9. (16 points) Let $f(x)=3 x-a x^{3}$,
$$
g(x)=b+3 x+c x^{2}+d x^{3},
$$

where $a, b, c, d \in \mathbf{R}$.
(1) Find the value of $a$ such that when $x \in[-1,1]$,
$$
|f(x)| \leqslant 1 \text{; }
$$
(2) Find the values of $b, c, d$ such that when $x \in[-1,1]$, $|g(x)| \leqslant 1$.
(1) From
$|f(1)|=|3-a| \leqslant 1 \Rightarrow 2 \leqslant a \leqslant 4$.
From $\left|f\left(\frac{1}{2}\right)\right|=\left|\frac{3}{2}-\frac{a}{8}\right| \leqslant 1 \Rightarrow 4 \leqslant a \leqslant 20$.
Thus, $a=4$.
Conversely, when $a=4$, let
$$
x=\sin \theta(0 \leqslant \theta<2 \pi) \text {. }
$$

Then $|f(x)|=\left|3 \sin \theta-4 \sin ^{3} \theta\right|$
$$
=|\sin 3 \theta| \leqslant 1 \text {. }
$$

Therefore, $a=4$.
(2) When $x \in[-1,1]$, $-x \in[-1,1]$, we have $|g(x)| \leqslant 1,|g(-x)| \leqslant 1$.

Then $\left|3 x+d x^{3}\right|=\frac{1}{2}|g(x)-g(-x)|$
$$
\leqslant \frac{1}{2}(|g(x)|+|g(-x)|) \leqslant 1 .
$$

From the conclusion of (1), we know $d=-4$.
Thus, for all $x \in[-1,1]$,
$$
\left|b+3 x+c x^{2}-4 x^{3}\right| \leqslant 1 \text {. }
$$

Taking $x=-1,1$, we get
$$
\begin{array}{l}
\left\{\begin{array} { l } 
{ | b + c + 1 | \leqslant 1 , } \\
{ | b + c - 1 | \leqslant 1 }
\end{array} \Rightarrow \left\{\begin{array}{l}
-2 \leqslant b+c \leqslant 0, \\
0 \leqslant b+c \leqslant 2
\end{array}\right.\right. \\
\Rightarrow b+c=0 .
\end{array}
$$

Taking $x=-\frac{1}{2}, \frac{1}{2}$, we get
$$
\begin{array}{l}
\Rightarrow b+\frac{c}{4}=0 \text {. } \\
\end{array}
$$

Thus, $b$ and $c$ satisfy
$$
\left\{\begin{array}{l}
b+c=0, \\
b+\frac{c}{4}=0
\end{array} \Rightarrow b=c=0 .\right.
$$

Therefore, $b=c=0, d=-4$.
Thus, $g(x)=3 x-4 x^{3}$ satisfies the conditions.