设$ {\cal A} $是$ * $-代数, $ \xi $是非零复数, $ A, B\in{\cal A} $的$ \xi $-Jordan $ * $-积定义为$ A\diamond_{\xi}B = AB+\xi BA^{*}, $$ 1 $-Jordan $ * $-积通常记为$ A\bullet B = AB+BA^{*}, $$ -1 $-Jordan $ * $- 积(斜Lie积)通常记为$ [A, B]_{*} = AB-BA^{*}. $近年来, 相关的研究吸引了许多作者的注意(参看文献[1-13]). 文献[1]中P. Šemrl在量子函数中首先引入并研究了$ -1 $-Jordan $ * $-积. 文献[2]研究了von Neumann代数(无中心交换投影)到$ {\cal B(H)} $上的非线性$ \xi $-Jordan $ * $-可导映射.

在以上文献研究的基础上, 下面将考虑更宽泛的结果: 设$ {\cal A} $是$ * $-代数, 对任意$ A, B, C\in{\cal A}, $定义$ A\diamond_{\xi}B\diamond_{\xi}C: = (A\diamond_{\xi}B)\diamond_{\xi}C $为$ \xi $-Jordan $ * $-三重积(其中$ \diamond_{\xi} $不具有结合性). $ \phi:{\cal A\rightarrow A} $是一个映射. 如果$ \phi(A\diamond_{\xi}B\diamond_{\xi}C) = \phi(A)\diamond_{\xi}B\diamond_{\xi}C+A\diamond_{\xi}\phi(B)\diamond_{\xi}C+A\diamond_{\xi}B\diamond_{\xi}\phi(C), $那么称$ \phi $是$ \xi $-Jordan $ * $-三重可导映射. 文献[3]得到了因子von Neumann代数上非线性$ -1 $-Jordan $ * $-三重可导映射是可加的$ * $-导子. 本文将要研究因子von Neumann代数上非线性$ \xi $-Jordan $ * $-三重可导映射.

设$ {\mathbb R} $和$ {\mathbb C} $分别是实数域和复数域, $ \text{i} $是虚数单位. von Neumann代数$ {\cal A} $是定义在Hilbert空间$ {\cal H} $上的包含单位算子$ I $的弱闭的, 自伴的算子代数. 记$ {\cal Z(A)} $为$ {\cal A} $的中心, 若$ {\cal Z(A)} = {\mathbb C}I, $则$ {\cal A} $是因子von Neumann代数. 因子von Neumann代数$ {\cal A} $是素的是指对$ A, B\in{\cal A}, $如果有$ A{\cal A}B = 0, $那么$ A = 0 $或$ B = 0. $

引理1.1^[4]  设$ {\cal A} $是因子von Neumann代数且$ A\in{\cal A}. $若对所有$ B\in{\cal A}, $有$ AB+BA^{*} = 0, $则$ A\in\text{i}{\mathbb R}I. $

定理2.1  设$ {\cal A} $是因子von Neumann代数, $ \xi $是非零复数. 非线性映射$ \phi:{\cal A\rightarrow A} $满足对所有$ A, B, C\in{\cal A}, $有$ \phi(A\diamond_{\xi}B\diamond_{\xi}C) = \phi(A)\diamond_{\xi}B\diamond_{\xi}C+A\diamond_{\xi}\phi(B)\diamond_{\xi}C+A\diamond_{\xi}B\diamond_{\xi}\phi(C) $当且仅当$ \phi $是可加的$ * $-导子且对所有$ A\in{\cal A}, $有$ \phi(\xi A) = \xi\phi(A). $

证  设$ P_{1}\in{\cal A} $为非平凡投影, 令$ P_{2} = I-P_{1}, {\cal A}_{ij} = P_{i}{\cal A}P_{j}, i, j = 1, 2, $则$ {\cal A} = \sum\limits_{i, j = 1}^{2}{\cal A}_{ij}. $所以对任意$ A\in{\cal A}, $记$ A = \sum\limits_{i, j = 1}^{2}A_{ij}, $其中$ A_{ij}\in{\cal A}_{ij}. $

当$ \xi = -1 $时, 文献[3]已证明并得到结论, 下面的证明中假设$ \xi\neq -1. $

断言1  $ \phi(0) = 0. $

取$ A = B = C = 0 $得

$ \phi(0) = \phi(0\diamond_{\xi}0\diamond_{\xi}0) = \phi(0)\diamond_{\xi}0\diamond_{\xi}0+0\diamond_{\xi}\phi(0)\diamond_{\xi}0+0\diamond_{\xi}0\diamond_{\xi}\phi(0) = 0. $

断言2  $ \phi(A_{11}+A_{22}) = \phi(A_{11})+\phi(A_{22}), \forall A_{11}\in{\cal A}_{11}, \forall A_{22}\in{\cal A}_{22}. $

设$ T = \phi(A_{11}+A_{22})-\phi(A_{11})-\phi(A_{22}). $由$ I\diamond_{\xi}\frac{P_{1}}{1+\xi}\diamond_{\xi}A_{22} = 0 $和断言1得

$ \begin{eqnarray*} &&\phi(I)\diamond_{\xi}\frac{P_{1}}{1+\xi}\diamond_{\xi}(A_{11}+A_{22})+I\diamond_{\xi}\phi(\frac{P_{1}}{1+\xi})\diamond_{\xi}(A_{11}+A_{22})\\ &&+I\diamond_{\xi}\frac{P_{1}}{1+\xi}\diamond_{\xi}\phi(A_{11}+A_{22})\\ & = &\phi(I\diamond_{\xi}\frac{P_{1}}{1+\xi}\diamond_{\xi}(A_{11}+A_{22})) = \phi(I\diamond_{\xi}\frac{P_{1}}{1+\xi}\diamond_{\xi}A_{11}) +\phi(I\diamond_{\xi}\frac{P_{1}}{1+\xi}\diamond_{\xi}A_{22})\\ & = &\phi(I)\diamond_{\xi}\frac{P_{1}}{1+\xi}\diamond_{\xi}(A_{11}+A_{22})+I\diamond_{\xi}\phi(\frac{P_{1}}{1+\xi})\diamond_{\xi}(A_{11}+A_{22})\\ &&+I\diamond_{\xi}\frac{P_{1}}{1+\xi}\diamond_{\xi}(\phi(A_{11})+\phi(A_{22})). \end{eqnarray*} $

