令$\mathbb{C}$为复平面,$\mathbb{D}$表示$\mathbb{C}$中单位圆盘,$\mathbb{T}$为$\mathbb{D}$的边界. ${\rm d}A(z)=\frac{1}{\pi}{\rm d}x{\rm d}y$为$\mathbb{D}$上正规化Lebesgue测度. 对$1\leq p<\infty$,定义

$$ L^{p,1}= \bigg\{f\bigg| \int_{{\Bbb D}}\bigg(\bigg|\frac{\partial f}{\partial z}\bigg|^p+ \bigg|\frac{\partial f}{\partial \overline{z}}\bigg|^p\bigg){\rm d}A(z)<\infty\bigg\}. $$

易知${\cal L}^{p,1}=\frac{L^{p,1}}{{\Bbb C}}$为Banach空间,其上范数为

$$ \|f\|_{L^{p,1}}=\bigg(\int_{{\Bbb D}}\bigg(\bigg|\frac{\partial f}{\partial z}\bigg|^p+ \bigg|\frac{\partial f}{\partial \overline{z}}\bigg|^p\bigg){\rm d}A(z)\bigg)^{1/p}. $$

调和Dirichlet空间${\cal D}^p_h\ (1\leq p<\infty)$是由${\cal L}^{p,1}$中 全体调和函数所成的闭子空间. $\mathbb{D}$上有界调和函数全体所成空间记为 $h^{\infty}$. 令$P_h$是从${\cal L}^{p,1}$到${\cal D}^p_h$上投影, 则$P_h$可以按如下方式表示为积分算子

$\begin{array}{l} {P_h}(f)(w) = {\langle f,{K^h}(z,w)\rangle _1} = \langle \frac{{\partial f}}{{\partial z}},\frac{{\partial K_w^h}}{{\partial z}}\rangle + \langle \frac{{\partial f}}{{\partial \bar z}},\frac{{\partial K_w^h}}{{\partial \bar z}}\rangle \\ = \int_D ( \frac{{\partial f}}{{\partial z}}\overline {\frac{{\partial K_w^h}}{{\partial z}}} + \frac{{\partial f}}{{\partial \bar z}}\overline {\frac{{\partial K_w^h}}{{\partial \bar z}}} ){\rm{d}}A(z), \end{array}$

其中$K^h_w(z)=K^h(z,w)$是${\cal D}^p_h$的再生核. 直接计算得

$$ K^h(z,w)=-\ln(1-z\overline{w})-\ln(1-\overline{z}w). $$

若$\varphi\in L^{p,1}(\mathbb{D},{\rm d}A)$,对$z,\overline{z}$的多项式$f(z,\overline{z})$, ${\cal D}_h^p$上以$\varphi$为符号的Toeplitz算子,小Hankel算子与Hankel算子分别稠密定义为

$$ T^h_{\varphi}f=P_h(\varphi f);\quad\Gamma^h_{\varphi}f=P_h(\varphi U f); \quad H^h_{\varphi}f=(I-P_h)(\varphi f), $$

其中$(Uf)(z)=f(\overline{z})$是${\cal D}_h^p$上的酉算子.

Hilbert函数空间上Toeplitz算子与Hankel算子的有界性,紧性, Fredholm指标已有许多算子论专家进行了深入广泛地研究,特别对Hardy空间$H^2$, Bergman空间$L^2_a$与Dirichlet空间${\cal D}^2$情形得到了丰富而深刻的结果. 对$1< p<\infty$,$H^p$,$L^p_a$与${\cal D}^p$上Toeplitz算子与 Hankel算子也有了很多有趣的结果,参见文献[1, 2]. Guo与Zheng在文献[3]中讨论了调和Bergman空间$L^2_h$上Toeplitz算子的性质, 得到了一些与解析Bergman空间不同的有趣结论. 但对$p=1$的极端情形却很少有相关成果, 参见文献[4, 5]. 本文讨论调和Dirichlet空间${\cal D}^1_h$上Toeplitz算子, 小Hankel算子与Hankel算子的相关性质.

易知${\cal D}^{p}_h\ (1< p<\infty)$上的调和Dirichlet投影是有界的. 于是,${\cal D}^p_h\ (1< p<\infty)$上符号在

$$ L^{\infty,1}=\bigg\{\varphi\bigg| \varphi,\frac{\partial \varphi}{\partial z},\frac{\partial \varphi} {\partial \overline{z}}\in L^{\infty}\bigg\} $$

中的Toeplitz算子,小Hankel算子与Hankel算子是有界的. 然而众所周知,调和Bergman空间 $L_h^1$上的Bergman投影是无界的,同样由$L^{1,1}$到${\cal D}^1_h$的调和Dirichlet投 影也 是无界的. 所以需要对${\cal D}^1_h$上的Toeplitz算子,小Hankel算子与Hankel算子有 界性进行更加深入地讨论.

Zhu第一个在文献[6]中研究了$L^1_a$上的Bergman投影,找出了一大类可以诱导$L^1_a$上 的有界Toeplitz算子的有界函数,他的结果为研究Dirichlet空间${\cal D}^1$上Toeplitz 的有界性给出很多好的思路.

Bergman空间$L^2_a$与Dirichlet空间${\cal D}^2$上Toeplitz算子的Fredholm性质已被广泛研究, 并得到了很多有意思的结果,参见文献[7, 8, 9]. Taskinen与 Virtanen讨论了Bergman空间$L^1_a$上Toeplitz算子的Fredholm性质, 计算了Fredholm指标,参见文献[5]. 对调和Dirichlet空间${\cal D}^1_h$ 上Toeplitz算子的Fredholm性质至今没有任何相关结果,是全新的领域. 后文将对

$$ C^1(\overline{\mathbb{D}})=\bigg\{\varphi\bigg|\varphi,\frac{\partial \varphi}{\partial z},\frac{\partial \varphi} {\partial \overline{z}}\in C(\overline{\mathbb{D}})\bigg\} $$

中某些函数诱导的Toeplitz算子建立Fredholm理论,这里的$C^1(\overline{\mathbb{D}})$是以

$$\|\varphi\|_*=\max\bigg\{\|\varphi\|_{\infty}, \bigg\|\frac{\partial \varphi}{\partial z}\bigg\|_{\infty}, \bigg\|\frac{\partial \varphi}{\partial \overline{z}} \bigg\|_{\infty}\bigg\} $$

为范数的Banach空间.

Dirichlet空间${\cal D}^2$上Toeplitz算子,小Hankel算子与Hankel算子的紧性相关结果参见 文献[9, 13]. 下文将讨论调和Dirichlet空间${\cal D}^1_h$上这些算子的紧性, 给出它们为紧算子的充分条件. 后面证明中将会用到文献[6]中关于对数有界均值震荡(BMO$_{\partial \log }$)函数的一些估计.

首先回顾一下文献[14]中得到的Bergman度量下有界均值震荡(BMO)的一些结论.

令$\rho(z,w)=|\frac{z-w}{1-\overline{z}w}|$,其中$z,w\in \mathbb{D}$, 则Bergman度量$\beta(z,w)$定义为

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

Bergman度量是M$\mathbb{D}dot{\mbox{o}}$bius不变的,即对任意$\phi\in $Aut$(\mathbb{D})$与$z,w\in \mathbb{D}$有

$\beta (\phi (z),\phi (w)) = \beta (z,w).$

对任意$z\in \mathbb{D}$与$r>0$,记

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

是以$z$为心,$r$为半径的Bergman圆盘. 众所周知$D(z,r)$也是欧式圆盘,其圆心和半径分别为

$ C=\frac{1-s^2}{1-s^2|z|^2}z,\quad R=\frac{1-|z|^2}{1-s^2|z|^2}s, $

这里的$s=\mbox{tanh}r\in (0,1)$. 若$\varphi\in L^1$满足对任意$z\in \mathbb{D}$有

${\rm{M}}{{\rm{O}}_r}(\varphi )(z) = {\left( {\widehat {|\varphi |_r^2}(z)-|{{\hat \varphi }_r}(z){|^2}} \right)^{1/2}} < M < \infty ,$

其中$\widehat{\varphi}_r(z)=\frac{1}{|D(z,r)|}\int_{D(z,r)}\varphi(w){\rm d}A(w)$, 则这些函数组成空间被称为有界均值震荡函数空间,记为BMO$_{\partial }$. 事实上,$\mbox{MO}_r(\varphi)$与$r$无关,对任意$r>0$,

$ \|\varphi\|_{{\rm BMO}_{\partial }}:=\sup\limits_{z\in\mathbb{D}}\mbox{MO}_r(\varphi)(z) \mbox{ }\mbox{等价于}\mbox{ } \sup _{z\in\mathbb{D}}\mbox{MO}(\varphi)(z)=({\widetilde{|\varphi|^2}(z)- |\widetilde{\varphi}(z)|^2})^{1/2}, $

其中$\widetilde{\varphi}$是Berezin变换.

消失均值震荡空间 VMO$_{\partial}$是BMO$_{\partial }$的闭子空间,由所有满足条件

$ \lim_{|z|\rightarrow 1^-}\mbox{MO}_r(\varphi)(z)=0 $

的函数构成,注意这里的极限过程与$r$无关.

下文将使用对数加权有界均值震荡空间BMO$_{\partial \log }$和 消失均值震荡空间VMO$_{\partial \log }$,它们的范数定义为

$$ \|\varphi\|_{{\rm BMO}_{\partial \log}}=\sup _{z\in \mathbb{D}} \log \frac{1}{1-|z|^2}\mbox{MO}_r(f)(z). $$

BMO$_{\partial \log }$与VMO$_{\partial \log }$在调和Bergman投影下的像 空间分别为对数加权调和Bloch空间${\cal LB}^h$与对数加权小调和Bloch空间${\cal LB}^h_0$.

