Abstract:
For a piecewise-continuous function $f$ on $[0,1]$ we denote by $\nu(f)$ the number of its sign changes. By $K_n[0,1]$ we denote the set of piecewise-continuous functions $f$ on $[0,1]$ such that $\nu(f)\le n$. We prove that for any $n\ge 2$ there are no integral transforms $\tilde{K}f(x)=\int_0^1 K(x,y)f(y)\,dy$ with a continuous kernel $K(x,y)$ such that $\nu(\tilde {K}f)=\nu(f)$, for every $f\in K_n[0,1]$. We give an example of a continuous kernel $K(x,y)$ such that $\nu(\tilde{K}f)=\nu(f)$, for every $f\in K_1[0,1]$.