We have a geometric Brownian motion
$$dS(u) = \mu S(u) dt +\sigma S(u) dZ(u), \quad S(0)=1,$$
where $Z$ is a standard Brownian motion. There are two nondecreasing continuous processes $L$ and $U$ with $L(0)=U(0)=0$. The process $L$ grows only when $B(t)/A(t) = \gamma{}S(t)$ and the process $U$ grows only when $B(t)/A(t) = S(t)/\gamma{}$, i.e.
$$L(t) = \int_0^t \unicode{x1D7D9}_{\{B(u)/A(u)=\gamma{}S(u) \}}dL(u), \quad U(t) = \int_0^t \unicode{x1D7D9}_{\{B(u)/A(u)=S(u)/\gamma{} \}}dU(u), \quad t \geq 0.$$
We have two finite-variation continuous processes $A$ and $B$ with $A(0)=B(0)=1$ given for $t\geq 0$ by
$$A(t) = 1 - \gamma{}\int_0^tA(u)dL(u) + \int_0^tA(u)dU(u),\\ B(t) = 1 + \int_0^tB(u)dL(u) - \gamma{}\int_0^tB(u)dU(u).$$
In addition, we have that $\gamma{}S(t) \leq \frac{B(t)}{A(t)} \leq \frac{1}{\gamma{}}S(t)$ for all $t \geq 0$. We believe that for each fixed $\gamma \in (0,1)$, the processes $L, U, A$ and $B$ exist and are uniquely determined by these conditions.
We want to find find the long-time limit for the process $A(t)*S(t)+B(t)$, i.e.
$$\mathbb{E}\left[\lim_{T \to \infty}\frac{1}{T} \log(A(T)S(T)+B(T))\right]$$