$\mathbb{D}$上的调和函数$f$属于调和Bloch空间${\cal B}^h$当且仅当

$$ \sup _{z\in\mathbb{D}}(1-|z|^2)\bigg(\bigg|\frac{\partial f}{\partial z}(z)\bigg|+\bigg|\frac{\partial f }{\partial \overline{z}}(z)\bigg|\bigg)<\infty; $$

$f$ 属于小调和Bloch空间$ {\cal B}^h_0$当且仅当

$$ |z|\rightarrow 1^-\mbox{ 时,}\ (1-|z|^2)\bigg(\frac{\partial f}{\partial z}(z)+\frac{\partial f} {\partial \overline{z}}(z)\bigg)\rightarrow 0. $$

若在上面两式乘上因子$-\log (1-|z|^2)$,则可得到${\cal LB}^h$与${\cal LB}_0^h$的定义.

下面考虑Toeplitz算子,小Hankel算子与Hankel算子的有界性.

定理 2.1 若$\varphi\in L^{2+\epsilon,1}\cap {\rm BMO}_{\partial \log }$, 其中$\epsilon$为任意正数, Toeplitz算子$T^h_{\varphi}$与Hankel算子$H^h_{\varphi}$在调和 Dirichlet空间${\cal D}^1_h$上有界,即存在常数$C,C_1$使得

${\left\| {T_\varphi ^h} \right\|_{L(D_h^1,D_h^1)}} \le C({\left\| \varphi \right\|_{{L^{2 + \varepsilon ,1}}}} + {\left\| \varphi \right\|_{{\rm{BM}}{{\rm{O}}_\partial }}}),$

${\left\| {H_\varphi ^h} \right\|_{L(D_h^1,{L^{1,1}})}} \le {C_1}({\left\| \varphi \right\|_{{L^{2 + \varepsilon ,1}}}} + {\left\| \varphi \right\|_{{\rm{BM}}{{\rm{O}}_\partial }}}),$

证 由定义知对任意$f\in {\cal D}^1_h$,

$\begin{array}{l} {\left\| {T_\varphi ^hf} \right\|_{{L^{1,1}}}} = \int_{\mathbb{D}} | \int_{\mathbb{D}} \varphi \frac{{\partial f}}{{\partial z}}\frac{1}{{{{(1-\bar zw)}^2}}}{\rm{d}}A(z) + \int_{\mathbb{D}} {\frac{{\partial \varphi }}{{\partial z}}} f\frac{1}{{{{(1-\bar zw)}^2}}}{\rm{d}}A(z)|{\rm{d}}A(w)\\ + \int_{\mathbb{D}} | \int_{\mathbb{D}} \varphi \frac{{\partial f}}{{\partial \bar z}}\frac{1}{{{{(1-z\bar w)}^2}}}{\rm{d}}A(z) + \int_{\mathbb{D}} {\frac{{\partial \varphi }}{{\partial \bar z}}} f\frac{1}{{{{(1-z\bar w)}^2}}}{\rm{d}}A(z)|{\rm{d}}A(w)\\ \le \int_{\mathbb{D}} | \int_{\mathbb{D}} \varphi \frac{{\partial f}}{{\partial z}}\frac{1}{{{{(1-\bar zw)}^2}}}{\rm{d}}A(z)|{\rm{d}}A(w) + \int_{\mathbb{D}} | \int_{\mathbb{D}} {\frac{{\partial \varphi }}{{\partial z}}} f\frac{1}{{{{(1-\bar zw)}^2}}}{\rm{d}}A(z)|{\rm{d}}A(w)\\ + \int_{\mathbb{D}} | \int_{\mathbb{D}} \varphi \frac{{\partial f}}{{\partial \bar z}}\frac{1}{{{{(1-z\bar w)}^2}}}{\rm{d}}A(z)|{\rm{d}}A(w) + \int_{\mathbb{D}} | \int_{\mathbb{D}} {\frac{{\partial \varphi }}{{\partial \bar z}}} f\frac{1}{{{{(1-z\bar w)}^2}}}{\rm{d}}A(z)|{\rm{d}}A(w). \end{array}$

因$L^{2+\epsilon,1}\hookrightarrow L^{\infty}$连续,故

$\begin{array}{l} \int_{\mathbb{D}} | \int_{\mathbb{D}} \varphi \frac{{\partial f}}{{\partial z}}\frac{1}{{{{(1-\bar zw)}^2}}}{\rm{d}}A(z)|{\rm{d}}A(w) \le C({\left\| \varphi \right\|_\infty } + {\left\| \varphi \right\|_{{\rm{BM}}{{\rm{O}}_{\partial \log }}}}){\left\| {\frac{{\partial f}}{{\partial z}}} \right\|_{{L^1}}}\\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; \le C({\left\| \varphi \right\|_{{L^{2 + \varepsilon ,1}}}} + {\left\| \varphi \right\|_{{\rm{BM}}{{\rm{O}}_{\partial \log }}}}){\left\| f \right\|_{{L^{1,1}}}}, \end{array}$

上面的第一个不等式来自文献[5,定理6]. 同理有

$\int_{\mathbb{D}} | \int_{\mathbb{D}} \varphi \frac{{\partial {\rm{ }}f}}{{\partial \bar z}}\frac{1}{{{{(1-\bar zw)}^2}}}{\rm{d}}A(z)|{\rm{d}}A(w) \le C({\left\| \varphi \right\|_{{L^{2 + \varepsilon ,1}}}} + {\left\| \varphi \right\|_{{\rm{BM}}{{\rm{O}}_{\partial \log }}}}){\left\| f \right\|_{{L^{1,1}}}}.$

因$L^{1,1}\hookrightarrow L^{2}$为连续,于是 $\frac{\partial\varphi}{\partial z}f\in L^{\frac{2(2+\epsilon)}{4+\epsilon}}$. 因此

$ \int_{\mathbb{D}}\bigg|\int_{\mathbb{D}}\frac{\partial\varphi}{\partial z} f \frac{1}{(1-\overline{z}w)^2} {\rm d}A(z)\bigg| {\rm d}A(w)\\ \leq C\bigg(\int_{\mathbb{D}}\bigg|\int_{\mathbb{D}}\frac{\partial\varphi}{\partial z} f \frac{1}{(1-\overline{z}w)^2} {\rm d}A(z)\bigg|^{\frac{2(2+\epsilon)}{4+\epsilon}} {\rm d}A(w)\bigg)^{\frac{4+\epsilon}{2(2+\epsilon)}}\\ \leq C\bigg\|\frac{\partial\varphi}{\partial z}f\bigg\|_{L^{\frac{2(2+\epsilon)}{4+\epsilon}}}\leq C \|\varphi\|_{L^{2+\epsilon,1}}\|f\|_{L^{2}}\leq C\|\varphi\|_{L^{2+\epsilon,1}}\|f\|_{L^{1,1}}, $

且

$$ \int_{\mathbb{D}}\bigg|\int_{\mathbb{D}}\frac{\partial\varphi}{\partial \overline{z}} f \frac{1}{(1-z\overline{w})^2} {\rm d}A(z)\bigg| {\rm d}A(w) \leq C\|\varphi\|_{L^{2+\epsilon,1}}\|f\|_{L^{1,1}}. $$

所以

$$\|T^h_{\varphi}f\|_{L^{1,1}}\leq C(\|\varphi\|_{L^{2+\epsilon,1}}+\|\varphi\|_{{\rm BMO}_{\partial \log}})\|f\|_{L^{1,1}}. $$

$H^h_{\varphi}$的有界性可由不等式

$ \|\varphi f\|_{L^{1,1}} =\int_{\mathbb{D}}\bigg(\bigg|\frac{\partial \varphi}{\partial z} f+\varphi \frac{\partial f}{\partial z}\bigg|+ \bigg|\frac{\partial \varphi}{\partial \overline{z}} f+\varphi\frac{\partial f}{\partial \overline{z}}\bigg|\bigg){\rm d}A(z)\\ \leq \int_{\mathbb{D}}\bigg(\bigg|\frac{\partial \varphi}{\partial z} f\bigg|+\bigg|\varphi \frac{\partial f}{\partial z}\bigg|+\bigg|\frac{\partial \varphi}{\partial \overline{z}} f\bigg|+ \bigg |\varphi\frac{\partial f}{\partial \overline{z}}\bigg|\bigg){\rm d}A(z)\\ \leq C_1(\|\varphi\|_{L^{2+\epsilon,1}}+\|\varphi\|_{{\rm BMO}_{\partial \log }})\|f\|_{L^{1,1}} $

与

$$ \|H^h_{\varphi}f\|_{L^{1,1}}\leq \|T^h_{\varphi}f\|_{L^{1,1}}+\|\varphi f\|_{L^{1,1}} $$

推得.

引理 2.1 算子$Uf(z)=f(\overline{z})$是${\cal D}^1_h$上的酉算子,其中$f\in {\cal D}^1_h$.

定理 2.2 若$\varphi\in L^{2+\epsilon,1}\cap {\rm BMO}_{\partial \log }$,其中$\epsilon$为任意正数,小Hankel算子$\Gamma^h_{\varphi}=T^h_{\varphi}U$在${\cal D}^1_h$上有界.

证 应用定理2.1与引理2.1可得.

定理 2.3 若$\varphi\in L^{2+\epsilon,1}(\overline{\mathbb{D}})\cap \mbox{VMO}_{\partial \log }$,其中$\epsilon$为任意正数,则$T^h_{\varphi}$在${\cal D}^{1}_h$上为紧算子当且仅当 $\varphi|_{\mathbb{T}}=0$.

