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Definition. Let $f$ be a Schwartz function. The Fourier transform of $f$ is defined to be $$\hat{f}(\xi) := \int_{\mathbb{R}^n} f(x) e^{-2\pi i\langle x,\xi\rangle}dx$$

Now let $\alpha$ be a multi-index. Then we have $$\widehat{\partial^\alpha f}(\xi) = (2\pi i\xi)^\alpha \hat{f}(\xi)$$ Now this identity is proven by integration by parts as far as my book tells. But somehow I do not quite see how. I also never heard this in the context of multi-dimensional integrals. This is from the book: $$\widehat{\partial^\alpha f}(\xi) = \int_{\mathbb{R}^n} (\partial^\alpha f)(x) e^{-2\pi i\langle x,\xi\rangle}dx = (-1)^{|\alpha|}\int_{\mathbb{R}^n}f(x)(-2\pi i \xi)^\alpha e^{-2\pi i\langle x,\xi\rangle}dx$$

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    It obviously suffices to prove this for a single partial derivative, in which case the integration by parts occurs just along a single copy of the real line, and the infinite integral is a limit of finite truncations. The boundary terms in integration by parts go to zero due to the rapid decay of the Schwartz function. – 2017-01-29
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    @paulgarrett Thanks a lot. Could you please write this out in an answer? I somehow get confused with the $x$ in the exponential (I do not quite see how to write this integration by parts formally). – 2017-01-29
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    Let $n=2$ then assuming everything converges nicely $\hat{f}(\xi) = \int_{-\infty}^\infty \int_{-\infty}^\infty f(x,y) e^{-2i \pi (\xi_x x+\xi_y y)}dx dy$ and $$ \int_{-\infty}^\infty \int_{-\infty}^\infty \partial_x [f(x,y) ]e^{-2i \pi (\xi_x x+\xi_y y)}dx dy = - \int_{-\infty}^\infty \int_{-\infty}^\infty f(x,y) \partial_x [e^{-2i \pi (\xi_x x+\xi_y y)}]dx dy = 2i \pi \xi_x \hat{f}(\xi)$$ – 2017-01-29

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As Paul Garrett pointed out it is enough to consider $\alpha = \underbrace{(0,\dots,0,1,0,\dots,0)}_{j}$ only, since any other multi-indices follow by induction. Integration by parts gives us $$\begin{align} \widehat{\partial_j f}(\xi) &= \int_{\mathbb{R}^n}(\partial_jf)(x)e^{-2\pi i \langle x,\xi\rangle}dx \\ &= \int_{\mathbb{R}^{n - 1}}\int_{- \infty}^\infty (\partial_jf)(x)e^{-2\pi i \langle x,\xi\rangle}dx_j dx'\\ &= \int_{\mathbb{R}^{n - 1}} f(x)e^{-2\pi i \langle x,\xi\rangle}\big|_{x_j = -\infty}^{x_j = \infty} dx' + 2\pi i\xi_j\int_{\mathbb{R}^n}f(x)e^{-2\pi i \langle x,\xi\rangle}dx\\ &= 2\pi i \xi_j \hat{f}(\xi) \end{align}$$

Since as $f$ is a Schwartz function we have $$|f(x)| \leq C (1 + |x|)^{-1}$$ for some $C > 0$ and thus $$\lim_{x \to \pm\infty}|f(x) e^{-2\pi i x \xi_j}| \leq \lim_{x \to \pm\infty} |f(x)| \leq \lim_{x \to \pm\infty}C (1 + |x|)^{-1} = 0$$