Timeline for Derivatives of delta function as a basis for distributions
Current License: CC BY-SA 4.0
5 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Aug 30, 2019 at 17:34 | vote | accept | fewfew4 | ||
| Aug 30, 2019 at 17:34 | comment | added | fewfew4 | Alright, I'm sorry for being imprecise. So the singular term "test function" denotes an entire set of functions? I was trying to show how that gaussian example could not be a distribution on the set of smooth test functions. Was my argument imprecise? Or did I just use the wrong language? | |
| Aug 30, 2019 at 9:19 | comment | added | Victor Ivrii | You need to stop juggling words and think (and read). One test function does not make much sense. You define what "test functions" are and get the space of distributions. If it is $\mathscr{D}:=C_0^\infty $ you get $\mathscr{D}'$, if it is $\mathscr{E}:=C^\infty $ you get $\mathscr{E}'$, if it is $\mathscr{S} $ you get $\mathscr{S}'$. Etc | |
| Aug 29, 2019 at 23:00 | comment | added | fewfew4 | So in other words, the "distribution" $\exp (-(x-y)^2/a^2)$ acting on the test function $\phi (y)=y^2$ gives a finite and smooth function. But if the test function were say $\phi(y)\sim\exp (y^4)$, the result would not make sense since the integral does not converge. If one wants to define distributions acting on a set functions, they better be well behaved for all test functions in that set. Is that roughly what you are saying? | |
| Aug 29, 2019 at 21:43 | history | answered | paul garrett | CC BY-SA 4.0 |