Convolution type form of infinitesimal generator
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1\. In this section we prove that under some conditions the infinitesimal generator $L$ can be represented in the special convolution type form $$Lf=\frac{d}{dx}S\frac{d}{dx}f,$$ where the operator $S$ is defined by the relation $$Sf=\frac{1}{2}Af+\int_{-\infty}^{\infty}k(y-x)f(y)dy,$$ and for arbitrary $M (0<M<\infty)$ we have $$\int_{-M}^{M}|k(t)|dt<\infty.$$ The representation $L$ in form (2.1) is convenient as the operator $L$ is expressed with the help of the classic differential and convolution operators.\
By $C_{c}$ we denote the set of functions $f(x){\in}C_{0}$ with a compact support.\
**Lemma 2.1.** *Let the following conditions be fulfilled.\
1. The function $\nu(x)$ is monotonically increasing, has the derivative when $x{\ne}0$ and* $$\int_{-\infty}^{\infty}\frac{x^{2}}{1+x^{2}}d\nu(x)=
\int_{-\infty}^{\infty}\frac{x^{2}}{1+x^{2}}{\nu}^{\prime}(x)dx<\infty,$$ $$\nu(x){\to}0, \quad x{\to}\infty.$$

2\. *For arbitrary $M \quad (0<M<\infty)$ we have* $$\int_{-M}^{M}|\nu(x)|dx<\infty,\quad
\int_{-M}^{M}|x|{\nu}^{\prime}(x)dx<\infty.$$ 3. $$x\nu(x){\to}0,\quad x{\to}0.$$ *Then the expression $$J=\int_{-\infty}^{\infty}[f(y+x)-f(x)]{\nu}^{\prime}(y)dy$$ can be represented in the convolution type form* $$J=\frac{d}{dx}\int_{-\infty}^{\infty}f^{\prime}(y)k(y-x)dy$$ *where* $f(x){\in}C_{0}^{2},\quad
k(x)=\int_{0}^{x}\nu(y)dy$.\
*Proof.* For every $f(x){\in}C_{c}$ there exists such $M \quad (0<M<\infty)$ that $$f(x)=0 ,\quad x{\notin}[-M,M].$$Let us introduce the following notations $$J_{1}=\frac{d}{dx}\int_{-\infty}^{x}f^{\prime}(y)k(y-x)dy,$$ $$J_{2}=\frac{d}{dx}\int_{x}^{\infty}f^{\prime}(y)k(y-x)dy.$$ Using (2.11) we have $$J_{1}=
-\frac{d}{dx}\int_{-M}^{x}[f(y)-f(x)+f(x)]k^{\prime}(y-x)dy.$$ From (2.11) and (2.13) we deduce the relation $$J_{1}=f(x)k^{\prime}(-M-x)+\int_{-M}^{x}[f(y)-f(x)]k^{\prime\prime}(y-x)dy.$$ When $M{\to}\infty$ we obtain the equality $$J_{1}=\int_{-\infty}^{0}[f(y+x)-f(x)]k^{\prime\prime}(y)dy.$$ In the same way we deduce the relation $$J_{2}=\int_{0}^{\infty}[f(y+x)-f(x)]k^{\prime\prime}(y)dy.$$ Relation (2.9) follows directly from formulas (2.15) 