1 Introduction
Let D be the unit disk in the complex plane
$\textbf{C}$
and
$H(D)$
be the class of functions analytic in D. Let
$dA$
be the area measure on D normalized in such a way that
${A(D)=1}$
. We write
$d\lambda $
for the Möbius-invariant measure on D, i.e.,
$d\lambda (z)= (1-|z|^2)^{-2}dA(z)$
.
Recall that the Bergman space
$A^2$
is defined by
$$ \begin{align*} A^2=\{f: f\in H(D),\; \|f\|_{A^2}^2=\int_D|f(z)|^2dA(z)<\infty\}. \end{align*} $$
The Hardy space
$H^2$
is the Hilbert space of analytic functions f on D such that
$$ \begin{align*} \|f\|_{H^2}^2=\sup_{0<r<1}\int_{0}^{2\pi}|f(re^{i\theta})|^2\frac{d\theta}{2\pi}<\infty. \end{align*} $$
Given an analytic function
$\varphi : D \rightarrow D$
, we have the composition operator
$C_{\varphi }$
on
$A^2$
or
$H^2$
defined by the formula
$C_{\varphi }(f)=f\circ \varphi $
. Recall that such a
$C_\varphi $
is always bounded [Reference Zhu5].
Let H be a separable Hilbert space. For any
$1\leq p<\infty $
, the Schatten p-class
$\mathcal {C}_p$
consists of bounded linear operators T on H satisfying the condition
$\|T\|_p<\infty $
, where the p-norm is defined by the formula
$$ \begin{align*} \|T\|_p=\{\text{tr}~(|T|^p)\}^{1/p}=\{\text{tr}~((T^*T)^{p/2})\}^{1/p}. \end{align*} $$
The membership of composition operator
$C_\varphi $
in the Schatten class
$\mathcal {C}_p$
has been a constant source of fascination for operator theorists. In the case of the Bergman space
$A^2$
, Luecking and Zhu showed that
$C_\varphi \in \mathcal {C}_p$
if and only if the function
$z \mapsto \{\log (1/|z|)\}^{-2}N_{\varphi ,2}(z)$
belongs to
$L^{p/2}(D,d\lambda )$
, where
$N_{\varphi ,2}$
is a counting function associated with
$\varphi $
[Reference Luecking and Zhu2].
On the other hand, it would be more desirable to obtain a criterion for the membership
$C_\varphi \in \mathcal {C}_p$
in which
$\varphi $
appears in a more explicit way. One such approach involves the Berezin transform. Let
$k_z$
be the normalized reproducing kernel for
$A^2$
. Recall that for
$T \in \mathcal {B}(A^2)$
, the Berezin transform of T is the function
$$ \begin{align*}z \mapsto \langle Tk_z,k_z\rangle \end{align*} $$
on D. It is known (see [Reference Zhu6]) that
$C_\varphi \in \mathcal {C}_p$
if and only if
$$ \begin{align} \int_{D} \langle C_\varphi ^\ast C_\varphi k_z,k_z\rangle ^{p/2}d\lambda (z) < \infty. \end{align} $$
In view of this, one naturally considers the condition
$$ \begin{align} \int_{D} \langle C_\varphi C_\varphi ^\ast k_z,k_z\rangle ^{p/2}d\lambda (z) < \infty. \end{align} $$
Compared with (1.1), (1.2) appears more desirable because, by an easy calculation,
$$ \begin{align*}\langle C_\varphi C_\varphi ^\ast k_z,k_z\rangle = \bigg({1-|z|^2\over 1 - |\varphi (z)|^2}\bigg)^2, \end{align*} $$
which involves
$\varphi $
in a very direct way. Thus the following problem arose.
