Throughout this work H denotes an arbitrarily fixed complex Hilbert space with the scalar product ⟨x|y⟩ which is linear in x and conjugate linear in y, giving rise to the norm ∥x∥ = ⟨x|x∥^{1/2}. For any vector c ∈ H, we shall write c^{∗} := [x ↦ ⟨x|a⟩] for its associated functional. The open ball of radius ρ > 0 and centered in 0 in H will be denoted with ρBall(H), and we write ρ\overline{Ball}(H)(= \{x : ∥x∥ ≤ ρ\}) for its closure.

Recall [3, Ch. VI] that the group of Aut Ball(H) all holomorphic automorphisms of Ball(H) admits a matrix representation in the following sense: Each element Ψ of Aut Ball(H) has the fractional linear form

(R)  Ψ(x) = \frac{Ax + b}{⟨x|c⟩ + d} , A ∈ L(H), b, c ∈ H, d ∈ ℂ

and we have

Ψ_{1}∘Ψ_{2}(x) = Ψ_{1}(Ψ_{2}(x)) = \frac{Ax + b}{⟨x|c⟩ + d} whenever [ \begin{matrix}A & b \\ c^{∗} & d \end{matrix} ] = [ \begin{matrix}A_{1} & b_{1} \\ c_{1}^{∗} & d_{1} \end{matrix} ] [ \begin{matrix}A_{2} & b_{2} \\ c_{2}^{∗} & d_{2} \end{matrix} ] .

This representation is unique up to a constant, since in (R) we necessarily have

[ \begin{matrix}A & b \\ c^{∗} & d \end{matrix} ] = d[\begin{matrix}(P_{a}+β\overline{P}_{a}) & a \\ a^{∗} & 1 \end{matrix} ][ \begin{matrix}U & 0 \\ 0 & 1 \end{matrix} ]  where  \begin{matrix}a:=Ψ(0), β =\sqrt{1−∥a∥^{2}}, \\ U = (β^{2}P_{a}+ β\overline{P}_{a})^{−1}Ψ′(0) \end{matrix}

in terms of the projections P_{a} onto ℂa resp. \overline{P}_{a} := 1 − P_{a} and the Fréchet derivative Ψ′. Any automorphism Ψ ∈ Aut Ball(H) extends holomorphically to (1 − ∥Ψ(0)∥)^{−1}Ball(H) and hence admits a continuous extension \overline{Ψ} to the closed unit ball \overline{Ball}(H). We write Aut \overline{Ball}(H) := \{\overline{Ψ} : ψ ∈ Aut Ball(H)\} for their collection.

Recall [3, Ch.VI] that any automorphism \overline{Ψ} ∈ Aut \overline{Ball}(H) is weakly continuous and preserves the family Aff(\overline{Ball}(H)) of all complex affine^{∗} closed subspaces intersected with \overline{Ball}(H). By Schauder’s fixed point theorem, Fix(\overline{Ψ}) ≠ ∅, since \overline{Ball}(H) is weakly compact. Moreover we have the following alternatives:

(1) Fix(\overline{Ψ}) ∈ Aff(\overline{Ball}(H)), (2) Fix(\overline{Ψ}) consists of two boundary points.

In case (2) from the proof of [3, Thm.VI.4.8] we see even that \overline{Ψ} = \overline{Φ} ∘ \overline{Θ}_{a} ∘ \overline{Φ}^{−1} with a suitable automorphism Φ ∈ Aut \overline{Ball}(H) and a Möbius shift

(M)  Θ_{a} : x ↦ \frac{P_{a}x + a + β_{a}\overline{P}_{a}x}{1 + ⟨x|a⟩},  β_{a} = \sqrt{1 − ∥a∥^{2}}

for some 0 ≠ a ∈ Ball(H) such that Fix(\overline{Θ})_{a} = \{−e, e\} where e := a/∥a∥.

The next result is an infinite dimensional extension for a simple special case of a far reaching theorem of Abate [4] established for finite dimensional uniformly convex domains. It

∗ If x = ∑_{k=1}^{2}λ_{k}x_{k} with λ_{1}, λ_{2} ∈ ℂ and ∑_{k=1}^{2}λ_{k} = 1 then \overline{Ψ}(x) = ∑_{k=1}^{2}α_{k}\overline{Ψ}(x_{k}) for some α_{1}, α_{2}∈ℂ with ∑_{k=1}^{2}α_{k} =1 (namely with α_{k} =λ_{k}[1+⟨x_{k}|U^{∗}a⟩]/[1+⟨λ_{1}x_{1}+λ_{2}x_{2}|U^{∗}a⟩]). seems that Dineen’s bidual embedding method [5] 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).

Lemma. Abelian subsets of Aut \overline{Ball}(H) admit common fixed points.

Proof. Assume \{\overline{Ψ}_{j} : j ∈ J \} ⊂ Aut \overline{Ball}(H) with \overline{Ψ}_{j} ∘ \overline{Ψ}_{k} = \overline{Ψ}_{k} ∘ \overline{Ψ}_{j} (j, k ∈ J). By weak continuity, for any index family K ⊂ J, the set ⋂_{k∈K} Fix(\overline{Ψ}_{k}) of common fixed points is weakly compact. Thus, according to Riesz’ intersection theorem, it suffices to see that ⋂_{k∈K} Fix(\overline{Ψ}_{k}) ≠ ∅ for finite index families K. By proceeding to contradiction, let K = \{k_{1}, …, k_{N}\} be a subset of J with minimal cardinality such that ⋂_{k∈K} Fix(\overline{Ψ}_{k}) = ∅. Necessarily N > 1 and S := ⋂_{n=1}^{N−1} Fix(\overline{Ψ}_{k_{n}}) ≠ ∅ is either a weakly compact convex subset of \overline{Ball}(H) or it consists of two boundary points. Since \overline{Ψ}_{k_{N}} commutes with all the maps \overline{Ψ}_{k_{n}} (n < N), we have \overline{Ψ}_{k_{N}} (S) ⊂ S. Indeed, if x ∈ S then \overline{Ψ}_{k_{n}}(\overline{Ψ}_{k_{N}} (x)) = \overline{Ψ}_{k_{N}} (\overline{Ψ}_{k_{n}}(x)) = \overline{Ψ}_{k_{N}} (x) ⇒ \overline{Ψ}_{k_{N}} (x) ∈ Fix(\overline{Ψ}_{k_{n}}) (n < N). Hence Schauder’s fixed point theorem excludes the case of S being convex. Suppose S = {p, q} ⊂ ∂Ball(H). Then necessarily Fix(\overline{Ψ}_{k_{M}}) = \{p, q\} for some index M < N and \overline{Ψ}_{k_{N}} : p ↔ q. However, in this case we can write \overline{Ψ}_{k_{N}} = \overline{Φ}∘\overline{Θ}_{a}∘\overline{Φ}^{−1} with a suitable automorphism \overline{Φ} ∈ Aut \overline{Ball}(H) and a Möbius shift \overline{Θ}_{a} of the form (M) where 0≠a∈Ball(H). Then, by setting \overline{Ω}:=\overline{Φ}^{−1}∘\overline{Ψ}_{k_{M}}∘\overline{Φ} and e:=a/|a∥, we have \{±e\}=Fix(\overline{Θ}_{a}) and \overline{Ω} : e↔−e. On the other hand, it is immediate that e = lim_{n→∞} \overline{Θ}_{a}^{n}(x) for every point x ∈ \overline{Ball}(H)\\{e\}. Taking any point f ∈ Fix(\overline{Ω}) we get the contradiction e = lim_{n→∞} \overline{Θ}_{a}^{n}(f) = lim_{n→∞} \overline{Θ}_{a}^{n}∘ \overline{Ω}(f) = lim_{n→∞} \overline{Ω}∘ \overline{Θ}_{a}^{n}(f) = \overline{Ω}(e) = −e. Q.e.d.

