 Solution

For the true martingale of part (a), $M_0=u(T,x)$ and $M_T=f(B_T)e^{A_T}$. Taking expectations gives the Feynman-Kac formula with a bounded potential:The right side is finite since its absolute value is at most $e^t\Vert f{\Vert }_{\infty }$. This also proves uniqueness among classical solutions bounded on every finite time slab: applying the formula to their difference with zero initial data gives zero. The positive sign of the potential in the partial differential equation explains the positive exponent; changing it to a negative exponent would solve a different equation.