that $ \psi(x,s){\geq}0$. Now it follows from (5.23) that $$\Phi(0,x)=\psi(x,0){\geq}0.$$ Let us consider the domains $\Delta_{1}$ and $\Delta_{2}$ which are connected by relation $\Delta_{2}=\Delta_{1}+\delta$. We denote the corresponding truncated generators by $L_{\Delta_{1}}$ and $L_{\Delta_{2}}$, we denote the corresponding quasi-potentials by $B_{1}$ and $B_{2}$ and the corresponding kernels by $\Phi_{1}(x,y)$ and $\Phi_{2}(x,y)$. We introduce the unitary operator $$Uf=f(x-\delta),$$ which maps the space $L^{2}(\Delta_{2})$ onto $L^{2}(\Delta_{1})$. At the beginning we suppose that the conditions of Theorem 2.1 are fulfilled. Using formulas (2.1) and (2.2) we deduce that $$L_{\Delta_{2}}=U^{-1}L_{\Delta_{1}}U.$$ Hence the equality $$B_{2}=U^{-1}B_{1}U$$ is valid. The last equality can be written in the terms of the kernels $$\Phi_{2}(x,y)=\Phi_{1}(x+\delta,y+\delta).$$According to (6.2) and (6.6) we have $$\Phi_{1}(\delta,y+\delta){\geq}0.$$ As $\delta$ is an arbitrary real number, relation (6.1) follows directly from (6.6). We remark that an arbitrary generator $L$ can be approximated by the operators of form (2.1) (see\[23\],Ch.2). Hence the proposition is proved.\
In view of (4.1), (4.5) and relation $Bf{\in}C_{\Delta}$ the following assertion is true.\
**Proposition 6.2.** *Let the quasi-potential $B$ satisfy the conditions of Proposition $6.2$, Then the equalities* $$\Phi(a_{k},y)=\Phi(b_{k},y)=0 \quad 1{\leq}k{\leq}n$$ *are valid*.\

Sectorial operators
===================

1\. We introduce the following notions.\
**Definition 7.1.** The bounded operator $B$ in the space $L^{2}(\Delta)$ is 