PRELIMINARY CONCEPTS
Remark. The authors have made an appendix of prerequisite knowledge; it contains,
notably, the following.
1. Riemann integral
Definition 1.1. A bounded real-valued function f : [a, b] → R is called Riemann
integrable if for every > 0 there exists a partition P such that U (P, f )−L(P, f ) < .
Rb
Definition 1.2. For f integrable on [a, b] we define a f (x) dx to be the common
value
Z b
(1)
f (x) dx = inf U (P, f ) = sup L(P, f ).
P
a
P
Remark. The following lemma will prove useful: that any integrable function can
be approximated (in some sense) by continuous functions. The proof also shows
that one can alternatively use step functions.
Lemma 1.3. Suppose f is integrable on the circle (i.e., a 2π periodic function on
R), bounded by B. Then there exists a sequence {fk } of continuous functions on
the circle, bounded by B, such that
Z π
(2)
|f (x) − fk (x)| dx → 0 as k → ∞
−π
2. Measure Zero
Remark. A notion of “smallness” for sets.
Definition 2.1. E ⊂ R is said to be of measure 0 if, given any > 0, there exists
a countable family of open intervals {Ik } such that
(1) P
E ⊂ ∪k Ik
∞
(2)
k=1 |Ik | < Lemma 2.2. The union of countably many sets of measure 0 is of measure 0.
Theorem 2.3. A bounded function f : [a, b] → R is integrable if and only if its set
of discontinuities is of measure 0.
Remark. The authors did not include the following knowledge in the appendix, but
it is also assumed. You will need to use this knowledge at various points, but not
prove it.
Elementary concepts: convergence of sequences, Cauchy sequences, Cauchy criterion, completeness, density.
2
PRELIMINARY CONCEPTS
3. Pointwise convergence
Remark. Just as one can consider the convergence of a sequence of points, one can
consider the convergence of a sequence {fn } of functions. The simplest type of
convergence is “pointwise convergence”, implying that for each fixed point, the sequence of points {fn (x)} converges. (Reference: Rudin, Principles of Mathematical
Analysis, chapter 7)
Definition 3.1. E ⊂ X a matrix space, {fn : E → C} a sequence of functions.
If limn→∞ fn (x) exists for each x ∈ E (call that limit f (x)), then we say {fn }
converges pointwise to f .
Definition 3.2 ( − N rephrasing). If, for each x ∈ E, we have that given any
> 0, there exists an N ∈ N such that n > N implies |fn (x) − f (x)| < , then we
say that {fn } converges pointwise to f on E.
Remark. There are certain important qualities that are not preserved by pointwise
convergence; that is, though all the {fn } might possess a certain property, their
pointwise limit f may not necessarily have that property.
Remarks.
(1) The pointwise limit of continuous functions need not be continuous.
(2) A pointwise convergent sum of continuous functions need not be continuous.
(3) Limit and derivative need not commute.
(4) Limit of a convergent sequence of integrable functions need not
be integrable.
(5) Limit and integration do not commute
4. Uniform Convergence
Remark. There is a stronger form of convergence, however: so-called uniform convergence, which turns out to preserve at least some of the above properties. (Reference: Rudin, Principles of Mathematical Analysis, chapter 7)
Definition 4.1 (uniform convergence). Let fn : E → C; n = 1, 2, 3, . . . . and f :
E → C. If, for all > 0, there exists an N such that n > N implies |fn (x)−f (x)| < for all x ∈ E, then we say {fn } converges to f uniformly.
Remark. Intuition: the tube.
Theorem 4.2 (Cauchy criterion). {fn } converges uniformly to f on E ⇐⇒ for all
> 0, there exists N such that m, n > N implies |fn (x) − fm (x)| < for all x ∈ E.
Proof. ⇒) Fix > 0. Since convergence is uniform on E, we know there exists
N > 0 such that if n > N then |fn (x) − f (x)| < /2 for all x ∈ E.
⇐) Note that for each fixed x ∈ E, {fn (x)} is a Cauchy sequence, and so
converges to some value, f (x). By hypothesis, given > 0, there exists N such
that n > N implies |fn (x) − fm (x)| < 2 for all n, m ≥ N . Let m → ∞; we get
|fn (x) − f (x)| ≤ 2 .
Theorem 4.3 (recharacterization). Say that limn→∞ fn (x) = f (x), and let Mn :=
supx∈E |fn (x)−f (x)|. Then fn → f uniformly on E if and only if limn→∞ Mn = 0.
PRELIMINARY CONCEPTS
3
Theorem 4.4 (Weierstrass M-test). Let {fn : E → C} be a set of functions.
P
If there exist constants
{Mn } such that |fn (x)| ≤ Mn for all x ∈ E and
Mn
P
converges, then
fn converges uniformly.
Proof. Use the Cauchy Criterion for sums.
5. Consequences of uniform convergence
Theorem 5.1. If {fn } are continuous functions that converge uniformly to some
f , then f must be continuous.
Proof. Exercise.
Theorem 5.2. Suppose fn : [a, b] → C are Riemann integrable functions. If fn →
f unifomly on [a, b], then f is itself Riemann integrable, and
Z b
Z b
(3)
f dx = lim
fn dx.
n→∞
a
a
Proof. Let n = sup[a,b] |fn (x) − f (x)|. Then fn − n ≤ f ≤ fn + n . Thus
Z
Z b
Z b
Z
(4)
(fn + n ) dx
(fn − n ) dx ≤ f ≤ f ≤
a
a
Z
(5)
⇒0≤
Z
f≤
Z
f≤
b
2n dx = 2n (b − a)
a
Let n → ∞; then n → 0, so the upper and lower integrals are equal.
Remark. It is not true that uniform convergence of fn → f implies that fn0 → f 0 .
However, one does have the following.
Theorem 5.3. {fn : [a, b] → C} be differentiable functions. If fn0 converge uniformly on [a, b], and the {fn } converge pointwise at some point x0 ∈ [a, b] then
(1) {fn } converge uniformly on [a, b] to some differentiable function f , and
(2) f 0 (x) = limn→∞ fn0 (x); x ∈ [a, b]
6. Fubini’s theorem
Remark. One theorem that will come in handy is the following. (Reference: Rudin,
Real and Complex Analysis, pp. 165ff)
Theorem 6.1. If f is a (measurable) function such that
Z Z
(6)
|f (x, y)| dx dy < ∞,
R R
R R
R R
then R RR fR (x, y) dx dy = R R f (x, y) dy dxthe order of integration in the double
integral R R f (x, y) dx dy may be reversed without changing the value of the integral.
Remark. Note that this is not true in general; see Rudin, p. 166 for examples.
THE WAVE EQUATION: D’ALEMBERT’S FORMULA
Initial Setting; Reduction of Problem
Suppose one has a string of length L > 0, fixed at both
ends. We let u(x, t) denote the displacement of the string
at position x at time t; e.g., u(x, 0) describes the string at
time t = 0, etc. Physical considerations imply that if u
is twice-differentiable, it satisfies the following (the “wave
equation”):
1 ∂ 2u ∂ 2u
(1)
= 2,
c2 ∂t2
∂x
for some constant c > 0.
Changing our units (let X = a1 x, T = 1b t), and letting
U (X, T ) = u(x, t), we get with appropriate choice (a =
L
L
π , b = cπ ) of constants that the wave equation (1) above is
equivalent to the following:
2
2
∂ U
∂ U
=
2
∂T
∂X 2
where 0 ≤ X ≤ π. That is, without loss of generality, we
may assume that the string is of length π, and that the
constant c = 1, i.e., that
The Wave
Equation
∂
∂T
∂
∂X
∂
= b ∂t
∂
= a ∂x
c2 π 2 ∂ 2 u
c2 L2 ∂T 2
=
π2 ∂ 2 u
L2 ∂X 2
(2)
(3)
∂ 2u ∂ 2u
= 2 on 0 ≤ x ≤ π, t ≥ 0.
∂t2
∂x
Solutions on R are all combinations of
traveling waves.
Goal: find
u(x, t)
satisfying
the above.
Book gives
two methods
both relevant
one more so
2
THE WAVE EQUATION: D’ALEMBERT’S FORMULA
First observation. Suppose we ignore the initial conditions (that u(0, t) = u(π, t) = 0 for all t > 0). Then, for
any twice differentiable function F , if we define either
(4)
u(x, t) = F (x + t) or u(x, t) = F (x − t),
it is an easy calculation that either solves the equation.
Such solutions are called traveling waves, for obvious reasons.
In fact, any twice-differentiable solution u to the
wave equation on R must be a combination of (opposing) traveling waves.
For let ξ = x + t, η = x − t, and notate
(5)
ν(ξ, η) = u(x, t).
Then, again by the Chain Rule,
∂ξ ∂
∂η ∂
∂
(6)
=
+
∂x ∂x ∂ξ ∂x ∂η
∂
∂
(7)
=
+
∂ξ ∂η
and, similarly,
∂
∂
∂
(8)
=
− ,
∂t ∂ξ ∂η
so the wave equation is equivalent to
∂ 2v
∂ 2v
∂ 2v
∂ 2v
∂ 2v
∂ 2v
(9)
+2
+
= 2 −2
+
,
∂ξ 2
∂ξ∂η ∂η 2
∂ξ
∂ξ∂η ∂η 2
i.e.,
∂ 2v
(10)
= 0.
∂ξ∂η
Thus
(11)
v(ξ, η) = F (ξ) + G(η),
or, switching back to x, t notation,
(12)
u(x, t) = F (x + t) + G(x − t),
a sum of two waves in opposite directions, as claimed.
LECTURE 2: D’ALEMBERT’S FORMULA (CONT’D);
STANDING WAVES; HEAT EQUATION
Remark. Finishing up Chapter I (motivating the problems).
From Chapter II onwards the presentation will be more rigorous; however this first chapter introduces some important
themes.
Method I: Traveling Waves
Recall: last time we showed that every solution to the wave
equation can be expressed as a sum of two travelling waves,
in opposing directions.
Returning to our string of length π, let u(x, 0) = f (x);
0 ≤ x ≤ π denote the initial configuration of the string; we
extend f first as an odd function to [−π, π], and then as a
2π-periodic function to all of R; do likewise for u(x, t) (also
set u(x, t) = u(x, −t) for t < 0). Then u is a solution for
the wave equation on all of R, so must be a sum of traveling
waves as above. What are F and G?
Well, we know
(1)
u(x, t) = F (x + t) + G(x − t),
so
(2)
f (x) = u(x, 0) = F (x) + G(x).
In fact, we haven’t imposed enough constraints to determine the functions uniquely; we impose an additional
(“initial velocity”) condition, that
∂u
(3)
(x, 0) = g(x),
∂t
where g is an odd, 2π-periodic function satisfying g(0) =
g(x) = 0 like f .
Q. What about fixedendpoint string motion?
Q. What are those
waves?
initial conditions
Not uniquely determined; we impose
Initial velocity condition
D’ALEMBERT, STANDING WAVES, HEAT EQUATION
2
Thus our system is
(4)
(5)
That’s enough; now
we can solve it
F (x) + G(x) = f (x)
F 0 (x) − G0 (x) = g(x),
which implies
2F 0 (x) = f 0 (x) + g(x)
2G0 (x) = f 0 (x) − g(x),
(6)
(7)
forcing
d’Alembert’s
formula
That’s one way; another which leads directly to the main
question of Fourier
analysis is
Z x
1
(8)
f (x) +
g(y) dy + C1
F (x) =
2
0
Z x
1
(9)
f (x) −
g(y) dy + C2
G(x) =
2
0
and thus (note C1 + C2 = 0)
Z
1
1 x+t
(10) u(x, t) = [f (x + t) + f (x − t)] +
g(y) dy.
2
2 x−t
Method II: superposition of standing waves
Remark. Motivation for the notion of Fourier series - the
simplest version of Fourier analysis.
Same problem
Standing waves
Again we examine the wave equation
∂ 2u ∂ 2u
(11)
= 2
∂t2
∂t
with initial condition u(x, 0) = f (x) and initial velocity
condition ∂u
∂t (x, 0) = g(x) as before.
Idea: consider solutions called “standing waves,” i.e., of
the simple form u(x, t) = φ(x)ψ(t). In that case, the
functions must satisfy the system
(12)
(13)
φ00 (x) − λφ(x) = 0
ψ 00 (t) − λψ(t) = 0,
D’ALEMBERT, STANDING WAVES, HEAT EQUATION
3
and thus (tossing out the non-oscillating solutions)
(14)
ψ(t) = A cos(mt) + B sin(mt)
(15)
φ(x) = Ã cos(mx) + B̃ sin(mx)
blah blah blah ....solutions are of the form
(16)
um (x, t) = (Am cos mt + Bm sin mt) sin mx
for m = 1, 2, 3, . . . (we combine negative indices with positive).
So we have an infinite number of solutions. Recall/observe
that any linear combination of solutions is still a solution,
and suppose u(x, t) is some linear combination of such solutions, i.e,.
∞
X
(17)
u(x, t) =
(Am cos mt + Bm sin mt) sin mx;
φ(0) = 0; so à = 0;
φ(π) = 0; so m ∈ Z
if B̃ 6= 0
infinite number of
solutions
m=1
What are the Am and Bm ? Recalling that u(x, 0) =
f (x), we see
∞
X
(18)
Am sin mx = f (x).
m=1
This motivates the question: given a sufficiently good
function f on [0, π] with f (0) =Pf (π) = 0, can we find
coefficients Am such that f (x) = ∞
m=1 Am sin mx?
More generally, the question we will really be concerned with is the following: given an arbitrary function
F on [−π, π], can we find coefficients am such that
∞
X
(19)
F (x) =
am eimx
m=−∞
(note that eimx = cos(mx) + i sin(mx)).
Well, it is an easy calculation that
Z π
1
0 if n 6= m
(20)
eimx e−inx dx =
1 if n = m
2π −π
The first question
(Note we also see
from the initial vel.
cond’n
that g(x) =
P∞
mB
m sin mx.)
m=1
The more general
question
4
“formally”: i.e., if
one ignores problems
of convergence
infinite dimensional
orthonormal basis
Main points: 1. introduce the terminology: heat equation,
Laplacian; Dirichlet
problem; polar coordinates, 2. observe
that the same problem arises.
Time-dependent
heat equation
∂u
∂t
=0
Dirichlet Problem
D’ALEMBERT, STANDING WAVES, HEAT EQUATION
so, at least formally, one would expect that
Z π
1
an =
(21)
F (x)e−inx dx.
2π −π
These coefficients are in fact called the Fourier coefficients
of F , and our job will be to determine in what sense, and
under what conditions, the above equality is true.
Remark. Recall from linear algebra: one may recognize this
as an inner product space; then we have an orthonormal
basis (infinite dimensional space). Fourier coefficients are
the coefficients...
The heat equation
Suppose one has an infinite plate (R2 ) with an initial
heat distribution. Let u(x, y, t) denote the temperature of
the place at time t, position (x, y). Physical considerations
yield an equation governing the evolution of that heat:
σ ∂u ∂ 2 u ∂ 2 u
(22)
= 2+ 2
κ ∂t
∂x
∂y
We’ll mainly be concerned with the steady-state heat
equation
∂ 2u ∂ 2u
(23)
∆u := 2 + 2 = 0
∂x
∂y
The Dirichlet problem (D some open domain (set), and
∂D its boundary):
∆u = 0 on D
(24)
u = f on ∂D
E.g., one can consider the Dirichlet problem on
D = {(x, y) ∈ R2 : x2 + y 2 < 1} = {(r, θ) ∈ R × S 1 : r ∈ [0, 1)}
Laplacian in polar
coordinates
the unit disk.
In polar coordinates, the condition ∆u = 0 becomes (exercise)
D’ALEMBERT, STANDING WAVES, HEAT EQUATION
Same approach as
before: separation of
variables
5
1 ∂ 2u
∂ 2 u 1 ∂u
(25)
+
= 0,
∆u = 2 +
∂r
r ∂r r2 ∂θ2
2
∂u
∂ 2u
2∂ u
i.e., r
(26)
+r
=− 2
∂r2
∂r
∂θ
We take the same “separation” approach as above: let
u(r, θ) = F (r)G(θ); then by a similar argument, we get
00
G (θ) + λG(θ) = 0
(27)
r2 F 00 (r) + rF 0 (r) − λF (r) = 0
As before, we obtain a family of solutions,
(28)
um (r, θ) = r|m| eimθ ; m ∈ Z
and, supposing that u be some linear combination
∞
X
(29)
am r|m| eimθ
u(r, θ) =
−∞
Different
setting,
same question
of those solutions, we see that, in this different setting, we
arrive at an identical question. For the boundary value
condition requires that
∞
X
(30)
u(1, θ) =
am eimθ = f (θ),
−∞
so, our question is, again: “Given any reasonable function
f on [0, 2π] with f (0) = f (2π), can we find coefficients am
such that
∞
X
(31)
f (θ) =
am eimθ ?”
−∞
LECTURE 3: INTRODUCTION TO FOURIER SERIES
Basic definitions: basic function classes (read pp. 31-33
yourself), Fourier series; Examples: calculations of Fourier
series, Dirichlet and Poisson kernels (pp. 34-39)
1. Basic knowledge
(1) Continuous functions on [0, L].
(2) Piecewise continuous functions (only finitely many
discontinuities)
(3) Riemann integrable functions (note bounded)
(4) Functions on the circle (correspondence with 2π periodic functions on R such that f (0) = f (2π).
Remark. We will assume that all of our functions are Riemann integrable
2. Definitions and examples
Definition 2.1. Given an integrable function f : [a, b] →
C, we define the Fourier series of f as
∞
X
2πinx
fˆ(n)e L ,
some definitions
Warning: only integrable functions
Important
defn.:
Fourier series
n=−∞
where
Z b
1
fˆ(n) :=
f (x)e−2πinx/L dx
L a
denotes the n-th Fourier coefficient of f for n ∈ N.
Example: the Fourier series of the 2π-periodic odd function defined on [0, π] by f (θ) = θ(π − θ).
Well, our function is f (θ) = θ(π − θ) for θ > 0 (with
derivative f 0 (θ) = π −2θ there), and f (θ) = θ(π +θ) for θ <
Fourier coefficient
Example of Fourier
series
2
Note we don’t know
if there’s any relation - think of Taylor
series
LECTURE 3: INTRODUCTION TO FOURIER SERIES
0 (with derivative π + 2θ there). Then, using integration
by parts, we get
Z π
1
ˆ
f (θ)e−inθ dθ
f (n) :=
2π −π
π
Z
1
−f (θ) inθ
1 π 0
=
e
f (θ)e−inθ dθ
+
2π
in
in −π
−π
Z 0
Z π
1
2
=
θe−inθ dθ −
θe−inθ dθ
2π in
0
Z π −π
1
θ(e−inθ + einθ ) dθ (C.O.V.: γ = −θ)
=−
inπ 0
π
Z
2
θ
1 π
=−
sin(nθ) −
sin(nθ) dθ
inπ
n
n
0
0
2
= 3 [(−1)n+1 + 1]
n πi
4
= 3
n πi
for n odd, and 0 otherwise. Thus the Fourier series is
f (θ) ∼
X 4
eikx .
3
k π
k odd
Simple observation
Remark. There is, a priori, no guarantee that the Fourier
series will converge at all; even if it does converge, it may
not converge to f . In fact, if f and g agree everywhere
excluding a finite set of points, their Fourier series will be
identical. So asking for pointwise convergence to the original function is in fact a priori futile.
partial sum
Definition 2.2. We define the N th partial sum of the Fourier
series of f , N ∈ N by
N
X
SN (f )(x) :=
fˆ(n)e2πinx/L .
n=−N
Define trigonometric
polynomials!
Only sketch it! Let
them fill in the details - it’s just calculus.
LECTURE 3: INTRODUCTION TO FOURIER SERIES
3
Remark. In this text, “convergence of the Fourier series to
f ” will always mean convergence of the above partial sums
to f .
3. Some important constructions
Dirichlet kernel: for
now, opaque
Definition 3.1. We define the Dirichlet kernel DN by
N
X
DN (x) =
einx
n=−N
where x ∈ [−π, π].
Lemma 3.2. DN (x) =
sin((N + 21 )x)
sin(x/2)
Leave to them.
Proof. Consider the geometric sums
N
X
−1
X
ix n
(e ) and
n=0
(eix )n .
n=−N
Question. (dumb) what are the Fourier coefficients of DN ?
Definition 3.3. We define the Poisson kernel Pr (θ) by
Pr (θ) =
∞
X
r|n| einθ
n=−∞
where θ ∈ [−π, π] and r ∈ [0, 1)
Remark. Note that the sum is both absolutely and uniformly convergent. (Obviously: notice the |n|.)
Question. (non-dumb?) what are the Fourier coefficients of
Pr (θ)? (Point: uniform convergence is necessary.)
Lemma 3.4. Pr (θ) =
1−r2
1−2r cos θ+r2
Poisson
arose
in
equation
kernel:
heat
4
LECTURE 3: INTRODUCTION TO FOURIER SERIES
Proof. Letting ω = reiθ , we have a sum of geometric series:
∞
∞
X
X
n
Pr (θ) =
ω +
ω̄ n
n=0
n=1
1
ω̄
1 − |ω|2
+
=
1 − ω 1 − ω̄
|1 − ω|2
1 − r2
=
1 − 2r cos θ + r2
=
4. Uniqueness of Fourier Series
Remark. The question: if the Fourier series did recover the
original function uniquely, then functions with the same
Fourier coefficients would have to be equal; in particular, if
fˆ(n) = 0 for all n ∈ Z, then f ≡ 0. This is of course false.
However one has the following.
Finally: first real
theorem!
i.e., on [−π, π], with
f (−π) = f (π)
Theorem 4.1 (Uniqueness). Let f be an integrable function on the circle. Suppose fˆ(n) = 0 for all n ∈ Z. If f is
continuous at θ0 , then f (θ0 ) = 0.
Remark. Thus f ≡ 0 a.e., since integrable functions are
continuous except on a set of measure zero.
Remark. Observe that if fˆ(n) = 0 for all n ∈ Z, then f
must be orthogonal to all finite linear combinations of the
blah, i.e., orthogonal to all trigonometric polynomials.
Proof. Proof by contradiction. Suppose f (0) > 0 (WLOG
assume θ0 = 0). By continuity, we know f (x) > 21 f (0) in
some neighborhood (−δ, δ) of 0.
We now create a sequence {pk (θ)} of trigonometric polynomials (i.e., finite linear combinations of the {einx : n ∈
Z}) as follows: Let
p(θ) = cos θ + ,
LECTURE 3: INTRODUCTION TO FOURIER SERIES
Draw a picture
5
where > 0 is chosen so small that outside of (−δ, δ), we
still have
|p(θ)| < 1 − .
2
Notice that by continuity of cos θ, we can choose a small
η > 0 such that inside of (−η, η), we have
p(θ) > 1 + .
2
k
Let pk (θ) = [p(θ)] (note that these are all trigonometric
polynomials).
Since the Fourier
coefficients fˆ(n) := hf, e2πinx/L i are all
P inx
0, we have hf, cn e i = 0 for all trigonometric polynomials. However, we have just created a sequence of trigonometric polynomials for which that does not happen....(to
be continued)
Point:
powers of
p will (uniformly)
shrink outside of the
δ neighborhood, but
grow to infinity uniformly inside the η
neighborhood
(The Idea)
LECTURE 4: UNIQUENESS, PART II
1. Uniqueness of Fourier Series, continued
Recall: we were in the midst of the following theorem:
Theorem 1.1 (Uniqueness). Let f be an integrable function on the circle. Suppose fˆ(n) = 0 for all n ∈ Z. If f is
continuous at θ0 , then f (θ0 ) = 0.
i.e., f integrable
on [−π, π], with
f (−π) = f (π)
Proof.
Proof by contradiction. Suppose f (0) > 0 (WLOG assume θ0 = 0). By continuity, we know f (x) > 21 f (0) in some neighborhood (−δ, δ) of 0.
We now create a sequence {pk (θ)} of trigonometric polynomials (i.e., finite linear combinations of the {einx : n ∈ Z}) as follows: Let
p(θ) = cos θ + ,
where > 0 is chosen so small that outside of (−δ, δ), we still have
|p(θ)| < 1 − .
2
Notice that by continuity of cos θ, we can choose a small η > 0 such that inside of (−η, η),
we have
p(θ) > 1 + .
2
Let pk (θ) = [p(θ)]k (note that these are all trigonometric polynomials).
P
Since the Fourier coefficients fˆ(n) := hf, e2πinx/L i are all 0, we have hf, cn einx i = 0
for all trigonometric polynomials. However, we have just created a sequence of trigonometric polynomials for which that does not happen....(to be continued)
Rπ
The details: We estimate −π f (θ)pk (θ) dθ in three parts:
(1) In the η-neighborhood, we have the crude estimate
Z
f (0)
(1 + )k
f (θ)pk (θ) dθ ≥ 2η
2
2
(−η,η)
which goes to infinity as k does.
(2) Outside of the δ neighborhood, we have the (again
crude) estimate
Z
f (θ)pk (θ) dθ ≤ 2πB(1 − )k
2
(−δ,δ)c
where B is the bound on (integrable) f ; this goes to
0 as k → ∞.
Draw a picture
Point:
powers of
p will (uniformly)
shrink outside of the
δ neighborhood, but
grow to infinity uniformly inside the η
neighborhood
(The Idea)
The details
2
LECTURE 4: UNIQUENESS, PART II
(3) Between the two, p and f are non-negative, so the
integral there is positive.
Together, the above prove that
Z
lim
f (θ)pk (θ) dθ = ∞,
k→∞
(−π,π)
a contradiction.
Some consequences
2. Consequences of Uniqueness theorem
Corollary 2.1. Suppose f is continuous on the circle. If
fˆ(n) = 0 for all n ∈ Z, then f ≡ 0.
Thus if two continuous functions have
the same Fourier coefficients, they must
be identical.
Recovering the function from the Fourier
series.
Proof. Obvious.
Question. At this point, do we know that if a function is
continuous, then its Fourier series converges back to the
function? (No. In fact, that is false.)
Corollary P
2.2. Suppose f is a continuous function on the
ˆ
circle, and ∞
n=−∞ |f (n)| < ∞ (i.e., the Fourier series of
f is absolutely convergent). Then
lim SN (f )(θ) = f (θ)
N →∞
uniformly on the circle.
Note we do not have
this for cts. fns. in
general.
Natural question
P
ˆ
Proof. (Trivial.) Since ∞
n=−∞ |f (n)| < ∞, we know that
the Fourier series converges uniformly to some continuous
function, which has Fourier coefficients fˆ(n); n ∈ Z. Thus
we have two continuous functions with the same Fourier
coefficients, so they must, by the previous lemma, be identical.
Question. When do we have absolute convergence of the
sum of the Fourier coefficients?
Introduce a useful
notation
LECTURE 4: UNIQUENESS, PART II
3
Definition 2.3 (Big O notation). We say f (x) = O(g(x))
as x → a if there exists a C > 0 such that
f (x)
lim
≤ C.
x→a g(x)
Corollary 2.4. Let f be a function on the circle. If f is
twice continuously differentiable (i.e., is of class C 2 ), then
fˆ(n) = O(1/|n|2 ) as |n| → ∞.
Proof. Proof is trivial: integration by parts.
Z 2π
2π fˆ(n) =
f (θ)e−inθ dθ
0
2π
Z
−e−inθ
1 2π 0
= f (θ)
+
f (θ)e−inθ dθ
in
in 0
0
Z 2π
1
f 0 (θ)e−inθ dθ
=
in 0
Continuing with another integration by parts,
Z 2π
−inθ 2π
1
−e
1
=
f 0 (θ)
+
f 00 (θ)e−inθ dθ
2
in
in
(in)
0
0
Z 2π
1
f 00 (θ)e−inθ dθ
=− 2
n 0
So
Z 2π
Z 2π
1
1
2
00
−inθ
ˆ
|f (n)||n | ≤
f (θ)e
dθ ≤
|f 00 | ≤ C,
2π 0
2π 0
as desired.
This is a good place to introduce the following important
notion:
Definition 2.5. Let f be a function for which there exists
a constant A such that for all x, h
|f (x + h) − f (x)| ≤ A|h|α .
ˆ
(n)|
lim|n|→∞ |f|n|
≤ C
2
for some C
And thus we can recover the function.
Stop there for a second! fb0 (n) = infˆ(n)
and so on and so
forth
4
LECTURE 4: UNIQUENESS, PART II
Then we say that f satisfies a Holder condition (or is Holder
continuous) of order α.
Remark. In fact (as you will show) if a function is Holder
continuous order α > 1/2, then the Fourier series converges
absolutely (and thus uniformly to f ).
LECTURE 5: CONVOLUTIONS AND GOOD KERNELS
1. Convolutions
Remark. Now for a seemingly simple, but important notion....
Definition 1.1. Let f, g : R → C be 2π-periodic functions.
The convolution f ∗ g of f and g is the function defined on
[−π, π] by
Z π
1
f (y)g(x − y) dy
(f ∗ g)(x) :=
2π −π
Remarks.
(1) It is an easy exercise (C.O.V.) to see that f ∗g = g ∗f .
(2) Convolution as weighted average.
(3) Turns out that many important constructs can be
expressed in terms of convolutions.
For example, consider f ∗ DN , the Dirichlet kernel :
1
(f ∗ DN )(x) :=
2π
Z
π
f (y)
−π
N
X
1
=
2π
=
=
−N
N
X
−N
N
X
e
ein(x−y) dy
−N
Z
inx
N
X
E.g., The Hilbert
Transform
“kernel” = “that
which one convolves
against”
π
f (y)ein(x−y) dy
−π
1
2π
Z
π
f (y)e−iny dy
−π
einx fˆ(n) =: SN (f )(x),
−N
the N th partial sum of the Fourier series.
The question about
convergence
of
Fourier series can
be thought of as
the
convergence
of a sequence of
particular weighted
averages.
2
LECTURE 5: CONVOLUTIONS AND GOOD KERNELS
2. Properties of Convolution
1
L (T) a Banach algebra.
Theorem 2.1 (Basic properties). Let f, g, h be 2π-periodic
integrable functions, and c ∈ C. Then
i.
ii.
iii.
iv.
v.
vi.
(Linearity I) f ∗ (g + h) = (f ∗ g) + (f ∗ h)
(Linearity II) (cf ) ∗ g = c(f ∗ g) = f ∗ (cg)
(Commutative) f ∗ g = g ∗ f
(Associative) (f ∗ g) ∗ h = f ∗ (g ∗ h)
(Continuity!) f ∗ g is continuous.
(Interaction with Fourier transform) f[
∗ g(n) = fˆ(n)ĝ(n)
What is f\
∗ DN (n)?
ˆ
f (n)χ[−N,N ] (n).
Obvious if continuous
Proof of (v), first for
cts. fns. (fairly standard argument)
Remark. If one assumes that f, g, h are continuous functions, then all of the properties, excluding the fifth, are
immediate calculations. E.g., (using Fubini’s theorem) one
can prove (vi) for continuous functions as follows.
Z π
1
(f ∗ g)(x)e−inx dx
f[
∗ g(n) :=
2π −π
Z π Z π
1
1
=
f (y)g(x − y) dy e−inx dx
2π −π 2π −π
Z π
Z π
1
1
=
f (y)e−iny
g(x − y)e−in(x−y) dx dy
2π −π
2π
Z−ππ
Z π
1
1
=
f (y)e−iny
g(x)e−inx dx dy
2π −π
2π −π
= fˆ(n)ĝ(n)
Proof. We will prove the second-to-last property, initially
in the case that f, g are continuous. Then we will show
how one obtains the result for integrable functions by approximating them with continuous ones.
Step I. Suppose f, g are continuous; we want to show
f ∗ g is also. I.e., we want to show that given any > 0,
there exists a δ such that |x1 − x2 | < δ implies |(f ∗ g)(x1 ) −
(f ∗ g)(x2 )| < .
This is actually
uniform continuity;
note
[−π, π]
is
compact.
LECTURE 5: CONVOLUTIONS AND GOOD KERNELS
“standard
estimates”
Key:
uniform
continuity of g
What if we don’t
have continuity?
Approximate
(in
L1 ) with continuous
functions.
Why does this suffice? (Dumb question.)
3
Well, we estimate crudely with the triangle inequality:
Z π
1
f (y)[g(x1 − y) − g(x2 − y)] dy
|(f ∗ g)(x1 ) − (f ∗ g)(x2 )| ≤
2π −π
Z π
1
≤
|f (y)| |g(x1 − y) − g(x2 − y)| dy.
2π −π
f , being integrable, is bounded by some B on [−π, π], so
we use the uniform continuity of g to choose δ > 0 such that
|a − b| < δ implies |g(a) − g(b)| < B . Then, if |x1 − x2 | < δ,
1
2πB B = . In other
we see the above is smaller than 2π
words, the convolution is (uniformly) continuous.
Question. Notice that we really needed the continuity of g
(but only boundedness of f ). How are we going to do this
without continuity?
Step II. Suppose f, g are integrable. By the lemma in
the appendix, any integrable function can be approximated
in the L1 sense by a sequence of continuous functions. So
let {fk } and {gk } be such sequences for f, g, respectively.
It suffices to show the following.
Claim: fk ∗ gk → f ∗ g uniformly on [−π, π].
Proof. First note that
f ∗ g − fk ∗ gk = (f − fk ) ∗ g + fk ∗ (g − gk ), so
|f ∗ g − fk ∗ gk | ≤ |(f − fk ) ∗ g| + |fk ∗ (g − gk )|
Now
Z π
1
|(f − fk ) ∗ g(x)| ≤
|f (x − y) − fk (x − y)| |g(y)| dy
2π −π
Z π
1
≤
sup |g(y)|
|f (y) − fk (y)| dy → 0
2π y
−π
as k → ∞. Similarly for |fk ∗(g −gk )|; thus the convergence
depends on k (and not on x); i.e., we have the desired
uniform convergence. (End of proof of claim.)
4
LECTURE 5: CONVOLUTIONS AND GOOD KERNELS
Then, by Step I, since fk ∗ gk are continuous, their uniform
limit f ∗ g is also.
LECTURE 6: CONVOLUTIONS AND GOOD KERNELS
1. Convolutions, continued
Recall: we were proving certain properties about the convolution of two (integrable) functions; more importantly,
we were showing examples of how one extends results on
continuous functions to integrable functions, using the “L1
approximation lemma.”
Claim: (vi) For integrable f, g, f[
∗ g = fˆĝ.
Proof. Let {fk }, {gk } be sequences of continuous functions
converging to f, g in L1 . In the previous example, we
showed that fk ∗ gk converges to f ∗ g uniformly, so, interchanging integral and limit,
[
f\
k ∗ gk (n) → f ∗ g(n)
for each fixed n ∈ Z.
Now, since fk and gk are continuous, we know already
b
that f\
k ∗ gk = fk gbk . That is, we actually have
fbk (n)gbk (n) → f[
∗ g(n)
Second example
of
extending
a
result from continuous functions to
integrable ones.
Key: uniform convergence of fk ∗ gk
We have the result
already for continuous functions.
as k → ∞. So STS that
fbk (n)gbk (n) → fˆ(n)ĝ(n).
That follows from the L1 statement: fbk (n) → fˆ(n), since
Z π
1
|fˆ(n) − fbk (n)| =
[f (x) − fk (x)]e−inx dx
2π −π
Z π
1
≤
|f (x) − fk (x)| dx
2π −π
which goes to 0 as k → ∞ (similarly for gk (n)) so we are
done.
Where L1 convergence comes into
play.
2
Convolution as multiplication: What is
1?
LECTURE 6: CONVOLUTIONS AND GOOD KERNELS
2. Good kernels (Approximations of the
Identity)
Remark. One thinks of convolution as a kind of multiplication; in fact, L1 (T) with convolution as multiplication is a
so-called Banach algebra.
Question. What’s the identity element (for this multiplication)?
Well, there isn’t one (it would be the Dirac delta function,
which is not a function). But although there isn’t an identity element, we can create a sequence of elements that
approximate the identity: a so-called good kernel.
An extremely useful
notion: an approximation of the identity
Definition 2.1. Let {Kn : T → R}∞
n=1 be a sequence of
functions on the circle. If
Rπ
1
i. (“sliding in” condition) 2π
−π Kn (x) dx = 1 for all n ∈
N,
Rπ
1
ii. there exists M > 0 such that 2π
−π |Kn (x)| dx ≤ M
for all n ∈ N, and
iii. given any δ > 0,
Z
|Kn (x)| dx → 0
δ≤|x|≤π
as n → ∞,
then we say {Kn }∞
n=1 is a family of good kernels or (more
commonly) an approximation to the identity.
What’s the point?
They approximate
the identity, i.e., the
Dirac delta function.
Theorem 2.2. Let f be an integrable function on the circle,
and {Kn } a family of good kernels. Then
lim (f ∗ Kn )(x0 ) = f (x0 )
n→∞
whenever f is continuous at x0 . If f is continuous everywhere, then the above limit is uniform.
Sliding in the function to take advantage of smoothness
(here, continuity)
Proof. Let > 0 be given. We want to control |f ∗Kn (x0 )−
f (x0 )|. Well,
LECTURE 6: CONVOLUTIONS AND GOOD KERNELS
Using ||Kn ||1 ≤ M
and continuity on
the first part
Using boundedness
of f and the shrinking of Kn
Q. What do the
bounds depend on?
3
Z π
1
|f ∗ Kn (x0 ) − f (x0 )| =
Kn (y)f (x0 − y) dy − f (x0 )
2π −π
Z π
1
=
|Kn (y)[f (x0 − y) − f (x0 )]| dy
2π −π
We know f is continuous at x0 , so choose δ > 0 such that
|y| < δ implies |f (x0 − y) − f (x0 )| < . Then, obviously, we
break the integral into two parts, the first over |y| < δ, the
second its complement. For the first integral we get:
Z π
|Kn (y)| dy
|Kn (y)[f (x0 − y) − f (x0 )]| dy ≤
2π −π
|y|<δ
≤
M.
2π
For the second integral,
Z
Z
1
2B
|Kn (y)[f (x0 − y) − f (x0 )]| dy ≤
|Kn (y)| dy
2π δ≤|y|≤π
2π δ≤|y|≤π
1
2π
Z
By the third property of good kernels, the second integral
can be made arbitrarily small; thus we are done.
Notice that the first bound is independent of x0 , but the
second depends on δ (which depends on x0 ). However, if
f were uniformly continuous, the second bound would not
depend on x0 ; in that case the convergence would be uniform.
Remark. Notice that if the Dirichlet kernel DN were an
approximation of the identity, then we’d immediately have
pointwise convergence of the Fourier series of an integrable
at every point of continuity (recall that we have no such
results for integrable functions). Unfortunately, the kernel
that governs convergence of the Fourier series is not an
approximation of the identity.
The Dirichlet kernel
is not good.
Later: a continuous
function
whose
Fourier
series
diverges
LECTURE 7: CESARO AND ABEL SUMMABILITY
Remark. As we are starting to see, even if the function is
continuous, the Fourier series may not recover the function.
On the other hand, there should be enough information in
the Fourier coefficients to recover the function, especially
for continuous functions: after all, we have the uniqueness
theorem. How can we recover the function from the Fourier
series, if not in the direct way?
A different way of
recovering the function.
There should be
enough information!
We shall present two different ways of averaging the Fourier
series to recover the function.
1. Cesaro summability
P
Definition 1.1. Given a sequence {cn }, let sn := nk=0 ck
be the sequence of partial sums. We define the N th Cesaro
th
mean σN of the
P∞sequence {sk } (a.k.a. the N Cesaro sum
of the series k=0 ck ) by
σN :=
Cesaro sum of the series
s0 + s1 + · · · + sN −1
.
N
Remark. Convergence implies Cesaro summability.
2. Fejer’s theorem
Definition 2.1. We define the Fejer kernel, FN , by letting FN (x) denote the N th Cesaro mean of the sequence
{Dk (x)}; i.e.,
N −1
1 X
FN (x) :=
Dk (x).
N
k=0
The Fejer kernel
2
Convolving with FN
gives the Cesaro sum
of the Fourier series
LECTURE 7: CESARO AND ABEL SUMMABILITY
Remark. Then consider:
N −1
1 X
f ∗ Dk (x)
f ∗ FN (x) :=
N
=
1
N
k=0
N
−1
X
Sk (f )(x),
k=0
the average of the first N partial sums of the Fourier series
of f .
The Fejer kernel is
an approximation of
the identity
Lemma 2.2. One has
1 sin2 (N x/2)
FN (x) =
;
N sin2 (x/2)
further, the Fejer kernel is a good kernel.
Since FN is a good
kernel, we immediately get
We can recover the
function!
We knew this already.
Proof. Exercise: prove the closed form. One can use the
closed form to prove theRgood kernel criteria. FN being
π
1
positive, the fact that 2π
−π FN = 1 gives us the first two
properties. The last follows from sin2 (x/2) ≥ cδ > 0 for
δ ≤ |x| ≤ π. For then the |FN (x)| ≤ N1cδ in that region, so
the integral converges to 0 as N → ∞.
Theorem 2.3. Let f be an integrable function on the circle.
If f is continuous at θ0 , then the Fourier series of f is
Cesaro summable to f at θ0 . Further, if f is continuous on
the entire circle, then the convergence of the Cesaro sums
is uniform.
Corollary 2.4. Let f be integrable on the circle. If fˆ(n) =
0 for all n ∈ Z, then f = 0 at all points of continuity of f .
Proof. Obvious: at points of continuity, the Fourier series
(namely, 0) is Cesaro summable to f .
Corollary 2.5. Any continuous function on the circle can
be uniformly approximated by trigonometric polynomials.
LECTURE 7: CESARO AND ABEL SUMMABILITY
3
Proof. The Cesaro sums are averages of the partial sums
(which are trigonometric polynomials) and so are polynomials themselves.
3. Abel means and summation
P
Definition 3.1. Let ∞
numbers.
k=0 be a series of complexP
i. We define the Abel means A(r) of the series
ck by
∞
X
A(r) :=
ck r k .
Abel means of a
series
Taking the terms
as coefficients of a
power series
k=0
ii. If, for every 0 ≤ r < 1, the Abel means A(r) converges,
and
lim A(r) = s
P
then we say the series
ck is Abel summable to s.
P
k
Example. Consider ∞
k=0 (−1) = 1 − 1 + 1 − 1 + 1 . . . . It
diverges, but is Abel summable to s = 12 :
∞
X
1
A(r) :=
(−1)k rk =
.
1+r
k=0
P
k
Example. (Done in text.) Consider ∞
k=0 (−1) (k + 1) =
1 − 2 + 3 − 4 + 5 . . . . It diverges, but is Abel summable to
s = 14 :
∞
X
1
A(r) :=
(−1)k (k + 1)rk =
.
(1 + r)2
Abel summable to s
r→1
Examples
k=0
4. The Abel means of a Fourier series
and the Poisson kernel
Definition 4.1. Suppose we know the Fourier series of a
function f :
∞
X
f (θ) ∼
an einθ .
n=−∞
Abel means of a
Fourier series
LECTURE 7: CESARO AND ABEL SUMMABILITY
4
We define the Abel means Ar (f )(θ) of the Fourier series of
the function f by
∞
X
r|n| an einθ
Ar (f )(θ) =
n=−∞
Remarks.
i. If we let c0 = a0 , cn = an einθ + a−n e−inθ , then the Abel
means of the P
Fourier series above equals the Abel means
of the series ∞
k=1 ck .
ii. For f integrable, |an | is uniformly bounded in n, so
Ar (f ) converges absolutely, and for each fixed 0 ≤ r <
1, uniformly.
This means can also
be recognized as a
convolution.
Lemma 4.2 (Abel means as a convolution).
Ar (f )(θ) = (f ∗ Pr )(θ)
Proof. Recall the Poisson kernel
∞
X
Pr (θ) :=
r|n| einθ
n=−∞
(We also saw that
2
Pr (θ) = 1−2r1−r
cos θ+r 2
for 0 ≤ r < 1.)
Then
Ar (f )(θ) :=
=
∞
X
r|n| an einθ
n=−∞
∞
X
r|n|
n=−∞
=
1
2π
Z
1
2π
π
f (φ)
−π
Z
π
f (φ)e−inφ dφ einθ
−π
∞
X
!
r|n| e−in(φ−θ)
dφ
n=−∞
= (f ∗ Pr )(θ).
(Note we needed
uniform
convergence.)
Pr is an approximation of the identity.
Lemma 4.3. The Poisson kernel is an approximation of
the identity (as r ↑ 1).
LECTURE 7: CESARO AND ABEL SUMMABILITY
5
Proof. Recall we observed that
∞
X
Pr (θ) :=
r|n| einθ
n=−∞
converges absolutely. We also saw that
Pr (θ) =
1 − |ω|2
1 − r2
=
|1 − ω|2
1 − 2r cos θ + r2
where ω = reiθ .
From the above, we note that Pr (θ) ≥ 0, and that
Z π
Z π X
∞
1
1
r|n| einθ dθ
Pr (θ) dθ =
2π −π
2π −π n=−∞
Z π
∞
X
1
=
r|n| einθ dθ = 1,
2π −π
n=−∞
Pr is nonnegative.
so the first two conditions for approximations of the identity
are satisfied.
Now, notice that
1 − 2r cos θ + r2 = (1 − r)2 + 2r(1 − cos θ).
In the region outside of |θ| < δ,
the denominator is
bounded below.
1
2
≤ r ≤ 1 and δ ≤ |θ| ≤ π we see that
2
1
1 − 2r cos θ + r2 = 1 −
+ (1 − cos θ) ≥ cδ > 0;
2
For
and thus
1 − r2
Pr (θ) ≤
cδ
in δ ≤ |θ| ≤ π; thus
Z
Pr (θ) dθ = 0,
lim
r↑1
δ≤|θ|≤π
the third condition is also satisfied.
Immediate
quence
conse-
6
LECTURE 7: CESARO AND ABEL SUMMABILITY
Corollary 4.4. Let f be an integrable function on the circle. Then the Abel means of the (Fourier series of ) f converges pointwise to f at every point of continuity. If, further, f is continuous on the circle, then the convergence is
uniform.
Again we can recover the function.
LECTURE 8: DIRICHLET PROBLEM ON THE UNIT
DISC
1. The Poisson kernel and
the Dirichlet Problem on the unit disc
Recall the Dirichlet problem:
∆u = 0 inside the unit disc D
u = f on ∂D
In Chapter I, we guessed that all solutions would be expressible as linear combinations of certain special functions;
precisely, that they would be of the form
∞
X
u(r, θ) =
fˆ(m)r|m| eimθ
Recall the Dirichlet problem
Guessed form of the
solution
which we now recognize as Poisson integrals
m=−∞
=: Ar (f )(θ)
Z π
1
=
f (φ)Pr (φ − θ)dθ
2π −π
Now we shall see that this is true.
Theorem 1.1. Let f be an integrable function defined on
the unit circle. Then the Poisson integral
u(r, θ) := (f ∗ Pr )(θ)
solves the heat equation on the disk. That is, u ∈ C 2 (D)
and satisfies:
i. ∆u = 0.
ii. If f is continuous at θ, then
lim u(r, θ) = f (θ)
r↑1
and if f is continuous everywhere then the convergence
is uniform.
The Poisson integral
does solve the heat
equation on the disc.
2
In fact, it is the
unique solution.
LECTURE 8: DIRICHLET PROBLEM ON THE UNIT DISC
iii. If f is continuous, then u(r, θ) is the unique solution to
the steady-state heat equation on the disc which satisfies
conditions (i) and (ii).
Proof. i. First off, notice that since f is integrable, again,
the Fourier series expansion
∞
X
u(r, θ) =
fˆ(m)r|m| eimθ
m=−∞
infinitely
tiable
differen-
Term-by-term
differentiation shows
Poisson integral is
harmonic.
of the Poisson integral converges absolutely and uniformly
on any disc 0 ≤ r < ρ, where ρ < 1. Because of (absolute
and) uniform convergence (of the derivatives of the partial
sums see Rudin, Thm 7.17), we know that u is differentiable
term-by-term, infinitely many times.
Using the polar form of the Laplacian,
∂ 2 u 1 ∂u
1 ∂ 2u
∆u = 2 +
+
∂r
r ∂r r2 ∂θ2
we get, by differentiating term-by-term,
∞ h
X
∆u(r, θ) =
|m|(|m| − 1)fˆ(m)r|m|−2 eimθ +
m=−∞
|m|fˆ(m)r|m|−2 eimθ + −m2 fˆ(m)r|m|−2 eimθ
i
= 0,
We do recover the
boundary function.
Uniqueness
so the Poisson integral is indeed harmonic.
(ii) is a restatement of the previous result.
To show (iii), suppose there is another such solution,
v(r, θ). It being twice-continuously differentiable, for each
fixed r ∈ (0, 1) we know v(r, ·) has a (uniformly convergent)
Fourier series
∞
X
an (r)einθ ,
n=−∞
where an (r) =
1
2π
Rπ
−π
v(r, θ)e−inθ dθ.
LECTURE 8: DIRICHLET PROBLEM ON THE UNIT DISC
3
Now, that series must satisfy the heat equation, ∆v = 0.
Plugging the above into the polar Laplacian, and assuming
that we can differentiate term-wise, we get
∞
X
[a00n (r)
n=−∞
Fake reasoning
1 0
n2
+ an (r) − 2 an (r)]einθ = 0
r
r
and thus
a00n (r)
1 0
n2
+ an (r) − 2 an (r) = 0
r
r
for all n ∈ Z.
The equation above implies (Exercise 11, Chapter 1) that
The coefficient must
be of a certain form
an (r) = An rn + Bn r−n
for some An and Bn . Now, for n ≥ 1, Bn = 0 since
Z π
1
an (r) :=
v(r, θ)e−inθ dθ
2π −π
is bounded; to find An , take
1
An = lim an (r) := lim
r↑1
r↑1 2π
Z
π
v(r, θ)e−inθ dθ.
−π
Since v(r, θ) converges uniformly to f , we get An = fˆ(n).
Thus for n ≥ 1,
an (r) = fˆ(n)rn ;
similarly for n < 0 (and for n = 0). All together, we have
v(r, θ) =
∞
X
fˆ(n)r|n| einθ
n=−∞
and so the solution is indeed unique.
The coefficient is actually the Fourier coefficient!
4
LECTURE 8: DIRICHLET PROBLEM ON THE UNIT DISC
2. Chapter III: L2 convergence of Fourier
Series
Definition 2.1. We define the L2 norm of a function f on
the circle by
Z π
1
|f (θ)|2 dθ.
||f ||2 :=
2π −π
L2 (T)
Review of basic linear algebra
Definitions: vector
space, inner product, positive definite, norm, Hermitian inner product....
Some review:
i. Vector space V over R, over C.
ii. Examples: Rd , Cd
iii. Inner product (X, Y ) on a vector space V over R. Symmetric (X, Y ) = (Y, X), (strictly) positive definite (X, X) ≥
0, linear in both variables.
iv. Given an inner product (·, ·), we can define a norm on
V by
||X|| := (X, X)1/2 .
E.g., the dot product on Rd gives rise to the standard
Euclidean norm.
v. Inner product (X, Y ) on a vector space V over C. Hermitian: (X, Y ) = (Y, X); linear in the first variable,
conjugate linear in the second; etc.
orthogonal vectors
Definition 2.2. Let V be a vector space over R or C with
inner product (·, ·) and associated norm ||·||. If (X, Y ) = 0,
we say that X and Y are orthogonal and write X ⊥ Y .
It turns out (see any good linear algebra book, e.g., Leon,
S. Linear Algebra) that any vector space with inner product then has the following properties (read the proofs yourselves).
Properties of (finitedimensional?) inner
product spaces
Theorem 2.3. Properties of inner product spaces
i. The Pythagorean Theorem: If X ⊥ Y , then ||X +
Y ||2 = ||X||2 + ||Y ||2 .
LECTURE 8: DIRICHLET PROBLEM ON THE UNIT DISC
5
ii. The Cauchy-Schwarz Inequality: Given any X, Y ∈ V ,
|(X, Y )| ≤ ||X|| ||Y ||
iii. The Triangle Inequality: Given any X, Y ∈ V , ||X +
Y || ≤ ||X|| + ||Y ||
Proof. The more complicated one is the Cauchy-Schwarz
inequality. First, if ||Y || = 0, then
0 ≤ ||X + tY ||2 for all t ∈ R
= (X + tY, X + tY ) = ||X||2 + 2t<(X, Y ) + 0.
If <(X, Y ) > 0, then taking t << 0 gives a contradiction;
similarly for <(X, Y ) < 0; thus <(X, Y ) = 0. Similarly for
=(X, Y ).
)
Now, if ||Y || =
6 0, let c = (X,Y
(Y,Y ) ; then (X − cY ) ⊥ Y . By
the Pythagorean theorem,
||X||2 = ||X − cY + cY ||2
= ||X − cY ||2 + ||cY ||2 ≥ |c|2 ||Y ||2 ,
i2
h
(X,Y )
2
i.e., ||X|| ≥ (Y,Y ) ||Y ||2 : the C-S inequality (squared).
Showing
that
(X, Y ) = 0 for all
X; note the inner
product is positive,
but not necessarily
strictly.
LECTURE 9: THE FOURIER SERIES
RECOVER THE FUNCTION IN THE L2 SENSE
1. Two important infinite-dimensional
vector spaces
1.1. `2 (Z).
Definition 1.1. Let {an }n∈Z denote a (two-sided) sequence of complex
numbers. We define the “little `2 norm” of {an }n∈Z by
X
|an |2
||{an }n∈Z ||`2 =
n∈Z
`2 (Z)
and let
is finite.
denote the vector space of all sequences whose little `2 norm
Definition 1.2. For A = {an } and B = {bn } in `2 (Z), we define the inner
product
X
(A, B) :=
an bn ;
By the two-sided
sum
we
mean
the limit of the
symmetric
partial
sums.
Inner
`2 (Z)
product
for
n∈Z
and let ||A|| := (A, A)1/2 denote the related norm.
Question
Question. Is
`2 (Z)
even a vector space?
Proof. We need to show that if A, B ∈ `2 (Z), then so is A + B. That is, we
need to show that
X
lim
|an + bn |2 < ∞.
N →∞
|n|≤N
However, this is obvious (why?): well, by the finite-dimensional triangle
inequality,

