Lemma 7.7.14521AF3E3E5C

Proof. Use the Hermitian product $\langle z,a\rangle =\sum z_ j\bar a_ j$, put $s_ a=\sqrt{1-|a|^2}$, and define

This expression depends smoothly on $(a,z)$ in the open set $\{ (a,z):|a|<1,\ |a||z|<1\} $, where the denominator does not vanish, and it is holomorphic in $z$; moreover $T_0$ is the identity. If $U$ is a unitary matrix, then $\langle Uz,Ua\rangle =\langle z,a\rangle $ and $s_{Ua}=s_ a$, so $T_{Ua}(Uz)=UT_ a(z)$. Given $a\ne 0$, choose $U$ with $Ua=(t,0,\ldots ,0)$, $t=|a|$; it therefore suffices to verify the following identities for $a=(t,0,\ldots ,0)$ with $0\le t<1$, and also for $-a=(-t,0,\ldots ,0)$. Put $s=s_ a=\sqrt{1-t^2}$. Then $\langle z,a\rangle =tz_1$, and the first component of the numerator of $T_ a(z)$ is $sz_1+t^2z_1/(1+s)-t=z_1-t$, because $s(1+s)+t^2=s+s^2+t^2=1+s$. Hence

the second formula by the same computation with $t$ replaced by $-t$. In particular $T_ a(a)=0$ and $T_{-a}(0)=a$. If $w=T_ a(z)$, then

so the first component of $T_{-a}(w)$ is $z_1$ and the others are $s\cdot sz_ k/(1-tz_1)\cdot (1-tz_1)/(1-t^2)=z_ k$; thus $T_{-a}\circ T_ a=\mathrm{id}$, and by symmetry $T_ a\circ T_{-a}=\mathrm{id}$. Finally,

because the terms $\mp 2t\operatorname{Re}z_1$ cancel; dividing by $|1-tz_1|^2$ gives

Hence $T_ a$ maps $B$ into $B$ and the sphere $\partial B$ into itself, and the same holds for $T_{-a}$; since $T_{-a}$ is inverse to $T_ a$, the map $T_ a$ is a biholomorphic automorphism of $B$ which maps $\partial B$ onto itself.

Set $S_{a,h}=T_{-(a+h)}\circ T_ a$ for $|a+h|<1$. It maps $a$ to $T_{-(a+h)}(0)=a+h$ and is the identity for $h=0$. The map $T_ a$ is holomorphic on the neighborhood $\{ z:|a||z|<1\} $ of $\overline B$, and, as we have just seen, it maps $\overline B$ into $\overline B$ and $\partial B$ onto $\partial B$; the same holds for $T_{-(a+h)}$. Hence $S_{a,h}$ is holomorphic on the neighborhood $\{ z:|a||z|<1,\ |a+h||T_ a(z)|<1\} $ of $\overline B$, and it maps $\partial B$ onto $\partial B$. Let $K\subset B$ be compact (we may assume $K\neq \varnothing $), put $t_ K=\max _ K|a|<1$, and choose $h_0>0$ with $t_ K+3h_0<1$. On the open set $O$ of all $(a,h,z)$ with $|a|<t_ K+h_0$, $|h|<2h_0$, $|a||z|<1$ and $(t_ K+3h_0)|T_ a(z)|<1$, we have $|a+h|<1$ and $|a+h||T_ a(z)|<1$, so $(a,h,z)\mapsto S_{a,h}(z)$ is defined and smooth on $O$, and holomorphic in $z$. Since $|T_ a(z)|\le 1$ for $|z|\le 1$, the set $O$ contains the compact set $K\times \overline{B(0,h_0)}\times \overline B$, on which the continuous first and second derivatives of $S_{a,h}(z)$ with respect to $h$ are therefore bounded by a constant $C$. $\square$