Henceforth we keep fixed the notation [Ψ^{t} : t ∈ ℝ] for an arbitrarily given strongly continuous one-parameter group in Aut Ball(H). That is

(P)  Ψ^{t}(x) = \frac{(P_{t} + β_{t}\overline{P}_{t})U^{t}x + a_{t}}{1 + ⟨U_{t}x|a_{t}⟩} = [Θ_{a_{t}} ∘ U_{t}](x)

where a_{t} =Ψ_{t}(0)∈Ball(H), U_{t} ∈L(H) is a unitary operator and, with the notations in (M), β_{t}:=β_{a_{t}}, P_{t}:=P_{a_{t}}, \overline{P}_{t}:=\overline{P}_{a_{t}} for short. In particular P_{t}x = [⟨x|a_{t}⟩/⟨a_{t}|a_{t}⟩]a_{t}, \overline{P}_{t} = I − P_{t}.

According to the previous lemma, the continuous extensions \overline{Ψ}^{t} (t ∈ ℝ) to \overline{Ball}(H) have a common fixed point which we shall denote with \overline{x}:

\overline{x} ∈ \overline{Ball}(H), \overline{Ψ}^{t}(\overline{x}) = \overline{x} (t ∈ ℝ).

Remark. In finite dimensions, it is customary to normalize (R) by requiring det \left[{A b \atop c^{∗}d}\right]=1. Thus, in case of dim(H) = N, in this manner one can establish a canonical identification of Aut Ball(H) with a subgroup of the classical matrix group 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.

Proposition. With the notations (P), the (R)-type representation matrices

Ψ̂^{t} := \frac{1}{1 + ⟨U_{t}\overline{x}|a_{t}⟩} [ \begin{matrix}(P_{t} + β_{t}\overline{P}_{t})U_{t} & a_{t} \\ [U_{t}^{∗}a_{t}]^{∗} & 1 \end{matrix} ] = \frac{1}{1 + ⟨U_{t}\overline{x}|a_{t}⟩} [ \begin{matrix}P_{t} + β_{t}\overline{P}_{t} & a_{t} \\ a_{t}^{∗} & 1 \end{matrix} ] [ \begin{matrix}U_{t} & 0 \\ 0 & 1 \end{matrix} ]

of [Ψ_{t} : t ∈ ℝ] form a strongly continuous one-parameter group of operators in H ⊕ ℂ.

Proof. Since Ψ^{t} ∘Ψ^{s} = Ψ^{t+s} (t, s ∈ ℝ), for the representation matrices we have Ψ̂^{t}Ψ̂^{s} = d_{t,s}Ψ̂^{t+s} with suitable constants d_{t,s} ∈ ℂ. The fixed point property \overline{Ψ}^{t}(\overline{x}) = \overline{x} implies

Ψ̂^{t} \left[{\overline{x} \atop 1}\right] = \left[{\overline{x} \atop 1}\right] (t ∈ ℝ).

Hence necessarily d_{t,s} =1 (t, s∈ℝ), thus [Ψ̂^{t} : t∈ℝ] is a one-parameter matrix group. By assumption, the function t ↦ a_{t} = Ψ^{t}(0) is norm-continuous ℝ → Ball(H). Hence we can deduce the strong continuity of the H-unitary operator valued function t ↦ U_{t}. Namely consider any vector x ∈ H. To establish the norm-continuity of the function t ↦ U_{t}, we may assume without loss of generality that x ∈ Ball(H). Then, by the aid of the Möbius shifts (M) we can write

U_{t}x = [Θ_{a_{t}}^{−1} ∘ Ψ_{t}](x) = Θ_{−a_{t}}(Ψ(x)) (t∈ℝ).

Observe that the norm continuity of t ↦ a_{t} implies the continuity of t ↦ ⟨x|a_{t}⟩ and t ↦ β_{t} ∈ [0, 1) entailing the norm-continuity of t ↦ P_{t} + β_{t}\overline{P}_{t} ∈ \overline{Ball}(L(H)). Hence the required norm-continuity of t ↦ U_{t}x = (1 − ⟨x|a_{t}⟩)^{−1}[(P_{t}x − a_{t} + β_{t}\overline{P}_{t}x] is immediate. In general, the product of two bounded strongly continuous linear operator valued functions ℝ → L(X) over a normed space X is strongly continuous. Hence we conclude that the entries (1, 1), (1, 2), (2, 1) resp. (2, 2) of the matrices Ψ̂^{t} are strongly continuous functions ℝ → L(H), ℝ → H, ℝ → H^{∗}≃H resp. ℝ → ℝ which completes the proof. Q.e.d.

Remark. It is worth to notice that the term ⟨U_{t}\overline{x}|a_{t}⟩ is actually independent of u_{t} as

(Ux)  ⟨U_{t}\overline{x}|a_{t}⟩ = \frac{⟨\overline{x} − a_{t}|a_{t}⟩}{1 − ⟨\overline{x}|a_{t}⟩},  Ψ̂^{t} = \frac{1 − ⟨\overline{x}|a_{t}⟩}{1 − ⟨a_{t}|a_{t}⟩} [ \begin{matrix}P_{t} + β_{t}\overline{P}_{t} & a_{t} \\ a_{t}^{∗} & 1 \end{matrix} ] [ \begin{matrix}U_{t} & 0 \\ 0 & 1 \end{matrix} ] .

