12. (GBR 4) $)^{\mathrm{MmO1}}$ Find all functions $f$ defined on the positive real numbers and taking positive real values that satisfy the following conditions:
(i) $f(x f(y))=y f(x)$ for all positive real $x, y$.
(ii) $f(x) \rightarrow 0$ as $x \rightarrow+\infty$.
12. Putting $y=x$ in (1) we see that there exist positive real numbers $z$ such that $f(z)=z$ (this is true for every $z=x f(x)$ ). Let $a$ be any of them.

Then $f\left(a^{2}\right)=f(a f(a))=a f(a)=a^{2}$, and by induction, $f\left(a^{n}\right)=a^{n}$. If $a>1$, then $a^{n} \rightarrow+\infty$ as $n \rightarrow \infty$, and we have a contradiction with (2). Again, $a=f(a)=f(1 \cdot a)=a f(1)$, so $f(1)=1$. Then, $a f\left(a^{-1}\right)=$ $f\left(\alpha^{-1} f(a)\right)=f(1)=1$, and by induction, $f\left(\alpha^{-n}\right)=a^{-n}$. This shows that $a<1$. Hence, $a=1$. It follows that $x f(x)=1$, i.e., $f(x)=1 / x$ for all $x$. This function satisfies (1) and (2), so $f(x)=1 / x$ is the unique solution.