The Constant of Interpolation in Bloch Type Spaces

Abstract

It is known that there exists a constant $0<\Delta _1 < 1$ such that any $\Delta _1$-separated sequence for the pseudohyperbolic distance in the open unit disk $\textbf{D}$ of $\mathbb {C}$ is interpolating for the classical Bloch space $\mathcal {B}$. We will prove that $0.8114< \Delta _1 < 0.9785$ and we will also generalize this result for Bloch type spaces $\mathcal {B}_{v_p}$ for $v_p(z)=(1-|z|^2)^p$. In particular, we will provide a construction to calculate an estimate of the lower and upper bounds for the corresponding constant of separation $\Delta _p$ for these spaces. We also prove that $\Delta _p$ tends to 1 when $p \rightarrow \infty$.

1 Introduction and Background

Let $\textbf{D}$ be the open unit disk of the complex plane $\mathbb {C}$. Recall that the classical Bloch space $\mathcal {B}$ is given by the space of analytic functions $f: \textbf{D}\rightarrow \mathbb {C}$ such that $\Vert f\Vert _B=\sup _{z\in \textbf{D}}(1-|z|^2)|f'(z)| <\infty$. The study of interpolating sequences for the classical Bloch space $\mathcal {B}$ was started by Attele in [1] and Madigan and Matheson in [9]. The study of interpolating sequences for Bloch type spaces $\mathcal {B}_v$ was introduced in [10], where the weight $1-|z|^2$ was substituted by a more general weight v(z). In this work, we will deal with weights $v_p(z)=(1-|z|^2)^p$ for $1 \le p < \infty$.

Recall that a sequence $(z_n)\subset \textbf{D}$ is said to be an interpolating sequence for $\mathcal {B}_{v}$ if for any $(a_n) \in \ell _\infty$ there exists $f\in \mathcal {B}_{v}$ such that $v(z_n) f'(z_n)=a_n \text{ for } \text{ any } n \in \mathbb {N}.$ The interpolating operator $T:\mathcal {B}_v \rightarrow \ell _\infty$ is defined by $T(f)=(v(z_n)f'(z_n))$. Notice that T is clearly linear and $(z_n)$ is interpolating for $\mathcal {B}_v$ if and only if T is surjective or, equivalently, if there exists a mapping $S: \ell _\infty \rightarrow \mathcal {B}_v$ such that $T \circ S=Id_{\ell _\infty }$. Given $\delta >0$, the sequence $(z_n)$ is said to be $\delta$-separated (or simply separated) if $\rho (z_k,z_j) \ge \delta$ for $k \ne j$, where the pseudohyperbolic distance $\rho$ is defined by:

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

It is well known that $\rho (z,w)=|\varphi _z(w)|$ where $\varphi _z$ is the automorphism from $\textbf{D}$ onto itself given by $\varphi _z(w)=(z-w)/(1-\bar{z}w)$. For any $z \in \textbf{D}$, we have that $\rho (\varphi _z(w_1),\varphi _z(w_2))=\rho (w_1,w_2)$, $w_1,w_2 \in \textbf{D}$. It is also well known that:

$$\begin{aligned} 1-\rho (z,w)^2=\frac{(1-|z|^2)(1-|w|^2)}{|1-\bar{z}w|^2} \end{aligned}$$

(1.1)

and these facts about $\rho$ will be used in the sequel. The constant of separation of $(z_n)$ is given by $S:=\inf _{n\ne k}\rho (z_n,z_k)$. It is well known that if $(z_n) \subset \textbf{D}$ is interpolating, then it is separated (see for instance Corollary 3 in [1]). On the other side, Attele (see Proposition 4 in [1]) and Madigan and Matheson (see Proposition 1 in [9]) proved that there exists a universal constant $\Delta _1$ such that any $\Delta _1$-separated sequence in $\textbf{D}$ is interpolating for $\mathcal {B}$. Making calculations from these works, it is not difficult to make an upper estimate for this constant. Nevertheless, this estimate is higher than 0.99. In Theorem 5 in [4], the authors improved this upper bound until 0.9882.

In this work, we will improve the upper estimate and we will calculate a lower estimate for $\Delta _1$, proving that $0.8114< \Delta _1 < 0.9785$. Furthermore, we will generalize the result for Bloch type spaces $\mathcal {B}_{v_p}$: we will prove that sufficiently separated sequences on the open unit disk are interpolating for $\mathcal {B}_{v_p}$ and will provide bounds for the corresponding universal constants $\Delta _p$.

2 Results

2.1 Lower Estimate

Let $a >1$ and $b >0$ and consider the sequence:

$$\begin{aligned} \Gamma (a,b)=\left\{ \frac{a^m(bn+i)-i}{a^m(bn+i)+i} \right\} _{m,n \in \mathbb {Z}}. \end{aligned}$$

(2.1)

K. Seip proved that $\Gamma (a,b)$ is interpolating for $\mathcal {B}_{v_p}$ if and only if $\frac{2 \pi }{b \log a} < \frac{1}{p}$. This is based on the study of the positive density $D^+$ of the sequence (see [11]). As Schuster observed [12], the constant of separation of $\Gamma (a,b)$ is given by:

$$\begin{aligned} S(a,b)=\min \left\{ \frac{a-1}{a+1},\frac{b}{\sqrt{b^2+4}}\right\} . \end{aligned}$$

Proposition 2.1

There exists a sequence $(z_n) \subset \textbf{D}$ which is $\delta$-separated for $\delta =0.811458$ and $(z_n)$ is not interpolating for the Bloch space $\mathcal {B}$. Hence, $\Delta _1 > 0.811458$.

Proof

Consider the sequence $\Gamma (a,b)$ in (2.1). Take b=$\frac{2\pi }{log a}$ for $a > 1$ and $b >0$. Then, we have $\frac{2 \pi }{b \log a}=1$ so the sequence is not interpolating for $\mathcal {B}$ by the comments above. Notice that functions:

$$\begin{aligned} \frac{a-1}{a+1}=1-\frac{2}{a+1} \ \text{ and } \frac{b}{\sqrt{b^2+4}}=\sqrt{1-\frac{4}{b^2+4}}=\sqrt{1-\frac{4}{(\frac{2 \pi }{\log a})^2+4}} \end{aligned}$$

are non-decreasing and non-increasing, respectively. Hence, the highest value of S(a, b) will be got when:

$$\begin{aligned} \frac{a-1}{a+1}=\frac{b}{\sqrt{b^2+4}} \end{aligned}$$

which is equivalent to:

$$\begin{aligned} g(a):=1-\frac{2}{a+1}-\sqrt{1-\frac{4}{b^2+4}}=0 \end{aligned}$$

and bearing in mind that $b=\frac{2 \pi }{\log a}$, the function g(a) is clearly non-decreasing continuous with respect to a. Since $\lim _{a \rightarrow 1^{+}}g(a)=-1$ and $\lim _{a \rightarrow +\infty } g(a)=1$, there exists a unique $a >1$ such that $g(a)=0$. Indeed, $a \approx 9.60773$ and $b \approx 2.77701$, which yields:

$$\begin{aligned} S(a,b) =\min \left\{ \frac{a-1}{a+1},\frac{b}{\sqrt{b^2+4}}\right\} \approx 0.811458. \end{aligned}$$

