令 ${\Bbb {D}}$ 记复平面 ${\Bbb C}$ 中的单位圆盘,${\Bbb T}$ 记单位圆周. 本文中,令 $n$ 是一个固定的大于或等于 2 的自然数, ${\Bbb {D}}^{n}$ 和 ${\Bbb T}^{n}$ 分别记 $n$ 个 ${\Bbb D}$ 和 ${\Bbb T}$ 的Cartesian乘积. 对 $1\leq p\leq\infty$,令 $L^{p}({\Bbb T}^{n})$ 是 ${\Bbb T}^{n}$ 上关于测度 ${\rm d}\sigma$ 的Lebesgue空间,其中 ${\rm d}\sigma$ 是 ${\Bbb T}^{n}$ 上的 Haar测度. Hardy空间 $H^{p}({\Bbb {D}}^{n})$ 定义为 $L^{p}({\Bbb T}^{n})$ 中解析多项式的闭包. 以 $P$ 记 $L^{2}({\Bbb T}^{n})$ 到 $H^{2}({\Bbb {D}}^{n})$ 上的正交投影. 对 $u\in L^{\infty}({\Bbb T}^{n})$,定义Toeplitz算子 $T_{u}$ 为 $$T_{u}(f)=P(uf),\ \ \ f \in H^{2}({\Bbb {D}}^{n}).$$ 众所周知,$T_{u}$ 是 $H^{2}({\Bbb D}^{n})$ 上的有界线性算子.

在单位圆盘的Hardy空间上,Brown和Halmos^[2]证明了两个Toeplitz 算子是交换的当且仅当它们的符号都是解析的,或者都是与解析的, 或者它们的一个非平凡线性组合为常数. 随后,关于这个问题出现大量后续研究. 例如,在单位圆盘的Bergman空间上,Axler和^[1] 刻画了两个调和符号的Toeplitz 算子的交换性. 在单位圆盘的Dirichlet空间上, Lee^[8] 得到两个调和符号的Toeplitz算子交换的充要条件. Chen和 Dieu^[4], Yu^[11]分别将Lee 的结论扩展到一般符号的Toeplitz算子. 在单位球的Bergman空间上,Zheng^[12]刻画了具有多调和符号的Toeplitz算子的交换性.

在多圆盘情形,Sun和Zheng研究了本性重交换的Toeplitz算子,参见文献^[10]. Choe,Koo和Lee^[3]刻画了多圆盘的的Bergman空间上多调和符号的交换Toeplitz算子. Lee^[9]研究了多圆盘的Hardy空间上两个Toeplitz算子的交换性,其中一个符号是多调和的.

最近,Ding,Sun和Zheng^[6]完全刻画了双圆盘的Hardy空间上Toeplitz算子的交换性. 本文中我们将其中一个结果推广到多圆盘,得到一个多圆盘的Hardy空间上Toeplitz交换的充要条件.

设 $S$ 是 $H^{2}({\Bbb D}^{n})$ 上一个有界线性算子,其Berezin变换 $\widetilde{S}$ 定义为

\begin{equation}\label{eq:11} \widetilde{S}(z)= \langle Sk_{z},k_{z}\rangle = \int_{{\Bbb T}^n} (Sk_{z})(\eta)\overline{k_{z}(\eta)}{\rm d}\sigma (\eta),\ \ z\in{\Bbb D}^{n}, \end{equation}

(1.1)

其中 $k_{z}$ 是 $H^{2}({\Bbb D}^{n})$ 的正规化再生核,定义见第二节.

对 $ 1 \leq k \leq n$,在本文中以 $ {z_{\scriptsize \overrightarrow{k}}}$ 和 ${z_{\scriptsize\overleftarrow k}}$ 分别记 $(z_{k},z_{k+1},\cdots,z_{n})$ 和 $(z_{1},z_{2},\cdots ,z_{k})$.

定理1.1 设 $f,g \in L^{\infty}({\Bbb T}^{n})$,则在 $H^{2}({\Bbb D}^{n})$ 上 $T_{f}T_{g}=T_{g}T_{f}$ 当且仅当

(a) 对 $1\leq k \leq n-1$ 和几乎所有的 $ {\xi_{\scriptsize\overleftarrow{k-1}}} \in {\Bbb T}^{k-1}$,Berezin变换

\begin{equation} \widetilde{[T_{f_{k+}(\xi_{\scriptsize\overleftarrow{k-1}},z_{k},\cdot)}T_{g_{k-} (\xi_{\scriptsize\overleftarrow{k-1}},z_{k},\cdot)}-T_{g_{k+}(\xi_{\scriptsize\overleftarrow{k-1}}, z_{k},\cdot)}T_{f_{k-}(\xi_{\scriptsize\overleftarrow{k-1}},z_{k},\cdot)}]} ({z_{\scriptsize\overrightarrow{k+1}}}) \end{equation}

(1.2)

对于 $z_{k}\in{\Bbb {D}}$ 是调和的;

(b) 对 $2\leq k\leq n$ 和几乎所有的 ${\xi_{\scriptsize\overrightarrow{k+1}}} \in {\Bbb T}^{n-k}$,Berezin变换

\begin{equation} \widetilde{[T_{f_{k+}(\cdot,z_{k},{\xi_{\scriptsize\overrightarrow{k+1}}})} T_{g_{k-}(\cdot,z_{k},{\xi_{\scriptsize\overrightarrow{k+1}}})}-T_{g_{k+} (\cdot,z_{k},{\xi_{\scriptsize\overrightarrow{k+1}}})}T_{f_{k-}(\cdot,z_{k}, {\xi_{\scriptsize\overrightarrow{k+1}}})}]}({z_{\scriptsize\overleftarrow{k-1}}}) \end{equation}

(1.3)

对于 $z_{k}\in{\Bbb {D}}$ 是调和的.

对 $i=1,2,\cdots ,n $,令 $K_{z_{i}}(\zeta_{i})$ 记Hardy空间 $H^{2}({\Bbb D})$ 的再生核 $\frac{1}{1-\overline{z_{i}}\zeta_{i}}$, $k_{z_{i}}(\zeta_{i})$ 记 $H^{2}({\Bbb D})$ 的正规化再生核 $\frac{\sqrt{1-|z_{i}|^{2}}}{1-\overline{z_{i}}\zeta_{i}}$. 则对于 $z=(z_{1},z_{2},\cdots,z_{n})\in {{\Bbb D}}^{n}$ 和 $ \zeta=(\zeta_{1},\zeta_{2},\cdots,\zeta_{n})\in {{\Bbb T}}^{n}$, $H^{2}({\Bbb D}^{n})$ 的再生核为 $$K_{z}(\zeta )= \prod_{i=1}^{n}K_{z_{i}}(\zeta_{i}),$$ 正规化再生核为 $$k_{z}(\zeta )= \prod_{i=1}^{n}k_{z_{i}}(\zeta_{i}).$$

众所周知,从 $L^{2}({\Bbb T}^{n})$ 到 $H^{2}({\Bbb {D}}^{n})$ 上的正交投影 $P$ 可表为一个积分算子

\begin{equation} P({f})(z)= \int_{{\Bbb T}^n}f(\xi)\overline{K_{z}(\xi )}{\rm d}\sigma (\xi),\ \ f\in L^{2}({\Bbb D}^{n}). \end{equation}

(2.1)

因此 $P$ 的定义域可扩展到 $L^{1}({\Bbb T}^{n})$.

设 $f\in L^{1}({\Bbb T}^{n})$,它的调和扩张 $\hat{f}$ 定义为

\begin{equation} \hat{f}(z) = \int_{{\Bbb T}^n}f(\xi)\prod_{i=1}^{n} \frac{1-|z_{i}|^{2}}{1-z_{i}\overline{\xi_{i}}}{\rm d}\sigma (\xi) = \int_{{\Bbb T}^n}f(\xi)|k_{z}(\xi)|^{2}{\rm d}\sigma (\xi) = \langle fk_z,k_z\rangle . \end{equation}

(2.2)

易见 Toeplitz算子 $T_{f}$ 的Berezin变换就是 $f$ 的调和扩张, 且 $\hat{f}(z)$ 在 ${\Bbb D}^{n}$ 中是 $n$ -调和的. 我们以后将 $f$ 的调和扩张仍记为 $f$.

对于 $f\in L^{1}({\Bbb T}^{n})$,$i=1,2,\cdots,n$, $z_{i}\in {\Bbb D}$,$\xi\in {\Bbb T}^{n}$,定义投影 $P_{i}$ 为 $$(P_{i}f)(\xi_{\scriptsize\overleftarrow{i-1}},z_i, \xi_{\scriptsize\overrightarrow{i+1}})= \int_{{\Bbb T}^n}f(\xi)K_{z_{i}}(\xi_{i}){\rm d}\sigma (\xi_{i}).$$ 相似于文献 ^[6],可以得到下述事实

$\bullet$~ 对 $i,j = 1,2,\cdots,n$,$P_{i}$ 与 $P_{j}$ 可交换, $P=P_{1}P_{2} \cdots P_{n}$,且对于 $q > 1$,$P$ 是从 $L^q({\Bbb T}^{n})$ 到 $H^q({\Bbb T}^{n})$ 上的有界线性算子.

$\bullet$~ 设 $f$ 和 $g$ 属于 $\bigcap\limits_{1 < q < \infty} L^{q}({\Bbb T}^{n})$, 则 $f+g$,$fg$ 也属于 $\bigcap\limits_{1 < q <\infty} L^{q}({\Bbb T}^{n})$. 事实上,$\bigcap\limits_{1 < q <\infty} L^{q}({\Bbb T}^{n})$ 是一个代数.

