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In Freedman's series of 3 books on Markov processes, I find that I keep on running into terms like:

P[$\max_{0 \leq s \leq 1, s \leq t \leq rs}$ | B(t) - B(s) | > $\epsilon$]

in the background of proofs I'm reading. As Freedman mentions in B+D (19), its easy to see that for all fixed $\epsilon > 0$ this goes to 0 as r goes to 1. Its not so easy for me to see how to get any even somewhat-reasonable bounds on this probability for fixed $\epsilon$ and $r$. Does anybody have any suggestions (or out-of-the-box theorems) for some bounds?

PS: I've tried only a few silly things - e.g. trying to approximate the probability for finite sums and use Donsker's principle, and asking some people around here about martingale tricks, but no luck so far.

EDIT: A clarification in response to the first answer. I know that this probability (call it p(r, $\epsilon$) ) goes to 0 as r goes to 1 for ever fixed $\epsilon$, but I'm interested in statements like p(0.04, 0.000000001) < 0.02, or (more optimistically) p(r, $\epsilon) < 500 \frac{\log(\log(r-1))}{\epsilon}$. I don't have any reason to believe those two bounds are true (though both seem plausible to me), they are merely illustrative.

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up vote 1 down vote accepted

If $B$ is a.s.-continuous (and, therefore, uniformly continuous) then the maximum in question converges to $0$ a.s.

Your statement means convergence in probability, and follows automatically.

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Hi Yuri, thanks for the comment. I was probably not clear enough. I agree that it is easy to show the probability converges to 0, and indeed your comment seems the best/most economical way. My question is (should have been), can we say anything about the rate? e.g. if epsilon = 0.1, r = 1.0000000000001, I guess that the probability is quite small, but I don't see an easy way to get a 'reasonable' upper bound on this number. – little_probabilist Aug 13 '10 at 4:09
these estimates should follow from estimates on the modulus of continuity of B which in turn follow from maximal inequalities. – Yuri Bakhtin Aug 13 '10 at 18:32

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