Let $X$ be a Banach space, $\psi \in C^k(X;\mathbb{R})$ and $M = \{ x \in X: \psi(x) = 0 \} \neq \emptyset$ and $\psi'(x) \neq 0$ on $M$. For $x \in X$ let $x^* := D_x \psi$ be the Frechét differential of $\psi$ at $x$ and $e \in X$ such that $x^*(e) = 1$.
I want to prove that for $x \in X$ we have $$x = u + x^*(x) e$$ where $u \in \text{Kernel}(x^*)$.
I know that the space is split in the direct sum of image and kernel, e.g. $$ X = \text{Kernel}(x^*) \oplus \text{Image}(x^*) = \text{Kernel}(x^*) \oplus \text{span}(e)$$ I'm rather curious about the term $x^*(x)$ in the equation.
If anybody would be able to help me, I'd greatly appreciate it.
$x^{*}$ is a linear functional from $X$ in to $\mathbb R$ and $x^{*} (x-x^{*}(x)e)=x^{*}(x)-x^{*}(x)x^{*}(e)=0$ since $x^{*}(e)=1$. Take $u=x-x^{*}(x)e$