Laplace Transform of a Brownian motion

Question

If $v(\omega,t) : \Omega \times [0,\infty) \to \mathbb{R}$ is a Standard Brownian motion, then for what values of $s,\omega$ does the Laplace transform $l(\omega,s) = \int_0^\infty e^{-st} v(\omega,t) dt$ converge (where $s$ is a complex number).

Answer 1

The Law of Iterated Logarithm for the Brownian motion implies that for a.e. $\omega$ $$v(\omega,t)\le C \sqrt{t\log\log (e^e+t)}\qquad \forall t\ge 0$$ Therefore, for a.e. $\omega$ the integral $\int_0^\infty e^{-st}v(\omega,t)$ converges for all $s$ with $\operatorname{Re}s>0$.

On the other hand, the integral diverges almost surely when $\operatorname{Re}s=0$. Indeed, let $s=iy$. The growth and continuity properties imply that there is an infinite sequence of disjoint time intervals $T_i=[t_i,t_i+0.1y^{-1}]$ in which $v(\omega,t)$ is contained in some ball $B(a,r)$ with $|a|\ge 2 r$. The absolute value of the integral $\int_{T_i} e^{-ity} v(\omega,t)$ is bounded below by a positive constant.

I don't think there can be a useful description of the exceptional values of $\omega$ for which the integral fails to converge. The pointwise structure of $\Omega$ is usually not important in probability.