Let $\varphi$ be a positive continuous function on $[0, 1)$, $\varphi$ is called a normal function if there are two constants $a$ and $b$: $0<a<b$ such that $\frac{\varphi(t)}{(1-t^2)^a}$ decreases and $\frac{\varphi(t)}{(1-t^2)^b}$ increases on $[0, 1)$. The simplest example is $\varphi(t)=(1-t^2)^{\beta}$, $\beta >0$.

Also, let $\mathbb{D}$ be the open unit disk in the complex plane $\mathbb{C}$ and $\mathbb{T}$ be its boundary, $dA(z)=\frac{r}{\pi}drd\theta$ be the normalized Lebesgue measure on $\mathbb{D}$. We write $dA_{\varphi}^p(z)$ $(1\leq p<\infty)$ for the weighted Lebesgue measure:$\frac{\varphi^p(|z|)}{1-|z|^2}dA(z)=\frac{\varphi^p(r)}{1-r^2}\frac{r}{\pi}drd\theta$, $z=re^{i\theta}$. A normal function $\varphi$ and $p\in[1, \infty)$ are often used to define a Banach space $L^p(\varphi)$ with norm

$ \|f\|_{p, \varphi}=\left(\int_{\mathbb{D}}|f(z)|^pdA_{\varphi}^p(z)\right)^{1/p}<\infty. $

Next for any $\alpha>b$ let $\psi(t)=\frac{(1-t^2)^{\alpha}}{\varphi(t)}$, then $\{\varphi, \psi\}$ is called a normal pair. Obviously $\psi$ is also a normal function with two constants $\alpha-b$ and $\alpha-a$. Of course, every $\psi$ can be used to give a Banach space $L^q(\psi)$ for $1\leq q<\infty$, relative to norm $\|\cdot\|_{q, \psi}$.

For $f\in L^p(\varphi)$ and $g\in L^q(\psi)$, we can define a dual form $\langle f, g\rangle$ as follows:

$ \langle f, g\rangle=\alpha\int_{\mathbb{D}}f(z)\overline{g(z)}(1-|z|^2)^{\alpha-1}dA(z). $

Let $L^{2, 1}_{\varphi}$ be the subspace of $L^2(\varphi)$ satisfying condition

$ \|u\|^2_{\frac{1}{2}}=\int_{\mathbb{D}}\big(\big|\frac{\partial u}{\partial z}(z)\big|^2+ \big|\frac{\partial u}{\partial \bar{z}}(z)\big|^2\big)dA^2_{\varphi}(z)<\infty. $

Let $\mathfrak{L}^{2, 1}_{\varphi}$ be the quotient space $L^{2, 1}_{\varphi}/\mathbb{C}$, where $\mathbb{C}$ is complex constant functions subspace of $L^{2, 1}_{\varphi}$, then $\mathfrak{L}^{2, 1}_{\varphi}$ is a Hilbert space with the inner product

$ \langle f, g\rangle_{\frac{1}{2}}=<\frac{\partial f}{\partial z}, \frac{\partial g}{\partial z}>_{L^2(\varphi)}+ \langle \frac{\partial f}{\partial \bar{z}}, \frac{\partial g}{\partial \bar{z}}\rangle_{L^2(\varphi)}. $

The weighted Dirichlet space $\mathcal {D}_{\varphi}$ is the subspace of all analytic function in $\mathfrak{L}^{2, 1}_{\varphi}$. Let $K_z(w)=\int_0^{\bar{z}}\int_0^w\frac{d\zeta d\eta}{(1-\zeta\eta)^{\alpha+1}}$ for $z, w\in\mathbb{D}$, then the linear operator $P$ is defined as follows:

$ (Pf)(z)=\alpha\int_{\mathbb{D}}\frac{\partial f(w)}{\partial w}\overline{\left(\frac{\partial K_z(w)}{\partial w}\right)}(1-|w|^2)^{\alpha-1}dA(w), \quad f\in \mathcal{D}_{\varphi}. $

