How can we find all functions g such that $g(B_t)$ is a martingale? (assume g(x) is twice continuously differentiable). Let $X_t$ = $g(B_t)$, by using the Ito formula, I get
$dX_t$ = $f_{B}'$($B_t$)$dB_t$ + $\frac{1}{2}$$f_{BB}''$($B_t$)dt, then by integrating both sides, I obtain X$_t$ = g(t) + $\frac{1}{2}$$g_{B}'$(B$_t$) $-$ $\frac{1}{2}$$g_{B}'$(0). It requires that $\textbf{E}$[X$_t$|$\mathcal{F}_s$] = X$_s$ for t $\geq$ s.
It follows that we need $\textbf{E}$[g(t) - g(s) + $\frac{1}{2}$[$g_B'$(B$_t$) - $g_B'$(B$_s$)] | $\mathcal{F}_s$] = 0.
Is it correct? Not sure how to continue to solve it. Any suggestion?
Your integration of both sides doesn't make any sense to me. You should get $$X_t = g(0) + \int_0^t g'(B_s)\,dB_s + \frac{1}{2} \int_0^t g''(B_s)\,ds$$ and then observe that a martingale has zero bounded variation part, so you will get a martingale only if the last term on the right side vanishes. You also need to ensure that the $dB_s$ term is a martingale, and not just a local martingale. That should give you an idea of an appropriate condition on $g$...