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In an old post, someone said that the Calderon-Zygmund decomposition "follows from a simple stopping time argument". I would like to see this probabilistic argument (should be something simple). I know that the Lebesgue differentiation theorem, wich is important in the usual proof, can be proved using probability theory and then... but is there a faster way?

Also (I do not know too much probability theory) Lebesgue theorem can be proved by a kind of "energy increment" method (it looks like a regularity result): some quantity is "incremented" until we get enough structure and the process can be stopped...(I am thinking also to Roth theorem). Is there a corespondence, or a probabilistic interpretation for this? That "stop" has something to do with "stopping time"? Another thing: is there a direct probabilistic proof of Roth theorem (not via Szemeredi theorem)?

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Eddy
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  • I've heard this description of this proof independently of that post... my assumption was that "stopping time" was being used in a deterministic form. I'm curious if anyone has insight here.

    I suppose predictability is still a valid and interesting concept even when the filtration comes from a deterministic process.

     – D. Kelleher Dec 08 '16 at 23:54
  • The stopping time is with respect to the dyadic filtration on Euclidean space (so, in particular, not a filtration on a probability space). See e.g. http://www.math.unm.edu/~crisp/papers/lectures.pdf or http://bigwww.epfl.ch/chaudhury/Dyadic_decompositions.pdf – Terry Tao Dec 09 '16 at 08:16
  • Reading very fast the proof from the first document... it seems to me that the Lebesgue theorem is still present. – Eddy Dec 09 '16 at 11:32

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