Proof: In general we have P_{t}y = ⟨y|a_{t}⟩⟨a_{t}|a_{t}⟩^{−2}a_{t} (0 ≠ a_{t}, y ∈ H). It follows ⟨P_{t}U_{t}\overline{x}|a_{t}⟩ = ⟨U_{t}\overline{x}|a_{t}⟩ with ⟨\overline{P}_{t}U_{t}\overline{x}|a_{t}⟩ = 0 for any t ∈ ℝ. Thus multiplying the fixed point equation \overline{x} = \overline{Ψ}^{t}(\overline{x}) = (1+⟨U_{t}\overline{x}|a_{t}⟩)^{−1}(P_{t}+β_{t}\overline{P}_{t})U_{t}\overline{x} with |a_{t}⟩, we get (1+⟨U_{t}\overline{x}|a_{t}⟩)^{−1}⟨U_{t}\overline{x}+a_{t}|a_{t}⟩ = ⟨\overline{x}|a_{t}⟩ whence the relations (Ux) are immediate.

Recalling the Hille–Yosida theorem [6, 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 A : D → H ⊕ ℂ (the infinitesimal generator of [Ψ̂^{t} : t∈ℝ]) such that

(Aa)

D = \{[{x \atop ξ}]∈H⊕ℂ : \frac{d}{dt}|_{t=0}Ψ̂^{t}[{x \atop ξ}] exists\},

\frac{d}{dt}Ψ̂^{t} = AΨ̂^{t} = Ψ̂^{t}A on D,

A\left[{x \atop ξ}\right] = \frac{d}{dt}|_{t=0}Ψ̂^{t}\left[{x \atop ξ}\right] for \left[{x \atop ξ}\right]∈D.

Henceforth we fix the notations A, D for the infinitesimal generator of [Ψ̂^{t} : t∈ℝ] and its domain, respectively. It is well-known that A has closed graph and D is a dense linear submanifold of H ⊕ ℂ being invariant under the maps Ψ̂^{t}. We shall write \{[{x \atop ξ}]\}_{H} := x resp. \{[{x \atop ξ}]\}_{ℂ} := ξ for the canonical projections in H ⊕ ℂ. We also introduce the notations

(N)

D_{0} := \{x ∈ H : \left[{x \atop 0}\right] ∈ D\} , B := (\overline{x} + D_{0}) ∩ \overline{Ball}(H) ;

Bz := \{A[{z \atop 0}]\}_{H} , Λz := \{A[{z \atop 0}]\}_{ℂ} for z ∈ D_{0} .

Proposition. D_{0} is a dense linear submanifold in H with D = [{D_{0} \atop 0} ] + ℂ[\overline{x} ; 1]. We have B = \{x ∈ Ball(H) : [{x \atop 1}] ∈ D\}. The set B is [Ψ^{t} : t ∈ ℝ]-invariant and dense in \overline{Ball}(H) such that

\frac{d}{dt}|_{t=0}\overline{Ψ}^{t}(x) = [Λ(\overline{x} − x)]x + B(x − \overline{x}) (x ∈ B).

Proof. The first statement in (Aa) implies [\overline{x} ; 1] ∈ D with A[\overline{x} ; 1] = 0 since Ψ̂^{t}[\overline{x} ; 1] ≡ [\overline{x} ; 1]. Since D is closed for linear combinations, it follows [{D_{0} \atop 0} ] + ℂ[\overline{x} ; 1] ⊂ D. Given any vector

[{x \atop ξ}] ∈ D, we have x − ξ\overline{x} ∈ D_{0} because [{x \atop ξ}] − ξ[\overline{x} ; 1] ∈ D. Hence also D ⊂ [{D_{0} \atop 0} ] + ℂ[\overline{x} ; 1].

As for the density of D_{0} in H and hence also the density of B in \overline{Ball}(H), consider an arbitrary vector x ∈ H. Since D is dense in H ⊕ ℂ, there is a sequence [{x_{1} \atop ξ_{1}} ], [{x_{2} \atop ξ_{2}} ], … ∈ D with [{x_{n} \atop ξ_{n}} ]→[{x \atop 0}] that is ξ_{n} →0, x_{n} →0 (in norm) whence D_{0} ∋ x_{n}−ξ_{n}\overline{x} → x. To see B = \{x ∈ \overline{Ball}(H) : [{x \atop 1}] ∈ D\} it suffices to notice that in general

(∗)  x ∈ D_{0} + \overline{x} ⟺ \left[{x \atop 1}\right] ∈ D  (x ∈ H).

Proof of (∗): x ∈ D_{0} ⟺ [{x−\overline{x} \atop 0}] ∈ D ⟺ [{x \atop 1}] − [\overline{x} ; 1] ∈ D ⟺ [{x \atop 1}] ∈ D.

Using (∗), we show the [\overline{Ψ}^{t} : t ∈ ℝ]-invariance of the set B as follows. Since Ψ̂^{t} : D ↔ D and \overline{Ψ}^{t} : \overline{Ball}(H) ↔ \overline{Ball}(H), we can deduce consecutively the equivalence of the statements (i) \overline{Ψ^{t}(x}∈ B, (ii) x∈ Ball(H) and [{Ψ^{t}(x) \atop 1}]∈ D, (iii) x∈ \overline{Ball}(H) and \{Ψ̂^{t}[{x \atop 1}]\}_{ℂ}^{−1}Ψ̂^{t}[{x \atop 1}]∈ D, (iv) x∈\overline{Ball}(H) and Ψ̂^{t}[{x \atop 1}]∈D, (v) x∈\overline{Ball}(H) and [{x \atop 1}]∈D, (vi) x∈B.

Finally, given any vector x ∈ B, by (∗) we have [{x \atop 1}] ∈ D whence A[{x \atop 1}] = \frac{d}{dt} |_{t=0}Ψ̂^{t}[{x \atop 1}]. Since

\overline{Ψ}^{t}(x) = \{Ψ̂^{t}[{x \atop 1}]\}_{ℂ}^{−1}\{Ψ̂^{t}[{x \atop 1}]\}_{H} and since Ψ̂^{0} = Id, A[\overline{x} ; 1] = \frac{d}{dt} |_{t=0}Ψ̂^{t}[\overline{x} ; 1] = \frac{d}{dt} |_{t=0}[\overline{x} ; 1] = 0,

hence we get

\frac{d}{dt}|_{t=0} \overline{Ψ}^{t}(x) = −\{Ψ̂^{0}[{x \atop 1}]\}_{ℂ}^{−2}[\frac{d}{dt}|_{t=0}\{Ψ̂^{0}[{x \atop 1}]\}_{ℂ}]\{Ψ̂^{0}[{x \atop 1}]\}_{H}+ \{Ψ̂^{0}[{x \atop 1}]\}_{ℂ}^{−1} \frac{d}{dt}|_{t=0}\{Ψ̂^{0}[{x \atop 1}]\}_{H}=

= −\{A[{x \atop 1}]\}_{ℂ}x + \{A[{x \atop 1}]\}_{H} = −\{A[{x − \overline{x} \atop 0}]\}_{ℂ}x + \{A[{x − \overline{x} \atop 0}]\}_{H}. Q.e.d.

