composition operators with closed range on the Dirichlet space

Guangfu Cao Cao: School of Mathematics and Information Science, Guangzhou University, Guangzhou 510006, China. [email protected] and Li He He: School of Mathematics and Information Science, Guangzhou University, Guangzhou 510006, China. [email protected]

Abstract.

It is well known that the composition operator on Hardy or Bergman space has a closed range if and only if its Navanlinna counting function induces a reverse Carleson measure. Similar conclusion is not available on the Dirichlet space. Specifically, the reverse Carleson measure is not enough to ensure that the range of the corresponding composition operator is closed. However, under certain assumptions, we in this paper set the necessary and sufficient condition for a composition operator on the Dirichlet space to have closed range.

Key words and phrases:

Dirichlet space, composition operator, closed range, counting function.

2010 Mathematics Subject Classification:

47B33, 47A53

G.C. was supported by NNSF of China (Grant No. 12071155), and L.H. was supported by NNSF of China (Grant No. 11871170).

*Corresponding author, email: [email protected]

1. Introduction

Let ${\mathbb{D}}$ be the unit disc in the complex plane ${\mathbb{C}},$ $\mathbb{T}$ be the unit circle, and let $dA$ denote area measure on ${\mathbb{D}}$ . The Dirichlet space, denoted by $\mathfrak{D}({\mathbb{D}})$ , consists of analytic functions $f$ on ${\mathbb{D}}$ such that

\|f\|^{2}=|f(0)|^{2}+\int_{\mathbb{D}}|f^{\prime}(z)|^{2}\,dA(z)<\infty.

Obviously, $\mathfrak{D}({\mathbb{D}})$ is a Hilbert space with the inner product

\langle f,g\rangle=f(0)\overline{g(0)}+\int_{\mathbb{D}}f^{\prime}(z)\overline{g^{\prime}(z)}\,dA(z).

Set $e_{0}(z)=1,\quad e_{n}(z)=\frac{1}{\sqrt{\pi}}\frac{z^{n}}{\sqrt{n}},\qquad(n=1,2,,\cdots),$ then $\{e_{n}\}$ is the standard orthonormal basis of $\mathfrak{D}({\mathbb{D}}).$ It follows that the reproducing kernel of $\mathfrak{D}({\mathbb{D}})$ is

K(z,w)=\sum_{n=0}^{\infty}e_{n}(z)\overline{e_{n}(w)}=1+\frac{1}{\pi}\log\frac{1}{1-z\overline{w}}.

Given an analytic self-map $\varphi:{\mathbb{D}}\to{\mathbb{D}}$ , for any $w\in\varphi({\mathbb{D}}),$ write $n_{\varphi}(w)$ is the cardinality of the set $\varphi^{-1}(w),$ $N_{\varphi}(w)$ is the (Nevanlinna) counting number function of $\varphi,$ and

\tau_{\varphi}(w)=\frac{N_{\varphi}(w)}{log\frac{1}{|w|}}.

In the past few decades, the closed-range composition operators on various function spaces has attracted wide attention, we refer to [1, 2, 3, 4, 6, 7, 8, 11, 12, 13, 14, 16, 17, 18] and the references therein. In 1974 Cima, Thompson, and Wogen determined when the composition operators on Hardy space $H^{2}$ have closed-ranges by utilizing the boundary behaviour of the inducing maps (see [2]). In the end of their paper, they posed the problem of characterizing the closed-range composition operators using the properties of the range of the inducing maps on the unit disc rather than the properties of the boundary. Zorboska [17] gave an answer to this question by using the Nevanlinna counting function and Luecking’s measure theoretic results on inequalities on Bergman spaces. He proved the following

Theorem Z.

([17]). Let $\varphi$ be an analytic self-mapping of ${\mathbb{D}}.$ Assume that $C_{\varphi}$ is a bounded composition operator on the Hardy space $H^{2}({\mathbb{D}})$ defined as

C_{\varphi}f(z)=(f\circ\varphi)(z),\hskip 14.22636pt\forall\hskip 5.69054ptf\in H^{2}({\mathbb{D}}).

Then $C_{\varphi}$ has closed range if and only if there exists a positive constant $c>0$ such that the set $G_{c}=\{z\in{\mathbb{D}}:\tau_{\varphi}(z)>c\}$ satisfies the condition:

There exists a constant $\delta>0$ such that

A(G_{c}\cap S(\zeta,r))>\delta A({\mathbb{D}}\cap S(\zeta,r))\hskip 28.45274pt(*)

for all $\zeta\in\mathbb{T}$ and $r>0,$ where $S(\zeta,r)=\{z\in{\mathbb{D}}:|z-\zeta|<r\}.$

In addition, Zorboska constructed a counterexample which shows that the range of $C_{\varphi}$ may not be closed even if $\varphi({\mathbb{D}})={\mathbb{D}}.$ It seems a little strange since $C_{\varphi}$ is invertible on $H^{2}({\mathbb{D}})$ if $\varphi\in Aut({\mathbb{D}}).$

For the case of the (weighted) Bergman space, one also obtained the necessary and sufficient conditions for the closed range composition operators (see [1, 8]). However, in the case of the Dirichlet space, it seems rather difficult.

Recall that the pseudo-hyperbolic metric on ${\mathbb{D}}$ is defined by

\rho(z,w)=|\frac{z-w}{1-\bar{z}w}|,\hskip 14.22636ptz,w\in{\mathbb{D}},

and the Bergman metric is defined by

\beta(z,w)=\frac{1}{2}log\frac{1+\rho(z,w)}{1-\rho(z,w)},\hskip 14.22636ptz,w\in{\mathbb{D}}.

For $0<\eta<1,$ $z\in{\mathbb{D}}$ and $r>0,$ write

D_{\eta}(z)=\{w\in{\mathbb{D}}|\rho(z,w)<\eta\}

and

D(z,r)=\{w\in{\mathbb{D}}|\beta(z,w)<r\}.

For $G\subset\mathbb{C},$ $A(G)$ denotes the Lebesgue measure of $G.$ Then for any fixed positive $r,$ we have

A(D(z,r))\sim(1-|z|^{2})^{2}.

In fact, $D(z,r)$ is the Euclidean disk with Euclidean center $(1-s^{2})z/(1-s^{2}|z|^{2})$ and Euclidean radius $(1-|z|^{2})s/(1-s^{2}|z|^{2}),$ where $s=tanh(r)\in(0,1).$ One may consult the book [15] by Kehe Zhu for details.

