Consider a Brownian motion $B$ and let $f(r)=\sqrt{2r \ln(\ln(r))}.$
Is it true that $\lim_{r \to +\infty}\frac{1}{f(r)}(B_r-B_{\left \lfloor{f(r)}\right \rfloor})= 0$ a.s. ?
If so, how to prove it? Otherwise what counter-example do you suggest?
$\lim_{r \to +\infty}\frac{1}{\sqrt{2r \ln(\ln(r))}}(B_r-B_{\left \lfloor{\sqrt{2r \ln(\ln(r))}}\right \rfloor})= 0$ a.s.?
Kurt.W.X
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