Problem 1.1 [Reference Luecking1, Reference Luecking and Zhu2, Reference Zhu5]
Let
$\varphi : D \rightarrow D$
be an analytic function. Is it true that for
${1 < p < \infty} $
, the composition operator
$C_\varphi : A^2 \rightarrow A^2$
is in the Schatten class
$\mathcal {C}_p$
if and only if
$$ \begin{align} \int_{D} \bigg({1-|z|^2\over 1 - |\varphi (z)|^2}\bigg)^pd\lambda (z) < \infty ? \end{align} $$
It is trivial that
$C_\varphi \in \mathcal {C}_2$
if and only if
$$ \begin{align*}\int_{D} \bigg({1-|z|^2\over 1 - |\varphi (z)|^2}\bigg)^2d\lambda (z) < \infty. \end{align*} $$
In [Reference Xia3], it was shown that when
$2 < p < \infty $
, (1.3) is not sufficient for the membership
$C_\varphi \in \mathcal {C}_p$
. That is, for each
$2 < p < \infty $
, there is an analytic
$\varphi : D \rightarrow D$
such that
$$ \begin{align*}\int_{D} \bigg({1-|z|^2\over 1 - |\varphi (z)|^2}\bigg)^pd\lambda (z) < \infty \end{align*} $$
and
$C_\varphi \not \in \mathcal {C}_p$
. In this article, we settle the remaining case, the case
$1 < p < 2$
. We will show that when
$1 < p < 2$
, (1.3) is not necessary for the membership
$C_\varphi \in \mathcal {C}_p$
. Here is our main result.
Theorem 1.1 For each
$1 < p < 2$
, there exists an analytic function
$\varphi : D \rightarrow D$
such that the composition operator
$C_{\varphi }:A^2\rightarrow A^2$
belongs to the Schatten class
$\mathcal {C}_p$
, and yet
$$ \begin{align} \int_{D} \left(\frac{1-|z|^2}{1-|\varphi(z)|^2}\right)^p d\lambda(z)=\infty. \end{align} $$
Together with the result in [Reference Xia3], Theorem 1.1 completes the contrast between (1.1) and (1.2). From the view point of operator theory, it is truly amazing that there is such a sharp contrast.
It will be interesting to consider what happens in the case of the Hardy space. Let
$k_z^{\text {Har}}$
denote the normalized reproducing kernel for the Hardy space
$H^2$
. By an easy calculation,
$$ \begin{align*}\langle C_\varphi C_\varphi ^\ast k_z^{\text{Har}},k_z^{\text{Har}}\rangle = {1-|z|^2\over 1 - |\varphi (z)|^2}. \end{align*} $$
Thus the Hardy-space equivalent of condition (1.3) is
$$ \begin{align} \int_{D} \bigg({1-|z|^2\over 1 - |\varphi (z)|^2}\bigg)^{p/2}d\lambda (z) < \infty. \end{align} $$
Recently, Yang and Yuan showed for each
$2<p<\infty $
, there is an analytic
$\varphi : D \rightarrow D$
such that
$$ \begin{align*} \int_{D} \left(\frac{1-|z|^2}{1-|\varphi(z)|^2}\right)^{p/2}d\lambda (z)<\infty \end{align*} $$
and such that the composition operator
$C_{\varphi } : H^2\rightarrow H^2$
does not belong to
$\mathcal {C}_p(H^2)$
[Reference Yang and Yuan4]. This settles the entire Hardy-space case. This is because (1.5) holds only if
$p> 2$
. For every
$p \leq 2$
, we have
$$ \begin{align*}\int_{D} (1-|z|^2)^{p/2}d\lambda (z) = \infty. \end{align*} $$
The remainder of the article consists of the proof of Theorem 1.1.
2 The proof of Theorem 1.1
The proof of Theorem 1.1 begins with a construction adapted from [Reference Xia3]. For
$n = 1,2,\ldots $
, define
$$ \begin{align*}T_n=\big(2^{-(n+1)},\; 2^{-n}\big] \quad \text{and} \quad S_n=\big((4/3)2^{-(n+1)},\; (5/3)2^{-(n+1)}\big]. \end{align*} $$
That is,
$S_n$
is the middle third of
$T_n$
. Denote
$t_n=(4/3)2^{-(n+1)}$
, the left end-point of
$S_n$
,
$n \in \textbf{N}$
.
