Use Blumenthal's 0-1 law to solve the following

Question

From Durett: $B(t)$ represents a Brownian motion,

"$(i)$ Suppose $f(t) > 0$ for all $t>0$. Use Blumenthal's $0-1$ law to conclude that $\lim \sup_{t \rightarrow 0^+} \frac{B(t)}{f(t)} = c$, $P_0$ a.s., where c is a constant in $[0,\infty]$.

$(ii)$ Show that if $f(t) = \sqrt{t}$ then $c=\infty$, so with probability one Brownian Paths are not Holder continuous of order $1/2$ at $0$"

I understand that the random variable $\lim \sup_{t \rightarrow 0^+} \frac{B(t)}{f(t)}$ is in the germ sigma field thus we can apply Blumenthal's $0-1$ law to all events involving it.

$(ii)$ We need to show that $P_0(\lim \sup_{t \rightarrow 0^+} \frac{B(t)}{\sqrt{t}} \leq s) = 0$, $\forall s \in \mathbb{R}$. By Blumenthal's $0-1$ law this is equivalent to $P_0(\lim \sup_{t \rightarrow 0^+} \frac{B(t)}{\sqrt{t}} \leq s) < 1$, $\forall s \in \mathbb{R}$

$P_0(\lim \sup_{t \rightarrow 0^+} \frac{B(t)}{\sqrt{t}} \leq s) = \lim_{n \rightarrow \infty} P_0(\sup_{t \leq 1/n} \frac{B(t)}{\sqrt{t}} \leq s) \leq P_0(\frac{B(1/n)}{\sqrt{1/n}} \leq s) = \Phi(s) < 1 $

Where the last equality comes from the fact that $\frac{B(t)}{\sqrt{t}} \sim N(0,1)$

(i) I can't find a similar way to deal with this one, any idea ?

Answer 1

If every set in a sigma field has probability 0 or 1 then any function measurable with respect to it is almost surely constant. To prove this just note that the distribution function $F$ of this random variable has the property that $F(x)$ is 0 or 1 for every $x$. It is fairly easy to show that if $a=\inf \{x:F(x)=1\}$ then $F(x)=0$ for $x<a$ and 1 for $x \geq a$. Hence the function is almost surely equal to $a$.

Answer 2

Here is an easy way of seeing that $c=+\infty$ almost surely, in the case that $f(t) = \sqrt t$.

Indeed, note that $B_t/\sqrt{t}$ has a $N(0,1)$ distribution for all $t$, hence $P(B_t/\sqrt t>M)=c>0$ for all $M>0$ and $t>0$.

By "reverse Fatou's lemma", it holds that $$P\bigg( \limsup_{t\to 0} \frac{B_t}{\sqrt{t}} >M\bigg) = E \big[ \limsup_{t \to 0} 1_{\{t^{-1/2}B_t >M\}}\big] \geq \limsup_{t \to 0} P\big( B_t/\sqrt{t}>M)=c>0. $$Now apply Blumenthal to conclude that the probability must be $1$.