Timeline for Is $C_b(Q,E)$ linearly isometrically isomorphic to $C(\beta Q,E)$ where $\beta Q$ is the Stone–Čech compactification of $Q$?
Current License: CC BY-SA 4.0
10 events
| when toggle format | what | by | license | comment | |
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| May 9, 2018 at 14:40 | comment | added | Parschallen | It will only work for functions with relatively compact range. However, you can use weak compactness to get a continuous extension, just that continuity will be with respect to the weak topology and not (in general) the norm. | |
| May 9, 2018 at 10:45 | comment | added | Idonknow | @Parschallen How about if we assume that $E$ is reflexive? | |
| May 8, 2018 at 23:06 | comment | added | Idonknow | @Parschallen Do you mean finite dimensional case? Because relative compactness always implies boundedness (as $A\subseteq \overline{A}$ and $\overline{A}$ is bounded). In finite dimensional case, bounded and totally bounded are equivalent notions and totally boundedness implies relative compact in a complete metric space. | |
| May 8, 2018 at 16:13 | comment | added | Parschallen | In the one dimensional case, boundedness and relative compactness are the same thing. | |
| May 8, 2018 at 15:43 | vote | accept | Idonknow | ||
| May 8, 2018 at 15:44 | |||||
| May 8, 2018 at 15:43 | comment | added | Idonknow | Then why would $C_b(Q)$ be linearly isometric isomorphic to $C(\beta Q),$ as the functions in latter set have compact range while the former may not. | |
| May 8, 2018 at 14:18 | comment | added | Jochen Wengenroth | Parschallen's answer means that the canonical map $\rho: C(\beta Q,E)\to C_b(Q,E)$, $f\mapsto f|_Q$ is hardly ever surjective: Continuous functions $\beta Q\to E$ have compact range but bounded continuous functions $f:Q\to E$ need not have compact range. | |
| May 8, 2018 at 13:49 | comment | added | Idonknow | Yes, you are right. Edited. | |
| May 8, 2018 at 13:28 | review | First posts | |||
| May 8, 2018 at 13:28 | |||||
| May 8, 2018 at 13:25 | history | answered | Parschallen | CC BY-SA 4.0 |