5Fourier transform

II Probability and Measure 5.4 Fourier transform in L
2
It turns out wonderful things happen when we take the Fourier transform of an
L
2
function.
Theorem
(Plancherel identity)
.
For any function
f L
1
L
2
, the Plancherel
identity holds:
k
ˆ
fk
2
= (2π)
d/2
kfk
2
.
As we are going to see in a moment, this is just going to follow from the
Fourier inversion formula plus a clever trick.
Proof.
We first work with the special case where
f,
ˆ
f L
1
, since the Fourier
inversion formula holds for f. We then have
kfk
2
2
=
Z
f(x)f(x) dx
=
1
(2π)
d
Z
Z
ˆ
f(u)e
i(u,x)
du
f(x) dx
=
1
(2π)
d
Z
ˆ
f(u)
f(x)e
i(u,x)
dx
du
=
1
(2π)
d
Z
ˆ
f(u)
f(x)e
i(u,x)
dx
du
=
1
(2π)
d
Z
ˆ
f(u)
ˆ
f(u) du
=
1
(2π)
d
k
ˆ
f(u)k
2
2
.
So the Plancherel identity holds for f .
To prove it for the general case, we use this result and an approximation
argument. Suppose that
f L
1
L
2
, and let
f
t
=
f g
t
. Then by our earlier
lemma, we know that
kf
t
k
2
kfk
2
as t 0.
Now note that
ˆ
f
t
(u) =
ˆ
f(u)ˆg
t
(u) =
ˆ
f(u)e
−|u|
2
t/2
.
The important thing is that e
−|u|
2
t/2
% 1 as t 0. Therefore, we know
k
ˆ
f
t
k
2
2
=
Z
|
ˆ
f(u)|
2
e
−|u|
2
t
du
Z
|
ˆ
f(u)|
2
du = k
ˆ
fk
2
2
as t 0, by monotone convergence.
Since f
t
,
ˆ
f
t
L
1
, we know that the Plancherel identity holds, i.e.
k
ˆ
f
t
k
2
= (2π)
d/2
kf
t
k
2
.
Taking the limit as t 0, the result follows.
What is this good for? It turns out that the Fourier transform gives as
a bijection from
L
2
to itself. While it is not true that the Fourier inversion
formula holds for everything in
L
2
, it holds for enough of them that we can just
approximate everything else by the ones that are nice. Then the above tells us
that in fact this bijection is a norm-preserving automorphism.
Theorem.
There exists a unique Hilbert space automorphism
F
:
L
2
L
2
such that
F ([f]) = [(2π)
d/2
ˆ
f]
whenever f L
1
L
2
.
Here [
f
] denotes the equivalence class of
f
in
L
2
, and we say
F
:
L
2
L
2
is
a Hilbert space automorphism if it is a linear bijection that preserves the inner
product.
Note that in general, there is no guarantee that
F
sends a function to its
Fourier transform. We know that only if it is a well-behaved function (i.e. in
L
1
L
2
). However, the formal property of it being a bijection from
L
2
to itself
will be convenient for many things.
Proof. We define F
0
: L
1
L
2
L
2
by
F
0
([f]) = [(2π)
d/2
ˆ
f].
By the Plancherel identity, we know F
0
preserves the L
2
norm, i.e.
kF
0
([f])k
2
= k[f]k
2
.
Also, we know that
L
1
L
2
is dense in
L
2
, since even the continuous functions
with compact support are dense. So we know
F
0
extends uniquely to an isometry
F : L
2
L
2
.
Since it preserves distance, it is in particular injective. So it remains to show
that the map is surjective. By Fourier inversion, the subspace
V = {[f] L
2
: f,
ˆ
f L
1
}
is sent to itself by the map
F
. Also if
f V
, then
F
4
[
f
] = [
f
] (note that
applying it twice does not suffice, because we actually have
F
2
[
f
](
x
) = [
f
](
x
)).
So
V
is contained in the image
F
, and also
V
is dense in
L
2
, again because it
contains all Gaussian convolutions (we have
ˆ
f
t
=
ˆ
f ˆg
t
, and
ˆ
f
is bounded and
ˆg
t
is decaying exponentially). So we know that F is surjective.