\magnification=\magstep1
\parindent=0pt

\font\tenopen=msbm10 \font\sevenopen=msbm7 \font\fiveopen=msbm5
\newfam\openfam
\def\openo{\fam\openfam\tenopen}
\textfont\openfam=\tenopen \scriptfont\openfam=\sevenopen
\scriptscriptfont\openfam=\fiveopen

\font\tenujsym=msam10 \font\sevenujsym=msam7 \font\fiveujsym=msam5
\newfam\ujsymfam
\def\ujsym{\fam\ujsymfam\tenujsym}
\textfont\ujsymfam=\tenujsym \scriptfont\ujsymfam=\sevenujsym
\scriptscriptfont\ujsymfam=\fiveujsym

\long\def\DEL#1{{}}

\def\RR{{\openo{R}}}
\def\CC{{\openo{C}}}

\def\eps{\varepsilon}

\def\qed{\hskip10mm Q.e.d.}

%References
\def\HS{1}       %Hayden-Suffridge
\def\GOEBEL{2}   %Goebel
\def\FRVE{3}     %Franzoni-Vesentini
\def\ABATE{4}    %Abate
\def\DINEEN{5}   %Dineen
\def\RSZ{6}      %Riesz-Sz-Nagy
\def\BKU{7}      %Brau-Kaup-Upmeier
\def\NAG{8}      %Engel-Nagel


Throughout this work ${\bf H}$ denotes an arbitrarily fixed complex Hilbert space
with the scalar product $\langle x \vert y \rangle$ which is linear in $x$ and conjugate linear in $y$,
giving rise to the norm $\Vert x\Vert = \langle x \vert x\Vert^{1/2}$.
For any vector $c\in{\bf H}$, we shall write $c^*:= [x\mapsto \langle x\vert a\rangle]$
for its associated functional. The open ball of radius $\rho>0$ and centered in $0$ in ${\bf H}$
will be denoted with $\rho {\rm Ball}({\bf H})$, and we write
$\rho \overline{{\rm Ball}}({\bf H}) (=\{ x:\ \Vert x\Vert \le \rho\})$ for its closure.

\bigskip

Recall [\FRVE, Ch. VI] that the group of
${\rm Aut}\,{\rm Ball}({\bf H})$
all holomorphic automorphisms of ${\rm Ball}({\bf H})$ admits a matrix
representation in the following sense:
Each element $\Psi$ of ${\rm Aut}\, {\rm Ball}({\bf H})$ has the fractional linear form
$$\Psi(x) = {{Ax + b}\over \langle x \vert c\rangle + d} \ , \quad
A\in{\cal L}({\bf H}), \ b,c\in {\bf H}, \ d\in\CC
\leqno(R)$$
and we have
$$\Psi_1 \!\circ\! \Psi_2(x) = \Psi_1\big(\Psi_2(x)\big) =
{Ax + b\over \langle x \vert c\rangle + d} \quad
\hbox{whenever} \quad
\left[ \matrix{A & b\cr c^* &d \cr} \right] =
\left[ \matrix{A_1 & b_1\cr c_1^* &d_1 \cr} \right]
\left[ \matrix{A_2 & b_2\cr c_2^* &d_2 \cr} \right].$$
This representation is unique up to a constant, since in (R) we necessarily have
$$%A \!=\! d \big( {\rm P}_a \!+\! \beta \overline{\rm P}_a \big) U, \ b \!=\! \lambda a, \ c \!=\!\overline{d} U^* a \ \
\left[ \matrix{ A &b \cr c^* & d\cr}\right] =
d \!\left[ \matrix{ \!({\rm P}_a \!\!+\! \beta \overline{\rm P}_a) & \!\!\!a \cr
a^* &\!\!\! 1 \cr}\right] \! \left[ \matrix{U & 0\cr 0 &1 \cr} \right]
\qquad \hbox{where} \qquad
\eqalign{ &a \!:=\! \Psi(0), \ \beta \!=\!\sqrt{1 \!-\!\Vert a\Vert^2}, \cr
&U \!=(\beta^2 {\rm P}_a \!+\beta\overline{\rm P}_a)^{-1}\Psi^\prime(0) \cr}$$
in terms of the %orthogonal
projections ${\rm P}_a$ onto ${\CC}a$ resp. $\overline{\rm P}_a:=1-{\rm P}_a$
and the Fr\'echet derivative $\Psi^\prime$.
Any automorphism $\Psi \in{\rm Aut}\,{\rm Ball}({\bf H})$
extends holomorphically to $(1-\Vert \Psi(0)\Vert)^{-1}{\rm Ball}({\bf H})$
and hence admits a continuous extension
$\overline{\Psi}$ to the closed unit ball $\overline{{\rm Ball}}({\bf H})$.
We write
${\rm Aut}\,\overline{{\rm Ball}}({\bf H}) :=
\{ \overline{\Psi}:\ \psi\in{\rm Aut}\,{\rm Ball}({\bf H})\}$
for their collection.

\medskip

Recall [\FRVE, Ch.VI] that any automorphism
$\overline{\Psi}\in{\rm Aut}\,\overline{{\rm Ball}}({\bf H})$ is weakly continuous
and
preserves the family ${\rm Aff}(\overline{{\rm Ball}}({\bf H}))$ of all
complex affine\footnote{$^*$}{If
$x \!=\!\sum_{k\!=\!1}^2 \!\lambda_k x_k$ with $\lambda_1,\lambda_2\!\!\in\!\CC$
and $\sum_{k\!=\!1}^2 \!\lambda_k \!=\!1$ then
$\overline{\Psi}(x) \!=\! \sum_{k\!=\! 1}^2 \!\alpha_k \overline{\Psi}(x_k)$ \
for some $\alpha_1,\alpha_2\!\!\in\!\CC$ with
$\sum_{k\!=\!1}^2 \!\alpha_k \!=\! 1$ $\big($namely with
$\alpha_k\! =\! \lambda_k [1 \!+\!\langle x_k\vert U^* a\rangle]/
[1 \!+\!\langle \lambda_1 x_1 \!+\! \lambda_2 x_2 \vert U^* a\rangle]\big)$.}
closed subspaces intersected with $\overline{{\rm Ball}}({\bf H})$.
By Schauder's fixed point theorem, ${\rm Fix}(\overline{\Psi})\ne\emptyset$, since
$\overline{{\rm Ball}}({\bf H})$ is weakly compact.
Moreover we have the following alternatives:

\smallskip
\hskip10mm (1) ${\rm Fix}(\overline{\Psi}) \in {\rm Aff}(\overline{{\rm Ball}}({\bf H}))$,\qquad
(2) ${\rm Fix}(\overline{\Psi})$ consists of two boundary points.
%$\{ p,q \}\subset\partial {\rm Ball}({\bf H})$.
\smallskip

In case (2) from the proof of [\FRVE, Thm.VI.4.8] we see even that
$\overline{\Psi} =
\overline{\Phi}\circ \overline{\Theta}_a \circ\overline{\Phi}\,\strut^{-1}$
with a suitable automorphism $\Phi\in{\rm Aut}\,\overline{{\rm Ball}}({\bf H})$ and a
M\"obius shift %translation
$$\Theta_a: x\mapsto {{\rm P}_a x+a+\beta_a \overline{{\rm P}}_a x \over 1+\langle x \vert a \rangle}, \qquad
\beta_a=\sqrt{1-\Vert a\Vert^2} \leqno(M)$$
for some $0\ne a\in{\rm Ball}({\bf H})$ such that
${\rm Fix}(\overline{\Theta})_a=\{-e,e\}$ where $e :=a/\Vert a\Vert$.


\bigskip

The next result is an infinite dimensional extension for a simple special case
of a far reaching theorem of Abate [\ABATE]
established for finite dimensional uniformly convex domains.
It seems that Dineen's bidual embedding method [\DINEEN] is suitable in proving a complete
infinite dimensional analogy with uniformly convex domains in Banach spaces.
Instead, below we give a short direct proof on the basis of the special algebraic form $(R)$.


\bigskip
\bf Lemma. \it  Abelian subsets of ${\rm Aut}\,\overline{{\rm Ball}}({\bf H})$
admit common fixed points.

\bigskip
\bf Proof. \rm
Assume $\{ \overline{\Psi}_j :\, j\!\in\!{\cal J}\} \!\subset\! {\rm Aut}\,\overline{{\rm Ball}}({\bf H})$
with $\overline{\Psi}_j \!\circ\! \overline{\Psi}_k = \overline{\Psi}_k \!\circ\! \overline{\Psi}_j$
$(j,k \!\in\! {\cal J})$.
By weak continuity, for any index family ${\cal K}\subset{\cal J}$, the set
$\bigcap_{k\in{\cal K}} {\rm Fix}(\overline{\Psi}_k)$ of common fixed points
is weakly compact.
Thus, according to Riesz' intersection theorem, it suffices to see that
$\bigcap_{k\in{\cal K}} {\rm Fix}(\overline{\Psi}_k) \ne \emptyset$
for finite index families ${\cal K}$.
By proceeding to contradiction,
let ${\cal K}=\{ k_1,\ldots,k_N\}$ be a subset of ${\cal J}$ with minimal cardinality such that
$\bigcap_{k\in{\cal K}} {\rm Fix}(\overline{\Psi}_k) = \emptyset$.
Necessarily $N>1$ and ${\bf S}:=\bigcap_{n=1}^{N-1} {\rm Fix}(\overline{\Psi}_{k_n}) \ne \emptyset$
is either a weakly compact convex subset of $\overline{{\rm Ball}}({\bf H})$ or it consists of two boundary points.
Since $\overline{\Psi}_{k_N}$ commutes with all the maps $\overline{\Psi}_{k_n}$ $(n<N)$,
we have $\overline{\Psi}_{k_N}({\bf S}) \subset {\bf S}$.
Indeed, if $x\in{\bf S}$ then
$\overline{\Psi}_{k_n}\big(\overline{\Psi}_{k_N}(x)\big) =
\overline{\Psi}_{k_N}\big(\overline{\Psi}_{k_n}(x)\big) =
\overline{\Psi}_{k_N}(x) \Rightarrow \overline{\Psi}_{k_N}(x)\in{\rm Fix}(\overline{\Psi}_{k_n})$ $(n<N)$.
Hence Schauder's fixed point theorem excludes the case of ${\bf S}$ being convex.
Suppose ${\bf S}=\{p,q\}\subset\partial{\rm Ball}({\bf H})$.
Then necessarily ${\rm Fix}(\overline{\Psi}_{k_M}) =\{ p,q\}$
for some index $M<N$ and $\overline{\Psi}_{k_N}: p\leftrightarrow q$.
However, in this case we can write
$\overline{\Psi}_{k_N} =
\overline{\Phi}\circ \overline{\Theta}_a \circ\overline{\Phi}\,\strut^{-1}$
with a suitable automorphism $\overline{\Phi}\in{\rm Aut}\,\overline{{\rm Ball}}({\bf H})$ and a
M\"obius shift
$\overline{\Theta}_a$
of the form $(M)$
%: x \!\mapsto\! (1 \!+\!\langle x \vert a \rangle)^{-1} ({\rm P}_a \!+\!\beta \overline{{\rm P}}_a) x$
%$\Theta: x\mapsto {{\rm P}_a x+\beta \overline{{\rm P}}_a x \over 1+\langle x \vert a \rangle}$
where $0 \!\ne\! a\!\in\!{\rm Ball}({\bf H})$.
% and $\beta \!=\!\sqrt{1 \!-\! \Vert a\Vert^2}$.
Then, by setting
$\overline{\Omega} \!:=\!\overline{\Phi}\,\strut^{-1}\!\circ\overline{\Psi}_{k_M}\!\circ\overline{\Phi}$
and $e \!:=\! a/\vert a\Vert$, we have
$\{ \pm e\} \!=\! {\rm Fix}(\overline{\Theta}_a)$ and $\overline{\Omega}: e\!\leftrightarrow\! -e$.
On the other hand, it is immediate that
$e=\lim_{n\to\infty}\overline{\Theta}_a^n (x)$ for every point $x\!\in\!\overline{\rm Ball}({\bf H})\!\setminus\!\{ e\}$.
Taking any point $f\in{\rm Fix}(\overline{\Omega})$ we get the contradiction
$e=\lim_{n\to\infty} \overline{\Theta}_a^n(f) =
\lim_{n\to\infty} \overline{\Theta}_a^n \!\!\circ \overline{\Omega}(f) =
\lim_{n\to\infty} \overline{\Omega} \!\circ \overline{\Theta}_a^n(f) = \overline{\Omega} (e) =-e$.
\qed

