47 Find all numbers $a, b, c, d$, such that (i) the function $f(x)=4 x^{3}-d x$ satisfies the inequality $|f(x)| \leqslant 1$ for $x \in[-1,1]$; (ii) the function $g(x)=4 x^{3}+a x^{2}+b x+c$ satisfies the inequality $|g(x)| \leqslant 1$ for $x \in[-1,1]$.
47 From the given
$$
\left\{\begin{array}{l}
|f(1)|=|4-d| \leqslant 1 \\
\left|f\left(\frac{1}{2}\right)\right|=\left|4 \cdot \frac{1}{8}-\frac{d}{2}\right| \leqslant 1,
\end{array}\right.
$$

we can solve to get $-1 \leqslant d \leqslant 3$ and $3 \leqslant d \leqslant 5$, hence $d=3$. Let $x=\cos \alpha$, then we know $d=3$ satisfies the condition. For $x \in[-1,1]$, we have
$$
|g(x)| \leqslant 1,
$$

thus
$$
|g(-x)| \leqslant 1
$$

Therefore,
$$
\begin{aligned}
\left|4 x^{3}+b x\right| & =\frac{1}{2}|g(x)-g(-x)| \\
& \leqslant \frac{1}{2}(|g(x)|+|g(-x)|) \\
& \leqslant \frac{1}{2} \times 2=1 .
\end{aligned}
$$

From the above, we know $b=-3$. Thus we have
$$
\left|4 x^{3}+a x^{2}-3 x+c\right| \leqslant 1
$$

Let $x=1, -1$ to get
$$
|1+a+c| \leqslant 1,|-1+a+c| \leqslant 1
$$

Thus $a+c=0, c=-a$.
Let $x=\frac{1}{2},-\frac{1}{2}$ to get
$$
\begin{array}{l}
-1 \leqslant-1+\frac{a}{4}-a \\
1+\frac{a}{4}-a \leqslant 1
\end{array}
$$

Thus $a=0, c=0$.