Elimination of the frequency shift by a holomorphic contraction retraction

The closed set lies inside the parameter domain. Fix ${\nu }_{+}\in O_{h\mathrm{/}4}(\mathrm{\Omega })$ and put $K:=\{\nu \in {\mathbb{C}}^n:\mathrm{\mid }\nu -{\nu }_{+}{\mathrm{\mid }}_{\mathrm{\infty }}\le 2B\}$, which is nonempty also when $B=0$. Every $\nu \in K$ satisfies ${\mathrm{dist}}_{\mathrm{\infty }}(\nu ,\mathrm{\Omega })\le {\mathrm{dist}}_{\mathrm{\infty }}({\nu }_{+},\mathrm{\Omega })+2B<h\mathrm{/}4+2\cdot h\mathrm{/}(16n)\le h\mathrm{/}4+h\mathrm{/}8<h\mathrm{/}2$, so $K\subseteq O_{h\mathrm{/}2}(\mathrm{\Omega })\subseteq O_h(\mathrm{\Omega })$.

The map and its contraction property. Define $T(\nu ):={\nu }_{+}-v(\nu )$ on $K$. For $\nu \in K$, $\mathrm{\mid }T(\nu )-{\nu }_{+}\mathrm{\mid }=\mathrm{\mid }v(\nu )\mathrm{\mid }\le B\le 2B$, so $T(K)\subseteq K$. For $\nu ,{\nu }^{\mathrm{\prime }}\in K$, the segment between them lies in $K\subseteq O_{h\mathrm{/}2}$. By [F2], each entry of $Dv$ has modulus at most $2B\mathrm{/}h$ there. Along the segment, the derivative of each realified complex component of $v$ is therefore bounded by $(2B\mathrm{/}h){\sum }_j\mathrm{\mid }{\nu }_j-{\nu }_j^{\mathrm{\prime }}\mathrm{\mid }\le (2nB\mathrm{/}h)\mathrm{\mid }\nu -{\nu }^{\mathrm{\prime }}{\mathrm{\mid }}_{\mathrm{\infty }}$. Applying [F7] componentwise gives $\mathrm{\mid }v(\nu )-v({\nu }^{\mathrm{\prime }}){\mathrm{\mid }}_{\mathrm{\infty }}\le (2nB\mathrm{/}h)\mathrm{\mid }\nu -{\nu }^{\mathrm{\prime }}{\mathrm{\mid }}_{\mathrm{\infty }}\le \frac{1}{8}\mathrm{\mid }\nu -{\nu }^{\mathrm{\prime }}{\mathrm{\mid }}_{\mathrm{\infty }}$, because $B\le h\mathrm{/}(16n)$. Thus $T:K\to K$ is a contraction with factor $1\mathrm{/}8$.

Fixed point. $K$ is complete by [F3], so [F4] gives exactly one $\nu \in K$ with $T(\nu )=\nu$, i.e. $\nu +v(\nu )={\nu }_{+}$, and then $\mathrm{\mid }\nu -{\nu }_{+}\mathrm{\mid }=\mathrm{\mid }v(\nu )\mathrm{\mid }\le B$. If ${\nu }_1,{\nu }_2\in O_{h\mathrm{/}2}(\mathrm{\Omega })$ both satisfy ${\nu }_i+v({\nu }_i)={\nu }_{+}$, then each lies in $K$ because $\mathrm{\mid }{\nu }_i-{\nu }_{+}\mathrm{\mid }=\mathrm{\mid }v({\nu }_i)\mathrm{\mid }\le B\le 2B$; their connecting segment lies in the convex ball $K$, so $\mathrm{\mid }{\nu }_1-{\nu }_2\mathrm{\mid }=\mathrm{\mid }v({\nu }_2)-v({\nu }_1)\mathrm{\mid }\le (1\mathrm{/}8)\mathrm{\mid }{\nu }_1-{\nu }_2\mathrm{\mid }$ by the same estimate of step 2.1, hence ${\nu }_1={\nu }_2$. This proves the uniqueness assertion of clause 1 and defines $\varphi$.

Holomorphy. On the open set $U:=O_{h\mathrm{/}4}(\mathrm{\Omega })$ define $f_0(x):=x$ and $f_{k+1}(x):=x-v(f_k(x))$. Steps 1.1 and 2.1 keep every $f_k(x)$ in the ball $K_x:=\{z:\mathrm{\mid }z-x{\mathrm{\mid }}_{\mathrm{\infty }}\le 2B\}\subseteq O_{h\mathrm{/}2}(\mathrm{\Omega })$, so induction and holomorphic composition make each $f_k:U\to {\mathbb{C}}^n$ holomorphic. The increments satisfy $\mathrm{\mid }{f}_k(x)-f_{k-1}(x){\mathrm{\mid }}_{\mathrm{\infty }}\le 8^{-(k-1)}B$ for $k\ge 1$, uniformly in $x$, hence $f_k\to \varphi$ uniformly on $U$ by step 3.1. Each scalar component of $\varphi$ is continuous: at a point, approximate it uniformly by a continuous component of $f_k$ within one third of a prescribed tolerance, and use continuity of that component for the remaining third. On any coordinate disc contained in $U$, the corresponding scalar components converge uniformly and are one-variable holomorphic, so the one-variable part of [F5] makes the limit holomorphic on that slice. Thus every component of $\varphi$ is continuous and separately holomorphic; the polydisc part of [F5], applied on a polydisc about each point of $U$, gives its local power-series expansion. This proves clause 2, including $B=0$ when every iterate is the identity; if $U$ is empty the assertion is vacuous.

Derivative bound. Differentiating the identity $\varphi ({\nu }_{+})+v(\varphi ({\nu }_{+}))={\nu }_{+}$ with respect to ${\nu }_{+}$ (both sides holomorphic in ${\nu }_{+}$ on $O_{h\mathrm{/}4}$) gives $(I_n+Dv(\varphi ({\nu }_{+})))D\varphi ({\nu }_{+})=I_n$, and $\mathrm{\parallel }Dv(\varphi ({\nu }_{+}))\mathrm{\parallel }\le 2nB\mathrm{/}h\le 1\mathrm{/}8$ by step 2.1, so [F6] gives $\mathrm{\parallel }D\varphi ({\nu }_{+})-I_n\mathrm{\parallel }=\parallel (I_n+Dv\circ \varphi {)}^{-1}-I_n\parallel \le \frac{\mathrm{\parallel }Dv\circ \varphi \mathrm{\parallel }}{1-\mathrm{\parallel }Dv\circ \varphi \mathrm{\parallel }}\le \frac{2nB\mathrm{/}h}{7\mathrm{/}8}=\frac{16n}{7}\frac{B}{h}\mathrm{.}$ This is clause 3.