Corollary. Given any vector v ∈ D_{0} we have

Re( − Λv + ⟨Bv|\overline{x}⟩ + ⟨Bv|v⟩) = 0 whenever ∥\overline{x} + v∥ = 1 .

Proof. Let v ∈ D_{0}, u := \overline{x} + v and assume ∥u∥ = 1. Since, as it is well-known [3], \overline{Ψ}(∂Ball(H)) = ∂Ball(H) for any holomorphic automorphism Ψ of the unit ball, we have ∥\overline{Ψ}^{t}(u)∥=1 (t∈ℝ). It follows

0 = \frac{d}{dt}|_{t=0}⟨\overline{Ψ}^{t}(u) | \overline{Ψ}^{t}(u)⟩ = 2 Re ⟨\frac{d}{dt}|_{t=0}\overline{Ψ}^{t}(u) | u⟩.

Applying the proposition with x:=u and v:=x−\overline{x}, this means that Re ⟨[−Λv]u+Bv | u⟩=0. Taking into account that ⟨u|u⟩ = 1, the statement is immediate. Q.e.d.

Proposition. For some constant λ ∈ ℝ we have

Λz = ⟨(B−2λI)z | \overline{x}⟩, Re⟨(B−λI)z | z⟩ = 0 (z ∈ D_{0}).

In particular, if ∥\overline{x}∥ < 1 then necessarily λ = 0 above.

Proof. Consider any vector z ≠ 0. Let ζ ∈ ℂ be the (unique) constant such that \overline{x} + ζz ⊥ z and define ϱ := \sqrt{1 − ∥\overline{x} + ζz∥^{2}}. Actually we have ζ = −⟨\overline{x}|z⟩/⟨z|z⟩ and 1 ≥ ∥\overline{x}∥^{2} = ∥\overline{x} + ζz∥^{2} + ∥ − ζz∥^{2} showing that both ζ and ϱ are well-defined. Consider the unit vectors

v_{φ} := \overline{x} + ζz + e^{iφ}ϱz (φ∈ℝ).

According to the last corollary,

Re((ζ+e^{iφ}ϱ)[ − Λz + ⟨Bz | \overline{x}⟩] + |ζ+e^{iφ}ϱ|^{2}⟨Bz | z⟩) = 0 (φ ∈ ℝ).

That is we have

Re(α + βe^{iφ} + γe^{−iφ}) = 0 (φ ∈ ℝ)
where α := ζ[ − Λz + ⟨Bz | \overline{x}⟩] + (|ζ|^{2} + ϱ^{2})⟨Bz | z⟩,
β := ϱ [ − Λz + ⟨Bz | \overline{x}⟩ + \overline{ζ}⟨Bz | z⟩], γ := ϱζ⟨Bz | z⟩.

Since 2 Re(α + βe^{iφ} + γe^{−iφ}) = 2 Re(α) + (β + \overline{γ})e^{iφ} + (\overline{β} + γ)e^{−iφ}, we have necessarily

Re(α) = β + \overline{γ} = 0 .

As β + \overline{γ} = 0, it follows

(B)  Λz − ⟨Bz | \overline{x}⟩ = 2 \overline{ζ} Re ⟨Bz | z⟩ = −2 ⟨z | \overline{x}⟩ \frac{Re ⟨Bz | z⟩}{⟨z | z⟩} ,

and substituting this into the relation 0 = Re(α), we get

(B′)  0 = (ϱ^{2} − |ζ|^{2}) Re ⟨Bz | z⟩ = (1 − ∥\overline{x}∥^{2}) Re ⟨Bz | z⟩ .

From (B) we see that

z ↦ \frac{Re⟨Bz | z⟩}{⟨z | z⟩} = −\frac{1}{2} \frac{Λz − ⟨Bz | \overline{x}⟩}{⟨z | \overline{x}⟩}

is a real valued Gâteaux holomorphic function on the algebraically open and in D_{0} algebraically dense domain \{z ∈ D_{0} : z ⊥̸ \overline{x}\} which is possible only if being constant on D_{0}. By writing λ for this constant value, the statement is immediate from (B) and (B′). Q.e.d.

Taking int account the obvious reversibility properties of the steps of the proof, we conclude immediately the following technical statement.^{∗}

Corollary. Given a possibly unbounded linear functional Λ : D_{0} → ℂ along with a possibly unbounded linear operator B : D_{0} → H, we have [Λ(\overline{x} − x)]x + B(x − \overline{x}) ⊥ x for any x ∈ [\overline{x} + D_{0}] ∩ ∂Ball(H) if and only if Λz =⟨(B−2λI)z | \overline{x}⟩, Re⟨(B−λI)z | z⟩=0 (z ∈D_{0}) with some real constant λ if ∥\overline{x}∥ = 1 and with λ = 0 in case of ∥\overline{x}∥ < 1.

Lemma. The operator B : D_{0} → D_{0} is closed.

Proof. Consider a convergent sequence D_{0} ∋ z_{n} → z with limit point in H such that Bz_{n} → w for some w ∈ H. We have to show that (∗) z ∈ D_{0} with Bz = w, For (∗), it suffices to observe that

Λz_{n} = ⟨(B − 2λI)z_{n}|\overline{x}⟩ → ⟨w − 2λz|\overline{x}⟩,

∗ Direct proof. Given any λ ∈ ℝ and a vector z ∈ H with ∥\overline{x} + z∥ = 1, we have the following chain of equivalences: 0 = Re⟨ − ⟨(iA − λ)z|\overline{x}⟩(\overline{x} + z) + (iA + λI)z|\overline{x} + z⟩ iff 0 = Re[ − ⟨(iA − λ)z|\overline{x}⟩ + ⟨(iA + λI)z|\overline{x} + z⟩] iff 0 = Re[2λ⟨z|\overline{x}⟩ + ⟨iAz|z⟩ + λ⟨z|z⟩] iff 0 = Re[λ(⟨\overline{x} + z|\overline{x} + z⟩ − ⟨\overline{x}|\overline{x}⟩ − ⟨\overline{x}|z⟩ + ⟨z|\overline{x}⟩) + ⟨iAz|z⟩] iff 0 = Re[λ(1 − ∥\overline{x}∥^{2}) + ⟨iAz|z⟩]. recalling that the not necessarily bounded linear operator A[{x \atop ξ}] = [{B(x−ξ\overline{x}) \atop Λ(x−ξ\overline{x})}] mapping the

linear manifold D := [{D_{0} \atop 0} ]+ℂ[\overline{x} ; 1] into itself is closed. Indeed we have

[{D_{0} \atop 0} ] ∋ [{z_{n} \atop 0} ] → [{z \atop 0}], A[{z_{n} \atop 0} ] = [{Bz_{n} \atop Λz_{n}} ] → [{w \atop ⟨w − 2λz|\overline{x}⟩}]

implying (∗) along with the relation Λz = ⟨w − 2λz|\overline{x}⟩ = ⟨(B − 2λI)z|\overline{x}⟩. Q.e.d.

