Norms of a Product of Integral and Composition Operators between Some Bloch-Type Spaces

Abstract

We present some formulas for the norm, as well as the essential norm, of a product of composition and an integral operator between some Bloch-type spaces of analytic functions on the unit ball, in terms of given symbols and weights.

1. Introduction

Let $\mathbb{B}$ be the open unit ball in ${\mathbb{C}}^n$, with the scalar product $\langle z,w\rangle ={\sum }_{k=1}^n{z}_k{\overline{w}}_k$ and the norm $\left|z\right|=\sqrt{\langle z,z\rangle }$ (here, as usual, $z=(z_1,\dots ,z_n)$, $w=(w_1,\dots ,w_n)$, and $\overline{z}=({\overline{z}}_1,\dots ,{\overline{z}}_n)$). We denote the space of analytic functions on $\mathbb{B}$ by $H\left(\mathbb{B}\right)$, whereas we denote the class of analytic self-maps of $\mathbb{B}$ by $S\left(\mathbb{B}\right)$ [1,2]. The linear operator $\Re f\left(z\right)={\sum }_{j=1}^n{z}_j{D}_{j}f\left(z\right)$, where $D_{j}f=\frac{\partial f}{\partial z_j}$, $j=\overline{1,n}$, is called a radial derivative.

We denote the set of all positive and continuous functions on $\mathbb{B}$ by $W\left(\mathbb{B}\right)$. A $w\in W\left(\mathbb{B}\right)$ is called a weight. Let $\mu \in W\left(\mathbb{B}\right)$. Then,

$$H_{\mu }^{\infty }\left(\mathbb{B}\right)={\{f\in H\left(\mathbb{B}\right):\parallel f\parallel }_{H_{\mu }^{\infty }}:=\underset{z\in \mathbb{B}}{sup}\mu \left(z\right)\left|f\left(z\right)\right|<+\infty \}$$

is called a weighted-type space. This space with the norm ${\parallel ·\parallel }_{H_{\mu }^{\infty }}$ is a Banach space. A little weighted-type space consists of $f\in H_{\mu }^{\infty }\left(\mathbb{B}\right)$ such that ${lim}_{\left|z\right|\to 1}\mu \left(z\right)\left|f\left(z\right)\right|=0.$ These spaces have been studied for a long time (see, e.g., [3,4,5,6,7,8,9]), as well as the operators acting on them (see, e.g., [10,11,12,13,14,15,16,17] and the references therein). If $\mu$ is a nonzero constant, we obtain the space $H^{\infty }\left(\mathbb{B}\right)$ with the norm ${\parallel f\parallel }_{\infty }={sup}_{z\in \mathbb{B}}\left|f\left(z\right)\right|$ (bounded analytic functions).

Let $\mu \in W\left(\mathbb{B}\right)$. Then, the space

$${\mathcal{B}}_{\mu }\left(\mathbb{B}\right)=\{f\in H\left(\mathbb{B}\right):b_{\mu }\left(f\right):=\underset{z\in \mathbb{B}}{sup}\mu \left(z\right)|\Re f\left(z\right)|<+\infty \},$$

is called a Bloch-type space. With the norm ${\parallel f\parallel }_{{\mathcal{B}}_{\mu }}=\left|f\left(0\right)\right|+b_{\mu }\left(f\right)$, it is a Banach space. A little Bloch-type space consists of $f\in {\mathcal{B}}_{\mu }\left(\mathbb{B}\right)$ such that ${lim}_{\left|z\right|\to 1}\mu \left(z\right)|\Re f\left(z\right)|=0.$ We obtain the Bloch space $\mathcal{B}$ and little Bloch space ${\mathcal{B}}_0$ for $\mu \left(z\right)=1-{\left|z\right|}^2$, whereas for $\mu \left(z\right)={(1-|z|}^2{)}^{\alpha },$ $\alpha >0,$ we obtain the $\alpha$-Bloch space ${\mathcal{B}}^{\alpha }$ and the little $\alpha$-Bloch space ${\mathcal{B}}_0^{\alpha }$. For

$$\mu \left(z\right)={\mu }_{{log}_k}\left(z\right)={(1-|z|}^2)\prod _{j=1}^k{ln}^{\left[j\right]}\frac{e^{\left[k\right]}}{1-{\left|z\right|}^2},$$

where $k\in \mathbb{N}$, $e^{\left[1\right]}=e,$ $e^{\left[l\right]}=e^{e^{[l-1]}}$, $l\in \mathbb{N}\backslash \left\{1\right\}$ and

$${ln}^{\left[j\right]}z=\underset{j\mathrm{times}}{\underbrace{ln\cdots ln}}z,$$

we obtain the iterated logarithmic Bloch space ${\mathcal{B}}_{{log}_k}\left(\mathbb{B}\right)={\mathcal{B}}_{{log}_k}$, which for $k=1$, reduces to ${\mathcal{B}}_{{log}_1}={\mathcal{B}}_{log}.$ The quantity

$$\begin{array}{c}\hfill {\parallel f\parallel }_{{\mathcal{B}}_{{log}_k}}^{\prime }=\left|f\left(0\right)\right|+\underset{z\in \mathbb{B}}{sup}{\mu }_{{log}_k}\left(z\right)|\nabla f\left(z\right)|,\end{array}$$

(1)

is a norm on ${\mathcal{B}}_{{log}_k}\left(\mathbb{B}\right)$. From $|\Re f(z\left)\right|\le |\nabla f(z\left)\right|$ and a known theorem ([18,19,20]), it follows that (1) is equivalent to the norm ${\parallel f\parallel }_{{\mathcal{B}}_{{log}_k}}=\left|f\left(0\right)\right|+{sup}_{z\in \mathbb{B}}{\mu }_{{log}_k}\left(z\right)|\Re f\left(z\right)|$ on ${\mathcal{B}}_{{log}_k}$.

Suppose $a\in [e^{\left[k\right]},+\infty )$. Then, for every $z\in \mathbb{B}$, we have