Hence, we have a sequence whose constant of separation is 0.811458 but fails to be interpolating for $\mathcal {B}$.$\square$

We can easily generalize Proposition 2.1 to $B_{v_p}$ spaces for $1 \le p < +\infty$ taking $b=2p \pi /\log a$ and following the same pattern:

Proposition 2.2

For any $1 \le p < +\infty$, there exist $a >1$ and $b>0$ such that the sequence $\Gamma (a,b)$ is separated but fails to be interpolating for $\mathcal {B}_{v_p}$. In particular, these values can be chosen such that the constant of separation tends to 1 when $p \rightarrow \infty$.

For some particular p, it is a straightforward calculation to determine the value of a and S(a, b) such that $b=\frac{2p \pi }{\log a}$ which yields examples of more and more separated sequences which are not interpolating for $\mathcal {B}_{v_p}$. For instance:

$$\begin{aligned} \begin{array}{ | c | c | c | } \hline {\textbf {p}} &{} {\textbf {a}} &{} {\textbf {S(a,b)}} \\ \hline 1 &{} 9.60773 &{} 0.811458 \\ \hline 2&{} 19.7151 &{} 0.903452 \\ \hline 3 &{} 31.707 &{} 0.938851 \\ \hline 4 &{} 45.3776 &{}0.956876 \\ \hline \end{array} \end{aligned}$$

Remark 2.3

There are $\Delta _p$-separated sequences $(z_n) \subset \textbf{D}$ which fail to be interpolating for $\mathcal {B}_{v_p}$ and $\Delta _p \rightarrow 1$ when $p \rightarrow \infty$. To show it, we will prove that $S(a,b) \rightarrow 1$ when $p \rightarrow \infty$. Indeed, from:

$$\begin{aligned} 1-\frac{2}{a+1}=\sqrt{1-\frac{4}{b^2+4}} \end{aligned}$$

and since $p \rightarrow \infty$ then $b \rightarrow \infty$ so the right term tends to 1. In order to get this, we need that the left term also tends to 1 which is only possible if $a \rightarrow \infty$ (since $b=2 p \pi /\log a$, we can take a bigger than p to get it and b also tends to infinity). Hence, $S(a,b) \rightarrow 1$ and we are done.

2.2 Upper Estimate

Given $a\in \textbf{D}$ and $0<r<1$, the pseudohyperbolic open disks centered at a and radius r are denoted by $D_\rho (a,r)=\{ z \in \textbf{D}: \rho (z,a) < r \}$. A standard calculation (see for instance [14]) shows that $D_\rho (a,r)$ is the euclidean disk with center $\frac{1-r^2}{1-r^2|a|^2}a$ and radius $\frac{1-|a|^2}{1-|a|^2r^2}r$. The following lemma is just an easy calculation:

Lemma 2.4

Let $z\in \textbf{D}$ and $0<r<1$. If $w\in D_{\rho }(z,r)$ then:

$$\begin{aligned} \frac{1-r^2}{4}(1-|z|^2)< (1-|w|^2) <\frac{4}{1-r^2}(1-|z|^2). \end{aligned}$$

Proof

An easy calculation shows that $w\in D_\rho (z,r)$ if and only if $\rho (z,w)^2<r^2$ if and only if $1-r^2<1-\rho (z,w)^2=\frac{(1-|z|^2)(1-|w|^2)}{|1-\overline{w}z|^2}$ which is equivalent to:

$$\begin{aligned}{} & {} \frac{1-r^2}{(1-|z|^2)(1-|w|^2)}<\frac{1}{|1-\overline{w}z|^2} \text{ if } \text{ and } \text{ only } \text{ if } \\{} & {} \frac{(1-r^2)^2|1-\overline{w}z|^2}{(1-|z|^2)^2(1-|w|^2)^2}<\frac{1}{|1-\overline{w}z|^2} \end{aligned}$$

and since $|1-\overline{w}z|\ge (1-|z|)$, it follows that:

$$\begin{aligned}{} & {} \frac{4}{(1-|z|^2)^2}>\frac{1}{(1-|z|)^2}\ge \frac{1}{|1-\overline{w}z|^2}>\frac{(1-r^2)^2|1-\overline{w}z|^2}{(1-|z|^2)^2(1-|w|^2)^2} \\{} & {} \quad> \frac{(1-r^2)^2(1-|z|)^2}{(1-|z|^2)^2(1-|w|^2)^2}=\frac{(1-r^2)^2}{(1+|z|)^2(1-|w|^2)^2}>\frac{(1-r^2)^2}{4(1-|w|^2)^2} \end{aligned}$$

and we are done. Changing z with w and repeating the proof we obtain the second inequality. $\square$

Lemma 2.5

Let $a \ne b \in \textbf{D}$, $\rho (a,b) \ge \delta$ and consider:

$$\begin{aligned} r_\delta =\frac{\delta }{1+\sqrt{1-\delta ^2}}. \end{aligned}$$

Then, $D_\rho (a,r_\delta ) \cap D_\rho (b,r_\delta ) = \emptyset$. In addition, if $\rho (a,b)=\delta$ then the disks $\overline{D_\rho (a,r_\delta )}$ and $\overline{D_\rho (b,r_\delta )}$ are externally tangent in the complex plane $\mathbb {C}$ so the radius $r_\delta$ is optimal to separate both disks.

Proof

It is an easy calculation that $r_\delta$ is the solution of the equation:

$$\begin{aligned} \frac{2r_\delta }{1+r_\delta ^2}=\delta . \end{aligned}$$

If $w \in D_\rho (a,r_\delta ) \cap D_\rho (b,r_\delta )$ then:

$$\begin{aligned} \rho (a,b) \le \frac{\rho (a,w)+\rho (b,w)}{1+\rho (a,w) \rho (b,w)} < \frac{2r_\delta }{1+r_\delta ^2}=\delta \end{aligned}$$

where last inequality is clear since given any $0 \le a < 1$ the real function $(x+a)/(1+ax)$ is non-decreasing for $0 \le x < 1$. This is a contradiction, so the disks are disjoint.

