Let $B$ stand for a brownian motion on a finite interval $[0,1]$. If i am not wrong, i think that there exists a positive constant $c$, such that almost surely, for h small enough , for all $0< t < 1- h$
\begin{align} |B(t+h)-B(t)| < c\sqrt{h\log(1/h)} \end{align} or something like this. As a result
\begin{align} \bigg|\frac{B(t+h)-B(t)}{h}\bigg| < K(h) \end{align}
Am i correct ?