Let
$1<p<2$
be given. We choose an
$\epsilon $
such that
$$ \begin{align*}0 < \epsilon < 1/p \end{align*} $$
and such that
$p\epsilon $
is a rational number. Thus
$p^{-1}> (p-1)\epsilon $
, and
$\lim _{k\rightarrow \infty } 2^{-(p^{-1}-(p-1)\epsilon )k} = 0$
. We can choose a strictly increasing sequence
$k(1)<\dots < k(n)<\dots $
of positive integers such that
$$ \begin{align} 2^{-(p^{-1}+\epsilon)k(n)}\cdot2\cdot2^{p\epsilon k(n)}=2^{-(p^{-1}-(p-1)\epsilon )k(n)+1}\leq (1/3)2^{-(n+1)} = |S_n| \end{align} $$
for every n and such that every
$p\epsilon k(n)$
is an integer. Note the difference between the choice of
$k(n)$
in this article and the choice in [Reference Xia3].
For integers
$n\geq 1$
and
$1\leq j\leq 2^{p\epsilon k(n)}$
, define the intervals
$$ \begin{align*} J_{n,j}&=(a_{n,j}, c_{n,j})=\big(t_n+2^{-(p^{-1}+\epsilon)k(n)}\cdot2\cdot(j-1),\; t_n+2^{-(p^{-1}+\epsilon)k(n)}\cdot2\cdot j\big),\\I_{n,j}&=(a_{n,j}, b_{n,j})=\big(t_n+2^{-(p^{-1}+\epsilon)k(n)}\cdot2\cdot(j-1),\; t_n+2^{-(p^{-1}+\epsilon)k(n)}\cdot(2j-1)\big). \end{align*} $$
It is easy to check that
$I_{n,j}$
is the left half of
$J_{n,j}$
and the
$J_{n,j}$
’s are pairwise disjoint. (2.1) ensures that
$$ \begin{align*} \bigcup_{j=1}^{2^{p\epsilon k(n)}}J_{n,j}\subset S_n. \end{align*} $$
We denote the length of the interval
$I_{n,j}$
by
$\rho _n$
. That is,
$$ \begin{align*} \rho_n=|I_{n,j}|=b_{n,j}-a_{n,j}=2^{-(p^{-1}+\epsilon)k(n)}. \end{align*} $$
We now define a measurable function u on the unit circle
$\textbf{T} = \{w \in \textbf{C}: |w|=1\}$
as follows:
$$ \begin{align*} u(e^{it})&=2^{-k(n)}\quad \text{if} \quad t\in \bigcup_{j=1}^{2^{p\epsilon k(n)}} I_{n,j},~~n\geq 1,\\ u(e^{it})&=1\quad \text{if} \quad t\in (-\pi, \pi]\setminus\Big\{\bigcup_{n=1}^{\infty} \bigcup_{j=1}^{2^{p\epsilon k(n)}} I_{n,j}\Big\}. \end{align*} $$
The harmonic extension of u to D will be denoted by the same symbol. Finally, define
$$ \begin{align*} \hspace{14pt} h(z)=\frac{1}{2\pi}\int_{-\pi}^{\pi}\frac{e^{it}+z}{e^{it}-z}u(e^{it})dt, \end{align*} $$
$$ \begin{align} \varphi(z)=\exp(-h(z)), \quad z\in D. \end{align} $$
Obviously,
$\text {Re}\{h(z)\}=u(z)>0$
, and consequently
$$ \begin{align*} |\varphi(z)|=e^{-\text{Re}\{h(z)\}} = e^{-u(z)} < 1 \end{align*} $$
for every
$z \in D$
. This implies
$\varphi (D) \subset D$
.
