# Polynomials non-negative on the integers

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.

• Yes, this follows from continuity (so $P$ has a minimum) and the fact that $P(x)$ tends to $+\infty$ as $|x|\to\infty$.
– user1688
Jul 20, 2017 at 7:51
• @WlodAA: I don't see how one can misunderstand this question which is well formulated (except for the missing implicit condition that at least one of the $\alpha_i$ should be nonzero). Also, I do not understand the downvote, the question looks reasonable and difficult to me. Jul 20, 2017 at 22:53
• The good news is that it looks loke there exists a universal (independent of $P$) sequence $\alpha_k={n\choose k}^2$ with this property. The bad news is that I can prove it only for $n=1$ (trivial) and $n=2$ (a bit less trivial). Jul 21, 2017 at 3:49
• If $\alpha_k$ are fixed, we may without loss of generality suppose that our polynomial has roots at $0,\dots,2k-1$ and $n-k$ pairs of double zeroes. If it helps anyhow. Jul 21, 2017 at 11:59
• by studying the extreme rays of the cone of polynomials which are non-negative at integer points Jul 21, 2017 at 12:54

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.
• Trivial. :-) ${}$ Jul 22, 2017 at 0:35
• The interesting part is how did you come up with this identity: $\sum_{k} \binom{n}{k}^{2}P(k) =\frac{(n!)^{2} \cos^{2}\lambda}{\pi^{2}} \sum_{x \in \Lambda}\frac{P(x)}{Q(x)^{2}}$ Jul 22, 2017 at 3:53
• @PaataIvanisvili The interesting part is actually why after teaching graduate complex analysis for the whole semester I didn't see it immediately :lol: It is clear (from the general duality mumbo-jumbo) that if the linear combination in question is, indeed, positive, then some identity like that must exist. Next, all such identities come from the residue theorem, so I need a function with poles at $\Lambda$ and extra poles at $0,\dots,n$ that is decaying at infinity faster than $P$ grows. I certainly know how to place poles with residues of the same sign at $\Lambda$ (continued) Jul 22, 2017 at 4:40
• That is what tangent is for. Extra poles and the decay dictate $Q^2$ in the denominator. Thus, I already get $\frac{\tan(\pi z+\lambda)}{Q(z)^2}P(z)$ forced upon me. Now, the second order poles are definitely not what I want. So that has to be corrected by offsetting the numerator by $\tan\lambda$ to make it vanish to the first order at integers. Then all that remains is to compute the residues. Jul 22, 2017 at 4:46