Upper bound of a quadratic, given the upper bound of variable Problem: Let $p(x)$ be a quadratic polynomial $p(x)$ = $ax^2+bx+c$ such that $|p(x)|$ $≤$ $1$ for |$x$| $≤$ $1$.
Prove that
$|cx^2+bx+a|$ $≤$ $2$ for |$x$| $≤$ $1$. (Source: Challenge and Thrill of Pre-College Mathematics, Page 201, Problem 16)
My attempt: I tried $|cx^2+bx+a|$ $≤$ |$a+b+c$| = $|p(1)|$ for $|x|$ $≤$ 1, but after that it gets complicated. More specifically, the upper bound for the quadratic asked actually comes out to be $|p(1)|$ $≤$ 1, and so the given statement is proved. However I'm thrown off by why the problem states $≤$ $2$ and not $1$.
Any hints would be appreciated, but please don't post a definite solution. In the case I solve it using the hints given, I'll post my own answer below. If I cannot, I'll edit this question asking for a solution.
 A: Employing a rather common strategy for such problems, from $p(x)=ax^2+bx+c$, we get $c=p(0), b = \frac12\left[p(1)-p(-1)\right]$ and $a = \frac12\left[p(1)+p(-1)\right]-p(0)$.
Hence with $q(x) = cx^2+bx+a$, we get
$$q(x) = p(0)x^2+\tfrac12\left[p(1) - p(-1)\right]x +\tfrac12\left[p(1)+p(-1)\right]-p(0)$$
$$\qquad = p(0)(x^2-1)+\tfrac12p(1)(x+1)-\tfrac12p(-1)(x-1)$$
$$\implies |q(x)| \leqslant |p(0)||x^2-1|+\tfrac12|p(1)||x+1|+\tfrac12|p(-1)||x-1|$$
$$\qquad \leqslant (1-x^2)+\tfrac12(1+x)+\tfrac12(1-x)=2-x^2\leqslant 2$$
--
P.S.  Take $p(x)=2x^2-1$ and you will notice this bound is tight.
A: Define the class of quadratic polynomials P2 whose members having sum of all abs coefficients being less than or equal to 1:
P2={p(x)=ax^2 +bx+c |  |a|+|b|+|c|<=1}
Show that

*

*If p(x) ε P2 then |p(x)|<=1 for all |x|<=1

Proof
|p(x)|<= |a||x|^2+|b| |x|+ |c|<= |a|+|b|+|c|<=1 for all |x|<=1


*If p ε P2 then q ε P2 where q(x)= cx^2 + bx +a then q ε P2

Proof
|q(x)|<= |c||x|^2 + |b| |x|+ |a|<= |c|+|b|+|a| <= 1 for all |x|<=1
It follows that |q(x)| <=2 for all |x| <=1
QED
