Sometimes additive means linear

Let $V$ and $W$ be topological vector spaces over the reals, and let $f : V \to W$ such that $f$ is continuous and additive, i.e.

$$\forall x, y \in V: f(x + y) = f(x) + f(y).$$

In this post I will show that $f$ is then also linear, i.e.

$$\forall x, y \in V: \forall \alpha, \beta \in \mathbb{R}: f(\alpha x + \beta y) = \alpha f(x) + \beta f(y).$$

Proof

Since $f(0) = f(0 + 0) = f(0) + f(0)$, it follows that $f(0) = 0$. Now

$$0 = f(0) = f(x + (-x)) = f(x) + f(-x),$$

from which it follows that $f(-x) = -f(x)$. Thus $f$ preserves negation. We now claim that if $n \in \mathbb{N}$, then $f(nx) = nf(x).$ The proof is by induction. Since $f(0x) = f(0) = 0 = 0f(0)$, the claim holds when $n = 0$. Assume that the claim holds for $n < k$, with $k> 0$. Then

$$f(kx) = f((k - 1)x + x) = f((k - 1)x) + f(x) = (k - 1)f(x) + f(x) = kf(x).$$

Thus $f$ preserves multiplication by natural numbers. Let $z \in \mathbb{Z}$. Then if $z < 0$,

$$f(zx) = f((-z)(-x)) = (-z)f(-x) = zf(x).$$

Thus $f$ preserves multiplication by integers. Let $p \in \mathbb{Z}$, with $p \neq 0$. Now

$$f(x) = f((p / p) x) = f(p (x / p)) = p f(x / p).$$

Thus $f(x / p) = f(x) /p$, and $f$ preserves multiplication by inverses of non-zero integers. Let $q \in \mathbb{Z}$. Then

$$f((q / p)x) = f(q (x / p)) = q f(x / p) = (q / p) f(x).$$

Thus $f$ preserves multiplication by rational numbers. Finally, let $(r_i)_{i \in \mathbb{N}}$ with $r_i \in \mathbb{Q}$ be a sequence of rational numbers converging to a real number $r \in \mathbb{R}$, i.e. $\lim_{i \to \infty} r_i = r$. Then

$$f(rx) = f\left(\lim_{i \to \infty} r_i x\right) = \lim_{i \to \infty} f(r_i x) = \lim_{i \to \infty} [r_i f(x)] =\left[\lim_{i \to \infty} r_i\right] f(x) =r f(x),$$

where the limit can be moved out of $f$ by continuity of $f$. The $f(x)$ can be moved out of the limit by the continuity of vector multiplication (which holds by definition for a topological vector space). Thus $f$ preserves multiplication by real numbers. Therefore, $f$ is both homogeneous and additive, and thus linear.

QED.