Definition. For convenience, we shall write A := −i(B + λI). Thus henceforth

(A)  Bz = (iA + λI)z , Λz = ⟨(iA − λI)z|\overline{x}⟩  (z ∈ D_{0})

where A : D_{0} → D_{0} is a densely defined closed symmetric linear operator and λ is some real number. We know also that any vector field of the form \overline{x}+D_{0} ∋ \overline{x}+z ↦ −[Λz](\overline{x}+z)+Bz is tangent to the unit sphere if and only if B and Λ are of the form (A) with a symmetric operator A and any real number λ if ∥\overline{x}∥ = 1 and with λ = 0 in case of ∥\overline{x}∥ = 1.

The case 0 ∈ B with ∥\overline{x}∥ = 1

Throughout this section we assume that a common fixed point \overline{x} of [\overline{Ψ}^{t} : t ∈ ℝ] lyes in the boundary of the unit ball and the origin belongs to the domain B of the non-linear generator that is we have

(Ab)  0 ∈ B, ∥\overline{x}∥ = 1.

Remark, Since B = \overline{x} + D_{0} ∩ \overline{Ball}(H) and since D_{0} is a complex linear submanifold of H, we have 0 ∈ B if and only if \overline{x} ∈ D_{0} which is equivalent to [{0 \atop 1}] = [\overline{x} ; 1] − [\overline{x} ; 0] ∈ D. In this case the terms B\overline{x}, Λ\overline{x} are also well-defined and in their terms we can write the infinitesimal generator A in a matrix form as

\begin{matrix}A[{0 \atop 1}] = A([\overline{x} ; 1] − [\overline{x} ; 0]) = −[{B\overline{x} \atop Λ\overline{x}}], \\ A[{x \atop ξ}] = [{B(x − ξ\overline{x}) \atop Λ(x − ξ\overline{x})}] = [{B −B\overline{x} \atop Λ −Λ\overline{x}}][{x \atop ξ}]. \end{matrix}

Notice that in case of [{0 \atop 1}] ∉ D we have only A[{x \atop ξ}] = [{B \atop Λ} ]([I − 1][{x \atop ξ}]) while the formal

product [{B \atop Λ} ][I − 1] cannot be defined.

Theorem. Under the hypothesis (Ab), B is the intersection of the dense complex linear submanifold D_{0} of H with the closed unit ball and V admits the Jordan form

(J)  V(x) = b − {xb^{∗}x} + iAx  (x ∈ B)

in terms of the Jordan triple product \{xc^{∗}y\} := \frac{1}{2}⟨x|c⟩y + \frac{1}{2}⟨y|c⟩x where

b := \frac{d}{dt}|_{t=0}Ψ^{t}(0) = lim_{t→0} a_{t}/t, Ax := lim_{t→0} \frac{1}{it}(U_{t} − I)x − Im⟨\overline{x}|b⟩x (x ∈ D_{0}).

Proof. By the definition of the domain D,

A[{0 \atop 1}] = lim_{t→0} \frac{1}{t}(Ψ̂^{t} − Ψ̂^{0})[{0 \atop 1}] = lim_{t→0} \frac{1}{t}\{\frac{1}{1 + ⟨U_{t}\overline{x}|a_{t}⟩}[{a_{t} \atop 1} ] − [{0 \atop 1}]\}.

Since lim_{t→0} a_{t} = 0 and ∥U_{t}\overline{x}∥ ≤ 1 (t ∈ ℝ), taking (Ux) into account, we see that the limit

b := lim_{t→0} \frac{1}{t}a_{t} = \frac{d}{dt}|_{t=0}Ψ^{t}(0)

is well-defined and

B\overline{x} = −\{A[{0 \atop 1}]\}_{H} = −b, Λ\overline{x} = −\{A[{0 \atop 1}]\}_{ℂ} = ⟨\overline{x} | b⟩.

As a consequence we also have

\begin{aligned}β_{t} & = \sqrt{1 − ⟨a_{t}|a_{t}⟩} = \sqrt{1 − ∥tb + o(t)∥^{2}} = \\ & = \sqrt{1 − t^{2}∥b∥^{2} + o(t^{2})} = 1 − \frac{1}{2}∥b∥^{2}t^{2} + o(t^{2}). \end{aligned}

Hence we deduce that 1 − β_{t} = o(t) and therefore, for any vector z ∈ D_{0}.

\begin{aligned}\frac{1}{t}(Ψ̂^{t} − I)[{z \atop 0}] & = \frac{1}{t} \left[{(P_{t} + β_{t}\overline{P}_{t})U_{t}z − ⟨U^{t}\overline{x}|a_{t}⟩z \atop tΛz + o(t)}\right] = \\ & = \left[{t^{−1}(U_{t} − I)z + t^{−1}(1 − β_{t})\overline{P}_{t}U_{t}z − ⟨\overline{x}|b⟩z \atop Λz}\right] + o(1). \end{aligned}

Since, by definition, [{Bz \atop Λz} ] = lim_{t→0} \frac{1}{t}(Ψ̂^{t} − I)[{z \atop 0}], we get

Bz = lim_{t→0} \frac{1}{t}(U_{t}−I)z − ⟨\overline{x} |b⟩z (z ∈ D_{0}).

In particular, from the relations (Ab) we obtain that for any unit vector z ∈ D_{0} we have i⟨Az|z⟩ + λ = lim_{t→0} t^{−1}⟨(U_{t} − I)z|z⟩ − ⟨\overline{x}|b⟩ whence the the symmetry of A along with the

properties 0 = \frac{d}{dt}_{t=0}∥U_{t}z∥^{2} = 2 Re lim_{t→0} t^{−1}⟨(U_{t} − I)z|z⟩ and λ ∈ ℝ implies

λ = −Re⟨b | \overline{x}⟩ , A = lim_{t→0}^{strong} \frac{1}{it}(U_{t} − I) − Im⟨\overline{x} | b⟩I .

