Timeline for Duality relations for Lebesgue spaces of sections of vector bundles
Current License: CC BY-SA 3.0
14 events
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| Jun 15, 2020 at 7:27 | history | edited | CommunityBot |
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| Aug 8, 2013 at 23:22 | comment | added | user14166 | @TaQ: good point, i overlooked all that! | |
| Aug 8, 2013 at 11:51 | comment | added | TaQ | However, for $p=1$, the duality does not generally hold without $\sigma$-finiteness. A simple counterexample is the counting measure on the Borel $\sigma$-algebra of the real line. | |
| Aug 8, 2013 at 11:50 | comment | added | TaQ | @ Alex Amenta: The base $X$ being Lindelöf indeed is sufficient to guarantee the bundle being globally measurably trivial but this does not relate in any way to the given Borel measure on $X$ since no regularity of the measure is inherent in it just being defined on the $\sigma$-algebra of Borel sets. So some additional assumptions are also needed in order for the Riesz representation theorem to be applicable to $L^p(X,\mathbb R^{\,n})$. In Dudley's statement of the theorem, $\sigma$-finiteness of the measure is assumed, but if I recall correctly, this is not necessary for $1<p<+\infty$. | |
| Aug 8, 2013 at 1:12 | comment | added | user14166 | @TaQ: it's sufficient for the space to be Lindelöf, and i think one could get away with less by using some kind of transfinite induction - i don't think any conditions on measures are needed here | |
| Aug 7, 2013 at 17:26 | comment | added | TaQ | @ Peter Michor: Obviously some additional assumptions on the base space are required to get such a global trivialization by orthonormal measurable sections. For example those given in my answer are sufficient. In view of this, your argument is simpler than mine provided that the scalar and hence the finite-dimensional vector valued case is assumed to be known. @ Alex Amenta: In view of the preceding, I will not waste my time to add the details for the proof of $\|y\|_{p'}\le\|u\|$ in my answer. | |
| Aug 7, 2013 at 13:52 | comment | added | user14166 | regarding your second edit: i risk overlooking something since it's currently late at night, but this looks great to me! thanks a lot, tonight i can sleep easy. | |
| Aug 7, 2013 at 13:50 | vote | accept | CommunityBot | ||
| Aug 7, 2013 at 13:38 | history | edited | Peter Michor | CC BY-SA 3.0 |
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| Aug 6, 2013 at 23:35 | comment | added | user14166 | regarding your edit: i can fill some of the gaps, but i still can't make it to the end of the proof. writing $\lambda_\alpha$ for the function in $L^{p^\prime}(U_\alpha;E)$ induced by the local action of $\lambda$, we need to prove that the sums $\sum^N \lambda_\alpha$ are bounded (uniformly in $N$) in $L^{p^\prime}$. i don't see how this follows from your argument - in particular i don't see how the partition of unity comes into play. any further tips are much appreciated! | |
| Aug 6, 2013 at 19:31 | history | edited | Peter Michor | CC BY-SA 3.0 |
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| Aug 6, 2013 at 9:53 | comment | added | user14166 | i'm not sure if this is applicable, but maybe i'm just a bit lost. doesn't Theorem 3.12 just show (when restricted to $L^p$ spaces) that $L^{p^\prime}$ is contained in $(L^p)^*$, but not the reverse containment? | |
| Aug 6, 2013 at 9:20 | comment | added | user14166 | thanks for the reference - a quick look hasn't made things any clearer to me, so I'll give it a longer look and see how it goes! | |
| Aug 6, 2013 at 8:11 | history | answered | Peter Michor | CC BY-SA 3.0 |