A quadratic polynomial proof

Question

Let $P(x)$ be a quadratic polynomial for which:

$|P(x)| \leq1$ for all $x$ in the set $\{-1, 0, 1\}$.

Prove that $|P(x)| \leq \frac{5}{4}$ for all $x$ in the interval $[-1, 1]$.

How might I go about showing this?

Answer 1

Note that $$P(x)=P(-1)\frac{x(x-1)}{2}+P(0)(1-x^2)+P(1)\frac{x(x+1)}{2}$$ So, for $x\in[-1,1]$ we have $$\eqalign{|P(x)|&\le \frac12\left(|x|(1-x)+2-2x^2+|x|(x+1)\right)\max(|P(-1)|,|P(0)|,|P(1)|)\cr &\le\left(1+|x|-x^2\right)\max(|P(-1)|,|P(0)|,|P(1)|)\cr &\le\left(\frac{5}{4}-\left(\frac{1}{2}-|x|\right)^2\right) \max(|P(-1)|,|P(0)|,|P(1)|)\cr &\le \frac{5}{4}\max(|P(-1)|,|P(0)|,|P(1)|) }$$ and the desired conclusion follows.

Answer 2

A quadratic polynomial $q(x)=ax^2+bx+c$ is fixed by its values at $-1,0,1$, namely $a-b+c,c,a+b+c$. The maximum of $|q(x)|$ over $[-1,1]$ is attained either at an endpoint or at the abscissa of the vertex, assuming it belongs to $[-1,1]$. We may assume without loss of generality that the vertex belongs to $[0,1]$ and check that with such assumptions the extremal polynomials are given by $\pm(x^2-x-1)$, with maximum modulus $\frac{5}{4}=\left|c-\frac{b^2}{4a}\right|$.