1.1 Backgrounds and motivations

Let $\mathbb {D}=\{z\in \mathbb {C} :|z|<1\}$ be the unit disk of the complex plane $\mathbb {C}$. Let $H(\mathbb {D})$ be the space of holomorphic functions on $\mathbb {D}$ and let $\varphi$ be a holomorphic function on $\mathbb {D}$ with $\varphi (\mathbb {D})\subset \mathbb {D}$. For $f\in H(\mathbb {D})$, the composition operator $C_{\varphi }$ is a linear operator defined by $C_{\varphi }(f)=f\circ \varphi$.

Recall that a positive $T$ on a separable Hilbert space $H$ is in the trace class if

\[ \mathrm{tr}(T)=\sum_{n=0}^\infty\langle Te_n,e_n\rangle_H<{+}\infty \]

for some (or all) orthonormal basis $\{e_n\}$ of $H$. For any $0< p<\infty$, the Schatten $p$-class $\mathcal {S}_p(H)$ of $H$ consists of bounded linear operators $T:H\to H$ such that $(T^\ast T)^{p/2}$ belongs to the trace class. In particular, $\mathcal {S}_1(H)$ is the trace class of $H$, and $\mathcal {S}_2(H)$ is called the Hilbert–Schmidt class. It is easy to check that $T\in \mathcal {S}_p(H)$ if and only if $T^*\in \mathcal {S}_p(H)$. For more details about Schatten $p$-class operators, we refer the readers to Zhu [Reference Zhu16].

The Hardy space $H^2$ is a Hilbert space of analytic functions $f$ on $\mathbb {D}$ such that

\[ \|f\|_{H^2}^2=\sup_{0< r<1}\int_0^{2\pi}|f(r e^{i\theta})|^2\frac{\mathrm{d}\theta}{2\pi} <\infty. \]

For $\alpha >-1$, the weighted Bergman space $A_\alpha ^2$ consists of holomorphic functions $f$ on $\mathbb {D}$ satisfying

\[ \|f\|_{A_\alpha^2}^2= \int_{\mathbb{D}}|f(z)|^2 {\mathrm{d} A_\alpha(z)} <\infty, \]

where $\mathrm {d} A_\alpha (z)=(\alpha +1)(1-|z|^2)^\alpha \mathrm {d} A(z)$ and $\mathrm {d} A(z)$ is the normalized area measure on $\mathbb {D}$. When $\alpha =0$, the space $A_0^2$ is usually denoted by $A^2$. Properties of composition operator on $A_\alpha ^2$ and $H^2$ has been widely investigated for decades, see e.g. [Reference Cowen and MacCluer3, Reference Shapiro8, Reference Zhu16]. In particular, conditions for $C_{\varphi }$ that belong to $\mathcal {S}_p(A_\alpha ^2)$ and $\mathcal {S}_p(H^2)$ are also characterized, see [Reference Benazzouz, El-Fallah, Kellay and Mahzouli1, Reference Bendaoud, Korrichi, Merghni and Yagoub2, Reference Luecking4–Reference Pau and Pérez7, Reference Wirths and Xiao9, Reference Xia10, Reference Yuan and Zhou12, Reference Zhu14].

It is well known (see e.g. Zhu [Reference Zhu15]) that $H^2$ can be viewed as the limit case of $A^2_{\alpha }$ as $\alpha \to -1^+$ in some sense. It is also known that for $0< p<\infty$, $C_{\varphi }\in \mathcal {S}_p(H^2)$ if and only if

\[ \int_{\mathbb{D}}\left(\frac{N_\varphi(z)}{\log\frac1{|z|}}\right)^{p/2}\mathrm{d} \lambda(z)<\infty, \]

where

\[ \mathrm{d}\lambda(z)=(1-|z|^2)^{{-}2}\mathrm{d} A(z) \]

is the Möbius invariant measure on $\mathbb {D}$, and

\[ N_\varphi(z)=\sum_{w\in \varphi^{{-}1}(z)}\log\frac1{|w|} \]