$\bullet$~ 对 $f\in\bigcap\limits_{1 < q <\infty} L^{q}({\Bbb T}^{n})$, 如果令 $f_{i+}=P_{i}(f)$ 和 $f_{i-}=(I-P_{i})(f)$,则

\begin{equation}\label{eq:23} f=f_{i+}+f_{i-}, \end{equation}

(2.3)

$i=1,2,\cdots ,n$ ,且 $f_{i+}$,$f_{i-}$ 属于 $\bigcap\limits_{1 < q <\infty} L^{q}({\Bbb T}^{n})$. 明显地 $f_{i+}$ 对第 $i$ 个变量解析,且 $f_{i-}$ 对第 $i$ 个变量余解析.

$\bullet$~ 如果 $f$ 和 $g$ 属于 $\bigcap\limits_{1 < q <\infty}L^{q}({\Bbb T}^{n})$, 则$T_{f}T_{g}$ 是 $H^{2}({\Bbb T}^{n})$ 上的稠定义算子.

对于 $1 \leq k \leq n$,本文中我们令 ${P_{\scriptsize\overrightarrow{k}}}$ 记 $P_{k}P_{k+1} \cdots P_{n}$. 给定 $f\in\bigcap\limits_{1 < q <\infty} L^{q}({\Bbb T}^{n})$ 和 $z_{\scriptsize\overleftarrow{k}}\in {\Bbb D}^{k}$, 令 $T_{f({z_{\scriptsize\overleftarrow{k}}},\cdot)}$ 是 $H^{2}({\Bbb D}^{n-k})$ ($0 \leq k \leq n-1$) 上Toeplitz算子 $$ T_{f(z_{\scriptsize\overleftarrow{k}},\cdot)}u= {P_{\scriptsize\overrightarrow{n-k+1}}}[f({z_{\scriptsize\overleftarrow{k}}},\cdot)u],\ \ u\in H^{\infty}({\Bbb D}^{n-k}). $$ 对 $f\in H^{2}({\Bbb T}^{n})$,有下述基本事实

\begin{equation}\label{eq:24} (T_fk_z)(w)=f(w)k_z(w), \end{equation}

(2.4)

\begin{equation}\label{eq:25} (T_{\bar f}k_z)(w)=\overline{f(z)}k_z(w). \end{equation}

(2.5)

引理2.1 设 $f$,$g$ $\in L^{\infty}({\Bbb T}^{n})$,$1\leq k\leq n-1$, 则对几乎所有的 ${\xi_{\scriptsize\overleftarrow{k-1}}} \in {\Bbb T}^{k-1}$,有

\begin{eqnarray} &&\langle [T_{f(\xi_{\scriptsize\overleftarrow{k-1}},\cdot)} T_{g(\xi_{\scriptsize\overleftarrow{k-1}},\cdot)} -T_{g(\xi_{\scriptsize\overleftarrow{k-1}},\cdot)} T_{f(\xi_{\scriptsize\overleftarrow{k-1}},\cdot)}]k_{z_{\scriptsize\overrightarrow{k}}}, k_{z_{\scriptsize\overrightarrow{k}}}\rangle\nonumber\\ &=&\langle \langle [T_{f_{k-}(\xi_{\scriptsize\overleftarrow{k}},\cdot)}T_{g_{k+} (\xi_{\scriptsize\overleftarrow{k}}, \cdot)}-T_{g_{k-}(\xi_{\scriptsize\overleftarrow{k}},\cdot)}T_{f_{k+} (\xi_{\scriptsize\overleftarrow{k}}, \cdot)}]k_{{z_{\scriptsize\overrightarrow{k+1}}}}, k_{{z_{\scriptsize\overrightarrow{k+1}}}}\rangle k_{z_{k}},k_{z_{k}}\rangle \nonumber\\ &&+ \langle [T_{f_{k+}(\xi_{\scriptsize\overleftarrow{k-1}},z_{k}, \cdot)}T_{g_{k+}(\xi_{\scriptsize\overleftarrow{k-1}},z_{k},\cdot)} -T_{g_{k+}(\xi_{\scriptsize\overleftarrow{k-1}},z_{k},\cdot)} T_{f_{k+}(\xi_{\scriptsize\overleftarrow{k-1}},z_{k},\cdot)}] k_{{z_{\scriptsize\overrightarrow{k+1}}}},k_{{z_{\scriptsize\overrightarrow{k+1}}}}\rangle \nonumber\\ &&+ \langle [T_{f_{k-}(\xi_{\scriptsize\overleftarrow{k-1}},z_{k},\cdot)}T_{g_{k-} (\xi_{\scriptsize\overleftarrow{k-1}},z_{k},\cdot)} -T_{g_{k-}(\xi_{\scriptsize\overleftarrow{k-1}},z_{k},\cdot)}T_{f_{k-}( \xi_{\scriptsize\overleftarrow{k-1}},z_{k},\cdot)}]k_{{z_{\scriptsize\overrightarrow{k+1}}}}, k_{{z_{\scriptsize\overrightarrow{k+1}}}}\rangle \nonumber\\ &&+ \langle [T_{f_{k+}(\xi_{\scriptsize\overleftarrow{k-1}},z_{k},\cdot)} T_{g_{k-}(\xi_{\scriptsize\overleftarrow{k-1}},z_{k},\cdot)} -T_{g_{k+}(\xi_{\scriptsize\overleftarrow{k-1}},z_{k},\cdot)}T_{f_{k-} (\xi_{\scriptsize\overleftarrow{k-1}},z_{k},\cdot)}] k_{{z_{\scriptsize\overrightarrow{k+1}}}}, k_{{z_{\scriptsize\overrightarrow{k+1}}}}\rangle . \end{eqnarray}

(2.6)

证 利用(2.3)式中的分解,对 $f\in L^{\infty}({\Bbb T}^{n})$,有 $$ f(\xi_{\scriptsize\overleftarrow{k-1}},\cdot)=f_{k+} (\xi_{\scriptsize\overleftarrow{k-1}},\cdot) + f_{k-}(\xi_{\scriptsize\overleftarrow{k-1}},\cdot) $$ 且对几乎所有的 ${\xi_{\scriptsize\overleftarrow{k-1}}} \in {\Bbb T}^{k-1}$,$f_{k+} (\xi_{\scriptsize\overleftarrow{k-1}},\cdot)$ 和 $f_{k-} (\xi_{\scriptsize\overleftarrow{k-1}},\cdot)$ 都属于 $\bigcap\limits_{1 < q <\infty} L^{q}({\Bbb T}^{n-k+1})$. 因此 \begin{eqnarray*} &&\langle [T_{f(\xi_{\scriptsize\overleftarrow{k-1}},\cdot)} T_{g(\xi_{\scriptsize\overleftarrow{k-1}},\cdot)} -T_{g(\xi_{\scriptsize\overleftarrow{k-1}},\cdot)}T_{f(\xi_{\scriptsize\overleftarrow{k-1}}, \cdot)}] k_{z_{\scriptsize\overrightarrow{k}}},k_{z_{\scriptsize\overrightarrow{k}}}\rangle \\ &=& \langle [T_{f_{k-}(\xi_{\scriptsize\overleftarrow{k-1}},\cdot)}T_{g_{k+} (\xi_{\scriptsize\overleftarrow{k-1}},\cdot)} -T_{g_{k-}(\xi_{\scriptsize\overleftarrow{k-1}},\cdot)}T_{f_{k+} (\xi_{\scriptsize\overleftarrow{k-1}},\cdot)}]k_{z_{\scriptsize\overrightarrow{k}}}, k_{z_{\scriptsize\overrightarrow{k}}}\rangle\\ && + \langle [T_{f_{k+}(\xi_{\scriptsize\overleftarrow{k-1}},\cdot)}T_{g_{k+} (\xi_{\scriptsize\overleftarrow{k-1}},\cdot)} -T_{g_{k+}(\xi_{\scriptsize\overleftarrow{k-1}},\cdot)}T_{f_{k+} (\xi_{\scriptsize\overleftarrow{k-1}},\cdot)}]k_{z_{\scriptsize\overrightarrow{k}}}, k_{z_{\scriptsize\overrightarrow{k}}}\rangle \\ &&+ \langle [T_{f_{k-}(\xi_{\scriptsize\overleftarrow{k-1}},\cdot)}T_{g_{k-} (\xi_{\scriptsize\overleftarrow{k-1}},\cdot)} -T_{g_{k-}(\xi_{\scriptsize\overleftarrow{k-1}},\cdot)}T_{f_{k-} (\xi_{\scriptsize\overleftarrow{k-1}},\cdot)}]k_{z_{\scriptsize\overrightarrow{k}}}, k_{z_{\scriptsize\overrightarrow{k}}}\rangle\\ & &+ \langle (T_{f_{k+}(\xi_{\scriptsize\overleftarrow{k-1}},\cdot)}T_{g_{k-} (\xi_{\scriptsize\overleftarrow{k-1}},\cdot)} -T_{g_{k+}(\xi_{\scriptsize\overleftarrow{k-1}},\cdot)}T_{f_{k-} (\xi_{\scriptsize\overleftarrow{k-1}},\cdot)})k_{z_{\scriptsize\overrightarrow{k}}}, k_{z_{\scriptsize\overrightarrow{k}}}\rangle . \end{eqnarray*} 对几乎所有的 ${\xi_{\scriptsize\overleftarrow{k-1}}} \in {\Bbb T}^{k-1}$ 和 $z_{\scriptsize\overrightarrow{k}},w_{\scriptsize\overrightarrow{k}} \in {\Bbb D}^{n-k+1}$, 由(2.4) 式可知 \begin{eqnarray*} T_{g_{k+}(\xi_{\scriptsize\overleftarrow{k-1}},\cdot)} k_{z_{\scriptsize\overrightarrow{k}}}(w_{\scriptsize\overrightarrow{k}}) &=&P_{\scriptsize\overrightarrow{k+1}}P_k [g_{k+} (\xi_{\scriptsize\overleftarrow{k-1}},\cdot)k_{z_{\scriptsize\overrightarrow{k}}}] (w_{\scriptsize\overrightarrow{k}}) \\ &=& P_{\scriptsize\overrightarrow{k+1}}[g_{k+}(\xi_{\scriptsize\overleftarrow{k-1}}, w_{k},\cdot) k_{{z_{\scriptsize\overrightarrow{k+1}}}}]({w_{\scriptsize\overrightarrow{k+1}}}) \cdot k_{z_{k}}(w_{k}), \end{eqnarray*} 因此

