If $$\left|ax^2+bx+c\right|\leq 2\quad \forall x\in[-1,1]$$ then find the maximum value of $$\left|cx^2+2bx+4a\right|\quad \forall x\in [-2,2].$$
My Attempt
Let $f(x)=ax^2+bx+c$, then
$$|cx^2+2bx+4a|=x^2|4\frac{a}{x^2}+2\frac{b}{x}+c|=x^2\left|f\left(\frac{2}{x}\right)\right|$$
But then I couldn't make any headway.
I also tried to express $a,b,c$ in terms of $f(-1),f(0),f(1)$ i.e $$c=f(0);b=\frac{f(1)-f(-1)}{2};a=\frac{f(-1)-2f(0)+f(1)}{2}$$ but again couldn't go further
On
We may also examine the behavior of the polynomials in absolute-value brackets as described in "vertex form", $$ p(x) \ \ = \ \ a·\left(x \ + \ \frac{b}{2a} \right)^2 \ + \ \left(c \ - \ \frac{b^2}{4a} \right) \ \ \ \text{and} \ \ \ q(x) \ \ = \ \ c·\left(x \ + \ \frac{b}{c} \right)^2 \ + \left(4a \ - \ \frac{b^2}{c} \right) \ \ . $$ It will be sufficient to look at the situation for $ \ p(x) \ $ with $ \ a \ > \ 0 \ \ , \ $ as the case with $ \ a \ < \ 0 \ \ $ is a "vertical reflection" which proves to produce identical results when the absolute-value of the polynomials is taken.
We can start from the symmetrical arrangement in which $ \ b \ = \ 0 \ \ , \ $ placing the vertices of the parabolas for both polynomials on the $ \ y-$axis. In order to have $ \ | \ p(x) \ | \ \le \ 2 \ $ on $ \ [-1 \ , \ 1] \ \ , \ $ it is necessary to have $ \ -2 \ \le \ p(x) \ \le \ 2 \ \ . \ $ Its vertex lies at $ \ (0 \ , \ c ) \ $ and the "endpoints" on the interval boundary are $ \ (\pm 1 \ , \ a + c) \ \ . \ $ This leads to $ \ -2 \ \le \ c \ \le \ 2 \ $ (vertex must be in the given range) and $ \ -2 \ \le \ a + c \ \le \ 2 \ $ (endpoints must be in the same range); this essentially "forces" $ \ c \ = \ -2 \ $ (vertex as "low" as permitted) and thus $ \ a \ = \ 4 \ \ $ is the largest value this coefficient may have. Taking the absolute-value of $ \ p(x) \ $ "folds up" its parabola at the $ \ x-$axis at $ \ x \ = \ \pm \frac{1}{\sqrt2} \ \ , \ $ so that the "vertex" of $ \ | \ p(x) \ | \ $ is at $ \ (0 \ , \ 2) \ \ $ but the endpoints remain at $ \ (\pm 1 \ , \ 2) \ \ . \ $
With $ \ p(x) \ = \ 4x^2 - 2 \ \ , \ $ we then have the corresponding polynomial $ \ q(x) \ = \ (-2)·x^2 + 4·4 $ $ = \ -2x^2 + 16 \ \ $ with its vertex at $ \ (0 \ , \ 16) \ \ . \ $ As this is a "downward-opening" parabola, the $ \ y-$coordinates of its endpoints on $ \ [-2 \ , \ 2] \ $ will be "lower" [ at $ \ (\pm 2 \ , \ 8) \ ] \ . \ $ Also, as $ \ q(x) \ \ge \ 0 \ $ on $ \ [-2\sqrt2 \ , \ 2 \sqrt2] \ \ , \ $ the parabola for $ \ | \ q(x) \ | \ $ is identical. Hence, the maximal value for $ \ | \ -2x^2 + 16 \ | \ $ on $ \ [-2 \ , \ 2] \ $ is $ \ 16 \ \ . $ (Apropos of AnCar's answer, taking the negatives of $ \ a \ $ and $ \ c \ $ inverts the two parabolas, but the curves of $ \ | \ p(x) \ | \ $ and $ \ | \ q(x) \ | \ $ are exactly the same.)
