274 CHAPTER 11. FUNDAMENTAL TRANSFORMS

Definition 11.5.1 α = (α1, · · · ,αn) for α1 · · ·αn nonnegative integers is called amulti-index. For α a multi-index, |α| ≡ α1 + · · ·+αn and if x ∈Rn,x = (x1, · · · ,xn) , and fa function, define xα ≡ xα1

1 xα22 · · ·xαn

n . A polynomial in n variables of degree m is a functionof the form

p(x) = ∑|α|≤m

aα xα .

Here α is a multi-index as just described and aα ∈ C. Also define for α = (α1, · · · ,αn) amulti-index

Dα f (x)≡ ∂ |α| f∂xα1

1 ∂xα22 · · ·∂xαn

n.

Definition 11.5.2 Define G1 to be the functions of the form p(x)e−a|x|2 where a> 0is rational and p(x) is a polynomial having all rational coefficients, aα being “rational”if it is of the form a+ ib for a,b ∈Q. Let G be all finite sums of functions in G1. Thus G isan algebra of functions which has the property that if f ∈ G then f ∈ G .

Thus there are countably many functions in G1. This is because, for each m, there arecountably many choices for aα for |α| ≤m since there are finitely many α for |α| ≤m andfor each such α, there are countably many choices for aα sinceQ+iQ is countable. (Why?)Thus there are countably many polynomials having degree no more than m. This is true foreach m and so the number of different polynomials is a countable union of countable setswhich is countable. Now there are countably many choices of e−α|x|2 and so there arecountably many in G1 because the Cartesian product of countable sets is countable.

Now G consists of finite sums of functions in G1. Therefore, it is countable because foreach m ∈ N, there are countably many such sums which are possible.

I will show now that G is dense in Lp (Rn) but first, here is a lemma which follows fromthe Stone Weierstrass theorem.

Lemma 11.5.3 G is dense in C0 (Rn) with respect to the norm,

∥ f∥∞≡ sup{| f (x)| : x ∈ Rn}

Proof: By the Weierstrass approximation theorem, it suffices to show G separates thepoints and annihilates no point. It was already observed in the above definition that f ∈ Gwhenever f ∈ G . If y1 ̸= y2 suppose first that |y1| ̸= |y2| . Then in this case, you can letf (x)≡ e−|x|

2. Then f ∈ G and f (y1) ̸= f (y2). If |y1|= |y2| , then suppose y1k ̸= y2k. This

must happen for some k because y1 ̸= y2. Then let f (x) ≡ xke−|x|2. Thus G separates

points. Now e−|x|2

is never equal to zero and so G annihilates no point of Rn. ■These functions are clearly quite specialized. Therefore, the following theorem is some-

what surprising.

Theorem 11.5.4 For each p ≥ 1, p < ∞,G is dense in Lp (Rn). Since G is count-able, this shows that Lp (Rn) is separable.

Proof: Let f ∈ Lp (Rn) . Then there exists g ∈ Cc (Rn) such that ∥ f −g∥p < ε . Now

let b > 0 be large enough that∫Rn

(e−b|x|2

)pdx < ε p. Then x→ g(x)eb|x|2 is in Cc (Rn)⊆

C0 (Rn) . Therefore, from Lemma 11.5.3 there exists ψ ∈ G such that∥∥∥geb|·|2 −ψ

∥∥∥∞

< 1.