\begin{eqnarray} &&\langle T_{f_{k+}(\xi_{\scriptsize\overleftarrow{k-1}},\cdot)}T_{g_{k+} (\xi_{\scriptsize\overleftarrow{k-1}},\cdot)}k_{z_{\scriptsize\overrightarrow{k}}}, k_{z_{\scriptsize\overrightarrow{k}}}\rangle \nonumber\\ &=& \langle f_{k+}(\xi_{\scriptsize\overleftarrow{k-1}},\cdot)P_{\scriptsize\overrightarrow{k+1}} [g_{k+}(\xi_{\scriptsize\overleftarrow{k-1}},w_{k},\cdot) k_{{z_{\scriptsize\overrightarrow{k+1}}}}]\cdot k_{z_{k}},k_{z_{k}}\cdot k_{{z_{\scriptsize\overrightarrow{k+1}}}}\rangle \nonumber\\ &=& \langle f_{k+}(\xi_{\scriptsize\overleftarrow{k-1}},z_{k},\cdot)P_{\scriptsize\overrightarrow{k+1}} (g_{k+}(\xi_{\scriptsize\overleftarrow{k-1}},z_{k},\cdot) k_{{z_{\scriptsize\overrightarrow{k+1}}}}),k_{{z_{\scriptsize\overrightarrow{k+1}}}} \rangle \nonumber\\ &=& \langle T_{f_{k+}(\xi_{\scriptsize\overleftarrow{k-1}},z_{k},\cdot)}T_{g_{k+} (\xi_{\scriptsize\overleftarrow{k-1}},z_{k},\cdot)}k_{{z_{\scriptsize\overrightarrow{k+1}}}}, k_{{z_{\scriptsize\overrightarrow{k+1}}}}\rangle , \end{eqnarray}

(2.7)

使用Fubini定理,可见

\begin{eqnarray} &&\langle T_{f_{k-}(\xi_{\scriptsize\overleftarrow{k-1}},\cdot)}T_{g_{k+} (\xi_{\scriptsize\overleftarrow{k-1}},\cdot)}k_{z_{\scriptsize\overrightarrow{k}}}, k_{z_{\scriptsize\overrightarrow{k}}}\rangle \nonumber\\ &=&\bigg\langle \int_{{\Bbb T}^{n-k}}f_{k-}(\xi_{\scriptsize\overleftarrow{k-1}}, \cdot,{\xi_{\scriptsize\overrightarrow{k+1}}})P_{\scriptsize\overrightarrow{k+1}}[g_{k+} (\xi_{\scriptsize\overleftarrow{k-1}},\cdot,{\xi_{\scriptsize\overrightarrow{k+1}}}) k_{{z_{\scriptsize\overrightarrow{k+1}}}}({\xi_{\scriptsize\overrightarrow{k+1}}})] \nonumber\\ &&\times \overline{k_{{z_{\scriptsize\overrightarrow{k+1}}}}({\xi_{\scriptsize\overrightarrow{k+1}}})} {\rm d}\sigma ({\xi_{\scriptsize\overrightarrow{k+1}}})\cdot k_{z_{k}}, k_{z_{k}}\bigg \rangle \nonumber\\ &=&\langle \langle (T_{f_{k-}(\xi_{\scriptsize\overleftarrow{k-1}},\cdot)} T_{g_{k+}(\xi_{\scriptsize\overleftarrow{k-1}}, \cdot)})k_{{z_{\scriptsize\overrightarrow{k+1}}}},k_{{z_{\scriptsize\overrightarrow{k+1}}}} \rangle k_{z_{k}},k_{z_{k}}\rangle. \end{eqnarray}

(2.8)

使用(2.5)式,可得 \begin{eqnarray*} T_{g_{k-}(\xi_{\scriptsize\overleftarrow{k-1}},\cdot)}k_{z_{\scriptsize\overrightarrow{k}}} (w_{\scriptsize\overrightarrow{k}}) &=& P_{\scriptsize\overrightarrow{k+1}}P_k [g_{k-}(\xi_{\scriptsize\overleftarrow{k-1}},\cdot) k_{z_k}k_{z_{\scriptsize\overrightarrow{k+1}}}](w_{\scriptsize\overrightarrow{k}}) \\ &=& P_{\scriptsize\overrightarrow{k+1}}[g_{k-}(\xi_{\scriptsize\overleftarrow{k-1}}, z_k,\cdot)k_{z_k}(w_k) k_{z_{\scriptsize\overrightarrow{k+1}}}](w_{\scriptsize\overrightarrow{k+1}}) \\ &=& P_{\scriptsize\overrightarrow{k+1}}[g_{k-}(\xi_{\scriptsize\overleftarrow{k-1}},z_k,\cdot) k_{{z_{\scriptsize\overrightarrow{k+1}}}}]({w_{\scriptsize\overrightarrow{k+1}}})\cdot k_{z_{k}}(w_{k}). \end{eqnarray*} 相似于 (2.7) 和 (2.8)式,我们有

\begin{equation} \langle T_{f_{k-}(\xi_{\scriptsize\overleftarrow{k-1}},\cdot)}T_{g_{k-} (\xi_{\scriptsize\overleftarrow{k-1}},\cdot)}k_{z_{\scriptsize\overrightarrow{k}}}, k_{z_{\scriptsize\overrightarrow{k}}}\rangle =\langle T_{f_{k-}(\xi_{\scriptsize\overleftarrow{k-1}},z_{k},\cdot)}T_{g_{k-} (\xi_{\scriptsize\overleftarrow{k-1}},z_{k},\cdot)}k_{{z_{\scriptsize\overrightarrow{k+1}}}}, k_{{z_{\scriptsize\overrightarrow{k+1}}}}\rangle; \end{equation}

(2.9)

\begin{equation} \langle T_{f_{k+}(\xi_{\scriptsize\overleftarrow{k-1}},\cdot)} T_{g_{k-}(\xi_{\scriptsize\overleftarrow{k-1}},\cdot)}k_{z_{\scriptsize\overrightarrow{k}}}, k_{z_{\scriptsize\overrightarrow{k}}}\rangle =\langle T_{f_{k+}(\xi_{\scriptsize\overleftarrow{k-1}},z_{k},\cdot)} T_{g_{k-}(\xi_{\scriptsize\overleftarrow{k-1}},z_{k},\cdot)}k_{{z_{\scriptsize\overrightarrow{k+1}}}}, k_{{z_{\scriptsize\overrightarrow{k+1}}}}\rangle. \end{equation}

(2.10)

联合 (2.7)-(2.10)式,易见引理的结论成立.

下面的结果(参见文献[7,p38,Corollary 2])显示了一个 $ L^{p}({\Bbb T})$ 中函数和它在单位圆盘的调和扩张的联系.

引理2.2 设 $f$ 是单位圆盘内的调和函数,$1<p<\infty$,满足当 $r\rightarrow 1^-$ 时 $$\int_{{\Bbb T}} |f(r\xi)|^p{\rm d}\sigma(\xi)$$ 是有界的. 则对几乎所有的 $\xi\in{\Bbb T}$,径向极限 $f^\ast(\xi)=\lim\limits_{r\rightarrow 1^-}f(r\xi)$ 存在, $f^\ast$ 属于 $ L^{p}({\Bbb T})$,且 $f$ 是 $f^\ast$ 的调和扩张.

引理2.3 设 $f,g \in\bigcap\limits_{1 < q < \infty} L^{q}({\Bbb T}^{n})$,如果 $$\langle T_{f(z_{1},\cdot)}T_{g(z_{1},\cdot)}k_{{z_{\vec{2}}}},k_{{z_{\vec{2}}}} \rangle$$ 对变量 $z_{1}$ 调和,则

\begin{equation} \langle T_{f(z_{1},\cdot)}T_{g(z_{1},\cdot)}k_{{z_{\vec{2}}}},k_{{z_{\vec{2}}}}\rangle =\langle \langle T_{f(\xi_{1},\cdot)}T_{g(\xi_{1},\cdot)}k_{{z_{\vec{2}}}}, k_{{z_{\vec{2}}}}\rangle k_{z_{1}},k_{z_{1}}\rangle . \end{equation}

(2.11)

证 一方面,取极坐标 $z_{1}=r\xi_{1}$, 应用Fubini定理和Cauchy-Schwarz不等式,有 \begin{eqnarray*} &&\int_{{\Bbb T}}|\langle T_{f(r\xi_{1},\cdot)}T_{g(r\xi_{1},\cdot)}k_{{z_{\vec{2}}}}(\cdot), k_{{z_{\vec{2}}}}(\cdot)\rangle |^{2}{\rm d}\sigma (\xi_{1})\\ &\leq &\int_{{\Bbb T}}\|f(r\xi_{1},\cdot)P_{\vec{2}}[g(r\xi_{1},\cdot)k_{{z_{\vec{2}}}}] \|^{2}{\rm d}\sigma (\xi_{1})\\ &\leq &\int_{{\Bbb T}^{n-1}}\int_{{\Bbb T}}|f(r\xi_{1},{\xi_{\vec{2}}})P_{\vec{2}}[g(r\xi_{1},{\xi_{\vec{2}}})k_{{z_{\vec{2}}}}( {\xi_{\vec{2}}})] |^{2} {\rm d}\sigma (\xi_{1}){\rm d}\sigma ( {\xi_{\vec{2}}})\\ &\leq &\prod_{i=2}^{n}c(z_{i})\cdot \int_{{\Bbb T}^{n}}|f(\xi)|^{4}{\rm d}\sigma (\xi) \cdot \int_{{\Bbb T}^{n}}|g(\xi)|^{4}{\rm d}\sigma (\xi), \end{eqnarray*} 其中 $c(z_{i})$ 是一个仅依赖于 $z_{i}$ 的常数.

