Key Concepts
1. Basic Properties of
Fourier Series
• Fourier Series
• Convolution
Outline
• Good kernel
I. Key Concepts
• Cesaro sum
II. Summary
• Abel sum
III. Convergence Theorems
Summary
IV. Uniqueness Theorems
V. More Convergence Theorems
Question 1. Suppose f and g are two Riemann integrable functions on the circle (2π-periodic) that
have the same Fourier coefficients. Are they equal?
VI. Approximation Theorems
VII. Special Kernels
Answer 1. If they are both continuous on the circle, then yes.
VIII. Other Results
Question 2. In what sense does the Fourier series of f converge to f ? Pointwise? Uniformly? In
the mean square sense?
To The Reader
These are the summaries of the notes I took from
the book Fourier Analysis by Elias Stein and Rami
Shakarchi (I am currently reading it!). The book
covers the basic facts about Fourier series, the elements of Fourier transforms, and finite Fourier
analysis. It also tackles several applications of
the theory to partial differential equations, number
theory, and other sciences. I figured I’d share this
in case someone needs notes to accompany them
in learning the subject. I have also included my
solutions to some exercises and problems given in
the book. So, feel free to email me or message me
(https://www.facebook.com/olej.azotra/) for any
corrections
or
questions.
For an arbitrary Riemann integrable function on
the circle, a complete answer is yet to be discussed
in the book. However, we did prove some results
about the uniform convergence with some added
assumptions to the function.
We did show that for all x ∈ [−π, π]
Z π
1
f (x)DN (x − y)dy
SN (f )(x) =
2π −π
Where SN is the N -th partial sum of the Fourier
series of f , and DN is the N -th Dirichlet kernel
defined by
∞
X
DN (x) =
einx .
n=−∞
We then studied the limiting behavior of the sequence of convolutions {f ∗ DN }∞
N =1 instead of the
1
• f is twice continuously differentiable.
sequence of partial sums {SN (f )}∞
N =1 .
From studying convolutions, we have seen that when
{Kn }∞
n=1 is a family of good kernels, then
Uniqueness Theorems
lim (f ∗ Kn )(θ) = f (θ)
• If fb(n) = 0 for all n ∈ Z, then f (θ) = 0
whenever θ is a continuity point of f .
n→∞
whenever θ is a continuity point of f , and the convergence is uniform when f is continuous on the
circle.
• If fb(n) = 0 for all n ∈ Z and f is continuous
everywhere, then f = 0.
This led to the question: Is {Dn }∞
n=1 a good kernel? Because if it is, then this would imply that
the Fourier series of f converges to f at continuity
points, and converges uniformly to f when f is a
continuous function on the circle.
More Convergence Theorems
• If {Kn }∞
n=1 is a good kernel, then
lim (f ∗ Kn )(θ) = f (θ)
Answer: NO :(
n→∞
whenever θ is a continuity point of f . The
limit is uniform if f is continuous.
Thus, for now, nothing can be said about the convergence of the Fourier series,
• The Fourier series of f is Cesaro summable,
and Abel summable to f at continuity points
of f
To conclude the section we studied different ways
of summing the Fourier series, and found that the
Cesaro and Abel sums behaved better than the
partial sums.
The Fourier series is uniformly Cesaro summable,
and uniformly Abel summable to f , if f is
continuous.
• Cesaro summation leads to Fejer’s theorem,
and the periodic analogue of the Weierstass
approximation theorem, which by the way
can be easily proven using good kernels.
Approximation Theorems
• Abel summation leads to a solution of the
Dirichlet problem for the steady-state heat
equation in the disc.
• Approximation by continuous functions on the
circle: For a given integrable function f defined on the circle, there exists a sequence
of continuous functions on the circle {fk }∞
k=1
such that
Convergence Theorems
sup fk (θ) ≤ sup f (θ)
The Fourier series of f converges uniformly to f if
f satisfies one of the following conditions:
θ∈[−π,π]
θ∈[−π,π]
for all k ∈ N, and that
Z π
|f (θ) − g(θ)|dθ → 0
• f is continuous and its
series absoP Fourier
b
lutely converges, i.e.,
|f (n)| < ∞.
−π
2
• The Poisson Kernel:
as k → ∞.
• Approximation by trigonometric polynomials:
A continuous function on the circle can be
uniformly approximated by trigonometric polynomials.
