Timeline for Is the space $C_0^{k}(\Omega)$ a Montel space?
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14 events
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| Mar 27 at 23:47 | comment | added | Pedro Lauridsen Ribeiro | @AymanMoussa your definition of (lc) inductive limit topology is correct, but that by itself doesn't guarantee that $E$ induces the topology of $E_n$ for all $n$. There are plenty of examples of (lc) inductive limits which are not strict. | |
| Mar 27 at 23:43 | comment | added | Pedro Lauridsen Ribeiro | @Math A closed subset $F$ of $E_n$ is closed in $E$ because it's a closed subset in the relative topology (for $E$ induces the topology of $E_n$ for all $n$ since it's a strict (lc) inductive limit), hence it's the intersection of $E_n$ (which is closed in $E$) with a closed subset of $E$. | |
| Mar 27 at 21:19 | vote | accept | Math | ||
| Mar 27 at 21:16 | comment | added | Ayman Moussa | I rephrased the answer, I hope it's clearer now ! | |
| Mar 27 at 21:16 | history | edited | Ayman Moussa | CC BY-SA 4.0 |
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| Mar 27 at 21:08 | comment | added | Math | @AymanMoussa Could you give some more details in the last statement of your answer? That is, how to justify if this unit ball is compact in $C_{0}^{k}(\Omega)$ then it is compact in $C_{K_n}^{k}(\Omega)$. | |
| Mar 27 at 18:58 | comment | added | Math | @PedroLauridsenRibeiro In your third comment, closed subsets of $E_n$, say $F$, are closed in $E$ because, $\overline{F}^{\tau_E}=\overline{F}^{\tau_{E_n}}=F$ ? Perhaps I'm missing something, but how can we conclcude that the unit ball is not compact in $C_0^k(\Omega)$? | |
| Mar 27 at 17:56 | comment | added | Ayman Moussa | Thanks @PedroLauridsenRibeirofor the precisions, I guess this strictness was included in the characterization as " finest topology satisfying [...] " or am I missing something? | |
| Mar 27 at 17:50 | comment | added | Pedro Lauridsen Ribeiro | With these provisos in mind, what Ayman shows is that the unit ball of $E_n$ is closed and bounded in $E$ but it cannot be compact there, for otherwise it would be compact in $E_n$ as well. Hence $E$ cannot be even semi-Montel, let alone Montel. | |
| Mar 27 at 17:47 | comment | added | Pedro Lauridsen Ribeiro | Not really, with the above remarks Ayman's proof is correct. Actually, the regularity of the inductive limit is not really needed (it only shows that any bounded subset of $E$ is a bounded subset of $E_n$ for some $n$, but we know that already for the unit ball of $E_n$ since continuous linear maps send bounded sets to bounded sets), but the closedness of $E_n$ in $E$ for all $n$ is directly needed to guarantee that a closed subset of $E_n$ is also closed in $E$. | |
| Mar 27 at 17:17 | comment | added | Math | @PedroLauridsenRibeiro So the conclusion is that there is problem in the above proof? | |
| Mar 27 at 17:13 | comment | added | Pedro Lauridsen Ribeiro | Otherwise you cannot argue that if the unit ball of $E_n$ is compact in $E$, then it's compact in $E_n$ as well. | |
| Mar 27 at 17:12 | comment | added | Pedro Lauridsen Ribeiro | A thing missing here is the fact that $E=C^k_0(\Omega)$ is a strict (lc) inductive limit and therefore it induces on $E_n=C^k_{K_n}(\Omega)$ its original topology for all $n$. This is only true for strict (lc) inductive limits. Moreover, since in this case (also not generally true for all (lc) inductive limits) $E_n$ is closed in $E$ for all $n$, it follows that $E$ is a regular inductive limit, that is, any bounded subset of $E$ is contained and bounded in $E_n$ for some $n$. All that together implies that a compact subset of $E$ must be contained and compact in $E_n$ for some $n$. | |
| Mar 27 at 16:31 | history | answered | Ayman Moussa | CC BY-SA 4.0 |