Problem 1136

[Proof 1] Let $f(x)=a x^{2}+b x+c, g(x)=c x^{2}+b x+a$, then
$$|g(1)|=|f(1)| \leqslant 1,|g(-1)|=|f(-1)| \leqslant 1 \text {. }$$

If $c=0$ or $c \neq 0$ but the maximum or minimum value of $g(x)$ is not reached within $(-1,1)$, then by the properties of linear or quadratic functions, we have
$$|g(x)|=\left|c x^{2}+b x+a\right| \leqslant 1 \text {, for any }-1 \leqslant x \leqslant 1 \text {. }$$

If $c \neq 0$ and the extremum of $g(x)$ is reached within $(-1,1)$, let $x_{0}$ be the extremum point, then
$$g(x)=c\left(x-x_{0}\right)^{2}+g\left(x_{0}\right) \text {. }$$

Thus, $\quad\left|g\left(x_{0}\right)\right| \leqslant|g(1)|+|c|\left|1-x_{0}\right|^{2} \leqslant 1+|c|$, and $|c|=|f(0)| \leqslant 1$.

Therefore, $\left|g\left(x_{0}\right)\right| \leqslant 2$. By the properties of quadratic functions, we have
$$|g(x)| \leqslant \max \left(\left|g\left(x_{0}\right)\right|,|g(1)|,|g(-1)|\right) \leqslant 2 \text {, for any }$$
$-1 \leqslant x \leqslant 1$.
[Proof 2] Let $f(x)=a x^{2}+b x+c$, by the assumption we have
$$\begin{array}{l} |f(0)|=|c| \leqslant 1, \\ |f(1)|=|a+b+c| \leqslant 1, \\ |f(-1)|=|a-b+c| \leqslant 1 . \end{array}$$

Since $c x^{2}+b x+a=c\left(x^{2}-1\right)+(a+b+c) \frac{1+x}{2}+(a-b+c)$
$$\frac{1-x}{2},$$

for any $x \in[-1,1]$ we have
$$\begin{aligned} \left|c x^{2}+b x+a\right| \leqslant & |c|\left|x^{2}-1\right|+|a+b+c|\left|\frac{1+x}{2}\right|+\mid a- \\ & \left.b+c|| \frac{1-x}{2} \right\rvert\, \\ \leqslant & \left|x^{2}-1\right|+\left|\frac{1+x}{2}\right|+\left|\frac{1-x}{2}\right| \\ & =1-x^{2}+\frac{1+x}{2}+\frac{1-x}{2} \\ & =2-x^{2} \\ & \leqslant 2 \end{aligned}$$