If we are free to break ties in any way we want, doesn't a regular (2N+1)-gon work? each vertex has two farthest points that are also vertices, and the trajectory that always chooses the one on the right will not repeat itself until it visits all vertices.

just to be sure, you mean 2D Browinan bridges both parametrized by $[0,1]$, right?

@MarcBerth, I want a differential inequality for a real-valued function $|f(\gamma(t))|$, so I'm using that $\partial_t|f(\gamma(t))|\leq |\partial_t f(\gamma(t))|$

@MarcBerth, it was a misprint. The argument bounds $|z|$ for a given $r$ from above, i.e., $r$ for a given $|z|$ from below.

@painday, I don't think this has anything to do with discontinuity though - a process with a drift non-symmetric about the origin has no particular reason to have zero equilibrium expectation.

@painday, I don't think there's any reason to expect it to be zero. Intuitively, the process will spend less time in a quadrant where drift towards the origin is stronger.

But then, one shouldn't expect the limit in the left-hand side to depend on $\mathbf{X}$ at all, since the process converges to its stationary distribution at large $t$.

What is $\mathbb{E}^{\mathbf{X}}$, and what is $\mathbf{X}$ in your last equation?

My comment was a bit hasty, so I wrote up an answer.

What do you mean by "inverse in some sense"? The map $g_\sharp$ is in general neither injective nor surjective.