Find all increasing functions $f: \mathbb{R}^{*} \rightarrow \mathbb{R}^{*}$ such that for all non-zero real numbers $x, y$, we have $f(x+y)(f(x)+f(y))=f(x) f(y)$.
Keys: change of function, return to a known equation.

Let $g(x)=1 / f(x)$ for all real $x$. The statement gives $g(x+y)=g(x)+g(y)$. Thus, $g$ satisfies the Cauchy equation...on $\mathbb{R}^{*}$, not on $\mathbb{R}$! We cannot directly conclude that $g$ is linear on $\mathbb{Q}$: a standard property of functions satisfying the Cauchy equation on a set containing at least $\mathbb{Q}$, but here, we cannot choose $x=0$ or $y=0$. Thus, we have fewer constraints on $g$. However, we can prove, as for the Cauchy equation on $\mathbb{Q}$, that $g$ is linear on $\mathbb{Q}_{+}^{*}$ (prove by induction on $\mathbb{N}^{*}$ that $g(n)=n g(1)$, and that $g(1)=n g(1 / n)$, then write that $g(p / q)=p g(1 / q)=p / q g(1)$ for all $\left.p, q \in \mathbb{N}^{*}\right)$ and similarly on $\mathbb{Q}_{-}^{*}$. Therefore, there exist $a, b \in \mathbb{R}^{*}$ such that $g(x)=a x$ if $x>0$ and $g(x)=b x$ if $x>0$. By setting $x=2$ and $y=-1$ in the equation $g(x+y)=g(x)+g(y)$, we get $a=2 a-b$ hence $a=b$.

Thus $g$ is linear on $\mathbb{Q}^{*}$. Moreover, it is increasing. Thus $a>0$, and for any non-zero real $r$, for all rationals $q \leqslant r \leqslant q^{\prime}$, we have $a q \leqslant g(r) \leqslant a q^{\prime}$. We can take $q$ and $q^{\prime}$ arbitrarily close to $r$ since $\mathbb{Q}$ is dense in $\mathbb{R}$. Thus, $g(r)=a r$ for any non-zero real $r$.

Therefore,

$$
f(x)=\frac{1}{a x}
$$

for all $x \in \mathbb{R}^{*}$. Conversely, all these functions (for $a>0$) are solutions to the problem.

## Remark 1.3.2.

The Cauchy equation $f(x+y)=f(x)+f(y)$ admits only linear functions as solutions on $\mathbb{Q}$. On $\mathbb{R}$, there exist pathological cases of non-linear solutions... but they cannot be constructed! You can consult this USAMO handout (in English) for more details.

## - Exercises -

In class, we covered exercises $1,4,5,8,9$.