Pr (x) =
∞
X
fb(n)r|n| einx
n=−∞
Properties:
• Approximation by polynomials (Weierstrass
approximation theorem): A continuous function can be uniformly approximated by polynomials.
1 − r2
– Pr (x) =
1 − 2r cos x + r2
– (f ∗ Pr )(x) = Ar (f )(x)
– Good kernel
– Key to solving Dirichlet’s problem for
the steady-state heat equation in the
unit disc.
The proofs of the last two theorems are simplified
thanks to good kernels.
Special Kernels
Other Results
• The Dirichlet Kernel
• fb0 (n) = infb(n)
N
X
DN (x) =
einx
• Convergence implies Cesaro summability implies Abel summability. The converses are
not always true.
n=−N
Properties:
• Basic properties of convolutions: For any Riemann integrable functions on the circle f, g
we have,
sin [(N + 21 )x]
– DN (x) =
.
sin x/2
– (f ∗ DN )(x) = SN (f )(x)
– (f ∗ g) + (f ∗ h) = f ∗ (g + h)
– Not a good kernel
– (cf ) ∗ g = f ∗ (cg)
• The Fejer Kernel (Average of the first
N Dirichlet kernels)
– f ∗g =g∗f
– (f ∗ g) ∗ h = f ∗ (g ∗ h)
FN (x) =
1
N
N
−1
X
– f ∗ g is continuous
Dn (x)
\
– (f
∗ g)(n) = fb(n)b
g (n)
n=0
Properties:
– FN (x) =
1 sin2 (N x/2)
.
N sin2 x/2
– (f ∗ FN )(x) = σN (f )(x)
– Good kernel
3
(Exercises and Problems)
Basic Properties of Fourier
Series
Well,
2h sin mp
m2 p(π − p)
2h
1
≤
,
p(π − p) m2
|Am | =
Exercise 1
and P
we know that for any constant C > 0 the series C/m2 converges. Hence, by the comparison
test, the sine series converges absolutely.
Let h > 0, and p be any point on [0, π]. Show
that the function f defined on the interval
[0, π] by

xh


p
f (x) = h(π
− x)


π−p
Exercise 2
0≤x≤p
On [−π, π] consider
(
0
f (θ) =
1 − |θ| /δ
p<x≤π
is equal to its Fourier sine series.
Show that
∞
X
2h sin mp
Am sin mx with Am = 2
m p(π − p)
m=1
∞
X
1 − cos nδ
δ
+2
cos nθ
f (θ) =
2π
n2 πδ
n=1
Solution. The Fourier series of a function on the
circle converges to the function provided that it
absolutely converges and that the function is continuous. So if we can prove that f is continuous
and that the sine series absolutely converges, then
we’re done.
Solution. Observe that f is even. So, its Fourier
series is a cosine series (we don’t have to worry
about the sine terms).
Find the Fourier coefficients of f :
Z π
1
b
f (0) =
f (θ)dθ
2π −π
Z
Z
1
=
f (θ)dθ +
f (θ)dθ
2π
|θ|≤δ
|θ|>δ
Prove that f is continuous: f is a polynomial of
degree 1 both on [0, p] and on (p, π]. So it is continuous everywhere on [0, π] except possibly on the
point p. But the left and right limit of f at p approaches h, and f (p) = h. So, p is also a continuity
point, and therefore f is continuous on [0, π].
f is 0 for |θ| > 0 so the second integral is also 0.
f (θ) = 1 − |θ|/δ when |θ| ≤ δ, so
Z
1
|θ|
b
f (0) =
1−
dθ
2π |θ|≤δ
δ
Z 0
Z δ
θ
θ
1
=
1 + dθ +
1 − dθ
2π
δ
δ
−δ
0
All that is left for us to do is to prove that
∞
X
|θ| > δ
|θ| ≤ δ
|Am | < ∞.
m=1
1
=
δ
2π
Compute the Fourier series of f :
X
f (θ) ∼ fb(0) +
2fb(n) cos nθ
For n 6= 0:
1
fb(n) =
2π
Z
π
f (θ)e−inθ dθ
−π
Z δ
1
|θ| −inθ
=
1−
e
dθ
2π −δ
δ
Z δ
Z
1 0 −inθ
1
−inθ
e
dθ +
=
θe
dθ
2π
δ −δ
−δ
Z
1 δ −inθ
θe
dθ
−
δ 0
=
δ
θe
0
Z
−inθ
Does the Fourier series converge absolutely? Let
{an }∞
n=0 be the sequence of absolute values of the
Fourier coefficients of f : That is, the sequence defined by
(
δ/2π
n=0
an =
2
2 |1 − cos nδ| /n πδ otherwise.
