Turan Inequalities

Question

A real entire function

$$\psi(x)=\sum_{k=0}^{\infty} \gamma_k\frac{x^k}{k!}$$

is said to be in the Laguerre-Polya class, denoted $\psi(x) \in \mathcal{LP}$, if it can be represented in the form \begin{eqnarray*} \psi(x)=c x^m e^{-\alpha x^{2}+\beta x} \prod_{k=1}^{\infty}\left(1+x/x_{k}\right)e^{- x/x_k}, \end{eqnarray*} where $c$, $\beta$, $x_{k}$ are real numbers, $\alpha \geq 0$, $m$ is a nonnegative integer and $\sum x^{-2}_{k}< \infty$.

It is well known that a necessary condition for a real entire function $\psi(x)$ to belong to $\mathcal{LP}$ class is that its Maclaurin coefficients satisfy the Turan inequalities

\begin{eqnarray} \gamma_{k}^2-\gamma_{k-1}\gamma_{k+1}\geq 0, \end{eqnarray}

for $k\geq 1$.

Question: Who proved the necessary condition? When? How?

Answer 1

These were first proved by Laguerre himself (They are called sometimes Laguerre's inequalities). The LP class can be characterized as the closure of real polynomials with real zeros, so it is enough to prove the inequalities for real polynomials whose all zeros are real, and this is sort of elementary.

First notice that $\gamma_k=\psi^{(k)}(0)$ and the inequalities follow from more general inequalities $$(\psi^{(k)})^2(x)-\psi^{(k-1)}(x)\psi^{(k+1)}(x)\geq 0.$$ As the class of polynomials with all zeros real is closed under differentiation, it is enough to prove this for $k=1$. As the class is also closed with respect to real shifts, what remains to prove is your inequality (at $x=0$) with $k=1$ for polynomials with all zeros real. Wlog $\psi(0)=1$, so $$\psi(x)=\prod\left(1+x/x_k\right),$$ $x_k$ are real, so $$\psi'(0)=\sum\frac{1}{x_k},\quad \psi''(0)=\sum\frac{1}{x_ix_j}$$ and our inequality becomes trivial.