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I'm trying to read this paper: Bourgain, J.; Jitomirskaya, S., Continuity of the Lyapunov exponent for quasiperiodic operators with analytic potential, J. Stat. Phys. 108, No. 5-6, 1203-1218 (2002), arXiv:math-ph/0110040. Zbl 1039.81019. CiteSeerX

But I don't understand the norm they are using in (1.3). They write down the condition $\vert\vert k\omega\vert\vert>C(\vert k\vert \log(1+\vert k\vert)^A)^{-1}$

($\omega$ is an irrational number, and I think $k$ is an integer), and say that this expression is a "strong Diophantine condition" on $\omega$.

Another line in the paper which may be illuminating is

"for $\vert k\vert\leq q/2, \vert kw-ka/q\vert<1/2q$ and hence $\vert\vert k\omega\vert\vert >1/2q$."

Here $a/q$ is a lowest terms fraction with property $\vert \omega-a/q\vert<1/q^2$.

Context suggests to me that the norm should be a related to how easy it is to approximate an irrational number with rationals, but they don't have a definition in the paper, and I don't think I understand Diophantine approximation well enough to guess what it is.

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    $\begingroup$ For people who don't want to download the paper, or don't have a subscription to the journal, please put some actual detail in this question. e.g. the formula for the norm, and write out the context yourself. Read mathoverflow.net/howtoask for reasons why. $\endgroup$
    – David Roberts ♦
    Commented Jan 23, 2012 at 22:42
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    $\begingroup$ Is it just the distance to the integers? $\endgroup$
    – Charles Matthews
    Commented Jan 23, 2012 at 23:11
  • $\begingroup$ Charles: I don't think so. Or at least that suggestion doesn't clarify the second example in particular. My guess is that the norm is related to c(x) in the wikipedia page en.wikipedia.org/wiki/Diophantine_approximation $\endgroup$
    – Darren Ong
    Commented Jan 23, 2012 at 23:40
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    $\begingroup$ Darren, I don't have access to Springer, but already your first borrowing makes clear that $k$ means and integer and $\|\ \|$ is the distance from a real number to the nearest integer (without the "strong" appearance of $\log(1+|k|)$, the estimate is guaranteed by Dirichlet's theorem. $\endgroup$
    – Wadim Zudilin
    Commented Jan 24, 2012 at 0:32
  • $\begingroup$ Wadim and Charles, you're right. After thinking through more clearly, it has to be the distance to the integers. Thanks a bunch! $\endgroup$
    – Darren Ong
    Commented Jan 24, 2012 at 0:35

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