Luecking [10] proved that a necessary condition for a composition operator $C_{\varphi}$ on the Dirichlet space $\mathcal{D}({\mathbb{D}})$ to have closed range is that $n_{\varphi}dA$ must be a reverse Carleson measure, that is, there is a $\delta>0$ and $\eta\in(0,1)$ such that

\int_{\varphi({\mathbb{D}})\cap D_{\eta}(z)}n_{\varphi}(w)dA(w)>\delta A(D_{\eta}(z))

for all $z\in\mathbb{D},$ or equivalently, there is a $\delta>0$ such that

\int_{\varphi({\mathbb{D}})\cap S(\zeta,r)}n_{\varphi}(w)dA(w)>\delta A(S(\zeta,r))

for all $\zeta\in\mathbb{T}.$ Note that

D(z,r)=\{w\in{\mathbb{D}}:\beta(z,w)<r\}=D_{\eta}(z)=\{w\in{\mathbb{D}}:\rho(z,w)<\eta\}

for $\eta=\frac{e^{2r}-1}{e^{2r}+1},$ and then that the following statements are equivalent:

(1) There is a $\delta>0$ and $r>0$ such that

\int_{\varphi({\mathbb{D}})\cap D(z,r)}n_{\varphi}(w)dA(w)>\delta A(D(z,r))

for all $z\in\mathbb{D}.$

(2) There is a $\delta>0$ and $\eta\in(0,1)$ such that

\int_{\varphi({\mathbb{D}})\cap D_{\eta}(z)}n_{\varphi}(w)dA(w)>\delta A(D_{\eta}(z))

for all $z\in\mathbb{D}.$

In 1999 Luecking [11] constructed a self-mapping $\varphi$ on ${\mathbb{D}}$ such that $n_{\varphi}dA$ is a reverse Carleson measure, but $C_{\varphi}$ has not closed range. Unfortunately, there is no necessary and sufficient condition for a composition operator to have closed range. In contrast to Hardy space, the composition operator with surjective symbol must have closed range on Dirichlet space. In fact, if $\varphi$ is an analytic self-mapping on ${\mathbb{D}}$ , and $\varphi({\mathbb{D}})={\mathbb{D}},$ then $n_{\varphi}(w)\geq 1$ for any $w\in{\mathbb{D}},$ and

$\displaystyle\\|C_{\varphi}f\\|^{2}_{\mathcal{D}}$	$\displaystyle=$	$\displaystyle\int_{{\mathbb{D}}}\|(f\circ\varphi)^{\prime}(z)\|^{2}dA(z)$
	$\displaystyle=$	$\displaystyle\int_{\varphi({\mathbb{D}})}\|f^{\prime}(w)\|^{2}n_{\varphi}(w)dA(w)$
	$\displaystyle=$	$\displaystyle\int_{{\mathbb{D}}}\|f^{\prime}(w)\|^{2}n_{\varphi}(w)dA(w)$
	$\displaystyle\geq$	$\displaystyle\int_{{\mathbb{D}}}\|f^{\prime}(w)\|^{2}dA(w)=\\|f\\|^{2}_{\mathcal{D}}.$

This implies that $C_{\varphi}$ has closed range.

The discussions above means that even if the measure $n_{\varphi}(w)dA(w)$ induced by counting function $n_{\varphi}(w)$ is a reverse Carleson measure, it also cannot guarantee that the composition operator $C_{\varphi}$ has closed range. However, if $\varphi$ is surjective, then the range of $C_{\varphi}$ must be closed. This phenomenon is not surprising, since the counting function $n_{\varphi}(w)$ may be unbounded when $w$ approaches the boundary of the domain, although the image of the symbol map cannot fill the neighborhood of $\mathbb{T},$ the counting function may introduce a reverse Carleson measure. It can show up from the counterexample of Luecking in [17]. This shows that the reverse Carleson measure is not a proper condition for composition operator with closed range on Dirichlet space $\mathcal{D}({\mathbb{D}})$ in some extreme cases.

Notice that

\int_{\varphi({\mathbb{D}})\cap D(z,r)}n_{\varphi}(w)A(w)\geq\int_{\varphi({\mathbb{D}})\cap D(z,r)}A(w)=A(\varphi({\mathbb{D}})\cap D(z,r)),

then the inequality

A(\varphi({\mathbb{D}})\cap D(z,r))\geq\delta A(D(z,r))

for all $z\in\mathbb{D}$ and some $r>0$ implies that $n_{\varphi}(w)A(w)$ is a reverse Carleson measure, but the contrary may not be true. In other words, the inequality

A(\varphi({\mathbb{D}})\cap D(z,r))\geq\delta A(D(z,r))

is stronger than that $n_{\varphi}(w)dA(w)$ is a reverse Carleson measure. Further, they are equivalent under the assumption that $n_{\varphi}(w)$ is bounded. In fact, if $n_{\varphi}(w)$ is bounded on ${\mathbb{D}},$ then for any $z\in\mathbb{D}$ and $r>0,$ there is a constant $M>0$ such that

$\displaystyle MA(\varphi({\mathbb{D}})\cap D(z,r))$	$\displaystyle\geq$	$\displaystyle\int_{\varphi({\mathbb{D}})\cap D(z,r)}n_{\varphi}(w)dA(w)$
	$\displaystyle\geq$	$\displaystyle\int_{\varphi({\mathbb{D}})\cap D(z,r)}dA(w)$
	$\displaystyle=$	$\displaystyle A(\varphi({\mathbb{D}})\cap D(z,r)).$

In this case, the inequality

A(\varphi({\mathbb{D}})\cap D(z,r))\geq\delta A(D(z,r))

is equivalent to that $n_{\varphi}(w)dA(w)$ is a reverse Carleson measure. It seems that a proper condition for composition operator with closed range is:

There exists a constant $\delta>0$ and $r>0$ such that

A(\varphi({\mathbb{D}})\cap D(z,r))\geq\delta A(D(z,r))\hskip 28.45274pt(**)

for all $z\in\mathbb{D}.$

Under some natural assumptions, $(**)$ is indeed the necessary and sufficient condition for composition operators to have closed range. Our main result is as follow

Main Theorem.

Let $\varphi$ be an analytic self-mapping function of ${\mathbb{D}}.$ If

lim_{k\rightarrow\infty}sup_{f\in(\mathcal{D}({\mathbb{D}}))_{1}}\int_{\varphi({\mathbb{D}})\cap\{w\in{\mathbb{D}}:n_{\varphi}(w)>k\}}|f^{\prime}(w)|^{2}n_{\varphi}(w)dA(w)=0,

where $(\mathcal{D}({\mathbb{D}}))_{1}$ denotes the unit sphere of $\mathcal{D}({\mathbb{D}}),$ then the following statements are equivalent:

(a)

$R(C_{\varphi})$ is closed on $\mathcal{D}({\mathbb{D}}).$
(b)

There is a $\delta>0$ and $r>0$ such that

$A(\varphi({\mathbb{D}})\cap D(z,r))\geq\delta A(D(z,r))$

for all $z\in{\mathbb{D}}$ .
(c)

$n_{\varphi}(w)dA(w)$ is a reverse Carleson measure.

2. Composition operator with closed range

Proposition 2.1.

Let $\varphi$ be an analytic self-mapping function of ${\mathbb{D}}.$ Assume $C_{\varphi}$ is a bounded composition operator on $\mathcal{D}({\mathbb{D}})$ defined as

C_{\varphi}f(z)=(f\circ\varphi)(z),\hskip 14.22636pt\forall\hskip 5.69054ptf\in\mathcal{D}({\mathbb{D}}).

If $R(C_{\varphi})=\{(f\circ\varphi|f\in\mathcal{D}({\mathbb{D}})\}$ , the range of $C_{\varphi},$ is closed, then for any $\zeta\in\mathbb{T}$ and any $r>0,$

S(\zeta,r)\cap\varphi({\mathbb{D}})\neq\phi.

Equivalently, $\mathbb{T}\subset\overline{\varphi({\mathbb{D}})}.$

Proof.

