Turán's inequalities

Remarks

At $x = \pm 1$ the left side is $0$, since $P_k(1) = 1$ and $P_k(-1) = (-1)^k$; so the strict inequality needs $\abs{x} < 1$. Check for $n = 1$: $x^2 - \frac{3x^2 - 1}{2} = \frac{1 - x^2}{2} > 0$ for $\abs{x} < 1$. The known proofs combine the three-term recurrence of the $P_n$ with their differential equation, and show that the left side cannot vanish inside $(-1, 1)$. The left side is called the Turán determinant, since it is $\det \begin{pmatrix} P_n & P_{n+1} \\ P_{n-1} & P_n \end{pmatrix}$. Analogues hold for other families: for the Chebyshev polynomials, $T_n(x)^2 - T_{n-1}(x) T_{n+1}(x) = 1 - x^2$. This follows from $\cos^2(n\theta) - \cos((n - 1)\theta) \cos((n + 1)\theta) = \sin^2 \theta$ with $x = \cos \theta$. For the Hermite polynomials one has $H_n(x)^2 - H_{n-1}(x) H_{n+1}(x) > 0$ for all real $x$.

Going deeper

For a student

Turán found the inequality for Legendre polynomials, and Szegő published several proofs of it in 1948. Similar Turán-type inequalities hold for ultraspherical, Laguerre and Hermite polynomials, and for Bessel functions in the order variable. Such inequalities for the Taylor coefficients of entire functions are related to the Laguerre–Pólya class. This link is the reason they appear in work on the Riemann hypothesis through the Jensen polynomials of the Riemann $\xi$ function.