For $(M_t)$ a continuous local martingale, and $M^\text{*}_t := \sup_{0 \leq s \leq t}M_s$ the running maximum of the process, then I'm seeking to prove $$ \int_0^T (M^\text{*}_s - M_s) \, \text{d}M^\text{*}_s = 0 $$ Intuitively, this is since $(M^\text{*}_s - M_s) \, \text{d}M^\text{*}_s$ is zero: if we're currently at our maximum then the bracket term is zero, and otherwise $M^\text{*}_s$ is constant near $s$ so $\text{d}M^\text{*}_s$ is $0$.
In a bit more detail, we'll write $P_{t,s} = P_t - P_s$ for $P$ one of the above processes. The integral is equal to the limit in probability of a sequence of partitions $\pi_n$ of $[0,T]$ with mesh tending to 0: $$ \lim_{n \to \infty} \sum_{[s,t]\in \pi_n} (M^\text{*}_{t,s} - M_{t,s})M^\text{*}_{t,s} $$ and the terms in the sum are 'at the limit' all zero: either we don't reach a new maximum in $[s,t]$, in which case the right term $M^\text{*}_{t,s} = 0$, or we do reach a maximum, in which case "for small $t-s$", we're actively changing the maximum, so that $M^\text{*}_t = M_t$ and $M^\text{*}_s = M_s$ and so the left difference is $0$ (since $M^\text{*}_{t,s} = M_{t,s}$).
But this last argument depend on choosing a partition after that the outcome has been determined, which can (as with quadratic variation and 2-variation) change the result of the limit. Is the result true? How do you deal with this partition choice?
(I have since also asked this question on MO here)