Example $\mathbf{3}$ Find the function $f: \mathbf{R}_{+} \rightarrow \mathbf{R}_{+}$, such that it satisfies $f(f(x))=6 x-f(x)$.
(30th IMO Canadian Training Problem)
Solution 1: Clearly, $f(x) \leqslant 6 x$, then
$$
6 x-f(x)=f(f(x)) \leqslant 6 f(x) \text {. }
$$

Therefore, $f(x) \geqslant \frac{6}{7} x$.
Thus, $6 x-f(x)=f(f(x)) \geqslant \frac{6}{7} f(x)$.
Hence, $f(x) \leqslant \frac{6 x}{1+\frac{6}{7}}$.
Therefore, $6 x-f(x)=f(f(x)) \leqslant \frac{6 f(x)}{1+\frac{6}{7}}$.
Solving, we get $f(x) \geqslant \frac{6 x}{1+\frac{6}{1+\frac{6}{7}}}$.
Let $a_{n}=\frac{6}{1+a_{n-1}}$, then $a_{1}=6$, and
$a_{2} x \leqslant f(x) \leqslant a_{1} x$,
$a_{4} x \leqslant f(x) \leqslant a_{3} x$,
$\qquad$
$a_{2 n} x \leqslant f(x) \leqslant a_{2 n-1} x$.
It is easy to prove that $\left\{a_{2 n-1}\right\}$ is decreasing, $\left\{a_{2 n}\right\}$ is increasing, and
$$
a_{2 n} < a_{2 n+2} < a_{2 n+1} < a_{2 n-1}.
$$

For $x>0$, define
$$
a_{0}=x, a_{1}=f(x), a_{n+1}=f\left(a_{n}\right)
$$

We get $a_{0}, a_{1}, a_{2}, \cdots$. By the condition
$$
a_{n+2}+a_{n+1}-6 a_{n}=0, n=0,1,2, \cdots,
$$

Solve the characteristic equation $\gamma^{2}+\gamma-6=0$. The characteristic roots are $-3,2$.
Then $a_{n}=c(-3)^{n}+k \times 2^{n}(c, k$ are constants $)$.
Thus, for $n \in \mathbf{N}, a_{n+1}=f\left(a_{n}\right)>0$.
Hence, $c=0$.
By $a_{0}=x$ we get $k=x$. Therefore,
$$
f(x)=f\left(a_{0}\right)=a_{1}=2 x \text {. }
$$

Substituting for verification
$$
f(f(x))=f(2 x)=4 x=6 x-f(x) \text {, }
$$

and for $x>0$, $2 x>0$.
Thus, $f(x)=2 x$ is the unique function that satisfies the conditions.