Let $U\in\mathbb{R}^n$ be open and $f:U\to\mathbb{R}^n$ be a $\mathcal{C}^1$ map. $\exists p\in U$ such that $\;f(p) = 0$ and $Df_{|p}$ is invertible. Define $\phi(x)= x-(Df_{|x})^{-1}f(x)$, show that $\exists U'$ an open neighborhood of $p$ such that $\phi(U')\subset U'$ and that $\forall x_0\in U$ the iteration $\phi^{\circ n}(x_0)\to p$.
I understand what I need to do, in fact proved the statement for a different iteration map $\psi(x) = x-(Df_{|p})^{-1}f(x)$. I show my proof here.
Let $A: = Df_{|p}$, since $A$ is invertible, $\|A^{-1}\|\ne 0$. By continuity of derivative
$$\exists \epsilon>0.\;\forall q\in N_\epsilon(q).\;|A-f'(q)|<\frac{1}{2}\|A^{-1}\|^{-1}$$
Now I try to bound the derivative
$$\forall x\in N_\epsilon(p).\;|\psi'(x)| = |I-A^{-1}f'(x)| = |A^{-1}(A-f'(x))|\le\|A^{-1}\|\cdot|A-f'(x)|<\frac{1}{2}$$
Since bounded derivative in convex open domain implies Lipschitz continuity, (Theorem 9.19 page 218 Rudin), we have
$$\forall x,y\in N_\epsilon(p).\;|\psi(x)-\psi(y)|\le \frac{1}{2} |x-y|$$
$\psi$ is a contraction, and $p$ is a fixed point of $\psi$.
$$\forall x\in N_{\epsilon}(p).\;x\ne o.\;\eta: = |x-p|.\;B: = N_\eta(p). \overline{B}\subsetneq U\subset\mathbb{R}^n$$
$$\forall x\in \overline{B}.\;|\psi(x)-p| = |\psi(x)-\psi(p)|\le\frac{1}{2}|x-p|\Longrightarrow \psi(x)\in \overline{B}$$
$\overline{B}$ is a compact subset of $\mathbb{R}^n$, it is complete. We have shown $\psi$ is a contraction of $\overline{B}$ into $\overline{B}$. Therefore by contraction fixed point theorem, there is a unique fixed point of $\psi$, which is $p$.
That is $\lim\limits_{n\to\infty} \psi^{\circ n}(x) = p$.
Now I try to do the same with $\phi$. I want to bound $\phi'(x)$. Due to the $Df_{|x}$ term, and that I don't know if $f$ is twice differentiable, I can't use $\phi'(x)$. So I try to show directly that $\phi(x)$ is a contraction.
$$|\phi(x)-\phi(y)| = |x-y-(Df_{|x})^{-1}\cdot f(x)+(Df_{|y})^{-1}\cdot f(y)|$$ $$ = |x-y-(Df_{|x})^{-1}\cdot f(x)+(Df_{|y})^{-1}\cdot f(y)-(Df_{|x})^{-1}\cdot f(y)+(Df_{|x})^{-1}\cdot f(y)|$$ $$\le |x-y|+|(Df_{|y})^{-1}-(Df_{|x})^{-1}|\cdot|f(y)|+\|(Df_{|x})^{-1}\|\cdot|f(x)-f(y)|$$ The last two terms can be made arbitrary small by continuity argument, I get $$|\phi(x)-\phi(y)|\le |x-y|+\epsilon$$ which is not as good as a contraction. Please help me show that $\phi$ is a contraction in some neighborhood of $p$. Thanks!
Ok, my friend Stefan helped me. It turned out that I don't need contraction fixed point theorem.
First observe $\phi(p) = p$.
By Taylor's Theorem (or definition or differentiability) $$0 = f(p) = f(x)+Df_{x}(p-x)+o(|p-x|)$$ $$\Longrightarrow\;\;\phi(x) = x-(Df_x)^{-1}f(x) = p+(Df_x)^{-1}o(|p-x|)$$ $$\Longrightarrow\;\;\frac{|\phi(x)-p|}{|x-p|}\le \|Df_x^{-1}\|\cdot \frac{o(|p-x|)}{|p-x|}$$ Choose $0<\epsilon$ so small that $$\forall x\in N_\epsilon(p).\;\frac{o(|p-x|)}{|p-x|}<\frac{1}{2\|Df_x^{-1}\|}$$ Therefore $$\forall x\in N_{\epsilon_2}(p).\;|\phi(x)-p|<\frac{1}{2}|x-p|$$ Thus $\phi(N_\epsilon(p))\subset N_\epsilon(p)$, further
$$|\phi(x)^{\circ n}-p|<\left(\frac{1}{2}\right)^n|x-p|$$ That is, $\forall x\in N_{\epsilon}(p).\;\lim\limits_{n\to\infty} \phi^{\circ n}(x) = p$