Potential Local Martingale property derived from its quadratic variation

Question

Suppose we have a continuous local martingale $M$ such that $\langle M \rangle_t =o(t)$ - i.e. $$\lim_{t \rightarrow 0} \frac{\langle M \rangle_t}{t} = 0$$ Does this imply that $$\lim_{t \rightarrow 0} \frac{M_t}{t} = 0 $$ as well? i.e. is $M$ also $o(t)$ about $0$? I am asking as a potential strategy to solve this question

Answer 1

"$ M $ may not be $ o(t) $", the following is an example. Let $ B=\{B_t,t\ge 0\} $ be the Brownian motion with $\langle B\rangle_t =t $ and \begin{equation*} M_t=B_{t^2}, \qquad t\ge 0. \end{equation*} Then $ M=\{M_t, t\ge 0\} $ is a continuous local martingale with $\langle M\rangle_t =t^2 $, hence \begin{equation*} \lim_{t\to0}\frac{\langle M\rangle_t}{t}=0. \end{equation*} Meanwhile \begin{equation*} \frac{|M_t|}{t}=\frac{|B_{t^2}|}{t} =\frac{|B_{t^2}|}{\sqrt{2t^2\log\log(1/t^2)}} \sqrt{2\log\log(1/t^2)}. \end{equation*} Using the LIL of BM at $ t=0 $ we have \begin{equation*} \varlimsup_{t\to0}\frac{|M_t|}{t}=+\infty. \end{equation*}