be substantial in Section \[sec:harmonic\] is the fact that the Green function for the positive half-line can be represented in terms of $V$ and $\widehat{V}$ in the following way: $$\label{eq:green_f}
G_{(0,\infty)}(x,y) = \int_0^x \widehat{V}'(u)V'(y-x+u)\,du, \quad y>x>0.$$ That identity is provided by [@Bertoin96 Theorem VI.20].

A number of helpful results concerning $\widehat{V}$, which is of greater importance and usage in the case of non-symmetric processes, are derived in [@TG19]. Below we provide a sharp estimate on the probability that the process, when exiting the interval $(0,R)$, chooses the right end. Such result seem to be interesting in and of itself, as we provide sharp two-sided bound, which is an analogue of the estimate for the symmetric case (see e.g. [@GR12 Proposition 3.7]).

\[prop:prob\_exit\_sharp\] Suppose that ${\mathbb{E}}X_1 = 0$ and ${\operatorname{Re}}\psi \in {{\rm WLSC}(\alpha,\chi)}$ for some $\alpha>1$ and $\chi \in (0,1]$. Then there is $c \in (0,1]$ such that for any $R>0$ and $0<x<R$, $$c \frac{\widehat{V}(x)}{\widehat{V}(R)} {\leqslant}{\mathbb{P}}^x \big( {\tau_{(0,R)}}< {\tau_{(0,\infty)}}\big) {\leqslant}\frac{\widehat{V}(x)}{\widehat{V}(R)}.$$

Fix $R>0$ and let $x \in (0,R)$. From the proof of [@TG19 Theorem 9] we get the following results. First, $${\mathbb{P}}^x \big( {\tau_{(0,R)}}< {\tau_{(0,\infty)}}\big) {\leqslant}\frac{\widehat{V}(x)}{\widehat{V}(R)}.$$ Next, in view of [@TG19 Proposition 4 and Theorem 6], there is $c_1$ such that $$c_1 