Assume $R(C_{\varphi})$ is closed, since $kerC_{\varphi}=\{0\},$ we know that there is a constant $c>0$ such that

\|C_{\varphi}f\|\geq c\|f\|,\hskip 14.22636pt\forall f\in\mathcal{D}({\mathbb{D}}).\hskip 56.9055pt(***)

Thus

$\displaystyle\\|C_{\varphi}f\\|^{2}$	$\displaystyle=$	$\displaystyle\int_{{\mathbb{D}}}\|(f\circ\varphi)^{\prime}(z)\|^{2}dA(z)+\|f\circ\varphi(0)\|^{2}$
	$\displaystyle=$	$\displaystyle\int_{\varphi({\mathbb{D}})}\|f^{\prime}(w)\|^{2}n_{\varphi}(w)dA(w)+\|f\circ\varphi(0)\|^{2}$
	$\displaystyle\geq$	$\displaystyle c^{2}[\int_{{\mathbb{D}}}\|f^{\prime}(z)\|^{2}dA(z)+\|f(0)\|^{2}].$

If there is a $\zeta\in\mathbb{T}$ and $r>0$ such that

S(\zeta,r)\cap\varphi({\mathbb{D}})=\phi,

let

f_{\zeta}(z)=\frac{1+\bar{\zeta}\cdot z}{2},\hskip 14.22636ptf_{k}(z)=f_{\zeta}^{k}(z),

then $f_{\zeta}(z)$ is the peak function at $\zeta$ on ${\mathbb{D}}.$ For arbitrary open neighborhood $U$ of $\zeta,$ it is easy to see that

			$\displaystyle lim_{k\rightarrow\infty}\sqrt[k]{\frac{\int_{{\mathbb{D}}-U}\|f^{\prime}_{k}(z)\|^{2}dA(z)}{\int_{{\mathbb{D}}}\|f^{\prime}_{k}(z)\|^{2}dA(z)}}$
		$\displaystyle=$	$\displaystyle\frac{max\|f_{\zeta}\|_{{\mathbb{D}}-U}\|}{max\|f_{\zeta}\|_{{\mathbb{D}}}\|}=max\|f_{\zeta}\|_{{\mathbb{D}}-U}\|<1.$

Hence

\frac{\int_{{\mathbb{D}}-U}|f^{\prime}_{k}(z)|^{2}dA(z)}{\int_{{\mathbb{D}}}|f^{\prime}_{k}(z)|^{2}dA(z)}\rightarrow 0\hskip 14.22636pt\mbox{as}\hskip 14.22636ptk\rightarrow\infty.

Direct calculation gives that

	$\displaystyle\frac{\\|C_{\varphi}f_{k}\\|^{2}}{\\|f_{k}\\|^{2}}$	$\displaystyle=$	$\displaystyle\frac{\int_{{\mathbb{D}}}\|(f_{k}\circ\varphi)^{\prime}(z)\|^{2}dA(z)+\|f_{k}\circ\varphi(0)\|^{2}}{\int_{{\mathbb{D}}}\|f^{\prime}_{k}(z)\|^{2}dA(z)+\|f_{k}(0)\|^{2}}$
		$\displaystyle=$	$\displaystyle\frac{\int_{\varphi({\mathbb{D}})}\|f^{\prime}_{k}(w)\|^{2}n_{\varphi}(w)dA(w)+\|f_{k}\circ\varphi(0)\|^{2}}{\int_{{\mathbb{D}}}\|f^{\prime}_{k}(z)\|^{2}dA(z)+\|f_{k}(0)\|^{2}}.$

By the boundedness of $C_{\varphi},$ there exists a positive constant $M$ such that

\int_{{\mathbb{D}}}|f^{\prime}(w)|^{2}n_{\varphi}(w)dA(w)\leq M\int_{{\mathbb{D}}}|f^{\prime}(w)|^{2}dA(w),\hskip 14.22636pt\forall f\in\mathcal{D}({\mathbb{D}}).

In particular,

\int_{{\mathbb{D}}}n_{\varphi}(w)dA(w)\leq M\int_{{\mathbb{D}}}dA(w)=MA({\mathbb{D}}).

Thus

lim_{k\rightarrow\infty}\sqrt[k]{\int_{\varphi({\mathbb{D}})}|f^{\prime}_{k}(z)|^{2}n_{\varphi}(z)dA(z)}=max|f_{\zeta}|_{\varphi({\mathbb{D}})}|.

Since there is a neighborhood $U$ of $\zeta$ such that $U\cap\overline{\varphi({\mathbb{D}})}=\phi,$ we have $\overline{\varphi({\mathbb{D}})}\subset\overline{{\mathbb{D}}}-U.$ Further,

			$\displaystyle lim_{k\rightarrow\infty}\sqrt[k]{\frac{\int_{\varphi({\mathbb{D}})}\|f^{\prime}_{k}(z)\|^{2}n_{\varphi}(z)dA(z)}{\int_{{\mathbb{D}}}\|f^{\prime}_{k}(z)\|^{2}dA(z)}}$
		$\displaystyle=$	$\displaystyle\frac{max\|f_{\zeta}\|_{\varphi({\mathbb{D}})}\|}{max\|f_{\zeta}\|_{{\mathbb{D}}}\|}=max\|f_{\zeta}\|_{\varphi({\mathbb{D}})}\|<1.$

With the fact that $f_{k}(0)\rightarrow 0$ and $f_{k}\circ\varphi(0)\rightarrow 0,$ we get

\frac{\|C_{\varphi}f_{k}\|^{2}}{\|f_{k}\|^{2}}=\frac{\int_{\varphi({\mathbb{D}})}|f^{\prime}_{k}(w)|^{2}n_{\varphi}(z)dA(w)+|f_{k}\circ\varphi(0)|^{2}}{\int_{{\mathbb{D}}}|f^{\prime}_{k}(z)|^{2}dA(z)+|f_{k}(0)|^{2}}\rightarrow 0,\hskip 8.53581pt\mbox{as}\hskip 8.53581ptk\rightarrow\infty.

This contradicts to $(***).$ It shows that

S(\zeta,r)\cap\varphi({\mathbb{D}})\neq\phi,

which implies $\mathbb{T}\subset\overline{\varphi({\mathbb{D}})}.$ ∎

Let $\alpha>-1$ and $1\leq p<\infty.$ We say that $G$ , a Borel subset of ${\mathbb{D}},$ satisfies the reverse Carleson condition on $A_{\alpha}^{p},$ the weighted Bergman space, if there exists positive constant $\eta$ such that

\eta\int_{G}|f(z)|^{p}(1-|z|^{2})^{\alpha}dA(z)\geq\int_{{\mathbb{D}}}|f(z)|^{p}(1-|z|^{2})^{\alpha}dA(z),\hskip 11.38109pt\forall f\in A_{\alpha}^{p}.

Theorem KRL.

([7, 9]) Let $G\subset{\mathbb{D}}$ be a Borel subset of ${\mathbb{D}}.$ Then $G$ satisfies the reverse Carleson condition if and only if the following condition holds:
There is a $\delta>0$ and $r>0$ such that

A(G\cap D(z,r))>\delta A(D(z,r))\hskip 14.22636pt\mbox{for each}\hskip 14.22636ptz\in{\mathbb{D}},

or equivalently, there is a $\delta>0$ and $\eta\in(0,1)$ such that

A(G\cap D_{\eta}(z))>\delta A(D_{\eta}(z))\hskip 14.22636pt\mbox{for each}\hskip 14.22636ptz\in{\mathbb{D}}.

Proposition 2.2.

