Proving that a certain continuous function is surjective.

Question

Let $f:\mathbb R \to \mathbb R$ be a continuous function such that $|f(x)-f(y)|≥|x-y| ,\forall x,y \in \mathbb R $ , then how do we prove that $f$ is surjective ?

Answer 1

$|f(x)-f(y)|≥|x-y| ,\forall x,y \in \mathbb R $

For $y=0=> |x|\leq |f(x)-f(0)|<=> -|f(x)-f(0)|\leq x\leq |f(x)-f(0)|,\forall x \in \mathbb R$

In a previous review we have shown that f is strictly monotone. We have two cases:

I. $f$ f is strictly increasing:

a) For $x>0=>f(x)-f(0)>0=>x\leq f(x)-f(0)=>\lim_{x\to\infty} f(x)=\infty;$

b) For $x<0=>f(x)-f(0)<0=>f(x)-f(0)\leq x=>\lim_{x\to-\infty} f(x)=-\infty.$

Applying intermediate value theorem it follows that $f$ is surjective.

II. $f$ f is strictly decreasing:

a) For $x>0=>f(x)-f(0)<0=>x\leq f(0)-f(x)=>f(x)\leq f(0)-x=>\lim_{x\to\infty} f(x)=-\infty;$

b) For $x<0=>f(x)-f(0)>0=>-f(x)+f(0)\leq x=>-x+f(0)\leq f(x)=>\lim_{x\to-\infty} f(x)=\infty.$

Applying intermediate value theorem it follows that $f$ is surjective.

Answer 2

WLOG let's say $f(0) = 0$, and $f$ is monotonically increasing (why?). Let $x$ be any positive real number - then $f(x)>x$. You can then use the Intermediate Value Theorem to show that $\exists y\in(0,x)$ such that $f(y) = x$. Can you fill in the rest of this proof?