Riemann–Lebesgue lemma

Remarks

Proof idea: for the indicator function of a box, $\hat{f}$ is a product of terms like $(e^{-2\pi i b \xi_k} - e^{-2\pi i a \xi_k}) / (-2\pi i \xi_k)$, which tends to $0$. Linear combinations of such indicators, the step functions, are dense in $L^1(\R^n)$. Since $\abs{\hat{f}(\xi) - \hat{g}(\xi)} \le \int \abs{f - g}$ for every $\xi$, the property passes to all integrable $f$.

Example: for $f = \mathbf{1}_{[-1/2, 1/2]}$ on $\R$, $\hat{f}(\xi) = \sin(\pi \xi) / (\pi \xi)$ for $\xi \ne 0$, which tends to $0$.

The same proof gives the version for Fourier coefficients. If $f$ is integrable on $[0, 2\pi]$, then $\int_0^{2\pi} f(x) e^{-ikx} \, dx \to 0$ as $\abs{k} \to \infty$ over the integers. The lemma gives no rate of decay in general.

Going deeper

For a student

The lemma says that the Fourier transform maps $L^1(\R^n)$ into the space $C_0(\R^n)$ of continuous functions vanishing at infinity. In Fourier series it is a first step in the study of pointwise convergence, through the Dini and Jordan criteria. The image of $L^1$ is a proper subset of $C_0$, so the decay at infinity alone does not characterise Fourier transforms of integrable functions.