1/2 
1/2 
1/2
X
X
X

|an + bn |2  ≤ 
|an |2  + 
|bn |2 
|n|≤N
|n|≤N
|n|≤N
≤ ||A|| + ||B||.
Taking the limit as N → ∞ shows that
||A + B|| ≤ ||A|| + ||B|| < ∞
so we’ve demonstrated that A + B ∈ `2 (Z) (and simultaneously that the
infinite-dimensional triangle inequality holds).
Just use the finite
dimensional C-S and
take the limit.
Example 2: a preHilbert space
L2 RECOVERY OF THE FUNCTION
2
Definition 1.3. An inner-product space with strictly positive-definite inner
product, which is complete with respect to the induced metric, is called a
Hilbert space.
R is not a Hilbert
space.
Let R denote the set of complex-valued Riemann integrable functions on
[0, 2π], with addition, scalar multiplication, inner product and norm defined
as usual. It turns out that R fails to be a Hilbert space in two senses:
first, the inner product is not strictly positive definite (||f || = 0 only implies
f = 0 a.e.), and second, the space is not complete.
1.2. Checking Cauchy-Schwarz in R.
Cute
proof
of
Cauchy-Schwarz for
R
Proof. Using the fact that 2AB ≤ (A2 + B 2 ), we see that for any λ > 0,
1
|f (x)g(x)| ≤ [λ|f (x)|2 + λ−1 |g(x)|2 ]
2
Then
Z 2π
1
|(f, g)| ≤
|f (x)g(x)| dx
2π 0
1
≤ [λ||f ||2 + λ−1 ||g||2 ];
2
taking λ =
||g||
||f ||
yields the Cauchy-Schwartz inequality.
2. Fourier series and orthogonality
Remark. We’ve already basically mentioned this before, but we can express
the notion of Fourier series more simply using the language of inner products
and orthogonality.
Fourier series in
the language of
infinite-dimensional
inner product spaces
i. Let R denote integrable functions on the circle.
ii. We define an inner product
Z 2π
1
(f, g) :=
f (θ)g(θ) dθ.
2π 0
with induced norm ||f ||22 = (f, f ) and (not-quite) metric d(f, g) : ||f −
g||2 .
iii. Let en (θ) := einθ . Easy to see that {en }n∈Z is an orthonormal set.
iv. Remark that the Fourier coefficients are precisely the coefficients of f
in terms of {en }n∈Z :
fˆ(n) = (f, en ).
v. For example, we can express the partial sum of the Fourier series as
X
SN (f ) =
(f, en )en .
Consequences
of
having an orthonormal set.
|n|≤N
A few observations from linear algebra:
P
i. f − |n|≤N (f, en )en is orthogonal to en for all |n| ≤ N .
Hilbert space
L2 RECOVERY OF THE FUNCTION
3
ii. Using the above, the Pythagorean theorem implies
||f ||2 = ||f − SN (f )||2 + ||SN (f )||2
X
|an |2 .
= ||f − SN (f )||2 +
|n|≤N
P
iii. Given any other approximation |n|≤N cn en , we have
X
X
(an − cn )en ||2 ,
cn en ||2 = ||f − SN (f )||2 + ||
||f −
The partial sum of
the Fourier series is
the “best approximation”
|n|≤N
|n|≤N
and thus SN (f ) is the best approximation of f in the L2 sense.
3. Main theorem: Recovery in the L2 sense
Theorem 3.1. Let f be an integrable function on the circle. Then
Z 2π
1
|f (θ) − SN (f )(θ)|2 dθ → 0 as N → ∞,
2π 0
i.e., the Fourier series converges to f “in the L2 sense.”
Proof. Suppose g is continuous. By the Weierstrass approximation theorem,
there exists a sequence of trigonometric polynomials pn which converge to g
uniformly on the circle, i.e., such that given any > 0, there exists N such
that n > N implies
|g(x) − pn (x)| < for all x ∈ [0, 2π].
Thus
1/2
Z 2π
1
2
||g − pn ||2 :=
|g(x) − pn (x)| dx
< .
2π 0
By the “best approximation lemma”, we see that then for some sufficiently
large N (not the same N as above), ||g − SN || ≤ as well.
Suppose now that f is an integrable function. By the L1 approximation
lemma, we can find a continuous function g such that g has the same bound
(B, say) as f and
Z 2π
1
2
|f (x) − g(x)| dx <
,
2π 0
2B
Proof
is
trivial
for
continuous
functions,
using
Weierstrass approximation.
For integrable functions, use the L1 approximation lemma.
The L2 difference is
Z 2π
1/2 1/2
Z
1
2B 2π
2
|f (x) − g(x)| dx
≤
|f (x) − g(x)| dx
2π 0
2π 0
≤ .
As before, we can find a trigonometric polynomial p approximating g s.t.
||g − p||2 < ; thus ||f − p||2 < 2 and, by the best approximation lemma, a
partial sum SN (f ) of the Fourier series of f , for N sufficiently large, must
approximate f with at most that error.
(Main work is in the
Weierstrass approximation theorem)
LECTURE 10:
L2 RECOVERY OF INTEGRABLE FUNCTIONS
AND CONSEQUENCES
1. Main theorem: Recovery in the L2 sense
Theorem 1.1. Let f be an integrable function on the circle. Then
Z 2π
1
|f (θ) − SN (f )(θ)|2 dθ → 0 as N → ∞,
2π 0
(Have them do the
second part as exercise.)
i.e., the Fourier series converges to f “in the L2 sense.”
Proof. Suppose g is continuous. By the Weierstrass approximation theorem, there exists
a sequence of trigonometric polynomials pn which converge to g uniformly on the circle,
i.e., such that given any > 0, there exists N such that n > N implies
|g(x) − pn (x)| < for all x ∈ [0, 2π].
Thus
||g − pn ||2 :=
1
2π
2π
Z
1/2
|g(x) − pn (x)|2 dx
< .
Proof
is
trivial
for
continuous
functions,
using
Weierstrass approximation.
0
By the “best approximation lemma”, we see that then for some sufficiently large N (not
the same N as above), ||g − SN || ≤ as well.
Suppose now that f is an integrable function. By the L1 approximation lemma, we
can find a continuous function g such that g has the same bound (B, say) as f and
Z 2π
2
1
,
|f (x) − g(x)| dx <
2π 0
2B
For integrable functions, use the L1 approximation lemma.
The L2 difference is
1/2 1/2
Z 2π
Z
1
2B 2π
|f (x) − g(x)|2 dx
≤
|f (x) − g(x)| dx
2π 0
2π 0
≤ .
As before, we can find a trigonometric polynomial p approximating g s.t. ||g − p||2 < ;
thus ||f − p||2 < 2 and, by the best approximation lemma, a partial sum SN (f ) of the
Fourier series of f , for N sufficiently large, must approximate f with at most that error.
Corollary 1.2 (Parseval’s Identity). Let f be an integrable
P
2
function, and an = fˆ(n). Then limN →∞ N
n=−N |an | converges to ||f ||2 .
Proof. Recall that, by the Pythagorean theorem,
X
2
2
||f || = ||f − SN (f )|| +
|an |2 .
|n|≤N
(Main work is in the
Weierstrass approximation theorem)
I.e., the map is an
isometry.
2
L2 RECOVERY OF THE FUNCTION
(R is incomplete.)
Comment
about
physical meaning sounds
Useful variant
Parseval:
Remark. In particular, notice that ||an ||`2 = ||f ||L2 ; and one
has a correspondence between `2 and R. However, there
exist sequences in `2 that do not arise as Fourier series of
functions in R
Corollary 1.3 (Riemann-Lebesgue Lemma). f integrable
on the circle. Then fˆ(n) → 0 as |n| → ∞.
of
Polarization identity
Lemma 1.4 (Polarized Parseval’s identity). Let f and g be
integrable on the circle, with Fourier coefficients {an } and
{bn }, respectively. Then
Z 2π
∞
X
1
f (θ)g(θ) dθ =
an bn .
2π 0
n=−∞
Proof. In any Hermitian inner product space, one has the
polarization identity
1
(f, g) = [||f + g||2 − ||f − g||2 + i(||f + ig||2 − ||f − ig||2 ].
4
Using this in R and Parseval’s identity, we get
i
X
X
X
1 hX
2
2
2
2
(f, g)2 =
|an + bn | −
|an − bn | + i(
|an + ibn | −
|an − ibn |
4
1
= [||an + bn ||2 − ||an − bn ||2 + i(||an + ibn ||2 − ||an − ibn ||2 ]
4
= (an , bn )`2 ,
by the polarization identity again...in `2 !
2. Return to pointwise convergence
Remark. Convergence in L2 does not guarantee that the
Fourier series converges for any θ. HUH? (blimps)
What about pointwise convergence?
We do get some
small results.
Theorem 2.1. Let f be an integrable function on the circle.
Suppose f is differentiable at θ0 . Then limN →∞ SN (f )(θ0 ) =
f (θ0 ).
L2 RECOVERY OF THE FUNCTION
3
Proof. Consider the “slope of the secant line” function
f (θ −t)−f (θ )
0
0
if t 6= 0 and |t| < π
t
F (t) :=
0
−f (θ0 )
if t = 0
F , being differentiable at 0, is bounded near 0. It is, further, integrable on δ < |t| for all δ > 0; thus F is integrable
on [−π, π] (see appendix).
Then
Z π
1
f (θ0 − t)DN (t) dt − f (θ0 )
SN (f )(θ0 ) − f (θ0 ) =
2π −π
Z π
1
=
[f (θ0 − t) − f (θ0 )]DN (t) dt
2π −π
Z π
1
=
F (t)tDN (t) dt.
2π −π
Now,
t
sin((N + 1/2)t)
sin(t/2)
t
=
[sin(N t) cos(t/2) + cos(N t) sin(t/2)]
sin(t/2)
tDN (t) =
so we get
1
2π
1
=
2π
Z
π
F (t)
−π
Z
π
F (t)tDN (t) dt
−π
t
[sin(N t) cos(t/2) + cos(N t) sin(t/2)] dt
sin(t/2)
which goes to 0 by the Riemann-Lebesgue lemma.
Remark. In fact, the conclusion is true even for f only Lipschitz at θ0 .
Alternate proof. (Chernoff, P. The American Mathematical
Monthly, vol. 87, No. 5, May 1980, pp. 399-400.)
Even cuter proof
4
(limit
exists
by
L’Hôpital’s rule)
L2 RECOVERY OF THE FUNCTION
WLOG assume that x0 = 0 and f (x0 ) = 0. Since f (0) =
0 and f 0 (0) exists, the function g(x) = f (x)/[eix − 1] is
bounded near 0 , and thus is integrable since f is.
Then
fˆ(n) = ĝ(n − 1) − ĝ(n),
a telescoping series, and
N
X
fˆ(n) = ĝ(−N − 1) − ĝ(N );
SN f (0) =
−N
which tends to 0 by the Riemann-Lebesgue lemma.
Convergence
of
SN (f )(θ0 depends
only on the behavior
of f
near θ0 .
(Shocking!?)
Corollary 2.2 (Localization principle of Riemann). Let f
and g be integrable on the circle. Suppose f ≡ g in some
neighborhood of a point θ0 . Then
lim SN (f )(θ0 ) − SN (g)(θ0 ) = 0.
N →∞
Remark. Note that neither f nor g need to be differentiable
at θ0 , and that this does not imply that the Fourier series
of either converges at θ0 , only that their convergence or divergence is connected (and, if they converge, they converge
to the same limit).
LECTURE 11: FOURIER SERIES NEED NOT
CONVERGE AT POINTS OF CONTINUITY
1. Continuous function with
Fourier series diverging at a point
1.1. A function which is not a Fourier series of a
function in R.
Consider the series
−1
X
einθ
.
n
n=−∞
Suppose the above is the Fourier series of some Riemann
integrable function, f . In that case, if we consider the Abel
means at 0, we get
∞
X
rn
|Ar (f )(0)| =
n
n=1
which diverges as r → 1. However, we should also have
Z π
1
|Ar (f )(0)| ≤
|f (θ)| Pr (θ) dθ ≤ sup |f (θ)|
2π −π
θ
a contradiction, so the above is not the Fourier series of a
Riemann integrable function.
1.2. A continuous function whose Fourier series diverges at a point.
Let
X einθ
X einθ
˜
and fN (θ) =
fN (θ) =
n
n
1≤|n|≤N
−N ≤n≤−1
Claim:
(i) |f˜N (0)| ≥ c log N
(ii) fN (θ) is uniformly bounded in N and θ
L2 RECOVERY OF THE FUNCTION
2
Recall Tauber’s
P theorem: if
cn is
Abel summable
to
P
s, then
cn actually converges to s if
cn = o(1/n).
(i) is evident. To prove (ii) will require a little machinery:
P
Lemma 1.1. Let ∞
n=1 cn be an infinite series. If
P
n
(i) the Abel means Ar = ∞
n=1 r cn are bounded as r ↑ 1,
and
(ii) cn = O(1/n)
P
then the partial sum sequence SN = N
n=1 is bounded.
Proof. We will control the difference between SN and Ar .
For
N
∞
X
X
n
SN − Ar =
(cn − r cn ) −
r n cn
n=1
n=N +1
so
|SN − Ar | ≤
N
X
n
|cn ||1 − r | +
n=1
Coarse bound
(1 − rn )
on
and the O(1/n) control
∞
X
|rn ||cn |
n=N +1
Now, let us make three observations:
(i) (1 − rn ) = (1 − r)(1 + r + · · · + rn−1 ) ≤ n(1 − r).
(ii) Since cn = O(1/n), we have n|cn | ≤ M for some M .
(iii) We also have that for n ≥ N + 1, |cn | ≤ M
N.
Using those two facts to bound the two sums, we continue:
N
∞
X
M X n
≤M
(1 − r) +
r
N
n=1
n=N +1
M 1
.
N 1−r
If we take r = 1 − 1/N , then we get
≤ M N (1 − r) +
Clever choice of r
|SN − Ar | ≤ 2M.
Since the Ar are bounded for large enough N , we see that
SN are also bounded.
Recall that we wanted to prove claim (ii), i.e.:
L2 RECOVERY OF THE FUNCTION
follows immediately
from the lemma
Claim: fN (θ) =
and θ.
P
1≤|n|≤N
einθ
n
3
is uniformly bounded in N
Proof. (of claim (ii)) fN (θ) is the partial sum of the Fourier
P
inθ
series n6=0 e n , the Fourier series of the sawtooth function
f . Since
(i) The Abel means are expressible as f ∗ Pr (θ), and since
f is bounded (and ||Pr ||1 ≤ M ) the Abel means are
bounded.
(ii) cn = einθ /n + e−inθ /n, which is certainly O(1/n).
we see SN (f )(θ) is uniformly bounded in N and θ.
Use the convolution
form of the Abel
means
1.3. Creating the example: the Heart of the Matter.
Recall
X einθ
and f˜N (θ) =
fN (θ) =
n
1≤|n|≤N
X
−N ≤n≤−1
einθ
,
n
Notice fN has no
n = 0 term.
trigonometric polynomials of degree N . We define frequencyshifted versions of fN and f˜N ,
frequency-shifted
versions of fN , f˜N
PN (θ) = ei2N θ fN (θ) and P̃N (θ) = ei2N θ f˜N (θ),
which are trigonometric polynomials of degree 3N and 2N −
1, respectively.
Then if we consider the partial sums of PN , we see
Lemma 1.2.

