1. Given
(1) $a>0$;
(2) When $-1 \leqslant x \leqslant 1$, it satisfies
$$
\left|a x^{2}+b x+c\right| \leqslant 1 \text {; }
$$
(3) When $-1 \leqslant x \leqslant 1$, $a x+\dot{b}$ has a maximum value of 2. Find the constants $a, b, c$.
1. From (1), we know that $y=a x^{2}+b x+c$ is a parabola opening upwards. From (1) and (3), we have
$$
a+b=2 \text{. }
$$

From (2), we have
$$
\begin{array}{l}
|a+b+c| \leqslant 1, \\
|c| \leqslant 1 .
\end{array}
$$

From (1) and (2), we have
$$
|2+c| \leqslant 1 \text{. }
$$

From (3) and (4), we get $c=-1$. Therefore, when $x=0$, $y=a x^{2}+b x+c$ reaches its minimum value.
Thus, $-\frac{b}{2 a}=0, b=0$.
From (1), we get $a=2$. Hence, $f(x)=2 x^{2}-1$.