Fourier Transform
László Erdős
Oct 22, 2008
The purpose of this note is to outline some basic material about Fourier series and transform.
1
Fourier series
In this Section we work on [0, 2π] where 0 and 2π are periodically identified. You can think
of if as the unit circle S 1 . All the Lp and C k spaces refer to the corresponding function space
on S 1 . The periodicity condition does not require any extra for Lp functions (f (0) = f (2π)
makes no sense), but it requires f (0) = f (2π) for continuous functions and f (j) (0) = f (j) (2π),
0 ≤ j ≤ k for C k functions.
1.1
L2 Theory of Fourier series
Define
einx
n∈Z
en (x) := √
2π
which is clearly an orthonormal set.
Let f ∈ L2 , then f ∈ L1 and we can define
Z 2π −inx
e
√ f (x)dx
cn = (en , f ) =
2π
0
P
The formal series
cn en (x) is called the Fourier series of f . The basic question is its
convergence (in which sense?) and its relation to f .
If we knew that {en } were complete, i.e. it would form an ONB, then
Pby the general
basis-representation theorem in abstract Hilbert spaces we would know that cn en converges
1
to f in the norm of L2 . In particular, the finite approximate sums,
N
X
SN (f )(x) :=
cn en (x)
n=−N
would converge to f (x) in L2 .
Note that SN (f ) is the projection of f onto the finite dimensional space spanned by
{en : |n| ≤ N}. This follows from the following lemma in abstract Hilbert spaces:
Lemma 1.1 Let M ⊂ H be a closed subspace of a Hilbert space. Let {en } be an ONB in M.
Then for any x ∈ H the projection of x onto M is given by
X
PM (x) =
(en , x)en
Proof. Since M ⊕ M ⊥ = H, we can uniquely write x = m + m′ with m ∈ M, m′ ∈ M ⊥ .
However it is easy to check that
X
X
x=
(en , x)en + x −
(en , x)en
exactly decomposes x into an element of M plusPan element of M ⊥ (the first term is in M
because M is closed, so the limit of the finite
x)en ∈ M sums lies in M. The
n≤N (en ,P
⊥
second term is in M , since for any em clearly em ⊥ x − (en , x)en by orthogonality). By
the uniqueness ofP
the x = m + m′ decomposition hence we know that the component of x in
M must be m = (en , x)en . 1.2
Pointwise Theory of Fourier series
So far we have seen that the L2 theory of F-series is fairly trivial (modulo the missing fact,
that en is complete). However, here we talk about functions and not just abstract elements
of abstract Hilbert spaces. There are several other questions one can ask about the relation
between F-series and the original function. For example: In which other sense does the
convergence hold? Can one differentiate/integrate Fourier series term-by-term? Etc.
The good analogy is the Taylor series, expcept that there essentially everything is trivial.
Recall that a Taylor series converges absolutely inside the interval of convergence (maybe not
at the endpoints), it converges uniformly on any compact subset away from the endpoints, it
can be integrated and differentiated arbitrary many times inside the interval of convergence
etc. The similar questions for F-series are much harder and the answers show greater variety.
I will show the most important things.
2
Theorem 1.2 Let f ∈ C(S 1 ) be continuous (you can think of it f ∈ C[0, 2π] with periodic
boundary condition, f (0) = f (2π)). Then the Fourier series is uniformly Cesaro summable.
It means that
sup |ΣN (f )(x) − f (x)| → 0
as N → ∞
x
where
1 S0 (f ) + S1 (f ) + . . . + SN (f )
N +1
P
is the so-called Cesaro sum of the series
cn en .
ΣN (f ) :=
Proof. Easy computation (summing up geometric series) gives
Z X
Z −iny
N
N
X
einx
1
e
√
√ f (y)dy =
SN (f )(x) =
ein(x−y) f (y)dy
2π n=−N
2π
2π
n=−N
=
where
Z
DN (x − y)f (y)dy =
1 sin
DN (z) :=
2π
Z
DN (y)f (x − y)dy
h
i
N + 21 z
sin z2
is the so-called Dirichlet kernel.
We would like to see that it behaves like an approximate
delta-function, i.e. its main
R
weight is supported around z ≈ 0, so in the integral DN (y)f (x − y)dy the main contribution
would come from the regime y ∼ x and it is then f (x) by continuity of f . It is clearly true
that
Z
Z X
N
1
DN (z)dz =
einz dz = 1
2π n=−N
(only the n = 0 contributes), but DN is not positive, and actually it is highly oscillatory and
not concentrated sufficiently strongly around the origin.
The Cesaro sum takes care of this. Again, a simple calculation (summing up geometric
series) shows that
h
i!