\bigskip
\bigskip

Henceforth we keep fixed the notation
$\big[ \Psi^t :\ t\in\RR \big]$
for an arbitrarily given
strongly continuous one-parameter group in ${\rm Aut}\, {\rm Ball}({\bf H})$.
That is
$$\Psi^t(x) = {(P_t+\beta_t \overline{P}_t)U^t x +a_t  \over 1+\langle U_t x \vert a_t \rangle }
= \big[\Theta_{a_t} \!\circ U_t \big] (x)
\leqno(P)$$
where
$a_t \!=\!\Psi_t(0)\!\in\! {\rm Ball}({\bf H})$, $U_t \!\in\!{\cal L}({\bf H})$ is a unitary operator
and, with the notations in $(M)$,
$\beta_t \!:=\! \beta_{a_t},  %\sqrt{1-\Vert a_t\Vert^2}$ and
P_t \!:=\!{\rm P}_{a_t}, \overline{P}_t \!:=\! \overline{{\rm P}}_{a_t}$
for short.
%denote the orthogonal projections onto
%the 1-dimensional subspace ${\CC} a_t$ resp. its complementary subspace ${\bf H}\ominus {\CC} a_t$
In particular
$P_t x = [\langle x \vert a_t\rangle / \langle a_t \vert a_t\rangle] a_t$, \
$\overline{P}_t =I-P_t
%\quad \beta_t=\sqrt{1-\langle a_t\vert a_t\rangle}
$.

\smallskip
According to the previous lemma,
the continuous extensions $\overline{\Psi}^t$ $(t\in\RR)$ to $\overline{{\rm Ball}}({\bf H})$
have a common fixed point which we shall denote with $\overline{x}$:
$$\overline{x}\in\overline{{\rm Ball}}({\bf H}), \qquad
\overline{\Psi}^t(\overline{x}) =\overline{x} \quad (t\in\RR). $$

\bigskip
\bf Remark. \rm In finite dimensions, it is customary to normalize (R) by requiring
${\rm det} \left[ {A \ b\atop c^* d} \right] \!=\! 1$.
Thus, in case of ${\rm dim}({\bf H})=N$, in this manner one can establish a canonical identification
of ${\rm Aut}\, {\rm Ball}({\bf H})$
with a subgroup of the classical matrix group ${\rm SL}(N+1)$.
Though in infinite dimensions such a normalization is not available,
for one-parameter groups with common fixed point there an alternative way as follows.

\bigskip
\bf Proposition. \it With the notations $(P)$, the $(R)$-type representation matrices
$$\widehat{\Psi}^t := {1\over 1+\langle U_t \overline{x}\vert a_t\rangle}
\left[ \matrix{ (P_t + \beta_t \overline{P}_t)U_t &a_t \cr [U_t^* a_t]^* &1 \cr}\right] =
{1\over 1+\langle U_t \overline{x}\vert a_t\rangle}
\left[ \matrix{ P_t + \beta_t \overline{P}_t &a_t \cr a_t^* &1 \cr}\right]
\left[ \matrix{ U_t  &0 \cr 0 &1 \cr}\right]$$
of $\big[\Psi_t:\ t\in\RR]$ form
a strongly continuous one-parameter group of operators in ${\bf H}\oplus\CC$.

\bigskip
\bf Proof. \rm
Since
 $\Psi^t\!\circ\!\Psi^s \!=\! \Psi^{t+s}$ $(t,s\!\in\!\RR)$,
for the representation matrices we have
$\widehat{\Psi}^t \widehat{\Psi}^s \!=\! d_{t,s} \widehat{\Psi}^{t+s}$
with suitable constants $d_{t,s}\in\CC$.
The fixed point property
$\overline{\Psi}^t (\overline{x})=\overline{x}$
implies
$$\widehat{\Psi}^t \left[ {\overline{x} \atop 1}\right] =
\left[ {\overline{x} \atop 1}\right] \qquad (t\in\RR).$$
Hence necessarily $d_{t,s}\!=\!1$ $(t,s\!\in\!\RR)$, thus
$\big[ \widehat{\Psi}^t: t\!\in\!\RR\big]$ is a one-parameter matrix group.

By assumption, the function $t\mapsto a_t=\Psi^t(0)$ is norm-continuous $\RR\to{\rm Ball}({\bf H})$.
Hence we can deduce the strong continuity of the ${\bf H}$-unitary operator valued
%$\RR\to{\cal L}({\f H})$
function $t\mapsto U_t$. Namely consider any vector $x\in{\bf H}$.
To establish the norm-continuity of the function $t\mapsto U_t$,
we may assume without loss of generality that $x\in{\rm Ball}({\bf H})$.
Then, by the aid of the M\"obius shifts $(M)$
we can write
$$U_t x = \big[\Theta_{a_t}^{-1} \circ \Psi_t \big](x) = \Theta_{-a_t} \big( \Psi(x)\big)
\qquad (t\!\in\!\RR)
%\quad \hbox{where} \ \
%\Theta_a : z\!\mapsto\! { [{\rm P}_a \!+\! \beta_a \overline{{\rm P}}_a]z \over 1 \!+\!\langle z\vert a\rangle}
%\ \ \hbox{with} \ \ \beta_a=\sqrt{1 \!-\!\Vert a \Vert^2}
.$$
Observe that the norm continuity of $t\mapsto a_t$ implies the
continuity of $t\mapsto \langle x \vert a_t\rangle$ and $t\mapsto\beta_t \in [0,1)$
entailing the norm-continuity of
$t\mapsto P_t +\beta_t \overline{P}_t \in\overline{{\rm Ball}}\big({\cal L}({\bf H})\big)$.
Hence the required norm-continuity of
$t\mapsto U_t x = (1-\langle x\vert a_t\rangle)^{-1} [(P_t x - a_t + \beta_t \overline{P}_t x]$
is immediate.
In general, the product of two bounded strongly continuous linear operator valued functions $\RR\to{\cal L}({\bf X})$
over a normed space ${\bf X}$
is strongly continuous. Hence we conclude that the entries $(1,1),(1,2),(2,1)$ resp. $(2,2)$
of the matrices $\widehat{\Psi}^t$ are strongly continuous functions
$\RR\to{\cal L}({\bf H})$, $\RR\to{\bf H}$, $\RR\to {\bf H}^* \!\!\simeq\!{\bf H}$ resp. $\RR\to\RR$
which completes the proof.
\qed

\bigskip
\bf Remark. \rm
It is worth to notice that the term $\langle U_t\overline{x}\vert a_t\rangle$ is
actually independent of $u_t$ as
$$\langle U_t\overline{x}\vert a_t\rangle =
{\langle \overline{x}-a_t\vert a_t\rangle  \over 1 - \langle \overline{x}\vert a_t \rangle} , \qquad
\widehat{\Psi}^t  =
{ 1-\langle \overline{x}\vert a_t\rangle \over 1 -\langle a_t\vert a_t\rangle}
\left[ \matrix{ P_t + \beta_t \overline{P}_t &a_t \cr a_t^* &1 \cr}\right]
\left[ \matrix{ U_t  &0 \cr 0 &1 \cr}\right] .
\leqno(Ux)$$
Proof: In general we have $P_t y = \langle y\vert a_t\rangle \langle a_t \vert a_t\rangle^{-2} a_t$
$(0\ne a_t,y\in{\bf H})$. It follows
$\langle P_t U_t \overline{x} \vert a_t\rangle = \langle U_t\overline{x}\vert a_t\rangle$
with $\langle \overline{P}_t U_t \overline{x} \vert a_t\rangle =0$ for any $t\in\RR$.
Thus multiplying the fixed point equation
$\overline{x} = \overline{\Psi}^t(\overline{x}) =
(1+\langle U_t\overline{x}\vert a_t\rangle)^{-1} (P_t+\beta_t\overline{P}_t)U_t\overline{x}$
with $\vert a_t\rangle$, we get
$(1+\langle U_t\overline{x}\vert a_t\rangle)^{-1} \langle U_t\overline{x} +a_t\vert a_t\rangle =
\langle \overline{x} \vert a_t\rangle$
whence the relations $(Ux)$ are immediate.

\bigskip
\bigskip
Recalling the Hille--Yosida theorem [\RSZ, Kap.10] on strongly continuous one-parameter groups
of bounded linear operators in Banach spaces, the above proposition ensures the existence
of a not necessarily bounded linear map ${\cal A}:{\cal D}\to{\bf H}\oplus\CC$
(the infinitesimal generator of $\big[ \widehat{\Psi}^t: t\!\in\!\RR\big]$)
such that
$$\eqalign{
&{\cal D} = \Big\{ \Big[ {x\atop \xi}\Big] \!\in\! {\bf H}\!\oplus\! \CC : \,
{d\over dt}\Big\vert_{t\!=\! 0} \!\!\widehat{\Psi}^t \Big[ {x\atop \xi}\Big] \ \hbox{exists}  \Big\}
%\supset
%\Big\{ {\scriptstyle{\int_0^\eps \!\widehat{\Psi}^t \left[ {x\atop\xi}\right] \, dt}} : \,
%\eps \!>\! 0, \, x \!\in\! {\bf H},\, \xi\!\in\!\CC \Big\}
 ,\cr
&{d\over dt} \widehat{\Psi}^t =
{\cal A}\widehat{\Psi}^t =\widehat{\Psi}^t {\cal A} \qquad {\rm on} \ {\cal D} ,\cr
&{\cal A} \!\left[ {x \atop \xi}\right] =
{d\over dt}\Big\vert_{t=0} \!\widehat{\Psi}^t \!\! \left[ {x \atop \xi}\right] \quad {\rm for} \ \
\left[ {x \atop \xi}\right]\!\!\in\!{\cal D} .\cr}
\leqno(Aa)$$
Henceforth we fix the notations ${\cal A},{\cal D}$ for the
infinitesimal generator of $\big[ \widehat{\Psi}^t: t\!\in\!\RR\big]$ and its domain, respectively.
It is well-known that ${\cal A}$ has closed graph
and
${\cal D}$ is a dense linear submanifold of ${\bf H}\oplus\CC$
being invariant under the maps $\widehat{\Psi}^t$.
We shall write $\big\{ \big[{x \atop\xi}\big]\big\}_{{\bf H}}:= x$ resp.
$\big\{ \big[{x \atop\xi}\big]\big\}_{\CC}:= \xi$
for the canonical projections in ${\bf H}\oplus\CC$.
We also introduce the notations
$$\eqalign{
&{\cal D}_0 := \Big\{ x\in{\bf H}:\ \left[ {x \atop 0}\right] \in {\cal D} \Big\} \, , \quad
{\cal B} := \big(\overline{x}+{\cal D}_0\big) \cap \overline{{\rm Ball}}({\bf H}) \ ;\cr
&Bz :=\Big\{ {\cal A}\Big[{z \atop 0}\Big]\Big\}_{{\bf H}} \ , \quad
\Lambda z :=\Big\{ {\cal A}\Big[{z \atop 0}\Big]\Big\}_{\CC} \quad
\hbox{for} \ \ z\in{\cal D}_0 \ .\cr}
\leqno(N)$$
\bf Proposition. \it ${\cal D}_0$ is a dense linear submanifold in ${\bf H}$ \
with \
${\cal D}= \big[ {{\cal D}_0 \atop 0}\big] + \CC \big[ {\overline{x} \atop 1}\big]$.
We have ${\cal B} =\big\{ x\in{\rm Ball}({\bf H}):\ \big[ {x\atop 1}\big]\in {\cal D} \big\}$.
The set ${\cal B}$ is
$\big[ \Psi^t: t\!\in\!\RR\big]$-invariant and dense in $\overline{{\rm Ball}}({\bf H})$
such that
$${d\over dt}\Big\vert_{t=0} \overline{\Psi}^t(x) =
%-\Big\{ {\cal A} \left[ {x\atop 1}\right] \Big\}_{\CC} x +
%\Big\{ {\cal A} \left[ {x\atop 1}\right] \Big\}_{{\bf H}}
[\Lambda(\overline{x} -x)]x + B(x-\overline{x})
\qquad (x\in{\cal B}).$$

