Continuous additive functions are linear

Remarks

Additivity gives $f(nx) = n f(x)$ for integers $n$, and then $f(q) = q f(1)$ for every rational $q = a/b$, since $b f(a/b) = f(a) = a f(1)$. The rationals are dense in $\R$, so every real $x$ is a limit of rationals $q_k$, and continuity gives $f(x) = \lim f(q_k) = \lim c q_k = c x$. For example, if a continuous $f$ satisfies $f(x + y) = f(x) + f(y)$ and $f(1) = 2$, then $f(x) = 2x$, so $f(\pi) = 2\pi$. Continuity cannot be dropped. A basis of $\R$ as a vector space over $\Q$ exists by the axiom of choice, and with it one builds additive functions that are not of the form $cx$. Weaker hypotheses suffice, such as continuity at a single point, monotonicity on an interval, or Lebesgue measurability. The same argument shows that a continuous additive map between real normed spaces is $\R$-linear.

Going deeper

For a student

This is the regular solution of Cauchy's functional equation, and many other classical functional equations reduce to it. For instance, a continuous $g \colon \R \to (0, \infty)$ with $g(x + y) = g(x) g(y)$ satisfies the hypothesis after taking logarithms, so $g(x) = a^x$ with $a = g(1)$. In linear algebra terms, the theorem says that a continuous homomorphism of the group $(\R, +)$ to itself is automatically $\R$-linear.