$$
\text { Three. ( } 20 \text { points) Let } f(x)=x^{2}+p x+q, p, q
$$
$\in \mathbf{R}$. If the maximum value of $|f(x)|$ on $[-1,1]$ is $M$, find the minimum value of $M$.
Three, let $M=\max _{-1 \leqslant x \leqslant 1}|f(x)|$, then
$M \geqslant|f(1)|, M \geqslant|f(0)|, M \geqslant|f(-1)|$,
i.e., $M \geqslant|1+p+q|, M \geqslant|q|, M \geqslant|1-p+q|$,
thus $4 M \geqslant|1+p+q|+2|-q|+|1-p+q|$
$$
\begin{array}{l}
\geqslant|(1+p+q)+(-2 q)+(1-p+q)| \\
=2 .
\end{array}
$$

Therefore, $M \geqslant \frac{1}{2}$.
When $f(x)=x^{2}-\frac{1}{2}$, $\max _{-1 \leqslant x \leqslant 1}|f(x)|=\frac{1}{2}$, then
the minimum value of $M$ is $\frac{1}{2}$.