$$\begin{array}{cc}\hfill & (1-|z\left|\right)\prod _{j=1}^k{ln}^{\left[j\right]}\frac{a}{1-\left|z\right|}=(1-|z\left|\right)\prod _{j=1}^k{ln}^{\left[j\right]}{e}^{\left[k\right]}\frac{a(1+|z\left|\right)}{e^{\left[k\right]}{(1-|z|}^2)}\hfill \\ \hfill \le & {(1-|z|}^2)\prod _{j=1}^k{ln}^{\left[j\right]}\left(\frac{2a}{e^{\left[k\right]}}\frac{e^{\left[k\right]}}{1-{\left|z\right|}^2}\right)={(1-|z|}^2)\prod _{j=1}^k{ln}^{[j-1]}\left(ln\frac{e^{\left[k\right]}}{1-{\left|z\right|}^2}+ln\frac{2a}{e^{\left[k\right]}}\right)\hfill \\ \hfill \le & {(1-|z|}^2)\prod _{j=1}^k{ln}^{[j-1]}\left(\left(1+ln\frac{2a}{e^{\left[k\right]}}\right)ln\frac{e^{\left[k\right]}}{1-{\left|z\right|}^2}\right)\hfill \\ \hfill =& {(1-|z|}^2)\left(1+ln\frac{2a}{e^{\left[k\right]}}\right)ln\frac{e^{\left[k\right]}}{1-{\left|z\right|}^2}\prod _{j=2}^k{ln}^{[j-2]}\left(ln\left(1+ln\frac{2a}{e^{\left[k\right]}}\right)+{ln}^{\left[2\right]}\frac{e^{\left[k\right]}}{1-{\left|z\right|}^2}\right)\hfill \\ \hfill \le & {(1-|z|}^2)\left(1+ln\frac{2a}{e^{\left[k\right]}}\right)ln\frac{e^{\left[k\right]}}{1-{\left|z\right|}^2}\prod _{j=2}^k{ln}^{[j-2]}\left(\left(1+ln\left(1+ln\frac{2a}{e^{\left[k\right]}}\right)\right){ln}^{\left[2\right]}\frac{e^{\left[k\right]}}{1-{\left|z\right|}^2}\right)\hfill \\ \hfill =& {(1-|z|}^2)\left(1+ln\frac{2a}{e^{\left[k\right]}}\right)ln\frac{e^{\left[k\right]}}{1-{\left|z\right|}^2}\left(1+ln\left(1+ln\frac{2a}{e^{\left[k\right]}}\right)\right){ln}^{\left[2\right]}\frac{e^{\left[k\right]}}{1-{\left|z\right|}^2}\hfill \\ \hfill & \times \prod _{j=3}^k{ln}^{[j-2]}\left(\left(1+ln\left(1+ln\frac{2a}{e^{\left[k\right]}}\right)\right){ln}^{\left[2\right]}\frac{e^{\left[k\right]}}{1-{\left|z\right|}^2}\right)\hfill \\ \hfill & ⋮\hfill \\ \hfill \le & {(1-|z|}^2)\left(1+ln\frac{2a}{e^{\left[k\right]}}\right)ln\frac{e^{\left[k\right]}}{1-{\left|z\right|}^2}\left(1+ln\left(1+ln\frac{2a}{e^{\left[k\right]}}\right)\right){ln}^{\left[2\right]}\frac{e^{\left[k\right]}}{1-{\left|z\right|}^2}\hfill \\ \hfill & \cdots \left(1+ln\left(1+\cdots +ln\left(1+ln\left(1+ln\frac{2a}{e^{\left[k\right]}}\right)\right)\cdots \right)\right){ln}^{\left[k\right]}\frac{e^{\left[k\right]}}{1-{\left|z\right|}^2}\hfill \\ \hfill =& c_a{(1-|z|}^2)\prod _{j=1}^k{ln}^{\left[j\right]}\frac{e^{\left[k\right]}}{1-{\left|z\right|}^2}\hfill \\ \hfill \le & 2c_a(1-|z\left|\right)\prod _{j=1}^k{ln}^{\left[j\right]}\frac{a}{1-\left|z\right|}.\hfill \end{array}$$

(2)

The consideration leading to (2) implies that, for $a\in [e^{\left[k\right]},+\infty )$, the quantity

$$\begin{array}{c}\hfill {\parallel f\parallel }_{{\mathcal{B}}_{{log}_k}}^{\left(a\right)}=\left|f\left(0\right)\right|+b_{{log}_k}^{\left(a\right)}\left(f\right):=\left|f\left(0\right)\right|+\underset{z\in \mathbb{B}}{sup}(1-|z\left|\right)\left(\prod _{j=1}^k{ln}^{\left[j\right]}\frac{a}{1-\left|z\right|}\right)|\nabla f\left(z\right)|,\end{array}$$

(3)

presents another equivalent norm on ${\mathcal{B}}_{{log}_k}$.

We define the corresponding little iterated logarithmic Bloch space ${\mathcal{B}}_{{log}_k,0}\left(\mathbb{B}\right)={\mathcal{B}}_{{log}_k,0}$ as the set of all $f\in H\left(\mathbb{B}\right)$ such that

$$\underset{\left|z\right|\to 1}{lim}(1-|z\left|\right)\left(\prod _{j=1}^k{ln}^{\left[j\right]}\frac{a}{1-\left|z\right|}\right)|\nabla f\left(z\right)|=0.$$

For some facts on logarithmic-type spaces, see, e.g., [10,14,21,22,23].

The product of the composition operator $C_{\phi }f\left(z\right)=f\left(\phi \left(z\right)\right)$ and an equivalent form of the integral operator in [24,25]

$$\begin{array}{c}\hfill P_{\phi }^g\left(f\right)\left(z\right)={\int }_0^{1}f\left(\phi \left(tz\right)\right)g\left(tz\right)\frac{dt}{t},z\in \mathbb{B},\end{array}$$

(4)

where $g\in H\left(\mathbb{B}\right)$, $g\left(0\right)=0$ and $\phi \in S\left(\mathbb{B}\right)$, was studied, e.g., in [22,26]. The introduction of the operators in [24,25] was motivated by some special cases mentioned therein (see also [27]). Many facts about this topic can be found in [28]. Operator (4), as well as some related ones, has been considerably studied (see, e.g., [29,30,31,32,33,34] and the cited references therein). Beside this product-type operator, many others have been studied during the last two decades. One can consult the following references: [10,14,15,35,36].

The essential norm of a linear operator $L:X\to Y$, where X and Y are Banach spaces and ${\parallel ·\parallel }_{X\to Y}$ denotes the operator norm, is the quantity

$${\parallel L\parallel }_{e,X\to Y}=inf\left\{{\parallel L+K\parallel }_{X\to Y}:K:X\to Y,K\mathrm{is} \mathrm{compact}\right\}.$$

One of the most popular topics in studying concrete linear operators is characterization of their operator-theoretic properties in terms of the induced symbols. One of the basic problems is the calculation of their norms and essential norms [18,19,20,37,38,39]. Some recent formulas for the norms can be found in [11,12,13,14,23,26,31].

Let $M_u\left(f\right)\left(z\right)=u\left(z\right)f\left(z\right)$, where $u\in H\left(\mathbb{B}\right)$. The following result was proved in [11].

Theorem 1.