证 若$\varphi|_{\mathbb{T}}= 0$, 对任意${\cal D}^{1}_h$中满足$\|f_{k}\|_{L^{1,1}}=1$且在${\cal D}^{1}_h$ 中$f_{k}$弱收敛到$ 0$的序列$\{f_{k}\}$有

$ \|T_{\varphi}f_{k}\|_{L^{1,1}}\\ =\int_{{\Bbb D}}\bigg|\frac{\partial P_h(\varphi f_{k})(w)}{\partial w}\bigg|{\rm d}A(w) +\int_{{\Bbb D}} \bigg|\frac{\partial P_h(\varphi f_{k})(w)}{\partial \overline{w}}\bigg|{\rm d}A(w)\\ =\int_{{\Bbb D}} \bigg|\int_{{\Bbb D}}\frac{\partial (\varphi f_{k})} {\partial z}\overline{\frac{\partial^{2} K_{w}(z)}{\partial z \partial \overline{w}}}{\rm d}A(z)\bigg|{\rm d}A(w)+\int_{{\Bbb D}} \bigg|\int_{{\Bbb D}}\frac{\partial (\varphi f_{k})}{\partial \overline{z}} \overline{\frac{\partial^{2} K_{w}(z)}{\partial z \partial \overline{w}}}{\rm d}A(z)\bigg|{\rm d}A(w)\\ =\int_{{\Bbb D}}\bigg|\int_{{\Bbb D}}\frac{\partial (\varphi f_{k})} {\partial z}\frac{1}{(1-\overline{z}w)^{2}}{\rm d}A(z)\bigg|{\rm d}A(w) +\int_{{\Bbb D}}\bigg|\int_{{\Bbb D}}\frac{\partial (\varphi f_{k})}{\partial \overline{z}}\frac{1}{(1-z\overline{w})^{2}}{\rm d}A(z)\bigg|{\rm d}A(w) \\ =\int_{{\Bbb D}}\bigg|\int_{{\Bbb D}}\frac{\partial \varphi}{\partial z} f_{k}\frac{1}{(1-\overline{z}w)^{2}}{\rm d}A(z)+ \int_{{\Bbb D}}\varphi \frac{\partial f_{k} }{\partial z}\frac{1} {(1-\overline{z}w)^{2}}{\rm d}A(z)\bigg|{\rm d}A(w)\\ +\int_{{\Bbb D}}\bigg|\int_{{\Bbb D}}\frac{\partial \varphi}{\partial \overline{z}}f_{k}\frac{1}{(1-z\overline{w})^{2}}{\rm d}A(z)+ \int_{{\Bbb D}}\varphi \frac{\partial f_{k} }{\partial \overline{z}} \frac{1}{(1-z\overline{w})^{2}}{\rm d}A(z)\bigg|{\rm d}A(w) \\ \leq \int_{{\Bbb D}}\bigg|\int_{{\Bbb D}}\frac{\partial \varphi}{\partial z} f_{k}\frac{1}{(1-\overline{z}w)^{2}}{\rm d}A(z)\bigg|{\rm d}A(w)+ \int_{{\Bbb D}}\bigg|\int_{{\Bbb D}}\varphi \frac{\partial f_{k} }{\partial z} \frac{1}{(1-\overline{z}w)^{2}}{\rm d}A(z)\bigg|{\rm d}A(w)\\ +\int_{{\Bbb D}}\bigg|\int_{{\Bbb D}}\frac{\partial \varphi}{\partial \overline{z}}f_{k}\frac{1}{(1-z\overline{w})^{2}}{\rm d}A(z)\bigg|{\rm d}A(w)+ \int_{{\Bbb D}}\bigg|\int_{{\Bbb D}}\varphi \frac{\partial f_{k} }{\partial \overline{z}}\frac{1}{(1-z\overline{w})^{2}}{\rm d}A(z)\bigg|{\rm d}A(w). $

因$L^{1,1}\hookrightarrow L^{2-\frac{\epsilon}{2}}$是紧的,故有$\|f_{k}\|_{L^{2-\frac{\epsilon}{2}}}\rightarrow 0$. 于是

$ \int_{{\Bbb D}}\bigg|\int_{{\Bbb D}}\frac{\partial \varphi}{\partial z}f_{k}\frac{1} {(1-\overline{z}w)^{2}}{\rm d}A(z)\bigg|{\rm d}A(w)\\ \leq C\bigg\|\frac{\partial \varphi}{\partial z}f_{k}\bigg\|_{L^{\frac{8+2\epsilon-\epsilon^2} {8+\epsilon}}} \leq C \bigg\|\frac{\partial \varphi}{\partial z}\bigg\|_{L^{2+\epsilon}} \|f_k\|_{L^{2-\frac{\epsilon}{2}}} =C\|\varphi\|_{L^{2+\epsilon,1}} \|f_k\|_{L^{2-\frac{\epsilon}{2}}}\rightarrow 0 $

且

$$ \int_{{\Bbb D}}\bigg|\int_{{\Bbb D}}\frac{\partial \varphi}{\partial \overline{z}}f_{k}\frac{1}{(1-z\overline{w})^{2}}{\rm d}A(z)\bigg|{\rm d}A(w) \leq C\|\varphi\|_{L^{2+\epsilon,1}}\|f_k\|_{L^{2-\frac{\epsilon}{2}}}\rightarrow 0. $$

注意到

$$ \int_{{\Bbb D}} \varphi \frac{\partial f_{k}}{\partial z}\frac{1}{(1-\overline{z}w)^{2}}{\rm d}A(z) = P^b\Big(\varphi \frac{\partial f_{k}}{\partial z}\Big)(w)=(\mathbb{D}dot{T}_{\varphi}f')(w), $$

其中$P^b$是解析Bergman投影,$\mathbb{D}dot{T}_{\varphi}$为Bergman空间$L_{a}^{1}$上以 $\varphi$为符号的Toeplitz算子. 因$L^{2+\epsilon,1}(\overline{\mathbb{D}})\hookrightarrow C(\overline{\mathbb{D}})$连续, $\varphi|_{\mathbb{T}}= 0$且$\varphi\in L^{2+\epsilon,1}(\overline{\mathbb{D}})\cap \mbox{VMO}_{\partial \log}$, 故$\mathbb{D}dot{T}_{\varphi}$为紧算子,参见文献[5]. 很明显$\frac{\partial f_{k}}{\partial z}$ 在$L_{a}^{1}$中弱收敛到0,因此有

$$ \int_{{\Bbb D}}\bigg|\int_{{\Bbb D}}\varphi \frac{\partial f_{k}}{\partial z} \frac{1}{(1-\overline{z}w)^{2}}{\rm d}A(z)\bigg|{\rm d}A(w) =\int_{{\Bbb D}}\bigg|\mathbb{D}dot{T}_{\varphi}\frac{\partial f_{k}}{\partial z}\bigg|{\rm d}A(w) =\bigg\|\mathbb{D}dot{T}_{\varphi}\frac{\partial f_{k}}{\partial z}\bigg\|_{L^1} \rightarrow 0. $$

同理知

$$ \int_{{\Bbb D}}\bigg|\int_{{\Bbb D}}\varphi \frac{\partial f_{k}}{\partial \overline{z}} \frac{1}{(1-z\overline{w})^{2}}{\rm d}A(z)\bigg|{\rm d}A(w) \rightarrow 0. $$

于是,$\|T^h_{\varphi}f_{k}\|_{L^{1,1}}\rightarrow 0$. 这说明 $T^h_{\varphi}$为紧算子.

反过来,设$T^h_{\varphi}$在${\cal D}^{1}_h$上紧. 下证 $\varphi|_{\mathbb{T}}= 0$. 若$\{f_{k}\}$为$L^{1}_{a} $中满足 $\|f_{k}\|_{L^1}=1$且$f_{k} $在$L^{1}_{a} $中弱收敛到0的序列,则$F_{k}(z)=\int f_{k}(z){\rm d}z$满足 $\|F_{k}\|_{L^{1,1}}=1$,$F_{k} $在${\cal D}^{1}_h$弱收敛到0且$\|F_{k}\|_{L^{2-\frac{\epsilon}{2}}}\rightarrow 0$. 由${\cal D}^{1}_h$上Toeplitz算子的定义知

$ \|\mathbb{D}dot{T}_{\varphi}f_{k}\|_{L^1} =\int_{{\Bbb D}}\bigg|\int_{{\Bbb D}}\frac{\varphi f_{k}}{(1-\overline{z}w)^{2}}{\rm d}A(z) \bigg|{\rm d}A(w) \\ =\int_{{\Bbb D}}\bigg|\int_{{\Bbb D}}\frac{\varphi f_{k}} {(1-\overline{z}w)^{2}}{\rm d}A(z)+ \int_{{\Bbb D}}\frac{\partial \varphi}{\partial z}\frac{F_{k}} {(1-\overline{z}w)^{2}}{\rm d}A(z)+\int_{{\Bbb D}}\frac{\partial \varphi} {\partial \overline{z}}\frac{F_{k}}{(1-z\overline{w})^{2}}{\rm d}A(z)\\ -\int_{{\Bbb D}}\frac{\partial \varphi}{\partial z}\frac{F_{k}} {(1-\overline{z}w)^{2}}{\rm d}A(z)-\int_{{\Bbb D}}\frac{\partial \varphi} {\partial \overline{z}}\frac{F_{k}}{(1-z\overline{w})^{2}}{\rm d}A(z)\bigg|{\rm d}A(w) \\ \leq \|T^h_{\varphi}F_k\|_{L^{1,1}}+\int_{{\Bbb D}}\bigg|\int_{{\Bbb D}} \frac{\partial \varphi}{\partial z}\frac{F_{k}}{(1-\overline{z}w)^{2}}{\rm d}A(z) \bigg|{\rm d}A(w)\\ +\int_{{\Bbb D}}\bigg|\int_{{\Bbb D}}\frac{\partial \varphi}{\partial \overline{z}}\frac{F_{k}}{(1-z\overline{w})^{2}}{\rm d}A(z)\bigg|{\rm d}A(w). $

由$T^h_{\varphi}$的紧性得

$$ \|T^h_{\varphi}F_k\|_{L^{1,1}}\rightarrow 0. $$

因$F_{k} $在${\cal D}^{1}_h$弱收敛到0,故$\|F_k\|_{L^{2-\frac{\epsilon}{2}}}\rightarrow 0$. 因此

$$ \int_{{\Bbb D}}\bigg|\int_{{\Bbb D}}\frac{\partial \varphi}{\partial z} \frac{F_{k}}{(1-\overline{z}w)^{2}}{\rm d}A(z)\bigg|{\rm d}A(w)\leq C\|\varphi\|_{L^{2+\epsilon,1}}\|F_{k}\|_{L^{2-\frac{\epsilon}{2}}}\rightarrow 0 $$

且

$$ \int_{{\Bbb D}}\bigg|\int_{{\Bbb D}}\frac{\partial \varphi}{\partial \overline{z}} \frac{F_{k}}{(1-z\overline{w})^{2}}{\rm d}A(z)\bigg|{\rm d}A(w)\leq C\|\varphi\|_{L^{2+\epsilon,1}}\|F_{k}\|_{L^{2-\frac{\epsilon}{2}}}\rightarrow 0. $$

于是, $\|\mathbb{D}dot{T}_{\varphi}f_{k}\|_{L^1}\rightarrow 0$,即 $\mathbb{D}dot{T}_{\varphi}$是紧算子. 故$\varphi|_{\mathbb{T}}=0$,参见文献[5].

