IMC 1998 · Day 2 · P8

Denote $x_0 = 1$, $x_1 = -1$, $x_2 = \frac{1}{2}$, $x_3 = -\frac{1}{2}$, $$w(x) = \prod_{i=0}^{3} (x - x_i),$$ $$w_k(x) = \frac{w(x)}{x - x_k}, \quad k = 0, \dots, 3,$$ $$l_k(x) = \frac{w_k(x)}{w_k(x_k)}.$$ Then for every $f \in P$ $$f''(x) = \sum_{k=0}^{3} l_k''(x) f(x_k),$$ $$|f''(x)| \le \sum_{k=0}^{3} |l_k''(x)|.$$ Since $f''$ is a linear function $\max\limits_{-1 \le x \le 1} |f''(x)|$ is attained either at $x = -1$ or at $x = 1$. Without loss of generality let the maximum point is $x = 1$. Then $$\sup_{f \in P} \max_{-1 \le x \le 1} |f''(x)| = \sum_{k=0}^{3} |l_k''(1)|.$$ In order to have equality for the extremal polynomial $f_*$ there must hold $$f_*(x_k) = \operatorname{sign} l_k''(1), \quad k = 0, 1, 2, 3.$$ It is easy to see that $\{l_k''(1)\}_{k=0}^{3}$ alternate in sign, so $f_*(x_k) = (-1)^{k-1}$, $k = 0, \dots, 3$. Hence $f_*(x) = T_3(x) = 4x^3 - 3x$, the Chebyshev polynomial of the first kind, and $f_*''(1) = 24$. The other extremal polynomial, corresponding to $x = -1$, is $-T_3$.