Timeline for Nice way to express $H^{-1}(\mathbb{S}^1)$
Current License: CC BY-SA 3.0
16 events
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| May 1, 2016 at 0:48 | history | edited | Willie Wong |
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| Apr 30, 2016 at 20:42 | vote | accept | Lewandowski | ||
| Apr 30, 2016 at 20:41 | answer | added | paul garrett | timeline score: 9 | |
| Apr 30, 2016 at 20:31 | comment | added | Deane Yang | And why does Johannes Hahn's situation imply the use of charts? You can define $H^{-1}$ as the space of bounded linear functionals on $H^1$. | |
| Apr 30, 2016 at 20:30 | comment | added | Deane Yang | I don't understand your question. Indeed, the $L^2$ inner product is undefined on $H^{-1}$, no matter what definition you use of the latter. | |
| Apr 30, 2016 at 20:23 | answer | added | Liviu Nicolaescu | timeline score: 4 | |
| Apr 30, 2016 at 20:21 | history | edited | Lewandowski | CC BY-SA 3.0 |
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| Apr 30, 2016 at 20:17 | comment | added | Lewandowski | @JohannesHahn it is just a question of taste, I guess and there are probably many people who disagree with me. I found this definition of $H^2$ very intuitive and direct whereas a characterization via charts somehow destroys my understanding of what kind of objects we have in $H^2(\mathbb{S}^1)$ and therefore I started wondering whether there is similar description of $H^{-1}$. | |
| Apr 30, 2016 at 20:10 | comment | added | Lewandowski | @DeaneYang well, which set would you take? Depending on what you choose, you might run into difficulties explaining what this $L^2$ inner-product is for $f \notin L^2$. | |
| Apr 30, 2016 at 20:08 | comment | added | Johannes Hahn | Is there a reason the standard interpretation of $H^{-1}$ being the dual of $H^1$ doesn't satisfy you? | |
| Apr 30, 2016 at 20:07 | comment | added | Deane Yang | There's no reason why Fourier series have to be restricted to ones in $L^2$. | |
| Apr 30, 2016 at 19:53 | comment | added | Lewandowski | @ChristianRemling thank you for pointing out the typo. Unfortunately, I don't really understand. You certainly want $L^2 \subset H^{-1},$ right? So if I would just adapt this notation, then I would still consider $\{f \in L^2; \text{some condition is satisfied}\}$ and there are no tempered distributions on $S^1$. | |
| Apr 30, 2016 at 19:50 | history | edited | Lewandowski | CC BY-SA 3.0 |
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| Apr 30, 2016 at 19:49 | comment | added | Christian Remling | I don't think what you wrote down is what one would usually denote by $H^2$ (two derivatives in $L^2$). In any event, the characterization via Fourier coefficients works for negative exponent also. | |
| Apr 30, 2016 at 19:43 | review | First posts | |||
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| Apr 30, 2016 at 19:41 | history | asked | Lewandowski | CC BY-SA 3.0 |