Timeline for Elliptic regularity and Sobolev spaces
Current License: CC BY-SA 4.0
10 events
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| Nov 2, 2023 at 22:11 | comment | added | TaQ | It holds because of the (well-known) norm (in)equalities $\|\,U\,\|_{L^2}=\|\,U\,\|_{H^0}$ and $\|\,U\,\|_{H^s}\le\|\,U\,\|_{H^t}$ for $-\infty<s<t<+\infty$. | |
| Nov 2, 2023 at 12:47 | comment | added | G. Blaickner | @TaQ Why does (strong) convergence in $(H^{s},\Vert\cdot\Vert_{H^{s}})$ (for $s\geq 0$) implies convergence in $(L^{2},\Vert\cdot\Vert_{L^{2}})$? Maybe I am missing something... | |
| Nov 2, 2023 at 11:37 | comment | added | TaQ | You write: "there is no relation between strong convergence in $L^2$ and $H^s$. This is false since convergence in $H^s$ implies that in $H^0=L^2$ for $0\le s$ and conversely when $s<0$ holds. Hence for $0\le s$ your problem is settled, as I wrote above, but for $s<0$ not. | |
| Nov 2, 2023 at 7:23 | comment | added | G. Blaickner | @TaQ I don't think it is so easy. If you take a differential operator $D:C^{\infty}_{c}\to C^{\infty}_{c}$, then in general, there is no relation between its closures in $H^{s}$ and $L^{2}$ in the sense that neither of the domains $\mathcal{D}(\overline{D}^{L^{2}})$ and $\mathcal{D}(\overline{D}^{H^{s}})$ is contained in each other (since there is no relation between strong convergence in $L^{2}$ and $H^{s}$). Of course, $\mathcal{D}(\overline{D}^{L^{2}})\cap \mathcal{D}(\overline{D}^{H^{s}})$ is non-empty and both operators agree on their common domain, but that is all we can say in general. | |
| Nov 1, 2023 at 22:45 | comment | added | TaQ | So, in other words, you refer to the former case in my comment above, and consequently, in the case $s\ge 0$ you have already implicitly answered "yes" to your own question because of the last sentence in my comment: closure in $H^s\times H^s$ is a subset of the one in $L^2\times L^2$, and referring to the inner product in $H^s$ just obscures matters. If the case $s<0$ is important to you, then it is a more complicated matter. | |
| Nov 1, 2023 at 21:56 | comment | added | G. Blaickner | @TaQ I mean closure in the standard functional analytic sense: If $A:D(A)\to H$ is a linear operator in a Hilbert space $H$, then $\overline{A}$ is the operator defined as follows: $x\in D(\overline{A})$ iff there exists a sequence $(x_n)_n$ in $D(A)$ converging to $x$ such that $(Ax_n)_n$ is convergent. In this case we set $\overline{A}x:=\lim_{n\to\infty} Ax_n$. In other words, $\overline{A}$ is the (unique) operator whose graph is the closure of the graph of $A$. | |
| Nov 1, 2023 at 20:08 | comment | added | TaQ | When you speak of taking "closure of $D$ in some space, say $E$, do you really mean taking closure of the set $D_0=\{(f,Df):f\in C_c^\infty\}$ in the space $E\times E$, or possibly taking the bipolar of the set $D_0$ with respect to the duality between the spaces $E\times E$ and $E'\times E'$? Speaking about the inner product of $H^s$ suggests that you mean the latter. I do not see what would be the purpose of this. In case you mean the former, I do not see any real problem here since the closure in $H^s\times H^s$ is a subset of the one in $L^2\times L^2$. | |
| Oct 30, 2023 at 14:07 | history | edited | G. Blaickner | CC BY-SA 4.0 |
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| Oct 30, 2023 at 10:57 | history | edited | G. Blaickner | CC BY-SA 4.0 |
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| Oct 30, 2023 at 10:29 | history | asked | G. Blaickner | CC BY-SA 4.0 |