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The Itô formula for $f(t,x)=e^{x+at}$ gives the stochastic differential equationWrite $N_t={\int }_0^t{Z}_{u}d{B}_u$ for its continuous local martingale part. Its clock is the quadratic variationBecause $Z_t>0$, $A$ is strictly increasing. For $s<A_{\infty }$, its inverse ${\tau }_s$ satisfies $A_{{\tau }_s}=s$ and $d{\tau }_s/ds=Z_{{\tau }_s}^{-2}$. The Dambis-Dubins-Schwarz theorem, or the Lévy characterization of Brownian motion after the optional time-change theorem, makes ${\beta }_s=N_{{\tau }_s}$ a Brownian motion for the time-changed filtration. With $Y_s=Z_{{\tau }_s}$,This is the Bessel process equation with dimension $\boxed{2+2a}$, on the positive stochastic interval. This establishes the Exponential Brownian-to-Bessel time change.

The lifetime convention is important: the inverse is defined only for $s<A_{\infty }$. If $a<0$, the strong law for Brownian motion gives $B_t/t\to 0$, so $Z_t\to 0$ exponentially and $A_{\infty }<\infty$. Then $Y_s\to 0$ as $s\uparrow A_{\infty }$, and this is its first hit of zero after continuous extension. If $a>0$ the clock diverges by the same strong law. If $a=0$ it diverges by infinite occupation time of one-dimensional Brownian motion on a bounded interval, since $e^{2B_t}$ is bounded below there. Thus the lifetime is infinite when $a\ge 0$. Negative dimensions are interpreted by the same positive-state equation killed at zero; no post-zero Bessel process behavior is asserted.

Under the corrected orthogonality hypothesis, the preceding quadratic variation calculation gives $[X{]}_t=A_t={\int }_0^t{e}^{2B_s}ds$. We justify the needed infinite occupation time of one-dimensional Brownian motion, rather than inferring an infinite integral from recurrent visits alone.

Let $T_0=0$, and recursively let $S_n$ be the first exit of $B$ from $(-1,1)$ after $T_n$, and $T_{n+1}$ its first return to zero after $S_n$. Recurrence of one-dimensional Brownian motion makes these stopping times finite almost surely. The Strong Markov property shows that $D_n=S_n-T_n$ are independent and identically distributed random variables, each strictly positive. Choose $\epsilon >0$ with $\mathbb{P}(D_n\ge \epsilon )>0$. The Borel-Cantelli lemma gives infinitely many such durations and hence ${\sum }_n{D}_n=\infty$ almost surely. These intervals are disjoint, and $|B_s|\le 1$ on each, soThis proves the intended conclusion. Without the orthogonality correction, the printed setup gives instead the more general quadratic variation in the previous part; it does not justify identifying that quantity with $A$. The proof here is explicitly for the corrected setup.

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.

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.