\bigskip
\bf Proof. \rm
The first statement in $(Aa)$ implies
$\big[ {\overline{x} \atop 1}\big] \!\in\!{\cal D}$ with ${\cal A} \big[ {\overline{x} \atop 1}\big]\!=\! 0$
since $\widehat{\Psi}^t \!\big[ {\overline{x} \atop 1}\big] \!\equiv\!\big[ {\overline{x} \atop 1}\big]$.
Since ${\cal D}$ is closed for linear combinations, it follows
$\big[ {{\cal D}_0 \atop 0}\big] + \CC \big[ {\overline{x} \atop 1}\big] \subset {\cal D}$.
Given any vector $\big[ {x \atop \xi}\big] \in{\cal D}$, we have
$x-\xi\overline{x} \in{\cal D}_0$  because
$\big[ {x \atop \xi}\big] - \xi \big[ {\overline{x} \atop 1}\big] \in{\cal D}$.
Hence also
${\cal D}\subset \big[ {{\cal D}_0 \atop 0}\big] + \CC \big[ {\overline{x} \atop 1}\big]$.

As for the density of ${\cal D}_0$ in ${\bf H}$
and hence also the density of ${\cal B}$ in $\overline{{\rm Ball}}({\bf H})$,
consider an arbitrary vector $x\in{\bf H}$.
Since ${\cal D}$ is dense in ${\bf H}\oplus\CC$, there is a sequence
$\big[ {x_1 \atop \xi_1}\big],\big[ {x_2 \atop \xi_2}\big],\ldots\in{\cal D}$ \
with \
$\big[ {x_n \atop \xi_n}\big] \!\to\! \big[ {x \atop 0}\big]$ that is
$\xi_n\!\to\! 0$, $x_n\!\to\! 0$  (in norm)
\ whence \
%${\cal D}\ni \left[ {x_n \atop \xi_n}\right] - \xi_n \left[ {\overline{x} \atop 1}\right] =
%\left[ {x_n-\overline{x} \atop 0}\right] \to \left[ {x \atop 0}\right]$ and
${\cal D}_0 \ni x_n \!-\! \xi_n\overline{x} \to x$.

To see ${\cal B}=\big\{ x\in\overline{{\rm Ball}}({\bf H}):\ \big[ {x\atop 1}\big]\in {\cal D} \big\}$
it suffices to notice that in general
$$x\in{\cal D}_0+\overline{x} \iff  \left[{ x\atop 1}\right] \in {\cal D}
\qquad (x\in{\bf H}).\leqno(*)$$
Proof of $(*)$: \ $x\in{\cal D}_0 \iff \big[ {x-\overline{x} \atop 0}\big] \in {\cal D}
\iff \big[ {x\atop 1}\big] - \big[ { \overline{x}\atop 1}\big] \in {\cal D}
\iff \big[ {x\atop 1}\big] \in{\cal D}$.

Using $(*)$, we show the $\big[ \overline{\Psi}^t: t\in\RR\big]$-invariance of the set ${\cal B}$ as follows.
Since
$\widehat{\Psi}^t : {\cal D}\leftrightarrow {\cal D}$ and
$\overline{\Psi}^t: \overline{{\rm Ball}}({\bf H}) \leftrightarrow \overline{{\rm Ball}}({\bf H})$,
we can deduce consecutively the equivalence of the statements
$(i) \,\overline{\Psi^t(x} \!\!\in\! {\cal B}$, \
$(ii)\, x\!\!\in\!{\rm Ball}({\bf H}) \!$ and $\!\big[ {\Psi^t(x) \atop 1}\big]\!\!\in\!{\cal D}$, \
$(iii)\, x\!\!\in\!\overline{{\rm Ball}}({\bf H}) \!$ and
$\!\big\{ \widehat{\Psi}^t \big[ {x\atop 1}\big] \big\}_{\!\CC}^{\!\!-1}
\widehat{\Psi}^t \big[ {x\atop 1}\big] \!\!\in{\cal D}$, \
$(iv)\, x\!\!\in\! \overline{{\rm Ball}}({\bf H})\!$ and
$\!\widehat{\Psi}^t \big[ { x\atop 1}\big] \!\!\in\! {\cal D}$, \
$(v)\, x\!\!\in\!\overline{{\rm Ball}}({\bf H})\!$ and
$\!\big[ {x\atop 1}\big] \!\!\in\! {\cal D}$, \
$(vi)\, x \!\!\in\! {\cal B}$.

Finally, given any vector $x\in{\cal B}$, by $(*)$ we have
%$x\!-\!\overline{x} \in {\cal D}_0$
$\big[ {x\atop 1} \big] \in{\cal D}$  whence
${\cal A}\big[ {x\atop 1} \big] = {d\over dt}\big\vert_{t=0} \widehat{\Psi}^t \big[ {x\atop 1} \big]$.
Since
$\overline{\Psi}^t (x) =
\big\{ \widehat{\Psi}^t \big[ {x\atop 1} \big] \big\}_{\!\CC}^{\!\!-1}
\big\{ \widehat{\Psi}^t \big[ {x\atop 1} \big] \big\}_{\!{\bf H}}$
and since
$\widehat{\Psi}^0={\rm Id}$,
${\cal A} \big[{\overline{x}\atop 1}\big] =
{d\over dt}\big\vert_{t=0} \widehat{\Psi}^t  \big[{\overline{x}\atop 1}\big] =
{d\over dt}\big\vert_{t=0} \big[{\overline{x}\atop 1}\big] =0 $,
hence we get
$$\eqalign{
{d\over dt}\Big\vert_{t=0} \!\!\!\!\overline{\Psi}^t(x) &=
-\Big\{ \!\widehat{\Psi}^0 \Big[ {x\atop 1} \Big] \!\Big\}_{\!\CC}^{\!\!-2} \!
\left[ {d\over dt}\Big\vert_{t=0} \! \Big\{ \!\widehat{\Psi}^0 \Big[ {x\atop 1} \Big] \!\Big\}_{\!\CC} \right] \!
\Big\{ \!\widehat{\Psi}^0 \Big[ {x\atop 1} \Big] \!\Big\}_{\!{\bf H}} \!\!+
\Big\{ \!\widehat{\Psi}^0 \Big[ {x\atop 1} \Big] \!\Big\}_{\!\CC}^{\!\!-1}
{d\over dt}\Big\vert_{t=0} \!\Big\{ \!\widehat{\Psi}^0 \Big[ {x\atop 1} \Big] \!\Big\}_{\!{\bf H}} \!\!=\cr
&= - \Big\{ {\cal A} \Big[ {x\atop 1} \Big] \Big\}_{\CC} x +
\Big\{ {\cal A} \Big[ {x\atop 1} \Big] \Big\}_{\!{\bf H}} =
 - \Big\{ {\cal A} \Big[ {x-\overline{x}\atop 0} \Big] \Big\}_{\CC} x +
\Big\{ {\cal A} \Big[ {x-\overline{x}\atop 0} \Big] \Big\}_{\!{\bf H}}. \ \
\hbox{Q.e.d.}
\cr}$$

\bigskip
\bf Corollary. \it Given any vector
$v\in{\cal D}_0$ we have
$${\rm Re} \Big( -\Lambda v +
\big\langle B v \big\vert \overline{x} \big\rangle +
\big\langle B v \big\vert v \big\rangle \Big)=0
\quad  \hbox{\rm whenever} \quad \big\Vert \overline{x}+v \big\Vert =1 \, .$$
\bf Proof. \rm
Let $v\!\in{\cal D}_0$, $u := \overline{x} \!+\! v$ and assume $\Vert u\Vert \!=\! 1$.
Since, as it is well-known [\FRVE],
$\overline{\Psi}\big( \partial {\rm Ball}({\bf H}) \big) =
\partial {\rm Ball}({\bf H})$ for any holomorphic automorphism
$\Psi$ of the unit ball, we have
$\big\Vert \overline{\Psi}^t(u) \big\Vert \!=\! 1$ \ $(t\!\in\!\RR)$.
It follows
$$0 \ = \ {d\over dt}\Big\vert_{t=0}
\Big\langle \overline{\Psi}^t(u) \,\Big\vert\, \overline{\Psi}^t(u) \Big\rangle \ = \
2 \,{\rm Re} \, \Big\langle {d\over dt}\Big\vert_{t=0} \!\!
\overline{\Psi}^t(u) \ \Big\vert \ u \Big\rangle .$$
Applying the proposition with $x \!:=\! u$ and $v \!:=\! x \!-\!\overline{x}$, this means that
${\rm Re}\,\big\langle [-\Lambda v] u \!+\! Bv \,\big\vert\, u \big\rangle \!=\!0$.
Taking into account that $\langle u\vert u\rangle =1$,
the statement is immediate. \qed

\bigskip
\bf Proposition. \it For some constant $\lambda\in\RR$
we have
$$\Lambda z = \big\langle (B \!-\! 2\lambda I) z \,\big\vert\, \overline{x} \big\rangle , \quad
{\rm Re} \big\langle \big(B \!-\!\lambda I \big)z \,\big\vert\, z \big\rangle =0 \qquad (z\in{\cal D}_0) .$$
In particular, if \ $\Vert \overline{x}\Vert <1$ then necessarily $\lambda=0$ above.

