(3)

For Ta f (x) := f (x ’ a), a ∈ Rd , we have Ta f (x) = e’i x a f (x).

T

(4)

For S± f (x) := f (x/±), ± > 0, we have S± f = ± d S1/± f .

(5)

If, in addition, x j f (x) ∈ L 1 (Rd ) then f is differentiable with respect to x j and

(6)

‚f

(x) = (’iy j f (y))§ (x).

‚x j

If ‚ f /‚x j is also in L 1 (Rd ) then

‚f

(x) = ix j f (x).

‚x j

Obviously the last item extends to higher-order derivatives in a natural way. The following

space will turn out to be the natural playground for Fourier transforms.

De¬nition 5.17 The Schwartz space S consists of all functions γ ∈ C ∞ (Rd ) that satisfy

|x ± D β γ (x)| ¤ C±,β,γ , x ∈ Rd ,

for all multi-indices ±, β ∈ Nd with a constant C±,β,γ that is independent of x ∈ Rd . The

0

functions of S are called test functions or good functions.

In other words, a good function and all of its derivatives decay faster than any polynomial.

∞

Of course all functions from C0 (Rd ) are contained in S but so also are the functions

γ (x) = e’±

2

, x ∈ Rd ,

x 2

for all ± > 0, and we are going to compute their Fourier transforms now. Because of the

scaling property of the Fourier transform it suf¬ces to pick one speci¬c ± > 0.

Theorem 5.18 The function G(x) := e’ 2 /2

2

satis¬es G = G.

x

Proof First note that, because

G(x) = (2π)’d/2 e’ y 2 /2 ’i x T y

e dy

2

Rd

∞

d

e’yj /2 e’ixj yj dy j ,

(2π)’1/2

2

=

’∞

j=1

the Fourier transform of the d-variate Gaussian G is the product of the univariate Fourier

transforms of the univariate Gaussian g(t) = e’t /2 , and it suf¬ces to compute this univariate

2

56 Auxiliary tools from analysis and measure theory

Fourier transform. Cauchy™s integral theorem yields

∞

g(r ) = (2π)’1/2 e’(t /2+ir t)

2

dt

’∞

∞

= (2π)’1/2 e’r /2

e’(t+ir ) /2

2 2

dt

’∞

∞

= (2π)’1/2 e’r /2

e’t /2

2 2

dt

’∞

= e’r /2

2

.

We need another class of functions. In a certain way they are the opposite of good

functions. Actually, they de¬ne continuous linear functionals on S, but this will not really

matter for us.

De¬nition 5.19 We say that a function f is slowly increasing if there exists a constant

m ∈ N0 such that f (x) = O( x m ) for x 2 ’ ∞.

2

Our next result concerns the approximation of functions by convolution.

Theorem 5.20 De¬ne gm (x) = (m/π)d/2 e’m

2

, m ∈ N, x ∈ Rd . Then the following hold

x 2

true:

gm (x)d x = 1;

(1) Rd

(2) gm (x) = (2π)’d/2 e’ x 2 /(4m)

,

2

(3) g m (x) = gm (x),

(x) = limm’∞ (ω)gm (ω ’ x)dω, provided that ∈ C(Rd ) is slowly increasing.

(4) Rd

Proof (1) follows from

gm (x)d x = (2π)’d/2 e’ y 2 /2

dy = 1.

2

Rd Rd

To prove (2) we remark that gm = (m/π)d/2 S1/√2m G. Thus Theorems 5.16 and 5.18 lead

to

gm = (m/π )d/2 (2m)’d/2 S√2m G = (2π )’d/2 S√2m G.

For (3) note that

√

g m = (2π )’d/2 (S√2m G )§ = (2π )’d/2 ( 2m )d S1/√2m G = gm .

For (4) we ¬rst restrict ourselves to the case x = 0. From (1) we see that

(ω)gm (ω)dω ’ (0) = [ (ω) ’ (0)]gm (ω)dω.

Rd Rd

Now choose an arbitrary > 0. Then there exists a δ > 0 such that | (ω) ’ (0)| < /2

for ω 2 ¤ δ. Furthermore, since is slowly increasing, there exists an ∈ N0 and an

5.2 Fourier transform and approximation by convolution 57

M > 0 such that | (ω)| ¤ M(1 + ω 2 ) , ω ∈ Rd . This means that we can ¬nd a generic

constant Cδ such that we can bound the integral as follows:

[ (ω) ’ (0)]gm (ω)dω ¤ | (ω) ’ (0)|gm (ω)dω

Rd ω 2 ¤δ

+ Cδ gm (ω) ω 2 dω