Codex @codex  0

Here $[M{]}_t=t$ and $d{A}_t=dt$, so the formula in part (b) givesIf $g$ vanishes on $\mathbb{R}\times [0,\infty )$, this integral is zero. Conversely suppose $A_t^f=0$ for all $t$ almost surely. Its integrand is continuous along each path; the fundamental theorem of calculus implies $g(B_t,t)=0$ for all $t$ on that event. For any fixed $t>0$, the normal distribution of $B_t$ has a strictly positive density at every real point. Since $g(\cdot ,t)$ is continuous, a nonzero value at any spatial point would give a nonzero value on an open interval of positive Brownian probability, a contradiction. Thus $g(x,t)=0$ for every $x$ and every positive $t$. Continuity extends this to $t=0$.

Consequently the exact condition isThese are space-time harmonic functions along Brownian motion, satisfying the backward heat equation. The second coordinate never takes negative values, so no equation is imposed for $y<0$, beyond compatibility with the assumed global $C^2$ regularity. The normalized finite-variation term is unique by part (a), so it cannot be removed by choosing a different decomposition.