I guess the following little problem is famous:
Suppose $f$ is a continuous group homomorphism from $(\mathbb{R},+)$ to $Gl_n(\mathbb{R})$ such that for all reals $x,y$ we have (*): $f(x+y)=f(x)f(y)$.
Show that there exists a matrix $M$ such that for all reals $t$, $f(t)=\exp(tM)$.
I would like to solve this by using the same famous argument with the density of rationals used to resolve Cauchy's functional equation. However I run into a problem with taking nth roots of a matrix, which are not well-defined.
A solution suggested to me by my professor begins by showing that $f$ is differentiable and then taking the derivative of (*) to get a differential equation whose solution is an exponential. I am not sure however hot to demonstrate derivability of $f$.
Note: $(\mathbb{R},+)$ denotes the group $\mathbb{R}$ under standard addition and $Gl_n(\mathbb{R})$ denotes the set of inversible matrices of order $n$ with real coefficients. $\exp(A)$ denotes the matrix exponential of $A$.[/quote]
Remark that the differential of the exponential $exp:M_n(\mathbb{R})\rightarrow Gl_n(\mathbb{R})$ is the identity, so it is a local diffeomorphism. This implies that there exists a neighborhood $[-c,c]$ of $0$ such that $f(t)=exp(M(t))$ where $M(t)$ is continuous.
For every $t,s\in [-c,c]$, such that $t+s\in [-c,c]$, $exp(M(s+t))=exp(M(s))exp(M(t))=exp(M(t+s))=exp(M(t))exp(M(s))$ implies that $exp(M(s+t)=exp(M(s)+M(t))$ and $M(s+t)=M(s)+M(t)$ since $M$ is continuous, we deduce that $M$ is linear.
$f'(0)=lim_{t\rightarrow 0}{{f(t)-f(0)}\over t}$
$=lim_{t\rightarrow 0}{{exp(M(t))-I}\over t}$
$=lim_{t\rightarrow 0}{{M(t)}\over t}exp(M(t))$, since $M(t)$ is linear, and $M(0)=0$, we deduce that $=lim_{t\rightarrow 0}{{M(t)}\over t}exp(M(t))$ exists and $f$ is differentiable at zero.
By using the fact that $f(x+y)=f(x)f(y)$, we deduce that $f$ is differentiable on $\mathbb{R}$.