respect to $x$, and for any $x \in {\mathbb{R}}$ we have $$\label{eq:1}
u^{\lambda}(x) = u^{\lambda}(0){\mathbb{E}}^{-x} e^{-\lambda T_0}.$$ Note that for recurrent processes we have $u(x)=\infty$ for all $x \in {\mathbb{R}}$. Instead one can define *the compensated $\lambda$-potential kernel* by setting for $\lambda >0$ $$K^{\lambda}(x) = u^{\lambda}(0)-u^{\lambda}(x), \quad x \in {\mathbb{R}}.$$ In view of we get that $K^{\lambda} {\geqslant}0$ for all $\lambda {\geqslant}0$.

The next natural move would be to pass with $\lambda$ to $0$ and define *the compensated potential kernel* $$K(x) = \lim_{\lambda \to 0^+} K^{\lambda}(x), \quad x \in {\mathbb{R}}.$$ In general, however, it is not easy to show even the existence of $K$, let alone its further properties. For some elaboration on that subject, including special cases when $K$ can be well-defined, see Section \[sec:asymptotics\]. *If* it does exist then one can express the probability of not hitting the origin in terms of $K$ and $\kappa$.

\[prop:5\] Suppose that $K$ exists. Then $${\mathbb{P}}^x (T_0=\infty)=\kappa K(-x).$$ If $\kappa=0$ then ${\mathbb{P}}^x (T_0<\infty)=1$ for all $x \in {\mathbb{R}}$.

It follows that if $K$ exists and ${\mathbb{P}}^x (T_0<\infty)=1$ then $\mathbf{X}$ is recurrent.

Observe that $$\lim_{\lambda \to 0^+} {\mathbb{E}}^x e^{-\lambda T_0} = {\mathbb{P}}^x (T_0 < \infty).$$ On the other hand, $${\mathbb{E}}^x e^{-\lambda T_0} = 