由于 $f,g \in\bigcap\limits_{1 < q < \infty} L^{q}({\Bbb T}^{n})$,可见 $$\limsup_{r\rightarrow 1^{-}}\int_{{\Bbb T}}|\langle T_{f(r\xi_{1},\cdot)}T_{g(r\xi_{1},\cdot)}k_{{z_{\vec{2}}}}(\cdot), k_{{z_{\vec{2}}}}(\cdot)\rangle |^{2}{\rm d}\sigma (\xi_{1}) < \infty.$$

另一方面,对几乎所有的 ${\xi_{\vec{2}}} \in {\Bbb T}^{n-1}$,令 $f$ 和 $g$ 的关于 $z_{1}$ Fourier级数分别为 $$f(z_{1},{\xi_{\vec{2}}})=\sum_{-\infty}^{+\infty}\widehat{f(n,{\xi_{\vec{2}}})}z_{1}^{n},\ \ \ g(z_{1},{\xi_{\vec{2}}})=\sum_{-\infty}^{+\infty}\widehat{g(m,{\xi_{\vec{2}}})}\overline{z_{1}}^{m}.$$ 由于$\langle T_{f(z_{1},\cdot)}T_{g(z_{1},\cdot)}k_{{z_{\vec{2}}}},k_{{z_{\vec{2}}}} \rangle$ 关于变量 $z_{1}$ 调和,应用引理 2.2,对 a.e. $\xi_{1}\in {\Bbb T}$,可得 \begin{eqnarray*} &&\lim_{r\rightarrow 1^{-}}\langle T_{f(r\xi_{1},\cdot)}T_{g(r\xi_{1},\cdot)}k_{{z_{\vec{2}}}}, k_{{z_{\vec{2}}}}\rangle \\ &=&\lim_{r\rightarrow 1^{-}}\langle f(r\xi_{1},\cdot)P_{\vec{2}}[g(r\xi_{1},\cdot)k_{{z_{\vec{2}}}}(\cdot)],k_{{z_{\vec{2}}}}(\cdot)\rangle \\ &=& \lim_{r\rightarrow 1^{-}}\sum_{-\infty}^{+\infty}\sum_{-\infty}^{+\infty}\langle \widehat{f(n,\cdot)}P_{\vec{2}}[\widehat{g(m,\cdot)}k_{{z_{\vec{2}}}}(\cdot)], k_{{z_{\vec{2}}}}(\cdot)\rangle (r\xi_{1})^{n}(r\overline{\xi_{1}})^{m}\\ &=& \sum_{-\infty}^{+\infty}\sum_{-\infty}^{+\infty}\langle \widehat{f(n,\cdot)}P_{\vec{2}}[\widehat{g(m,\cdot)}k_{{z_{\vec{2}}}}(\cdot)],k_{{z_{\vec{2}}}}(\cdot)\rangle (\xi_{1})^{n}(\overline{\xi_{1}})^{m}\\ &=& \langle f(\xi_{1},\cdot)P_{\vec{2}}[g(\xi_{1},\cdot)k_{{z_{\vec{2}}}}(\cdot)],k_{{z_{\vec{2}}}}(\cdot)\rangle \\ &=& \langle T_{f(\xi_{1},\cdot)}T_{g(\xi_{1},\cdot)}k_{{z_{\vec{2}}}}, k_{{z_{\vec{2}}}}\rangle . \end{eqnarray*} 再次应用引理 2.2,我们有

\begin{equation} \langle T_{f(z_{1},\cdot)}T_{g(z_{1},\cdot)k_{{z_{\vec{2}}}}}, k_{{z_{\vec{2}}}}\rangle =\langle \langle T_{f(\xi_{1},\cdot)}T_{g(\xi_{1},\cdot)}k_{{z_{\vec{2}}}}, k_{z_{\vec{2}}} \rangle k_{z_{1}},k_{z_{1}}\rangle . \end{equation}

(2.12)

引理得证.

在本节我们给出主要结果的证明.

证 首先假设在 $H^{2}({\Bbb D}^{n})$上 $T_{f}T_{g}=T_{g}T_{f}$. 注意到对 $z= (z_{1},z_{2},\cdots,z_{n}) \in {\Bbb D}^{n}$,$f=f_{1+} + f_{1-}$ 和 $g=g_{1+} + g_{1-}$,由引理 2.1,有

\begin{eqnarray} &&-\langle (T_{f_{1+}(z_{1},\cdot)}T_{g_{1-}(z_{1},\cdot)} -T_{g_{1+}(z_{1},\cdot)}T_{f_{1-}(z_{1},\cdot)})k_{{z_{\vec{2}}}}, k_{{z_{\vec{2}}}}\rangle \nonumber\\ &=&\langle \langle (T_{f_{1-}(\xi_{1},\cdot)}T_{g_{1+}(\xi_{1}, \cdot)}-T_{g_{1-}(\xi_{1},\cdot)}T_{f_{1+}(\xi_{1}, \cdot)})k_{{z_{\vec{2}}}},k_{{z_{\vec{2}}}}\rangle k_{z_{1}}, k_{z_{1}}\rangle \nonumber\\ &&+ \langle (T_{f_{1+}(z_{1},\cdot)}T_{g_{1+}(z_{1},\cdot)}-T_{g_{1+}(z_{1}, \cdot)}T_{f_{1+}(z_{1},\cdot)})k_{{z_{\vec{2}}}},k_{{z_{\vec{2}}}}\rangle \nonumber\\ &&+ \langle (T_{f_{1-}(z_{1},\cdot)}T_{g_{1-}(z_{1},\cdot)} -T_{g_{1-}(z_{1},\cdot)}T_{f_{1-}(z_{1},\cdot)})k_{{z_{\vec{2}}}}, k_{{z_{\vec{2}}}}\rangle . \end{eqnarray}

(3.1)

因为上式右端的每一项都关于 $z_{1}$ 调和,如果令 $$ u_{1}(z)=\langle [T_{f_{1+}(z_{1},\cdot)}T_{g_{1-}(z_{1},\cdot)} -T_{g_{1+}(z_{1},\cdot)}T_{f_{1-}(z_{1},\cdot)}]k_{{z_{\vec{2}}}}, k_{{z_{\vec{2}}}}\rangle, $$ 则 $u_{1}(z)$ 关于 $z_{1}$ 调和. 使用引理 2.3,可见

\begin{equation} u_{1}(z)=\langle \langle [T_{f_{1+}(\xi_{1},\cdot)}T_{g_{1-} (\xi_{1},\cdot)}-T_{g_{1+}(\xi_{1},\cdot)}T_{f_{1-}(\xi_{1}, \cdot)}]k_{{z_{\vec{2}}}},k_{{z_{\vec{2}}}}\rangle k_{z_{1}},k_{z_{1}}\rangle . \end{equation}

(3.2)

因此,应用引理 2.1,有 \begin{eqnarray*} &&\langle (T_{f}T_{g}-T_{g}T_{f})k_{z},k_{z}\rangle \\ &=&\langle \langle[T_{f_{1-}(\xi_{1},\cdot)}T_{g_{1+}(\xi_{1},\cdot)} -T_{g_{1-}(\xi_{1},\cdot)}T_{f_{1+}(\xi_{1},\cdot)}]k_{{z_{\vec{2}}}}, k_{{z_{\vec{2}}}}\rangle k_{z_{1}},k_{z_{1}} \rangle \\ &&+ \langle [T_{f_{1+}(z_{1},\cdot)}T_{g_{1+}(z_{1},\cdot)}-T_{g_{1+}(z_{1}, \cdot)}T_{f_{1+}(z_{1},\cdot)}]k_{{z_{\vec{2}}}},k_{{z_{\vec{2}}}}\rangle \\ &&+ \langle [T_{f_{1-}(z_{1},\cdot)}T_{g_{1-}(z_{1},\cdot)}-T_{g_{1-}(z_{1}, \cdot)}T_{f_{1-}(z_{1},\cdot)}]k_{{z_{\vec{2}}}},k_{{z_{\vec{2}}}}\rangle \\ &&+ \langle [T_{f_{1+}(z_{1},\cdot)}T_{g_{1-}(z_{1},\cdot)} -T_{g_{1+}(z_{1},\cdot)}T_{f_{1-}(z_{1},\cdot)}]k_{{z_{\vec{2}}}}, k_{{z_{\vec{2}}}}\rangle \\ &=& \langle \langle [T_{f(\xi_{1},\cdot)}T_{g(\xi_{1},\cdot)} -T_{g(\xi_{1},\cdot)}T_{f(\xi_{1},\cdot)}]k_{{z_{\vec{2}}}}, k_{{z_{\vec{2}}}}\rangle k_{z_{1}},k_{z_{1}}\rangle. \end{eqnarray*} 对 $1\leq k\leq n-1$,令 $u_{k}(z)$ 记 $$ \langle \langle [T_{f_{k+}(\xi_{\scriptsize\overleftarrow{k-1}}, z_{k},\cdot)}T_{g_{k-}(\xi_{\scriptsize\overleftarrow{k-1}},z_{k},\cdot)} -T_{g_{k+}(\xi_{\scriptsize\overleftarrow{k-1}},z_{k},\cdot)}T_{f_{k-} (\xi_{\scriptsize\overleftarrow{k-1}},z_{k},\cdot)}]k_{{z_{\scriptsize\overrightarrow{k+1}}}}, k_{{z_{\scriptsize\overrightarrow{k+1}}}}\rangle k_{{z_{\scriptsize\overleftarrow{k-1}}}}, k_{{z_{\scriptsize\overleftarrow{k-1}}}}\rangle. $$ 我们将使用归纳法证明如下.

