Feynman-Kac formula with a bounded potential ($u_t=\frac{1}{2}{u}_{xx}+Vu$)

For bounded $V$ and bounded initial data $f$, a sufficiently regular classical solution bounded on every finite time slab satisfiesTo prove this, the Itô formula and Itô product rule make $u(T-t,B_t)e^{{\int }_0^{t}V(B_s)ds}$ a local martingale. Its deterministic bound on $[0,T]$ makes it a true martingale; taking expectations at the two endpoints gives the formula. This also gives uniqueness in the finite-horizon bounded class. The exponent has the same sign as the potential in the equation.

Let $\varphi$ be continuous, nonnegative, zero on $[0,\infty )$, and strictly positive on $(-\infty ,0)$. By continuity of Brownian paths, its occupation integral vanishes exactly on nonnegative paths. Dominated convergence gives the displayed limit. The Brownian reflection principle makes it $1-2{\mathbb{P}}_x(B_t\le 0)$ for $x\ge 0$ and zero for $x<0$. In particular the limit at $x=0$ is zero for every positive time. The choice $\varphi (x)=\min (1,\max (0,-x))$ is a bounded soft killing potential.

For globally Lipschitz drift $b$, couple $d{X}^{\epsilon }=b(X^{\epsilon })dt+\epsilon dW$ and $\dot{x}=b(x)$ from the same point. The Gronwall inequality gives ${\sup }_{s\le t}|X_s^{\epsilon }-x_s|\le \epsilon e^{Lt}{\sup }_{s\le t}|W_s|$. Thus convergence is uniform on each finite horizon. The Feynman-Kac formula for generator ${\epsilon }^2\Delta /2+b\cdot \nabla +c$ has weight $\exp \int c$, with a positive sign. Bounded continuous $c$ is uniformly continuous on a compact neighbourhood of the deterministic trajectory, and bounded continuous $f$ supplies a deterministic integrable bound. The dominated convergence theorem proves the displayed limit without requiring global uniform continuity of $c$.

Suppose bounded $V\ge 0$ is bounded below by a positive constant on some nonempty open interval, and $f\ge C>0$. The Feynman-Kac formula with a bounded potential gives $u(t,x)\ge C{\mathbb{E}}_x{e}^{{\int }_0^{t}V(B_s)ds}$. Infinite occupation time of one-dimensional Brownian motion makes that integral tend to infinity almost surely. The monotone convergence theorem then gives $u(t,x)\to \infty$ for every fixed $x$. Thus no globally bounded classical solution exists under these hypotheses, although boundedness on every finite time slab is compatible with the conclusion.