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Aug 30, 2019 at 0:32 comment added Victor Ivrii Yes, it converges on these functions, as I wrote "on some analytic functions". However, if you consider the remainder it will be very large for $|p|\ge \sqrt{N}$
Aug 29, 2019 at 22:47 comment added fewfew4 Well that series is certainly convergent on $\exp{(ip(x-y))}$, right? So as far as my example is concerned $\exp{a^2\partial^2}$ (yes that is what I meant, thank you for correcting) can be regarded as that series. What does this not contradict? I never specified that the distribution was a differential operator.
Aug 29, 2019 at 21:54 comment added Victor Ivrii First $\exp(-a^2\partial ^2)$ is really bad, since it corresponds to a heat equation with negative time. You meant probably $\exp(a^2\partial ^2)$. This is not a differential operator, so it does not contradicts, and it is definitely not $\sum_{n=0}^\infty \frac{1}{n!} (a\partial)^{2n}$ since this series converges only on some (but not all!) analytic functions .
Aug 29, 2019 at 21:52 history edited Victor Ivrii CC BY-SA 4.0
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Aug 29, 2019 at 17:42 comment added fewfew4 @paul garrett yes, that is what I am referring to. I simply expressed the delta function as $\delta (x-y)=\int_{-\infty}^{\infty}\frac{dp}{2\pi}\exp(ip(x-y))$ and performed the $p$ integral.
Aug 29, 2019 at 17:24 comment added paul garrett @LucashWindowWasher, assumming that your $\partial$ means derivative: any differentiation of a distribution cannot increase support, quite generally. Can you clarify your comment?
Aug 29, 2019 at 17:22 comment added fewfew4 $\exp{(-a^2\partial^2)}\delta (x-y)\propto \exp{(x-y)^2/4a^2}$ is an example of the sort of sum mentioned that is not supported on the diagonal.
Aug 29, 2019 at 17:07 history answered Victor Ivrii CC BY-SA 4.0