I am having a question on the following proof of the Inverse function Theorem in Deimling's book. It is stated that if $G$ is continuously differentiable, then $G_{|U}^{-1}$ is also continuosly differentable. However in the proof Demling proceeds to show that $T=G_{|U}^{-1}$ is differentiable (locally) and $T'y = G'(Ty)^{-1}$ holds. However it doesn't explain that $T'(\cdot)$ i.e. $G'(\cdot)^{-1}$ is continuos?
So my question: how does continuity of $T'(\cdot)$ follows from continuity of $G'(\cdot)?$
Because he proceeds to prove for higher degrees of smoothnes, by induction. Induction however is not a proper thing to do if he hasn't proven the case $m=1$ in his notations. If it's obvious from the circled equality that in addition to being
P.S. I am aware of my question here my question her and my current question has the same principle in core I think. Namely that the inversion map $A \to A^{-1}$ is smooth. In particular continuous which would probably explain what Deimling has in mind.
How do I prove that indeed the inversion map is smooth? And doe this , like I suspect, resolve my problem?
Here I need only the continuity of $Inv$ and its even Lipchitz continous since $||Inv(A+H)-Inv(A)||\leq ||H||.\sum_{k=1}^{\infty}||A^{-1}||^{k+1}||H||^{k-1}=c||H||.$ (Provided $A$ is invertiable and $||A-H|| < 1/||A^{-1}||.$) And now $T'(\cdot) = G'(\cdot)^{-1} = Inv \circ G'(\cdot)$ and both $Inv$ and $G'(\cdot)$ are continuous. So indeed...:) Makes sense..
– Petar Jun 11 '23 at 23:46