TITLE: Any continuous group homomorphism $\mathbb{R}\to \mathbb{R}^n$ is $C^\infty$
QUESTION [7 upvotes]: Show that any continuous homomorphism $\mathbb{R}\to \mathbb{R}^n$, with respect to the usual abelian group structure, is actually $C^\infty$.
My attempt: Let $\varphi$ be such a map.
$$\lim_{h\to 0}\frac{\varphi(x+h) - \varphi(x)} {h} = \lim_{h \to 0} \frac{\varphi (h)}{h},$$
so if $\varphi'$ exists, it is constant on $\mathbb{R}$, and it is sufficient for it to exist at zero. Now note that $h\mapsto \frac{\varphi(h)}{h}$ is continuous for $h \neq 0$ since it is the product of a continuous scalar-valued function and a continuous vector-valued function. Now
$$ \frac {\varphi \left(\frac{h}{m/n} \right ) } {\frac{h}{m/n}} = \frac{(n/m)\varphi(h)}{\frac{h}{m/n}} = \frac{\varphi(h)}{h}, \;\;\; m,n \in \mathbb{Z}$$
so $h\mapsto \frac{\varphi(h)}{h}$ is constant on $\mathbb{Q}$. Since it is continuous, it is constant on $\mathbb{R}$, and it has a limit at zero. 
I would appreciate it if you could let me know if there are parts of the argument I could explain better, or point out any errors.

REPLY [4 votes]: You can even show that $\varphi$ is linear (in particular $C^\infty$), i.e. there is $x_0 \in \mathbb{R}^n$ such that $\varphi(t)=tx_0$ for all $t\in \mathbb{R}$. 
The argument is quite similar to yours: Let $x_0 = \varphi(1)$. By additivity $\varphi(m)=mx_0$ for all integers $m \ge 0$. Let $q=m/n$ with integers $m,n> 0$. Form $mx_0=\varphi(m)=\varphi(nq)=n\varphi(q)$ coclude $\varphi(q)=m/nx_0=qx_0$ for all rationals $q \ge 0$. Since group homomorphism map inverses to inverses, we obtain $\varphi(-q)=-\varphi(q)=(-q)x_0$. Hence $\varphi(q)=qx_0$ for all rationals $q$. So $\varphi$ is linear on the rationals and by continuity the claim follows.