We wish to find a convergent series that dominates
P
the Fourier series, i.e., a convergent series
bn
such that |an | ≤ bn for all sufficiently large n.
P
And this series that we’re looking for is
2/πδn2
because for n > 0 we have
δe−inδ e−inδ
1
dθ = −
+ 2 − 2
in
n
n
0
θe−inθ dθ = −
−δ
1 − cos nδ
δ
+2
cos nθ
2π
n2 πδ
n=1
Is f continuous on [−π, π]?: f is a constant function on [−π, −δ) ∪ (δ, π] and a polynomial on both
[−δ, 0] and [0, δ]. Furthermore, f (x) approaches 0
as x → δ and f (δ) = 0. So, yep. f is continuous
everywhere on [−π, π].
So we’ve broken down the integral into three integrals. We evaluate each of them separately:
Z δ
e−inδ einδ
e−inθ dθ = −
+
in
in
−δ
2
= sin nδ
n
Z
n≥1
∞
X
δeinδ einδ
1
− 2 + 2
in
n
n
2|1 − cos nδ|
2
≤
πδn2
πδn2
Hence,
Z δ
2
2 cos nδ 2δ sin nδ
+
,
e−inθ |θ| dθ = − 2 +
n
n2
n
−δ
and hence
Z δ
Z
1 δ −inθ
2
2 cos nδ
−inθ
e
dθ −
e
|θ| dθ = 2 −
.
δ −δ
δn
δn2
−δ
So for all n 6= 0 we have
1 − cos nδ
fb(n) =
n2 πδ
2
2. Let > 0 be given, and M > 0 be a constant
such that |Bk | ≤ M for all k ∈ N (Such a constant
exists because the partial
P sums are bounded). We
claim that the series
an bn converges. Choose
N ∈ N so large that
Exercise 3
N
Suppose {an }N
n=1 and {bn }n=1 are two finite
sequences
of complex numbers. Let Bk =
Pk
Pn=1 bn denote the partial sums of the series
bn with the convention B0 = 0.
|aN | <
1. Prove the summation by parts formula:
N
X
Then n ≥ m ≥ N implies
an b n
n
X
n=M
= aN BN − aM BM −1 −
N
−1
X
(an+1 − an )Bn
ak b k
k=m
n=M
= an Bn − am Bm−1 −
≤B
|an | + |am | +
n−1
X
!
|ak+1 − ak |
Because {ak } decreases monotonically to 0, it follows that aN ≥ an ≥ am ≥ 0 and ak − ak+1 ≥ 0.
Hence,
n
X
ak bk ≤ B(2aN )
k=m
< 2B ·
an b n
n=M
=
(ak+1 − ak )Bk
k=m
Solution. 1. The key idea is to use the fact that
Bn − Bn−1 = bn to break apart the sum on the left:
=
n−1
X
k=m
2. Deduce from this formula Dirichlet’s test
for convergence of P
a series: If the partial
sums of the series
bn are bounded, and
{an } is a sequence of real numbers
P that decreases monotonically to 0, then an bn converges.
N
X
.
2B
N
X
n=M
N
X
Hence, the series
converges.
an (Bn − Bn−1 )
an Bn −
n=M
= aN BN +
N
X
an Bn −
n=M
= aN BN − aM BM −1 +
= aN BN − aM BM −1 −
N
X
an Bn−1 − aM BM −1
n=M +1
N
−1
X
n=M
N
−1
X
ak bk is Cauchy, and therefore
(I’ll be adding more problems. A lot more!)
an Bn−1
n=M
N
−1
X
P
= .
2B
an Bn −
N
−1
X
an+1 Bn
n=M
(an+1 − an )Bn .
n=M
3
Basic Results From Orthogonality
2. Convergence of
Fourier Series
The following three results are valid in any finitedimensional inner product space, and the infinitedimensional spaces `2 (Z), and R.
