11.6. FOURIER TRANSFORMS OF JUST ABOUT ANYTHING 281

Proof: From the definition and Fubini’s theorem,

F f (φ) ≡∫Rn

f (t)Fφ (t)dt =∫Rn

f (t)(

12π

)n/2 ∫Rn

e−it·xφ (x)dxdt

=∫Rn

((1

2π

)n/2 ∫Rn

f (t)e−it·xdt

)φ (x)dx.

Since φ ∈ G is arbitrary, it follows from Theorem 11.6.8 that F f (x) is given by the claimedformula. The case of F−1 is identical. ■

Here are interesting properties of these Fourier transforms of functions in L1.

Theorem 11.6.11 If f ∈ L1 (Rn) and ∥ fk − f∥1 → 0, then F fk and F−1 fk con-verge uniformly to F f and F−1 f respectively. If f ∈ L1 (Rn), then F−1 f and F f are bothcontinuous and bounded. Also,

lim|x|→∞

F−1 f (x) = lim|x|→∞

F f (x) = 0. (11.8)

Furthermore, for f ∈ L1 (Rn) both F f and F−1 f are uniformly continuous.

Proof: The first claim follows from the following inequality.

|F fk (t)−F f (t)| ≤ (2π)−n/2∫Rn

∣∣e−it·x fk(x)− e−it·x f (x)∣∣dx

= (2π)−n/2∫Rn| fk (x)− f (x)|dx = (2π)−n/2 ∥ f − fk∥1 .

which a similar argument holding for F−1.Now consider the second claim of the theorem.∣∣F f (t)−F f

(t′)∣∣≤ (2π)−n/2

∫Rn

∣∣∣e−it·x− e−it′·x∣∣∣ | f (x)|dx

The integrand is bounded by 2 | f (x)|, a function in L1 (Rn) and converges to 0 as t′ → tand so the dominated convergence theorem implies F f is continuous. To see F f (t) isuniformly bounded,

|F f (t)| ≤ (2π)−n/2∫Rn| f (x)|dx < ∞.

A similar argument gives the same conclusions for F−1.

Let ∥tk∥∞→ ∞. Then for g ∈ G , we see that for some ik where,

∣∣∣t ikk

∣∣∣= ∥tk∥∞

tk ≡(t11 , · · · , tn

k), lim

k→∞

∣∣∣t ikk

∣∣∣= ∞

since otherwise we could not have ∥tk∥∞→ ∞. Then integrating by parts,

|Fg(tk)|=∣∣∣∣(2π)−n/2

∫Rn

e−itk·xg(x)dx∣∣∣∣≤Cn

∫Rn

1∣∣∣t ikk

∣∣∣ |Dtig(x)|dx