2
$\begingroup$

I have seen an author use the Implicit Function Theorem for a map whose second partial derivative has a bounded inverse, but is unbounded. The map itself is not defined on an open set, but only on a domain. I failed to find a convincing reference, so I ask here.

To be more precise, assume $X$ is a Banach space, $D$ a linear subspace (dense if needed) and $A$ an unbounded closed linear operator of $X$, with domain $D$. Let $b:U\subset X\to X$ be an analytic map where $U$ is an open set. Consider the map $\Phi:U\cap D\to X$ defined by $\Phi(x)=Ax+b(x)$.

Assume that at some point $x_0\in U\cap D$, the linear unbounded operator $A+Db_{x_0}:D\to X$ is injective and has a bounded right inverse (which thus takes its values in $D$). Is it true that there is an analytic map $\Psi:V\to X$, defined in a neighborhood $V$ of $\Phi(x_0)$, taking its values in $D\cap U$, such that for all $x\in V$ it holds $\Phi(\Psi(x))=x$?

Any similar statement, pointers to literature, Implicit Function Theorem variants would be helpful. Maybe the usual proof only needs some slight adaptation, but right know I don't see it.

Edit: the question is basically answered by Michael Renardy comment.

I also wonder what would be the general definition of an analytic map in such context, where the map is only define on a dense subspace.

Improve this question
$\endgroup$
9
  • 3
    $\begingroup$ 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? $\endgroup$
    – Michael Renardy
    Commented Nov 5, 2016 at 15:22
  • $\begingroup$ @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. $\endgroup$
    – Benoît Kloeckner
    Commented Nov 6, 2016 at 11:42
  • $\begingroup$ 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. $\endgroup$
    – TaQ
    Commented Nov 8, 2016 at 0:21
  • $\begingroup$ @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. $\endgroup$
    – Benoît Kloeckner
    Commented Nov 8, 2016 at 8:06
  • $\begingroup$ @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. $\endgroup$
    – Benoît Kloeckner
    Commented Nov 8, 2016 at 8:13

0

Reset to default

You must log in to answer this question.