1
Z
Z
N sin
N
+
z
X
2
1
1
f (x − y)dy = KN (y)f (x − y)dy
ΣN (f )(x) =
N + 1 2π
sin 2z
n=0
3
with
KN (z) :=
sin N +1 z 2
1
2
2π(N + 1)
sin 2z
which is called the Fejér-kernel. Note that KN ≥ 0 (it is a “miracle”), and it enjoys the
following properties:
R
(i) KN = 1
(ii) For any δ > 0, KN (z) → 0 uniformly on [δ, 2π − δ].
The first one is trivial, since KN is just the average of DN ’s,
KN =
1
(D0 + D1 + . . . + DN )
N +1
R
and Dk = 1 for any k.
The second one follows from the fact that if z ∈ [δ, 2π − δ], then | sin(z/2)| ≥ δ/4 (draw
the graph), so for such z’s
16
KN (z) ≤ 2
→0
δ (N + 1)
as N → ∞.
Now we can show that ΣN (f ) converges uniformly to f for continuous f . Since f is
continuous on a compact set (S 1 ), it is uniformly continuous, so for any ε > 0, ∃δ s.t. |f (z) −
f (x)| < ε if |z − x| < δ. Then we have
Z
Z
|ΣN (f )(x) − f (x)| = KN (x − z)f (z)dz − f (x) = KN (x − z)(f (z) − f (x))dz (notice how f (x) was smuggled in the integral using
Z
Z
≤ KN (x − z)|f (z) − f (x)|dz =
R
KN = 1)
(. . .)dz +
|z−x|≥δ
≤ 2kf k∞
Z
|z−x|≥δ
KN (z − x)dz + ε
Z
(. . .)dz
|z−x|<δ
Z
KN
16
+ε
+ 1)
In the first term we estimated |f (z) − f (x)| ≤ 2kf k∞ , in the second we used |f (z) − f (x)| < ε
if |z − x| < δ. Note that KN is positive, so no need to take absolute value.
Now this can be made arbitrarily small for large N, uniformly in x. First you choose the ε
sufficiently small, then fix δ and choose N sufficiently large. [THINK THIS OVER, it is very
important!] ≤ 2kf k∞ 2π
δ 2 (N
4
Corollary 1.3 For any f ∈ L2 [0, 2π] we have SN (f ) → f in L2 . Moreover, {en }n∈Z is
complete
Proof. Clearly
kf − SN (f )k2 ≤ kf − ΣN (f )k2
(1.1)
because SN (f ) is the projection of f onto the space spanned by en , |n| ≤ N, and clearly
ΣN (f ) is also in this space. Since the projection gives the closest element, this inequality is
trivial.
Now let f ∈ L2 . We choose a continuous function g such that kf − gk < ε, this is possible,
since C is dense in L2 . We write
kf − SN (f )k2 ≤ kf − gk2 + kg − SN (g)k2 + kSN (g) − SN (f )k2 ≤ 2ε + kg − SN (g)k2
using that kSN (g) − SN (f )k2 = kSN (g − f )k2 ≤ kg − f k2 by the fact that any projection has
norm at most 1 and SN is a projection.
Since g ∈ C, using (1.1) for g and Theorem 1.2, we see that kg − SN (g)k2 ≤ ε if choosing
N big enough, so the whole thing is smaller than 3ε.
The completeness of {en }n∈Z follows immediately; if Span{en } were not the whole H, then
it would be a proper closed subspace of it, and then there would be a nonzero element h ∈ H
in the orthogonal complement, in particular h ⊥ en . Therefore all Fourier coefficients were
zero, SN (h) ≡ 0 for any N, and hence lim SN (h) = 0. On the other hand, lim SN (h) = h for
any h ∈ H, contradiction. The completeness also implies that the general statement about unitary equivalence with
ℓ2 holds. The map
F : L2 ([0, 2π]) 7→ ℓ2 (Z)
f 7→ {(en , f )}n∈Z
is isometrically isomorphic, (in other words: unitary) between the two spaces. It is called the
Fourier transform.
Note that for convenience we work with ℓ2 (Z) instead of the usual ℓ2 = ℓ2 (N), but of
course these two spaces can be easily mapped into each other by relabelling all integers by
natural numbers.
Having completed the L2 theory, the next question is in which other senses does the
Fourier series converge. Pointwise convergence for a general L2 function does not make sense.
Unfortunately, pointwise convergence does not hold even for arbitrary continuous functions,
where it could make sense:
5
Example. There exists f ∈ C(S 1 ) such that its F-series does not converge pointwise.
The construction is not easy, and later we’ll give a proof of the existence of such function without really constructing it, but it will require an extra theoretical material (BanachSteinhaus theorem).