Finally we calculate V = \frac{d}{dt} |_{t=0}Ψ^{t} by substituting the previous results int the form V(x) = [Λ(\overline{x} − x)]x + B(x − \overline{x}). Namely, given any vector x ∈ B, taking into account the antisymmetry of the operator B−λI, we can write

V(x) = [Λ\overline{x}]x − [Λx]x + Bx − B\overline{x}) =
= ⟨\overline{x}|b⟩ − ⟨(B − 2λI)x|\overline{x}⟩x + Bx + b =
= ⟨\overline{x}|b⟩ − ⟨(B − λI)x|\overline{x}⟩x + ⟨(B − λI)x|\overline{x}⟩x + Bx + b =
= b + [⟨\overline{x}|b⟩I + B]x + ⟨x|(B + λI)\overline{x}⟩x =
= b + [⟨\overline{x}|b⟩I + B]x + ⟨x|(−b + 2λ\overline{x})⟩x =
= b − ⟨x|b⟩x + [(⟨\overline{x}|b⟩ + λ + (B − λI)]x =
= b − ⟨x|b⟩x + [i⟨A\overline{x}|\overline{x}⟩I + iA]x

whence the Jordan form (J) is immediate. Q.e.d.

Lemma. We have (1 + ⟨U_{t}\overline{x}|a_{t}⟩)^{−1}[{P_{t}+β_{t}\overline{P}_{t} a_{t} \atop a_{t}^{∗} 1} ] = [{I 0 \atop 0 1}] + t[{−⟨\overline{x}|b⟩I b \atop b^{∗} −⟨\overline{x}|b⟩}] + o^{norm}(t).

Proof. This is a direct consequence of the facts that a_{t} = tb + o^{norm}(t), β_{t} = 1 + o(t), (1 + ⟨U_{t}\overline{x}|a_{t}⟩)^{−1} = (1 − ⟨\overline{x}|a_{t}⟩)(1 − ⟨a_{t}|a_{t}⟩)^{−1} = (1 − t⟨\overline{x}|b⟩ + o(t))β^{−2} = 1 − t⟨\overline{x}|b⟩ + o(t) and P_{t} + β_{t}\overline{P}_{t} = (1 − β_{t})P_{t} + β_{t}(P_{t} + \overline{P}_{t}) = I + o^{norm}(t) as t → 0. Q.e.d.

Corollary. We have D_{0} = dom( lim_{t→0} t^{−1}(U_{t} − I)), that is x ∈ D_{0} if and only if the limit

lim_{t→0} t^{−1}(U^{t}x − x) exists.

Proof. Recall that D_{0} = \{x ∈ H : \frac{d}{dt} |_{t=0}Ψ̂^{t}[{x \atop 0}] is well-defined\}. From the theorem we know that iAx = lim_{t→0} t^{−1}(U_{t} − I)x is well-defined for every vector x ∈ D_{0}. Conversely,

suppose u = lim_{t→0} t^{−1}(U_{t}x − x) is well-defined. Then U_{t}x = x + tu + o^{norm}(t) and

Ψ̂^{t}[{x \atop 0}] = (1 + ⟨U_{t}\overline{x}|a_{t}⟩)[{P_{t} + β_{t}\overline{P}_{t} a_{t} \atop a_{t}^{∗} 1} ] [{U_{t} 0 \atop 0 1}] [{x \atop 0}] =

= ([{I 0 \atop 0 1}] − t[{⟨\overline{x}|a_{t}⟩ − b \atop −b^{∗} ⟨\overline{x}|a_{t}⟩}] + o^{norm}(1)) [{x + tu + o^{norm}(t) \atop 0}] =

= [{x \atop 0}] + t[{u − ⟨\overline{x}|a_{t}⟩x \atop ⟨x|b⟩}] + o^{norm}(t). Q.e.d.

Lemma. We have U_{−t} = U_{t}^{−1} = U_{t}^{∗}, a_{−t} = −U_{t}^{∗}a_{t} (t ∈ ℝ).

Proof. Given any parameter t ∈ ℝ, we have Ψ^{−t} = Ψ_{t}^{−1} that is Θ_{a_{−t}}U_{−t} = [Θ_{a_{t}}U_{t}]^{−1} = U_{t}^{−1}Θ_{a_{t}}^{−1} = U_{t}^{−1}Θ_{−a_{t}} = [U_{t}^{−1}Θ_{−a_{t}}U_{t}]U_{t}^{−1} = Θ_{U_{t}^{−1}(−a_{t})}U_{t}^{−1}. Using the uniqueness of the decomposability of holomorphic automorphisms of circular domains into Möbius and unitary parts [7], hence we deduce that Θ_{a_{−t}} = Θ_{−U_{t}^{−1}a_{t}} and U_{−t} = U_{t}^{−1}. Q.e.d.

Proposition. The operator A in (J) is self-adjoint.

Proof. In view of the previous lemmas we can conclude that for any vector [{x \atop ξ}] ∈ D,

\frac{d}{dt}|_{t=0}Ψ̂^{t} = \frac{d}{dt}|_{t=0}(1 + ⟨U_{t}\overline{x}|a_{t}⟩)^{−1}[{P_{t} + β_{t}\overline{P}_{t} a_{t} \atop a_{t}^{∗} 1} ][{U_{t} 0 \atop 0 1}][{x \atop ξ}] =

= \frac{d}{dt}|_{t=0}(1 + ⟨U_{t}\overline{x}|a_{t}⟩)^{−1}[{P_{t} + β_{t}\overline{P}_{t} a_{t} \atop a_{t}^{∗} 1} ][{x \atop ξ}] + \frac{d}{dt}|_{t=0}[{U_{t} 0 \atop 0 1}][{x \atop ξ}] =

= [ {iA−⟨\overline{x}|b⟩I b \atop b^{∗} −⟨\overline{x}|b⟩}][{x \atop ξ}]

The linear operator in L(H⊕ℂ) with matrix [ {−⟨\overline{x}|b⟩I b \atop b^{∗} −⟨\overline{x}|b⟩}] is bounded. Since A = \frac{d}{dt} |_{t=0}Ψ̂^{t} with domain |D is the generator of a strongly continuous semigroup in L(H ⊕ ℂ), by the theorem of bounded perturbations [8], also the operator with matrix [{iA 0 \atop 0 0}] with domain D is the generator of a strongly continuous group in L(H ⊕ ℂ) strongly continuous oneparameter subgroup of L(H) entailing that iA in (J) is the generator of a strongly continuous group [V_{t} : t ∈ ℝ] in L(H). Since U_{−t} = U_{t}^{−1} = U_{t}^{∗}, the arguments on sun adjoint semigroups in [8, p. 69] show that lim_{t→0} t^{−1}(U_{t}^{∗} − I) = −iA is the generator of the sun adjoint group [V_{t}^{∗} : t ∈ ℝ] = [V_{−t} : t ∈ ℝ] and we have −iA = (iA)^{∗}. Q.e.d.

