Find all functions $f: \mathbb{R} \mapsto \mathbb{R}$ such that for all $x, y \in \mathbb{R}$,

$$
f(x+y)=x f(x)+y f(y)
$$
We guess (and at the same time verify) that the null function is a solution. Let's show that it is the only one. By setting $y=f(0)$ we have $f(x)=x f(x)$. In other words, for all $x \in \mathbb{R}$,

$$
(x-1) f(x)=0 \text {. }
$$

Thus, by substituting $x$ with any value different from 1, we obtain that the function is zero except possibly at 1. We are not done yet! Indeed, there still exists an infinity of such functions. Let's show that $f(1)=0$. For example, with $x=y=\frac{1}{2}$ we have

$$
f(1)=f\left(\frac{1}{2}+\frac{1}{2}\right)=\frac{1}{2} f\left(\frac{1}{2}\right)+\frac{1}{2} f\left(\frac{1}{2}\right)=0
$$

since $f\left(\frac{1}{2}\right)=0$. Therefore, 0 is the only function solution.