Find all functions $f:\Bbb R \rightarrow \Bbb R$ such that for every $x,y \in \Bbb R$: i) $f(x)\leq x$ ii) $f(x+y)\leq f(x)+f(y) $

Question

Find all functions $f:\Bbb R \rightarrow \Bbb R$ such that for every $x,y \in \Bbb R$:
i) $f(x)\leq x$
ii) $f(x+y)\leq f(x)+f(y) $

In functional equations, I have trouble when inequality has a role...

Answer 1

Let $x=y=0$. Then $f(0+0)\le f(0)+f(0)$, whence $f(0)\ge 0$. Put $y=-x$. Then $0\le f(0)\le f(x)+f(-x)\le x+(-x)=0$ and $f$ is an odd function. Now $f(x)\le x$. But also $f(-x)\le -x$, so $-f(x)\le -x$ (since $f$ was odd). Then $f(x)\ge x$ and as a consequence we have $f(x)=x$ for all $x\in\Bbb R$.

Answer 2

Setting y=0 in ii), you get $f(x) \leq f(x)+f(0)$, so $0 \leq f(0)$. On the other hand, from i) $f(0) \leq 0$, therefore you get $f(0)=0$.

Now set y=-x in ii). The inequality becomes $0 \leq f(x)+f(-x)$. However, from i) $f(x) \leq x, f(-x) \leq -x$, so $f(x)+f(-x) \leq 0$. Therefore from the two inequalities you get $f(x)=x$.