Suppose f from R to R satisfies f(x + y) = f(x) + f(y) for all real x and y. Prove f(qx) = q f(x) for rational q, explain why that fails to force f(x) = cx over the reals, and give a condition that does.

Suppose f from R to R satisfies f(x + y) = f(x) + f(y) for all real x and y. Prove f(qx) = q f(x) for rational q, explain why that fails to force f(x) = cx over the reals, and give a condition that does.

Approach: Build up from integers to rationals by repeated addition and division, then consider what the equation leaves undetermined between the rationals and the reals when no continuity is assumed.

f(x) = cx follows once f is assumed measurable, monotone on an interval, or bounded on a set of positive measure, and rational homogeneity alone is too weak. Setting y = 0 gives f(0) = 0, and y = -x gives f(-x) = -f(x). Induction on the Cauchy functional equation gives f(nx) = n f(x) for positive integers, and applying it to x/n gives f(x/n) = f(x)/n, so f(qx) = q f(x) for every rational q, and in particular f(q) = q f(1) on the rationals. Over the reals this pins nothing further down, because R is a vector space over Q and a Hamel basis lets you assign values independently on each basis element, producing additive functions whose graph is dense in the plane. Any regularity kills those: with a continuity assumption, f agrees with c x on a dense set and both sides are continuous, so they agree everywhere, and boundedness on any interval upgrades to continuity by the additivity. Every pathological solution needs the axiom of choice, so none can be written down explicitly.

Follow-up: What are the solutions of f(x + y) = f(x) f(y) and of f(xy) = f(x) + f(y) under the same continuity assumption?

Key concepts: cauchy functional equation, rational homogeneity, continuity assumption, hamel basis.