Continuous bijective function If $f:\mathbb R\to \mathbb R$ is a continuous function satisfying $\vert f(x)-f(y)\vert\geq \dfrac{1}{2}\vert x-y\vert$ for all $x,y\in \mathbb R$, then is $f$ bijective?
I believe that $f$ is one-one, as $f(x)=f(y)\Rightarrow x=y$. But is $f$ onto? Please help!
 A: $f$ is injective since $x\ne y$ means that $|f(x)-f(y)|\ge|x-y|>0$. Every injective and continuous function $\Bbb R\to\Bbb R$ is either monotonically increasing or decreasing. Let's assume that $f$ is increasing (otherwise replace $f$ by $-f$).
If the range of $f$ has a supremum $s$ and there is some $x$ such that $s-f(x)<\infty$, but for each $y>x+s-f(x)$ we have $f(y)-f(x)>y-x>s-f(x)$, thus $f(y)>s$. So the supremum must be $\infty.$
On the other hand, if $i$ is the infimum of the range of $f$ and $f(x)-i<\infty$, then for $y<x-f(x)+i$ we have $f(x)-f(y)>x-y>f(x)-i$, so $f(y)<i$, hence $i$ must be $-\infty$.
A: Yes, it must be a bijection. Assume not, and assume wlog, that it misses points in the right-half of the plane, i.e., f does not hit every positive Real. Then f must approach a finite limit M as $x \rightarrow \infty$ . This means, for given $\epsilon>0$, there exists some $N>0$ so that for all $x>N$, we must have $$|f(a)-f(b)|<\epsilon$$ for all $a,b>N$ . But you can make the difference $|a-b|$ indefinitely large  (using $a,b>N$).... 