所以$ I\diamond_{\xi}\frac{P_{1}}{1+\xi}\diamond_{\xi}T = 0, $由此可得$ (1+\xi)T_{11}+T_{12}+\xi T_{21} = 0. $由$ \xi\neq 0, -1 $得$ T_{11} = T_{12} = T_{21} = 0. $类似可得$ T_{22} = 0. $因此$ \phi(A_{11}+A_{22}) = \phi(A_{11})+\phi(A_{22}). $

断言3  $ \phi(A_{12}+A_{21}) = \phi(A_{12})+\phi(A_{21}), \forall A_{12}\in{\cal A}_{12}, \forall A_{21}\in{\cal A}_{21}. $

令$ T = \phi(A_{12}+A_{21})-\phi(A_{12})-\phi(A_{21}). $由$ I\diamond_{\xi}\frac{P_{1}-\frac{1}{\overline{\xi}}P_{2}}{1+\xi}\diamond_{\xi}A_{12} = 0 $和断言1得

$ \begin{eqnarray*} &&\phi(I)\diamond_{\xi}\frac{P_{1}-\frac{1}{\overline{\xi}}P_{2}}{1+\xi}\diamond_{\xi}(A_{12}+A_{21})+I\diamond_{\xi}\phi(\frac{P_{1}-\frac{1}{\overline{\xi}}P_{2}}{1+\xi})\diamond_{\xi}(A_{12}+A_{21})\\ &&+I\diamond_{\xi}\frac{P_{1}-\frac{1}{\overline{\xi}}P_{2}}{1+\xi}\diamond_{\xi}\phi(A_{12}+A_{21}) \\ & = &\phi(I\diamond_{\xi}\frac{P_{1}-\frac{1}{\overline{\xi}}P_{2}}{1+\xi}\diamond_{\xi}(A_{12}+A_{21}))\\ & = &\phi(I\diamond_{\xi}\frac{P_{1}-\frac{1}{\overline{\xi}}P_{2}}{1+\xi}\diamond_{\xi}A_{12}) +\phi(I\diamond_{\xi}\frac{P_{1}-\frac{1}{\overline{\xi}}P_{2}}{1+\xi}\diamond_{\xi}A_{21})\\ & = &\phi(I)\diamond_{\xi}\frac{P_{1}-\frac{1}{\overline{\xi}}P_{2}}{1+\xi}\diamond_{\xi}(A_{12}+A_{21})+I\diamond_{\xi}\phi(\frac{P_{1}-\frac{1}{\overline{\xi}}P_{2}}{1+\xi})\diamond_{\xi}(A_{12}+A_{21})\\ &&+I\diamond_{\xi}\frac{P_{1}-\frac{1}{\overline{\xi}}P_{2}}{1+\xi}\diamond_{\xi}(\phi(A_{12})+\phi(A_{21})). \end{eqnarray*} $

所以$ I\diamond_{\xi}\frac{P_{1}-\frac{1}{\overline{\xi}}P_{2}}{1+\xi}\diamond_{\xi}T = 0, $由此可得$ (1+\xi)T_{11}-(1+\frac{1}{\overline{\xi}})T_{22}+(\xi-\frac{1}{\overline{\xi}})T_{21} = 0. $由$ \xi\neq 0, -1 $得$ T_{11} = T_{22} = 0. $

又由$ A_{12}\diamond_{\xi}P_{1}\diamond_{\xi}I = 0 $得

$ \begin{eqnarray*} &&\phi(A_{12}+A_{21})\diamond_{\xi}P_{1}\diamond_{\xi}I+(A_{12}+A_{21})\diamond_{\xi}\phi(P_{1})\diamond_{\xi}I+(A_{12}+A_{21})\diamond_{\xi}P_{1}\diamond_{\xi}\phi(I)\\ & = &\phi((A_{12}+A_{21})\diamond_{\xi}P_{1}\diamond_{\xi}I) = \phi(A_{12}\diamond_{\xi}P_{1}\diamond_{\xi}I) +\phi(A_{21}\diamond_{\xi}P_{1}\diamond_{\xi}I)\\ & = &(\phi(A_{12})+\phi(A_{21}))\diamond_{\xi}P_{1}\diamond_{\xi}I+(A_{12}+A_{21})\diamond_{\xi}\phi(P_{1})\diamond_{\xi}I+(A_{12}+A_{21})\diamond_{\xi}P_{1}\diamond_{\xi}\phi(I). \end{eqnarray*} $

所以$ T\diamond_{\xi}P_{1}\diamond_{\xi}I = 0, $由此可得$ (1+|\xi|^{2})T_{21}+2\xi T_{21}^{*} = 0, $进而$ T_{21} = 0. $类似可得$ T_{12} = 0. $因此$ \phi(A_{12}+A_{21}) = \phi(A_{12})+\phi(A_{21}). $

断言4  设$ i, j, k\in\{1, 2\}, $若$ A_{kk}\in {\cal A}_{kk}, A_{ij}\in {\cal A}_{ij}, i\neq j, $则$ \phi(A_{kk}+A_{ij}) = \phi(A_{kk})+\phi(A_{ij}). $

下面只证明$ i = k = 1, j = 2, $其他情形同理可得. 令$ T = \phi(A_{11}+A_{12})-\phi(A_{11})-\phi(A_{12}). $由$ I\diamond_{\xi}\frac{P_{2}}{1+\xi}\diamond_{\xi}A_{11} = 0 $和断言1得

$ \begin{eqnarray*} &&\phi(I)\diamond_{\xi}\frac{P_{2}}{1+\xi}\diamond_{\xi}(A_{11}+A_{12})+I\diamond_{\xi}\phi(\frac{P_{2}}{1+\xi})\diamond_{\xi}(A_{11}+A_{12})\\ &&+I\diamond_{\xi}\frac{P_{2}}{1+\xi}\diamond_{\xi}\phi(A_{11}+A_{12})\\ & = &\phi(I\diamond_{\xi}\frac{P_{2}}{1+\xi}\diamond_{\xi}(A_{11}+A_{12})) = \phi(I\diamond_{\xi}\frac{P_{2}}{1+\xi}\diamond_{\xi}A_{11}) +\phi(I\diamond_{\xi}\frac{P_{2}}{1+\xi}\diamond_{\xi}A_{12})\\ & = &\phi(I)\diamond_{\xi}\frac{P_{2}}{1+\xi}\diamond_{\xi}(A_{11}+A_{12})+I\diamond_{\xi}\phi(\frac{P_{2}}{1+\xi})\diamond_{\xi}(A_{11}+A_{12})\\ &&+I\diamond_{\xi}\frac{P_{2}}{1+\xi}\diamond_{\xi}(\phi(A_{11})+\phi(A_{12})). \end{eqnarray*} $

