Given that f is additive and f is continous at 0, prove f(x)=kx where k=f(1). My first response for this problem would be to use the straight definition of continuity at 0, using epsilon and delta, but I'm getting stuck. 
I also tried, let k=f(1) and g(x) = f(x) -kx  and then somehow show that g(x)=0 but this is not working either. 
Any help would be appreciated :)
 A: You definitely want to start with $f(1)=k$.  You can then show by induction that $f(n)=kn$ for all $n\in\Bbb N^+$ by induction and additivity.  This can be easily extended to $f(n)=kn$ for all $n\in\Bbb Z$.
Next use additivitity, to show that $f(1/n)=k/n$ for all $n\in\Bbb N^+$, and then you can extended to all rationals.  You will be able to extend continuity to $\Bbb Q$ also. Finally use density and continuity to get the result for $\Bbb R$.
Edit: Actually $f$ additive and continuous at $0$ implies $f$ continuous everywhere.
A: $f(0)=f(0+0)=2f(0), f(2)=f(1+1)=2f(1)$, so, over the integers, $f(n)=nf(1)=nk$ Then work over the Rationals : $f( a/b)= f(a(1/b))=af(1/b)$, then use $f(1)=k$.
e.g., $f(1)=f(1/2+1/2)=2f(1/2)=k $, so $f(1/2)=k/2$ . Extend this argument to all the Rationals, to show that $f(p/q)= k(p/q)$. Then use the fact that the Rationals are dense in the Reals , and that f(x)=kx is uniformly continuous on the Rationals and then use that every uniformly-continuous function defined on a dense subset of a set S extends uniquely into the set S.
Summarizing:
1) For $n \in \mathbb Z $, we see that $f(n)=kf(1)$
2) Use base case $f(1)= f(n(1/n)$ to show $f(a/b) =f(1)(a/b)$
3) $f(x)=kx$ is uniformly continuous, and the Rationals are dense on the Reals. Uniformly-continuous functions defined on dense subsets extend uniquely into the space; this can be done using sequential continuity ( use that uniform continuous preserves Cauchy sequences ).
