14. (S) Find the Fourier transform of the function f (x) = xe−x
√
[Ans: −iξ 2π exp(−ξ 2 /2)]
Solution: Set g(x) = e−x
2
/2
√
. By Problem 13, ĝ(ξ) =
2πe−ξ
2
2
/2
/2
. [Hint: f is the derivative of ...]
Q.14 is part
hand-in for
Week 4
. Observe that f (x) = −g ′ (x) therefore
fˆ(ξ) = − c
g ′ (ξ)
= −iξĝ(ξ)
√
2
= −iξ 2π e−ξ /2
15. (S) Show that the Fourier transform of the function
f (x) = √
Q.15 is
completion of
hand-in for
Week 4
2
1
e−x /(2a)
2πa
where a is a positive constant, is
2
fˆ(ξ) = e−aξ /2 .
[Hint: Combine the results of problems 13 and 8]. What happens to f and fˆ as a → 0?
Solution: By the result of Problem 13 the Fourier transform of φ(x) =
λ > 0 define φλ (x) = φ(λx) =
Set λ =
√1 .
a
2 2
√ 1 e−λ x /2 .
2πa
2
√ 1 e−x /2
2πa
is φ̂(ξ) =
2
√1 e−ξ /2 .
a
For
Then (Problem 8)
ξ2
cλ (ξ) = 1 φ̂( ξ ) = 1 √1 e− 2λ2 .
φ
λ λ
λ a
Then φλ (x) =
2
√ 1 e−x /(2a)
2πa
is the function we are interested in, and its Fourier transform is
φ̂λ (ξ) =
2
1 1 − ξ22
√ e 2λ = e−aξ /2 .
λ a
As a → 0 the function f peaks at the origin while its Fourier transform gets ‘flatter’. This means that we
can’t localize both f and fˆ. We’ll see later that this is a general phenomenon called the Heisenberg uncertainty
principle.
16. (S) Let f, g ∈ S(R). Show that
Z ∞
−∞
Solution: Set If =
x 7→ x′ = x − y)
R∞
−∞
(f ∗ g)(x)dx =
f (x)dx and Ig =
Z
∞
−∞
R∞
−∞
Z
∞
−∞
f (x)dx ·
Z
∞
g(x)dx .
−∞
g(x)dx. Observe that for any y ∈ R we have (change variables
f (x − y)dx =
Z
∞
f (x′ )dx′ = If
−∞
Now
Z
∞
−∞
(f ∗ g)(x)dx =
=
=
=
Z
∞
−∞
Z ∞
−∞
Z ∞
−∞
∞
Z
−∞
= If
Z
Z
∞
−∞
Z ∞
f (x − y)g(y) dx dy
Z ∞
g(y)
f (x − y) dx dy
−∞
−∞
g(y)If dy
∞
−∞
= If Ig
f (x − y)g(y) dy dx
g(y) dy
17. (P) Let f, g ∈ S(R). Recall that the convolution f ∗ g of f and g is defined by
Z ∞
(f ∗ g)(x) =
f (x − y)g(y)dy
−∞
a. Show that the generalised Riemann integral above exists (hence f ∗ g is well defined).
b. Show that f ∗ g = g ∗ f .
c. Assuming that differentiation under the integral is valid for integrands in S(R), show that f ∗ g is differentiable and that (f ∗ g)′ = f ′ ∗ g = g ∗ f ′ . Conclude that f ∗ g is actually infinitely differentiable. (Actually
f ∗ g ∈ S(R).)
Solution:
a. f is bounded, say |f | ≤ M . Therefore |f (x − y)g(y)| ≤ M |g(y)|. Since
R∞
−∞ f (x − y)g(y)dy.
b. Changing variables y 7→ y ′ := x − y we have
Z ∞
Z
(f ∗ g)(x) =
f (x − y)g(y)dy =
−∞
∞
−∞
R∞
−∞
|g(y)|dy exists, so does
f (y ′ )g(x − y ′ )dy = (g ∗ f )(x)
c. Differentiating under the integral sign we get
(f ∗ g)′ (x) =
d
dx
Z
∞
−∞
f (x − y)g(y)dy =
Z
∞
−∞
d
[f (x − y)g(y)] dy
dx
=
Z
∞
−∞
f ′ (x − y)g(y)dy = (f ′ ∗ g)(x)
Using this and the result of part 2 we get
(f ∗ g)′ = (g ∗ f )′ = g ′ ∗ f = f ∗ g ′
Now f ′ ∗ g is a convolutions of two members of S(R). By what we have just shown it is itself differentiable
with
(f ′ ∗ g)′ = f ′′ ∗ g
This shows that f ∗ g is twice differentiable with
(f ∗ g)′′ = f ′′ ∗ g
An easy induction argument shows that f ∗ g is infinitely differentiable with
(f ∗ g)(k) = f (k) ∗ g
18. (S) Let f ∈ S(R). Show that f is real valued if and only if for all ξ ∈ R, fˆ(−ξ) = fˆ(ξ). [Hint for the
converse: find the Fourier transform of f − f .]