所以$ I\diamond_{\xi}\frac{P_{2}}{1+\xi}\diamond_{\xi}T = 0, $由此可得$ T_{12} = T_{21} = T_{22} = 0. $

又由$ I\diamond_{\xi}\frac{P_{1}-\frac{1}{\overline{\xi}}P_{2}}{1+\xi}\diamond_{\xi}A_{12} = 0 $得

$ \begin{eqnarray*} &&\phi(I)\diamond_{\xi}\frac{P_{1}-\frac{1}{\overline{\xi}}P_{2}}{1+\xi}\diamond_{\xi}(A_{11}+A_{12})+I\diamond_{\xi}\phi(\frac{P_{1}-\frac{1}{\overline{\xi}}P_{2}}{1+\xi})\diamond_{\xi}(A_{11}+A_{12})\\ &&+I\diamond_{\xi}\frac{P_{1}-\frac{1}{\overline{\xi}}P_{2}}{1+\xi}\diamond_{\xi}\phi(A_{11}+A_{12})\\ & = &\phi(I\diamond_{\xi}\frac{P_{1}-\frac{1}{\overline{\xi}}P_{2}}{1+\xi}\diamond_{\xi}(A_{11}+A_{12}))\\ & = &\phi(I\diamond_{\xi}\frac{P_{1}-\frac{1}{\overline{\xi}}P_{2}}{1+\xi}\diamond_{\xi}A_{11}) +\phi(I\diamond_{\xi}\frac{P_{1}-\frac{1}{\overline{\xi}}P_{2}}{1+\xi}\diamond_{\xi}A_{12})\\ & = &\phi(I)\diamond_{\xi}\frac{P_{1}-\frac{1}{\overline{\xi}}P_{2}}{1+\xi}\diamond_{\xi}(A_{11}+A_{12})+I\diamond_{\xi}\phi(\frac{P_{1}-\frac{1}{\overline{\xi}}P_{2}}{1+\xi})\diamond_{\xi}(A_{11}+A_{12})\\ &&+I\diamond_{\xi}\frac{P_{1}-\frac{1}{\overline{\xi}}P_{2}}{1+\xi}\diamond_{\xi}(\phi(A_{11})+\phi(A_{12})). \end{eqnarray*} $

所以$ I\diamond_{\xi}\frac{P_{1}-\frac{1}{\overline{\xi}}P_{2}}{1+\xi}\diamond_{\xi}T = 0, $由此可得$ T_{11} = 0. $因此$ \phi(A_{11}+A_{12}) = \phi(A_{11})+\phi(A_{12}). $

断言5  $ \phi(A_{11}+A_{12}+A_{21}) = \phi(A_{11})+\phi(A_{12})+\phi(A_{21}), \phi(A_{12}+A_{21}+A_{22}) = \phi(A_{12})+\phi(A_{21})+\phi(A_{22}), \forall A_{11}\in{\cal A}_{11}, \forall A_{12}\in{\cal A}_{12}, \forall A_{21}\in{\cal A}_{21}, \forall A_{22}\in{\cal A}_{22}. $

令$ T = \phi(A_{11}+A_{12}+A_{21})-\phi(A_{11})-\phi(A_{12})-\phi(A_{21}). $由断言3得

$ \begin{eqnarray*} &&\phi(I)\diamond_{\xi}\frac{P_{2}}{1+\xi}\diamond_{\xi}(A_{11}+A_{12}+A_{21})+I\diamond_{\xi}\phi(\frac{P_{2}}{1+\xi})\diamond_{\xi}(A_{11}+A_{12}+A_{21})\\ &&+I\diamond_{\xi}\frac{P_{2}}{1+\xi}\diamond_{\xi}\phi(A_{11}+A_{12}+A_{21})\\ & = &\phi(I\diamond_{\xi}\frac{P_{2}}{1+\xi}\diamond_{\xi}(A_{11}+A_{12}+A_{21}))\\ & = &\phi(I\diamond_{\xi}\frac{P_{2}}{1+\xi}\diamond_{\xi}A_{11})+\phi(I\diamond_{\xi}\frac{P_{2}}{1+\xi}\diamond_{\xi}(A_{12}+A_{21}))\\ & = &\phi(I)\diamond_{\xi}\frac{P_{2}}{1+\xi}\diamond_{\xi}(A_{11}+A_{12}+A_{21})+I\diamond_{\xi}\phi(\frac{P_{2}}{1+\xi})\diamond_{\xi}(A_{11}+A_{12}+A_{21})\\ &&+I\diamond_{\xi}\frac{P_{2}}{1+\xi}\diamond_{\xi}(\phi(A_{11})+\phi(A_{12})+\phi(A_{21})). \end{eqnarray*} $

所以$ I\diamond_{\xi}\frac{P_{2}}{1+\xi}\diamond_{\xi}T = 0, $由此可得$ T_{12} = T_{21} = T_{22} = 0. $

又由$ I\diamond_{\xi}\frac{-\frac{1}{\overline{\xi}}P_{1}+P_{2}}{1+\xi}\diamond_{\xi}A_{21} = 0 $和断言4得

$ \begin{eqnarray*} &&\phi(I)\diamond_{\xi}\frac{-\frac{1}{\overline{\xi}}P_{1}+P_{2}}{1+\xi}\diamond_{\xi}(A_{11}+A_{12}+A_{21})+I\diamond_{\xi}\phi(\frac{-\frac{1}{\overline{\xi}}P_{1}+P_{2}}{1+\xi})\diamond_{\xi}(A_{11}+A_{12}+A_{21})\\ &&+I\diamond_{\xi}\frac{-\frac{1}{\overline{\xi}}P_{1}+P_{2}}{1+\xi}\diamond_{\xi}\phi(A_{11}+A_{12}+A_{21})\\ & = &\phi(I\diamond_{\xi}\frac{-\frac{1}{\overline{\xi}}P_{1}+P_{2}}{1+\xi}\diamond_{\xi}(A_{11}+A_{12}+A_{21}))\\ & = &\phi(I\diamond_{\xi}\frac{-\frac{1}{\overline{\xi}}P_{1}+P_{2}}{1+\xi}\diamond_{\xi}(A_{11}+A_{12}))+\phi(I\diamond_{\xi}\frac{-\frac{1}{\overline{\xi}}P_{1}+P_{2}}{1+\xi}\diamond_{\xi}A_{21})\\ & = &\phi(I)\diamond_{\xi}\frac{-\frac{1}{\overline{\xi}}P_{1}+P_{2}}{1+\xi}\diamond_{\xi}(A_{11}+A_{12}+A_{21})+I\diamond_{\xi}\phi(\frac{-\frac{1}{\overline{\xi}}P_{1}+P_{2}}{1+\xi})\diamond_{\xi}(A_{11}+A_{12}+A_{21})\\ &&+I\diamond_{\xi}\frac{-\frac{1}{\overline{\xi}}P_{1}+P_{2}}{1+\xi}\diamond_{\xi}(\phi(A_{11})+\phi(A_{12})+\phi(A_{21})). \end{eqnarray*} $

