Measure, Integration & Real Analysis, 2021a

366 Chapter 11 Fourier Analysis
Note that f ′ (t) =−2πte −πt2 = −2πtf(t). Combining this equation with 11.52
shows that
( ̂f
f
) ′(t)
=
f (t)( ̂f ) ′ (t) − f ′ (t) ̂f (t)
(
f (t)
) 2
= −2πt f (t) ̂f (t) − f (t) ̂f (t)
(
f (t)
) 2
= 0
for all t ∈ R. Thus ̂f / f is a constant function. In other words, there exists c ∈ C
such that ̂f = cf. To evaluate c, note that
11.53 ̂f (0) =
∫ ∞
−∞
e −πx2 dx = 1 = f (0),
where the integral above is evaluated by writing its square as the integral times the
same integral but using y instead of x for the dummy variable and then converting to
polar coordinates (dx dy = r dr dθ).
Clearly 11.53 implies that c = 1. Thus ̂f = f .
The next result gives a formula for the Fourier transform of a derivative. See
Exercise 9 for a formula for the Fourier transform of the n th derivative.
11.54 Fourier transform of a derivative
Suppose f ∈ L 1 (R) is a continuously differentiable function and f ′ ∈ L 1 (R).
If t ∈ R, then
( f ′ )̂(t) =2πit ̂f (t).
Proof
Suppose ε > 0. Because f and f ′ are in L 1 (R), there exists a ∈ R such that
∫ ∞
| f ′ (x)| dx < ε and | f (a)| < ε.
Now if b > a then
| f (b)| = ∣
∫ b
a
a
f ′ (x) dx + f (a) ∣ ≤
∫ ∞
Hence lim x→∞ f (x) =0. Similarly, lim x→−∞ f (x) =0.
If t ∈ R, then
( f ′ )̂(t) =
∫ ∞
−∞
f ′ (x)e −2πitx dx
= f (x)e −2πitx] x=∞ ∫ ∞
+ 2πit
x=−∞
= 2πit ̂f (t),
a
| f ′ (x)| dx + | f (a)| < 2ε.
−∞
f (x)e −2πitx dx
where the second equality comes from integration by parts and the third equality
holds because we showed in the paragraph above that lim x→±∞ f (x) =0.
The next result gives formulas for the Fourier transforms of some algebraic
transformations of a function. Proofs of these formulas are left to the reader.
Measure, Integration & Real Analysis, by Sheldon Axler