Let $u\in H\left(\mathbb{B}\right)$, $\phi \in S\left(\mathbb{B}\right)$, $\mu \in W\left(\mathbb{B}\right)$ and $M_u{C}_{\phi }:X\to H_{\mu }^{\infty }$ be bounded, where $X\in \{\mathcal{B},{\mathcal{B}}_0).$ Then,

$$\begin{array}{c}\hfill \parallel M_u{C}_{\phi }{\parallel }_{X\to H_{\mu }^{\infty }}=max\left\{{\parallel u\parallel }_{H_{\mu }^{\infty }},\frac{1}{2}\underset{z\in B}{sup}\mu \left(z\right)\left|u\left(z\right)\right|ln\frac{1+\left|\phi \right(z\left)\right|}{1-\left|\phi \right(z\left)\right|}\right\},\end{array}$$

(5)

where the norm on $\mathcal{B}$ is given by ${\parallel f\parallel }_{\mathcal{B}}=\left|f\left(0\right)\right|+{sup}_{z\in \mathbb{B}}{(1-|z|}^2\left)\right|\nabla f\left(z\right)|.$

One can try to calculate the norm of $M_u{C}_{\phi }:{\mathcal{B}}^{\alpha }\to H_{\mu }^{\infty }$. To solve it, in [13], we had to change the weight ${(1-|z|}^2{)}^{\alpha }$. The method also works in some other situations [23]. Here, we employ this idea to calculate the norm of $P_{\phi }^g:{\mathcal{B}}_{{log}_k}\left(\mathrm{o}\mathrm{r}{\mathcal{B}}_{{log}_k,0}\right)\to {\mathcal{B}}_{\mu }\left(\mathrm{o}\mathrm{r}{\mathcal{B}}_{\mu ,0}\right)$. Beside this, we present a formula for its essential norm, extending the results in [23]. We use some of the methods and ideas in [13,14,23,26].

2. Auxiliary Results

Our first auxiliary result is a nontrivial technical lemma.

Lemma 1.

Assume that $k\in \mathbb{N}$, $a\in [e^{\left[k\right]},+\infty ).$ Then,

$$\begin{array}{c}\hfill h_k\left(x\right)=x\prod _{j=1}^k{ln}^{\left[j\right]}\frac{a}{x},\end{array}$$

(6)

is a nonnegative and increasing function on $(0,\frac{a}{e^{\left[k\right]}}].$

Proof. 

The case $k=1$ is simple [23]. So, assume $k\in \mathbb{N}\backslash \left\{1\right\}.$ We have

$$\begin{array}{c}\hfill h_k\left(x\right)=h_{k-1}\left(x\right){ln}^{\left[k\right]}\left(\frac{a}{x}\right).\end{array}$$

(7)

From (7), it follows that

$$\begin{array}{c}\hfill h_k^{\prime }\left(x\right)=h_{k-1}^{\prime }\left(x\right){ln}^{\left[k\right]}\left(\frac{a}{x}\right)-1.\end{array}$$

(8)

The recursive relation in (8) implies

$$\begin{array}{c}\hfill h_k^{\prime }\left(x\right)=\left(\left(\cdots \left(\left(ln\left(\frac{a}{x}\right)-1\right){ln}^{\left[2\right]}\left(\frac{a}{x}\right)-1\right)\cdots \right){ln}^{[k-1]}\left(\frac{a}{x}\right)-1\right)·{ln}^{\left[k\right]}\left(\frac{a}{x}\right)-1.\end{array}$$

(9)

From (9), it follows that $h_k^{\prime }\left(x\right)$ is decreasing on the interval $(0,\frac{a}{e^{[k-1]}})$ (here, we regard that $e^{\left[0\right]}=1$). Hence,

$$\begin{array}{cc}\hfill h_k^{\prime }\left(x\right)\ge & h_k^{\prime }\left(\frac{a}{e^{\left[k\right]}}\right)\hfill \\ \hfill =& \left(\left(\cdots \left(\left(ln{e}^{\left[k\right]}-1\right){ln}^{\left[2\right]}{e}^{\left[k\right]}-1\right)\cdots \right){ln}^{[k-1]}{e}^{\left[k\right]}-1\right)·{ln}^{\left[k\right]}{e}^{\left[k\right]}-1\hfill \\ \hfill =& \left(\left(\cdots \left(\left(e^{[k-1]}-1\right)e^{[k-2]}-1\right)\cdots \right)e^{\left[1\right]}-1\right)-1>0,\hfill \end{array}$$

for $x\in (0,\frac{a}{e^{\left[k\right]}}]$, from which the lemma follows. □

Now, we present some point evaluation estimates for the functions in ${\mathcal{B}}_{{log}_k}\left(\mathbb{B}\right)$.

Lemma 2.

Assume that $k\in \mathbb{N}$, $a\in [e^{\left[k\right]},+\infty )$, $f\in {\mathcal{B}}_{{log}_k}\left(\mathbb{B}\right)$, $z\in \mathbb{B},$ and $r\in [0,1).$ Then,

$$\begin{array}{c}\hfill |f\left(z\right)-f\left(rz\right)|\le b_{{log}_k}^{\left(a\right)}\left(f\right)\left({ln}^{[k+1]}\frac{a}{1-\left|z\right|}-{ln}^{[k+1]}\frac{a}{1-r\left|z\right|}\right),\end{array}$$

(10)

and

$$\begin{array}{c}\hfill \left|f\left(z\right)\right|\le {\parallel f\parallel }_{{\mathcal{B}}_{{log}_k}}^{\left(a\right)}max\left\{1,{ln}^{[k+1]}\frac{a}{1-\left|z\right|}-{ln}^{[k+1]}a\right\}.\end{array}$$

(11)

Proof. 

Let $\nabla f=(D_{1}f,\dots ,D_{n}f)$. Then,

$$\begin{array}{cc}\hfill \left|f\right(z)-f(rz\left)\right|=& \left|{\int }_r^1\langle \nabla f\left(tz\right),\overline{z}\rangle dt\right|\hfill \\ \hfill \le & b_{{log}_k}^{\left(a\right)}\left(f\right){\int }_r^1\frac{\left|z\right|dt}{(1-|z\left|t\right){\prod }_{j=1}^k{ln}^{\left[j\right]}\frac{a}{1-\left|z\right|t}}\hfill \\ \hfill =& b_{{log}_k}^{\left(a\right)}\left(f\right)\left({ln}^{[k+1]}\frac{a}{1-\left|z\right|}-{ln}^{[k+1]}\frac{a}{1-r\left|z\right|}\right).\hfill \end{array}$$

(12)

From (12), for $r=0$, it follows that

$$\begin{array}{c}\hfill |f\left(z\right)-f\left(0\right)|\le b_{{log}_k}^{\left(a\right)}\left(f\right)\left({ln}^{[k+1]}\frac{a}{1-\left|z\right|}-{ln}^{[k+1]}a\right).\end{array}$$

(13)

Relation (13), along with the definition of ${\parallel ·\parallel }_{{\mathcal{B}}_{{log}_k}}^{\left(a\right)}$ and the triangle inequality for numbers, implies (11). □

For the next lemma, see [22].

Lemma 3.

Let $f,g\in H\left(\mathbb{B}\right)$ and $g\left(0\right)=0$. Then,

$$\begin{array}{c}\hfill \Re P_{\phi }^g\left(f\right)\left(z\right)=f\left(\phi \left(z\right)\right)g\left(z\right),z\in \mathbb{B}.\end{array}$$

(14)

The following result is closely related to the corresponding one in [40], because of which the proof is omitted.

Lemma 4.

