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 26.2.20f ∈S, the Schwartz class, if f ∈ C^{∞}(ℝ^{n}) and for all positiveintegers N,
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 } < ∞ (26.12)
(26.12)
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
S ⊆ Lp (ℝn )∩ L∞ (ℝn )
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}^{∞}
(ℝn )
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 e−it⋅xf (x)dx.
ℝn
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 26.2.21If 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.
F− 1f (t +hej) − F −1f(t) ∫ eihxj − 1
----------h-----------= (2π)−n∕2 n eit⋅xf(x)(--h---)dx. (26.13)
ℝ
where the boundary term vanishes because f ∈S. Returning to 26.14, use the fact that
|e^{ia}| = 1 to conclude
∫ a
|tβD αF− 1f(t)| ≤C n |D β((ix) f(x))|dx < ∞.
ℝ
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 26.2.22Let ψ ∈S. Then (F ∘F^{−1})(ψ) = ψ and (F^{−1}∘F)(ψ) = ψwhenever 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.
■