Define \[ F(x) &\da \qty{ \sin(\pi x) \over \pi x}^2 \\ G(x) &\da \begin{cases} 1 - \abs{x} & \abs{x} \leq 1 \\ 0 & \text{else}. \end{cases} \]
Show that \(\fourier{G}(\xi) = F(\xi)\)
Compute \(\fourier{F}\).
Give an example of a function \(g\not \in L^1(\RR)\) which is the Fourier transform of an \(L^1\) function.
Hint: write \(\fourier{G}(\xi) = H(\xi) + H(-\xi)\) where \[ H(\xi) \da e^{2\pi i \xi} \int_0^1 y e^{2\pi i y \xi }\dy .\]
Solution
Use \(\hat g(\xi) = \int g(x)e^{-2\pi i x\xi}\,dx\).
1. \(\hat G(\xi) = F(\xi)\).
\(G\) is even, so the imaginary part of \(\int_{-1}^1(1-|x|)e^{-2\pi ix\xi}\,dx\) vanishes and \(\hat G(\xi) = 2\int_0^1(1-x)\cos(2\pi x\xi)\,dx\). For \(a = 2\pi\xi \neq 0\), integration by parts gives \(\int_0^1(1-x)\cos(ax)\,dx = \left[(1-x)\frac{\sin ax}{a}\right]_0^1 + \frac1a\int_0^1\sin(ax)\,dx = \frac{1-\cos a}{a^2}\). With \(1 - \cos 2\theta = 2\sin^2\theta\) for \(\theta = \pi\xi\), \(\hat G(\xi) = \frac{4\sin^2(\pi\xi)}{4\pi^2\xi^2} = F(\xi)\). At \(\xi = 0\) both sides equal \(1\).
2. \(\hat F = G\).
\(G\) is continuous and in \(L^1\), and \(\hat G = F \in L^1\) since \(F(\xi) \le \min(1, (\pi\xi)^{-2})\). The Fourier inversion theorem gives \(G(x) = \int F(\xi)e^{2\pi ix\xi}\,d\xi = \hat F(-x)\) for every \(x\). Since \(G\) is even, \(\hat F = G\).
3. \(g(\xi) = \frac{\sin(2\pi\xi)}{\pi\xi}\) is the Fourier transform of \(\chi_{[-1,1]} \in L^1\), and \(g \notin L^1(\RR)\).
\(\hat\chi_{[-1,1]}(\xi) = \int_{-1}^1 e^{-2\pi i x\xi}\,dx = \frac{e^{2\pi i\xi} - e^{-2\pi i\xi}}{2\pi i\xi} = \frac{\sin(2\pi\xi)}{\pi\xi}\). For \(k \geq 1\) and \(\xi \in [k/2 + 1/8, k/2 + 3/8]\), \(|\sin 2\pi\xi| \ge 1/\sqrt2\) and \(\xi \le k/2 + 1\), so \(\int|g| \ge \sum_{k\ge1}\frac{1}{4\sqrt2\,\pi(k/2 + 1)} = \infty\).