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Jun 30, 2014 at 0:29 comment added Christian Remling @M Lemm: You're right, I was confused. In my "example", we would have to make $\varphi_n\sim 2^{-n}$ on $2^n<x<2^{n+1}$, and this part is $\sim 1/x$, so does not go to zero in $\mathcal S$. (The DC argument is fine, of course.)
Jun 30, 2014 at 0:20 comment added username @Christian Remling:It's not clear to me that the approximation will work near the boundary of $B_{2^n}$ in the very strong Schwartz topology. What about using dominated convergence to prove that it is closed? Pointwise convergence is trivial and from uniform convergence of $(1+x^2)^{N/2+1}|f_n-f|$ one gets a dominating function.
Jun 3, 2014 at 21:48 history edited username CC BY-SA 3.0
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May 17, 2014 at 2:55
May 17, 2014 at 0:30 answer added Christian Remling timeline score: 6
May 17, 2014 at 0:16 history asked username CC BY-SA 3.0