Assume $f:\mathbb{R}_{>0}\to\mathbb{R}_{>0}$ is a continuous function such that $xf(f(f(x)))=1$ for all $x>0$.

I found that $f(x)=1/x$ is a solution. Could we find another such function? And why?


1 Answer 1


No, $f(x)=1/x$ is the only solution.

The proof consists of many simple steps.

part 1: $f$ is strictly decreasing:

First, we show that $f$ is surjective. Let $x>0$. Then $$ f(f(f(1/x)))(1/x)=1 \quad\Rightarrow\quad f(f(f(1/x)))=x. $$ Thus $f$ is surjective. Next, we show that $f$ is injective. Let $x,y>0$ with $f(x)=f(y)$. Then $$ x f(f(f(y))) = x f(f(f(x))) = 1 = y f(f(f(y))) $$ which implies $x=y$. Thus $f$ is injective and therefore bijective. Since $f$ is bijective and continuous, it has to be either strictly increasing or strictly decreasing. Assume that $f$ is strictly increasing. Since the composition of strictly increasing functions is strictly increasing, it follows that $f\circ f\circ f$ is strictly increasing. Since $x\mapsto x$ is strictly increasing and the multiplication of strictly increasing nonnegative functions is strictly increasing, it follows that $ x \mapsto x f(f(f(x)))=1$ is strictly increasing. This is a contradiction. Thus, $f$ has to be strictly decreasing.

part 2: an indirect proof:

Assume that for some $x>0$ we have $f(x)>1/x$. In the following calculation, each step is either a simple application of $f$ or a multiplication/division by $x$. Since $f$ is strictly decreasing, it means we have to switch the sign each time we apply $f$. We obtain $$ \begin{align} && f(x) &>1/x \\ \Rightarrow && f(f(x)) &< f(1/x) \\ \Rightarrow && f(f(f(x))) &> f(f(1/x)) \\ \Rightarrow && 1 = xf(f(f(x))) &> xf(f(1/x)) \\ \Rightarrow && 1/x &> f(f(1/x)) \\ \Rightarrow && f(1/x) &< f(f(f(1/x))) \\ \Rightarrow && (1/x)f(1/x) &< (1/x)f(f(f(1/x))) = 1 \\ \Rightarrow && f(1/x) &<x \end{align} $$ With this information we can start again. $$ \begin{align} && f(x) &>1/x \\ \Rightarrow && f(f(x)) &< f(1/x) \\ \Rightarrow && f(f(x)) &< x \\ \Rightarrow && f(f(f(x))) &> f(x) > 1/x \\ \Rightarrow && 1=xf(f(f(x))) & > xf(x) > x/x = 1 \end{align} $$ which is a contradiction.

Now assume $f(x)<1/x$ for some $x>0$. Then we can obtain a contradiction with the same argument as above, but exchanging the signs $<$ and $>$.

Thus the only possible option is that $f(x)=1/x$ for all $x$.

  • $\begingroup$ could someone explain why this was downvoted? I would like to know about possible mistakes. $\endgroup$
    – supinf
    Mar 29, 2019 at 15:41
  • 1
    $\begingroup$ For me, the solution is good. Bravo! There are some parenthesis missing (that you can edit), but it is readable and correct. You have proved the surjectivity of $f(f(f(x)))$, so a comment should be added for completeness. For the rest, I am pleased by your proof :) $\endgroup$ Apr 1, 2019 at 11:58

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