Lipschitz continuity of the dilation of Bloch functions on the unit ball of a Hilbert space and applications

Abstract

Let $B_E$ be the open unit ball of a complex finite- or infinite-dimensional Hilbert space. If f belongs to the space $\mathcal {B}(B_E)$ of Bloch functions on $B_E$, we prove that the dilation map 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, is Lipschitz continuous with respect to the pseudohyperbolic distance $\rho _E$ in $B_E$, which extends to the finite- and infinite-dimensional setting the result given for the classical Bloch space $\mathcal {B}$. To provide this result, we will need to prove that $\rho _E(zx,zy) \le |z| \rho _E(x,y)$ for $x,y \in B_E$ under some conditions on $z \in \mathbb {C}$. Lipschitz continuity of $x \mapsto (1-\Vert x\Vert ^2) \mathcal {R}f(x)$ will yield some applications on interpolating sequences for $\mathcal {B}(B_E)$ which also extends classical results from $\mathcal {B}$ to $\mathcal {B}(B_E)$. Indeed, we show that it is necessary for a sequence in $B_E$ to be separated to be interpolating for $\mathcal {B}(B_E)$ and we also prove that any interpolating sequence for $\mathcal {B}(B_E)$ can be slightly perturbed and it remains interpolating.

1 Introduction and background

Let $\textbf{D}$ be the open unit disk of the complex plane $\mathbb {C}$ and $\rho$ the pseudohyperbolic distance defined by $\rho (z,w)=\left| \frac{z-w}{1-z {\bar{w}}} \right|$. Consider a function f of the classical Bloch space $\mathcal {B}$. The study of the Lipschitz continuity of the dilation map $z \mapsto (1-|z|^2)f'(z)$ with respect to $\rho$ was first started by Attele to study conditions for a sequence on $\textbf{D}$ to be interpolating for $\mathcal {B}$ (see [2]). Ghatage, Yan, and Zheng also discussed a similar result providing a different proof [8]. Then, Xiong improved this result in [14] and other authors extended it for high-dimensional holomorphic functions [6] and planar harmonic mappings [7]. For bounded symmetric domains of $\mathbb {C}^n$, see also [10].

The aim of this paper is to extend the previous result for the infinite-dimensional setting and give some applications. Other applications on bounded below composition operators can be found in [12]. Along this work, E will denote a finite- or infinite-dimensional complex Hilbert space and its open unit ball will be denoted by $B_E$. In Sect. 2, we will study the boundness of

$$\begin{aligned} \frac{\rho _E(zx,zy)}{|z|\rho _E(x,y)} \end{aligned}$$

(1.1)

for $z \in \mathbb {C}$ and $x,y \in B_E$, such that $zx,zy \in B_E$ proving that, in general, this expression is unbounded. Nevertheless, we will show that if |z| is bounded above by

$$\begin{aligned} \frac{1+\max \{\Vert x\Vert ,\Vert y\Vert \}}{2 \max \{\Vert x\Vert ,\Vert y\Vert \}}, \end{aligned}$$

then the expression above is bounded by 2 and this bound is the best possible. We first prove the case when we deal with $E=\mathbb {C}$ and then with any finite- or infinite-dimensional Hilbert space E. At the end of this section, we will extend this result to the case when we deal with the Banach space $C_0(S)$.

In Sect. 3, we will consider the space $\mathcal {B}(B_E)$ of Bloch functions f on $B_E$. Recall that for $f \in \mathcal {B}(B_E)$ and $x \in B_E$, the radial derivative Rf(x) is given by $Rf(x)=\langle x,f'(x) \rangle .$ As a consequence of the boundness of (1.1), we show that the dilation map $x \mapsto (1-\Vert x\Vert ^2) \mathcal {R}f(x)$ for $x \in B_E$ is Lipschitz continuous with respect to the pseudohyperbolic distance $\rho _E$ in Sect. 3.1, extending to the finite- and infinite-dimensional setting the results mentioned above. Hence, we derive some results about interpolating sequences for $\mathcal {B}(B_E)$ in Sect. 3.2. Indeed, we supply a proof that these sequences are separated for the pseudohyperbolic distance. We also prove that interpolating sequences can be slightly perturbed and they remain interpolating, which also extends the result for $\mathcal {B}$ given in [2].

1.1 The pseudohyperbolic and hyperbolic distance

Let $\textbf{D}$ be the open unit disk of the complex plane $\mathbb {C}$. As we have mentioned, the pseudohyperbolic distance for $z,w \in \textbf{D}$ is given by

$$\begin{aligned} \rho (z,w)=\left| \frac{z-w}{1-{\bar{z}}w} \right| . \end{aligned}$$

If we deal with a complex Banach space X with open unit ball $B_X$, recall that $f: B_X \rightarrow \mathbb {C}$ is said to be holomorphic (or analytic) if it is Fréchet differentiable for any $x \in B_X$ (see [12] for further information). For any $x,y \in B_X$, the pseudohyperbolic distance $\rho _X(x,y)$ is given by

$$\begin{aligned} \rho _X (x,y)=\sup \{ \rho (f(x),f(y)): f \in H^{\infty }(B_X), \Vert f\Vert _{\infty } \le 1 \}, \end{aligned}$$

where $H^{\infty }(B_X)$ is the space of bounded holomorphic functions on $B_X$ which become a Banach space (a uniform algebra, indeed) endowed with the sup-norm. The hyperbolic distance for $x,y \in B_X$ is given by

$$\begin{aligned} \beta _X(x,y)=\frac{1}{2} \log \left( \frac{1+\rho _X(x,y)}{1-\rho _X(x,y)} \right) . \end{aligned}$$

1.2 Automorphisms and pseudohyperbolic distance on $B_E$

If we deal with a complex Hilbert space E, we will denote by $\textrm{Aut}(B_E)$ the space of automorphisms of $B_E$, that is, the maps $\varphi : B_E \rightarrow B_E$ which are bijective and bianalytic (see [9]). For any $x \in B_E$, the automorphism $\varphi _x: B_{E} \longrightarrow B_{E}$ is defined according to

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

(1.2)

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 the orthogonal projection along the one-dimensional subspace spanned by x, that is

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

and $Q_x: E \longrightarrow E$ is the orthogonal complement $Q_x=\textrm{Id}_E-P_x$, where $\textrm{Id}_E$ denotes the identity operator on E. It is clear that $\varphi _x(0)=x$ and $\varphi _x(x)=0$. The automorphisms of the unit ball $B_{E}$ turn to be compositions of these $\varphi _x$ with unitary transformations U of E.

