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.?
no, the second term is basically $B\_t/t$ which (proabably) does go to 0, but the first is governed by the law of the iterated logarithm and does not. In fact, you know that limsup of that term is 1.


