Differentiation and the Fourier transform in Schwartz space I was trying to prove the following ($\mathscr S$ is the Schwartz space):

Let $f\in \mathscr S$. Then, $\hat f \in \mathscr S$ and $\forall \beta \in \mathbb N$ we have
  $$\begin{cases} \hat f ^ {(\beta)} (\zeta) = (-i)^ {\beta} \widehat{x^ {\beta}f}(\zeta)
\\ \widehat{f^ {(\beta)}}(\zeta)=i^ {\beta}\zeta^ {\beta}\hat f(\zeta)
\end{cases}$$

I encountered several obstacles.
1 - First, notice that $f\in\mathscr S \subseteq L^1$, so it makes sense to talk of its Fourier transform. We want to prove that $\forall \alpha,\beta \in\mathbb N_0, p_{\alpha,\beta}(\hat f) = \sup\vert x^ {\alpha} \hat f ^ {(\beta)}(x)
 \vert < \infty$.
We have $\vert x^ {\alpha} \hat f ^ {(\beta)}(x) \vert =  \vert \int x^ {\alpha} f (w) \partial_x^\beta(e^{-iwx}) dw \vert \le  \int \vert x^ {\alpha} f (w) w^\beta e^{-iwx}\vert dw = \int \vert x^ {(\alpha)} f (w) w^\beta\vert dw $. But what about now? Also, I passed the differentiation into the integral, and I do not know if I can.
2 - I was able to prove the first case, but again passing the differentiation into the integral...
3 - To prove the second case, I can prove for $\beta=1$ and then use induction.
 I want to prove that $\widehat{f'}(\zeta)=i\zeta\hat f(\zeta)$. We have 
$$
\widehat{f'}(\zeta) = \int f'(x)e^{-ix\zeta}dx = f(x)e^{-ix\zeta}\Big|^{+\infty}_{-\infty}-\int f(x) i\zeta e^{-ix\zeta} dx
$$
So if $\underset {x\rightarrow-\infty} \lim f(x)e^{-ix\zeta}=0 $, it is done, but I do not see why it should be the case.
 A: One thing you want to do is the following:
$$\partial_w \int_{\Bbb R}f(x) e^{-iwx}\,dx\overset?=\int_{\Bbb R}f(x) \partial_we^{-iwx}\,dx=\int_{\Bbb R}f(x) \, (-ix) e^{-iwx}\,dx\tag{1}$$
Why can you do this? It is called differentiation under the integral and follows from the Lebesgue theorem:
Let $w',w$ be two numbers, then there exists some $w''\in[w',w]$ so that $(e^{-i w' x}-e^{-iw x})/(w'-w) = (-ix)\,e^{i w'' x}$, in other words the difference is bounded by $|x|$. Now as you know, for Schwartz functions $f$ the function $x\mapsto |x|\,|f(x)|$ is also integrable. But this function is a bound of
$$f_n(x;w):=f(x)\frac{e^{-iw_n x}-e^{-i wx}}{w_n-w}$$
which for $w_n\to w$ converges pointwise to the integrand of the right hand side of $(1)$. So one finds by the Lebesgue theorem:
$$\partial_w \int_{\Bbb R} f(x) e^{-iwx}\,dx = \lim_{n\to\infty}\int_{\Bbb R}f_n(x;w)=\int_{\Bbb R}\lim_{n\to\infty}f_n(x;w)=\int_{\Bbb R}f(x)\,(ix)e^{-iwx}\,dx $$
Doing an induction will give you your first statement.
The second statement actually follows directly from partial integration and an induction:
$$w\int_{\Bbb R}f(x)e^{-iwx}\,dx = \int_{\Bbb R}f(x)\, w e^{-iwx}\,dx=\int_{\Bbb R}f(x)\, (-i\partial_x e^{-iwx})\,dx=\int_{\Bbb R}(i\partial _xf(x))\,e^{-iwx}\,dx$$
(Schwartz functions vanish at $\pm\infty$ because this is part of their definition.)