We note that the operator $P$ is bounded from $L^{2, 1}_{\varphi}$ onto $\mathcal{D}_{\varphi}$. Moreover, $(Pf)(z)=f(z)$ for $f\in\mathcal{D}_{\varphi}$, so $K_z$ is called the reproducing kernel (see ^[1-3]). Let $G$ be a domain in $\mathbb{C}$, and define

$ L^{\infty, 1}(G)=\{u: u, \frac{\partial u}{\partial z}, \frac{\partial u}{\partial \bar{z}}\in L^{\infty}(G)\} $

for $u\in L^{\infty, 1}(G)$, $\|u\|_{1, \infty}= \mathop{{\rm esssup}}\limits_{z\in G}\max\{|u(z)|, |\frac{\partial u}{\partial z}(z)|, |\frac{\partial u}{\partial \bar{z}}(z)|\}$.

Definition 1.1  Suppose $u\in L^{2, 1}_{\varphi}$, the operator

$ T_uf(z)=P(uf)(z)=\alpha\int_{\mathbb{D}}\frac{\partial (u(w)f(w))}{\partial w}\overline{\left(\frac{\partial K_z(w)}{\partial w}\right)} (1-|w|^2)^{\alpha-1}dA(w) \quad \left(f\in \mathcal{D}_{\varphi}\right) $

is said to be the Toeplitz operator with symbol $u$, this operator is densely defined.

In the case of Hardy space, it is well known that $T_{\varphi}$ is bounded if and only if $\varphi$ is essentially bounded, and $T_{\varphi}$ is compact if and only if $\varphi=0$ (see ^{[4, 5]}). However, there are indeed bounded and compact Toeplitz operators with unbounded symbols, in fact, Miao and Zheng ^[6] introduced a class of functions, called $BT$, which contains $L^{\infty}$, for $\varphi\in BT$, $T_{\varphi}$ is compact on Bergman space $L^2_a(\mathbb{D})$, if and only if the Berezin transform of $\varphi$ vanishes on the unit circle $\mathbb{T}$. Zorboska ^[7] proved that if $\varphi$ belongs to the hyperbolic BMO space, then $T_{\varphi}$ is compact if and only if the Berezin transform of $\varphi$ vanishes on the unit circle. Cima and Cuckovic ^[8] constructed a class of unbounded functions built over Cantor set, the Toeplitz operator with these functions are compact. Essentially, if the value of the function $\varphi$ vanishes rapidly near the unit circle in the sense of measure $dA$, the $T_{\varphi}$ will be compact. Cao ^[9] constructed compact Toeplitz operators on Bergman space $L^2_a(\mathbb{B}_n, d\nu)$ with unbounded symbols. Wang, Xia and Cao ^[10] constructed a trass class Toeplitz operator $T_{\varphi}$ on Dirichlet space $\mathcal {D}$ with unbounded symbols.

In this paper, we construct a class of unbounded function on $\mathbb{D}$, the Toeplitz operators with these symbols are compact. We also construct a function $\phi$ on any countable dense subset in $\mathbb{T}$ which has nontagential limit infinity everywhere, such that $T_{\phi}$ is trace class.

For $\delta>0, \xi\in\mathbb{T}$, set

$ \Omega(\xi, \delta)=\{z\in\mathbb{D}: [1-(1-|z|)^{\delta}]^{\frac{1}{2}}|z-\xi|<|{\rm Re}(\xi\overline{(z-\xi)})|, {\rm Re}(z\bar{\xi})>0\}. $

Then $\Omega(\xi, \delta)$ is an open subset of $\mathbb{D}$ and this domain is said to be circle cone like with vertex $\xi$. For any $0<r<1$, let $\mathbb{D}_r=\{z: |z|<r\}$ be the disc with center 0 and radius $r$, $\mathbb{T}_r$ its boundary. We denote the Lebesgue measure on $\mathbb{T}$ by $d\theta$. Assume $b$ is an arbitrary positive number, it is obvious that we may choose a suitable $\delta=\delta(b)>0$ such that for arbitrary $0<r<1$,

$ \theta[\Omega(\xi, \delta)\cap\mathbb{T}_r]<d(1-r^2)^b, $

where $d$ is a constant which is independent of $\xi$ and $r$. For convenience, we write $\Omega_b(\xi)=\Omega(\xi, \delta(b))$.

