# Why $(2\pi x)^\alpha f(x)(2\pi i\xi)^\beta e^{-2i\pi x\cdot \xi}=(\partial _x)^\beta [(2\pi x)^\alpha f(x)]e^{-2i\pi x\cdot \xi}$?

Why $$(2\pi x)^\alpha f(x)(2\pi i\xi)^\beta e^{-2i\pi x\cdot \xi}=(\partial _x)^\beta [(2\pi x)^\alpha f(x)]e^{-2i\pi x\cdot \xi} ?$$

Indeed, to prove that the fourier transform is in the Schwarz space, we do as following \begin{align*} (2\pi i\xi)^\beta (i\partial _\xi)^\alpha \hat f(\xi)&=\int_{\mathbb R^n}(2\pi x)^\alpha f(x)(2i\pi i\xi)^\beta e^{-2\pi ix\cdot \xi}dx\\ &=\int_{\mathbb R^n}(2\pi x)^\alpha f(x)(-\partial _x)^\beta e^{-2\pi ix\cdot \xi}dx\\ &\underset{(*)}{=}\int_{\mathbb R^n}(\partial _x)^\beta [(2\pi x)^\alpha f(x)]e^{-2i\pi x\cdot \xi}dx \end{align*} but I don't understand the equality $(*)$. I recall that $\alpha,\beta \in \mathbb N^n$ and that we used $$x^\alpha =\prod_{i=1}^n x_i^{\alpha _i}.$$

• Are you sure $x^\alpha = \sum_i x_i^{\alpha_i}$ and not $x^\alpha = x_1^{\alpha_1}x_2^{\alpha_2}\cdots x_n^{\alpha_n}$? – Neal Jul 3 '16 at 17:33
• @Neal: For sure you right :) I corrected it. – MSE Jul 3 '16 at 17:35
• Why the downvote ? – MSE Jul 3 '16 at 18:13
• It wasn't me, but the downvote may be because the equation you ask about is pretty obviously false; one side involves derivatives of $f$ while the other side doesn't. Showing the Fourier transform maps the Schwarz space to itself isn't deep or all that hard but it's slightly trickier than the above. Gotta go now... – David C. Ullrich Jul 3 '16 at 18:52
• @DavidC.Ullrich: If you are inetrested, here is the proof of my course (proposition 2.1). There is maybe a subtlety I don't see. Thanks, jahia-prod.epfl.ch/files/content/sites/pde/files/cours/… – MSE Jul 3 '16 at 19:35