What happens if we remove this symmetry by giving $ \ b \ $ a non-zero value? (We'll just consider $ \ b \ > \ 0 \ \ $ since a "rightward" rather than a "leftward" shift has the same effect here.) We now have the tricky task of keeping the vertex $ \ \left(b \ , \ c \ - \ \frac{b^2}{4a} \right) \ $ and the "higher endpoint" $ \ (1 \ , \ a + b + c) \ $ within the allowed range. But increasing $ \ b \ > \ 0 \ $ "lowers" the vertex of $ \ p(x) \ $ unless we increase $ \ a \ > \ 0 \ $ or increase $ \ c \ $ from its negative value (or both); both of these alterations act to increase $ \ a + b + c \ \ , \ $ which is undesired. For example, if we shift the vertex to $ \ x \ = \ -1 \ \ , \ $ we require $ \ b \ = \ 2a \ \ ; \ $ to accommodate the requirement on the endpoint at $ \ x \ = \ 1 \ \ , \ $ we then need to have $ \ 3a + c \ \le \ 2 \ \ . $ The restrictions can all be met by $ \ p(x) \ = \ x^2 + 2x - 1 \ \ \ , \ $ which maintains $ \ | \ p(x) \ | \ \le \ 2 \ \ . \ $ However, the corresponding $ \ q(x) \ = \ -x^2 + 4x + 4 \ $ has vertex and right endpoint $ \ (2 \ , \ 4·[a + b + c] = 8) \ $ and left endpoint $ \ (-2 \ , \ 4·[a - b + c] = -8 ) \ \ . \ $ So $ \ c \ $ is increased (less negative) but $ \ a \ $ is reduced, which in turn reduced the range of $ \ q(x) \ \ ; \ $ we now have $ \ | \ q(x) \ | \ \le \ 8 \ \ $ on $ \ [-2 \ , \ 2] \ \ . $
Moving the vertex of $ \ p(x) \ $ outside of $ \ [-1 \ , \ 1] \ $ to $ \ -X \ < \ -1 \ $ does not improve matters: with $ \ b \ = \ 2Xa \ \ , \ $ the endpoints must satisfy $ \ a - 2Xa + c \ = \ -2 \ $ and $ \ a + 2Xa + c \ = \ 2 \ \ , \ $ so $$ \ a \ \ = \ \ \frac{1}{X} \ \ \Rightarrow \ \ \frac{1}{X} \ \pm 2X·\frac{1}{X} \ + \ c \ \ = \ \ \pm 2 \ \ \Rightarrow \ \ c \ \ = \ \ -\frac{1}{X} \ \ . $$ This has the effect of "driving" the vertex of $ \ q(x) \ $ outside of $ \ [-2 \ , \ 2] \ $ "to the right" and to larger and larger positive values of $ \ y \ \ . \ $ But its endpoint values are still $ \ (2 \ , \ 4·[a + b + c] = 8) \ $ and $ \ (-2 \ , \ 4·[a - b + c] = -8 ) \ \ . \ $ The value of $ \ q(x) \ $ at its vertex has become irrelevant: its range on $ \ [-2 \ , \ 2] \ $ remains $ \ -8 \ \le \ q(x) \ \le \ 8 \ \ , \ $ and so $ \ | \ q(x) \ | \ \le \ 8 \ \ . $
We conclude that the symmetrical case of $ \ b \ = \ 0 \ \ , \ \ a \ = \ \pm 4 \ \ , \ \ c \ = \ \mp 2 \ $ offers the maximum possible value for $ \ | \ q(x) \ | \ $ of $ \ | \ q(0) \ | \ = \ 16 \ \ . $
On
We have \begin{align*} |a - b + c| &\le 2, \tag{1}\\ |a + b + c| &\le 2, \tag{2}\\ |c| &\le 2. \tag{3} \end{align*}
Using (1) and (2), we have $$|a + c| + |b| \le 2. \tag{4}$$ (Note: If $(a + c)b \ge 0$, then $|a + c| + |b| = |a + c + b| \le 2$. If $(a + c)b < 0$, then $|a + c| + |b| = |a + c - b| \le 2$.)
Using (1) and (3), we have $$|a - b| \le |a - b + c| + |-c| \le 4. \tag{5}$$
Using (2) and (3), we have $$|a + b| \le |a + b + c| + |-c| \le 4. \tag{6}$$
Using (5) and (6), we have $$|a| + |b| \le 4. \tag{7}$$ (Note: If $ab \ge 0$, then $|a| + |b| = |a + b| \le 4$. If $ab < 0$, then $|a| + |b| = |a - b| \le 4$.)
Using (4) and (7), we have, for all $x\in [-2, 2]$, \begin{align*} |cx^2 + 2bx + 4a| &\le |cx^2 + 4a| + |2bx| \\ &\le \max(|4c + 4a|, |4a|) + 4|b| \tag{8}\\ &\le 16 \end{align*} where in (8) we have used $$\max_{x\in [-2, 2]} |cx^2 + 4a| = \max(|c\cdot (-2)^2 + 4a|, |c \cdot 2^2 + 4a|, |c \cdot 0^2 + 4a|).$$
On the other hand, when $a = -4, b = 0, c = 2$, we have $|ax^2 + bx + c| = |-4x^2 + 2|\le 2$ for all $x\in [-1, 1]$, and the maximum of $|cx^2 + 2bx + 4a|= |2x^2 - 16|$ on $[-2, 2]$ is $16$.