Now notice that $z \in \overline{D_\rho (a,r_\delta )} \cap \overline{D_\rho (b,r_\delta )}$ if and only if $\rho (a,z) \le r_\delta$ and $\rho (b,z) \le r_\delta$. Since $\rho$ is invariant by automorphisms and $\varphi _a(a)=0$, this is equivalent to: $\rho (0,\varphi _a(z)) \le r_\delta$ and $\rho (\varphi _a(b),\varphi _a(z)) \le r_\delta$ if and only if $\varphi _a(z) \in \overline{D_\rho (0,r_\delta )} \cap \overline{D_\rho (\varphi _a(b),r_\delta )}$. Hence z belongs or not to $\overline{D_\rho (a,r_\delta )}$ and $\overline{D_\rho (b,r_\delta )}$ if and only if $\varphi _a(z)$ belongs or not to $\overline{D_\rho (0,r_\delta )}$ and $\overline{D_\rho (\varphi _a(b),r_\delta )}$ respectively. Notice that:

$$\begin{aligned} \overline{D_\rho (0,r_\delta )}= & {} \overline{D_{| \cdot |} (0,r_\delta )} \ \ \text{ and } \\ \overline{D_\rho (\varphi _a(b),r_\delta )}= & {} \overline{D_{| \cdot |} \left( \frac{1-r_\delta ^2}{1-r_\delta ^2 |\varphi _a(b)|^2} \varphi _a(b),\frac{1-|\varphi _a(b)|^2}{1-r_\delta ^2 |\varphi _a(b)|^2}r_\delta \right) } \end{aligned}$$

where $D_{| \cdot |}$ denotes the corresponding euclidean disks, so we will prove that these disks are externally tangent. For this, it is sufficient to prove that that the euclidean distance between both centers equals to the sum of both radius, that is:

$$\begin{aligned} \frac{1-r_\delta ^2}{1-r_\delta ^2 |\varphi _a(b)|^2} |\varphi _a(b)|= & {} r_\delta +\frac{1-|\varphi _a(b)|^2}{1-r_\delta ^2 |\varphi _a(b)|^2}r_\delta \text{ if } \text{ and } \text{ only } \text{ if } \\ \frac{1-r_\delta ^2}{1-r_\delta ^2 |\varphi _a(b)|^2} |\varphi _a(b)|= & {} r_\delta \left( 1+\frac{1-|\varphi _a(b)|^2}{1-r_\delta ^2 |\varphi _a(b)|^2}\right) \\= & {} r_\delta \frac{2- r_\delta ^2 |\varphi _a(b)|^2-|\varphi _a(b)|^2}{1-r_\delta ^2 |\varphi _a(b)|^2} \end{aligned}$$

and bearing in mind that $|\varphi _a(b)|=\rho (a,b)=\delta$, this is equivalent to:

$$\begin{aligned} (1-r^2) \delta =r_\delta (2- r_\delta ^2 |\varphi _a(b)|^2-|\varphi _a(b)|^2)=r_\delta (2-(r_\delta ^2+1)\delta ^2) \end{aligned}$$

and dividing by $1+r_\delta ^2$ we have:

$$\begin{aligned} \frac{1-r_\delta ^2}{1+r_\delta ^2} \delta = \frac{2r_\delta }{1+r_\delta ^2}- r_\delta \delta ^2 \end{aligned}$$

and since $\frac{1-r_\delta ^2}{1+r_\delta ^2}=\sqrt{1-\delta ^2}$ because $\left( \frac{2r_\delta }{1+r_\delta ^2}\right) ^2+\left( \frac{1-r_\delta ^2}{1+r_\delta ^2}\right) ^2=1$, the equality is equivalent to:

$$\begin{aligned} \sqrt{1-\delta ^2} \delta = \delta -r_\delta \delta ^2 \end{aligned}$$

and dividing by $\delta$ and solving for $r_\delta$ we obtain:

$$\begin{aligned} r_\delta =\frac{1-\sqrt{1-\delta ^2}}{\delta }=\frac{\delta }{1+\sqrt{1-\delta ^2}} \end{aligned}$$

and we are done. $\square$

The following result is an easy consequence of the mean value for complex analytic functions f on $\textbf{D}$:

Proposition 2.6

If $f: \textbf{D}\rightarrow \mathbb {C}$ is analytic then for any $a\in \textbf{D}$ and $0<r<1$ we have:

$$\begin{aligned} |f(a)|(1-|a|^2)^2\le \frac{1}{\pi r^2}\int _{D_{\rho }(a,r)} |f(z)| \mathrm{{d}}z. \end{aligned}$$

Proof

By Cauchy’s integral formula, we have for any $0<s<1$ that

$$\begin{aligned} f(0)=\frac{1}{2\pi }\int _{0}^{2\pi } f(se^{i\sigma }) \mathrm{{d}}\sigma \Rightarrow |f(0)|\le \frac{1}{2\pi }\int _{0}^{2\pi } |f(se^{i\sigma })|\mathrm{{d}}\sigma \end{aligned}$$

and using polar coordinates:

$$\begin{aligned} \frac{1}{\pi r^2}\int _{D(0,r)} |f(z)|\mathrm{{d}}z=\frac{1}{\pi r^2}\int _{0}^r\int _{0}^{2\pi } s|f(se^{i\sigma })|\mathrm{{d}}\sigma \mathrm{{d}}s \ge \frac{2}{ r^2} |f(0)|\int _{0}^r s\mathrm{{d}}s=|f(0)|. \end{aligned}$$

Applying this inequality to the function $h:=(f\circ \varphi _{a})(\varphi '_{a})^2$, we obtain as a consequence of the change of variable formula and the Cauchy-Riemman equations that

$$\begin{aligned} |f(a)|(1-|a|^2)^2= & {} |h(0)|\le \frac{1}{\pi r^2}\int _{D(0,r)} |h(z)| \mathrm{{d}}z\\= & {} \frac{1}{\pi r^2}\int _{D(0,r)} |f(\varphi _{a}(z))||\varphi '_{a}(z)|^2 \mathrm{{d}}z= \\= & {} \frac{1}{\pi r^2}\int _{\varphi _{a}(D(0,r))} |f(z)| \mathrm{{d}}z = \frac{1}{\pi r^2}\int _{D_{\rho }(a,r)} |f(z)| \mathrm{{d}}z \end{aligned}$$

and we are done.

The following results are well-known (see Theorem 1.7 in [8] or Lemma 3.10 in [14]). The measure A will denote the classical Borel measure defined on $\mathbb {C}$. $\square$

Proposition 2.7

We have the following results:

• a) If $u >-1$ and $v \in \mathbb {R}$ such that v/2 is neither 0 nor a negative integer then:

  $$\begin{aligned} \int _{\textbf{D}} \frac{(1-|w|^2)^u}{|1-\bar{z} w|^{v}}\mathrm{{d}}A(w)=\frac{\pi \Gamma (u+1)}{\Gamma (\frac{v}{2})^2} \sum _{m=0}^\infty \frac{\Gamma (m+\frac{v}{2})^2}{m! \Gamma (m+u+2)} |z|^{2m}. \end{aligned}$$

• b) For any $0 \le x < 1$ and $s \in \mathbb {R}$, s neither 0 nor a negative integer, we have:

  $$\begin{aligned} \frac{1}{(1-x)^{s}}=\sum _{m=0}^\infty \frac{\Gamma (m+s)}{m! \Gamma (s)} x^m \end{aligned}$$

Proposition 2.8

Let $p,q,r \in \mathbb {R}$ such that $p > 0$, $p \ge r-2 \ge \frac{q-1}{2} >-1$ and $r-2 >0$. If $(z_n)\subset \textbf{D}$ is a separated sequence then:

$$\begin{aligned} E(z)=\sup _{z\in \textbf{D}}\left\{ \sum _{n=1}^{\infty }\frac{(1-|z_n|^2)^r (1-|z|^2)^p}{|1-\overline{z_n}z|^{q+1}} \right\} \end{aligned}$$

is bounded on $\textbf{D}$.