{断言}: $u_{k}$ 关于 $z_k$ 调和,且

\begin{equation} \langle (T_{f}T_{g}-T_{g}T_{f})k_{z},k_{z}\rangle = \langle \langle [T_{f(\xi_{\scriptsize\overleftarrow{k}},\cdot)} T_{g(\xi_{\scriptsize\overleftarrow{k}},\cdot)} -T_{g(\xi_{\scriptsize\overleftarrow{k}},\cdot)}T_{f(\xi_{\scriptsize\overleftarrow{k}}, \cdot)}]k_{z_{\scriptsize\overrightarrow{k+1}}},k_{z_{\scriptsize\overrightarrow{k+1}}}\rangle k_{z_{\scriptsize\overleftarrow{k}}},k_{z_{\scriptsize\overleftarrow{k}}}\rangle. \end{equation}

(3.3)

假设上面的断言对 $k$ ($1\leq k\leq n-2$) 成立. 由 引理 2.1,有 \begin{eqnarray*} &&\langle (T_{f}T_{g}-T_{g}T_{f})k_{z},k_{z}\rangle \\ &=& \langle \langle [T_{f(\xi_{\scriptsize\overleftarrow{k}},\cdot)} T_{g(\xi_{\scriptsize\overleftarrow{k}},\cdot)} -T_{g(\xi_{\scriptsize\overleftarrow{k}},\cdot)}T_{f(\xi_{\scriptsize\overleftarrow{k}}, \cdot)}]k_{z_{\scriptsize\overrightarrow{k+1}}},k_{z_{\scriptsize\overrightarrow{k+1}}}\rangle k_{z_{\scriptsize\overleftarrow{k}}},k_{z_{\scriptsize\overleftarrow{k}}}\rangle \\ &=& \langle \langle \langle [T_{f_{(k+1)-}(\xi_{\scriptsize\overleftarrow{k}}, \xi_{k+1},\cdot)}T_{g_{(k+1)+}(\xi_{\scriptsize\overleftarrow{k}},\xi_{k+1},\cdot)}\\ && -T_{g_{(k+1)-}(\xi_{\scriptsize\overleftarrow{k}},\xi_{k+1},\cdot)} T_{f_{(k+1)+}(\xi_{\scriptsize\overleftarrow{k}},\xi_{k+1},\cdot)}] k_{z_{\scriptsize\overrightarrow{k+2}}},k_{z_{\scriptsize\overrightarrow{k+2}}}\rangle k_{z_{k+1}},k_{z_{k+1}}\rangle k_{z_{\scriptsize\overleftarrow{k}}}, k_{z_{\scriptsize\overleftarrow{k}}}\rangle \\ &&+ \langle \langle [T_{f_{(k+1)+}(\xi_{\scriptsize\overleftarrow{k}},z_{k+1}, \cdot)}T_{g_{(k+1)+}(\xi_{\scriptsize\overleftarrow{k}},z_{k+1},\cdot)}\\ && -T_{g_{(k+1)+}(\xi_{\scriptsize\overleftarrow{k}},z_{k+1},\cdot)} T_{f_{(k+1)+}(\xi_{\scriptsize\overleftarrow{k}},z_{k+1},\cdot)}] k_{z_{\scriptsize\overrightarrow{k+2}}},k_{z_{\scriptsize\overrightarrow{k+2}}}\rangle k_{z_{\scriptsize\overleftarrow{k}}},k_{z_{\scriptsize\overleftarrow{k}}}\rangle \\ &&+ \langle \langle [T_{f_{(k+1)-}(\xi_{\scriptsize\overleftarrow{k}},z_{k+1}, \cdot)}T_{g_{(k+1)-}(\xi_{\scriptsize\overleftarrow{k}},z_{k+1},\cdot)}\\ &&-T_{g_{(k+1)-}(\xi_{\scriptsize\overleftarrow{k}},z_{k+1},\cdot)} T_{f_{(k+1)-}(\xi_{\scriptsize\overleftarrow{k}},z_{k+1},\cdot)}] k_{z_{\scriptsize\overrightarrow{k+2}}},k_{z_{\scriptsize\overrightarrow{k+2}}}\rangle k_{z_{\scriptsize\overleftarrow{k}}},k_{z_{\scriptsize\overleftarrow{k}}}\rangle \\ &&+ \langle \langle [T_{f_{(k+1)+}(\xi_{\scriptsize\overleftarrow{k}},z_{k+1},\cdot)} T_{g_{(k+1)-}(\xi_{\scriptsize\overleftarrow{k}},z_{k+1},\cdot)}\\ &&-T_{g_{(k+1)+}(\xi_{\scriptsize\overleftarrow{k}},z_{k+1},\cdot)}T_{f_{(k+1)-} (\xi_{\scriptsize\overleftarrow{k}},z_{k+1},\cdot)}]k_{z_{\scriptsize\overrightarrow{k+2}}}, k_{z_{\scriptsize\overrightarrow{k+2}}}\rangle k_{z_{\scriptsize\overleftarrow{k}}}, k_{z_{\scriptsize\overleftarrow{k}}}\rangle. \end{eqnarray*} 由于 $T_{f}T_{g}-T_{g}T_{f}=0$,得 \begin{eqnarray*} -u_{k+1}(z) &=&\langle \langle \langle [T_{f_{(k+1)-}(\xi_{\scriptsize\overleftarrow{k}},\xi_{k+1}, \cdot)}T_{g_{(k+1)+}(\xi_{\scriptsize\overleftarrow{k}},\xi_{k+1},\cdot)}\\ && -T_{g_{(k+1)-}(\xi_{\scriptsize\overleftarrow{k}},\xi_{k+1},\cdot)}T_{f_{(k+1)+} (\xi_{\scriptsize\overleftarrow{k}},\xi_{k+1},\cdot)}] k_{z_{\scriptsize\overrightarrow{k+2}}}, k_{z_{\scriptsize\overrightarrow{k+2}}}\rangle k_{z_{k+1}},k_{z_{k+1}}\rangle k_{z_{\scriptsize\overleftarrow{k}}},k_{z_{\scriptsize\overleftarrow{k}}}\rangle \\ &&+ \langle \langle [T_{f_{(k+1)+}(\xi_{\scriptsize\overleftarrow{k}},z_{k+1},\cdot)} T_{g_{(k+1)+}(\xi_{\scriptsize\overleftarrow{k}},z_{k+1},\cdot)}\\ && -T_{g_{(k+1)+}(\xi_{\scriptsize\overleftarrow{k}},z_{k+1},\cdot)}T_{f_{(k+1)+} (\xi_{\scriptsize\overleftarrow{k}},z_{k+1},\cdot)}]k_{z_{\scriptsize\overrightarrow{k+2}}}, k_{z_{\scriptsize\overrightarrow{k+2}}}\rangle k_{z_{\scriptsize\overleftarrow{k}}}, k_{z_{\scriptsize\overleftarrow{k}}}\rangle \\ &&+ \langle \langle [T_{f_{(k+1)-}(\xi_{\scriptsize\overleftarrow{k}},z_{k+1},\cdot)} T_{g_{(k+1)-}(\xi_{\scriptsize\overleftarrow{k}},z_{k+1},\cdot)}\\ &&-T_{g_{(k+1)-}(\xi_{\scriptsize\overleftarrow{k}},z_{k+1},\cdot)}T_{f_{(k+1)-} (\xi_{\scriptsize\overleftarrow{k}},z_{k+1},\cdot)}]k_{z_{\scriptsize\overrightarrow{k+2}}}, k_{z_{\scriptsize\overrightarrow{k+2}}}\rangle k_{z_{\scriptsize\overleftarrow{k}}}, k_{z_{\scriptsize\overleftarrow{k}}}\rangle. \end{eqnarray*} 由上式右端的每项都关于 $z_{k+1}$ 调和,可见 $u_{k+1}(z)$ 关于 $z_{k+1}$ 调和. 因此由引理 2.3 和Fubini定理,可得 \begin{eqnarray*} &&\langle (T_{f}T_{g}-T_{g}T_{f})k_{z},k_{z}\rangle \\ &=&\langle \langle [T_{f_{(k+1)-}(\xi_{\scriptsize\overleftarrow{k+1}},\cdot)} T_{g_{(k+1)+}(\xi_{\scriptsize\overleftarrow{k+1}},\cdot)} -T_{g_{(k+1)-}(\xi_{\scriptsize\overleftarrow{k+1}},\cdot)}T_{f_{(k+1)+} (\xi_{\scriptsize\overleftarrow{k+1}},\cdot)}]\\ &&\times k_{z_{\scriptsize\overrightarrow{k+2}}}, k_{z_{\scriptsize\overrightarrow{k+2}}}\rangle k_{z_{\scriptsize\overleftarrow{k+1}}}, k_{z_{\scriptsize\overleftarrow{k+1}}}\rangle \\ &&+ \langle \langle [T_{f_{(k+1)+}(\xi_{\scriptsize\overleftarrow{k+1}},\cdot)} T_{g_{(k+1)+}(\xi_{\scriptsize\overleftarrow{k+1}},\cdot)} -T_{g_{(k+1)+}(\xi_{\scriptsize\overleftarrow{k+1}},\cdot)}T_{f_{(k+1)+} (\xi_{\scriptsize\overleftarrow{k+1}},\cdot)}] \\ &&\times k_{z_{\scriptsize\overrightarrow{k+2}}}, k_{z_{\scriptsize\overrightarrow{k+2}}}\rangle k_{z_{\scriptsize\overleftarrow{k+1}}}, k_{z_{\scriptsize\overleftarrow{k+1}}}\rangle \\ &&+ \langle \langle [T_{f_{(k+1)-}(\xi_{\scriptsize\overleftarrow{k+1}},\cdot)} T_{g_{(k+1)-}(\xi_{\scriptsize\overleftarrow{k+1}},\cdot)} -T_{g_{(k+1)-}(\xi_{\scriptsize\overleftarrow{k+1}},\cdot)}T_{f_{(k+1)-} (\xi_{\scriptsize\overleftarrow{k+1}},\cdot)}] \\ &&\times k_{z_{\scriptsize\overrightarrow{k+2}}}, k_{z_{\scriptsize\overrightarrow{k+2}}}\rangle k_{z_{\scriptsize\overleftarrow{k+1}}}, k_{z_{\scriptsize\overleftarrow{k+1}}}\rangle\\ &&+\langle \langle [T_{f_{(k+1)+}(\xi_{\scriptsize\overleftarrow{k+1}}, \cdot)}T_{g_{(k+1)-}(\xi_{\scriptsize\overleftarrow{k+1}},\cdot)} -T_{g_{(k+1)+}(\xi_{\scriptsize\overleftarrow{k+1}},\cdot)}T_{f_{(k+1)-} (\xi_{\scriptsize\overleftarrow{k+1}},\cdot)}]\\ &&\times k_{z_{\scriptsize\overrightarrow{k+2}}}, k_{z_{\scriptsize\overrightarrow{k+2}}}\rangle k_{z_{\scriptsize\overleftarrow{k}}}, k_{z_{\scriptsize\overleftarrow{k}}}\rangle\\ &=&\langle \langle [T_{f(\xi_{\scriptsize\overleftarrow{k+1}},\cdot)} T_{g(\xi_{\scriptsize\overleftarrow{k+1}},\cdot)} -T_{g(\xi_{\scriptsize\overleftarrow{k+1}},\cdot)}T_{f(\xi_{\scriptsize\overleftarrow{k+1}}, \cdot)}] k_{z_{\scriptsize\overrightarrow{k+2}}},k_{z_{\scriptsize\overrightarrow{k+2}}}\rangle k_{z_{\scriptsize\overleftarrow{k+1}}},k_{z_{\scriptsize\overleftarrow{k+1}}}\rangle. \end{eqnarray*} 断言得证.