is the Nevanlinna counting function of $\varphi$. Similarly, $C_{\varphi }\in \mathcal {S}_p(A_\alpha ^2)$ if and only if

\[ \int_{\mathbb{D}}\left(\frac{N_{\varphi,\alpha+2}(z)}{(\log \frac1{|z|})^{\alpha+2}}\right)^{p/2}\mathrm{d} \lambda(z)<\infty, \]

where $N_{\varphi,\alpha +2}(z)$ is a generalized Nevanlinna counting function of $\varphi$ given by

\[ N_{\varphi,\alpha+2}(z)=\sum_{w\in \varphi^{{-}1}(z)}\left(\log\frac1{|w|}\right)^{\alpha+2}. \]

See Luecking-Zhu [Reference Luecking and Zhu5].

1.2 Main results

A holomorphic map $\varphi :\mathbb {D}\to \mathbb {D}$ is of bounded valence if there is a positive integer $N$ such that for each $z\in \mathbb {D}$, the set $\varphi ^{-1}(z)$ contains at most $N$ points. Zhu [Reference Zhu14] shows that if $\alpha >-1$, $2\le p<\infty$ and $\varphi :\mathbb {D}\to \mathbb {D}$ is an analytic function of bounded valence, then $C_{\varphi }$ is in the Schatten class $\mathcal {S}_p$ of $A_\alpha ^2$ if and only if

\[ \int_{\mathbb{D}} \left(\frac{1-|z|^2}{1-|\varphi(z)|^2}\right)^{p(\alpha+2)/2} {\mathrm{d} \lambda(z)} <\infty. \]

Meanwhile, Zhu [Reference Zhu16, Exercise 11.6.7] says that if $p>2$ and $C_{\varphi }\in \mathcal {S}_p(H^2)$, then

\[ \int_{\mathbb{D}} \left(\frac{1-|z|^2}{1-|\varphi(z)|^2}\right)^{p/2} {\mathrm{d} \lambda(z)} <\infty. \]

These observations hint us to give the following result.

Theorem 1.1 If $2< p<\infty,$ $\varphi$ has bounded valence and

(1.1)\begin{equation} \int_{\mathbb{D}} \left(\frac{1-|z|^2}{1-|\varphi(z)|^2}\right)^{p/2} \mathrm{d}\lambda(z)<\infty, \end{equation}

then $C_{\varphi }\in \mathcal {S}_p(H^2)$.

For $p>2$, Xia [Reference Xia10] constructs a holomorphic map $\varphi :\mathbb {D}\to \mathbb {D}$ such that

\[ \int_{\mathbb{D}} \left(\frac{1-|z|^2}{1-|\varphi(z)|^2}\right)^p \mathrm{d}\lambda(z)<\infty \]

and such that $C_{\varphi }: A^2\to A^2$ does not belong to the Schatten class $\mathcal {S}_p(A^2)$. Motivated by Xia [Reference Xia10], we prove the following theorem:

Theorem 1.2 For any $2< p<\infty,$ there exists a holomorphic function $\varphi :\mathbb {D}\to \mathbb {D}$ such that

(1.2)\begin{equation} \int_{\mathbb{D}} \left(\frac{1-|z|^2}{1-|\varphi(z)|^2}\right)^{p/2} \mathrm{d}\lambda(z)<\infty, \end{equation}

but $C_{\varphi }:H^2\to H^2$ does not belong to the Schatten class $\mathcal {S}_p(H^2)$.

The proof of theorem 1.1 is based on Wirths-Xiao [Reference Wirths and Xiao9] and Zhu [Reference Zhu14]. The proof of theorem 1.2 is modified from Xia [Reference Xia10]. Although the idea of the proof of theorem 1.2 is coming from [Reference Xia10], there are several technical barriers we need to overcome. Thus, we need to adapt Xia's construction for our situation.

