On the Brownian first-passage time over a one-sided stochastic boundary

Let $B=(B_t)_{t \ge 0}$ be standard Brownian motion started at $0$ under $P$, let $S_t=\max_{ 0 \l r \l t} B_r$ be the maximum process associated with $B$, and let $g\colon\mathbf{R}_+\to\mathbf{R}$ be a (strictly) monotone continuous function satisfying $g(s) < s$ for all $s \ge 0 $. Let $ \tau $ be the first-passage time of $B$ over $t \mapsto g(S_t)$:
$$ \tau=\inf\{t>0\mid B_t\le g(S_t)\}. $$
Let $G$ be the function defined by
$$ G(y)=\exp(-\int_0^{g^{-1}(y)}\frac{ds}{s-g(s)}) $$
for $y \in \bf R$ in the range of $g$. Then, if $g$ is increasing, we have
$$ \lim_{t\to\infty}\sqrt{t}\mathsf{P}\{\tau\ge t\}=\sqrt{\frac2{\pi}}(-g(0)-\int_{g(0)}^{g(\infty)}G(y) dy), $$
and this number is finite. Similarly, if $g$ is decreasing, we have
$$ \lim_{t\to\infty}\sqrt{t}\mathsf{P}\{\tau\ge t\}=\sqrt{\frac2{\pi}}(-g(0)+\int_{g(\infty)}^{g(0)}G(y) dy\} $$
and this number may be infinite. These results may be viewed as a stochastic boundary extension of some known results on the first-passage time over deterministic boundaries. The method of proof relies on the classical Tauberian theorem and certain extensions of the Novikov-Kazamaki criteria for exponential martingales.