Local times and quadratic variation Suppose we have a continuous local martingale $(M_t)_{t \geq 0}$, where $M_0 = 0$. Let $L_{t}^{0}$ denote the local time of this process at zero. I would like to prove the simple statement that:
\begin{equation}
\inf \{t : L_{t}^{0} > 0 \} = \inf \{t : \langle M \rangle_t > 0 \},
\end{equation}
where the bracket denotes the quadratic variation process.
My initial idea was to use the occupation density formula, which relates local times and the quadratic variation through the integrals involving non-negative Borel measurable functions. However, the formula integrates against the superscript of $L$ rather than the subscript (that is, one measures the change with respect to the position of the process, not with respect to the time for a fixed position). Therefore, I do not hope that this could be useful.
I am not aware of any other (useful) results connecting the quadratic variation and the local times, so I would appreciate any help. I appreciate the fact that (heuristically) the local time measures the time a process spends at a particular point, with the clock given by the quadratic variation rather than the time indexing the process, but I can't quite get  a rigorous proof of this.
 A: By the theorem of Dubins & Schwartz, there is a Brownian motion $B$ such that $M_t=B_{\langle M\rangle_t}$ for all $t\ge 0$, a.s.
The  local times of $B$ and $M$ are related in the same way: $L^x(M)_t = L^x(B)_{\langle M\rangle_t}$.
In particular, taking $x=0$ and defining $A(u):=\inf\{t:\langle M\rangle_t>u\}$, you have
$$
\inf\{t:L^0(M)_t>0\}=A(\inf\{t:L^0_t(B)>0\}).
$$
But as $M_0=0$, you have $B_0=0$ as well, and so $\inf\{t:L^0_t(B)>0\}=0$ a.s. Therefore
$$
\inf\{t:L^0(M)_t>0\}=A(0)=\inf\{t:\langle M\rangle_t>0\},
$$
a.s.
A: I'd just like to add an alternative solution, via the occupation and Tanaka formulas, which I found to be the most natural tools to use in the first place. We define:
$$
T_1 = \inf \left\{ t \geq 0 : L_{t}^{0} > 0  \right\}, \\
T_2 = \inf \left\{ t \geq 0 : \langle M \rangle_{t} > 0  \right\}
$$
to be the stopping times in question.

*

*To see that $T_1 \geq T_2$, we recall that by taking an appropriate modification of $L_{t}^{0}$, we have via the occupation formula:
$$
L_{T_2}^{0} =  \lim_{\epsilon \to 0} \frac{1}{2 \epsilon} \int_{0}^{T_2} \mathbb{1}_{\{- \epsilon < M_s < \epsilon\}} d \langle M \rangle_s \leq \lim_{\epsilon \to 0} \frac{1}{2 \epsilon} \langle M \rangle _{T_2} = 0,
$$
by continuity of $M$. Hence, the local time can start changing only after the quadratic variation.

*To see that $T_2 \geq T_1$, by Tanaka's formula:
$$
|M_{t \wedge T_1}| = \int_{0}^{t \wedge T_1} sgn(M_s)dM_s,
$$
is a local martingale. By localizing, we get that $\mathbb{E} [|M_{T_1 \wedge \tau_n}|]$ for an appropriate sequence of stopping times $(\tau_n)_{n \in \mathbb{N}}$ increasing to infinity. Via Fatou's lemma and and continuity, this tells us that $M_t \equiv 0$ on $[0, T_1]$, and so certainly $\langle M \rangle_{T_1} = 0$, which means $T_2 \geq T_1.$
