the potential $Q$ are ( in general) unbounded operators. Therefore the operators $L$ and $Q$ are defined not in the whole space $L^{2}(-\infty.\infty)$ but only in the subsets $D_{L}$ and $D_{Q}$ respectively. We use the following property of the potential $Q$ (see\[23\]).\
**Proposition 3.1.** *If $f=Q\,g,\quad (g{\in}D_{Q})$ then $f{\in}D_{L}$ and* $$-Lf=g.$$ **Example 3.1.** Compound Poisson process.\
Let the generator $L$ has form (2.50) where $$M=\int_{-\infty}^{\infty}\nu^{\prime}(x)dx<\infty,\quad
\int_{-\infty}^{\infty}[\nu^{\prime}(x)]^{2}dx<\infty.$$ We introduce the functions $$K(u)=-\frac{1}{M\sqrt{2\pi}}\int_{-\infty}^{\infty}\nu^{\prime}(x)e^{-iux}dx,$$ $$N(u)=\frac{K(u)}{1-\sqrt{2\pi}K(u)}.$$Let us note that $$|K(u)|<\frac{1}{\sqrt{2\pi}},\quad u{\ne}0;\quad K(0)=-\frac{1}{\sqrt{2\pi}}.$$ It means that $N(u){\in}L^{2}(-\infty,\infty).$ Hence the function $$n(x)=-\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}N(u)e^{-iux}du$$ belongs to $L^{2}(-\infty,\infty)$ as well. It follows from (2.50),(3.2)and (3.7) that the corresponding potential $Q$ has the form (see \[23\], Ch.11)$$Qf=\frac{1}{M}[f(x)+\int_{-\infty}^{\infty}f(y)n(x-y)dy].$$ **Proposition 3.2.** *Let conditions $(3.3)$ be fulfilled. Then the operators $L$ and $Q$ are bounded in the space* $L^{2}(-\infty,\infty).$\
Now we shall give an example when the kernel $n(x)$ can be written in an explicit form.\
**Example 3.2.** We consider the case when $${\nu}^{\prime}(x)=e^{-|x|},\quad -\infty<x<\infty.$$ In this case $M=2$ and the operator $L$ takes the form $$Lf=-2f(x)+\int_{-\infty}^{\infty}f(y)e^{-|x-y|}dy.$$ Formulas (3.4)-(3.7) imply that $$Qf=\frac{1}{2}f(x)-\frac{1}{4\sqrt{2}}\int_{-\infty}^{\infty}f(y)e^{-|x-y|\sqrt{2}}dy.$$

Truncated generators and quasi-potentials
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Let us denote by $\Delta$ the set of segments $[a_{k},b_{k}]$ such that\
$a_{1}<b_{1}<a_{2}<b_{2}<...<a_{n}<b_{n},\quad 1{\leq}k{\leq}n.$ By $C_{\Delta}$ we denote the set of functions $g(x)$ on $L^{2}(\Delta)$ such 