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Benoît Kloeckner
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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.

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 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 $A+Db_{x_0}:D\to X$ 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.

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.

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.

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Benoît Kloeckner
  • 14.4k
  • 1
  • 60
  • 106

Inverse and implicit function theorems with domain

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 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 $A+Db_{x_0}:D\to X$ 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.

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.