It is well known (see [9]) that the pseudohyperbolic distance on $B_E$ is given by

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

(1.3)

and

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

(1.4)

1.3 The Bloch space

The classical Bloch space $\mathcal {B}$ is the set of holomorphic functions $f: \textbf{D}\rightarrow \mathbb {C}$, such that $\Vert f\Vert _B=\sup _{z \in \textbf{D}} (1-|z|^2)|f'(z)|$ is bounded. This supremum defines a semi-norm which becomes a norm by adding up a constant: $|f(0)|+\sup _{z \in \textbf{D}} (1-|z|^2)|f'(z)|$. Hence, $\mathcal {B}$ becomes a complex Banach space. The semi-norm $\Vert \cdot \Vert _B$ is invariant by automorphisms, that is, $\Vert f \circ \varphi \Vert _B=\Vert f\Vert _B$ for any $f \in \mathcal {B}$ and $\varphi : \textbf{D}\rightarrow \textbf{D}$ an automorphism of $\textbf{D}$.

Timoney extended Bloch functions if we deal with a finite-dimensional Hilbert space (see [13]). Blasco, Galindo, and Miralles extended them to the infinite-dimensional setting (see [5]). If we deal with a complex finite- or infinite-dimensional Hilbert space E, the analytic function $f: B_E \rightarrow \mathbb {C}$ is said to belong 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 it is clear that $\nabla f(x)$ is the derivative $f'(x)$ or, equivalently, if

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

where $\mathcal {R}f(x)$ is the radial derivative of f at x given by $\mathcal {R}f(x)=\langle x, \overline{\nabla f (x)}\rangle$. These semi-norms are equivalent to the following one:

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

(1.5)

where ${\widetilde{\nabla }} f (x)$ denotes the invariant gradient of f at x which is given by ${\widetilde{\nabla }} f (x) =\nabla (f \circ \varphi _x)(0)$, where $\varphi _x$ is the automorphism given in (1.2).

The three semi-norms $\Vert \cdot \Vert _\mathcal {B}, \Vert \cdot \Vert _\mathcal {R}$ and $\Vert \cdot \Vert _\mathcal {I}$ define equivalent Banach space norms-modulo the constant functions- in $\mathcal {B}(B_E)$ (see [5]). In particular, there exists a constant $A_0 >0$, such that

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

(1.6)

Hence, the space $\mathcal {B}(B_E)$ can be endowed with any of the norms $\Vert \cdot \Vert _{\mathcal {B}-Bloch}=|f(0)|+ \Vert \cdot \Vert _\mathcal {B}$ or $\Vert \cdot \Vert _{\mathcal {R}-Bloch}=|f(0)|+ \Vert \cdot \Vert _\mathcal {R}$ or $\Vert \cdot \Vert _{\mathcal {I}-Bloch}=|f(0)|+ \Vert \cdot \Vert _\mathcal {I}$ and $\mathcal {B}(B_E)$ becomes a Banach space. We will make use of these three semi-norms and norms along the sequel. We will also make use of this result, which states that Bloch functions on $B_E$ are Lipschitz with respect to the hyperbolic distance (see [3]):

Proposition 1.1

Let E be a complex Hilbert space and let $f \in \mathcal {B}(B_E)$. Then, for any $x,y \in B_E$

$$\begin{aligned} |f(x) -f(y)| \le \Vert f\Vert _\mathcal {I}\beta _E(x,y). \end{aligned}$$

2 Inequalities with the pseudohyperbolic distance

Let X be a complex Banach space. If $\varphi : B_X \rightarrow B_X$ is an analytic self-map, it is well known that $\rho _X(\varphi (x),\varphi (y)) \le \rho _X (x,y)$ for any $x,y \in B_X$ and the equality is attained if and only if $\varphi$ is an automorphism of $B_X$. Hence, if we consider $z \in \mathbb {C}$, $|z| \le 1$ and $x,y \in B_X$, it is clear that

$$\begin{aligned} \frac{\rho _X(zx,zy)}{\rho _X(x,y)} \le 1, \end{aligned}$$

since the map $\varphi : B_X \rightarrow B_X$ given by $\varphi (x)=z x$ is analytic on $B_X$. However, this situation changes dramatically if we consider the expression

$$\begin{aligned} \frac{\rho _X(zx,zy)}{|z|\rho _X(x,y)} \end{aligned}$$

(2.1)

for any $z \in \mathbb {C}$, such that $zx,zy \in B_X$. We show that, in general, this expression is unbounded. Anyway, if we deal with X a complex Hilbert space or $C_0(S)$ and $z \in \mathbb {C}$ satisfies

$$\begin{aligned} |z| \le \frac{1+\max \{\Vert x\Vert ,\Vert y\Vert \}}{2 \max \{\Vert x\Vert ,\Vert y\Vert \}}, \end{aligned}$$

then expression (2.1) is bounded by 2 and this will permit us to provide several applications in Sect. 3.

2.1 Unboundness

In this section, we prove that expression (2.1) is unbounded in general.

Proposition 2.1

Let E be a complex Hilbert space. There exist a sequence $(z_n) \subset \mathbb {C}$, $x \in B_E$ and a sequence $(y_n) \subset B_E$, such that $z_n x,z_n y_n \in B_E$ but

$$\begin{aligned} \frac{\rho _E(z_n x,z_n y_n)}{|z_n| \rho _E(x,y_n)} \end{aligned}$$

is unbounded.

Proof

We prove it for $E=\mathbb {C}$. Take for instance $x=1/2$, $y_n=1/2-\frac{1}{n}$ and $z_n =2-\frac{1}{n}$. It is clear that $|z_n x|<1$ and $|z_n y_n| <1$. However

$$\begin{aligned} \rho (z_n x,z_ny_n)=\frac{\frac{1}{n}\left( 2-\frac{1}{n} \right) }{1-\frac{1}{2}\left( \frac{1}{2}-\frac{1}{n}\right) \left( 2-\frac{1}{n}\right) ^2} =\frac{\frac{2n-1}{n^2}}{\frac{2-9n+12n^2}{4n^3}}=\frac{4n(2n-1)}{12n^2-9n+2} \end{aligned}$$

and

$$\begin{aligned} |z_n|\rho ( x,y_n)=\left( 2-\frac{1}{n}\right) \frac{\frac{1}{n}}{1-\frac{1}{2} \left( \frac{1}{2}-\frac{1}{n}\right) }=\frac{\frac{2n-1}{n^2}}{\frac{3n+2}{4n}}=\frac{4(2n-1)}{3n^2+2n}. \end{aligned}$$

