Timeline for Fourier transform of a translation invariant operator on $l^2(\mathbb{Z}) \otimes l^2(\mathbb{Z})$
Current License: CC BY-SA 4.0
10 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Mar 21, 2020 at 9:48 | vote | accept | Frederik Ravn Klausen | ||
| Mar 19, 2020 at 12:53 | answer | added | Nik Weaver | timeline score: 1 | |
| S Mar 19, 2020 at 11:55 | history | suggested | gmvh | CC BY-SA 4.0 |
Corrected obvious typos and added some punctuation
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| Mar 19, 2020 at 11:46 | review | Suggested edits | |||
| S Mar 19, 2020 at 11:55 | |||||
| Mar 19, 2020 at 10:45 | comment | added | Frederik Ravn Klausen | Yes, we complete with respect to the norm coming from the inner product in the tensor product. | |
| Mar 18, 2020 at 20:28 | comment | added | LSpice | I believe that $\mathcal F^{-1}\phi$ lies in $\ell^2(\mathbb Z \times \mathbb Z)$, but that properly contains the ordinary vector-space product of $\ell^2(\mathbb Z) \otimes \ell^2(\mathbb Z)$. Did you mean some kind of completed vector product? | |
| Mar 18, 2020 at 20:27 | comment | added | Frederik Ravn Klausen | And this is supposed to be the norm of $\phi$ in the space $\otimes_{r \in \mathbb{Z}} L^2( \lbrack 0, 2 \pi \rbrack) $. | |
| Mar 18, 2020 at 20:25 | comment | added | Frederik Ravn Klausen | Well, I might be wrong but here is my take. The norm of a vector in $l^2(\mathbb{Z}) \otimes l^2(\mathbb{Z}) $ I calculate as $\vert\vert a \vert \vert = \sum_{k,j} \vert a_{k,j} \vert^2$ and hence $ \vert \vert \mathcal{F}^{-1} \phi \vert \vert = \sum_{x,y} \vert \int_0^{2 \pi} e^{ikx} \phi(k, x-y) dk \vert^2 = \sum_{x,y} \int_0^{2\pi} \int_0^{2\pi} dk dk' e^{i(k-k')x} \vert \phi(k,x-y) \vert^2 $. Now $\sum_{y} \vert \phi(k,x-y) \vert^2 $ is independent of $x$ and then the sum over $x$ of $e^{i(k-k')x}$ collapses one into to yield $\int_0^{2\pi} dk \sum_{n} \vert \phi(k,n) \vert^2 $. | |
| Mar 18, 2020 at 19:52 | comment | added | LSpice | Why does $\mathcal F^{-1}\phi$ live in $\ell^2(\mathbb Z) \otimes \ell^2(\mathbb Z)$? Is that meant to be some sort of completed tensor product? (This probably has to do with what you mean by 'basis', too.) | |
| Mar 18, 2020 at 19:41 | history | asked | Frederik Ravn Klausen | CC BY-SA 4.0 |