30.3.4 The Schwartz Class
The problem with G is that it does not contain Cc∞
. I have used it in presenting the
Fourier transform because the functions in
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 Cc∞
also has the same nice properties as
. 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
. To describe this precisely requires some
Definition 30.3.20 f ∈S, the Schwartz class, if f ∈ C∞(ℝn) and for all positive
Thus f ∈S if and only if f ∈ C∞(ℝn) and
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
for any p ≥ 1. To see this assertion about the p
, it suffices to consider the case of the
product of two elements of the Schwartz class. If
, then Dα
is a finite sum of
times derivatives of g
. Therefore, ρN
for all N
may wonder about examples of things in S
. Clearly any function in Cc∞
. However there are other functions in S
. For example e−
is in S
as you can verify for yourself and so is any function from G
. Note also that
the density of Cc
is dense in Lp
Recall the Fourier transform of a function in L1
is given by
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 30.3.21 If f ∈S, then Ff and F−1f are also in S.
Proof: To begin with, let α = ej = (0,0,
0), the 1 in the jth
Consider the integrand in 30.3.14.
and this is a function in L1
) because f ∈S
. Therefore by the Dominated
and so one can continue in this way and take derivatives indefinitely.
Thus F−1f ∈ C∞
) and from the above argument,
To complete showing F−1f ∈S,
Integrate this integral by parts to get
Here is how this is done.
where the boundary term vanishes because f ∈S
. Returning to 30.3.15
, use the fact that
= 1 to conclude
It follows F−1f ∈S. Similarly Ff ∈S whenever f ∈S. ■
Of course S can be considered a subset of G∗ as follows. For ψ ∈S,
Theorem 30.3.22 Let 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
. If Fψ
then do F−1
to both sides to
is one to one and onto. Similarly, F−1
is one to one and onto.