Hence

$$\begin{aligned} \frac{\rho (z_nx,z_ny_n)}{|z_n|\rho (x,y_n)}=\frac{\frac{4n(2n-1)}{12n^2-9n+2}}{\frac{4(2n-1)}{3n^2+2n}}=\frac{3n^3+2n^2}{12n^2-9n+2}, \end{aligned}$$

which is clearly unbounded, since it tends to $\infty$ when $n \rightarrow \infty$. The result remains true if we deal with any complex Hilbert space E, since we can take $x_0 \in E$, such that $\Vert x_0\Vert =1$ and take $u=\frac{1}{2} x_0$, $v_n= y_n x_0$. We have that

$$\begin{aligned} \rho _E(z_n u,z_n v_n)^2= & {} 1-\frac{(1-\Vert u_n\Vert ^2)(1-\Vert v_n\Vert ^2)}{|1-\langle u_n,v_n \rangle |^2} \\ {}= & {} 1-\frac{(1-\Vert z_n x\Vert ^2)(1-\Vert z_n y_n\Vert ^2)}{|1-\langle z_n x,z_n y_n \rangle |^2}=\rho _E(z_n x, z_n y_n)^2, \end{aligned}$$

and similarly, we have $\rho _E(u,v_n)=\rho _E(x,y_n)$, so we apply the case $E=\mathbb {C}$ and we are done. $\square$

An easy consequence is a well-known result: the pseudohyperbolic distance cannot be extended to a norm on E, since

$$\begin{aligned} \frac{\rho _E(zx,zy)}{|z|\rho _E(x,y)} \end{aligned}$$

is unbounded, so $\rho _E(zx,zy) \ne |z| \rho _E(x,y)$.

2.2 Boundness

The main result of this section is Theorem 2.7 which states that under condition (2.1), then the expression (1.1) is bounded and the best bound possible is given by 2. The following lemma will be used to prove this result.

Lemma 2.2

Let E be a finite- or infinite-dimensional Hilbert space, $z \in \mathbb {C}$ and $x,y \in B_E$, such that

$$\begin{aligned} \displaystyle |z| \le \frac{1+\max \{\Vert x\Vert ,\Vert y\Vert \}}{2 \max \{\Vert x\Vert ,\Vert y\Vert \}}. \end{aligned}$$

Then, $|1-p| \le 2 |1-|z|^2 p|$ where p denotes the scalar product $\langle x,y \rangle$.

Proof

Suppose without loss of generality that $\Vert x\Vert \ge \Vert y\Vert$ and $x \ne 0$. Otherwise, the inequality is clearly true for any $z \in \mathbb {C}$. Notice that

$$\begin{aligned} |1-p|=|1-|z|^2p+|z|^2p-p| \le |1-|z|^2p|+||z|^2-1||p|, \end{aligned}$$

so it is sufficient to prove that $|1-|z|^2p|+||z|^2-1||p| \le 2|1-|z|^2p|$, which is equivalent to $|1-|z|^2p| \ge ||z|^2-1||p|$. We consider two cases:

i) if $|z|^2 \le 1$, then $1-|z|^2 \ge 0$, so we need to prove $|1-|z|^2 p| \ge (1-|z|^2)|p|$ which is clearly satisfied, since

$$\begin{aligned} |1-|z|^2p| \ge 1-|z|^2 |p| \ge |p|-|z|^2 |p|=(1-|z|^2)|p|, \end{aligned}$$

where second inequality is true, because $|p| \le \Vert x\Vert \Vert y\Vert < 1$.

ii) On the other hand, suppose that $|z|^2 >1$: we need to prove that $|1-|z|^2p| \ge (|z|^2-1)|p|$. Since $|1-|z|^2p| \ge 1-|z|^2 |p|$, it is sufficient to prove that

$$\begin{aligned} 1-|z|^2 |p| \ge (|z|^2-1)|p|. \end{aligned}$$

(2.2)

Notice that $1-|z|^2|p| >0$, since $|z|^2 |p| \le \Vert zx\Vert \Vert zy\Vert < 1$, because $zx,zy \in B_E$. Inequality (2.2) is equivalent to $2|z|^2 |p| <1+|p|$ which is true, since

$$\begin{aligned} 2|z|^2|p| \le 2\left( \frac{1+\Vert x\Vert }{2\Vert x\Vert }\right) ^2 |p|, \end{aligned}$$

so we need to prove that

$$\begin{aligned} 2\left( \frac{1+\Vert x\Vert }{2\Vert x\Vert }\right) ^2 |p| \le 1+|p| \leftrightarrow \left( \frac{(1+\Vert x\Vert )^2}{2\Vert x\Vert ^2}-1 \right) |p| \le 1. \end{aligned}$$

However

$$\begin{aligned} \left( \frac{(1+\Vert x\Vert )^2}{2\Vert x\Vert ^2}-1 \right) |p| = \frac{1+(2-\Vert x\Vert )\Vert x\Vert }{2\Vert x\Vert ^2}|p| \le \frac{1+1}{2\Vert x\Vert ^2}\Vert x\Vert ^2=1, \end{aligned}$$

where last inequality is true because of the arithmetic mean-geometric mean inequality and since $|p| \le \Vert x\Vert \Vert y\Vert \le \Vert x\Vert ^2$. $\square$

2.2.1 The case $E=\mathbb {C}$

If we deal with $E=\mathbb {C}$, it is easy to prove that (2.1) is bounded:

Proposition 2.3

Let $x,y \in \textbf{D}$ and $z \in \mathbb {C}$, such that

$$\begin{aligned} |z| \le \frac{1+\max \{|x|,|y|\}}{2 \max \{|x|,|y|\}}. \end{aligned}$$

Then, $zx,zy \in \textbf{D}$ and

$$\begin{aligned} \rho (zx,zy)\le 2 |z|\rho (x,y). \end{aligned}$$

Proof

Suppose without loss of generality that $|x| \ge |y|$ and take $z \ne 0$ (otherwise, it is clear). Notice that $zx,zy \in \textbf{D}$, since

$$\begin{aligned} |zy| \le |zx| \le \frac{1+|x|}{2|x|} |x| = \frac{1+|x|}{2} < 1. \end{aligned}$$

We have

$$\begin{aligned} \rho (zx,zy)=\left| \frac{zx-zy}{1-zx{\overline{zy}}}\right| =|z|\left| \frac{x-y}{1-|z|^2x{\overline{y}}}\right| =|z| \frac{|x-y|}{|1-|z|^2x{\overline{y}}|} \end{aligned}$$

and

$$\begin{aligned} |z| \rho (x,y)=|z| \left| \frac{x-y}{1-x{\overline{y}}}\right| =|z| \frac{|x-y|}{|1-x{\overline{y}}|}, \end{aligned}$$

so the inequality is equivalent to

$$\begin{aligned} |z| \frac{|x-y|}{|1-|z|^2x{\overline{y}}|} \le 2 |z| \frac{|x-y|}{|1-x{\overline{y}}|} \leftrightarrow |1-x{\overline{y}}| \le 2 |1-|z|^2x{\overline{y}}|. \end{aligned}$$

