Timeline for Orthonormal Basis of Multi-Dimensional Sobolev Space of Different Orders without Reproducing Kernel
Current License: CC BY-SA 4.0
10 events
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Dec 25, 2018 at 22:18 | history | edited | Minkov |
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Dec 24, 2018 at 2:13 | history | edited | Minkov | CC BY-SA 4.0 |
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Dec 23, 2018 at 12:08 | comment | added | Alex Gavrilov | Even this is not so ``natural'' a choice. What is the big reason to take all of this terms with weight one? (It is not invariant with respect to scaling, for example.) I admit that in this case the question actually has an answer, but I also admit that, personally, I totally would not care about it. | |
Dec 23, 2018 at 10:12 | history | edited | Minkov | CC BY-SA 4.0 |
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Dec 23, 2018 at 9:58 | history | edited | Minkov | CC BY-SA 4.0 |
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Dec 23, 2018 at 9:43 | comment | added | Minkov | @AlexGavrilov I totally agree with you on that the norm and inner product are not unique. We are interested in the "natural" choice of Sobolev norm and inner product $<u, v> = \sum_{i=1}^k \langle D^i u, D^i v \rangle$, especially how to construct the corresponding basis, for example, the eigenvector of the reproducing kernel. Also, for $0 \leq s_1, s_2 < d/2$, is it possible to align the orthonormal bases of $H^{s_1}(\Omega)$ and $H^{s_2}(\Omega)$ so that they only differ up to a rescaling of magnitudes? | |
Dec 23, 2018 at 9:34 | history | edited | Minkov | CC BY-SA 4.0 |
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Dec 23, 2018 at 9:32 | comment | added | Alex Gavrilov | I believe that this is not a meaningful question. You see, even though $H^s(\Omega)$ is called a Hilbert space it is technically not, in the sense that the actual Hilbert norm on it is not fixed. It is a sort of infinite dimensional analog of a linear space of finite dimension: you are free to chose any of the (equivalent) scalar products. Naturally, an orthonormal basis would very much depend on the choice. | |
Dec 23, 2018 at 9:17 | history | edited | Minkov | CC BY-SA 4.0 |
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Dec 23, 2018 at 9:08 | history | asked | Minkov | CC BY-SA 4.0 |