定理 2.4 若$\varphi \in L^{2+\epsilon,1}(\overline{\mathbb{D}})\cap \mbox{VMO}_{\partial \log }$,其中$\epsilon$为任意正数,则$H^h_{\varphi}:{\cal D}_h^{1}\rightarrow L^{1,1}$是紧算子.

证 由定义知对任意${\cal D}^{1}_h$中满足$\|f_{k}\|_{L^{1,1}}=1$且在${\cal D}^{1}_h$中$f_{k}$弱收敛到$ 0$的序列$\{f_{k}\}$有

$ \|H_{\varphi}f_{k}\|_{L^{1,1}} \\ =\int_{{\Bbb D}}\bigg(\bigg|\frac{\partial \varphi}{\partial w}f_{k}+\varphi \frac{\partial f_{k} }{\partial w} -\int_{{\Bbb D}}\frac{\partial \varphi}{\partial z}\frac{f_{k}} {(1-w\overline{z})^{2}}{\rm d}A(z) -\int_{{\Bbb D}}\varphi \frac{\partial f_{k} }{\partial z} \frac{1}{(1-w\overline{z})^{2}}{\rm d}A(z)\bigg|\\ +\bigg|\frac{\partial \varphi}{\partial \overline{w}}f_{k}+\varphi \frac{\partial f_{k} }{\partial \overline{w}} -\int_{{\Bbb D}}\frac{\partial \varphi}{\partial \overline{z}}\frac{f_{k}} {(1-z\overline{w})^{2}}{\rm d}A(z) -\int_{{\Bbb D}}\varphi \frac{\partial f_{k} }{\partial \overline{z}} \frac{1}{(1-z\overline{w})^{2}}{\rm d}A(z)\bigg|\bigg){\rm d}A(w)\\ \leq \int_{{\Bbb D}}\bigg(\bigg|\frac{\partial \varphi}{\partial w}f_{k}\bigg|+ \bigg|\int_{{\Bbb D}}\frac{\partial \varphi}{\partial z} \frac{f_{k}}{(1-w\overline{z})^{2}}{\rm d}A(z)\bigg| +\bigg|\varphi \frac{\partial f_{k} } {\partial w}-\int_{{\Bbb D}}\varphi \frac{\partial f_{k} }{\partial z} \frac{1}{(1-w\overline{z})^{2}}{\rm d}A(z)\bigg|\\ +\bigg|\frac{\partial \varphi}{\partial \overline{w}}f_{k}\bigg| +\bigg|\int_{{\Bbb D}} \frac{\partial \varphi}{\partial \overline{z}} \frac{f_{k}}{(1-z\overline{w})^{2}}{\rm d}A(z)\bigg|+ \bigg|\varphi \frac{\partial f_{k} } {\partial \overline{w}}-\int_{{\Bbb D}}\varphi \frac{\partial f_{k} }{\partial \overline{z}} \frac{1}{(1-z\overline{w})^{2}}{\rm d}A(z)\bigg|\bigg){\rm d}A(w). $

因$\|f_{k}\|_{L^{2-\frac{\epsilon}{2}}}\rightarrow 0$,故有

$$ \int_{{\Bbb D}}\bigg|\frac{\partial \varphi}{\partial w}f_{k}\bigg|{\rm d}A(w) \leq \bigg\|\frac{\partial \varphi}{\partial w}f_{k}\bigg\|_{L^{\frac{8+2\epsilon-\epsilon^2} {8+\epsilon}}}\leq \bigg\|\frac{\partial \varphi}{\partial w}\bigg\|_{L^{2+\epsilon}}\|f_k\|_{L^{2-\frac{\epsilon}{2}}}=\|\varphi\|_{L^{2+\epsilon,1}}\|f_k\|_{L^{2-\frac{\epsilon}{2}}}\rightarrow 0 $$

且

$$ \int_{{\Bbb D}}\bigg|\frac{\partial \varphi}{\partial \overline{w}}f_{k}\bigg|{\rm d}A(w) \leq C\|\varphi\|_{L^{2+\epsilon,1}}\|f_k\|_{L^{2-\frac{\epsilon}{2}}}\rightarrow 0; $$

还有

$ \int_{{\Bbb D}}\bigg|\int_{{\Bbb D}}\frac{\partial \varphi}{\partial z} \frac{f_{k}}{(1-w\overline{z})^{2}}{\rm d}A(z)\bigg|{\rm d}A(w)\\ \leq C\bigg \|\frac{\partial \varphi}{\partial w}f_{k}\bigg\|_{L^{\frac{8+2\epsilon-\epsilon^2} {8+\epsilon}}} \leq C\bigg \|\frac{\partial \varphi}{\partial w}\bigg\|_{L^{2+\epsilon}}\|f_k\|_{L^{2-\frac{\epsilon}{2}}}= C\|\varphi\|_{L^{2+\epsilon,1}}\|f_k\|_{L^{2-\frac{\epsilon}{2}}}\rightarrow 0 $

且

$$ \int_{{\Bbb D}}\bigg|\int_{{\Bbb D}}\frac{\partial \varphi}{\partial \overline{z}} \frac{f_{k}}{(1-z\overline{w})^{2}}{\rm d}A(z)\bigg|{\rm d}A(w) \leq C\|\varphi\|_{L^{2+\epsilon,1}}\|f_k\|_{L^{2-\frac{\epsilon}{2}}}\rightarrow 0. $$

注意到

$$\int_{{\Bbb D}}\bigg |\varphi \frac{\partial f_{k} }{\partial w}- \int_{{\Bbb D}}\varphi \frac{\partial f_{k} }{\partial z}\frac{1} {(1-\overline{z}w)^{2}}{\rm d}A(z)\bigg| {\rm d}A(w)=\bigg\|\mathbb{D}dot{H}_{\varphi} \frac{\partial f_{k} }{\partial z}\bigg\|_{L^1}, $$

其中$\mathbb{D}dot{H}_{\varphi}$ 是$L^1_a\rightarrow L^1$的Hankel算子. 因对任意$\varphi\in L^{2+\epsilon,1}(\overline{\mathbb{D}})\cap \mbox{VMO}_{\partial \log }$, $\mathbb{D}dot{H}_{\varphi}$是紧算子,且$\frac{\partial f_{k}}{\partial z}$在$L_{a}^{1}$中弱收敛到0, 故

$$\int_{{\Bbb D}} \bigg|\varphi \frac{\partial f_{k} }{\partial w}-\int_{{\Bbb D}}\varphi \frac{\partial f_{k} }{\partial z}\frac{1}{(1-\overline{z}w)^{2}}{\rm d}A(z)\bigg| {\rm d}A(w) =\int_{{\Bbb D}}\bigg|\mathbb{D}dot{H}_{\varphi}\frac{\partial f_{k}}{\partial z}\bigg|{\rm d}A(w)\rightarrow 0. $$

同理可得

$$ \int_{{\Bbb D}} \bigg|\varphi \frac{\partial f_{k} }{\partial \overline{w}} -\int_{{\Bbb D}}\varphi \frac{\partial f_{k} }{\partial \overline{z}} \frac{1}{(1-z\overline{w})^{2}}{\rm d}A(z)\bigg|{\rm d}A(w)\rightarrow 0. $$

因此$\|H^h_{\varphi}f_{k}\|_{L^{1,1}}\rightarrow 0$. 这说明$H^h_{\varphi}$为紧算子.

下面得到的调和Dirichlet空间上小Hankel算子的紧性充分条件与解析Dirichlet空间情形有很大区别.

定理 2.5 若$\varphi\in L^{2+\epsilon,1}(\overline{\mathbb{D}})\cap \mbox{VMO}_{\partial \log}$, 其中$\epsilon$为任意正数,则$\Gamma^h_{\varphi}$是${\cal D}^{1}_h$ 上紧算子当且仅当$\varphi|_{\mathbb{T}}=0$.

证 应用定理2.3与引理2.1.

命题 2.1 若$\varphi\in L^{2+\epsilon,1}\cap {\rm BMO}_{\partial \log}$且 $\psi\in L^{2+\epsilon,1}\cap \mbox{VMO}_{\partial \log}$,其中$\epsilon$为任意正数, $T^h_{\varphi}T^h_{\psi}-T^h_{\varphi\psi}$为紧算子.

