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Give an example of a function sequence in the Schwartz space $\mathcal S(\Bbb R)$ which does not converge.
That is, for any $a,b \in \Bbb Z_+$, $$ \|f_n\|_{a,b} < \infty, $$ but $$ \|f\|_{u,v}=\infty, $$ for some $u,v \in \Bbb Z_+$, when $\{f_n\} \subset \mathcal S$ converges to $f$ pointwise, that is $$ \lim_n f_n(x)=f(x), \forall x \in \Bbb R. $$
Use the standard $\mathcal S$-norms $$ \|f\|_{a,b}=\sup_{x \in \Bbb R} \left| x^a f^{(b)}(x) \right|, \, a,b \in \Bbb Z_+. $$

The following function does not belongs to $\mathcal S$ \begin{align} f(x)&=e^{-x^2}sin\left(e^{x^2}\right), \\ f(x)&=\frac{1}{1+x^2}. \end{align} But I don't see how to make an $\mathcal S$-sequence converging to them pointwise.

This question arised when considering this post.

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    I think you should clarify which mode of convergence is involved in your question. For example, if the convergence is induced by the $\mathcal{S}$-norms, then $f$ also lie in $\mathcal{S}$ (since the Schwartz space is a Fréchet-Montel space.) – Sangchul Lee Nov 13 '12 at 00:26
  • @sos440 Good point. I choosed pointwise convergence of $f_n$ to $f$. –  Nov 13 '12 at 00:58
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    Then how about a sequence of Gaussian kernels? For example $$ f_n(x) = \exp\left( -\frac{x^2}{n} \right) $$ is a sequence of Schwartz functions while converging to $1$ pointeise. – Sangchul Lee Nov 13 '12 at 01:30
  • @sos440 A perfectly fine example. Thank you. –  Nov 13 '12 at 01:45
  • @sos440 Your can post your comment as an answer. – Davide Giraudo Nov 16 '12 at 21:51

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