The following is a theorem from Fischer and Marsden's 1975's paper: Linearization stability of nonlinear PDEs.


Let $X, Y$ be Banach manifolds and $\Phi: X \rightarrow Y$ be $C^1$. Let $x_0 \in X$ and $\Phi\left(x_0\right)=y_0$. Suppose that $T \Phi\left(x_0\right)$ is surjective with the splitting kernel. Then the equation $\Phi(x)=y_0$ is linearization stable about $x_0$.

Proof. From the implicit function theorem, the set $\Phi^{-1}\left(y_0\right)$ is a $C^1$ submanifold near $x_0$ with tangent space the kernel of $T \Phi\left(x_0\right)$. Thus $h \in T_{x_0} X$ is a first order deformation iff $h \in \operatorname{ker} T \Phi\left(x_0\right)$ iff $h \in T_{x_1}\left(\Phi^{-1}\left(y_0\right)\right)$, and since $\Phi^{-1}\left(y_0\right)$ is a submanifold, there exists a curve $x(\lambda) \in \Phi^{-1}\left(y_0\right)$ which is actually tangent to $h$. (QED)

Can someone please explain the following sentence? From the implicit function theorem, the set $\Phi^{-1}\left(y_0\right)$ is a $C^1$ submanifold near $x_0$ with tangent space the kernel of $T \Phi\left(x_0\right)$.

How the implicit function theorem (Banach space version) gives us $T_{x_{0}} \phi^{-1}(y_{0})$ = kernel $(T \phi (x_{0})$? The remaining part of the proof is straightforward.

Most importantly, what is splitting kernel in this context? I worked with some examples that showed that the surjectivity of the map $\phi$ is required; otherwise, we can create counterexamples. Please note that I posted this problem here, and I'm sorry if this problem is inappropriate for this site. Thanks so much.

Edit: Thanks again. I think now I understand the proof.

  • 1
    $\begingroup$ The sentence you highlighted in bold is completely analogous to the finite dimensional implicit function theorem, I don't see what is there to explain. The terminology "splitting kernel" I guess is not the only way to phrase it. I believe it is simply a necessary hypothesis to apply the Banach space version of the implicit function theorem, that $X = T_{x_0}X \cong \ker T\Phi(x_0) \times Z$, where $Z\subset T_{x_0}X$ complements $\ker T\Phi(x_0)$. In general Banach spaces, not all subspaces are complemented. $\endgroup$ Nov 20, 2022 at 16:52
  • $\begingroup$ Thanks so much. $\endgroup$
    – Boka Peer
    Nov 20, 2022 at 20:15
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    $\begingroup$ Instead of surjective with splitting kernel you can also say that $T\phi(x_0)$ has a continuous linear right inverse. $\endgroup$ Nov 21, 2022 at 8:01


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