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Sep 25, 2019 at 14:40 comment added Yidong Luo @JochenGlueck Can this way yield an explicit expression for one infinite-dimensional range?
Sep 25, 2019 at 14:24 comment added Jochen Glueck You can, for instance, integrate the spectral measure of your operator over the constant function with value $1$ over any subset of the spectrum that does not contain $0$ in its closure; the result will be a spectral projection $P$ such that the restriction of your operator to the range of $P$ is countinuously invertible. Hence, $\mathcal{B}_i$ can be chosen as the range of any such projection $P$.
Sep 25, 2019 at 14:21 history edited Arun Debray CC BY-SA 4.0
\"{o} -> ö hopefully makes this easier to search for
Sep 25, 2019 at 13:14 comment added Yidong Luo @JochenGlueck This is a naive example i ignore. Ok, now the focus is on the infinite-dimensional case. Is there some known research on it?
Sep 25, 2019 at 13:09 comment added Jochen Glueck Well, every linear operator from a finite dimensional space to any normed space is continuous, so you can choose $\mathcal{B}_i$ to be any finite dimensional subspace of $\mathcal{R}(A_i)$ and the restriction of $A_i^{-1}$ to $\mathcal{B}_i$ will be continuous.
Sep 25, 2019 at 11:51 comment added Yidong Luo @JochenGlueck Now i have no mature thoughts. Maybe you can give me some examples to feel? Even the finite dimensional case will be welcome. Of course the bigger and infinite-dimensional will be better.
Sep 25, 2019 at 11:19 comment added Jochen Glueck What kind if conditions do you want $\mathcal{B}_i$ to satisfy? For instance, the situation becomes trivial if you allow finite-dimensional $\mathcal{B}_i$.
Sep 25, 2019 at 10:11 history edited Yidong Luo CC BY-SA 4.0
added 2 characters in body
Sep 25, 2019 at 9:36 history asked Yidong Luo CC BY-SA 4.0