Doob-Meyer Decomposition of the range of a standard Brownian Motion I am curious about whether there exists a Doob-Meyer decomposition for the range process $R_t$, or the squared range $R_t^2$ of a standard Brownian motion $B_t$, defined as:
$$ R_t = M_t-m_t, $$
where $ M_t := \sup_{0\leq s\leq t} B_s$  and $m_t := \inf_{0\leq s\leq t} B_s$. Clearly, $R_t$ and hence $R_t^2$ is monotonically increasing and continuous in $t$, thus they are by design submartingales which should possess unique Doob-Meyer decompositions.
I know that by the Tanaka's equation, we have:
$$ |B_t| = \int_0^t \mathrm{sgn}(B_s)dB_s + L_t,$$
where $L_t$ is the Brownian local time at zero. Consequently, this provides the Doob-Meyer decomposition for $|B_t|$. Also, for $B_t^2$ this is even simpler:
$$B_t^2 = 2\int_0^t B_s dB_s + t,$$
directly from Ito's lemma. Therefore, I was thinking whether these results can be easily extended to describe $R_t$ or $R_t^2$ in light of the well-known relation that $M_t\overset{d}{=}  |B_t| \overset{d}{=}  M_t-B_t \overset{d}{=}-m_t \overset{d}{=} -B_t-m_t$? I spent hours trying to find relevant results in the literature but could not find anything relevant. Any suggestions or hints are highly appreciated.

Since $m_t=-\max_{0\le s\le t}(-B_s)$ ($-m_t$ is a disguised max) it is enough to find the Dooob-Meyer decomposition of $M_t=\max_{0\le s\le t}B_s\,.$ But that's trivial. Since $M_t$ is inreasing its decomposition is $M_t=0+A_t$ where the increasing process $A_t$ is $M_t$ itself and the martingale part is zero.
