I am studying Lie algebra and Lie groups from the lecture notes given by our instructor. Here I find a definition of differentiable Lie algebra homomorphism. What it says is as follows $:$
Let $E$ be a Banach space and $G,H \subseteq GL(E)$ be two linear Banach Lie groups. Then a continuous group homomorphism $\Phi : G \longrightarrow H$ is said to be differentiable if for any given $\varepsilon \gt 0$ the following conditions hold $:$
$(1)$ For any differentiable curve $\gamma : (-\varepsilon, \varepsilon) \longrightarrow G$ the composition $\Phi \circ \gamma : (-\varepsilon, \varepsilon) \longrightarrow H$ is also differentiable.
$(2)$ If $\alpha, \beta : (-\varepsilon, \varepsilon) \longrightarrow G$ be two differentiable curves with $\alpha (0) = \beta (0)$ and $\alpha'(0) = \beta'(0)$ then $$(\Phi \circ \alpha)'(0) = (\Phi \circ \beta)'(0).$$
Then a theorem has been left as an exercise which states that Any continuous Lie group homomorphism is differentiable. I have tried that and able to prove that there exists a unique bounded linear map $\phi$ between the corresponding Lie algebras $\mathfrak g$ and $\mathfrak h$ such that for all $t \in \Bbb R$ and for any $X \in \mathfrak g$ we have $$\Phi (e^{tX}) = e^{t \phi (X)} \ \ \ \ (1)$$ Does it help me anyway to prove the differentiability of $\Phi\ $? Would anybody please help me in this regard?
Thank you so much for your time.
EDIT $:$ If $\gamma$ is a differentiable curve such that for all $t,s \in (-\varepsilon, \varepsilon)$ with $t + s \in (-\varepsilon, \varepsilon)$ we have $\gamma (t + s) = \gamma (t)\ \gamma (s)$ then we are through. Because I have come across a theorem which states that for such a curve $\gamma$ there always exist a unique $X \in \mathfrak g$ such that for all $t \in \Bbb R$ we have $\gamma (t) = e^{tX}.$ Since the right hand side of $(1)$ is always differentiable it follows that $\Phi \circ \gamma$ is also differentiable. Also for two such curves $\alpha$ and $\beta$ if $\alpha '(0) = \beta'(0)$ then $\alpha \equiv \beta.$ So we have $(\Phi \circ \alpha)' (0) = (\Phi \circ \beta)'(0).$
