32.3. FOURIER TRANSFORMS OF JUST ABOUT ANYTHING 1105

Theorem 32.3.10 If f ∈ L1 (Rn) and || fk− f ||1 → 0, then F fk and F−1 fk converge uni-formly to F f and F−1 f respectively. If f ∈ L1 (Rn), then F−1 f and F f are both continuousand bounded. Also,

lim|x|→∞

F−1 f (x) = lim|x|→∞

F f (x) = 0. (32.3.3)

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.It remains to verify 32.3.3 and the claim that F f and F−1 f are uniformly continuous.

|F f (t)| ≤∣∣∣∣(2π)−n/2

∫Rn

e−it·x f (x)dx∣∣∣∣

Now let ε > 0 be given and let g ∈C∞c (Rn) such that (2π)−n/2 ||g− f ||1 < ε/2. Then

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

+

∣∣∣∣(2π)−n/2∫Rn

e−it·xg(x)dx∣∣∣∣

≤ ε/2+∣∣∣∣(2π)−n/2

∫Rn

e−it·xg(x)dx∣∣∣∣ .

Now integrating by parts, it follows that for ||t||∞≡max

{∣∣t j∣∣ : j = 1, · · · ,n

}> 0

|F f (t)| ≤ ε/2+(2π)−n/2

∣∣∣∣∣ 1||t||

∞

∫Rn

n

∑j=1

∣∣∣∣∂g(x)∂x j

∣∣∣∣dx

∣∣∣∣∣ (32.3.4)