390 CHAPTER 13. FOURIER TRANSFORMS

13.3 Exercises1. For f ∈ L1 (Rn), show that if F−1 f ∈ L1 or F f ∈ L1, then f equals a continuous

bounded function a.e.

2. Suppose f ,g ∈ L1(R) and F f = Fg. Show f = g a.e.

3. Show that if f ∈ L1 (Rn) , then lim|x|→∞ F f (x) = 0.

4. ↑ Suppose f ∗ f = f or f ∗ f = 0 and f ∈ L1(R). Show f = 0.

5. For this problem define∫

∞

a f (t)dt ≡ limr→∞

∫ ra f (t)dt. Note this coincides with the

Lebesgue integral when f ∈ L1 (a,∞). Show

(a)∫

∞

0sin(u)

u du = π

2

(b) limr→∞

∫∞

δ

sin(ru)u du = 0 whenever δ > 0.

(c) If f ∈ L1 (R), then limr→∞

∫R sin(ru) f (u)du = 0.

Hint: For the first two, use 1u =

∫∞

0 e−utdt and then, using this, apply Fubini’stheorem to

∫ R0 sinu

∫R e−utdtdu. For the last part, first establish it for f ∈C∞

c (R) andthen use the density of this set in L1 (R) to obtain the result. This is sometimes calledthe Riemann Lebesgue lemma.

6. ↑Suppose that g∈ L1 (R) and that at some x > 0, g is locally Holder continuous fromthe right and from the left. This means

limr→0+

g(x+ r)≡ g(x+) , limr→0+

g(x− r)≡ g(x−)

exists and there exist constants K,δ > 0 and r ∈ (0,1] such that for |x− y|< δ ,

|g(x+)−g(y)|< K |x− y|r , |g(x−)−g(y)|< K |x− y|r

for y > x and y < x respectively. Show that under these conditions,

limr→∞

2π

∫∞

0

sin(ur)u

(g(x−u)+g(x+u)

2

)du =

g(x+)+g(x−)2

.

7. Let g ∈ L1 (R) and suppose g is locally Holder continuous from the right and fromthe left at x. Show that then

limR→∞

12π

∫ R

−Reixt∫

∞

−∞

e−ityg(y)dydt =g(x+)+g(x−)

2.

This is very interesting. If g ∈ L2 (R), this shows F−1 (Fg)(x) = g(x+)+g(x−)2 , the

midpoint of the jump in g at the point, x. In particular, if g ∈ G , F−1 (Fg) = g. Hint:Show the left side of the above equation reduces to

2π

∫∞

0

sin(ur)u

(g(x−u)+g(x+u)

2

)du

and then use Problem 6 to obtain the result.