Timeline for When is the periodisation of a function continuous?
Current License: CC BY-SA 4.0
7 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Aug 20, 2020 at 16:43 | comment | added | JensVF | Case 1: Let f be a function of compact support on [0,T]. Because f is already continuous, the only additional condition required is that f(T) = f(0). Case 2: Let f be a function of compact support but not on [0,T]. This case depends on "how often" the given interval "fits into" the interval [0,T] or vice versa. It's a matter of divisibility. In the worst case, the discontinuity is "sliding" such that only Schwartz functions fulfill the condition. Case 3: Infinite support: Only Schwartz functions fulfill the condition. | |
| Aug 19, 2020 at 9:15 | comment | added | MatthieuMeo | Hello, thank you very much for this very detailed and pedagogical answer. I agree with your statement that Schwartz functions are good candidates. However, as you point out yourself with the example of the rectangular function, there exists some functions much less regular whose periodisation is continuous. And this is precisely what I am interested in: I want to find the subspace of all functions whose periodisation is continuous. More precisely, I am interested in characterising the space: $$ \left\{f\in\mathcal{C}_0(\mathbb{R}): \Delta\Delta\Delta_T f \in \mathcal{C}([0,T])\right\}$$ | |
| Aug 17, 2020 at 21:44 | history | edited | JensVF | CC BY-SA 4.0 |
added 844 characters in body
|
| Aug 17, 2020 at 20:48 | comment | added | LSpice | Your first reference is to the paper that @MatthieuMeo mentions in their post. | |
| Aug 17, 2020 at 20:48 | history | edited | LSpice | CC BY-SA 4.0 |
Names of "these"
|
| Aug 17, 2020 at 20:20 | review | First posts | |||
| Aug 17, 2020 at 22:48 | |||||
| Aug 17, 2020 at 20:18 | history | answered | JensVF | CC BY-SA 4.0 |