Show the Fourier transform is continuous in the Schwartz space $\mathcal S(\Bbb R)$ Show the Fourier transform $\mathcal F$ is continuous in the Schwartz space $\mathcal S(\Bbb R)$.
Use the standard $\mathcal S$-norms
$$
\|f\|_{a,b}=\sup_{x \in \Bbb R} \left| x^af^{(b)}(x)\right|, \, a,b \in \Bbb Z_+.
$$ 
Let $\{f_n\}$ be a sequence converging to $f$ in $\mathcal S$.
To get the result, is it sufficient to show that
$$
\lim_n \|\hat f_n\|_{a,b} =  \|\hat f\|_{a,b}
$$
for any $a,b \in \Bbb Z_+$? 
This post prooves the result in another way. 
 A: You can check easily that the Schwartz space is locally convex because it is metrizable. And you can also check easily that a Cacuhy sequence must converge to a element in the Schwartz space. The argument above proves that Schwartz space is a Frechet space. Now to prove Fourier transform is continuous, you only have to prove that this operator is closed, i.e. the graph of this operator is closed(By closed graph theorem of Frechet space.). Let $P$ be an arbitrary polynomial and $D$ be the derivative operator. By the properties of Fourier tranform: $\mathcal{F}(P(D)f)$ is $P(\cdot)\mathcal{F}f$ and $\mathcal{F}(P(\cdot)f)$ is $P(-D)\mathcal{F}$. So you only have to prove that:
(i) taking a derivative,

(ii) multiplying by a polynomial

are continuous mapping from Schwartz space to Schwartz space (by Leibniz rule and closed graph theorem). With the aid of the above argument, you only have to prove that if $f_i$ tends to f in Schwartz space, then $\mathcal{F}f_i$ tends to $\mathcal{F} f$ pointwise which is an easy application of the Lebesgue dominated convergence theorem. The argument above shows that Fourier transformation is a closed operator. The continuity of the Fourier transform thus follows.
