Let $X,Y$ be complete metric spaces, and let $\Sigma:X\times Y\rightarrow Y$ be a continous mapping which satisfies the following property: there exists a $C<1$, such that for all $x\in X$ and $y_{1},y_{2}\in Y$ one has $d(\Sigma(x,y_{1}),\Sigma(x,y_{2}))\leq Cd(y_{1},y_{2})$. The fixed point theorem for complete metric spaces then implies that for all $x\in X$, there exists a unique $y(x)\in Y$ such that $\Sigma(x,y(x))=y(x)$. It is possible to prove that the correspondence $x\mapsto y(x)$ is a continuous map of metric spaces $X\rightarrow Y$.

So far I stated what I know about general metric spaces. Now furtherly assume that in particular, $X,Y$ are Fréchet spaces, and that $\Sigma$ is a smooth map of Fréchet manifolds.

My question is this: can we conclude under these additional assumptions that the correspondence $x\mapsto y(x)$ is actually a smooth map between Fréchet manifolds? I haven't found any reference for this, nor could I find a way to extend the continuous-metric proof to the smooth setting.

I will be extremely thankful for guidance or reference.

  • 2
    $\begingroup$ This is not enough to conclude that the map is smooth. You need an assumption on the differential of $\Sigma$. You can see this when the dimension is finite or even just $1$. The smooth dependence is in general proved using an implicit function theorem. For Fréchet maps, the appropriate one is the Nash implicit function theorem. $\endgroup$
    – Deane Yang
    Jul 18 at 15:56

Deane Yang's comment shows that the premise of my question naively fails even in finite dimension. However, thanks to his insight, I think that I managed to figure it out in the Fréchet setting as well. For reference's sake, here is how I think my question falls into the setting of Theorem 3.3.1 in Richard Hamilton's article on the Nash-Moser inverse function theorem (1982):

Suppose that $X,Y$ are tame Fréchet spaces (which they are in my original setting, since they are both section spaces of vector bundles over a compact Riemannian manifold with boundary, equipped with the $C^{\infty}$-topology. More details are in Hamilton's article) and $\Sigma$ is a tame map (in my original setting, a nonlinear differential operator between these spaces). Consider the map $A=\Sigma-I_{Y}$, where $I_{Y}(x,y)=y$. Then the correspondence $x\mapsto y(x)$ in the above becomes the condition $A(x,y(x))=0$. Here is where Deane's comment about the assumption on the differential of $\Sigma$ enters into play: we need to assume that the partial directional derivative (also known as partial Gateaux derivative) $D_{x}A(x,y)=D_{x}\Sigma(x,y):X\rightarrow X$ is a surjective linear map whenever $A(x,y)=0$. Under this assumption, and a certain "good guess" for an inverse which Hamilton explains about in his article, if $A(x_{0},y_{0})=0$ for some $(x_{0},y_{0})$, we can find a neighbourhood of $(x_{0},y_{0})$ for which every $x$ has a $y$ such that $A(x,y)=0$, and the solution map $x\mapsto y(x)$ restricted to this neighbourhood is a smooth tame map.

Of course the situation in the case where $X,Y$ are Banach spaces (which includes the finite dimensional case) is much simpler as we only need to use the standard implicit function theorem. Thank you Deane!


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