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02.03.2012

## Claim

Let $V$ and $W$ be topological vector spaces over $\mathbb{R}$, and $f : V \to W$ be continuous and additive, i.e.

Then $f$ is linear, i.e.

## Proof

It follows that $f(0) = 0$. Let $x \in V$. By additivity,

It follows that $f(-x) = -f(x)$. Thus $f$ preserves negation. We claim that $f(nx) = nf(x)$, for all $n \in \mathbb{N}$. Since

the claim holds for $n = 0$. Assume 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 $p \neq 0$. Then

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. Let $\{r_i \in \mathbb{Q}\}_{i \in \mathbb{N}}$ 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.