After pondering this MO question > [Location of maximum of Brownian motion with rough drift](https://mathoverflow.net/questions/193848) <, 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?



Can a Brownian motion be fast at its extrema?
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, 


1. This is only a heuristic, and
2. 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 $|s-t|\le 1/n$.
Since $M\_n$ is countable, the packing dimension $d\_P(M\_n)=0$.


So by Theorem 10.22 of <http://research.microsoft.com/en-us/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 $W$.


