Example 6 Find all functions $f$, whose domain is the set of all positive real numbers, and whose values are positive real numbers, satisfying the following conditions: (1) $f[x f(y)]=y f(x)$, for all $x, y \in \mathbf{R}^{+}$; (2) as $x \rightarrow \infty$, $f(x) \rightarrow 0$.
(IMO - 24 Problem)
Analysis When $x=y>0$, we have $f[x f(x)]=x f(x)$, which means $x f(x)$ is a fixed point.
Solution First, we prove that $x=1$ is a fixed point of $f(x)$, i.e., $f(1)=1$. In condition (1), take $x=1, y=\frac{1}{f(1)}$, then $f\left[1 \cdot f\left(\frac{1}{f(1)}\right)\right]=\frac{1}{f(1)} f(1)=1$.

Next, in condition (1), take $y=1, x=\frac{1}{f(1)}$ to get $f\left[\frac{1}{f(1)} \cdot f(1)\right]=1 \cdot f\left(\frac{1}{f(1)}\right)$, i.e., $f(1)=f\left(\frac{1}{f(1)}\right)$.
Comparing (1) and (2), we get $f[f(1)]=1$.
Again, in condition (1), let $x=1, y=1$, then
$f[1 \cdot f(1)]=1 \cdot f(1)$, so $f(1)=1$.
Next, we prove that $x=1$ is the only fixed point.
Assume $x=a$ is a fixed point, i.e., $f(a)=a$.
(i) When $a>1$, then $f\left(a^{n}\right)=f\left(a^{n-1} \cdot a\right)=f\left[a^{n-1} \cdot f(a)\right]=a f\left(a^{n-1}\right)$
$$
=a \cdot f\left(a^{n-2} \cdot a\right)=a \cdot f\left[a^{n-2} \cdot f(a)\right]=\cdots=a^{n},
$$
i.e., $x=a^{n}$ is also a fixed point. And $\lim _{n \rightarrow \infty} f\left(a^{n}\right)=\lim _{n \rightarrow \infty} a^{n}=\infty$, which contradicts condition (2), so $a>1$ is impossible.
(ii) When $0<a<1$, then $f\left(a^{-n}\right)=f\left(a^{-n+1} \cdot a^{-1}\right)=f\left[a^{-n+1} \cdot f(a^{-1})\right]=a^{-1} f\left(a^{-n+1}\right)$
$$
=a^{-1} \cdot f\left(a^{-n+2} \cdot a^{-1}\right)=a^{-1} \cdot f\left[a^{-n+2} \cdot f(a^{-1})\right]=\cdots=a^{-n},
$$
i.e., $x=a^{-n}$ is also a fixed point. And $\lim _{n \rightarrow \infty} f\left(a^{-n}\right)=\lim _{n \rightarrow \infty} a^{-n}=\infty$, which again contradicts condition (2), so $0<a<1$ is also impossible.
Combining (i) and (ii), we conclude that $a=1$, i.e., $x=1$ is the only fixed point.
Finally, let $x=y$, from the given condition (1), we get $f[x f(x)]=x f(x)$, i.e., for any $x>0, x f(x)$ is a fixed point, and since $x=1$ is the only fixed point, we have $x f(x) \equiv 1$, i.e., $f(x)=\frac{1}{x}$.