Timeline for Riesz Representation Theorem for $L^2(\mathbb{R}) \oplus L^2(\mathbb{T})$?
Current License: CC BY-SA 4.0
18 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Aug 21, 2020 at 15:38 | comment | added | Goulifet | @LSpice Sure, I will remember that. | |
| Aug 21, 2020 at 15:36 | comment | added | LSpice | Just so you are aware, any edit, even for a change of notation like $\mathcal L^2 \rightarrow L^2$, still bumps a question. | |
| Aug 21, 2020 at 14:47 | history | edited | Goulifet | CC BY-SA 4.0 |
deleted 280 characters in body
|
| Aug 15, 2020 at 11:29 | history | edited | YCor |
edited tags
|
|
| Aug 14, 2020 at 20:55 | comment | added | Goulifet | @MichaelRenardy This seems reasonable: it would be all the square-integrable functions such that the periodization is well-defined and specifies a square-integrable periodic function. The big question is then to clarify which functions can be periodized in this way. | |
| Aug 14, 2020 at 19:10 | comment | added | Michael Renardy | A good candidate would be the set of all functions $g\in L^2(R)$ for which $\sum_{k\in Z} g(\cdot+2\pi k)$ is in $L^2(0,2\pi)$. | |
| Aug 14, 2020 at 17:24 | comment | added | Goulifet | @ChristianRemling I completely agree and this is part of the problem. The first displayed equation specifies a linear functional, but the $g$ in the second one is definitely not in the direct sum. I hope to find the class of functions such that the second displayed equation "works" (gives a continuous linear functional over $L^2(\mathbb{R}) \oplus L^2(\mathbb{T})$). | |
| Aug 14, 2020 at 17:15 | comment | added | Christian Remling | @Goulifet: Thanks for clarifying. You can certainly not get all functionals $F$ on $L^2(\mathbb R)\oplus L^2(\mathbb T)$ in this way because $g$ is already determined by $F$ restricted to the first summand. | |
| Aug 14, 2020 at 16:56 | history | edited | Goulifet | CC BY-SA 4.0 |
minor
|
| Aug 14, 2020 at 16:52 | comment | added | Goulifet | @MarkWildon It is not a typo but my way of writing is not very good, as Christian Remling also pointed out. I try to update for a better version. | |
| Aug 14, 2020 at 16:51 | comment | added | Goulifet | @ChristianRemling Good point, $f_1 + f_2$ is not in $\mathcal{L}^2(\mathbb{R})$ but still, so the notation $\langle g, f_1+f_2\rangle$ is a bit ambiguous, but I am considering the integration of the product $g \times (f_1+f_2)$. I am updating accordingly. | |
| Aug 14, 2020 at 16:49 | history | edited | Goulifet | CC BY-SA 4.0 |
precision on periodic square-integrable
|
| Aug 14, 2020 at 16:48 | comment | added | Goulifet | @LSpice Yes, that's right, I will had a precision about that. | |
| Aug 14, 2020 at 14:23 | history | edited | Goulifet | CC BY-SA 4.0 |
deleted 7 characters in body
|
| Aug 14, 2020 at 14:20 | comment | added | Christian Remling | What do you mean by $f_1+f_2$ in the last displayed equation? There doesn't seem to be an obvious way to define this sum that gives an element of $L^2(\mathbb R)$ (which is what you need here). | |
| Aug 14, 2020 at 14:16 | comment | added | Robert Israel | $\mathcal L^2(\mathbb R) \oplus \mathcal L^2(\mathbb T)$ can be considered as $\mathcal L^2$ of the disjoint union of a line and a circle, with Lebesgue measure on each. | |
| Aug 14, 2020 at 14:10 | comment | added | LSpice | To regard $\mathcal L^2(\mathbb T)$ as a subspace of the tempered distributions, you pull back to periodic functions on $\mathbb R$ and then integrate them against Schwartz functions? | |
| Aug 14, 2020 at 13:35 | history | asked | Goulifet | CC BY-SA 4.0 |