= Solution
The <infinite occupation time of one-dimensional Brownian motion> implies
$$
A_\infty=\int_0^\infty\frac{ds}{1+B_s^2}\geq\frac12\int_0^\infty\mathbf1_{\{|B_s|\leq1\}}ds=\infty\quad\text{almost surely}.
$$
For a start $x\ne0$, 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 $f\geq C>0$, the <Feynman-Kac formula> gives $u(t,x)\geq C\mathbb E_x e^{A_t}$. 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>:
$$
\boxed{\lim_{t\to\infty}u(t,x)=+\infty\quad\text{for every fixed }x\in\mathbb R.}
$$
\b[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 $f\geq C>0$ 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 $f\equiv1$ 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.
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