 PN
SM (PN ) =
P̃
 N
0
if M ≥ 3N
if M = 2N
if M < N
(obvious)
Now choose any convergent positive series
quence of integers {Nk } such that
(i) Nk+1 > 3Nk
(ii) limk→∞ αk log Nk = ∞
P
αk and se-
4
L2 RECOVERY OF THE FUNCTION
and let
f (θ) =
∞
X
Creating the
function!
αk PNk (θ).
k=1
Since |PN (θ) = |fN (θ)|, which were uniformly bounded, the
above series converges uniformly to a continuous periodic
function.
On the other hand,
Key claim:
It’s continuous.
However....
|S2Nm (f )(0)| ≥ cαm log Nm + O(1)
Proof. (of this last claim)
X
S2Nm (f )(0) = S2Nm (
αk PNk )(0)
Use the lemma
k
=
X
αk S2Nm PNk (0)
= αm S2Nm PNm (0) +
X
αk PNk (0) +
m<k
= αm P̃Nm (0) + 0 +
X
X
αk PNk (0)
m>k
αk PNk (0)
m>k
Here’s how
two claims
into play.
those
come
by the lemma.
Now, P̃Nm (0) = f˜Nm (0) which, recall, is of modulus ≈
c log Nm , and
X
X
|
αk PNk (0)| ≤ B|
αk | ≤ BA
m>k
P
(Recall the
αk =
A was a convergent
positive series.)
k
since the PNk (0) = fNk (0), which were uniformly bounded
(by B, say) in N .
Thus the partial sums of the Fourier series diverge at 0,
despite the function being continuous everywhere.
Remark. It is possible to see, using the Baire category theorem and Banach-Steinhaus theorem, that the set of continuous functions whose Fourier series diverge at a point
is actually dense in the set of continuous functions (e.g.,
L2 RECOVERY OF THE FUNCTION
5
http://www.math.uchicago.edu/ may/VIGRE/VIGRE2010/
REUPapers/Stratmann.pdf for an undergraduate’s REU
summary of the argument).
LECTURE 12: FOURIER SERIES AND THE
ISOPERIMETRIC INEQUALITY
1. Basic knowledge about curves.
Definition 1.1. a C 1 mapping
γ : [a, b] → R2 ,
such that γ 0 (s) 6= 0, is called a parametrized curve. We call
the image of γ a curve (denoted by Γ, say). If γ is 1 − 1,
we call Γ simple; if γ(a) = γ(b) we call Γ closed.
parametrized curve
Remarks.
(1) Can extend γ to a b − a periodic function on R.
(2) The smoothness conditions ensure the existence of a
continuous tangent vector.
Definition 1.2. We define the length of the curve Γ, parametrized
by γ(s) = (x(s), y(s)) by
length of a curve
Z b
Z b
`=
|γ 0 (s)| ds =
(x0 (s)2 + y 0 (s)2 )1/2 ds
a
a
Definition 1.3. Let s : [c, d] → [a, b] be a bijective, C 1
mapping. Then we call
η(t) := γ ◦ s(t)
a (re-)parametrization of Γ. [Note length is independent of
parametrization.] If, further,
|η 0 (t)| = 1 for all t
we call η the arclength parametrization of Γ.
Definition 1.4. We define, as the area of the region enclosed by the simple closed curve Γ, the value
reparametrization of
a curve
arclength
parametrization
area enclosed by a
curve
2
ISOPERIMETRIC INEQUALITY, WEYL’S EQUIDISTRIBUTION THEOREM
Z
1
A=
(x dy − y dx)
2 Γ
Z b
1
=
x(s)y 0 (s) − y(s)x0 (s) ds
2 a
2. The Isoperimetric inequality and
Parseval’s identity
Theorem 2.1. Let Γ ⊂ R2 be a simple closed curve. Let
` denote the length of Γ, A the area of its enclosed region.
Then
`2
A≤
,
2π
with equality if and only if Γ is a circle.
Statement
inequality
of
The parametrization
is the arclength
parametrization.
Translation of this
fact in terms of
Fourier series (via
Parseval).
Proof. Part I. WLOG (by scaling the plane by (x, y) →
(δx, δy) one may assume that ` = 2π.
Take the arclength parametrization γ(s) = (x(s), y(s)) =
of Γ; then
Z 2π
1
[x0 (s)2 + y 0 (s)2 ] ds = 1
2π 0
i.e., ||x0 ||22 + ||y 0 ||22 = 1.
x(s) and y(s) are 2π periodic functions and have Fourier
coefficients {an } and {bn }, respectively; their derivatives
have coefficients {inan } and {inbn }, respectively. (Note
that an = a−n and bn = b−n since x, y are real-valued.)
Using Parseval’s identity on the above, we get
(1)
||{inan }||2`2 + ||{inbn }||2`2 = 1,
∞
X
i.e.,
(2)
|n|2 [|an |2 + |bn |2 ] = 1.
n=−∞
The area is what.
Now, the area is by “definition”
Z b
1
1
A = 2π
x(s)y 0 (s) − y(s)x0 (s) ds
2
2π a
ISOPERIMETRIC INEQUALITY, WEYL’S EQUIDISTRIBUTION THEOREM
3
Using Parseval again, we get
(3)
A=π
X
Translation of above
via Parseval.
an (−inbn ) − bn (−inan )
n∈Z
(4)
≤π
X
|n||an bn − an bn |
n∈Z
Now, let’s bound this using the previous observation. Since
(5)
trivial quadratic inequality
|an bn − an bn | ≤ 2|an | |bn | ≤ |an |2 + |bn |2
we see that (observe |n| ≤ |n|2 )
X
(6)
A≤π
|n|(|an |2 + |bn |2 )
(with equality
|an | = |bn |)
n∈Z
(7)
≤π
X
|n|2 (|an |2 + |bn |2 ) = π
n∈Z
using the statement (2) above about the arclength parametrization for the final equality. So we’ve shown the isoperimetric
inequality.
the case A = π
Part II. Why can A = π only in the case of a circle? We
make the following observations:
i. First: equality in (6, 7) could only occur if we have no
terms for |n| ≤ 2, i.e,
x(s) = a−1 e−is + a0 + a1 eis and
y(s) = b−1 e−is + b0 + b1 eis .
ii. Further, since a−1 = a1 and b−1 = b1 , (7) implies that
2(|a1 |2 + |b1 |2 ) = 1.
iii. Now, by (5), equality can hold only if
|a1 b1 − a1 b1 | = 2|a1 | |b1 | = |a1 |2 + |b1 |2 =
1
2
iff
4
ISOPERIMETRIC INEQUALITY, WEYL’S EQUIDISTRIBUTION THEOREM
and thus, since (|an | − |bn |)2 = 0 implies |an | = |bn |,
|a1 | = |b1 | = 1/2. So
1
1
a1 = eiα and b1 = eiβ
2
2
for some α, β.
iv. In this notation, 1 = 2|a1 b1 − a1 b1 | (from (iii) above) is
equivalent to
1 i(α−β)
[e
− e−i(α−β) ] = 1
2
i.e., | sin(α − β)| = 1.
so α − β = kπ/2 for some k ∈ Z.
So
1
1
x(s) = e−iα e−is + a0 + eiα eis = a0 + cos(α + s)
2
2
1
1
and y(s) = e−iβ e−is + b0 + eiβ eis = b0 + cos(β + s)
2
2
kπ
= b0 + cos(α +
+ s) = b0 ± sin(α + s)
2
- i.e., Γ is a circle, as desired.
LECTURE 13: WEYL’S EQUIDISTRIBUTION
THEOREM
1. Number theory:
Weyl’s equidistribution theorem
1.1. Basic knowledge.
Definition 1.1. Let x be a real number. Then
(1) Let [x], the integer part of x, denote the greatest integer less than or equal to x.
(2) Let hxi := x − [x] denote the fractional part of x.
(3) Given x, y ∈ R, if x − y ∈ Z we say
x ≡ y mod Z or x ≡ y mod 1.
(Another
easilyattained result)
basic definitions
Of course x ≡ y mod Z iff hxi = hyi.
1.2. Main theorem.
The problem: consider the collection {hnγi : n ∈ N}: is
it dense in [0, 1)? Kronecker’s theorem: yes, if γ ∈
/ Q.
Definition 1.2 (Definition of equidistributed sequence).
If, for every interval (a, b) ⊂ [0, 1), then we call {ξn } an
equidistributed sequence.
#{n ∈ {1, 2, . . . , N } : ξn ∈ (a, b)}
lim
=b−a
N →∞
N
Theorem 1.3 (Weyl’s Equidistribution theorem). If γ ∈
/Q
then {hnγi} is equidistributed in [0, 1).
The question:
{hnγi} dense?
is
equidistributed
sequence
Probability that an
element of the (first
N points of the) sequence lies in the interval
main theorem
Corollary 1.4 (Kronecker’s theorem).
1.3. Translation from number theory to analysis.
We rephrase the main theorem as follows. Let χ(a,b) denote the characteristic function of (a, b) ⊂ [0, 1), extended
as a 1-periodic function. Then we observe that
Rephrasal of problem as an analysis
question
2
WEYL’S EQUIDISTRIBUTION THEOREM
#{1 ≤ n ≤ N : hnγi ∈ (a, b)} =
N
X
χ(a,b) (nγ).
n=1
So the theorem can be rephrased as follows:
Theorem 1.5. Given any γ ∈
/ Q, and any (a, b) ⊂ [0, 1),
Z 1
N
1 X
lim
χ(a,b) (nγ) =
χ(a,b) (x) dx
N →∞ N
0
n=1
We show the cubature formula for continuous functions.
1.4. Main lemma.
Lemma 1.6 (Main lemma). f continuous and periodic
on [0, 1). If γ ∈
/ Q, then
Z 1
N
1 X
f (nγ) →
f (x) dx.
N n=1
0
Technique:
statement trivial for
trigonometric
polynomials, which
are L∞ dense in the
continuous periodic
functions.
(geometric sum)
(by the uniform
convergence
of
the Cesaro means
for
continuous
functions, i.e,. the
goodness of the
Fejer kernel - here’s
where the Fourier
analysis enters)
Proof. (trivial)
i. Case: f ∈ {1, e2πix , . . . , e2πikx , . . . }.
R1
f = 1 is obvious. Otherwise, the integral 0 f = 0, so
we need to see that the average tends to 0. But, since
e2πikγ 6= 1,
N
N
1 X
1 X 2πiknγ
f (nγ) :=
e
N n=1
N n=1
e2πikγ 1 − e2πikN γ
=
N 1 − e2πikγ
which goes to 0 as N → ∞.
ii. Case: trigonometric polynomials.
The problem is linear, so the lemma holds for linear
combinations of exponentials.
iii. Case: f a continuous periodic function. We know that
there exists a trigonometric polynomial P such that
||f (x) − P (x)||∞ < /3 and, by step i, that the lemma
WEYL’S EQUIDISTRIBUTION THEOREM
3
holds for P . Then
Z 1
N
N
1 X
1 X
f (nγ) −
f (x) dx ≤
|f (nγ) − P (nγ)|+
N n=1
N
0
n=1
Z 1
Z
N
1
1 X
|P (x) − f (x)| dx.
P (nγ) −
P (x) dx +
N n=1
0
0
The first and last are average differences between f
and P (and thus are controlled by 3 ); the middle term
is smaller than 3 for sufficiently large N by part i.
1.5. Proof of main theorem.
Proof of Weyl’s equidistribution theorem.
For each > 0, let f+ and f− be continuous, 1-periodic
functions which
(i) approximate χ(a,b) from above and below
(ii) are bounded by 1
(iii) agree with χ(a,b) except on intervals of total length 2.
Then the averages of f− , χ(a,b) , and f+ have the following
relation:
N
N
N
1 X −
1 X +
1 X
f (nγ) ≤
χ(a,b) (nγ) ≤
f (nγ)
N n=1 N n=1
N n=1 Proof of Weyl’s
equidistribution
theorem
(draw picture)
Now, the lemma implies (taking the limit as N → ∞)
lim inf and lim sup!
Z 1
Z 1
N
N
X
X
f− (x) dx ≤ lim inf
χ(a,b) (nγ) ≤ lim sup
χ(a,b) (nγ) ≤
f+ (x) dx
N →∞
0
N →∞
n=1
n=1
Since
Z
b − a − 2 ≤
0
1
f− (x) dx
Z
and
1
f+ (x) dx ≤ b − a + 2
0
and the above is true for all , we see that the desired limit
exists, and equals the desired b − a.
0
4
WEYL’S EQUIDISTRIBUTION THEOREM
1.6. Observations.
Corollary 1.7. In fact, the main lemma holds even if f is
merely Riemann integrable.
Proof. Approximate f such a step function s such that ||f −
s||∞ < /3 (see the proof of the L1 approximation lemma).
Then
Z 1
N
N
N
1 X
1 X
1 X
f (nγ) −
f (x) dx ≤
f (nγ) −
s(nγ)
N n=1
N n=1
N n=1
0
Z 1
Z 1
Z 1
N
1 X
+
s(nγ) −
s(x) dx +
s(x) dx −
f (x) dx
N n=1
0
0
0
Since s is a finite linear combination of characteristic functions of intervals, the middle term can be made smaller
than /3 by taking N sufficiently large; the other terms are
both also smaller than /3.
Remark. Connection with dynamical systems: the system
is “ergodic”; that is, for all irrational γ, denoting
ρ(θ) = θ + 2πγ
mod 2π,
the “time average”
N
1 X
f (ρn (θ))
lim
N →∞ N
n=1
exists for each θ, and equals the “space average”
Z 2π
1
f (θ) dθ
2π 0
Remark. Notice that along the way, we proved the forward
direction of the following statement:
Theorem 1.8. Weyl’s criterion: A sequence of real numbers {ξi } in (0, 1) is equidistributed if and only if for all
WEYL’S EQUIDISTRIBUTION THEOREM
k ∈ Z,
1 X 2πikξn
lim
e
→ 0.
N →∞ N
5
LECTURE 14: A CONTINUOUS, NOWHERE
DIFFERENTIABLE FUNCTION
1. Cool examples
i. Weierstrass’s example: Let a ∈ N, a > 1; b ∈ (0, 1).
Then if ab > 1 + 3π
2 ,
W (x) :=
∞
X
bn cos(an x)
n=1
is a nowhere differentiable function (!).
ii. Riemann’s near-example:
R(x) :=
∞
X
sin(n2 x)
n=1
n2
is differentiable att the points
integers.
πp
q
where p, q are odd
P
−nα i2n x
Theorem 1.1. fα (x) = ∞
e , for α ∈ (0, 1) is a
n=0 2
continuous, nowhere differentiable function.
“at and only at”
Using a lacunary
Fourier series to
create a cts nowhere
differentiable
function
2. The main idea: delayed means
Recall that we have a couple of ways of summing the
Fourier series: first, the standard way, taking SN (g) = g ∗
DN with DN the Dirichlet kernel; second, the Cesaro way,
taking σN (g) = g ∗ FN , with FN the Fejer kernel.
If we look at the partial sum SN on the Fourier coefficient
side, we see that
\
d
S
N (g)(n) = ĝ(n)DN (n)
How things look
on the Fourier coefficient side
A CONTINUOUS, NOWHERE DIFFERENTIABLE FUNCTION
2
where
d
D
N (n) =
Partial sums: truncation
1 if |n| ≤ N
0 if |n| > N
(Draw this!) I.e., we “chop off” the Fourier series for the
terms with indices |n| > N .
Doing the same thing for the Cesaro sums, we see that
on the Fourier coefficient side one weights them as follows:
Cesaro
sums:
weighted truncation
\
σ\
N (g)(n) := g ∗ FN (n)
1 c
c
\
= ĝ(n) [D
0 (n) + D1 (n) + · · · + DN −1 (n)]
N 1
ĝ(n) N [N − |n|] for |n| ≤ N
=
0
for |n| > N.
Equivalently,
S0 (g)(x) + S1 (g)(x) + · · · + SN −1 (g)(x)
N
N
−1
1 XX
ak eikx
=
N
`=0 |kl≤`
1 X
=
(N − |n|)an einx
N
|n|≤N
X |n|
an einx
=
1−
N
σN (g)(x) :=
|n|≤N
Definition 2.1. We define the delayed means ∆N (g) of the
Fourier series of g by
∆N (g) := 2σ2N (g) − σN (g)
= 2g ∗ F2N − g ∗ FN = g ∗ (2F2N − FN ).
Delayed means: Big
triangle - little triangle = trapezoid
On the Fourier coefficient side, it is easy to see that
A CONTINUOUS, NOWHERE DIFFERENTIABLE FUNCTION

