Theorem 1

Let E be a complete normed algebra over the field $\mathbf{R}$, with a unit element $\mathbf{e}$, and let $\mathbf{f}$ be a continuous group homomorphism of the additive group $\mathbf{R}$ into the multiplicative group G of invertible elements of E. Then the map $\mathbf{f}$ is differentiable at every $x \in \mathbf{R}$, and

First we note that, E being a complete algebra, G is open in E (Gen. Top., IX, p. 179, prop. 14). Consider the function $g(x) = \int_0^a \mathbf{f}(x + t)\,dt$, where $a > 0$ is a number which we shall choose later; since $\mathbf{f}(x + t) = \mathbf{f}(x)\mathbf{f}(t)$ by hypothesis, we have $g(x) = \int_0^a \mathbf{f}(x)\mathbf{f}(t)\,dt = \mathbf{f}(x) \int_0^a \mathbf{f}(t)\,dt$ (I, p. 6, prop. 3). Let $\alpha > 0$ be such that the ball $\|x - \mathbf{e}\| \leq \alpha$ is contained in G; since $\mathbf{f}(0) = \mathbf{e}$ and $\mathbf{f}$ is continuous by hypothesis, one can assume that $a$ is small enough so that $\|\mathbf{f}(t) - \mathbf{e}\| \leq \alpha$ on $[0, a]$; consequently (II, p. 61, formula (16)) one has

and $\frac{1}{a} \int_0^a f(t) \, dt$ belongs to G; in other words, is invertible; so too is $b = \int_0^a f(t) \, dt$ and one can write $f(x) = g(x)b^{-1}$; it is therefore enough to show that g is differentiable; now, by the change of variable $x + t = u$ we have $g(x) = \int_x^{x+a} f(u) \, du$; since f is continuous, g is differentiable for all $x \in \mathbf{R}$ (II, p. 56, prop. 3), and

Hence $f'(x) = g'(x)b^{-1} = f(x)c$, where $c = (f(a) - e)b^{-1}$, and clearly $f'(0) = c$.

Conversely, one can show, either directly (III, p. 115, exerc. 1), or by means of the theory of linear differential equations (IV, p. 188), that every differentiable map f of $\mathbf{R}$ into a complete normed algebra E, such that $f'(x) = f(x)c$ and $f(0) = e$, is a homomorphism of the additive group $\mathbf{R}$ into the multiplicative group G.