Outline
• The Pythagorean Theorem: If X and Y are
orthogonal, then
I. Key Concepts
II. Summary
kX + Y k2 = kXk2 + kY k2
III. Basic Results From Orthogonality
• The Cauchy-Schwartz Inequality:
IV. Special Infinite-Dimensional Vector Spaces
|hX, Y i| = kXkkY k
V. Approximation Theorems
VI. Mean-Square Convergence
• The Triangle Inequality
VII. Proof of Mean-Square Convergence
kX + Y k ≤ kXk + kY k
VIII. Other Results
Special Infinite Dimensional Vector
Spaces
Key Concepts
• Orthogonality
• The space of all (double-sided) sequences
• Completeness
(. . . , a−n , . . . , a−1 , a0 , a1 , . . . , an , . . . )
• Hilbert Space
such that
Summary
∞
X
|an |2 < ∞
n=−∞
– Addition & Scalar Multiplication:
The mean-square of a Riemann integrable function
defined on [a, b] is the integral
Z b
1
|f (t)|2 dt
b−a a
{an } + c{bn } = {an + cbn }
– Inner Product & Norm:
This short chapter is devoted to answering the
question of whether the Fourier series of an integrable function converges in the mean-square sense:
Z π
1
|f (t) − SN (f )(t)|2 dt → 0
2π −π
han , bn i =
∞
X
an b n
n=−∞
kan k =
∞
X
n=−∞
– Denoted by `2 (Z)
as N → ∞.
1
!1/2
2
|an |
Mean-Square Convergence
– A Hilbert space
• The space of all complex-valued Riemann integrable functions on the circle
Let f be a Riemann integrable
function on
P∞
the circle with f (θ) ∼ n=−∞ an e−inθ . Then
we have:
– Addition & Scalar Multiplication:
(f + cg)(x) = f (x) + cg(x)
(a) Mean-square convergence of the
Fourier series:
Z π
1
|f (t) − SN (f )(t)|2 dt → 0
2π −π
– Inner Product & Norm:
Z 2π
1
f (θ)g(θ) dθ
hf, gi =
2π 0
Z 2π
1/2
1
2
kf k =
kf (θk
2π 0
as N → ∞.
(b) Parseval’s identity:
– Denoted by R
1
2π
– A pre-Hilbert space
Z
π
2
|f (θ)| dθ =
−π
∞
X
|an |2
n=−∞
Parseval’s identity is a consequence of the meansquare convergence of the Fourier series.
Approximation Theorems
• Best Approximation:
Proof of Mean-Square Convergence
For any sequence of complex numbers
{cn }|n|≤N , we have
Fix > 0. Since f is integrable on the circle, there
exists a continuous function g such that
kf − SN (f )k ≤ f −
X
cn einθ
sup |g(θ| ≤ sup |f (θ| = B
|n|≤N
θ∈[0,2π]
and
– Proved from the observation that for all n ∈ Z
[f − SN (f )] ⊥ en .
(Here en (x) = e
inx
θ∈[0,2π]
π
2 π
|f (θ) − g(θ)| dθ <
9B
−π
Z
This continuous function will be the key to the
proof. We want to prove that there exists N ∈ N
such that n ≥ N implies kf − SN k < . The answer is the degree of the trigonometric polynomial
P such that kg − P k < /3. Since g is continuous
)
2
on the circle, such a polynomial exists:
Z 2π
1
2
kf − gk =
|f (θ) − g(θ)|2 dθ
2π 0
Z
B 2π
|f (θ) − g(θ)| dθ
≤
π 0
2
<
9
So, if M is the degree of P , then N > M implies
eiN θ + P
kf − SN k ≤ f −
3
≤ kf − P k +
3
≤ kf − gk + kg − P k +
3
≤ + + =
3 3 3
Notice how I played with the epsilons! The best
approximation lemma tells us that the first equality holds if P and SN have the same degree, which
isn’t the case. So I had to add that extra (/3)eiN θ
to make the degrees match and be able to use the
lemma. But I couldn’t have done this without g
approximating f and P approximating g. It feels
amazing being able to approximate a function. I
feel like a god. This concludes the proof.
Other Results
• Riemann-Lebesgue Lemma
• Generalized Parseval’s Identity
There’s actually two more sections in the chapter.
It’s about pointwise convergence, and an example
of a continuous function with a diverging Fourier
series. I haven’t read them yet tho.
3
Now, use the identity sin ax = (eiax − e−iax )/2i.