The following is a major theorem despite its innocent look:
Theorem 1.4 (Carleson) If f ∈ L2 [0, 2π], then SN (f ) → f almost everywhere pointwise.
1.3
Regularity and Fourier series
If one assumes extra regularity on the function, then one gets better convergence:
1
Theorem 1.5 If f ∈ C 1 (S 1 ) = Cper
([0, 2π]) (periodicity in the sense that f (0) = f (2π) and
′
′
f (0) = f (2π)), then SN (f ) → f uniformly
P
P
Since f ′ ∈ C ⊂ L2 , it has a F-series, f ′ =
bn en that converges in L2 . Let f =
cn en
be the F-series of f (that is also L2 ). We have
Z −inx
Z −inx
e
e
′
√ f (x)dx = in
√ f (x) = incn
bn =
2π
2π
1
by integration by parts. This shows that the
P F-seriesPof a C functionPcan2 be 2differentiated
P
termwise (clearly, the termwise derivative of cn en is incn en .) Since n |cn | = |bn |2 <
∞, we get
X
1/2 X
1/2
X
X
1
|cn | =
|cn ||n| ·
≤
|cn |2 n2
n−2
<∞
(1.2)
|n|
P
P
P
Therefore the series
cn en converges uniformly since
|cn |ken k∞ =
|cn | is summable.
This is exactly the uniform convergence of SN (f ) → f .
P
In particular, we have seen that if f ∈ C 1 , then
|cn |2 n2 < ∞. P
It is natural to ask if the converse is true, i.e. is it true that
|cn |2 n2 < ∞ implies
differentiability. The answer is no, but it is true under a stronger condition
Theorem 1.6 Suppose that
Then f ∈ C 1 .
P
|cn ||n| < ∞ holds for the Fourier series of an f ∈ L2 function.
6
P
P
Remark.
Note
that
the
condition
|c
||n|
<
∞
is
stronger
than
|cn |2 n2 < ∞. If
n
P
|cn |n < ∞, then in particular
|cn ||n| ≤ P
K for some constant (every convergent series have
P
2 2
bounded terms), therefore
|cn | n ≤ K |cn ||n|, so if the latter is finite, so is the former.
Proof. Recall a standard theorem in analysis: If a sequence of functions fN converges to
f pointwise (actually convergence at one point is enough) and fN′ converges to some function
h uniformly,
then f is differentiable and f ′ = h. Use this theorem for fN := SN (f ) and check
P
that fN′ = N
−N incn en converges uniformly. But this is clear from the condition. Similar theorem is true for higher derivatives (The proof is by induction on k):
P
Theorem 1.7 If
|cn ||n|k < ∞, then f ∈ C k .
1.4
Diagonalization of derivatives
The F-transform diagonalizes the differentiation operator D :=
vectors:
Den = inen
d
dx
with en being the eigen-
(compare it with Av = λv eigenvalue equation; view D as a linear transformation, i.e. as an
“infinite dimensional” matrix, and then en ’s are the eigenvectors with eigenvalues n).
Instead of the operator D it is better to look at −iD, the formulas are nicer. Note that
using integration by parts, −iD is a symmetric operator
Z
Z
Z
′
′
(f, (−iD)g) = −i f¯(x)g (x)dx = i f¯ (x)g(x)dx = −if ′ (x)g(x)dx = ((−iD)f, g)
while D itself would be antisymmetric. This is the main reason we like −iD better than D.
If you consider the differentiation map under F-transform, then
F (−iDf ) = {ncn }
if F (f ) = {cn }. I.e. the action of the −iD in the Fourier representation is simply multiplication
by n.
Again, higher derivatives are similar, for example −∆ = (−iD)2 in the Fourier picture is
just multiplication by n2 .
1.5
Sobolev spaces
We have seen at Theorem 1.6 and in the Remark afterwards that differentiable functions
cannot be nicely characterized by F-transform. However, we can use the F-transform to
extend the concept of differentiability. The following definition is fundamental:
7
Definition 1.8 (Sobolev spaces) Let k be a natural number. We define
X
H k := {f ∈ L2 [0, 2π] :
n2k |cn |2 < ∞}
P
to be the Sobolev space of order k. For f ∈ H k , the function (in)k cn en (that makes sense as
an L2 function) is called the weak k-th order derivative of f . If f ∈ C k , then this notion
coincides with the usual k-th order derivative.
P
P
We have seen that if
|cn ||n|k < ∞, then f ∈ C k . The condition
|cn |2 |n|2k < ∞ is
2
weaker, nevertheless we could define the k-th derivative of f , as an L function. We did not
define it as the limit of the difference quotient since that does not exist. We defined it via
Fourier transform, but one can check that every properties of the derivatives hold and it is
indeed an extension of the usual derivative. For example:
Homework 1.9 (Leibniz rule for H 1 in 1 dimensions.) . Let f, g ∈ H 1 , then f g ∈ H 1
and (f g)′ = f ′ g + f g ′ (all derivatives are in the weak sense).