Let $\varphi$ be an analytic self-mapping of ${\mathbb{D}}$ and $C_{\varphi}$ be bounded on $\mathcal{D}({\mathbb{D}}).$ Then the following statements are equivalent:

(a)

There is a $\delta>0$ and $r>0$ such that

$A(\varphi({\mathbb{D}})\cap D(z,r))\geq\delta A(D(z,r))$

for all $z\in{\mathbb{D}}$ .
(b)

For arbitrary $\alpha\in(0,1)$ , there is a $\delta>0$ and $r>0$ such that

$\int_{\varphi({\mathbb{D}})\cap D(z,r)}n_{\varphi}^{\alpha}(w)dA(w)\geq\delta A(D(z,r))$

for all $z\in{\mathbb{D}}.$

(c)

For arbitrary $\alpha\in(0,1)$ , there is a $\delta>0$ such that

\int_{\varphi({\mathbb{D}})}|f^{\prime}(w)|^{2}n_{\varphi}^{\alpha}(w)dA(w)\geq\delta\int_{{\mathbb{D}}}|f^{\prime}(w)|^{2}dA(w)

for all $f\in\mathcal{D}({\mathbb{D}}).$

Proof.

Note $n_{\varphi}^{\alpha}(w)\geq 1$ for any $w\in\varphi({\mathbb{D}})$ and $\alpha\in(0,1),$ we have

\int_{\varphi({\mathbb{D}})}|f^{\prime}(w)|^{2}n_{\varphi}^{\alpha}(w)dA(w)\geq\int_{\varphi({\mathbb{D}})}|f^{\prime}(w)|^{2}dA(w)

for all $f\in\mathcal{D}({\mathbb{D}}).$ Hence $(a)\Longrightarrow(c)$ is obvious by Theorem KRL.

Conversely, for arbitrary $\alpha\in(0,1)$ and $\epsilon>0,$ there is a $N\in\mathbb{N}$ such that

\frac{n^{\alpha}}{n}=\frac{1}{n^{1-\alpha}}<\epsilon

for all $n\geq N.$ Thus

			$\displaystyle\int_{\varphi({\mathbb{D}})\cap\{w:n_{\varphi}(w)>N\}}\|f^{\prime}(w)\|^{2}n_{\varphi}^{\alpha}(w)dA(w)$
		$\displaystyle=$	$\displaystyle\int_{\varphi({\mathbb{D}})\cap\{w:n_{\varphi}(w)>N\}}\|f^{\prime}(w)\|^{2}\frac{n_{\varphi}^{\alpha}(w)}{n_{\varphi}(w)}n_{\varphi}(w)dA(w)$
		$\displaystyle\leq$	$\displaystyle\epsilon\int_{\varphi({\mathbb{D}})\cap\{w:n_{\varphi}(w)>N\}}\|f^{\prime}(w)\|^{2}n_{\varphi}(w)dA(w)$
		$\displaystyle\leq$	$\displaystyle\epsilon\int_{\varphi({\mathbb{D}})}\|f^{\prime}(w)\|^{2}n_{\varphi}(w)dA(w)$
		$\displaystyle\leq$	$\displaystyle\epsilon\\|C_{\varphi}\\|^{2}\cdot\int_{{\mathbb{D}}}\|f^{\prime}(w)\|^{2}dA(w)$

and

			$\displaystyle\int_{\varphi({\mathbb{D}})\cap\{w:n_{\varphi}(w)\leq N\}}\|f^{\prime}(w)\|^{2}n_{\varphi}^{\alpha}(w)dA(w)$
		$\displaystyle\leq$	$\displaystyle N^{\alpha}\int_{\varphi({\mathbb{D}})\cap\{w:n_{\varphi}(w)\leq N\}}\|f^{\prime}(w)\|^{2}dA(w)$
		$\displaystyle\leq$	$\displaystyle N^{\alpha}\int_{\varphi({\mathbb{D}})}\|f^{\prime}(w)\|^{2}dA(w)$

for all $f\in\mathcal{D}({\mathbb{D}}).$ Further,

			$\displaystyle\int_{\varphi({\mathbb{D}})}\|f^{\prime}(w)\|^{2}n_{\varphi}^{\alpha}(w)dA(w)$
		$\displaystyle\leq$	$\displaystyle\epsilon\\|C_{\varphi}\\|^{2}\cdot\int_{{\mathbb{D}}}\|f^{\prime}(w)\|^{2}dA(w)+N^{\alpha}\int_{\varphi({\mathbb{D}})}\|f^{\prime}(w)\|^{2}dA(w)$

for all $f\in\mathcal{D}({\mathbb{D}}).$ If there is a $\delta>0$ such that

\delta\int_{{\mathbb{D}}}|f^{\prime}(w)|^{2}dA(w)\leq\int_{\varphi({\mathbb{D}})}|f^{\prime}(w)|^{2}n_{\varphi}^{\alpha}(w)dA(w)

for all $f\in\mathcal{D}({\mathbb{D}}),$ then

$\displaystyle\delta\int_{{\mathbb{D}}}\|f^{\prime}(w)\|^{2}dA(w)$	$\displaystyle\leq$	$\displaystyle\int_{\varphi({\mathbb{D}})}\|f^{\prime}(w)\|^{2}n_{\varphi}^{\alpha}(w)dA(w)$
	$\displaystyle\leq$	$\displaystyle\epsilon\\|C_{\varphi}\\|\cdot\int_{{\mathbb{D}}}\|f^{\prime}(w)\|^{2}dA(w)$
	$\displaystyle+$	$\displaystyle N^{\alpha}\int_{\varphi({\mathbb{D}})}\|f^{\prime}(w)\|^{2}dA(w).$

Choose $\epsilon_{0}>0$ such that $\epsilon_{0}\|C_{\varphi}\|^{2}<\frac{1}{2}\delta$ and $N_{0}\in\mathbb{N}$ such that $\frac{n^{\alpha}}{n}<\epsilon_{0}$ for all $n\geq N_{0}.$ Then

\frac{\delta}{2N_{0}^{\alpha}}\int_{{\mathbb{D}}}|f^{\prime}(w)|^{2}dA(w)\leq\int_{\varphi({\mathbb{D}})}|f^{\prime}(w)|^{2}dA(w).

By Theorem KRL again, we have $(c)\Longrightarrow(a)$ .

$(a)\Longrightarrow(b)$ is obvious since $n_{\varphi}(w)\geq 1$ for all $w\in\varphi({\mathbb{D}}).$ To prove $(b)\Longrightarrow(a),$ assume for arbitrary $\alpha\in(0,1)$ , there is a $\delta>0$ and $r>0$ such that

\int_{\varphi({\mathbb{D}})\cap D(z,r)}n_{\varphi}^{\alpha}(w)dA(w)\geq\delta A(D(z,r))

for all $z\in{\mathbb{D}}.$ Write $E_{N}=\varphi({\mathbb{D}})\cap D(z,r)\cap\{w:n_{\varphi}(w)>N\}$ for any $N.$ Then for any $\alpha\in(0,1),$

	$\displaystyle\int_{E_{N}}n_{\varphi}^{\alpha}(w)dA(w)$	$\displaystyle\leq$	$\displaystyle\frac{1}{N^{1-\alpha}}\int_{E_{N}}n_{\varphi}dA(w)$
		$\displaystyle\leq$	$\displaystyle\frac{1}{N^{1-\alpha}}\int_{D(z,r)}n_{\varphi}dA(w).$

Since $C_{\varphi}$ is bounded on $\mathcal{D}({\mathbb{D}}),$ there is a $M>0$ such that

\int_{D(z,r)}n_{\varphi}dA(w)\leq MA(D(z,r))

(see [11]). Hence

\int_{E_{N}}n_{\varphi}^{\alpha}(w)dA(w)\leq\frac{M}{N^{1-\alpha}}A(D(z,r)).

