TITLE: If $|x|\leq 1$, $|ax^2+bx+c|\leq 1$, find the max possible value of $|2ax+b|$
QUESTION [5 upvotes]: Given $a,b,c\in \mathbb{R}$ such that  $|x|\leq 1$ and $|ax^2+bx+c|\leq 1$, find the maximum value of
$$
|2ax+b|
$$

My Attempt:
Set $x=1$ in $|ax^2+bc+c|\leq 1$ to get
$$
\tag{$\star$}|a+b+c|\leq 1
$$
Similarly, set $x=-1$ in $|ax^2+bc+c|\leq 1$ to get
$$
\tag{$\star\star$}|a-b+c|\leq 1
$$
Similarly, set $x=0$ in $|ax^2+bx+c|\leq 1$ to get
$$
\tag{$\star\star\star$}|c|\leq 1
$$
Now, adding  $(\star)$ and $(\star\star)$ gives
$$
|a+c|\leq 1
$$
Subtracting $(\star)$ from $(\star\star)$ gives
$$
|b|\leq 1
$$
Now, we have
$$
-1\leq (a+c)\leq 1\\
-1-c\leq a \leq 1-c
$$
Since $-1\leq c\leq 1$, we get
$$
-2 \leq a \leq 2,\; -1 \leq b\leq 1
$$
How can I complete the solution from this point?

REPLY [3 votes]: By the Markov brothers' inequality, for a polynomial $p$ of degree $n$ we have
$$ \max_{-1\le x\le 1} |p'(x)| \le n^2\max_{-1\le x\le 1} |p(x)| $$
with equality for Chebyshev polynomials.  Taking $p(x)=ax^2+bx+c$ we get $|2ax+b|\le 4$.