Theorem. Any vector field of the form (J) where A is a not necessarily bounded selfadjoint operator with dense domain D_{0} ⊂ H, is the infinitesimal generator defined on B := D_{0} ∩ \overline{Ball}(H) of a pointwise continuous one-parameter group Φ^{t} : t ∈ ℝ] of holomorphic automorphisms of \overline{Ball}(H).^{∗}

Proof. It suffices to see that there is a strongly (i.e. pointwise) continuous one-parameter group [V^{t} : t ∈ ℝ] of bounded linear operators of the space H ⊕ ℂ such that

\frac{d}{dt}|_{t=0}V^{t}[{x \atop ξ}] = [{iA b \atop b^{∗} 0}][{x \atop ξ}] (x ∈ D_{0}, ξ ∈ ℂ), V^{t}K ⊂ K := \{[{x \atop ξ}] : ∥x∥^{2} ≥ |ξ|^{2}\}.

Namely, in this case the maps

Φ^{t}(x) := \{V^{t}[{x \atop 1}]\}_{ℂ}^{−1}\{V^{t}[{x \atop 1}]\}_{H} (t ∈ ℝ, x ∈ B)

suit the requirements of the theorem since x ∈ B ⇒ [{x \atop 1}] ∈ K ⇒ V^{t}[{x \atop 1}] ⇒ Φ^{t}(x) ∈ B and x∈B ⇒ \frac{d}{dt} |_{t=0}Φ^{t}(x)=−\{V^{0}[{x \atop 1}]\}_{ℂ}^{−2} \frac{d}{dt} |_{t=0}\{V^{t}[{x \atop 1}]\}_{ℂ}\{V^{0}[{x \atop 1}]\}_{H}+\{V^{0}[{x \atop 1}]\}_{ℂ}^{−1} \frac{d}{dt} |_{t=0}\{V^{t}[{x \atop 1}]\}_{H} = −\{[{iA b \atop b^{∗} 0}][{x \atop 1}]\}_{ℂ}x+\{[{iA b \atop b^{∗} 0}][{x \atop 1}]\}_{H} = −⟨x|b⟩x + iAx + b = V(x).

∗ That is, for all x ∈ B := D_{0} ∩ \overline{Ball}(H), the functions t ↦ Φ^{t}(x) range in D_{0}, they are differentiable and satisfy the identity \frac{d}{dt}Φ^{t}(x) = V(Φ^{t}(x)) (t ∈ ℝ).

Notice that a strongly continuous one parameter group of linear operator leaves the cone K invariant if all its members map the boundary ∂K=\{[{x \atop ξ}]: ∥x∥=|ξ|\}=\{[{x \atop e^{iτ} ∥x∥}]: x∈H, τ∈ℝ\} into itself.^{∗} Therefore it suffices to check that there is a (necessarily unique)strongly continuous one-parameter group in L(H ⊕ ℂ) with domain D_{0} ⊕ ℂ such that

\frac{d}{dt}V^{t}[{x \atop ξ}] = [{iA b \atop b^{∗} 0}]V^{t}[{x \atop ξ}], ∥\{V^{t}[{x \atop ∥x∥}]\}_{H}∥ = |\{V^{t}[{x \atop ∥x∥}]\}_{ℂ}| (x ∈ D_{0}, t ∈ ℝ).

By Stone’s theorem, the H ⊕ ℂ-unitary operators W^{t}[{x \atop ξ}] := [{exp(itA)x \atop ξ}] form a strongly continuous one-parameter group whose infinitesimal generator is defined on dom(A)⊕ℂ = D_{0} ⊕ ℂ with the diagonal matrix [{iA 0 \atop 0 0}]. Since the matrix [{0 b \atop b^{∗} 0}] represents a bounded linear operator in H ⊕ ℂ, by the theorem of bounded perturbations 8], there is a strongly continuous one-parameter group [V^{t} : t ∈ ℝ] whose generator is defined on D_{0} ⊕ ℂ with the matrix [{iA b \atop b^{∗} 0}]. In particular \frac{d}{dt}V^{t}[{x \atop ξ}] = [{iA b \atop b^{∗} 0}]V^{t}[{x \atop ξ}] (t ∈ ℝ, x ∈ D_{0}). To complete the proof, we show that necessarily

\frac{d}{dt} [∥\{V^{t}[{x \atop ∥x∥}]\}_{H}∥^{2} − |\{V^{t}[{x \atop ∥x∥}]\}_{ℂ}|^{2}] = 0 (t ∈ ℝ, x ∈ D_{0}).

Consider any vector x ∈ D_{0} and write [{x_{t} \atop ξ_{t}} ] := V^{t}[{x \atop ∥x∥}] for all parameters t ∈ ℝ. Then

\frac{d}{dt}[∥x_{t}∥^{2} − |ξ_{t}|^{2}] = 2 Re[⟨dx_{t}/dt|x_{t}⟩ − (dξ_{t}/dt)\overline{ξ_{t}}] =

= 2 Re [⟨\{[{iA b \atop b^{∗} 0}][{x_{t} \atop ξ_{t}} ]\}_{H}|x_{t}⟩ − \{[{iA b \atop b^{∗} 0}][{x_{t} \atop ξ_{t}} ]\}_{ℂ}(dξ_{t}/dt)\overline{ξ_{t}}] =

= 2 Re [⟨iAx_{t} + ξ_{t}b|x_{t}⟩ − ⟨x_{t}|b⟩\overline{ξ_{t}}] =

= −2 Im⟨Ax_{t}|x_{t}⟩ + 2 Im(⟨ξb|x_{t}⟩ − ⟨x_{t}|ξb⟩) = 0. Q.e.d.

