Timeline for Calderon-Zygmund decomposition on manifolds?
Current License: CC BY-SA 4.0
9 events
when toggle format | what | by | license | comment | |
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Nov 23, 2021 at 16:51 | vote | accept | Yuval | ||
Nov 22, 2021 at 18:36 | history | edited | YCor | CC BY-SA 4.0 |
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Nov 22, 2021 at 18:32 | answer | added | Terry Tao | timeline score: 11 | |
Nov 22, 2021 at 17:19 | comment | added | mlk | @Yuval I think you might even be able to get away with less. You'll only at some point have to relax your definition of $B_j$ a bit to include slightly less nice sets. E.g. if you glue a tiny handle to the plane and concentrate all of $f$ on there, then the only way out is to use one or more large sets that include parts or all of the handle as well as a good chunk of the plane to draw out the average. | |
Nov 22, 2021 at 16:18 | comment | added | Quarto Bendir | My impression is that this is done on page 628 of Coifman and Weiss "Extensions of Hardy spaces and their use in analysis," but I'm not familiar with any details | |
Nov 22, 2021 at 11:10 | comment | added | Yuval | @mlk It will be ideal if one only assumes a volume doubling condition, i.e. $\nu(B_{2R}(x))\leq 2^\kappa \nu(B_R(x))$, and perhaps the Poincare inequality. Assuming a lower bound in injectivity radius would be too strong in many situations... | |
Nov 22, 2021 at 9:31 | comment | added | mlk | On a compact manifold where the measure is induced by the Riemannian metric, could you not just apply the classic result in local coordinates? You get some additional multiplicative constants from this, but since you only need finitely many coordinate charts these are bounded and the result would probably still be enough for most estimates that need the decomposition. | |
Nov 22, 2021 at 9:24 | comment | added | Z. M | Do you endow the manifold with a Radon measure? It is unclear to me whether these volume conditions would depend on the choice of measure. | |
Nov 22, 2021 at 5:14 | history | asked | Yuval | CC BY-SA 4.0 |