Bounded below composition operators on the space of Bloch functions on the unit ball of a Hilbert space

Abstract

Let $B_E$ be the open unit ball of a complex finite or infinite dimensional Hilbert space E and consider the space $\mathcal {B}(B_E)$ of Bloch functions on $B_E$. Using Lipschitz continuity of the dilation map on $B_E$ given by $x \mapsto (1-\Vert x\Vert ^2) \mathcal {R}f(x)$ for $x \in B_E$, where $\mathcal {R}f$ denotes the radial derivative of $f \in \mathcal {B}(B_E)$, we study when a composition operator on $\mathcal {B}(B_E)$ is bounded below.

1 Introduction and background

Let E be a complex Hilbert space and consider its open unit ball $B_E$. The space of Bloch functions f on $B_E$ will be denoted by $\mathcal {B}(B_E)$. We will study various properties of automorphisms of the unit ball $B_E$ which will allow us to supply conditions for a composition operator to be bounded below on $\mathcal {B}(B_E)$, extending the one-dimensional results given [4]. The study of operators on Bloch spaces on an infinite dimensional setting can be found in [3], where the boundness and compactness of composition operators are studied. Hamada also deals with bounded below composition operators on the space of Bloch functions on bounded symmetric domains [10]. The author also studies properties of extended Cesàro operators on spaces of Bloch-type functions in [9]. The results given in this work are presented as a preprint in [13].

For our purpose, it will be needed that for any $f \in \mathcal {B}(B_E)$, the map on $x \in B_E$ given by $x \mapsto (1-\Vert x\Vert ^2) \mathcal {R}f(x)$ is Lipschitz with respect to $\rho _E$, where $\rho _E$ denotes the pseudohyperbolic distance (see [13]) and bearing in mind that Rf is the radial derivative of the function f.

1.1 Automorphisms on $B_E$. The pseudohyperbolic distance

If X is a complex Banach space and we denote by $B_X$ its open unit ball, the function $f: B_X \rightarrow \mathbb {C}$ is said to be analytic (or holomorphic) if f is Fréchet differentiable for all $x \in B_X$ (see [15] for further information). The pseudohyperbolic distance $\rho _X(x,y)$ for $x,y \in B_X$ is described by:

$$\begin{aligned} \rho _X (x,y)=\sup \{ \rho (f(x),f(y)): f \in H^{\infty }(B_X), \ |f| < 1 \text{ on } \mathbb {D}\} \end{aligned}$$

where we denote by $H^{\infty }(B_X)$ the space of analytic functions on $B_X$ which are bounded. This space is endowed with the sup-norm and the pseudohyperbolic distance on $\mathbb {D}$ is given by:

$$\begin{aligned} \rho (z,w)=\left| \frac{z-w}{1-\bar{z}w} \right| \ \ \text{ for } \text{ all } z,w \in \mathbb {D}. \end{aligned}$$

Now consider a complex Hilbert space E and denote by $\langle \cdot , \cdot \rangle$ the natural inner product of E. We will denote by ${\text {Aut}}(B_E)$ the space of automorphisms of $B_E$, that is, the bijective maps $\varphi : B_E \rightarrow B_E$ which are bianalytic. We will use these automorphisms several times in this work (see [8] for further information). For every $x \in B_E$, we will denote the automorphism $\varphi _x: B_{E} \longrightarrow B_{E}$ by:

$$\begin{aligned} \varphi _{x}(y)= (s_x Q_x+P_x)(m_x(y)) \end{aligned}$$

(1.1)

where $s_x=\sqrt{1-\Vert x\Vert ^2},$ $m_x: B_{E} \longrightarrow B_{E}$ is the analytic self-map:

$$\begin{aligned} m_x(y)=\frac{x-y}{1- \langle y,x \rangle }, \end{aligned}$$

$P_x: E \longrightarrow E$ is given by:

$$\begin{aligned} P_x(y)=\frac{\langle y, x \rangle }{\langle x,x \rangle } x \end{aligned}$$

and $Q_x: E \longrightarrow E$ is defined by $Q_x=Id_E-P_x$, where $Id_E$ is the identity on E. Notice that $\varphi _x(0)=x$ and also $\varphi _x(x)=0$. It is well-known that the space of automorphisms of $B_{E}$ is given by compositions of $\varphi _x$ for some $x \in B_E$ with unitary transformations U of E. In addition, this space acts transitively on $B_E$.

The pseudohyperbolic distance on $B_E$ is given by (see [8]):

$$\begin{aligned} \rho _{E}(x,y) = \Vert \varphi _{y}(x)\Vert \text{ for } \text{ any } x, y \in B_{E}. \end{aligned}$$

(1.2)

and using the definition of $\varphi _x$ it is easy to conclude that:

$$\begin{aligned} \rho _E(x,y)^2=1-\frac{(1-\Vert x\Vert ^2)(1-\Vert y\Vert ^2)}{|1-\langle x,y \rangle |^2}. \end{aligned}$$

(1.3)

1.2 The space of Bloch functions

Let $\mathbb {C}$ be the space of complex numbers and $\mathbb {D}$ the open disk of radius 1 centered at 0. The classical Bloch space $\mathcal {B}$ is given by the set of holomorphic functions $f: \mathbb {D}\rightarrow \mathbb {C}$ such that $\Vert f\Vert _B=\sup _{z \in \mathbb {D}} (1-|z|^2)|f'(z)| < +\infty$. Timoney extended this space by considering domains of finite dimensional Hilbert spaces (see [17]) and in [2] the authors extended these functions to an infinite dimensional context. When we deal with a complex Hilbert space E, the holomorphic function $f: B_E \rightarrow \mathbb {C}$ belongs to the Bloch space $\mathcal {B}(B_E)$ if:

$$\begin{aligned} \Vert f\Vert _\mathcal {B}=\sup _{x \in B_E} (1-\Vert x\Vert ^2) \Vert \nabla f(x)\Vert < +\infty , \end{aligned}$$

where the gradient $\nabla f(x)$ denotes the Fréchet derivative $f'(x)$ of f at x or, equivalently, if:

$$\begin{aligned} \Vert f\Vert _\mathcal {R}=\sup _{x \in B_E} (1-\Vert x\Vert ^2) \Vert \mathcal {R}f(x)\Vert < +\infty , \end{aligned}$$

where $\mathcal {R}f(x)=\langle x, \overline{\nabla f (x)}\rangle$. Both semi-norms are also equivalent to:

$$\begin{aligned} \Vert f\Vert _\mathcal {I}=\sup _{x \in B_E} \Vert \widetilde{\nabla } f (x) \Vert , \end{aligned}$$

(1.4)

where $\widetilde{\nabla } f (x)$ is the invariant gradient of f at x, that is, $\widetilde{\nabla } f (x) =\nabla (f \circ \varphi _x)(0)$ and bearing in mind that $\varphi _x$ is the automorphism described in (1.1).

