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I am looking at the following proof of the Inverse function theorem in "Nonlinear Functional Analysis" by Deimling. Here in the proof, he uses the following argument which seems extremely far from trivial and longer than the proof itself and he mentions nothing what so ever of its validance: Knowing that $G'(x_0)$ is homeomorphism and $G'(.)$ is $||\cdot||_{X}--||\cdot||_{L(X,Y)}$ continous and $T$ is continuous, then $G'(Ty) is alswo homeomorphims.

Essentially why is that $G'(x)$ is also homeomorphism for x sufficiently close to $x_0?$

I tried a few things, among which are related to the answer of wdacda here : injective-linear-operators

So if i could prove that $\inf\{||G'(x)\overline{x}|| : \overline{x} \in S_X\} > 0$ is a possitive number then it would follow that $G'(x)$ is injective and the inverse is bounded.

Petar
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    The group of invertible operators is open, and $G':U_0\rightarrow L(X,Y)$ is continuous at $x_0$. This, in a small neighborhood of $B$ of $x_0$, $G:B\rightarrow L(X,Y)$ is invertible. – Mittens Jun 10 '23 at 21:16
  • The invertibility of the map $G'$ follows from the implicit function theorem. In a different posting of your I derived in my answer the implicit function theorem via a uniform contraction principle that solves twi fixed point equations, one for the implicit function and another one for its derivative (here it can be seen that invertibility holds). – Mittens Jun 10 '23 at 23:16

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The set of bound, invertible linear operators is open in $L(X,Y)$. Because $G'(.)$ is continuous, then the set $\{x: \text{$G'(x)$ is invertible}\}$ is open in $X$. So if $G(x_{0})$ is invertible, there must be a neighborhood of $x_{0}$ so that $G'(x)$ is invertible for all $x$ in there.

FZan
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  • Is $L(X,Y)$ even a Banach algebra? Since we can't compose two $X\to Y$ maps? And the proof there is for $L(X,X)?$ – Petar Jun 10 '23 at 21:06
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    Ah yes sorry about that, by the answer I linked there's some neighborhood of $x_{0}$ so that $G'^{-1}(x_{0})\circ G'(x)$ is invertible so the result follow by composing with $G'(x_{0})$ again – FZan Jun 10 '23 at 22:05
  • Thank you! This was very helpful. – Petar Jun 10 '23 at 23:09