After pondering this MO question > Location of maximum of Brownian motion with rough drift <, I wonder whether a Brownian motion can be fast (i.e. beats the law of the iterated logarithm) at its extrema? Is it necessarily fast?

1$\begingroup$ Can you ask the question in a more precise way? $\endgroup$ – Kurisuto Asutora Jan 29 '15 at 14:01
Heuristically, a point $(t,B_t)$ being a extremum is antithetical to fast oscillation, since there is no oscillation on one side of (above/below) $B_t$.
However,
 This is only a heuristic, and
 One may wonder whether there are so many fast times and so many maxima as to force an overlap.
At least we can rule out (2) as follows:
Theorem. If $W$ and $B$ are two independent Brownian motions then the set of maxima of $W$ is disjoint from the set of fast times of $B$.
Proof: Let $M$ be the set of local maxima of $W$ on a given closed interval. Then $M$ is a union of finite discrete, hence closed, sets $M_n$, where $M_n$ consists of those $t$ such that $B_s\le B_t$ for all $s$ with $st\le 1/n$. Since $M_n$ is countable, the packing dimension $d_P(M_n)=0$.
So by Theorem 10.22 of http://research.microsoft.com/enus/um/people/peres/brbook.pdf we have almost surely that $$\sup_{t\in M_n} \limsup_{h\downarrow 0} \frac{B_{t+h}B_t}{\sqrt{2h\log(1/h)}}=\sqrt{d_P(M_n)}=0.$$ Given $a>0$ we call a time $t\in [0,1]$ an $a$fast time if $$\limsup_{h\downarrow 0} \frac{B_{t+h}B_t}{\sqrt{2h\log(1/h)}}\ge a.$$ And a time is fast if it is $a$fast for some $a>0$. Thus, a.s. no fast time of $B$ is a local maximum of $W$.

$\begingroup$ Thanks for the update. I did consider that heuristic but on the other hand also thought that the Brownian motion has to get away from the local minimum, and it might even be the fast movement that lead to the local extrema. My intuition for these things isn't great! $\endgroup$ – P.Windridge Feb 16 '15 at 22:40