1. Find all functions $f: \mathbb{R} \rightarrow \mathbb{R}$ such that for all $x \in \mathbb{R}$, $f(x) \leqslant x$ and for all $x, y \in \mathbb{R}$, $f(x+y) \leqslant f(x)+f(y)$.
1. For $x=0$ we get from the first condition $f(0) \leqslant 0$, and from the second condition $f(y) \leqslant$ $f(0)+f(y)$, i.e., $0 \leqslant f(0)$, so together we have $f(0)=0$. Now for $y=-x$, using both given conditions, we get

$$
0=f(0)=f(x+(-x)) \leqslant f(x)+f(-x) \leqslant x+(-x)=0
$$

from which we conclude $f(x)=-f(-x)$ for all $x \in \mathbb{R}$. Since from the condition in the problem statement it follows that $f(-x) \leqslant -x$, we get

$$
x \leqslant -f(-x)=f(x) \leqslant x
$$

(at the end we again used the condition from the problem statement). From this, it directly follows that $f(x)=x$ for all $x \in \mathbb{R}$.