On the other hand,

	$\displaystyle\int_{\varphi({\mathbb{D}})\cap D(z,r)-E_{N}}n_{\varphi}^{\alpha}(w)dA(w)$	$\displaystyle\leq$	$\displaystyle N^{\alpha}\int_{\varphi({\mathbb{D}})\cap D(z,r)-E_{N}}dA(w)$
		$\displaystyle\leq$	$\displaystyle N^{\alpha}A(\varphi({\mathbb{D}})\cap D(z,r)).$

Thus,

$\displaystyle\int_{\varphi({\mathbb{D}})\cap D(z,r)}n_{\varphi}^{\alpha}(w)dA(w)$	$\displaystyle=$	$\displaystyle\int_{E_{N}}n_{\varphi}^{\alpha}(w)dA(w)$
	$\displaystyle+$	$\displaystyle\int_{\varphi({\mathbb{D}})\cap D(z,r)-E_{N}}n_{\varphi}^{\alpha}(w)dA(w)$
	$\displaystyle\leq$	$\displaystyle\frac{M}{N^{1-\alpha}}A(D(z,r))$
	$\displaystyle+$	$\displaystyle N^{\alpha}A(\varphi({\mathbb{D}})\cap D(z,r)).$

Choose $N_{0}>0$ such that $\frac{M}{N_{0}^{1-\alpha}}<\frac{\delta}{2},$ then with the inequality $(b)$ we have that

A(\varphi({\mathbb{D}})\cap D(z,r))\geq\frac{\delta}{2N_{0}^{\alpha}}A(D(z,r)).

Hence $(b)\Longrightarrow(a).$ We complete the proof. ∎

Theorem 2.3.

Let $\varphi$ be an analytic self-mapping of ${\mathbb{D}}.$ Assume that $C_{\varphi}$ is a bounded composition operator on $\mathcal{D}({\mathbb{D}}).$ If for arbitrary $\alpha\in(0,1)$ , there is a $\delta>0$ and $r>0$ such that

\int_{\varphi({\mathbb{D}})\cap D(z,r)}n_{\varphi}^{\alpha}(w)dA(w)\geq\delta A(D(z,r))

for all $z\in{\mathbb{D}}.$ Or equivalently, $n_{\varphi}^{\alpha}(w)dA(w)$ is a reverse Carleson measure, then $R(C_{\varphi})$ is closed.

Proof.

Since for arbitrary $\alpha\in(0,1)$ , there is a $\delta>0$ and $r>0$ such that

\int_{\varphi({\mathbb{D}})\cap D(z,r)}n_{\varphi}^{\alpha}(w)dA(w)\geq\delta A(D(z,r))

for all $z\in{\mathbb{D}},$ we know that there is a $\tilde{\delta}>0$ such that

\int_{\varphi({\mathbb{D}})}|f^{\prime}(w)|^{2}dA(w)\geq\tilde{\delta}\int_{{\mathbb{D}}}|f^{\prime}(w)|^{2}dA(w)

for all $f\in\mathcal{D}({\mathbb{D}})$ by Proposition 2.2 and Theorem KRL. Thus,

$\displaystyle\\|C_{\varphi}f\\|^{2}$	$\displaystyle=$	$\displaystyle\|f(\varphi(0))\|^{2}+\int_{{\mathbb{D}}}\|(f\circ\varphi)^{\prime}(z)\|^{2}dA(z)$
	$\displaystyle=$	$\displaystyle\|f(\varphi(0))\|^{2}+\int_{{\mathbb{D}}}\|f^{\prime}(\varphi(z))\|^{2}\|\varphi^{\prime}(z)\|^{2}dA(z)$
	$\displaystyle=$	$\displaystyle\|f(\varphi(0))\|^{2}+\int_{\varphi({\mathbb{D}})}\|f^{\prime}(w)\|^{2}n_{\varphi}(w)dA(w)$
	$\displaystyle\geq$	$\displaystyle\|f(\varphi(0))\|^{2}+\int_{\varphi({\mathbb{D}})}\|f^{\prime}(w)\|^{2}dA(w)$
	$\displaystyle\geq$	$\displaystyle\|f(\varphi(0))\|^{2}+\tilde{\delta}\int_{{\mathbb{D}}}\|f^{\prime}(w)\|^{2}dA(w).$

It is easy to see that $R(C_{\varphi})$ has closed range. In fact, if $R(C_{\varphi})$ is not closed, then there is a sequence $\{g_{m}\}\subset\mathcal{D}({\mathbb{D}})$ with $\|g_{m}\|=1$ such that

\|C_{\varphi}g_{m}\|\rightarrow 0\hskip 11.38109pt\mbox{as}\hskip 11.38109ptm\rightarrow\infty.

Without loss of generality, assume $g_{m}\xrightarrow{w}g.$ Then

C_{\varphi}g_{m}\xrightarrow{w}C_{\varphi}g.

By $\|C_{\varphi}g_{m}\|\rightarrow 0,$ we see that $C_{\varphi}g=0.$ Thus $g=0.$ That is, $g_{m}\xrightarrow{w}0.$ In particular, $g_{m}(0)\rightarrow 0$ and $g_{m}(\varphi(0))\rightarrow 0.$ Further,

	$\displaystyle\\|C_{\varphi}g_{m}\\|^{2}$	$\displaystyle\geq$	$\displaystyle\|g_{m}(\varphi(0))\|^{2}+\tilde{\delta}\int_{{\mathbb{D}}}\|g_{m}^{\prime}(w)\|^{2}dA(w)$
		$\displaystyle\geq$	$\displaystyle\tilde{\delta}\int_{{\mathbb{D}}}\|g_{m}^{\prime}(w)\|^{2}dA(w).$

Note $\|g_{m}\|=1$ and $g_{m}(0)\rightarrow 0,$ we see that

\int_{{\mathbb{D}}}|g_{m}^{\prime}(w)|^{2}dA(w)\rightarrow 1\hskip 14.22636pt\mbox{as}\hskip 14.22636ptm\rightarrow\infty.

This contradicts to $\|C_{\varphi}g_{m}\|\rightarrow 0$ as $m\rightarrow\infty.$ Hence, $R(C_{\varphi})$ is closed. This completes the proof. ∎

Remark 1.

The operator $T$ on the Hilbert or Banach space $H$ is called semi-Fredholm operator if $R(T),$ the range of $T,$ is closed and at least one of $KerT$ and $KerT^{*}$ is a finite dimensional space of $H.$ Since $KerC_{\varphi}$ is always trivial, we see that $C_{\varphi}$ is semi-Fredholm operator if and only if $C_{\varphi}$ has closed range.

Corollary 2.4.

Let $\varphi$ be the analytic function on ${\mathbb{D}}$ which maps ${\mathbb{D}}$ into ${\mathbb{D}}.$ $C_{\varphi}$ is the bounded composition operator on $\mathcal{D}({\mathbb{D}}).$ If for arbitrary $\alpha\in(0,1)$ , there is a $\delta>0$ and $r>0$ such that

\int_{\varphi({\mathbb{D}})\cap D(z,r)}n_{\varphi}^{\alpha}(w)dA(w)\geq\delta A(D(z,r))

for all $z\in{\mathbb{D}}.$ Or equivalently, $n_{\varphi}^{\alpha}(w)dA(w)$ is a reverse Carleson measure, then $C_{\varphi}$ is a semi-Fredholm operator.

Remark 2.

Proposition 2.2 seems to mean that the condition in Theorem 2.3 is also necessary. We have the following conjecture

Conjecture 1.