For
$z \in D$
and
$e^{it} \in \textbf{T}$
, let
$P(z, e^{it})=\frac {1-|z|^2}{|e^{it}-z|^2}$
be the Poisson kernel. It was shown in [Reference Xia3, p. 2508] that if
$1/2\leq r<1$
and
$|\theta -t|\leq 5$
, then there exist constants
$0<\alpha <\beta <\infty $
such that
$$ \begin{align} \frac{\alpha (1-r)}{(1-r)^2+(\theta-t)^2}\leq \frac{1}{2\pi}P(re^{i\theta}, e^{it})\leq \frac{\beta (1-r)}{(1-r)^2+(\theta-t)^2}. \end{align} $$
For any
$n \in \textbf{N}$
and
$1\leq j\leq 2^{p\epsilon k(n)}$
, define
$$ \begin{align} G_{n,j}=\{re^{i\theta}: \theta\in I_{n,j},\; 0<1-r\leq \rho_n\}. \end{align} $$
Given such a pair of
$n,j$
, we have
$$ \begin{align*} G_{n,j}=\bigcup_{\nu =0}^{k(n)}G_{n,j}^\nu , \end{align*} $$
where
$$ \begin{align*} G_{n,j}^0&=\{re^{i\theta}: \theta\in I_{n,j}, \;0<1-r\leq \rho_n\cdot 2^{-k(n)}\},\\G_{n,j}^\nu &=\{re^{i\theta}: \theta\in I_{n,j},\; \rho_n\cdot 2^{-k(n)}\cdot 2^{\nu -1}<1-r\leq \rho_n\cdot 2^{-k(n)}\cdot2^\nu \} \end{align*} $$
for
$1\leq \nu \leq k(n)$
. By [Reference Xia3, (2.6) and (2.7)], there is a constant
$0<c<1$
independent of
$n,j$
such that
$$ \begin{align} u(z)\geq c 2^{-k(n)+\nu } \quad \text{if} \ z\in G_{n,j}^\nu , \ 0\leq \nu \leq k(n). \end{align} $$
Recalling [Reference Xia3, (2.10)], we have
$$ \begin{align} A(G_{n,j}^\nu ) \leq \rho_n^2\cdot 2^{-k(n)}\cdot 2^\nu , \quad 0\leq \nu \leq k(n). \end{align} $$
The following two lemmas are quoted from [Reference Xia3, Lemma 7] and [Reference Xia3, Lemma 5], respectively.
Lemma 2.1 There is a
$c_1>0$
such that
$$ \begin{align*} u(z)\geq c_1 \quad \text{for every} \quad z\in D\Big\backslash \Big\{\bigcup_{n=1}^{\infty} \bigcup_{j=1}^{2^{p\epsilon k(n)}} G_{n,j}\Big\} \end{align*} $$
where
$G_{n,j}$
is defined by (2.4).
Lemma 2.2 For any
$n\geq 1$
and
$1\leq j\leq 2^{p\epsilon k(n)}$
, let
$B_{n,j}$
be the middle third of
$I_{n,j}$
. That is,
$B_{n,j}=(3^{-1}(b_{n,j}+2a_{n,j}),\; 3^{-1}(2b_{n,j}+a_{n,j}))$
, where
$a_{n,j}<b_{n,j}$
are the end-points of
$I_{n,j}$
. Furthermore, for such n and j, define
$$ \begin{align*} E_{n,j}=\{re^{it}: t\in B_{n,j},\; 0<1-r\leq \rho_n\cdot 2^{-k(n)}\}. \end{align*} $$
Then
$\sup _{z\in E_{n,j}} u(z)\leq (1+6\beta )2^{-k(n)}$
, where
$\beta $
is the constant that appears in (2.3).
We need one more lemma.