These three semi-norms describe norms of Banach spaces which are equivalent-modulo constant functions- in $\mathcal {B}(B_E)$ [2]. Indeed, there is a positive constant $A_0$ satisfying:

$$\begin{aligned} \Vert f\Vert _\mathcal {R}\le \Vert f\Vert _\mathcal {B}\le \Vert f\Vert _\mathcal {I}\le A_0 \Vert f\Vert _R, \end{aligned}$$

(1.5)

so we obtain a Banach space if we endow $\mathcal {B}(B_E)$ with one of the norms $\Vert \cdot \Vert _{\mathcal {B}-{\text {Bloch}}}=|f(0)|+ \Vert \cdot \Vert _\mathcal {B}$ or $\Vert \cdot \Vert _{\mathcal {R}-{\text {Bloch}}}$ and $\Vert \cdot \Vert _{\mathcal {I}-{\text {Bloch}}}$ which are defined with the corresponding semi-norms $\Vert \cdot \Vert _{\mathcal {R}}$ and $\Vert \cdot \Vert _{\mathcal {I}}$. These semi-norms will be used along our work.

1.3 The function $(1-\Vert x\Vert ^2) |\mathcal {R}f(x)|$ is Lipschitz continuous

Let $f \in \mathcal {B}(B_E)$. Recall that the function defined on $x \in B_E$ and given by $x \mapsto (1-\Vert x\Vert ^2) |\mathcal {R}f(x)|$ is Lipschitz with respect to $\rho _E$. We recall several results which can be found in [13].

Theorem 1.1

If f belongs to $\mathcal {B}(B_E)$ then:

$$\begin{aligned} |(1-\Vert x\Vert ^2) \mathcal {R}f(x)-(1-\Vert y\Vert ^2) \mathcal {R}f(y)| \le 14 \Vert f\Vert _\mathcal {I}\rho _E(x,y) \ \text{ for } \text{ all } x,y \in B_E. \end{aligned}$$

As a consequence, we obtain a corollary which extends results given in [1] by Attele and improved in [18] by Xiong on the Bloch space $\mathcal {B}$:

Corollary 1.2

Consider a complex Hilbert space E. The function defined for $x \in B_E$ and given by $x \mapsto (1-\Vert x\Vert ^2) |\mathcal {R}f(x)|$ is Lipschitz with respect to the pseudohyperbolic distance $\rho _E$. In addition, we have:

$$\begin{aligned} |(1-\Vert x\Vert ^2) |\mathcal {R}f(x)|-(1-\Vert y\Vert ^2) |\mathcal {R}f(y)| | \le 14 \Vert f\Vert _\mathcal {I}\rho _E(x,y). \end{aligned}$$

This will allow us to provide conditions for a composition operator on the Bloch space $\mathcal {B}(B_E)$ to be bounded below.

2 Composition operators on $\mathcal {B}(B_E)$ which are bounded below

Let X and Y be Banach spaces. A linear operator $T: X \rightarrow Y$ is said to be bounded below if there is a positive constant $k >0$ satisfying $\Vert x\Vert \le k \Vert T(x)\Vert$. A linear continuous operator T is bounded below if and only if T has closed range and it is injective.

If $\varphi : \mathbb {D}\rightarrow \mathbb {D}$ denotes an analytic map, the composition operator $C_{\varphi }: \mathcal {B}\rightarrow \mathcal {B}$ is defined by $C_{\varphi }(f)=f \circ \varphi$ and it is continuous for any $\varphi$. Define:

$$\begin{aligned} \tau _{\varphi }(z)=\frac{1-|z|^2}{1-|\varphi (z)|^2} \varphi '(z). \end{aligned}$$

(2.1)

In [7], it was investigated when $\varphi$ induces a composition operator which has closed range on $\mathcal {B}$. They proved:

Proposition 2.1

Let $C_{\varphi }$ be bounded below. Then there are $\varepsilon , r >0$ such that $r < 1$ satisfying that for all $z \in \mathbb {D}$ we have $\rho (\varphi (w),z) \le r$ for all $w \in \mathbb {D}$ satisfying $|\tau _{\varphi }(w)| > \varepsilon$.

The authors also studied the map defined for $z \in \mathbb {D}$ by $z \mapsto (1-|z|^2)|f'(z)|$, proving that it is Lipschitz with respect to $\rho$ if f belongs to the Bloch space $\mathcal {B}$. Indeed, for any $f \in \mathcal {B}$ and $z,w \in \mathbb {D}$ we have:

$$\begin{aligned} |(1-|z|^2)|f'(z)|-(1-|w|^2)|f'(w)|| \le 3.31 \Vert f\Vert _\mathcal {B}\rho (z,w). \end{aligned}$$

(2.2)

This result refines a result of Attele (see [1]) who provided the constant 9 instead of 3.31. Xiong improved the constant in [18], giving $3 \sqrt{3}/2 \approx 2.6$. From (2.2) we have the following sufficient condition for $C_{\varphi }$ to be bounded below (see [7]):

Theorem 2.2

Consider an analytic self-map $\varphi$ on $\mathbb {D}$ and suppose that there is $0< r < \frac{1}{4}$ and $\varepsilon >0$ satisfying that for any $w \in \mathbb {D}$ there exists $z_w \in \mathbb {D}$ such that $\rho (\varphi (z_w),w) < r$ and $|\tau _{\varphi }(z_w)| > \varepsilon$. Then the operator $C_\varphi : \mathcal {B}\rightarrow \mathcal {B}$ is bounded below.

F. Deng, L. Jiang and C. Ouyang [6] and H. Chen [4] considered self-maps $\varphi$ on $B_n$, where $B_n$ denotes the open unit ball of a finite dimensional Hilbert space, extending these results from the one-dimensional case. However, they replaced $\tau _\varphi (z)$ by:

$$\begin{aligned} \left( \frac{1-\Vert z\Vert ^2}{1-\Vert \varphi (z)\Vert ^2}\right) ^{(n+1)/2} |det(J_{\varphi }(z))| \end{aligned}$$

(2.3)

where $J_{\varphi }(z)$ is the Jacobian matrix of $\varphi$. If $\varphi$ is an automorphism of $B_n$ then it is easy that $\tau _\varphi (z)=1$. Moreover, the proofs of these results used the definition given by Timoney of Bloch function on $B_n$ depending on the Bergman metric [17].