所以$ I\diamond_{\xi}\frac{-\frac{1}{\overline{\xi}}P_{1}+P_{2}}{1+\xi}\diamond_{\xi}T = 0, $计算可得$ T_{11} = 0. $因此$ \phi(A_{11}+A_{12}+A_{21}) = \phi(A_{11})+\phi(A_{12})+\phi(A_{21}), $同理$ \phi(A_{12}+A_{21}+A_{22}) = \phi(A_{12})+\phi(A_{21})+\phi(A_{22}). $

断言6  $ \phi(A_{ij}+B_{ij}) = \phi(A_{ij})+\phi(B_{ij}), \forall A_{ij}, B_{ij}\in{\cal A}_{ij}, 1\leq i\neq j\leq 2. $

因为$ I\diamond_{\xi}\frac{P_{i}+A_{ij}}{1+\xi}\diamond_{\xi}(P_{j}+B_{ij}) = A_{ij}+B_{ij}+\xi A_{ij}^{*}+\xi B_{ij}A_{ij}^{*}, $由断言3, 断言4和断言5得

$ \begin{eqnarray*} &&\phi(A_{ij}+B_{ij})+\phi(\xi A_{ij}^{*})+\phi(\xi B_{ij}A_{ij}^{*})\\ & = &\phi(I\diamond_{\xi}\frac{P_{i}+A_{ij}}{1+\xi}\diamond_{\xi}(P_{j}+B_{ij}))\\ & = &\phi(I)\diamond_{\xi}\frac{P_{i}+A_{ij}}{1+\xi}\diamond_{\xi}(P_{j}+B_{ij})+I\diamond_{\xi}\phi(\frac{P_{i}+A_{ij}}{1+\xi})\diamond_{\xi}(P_{j}+B_{ij})\\ &&+I\diamond_{\xi}\frac{P_{i}+A_{ij}}{1+\xi}\diamond_{\xi}\phi(P_{j}+B_{ij})\\ & = &\phi(I)\diamond_{\xi}\frac{P_{i}+A_{ij}}{1+\xi}\diamond_{\xi}(P_{j}+B_{ij})+I\diamond_{\xi}(\phi(\frac{P_{i}}{1+\xi})+\phi(\frac{A_{ij}}{1+\xi}))\diamond_{\xi}(P_{j}+B_{ij})\\ &&+I\diamond_{\xi}\frac{P_{i}+A_{ij}}{1+\xi}\diamond_{\xi}(\phi(P_{j})+\phi(B_{ij}))\\ & = &\phi(I\diamond_{\xi}\frac{P_{i}}{1+\xi}\diamond_{\xi}P_{j})+\phi(I\diamond_{\xi}\frac{A_{ij}}{1+\xi}\diamond_{\xi}P_{j})+\phi(I\diamond_{\xi}\frac{P_{i}}{1+\xi}\diamond_{\xi}B_{ij})+\phi(I\diamond_{\xi}\frac{A_{ij}}{1+\xi}\diamond_{\xi}B_{ij})\\ & = &\phi(A_{ij}+\xi A_{ij}^{*})+\phi(B_{ij})+\phi(\xi B_{ij}A_{ij}^{*})\\ & = &\phi(A_{ij})+\phi(B_{ij})+\phi(\xi A_{ij}^{*})+\phi(\xi B_{ij}A_{ij}^{*}). \end{eqnarray*} $

所以$ \phi(A_{ij}+B_{ij}) = \phi(A_{ij})+\phi(B_{ij}). $

断言7  $ \phi(X_{12}A_{21}+X_{12}B_{21}+\xi A_{12}X_{12}^{*}+\xi B_{12}X_{12}^{*}) = \phi(X_{12}A_{21})+\phi(X_{12}B_{21})+\phi(\xi A_{12}X_{12}^{*})+\phi(\xi B_{12}X_{12}^{*}), \forall X_{12}, A_{12}, B_{12}\in {\cal A}_{12}, \forall A_{21}, B_{21}\in {\cal A}_{21}. $

由断言3和断言6得

$ \begin{eqnarray*} &&\phi(X_{12}A_{21}+X_{12}B_{21}+\xi A_{12}X_{12}^{*}+\xi B_{12}X_{12}^{*})\\ & = &\phi(I\diamond_{\xi}\frac{X_{12}}{1+\xi}\diamond_{\xi}(A_{21}+B_{21}+A_{12}+B_{12}))\\ & = &\phi(I)\diamond_{\xi}\frac{X_{12}}{1+\xi}\diamond_{\xi}(A_{21}+B_{21}+A_{12}+B_{12})+I\diamond_{\xi}\phi(\frac{X_{12}}{1+\xi})\diamond_{\xi}(A_{21}+B_{21}+A_{12}+B_{12})\\ &&+I\diamond_{\xi}\frac{X_{12}}{1+\xi}\diamond_{\xi}\phi(A_{21}+B_{21}+A_{12}+B_{12})\\ & = &\phi(I)\diamond_{\xi}\frac{X_{12}}{1+\xi}\diamond_{\xi}(A_{21}+B_{21}+A_{12}+B_{12})+I\diamond_{\xi}\phi(\frac{X_{12}}{1+\xi})\diamond_{\xi}(A_{21}+B_{21}+A_{12}+B_{12})\\ &&+I\diamond_{\xi}\frac{X_{12}}{1+\xi}\diamond_{\xi}(\phi(A_{21}+B_{21})+\phi(A_{12}+B_{12}))\\ & = &\phi(I)\diamond_{\xi}\frac{X_{12}}{1+\xi}\diamond_{\xi}(A_{21}+B_{21}+A_{12}+B_{12})+I\diamond_{\xi}\phi(\frac{X_{12}}{1+\xi})\diamond_{\xi}(A_{21}+B_{21}+A_{12}+B_{12})\\ &&+I\diamond_{\xi}\frac{X_{12}}{1+\xi}\diamond_{\xi}(\phi(A_{21})+\phi(B_{21})+\phi(A_{12})+\phi(B_{12}))\\ & = &\phi(I\diamond_{\xi}\frac{X_{12}}{1+\xi}\diamond_{\xi}A_{21})+\phi(I\diamond_{\xi}\frac{X_{12}}{1+\xi}\diamond_{\xi}\diamond_{\eta}B_{21})\\ &&+\phi(I\diamond_{\xi}\frac{X_{12}}{1+\xi}\diamond_{\xi}A_{12})+\phi(I\diamond_{\xi}\frac{X_{12}}{1+\xi}\diamond_{\xi}B_{12})\\ & = &\phi(X_{12}A_{21})+\phi(X_{12}B_{21})+\phi(\xi A_{12}X_{12}^{*})+\phi(\xi B_{12}X_{12}^{*}). \end{eqnarray*} $

