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Show there exists a sequence $\{f_n\}$ in the Schwartz space $S(\mathbb R)$ with limit $f$ for which $$ \lim \|f_n\|_{u,v} \text{ induced that } f \not\in S(\mathbb R) \text{ for some } u,v. $$ But $$ \lim \|f_n\|_{a,b} \text{ does not converges when } a \ne u,b\ne v. $$
Use the standard norm on $S(\mathbb R)$ $$ \|f\|_{a,b}= \sup_{x \in \mathbb R} |x^af^{(b)}(x)| ,\, a,b \in \mathbb Z_+. $$

This post shows $\sqrt{f(x)}$ does not belongs to $S(\mathbb R)$ when $f(x)=e^{-x^2} \left(e^{-x^2}+\sin ^2(x)\right)$.
I guess one way is to make $\|f_n\|_{u,v}$ unbounded, while keeping the other derivative bounded.
But I don't see how to do that.

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It $\{f_n\}$ which have to converge to $f$ in $\lVert\cdot\rVert_{u,v}$, not the sequence of norms. –  Davide Giraudo Nov 3 '12 at 13:20
    
@DavideGiraudo I meant what you said, but didn't express it well. Thank you. –  Nicolas Essis-Breton Nov 3 '12 at 16:12
    
@DavideGiraudo I adjusted the question, to reflect your point. Thank you. –  Nicolas Essis-Breton Nov 3 '12 at 17:09
    
What does "induced" mean here? –  joriki Nov 5 '12 at 22:20
    
@joriki I meant that $\lim \|f_n\|_{u,v}$ breaks the rule for $f$ to be $S$, that is $\lim \|f\|_{u,v}=\infty$. I think the question highlight the completeness of $S$, with respect to its topology. As $\lim \|f\|_{u,v}=\infty$ should implies that $\{f_n\} \subset S$ does not converges in $S$. Whence, the other limits, $\lim\|f_n\|_{a,b}$, should not exist. –  Nicolas Essis-Breton Nov 6 '12 at 0:19

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