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

$$
f(\lfloor x\rfloor y)=f(x)\lfloor f(y)\rfloor
$$
For $x=y=0$, we have:

$$
f(0)=f(0)\lfloor f(0)\rfloor
$$

Thus, $f(0)=0$ or $\lfloor f(0)\rfloor=1$.

Suppose $\lfloor f(0)\rfloor=1$. With $y=0$, we get $f(x)=f(0)$, so $f$ is constant. Since $\lfloor f(0)\rfloor=1$, we have $f=c$ with $c \in[1,2[$. The initial equation is then satisfied.

Now suppose $f(0)=0$. Taking $x=y=1$, we similarly have

$$
f(1)=f(1)\lfloor f(1)\rfloor
$$

and thus $f(1)=0$ or $\lfloor f(1)\rfloor=1$.

If $\lfloor f(1)\rfloor=1$, with $y=1$ and $0 \leqslant x<1$, we have:

$$
f(0)=f(x)\lfloor f(1)\rfloor
$$

hence $f(x)=0$. Taking now $x=2$ and $y=\frac{1}{2}$, we get $f(1)=0$. This is absurd.

Therefore, $f(1)=0$ and with $x=1$, we have $f=0$ and the initial equation is satisfied.

The solutions are therefore the constant functions $f=c$ with $c=0$ or $c \in[1,2[$.

## 7 Polynomials (Jean)