4.1 Construction of $\varphi$

The construction is modified from Xia [Reference Xia10]. We adapt some parameters for our argument. For $n=1,\,2,\,\dots$, let

\[ T_n=\left(2^{-(n+1)},2^{{-}n}\right]\quad \hbox{ and }\quad S_n=\left(( 4/3) 2^{-(n+1)}, ( 5/3) 2^{-(n+1)}\right]. \]

That is, $S_n$ is the middle third of $T_n$. Let $t_n= ( 4/3) 2^{-(n+1)}$ be the left end-point of $S_n$.

For fixed $p\in (2,\,\infty )$, let $\varepsilon$ be a fixed rational number such that

\[ 0<\varepsilon<\frac2p<1. \]

We can choose a strictly increasing sequence $k(1)<\dots < k(n)<\dots$ of positive integers such that

\[ 2^{-(\frac2p+\varepsilon)k(n)} \cdot 2\cdot 2^{ \varepsilon k(n) }=2^{- \frac2p k(n)+1} \le (1/3) 2^{-(n+1)}=|S_n| \]

for all $n$ and such that every $\varepsilon k(n)$ is an integer.

For integers $n\ge 1$ and $1\le j\le 2^{ \varepsilon k(n)}$, recall that $t_n$ is the left end-point of $S_n$. Define the intervals

\[ J_{n,j}=(a_{n,j},c_{n,j})=\left(t_n+2^{-(\frac2p+\varepsilon)k(n)} \cdot2\cdot (j-1), t_n+2^{-(\frac2p+\varepsilon)k(n)} \cdot2\cdot j\right) \]

and

\[ I_{n,j}=(a_{n,j},b_{n,j})=\left(t_n+2^{-(\frac2p+\varepsilon)k(n)} \cdot2\cdot (j-1), t_n+2^{-(\frac2p+\varepsilon)k(n)} \cdot (2j-1)\right). \]

It is easy to check that $I_{n,j}$ is the left half of $J_{n,j}$, $J_{n,j}$'s are pairwise disjoint,

\[ \bigcup_{j=1}^{2^{ \varepsilon k(n)}}J_{n,j}\subset S_n, \]

and the length of the interval $I_{n,j}$ is denoted by $\rho _n$, that is

(4.1)\begin{equation} \rho_n=|I_{n,j}|=b_{n,j}- a_{n,j}=2^{-(\frac2p+\varepsilon)k(n)}. \end{equation}

We now define a measurable function $u$ on the unit circle $\mathbb {T}=\{w\in \mathbb {C}:|w|=1\}$ as follows:

\begin{align*} u(\mathrm{e}^{\mathrm{i} t}) & =2^{{-}k(n)} \quad \hbox{ if } t\in \bigcup_{j=1}^{2^{ \varepsilon k(n)}}I_{n,j},n\ge1 ,\\ u(\mathrm{e}^{\mathrm{i} t}) & =1 \quad \hbox{ if } t\in (-\pi,\pi]\setminus\left(\bigcup_{n=1}^\infty \bigcup_{j=1}^{2^{ \varepsilon k(n)}}I_{n,j}\right). \end{align*}

The harmonic extension of $u$ to $\mathbb {D}$ is also denoted by $u$. Let

\[ h(z)=\frac1{2\pi}\int_{-\pi}^\pi\frac{\mathrm{e}^{\mathrm{i} t}+z}{\mathrm{e}^{\mathrm{i} t}-z} \,u(\mathrm{e}^{\mathrm{i} t})\,\mathrm{d} t \]

and

(4.2)\begin{equation} \varphi(z)=\exp({-}h(z)) \end{equation}

for all $z\in \mathbb {D}$. Then, $\mathrm {Re} (h(z))=u(z)>0$ for each $z\in \mathbb {D}$, and thus,

\[ |\varphi(z)|=\mathrm{e}^{\mathrm{Re} (h(z))}=\mathrm{e}^{{-}u(z)}<1. \]

This implies $\varphi (\mathbb {D})\subset \mathbb {D}$. We will need the fact that $\varphi \in H^2$ with