To extend the results given for the classical Bloch space $\mathcal {B}$ to a more general setting (finite or infinite dimensional), we will give sufficient and necessary conditions which avoid the Bergman metric and expression (2.3). Hence, consider a complex Hilbert space E and an analytic map $\psi : B_E \rightarrow B_E$. We define for $x \in B_E$ the expressions $\tau _\psi (x)$ and $\widetilde{\tau _\psi }(x)$ which are given by:

$$\begin{aligned} {\tau _\psi }(x) =\frac{1-\Vert x\Vert ^2}{1-\Vert \psi (x)\Vert ^2} \Vert \psi '(x)\Vert \end{aligned}$$

(2.4)

and:

$$\begin{aligned} \widetilde{\tau _\psi }(x)=\frac{\sqrt{1-\Vert x\Vert ^2}}{1-\Vert \psi (x)\Vert ^2} \Vert \psi '(x)\Vert . \end{aligned}$$

(2.5)

It is easy that $\widetilde{\tau _\psi }(x) \ge {\tau _\psi }(x)$.

In [3] the authors studied the boundness and also the compactness of $C_{\psi }: \mathcal {B}(B_E) \rightarrow \mathcal {B}(B_E)$ which is the composition operator defined by $C_{\psi }(f)=f \circ \psi$. It was proved that for any analytic self map $\psi$ on $B_E$, the operator $C_{\psi }$ is bounded. Furthermore, they proved the inequality $\Vert f \circ \psi \Vert _\mathcal {I}\le \Vert f\Vert _\mathcal {I}$ where $\Vert \cdot \Vert _\mathcal {I}$ is the semi-norm defined in Sect. 1.2.

This Lemma will be useful for Lemma 2.4:

Lemma 2.3

Consider a complex Hilbert space E and $f \in \mathcal {B}(B_E)$. Then:

$$\begin{aligned} |f(x)-f(0)| \le \Vert x\Vert \frac{\Vert f\Vert _\mathcal {B}}{1-\Vert x\Vert ^2} \ \ \text{ for } \text{ any } x \in B_E. \end{aligned}$$

Proof

Note that:

$$\begin{aligned} |f(x)-f(0)|= & {} \left| \left( \int _{0}^{1} f'(xt) {\text {d}}t \right) (x) \right| \le \Vert x\Vert \left\| \int _{0}^{1} \frac{f'(xt) (1-\Vert tx\Vert ^2)}{1-\Vert tx\Vert ^2} {\text {d}}t \right\| \\\le & {} \Vert x\Vert \Vert f\Vert _\mathcal {B}\int _{0}^{1} \left| \frac{1}{1-\Vert tx\Vert ^2} \right| {\text {d}}t \\\le & {} \Vert x\Vert \Vert f\Vert _\mathcal {B}\int _{0}^{1} \frac{1}{1-\Vert x\Vert ^2} {\text {d}}t = \Vert x\Vert \frac{\Vert f\Vert _\mathcal {B}}{1-\Vert x\Vert ^2} \end{aligned}$$

so the result is clear. $\square$

Recall that $\Vert \cdot \Vert _\mathcal {R}$, $\Vert \cdot \Vert _\mathcal {I}$ and $\Vert \cdot \Vert _\mathcal {B}$ are equivalent, so they can be used interchangeably when studying if $C_\psi$ is bounded below.

The following Lemma was given in [10, Lemma 2.14] with a different proof for the general case when $B_E$ is the unit ball of a $JB^*$-triple. For completeness, we give a direct proof:

Lemma 2.4

Consider a complex Hilbert space E and an analytic map $\psi : B_E \rightarrow B_E$. The composition operator $C_{\psi }: \mathcal {B}(B_E) \rightarrow \mathcal {B}(B_E)$ is bounded below if and only if there is $k >0$ such that:

$$\begin{aligned} \Vert C_\psi (f)\Vert _\mathcal {I}\ge k \Vert f\Vert _\mathcal {I}\ \text{ for } \text{ all } f \in \mathcal {B}(B_E). \end{aligned}$$

Proof

If $C_{\psi }$ is bounded below then there exists $k >0$ such that $\Vert C_{\psi }(f)\Vert _{\mathcal {I}-{\text {Bloch}}} \ge k \Vert f\Vert _{\mathcal {I}-{\text {Bloch}}}$ for $f \in \mathcal {B}(B_E)$. We define $g(x)=f(x)-f(\psi (0))$ and clearly $g(\psi (0))=0$. We have:

$$\begin{aligned} \Vert C_{\psi }(f)\Vert _{\mathcal {I}}= & {} \Vert f \circ \psi \Vert _\mathcal {I}= \Vert g \circ \psi \Vert _\mathcal {I}= \Vert g \circ \psi \Vert _{\mathcal {I}-{\text {Bloch}}} \\\ge & {} k \Vert g\Vert _{\mathcal {I}-{\text {Bloch}}} \ge k \Vert g\Vert _\mathcal {I}=k \Vert f\Vert _\mathcal {I}. \end{aligned}$$

Now consider $\Vert C_\psi (f)\Vert _\mathcal {I}\ge k \Vert f\Vert _\mathcal {I}$ for some constant $0 < k \le 1$. We will find $k' >0$ satisfying $\Vert C_\psi (f)\Vert _{\mathcal {I}-{\text {Bloch}}} \ge k' \Vert f\Vert _{\mathcal {I}-{\text {Bloch}}}.$ Using Lemma 2.3 we obtain:

$$\begin{aligned} |f(\psi (0))-f(0)| \le \Vert \psi (0)\Vert \frac{\Vert f\Vert _\mathcal {B}}{1-\Vert \psi (0)\Vert ^2} \end{aligned}$$

so we have:

$$\begin{aligned} |f(\psi (0))| \ge |f(0)|-\Vert \psi (0)\Vert \frac{\Vert f\Vert _\mathcal {B}}{1-\Vert \psi (0)\Vert ^2} \ge |f(0)|-\frac{\Vert f\Vert _\mathcal {I}}{1-\Vert \psi (0)\Vert ^2}. \end{aligned}$$

and we obtain:

$$\begin{aligned} |f(\psi (0))|+\frac{1}{(1-\Vert \psi (0)\Vert ^2)} \Vert f\Vert _\mathcal {I}\ge |f(0)|. \end{aligned}$$