Lemma 2.1  Suppose $\{f_k\}\subset \mathcal {D}_{\varphi}$ and $\|f_k\|_{L^{2, 1}_{\varphi}}=1$, $f_k\longrightarrow 0$ weakly in $\mathcal {D}_{\varphi}$, then $\|f_k\|_{L^2(\varphi)}\leq A$ and $\|f_k\|_{L^2(\varphi)}\longrightarrow 0$, where $A$ is a constant.

Theorem 2.2  Suppose $c>0$, $U_c(z)=(1-|z|^2)^{-c}$, $z\in \mathbb{D}$. For any $\xi\in\mathbb{T}$, let $b\geq2c+4$ and $\chi_{\Omega_b(\xi)}$ be the characteristic function of $\Omega_b(\xi)$, then $\phi=\chi_{\Omega_b(\xi)}U_c(z)$ induces a compact Toeplitz operator on weighted Dirichlet space $\mathcal {D}_{\varphi}$.

Proof  Suppose $\{f_k\}\subset\mathcal {D}_{\varphi}$ with $\|f_k\|_{L^{2, 1}_{\varphi}}=1$ is a sequence which weakly converges to zero, it is enough to prove that $\|T_{\phi}f_k\|_{L^{2, 1}_{\varphi}}\rightarrow0$ when $k\rightarrow\infty$. Note

$ \begin{eqnarray*} T_{\phi}f_k(z)&\;=&\;\alpha\int_{\mathbb{D}}\frac{\partial (\phi(w)f_k(w))}{\partial w}\overline{\left(\frac{\partial K_z(w)}{\partial w}\right)} (1-|w|^2)^{\alpha-1}dA(w)\\ &=&\;\alpha\int_{\mathbb{D}}\frac{\partial (\phi(w))}{\partial w}f_k(w)\overline{\left(\frac{\partial K_z(w)}{\partial w}\right)} (1-|w|^2)^{\alpha-1}dA(w)\\ &&\;+\alpha\int_{\mathbb{D}}\phi(w)f^{'}_k(w)\overline{\left(\frac{\partial K_z(w)}{\partial w}\right)} (1-|w|^2)^{\alpha-1}dA(w). \end{eqnarray*} $

Then

$ \begin{eqnarray*} \frac{\partial T_{\phi}f_k(z)}{\partial z}&\;=&\;\alpha\int_{\mathbb{D}}\frac{\partial (\phi(w))} {\partial w}f_k(w)\frac{(1-|w|^2)^{\alpha-1}}{(1-z\bar{w})^{\alpha+1}}dA(w)\\ &&+\;\alpha\int_{\mathbb{D}}\phi(w)f^{'}_k(w)\frac{(1-|w|^2)^{\alpha-1}}{(1-z\bar{w})^{\alpha+1}}dA(w), \end{eqnarray*} $

we see that

$ \begin{eqnarray*} \|T_{\phi}f_k\|^2_{L^{2, 1}_{\varphi}}&\;=&\;\alpha\int_{\mathbb{D}}\big|\frac{\partial T_{\phi}f_k(z)}{\partial z}\big|^2(1-|z|^2)^{\alpha-1}dA(z)\\ &\leq&\;C\alpha\int_{\mathbb{D}}\big(\big|\alpha\int_{\Omega_b(\xi)}\bar{w}(1-|w|^2)^{-c-1}f_k(w)\frac{(1-|w|^2)^{\alpha-1}}{(1-z\bar{w})^{\alpha+1}}dA(w)\big|^2\\ &&+\big|\alpha\int_{\Omega_b(\xi)}f'_k(w)(1-|w|^2)^{-c}\frac{(1-|w|^2)^{\alpha-1}}{(1-z\bar{w})^{\alpha+1}}dA(w)\big|^2\big)(1-|z|^2)^{\alpha-1}dA(z). \end{eqnarray*} $

