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The Littlewood Conjecture asserts that for all real numbers $r$ and $s$, and for every $\epsilon > 0$, the inequality $|x(rx-y)(sx-z)| < \epsilon$ is solvable in integers $x, y, z$ with $x > 0$.

The Littlewood conjecture is clearly a consequence of the following:

For all real numbers $r$ and $s$, and for every $\epsilon > 0$ the inequalities

$|x(rx-y)| < 1, \,\,|sx-z| < \epsilon$ are solvable in integers $x, y, z$ with $x > 0$.

Does anyone know a counter-example to the latter statement? Does anyone know of any references to it in the literature?

Note that the inequality $|x(rx-y)| < 1$ always has infinitely many solutions $(x,y)$ with $x > 0$. This is a consequence of Dirichlet's Approximation Theorem. So it is natural to ask: "How does $sx$ behave mod 1 as $(x,y)$ runs through the solutions of $|x(rx-y)| < 1$?" For example, can the closure of the $sx$ mod 1 contain some non-empty open set and be disjoint from another?

Experiments with Sage seem to "suggest" that the numbers $sx$ are either dense mod 1 or have just finitely many limit points mod 1, depending on whether the numbers $1,r,s$ are linearly independent over the rationals. Again, any counter-examples or references relevant to this statement would be appreciated.

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3 Answers 3

$ \vert x(r x-y)\vert <1$ implies that $y/x$ is a convergent of the continued fraction expansion of $r$. This can be used to construct a counter-example as follows. Consider for $r$ for example a fairly large irrational quadratic real number with constant continued fraction $[l,l,l,l,\dots]$. Let $d_1 < d_2 < d_3,\dots$ be the sequence of denominators appearing in the convergents of $r$. Consider $s$ of the form $s=1/2\sum_{n=1}^\infty \alpha_n/d_n$ with $\alpha_i\in\{0,1}$ recursively defined such that the distance of $d_i(1/2\sum_{n=1}^i\alpha_i/d_i)$ to the nearest integer is $\geq 1/4$. This implies that the distance of $d_i(1/2\sum_{n=1}^\infty \alpha_n/d_n)$ to the nearest integer is $>1/4-\epsilon$ for $l$ large enough. Indeed, the sequence $d_1,d_2,\dots$ grows roughly like a geometric sequence of argument $l$. This implies that $d_i(1/2\sum_{n=i+1}^\infty \alpha_n/d_n)$ is at most of absolute value roughly given by $1/(2l(1-1/l))=1/(2l-2)$ which can be made arbitrarily small by choosing $l$ large.

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Roland, It is not quite true that the convergents of the continued fraction expansion of $r$ coincide with the solutions of $x(rx-y)| < 1$ in the way you describe. Non-convergents can satisfy such an inequality and moreover there can be solutions $(x,y)$ with $x$ and $y$ not relatively prime. See Rockett and Szusz, Continued Fractions, Section II.5. Does this fatally break the estimate given in the last two sentences of your argument? I'm not sure.... but it needs another look . (Note: This is a corrected version of my initial comment) –  SJR Apr 22 '10 at 6:08
You are of course right. My attempts to fix the resulting complications were unsuccessful. Perhaps it is more promising to consider for $r$ the golden number for which my claim (without error of my part) is indeed correct. Unfortunately it is to small to give useful error-terms and it is difficult to push through the end of the argument. –  Roland Bacher Apr 22 '10 at 15:14
I can't quite follow what your argument is, but I would point out that it is a known fact that if a sequence $d_1,d_2,...$ grows geometrically (in the strong sense that there is a positive constant c such that all ratios are at least 1+c) then there must exist a real number s and a positive epsilon such that for every n the distance from $sd_n$ to the nearest integer is at least epsilon. –  gowers Apr 22 '10 at 16:01
But that settles it! As I think Roland was suggesting, if $r=(1+5^{1/2})/2$, then the positive values of $x$ that appear in solutions to the inequality $|x(rx-y)| < 1$ are exactly the Fibonacci numbers, which grow exponentially in the sense you describe. Then the $s$ of your known fact gives a counter-example to my generalization of Littlewood's conjecture. Maybe for ANY irrational $r$ the $x$'s will grow exponentially, but I'm not sure about this. I think I see how to attack the proof of your "known fact". Is there a name attached to it? Thanks. –  SJR Apr 22 '10 at 17:26
@gowers: Do you have a reference for the "known fact" of your last comment? I can prove it if the ratio $d_{n+1}/d_n$ is at least $2+c$, but "$1+c$" is giving me trouble. –  SJR Oct 11 '10 at 15:11

This answer pieces together the various comments made by Roland Bacher, SJR and gowers previously. The proposed generalization of Littlewood's conjecture is false.

As suggested by Roland and SJR, take $r$ to be the golden ratio $(1+\sqrt{5})/2$.
Then $x|rx-y| \le 1$ only when $x$ runs over the Fibonacci numbers $F_k$. Now we want to show that there is an irrational number $s$ such that $F_k s$ is bounded away from integers.

As suggested by gowers there do exist such $s$ for any lacunary sequence $n_k$ (that is a sequence with $n_{k+1}/n_k \ge 1+ c>1$ for all large $k$), and so in particular for the Fibonacci numbers. This is related to a conjecture of Erdos, that for any lacunary sequence $n_k$ there exist irrational numbers $\alpha$ with $n_k \alpha$ not being dense $\mod 1$. Erdos's problem was settled independently by de Mathan and Pollington in a stronger form, showing that there exist many such $\alpha$. We need in fact that the values $\mod 1$ are bounded away from $0$. This is worked out in detail using Pollington's argument in a recent nice preprint of Haynes and Munday (see Lemma 1 of http://arxiv.org/abs/1308.0208 ).

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"This answer pieces together the various comments made by Roland Bacher, SJR and gowers above." For some value of the word, "above". –  Gerry Myerson Aug 18 '13 at 23:35

I can't find the paper online, but this looks rather like a question that is answered by a paper of Pollington and Vellani. Here is a link to an abstract of the paper. (It may be clear from the abstract that they answer your question -- I am feeling lazy and so have not checked.)


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@gowers: Thanks for the reference. The theorem mentioned in the abstract doesn't seem to settle my question, at least directly... Maybe I can use the result mentioned in the abstract to push through an argument like the one that Roland gave? –  SJR Apr 22 '10 at 3:30
Hmm, I have now thought a bit more and I think this paper is irrelevant -- though it is interesting in its own right. I like your question too. –  gowers Apr 22 '10 at 11:58

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