Proposition 2

Let $A$ be a parallelopte in $\mathbf{R}^n$ which contains $o$; and let $f$ be a continuous mapping of $A$ into a topological group $G$ (written multiplicatively) such that $f(x + y) = f(x)f(y)$ for each pair of points $x, y$ such that $x \in A,\ y \in A$ and $x + y \in A$. Then there exists a unique continuous homomorphism of $\mathbf{R}^n$ into $G$ which extends $f$.

By the same reasoning as in Proposition 6 of Chapter V, § i, no. 4 we show that the homomorphism extending $f$, if it exists, is unique. Next, the subgroup $G_1$ of $G$, generated by $f(A)$, is commutative; for if $x$ and $y$ are any two points of $A$, then $\frac{1}{2} x, \frac{1}{2} y$ and $\frac{1}{2} (x + y)$ belong to $A$, therefore

which shows that $f \left( \frac{1}{2} x \right)$ and $f \left( \frac{1}{2} y \right)$ commute; therefore so do $f(x) = (f \left( \frac{1}{2} x \right))^2$ and $f(y) = (f \left( \frac{1}{2} y \right))^2$, which proves the assertion.

Let $a_1, a_2, \ldots, a_n$ be $n$ non-zero vectors contained in $A$ and proportional to the basis vectors of the parallelootope, and for each index $i$ let $D_i$ be the line through $o$ and $a_i$, that is, the set of points $ta_i$ as $t$ runs through $\mathbf{R}$. Let $A_i$ be the set of all $t \in \mathbf{R}$ such that $ta_i \in A$; then $A_i$ is an interval of $\mathbf{R}$ containing $[0, 1]$, and the function

is defined and continuous on $A_i$ and satisfies the relation

whenever $t, t'$ and $t + t'$ all belong to $A_i$. By Proposition 6 of Chapter V, § 1, no. 4, there exists a continuous homomorphism $\bar{f}_i$ of $\mathbf{R}$ into $G$ which extends $f_i$. Since $\mathbf{R}^n$ is the direct sum of the subgroups $D_i$, we can define a homomorphism $\bar{f}$ of $\mathbf{R}^n$ into the abelian group $G_1$ by the rule $\bar{f}(x) = \prod_{i=1}^n \bar{f}_i(t_i)$ where $x = \sum_{i=1}^n t_i a_i$; $\bar{f}$ is an extension of $f$ since, if $x \in A$, all the components $x_i$ of $x$ on the lines $D_i$ also belong to $A$, by reason of the choice of the $a_i$; moreover, $\bar{f}$ is continuous on $\mathbf{R}^n$, because it is continuous on each of the lines $D_i$, and $x_i$ is a linear function of $x$ (and therefore continuous).