The law of iterated logarithms for the standard Brownian motion asserts that
$(\ast) \limsup\limits_{h \downarrow 0} \frac{B(h)}{\sqrt{2h\log\log(\frac{1}{h})}} = 1$
I'm trying to prove the following: if $T_a = \inf\{t \geq 0 \ | B(t) = a\}$ then
$\limsup\limits_{h \downarrow 0} \frac{B(T_1)-B(T_1-h)}{\sqrt{2h\log\log(\frac{1}{h})}} \leq 1$
I tried to use $(\ast)$ for the right stopping time and the right Brownian motion, but I couldn't use one of the form $(B(T_1+h)-B(T_1) \ | \ h \geq 0)$ because in the question have an expression of the form $B(T_1-h)$ (and the strong Markov property does not work back in time).
The book (http://research.microsoft.com/en-us/um/people/peres/brbook.pdf) hints to bound $\psi(T_1-T_{1-q^{-n}})$ (where $\psi(h)=\sqrt{2h\log\log(\frac{1}{h}}$)) from below using Borel Cantelli. Since $T_1-T_{1-q^{-n}} = T_{q^{-n}}$ in distribution, I suppose this could be done, be I don't see how this is helpful (in particular I don't know how to use $T_{1-q^{-n}}$).
Any help would be much appreciated, thanks!