Source: Challenge and Thrill of Pre-College Mathematics
"Let $$p(x)=ax^2+bx+c$$ be a polynomial in $ℝ(x)$ such that $|p(α)|≤1$ for $|α|≤1$. Prove that $|2aα+b|≤4$ for $|α|<1$."
First of all, it is interesting to note that the exercise has multiple questions of the same pattern, and while I was able to do all other such questions, this one seems slightly more trickier. An answer to one of the other questions may be found here, for context/reference: Prove that $|a|+|b|+|c|\le17$ if $p(x)=ax^2+bx+c$ is a real polynomial with $|p|\le1$ for $0\le x\le1$
My attempt: Expressing $a,b,c$ in terms of $p(0.5), p(0)$ and $p(1)$ as in the answer given in the link I just shared. This can be done as follows:
$1) a=2(a+b+c)-4(0.25a+0.5b+c)+2c =2p(1)-4p(0.5)+2p(0)$ ⇒ $max (a)=2+4+2=8$
Similarly, it can be found that $max(b)=8$, but this in-fact doesn't satisfy our proof.
The problem with my approach is that I considered my proof to assume $|α|≤1$, and not $|α|<1$ as asked by the question.
Could anyone help me on how to proceed with $|α|<1$ ? I can't quite get my finger on it.
Hint:
Show both $f'(1) = \frac32f(1)+\frac12f(-1)-2f(0)$ and $f'(-1) = -\frac12f(1)-\frac32f(-1)+2f(0)$ are in $[-4, 4]$. Further, $f'$ is linear.