\bigskip
\bf Proof. \rm
Consider any vector $z\ne 0$.
Let $\zeta\in \CC$ be the (unique) constant such that
$\overline{x}+\zeta z \perp z$ and define
$\varrho :=\sqrt{1-\Vert \overline{x}+\zeta z\Vert^2}$.
Actually we have
$\zeta = -\langle \overline{x} \vert z\rangle / \langle z\vert z\rangle$ and
$1\ge \Vert\overline{x} \Vert^2 = \Vert \overline{x}+\zeta z\Vert^2 + \Vert -\zeta z\Vert^2$
showing that both $\zeta$ and $\varrho$ are well-defined.
Consider the unit vectors
$$v_\varphi:= \overline{x}+\zeta z + e^{i\varphi} \varrho z \qquad (\varphi\!\in\!\RR) .$$
According to the last corollary,
$${\rm Re} \Big( (\zeta \!+\! e^{i\varphi} \varrho) \big[- \Lambda z +
\big\langle Bz \,\big\vert\, \overline{x}\big\rangle \big] +
\vert \zeta \!+\! e^{i\varphi}\varrho\vert^2 \big\langle Bz\,\big\vert\, z \big\rangle \Big) =0
\qquad (\varphi\in\RR).$$
That is we have
$$\eqalign{
&{\rm Re} \big( \alpha + \beta e^{i\varphi} +\gamma  e^{-i\varphi}\big)=0
\quad  (\varphi\in\RR) \cr
&{\rm where} \ \ \alpha:=
\zeta\big[ -\Lambda z + \big\langle Bz \,\big\vert\, \overline{x}\big\rangle \big] +
(\vert\zeta\vert^2 + \varrho^2) \big\langle Bz \,\big\vert\, z\big\rangle,\cr
&\hskip11mm
\beta:= \varrho\, \big[ -\Lambda z + \big\langle Bz \,\big\vert\, \overline{x} \big\rangle +
\overline{\zeta} \big\langle Bz \,\big\vert\, z \big\rangle\big],
\quad \gamma:= \varrho\zeta \big\langle Bz \,\big\vert\, z \big\rangle .\cr}$$
Since \
$2\, {\rm Re} \big( \alpha + \beta e^{i\varphi} +\gamma  e^{-i\varphi} \big) =
2\, {\rm Re} (\alpha) + (\beta+\overline{\gamma})
e^{i\varphi} + (\overline{\beta}+\gamma)e^{-i\varphi}$,
we have necessarily
$${\rm Re}(\alpha) = \beta+\overline{\gamma}=0 \ .$$
%The identity $\beta+\overline{\gamma}=0$ corresponds to our statement.
As \ $\beta+\overline{\gamma}=0,$ \ it follows
$$\Lambda z - \langle Bz \,\vert\, \overline{x} \rangle =
2\, \overline{\zeta}\, {\rm Re}\, \langle Bz \,\vert\, z \rangle =
-2\, \langle z \,\vert\, \overline{x} \rangle \,
{{\rm Re}\, \langle Bz \,\vert\, z \rangle \over \langle z \,\vert\, z \rangle} \ ,
\leqno(B)$$
and substituting this into the relation
$0={\rm Re} (\alpha)$, we get
$$0=(\varrho^2 - \vert\zeta\vert^2) \, {\rm Re}\, \big\langle Bz \,\big\vert\, z\big\rangle =
\big(1-\Vert \overline{x}\Vert^2 \big) \, {\rm Re}\, \big\langle Bz \,\big\vert\, z\big\rangle \ .
\leqno(B^\prime)$$
From $(B)$ we see that
$$z\mapsto{ {\rm Re} \big\langle Bz \,\big\vert\, z\big\rangle
\over  \big\langle z \,\big\vert\, z\big\rangle } =
-{1\over 2}\,
{\Lambda z - \big\langle Bz \,\big\vert\, \overline{x} \big\rangle
\over\big\langle z \,\big\vert\, \overline{x} \big\rangle }$$
is a
real valued G\^ateaux holomorphic function on the algebraically open
and in ${\cal D}_0$ algebraically dense
domain $\{ z\!\in\!{\cal D}_0 : \ z\not\perp \overline{x} \}$ which
is possible only if being constant on ${\cal D}_0$.
By writing $\lambda$ for this constant value, the statement is immediate from $(B)$ and $(B^\prime)$.
\qed



\bigskip

Taking int account the obvious reversibility properties of the steps of the proof,
we conclude immediately the following technical statement.\footnote{$^*$}{Direct proof.
Given any $\lambda\in\RR$ and a vector $z\in{\bf H}$ with $\Vert \overline{x}+z \Vert=1$,
we have the following chain of equivalences:
$0= {\rm Re} \big\langle -\langle (iA-\lambda)z \vert \overline{x} \rangle (\overline{x}+z) +
(iA+\lambda I)z \big\vert \overline{x}+z \big\rangle$
iff
$0= {\rm Re} \big[ -\langle (iA-\lambda)z \vert \overline{x} \rangle +
\langle (iA+\lambda I)z \vert \overline{x}+z \rangle \big]$
iff
$0= {\rm Re} \big[ 2\lambda \langle z \vert \overline{x} \rangle +
\langle iA z \vert z \rangle + \lambda \langle z \vert z\rangle \big]$
iff
$0= {\rm Re} \big[ \lambda \big( \langle \overline{x}+z \vert \overline{x} +z \rangle
- \langle \overline{x} \vert \overline{x}\rangle
- \langle \overline{x} \vert z\rangle + \langle z \vert \overline{x} \rangle \big) +
\langle iA z \vert z \rangle \big]$
iff
$0= {\rm Re} \big[ \lambda ( 1 - \Vert \overline{x} \Vert^2 ) +
\langle iA z \vert z \rangle \big]$.
}


\bigskip
\bf Corollary. \it Given a possibly unbounded linear functional
$\Lambda: {\cal D}_0\to\CC$ along with a possibly unbounded linear operator
$B:{\cal D}_0\to {\bf H}$, we have
$\big[\Lambda(\overline{x}-x)\big]x + B(x-\overline{x}) \perp x$ for any
$x\in [\overline{x}+{\cal D}_0]\cap\partial{\rm Ball}({\bf H})$
if and only if
$\Lambda z \!=\! \big\langle (B \!-\! 2\lambda I) z \,\big\vert\, \overline{x} \big\rangle , \
{\rm Re} \big\langle \big(B \!-\!\lambda I \big)z \,\big\vert\, z \big\rangle \!=\!0$ $(z\!\in\!{\cal D}_0)$
with some real constant $\lambda$ if $\Vert\overline{x}\Vert=1$ and
with $\lambda=0$ in case of $\Vert\overline{x}\Vert<1$.

\bigskip
\bf Lemma. \it
The operator $B:{\cal D}_0\to{\cal D}_0$ is closed.

\bigskip
\bf Proof. \rm
Consider a convergent sequence
${\cal D}_0 \ni z_n \to z$ with limit point in ${\bf H}$
such that
$B z_n \to w$ for some $w\in{\bf H}$.
We have to show that \ $(*)$ $z\!\in\!{\cal D}_0$ with $Bz \!=\! w$,
For $(*)$, it suffices to
observe that
$$\Lambda z_n = \big\langle (B-2\lambda I) z_n \big\vert \overline{x} \big\rangle
\to \big\langle w -2\lambda z \big\vert \overline{x} \big\rangle ,$$
recalling that
%, by the Hille-Yosida theorem,
the not necessarily bounded linear operator
${\cal A} \big[ { x \atop \xi} \big] =
\big[ { B(x-\xi\overline{x}) \atop \Lambda(x-\xi\overline{x})} \big]$ mapping
the linear manifold
${\cal D} := \big[ { {\cal D}_0 \atop 0} \big] \!\!+\! \CC \big[ { \overline{x} \atop 1} \big]$
into itself is closed.
Indeed we have
$$\Big[ { {\cal D}_0 \atop 0} \Big] \ni \Big[ { z_n \atop 0} \Big] \to \Big[ { z \atop 0} \Big] , \quad
{\cal A} \Big[ { z_n \atop 0} \Big] = \Big[ { B z_n \atop \Lambda z_n} \Big] \to
\Big[ { w \atop \langle w -2\lambda z \vert \overline{x} \rangle} \Big]$$
implying $(*)$ along with the relation
$\Lambda z= \langle w -2\lambda z \vert \overline{x} \rangle =
\langle (B-2\lambda I) z \vert \overline{x} \rangle$.
\qed

\bigskip

\bf Definition. \rm
For convenience, we shall write
$A:= -i(B+\lambda I)$. Thus henceforth
$$Bz=(iA +\lambda I)z \, , \quad
\Lambda z = \big\langle (iA-\lambda I)z \big\vert \overline{x} \big\rangle
\qquad (z\in{\cal D}_0)
\leqno(A)$$
where $A:{\cal D}_0\to{\cal D}_0$ is a densely defined closed symmetric linear operator
and $\lambda$ is some real number.
We know also that any vector field of the form
$\overline{x}+{\cal D}_0 \ni \overline{x} + z \mapsto
-[\Lambda z] (\overline{x}+z)+Bz$ is tangent to the unit sphere if and only if
$B$ and $\Lambda$ are of the form $(A)$ with a symmetric operator $A$
and any real number $\lambda$ if $\Vert \overline{x}\Vert =1$ and with $\lambda=0$
in case of $\Vert\overline{x}\Vert=1$.


\DEL{
In order to pass to symmetric operators. Thus for any vector
$\big[ {x\atop\xi}\big] \in{\cal D}$ we have
$${\cal A} \Big[ {x\atop\xi} \Big]= {d\over dt}\Big\vert_{t=0} \widehat{\Psi}^t \Big[ {x\atop\xi} \Big] =
\left[{ (iA+\lambda I) (x-\xi\overline{x}) \atop \langle (iA-\lambda I) (x-\xi\overline{x}) \vert \overline{x} \rangle} \right]
$$
and for any point $x \in [\overline{x}+{\cal D}_0]\cap\overline{\rm Ball}({\bf H})$ we have
$${d\over dt}\Big\vert_{t=0} \Psi^t (x) =
-\big\langle (iA-\lambda I)(x-\overline{x}) \big\vert \overline{x} \big\rangle x +
(iA+\lambda I) (x-\overline{x})$$
where $A:{\cal D}_0\to{\cal D}_0$ is a densely defined closed symmetric linear operator
and $\lambda$ is some real number.
}%ENDDEL


\DEL{

\bigskip\bf Proposition. \it
There exists a constant $\rho\in\RR$ such that
$$\Lambda v = \big\langle (B \!+\! \rho I)v \,\big\vert\, \overline{x} \big\rangle, \quad
2\,{\rm Re}\, \big\langle Bv \big\vert v \big\rangle = \rho
\big\langle v \big\vert v \big\rangle \qquad (v \in {\cal D}_0).$$
\bf Proof. \rm Consider any vector $0\ne v\in{\cal D}_0$ and let
$$z:= \zeta v \quad \hbox{where} \quad
\zeta := -{ \langle \overline{x} \vert v \rangle \over \langle v
\vert v \rangle } .$$ Observe that \ $\overline{x} \!+\! z \perp z
\in{\cal D}_0$. Hence we may apply the lemma with the conclusion
$\zeta\, \Lambda v = \big\langle \zeta Bv \,\big\vert\, \overline{x}
\big\rangle + 2\,{\rm Re} \big\langle \zeta Bv \,\big\vert\, \zeta
v\big\rangle$ that is $\Lambda v = \big\langle Bv \,\big\vert\,
\overline{x} \big\rangle + 2 \,\overline{\zeta}\, {\rm Re}
\big\langle Bv \,\big\vert\, v\big\rangle$ implying
$$\Lambda v = \big\langle Bv \,\big\vert\, \overline{x} \big\rangle +
2 \,{ {\rm Re} \big\langle Bv \,\big\vert\, v\big\rangle \over
\big\langle v \,\big\vert\, v\big\rangle } \, \big\langle v
\,\big\vert\, \overline{x} \big\rangle \ .$$ Hence we see that
$$v\mapsto 2 \,{ {\rm Re} \big\langle Bv \,\big\vert\, v\big\rangle
\over  \big\langle v \,\big\vert\, v\big\rangle } = {\Lambda v -
\big\langle Bv \,\big\vert\, \overline{x} \big\rangle
\over\big\langle v \,\big\vert\, \overline{x} \big\rangle }$$ is a
real valued G\^ateaux holomorphic function on the algebraically open
domain $\{ v\!\in\!{\cal D}_0 : \ v\not\perp \overline{x} \}$ which
is possible only if being constant. Hence the statement is
immediate. \qed

