Local Maximum of Brownian motion.

Question

Given two positive rational number $a,b$. How to show that almost surely Brownian motion attains a local maximum at some time in $(a,b)$?

Answer 1

From Bluemental's 0-1 law and the symmetry of the Brownian motion we can see that for $B_0=0$ almost surely there $t_n \rightarrow 0$ and $s_n\rightarrow 0$ such that $B_{t_n}>0$ and $B_{s_n}<0$. We choose $0<t_m<s_k$.

Now on the interval $[0,s_k]$, $B_t$ cannot attain its maximum on the boundary hence it is somewhere in the interior, which gives exactly the existence of a local minimum.

Answer 2

Without loss of generality, we can assume $a=0$ (since the process $W_t := B_{t+a}-B_a$ is again a Brownian motion). Denote by $\Omega_{\max}$ the set of $w \in \Omega$ such that the path $(0,b) \ni t \mapsto B_t(w)$ does not attain a local maximum in $(0,b)$.

Let $w \in \Omega_{\max}$. Since the Brownian motion has continuous paths there are two possibilities:

• $(0,b) \ni t \mapsto B_t(w)$ does not attain a local minimum. This implies the monotonicity of the mapping.
• $(0,b) \ni t \mapsto B_t(w)$ does attain a local minimum. In this case, we can find a partition $0=s_0<\ldots<s_n=b$ such that $B_{\cdot}(w)|_{[s_{j-1},s_j]}$ is monotone for $j=1,\ldots,n$.

Thus, we conclude that the path is of bounded variation, i.e. $$\text{VAR}(B_{\cdot}(w),b) := \sup_{\Pi} \sum_{t_j-t_{j-1} \in \Pi} |B_{t_j}(w)-B_{t_{j-1}}(w)| \leq \sum_{j=1}^m |B_{s_j}(w)-B_{s_{j-1}}(w)| < \infty$$ where the supremum is taken over partitions $\Pi=\{0=t_0<\ldots<t_m=b\}$.

On the other hand, it is known that the total variation $\text{VAR}(B_{\cdot},b)$ is equal to $\infty$ a.s., so we conclude $\mathbb{P}(\Omega_{\max})=0$.