Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
Schwarts space is not first countable, so apprach with sequential continuity is not equivalent to the continuiuty. – Norbert Nov 12 '12 at 19:44
Notice the Schwartz space is a Frechet space, so apply the closed graph theorem. – lee Jan 2 '13 at 13:51
@Davide Giraudo I'm very bad at editting the answer because I don't know how to use Latex. So I can only do my best with the aid of language to prove it. – lee Jan 15 '13 at 7:49
@Norbert I don't understand your comment : Schwartz space is metrisable, so I don't see any issue to use sequential continuity. And even if it wasn't metrisable, there wouldn't be necessarily an issue, as with $\mathcal{D}$, because on a lot of topological vector spaces, sequential continuity is equivalent to continuity, for linear maps ! – Ahriman Jan 15 '13 at 8:41
up vote 3 down vote accepted

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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.