Thus, the maximum of $|cx^2 + 2bx + 4a|$ on $[-2, 2]$ is $16$.
A solution using elementary calculus.
First observe by changing variables $x\mapsto\frac{x}{2}$ that the question becomes finding the maximum value of $4|cx^2+bx+a|$ on $[-1,1]$. Let's set $f(x)=ax^2+bx+c$ and $g(x)=cx^2+bx+a$. Also note that by swapping between $f$ and $-f$ (and $g$ and $-g$ respectively), we can assume wlog that $c\geq 0$. Also wlog, since the domains are symmetric around the origin, we can assume $b\geq 0$.
Now there are four main ingredients to the elementary proof.
Step 1: As Maverick pointed out, one can get bounds on the coefficients. Indeed, $a=\frac{f(-1)-2f(0)+f(1)}{2}$, $b=\frac{f(1)-f(-1)}{2}$, and $c=f(0)$ implying with the above that $b,c\in[0,2]$ and $|a|\leq4$.
Step 2: The extremal values a parabola attains on a closed interval are either at the endpoints of the interval or at the global extremum of the parabola.
Therefore \begin{equation*} \max_{[-1,1]}|g(x)|\in\{|g(1)|,|g(-1)|,|g(-\frac{b}{2c})|\}=\{|c+b+a|,|c-b+a|,|a-\frac{b^2}{4c}|\}=\{|f(1)|,|f(-1)|,|a-\frac{b^2}{4c}|\}. \end{equation*}
So $\max_{[-1,1]}|g(x)|\leq \max(2,|a-\frac{b^2}{4c}|)$. As a technicality, if $c=0$, $g$ is linear, so it will attain extremal values only at the endpoints of the interval, so we don't need to worry about $g(-\frac{b}{2c})$.
Step 3: Note that by taking $a=-4,b=0,c=2$, we have that $|g(0)|=4$, so we are only further interested in looking at the case when the extremal value of $|g|$ is attained at the global extremal value of $g$ and when this value is attained inside the interval $[-1,1]$.
This implies $c>0$ and $-1\leq -\frac{b}{2c}\leq 0$. Observe that if $g(-\frac{b}{2c})\geq 0$, since we assumed $c\geq0$, $g$ increases away from $-\frac{b}{2c}$ and therefore $g(1)$ and $g(-1)$ will both be bigger than $g(-\frac{b}{2c})$. Therefore we can restrict our analysis to the case when $g(-\frac{b}{2c})<0$.
Step 4: Arguing that $g$ is bounded below on $[-1,1]$ by $-4$.
Assume this is not the case. Then necessarily $g(-\frac{b}{2c})<-4$. We now apply the mean value theorem on the interval $[-1,-\frac{b}{2c}]\subset [-1,0]$ to find some $t\in[-1,0]$ such that
\begin{equation*} g'(t)=\frac{g(-\frac{b}{2c})-g(-1)}{-\frac{b}{2c}-(-1)}\leq g(-\frac{b}{2c})-g(-1)<- 2, \end{equation*} because $0\geq-\frac{b}{2c}\geq -1, g(-1)=f(-1)\geq -2$ and because we assumed $g(-\frac{b}{2c})<-4$.
But this implies that $2ct+b<-2$, or, since $t\in[-1,0]$, that $c>\frac{b}{2|t|}+\frac{1}{|t|}\geq\frac{1}{|t|}\geq 1$.
Now since we assumed that $g(-\frac{b}{2c})<-4$, we have
\begin{equation*} -4>a-\frac{b^2}{4c}=(a-b+c)+b-c-\frac{b^2}{4c}\geq -2 +b-c-\frac{b^2}{4c}. \end{equation*}
Rearranging, this means $\frac{1}{4c}b^2-b+c-2>0$. Viewing this as a degree $2$ inequality in $b$, we can compute the discriminant $\Delta=\sqrt{\frac{2}{c}}$ and roots $b_{1,2}=\frac{1\pm\sqrt{\frac{2}{c}}}{\frac{1}{2c}}=2c\pm 2\sqrt{2c}$. For the inequality to hold, we must therefore have that either $b<2c-2\sqrt{2c}=2\sqrt{c}(\sqrt{c}-\sqrt{2})<0$ (since $c\in(1,2]$), impossible since $b\geq 0 $, or that $b>2c+2\sqrt{2c}>2c>2$ (since $c>1$), impossible as $b\leq 2$. So we have a contradiction, concluding Step 4.
This means that the required maximum is $16$ and it can be achieved by taking $a=-4,b=0,c=2$.