Timeline for A question of Erdős on equidistribution
Current License: CC BY-SA 2.5
11 events
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Jun 7, 2010 at 22:15 | vote | accept | Brad Rodgers | ||
Jun 6, 2010 at 1:38 | answer | added | George Lowther | timeline score: 23 | |
Apr 30, 2010 at 23:39 | comment | added | Brad Rodgers | TonyK, that's the right interpretation. Gjergji's link would seem to be pretty strong evidence that the problem is still unsolved... | |
Apr 29, 2010 at 21:37 | comment | added | Gjergji Zaimi | This is still referred to as an open question in arxiv.org/abs/math/0312440 | |
Apr 29, 2010 at 18:05 | comment | added | TonyK | Ah, so implicit in the conjecture is that the LHS takes the same value, independent of alpha, for almost all alpha. Is that right? | |
Apr 29, 2010 at 17:58 | history | edited | Brad Rodgers | CC BY-SA 2.5 |
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Apr 29, 2010 at 17:54 | comment | added | Brad Rodgers | @TonyK: I'm using $\alpha$ as both a dummy variable on the RHS, and an actual variable on the LHS. In fairness, I'm following Harman (which is otherwise a great book). I'll edit this above. @gowers: There is a counterexample due to JM Marstrand of Khintchine's conjecture given in Harman's book, for f an indicator function of some measurable set. It runs a few pages and is mainly arithmetical, rather than analytic. I don't know if it's the only counterexample known though. According to Harman it does not work in Erdős's question. (I haven't yet checked this myself.) | |
Apr 29, 2010 at 17:52 | history | edited | Brad Rodgers | CC BY-SA 2.5 |
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Apr 29, 2010 at 8:23 | comment | added | TonyK | The integrand should be f(\alpha x) dx, I think. | |
Apr 29, 2010 at 6:55 | comment | added | gowers | No idea what the answer is, but thanks for drawing my attention to this nice problem. (I'm hoping someone will say that it's still open.) How easy is the counterexample to Khintchine's conjecture? | |
Apr 29, 2010 at 4:00 | history | asked | Brad Rodgers | CC BY-SA 2.5 |