Riemann-Lebesgue Lemma, higher dimensions Let $f\in L^1(\mathbb{R}^n)$ then $\lim_{|\xi|\rightarrow \infty} \hat{f}(\xi)=0$, where
$$\hat{f}(\xi)=\int_{\mathbb{R}^n} f(x) e^{-2\pi ix \cdot \xi} dx$$
What is the proof in higher dimensions: $n>1$? In most references only one dimensional case is presented. Can somebody please give the rigorous argument?
 A: Notice that if $f\in L^1$ then $||\widehat{f}||_\infty \leq ||f||_1$.
You can take the Schwartz class $$\mathscr S=\{\phi\in\mathscr{C}^\infty(\mathbb R^n) : ||x^\beta \partial^\alpha \phi||_\infty < \infty \;\forall \alpha,\beta \text{ multi-indices}\}$$
One can prove that $\widehat{\partial^\alpha f}(x) = (2\pi i x)^\alpha \widehat{f}(x)$ and $\partial^\alpha \widehat{f}(x) = \widehat{(-2\pi i x)^\alpha f(x)}$ and we can easily check that if $\phi\in\mathscr S$ then $\widehat{\phi}\in\mathscr{S}$. Also, the Schwartz class is dense in $L^1$ (with the $L^1$ norm) since the infinitely differentiable functions of compact support are included and those are dense in $L^1$.
Let $f\in L^1(\mathbb R^n)$ and take $\phi_n\in\mathscr S$ such that $\phi_n\to f$ in $L^1$. Therefore $$||\widehat{\phi_n}-\widehat{f}||_\infty = ||\widehat{\phi_n - f}||_\infty \leq ||\phi_n - f||_{L^1} \to 0$$
Thus $\widehat{\phi_n}$ converges uniformly to $\widehat{f}$. Since $\widehat{\phi_n}\in\mathscr S$ and the convergence is uniform we get that $\displaystyle\lim_{\xi\to\infty}\widehat{f}(\xi)=0$ as desired.
