Nondegenerate local Hölder coordinates for a Hölder coefficient

Replace $U$ by $U-p$ and $\mu (z)$ by $\mu (p+z)$; translation preserves the Hölder norms and equation, and we undo it at the end. Now $p=0$. Put $a:=\mu (0)$ and $A(z):=z+a\bar{z}$. Its real Jacobian is $1-\mathrm{\mid }a{\mathrm{\mid }}^2>0$ and $A^{-1}(w)=(w-a\bar{w})\mathrm{/}(1-\mathrm{\mid }a{\mathrm{\mid }}^2)$. On $A(U)$ define $\nu (w):=(\mu (A^{-1}w)-a)\mathrm{/}(1-\bar{a}\mu (A^{-1}w))$. The denominator has modulus at least $1-k_0^2>0$. If $g_{\bar{w}}=\nu g_w$ and $\mathrm{\Phi }=g\circ A$, [F3] and the chain rule [F11] give ${\mathrm{\Phi }}_{\bar{z}}=(g_w\circ A)(a+\nu \circ A)$ and ${\mathrm{\Phi }}_z=(g_w\circ A)(1+\bar{a}\nu \circ A)$, so ${\mathrm{\Phi }}_{\bar{z}}=\mu {\mathrm{\Phi }}_z$ by the definition of $\nu$.

We first record the finite product bound used below. Repeated coordinatewise product rules give $D^{\beta }(uv)={\sum }_{\gamma \le \beta }\left(\genfrac{}{}{0px}{}{\beta }{\gamma }\right)D^{\gamma }u{D}^{\beta -\gamma }v$. For $\mathrm{\mid }\gamma \mathrm{\mid }<k$, the mean-value theorem bounds the $\alpha$-seminorm of $D^{\gamma }u$ by its next derivative when $\mathrm{\mid }x-y\mathrm{\mid }\le 1$, and the sup norm does so when $\mathrm{\mid }x-y\mathrm{\mid }\ge 1$; for $\mathrm{\mid }\gamma \mathrm{\mid }=k$ the top seminorm is part of the norm. Since $[fg{]}_{0,\alpha }\le \mathrm{\parallel }f{\mathrm{\parallel }}_{\mathrm{\infty }}[g{]}_{0,\alpha }+\mathrm{\parallel }g{\mathrm{\parallel }}_{\mathrm{\infty }}[f{]}_{0,\alpha }$, the Leibniz sum gives $P_{k,\alpha }<\mathrm{\infty }$ with $\mathrm{\parallel }uv{\mathrm{\parallel }}_{C^{k,\alpha }}\le P_{k,\alpha }\mathrm{\parallel }u{\mathrm{\parallel }}_{C^{k,\alpha }}\mathrm{\parallel }v{\mathrm{\parallel }}_{C^{k,\alpha }}$ on ${\mathbb{R}}^2$.

The identity $\mathrm{\mid }1-\bar{a}t{\mathrm{\mid }}^2-\mathrm{\mid }t-a{\mathrm{\mid }}^2=(1-\mathrm{\mid }a{\mathrm{\mid }}^2)(1-\mathrm{\mid }t{\mathrm{\mid }}^2)$ shows that $\nu (0)=0$ and $\mathrm{\mid }\nu (w)\mathrm{\mid }<1$. The denominator bound and repeated chain and product rules show that $\nu$ is $C^{k,\alpha }$ near $0$: the rational map $t\mapsto (t-a)\mathrm{/}(1-\bar{a}t)$ has bounded derivatives on $\mathrm{\mid }t\mathrm{\mid }\le k_0$, and composition with the affine map $A^{-1}$ preserves the finite-order derivative and top Hölder bounds.

