= Solution
Set $V(x)=(1+x^2)^{-1}$ and $A_t=\int_0^tV(B_s)ds$. On a fixed horizon $T$, the <Itô formula> and <Itô product rule> give
$$
d\bigl(e^{A_t}u(T-t,B_t)\bigr)=e^{A_t}\bigl(-u_t+\tfrac12u_{xx}+Vu\bigr)(T-t,B_t)dt+e^{A_t}u_x(T-t,B_t)dB_t.
$$
The drift vanishes by the <partial differential equation>. This makes $M$ a <continuous local martingale>. Since $0\leq A_t\leq T$,
$$
\boxed{|M_t|\leq e^T\sup_{0\leq s\leq T,\,y\in\mathbb R}|u(s,y)|\quad(0\leq t\leq T).}
$$
The right side is a finite deterministic bound under the stated boundedness assumption, so the <dominated convergence theorem> removes a <localizing sequence> and $M$ is a true <martingale>. If the derivatives are also bounded, the <Itô isometry> alternatively proves the stochastic integral is <square-integrable> because $\mathbb E\int_0^Te^{2A_s}|u_x(T-s,B_s)|^2ds\leq Te^{2T}\|u_x\|_\infty^2$.
For the <bounded parabolic differentiability class>, distinguish global bounds from bounds on each finite time slab. Only the latter are needed here; a single global bound on the <classical solution> $u$ would conflict with its requested long-time growth in part (c).
Back to article page
