1983 IMO Problems/Problem 1

Find all functions $f$ defined on the set of positive reals which take positive real values and satisfy: $f(xf(y))=yf(x)$ for all $x,y$ ; and $f(x)\to0$ as $x\to \infty$ .

Let $x=y=1$ and we have $f(f(1))=f(1)$ . Now, let $y=f(1)$ and we have $f(xf(f(1)))=f(1)f(x)$ which simplifies to $f(x)=f(1)f(x)$ and since $f(x)\ne0$ we have $f(x)=1$ .

Plug in $y=x$ and we have $f(xf(x))=xf(x)$ . If $a=1$ is the only solution to $f(a)=a$ then we have $xf(x)=1\Rightarrow f(x)=\frac{1}{x}$ . We prove that this is the only function by showing that there does not exist any other $a$ :

Suppose there did exist such an $a\ne1$ . Then, letting $y=a$ in the functional equation yields $f(xa)=af(x)$ . Then, letting $x=\frac{1}{a}$ yields $f(\frac{1}{a})=\frac{1}{a}$ . Notice that since $a\ne1$ , one of $a,\frac{1}{a}$ is greater than $1$ . Let $b$ equal the one that is greater than $1$ . Then, we find similarly (since $f(b)=b$ ) that $f(xb)=bf(x)$ . Putting $x=b$ into the equation, yields $f(b^2)=b^2$ . Repeating this process we find that $f(b^{2^k})=b^{2^k}$ for all natural $k$ . But, since $b>1$ , as $k\to \infty$ , we have that $b^{2^k}\to\infty$ which contradicts the fact that $f(x)\to0$ as $x\to \infty$ .

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1983 IMO Problems/Problem 1