所以$ \phi(X_{12}A_{21}+X_{12}B_{21}+\xi A_{12}X_{12}^{*}+\xi B_{12}X_{12}^{*}) = \phi(X_{12}A_{21})+\phi(X_{12}B_{21})+\phi(\xi A_{12}X_{12}^{*})+\phi(\xi B_{12}X_{12}^{*}). $

断言8  $ \phi $是可加的.

令$ A = \sum\limits_{i, i = 1}^{2}A_{ij}, B = \sum\limits_{i, i = 1}^{2}B_{ij}, T = \phi(A+B)-\phi(A)-\phi(B). $任取$ X_{12}\in {\cal A}_{12}, $由断言5, 断言6和断言7得

$ \begin{eqnarray*} &&\phi(I)\diamond_{\xi}\frac{X_{12}}{1+\xi}\diamond_{\xi}(A+B)+I\diamond_{\xi}\phi(\frac{X_{12}}{1+\xi})\diamond_{\xi}(A+B)+I\diamond_{\xi}\frac{X_{12}}{1+\xi}\diamond_{\xi}\phi(A+B)\\ & = &\phi(I\diamond_{\xi}\frac{X_{12}}{1+\xi}\diamond_{\xi}(A+B))\\ & = &\phi(X_{12}A_{22}+X_{12}B_{22}+X_{12}A_{21}+X_{12}B_{21}\\ &&+\xi A_{12}X_{12}^{*}+\xi B_{12}X_{12}^{*}+\xi A_{22}X_{12}^{*}+\xi B_{22}X_{12}^{*})\\ & = &\phi(X_{12}A_{22}+X_{12}B_{22})+\phi(X_{12}A_{21}+X_{12}B_{21}+\xi A_{12}X_{12}^{*}+\xi B_{12}X_{12}^{*})\\ &&+\phi(\xi A_{22}X_{12}^{*}+\xi B_{22}X_{12}^{*})\\ & = &\phi(X_{12}A_{22})+\phi(X_{12}B_{22})+\phi(X_{12}A_{21})+\phi(X_{12}B_{21})\\ &&+\phi(\xi A_{12}X_{12}^{*})+\phi(\xi B_{12}X_{12}^{*})+\phi(\xi A_{22}X_{12}^{*})+\phi(\xi B_{22}X_{12}^{*})\\ & = &\phi(X_{12}A_{22})+\phi(X_{12}A_{21}+\xi A_{12}X_{12}^{*})+\phi(\xi A_{22}X_{12}^{*})\\ &&+\phi(X_{12}B_{22})+\phi(X_{12}B_{21}+\xi B_{12}X_{12}^{*})+\phi(\xi B_{22}X_{12}^{*})\\ & = &\phi(X_{12}A_{22}+X_{12}A_{21}+\xi A_{12}X_{12}^{*}+\xi A_{22}X_{12}^{*})\\ &&+\phi(X_{12}B_{22}+X_{12}B_{21}+\xi B_{12}X_{12}^{*}+\xi B_{22}X_{12}^{*})\\ & = &\phi(I\diamond_{\xi}\frac{X_{12}}{1+\xi}\diamond_{\xi}A)+\phi(I\diamond_{\xi}\frac{X_{12}}{1+\xi}\diamond_{\xi}B)\\ & = &\phi(I)\diamond_{\xi}\frac{X_{12}}{1+\xi}\diamond_{\xi}(A+B)+I\diamond_{\xi}\phi(\frac{X_{12}}{1+\xi})\diamond_{\xi}(A+B) +I\diamond_{\xi}\frac{X_{12}}{1+\xi}\diamond_{\xi}(\phi(A)+\phi(B)). \end{eqnarray*} $

由此可得$ I\diamond_{\xi}\frac{X_{12}}{1+\xi}\diamond_{\xi}T = 0, $即

(2.1) $ \begin{equation} X_{12}T+\xi TX_{12}^{*} = 0 \end{equation} $

对所有$ X_{12}\in {\cal A}_{12} $成立. (2.1) 式中$ X_{12} $用$ \text{i}X_{12} $代替得

(2.2) $ \begin{equation} X_{12}T-\xi TX_{12}^{*} = 0. \end{equation} $

(2.1) 式和(2.2) 式相加得对所有$ X_{12}\in {\cal A}_{12}, $有$ X_{12}T = 0 $成立, 由$ {\cal A} $是素的可知$ T = 0, $即$ \phi $是可加的.

断言9  $ \phi $是可加的$ * $-导子且$ \phi(\xi A) = \xi\phi(A), \forall A\in{\cal A}. $

分以下三种情形讨论.