Hence:

$$\begin{aligned}{} & {} k(1-\Vert \psi (0)\Vert ^2)|f(\psi (0))|+\Vert C_{\psi }(f)\Vert _\mathcal {I}\\{} & {} \quad \ge k(1-\Vert \psi (0)\Vert ^2) |f(\psi (0))|+k \Vert f\Vert _\mathcal {I}\ge k (1-\Vert \psi (0)\Vert ^2) |f(0)| \end{aligned}$$

so we have that:

$$\begin{aligned} 2(|f(\psi (0))|+\Vert C_\psi (f)\Vert _\mathcal {I})= & {} 2|f(\psi (0))| + \Vert C_\psi (f)\Vert _\mathcal {I}+\Vert C_\psi (f)\Vert _\mathcal {I}\\\ge & {} k(1-\Vert \psi (0)\Vert ^2)|f(\psi (0))|+ \Vert C_\psi (f)\Vert _\mathcal {I}+\Vert C_\psi (f)\Vert _\mathcal {I}\\\ge & {} k(1-\Vert \psi (0)\Vert ^2)|f(0)|+\Vert C_\psi (f)\Vert _\mathcal {I}\\\ge & {} k(1-\Vert \psi (0)\Vert ^2)(|f(0)|+\Vert f\Vert _\mathcal {I}) \end{aligned}$$

and we can conclude:

$$\begin{aligned} \Vert C_\psi (f))\Vert _{\mathcal {I}-{\text {Bloch}}} \ge \frac{k(1-\Vert \psi (0)\Vert ^2)}{2} \Vert f\Vert _{\mathcal {I}-{\text {Bloch}}} \end{aligned}$$

so taking $k'=k(1-\Vert \psi (0)\Vert ^2)/2$ we obtain that $C_\psi$ is a bounded below operator. $\square$

2.1 The automorphisms $\varphi _x$ on $B_E$

In this section we will give some calculations related to the automorphisms $\varphi _x$ of $B_E$ given in (1.1) which will permit us to study conditions for $C_\varphi$ to be bounded below. If E is finite dimensional, then it is well-known that $\varphi _x$ is an involution (see [16]). Since the proof uses the Cartan’s uniqueness theorem, we first give a new proof of this assertion, extending the result for infinite dimensional spaces:

Lemma 2.5

If E is a complex Hilbert space and $x \in B_E$, then $\varphi _x \circ \varphi _x=Id_E$, that is, $\varphi _x$ is an involution.

Proof

Using (1.1), we have:

$$\begin{aligned} \varphi _x(\varphi _x(y))=(s_x Q_x+P_x)(m_x(\varphi _x(y))= (s_x Q_x+P_x) \left( \frac{x-\varphi _x(y)}{1-\langle \varphi _x(y),x \rangle }\right) \end{aligned}$$

and using the following result (it can be found as Lemma 3.6 in [14]):

$$\begin{aligned} 1-\langle \varphi _x(y),x\rangle =1-\langle \varphi _x(y),\varphi _x(0)\rangle =\frac{1-\Vert x\Vert ^2}{1-\langle y,x \rangle } \end{aligned}$$

we obtain:

$$\begin{aligned} \varphi _x(\varphi _x(y))= & {} \frac{1-\langle y,x \rangle }{1-\Vert x\Vert ^2} (s_x Q_x+P_x) \left( x-\varphi _x(y)\right) \\= & {} \frac{1-\langle y,x \rangle }{1-\Vert x\Vert ^2} \left( (s_x Q_x+P_x)(x)-(s_x Q_x +P_x)((s_x Q_x+P_x)(m_x(y)))\right) . \end{aligned}$$

Using $P_x \circ Q_x=Q_x \circ P_x=0$, $P_x+Q_x=Id_E$, $P_x^2=P_x$ and $Q_x^2=Q_x$ we have:

$$\begin{aligned} \varphi _x(\varphi _x(y))= & {} \frac{1-\langle y,x \rangle }{1-\Vert x\Vert ^2} \left( x-(s_x^2 Q_x +P_x)\left( \frac{x-y}{1-\langle y,x \rangle } \right) \right) \\= & {} \frac{1-\langle y,x \rangle }{(1-\Vert x\Vert ^2)(1-\langle y,x \rangle )} \left( (1-\langle y,x \rangle ) x-(s_x^2 Q_x +P_x)\left( x-y \right) \right) \\= & {} \frac{1}{(1-\Vert x\Vert ^2)} \left( (x-\Vert x\Vert ^2 P_x(y)-x+(1-\Vert x\Vert ^2) Q_x(y)+P_x(y)\right) \\= & {} \frac{1}{(1-\Vert x\Vert ^2)}(1-\Vert x\Vert ^2)(P_x(y)+Q_x(y))=y \end{aligned}$$

so we obtain the result. $\square$

Lemma 2.6

For any $x \in B_E$ we have that the operator $\varphi _x'(0)$ is invertible and $\varphi _x'(0)^{-1}=\varphi _x'(x)$.

Proof

Using Lemma 2.5, we have $(\varphi _x \circ \varphi _x)'(0)=Id_E'(0)=Id_E$ so:

$$\begin{aligned} \varphi _x'(\varphi _x(0)) \circ \varphi _x'(0)=\varphi _x'(x) \circ \varphi _x'(0)=Id_E \end{aligned}$$

and we are done. $\square$

Recall that $\Vert f\Vert _\mathcal {I}=\sup _{x \in B_E} \Vert \widetilde{\nabla } f(x) \Vert$ by (1.4). For all $x \in B_E$ we have:

$$\begin{aligned} \Vert \widetilde{\nabla } f(x) \Vert = \sup _{u \in \overline{B_E}} \Vert f'(\varphi _x(0)) \circ \varphi _x'(0)(u)\Vert =\sup _{w \in E {\setminus } \{0\}} \frac{| f'(x)(w)|}{\Vert \varphi _x'(0)^{-1}(w)\Vert } \end{aligned}$$

(2.6)

and for all $w \in E$ we have that (see [2]):

$$\begin{aligned} \Vert \varphi _x'(0)^{-1}(w)\Vert ^2=\frac{(1-\Vert x\Vert ^2) \Vert w\Vert ^2+| \langle w,x \rangle |^2}{(1-\Vert x\Vert ^2)^2}. \end{aligned}$$

(2.7)

In [2] the following equality was also given:

$$\begin{aligned} \Vert \widetilde{\nabla } f(x) \Vert ^2= (1-\Vert x\Vert ^2)\left( \Vert \nabla f (x)\Vert ^2 -|\mathcal {R}f(x)|^2 \right) . \end{aligned}$$

(2.8)

For an analytic map $\psi : B_E \rightarrow B_E$, $x \in B_E$ and $w \in E$ we will use the infinitesimal Kobayashi metric described in [10]. For a complex Hilbert space E, this metric can be described in terms of the automorphisms $\varphi _x$ by:

$$\begin{aligned} \kappa _E(x,w)=\Vert \varphi _{x}'(0)^{-1}(w)\Vert \ \text{ for } x \in B_E \text{ and } w \in E. \end{aligned}$$