Proof

Suppose that $\rho (z_n,z_k) \ge \delta >0$ for $n\ne k$ and consider the radius $r_\delta$ given in Lemma 2.5. Fix $z\in \textbf{D}$ and define the function $f_z(w):=\frac{1}{(1-\overline{z}w)^{q+1}}$, which is holomorphic on $\textbf{D}$. Applying Proposition 2.6 and evaluating at $z_n$ we have:

$$\begin{aligned} \frac{(1-|z_n|^2)^2}{|1-\overline{z}z_n|^{q+1}}=|f_z(z_n)|(1-|z_n|^2)^2\le \frac{1}{\pi r_\delta ^2}\int _{D_{\rho }(z_n,r_\delta )} |f_z(w)|\ \mathrm{{d}}w. \end{aligned}$$

By hypothesis $D_{\rho }(z_n,r_\delta )\cap D_{\rho }(z_k,r_\delta )=\emptyset$ for $n\ne k$, so:

$$\begin{aligned} E(z):= & {} \sum _{n=1}^{\infty }\frac{(1-|z_n|^2)^r(1-|z|^2)^p}{|1-\overline{z}z_n|^{q+1}} \le (1-|z|^2)^p\sum _{n=1}^{\infty }\left( \frac{1}{\pi r_\delta ^2}(1-|z_n|^2)^{r-2}\right. \\{} & {} \left. \int _{D_{\rho }(z_n,r_\delta )} |f_z(w)|\mathrm{{d}}w\right) \\\le & {} (1-|z|^2)^p \sum _{n=1}^{\infty }\left( \frac{4^{r-2}}{(1-r_\delta ^2)^{r-2} \pi r_\delta ^2}\int _{D_{\rho }(z_n,r_\delta )} (1-|w|^2)^{r-2}|f_z(w)|\mathrm{{d}}w\right) \\= & {} \frac{4^{r-2}(1-|z|^2)^p}{(1-r_\delta ^2)^{r-2}\pi r_\delta ^2} \int _{\bigcup _{n=1}^\infty D_{\rho }(z_n,r_\delta )} \frac{(1-|w|^2)^{r-2}}{|1-\overline{z}w|^{q+1}}\mathrm{{d}}w \\\le & {} \frac{4^{r-2}(1-|z|^2)^{p}}{(1-r_\delta ^2)^{r-2}\pi r_\delta ^2}\int _{\textbf{D}} \frac{(1-|w|^2)^{r-2}}{|1-\overline{z}w|^{q+1}}\mathrm{{d}}w \end{aligned}$$

being the second inequality a consequence of Lemma 2.4. By Proposition 2.7 a), we have:

$$\begin{aligned} E(z) \le \frac{4^{r-2}(1-|z|^2)^{p}}{(1-r_\delta ^2)^{r-2} \pi r_\delta ^2}\frac{\pi \Gamma (r-1)}{\Gamma (\frac{q+1}{2})^2} \sum _{m=0}^\infty \frac{\Gamma (m+\frac{q+1}{2})^2}{m! \Gamma (m+r)} |z|^{2m}. \end{aligned}$$

Therefore, for any $z\in \textbf{D}$, we have that:

$$\begin{aligned} E(z)\le & {} \frac{4^{r-2}(1-|z|^2)^{p}}{(1-r_\delta ^2)^{r-2} r_\delta ^2}\frac{(r-2)\Gamma (r-2)^2}{\Gamma (\frac{q+1}{2})^2 \Gamma (r-2)} \\{} & {} \times \left( \sum _{m=0}^2 \frac{\Gamma (m+\frac{q+1}{2})^2}{m! \Gamma (m+r)} |z|^{2m}\right. \\{} & {} \quad \left. +\sum _{m=3}^\infty \frac{(m+\frac{q-1}{2})^2}{(m+r-1)(m+r-2)}\frac{\Gamma (m+\frac{q-1}{2})^2}{m! \Gamma (m+r-2)} |z|^{2m}\right) . \end{aligned}$$

$\bullet$ Case $r=2$. We obtain:

$$\begin{aligned} E(z)\le & {} \frac{(1-|z|^2)^p}{\pi r_\delta ^2} \frac{1}{\Gamma (\frac{q+1}{2})^2} \left( \sum _{m=0}^2 \frac{\Gamma (m+\frac{q+1}{2})^2}{m! \Gamma (m+2)} |z|^{2m}\right. \\{} & {} \left. +\sum _{m=3}^\infty \frac{(m+\frac{q-1}{2})^2}{(m+1)m}\frac{\Gamma (m+\frac{q-1}{2})^2}{m! \Gamma (m)} |z|^{2m}\right) . \end{aligned}$$

Since $0 \ge \frac{q-1}{2}$ then:

$$\begin{aligned} \frac{(m+\frac{q-1}{2})^2}{(m+1)m} \le 1 \end{aligned}$$

and since the function $\Gamma (x)$ is non-decreasing for $x >1.462$, we have for any $m \ge 3$:

$$\begin{aligned} \frac{\Gamma (m+\frac{q-1}{2})^2}{m! \Gamma (m)} \le \frac{\Gamma (m)^2}{m! \Gamma (m)}=\frac{\Gamma (m)}{m!}=\frac{1}{m} \end{aligned}$$

and bearing in mind that $\sum _{m=1}^{\infty } \frac{1}{m} |z|^{2m}=-\log (1-|z|^2)$, we obtain:

$$\begin{aligned} E(z) \le \frac{(1-|z|^2)^p}{\pi r_\delta ^2} \frac{1}{\Gamma (\frac{q+1}{2})^2} (-\log (1-|z|^2)+P_1(z)) \end{aligned}$$

where $P_1(z)$ is a complex polynomial on |z|. Since $(1-|z|^2) \log (1-|z|^2)$ is bounded on $\textbf{D}$, we are done.

