Definition 5

If E is a (left) vector space over a non-discrete valued division ring K, a norm on E is a mapping $x \to p(x)$ of E into $\mathbf{R}_+$ which satisfies the following axioms:

(NO_I) $p(x) = 0$ if and only if $x = 0$; (NO_{II}) $p(x + y) \leq p(x) + p(y)$ for all $x, y$ in E; (NO_{III}) $p(tx) = |t|p(x)$ for all $t \in K$ and all $x \in E$.

The normed spaces most frequently met with have either $\mathbf{R}$ or $\mathbf{C}$ as field of scalars (with the usual absolute value).

From (NO_{III}) it follows in particular that $p(-x) = p(x)$; hence if we put $d(x, y) = p(x - y)$, $d$ is an invariant metric on the additive group E, and defines a metric space topology compatible with the additive group structure of E (no. 1, Proposition 3); moreover, the mapping

is continuous on $K \times E$; for we have

and therefore

which shows that the left-hand side can be made as small as we please by taking $|t - t_0|$ and $p(x - x_0)$ sufficiently small.

Let E be a subset of K. A function $f : E \to K$ is called a normed space over K.

A normed space is endowed with the topology and the uniformity defined by its norm.

Consider a finite-dimensional division ring K. Considered as a real vector space, the absolute value $|x|$ is a norm.

The norm $\|x\| = \sqrt{\sum x_i^2}$, which we have called the Euclidean norm on the space $\mathbf{R}^n$ (Chapter VI, § 2) is evidently a norm in the sense of Definition 3. So are the functions $\sup |x_i|$ and $\sum |x_i|$.

Let $S(E; K')$ be the set of all functions $f$ on a set E which take their values in a division ring $K'$ and are such that the real-valued function $x \mapsto f(x)$ is bounded on E. This set is clearly a vector space over $K'$. Let us denote by $p(f) = \sup_{x \in E} f(x)$, then $p$ is a norm on the vector space $S(E; K')$ (cf. Chapter X, § 1).

\* 4) On the vector space $C(I; \mathbf{R})$ of all finite continuous real-valued functions defined on the interval $I = [0, 1]$ of $\mathbf{R}$, the function

is a norm.

In a normed space E, the closed ball B with centre o and radius 1, that is the set of all $x$ such that $p(x) \leq 1$, will be called the unit ball of E. We shall show that a fundamental system of neighbourhoods of o is formed by the translations of the unit ball by the elements of K. The open ball with centre o and radius 1 is the closed ball with centre o and radius 1; hence it is enough to show that for each real number $r > 0$ there exists $k \in \mathbf{N}$ such that $0 < |k| < r$; it is sufficient to take $k = \lfloor c/r \rfloor$, where $c$ is a sufficiently large integer, in order that $|k|^2 < r$.

Let E be a normed space over a non-discrete valued field $K$.