May 8, 2015
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10-18
CHAPTER 10. FOURIER SERIES
p
p
p
p
2
2
-p
-
p
p
2
2
-
2 p -p
p
-
p
p
2
p
2
-
2
-p
p
2p
p
2
-p
Figure 9. These plots illustrate the functions of Example 10.4. On
the left are shown f (x), s8 (x) and 8 (x). On the right are shown f (x),
3 (x), 5 (x) and 20 (x). Compare this with Figure 4.
Proof. By assumption,
0=
n (f, t)
=
n (f, t)
n (g, t)
n (g, t)
for all n and Theorem 10.14 implies
!f
g.
⇤
In the case of continuous functions, the convergence is uniform, rather than
pointwise.
Theorem 10.16 (Fej´er). If f : R ! R is 2⇡-periodic and continuous, then
(x)
◆ f (x).
n
Proof. By Exercise 6.32, f is uniformly continuous. This can be used to show
the calculation within the proof of Theorem 10.14 does not depend on x. The details
are left as Exercise 10.12.
⇤
A perspicacious reader will have noticed the similarity between Theorem 10.16
and the Weierstrass approximation theorem, Theorem 9.13. In fact, the Weierstrass
approximation theorem can be proved from Theorem 10.16 using power series and
Theorem 9.24. (See Exercise 10.13.)
Example 10.4. As in Example 10.2, let f (x) = x for ⇡ < x  ⇡ and extend
f to be periodic on R with period 2⇡. Figure 9 shows the di↵erence between the
Fej´er and classical methods of summation. Notice that the Fej´er sums remain much
more smoothly affixed to the function and do not show Gibbs phenomenon.
8. Exercises
10.1. Prove Theorem 10.2.
10.2. Suppose f : R ! R is 2⇡-periodic and integrable on [ ⇡, ⇡]. If b
Rb
R⇡
then a f = ⇡ f .
a = 2⇡,
10.3. Let f : [ ⇡, ⇡) be a function with Fourier coefficients an and bn . If f is odd,
then an = 0 for all n 2 !. If f is even, then bn = 0 for all n 2 N.
10.4. If f (x) = sgn(x) on [ ⇡, ⇡), then find the Fourier series for f .
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8. EXERCISES
10.5. Is
1
X
10-19
sin nx the Fourier series of some function?
n=1
10.6. Let f (x) = x2 when ⇡  x < ⇡ and extend f to be 2⇡-periodic on all of R.
Use Dini’s Test to show sn (f, x) ! f (x) everywhere.
10.7. Use Exercise 10.6 to prove
1
X
⇡2
1
=
.
2
6
n
n=1
10.8. Suppose f is integrable on [ ⇡, ⇡]. If f is even, then the Fourier series for f
has no sine terms. If f is odd, then the Fourier series for f has no cosine terms.
10.9. If f : R ! R is periodic with period p and continuous on [0, p], then f is
uniformly continuous.
10.10. Prove
n
X
cos(2k
k=1
for t 6= k⇡, k 2 Z and n 2 N. (Hint: 2
1)t =
Pn
k=1
sin 2nt
2 sin t
cos(2k
1)t = D2n (t)
Dn (2t).)
10.11. The function g(t) = t/ sin(t/2) is undefined whenever t = 2n⇡ for some
n 2 Z. Show that it can be redefined on the set {2n⇡ : n 2 Z} to be periodic and
uniformly continuous on R.
10.12. Prove Theorem 10.16.
10.13. Prove the Weierstrass approximation theorem using Fourier series and
Taylor’s theorem.
10.14. If f (x) = x for ⇡  x < ⇡, can the Fourier series of f be integrated
term-by-term on ( ⇡, ⇡) to obtain the correct answer?
10.15. If f (x) = x for ⇡  x < ⇡, can the Fourier series of f be di↵erentiated
term-by-term on ( ⇡, ⇡) to obtain the correct answer?
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