Exercise 9. Let $\mathbb{R}_{+}^{*}$ be the set of strictly positive real numbers. Find all functions $f: \mathbb{R}_{+}^{*} \longrightarrow \mathbb{R}_{+}^{*}$ such that for all $x, y>0$, we have

$$
f\left(\frac{f(x)}{y f(x)+1}\right)=\frac{x}{x f(y)+1}
$$
Solution to Exercise 9 First, we note that for all $z>0$, the function $x \in \mathbb{R}^{+*} \longmapsto \frac{x}{x z+1} \in ]0 ; z^{-1}[$ is a strictly increasing bijection.

Now, if $f(x)=f\left(x^{\prime}\right)$, we find that

$$
\frac{x}{x f(1)+1}=f\left(\frac{f(x)}{1 \times f(x)+1}\right)=f\left(\frac{f\left(x^{\prime}\right)}{f\left(x^{\prime}\right) \times 1+1}\right)=\frac{x^{\prime}}{x^{\prime} f(1)+1}
$$

thus $x=x^{\prime}$ and $f$ is injective.

On the other hand, fix $y>0$. Each $z$ between 0 and $f(y)^{-1}$ (strictly) can be written as $\frac{x}{x f(y)+1}$ for some $x>0$, for all $00$. We verify that $g$ satisfies the same functional equation as $f$: by applying the initial functional equation to $g(x)$ and $g(y)$, we get

$$
f\left(\frac{x}{x g(y)+1}\right)=\frac{g(x)}{y g(x)+1}
$$

Applying $g$ to both sides, we find that $g$ satisfies the same equation as $f$.

Now let $x, x^{\prime}>0$ such that $x^{\prime}0$ such that $x^{\prime}=$ $\frac{f(x)}{y f(x)+1}$ (it suffices to take $y=\frac{1}{x^{\prime}}-\frac{1}{f(x)}$, so

$$
f\left(x^{\prime}\right)=f\left(\frac{f(x)}{y f(x)+1}\right)=\frac{x}{x f(y)+1}0, f(x) \leqslant x$.

However, we have seen that $g$ is also a solution to the equation, so for all $x>0$, we have $g(x) \leqslant x$. In particular, if $x>0, x=g(f(x)) \leqslant f(x)$. Therefore, we must have $f(x)=x$, so $f$ is the identity, and it is easily verified that the identity is indeed a solution.