Assume $A$ is a real Banach algebra (which need not necessarily be commutative or finite-dimensional) with unit and the function $f: \mathbb{R}\to A$ satisfies the differential equation $$\frac{df\left( t \right)}{dt}=f\left( t \right)\cdot s\left( t \right)$$ for all $t\in \mathbb{R}$ with a given continuous function $s: \mathbb{R}\to A$. How can you prove that $f\left( t \right)$ is invertible for all $t\in \mathbb{R}$ if there is (at least) one $t_{0}\in \mathbb{R}$ for which $f\left( t_{0} \right)$ is invertible?
In case the Banach algebra is just $\mathbb{R}$ everybody knows the answer: the only function $f:\mathbb{R}\to \mathbb{R}$ which then satisfies the differential equation and $f\left( t_{0} \right)=r_{0}$ with a given $r_{0}\in \mathbb{R}$ is defined by $$f\left( t \right)=r_{0}\cdot \exp\left( \int_{t_{0}}^{t}s\left( u \right)du \right)$$ for all $t\in \mathbb{R}$ and thus the assertion ensues immediately. Virtually the same reasoning is valid if the Banach algebra is commutative (no matter whether it is finite-dimensional), only the real exponential function has to be replaced by the exponential on the Banach algebra which is defined by the power series $$\exp\left( a \right)=\sum_{i=0}^{\infty}\frac{1}{i!}\cdot a^{i}$$which is convergent for all $a\in A$, and once again, the uniquely determined function $f:A\to A$ which satisfies the differential equation and $f\left( t_{0} \right)=r_{0}$ with a given $r_{0}\in A$ is defined by $f\left( t \right)=r_{0}\cdot \exp\left( \int_{t_{0}}^{t}s\left( u \right)du \right)$ for all $t\in \mathbb{R}$, where it is well known how to define this integral over the continuous function $s$ with values in $A$. So, if $f\left( t_{0} \right)=r_{0}$ is invertible then $f\left( t \right)$ is invertible for all $t\in \mathbb{R}$ as this is true for $\exp\left( \int_{t_{0}}^{t}s\left( u \right)du \right)$.
Unfortunately this does not work if the Banach algebra A is not commutative because then a solution of the differential equation cannot be given explicitly using the exponential function on A.
Best Answer
Let $f'(t)=f(t)s(t)$, $a:=f(t_0)$ invertible. Consider the IVP $g'(t)=-s(t)g(t)$, $g(t_0)=a^{-1}$. Then for all $t$ $$ (fg)'(t)=f'(t)g(t)+f(t)g'(t)= f(t)s(t)g(t)-f(t)s(t)g(t)=0 $$ and $f(t_0)g(t_0)=1$ ($1$ the unit in $A$). Thus $f(t)g(t)=1$ for all $t$. So we have a right inverse for all time. Consider $f$ for example to the right $(t \ge t_0)$: The set of invertible elements is open hence $f(t)$ is invertible on an interval $[t_0,T)$. Then $g(t)=f(t)^{-1}$ $(t \in [t_0,T))$. Assume $f(T)$ is not invertible. Then $\|f^{-1}(t)\|=\|g(t)\|$ is unbounded on $[t_0,T)$, a contradiction. Thus $f$ stays invertible on $t \ge t_0$. The case $t \le t_0$ should by similar.