Remark: in higher dimensions the Leibniz rule is more refined, one needs more conditions
on f , g. One possible version is that if f ∈ C0∞ ([0, 2π]d ) and g ∈ H 1 ([0, 2π]d ), then f g ∈
H 1 ([0, 2π]d ) and (f g)′ = f ′ g + f g ′ holds. There are more complicated versions where both
f, g are in certain Sobolev spaces.
Homework 1.10 (Weak derivative as L2 -limit of difference quotient) Let f ∈ H 1 . Prove
that the limit
f (x + ε) − f (x)
lim
ε→0
ε
2
′
exists in L -sense and the limit is f (x).
The space H k has a major advantage over C k , it is a (complete) Hilbert space with the
scalar product
X
(1 + |n|2k )cn dn
(f, g)H k :=
P
P
where f =
cn en and g = dn en . The norm is called the k-th order Sobolev norm:
X
1/2
kf kH k :=
(1 + |n|2k )|cn |2
This concept perfectly makes sense for any k ≥ 0, not just for integers. It is clearly a Hilbert
space, because it is an L2 space on Z with the weigthed counting measure
µ({n}) = 1 + |n|2k
8
See also a remark later in Section 1.7 on fractional exponents, k 6∈ N.
Actually, one can also define Sobolev spaces with negative exponents, k < 0, but these are
in general not functions but distributions and the definition is slightly different, so we do not
touch them here.
There are several alternative definitions of the Sobolev norm(s). The following norm
X
1/2
kf k′H k :=
(1 + n2 )k |cn |2
is not the same as kf kH k , but it is equivalent to it in the following sense: there exist positive
constants, K1 , K2 , depending only on k, such that
K1 kf kH k ≤ kf k′H k ≤ K2 kf kH k
In particular the two norms define the same topology (i.e. the concept of convergence is the
same), so for every practical purpose it does not matter which one you use. (Most books
actually prefer kf k′H k ).
Note that
C k ⊂ H k ⊂ L2
The set C k is dense in L2 in the usual L2 -norm, in other words, if you close C k in the k · k2 ,
then you get L2 . But if you close C k in the stronger k · kH k norm, then the closure is H k .
The major advantage is that one can interchange derivatives and limits under a control
given by a scalar product. We can interchange derivatives and limits if uniform control is
given. But typically it is hard to check if fn converges uniformly. It is much easier to check
convergence in k · kH k norms because you can compute on the Fourier side.
For example, suppose that fn ∈ C 1 and fn is Cauchy in k · kH 1 . This is a checkable
condition in many cases. Then fn converges to f in k · kH 1 , and f will have weak derivative,
namely fn′ → f ′ in L2 sense. It may be that f 6∈ C 1 , but still we can talk about its derivative,
and having introduced this concept, the limit and differentiation can be interchanged.
Another interesting point is the following. We have seen that C 1 ⊂ H 1 ⊂ L2 , and it is
also easy to check that H k ⊂ H ℓ whenever k ≥ ℓ.
We also know that C 1 ⊂ C ⊂ L2 . Question: where is C “nested” in the Sobolev space hierarchy. These questions are answered by various Sobolev-embedding theorems and are nontrivial
for general spaces. On the [0, 2π] the answer is the following:
Theorem 1.11 (Sobolev) For any ε > 0
but H 1/2 [0, 2π] 6⊂ C[0, 2π].
H 1/2+ε [0, 2π] ⊂ C[0, 2π]
9
Proof. The P
proof of the first statement P
is similar to the proof of Theorem (1.5). The key is
1+2ε
2
to notice that
|n|
|cn | < ∞ implies n |cn | < ∞ which isPa similar Schwarz inequality
used in the proof of Theorem (1.5), together with the fact that n |n|−1−2ε < ∞. The second
statement requires to construct a non-continuous H 1/2 function. Hint: try a function that is
very mildly singular.
1.6
Solution to the heat equation on S 1
We show the power of the F-transform by solving the simplest PDE (partial diff. equation).
Let f (x, t) be periodic function (imagine it is the temperature at point x ∈ S 1 at time t),
satisfying
∂t f = ∆x f
where ∆x = d2 /dx2 . we set the initial condition f (x, 0) = f0 (x), where f0 ∈ L2 (S 1 ) is given.
