sororak wrote: Let $ p(x) $ be a quadratic polynomial for which : \[ |p(x)| \leq 1 \qquad \forall x \in \{-1,0,1\} \] Prove that: \[ \ |p(x)|\leq\frac{5}{4} \qquad \forall x \in [-1,1]\] Spain 96!

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$P(x) = ax^2 + bx + c$. $P(0) = c$,$P(1) = a + b + c$ and $P(-1) = a - b + c$. this are all we have so let rewrite $P(x)$ using $P(1),P(-1),P(0)$. $P(x) = \frac{x^2+x}{2}P(1) + \frac{x^2-x}{2}P(-1) + (1-x^2)P(0)$. Case $1 : 0 \le x \le 1$ $|P(x)| = |\frac{x^2+x}{2}P(1) + \frac{x-x^2}{2}P(-1) + (1-x^2)P(0)| \implies |P(x)| \le |\frac{x^2+x}{2} + \frac{x-x^2}{2} + (1-x^2)| = |x + 1 - x^2| \le \frac{5}{4}$. Case $2 : -1 \le x \le 0$ $|P(x)| = |\frac{x^2+x}{2}P(1) + \frac{x-x^2}{2}P(-1) + (1-x^2)P(0)| \implies |P(x)| \le |-\frac{x^2+x}{2} - \frac{x-x^2}{2} - (1-x^2)| = |-x - 1 + x^2| \le \frac{5}{4}$. equality holds for $\pm \frac{1}{2}$.