10.5.1 Fourier Transforms in G∗
It turns out you can make sense of the Fourier transform of any linear map defined on G. This is a very
abstract way to look at things but if you want ultimate generality, you must do something like this. Part of
the problem is that it is desired to take Fourier transforms of functions which are not in L1
. Thus the
integral which defines the Fourier transform in the above will not make sense. You run into this problem as
soon as you try to take the Fourier transform of a function in
because such functions might not be in
if they are defined on ℝ
. However, it was realized long ago that if a function is in
L1 ∩ L2,
then the L2
norm of the function is equal to the L2
norm of the Fourier transform of the
function. Thus there is an obvious question about whether you can get a definition which will
allow you to directly deal with the Fourier transform on L2
. If you solve this, perhaps by using
density of L1 ∩ L2
you are still faced with the problem of taking the Fourier transform
of an arbitrary function in Lp
. The method developed here removes all these difficulties at
Definition 10.5.1 Let G∗ denote the vector space of linear functions defined on G which have values in ℂ.
Thus T ∈G∗ means T : G→ℂ and T is linear,
Let ψ ∈G. Then we can regard ψ as an element of G∗ by defining
This implies the following important lemma.
Lemma 10.5.2 The following is obtained for all ϕ,ψ ∈G.
Also if ψ ∈G and ψ = 0 in G∗ so that ψ
for all ϕ ∈G, then ψ
= 0 as a function.
The other claim is similar.
Suppose now ψ
= 0 for all
for all ϕ ∈G. Therefore, this is true for ϕ = ψ and so ψ = 0. ■
This lemma suggests a way to define the Fourier transform of something in G∗.
Definition 10.5.3 For T ∈G∗, define FT,F−1T ∈G∗ by
Lemma 10.5.4 F and F−1 are both one to one, onto, and are inverses of each other.
Proof: First note F and F−1 are both linear. This follows directly from the definition. Suppose now
FT = 0. Then FT
= 0 for all
. But F
because if ψ ∈G
, then as
shown above, ψ
= 0 and so F
is one to one. Similarly F−1
is one to one.
Therefore, F−1 ∘F
Similarly, F ∘F−1
Thus both F
are one to one and onto and
are inverses of each other as suggested by the notation. ■
Probably the most interesting things in G∗ are functions of various kinds. The following lemma will be
useful in considering this situation.
Definition 10.5.5 A function f defined on ℝn is in Lloc1
if fXB ∈ L1
for every ball B.
Such functions are termed locally integrable.
Lemma 10.5.6 If f ∈ Lloc1
= 0 for all ϕ ∈ Cc
, then f
Proof: Let E be bounded and Lebesgue measurable. By regularity, there exists a compact set Kk ⊆ E
and an open set V k ⊇ E such that mn
. Let hk
equal 1 on Kk
, vanish on V kC,
values between 0 and 1. Then hk
converges to XE
a set of measure zero.
Hence, by the dominated convergence theorem,
It follows that for E an arbitrary Lebesgue measurable set,
By Corollary 6.7.6, there exists
, a sequence of simple functions converging pointwise to
Then by the dominated convergence theorem again,
Since R is arbitrary,
= 0 a.e.
Corollary 10.5.7 Let f ∈ L1
for all ϕ ∈G. Then f = 0 a.e.
Proof: Let ψ ∈ Cc
Then by the Stone Weierstrass approximation theorem, there exists a
sequence of functions,
such that ϕk → ψ
uniformly. Then by the dominated convergence
By Lemma 10.5.6 f = 0. ■
The next theorem is the main result of this sort.
Theorem 10.5.8 Let f ∈ Lp
, p ≥
1, or suppose f is measurable and has polynomial
for some m ∈ ℕ. Then if
for all ψ ∈G, then it follows f = 0.
Proof: First note that if f ∈ Lp
or has polynomial growth, then it makes sense to write the
described above. This is obvious in the case of polynomial growth. In the case where
f ∈ Lp
it also makes sense because
due to the fact mentioned above that all these functions in G are in Lp
1. Suppose now
that f ∈ Lp,p ≥
The case where f ∈ L1
was dealt with in Corollary
. Suppose f ∈ Lp
and by density of G in Lp′
), there exists a sequence
which converges to 0. Hence f
It remains to consider the case where f has polynomial growth. Thus x → f
Therefore, for all ψ ∈G
. Therefore, by the first part, f
= 0 a.e. ■
Note that “polynomial growth” could be replaced with a condition of the form
and the same proof would yield that these functions are in G∗. The main thing to observe is that almost all
functions of interest are in G∗.
Theorem 10.5.9 Let f be a measurable function with polynomial growth,
or let f ∈ Lp
for some p ∈
. Then f ∈G∗ if
Proof: Let f have polynomial growth first. Then the above integral is clearly well defined and so in
this case, f ∈G∗.
Next suppose f ∈ Lp
∞ > p ≥
1. Then it is clear again that the above integral is well defined
because of the fact that ϕ
is a sum of polynomials times exponentials of the form e−c
and these are in
ϕ → f
is clearly linear in both cases.
This has shown that for nearly any reasonable function, you can define its Fourier transform as
described above. You could also define the Fourier transform of a finite Borel measure μ because for such a
is a linear functional on G. This includes the very important case of probability distribution