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



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


**EDIT:** Sorry, I corrected my mistake.



Standard Brownian motion, Hölder continuous with exponent $\gamma$ for any $\gamma < 1/2$, not for any $\gamma \ge 1/2$
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 $\sup\_{0 < t < 1} B\_t/\sqrt{t} = +\infty$.