证 由定义知对任意${\cal D}^{1}_h$中满足$\|f_{k}\|_{L^{1,1}}=1$且在 ${\cal D}^{1}_h$中$f_{k}$弱收敛到$ 0$的序列$\{f_{k}\}$有

$ \|(T^h_{\varphi}T^h_{\psi}-T^h_{\varphi\psi})f_{k}\|_{L^{1,1}}\\ =\int_{{\Bbb D}}\bigg(\bigg|\int_{{\Bbb D}}\frac{\partial \varphi}{\partial z}\frac{(T^h_{\psi}f_{k})}{(1-\overline{z}w)^{2}}{\rm d}A(z) -\int_{{\Bbb D}}\frac{\partial \varphi}{\partial z}\frac{(\psi f_{k})}{(1-\overline{z}w)^{2}}{\rm d}A(z) +\int_{{\Bbb D}}\frac{\partial(-H^h_{\psi}f_{k})}{\partial z}\frac{\varphi}{(1-\overline{z}w)^{2}}{\rm d}A(z) \bigg|\\ +\bigg|\int_{{\Bbb D}}\frac{\partial \varphi}{\partial \overline{z}} \frac{(T^h_{\psi}f_{k})}{(1-z\overline{w})^{2}}{\rm d}A(z) -\int_{{\Bbb D}}\frac{\partial \varphi}{\partial z}\frac{(\psi f_{k})} {(1-z\overline{w})^{2}}{\rm d}A(z) \\ +\int_{{\Bbb D}}\frac{\partial(-H^h_{\psi}f_{k})}{\partial \overline{z}} \frac{\varphi}{(1-z\overline{w})^{2}}{\rm d}A(z)\bigg|\bigg){\rm d}A(w)\\ \leq \int_{{\Bbb D}} \bigg|\int_{{\Bbb D}}\frac{\partial \varphi}{\partial z} \frac{(T^h_{\psi}f_{k})}{(1-\overline{z}w)^{2}}{\rm d}A(z)\bigg|{\rm d}A(w) +\int_{{\Bbb D}}\bigg|\int_{{\Bbb D}}\frac{\partial \varphi}{\partial \overline{z}}\frac{(T^h_{\psi}f_{k})}{(1-z\overline{w})^{2}}{\rm d}A(z)\bigg|{\rm d}A(w)\\ + \int_{{\Bbb D}}\bigg|\int_{{\Bbb D}}\frac{\partial \varphi}{\partial z} \frac{(\psi f_{k})}{(1-\overline{z}w)^{2}}{\rm d}A(z)\bigg|{\rm d}A(w) +\int_{{\Bbb D}}\bigg|\int_{{\Bbb D}}\frac{\partial \varphi}{\partial \overline{z}} \frac{(\psi f_{k})}{(1-z\overline{w})^{2}}{\rm d}A(z)\bigg|{\rm d}A(w)\\ + \int_{{\Bbb D}}\bigg|\int_{{\Bbb D}}\frac{\partial(-H^h_{\psi}f_{k})} {\partial z}\frac{\varphi}{(1-\overline{z}w)^{2}}{\rm d}A(z)\bigg|{\rm d}A(w) +\int_{{\Bbb D}} \bigg|\int_{{\Bbb D}}\frac{\partial(-H^h_{\psi}f_{k})} {\partial \overline{z}}\frac{\varphi}{(1-z\overline{w})^{2}}{\rm d}A(z)\bigg|{\rm d}A(w). $

因$T^h_{\psi}$在$ {\cal D}_h^{1} $上有界,故$\|f_{k}\|_{L^{2-\frac{\epsilon}{2}} }\rightarrow 0$且$\|T^h_{\psi}f_{k}\|_{L^{2-\frac{\epsilon}{2}}}\rightarrow 0$,于是有

$$ \int_{{\Bbb D}}\bigg|\int_{{\Bbb D}}\frac{\partial \varphi}{\partial z} \frac{(T^h_{\psi}f_{k})}{(1-\overline{z}w)^{2}}{\rm d}A(z)\bigg|{\rm d}A(w)\leq C\|\varphi\|_{L^{2+\epsilon,1}}\|T^h_{\psi}f_{k}\|_{L^{2-\frac{\epsilon}{2}}}\rightarrow 0 $$

且

$$ \int_{{\Bbb D}}\bigg|\int_{{\Bbb D}}\frac{\partial \varphi}{\partial \overline{z}}\frac{(T^h_{\psi}f_{k})}{(1-z\overline{w})^{2}}{\rm d}A(z)\bigg|{\rm d}A(w)\leq C\|\varphi\|_{L^{2+\epsilon,1}}\|T^h_{\psi}f_{k}\|_{L^{2-\frac{\epsilon}{2}}}\rightarrow 0; $$

还有

$ \int_{{\Bbb D}}\bigg|\int_{{\Bbb D}}\frac{\partial \varphi}{\partial z} \frac{(\psi f_{k})}{(1-\overline{z}w)^{2}}{\rm d}A(z)\bigg|{\rm d}A(w)\\ \leq C\|\varphi\|_{L^{2+\epsilon}}\|\psi\|_{L^{\infty}} \|f_{k}\|_{L^{2-\frac{\epsilon}{2}}} \leq C\|\varphi\|_{L^{2+\epsilon}}\|\psi\|_{L^{2+\epsilon}} \|f_{k}\|_{L^{2-\frac{\epsilon}{2}}}\rightarrow 0 $

且

$$ \int_{{\Bbb D}}\bigg|\int_{{\Bbb D}}\frac{\partial \varphi}{\partial \overline{z}} \frac{(\psi f_{k})}{(1-z\overline{w})^{2}}{\rm d}A(z)\bigg|{\rm d}A(w)\leq C\|\varphi\|_{L^{2+\epsilon}}\|\psi\|_{L^{2+\epsilon}} \|f_{k}\|_{L^{2-\frac{\epsilon}{2}}}\rightarrow 0. $$

注意到$H^h_{\psi}$为紧且在${\cal D}^{1}_h$中$f_{k}$弱收敛到$ 0$, 故知$\|H^h_{\psi}f_{k}\|_{L^{1,1}}\rightarrow 0$,所以

$ \int_{{\Bbb D}}\bigg|\int_{{\Bbb D}}\frac{\partial(-H^h_{\psi}f_{k})} {\partial z}\frac{\varphi}{(1-\overline{z}w)^{2}}{\rm d}A(z)\bigg|{\rm d}A(w)\\ \leq C\|\varphi\|_{L^{\infty}}\bigg\|\frac{\partial(-H^h_{\psi}f_{k})} {\partial z}\bigg\|_{L^{1,1}} \leq C \|\varphi\|_{L^{2+\epsilon,1}}\|H^h_{\psi }f_{k}\|_{L^{1,1}}\rightarrow 0. $

且

$$ \int_{{\Bbb D}}\bigg|\int_{{\Bbb D}}\frac{\partial(-H_{\psi}f_{k})} {\partial \overline{z}}\frac{\varphi}{(1-z\overline{w})^{2}}{\rm d}A(z)\bigg|{\rm d}A(w) \leq C \|\varphi\|_{L^{2+\epsilon,1}}\|H^h_{\psi }f_{k}\|_{L^{1,1}}\rightarrow 0. $$

于是$\|(T^h_{\varphi}T^h_{\psi}-T^h_{\varphi \psi})f_{k}\|_{L^{1,1}}\rightarrow 0$, 即$T^h_{\varphi}T^h_{\psi}-T^h_{\varphi\psi}$为紧算子.

${\cal D}^1_h$的对偶空间${\cal B}^{h,1}$定义为

$$ {\cal B}^{h,1} =\frac{\{f|f\in h,\mbox{ }\frac{\partial f}{\partial z},\frac{\partial f} {\partial \overline{z}}\in {\cal B}\}}{ \mathbb{C} }, $$

其中$h$表示$\mathbb{D}$上调和函数所成空间. 很明显$f\in {\cal B}^{h,1}$的范数为

$$ \|f\|_{{\cal B}^{h,1}}=\bigg\|\frac{\partial f}{\partial z}\bigg\|_{{\cal B}}+ \bigg\|\overline{\frac{\partial f}{\partial \overline{z}}}\bigg\|_{{\cal B}}. $$

命题 2.2 若$\varphi\in L^{\infty,1}\cap {\rm BMO}_{\partial \log}$, 则$(T^h_{\varphi})^{*}-T^h_{\overline{\varphi}}$为${\cal B}^{h,1}$上紧算子.

证 由定义知对${\cal B}^{h,1}$中满足$\|g_{k}\|_{{\cal B}^{h,1}}=1$且$g_{k}$在 ${\cal B}^{h,1}$中弱*收敛到0的序列$g_{k}$有

$ \|((T^h_{\varphi})^{*}-T^h_{\overline{\varphi}})g_{k}\|_{{\cal B}^{h,1}}\\ =\sup\limits_{w\in\mathbb{D}}(1-|w|^2) \bigg(\bigg| \frac{\partial^2 \langle (T^{*}_{\varphi} -T_{\overline{\varphi}})g_{k},K_{w}(z)\rangle_{1}}{\partial ^2 w}\bigg|+ \bigg| \frac{\partial^2 \langle (T^{*}_{\varphi}-T_{\overline{\varphi}}) g_{k},K_{w}(z)\rangle_{1}}{\partial^2 \overline{w}}\bigg|\bigg)\\ \leq \sup\limits_{w\in\mathbb{D}}(1-|w|^2)\bigg|\bigg\langle \frac{\partial g_k}{\partial z}, \frac{\partial \varphi}{\partial z}\frac{z^2}{(1-z\overline{w})^2}\bigg\rangle\bigg| +\sup_{w\in\mathbb{D}}(1-|w|^2)\bigg|\bigg\langle \frac{\partial \overline{\varphi}}{\partial z}g_k, \frac{z}{(1-z\overline{w})^3}\bigg\rangle\bigg|\\ +\sup\limits_{w\in\mathbb{D}}(1-|w|^2)\bigg|\bigg\langle \frac{\partial g_k}{\partial \overline{z}}, \frac{\partial \varphi}{\partial \overline{z}}\frac{\overline{z}^2}{(1-\overline{z}w)^2} \bigg\rangle\bigg|+\sup_{w\in\mathbb{D}}(1-|w|^2) \bigg|\bigg\langle \frac{\partial \overline{\varphi}} {\partial \overline{z}}g_k,\frac{\overline{z}}{(1-\overline{z}w)^3}\bigg\rangle\bigg|. $