 ĝ(n)
\
∆
ĝ(n)2(1 −
N (g)(n) =

0
|n|
2N )
3
if |n| ≤ N
if N ≤ |n| ≤ 2N
if |n| > 2N
3. Getting the contradiction
Let’s recall our function:
∞
X
n
fα (x) =
2−nα ei2 x , with α ∈ (0, 1)
n=0
Observation: For fixed N , if we choose the largest k for
which 2k ≤ N , then
∆2k (fα ) = SN (fα )
Even though 2k ≤ N , there are no frequencies between 2k
and N anyway, because of the lacunary nature of the series.
the frequencies are of
the form 2n , n ∈ N
We don’t miss anything, because of the
lacuna.
Dumb observation: If 2N = 2n , then
n
∆2N (f ) − ∆N (f ) = 2−nα ei2 x
Question. Why not do this using the partial sum operator?
Our contradiction will be obtained as follows. By the
above dumb observation, for any point x0 ,
n
(1) |∆2N (f )0 (x0 ) − ∆N (f )0 (x0 )| = |i2n 2−nα ei2 x0 |
(2)
= 2n(1−α) = (2N )1−α .
However, we shall prove the following
Lemma 3.1. Let g be continuous. If g is differentiable at
x0 , then
σN (g)0 (x0 ) = O(log N ).
Since
∆N (g)0 (x0 ) = 2σ2N (g)0 (x0 ) − σN (g)0 (x0 ) = O(log N ),
if fα is differentiable at some point x0 we would have a
contradiction with (??) above.
(We can catch the
top term in an obvious way.)
4
A CONTINUOUS, NOWHERE DIFFERENTIABLE FUNCTION
Proof of Lemma. Recall: σn g(x0 ) := FN ∗ g(x0 ) (with FN
the Fejer kernel), so
Z π
σN (g)0 (x0 ) =
FN0 (x0 − t)g(t) dt
Z−ππ
=
FN0 (t)g(x0 − t) dt
−π
Take the derivative
inside.
Standard trick: use
cancellation to slide
in a constant to
take advantage of
smoothness
Rπ
by change of variables. Using the fact that −π FN0 (t) dt =
0, we see that we can slide in a constant to the above: and
thus, using the differentiability of g at x0 ,
Z π
=
FN0 (t)[g(x0 − t) − g(x0 )] dt
−π
0
Z
π
|σN (g) (x0 )| ≤ C
−π
Useful estimates
|FN0 (t)| |t| dt.
FACTS: Useful estimates on FN0
i. |FN0 (t)| ≤ AN 2
ii. |FN0 (t)| ≤ |t|A2
Break up the integral
Putting it all together, we see
Z π
0
|σN (g) (x0 )| ≤ C
|FN0 (t)| |t| dt
Z−π
Z
0
≤C
|FN (t)| |t| dt + C
|FN0 (t)| |t| dt
|t|≥ 1
|t|≤ N1
Z N
Z
A
2 1
≤C
|t|
dt
+
C
AN
dt
2
N
|t|≥ N1 |t|
|t|≤ N1
The first term is O(log N ), and the second O(1).
A CONTINUOUS, NOWHERE DIFFERENTIABLE FUNCTION
Proof of the second fact. WTS that |FN0 (t)| ≤
Well
1 sin2 (N t/2)
FN (t) =
,
N sin2 (t/2)
5
problem
A
|t|2 .
sin(N t/2) cos(N t/2)
1 cos(t/2) sin2 (N t/2)
so
=
−
.
N
sin2 (t/2)
sin3 (t/2)
1
sin(N t/2) cos(N t/2)
1 cos(t/2) sin2 (N t/2)
=
−
1
N
sin(t/2)
sin2 (t/2)
Now | sin(N t/2)| ≤ CN |t| and | sin(t/2)| ≤ c|t| for |t| ≤ π
so we get
CN |t| cos(N t/2)
1 cos(t/2)C 2 N 2 |t|2
0
so |FN (t)| ≤
−
....
c2 |t|2
N
c3 |t|3
FN0 (t)
Remark. Something seems wrong here.
LECTURE 15: FINALLY, THE FOURIER TRANSFORM!
Remark. Normally, one defines the Fourier transform on
L2 (R). However, we cannot define this space (without first
defining the Lebesgue integral). Instead, we’ll work on the
Schwartz class S(R). When you are older (Book III) you’ll
see that this space is dense in L2 (R), and that one can extend uniquely our Fourier transform to L2 (R). By the way,
restricting ourselves to the Schwartz space “is a device that
allows us to come quickly to the main conclusions, formulated in a direct and transparent fashion” (but in some
sense oversimplifies the matter).
1. Basic definitions
improper integrals
Definition 1.1. We define (if the limit exists)
Z
∞
Z
N
f (x) dx = lim
N →∞
−∞
f (x) dx
−N
Definition 1.2. Let f : R → C be continuous. if there
exists A > 0 such that
|f (x)| ≤
A
1 + x2
for all x ∈ R, then we say f is of moderate decrease, and
denote by M (R) the (vector) space of such functions.
Remark. It is easy to see that the improper integral
converges for f ∈ M (R) R
(Exercise: show that then {IN :=
N
−N
f } forms a Cauchy sequence.
R
Rf
Vector space M (R)
of Functions of Moderate Decrease
Could use 1 + instead of 2.
2
LECTURE 15: FINALLY, THE FOURIER TRANSFORM!
Well, given > 0, we see that
Z
|IM − IN | ≤
f
N ≤|x|≤M
Z
≤A
N ≤|x|≤M
1
1
1
dx = 2A
−
x2
N
M
2A
≤
,
N
which we can make smaller than as long as N is sufficiently large.)
Basic properties of
improper integration
Proposition 1.3. (Properties of the improper integral) Let
f, g ∈ M (R), and α, β ∈ C. Then
i. (Linearity)
Z ∞
Z
Z
[αf + βg] = α f + β
g.
−∞
ii. (Translation invariance)
Z ∞
Z
f (x − α) dx =
−∞
⇒ L1 norm doesn’t
change under dilation
∞
f (x) dx.
−∞
iii. (Scaling under dilations) Given any δ > 0,
Z
Z ∞
δ f (δx) dx =
f (x) dx.
R
iv. (Continuity)
Z
lim
h→0
Stein will start to
leave out details
now...but this is
nothing.
R
R
−∞
∞
|f (x − h) − f (x)| dx = 0.
−∞
Proof. The proofs are straightforward.
Proof of ii. It suffices to show that
Z N
Z N
lim
f (x − α) dx −
f (x) dx = 0.
N →∞
−N
−N
Well, via change of variables,
Z N
Z N
Z N −α
Z N
f (x − α) dx −
f (x) dx =
f (x) dx −
f (x) dx
−N
−N
−N −α
−N
Z −N
Z N
f
=
f+
−N −α
N −α
LECTURE 15: FINALLY, THE FOURIER TRANSFORM!
Now, take N > 2h. Then
Z −N
Z N
Z
|f | +
|f | ≤
−N −α
1
2 N ≤|x|≤2N
N −α
3
A
3A
dx
=
,
x2
N
which goes to 0 as N → ∞.
Proof of iv. Given > 0, we want to show there exists
H > 0 such that if |h| < H, then
Z N
|f (x − h) − f (x)| dx < .
lim
N →∞
−N
WLOG take |h| ≤ 1 and N0 large enough that
Z
Z
|f | ≤ and
|f (x − h)| dx ≤
4
4
|x|≥N0
|x|≥N0
By the uniform continuity of f , we can choose H such that
for h < H,
sup |f (x − h) − f (x)| <
4N0
|x|≤N0
Then for any N > N0 , we see that
Z N
Z N0
|f (x − h) − f (x)| dx ≤
|f (x − h) − f (x)| dx
−N
−N0
Z
Z
+
|f | +
|f (x − h)| dx
|x|≥N0
Z N0
≤
|x|≥N0
|f (x − h) − f (x)| dx +
−N0
≤
2
+ = .
2 2
Thus
Z
∞
|f (x − h) − f (x)| dx ≤ −∞
for all h < H.
Point: f ∈ M (R)
implies f lives inside
a lemon.
4
LECTURE 15: FINALLY, THE FOURIER TRANSFORM!
2. The Fourier Transform
Definition 2.1. Given f ∈ M (R), we define the Fourier
transform fˆ of f by
Z ∞
fˆ(ξ) :=
f (x)e−2πixξ dx
−∞
Fourier transform
Remark. Easy to see that fˆ is a bounded, continuous function that vanishes at ∞. However, fˆ is not necessarily in
M (R) itself. This lack motivates the definition of the following class of functions.
Rapidly decreasing
function
Definition 2.2. We call a function rapidly decreasing if for
every k ≥ 0, we have
sup |x|k |f (x)| < ∞,
x∈R
i.e., the function shrinks faster than the reciprocal of any
polynomial function.
S(R)
Definition 2.3. Let f be an infinitely differentiable (C ∞ )
function. If f and all of its derivatives are rapidly decreasing, we call f a Schwartz class function and write f ∈ S(R).
Remark. Observe that S(R) is a vector space (over C) and
is closed under both differentiation and multiplication by
polynomials.
Principle: decay of
the Fourier transform corresponds to
the smoothness of f
(connects with 2.4 in
chapter II).
3. Fourier transform on S(R)
Definition 3.1. For f ∈ S(R) we define the Fourier transform of f by
Z ∞
fˆ(ξ) =
f (x)e−2πixξ dx
−∞
Proposition 3.2. Let f ∈ S(R), h ∈ R, and δ > 0. Then
(the Fourier transform maps the following):
LECTURE 15: FINALLY, THE FOURIER TRANSFORM!
5
i. f (x + h) −→ fˆ(ξ)e2πihξ (translation becomes modulation)
ii. f (x)e−2πixh −→ fˆ(ξ + h)
iii. f (δx) −→ δ −1 fˆ(δ −1 ξ) (dilation)
iv. f 0 (x) −→ 2πiξ fˆ(ξ) (differentiation becomes polynomial
multiplication)
v. −2πixf (x) −→ dξd fˆ(ξ)
Proof. The only one of interest is (v).
We want to show that
d ˆ
\ (ξ) → 0
f (ξ) + 2πixf
dξ
Fix > 0. Consider:
fˆ(ξ + h) − fˆ(ξ) \
+ 2πixf (ξ)
h
Z
Z
1 −2πix(ξ+h)
− e−2πixξ ] dx + 2πixf (x)e−2πixξ dx
=
f (x) [e
h
R
−2πixh
ZR
e
−1
+ 2πix dx
=
f (x)e2πixξ
h
R
...(to be continued).
LECTURE 16: BASIC PROPERTIES
OF THE FOURIER TRANSFORM
Question (Badgering). What’s the difference between the
theory we’re developing now and the theory we developed
before? What was the “best” result we had for convergence
of Fourier series? Do we have anything parallel here?
1. Fourier transform on S(R)
Recall the Schwartz class S(R). Notice that it is closed
under differentiation and under multiplication by polynomials.
Definition 1.1. For f ∈ S(R) we define the Fourier transform of f by
Z ∞
fˆ(ξ) =
f (x)e−2πixξ dx
−∞
Proposition 1.2. Let f ∈ S(R), h ∈ R, and δ > 0. Then
(the Fourier transform maps the following):
i. f (x + h) −→ fˆ(ξ)e2πihξ (translation becomes modulation)
ii. f (x)e−2πixh −→ fˆ(ξ + h)
iii. f (δx) −→ δ −1 fˆ(δ −1 ξ) (dilation)
iv. f 0 (x) −→ 2πiξ fˆ(ξ) (differentiation becomes polynomial
multiplication)
v. −2πixf (x) −→ dξd fˆ(ξ)
Proof. The main proof of interest is (v).
We want to show that
d ˆ
\ (ξ) → 0
f (ξ) + 2πixf
dξ
definition of the
Fourier transform:
analogous to definition of Fourier
coefficients
interactions with the
Fourier transform
2
BASIC PROPERTIES
Fix > 0. Consider:
fˆ(ξ + h) − fˆ(ξ)
h
\ (ξ)
+ 2πixf
Z
Z
1 −2πix(ξ+h)
f (x) [e
=
− e−2πixξ ] dx + 2πixf (x)e−2πixξ dx
h
−2πixh
R
ZR
e
−1
f (x)e−2πixξ
=
+ 2πix dx
h
R
We do need the second estimate
Now, f (x) and xf (x) are both rapidly decreasing, so
there exists
Z N ∈ N such that
Z
|f | < and
|x|≥N
For each fixed x0 ,
we can find an h.
By continuity, that h
will will in a small
neighborhood of x0 .
Cover [−N, N ] with
such neighborhoods;
choose the minimum
h, which is independent of x.
eix − 1 = 2| sin x2 |
|x||f (x)| dx < |x|≥N
and, by L’Hôpital’s Rule, for sufficiently small h we have,
for the compact set |x| ≤ N ,
e−2πixh − 1
+ 2πix ≤
h
N
Outside of |x| ≤ N , we have the bound
2 |sin(−πixh)|
e−2πixh − 1
+ 2πix =
+ 2πix
h
h
≤ A + 2π|x|
since
sin h
h
is bounded. Thus we have
Z
N
≤
−N
fˆ(ξ + h) − fˆ(ξ) \
+ 2πixf (ξ)
h
−2πixh
e
−
1
+ 2πix dx + f (x)e−2πixξ
h
≤ C
Corollary 1.3. f ∈ S(R) implies fˆ ∈ S(R).
BASIC PROPERTIES
3
Proof. Recall that S(R) is closed under differentiation and
multiplication by polynomials; thus if f ∈ S(R), then so is
k
d
1
[(−2πix)l `f (x)].
k
(2πi) dx
Thus the Fourier transform of the above, namely (by the
previous proposition)
`
d
k
fˆ(ξ)
ξ
dξ
is bounded (since the Fourier transform of any Schwartz
class function is bounded). Thus fˆ and all its derivatives
are rapidly decreasing, as desired.
Immediate
consequence of the above
and the closure of
S(R)
2. Creating an approximation of the identity
using dilated Gaussians
2
Definition 2.1. We call f (x) = e−x the Gaussian.
Remark. The Gaussian is a Schwartz class function; in fact
2
e−ax is in S(R) for all a > 0. The choice a = π is particular
because
Z ∞
2 Z ∞ Z ∞
2
2
−πx2
e−π(x +y ) dx dy
e
dx =
−∞
−∞ −∞
Z 2π Z ∞
2
e−πr r dr dθ
=
Z0 ∞ 0
2
2
=
2πre−πr dr = [−e−πr ]∞
0 = 1.
0
2
Theorem 2.2. Let f (x) = e−πx . Then fˆ = f .
Proof. We shall show that fˆ satisfies a certain boundary
value problem. Let
Z ∞
2
F (ξ) := fˆ(ξ) =
e−πx e−2πixξ dx
−∞
(You probably saw
this in multivariable
calculus.)
4
BASIC PROPERTIES
Then by property (v) in Proposition 1.2, we have
Z ∞
0
F (ξ) =
[−2πixf (x)]e−2πixξ dx, which
−∞
Z
∞
=i
f 0 (x)e−2πixξ dx since f 0 (x) = −2πxf (x)
−∞
(the Fourier transform of f 0 (x) is
2πiξ fˆ(ξ))
= i2πiξ fˆ(ξ) = −2πξF (ξ).
Elementary differential equations (separation of variables)
2
implies that F (ξ) = Ce−πξ
and, plugging in ξ = 0 we
R ∞ , −πx
2
ˆ
see C = F (0) = f (0) = −∞ e
dx = 1 by the previous
calculation.
LECTURE 17: FOURIER INVERSION
1. Creating an approximation of the identity
using dilated Gaussians
Recall:
2
Definition 1.1. We call f (x) = e−x the Gaussian.
2
Remark. The Gaussian is a Schwartz class function; in fact e−ax is in S(R) for all a > 0.
The choice a = π is particular because
Z ∞
2 Z ∞ Z ∞
2
2
2
e−πx dx
=
e−π(x +y ) dx dy
−∞
−∞ −∞
2π Z ∞
Z
=
2
e−πr r dr dθ
=
Z0 ∞
(You probably saw
this in multivariable
calculus.)
0
2
2
2πre−πr dr = [−e−πr ]∞
0 = 1.
0
−πx2
Theorem 1.2. Let f (x) = e
. Then fˆ = f .
Proof. We shall show that fˆ satisfies a certain boundary value problem. Let
Z ∞
2
F (ξ) := fˆ(ξ) =
e−πx e−2πixξ dx
−∞
Then by property (v) in Proposition ??, we have
Z ∞
F 0 (ξ) =
[−2πixf (x)]e−2πixξ dx, which
−∞
Z ∞
=i
f 0 (x)e−2πixξ dx since f 0 (x) = −2πxf (x)
−∞
= i2πiξ fˆ(ξ) = −2πξF (ξ).
2
Elementary differential equations (separation of variables) implies that F (ξ) = Ce−πξ ,
R∞
2
and, plugging in ξ = 0 we see C = F (0) = fˆ(0) = −∞ e−πx dx = 1 by the previous
calculation.
(the Fourier transform of f 0 (x) is
2πiξ fˆ(ξ))
1.1. Creating the approximation of the identity.
√
Now, let, for δ > 0, Kδ denote a dilated (by δ) Gaussian:
Kδ (x) = δ −1/2 f (δ −1/2 x)
= δ −1/2 e−πx
2
/δ
Then, by the interaction of dilation and Fourier transform
Miraculously (?), dilations of this eigenfunction form an approximation of the
identity.
2
LECTURE 17: FOURIER INVERSION
(function dilated by ρ maps to Fourier transform scaled by
ρ−1 ) we see
We do get an approximation of the
identity.
cδ (ξ) = fˆ(δ 1/2 ξ) = e−πδξ 2 .
Corollary 1.3. K
Theorem 1.4. {Kδ }δ>0 is, as δ ↓ 0, an approximation of
the identity.
Proof. Easy. Kδ is a dilation of the Gaussian (a positive
function), which has L1 norm of 1, so the first two conditions are satisfied. For the last condition,
Z
Z
1
2
√ e−πx /δ dx
|Kδ (x)| dx =
δ
|x|>η
Z|x|>η
√
2
=
e−πy dy (let x = δy)
|y|> √ηδ
which obviously (the integrand is rapidly decreasing) goes
to 0 as δ does.
Remark. Consider: what would be the Fourier transform of
the Dirac delta function? As expected (?), on the Fourier
cδ (ξ) = e−πδξ 2 , which converges
transform side, we see that K
pointwise to the constant (≡ 1) function.
Convolution of S(R)
functions
Definition 1.5. Given f, g ∈ S(R), we define the convolution f ∗ g by
Z ∞
(f ∗ g)(x) :=
f (x − t)g(t) dt.
−∞
Corollary 1.6. Given any f ∈ S(R),
lim+ f ∗ Kδ (x) = f (x)
δ→0
uniformly.
The approximation
of the identity works
as before.
Proof. First, we note that any Schwartz class function f ∈
S(R) is uniformly continuous on the real line. For given
any > 0, we can choose an interval [−R, R] outside of
LECTURE 17: FOURIER INVERSION
Point: we needed
uniform continuity
(first on the circle,
now on R) and the
decay of the good
kernel.
3
which |f (x)| ≤ /4. f being continuous, it is uniformly
continuous on [−R, R], and thus onR all of R.
f ∈ S(R) ⇒ f is
∞
continNow then, using the fact that −∞ Kδ = 1 and that uniformly
uous.
(Recall the
Kδ (x) ≥ 0,
Fourier
coefficient
result.)
Z ∞
Kδ (t)[f (x − t) − f (x)] dt
|(f ∗ Kδ )(x) − f (x)| =
−∞
Z
Z
≤
+
Kδ (t)|f (x − t) − f (x)| dt
|t|≥η
|t|≤η
Both of these can be made small, independently of x (exercise). (Away from the origin, use the vanishing of the
approximation of the identity; near the origin, use the uniform continuity of f . The end.)
2. Fourier inversion formula
Proposition 2.1 (“Multiplication formula”). Let f, g ∈
S(R). Then
Z ∞
Z ∞
f (x)ĝ(x) dx =
fˆ(y)g(y) dy
−∞
−∞
Remark. Recall Fubini’s theorem: If F (x, y) is a continuous
function on R2 satisfying the condition
|F (x, y)| ≤
Multiplication
formula: forerunner
to
Plancherel’s
theorem
A
(1 + x2 )(1 + y 2 )
then
Z
∞
Z
∞
Z
∞
Z
∞
F (x, y) dy dx =
−∞
−∞
F (x, y) dx dy.
−∞
−∞
Proof. Let F (x, y) = f (x)g(y)e−2πixy . Then certainly F
satisfies the decay conditions for Fubini’s theorem, and
Just a consequence
of Fubini’s theorem
LECTURE 17: FOURIER INVERSION
4
Z
∞
Z
∞
Z
−2πixy
∞
∞
Z
f (x)g(y)e
dy dx =
f (x)g(y)e−2πixy dx dy,
−∞ −∞
−∞
Z ∞Z ∞
Z−∞
∞ Z ∞
fˆ(y)g(y) dy,
i.e.,
f (x)ĝ(x) dx =
−∞
−∞
−∞
−∞
which is what we wanted to show.
Theorem 2.2 (Fourier inversion). Let f ∈ S(R). Then
Z ∞
fˆ(ξ)e2πixξ dξ.
f (x) =
−∞
Use i. multiplication
formula
ii. that the Gaussian
behaves nicely,
and iii. that we have
an approximation of
the kernel.
Proof. We first prove this for x = 0.
Let Gδ (x) be what should be the inverse Fourier trans2
cδ (ξ) = Kδ (ξ)
form of Kδ : i.e., let Gδ (x) = e−πδx ; then G
(see scratchwork below). Then the multiplication formula
says
Z ∞
Z ∞
f (x)Kδ (x) dx =
fˆ(ξ)Gδ (ξ) dξ.
−∞
−∞
Let δ → 0 on both sides. The LHS is f ∗ Kδ (0), which
converges to f (0) (since the family is an approximation
of
R∞
the identity). The RHS “clearly” converges to −∞ fˆ(ξ) dξ
as δ ↓ 0 (exercise - recall fˆ ∈ S(R)), so the statement is
proven for x = 0.
For a general x, let F (y) = f (y+x). Then by the previous
case,
Z ∞
Z ∞
f (x) = F (0) =
F̂ (ξ) dξ =
fˆ(ξ)e2πixξ dξ.
−∞
−∞
−πx2
Scratchwork: Let γ(x) = e√ , i.e., the normalized Gaussian. Then our
above choice of Gδ (x) = γ( δx), and thus (scaling becomes dilation)
1
1
ξ
2
c
Gδ (ξ) = √ γ̂ √
= √ e−πξ /δ = Kδ (ξ)
δ
δ
δ
Inverse
transform
Fourier
Definition 2.3. Given g ∈ S(R), we define the inverse
Fourier transform ǧ of g by
LECTURE 17: FOURIER INVERSION
Z
∗
5
∞
g(ξ) e2πixξ dξ
F (g)(x) = ǧ(x) :=
−∞
Thus the Fourier inversion theorem can be written: for
f ∈ S(R),
ˇ
f (x) = fˆ(x).
Note that F(f )(y) =
F ∗ (f )(−y).
It is easy to see that for g ∈ S(R), ǧˆ(ξ) = g(ξ) (i.e., F ◦F ∗ =
F ∗ ◦ F = I) and thus the Fourier transform is bijective on
S(R).
3. Plancherel’s theorem
Theorem 3.1 (Plancherel’s theorem). For any f ∈ S(R),
we have ||fˆ||L2 (R) = ||f ||L2 (R) .
Actually we don’t really need the above properties: Plancherel
can be obtained directly from the multiplication and inversion formulae as follows.
Simple
Proof. The multiplication theorem says that for f, g ∈ S(R),
Z ∞
Z ∞
f (x)ĝ(x) dx =
fˆ(y)g(y) dy
−∞
−∞
So that g such that ĝ = f¯, i.e., g = fˇ¯. Then we get
Z ∞
Z ∞
f (x)f¯(x) dx =
fˆ(y)fˇ¯(y) dy
−∞
Z−∞
∞
¯
=
fˆ(y)fˆ(y) dy,
ˆ
¯ = f¯
ˆ
fˇ¯ = f¯; fˇ
−∞
as desired.
proof of
Plancherel’s
theorem:
basically
just multiplication
theorem
LECTURE 18: WEIERSTRASS APPROXIMATION;
HEAT EQUATION ON THE LINE
1. F and convolutions: Plancherel’s theorem
Some further properties of the Fourier transform:
Proposition 1.1 (Fourier transform and convolutions).
Let f, g ∈ S(R). Then
i. f ∗ g ∈ S(R)
ii. f ∗ g = g ∗ f .
\
iii. (f
∗ g)(ξ) = fˆ(ξ)ĝ(ξ)
Convolutions
of
S(R)
functions;
plus the Fourier
transform
(Please read the proofs yourselves.)
These properties, plus the inversion formula, can be used
to give us the following useful theorem:
Theorem 1.2 (Plancherel’s theorem). For any f ∈ S(R),
we have ||fˆ||L2 (R) = ||f ||L2 (R) .
Proof. Let f [ (x) := f (−x), and h = f ∗ f [ (which is in
S(R), by the above proposition).
Applying the inversion formula to h at x = 0, we get
Z ∞
h(0) =
ĥ(ξ) dξ
Z−∞
∞
[
i.e., (f ∗ f )(0) =
|fˆ(ξ)|2 dξ
Z ∞
Z−∞
∞
2
i.e.,
|f (x)| dx =
|fˆ(ξ)|2 dξ
−∞
−∞
The moral: always
make your mistakes
in pairs, so that
one might cancel the
other one out.
2
WEIERSTRASS APPROX’N; HEAT EQUATION
2. Application: Weierstrass Approximation
Theorem
Theorem 2.1. Let f : [a, b] → C. If f is continuous, then
given any > 0, there exists a polynomial P such that
sup |f (x) − P (x)| < .
x∈[a,b]
Create an extension
to all of R
Approximate the extension.
Approximate the approximation.
Actually we approximate the kernel using a polynomial
and then notice that
the convolution with
the polynomial approximates the convolution with the
kernel
Any
continuous
function on a closed
and bounded interval can be uniformly
approximated by a
polynomial.
Proof. Let [−M, M ] be an interval containing [a, b]; let g
be a continuous function on R that agrees with f on [a, b]
and vanishes outside of [−M, M ]; let B denote a bound for
g.
Since g is uniformly continuous, we can choose δ0 such
that for all x ∈ R,
|g(x) − (g ∗ Kδ0 )(x)| <
2
Now,
∞
2
X
1
1 −πx
(−πx20 /δ0 )n
δ
0
=√
Kδ0 (x) = √ e
.
n!
δ0
δ0 n=0
The series expression converges uniformly on every compact
interval of R, so there exists an N0 such that
N0
(−πx2 /δ0 )n
1 X
≤
Kδ0 (x) − √
n!
4M B
δ0 n=0
for all x ∈ [−2M, 2M ]; denote the finite sum by R(x).
Then, for x ∈ [−M, M ] (so that x − t ∈ [−2M, 2M ] for
t ∈ [−M, M ]), using the above bound we see
Z M
|(g ∗ Kδ0 )(x) − (g ∗ R)(x)| =
g(t)[Kδ0 (x − t) − R(x − t)] dt
−M
Z M
≤
|g(t)| |Kδ0 (x − t) − R(x − t)| dt
−M
≤ 2M B
sup
z∈[−2M,2M ]
< /2
|Kδ0 (z) − R(z)|
WEIERSTRASS APPROX’N; HEAT EQUATION
3
and thus g is uniformly approximated by g∗R on [−M, M ].
g is, of course, f on [a, b] ⊂ [−M, M ].
The last thing we need to show is that g ∗ R is actually
a polynomial. Well,
Z M
(g ∗ R)(x) =
g(t)R(x − t) dt
Putting the two
bounds together,
Was that thing a
polynomial?
−M
2
(−π(x−t)
and R(x−t) = √1δ
n=0
n!
P2N00
n
(of the form n an (t)x ) in x
PN0
/δ0 )n
which is a polynomial
3. Application to PDEs
3.1. Time-dependent Heat equation on the line.
Remark. Crucial property of Fourier Transform: interchanges
differentiation with multiplication by polynomials.
The problem: given an infinite rod and an initial temperature distribution f (x) at t = 0, what is u(x, t), the
temperature at point x ∈ R at time t > 0? Physical considerations imply that
∂u ∂ 2 u
= 2.
∂t
∂x
The heat equation
3.2. Finding the solution via the Fourier transform.
Taking (formally - assuming that, in particular, a solution
exists and that it is in S(R)) the Fourier transform (in the
first variable) of both sides, we get
∂ û
(ξ, t) = −4π 2 ξ 2 û(ξ, t)
∂t
and thus, fixing ξ, one gets a trivial differential equation,
viz.
(formally = assuming that everything
works)
4
WEIERSTRASS APPROX’N; HEAT EQUATION
∂ û
∂t (ξ, t)
û(ξ, t)
= −4π 2 ξ 2 ,
⇒ û(ξ, t) = A(ξ)e−4π
2 2
ξ t
.
Taking then the Fourier transform of the initial condition,
we get û(ξ, 0) = fˆ(ξ), so A(ξ) = fˆ(ξ). Thus
2 2
û(ξ, t) = fˆ(ξ)e−4π ξ t ,
The solution is convolution with a particular kernel (which
we call the heat kernel).
and (taking the inverse Fourier transform of both sides)
we see that a solution, if it exists, must be of the form
u(x, t) = f ∗ Ht (x), where
Ht (x) = K4πt (x).
We call Ht (x) the heat kernel of the line.
Theorem 3.1. Let f ∈ S(R), and let u(x, t) := (f ∗Ht )(x).
Then
i. u(x, t) is C 2 (R2+ ) and solves the heat equation,
ii. u(x, t) → f (x) uniformly in x as t → 0 (and thus is
continuous on R2+ )
iii. u(x, t) → f (x) in L2 as t → 0.
Proof. Using the Fourier inversion formula, we see that
Z ∞
ct (ξ)e−2πξx dξ;
u(x, t) =
fˆ(ξ)H
−∞
Main part: showing
convergence in L2 ,
using Plancherel
differentation under the integral sign proves that it (is not
only infinitely differentiable but also) solves the heat equation. (ii) is immediate, since {Kδ } is an approximation of
the identity.
WEIERSTRASS APPROX’N; HEAT EQUATION
5
How do we show L2 convergence? By Plancherel’s theorem,
Z ∞
Z ∞
2
ct (ξ) − fˆ(ξ)|2 dξ
|u(x, t) − f (x)| dx =
|fˆ(ξ)H
−∞
Z−∞
∞
2 2
=
|fˆ(ξ)|2 |e−4π tξ − 1| dξ.
−∞
Let > 0 be fixed. Using the rapid decrease of f ∈ S(R),
choose N such
Z that
2 2
|fˆ(ξ)|2 |e−4π tξ − 1| dξ < .
(1)
Bound the part away
from the origin
|ξ|≥N
Then for all t small enough (note that fˆ is bounded), we
have
2 2
(2)
.
sup |fˆ(ξ)|2 |e−4π tξ − 1| <
2N
|ξ|≤N
So
Z
(3)
|fˆ(ξ)|2 |e−4π
2
tξ 2
− 1| dξ < |ξ|≤N
for all small t, so the L2 difference is smaller than /2. Bound the part near
the origin
LECTURE 19: THE FOURIER TRANSFORM AND
PARTIAL DIFFERENTIAL EQUATIONS, CT’D.
1. The Heat Equation, ct’d.
Recall: last time we proved the following. We defined the
heat kernel by Ht (x) := K4πt (x), i.e.,
1
−x2 /4t
Ht (x) =
e
(4πt)1/2
ct (ξ) = e−4π2 tξ 2
H
We solved the heat
equation using the
appropriate kernel.
Theorem 1.1. Let f ∈ S (R), and let u(x, t) := (f ∗
Ht )(x). Then
i. u(x, t) is C 2 (R2+ ) and solves the heat equation,
ii. u(x, t) → f (x) uniformly in x as t → 0 (and thus is
continuous on R2+ )
iii. u(x, t) → f (x) in L2 as t → 0.
We also saw that the solution could be expressed as:
Z ∞
ct (ξ)e−2πξx dξ;
u(x, t) =
fˆ(ξ)H
−∞
and noted that for each fixed t, the convolution f ∗ Ht was
in S (R). In fact, we have something stronger:
Corollary 1.2. u(·, t) ∈ S (R) uniformly in t in the sense
that given any T > 0,
`
k ∂
sup |x|
u(x, y) < ∞
∂x`
x∈R;t∈(0,T )
For fixed t, f ∗
Ht ∈ S (R) (convolution of Schwartz
functions).
In fact, uniformly in
S (R).
for each k, ` ≥ 0.
Proof. We shall show that u(x, t) is rapidly decreasing, uni∂`
formly for t ∈ (0, T ); the argument is identical for the ∂x
` u.
We bound |u| as follows:
Do it for u.
Break into parts
where |x−y| ≈ |x|....
2
FOURIER TRANSFORM AND PDES
Z
Z
|u(x, t)| ≤
|y|≤ |x|
2
Note
that
we
couldn’t have used
this bound for all y.
|f (x − y)|Ht (y) dy +
same
|y|≥ |x|
2
So now, think: |y| ≤ |x|
2 , so |x − y| ≈ |x|. Thus, using the
rapid decay of f , we see that for the first integral,
Z
Z
CN
|f (x − y)|Ht (y) dy ≤
Ht (y) dy
|x| (1 + |x|)N
|y|≤
|y|≤ |x|
2
2
CN
≤
;
(1 + |x|)N
that is, for any N ∈ N the first integral shrinks faster than
1
(1+|x|)N (is rapidly decreasing).
For the second integral, we see that (recall Ht (y) :=
2
−y
1
4t
e
1/2
(4πt)
) for |y| ≥
|x|
2 ,
we have, for t ∈ (0, T ),
Ht (y) ≤
Note
that
we
couldn’t have used
this bound for all y
either.
C −cx2
e t ,
t1/2
thus
Z
Z
C −cx2
|f (x − y)|Ht (y) dy ≤ 1/2 e t
|f (x − y)| dy
|x|
t
|y|≥ 2
|y|≥ |x|
2
−cx2
1
≤ C 0 1/2 e t
t
is rapidly decreasing in x....
Theorem 1.3 (Uniqueness of solution). Suppose u(x, t)
satisfies the following conditions:
i. u solves the heat equation on R2+
ii. u(x, 0) = 0
iii. u is continuous on R2+
iv. u(·, t) ∈ S (R) uniformly in t (in the sense of Corollary
(1.2))
then u ≡ 0.
FOURIER TRANSFORM AND PDES
3
Proof. Define the energy of the solution u(x, t) at time t by
Z
E(t) := ||u(x, ·)||L2 (R) :=
|u(x, t)|2 dx
R
We notice that E(0) = 0 and that E ≥ 0; now we claim
that E 0 (t) ≤ 0. Differentiating under the integral sign, we
get
Z
dE
= [∂t u(x, t)ū(x, t) + u(x, t)∂t ū(x, t)] dx
dt
R
Since u solves the heat equation, we see that ∂t u = ∂x2 u; so
(passing the derivatives over using integration by parts)
Z
dE
= [∂x2 u(x, t)ū(x, t) + u(x, t)∂x2 ū(x, t)] dx
dt
RZ
=−
This is so cool.
[∂x u(x, t)∂x ū(x, t) + ∂x u(x, t)∂x ū(x, t)] dx
ZR
= −2
|∂x u(x, t)|2 dx ≤ 0.
R
The end.
2. Steady-state heat equation in UHP
The boundary value problem we will examine now is the
following.
(
2
∂2
∂
∆u := ∂x2 + ∂y2 u = 0 on R2+
u(x, 0) = f (x)
As before, we proceed formally. Taking the Fourier transform in the first variable of the data above, we get
∂2
2 2
−4π ξ û(ξ, y) + 2 û(ξ, y) = 0
∂y
û(ξ, 0) = fˆ(ξ)
As before, elementary ODE theory gives that the solution
must be of the form
û(ξ, y) = A(ξ)e−2π|ξ|y + B(ξ)e2π|ξ|y .
Steady-state
heat
equation in R2+ .
4
FOURIER TRANSFORM AND PDES
for some functions A(ξ), B(ξ). We assume B(ξ) ≡ 0, since
otherwise we’d have to take the inverse Fourier transform
of an exponentially growing function.
The boundary condition û(ξ, 0) = fˆ(ξ) implies then that
A(ξ) = fˆ(ξ), i.e.,
û(ξ, y) = fˆ(ξ)e−2π|xi|y
i.e., u(x, y) = f ∗ Py (x)
cy (ξ) = e−2π|ξ|y .
for Py satisfying P
The Poisson kernel
on R2+ (is there some
relation?)
Definition 2.1. We define the Poisson kernel on the UHP
by
1 y
Py (x) :=
π x2 + y 2
LECTURE 20: THE STEADY-STATE HEAT EQUATION
IN THE UPPER HALF PLANE.
Recall the problem:
Steady-state
heat
equation in R2+ .
∆u = 0 on R2+
u(x, 0) = f (x)
Taking the Fourier transform in the first variable, we ultimately saw that
û(ξ, y) = fˆ(ξ)e−2π|xi|y
i.e., u(x, y) = f ∗ Py (x)
cy (ξ) = e−2π|ξ|y .
for Py satisfying P
Definition 0.1. We define the Poisson kernel on the UHP
by
1 y
Py (x) :=
π x2 + y 2
Lemma 0.2 (Py (x) is what we would want it to be).
Z ∞
e−2π|ξ|y e2πiξx dξ = Py (x)
Z −∞
∞
Py (x)e−2πixξ dx = e2π|ξ|y .
The Poisson kernel
on R2+ (Obvious
question: is there
some
relation
between the Poisson
kernel on the circle
and on the line?)
−∞
Proof. RJust a calculation.
Dividing the integral into two
R∞
0
parts, −∞ and 0 , we see:
Z ∞
Z ∞
−2πξy 2πiξx
e
e
dξ =
e2πi(x+iy)ξ dξ
0
0 2πi(x+iy)ξ ∞
1
e
=−
=
2πi(x + iy) 0
2πi(x + iy)
Inverse
Fourier
transfprm
of
e−2π|ξ|y is Py (x),
and tautologically
equivalent statement
2
FOURIER TRANSFORM AND PDES
Similarly,
Calculus I exercise.
Z
0
1
;
2πi(x − iy)
−∞
together these two calculations give the first identity. The
second identity is equivalent to the first via the Fourier
inversion theorem (which we can use since e−2π|ξ|y and Py
are both of moderate decrease).
The question is:
for what PDEs is
one guaranteed this
phenomenon?
Cf.
Green’s functions.
More calculus.
e2πξy e2πiξx dξ =
As before, we are extremely lucky: the kernel with which
we convolve to obtain a solution of our equation is again
an approximation of the identity:
Lemma 0.3. Py is an approximation of the identity on R
as y ↓ 0.
Proof. Trivial. We have already shown above that ||Py ||1 =
1, and the Poisson kernel is positive, so the only property we
need prove is the third, which is an(other) integral calculus
problem:
Z ∞
y
π
δ
−1
dx
=
·
·
·
=
−
tan
x2 + y 2
2
y
δ
which goes to 0 as y → 0.
Now let us show that our intuited solution actually is one:
Theorem 0.4. Let f ∈ S (R); let u(x, y) := (f ∗ Py )(x)
for (x, y) ∈ R2+ . Then
i. u(x, y) ∈ C 2 (R2+ ) and ∆u = 0
ii. u(x, y) → f (x) uniformly as y → 0.
iii. u(·, y) converges to f in L2 as y → 0.
iv. Letting u(x, 0) = f (x), then u is continuous on R2+ and
vanishes at infinity in the sense that
u(x, y) → 0
as |x| + y → ∞.
FOURIER TRANSFORM AND PDES
3
Proof of (iv). (The rest are similar to the proof in the case
of the heat equation on the line, and left as exercises).
We note that if f is any function of moderate decrease,
then we have the following bound (first estimate):
y
1
|(f ∗ Py )(x)| ≤ C
+
.
1 + x2 x2 + y 2
For
Z
Z
Z ∞
f (x − t)Py (t) dt =
+
.
−∞
|t|< |x|
2
First estimate
proof of first estimate
|t|> |x|
2
For the first integral, |x − t| ≈ |x|, so
C
C
≈
.
|f (x − t)| ≤
1 + (x − t)2
1 + x2
For the second integral, |t| > |x|
2 , so
y
y
Py (t) = 2
<
C
.
t + y2
x2 + y 2
We also have the second estimate
C
|(f ∗ Py )(x)| ≤ .
y
Second estimate
since supx Py (x) ≤ Cy (and we are “averaging” Py with a
function of moderate decrease - exercise).
To show the vanishing at infinity, then: when |x| ≥ |y|,
as |x| + y → ∞, we see that we have the bound
1
|y|
1
|x|
+
≤
+
1 + x2 x2 + y 2
1 + x2 x2 + y 2
1
|x|
≤
+
→ 0.
1 + x2 x2
And if |x| ≤ |y|, then
C
|f ∗ Py (x)| ≤ → 0.
y
|(f ∗ Py )(x)| ≤
4
FOURIER TRANSFORM AND PDES
Uniqueness of the solution will follow from the following important (and in fact, characterizing) property of harmonic functions:
MVP
Point:
a consequence of the
Laplacian
being
equal to 0.
Lemma 0.5 (Mean Value Property). Let Ω ⊂ R2 be open;
let u ∈ C 2 (Ω). if ∆u = 0 in Ω, then given any closed disc
BR (x, y) ⊂ Ω, one has, for all r ∈ [0, R],
Z 2π
1
u(x + r cos θ, y + r sin θ) dθ.
u(x, y) =
2π 0
Proof. Let U (r, θ) = u(x+r cos θ, y+r sin θ). Then ∆u = 0,
expressed in polar form, is equivalent to (easy exercise)
∂
∂U
∂ 2U
+r
r
.
0=
∂θ2
∂r
∂r
Average
the above over the circle; then letting F (r) =
R 2π
1
2π 0 U (r, θ) dθ, we get
Z 2π
∂
∂F
1
∂ 2U
r
r
=
− 2 (r, θ) dθ
∂r
∂r
2π 0
∂θ
Since ∂U
∂θ is periodic, the RHS of the above is 0; thus the
continuous function r ∂F
∂r is constant, and thus (take r = 0)
∂F
∂F
r ∂r = 0 (implying ∂r = 0); i.e., F is constant. Since
F (0) = u(x, y), we see that F (r) = u(x, y) for all r ∈ [0, R].
The end.
LECTURE 21: THE STEADY-STATE HEAT EQUATION
IN THE UPPER HALF PLANE (END); POISSON
SUMMATION FORMULA
1. Mean Value Property and Uniqueness of
Solutions
Recall: we were about to prove the uniqueness of solutions for the steady-state heat equation in the UHP using
the following important property of harmonic functions.
Lemma 1.1 (Mean Value Property). Let Ω ⊂ R2 be open;
let u ∈ C 2 (Ω). if ∆u = 0 in Ω, then given any closed disc
BR (x, y) ⊂ Ω, one has, for all r ∈ [0, R],
Z 2π
1
u(x, y) =
u(x + r cos θ, y + r sin θ) dθ.
2π 0
MVP gives
Remark. In fact, if a function is continuous on an (open,
connected) domain in Rn and satisfies the Mean Value
Property, then it is harmonic and in C ∞ .
Theorem 1.2 (Uniqueness of solutions to the steady-state
heat equation on R2+ ). Let u be a solution to ∆u = 0 on
R2+ . If
i. u is continuous on R2+ ,
ii. u(x, 0) = 0, and
iii. u(x, y) vanishes at infinity
then u ≡ 0.
Proof. By contradiction. Suppose u(x, y) is (WLOG) realvalued and u(x0 , y0 ) > 0 for some (x0 , y0 ) ∈ R2+ .
+
Choose a semi-disc DR
:= DR ∩ R2+ with R sufficiently
large that u(x, y) ≤ 12 u(x0 , y0 ) in the complement (note
+
+
(x0 , y0 ) ∈ DR
). Since u is continuous on DR
, it attains
Uniqueness of solutions
2
Observe: where a
max is attained, by
the MVP the function must attain that
max on every circle centered at that
point.
Other properties of
harmonic functions
Connection between
analysis on circle
and R.
Periodization of a
function
POISSON SUMMATION FORMULA
+
. Notice that
its maximum M at some point (x1 , y1 ) ∈ DR
u ≤ M throughout R2+ . Now, by the MVP, we know that
Z 2π
1
u(x1 + ρ cos θ, y1 + ρ sin θ) dθ
u(x1 , y1 ) =
2π 0
for, in particular, ρ ∈ (0, y1 ). Since u(x1 , y1 ) = M , u ≡ M
on that entire circle. Let ρ → y1 ; we see that then (since u
is continuous on R2+ ) u(x1 , 0) = M also: ※.
Remarks (Other properties of harmonic functions).
i. Maximum principle: If u is continuous on the closure
of a bounded domain D and harmonic on the interior,
then the maximum must be attained on the boundary
(unless u is constant).
ii. Liouville’s theorem: A harmonic function on Rn which
is bounded must be constant.
2. Poisson summation formula
Definition 2.1. Let f ∈ S (R). We define the periodization of f to be the (continuous) 1-periodic function F1 :
[0, 1] → C defined by
F1 (x) :=
∞
X
f (x + n)
n=−∞
Remark. The sum converges absolutely and uniformly on
every compact subset of R, so converges to a continuous
function.
Poisson summation
Theorem 2.2 (Poisson summation formula). Let f ∈ S (R).
Then
X
X
f (x + n) =
fˆ(n)e2πinx ,
n∈Z
The periodic function is the one we get
via a discrete version
of the Fourier transform.
n∈Z
POISSON SUMMATION FORMULA
Remark. In particular,
X
f (n) =
n∈Z
X
3
fˆ(n).
n∈Z
Proof. Call the second function F2 . Notice that F2 is (because fˆ ∈ S (R) also) again absolutely and uniformly converging (on all of R), and so is continuous.
Since F1 and F2 are continuous functions on the circle,
showing that they have the same Fourier coefficients would
force their difference to be 0 everywhere. So we calculate
the Fourier coefficients of F1 : using uniform convergence,
we see
!
Z 1 X
XZ 1
−2πimx
f (x + n)e−2πimx dx
f (x + n) e
=
0
n∈Z
n∈Z
=
0
XZ
n∈Z
Z
∞
=
Continuous ⇒ STS
Fourier coefficients
equal.
i.e., Fourier coefficients as functions
on [0,1]
n+1
f (y)e−2πimy dy
n
f (y)e−2πimy dy = fˆ(m),
−∞
which is the m-th Fourier coefficient of F2 .
Remark. Theorem (and proof) holds if f, fˆ are of moderate
decrease.
3. Applications of Poisson summation
3.1. Heat Kernels on the Circle and on the Line.
Recall from Chapter IV: given a function u(x, t) where
t ≥ 0, x ∈ [0, 1] describing the temperature distribution on
a ring (with initial distribution f (x)), it can be shown that
u must satisfy the following problem:
∂u ∂ 2 u
∂t = ∂x2
u(x, 0) = f (x)
Using Poisson summation formula
Recall heat kernel on
the circle.
4
POISSON SUMMATION FORMULA
As usual, we look for standing wave solutions: u(x, t) =
A(x)B(t), resulting in (an infinite superposition)
X
2 2
u(x, t) =
an e−4π n t e2πinx .
n∈Z
Setting t = 0 shows us that an = fˆ(n).
Definition 3.1. We notate
X
2 2
Ht (x) =
e−4π n t e2πinx
n∈Z
heat kernel on the
circle
Heat kernel on the
circle is the periodization of the heat
kernel on the line
and call Ht the heat kernel for the circle.
Then the solution for the heat equation on [0, 1] with initial
data f can be written as u(x, t) = (f ∗ Ht )(x) (where the
convolution is on [0,1]).
Recalling that the heat kernel on the line was given by
1
−x2 /4t
ct (ξ) = e−4π2 tξ 2 ,
Ht (x) =
e
;
i.e.,
H
(4πt)1/2
we note the following.
Theorem 3.2.
Ht (x) =
X
Ht (x + n).
n∈Z
How cool!
Proof. This is exactly the Poisson summation formula:
X
X
ct (n)e2πinx
Ht (x + n) =
H
n∈Z
n∈Z
=
X
e−4π
2
tn2 2πinx
e
=: Ht (x).
n∈Z
LECTURE 21: THE STEADY-STATE HEAT EQUATION
IN THE UPPER HALF PLANE (END); POISSON
SUMMATION FORMULA
Theorem 0.1 (Uniqueness of solutions to the steady-state
heat equation on R2+ ). Let u be a solution to ∆u = 0 on
R2+ . If
i. u is continuous on R2+ ,
ii. u(x, 0) = 0, and
iii. u(x, y) vanishes at infinity
then u ≡ 0.
Proof. By contradiction. Suppose u(x, y) is (WLOG) realvalued and u(x0 , y0 ) > 0 for some (x0 , y0 ) ∈ R2+ .
+
Choose a semi-disc DR
:= DR ∩ R2+ with R sufficiently
large that u(x, y) ≤ 12 u(x0 , y0 ) in the complement (note
+
+
(x0 , y0 ) ∈ DR
). Since u is continuous on DR
, it attains
+
its maximum M at some point (x1 , y1 ) ∈ DR . Notice that
u ≤ M throughout R2+ . Now, by the MVP, we know that
Z 2π
1
u(x1 , y1 ) =
u(x1 + ρ cos θ, y1 + ρ sin θ) dθ
2π 0
for, in particular, ρ ∈ (0, y1 ). Since u(x1 , y1 ) = M , u ≡ M
on that entire circle. Let ρ → y1 ; we see that then (since u
is continuous on R2+ ) u(x1 , 0) = M also: ※.
Recall:
1. Poisson summation formula
Theorem 1.1 (Poisson summation formula). Let f ∈
Then
X
X
f (x + n) =
fˆ(n)e2πinx ,
n∈Z
n∈Z
Observe: where a
max is attained, by
the MVP the function must attain that
max on every circle centered at that
point.
Maximum principle:
If u is continuous
on the closure of
a bounded domain
D and harmonic on
S (R). the interior, then
the maximum must
be attained on the
boundary (unless u
is constant).
Connection between
analysis on circle
and R.
Poisson summation
The periodic function is the one we get
via a discrete version
of the Fourier trans-
2
Using Poisson summation formula
Recall heat kernel on
the circle.
POISSON SUMMATION FORMULA
2. Applications of Poisson summation
2.1. Heat Kernels on the Circle and on the Line.
Recall from Chapter IV: given a function u(x, t) where t ≥ 0, x ∈ [0, 1] describing the
temperature distribution on a ring (with initial distribution f (x)), it can be shown that
u must satisfy the following problem:
∂u
2
= ∂∂xu2
∂t
u(x, 0) = f (x)
As usual, we look for standing wave solutions: u(x, t) = A(x)B(t), resulting in (an
infinite superposition)
X
2 2
u(x, t) =
an e−4π n t e2πinx .
n∈Z
Setting t = 0 shows us that an = fˆ(n).
Definition 2.1. We notate
X
2 2
Ht (x) =
e−4π n t e2πinx
n∈Z
heat kernel on the
circle
and call Ht the heat kernel for the circle.
Then the solution for the heat equation on [0, 1] with initial data f can be written as
u(x, t) = (f ∗ Ht )(x) (where the convolution is on [0,1]).
Recalling that the heat kernel on the line was given by
1
−x2 /4t
ct (ξ) = e−4π2 tξ 2 ,
Ht (x) =
e
;
i.e.,
H
(4πt)1/2
Heat kernel on the
circle is the periodization of the heat
kernel on the line
we note the following.
Theorem 2.2.
Ht (x) =
X
Ht (x + n).
n∈Z
A cool consequence of the above expression:
Corollary 2.3. The heat kernel {Ht } on the circle is an
approximation of the identity (on the circle) as t ↓ 0.
Proof was too hard
to do until now.
Proof. Using uniform convergence, it is immediate that
Z 1/2
Ht (x) dx = 1.
−1/2
Since Ht ≥ 0, the above theorem implies that Ht ≥ 0
(not at all obvious otherwise); so the first two properties of
POISSON SUMMATION FORMULA
3
good kernels are satisfied. It remains to see that given any
η < 1/2,
Z
|Ht (x)| dx → 0 as t → 0.
η<|x|≤ 12
Well, consider:
Ht (x) =
X
Ht (x + n)
n∈Z
= Ht (x) +
X
Ht (x + n) =: Ht (x) + Et (x).
n∈Z∗
Since {Ht } is a good kernel, it suffices to show that
Z
|Et (x)| dx → 0
|x|≤ 21
as t → 0. We shall see that (claim:)
c
|Et (x)| ≤ Ce− t .
For, consider:
1 X −(x+n)2
e 4t
4πt n∈Z ∗
C X −cn2
√
≤
e t .
t n∈Z∗
Et (x) := √
since |x| ≤ 21 . Now, for t ∈ (0, 1] (which we can assume) we
see that
n2
1 1
2
≥
+n
t
2 t
This is so cool...we
can estimate the difference between the
(good) heat kernel
on the line and its
periodization.
4
POISSON SUMMATION FORMULA
(the n2 /t is greater than either of the terms averaged) and
so
C X − cn2
|Et (x)| ≤ √
e t
t n∈Z∗
C − c X − cn2
≤ √ e 2t
e 2
t
n∈Z∗
c
≤ Ce− t .
R
This bound implies the desired control on |x|≤ 1 |Et (x)| dx
2
and thus the third property of good kernels.
Showing
Poisson
kernels related
2.2. Poisson kernels on the disc and upper half plane.
Recall the Poisson kernels on the disc and upper half plane:
1 y
1 − r2
and
P
(x)
=
.
Pr (ϑ) =
y
1 − 2r cos ϑ + r2
π y 2 + x2
Corollary 2.4. With r = e2πy ,
X
Pr (2πx) =
Py (x + n)
n∈Z
Don’t need to mention this in class.
Proof. Use the Poisson summation formula.
3. Digression into analytic number theory
(Reference: Whittaker, E.T., and G.N. Watson, A Course
of Modern Analysis: An Introduction to the General Theory
of Infinite Processes and of Analytic Functions; With an
Account of the Principal Transcendental Functions, Cambridge University Press, 1902.)
theta function
Definition 3.1. For s > 0, the theta function ϑ(s) is defined
∞
X
2
ϑ(s) :=
e−πn s
n=−∞
Functional relation:
consequence of Poisson summation
POISSON SUMMATION FORMULA
5
Theorem 3.2 (Functional relation for ϑ). For s > 0,
1
s−1/2 ϑ
= ϑ(s).
s
2
Proof. Consider the function f (x) = e−πsx ; its Fourier
transform (exercise) is
2
πξ
1
fˆ(ξ) = s− 2 e− s .
Then, by Poisson summation,
∞
∞
X
X
1
πn2
−πs(x+n)2
e
=
s− 2 e− s e2πinx .
n=−∞
n=−∞
Evaluating at x = 0 yields the desired relation.
Another theta function
Definition 3.3. We define the theta function Θ(z|τ ) for
z ∈ C, =(τ ) > 0 by
∞
X
2
Θ(z|τ ) :=
eiπn τ e2πinz .
n=−∞
Remarks.
i. Θ(0|is) = ϑ(s).
ii. Θ(x|4πit) = Ht (x)
Definition 3.4. For s ∈ C such that <(s) > 1, we define
the celebrated Riemann zeta function by
∞
X
1
ζ(s) =
ns
n=1
It can be shewn that ϑ, ζ, and Γ are related by
Z ∞
s
1
π −s/2 Γ(s/2)ζ(s) =
t 2 −1 (ϑ(s) − 1) dt.
2 0
Remark. This will become more relevant later (in your life).
Digressing even further....
The Riemann zeta
function
6
POISSON SUMMATION FORMULA
4. The Heisenberg Uncertainty Principle
state function
expected position
Remark (Motivation). To what extent can one simultaneously locate the position and momentum of a particle?
In quantum mechanics, a particle has associate with it a
state function ψ (of L2 norm 1) which governs the position
in the sense that the probability that the particle lies in
aR particular region (a, b) ∈ R (one-dimensional space) is
2
(a,b) |ψ| . Then the expectation (expected position) is given
by
Z ∞
x|ψ(x)|2 dx,
x :=
−∞
variance of position
and the variance (uncertainty of the expectation) is given
by
Z ∞
(x − x)2 |ψ(x)|2 dx.
−∞
Heisenberg Uncertainty Principle
One has an analogous function describing the momentum
of the particle. Importantly, it turns out that the probability
of the momentum belonging to an interval (a, b) is
R
2
(a,b) |ψ̂(ξ)| dξ. We shall now see the Heisenberg Uncertainty Principle, i.e., that
1
Variance of position × Variance of momentum &
.
16π 2
Theorem 4.1. Let ψ ∈ S (R), and suppose ||ψ||2 = 1.
Then
Z ∞
Z ∞
1
x2 |ψ(x)|2 dx
ξ 2 |ψ̂(ξ)|2 dξ ≥
16π 2
−∞
−∞
2
−Bx
and
where B > 0, A2 =
p equality holds iff ψ(x) = Ae
2B/π.
In fact, we have, for every x0 , ξ0 ∈ R, blahblahblah (with
the individual terms minimized when x0 = x, ξ0 = ξ.
Easy calculation: integration by parts
Proof. Integration by parts implies the following.
POISSON SUMMATION FORMULA
7
∞
Z
|ψ(x)|2 dx
1=
−∞
Z
∞
d
x |ψ(x)|2 dx
dx
Z−∞
∞ 0
0
=−
xψ (x)ψ(x) + xψ (x)ψ(x) dx.
=−
−∞
Thus Z
1≤2
∞
|x||ψ(x)||ψ 0 (x)| dx
−∞
∞
Z
≤2
−∞
Z ∞
=2
1/2 Z
x |ψ(x)| dx
2
∞
0
2
2
1/2
|ψ (x)| dx
−∞
x2 |ψ(x)|2 dx
1/2 4π 2
−∞
Z
∞
ξ 2 |ψ̂(ξ)|2 dξ
1/2
,
−∞
using the Plancherel theorem (and the basic properties of
the Fourier transform) for the equality in the last line. Now,
equality can hold only if equality held in the application of
the Cauchy-Schwartz inequality. which implies that the
functions must be scalar multiples of each other:
ψ 0 (x) = βxψ(x)
for some scalar β. Again, elementary ODE theory implies
2
ψ(x) = Aeβx /2 .
To ensure the function is in S (R), we requirep
β = −2B for
2
some positive B; then ||ψ||2 = 1 forces A = 2B/π. Followed
by
Cauchy-Schwartz
and
Plancherel’s
theorem
LECTURE 21: THE STEADY-STATE HEAT EQUATION
IN THE UPPER HALF PLANE (END); POISSON
SUMMATION FORMULA
Recall:
1. Poisson summation formula
Theorem 1.1 (Poisson summation formula). Let f ∈
Then
X
X
f (x + n) =
fˆ(n)e2πinx ,
n∈Z
Connection between
on circle
S (R). analysis
and R.
Poisson summation
n∈Z
2. Applications of Poisson summation
2.1. Heat Kernels on the Circle and on the Line.
Recall from Chapter IV: given a function u(x, t) where t ≥ 0, x ∈ [0, 1] describing the
temperature distribution on a ring (with initial distribution f (x)), it can be shown that
u must satisfy the following problem:
∂u
2
= ∂∂xu2
∂t
u(x, 0) = f (x)
The periodic function is the one we get
via a discrete version
of the Fourier transform.
Using Poisson summation formula
Recall heat kernel on
the circle.
As usual, we look for standing wave solutions: u(x, t) = A(x)B(t), resulting in (an
infinite superposition)
X
2 2
u(x, t) =
an e−4π n t e2πinx .
n∈Z
Setting t = 0 shows us that an = fˆ(n).
Definition 2.1. We notate
X
2 2
Ht (x) =
e−4π n t e2πinx
n∈Z
and call Ht the heat kernel for the circle.
heat kernel on the
circle
Then the solution for the heat equation on [0, 1] with initial data f can be written as
u(x, t) = (f ∗ Ht )(x) (where the convolution is on [0,1]).
Recalling that the heat kernel on the line was given by
1
−x2 /4t
ct (ξ) = e−4π2 tξ 2 ,
e
;
i.e.,
H
Ht (x) =
(4πt)1/2
we note the following.
Heat kernel on the
circle is the periodization of the heat
kernel on the line
2
POISSON SUMMATION FORMULA
Theorem 2.2.
Ht (x) =
X
Ht (x + n).
n∈Z
A cool consequence of the above expression:
Corollary 2.3. The heat kernel {Ht } on the circle is an
approximation of the identity (on the circle) as t ↓ 0.
Proof was too hard
to do until now.
Proof. Using uniform convergence, it is immediate that
Z 1/2
Ht (x) dx = 1.
−1/2
Since Ht ≥ 0, the above theorem implies that Ht ≥ 0
(not at all obvious otherwise); so the first two properties of
good kernels are satisfied. It remains to see that given any
η < 1/2,
Z
|Ht (x)| dx → 0 as t → 0.
η<|x|≤ 21
Well, consider:
X
Ht (x) =
Ht (x + n)
n∈Z
= Ht (x) +
X
Ht (x + n) =: Ht (x) + Et (x).
n∈Z∗
This is so cool...we
can estimate the difference between the
(good) heat kernel
on the line and its
periodization.
Since {Ht } is a good kernel, it suffices to show that
Z
|Et (x)| dx → 0
|x|≤ 21
as t → 0. We shall see that (claim:)
c
|Et (x)| ≤ Ce− t .
POISSON SUMMATION FORMULA
3
For, consider:
1 X −(x+n)2
e 4t
Et (x) := √
4πt n∈Z ∗
C X −cn2
e t .
≤√
t n∈Z∗
since |x| ≤ 21 . Now, for t ∈ (0, 1] (which we can assume) we
see that
1 1
n2
≥
+ n2
t
2 t
(the n2 /t is greater than either of the terms averaged) and
so
C X − cn2
|Et (x)| ≤ √
e t
t n∈Z∗
X cn2
c
C
≤ √ e− 2t
e− 2
t
n∈Z∗
c
≤ Ce− t .
R
This bound implies the desired control on |x|≤ 1 |Et (x)| dx
2
and thus the third property of good kernels.
2.2. Poisson kernels on the disc and upper half plane.
Recall the Poisson kernels on the disc and upper half plane:
1 − r2
1 y
and
P
(x)
=
.
Pr (ϑ) =
y
1 − 2r cos ϑ + r2
π y 2 + x2
Showing
Poisson
kernels related
Corollary 2.4. With r = e2πy ,
X
Pr (2πx) =
Py (x + n)
n∈Z
Don’t need to mention this in class.
Proof. Use the Poisson summation formula.
4
POISSON SUMMATION FORMULA
3. Digression into analytic number theory
(Reference: Whittaker, E.T., and G.N. Watson, A Course
of Modern Analysis: An Introduction to the General Theory
of Infinite Processes and of Analytic Functions; With an
Account of the Principal Transcendental Functions, Cambridge University Press, 1902.)
theta function
Definition 3.1. For s > 0, the theta function ϑ(s) is defined
∞
X
2
ϑ(s) :=
e−πn s
n=−∞
Functional relation:
consequence of Poisson summation
Theorem 3.2 (Functional relation for ϑ). For s > 0,
1
−1/2
= ϑ(s).
s
ϑ
s
2
Proof. Consider the function f (x) = e−πsx ; its Fourier
transform (exercise) is
2
πξ
1
fˆ(ξ) = s− 2 e− s .
Then, by Poisson summation,
∞
∞
X
X
1
πn2
−πs(x+n)2
e
=
s− 2 e− s e2πinx .
n=−∞
Another theta function
n=−∞
Evaluating at x = 0 yields the desired relation.
Definition 3.3. We define the theta function Θ(z|τ ) for
z ∈ C, =(τ ) > 0 by
∞
X
2
Θ(z|τ ) :=
eiπn τ e2πinz .
n=−∞
Remarks.
i. Θ(0|is) = ϑ(s).
ii. Θ(x|4πit) = Ht (x)
Digressing even further....
POISSON SUMMATION FORMULA
5
Definition 3.4. For s ∈ C such that <(s) > 1, we define
the celebrated Riemann zeta function by
∞
X
1
ζ(s) =
ns
n=1
The Riemann zeta
function
It can be shewn that ϑ, ζ, and Γ are related by
Z ∞
s
1
π −s/2 Γ(s/2)ζ(s) =
t 2 −1 (ϑ(s) − 1) dt.
2 0
Remark. This will become more relevant later (in your life).
4. The Heisenberg Uncertainty Principle
Remark (Motivation). To what extent can one simultaneously locate the position and momentum of a particle?
In quantum mechanics, a particle has associate with it a
state function ψ (of L2 norm 1) which governs the position
in the sense that the probability that the particle lies in
region (a, b) ∈ R (one-dimensional space) is
Ra particular
2
(a,b) |ψ| . Then the expectation (expected position) is given
by
Z ∞
x :=
x|ψ(x)|2 dx,
state function
expected position
−∞
and the variance (uncertainty of the expectation) is given
by
Z ∞
(x − x)2 |ψ(x)|2 dx.
variance of position
−∞
One has an analogous function describing the momentum
of the particle. Importantly, it turns out that the probability
of the momentum belonging to an interval (a, b) is
R
2
(a,b) |ψ̂(ξ)| dξ. We shall now see the Heisenberg Uncertainty Principle, i.e., that
1
Variance of position × Variance of momentum &
.
16π 2
Heisenberg Uncertainty Principle
6
POISSON SUMMATION FORMULA
Theorem 4.1. Let ψ ∈ S (R), and suppose ||ψ||2 = 1.
Then Z ∞
Z ∞
1
x2 |ψ(x)|2 dx
ξ 2 |ψ̂(ξ)|2 dξ ≥
16π 2
−∞
−∞
2
−Bx
and
where B > 0, A2 =
p equality holds iff ψ(x) = Ae
2B/π.
In fact, we have, for every x0 , ξ0 ∈ R, blahblahblah (with
the individual terms minimized when x0 = x, ξ0 = ξ.
Easy calculation: integration by parts
Proof. Integration by parts implies the following.
Z ∞
1=
|ψ(x)|2 dx
−∞
Z ∞
d
=−
x |ψ(x)|2 dx
dx
Z−∞
∞ 0
0
xψ (x)ψ(x) + xψ (x)ψ(x) dx.
=−
−∞
Thus Z
1≤2
∞
|x||ψ(x)||ψ 0 (x)| dx
−∞
∞
Z
≤2
−∞
Z ∞
=2
1/2 Z
x |ψ(x)| dx
2
∞
0
2
1/2
|ψ (x)| dx
−∞
x2 |ψ(x)|2 dx
1/2 4π 2
Z
−∞
Followed
by
Cauchy-Schwartz
and
Plancherel’s
theorem
2
∞
ξ 2 |ψ̂(ξ)|2 dξ
1/2
,
−∞
using the Plancherel theorem (and the basic properties of
the Fourier transform) for the equality in the last line. Now,
equality can hold only if equality held in the application of
the Cauchy-Schwartz inequality. which implies that the
functions must be scalar multiples of each other:
ψ 0 (x) = βxψ(x)
for some scalar β. Again, elementary ODE theory implies
2
ψ(x) = Aeβx
/2
.
POISSON SUMMATION FORMULA
7
To ensure the function is in S (R), we requirep
β = −2B for
2
some positive B; then ||ψ||2 = 1 forces A = 2B/π. LECTURE 22: POISSON SUMMATION FORMULA,
CONTINUED
1. Review from last lecture
Recall:
Theorem 1.1 (Poisson summation formula). Let f ∈ S (R).
Then
X
X
f (x + n) =
fˆ(n)e2πinx ,
n∈Z
n∈Z
Definition 1.2. We notate
X
2 2
Ht (x) =
e−4π n t e2πinx
Connection between
analysis on circle
and R.
Poisson summation
The periodic function is the one we get
via a discretization
of the Fourier transform.
n∈Z
and call Ht the heat kernel for the circle.
Then the solution for the heat equation on [0, 1] with initial
data f can be written as u(x, t) = (f ∗ Ht )(x) (where the
convolution is on [0,1]).
Recalling that the heat kernel on the line was given by
1
2
ct (ξ) = e−4π2 tξ 2 ,
Ht (x) =
e−x /4t ; i.e., H
1/2
(4πt)
we note the following.
Theorem 1.3.
Ht (x) =
X
Ht (x + n).
n∈Z
2. The Heat Kernel on the Line is Good
A cool consequence of the above expression:
Corollary 2.1. The heat kernel {Ht } on the circle is an
approximation of the identity (on the circle) as t ↓ 0.
heat kernel on the
circle
Heat kernel on the
circle is the periodization of the heat
kernel on the line
2
POISSON SUMMATION FORMULA
Proof. Using uniform convergence, it is immediate that
Z 1/2
Ht (x) dx = 1.
−1/2
Since Ht ≥ 0, the above theorem implies that Ht ≥ 0
(not at all obvious otherwise); so the first two properties of
good kernels are satisfied. It remains to see that given any
η < 1/2,
Z
|Ht (x)| dx → 0 as t → 0.
η<|x|≤ 21
Well, consider:
X
Ht (x) =
Ht (x + n)
n∈Z
= Ht (x) +
X
Ht (x + n) =: Ht (x) + Et (x).
n∈Z∗
This is so cool...we
can estimate the difference between the
(good) heat kernel
on the line and its
periodization.
Since {Ht } is a good kernel, it suffices to show that
Z
|Et (x)| dx → 0
|x|≤ 21
as t → 0. We shall see that (claim:)
c
|Et (x)| ≤ Ce− t .
Error here?
[Proof of claim.] Consider:
1 X −(x+n)2
e 4t
4πt n∈Z ∗
C X −cn2
e t .
≤√
t n∈Z∗
Et (x) := √
since |x| ≤ 21 . Now, for t ∈ (0, 1] (which we can assume) we
see that
n2
1 1
2
≥
+n
t
2 t
Proof was too hard
to do until now.
POISSON SUMMATION FORMULA
3
(the n2 /t is greater than either of the terms averaged) and
so
C X − cn2
|Et (x)| ≤ √
e t
t n∈Z∗
C − c X − cn2
≤ √ e 2t
e 2
t
n∈Z∗
c
≤ Ce− t .
R
This bound implies the desired control on |x|≤ 1 |Et (x)| dx
2
and thus the third property of good kernels.
2.1. Poisson kernels on the disc and upper half plane.
Recall the Poisson kernels on the disc and upper half plane:
1 y
1 − r2
and
P
(x)
=
.
Pr (ϑ) =
y
1 − 2r cos ϑ + r2
π y 2 + x2
(Seems to be some
error
here
(in
the last inequality), but showing
c
|Et (x)| ≤ √Ct e− 2t
gives the desired
bound anyway.)
Showing
Poisson
kernels related
Corollary 2.2. With r = e2πy ,
X
Pr (2πx) =
Py (x + n)
n∈Z
Proof. Use the Poisson summation formula.
Don’t need to mention this in class.
3. Digression into analytic number theory
(Reference: Whittaker, E.T., and G.N. Watson, A Course
of Modern Analysis: An Introduction to the General Theory
of Infinite Processes and of Analytic Functions; With an
Account of the Principal Transcendental Functions, Cambridge University Press, 1902.)
Definition 3.1. For s > 0, the theta function ϑ(s) is defined
∞
X
2
ϑ(s) :=
e−πn s
theta function
n=−∞
Functional relation:
consequence of Poisson summation
4
POISSON SUMMATION FORMULA
Theorem 3.2 (Functional relation for ϑ). For s > 0,
1
s−1/2 ϑ
= ϑ(s).
s
2
Proof. Consider the function f (x) = e−πsx ; its Fourier
transform (exercise) is
2
πξ
1
fˆ(ξ) = s− 2 e− s .
Then, by Poisson summation,
∞
∞
X
X
πn2
1
−πs(x+n)2
e
=
s− 2 e− s e2πinx .
n=−∞
Another theta function
n=−∞
Evaluating at x = 0 yields the desired relation.
Definition 3.3. We define the theta function Θ(z|τ ) for
z ∈ C, =(τ ) > 0 by
∞
X
2
Θ(z|τ ) :=
eiπn τ e2πinz .
n=−∞
Remarks.
i. Θ(0|is) = ϑ(s).
ii. Θ(x|4πit) = Ht (x)
Digressing even further....
The Riemann zeta
function
Definition 3.4. For s ∈ C such that <(s) > 1, we define
the celebrated Riemann zeta function by
∞
X
1
ζ(s) =
ns
n=1
It can be shewn that ϑ, ζ, and Γ are related by
Z ∞
s
1
π −s/2 Γ(s/2)ζ(s) =
t 2 −1 (ϑ(s) − 1) dt.
2 0
Remark. This will become more relevant later (in your life).
LECTURE 23: HEISENBERG UNCERTAINTY;
BACKGROUND FOR F ON Rd
1. The Heisenberg Uncertainty Principle
Remark (Motivation). To what extent can one simultaneously specify the position and momentum of a particle?
In quantum mechanics, a particle has associated with it a
state function ψ (of L2 norm 1) which governs the position
in the sense that the probability that the particle lies in
region (a, b) ∈ R (one-dimensional space) is
Ra particular
2
(a,b) |ψ| . Then the expectation (expected position) is given
by
Z ∞
x :=
x|ψ(x)|2 dx,
(Or, phrase in terms
of time-frequency localization.)
state function
expected position
−∞
and the variance (uncertainty of the expectation) is given
by
Z ∞
(x − x)2 |ψ(x)|2 dx.
variance of position
from the expected
position
−∞
One has an analogous function describing the momentum
of the particle. Importantly, it turns out that the probability
of the momentum belonging to an interval (a, b) is
R
2
(a,b) |ψ̂(ξ)| dξ. We shall now see the Heisenberg Uncertainty Principle, i.e., that
1
Variance of position × Variance of momentum &
.
16π 2
Theorem 1.1. Let ψ ∈ S (R), and suppose ||ψ||2 = 1.
Then
Z ∞
Z ∞
1
ξ 2 |ψ̂(ξ)|2 dξ ≥
x2 |ψ(x)|2 dx
16π 2
−∞
−∞
Heisenberg Uncertainty Principle
HEISENBERG UNCERTAINTY; F ON RD
2
2
−Bx
and
where B > 0, A2 =
p equality holds iff ψ(x) = Ae
2B/π.
In fact, we have, for every x0 , ξ0 ∈ R,
Z ∞
Z ∞
1
(x − x0 )2 |ψ(x)|2 dx
,
(ξ − ξ0 )2 |ψ̂(ξ)|2 dξ &
2
16π
−∞
−∞
with the individual terms (and, subsequently, the product)
minimized when x0 = x, ξ0 = ξ.
Easy calculation: integration by parts
Proof. Integration by parts implies the following.
Z ∞
1=
|ψ(x)|2 dx
−∞
Z ∞
d
=−
x |ψ(x)|2 dx
dx
Z−∞
∞ 0
0
xψ (x)ψ(x) + xψ (x)ψ(x) dx.
=−
−∞
Thus
Z
∞
|x||ψ(x)||ψ 0 (x)| dx
1≤2
−∞
∞
Z
≤2
Z
1/2 Z
x2 |ψ(x)|2 dx
−∞
∞
|ψ 0 (x)|2 dx
2
2
1/2 4π
2
Z
∞
2
2
ξ |ψ̂(ξ)| dξ
−∞
Followed
by
Cauchy-Schwartz
and
Plancherel’s
theorem
1/2
−∞
x |ψ(x)| dx
=2
∞
1/2
,
−∞
using the Plancherel theorem (and the basic properties of
the Fourier transform) for the equality in the last line. Now,
equality can hold only if equality held in the application of
the Cauchy-Schwartz inequality. which implies that the
functions must be scalar multiples of each other:
ψ 0 (x) = βxψ(x)
for some scalar β. Again, elementary ODE theory implies
2
ψ(x) = Aeβx
/2
.
HEISENBERG UNCERTAINTY; F ON Rd
3
To ensure the function is in S (R), we requirep
β = −2B for
2
some positive B; then ||ψ||2 = 1 forces A = 2B/π.
To get the second part of the theorem, apply the first to
−2πixξ0
e
ψ(x + x0 ).
2. Fourier Transform on Rd : Background
Remark. Not much content in this section: basically just
a re-hash of the one-dimensional theory. Only thing of
interest is to understand the details of how one makes that
extension.
Basic terms
d
Let x = (x1 , · · · , xd ) denote a vector in R ; let |x| denote
its magnitude, and x · y denote the standard inner product
on Rn .
Definition 2.1. Let x ∈ Rd , and let α ∈ Zd be a multiindex, i.e., a d-tuple of non-negative integers. We notate
xα := xα1 1 xα2 2 · · · xαd d .
and
α1 α2
αd
α
∂
∂
∂
∂
:=
···
∂x
∂x1
∂x2
∂xd
∂ |α|
= α1
∂x1 · · · ∂xαd d
Definition 2.2. Let R : Rd → Rd be a linear transformation. If
R(x) · R(y) = x · y for all x, y ∈ Rd
(or, equivalently: if Rt = R−1 ) then we call R a rotation.
Of course | det(R)| = 1; if det(R) = 1, we say R is a proper
rotation; otherwise is it called improper.
Remark. Given any orthonormal basis {e1 , . . . , ed }, then
{R(e1 ), . . . , R(ed )} is another orthonormal basis. Conversely,
given any two orthonormal bases {ei }, {e0i }, we can define
a rotation R by letting R(ei ) = e0i .
multi-index
rotations
HEISENBERG UNCERTAINTY; F ON RD
4
Rapidly decreasing
functions
Integral on R
d
Definition 2.3. Let f : Rd → C. If
i. f is continuous, and
ii. for all α, |xα f (x)| is bounded,
then f is called rapidly decreasing. (If |x|α |f (x)| is bounded
for α = d + 1, we call the function of moderate decrease.)
Definition 2.4. Given f a function of rapid (or merely
moderate) decrease, we define
Z
Z
f = lim
f (x) dx.
N →∞
Rd
Polar coordinate integration on Rd
QN
2.1. Polar coordinates.
2.1.1. R2 .
Recall: one can, using a change to polar coordinates,
express the integral of a function over the plane as
Z
Z 2π Z ∞
f (r cos θ, r sin θ) r dr dθ.
f (x) dx =
R2
bogus
0
0
With the notation
Z
Z
g(γ) dσ(γ) :=
S1
2π
g(cos θ, sin θ) dθ,
0
we can rewrite the above as
Z
Z Z
f (x) dx =
R2
S1
∞
f (rγ)r dr dσ(γ)
0
2.1.2. R3 .
In R3 , recall, using spherical coordinates, we can do a
change-of-variables to show that
Z
Z 2π Z π Z ∞
f=
f (r sin θ cos φ, r sin θ sin φ, r cos θ)r2 dr sin θ dθ dφ.
R3
0
0
0
As above, if we abbreviate (or more honestly, define)
Z
Z 2 Z π
g(γ) dσ(γ) =
π
g(sin θ cos φ, sin θ sin φ, cos θ) sin θ dθ dφ,
S2
0
0
HEISENBERG UNCERTAINTY; F ON Rd
5
then the formula immediately above can be more concisely
expressed as
Z Z ∞
Z
f=
f (rγ)r2 dr dσ(γ).
R3
S2
0
2.1.3. Rd .
In general, one has the following formula:
Z
Z ∞
Z
f=
f (rγ)rd−1 drdσ(γ).
Rd
S d−1
General formula
0
3. Elementary Theory of the Fourier
Transform in Rd
Definition 3.1. Let f : Rd → C be an infinitely differentiable function. If, for each pair of multi-indices α and
β,
β
∂
sup xα
f (x) < ∞
∂x
d
x∈R
then we say that f ∈ S (Rd ).
Definition 3.2. Given f ∈ S (Rd ), we define its Fourier
transform fˆ : Rd → C by
Z
fˆ(ξ) :=
f (x)e−2πix·ξ dx.
Rd
Proposition 3.3. Let f ∈ S (Rd ), h ∈ Rd , and δ > 0.
Then
i. f (x + h) → fˆ(ξ)e2πiξ·h
ˆ
ii. f (x)e−2πix·h →
f(x + h)
iii. f (δx) → δ1d fˆ ξδ
∂ α
αˆ
iv. ∂x
f (x) → (2πiξ)
fα(ξ)
∂
v. (−2πix)α f (x) → ∂ξ
fˆ(ξ)
vi. f (Rx) → fˆ(Rξ) for all rotations R.
Just as in the
one-dimensional
case.
Interchange
of
differentiation
and multiplication
by
corresponding
monomials, etc.
6
HEISENBERG UNCERTAINTY; F ON RD
Proof of (vi). A calculation:
Z
f (R(x))e−2πix·ξ dx
F [f (Rx)](ξ) :=
Z Rd
−1
f (y)e−2πi(R y)·ξ |det(R−1 )|dy (let y = Rx)
=
ZRd
f (y)e−2πiy·(Rξ) dy = fˆ(Rξ), as desired.
=
Rd
As before, we have the following:
Corollary 3.4. F maps S (Rd ) to itself.
3.1. Slight digression: Radial functions.
Definition 3.5. A function f on Rd is called radial if its
value is constant on spheres about the origin, i.e., there
exists f0 : R≥0 → C such that f (x) = f0 (|x|).
Remark. Obviously, f is radial ⇐⇒ f (Rx) = f (x) for all
rotations R.
Corollary 3.6. If f is radial, then fˆ is also.
Proof. WTS fˆ(Rξ) = fˆ(ξ) for all rotations R. By the
above,
fˆ(Rξ) = F [f (Rx)](ξ) = F (f )(ξ) = fˆ(ξ).
3.2. Fourier inversion and Plancherel on Rd .
Theorem 3.7. Let f ∈ S (Rd ). Then the Fourier inversion formula
Z
f (x) =
fˆ(ξ)e2πix·ξ dξ
Rd
holds, as does the Plancherel theorem, ||f ||2 = ||fˆ||2 .
HEISENBERG UNCERTAINTY; F ON Rd
7
Proof. The proof is basically analogous to that of the onedimensional case:
i. First one shows (via iteration) that the d-dimensional
Gaussian is an eigenfunction of F , i.e.,
2
2
F (e−π|x| )(ξ) = e−π|ξ| .
ii. Using the interaction of dilation and F , we next see
that
π|ξ|2
1
2
F[e−πδ|x| ](ξ) = d/2 e− δ .
δ
π|x|2
1 − δ
e
, one shows that {Kδ }
iii. Then, letting Kδ (x) = δd/2
is an approximation of the identity.
iv. Multiplication formula: for f, g ∈ S (Rd ),
Z
Z
f ĝ =
fˆg
Rd
Rd
(proof obtained via Fubini’s theorem exactly as in the
one-dimensional result).
Once one has (iii) and (iv) above, the proof of the inversion
formula follows exactly as in the one-dimensional case; once
one has (iv) and the inversion formula, one can analogously
play the “hats game” to get Plancherel’s theorem (or go via
convolutions).
Remark. Not all that interesting. However, what we will
explore next, the wave equation in Rd × R and how the
theory differs in odd and even dimensions, is quite subtle.
LECTURE 24:
THE WAVE EQUATION ON Rd × R
1. Radial functions and the Fourier transform
Definition 1.1. A function f on Rd is called radial if its
value is constant on spheres about the origin, i.e., there
exists f0 : R≥0 → C such that f (x) = f0 (|x|).
Radial functions
Remark. Obviously, f is radial ⇐⇒ f (Rx) = f (x) for all
rotations R.
Corollary 1.2. If f is radial, then fˆ is also.
Fourier transform of
radial function is radial.
Proof. WTS fˆ(Rξ) = fˆ(ξ) for all rotations R. Well,
Z
fˆ(Rξ) =
f (x)e−2πiRξ·x dx
ZRd
−1
=
f (x)e−2πiξ·R x dx
ZRd
=
f (Ry)e−2πiξ·y |det(R−1 )| dy
ZRd
=
f (y)e−2πiξ·y dy = fˆ(ξ).
Rd
2. The Wave Equation
Definition 2.1. We define the d-dimensional Laplacian by
Laplacian
∆=
∂2
∂2
+
·
·
·
+
.
∂x21
∂x2d
LECTURE 24: THE WAVE EQUATION ON RD × R
Definition 2.2. The Cauchy problem for the wave equation
is the following:

2
 ∆u = ∂∂tu2
u(x, 0) = f (x)
 ∂u
∂t (x, 0) = g(x)
2
The
formal
argument to find a
solution:
As usual, we first run a formal argument. Taking the
Fourier transform in the first variable, we see that the wave
equation becomes
∂ 2 û
2 2
2
−4π (ξ1 + · · · + ξd )û(ξ, t) = 2 (ξ, t)
∂t
where in the above theˆindicates the Fourier transform in
the first variable. For each fixed ξ, then, one obtains an
ordinary differential equation in t with solution
û(ξ, t) = A(ξ) cos(2π|ξ|t) + B(ξ) sin(2π|ξ|t).
Fourier transform of
initial conditions
As usual, taking the Fourier transforms of the initial conditions yields
û(ξ, 0) = fˆ(ξ)
∂ û
∂t (ξ, 0) = ĝ(ξ);
then, of course
A(ξ) = fˆ(ξ) and 2π|ξ|B(ξ) = ĝ(ξ).
The guessed solution.
It actually is a solution.
Thus the solution is expected to be
ˆ
g(ξ)
ˆ
û(ξ, t) = f (ξ) cos(2π|ξ|t) +
sin(2π|ξ|t);
2π|ξ|
and, in fact, it is:
Theorem 2.3. Given the Cauchy problem for the wave
equation

