Show that $X_t=|B_t|$ solves the martingale problem for the operator $\mathcal{L}f=\frac12 f''$ with domain $$\mathcal{A}=\{f\in C_b^2([0,\infty)):f'(0)=0\}.$$ Hint: Functions in $\mathcal{A}$ can be extended to symmetric functions.
First of all I don't see how the hint helps as we consider non-negative time…
Call the filtration $F_t$. I need to show $$ M_t^{f}=f(X_t)-\int_0^t \mathcal{L}_sf(X_s) ds $$ is an $(F_t)$ martingale, i.e $$E(M_t^f|F_s)=M_s.$$
I think adaptedness and integrability is clear. Since $f\in C_b^2([0,\infty))$ and $X_t$ is a Markov process, $M_t$ is $F_t$ adapted. It is also in $L^1$ since $X_t$ is.
I want to prove $$E(f(X_t)-\frac12 \int_0^t f''(X_r) |\mathcal{F_s})=f(X_s)-\frac12 \int_0^s f''(X_r)$$
Using a continuous function of $X_s$ is $\mathcal{F_s}$-measurable, I write LHS as $$E(f((X_t-X_s)+X_s)-1/2\int_s^t f''(X_r) dr|F_s)-1/2\int_0^s f''(X_s). $$ My problem is how to write $f(X_t)$ as $f(X_s)$ and something that vanishes when you condition on it. I also have an integral from $s$ to $t$ which needs to vanish. Maybe you can use independent increments of the Brownian motion somehow? And can use the fact $f'(0)=0$ somehow? Since $X_t$ is a Markov process I have in my notes $$E(f((X_t)_{t\geq s})|\mathcal{F_s})=E_{X_s}(f)$$ but I am not sure how to work with that.
Any ideas how to proceed?
I think the Hint is driving at this: If $f\in \mathcal A$, define $g\in C^2_b(\Bbb R)$ by $g(x) :=f(|x|)$. Then $g(B_t)-(1/2)\int_0^t g''(B_s)\,ds$ is a martingale. (I'm guessing you've already seen the martingale problem for standard Brownian motion.) Of course you need to check that $g$ really is twice continuously differentiable, and then see how $g''$ is related to $f''$.