Assume that $g\in H\left(\mathbb{B}\right)$, $g\left(0\right)=0$, $\phi \in S\left(\mathbb{B}\right)$ and $\mu \in W\left(\mathbb{B}\right)$. Then, $P_{\phi }^g:{\mathcal{B}}_{{log}_k}\left(or{\mathcal{B}}_{{log}_k,0}\right)\to {\mathcal{B}}_{\mu }$ is compact if and only if it is bounded and for any bounded sequence ${\left(f_k\right)}_{k\in \mathbb{N}}\subset {\mathcal{B}}_{{log}_k}\left(or{\mathcal{B}}_{{log}_k,0}\right)$ converging to zero uniformly on compacts of $\mathbb{B}$, we have ${lim}_{k\to +\infty }{\parallel P_{\phi }^g{f}_k\parallel }_{{\mathcal{B}}_{\mu }}=0.$

3. Main Results

Now, we are in a position to state and prove our main results.

Theorem 2.

Suppose that $k\in \mathbb{N},$ $a\in [2e^{\left[k\right]},+\infty )$, $g\in H\left(\mathbb{B}\right)$, $g\left(0\right)=0$, $\phi \in S\left(\mathbb{B}\right)$, $\mu \in W\left(\mathbb{B}\right)$ and that $P_{\phi }^g:X\to {\mathcal{B}}_{\mu }$ is bounded, where $X\in \{{\mathcal{B}}_{{log}_k},{\mathcal{B}}_{{log}_k,0}\}.$ Then,

$$\begin{array}{c}\hfill \parallel P_{\phi }^g{\parallel }_{X\to {\mathcal{B}}_{\mu }}=max\left\{{\parallel g\parallel }_{H_{\mu }^{\infty }},\underset{z\in \mathbb{B}}{sup}\mu \left(z\right)\left|g\left(z\right)\right|\left({ln}^{[k+1]}\frac{a}{1-\left|\phi \right(z\left)\right|}-{ln}^{[k+1]}a\right)\right\}.\end{array}$$

(15)

Proof. 

From (14) and (11), it follows that, for $f\in {\mathcal{B}}_{{log}_k}$, we have

$$\begin{array}{cc}\hfill \parallel P_{\phi }^g{f\parallel }_{{\mathcal{B}}_{\mu }}=& \underset{z\in \mathbb{B}}{sup}\mu \left(z\right)\left|g\left(z\right)f\left(\phi \left(z\right)\right)\right|\hfill \\ \hfill \le & {\parallel f\parallel }_{{\mathcal{B}}_{{log}_k}}^{\left(a\right)}\underset{z\in \mathbb{B}}{sup}\mu \left(z\right)\left|g\left(z\right)\right|max\left\{1,{ln}^{[k+1]}\frac{a}{1-\left|\phi \right(z\left)\right|}-{ln}^{[k+1]}a\right\},\hfill \end{array}$$

(16)

Hence,

$$\begin{array}{c}\hfill \parallel P_{\phi }^g{\parallel }_{X\to {\mathcal{B}}_{\mu }}\le max\left\{{\parallel g\parallel }_{H_{\mu }^{\infty }},\underset{z\in \mathbb{B}}{sup}\mu \left(z\right)\left|g\left(z\right)\right|\left({ln}^{[k+1]}\frac{a}{1-\left|\phi \right(z\left)\right|}-{ln}^{[k+1]}a\right)\right\}.\end{array}$$

(17)

If $P_{\phi }^g:X\to {\mathcal{B}}_{\mu }$ is bounded, then for $f_0\left(z\right)\equiv 1\in {\mathcal{B}}_{{log}_k,0}$, we have $\parallel f_0{\parallel }_{{\mathcal{B}}_{{log}_k}}=1$, from which together with the boundedness, it follows that

$$\begin{array}{c}\hfill \parallel P_{\phi }^g{\parallel }_{X\to {\mathcal{B}}_{\mu }}\ge \parallel P_{\phi }^g{f}_0{\parallel }_{{\mathcal{B}}_{\mu }}=\underset{z\in \mathbb{B}}{sup}\mu \left(z\right)\left|g\left(z\right)\right|.\end{array}$$

(18)

Let

$$\begin{array}{c}\hfill h_w\left(z\right)={ln}^{[k+1]}\frac{a}{1-\langle z,w\rangle }-{ln}^{[k+1]}a,\end{array}$$

(19)

and $w\in \mathbb{B}$.

Then,

$$\begin{array}{c}\hfill 1-\left|z\right|\le |1-\langle z,w\rangle |<2.\end{array}$$

(20)

for $z,w\in \mathbb{B}$. Relation (20) together with Lemma 1 implies

$$\begin{array}{cc}\hfill (1-|z\left|\right)\left(\prod _{j=1}^k{ln}^{\left[j\right]}\frac{a}{1-\left|z\right|}\right)|\nabla h_w\left(z\right)|=& \frac{\left|w\right|(1-\left|z\right|){\prod }_{j=1}^k{ln}^{\left[j\right]}\frac{a}{1-\left|z\right|}}{|1-\langle z,w\rangle |\left|{\prod }_{j=1}^k{ln}^{\left[j\right]}\frac{a}{1-\langle z,w\rangle }\right|}\hfill \end{array}$$

(21)

$$\begin{array}{cc}\hfill \hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}& \le \frac{\left|w\right|(1-\left|z\right|){\prod }_{j=1}^k{ln}^{\left[j\right]}\frac{a}{1-\left|z\right|}}{|1-\langle z,w\rangle |{\prod }_{j=1}^k{ln}^{\left[j\right]}\frac{a}{|1-\langle z,w\rangle |}}<1.\hfill \end{array}$$

(22)

Inequality (22) along with the fact that $h_w\left(0\right)=0$ implies

$$\begin{array}{c}\hfill \underset{w\in \mathbb{B}}{sup}{\parallel h_w\parallel }_{{\mathcal{B}}_{{log}_k}}^{\left(a\right)}\le 1.\end{array}$$

(23)

Let $\left|z\right|\to 1$ in (21); then, we have $h_w\in {\mathcal{B}}_{{log}_k,0}$, $w\in \mathbb{B}.$

If $\phi \left(w\right)\ne 0$ and $t\in (0,1)$, then from the boundedness of $P_{\phi }^g:X\to {\mathcal{B}}_{\mu }$ and (23), we have

$$\begin{array}{cc}\hfill \parallel P_{\phi }^g{\parallel }_{X\to {\mathcal{B}}_{\mu }}\ge & \parallel P_{\phi }^g{h}_{t\phi \left(w\right)/\left|\phi \right(w\left)\right|}{\parallel }_{{\mathcal{B}}_{\mu }}\hfill \\ \hfill =& \underset{z\in \mathbb{B}}{sup}\mu \left(z\right)\left|g\left(z\right)\right||{ln}^{[k+1]}\frac{a}{1-t\langle \phi \left(z\right),\phi \left(w\right)/\left|\phi \right(w\left)\right|\rangle }-{ln}^{[k+1]}a|\hfill \\ \hfill \ge & \mu \left(w\right)\left|g\left(w\right)\right|\left({ln}^{[k+1]}\frac{a}{1-t\left|\phi \right(w\left)\right|}-{ln}^{[k+1]}a\right).\hfill \end{array}$$

(24)