We will use $\kappa (x,w)$ and $\kappa (\psi (x),\psi '(x)(w))$ for an analytic self-map $\psi : B_E\rightarrow B_E$ several times in the sequel. Notice that:

$$\begin{aligned} \kappa (\psi (x),\psi '(x)(w))=\Vert \varphi _{\psi (x)}'(0)^{-1}(\psi '(x)(w))\Vert . \end{aligned}$$

(2.9)

Lemma 2.7

If $\psi : B_E \rightarrow B_E$ is analytic and $x \in B_E$ then:

• (a) For $w \in E$:

  $$\begin{aligned} \frac{\Vert w\Vert ^2}{1-\Vert x\Vert ^2} \le \kappa (x,w)^2 \le \frac{\Vert w\Vert ^2}{(1-\Vert x\Vert ^2)^2} \end{aligned}$$

  (2.10)

  and:

  $$\begin{aligned} \frac{\Vert \psi '(x)(w)\Vert ^2}{1-\Vert \psi (x)\Vert ^2} \le \kappa (\psi (x),\psi '(x)(w))^2 \le \frac{\Vert \psi '(x)(w)\Vert ^2}{(1-\Vert \psi (x)\Vert ^2)^2}. \end{aligned}$$

  (2.11)

• (b) If there is $w_x \in E$ satisfying $\psi '(x)(w_x)=\Vert \psi '(x)\Vert \psi (x)$ then:

  $$\begin{aligned} \frac{\Vert \psi '(x)\Vert \Vert \psi (x)\Vert }{1-\Vert \psi (x)\Vert ^2} = \kappa (\psi (x),\psi '(x)(w_x)) \le \frac{\Vert \psi '(x)\Vert }{1-\Vert \psi (x)\Vert ^2} \end{aligned}$$

  (2.12)

  and under the condition $w_x \ne 0$, then:

  $$\begin{aligned} \frac{\kappa (\psi (x),\psi '(x)(w_x))}{\kappa (x,w_x)} \ge \tau _{\psi }(x) \frac{\Vert \psi (x)\Vert }{\Vert w_x\Vert }. \end{aligned}$$

  (2.13)

Proof

We will prove a). By (2.7) and (2.9) we obtain:

$$\begin{aligned} \kappa (x,w)^2=\frac{(1-\Vert x\Vert ^2) \Vert w\Vert ^2+| \langle w,x\rangle |^2}{(1-\Vert x\Vert ^2)^2}. \end{aligned}$$

Hence:

$$\begin{aligned} \frac{\Vert w\Vert ^2}{(1-\Vert x\Vert ^2)} \le \kappa (x,w)^2 \le \frac{\Vert w\Vert ^2}{(1-\Vert x\Vert ^2)^2} \end{aligned}$$

where last inequality is true because $|\langle w,x \rangle | \le \Vert w \Vert \Vert x\Vert$, so we conclude (2.10). Following the same pattern, we obtain a proof for (2.11).

Now we prove b). We have:

$$\begin{aligned} \kappa (\psi (x),\psi '(x)(w_x))^2= & {} \frac{(1-\Vert \psi (x)\Vert ^2) \Vert \psi '(x)(w_x)\Vert ^2+| \langle \psi '(x)(w_x),\psi (x)\rangle |^2}{(1-\Vert \psi (x)\Vert ^2)^2} \\= & {} \frac{(1-\Vert \psi (x)\Vert ^2)\Vert \psi '(x)\Vert ^2 \Vert \psi (x)\Vert ^2+\Vert \psi (x)\Vert ^4 \Vert \psi '(x)\Vert ^2}{(1-\Vert \psi (x)\Vert ^2)^2}\\= & {} \frac{\Vert \psi '(x)\Vert ^2 \Vert \psi (x)\Vert ^2}{(1-\Vert \psi (x)\Vert ^2)^2} \end{aligned}$$

and we obtain inequality (2.12). Together with inequality (2.10) results in (2.13) since:

$$\begin{aligned} \frac{\kappa (\psi (x),\psi '(x)(w_x))}{\kappa (x,w_x)} \ge \frac{1-\Vert x\Vert ^2}{1-\Vert \psi (x)\Vert ^2} \frac{\Vert \psi '(x)\Vert \Vert \psi (x)\Vert }{\Vert w_x\Vert } \end{aligned}$$

and we conclude the result. $\square$

From Lemma 2.7 we have:

Lemma 2.8

For any $x \in B_E$ and $w \in E {\setminus } \{0\}$:

$$\begin{aligned} \frac{\kappa (\psi (x),\psi '(x)(w))}{\kappa (x,w)} \le \frac{\sqrt{1-\Vert x\Vert ^2}}{1-\Vert \psi (x)\Vert ^2} \left\| \psi '(x) \left( \frac{w}{\Vert w\Vert }\right) \right\| \end{aligned}$$

(2.14)

and:

$$\begin{aligned} \frac{\kappa (\psi (x),\psi '(x)(w))}{\kappa (x,w)} \ge \frac{1-\Vert x\Vert ^2}{\sqrt{1-\Vert \psi (x)\Vert ^2}} \left\| \psi '(x) \left( \frac{w}{\Vert w\Vert }\right) \right\| . \end{aligned}$$

(2.15)

The following lemma is just a contractive property of the infinitesimal Kobayashi metric. We omit the proof:

Lemma 2.9

If $\psi$ is an analytic self-map on $B_E$, then for any $x \in B_E$ and $w \in E {\setminus } \{0\}$ we have:

$$\begin{aligned} \frac{\kappa (\psi (x),\psi '(x)(w))}{\kappa (x,w)}=\frac{\Vert \varphi _{\psi (x)}'(0)^{-1}(\psi '(x)(w))\Vert }{\Vert \varphi _{x}'(0)^{-1}(w)\Vert } \le 1. \end{aligned}$$

The following extension of the Schwarz-Pick lemma generalizes a result of Kalaj [12] when we deal with an infinite dimensional space. The same result for bounded symmetric domains can be found in [5].

Corollary 2.10

Consider an analytic self map $\psi$ on $B_E$. Then:

$$\begin{aligned} \frac{1-\Vert x\Vert ^2}{\sqrt{1-\Vert \psi (x)\Vert ^2}} \Vert \psi '(x)\Vert \le 1 \ \text{ for } \text{ all } x \in B_E. \end{aligned}$$

Proof

Applying Lemma 2.9 and using inequality (2.15) in Lemma 2.7 we are done. $\square$

Remark 2.11

Hamada and Kohr [11] proved that Corollary 2.10 is sharp. Kalaj [12] also proved this sharpness by considering for all $t \in (0,\pi /2)$ the self-map $\psi _t: B_2 \rightarrow B_2$ defined by $\psi _t(z,w)=( z \sin t, \cos t)$.