由于 \begin{eqnarray*} u_{k}(z) &=&\langle \langle [T_{f_{k+}(\xi_{\scriptsize\overleftarrow{k-1}},z_{k}, \cdot)}T_{g_{k-}(\xi_{\scriptsize\overleftarrow{k-1}},z_{k},\cdot)}\\ &&-T_{g_{k+}(\xi_{\scriptsize\overleftarrow{k-1}},z_{k},\cdot)}T_{f_{k-} (\xi_{\scriptsize\overleftarrow{k-1}},z_{k},\cdot)}]k_{z_{\scriptsize\overrightarrow{k+1}}}, k_{z_{\scriptsize\overrightarrow{k+1}}}\rangle k_{z_{\scriptsize\overleftarrow{k-1}}}, k_{z_{\scriptsize\overleftarrow{k-1}}}\rangle \end{eqnarray*} 关于 $z_{k}$ 调和, 对方程两边作用Laplace算子 $\triangle_{k}=\frac{\partial^{2}} {\partial z_{k}\partial{\bar z_{k}}}$,可见 \begin{eqnarray*} 0 &=&\langle \triangle_{k}\{\langle [T_{f_{k+}(\xi_{\scriptsize\overleftarrow{k-1}}, z_{k},\cdot)}T_{g_{k-}(\xi_{\scriptsize\overleftarrow{k-1}},z_{k},\cdot)}\\ &&-T_{g_{k+}(\xi_{\scriptsize\overleftarrow{k-1}},z_{k},\cdot)}T_{f_{k-} (\xi_{\scriptsize\overleftarrow{k-1}},z_{k},\cdot)}]k_{z_{\scriptsize\overrightarrow{k+1}}}, k_{z_{\scriptsize\overrightarrow{k+1}}}\rangle\} k_{z_{\scriptsize\overleftarrow{k-1}}}, k_{z_{\scriptsize\overleftarrow{k-1}}}\rangle. \end{eqnarray*} 由于Berezin变换是一对一的(参见文献[13,Proposition 6.2]), 对于几乎所有的 $\xi_{\scriptsize\overleftarrow{k-1}}\in {\Bbb T}^{k-1} $,有 $$ \triangle_{k}\{\langle [T_{f_{(k)+}(\xi_{\scriptsize\overleftarrow{k-1}},z_{k}, \cdot)}T_{g_{(k)-}(\xi_{\scriptsize\overleftarrow{k-1}},z_{k},\cdot)} -T_{g_{(k)+}(\xi_{\scriptsize\overleftarrow{k-1}},z_{k},\cdot)}T_{f_{(k)-} (\xi_{\scriptsize\overleftarrow{k-1}},z_{k},\cdot)}]k_{z_{\scriptsize\overrightarrow{k+1}}}, k_{z_{\scriptsize\overrightarrow{k+1}}}\rangle\}=0. $$ 因此,对于几乎所有的 $\xi_{\scriptsize\overleftarrow{k-1}}\in {\Bbb T}^{k-1} $, \begin{eqnarray*} &&\widetilde{[T_{f_{(k)+}(\xi_{\scriptsize\overleftarrow{k-1}},z_{k},\cdot)} T_{g_{(k)-}(\xi_{\scriptsize\overleftarrow{k-1}},z_{k},\cdot)}-T_{g_{(k)+} (\xi_{\scriptsize\overleftarrow{k-1}},z_{k},\cdot)}T_{f_{(k)-}(\xi_{\scriptsize\overleftarrow{k-1}}, z_{k},\cdot)}]}(z_{\scriptsize\overrightarrow{k+1}})\\ &=& \langle [T_{f_{(k)+}(\xi_{\scriptsize\overleftarrow{k-1}},z_{k},\cdot)} T_{g_{(k)-}(\xi_{\scriptsize\overleftarrow{k-1}},z_{k},\cdot)}-T_{g_{(k)+} (\xi_{\scriptsize\overleftarrow{k-1}},z_{k},\cdot)}T_{f_{(k)-} (\xi_{\scriptsize\overleftarrow{k-1}}, z_{k},\cdot)}]k_{z_{\scriptsize\overrightarrow{k+1}}}, k_{z_{\scriptsize\overrightarrow{k+1}}}\rangle \end{eqnarray*} 关于 $z_{k}$ 调和,条件 (a) 得证.

类似地可证条件 (b) 成立.

另一方面,设条件 (a) 和 (b) 成立. 由引理 2.1,有 \begin{eqnarray*} \langle [T_{f}T_{g}-T_{g}T_{f}]k_{z},k_{z}\rangle &=& \langle \langle [T_{f_{1-}(\xi_{1},\cdot)}T_{g_{1+}(\xi_{1}, \cdot)}-T_{g_{1-}(\xi_{1},\cdot)}T_{f_{1+}(\xi_{1}, \cdot)}]k_{{z_{\vec{2}}}},k_{{z_{\vec{2}}}}\rangle k_{z_{1}},k_{z_{1}}\rangle \\ &&+ \langle [T_{f_{1+}(z_{1},\cdot)}T_{g_{1+}(z_{1},\cdot)}-T_{g_{1+} (z_{1},\cdot)}T_{f_{1+}(z_{1},\cdot)}]k_{{z_{\vec{2}}}},k_{{z_{\vec{2}}}}\rangle \\ &&+ \langle [T_{f_{1-}(z_{1},\cdot)}T_{g_{1-}(z_{1},\cdot)} -T_{g_{1-}(z_{1},\cdot)}T_{f_{1-}(z_{1},\cdot)}]k_{{z_{\vec{2}}}}, k_{{z_{\vec{2}}}}\rangle \\ &&+ \langle [T_{f_{1+}(z_{1},\cdot)}T_{g_{1-}(z_{1},\cdot)} -T_{g_{1+}(z_{1},\cdot)}T_{f_{1-}(z_{1},\cdot)}]k_{{z_{\vec{2}}}}, k_{{z_{\vec{2}}}}\rangle . \end{eqnarray*} 由条件 (a),上式右端的后三项关于 $z_1$ 调和,应用引理 2.3,可得

\begin{eqnarray} \langle [T_{f}T_{g}-T_{g}T_{f}]k_{z},k_{z}\rangle = \langle \langle [T_{f(\xi_{1},\cdot)}T_{g(\xi_{1},\cdot)} -T_{g(\xi_{1},\cdot)}T_{f(\xi_{1},\cdot)}]k_{{z_{\vec{2}}}}, k_{{z_{\vec{2}}}}\rangle k_{z_{1}},k_{z_{1}}\rangle. \end{eqnarray}