}%ENDDEL

\bigskip\bigskip
\bf The case %$\overline{x} \in{\cal D}_0$.
$0 \in{\cal B}$ with $\Vert \overline{x} \Vert=1$
\rm
\bigskip

Throughout this section we assume
that
a common fixed point $\overline{x}$ of $\big[ \overline{\Psi}^t:\, t\!\in\!\RR\big]$
lyes in the boundary of the unit ball and the origin belongs to the domain
${\cal B}$ of the non-linear generator
that is we have
$$0\in {\cal B}, \quad \Vert \overline{x} \Vert =1 .
\leqno(Ab)$$

\bf Remark, \rm
Since ${\cal B}=\overline{x}+{\cal D}_0 \cap \overline{{\rm Ball}}({\bf H})$
and since ${\cal D}_0$ is a complex linear submanifold of ${\bf H}$,
we have $0\in{\cal B}$ if and only if
$\overline{x} \in{\cal D}_0$
which is equivalent to
$\big[{0\atop 1}\big]= \big[{\overline{x}\atop 1}\big] - \big[{\overline{x}\atop 0}\big] \in{\cal D}$.
In this case the terms $B\overline{x},\Lambda\overline{x}$
are also well-defined
and in their terms we can write the infinitesimal generator ${\cal A}$ in a matrix form

as
$$\eqalign{
&{\cal A} \Big[ {0\atop 1}\Big]
= {\cal A} \Big( \Big[ {\overline{x}\atop 1}\Big] - \Big[ {\overline{x}\atop 0}\Big] \Big) =
- \Big[ {B\overline{x}\atop \Lambda \overline{x}} \Big] ,\cr
&{\cal A} \Big[ {x\atop \xi}\Big] =
\Big[ {B(x-\xi \overline{x}) \atop \Lambda(x-\xi \overline{x})} \Big] =
\Big[ { B \ -\!B\overline{x} \atop \Lambda \ -\!\Lambda\overline{x} }\Big] \Big[ {x\atop \xi}\Big] .\cr}$$
Notice that in case of
$\big[{0\atop 1}\big] \not\in{\cal D}$ we have only
${\cal A}\big[{x\atop \xi}\big] =
\big[ {B \atop \Lambda} \big]\Big( [I \ -1]\big[{x\atop \xi}\big] \Big)$
while the formal product $\big[ {B \atop \Lambda} \big] [I \ -1]$ cannot be defined.

\bigskip
\bf Theorem. \it
Under the hypothesis $(Ab)$,
${\cal B}$ is the intersection of the dense complex linear submanifold ${\cal D}_0$ of ${\bf H}$
with the closed unit ball and
${\cal V}$ admits the Jordan form
$${\cal V}(x) = b-\{ xb^*x \} + iA x \qquad (x\in{\cal B})
\leqno(J)$$
in terms of the Jordan triple product
$\{ xc^*y\} := {1\over 2} \langle x\vert c\rangle y + {1\over 2} \langle y\vert c\rangle x$
where
$$b :={d\over dt}\big\vert_{t=0} \Psi^t(0)= \lim\limits_{t\to 0} a_t/t , \quad
Ax := \lim_{t\to 0} {1\over it} (U_t-I)x - {\rm Im} \big\langle \overline{x} \big\vert b \big\rangle x \quad (x\in {\cal D}_0).$$
%Here $A$ is a closed symmetric not necessarily bounded
%linear operator such that $iA$ is the generator of a strongly continuous one-parameter subgroup of ${\cal L}({\bf H})$.

\bigskip
\bf Proof. \rm
By the definition of the domain ${\cal D}$,
$${\cal A} \Big[ {0\atop 1}\Big] =
\lim_{t\to 0} {1\over t} \big( \widehat{\Psi}^t - \widehat{\Psi}^0\big) \Big[ {0\atop 1}\Big] =
\lim_{t\to 0} {1\over t} \Big\{ {1\over 1+\langle U_t \overline{x} \vert a_t \rangle}
\Big[ {a_t \atop 1} \Big] - \Big[ {0 \atop 1}\Big] \Big\} .$$
Since $\lim_{t\to 0} a_t =0$
and $\Vert U_t \overline{x} \Vert\le 1$ $(t\in\RR)$,
taking $(Ux)$ into account, we see
that the limit
$$b:=\lim_{t\to 0} {1\over t} a_t = {d\over dt} \Big\vert_{t=0} \Psi^t(0)$$
is well-defined and
$$B \overline{x} = -\Big\{ {\cal A} \Big[ {0\atop 1}\Big] \Big\}_{{\bf H}} =-b ,
\quad
\Lambda \overline{x} = -\Big\{ {\cal A} \Big[ {0\atop 1}\Big] \Big\}_{\CC} =
\big\langle \overline{x} \,\big\vert\, b \big\rangle .
%\langle U_t \overline{x} \vert a_t \}rangle = t \Lambda \overline{x} + o(t) , \quad
%a_t= t b + o(t) .
$$
As a consequence we also have
$$\eqalign{
\beta_t &=\sqrt{1-\langle a_t\vert a_t\rangle} =
\sqrt{1-\Vert tb+o(t)\Vert^2} =\cr
&= \sqrt{1- t^2 \Vert b\Vert^2 +o(t^2)} =
1 - {1\over 2} \Vert b\Vert^2 t^2 + o(t^2) .\cr}$$
Hence we deduce that
$1-\beta_t = o(t)$ and therefore, for any vector $z\in{\cal D}_0$.
$$\eqalign{
{1\over t} \big( \widehat{\Psi}^t - I \big) \Big[ {z\atop 0} \Big] &=
{1\over t} \left[ {(P_t + \beta_t \overline{P}_t) U_t z -
\langle U^t\overline{x} \vert a_t\rangle z
\atop
t \Lambda z + o(t) } \right] =\cr
&= \left[ {
t^{-1} (U_t-I) z + t^{-1} (1-\beta_t) \overline{P}_t U_t z -
\langle \overline{x} \vert b \rangle z
\atop
\Lambda z} \right] + o(1) .\cr}$$
Since, by definition,
$\big[ {Bz  \atop \Lambda z} \big]= \lim\limits_{t\to 0}
{1\over t} \big( \widehat{\Psi}^t - I \big) \big[ {z\atop 0} \big]$,
we get
$$Bz = \lim_{t\to 0} {1\over t} (U_t \!-\! I) z -
\big\langle \overline{x} \, \big\vert  b \big\rangle z
\qquad (z\in {\cal D}_0).  $$
In particular, from the relations $(Ab)$
we obtain that
for any unit vector $z\in{\cal D}_0$ we have
$i \langle Az \vert z \rangle + \lambda = \lim\limits_{t\to 0} t^{-1} \langle (U_t-I)z \vert z\rangle
- \langle \overline{x} \vert b \rangle$
whence the the symmetry of $A$ along with the properties
$0= {d\over dt}_{t=0} \Vert U_t z\Vert^2 = 2\,{\rm Re} \lim\limits_{t\to 0} t^{-1}\langle (U_t-I)z \vert z\rangle$
and $\lambda\in\RR$ implies
$$\lambda= -{\rm Re} \big\langle b \,\big\vert\, \overline{x} \big\rangle \, , \quad
A = \lim_{t\to 0} \hskip-4mm{\phantom{\big\vert}}^{{\rm strong}} \
{1\over it} \big( U_t -I\big) - {\rm Im} \big\langle \overline{x} \,\big\vert\, b \big\rangle I \, .$$
Finally we calculate ${\cal V} ={d\over dt}\big\vert_{t=0} \Psi^t$ by substituting the
previous results int the form ${\cal V}(x)= \big[\Lambda(\overline{x}-x)\big]x + B(x-\overline{x})$.
Namely, given any vector $x\in{\cal B}$,
taking into account the antisymmetry of the operator $B \!-\!\lambda I$,
we can write
$$\eqalign{
{\cal V}(x) &= \big[\Lambda \overline{x}\big]x - \big[\Lambda x\big]x+ Bx - B\overline{x}) =\cr
&= \langle \overline{x} \vert b \rangle - \langle (B-2\lambda I)x \vert \overline{x}\rangle x + Bx + b = \cr
&= \langle \overline{x} \vert b \rangle - \langle (B-\lambda I)x \vert \overline{x}\rangle x
+ \langle (B-\lambda I)x \vert \overline{x}\rangle x + Bx + b = \cr
&= b + \big[ \langle \overline{x} \vert b\rangle I + B \big] x +
\langle x \vert (B + \lambda I)\overline{x} \rangle x =\cr
&= b + \big[ \langle \overline{x} \vert b\rangle I + B \big] x +
\langle x \vert (-b + 2\lambda \overline{x}) \rangle x =\cr
&= b - \langle x\vert b\rangle x + \big[ (\langle \overline{x} \vert b\rangle + \lambda + (B-\lambda I)\big] x =\cr
&= b - \langle x\vert b\rangle x + \big[ i \langle A \overline{x} \vert \overline{x} \rangle I + iA \big] x\cr}$$
whence the Jordan form $(J)$ is immediate.
\qed

\bigskip
\bf Lemma. \it We have
$(1+\langle U_t \overline{x}\vert a_t\rangle)^{-1}
\Big[  {P_t+\beta_t \overline{P}_t \ a_t \atop \ \ \ a_t^* \ \ \ 1} \Big] =
\Big[ {I \ 0 \atop 0 \ 1 }\Big] +
t \Big[ \!\!\!\!
{ -\langle \overline{x} \vert b\rangle I\ b \atop \quad b^* \ -\langle \overline{x} \vert b\rangle} \Big]
+ o^{{\rm norm}}(t).$

\bigskip
\bf Proof. \rm
This is a direct consequence of the facts that
$a_t = tb+ o^{{\rm norm}}(t)$, $\beta_t=1+o(t)$,
$(1+\langle U_t \overline{x}\vert a_t\rangle)^{-1} =
(1-\langle \overline{x} \vert a_t\rangle)(1-\langle a_t \vert a_t\rangle)^{-1} =
(1-t\langle \overline{x} \vert b\rangle +o(t)) \beta^{-2} = 1-t\langle \overline{x} \vert b\rangle +o(t)$
and $P_t +\beta_t \overline{P}_t = (1-\beta_t)P_t +\beta_t (P_t+\overline{P}_t) = I +o^{{\rm norm}}(t)$
as $t\to 0$.
\qed


\bigskip
\bf Corollary. \it We have ${\cal D}_0 = {\rm dom} \big( \lim\limits_{t\to 0} t^{-1} (U_t-I) \big)$,
that is $x\in{\cal D}_0$ if and only if the limit $\lim\limits_{t\to 0} t^{-1}(U^tx-x)$ exists.