Calling $p=x{\overline{y}}$, we have to prove that $|1-p| \le 2 |1-|z|^2 p|$. Apply Lemma 2.2 for $E=\mathbb {C}$ and we are done. $\square$

Remark 2.4

Notice that the bound 2 is the best possible. Indeed, take $x_n,y_n \in \textbf{D}$, such that $x_n \rightarrow 1$ and $y_n \rightarrow -1$. It is clear that for $z_n \rightarrow 0$, the expression $|1-\overline{x_n}y_n|$ tends to 2 when $n \rightarrow \infty$ and the expression $2 |1-|z_n|^2\overline{x_n}y_n|$ also tends to 2, so the inequality above is sharp. $\square$

2.2.2 The case when E is any complex Hilbert space

We will deal with $x,y \in \mathcal {B}_E$ and $z \in \mathbb {C}$, such that $zx,zy \in B_E$, we will denote by $r=\Vert x\Vert$, $s=\Vert y\Vert$ and $u=\Vert x\Vert ^2+\Vert y\Vert ^2=r^2+s^2$. We will also denote by p the scalar product $\langle x,y \rangle$ and $m=\Re p$. This notation will be used in Lemma 2.5 and Theorem 2.7.

Lemma 2.5

Let E be a complex Hilbert space and $x,y \in B_E$. Then

$$\begin{aligned} \frac{\Vert x-y\Vert ^2|1-p|^2}{\Vert x-y\Vert ^2-(\Vert x\Vert ^2 \Vert y\Vert ^2-|p|^2)} \le 4. \end{aligned}$$

Proof

The inequality is equivalent to

$$\begin{aligned}{} & {} (1-2\Re p +|p|^2) \Vert x-y\Vert ^2 \le 4\Vert x-y\Vert ^2 -4r^2s^2+4|p|^2 \text{ if } \text{ and } \text{ only } \text{ if: } \\{} & {} \quad (1-2\Re p +|p|^2) (r^2+s^2-2\Re p) \le 4(r^2+s^2)-8 \Re p -4r^2s^2+4|p|^2. \end{aligned}$$

Bearing in mind that $m=\Re p$ and $u=r^2+s^2$, we need to prove

$$\begin{aligned}{} & {} (1-2m +|p|^2) (u-2m) \le 4u-8m -4r^2s^2+4|p|^2 \\{} & {} \quad \leftrightarrow u-2um+u |p|^2-2m+4m^2-2m|p|^2 \le 4u-8m-4r^2s^2+4|p|^2 \\{} & {} \quad \leftrightarrow 4u-8m-4r^2s^2+4|p|^2-u+2um-u|p|^2+2m-4m^2+2m|p|^2 \ge 0. \end{aligned}$$

Notice that $|p| \ge |m|$ so $(4+2\,m-u) |p|^2 \ge ~(4+m-u) m^2$ since $4+2\,m-u \ge 0$. Hence, it is sufficient to prove

$$\begin{aligned} 4u-8m-4r^2s^2+(4+2m-u)m^2-u+2um+2m-4m^2 \ge 0. \end{aligned}$$

It is also clear that $u^2 \ge 4r^2s^2$, so last inequality is equivalent to

$$\begin{aligned}{} & {} 4u-8m-u^2+(4+2m-u)m^2-u+2um+2m-4m^2 \ge 0 \\{} & {} \quad \leftrightarrow 2m^3-u m^2+2(u-3)m+3u-u^2 \ge 0. \end{aligned}$$

The expression at left can be easily factorized and is equal to

$$\begin{aligned} (u-2m)(3-u-m^2), \end{aligned}$$

where both factors are clearly greater or equal to 0 and we are done. $\square$

This lemma will be used at the end of the proof of Theorem 2.7:

Lemma 2.6

Let $f(a,b,c)=(3-b^2)(a-c)-(a^2-b^2)(2-c)$. Then, $f(a,b,c) \ge 0$ for any $0 \le c \le b \le a \le 1$.

Proof

Notice that $f(a,b,c)=(3-b^2)a-2(a^2-b^2)-(3-a^2)c$, so f is affine with respect to c. Hence, it is enough to prove the inequality for $c=b$. The function becomes $f(a,b,b)=(3-b^2)(a-b)-(a^2-b^2)(2-b)=(a-b)((3-b^2)-(a+b)(2-b))$, and since $a-b \ge 0$, it is enough to prove that $(3-b^2)-(a+b)(2-b) \ge 0$. The expression $g(a,b)=(3-b^2)-(a+b)(2-b)$ is affine with respect to a, so it is enough to prove it for $a=1$. Notice that $g(1,b)=(3-b^2)-(1+b)(2-b)=1-b$ which is clearly greater or equal to 0, so we are done. $\square$

Theorem 2.7

Let E be a finite- or infinite-dimensional complex Hilbert space, $z \in \mathbb {C}$ and $x,y \in B_E$. If

$$\begin{aligned} |z| \le \frac{1+\max \{\Vert x\Vert ,\Vert y\Vert \}}{2 \max \{\Vert x\Vert ,\Vert y\Vert \}}, \end{aligned}$$

then $zx,zy \in B_E$ and

$$\begin{aligned} \frac{\rho _E(zx,zy)}{|z|\rho _E(x,y)} \le 2. \end{aligned}$$

Proof

Suppose without loss of generality that $\Vert x\Vert \ge \Vert y\Vert$ and $z \ne 0$. We will denote $\rho =\rho _E(x,y)$ and $\rho _z=\rho _E(zx,zy)$. If $\frac{1}{2} \le |z| < 1$, then the result is clear, since

$$\begin{aligned} \rho _E(zx,zy) \le \rho _E(x,y) \le 2 |z| \rho _E(x,y), \end{aligned}$$

where first inequality is true because of the contractivity of the pseudohyperbolic distance for the function $g: B_E \rightarrow B_E$ given by $g(x)=z x$.

