Differentiation and multiplication under the Fourier transform

Remarks

For part 1, differentiate under the integral sign. The difference quotients of $e^{-2\pi i x \cdot \xi}$ in $\xi_j$ are bounded by $2\pi \abs{x_j}$, so dominated convergence applies with dominating function $2\pi \abs{x_j f(x)}$. For part 2, integrate by parts in $x_j$. On almost every line parallel to the $j$-th axis, $f$ and $\partial_j f$ are integrable, so $f$ has limits at both ends, and these limits are $0$; the boundary terms vanish.

Example: for $f(x) = e^{-\pi x^2}$ on $\R$, part 2 gives $\widehat{f'}(\xi) = 2\pi i \xi e^{-\pi \xi^2}$, and part 1 gives $\frac{d}{d\xi} e^{-\pi \xi^2} = -2\pi \xi e^{-\pi \xi^2}$; both agree with $f' = -2\pi x f$.

Iterating, if $f$ lies in the Schwartz space then $\widehat{\partial^\alpha f}(\xi) = (2\pi i \xi)^\alpha \hat f(\xi)$ and $\partial^\alpha \hat f$ is the transform of $(-2\pi i x)^\alpha f$ for every multi-index $\alpha$. The integrability hypotheses cannot be dropped: if $x_j f$ is not integrable, $\hat f$ need not be differentiable, as for $f(x) = 1/(1 + x^2)$ on $\R$, whose transform $\pi e^{-2\pi \abs{\xi}}$ has a corner at $0$.

Going deeper

For a student

These rules turn a linear differential operator with constant coefficients $P(\partial)$ into multiplication by the polynomial $P(2\pi i \xi)$. This is how the transform solves the heat, wave and Schrödinger equations on $\R^n$. They also give the duality between smoothness and decay: the more derivatives $f$ has in $L^1$, the faster $\hat f$ decays, and the faster $f$ decays, the smoother $\hat f$ is. This duality is the reason the Schwartz space is mapped onto itself.