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Let $\{U(t)\;|\;t \in \mathbb{R}\}$ be a subgroup of $\text{GL}_n(\mathbb{R})$ such that $t \mapsto U(t)$ is continuous and satisfies $U(0) = E_n$ and $$U(t+s) = U(s)U(t)$$

Is it true then that $t \mapsto U(t)$ is differentiable at $0$ (and hence for all $t \in\mathbb{R}$)?

If yes, I would appreciate a hint on how to prove this since it is clear by definition that $$U(h) - U(0) = o(1) \text{ for } h \to 0$$ but not $$U(h) - U(0) = o(||h||)\text{ for } h \to 0$$

Andrei Kh
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2 Answers2

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Yes. Since $I$ is non-singular, we may define in its small neighbourhood a matrix logarithm. It follows that $f(t) = \log U(t)$ is well defined in some neighbourhood of zero because $U$ is continuous and $U(0)=I$.

By definition, the subgroup $\{U(t): t\in\mathbb R\}$ is Abelian. Hence $\log\left(U(s)U(t)\right)=\log U(s)+\log U(t)$. Consequently, if we denote by $f_{ij}$ the $(i,j)$-th component of $f$, then we can reduce the original functional equation to $$ f(s+t)=f(s)+f(t) $$ or $f_{ij}(s+t)=f_{ij}(s)+f_{ij}(t)$ for each $(i,j)$. This is basically Cauchy's functional equation, with the exception that the domain of $f_{ij}$ is a neighbourhood of zero instead of the whole real line. However, the same solution method applies: using the additive property, one can show that there exists a real number $c_{ij}$ such that $f_{ij}(q) = c_{ij}q$ for every rational number $q$ inside the domain of $f_{ij}$. Then, using the continuity of $f_{ij}$ and the denseness of rational numbers in $\mathbb R$, we conclude that $f_{ij}(t)=c_{ij}t$.

Hence $f(t)=tC$ for some constant matrix $C$. That is, $U(t)=e^{tC}$ on some neighbourhood of zero. Thus $U$ is differentiable at zero.

Community
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user1551
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  • Great answer! Is it possible to prove this without using the multiplicativity of the matrix logarithm? (Since at this point in the lecture, we learned only that $||\log(A+E)-A|| \leq c||A||^2$ for some constant $c > 0$, i.e $\log$ is differentiable at $E$ with Frechet derivative $E$.) – Andrei Kh Mar 31 '17 at 16:44
  • @AndreiKh I don't know, but what is being used here is actually not the multiplicativity, but that (a) the logarithm of a matrix is a polynomial in the matrix, and (b) the matrix exponential function is locally bijective and its local inverse is the matrix logarithm function. Now, as $U(s)$ and $U(t)$ commute, $\log U(s)$ and $\log U(t)$ also commute. Therefore $$\exp(\log U(s)+\log U(t))=\exp(\log(U(s)))\exp(\log(U(t)))=U(s)U(t)=\exp(\log(U(s)U(t))).$$ Hence $\log U(s)+\log U(t)=\log(U(s)U(t))$. – user1551 Mar 31 '17 at 17:08
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Let $s(x) =\sum_k \frac{(-1)^k}{k} x^k $ then it is easy to see that $s\circ U $ is differentiable and thus so is $U.$