5Fourier transform
II Probability and Measure
5.3 Fourier inversion formula
We now want to work towards proving the Fourier inversion formula:
Theorem (Fourier inversion formula). Let f,
ˆ
f ∈ L
1
. Then
f(x) =
1
(2π)
d
Z
ˆ
f(u)e
−i(u,x)
du a.e.
Our strategy is as follows:
(i)
Show that the Fourier inversion formula holds for a Gaussian distribution
by direct computations.
(ii)
Show that the formula holds for Gaussian convolutions, i.e. the convolution
of an arbitrary function with a Gaussian.
(iii)
We show that any function can be approximated by a Gaussian convolution.
Note that the last part makes a lot of sense. If
X
is a random variable, then
convolving with a Gaussian is just adding
X
+
√
tZ
, and if we take
t →
0, we
recover the original function. What we have to do is to show that this behaves
sufficiently well with the Fourier transform and the Fourier inversion formula
that we will actually get the result we want.
Gaussian densities
Before we start, we had better start by defining the Gaussian distribution.
Definition (Gaussian density). The Gaussian density with variance t is
g
t
(x) =
1
2πt
d/2
e
−x
2
/2t
.
This is equivalently the density of
√
tZ
, where
Z
= (
Z
1
, ··· , Z
d
) with
Z
i
∼
N(0, 1) independent.
We now want to compute the Fourier transformation directly and show that
the Fourier inversion formula works for this.
We start off by working in the case
d
= 1 and
Z ∼ N
(0
,
1). We want to
compute the Fourier transform of the law of this guy, i.e. its characteristic
function. We will use a nice trick.
Proposition. Let Z ∼ N(0, 1). Then
φ
Z
(a) = e
−u
2
/2
.
We see that this is in fact a Gaussian up to a factor of 2π.
Proof. We have
φ
Z
(u) = E[e
iuZ
]
=
1
√
2π
Z
e
iux
e
−x
2
/2
dx.
We now notice that the function is bounded, so we can differentiate under the
integral sign, and obtain
φ
0
Z
(u) = E[iZe
iuZ
]
=
1
√
2π
Z
ixe
iux
e
−x
2
/2
dx
= −uφ
Z
(u),
where the last equality is obtained by integrating by parts. So we know that
φ
Z
(u) solves
φ
0
Z
(u) = −uφ
Z
(u).
This is easy to solve, since we can just integrate this. We find that
log φ
Z
(u) = −
1
2
u
2
+ C.
So we have
φ
Z
(u) = Ae
−u
2
/2
.
We know that A = 1, since φ
Z
(0) = 1. So we have
φ
Z
(u) = e
−u
2
/2
.
We now do this problem in general.
Proposition.
Let
Z
= (
Z
1
, ··· , Z
d
) with
Z
j
∼ N
(0
,
1) independent. Then
√
tZ
has density
g
t
(x) =
1
(2πt)
d/2
e
−x
2
/(2t)
.
with
ˆg
t
(u) = e
−u
2
t/2
.
Proof. We have
ˆg
t
(u) = E[e
i(u,
√
tZ)
]
=
d
Y
j=1
E[e
i(u
j
,
√
tZ
j
)
]
=
d
Y
j=1
φ
Z
(
√
tu
j
)
=
d
Y
j=1
e
−tu
2
j
/2
= e
−u
2
t/2
.
Again,
g
t
and
ˆg
t
are almost the same, apart form the factor of (2
πt
)
−d/2
and
the position of t shifted. We can thus write this as
ˆg
t
(u) = (2π)
d/2
t
−d/2
g
1/t
(u).
So this tells us that
ˆ
ˆg
t
(u) = (2π)
d
g
t
(u).
This is not exactly the same as saying the Fourier inversion formula works,
because in the Fourier inversion formula, we integrated against
e
−i(u,x)
, not
e
i(u,x)
. However, we know that by the symmetry of the Gaussian distribution,
we have
g
t
(x) = g
t
(−x) = (2π)
−d
ˆ
ˆg
t
(−x) =
1
2π
d
Z
ˆg
t
(u)e
−i(u,x)
du.
So we conclude that
Lemma.
The Fourier inversion formula holds for the Gaussian density function.
Gaussian convolutions
Definition
(Gaussian convolution)
.
Let
f ∈ L
1
. Then a Gaussian convolution
of f is a function of the form f ∗ g
t
.
We are now going to do a little computation that shows that functions of
this type also satisfy the Fourier inversion formula.
Before we start, we make some observations about the Gaussian convolution.
By general theory of convolutions, we know that we have
Proposition.
kf ∗ g
t
k
1
≤ kfk
1
.
We also have a pointwise bound
f ∗ g
t
(x) =
Z
f(x − y)e
−y
2
/(2t)
1
2πt
d/2
dy
≤ (2πt)
−d/2
Z
f(x − y) dx
≤ (2πt)
−d/2
kfk
1
.
This tells us that in fact
Proposition.
kf ∗ g
t
k
∞
≤ (2πt)
−d/2
kfk
1
.
So in fact the convolution is pointwise bounded. We see that the bound gets
worse as
t →
0, and we will see that this is because as
t →
0, the convolution
f ∗ g
t
becomes a better and better approximation of
f
, and we did not assume
that f is bounded.
Similarly, we can compute that
Proposition.
k
\
f ∗ g
t
k
1
= k
ˆ
f ˆg
t
k
1
≤ (2π)
d/2
t
−d/2
k
ˆ
fk
1
,
and
k
\
f ∗ g
t
k
∞
≤ k
ˆ
fk
∞
.