情形1  $ \xi = 1. $

任取$ \lambda\in{\mathbb R}, A\in{\cal A}, $由断言1得

$ \begin{eqnarray*} 0& = &\phi(\text{i}\lambda I\bullet A\bullet I) = \phi(\text{i}\lambda I)\bullet A\bullet I+\text{i}\lambda I\bullet \phi(A)\bullet I+\text{i}\lambda I\bullet A\bullet\phi(I)\\ & = &\phi(\text{i}\lambda I)(A+A^{*})+(A+A^{*})\phi(\text{i}\lambda I)^{*}. \end{eqnarray*} $

所以任取$ B = B^{*}\in{\cal A} $有$ \phi(\text{i}\lambda I)B+B\phi(\text{i}\lambda I)^{*} = 0. $设$ B_{1} = \frac{B+B^{*}}{2}, B_{2} = \frac{B-B^{*}}{2\text{i}} $是自伴元, 则对所有$ B\in{\cal A} $有$ B = B_{1}+\text{i}B_{2}, $因此$ \phi(\text{i}\lambda I)B+B\phi(\text{i}\lambda I)^{*} = 0. $由引理1.1得$ \phi(\text{i}\lambda I)\subseteq\text{i}{\mathbb R}I, $所以

(2.3) $ \begin{equation} \phi(\text{i}{\mathbb R}I)\subseteq\text{i}{\mathbb R}I. \end{equation} $

对所有$ \lambda\in{\mathbb C}, A\in{\cal A}, $由(2.3) 式得

$ \begin{eqnarray*} 0& = &\phi(A\bullet\text{i}I\bullet\lambda I) = \phi(A)\bullet\text{i}I\bullet\lambda I+A\bullet\phi(\text{i}I)\bullet\lambda I+A\bullet\text{i}I\bullet\phi(\lambda I)\\ & = &\text{i}(A+A^{*})\phi(\lambda I)-\text{i}\phi(\lambda I)(A+A^{*}). \end{eqnarray*} $

由此可得, 对所有$ B = B^{*}\in{\cal A}, $有$ B\phi(\lambda I) = \phi(\lambda I)B. $设$ B_{1} = \frac{B+B^{*}}{2}, B_{2} = \frac{B-B^{*}}{2\text{i}} $是自伴元, 则对所有$ B\in{\cal A}, $有$ B = B_{1}+\text{i}B_{2}, $因此$ B\phi(\lambda I) = \phi(\lambda I)B. $则$ \phi(\lambda I)\subseteq{\mathbb C}I. $所以

(2.4) $ \begin{equation} \phi({\mathbb C}I)\subseteq{\mathbb C}I. \end{equation} $

由(2.4) 式可得, 存在$ \lambda\in{\mathbb C} $使得$ \phi(I) = \lambda I. $由

$ \begin{eqnarray*} 4\lambda I& = &4\phi(I) = \phi(I\bullet I\bullet I) = \phi(I)\bullet I\bullet I+I\bullet \phi(I)\bullet I+I\bullet I\bullet \phi(I)\\ & = &4(\lambda+\overline{\lambda})I+4\lambda I \end{eqnarray*} $

得$ \lambda+\overline{\lambda} = 0, $则$ \lambda\in\text{i}{\mathbb R}I, $于是存在$ \lambda_{1}\in{\mathbb R}, $使得

(2.5) $ \begin{equation} \phi(I) = \text{i}\lambda_{1}I. \end{equation} $

设$ P_{i}\in{\cal A}, i = 1, 2 $为非平凡投影, 由(2.5) 式得

$ \begin{eqnarray*} 4\phi(P_{i})& = &\phi(I\bullet P_{i}\bullet I) = \phi(I)\bullet P_{i}\bullet I+I\bullet\phi(P_{i})\bullet I+I\bullet P_{i}\bullet\phi(I) \\ & = &2(\phi(P_{i})+\phi(P_{i})^{*})+4\text{i}\lambda_{1}P_{i}. \end{eqnarray*} $

即

(2.6) $ \begin{equation} \phi(P_{i})^{*} = \phi(P_{i})-2\text{i}\lambda_{1}P_{i}, i = 1, 2. \end{equation} $

设$ i, j = 1, 2, i\neq j, $由(2.6) 式可得

(2.7) $ \begin{eqnarray} 0& = &\phi(P_{i}\bullet P_{j}\bullet P_{i}){}\\ & = &\phi(P_{i})\bullet P_{j}\bullet P_{i}+P_{i}\bullet\phi(P_{j})\bullet P_{i}+P_{i}\bullet P_{j}\bullet \phi(P_{i}){}\\ & = &P_{j}\phi(P_{i})P_{i}+P_{i}\phi(P_{i})P_{j}+2P_{i}\phi(P_{j})P_{i}+\phi(P_{j})P_{i}+P_{i}\phi(P_{j}). \end{eqnarray} $

(2.7) 式两边同乘$ P_{i} $得

(2.8) $ \begin{equation} P_{i}\phi(P_{j})P_{i} = 0. \end{equation} $

(2.7) 式左乘$ P_{i} $右乘$ P_{j} $得

(2.9) $ \begin{equation} P_{i}\phi(P_{i})P_{j}+P_{i}\phi(P_{j})P_{j} = 0. \end{equation} $

(2.7) 式左乘$ P_{j} $右乘$ P_{i} $得

(2.10) $ \begin{equation} P_{j}\phi(P_{i})P_{i}+P_{j}\phi(P_{j})P_{i} = 0. \end{equation} $

对所有$ A_{ji}\in{\cal A}_{ji}, $由(2.5) 式和(2.6) 式得

$ \begin{eqnarray*} 2\phi(A_{ji})& = &\phi(I\bullet P_{j}\bullet A_{ji})\\ & = &\phi(I)\bullet P_{j}\bullet A_{ji}+I\bullet\phi(P_{j})\bullet A_{ji}+I\bullet P_{j}\bullet\phi(A_{ji})\\ & = &2\phi(P_{j})A_{ji}+2A_{ji}\phi(P_{j})^{*}+2P_{j}\phi(A_{ji})+2\phi(A_{ji})P_{j}\\ & = &2\phi(P_{j})A_{ji}+2A_{ji}\phi(P_{j})+2P_{j}\phi(A_{ji})+2\phi(A_{ji})P_{j}. \end{eqnarray*} $

上式左乘$ P_{j} $右乘$ P_{i} $得$ P_{j}\phi(P_{j})A_{ji}+A_{ji}\phi(P_{j})P_{i} = 0. $结合(2.8) 式可得$ P_{j}\phi(P_{j})A_{ji} = 0. $由于$ {\cal A} $是素的, 所以

(2.11) $ \begin{equation} P_{j}\phi(P_{j})P_{j} = 0. \end{equation} $

由(2.8)–(2.11) 式可得

(2.12) $ \begin{equation} \phi(I) = \phi(P_{i})+\phi(P_{j}) = P_{i}\phi(P_{i})P_{j}+P_{j}\phi(P_{i})P_{i}+P_{i}\phi(P_{j})P_{j}+P_{j}\phi(P_{j})P_{i} = 0. \end{equation} $

对所有$ A, B\in{\cal A}, $由(2.12) 式得

$ \phi(A\bullet B) = \phi(\frac{I}{2}\bullet A\bullet B) = \phi(A)\bullet B+A\bullet\phi(B). $

由文献[6]的主要定理得, $ \phi $是可加的$ * $-导子.