Therefore, let us prove it for $|z| < 1/2$ or $|z| \ge 1$. Taking squares, the inequality is equivalent to prove

$$\begin{aligned} \frac{\rho _z^2}{|z|^2\rho ^2} \le 4. \end{aligned}$$

(2.3)

Bear in mind the expression (1.4) for the pseudohyperbolic distance and call $t=|z|^2$ which is different from 0, since $z \ne 0$. We have

$$\begin{aligned} \displaystyle \frac{\rho _z^2}{|z|^2\rho ^2}= & {} \frac{\frac{|1-t p|^2-(1-t r^2)(1-ts^2)}{|1-tp|^2}}{\frac{t(|1-p|^2-(1-r^2)(1-s^2))}{|1-p|^2}} = \frac{(|1-t p|^2-(1-t r^2)(1-ts^2))|1-p|^2}{t(|1-p|^2-(1-r^2)(1-s^2))|1-tp|^2} \\= & {} \frac{(1+t^2|p|^2-2t\Re p-1-t^2 r^2s^2+t(r^2+s^2))|1-p|^2}{t(1+|p|^2-2\Re p-1-r^2s^2+r^2+s^2)|1-tp|^2}\\= & {} \frac{(t^2|p|^2-2t\Re p-t^2 r^2s^2+t(r^2+s^2))|1-p|^2}{t(|p|^2-2\Re p-r^2s^2+r^2+s^2)|1-tp|^2} \\= & {} \frac{(\Vert x-y\Vert ^2-t(r^2s^2-|p|^2))|1-p|^2}{(\Vert x-y\Vert ^2-(r^2s^2-|p|^2))|1-tp|^2}. \end{aligned}$$

We will introduce the following notation:

$$\begin{aligned} A:= & {} \Vert x-y\Vert ^2=r^2+s^2-2 \Re p=u-2m \end{aligned}$$

(2.4)

$$\begin{aligned} B:= & {} r^2s^2-|p|^2. \end{aligned}$$

(2.5)

Notice that $A-B \ge 0$, since

$$\begin{aligned}{} & {} A-B=r^2+s^2-2m-r^2s^2+|p|^2=|1-p|^2-(1-r^2)(1-s^2) \ge 0 \\{} & {} \quad \leftrightarrow |1-p|^2 \ge (1-r^2)(1-s^2) \leftrightarrow 1-\frac{(1-r^2)(1-s^2)}{|1-p|^2}=\rho (x,y)^2 \ge 0. \end{aligned}$$

Using this notation, inequality (2.3) is equivalent to

$$\begin{aligned} \frac{(A-tB) |1-p|^2}{(A-B)|1-tp|^2} \le 4, \end{aligned}$$

(2.6)

so bearing in mind that $t=|z|^2$, we only need to prove (2.6) for $t \ge 1$ and $t \le 1/4$. If $t \ge 1$, the result is clear, since

$$\begin{aligned} \frac{(A-tB) |1-p|^2}{(A-B)|1-tp|^2} \le \left( \frac{A-B}{A-B}\right) \frac{|1-p|^2}{|1-tp|^2} = \frac{|1-p|^2}{|1-tp|^2} \le 4, \end{aligned}$$

where last inequality is true by Lemma 2.2. Therefore, it remains to prove it for $0 \le t \le 1/4$. Inequality 2.6 is equivalent to

$$\begin{aligned}{} & {} 4 (A-B)|1-tp|^2 \ge (A-tB)|1-p|^2 \\{} & {} \quad \leftrightarrow 4 (A-B)(1-2mt+t^2|p|^2) \ge (A-tB)|1-p|^2 \\{} & {} \quad \leftrightarrow 4 (A-B)|p|^2t^2 +(B|1-p|^2-8m(A-B))t+ 4(A-B)-A|1-p|^2 \ge 0. \end{aligned}$$

Since $B,t \ge 0$ and $4(A-B)-A|1-p|^2 \ge 0$ by Lemma 2.5, this inequality is clearly true if $m <0$. Therefore, we can suppose without loss of generality that $m \ge 0$. The inequality is equivalent to

$$\begin{aligned} 4 (A-B)|1-tp|^2 - (A-tB)|1-p|^2 \ge 0, \end{aligned}$$

so we will prove last inequality. Notice that

$$\begin{aligned}{} & {} 4 (A-B)|1-tp|^2 - (A-tB)|1-p|^2 \\{} & {} \quad =4(A-B)(1-2mt+t^2 |p|^2)-(A-B)(1-2m+|p|^2)-B(1-t)|1-p|^2\\{} & {} \quad =4(A-B)(1-2m+|p|^2+2m(1-t)-(1-t^2)|p|^2)\\{} & {} \qquad -(A-B)(1-2m+|p|^2)-B(1-t)|1-p|^2\\{} & {} \quad =3(A-B)(1-2m+|p|^2)+8m(1-t)(A-B)\\{} & {} \qquad -4(1-t^2)|p|^2(A-B)-B(1-t)|1-p|^2. \end{aligned}$$

Since $0 \le t \le 1/4$, we have that $3/4 \le 1-t \le 1$, so

$$\begin{aligned}{} & {} 4 (A-B)|1-tp|^2 - (A-tB)|1-p|^2 \\{} & {} \quad \ge 3(A-B)(1-2m+|p|^2)+8m(1-t)(A-B) \\{} & {} \qquad -4(1-t^2)|p|^2(A-B)-B(1-t)|1-p|^2\\{} & {} \quad \ge 3(A-B)-6m(A-B)+3|p|^2(A-B)+8m \cdot \frac{3}{4}(A-B)\\{} & {} \qquad -4|p|^2(A-B)-B|1-p|^2 \\{} & {} \quad =3(A-B)-6m(A-B)+6m(A-B)-|p|^2(A-B)-B|1-p|^2 \\{} & {} \quad =(3-|p|^2)(A-B)-B|1-p|^2. \end{aligned}$$

Bearing in mind (2.4) and (2.5), notice that

$$\begin{aligned}{} & {} (3-|p|^2)(A-B)-B|1-p|^2 = (3-|p|^2)A-B(3-|p|^2+1-2m+|p|^2)\\{} & {} \quad =(3-|p|^2)A-B(4-2m) = (3-|p|^2)(r^2+s^2-2m)-(4-2m)(r^2s^2-|p|^2) \\{} & {} \quad \ge (3-|p|^2)(2rs-2m)-(4-2m)(r^2s^2-|p|^2) \\{} & {} \quad =2(3-|p|^2)(rs-m)-2(2-m)(r^2s^2-|p|^2)=2f(rs,|p|,m), \end{aligned}$$

where f is the function defined in Lemma 2.6 and since $0 \le m \le |p| \le rs \le 1$, using the lemma, we are done, since

$$\begin{aligned} (3-|p|^2)(A-B)-B|1-p|^2 \ge 2f(rs,|p|,m)\ge 0. \end{aligned}$$

$\square$

2.2.3 Results for $X=C_0(S)$

Let S be a locally compact topological space and consider $X=C_0(S)$ given by the space of continuous functions $f: S \rightarrow \mathbb {C}$, such that for any $\varepsilon >0$, there exists a closed compact subset $K \subset S$, such that $|f(x)| < \varepsilon$ for any $x \in S {\setminus } K$. Endowed with the sup-norm, $C_0(S)$ becomes a Banach space and the pseudohyperbolic distance for $x,y \in C_0(S)$ is well known (see [1]) and it is given by

$$\begin{aligned} \rho _X(x,y)=\sup _{t \in S} \rho (x(t),y(t)). \end{aligned}$$

(2.7)

