# Standard Brownian motion, Hölder continuous with exponent $\gamma$ for any $\gamma < 1/2$, not for any $\gamma \ge 1/2$

In some results on Hölder continuity with regards to standard Brownian motion, the following is asserted without proof.

It is not hard to see that for every $k < \infty$, and every $\epsilon > 0$,$$\mathbb{P}\left\{ \sup_{0 < s < t < 1} {{|B_t - B_s|}\over{\sqrt{t - s}}} \ge k\right\} > 1 - \epsilon,$$where $B_t$ is a standard Brownian motion.

To me, this is not all that trivial. Could anyone explain why this is the case?

EDIT: Sorry, I corrected my mistake.

• What does "is positive" refer to? The display is an inequality, not a quantity. Oct 7, 2015 at 3:48
• In fact can you please proofread the display? Written as "for all $k$ and for all $\epsilon$" it doesn't seem to make sense. Oct 7, 2015 at 3:55

## 2 Answers

Perhaps this is what you're looking for?

Fix a sequence $t_n \downarrow 0$ and some $k > 0$ and let $A_n = \{ \sup_{0 < t < t_n} B_t/\sqrt{t} \ge k\}$. Since $B_{t_n}/\sqrt{t_n}$ has a standard normal distribution, we have $$\mathbb{P}(A_k) \ge \mathbb{P}(B_{t_n}/\sqrt{t_n} \ge k) = 1-\Phi(k) > 0$$ where $\Phi$ is the normal cdf.

Now $A_1 \supseteq A_2 \supseteq \cdots$ so if $A = \bigcap_n A_n$, then by "continuity from above" we have $$\mathbb{P}(A) = \lim_{n \to \infty} \mathbb{P}(A_n) \ge 1-\Phi(k) > 0.$$ But $A$ is in the "germ field" $\mathcal{F}_0^+$ so the Blumenthal 0-1 law says we must have $\mathbb{P}(A) = 1$. On the event $A$ we have $\sup_{0 < t < 1} B_t/\sqrt{t} \ge k$, and $k$ was arbitrary, so almost surely we have $\sup_{0 < t < 1} B_t/\sqrt{t} = +\infty$.

Also, this immediately follows from the local law of the iterated logarithm for the Brownian motion; see e.g. Corollary 5.3 on page 123.

• Even though I agree with this answer and upvoted it, one should always provide full proof in a self contained manner on MathOverflow :) Aug 7, 2021 at 0:14