IMO 1983 · Problem 1

Find all functions \(f\) defined on the set of positive real numbers which take positive real values and satisfy the conditions:

(i) \(f(xf(y))=yf(x)\) for all positive \(x,y\);

(ii) \(f(x)\to0\) as \(x\to\infty\).

Official text of the problem. Official problem paper (PDF)

Answer

\[ \boxed{f(x)=\frac1x\quad\text{for every }x>0.} \]

Solution

First, we prove that any fixed point of \(f\) must equal \(1\). Suppose \(a>0\) and \(f(a)=a\). Setting \(y=a\) in (i) gives \[ f(ax)=a f(x)\qquad(x>0). \] Taking \(x=1\), we obtain \(a=a f(1)\), so \(f(1)=1\). Also, replacing \(x\) by \(x/a\) yields \[ f(x/a)=\frac{f(x)}a. \] Iterating these two identities from \(f(1)=1\), we obtain \[ f(a^n)=a^n\qquad\text{for every integer }n. \]

If \(a>1\), then \(a^n\to\infty\) as \(n\to\infty\), so condition (ii) requires \(f(a^n)\to0\), contradicting \(f(a^n)=a^n\to\infty\). If \(0<a<1\), the same contradiction follows from the sequence \(a^{-n}\). Therefore \(a=1\).

Now let \(x>0\). Setting \(y=x\) in (i) gives \[ f(xf(x))=xf(x). \] Thus the positive number \(xf(x)\) is a fixed point of \(f\), and hence \(xf(x)=1\). Therefore \[ f(x)=\frac1x\qquad(x>0). \]

Conversely, this function takes positive values and satisfies \[ f(xf(y))=f(x/y)=\frac yx=yf(x), \] as well as \(f(x)\to0\) as \(x\to\infty\). It is therefore the unique solution.

The solution was checked against John Scholes' IMO solution archive (Kalva).