We prove that expression (2.1) is also bounded by 2 when we deal with the space $X=C_0(S)$:

Proposition 2.8

Let $X=C_0(S)$ and $x,y \in X$. If $z \in \mathbb {C}$ satisfies

$$\begin{aligned} |z| \le \frac{1+\max \{\Vert x\Vert ,\Vert y\Vert \}}{2 \max \{\Vert x\Vert ,\Vert y\Vert \}}, \end{aligned}$$

then

$$\begin{aligned} \frac{\rho _X(zx,zy)}{|z| \rho _X(x,y)} \le 2. \end{aligned}$$

Proof

Suppose without loss of generality that $\Vert x\Vert \ge \Vert y\Vert$. For any $t \in S$, we have that $x(t),y(t) \in \textbf{D}$, since $\Vert x\Vert =\sup _{t \in S}|x(t)| <1$ and $\Vert y\Vert =\sup _{t \in S} |y(t)|<1$. The result is clear, since

$$\begin{aligned} \rho _X(zx,zy)= & {} \sup _{t \in S} \rho (z x(t),zy(t)) \le \sup _{t \in S} 2|z| \rho (x(t),y(t))\\ {}= & {} 2|z| \sup _{t \in S} \rho (x(t),y(t))=2 |z| \rho _X(x,y), \end{aligned}$$

where first inequality is clear because of Proposition 2.3 and because for any $t \in X$, we have that

$$\begin{aligned} |z| \le \frac{1+\Vert x\Vert }{2\Vert x\Vert }=\frac{\frac{1}{\Vert x\Vert }+1}{2} \le \inf _{t \in X} \left\{ \frac{1+|x(t)|}{2|x(t)|}\right\} \le \frac{1+|x(t)|}{2|x(t)|}, \end{aligned}$$

and we are done. $\square$

3 Applications

Theorem 2.7 yields several applications. As we have mentioned, we first show that for a Bloch function $f: B_E \rightarrow \mathbb {C}$, the function $x \mapsto (1-\Vert x\Vert ^2) |\mathcal {R}f(x)|$ for $x \in B_E$ is Lipschitz continuous with respect to the pseudohyperbolic distance $\rho _E$. Hence, we derive some results about interpolating sequences for $\mathcal {B}(B_E)$ in Sect. 3.2. Indeed, we provide a new proof that these sequences are separated for the pseudohyperbolic distance. We also prove that these sequences can be slightly perturbed and they remain interpolating.

3.1 The Lipschitz continuity of $(1-\Vert x\Vert ^2) |\mathcal {R}f(x)|$

We will denote by $\Pi$ the unit circle of the complex plane $\mathbb {C}$, that is, the set of complex numbers u, such that $|u|=1$.

Lemma 3.1

Let $f \in \mathcal {B}(B_E)$. Fix $\varepsilon >0$ and $x,y \in B_E$. If $(1+\varepsilon u)x$ and $(1+\varepsilon u)y$ belongs to $B_E$ for any $u \in \Pi$, then there exists $u_0 \in \Pi$, such that

$$\begin{aligned} |\mathcal {R}f(x)-\mathcal {R}f(y)| \le \frac{1}{\varepsilon } \Vert f\Vert _\mathcal {I}\beta ((1+\varepsilon u_0) x,(1+\varepsilon u_0)y). \end{aligned}$$

(3.1)

Proof

Fix $x,y \in B_E$ and $\varepsilon >0$. Notice that the function $f(x+\varepsilon u x)-f(y+\varepsilon u y)$ defined for $u \in \Pi$ is continuous. Since $\Pi$ is a compact set, there exists $u_0 \in \Pi$, such that

$$\begin{aligned} f(x+\varepsilon u_0 x)-f(y+\varepsilon u_0 y)= \max \{f(x+\varepsilon u x)-f(y+\varepsilon u y): u \in \Pi \}. \end{aligned}$$

Consider $g(u)=f(x+\varepsilon u x)$ for u defined on an open disk of the complex plane $\mathbb {C}$ which contains $\Pi$. It is clear that

$$\begin{aligned} g'(u)=\nabla f(x+\varepsilon u x)(\varepsilon x), \end{aligned}$$

so $g'(0)=\varepsilon \mathcal {R}f(x)$. Similarly, if $h(u)=f(y+\varepsilon u y)$, then $h'(0)=\varepsilon R f(y)$. By the Cauchy’s integral formula, we have

$$\begin{aligned} |\mathcal {R}f(x)-\mathcal {R}f(y)|= & {} |\langle x,\overline{\nabla f(x)} \rangle -\langle y,\overline{\nabla f(y)} \rangle |\\= & {} \left| \frac{1}{\varepsilon } \frac{1}{2 \pi i} \int _{|u|=1} f(x+\varepsilon u x)-f(y+\varepsilon u y) \frac{du}{u^2}\right| \\\le & {} \frac{2\pi }{2 \pi \varepsilon } |f(x+\varepsilon u_0 x)-f(y+\varepsilon u_0 y) |\\ {}\le & {} \frac{1}{\varepsilon } \Vert f\Vert _\mathcal {I}\beta ((1+\varepsilon u_0) x,(1+\varepsilon u_0)y), \end{aligned}$$

where last inequality is true by Proposition 1.1. $\square$

The proof of the following lemma is an easy calculation. It will be used in Lemma 3.3.

Lemma 3.2

For any $0 \le t < 1$, we have

$$\begin{aligned} \frac{1}{2} \log \left( \frac{1+t}{1-t}\right) \le \frac{t}{1-t}. \end{aligned}$$

Lemma 3.3

Let $f \in \mathcal {B}(B_E)$ and $x,y \in B_E$, such that $\Vert x\Vert \ge \Vert y\Vert$. Then

$$\begin{aligned} (1-\Vert x\Vert ^2) |\mathcal {R}f(x)-\mathcal {R}f(y)| \le 12 \Vert f\Vert _\mathcal {I}\rho _E(x,y). \end{aligned}$$

Proof

Take

$$\begin{aligned} \varepsilon =\frac{1-\Vert x\Vert }{2\Vert x\Vert } >0. \end{aligned}$$

