$F$ we define *the first hitting time* of $F$ as the first exit time from its complement $F^c$, that is $T_F = \tau_{F^c}$. If $F=\{b\}$ is a singleton, then slightly abusing the notation, we write $T_b = T_{\{b\}}$. For $\lambda>0$ we let $u^{\lambda}$ be *the $\lambda$-potential kernel*, that is the Laplace transform of $p(\cdot,x)$: $$u^{\lambda}(x) = \int_0^{\infty} e^{-\lambda t} p(t,x)\,dt.$$ If $u^{\lambda}$ exists for $\lambda=0$ then we set $u^0=u$ and then the process is transient (see [@Bertoin96 Theorem I.17]). In general, this needs not be the case (take for example the classical stable process with stability index $\alpha>1$, which is recurrent). Nevertheless, the condition $$\int_0^{\infty} \frac{d\xi}{1+{\operatorname{Re}}\psi(\xi)}<\infty,$$ together with [@Bretagnolle71 Theoreme 7 and 8], implies that $0$ is regular for itself, that is $${\mathbb{P}}^0 \big( T_0=0 \big)=1.$$ That in turn, combined with [@Kesten69 Theorem 1], yield that ${\mathbb{P}}^x (T_0<\infty)>0$ for any $x \in {\mathbb{R}}$.

Let us also set $$\kappa = \lim_{\lambda \to 0^+} \frac{1}{u^{\lambda}(0)} = \lim_{\lambda \to 0^+} \frac{1}{2\pi} \Bigg( \int_{{\mathbb{R}}} {\operatorname{Re}}\bigg[ \frac{1}{\lambda+\psi(\xi)} \bigg]\,d\xi \Bigg)^{-1}.$$ Clearly, $\kappa \in [0,\infty)$. Moreover, from [@Bertoin96 Theorem I.17] it follows that the process is transient if $\kappa>0$.

By [@Bertoin96 Corollary II.19 and Theorem II.19] we get that $x \mapsto {\mathbb{E}}^x e^{-\lambda T_0}$ is continuous with 