Solution: Suppose first that f (x) is real valued. Then f (x) = f (x). Therefore
Z ∞
fˆ(−ξ) =
f (x)e+ixξ dx
−∞
Z ∞
=
f (x) e−ixξ dx
=
−∞
∞
Z
f (x)e−ixξ dx
−∞
= fˆ(ξ)
Conversely, suppose that
fˆ(−ξ) = fˆ(ξ)
(2)
Consider the function g(x) = f (x) − f (x). Its Fourier transform is
Z ∞
g(x)e−ixξ dx
ĝ(ξ) =
−∞
Z ∞ =
f (x) − f (x) e−ixξ dx
−∞
Z ∞
Z ∞
=
f (x)e−ixξ dx −
f (x)e−ixξ dx
=
−∞
∞
Z
−∞
f (x)e+ixξ dx −
= fˆ(−ξ) − fˆ(ξ)
=0
Z
−∞
∞
f (x)e−ixξ dx
−∞
(replace ξ by − ξ in (2))
Therefore g(x) = 0, which gives f (x) = f (x). This means that f is real valued.
19. (S)Show that if f ∈ S(R) then for all integers k ≥ 0,
Z ∞
Z ∞ 2
1
(k) 2
|ξ|2k fˆ(ξ) dξ
f (x) dx =
2π −∞
−∞
[Hint: Apply Plancherel’s theorem to an appropriate function.]
Solution: Apply Plancherel’s theorem to f (k) .
Z ∞ Z ∞ 2
1
(k) 2
d
f (x) dx =
f (k) (ξ) dξ
2π −∞
−∞
Z ∞ 1
ˆ 2
=
f (ξ) dξ
2π −∞
Z ∞
2
1
2k =
|ξ| fˆ(ξ) dξ
2π −∞
ℜ
R∞
−∞
f (x)ḡ(x)dx
. Show that
kf k2 kgk2
| cos θ| ≤ 1. Show that Fourier transformation preserves angles, i.e. the angle between f and g is the same as
the angle between fˆ and ĝ.
R∞
R∞
R∞
Solution: By Schwarz’s inequality, −∞ |f ||g|dx ≤ ( −∞ |f |2 dx)1/2 ( −∞ |g|2 dx)1/2 . Hence
R ∞
1 −∞ (f (x)ḡ(x) + f¯(x)g(x))dx | cos θ| =
2
kf k2 kgk2
R∞
1 −∞ (|f (x)|ḡ(x)| + |f¯(x)||g(x)|)dx
≤
2
kf k2 kgk2
R∞
1 −∞ (|f (x)|g(x)| + |f (x)||g(x)|)dx
=
2
kf k2 kgk2
≤ 1.
20. (S) For f, g ∈ S(R), define the angle θ between the two functions by cos θ =
Then
2
kf + gk2
=
=
=
=
=
Z
∞
−∞
Z ∞
−∞
2
kf k2
(f + g)(f + g)dx
f f¯ + f ḡ + g f¯ + gḡ dx
2
+ 2 kf k2 kgk2 cos θ + kgk2
2
1 ˆ
by Plancherel
f + ĝ
2π 2
1 ˆ2
2
f + 2 fˆ kĝk2 cos θ̂ + kĝk2
2π
2
2
by a similar calculation, where θ̂ is the angle between fˆ and ĝ. Applying Plancherel again to f and g separately,
we deduce that cos θ = cos θ̂. Hence result.
21. (S) By using the result of Q4c and the inversion theorem, find the Fourier Transform of f (x) = 1/(1 + x2 ).
Does either f or fˆ belong to S(R)?
[Ans: π exp(−|ξ|), no, no]
Solution: Applying the inversion theorem to exp(−|x|), we have
Z ∞
1
2
exp(−|x|) =
exp(ixξ)dξ.
2π −∞ 1 + ξ 2
Interchanging x, ξ, we deduce that
1
exp(−|ξ|) =
π
Z
∞
−∞
1
exp(ixξ)dx.
1 + x2
Replacing ξ by −ξ gives the required result.
/ S(R).
f is not differentiable at x = 0, so f ∈
/ S(R). |x|3 fˆ(x) is not bounded, and so fˆ ∈
Z ∞
p
p
cos t
√ dt. (Hint: use the fact
22. (S) Show that the Fourier Transform of 1/ |x| is C/ |ξ|, where C = 2
t
0
that the function to be transformed is even.) By using the Inversion Theorem, show that C 2 = 2π.
Solution:
fˆ =
=
=
=
=
Z
∞
exp(−ixξ)
p
dx
|x|
−∞
Z 0
Z ∞
exp(−ixξ)
p
dx
+
|x|
0
−∞
Z ∞
Z ∞
exp(ixξ)
exp(−ixξ)
p
p
dx +
dx
|x|
|x|
0
0
Z ∞
cos xξ
√ dx
2
x
0
Z ∞
2
cos t
p
√ dt.
t
|ξ| 0
(replacing x → −x in the first integral)
using the change of variable x = t/|ξ|. (Make sure you understand why |ξ| is needed, and not just ξ.)
By a virtually identical calculation, the Inversion Theorem gives
Z ∞
1
C
1
p
p exp(ixξ)dξ
=
2π −∞ |ξ|
|x|
=
Hence result.
C2 1
p .
2π |x|