2.2 Results on bounded below composition operators

We will apply the study on the automorphisms $\varphi _x$ to study bounded below composition operators. Hamada [10] provided a necessary condition in the context of the unit ball of a $JB^*-$triple by considering the existence of $\varepsilon >0$ and $0< r < 1$ such that if $y \in B_E$ then $\rho (\psi (x_y),y) \le r$ for any $x_y \in B_E$ satisfying $\tau _{\psi }^*(x_y) \ge \varepsilon$ where:

$$\begin{aligned} \tau _{\psi }^*(x_y)=\sup \left\{ \frac{\kappa _E(\psi (x_y),\psi '(x_y)(y))}{\kappa _E(x_y,y)}: w \in E {\setminus } \{0\} \right\} \end{aligned}$$

We provide a necessary condition for the Hilbert case by adapting the proof of Theorem 2 in [6] and using $\widetilde{\tau _\psi }(x_y)$ instead of $\tau _{\psi }^*(x_y)$:

Theorem 2.12

Consider an analytic self map $\psi$ on $B_E$ and suppose that $C_{\psi }: \mathcal {B}(B_E) \rightarrow \mathcal {B}(B_E)$ is a bounded below operator. Then there are $\varepsilon >0$ and $0< r < 1$ such that if $y \in B_E$ we have $\rho (\psi (x_y),y) \le r$ for some $x_y \in B_E$ satisfying $\widetilde{\tau _\psi }(x_y) \ge \varepsilon$.

Proof

If $C_\psi$ is a bounded below operator, consider $y \in B_E$ and let $f: B_E \rightarrow \mathbb {C}$ be an analytic function given by $f_y(x)=1/(1-\langle x,y \rangle ).$

We have:

$$\begin{aligned} f_y'(x)=\frac{\langle \cdot , y \rangle }{(1-\langle x,y \rangle )^2} \end{aligned}$$

so we have:

$$\begin{aligned} \Vert f_y\Vert _\mathcal {B}= & {} \sup _{x \in B_E} (1-\Vert x\Vert ^2) \Vert f_y'(x)\Vert =\sup _{x \in B_E} (1-\Vert x\Vert ^2)\frac{\Vert y\Vert }{|1-\langle x,y \rangle |^2} \\= & {} \sup _{x \in B_E} \Vert y\Vert \frac{1-\Vert \varphi _y(x)\Vert ^2}{1-\Vert y\Vert ^2}= \frac{\Vert y\Vert }{1-\Vert y\Vert ^2}. \end{aligned}$$

Define $g_y: B_E \rightarrow \mathbb {C}$ by $g_y(x)=\displaystyle f_y(x)/\Vert f_y\Vert _\mathcal {B}$ which is analytic and it is satisfied that $\Vert g_y\Vert _\mathcal {I}\ge \Vert g_y\Vert _\mathcal {B}=1$. Using Lemma 2.4, there is a positive number k satisfying $\Vert g_y \circ \psi \Vert _\mathcal {I}\ge k \Vert g_y\Vert _\mathcal {I}$ so since:

$$\begin{aligned} \Vert g_y \circ \psi \Vert _\mathcal {I}=\sup _{x \in B_E} \Vert \widetilde{\nabla } (g_y \circ \psi )(x)\Vert , \end{aligned}$$

there exists $x_y \in B_E$ which satisfies $\Vert \widetilde{\nabla } (g_y \circ \psi )(x_y) \Vert \ge k/2$. Hence:

$$\begin{aligned} \frac{k}{2} \le \Vert \widetilde{\nabla } (g_y \circ \psi )(x_y) \Vert= & {} \sup _{w \in E {\setminus } \{0\} } \frac{\Vert \widetilde{\nabla } (g_y \circ \psi )(x_y) (w) \Vert }{\Vert w\Vert }\nonumber \\= & {} \sup _{w \in E {\setminus } \{0\} } \frac{|g_y'(\psi (x_y))(\psi '(x_y)(w))|}{\Vert \varphi _{x_y}'(0)^{-1}(w)\Vert }\nonumber \\= & {} \sup _{w \in E {\setminus } \{0\} } \frac{|g_y'(\psi (x_y))(\psi '(x_y)(w))|}{\kappa (\psi (x_y),\psi '(x_y)(w))} \frac{\kappa (\psi (x_y),\psi '(x_y)(w))}{\kappa (x_y,w)}\nonumber \\\le & {} \Vert \widetilde{\nabla } g_y (\psi (x_y))\Vert \widetilde{\tau _{\psi }}(x_y) \end{aligned}$$

(2.16)

where using (2.6) and (2.14) in Lemma 2.8 it is clearly deduced last inequality. By (2.8) we conclude:

$$\begin{aligned} \Vert \widetilde{\nabla } g_y (\psi (x_y))\Vert ^2= & {} (1-\Vert \psi (x_y)\Vert ^2)( \Vert \nabla g_y (\psi (x_y))\Vert ^2-|\mathcal {R}g_y (\psi (x_y))|^2)\\= & {} (1-\Vert \psi (x_y)\Vert ^2) \frac{(1-\Vert y\Vert ^2)^2}{\Vert y\Vert ^2}\\{} & {} \quad \left( \frac{\Vert y\Vert ^2}{|1-\langle \psi (x_y),y \rangle |^4}-\frac{|\langle \psi (x_y),y\rangle |^2}{|1-\langle \psi (x_y),y \rangle |^4}\right) \\= & {} (1-\Vert \psi (x_y)\Vert ^2)(1-\Vert y\Vert ^2)^2 \frac{1-\left| \Big \langle \psi (x_y),\frac{y}{\Vert y\Vert } \Big \rangle \right| ^2}{|1-\langle \psi (x_y),y \rangle |^4}. \end{aligned}$$

The inequality $|1-\langle c,d/\Vert d\Vert \rangle | \le 2 |1-\langle c,d \rangle |$ for any $c,d \in B_E$ is clear since:

$$\begin{aligned} |1-\langle c,d/\Vert d\Vert \rangle |\le & {} |1-\langle c,d \rangle |+|\langle c,d-d/\Vert d\Vert \rangle | \\\le & {} |1-\langle c,d \rangle |+ 1-\Vert d\Vert \le |1-\langle c,d \rangle |+1-|\langle c,d \rangle | \\= & {} 2 |1-\langle c,d \rangle |. \end{aligned}$$