Suppose that ft = f (·, t) ∈ C 2 for any t > 0 (we’ll prove it later) and let its F-transform
be
X
ft =
cn (t)en
(note that the F-coefficients depend on t). Then
X
∆x f =
(−n2 )cn (t)en
and
∂t ft =
X
c′n (t)en
(we’ll see in a moment, than c′n (t) decays sufficiently fast as n → ∞ so that you can interchange
the infinite sum and the t-derivative).
Equating the F-coefficients, we have
c′n (t) = −n2 cn (t)
P
and the initial condition is cn (0) = cn , where f0 = cn en . Therefore
2
cn (t) = e−tn cn
So we can summarize
Theorem 1.12 Let f0 ∈ L2 (S 1 ), f0 =
P
cn en . The the function
X
2
f (t, x) =
e−tn cn en (x)
10
is smooth (C ∞ ), for any t > 0 it solves
∂t f = ∆x f
and as t → 0
lim f (t, ·) = f0
t→0
where the limit is in L2
Note that the time direction is important: the solution works for t > 0. If t < 0, then the
F-series of the presumed solution would blow up (unless only finitely many cn ’s are nonzero).
Proof. We just have to check the conditions we imposed in the calculation above the
theorem. If t > 0, then
X
1/2 X
X
2
2 1/2
nk · e−tn |cn | ≤
|cn |2
n2k e−2tn
<∞
since the Gaussian function decays much faster than any polynomial. Using Theorem 1.7, we
obtain that ft ∈ C k for any k.
Similarly one can check thatP∂t can be brought
inside the summation, because the derivative
P 2 −tn
2
′
series uniformly converges by
|cn (t)| = n e
< ∞.
Finally we can check that
X
2
kf (t, ·) − f0 k2 =
(e−tn − 1)2 |cn |2 → 0
as t → 0 by monotone (or dominated) convergence.
Note that the formal solution to the heat equation is
ft = et∆ f0
so the question is how to define et∆ . More general, one can consider a function G : R → R
and ask how to define G(∆). The answer is simply multiplication on the Fourier side, i.e.
X
[G(∆)f ](x) :=
G(−n2 )cn en (x)
n
P
if f (x) = n cn en (x) and f ∈ L2 . But one has to make sure that the RHS makes sense, i.e.
you can define G(∆) on the function f only if
X
|G(−n2 )|2 |cn |2 < ∞
n
11
1.7
Remarks on fractional Sobolev spaces and several dimensions
The fractional Sobolev spaces (i.e. when the order k is not integer) give rise to the possibility
to define fractional derivatives. The natural definition would be
X
(−iD)k f =
nk cn en (x)
n
for functions where the RHS is well defined, but this has a major flaw for noninteger k. When
k is not integer and n is √
a negative number, then the definition of nk is not unique (think of
k = 1/2, n = −1, then −1 = ±i). To avoid the problem of choosing the “right” version,
one defines only (−∆)k/2 instead of (−iD)k , i.e.
X
(−∆)k/2 f :=
|n|k cn en (x)
n
Note that the absolute value removes the ambiguity, but this operator is not the same as
(−iD)k (although the identity [(−iD)2 ]k/2 = (−∆)k/2 looks tempting, one should remember
that taking fractional powers even for complex numbers is a multivalued operation. Without
keeping this mind, one would run into the
i = (−1)1/2 = [(−1)2 ]1/4 = 11/4 = 1
nonsense).
It is also clear for integer k’s, e.g.
X
X
|n|k cn en (x) 6=
nk cn en (x) = (−iD)f
(−∆)1/2 f :=
n
n
despite the fact that the square of both operators are the same:
(−∆)1/2 (−∆)1/2 = −∆ = (−iD)(−iD)
But this should not surprise us, since (−1)(−1) = 1 · 1 but −1 6= 1. As usual,
√ taking
the absolute value (which is the nonnegative square root of the square: |n| = n2 ) loses
information; in this way both (−iD)k and (iD)k are translated into the same operator.
Finally, I’d like to mention, that we dealt with one-dimensional setup for mere simplicity.
Everything is true in higher dimensions. Consider T d = [0, 2π]d the d-dimensional torus,
which is the cube [0, 2π]d with its opposite sides identified. This in particular means that for
continuous functions on T d we require periodicity in all variables:
f (x1 , x2 , . . . , xk−1 , 0, xk+1 , . . .) = f (x1 , x2 , . . . , xk−1 , 2π, xk+1, . . .)
12
(for all x1 , x2 , . . . without xk ). For spaces with higher derivatives we also require the periodicity
of the corresponding derivatives.
