Let $B(t)$ be a Brownian Motion. My goal is to prove that $B(t)$ is not Hölder continuous with exponent $\gamma$ at any point in $[0,1]$ if $\gamma>1/2$. The solutions I found is as follows:
Fix $k\in \mathbb N$, $\gamma > 1/2 + 1/k$ and $C<\infty$. For $n\geq k$ , let $$ A_n = \{\omega : \text{there is an }s \in [0, 1] \text{ so > that }|B(t) - B(s)| \leq C|t - s|^\gamma \text{ when }|t - s| \leq k/n\}. $$ Thus $\cup_{C=1}^\infty \cup_{n=1}^\infty A_n(C)$ is the set that there exists $s\in [0,1]$ such that $B$ is locally $\gamma$-H\"older continuous at $s$. For $1 \leq i \leq n - k + 1$, let
$$ Y_{i,n}=\max\left\{\left|B\left(\frac {i+j}n\right)- B\left(\frac {i+j-1}{n} \right) \right|:j=0,1,\dots,k-1 \right\}, $$
This is the maximum increment among $k$ consecutive intervals, as $i$ travels from left to right. Also let $$ B_n:=\left\{\text{at least one }Y_{i,n} \leq \frac {C (2k-1)}{n^\gamma}\right\} $$ which is a possible interval of small increment.
Triangle inequality gives $A_n\subseteq B_n$.
The triangle inequality is where I don't understand. Where does the number $2k-1$ come from? Thanks!