Let $P$ be a real polynomial of exact degree $2n$ ($n \geq 1$) whose zeros are real numbers and such that 
\begin{equation\*}
P(j) \geq 0 
\quad \text{for any} \quad 
j \in \mathbb{Z}. 
\end{equation\*}  

Does there exist non-negative real numbers $\alpha\_0,\alpha\_1,\ldots,\alpha\_n,$ with at least one of the $\alpha\_i$ non-zero, such that the polynomial 
\begin{equation\*}
Q(x) = \sum\_{k=0}^{n} \alpha\_k P(x+k)
\end{equation\*}
is non-negative on the whole real line, i.e.; $Q(x) \geq 0$ for any $x \in \mathbb{R}$ ?


I would like to add that this question is not merely a mathematical curiosity but pops up naturally while working on spectral transformations of discrete measures.



Polynomials non-negative on the integers
Occasionally I wish somebody could give me a good whack on the head to keep my brains running and the older I get, the more frequently I need it. The problem is actually trivial.


I will prefer to think that $P$ is non-negative on some disjoint with integers arithmetic progression $\Lambda$ with step $1$ . Then we need to show that 
$$
\sum\_{k=0}^n {n\choose k}^2 P(k)\ge 0\,.
$$
Let $\lambda$ be a number such that $\Lambda=\{x:\cos(\pi x+\lambda)=0\}$ and put $Q(x)=x(x-1)\dots(x-n)$. Consider the meromorphic function 
$$
F(z)=\frac{\tan(\pi z+\lambda)-\tan\lambda}{Q(z)^2}P(z)\,.
$$
Note that the poles of $F$ are simple and $F(z)$ decays like $|z|^{-2}$ on big circles between the poles of the tangent, so the sum of the residues converges to $0$.
Now the residue at the zero $k$ of $Q$ is $\frac{\pi}{(n!)^2\cos^2\lambda}{n\choose k}^2P(k)$ while the residues at the poles $x\in\Lambda$ of the tangent are $-\frac 1{\pi Q(x)^2}P(x)\le 0$. The end.