$\bullet$ Case $r \ne 2$. If $r-2 \ge \frac{q-1}{2}$ then:

$$\begin{aligned} \frac{(m+\frac{q-1}{2})^2}{(m+r-1)(m+r-2)} \le 1 \end{aligned}$$

and since the function $\Gamma (x)$ is non-decreasing for $x >1.462$, we have for any $m \ge 3$:

$$\begin{aligned} \frac{\Gamma (m+\frac{q-1}{2})^2}{\Gamma (m+r-2)} \le \frac{\Gamma (m+r-2)^2}{\Gamma (m+r-2)}=\Gamma (m+r-2). \end{aligned}$$

So:

$$\begin{aligned} E(z) \le \frac{4^{r-2}(1-|z|^2)^{p}}{(1-r_\delta ^2)^{r-2} r_\delta ^2}\frac{(r-2) \Gamma (r-2)^2}{\Gamma (\frac{q+1}{2})^2} G(z) \end{aligned}$$

where:

$$\begin{aligned} G(z){} & {} =\sum _{m=0}^\infty \frac{\Gamma (m+r-2)}{m! \Gamma (r-2)}|z|^{2m}-\frac{\Gamma (r-2)}{0! \Gamma (r-2)}|z|^{0}-\frac{\Gamma (1+r-2)}{1! \Gamma (r-2)}|z|^{2}\\{} & {} \quad - \frac{\Gamma (2+r-2)}{2! \Gamma (r-2)}|z|^{4}+\frac{\Gamma (\frac{q+1}{2})^2}{0! \Gamma (r) \Gamma (r-2)}+\frac{\Gamma (\frac{q+3}{2})^2}{1! \Gamma (r+1) \Gamma (r-2)}|z|^2\\{} & {} \quad +\frac{\Gamma (\frac{q+5}{2})^2}{2! \Gamma (r+2) \Gamma (r-2)}|z|^4 \end{aligned}$$

and since $r-2 >0$, by Proposition 2.7 b), we have:

$$\begin{aligned} \sum _{m=0}^\infty \frac{\Gamma (m+r-2)}{m! \Gamma (r-2)}|z|^{2m}=\frac{1}{(1-|z|^2)^{r-2}}. \end{aligned}$$

Considering $p \ge r-2$, we have:

$$\begin{aligned} \frac{(1-|z|^2)^p}{(1-|z|^2)^{r-2}} \le 1 \end{aligned}$$

so:

$$\begin{aligned} E(z) \le \frac{4^{r-2}}{(1-r_\delta ^2)^{r-2} r_\delta ^2}\frac{(r-2) \Gamma (r-2)^2}{\Gamma (\frac{q+1}{2})^2} \left( 1+(1-|z|^2)^{p}P_2(|z|)\right) \end{aligned}$$

where $P_2(|z|)$ is a polynomial depending on |z| so E(z) is uniformly bounded for $z \in \textbf{D}$. $\square$

Hence, we obtain a well-known result about separated sequences:

Corollary 2.9

If $(z_n)\subset \textbf{D}$ is a separated sequence then for every $r >1$ we have:

$$\begin{aligned} \sum _{n=1}^{\infty }(1-|z_n|)^{r}<\infty . \end{aligned}$$

Proof

Take $r >1$, $p >0$ such that $p \ge r-2 \ge \frac{q-1}{2} > -1$. Apply Proposition 2.8 evaluated at $z=0$ and we are done. $\square$

Our main result refines the Madigan and Matheson result to get an upper estimate for $\Delta _1$ (see [9]). In addition, we generalize it for Bloch type spaces $\mathcal {B}_{v_p}$ and provide a procedure to estimate $\Delta _p$ for any $1 \le p < +\infty$.

Theorem 2.1

Let $p \ge 1$. There exists a constant $0<\Delta _p<1$ such that if $(z_n)\subset \textbf{D}$ is $\Delta _p$-separated then it is interpolating for the Bloch space $\mathcal {B}_{v_p}$.

Proof

Let $p \ge 1$ and consider r, q such that $p \ge r-2 \ge \frac{q-1}{2} > -1$ and $p+r=q+1$. For any $\lambda =(\lambda _n)\in \ell _\infty$, define:

$$\begin{aligned} T_\lambda (z):=\sum _{k=1}^\infty \lambda _k\frac{1}{q \overline{z_k}}\frac{(1-|z_k|^2)^r}{(1-\overline{z_k}z)^q}. \end{aligned}$$

If $z_{k_0}=0$ for any $k_0$ (there could only be one term because of the separability) then substitute the corresponding $k_0$-term of the previous series by $\lambda _{k_0} z$. Notice that $T_\lambda$ is well-defined since the series is absolutely and uniformly convergent on compact sets of $\textbf{D}$. To show this, set $\eta :=\inf _{k\in \mathbb {N}}\{|z_n|\mid z_k\ne 0 \}$. Since $(z_k)\subset \textbf{D}$ is separated it does not have accumulation points in $\textbf{D}$ and thus $\eta >0$. Now for any $|z|\le s<1$, we have:

$$\begin{aligned} |T_\lambda (z)|\le \sum _{k=1}^\infty |\lambda _k|\frac{1}{q |z_k|}\frac{(1-|z_k|^2)^r}{(1-|z_k|s)^q}\le \frac{||\lambda ||_\infty }{q\eta }\frac{1}{(1-s)^q}\sum _{k=1}^\infty (1-|z_k|^2)^r<\infty \end{aligned}$$

because of Corollary 2.9. Adding $\lambda _k z$ for a possible $z_k=0$ does not affect to the convergence.

Thus, $T_\lambda$ is holomorphic for any $\lambda \in \ell _\infty$ and $T'_\lambda (z)=\sum _{k=1}^\infty \lambda _k \frac{(1-|z_k|^2)^r}{(1-\overline{z_k}z)^{q+1}}$, which implies together with Proposition 2.8 that:

$$\begin{aligned} \sup _{z\in \textbf{D}} \{ (1-|z|^2)^p|T'_\lambda (z)| \}\le ||\lambda ||_\infty \sup _{z\in \textbf{D}} \left\{ \sum _{k=1}^{\infty }\frac{(1-|z_k|^2)^r(1-|z|^2)^p}{|1-\overline{z_k}z|^{q+1}} \right\} <\infty , \end{aligned}$$

so $T_\lambda \in \mathcal {B}_{v_p}$. Consider $T: \ell _\infty \rightarrow \mathcal {B}_{v_p}$ given by $T(\lambda )=T_{\lambda }(z)$. By the comments above, it is clear that T is a well-defined linear bounded operator. Let $L: \mathcal {B}_{v_p} \rightarrow \ell _\infty$ given by $L(f)=(f'(z_n)(1-|z_n|^2)^p)$. We will prove that if $\Delta _p$ is close enough to 1, then the operator $L\circ T: \ell _\infty \rightarrow \ell _\infty$ is invertible, so in particular L is surjective. We have that:

$$\begin{aligned} L\circ T(\lambda )= & {} \left( (1-|z_n|^2)^p T'_\lambda (z_n)\right) _n\\= & {} \left( \lambda _n (1-|z_n|^2)^{r+p-q-1} +\sum _{k=1,k\ne n }^\infty \lambda _k\frac{(1-|z_k|^2)^r(1-|z_n|^2)^p}{(1-\overline{z_k}z_n)^{q+1}} \right) _n \end{aligned}$$

and since $r+p=q+1$ therefore:

$$\begin{aligned} (L\circ T-Id_{\ell _\infty })(\lambda )=\left( \sum _{k=1,k\ne n }^\infty \lambda _k\frac{(1-|z_k|^2)^r(1-|z_n|^2)^p}{(1-\overline{z_k}z_n)^{q+1}} \right) _n=:(F_n(\lambda )). \end{aligned}$$

