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Nov 11, 2016 at 11:45 comment added Benoît Kloeckner @TaQ I see where the misunderstanding is: I wonder about analyticity in the hypothesis (if we were to state a more general result than above), not about the analyticity in the conclusion which is just fine.
Nov 11, 2016 at 0:40 comment added TaQ So, if I understand you correctly, you are wondering what it means for the map $\Psi$ in your question to be analytic? Its domain, however, is an open set $V$ in the Banach space $X$ and its target space $D$ with the graph norm topology is also a Banach space when $A$ is assumed closed. Then $\Psi$ has, locally, a convergent power series representation in any reasonable sense around every point in $V$. Since the series converges in the graph norm topology, it also converges in the weaker original topology induced from $X$.
Nov 10, 2016 at 22:06 comment added Benoît Kloeckner I don't know what would be my problem since I don't think I see which definition you are talking about. I usually define an analytic map as one having a Taylor series expansion whose terms have operator norms defining a Taylor series with a positive radius of convergence. This implies the map is well-defined on an open set (and for absolute convergence to imply convergence I need completeness).
Nov 10, 2016 at 21:29 comment added TaQ What is your problem with the definition of analyticity for a map defined on (the whole of?) a dense subspace? A dense subspace of a Banach space is just a possibly incomplete normed/normable space. Further, the "power series approach" already subsumes several possibilities according to what kind of convergence one wishes to have. Moreover, the addition of the injectivity assumption seems to be superfluous.
Nov 8, 2016 at 8:14 history edited Benoît Kloeckner CC BY-SA 3.0
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Nov 8, 2016 at 8:13 comment added Benoît Kloeckner @TaQ: concerning real analyticity, I am interested in the power series approach, but my specific problem would be to define analyticity for maps defined on a dense subspace while it is usually done for maps defined on an open set.
Nov 8, 2016 at 8:06 comment added Benoît Kloeckner @TaQ: of course, I'll add that $A$ is closed (I had this in mind but forgot to write it) and that $A+Db_{x_0}$ is injective.
Nov 8, 2016 at 0:21 comment added TaQ I do not have an explicit counterexample but it seems to me that at least one should add an assumption like $A$ being closed (or something similar) for the suggested result to hold true. For the various possibilities of analyticity, see e.g. the references here on page 1.
Nov 6, 2016 at 11:42 comment added Benoît Kloeckner @MichaelRenardy: I feel silly, but I guess this is what MO is for: question that are not so easily dealt with by oneself for an outsider, but that only take seconds to answer for someone in the field. I am surprised my tentative bibliographic searches did not point me to this! You should make this an answer.
Nov 5, 2016 at 15:22 comment added Michael Renardy The usual way of handling this is to make D a Banach space (with the graph norm of A) and then apply the usual inverse function theorem to the mapping from D to X. Is there any reason why your example does not fit this?
Nov 5, 2016 at 14:47 history asked Benoît Kloeckner CC BY-SA 3.0