Interpretation of the resolvent of Brownian motion

$(P_t)_{t\geq 0}$ denotes the semigroup of Brownian motion, i.e. $$P_tu(x) = E^x[u(B_t)]$$ where $u$ is a bounded and measurable function. The following operator is called the $\alpha$-potential. $$U_{\alpha}u(x) = \int_0^{\infty}e^{-\alpha t}P_tu(x)\,dt$$ Here $\alpha > 0$. What I am trying to understand is what the $\alpha$-potential of the indicator function $\mathbb{1}_{C}(x)$ means in a probabilistic sense. Here $C$ is some Borel measurable set. By the definition above, \begin{align} U_{\alpha}\mathbb{1}_{C}(x) &= \int_0^{\infty}e^{-\alpha t}P_tu(x)\,dt \\ &= \int_0^{\infty}e^{-\alpha t} E^x[\mathbb{1}_{C}(B_t)]\,dt \\ &= E^x\left[\int_0^{\infty}e^{-\alpha t} \mathbb{1}_{C}(B_t)\,dt \right] \end{align} By the last expression, I first thought $U_{\alpha}\mathbb{1}_{C}(x)$ is the Laplace transform (i.e. moment generating function) of the time Brownian motion spends in $C$. But that is not true because the time BM spends in $C$ is $$\int_0^{\infty}\mathbb{1}_{C}(B_t)\,dt$$ and the Laplace transform of this random variable would be $$E^x\left[e^{-\alpha \int_0^{\infty} \mathbb{1}_{C}(B_t)\,dt} \right]$$ Defining the hitting time $$\tau = \inf\{t \geq 0: B_t \in C\}$$ I can write $\{B_t \in C\} \subset \{\tau \leq t\}$. But that does not lead to some insightful interpretation of the $\alpha$-potential.

A related question is what $\lim_{\alpha \to 0}U_{\alpha}\mathbb{1}_{C}$ is. I guess by monotone convergence the limit is the average time BM spends in $C$. This interpretation is correct, right?

• $\alpha \mapsto U_{\alpha} 1_C(x)$ is the Laplace transform of $t \mapsto \mathbb{P}^x(B_t \in C)$ ... I don't think that there is more about it. And yes, you are right about the limit $\alpha \to 0$ (you can apply monotone convergence theorem). – saz Apr 14 '18 at 9:24
• @saz Thank you . – Calculon Apr 14 '18 at 9:29