to the measure $\mu$, if its roots are all simple, lying in $\displaystyle \overset{\circ}{\Delta}$, with at most one zero into each interval of $\displaystyle \Delta \setminus \mbox{supp}(\mu)$. The system of nodes $(x_{1},\ldots,x_{n})$ is also said to be admissible with respect to $\mu$.

\[interwendroff\] Let $\displaystyle P_n(x)=\prod_{j=1}^n \left(x-x_{j}\right)$ and $\displaystyle \widetilde{P}_{n}(x)=\prod_{j=1}^{n-1} \left(x-\widetilde{x}_{j}\right)$ be two admissible polynomials with respect to $\mu$ that satisfy $\displaystyle x_1<\widetilde{x}_1<x_2< \cdots < \widetilde{x}_{n-1}<x_n.$ Then there exists a positive integrable function $\rho_n$ with respect to $\mu$ ($\rho_n$ is a weight function for $\mu$) such that for the measure $\mu_n$ which differential form $\displaystyle \mbox{d} \mu_n(x)=\rho_n(x) \, \mbox{d} \mu(x),$ $x \in \mbox{supp} (\mu)$, $P_n\equiv q_{\mu_n,n}$ and $\widetilde{P}_{n}\equiv q_{\mu_n, n-1}$ are the $n$-th and $n-1$-th monic orthogonal polynomials with respect to $\mu_n$, respectively.

In the proof we follow techniques used in [@kroo].

Consider $\Phi$ a set of weight functions such that for every constant $\alpha >0$ it satisfies:

-   $\rho \in \Phi \implies \alpha \rho \in \Phi.$

-   $(\rho, \widetilde{\rho})\in \Phi^2= \Phi \times \Phi \implies \alpha \rho +(1-\alpha)\widetilde{\rho} \in \Phi,$ $\alpha \leq 1.$

-   If a polynomial $Q$ satisfies $\displaystyle \int Q(x) \rho(x) \, \mbox{d} \mu(x) >0$ for all $\rho \in \Phi,$ then $Q \geq 0$ in $\mbox{supp} (\mu)$.

Two 