Example 1 (2005 China Southeast Mathematical Olympiad Winter Camp Pre-competition Training $A$ Paper) Find the function $f: \mathbf{R} \rightarrow \mathbf{R}$, satisfying:
$$
x[f(x+1)-f(x)]=f(x), \forall x \in \mathbf{R} \text { and }|f(x)-f(y)| \leqslant|x-y|, \forall x, y \in \mathbf{R} \text {. }
$$
Solving for $x=0$ gives $f(0)=0 ;$ when $x \neq 0,-1$, it can be rewritten as: $\frac{f(x+1)}{x+1}=\frac{f(x)}{x}$. In particular, for any $x>0$, and $n \in \mathbf{N}^{+}$, we have $\frac{f(x+n)}{x+n}=\frac{f(x)}{x}$,

Then
$$
\begin{aligned}
|x-y| & \geqslant|f(x+n)-f(y+n)| \\
& =\left|\frac{x+n}{x} f(x)-\frac{y+n}{y} f(y)\right| \\
& =\left|f(x)-f(y)+n\left[\frac{f(x)}{x}-\frac{f(y)}{y}\right]\right| .
\end{aligned}
$$

Thus $\frac{|x-y|}{n} \geqslant\left|\frac{f(x)-f(y)}{n}+\left[\frac{f(x)}{x}-\frac{f(y)}{y}\right]\right|$,
letting $n \rightarrow+\infty$, we get $\frac{f(x)}{x}=\frac{f(y)}{y}, \forall x, y>0$, i.e., $f(x)=k x, x>0$.