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 L^{1}
n
(ℝ )
. 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 L^{2} because such functions might not be in
L^{1} if they are defined on ℝ or ℝ^{n}. However, it was realized long ago that if a function is in
L^{1}∩ L^{2}, then the L^{2} norm of the function is equal to the L^{2} 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 L^{2}. If you solve this, perhaps by using
density of L^{1}∩ L^{2} in L^{2}, you are still faced with the problem of taking the Fourier transform
of an arbitrary function in L^{p}. The method developed here removes all these difficulties at
once.
Definition 10.5.1Let 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,
T (aψ +bϕ) = aT (ψ) + bT (ϕ) for all a,b ∈ ℂ, ψ,ϕ ∈ G
Let ψ ∈G. Then we can regard ψ as an element of G^{∗}by defining
∫
ψ (ϕ) ≡ n ψ (x)ϕ(x)dx.
ℝ
This implies the following important lemma.
Lemma 10.5.2The following is obtained for all ϕ,ψ ∈G.
, p ≥ 1, or suppose f is measurable and has polynomialgrowth,
( 2)m
|f (x)| ≤ K 1+ |x|
for some m ∈ ℕ. Then if
∫
fψdx = 0
for all ψ ∈G, then it follows f = 0.
Proof: First note that if f ∈ L^{p}
(ℝn )
or has polynomial growth, then it makes sense to write the
integral ∫fψdx described above. This is obvious in the case of polynomial growth. In the case where
f ∈ L^{p}
∫ p ∫ ( p−2-- ) ∫
ℝn |f |dx = ℝn f |f| f − gk dx+ ℝn fgkdx
∫ ( p−2-- )
= n f |f| f − gk dx
ℝ |||| -||||
≤ ||f||Lp||gk − |f|p−2f||′
p
which converges to 0. Hence f = 0.
It remains to consider the case where f has polynomial growth. Thus x → f
(x)
e^{−|x|
2
}∈ L^{1}
n
(ℝ )
.
Therefore, for all ψ ∈G,
∫ 2
0 = f (x)e− |x|ψ (x)dx
because e^{−}
|x|
^{2}ψ
(x )
∈G. Therefore, by the first part, f
(x)
e^{−|x|
2
} = 0 a.e. ■
Note that “polynomial growth” could be replaced with a condition of the form
|f (x )| ≤ K (1 + |x|2)m ek|x|α, α < 2
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 10.5.9Let f be a measurable function with polynomial growth,
( 2)N
|f (x )| ≤ C 1 + |x| for some N,
or let f ∈ L^{p}
(ℝn)
for some p ∈
[1,∞ ]
. Then f ∈G^{∗}if
∫
f (ϕ ) ≡ fϕdx.
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 ∈ L^{p}
n
(ℝ )
with ∞ > 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|x|
2
} and these are in
L^{p′
}
n
(ℝ )
. Also ϕ → f
(ϕ)
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
∫
ψ → n ψdμ
ℝ
is a linear functional on G. This includes the very important case of probability distribution
measures.