This is an interesting problem, which has however already been treated elsewhere (Try to get your hands on Natanson's book "Theory of Functions of a Real Variable" or on Kolmogorov/Fomin). You can prove that any additive function that is alsoOriginally Posted byThePerfectHackercontinuous at 0must be of the form f(x) = ax, as follows.

Firstly, f(0) = f(0+0) = f(0) + f(0), so f(0) = 0. If f is continuous at x = 0, then , so f is continuous everywhere.

Next set a = f(1). By induction you can prove that f(n) = an for all positive integers. Also, 0 = f(0) = f(n-n) = an + f(-n), so this is true for all integers n.

Then consider a rational number x = m/n, with n > 0. Then nx = m = x + ... + x, so by induction , thus .

Finally for general x, .

Now if f is not continuous at 0, things are different. Your instinct that an additive function need not be of the form f(x) = ax is indeed correct. Counterexamples are tricky and require the Axiom of Choice. Namely, consider a Hamel basis H, that is a basis for theset of real numbers viewed as a vector space over the rationals. For each , define , where is an arbitrary real number. Then if where the are rational and the , one extends f as . This is an additive function that is not of the simple form f(x) = ax (and not continuous at 0).