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Jul 27, 2019 at 21:42 comment added reuns If $T$ is a distribution then $T_n = T \ast n e^{-\pi n^2 x^2}$ is smooth, if $T$ was a probability density (ie. non-negative with $<T,1> =1$) then $T_n$ is a probability density so $T_n$ is smooth and $L^1$ and we can approximate $T_n \ast n e^{-\pi n^2 x^2}$ with the mixture of gaussian $\sum_{k=-K}^K \frac1K T_n(k/K) n e^{-\pi n^2 (x-k/K)^2}$. Finally $T_n \ast n e^{-\pi n^2 x^2}$ is an approximation of $T$.
Jul 27, 2019 at 18:23 history edited András Bátkai CC BY-SA 4.0
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Jul 27, 2019 at 17:25 review Late answers
Jul 27, 2019 at 18:28
Jul 27, 2019 at 17:10 review First posts
Jul 27, 2019 at 18:23
Jul 27, 2019 at 17:07 history answered Mattia Villani CC BY-SA 4.0