11.5. FOURIER TRANSFORMS IN Rn 275

Therefore, letting φ (x) ≡ ψ (x) it follows that φ ∈ G and for all x ∈ Rn, |g(x)−φ (x)| <e−b|x|2 Therefore,(∫

Rn|g(x)−φ (x)|p dx

)1/p

≤(∫

Rn

(e−b|x|2

)pdx)1/p

< ε .

It follows ∥ f −φ∥p ≤ ∥ f −g∥p +∥g−φ∥p < 2ε . ■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 functionswhich 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 11.5.5 For ψ ∈ G , p a polynomial, and α,β multi-indices, Dα ψ ∈ G andpψ ∈ G . Also

sup{|xβ Dαψ(x)| : x ∈ Rn}< ∞

Thus these special functions are infinitely differentiable (smooth). They also have theproperty that they and all their partial derivatives vanish as |x| → ∞. This is because everymixed partial derivative of one of these will be a finite sum of polynomials multiplied bye−b|x|2 for some positive b.

The idea is to first understand the Fourier transform on these very specialized functionsin G .

Definition 11.5.6 For ψ ∈ G , define the Fourier transform F and the inverseFourier transform F−1 by

Fψ(t)≡ (2π)−n/2∫Rn

e−it·xψ(x)dx,

F−1ψ(t)≡ (2π)−n/2

∫Rn

eit·xψ(x)dx.

where t ·x≡∑ni=1 tixi. Note there is no problem with this definition because ψ is in L1 (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 Fouriertransform of these functions. The first thing to do is to verify F and F−1 map G to G andthat F−1 ◦F (ψ) = ψ.

Lemma 11.5.7 The following holds. (c > 0)(1

2π

)n/2 ∫Rn

e−c|t|2e−is·tdt =(

12π

)n/2 ∫Rn

e−c|t|2eis·tdt

=

(1

2π

)n/2

e−|s|24c

(√π√c

)n

=

(12c

)n/2

e−14c |s|

2. (11.7)

Proof: Consider first the case of one dimension. Let H (s) be given by

H (s)≡∫R

e−ct2e−istdt =

∫R

e−ct2cos(st)dt