1. Find all functions $f: \mathbf{R} \rightarrow \mathbf{R}$, such that the equation
$$
f([x] y)=f(x)[f(y)]
$$

holds for all $x, y \in \mathbf{R}$ (where $[z]$ denotes the greatest integer less than or equal to the real number $z$).
(French contribution)
1. $f(x)=c$ (constant), where, $c=0$ or $1 \leqslant c<2$.

Let $x=0$ substitute into equation (1) to get
$$
f(0)=f(0)[f(y)]
$$

for all $y \in \mathbf{R}$.
Thus, there are two cases.
(1) When $f(0) \neq 0$, from equation (2) we know $[f(y)]=1$ for all $y \in \mathbf{R}$.
Therefore, equation (1) $\Leftrightarrow f([x] y)=f(x)$.
Let $y=0$, we get $f(x)=f(0)=c \neq 0$.
From $[f(y)]=1=[c]$, we know $1 \leqslant c<2$.
(2) When $f(0)=0$, if there exists $0<\alpha<1$ such that $f(\alpha) \neq 0$, let $x=\alpha$ substitute into equation (1) to get
$$
0=f(0)=f(\alpha)[f(y)]
$$

for all $y \in \mathbf{R}$.
Thus, $[f(y)]=0$ for all $y \in \mathbf{R}$.
Let $x=1$ substitute into equation (1) to get $f(y)=0$ for all $y \in \mathbf{R}$, which contradicts $f(\alpha) \neq 0$.
Therefore, $f(\alpha)=0(0 \leqslant \alpha<1)$.
For any real number $z$, there exists an integer $N$, such that
$$
\alpha=\frac{z}{N} \in[0,1) \text {. }
$$

From equation (1) we have
$$
f(z)=f([N] \alpha)=f(N)[f(\alpha)]=0
$$

for all $z \in \mathbf{R}$.
Upon verification, $f(x)=c$ (constant), where, $c=0$ or $1 \leqslant c<2$ satisfies the problem's conditions.