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Jan 25, 2010 at 14:20 comment added fedja No, you cannot take $g=H_0$. It is not Schwartz. The question as I understand it is the following. Let $g\in S$. Take the Hermite polynomial expansion $g\asymp \sum_n c_nH_n$ in $L^2(e^{-x^2})$ and multiply the $n$-th coefficient by $e^{-nt}$. Will the resulting function be Shwartz again? This question makes sense and is not immediately obvious to me. Still, I prefer to think of other things before unknown(yahoo) confirms that it is what he meant (or denies it)
Jan 25, 2010 at 10:01 history answered S. Carnahan CC BY-SA 2.5