1106 CHAPTER 32. FOURIER TRANSFORMS

and this last expression converges to zero as ||t||∞→∞. The reason for this is that if t j ̸= 0,

integration by parts with respect to x j gives

(2π)−n/2∫Rn

e−it·xg(x)dx = (2π)−n/2 1−it j

∫Rn

e−it·x ∂g(x)∂x j

dx.

Therefore, choose the j for which ||t||∞=∣∣t j∣∣ and the result of 32.3.4 holds. Therefore,

from 32.3.4, if ||t||∞

is large enough, |F f (t)| < ε . Similarly, lim||t||→∞ F−1 (t) = 0. Con-sider the claim about uniform continuity. Let ε > 0 be given. Then there exists R suchthat if ||t||

∞> R, then |F f (t)|< ε

2 . Since F f is continuous, it is uniformly continuous onthe compact set [−R−1,R+1]n. Therefore, there exists δ 1 such that if ||t− t′||

∞< δ 1 for

t′, t ∈ [−R−1,R+1]n, then ∣∣F f (t)−F f(t′)∣∣< ε/2. (32.3.5)

Now let 0 < δ < min(δ 1,1) and suppose ||t− t′||∞< δ . If both t, t′ are contained in

[−R,R]n, then 32.3.5 holds. If t ∈ [−R,R]n and t′ /∈ [−R,R]n, then both are contained in[−R−1,R+1]n and so this verifies 32.3.5 in this case. The other case is that neither pointis in [−R,R]n and in this case,∣∣F f (t)−F f

(t′)∣∣ ≤ |F f (t)|+

∣∣F f(t′)∣∣

<ε

2+

ε

2= ε.

There is a very interesting relation between the Fourier transform and convolutions.

Theorem 32.3.11 Let f ,g ∈ L1(Rn). Then f ∗g ∈ L1 and F( f ∗g) = (2π)n/2 F f Fg.

Proof: Consider ∫Rn

∫Rn| f (x−y)g(y)|dydx.

The function, (x,y)→ | f (x−y)g(y)| is Lebesgue measurable and so by Fubini’s theorem,∫Rn

∫Rn| f (x−y)g(y)|dydx =

∫Rn

∫Rn| f (x−y)g(y)|dxdy = || f ||1 ||g||1 < ∞.

It follows that for a.e. x,∫Rn | f (x−y)g(y)|dy < ∞ and for each of these values of x, it

follows that∫Rn f (x−y)g(y)dy exists and equals a function of x which is in L1 (Rn) , f ∗

g(x). Now

F( f ∗g)(t)≡ (2π)−n/2∫Rn

e−it·x f ∗g(x)dx

= (2π)−n/2∫Rn

e−it·x∫Rn

f (x−y)g(y)dydx

= (2π)−n/2∫Rn

e−it·yg(y)∫Rn

e−it·(x−y) f (x−y)dxdy

= (2π)n/2 F f (t)Fg(t) .

There are other considerations involving Fourier transforms of functions in L1 (Rn).