Notice that for any $u \in \Pi$, we have that $(1+\varepsilon u)x$ and $(1+\varepsilon u) y$ belongs to $B_E$, since

$$\begin{aligned} (1+\varepsilon ) \Vert x\Vert \le \left( 1+\frac{1-\Vert x\Vert }{2\Vert x\Vert } \right) \Vert x\Vert =\frac{1+\Vert x\Vert }{2 \Vert x\Vert } \Vert x\Vert = \frac{1+\Vert x\Vert }{2} < 1, \end{aligned}$$

so clearly $\Vert (1+\varepsilon u) x\Vert \le (1+\varepsilon ) \Vert x\Vert < 1$ and since $\Vert y\Vert \le \Vert x\Vert$

$$\begin{aligned} \Vert (1+\varepsilon u) y\Vert \le (1+\varepsilon ) \Vert y\Vert \le (1+\varepsilon ) \Vert x\Vert < 1. \end{aligned}$$

By Lemma 3.1, there exists $u_0 \in \Pi$, such that:

$$\begin{aligned} |\mathcal {R}f(x)-\mathcal {R}f(y)| \le \frac{2\Vert x\Vert }{1-\Vert x\Vert } \Vert f\Vert _\mathcal {I}\beta _E((1+\varepsilon u_0) x,(1+\varepsilon u_0)y). \end{aligned}$$

Take $z_0=1+\varepsilon u_0$ which satisfies

$$\begin{aligned} |z_0| \le 1+\varepsilon =1+\frac{1-\Vert x\Vert }{2\Vert x\Vert }= \frac{1+\Vert x\Vert }{2\Vert x\Vert }. \end{aligned}$$

By Theorem 2.7, we have that $z_0 x,z_0 y \in B_E$ and:

$$\begin{aligned} \rho _E(z_0 x,z_0y) \le 2 |z_0| \rho _E(x,y). \end{aligned}$$

(3.2)

Denote

$$\begin{aligned} C:=(1-\Vert x\Vert ^2) |\mathcal {R}f(x)-\mathcal {R}f(y)|, \end{aligned}$$

(3.3)

so we have

$$\begin{aligned} C \le (1-\Vert x\Vert ^2) \frac{2\Vert x\Vert }{1-\Vert x\Vert } \Vert f\Vert _\mathcal {I}\beta _E(z_0x,z_0y). \end{aligned}$$

Notice that using Lemma 3.2, we have

$$\begin{aligned} \beta _E (z_0 x,z_0y) \le \frac{\rho _E(z_0x,z_0y)}{1-\rho _E(z_0 x,z_0y)}, \end{aligned}$$

so we obtain

$$\begin{aligned} C\le & {} (1+\Vert x\Vert )(1-\Vert x\Vert ) \frac{2\Vert x\Vert }{1-\Vert x\Vert } \Vert f\Vert _\mathcal {I}\frac{\rho _E(z_0x,z_0y)}{1-\rho _E(z_0 x,z_0y)} \\\le & {} 4 \Vert x\Vert \Vert f\Vert _\mathcal {I}\frac{\rho _E(z_0x,z_0y)}{1-\rho _E(z_0 x,z_0y)} = \frac{4 \Vert x\Vert \Vert f\Vert _\mathcal {I}}{\frac{1}{\rho _E(z_0 x,z_0y)}-1}, \end{aligned}$$

so

$$\begin{aligned} \left( \frac{1}{\rho _E(z_0 x,z_0y)}-1\right) C \le 4 \Vert x\Vert \Vert f\Vert _\mathcal {I}\leftrightarrow \frac{C}{\rho _E(z_0 x,z_0y)} -C \le 4 \Vert x\Vert \Vert f\Vert _\mathcal {I}. \end{aligned}$$

Bearing in mind that $\Vert f\Vert _\mathcal {B}\le \Vert f\Vert _\mathcal {I}$ (see (1.6)), we have

$$\begin{aligned} C\le & {} (1-\Vert x\Vert ^2) |\mathcal {R}f(x)|+(1-\Vert x\Vert ^2) |\mathcal {R}f(y)|\\\le & {} (1-\Vert x\Vert ^2) \Vert \nabla f(x) \Vert \Vert x\Vert +(1-\Vert y\Vert ^2) \Vert \nabla f(y)\Vert \Vert y\Vert \le 2\Vert f\Vert _\mathcal {B}\Vert x\Vert \le 2 \Vert x\Vert \Vert f\Vert _\mathcal {I}, \end{aligned}$$

so

$$\begin{aligned} \frac{C}{\rho _E(z_0 x,z_0y)} \le 4 \Vert x\Vert \Vert f\Vert _\mathcal {I}+ C \le 4 \Vert x\Vert \Vert f\Vert _\mathcal {I}+ 2 \Vert x\Vert \Vert f\Vert _\mathcal {I}= 6\Vert x\Vert \Vert f\Vert _\mathcal {I}, \end{aligned}$$

and we conclude $C \le 6 \Vert x\Vert \Vert f\Vert _\mathcal {I}\rho _E(z_0 x,z_0y)$.

Finally, we apply inequality (3.2), and since $|z_0| \le \frac{1+\Vert x\Vert }{2\Vert x\Vert }$, we obtain

$$\begin{aligned} C\le & {} 6 \Vert x\Vert \Vert f\Vert _\mathcal {I}2 |z_0| \rho _E(x,y) \le 12 \Vert x\Vert \Vert f\Vert _\mathcal {I}\frac{1+\Vert x\Vert }{2\Vert x\Vert } \rho _E(x,y) \\= & {} 12 \Vert f\Vert _\mathcal {I}\frac{1+\Vert x\Vert }{2} \rho _E(x,y) \le 12 \Vert f\Vert _\mathcal {I}\rho _E(x,y), \end{aligned}$$

and we are done. $\square$

Theorem 3.4

Let $f \in \mathcal {B}(B_E)$ and $x,y \in B_E$. Then

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

Proof

We call

$$\begin{aligned} F:=|(1-\Vert x\Vert ^2) \mathcal {R}f(x)-(1-\Vert y\Vert ^2) \mathcal {R}f(y)|, \end{aligned}$$

(3.4)

and suppose without loss of generality that $\Vert x\Vert \ge \Vert y\Vert$. We have that

$$\begin{aligned} F= & {} |(1-\Vert x\Vert ^2) (\mathcal {R}f(x)-\mathcal {R}f(y))-(\Vert x\Vert ^2-\Vert y\Vert ^2) \mathcal {R}f(y)| \nonumber \\\le & {} (1-\Vert x\Vert ^2) |\mathcal {R}f(x)-\mathcal {R}f(y)| +(\Vert x\Vert ^2-\Vert y\Vert ^2) |\mathcal {R}f(y)|. \end{aligned}$$

(3.5)

Since $\Vert x\Vert ^2-\Vert y\Vert ^2=(\Vert x\Vert +\Vert y\Vert )(\Vert x\Vert -\Vert y\Vert ) \le 2 (\Vert x\Vert -\Vert y\Vert )$ and bearing in mind that $\rho _E(\Vert x\Vert ,\Vert y\Vert ) \le \rho _E(x,y)$, we obtain

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

By Lemma 3.3, we know that

$$\begin{aligned} (1-\Vert x\Vert ^2) |\mathcal {R}f(x)-\mathcal {R}f(y)| \le 12 \Vert f\Vert _\mathcal {I}\rho _E(x,y), \end{aligned}$$

so from (3.5), we conclude

$$\begin{aligned} F \le 12 \Vert f\Vert _\mathcal {I}\rho _E(x,y)+2\Vert f\Vert _\mathcal {I}\rho _E(x,y)=14 \Vert f\Vert _\mathcal {I}\rho _E(x,y), \end{aligned}$$

and we are done. $\square$

Hence, we obtain the result which proves the Lipschitz continuity of the mapping $x \mapsto (1-\Vert x\Vert ^2)|Rf(x)|$ for $x \in B_E$.

