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For Proposition 2.5 of Chapter I in Lang's Fundamentals of Differential Geometry, it is stated

Let $E$, $F$ be Banach spaces. Then the set of toplinear (that is, bounded linear) isomorphisms ${\rm Lis}(E,F)$ is open in $L(E,F)$ (the set of bounded linear maps).

And the proof is as follows:

If Lis($E,F$) is not empty, one is immediately reduced to proving that ${\rm Laut}(E)$ is open in $L(E,E)$. We then remark that if $u \in L(E,E)$ and $|u| < 1$, then the series \begin{equation} 1 + u + u^2 + \ldots \end{equation} converges. Given any toplinear automorphism $w$ of $E$, we can find an open neighborhood by translating the open unit ball multiplicatively from $1$ to $w$.

I am not able to follow this argument very well.

The reduction to the case of endomorphisms and automorphisms is believable enough, but after that, I cannot even tell how the following statements are related to each other, unfortunately.

It seems to me that they are indicating that for a bounded linear map $u$ of operator norm less than 1, that the operator $(1-u)^{-1} = \sum_{i \geq 0} u^i$ is well defined.

However, I have never heard the phrase "translate the open unit ball multiplicatively" before, and it is unclear to me what is meant.

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    $\begingroup$ To translate a set $A$ multiplicatively by the multiplier $k$ on the left, you replace $A$ with $kA:=\{kx\;|\;x\in A\}$. (You can also translate on the right to get $Ak:=\{xk\;|\;x\in A\}$.) In Lang's case, $A$ is the open unit ball in $\mathrm L(E,E)$ and $k$ is $w$, because $w$ is the multiplier that will translate $1$ to $w$ (since $w1=w$). Lang specifies that we're translating multiplicatively because it's more common to translate subsets of vector spaces additively by replacing $A$ with $A+v:=\{x+v\;|\;x\in A\} $, but Lang doesn't want that. $\endgroup$ Commented yesterday

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Do you agree that this argument shows there is an open neighborhood of the identity endomorphism consisting entirely of invertible elements… or in other words, the identity lies in the interior of $Laut(E)$ (also written $GL(E)$)?

The next part of the argument is telling us how to get from a neighborhood of the identity to a neighbourhood of a general invertible element $T$. And the claim is that $u\mapsto T\circ u$ is a linear homeomorphism (why is it continuous, and why is the inverse continuous?) on the space of endomorphisms, so maps a neighbourhood of the identity to a neighbourhood of $T$. The former neighborhood consists entirely of invertible elements, hence so does the latter (why?). Thus every point of $GL(E)$ is an interior point.

This is a general proof strategy when dealing with groups: prove something near the identity, then you can use the group translations to consider a general point.

Anyway, if this sounds too abstract, just consider a general invertible element $T$, and write $T-v=T\circ (I- T^{-1}\circ v)$. Do you see based on what you have that if $v$ has sufficiently small (how small?) operator norm, then the bracketed term is invertible, and hence so is $T-v$?

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The wording in the proof seems unecessairily complicated and abstract.

Just let $p$ be an arbitrary perturbation of $w$, and set $u=w^{-1}p$. We have $$ w - p = w - w w^{-1} p = w(1-u). $$ By the submultiplicative property of the operator norm, we have $$ |u| \le |w^{-1}||p|. $$

If $|p| < |w^{-1}|^{-1}$, then $|u| < 1$, so $(1-u)$ is invertible, so $w(1-u)$ is invertible.

Therefore, we have found an open ball of invertible operators around $w$ (and the ball has radius $|w^{-1}|^{-1}$).

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