MHB How can we prove $|Cx^2+Bx+A|\le 2$ for the given condition?

[anemone]

The real numbers $A,\,B,\,C$ satisfy the condition such that for $-1\le x\le 1$, the inequality $|Ax^2+Bx+C|\le 1$ holds.

Prove that $|Cx^2+Bx+A|\le 2$ for the same $x$.

 

[Albert1]

The real numbers $A,\,B,\,C$ satisfy the condition such that for $-1\le x\le 1$, the inequality $|Ax^2+Bx+C|\le 1$ holds.

Prove that $|Cx^2+Bx+A|\le 2$ for the same $x$.

Spoiler

let :$f(x)=Ax^2+Bx+C,\,\,\,g(x)=Cx^2+Bx+A$
then we have :$-1\leq f(0)=C\leq 1----(1)$
$-1\leq f(1)=A+B+C\leq 1---(2)$
$-1\leq f(-1)=A-B+C\leq 1---(3)$
now $-2<-1\leq g(1)=C+B+A=A+B+C\leq 1<2---(4)$
$-2<-1\leq g(-1)=C-B+A=A-B+C\leq 1<2---(5)$
as for the domain of $A$: to satisfy $|Cx^2+Bx+A|\le 2$
we get: $-2\leq g(0)=A\le 2$
for one case :$f(x)=2x^2-1,\,\,\, g(x)=-x^2+2$

 

[Opalg]

[sp]I'll use Albert's notation $f(x) = Ax^2 + Bx + C$, $g(x) = Cx^2 + Bx + A$. Notice that $g(-1) = f(-1) = A-B+C$ and $g(1) = f(1) = A+B+C$. Therefore $|g(-1)| \leqslant 1$ and $|g(1)| \leqslant 1.$ Also, $C = f(0)$ and so $|C| \leqslant 1$.

Suppose that for some $x$ with $-1< x< 1$, $g(x) >2$. Then $g(x) - g(1) > 2-1 = 1$, so that $\dfrac{g(x) - g(1)}{1-x} > \dfrac1{1-x}.$ Similarly, $g(x) - g(-1) >1$ so that $\dfrac{g(x) - g(-1)}{1+x} > \dfrac1{1+x}.$ Therefore $$\dfrac{g(x) - g(1)}{1-x} + \dfrac{g(x) - g(-1)}{1+x} > \dfrac1{1-x} + \dfrac1{1+x} = \dfrac2{1-x^2} >2.$$
On the other hand, $g(x) - g(1) = Cx^2 + Bx + A - (C+B+A) = C(x^2-1) + B(x-1)$, so that $\dfrac{g(x) - g(1)}{1-x} = -C(x+1) - B.$

Similarly, $g(x) - g(-1) = C(x^2-1) + B(x+1)$, so that $\dfrac{g(x) - g(-1)}{1+x} = C(x-1) + B.$

Therefore $\dfrac{g(x) - g(1)}{1-x} + \dfrac{g(x) - g(-1)}{1+x} = -2C \leqslant 2$ (because $|C|\leqslant 1).$ That contradicts the previous inequality and shows that $g(x)$ can never be greater than $2$.

A similar argument (with most of the signs changed) shows that $g(x)$ can never be less than $-2$. In conclusion, $|g(x)| \leqslant 2$ whenever $-1\leqslant x\leqslant 1.$

Albert's example $f(x) = 2x^2-1$ shows that $|g(x)|$ can attain the value $2$, so $2$ is the best possible constant in this result.

The basic idea in the above proof is that the constant $C$ controls the curvature of the function $g$. Since $|C|\leqslant 1$, the curve cannot bend too quickly. If $g(x)$ takes a value $\leqslant 1$ when $x=-1$, then increases to a value greater than $2$ (at some point $x$) then it cannot bend round sharply enough to get back down to a value $\leqslant 1$ by the time that $x=1$.[/sp]

 

[Albert1]

Spoiler

let :$f(x)=Ax^2+Bx+C,\,\,\,g(x)=Cx^2+Bx+A$
then we have :$-1\leq f(0)=C\leq 1----(1)$
$-1\leq f(1)=A+B+C\leq 1---(2)$
$-1\leq f(-1)=A-B+C\leq 1---(3)$
now $-2<-1\leq g(1)=C+B+A=A+B+C\leq 1<2---(4)$
$-2<-1\leq g(-1)=C-B+A=A-B+C\leq 1<2---(5)$
as for the domain of $A$: to satisfy $|Cx^2+Bx+A|\le 2$
we get: $-2\leq g(0)=A\le 2$
for one case :$f(x)=2x^2-1,\,\,\, g(x)=-x^2+2$

another proof for $-2\leq A\leq 2$

Spoiler

$-1\leq A+B+C\leq 1---(1)$
$-1\leq A-B+C\leq 1---(2)$
$-1\leq C\leq 1---(3)$
$(1)+(2)$ we have :
$-1\leq A+C\leq 1---(4)$
fom $(3),(4)$
$A-1\leq A+C\leq 1, \rightarrow A\leq 2$
$-1\leq A+C\leq A+1,\rightarrow -2\leq A$

 

[anemone]

Thank you Albert and Opalg for participating in this challenge problem!

Another solution of other that I wish to share with MHB:

Spoiler

Using also Albert's notation, we have

$|A+B+C|=|f(1)|\le 1$, $|A-B+C|=|f(-1)|\le 1$ and $|C|=|f(0)|<1$

Hence,

$\begin{align*}|g(x)|&=\left|C\cdot(x^2-1)+(A+B+C)\cdot\left(\dfrac{1+x}{2}\right)+(A-B+C)\cdot\left(\dfrac{1-x}{2}\right)\right|\\&\le |x^2-1|+\dfrac{|1+x|}{2}+\dfrac{|1-x|}{2}\\&=1-x^2+\dfrac{1+x}{2}+\dfrac{1-x}{2}\\&=2-x^2\\&\le 2\end{align*}$

for $-1\le x\le 1$, and we're done.