Lemma 2.3 There is a
$c_3> 0$
such that
$$ \begin{align*}\int _{E_{n,j}}{(1-|z|^2)^{p-2}\over (1 - |\varphi (z)|^2)^p}dA(z) \geq c_32^{-p\epsilon k(n)} \end{align*} $$
for all
$n \geq 1$
and
$1 \leq j \leq 2^{p\epsilon k(n)}$
.
Proof Denote
$\varphi _{n,j} = \inf \{|\varphi (z)| : z \in E_{n,j}\}$
,
$n \geq 1$
and
$1 \leq j \leq 2^{p\epsilon k(n)}$
. Then
$\varphi _{n,j} \geq e^{-C2^{-k(n)}}$
by Lemma 2.2. Writing
$\sigma = \sup _{0<x\leq C}(1 - e^{-x})/x$
, we have
$$ \begin{align*}{1\over 1 - |\varphi (z)|} \geq {1\over 1 - \varphi _{n,j}} \geq {1\over \sigma C2^{-k(n)}} = {2^{k(n)}\over \sigma C} \quad \text{for} \ z \in E_{n,j}. \end{align*} $$
Let
$c_2 = 2^{-2}(\sigma C)^{-p}$
. Then
$$ \begin{align*} \int _{E_{n,j}}{(1-|z|^2)^{p-2}\over (1 - |\varphi (z)|^2)^p}dA(z) &\geq c_22^{pk(n)}\int _{E_{n,j}}(1-|z|)^{p-2}dA(z) \\ &\geq c_32^{pk(n)}\cdot (\rho _n2^{-k(n)})^{p-1}\cdot \rho _n = c_32^{-p\epsilon k(n)}. \end{align*} $$
This completes the proof.
Proof of Theorem 1.1
We must show that the analytic function
$\varphi : D\rightarrow D$
defined by (2.2) has the property that
$C_{\varphi }\in \mathcal {C}_p$
and satisfies (1.4). Let us first verify
$C_{\varphi }\in \mathcal {C}_p$
.
To show that
$C_{\varphi }\in \mathcal {C}_p$
, we need the following inequality: For any
$0<\rho <1$
and
$0<x<1$
, using Hölder’s inequality with conjugate exponents
$1/\rho $
and
$1/(1-\rho )$
, we have
$$ \begin{align}\nonumber \sum_{l=0}^{\infty}(l+1)^\rho x^l &=\sum_{l=0}^{\infty}(l+1)^\rho\cdot x^{l\rho}\cdot x^{l(1-\rho)} \\\nonumber &\leq \Big\{ \sum_{l=0}^{\infty}\big((l+1)^\rho x^{l\rho}\big)^{1/\rho}\Big\}^\rho\cdot \Big\{ \sum_{l=0}^{\infty}\big(x^{l(1-\rho)}\big)^{1/(1-\rho)}\Big\}^{1-\rho} \\\nonumber &=\Big(\frac{1}{(1-x)^2}\Big)^\rho\Big(\frac{1}{1-x}\Big)^{1-\rho} \\ &= \frac{1}{(1-x)^{\rho+1}}. \end{align} $$
Let
$e_{l}(z)=(l+1)^{1/2}z^l, l=0,1,2,\dots $
. Recall that
$ \{e_l: l\geq 0\}$
is the standard orthonormal basis for the Bergman space
$A^2$
. Because
$1<p<2$
and
$\|e_l\|=1$
, it follows that
