If $ f $ is a continuous function and $ f ( a + b ) = f ( a ) + f ( b ) $, how do I prove that $ f ( x ) = m x $ for any $ x $ in real numbers, where $ m = f ( 1 ) $?
I know that I have to start by showing $ f ( x ) = m x $ for any rational $ x $ and then extend that to any real number with continuity. However, I do not know how to go about it.
First use induction to show $f(n)=nm$ for $n\in\Bbb N$. Next, $f(0)=f(0)+f(0)$ implies $f(0)=0$, so $0=f(1-1)=m+f(-1)$. This tells us $f(-1)=-m$, and now we can show by induction $f(n)=mn$ for any $n\in\Bbb Z$.
Next, we claim that, for any rational number $p/q$, $f(p/q)=mp/q$. We can assume $q$ is a positive integer, and then
$$mp=f(p)=f(qp/q)=f(p/q+p/q+\cdots+p/q)=qf(p/q),$$
which gives the result. Now use the fact that $f$ is continuous to extend the result to every real number.