因$\frac{z^2(1-|w|^2)}{(1-z\overline{w})^2},\frac{z^2(1-|w|^2)} {(1-\overline{z}w)^2}\in L^2$且满足$\|\frac{z^2(1-|w|^2)}{(1-z\overline{w})^2}\|_{L^2}\leq 1$与$\|\frac{z^2(1-|w|^2)}{(1-\overline{z}w)^3}\|_{L^2}\leq 1$,故

$$ \sup\limits_{w\in\mathbb{D}}(1-|w|^2)\bigg|\bigg\langle \frac{\partial g_k}{\partial z},\frac{\partial \varphi} {\partial z}\frac{z^2}{(1-z\overline{w})^2}\bigg\rangle\bigg|\rightarrow 0,\quad \sup\limits_{w\in\mathbb{D}}(1-|w|^2)\bigg|\bigg\langle \frac{\partial g_k}{\partial \overline{z}}, \frac{\partial \varphi}{\partial \overline{z}}\frac{z^2}{(1-z\overline{w})^2}\bigg\rangle\bigg|\rightarrow 0. $$

注意到$g_{k}$在${\cal B}^{h,1}$中弱*收敛到0且$\|g_{k}\|_{{\cal B}^{h,1}}=1$,于是存在充分小的$\delta>0$,对任意$1-\delta < r < 1$,在$\mathbb{D}-\overline{\mathbb{D}_r}$中有$|\frac{\partial g_k}{\partial z}(z)|\leq C(-\log(1-|z|^2)),|\frac{\partial g_k}{\partial \overline{z}}(z)|\leq C(-\log(1-|z|^2))$,且在$\overline{\mathbb{D}_r}=\{z|\mbox{ }|z|\leq 1-\delta\}$一致地有$\frac{\partial g_k}{\partial z},\frac{\partial g_k}{\partial \overline{z}}\rightarrow 0$. 因此,对任意$\epsilon>0$,存在$K$,当$k>K$,对任意$1-\delta < r < 1$,其中充分小$\delta>0$,有

$ \sup\limits_{z\in\overline{\mathbb{D}_r}}(1-|w|^2) \bigg|\frac{\partial g_k}{\partial \overline{w}}(w)\bigg|+\sup\limits_{z\in\overline{\mathbb{D}_r}}(1-|w|^2) \bigg|\frac{\partial g_k}{\partial \overline{w}}(w)\bigg|<\frac{\epsilon}{2};\\ \sup\limits_{z\in\mathbb{D}-\overline{\mathbb{D}_r}}(1-|w|^2) \bigg|\frac{\partial g_k}{\partial \overline{w}}(w)\bigg|+\sup\limits_{z\in\mathbb{D}-\overline{\mathbb{D}_r}}(1-|w|^2) \bigg|\frac{\partial g_k}{\partial \overline{w}}(w)\bigg|\\ \leq C \sup\limits_{z\in\mathbb{D}-\overline{\mathbb{D}_r}}(1-|w|^2)(-\log(1-|w|^2))<\frac{\epsilon}{2}. $

于是

$ \|g_k\|_{{\cal B}^h} \leq \sup\limits_{z\in\overline{\mathbb{D}_r}}(1-|w|^2) \bigg(\bigg|\frac{\partial g_k}{\partial \overline{w}}(w)\bigg|+\bigg|\frac{\partial g_k}{\partial \overline{w}}(w)\bigg|\bigg)\\ +\sup\limits_{z\in\mathbb{D}-\overline{\mathbb{D}_r}}(1-|w|^2)\bigg(\bigg|\frac{\partial g_k}{\partial \overline{w}}(w)\bigg| +\bigg|\frac{\partial g_k}{\partial \overline{w}}(w)\bigg|\bigg)\rightarrow 0. $

又注意到$\frac{\partial \overline{\varphi}}{\partial z}\in L^{\infty}$且 $\frac{(1-|w|^2)z}{(1-z\overline{w})^3}\in L^1$,故

$ \sup\limits_{w\in\mathbb{D}}(1-|w|^2)\bigg(\bigg|\bigg\langle \frac{\partial \overline{\varphi}}{\partial z}g_k, \frac{z}{(1-z\overline{w})^3}\bigg\rangle\bigg|+\bigg|\bigg\langle \frac{\partial \overline{\varphi}} {\partial \overline{z}}g_k,\frac{\overline{z}}{(1-\overline{z}w)^3}\bigg\rangle\bigg|\bigg)\\ \leq C\|g_k\|_{{\cal B}^h}\bigg(\bigg\|\frac{\partial \overline{\varphi}}{\partial z}\bigg\|_{L^{\infty}} +\bigg\|\frac{\partial \overline{\varphi}}{\partial \overline{z}}\bigg\|_{L^{\infty}}\bigg) \bigg\| \frac{(1-|w|^2)z}{(1-z\overline{w})^2}\bigg\|_{L^1}\rightarrow 0. $

因此,

$$ \|(T^{*}_{\varphi}-T_{\overline{\varphi}})g_{k}\|_{{\cal B}^{h,1}}\rightarrow 0, $$

即 $T_{\varphi}^{*}-T_{\overline{\varphi}}$是紧算子.

Banach空间$X$上有界算子$A$为Fredholm算子当且仅当$A$的核与余核是有限维; Fredholm指标定义如下

$$ \mbox{Ind}\mbox{ }A=\mbox{dim}\mbox{ }\mbox{ker}\mbox{ }A-\mbox{dim}\mbox{ }\mbox{coker}\mbox{ }A. $$

下面定理中得到了${\cal D}^1_h$上Toeplitz算子$T^h_{\varphi}$为Fredholm算子的充要条件.

定理 3.1 设$\varphi \in L^{2+\epsilon,1}(\overline{\mathbb{D}})\cap \mbox{VMO}_{\partial \log }$, 则$T_{\varphi}$是Fredholm算子当且仅当对任意$z\in \mathbb{T}$,$\varphi(z)\neq 0$.

证 不失一般性,设$0\in \varphi(\mathbb{T})$, 则$\mathbb{D}dot{T}_{\varphi}$不是$L^{1}_{a}({\Bbb D})$上Fredholm算子, 故对$\{f_{k}\}\subset L^{1}_{a} $ 满足$\|f_{k}\|_{L^1}=1$,且在$L^{1}_{a} $中$f_{k}$弱收敛到0有 $\|\mathbb{D}dot{T}_{\varphi}f_{k}\|_{L^1}\rightarrow 0$,或 对$\{g_{k}\}\subset {\cal B}$,其中${\cal B}$为解析Bloch空间, 满足$\|g_{k}\|_{{\cal B}_h}=1$,$g_{k} $在${\cal B}$中弱$*$收敛到0有 $\|\mathbb{D}dot{T}^{*}_{\varphi}g_{k}\|_{{\cal B}}\rightarrow 0$.

记$F_{k}=\int f_{k}{\rm d}z$,则$\{F_{k}\}\subset {\cal D}^{1}_h $, $\|F_{k}\|_{L^{1,1}}=1$,且$F_{k}$在${\cal D}^{1}_h $中弱收敛到0. 注意到

$ \|T^h_{u} F_{k}\|_{L^{1,1}}\\ =\int_{{\Bbb D}}\bigg|\frac{\partial P_h(\varphi F_{k})(w)}{\partial w} \bigg|{\rm d}A(w)+\int_{{\Bbb D}}\bigg|\frac{\partial P_h(\varphi F_{k})(w)} {\partial \overline{w}}\bigg|{\rm d}A(w)\\ =\int_{{\Bbb D}}\bigg|\int_{{\Bbb D}}\frac{\partial (\varphi F_{k})} {\partial z}\frac{1}{(1-\overline{z}w)^{2}}{\rm d}A(z)\bigg|{\rm d}A(w)+ \int_{{\Bbb D}}\bigg|\int_{{\Bbb D}}\frac{\partial (\varphi F_{k})} {\partial \overline{z}}\frac{1}{(1-z\overline{w})^{2}}{\rm d}A(z)\bigg|{\rm d}A(w)\\ =\int_{{\Bbb D}}\bigg|\int_{{\Bbb D}}\frac{\partial \varphi} {\partial z}F_{k}\frac{1}{(1-\overline{z}w)^{2}}{\rm d}A(z) +\int_{{\Bbb D}}\varphi f_{k}\frac{1}{(1-\overline{z}w)^{2}}{\rm d}A(z)\bigg|{\rm d}A(w)\\ +\int_{{\Bbb D}}\bigg|\int_{{\Bbb D}}\frac{\partial \varphi}{\partial \overline{z}}F_{k}\frac{1}{(1-z\overline{w})^{2}}{\rm d}A(z)\bigg|{\rm d}A(w)\\ \leq \int_{{\Bbb D}}\bigg|\int_{{\Bbb D}}\frac{\partial \varphi} {\partial z}F_{k}\frac{1}{(1-\overline{z}w)^{2}}{\rm d}A(z)\bigg|{\rm d}A(w)+ \int_{{\Bbb D}}\bigg|\int_{{\Bbb D}}\varphi f_{k}\frac{1} {(1-\overline{z}w)^{2}}{\rm d}A(z)\bigg|{\rm d}A(w)\\ + \int_{{\Bbb D}}\bigg|\int_{{\Bbb D}}\frac{\partial \varphi}{\partial \overline{z}}F_{k}\frac{1}{(1-z\overline{w})^{2}}{\rm d}A(z)\bigg|{\rm d}A(w). $

