Example 2 Find the largest real number $m$, such that as long as the quadratic function $f(x)=a x^{2}+b x+c$ satisfies the conditions
(1) $a>0$;
(2) $|f(x)| \leqslant 1$, when $x \in[-1,1]$;
(3) $x_{0}=-\frac{b}{2 a} \in[-2,-1) \cup(1,2]$,

it must be true that $f(x) \geqslant m(x \in[-2,2])$.
Analysis: $m_{\text {max }}=f(x)_{\text {min }}=f\left(x_{0}\right)$, so, the quadratic function can be written in vertex form. Without loss of generality, we can assume $x_{0}=-\frac{b}{2 a} \in(1,2]$. From the graph of $f(x)=a x^{2}+b x+c$, we know that $f(x)$ is monotonically decreasing on $[-1,1]$.
Therefore, condition $(2) \Leftrightarrow\left\{\begin{array}{l}f(1) \geqslant-1, \\ f(-1) \leqslant 1 .\end{array}\right.$
Solution: Let $f(x)=a\left(x-x_{0}\right)^{2}+d$. Without loss of generality, assume $x_{0}=-\frac{b}{2 a} \in(1,2]$. From (2), we have
$$
\left\{\begin{array}{l}
1 \geqslant f(-1)=a\left(1+x_{0}\right)^{2}+d, \\
-1 \leqslant f(1)=a\left(1-x_{0}\right)^{2}+d .
\end{array}\right.
$$

Thus, $1-a\left(1+x_{0}\right)^{2} \geqslant d \geqslant-1-a\left(1-x_{0}\right)^{2}$. From $1-a\left(1+x_{0}\right)^{2} \geqslant-1-a\left(1-x_{0}\right)^{2}$, we get $a \leqslant \frac{1}{2 x_{0}}$. Therefore,
$$
\begin{array}{l}
d \geqslant-1-a\left(1-x_{0}\right)^{2} \geqslant-1-\frac{1}{2 x_{0}}\left(1-x_{0}\right)^{2} \\
=-1-\frac{x_{0}}{2}\left(1-\frac{1}{x_{0}}\right)^{2},
\end{array}
$$

where $x_{0} \in(1,2]$.
The function $\varphi(x)=-1-\frac{x}{2}\left(1-\frac{1}{x}\right)^{2}$ is a monotonically decreasing function on $(1,2]$, so
$$
\varphi\left(x_{0}\right) \geqslant \varphi(2)=-\frac{5}{4} .
$$

Therefore, $d \geqslant-\frac{5}{4}$.
The function $f(x)=\frac{1}{4}(x-2)^{2}-\frac{5}{4}$ satisfies conditions (1), (2), and (3), and its minimum value on $[-2,2]$ is $-\frac{5}{4}$, hence $m_{\max }=-\frac{5}{4}$.

Note: The solution cleverly uses the monotonicity of the function $\varphi(x)=-1-\frac{x}{2}\left(1-\frac{1}{x}\right)^{2}$ on $(1,2]$.