relations $$V(x)f(x)=0,\quad -\infty<x<\infty,$$ $$\frac{1}{2\pi}\mathrm{lim}\int_{-N}^{N}\int_{\Delta}e^{-ixp}f(x)dxdp=f(0),\quad N{\to}\infty.$$ Since the function $Q(x,t,u)$ monotonically decreases with respect to $"u"$ this is also true for the function $\psi(x,s,u)$ according to (5.10). Hence there exists the limit $$\psi(x,s)=\mathrm{\lim}\psi(x,s,u),\quad u{\to}\infty,$$where $$\psi(x,s)=0,\quad x{\notin}\Delta.$$ The probabilistic meaning of $\psi(x,s)$ follows from the equality $$\int_{0}^{\infty}e^{-st}p(t,\Delta)dt=\int_{\Delta}\psi(x,s)dx.$$ Using the properties of the Fourier transform and conditions (5.19) , (5.20) we deduce from (5.16) the following assertion.\
**Theorem 5.1.** *Let the considered Levy process have the continuous density. Then the relation $$((sI-L_{\Delta})f,\psi(x,s))_{\Delta}=f(0)$$ is true.*\
**Remark 5.1** For symmetric stable processes relation (5.22) was deduced by M.Kac \[11\].\
**Remark 5.2** As it is known ,the stable processes, the variance damped Levy processes , the variance gamma processes, the normal inverse Gaussian process, the Meixner process have continuous densities (see(\[24\],\[31\]).\
**Remark 5.3.** So we have obtained the formula (5.21) for Laplace transform of $p(t, \Delta)$ in terms of $\psi(x,s)$. The double Laplace transform of $p(t,\Delta)$ was obtained by G.Baxter and M.D.Donsker \[3\] for the case when $\Delta=(-\infty,a].$\
We express the important function $\psi(x,s)$ with the help of the quasi-potential $B$.\
**Theorem 5.2.** *Let the considered Levy process have the continuous density and let the quasi-potential $B$ be regular . Then in the space $L^{p}(\Delta)\quad (p>1)$ 