Prove that $|a|+|b|+|c|\le17$ if $p(x)=ax^2+bx+c$ is a real polynomial with $|p|\le1$ for $0\le x\le1$

Question

Let $ax^2+bx+c$ be a quadratic polynomial with real coefficients such that $$|ax^2+bx+c| \leq 1,$$ for $ 0\leq x\leq 1$. Prove that $$|a|+|b|+|c|\leq 17$$

How to proceed in this particular question. Sorry I can't show any work because I really not getting how to initiate.

Answer 1

Let $f(x) = ax^2 + bx + c$ We know that $$ \left|\frac{a}2\right| = |f(0) + f(1) - 2f(0.5)| = |[f(0) - f(0.5)] - [f(0.5) - f(1)]|\\ \leq |f(0) - f(0.5)| + |f(0.5) - f(1)| \leq 2+2 = 4 $$ so $|a| \leq 8$. Clearly, $|c| = |f(0)| \leq 1$. That leaves $b$. We get $$ |b| = |4f(0.5) - f(1) - 3f(0)| \leq 3|f(0.5)-f(0)| + |f(0.5) - f(1)| \leq 3\cdot 2 + 2 = 8 $$ which is what we need.

It's also worth noting that $$ f(x) = 8x^2 - 8x + 1 $$ demonstrates that $17$ is a strict bound, so we cannot do any better.

Answer 2

Hint: Let $P$ your polynomial. Put $P(0)=u$, $P(1/2)=v$ and $P(1)=w$. Then $|u|, |v|, |w|$ are $\leq 1$. Find $a,b,c$ in function of $u,v,w$, and bound them.