14. Given $a, b, c \in \mathbf{R}$, functions $f(x)=a x^{2}+b x+c, g(x)=a x+b$, when $|x| \leqslant$ 1, $|f(x)| \leqslant 1$.
（1）Prove: $|c|0$, when $|x| \leqslant 1$, the maximum value of $g(x)$ is 2, find $f(x)$.
14. (1) Let $x=0,|f(0)| \leqslant 1$, i.e., $|c| \leqslant 1$.
(2) $\left\{\begin{array}{l}f(0)=c \\ f(-1)=a-b+c \\ f(1)=a+b+c,\end{array}\right.$ we get $\left\{\begin{array}{l}a=\frac{1}{2}[f(1)+f(-1)-2 f(0)] \\ b=\frac{1}{2}[f(1)-f(-1)] .\end{array}\right.$ When $|x| \leqslant 1$,
$$
\begin{aligned}
|g(x)| & =|a x+b|=\left|\frac{1}{2}[f(1)+f(-1)-2 f(0)] x+\frac{1}{2}[f(1)-f(-1)]\right| \\
& =\frac{1}{2}|(x+1) f(1)+(x-1) f(-1)-2 x f(0)| \\
& \leqslant \frac{1}{2}(|x+1||f(1)|+|x-1||f(-1)|+2|x||f(0)|) \\
& \leqslant \frac{1}{2}(x+1+1-x+2)=2 .
\end{aligned}
$$
(3) From the problem, we have $2=g(1)=a+b=f(1)-f(0)$, then $-1 \leqslant f(0)=f(1)-2 \leqslant 1-2 \leqslant$ -1, so $c=f(0)=-1$. Also, when $|x| \leqslant 1$, $|f(x)| \leqslant 1$, thus $f(x) \geqslant-1=c=f(0)$, which means $f(0)$ is the minimum value of $f(x)$. $-\frac{b}{2 a}=0, b=0$. Therefore, $a=2, f(x)=2 x^{2}-1$.