Note that (24) also holds when $\phi \left(w\right)=0.$

Let $t\to 1^{-}$ in (24), and taking the supremum over $\mathbb{B}$, we obtain

$$\begin{array}{c}\hfill \parallel P_{\phi }^g{\parallel }_{X\to {\mathcal{B}}_{\mu }}\ge \underset{z\in \mathbb{B}}{sup}\mu \left(z\right)\left|g\left(z\right)\right|\left({ln}^{[k+1]}\frac{a}{1-\left|\phi \right(z\left)\right|}-{ln}^{[k+1]}a\right).\end{array}$$

(25)

Relations (18) and (25) imply

$$\begin{array}{c}\hfill \parallel P_{\phi }^g{\parallel }_{X\to {\mathcal{B}}_{\mu }}\ge max\left\{{\parallel g\parallel }_{H_{\mu }^{\infty }},\underset{z\in \mathbb{B}}{sup}\mu \left(z\right)\left|g\left(z\right)\right|\left({ln}^{[k+1]}\frac{a}{1-\left|\phi \right(z\left)\right|}-{ln}^{[k+1]}a\right)\right\}.\end{array}$$

(26)

Combining the inequalities in (17) and (26), the formula in (15) immediately follows. □

Using the test function $f_0\left(z\right)\equiv 1$ and the fact that the set of polynomials is dense in ${\mathcal{B}}_{{log}_k,0}$, the following theorem is easily proved. We omit the standard proof.

Theorem 3.

Suppose that $k\in \mathbb{N},$ $g\in H\left(\mathbb{B}\right)$, $g\left(0\right)=0$, $\phi \in S\left(\mathbb{B}\right)$, and $\mu \in W\left(\mathbb{B}\right)$. Then, $P_{\phi }^g:{\mathcal{B}}_{{log}_k,0}\to {\mathcal{B}}_{\mu ,0}$ is bounded if and only if $P_{\phi }^g:{\mathcal{B}}_{{log}_k,0}\to {\mathcal{B}}_{\mu }$ is bounded and $g\in H_{\mu ,0}^{\infty }.$

The following result is a consequence of the previous two theorems.

Corollary 1.

Suppose that $k\in \mathbb{N},$ $g\in H\left(\mathbb{B}\right)$, $g\left(0\right)=0$, $\phi \in S\left(\mathbb{B}\right)$, $\mu \in W\left(\mathbb{B}\right)$ and that $P_{\phi }^g:{\mathcal{B}}_{{log}_k,0}\to {\mathcal{B}}_{\mu ,0}$ is bounded. Then,

$$\parallel P_{\phi }^g{\parallel }_{{\mathcal{B}}_{{log}_k,0}\to {\mathcal{B}}_{\mu ,0}}=max\left\{{\parallel g\parallel }_{H_{\mu }^{\infty }},\underset{z\in \mathbb{B}}{sup}\mu \left(z\right)\left|g\left(z\right)\right|\left({ln}^{[k+1]}\frac{a}{1-\left|\phi \right(z\left)\right|}-{ln}^{[k+1]}a\right)\right\}.$$

Theorem 4.

Suppose that $k\in \mathbb{N}$, $a\in [2e^{\left[k\right]},+\infty )$, $g\in H\left(\mathbb{B}\right)$, $g\left(0\right)=0$, $\phi \in S\left(\mathbb{B}\right)$, $\mu \in W\left(\mathbb{B}\right)$ and $P_{\phi }^g:X\to {\mathcal{B}}_{\mu }$ is bounded, where $X\in \{{\mathcal{B}}_{{log}_k},{\mathcal{B}}_{{log}_k,0}\}.$ Then,

(a) 

If ${\parallel \phi \parallel }_{\infty }=1$, we have

$$\begin{array}{c}\hfill \parallel P_{\phi }^g{\parallel }_{e,X\to {\mathcal{B}}_{\mu }}=\underset{\left|\phi \right(z\left)\right|\to 1}{lim\; sup}\mu \left(z\right)\left|g\left(z\right)\right|\left({ln}^{[k+1]}\frac{a}{1-\left|\phi \right(z\left)\right|}-{ln}^{[k+1]}a\right);\end{array}$$

(27)

(b) 

If ${\parallel \phi \parallel }_{\infty }<1$, we have

$$\begin{array}{c}\hfill \parallel P_{\phi }^g{\parallel }_{e,X\to {\mathcal{B}}_{\mu }}=0.\end{array}$$

(28)

Proof. 

(a) Let $\epsilon >0$ and $w\in \mathbb{B}\backslash \left\{0\right\}$ be fixed, and

$$h_{w,\epsilon }\left(z\right)={\left({ln}^{[k+1]}\frac{a}{1-\left|w\right|}-{ln}^{[k+1]}a\right)}^{-\epsilon }{\left({ln}^{[k+1]}\frac{a(1+|w\left|\right)}{1-\langle z,w\rangle }-{ln}^{[k+1]}a\right)}^{\epsilon +1},z\in \mathbb{B}.$$

Then,

$$\begin{array}{cc}\hfill & (1-|z\left|\right)\left(\prod _{j=1}^k{ln}^{\left[j\right]}\frac{a}{1-\left|z\right|}\right)|\nabla h_{w,\epsilon }\left(z\right)|\hfill \\ \hfill =& (\epsilon +1)\frac{\left|w\right|(1-\left|z\right|){\prod }_{j=1}^k{ln}^{\left[j\right]}\frac{a}{1-\left|z\right|}}{|1-\langle z,w\rangle |\left|{\prod }_{j=1}^k{ln}^{\left[j\right]}\frac{a(1+|w\left|\right)}{1-\langle z,w\rangle }\right|}\hfill \\ \hfill & \times |{ln}^{[k+1]}\frac{a(1+|w\left|\right)}{1-\langle z,w\rangle }-{ln}^{[k+1]}a{|}^{\epsilon }{\left({ln}^{[k+1]}\frac{a}{1-\left|w\right|}-{ln}^{[k+1]}a\right)}^{-\epsilon }\hfill \\ \hfill \le & (\epsilon +1)\frac{\left|w\right|(1-\left|z\right|){\prod }_{j=1}^k{ln}^{\left[j\right]}\frac{a}{1-\left|z\right|}}{|1-\langle z,w\rangle |\left|{\prod }_{j=1}^k{ln}^{\left[j\right]}\frac{a(1+|w\left|\right)}{1-\langle z,w\rangle }\right|}{\left({ln}^{[k+1]}\frac{a}{1-\left|w\right|}-{ln}^{[k+1]}a\right)}^{-\epsilon }\hfill \\ \hfill & \times {\left(ln\left(ln\left(\cdots \left(ln\frac{a(1+|w\left|\right)}{|1-\langle z,w\rangle |}+2\pi \right)\cdots \right)+2\pi \right)+2\pi -{ln}^{[k+1]}a\right)}^{\epsilon }\hfill \\ \hfill \le & (\epsilon +1)\left|w\right|{\left({ln}^{[k+1]}\frac{a}{1-\left|w\right|}-{ln}^{[k+1]}a\right)}^{-\epsilon }\hfill \end{array}$$