We define
ein·x
en (x) :=
x ∈ [0, 2π]d
(2π)d/2
for any n = (n1 , . . . nd ) ∈ Zd . This is an ONB. For any f ∈ L2 ([0, 2π]d ) we define
Z
e−in·x
cn := (en , f ) =
f (x)dx
d/2
T d (2π)
to be the F-coefficient. Then
f=
X
cn en
n∈Zd
in L2 ([0, 2π]d)-sense, i.e.
in L2 where
SN (f ) → f
SN (f ) :=
X
cn en
n∈Zd
|ni |≤N
All one-dimensional theorems goes through essentially directly.
Homework 1.13 Think over all statements mentioned in this notes up to now on Fourier
series and generalize them to several dimensions. Most of them go through without changes,
but the relations between C k and H k spaces gets modified. Notice that the
was
P dimension
−2
essentially used only at one point, namely in (1.2), the summability of
is a onenn
dimensional statement. Find out its replament in d-dimensions.
2
Fourier transform in Rd
The presentation basically Lieb-Loss Chapter 5, but with the convention used in Reed-Simon.
All spaces are over Rd .
2.1
L1 Fourier transform
Let f ∈ L1 (Rd ), then its F-transform is defined as
Z
1
ˆ
f (k) :=
e−ikx f (x)dx
(2π)d/2
13
Note that the integral makes sense. In the exponent we have kx = k · x scalar product of two
n-vectors, but for simplicity we’ll just write it as kx.
ˆ
Sometimes we’ll denote the F-transform by F (f ) := f.
Remark. Whenever you do F-transform, there is a constant hassle with the 2π’s. Different
−n/2
ˆ
books use different conventions, sometimes
prefactor, someR −2πikxf is defined without the (2π)
times they put it in the exponent: e
f (x)dx etc. I use the convention of Reed-Simon.
But whenever you read another book and the 2π’s don’t come out right, then you should
check at the beginning how that book defined the F-transform...
One can view the F-transform as a map: ˆ : f 7→ fˆ between L1 7→ L∞ . It is clear that
kfˆk∞ ≤ (2π)−d/2 kf k1 .
Proposition 2.1 (Properties of F-transform) (i) [Convolution] Let f, g ∈ L1 , then
\
(f
⋆ g) = (2π)d/2 fˆĝ
i.e. the convolution goes into product under F-transform
(ii) [Translation]
f\
(· − h) = e−ikh fˆ(k)
(iii) [Scaling]
f\
(·/λ) = λd fˆ(λk).
Proof. Calculation. The only point is in (i) to use Fubini properly:
Z
Z
1
−ikx
ˆ
(f ⋆ g)(k) =
f (x − y)g(y)dydx
e
(2π)d/2
Z
Z
1
−iky
−ik(x−y)
ˆ
=
e
g(y)
e
f
(x
−
y)dx
dy = (2π)d/2 f(k)ĝ(k)
(2π)d/2
where the interchange of integration is guaranteed because f, g ∈ L1 , so the integrand is
always in L1 (dxdy).
14
2.2
Fourier transform of Gaussian function.
Let
f (x) =
then
R
1
2
e−x /2
d/2
(2π)
f = 1. We compute
1
fˆ(k) =
(2π)d
Z
e−ikx e−x
with
g(k) :=
Z
2 /2
=
1
1
−k 2 /2
g(k)e
(2π)d
2
e− 2 (x+ik) dx
(integral is convergent). One can easily check by integration by parts, that ∇g(k) = 0,
therefore g(k) = g(0) = (2π)d/2 . Hence we obtain
fˆ = f
(The fact that g is constant also follows from the Cauchy Theorem in complex analysis by inte2
grating the analytic function e−z and moving the integration contour from the line Im(z) = k
to the real axis.)
Actually the Gaussian is the only nonnegative function with the property that the Ftransform leaves it invariant.
2.3
Fourier transform of L2 functions
Theorem 2.2 Let f ∈ L1 ∩ L2 . Then
ˆ 2 (also called Plancherel formula, similarly to Parseval identity
(i) fˆ ∈ L2 and kf k2 = kfk
for F-series).
(ii) The F-transform as a map f → fˆ can be extended as a bounded map on L2 from
f ∈ L1 ∩ L2 to f ∈ L2 .
ˆ ĝ). I.e. the Fourier transform is isometry on L2 .
(iii) If f, g ∈ L2 , then (f, g) = (f,
Remark: We do not know yet that the F-transform is bijection, isometry is weaker. It is,
but this will come later.
