In this section is a general treatment of Fourier transforms. It turns out you can take the Fourier transform
of almost anything you like. First is a definition of a very specialized set of functions. Here the measure
space will be
(ℝn,mn,ℱn )
, m_{n} Lebesgue measure on ℝ^{n}.
First is the definition of a polynomial in many variables.
Definition 10.4.1α = (α_{1},
⋅⋅⋅
,α_{n}) for α_{1}
⋅⋅⋅
α_{n}nonnegative integers is called a multi-index.For α amulti-index, |α|≡ α_{1} +
⋅⋅⋅
+ α_{n}and if x ∈ ℝ^{n},
x = (x1,⋅⋅⋅,xn ),
and f a function, define
xα ≡ xα11xα22 ⋅⋅⋅xαnn.
A polynomial in n variables of degree m is a function of the form
∑ α
p(x) = aαx .
|α|≤m
Here α is a multi-index as just described and a_{α}∈ ℂ.Also define for α = (α_{1},
⋅⋅⋅
,α_{n}) a multi-index
∂ |α|f
D αf (x) ≡ --α1--α2-----αn.
∂x1 ∂x2 ⋅⋅⋅∂xn
Definition 10.4.2Define G_{1}to be the functions of the form p
(x)
e^{−a|x|
2
}where a > 0 is rationaland p
(x)
is a polynomial having all rational coefficients, a_{α}being “rational” if it is of the form a+ibfor a,b ∈ ℚ. Let G be all finite sums of functions in G_{1}. Thus G is an algebra of functions which hasthe property that if f ∈G then f∈G.
Thus there are countably many functions in G_{1}. This is because, for each m, there are countably many
choices for a_{α} for
|α|
≤ m since there are finitely many α for
|α|
≤ m and for each such α, there are
countably many choices for a_{α} since ℚ+iℚ is countable. (Why?) Thus there are countably many
polynomials having degree no more than m. This is true for each m and so the number of different
polynomials is a countable union of countable sets which is countable. Now there are countably many
choices of e^{−α|x|
2
} and so there are countably many in G_{1} because the Cartesian product of countable sets is
countable.
Now G consists of finite sums of functions in G_{1}. Therefore, it is countable because for each m ∈ ℕ,
there are countably many such sums which are possible.
I will show now that G is dense in L^{p}
(ℝn)
but first, here is a lemma which follows from the Stone
Weierstrass theorem.
Lemma 10.4.3G is dense in C_{0}
(ℝn)
with respect to the norm,
n
||f ||∞ ≡ sup{|f (x)| : x ∈ ℝ }
Proof: By the Weierstrass approximation theorem, it suffices to show G separates the points and
annihilates no point. It was already observed in the above definition that f∈G whenever f ∈G. If y_{1}≠y_{2}
suppose first that
|y1|
≠
|y2|
. Then in this case, you can let f
(x)
≡ e^{−|x|
2
}. Then f ∈G and f
(y1)
≠f
(y2)
.
If
|y1|
=
|y2|
, then suppose y_{1k}≠y_{2k}. This must happen for some k because y_{1}≠y_{2}. Then let
f
(x )
≡ x_{k}e^{−|x|
2
}. Thus G separates points. Now e^{−|x|
2
} is never equal to zero and so G annihilates no point
of ℝ^{n}. ■
These functions are clearly quite specialized. Therefore, the following theorem is somewhat
surprising.
Theorem 10.4.4For each p ≥ 1,p < ∞,G is dense in L^{p}
(ℝn)
. Since G is countable, this showsthat L^{p}
(ℝn )
is separable.
Proof: Let f ∈ L^{p}
(ℝn)
. Then there exists g ∈ C_{c}
(ℝn)
such that
||f − g||
_{p}< ε. Now let b > 0 be large
enough that
∫ ( 2)p
n e−b|x| dx < εp.
ℝ
Then x → g
(x )
e^{b|x|
2
} is in C_{c}
n
(ℝ )
⊆ C_{0}
n
(ℝ )
. Therefore, from Lemma 10.4.3 there exists ψ ∈G such
that
From now on, drop the restriction that the coefficients of the polynomials in G are rational. Also drop
the restriction that a is rational. Thus G will be finite sums of functions which are of the form p
(x)
e^{−a|x|
2
}
where the coefficients of p are complex and a > 0.
The following lemma is also interesting even if it is obvious.
Lemma 10.4.5For ψ ∈G , p a polynomial, and α,β multi-indices, D^{α}ψ ∈G and pψ ∈G.Also
sup{|xβD αψ(x)| : x ∈ ℝn} < ∞
Thus these special functions are infinitely differentiable (smooth). They also have the property that
they and all their partial derivatives vanish as
|x|
→∞.
Let G be the functions of Definition 10.4.2 except, for the sake of convenience, remove all references to
rational numbers. Thus G consists of finite sums of polynomials having coefficients in ℂ times e^{−a}
|x|
^{2}
for
some a > 0. The idea is to first understand the Fourier transform on these very specialized
functions.
Definition 10.4.6For ψ ∈G Define the Fourier transform, F and the inverse Fourier transform, F^{−1}by
−n∕2∫ −it⋅x
Fψ (t) ≡ (2π) n e ψ (x )dx,
ℝ
∫
F−1ψ (t) ≡ (2π)−n∕2 eit⋅xψ(x)dx.
ℝn
wheret ⋅ x ≡∑_{i=1}^{n}t_{i}x_{i}. Note there is no problem with this definition because ψ is in L^{1}
(ℝn )
andtherefore,
| |
|eit⋅xψ(x)| ≤ |ψ (x)|,
an integrable function.
One reason for using the functions G is that it is very easy to compute the Fourier transform of these
functions. The first thing to do is to verify F and F^{−1} map G to G and that F^{−1}∘ F
(ψ )
= ψ.
Lemma 10.4.7The following hold.
(c > 0)
( 1)n ∕2∫ −c|t|2 −is⋅t ( 1 )n∕2∫ −c|t|2 is⋅t
2π- n e e dt = 2π- n e e dt
ℝ ℝ
This proves the formula in the case of one dimension. The case of the inverse Fourier transform is similar.
The n dimensional formula follows from Fubini’s theorem. ■
With these formulas, it is easy to verify F,F^{−1} map G to G and F ∘ F^{−1} = F^{−1}∘ F = id.
Theorem 10.4.8Each of F and F^{−1}map G to G. Also F^{−1}∘ F
(ψ )
= ψ and F ∘ F^{−1}
(ψ)
= ψ.
Proof: To make the notation simpler, ∫
will symbolize
--1n∕2-
(2π)
∫_{ℝn}. Also, f_{b}
(x)
≡ e^{−b|x|
2
}. Then from
the above
Ffb = (2b)−n∕2f(4b)−1
The first claim will be shown if it is shown that Fψ ∈G for ψ
(x)
= x^{α}e^{−b|x|
2
} because an arbitrary
function of G is a finite sum of scalar multiples of functions such as ψ. Using Lemma 10.4.7,
∫
F ψ(t) ≡ e−it⋅xxαe−b|x|2dx
( )
−|α| α ∫ −it⋅x − b|x|2
= (− i) D t e e dx , (Differentiating under integral)
( |t|2 (√ π)n )
= (− i)−|α|D αt e− 4b √--
b
and this is clearly in G because it equals a polynomial times e^{−|t|2
4b
}. Similarly, F^{−1} : G→G. Now
consider F^{−1}∘ F
(ψ)
(s)
where ψ was just used. From the above, and integrating by parts,