Let $||\lambda ||_\infty \le 1$ and consider $\alpha \ge 0$ such that $\alpha < p$, so $p-\alpha \ge r-2-\alpha$ and $r-2-\alpha \ge \frac{q-1}{2}-\alpha$ and $r-2-\alpha , \frac{q+1}{2}-\alpha> 1-\alpha >0$. Since $\rho (z_n,z_k)\ge \Delta _p$ for $n\ne k$ we have for any $0 \le \alpha < p$:

$$\begin{aligned} \frac{(1-|z_k|^2)^{\alpha }(1-|z_n|^2)^{\alpha }}{|1-\overline{z_k}z_n|^{2\alpha }}=(1-\rho ^2(z_n,z_k))^\alpha \le (1-\Delta _p^2)^\alpha \end{aligned}$$

and thus:

$$\begin{aligned} |F_n(\lambda )|{} & {} = \left| \sum _{k=1,k\ne n }^\infty \lambda _k\frac{(1-|z_k|^2)^{r}(1-|z_n|^{2})^p}{(1-\overline{z_k}z_n)^{q+1}}\right| \\{} & {} \le \sum _{k=1,k\ne n }^\infty \frac{(1-|z_k|^2)^r(1-|z_n|^2)^p}{|1-\overline{z_k}z_n|^{q+1}}\\{} & {} \le (1-\Delta _p^2)^\alpha \sum _{k=1,k\ne n }^\infty \frac{(1-|z_k|^2)^{r-\alpha }(1-|z_n|^2)^{p-\alpha }}{|1-\overline{z_k}z_n|^{q+1-2\alpha }}. \end{aligned}$$

If we apply Proposition 2.6 to the function $f_n(z):=\frac{1}{(1-\overline{z_n}z)^{q+1-2\alpha }}$ at the point $z=z_k$ for $k\ne n$ and taking $r_p=\frac{\Delta _p}{1+\sqrt{1-\Delta _p^2}}$ from Proposition 2.5 we obtain:

$$\begin{aligned}{} & {} \frac{(1-|z_k|^2)^{r-\alpha }(1-|z_n|^2)^{p-\alpha }}{|1-\overline{z_k}z_n|^{q+1-2 \alpha }}\\{} & {} \quad = |f_n(z_k)|(1-|z_k|^2)^2(1-|z_k|^2)^{r-2-\alpha }(1-|z_n|^2)^{p-\alpha }\\{} & {} \quad \le \frac{(1-|z_n|^2)^{p-\alpha }}{\pi r_p^2}\int _{D_{\rho }(z_k,r_p)} \frac{(1-|z_k|^2)^{r-2-\alpha }}{|1-\overline{z_n}w|^{q+1-2 \alpha }}\mathrm{{d}}w \\{} & {} \quad \le \frac{(1-|z_n|^2)^{p-\alpha }}{\pi r_p^2}\int _{D_{\rho }(z_k,r_p)} \frac{4^{r-2-\alpha }(1-|w|^2)^{r-2-\alpha }}{ (1-r_p^2)^{r-2-\alpha }|1-\overline{z_n}w|^{q+1-2\alpha }}\mathrm{{d}}w \end{aligned}$$

the last inequality being a consequence of Lemma 2.4. Since the disks $D_\rho (r_k,r_p)$ are disjoint, we obtain:

$$\begin{aligned} S:= & {} \sum _{k=1,k\ne n }^\infty \frac{(1-|z_k|^2)^{r-\alpha }(1-|z_n|^2)^{p-\alpha }}{|1-\overline{z_k}z_n|^{q+1-2\alpha }}\\\le & {} \frac{4^{r-2-\alpha }(1-|z_n|^2)^{p-\alpha }}{\pi r_p^2(1-r_p^2)^{r-2-\alpha }} \int _{\textbf{D}} \frac{(1-|w|^2)^{r-2-\alpha }}{ |1-\overline{z_n}w|^{q+1-2\alpha }}\mathrm{{d}}w \end{aligned}$$

where $r-2-\alpha$ and $q+1-2\alpha >0$. Bearing in mind Proposition 2.7 a) we obtain:

$$\begin{aligned} S \le \frac{4^{r-2-\alpha }(1-|z_n|^2)^{p-\alpha }}{\pi r_p^2(1-r_p^2)^{r-2-\alpha }} \frac{\pi \Gamma (r-1-\alpha )}{\Gamma (\frac{q+1}{2}-\alpha )^2}\sum _{m=0}^\infty \frac{\Gamma (m+\frac{q+1}{2}-\alpha )^2}{m! \Gamma (m+r-\alpha )} |z_n|^{2m}. \end{aligned}$$

Therefore, for any $z\in \textbf{D}$, we have that:

$$\begin{aligned} S\le & {} \frac{4^{r-2-\alpha }(1-|z_n|^2)^{p-\alpha }}{(1-r_p^2)^{r-2-\alpha } r_p^2}\frac{\Gamma (r-1-\alpha )}{\Gamma (\frac{q+1}{2}-\alpha )^2} \left( \sum _{m=0}^2 \frac{\Gamma (m+\frac{q+1}{2}-\alpha )^2}{m! \Gamma (m+r-\alpha )} |z_n|^{2m}\right. \\{} & {} \left. + \sum _{m=3}^\infty \frac{(m+\frac{q-1}{2}-\alpha )^2}{(m+r-1-\alpha )(m+r-2-\alpha )}\frac{\Gamma (m+\frac{q-1}{2}-\alpha )^2}{m! \Gamma (m+r-2-\alpha )} |z_n|^{2m}\right) . \end{aligned}$$

If $r-2 \ge \frac{q-1}{2}$ and $m \ge 3$ then:

$$\begin{aligned} \frac{(m+\frac{q-1}{2}-\alpha )^2}{(m+r-1-\alpha )(m+r-2-\alpha )} \le 1 \end{aligned}$$

and for any $m \ge 3$, we also have:

$$\begin{aligned} \frac{\Gamma (m+\frac{q-1}{2}-\alpha )^2}{\Gamma (m+r-2-\alpha )} \le \frac{\Gamma (m+r-2-\alpha )^2}{\Gamma (m+r-2-\alpha )}=\Gamma (m+r-2-\alpha ). \end{aligned}$$