Let $\varphi$ be an analytic self-mapping of ${\mathbb{D}}.$ Assume that $C_{\varphi}$ is a bounded composition operator on $\mathcal{D}({\mathbb{D}}).$ Then $R(C_{\varphi})$ is closed if and only if for arbitrary $\alpha\in(0,1)$ , there is a $\delta>0$ and $r>0$ such that

\int_{\varphi({\mathbb{D}})\cap D(z,r)}n_{\varphi}^{\alpha}(w)dA(w)\geq\delta A(D(z,r))

for all $z\in{\mathbb{D}}.$ Or equivalently, $n_{\varphi}^{\alpha}(w)dA(w)$ is a reverse Carleson measure.

If the counting function $n_{\varphi}$ of the mapping $\varphi$ is unbounded, as long as the integrals, relative to the Carleson measure of functions in the unit sphere of the Dirichlet space, are equally absolute continuous, then the reverse Carleson condition can ensure that the composition operator has closed range. That is, we have MainTheorem.

Proof of Main Theorem.

By the condition in the main Theorem, we see easily that $C_{\varphi}$ is bounded on $\mathcal{D}({\mathbb{D}}).$ $(a)\Longrightarrow(c)$ and $(b)\Longrightarrow(c)$ are obvious. Also, Theorem KRL implies that $(b)\Longrightarrow(a)$ . It remains to show $(c)\Longrightarrow(b).$ Assume $(c)$ holds, that is, there is a $\delta>0$ and $r>0$ such that

\int_{\varphi({\mathbb{D}})\cap D(z,r)}n_{\varphi}(w)dA(w)\geq\delta A(D(z,r))

for all $z\in{\mathbb{D}}.$ We are to prove that there is a $\delta>0$ and $r>0$ such that

A(\varphi({\mathbb{D}})\cap D(z,r))\geq\delta A(D(z,r))

for all $z\in{\mathbb{D}}$ . For any $z\in{\mathbb{D}}$ and fixed $r>0,$ write

k_{z}(w)=\frac{1-|z|^{2}}{(1-\bar{z}w)^{2}},

then by the equivalence of $1-|z|^{2}$ and $|1-\bar{z}w|$ on $D(z,r),$ together with assumption $(c),$ we get that there exist positive constants $M_{1}$ , $M_{2}$ such that

			$\displaystyle\delta\leq\frac{1}{A(D(z,r))}\int_{D(z,r)}n_{\varphi}(w)dA(w)$
		$\displaystyle\leq$	$\displaystyle M_{1}\int_{D(z,r)}\|k_{z}(w)\|^{2}n_{\varphi}(w)dA(w)$
		$\displaystyle=$	$\displaystyle M_{1}[\int_{D(z,r)\cap\{w\in{\mathbb{D}}:n_{\varphi}(w)>k\}}\|k_{z}(w)\|^{2}n_{\varphi}(w)dA(w)$
		$\displaystyle+$	$\displaystyle\int_{D(z,r)\cap\{w\in{\mathbb{D}}:n_{\varphi}(w)\leq k\}}\|k_{z}(w)\|^{2}n_{\varphi}(w)dA(w)]$
		$\displaystyle\leq$	$\displaystyle M_{1}[sup_{f\in(\mathcal{D}({\mathbb{D}}))_{1}}\int_{\varphi({\mathbb{D}})\cap\{w\in{\mathbb{D}}:n_{\varphi}(w)>k\}}\|f^{\prime}(w)\|^{2}n_{\varphi}(w)dA(w)$
		$\displaystyle+$	$\displaystyle\int_{D(z,r)\cap\{w\in{\mathbb{D}}:n_{\varphi}(w)\leq k\}}\|k_{z}(w)\|^{2}n_{\varphi}(w)dA(w)]$
		$\displaystyle\leq$	$\displaystyle M_{1}sup_{f\in(\mathcal{D}({\mathbb{D}}))_{1}}\int_{\varphi({\mathbb{D}})\cap\{w\in{\mathbb{D}}:n_{\varphi}(w)>k\}}\|f^{\prime}(w)\|^{2}n_{\varphi}(w)dA(w)$
		$\displaystyle+$	$\displaystyle\frac{M_{2}}{A(D(z,r))}\int_{D(z,r)\cap\{w\in{\mathbb{D}}:n_{\varphi}(w)\leq k\}}n_{\varphi}(w)dA(w).$

Note that

lim_{k\rightarrow\infty}sup_{f\in(\mathcal{D}({\mathbb{D}}))_{1}}\int_{\varphi({\mathbb{D}})\cap\{w\in{\mathbb{D}}:n_{\varphi}(w)>k\}}|f^{\prime}(w)|^{2}n_{\varphi}(w)dA(w)=0,

then for $\epsilon=\frac{\delta}{2(M_{1}+1)},$ there is a $k_{0}>0$ such that

sup_{f\in(\mathcal{D}({\mathbb{D}}))_{1}}\int_{\varphi({\mathbb{D}})\cap\{w\in{\mathbb{D}}:n_{\varphi}(w)>k\}}|f^{\prime}(w)|^{2}n_{\varphi}(w)dA(w)<\epsilon

for all $k\geq k_{0}.$ This makes that

			$\displaystyle M_{1}sup_{f\in(\mathcal{D}({\mathbb{D}}))_{1}}\int_{\varphi({\mathbb{D}})\cap\{w\in{\mathbb{D}}:n_{\varphi}(w)>k_{0}\}}\|f^{\prime}(w)\|^{2}n_{\varphi}(w)dA(w)$
		$\displaystyle+$	$\displaystyle\frac{M_{2}}{A(D(z,r))}\int_{D(z,r)\cap\{w\in{\mathbb{D}}:n_{\varphi}(w)\leq k_{0}\}}n_{\varphi}(w)dA(w)$
		$\displaystyle\leq$	$\displaystyle M_{1}\epsilon+\frac{k_{0}M_{2}}{A(D(z,r))}\int_{\varphi({\mathbb{D}})\cap D(z,r)}dA(w)$
		$\displaystyle\leq$	$\displaystyle\frac{\delta}{2}+\frac{k_{0}M_{2}}{A(D(z,r))}A(\varphi({\mathbb{D}})\cap D(z,r)).$

Hence,

A(\varphi({\mathbb{D}})\cap D(z,r))=\int_{\varphi({\mathbb{D}})\cap D(z,r)}dA(w)\geq\frac{\delta}{2k_{0}M_{2}}A(D(z,r)).

That is, $(c)\Longrightarrow(b).$ The proof is thus complete. ∎

Corollary 2.5.

Let $\varphi$ be an analytic self-mapping of ${\mathbb{D}}$ and $n_{\varphi}(w)$ be bounded on ${\mathbb{D}}.$ Then the following statements are equivalent:

(a)

$R(C_{\varphi})$ is closed on $\mathcal{D}({\mathbb{D}}).$
(b)

There is a $\delta>0$ and $r>0$ such that

$A(\varphi({\mathbb{D}})\cap D(z,r))\geq\delta A(D(z,r))$

for all $z\in{\mathbb{D}}$ .
(c)

$n_{\varphi}(w)dA(w)$ is a reverse Carleson measure.

Proof.

If $n_{\varphi}$ is bounded on ${\mathbb{D}},$ then $C_{\varphi}$ is obviously bounded on $\mathcal{D}({\mathbb{D}})$ and $n_{\varphi}$ satisfies the condition in Theorem B. Hence $(a)\Longleftrightarrow(b)\Longleftrightarrow(c)$ . ∎

3. Composition operators with some special symbols

If $\varphi$ is analytic on the closed unit disc, then we have the following

Proposition 3.1.

