Timeline for Find an element with given periodicity
Current License: CC BY-SA 4.0
37 events
| when toggle format | what | by | license | comment | |
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| S Dec 8, 2020 at 15:13 | history | bounty ended | Kung Yao | ||
| S Dec 8, 2020 at 15:13 | history | notice removed | Kung Yao | ||
| Dec 7, 2020 at 3:22 | answer | added | Noam D. Elkies | timeline score: 7 | |
| Dec 2, 2020 at 20:07 | comment | added | Fedor Petrov | Start with any smooth function having a compact support inside $(0,1)^2$ and extend to the whole plane using your relations. | |
| Dec 2, 2020 at 19:54 | vote | accept | Kung Yao | ||
| Dec 2, 2020 at 19:43 | comment | added | LSpice | Your latest edit is confusing, because it only makes sense when one knows the original version and reads all the comments. I think it is better just to edit down to the actual question, or at least to put in enough context that one can understand just by reading the question. | |
| Dec 2, 2020 at 19:42 | answer | added | Bertram Arnold | timeline score: 9 | |
| Dec 2, 2020 at 19:36 | history | edited | Kung Yao | CC BY-SA 4.0 |
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| Dec 2, 2020 at 19:20 | history | edited | Kung Yao | CC BY-SA 4.0 |
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| Dec 2, 2020 at 19:04 | comment | added | Fedor Petrov | Well, I proceed being confused. In $\mathbb{R}/\mathbb{Z}$ we have $x_1+1=x_1$, do not we? | |
| Dec 2, 2020 at 19:02 | comment | added | Kung Yao | @FedorPetrov absolutely | |
| Dec 2, 2020 at 19:02 | history | edited | Kung Yao | CC BY-SA 4.0 |
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| Dec 2, 2020 at 19:01 | comment | added | Fedor Petrov | You mean $L^2(\mathbb{R}^2/\mathbb{Z}^2)$, right? | |
| Dec 2, 2020 at 18:30 | history | edited | Kung Yao | CC BY-SA 4.0 |
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| Dec 2, 2020 at 17:04 | comment | added | juan | @KungYao Your use of the word boundary is not clear to me. Are you taking the elements in $[0,1]^2$ mod 1? It is your first condition, for example, equivalent to $f(0, x_2)=e^{-\pi i x_2} f(1,x_2)$? This is what I will have called a boundary condition. Or can we use $x_1=1/3$? In this case what is the meaning of $x_1+1$? | |
| Dec 2, 2020 at 15:22 | comment | added | Kung Yao | @BertramArnold does it really satisfy the first boundary condition? | |
| Dec 2, 2020 at 14:22 | comment | added | Kung Yao | @RaphaelB4 perhaps it does not quite respect the different signs in the boundary condition, i.e. it fulfils the first one but not the second one? | |
| Dec 2, 2020 at 14:12 | comment | added | RaphaelB4 | $f(x_1,x_2)=e^{i\pi x_1x_2}$ ? | |
| Dec 2, 2020 at 14:01 | history | edited | Kung Yao | CC BY-SA 4.0 |
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| Dec 2, 2020 at 13:51 | history | edited | Kung Yao | CC BY-SA 4.0 |
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| S Dec 2, 2020 at 13:39 | history | bounty started | Kung Yao | ||
| S Dec 2, 2020 at 13:39 | history | notice added | Kung Yao | Authoritative reference needed | |
| Nov 30, 2020 at 19:47 | comment | added | Kung Yao | @AnthonyQuas yes, thank you for your comment though. I was somehow not writing what I intended before though | |
| Nov 30, 2020 at 19:42 | comment | added | Anthony Quas | So there is no smooth function (or even continuous) function $g$ of absolute value 1 everywhere satisfying your boundary conditions. This means the approach I suggested previously cannot give you a basis of smooth functions of the type that you now say you want. | |
| Nov 30, 2020 at 15:54 | comment | added | Kung Yao | @MikaeldelaSalle thank you, I actually now understood I was asking the wrong question. | |
| Nov 30, 2020 at 15:54 | history | edited | Kung Yao | CC BY-SA 4.0 |
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| Nov 30, 2020 at 15:38 | comment | added | Mikael de la Salle | Ok. Then the restriction map is an isometry from your space onto $L^2([0,1]^2)$, so any orthonormal basis of $L^2([0,1]^2)$ (for examplet the usual Fourier basis) gives an orthonormal of you space. | |
| Nov 30, 2020 at 15:34 | comment | added | Kung Yao | the inner product is the standard $L^2$ inner product with Lebesgue measure on $[0,1]^2$ | |
| Nov 30, 2020 at 15:32 | comment | added | Mikael de la Salle | You talk about orthonormality, but you did not tell us what scalar product you consider. | |
| Nov 30, 2020 at 15:29 | comment | added | Anthony Quas | It will be orthogonal if $g$ had absolute value 1 everywhere | |
| Nov 30, 2020 at 15:19 | comment | added | Kung Yao | @AnthonyQuas how will this be orthogonal? | |
| Nov 30, 2020 at 15:18 | history | edited | Kung Yao | CC BY-SA 4.0 |
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| Nov 30, 2020 at 6:34 | history | edited | Kung Yao | CC BY-SA 4.0 |
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| Nov 30, 2020 at 6:33 | comment | added | Anthony Quas | Do you know any reasonable function in that space? If so, call it $g$, and consider the functions $f_{\mathbf n}=(x_1,x_2)=g(x_1,x_2)e^{2\pi i(n_1x_1+n_2x_2)}$. | |
| Nov 30, 2020 at 6:07 | comment | added | Kung Yao | I think this is more of theoretical relevance, since nobody is ever going to compute that projection explicitly. | |
| Nov 30, 2020 at 6:03 | comment | added | Thomas Kojar | One idea is to use the Hermite polynomials basis (math.stackexchange.com/questions/1927614/…), which has a nice Fourier interpretation, and then projecting them to your particular subspace. | |
| Nov 30, 2020 at 5:57 | history | asked | Kung Yao | CC BY-SA 4.0 |