Theorem 1

Let $E_i (1 \leq i \leq n)$ and $F$ be normed spaces over a non-discrete valued division ring $K$, and let $f$ be a multilinear mapping of $\prod_{i=1}^n E_i$ into $F$. Then $f$ is continuous on $\prod_{i=1}^n E_i$ if and only if there exists a real number $a > 0$ such that, for all $x_i \in E_i \ (1 \leq i \leq n)$, we have

(3)

The condition is necessary. For if $f$ is continuous at the point $(0, 0, \ldots, 0)$ there exists a number $b > 0$ such that the relations $||x_i|| \leq b \ (1 \leq i \leq n)$ imply $||f(x_1, \ldots, x_n)|| \leq 1$. Let $t_0$ be an element of $K$ such that $0 < |t_0| < 1$; then for every point $(x_i) \in \prod_{i=1}^n E_i$ such that none of the $x_i$ is zero, there exist $n$ rational integers $k_i$ such that $b|t_0| < ||t_0^{k_i} x_i|| \leq b$; consequently we have

on the other hand we have $\frac{1}{|t_0|^{k_i}} \leq \frac{1}{b|t_0|} ||x_i||$, and the relation (3) therefore follows, with $a = (1/b|t_0|)^n$. This relation is evidently still valid when one of the $x_i$ is zero.

The condition is sufficient. We shall show that, if it is satisfied, $f$ is continuous at every point $(a_i)$ of $\prod_{i=1}^n E_i$. We can write

Now, using (3), the conditions $||x_i - a_i|| \leq r$ ($1 \leq i \leq n$) imply that

hence, if $c$ is the maximum of the numbers $||a_i||$ ($1 \leq i \leq n$), we have

Since the right-hand side of this inequality is a polynomial in $r$ with zero constant term, it tends to 0 as $r$ tends to 0; hence $f$ is continuous.