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.

Thanks in advance.

  • 4
    $\begingroup$ Step 1: Prove $f(0)=0$. Say if you need further or detailed hints. $\endgroup$ – Ingix Jan 23 at 13:05
  • $\begingroup$ Very helpful indeed. I will post an answer. Thank you. $\endgroup$ – Nikolaos Skout Jan 23 at 13:28

Following the useful hint by @Ingix, I post an answer of the exercise.

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 $x.$


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