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Jul 7, 2014 at 21:38 comment added Christian Remling Yes, Oseledec sounds like the kind of ingredient that's needed here.
Jul 7, 2014 at 19:08 comment added Ben Great. I'll take a look at that reference. I'm also just wondering if it's a quick step from knowing that for a fixed eigenvalue $E$, a positive Lyapunov exponent implies that the corresponding eigenfunction is exponentially decaying. Is this just the Ruelle-Osceledec theorem?
Jul 7, 2014 at 18:57 comment added Ben Right. This would just give that there is no absolutely continuous part of the spectral measure.
Jul 7, 2014 at 18:57 comment added Christian Remling A reference you might useful (or not, it's not exactly what you asked) is math.caltech.edu/SimonPapers/250.pdf (see Section 7 especially).
Jul 7, 2014 at 18:56 comment added Christian Remling No, positive Lyapunov exponent will not give this. By Fubini, you obtain that almost surely, for almost all $E$ you have exponential decay, but the (singular) spectral measure could be supported on the exceptional set.
Jul 7, 2014 at 18:50 comment added Ben Thanks! Does this follow just from the fact that the Lyapunov exponent of the transfer matrix is non-zero? In other words, if I knew that with probability one, the Lyapunov exponent is non-zero Lebesgue almost everywhere, would this imply the exponential decay for all eigenvectors?
Jul 7, 2014 at 18:45 comment added Christian Remling Your last paragraph is correct, in the following sense: Fix $E$. Then $|\psi(n)|\lesssim e^{-\gamma |n|}$ almost surely. Note, however, that this on its own does not imply that almost surely, we have exponential decay for eigenvalues $E$ (this needs to be proved separately).
Jul 7, 2014 at 18:39 history asked Ben CC BY-SA 3.0