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Let $\mathbf{E}$ be a real banach space and $\mathbf{L} = L(\mathbf{E})$ the resultant banach space of bounded linear operators $T:\mathbf{E} \to \mathbf{E}$ equipped with the operator norm $\| T \| = \sup_{\| x \| \leqslant 1 } \|Tx\|$.

From https://en.wikipedia.org/wiki/Neumann_series I can see why the subset $\Omega \subset \mathbf{L}$ consisting of all invertible operators is an open set of $\mathbf{L}$. Moreover, from the accepted answer to differential inverse matrix I can see why the map $\Phi: \Omega \mapsto \mathbf{L} \, \,; \,\, T \mapsto T^{-1}$ is frechet differentiable, and specifically $D\Phi(T)S = -T^{-1} S T^{-1}$.

How do I use this knowledge to prove the stronger result that $\Phi$ is smooth?

Regarding finite dimensional spaces $\mathbf{E}$, this question has been asked before on stackexchange, but the answers almost always employ Cramer's rule for matrices and partial derivatives as ratios of smooth functions. I am not assuming $\mathbf{E}$ is finite dimensional, so Cramer's rule doesn't apply. I'm trying to find an answer that doesn't deal with partial derivatives at all.

Community
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joeb
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1 Answers1

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As seen from your link to the Neumann series, the inversion map $\Phi$ can be given by a power series at $T$ as: $$ \Phi(T+S) = \sum_{n \ge 0} (-T^{-1} S)^n T^{-1}. $$ This power series has radius of convergence $\rho = \frac{1}{\|T^{-1}\|}>0$, so it absolutely converges inside that, and hence $\Phi$ is analytic.

WillG
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Jaap Eldering
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    Assuming $T$ is invertible, for $S$ such that $| S | < \frac{1}{| T^{-1} |}$ we have $| -T^{-1} S | \leqslant | T^{-1} | \cdot | S | < 1$, and so the Neumann series $\sum_{n=0}^\infty (-T^{-1}S)^n $ converges absolutely. Therefore, $\mathrm{Id} - (-T^{-1}S) = \mathrm{Id} + T^{-1}S$ is invertible, and its inverse equals the limit of said Neumann series. But then $\Phi(T + S) = (T + S)^{-1} = (T(\mathrm{Id} +T^{-1}S))^{-1} = (\mathrm{Id} + T^{-1}S)^{-1}T^{-1} = \left( \sum_{n=0}^\infty (-T^{-1}S)^n \right) T^{-1} = \sum_{n=0}^\infty (-T^{-1}S)^nT^{-1}$. – joeb Feb 15 '17 at 23:30
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    Thus if $\tilde{S}$ is such that $| \tilde{S} - T | < \frac{1}{| T^{-1} |}$, upon letting $S = \tilde{S} - T$, it follows that $\Phi(\tilde{S}) = \Phi( T + S) = \sum_{n=0}^\infty (-T^{-1}S)^nT^{-1} = \sum_{n=0}^\infty (-T^{-1}( \tilde{S} - T))^nT^{-1}$. Is this a power series expansion at $T$? What would the coefficients be? Also, I'm not familiar with power series in Banach algebras. Is there a resource that explains why such a series is smooth on its domain of convergence? I'm not really that confident with what analytic means outside of complex functions of one complex variable. – joeb Feb 15 '17 at 23:39
  • Yes, that's the power series expansion. It's not easy to talk about coefficients, because these would be infinite-dimensional matrices and higher order objects. A book I've used as reference is: J. Mujica. Complex analysis in Banach spaces. Vol. 120. North-Holland Mathematics Studies. Amsterdam: North-Holland Publishing Co., 1986, pp. xii+434. ISBN: 978-0-444-87886-1 – Jaap Eldering Feb 16 '17 at 01:43