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.

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

## Proof

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

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

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

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

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

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

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.

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