for a fixed $t \in [0, \infty)$, I have to show that
$ \mathbb{P} (D^+W_t = + \infty$ and $D_+W_t = -\infty )$, where $D^+$ (and $D_+$) denotes the upper right-hand derivative (and respectively the lower right-hand derivative ).
I construct a further Brownian motion X by time-inversion of a Brownian motion B, then:
$D^+X(0) \geq \limsup\limits_{n \to \infty} \frac{X(1/n)-X(0)}{1/n} \geq \limsup\limits_{n \to \infty} \sqrt n X(1/n) = \limsup\limits_{n \to \infty} \frac{1}{\sqrt n}B(n)$
Now all I need to prove is that $\limsup\limits_{n \to \infty} \frac{1}{\sqrt n}B(n) = +\infty$ because then I define $X(s) = B(t+s) – B(t)$
which is a Brownian motion and differentiability of $X$ at $0$ is equivalent to that of $B$ at $t$.
I thought about using Hewitt-Savage $ 0$-$1 $ law to prove the last claim but I can't conclude.
I think we could also use Borel-cantelli lemma but I don't know how.
If you see another more direct approach I will be glad to take it.
Best Answer
Note that the event $A=[\limsup\limits_{n\to\infty}B_n/\sqrt{n}=+\infty]$ is also $A=[\limsup\limits_{n\to\infty}(B_{n+k}-B_k)\sqrt{n}=+\infty],$ for every $k$, hence $A$ is in the tail sigma-algebra of the i.i.d. sequence $(B_n-B_{n-1})$.
Kolmogorov (independent) zero-one law shows that $P[A]$ is $0$ or $1$. By symmetry, $P[A]=P[A']$ where $A'=[\liminf\limits_{n\to\infty}B_n/\sqrt{n}=-\infty]$ hence $A$ is almost sure as soon as $[\limsup\limits_{n\to\infty}|B_n|/\sqrt{n}=+\infty]$ is almost sure.
Let $c\gt0$. Note that, if $|x|\leqslant c\sqrt{2^n}$ and $|x+y|\leqslant c\sqrt{2^{n+1}}$ then $|y|\leqslant3c\sqrt{2^n}$. For every $n$, define $A_n=[|B_{2^n}|\leqslant c\sqrt{2^n}]$ then the preceding remark and the independence of the increments of the Brownian motion show that $$ P[A_{n+1}\mid A_n]\leqslant P[|B_{2^{n+1}}-B_{2^n}|\leqslant3c\sqrt{2^n}]=P[|B_1|\leqslant3c]. $$ The last equal sign uses the Brownian scaling property. Let $p_c$ denote the RHS, thus $p_c\lt1$. For every $k\leqslant\ell$, $$ P\left[\bigcap_{n=k}^{\ell+1}A_n\right]\leqslant p_cP\left[\bigcap_{n=k}^{\ell}A_n\right], $$ hence, for every $k$, $$ P\left[\bigcap_{n\geqslant k}A_n\right]=0, $$ thus, the event $\liminf\limits_{n\to\infty}A_n$ has probability zero, that is, almost surely, $|B_{2^n}|\gt c\sqrt{2^n}$ for infinitely many $n$, thus, $$ P[\limsup\limits_{n\to\infty}|B_n|/\sqrt{n}\geqslant c]=1. $$ This holds for every $c$ hence the claim is proved.