32.4. EXERCISES 1115

32.4 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 apply Fubini’s theorem to∫ R

0sinu

∫R

e−utdtdu

For the last part, first establish it for f ∈C∞c (R) and then use the density of this set in

L1 (R) to obtain the result. This is sometimes called the 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+)

exists,lim

r→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

for y > x and|g(x−)−g(y)|< K |x− y|r

for y < x. Show that under these conditions,

limr→∞

∫∞

0

sin(ur)u

(g(x−u)+g(x+u)

2

)du

=g(x+)+g(x−)

2.

32.4. EXERCISES 111532.4 Exercises1. For f € L'(R"), show that if F~'f ¢ L' or Ff € L!, then f equals a continuousbounded function a.e.2. Suppose f,g € L'(R) and Ff = Fg. Show f = gae.3. Show that if f € L' (R"), then limyy)_,.. Ff (x) = 0.4. + Suppose f * f = f or f * f =O and f € L'(R). Show f =0.5. For this problem define f'” f (t) dt =lim,-,.. [7 f (t) dt. Note this coincides with theLebesgue integral when f € L! (a,c°). Show(a) fo du = §(b) lim, [5° sin) 4 = 0 whenever 5 > 0.(c) If feL'(R \. then lim,—y.0 fig sin (ru) f (u) du = 0.Hint: For the first two, use i = fy e “dt and apply Fubini’s theorem toR[ sinu [ e “dtdu0 RFor the last part, first establish it for f € Cz (IR) and then use the density of this set inL! (R) to obtain the result. This is sometimes called the Riemann Lebesgue lemma.6. tSuppose that g € L! (IR) and that at some x > 0, g is locally Holder continuous fromthe right and from the left. This meanslim g(x+r) = g(x+)r>0+exists,jim g(x r) =g(x-)exists and there exist constants K,6 > 0 and r € (0, 1] such that for |x—y| <6,ig(x+)—8(y)|<K|x—y/"for y > x andIg(x—) —g(y)| <K|x—yI"for y < x. Show that under these conditions,2 sin (ur) (Se tulim —roo TT JO Uu