(29)

$$\begin{array}{cc}\hfill & {\left(ln\left(ln\left(\cdots \left(ln\frac{a(1+|w\left|\right)}{1-\left|w\right|}+2\pi \right)\cdots \right)+2\pi \right)+2\pi -{ln}^{[k+1]}a\right)}^{\epsilon }.\hfill \end{array}$$

(30)

Relation (29) implies $h_{w,\epsilon }\in {\mathcal{B}}_{{log}_k,0},$ for $w\in \mathbb{B}\backslash \left\{0\right\}$, while by taking limit in relation (30), we obtain

$$\begin{array}{c}\hfill \underset{\left|w\right|\to 1}{lim\; sup}{b}_{{log}_k}^{\left(a\right)}\left(h_{w,\epsilon }\right)\le \epsilon +1.\end{array}$$

(31)

Note also that

$$\begin{array}{c}\hfill \underset{\left|w\right|\to 1}{lim}\left|h_{w,\epsilon }\left(0\right)\right|=0.\end{array}$$

(32)

From (31) and (32), it follows that

$$\begin{array}{c}\hfill \underset{\left|w\right|\to 1}{lim\; sup}{\parallel h_{w,\epsilon }\parallel }_{{\mathcal{B}}_{{log}_k}}^{\left(a\right)}\le \epsilon +1.\end{array}$$

(33)

If ${\left(\phi \left(z_k\right)\right)}_{k\in \mathbb{N}}\subset \mathbb{B}$ satisfies the condition $\left|\phi \right(z_k\left)\right|\to 1$ as $k\to +\infty$, then (33) for

$$f_k\left(z\right):=h_{\phi \left(z_k\right),\epsilon }\left(z\right),k\in \mathbb{N},$$

implies

$$\begin{array}{c}\hfill \underset{k\to +\infty }{lim\; sup}{\parallel f_k\parallel }_{{\mathcal{B}}_{{log}_k}}^{\left(a\right)}\le \epsilon +1.\end{array}$$

(34)

The assumption $f_k\to 0$ on compacts of $\mathbb{B}$ implies that $f_k\to 0$ weakly in ${\mathcal{B}}_{{log}_k,0}$ as $k\to +\infty$. Indeed, the operator $L\left(f\right)=f^{\prime }$ is an isometric isomorphism between ${\mathcal{B}}_{{log}_k,0}/\mathbb{C}$ and $H_{{log}_k,0}^{\infty }$. On the other hand, a bounded sequence converges weakly to zero in $H_{{log}_k,0}^{\infty }$ if and only if it converges to zero uniformly on compacts of $\mathbb{B}$ (see, e.g., some reasoning in [3] and the estimate in (11), and note that the unit ball in $H_{{log}_k,0}^{\infty }$ is a normal family).

Hence, if $K:{\mathcal{B}}_{{log}_k,0}\to {\mathcal{B}}_{\mu }$ is compact, then ${lim}_{k\to +\infty }{\parallel K{f}_k\parallel }_{{\mathcal{B}}_{\mu }}=0.$ This fact, (34), and the estimate

$$\begin{array}{cc}\hfill \parallel f_k{\parallel }_{{\mathcal{B}}_{{log}_k}}^{\left(a\right)}{\parallel P_{\phi }^g+K\parallel }_{{\mathcal{B}}_{{log}_k,0}\to {\mathcal{B}}_{\mu }}\ge & \parallel (P_{\phi }^g+K)\left(f_k\right){\parallel }_{{\mathcal{B}}_{\mu }}\hfill \\ \hfill \ge & \parallel P_{\phi }^g{f}_k{\parallel }_{{\mathcal{B}}_{\mu }}-{\parallel K{f}_k\parallel }_{{\mathcal{B}}_{\mu }},\hfill \end{array}$$

imply

$$\begin{array}{cc}\hfill \frac{\parallel P_{\phi }^g{+K\parallel }_{{\mathcal{B}}_{{log}_k,0}\to {\mathcal{B}}_{\mu }}}{{(\epsilon +1)}^{-1}}\ge & \underset{k\to \infty }{lim\; sup}\parallel f_k{\parallel }_{{\mathcal{B}}_{{log}_k}}^{\left(a\right)}{\parallel P_{\phi }^g+K\parallel }_{{\mathcal{B}}_{{log}_k,0}\to {\mathcal{B}}_{\mu }}\hfill \\ \hfill \ge & \underset{k\to \infty }{lim\; sup}(\parallel P_{\phi }^g{f}_k{\parallel }_{{\mathcal{B}}_{\mu }}-\parallel K{f}_k{\parallel }_{{\mathcal{B}}_{\mu }})\hfill \\ \hfill =& \underset{k\to \infty }{lim\; sup}{\parallel P_{\phi }^g{f}_k\parallel }_{{\mathcal{B}}_{\mu }}\hfill \\ \hfill =& \underset{k\to \infty }{lim\; sup}\underset{z\in \mathbb{B}}{sup}\mu \left(z\right)\left|g\left(z\right)\right||f_k\left(\phi \left(z\right)\right)|\hfill \\ \hfill \ge & \underset{k\to \infty }{lim\; sup}\mu \left(z_k\right)\left|g\left(z_k\right)f_k\left(\phi \left(z_k\right)\right)\right|\hfill \\ \hfill =& \underset{k\to \infty }{lim\; sup}\mu \left(z_k\right)\left|g\left(z_k\right)\right|\left({ln}^{[k+1]}\frac{a}{1-\left|\phi \right(z_k\left)\right|}-{ln}^{[k+1]}a\right).\hfill \end{array}$$

(35)

From (35) and since $K:{\mathcal{B}}_{{log}_k,0}\to {\mathcal{B}}_{\mu }$ is an arbitrary compact operator, by letting $\epsilon \to +0$, we have

$$\parallel P_{\phi }^g{\parallel }_{e,{\mathcal{B}}_{{log}_k,0}\to {\mathcal{B}}_{\mu }}\ge \underset{k\to \infty }{lim\; sup}\mu \left(z_k\right)\left|g\left(z_k\right)\right|\left({ln}^{[k+1]}\frac{a}{1-\left|\phi \right(z_k\left)\right|}-{ln}^{[k+1]}a\right).$$

Hence,

$$\begin{array}{c}\hfill \parallel P_{\phi }^g{\parallel }_{e,{\mathcal{B}}_{{log}_k,0}\to {\mathcal{B}}_{\mu }}\ge \underset{\left|\phi \right(z\left)\right|\to 1}{lim\; sup}\mu \left(z\right)\left|g\left(z\right)\right|\left({ln}^{[k+1]}\frac{a}{1-\left|\phi \right(z\left)\right|}-{ln}^{[k+1]}a\right).\end{array}$$

(36)

Let ${\rho }_m\in (0,1)$, $m\in \mathbb{N}$, ${\rho }_m\nearrow 1$ as $m\to +\infty$, and

$$\begin{array}{c}\hfill P_{{\rho }_m\phi }^g\left(f\right)\left(z\right)={\int }_0^{1}f\left({\rho }_m\phi \left(tz\right)\right)g\left(tz\right)\frac{dt}{t},m\in \mathbb{N}.\end{array}$$

(37)

Suppose that ${\left(h_k\right)}_{k\in \mathbb{N}}\subset X$ is bounded and $h_k\to 0$ uniformly on compacts of $\mathbb{B}$. We have $P_{\phi }^g\left(f_0\right)=g\in H_{\mu }^{\infty },$ so

$$\mu \left(z\right)|\Re P_{{\rho }_m\phi }^g\left(h_k\right)\left(z\right)|=\mu \left(z\right)|g\left(z\right)h_k\left({\rho }_m\phi \left(z\right)\right){|\le \parallel g\parallel }_{H_{\mu }^{\infty }}\underset{\left|w\right|\le {\rho }_m}{sup}\left|h_k\left(w\right)\right|\to 0,$$

as $k\to +\infty .$

Thus, Lemma 4 implies the compactness of $P_{{\rho }_m\phi }^g:X\to {\mathcal{B}}_{\mu }$, for each $m\in \mathbb{N}$.

