# composition operators with closed range on the Dirichlet space

Guangfu Cao  Cao: School of Mathematics and Information Science, Guangzhou
University, Guangzhou 510006, China<EMAIL_ADDRESS>and  Li He  He:
School of Mathematics and Information Science, Guangzhou University, Guangzhou
510006, China<EMAIL_ADDRESS>
###### 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_ADDRESS>
##  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:

  1. (a)

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

  2. (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 $z\in{\mathbb{D}}$.

  3. (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:

  1. (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 $z\in{\mathbb{D}}$.

  2. (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 $z\in{\mathbb{D}}.$

  3. (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:

  1. (a)

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

  2. (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 $z\in{\mathbb{D}}$.

  3. (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:

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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