Let $t\mapsto X_t(\omega)$ be the sample path of a Brownian motion $X$. Certainly it is possible for the sample path to be Hölder continuous on a bounded interval $[0,K]$ but can it be Hölder continuous on all of $[0,\infty)$?


No, Brownian motion is a.s. not Holder continuous of any exponent on all of $[0,\infty)$.

For a short proof, we recall two facts about Brownian motion:

  1. On compact intervals, it is Holder-continuous of every exponent $\alpha<1/2$, but not $\alpha \geq 1/2$.

  2. As $t\to \infty$, the random variables $t^{-1/2}B_t$ converge in distribution to $N(0,1)$, since trivially, $t^{-1/2}B_t\sim N(0,1)$ for all $t$.

Now for contradiction, suppose that (with positive probability) Brownian motion is Holder-$\gamma$ on all $[0,\infty)$. Then by (1), we necessarily have that $\gamma<1/2$, which contradicts (2) since $t^{-1/2}B_t$ would then converge to $0$ with positive probability.

  • 2
    $\begingroup$ Moreover, the law of iterated logarithm says that $t^{-1/2} B_t$ grows like $\sqrt{2 \log\log t}$ instead of remaining bounded. $\endgroup$ – Nate Eldredge Jan 19 '18 at 23:41

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