∗ In general, if a Hilbert space W is the orthogonal sum of the subspaces W_{k} (k =1, 2) and C is the infinitesimal generator of a strongly continuous one-parameter group [T^{t} : t∈ℝ] in L(W) then we have T^{t}C=C (t∈ℝ) for the cone C:={w_{1}⊕w_{2} : w_{k} ∈W_{k}, ∥w_{1}∥>∥w_{2}∥} if and only if C is tangent to the boundary of C that is if (∗) Re⟨C(w_{1}⊕w_{2}),w_{1}⟩=Re⟨C(w_{1}⊕w_{2}),w_{2}⟩ whenever w_{1}⊕w_{2} ∈dom(C) with ∥w_{1}∥=∥w_{2}∥. Proof. It is immediate that T^{t}C⊂C (t∈ℝ) ⇒ T^{t}∂C⊂∂C (t∈ℝ) ⇒ \frac{d}{dt}|_{t=0}T^{t}(w_{1} ⊕ w_{2}) ∈ Tan_{w_{1}⊕w_{2}}(C) for w_{1} ⊕ w_{2} ∈ dom(C) ⇒ (∗). Assume (∗) and let P denote the canonical projection of W onto W_{1} and define T^{t,s} := exp(tC + sP) (s, t ∈ ℝ). By the theorem of bounded perturbations [8, p.158] the operators T^{t,s} are all well-defined. Moreover, by [8, Corollary 1.7 p. 161] (applied with B := sP and A := C there) we have lim_{s→0} T^{t,s}w = T^{t}w (w ∈ dom(C), t ∈ ℝ). Therefore, to establish that T^{t}C ⊂ C (t ∈ ℝ), it suffices to see only that (∗∗) T^{t,s}w∈C whenever w∈dom(C)∩C and t, s>0. To proceed to contradiction, let s, t > 0 and w := w_{1} ⊕ w_{2} ∈ dom(C) with ∥w_{1}∥ > ∥w_{2}∥ but ∥[T^{t,s}w]_{1}∥ ≤ ∥[T^{t,s}w]_{2}∥. The function δ(τ) := ∥[T^{τ,s}w]_{1}∥^{2} − ∥[T^{τ,s}w]_{2}∥^{2} is differentiable in τ on the whole ℝ and δ(0) > 0 ≥ δ(t). Thus there exist a point t_{∗} ∈ (0, t] such that δ(τ) > 0 = δ(t_{∗}) (0 ≤ τ < t_{∗}). Since δ(t_{∗}) = 0, w_{∗} := T^{t_{∗},s}w ∈ ∂C and hence Re⟨Cw_{∗}, [w_{∗}]_{1}⟩ = Re⟨Cw_{∗}, [w_{∗}]_{2}⟩. We get the contradiction 0 ≥ δ′(t_{∗}) = 2Re⟨(C + sP)w_{∗}, [w_{∗}]_{1}⟩ − 2Re⟨(C + sP)w_{∗}, [w_{∗}]_{2}⟩ = 2s∥[w_{∗}]_{1}∥^{2} > 0.

Reduction to the Jordan case 0 ∈ B

In accordance with the previous standard notations, throughout this section [\overline{Ψ}^{t} : t ∈ ℝ] denotes a pointwise continuous one-parameter subgroup of Aut(\overline{Ball}(H)) with infinitesimal generator X : B → H where B := \{x ∈ \overline{Ball}(H) : t ↦ \overline{Ψ}^{t}(x) is differentiable on ℝ\}. We regard X as a vector field defined on the intersection of a dense complex-affine submanifold with the unit ball of H. Let us fix any point a ∈ B with ∥a∥ < 1 and, in terms of the Möbius transformations (M), consider the transformations

\overline{Φ}^{t} := Θ_{−a}^{#}\overline{Ψ}^{t} = Θ_{−a} ∘ \overline{Ψ}^{t} ∘ Θ_{a}

given by the adjoint action of Θ_{−a} = Θ_{a}^{−1}. Clearly [\overline{Φ}^{t} : t ∈ ℝ] is a pointwise continuous one-parameter subgroup of Aut(\overline{Ball}(H)) whose infinitesimal generator is the vector field

V := Θ_{−a}^{#}X : y ↦ \frac{d}{dt}|_{t=0}Θ_{−a}(\overline{Ψ}^{t}(Θ_{a}(y)))

defined on the domain

B_{1} := Θ_{−a}B = D_{1} ∩ \overline{Ball}(H).

Recall that 0 ∈ B_{1} and D_{1} is some dense complex-linear submanifold in H and, as a consequence of the results of the previous section, we can write V in the Jordan form (J) with some vector b ∈ H and a suitable not necessarily bounded self-adjoint linear operator A : D_{1} → H. By the aid of the inverse adjoint action Θ_{a}^{#}, hence we get X in the form

X (x) = [Θ_{a}^{#}V](x) = \frac{d}{dt}|_{t=0}Θ_{a}(\overline{Φ}^{t}(Θ_{−a}(x))) =

= \frac{d}{dt}|_{t=0}Θ_{a}(Θ_{−a}(x) + tV(Θ_{−a}(x)) + o(t)) =

= \frac{d}{dt}|_{t=0}(1 + ⟨y + tk|a⟩)^{−1}([P_{a} + β_{a}\overline{P}_{a}](y + tk) + a)|_{{y=Θ_{−a}(x) \atop k=b−⟨y|b⟩y+iAy}}

for the locations x ∈ B. Next we calculate the a detailed algebraic form for X. Notice that

\frac{d}{dt}|_{t=0}(1 + ⟨y + tk|a⟩)^{−1} = −⟨k|a⟩(1 + ⟨y|a⟩)^{−2}

!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

The classical case ∥\overline{x}∥ < 1. Since the Möbius transformation Θ\overline{x} takes the origin into \overline{x} and since its inverse coincides with Θ_{−\overline{x}}, the automorphisms of the form T^{t} := Θ_{−\overline{x}}∘Ψ^{t}∘Θ\overline{x} leave the origin fixed and hence [T^{t} : t ∈ ℝ] is a strongly continuous one-parameter group consisting of restrictions of linear unitary operators to Ball(H). By Stone’s classical theorem, there is a not necessarily bounded self-adjoint operator A with dense domain in H such that T^{t} = exp(itA) (t ∈ ℝ). It follows Ψ^{t} = Θ\overline{x} ∘ exp(itA) ∘ Θ_{−\overline{x}} (t ∈ ℝ) entailing that the (non-linear) infinitesimal generator of [\overline{Ψ}^{t} : t ∈ ℝ] is

\frac{d}{dt}|_{t=0}\overline{Ψ}^{t}(x) = i[Θ′ ; \overline{x}(Θ_{−\overline{x}}(x))]AΘ_{−\overline{x}}(x)

for x in the domain Θ\overline{x}(dom(A) ∩ \overline{Ball}(H)) = (\overline{x} + D_{0}) ∩ \overline{Ball}(H) = B.

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