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?


  • 1
    $\begingroup$ see this math.stackexchange.com/questions/592727/… $\endgroup$ – Seyhmus Güngören Jan 28 '14 at 12:09
  • $\begingroup$ @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. $\endgroup$ – Moritzplatz Jan 28 '14 at 12:17
  • $\begingroup$ @SeyhmusGüngören But then the last hint doesn't really make sense, it should have been without modulus, or am I mistaken once more? $\endgroup$ – Moritzplatz Jan 28 '14 at 12:18
  • $\begingroup$ 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. $\endgroup$ – Seyhmus Güngören Jan 28 '14 at 12:48

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.