因$\|F_{k}\|_{L^{2-\frac{\epsilon}{2}}}\rightarrow 0$,故

$$ \int_{{\Bbb D}}\bigg|\int_{{\Bbb D}}\frac{\partial \varphi}{\partial z}F_{k} \frac{1}{(1-\overline{z}w)^{2}}{\rm d}A(z)\bigg|{\rm d}A(w)\leq C\|\varphi\|_{L^{2+\epsilon,1}}\|f\|_{L^{2-\frac{\epsilon}{2}}}\rightarrow 0 $$

且

$$ \int_{{\Bbb D}}\bigg|\int_{{\Bbb D}}\frac{\partial \varphi}{\partial \overline{z}}F_{k}\frac{1}{(1-z\overline{w})^{2}}{\rm d}A(z)\bigg|{\rm d}A(w)\leq C\|\varphi\|_{L^{2+\epsilon,1}}\|f\|_{L^{2-\frac{\epsilon}{2}}}\rightarrow 0. $$

又因

$$\int_{{\Bbb D}}\bigg|\int_{{\Bbb D}}\varphi f_{k}\frac{1}{(1-\overline{z}w)^{2}} {\rm d}A(z)\bigg|{\rm d}A(w)=\int_{{\Bbb D}}|P^b(\varphi f_{k})(w)|{\rm d}A(w) =\int_{{\Bbb D}}|(\mathbb{D}dot{T}_{u}f_{k})(w)|{\rm d}A(w)\rightarrow 0, $$

其中$P^b$为 $L^1$到$L^1_a$上的投影,$\mathbb{D}dot{T}_{u}$是$L^1_a$上的Toeplitz算子, 则$\|T^h_{\varphi}f_{k}\|_{L^{1,1}}\rightarrow 0$. 这说明$ T_{\varphi}$不是Fredholm算子.

记$G_{k}=\int g_{k}{\rm d}z$,则有$\{G_{k}\}\subset {\cal B}^{h,1} $. 很明显$f\in {\cal B}^1$满足$\|f\|_{{\cal B}^{h,1}}=\|f'\|_{{\cal B}}$. 故有$\|G_{k}\|_{{\cal B}^{h,1}}=1$,且$G_{k}$在${\cal B}^{h,1} $中弱*收敛到0. 注意

$ \|(T^h_{\varphi})^{*} G_{k}\|_{{\cal B}^{h,1}} =\sup\limits_{w\in\mathbb{D}}(1-|w|^2) \bigg(\bigg|\frac{\partial^2 \langle (T^h_{\varphi})^{*} G_k,K^h_w(z)\rangle_1}{\partial^2 w} \bigg|+ \bigg|\frac{\partial^2 \langle(T^h_{\varphi})^{*} G_k,K^h_w(z)\rangle_1}{\partial^2 \overline{w}}\bigg|\bigg)\\ =\sup\limits_{w\in\mathbb{D}}(1-|w|^2) \bigg(\bigg|\frac{\partial^2 \langle G_k,T^h_{\varphi} K^h_w(z)\rangle_1}{\partial^2 w}\bigg| +\bigg|\frac{\partial^2 \langle G_k,T^h_{\varphi} K^h_w(z)\rangle_1}{\partial^2 \overline{w}}\bigg|\bigg)\\ =\sup\limits_{w\in\mathbb{D}}(1-|w|^2) \bigg(\bigg|\frac{\partial^2 \langle G_k,\varphi K^h_w(z)\rangle_1}{\partial^2 w}\bigg| +\bigg|\frac{\partial^2 \langle G_k,\varphi K^h_w(z)\rangle_1}{\partial^2 \overline{w}}\bigg|\bigg)\\ \leq \sup\limits_{w\in\mathbb{D}}(1-|w|^2)\bigg| \bigg\langle g_k,\frac{\partial \varphi}{\partial z} \frac{z^2} {(1-z\overline{w})^2}\bigg\rangle\bigg|+\sup\limits_{w\in\mathbb{D}}(1-|w|^2) \bigg|\frac{d \langle g_k,\varphi \frac{1}{(1-z\overline{w})^2} \rangle}{d w}\bigg|. $

因$g_k\rightarrow 0$在${\cal B}$中弱*收敛到0, $\frac{\partial \varphi}{\partial z}\in L^{2+\epsilon}$, $\frac{(1-|w|^2)z^2}{(1-z\overline{w})^2}\in L^2_a$且满足 $\|\frac{(1-|w|^2)z^2}{(1-z\overline{w})^2}\|_{L^{2}}\leq 1$,故

$$ \sup\limits_{w\in\mathbb{D}}(1-|w|^2)\bigg|\bigg \langle g_k,\frac{\partial \varphi}{\partial z} \frac{z^2}{(1-z\overline{w})^2}\bigg\rangle\bigg| \rightarrow 0. $$

又因

$$ \|(\mathbb{D}dot{T}_{\varphi})^* g_k\|_{{\cal B}}= \sup\limits_{w\in\mathbb{D}}(1-|w|^2)\bigg|\frac{d \langle g_k,\varphi \frac{1}{(1-z\overline{w})^2} \rangle}{d w}\bigg|\rightarrow 0, $$

所以

$$ \|(T^h_{\varphi})^{*} G_{k}\|_{{\cal B}^1}\rightarrow 0. $$

这说明$ T^h_{\varphi}$不是Fredholm算子.

反过来,若$T^h_{\varphi}$不是Fredholm算子,则 存在序列$\{f_{k}\}\subset {\cal D}^{1}_h$ 满足$\|f_{k}\|_{L^{1,1}}=1$,且$f_{k}$在${\cal D}^{1}_h$中弱收敛到0,使得 $\|T^h_{\varphi}f_{k}\|_{L^{1,1}}\rightarrow 0$,或存在 $\{g_{k}\}\subset {\cal B}^{h,1}$满足$\|g_{k}\|_{{\cal B}^{h,1}}=1$,且$g_{k}$在$ {\cal B}^{h,1}$中弱*收敛到0,使得 $\|(T^{h}_{\varphi})^* g_{k}\|_{{\cal B}^{h,1}}\rightarrow 0$. 注意到$\|\frac{\partial f_{k}}{\partial z}\|_{L^{1}}+\|\frac{\partial f_{k}}{\partial \overline{z}}\|_{L^{1}}=1$,选取恰当子列得$\frac{\partial f_{k}}{\partial z}$在$L^{1}_a({\Bbb D})$中弱收敛到0,但$\|\frac{\partial f_{k}}{\partial z}\|_{L^1}\geq \delta>0$. 又注意到$\|\frac{\partial g_{k}}{\partial z}\|_{{\cal B}}+\|\overline{\frac{\partial g_{k}}{\partial \overline{z}}}\|_{{\cal B}}=1$,选取恰当子列得$\frac{\partial g_{k}}{\partial z}$在${\cal B}$中弱*收敛到0,但$\|\frac{\partial g_{k}}{\partial z}\|_{{\cal B}}\geq \delta>0$.

若$\|T^h_{\varphi}f_{k}\|_{L^{1,1}}\rightarrow 0$,因

$ \|T^h_{\varphi} f_{k}\|_{L^{1,1}} =\int_{{\Bbb D}}\bigg|\frac{\partial P_h( \varphi f_{k})(w)}{\partial w} \bigg|{\rm d}A(w)+\int_{{\Bbb D}}\bigg|\frac{\partial P_h( \varphi f_{k})(w)} {\partial \overline{w}}\bigg|{\rm d}A(w)\\ =\int_{{\Bbb D}}\bigg|\int_{{\Bbb D}}\frac{\partial (\varphi f_{k})} {\partial z}\frac{1}{(1-\overline{z}w)^{2}}{\rm d}A(z)\bigg|{\rm d}A(w)\\ + \int_{{\Bbb D}}\bigg|\int_{{\Bbb D}}\frac{\partial (\varphi f_{k})}{\partial \overline{z}}\frac{1}{(1-z\overline{w})^{2}}{\rm d}A(z)\bigg|{\rm d}A(w)\\ =\int_{{\Bbb D}}\bigg|\int_{{\Bbb D}}\frac{\partial \varphi}{\partial z} f_{k}\frac{1}{(1-\overline{z}w)^{2}}{\rm d}A(z)+ \int_{{\Bbb D}} \varphi \frac{\partial f_{k}}{\partial z} \frac{1}{(1-\overline{z}w)^{2}}{\rm d}A(z)\bigg|{\rm d}A(w)\\ +\int_{{\Bbb D}}\bigg|\int_{{\Bbb D}}\frac{\partial \varphi}{\partial \overline{z}}f_{k}\frac{1}{(1-\overline{z}w)^{2}}{\rm d}A(z)+ \int_{{\Bbb D}} \varphi \frac{\partial f_{k}}{\partial \overline{z}} \frac{1}{(1-\overline{z}w)^{2}}{\rm d}A(z)\bigg|{\rm d}A(w) $

且

$$ \int_{{\Bbb D}}\bigg|\int_{{\Bbb D}}\frac{\partial\varphi}{\partial z}f_{k}\frac{1} {(1-\overline{z}w)^{2}}{\rm d}A(z)\bigg|{\rm d}A(w)\leq C\|\varphi \|_{L^{2+\epsilon,1}}\|f\|_{L^{2-\frac{\epsilon}{2}}}\rightarrow 0, $$

则

$$\bigg\| \mathbb{D}dot{T}_{\varphi}\frac{\partial f_{k}}{\partial z}\bigg\|_{L^{1}}=\int_{{\Bbb D}} \bigg|\int_{{\Bbb D}}\varphi \frac{\partial f_{k}}{\partial z}\frac{1}{(1-\overline{z}w)^{2}}{\rm d}A(z) \bigg|{\rm d}A(w)\rightarrow 0. $$

这说明$ \mathbb{D}dot{T}_{\varphi}$不是Fredholm算子,故 $0\in \varphi|_{\mathbb{T}}$.

