# $xf(f(f(x)))=1$ continuous

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?

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) & 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 $$x$$.

• could someone explain why this was downvoted? I would like to know about possible mistakes. Mar 29, 2019 at 15:41
• 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 :) Apr 1, 2019 at 11:58