For $m\in(0, 1)$, set $\Omega_b(\xi, m)=\{z\in\Omega_b(\xi): |z|>m\}$, then

$ \begin{eqnarray*} \|T_{\phi}f_k\|^2_{L^{2, 1}_{\varphi}}&\;\leq&\;2C\alpha\int_{\mathbb{D}}\big[\big|\alpha\int_{\Omega_b(\xi)-\Omega_b(\xi, m)}\bar{w}(1-|w|^2)^{-c-1} f_k(w)\frac{(1-|w|^2)^{\alpha-1}}{(1-z\bar{w})^{\alpha+1}}dA(w)\big|^2\\ &&+\big|\alpha\int_{\Omega_b(\xi, m)}\bar{w}(1-|w|^2)^{-c-1}f_k(w)\frac{(1-|w|^2)^{\alpha-1}}{(1-z\bar{w})^{\alpha+1}}dA(w)\big|^2\\ &&+\big|\alpha\int_{\Omega_b(\xi)-\Omega_b(\xi, m)}f'_k(w)(1-|w|^2)^{-c}\frac{(1-|w|^2)^{\alpha-1}}{(1-z\bar{w})^{\alpha+1}}dA(w)\big|^2\\ &&+\big|\alpha\int_{\Omega_b(\xi, m)}f'_k(w)(1-|w|^2)^{-c}\frac{(1-|w|^2)^{\alpha-1}}{(1-z\bar{w})^{\alpha+1}}dA(w)\big|^2\big](1-|z|^2)^{\alpha-1}dA(z). \end{eqnarray*} $

Note

$ \begin{eqnarray*} &&\big|\alpha\int_{\Omega_b(\xi, m)}\bar{w}(1-|w|^2)^{-c-1}f_k(w)\frac{(1-|w|^2)^{\alpha-1}}{(1-z\bar{w})^{\alpha+1}}dA(w)\big|^2\\ &\leq&\;\alpha\int_{\Omega_b(\xi, m)}|f_k(w)|^2(1-|w|^2)^{\alpha-1}dA(w)\alpha\int_{\Omega_b(\xi, m)} \frac{(1-|w|^2)^{-2c-2}(1-|w|^2)^{\alpha-1}}{|1-z\bar{w}|^{2(\alpha+1)}}dA(w), \end{eqnarray*} $

and

$ \begin{eqnarray*} \alpha\int_{\mathbb{D}}\frac{(1-|z|^2)^{\alpha-1}}{|1-z\bar{w}|^{2(\alpha+1)}}dA(z) =\frac{1}{(1-|w|^2)^{\alpha+1}}. \end{eqnarray*} $

Since $\|f_k(w)\|_{L^2(\varphi)}\leq A\|f_k(w)\|_{L^{2, 1}_{\varphi}}\leq A$, thus

$ \begin{eqnarray*} &&\alpha\int_{\mathbb{D}}\big|\alpha\int_{\Omega_b(\xi, m)}\bar{w}(1-|w|^2)^{-c-1}f_k(w) \frac{(1-|w|^2)^{\alpha-1}}{(1-z\bar{w})^{\alpha+1}}dA(w)\big|^2(1-|z|^2)^{\alpha-1}dA(z)\\ &\leq&\;A\alpha\int_{\mathbb{D}}\alpha\int_{\Omega_b(\xi, m)}\frac{(1-|w|^2)^{-2c-2}(1-|w|^2)^{\alpha-1}}{|1-z\bar{w}|^{2(\alpha+1)}} (1-|z|^2)^{\alpha-1}dA(w)dA(z)\\ &=&\;A\alpha\int_{\Omega_b(\xi, m)}\frac{(1-|w|^2)^{-2c-2}(1-|w|^2)^{\alpha-1}}{(1-|w|^2)^{\alpha+1}}dA(w)\\ &\leq&\;A_0\int^1_m(1-r^2)^{b-2c-4}dr^2\\ &=&\;\frac{A_0}{b-2c-3}(1-m^2)^{b-2c-3}\leq A_0(1-m^2), \end{eqnarray*} $

where $A_0$ is a constant. It is obvious that for any $\varepsilon>0$, there is an $m_1\in(0, 1)$ such that

