I am reading Peter Morters and Yuval Peres's book Brownian Motion. In theorem 1.27, it is proved that for a fixed $t \ge 0,$ almost surely, the Brownian motion $B(t)$ is not differentiable at $t$ and that $D^*B(t)=+ \infty$ and $D_*B(t)=-\infty.$ First we consider the Brownian motion $X(t) := tB(1/t)$ obtained from the time inversion of $B(t).$ Then $$D^*X(0) = \limsup_{h \downarrow 0} \frac {X(0 + h)-X(0)}{h} \ge \limsup_{n \to \infty} \frac {X(1/n)-X(0)}{1/n} \ge \limsup_{n \to \infty} \sqrt n X(1/n) = \limsup_{n \to \infty} B(n)/\sqrt n=+\infty.$$
(It is already stated before in the book that $\limsup_{n \to \infty} B(n)/ \sqrt n = + \infty$ and $\liminf_{n \to \infty} B(n)/ \sqrt n = - \infty.$)
If I employ the same strategy for $D_*X(0)$, then I want the inequalities as obtained above to be reversed for $\liminf$. But it can be seen that the second inequality doesn't reverse! (it is only based on the inequality $n \gt \sqrt n$). So in this way I can't show that $D_*X(0) = \liminf_{h \downarrow 0} \frac {X(0 + h)-X(0)}{h} \le \liminf_{n \to \infty} B(n)/ \sqrt n.$ So what it is it that I am missing here?
Moreover, we want $D_*X(0) = D_*B(t).$ In the book, for a fixed $t \ge 0$, we define $X(s)=B(t+s)-B(t)$. But how is that possible? Isn't $X(s) := sB(1/s)?$(I have this very same doubt for showing $D^*X(0)=D^*B(t).$)
The (right) derivates of $B$ at time $t$ coincide with the right derivates of $B^{(t)}$, defined by $B^{(t)}_s:=B_{t+s}-B_t$, at time $0$. Because $B^{(t)}$ is again a Brownian motion, this observation permits one to reduce to the case $t=0$.
Once you have shown (using $X$ and the above reduction) that $D^*B(t)=+\infty$, a.s., you can apply this to the Brownian motion $-B$ to see that (because $-B$ has the same distribution as $B$) $$ D_*B(t)\stackrel{d}{=}D_*(-B)(t) = -D^*B(t) = -\infty, $$ a.s.