Fourier inversion theorem

Remarks

Proof idea: insert the Gaussian factor $e^{-\pi \varepsilon^2 \abs{\xi}^2}$ into the inversion integral. With this factor the double integral converges absolutely. Since the Gaussian is its own Fourier transform, Fubini's theorem turns it into the convolution of $f$ with a Gaussian of width $\varepsilon$ and total mass $1$. As $\varepsilon \to 0$, the left side tends to the inversion integral by dominated convergence, since $\hat f$ is integrable. The convolutions tend to $f$ in $L^1$, and hence almost everywhere along a subsequence.

Example: $f(x) = e^{-\abs{x}}$ on $\R$ has $\hat f(\xi) = \frac{2}{1 + 4\pi^2 \xi^2}$, which is integrable. Since $f$ is continuous, $e^{-\abs{x}} = \int_\R \frac{2 e^{2\pi i x \xi}}{1 + 4\pi^2 \xi^2} \, d\xi$ for every $x$.

The integrability of $\hat f$ is needed for the integral to make sense. For $f = \mathbf{1}_{[-1/2, 1/2]}$, $\hat f(\xi) = \frac{\sin \pi \xi}{\pi \xi}$ is not integrable, so the theorem does not apply as stated. Then $f$ is recovered only by the symmetric limits $\lim_{R \to \infty} \int_{-R}^{R}$, which give $f(x)$ for $\abs{x} \ne 1/2$ and $1/2$ at the jumps. A consequence of the theorem is uniqueness: if $f$ is integrable and $\hat f = 0$, then $f = 0$ almost everywhere.

Going deeper

For a student

The theorem lets one solve problems on the frequency side, where derivatives become multiplications, and then return. It is the basis of the Fourier method for linear differential equations with constant coefficients. On the Schwartz space the transform is a bijection and inversion holds everywhere without extra hypotheses. The Plancherel theorem gives the corresponding statement in $L^2$, where the inverse transform is the adjoint of the transform.