Additive function continuous at one point is linear ($f(x)=xf(1)$)

A real additive function $f:\mathbb{R}\to \mathbb{R}$ that is continuous at one point $x_0$ is continuous everywhere and has form $f(x)=xf(1)$. Indeed $f(0)=0$, and $f(h)=f(x_0+h)-f(x_0)\to 0$ as $h\to 0$. Then $f(x+h)-f(x)=f(h)\to 0$ for every $x$. Additivity gives $f(q)=qf(1)$ for every rational number; continuity and the density of the rational numbers extend this identity to every real argument.

The step from the rational numbers to the real numbers can be made explicit. Put $c=f(1)$. For an arbitrary real number $x$, the density of the rational numbers supplies a sequence of rational numbers $q_n\to x$. Because continuity at the original point already implied continuity everywhere,This is where continuity enters: knowing all the rational values alone does not justify the first equality.

For an original numerical example, use the floor function to define the rational lower approximants $q_n=\lfloor 10^n\sqrt{2}\rfloor /10^n$. They obey $0\le \sqrt{2}-q_n<10^{-n}$. When $f(1)=3$ and $f$ is continuous at one point, rational homogeneity of an additive function and the preceding argument giveThe error bound here follows from the resulting linear map, rather than from a quantitative bound on the initially assumed continuity.