\bigskip
\bf Proof. \rm
Recall that ${\cal D}_0 = \big\{ x\in{\bf H} :\ {d\over dt}\big\vert_{t=0} \widehat{\Psi}^t \big[ {x\atop 0}\big]$
{\rm is well-defined}$\big\}$.
From the theorem we know that
$iAx =\lim\limits_{t\to 0} t^{-1}(U_t-I)x$ is well-defined for every vector $x\in{\cal D}_0$.
Conversely, suppose $u=\lim\limits_{t\to 0} t^{-1}(U_tx-x)$ is well-defined. Then
$U_tx= x+tu+o^{{\rm norm}}(t)$ and
$$\eqalign{
\widehat{\Psi}^t \Big[ {x\atop 0} \Big] &=
(1+\langle U_t \overline{x} \vert a_t\rangle )
\Big[ {P_t +\beta_t \overline{P}_t \ a_t \atop \ \ \ a_t^* \ \ \ 1 } \Big] \,
\Big[ {U_t \ 0 \atop 0 \ 1} \Big] \, \Big[ {x\atop 0} \Big] =\cr
&= \Big( \Big[ { I \ 0 \atop 0 \ 1} \Big] -
t \Big[ { \langle \overline{x} \vert a_t\rangle \ -b \atop -b^* \ \langle \overline{x} \vert a_t\rangle} \Big]
+ o^{{\rm norm}}(1) \Big) \,
\Big[ {x+t u+o^{{\rm norm}}(t) \atop 0} \Big] =\cr
&= \Big[ {x\atop 0} \Big] + t \Big[ {u - \langle \overline{x} \vert a_t\rangle x \atop
\langle x \vert b \rangle} \Big] + o^{{\rm norm}}(t) . \hbox{\qed}\cr}
$$

\bigskip
\bf Lemma. \it We have \ \ $U_{-t} = U_t^{-1} = U_t^*$, \ $a_{-t} = -U_t^* a_t$ \ \ $(t\in\RR)$.

\bigskip
\bf Proof. \rm
Given any parameter $t\in\RR$, we have $\Psi^{-t} = \Psi_t^{-1}$ that is
$\Theta_{a_{-t}} U_{-t} = \big[ \Theta_{a_t} U_t \big]^{-1} =
U_t^{-1} \Theta_{a_t}^{-1} = U_t^{-1} \Theta_{-a_t} = \big[ U_t^{-1} \Theta_{-a_t} U_t\big] U_t^{-1} =
\Theta_{U_t^{-1} (-a_t)} U_t^{-1}$.
Using the uniqueness of the decomposability of holomorphic automorphisms of circular domains into
M\"obius and unitary parts [\BKU], hence we deduce that
$\Theta_{a_{-t}} = \Theta_{-U_t^{-1} a_t}$ and $U_{-t} = U_t^{-1}$.
\qed

\bigskip

\bf Proposition. \it The operator $A$ in $(J)$ is self-adjoint.

\bigskip
\bf Proof. \rm
In view of the previous lemmas
we can conclude that
for any vector $\big[ {x\atop \xi}\big] \in{\cal D}$,
$$\eqalign{
{d\over dt}\Big\vert_{t=0} \widehat{\Psi}^t &=
{d\over dt}\Big\vert_{t=0} (1+\langle U_t \overline{x}\vert a_t\rangle)^{-1}
\Big[  {P_t+\beta_t \overline{P}_t \ a_t \atop \ \ \ a_t^* \ \ \ 1} \Big]
\Big[ {U_t \ 0 \atop 0 \ \ 1} \Big] \Big[ { x\atop \xi}\Big] =\cr
&= {d\over dt}\Big\vert_{t=0} (1+\langle U_t \overline{x}\vert a_t\rangle)^{-1}
\Big[  {P_t+\beta_t \overline{P}_t \ a_t \atop \ \ \ a_t^* \ \ \ 1} \Big] \Big[ { x\atop \xi}\Big] +
{d\over dt}\Big\vert_{t=0} \Big[ {U_t \ 0 \atop 0 \ \ 1} \Big] \Big[ { x\atop \xi}\Big] =\cr
&= \Big[ { iA \!-\! \langle \overline{x} \vert b\rangle I \ \ b \atop \quad b^*  \quad -\!\langle \overline{x} \vert b\rangle} \Big]
\Big[ { x\atop \xi}\Big] \cr}$$
The linear operator in ${\cal L}({\bf H}\oplus\CC)$ with matrix
$\big[ \!\!\!\! {-\!
\langle \overline{x} \vert b\rangle I \ \ b \atop \quad b^* \quad -\langle \overline{x} \vert b\rangle} \big] $
is bounded. Since ${\cal A} ={d\over dt}\big\vert_{t=0} \widehat{\Psi}^t$ with domain ${|\cal D}$
is the generator of a strongly continuous semigroup in ${\cal L}({\bf H}\oplus\CC)$,
by the theorem of bounded perturbations [\NAG], also
the operator with matrix
$\big[ {iA \ 0 \atop \ 0 \ 0}\big]$ with domain ${\cal D}$ is the
generator of a strongly continuous group in ${\cal L}({\bf H}\oplus\CC)$
strongly continuous one-parameter subgroup of ${\cal L}({\bf H})$
entailing that $iA$ in $(J)$ is the
generator of a strongly continuous group $\big[ V_t: t\in\RR\big]$ in ${\cal L}({\bf H})$.
Since $U_{-t}=U_t^{-1}=U_t^*$, the arguments on sun adjoint semigroups in [\NAG, p. 69]
show that $\lim_{t\to 0} t^{-1}(U_t^* -I) =-iA$ is the generator of the sun adjoint group
$\big[ V_t^*: t\in\RR\big]= \big[ V_{-t}: t\in\RR\big]$
and we have $-iA=(iA)^*$.
\qed

\bigskip
\bf Theorem. \it Any vector field of the form $(J)$
where $A$ is a not necessarily bounded self-adjoint operator with dense domain ${\cal D}_0 \subset {\bf H}$,
is the infinitesimal generator defined on ${\cal B}:={\cal D}_0 \cap\overline{{\rm Ball}}({\bf H})$
of a pointwise continuous one-parameter group
$\Phi^t:\ t\in\RR]$ of holomorphic automorphisms of $\overline{{\rm Ball}}({\bf H})$.\footnote{$^*$}{\rm
That is, for all $x\in {\cal B}:={\cal D}_0\cap\overline{{\rm Ball}}({\bf H})$,
the functions $t\mapsto \Phi^t(x)$ range in ${\cal D}_0$, they are differentiable and satisfy
the identity ${d\over dt} \Phi^t(x)={\cal V}\big(\Phi^t(x)\big)$ $(t\in\RR)$.}

\bigskip
\bf Proof. \rm
It suffices to see that
there is a strongly (i.e. pointwise) continuous one-parameter group
$\big[ V^t:\ t\in\RR\big]$ of
bounded linear operators of the space ${\bf H}\oplus \CC$ such that
$${d\over dt}\Big\vert_{t=0} V^t \Big[ {x \atop \xi} \Big] =
\Big[ {iA \ b \atop b^* \ \ 0}\Big] \Big[ {x \atop \xi} \Big] \quad (x\in{\cal D}_0,\ \xi\in\CC),
\qquad V^t {\bf K} \subset {\bf K} := \Big\{ \Big[ {x\atop \xi}\Big] : \ \Vert x\Vert^2 \ge \vert \xi\vert^2 \Big\}. $$
Namely, in this case the maps
$$\Phi^t (x) := \Big\{ V^t \Big[ {x\atop 1} \Big] \Big\}_{\CC}^{-1} \Big\{ V^t \Big[ {x\atop 1} \Big] \Big\}_{{\bf H}}
\qquad (t\in\RR,\ x\in{\cal B})$$
suit the requirements of the theorem since $x\in{\cal B} \Rightarrow \big[ {x\atop 1}\big] \in{\bf K}
\Rightarrow V^t \big[ {x\atop 1}\big] \Rightarrow \Phi^t(x)\in{\cal B}$ and
$x\!\in\!{\cal B} \!\Rightarrow\! {d\over dt}\big\vert_{t=0} \! \Phi^t(x) \!=\!
- \!\big\{ V^0 \big[ {x\atop 1} \big] \big\}_{\!\CC}^{\!-2}
{d\over dt}\big\vert_{t=0} \big\{ V^t \big[ {x\atop 1} \big] \big\}_{\!\CC}
\big\{ V^0 \big[ {x\atop 1} \big] \big\}_{\!{\bf H}} +
\big\{ V^0 \big[ {x\atop 1} \big] \big\}_{\!\CC}^{\!-1}
{d\over dt}\big\vert_{t=0} \big\{ V^t \big[ {x\atop 1} \big] \big\}_{\!{\bf H}}$
$=
- \big\{ \big[ {iA \ b \atop b^* \ 0}\big] \big[ {x\atop 1}\big] \big\}_{\CC} x \!+\!
\big\{ \big[ {iA \ b \atop b^* \ 0}\big] \big[ {x\atop 1} \big] \big\}_{\!{\bf H}} =
-\langle x\vert b\rangle x + iAx +b ={\cal V}(x)$.

