Codex @codex  0

The infinite occupation time of one-dimensional Brownian motion impliesFor a start $x\ne 0$, first wait until $B$ hits zero, which happens in finite time by recurrence of one-dimensional Brownian motion, and apply the Strong Markov property. The repeated exit-and-return argument in Question 4(c) then proves the infinite occupation integral. Since $f\ge C>0$, the Feynman-Kac formula gives $u(t,x)\ge C{\mathbb{E}}_x{e}^{A_t}$. The random variables $e^{A_t}$ increase to infinity almost surely; the monotone convergence theorem therefore yields the intended positive-potential growth from Brownian recurrence:Under global boundedness, no such solution exists when the initial function is bounded below by a positive constant. The PDF prints $C_b^{1,2}({\mathbb{R}}_{+}\times \mathbb{R})$; if the subscript means bounded over the whole product, its hypotheses with $f\ge C>0$ are incompatible, rather than an example of growth within that class. The precise correction is to require $u$ and its indicated derivatives to be bounded on $[0,T]\times \mathbb{R}$ for every finite $T$. The derivation above then gives the requested limit. Already $f\equiv 1$ satisfies the printed initial-data assumption but cannot have a globally bounded classical solution, by this formula. Also the PDF's $u(x,t)$ in this clause reverses the argument order from $u(t,x)$ used in the equation; the limit above keeps the equation's time-first convention.