(4.3)\begin{equation} \|\varphi\|_{H^2}^2=\sup_{0< r<1}\frac1{2\pi}\int_{-\pi}^\pi\left| \varphi(r\mathrm{e}^{ \mathrm{i}\theta})\right|^2\mathrm{d}\theta= \frac1{2\pi}\int_{-\pi}^\pi\left| \varphi( \mathrm{e}^{ \mathrm{i}\theta})\right|^2\mathrm{d}\theta. \end{equation}

4.2 Estimates

For $z\in \mathbb {D}$ and $\mathrm {e}^{\mathrm {i} t}\in \mathbb {T}$, let

\[ P(z,\mathrm{e}^{\mathrm{i} t})=\frac{1-|z|^2}{|\mathrm{e}^{\mathrm{i} t}-z|^2} \]

be the Poisson kernel. It is shown in [Reference Xia10, p. 2508] that if $1/2\le r<1$ and $|\theta -t|\le 5$, then there exist constants $0<\alpha <\beta <\infty$ such that

(4.4)\begin{equation} \frac{\alpha(1-r)}{(1-r)^2+(\theta-t)^2}\le \frac1{2\pi} P(r\mathrm{e}^{\mathrm{i}\theta},\mathrm{e}^{\mathrm{i} t})\le \frac{\beta(1-r)}{(1-r)^2+(\theta-t)^2} . \end{equation}

We have the following lemma modified from [Reference Xia10, Lemma 4].

Lemma 4.1 For any positive integer $n$ and $1\le j\le 2^{ \varepsilon k(n)}$, let $G_{n,j}$ be the Carleson box based on $I_{n,j}$, i.e.

(4.5)\begin{equation} G_{n,j}=\left\{r\mathrm{e}^{\mathrm{i}\theta}: \theta\in I_{n,j},0<1-r\le\rho_n\right\}\!. \end{equation}

Then there is a constant $C_1$ independent of $n,\,j$ such that

(4.6)\begin{equation} \int_{G_{n,j}}\left(\frac{1-|z|}{1-|\varphi(z)|}\right)^{p/2}\mathrm{d} \lambda(z)\le C_1 2^{-\frac{p\varepsilon}2 k(n) }. \end{equation}

Proof. Given such a pair of $n,\,j$, we write

\[ G_{n,j}=\bigcup_{\nu=0}^{k(n)} G_{n,j}^\nu, \]

where

\[ G_{n,j}^0=\left\{r\mathrm{e}^{\mathrm{i}\theta}: \theta\in I_{n,j}, 0<1-r\le\rho_n\cdot 2^{{-}k(n)}\right\}\!, \]

and

\[ G_{n,j}^\nu=\left\{r\mathrm{e}^{\mathrm{i}\theta}: \theta\in I_{n,j}, \rho_n\cdot 2^{{-}k(n)}\cdot 2^{\nu-1}<1-r\le\rho_n\cdot 2^{{-}k(n)}\cdot 2^{\nu }\right\}\!, \]

for $1\le \nu \le k(n)$.

It is shown in [Reference Xia10, p. 2509] that there is a constant $0< c<1$ independent of $n,\,j$ such that

\[ 1-|\varphi(z)|=1-\mathrm{e}^{{-}u(z)}\ge 1-\exp({-}c 2^{{-}k(n)+\nu}) \]

if $z\in G_{n,j}^\nu$ and $0\le \nu \le k(n)$. Let $\delta =\inf _{0< x\le 1}x^{-1}(1-\mathrm {e}^{-x})$. Then,

(4.7)\begin{align} \inf_{z\in G_{n,j}^\nu}\left(1-|\varphi(z)|\right)^{p/2}\ge (\delta c)^{p/2}\cdot 2^{{-}p/2k(n)}\cdot 2^{p/2\nu}, \quad 0\le\nu \le k(n). \end{align}