Notice that a strongly continuous one parameter group of linear operator
leaves the cone ${\bf K}$ invariant if all its members map the boundary
$\partial {\bf K} \!\!=\! \big\{ \!\big[ {x\atop \xi} \big] \!\! : \Vert x\Vert \!\!=\!\!\vert \xi\vert\big\} \!\!=\!
\big\{ \!\big[ {x\atop e^{i\tau}\Vert x\Vert} \big] \!\! :  x\!\!\in\!\!{\bf H},\, \tau\!\!\in\!\!\RR \big\}$
\break
into itself.\footnote{$^*$}{In general, if a Hilbert space ${\bf W}$ is the orthogonal 
%direct 
sum
of the subspaces ${\bf W_k}$ $(k\!=\!1,2)$ and $C$ is the infinitesimal generator of a 
strongly continuous one-parameter group $[T^t: t\!\in\!\RR ]$ in ${\cal L}({\bf W})$
then we have $T^t {\bf C} \!=\!{\bf C}$ $(t\!\in\!\RR)$ for the cone
${\bf C} \!:=\!\big\{ {\bf w}_1\!\oplus\!{\bf w}_2: {\bf w}_k\!\in\!{\bf W}_k, \, 
\Vert{\bf w}_1\Vert \!>\! \Vert{\bf w}_2\Vert \big\}$
if and only if $C$ is tangent to the boundary of ${\bf C}$
that is if 
\hfill\break 
$(*)$ ${\rm Re}\big\langle\! C({\bf w}_1\!\oplus\!{\bf w}_2),\!{\bf w}_1 \!\big\rangle \!=\!
{\rm Re}\big\langle\! C({\bf w}_1\!\oplus{\bf w}_2),\!{\bf w}_2 \!\big\rangle$ whenever
${\bf w}_1\!\oplus\!{\bf w}_2 \!\in\!{\rm dom}(C)$ with
$\Vert{\bf w}_1\Vert \!=\! \Vert{\bf w}_2\Vert$.
\hfill\break
Proof. It is immediate that $T^t{\bf C\!}\!\subset\! {\bf C}$ $(t\!\in\!\RR)$
$\Rightarrow T^t\partial{\bf C}\!\subset\! \partial{\bf C}$ $(t\!\in\!\RR)$
$\Rightarrow {d\over dt}\vert_{t=0} T^t({\bf w}_1\oplus{\bf w}_2) \in 
{\rm Tan}_{{\bf w}_1\oplus{\bf w}_2}({\bf C})$ for ${\bf w}_1\oplus{\bf w}_2\in{\rm dom}(C)$
$\Rightarrow (*)$.
Assume $(*)$ and
let $P$ denote the canonical projection of ${\bf W}$ onto ${\bf W}_1$
and define
$T^{t,s}:=\exp(tC+sP)$ $(s,t\in\RR)$. 
By the theorem of bounded perturbations [\NAG, p.158]
the operators $T^{t,s}$ are all well-defined.
Moreover, by [\NAG, Corollary 1.7 p. 161]
(applied with $B:=sP$ and $A:=C$ there) 
we have $\lim_{s\to 0} T^{t,s}{\bf w} = T^t{\bf w}$ $({\bf w}\in{\rm dom}(C), t\in\RR)$.
Therefore, to establish that $T^t{\bf C}\subset {\bf C}$ $(t\in\RR)$, 
it suffices to see only that \ $(**)$
$T^{t,s}{\bf w} \!\in\!{\bf C}$ whenever 
${\bf  w}\!\in\!{\rm dom}(C)\!\cap\! {\bf C}$ and $t,s\!>\! 0$.
To proceed to contradiction, let 
$s,t>0$ and ${\bf w}:={\bf w}_1\oplus {\bf w}_2 \in{\rm dom}(C)$ with
$\Vert{\bf w}_1\Vert > \Vert{\bf w}_2\Vert$ but
$\Vert [T^{t,s}{\bf w}]_1\Vert \le \Vert [T^{t,s}{\bf w}]_2\Vert$.
The function 
$\delta(\tau):=\Vert [T^{\tau,s}{\bf w}]_1\Vert^2 - \Vert [T^{\tau,s}{\bf w}]_2\Vert^2$ 
is differentiable in $\tau$ on the whole $\RR$ and
$\delta(0)> 0 \ge \delta(t)$. Thus there exist a point $t_*\in(0,t]$
such that $\delta(\tau)>0=\delta(t_*)$ $(0\le\tau<t_*)$.
Since $\delta(t_*)=0$, 
${\bf w}_*:=T^{t_*,s}{\bf w} \in\partial{\bf C}$
and hence 
${\rm Re}\langle C{\bf w}_*,[{\bf w}_*]_1\rangle=
{\rm Re}\langle C{\bf w}_*,[{\bf w}_*]_2\rangle$. 
We get the contradiction $0\ge \delta^\prime(t_*) =
2{\rm Re}\langle (C+sP){\bf w}_*,[{\bf w}_*]_1\rangle -
2{\rm Re}\langle (C+sP){\bf w}_*,[{\bf w}_*]_2\rangle =
2s \Vert[{\bf w}_*]_1\Vert^2 >0$.      
}  
Therefore it suffices to check that
there is a (necessarily unique)strongly continuous one-parameter group
in ${\cal L}({\bf H}\oplus\CC)$ with domain ${\cal D}_0\oplus\CC$ such that
$${d\over dt} V^t \Big[ {x \atop \xi} \Big] =
\Big[ {iA \ b \atop b^* \ \ 0}\Big] V^t \Big[ {x \atop \xi} \Big] ,
\quad \left\Vert \Big\{ V^t \Big[ { x\atop \Vert x\Vert } \Big] \Big\}_{{\bf H}} \right\Vert=
\left\vert \Big\{ V^t \Big[ { x\atop \Vert x\Vert } \Big] \Big\}_{\CC} \right\vert
\qquad (x\in{\cal D}_0,\ t\in\RR) . $$
By Stone's theorem, the ${\bf H}\oplus\CC$-unitary operators
$W^t \big[{x\atop\xi}\big] := \big[{\exp(itA)x \atop \xi}\big]$
form a strongly continuous one-parameter group whose infinitesimal generator
is defined on ${\rm dom}(A) \oplus\CC ={\cal D}_0\oplus\CC$ with the diagonal matrix
$\big[ {iA \ 0 \atop 0 \ \ 0}\big]$.
Since the matrix $\big[ {0 \ b \atop b^* \ 0} \big]$ represents a bounded linear operator
in ${\bf H}\oplus\CC$,
by the theorem of bounded perturbations {\NAG],
there is a strongly continuous one-parameter group
$\big[ V^t: \, t\in\RR\big]$
whose generator is defined on
${\cal D}_0\oplus\CC$ with the matrix $\big[ {iA \ b \atop b^* \ 0}\big]$.
In particular
${d\over dt} V^t \big[ {x\atop \xi}\big] =
\big[ {iA \ b \atop b^* \ 0}\big] V^t \big[ {x\atop \xi}\big]$ $(t\in\RR,\ x\in{\cal D}_0)$.
To complete the proof, we show that necessarily
$${d\over dt} \left[ \Big\Vert\Big\{ V^t \Big[ { x\atop \Vert x\Vert } \Big] \Big\}_{{\bf H}} \Big\Vert^2 -
\Big\vert \Big\{ V^t \Big[ { x\atop \Vert x\Vert } \Big] \Big\}_{\CC} \Big\vert^2 \right] =0
\qquad (t\in\RR,\ x\in{\cal D}_0). $$
Consider any vector
$x\in{\cal D}_0$ and write $\big[ {x_t \atop \xi_t} \big] :=  V^t \big[ {x\atop \Vert x\Vert }\big]$
for all parameters $t\in\RR$.
Then
$$\eqalign{
{d\over dt} \Big[ \Vert x_t \Vert^2 - \vert \xi_t\vert^2 \Big] &=
2\,{\rm Re} \Big[ \big\langle dx_t/dt \big\vert x_t \big\rangle - \big( d\xi_t/dt) \overline{\xi_t} \Big] =\cr
&= 2\,{\rm Re} \left[ \Big\langle \Big\{ \Big[ {iA \ b \atop b^* \ \ 0}\Big]\Big[ {x_t\atop\xi_t}\Big] \Big\}_{\!{\bf H}}
\Big\vert x_t \Big\rangle -
\Big\{ \Big[ {iA \ b \atop b^* \ \ 0}\Big]\Big[ {x_t\atop\xi_t}\Big] \Big\}_{\!\CC}
\big( d\xi_t/dt) \overline{\xi_t} \right] =\cr
&= 2\,{\rm Re} \left[ \big\langle iAx_t + \xi_t b \big\vert x_t \big\rangle -
\big\langle x_t \big\vert b\big\rangle \overline{\xi_t} \right] =\cr
&= -2\,{\rm Im} \big\langle Ax_t \big\vert x_t \big\rangle +
2\,{\rm Im}\big( \big\langle \xi b\big\vert x_t\big\rangle - \big\langle x_t \big\vert \xi b\big\rangle \big)
=0 . \hbox{\qed}
\cr}$$

\bigskip
\bigskip

\bf Reduction to the Jordan case $0\in{\cal B}$
\rm

\bigskip

In accordance with the previous standard notations,
throughout this section $\big[ \overline{\Psi}^t:\ t\in\RR\big]$ denotes
a pointwise continuous one-parameter subgroup of ${\rm Aut}\big(\overline{{\rm Ball}}({\bf H})\big)$
with infinitesimal generator ${\cal X}: {\cal B}\to {\bf H}$
where ${\cal B}:=\big\{ x\in\overline{{\rm Ball}}({\bf H}):\
t\mapsto\overline{\Psi}^t(x)$ is differentiable on $\RR\big\}$.
We regard ${\cal X}$ as a vector field defined on the intersection of a dense complex-affine
submanifold with the unit ball of ${\bf H}$.

Let us fix any point $a\in{\cal B}$ with $\Vert a\Vert<1$ and,
in terms of the M\"obius transformations $(M)$, consider the transformations
$$\overline{\Phi}^t := \Theta_{-a}^\# \overline{\Psi}^t =
\Theta_{-a} \circ \overline{\Psi}^t\circ \Theta_a$$
given by the adjoint action of $\Theta_{-a} =\Theta_a^{-1}$.
Clearly $\big[ \overline{\Phi}^t:\ t\in\RR\big]$ is a
pointwise continuous one-parameter subgroup of
${\rm Aut}\big(\overline{{\rm Ball}}({\bf H})\big)$
whose infinitesimal generator is the vector field
$${\cal V} := \Theta_{-a}^\# {\cal X} :  y\mapsto
{d\over dt} \Big\vert_{t=0} \Theta_{-a} \big( \overline{\Psi}^t \big( \Theta_a(y)\big)\big)$$
defined on the domain
$${\cal B}_1 := \Theta_{-a} {\cal B} = {\cal D}_1 \cap \overline{{\rm Ball}}({\bf H}).$$
Recall that $0\in{\cal B}_1$ and ${\cal D}_1$ is some dense complex-linear submanifold in ${\bf H}$
and, as a consequence of the results of the previous section,
we can write ${\cal V}$ in the Jordan form $(J)$ with some vector $b\in{\bf H}$ and a suitable
not necessarily bounded self-adjoint linear operator $A:{\cal D}_1\to{\bf H}$.
By the aid of the inverse adjoint action $\Theta_a^\#$, hence we get ${\cal X}$ in the form
$$\eqalign{
{\cal X}(x) &= \big[\Theta_a^\# {\cal V}\big] (x) =
{d\over dt}\Big\vert_{t=0} \Theta_a \Big(\overline{\Phi}^t\big( \Theta_{-a}(x)\big)\Big) =\cr
&= {d\over dt}\Big\vert_{t=0} \Theta_a \Big(
\Theta_{-a}(x) + t {\cal V}\big(\Theta_{-a}(x)\big) +o(t) \Big) =\cr
&= {d\over dt}\Big\vert_{t=0} \big(1+\langle y+tk \vert a\rangle \big)^{-1}
\Big( \big[ {\rm P}_a + \beta_a \overline{{\rm P}}_a \big] (y+tk) +a\Big)
\Big\vert\strut_{\hskip-7mm{y=\Theta_{-a}(x)\atop \hskip7mm k=b-\langle y\vert b\rangle y +iAy}} \cr}$$
for the locations $x\in{\cal B}$.
Next we calculate the a detailed algebraic form for ${\cal X}$.
Notice that
$${d\over dt}\Big\vert_{t=0} \big(1+\langle y+tk \vert a\rangle \big)^{-1} =
-\langle k\vert a\rangle \big( 1 + \langle y\vert a\rangle\big)^{-2} $$








\bigskip\bigskip\bigskip
!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
\bigskip


\bigskip\bigskip
\bf The classical case $\Vert \overline{x} \Vert <1$.
\rm
Since the M\"obius transformation
$\Theta_{\overline{x}}$ takes the origin into $\overline{x}$ and since
its inverse coincides with $\Theta_{-\overline{x}}$,
the automorphisms of the form $T^t:=\Theta_{-\overline{x}} \circ \Psi^t \circ \Theta_{\overline{x}}$
leave the origin fixed and hence
$\big[ T^t: t\in\RR \big]$ is a strongly continuous one-parameter group
consisting of restrictions of linear unitary operators
to ${\rm Ball}({\bf H})$.
By Stone's classical theorem,
there is a not necessarily bounded self-adjoint operator
$A$ with dense domain in ${\bf H}$ such that
$T^t= \exp(itA)$ $(t\!\in\!\RR)$.
It follows
$\Psi^t =\Theta_{\overline{x}} \circ \exp(itA) \circ \Theta_{-\overline{x}}$
$(t\!\in\!\RR)$ entailing that
the (non-linear) infinitesimal generator of
$\big[ \overline{\Psi}^t: t\in\RR \big]$
is
$${d\over dt}\Big\vert_{t=0}  \overline{\Psi}^t (x) =
i \Big[\Theta_{\overline{x}}^\prime\big(  \Theta_{-\overline{x}}(x)\big)\Big]
A \Theta_{-\overline{x}}(x)$$
for $x$ in the domain
$\Theta_{\overline{x}} \big( {\rm dom}(A)\cap \overline{{\rm Ball}}({\bf H}) \big) =
(\overline{x}+{\cal D}_0)\cap\overline{{\rm Ball}}({\bf H}) ={\cal B} $.

\DEL{

\bigskip\bigskip
\bf The case of boundary fixed points:
$\overline{x} \in {\rm Fix}\{ \overline{\Psi}^t : t\in\RR\}\cap \partial{\rm Ball}({\bf H})$.
\rm

Henceforth we assume $\Vert \overline{x} \Vert=1$.