So:

$$\begin{aligned} S{} & {} \le \frac{4^{r-2-\alpha }(1-|z_n|^2)^{p-\alpha }}{(1-r_p^2)^{r-2-\alpha } r_p^2}\frac{(r-2-\alpha )\Gamma (r-2-\alpha )}{\Gamma (\frac{q+1}{2}-\alpha )^2} \\{} & {} \quad \times \left( \sum _{m=0}^2 \frac{\Gamma (m+\frac{q+1}{2}-\alpha )^2}{m! \Gamma (m+r-\alpha )} |z_n|^{2m} + \sum _{m=3}^\infty \frac{\Gamma (m+r-2-\alpha )}{m!} |z_n|^{2m}\right) \\{} & {} \le \frac{4^{r-2-\alpha }(1-|z_n|^2)^{p-\alpha }}{(1-r_p^2)^{r-2-\alpha } r_p^2}\frac{(r-2-\alpha )\Gamma (r-2-\alpha )^2}{\Gamma (\frac{q+1}{2}-\alpha )^2} \\{} & {} \quad \times \left( \sum _{m=0}^2 \frac{\Gamma (m+\frac{q+1}{2}-\alpha )^2}{m! \Gamma (m+r-\alpha ) \Gamma (r-2-\alpha )} |z_n|^{2m}\right. \\{} & {} \quad \left. + \sum _{m=3}^\infty \frac{\Gamma (m+r-2-\alpha )}{m! \Gamma (r-2-\alpha )} |z_n|^{2m}\right) . \end{aligned}$$

Calling:

$$\begin{aligned} M= & {} \left( \sum _{m=0}^2 \frac{\Gamma (m+\frac{q+1}{2}-\alpha )^2}{m! \Gamma (m+r-\alpha ) \Gamma (r-2-\alpha )} |z_n|^{2m}+ \sum _{m=3}^\infty \frac{\Gamma (m+r-2-\alpha )}{m! \Gamma (r-2-\alpha )} |z_n|^{2m}\right) \end{aligned}$$

we have:

$$\begin{aligned} M= & {} \sum _{m=0}^\infty \frac{\Gamma (m+r-2-\alpha )}{m! \Gamma (r-2-\alpha )}|z_n|^{2m}- \frac{\Gamma (r-2-\alpha )}{0! \Gamma (r-2-\alpha )}|z_n|^{0}- \frac{\Gamma (r-1-\alpha )}{1! \Gamma (r-2-\alpha )}|z_n|^{2} \\{} & {} - \frac{\Gamma (r-\alpha )}{2! \Gamma (r-2-\alpha )}|z_n|^{4} + \frac{\Gamma (\frac{q+1}{2}-\alpha )^2}{0! \Gamma (r-\alpha ) \Gamma (r-2-\alpha )} \\{} & {} + \frac{\Gamma (\frac{q+3}{2}-\alpha )^2}{1! \Gamma (r+1-\alpha ) \Gamma (r-2-\alpha )} |z_n|^2 +\frac{\Gamma (\frac{q+5}{2}-\alpha )^2}{2! \Gamma (r+2-\alpha ) \Gamma (r-2-\alpha )} |z_n|^4 \end{aligned}$$

and by Proposition 2.7 b) we have:

$$\begin{aligned} \sum _{m=0}^\infty \frac{\Gamma (m+r-2-\alpha )}{m! \Gamma (r-2-\alpha )}|z_n|^{2m}=\frac{1}{(1-|z|^2)^{r-2-\alpha }} \end{aligned}$$

we have:

$$\begin{aligned} M= & {} \frac{1}{(1-|z|^2)^{r-2-\alpha }}-\left( 1- \frac{\Gamma (\frac{q+1}{2}-\alpha )^2}{0! \Gamma (r-\alpha ) \Gamma (r-2-\alpha )} \right) \\{} & {} - \left( r-2-\alpha - \frac{\Gamma (\frac{q+3}{2}-\alpha )^2}{1! \Gamma (r+1-\alpha ) \Gamma (r-2-\alpha )} \right) |z_n|^{2} \\{} & {} - \left( \frac{(r-1-\alpha )(r-2-\alpha )}{2} -\frac{\Gamma (\frac{q+5}{2}-\alpha )^2}{2! \Gamma (r+2-\alpha ) \Gamma (r-2-\alpha )} \right) |z_n|^{4} \end{aligned}$$

Since $p \ge r-2$ we obtain:

$$\begin{aligned} \frac{(1-|z_n|^2)^{p-\alpha }}{(1-|z_n|^2)^{r-2-\alpha }} \le 1 \end{aligned}$$

so we conclude:

$$\begin{aligned} S \le \frac{4^{r-2-\alpha }}{(1-r_p^2)^{r-2-\alpha } r_p^2}\frac{(r-2-\alpha ) \Gamma (r-2-\alpha )^2}{\Gamma (\frac{q+1}{2}-\alpha )^2} \left( 1+(1-|z_n|^2)^{p-\alpha }Q(|z_n|)\right) \end{aligned}$$

where Q(|z|) is the polynomial given by:

$$\begin{aligned}{} & {} \left( -1+ \frac{\Gamma (\frac{q+1}{2}-\alpha )^2}{0! \Gamma (r-\alpha ) \Gamma (r-2-\alpha )} \right) \\{} & {} \quad - \left( r-2-\alpha - \frac{\Gamma (\frac{q+3}{2}-\alpha )^2}{1! \Gamma (r+1-\alpha ) \Gamma (r-2-\alpha )} \right) |z_n|^{2}\\{} & {} \quad -\left( \frac{(r-1-\alpha )(r-2-\alpha )}{2} -\frac{\Gamma (\frac{q+5}{2}-\alpha )^2}{2! \Gamma (r+2-\alpha ) \Gamma (r-2-\alpha )} \right) |z_n|^{4} \end{aligned}$$

so the expression $1+(1-|z_n|^2)^{p-\alpha }Q(|z|)$ is uniformly bounded for $z \in \textbf{D}$ by a constant C. Hence:

$$\begin{aligned} ||Id_{\ell _\infty }-L\circ T||\le \frac{4^{r-2-\alpha }(1-\Delta _p^2)^\alpha }{ r_p^2(1-r_p^2)^{r-2-\alpha }} \frac{(r-2-\alpha ) \Gamma (r-2-\alpha )^2}{\Gamma (\frac{q+1}{2}-\alpha )^2} C \end{aligned}$$

so we will be done if we prove that:

$$\begin{aligned} \frac{4^{r-2-\alpha }(1-\Delta _p^2)^\alpha }{ r_p^2(1-r_p^2)^{r-2-\alpha }} \frac{(r-2-\alpha ) \Gamma (r-2-\alpha )^2}{\Gamma (\frac{q+1}{2}-\alpha )^2} \rightarrow 0 \end{aligned}$$

as $\Delta _p\rightarrow 1$ for $\alpha > \frac{r-2}{3}$. Indeed:

$$\begin{aligned} 1-r_p^2=1-\frac{\Delta _p^2}{(1+\sqrt{1-\Delta _p^2})^2}=2 \frac{\sqrt{1-\Delta _p^2}}{1+\sqrt{1-\Delta _p^2}} \end{aligned}$$

