Theorem 1

Let G and H be two Lie group germs over $\mathbf{R}$ or $\mathbf{Q}_p$. Let f be a continuous morphism of G into H. Then f is analytic.

We give $L(G)$ and $L(H)$ norms which define their topologies and such that $\| [x, y] \| \leq \|x\| \|y\|$ for all $x, y$. There exist an open ball V of centre 0 in $L(G)$ and an exponential mapping $\phi$ of G defined on V such that: (1) $\phi(V)$ is an open neighbourhood of $e$ in G; (2) $\phi$ is an isomorphism of the analytic manifold V onto the analytic manifold $\phi(V)$; (3) $\phi(nx) = \phi(x)^n$ for all $x \in V$ and all $n \in \mathbf{Z}$ such that $nx \in V$. We define similarly W and $\psi$ for H. By shrinking V if necessary, it can be assumed that $f(\phi(V)) \subset \psi(W)$. Then $g = \psi^{-1} \circ f \circ \phi$ is a continuous mapping of V into W.

We show that

It can be assumed that $\lambda \neq 0$. Let $\lambda = \frac{r}{q}$ with $r, q$ in $\mathbf{Z} - \{0\}$. Let $y = \frac{r}{q} x$.

If $K = \mathbf{R}$, we write $z = \frac{x}{q} = \frac{y}{r} \in V$. Then $x = qz, y = rz$, whence

We show that $\psi^{-1}(f(\phi(z))^q) = q \psi^{-1}(f(\phi(z))) = qg(z)$. It suffices to verify that, if $u \in \psi(W)$ is such that $u^q \in \psi(W)$, then $\psi^{-1}(u^q) = q \psi^{-1}(u)$; but if $u = \psi(v)$, $u^q = \psi(v^q)$, then $v^1/q \in W$ and $(\psi(v^1/q))^q = u^q$, hence $\psi(v^1/q) = u = \psi(v)$ and therefore $v^1 = qv$.

Similarly, $g(y) = rg(z)$, whence (1).

If $K = \mathbf{Q}_p$, we write $z = rx = qy \in V$, whence $g(z) = rg(x) = qg(y)$, whence again (1).

As $\mathbf{Q}$ is dense in $K$, (1) implies that

Let $x \in L(G)$ and $\lambda, \lambda'$ be elements of $K^*$ such that $\lambda x \in V, \lambda' x \in V$. Then by (2) and hence $\frac{1}{\lambda} g(\lambda x) = \frac{1}{\lambda'} g(\lambda' x)$. Thus an extension $h$ of $g$ to $L(G)$ is defined by writing $h(x) = \frac{1}{\lambda} g(\lambda x)$ for all $\lambda$ such that $\lambda x \in V$. Clearly $h$ is continuous. We show that Let $\lambda' \in K^*$ be such that $\lambda' x \in V$ and $\lambda' \lambda x \in V$. Then Let $x, y$ be in $L(G)$. Then, by Proposition 4 of § 4, no. 3, For $|\lambda|$ sufficiently small, $\lambda x \in V$ and $\lambda y \in V$ and hence the above expression is equal to Thus $h$ is continuous and linear, hence $g = h|V$ is analytic, hence $f$ is analytic on $\phi(V)$ and hence $f$ is analytic (§ 1, no. 10).