, p ≥ 1, or suppose f is measurable and haspolynomial growth,
( )m
|f (x)| ≤ K 1+ |x|2
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}
∫ ∫ ∫
|f|pdx = f (|f|p− 2f − g ) dx+ f gdx
ℝn ℝn k ℝn k
∫ ( p− 2-- )
= ℝn f |f| f − gk dx
|||| p− 2-||||
≤ ||f||Lp||gk − |f| f||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}
(ℝn)
. Therefore, for all ψ ∈G,
∫
0 = f (x)e−|x|2ψ (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 26.2.8Let f be a measurable function with polynomial growth,
( )
|f (x)| ≤ C 1 +|x|2 N 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 → ψdμ
ℝn
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
later.