$$ \begin{align*} \langle (C_{\varphi}^*C_{\varphi})^{p/2}e_{l}, e_{l} \rangle &\leq\{\langle C_{\varphi}^*C_{\varphi}e_{l}, e_{l} \rangle\}^{p/2}=\|C_{\varphi}e_{l}\|_{A^2}^p=(l+1)^{p/2}\|\varphi^l\|_{A^2}^p\\ &=(l+1)^{p/2}\Big\{\int_D|\varphi(z)|^{2l} dA(z)\Big\}^{p/2}. \end{align*} $$
Let
$$ \begin{align*} G=\bigcup_{n=1}^{\infty} \bigcup_{j=1}^{2^{p\epsilon k(n)}} G_{n,j}, \end{align*} $$
where
$G_{n,j}$
is given by (2.4). For
$z\in D\backslash G$
, Lemma 2.1 implies that
$$ \begin{align} |\varphi(z)|=e^{-\text{Re}(h(z))}=e^{-u(z)}\leq e^{-c_1}. \end{align} $$
We have
$$ \begin{align*} \text{tr}~(({C_{\varphi}}^*C_{\varphi})^{p/2})&=\sum_{l=0}^{\infty}\langle (C_{\varphi}^*C_{\varphi})^{p/2}e_{l}, e_{l} \rangle\leq \sum_{l=0}^{\infty}(l+1)^{p/2}\Big\{\int_D|\varphi(z)|^{2l} dA(z)\Big\}^{p/2} \\&=\sum_{l=0}^{\infty}(l+1)^{p/2}\Big\{\int_{D\setminus G}|\varphi(z)|^{2l} dA(z)+\int_{G}|\varphi(z)|^{2l} dA(z)\Big\}^{p/2} \\&\leq I+J, \end{align*} $$
where
$$ \begin{align*} I &= \sum_{l=0}^{\infty}(l+1)^{p/2}\Big\{\int_{D\setminus G}|\varphi(z)|^{2l} dA(z)\Big\}^{p/2} \quad \text{and} \\ J &=\sum_{l=0}^{\infty}(l+1)^{p/2}\Big\{\int_{G}|\varphi(z)|^{2l} dA(z)\Big\}^{p/2}. \end{align*} $$
Applying (2.8), we obtain
$$ \begin{align*} I&=\sum_{l=0}^{\infty}(l+1)^{p/2}\Big\{\int_{D\setminus G}|\varphi(z)|^{2l} dA(z)\Big\}^{p/2} \leq \sum_{l=0}^{\infty}(l+1)^{p/2}\Big\{(e^{-c_1})^{2l}\int_{D\setminus G} dA(z)\Big\}^{p/2} \\&\leq \sum_{l=0}^{\infty}(l+1)^{p/2}\big\{(e^{-c_1})^{2l}\big\}^{p/2}=\sum_{l=0}^{\infty}(l+1)^{p/2} (e^{-p c_1})^{l} \leq \frac{1}{(1-e^{-p c_1})^{(p/2) +1}}, \end{align*} $$
where the last
$\leq $
follows from the condition
$p/2 < 1$
and (2.7).
Next we show that
$J < \infty $
. Note that
$$ \begin{align*} \Big(\sum_{n}a_n\Big)^s\leq \sum_{n}a_n^s \end{align*} $$
if
$s\leq 1$
and
$a_n\geq 0$
. Applying (2.5), (2.6), and (2.7), we obtain
$$ \begin{align*} J &= \sum_{l=0}^{\infty}(l+1)^{p/2}\Big\{ \sum_{n=1}^{\infty} \sum_{j=1}^{2^{p\epsilon k(n)}} \sum _{\nu =0}^{k(n)}\int_{G_{n,j}^\nu }|\varphi(z)|^{2l} dA(z) \Big\}^{p/2} \\&\leq \sum_{l=0}^{\infty}(l+1)^{p/2}\Big\{ \sum_{n=1}^{\infty} \sum_{j=1}^{2^{p\epsilon k(n)}} \sum_{\nu =0}^{k(n)}(e^{-c 2^{-k(n)+\nu }})^{2l}\rho _n^22^{-k(n)+\nu }\Big\}^{p/2} \\ &= \sum_{l=0}^{\infty}(l+1)^{p/2}\Big\{ \sum_{n=1}^{\infty}2^{p\epsilon