If R->R satisfies f(x+y)=f(x)+f(y) for all x,y in R and 0 is in C(f), then f is continuous.
show me a great proof...
thanks in advance
A great one huh?
This is obvious since $\displaystyle f(0)=f(0)+f(0)\implies f(0)=0$ and thus $\displaystyle 0=f(0)=f(x-x)=f(x)+f(-x)\implies f(-x)=-f(x)$. Thus, $\displaystyle |f(x)-f(y)|=|f(x-y)|$ now there exists some $\displaystyle \delta>0$ such that $\displaystyle |z|<\delta\implies |f(z)|<\varepsilon$. Make $\displaystyle |x-y|<\delta$
Let z=x-y. Then $\displaystyle |z|=|x-y|<\delta\implies |f(z)| = |f(x-y)| = |f(x)-f(y)|<\varepsilon$.
Another way to see it is: f is continuous at x iff f(x+h)=f(x)+o(1) as $\displaystyle h\to0$. So in this case, f(x+h) = f(x) + f(h) by linearity and f(h)=o(1) by continuity at 0.