For each $T > 0$, let $B^T$ be a Brownian bridge on $[0, T]$, conditioned to start and end at $0$.


**Question:** Is it true that $\mathbb E[|\text{exp}\, (\sup\_{0 \leq t \leq T} B^T\_t) - 1|] \to 0$ as $T \to 0^+$?



Exponential of supremum of Brownian bridge on short time frame
Without loss of generality, $B\_t^T=B\_t-\frac tT\,B\_T$, where $B\_\cdot$ is a standard Brownian motion. So,
$$0\le\sup\_{t\in[0,T]}B\_t^T\le M\_T+|B\_T|,$$
where $M\_T:=\sup\_{t\in[0,T]}B\_t$. So, in view of the Cauchy-Scwarz inequality,
$$E|\exp\sup\_{t\in[0,T]}B\_t^T-1| 
=E\exp\sup\_{t\in[0,T]}B\_t^T-1 \\ 
\le\sqrt{E\exp(2M\_T)} \sqrt{E\exp(2|B\_T|)}-1.$$
By the  [reflection principle](https://en.wikipedia.org/wiki/Reflection_principle_(Wiener_process)#Statement), $M\_T$ equals $|B\_T|$ in distribution. So,
$$E|\exp\sup\_{t\in[0,T]}B\_t^T-1| 
\le E\exp(2|B\_T|)-1=E\exp(2\sqrt T\,|B\_1|)-1\to0$$
as $T\downarrow0$, by dominated convergence. $\quad\Box$


