# Give an example a function sequence in the Schwartz space $\mathcal S(\Bbb R)$ which does not converge

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.) –  sos440 Nov 13 '12 at 0:26
@sos440 Good point. I choosed pointwise convergence of $f_n$ to $f$. –  Nicolas Essis-Breton Nov 13 '12 at 0:58
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. –  sos440 Nov 13 '12 at 1:30
@sos440 A perfectly fine example. Thank you. –  Nicolas Essis-Breton Nov 13 '12 at 1:45
@sos440 Your can post your comment as an answer. –  Davide Giraudo Nov 16 '12 at 21:51