Example 9 Let $f(x)=\left\{\begin{array}{cc}1 & 1 \leqslant x \leqslant 2 \\ x-1 & 2<x \leqslant 3\end{array}\right.$, for any $a(a \in \mathrm{R})$, denote $v(a)=\max \{f(x)-a x \mid x \in[1,3]\}-\min \{f(x)-a x \mid x \in[1,3]\}$, try to draw the graph of $v(a)$, and find the minimum value of $v(a)$.
When $a \leqslant 0$, $y=-a x$ and $y=f(x)$ are not decreasing functions, in this case we have
$$
\begin{array}{l}
\max \{f(x)-a x \mid x \in[1,3]\}=2-3 a, \\
\min \{f(x)-a x \mid x \in[1,3]\}=1-a .
\end{array}
$$

For the case when $a>0$, since the graph of $f(x)$ undergoes a fundamental change at the point $(2,1)$, we consider two scenarios:
When $0 < a \leqslant \frac{1}{2}$, we have
$$
\max \{f(x)-a x \mid x \in[1,3]\}=
$$
$2-3a$,
$$
\min \{f(x)-a x \mid x \in[1,3]\}=
$$
$1-2a$.
When $a > \frac{1}{2}$, similarly we have
$$
\max \{f(x)-a x \mid x \in[1,3]\}=
$$
$1-a$,
$$
\min \{f(x)-a x \mid x \in[1,3]\}=
$$
$1-2a$.
From the above discussion, we know
$$
v(a)=\left\{\begin{array}{cc}
1-2 a & a \leqslant 0 \\
1-a & 0 < a \leqslant \frac{1}{2} \\
1-2 a & a > \frac{1}{2}
\end{array}\right.
$$

Its graph is shown in Figure 3-2, $v(a)_{\min }=$ $\frac{1}{2}$.