26.2.1 Fourier Transforms Of G∗
Definition 26.2.1 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
Then we have the following important lemma.
Lemma 26.2.2 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
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
Definition 26.2.3 For T ∈G∗, define FT,F−1T ∈G∗ by
Lemma 26.2.4 F and F−1 are both one to one, onto, and are inverses of each
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
to one. Similarly F−1
is one to one. Now
Therefore, F−1 ∘ F
Similarly, F ∘ F−1
Thus both F
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.
Lemma 26.2.5 If f ∈ Lloc1
= 0 for all ϕ ∈ Cc
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
, vanish on V kC,
and take values between 0 and 1. Then hk
converges to XE
a set of measure zero. Hence, by the dominated convergence
It follows that for E an arbitrary Lebesgue measurable set,
By Theorem 22.2.1, there exists
, a sequence of simple functions converging
Then by the dominated convergence theorem
Since R is arbitrary,
= 0 a.e.
Corollary 26.2.6 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
By Lemma 26.2.5 f = 0. ■
The next theorem is the main result of this sort.
Theorem 26.2.7 Let f ∈ Lp
, p ≥
1, or suppose f is measurable and has
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 integral
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
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 26.2.8 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
and these are in Lp′
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 measure
is a linear functional on G. This includes the very important case of probability
distribution measures. The theoretical basis for this assertion will be given a little