(3.4)

对 $\xi_{1}\in {\Bbb T}$ 和 $z_{\vec{2}}\in {\Bbb D}^{n-1}$,令 $$ \Phi_{z_{\vec{2}}}(\xi_{1})=\langle [T_{f(\xi_{1},\cdot)}T_{g(\xi_{1}, \cdot)}-T_{g(\xi_{1},\cdot)}T_{f(\xi_{1},\cdot)}]k_{{z_{\vec{2}}}}, k_{{z_{\vec{2}}}}\rangle. $$ 则

\begin{equation} \langle [T_{f}T_{g}-T_{g}T_{f}]k_{z},k_{z}\rangle=\langle \Phi_{z_{\vec{2}}}k_{z_{1}},k_{z_{1}}\rangle. \end{equation}

(3.5)

应用条件 (a),对几乎所有的 $\xi_{1}\in {\Bbb T}$,有 $$\langle [T_{f_{2+}(\xi_{1},z_{2},\cdot)}T_{g_{2-}(\xi_{1},z_{2}, \cdot)}-T_{g_{2+}(\xi_{1},z_{2},\cdot)}T_{f_{2-}(\xi_{1},z_{2}, \cdot)}]k_{z_{\scriptsize\overrightarrow{3}}},k_{z_{\scriptsize\overrightarrow{3}}}\rangle $$ 关于 $z_{2}$ 调和. 因此由引理 2.3,有 \begin{eqnarray*} &&\langle [T_{f_{2+}(\xi_{1},z_{2},\cdot)}T_{g_{2-}(\xi_{1},z_{2}, \cdot)}-T_{g_{2+}(\xi_{1},z_{2},\cdot)}T_{f_{2-}(\xi_{1},z_{2}, \cdot)}]k_{z_{\scriptsize\overrightarrow{3}}},k_{z_{\scriptsize\overrightarrow{3}}}\rangle \\ &=& \langle \langle [T_{f_{2+}(\xi_{1},\xi_{2},\cdot)}T_{g_{2-}(\xi_{1},\xi_{2}, \cdot)}-T_{g_{2+}(\xi_{1},\xi_{2},\cdot)}T_{f_{2-}(\xi_{1},\xi_{2}, \cdot)}]k_{z_{\scriptsize\overrightarrow{3}}},k_{z_{\scriptsize\overrightarrow{3}}}\rangle k_{z_{2}},k_{z_{2}}\rangle . \end{eqnarray*} 使用引理 2.1 和 2.3,对几乎所有的 $\xi_{1}\in {\Bbb T}$,有 \begin{eqnarray*} \Phi_{z_{\vec{2}}}(\xi_{1}) &=& \langle \langle [T_{f_{2-}(\xi_{1},\xi_{2},\cdot)}T_{g_{2+}(\xi_{1},\xi_{2}, \cdot)}-T_{g_{2-}(\xi_{1},\xi_{2},\cdot)}T_{f_{2+}(\xi_{1},\xi_{2}, \cdot)}]k_{z_{\scriptsize\overrightarrow{3}}}, k_{z_{\scriptsize\overrightarrow{3}}}\rangle k_{z_{2}},k_{z_{2}}\rangle \\ &&+ \langle [T_{f_{2+}(\xi_{1},z_{2},\cdot)}T_{g_{2+}(\xi_{1},z_{2}, \cdot)}-T_{g_{2+}(\xi_{1},z_{2},\cdot)}T_{f_{2+}(\xi_{1},z_{2}, \cdot)}]k_{z_{\scriptsize\overrightarrow{3}}},k_{z_{\scriptsize\overrightarrow{3}}}\rangle \\ &&+ \langle [T_{f_{2-}(\xi_{1},z_{2},\cdot)}T_{g_{2-}(\xi_{1},z_{2}, \cdot)}-T_{g_{2-}(\xi_{1},z_{2},\cdot)}T_{f_{2-}(\xi_{1},z_{2}, \cdot)}]k_{z_{\scriptsize\overrightarrow{3}}},k_{z_{\scriptsize\overrightarrow{3}}}\rangle \\ &&+ \langle [T_{f_{2+}(\xi_{1},z_{2},\cdot)}T_{g_{2-}(\xi_{1},z_{2}, \cdot)}-T_{g_{2+}(\xi_{1},z_{2},\cdot)}T_{f_{2-}(\xi_{1},z_{2}, \cdot)}]k_{z_{\scriptsize\overrightarrow{3}}},k_{z_{\scriptsize\overrightarrow{3}}}\rangle \\ &=& \langle \langle [T_{f(\xi_{1},\xi_{2},\cdot)}T_{g(\xi_{1},\xi_{2},\cdot)} -T_{g(\xi_{1},\xi_{2},\cdot)}T_{f(\xi_{1},\xi_{2},\cdot)}] k_{z_{\scriptsize\overrightarrow{3}}},k_{z_{\scriptsize\overrightarrow{3}}}\rangle k_{z_{2}},k_{z_{2}}\rangle. \end{eqnarray*} 对 $\xi_{\scriptsize\overleftarrow{2}}\in {\Bbb T}^2$ 和 $z_{\scriptsize\overrightarrow{3}}\in {\Bbb D}^{n-2}$,令 $$ \Phi_{z_{\scriptsize\overrightarrow{3}}}(\xi_{\scriptsize\overleftarrow{2}})= \langle [T_{f(\xi_{\scriptsize\overleftarrow{2}}, \cdot)}T_{g(\xi_{\scriptsize\overleftarrow{2}}, \cdot)}-T_{g(\xi_{\scriptsize\overleftarrow{2}}, \cdot)}T_{f(\xi_{\scriptsize\overleftarrow{2}}, \cdot)}]k_{z_{\scriptsize\overrightarrow{3}}}, k_{z_{\scriptsize\overrightarrow{3}}}\rangle , $$ 则 $$ \langle [T_{f}T_{g}-T_{g}T_{f}]k_{z},k_{z}\rangle=\langle\langle \Phi_{z_{\scriptsize\overrightarrow{3}}} k_{z_{2}},k_{z_{2}}\rangle k_{z_{1}}, k_{z_{1}}\rangle=\langle \Phi_{z_{\scriptsize\overrightarrow{3}}} k_{z_{\scriptsize\overleftarrow{2}}}, k_{z_{\scriptsize\overleftarrow{2}}}\rangle. $$ 对 $\xi_{\scriptsize\overleftarrow{n-2}}\in {\Bbb T}^{n-2}$ 和 $z_{\scriptsize\overrightarrow{n-1}} \in {\Bbb D}^{2}$,如果令 $$ \Phi_{z_{\scriptsize\overrightarrow{n-1}}}(\xi_{\scriptsize\overleftarrow{n-2}})=\langle [T_{f(\xi_{\scriptsize\overleftarrow{n-2}},\cdot)} T_{g(\xi_{\scriptsize\overleftarrow{n-2}},\cdot)} -T_{g(\xi_{\scriptsize\overleftarrow{n-2}},\cdot)} T_{f(\xi_{\scriptsize\overleftarrow{n-2}},\cdot)}] k_{z_{\scriptsize\overrightarrow{n-1}}},k_{z_{\scriptsize\overrightarrow{n-1}}}\rangle , $$ 重复上面的程序,可得

\begin{equation} \langle [T_{f}T_{g}-T_{g}T_{f}]k_{z},k_{z}\rangle=\langle \Phi_{z_{\scriptsize\overrightarrow{n-1}}} k_{z_{\scriptsize\overleftarrow{n-2}}}, k_{z_{\scriptsize\overleftarrow{n-2}}}\rangle. \end{equation}

(3.6)

再次使用引理 2.1 和 2.3,可得对几乎所有的 $\xi_{\scriptsize\overleftarrow{n-1}} \in {\Bbb T}^{n-1}$,

\begin{eqnarray} \Phi_{z_{\scriptsize\overrightarrow{n-1}}}(\xi_{\scriptsize\overleftarrow{n-2}}) &=& \int_{{\Bbb T}}\langle [T_{f(\xi_{\scriptsize\overleftarrow{n-1}},\cdot)} T_{g(\xi_{\scriptsize\overleftarrow{n-1}},\cdot)}-T_{g(\xi_{\scriptsize\overleftarrow{n-1}}, \cdot)}T_{f(\xi_{\scriptsize\overleftarrow{n-1}},\cdot)}] \nonumber\\ &&\times k_{z_{n}},k_{z_{n}}\rangle |k_{z_{n-1}}(\xi_{n-1})|^{2}{\rm d}\sigma (\xi_{n-1}). \end{eqnarray}

(3.7)

