analysis problem of continity

Question

Let $f:\mathbb R\to \mathbb R$ be a continuous function such that $|f(x)-f(y)|\ge |x-y|$.

Show that $f$ is onto.

Answer 1

I will assume you meant $|f(x)-f(y)|\ge|x-y|$.

If $f(x)=f(y)$, we have that $0\ge|x-y|\ge0$ so that $x=y$ and f is injective, so it is always increasing or decreasing. Assume it is increasing. Then for any $x>0$ you have $f(x)-f(0)\ge0$ So it is equal to it's absolute value and by your assumption $f(x)-f(0)\ge x$, so that $[f(0),\infty)$ is in the range of f by the intermediate value theorem. Similarly, you can prove that $(-\infty,f(0)]$ is in the range of f.

Answer 2

Assume (without loss of generality) that there exist values of $x,y$ such that $x>y$ and $f(x)>f(y)$. Now, using continuity of $f$ and your condition, you can show that $f$ is always increasing, so you now have a function where, for every $x>y$, you know that

$$f(x)-f(y) > x-y$$

Fixing $y=0$, for example, gives you $f(x) > x + f(0)$, meaning that $f$ covers all values from $f(0)$ to $\infty$. Similarly, fixing $x=0$, you have $f(y) < f(0)+ y$, meaning that $f$ covers all values from $-\infty$ to $f(0)$.

Answer 3

Since $x\neq y$ implies $f(x)\neq f(y)$, $f$ is injective, thus monotonic. So $\lim_{x\to+\infty}f(x)$ and $\lim_{x\to-\infty}f(x)$ must exist. You need only this two limits cannot be finite, then by the continuity, the image of $f$ is $\mathbb R$.