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.
Solution
Use the convention \(\hat f(\xi) = \int_\RR f(x) e^{-ix\xi}\,dx\), for which the inversion formula reads \(f(x) = \frac{1}{2\pi}\int_\RR \hat f(\xi) e^{ix\xi}\,d\xi\).
1. \(f \in L^1\) does not imply \(\hat f \in L^1\).
1.1. \(f = \chi_{[-1,1]}\) is in \(L^1(\RR)\) and \(\hat f(\xi) = \frac{2\sin\xi}{\xi}\).
\(\|f\|_1 = 2\), and \(\hat f(\xi) = \int_{-1}^{1} e^{-ix\xi}\,dx = \frac{e^{i\xi} - e^{-i\xi}}{i\xi} = \frac{2\sin\xi}{\xi}\).
1.2. \(\frac{\sin\xi}{\xi} \notin L^1(\RR)\).
For \(k \geq 1\), \(|\sin\xi| \geq 1/2\) on \([k\pi + \pi/6, k\pi + 5\pi/6]\), an interval of length \(2\pi/3\) on which \(|\xi| \leq (k+1)\pi\). So \(\int |\sin\xi/\xi|\,d\xi \geq \sum_{k\geq1} \frac{1}{3(k+1)} = \infty\).
1.3. Q.E.D.
Steps 1 and 2 exhibit an \(L^1\) function whose Fourier transform is not in \(L^1\).
2. Let \(f, \hat f \in L^1\) and \(g(x) \coloneqq \frac{1}{2\pi}\int \hat f(\xi) e^{ix\xi}\,d\xi\). Then \(f = g\) a.e.
This is the Fourier inversion theorem, which applies because \(f\) and \(\hat f\) are integrable.
3. \(g\) is bounded by \(\frac{1}{2\pi}\|\hat f\|_1\) and is uniformly continuous.
\(|g(x)| \leq \frac{1}{2\pi}\int |\hat f(\xi)|\,d\xi\). For \(h \in \RR\), \(|g(x+h) - g(x)| \leq \frac{1}{2\pi}\int |\hat f(\xi)|\,|e^{ih\xi} - 1|\,d\xi\), a bound independent of \(x\) that tends to \(0\) as \(h \to 0\) by dominated convergence with dominating function \(2|\hat f|\).
4. \(g(x) \to 0\) as \(|x| \to \infty\).
\(g(x) = \frac{1}{2\pi}\widehat{\hat f}(-x)\) is a Fourier transform of the \(L^1\) function \(\hat f\), so the Riemann–Lebesgue lemma applies.
5. Q.E.D.
Step 1 answers the first question negatively. By step 2, \(f\) agrees a.e. with \(g\), and steps 3 and 4 show that \(g\) is bounded, uniformly continuous, and vanishes at infinity; so \(f\) has these properties after modification on a null set.