# Linearization stability condition

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

Theorem.

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.

• 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. Nov 20, 2022 at 16:52
• Thanks so much. Nov 20, 2022 at 20:15
• Instead of surjective with splitting kernel you can also say that $T\phi(x_0)$ has a continuous linear right inverse. Nov 21, 2022 at 8:01