Proposition 6

Let $I$ be an interval of $\mathbf{R}$ which contains $o$ and at least one other point; let $f$ be a continuous mapping of $I$ 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 $x \in I, y \in I$ and $x + y \in I$. Then there is a unique continuous homomorphism of $\mathbf{R}$ into $G$ which extends $f$..

The uniqueness of the extension (if it exists) follows from the preceding remarks, because $I$ generates the group $\mathbf{R}$. We have to establish the existence of such an extension.

If $n$ is an integer $> 0$ and if $x \in I$ and $nx \in I$, we have $f(nx) = (f(x))^n$, by induction on $n$ (since $mx \in I$ for all integers $m$ such that $1 \leq m \leq n$). Put $J = \bigcup_{n \in \mathbf{N}} nI$; $J$ is either the whole line $\mathbf{R}$, or one of the intervals $[0, +\infty[$ or $[-\infty, 0]$, according as $o$ is or is not an interior point of $I$. If $x \in J$ we have $x/n \in I$ whenever $n$ is a sufficiently large integer $> 0$. Let $x \in J$, and let $m, n$ be two integers $> 0$ such that $x/n \in I$ and $x/m \in I$; then $x/mn \in I$, and therefore

in other words, the element $(f(x/n))^n$ of $G$ is the same for all integers $n > 0$ such that $x/n \in I$. Let us denote this element by $f_1(x)$; $f_1$ is a mapping of $J$ into $G$, which agrees with $f$ on $I$ and is therefore continuous at the point $o$ (with respect to $J$). Let $x, y$ be two elements of J and let n be a sufficiently large integer > 0 such that $\frac{x}{n} \in I, \frac{y}{n} \in I$ and $\frac{x+y}{n} \in I$; then

which shows that $f\left( \frac{x}{n} \right)$ and $f\left( \frac{y}{n} \right)$ commute; by the definition of $f_1$, we have therefore $f_1(x+y) = f_1(x) f_1(y)$. If $J = \mathbf{R}$, the proof is complete; if not, say $J = [0, +\infty[$, and for each $x < 0$ define $f_1(x)$ to be $(f_1(-x))^{-1}$. Then the relation $f_1(x+y) = f_1(x) f_1(y)$ remains valid for all $x \in \mathbf{R}$ and all $y \in \mathbf{R}$. This is clear if $x < 0$ and $y < 0$; if $x \geqslant 0$, $y < 0$ and $x + y \geqslant 0$ it follows from $f_1(x) = f_1(x+y) f_1(-y)$; similarly if $x \geqslant 0, y < 0$ and $x + y < 0$, for then we have

analogous proofs for $x < 0$ and $y \geqslant 0$. We see therefore that $f_1$ is a homomorphism of $\mathbf{R}$ into G, so that $f_1(0) = e$, the identity element of G; and since $f_1$ is continuous with respect to J, it has a limit on the right at 0, equal to $e$; since $f_1(-x) = (f_1(x))^{-1}$, $f_1$ also has a limit at 0 on the left, equal to $e$; thus $f_1$ is continuous at 0, and the proof is complete.