From:

$$\begin{aligned} 1-\left| \Big \langle \psi (x_y),\frac{y}{\Vert y\Vert } \Big \rangle \right| ^2 \le \left( 1+\left| \Big \langle \psi (x_y),\frac{y}{\Vert y\Vert } \Big \rangle \right| \right) \left( 1-\left| \Big \langle \psi (x_y),\frac{y}{\Vert y\Vert } \Big \rangle \right| \right) \end{aligned}$$

we conclude:

$$\begin{aligned} \Vert \widetilde{\nabla } g (\psi (x_y))\Vert ^2\le & {} 4 (1-\Vert \psi (x_y)\Vert ^2)(1-\Vert y\Vert ^2) \frac{1}{|1-\langle \psi (x_y),y \rangle |^2} \\= & {} 4 (1-\Vert \varphi _y (\psi (x_y))\Vert ^2)= 4(1-\rho (y,\psi (x_y))^2) \end{aligned}$$

so:

$$\begin{aligned} \frac{k}{2} \le 2 (1-\rho (y,\psi (x_y))^2)^{1/2} \widetilde{\tau _{\psi }}(x_y) \end{aligned}$$

which is true if and only if $\displaystyle \frac{k}{4} \le (1-\rho (y,\psi (x_y))^2)^{1/2} \widetilde{\tau _{\psi }}(x_y)$

and we have $\widetilde{\tau _{\psi }}(x_y) \ge \frac{k}{4}$.

Using (2.16) we have:

$$\begin{aligned} \frac{k}{2} \le 2 (1-\rho (y,\psi (x_y))^2)^{1/2} \sup _{w \in E {\setminus } \{0\}}\frac{\kappa (\psi (x_y),\psi '(x_y)(w))}{\kappa (x_y,w)} \end{aligned}$$

so applying Lemma 2.9:

$$\begin{aligned} \sqrt{1-\rho (y,\psi (x_y))^2} \ge k/4 \end{aligned}$$

and this expression is equivalent to:

$$\begin{aligned} \rho (y,\psi (x_y)) \le \sqrt{1-k^2/16}. \end{aligned}$$

Taking $r=\sqrt{1-k^2/16}$ and $\varepsilon =k/4$ we conclude the result. $\square$

Hamada [10] provided a sufficient condition for a composition operator to be bounded below when we deal with unit balls of $JB^*-$triples. We will provide a new condition by extending the result given in Theorem 2.2. Hence we will consider the following condition: we will suppose that $\psi (x_y)$ belongs to the range of $\psi '(x_y)$. Recall that, as we have mentioned in (1.5), there is a positive constant $A_0$ satisfying:

$$\begin{aligned} \Vert f\Vert _\mathcal {R}\le \Vert f\Vert _\mathcal {B}\le \Vert f\Vert _\mathcal {I}\le A_0 \Vert f\Vert _R \ \text{ for } \text{ any } f \in \mathcal {B}(B_E). \end{aligned}$$

Theorem 2.13

Let $\psi$ be an analytic self-map on $B_E$. Suppose there are constants $r,\varepsilon$ satisfying $0< r < \frac{1}{15 A_0}$ and $\varepsilon > 0$ which also satisfies that for any $y \in \mathcal {B}_E$ there exists $x_y \in B_E$ such that $\rho (\psi (x_y),y)<r$ and ${\tau _\psi }(x_y) > \varepsilon$. Suppose also that $\psi (x_y) = \psi '(x_y)(w_{x_y})$ for some point $w_{x_y} \in E$ satisfying $\sup _{y \in B_E} \Vert w_{x_y}\Vert < +\infty$. Then we have that $C_{\psi }: \mathcal {B}(B_E) \rightarrow \mathcal {B}(B_E)$ is bounded below.

Proof

Consider a function $f \in \mathcal {B}(B_E)$ satisfying $\Vert f\Vert _\mathcal {I}=1$. We show the existence of $k >0$ which satisfies that $\Vert f \circ \psi \Vert _\mathcal {I}\ge k$. We have that $\Vert f\Vert _\mathcal {R}\ge \Vert f\Vert _\mathcal {I}/A_0$ by (1.5) so $\Vert f\Vert _\mathcal {R}\ge 1/A_0$. Taking $y \in B_E$ satisfying $| \mathcal {R}f(y)|(1-\Vert y\Vert ^2) \ge 14/(15 A_0)$, there exists $x_y \in B_E$ such that $\rho (y,\psi (x_y)) < r$ and $\tau _\psi (x_y) > \varepsilon$. Using (1.4) and (2.6) and also by (2.9), we have for any $w \in E {\setminus } \{0\}$:

$$\begin{aligned} \Vert f \circ \psi \Vert _\mathcal {I}= & {} \sup _{x \in B_E} \Vert \widetilde{\nabla } (f \circ \psi )(x)\Vert \\\ge & {} \frac{|(f \circ \psi )'(x_y)(w)|}{\Vert \varphi _{x_y}'(0)^{-1}(w)\Vert }= \frac{|f'(\psi (x_y))(\psi '(x_y)(w))|}{\kappa (\psi (x_y),\psi '(x_y)(w))} \frac{\kappa (\psi (x_y),\psi '(x_y)(w))}{\kappa (x_y,w)}. \end{aligned}$$

Since $\psi (x_y) \in \psi '(x_y)(E)$, there exists $w_{x_y} \in E$ such that $\psi '(x_y)(w_{x_y})=\Vert \psi '(x_y)\Vert \psi (x_y)$ so the inequality above is clearly true taking $w_{x_y}$. Using (2.12) from Lemma 2.7 we obtain:

$$\begin{aligned} \frac{|f'(\psi (x_y))(\psi '(x_y)(w_{x_y}))|}{\kappa (\psi (x_y),\psi '(x_y)(w_{x_y}))}= & {} \frac{|f'(\psi (x_y))(\Vert \psi '(x_y)\Vert \psi (x_y))|}{\kappa (\psi (x_y),\psi '(x_y)(w_{x_y}))}\\= & {} \frac{\Vert \psi '(x_y)\Vert |f'(\psi (x_y))( \psi (x_y))|(1-\Vert \psi (x_y)\Vert ^2)}{\Vert \psi '(x_y)\Vert \Vert \psi (x_y)\Vert }\\= & {} \frac{|\mathcal {R}f( \psi (x_y))| (1-\Vert \psi (x_y)\Vert ^2)}{\Vert \psi (x_y)\Vert } \end{aligned}$$

so:

$$\begin{aligned} \Vert f \circ \psi \Vert _\mathcal {I}\ge \frac{\mathcal {R}f(\psi (x_y)) (1-\Vert \psi (x_y)\Vert ^2)}{\Vert \psi (x_y)\Vert }\frac{\kappa (\psi (x_y),\psi '(x_y)(w_{x_y}))}{\kappa (x_y,w_{x_y})} \end{aligned}$$

and using (2.13) from Lemma 2.7 we have:

$$\begin{aligned} \Vert f \circ \psi \Vert _\mathcal {I}\ge & {} \frac{|\mathcal {R}f(\psi (x_y))| (1-\Vert \psi (x_y)\Vert ^2)}{\Vert \psi (x_y)\Vert } \frac{\Vert \psi (x_y)\Vert \tau _\psi (x_y)}{\Vert w_{x_y}\Vert } \\\ge & {} |\mathcal {R}f(\psi (x_y))| (1-\Vert \psi (x_y)\Vert ^2) \frac{\varepsilon }{\Vert w_{x_y}\Vert }. \end{aligned}$$