This implies that

(4.8)\begin{align} & \int_{G_{n,j}}\left(\frac{1-|z|}{1-|\varphi(z)|}\right)^{p/2}\mathrm{d} \lambda(z)\nonumber\\& \quad=\int_{G^0_{n,j}}\left(\frac{1-|z|}{1-|\varphi(z)|}\right)^{p/2}\mathrm{d} \lambda(z)+\sum_{\nu=1}^{k(n)} \int_{G^\nu_{n,j}}\left(\frac{1-|z|}{1-|\varphi(z)|}\right)^{p/2}\mathrm{d} \lambda(z)\nonumber\\ & \quad\le\frac{2^{p/2k(n)}}{(\delta c)^{p/2}}\int_{G^0_{n,j}}(1-|z|^2)^{p/2-2}\mathrm{d} A(z)\nonumber\\& \qquad+ \sum_{\nu=1}^{k(n)}\frac{2^{p/2k(n)}}{(\delta c)^{p/2} \cdot 2^{p/2\nu}} \int_{G^\nu_{n,j}} (1-|z| )^{p/2-2}\mathrm{d} A(z). \end{align}

Notice that $p/2-2>-1$. Straightforward computation shows that

(4.9)\begin{align} \int_{G^0_{n,j}}(1-|z|^2)^{p/2-2}\mathrm{d} A(z) & =\frac1\pi \int_{I_{n,j}} \mathrm{d} \theta\int_{1-\rho_n\cdot2^{{-}k(n)}}^1(1-r^2)^{p/2-2} r\, \mathrm{d} r\nonumber\\& \le C_2 \rho_n^{p/2}\cdot 2^{-(p/2-1)k(n)} \end{align}

for some $C_2>0$, and

(4.10)\begin{align} \int_{G^\nu_{n,j}}(1-|z| )^{p/2-2}\mathrm{d} A(z) & =\frac1\pi \int_{I_{n,j}} \mathrm{d} \theta\int_{1-\rho_n\cdot2^{{-}k(n)}\cdot 2^\nu}^{1-\rho_n\cdot 2^{{-}k(n)}\cdot 2^{\nu-1}}(1-r )^{p/2-2}r\, \mathrm{d} r\nonumber\\& \le C_3 \rho_n^{p/2}\cdot 2^{-(p/2-1)k(n)} \cdot 2^{(p/2-1)\nu} \end{align}

for some $C_3>0$. Put (4.8), (4.9) and (4.10) together, we have

\begin{align*} & \int_{G_{n,j}}\left(\frac{1-|z|}{1-|\varphi(z)|}\right)^{p/2}\mathrm{d} \lambda(z)\\& \quad \le\frac{C_2 \cdot 2^{ k(n)} \cdot \rho_n^{p/2}}{(\delta c)^{p/2}} + \sum_{\nu=1}^{k(n)}\frac{2^{p/2k(n)}\cdot C_3 \rho_n^{p/2}\cdot 2^{-(p/2-1)k(n)} \cdot 2^{(p/2-1)\nu}}{(\delta c)^{p/2} \cdot 2^{p/2\nu}}\\ & \quad= 2^{ k(n)} \cdot \rho_n^{p/2}\cdot\left( \frac{C_2 }{(\delta c)^{p/2}}+ \frac{C_3 }{(\delta c)^{p/2}} \sum_{\nu=1}^{k(n)} 2^{-\nu}\right)\!. \end{align*}

Recall the inequality (4.1), we get the desired inequality (4.6) by letting

\[ C_1= \frac{C_2 }{(\delta c)^{p/2}}+ \frac{C_3 }{(\delta c)^{p/2}} \sum_{\nu=1}^{\infty} 2^{-\nu}= \frac{C_2 +C_3 }{(\delta c)^{p/2}}. \]

The following lemma is quoted from [Reference Xia10, Lemma 7].

