First is a definition of a very specialized set of functions. Here the measure space will be
(ℝn, mn,ℱn )
, familiar Lebesgue measure.
First recall the following definition of a polynomial.
Definition 26.1.1α = (α_{1},
⋅⋅⋅
,α_{n}) for α_{1}
⋅⋅⋅
α_{n}nonnegative integers is called amulti-index.For α a multi-index, |α|≡ α_{1} +
⋅⋅⋅
+ α_{n}and if x ∈ ℝ^{n},
x = (x1,⋅⋅⋅,xn),
and f a function, define
x α ≡ 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}) amulti-index
∂|α|f
D αf (x) ≡--α1--α2-----αn.
∂x1 ∂x2 ⋅⋅⋅∂xn
Definition 26.1.2Define G_{1}to be the functions of the form p
(x)
e^{−a|x|
2
}where a > 0 is rational and p
(x)
is a polynomial having all rational coefficients, a_{α}being “rational” if it is of the form a + ib for a,b ∈ ℚ. Let G be all finite sums offunctions in G_{1}. Thus G is an algebra of functions which has the property that if f ∈Gthen 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 26.1.3G is dense in C_{0}
(ℝn )
with respect to the norm,
||f|| ≡ sup {|f (x)| : x ∈ ℝn}
∞
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 26.1.4For each p ≥ 1,p < ∞,G is dense in L^{p}
(ℝn )
. Since G iscountable, this shows that 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
∫
( −b|x|2)p p
ℝn e dx < ε .
Then x → g
(x )
e^{b|x|
2
} is in C_{c}
(ℝn)
⊆ C_{0}
(ℝn)
. Therefore, from Lemma 26.1.3 there exists
ψ ∈G such that
From now on, we can drop the restriction that the coefficients of the polynomials in G are
rational. We 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 26.1.5For ψ ∈G , p a polynomial, and α,β multi-indices, D^{α}ψ ∈G andpψ ∈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 26.1.2 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 26.1.6For ψ ∈G Define the Fourier transform, F and the inverseFourier transform, F^{−1}by
∫
Fψ (t) ≡ (2π)−n∕2 e− it⋅xψ (x)dx,
ℝn
−1 −n∕2∫ it⋅x
F ψ(t) ≡ (2π) ℝn e ψ(x)dx.
wheret ⋅ x ≡∑_{i=1}^{n}t_{i}x_{i}. Note there is no problem with this definition because ψ is inL^{1}
(ℝn )
and therefore,
| |
|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 26.1.7The 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 26.1.8Each of F and F^{−1}map G to G. Also F^{−1}∘F
(ψ)
= ψ andF ∘ F^{−1}
(ψ)
= ψ.
Proof: Proof: To make the notation simpler, ∫
will symbolize
1
(2π)n∕2-
∫_{ℝn}. Also,
f_{b}
(x)
≡ e^{−b|x|
2
}. Then from the above
−n∕2
Ffb = (2b) f(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
26.1.7,
∫
F ψ(t) ≡ e−it⋅xxαe−b|x|2dx
( )
− |α| α ∫ − it⋅x −b|x|2
= (− i) Dt e e dx , (Differentiating under integral)
( |t|2( √ π)n)
= (− i)− |α|Dαt 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,
∫ ( ∫ )
F−1 ∘F (ψ)(s) = (− i)−|α| eis⋅tD α e−it⋅xe−b|x|2dx dt
t
−|α| |α| α∫ is⋅t(∫ − it⋅x −b|x|2 )
= (− i) (− i) s e e e dx dt
α −1
= s F (F (fb))(s)
−1 −1( − n∕2 ) − n∕2 −1( )
F (F (fb))(s) = F (2b) f(4b)−1 (s) = (2b) F f(4b)−1 (s)
− n∕2( −1)−n∕2
= (2b) 2(4b) f(4(4b)−1)−1 (s) = fb(s)