Example 3

The set $\mathscr{B}(X; K)$ of all functions $f$ on a set $X$ with values in a non-discrete valued field $K$, such that $x \to |f(x)|$ is bounded on $X$, is an algebra over $K$; the norm $\|f\| = \sup_{x \in X} |f(x)|$ is compatible with the ring structure of $\mathscr{B}(X; K)$, because we have $\|fg\| \leq \|f\|\cdot\|g\|$ (cf. Chapter X, § 1).

Let $a$ be a closed two-sided ideal in a normed algebra $A$. If in the quotient algebra $A/a$ we put $||\dot{x}|| = \inf_{x \in X} ||x||$, we get a norm on $A/a$ which defines the topology which is the quotient by $a$ of the topology of $A$ (Proposition 9); since this quotient topology is compatible with the quotient ring structure of $A/a$ (Chapter III, § 6, no. 4) it follows that the quotient algebra $A/a$, with the norm $||x||$, is a normed algebra.

Likewise, if $(A_i)_{1 \leq i \leq n}$ is a family of $n$ normed algebras over a valued field $K$, and if in the product algebra $A = \prod_{i=1}^n A_i$ we put $||x|| = \sup_i ||x_i||$, where $x = (x_i)$, we have a norm on $A$ which defines the product of the topologies of the $A_i$ (Proposition 10); since this topology is compatible with the ring structure of $A$ (Chapter III, § 6, no. 4), it follows that the product algebra $A$, with the norm $||x||$, is a normed algebra.

Let $A$ be a normed algebra over a valued field $K$. The completion $\hat{A}$ of $A$ (Chapter III, § 6, no. 5, Proposition 6) is also endowed with a $\hat{K}$-vector space structure (no. 3, Proposition 8), and it is clear from the principle of extension of identities that we have $t(xy) = (tx)y = x(ty)$ for all $t \in \hat{K}$ and all $x, y \in \hat{A}$. Hence $\hat{A}$ is an algebra over $\hat{K}$; on the other hand (no. 3, Proposition 8), the norm on $A$ extends by continuity to a norm which defines the topology of $\hat{A}$, and therefore $\hat{A}$, endowed with this norm, is a normed algebra over the field $\hat{K}$.

If $(x_\lambda)_{\lambda \in L}$ and $(y_\mu)_{\mu \in M}$ are two absolutely summable families in a normed algebra $A$, the family $(x_\lambda y_\mu)_{(\lambda, \mu) \in L \times M}$ is absolutely summable, because $||x_\lambda y_\mu|| \leq ||x_\lambda|| \cdot ||y_\mu||$ (Chapter IV, § 7, no. 3, Proposition 1); if in addition $A$ is complete, all three families are summable and we have

by the associativity of the sum on the left-hand side (Chapter III, § 5, no. 3, formula (2)).

If the normed algebra $A$ has an identity element $e \neq 0$, the mapping $t \to te$ is an isomorphism of the field structure of $K$ onto that of the subfield $Ke$ of $A$; this isomorphism is also an isomorphism of the topological field structure of $K$ onto that of $Ke$ (the topology of the latter being induced by that of $A$), for the restriction $||te||$ of the norm of $A$ to $Ke$ is a norm equivalent to the absolute value

If $||e|| = 1$ we have $||te|| = |t|$, and we can then identify the valued field $K$ with the normed subfield $Ke$ of $A$, and in particular we may denote the identity element of $A$ by the symbol 1.

In what follows we shall be concerned only with normed algebras which have an identity element $e$, and in which the norm satisfies the inequality (5); putting $x = y = e$ in this inequality, it follows that $||e|| \geq 1$.