Choose $r>0$ so small that ${\bar{B}}_{2r}(0)\subset A(U)$ and define $b_r(\xi ):=\chi (\xi )\nu (r\xi )$ on $B_2(0)$, extended by zero outside. Since $\chi$ is supported in a compact subset of $B_2$, this extension is $C^{k,\alpha }$ and supported in ${\bar{D}}_2$. Its norm tends to zero as $r\downarrow 0$. For $k=0$, $\nu (0)=0$ gives ${\sup }_{B_2}\mathrm{\mid }\nu (r\cdot )\mathrm{\mid }\le 2^{\alpha }{r}^{\alpha }[\nu {]}_{0,\alpha ;B_{2r}}$ and $[\nu (r\cdot ){]}_{0,\alpha ;B_2}=r^{\alpha }[\nu {]}_{0,\alpha ;B_{2r}}$. For $k\ge 1$, the zeroth-order supremum is $O(r)$, the order-$j$ derivative suprema are $O(r^j)$ for $1\le j\le k$, and the top seminorm is $r^{k+\alpha }[D^k\nu {]}_{0,\alpha ;B_{2r}}$. The product bound of step 1.2 with the fixed cutoff proves $\mathrm{\parallel }{b}_r{\mathrm{\parallel }}_{C^{k,\alpha }({\mathbb{R}}^2)}\to 0$. Decrease $r$ so also $P_{k,\alpha }{M}_{k,\alpha }\mathrm{\parallel }{b}_r{\mathrm{\parallel }}_{C^{k,\alpha }}<1\mathrm{/}2$, $2(1+M_{k,\alpha })\mathrm{\parallel }{b}_r{\mathrm{\parallel }}_{C^{k,\alpha }}\le 1\mathrm{/}4$, and $\mathrm{\parallel }{b}_r{\mathrm{\parallel }}_{\mathrm{\infty }}<1$.

Let $X:=\{q\in C_b^{k,\alpha }({\mathbb{R}}^2;\mathbb{C}):\mathrm{supp}q\subseteq {\bar{D}}_2\}$. It is nonempty and closed in the Banach space of [F6], since norm convergence implies uniform convergence and a uniform limit of functions vanishing outside ${\bar{D}}_2$ still vanishes there. Hence $X$ is complete by [F7]. With the radius from step 3.1, define $\mathcal{T}(q):=b_r(1+Sq)$. It maps $X$ to $X$, and steps 1.2 and [F14] give $\mathrm{\parallel }\mathcal{T}(q_1)-\mathcal{T}(q_2){\mathrm{\parallel }}_{C^{k,\alpha }}\le P_{k,\alpha }{M}_{k,\alpha }\mathrm{\parallel }{b}_r{\mathrm{\parallel }}_{C^{k,\alpha }}\mathrm{\parallel }{q}_1-q_2{\mathrm{\parallel }}_{C^{k,\alpha }}$. By [F8] there is a fixed point $q\in X$; its equation and the same bound give $\mathrm{\parallel }q{\mathrm{\parallel }}_{C^{k,\alpha }}\le 2\mathrm{\parallel }{b}_r{\mathrm{\parallel }}_{C^{k,\alpha }}$.

Put $\phi (\xi ):=\xi +Tq(\xi )$. By [F5], ${\phi }_{\bar{\xi }}=q=b_r(1+Sq)=b_r{\phi }_{\xi }$. The real operator norm of $D(Tq)$ is $\mathrm{\mid }Sq\mathrm{\mid }+\mathrm{\mid }q\mathrm{\mid }$, since its action is $h\mapsto (Sq)h+q\bar{h}$ and the argument of $h$ can align the two summands. Hence $\mathrm{\parallel }D(Tq){\mathrm{\parallel }}_{\mathrm{\infty }}\le (1+M_{k,\alpha })\mathrm{\parallel }q{\mathrm{\parallel }}_{C^{k,\alpha }}\le 1\mathrm{/}4$ by steps 3.1 and 4.1. Thus $\mathrm{\mid }{\phi }_{\xi }-1\mathrm{\mid }=\mathrm{\mid }Sq\mathrm{\mid }\le 1\mathrm{/}4$, and $J_{\phi }=\mathrm{\mid }{\phi }_{\xi }{\mathrm{\mid }}^2-\mathrm{\mid }{\phi }_{\bar{\xi }}{\mathrm{\mid }}^2=\mathrm{\mid }{\phi }_{\xi }{\mathrm{\mid }}^2(1-\mathrm{\mid }{b}_r{\mathrm{\mid }}^2)>0$. For the two real components of $Tq$, [F9] on the segment from $x$ to $y$ gives $\mathrm{\mid }Tq(x)-Tq(y)\mathrm{\mid }\le \sqrt{2}\mathrm{\parallel }D(Tq){\mathrm{\parallel }}_{\mathrm{\infty }}\mathrm{\mid }x-y\mathrm{\mid }\le (\sqrt{2}\mathrm{/}4)\mathrm{\mid }x-y\mathrm{\mid }$. Hence $\mathrm{\mid }\phi (x)-\phi (y)\mathrm{\mid }\ge (1-\sqrt{2}\mathrm{/}4)\mathrm{\mid }x-y\mathrm{\mid }>\mathrm{\mid }x-y\mathrm{\mid }\mathrm{/}2$: $\phi$ is injective and its inverse is Lipschitz. Since $J_{\phi }>0$, [F10] makes it a local $C^1$ diffeomorphism; injectivity then makes it a global diffeomorphism onto its open image.