From Corollary 1.2, we obtain:

$$\begin{aligned} | |\mathcal {R}f(\psi (x_y))| (1-\Vert \psi (x_y)\Vert ^2)-|\mathcal {R}f(y)| (1-\Vert y\Vert ^2)| \le 14 \Vert f\Vert _\mathcal {I}\rho _E(\psi (x_y),y) \end{aligned}$$

and using $\Vert f\Vert _\mathcal {I}=1$, we conclude:

$$\begin{aligned} \Vert f \circ \psi \Vert _\mathcal {I}\ge & {} (|\mathcal {R}f(y)| (1-\Vert y\Vert ^2)| - 14 \rho (\psi (x_y),y)) \frac{\varepsilon }{\Vert w_{x_y}\Vert } \\\ge & {} \left( \frac{14}{15 A_0} - 14 r \right) \frac{\varepsilon }{\sup _{y \in B_E}\Vert w_{x_y}\Vert } \end{aligned}$$

so we can take:

$$\begin{aligned} k=\displaystyle 14\left( \frac{1}{15 A_0} - r \right) \frac{\varepsilon }{\sup _{y \in B_E}\Vert w_{x_y}\Vert } >0 \end{aligned}$$

and we finally conclude $\Vert C_\psi (f)\Vert _\mathcal {I}\ge k$. $\square$

Now we check that the automorphism $\varphi _a$ of $B_E$ for any $a \in B_E$ satisfies the conditions of Theorem 2.13. We will need this result, which shows $\tau _{\varphi _a}(x) \ge 1$ for all $x \in B_E$.

Lemma 2.14

For all $a \in B_E$ we have $\tau _{\varphi _a}(x) \ge 1$ if $x \in B_E$.

Proof

Notice that by (1.3) we have:

$$\begin{aligned} \frac{1-\Vert x\Vert ^2}{1-\Vert \varphi _a(x)\Vert ^2}= \frac{|1-\langle x,a \rangle |^2}{1-\Vert a\Vert ^2} \end{aligned}$$

and since: $\varphi _a(x)=(P_a +s_a Q_a) \left( m_a(x)\right)$, then we obtain:

$$\begin{aligned} \varphi _a'(x)=(P_a+s_a Q_a)'(m_a(x)) \circ m_a'(x)=(P_a+s_a Q_a)(m_a'(x)) \end{aligned}$$

so we have:

$$\begin{aligned} \Vert \varphi _a'(x)\Vert ^2= \Vert P_a(m_a'(x))\Vert ^2+s_a^2 \Vert Q_a(m_a'(x))\Vert ^2 \ge \Vert P_a(m_a'(x))\Vert ^2. \end{aligned}$$

It is easy that:

$$\begin{aligned} m_a'(x)(y)=\frac{-(1-\langle x,a \rangle )y+ \langle y,a \rangle (a-x)}{(1-\langle x, a \rangle )^2} \end{aligned}$$

so:

$$\begin{aligned} \Vert \varphi _a'(x)\Vert \ge \Vert P_a(m_a'(x))\Vert= & {} \sup _{y \in \overline{B}_E} \Vert P_a(m_a'(x))(y))\Vert \\\ge & {} \left\| P_a \left( m_a'(x) \left( \frac{a}{\Vert a\Vert }\right) \right) \right\| \\= & {} \left\| P_a \left( \frac{-(1-\langle x,a \rangle ) \frac{a}{\Vert a\Vert }+ \langle \frac{a}{\Vert a\Vert },a \rangle (a-x)}{(1-\langle x, a \rangle )^2} \right) \right\| \end{aligned}$$

so we obtain:

$$\begin{aligned} \tau _{\varphi _a}(x)\ge & {} \frac{|1-\langle x,a \rangle |^2}{1-\Vert a\Vert ^2} \left\| P_a \left( \frac{-(1-\langle x,a \rangle ) \frac{a}{\Vert a\Vert }+ \langle \frac{a}{\Vert a\Vert },a \rangle (a-x)}{(1-\langle x, a \rangle )^2} \right) \right\| \\= & {} \frac{1}{1-\Vert a\Vert ^2} \left\| P_a \left( -(1-\langle x,a \rangle ) \frac{a}{\Vert a\Vert }+ \langle \frac{a}{\Vert a\Vert },a \rangle (a-x) \right) \right\| \\= & {} \frac{1}{1-\Vert a\Vert ^2} \left\| \left( -(1-\langle x,a \rangle ) \frac{a}{\Vert a\Vert }+ \Vert a\Vert a- \frac{\langle x,a \rangle }{\Vert a\Vert ^2} \Vert a\Vert a \right) \right\| \\= & {} \frac{1}{1-\Vert a\Vert ^2} \left\| -\frac{1-\Vert a\Vert ^2}{\Vert a\Vert } a \right\| =1 \end{aligned}$$

and we have $\tau _{\varphi _a}(x) \ge 1$ so we are done. $\square$

Remark 2.15

Conditions of Theorem 2.13 are satisfied by the automorphisms $\varphi _a$ for any $a \in B_E$ since by Lemma 2.14 we have:

$$\begin{aligned} \frac{1-\Vert x\Vert ^2}{1-\Vert \varphi _a(x)\Vert ^2} \Vert \varphi _a'(x)\Vert \ge 1 \end{aligned}$$

so choose $\varepsilon =1$, $r=0$ and for any $y \in B_E$ take $x_y= \varphi _a(y)$. Furthermore, $\varphi _a(x_y)=\varphi _a(\varphi _a(y))=y = \varphi _a'(x_y)(w_{x_y})$ for some $w_{x_y}$ belonging to E which satisfies $\sup _{y \in B_E} \Vert w_{x_y}\Vert < +\infty$ since the operator $\varphi _a'(x_y)$ is invertible on the space E.