Timeline for Does a measurable $F :[0, T] \to L^p (\mathbb R^d; \mathbb R_{\ge 0})$ have a "flattened" measurable version?
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8 events
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| Aug 19, 2023 at 8:43 | answer | added | Akira | timeline score: 0 | |
| Aug 11, 2023 at 20:20 | comment | added | Martin Väth | Yes, it holds also for $p=\infty$ and, more general, when $L^p$ is replaced by an ideal space and when $\mathbb R_{\ge0}$ is replaced by a Banach space. For instance, it also holds for every Orlicz space (which in general is non-separable if the generating Young function fails to satisfy the $\Delta_2$ condition). Also $\mathbb R^d$ can be replaced by any $\sigma$-finite measure space. A proof is given in the reference from the link. Note, however, that "measurable" needs to be defined as "Bochner measurable" (which implies essentially ("a.e.") separable range). | |
| Aug 10, 2023 at 11:22 | comment | added | Akira | @MartinVäth Below answer confirms that the result holds for $p \in [1, \infty)$. Could you please confirm if the result holds when we replace $L^p (\mathbb R^d; \mathbb R_{\ge 0})$ with a non-separable Banach space? | |
| Aug 9, 2023 at 22:38 | vote | accept | Akira | ||
| Aug 9, 2023 at 22:38 | comment | converted from answer | Martin Väth | Yes, see mathoverflow.net/questions/67434/… | |
| Aug 9, 2023 at 21:59 | answer | added | Iosif Pinelis | timeline score: 2 | |
| Aug 9, 2023 at 21:52 | history | edited | Akira | CC BY-SA 4.0 |
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| Aug 9, 2023 at 20:52 | history | asked | Akira | CC BY-SA 4.0 |