1
fb(n) = sin πα
(sin πα)(n + α)
1
=
n+α
Hence,
∞
X
π
i(π−x)α
e
∼
fb(n)einx
sin πα
n=−∞
(Exercises and Problems)
Convergence of Fourier Series
Exercise 1
Show that for α not an integer, the Fourier
series of
π
ei(π−x)α
sin πα
on [0, 2π] is given by
∞
X
einx
=
n+α
n=−∞
∞
X
einx
n+α
n=−∞
This concludes the first part of the problem. Next,
obtain the mean of the square of the norm/absolute value of the given function: The square of the
norm is
2
π2
π
2
ei(π−x)α =
ei(π−x)α
2
sin πα
(sin πα)
π2
=
(sin πα)2
Apply Parseval’s identity to show that
∞
X
1
π2
=
(n + α)2
(sin πα)2
n=−∞
Solution. Find the Fourier coefficients of f : For all
n ∈ Z, we have
Z 2π
1
π
b
f (n) =
ei(π−x)α e−inx dx
2π 0 sin πα
Z 2π
eiπα
=
e−i(n+α)x
2 sin πα 0
eiπα e−2πi(n+α) − 1
=
2 sin πα −i(n + α)
eiπα (e−2πiα (eiπ )2n − 1)
=
−2i(sin πα)(n + α)
2
ei(π−x)α = 1 by the definition of the absolute
value of a complex number, and the identities eit =
cos t + i sin t and cos2 t + sin2 t = 1:
ei(π−x)α
= Re[ei(π−x)α ]2 + Im[ei(π−x)α ]2
= cos2 (π − x)α + sin2 (π − x)α
= 1.
Hence, the mean we’re looking for is
Z 2π
Z 2π
2
π
π2
1
1
i(π−x)α
e
dx =
dx
2π 0
sin πα
(sin πα)2 2π 0
π2
=
(1)
(sin πα)2
P
Finally, find ∞
|fb(n)|2 :
For the last line use the fact that eiπ = −1, which
gives (eiπ )2n = (−1)2n = 1, and then distribute the
factor eiπα in the numerator. Doing this, we get
−πiα
2
n=−∞
∞
X
iπα
(e
−e )
−2i(sin πα)(n + α)
(eiπα − e−πiα )
1
=
2i
(sin πα)(n + α)
fb(n) =
|fb(n)|2 =
n=−∞
=
1
∞
X
n=−∞
∞
X
1
n+α
2
1
(n + α)2
n=−∞
(2)
Parseval’s identity tells us that (1) and (2) are
equal. Therefore,
(c) For any compact interval [a, b] and any
continuously differentiable function f
with f (a) = f (b) = 0, show that
∞
X
1
π2
=
(n + α)2
(sin πα)2
n=−∞
Z
a
The inequalities of Wirtinger and Poincare
establish a relationship between the norm of
a function and that of its derivative.
(a) If f is T -periodic, continuous, and
RT
piecewise C 1 with 0 f (t)dt = 0, show
that
Z T
Z
T2 T 0 2
2
|f (t)| dt ≤ 2
|f (t)| dt,
4π 0
0
Z
b
2
|f 0 (t)| dt
a
Partial solution.
(a) Find the Fourier series of f and f 0 : Because f is C 1 , the Fourier coefficients Rof f and
T
f 0 are defined. Furthermore, because 0 f (t)dt =
0, it follows that fb(0) = 0. So we have
X
fb(n)e2πint/T
f (t) ∼
with equality if and only if f (t) =
A sin 2πt/T + B cos 2πt/T . [Hint: Apply Parseval’s Identity]
(b) If f is as above and g is just C 1 and
T -periodic, prove that
T
(b − a)2
|f (t)| dt ≤
π2
2
Discuss the case of equality, and prove
that the constant (b − a)2 /π 2 cannot be
improved. [Hint: Extend f to be odd
with respect to a and periodic of period
T = 2(b − a) so that its integral over
an interval of length T is 0. Apply part
(a) to get the inequality, and conclude
that equality holds if and only if f (t) =
t−a
].
A sin π b−a
Exercise 2
Z
b
2
n6=0
f (t)g(t)dt
Z
Z T
T2 T
2
2
≤ 2
|f (t)| dt
|g 0 (t)| dt
4π 0
0
and
0
f 0 (t) ∼
∞
X
fb0 (n)e2πint/T
n=−∞
The first series extends over all non-zero integers; the second extends over all integers
because fb0 (0) need not be zero.