Since $g\in H_{\mu }^{\infty }$, by Lemmas 2 and 3, we have that, for $r\in (0,1)$,

$$\begin{array}{cc}\hfill \parallel P_{\phi }^g-P_{{\rho }_m\phi }^g{\parallel }_{{\mathcal{B}}_{{log}_k}\to {\mathcal{B}}_{\mu }}=& \underset{{\parallel f\parallel }_{{\mathcal{B}}_{{log}_k}}^{\left(a\right)}\le 1}{sup}\underset{z\in \mathbb{B}}{sup}\mu \left(z\right)\left|g\left(z\right)\right||f\left(\phi \left(z\right)\right)-f\left({\rho }_m\phi \left(z\right)\right)|\hfill \\ \hfill \le & \underset{{\parallel f\parallel }_{{\mathcal{B}}_{{log}_k}}^{\left(a\right)}\le 1}{sup}\underset{\left|\phi \right(z\left)\right|\le r}{sup}\mu \left(z\right)\left|g\left(z\right)\right||f\left(\phi \left(z\right)\right)-f\left({\rho }_m\phi \left(z\right)\right)|\hfill \\ \hfill & +\underset{{\parallel f\parallel }_{{\mathcal{B}}_{{log}_k}}^{\left(a\right)}\le 1}{sup}\underset{\left|\phi \right(z\left)\right|>r}{sup}\mu \left(z\right)\left|g\left(z\right)\right||f\left(\phi \left(z\right)\right)-f\left({\rho }_m\phi \left(z\right)\right)|\hfill \\ \hfill \le & {\parallel g\parallel }_{H_{\mu }^{\infty }}\underset{{\parallel f\parallel }_{{\mathcal{B}}_{{log}_k}}^{\left(a\right)}\le 1}{sup}\underset{\left|\phi \right(z\left)\right|\le r}{sup}|f\left(\phi \left(z\right)\right)-f\left({\rho }_m\phi \left(z\right)\right)|\hfill \\ \hfill & +\underset{\left|\phi \right(z\left)\right|>r}{sup}\mu \left(z\right)\left|g\left(z\right)\right|\left({ln}^{[k+1]}\frac{a}{1-\left|\phi \right(z\left)\right|}-{ln}^{[k+1]}\frac{a}{1-{\rho }_m|\phi \left(z\right)|}\right)\hfill \\ \hfill \le & {\parallel g\parallel }_{H_{\mu }^{\infty }}\underset{{\parallel f\parallel }_{{\mathcal{B}}_{{log}_k}}^{\left(a\right)}\le 1}{sup}\underset{\left|\phi \right(z\left)\right|\le r}{sup}|f\left(\phi \left(z\right)\right)-f\left({\rho }_m\phi \left(z\right)\right)|\hfill \\ \hfill & +\underset{\left|\phi \right(z\left)\right|>r}{sup}\mu \left(z\right)\left|g\left(z\right)\right|\left({ln}^{[k+1]}\frac{a}{1-\left|\phi \right(z\left)\right|}-{ln}^{[k+1]}a\right).\hfill \end{array}$$

(38)

Furthermore,

$$\begin{array}{cc}\hfill & \underset{m\to +\infty }{lim\; sup}\underset{{\parallel f\parallel }_{{\mathcal{B}}_{{log}_k}}^{\left(a\right)}\le 1}{sup}\underset{\left|\phi \right(z\left)\right|\le r}{sup}|f\left(\phi \left(z\right)\right)-f\left({\rho }_m\phi \left(z\right)\right)|\hfill \\ \hfill \le & \underset{m\to +\infty }{lim\; sup}\underset{{\parallel f\parallel }_{{\mathcal{B}}_{{log}_k}}^{\left(a\right)}\le 1}{sup}\underset{\left|\phi \right(z\left)\right|\le r}{sup}(1-{\rho }_m)|\phi \left(z\right)|\underset{\left|w\right|\le r}{sup}|\nabla f\left(w\right)|\hfill \\ \hfill \le & \underset{m\to +\infty }{lim\; sup}\frac{(1-{\rho }_m)r}{(1-r){\prod }_{j=1}^k{ln}^{\left[j\right]}\frac{a}{1-r}}\underset{{\parallel f\parallel }_{{\mathcal{B}}_{{log}_k}}^{\left(a\right)}\le 1}{sup}{\parallel f\parallel }_{{\mathcal{B}}_{{log}_k}}^{\left(a\right)}=0.\hfill \end{array}$$

(39)

Letting $m\to +\infty$ in (38), using (39), then letting $r\to 1$, it follows that

$$\begin{array}{c}\hfill \parallel P_{\phi }^g{\parallel }_{e,{\mathcal{B}}_{{log}_k}\to {\mathcal{B}}_{\mu }}\le \underset{\left|\phi \right(z\left)\right|\to 1}{lim\; sup}\mu \left(z\right)\left|g\left(z\right)\right|\left({ln}^{[k+1]}\frac{a}{1-\left|\phi \right(z\left)\right|}-{ln}^{[k+1]}a\right).\end{array}$$

(40)

Relations (36), (40), and the obvious inequality

$$\parallel P_{\phi }^g{\parallel }_{e,{\mathcal{B}}_{{log}_k}\to {\mathcal{B}}_{\mu }}\ge {\parallel P_{\phi }^g\parallel }_{e,{\mathcal{B}}_{{log}_k,0}\to {\mathcal{B}}_{\mu }},$$

imply (27).

(b) From this assumption, the compactness of $P_{\phi }^g:X\to {\mathcal{B}}_{\mu }$ follows, similar to the operator in (37). So, (28) holds. □

Theorem 5.

