Show that if \(f\in L^1\) then \(\hat f\) is bounded and uniformly continuous.
Is it the case that \(f\in L^1\) implies \(\hat f\in L^1\)?
Show that if \(f, \hat f \in L^1\) then \(f\) is bounded, uniformly continuous, and vanishes at infinity.
- Show that this is not true for arbitrary \(L^1\) functions.
Show that if \(f\in L^1\) and \(\hat f = 0\) almost everywhere then \(f = 0\) almost everywhere.
- Prove that \(\hat f = \hat g\) implies that \(f=g\) a.e.
Show that if \(f, g \in L^1\) then \[\int \hat f g = \int f\hat g.\]
- Give an example showing that this fails if \(g\) is not bounded.
Show that if \(f\in C^1\) then \(f\) is equal to its Fourier series.
Solution
The Fourier transform is normalized as \(\hat f(\xi) = \int f(x)e^{-2\pi i x\xi}\,dx\). Related solutions are E-WURI3, E-FZXFR, E-O742O, and E-TF33D.
1. \(\hat f\) is bounded and uniformly continuous.
boundedness: \(|\hat f(\xi)| \le \int|f| = \norm{f}_1\).
2. \(f \in L^1\) does not imply \(\hat f \in L^1\).
\(f = \chi_{[-1,1]}\) has \(\hat f(\xi) = \frac{\sin 2\pi\xi}{\pi\xi}\), which is not integrable (its absolute value integrates like \(1/|\xi|\) at infinity).
3. If \(f, \hat f \in L^1\), then \(f\) is bounded, uniformly continuous, and vanishes at infinity.
the Fourier inversion formula \(f(x) = \int \hat f(\xi) e^{2\pi i x\xi}\,d\xi\) holds (valid a.e., and here the right side is a continuous function, so \(f\) has a continuous representative).
4. This fails for arbitrary \(L^1\) functions.
\(\chi_{[-1,1]}\) is bounded and uniformly continuous but does not vanish at infinity.
5. \(\hat f = 0\) a.e. implies \(f = 0\) a.e.; hence \(\hat f = \hat g\) implies \(f = g\) a.e.
6. \(\int \hat f\,g = \int f\,\hat g\) for \(f, g \in L^1\).
by Tonelli–Fubini, \[ \int \hat f(\xi) g(\xi)\,d\xi = \int\!\!\int f(x) g(\xi) e^{-2\pi i x\xi}\,dx\,d\xi = \int f(x) \Big(\int g(\xi) e^{-2\pi i x\xi}\,d\xi\Big)\,dx = \int f\,\hat g, \] since \(|f(x)g(\xi)|\) is integrable on the product space.
7. The identity fails if \(g\) is not bounded.
take \(f(x) = e^{-|x|}\), so \(\hat f(\xi) = \frac{2}{1 + (2\pi\xi)^2}\), and \(g(\xi) = 1 + (2\pi\xi)^2\), unbounded and not in \(L^1\).
8. If \(f \in C^1\), then \(f\) equals its Fourier series.
for \(f \in C^1\), \(\widehat{f'}(n) = 2\pi i n \hat f(n)\), so \(\hat f(n) = \widehat{f'}(n)/(2\pi i n)\) for \(n \neq 0\); by Parseval and Cauchy–Schwarz, \[ \sum_{n \neq 0}|\hat f(n)| \le \Big(\sum_{n\neq 0}\frac{1}{(2\pi n)^2}\Big)^{1/2}\Big(\sum_n|\widehat{f'}(n)|^2\Big)^{1/2} < \infty, \] so the Fourier series converges absolutely, hence uniformly, to \(f\) (Fejér’s theorem gives the pointwise/uniform limit \(f\) since the series converges).
9. Q.E.D.