# Find $f$ if $f(x)\leq x$ and $f(x+y)\leq f(x)+f(y)$ for all $x,~y\in \mathbb{R}.$

Find the formula of function $$f:\mathbb{R}\to \mathbb{R}$$ if: $$f(x)\leq x$$ and $$f(x+y)\leq f(x)+f(y)$$ for all $$x,~y\in \mathbb{R}.$$

Attempt. Identity function $$I(x)=x$$ satisfies the needed properties. I suspect that is the only one. In that case we need only to show that $$f(x)\geq x$$ for all $$x$$. At this point though, I couldn't use sulinearity of $$f$$ to prove my statement. Any help is appreciated.

• Step 1: Prove $f(0)=0$. Say if you need further or detailed hints. – Ingix Jan 23 at 13:05
Since $$f(0)=f(0+0)\leq f(0)+f(0)$$ we get $$f(0)\geq 0$$.
Since $$f(0)\leq 0$$ by hypothesis, we get $$f(0)=0.$$ So for all $$x$$: $$0=f(0)=f(x+(-x))\leq f(x)+f(-x),$$ so: $$f(x)\geq -f(-x)\geq -(-x)=x.$$ Since $$f(x)\leq x$$ by hypothesis, we get $$f(x)=x$$ for all $$x.$$