Now given these bounds, it makes sense to write down the Fourier inversion
formula for a Gaussian convolution.
Lemma. The Fourier inversion formula holds for Gaussian convolutions.
We are going to reduce this to the fact that the Gaussian distribution itself
satisfies Fourier inversion.
Proof. We have
f ∗ g
t
(x) =
Z
f(x − y)g
t
(y) dy
=
Z
f(x − y)
1
(2π)
d
Z
ˆg
t
(u)e
−i(u,y)
du
dy
=
1
2π
d
ZZ
f(x − y)ˆg
t
(u)e
−i(u,y)
du dy
=
1
2π
d
Z
Z
f(x − y)e
−i(u,x−y)
dy
ˆg
t
(u)e
−i(u,x)
du
=
1
2π
d
Z
ˆ
f(u)ˆg
t
(u)e
−i(u,x)
du
=
1
2π
d
Z
\
f ∗ g
t
(u)e
−i(u,x)
du
So done.
The proof
Finally, we are going to extend the Fourier inversion formula to the case where
f,
ˆ
f ∈ L
2
.
Theorem (Fourier inversion formula). Let f ∈ L
1
and
f
t
(x) = (2π)
−d
Z
ˆ
f(u)e
−u
2
t/2
e
−i(u,x)
du = (2π)
−d
Z
\
f ∗ g
t
(u)e
−i(u,x)
du.
Then
kf
t
−fk
1
→
0, as
t →
0, and the Fourier inversion holds whenever
f,
ˆ
f ∈ L
1
.
To prove this, we first need to show that the Gaussian convolution is indeed
a good approximation of f:
Lemma.
Suppose that
f ∈ L
p
with
p ∈
[1
, ∞
). Then
kf ∗ g
t
− fk
p
→
0 as
t → 0.
Note that this cannot hold for
p
=
∞
. Indeed, if
p
=
∞
, then the
∞
norm
is the uniform norm. But we know that
f ∗ g
t
is always continuous, and the
uniform limit of continuous functions is continuous. So the formula cannot hold
if f is not already continuous.
Proof.
We fix
ε >
0. By a question on the example sheet, we can find
h
which
is continuous and with compact support such that kf − hk
p
≤
ε
3
. So we have
kf ∗ g
t
− h ∗ g
t
k
p
= k(f − h) ∗ g
t
k
p
≤ kf − hk
p
≤
ε
3
.
So it suffices for us to work with a continuous function
h
with compact support.
We let
e(y) =
Z
h(x − y) − h(x)
p
dx.
We first show that e is a bounded function:
e(y) ≤
Z
2
p
(h(x − y)
p
+ h(x)
p
) dx
= 2
p+1
khk
p
p
.
Also, since
h
is continuous and bounded, the dominated convergence theorem
tells us that e(y) → 0 as y → 0.
Moreover, using the fact that
R
g
t
(y) dy = 1, we have
kh ∗ g
t
− hk
p
p
=
Z
Z
(h(x − y) − h(x))g
t
(y) dy
p
dx
Since
g
t
(
y
) d
y
is a probability measure, by Jensen’s inequality, we can bound
this by
≤
ZZ
h(x − y) − h(x)
p
g
t
(y) dy dx
=
Z
Z
h(x − y) − h(x)
p
dx
g
t
(y) dy
=
Z
e(y)g
t
(y) dy
=
Z
e(
√
ty)g
1
(y) dy,
where we used the definition of
g
and substitution. We know that this tends to 0
as
t →
0 by the bounded convergence theorem, since we know that
e
is bounded.
Finally, we have
kf ∗ g
t
− fk
p
≤ kf ∗ g
t
− h ∗ g
t
k
p
+ kh ∗ g
t
− hk
p
+ kh − fk
p
≤
ε
3
+
ε
3
+ kh ∗ g
t
− hk
p
=
2ε
3
+ kh ∗ g
t
− hk
p
.
Since we know that
kh ∗g
t
− hk
p
→
0 as
t →
0, we know that for all sufficiently
small t, the function is bounded above by ε. So we are done.
With this lemma, we can now prove the Fourier inversion theorem.
Proof of Fourier inversion theorem.
The first part is just a special case of the
previous lemma. Indeed, recall that
\
f ∗ g
t
(u) =
ˆ
f(u)e
−u
2
t/2
.
Since Gaussian convolutions satisfy Fourier inversion formula, we know that
f
t
= f ∗ g
t
.
So the previous lemma says exactly that kf
t
− fk
1
→ 0.
Suppose now that
ˆ
f ∈ L
1
as well. Then looking at the integrand of
f
t
(x) = (2π)
−d
Z
ˆ
f(u)e
−u
2
t/2
e
−i(u,x)
du,
we know that
ˆ
f(u)e
−u
2
t/2
e
−i(u,x)
≤ 
ˆ
f.
Then by the dominated convergence theorem with dominating function

ˆ
f
, we
know that this converges to
f
t
(x) → (2π)
−d
Z
ˆ
f(u)e
−i(u,x)
du as t → 0.
By the first part, we know that
kf
t
− fk
1
→
0 as
t →
0. So we can find a
sequence (
t
n
) with
t
n
>
0,
t
n
→
0 so that
f
t
n
→ f
a.e. Combining these, we
know that
f(x) =
Z
ˆ
f(u)e
−i(u,x)
du a.e.
So done.