Proving the continuity of Fourier Transform between Schwartz spaces via sequences Prove that if $f_k \rightarrow f$ in the Schwartz space $\mathcal{S}(\mathbb{R}^n)$, then $\hat{f_k} \rightarrow \hat{f}$ in $\mathcal{S}(\mathbb{R}^n)$.
This is the Exercise 2.2.2  in Loukas Grafakos's book named Classical Fourier Analysis (3º edition) and it's used to prove the Corollary 2.2.15 which says the Fourier transform is a homeomorphism between Schwartz spaces.
The convergence in $\mathcal{S}(\mathbb{R}^n)$ is defined by: $f_k \rightarrow f$ in $\mathcal{S}(\mathbb{R}^n)$ if $\rho_{\alpha,\beta}(f_k-f) \rightarrow 0$ as $k \rightarrow \infty$, $\forall\alpha, \beta$  multi-index, where
$$\rho_{\alpha,\beta}(f) = \sup_{x \in \mathbb{R}^n} |x^\alpha \partial^\beta(f)(x)|.$$
My idea is to prove the convergence $\hat{f_k} \rightarrow \hat{f}$ by definition using the folowing identity: 
$$\xi^\alpha \partial^\beta \hat{g}(\xi) = \frac{(-2\pi i)^{|\beta|}}{(2\pi i)^{|\alpha|}} (\partial^\alpha(x^\beta g(x)))^\wedge(\xi), \;\; \forall g \in\mathcal{S}(\mathbb{R}^n),$$
but I'm can't get the convergence I want.

$\hat{g} = x^\alpha \partial^\beta(\hat{f})(x)$ $$ \sup |\hat{g}(x)|\le \|g\|_{L^1}\le  \pi\|(1+x^2)g\|_{\infty} \le \pi\|g\|_{\infty}+\pi\|x^2g\|_{\infty}$$
