1112 CHAPTER 32. FOURIER TRANSFORMS

Proof: To begin with, let α = e j = (0,0, · · · ,1,0, · · · ,0), the 1 in the jth slot.

F−1 f (t+he j)−F−1 f (t)h

= (2π)−n/2∫Rn

eit·x f (x)(eihx j −1

h)dx. (32.3.14)

Consider the integrand in 32.3.14.∣∣∣∣eit·x f (x)(eihx j −1

h)

∣∣∣∣ = | f (x)|

∣∣∣∣∣(ei(h/2)x j − e−i(h/2)x j

h)

∣∣∣∣∣= | f (x)|

∣∣∣∣ isin((h/2)x j)

(h/2)

∣∣∣∣≤ | f (x)| ∣∣x j∣∣

and this is a function in L1(Rn) because f ∈S. Therefore by the Dominated ConvergenceTheorem,

∂F−1 f (t)∂ t j

= (2π)−n/2∫Rn

eit·xix j f (x)dx

= i(2π)−n/2∫Rn

eit·xxe j f (x)dx.

Now xe j f (x) ∈ S and so one can continue in this way and take derivatives indefinitely.Thus F−1 f ∈C∞(Rn) and from the above argument,

Dα F−1 f (t) =(2π)−n/2∫Rn

eit·x(ix)α f (x)dx.

To complete showing F−1 f ∈S,

tβ Dα F−1 f (t) =(2π)−n/2∫Rn

eit·xtβ (ix)a f (x)dx.

Integrate this integral by parts to get

tβ Dα F−1 f (t) =(2π)−n/2∫Rn

i|β |eit·xDβ ((ix)a f (x))dx. (32.3.15)

Here is how this is done.∫R

eit jx j tβ jj (ix)α f (x)dx j =

eit jx j

it jtβ jj (ix)α f (x) |∞−∞ +

i∫R

eit jx j tβ j−1j De j((ix)α f (x))dx j

where the boundary term vanishes because f ∈S. Returning to 32.3.15, use the fact that|eia|= 1 to conclude

|tβ Dα F−1 f (t)| ≤C∫Rn|Dβ ((ix)a f (x))|dx < ∞.

It follows F−1 f ∈S. Similarly F f ∈S whenever f ∈S.Of course S can be considered a subset of G ∗ as follows. For ψ ∈S,

ψ (φ)≡∫Rn

ψφdx