$ \alpha\int_{\mathbb{D}}\big|\alpha\int_{\Omega_b(\xi, m)}\bar{w}(1-|w|^2)^{-c-1}f_k(w)\frac{(1-|w|^2)^{\alpha-1}}{(1-z\bar{w})^{\alpha+1}} dA(w)\big|^2(1-|z|^2)^{\alpha-1}dA(z) \leq\\ A_0(1-m^2)<\varepsilon $

for $m\in[m_1, 1)$. For the same reason, there exists an $m_2\in(0, 1)$ such that

$ \alpha\int_{\mathbb{D}}\big|\alpha\int_{\Omega_b(\xi, m)}f'_k(w)(1-|w|^2)^{-c}\frac{(1-|w|^2)^{\alpha-1}}{(1-z\bar{w})^{\alpha+1}}dA(w)\big|^2 (1-|z|^2)^{\alpha-1}dA(z)\leq A_0(1-m^2)<\varepsilon $

for $m\in[m_2, 1)$. We write $m_0=\max\{m_1, m_2\}$, then both inequalities hold for $m\in[m_0, 1)$. On the other hand, since $\Omega_b(\xi)-\Omega_b(\xi, m_0)\subset\{z\in\mathbb{D}: |z|\leq m_0\}$, we know that $f_k(w)\rightarrow0$ and $f'_k(w)\rightarrow0$ uniformly on $\Omega_b(\xi)-\Omega_b(\xi, m_0)$. Hence for any $\varepsilon>0$, there is a $K_0$, such that for $k>K_0$, $|f_k(w)|<\varepsilon$ and $|f'_k(w)|<\varepsilon$ for any $w\in\Omega_b(\xi)-\Omega_b(\xi, m_0)$. Therefore

$ \begin{eqnarray*} &&\alpha\int_{\mathbb{D}}\big|\alpha\int_{\Omega_b(\xi)-\Omega_b(\xi, m)}\bar{w}(1-|w|^2)^{-c-1}f_k(w) \frac{(1-|w|^2)^{\alpha-1}}{(1-z\bar{w})^{\alpha+1}}dA(w)\big|^2(1-|z|^2)^{\alpha-1}dA(z)\\ &\leq&\;\varepsilon \alpha\int_{\Omega_b(\xi)-\Omega_b(\xi, m)}\frac{1}{(1-|w|^2)^{2c+4}}dA(w)\leq\frac{\varepsilon \alpha}{(1-m_0^2)^{2c+4}}, \end{eqnarray*} $

and

$ \begin{eqnarray*} &&\alpha\int_{\mathbb{D}}\big|\alpha\int_{\Omega_b(\xi)-\Omega_b(\xi, m)}(1-|w|^2)^{-c}f'_k(w) \frac{(1-|w|^2)^{\alpha-1}}{(1-z\bar{w})^{\alpha+1}}dA(w)\big|^2(1-|z|^2)^{\alpha-1}dA(z)\\ &\leq&\;\varepsilon \alpha\int_{\Omega_b(\xi)-\Omega_b(\xi, m)}\frac{1}{(1-|w|^2)^{2c+2}}dA(w)\leq\frac{\varepsilon \alpha}{(1-m_0^2)^{2c+2}} \end{eqnarray*} $

consequently $\|T_{\varphi}f_k\|_{L^{2, 1}_{\varphi}}\rightarrow0$.