On $D_r(0)$ define $g(w):=\phi (w\mathrm{/}r)$. Since $\chi =1$ on ${\bar{D}}_1$, step 5.1 gives $g_{\bar{w}}=\nu (w)g_w$ there. Put $V:=A^{-1}(D_r(0))\subset U$ and $\mathrm{\Phi }:=g\circ A$ on $V$. By step 1.1 and the real chain rule, ${\mathrm{\Phi }}_{\bar{z}}=\mu {\mathrm{\Phi }}_z$ at every $z\in V$, and $J_{\mathrm{\Phi }}(z)=\mathrm{\mid }{g}_w(Az){\mathrm{\mid }}^2(\mathrm{\mid }1+\bar{a}\nu (Az){\mathrm{\mid }}^2-\mathrm{\mid }a+\nu (Az){\mathrm{\mid }}^2)=\mathrm{\mid }{g}_w(Az){\mathrm{\mid }}^2(1-\mathrm{\mid }a{\mathrm{\mid }}^2)(1-\mathrm{\mid }\nu (Az){\mathrm{\mid }}^2)>0$. Here $g_w=r^{-1}{\phi }_{\xi }(w\mathrm{/}r)$ is nonzero by step 5.1. The maps $A$ and $g{\mathrm{\mid }}_{D_r}$ are injective, so $\mathrm{\Phi }$ is injective; $\mathrm{\Phi }(V)=\phi (D_1)$ is open by step 5.1. The affine changes and the local bound in [F5] give $\mathrm{\Phi }\in C^{k+1,\alpha }(V)$. Undoing the translation gives the desired neighborhood and map at the original $p$.

It remains to check the full Hölder regularity of the inverse. The real derivative field $D\phi$ has bounded $C^{k,\alpha }$ norm, since its Wirtinger components are $1+Sq$ and $q$. The bound $\mathrm{\parallel }D\phi -I{\mathrm{\parallel }}_{\mathrm{\infty }}\le 1\mathrm{/}4$ keeps these matrices in a bounded subset of $GL(2,\mathbb{R})$ with inverses uniformly bounded. The explicit cofactor-over-determinant formula and the product and chain rules therefore give $M:=(D\phi {)}^{-1}\in C_b^{k,\alpha }({\mathbb{R}}^2)$. Let $H:={\phi }^{-1}$ on $\phi ({\mathbb{R}}^2)$. By [F10], $DH=M\circ H$, and step 5.1 makes $H$ globally Lipschitz. Thus $DH$ is $\alpha$-Hölder with a uniform bound when $k=0$. For $k\ge 1$, induction on $m=1,\dots ,k$ applies the finite chain/product formulas to $M\circ H$: if $H$ has derivatives through order $m$ with bounded suprema and top $\alpha$-seminorm, then $M\circ H$ has the same regularity, so $DH=M\circ H$ gives the next derivative of $H$ with bounded suprema and the required top seminorm. The top composition term is $D^{k}M\circ H$, which is $\alpha$-Hölder because $H$ is Lipschitz; all other factors are covered by the product estimate of step 1.2. Hence $H$ has bounded derivatives through order $k+1$ and bounded top $\alpha$-seminorm on $\phi (D_1)$. Since $H(\phi (D_1))=D_1$, its function supremum is finite there as well. For $\eta \in \mathrm{\Phi }(V)=\phi (D_1)$, ${\mathrm{\Phi }}^{-1}(\eta )=A^{-1}(rH(\eta ))$, so the inverse is $C^{k+1,\alpha }$ on $\mathrm{\Phi }(V)$.