I read in these notes, in the remark to corollary 4.3., that with probability one, the set of locally $1/2$ Hölder continuous times is of measure zero in $[0,1]$. I would like to use this fact for a problem I'm working on, but I do not know how to prove it. Does anyone have any references or ideas? I'm tried a proof by contradiction, but I'm not sure wherein the contradiction would lie. Obviously the Brownian motion can't be Hölder continuous of order $1/2$ on any interval, but all I've been able to get by supposing a contradiction is the existence of a closed, positive measure set on which it's Hölder continuous of order $1/2$. Is this impossible?
Thanks!
Consider the mapping
$$(t,\omega) \mapsto L(t,\omega) := \sup_{h \in (0,1)} \left| \frac{B(t+h,\omega)-B(t,\omega)}{\sqrt{h}} \right|.$$
Since the Brownian motion has continuous sample paths, we have
$$L(t,\omega) = \sup_{h \in (0,1) \cap \mathbb{Q}} \left| \frac{B(t+h,\omega)-B(t,\omega)}{\sqrt{h}} \right|,$$
and using that the mapping $(t,\omega) \mapsto B(t,\omega)$ is measurable, it follows that $(t,\omega) \mapsto L(t,\omega)$ is measurable. In particular,
$$(t,\omega) \mapsto 1_{\{|L_t(\omega)|<\infty\}}$$
is measurable. The law of the iterated logarithm shows that $\mathbb{P}(|L_t|<\infty)=0$ for each $t>0$, and therefore
$$\int_0^1 \mathbb{P}(L_t<\infty) \, dt = 0.$$
An application of Tonelli's theorem shows that
$$0=\int_0^1 \mathbb{P}(L_t<\infty) \, dt = \mathbb{E} \left( \int_0^1 1_{\{|L_t|<\infty\}} \, dt \right) = \mathbb{E}(\lambda(\{t \in [0,1]; |L(t,\cdot)|<\infty\})).$$
Hence,
$$\lambda(\{t \in [0,1]; |L(t,\cdot)|<\infty\})=0 \quad \text{almost surely}$$
Remark: The points $t$ for which $|L(t,\omega)|$ is finite are called slow points. They have been studied quite intensively in the 80s, e.g. by Kahane, Perkins et al. ; they obtained quite interesting results on the Hausdorff dimensions of slow points as well as possible values for the Hölder constant. You can find references to some of the papers e.g. in the monographs by Karatzas & Shreve and Mörters & Peres.