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Suppose $E$ is a banach space and $X$ is the banach space of continuous linear maps $E \to E$ (using the operator norm for $X$).

Let $\Omega = \{ \lambda \in X : \lambda \,\, \mathrm{invertible} \}$. Why is the map

$\omega:\Omega \to X \, ; \, \lambda \mapsto \lambda^{-1}$

$C^\infty$? I can prove this for finite dimensional spaces $E$, but don't see what I should do for infinite dimensional spaces.

EDIT: I know why $\Omega$ is an open subset of $X$, from this wikipedia article: https://en.wikipedia.org/wiki/Neumann_series

Also, I looked at the answer to this post differential inverse matrix and I think I see why the answer works perfectly well in infinite dimensional spaces, too, as the operator norm on $X$ (as I have defined $X$) is complete and sub-multiplicative. This gets me to the point where $\omega$ is differentiable and for a given $\lambda \in \Omega$ the derivative $D\omega(\lambda)$ is the continuous linear map $X \to X \, ; \mu \mapsto -\lambda^{-1} \circ \mu \circ \lambda^{-1}$.

How do I get from this description of $D\omega$ to proving that $\omega$ is smooth?

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joeb
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  • Use a local taylor expansion for the inverse, using the formula $\lambda^{-1} = \sum_0^{\infty} (1-\lambda)^n$, which holds in any unital Banach algebra if $| \lambda-1|<1$. Then use the fact that analytic functions are necessarily smooth in any Banach algebra. – shalin Jan 31 '17 at 03:10
  • Thanks for the tip but I don't know anything about taylor expansion in Banach algebras. I think I'd be able to prove that $\lambda^{-1} = \sum_0^\infty (1 - \lambda)^n$, but not much more. I can barely follow the answer by @copper.hat in http://math.stackexchange.com/questions/128556/inversion-of-matrices-is-a-diffeomorphism concerning first order differentiation. Is there a resource that you can recommend? – joeb Jan 31 '17 at 03:34

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