Kenneth Kuttler

282 CHAPTER 11. FUNDAMENTAL TRANSFORMS

which converges to 0 as k→ ∞. This shows 11.8 in case the function is in G . Now inthe general case, let ∥g− f∥L1(Rn) < ε where g ∈ G and f ∈ L1 (Rn). Then it follows that|F ( f )(t)−Fg(t)| ≤Cn ∥ f −g∥L1(Rn) and so if ∥tk∥∞

→ ∞,

lim supk→∞

|F ( f )(tk)| ≤ lim supk→∞

|F ( f )(tk)−Fg(tk)|+=0

limk→∞

|Fg(tk)| ≤ ε

if g is chosen to make Cn ∥ f −g∥L1(Rn) < ε. Since ε is arbitrary, this shows 11.8 in general.It remains to verify the claim that F f and F−1 f are uniformly continuous. Let ε > 0 be

given. Then there exists R such that if ∥t∥∞> R, then |F f (t)|< ε

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

∞< δ 1 for t′, t ∈ [−R−1,R+1]n, then∣∣F f (t)−F f

(t′)∣∣< ε/2. (11.9)

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

[−R,R]n, then 11.9 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 11.9 in this case. The other case is that neither point isin [−R,R]n and in this case,∣∣F f (t)−F f

(t′)∣∣≤ |F f (t)|+

∣∣F f(t′)∣∣< ε

2= ε. ■

11.6.3 Convolutions in L1 (Rn)

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

f ∗g(x)≡∫Rn

f (x−y)g(y)dy

and part of the problem is in showing that this even makes sense. This is dealt with in thefollowing theorem.

Theorem 11.6.12 Suppose that f ,g ∈ L1(Rn). Then also f ∗g ∈ L1 and it followsthat F( f ∗g) = (2π)n/2 F f Fg.

Proof: Assume both f and g are Borel measurable representatives. Consider∫Rn

∫Rn| f (x−y)g(y)|dydx.

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) . ■

282 CHAPTER 11. FUNDAMENTAL TRANSFORMSwhich converges to 0 as k —> 0. This shows 11.8 in case the function is in Y. Now inthe general case, let |g — f|;1 (qn) < € where g € Y and f € L! (R"). Then it follows thatIF (f) (t) — Fg ()| <Grllf— sllicen) and so if ||t||.. +,=0lim sup |F (f) (te)| < lim sup |F (f) (tk) — Fg (tx) | + lim |F'g (t)| < €k-yoo kyo k-y00if g is chosen to make C, || f — g|l1 (Rn) < €. Since € is arbitrary, this shows 11.8 in general.It remains to verify the claim that F f and F~! f are uniformly continuous. Let € > 0 begiven. Then there exists R such that if ||t||,, > R, then |F f (t)| < 4. Since Ff is continuous,it is uniformly continuous on the compact set [-R—1,R+1]". Therefore, there exists 6,such that if ||t—t'||,, < 6) for t’,t € [-R—1,R+1]", then|Ff (t)-Ff (t’)| <e/2. (11.9)Now let 0 < 6 < min(6;,1) and suppose ||t—t’||,, < 6. If both t,t’ are contained in[—R,R]", then 11.9 holds. If t € [—R,R]" and t’ ¢ [—R,R]", then both are contained in[—R—1,R+1]" and so this verifies 11.9 in this case. The other case is that neither point isn [—R,R]" and in this case,IFF(Q)-FSF(€)| SIFFOI+/FL(C)| <5Ile.an €=e.27 211.6.3 Convolutions in L! (R”)There is a very interesting relation between the Fourier transform and convolutions. Recallfae(s)= |. £x-y)e(ydyand part of the problem is in showing that this even makes sense. This is dealt with in thefollowing theorem.Theorem 11.6.12 Suppose that f,g € L'(R"). Then also f *g € L! and it followsthat F(f *g) = (20)" F fF.Proof: Assume both f and g are Borel measurable representatives. ConsiderJ [ites lavas.n R”By Fubini’s theorem,[ Llte-vewlaax= [Of ey) @)laxdy = Il lish <=.It follows that for a.e. x, fn |f(k—y)g(y)|dy < o and for each of these values of x, itfollows that fin f (x —y) g (y) dy exists and equals a function of x which is in L' (R"), f *g(x). NowF(fxs) (t) = (2)-" [ep g(x) dx(27) ye fe ew f (x—y)g(y) dydx= any"? | eM e(y) [ chebJR" JR"= (20)"? Ff (t)Fe(t).