Suppose $X(t)\in M_n(\Bbb R), t\in\Bbb R$ and is continuous, invertible at every point on the real line, if the equation $$X(t)X(s)=X(t+s)$$ holds for all $t,s\in\Bbb R$, prove that there exists a constant square matrix $A\in M_n(\Bbb R)$ such that $$X(t)=\exp(At)$$
Now I've done most of the major work: I have worked out $X(0)=I$, if $X'(0)$ exists, then $X$ must fit this DE for $X$: $$X'=X'(0)X$$ Note that all solutions to this DE can be represented as $$X=\exp(X'(0)t)C$$, so $C=I$ will be determined by $X(0)=I$, which yields the desired result.
The only problem is I do not know how to show $X$ is differentiable on the real line, or in particular, at any given point, say $0$.
The only way seems to be checking by definition, so I must show, for instance, that $$\lim_{\Delta t\to 0}\frac{X(\Delta t)-I}{\Delta t}$$ converges.
But since I don't have any information about what particular form $X$ may take, so there's noting I can do with this expression.
Can you help me? Or is there any alternative methods that can avoid differentiation? Best regards!
Averaging a function makes it smoother. By the fundamental theorem of calculus, we know that for every $a > 0$ the function
$$Y(t) = \frac{1}{a}\int_t^{t+a} X(s)\,ds = \frac{1}{a}\int_0^a X(s+t)\,ds\tag{1}$$
is differentiable. Since $X$ is a homomorphism, we can write $(1)$ as
$$Y(t) = \frac{1}{a} \int_0^a X(s+t)\,ds = \frac{1}{a}\int_0^a X(s)X(t)\,ds = \underbrace{\Biggl(\frac{1}{a}\int_0^a X(s)\,ds\Biggr)}_{B_a}\cdot X(t).\tag{2}$$
If we can choose $a > 0$ so that $B_a$ is invertible, then we can write $(2)$ as
$$X(t) = B_a^{-1}\cdot Y(t),\tag{3}$$
which then shows the differentiability of $X$.
Why can we choose $a > 0$ so that $B_a$ is invertible?