情形2  $ |\xi| = 1 $且$ \xi\neq 1, -1. $

因为

$ \begin{eqnarray*} 0& = &\phi(I\diamond_{\xi}\text{i}I\diamond_{\xi}\text{i}I)\\ & = &\phi(I)\diamond_{\xi}\text{i}I\diamond_{\xi}\text{i}I+I\diamond_{\xi}\phi(\text{i}I)\diamond_{\xi}\text{i}I+I\diamond_{\xi}\text{i}I\diamond_{\xi}\phi(\text{i}I)\\ & = &\text{i}(1+\xi)(\phi(\text{i}I)+\phi(\text{i}I)^{*}). \end{eqnarray*} $

所以

(2.13) $ \begin{equation} \phi(\text{i}I)^{*} = -\phi(\text{i}I). \end{equation} $

由(2.13) 式得

(2.14) $ \begin{eqnarray} 2\phi(\text{i}(1-\xi)I)& = &\phi(\text{i}I\diamond_{\xi}I\diamond_{\xi}I){}\\ & = &\phi(\text{i}I)\diamond_{\xi}I\diamond_{\xi}I+\text{i}I\diamond_{\xi}\phi(I)\diamond_{\xi}I+\text{i}I\diamond_{\xi}I\diamond_{\xi}\phi(I){}\\ & = &2(1-\xi)\phi(\text{i}I)+\text{i}(1-\xi)\phi(I)+\text{i}(1-\xi)\phi(I)^{*}+2\text{i}(1-\xi)\phi(I). \end{eqnarray} $

另一方面

(2.15) $ \begin{eqnarray} 2\phi(\text{i}(1+\xi)I)& = &\phi(I\diamond_{\xi}I\diamond_{\xi}\text{i}I){}\\ & = &\phi(I)\diamond_{\xi}I\diamond_{\xi}\text{i}I+I\diamond_{\xi}\phi(I)\diamond_{\xi}\text{i}I+I\diamond_{\xi}I\diamond_{\xi}\phi(\text{i}I){}\\ & = &\text{i}(3+\xi)\phi(I)+\text{i}(1+3\xi)\phi(I)^{*}+2(1+\xi)\phi(\text{i}I). \end{eqnarray} $

(2.14) 式加(2.15) 式得

(2.16) $ \begin{equation} \phi(I)^{*} = \frac{\xi-3}{\xi+1}\phi(I). \end{equation} $

(2.16) 式两边取$ * $得

$ \phi(I) = \frac{\overline{\xi}-3}{\overline{\xi}+1}\phi(I)^{*} = \frac{\overline{\xi}-3}{\overline{\xi}+1}\cdot\frac{\xi-3}{\xi+1}\phi(I). $

若$ \phi(I)\neq 0, $则$ \xi = 1, $与题设矛盾, 所以

(2.17) $ \begin{equation} \phi(I) = 0. \end{equation} $

任取$ A\in{\cal A}, $由(2.17) 式有

$ 2\phi((1+\xi)A) = \phi(I\diamond_{\xi}I\diamond_{\xi}A) = I\diamond_{\xi}I\diamond_{\xi}\phi(A) = 2(1+\xi)\phi(A). $

由此可得

(2.18) $ \begin{equation} \phi((1+\xi)A) = (1+\xi)\phi(A), \forall A\in{\cal A}. \end{equation} $

任取$ A, B\in{\cal A}, $有

$ \begin{eqnarray*} \phi((1+\xi)A\bullet B)& = &\phi(I\diamond_{\xi}A\diamond_{\xi}B)\\ & = &I\diamond_{\xi}\phi(A)\diamond_{\xi}B+I\diamond_{\xi}A\diamond_{\xi}\phi(B) \\ & = &(1+\xi)(\phi(A)\bullet B+A\bullet\phi(B)). \end{eqnarray*} $

结合(2.18) 式可得

$ \phi(A\bullet B) = \phi(A)\bullet B+A\bullet\phi(B). $

由文献[6]的主要定理得, $ \phi $是可加的$ * $-导子. 再由(2.18) 式得$ \phi(\xi A) = \xi\phi(A), \forall A\in{\cal A}. $

情形3  $ |\xi|\neq 1. $

由$ \phi(I\diamond_{\xi}\text{i}I\diamond_{\xi}\text{i}I) = \phi(\text{i}I\diamond_{\xi}\text{i}I\diamond_{\xi}I) $得

$ \begin{eqnarray*} &&2\text{i}\phi(\text{i}I)-|\xi|^{2}\text{i}\phi(\text{i}I)+|\xi|^{2}\text{i}\phi(\text{i}I)^{*}+\xi\text{i}\phi(\text{i}I)+\xi\text{i}\phi(\text{i}I)^{*}-\phi(I)+|\xi|^{2}\phi(I)\\ & = &-\phi(I)+|\xi|^{2}\phi(I)+2\text{i}\phi(\text{i}I)-\xi\text{i}\phi(\text{i}I)-\xi\text{i}\phi(\text{i}I)^{*}-|\xi|^{2}\text{i}\phi(\text{i}I)+|\xi|^{2}\text{i}\phi(\text{i}I)^{*}. \end{eqnarray*} $

所以$ 2\xi\text{i}(\phi(\text{i}I)+\phi(\text{i}I)^{*}) = 0. $由此可得

(2.19) $ \begin{equation} \phi(\text{i}I)^{*} = -\phi(\text{i}I). \end{equation} $

另一方面, 由$ \phi(\text{i}I\diamond_{\xi}\text{i}I\diamond_{\xi}\text{i}I) = -\phi(I\diamond_{\xi}\text{i}I\diamond_{\xi}I) $和(2.19) 式得

$ (1-|\xi|^{2})\phi(\text{i}I) = (1-|\xi|^{2})\text{i}\phi(I). $

由此可得

(2.20) $ \begin{equation} \phi(\text{i}I) = \text{i}\phi(I). \end{equation} $

所以

(2.21) $ \begin{equation} \phi(I)^{*} = \phi(I). \end{equation} $

由(2.19) 式和(2.21) 式可得

(2.22) $ \begin{eqnarray} \phi(\text{i}(1-2\xi+|\xi|^{2})I) & = &\phi(\text{i}I\diamond_{\xi}I\diamond_{\xi}I){}\\ & = &\phi(\text{i}I)\diamond_{\xi}I\diamond_{\xi}I+\text{i}I\diamond_{\xi}\phi(I)\diamond_{\xi}I+\text{i}I\diamond_{\xi}I\diamond_{\xi}\phi(I){}\\ & = &\phi(\text{i}I)-2\xi\phi(\text{i}I)+|\xi|^{2}\phi(\text{i}I)+2\text{i}\phi(I)-4\text{i}\xi\phi(I)+2\text{i}|\xi|^{2}\phi(I). \end{eqnarray} $

