Local maximum of brownian motions Let $B=(B_t)_{t\geq 0}$ be the standard Brownian motion.
I want to show that 
for every $t_0 \geq 0$
$\mathbb{P}$($B$ has a local maximum in $t_0$)=0.
I've already shown that for every $0<a<b<\infty$ $B$ is $\mathbb{P}$-a.s. not monotone on the interval [$a,b$].
My ideas were the following:
Suppose $B_t$ attains a local maximum in $t_0$. Can I assume that since $B$ has a.s. continuous paths, $t_0$ is preceded by an interval ($t_0-\epsilon, t_0$) where $B_t$ increases and is followed by an interval ($t_0,t_0+\epsilon$) where $B_t$ decreases? ($\epsilon > 0$) Then I would have two intervals where $B$ is monotone and since $B$ is not monotone on these intervals $\mathbb{P}$-a.s., I get
$\mathbb{P}$($B$ has a local maximum in $t_0$) = $\mathbb{P}$($B$ is monotone on ($t_0-\epsilon,t_0$) and $(t_0,t_0+\epsilon)$)=0.
Is that correct or do I have to argue in a different way? Can I always find those non-empty increasing and decreasing intervals "before" the next extremum?
Thanks in advance. 
 A: Here is one argument that would work, assuming that one already knows the following facts:


*

*For any $s>0$, the process $(B_{s+t}-B_s)_{t \geq 0}$ is a Brownian motion.

*The time inversion $(tB_{\frac{1}{t}})$ of a Brownian motion is a Brownian motion (defined to start at $0$ at $t=0$).

*$\limsup_{t \to \infty} B_t >0$ (in fact, the lim sup is $+\infty$).
There exist very succinct proofs of each of these statements, each of which can be derived directly from the basic axioms of Brownian motion. See theorem 1.9, proposition 1.23, and theorem 2.3 in the book by Morters and Peres, for example.
We can use these three things to get a short proof of your statement:
Fix $t_0 \geq 0$.
By $(1)$, we know that the process $Y_t = B_{t+t_0}-B_{t_0}$ is a Brownian motion.
By $(2)$, we know that the process $(X_t)$, defined by $X_t = tY_{\frac{1}{t}}$ when $t>0$ (and $X_0=0$), is a Brownian motion.
Now convince yourself that the following inclusion of events is true: $$\{ t_0 \text{ is a local maximum of } (B_t)\} \subseteq \{ \limsup_{t\to \infty} X_t \leq 0\}$$ The reason that this is true, is that if $t_0$ is a local maximum of $(B_t)$, then $Y_t \leq 0$ for small enough $t$, and thus $X_t \leq 0$ for large enough $t$.
By statement $(3)$, we see that the event on the right has probability $0$, and so the event on the left has probability $0$.
(I do concede, however, that there may be a shorter proof which relies on less statements. But I don't know one off the top of my head...)