Let $H(\bar{{\mathbb{D}}})$ be the space of analytic functions on $\bar{{\mathbb{D}}},$ $\varphi\in H(\bar{{\mathbb{D}}})$ maps ${\mathbb{D}}$ into ${\mathbb{D}}.$ Then $R(C_{\varphi})=\{f\circ\varphi|f\in\mathcal{D}({\mathbb{D}})\}$ , the range of $C_{\varphi},$ is closed if and only if

\mathbb{T}\subset\varphi(\bar{{\mathbb{D}}}).

Proof.

Assume $R(C_{\varphi})$ is closed, since $kerC_{\varphi}=\{0\},$ we know that there is a constant $c>0$ such that for any $f\in\mathcal{D}({\mathbb{D}}),$

\|C_{\varphi}f\|\geq c\|f\|.

Thus

$\displaystyle\\|C_{\varphi}f\\|^{2}$	$\displaystyle=$	$\displaystyle\int_{{\mathbb{D}}}\|(f\circ\varphi)^{\prime}(z)\|^{2}dA(z)$
	$\displaystyle=$	$\displaystyle\int_{\varphi({\mathbb{D}})}\|f^{\prime}(w)\|^{2}n_{\varphi}(w)dA(w)$
	$\displaystyle\geq$	$\displaystyle c^{2}\int_{{\mathbb{D}}}\|f^{\prime}(z)\|^{2}dA(z).$

Note $\varphi\in H(\overline{{\mathbb{D}}}),$ we see that $C_{\varphi}$ is bounded on $\mathcal{D}({\mathbb{D}})$ since $n_{\varphi}(w)$ is bounded on ${\mathbb{D}}.$ If $\mathbb{T}-\varphi(\bar{{\mathbb{D}}})\neq\phi,$ choose $\zeta\in\mathbb{T}-\varphi(\bar{{\mathbb{D}}}).$ Let

f_{\zeta}(z)=\frac{1+\bar{\zeta}\cdot z}{2},\hskip 14.22636ptf_{k}(z)=f_{\zeta}^{k}(z),

then $f_{\zeta}(z)$ is the peak function at $\zeta$ on ${\mathbb{D}}.$ For any open neighborhood $U$ of $\zeta,$ it is easy to see that

			$\displaystyle lim_{k\rightarrow\infty}\sqrt[k]{\frac{\int_{{\mathbb{D}}-U}\|f^{\prime}_{k}(z)\|^{2}dA(z)}{\int_{{\mathbb{D}}}\|f^{\prime}_{k}(z)\|^{2}dA(z)}}$
		$\displaystyle=$	$\displaystyle\frac{max\|f_{\zeta}\|_{{\mathbb{D}}-U}\|}{max\|f_{\zeta}\|_{{\mathbb{D}}}\|}=max\|f_{\zeta}\|_{{\mathbb{D}}-U}\|<1.$

Hence

\frac{\int_{{\mathbb{D}}-U}|f^{\prime}_{k}(z)|^{2}dA(z)}{\int_{{\mathbb{D}}}|f^{\prime}_{k}(z)|^{2}dA(z)}\rightarrow 0.

Note

$\displaystyle\frac{\\|C_{\varphi}f_{k}\\|^{2}}{\\|f_{k}\\|^{2}}$	$\displaystyle=$	$\displaystyle\frac{\int_{{\mathbb{D}}}\|(f_{k}\circ\varphi)^{\prime}(z)\|^{2}dA(z)}{\int_{{\mathbb{D}}}\|f^{\prime}_{k}(z)\|^{2}dA(z)}$
	$\displaystyle=$	$\displaystyle\frac{\int_{\varphi({\mathbb{D}})}\|f^{\prime}_{k}(w)\|^{2}n_{\varphi}(w)dA(w)}{\int_{{\mathbb{D}}}\|f^{\prime}_{k}(z)\|^{2}dA(z)}$
	$\displaystyle\leq$	$\displaystyle M\frac{\int_{\varphi({\mathbb{D}})}\|f^{\prime}_{k}(w)\|^{2}dA(w)}{\int_{{\mathbb{D}}}\|f^{\prime}_{k}(z)\|^{2}dA(z)},$

and there is a neighborhood $U$ of $\zeta$ such that $U\cap\varphi(\bar{{\mathbb{D}}})=\phi,$ thus $\varphi(\bar{{\mathbb{D}}})\subset\bar{{\mathbb{D}}}-U.$ Further,

\frac{\|C_{\varphi}f_{k}\|^{2}}{\|f_{k}\|^{2}}\leq M\frac{\int_{\varphi({\mathbb{D}})}|f^{\prime}_{k}(w)|^{2}dA(w)}{\int_{{\mathbb{D}}}|f^{\prime}_{k}(z)|^{2}dA(z)}\rightarrow 0.

This contradicts to $\|C_{\varphi}f\|\geq c\|f\|.$ It shows that $\mathbb{T}\subset\varphi(\bar{{\mathbb{D}}}).$

Conversely, assume $\mathbb{T}\subset\varphi(\bar{{\mathbb{D}}}),$ we are to prove that $R(C_{\varphi})$ is closed. Assume the contrary, $R(C_{\varphi})$ is not closed, then there is a sequence $\{g_{k}\}\subset\mathcal{D}({\mathbb{D}})$ with $\|g_{k}\|=1$ such that $\|C_{\varphi}g_{k}\|\rightarrow 0.$ Without loss of generality, assume $g_{k}\xrightarrow{w}g.$ Then

C_{\varphi}g_{k}\xrightarrow{w}C_{\varphi}g.

By $\|C_{\varphi}g_{k}\|\rightarrow 0,$ we see that $C_{\varphi}g=0.$ Thus $g=0.$ That is, $g_{k}\xrightarrow{w}0.$ Further, $g_{k}\rightarrow 0$ uniformly on any compact subset of ${\mathbb{D}}.$

Since $\varphi\in H(\bar{{\mathbb{D}}}),$ for any $\zeta\in\mathbb{T},$ there is an open neighborhood $U(\zeta)$ such that $n_{\varphi}(w)\geq 1$ on $U(\zeta).$ By finite covering thoreom, there are finte $\zeta_{i},i=1,\cdots,m$ such that

\mathbb{T}\subset\cup_{i=1}^{m}U(\zeta_{i}),

and $n_{\varphi}(w)\geq 1$ on $U(\zeta_{i}).$ Note $\cup_{i=1}^{m}U(\zeta_{i})$ is open, we may choose $r\in(0,1)$ such that

\overline{{\mathbb{D}}-{\mathbb{D}}_{r}}\subset\cup_{i=1}^{m}U(\zeta_{i}).

Clearly, $g_{k}\rightarrow 0$ uniformly on $\overline{{\mathbb{D}}_{r}},$ for any $\epsilon>0,$ there is a $k_{0}$ such that

\int_{{\mathbb{D}}_{r}}|g^{\prime}_{k}|^{2}(z)dA(z)<\epsilon\hskip 14.22636pt\mbox{for}\hskip 11.38109ptk>k_{0}.