2
 ∆u = ∂∂tu2
u(x, 0) = f (x)
 ∂u
∂t (x, 0) = g(x),
The Cauchy problem
for the wave equation
LECTURE 24: THE WAVE EQUATION ON Rd × R 3
the following is a solution:
Z ĝ(ξ)
fˆ(ξ) cos(2π|ξ|t) +
u(x, t) :=
sin(2π|ξ|t) e2πix·ξ dξ.
2π|ξ|
Rd
Proof is an exercise in differentiation
(using the fact that
we can differentiate
under the integral
sign.)
Proof. Differentiate under the integral signs. First, differentiating with respect to x1 , . . . , xd , we get
!
Z d
X
ĝ(ξ)
ξk2 e2πix·ξ dξ.
fˆ(ξ) cos(2π|ξ|t) +
sin(2π|ξ|t) (2πi)2
∆u(x, t) =
2π|ξ|
Rd
k=1
Z ĝ(ξ)
=
sin(2π|ξ|t) (−4π 2 i|ξ|2 )e2πix·ξ dξ.
fˆ(ξ) cos(2π|ξ|t) +
2π|ξ|
Rd
Diffeentiating u(x, t) with respect to t also yields the same
thing, obviously; so our guessed solution does solve the
wave equation. Checking the initial conditions is equally
clear:
Z
u(x, 0) :=
fˆ(ξ)e2πix·ξ dξ = f (x)
Rd
by Fourier inversion. Similarly,
Z ∂u
ĝ(ξ)
(x, t) =
cos(2π|ξ|t) e2πix·ξ dξ;
fˆ(ξ)(−2π|ξ|) sin(2π|ξ|t) + (2π|ξ|)
∂t
2π|ξ|
Rd
evaluating at t = 0 shows the initial velocity condition
∂u
(x, 0) = g(x)
∂t
is also satisfied.
Remark. In fact, the solution is unique. This fact can be
shown via a conservation of energy argument.
3. Conservation of Energy
Definition 3.1. Let u be a solution of the wave equation.
We define the (total = kinetic + potential)) energy of the
Uniqueness of solution is a conservation
of energy argument.
LECTURE 24: THE WAVE EQUATION ON RD × R
solution as
Z
2
2
2
∂u
∂u
∂u
+
+ ··· +
dx.
E(t) :=
∂x1
∂xd
Rd ∂t
4
Definition of energy
of a solution
Lemma 3.2. Let a, b ∈ C; let α ∈ R. Then
|a cos α + b sin α|2 + | − a sin α + b cos α|2 = |a|2 + |b|2
Pythagorean
theorem again.
Proof. (cos α, sin α) and (− sin α, cos α) are orthonormal in
C2 ; so by the Pythagorean Theorem,
|(a, b) · (cos α, sin α)|2 + |(a, b) · (− sin α, cos α)|2 = |(a, b)|2 .
Theorem 3.3. For the aforementioned solution u of the
wave equation, E(t) is constant.
Proof. Recall
ĝ(ξ)
û(ξ, t) = fˆ(ξ) cos(2π|ξ|t) +
sin(2π|ξ|t);
2π|ξ|
thus, using Plancherel’s theorem,
Z
Z X
d
2
∂u
dx =
2π|ξ|fˆ(ξ) cos(2π|ξ|t) + ĝ(ξ) sin(2π|ξ|t)
∂x
d
d
j
R
R j=1
Note:
tives
first deriva-
and
Z
Rd
∂u
∂t
2
Z
dx =
−2π|ξ|fˆ(ξ) sin(2π|ξ|t) + ĝ(ξ) cos(2π|ξ|t)
Rd
Then, by the lemma (a = 2π|ξ|fˆ(ξ), b = ĝ(ξ)), we get
Z
2
2
2
∂u
∂u
∂u
+ ··· +
dx
E(t) =
+
∂t
∂x
∂x
d
1
d
ZR
=
(4π 2 |ξ|2 |fˆ(ξ)|2 + |ĝ(ξ)|2 ) dξ,
Rd
which is independent of t.
2
dξ.
2
dξ.
LECTURE 24: THE WAVE EQUATION ON Rd × R
4. Wave Equation in R3 × R
motivation
5
Recall d’Alembert’s solution to the wave equation:
Z
u(x + t) + u(x − t) 1 x+t
u(x, t) =
+
g(y) dy.
2
2 x−t
“This suggests a generalization to higher dimensions, where
we might expect to write the solution of our problem as
averages of the initial data.”
Definition 4.1. Let f : R3 → C. We define the spherical
mean of f at x with radius t by
Z
1
Mt (f )(x) =
f (x − tγ)dσ(γ),
4π S 2
that is, the average of f over the sphere of radius t centered
at x.
Some useful lemmata:
Lemma 4.2. Let f ∈ S(R3 ). Then for each fixed t, Mt (f ) ∈
S(R3 ). Further, Mt (f ) is infinitely differentiable in t, and
each t-derivative is in S(R3 ).
Lemma 4.3 (Fourier transform of the surface measure).
Z
1
sin(2π|ξ|)
e−2πiξ·γ dσ(γ) =
.
4π S 2
2π|ξ|
c
Remark. Denote the LHS by dσ(ξ).
We notice that (it is
the Fourier transform of a radial “function.”): the above
lemma shows that it is then radial.
Proof. We first observe the formula is true for ξ = (0, 0, ρ)
for any ρ > 0 (ρ = 0 is immediate); then we show the left
hand side is a radial function. By definition,
Z
Z 2π Z π
1
1
−2πiξ·γ
e
dσ(γ) :=
e−2πiξ·γ sin θ dθ dφ
4π S 2
4π 0
0
spherical mean: first
example of a convolution of a function
with a measure.
LECTURE 24: THE WAVE EQUATION ON RD × R
(where γ = (sin θ cos φ, sin θ sin φ, cos θ) in the right-hand
side).
Z 2π Z π
1
=
e−2πiρ cos θ sin θ dθ dφ
4π 0
0
Z π
1
e−2πiρ cos θ sin θ dθ
=
2 0
Z
1 1 2πiρu
=
e
du (C.O.V.: u = − cos θ)
2 −1
sin(2πρ)
1 2πiρu 1
=
;
e
=
−1
4πiρ
2πρ
so the lemma is proven for ξ = (0, 0, ρ) for ρ ≥ 0.
(To be continued....)
6
LECTURE 25:
THE WAVE EQUATION ON R3 × R
In order to solve the initial value for the wave equation in dimensions bigger than one, we shall use the method of spherical means, due to Hadamard. The intuitive grounds for this
methodology lies in our conception of waves as being made
of a superposition of spherically symmetric fronts emanating
from point sources. Huygens was a pioneer of this view of
waves, which he used to show that the laws of optics suggested that light was a wave phenomenon.
(http://www.math.nyu.edu/faculty/tabak/PDEs/WE.pdf)
1. Wave Equation in R3 × R
Recall d’Alembert’s solution to the wave equation:
Z
u(x + t) + u(x − t) 1 x+t
+
g(y) dy.
u(x, t) =
2
2 x−t
“This suggests a generalization to higher dimensions, where
we might expect to write the solution of our problem as
averages of the initial data.”
Definition 1.1. Let f : R3 → C. We define the spherical
mean of f at x with radius t by
Z
1
f (x − tγ)dσ(γ),
Mt (f )(x) =
4π S 2
that is, the average of f over the sphere of radius t centered
at x.
Some useful lemmata:
Lemma 1.2. Let f ∈ S(R3 ). Then for each fixed t, Mt (f ) ∈
S(R3 ). Further, Mt (f ) is infinitely differentiable in t, and
each t-derivative is in S(R3 ).
motivation
spherical mean: first
example of a convolution of a function
with a measure.
LECTURE 25: THE WAVE EQUATION ON R3 × R
Lemma 1.3 (Fourier transform of the surface measure).
Z
sin(2π|ξ|)
1
e−2πiξ·γ dσ(γ) =
.
4π S 2
2π|ξ|
2
c
Remark. Denote the LHS by dσ(ξ).
We notice that (it is
the Fourier transform of a radial “function.”): the above
lemma shows that it is then radial.
Formula is true for a
single ξ = (0, 0, ρ).
Proof. We first observe the formula is true for ξ = (0, 0, ρ)
for any ρ > 0 (ρ = 0 is immediate); then we show the left
hand side is a radial function. By definition,
Z
Z 2π Z π
1
1
e−2πiξ·γ sin θ dθ dφ
e−2πiξ·γ dσ(γ) :=
4π S 2
4π 0
0
(where γ = (sin θ cos φ, sin θ sin φ, cos θ) in the right-hand
side).
Z 2π Z π
1
=
e−2πiρ cos θ sin θ dθ dφ
4π 0
0
Z
1 π −2πiρ cos θ
=
e
sin θ dθ
2 0
Z
1 1 2πiρu
=
e
du (C.O.V.: u = − cos θ)
2 −1
1 2πiρu 1
sin(2πρ)
=
e
=
;
−1
4πiρ
2πρ
Both sides of the
equation are radial
functions; so proving
it for a single vector
proves it for the entire sphere.
so the lemma is proven for ξ = (0, 0, ρ) for ρ ≥ 0.
START FROM HERE
At this point, it suffices to show that the Fourier transform of the surface measure is a radial function. We recall
the fact (see last page of appendix) that given any function
f of moderate decrease,
Z
Z
f (R(γ))dσ(γ) =
f (γ) dσ(γ).
S d−1
S d−1
LECTURE 25: THE WAVE EQUATION ON R3 × R
Then, given any rotation R,
Z
1
c
e−2πi(Rξ)·γ dσ(γ)
dσ(Rξ)
:=
4πZ S 2
1
−1
=
e−2πiξ·R γ dσ(γ)
4π ZS 2
1
c
e−2πiξ·γ dσ(γ) = dσ(ξ).
=
4π S 2
3
The next lemma basically says that the Fourier transform
of the spherical averaging operator (which, for t = 1 is
convolution f ∗ dσ) is the product of Fourier transforms.
c
(fˆ(ξ)dσ(ξt))
Lemma 1.4.
\
ˆ sin(2π|ξ|t) .
M
t (f )(ξ) = f (ξ)
2π|ξ|t
Proof. By definition,
\
M
e−2πix·ξ
f (x − γt) dσ(γ) dx
t (f )(ξ) :=
R3
S2
Z Z
1
−2πix·ξ
=
f (x − γt)e
dx dσ(γ)
4π S 2 R3
Z Z
1
=
f (y)e−2πi(y+γt)·ξ dy dσ(γ) (let y = x − γt)
4π S 2 R3
Z
1
= fˆ(ξ)
e−2πi(γt)·ξ dσ(γ)
4π S 2
Z
1
sin(2π|ξ|t)
= fˆ(ξ)
e−2πiγ·tξ dσ(γ) = fˆ(ξ)
4π S 2
2π|ξ|t
by the previous lemma.
Z
Fourier transform of
the spherical averaging operator: calculation using the previous lemma
1
4π
Z
Once we have the above lemmata, the solution becomes
clear:
4
The explicit formula
of the solution
Easy once we have
the above lemmata
Calculus I trick
Elementary school
arithmetic trick
LECTURE 25: THE WAVE EQUATION ON R3 × R
Theorem 1.5. In R3 × R, the solution to the Cauchy problem for the wave equation is
∂
u(x, t) = (tMt (f )(x)) + tMt (g)(x)
∂t
Proof. We break the problem into two subproblems: the
case when g = 0, and the case when f = 0. Using the
Fourier inversion expression of the solution that we worked
out before, we see in the first case,
Z h
i
u(x, t) =
fˆ(ξ) cos(2π|ξ|t) e2πix·ξ dξ
R3
Z sin(2π|ξ|t)
∂
2πix·ξ
t
e
dξ
fˆ(ξ)
=
∂t
2π|ξ|t
3
R
∂
= (tMt (f )(x)),
∂t
and in the second case,
Z sin(2π|ξ|t) 2πix·ξ
u(x, t) =
e
dξ
ĝ(ξ)
2π|ξ|
R3
Z sin(2π|ξ|t) 2πix·ξ
=t
e
dξ
ĝ(ξ)
2π|ξ|t
3
R
= tMt (g)(x).
The solution to the general problem (for general f, g ∈
S(R3 )) is then the superposition of these two cases.
2. Cool observation about the solution:
Huygen’s Principle
Huygen’s Principle
Considering the form of the solution given above, i.e.,
that
∂
u(x, t) = (tMt (f )(x)) + tMt (g)(x)
∂t
we see that the solution at (x, t) depends on the averages of
f and g (that is, data on the boundary t = 0) over spheres
(in R3 ) centered at x of radius t; equivalently, “the data at
a point x0 in the plane t = 0 influences the solution on the
LECTURE 25: THE WAVE EQUATION ON R3 × R 5
boundary of a forward light cone originating at x0 . (See
http://en.wikipedia.org/wiki/Huygens-Fresnel principle)
3. The Wave Equation in R2 × R:
Hadamard’s Method of Descent
Definition 3.1. We define, for F : R2 → C, a weighted
ft (F )(x) over the disk of radius t centered at x ∈
average M
2
R by
Z
1
1
ft (F )(x) :=
F (x − ty)
dy.
M
2π| |y|≤1
(1 − |y|2 )1/2
Relevant weighted
averaging operator
Theorem 3.2. Let f, g ∈ S(R2 ) be initial data for the
Cauchy problem for the wave equation on R2 × R. Then
a solution is
∂ f
f
u(x, t) = (tM
t (f )(x)) + tMt (g)(x).
∂t
Proof. We use f, g ∈ S(R2 ) to create a Cauchy problem for
the wave equation in R3 × R as follows. Fix some T > 0,
and let η ∈ S(R) be a (bump) function such that
η(x) = 1 whenever |x| ≤ 3T.
We
turn
the
two-dimensional
problem
into
a
three-dimensional
one.
We create f [ , g [ ∈ S(R3 ) by defining
f [ (x1 , x2 , x3 ) := f (x1 , x2 )η(x3 )
g [ (x1 , x2 , x3 ) := g(x1 , x2 )η(x3 ).
Let, now, u[ be the solution to the Cauchy problem for
the wave equation on R3 × R with initial data f [ , g [ . By
Huygen’s principle, we see that for |t| ≤ T , u[ (x, t) is constant in x3 for all |x3 | ≤ T : after all, the backwards light
cone for such (x, t) is contained in R2 × [−3T, 3T ], over
which region the initial data f [ , g [ is constant in x3 . (Draw
picture: the solution depends on data which is constant in
x3 .)
We already can solve
the R3 × R equation;
ignoring the third
variable (it’s constant in that variable) gives the R2 ×
R solution.
LECTURE 25: THE WAVE EQUATION ON R3 × R
Now, define
u(x1 , x2 , t) := u[ (x1 , x2 , 0, t).
u solves the 2-dimensional Cauchy problem for |t| < T ; let’s
call it uT . Notice now that if we take a T2 > T1 , then uT2
agrees with uT1 for |t| < T1 . Since T was arbitrary, we thus
obtain a well-defined solution u(x1 , x2 , t) for all t > 0.
Now we need to show that the solution actually has the
desired form.
6
Now we need to
show
that
the
spherical
averages
of the “extended to
R3 data” are the
weighted averages of
the original R2 data.
Calculus II
Lemma 3.3. Let H be a function on the sphere S 2 . If there
exists some two-variable function h such that H(x1 , x2 , x3 ) =
h(x1 , x2 ), then
ft (h)(x1 , x2 ).
Mt (H)(x1 , x2 , 0) = M
Proof of lemma. By definition, Mt (H)(x1 , x2 , 0)
Z 2π Z π
1
h(x1 − t sin θ cos φ, x2 − t sin θ sin φ) sin θ dθ dφ
=
4π 0
0
Z 2π Z π/2
1
=
h(x1 − t sin θ cos φ, x2 − t sin θ sin φ) sin θ dθ dφ
4π 0
0
Z 2π Z π
1
+
h(x1 − t sin θ cos φ, x2 − t sin θ sin φ) sin θ dθ dφ
4π 0
π/2
Letting r = sin θ, we get
Z 2π Z π/2
1
=
h(x1 − t sin θ cos φ, x2 − t sin θ sin φ) sin θ dθ dφ
4π 0
Z 2π Z 0π
1
+
h(x1 − t sin θ cos φ, x2 − t sin θ sin φ) sin θ dθ dφ
4π 0
π/2
Z 2π Z 1
1
1
=
h(x1 − tr cos φ, x2 − tr sin φ) √
rdr dφ
2
2π 0
1
−
r
0
Z
1
1
ft (h)(x)
=
h(x − ty) p
dy = M
2
2π |y|≤1
1 − |y|
as desired (taking y = (r cos φ, r sin φ) in the last change of
variables).
LECTURE 25: THE WAVE EQUATION ON R3 × R
Once we have the above lemma, since
u(x1 , x2 , t) := u[ (x1 , x2 , 0, t)
∂
= (tMt (f [ )(x1 , x2 , 0) + tMt (g [ )(x1 , x2 , 0)
∂t
∂ f
f
= (tM
t (f )(x1 , x2 ) + tMt (g(x1 , x2 ),
∂t
our solution indeed has the form that we claimed.
4. Comments about the solutions
Since the propagation of light is governed by the
three-dimensional wave equation, if at t = 0 a
point of light flashes at the origin, after a finite
amount of time an observer will see the flash
only for an instant. However, if we drop a stone
in a lake, after a finite amount of time any point
on the surface will begin to undulate and will
continue to do so indefinitely (in principle).
7
Difference between
odd
and
even
dimensions
LECTURE 26: RADIAL SYMMETRY, THE FOURIER
TRANSFORM, AND BESSEL FUNCTIONS; THE
RADON TRANSFORM
1. f0 and F0 , Bessel Functions, and Parity
d
Question. Recall: if f is a radial function on R (i.e., f (x) =
f0 (|x|), then so is fˆ (fˆ(ξ) = F0 (ξ)). What is the relation
between f0 and F0 ?
1.1. Case R. In one dimension, “radial” is the same as
even; so let |ξ| = ρ and consider:
Z ∞
F0 (ρ) := fˆ(ξ) :=
f (x)e−2πix|ξ| dx
−∞
Z ∞
−2πir|ξ|
2πir|ξ|
=
f0 (r) e
+e
dr
0Z
∞
=2
cos(2πρr)f0 (r) dr
0
1.2. Case R3 . In R3 , we (obviously) use the polar integration formula, and then the formula for the Fourier trans2
form of the surface element
Z on S .
F0 (ρ) = fˆ(ξ) :=
f (x)e−2πix·ξ dx
RZ3
Z ∞
=
f0 (r)
e−2πirγ·ξ dσ(γ)r2 dr
S2
Z0 ∞
2
c
=
f0 (r)4π dσ(rξ)r
dr
0
Z ∞
4π sin(2πρr) 2
r dr
=
f0 (r)
2πρr
0Z
2 ∞
=
sin(2πρr)f0 (r)r dr.
ρ 0
1
Seems sort of a digression for now....
A trivial calculation
using the evenness of
the function.
2
RADIAL SYMMETRY; RADON TRANSFORM
1.3. Case R2 .
Bessel function: you
might have seen
them in ODEs
Definition 1.1. For each n ∈ Z, let the nth Bessel function
Jn (ρ) denote the nth Fourier coefficient of eiρ sin θ ; that is,
Z 2π
1
eiρ sin θ e−inθ dθ, or
Jn (ρ) =
2π 0
∞
X
iρ sin θ
Jn (ρ)einθ .
e
=
n=−∞
Polar coordinates:
x = (r cos θ, r sin θ)
Then we have the following relation:
Z
ˆ
f (ξ) =
f (x)e−2πix·(0,−ρ) dx
2
ZR2π Z ∞
=
f0 (r)e2πirρ sin θ r dr dθ
0 Z 0
∞
= 2π
J0 (2πrρ)f0 (r)r dr.
0
Fubini’s theorem
In general, the relation between f0 and F0 involves the
Bessel function of order d2 − 1 (one needs a more general
definition in the odd dimensions for the Bessel functions of
fractional order).
2. The Radon Transform
2.1. Various forms of the Radon Transform.
The so-called X-ray
transform
Definition 2.1. Let ρ be a function on R2 . For each line
L ⊂ R2 , we define the Radon transform X of ρ on L by
Z
X(ρ)(L) :=
ρ.
L
Definition 2.2. Let G2,3 denote the Grassmannian (manifold) of two-dimensional affine planes in R3 .
The Radon Transform in R3
Definition 2.3. Let f be a function on R3 (f ∈ S(R3 ),
say). We define the Radon transform R(f ) on G2,3 by
RADIAL SYMMETRY; RADON TRANSFORM
3
Z
R(f )(P) =
f
P
Remark. Usually Radon transform means integration over
planes of co-dimension 1. One uses the term k-plane Radon
transform for integrals over k-planes, and the term X-ray
transform for integrals over lines.
2.2. Calculation of the Radon Transform in R3 .
Remark. One can parametrize the elements ot G2,3 as follows: for γ ∈ S 2 and t ∈ R, let Pt,γ denote the plane
{x ∈ R3 : x · γ = t}. I.e., Pt,γ is the plane orthogonal to γ
and passing through (the terminal point of) tγ.
Note that Pt,γ
P−t,−γ .
=
Definition 2.4. Given f ∈ S(R3 ), we define its integral
over Pt,γ by
Z
Z
f :=
f (tγ + u1 e1 + u2 e2 ) du1 du2
R2
Pt,γ
where {γ, e1 , e2 } is an orthonormal basis of R3 .
Proposition
2.5. Let f ∈ S(R3 ). Then the above definiR
tion of Pt,γ f is independent of the choice of e1 and e2 .
Proof. Trivial.
That is, {e1 , e2 } is
an ONB of the plane
P0,γ .
Lemma 2.6.
Z
∞
!
Z
f
−∞
Pt,γ
Z
dt =
f (x) dx
R3
Proof. Just a calculation. Let R be the rotation taking the
standard basis vectors of R3 to γ, e1 , e2 . Then
Intuitively obvious
4
RADIAL SYMMETRY; RADON TRANSFORM
Z
Z
f (Rx) dx
f (x) dx =
R3
R3
Z
=
R3
Z
f (x1 γ + x2 e1 + x3 e2 ) dx1 dx2 dx3
!
Z
∞
=
f
−∞
dt.
Pt,γ
Definition 2.7 (Alternate definition of Radon Transform).
Let f ∈ S(R3 ). Then for (t, γ) ∈ R × S 2 we define the
Radon Transform of f by
Z
R(f )(t, γ) :=
f.
Again,
we note
that R(f )(t, γ) =
R(f )(−t, −γ); that
is, really
Relevant Schwartz
class S(R × S 2 )
Pt,γ
We shall need an appropriately-defined Schwartz space.
Definition 2.8. Let F be a continuous function on R2 ×S 2
that is infinitely differentiable in t. If
`
k ∂ F
(t, γ) < ∞
sup |t|
∂t`
t∈R,γ∈S 2
for all nonnegative k, ` ∈ Z (i.e., F (·, γ) is in S(R) uniformly in γ) then we say F ∈ S(R × S 2 ).
THE key lemma:
The Fourier Slice
theorem
Lemma 2.9 (Fourier Slice, or Projection Theorem). If f ∈
S(R3 ), then R(f )(t, γ) ∈ S(R) for each γ, and
b )(s, γ) = fˆ(sγ)
R(f
where ˆ denotes the one-dimensional Fourier transform in
the first variable in the LHS, and the three-dimensional
Fourier transform on the RHS.
Remark. In other words, if you project f onto the line {tγ}
and then take the 1-D Fourier transform, it’s the same as
taking the slice of the 2-D Fourier transform parallel to γ.
LECTURE 27: FOURIER SLICE THEOREM AND
RADON INVERSION FORMULA
1. Fourier Slice Theorem
Recall:
Definition 1.1 (Definition of Radon Transform). Let f ∈
S(R3 ). Then for (t, γ) ∈ R×S 2 we define the Radon Transform of f by
Z
R(f )(t, γ) :=
f.
Pt,γ
We shall need an appropriately-defined Schwartz space.
Definition 1.2. Let F be a continuous function on R2 ×S 2
that is infinitely differentiable in t. If
`
k ∂ F
(t, γ) < ∞
sup |t|
∂t`
t∈R,γ∈S 2
Again,
we note
that R(f )(t, γ) =
R(f )(−t, −γ); that
is, really
Relevant Schwartz
class S(R × S 2 )
for all nonnegative k, ` ∈ Z (i.e., F (·, γ) is in S(R) uniformly in γ) then we say F ∈ S(R × S 2 ).
Lemma 1.3 (Fourier Slice, or Projection Theorem). If f ∈
S(R3 ), then R(f )(t, γ) ∈ S(R) for each γ, and
b )(s, γ) = fˆ(sγ)
R(f
where ˆ denotes the one-dimensional Fourier transform in
the first variable in the LHS, and the three-dimensional
Fourier transform on the RHS.
Remark. In other words, if you project f onto the line {tγ}
and then take the 1-D Fourier transform, it’s the same as
taking the slice of the 3-D Fourier transform parallel to γ.
1
THE key lemma:
The Fourier Slice
theorem
2
FOURIER SLICE THEOREM AND RADON TRANSFORM INVERSION
Proof of the formula.
Z ∞ Z
b )(s, γ) :=
R(f
−∞
∞ Z
!
f
e−2πist dt
Pt,γ
Z
f (tγ + u1 e1 + u2 e2 ) du1 du2 e−2πist dt
:=
2
Z −∞ R
f (tγ + u)e−2πist du dt
=
ZR3
=
f (tγ + u)e−2πisγ·(tγ+u) du dt
ZR3
=
f (x)e−2πisγ·x dx (C.O.V.: x = tγ + u)
R3
= fˆ(sγ).
Remark. The Fourier Slice theorem plus the (one-dimensional)
inversion formula yields the following:
Z ∞
R(f )(t, γ) =
fˆ(sγ)e2πits ds;
−∞
in other words, the Radon transform of f at (t, γ) is actually the inverse along the line {sγ : s ∈ R} of the (3dimensional) Fourier transform of f , evaluated at t.
Corollary 1.4. Let f, g ∈ S(R3 ). If R(f ) = R(g) then
f = g.
Uniqueness theorem
(not at all so simple
in general; a topic of
serious study)
Proof. Since R(f ) = R(g), R(f − g)(t, γ) ≡ 0. But then
Fourier slice implies f[
− g(sγ) = 0 for all s, γ; so f[
− g(ξ) ≡
0. Fourier inversion implies f − g ≡ 0, so f = g.
2. Filtered Backprojection Inversion formula
2.1. The dual Radon transform.
Dual Radon Transform
Definition 2.1. Let F : R × S 2 → R. We define the dual
Radon transform R∗ (F ) : R3 → R of F by
FOURIER SLICE THEOREM AND RADON TRANSFORM INVERSION
3
Z
∗
F (x · γ, γ) dσ(γ).
R (F )(x) :=
S2
Remark. If F (t, γ) is the Radon transform R(f )(t, γ) then
F (x · γ, γ) = R(f )(x · γ, γ)
Z
=
f,
What is the dual
Radon
transform
of
the
Radon
transform?
Px·γ,γ
the integral over the plane with normal vector γ of distance
x · γ from the origin: in other words, the plane passing
through x with normal vector γ. Thus R∗ (F )(x) is the
integral of f over all planes passing through x: the so-called
backprojection of f .
2.2. Why is it called the dual? Let V1 = S(R3 ), with
inner product
Z
(f, g)1 =
f g;
The backprojection
operator
R3
2
let V2 = S(R × S ) with inner product
Z Z
(F, G)2 =
F (t, γ)G(t, γ) dσ(γ) dt.
R
S2
Then
Duality formula
∗
(R(f ), F )2 = (f, R (F ))1 .
2.3. The inversion formula.
Theorem 2.2. Let f ∈ S(R3 ). Then
∆(R∗ R(f )) = −8π 2 f.
Proof. As we noted earlier, the Fourier slice theorem (plus
the one-dimensional inversion formula) imply
Z ∞
R(f )(t, γ) =
fˆ(sγ)e2πits ds;
−∞
so
∗
Z Z
∞
R R(f )(x) =
S2
−∞
fˆ(sγ)e2πix·γs ds dσ(γ).
a.k.a.
the filtered
backprojection
inversion formula
1. Fourier slice theorem
4
2. differentiation under the integral sign
FOURIER SLICE THEOREM AND RADON TRANSFORM INVERSION
Then, differentiation under the integral sign (plus the fact
that γ ∈ S 2 ) imply
Z Z ∞
∗
fˆ(sγ)(−4π 2 s2 )e2πix·γs ds dσ(γ)
∆(R R(f ))(x) =
S 2 −∞
Z Z ∞
2
= −4π
fˆ(sγ)e2πix·γs s2 ds dσ(γ)
2
ZS Z−∞
∞
2
= −4π
fˆ(sγ)e2πix·γs s2 ds dσ(γ)
2
Z SZ 0 0
fˆ(sγ)e2πix·γs s2 ds dσ(γ)
− 4π 2
S 2 −∞
Z Z ∞
= −8π 2
fˆ(sγ)e2πix·γs s2 ds dσ(γ)
S2
0
2
= −8π f (x).
invariance under rotation
Notes on the above calculation:
1. The second-to-last inequality follows since
Z Z 0
−4π 2
fˆ(sγ)e2πix·γs s2 ds dσ(γ)
S 2 −∞
Z Z ∞
2
= −4π
fˆ(−sγ)e−2πix·γs s2 ds dσ(γ)
2
ZS Z0 ∞
= −4π 2
fˆ(sγ)e2πix·γs s2 ds dσ(γ)
S2
3. polar integration
0
where one rotates γ to −γ in the last step.
2. The last equality follows from (reverting) the polar
integration formula:
Z Z ∞
Z
2πix·γs
2
fˆ(sγ)e
s ds dσ(γ) =
fˆ(ξ)e2πix·ξ dξ
S2
0
R3
= f (x).
Reconstruction formula in Rd
Remark. In general, the reconstruction formula for the Radon
transform is as follows:
FOURIER SLICE THEOREM AND RADON TRANSFORM INVERSION
5
(d−1)
(2π)1−d
(−∆) 2 R∗ (R(f )) = f,
2
where the fractional Laplacian is defined by
Z
(2π|ξ|)2α fˆ(ξ)e2πiξ·x dξ
(−∆)α f (x) :=
Fractional
Laplacian,
or inverse
Riesz transform
Rd
d
for f ∈ S(R ). Note that the formula is (again) more simple
for the odd dimensions than for the even.
Remark. In using the Schwartz class, we have actually swept
an entire world of details under a rug. For the interested
reader, please see, for example, http://equinto.math.tufts.edu/
research/sc-article.pdf, an introductory article to the field Reference
to
Quinto’s
article
by Todd Quinto of Tufts (whose thesis advisor was Cor- from the workshop
mack...wait, no, that’s false (http://equinto.math.tufts.edu/
CV/vitatuft.pdf)).
3. Wave Equation and Radon Transform
Definition 3.1. Let F : R → R be a C 2 (R) function, and
u a function on Rd × R. If there exists a vector γ ∈ S d−1
such that
u(x, t) = F ((x · γ) − t),
then we call F a plane wave.
Remarks. (Plane waves and the wave equation.)
i. Such functions are solutions of the wave equation in Rd .
ii. Such u are constant on planes perpendicular to γ.
iii. As t increases, the wave travels in the γ direction.
iv. In fact, for d > 1 the solution of the wave equation can
be expressed as an integral of plane waves (by using the
Radon transform).
Reference: Helgason, Sigurdur. Radon Transforms and
Wave Equations, Springer Lecture Notes in Mathematics,
1996. Full text is available at http://www.springerlink.com/
content/d664n43145015772/fulltext.pdf.
Plane wave
Plane waves and the
wave equation
LECTURE 28: FINITE FOURIER ANALYSIS:
BASIC DEFINITIONS
1. Background knowledge: The Group Z/N Z
Definition 1.1. Let z ∈ C, N ∈ N. If z N = 1, we say z is
an N th root of unity.
1.1. Z(N ).
Note that the set, which we’ll notate as Z(N ), of N th
roots of unity is exactly
roots of unity
Z(N )
{1, e2πi/N , e2πi2/N , . . . , e2πi(N −1)/N }.
Also note that Z(N ) is, with complex multiplication as its
group law, an abelian group, i.e., it is:
i. (Closed under group law:) If z, w ∈ Z(N ) then zw ∈
Z(N )
ii. (Abelian:) If z, w ∈ Z(N ) then zw = wz
iii. (Identity) 1 ∈ Z(N )
iv. (Inverses) If z ∈ Z(N ) then there exists z −1 ∈ Z(N )
such that zz −1 = 1.
1.2. Z/N Z.
Definition of
abelian group
an
Z/N Z
Definition 1.2. Let N ∈ N, and x, y ∈ Z. If x − y is
divisible by N , we say x is congruent to y mod N :
x≡y
mod N.
It is obvious that the relation is
i. (reflexive) x ≡ x mod N for all x ∈ Z
ii. (symmetric) x ≡ y mod N implies y ≡ x mod N
iii. (transitive) x ≡ y mod N and y ≡ z mod N imply
x ≡ z mod N ;
in other words, an equivalence relation on Z.
1
Equivalence relation
2
Z/N Z
Additive group law
for Z/N Z
Isomorphism
THE FAST FOURIER TRANSFORM
Definition 1.3. One calls the set of equivalence classes on
Z modulo this relation the integers modulo N , denoted by
Z/N Z.
Now, for each x ∈ Z, let R(x) denote the equivalence class
corresponding to x. It is easy to see that one can define
a group law (addition) on the set of equivalence classes by
defining
R(x) + R(y) = R(x + y).
It is easy to see that if x0 ∈ R(x) and y 0 ∈ R(y) (i.e.,
x0 ≡ x mod N , y 0 ≡ y mod N ) then x0 + y 0 ∈ R(x + y)
(i.e., x0 + y 0 ≡ x + y mod N ).
Proposition 1.4. The association R(k) ↔ e2πik/N gives
a correspondence (in fact, a group isomorphism) between
Z/N Z and Z(N ).
Remark. In the same way, we can create an identification
between the functions on Z/N Z and Z(N ).
The space of (cts.)
functions on Z(N )
2. The characters on Z(N )
Notation 2.1. We let V denote the (N -dimensional) inner product space of functions F : Z(N ) → C, with inner
product
(F, G) :=
N
−1
X
F (k)G(k).
k=0
Defining the (obvious) inner product
and the associated norm.
Question. What should be the analogues, for Fourier analysis on Z(N ), of the functions en (x) = e2πinx on the circle?
Desirable properties
of characters
The key properties of those functions are the following:
i. {en }n∈Z is an orthonormal set, with respect to the inner
product
THE FAST FOURIER TRANSFORM
Note that if we had
a set whose span was
dense in V , then the
span would have to
be all of V .
3
ii. The collection of finite linear combinations of the {en }
is dense in the space of continuous functions on the
circle
iii. en (x + y) = en (x)en (y)
Notation 2.2. Let ζ = e2πi/N . We define, for ` = 0, . . . , N −
1, the functions e` on Z(N ) by
The
characters...?
(Not quite.)
k
e` (k) := ζ ` = e2πi`k/N .
Lemma 2.3. The set {e0 , . . . , eN −1 } is orthogonal (and
thus a basis of V ).
Proof. A calculation:
(em , e` ) :=
N
−1
X
ζ
mk −`k
k=0
ζ
=
N
−1
X
ζ (m−`)k ,
k=0
which, if m 6= ` yields (a geometric sum)
1 − (ζ m−` )N
= 0,
1 − ζ m−`
and yields N if m = `.
Thus if we normalize the {e` }, we obtain an orthonormal
basis of V :
Notation 2.4 (Orthonormal basis of V ). Let e∗` =
√1 e` .
N
The ONB
3. Finite Fourier Analysis
Definition 3.1. Give F ∈ V , we define the nth Fourier
coefficient of F by
1
an = √ (F, e∗n )
N
N −1
1 X
=
F (k)e−2πink/N
N
k=0
Fourier coefficients
on Z(N )
LECTURE 29: FINITE FOURIER ANALYSIS:
THE FAST FOURIER TRANSFORM
1. Finite Fourier Analysis
Definition 1.1. Give F ∈ V , we define the nth Fourier
coefficient of F by
1
an = √ (F, e∗n )
N
N −1
1 X
=
F (k)e−2πink/N
N
Fourier coefficients
on Z(N )
k=0
Finite Fourier inversion and Plancherel are (obviously)
linear algebra statements about orthonormal bases.
Theorem 1.2 (Fourier inversion). Let F ∈ V . Then
N
−1
X
F (k) =
an e2πink/N .
“Obviously” because
Fourier analysis =
infinite-dimensional
linear algebra; so
finite-dimensional
Fourier analysis =
linear algebra.
n=0
Fourier inversion on
Z(N )
Proof. (Do this backwards.)
N
−1
N
−1
X
X
1
2πink/N
√ (F, e∗n )e2πink/N
an e
:=
N
n=0
n=0
=
N
−1
X
(F, e∗n )e∗n (k) = F (k)
n=0
since
{e∗n }
is an ONB of V .
Theorem 1.3 (Parseval-Plancherel formulae).
N −1
N
−1
X
1 X
2
|F (n)|2 .
|an | =
N n=0
n=0
1
Again, just properties of ONBs.
THE FAST FOURIER TRANSFORM
2
Proof.
N −1
1 X
1
|F (n)|2 =: ||F ||2
N n=0
N
N −1
1 X
=
|(F, e∗n )|2
N n=0
=
N
−1
X
n=0
N
−1
2
X
1
∗
√ (F, en ) =:
|an |2 .
N
n=0
2. FFT (the Fast Fourier Transform)
Question. How does one best calculate the Fourier coefficients of a function F on Z(N )?
The Question
2.1. Naive calculation.
It is easy to see that if one is given the values of F (0), . . . , F (N −
1) and ωN = e−2πi/N , then calculating the N Fourier coeffiN −1
2
cients {aN
k (F )}k=0 of F on Z(N ) requires at most 2N + N
operations. After all, by definition,
aN
k (F )
Straightforward
counting
N −1
1 X
kr
:=
F (r)ωN
.
N r=0
Then
N −1
2
i. Calculating ωN
, . . . , ωN
takes at most N − 2 multiplications
ii. For each aN
k (F ), one needs at most 1+N multiplications
1
(first by N , then the N products inside the sum) and
N − 1 additions (the sum).
iii. Thus we have N − 2 operations at the beginning, and
then N × 2N operations afterwards, totalling 2N 2 +
N − 2.
However, one can actually do much better than this.
THE FAST FOURIER TRANSFORM
3
2.2. The fast Fourier transform.
Theorem 2.1 (FFT). Given ωN = e−2πi/N (with N = 2n ),
it takes at most
4 · 2n n = 4N log2 (N ) = O(N log N )
operations to calculate the Fourier coefficients of a function
on Z(N ).
Notation 2.2. Let #(M ) denote the minimum number
of operations needed to calculate all the Fourier coefficients of any function on Z(M ).
Much better version.
Notation:
minimum number of
operations
Lemma 2.3 (The key lemma). If we are given ω2M =
e−2πi/(2M ) then
#(2M ) ≤ 2#(M ) + 8M.
2M
Proof of lemma. First note that to calculate ω2M , . . . , ω2M
2
takes no more than 2M operations; also observe that ω2M
=
−2πi/M
e
=: ωM .
Now, given any function F on Z(2M ) define F0 and F1
on Z(M ) by
F0 (n) = F (2n), F1 (n) = F (2n + 1);
by definition, it is possible to calculate the Fourier coefficients of F0 and F1 with at most #(M ) operations each.
Notating Fourier coefficients corresponding to Z(2M ) and
Z(M ) as a2M
and aM
k
k , we claim:
1 M
M
k
a2M
(F
)
=
a
(F
)
+
a
(F
)ω
0
1
k
2M ;
k
2 k
thus after obtaining the Fourier coefficients for F0 and F1 ,
each a2M
k (F ) can be obtained in three operations (one multiplication, one addition, one multiplication) and thus
#(2M ) ≤ 2M = 2#(M ) + 3 × 2M = 2#(M ) + 8M,
k
(steps to calculate the {ω2M
}2M
k=1 , the Fourier coefficients
of F0 and F1 , and then finally the coefficients a2M
k (F )) as
desired.
The
main
idea:
to divide F into
even and odd parts,
whose
Fourier
coefficients can be
obtained in ≤ #(M )
steps
and notice that the
Fourier coefficients
of F can be obtained
from the Fourier
coefficients of the
even and odd parts.
THE FAST FOURIER TRANSFORM
4
The first step we do
nothing; the second
we do even less.
So the only thing left to do is to prove the claim. But
this is basically a tautology: just break the sum into its
even and odd terms:
2M −1
1 X
kr
2M
F (r)ω2M
ak (F ) :=
2M r=0
=
=:
1
2
1
2
1
M
1
M
M
−1
X
`=0
M
−1
X
`=0
k(2`)
F (2`)ω2M +
k(2`)
F0 (`)ω2M +
1
M
1
M
M
−1
X
m=0
M
−1
X
!
k(2m+1)
F (2m + 1)ω2M
!
k(2m+1)
F1 (m)ω2M
.
m=0
Noticing that
k(2`)
k`
ω2M = ωM
and
mk k
k(2m+1)
2
mk k
ω2M
= ω2M
ω2M = ωM
ω2M
by the observation in the first paragraph of the proof finishes the claim.
Remark. Notice that the algorithm is built into the proof
of the lemma. The FFT was discovered by Cooley and
Tukey in 1965; however, in 1984 it was discovered that it
(as usual) had already been known to Gauss around 1805.
LECTURE 30: FINITE FOURIER ANALYSIS:
THE FAST FOURIER TRANSFORM
1. Proof of FFT
Theorem 1.1 (FFT). Given ωN = e−2πi/N (with N = 2n ),
it takes at most
4 · 2n n = 4N log2 (N ) = O(N log N )
operations to calculate the Fourier coefficients of a function
on Z(N ).
Much better version.
Recall: the key was that calculation of the coefficients
on Z(2M ) could be done by calculating the coefficients of
related functions (the odd and even parts) on Z(M ):
Lemma 1.2 (The key lemma). If we are given ω2M =
e−2πi/(2M ) then
#(2M ) ≤ 2#(M ) + 8M.
Once we have the key lemma, the theorem follows immediately:
Proof of Theorem. Let N = 2n ; we induct on n. In the case
n = 1 (N = 2), by definition
1
1
N
aN
0 (F ) = [F (1) + F (−1)] and a1 (f ) = [F (1) − F (−1)],
2
2
which requires 5 < 8 operations; so case n = 1 is verified.
For the inductive step, we assume that the theorem is
true for N = 2n−1 ; i.e., that
#(N ) ≤ 4 · 2n−1 (n − 1).
Then, by the lemma,
#(2N ) ≤ 2 · 4 · 2n−1 (n − 1) + 8 · 2n−1 = 4 · 2n n,
as desired.
1
Proof of theorem follows from lemma by
induction trivially.
THE FAST FOURIER TRANSFORM
2
2. Fourier Analysis on (finite) Abelian Groups
Remark. We’re just going to set up the background knowledge necessary. See how we extend the notions of Fourier
analysis; what’s needed, etc.
In fact one can
create a theory for
locally
compact
(i.e.,
each point
has a neighborhood
contained
in
a
compact set) abelian
groups.
Definition of abelian
group
Definition 2.1. An abelian group is a set with a binary
operation (“group law”) on pairs of elements of G,
G×G→G
(a, b) 7−→ a · b,
that satisfies the follwing properties:
i. (commutativity) a · b = b · a for all a, b ∈ G;
ii. (associativity) a · (b · c) = (a · b) · c for all a, b, c ∈ G;
iii. (identity) there exists an element u ∈ G such that a·u =
a for all a ∈ G; and
iv. (inverses) given any a ∈ G, there exists an element
a−1 ∈ G such that aa−1 = u.
Examples and non-examples: (R, ·), (R∗ , ·), (R, +), SL2 (R),
SO2 (R), etc.
Example: Z∗ (q)
Definition 2.2. We say n ∈ Z(q) (q ∈ N) is a unit if there
exists an m ∈ Z(q) such that
(1)
Remark
that
multiplication
is
well-defined on the
equivalence classes
of Z(n)
order
nm ≡ 1 mod q.
The set of all units in Z(q) is denoted Z∗ (q); it is an abelian
group under multiplication mod q.
Definition 2.3. The number #(G) of elements in a group
G is called the order of G, and denoted |G|.
Definition 2.4. Let G, H be (abelian) groups. If a map
f : G → H “preserves the group law” i.e., satisfies
f (a · b) = f (a) · f (b),
homomorphism
then we call f a (group) homomorphism.
THE FAST FOURIER TRANSFORM
3
Definition 2.5. A homomorphism f : G → H, if one-toone and onto, is called an isomorphism; and in that case
one says “G is isomorphic to H” and writes G ≈ H.
isomorphism
Example: the exponential function exp : R → R+ (R
equipped with addition and R+ with multiplication, respectively, as the group laws) is a group isomorphism.
Remark. The existence of an isomorphism is equivalent to
the existence of an inverse homomorphism.
2.1. Structure Theorem for Finite Abelian Groups.
Definition 2.6. Given two finite abelian groups G1 and
G2 , we define the direct product G1 × G2 to be the set of
cartesian pairs
direct product
{(g1 , g2 ) : g1 ∈ G1 , g2 ∈ G2 }
with the group law given by
(g1 , g2 ) · (g10 , g20 ) := (g1 · g10 , g2 · g20 ),
with which the set becomes (check) itself a finite abelian
group.
Theorem 2.7 (Structure theorem for finite abelian groups).
Any finite abelian group is isomorphic to a direct product
of groups of the form Z(N ).
Example. Consider Z∗ (8), the multiplicative group of
units of Z/8Z; namely, Z∗ (8) = {1, 3, 5, 7}. Z∗ (8) can
be shown to be isomorphic to Z(2) × Z(2) via, for example, the mapping under which {1, 3, 5, 7} correspond to
{(0, 0), (1, 0), (0, 1), (1, 1)} respectively.
3. Characters
Notation 3.1. Let S 1 denote the unit circle in C, equipped
with complex multiplication as the group law.
Example of structure theorem
4
character
dual group
THE FAST FOURIER TRANSFORM
Definition 3.2. Let G be a finite abelian group, and e :
G → S 1 . If for all a, b ∈ G,
e(a · b) = e(a)e(b),
(in other words, if e is a homomorphism) then we call e a
character.
b denote the set of all characters of G.
Lemma 3.3. Let G
b the product e1 · e2 by
If we define, for e1 , e2 ∈ G,
(e1 · e2 )(a) := e1 (a)e2 (a) for all a ∈ G,
b becomes itself an abelian group, called the dual group.
then G
Useful lemma:
Useful lemma: any
multiplicative, nonvanishing map is a
character
Lemma 3.4. Let G be a finite abelian group, and e : G →
C\{0}. If e is multiplicative, i.e.,
e(a · b) = e(a)e(b) for all a, b ∈ G,
then e is a character.
Proof. Notice that the function |e| is bounded both above
and below (away from 0) on G (since G is finite). If |e(g)| >
1, then |e(g n )| = |e(g)|n would go to infinity with n: impossible. Similarly for |e(g)| < 1. Thus e maps into S 1 and
is a character.
LECTURE 31:
FOURIER ANALYSIS ON FINITE ABELIAN GROUPS
Recall that we had, in the distant past, just introduced
b of
the notion of characters and defined the dual group G
characters on G.
1. Characters
Definition 1.1. Let G be a finite abelian group, and e : G → S 1 . If for all
a, b ∈ G,
e(a · b) = e(a)e(b),
(in other words, if e is a homomorphism) then we call e a character.
b denote the set of all characters of G. If we define, for
Lemma 1.2. Let G
b
e1 , e2 ∈ G, the product e1 · e2 by
character
dual group
(e1 · e2 )(a) := e1 (a)e2 (a) for all a ∈ G,
b becomes itself an abelian group, called the dual group.
then G
Lemma 1.3. Let G be a finite abelian group, and e : G → C\{0}. If e is
multiplicative, i.e.,
e(a · b) = e(a)e(b) for all a, b ∈ G,
then e is a character.
Useful lemma: any
multiplicative, nonvanishing map is a
character
2. The characters are what we want
Definition 2.1. Given a finite abelian group G, let V denote the |G|-dimensional (complex) vector space of functions f : G → C with inner product given by
1 X
(f, g) :=
f (a)g(a) for f, g, ∈ V.
|G|
a∈G
b is an orTheorem 2.2. With the above inner product, G
thonormal family.
1
The inner product
space of (continuous) functions.
FINITE ABELIAN GROUPS
2
b be a non-trivial character. Then
Lemma 2.3. Let e ∈ G
X
e(a) = 0.
a∈G
Proof. Since e is non-trivial, there exists an element b such
that e(b) 6= 1. Then
X
X
X
e(b)
e(a) =
e(ba) =
e(a),
a∈G
a∈G
a∈G
the last following since multiplying
by b is an invertible
P
(and thus 1-1) map. Thus a∈G e(a) = 0.
That the norms are
1 is obvious.
b (e, e) =
Proof of the theorem. Let’s first show that for e ∈ G,
1.
1 X
1 X
(e, e) :=
e(a)e(a) =
|e(a)|2 .
|G|
|G|
a∈G
a∈G
1
since e : G → S the first step is done.
Now it remains to be seen that (e, e0 ) = 0 for e 6= e0 ; that
is,
X
e(a)e0 (a) = 0.
a∈G
1
Well, in the group S , complex conjugation of an element
corresponds to taking the inverse of that element (since
zz = |z|2 = 1 for z ∈ S 1 ), so we can rewrite the above
equivalently as:
X
e(a)(e0 (a))−1 = 0.
a∈G
b - we
I.e. - expressing the above using the group law for G
want to show that
X
[e · (e0 )−1 ](a) = 0.
a∈G
Since e · (e0 )−1 is by construction (e =
6 e0 ) a non-trivial
character, the previous lemma finishes the proof.
Key lemma: showing the cancellation
property (moment
condition)
for
characters
FINITE ABELIAN GROUPS
3
Remark. Thus the characters are linearly independent elb ≤ G. In
ements of V . Since dim(V ) = |G|, we see |G|
fact....
b is an ONB of V
3. G
b is a
Theorem 3.1. Let G be a finite abelian group. G
(orthonormal) basis for the vector space of functions on G.
3.1. Some linear algebra: the Spectral Theorem.
unitary transformations
Definition 3.2. Let V be a d-dimensional inner product
space and T : V → V be a linear transformation. If, for all
v, w ∈ V , (T v, T w) = (v, w) then we call T unitary.
Theorem 3.3 (The Spectral Theorem). Given any unitary
transformation T : V → V , there exists a basis {v1 , . . . , vd }
of V of eigenvectors of T .
Corollary 3.4 (Simultaneous diagonalization). Let V be a
finite-dimensional inner product space, and {T1 , . . . , Tk } be
a family of linear transformations on V . If all {Ti } commute, then there exists a basis for V consisting of eigenvectors for every Ti .
Proof. By induction. Suppose the lemma is true for any
family of k − 1 commuting unitary transformations.
Consider the family {T1 , . . . , Tk }. By the spectral theorem, we can decompose V into a direct sum of eigenspaces
of the last transformation Tk ,
V = Vλ1 ⊕ · · · ⊕ Vλs .
Now, given any Tj , 1 ≤ j ≤ k − 1, we notice that for any
v ∈ Vλi , and any 1 ≤ i ≤ s,
Tk Tj (v) = Tj Tk (v) = Tj (λi v) = λi Tj (v);
That is, Tj (v) ∈ Vλi . In other words, each of the Tj preserves each of the eigenspaces.
The Spectral Theorem: unitary ⇒ diagonalizable
Consequence: simultaneous diagonalization of commuting
unitary transformations
(Base case is the
spectral theorem.)
Decompose V into
eigenspaces of Tk .
Notice (by
mutation)
Tj preserve
eigenspaces
comthat
those
4
FINITE ABELIAN GROUPS
Now, by induction hypothesis, the family {T1 , . . . , Tk−1 }
is simultaneously diagonalizable on each subspace Vλi ; i.e.,
each Vλi is decomposable into a direct sum of subspaces
which are eigenspaces for all of the {T1 , . . . , Tk−1 }. Those
eigenspaces are of course contained in Vλi , the eigenspace
for Tk ; so they provide a decomposition of V into eigenspaces
for the entire family.
Thus ends the linear algebra portion of this show.
Then
we
can
decompose
those
eigenspaces
(of
Tk )
into
further
eigenspaces
of
the
family
{T1 , . . . , Tk−1 }
LECTURE 32:
FOURIER ANALYSIS ON FINITE ABELIAN GROUPS
b is an ONB of V
1. G
b is a
Theorem 1.1. Let G be a finite abelian group. G
(orthonormal) basis for the vector space of functions on G.
Corollary 1.2 (Simultaneous diagonalization). Let V be
a finite-dimensional inner product space, and {T1 , . . . , Tk }
be a family of linear transformations on V . If all {Ti }
are unitary and commute, then there exists a basis for V
consisting of eigenvectors for every Ti .
Proof of theorem. Recall that dim(V ) = |G|, and that the
b of characters has order |G|
b ≤ |G|. Recall also
dual group G
that the characters form an orthonormal family; to show
b = |G|.
they form a basis, it suffices to show that |G|
Sketch: we consider the (finite) family of linear transformations (“left-translation”) on V which are not only unitary, but also commute (because G is abelian). We shall see
that the basis of V which simultaneously diagonalizes the
family consists (essentially) of characters: thus the number
of characters equals dim(V ) = |G|, as desired.
To each a ∈ G, associate a linear transformation Ta :
V → V defined by, for f ∈ V ,
(Ta f )(x) = f (a · x) for all x ∈ G.
With the inner product defined on V , we note that
1
Consequence: simultaneous diagonalization of commuting
unitary transformations
Need to check that
there are sufficient
characters to form a
basis of V
Create commuting,
unitary transformations: “translation”
operators
Why unitary: if we
shift both functions
over by the same
translation,
then
the inner product
doesn’t change.
2
b IS AN ORTHONORMAL BASIS
G
1 X
Ta f (b)Ta g(b)
|G|
b∈G
1 X
Ta f (b)Ta g(b)
=
|G|
b∈G
X
1
f (a · b)g(a · b)
=
|G|
b∈G
1 X
f (b)g(b) = (f, g);
=
|G|
(Ta f, Ta g) :=
b∈G
The previous corollary yields a basis (of
V ) of eigenfunctions.
In
fact,
those
|G| eigenfunctions,
properly normalized,
are characters; so
we have enough
characters to form a
basis.
i.e., the Ta are all unitary transformations. Further, they
commute (since G is abelian); so by the previous lemma,
they are simultaneously diagonalizable.
Thus we have a basis for V of functions {vb : G → C}b∈G ,
each of which is an eigenfunction for all of the {Ta }a∈G . We
now
Claim: For each of the {vb : G → C}b∈G , let
vb (x)
wb (x) :=
vb (u)
where u denotes the identity element of G. Then
wb is a character.
Remark. Why was this the obvious thing to do? Well, think
of what a character of G must do: it must, first, preserve
the group law:
χ(a · x) = χ(a)χ(x);
second, it must preserve the identity:
χ(uG ) = uS1 = 1.
The first statement is equivalent to saying that χ must be
an eigenfunction for the family of translations Ta (for all
a ∈ G); the second forces the normalization required of w
above. Can we get such eigenfunctions? Yes, since the Ta
are unitary and commute.
b IS AN ORTHONORMAL BASIS
G
3
Proof of claim: Let’s use v to denote one of the basis eigenfunctions, and w for the corresponding normalized
eigenfunction. We first notice that v(u) 6= 0: if v(u) = 0,
then
v(a) = v(a · u) = Ta v(u) = λa v(u) = 0
and then v(a) ≡ 0 for all a ∈ G. In fact, v can never vanish,
for if v(a) = 0, then
w never vanishes.
v(u) = v(a−1 a) = Ta−1 v(a) = λa−1 v(a) = 0
Similarly, w can never vanish.
It thus suffices to show that w : G → C\{0} is multiplicative; the lemma at the beginning of this lecture would
then imply that it must be a character. Well, letting λa
denote the eigenvalue of v for Ta , we observe
w is multiplicative.
v(a · b)
v(u)
Ta v(b) λa v(b) λa λb v(u)
=
=
=
v(u)
v(u)
v(u)
= λa λb = w(a)w(b),
w(a · b) :=
using the fact that
w(x) :=
v(x) Tx v(u)
=
= λx ;
v(u)
v(u)
so we’re done. Thus there are in fact dim(V ) = |G| characters, so they do form an ONB of V )
Once we have that the characters form an orthonormal
basis of V , our Fourier analysis of functions on G follows
as follows.
2. Fourier Analysis on Finite Abelian Groups
Definition 2.1. Let G be a finite abelian group. Given
b we define the
any function f : G → C and character e ∈ G,
Fourier coefficients
on (finite) abelian
groups
b IS AN ORTHONORMAL BASIS
G
4
Fourier series
Fourier
inversion:
just the statement
b is an ONB of
that G
V.
Fourier coefficient of f w.r.t. e as
1 X
fˆ(e) := (f, e) :=
f (a)e(a).
|G|
a∈G
P
Further, we define the Fourier series of f as e∈Gb fˆ(e)e.
Of course, since the characters form an ONB, we have
immediately that f actually equals its Fourier series:
Theorem 2.2 (Fourier inversion). Let G be a finite abelian
group. Then given any function f : G → C, we have that
X
f=
fˆ(e)e.
b
e∈G
As before we also have a Plancherel/Parseval theorem:
Theorem 2.3 (Plancherel/Parseval). Let G be a finite abelian
group. Then given any function f : G → C, we have that
X
||f ||2 =
|fˆ(e)|2 .
b
e∈G
Proof. Using Fourier inversion, we see that


