Show that if \(f\in L^1(\RR)\) and \(\hat f = 0\), then \(f = 0\) almost everywhere.
Show that Fourier inversion formula \(f(x) = \int_\RR \hat f(\xi) e^{2\pi i x \xi}\, d\xi\) does not necessarily hold pointwise for arbitrary \(f \in L^1(\RR)\) (since \(\hat f\) need not be in \(L^1(\RR)\)).
Solution
(1) Injectivity. Let \(f \in L^1(\RR)\) with \(\hat f(\xi) = \int_\RR f(x) e^{-2\pi i \xi x}\, dx = 0\) for all \(\xi \in \RR\). For \(\varepsilon > 0\) let \(\Phi_\varepsilon\) be the Gaussian approximate identity \[ \Phi_\varepsilon(x) = \frac{1}{\varepsilon} e^{-\pi (x/\varepsilon)^2}, \qquad \text{with } \widehat{\Phi_\varepsilon}(\xi) = e^{-\pi \varepsilon^2 \xi^2}. \] Fourier inversion holds for the Gaussian: \(\Phi_\varepsilon(x-y) = \int_\RR \widehat{\Phi_\varepsilon}(\xi) e^{2\pi i (x-y)\xi}\, d\xi\). Substituting this into the convolution and applying Fubini’s theorem, which is justified because \(\abs{f(y)}\widehat{\Phi_\varepsilon}(\xi)\) is integrable on \(\RR^2\), gives \[ (f * \Phi_\varepsilon)(x) = \int_\RR f(y) \Phi_\varepsilon(x - y)\, dy = \int_\RR \hat f(\xi) e^{2\pi i x \xi} \widehat{\Phi_\varepsilon}(\xi)\, d\xi. \] Since \(\hat f(\xi) = 0\) everywhere, the right-hand side is identically \(0\), so: \[ (f * \Phi_\varepsilon)(x) = 0 \quad \text{for all } x \in \RR \text{ and all } \varepsilon > 0. \] Since \(\Phi_\varepsilon\) is an approximate identity on \(L^1(\RR)\), as \(\varepsilon \to 0^+\): \[ \|f * \Phi_\varepsilon - f\|_{L^1} \to 0. \] Because \(f * \Phi_\varepsilon = 0\) for all \(\varepsilon > 0\), we have: \[ \|0 - f\|_{L^1} = \|f\|_{L^1} = 0. \] Thus \(f(x) = 0\) almost everywhere on \(\RR\).
(2) Failure of pointwise inversion. The inversion integral \(\int_\RR \hat f(\xi) e^{2\pi i x \xi}\, d\xi\) is a Lebesgue integral only when \(\hat f \in L^1(\RR)\). For \(f \in L^1(\RR)\), \(\hat f\) is continuous and vanishes at infinity by the Riemann–Lebesgue lemma, but \(\hat f\) need not be in \(L^1(\RR)\).
Let \(f = \mathbf{1}_{[-1, 1]} \in L^1(\RR)\). Its Fourier transform is \[ \hat f(\xi) = \int_{-1}^1 e^{-2\pi i x \xi}\, dx = \frac{e^{2\pi i \xi} - e^{-2\pi i \xi}}{2\pi i \xi} = \frac{\sin(2\pi \xi)}{\pi \xi}. \] For \(k \geq 1\) and \(\xi \in [k/2 + 1/8, k/2 + 3/8]\), \(\abs{\sin(2\pi\xi)} \geq 1/\sqrt 2\) and \(\abs{\hat f(\xi)} \geq \frac{1}{\sqrt 2\, \pi (k/2+1)}\); each such interval has length \(1/4\), so comparison with the harmonic series gives \[ \int_\RR |\hat f(\xi)|\, d\xi = \int_\RR \left| \frac{\sin(2\pi \xi)}{\pi \xi} \right| d\xi = \infty. \] Hence the classical inversion integral \(\int_\RR \hat f(\xi) e^{2\pi i x \xi}\, d\xi\) does not exist in the Lebesgue sense for arbitrary \(L^1\) functions. Furthermore, at points of discontinuity (such as \(x = \pm 1\)), the principal value limit equals \(1/2 \neq f(\pm 1) = 1\).