Let $M$ be an almost surely continuous martingale that is not almost surely constant in time - that is, it is not the case that almost surely, $M\_t = M\_0$ for all $t$.


Assume further that $M$ is a time homogeneous Markov process, and that it is transitive, in the sense that for any measurable subset $U$ of $\mathbb R$ with nonzero Lebesgue measure, we have


$$\mathbb E\big[\int\_0^\infty \mathbf 1\_U (M\_t) \, dt \big] > 0.$$


Suppose $f: \mathbb R \to \mathbb R$ is a continuous function such that $f(M\_t)$ is a martingale.


**Question:** Does if follow that $f$ is necessarily an affine linear function? That is, $f(x) = a + bx$ for some $a, b \in \mathbb R$.



If a continuous function of a Markov martingale is a martingale, does the function have to be affine linear?
If you allow for an arbitrary starting point, then just use the optional stopping theorem for $f(M\_t)$: $$\begin{aligned} f(x) & = \mathbb E^x f(M(\tau\_{(a,b)})) \\ & = \mathbb P^x(M(\tau\_{(a,b)}) = b) f(b) + P^x(M(\tau\_{(a,b)}) = a) f(a) \\ & = \frac{x-a}{b-a} f(b) + \frac{b-x}{b-a} f(a) , \end{aligned}$$
as desired. Here $\tau\_{(a,b)}$ is the first hitting time of $a$ or $b$; it is finite almost surely because $M$ is transitive.


If the starting point (or distribution) is fixed, follow the same argument after the first hitting time of any given $x$ (the latter is almost surely finite because the process is transitive) and use the strong Markov property.


Note: one-dimensional diffusions have been studied by E.B. Dynkin and others, and an essentially complete description is available in terms of the speed measure and the scale function. I think the above argument belongs to this theory (although it is very simple, of course).


