# Almost sure non differentiability of Brownian Motion

Problem: Let $t>0$, show that the standard Brownian motion is almost surely not differentiable a $t$

Now, through a Borel Cantelli argument I proved that, almost surely

$$\limsup_{\epsilon \rightarrow 0^+} \frac{|B_{t+\epsilon}-B_t|}{\epsilon} = \infty$$

Isn't this enough to prove non differentiability?

I ask this because I am given two more hints:

-Recall that by Blumenthal's 0-1 law, $\forall \epsilon > 0$ $B_{t+u}-B_{t}$ almost surely attains both a negative and a positive value for $u\in [t,t+\epsilon]$.

-Conclude considering behaviours of $\limsup$ and $\liminf$ of $|B_{t+\epsilon}-B_t|/\epsilon$ as $\epsilon \rightarrow 0$

I also think that there shouldn't be a modulus sign in the last hint, it doesn't make much sense to me otherwise, what do you think?

Thanks.

• – Seyhmus Güngören Jan 28 '14 at 12:09
• @SeyhmusGüngören Thanks, If I understand correctly it is about the first part of Did's answer: proving that the limsup of the modulus is +infinity implies that either limsup or liminf of the same thing without modulus is infinite. Since these events have the same probability by symmetry and it is either 0 or 1 by Blumenthal's 0-1 law, they both must occur almost surely. – Moritzplatz Jan 28 '14 at 12:17
• @SeyhmusGüngören But then the last hint doesn't really make sense, it should have been without modulus, or am I mistaken once more? – Moritzplatz Jan 28 '14 at 12:18
• According to what I read from Did, no modulus is required. But it is when $k$ is constant and only when one single $n$ goes to infinity. In your case you have two arguments which are both in $n$. For such Did also uses the modulus. I am not a expert here. – Seyhmus Güngören Jan 28 '14 at 12:48