The limit supremum of a function involving Brownian motion

Question

I would like, for some $\delta>0$ and a Brownian motion $B$, to calculate

$\displaystyle\limsup_{t\to\infty}\left(\exp\left( (1+\delta)t\right)\cdot\exp\left(-B_t-\frac{t}{2}\right)\right)$

which, upon calculation, is equal to

$\displaystyle\limsup_{t\to\infty}\left(\exp\left(t\left(\frac12+\delta\right)\right)\exp\left(-B_t\right)\right)$.

I reckon this is $\infty$, simply by looking at the function and noting that the least upper bound for the Brownian motion must be infinite. This doesn't seem to be a satisfactory argument. I tried arguing via the law of the iterated logarithm and similar results but very quickly got stuck. If I am right in saying that the value is infinite, surely there must be a short argument which doesn't fall into the category of "proof by handwaving"?

EDIT: Just a small note: I have now turned to MATLAB for help, and the simulation I ran all but confirmed that the limit superior is in fact infinite.

Answer 1

The Brownian motion is recurrent, in particular the random set of times $T=\{t\geqslant0\mid B_t\leqslant0\}$ is almost surely unbounded. If $t$ is in $T$ the quantity of interest is $\geqslant\exp\left((\delta+1/2)t\right)$ hence its limsup is $+\infty$, almost surely.

The law of the iterated logarithm also yields the result immediately.