Let $B$ be a $d$-dimensional standard Brownian motion. Fix $t \geq 0$. I have to show the following:
There is a $c>0$ such that for all $\varphi \in (\mathbb{R}^d)^*$ with $|\varphi|=1$, $$ \mathbb{P}\left( \limsup_{h \downarrow 0}\frac{|\varphi(B_{t,t+h})|}{\sqrt{h \ln\ln (1/h)}} \geq c\right)=1$$
By the law of iterated logarithm one would expect to show this with $c=\sqrt{2}$, but all my efforts have been futile.
I tried to reason as follows: Since $|\varphi|=1$, there is a sequence in the unit sphere in $\mathbb{R}^d$, say $\{x_n\}$, such that $|\varphi(x_n)| \uparrow 1$ as $n \uparrow \infty$. Now, if I could write $x_n = \frac{B_{t,t+h_n}}{|B_{t,t+h_n}|}$ for some sequence $h_n \downarrow 0$ and if (perhaps through a subsequence) $\frac{|B_{t,t+h_n}|}{\sqrt{h_n \ln\ln (1/h_n)} }\uparrow \sqrt{2}$ as $n \uparrow \infty$, I would have, $$ \limsup_{h\downarrow 0} \frac{|\varphi(B_{t,t+h})|}{\sqrt{h \ln\ln (1/h)}} \geq \lim_{n\uparrow \infty} \left|\varphi\left(\frac{B_{t,t+h_n}}{|B_{t,t+h_n)}|}\right)\right|\frac{|B_{t,t+h_n}|}{\sqrt{h_n \ln\ln (1/h_n)}} =\sqrt{2}.$$
But there is no guarantee that this will happen; in particular, there is no guarantee that the range of $\frac{B_{t,t+h}}{|B_{t,t+h}|}$ will even intersect the set $\{x_n\}$.