k(n)} \sum_{\nu =0}^{k(n)}(e^{-c 2^{-k(n)+\nu }})^{2l}\rho _n^22^{-k(n)+\nu }\Big\}^{p/2} \\ &\leq \sum_{l=0}^{\infty}(l+1)^{p/2} \sum_{n=1}^{\infty} \sum_{\nu =0}^{k(n)}e^{-C_1l2^{-k(n)+\nu }}\cdot \rho _n^p\cdot 2^{(p^2/2)\epsilon k(n)}\cdot (2^{-k(n)+\nu })^{p/2} \\ & = \sum_{n=1}^{\infty}\sum_{\nu =0}^{k(n)} \bigg(\sum_{l=0}^{\infty}(l+1)^{p/2}e^{-C_1l2^{-k(n)+\nu }}\bigg)\rho _n^p\cdot 2^{(p^2/2)\epsilon k(n)}\cdot (2^{-k(n)+\nu })^{p/2} \\ &\leq \sum_{n=1}^{\infty}\sum_{\nu =0}^{k(n)}\frac{\rho _n^p\cdot 2^{(p^2/2)\epsilon k(n)}\cdot (2^{-k(n)+\nu })^{p/2}}{(1-e^{-C_12^{-k(n)+\nu }})^{(p/2) +1}}. \end{align*} $$
Let
$\delta =\inf _{0<x\leq C_1} x^{-1}(1-e^{-x}).$
Continuing with the above, we obtain
$$ \begin{align*} J &\leq \sum_{n=1}^{\infty} \sum_{\nu =0}^{k(n)}\frac{\rho _n^p\cdot 2^{(p^2/2)\epsilon k(n)}\cdot (2^{-k(n)+\nu })^{p/2}}{(\delta C_12^{-k(n)+\nu })^{(p/2) +1}} \\&= {1\over (\delta C_1)^{(p/2)+1}}\sum_{n=1}^{\infty} \sum_{\nu =0}^{k(n)}\frac{\rho _n^p\cdot 2^{(p^2/2)\epsilon k(n)}}{2^{-k(n)+\nu }} \\&\leq {2\over (\delta C_1)^{(p/2)+1}}\sum_{n=1}^{\infty}\frac{2^{-(1+p\epsilon )k(n)}\cdot 2^{(p^2/2)\epsilon k(n)}}{2^{-k(n)}} \\&= {2\over (\delta C_1)^{(p/2)+1}}\sum_{n=1}^{\infty} 2^{-(1-(p/2))p\epsilon k(n)} < \infty , \end{align*} $$
where the last step again uses the condition
$p/2 < 1$
. Therefore
$$ \begin{align*} \text{tr}~(({C_{\varphi}}^*C_{\varphi})^{p/2})\leq I+J <\infty. \end{align*} $$
This implies that
$C_{\varphi }\in \mathcal {C}_p$
.
It remains to verify that
$$ \begin{align*} \int_D\left(\frac{1-|z|^2}{1-|\varphi(z)|^2}\right)^p d\lambda(z)=\infty. \end{align*} $$
Obviously,
$$ \begin{align*}\int_D\left(\frac{1-|z|^2}{1-|\varphi(z)|^2}\right)^p d\lambda(z) \geq \sum _{n=1}^\infty \sum _{j=1}^{2^{p\epsilon k(n)}}\int_{E_{n,j}}\frac{(1-|z|^2)^{p-2}}{(1-|\varphi(z)|^2)^p}dA(z). \end{align*} $$
Applying Lemma 2.3, we have
$$ \begin{align*}\int_D\left(\frac{1-|z|^2}{1-|\varphi(z)|^2}\right)^p d\lambda(z) \geq \sum _{n=1}^\infty \sum _{j=1}^{2^{p\epsilon k(n)}}c_32^{-p\epsilon k(n)} = c_3\sum _{n=1}^\infty 2^{p\epsilon k(n)}\cdot 2^{-p\epsilon k(n)} = \infty. \end{align*} $$
This completes the proof.
