The problem with G is that it does not contain C_{c}^{∞}
n
(ℝ )
. I have used it in presenting the Fourier
transform because the functions in G have a very specific form which made some technical details
work out easier than in any other approach I have seen. The Schwartz class is a larger class
of functions which does contain C_{c}^{∞}
n
(ℝ )
and also has the same nice properties as G. The
functions in the Schwartz class are infinitely differentiable and they vanish very rapidly as
|x|→∞ along with all their partial derivatives. This is the description of these functions, not
a specific form involving polynomials times e^{−α|x|
2
}. To describe this precisely requires some
notation.
Definition 10.5.21f ∈S, the Schwartz class, if f ∈ C^{∞}(ℝ^{n}) and for all positive integersN,
ρN (f) < ∞
where
ρN(f) = sup{(1+ |x|2)N |D αf(x)| : x ∈ ℝn,|α| ≤ N }.
Thus f ∈S if and only if f ∈ C^{∞}(ℝ^{n}) and
sup{|xβD αf(x)| : x ∈ ℝn} < ∞ (10.17)
(10.17)
for all multi indices α and β.
Also note that if f ∈S, then p(f) ∈S for any polynomial,p with p(0) = 0 and that
p n ∞ n
S ⊆ L (ℝ )∩ L (ℝ )
for any p ≥ 1. To see this assertion about the p
(f)
, it suffices to consider the case of the product of two
elements of the Schwartz class. If f,g ∈S, then D^{α}
(fg)
is a finite sum of derivatives of f
times derivatives of g. Therefore, ρ_{N}
(fg)
< ∞ for all N. You may wonder about examples of
things in S. Clearly any function in C_{c}^{∞}
n
(ℝ )
is in S. However there are other functions in
S. For example e^{−|x|
2
} is in S as you can verify for yourself and so is any function from G.
Note also that the density of C_{c}
(ℝn )
in L^{p}
(ℝn)
shows that S is dense in L^{p}
(ℝn )
for every
p.
Recall the Fourier transform of a function in L^{1}
(ℝn )
is given by
∫
F f(t) ≡ (2π )− n∕2 n e−it⋅xf(x )dx.
ℝ
Therefore, this gives the Fourier transform for f ∈S. The nice property which S has in common with G is
that the Fourier transform and its inverse map S one to one onto S. This means I could have presented
the whole of the above theory in terms of S rather than in terms of G. However, it is more
technical.
Theorem 10.5.22If f ∈S, then Ff and F^{−1}f are also in S.
Proof: To begin with, let α = e_{j} = (0,0,
⋅⋅⋅
,1,0,
⋅⋅⋅
,0), the 1 in the j^{th} slot.
− 1 −1 ∫ ihxj
F--f-(t-+-hej)-− F--f(t)= (2π)−n∕2 eit⋅xf(x)(e---−-1)dx. (10.18)
h ℝn h
where the boundary term vanishes because f ∈S. Returning to 10.19, use the fact that |e^{ia}| = 1 to
conclude
∫
|tβD αF −1f(t)| ≤C |D β((ix)af(x))|dx < ∞.
ℝn
It follows F^{−1}f ∈S. Similarly Ff ∈S whenever f ∈S. ■
Of course S can be considered a subset of G^{∗} as follows. For ψ ∈S,
∫
ψ(ϕ) ≡ ψϕdx
ℝn
Theorem 10.5.23Let ψ ∈S. Then (F ∘ F^{−1})(ψ) = ψ and (F^{−1}∘ F)(ψ) = ψ whenever ψ ∈S.Also F and F^{−1}map S one to one and onto S.
Proof: The first claim follows from the fact that F and F^{−1} are inverses of each other on G^{∗} which was
established above. For the second, let ψ ∈S. Then ψ = F
( )
F−1ψ
. Thus F maps S onto S. If Fψ = 0,
then do F^{−1} to both sides to conclude ψ = 0. Thus F is one to one and onto. Similarly, F^{−1} is one to one
and onto. ■