Thus for $k>k_{0},$

$\displaystyle\\|C_{\varphi}g_{k}\\|^{2}$	$\displaystyle=$	$\displaystyle\int_{{\mathbb{D}}}\|(g_{k}\circ\varphi)^{\prime}\|^{2}(z)dA(z)$
	$\displaystyle=$	$\displaystyle\int_{\varphi({\mathbb{D}})}\|g^{\prime}_{k}\|^{2}(w)n_{\varphi}(w)dA(w)$
	$\displaystyle=$	$\displaystyle\int_{\varphi({\mathbb{D}})-({\mathbb{D}}-{\mathbb{D}}_{r})}\|g^{\prime}_{k}\|^{2}(w)n_{\varphi}(w)dA(w)+\int_{{\mathbb{D}}-{\mathbb{D}}_{r}}\|g^{\prime}_{k}\|^{2}(w)n_{\varphi}(w)dA(w)$
	$\displaystyle\geq$	$\displaystyle\int_{{\mathbb{D}}-{\mathbb{D}}_{r}}\|g^{\prime}_{k}\|^{2}(w)n_{\varphi}(w)dA(w)$
	$\displaystyle\geq$	$\displaystyle\int_{{\mathbb{D}}-{\mathbb{D}}_{r}}\|g^{\prime}_{k}\|^{2}(w)dA(w)$
	$\displaystyle=$	$\displaystyle\int_{{\mathbb{D}}}\|g^{\prime}_{k}\|^{2}(w)dA(w)-\int_{{\mathbb{D}}_{r}}\|g^{\prime}_{k}\|^{2}(w)dA(w)$
	$\displaystyle\geq$	$\displaystyle\int_{{\mathbb{D}}}\|g^{\prime}_{k}\|^{2}(w)dA(w)-\epsilon$
	$\displaystyle=$	$\displaystyle\\|g_{k}\\|^{2}-\epsilon$
	$\displaystyle=$	$\displaystyle 1-\epsilon.$

This contradicts to $\|C_{\varphi}g_{k}\|\rightarrow 0.$ It shows that $R(C_{\varphi})$ must be closed. ∎

B. Hou and Ch.L. Jiang prove recently a similar result in the case of weighted Hardy space of polynomial growth (see [5]). In general, $\mathbb{T}\subset\overline{\varphi({\mathbb{D}})}$ is not enough to ensure that $C_{\varphi}$ has a closed range on $\mathcal{D}({\mathbb{D}})$ even if $n_{\varphi}$ is bounded on ${\mathbb{D}}.$ The following example illustrates this conclusion.

Example 1.

As shown figure 1:

Let $\zeta\in\mathbb{T}$ and ${\mathbb{D}}_{\zeta}$ is the disk inside the unit disk and is tangent to $\mathbb{T}$ at point $\zeta.$ Write $\Omega={\mathbb{D}}-\overline{{\mathbb{D}}_{\zeta}},$ $\varphi$ is the Riemann map from ${\mathbb{D}}$ to $\Omega\subset{\mathbb{D}}.$ Then $C_{\varphi}$ is a contraction operator on $\mathcal{D}({\mathbb{D}})$ since $n_{\varphi}(w)\leq 1$ for all $w\in{\mathbb{D}}.$ By Corollary 2.6, we know that $R(C_{\varphi})$ is not closed, although $\mathbb{T}\subset\overline{\varphi({\mathbb{D}})}.$

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$\displaystyle\\|C_{\varphi}f\\|^{2}_{\mathcal{D}}$	$\displaystyle=$	$\displaystyle\int_{{\mathbb{D}}}\|(f\circ\varphi)^{\prime}(z)\|^{2}dA(z)$
	$\displaystyle=$	$\displaystyle\int_{\varphi({\mathbb{D}})}\|f^{\prime}(w)\|^{2}n_{\varphi}(w)dA(w)$
	$\displaystyle=$	$\displaystyle\int_{{\mathbb{D}}}\|f^{\prime}(w)\|^{2}n_{\varphi}(w)dA(w)$
	$\displaystyle\geq$	$\displaystyle\int_{{\mathbb{D}}}\|f^{\prime}(w)\|^{2}dA(w)=\\|f\\|^{2}_{\mathcal{D}}.$

$\displaystyle\\|C_{\varphi}f\\|^{2}$	$\displaystyle=$	$\displaystyle\int_{{\mathbb{D}}}\|(f\circ\varphi)^{\prime}(z)\|^{2}dA(z)+\|f\circ\varphi(0)\|^{2}$
	$\displaystyle=$	$\displaystyle\int_{\varphi({\mathbb{D}})}\|f^{\prime}(w)\|^{2}n_{\varphi}(w)dA(w)+\|f\circ\varphi(0)\|^{2}$
	$\displaystyle\geq$	$\displaystyle c^{2}[\int_{{\mathbb{D}}}\|f^{\prime}(z)\|^{2}dA(z)+\|f(0)\|^{2}].$

	$\displaystyle\frac{\\|C_{\varphi}f_{k}\\|^{2}}{\\|f_{k}\\|^{2}}$	$\displaystyle=$	$\displaystyle\frac{\int_{{\mathbb{D}}}\|(f_{k}\circ\varphi)^{\prime}(z)\|^{2}dA(z)+\|f_{k}\circ\varphi(0)\|^{2}}{\int_{{\mathbb{D}}}\|f^{\prime}_{k}(z)\|^{2}dA(z)+\|f_{k}(0)\|^{2}}$
		$\displaystyle=$	$\displaystyle\frac{\int_{\varphi({\mathbb{D}})}\|f^{\prime}_{k}(w)\|^{2}n_{\varphi}(w)dA(w)+\|f_{k}\circ\varphi(0)\|^{2}}{\int_{{\mathbb{D}}}\|f^{\prime}_{k}(z)\|^{2}dA(z)+\|f_{k}(0)\|^{2}}.$

			$\displaystyle\int_{\varphi({\mathbb{D}})\cap\{w:n_{\varphi}(w)>N\}}\|f^{\prime}(w)\|^{2}n_{\varphi}^{\alpha}(w)dA(w)$
		$\displaystyle=$	$\displaystyle\int_{\varphi({\mathbb{D}})\cap\{w:n_{\varphi}(w)>N\}}\|f^{\prime}(w)\|^{2}\frac{n_{\varphi}^{\alpha}(w)}{n_{\varphi}(w)}n_{\varphi}(w)dA(w)$
		$\displaystyle\leq$	$\displaystyle\epsilon\int_{\varphi({\mathbb{D}})\cap\{w:n_{\varphi}(w)>N\}}\|f^{\prime}(w)\|^{2}n_{\varphi}(w)dA(w)$
		$\displaystyle\leq$	$\displaystyle\epsilon\int_{\varphi({\mathbb{D}})}\|f^{\prime}(w)\|^{2}n_{\varphi}(w)dA(w)$
		$\displaystyle\leq$	$\displaystyle\epsilon\\|C_{\varphi}\\|^{2}\cdot\int_{{\mathbb{D}}}\|f^{\prime}(w)\|^{2}dA(w)$

			$\displaystyle\int_{\varphi({\mathbb{D}})\cap\{w:n_{\varphi}(w)\leq N\}}\|f^{\prime}(w)\|^{2}n_{\varphi}^{\alpha}(w)dA(w)$
		$\displaystyle\leq$	$\displaystyle N^{\alpha}\int_{\varphi({\mathbb{D}})\cap\{w:n_{\varphi}(w)\leq N\}}\|f^{\prime}(w)\|^{2}dA(w)$
		$\displaystyle\leq$	$\displaystyle N^{\alpha}\int_{\varphi({\mathbb{D}})}\|f^{\prime}(w)\|^{2}dA(w)$