X
X
2
ˆ

||f || := (f, f ) =
f (e)e,
fˆ()
b
e∈G
=
XX
b
∈G
fˆ(e)fˆ()(e, ) =
b ∈G
b
e∈G
X
|fˆ(e)|2 ,
b
e∈G
since the characters are orthonormal.
A result that shows
the ultimate power
of Fourier analysis
3. Dirichlet’s theorem
Theorem 3.1. Let q, ` ∈ N. If q and ` have no common
factor, then the sequence
`, ` + q, ` + 2q, . . . , ` + kq, . . .
contains infinitely many primes.
b IS AN ORTHONORMAL BASIS
G
5
4. Background knowledge:
The Fundamental Theorem of Arithmetic
Theorem 4.1 (Euclid’s algorithm). Let a, b ∈ Z; b > 0.
Then there exists unique integers q and r, with 0 ≤ r < b
such that a = qb + r.
Proof. Consider
S := {a − qb : q ∈ Z; a − qb ≥ 0}.
S is non-empty; let r = min S. Of course a = qb + r and
r ≥ 0; we need to show r < b.
If r ≥ b, then r = b + s for some s ≥ 0; so
b + s = a − qb,
and thus 0 ≤ s = a − (q + 1)b < a − qb = r.
But then s ∈ S and s < r; contradiction.
Proving uniqueness: suppose that a = qb + r = q1 b + r1
with 0 ≤ r, r1 < b. Then
(q − q1 )b = r1 − r.
Since |LHS| is (nonnegative) integral multiple of b, and
|RHS| < b, |LHS| = |RHS| = 0b = 0.
The basis of long division.
Slight error in text:
need not assume s <
r.
LECTURE 33:
ELEMENTARY NUMBER THEORY
1. Background knowledge:
The Fundamental Theorem of Arithmetic
Theorem 1.1 (Euclid’s algorithm). Let a, b ∈ Z; b > 0.
Then there exists unique integers q and r, with 0 ≤ r < b
such that a = qb + r.
Proof. Consider
S := {a − qb : q ∈ Z; a − qb ≥ 0}.
S is non-empty; let r = min S. Of course a = qb + r and
r ≥ 0; we need to show r < b.
If r ≥ b, then r = b + s for some s ≥ 0; so
b + s = a − qb,
and thus 0 ≤ s = a − (q + 1)b < a − qb = r.
But then s ∈ S and s < r; contradiction.
Proving uniqueness: suppose that a = qb + r = q1 b + r1
with 0 ≤ r, r1 < b. Then
(q − q1 )b = r1 − r.
Since |LHS| is (nonnegative) integral multiple of b, and
|RHS| < b, |LHS| = |RHS| = 0b = 0.
Definitions. Let a, b ∈ Z.
i. If there exists c ∈ Z such that ac = b, then we say a
divides b (a is a divisor of b) and write a|b.
ii. A prime number is a positive integer, greater than 1,
which has no divisors besides 1 and itself.
iii. Say a, b ∈ N. The greatest common divisor gcd(a, b) of
a and b is the largest integer that divides both a and b.
iv. If gcd(a, b) = 1, we say a and b are relatively prime
1
The basis of long division.
Slight error in text:
need not assume s <
r.
Elementary concepts
from number theory
(i.e., they have no
nontrivial common
divisor)
2
gcd(a, b) is in the
(Z)-span of {a, b}.
Let’s show that s|a.
We want to show
that r = 0.
ELEMENTARY NUMBER THEORY
Theorem 1.2. Let a, b, ∈ N. Then there exist x, y ∈ Z
such that
xa + yb = gcd(a, b).
Proof. Consider the (positive) span of a and b:
S := {xa + yb : x, y ∈ Z; xa + yb > 0}.
Let s = min S; then in fact we claim that s = gcd(a, b).
Why? Well, obviously, any divisor of both a and b will
divide s; so s ≥ gcd(a, b). So STS s|a and s|b.
By Euclid’s algorithm,
a = qs + r with 0 ≤ r < s.
Let’s think about qs. Since s ∈ S,
xa + yb = s for some x, y ∈ Z
and thus qxa + qyb = qs.
Together, that gives us
qax + qby = a − r;
i.e. r = (1 − qx)a + (−qy)b;
in other words, 0 ≤ s − r ≤ s and s − r ∈ S. Since s =
min S, this forces r = 0. Similarly, s|b; so s = gcd(a, b). Corollary 1.3. Let a, b ∈ N. a and b are relatively prime
if and only if there exist x, y ∈ Z such that ax + by = 1.
Characterization of
relative prime-ness
Proof. Obvious: if relatively prime, then the theorem implies there exist such integers. If such integers exist, then
any divisor of both a and b must divide 1; thus the only
divisor is 1.
Corollary 1.4. Let a, c be relatively prime. If c|ab, then
c|b.
Useful property of
relatively
prime
numbers
Proof. gcd(a, c) = 1 means there exist x, y ∈ Z such that
xa+yc = 1. Thus xab+ycb = b, and since c|ab and (clearly)
c|ycb, c also divides b.
ELEMENTARY NUMBER THEORY
3
Corollary 1.5. Let p be prime. If p|a1 · · · ar , then p|ai for
some i.
Corollary of the previous corollary
Proof. Suppose p - a1 . Then p|a2 · · · ar . Eventually p must
divide one of the ai .
Theorem 1.6 (Fundamental Theorem of Arithmetic). Every positive integer greater than 1 can be factored uniquely
into a product of primes.
hilarious proof
Proof. First we note that all positive integers have prime
factorizations. For consider the set of positive integers without prime factorizations; and let n = min S. Since n is not
a prime, n = ab for some integers 1 < a, b < n. Since n
was minimal, a, b both have prime factorizations; and so n
does: ※.
Now suppose n has two prime factorizations, n = p1 p2 · · · pr
and n = q1 q2 · · · qs . Since p1 |n = q1 q2 · · · qs , we know p1
must divide one of the qi (relabeling, call it q1 ); thus p1 = q1 .
In this manner we see that r = s and that the factorizations
are identical up to permutation.
2. Euclid’s proof of the infinitude of primes
Theorem 2.1. There are infinitely many primes
Proof. By contradiction: suppose there are finitely many,
which we
Qn label {p1 , . . . , pn }. Then consider the number
N := ( i=1 pi ) + 1. It is larger than any pi , so must be
non-prime; thus has a prime factorization.
Qn However, if the
prime pk divides N , then it divides N −( i=1 pi ) = 1, which
is impossible.
Notice that we do
need the Fundamental Theorem of
Arithmetic for this
proof.
Corollary 2.2. The number of primes of the form 4k + 3
is infinite.
Variation
theme
on
the
4
ELEMENTARY NUMBER THEORY
Proof. Suppose there are only finitely many such primes:
{p1 = 7, p2 = 11, . . . , pn } (leave 3 out of the list for now).
Let
N = 4p1 p2 · · · pn + 3.
Observe the odd
primes are either of
the form 4k + 1 or
4k + 3
Since N is of the form 4k +3, and further, N > pn , we know
N cannot be prime, and thus has a prime factorization (not
containing 2, since N is odd). Observe that (4m + 1)(4n +
1) = 4(mn + m + n) + 1; so N ’s prime factors must include
a prime of the form 4k + 3. But, just as in the previous
argument, none of the primes in our finite list divide evenly
into N .
3. One result about infinite products
infinite product
∞
Definition
Q 3.1. Given {An ∈ R}n=1 , we define the infinite
product An by
∞
Y
An = lim
N →∞
n=1
condition implying
convergence
of
product
Proof of lemma is
via power series expansion of ln(1 + x)
around 0
An .
n=1
The main result we shall need is the following:
P
Theorem 3.2.
Let
A
=
1
+
a
.
If
an converges abson
n
Q
lutely, then An converges.
Lemma 3.3. For |x| < 21 , | ln(1 + x)| ≤ 2|x|.
P
Proof of theorem. Since
an converges, WLOG we may
1
assume |an | < 2 for all n. Then
N
Y
n=1
An :=
N
Y
n=1
PN
=e
Here’s
where
we use absolute
convergence
N
Y
eln(1+an )
n=1
ln(1+an )
.
By the lemma, | ln(1 + an )| ≤ 2|an |; so
PN
n=1 ln(1
+ an )
ELEMENTARY NUMBER THEORY
5
converges to a limit, B. Thus, by continuity of the exponential function,
N
Y
PN
lim
An = lim e n=1 ln(1+an ) = eB .
N →∞
n=1
N →∞
Remarks.
i. Notice that if the product vanishes, then one of the
initial factors (i.e., before |an | < 1/2) must have been
0, since the infinite tail is eB 6= 0.
ii. Notice that if in addition an 6= 1 for all n, then
N
Y 1
Y
1
1
= lim
= lim QN
N
N
1 − an
1 − an
n (1 − an )
n
n
also converges.
LECTURE 34:
EULER’S PRODUCT FORMULA; FOURIER SERIES
FOR CHARACTERISTIC FUNCTIONS OF POINTS
1. Euler’s product formula for the Riemann
zeta function
Definition 1.1. Let s > 1. We define the zeta function by
zeta function
∞
X
1
ζ(s) :=
.
s
n
n=1
Remark. That this series converges can be seen via the integral test:
Z ∞
∞
∞ Z n
X
X
1
dx
dx
1
≤
1
+
=
1
+
=
1
+
.
s
s
s
n
x
x
s
−
1
n−1
1
n=1
n=2
Theorem 1.2 (Euler’s product formula). Let s > 1, and
let ℘ denote the collection of primes. Then
Y
1
ζ(s) =
.
s
1
−
1/p
p∈℘
Remark. The intuition: each of the terms in the product
can be realized as a geometric series:
1
1
1
=
1
+
+
+ ··· ;
ps p2s
1 − p1s
Y
Y
1
1
1
so
=
1 + s + 2s + · · ·
s
1
−
1/p
p
p
p∈℘
p
1
Converges because
of the integral test
Analytic
version
of the fundamental
theorem
of
arithmetic
2
DIRICHLET CHARACTERS AND REDUCTION OF THEOREM
Now, by the fundamental theorem of arithmetic, each n ∈
N can be written as a product n = pk11 pk22 · · · pkmm , so
1
1
=
.
ns
(pk11 pk22 · · · pkmm )s
“Thus” (at least formally)
∞
Y
X
1
1
1
=
1 + s + 2s + · · ·
ζ(s) :=
s
n
p
p
p
n=1
Proof. Let N ∈ N, and consider
N
X
1
n=1
ns
.
Now, pick any M > N . Each integer n = 1, 2, . . . , N has
a prime factorization in terms of primes p < N , where p
occurs fewer than M times in that factorization. Thus
N
X
Y
1
1
1
1
≤
1 + s + 2s + · · · + M s
s
n
p
p
p
n=1
p≤N
Y 1 Y 1 ≤
≤
1 − p−s
1 − p−s
p
p≤N
Letting N → ∞ gives
ζ(s) ≤
Y
p
1
1 − p−s
.
Conversely, by the fundamental theorem of arithmetic, we
have for any N ,
X
∞
Y
1
1
1
1
1 + s + 2s + · · · + M s ≤
p
p
p
ns
n=1
p≤N
Thus, letting M → ∞,
Y
p≤N
1
1 − p−s
∞
X
1
≤
ns
n=1
DIRICHLET CHARACTERS AND REDUCTION OF THEOREM
3
and thus, letting N → ∞,
∞
Y 1 X
1
.
≤
−s
s
1
−
p
n
p
n=1
Theorem 1.3 (Euler’s analytic version of the infinitude of
primes).
X1
diverges.
The series
p
p∈℘
Take that, Euclid!
Lemma 1.4. ln(1 + x) = x + E(x) where |E(x)| ≤ x2 for
|x| < 1/2.
Proof of lemma. Using the power series expansion,
E(x) = ln(1 + x) − x = −
x2 x3 x4
+
−
+ −···
2
3
4
x2
so |E(x)| ≤ (1 + |x| + |x|2 + · · · )
2 x2
1
1
≤
1 + + 2 + · · · = x2
2
2 2
for |x| ≤ 12 .
Proof of theorem. By Euler’s formula
Y 1 ζ(s) =
, so
−s
1
−
p
p∈℘
X 1
ln ζ(s) = −
ln 1 − s
p
p∈℘
By the lemma,
Slight error in book:
big O mistake
4
DIRICHLET CHARACTERS AND REDUCTION OF THEOREM
X 1
1
− s +E
ln ζ(s) = −
p
ps
p∈℘
X 1
+ C,
=
s
p
p∈℘
since
X
E
p∈℘
1
ps
∞
X 1
X
1
π2
≤
≤
= .
2s
2
p
n
6
p∈℘
n=1
I.e.,
X 1
= ln ζ(s) − C.
s
p
p∈℘
Then consider:
∞
X
1
lim inf
ζ(s)
:=
lim
inf
s→1+
s→1+
ns
n=1
M
M
X
X
1
1
≥ lim inf
=
,
s
s→1+
n
n
n=1
n=1
for every M ∈ N. So then
X 1
≥ lim inf
ln ζ(s) = ∞
lim inf
s
s→1+
s→1+
p
p∈℘
P
P
and thus p∈℘ p1 ≥ lim inf s→1+ p∈℘ p1s = ∞ as well.
2. Dirichlet characters and
the reduction of the theorem
The above approach showing that there are infinitely
many primes was extended by Dirichlet to show that there
are an infinitude of primes of the form p ≡ ` mod q. It
suffices to show that
X
1
diverges.
p
p∈℘: p≡` mod q
DIRICHLET CHARACTERS AND REDUCTION OF THEOREM
5
The proof will involve using the Fourier series for functions
on Z∗ (q), the units (invertible elements) of Z(q).
Question. What is Z∗ (q)? a ∈ Z(q) being invertible means
of course that there exists x ∈ Z(q) such that xa ≡ 1
mod q, i.e., that gcd(a, q) = 1. In other words, Z∗ (q) is
precisely the set of (equivalence classes of) numbers relatively prime to q.
Definition 2.1. Define the Euler ϕ-function by
ϕ(q) := | Z∗ (q)|.
Let δ` : Z∗ (q) → R denote the characteristic function χ{`}
of the singleton ` ∈ Z∗ (q), i.e.,
1 if n ≡ ` mod q
δ` (n) :=
0 otherwise.
Dumb comment: we extend δ` to all of Z(q) (and thus all
of Z) by defining it as 0 on the non-units of Z(q), i.e.,
δ` (n) =
χ{`} ([n])
0
minor notation: Euler phi function
Characteristic functions of singletons in
Z∗ (q):
Extending δ` to all of
Z - this is just a periodic extension
if n and q are relatively prime
otherwise.
b be a character. We define the
Definition 2.2. Let e ∈ G
Dirichlet character modulo q extending e, denoted χ = χ(e) :
Z → C by
e([m]) if m and q are relatively prime
χ(m) :=
0
otherwise.
We denote the extension of the trivial character of G by χ0 .
Remark. Observe that the Dirichlet characters are multiplicative on all of Z.
Lemma 2.3.
X
1
δ` (n) :=
χ(`)χ(n).
ϕ(q)
b
χ(e) :e∈G
Extending
characters to all of Z,
again periodically
Expressing δ` in
terms of characters
6
DIRICHLET CHARACTERS AND REDUCTION OF THEOREM
Proof. Consider the Fourier coefficients on G of δ` :
1 X
1
b
δ` (e) :=
δ` (m)e(m) =
e(`).
|G|
|G|
m∈G
Fourier
series
of
characteristic
function
of
a
singleton
Thus the Fourier inversion formula yields
X
δ` (n) :=
δb` (e)e(n)
b
e∈G
=
1 X
e(`)e(n).
|G|
b
e∈G
Rewriting the above in terms of the extensions of the functions to Z, we have the desired result.
LECTURE 35:
EULER’S PRODUCT FORMULA; FOURIER SERIES
FOR CHARACTERISTIC FUNCTIONS OF POINTS
1. Reduction of the problem
Recall: our goal is to show Dirichlet’s theorem, i.e., that
for q, ` which are relatively prime,
X
1
diverges.
p
p∈℘: p≡` mod q
We reduced the problem as follows. First, we took the
characteristic function δ` of {`} as a function on Z∗ (q), and
∗ (q). By our work in the previous
the characters e ∈ Z[
chapter (recall by definition ϕ(q) := | Z∗ (q)|), we have the
following Fourier series expansion:
Lemma 1.1.
δ` (n) =
1 X
e(`)e(n).
ϕ(q)
b
e∈G
We then extended the definitions of δ` and e to all of Z
periodically (defining them as 0 whenever n was not relatively prime with q), and letting χ denote the extension of e
to Z. Then by the above inversion formula, since δ` (p) = 1
iff p ≡ ` mod q,
X
X δ` (p)
1
=:
ps
ps
p∈℘
p∈℘ s.t. p≡` mod q
X χ(p)
1 X
=
χ(`)
.
s
ϕ(q)
p
p∈℘
b
χe :e∈G
Breaking the first sum into the (Dirichlet extension of the)
trivial character and the non-trivial ones, we get that the
1
Perhaps the best
way is to present
this
backwards,
i.e.,
realize the
sum in terms of
the
characteristic
function which can
be thought of
as a function on
Z∗ (q) and as thus
possessing a Fourier
series expansion.
2
DIRICHLET CHARACTERS AND REDUCTION OF THEOREM
above equals
X χ(p)
1 X χ0 (p)
1 X
=
+
χ(`)
ϕ(q) p∈℘ ps
ϕ(q)
ps
p∈℘
χe 6=χ0
=
X χ(p)
1 X 1
1 X
+
χ(`)
,
s
ϕ(q)
ps ϕ(q)
p
p∈℘
χe 6=χ0
p∈℘: p-q
the last because χ0 (p) is the extension of the trivial character (taking 1 on all of Z∗ (q)); i.e., it indicates whether or
not p ∈ ℘ is relatively prime with q, which is equilvalent to
saying that p does not divide q.
Now let’s consider the first sum,
X 1
.
ps
p∈℘:p-q
Snce
but finitely many primes do not divide q, and since
P all
1
p∈℘ p diverges (Euler’s theorem), we see that this sum
diverges as s → 1+ .
Thus to show that the sum diverges, it suffices to show
that
X χ(p)
1 X
, < ∞,
χ(`)
s
ϕ(q)
p
p∈℘
χe 6=χ0
as s ↓ 1, which we shall do by proving the following.
Theorem 1.2. Let χ be any nontrivial Dirichlet character.
Then
X
χ(p)
ps
p∈℘:p≡` mod q
is bounded as s → 1+ .
2. The strategy
The key to proving the reduced problem will be the following.
DIRICHLET CHARACTERS AND REDUCTION OF THEOREM
3
Definition 2.1. For s > 1, χ a Dirichlet character (modulo
q), we define the L-function L by
∞
X
χ(n)
L(s, χ) :=
.
s
n
n=1
Theorem 2.2 (Product formula for L-functions). For s >
1,
∞
X
χ(n) Y
1
L(s, χ) :=
.
=
s
χ(p)
n
1
−
s
p∈℘
n=1
p
Then, formally,
χ(p)
ln L(s, χ) = −
ln 1 − s
p
p∈℘
X
χ(p)
1
=−
− s +E
p
ps
p∈℘
X χ(p)
=
+ C;
s
p
p∈℘
P
we use this relation to show that lims↓1 p∈℘ χ(p)
ps is finite.
To make the above argument rigorous, we need to do the
following.
i. Extend the logarithm to complex numbers,
ii. Interpret properly the (complex, multi-valued) logarithm of the product,
iii. Show that L(s, χ) is continuous at s = 1, and
iv. Show that L(1, χ) 6= 0.
X
Again, |C| ≤
π2
6 .
3. Logarithms and infinite products on C
Question. How do we extend the logarithm to C?
Recall that for |x| < 1 one has the power series
∞
X
(−1)n+1 n
x
log(1 + x) =
n
n=1
Via power series expansion of the logarithm.
4
DIRICHLET CHARACTERS AND REDUCTION OF THEOREM
which implies that (again for |x| < 1)
X
∞
1
1 n
log
=
x .
1−x
n
n=1
Extension of the
natural
logarithm
to....somewhere.
Definition 3.1. For z ∈ C, |z| < 1, we define
∞
X
zk
1
:=
.
log1
1−z
k
k=1
Question. Consider {w =
of points?
Question: For what
set of points are we
actually defining the
log?
Claim: {w =
1
1−z
1
1−z
: |z| < 1}. What is this set
: |z| < 1} = {w ∈ C : <(w) > 1/2}.
Proof. ⊂:
1
1 + re−iθ
1
w=
·
=
1−z
1 − reiθ 1 + re−iθ
1 + re−iθ
=
1 − r2 r
1
+
eiθ
=
2
2
1−r
1−r
1
r
r
=
+
cos θ +
i sin θ
1 − r2
1 − r2
1 − r2
Considering the real part of the above, we see that
1−r
1
<(w) ≥
=
;
1 − r2
1+r
since 0 ≤ r < 1, <(w) > 1/2.
1
⊃: Conversely, if <(w) > 1/2, and we write w = 1−z
, we
see that z = w−1
w . Thus
(w − 1)(w − 1)
|z|2 = zz =
ww
1 − (w + w)
=1+
.
ww
Since w + w = 2<(w) > 1, the second term above is negative; so 0 ≤ |z| < 1.
DIRICHLET CHARACTERS AND REDUCTION OF THEOREM
5
Remark. So our above definition is a generalization of the
natural logarithm for x > 1/2.
Question. What happens if we extend the logarithm in a
different way? Do we get the same function?
Proposition 3.2 (Properties of log1 ).
i. If |z| < 1, then
elog1 ( 1−z ) =
1
What
properties
does
this
new
logarithm have?
1
.
1−z
ii. If |z| < 1, then
log1
1
1−z
= z + E1 (z),
where |E1 (z)| ≤ |z|2 for |z| ≤ 1/2.
iii. If |z| < 1/2, then
1
log1
≤ 2|z|.
1−z
Proof of (i). Using polar coordinates, let z = reiθ (0 ≤ r <
1). We want to show that
P∞
(1 − reiθ )e
k=1
(reiθ )k
k
= 1.
Well, when r = 0, the left hand side does equal 1, so it
d
(LHS) = 0. But this is easy (using
suffices to show that dr
the fact that we can differentiate term-by-term in the disk
of convergence):
!0
∞
iθ k
iθ )k
X
P
P∞ (reiθ )k
∞
(re
(re )
d
(LHS) = (1 − reiθ )e k=1 k ·
+ (−eiθ )e k=1 k
dr
k
k=1
!0 #
"
∞
iθ k
X
P∞ (reiθ )k
(re )
= −eiθ + (1 − reiθ )
e k=1 k
k
k=1
6
DIRICHLET CHARACTERS AND REDUCTION OF THEOREM
Differentiating term-by-term, we see that
"
!0 #
∞
iθ k
X
(re
)
−eiθ + (1 − reiθ )
k
= −eiθ + (1 − reiθ )e
k=1
∞
X
iθ
(reiθ )k−1
k=1
= −eiθ + (1 − reiθ )eiθ
With the logarithm
in hand, we can now
prove an analogous
theorem for infinite
products of complex
numbers.
The proof is completely the same; no
point in showing it.
1
= 0.
1 − reiθ
P
Proposition 3.3 (Condition forP
convergence). Let an be
a series of complex numbers. If
an converges absolutely,
and an 6= 1 for all n, then
∞ Y
1
converges.
1
−
a
n
n=1
Proof. As before, WLOG assume |an | < 1/2 for all n. Using
the properties of the log1 , and then the exponential function
(note that we use the complex exponential function from
chapter 1 as well)
Y
N N
Y
1
1
=
elog1 ( 1−an )
1 − an
n=1
n=1
PN
1
= e n=1 log1 ( 1−an ) .
P
Just like in the earlier proof, we note that since
|an |
converges, and since
1
log1
≤ 2|an |,
1 − an
the sum in the exponent converges, and thus the limit exists.
LECTURE 35:
EULER’S PRODUCT FORMULA; FOURIER SERIES
FOR CHARACTERISTIC FUNCTIONS OF POINTS
1. Reduction of the problem
Recall: our goal is to show Dirichlet’s theorem, i.e., that
for q, ` which are relatively prime,
X
1
diverges.
p
p∈℘: p≡` mod q
We reduced the problem as follows. First, we took the
characteristic function δ` of {`} as a function on Z∗ (q), and
∗ (q). By our work in the previous
the characters e ∈ Z[
chapter (recall by definition ϕ(q) := | Z∗ (q)|), we have the
following Fourier series expansion:
As before
P 1 we’ll consider
ps as s ↓ 1.
Lemma 1.1.
1 X
δ` (n) =
e(`)e(n).
ϕ(q)
b
e∈G
We then extended the definitions of δ` and e to all of Z
periodically (defining them as 0 whenever n was not relatively prime with q), and letting χ denote the extension of
e to Z. Then since δ` (p) = 1 iff p ≡ ` mod q,
X
X δ` (p)
1
=:
ps
ps
p∈℘
p∈℘ s.t. p≡` mod q
X χ(p)
1 X
=
χ(`)
.
s
ϕ(q)
p
p∈℘
Perhaps the best way
is to present this
backwards, i.e., realize the sum in terms
of the characteristic
function which can
be realized as a
function on Z∗ (q)
and as thus possessing a Fourier series
expansion.
b
χe :e∈G
the second equality being the above Fourier series expansion. Breaking the first sum into the (Dirichlet extension of
1
Step 1: Fourier series expansion
2
Step 2: break sum
into trivial and nontrivial characters
DIRICHLET CHARACTERS AND REDUCTION OF THEOREM
the) trivial character and the non-trivial ones, we get that
the above equals
X χ(p)
1 X χ0 (p)
1 X
χ(`)
=
+
ϕ(q) p∈℘ ps
ϕ(q)
ps
p∈℘
χe 6=χ0
=
X χ(p)
1 X
1 X 1
χ(`)
+
,
s
ϕ(q)
ps ϕ(q)
p
p∈℘
χe 6=χ0
p∈℘: p-q
What’s χ0 ?
It
will take the value
1 at numbers relatively prime to q;
that is, primes that
do not divide q.
Observe that almost
all primes do not divide q
the last because χ0 (p) is the extension of the trivial character (taking 1 on all of Z∗ (q)); i.e., it indicates whether or
not p ∈ ℘ is relatively prime with q, which is equilvalent to
saying that p does not divide q.
Now let’s consider the first sum,
X 1
.
ps
p∈℘:p-q
Since P
all but finitely many primes do not divide q, and
since p∈℘ p1 diverges (Euler’s theorem), we see that this
sum diverges as s → 1+ .
Thus to show that the sum diverges, it suffices to show
that
X χ(p)
1 X
, < ∞,
χ(`)
s
ϕ(q)
p
p∈℘
χe 6=χ0
as s ↓ 1, which we shall do by proving the following.
Reduction
problem
of
the
Theorem 1.2. Let χ be any nontrivial Dirichlet character.
Then
X
χ(p)
p∈℘:p≡` mod q
is bounded as s → 1+ .
ps
DIRICHLET CHARACTERS AND REDUCTION OF THEOREM
Dirichlet L-function
(Generalization of
zeta function)
3
2. Dirichlet L-functions; Product formula
Definition 2.1. For s > 1, χ a Dirichlet character (modulo
q), we define the L-function by
∞
X
χ(n)
.
L(s, χ) :=
s
n
n=1
The key to proving the reduced problem will be the following product formula, which will play a role analogous to
that of Euler’s product formula for the zeta function.
Statement of product
formula
Theorem 2.2 (Product formula for L-functions). For s >
1,
∞
X
χ(n) Y
1
L(s, χ) :=
=
.
s
χ(p)
n
p∈℘ 1 − s
n=1
p
To accomplish this theorem will involve extending the notions of logarithm and infinite products to complex numbers.
What we need isQnot
much: logs and .
3. Logarithms and infinite products on C
Question. How do we extend the logarithm to C?
Recall that for |x| < 1 one has the power series
∞
X
(−1)n+1 n
log(1 + x) =
x
n
n=1
Motivation: power
series expansion of
the logarithm.
which implies that (again for |x| < 1)
∞
X
1
1 n
x .
log
= − log(1 − x) =
1−x
n
n=1
Definition 3.1. For z ∈ C, |z| < 1, we define
∞
X
1
zk
log1
:=
.
1−z
k
k=1
Extension of the
natural
logarithm
to....somewhere.
4
Question: For what
set of points are we
actually defining the
log?
Some “boring” calculations (actually
they were slightly
fun).
DIRICHLET CHARACTERS AND REDUCTION OF THEOREM
Question. Consider {w =
of points?
Claim: {w =
1
1−z
1
1−z
: |z| < 1}. What is this set
: |z| < 1} = {w ∈ C : <(w) > 1/2}.
Proof. ⊂:
1
1
1 + re−iθ
w=
=
·
1−z
1 − reiθ 1 + re−iθ
1 + re−iθ
=
1 − r2 1
r
=
+
eiθ
2
2
1−r
1−r
1
r
r
=
+
cos θ +
i sin θ
1 − r2
1 − r2
1 − r2
Considering the real part of the above, we see that
1
1−r
=
;
<(w) ≥
1 − r2
1+r
since 0 ≤ r < 1, <(w) > 1/2.
1
⊃: Conversely, if <(w) > 1/2, and we write w = 1−z
, we
w−1
see that z = w . Thus
(w − 1)(w − 1)
|z|2 = zz =
ww
1 − (w + w)
.
=1+
ww
Since w + w = 2<(w) > 1, the second term above is negative; so 0 ≤ |z| < 1.
Remark. So our above definition is a generalization of the
natural logarithm for x > 1/2.
multi-valued
logarithm
Question. What happens if we extend the logarithm in a
different way? Do we get the same function? Consider
the complex exponential function. Since ew+2kπi = ew , we
see that e is not a 1-1 function; thus there are multiple
logarithms possible: log1 is one of them.
DIRICHLET CHARACTERS AND REDUCTION OF THEOREM
5
Does this logarithm
behave as it ought
to? What properties
does this new logarithm have?
Proposition 3.2 (Properties of log1 ).
i. If |z| < 1, then
elog1 ( 1−z ) =
1
1
.
1−z
ii. If |z| < 1, then
log1
1
1−z
= z + E1 (z),
where |E1 (z)| ≤ |z|2 for |z| ≤ 1/2.
iii. If |z| < 1/2, then
1
log1
≤ 2|z|.
1−z
Proof of (i). (The other two facts follow as before.)
Using polar coordinates, let z = reiθ (0 ≤ r < 1). We
want to show that
iθ
P∞
(1 − re )e
Note
that
we
can’t
differentiate
term-by-term when
|z| ≥ 1: otherwise,
this would be true
for all z.
k=1
(reiθ )k
k
Use polar coordinates; see that
f (r) = 1, f 0 (r) = 0,
so f (r) ≡ 1.
= 1.
Well, when r = 0, the left hand side does equal 1, so it
d
(LHS) = 0. But this is easy (using
suffices to show that dr
the fact that we can differentiate term-by-term in the disk
of convergence):
d
(LHS) = (1 − reiθ )e
dr
"
(reiθ )k
k=1
k
P∞
iθ
·
∞
X
(reiθ )k
= −e + (1 − re )
∞
X
(reiθ )k
k=1
k
P∞
+ (−eiθ )e
k
!0 #
k=1
iθ
!0
e
P∞
k=1
(reiθ )k
k
k=1
(reiθ )k
k
6
DIRICHLET CHARACTERS AND REDUCTION OF THEOREM
Differentiating term-by-term, we see that
"
!0 #
∞
iθ k
X
(re
)
−eiθ + (1 − reiθ )
k
= −eiθ + (1 − reiθ )e
k=1
∞
X
iθ
(reiθ )k−1
k=1
= −eiθ + (1 − reiθ )eiθ
With the logarithm
in hand, we can now
prove an analogous
theorem for infinite
products of complex
numbers.
The proof is completely the same;
almost no point in
showing it.
1
= 0.
1 − reiθ
P
Proposition 3.3 (Condition forP
convergence). Let an be
a series of complex numbers. If
an converges absolutely,
and an 6= 1 for all n, then
∞ Y
1
converges.
1
−
a
n
n=1
Proof. As before, WLOG assume |an | < 1/2 for all n. Using
the properties of the log1 , and then the exponential function
(note that we use the complex exponential function from
chapter 1 as well)
Y
N N
Y
1
1
=
elog1 ( 1−an )
1 − an
n=1
n=1
PN
1
= e n=1 log1 ( 1−an ) .
P
Just like in the earlier proof, we note that since
|an |
converges, and since
1
log1
≤ 2|an |,
1 − an
the sum in the exponent converges, and thus the limit exists.
LECTURE 35: PRODUCT FORMULA FOR THE
L-FUNCTIONS
1. Proof of the product formula
Recall: we hope to obtain a product formula for L-functions
analogous to that for the zeta function, i.e.:
Theorem 1.1. For s > 1,
L(s, χ) :=
∞
X
χ(n)
ns
n=1
Y
=
1
p∈℘ 1 −
χ(p)
ps
.
Proof. P
First observe both sides converge: the left since |χ| ≤
1
1 and ∞
n=1 ns converges for s > 1, and the right side converges by the previous proposition (on the convergence of
infinite products), since (letting pn denote the n-th prime)
for s > 1
∞
X
χ(pn )
n=1
psn
converges absolutely,
again in comparison with
P∞
1
n=1 ns .
Now, we want to show that
∞
X
χ(n)
n=1
Approximate
sides with
sums
both
finite
ns
−
Y
p∈℘
1
1−
χ(p)
ps
=0
i.e., that the difference is smaller than any > 0. Well, by
the triangle inequality, for any N, M ,
1
Proof of product formula
PROOF OF DIRICHLET’S THEOREM
2
∞
X
χ(n)
n=1
∞
X
n=1
ns
−
Y
1
p∈℘ 1 −
χ(p)
ps
≤
χ(n) X χ(n)
+
−
ns
ns
n≤N
X χ(n) Y χ(p)
χ(pM )
+
−
1 + s + · · · + Ms
ns
p
p
n≤N
Y
p≤N
Y
p≤N
p≤N
M
χ(p)
χ(p )
+
·
·
·
+
−
ps
pM s
p≤N
!
Y
1
1
−
χ(p)
1 − χ(p)
p∈℘ 1 − ps
ps
1+
1
Y
1−
χ(p)
ps
!
+
=: |L − SN | + |SN − ΠN,M | + |ΠN,M − ΠN | + |ΠN − Π|
= I + II + III + IV.
Fix > 0. (I and IV:) By convergence, we can choose N
such that for all numbers greater,
|SN − L| < and |ΠN − Π| < .
Next we claim that we can choose M large enough that
We can control
first and fourth
ferences: those
just the finite
proximations
the
difare
ap-
II: |SN − ΠN,M | < and
III: |ΠN,M − ΠN | < .
III is clear since, using the multiplicativity of the Dirichlet
characters,
Third difference
obvious. χ(p)
χ(pM )
χ(p)
[χ(p)]M
lim 1 + s + · · · + M s
= lim 1 + s + · · · +
M →∞
M →∞
p
p
p
pM s
∞
X
χ(p)n
1
=
=
ns
p
1 − χ(p)
n=1
ps
is
PROOF OF DIRICHLET’S THEOREM
3
The only non-trivial inequality is II.
X χ(n) Y χ(p)
χ(pM )
|SN − ΠN,M | :=
1 + s + · · · + Ms
−
;
ns
p
p
n≤N
p≤N
Of course, for n ≤ N , n’s prime factorization is composed of
primes p ≤ N ; further, there cannot be more than N primes
involved. So take M > N . Then, using multiplicativity,
we see that the second term contains all thePterms in the
first. The sum of the terms not contained in n≤N χ(n)
ns are
P
contained in n>N χ(n)
ns , which has magnitude smaller than
by our choice of N .
Proposition 1.2 (Corollary of the the product formulae).
Recall the trivial Dirichlet character modulo q:
1 if n and q are relatively prime
χ0 (n) :=
0 otherwise.
Minor observation
along
the
way:
L(s, χ0 ) is basically
ζ(s)!
If q = pa11 · · · paNN is the prime factorization of q, then
−s
−s
L(s, χ0 ) = (1 − p−s
1 )(1 − p2 ) · (1 − pN )ζ(s).
Proof. By the product formula just proven,
X χ0 (n) Y
1
=
χ0 (p)
ns
s
p∈℘ 1 −
n
Proof: use the product formulae.
and by Euler’s product formula for ζ,
Y 1
ζ(s) =
1 − p1s
p∈℘
χ0 (p) = 0 when p|q.
p
The terms that are missing are those which correspond to
the primes that divide q; i.e., p1 , . . . , pN .
2. Logarithms of L-functions
(Introduce “second reduction” as “corollary” here.)
Proposition 2.1 (The technical proposition). Let χ be a
non-trivial Dirichlet character. Then
We’d like to be able
to take the log of
both sides of the
above product formula. But we’ll need
to do a lot of work
first....
The technical proposition: allowing us to
define the log of L.
Where do we need
0 < s < 1?
PROOF OF DIRICHLET’S THEOREM
4
i.
L(s, χ) :=
∞
X
χ(n)
n=1
Obtain controls on
the growth of L and
d
ds L
Simple but crucial
observation
in demonstrating
absolute and uniform convergence
in s > 0
ns
converges for s > 0.
ii. L(s, χ) is continuously differentiable in s.
iii. For some constants c, c0 > 0, we have
L(s, χ) = 1 + O(e−cs ) as s → ∞
d
0
L(s, χ) = O(e−c s ) as s → ∞.
ds
Lemma 2.2 (The key lemma). If χ is a non-trivial Dirichlet character modulo q, then
k
X
χ(n) ≤ q for any k.
n=1
Proof. First recall that since χ is non-trivial,
q
X
χ(n) = 0.
n=1
Q. Why do they
prove this again?
Using the Euclidean algorithm, write k = aq + b with
0 ≤ b < q; then
aq
aq+b
aq+b
k
X
X
X
X
χ(n) =
+
χ(n) =
χ(n).
n=1
Characters map into
S 1 ; Dirichlet characters can equal 0.
n=1
n=aq+1
n=aq+1
By the triangle inequality (since |χ(n)| ∈ {0, 1}) the last
term is less than or equal to q.
3. Proof of the technical proposition
Proof of the technical proposition. It is easy to see that the
series defining L(s, χ) converges uniformly and absolutely
d
for s > 1 as does the term-by-term derivative ds
L(s, χ).
We use some elementary manipulation to
rewrite the series....
(Note that the book
is incorrect to say
that (p.263) one
uses summation by
parts.)
Proof of (i): To show convergence for s > 0, we rewrite the
P
series as follows. Let Sk denote the partial sum kn=1 χ(n)
(and S0 := 0). Then
The L(·, χ) functions
make sense for 0 <
s ≤ 1 - as long as χ
is non-trivial.
PROOF OF DIRICHLET’S THEOREM
N
X
χ(k)
k=1
ks
5
N
X
Sk − Sk−1
=
k=1
N
X
=
k=1
N
X
ks
N
Sk X Sk−1
−
(since S0 = 0)
ks
ks
k=2
N
−1
X
Sk
Sk
−
ks
(k + 1)s
k=1
k=1
N
−1
X
1
1
SN
+
Sk s −
=
k
(k + 1)s
Ns
=
k=1
N
−1
X
SN
Ns
k=1
P
Consider: the convergence of fk (s): by the key lemma,
1
1
|fk (s)| := Sk s −
k
(k + 1)s
d −s
≤ q max
x = −sx−s−1
x∈[k,k+1] ds
1
= qs s+1 ;
k
P
thus k fk (s) converges absolutely and uniformly for s > 0.
Differentiating the series term-by-term, we see by an argument similar to that above that the differentiated series
converges uniformly for s > 0, and thus converges to a
continuous function.
=:
fk (s) +
Proof of ii: Consider: for s > 1 + (say),
|L(s, χ) − 1| :=
∞
X
χ(n)
Z n=2
∞
≤
ns
−s
x
2
−s
dx = 2
2
≤ 2−s O(1).
s−1
...as an absolutely
(and uniformly) converging series! Thus
the original series
also converges absolutely uniformly.
Control
of
growth is easy
the
6
PROOF OF DIRICHLET’S THEOREM
Taking c = log 2, gives the desired result:
L(s, χ) = 1 + O(e−cs ) as s → ∞.
iii is proved similarly.
4. What was the point of that?
To define the log of L(s, χ)
LECTURE 38: NON-VANISHING OF L(1, χ) FOR
NON-TRIVIAL REAL DIRICHLET CHARACTERS
1. Recall
Our goal is to prove the following.
Theorem 1.1 (Partial theorem, part II). For non-trivial,
real Dirichlet characters χ, L(1, χ) 6= 0.
We stated (without proof) the following propositions.
Proposition 1.2. There exists a number γ (the EulerMascheroni constant) satisfying
N
X
1
n=1
n
Some simple results
on “p-series”.
Of
course they diverge,
but we will need
more precise estimates on the partial
sums.
− log N = γ + O(1/N ).
(See en.wikipedia.org/wiki/Euler-Mascheroni constant)
Proposition 1.3. For N ∈ N,
N
X
1
− 2N 1/2 = c + O(1/N 1/2 ).
1/2
n
n=1
2. Hyperbolic sums
Let F : N × N → C, N ∈ N, and AN := {(m, n) ∈ N × N :
mn ≤ N }. Let SN denote the finite sum
X
SN :=
F (m, n).
Trivial observations
on particular finite
sums: that we can
sum them in different ways.
(m,n)∈AN
Now observe that we can express SN in three different
ways:
1
Completely trivial,
but yet so useful....
PROOF OF DIRICHLET’S THEOREM
2
SN =
N
X
X
F (m, n) (vertically)
m=1 1≤n≤N/m
=
N
X
X
F (m, n) (horizontally).
n=1 1≤m≤N/n
=
N X
X
F (m, n) (i.e., along hyperbolae indexed by k)
k=1 nm=k
3. Return to Dirichlet’s theorem
To finish the proof of Dirichlet’s theorem we will show
that L(1, χ) is well-approximated by ( √1N times) a particular hyperbolic sum, namely
X
SN :=
F (m, n)
(m,n)∈AN
where
Q. Why is it obvious that this hyperbolic sum
√ approximates 2 N L(1, χ)?
Recall L(1, χ) =
P χ(n)
n .
Thus if we prove
the proposition, the
proof of Dirichlet’s
theorem is over!
Analysis of the sum
of χ over divisors of
k
χ(n)
;
F (m, n) := √
nm
we’ll also show that the sum grows faster than the log N .
Precisely,
Proposition 3.1. Let χ be a non-trivial real Dirichlet character. With the above definitions,
i. ∃ c > 0 such that SN ≥ c log N .
ii. SN = 2N 1/2 L(1, χ) + O(1).
Remark. The above proposition finishes the theorem for, if
L(1, χ) = 0 then (ii) would imply that SN = O(1), while
(i) states that SN ≥ c log N ; a contradiction.
To prove the two parts of the proposition we will need
the following two lemmas.
Lemma 3.2. Let k ∈ N. Then
PROOF OF DIRICHLET’S THEOREM
X
χ(n) ≥
n|k
3
0 for all k
1 if k = `2 for some ` ∈ Z
Proof of lemma.
Case I: k = pα for some p prime. Then
X
χ(n) = χ(1) + χ(p) + χ(p2 ) + · · · + χ(pα )
Simple case first: k a
prime power.
n|k
= χ(1) + χ(p) + χ(p)2 + · · · + χ(p)α .
Now, χ being a real Dirichlet character, χ(p) ∈ {−1, 0, 1};
so