(2.2)

and:

$$\begin{aligned} \frac{4^{r-2-\alpha }(1-\Delta _p^2)^\alpha }{ r_p^2(1-r_p^2)^{r-2-\alpha }}&=\frac{4^{r-2-\alpha }\left( \sqrt{1-\Delta _p^2}\right) ^{2\alpha }}{ r_p^2 \left( \frac{2\sqrt{1-\Delta _p^2}}{1+\sqrt{1-\Delta _p^2}} \right) ^{r-2-\alpha }}\\&= \frac{2^{r-2-\alpha } \left( \sqrt{1-\Delta _p^2}\right) ^{3\alpha -r+2} \left( 1+\sqrt{1-\Delta _p^2}\right) ^{r-2-\alpha }}{ r_p^2} \end{aligned}$$

so the proof is completed since $r_p \rightarrow 1$ when $\Delta _p \rightarrow 1$ and $3 \alpha -r+2 >0$. $\square$

Remark 2.10

Theorem gives us a procedure to calculate an upper estimate for the constant of separation $\Delta _p$ for $\mathcal {B}_{v_p}$ if $p \ge 1$.

Lemma 2.11

If $a,b,c \ge 0$ and $0 \le \alpha < 1$, the function $f_\alpha (x)=(1-x^2)^{1-\alpha }(-a-bx^2-cx^4)$ satisfies:

$$\begin{aligned} (1-x^2)^{\alpha }f_{\alpha }'(x)=2(a(1-\alpha )-b)x+2(b(2-\alpha )-2c)x^3+2c(3-\alpha )x^5. \end{aligned}$$

Proposition 2.12

There exists $0.811< \Delta _1 < 0.9785$ such that for any separated sequence $(z_n) \subset \textbf{D}$ such that $\rho (z_k,z_j) \ge \Delta _1$, then $(z_n)$ is interpolating for the classical Bloch space $\mathcal {B}$ and there are examples of non-interpolating sequences which are $\delta$-separated for values $\delta < \Delta _1$.

Proof

Inequality $0.81 < \Delta _1$ is straightforward from Proposition 2.1. For the other inequality, use proof of Theorem 2.1. Take $p=1, r=3$ and $q=3$. Notice that $0 \le |z_n| < 1$ and take $0 \le \alpha < 1$. We have:

$$\begin{aligned} S \le \frac{4^{1-\alpha } (1-\Delta _1^2)^{\alpha } }{(1-\alpha )r_1^2(1-r_1^2)^{1-\alpha }} \left( 1+(1-|z_n|^2)^{1-\alpha } Q(|z_n|)\right) \end{aligned}$$

where the polynomial $Q(|z_n|)$ is given by:

$$\begin{aligned}{} & {} - \left( 1-\frac{1-\alpha }{2-\alpha }\right) -\left( 1-\alpha -\frac{(2-\alpha )(1-\alpha )}{3-\alpha }\right) |z_n|^2\\{} & {} \quad -\left( \frac{(2-\alpha )(1-\alpha )}{2}-\frac{(3-\alpha )(2-\alpha )(1-\alpha )}{2(4-\alpha )}\right) |z_n|^4\\{} & {} =-\frac{1}{2-\alpha }-\frac{1-\alpha }{3-\alpha } |z_n|^2-\frac{(2-\alpha )(1-\alpha )}{2(4-\alpha )} |z_n|^4 \end{aligned}$$

so the expression $f_\alpha (|z_n|):=1+(1-|z_n|^2)^{1-\alpha }Q(|z_n|)$ becomes:

$$\begin{aligned} f_\alpha (|z_n|)=1+(1-|z_n|^2)^{1-\alpha }\left( -\frac{1}{1-\alpha }-\frac{1-\alpha }{3-\alpha } |z_n|^2-\frac{(2-\alpha )(1-\alpha )}{2(4-\alpha )} |z_n|^4 \right) . \end{aligned}$$

We will prove that for some $0 \le \alpha < 1$, we have $f_\alpha '(|z_n|) \ge 0$ for $0 \le |z_n| \le 1$ so $f_\alpha (|z_n|)$ will be non-decreasing for $|z_n|$. Notice that this is equivalent to proving that $(1-|z_n|^2)^{\alpha }f_\alpha '(|z_n|) \ge 0$. Taking $\alpha =4/5$ and bearing in mind Lemma 2.11 we have for any $0 \le |z_n| \le 1$:

$$\begin{aligned} (1-|z_n|^2)^{4/5}f'(|z_n|)=\frac{5}{33}|z_n|+\frac{3}{44}|z_n|^3+\frac{33}{200}|z_n|^5 \ge 0. \end{aligned}$$

Hence the expression S is a non-decreasing function of $|z_n|$ and we have:

$$\begin{aligned} S\le & {} \frac{4^{1-\alpha } (1-\Delta _1^2)^{\alpha } }{(1-\alpha )r_1^2(1-r_1^2)^{1-\alpha }} \lim _{|z_n| \rightarrow 1^{-}} \left( 1+(1-|z_n|^2)^{1-\alpha } Q(|z_n|)\right) \\= & {} \frac{4^{1-\alpha } (1-\Delta _1^2)^{\alpha } }{(1-\alpha )r_1^2(1-r_1^2)^{1-\alpha }} \end{aligned}$$

so calling $t=\sqrt{1-\Delta _1^2} < 1$ and bearing in mind (2.2) we have:

$$\begin{aligned} S \le \frac{4^{1-\alpha } t^{2\alpha }}{(1-\alpha )\left( \frac{2t}{1+t}\right) ^{1-\alpha } \frac{1-t}{1+t}}=\frac{2^{1-\alpha } t^{3\alpha -1} (1+t)^{2-\alpha }}{(1-\alpha ) (1-t)} \end{aligned}$$

and for $\alpha =4/5$, we have:

$$\begin{aligned} \frac{2^{1/5} t^{7/5} (1+t)^{6/5}}{1/5 (1-t)} < 1 \end{aligned}$$

if and only if $2^{1/5} t^{7/5} (1+t)^{6/5}-\frac{1}{5}(1-t) < 0$.

which clearly has real solutions on $0< t < 1$ by Bolzano’s Theorem. A root of $2^{1/5} t^{7/5} (1+t)^{6/5}-\frac{1}{5}(1-t) = 0$ is $t \approx 0.2069$ so $\Delta _1 =\sqrt{1-t^2} \approx 0.9784$ and we are done. $\square$