若$\|(T^{h}_{u})^* g_{k}\|_{{\cal B}^{h,1}}\rightarrow 0$,因

$ \|(T^{h}_{u})^* g_{k}\|_{{\cal B}^{h,1}} =\sup\limits_{w\in\mathbb{D}}(1-|w|^2) \bigg(\bigg|\frac{\partial^2 \langle (T^{h}_{u})^* g_k,K^h_w(z)\rangle_1} {\partial^2 w}\bigg|+\bigg|\frac{\partial^2 \langle (T^{h}_{u})^* g_k, K^h_w(z)\rangle_1}{\partial^2 \overline{w}}\bigg|\bigg)\\ =\sup\limits_{w\in\mathbb{D}}(1-|w|^2) \bigg(\bigg|\frac{\partial^2 \langle g_k,T_{\varphi} K^h_w(z)\rangle_1} {\partial^2 w}\bigg|+\bigg|\frac{\partial^2 \langle g_k,T_{\varphi} K^h_w(z)\rangle_1}{\partial^2 \overline{w}}\bigg|\bigg)\\ =\sup\limits_{w\in\mathbb{D}}(1-|w|^2)\bigg|\frac{\partial^2 \langle \frac{\partial g_k}{\partial z}, \frac{\partial \varphi}{\partial z} K^h_w(z)\rangle}{\partial^2 w}+\frac{d \langle \frac{\partial g_k}{\partial z}, \varphi\frac{1}{(1-z\overline{w})^2} \rangle}{d w}\bigg|\\ +\sup\limits_{w\in\mathbb{D}}(1-|w|^2)\bigg|\frac{\partial^2 \langle \frac{\partial g_k}{\partial \overline{z}}, \frac{\partial \varphi}{\partial \overline{z}} K^h_w(z)\rangle}{\partial^2 \overline{w}}+ \frac{d \langle \frac{\partial g_k}{\partial \overline{z}},\varphi\frac{1}{(1-\overline{z}w)^2} \rangle}{d \overline{w}}\bigg|. $

再因$\frac{\partial g_k}{\partial \overline{z}} $在${\cal B}$中弱*收敛到0, $\frac{\partial \varphi}{\partial z}\in L^{2+\epsilon}$, $\frac{(1-|w|^2)z^2}{(1-z\overline{w})^2}\in L^2_a$且 $\|\frac{(1-|w|^2)z^2}{(1-z\overline{w})^2}\|_{L^{2}}\leq 1$,得

$$ \sup\limits_{w\in\mathbb{D}}(1-|w|^2)\bigg|\bigg \langle \frac{\partial g_k}{\partial \overline{z}}, \frac{\partial \varphi}{\partial z} \frac{z^2}{(1-z\overline{w})^2}\bigg\rangle\bigg| \rightarrow 0, $$

所以

$$\bigg\|\mathbb{D}dot{T}^{*}_{\varphi}\frac{\partial g_k}{\partial \overline{z}}\bigg\|_{{\cal B}} =\sup\limits_{w\in\mathbb{D}}(1-|w|^2)\bigg|\frac{d \langle \frac{\partial g_k}{\partial \overline{z}}, \varphi\frac{1}{(1-z\overline{w})^2} \rangle}{d w}\bigg|\rightarrow 0. $$

这说明$ \mathbb{D}dot{T}^{*}_{\varphi}$不是Fredholm算子,于是 $0\in \varphi|_{\mathbb{T}}$.

引理3.1 若$m$为正整数,则

$$\mathbb{D}im (\ker T^h_{z^m})=\mathbb{D}im (\ker (T^h_{z^m})^*)=1. $$

证 首先证明$(T^h_{z^m})^*z^m=0$. 对$l>0$,有

$$ \langle (T^h_{z^m})^*z^m,z^l\rangle_1=\langle z^m,z^m z^l\rangle_1=0,\mbox{ }\mbox{且}\mbox{ } \langle (T^h_{z^m})^*z^m,\overline{z}^l\rangle_1=\langle z^m,z^m \overline{z}^l\rangle_1=0, $$

故

$$(T^h_{z^m})^*z^m=0. $$

于是,记

$$h=\sum\limits_{k=1,k\neq m}^{\infty}a_k z^k+\sum\limits_{k=1}^{\infty}\mathbb{T}ilde{a}_k \overline{z}^k\in \mbox{ker} (T^h_{z^m})^*. $$

对任意$f\in {\cal D}^1$,有

$$ \langle z^m f,h\rangle_1= \langle f,(T^h_{z^m})^* h\rangle_1=0. $$

注意到

$$\langle z^m f,h\rangle_1=\bigg\langle z^{m-1}(m f+z f'), \frac{\partial h}{\partial z}\bigg\rangle, $$

故

$$h=\sum\limits_{k=1}^{m-1}a_kz^k+\sum\limits_{k=1}^{\infty}\mathbb{T}ilde{a}_k \overline{z}^k. $$

因对$k< m$与$l>0$,

$$ \langle (T^h_{z^m})^* z^k,z^l\rangle_1=\langle z^k,z^m z^l\rangle_1=0 \mbox{ } \mbox{ 且} \mbox{ } \langle (T^h_{z^m})^* z^k,\overline{z}^l\rangle_1=\langle z^k,z^m\overline{z}^l\rangle_1=\langle k z^{k-1},m z^{m-1}\overline{z}^l\rangle. $$

故

$$ (T^h_{z^m})^* z^k=\frac{k}{m-k}\overline{z}^{m-k}. $$

又因对$k>0$与$l>0$,

$$ \langle (T^h_{z^m})^* \overline{z}^k,z^l\rangle_1=0, \mbox{ } \mbox{ 且} \mbox{ } \langle (T^h_{z^m})^* \overline{z}^k,\overline{z}^l\rangle_1=\langle \overline{z}^k,z^m \overline{z}^l\rangle_1=\langle k\overline{z}^{k-1},z^m l\overline{z}^{l-1}\rangle. $$

所以

$$ (T^h_{z^m})^* \overline{z}^k=\frac{k}{m+k}\overline{z}^{m+k}. $$

于是

$$ (T^h_{z^m})^* h=\sum\limits_{k=1}^{m-1}a_k\frac{k}{m-k}\overline{z}^{m-k}+\sum\limits_{k=1}^{\infty}\mathbb{T}ilde{a}_k \frac{k}{m+k}\overline{z}^{m+k}=0. $$

这说明$h=0$且$\mbox{dim}(\mbox{ker}(T^h_{z^m})^*)=1$. 同理可得$\mbox{dim}(\mbox{ker}T^h_{z^m})=1$.

应用上述结论可得对$m\geq 0$有Ind $T^h_{z^m}=0$,其中Ind $T^h_{z^m}$表示$T^h_{z^m}$ 的Fredholm指标,同理对$m> 0$,有Ind $T^h_{\overline{z}^m}=0$.

为证明下面结论,这里引入无零点连续函数$u(z)$的绕数,其定义为

$$ \mbox{wind}\mbox{ }u|_{{\Bbb T}}=\frac{[\mbox{arg}\mbox{ }u]_{{\Bbb \partial {\Bbb D}}}}{2\pi}, $$

这里的$[\mbox{arg}\mbox{ }u]_{{\Bbb \partial {\Bbb D}}}$表示当自变量在 ${\Bbb T}=\partial{\Bbb D}$中取遍一周时,辐角arg $u(t)$的全增量.

定理 3.2 设$u\in L^{2+\epsilon,1}(\overline{\mathbb{D}})\cap \mbox{VMO}_{\partial \log}$, 对任意$z\in \mathbb{T}$满足$u(z)\neq 0$,则

$$ \mbox{Ind} T^h_{u}=0. $$

证 易知存在函数$v\in C^{2}(\overline{{\Bbb D}})$ 在${\Bbb T}$上无零点且满足

$$ \mbox{Ind}\mbox{ }T^h_{u}=\mbox{Ind}\mbox{ }T^h_{v}\mbox{ }\mbox{ } \mbox{和}\mbox{ }\mbox{ }\mbox{wind}\mbox{ }u|_{{\Bbb T}}=\mbox{wind}\mbox{ }v|_{{\Bbb T}}=:k\geq 0. $$

对$s\in [0,1]$,定义

$$ F_{s}(t)=t^{k}\exp (s\log g(t)) \mbox{ }\mbox{ }\mbox{ }(t\in {\Bbb T}), $$

其中$g(t)=t^{-k}v(t)$. 因$v\in C^{2}({\Bbb T})$,wind $v|_{{\Bbb T}}=k$, 且$F_{s}$是$C^{2}({\Bbb T})$中的同伦映射,在 ${\Bbb T}$上无零点,故可以延拓$F_{s}:{\Bbb T}\rightarrow {\Bbb C}$为 $C^{1}(\overline{{\Bbb D}})$中映射. 于是对任意$s\in [0,1]$,$T^h_{F_{s}}$为Fredholm算子, 又由 Fredholm指标的连续性得

$$ \mbox{Ind}\mbox{ }T^h_{z^{k}}=\mbox{Ind}\mbox{ }T^h_{F_{0}}=\mbox{Ind}\mbox{ }T^h_{F_{1}}=\mbox{Ind}\mbox{ }T^h_{v}=\mbox{Ind}\mbox{ }T^h_{u}. $$

当$k\geq 0$,不难得到

$$ 0=\mbox{Ind}\mbox{ }T^h_{z^{k}}=\mbox{Ind}\mbox{ }T^h_{u}. $$

当$k<0$,同样讨论可知

$$ \mbox{wind}\mbox{ }u|_{{\Bbb T}}=\mbox{wind}\mbox{ }\overline{z}^{|k|}|_{{\Bbb T}}=:k<0. $$

于是,

$$ \mbox{Ind}\mbox{ }T^h_{u}=\mbox{Ind}\mbox{ }T^h_{\overline{z}^{|k|}}=0. $$