290 CHAPTER 11. FUNDAMENTAL TRANSFORMS

11.7 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. This is Theorem 11.6.11 but review it.

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

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

4. Let h(x) =(∫ x

0 e−t2dt)2

+

(∫ 10

e−x2(1+t2)1+t2 dt

). Show that h′ (x) = 0 and h(0) = π/4.

Then let x→∞ to conclude that∫

0 e−t2dt =

√π/2. Show that

∫∞

−∞e−t2

dt =√

π and

that∫

−∞e−ct2

dt =√

π√c .

5. Let h(x) =(∫ x

0 e−t2dt)2

. Then

h′ (x) = 2(∫ x

0e−t2

dt)

e−x2= 2xe−x2

(∫ 1

0e−(xu)2

du).

Now h(x) =∫ x

0 h′ (t)dt. Do integration by parts to obtain

−e−t2∫ 1

0e−(tu)

2du|x0−

∫ x

0e−t2

∫ 1

0e−(tu)

22tu2dudt

= −e−x2∫ 1

0e−(xu)2

du +1−∫ x

0

∫ 1

0e−t2(1+u2)2tu2dudt

= −e−x2∫ 1

0e−(xu)2

du +1−∫ 1

0u2∫ x

0e−t2(1+u2)2tdtdu

= e(x)+1−∫ 1

0u2

(−e−x2(1+u2)−1

1+u2

)du

= e(x)+1−∫ 1

0

u2

1+u2 du

where limx→∞ e(x) = 0. Now explain why 1−∫ 1

0u2

1+u2 du = 14 π. Hence

∫∞

0 e−t2dt =

√π

2 .

6. Recall that for f a function, fy (x) = f (x−y) . Find a relationship between F fy (t)and F f (t) given that f ∈ L1 (Rn).

7. For f ∈ L1 (Rn) , simplify F f (t+y) .

8. For f ∈ L1 (Rn) and c a nonzero real number, show F f (ct) = Fg(t) where g(x) =f( x

c

).

9. Suppose that f ∈ L1 (R) and that∫|x| | f (x)|dx < ∞. Find a way to use the Fourier

transform of f to compute∫

x f (x)dx.

10. Suppose f ∈ G . Go over why F( fx j)(t) = it jF f (t).

290 CHAPTER 11. FUNDAMENTAL TRANSFORMS11.7. Exercises1. For f € L'(R"), show that if F~'f €¢ L' or Ff € L', then f equals a continuousbounded function a.e. This is Theorem 11.6.11 but review it.2. Suppose f,g € L'(R) and Ff = Fg. Show f = g ae.3. + Suppose f« f = f or fx f =O and f € L!(R). Show f =0.(14. Leth(x) = ( 6 e Par) + ( 0 oe) . Show that h’ (x) = 0 and h(0) = 27/4.Then let x > to conclude that {ye “dt = 1/2. Show that et dt= /m andthat [*, edt = x5. Let h(x) = (ie -1 *dt)” Thenx 1h' (x) =2 (/ ear) e* =2xe* (/ eau) .0 0Now h(x) = fo hr’ (t) dt. Do integration by parts to obtain-et [- ~( duls — [e + * [re (0) 2? dudt= ve [e ~(u)"qy 41 — [ fe (140?) 942 dudt~ [re 0H) gy 41 — [iu aa (1) o¢dtdu1 e *(14+u?) 4= 1— | w {| -———.—_ ]de(x)+ [« ( Tew u1 uwwhere lim, 4..e (x) = 0. Now explain why 1 — {; jodu = 4m. Hence fy et dt=va56. Recall that for f a function, fy (x) = f (x—y). Find a relationship between F fy (t)and Ff (t) given that f € L' (R").7. For f € L' (R"), simplify Ff (t+y).8. For f € L'(R”) and c a nonzero real number, show Ff (ct) = Fg(t) where g(x) =FE).9. Suppose that f € L' (IR) and that f |x| | f (x)|dx < co. Find a way to use the Fouriertransform of f to compute [xf (x) dx10. Suppose f € Y. Go over why F'(f;;)(t) = itjF f(t).