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Jul 17, 2021 at 5:47 vote accept Raphael J.F. Berger
Jul 16, 2021 at 22:11 comment added Will Sawin @RaphaelJ.F.Berger Actually I expect the function will behave like $f_n(x) = \begin{cases} 0 & x\leq0 \\ nx & 0 \leq x \leq 1/n \\ 2-nx & 1/n \leq x \leq 2/n \\ 0 & x \geq 2/n \end{cases}$ which pointwise converges to $0$ everywhere but has integral $1$.
Jul 16, 2021 at 22:08 comment added Raphael J.F. Berger The other alternative would be to divide the intervalls simply in $k$ pieces, such that all functions have $\delta(1)\ne0$ in that case it's immediately clear that it will converge to constant $1$ since all "combinations exist".
Jul 16, 2021 at 22:03 comment added Raphael J.F. Berger Very nice - thank you very much! I suppose then it will have poles at $\frac{1}{d}$? Can we say anything on the order of these poles? After all the integral has to be $1$ as well, in this case, I suppose.
Jul 16, 2021 at 21:56 history answered Will Sawin CC BY-SA 4.0