product (${\bf a}\cdot {\bf u}=0$, for all ${\bf u}\in \mathcal{A}$), and for each ${\bf v}_{\rho}\in \mathcal{K}$, ${\bf v}_{\rho} \cdot {\bf a}>0$. So the polynomials $$p_{n-1}(x)=a_{0,n-1}+a_{1,n-1}x+ \ldots+a_{n-2,n-1}x^{n-2}$$ and $$p_{n}(x)=a_{0,n}+a_{1,n}x+ \ldots+a_{n-1,n}x^{n-1}$$ satisfy that $$0< \int \left[p_{n-1}(x)\widetilde{P}_{n}(x)+p_{n}P_{n}(x)\right]\, \rho(x) \, \mbox{d} \mu(x), \quad \mbox{for all} \quad \rho \in \Phi.$$ According to condition (iii) the polynomial $\mathcal{P}(x)=p_{n-1}(x)\widetilde{P}_n(x)+p_{n}P_{n}(x)$, with real coefficients, must be non-negative in $\mbox{supp}(\mu)$. However we shall prove that this is impossible, arriving then to a contradiction.

Assume that $\mathcal{P}(x)=p_{n-1}(x)\widetilde{P}_n(x)+p_{n}P_{n}(x)$ does not change sign in $\mbox{supp}(\mu)$. Suppose that there is a point $t\in \mbox{supp}(\mu)$, such that $t=x_k$ $\displaystyle k\in \left\{1, \ldots, n\right\}$ satisfying that $\mathcal{P}(x_k)=0$, then taking into account that $\widetilde{P}_{n}$ and $P_{n}$ interlace zeros, we have that $p_{n-1}(x_k)=0$. Also, since $\mathcal{P}$ is non-negative on $\mbox{supp}(\mu)$, we have that $t=x_k$ is a zero of multiplicity even for $\mathcal{P}$. Consider $\displaystyle S=\left\{t_1, \ldots, t_{\ell}\right\}\subset \left\{x_1,\ldots,x_{n}\right\}$ the set of all points where $P_{n}$ and $\mathcal{P}$ vanishe at same time. Then we can write $$\label{positivepol}
\mathcal{P}=q(x)\prod_{i=1}^{\ell} \left(x-t_{i}\right)^{2 d_i}, \quad d_i\in \mathbb{N}, \quad i=1,\ldots,\ell,$$ where $q$ is a polynomial with positive values at every root of $P_{n}$. We also write $$\label{pnexpand}
p_{n-1}(x)=\widetilde{p}(x)\prod_{i=1}^{\ell} \left(x-t_{i}\right).$$ The polynomial $\widetilde{p}$ has degree $\deg p_n-\ell$. Since $\widetilde{P}_{n}$ and $P_n$ interlace zeros, we have 