5. determine all functions $f: \mathbb{R}_{\geq 0} \rightarrow \mathbb{R}_{\geq 0}$ with the following properties:

(a) $f(1)=0$,

(b) $f(x)>0$ for all $x>1$,

(c) For all $x, y \geq 0$ with $x+y>0$ applies

$$
f(x f(y)) f(y)=f\left(\frac{x y}{x+y}\right) .
$$
## Solution

With $y=1$ it follows from (a) and (c) for all $x \geq 0$

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

The expression $\frac{x}{x+1}$ takes on all values in the interval $[0,1[$ if $x$ passes through all non-negative real numbers. Thus, together with (a)

$$
f(x)=0, \quad 0 \leq x \leq 1
$$

For the following, we always assume $y \geq 1$. Because $f(y) \neq 0$, $f(x f(y))=0$ applies if and only if $f\left(\frac{x y}{x+y}\right)=0$. Because of (1), the following also applies

$$
x f(y) \leq 1 \quad \Longleftrightarrow \quad \frac{x y}{x+y} \leq 1 \quad \Longleftrightarrow \quad x \leq \frac{y}{y-1}
$$

In other words

$$
f(y)\left\{\begin{array}{l}
\leq \\
>
\end{array}\right\} z \quad \text { for all } z \text { with } \quad z\left\{\begin{array}{l}
\geq \\
1$. The only function that can fulfill the equation is therefore

$$
f(x)= \begin{cases}0 & \text { for } 0 \leq x \leq 1 \\ 1-\frac{1}{x} & \text { for } x>1\end{cases}
$$

We now show that this really is a solution and distinguish between two cases.

(1) Let $\frac{1}{x}+\frac{1}{y} \geq 1$, then $\frac{x y}{x+y} \leq 1$ and therefore $f\left(\frac{x y}{x+y}\right)=0$. If $y \leq 1$, then $f(y)$ also disappears and the left-hand side of the equation is also 0 . If $y>1$, then $x f(y)=x\left(1-\frac{1}{y}\right) \leq x \cdot \frac{1}{x}=1$, so this time $f(x f(y))=0$ and the left-hand side of the equation disappears again.

(2) Now let $\frac{1}{x}+\frac{1}{y}1$ and the same calculations as in the first case show that $x f(y)>1$. Consequently, the following applies

$$
\begin{aligned}
f(x f(y)) f(y) & =\frac{x f(y)-1}{x f(y)} \cdot \frac{y-1}{y}=\frac{x \frac{y-1}{y}-1}{x \frac{y-1}{y}} \cdot \frac{y-1}{y}=\frac{x y-x-y}{x y} \\
f\left(\frac{x y}{x+y}\right) & =\frac{x y-x-y}{x y}
\end{aligned}
$$

as desired. That shows everything.