Proof. (i) Since fˆ ∈ L∞ , for any ε > 0 we have
Z
ˆ 2 e−εk2 dk < ∞
|f(k)|
15
We compute it
Z
2
|fˆ(k)|2 e−εk dk =
Z Z
Z
1
2
ikx
−iky
f
(x)e
dx
f
(y)e
dy
e−εk dk
d
(2π)
Z
Z
1
ik(x−y) −εk 2
f
(x)f
(y)
e
e
dkdxdy
=
(2π)d
2
The integrals can be interchanged since f (x)f (y)e−εk is in the L1 space of dxdydk. The last
Gaussian integral can be explicitly computed, we obtain
Z
= f (x)f (y)jε (x − y)dxdy
where
jε (x − y) :=
therefore, in summary
Z
(x−y)2
1
− 4ε
e
(4επ)d/2
2
ˆ 2 e−εk dk = (f, f ⋆ jε )
|f(k)|
(2.3)
Recall an old theorem, that we proved (Theorem 2.16 in [Lieb-Loss])
R
R
Theorem 2.3 Let j(x) be such that j = 1 and |j| < ∞. Let jδ (x) := δ −n j(x/δ). Then
for any f ∈ Lp (p < ∞) we have f ⋆ jδ → f in Lp as δ → 0.
Using this theorem, we recognize that the right hand side of (2.3) goes to kf k2 as ε → 0. In
particular the right hand side is bounded, uniformly in ε. Therefore by monotone convergence,
ˆ 2 , so we get (i).
the LHS converges to kfk
(ii) The extension is a standard argument. Let f ∈ L2 , take an approximating sequence
2
2
fn ∈ L1 ∩ L2 such that fn → f in L2 (e.g. fn (x) := f (x)χ(|x| ≤ n or fn (x) := f (x)e−x /n ).
Then
kfˆn − fˆm k2 = kfn − fm k2
by part (i), but this is a Cauchy sequence since fn → f . Therefore fˆn is also Cauchy and its
ˆ One has to
limit exists. This limit is defined to be the F-transform of f , again denoted by f.
check that this definition does not depend on the approximating sequence. But if fˆn → F and
ĝn → G, where fn → f and gn → f , then clearly kfˆn −ĝn k = kfn −gn k ≤ kfn −f k+kgn −f k →
ˆ holds
0, so F = G. Again, by limiting argument we see that the Parseval identity, kf k = kfk
2
for f ∈ L as well.
16
The following formula gives the Fourier transform explicitly (using the fn (x) := f (x)e−x
cutoff) for a function f ∈ L2 :
Z
1
2
2
ˆ
f(k) := lim √
e−ikx e−x /n f (x)dx
n→∞
2π
2 /n2
where the limit is in L2 -sense (not pointwise!). Note that one cannot use dominated convergence to interchange the limit and integrals; the limit does not exists for all k, but it exists
for almost all k. The integral above if well defined for any finite n, because f ∈ L2 implies
2
2
that e−x /n f (x) ∈ L1 (WHY?).
ˆ ĝ) because in any complex
(iii) From kf k = kfˆk we easily conclude that (f, g) = (f,
H-space the following, so called polarization identity holds:
i
1h
(x, y) = kx + yk2 − kx − yk2 − ikx + iyk2 + ikx − iyk2
4
(it is amusing to check it). This ensures that the norm determines the scalar product (not
just the other way around).
2.4
Inversion formula
Theorem 2.4 For any g ∈ L2 (Rd ) we define
1
ǧ(x) := ĝ(−x) =
(2π)d/2
Then
Z
eikx g(k)dk
(fˆ)ˇ= f
In other words, the Fourier transform is invertible, the inverse transform is almost the same,
but note that the sign in the exponent is different. In particular, we have
Corollary 2.5 The F-transform f → fˆ on L2 is unitary.
This directly follows from the isometry and the invertibility.
Before the rigorous proof, let me show you how a physicist think about it. Let me do
n = 1 dimension. We would like to check that
Z
Z
1
ikx
e−iky f (y)dydk = f (x)
e
2π
17
If you disregard the question of validity of interchanging integrals, then we have to show that
Z
1
eik(x−y) dk = δ(x − y)
(2.4)
2π
where δ(z) is the delta function at the origin.
There are some confusion about the delta function, it looks like nonrigorous mathematics.
The way to think about it is that it is not a function but really a measure that is written like
δ(z)dz. This is the Lebesgue-Stieltjes measure associated to the absolutely honest monotonic
function H(x) = χ(x ≥ 1) (which is by the way called the Heaviside function). The measure
assigns the value 1 to any set that contains the origin and 0 otherwise. The integration with
respect to this measure is
Z
f (z)δ(z)dz = f (0)
So the real meaning of (2.4) is exactly
Z Z
1
eik(x−y) dk f (y)dy = f (x)
2π
The rigorous proofs have to do some regularization of the divergent integral in (2.4). This
is somewhat delicate and one can get the wrong result if one is not careful. The physicists
have an enormous experience and instinct to do it always (mostly) right without going through
rigorous steps.