Lemma 4.2 There is a $C_4>0$ such that

\[ u(z)\ge C_4 \quad\hbox{ for every }\quad z\in\mathbb{D}\setminus\left( \bigcup_{n=1}^\infty \bigcup_{j=1}^{2^{ \varepsilon k(n)}} G_{n,j}\right)\!, \]

where $G_{n,j}$ is defined by (4.5).

4.3 Proof of theorem 1.2

Let $\varphi$ be the holomorphic self-map of $\mathbb {D}$ given by (4.2). It is sufficient to check the inequality (1.2) for this $\varphi$, and $C_{\varphi }\notin \mathcal {S}_p(H^2)$.

Let

\[ G=\bigcup_{n=1}^\infty \bigcup_{j=1}^{2^{ \varepsilon k(n)}} G_{n,j}, \]

where $G_{n,j}$ is given by (4.5). For $z\in \mathbb {D}\setminus G$, lemma 4.2 implies that

\[ |\varphi(z)|=\mathrm{e}^{-\mathrm{Re}(h(z))}=\mathrm{e}^{{-}u(z)}\le \mathrm{e}^{{-}C_4}. \]

Since $p/2-2>-1$, we have

(4.11)\begin{align} \int_{\mathbb{D}\setminus G}\left(\frac{1-|z|^2}{1-|\varphi(z)|^2}\right)^{p/2}\mathrm{d} \lambda(z)& \le \frac1{(1-\mathrm{e}^{{-}C_4})^{p/2}} \int_{\mathbb{D}\setminus G} (1-|z|^2)^{p/2-2}\mathrm{d} A(z) \nonumber\\ & \le \frac1{(1-\mathrm{e}^{{-}C_4})^{p/2}} \int_{\mathbb{D}} (1-|z|^2)^{p/2-2}\mathrm{d} A(z)<\infty. \end{align}

Meanwhile, lemma 4.1 implies that

(4.12)\begin{align} \int_G \left(\frac{1-|z|^2}{1-|\varphi(z)|^2}\right)^{p/2}\mathrm{d} \lambda(z)& \approx \int_G \frac{(1-|z|)^{p/2-2} }{(1-|\varphi(z)|)^{p/2} }\, \mathrm{d} A(z)\nonumber\\& =\sum_{n=1}^\infty \sum_{j=1}^{2^{\varepsilon k(n)}}\int_{G_{n,j}} \frac{(1-|z|)^{p/2-2} }{(1-|\varphi(z)|)^{p/2} }\, \mathrm{d} A(z) \nonumber\\ & \le C_1 \sum_{n=1}^\infty 2^{\varepsilon k(n)}\cdot 2^{-\frac{p\varepsilon}2 k(n) } \le C_1 \sum_{n=1}^\infty 2^{-(p/2-1)\varepsilon k(n)}<\infty, \end{align}

where the last inequality is following from the fact that $p/2-1>0$. Now (1.2) follows from (4.11) and (4.12) easily.

It remains to check that $C_{\varphi }\notin \mathcal {S}_p(H^2)$, or equivalently, $\mathrm {tr}((C_{\varphi }^\ast C_{\varphi })^{\frac p2})=\infty$. Let $e_\ell (z)=z^\ell$, $\ell =0,\,1,\,2,\,\dots$. It is well known that $\{e_\ell :\ell \ge 0\}$ is an orthonormal basis for $H^2$. Since $p/2>1$, we have

\begin{align*} & \left\langle \left(C_{\varphi}^\ast C_{\varphi}\right)^{p/2} e_\ell,e_\ell\right \rangle_{H^2}\ge \left(\left\langle C_{\varphi}^\ast C_{\varphi} e_\ell, e_\ell\right\rangle_{H^2}\right)^{p/2}\\ & \quad =\left\| C_{\varphi} e_\ell\right\|_{H^2}^p= \left\| \varphi^l \right\|_{H^2}^p=\left(\frac1{2\pi}\int_{-\pi}^\pi \left|\varphi(\mathrm{e}^{ \mathrm{i}\theta})\right|^{2\ell}\mathrm{d}\theta\right)^{p/2}. \end{align*}

