I have trouble understanding why the Brownian motion is nowhere differentiable, and I found somewhere that after showing the total variation of the Brownian motion is $+\infty$, the author claims that "as a result (of the infinite total variation), the path of the Brownian motion $B$ has an infinite number of local minima and maxima on any interval $[0,T]$, where $T>0$. "
I don't get how to show the part that there exists an infinite number of local minima and maxima on the interval $[0,T]$.
Any help will be extremely appreciated!
Best and regards
Best Answer
The first step is to prove that for arbitrary $0<a<b$, the Brownian motion $B$ attains almost surely a local maximum (minimum) in $(a,b)$ (see this question for a proof).
Now choose a sequence $(t_n)_{n \in \mathbb{N}}$ such that $t_n < t_{n+1}$ and $t_n \to T$ as $n \to \infty$, e.g. $t_n := T- \frac{1}{n}$. By the first part, $(B_t)_{t \geq 0}$ attains almost surely a local maximum at some time $s_n \in (t_n,t_{n+1})$. Consequently, the set $\{s_n; n \in \mathbb{N}\}$ consists of local maxima in the interval $[0,T]$.