易见 $$ \langle T_{f_{n+}(\xi_{\scriptsize\overleftarrow{n-1}},\cdot)} T_{g_{n+}(\xi_{\scriptsize\overleftarrow{n-1}},\cdot)}k_{z_{n}},k_{z_{n}}\rangle = f_{n+}(\xi_{\scriptsize\overleftarrow{n-1}},z_{n})g_{n+} (\xi_{\scriptsize\overleftarrow{n-1}},z_{n}), $$ $$ \langle T_{f_{n-}(\xi_{\scriptsize\overleftarrow{n-1}},\cdot)} T_{g_{n-}(\xi_{\scriptsize\overleftarrow{n-1}},\cdot)}k_{z_{n}},k_{z_{n}}\rangle = f_{n-}(\xi_{\scriptsize\overleftarrow{n-1}},z_{n})g_{n-}(\xi_{\scriptsize\overleftarrow{n-1}},z_{n}), $$ $$ \langle T_{f_{n+}(\xi_{\scriptsize\overleftarrow{n-1}},\cdot)}T_{g_{n-} (\xi_{\scriptsize\overleftarrow{n-1}},\cdot)}k_{z_{n}},k_{z_{n}}\rangle = f_{n+}(\xi_{\scriptsize\overleftarrow{n-1}},z_{n})g_{n-}(\xi_{\scriptsize\overleftarrow{n-1}},z_{n}), $$ $$ \langle T_{f_{n-}(\xi_{\scriptsize\overleftarrow{n-1}},\cdot)}T_{g_{n+} (\xi_{\scriptsize\overleftarrow{n-1}},\cdot)}k_{z_{n}},k_{z_{n}}\rangle = \int_{{\Bbb T}}f_{n-}(\xi_{\scriptsize\overleftarrow{n-1}},\xi_{n})g_{n+} (\xi_{\scriptsize\overleftarrow{n-1}},\xi_{n})|k_{z_{n}}( \xi_{n})|^{2}{\rm d}\sigma (\xi_{n}). $$ 注意到 $$ f(\xi_{\scriptsize\overleftarrow{n-1}},\cdot)=f_{n+}(\xi_{\scriptsize\overleftarrow{n-1}}, \cdot)+f_{n-}(\xi_{\scriptsize\overleftarrow{n-1}},\cdot), $$ $$ g(\xi_{\scriptsize\overleftarrow{n-1}},\cdot)=g_{n+}(\xi_{\scriptsize\overleftarrow{n-1}}, \cdot)+g_{n-}(\xi_{\scriptsize\overleftarrow{n-1}},\cdot). $$ 因此有

\begin{eqnarray} &&\Phi_{z_{\scriptsize\overrightarrow{n-1}}}(\xi_{\scriptsize\overleftarrow{n-2}}) \nonumber\\ &=&\int_{{\Bbb T}}[f_{n+}(\xi_{\scriptsize\overleftarrow{n-1}},z_{n})g_{n-} (\xi_{\scriptsize\overleftarrow{n-1}},z_{n})- g_{n+}(\xi_{\scriptsize\overleftarrow{n-1}}, z_{n})f_{n-}(\xi_{\scriptsize\overleftarrow{n-1}},z_{n})]|k_{z_{n-1}} (\xi_{n-1})|^{2}{\rm d}\sigma (\xi_{n-1}) \nonumber\\ &&+ \int_{{\Bbb T}}\int_{{\Bbb T}}[f_{n-}(\xi_{\scriptsize\overleftarrow{n-1}}, \xi_{n})g_{n+}(\xi_{\scriptsize\overleftarrow{n-1}},\xi_{n})-g_{n-} (\xi_{\scriptsize\overleftarrow{n-1}},\xi_{n})f_{n+}(\xi_{\scriptsize\overleftarrow{n-1}}, \xi_{n})]\nonumber\\ && \times | k_{z_{n}}(\xi_{n})|^{2} {\rm d}\sigma (\xi_{n})|k_{z_{n-1}}(\xi_{n-1})|^{2}{\rm d}\sigma (\xi_{n-1}). \end{eqnarray}

(3.8)

由条件 (b), $$ \langle [T_{f_{n+}( \cdot,z_{n})}T_{g_{n-}( \cdot,z_{n})}-T_{g_{n+} ( \cdot,z_{n})}T_{f_{n-}( \cdot,z_{n})}]k_{z_{\scriptsize\overleftarrow{n-1}}}, k_{z_{\scriptsize\overleftarrow{n-1}}}\rangle $$ 关于 $z_{n}$ 调和. 因此使用 (3.5) 式中同样的讨论,可得 \begin{eqnarray*} &&\langle [T_{f}T_{g}-T_{g}T_{f}]k_{z},k_{z}\rangle \\ &=&\langle \langle [T_{f(\cdot,\xi_{n})}T_{g(\cdot,\xi_{n})} -T_{g(\cdot,\xi_{n})}T_{f(\cdot,\xi_{n})}]k_{z_{\scriptsize\overleftarrow{n-1}}}, k_{z_{\scriptsize\overleftarrow{n-1}}}\rangle k_{z_{n}},k_{z_{n}}\rangle \\ &=&\int_{{\Bbb T}}\langle [T_{f(\cdot,\xi_{n})}T_{g(\cdot,\xi_{n})} -T_{g(\cdot,\xi_{n})}T_{f(\cdot,\xi_{n})}]k_{z_{\scriptsize\overleftarrow{n-1}}}, k_{z_{\scriptsize\overleftarrow{n-1}}}\rangle |k_{z_{n}}(\xi_{n})|^{2}{\rm d}\sigma (\xi_{n}). \end{eqnarray*} 因此有 $$ \langle [T_{f}T_{g}-T_{g}T_{f}]k_{z},k_{z}\rangle $$ 关于 $z_{n}$ 调和. 在 (3.6)式 两端作用Laplace算子 $\triangle_{n}=\frac{\partial^{2}} {\partial z_{n}\partial{\bar z_{n}}}$,有 $$ 0=\triangle_{n}\langle \Phi_{z_{\scriptsize\overrightarrow{n-1}}} k_{z_{\scriptsize\overleftarrow{n-2}}}, k_{z_{\scriptsize\overleftarrow{n-2}}}\rangle =\langle \triangle_{n}(\Phi_{z_{\scriptsize\overrightarrow{n-1}}}) k_{z_{\scriptsize\overleftarrow{n-2}}}, k_{z_{\scriptsize\overleftarrow{n-2}}}\rangle. $$ 由于Berezin变换是一对一的,$\Phi_{z_{\scriptsize\overrightarrow{n-1}}}$ 关于 $z_{n}$ 调和. 对几乎所有的 $\xi_{\scriptsize\overleftarrow{n-2}}\in {\Bbb T}^{n-2}$,由于 (3.8) 式 右端的第二项是一个可积函数的Poisson积分,故关于 $z_{n}$ 调和. 因此对几乎所有的 $\xi_{\scriptsize\overleftarrow{n-2}}\in {\Bbb T}^{n-2}$, $$ \int_{{\Bbb T}}[f_{n+}(\xi_{\scriptsize\overleftarrow{n-1}}, z_{n})g_{n-}(\xi_{\scriptsize\overleftarrow{n-1}},z_{n})- g_{n+}(\xi_{\scriptsize\overleftarrow{n-1}}, z_{n})f_{n-}(\xi_{\scriptsize\overleftarrow{n-1}},z_{n})]|k_{z_{n-1}} (\xi_{n-1})|^{2}{\rm d}\sigma (\xi_{n-1}) $$ 也关于 $z_{n}$ 调和.

对上面函数使用Laplace算子 $\triangle_{n}=\frac{\partial^{2}} {\partial z_{n}\partial {\bar z_{n}}}$,由积分号下求导公式,可得对几乎所有的 $\xi_{\scriptsize\overleftarrow{n-2}}\in {\Bbb T}^{n-2}$, $$ \int_{{\Bbb T}}\triangle_{n}[f_{n+}(\xi_{\scriptsize\overleftarrow{n-1}},z_{n})g_{n-} (\xi_{\scriptsize\overleftarrow{n-1}},z_{n})- g_{n+}(\xi_{\scriptsize\overleftarrow{n-1}}, z_{n})f_{n-}(\xi_{\scriptsize\overleftarrow{n-1}},z_{n})]|k_{z_{n-1}} (\xi_{n-1})|^{2}{\rm d}\sigma (\xi_{n-1}) =0. $$ 由于Berezin变换是一对一的,故对几乎所有的 $\xi_{\scriptsize\overleftarrow{n-1}}\in {\Bbb T}^{n-1}$, $$ f_{n+}(\xi_{\scriptsize\overleftarrow{n-1}},z_{n})g_{n-}(\xi_{\scriptsize\overleftarrow{n-1}}, z_{n})- g_{n+}(\xi_{\scriptsize\overleftarrow{n-1}},z_{n})f_{n-}(\xi_{\scriptsize\overleftarrow{n-1}},z_{n}) $$ 关于 $z_{n}$ 调和.

应用文献 ^[6] 中的定理2.2,可得对几乎所有的 $\xi_{\scriptsize\overleftarrow{n-1}}\in {\Bbb T}^{n-1}$, $$ T_{f(\xi_{\scriptsize\overleftarrow{n-1}},\cdot)}T_{g(\xi_{\scriptsize\overleftarrow{n-1}}, \cdot)} =T_{g(\xi_{\scriptsize\overleftarrow{n-1}},\cdot)}T_{f(\xi_{\scriptsize\overleftarrow{n-1}}, \cdot)} $$ 在Hardy空间 $H^2({\Bbb D})$ 上成立. 因此由 (3.7)式 可得 $$ \Phi_{z_{\scriptsize \overleftarrow{n-1}}}(\xi_{\scriptsize\overleftarrow{n-2}})=0. $$ 使用 (3.6)式,有 $$ \langle [T_{f}T_{g}-T_{g}T_{f}]k_{z},k_{z}\rangle=0.$$ 因此在 $H^2({\Bbb D}^{n})$ 上 $$ T_{f}T_{g}=T_{g}T_{f}. $$ 定理得证.