Let $0 < a < 1$ be an irrational number. Is it true that
$$\liminf_{n \in \mathbb N, n \to \infty} n \{na\} = 0?$$
Note: Here $\{\cdot\}$ denotes the fractional part.
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7$\begingroup$ This is equivalent to the unboundedness of the terms in the continued fraction of $a$ (up to some technicalities), so in particular it is false for the quadratic irrationals such as $\phi$ or $\sqrt{2}$. But I think this question is more appropriate to math stackexchange. $\endgroup$– Aleksei KulikovCommented May 16, 2023 at 20:58
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$\begingroup$ @AlekseiKulikov Right, please feel free to migrate the question. Do you think the corresponding question asking if it’s true for almost all irrationals is worth asking here? (I will make a new post, of course.) $\endgroup$– Nate RiverCommented May 16, 2023 at 21:38
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$\begingroup$ @NateRiver a generic geodesic on $SL_{2}(\mathbb{R})/SL_{2}(\mathbb{Z})$ is equidistributed and in particular unbounded. For a number to be badly approximable, one needs the geodesic to be bounded. $\endgroup$– AsafCommented May 16, 2023 at 21:51
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4$\begingroup$ There are countably many $a\in\mathbb{R}$ for which $\liminf n\{na\}$ exceeds $1/3$, and there are continuum many $a\in\mathbb{R}$ for which $\liminf n\{na\}$ equals $1/3$. See Cassels: An introduction to diophantine approximation (1957). Keyword: Markov spectrum (aka Markoff spectrum). $\endgroup$– GH from MOCommented May 16, 2023 at 22:12
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1$\begingroup$ @GHfromMO Thanks for the reference! Very interesting result. $\endgroup$– Nate RiverCommented May 16, 2023 at 22:25
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1 Answer
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The answer is no, because for some real $c>0$ and all integers $q>0$ we have $$q\{q\sqrt2\}=q(q\sqrt2-\lfloor q\sqrt2\rfloor) =q|q\sqrt2-\lfloor q\sqrt2\rfloor| \\ \ge q\,\inf_{p\in\mathbb Z}|q\sqrt 2-p| =q^2\,\inf_{p\in\mathbb Z}|\sqrt 2-p/q|\ge c.$$ So, $\liminf\limits_{q\to\infty} q\{q\sqrt2\}\ge c>0$.
More specifically, letting $p:=p_q:=\lfloor q\sqrt2\rfloor$ and $q\to\infty$, we have $p<q\sqrt2$, $p\sim q\sqrt2$, and hence $$q\{q\sqrt2\}=q(q\sqrt2-p) =q\frac{2q^2-p^2}{q\sqrt2+p} \ge q\frac1{q\sqrt2+p}\sim\frac1{2\sqrt2}.$$ So, $$\liminf_{q\to\infty} q\{q\sqrt2\}\ge\frac1{2\sqrt2}>0.$$
answered May 16, 2023 at 21:05
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$\begingroup$ Sorry, why is $q$ times the infimum bounded below? Is this to do with the irrationality measure being $2$? $\endgroup$ Commented May 16, 2023 at 21:08
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1$\begingroup$ @NateRiver See my comment below the original post. $\endgroup$ Commented May 16, 2023 at 22:13
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1$\begingroup$ @NateRiver Ok, I was wrong. Was thinking of the reverse. Look up the proof of Duffin-Schaeffer by Koukoulopoulos-Maynard for background. $\endgroup$ Commented May 17, 2023 at 0:09
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1$\begingroup$ @mathworker21 It's true that this property holds for almost all real. Khinchin proved in 1926, that if $\psi(x)$ is a positive decreasing real-valued function such that $\sum \psi(n)$ diverges, then for almost all real $\theta$ the inequality $\{q\theta\}<\psi(q)$ has infinitely many solutions. See Cassels's book mentioned by GH, Chapter 7. $\endgroup$ Commented May 17, 2023 at 0:55