Corollary 3.5

Let E be a complex Hilbert space. The function $x \mapsto (1-\Vert x\Vert ^2) |\mathcal {R}f(x)|$ for $x \in B_E$ is Lipschitz with respect to the pseudohyperbolic distance and the following inequality holds:

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

Proof

Applying Theorem 3.4, it is clear that

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

and we are done. $\square$

3.2 Results on interpolating sequences for the Bloch space

Recall that a sequence $(x_n) \subset B_E {\setminus } \{0\}$ is said to be interpolating for the Bloch space $\mathcal {B}(B_E)$ if, for any bounded sequence $(a_n)$ of complex numbers, there exists $f \in \mathcal {B}(B_E)$ such that $(1-\Vert x_n\Vert ^2) \mathcal {R}f(x_n)=a_n$. Attele studied in [2] this kind of interpolation for the classical Bloch space $\mathcal {B}$ and the finite- and infinite-dimensional setting was studied in [4]. We provide a new approach to prove that a necessary condition for a sequence $(x_n) \subset B_E$ to be interpolating for $\mathcal {B}(B_E)$ is to be separated for the pseudohyperbolic distance $\rho _E$.

Proposition 3.6

Let E be a complex Hilbert space. If $(x_n) \subset B_E {\setminus } \{0\}$ is interpolating for $\mathcal {B}(B_E)$, then there exists $C>0$, such that $\rho (x_k,x_j) \ge C$ for any $k \ne j$, $k,j \in \mathbb {N}$.

Proof

Since $(x_n) \subset B_E {\setminus } \{0\}$ is interpolating, there exists a sequence $(f_n) \subset \mathcal {B}(B_E)$, such that

$$\begin{aligned} (1-\Vert x_n\Vert ^2) \mathcal {R}f_n(x_n) =1 \ \text{ and } \ (1-\Vert x_k\Vert ^2) \mathcal {R}f_n(x_k) =0 \ \text{ if } \ k \ne n. \end{aligned}$$

The operator $T: \mathcal {B}(B_E) \rightarrow \ell _{\infty }$ given by $T(f)=((1-\Vert x_n\Vert ^2) \mathcal {R}f(x_n))$ is surjective, so by the Open Mapping Theorem, there exists $M >0$, such that $\Vert f\Vert _R \le M \sup _{j \in \mathbb {N}} (1-\Vert x_j\Vert ^2) |\mathcal {R}f(x_j)|$, so $\Vert f_n\Vert _R \le M$ for any $n \in \mathbb {N}$. Applying Theorem 3.4, we have

$$\begin{aligned}{} & {} |(1-\Vert x_n\Vert ^2) \mathcal {R}f_n(x_n)-(1-\Vert x_k\Vert ^2) \mathcal {R}f_n(x_k)| \le 14 \Vert f_n\Vert _\mathcal {I}\rho _E(x_k,x_j) \\{} & {} \quad \le 14 A_0 \Vert f_n\Vert _\mathcal {R}\rho _E(x_k,x_j) \le 14 A_0 M \rho _E(x_k,x_j). \end{aligned}$$

Hence, $1-0 \le 14 A_0 M \rho _E(x_k,x_j)$, and we conclude that

$$\begin{aligned} \rho _E(x_k,x_j) \ge \frac{1}{14 A_0 M}, \end{aligned}$$

so we are done. $\square$

Attele (see [2]) also proved that any interpolating sequence $(z_n) \subset \textbf{D}$ for $\mathcal {B}$ can be slightly perturbed and the sequence remains interpolating. By means of Theorem 3.4, we adapt his proof and generalize the result to the case when we deal with any complex Hilbert space E.

Theorem 3.7

If $(x_n) \subset B_E {\setminus } \{ 0 \}$ is an interpolating sequence for $\mathcal {B}(B_E)$, then there exists $\delta >0$, such that if $(y_n) \subset B_E$ satisfies that $\sup _{n \in \mathbb {N}} \rho _E(x_n,y_n)<\delta$, then $(y_n)$ is also an interpolating sequence for $\mathcal {B}(B_E)$.

Proof

Since $(x_n)$ is interpolating, the operator $T: \mathcal {B}(B_E) \rightarrow \ell _\infty$ given by $T(f)=((1-\Vert x_n\Vert ^2) \mathcal {R}f(x_n))$ is surjective. Hence, its adjoint $T^*: \ell _{\infty }^* \rightarrow (\mathcal {B}(B_E))^*$ is injective and it has closed range. In particular, $T^*$ is left-invertible. The set of left-invertible elements is open in the Banach algebra of linear operators from $\ell _{\infty }^*$ to $(\mathcal {B}(B_E))^*$. Therefore, there exists $\delta$, such that if $\Vert T^*-R\Vert < 14 A_0 \delta$, then R is left-invertible. If we consider $S(f)=((1-\Vert y_n\Vert ^2) \mathcal {R}f(y_n))$, then by Theorem 3.4

$$\begin{aligned} \Vert (T-S)(f)\Vert _{\infty }= & {} \sup _{n \in \mathbb {N}} |(1-\Vert x_n\Vert ^2) \mathcal {R}f(x_n) -(1-\Vert y_n\Vert ^2) \mathcal {R}f(y_n)| \\\le & {} 14 \rho _E(x_n,y_n) \Vert f\Vert _\mathcal {I}\le 14 A_0 \Vert f\Vert _\mathcal {R}\rho _E(x_n,y_n) < 14 A_0 \delta \Vert f\Vert _\mathcal {R}, \end{aligned}$$

so $\Vert T-S\Vert < 14 A_0 \delta$, and hence, $\Vert T^*-S^*\Vert =\Vert T-S\Vert < 14 A_0 \delta$. We conclude that $S^*$ is left-invertible, and hence, S is surjective, as we wanted. $\square$