I had a conversation today as to whether or not $$M_t = \max_{0 \le s \le t}(B_s)$$ is an Itô process? My intuition is that it is not an Itô process since it needs to remember an arbitrarily ancient event from the past, like the max value, so it doesn't have the Markov property, but it may not be as simple as that.
The idea is that if it is an Itô process then it can be written as some SDE in general form.
On
I agree that it may not be that simple. It doesn't need to "remember" the max value from the past because the max value is given by $M_t$. If we let $W_t$ be a standard Brownian motion, then $B_t := \int_0^t \text{sgn}(W_s)dW_s$ is as well by Levy's characterization. From Tanaka's formula, $|W|_t = B_t + L_t$ where $L_t$ is the local time of $W$ at $0$. The local time is absolutely continuous with respect to Lebesgue measure and is progressively measurable, so $L_t = \int_0^t K_s ds$ for some progressively measurable process $K$ by the Radon-Nikodym theorem. Now thanks to a lemma from Skorohod, we know $L_t = \sup_{s \le t}(-B_s)$ (see section 3.5 in this paper by Bjork). If we define $\hat B_t := -B_t$ then $\hat B_t$ is a Brownian motion, and if $M_t := \sup_{s \le t} \hat B_t$ then we have $M_t = L_t$ and is therefore an Ito process.
This shows an example where $M_t$ is defined to be the running max of a specific Brownian motion and is an Ito process. I'm not certain if that is the case whenever $M_t$ is defined to be the running max of any Brownian motion (here the filtration is larger than the one generated by the Brownian motion we are looking at the running max of), but it certainly shows it is not quite that simple.
The running maximum of a Brownian motion $(B_t)$ has the following martingale representation: $$ M_T=\sqrt{\frac{2T}{\pi}}+\int_0^T 2\bar{\Phi}\!\left(\frac{M_t-B_t}{\sqrt{T-t}}\right)\,dB_t. $$ (See, e.g., Example 41.13 on page 92 here.)