Suppose that $k\in \mathbb{N}$, $a\in [2e^{\left[k\right]},+\infty )$, $g\in H\left(\mathbb{B}\right)$, $g\left(0\right)=0$, $\phi \in S\left(\mathbb{B}\right)$, $\mu \in W\left(\mathbb{B}\right)$, and $P_{\phi }^g:X\to {\mathcal{B}}_{\mu ,0}$ is bounded, where $X\in \{{\mathcal{B}}_{{log}_k},{\mathcal{B}}_{{log}_k,0}\}.$ Then,

$$\begin{array}{c}\hfill \parallel P_{\phi }^g{\parallel }_{e,X\to {\mathcal{B}}_{\mu ,0}}=\underset{\left|z\right|\to 1}{lim\; sup}\mu \left(z\right)\left|g\left(z\right)\right|\left({ln}^{[k+1]}\frac{a}{1-\left|\phi \right(z\left)\right|}-{ln}^{[k+1]}a\right).\end{array}$$

(41)

Proof. 

Since $P_{\phi }^g:X\to {\mathcal{B}}_{\mu ,0}$ is bounded, we have $P_{\phi }^g{f}_0=g\in H_{\mu ,0}^{\infty }.$

Assume that ${\parallel \phi \parallel }_{\infty }=1.$ Then,

$$\begin{array}{cc}& \underset{\left|z\right|\to 1}{lim\; sup}\mu \left(z\right)\left|g\left(z\right)\right|\left({ln}^{[k+1]}\frac{a}{1-\left|\phi \right(z\left)\right|}-{ln}^{[k+1]}a\right)\hfill \\ \hfill \ge & \underset{\left|\phi \right(z\left)\right|\to 1}{lim\; sup}\mu \left(z\right)\left|g\left(z\right)\right|\left({ln}^{[k+1]}\frac{a}{1-\left|\phi \right(z\left)\right|}-{ln}^{[k+1]}a\right).\hfill \end{array}$$

(42)

Choose ${\left(z_k\right)}_{k\in \mathbb{N}}\subset \mathbb{B}$ so that the following relation holds

$$\begin{array}{cc}& \underset{\left|z\right|\to 1}{lim\; sup}\mu \left(z\right)\left|g\left(z\right)\right|\left({ln}^{[k+1]}\frac{a}{1-\left|\phi \right(z\left)\right|}-{ln}^{[k+1]}a\right)\hfill \\ \hfill =& \underset{k\to \infty }{lim}\mu \left(z_k\right)\left|g\left(z_k\right)\right|\left({ln}^{[k+1]}\frac{a}{1-\left|\phi \right(z_k\left)\right|}-{ln}^{[k+1]}a\right).\hfill \end{array}$$

(43)

If ${sup}_{k\in \mathbb{N}}\left|\phi \left(z_k\right)\right|<1$, then the fact that $g\in H_{\mu ,0}^{\infty },$ implies

$$\underset{k\to \infty }{lim}\mu \left(z_k\right)\left|g\left(z_k\right)\right|\left({ln}^{[k+1]}\frac{a}{1-\left|\phi \right(z_k\left)\right|}-{ln}^{[k+1]}a\right)=0.$$

Thus, (42) and (43) imply

$$\underset{\left|\phi \right(z\left)\right|\to 1}{lim\; sup}\mu \left(z\right)\left|g\left(z\right)\right|\left({ln}^{[k+1]}\frac{a}{1-\left|\phi \right(z\left)\right|}-{ln}^{[k+1]}a\right)=0.$$

If ${sup}_{k\in \mathbb{N}}\left|\phi \left(z_k\right)\right|=1$, then $\left|\phi \right(z_{k_m}\left)\right|\to 1$ as $m\to +\infty$, for a subsequence ${\left(\phi \left(z_{k_m}\right)\right)}_{m\in \mathbb{N}}$. Hence,

$$\begin{array}{cc}& \underset{\left|z\right|\to 1}{lim\; sup}\mu \left(z\right)\left|g\left(z\right)\right|\left({ln}^{[k+1]}\frac{a}{1-\left|\phi \right(z\left)\right|}-{ln}^{[k+1]}a\right)\hfill \\ \hfill =& \underset{\left|\phi \right(z\left)\right|\to 1}{lim\; sup}\mu \left(z\right)\left|g\left(z\right)\right|\left({ln}^{[k+1]}\frac{a}{1-\left|\phi \right(z\left)\right|}-{ln}^{[k+1]}a\right).\hfill \end{array}$$

This, along with Theorem 4, implies the theorem in this case.

If ${\parallel \phi \parallel }_{\infty }<1,$ then $P_{\phi }^g:X\to {\mathcal{B}}_{\mu ,0}$ is compact, so that $\parallel P_{\phi }^g{\parallel }_{e,X\to {\mathcal{B}}_{\mu ,0}}=0.$ Since $g\in H_{\mu ,0}^{\infty },$ we have

$$\begin{array}{cc}\hfill & \underset{\left|z\right|\to 1}{lim\; sup}\mu \left(z\right)\left|g\left(z\right)\right|\left({ln}^{[k+1]}\frac{a}{1-\left|\phi \right(z\left)\right|}-{ln}^{[k+1]}a\right)\hfill \\ \hfill \le & \left({ln}^{[k+1]}\frac{a}{1-{\parallel \phi \parallel }_{\infty }}-{ln}^{[k+1]}a\right)\underset{\left|z\right|\to 1}{lim}\mu \left(z\right)\left|g\left(z\right)\right|=0.\hfill \end{array}$$

Hence, in this case, (41) holds. □

Corollary 2.

Suppose that $k\in \mathbb{N}$, $a\in [2e^{\left[k\right]},+\infty )$, $g\in H\left(\mathbb{B}\right)$, $g\left(0\right)=0$, $\phi \in S\left(\mathbb{B}\right)$, $\mu \in W\left(\mathbb{B}\right)$, and $X\in \{{\mathcal{B}}_{{log}_k},{\mathcal{B}}_{{log}_k,0}\}.$ Then, the following claims hold.

(a) 

$P_{\phi }^g:X\to {\mathcal{B}}_{\mu }$ is bounded if and only if

$$max\left\{{\parallel g\parallel }_{H_{\mu }^{\infty }},\underset{z\in \mathbb{B}}{sup}\mu \left(z\right)\left|g\left(z\right)\right|\left({ln}^{[k+1]}\frac{a}{1-\left|\phi \right(z\left)\right|}-{ln}^{[k+1]}a\right)\right\}<+\infty .$$

(b) 

If $P_{\phi }^g:X\to {\mathcal{B}}_{\mu }$ is bounded, then $P_{\phi }^g:X\to {\mathcal{B}}_{\mu }$ is compact if and only if

$$\underset{\left|\phi \right(z\left)\right|\to 1}{lim}\mu \left(z\right)\left|g\left(z\right)\right|\left({ln}^{[k+1]}\frac{a}{1-\left|\phi \right(z\left)\right|}-{ln}^{[k+1]}a\right)=0.$$

(c) 

If $P_{\phi }^g:X\to {\mathcal{B}}_{\mu ,0}$ is bounded, then $P_{\phi }^g:X\to {\mathcal{B}}_{\mu ,0}$ is compact if and only if

$$\underset{\left|z\right|\to 1}{lim}\mu \left(z\right)\left|g\left(z\right)\right|\left({ln}^{[k+1]}\frac{a}{1-\left|\phi \right(z\left)\right|}-{ln}^{[k+1]}a\right)=0.$$