Timeline for Does equidistribution of zero average, due to irrationality, imply boundedness?
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| Dec 8, 2013 at 9:09 | vote | accept | smyrlis | ||
| Dec 5, 2013 at 15:59 | comment | added | Will Sawin | Actually choose an interval of appropriate $\alpha$s. Then we can repeat the process arbitrarily often, making a bigger $s_n$ each time. The only difficulty is making sure that later $s_n$s don't mess up earlier $s_n$s. But we control the contribution from $m$ that come just after our $m$, because $e^{2 \pi i m\alpha]-1$ will be large for there, and we can control the contribution from far-away terms because the trigonometric sum will be at most $n$ and $a_m$ will be very small. So it should be possible to do this, but the calculation will probably be messy. | |
| Dec 5, 2013 at 15:31 | comment | added | Will Sawin | I believe so, by restricting to successive intervals. Suppose we have already chosen $\alpha$ to lie in some very small interval such that for some number of $m$, $e^{2 \pi i m \alpha} - 1$ is quite small, and for some $n$, $s_n$ is very large. Let $w$ be the width of the interval. Then for all $m> 1/w $, we can choose an appropriate $\alpha$ such that $e^{ 2\pi m \alpha}$ is any point on the unit circle we want. So fix some $m>1/w$ such that $a_m\neq 0$, and choose an $\alpha$ such that $e^{2 \pi m \alpha}$ is incredibly close to $1$, making some appropriate $s_n$ arbitrarily large. | |
| Dec 5, 2013 at 12:14 | comment | added | smyrlis | Very good counterexample. I am wondering whether the following is true: Is it possible for every smooth 1-periodic and 0-average function (which is not a trigonometric polynomial) to find an irrational, such that the sequence in my question is not bounded? | |
| Dec 5, 2013 at 4:37 | history | edited | Will Sawin | CC BY-SA 3.0 |
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| Dec 5, 2013 at 2:30 | history | edited | Will Sawin | CC BY-SA 3.0 |
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| Dec 4, 2013 at 23:52 | history | edited | Will Sawin | CC BY-SA 3.0 |
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| Dec 4, 2013 at 23:40 | history | answered | Will Sawin | CC BY-SA 3.0 |