How use Lagrange interpolation prove this inequality

Question

How can I use Lagrange interpolation to solve the following problem?

Let $a,b,c,d\in \mathbb{R}$ such that $$|ax^3+bx^2+cx+d|\le 1 $$ for every $x\in[-1,1]$. Show that $$|a|+|b|+|c|+|d|\le 7$$

Answer 1

Let $f(x)=ax^3+bx^2+cx+d$.

From Lagrange Interpolation (or solving a system of linear equations), we find that $$f(x)=\frac{-2f(-1)+4f(-\frac{1}{2})-4f(\frac{1}{2})+2f(1)}{3}x^3+\frac{2f(-1)-2f(-\frac{1}{2})-2f(\frac{1}{2})+2f(1)}{3}x^2+\frac{\frac{1}{2}f(-1)-4f(-\frac{1}{2})+4f(\frac{1}{2})-\frac{1}{2}f(1)}{3}x+\frac{-\frac{1}{2}f(-1)+2f(-\frac{1}{2})+2f(\frac{1}{2})-\frac{1}{2}f(1)}{3}$$

Now, we divide cases. The four polynomials in the form of $\pm f(\pm x)$ satisfy the exact same conditions as $f(x)$.

It is safe to assume that $a,b \ge 0$.

Case 1. $c \ge 0$.

If $d \ge 0$, we are done because $|a|+|b|+|c|+|d| =a+b+c+d=f(1) \le 1$.

If $d \le 0$, we have $|a|+|b|+|c|+|d|=a+b+c-d=f(1)-2f(0) \le 3$.

Case 2. $c \le 0$.

If $d \ge 0$, we have $$|a|+|b|+|c|+|d|=a+b-c+d=\frac{4}{3}f(1)-\frac{1}{3}f(-1)-\frac{8}{3}f(\frac{1}{2})+\frac{8}{3}f(-\frac{1}{2}) \le \frac{4}{3}+\frac{1}{3}+\frac{8}{3}+\frac{8}{3}=7$$

If $d \le 0$, we have $$|a|+|b|+|c|+|d|=a+b-c-d=\frac{5}{3}f(1)-4f(\frac{1}{2})+\frac{4}{3}f(-\frac{1}{2}) \le \frac{5}{3}+4+\frac{4}{3} = 7$$

This completes the proof. The bound is tight since $f(x)=4x^3-3x$ reaches equality.

EDIT: Note how $4x^3-3x$ is a Chebyshev Polynomial, which appears a lot in Lagrange Interpolation problems.