Timeline for Elementary curiosity on the dual of $H^{-1}(\Omega)$
Current License: CC BY-SA 4.0
9 events
| when toggle format | what | by | license | comment | |
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| Aug 15, 2020 at 6:16 | history | edited | Michael Hardy | CC BY-SA 4.0 |
added 12 characters in body
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| Aug 14, 2020 at 14:54 | comment | added | Kosh | @WillieWong we are not writing $g$ as the gradient of a function but as the inverse image of the operator $I-\Delta$, which is nothing but the Riesz isomorphism in $H^{-1}$. | |
| Aug 14, 2020 at 13:10 | comment | added | Kosh | Maybe, an advantage is that if one derives estimates in terms of $g_0$ and $g$, then one can immediately understand what happens for "source" in the more regular space $L^2$ by simply forgetting about $g$. | |
| Aug 14, 2020 at 9:02 | comment | added | Kosh | So, it seems like the 2nd one reveals more structure. Also, it is the gradient of the same function $f$ so that even from the notational point of view, one does not need to introduce two functions. If one decides to write a book, I cannot find good reasons to use the first one instead of the second one, other than the willingness of following the tradition. | |
| Aug 14, 2020 at 9:01 | comment | added | Kosh | Note sure to get your point. Let me try to explain better what I meant with the post. If one is interested in the properties of equations of the type $L u = f$ with $L$ the classical 2nd order linear elliptic operator then, from my point of view, the first RHS tells that the source is the sum of an $L^2$ function and the divergence of an $L^2$ function, while the second tells that it is the sum of a "more regular" $H_0^1$ function with the divergence of an $L^2$ function which one knows being "regular" because is the gradient of an $H_0^1$ function. | |
| Aug 14, 2020 at 8:37 | comment | added | Piero D'Ancona | If you use the second representation, it gets awkward to prove that a distribution is in $H^{-1}$. The first one tells you that any derivative of an $L^2$ function is in $H^{-1}$, no special structure required | |
| Aug 14, 2020 at 1:13 | comment | added | Kosh | By Riesz representation theorem, both expressions span the whole of $H^{-1}$. There is no difference between the two in terms of generality. They give two equivalent representations of $H^{-1}$. The second one is exactly the one that one gets by Riesz representation theorem. | |
| Aug 14, 2020 at 0:58 | review | Close votes | |||
| Aug 14, 2020 at 21:24 | |||||
| Aug 13, 2020 at 22:37 | history | asked | Kosh | CC BY-SA 4.0 |