let $(\Omega, \mathcal{A}, P)$ a probability space, $B=\left(B_{t}\right)_{t \geq 0}$ a standard brownian motion, $\left(\mathcal{F}_{t}\right)_{t \geq 0}$ its natural filtration, $\mathcal{F}_{\infty}=\sigma\left(\mathcal{F}_{t}, t \geq 0\right)$ and $S_{t}=\sup _{0 \leq s \leq t} B_{s}$
let $a \in \mathbb{R}, x \leq a$ and $f:[a,+\infty[ \rightarrow \mathbb{R}$ be an $L^1([a,+\infty[)$ function.
show that for big values of $u$ we have : $$\mathbb{E}[f(\max(a, x + S_u))] \sim \sqrt{\frac{2}{\pi u}}\left((a-x) f(a)+\int_{a}^{\infty} f(y) d y\right)$$
so for one, I know that $S_t$ is equal to $|B_t|$ in distribution.
if I write the expectation as follows :
$$\mathbb{E}[f(\max(a, x + |B_u|))] = \sqrt{\frac{1}{2 \pi u}}\int_{-\infty}^{+\infty} f(\max(a, x + |y|)) \exp{(-\frac{y^2}{2u})} dy$$
since the integrand is even, then :
$$\mathbb{E}[f(\max(a, x + |B_u|))] = \sqrt{\frac{2}{\pi u}}\int_{0}^{+\infty} f(\max(a, x + y)) \exp{(-\frac{y^2}{2u})} dy$$
on the other hand we have :
$a \geq x + y \iff y \leq a -x $
so :
$$\begin{align} \mathbb{E}[f(\max(a, x + |B_u|))] &= \sqrt{\frac{2}{\pi u}}\int_{0}^{+\infty} f(\max(a, x + y)) \exp{(-\frac{y^2}{2u})} dy \\ &= \sqrt{\frac{2}{\pi u}} \left( f(a) \int_{0}^{a-x} \exp{(-\frac{y^2}{2u})} dy + \int_{a-x}^{\infty} f(x+y) \exp{(-\frac{y^2}{2u})} dy \right) \\ &= \sqrt{\frac{2}{\pi u}} \left( f(a) \int_{0}^{a-x} \exp{(-\frac{y^2}{2u})} dy + \int_{a}^{\infty} f(y) \exp{(-\frac{(y-x)^2}{2u})} dy \right) \end{align}$$
and so what I can conclude from my attempt is :
$$\forall u \geq 0, \,\,\, \mathbb{E}[f(\max(a, x + S_u))] \leq \sqrt{\frac{2}{\pi u}}\left((a-x) f(a)+\int_{a}^{\infty} f(y) d y\right)$$
is there way to reach the result in the title from this or should I take another route ?
EDIT : it's actually really simple, it's just an application of the dominated convergence theorem, and we use the fact that as $u \to +\infty$ we have that $\exp(-\frac{\alpha^2}{2u}) \to 1$