Find a relationship between fb, and fb0 :
Simply do integration-by-parts to the integral
2
defining fb0 . For each n 6= 0, we have
(b)
fb0 (n)
Z T
f 0 (t)e−2πint/T dt
=
Z
2
T
f (t)g(t)dt
0
2
0
=
T
f (t)e−2πint/T 0
2πin
+
T
Z
= T
T
f (t)e
−2πint/T
dt
fb(n)b
g (n)
n6=0
0
≤ T2
Z
2πin T
=
f (t)e−2πint/T dt
T
0
2πin b
f (n)
=
T
X
2
fb(n)
n6=0
X
= T2
n6=0

2
0
X
2
T
gb (n) 
fb(n)  2
4π n6=0 n
∞
X
2
∞
2
T 2 X b0
2
b
≤T
f (n)
g
(n)
4π 2 n=−∞
n=−∞
Z T
Z T
T
2
2
0
=T
|f (t)| dt
|g (t)| dt
4π 2 0
0
Z T
Z
T2 T
2
2
|g 0 (t)| dt
|f (t)| dt
= 2
4π 0
0
n6=0
T 3 X fb0 (n)
= 2
4π n6=0 n
X
|b
g (n)|2
n6=0

We can now obtain the desired inequality:
Z T
X
2
fb(n)
|f (t)|2 dt = T
0
X
2
∞
2
T 3 X b0
≤ 2
f (n)
4π n=∞
Z
T2 T 0 2
|f (t)| dt
=
4π 0
2
!
(2) follows from the generalized Parseval’s identity, and (3) from the Cauchy-Schwarz inequality
(c) hehe ala pa ih. I’m working on it tho.
The first and last equalities hold by Parseval’s
identity. To see why equality holds if and only
if f (t) = A sin 2πt/T + B cos 2πt/T , the key
observation is that the equality holds only if
fb0 (n) = 0 for every integer n with |n| ≥ 2. So
if the equality holds and the Fourier series of
f converges to f , then
Questions:
• Does the Fourier series of f converge? (the f
I’m talking about here is the one in the problem)
[Unless I know this is true, my solution to (a) is
incomplete]
f (t) = Ae2πt/T + Be−2πt/T
= A sin 2πt/T + B cos 2πt/T
(I’ll be adding more problems. A lot more!)
where A = fb(1) and B = fb(−1)
3
(Exercises and Problems)
The Fourier Transform on R
for all non-zero integer n, we have
1b
f (n/L)
L
AL
≤
|n|2
|an (L)| =
(Kinocomputerize ko pa yung summary ko for this
chapter. For now, these are my answers to some
of the exercises from the book.)
for some constant A > 0 (which exists because fb is
of moderate decrease). So, f is continuous and periodic, and its Fourier
converges absolutely.
P bseries 2πinδx
Therefore f (x) = δ f (nδ)e
with δ = 1/L
Exercise 1
Suppose f is a continuous function supported on an interval [−M, M ], whose
Fourier transform fb is of moderate decrease.
(b) (oof this is the trickiest part) We want to prove
that for every > 0, there exists δ0 > 0 such that
δ < δ0 implies:
Z ∞
∞
X
F (ξ)dξ − δ
F (nδ) < (a) Fix L with
P L/2 > M , and show that
f (x) = an (L)e2πinx/L where
Z
1 L/2
an (L) =
f (x)e−2πinx /L dx
L −L/2
1
= fb(n/L)
L
−∞
Here’s what we’re gonna do. Let > 0. Choose
M, N ∈ N so large that
Z N
Z ∞
F (ξ)dξ <
F (ξ)dξ −
4
−N
−∞
Alternatively, we may write f (x) =
P
δ fb(nδ)e2πiδnx with δ = 1/L
and for all δ > 0 we have
(b) Prove that if F is of moderate decrease, then
Z
∞
F (ξ)dξ = lim δ
δ→0
−∞
∞
X
δ
n=−∞
X
F (nδ) − δ
F (nδ) <
|n|≤M/δ
4
Then for all δ > 0,
Z ∞
∞
X
F (ξ)dξ − δ
F (nδ)
∞
f (x) =
∞
X
n=−∞
F (nδ)
(c) Conclude that
Z
n=−∞
−∞
fb(ξ)e2πixξ dξ
−∞
n=−∞
Z
N
≤
F (ξ)dξ − δ
−N
Partial Solution. (a) Since L/2 > M , f (−L/2) =
f (L/2) = 0. So, f is L/2-periodic. Furthermore,
X
|n|≤M/δ
F (nδ) +
2
And this is the point where I’m STUCK! I can expand
R N that difference by approximating the integral
F with Riemann sums. Let’s call that Rie−N
mann sum R. Then this gives me a new problem.