Write

\[ I_n=\bigcup_{j=1}^{2^{\varepsilon k(n)}}I_{n,j}. \]

Then,

\[ |I_n|=2^{\varepsilon k(n)}\rho_n= 2^{- \frac2p k(n)}, \]

and

\[ \left|\varphi(\mathrm{e}^{\mathrm{i}\theta})\right|= \exp({-u(\mathrm{e}^{\mathrm{i}\theta})}) =\exp({-}2^{{-}k(n)}) \]

for almost every $\theta \in I_n$. Thus,

\[ \int_{-\pi}^\pi \left|\varphi(\mathrm{e}^{ \mathrm{i}\theta})\right|^{2\ell}\mathrm{d}\theta \ge\sum_{n=1}^\infty \int_{I_{n }} \left|\varphi(\mathrm{e}^{ \mathrm{i}\theta})\right|^{2\ell}\mathrm{d} \theta= \sum_{n=1}^\infty \mathrm{e}^{{-}2\ell\cdot 2^{{-}k(n)}}\cdot 2^{- \frac2p k(n)}. \]

Notice that

\[ \left(\sum_n a_n\right)^s\ge \sum_n a_n^s \]

if $s\ge 1$ and $a_n\ge 0$. We get

\[ \left( \int_{-\pi}^\pi \left|\varphi(\mathrm{e}^{ \mathrm{i}\theta})\right|^{2\ell}\mathrm{d}\theta \right)^{p/2}\ge \left(\sum_{n=1}^\infty \mathrm{e}^{{-}2\ell\cdot 2^{{-}k(n)}}\cdot 2^{- \frac2p k(n)}\right)^{p/2}\ge \sum_{n=1}^\infty \mathrm{e}^{{-}p\ell\cdot 2^{{-}k(n)}}\cdot 2^{- k(n)}. \]

This gives that

\begin{align*} \mathrm{tr}\left((C_{\varphi}^\ast C_{\varphi})^{p/2}\right)& =\sum_{\ell=0}^\infty \left\langle (C_{\varphi}^\ast C_{\varphi})^{p/2} e_\ell,e_\ell\right\rangle_{H^2}\ge \sum_{\ell=0}^\infty \left(\frac1{2\pi}\int_{-\pi}^\pi \left|\varphi(\mathrm{e}^{ \mathrm{i}\theta})\right|^{2\ell}\mathrm{d}\theta\right)^{p/2}\\ & \ge\frac1{(2\pi)^{p/2}} \sum_{\ell=0}^\infty \sum_{n=1}^\infty \mathrm{e}^{{-}p\ell\cdot 2^{{-}k(n)}}\cdot 2^{- k(n)}\\ & =\frac1{(2\pi)^{p/2}}\sum_{n=1}^\infty \left( 2^{- k(n)} \sum_{\ell=0}^\infty \mathrm{e}^{{-}p\ell\cdot 2^{{-}k(n)}}\right)\\ & = \frac1{(2\pi)^{p/2}}\sum_{n=1}^\infty 2^{- k(n)} \cdot\frac1{1-\mathrm{e}^{{-}p \cdot 2^{{-}k(n)}}}. \end{align*}

Since

\[ \sup_{x>0}\frac{1-\mathrm{e}^{{-}x}}x\le1. \]

We have

\[ \frac1{1-\mathrm{e}^{{-}p \cdot 2^{{-}k(n)}}}\ge \frac1{p \cdot 2^{{-}k(n)}}. \]

Then,

\[ \sum_{n=1}^\infty 2^{- k(n)} \cdot\frac1{1-\mathrm{e}^{{-}p \cdot 2^{{-}k(n)}}}\ge \sum_{n=1}^\infty 2^{- k(n)} \cdot\frac1{p \cdot 2^{{-}k(n)}}=\sum_{n=1}^\infty\frac1p=\infty. \]

This implies that $C_{\varphi }\notin \mathcal {S}_p(H^2)$ and the proof is complete.