Theorem 2.3  There is a function $\phi\in L^{2, 1}_{\varphi}$ which is unbounded on any neighborhood of each boundary point of $\mathbb{D}$ (i.g. for any $\xi\in\mathbb{T}$, and $r>0$, $ \mathop{{\rm esssup}}\limits_{z\in\mathbb{D}\cap\mathbb{D}(\xi, r)}|\phi(z)|=\infty$, where $\mathbb{D}(\xi, r)=\{z:|z-\xi|<r\}$) such that $T_{\phi}$ is a compact operator on $\mathcal {D}_{\varphi}$.

Proof  Set $c>0$, $b\geq4c+5$ and let $U_c(z)$ be the function as in Theorem 2.2. Choose a countable dense subset $\{\xi_i\big|i=1, 2, \cdots\}$ of $\mathbb{T}$. For each $\xi_i$, write $\phi_i=\chi_{\Omega_b(\xi_i)}U_c$ then $T_{\phi_i}$ is a compact operator by Theorem 2.2. For any $f\in\mathcal {D}_{\varphi}$,

$ \begin{eqnarray*} &&\|T_{\phi_i}f\|^2_{L^{2, 1}_{\varphi}}\\ &\leq&\;\alpha\int_{\mathbb{D}}\big(\big|\alpha\int_{\Omega_b(\xi_i)}\frac{\partial \phi_i(w)}{\partial w}f(w) \frac{(1-|w|^2)^{\alpha-1}}{(1-z\bar{w})^{\alpha+1}}dA(w)\big|^2\\ &&+\big|\alpha\int_{\Omega_b(\xi_i)}\phi_i(w)f'(w)\frac{(1-|w|^2)^{\alpha-1}}{(1-z\bar{w})^{\alpha+1}}dA(w)\big|^2(1-|z|^2)^{\alpha-1}\big)dA(z)\\ &\leq&\;\|f\|^2_{L^{2, 1}_{\varphi}}\big(\alpha\int_{\mathbb{D}}\alpha\int_{\Omega_b(\xi_i)} \frac{(1-|w|^2)^{-2c-2}(1-|w|^2)^{\alpha-1}}{|1-z\bar{w}|^{2(\alpha+1)}}dA(w)(1-|z|^2)^{\alpha-1}dA(z)\\ & &+\alpha\int_{\mathbb{D}}\alpha\int_{\Omega_b(\xi_i)} \frac{(1-|w|^2)^{-2c}(1-|w|^2)^{\alpha-1}}{|1-z\bar{w}|^{2(\alpha+1)}}dA(w)(1-|z|^2)^{\alpha-1}dA(z)\big)\\ &=&\;\|f\|^2_{L^{2, 1}_{\varphi}}\left(\alpha\int_{\Omega_b(\xi_i)}\left((1-|w|^2)^{-2c-2}+(1-|w|^2)^{-2c}\right)\frac{1}{(1-|w|^2)^2}dA(w)\right) \\ &\leq&\;\|f\|^2_{L^{2, 1}_{\varphi}}\alpha\int^1_0\int_{\Omega_b(\xi_i)\cap\mathbb{T}_r}\big((1-r^2)^{-2c-2}+(1-r^2)^{-2c}\big)\frac{r}{(1-r^2)^2}drd\theta\\ &\leq&\;\frac{d\alpha}{2\pi}\|f\|^2_{L^{2, 1}_{\varphi}}\int^1_0\big((1-r^2)^{-2c-2}+(1-r^2)^{-2c}\big)(1-r^2)^b\frac{1}{(1-r^2)^2}dr^2\\ &\leq&\;K\|f\|^2_{L^{2, 1}_{\varphi}}, \end{eqnarray*} $

where $K$ is a constant, hence $\|T_{\phi_i}\|\leq K$. Write $T_N=\sum^N\limits_{i=1}\frac{1}{2^i}T_{\phi_i}$, then $T_N$ is a compact, and for any $M$, $N$ and $f\in\mathcal {D}_{\varphi}$,

