Sometimes additive means linear

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.

$\begin{equation} \begin{split} \forall x, y \in V: f(x+y) = f(x) + f(y). \end{split} \end{equation}$

Then $f$ is linear, i.e.

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

Proof

By additivity,

$\begin{equation} \begin{split} f(0) & = f(0 + 0) \\ {} & = f(0) + f(0). \end{split} \end{equation}$

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

$\begin{equation} \begin{split} 0 & = f(0) \\ {} & = f(x + (-x)) \\ {} & = f(x) + f(-x). \end{split} \end{equation}$

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

$\begin{equation} \begin{split} f(0 x) & = f(0) \\ {} & = 0 \\ {} & = 0 f(0), \end{split} \end{equation}$

the claim holds for $n = 0$ . Assume the claim holds for $n < k$ , with $k > 0$ . Then

$\begin{equation} \begin{split} f(k x) & = f((k-1)x + x) \\ {} & = f((k-1)x) + f(x) \\ {} & = (k-1) f(x) + f(x) \\ {} & = kf(x). \end{split} \end{equation}$

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

$\begin{equation} \begin{split} f(zx) & = f((-z)(-x)) \\ {} & = (-z)f(-x) \\ {} & = zf(x). \end{split} \end{equation}$

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

$\begin{equation} \begin{split} f(x) & = f((p/p)x) \\ {} & = f(p(x/p)) \\ {} & = pf(x/p). \end{split} \end{equation}$

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

$\begin{equation} \begin{split} f((q/p)x) & = f(q(x/p)) \\ {} & = qf(x/p) \\ {} & = (q/p)f(x). \end{split} \end{equation}$

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

$\begin{equation} \begin{split} f(rx) & = f\left(\left[\lim_{i \to \infty} r_i \right] x\right) \\ {} & = f\left(\lim_{i \to \infty} r_i x \right) \\ {} & = \lim_{i \to \infty} f(r_i x) \\ {} & = \lim_{i \to \infty}[r_i f(x)] \\ {} & = [\lim_{i \to \infty} r_i] f(x) \\ {} & = rf(x), \end{split} \end{equation}$

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.