Show that $f$ is unbounded below 
Let $f:\Bbb R\to \Bbb R$ be continuous and satisfies $|f(x)|\ge |x| $ for all $x$.Also $f(x+y)=f(x)+f(y)$ for all $x,y$.Show that $f$ is bijective.

My try:
$f$ is injective ;since $f(x)=f(y)\implies f(x-y)=0\implies |x-y|\le 0\implies x=y$.
To show that $f$ is onto.
Every continuous injection is either strictly increasing or decreasing.Hence if I can show that $f$ is both unbounded above and below then by the IVP we can show that $f$ is surjective.
Now since $|f(x)|\ge |x|>x$ forall $x$ so $f$ is unbounded above.However I am failing to show that $f$ is unbounded below also.
Please help me out here.
 A: From $f(x+y)=f(x)+f(y)$ and continuity you get
$f(x)=cx$
https://en.wikipedia.org/wiki/Cauchy's_functional_equation
From $|f(x)|\ge |x|$ you get $c\ne 0$
So $f$ is bijective.
A: First an easy induction argument shows that $f(nx)=nf(x)$ for all $n \in \Bbb N$ and $x \in \Bbb R$. Next, for $r =p/q\in \Bbb Q$, one has for all $x\in \Bbb R$,
$$f(rx)= f\left(p\times \frac{x}{q}\right)= pf(x/q)$$
But:
$$f(x) = f(q \times x/q) = qf(x/q) \implies f(x/q) = \frac1q f(x)$$
Then $f(rx) = rf(x)$. 
Now for $x\in \Bbb R$ and $a\in \Bbb R$, by density of $\Bbb Q$ there is a rational sequence $a_n \to a$. For all $n$, $f(a_n x) = a_n f(x)$. By continuity of $f$, $f(ax) = af(x)$. 
Then for $x\in \Bbb R$, $f(x) = f(x \times 1) = x f(1) = cx$. As $|c| =|f(1)|\ge 1$, $c \neq 0$. So $f$ is bijective.
A: Note that $f(0)=f(0)+f(0)$, so $f(0)=0$.
Also, $f(-x)+f(x)=f(0)=0$, so $f(-x)=-f(x)$.
Since you've already showed that $f$ is unbounded above, it follows that it is also unbounded below.
A: To complete your proof, you can show that $f(n)=nf(1)$ for $n\in \mathbb Z$
$f(n) = f(1+1+...+1) = nf(1)$ when $n\ge 0$
$f(0)=f(0)+f(0)$ so $f(0)=0$
$f(0)=f(n)+f(-n)$ so $f(-n)=-f(n)$
Now: $|f(1)|\ge 1$ from where $f(1)\ne 0$
From where you can prove unboundness below and above.
A: Hint: the given function has to be linear, continuous, passing through zero, with slope greater than $1$.