$ \|\sum^M\limits_{i=N}\frac{1}{2^i}T_{\phi_i}f\|_{L^{2, 1}_{\varphi}}\leq C\sum^M\limits_{i=N}\frac{1}{2^i}\|f\|_{L^{2, 1}_{\varphi}}. $

It follows that

$ \|\sum^M\limits_{i=N}\frac{1}{2^i}T_{\phi_i}\|_{L^{2, 1}_{\varphi}}\leq C\sum^M\limits_{i=N}\frac{1}{2^i}, $

hence $T=\sum^{\infty}\limits_{i=1}\frac{1}{2^i}T_{\phi_i}$ converges in norm. Furthermore, $T$ is a compact operator. It is easy to check that $\phi_i\in L^{2, 1}_{\varphi}$ and $\|\phi_i\|_{L^{2, 1}_{\varphi}}\leq C$. Thus $\sum^{\infty}\limits_{i=1}\frac{1}{2^i}\phi_i$ converges to a function $\phi\in L^{2, 1}_{\varphi}$. It is not difficult to see that for each polynomial $P(w)$,

$ \|(T_{\phi}-T_N)P\|_{L^{2, 1}_{\varphi}}=\|T_{\sum^{\infty}\limits_{i=N+1}\frac{1}{2^i}\phi_i}P\|\leq C\|P\|_{L^{2, 1}_{\varphi}}\sum^{\infty}\limits_{i=N+1}\frac{1}{2^i}\rightarrow0. $

Therefore $T=T_{\phi}$, namely, $T$ is a Toeplitz operator with symbol $\phi=\sum^{\infty}\limits_{i=1}\frac{1}{2^i}\phi_i$. Since $\{\xi_i:i=1, 2, \cdots\}$ is dense in $\mathbb{T}$, it is obvious that for any $\xi_i\in\mathbb{T}$ and $r>0$, $\mathop{{\rm esssup}}\limits_{z\in\mathbb{D}\cap\mathbb{D}(\xi, r)}|\phi(z)|=\infty$.

In this section, we prove that there is a family of trace class of Toeplitz operators with unbounded symbols.

Theorem 3.1  There is a function $\phi\in L^{2, 1}_{\varphi}$ which is unbounded on any neighborhood of each boundary point of $\mathbb{D}$ (that is, for arbitrary $\xi\in\mathbb{T}$, and $r>0$, $\mathop{\rm {esssup}}\limits_{z\in\mathbb{D}\cap\mathbb{D}(\xi, r)}|\phi(z)|=\infty$, where $\mathbb{D}(\xi, r)$ is same as in Theorem 2.2) such that $T_{\phi}$ is a trace class operator.

Proof  Let $e_k(w)=w^k/\sqrt{C_{\alpha, k}} (k\in\mathbb{Z^+})$ be the standard orthonormal basis of $\mathcal {D}_{\varphi}$, where $C_{\alpha, k}=k\frac{\Gamma(k+1)\Gamma(\alpha+1)}{\Gamma(\alpha+k)}$. Let $\{\xi_i: i=1, 2, \cdots\}$ be a countable dense subset of $\mathbb{T}$, and $U_c=(1-|w|^2)^{-c}(c>0)$. We must show that

$ \sum^{\infty}\limits_{k=1}\langle|T|e_k, e_k\rangle_{L^{2, 1}_{\varphi}}<\infty. $

For any $m\in(0, 1)$ set $\phi_i(w)=\chi_{\Omega_b(\xi_i, m)}U_c(w)$, where $b\geq c+5$, then

