Timeline for Fourier transform of periodic distributions
Current License: CC BY-SA 4.0
11 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| May 26 at 23:52 | comment | added | eraldcoil | A question: $\mathcal{F}_{T^n}:D'(T^n)\to S'(\mathbb{Z}^n)$ is an isomorphism? | |
| Jul 14, 2021 at 12:57 | comment | added | spaceman | @AlexanderKalmynin That's perfect, thank you for your help. | |
| Jul 14, 2021 at 12:49 | comment | added | Alexander Kalmynin | The last identity holds due to Fourier inversion theorem, which in this case is essentially the representation of a smooth function as a Fourier series | |
| Jul 14, 2021 at 12:42 | comment | added | Alexander Kalmynin | Yeah, this looks correct to me: in your definition, $$ \langle \mathcal F_{\mathbb T^n}^{-1}\mathcal F_{\mathbb T^n}u,\varphi\rangle=\langle \mathcal F_{\mathbb T^n}u,\iota\circ\mathcal F_{\mathbb T^n}\varphi\rangle=\langle u,\iota \circ\mathcal F_{\mathbb T^n}^{-1}\circ \iota \circ\mathcal F_{\mathbb T^n}\varphi\rangle=\langle u,\varphi\rangle $$ | |
| Jul 14, 2021 at 12:32 | comment | added | spaceman | Would it be given by $\mathcal{F}_{\mathbb{T}^n}^{-1} : \mathcal{S}'(\mathbb{Z}^n) \to \mathcal{D}'(\mathbb{T}^n)$: where for $v \in \mathcal{S}'(\mathbb{Z}^n)$ $$ \langle \mathcal{F}_{\mathbb{T}^n}^{-1} v, \varphi\rangle = \langle v, \iota \circ \mathcal{F}_{\mathbb{T}_n} \varphi \rangle $$ for $\varphi \in C^{\infty}(\mathbb{T}^n)$? | |
| Jul 14, 2021 at 12:23 | vote | accept | spaceman | ||
| Jul 14, 2021 at 12:23 | comment | added | spaceman | Given this definition, out of curiosity how would one define the inverse Fourier transform of these periodic distributions? | |
| Jul 14, 2021 at 12:21 | comment | added | spaceman | Ah yes, I see! And indeed, following the representation of continuous linear functionals on $\mathcal{S}(\mathbb{Z}^n$ from Exercise 3.1.7, i.e. as $\varphi \mapsto \langle u, \varphi\rangle = \sum_{\xi \in \mathbb{Z}^n} \varphi(\xi) u(\xi)$, we obtain that $\mathcal{F}_{\mathbb{T}^n}\delta_0 = 1$ in the sense of distributions. | |
| Jul 14, 2021 at 12:00 | comment | added | Gerald Edgar | The fact that the dual group for $\mathbb R^n$ is isomorphic to $mathbb R^n$ has caused confusion. | |
| Jul 14, 2021 at 11:35 | history | edited | Alexander Kalmynin | CC BY-SA 4.0 |
deleted 126 characters in body
|
| Jul 14, 2021 at 11:26 | history | answered | Alexander Kalmynin | CC BY-SA 4.0 |