Rigorous proof of the Inversion formula. We need the following
Lemma 2.6 Let f ∈ L2 , λ > 0 then
Z
Z
2
(x−y)2
1
1
− 2λ
− λk2 ikx ˆ
e
e
f
(y)dy
=
e f (k)dk
(2πλ)d/2
(2π)d/2
(2.5)
Proof: First we check it for f ∈ L1 . Then the RHS is
Z
2
1
− λk2 ikx −iky
e e
f (y)dkdy
RHS =
e
(2π)d
and the integrals can be freely interchanged since we have L1 control in both k and y. The
Gaussian dk integral can be computed and one gets the LHS.
If f ∈ L2 , then we approximate it by fn ∈ L1 , fn → f (in L2 ). By Plancherel (extended to
2
L ) we obtain fˆn → fˆ, and plugging these limits into (2.5), checking that the Gaussian factors
on both sides are in L2 , the convergence follows. (Another way to argue is to use the second
18
part of Riesz-Fischer theorem, i.e. one can choose a subsequence fnj so that fnj (x) → f (x)
and fbnj (k) → fb(k) almost everywhere, and both of these sequences are dominated by an L2
function, |fnj (x)| ≤ F (x) ∈ L2 and |fbnj (k)| ≤ G(k) ∈ L2 . Together with the Gaussian factors,
these L2 functions are also in L1 , giving the necessary domination, the claim also follows from
dominated convergence). Once (2.5) is established, we take λ → 0 limit. The LHS goes to f (x) (in L2 ) by Theorem
(2.3) (one has to check that the Gaussian on the LHS with the given normalization has integral
2
ˆ pointwise convergence (as λ → 0) and as fˆ ∈ L2
1). On the RHS we have e−λk /2 fˆ(k) → f(k)
we have also L2 convergence (dominated convergence). Using Plancherel for the inverse Ftransform (which is clearly valid since the F-transform and its inverse differ only by a reflection
2
ˆ
x 7→ −x) we get that the RHS of (2.5), that is the inverse F-transform of e−λk /2 f(k)
also
converges to the inverse F-transform of fˆ and this was to be proved. 2.5
Fourier transform in Lp
From the original definition of the Fourier transform it is clear that
kfbk∞ ≤ (2π)−d/2 kf k1
i.e. the Fourier transform is a bounded linear transformation from L1 (Rd ) to L∞ (Rd ).
From the L2 -theory we have seen that the Fourier transform is a linear isomorphism, in
particular a bounded map (with bound 1) from L2 to L2
b 2 = kf k2
kfk
From these two properties it follows that the Fourier transform is also a bounded map from
Lp to Lq where p−1 + q −1 = 1 and 1 ≤ p ≤ 2. This is:
Theorem 2.7 (Hausdorff-Young) There exists a constant C, depending only on the dimension, such that
b Lq (Rd ) ≤ Ckf kLp (Rd )
kfk
for any 1 ≤ p ≤ 2.
The best possible constant (that also depends on p) can be found in Theorem 5.7 of
Lieb-Loss.
The standard proof of this theorem relies on an interpolation theorem for linear maps,
called the Riesz-Thorin interpolation theorem. This one says that if a linear map T acting on
functions is bounded from Lp1 to Lq1 and from Lp2 to Lq2 then it is also bounded from any
Lp3 to Lq3 as long as p3 ∈ [p1 , p2 ]. (Here pj and qj are dual exponents). The precise statement
of Riesz-Thorin and its proof is found, e.g., in Reed-Simon, Vol II, Theorem IX.17.
19
2.6
Heat equation on Rd
Let f ∈ C 1 ∩ L2 , then a simple integration by parts shows that
c
∇f(k)
= ik fˆ(k)
(note that k is a vector and ∇ is also a vector). Similarly for higher derivatives, for example
c (k) = −k 2 fˆ(k).
∆f
Consider the heat equation
∂t ft = ∆ft
for the unknown function ft (x) = f (x, t) with initial data f (x, 0) = f0
Suppose that ft ∈ C 2 , then we can take the F-transform of the equation:
∂ fˆt = −k 2 fˆt (k)
so the solution is
2
fˆt (k) = e−tk fˆ0 (k)
Back to x-space
2
ft (x) = e−tk fˆ0 ˇ=
1 −tk2 1
e
ˇ⋆ f0 =
d/2
(2π)
(4πt)d/2
Z
e−
(x−y)2
4t
f0 (y)dy
where we used how the F-transform (and its inverse) behaves under convolution. Once this
representation is given, it is easy to check that indeed ft ∈ C 2 (even more: ft ∈ C ∞ ) for t > 0.
Finally one checks that ft → f0 in L2 as t → 0 + 0 exactly as in the case of S 1 .
20