Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

In his book Metric Number Theory, Glyn Harman mentions the following problem he attributes to Erdős:

Let $f(\alpha)$ be a bounded measurable function with period 1. Is it true that

$$\lim_{N\rightarrow\infty} \frac{1}{\log N} \sum_{n=1}^N \frac{1}{n}f(n\alpha) = \int_0^1 f(x) dx$$

for almost all $\alpha$,

writing "so far as the author is aware, this question remains open."

Harman's book is from 1997. Does anyone know the current status of the problem?

Motivation, for the curious

We lose no generality in assuming $f$ has mean $0$. The rough idea is that for almost all $\alpha$, $n\alpha$ will be equidistributed $(\mod 1)$ in a strong enough way to cause a great deal of cancellation in the sum, so in particular we might guess the sum is $o(\log N)$. It is a weaker version of a more classical conjecture of Khintchine that

$$\lim_{N\rightarrow\infty} \frac{1}{N} \sum_{n=1}^N f(n\alpha) = \int_0^1 f(x) dx$$

for almost all $\alpha$, where $f$ is as above. This is known to be false. (Of course, if $f$ is continuous it is true, for all irrational $\alpha$ even.)

share|improve this question
    
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? –  gowers Apr 29 '10 at 6:55
    
The integrand should be f(\alpha x) dx, I think. –  TonyK Apr 29 '10 at 8:23
    
@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.) –  Brad Rodgers Apr 29 '10 at 17:54
    
Ah, so implicit in the conjecture is that the LHS takes the same value, independent of alpha, for almost all alpha. Is that right? –  TonyK Apr 29 '10 at 18:05
    
This is still referred to as an open question in arxiv.org/abs/math/0312440 –  Gjergji Zaimi Apr 29 '10 at 21:37
show 1 more comment

1 Answer

up vote 12 down vote accepted

The statement was shown to be false by J. Bourgain in a paper published in 1988 (Almost Sure Convergence and Bounded Entropy, doi:10.1007/BF02765022). Well before either Harman's 1997 book and the 2003 paper Gjergji mentioned in the comment, which both say that it is an open problem!

From page 2 of Bourgain's paper

As further application of our method, a problem due to A. Bellow and a question raised by P. Erdős are settled.

And, further down (using ${\bf T}={\bf R}/{\bf Z}$),

The problem of Erdős mentioned above deals with weaker versions of the Khintchine problem. In particular he raised the question whether given a measurable subset $A$ of $\bf T$, then for almost all $x$ the set $\lbrace j\in{\bf Z }\_+\mid jx\in A\rbrace$ has a logarithmic density, i.e. $$ \frac{1}{\log n}\sum_{\substack{j\le n\\\\ jx\in A}}\frac1j\to\vert A\vert. $$ We will disprove this fact.

I can't vouch that Bourgain's paper is free of errors, as I have only just found it now and haven't read through it all in detail. However, Bourgain's result is also (very briefly) referred to in this 2004 paper, http://arxiv.org/abs/math/0409001v1, so I assume it is considered to be valid.

share|improve this answer
    
Bourgain's paper does seem to be quite well known. See also arxiv.org/abs/math/0611621v2 (Bourgain's Entropy Estimates Revisited) which re-proves of Bourgain's main results, although it doesn't treat the example asked for here. –  George Lowther Jun 6 '10 at 2:35
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.