# The additive function is bounded

1. Jan 4, 2012

### LikeMath

Hi,
If I have an additive function which is $f(x+y)=f(x)+f(y)$,
the question is
how can we prove that if this function has a limit at each real number then there is a number a greater than zero and M greater than zero
such that
$|f(x)|\leq M$, for all $x\in[-a,a]$,

2. Jan 4, 2012

### HallsofIvy

Can you show that, for any rational number, r, f(rx)= rf(x)? From that, you should be able to show that if the limit exists at any x, then f is continuous there and so is continuous for all x. From that it follows that f(x)= cf(x) where c is a real constant.