This is the million-dollar question:
1
ous). Then |ξ − ζ| < δ implies:
Z ∞
|f (x)| e−2πix eξ − eζ dx
−∞
Z N
|f (x)| e−2πix eξ − eζ dx
≤
Z −N
|f (x)| e−2πix eξ − eζ dx
+
|x|≥N
Z
ξ
ζ
|f (x)| dx
≤ 2BN e − e + 2
Can we find a constant δ0 > 0 such that
δ > δ0 implies
R−δ
X
F (nδ) <
|n|≤M/δ
4
I’ll return to this problem later. For now, let’s
answer part (c).
(c) For all x, fb(ξ)e2πinxξ is of moderate decrease,
so
Z ∞
∞
X
2πixξ
b
fb(ξ)e2πinδx
f (ξ)e
dξ = lim δ
−∞
δ→0
|x|≥N
≤ + = .
2 2
Hence fb is continuous. For the second part of (a),
observe that for any ξ ∈ R, we have
Z ∞
1
1
fb(ξ) =
f (x) − f x −
e−2πixξ dx
2 −∞
2ξ
n=−∞
= f (x)
Exercise 2
Suppose f is of moderate decrease.
From this, it is clear to see that f (ξ) → 0 as
|ξ| → ∞. Here’s a complete verification of the observation: distribute the factor e−2πixξ inside the
integrand. Then we’ll get
Z
Z ∞ 1
1 ∞
−2πixξ 1
b
f (x)e
−
f x−
e−2πixξ dx
f (ξ) =
2 −∞
2 −∞
2ξ
R∞
Now use the invariance of the integral −∞ f (x −
1/2ξ)e−2πixξ dx
R ∞under translation to show that it
is equal to − −∞ f (x)e−2πixξ dx:
Z ∞
f (x − 1/2ξ)e−2πixξ dx
−∞
Z ∞
1
=
f (x)e−2πi(x+ 2ξ )ξ dx
Z−∞
∞
1
=
f (x)e−2πi(x+ 2ξ )ξ dx
Z−∞
∞
=
f (x)e−2πixξ e−πi dx
−∞
Z ∞
=−
f (x)e−2πixξ dx
(a) Prove that fb is continuous and fb(ξ) →
0 as |ξ| → ∞.
(b) Show that if fb(ξ) = 0 for all ξ, then f
is identically 0.
Solution.
(a) Let’s first prove that fb is continuous. Let >
0 be given, and ξ ∈ R. Since f is of moderate
decrease, kf k ≤ B for some constant B > 0, and
we can choose N ∈ N so large that
Z
|f (x)| dx < .
4
|x|≥N
Now choose a constant δ > 0 so small that |ξ − ζ| <
δ implies eξ − eζ < /4BN (such a constant exists because the exponential function is continu-
−∞
2
(b) The proof relies on the fact that the multiplication formula:
Z ∞
Z ∞
f (x)b
g (x) dx =
fb(y)g(y) dy
−∞
−∞
still holds for functions f and g of moderate decrease. Now all we have to do is to combine this
with the fact that there exists a family {gδ }δ>0 of
functions in S(R) such that for all δ > 0, and all
x, t ∈ R, we have gbδ (t) = Kδ (x − t):
Z ∞
Z ∞
b
gδ (y)f (y) dy =
Kδ (x − t)f (t) dt
−∞
−∞
= (f ∗ Kδ ) (x) → f (x)
as δ → 0.
Some verifications to do:
• The multiplication formula holds for functions of moderate decrease
• There exists a family {gδ }δ>0 of functions in
S(R) such that for all δ > 0, and all x, t ∈ R,
we have gbδ (t) = Kδ (x − t) (See Corollary
4.1.10 in the book)
(I’ll be adding more problems. A lot more!)
3