Finding the Derivative of Fourier Transform

Question

Suppose $f$ is in the Schwartz space $S(\mathbb R)$. i.e $f^n=O(\frac{1}{x^k})$ for any $n, k\in \mathbb N,$.

I am trying to show $\frac{d}{dx}\hat f(\xi)$ is the fourier transform of $-2\pi ixf$. i.e $\hat{-2\pi ixf(\xi)}$.

I computed $\frac{\hat f(\xi+h)-\hat f(\xi)}{h}-(\hat{-2\pi ixf})(\xi)=\int_{-\infty}^\infty f(x)e^{-2\pi ix \xi}(\frac{e^{-2\pi i xh}-1}{h}+2\pi ix).$ Fix $\epsilon>0 $, Since $f, xf$ decreaces rapidly, we do not need to worry about the integral on $[-\infty,N]\cup[N,\infty]$. What remains is to estimate $\frac{e^{-2\pi i xh}-1}{h}+2\pi ix$ when $|x|<N$. Here I noticed that $\frac{d}{dh}(e^{-2\pi i xh})(0)=2\pi ix$. So for each $|x|<N$, tehere exists $h_x>0$ such that if $|h|<h_x$, then $|\frac{e^{-2\pi i xh}-1}{h}+2\pi ix|<\frac{\epsilon}{N}$. if $h_x$ is independent of $x$, then we can conclude $|\int_{-N}^N f(x)e^{-2\pi ix \xi}(\frac{e^{-2\pi i xh}-1}{h}+2\pi ix)|<A\epsilon$ for some $A>0$. But I am stuck at removing the dependency of $h_x$ on $x$. Any suggestion?

Answer 1

I think you can't prove it this way, because $h_x$ is not independent of $x$. For any fixed $h$, the quantity $|\frac{e^{-2\pi i xh}-1}{h}+2\pi ix|$ explodes as $x\to\pm\infty$. Instead, if you are using Lebesgue integrals:

$$\frac d{d\xi}\int_{\mathbb R} f(x)e^{-2\pi i x\xi}dx = \int_{\mathbb R} \frac d{d\xi}(f(x)e^{-2\pi i x\xi}) dx = \int_{\mathbb R} f(x)\frac d{d\xi}e^{-2\pi i x\xi} dx = (-2\pi i x f)^{\wedge}(\xi), $$ where (here's your hint) the first equal sign is justified by DCT, and the rapid decay of $f$. The application of DCT to allow the interchange of limits is e.g. as in this MSE post.