\bigskip
\bf Lemma. \it
Given any vector $z\in{\cal D}_0$
such that $\overline{x}\!+\! z \perp z$, with the notations $(N)$ we have
$$\Lambda z = \big\langle Bz \,\big\vert\, \overline{x} \big\rangle +
2\, {\rm Re} \big\langle Bz \,\big\vert\, z \big\rangle \ .$$
\bf Proof. \rm
For any parameter $\varphi\!\in\!\RR$,
$\Vert \overline{x} \!+\! z \!+\! e^{i\varphi}z \Vert^2 \!=\!
\Vert \overline{x} \!+\! z \Vert^2 + \Vert (-z) \Vert^2 \!=\!
\Vert (\overline{x} \!+\! z) \!-\! z \Vert^2 \!=\!
\Vert \overline{x} \Vert^2 \!\!=\! 1$.
Thus, according to the last corollary,
$${\rm Re} \Big( -(1 \!+\! e^{i\varphi}) \Lambda z +
(1 \!+\! e^{i\varphi}) \big\langle Bz \,\big\vert\, \overline{x} \big\rangle +
\vert 1 \!+\! e^{i\varphi}\vert^2 \big\langle Bz \,\big\vert\, z \big\rangle \Big)
=0 \qquad (\varphi\in\RR).$$
That is we have
$$\eqalign{
&{\rm Re} \big( \alpha + \beta e^{i\varphi} +\gamma  e^{-i\varphi} \big)=0
\quad  (\varphi\in\RR) \cr
&{\rm where} \ \
\alpha:= -\Lambda z + \big\langle Bz \,\big\vert\, \overline{x} +2z \big\rangle, \
\beta:= -\Lambda z + \big\langle Bz \,\big\vert\, \overline{x} +z \big\rangle, \
\gamma:= \big\langle Bz \,\big\vert\, z \big\rangle .\cr}$$
Since \
$2\, {\rm Re} \big( \alpha + \beta e^{i\varphi} +\gamma  e^{-i\varphi} \big) =
2\, {\rm Re} (\alpha) + (\beta+\overline{\gamma}) e^{i\varphi} +
(\overline{\beta}+\gamma) e^{-i\varphi}$,
we have necessarily ${\rm Re}(\alpha) = \beta+\overline{\gamma}=0$.
The identity $\beta+\overline{\gamma}=0$ corresponds to our statement.
\qed

\bigskip
\bf Remark. \rm
In course of the proof, we cannot gain more information
by investigating the relation ${\rm Re} (\alpha)=0$ because
$\beta+\overline{\gamma }=0 \Rightarrow {\rm Re} (\alpha)=0$.

\bigskip\bf Proposition. \it
There exists a constant $\rho\in\RR$ such that
$$\Lambda v = \big\langle (B \!+\! \rho I)v \,\big\vert\, \overline{x} \big\rangle, \quad
2\,{\rm Re}\, \big\langle Bv \big\vert v \big\rangle
= \rho \big\langle v \big\vert v \big\rangle
\qquad (v \in {\cal D}_0).$$
\bf Proof. \rm
Consider any vector $0\ne v\in{\cal D}_0$
and let
$$z:= \zeta v \quad \hbox{where} \quad
\zeta := -{ \langle \overline{x} \vert v \rangle \over \langle v \vert v \rangle } .$$
Observe that \ $\overline{x} \!+\! z \perp z \in{\cal D}_0$.
Hence we may apply the lemma with the conclusion
$\zeta\, \Lambda v =
\big\langle \zeta Bv \,\big\vert\, \overline{x} \big\rangle +
2\,{\rm Re} \big\langle \zeta Bv \,\big\vert\, \zeta v\big\rangle$
that is
$\Lambda v =
\big\langle Bv \,\big\vert\, \overline{x} \big\rangle +
2 \,\overline{\zeta}\, {\rm Re} \big\langle Bv \,\big\vert\, v\big\rangle$
implying
$$\Lambda v = \big\langle Bv \,\big\vert\, \overline{x} \big\rangle +
2 \,{ {\rm Re} \big\langle Bv \,\big\vert\, v\big\rangle
\over  \big\langle v \,\big\vert\, v\big\rangle } \,
\big\langle v \,\big\vert\, \overline{x} \big\rangle \ .$$
Hence we see that
$$v\mapsto 2 \,{ {\rm Re} \big\langle Bv \,\big\vert\, v\big\rangle
\over  \big\langle v \,\big\vert\, v\big\rangle } =
{\Lambda v - \big\langle Bv \,\big\vert\, \overline{x} \big\rangle
\over\big\langle v \,\big\vert\, \overline{x} \big\rangle }$$
is a real valued Gateaux holomorphic function on the
algebraically open domain $\{ v\!\in\!{\cal D}_0 : \ v\not\perp \overline{x} \}$
which is possible only if being constant. Hence the statement is immediate.
\qed







\vskip10mm
!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
\vskip10mm


\bigskip
\bf Remark. \rm In general, if $t\mapsto f_t$ is a function from $\RR$ into a normed space ${\bf X}$ and
we have
${1\over t} \Big( (1+o(1) f_t \Big) = g +o(1)$ for some $g\in{\bf X}$ as $t\to 0$ then also
${1\over t} f_t = g+o(1)$ \quad Proof: $f_t/t =[g+o(1)]/[1+o(1)]$.

\bigskip
\bf Lemma. \it We have $a_t\to 0$ and $\langle U_t \overline{x} \vert a_t\rangle \to 0$.
Furthermore there is a $($not necessarily bounded$)$
linear functional $\lambda: {\cal D}_0\to\CC$ such that
$$\lambda(x) =\lim_{t\to 0} {1\over t} \langle U_tx \vert a_t\rangle \qquad (x\in {\cal D}_0).
\leqno(L)$$
\bf Proof. \rm
By assumption $a_t=\Psi^t(0)\to 0$. Hence also
$\big\vert \langle U_t \overline{x} \vert a_t\rangle \big\vert \le \Vert \overline{x}\Vert \Vert a_t\Vert \to 0$.
Let $x\in{\cal D}_0$. Then
$${1\over t} \big[ \widehat{\Psi}^t -I\big] \left[ {x \atop 0}\right] =
{1\over t} \left\{ {1 \over 1+\langle U_t\overline{x} \vert a_t\rangle }
\left[ \matrix{(P_t+\beta_t \overline{P}_t)U_tx \cr
\langle U_tx \vert a_t\rangle }\right] -
\left[ {x \atop 0}\right] \right\}
$$
\qed


$$A_t=P_t+\beta_t \overline{P}_t$$

$${1\over t} \left[
{1\over 1+\langle \overline{x}\vert a_t\rangle}
\left[\matrix{A_tU_t &a_t\cr a^*_tU_t &1\cr}\right]
\left[\matrix{x+\overline{x}\cr 1\cr}\right]-
\left[\matrix{x+\overline{x}\cr 1\cr}\right]
\right] \to {\cal A}(x+\overline{x}) \qquad (x\in{\cal D}_0)$$

$$\left[\matrix{A_tU_t &a_t\cr a^*_tU_t &1\cr}\right]
\left[\matrix{\overline{x}\cr 1\cr}\right]=
\left[\matrix{\overline{x}\cr 1\cr}\right]$$

$$A_tU_t \overline{x}+a_t = \big(1+\langle U_t\overline{x} \vert a_t\rangle \big) \overline{x} , \quad
\langle U_t\overline{x} \vert a_t\rangle =o(1) $$

$${1\over t} \langle U_t x \vert a_t\rangle \to \lambda(x), \ \
{1\over t} \big[ A_tU_t x- \big(1+\langle U_t\overline{x} \vert a_t\rangle \big)x \to B(x)
\ \ (x\in{\cal D}_0) $$

$$\eqalign{
{\cal A}(x+\overline{x} &=
\lim_{t\to 0} \left[
{1/t\over 1+\langle \overline{x}\vert a_t\rangle}
\left[\matrix{A_tU_t &a_t\cr a^*_tU_t &1\cr}\right]
\left[\matrix{x\cr 0\cr}\right]-
\left[\matrix{x\cr 0\cr}\right] \right] =\cr
&=\lim_{t\to 0} {1\over t}\left[
\matrix{A_tU_t x - (1+\langle U_t\overline{x} \vert a_t\rangle)x\cr
\langle U_t x \vert a_t\rangle \cr}\right]
\cr}$$

$$\eqalign{
&{1\over t} \big[ \Psi_t (x+\overline{x}) - (x+\overline{x}) \big] =
{1\over t} \left[{A_tU_t (x+\overline{x})+a_t \over
1+\langle U_t(x+\overline{x}) \vert a_t\rangle} - (x+\overline{x}) \right] =\cr
&= {1\over t} {A_tU_t x - (1+\langle U_t\overline{x} \vert a_t\rangle) x \over
(1+\langle U_t(x+\overline{x}) \vert a_t\rangle} +
{1\over t}{A_tU_t \overline{x} - (1+\langle U_t\overline{x} \vert a_t\rangle) \overline{x} \over
(1+\langle U_t(x+\overline{x}) \vert a_t\rangle} =\cr
&= o(1) + {1\over t} \Big\{ A_t U_t x - (1 \!+\! \langle U_t\overline{x} \vert a_t\rangle) x -
\lambda(x)x + (1 \!+\! \langle U_t\overline{x} \vert a_t\rangle) \overline{x} -
(1 \!+\!\langle U_t(x \!+\!\overline{x}) \vert a_t\rangle) \overline{x} \Big\} =\cr
&= o(1) + B(x) - \lambda(x)x - \lambda(x)\overline{x}  \qquad (x\in{\cal D}_0)\cr}$$

}%ENDDEL

\vfill
\eject

\bf References
\rm

\medskip
\item{[\HS]} T.L. Hayden -- T.J. Suffridge, \it Biholomorphic maps in Hilbert space have a fixed point,
\rm Pacific J. Math., \bf 38 \rm (1971) 419-422.

\medskip
\item{[\GOEBEL]} K. Goebel, \it Fixed points and invariant domains of holomorphic mappings of the Hilbert ball,
\rm Nonlinear Analysis, Theory, Methods and Applications, \bf 6/12 \rm (1982) 1327-1334.


\medskip
\item{[\FRVE]} T. Franzoni -- E. Vesentini, Holomorphic Maps and Invariant Distances,
North Holland Math. Studies 40, North Holland Publ. Co., Amsterdam-New York-Oxford, 1980.

\medskip
\item{[\ABATE]} M. Abate, \it Common fixed points of commuting holomorphic maps, \rm
Math. Ann. \bf 283 \rm (1989) 645-655.

\medskip
\item{[\DINEEN]} S. Dineen, \it Complete holomorphic vector fields on the second dual of a Banach space,
\rm Math. Scand. \bf 59 \rm (1986) 131-142.

\medskip
\item{[\RSZ]} F. Riesz -- B. Sz.-Nagy,
Vorlesungen \"uber Funktionalanalysis,
VEB Deutscher Verlag der Wissenschaften, Berlin, 1968.

\medskip
\item{[\BKU]} H. Braun -- W. Kaup -- H. Upmeier, \it On the automorphisms of
circular and Reinhardt domains in complex Banach spaces,
\rm Manuscripta Math., \bf 25 \rm (1978), 97-133.

\medskip
\item{[\NAG]} K.-J. Engel -- R. Nagel,
One-Parameter Semigroups for Linear Evolution Equations,
Graduate Texts in Mathematics Vol. 194, Springer-Verlag, New York, 2000.


\bye