另一方面

(2.23) $ \begin{eqnarray} \phi(\text{i}(1+2\xi+|\xi|^{2})I)& = &\phi(I\diamond_{\xi}I\diamond_{\xi}\text{i}I){}\\ & = &\phi(I)\diamond_{\xi}I\diamond_{\xi}\text{i}I+I\diamond_{\xi}\phi(I)\diamond_{\xi}\text{i}I+I\diamond_{\xi}I\diamond_{\xi}\phi(\text{i}I){}\\ & = &2\text{i}\phi(I)+4\text{i}\xi\phi(I)+2\text{i}|\xi|^{2}\phi(I)+\phi(\text{i}I)+2\xi\phi(\text{i}I)+|\xi|^{2}\phi(\text{i}I). \end{eqnarray} $

(2.22) 式加(2.23) 式得

(2.24) $ \begin{equation} \phi(\text{i}(1+|\xi|^{2})I) = \phi(\text{i}I)+2\text{i}\phi(I)+|\xi|^{2}\phi(\text{i}I)+2\text{i}|\xi|^{2}\phi(I). \end{equation} $

又由(2.19) 式和(2.21) 式得

(2.25) $ \begin{eqnarray} \phi(\text{i}(1-|\xi|^{2})I)& = &\phi(I\diamond_{\xi}\text{i}I\diamond_{\xi}I){}\\ & = &\phi(I)\diamond_{\xi}\text{i}I\diamond_{\xi}I+I\diamond_{\xi}\phi(\text{i}I)\diamond_{\xi}I+I\diamond_{\xi}\text{i}I\diamond_{\xi}\phi(I){}\\ & = &\text{i}\phi(I)-\text{i}|\xi|^{2}\phi(I)+\phi(\text{i}I)-|\xi|^{2}\phi(\text{i}I)+\text{i}\phi(I)-\text{i}|\xi|^{2}\phi(I). \end{eqnarray} $

(2.24) 式加(2.25) 式得$ 4\text{i}\phi(I) = 0, $则

(2.26) $ \begin{equation} \phi(I) = 0. \end{equation} $

由(2.20) 式得

(2.27) $ \begin{equation} \phi(\text{i}I) = 0. \end{equation} $

任取$ A\in{\cal A}, $由(2.27) 式得

$ \phi(A\diamond_{\xi}\text{i}I\diamond_{\xi}\text{i}I) = \phi(A)\diamond_{\xi}\text{i}I\diamond_{\xi}\text{i}I. $

计算可得

(2.28) $ \begin{equation} \phi(|\xi|^{2}A) = |\xi|^{2}\phi(A). \end{equation} $

由(2.26) 式得$ \phi(A\diamond_{\xi}I\diamond_{\xi}I) = \phi(A)\diamond_{\xi}I\diamond_{\xi}I, $结合(2.28) 式得

(2.29) $ \begin{equation} \phi(\xi A^{*}) = \xi\phi(A)^{*}. \end{equation} $

另一方面$ \phi(I\diamond_{\xi}I\diamond_{\xi}A^{*}) = I\diamond_{\xi}I\diamond_{\xi}\phi(A^{*}). $计算可得

(2.30) $ \begin{equation} \phi(\xi A^{*}) = \xi\phi(A^{*}), \forall A\in{\cal A}. \end{equation} $

结合(2.29) 式和(2.30) 式得

(2.31) $ \begin{equation} \phi(A^{*}) = \phi(A)^{*}, \forall A\in{\cal A}. \end{equation} $

任取$ A\in{\cal A}, $由(2.28) 式得

$ \begin{eqnarray*} (1-|\xi|^{2})\phi(\text{i}A)& = &\phi((1-|\xi|^{2})\text{i}A) = \phi(I\diamond_{\xi}\text{i}I\diamond_{\xi}A)\\ & = &I\diamond_{\xi}\text{i}I\diamond_{\xi}\phi(A) = (1-|\xi|^{2})\text{i}\phi(A). \end{eqnarray*} $

所以

(2.32) $ \begin{equation} \phi(\text{i}A) = \text{i}\phi(A), \forall A\in{\cal A}. \end{equation} $

任取$ A, B\in{\cal A}, $有

(2.33) $ \begin{eqnarray} &&\phi(AB+\xi AB+\xi BA^{*}+|\xi|^{2}BA^{*}){}\\ & = &\phi(I\diamond_{\xi}A\diamond_{\xi}B){}\\ & = &I\diamond_{\xi}\phi(A)\diamond_{\xi}B+I\diamond_{\xi}A\diamond_{\xi}\phi(B){}\\ & = &\phi(A)B+\xi\phi(A)B+\xi B\phi(A)^{*}+|\xi|^{2}B\phi(A)^{*}{}\\ &&+A\phi(B)+\xi A\phi(B)+\xi\phi(B)A^{*}+|\xi|^{2}\phi(B)A^{*}. \end{eqnarray} $

另一方面, 由(2.32) 式得

(2.34) $ \begin{eqnarray} &&\phi(AB+\xi AB-\xi BA^{*}-|\xi|^{2}BA^{*}){}\\ & = &\phi(I\diamond_{\xi}\text{i}A\diamond_{\xi}(-\text{i}B)){}\\ & = &I\diamond_{\xi}\phi(\text{i}A)\diamond_{\xi}(-\text{i}B)+I\diamond_{\xi}\text{i}A\diamond_{\xi}\phi(-\text{i}B){}\\ & = &\phi(A)B+\xi\phi(A)B-\xi B\phi(A)^{*}-|\xi|^{2}B\phi(A)^{*}{}\\ &&+A\phi(B)+\xi A\phi(B)-\xi\phi(B)A^{*}-|\xi|^{2}\phi(B)A^{*}. \end{eqnarray} $

结合(2.33) 式和(2.34) 式得

$ \phi((1+\xi)AB) = (1+\xi)\phi(A)B+(1+\xi)A\phi(B). $

又由(2.30) 式得$ (1+\xi)\phi(AB) = (1+\xi)(\phi(A)B+A\phi(B)). $所以$ \phi(AB) = \phi(A)B+A\phi(B). $定理2.1证毕.