2.$$r=\int_{-c}^{c}N_{1}(x)dx{\ne}0$$* *Then the corresponding operator $B$ has the form $$Bf=\int_{-c}^{c}\Phi(x,y,c)f(y)dy$$ where* $$\Phi(x,y,c)=\frac{1}{2}\int_{x+y}^{2c-|x-y|}q[(s+x-y)/2,(s-x+y)/2]ds,$$ $$q(x,y)=[N_{1}(-y)N_{2}(x)-N_{2}(-y)N_{1}(x)]/r.$$ It follows from (8.4) and (8.5) that $$\Phi({\pm}c,y)=\Phi(x,{\pm}c)=0.$$ Here we use the following relation\
$q[(s+x-y)/2,(s-x+y)/2]=$ $$/r.$$Thus $$q[(s+x-y)/2,(s-x+y)/2]=-q[(-s+x-y)/2,(-s-x+y)/2].$$ From formulas (8.4) and (8.5) we deduce the following statement.\
**Proposition 8.1.** *Let the conditions of Theorem $8.1$ be fulfilled. There exists a function $\phi(x)$ such that* $$|\Phi(x,y,c)|{\leq}\phi(x-y),$$ $$\int_{-R}^{R}\phi(x)dx<\infty \quad if \quad
0<R<\infty.$$ *Proof.* Relation (8.4) can be written in the form $$\Phi(x,y,c)=\int_{x}^{c+(x-y-|x-y|)/2}q(t,t-x+y)dt.$$ By relations $$N_{k}(x)=0,\quad x{\notin}[-c,c], \quad k=1,2$$ we extend the functions $N_{k}(x)$ from the segment $[-c,c]$ to the segment $[-2c,2c].$ It follows from (8.11) and (8.12) that inequality (8.9) is valid , if $$\phi(x)=\int_{-c}^{c}[|N_{1}(t)N_{2}(t-x)|+|N_{2}(t)N_{1}(t-x)|]dt/|r|.$$ Equality (8.13) imply that $\phi(x){\in}L^{p}[-2c,2c].$ The proposition is proved.\
It follows from Proposition 8.1 that the operator $B$ is bounded in all the spaces $L^{p}(-c,c),\quad
p{\geq}1$. We shall prove that the operator $B$ is compact.\
**Proposition 8.2.** *Let the conditions of Theorem $8.1$ be fulfilled. Then the operator $B$ is compact in all the spaces* $L^{p}(-c,c),\quad p{\geq}1.$\
*Proof.* Let us consider the operator $B^{\star}$ in the space $L^{q}(-c,c),\quad 1/p+1/q=1.$ Using relation (8.3) we have $$B^{\star}f_{n}=\int_{-c}^{c}\Phi(y,x,c)f_{n}(y)dy$$ where the functions $f_{n}(x){\to}0$ in the weak sense. Relation (8.14) can be represented in the following form $$B^{\star}f_{n}=\int_{-c}^{c}f_{n}(y)\int_{y}^{c+(y-x-|x-y|)/2}q(t,t-y+x)dtdy.$$By interchanging 