α + 1 if χ(p) = 1



X
1
if χ(p) = −1 and α is even
χ(n) =
0
if χ(p) = −1 and α is odd


n|k

1
if χ(p) = 0, i.e., p|q.
P
i.e., as desired, n|k χ(n) is greater than or equal to 0, and
only equals zero when α is odd (i.e., k is not a square).
Case II: General k. If k = pα1 1 · · · pαMM , then the divisors
of k consist of the set
{pβ1 1 · · · pβMM : 0 ≤ βj ≤ αj ; j = 1, . . . , M }.
Thus every divisor has exactly M prime factors of powers
0 through αj (and every such product is a divisor); so
X
n|k
χ(n) =
M
Y
α χ(p0j ) + χ(p1j ) + χ(p2j ) + · · · + χ(pj j ) .
j=1
As before, the only possibility of getting a 0 out of this is
if one of the αj is odd, in which case k is not a square. Lemma 3.3 (Second lemma). Let a, b ∈ N. If a < b then
P
√ = O(a−1/2 ) and
i. bn=a χ(n)
n
Pb χ(n)
ii. n=a n = O(a−1 ).
Remark. The proofs of these two facts are relatively straightforward; so we leave them to you.
The general case
follows immediately
from the first.
(Bad notation in
text: using N for
two purposes)
4
PROOF OF DIRICHLET’S THEOREM
Now we can prove the proposition, which we restate again
for convenience.
Proposition 3.4. Let χ be a non-trivial real Dirichlet character. With the above definitions,
i. ∃ c > 0 such that SN ≥ c log N .
ii. SN = 2N 1/2 L(1, χ) + O(1).
Proof. (i) Summing along hyperbolae, we see
X
SN :=
(m,n)∈AN
χ(n)
√
nm
N X
X
χ(n)
√
=
nm
=
k=1 nm=k
N
X
X
k=1
1
√
k
χ(n).
n|k
By the first lemma, then, we can obtain the desired lower
bound on SN as follows:
N
X
1 X
√
SN =
χ(n)
k
k=1
n|k
X 1
X
1
√ =
≥
= log N 1/2 + O(1).
k `≤N 1/2 `
k=`2 ,`≤N 1/2
(ii) We want to get a precise estimate of
SN :=
X
(m,n)∈AN
χ(n)
√
.
nm
PROOF OF DIRICHLET’S THEOREM
5
We separate SN into the sum SI + SII + SIII , where the
indices in the respective sums lie in the regions
√
√
N
N, N < n ≤ }
√
√ m
II := {(n, m) ∈ N : 1 ≤ m ≤ N , 1 ≤ n ≤ N }
√
√
N
III := {(n, m) ∈ N : N < m ≤ , 1 ≤ n < N }.
n
I := {(n, m) ∈ N : 1 ≤ m <
Consider SI : summing vertically, we get
SI :=
X
√ √
N
1≤m< N , N <n≤ m
χ(n)
√
nm

=
X
√
m< N

X
1
√ 
m √
N
N <n≤ m
χ(n) 
√
n
By the second lemma, the term in parentheses is of the order O((N 1/2 )−1/2 ) = O(N −1/4 ); using the 21 -series estimate
X
1≤n≤M
1
n1/2
= 2M 1/2 + c + O
1
√
M
we get
1
1/4
SI = 2N + c + O
O(N −1/4 ) = O(1)
1/4
N
as N → ∞.
6
PROOF OF DIRICHLET’S THEOREM
For the other two terms, we sum horizontally, again using
the 12 -series estimate:


X χ(n)
X
1 

SII + SIII =
1/2
m1/2
√ n
1≤m≤ N
1≤n< N
( n
)
1/2
X χ(n)
N
n 1/2
=
2
+
c
+
O
1/2
n
N
√ n
1≤n< N

X χ(n)
X χ(n)
 1
+c
+
O
= 2N 1/2
n
n1/2
N 1/2
1/2
1/2
1≤n≤N
the last term from
X χ(n) n1/2
1
=
n1/2 N 1/2
N 1/2
1/2
1≤n<N
1≤n≤N
X
χ(n).
1≤n≤N 1/2
Let’s call these three terms A, B, and C.
P
χ(n)
Now, since L(1, χ) = ∞
n=1 n , and since
∞
X
χ(n)
1/2
−1/2
1/2
=N O N
,
N
n
1/2
n=N
we see that A = 2N 1/2 L(1/χ) + O(1). Further, part (i) of
the lemma implies that
X χ(n)
= O(1);
1/2
n
1/2
1≤n≤N
thus B = O(1). Finally, obviously C = O(1), and so
SN = 2N 1/2 L(1/χ) + O(1),
and the proposition is proved.

X
1≤n≤N 1/2
1 ,
LECTURE 37: PRODUCT FORMULA FOR THE
L-FUNCTIONS
We’d like to be able
to take the log of
both sides of the
above product formula. But we’ll need
to do a lot of work
first....
The technical proposition: allowing us to
define the log of L.
Where do we need
0 < s < 1?
The L(·, χ) functions
make sense for 0 <
s ≤ 1 - as long as χ
is non-trivial.
Obtain controls on
the growth of L and
d
ds L
Notice we don’t define log2 in general.
The point is that this
integral turns out,
for a non-trivial
Dirichlet character to be the primary branch of the
log.
We need control of
the growth to justify
this definition.
1. Recall the technical proposition
Proposition 1.1 (The technical proposition). Let χ be a
non-trivial Dirichlet character. Then
i.
∞
X
χ(n)
L(s, χ) :=
converges for s > 0.
s
n
n=1
ii. L(s, χ) is continuously differentiable in s.
iii. For some constants c, c0 > 0, we have
L(s, χ) = 1 + O(e−cs ) as s → ∞
d
0
L(s, χ) = O(e−c s ) as s → ∞.
ds
2. What was the point of that?
To define the log of L(s, χ)
Definition 2.1. Let s > 1, χ a non-trivial Dirichlet character. We define the “logarithm of L(s, χ)” by
Z ∞ 0
L (t, χ)
dt.
log2 L(s, χ) := −
L(t, χ)
s
Notice that the integral converges, since L(t, χ) can be made
arbitrarily close to 1, and L0 (t, χ) = O(e−ct ), which together
imply
L0 (t, χ)
= O(e−ct ) as t → ∞.
L(t, χ)
Proposition 2.2. Let s > 1. Then
i. elog2 L(s,χ) = L(s, χ) and (most importantly for us,)
1
PROOF OF DIRICHLET’S THEOREM
2
ii. log2 L(s, χ) =
P
p∈℘ log1
1
1− χ(p)
ps
Getting the logarithmic version of
Dirichlet’s product
formula
for
the
L-functions
Proof. (i) We want to show that
e− log2 L(s,χ) L(s, χ) = 1.
Well, the derivative of the LHS w.r.t s is
L0 (s, χ)
− log2 L(s,χ) 0
− log2 L(s,χ)
L(s, χ) = 0,
e
L (s, χ) + e
−
L(s, χ)
so the LHS is constant with respect to s. Now consider
R∞
lim e− log2 L(s,χ) L(s, χ) := lim e
s→∞
s→∞
s
L0 (t,χ)
L(t,χ) dt
The derivative trick
again.
L(s, χ).
The integral, being convergent, vanishes as s → ∞, so the
first term tends to 1 as s → ∞; since L(s, χ) = 1 + O(e−cs ),
so does the second.
(ii) Now let’s show
!
X
1
log2 L(s, χ) =
log1
.
χ(p)
1 − ps
p∈℘
LHS
Well, consider the exponential of both sides. e
e
= L(s, χ), as we just saw. As for the RHS,
log2 L(s,χ)
=
The crucial second step in the
second reduction
We’ll show eA = eB ,
which almost implies
A = B...
!
P
log2 L(s,χ)
e
=e
p
log1
1
χ(p)
1− s
p
!
=
Y
log1
e
1
χ(p)
1− s
p
p∈℘
=
Y
p∈℘
1
1−
χ(p)
ps
!
= L(s, χ) also.
We’re not done! This only means that
!
X
1
log2 L(s, χ) −
log1
= 2πiM (s)
χ(p)
1
−
p∈℘
ps
...but not quite.
PROOF OF DIRICHLET’S THEOREM
3
for some integer-valued function M . It is left as an exercise
to show that M is continuous and that lims→∞ M (s) = 0
which forces M ≡ 0 (since M (s) ∈ Z).
What was the point
of all that?
Following the same
lines of the argument of Euler, but
to show finiteness
rather than infiniteness.
Here’s
why
we
needed the integral
representation of the
log
Now comes the deep
part....
My God, what a
wonderful piece of
work.
3. Second reduction of the problem
By the proposition just proved and subsequently the properties of log1 proven earlier,
!
X
1
log2 L(s, χ) =
log1
1 − χ(p)
p∈℘
ps
X χ(p)
1
=
+E
s
p
ps
p∈℘
X χ(p)
=
+ C.
s
p
p∈℘
Now, if we can show that L(1, χ) 6= 0 for χ non-trivial, then
since L(s, χ) is continuously differentiable for 0 < s < ∞,
L0 (s, χ) is bounded near s = 1, so
Z ∞ 0
L (t, χ)
lim+ log2 L(s, χ) := lim+
dt is bounded.
s→1
s→1
L(t, χ)
s
P
In that case, the above equality implies lims↓1 p∈℘ χ(p)
ps is
also bounded; that is, the reduced problem would be solved.
Thus it suffices to prove the following claim:
Theorem 3.1. L(1, χ) 6= 0 for non-trivial χ.
We will do this in two cases, that of complex Dirichlet characters and that of real ones. The former is the easier.
4. Non-vanishing of the L-function
Case I: complex Dirichlet characters
Theorem 4.1 (Partial theorem). For non-trivial, complex
Dirichlet characters χ, L(1, χ) 6= 0.
4
PROOF OF DIRICHLET’S THEOREM
The proof will be by contradiction and rely on the following
two lemmas.
Q
Lemma 4.2. Let s > 1. Then χ L(s, χ) ≥ 1; in particular, the product is real-valued.
Proof. Using the product formula for the L-function and
the definition of log1 , we see
!!
Y
XX
1
L(s, χ) = exp
log1
1 − χ(p)
χ
χ p∈℘
ps
!!
XX
1
= exp
log1
1 − χ(p)
χ p∈℘
ps
!
∞
XXX
1 χ(pk )
= exp
k pks
χ p∈℘ k=1
!
∞
X
XX
1 1
χ(pk ) .
= exp
ks
kp χ
p∈℘
Product over all χ is
real-valued.
k=1
Now, by Fourier series expansion
characteristic funcP of the
1
k
k
tions of points, δ1 (p ) = ϕ(q) χ χ(p ), so the above equals
!
∞
XX
1 1
exp
ϕ(q)δ1 (pk ) ≥ 1,
ks
kp
p∈℘
(Another cool application of Fourier series.)
Error in text: δ1 ,
not δ0 . (Unless by 0
they mean 1.)
k=1
since the term in the parentheses is non-negative.
Lemma 4.3.
i. If L(1, χ) = 0, the L(1, χ) = 0 (obvious).
ii. If L(1, χ) = 0 and χ is non-trivial, then
|L(s, χ)| ≤ C|s − 1| for 1 ≤ s ≤ 2.
iii. In the case of χ = χ0 , we have
C
|L(s, χ0 )| ≤
for 1 < s ≤ 2.
|s − 1|
On the behavior of
L(s, χ) as s ↓ 1: how
it vanishes; how it
blows up.
PROOF OF DIRICHLET’S THEOREM
5
Proof. i) L(1, χ) = L(1, χ).
ii) For χ non-trivial, L(s, χ) is continuously differentiable
for s > 0. Thus, by the Mean Value Theorem,
L(s, χ) − L(1, χ)
= L0 (s0 , χ)
s−1
for some s0 ∈ (1, s). Since L0 (s, χ) is continuous, it has a
bound C on [1, 2].
1
s
iii) Recall that ζ(s) ≤ 1 + s−1
= s−1
. For 1 < s ≤ 2,
then,
2
|ζ(s)| ≤
.
s−1
Since L(s, χ0 ) = Cζ(s), we are done.
Theorem 4.4 (Partial theorem). For non-trivial, complex
Dirichlet characters χ, L(1, χ) 6= 0.
The point:
for
complex
Dirichlet
characters,
we’re
guaranteed
that
they occur in pairs;
so if L(1, χ) = 0,
then L(1, χ) = 0,
overpowering
the
blowing-up L(1, χ0 ).
Proof of the partial theorem. By contradiction: suppose L(1, χ1 ) =
0; then by the previous lemma,
i. L(1, χ1 ) = 0 and
ii. For χ = χ1 and χ1 ,
|L(s, χ)| ≤ C|s − 1| for 1 ≤ s < 2.
Q
If we consider the product χ L(s, χ), then, we see that
there are at least two Dirichlet characters (χ1 and χ1 ) that
vanish like |s − 1| as s ↓ 1. L(s, χ) being continuously
differentiable on s > 0 for χ non-trivial, we see the only
term in this product that could (and does) go to infinity as
1
s ↓ 1 is L(s, χ0 ). But its growth is of order O( s−1
), so
Y
lim
L(s, χ) = 0, in contradiction with the earlier lemma.
s→1+
χ
LECTURE 37: NON-VANISHING OF L(1, χ) FOR
COMPLEX DIRICHLET CHARACTERS
We’d like to be able
to take the log of
both sides of the
above product formula. But we’ll need
to do a lot of work
first....
The technical proposition: allowing us to
define the log of L.
Where do we need
0 < s < 1?
The L(·, χ) functions
make sense for 0 <
s ≤ 1 - as long as χ
is non-trivial.
Obtain controls on
the growth of L and
d
ds L
Notice we don’t define log2 in general.
The point is that this
integral turns out,
for a non-trivial
Dirichlet character to be the primary branch of the
log.
We need control of
the growth to justify
this definition.
1. Recall the Technical Proposition
Proposition 1.1 (The technical proposition). Let χ be a
non-trivial Dirichlet character. Then
i.
∞
X
χ(n)
converges for s > 0.
L(s, χ) :=
s
n
n=1
ii. L(s, χ) is continuously differentiable in s.
iii. For some constants c, c0 > 0, we have
L(s, χ) = 1 + O(e−cs ) as s → ∞
d
0
L(s, χ) = O(e−c s ) as s → ∞.
ds
2. What was the point of that?
To define the log of L(s, χ)
Definition 2.1. Let s > 1, χ a non-trivial Dirichlet character. We define the “logarithm of L(s, χ)” by
Z ∞ 0
L (t, χ)
log2 L(s, χ) := −
dt.
L(t,
χ)
s
Remark. Notce that this integral converges since L(t, χ) can
be made arbitrarily close to 1, and L0 (t, χ) = O(e−ct ), which
together imply
L0 (t, χ)
= O(e−ct ).
L(t, χ)
Proposition 2.2. Let s > 1. Then
i. elog2 L(s,χ) = L(s, χ) and, most importantly for us,
1
PROOF OF DIRICHLET’S THEOREM
2
ii. log2 L(s, χ) =
P
p∈℘ log1
1
1− χ(p)
ps
Getting the logarithmic version of
Dirichlet’s product
formula
for
the
L-functions
Proof. i) We want to show that
e− log2 L(s,χ) L(s, χ) = 1.
Well, the derivative of the LHS w.r.t s is
L0 (s, χ)
− log2 L(s,χ) 0
− log2 L(s,χ)
L(s, χ) = 0,
e
L (s, χ) + e
−
L(s, χ)
so the LHS is constant with respect to s. Now consider
R∞
lim e− log2 L(s,χ) L(s, χ) := lim e
s→∞
s
L0 (t,χ)
L(t,χ) dt
s→∞
L(s, χ).
The integral, being convergent, vanishes as s → ∞, so the
first term tends to 1 as s → ∞; since L(s, χ) = 1 + O(e−cs ),
so does the second.
ii) Now let’s show
!
X
1
log2 L(s, χ) =
log1
.
χ(p)
1 − ps
p∈℘
LHS
Well, consider the exponential of both sides. e
e
= L(s, χ), as we just saw. As for the RHS,
log2 L(s,χ)
=
The crucial second step in the
second reduction
We’ll show eA = eB ,
which almost implies
A = B...
!
P
e
log2 L(s,χ)
=e
p
log1
1
χ(p)
1− s
p
log1
1
χ(p)
1− s
p
!
P
=e
p
!
=
Y
log1
e
1
χ(p)
1− s
p
p∈℘
=
Y
p∈℘
1
1−
χ(p)
ps
!
= L(s, χ) also.
We’re not done! This only means that
!
X
1
log2 L(s, χ) −
log1
= 2πiM (s)
χ(p)
1
−
p∈℘
ps
...but not quite.
PROOF OF DIRICHLET’S THEOREM
3
for some integer-valued function M . It is left as an exercise
to show that M is continuous and that lims→∞ M (s) = 0
which forces M ≡ 0 (since M (s) ∈ Z).
What was the point
of all that?
Following the same
lines of the argument of Euler, but
to show finiteness
rather than infiniteness.
(Here’s why we
needed the integral
representation
of
log L.)
Now comes the deep
part....
My God, what a
wonderful piece of
work.
3. Second reduction of the problem
By the proposition just proved and subsequently the properties of log1 proven earlier,
!
X
1
log2 L(s, χ) =
log1
1 − χ(p)
p∈℘
ps
X χ(p)
1
=
+E
s
p
ps
p∈℘
X χ(p)
=
+ C.
s
p
p∈℘
Now, if we can show that L(1, χ) 6= 0 for χ non-trivial, then
since L(s, χ) is continuously differentiable for 0 < s < ∞,
L0 (s, χ) is bounded near s = 1, so
Z ∞ 0
L (t, χ)
lim log2 L(s, χ) := lim+
dt is bounded.
s↓1
s→1
L(t, χ)
s
P
In that case, by the above equality, lims↓1 p∈℘ χ(p)
ps is also
bounded; that is, the reduced problem would be solved.
Thus it suffices to show the following claim:
Theorem 3.1. L(1, χ) 6= 0 for non-trivial χ.
We will do this in two cases, that of complex Dirichleet
characters and that of real ones. The former is the easier.
4. Non-vanishing of the L-function
Case I: complex Dirichlet characters
Theorem 4.1 (Partial theorem). For non-trivial, complex
Dirichlet characters χ, L(1, χ) 6= 0.
4
PROOF OF DIRICHLET’S THEOREM
The proof will be by contradiction and rely on the following
two lemmas.
Q
Lemma 4.2. Let s > 1. Then χ L(s, χ) ≥ 1; in particular, the product is real-valued.
Proof. Using the product formula for the L-function and
the definition of log1 , we see
!!
Y
XX
1
L(s, χ) = exp
log1
1 − χ(p)
χ
χ p∈℘
ps
!!
XX
1
= exp
log1
1 − χ(p)
χ p∈℘
ps
!
∞
XXX
1 χ(pk )
= exp
k pks
χ p∈℘ k=1
!
∞
XX
1 1 X
= exp
χ(pk ) .
ks
kp χ
p∈℘
k=1
Now, by Fourier series expansion
characteristic funcP of the
1
k
k
tions of points, δ1 (p ) = ϕ(q) χ χ(p ), so the above equals
!
∞
XX
1 1
ϕ(q)δ1 (pk ) ≥ 1,
exp
ks
kp
p∈℘
k=1
since the term in the parentheses is non-negative.
Lemma 4.3.
i. If L(1, χ) = 0, the L(1, χ) = 0 (obvious).
ii. If L(1, χ) = 0 and χ is non-trivial, then
|L(s, χ)| ≤ C|s − 1| for 1 ≤ s ≤ 2.
iii. In the case of χ = χ0 , we have
C
|L(s, χ0 )| ≤
for 1 < s ≤ 2.
|s − 1|
(Another cool application of Fourier series. Recall δ` (m) =
P
1
χ χ(`)χ(m).)
ϕ(q)
Error in text: δ1 ,
not δ0 . (Unless by 0
they mean 1.)
On the behavior of
L(s, χ) as s ↓ 1: how
it vanishes (if it vanishes); how it blows
up (if χ = χ0 ).
PROOF OF DIRICHLET’S THEOREM
P
1
ns
≤ 1+
1
.
1 + s−1
R∞
1
dx
xs
=
5
Proof. i) L(1, χ) = L(1, χ)
ii) For χ non-trivial, L(s, χ) is continuously differentiable
for s > 0. Thus, by the Mean Value Theorem,
L(s, χ) − L(1, χ)
(1)
= L0 (s0 , χ)
s−1
for some s0 ∈ (1, s). Since L0 (s, χ) is continuous, it has a
bound C on [1, 2].
1
s
iii) Recall that ζ(s) ≤ 1 + s−1
= s−1
. For 1 < s ≤ 2, then,
2
|ζ(s)| ≤
(2)
.
s−1
Since L(s, χ0 ) = cζ(s), we are done.
Theorem 4.4 (Partial theorem). For non-trivial, complex
Dirichlet characters χ, L(1, χ) 6= 0.
The point:
for
complex
Dirichlet
characters,
we’re
guaranteed
that
they occur in pairs;
so if L(1, χ) = 0,
then L(1, χ) = 0,
overpowering
the
blowing-up
of
L(1, χ0 ) as s ↓ 1.
Proof of the partial theorem.
By contradiction: suppose L(1, χ1 ) 6= 0; then, by the
previous lemma, L(1, χ1 ) = 0 also, and for both χ = χ1 , χ1 ,
we have
(3)
|L(s, χ)| ≤ C|s − 1| for 1 ≤ s < 2.
Q
Now consider χ L(s, χ). For non-trivial χ, L(s, χ) is
continuously differentiable on s > 0 and thus finite at
L(1, χ) (i.e., when s = 1; among those Dirichlet characters, there are at least two (viz., χ1 and χ1 ) that vanish
(like O(|s − 1|) as s ↓ 1). The only term that can (and
does) go
to infinity as s ↓ 1 is L(s, χ0 ), which grows like
1
O s−1 . Thus
Y
(4)
lim+
L(s, χ) = 0,
s→1
χ
in contradiction with the fact that
Q
χ L(s, χ)
≥ 1.
LECTURE 38: NON-VANISHING OF L(1, χ) FOR
NON-TRIVIAL REAL DIRICHLET CHARACTERS
1. Recall the Goal; Behavior of p-series
Our goal is to prove the following.
Theorem 1.1 (Partial theorem, part II). For non-trivial,
real Dirichlet characters χ, L(1, χ) 6= 0.
We stated (without proof) the following propositions.
Proposition 1.2. There exists a number γ (the EulerMascheroni constant) satisfying
N
X
1
− log N = γ + O(1/N ).
n
n=1
Precise estimates on
the partial sums of
certain divergent pseries.
(See en.wikipedia.org/wiki/Euler-Mascheroni constant)
Proposition 1.3. For N ∈ N,
N
X
1
− 2N 1/2 = c + O(1/N 1/2 ).
1/2
n
n=1
2. Approximating L(1, χ) with hyperbolic sums
To finish the proof of Dirichlet’s theorem we will show
that L(1, χ) is well-approximated by a particular hyperbolic
sum, namely
X χ(n)
√
SN :=
nm
(m,n)∈AN
where AN := {(m, n) ∈ N × N : mn ≤ N }. We’ll see that
∞
X
X χ(n)
χ(n)
1
1
√
√
L(1, χ) :=
≈
=: √ SN
n
mn
2 N (m,n)∈A
2 N
n=1
N
1
The trick: approximating L(1, χ) with
a particular hyperbolic sum.
PROOF OF DIRICHLET’S THEOREM
2
Q. Why is it obvious that this hyperbolic sum
√ approximates 2 N L(1, χ)?
Recall L(1, χ) =
P χ(n)
n .
Thus if we prove
the proposition, the
proof of Dirichlet’s
theorem is over!
Analysis of the sum
of χ over divisors of
k
and, further, that the sum grows faster than log N . Precisely:
Proposition 2.1. Let χ be a non-trivial real Dirichlet character. With the above definitions,
i. ∃ c > 0 such that SN ≥ c log N .
ii. SN = 2N 1/2 L(1, χ) + O(1).
Remark. The above proposition finishes the theorem for, if
L(1, χ) = 0 then (ii) would imply that SN = O(1), while
(i) states that SN ≥ c log N ; a contradiction.
To prove the two parts of the proposition we will need
the following two lemmas.
Lemma 2.2. Let k ∈ N. Then
X
0 for all k
χ(n) ≥
1 if k = `2 for some ` ∈ Z
n|k
Simple case first: k a
prime power.
Proof of lemma.
Case I: k = pα for some p prime. Then
X
χ(n) = χ(1) + χ(p) + χ(p2 ) + · · · + χ(pα )
n|k
The general case
follows immediately
from the first.
(Bad notation in
text: using N for
two purposes)
= χ(1) + χ(p) + χ(p)2 + · · · + χ(p)α .
Now, χ being a real Dirichlet character, χ(p) ∈ {−1, 0, 1};
so

α + 1 if χ(p) = 1



X
1
if χ(p) = −1 and α is even
χ(n) =
0
if χ(p) = −1 and α is odd


n|k

1
if χ(p) = 0, i.e., p|q.
P
i.e., as desired, n|k χ(n) is greater than or equal to 0, and
only equals zero when α is odd (i.e., k is not a square).
Case II: General k. If k = pα1 1 · · · pαMM , then the divisors
of k consist of the set
{pβ1 1 · · · pβMM : 0 ≤ βj ≤ αj ; j = 1, . . . , M }.
PROOF OF DIRICHLET’S THEOREM
3
Thus every divisor has exactly M prime factors of powers
0 through αj (and every such product is a divisor); so
X
n|k
χ(n) =
M
Y
α χ(p0j ) + χ(p1j ) + χ(p2j ) + · · · + χ(pj j ) .
j=1
As before, the only possibility of getting a 0 out of this is
if one of the αj is odd, in which case k is not a square. Lemma 2.3 (Second lemma). Let a, b ∈ N. If a < b then
P
√ = O(a−1/2 ) and
i. bn=a χ(n)
n
Pb χ(n)
ii. n=a n = O(a−1 ).
The product can
equal
zero
only
when one of the
sums is 0; by the
previous case, αj
must be odd
Remark. The proofs of these two facts are relatively straightforwardl; we leave them to you.
Now we can prove the proposition, which we restate again
for convenience.
Proposition 2.4. Let χ be a non-trivial real Dirichlet character. With the above definitions,
i. ∃ c > 0 such that SN ≥ c log N .
ii. SN = 2N 1/2 L(1, χ) + O(1).
Proof. (i) Summing along hyperbolae, we see
X χ(n)
√
SN :=
nm
Summing
along
hyperbolae gives
the lower bound
result.
(m,n)∈AN
N X
X
χ(n)
√
=
nm
k=1 nm=k


N
X 1
X

√
=
χ(n) .
k n|k
k=1
By the first lemma, then, since the term in parentheses is
(The pair (m, n)
such that nm = k is
determined by the
first entry.)
4
PROOF OF DIRICHLET’S THEOREM
≥ 1 when k is square,
1
√
k
k=`2 ,`≤N 1/2
X 1
=
= log N 1/2 + O(1),
`
1/2
SN ≥
X
`≤N
Summing
vertically
and
horizontally gives
L(1, χ).
which is the desired lower bound on SN (the last equality
follows from the estimate of the partial sum of the harmonic
series).
(ii) Consider:
X χ(n)
√
SN :=
.
nm
(m,n)∈AN
Break
the
sum
into various regions.
(Draw the hyperbola
correctly!)
We separate SN into the sum SN = SI + SII + SIII , where
the indices for the respective sums lie in the regions
√ √
N
I := {(n, m) ∈ N : 1 ≤ m < N , N < n ≤ }
√
√ m
II := {(n, m) ∈ N : 1 ≤ m ≤ N , 1 ≤ n ≤ N }
√
√
N
III := {(n, m) ∈ N : N < m ≤ , 1 ≤ n < N }.
n
Consider SI : summing vertically, we get
X
χ(n)
√
SI :=
nm
√ √
N
1≤m< N , N <n≤ m


X χ(n)
X 1
√ 
√ 
=
m √
n
√
N
m< N
Use
the bound on
P χ(n)
√
n
and the bound on the
p = 12 -series
N <n≤ m
By the second lemma, the term in parentheses is of the order O((N 1/2 )−1/2 ) = O(N −1/4 ); using the 21 -series estimate
we get
PROOF OF DIRICHLET’S THEOREM
5
1
1/4
SI = 2N + c + O
O(N −1/4 ) = O(1)
1/4
N
as N → ∞.
For the other two terms, we sum horizontally. Using the
1
SII + SIII : use the
2 -series estimate,
1
2 -series bound again


X χ(n)
X
1 

SII + SIII =
1/2
m1/2
√ n
N
1≤m≤
1≤n< N
( n
)
1/2
X χ(n)
n 1/2
N
=
2
+
c
+
O
1/2
n
N
√ n
1≤n< N


X χ(n)
X
X χ(n)
 1
+c
+
O
1 ,
= 2N 1/2
1/2
1/2
n
n
N
1/2
1/2
1/2
1≤n≤N
1≤n≤N
1≤n≤N
the last term from
X
1≤n<N
χ(n) n1/2
1
=
n1/2 N 1/2
N 1/2
1/2
X
χ(n).
1≤n≤N 1/2
Let’s call these three terms A, B, and C.
We’ll deal with them in reverse order. Obviously C =
O(1). Further, part (i) of the lemma implies that
X
1≤n≤N
χ(n)
= O(1);
1/2
n
1/2
thus B = O(1).
P
χ(n)
Now, since L(1, χ) = ∞
n=1 n , and since
∞
X
χ(n)
−1/2
=O N
,
n
1/2
n=N
C and B are O(1)
A: we basically have
L(1, χ)
6
PROOF OF DIRICHLET’S THEOREM
we see that
χ(n)
n
1≤n≤N 1/2
h
i
1/2
−1/2
= 2N
L(1, χ) − O N
A := 2N 1/2
X
= 2N 1/2 L(1, χ) + O(1).
giving the desired estimate for SN .