$ \begin{eqnarray*} &&|\langle T_{\phi_i}e_k, e_k\rangle_{L^{2, 1}_{\varphi}}|\\ &=&\;\frac{1}{C_{\alpha, k}}\big|\alpha\int_{\mathbb{D}}\big(\alpha\int_{\mathbb{D}} \frac{\partial \phi_i(w)}{\partial w}\frac{w^kk\bar{z}^{k-1}(1-|w|^2)^{\alpha-1}}{(1-z\bar{w})^{\alpha+1}}dA(w)\\ & &+\alpha\int_{\mathbb{D}}\phi_i(w)\frac{w^{k-1}k^2\bar{z}^{k-1}(1-|w|^2)^{\alpha-1}}{(1-z\bar{w})^{\alpha+1}} dA(w)\big)(1-|z|^2)^{\alpha-1}dA(z)\big|\\ &=&\;\frac{1}{C_{\alpha, k}}\big|\alpha\int_{\mathbb{D}}\frac{\partial \phi_i(w)}{\partial w}w_k\bar{w}^{k-1} (1-|w|^2)^{\alpha-1}dA(w)\end{eqnarray*} $

$ \begin{eqnarray*} & &+\alpha\int_{\mathbb{D}}\phi_i(w)k^2|w|^{k-1}(1-|w|^2)^{\alpha-1}dA(w)\big|\\ &=&\;\big|\frac{c\alpha k}{C_{\alpha, k}}\int_{\Omega_b(\xi_i, m)}(1-|w|^2)^{\alpha-c-2}|w|^{2k}dA(w)\\ & &+\frac{\alpha k^2}{C_{\alpha, k}}\int_{\Omega_b(\xi_i, m)}(1-|w|^2)^{\alpha-c-1}|w|^{2(k-1)}dA(w)\\ &\leq&\;\frac{Ak\alpha}{C_{\alpha, k}}\big|\int^1_m(1-r^2)^{b+\alpha-c-2}r^{2k}dr^2+k\int^1_m(1-r^2)^{b+\alpha-c-1}r^{2(k-1)}dr^2\big|. \end{eqnarray*} $

Since

$ \begin{eqnarray*} &&\int^1_m(1-r^2)^{b+\alpha-c-2}r^{2k}dr^2+k\int^1_m(1-r^2)^{b+\alpha-c-1}r^{2(k-1)}dr^2\\ &\leq&\int^1_{m^2}(1-r)^{\alpha+3}r^kdr+k\int^1_{m^2}(1-r)^{\alpha+4}r^{k-1}dr\\ &\leq&2\frac{\Gamma(k+1)\Gamma(\alpha+5)}{\Gamma(\alpha+k+5)}, \end{eqnarray*} $

then

$ |\langle T_{\phi_i}e_k, e_k\rangle_{L^{2, 1}_{\varphi}}|\leq\frac{Ak\alpha}{C_{\alpha, k}}\frac{2\Gamma(k+1)\Gamma(\alpha+5)}{\Gamma(\alpha+k+5)} \leq\frac{B(\alpha)}{k^5}, $

where $B(\alpha)$ is a constant only depending on $\alpha$. Set $T=\sum^{\infty}\limits_{i=1}\frac{1}{2^i}T_{\phi_i}$, then $T$ is a compact operator. Note $T$ is positive, thus

$ \begin{eqnarray*} &&\sum^{\infty}_{k=1}\big|\langle Te_k, e_k\rangle_{L^{2, 1}_{\varphi}}\big|\\ &=&\;\sum^{\infty}_{i=1}\sum^{\infty}_{k=1}\frac{1}{2^i}\langle T_{\phi_i}e_k, e_k\rangle_{L^{2, 1}_{\varphi}}\\ &\leq&\;\sum^{\infty}_{i=1}\sum^{\infty}_{k=1}\frac{1}{2^i}\frac{B(\alpha)}{k^5} \leq\sum^{\infty}_{k=1}\frac{B(\alpha)}{k^5}<\infty. \end{eqnarray*} $

Hence $T$ is a trace class operator. In the same manner we can also construct Hilbert-Schmidt Toeplitz operator $T_{\phi}$ on $\mathcal {D}_{\varphi}$ with unbounded symbols $\phi\in L^{2, 1}_{\varphi}$.