Measure, Integration & Real Analysis, 2021a

Section 11C Fourier Transform 365
11.50 derivative of a Fourier transform
Suppose f ∈ L 1 (R). Define g : R → C by g(x) =xf(x). Ifg ∈ L 1 (R), then
̂f is a continuously differentiable function on R and
for all t ∈ R.
( ̂f ) ′ (t) =−2πiĝ(t)
Proof
Fix t ∈ R. Then
̂f (t + s) − ̂f (t)
lim
s→0
s
∫ ∞
= lim
s→0 −∞
=
∫ ∞
−∞
f (x)e −2πitx( e −2πisx − 1
)
dx
s
f (x)e −2πitx( e −2πisx − 1
)
lim
dx
s→0 s
∫ ∞
= −2πi xf(x)e −2πitx dx
−∞
= −2πiĝ(t),
where the second equality is justified by using the inequality |e iθ − 1| ≤|θ| (valid
for all θ ∈ R, as the reader should verify) to show that |(e −2πisx − 1)/s| ≤2π|x|
for all s ∈ R \{0} and all x ∈ R; the hypothesis that xf(x) ∈ L 1 (R) and the
Dominated Convergence Theorem (3.31) then allow for the interchange of the limit
and the integral that is used in the second equality above.
The equation above shows that ̂f is differentiable and that ( ̂f ) ′ (t) =−2πiĝ(t)
for all t ∈ R. Because ĝ is continuous on R (by 11.49), we can also conclude that ̂f
is continuously differentiable.
11.51 Example e −πx2 equals its Fourier transform
Suppose f ∈ L 1 (R) is defined by f (x) =e −πx2 . Then the function g : R → C
defined by g(x) =xf(x) =xe −πx2 is in L 1 (R). Hence 11.50 implies that if t ∈ R
then
( ̂f
∫ ∞
) ′ (t) =−2πi xe −πx2 e −2πitx dx
−∞
= ( ie −πx2 e −2πitx)] x=∞ ∫ ∞
− 2πt e −πx2 e −2πitx dx
x=−∞
−∞
11.52
= −2πt ̂f (t),
where the second equality follows from integration by parts (if you are nervous about
doing an integration by parts from −∞ to ∞, change each integral to be the limit as
M → ∞ of the integral from −M to M).
Measure, Integration & Real Analysis, by Sheldon Axler