设$ {\cal A} $是复数域$ {\Bbb C} $上的$ * $- 代数, $ \xi $是非零复数, $ A, B\in{\cal A} $的$ \xi $-Jordan $ * $- 积定义为$ A\diamond B=AB+\xi BA^{*}, $$ 1 $-Jordan $ * $- 积通常记为$ A\bullet B=AB+BA^{*}, $$ (-1) $-Jordan $ * $- 积(Lie $ 1 $-$ * $- 积)通常记为$ [A, B]_{*}=AB-BA^{*}. $近年来, 相关的研究吸引了许多作者的注意^[1-19]. P. Šemrl ^[1]在量子函数中首先引入并研究了$ (-1) $-Jordan $ * $- 积. 霍等^[2]刻画了von Neumann代数上保持三重$ \xi $-Jordan $ * $- 积的非线性映射.

设$ \phi:{\cal A\rightarrow A} $是一个映射(不必可加或线性). 如果对任意的$ A, B\in{\cal A}, $有$ \phi(AB)=\phi(A)B+A\phi(B) $和$ \phi(A^{*})=\phi(A)^{*}, $那么称$ \phi $是$ * $- 导子. 设$ n\geq 2 $为固定的正整数, 如果对所有$ A_{1}, A_{2}, \ldots , A_{n}\in{\cal A}, $有$ \phi(A_{1}\bullet A_{2}\bullet\ldots\bullet A_{n})=\Sigma_{k=1}^{n}A_{1}\bullet\ldots\bullet A_{k-1}\bullet\phi(A_{k})\bullet A_{k+1}\bullet\ldots\bullet A_{n}, $其中$ A_{1}\bullet A_{2}\bullet\ldots\bullet A_{n}=(\ldots ((A_{1}\bullet A_{2})\bullet A_{3})\bullet\ldots\bullet A_{n}), $那么称$ \phi $是$ * $-Jordan- 型导子; 如果有$ \phi(A_{1}\diamond A_{2}\diamond\ldots\diamond A_{n})=\Sigma_{k=1}^{n}A_{1}\diamond\ldots\diamond A_{k-1}\diamond\phi(A_{k})\diamond A_{k+1}\diamond\ldots\diamond A_{n}, $其中$ A_{1}\diamond A_{2}\diamond\ldots\diamond A_{n}=(\ldots ((A_{1}\diamond A_{2})\diamond A_{3})\diamond\ldots\diamond A_{n}), $那么称$ \phi $是$ \xi $-$ * $-Jordan- 型导子. 李等^[3]得到下面结论: 设$ {\cal A} $是含单位元$ I $和非平凡投影$ P $的$ * $- 代数, 满足: (1) 若$ X{\cal A} P=0, $则$ X=0; $ (2) 若$ X{\cal A}(I-P)=0, $则$ X=0. $如果$ \phi $是$ {\cal A} $上的非线性$ * $-Jordan- 型导子当且仅当$ \phi $是可加的$ * $- 导子. 在文献[3] 中作者还给出了下面的猜想: 设$ {\cal A} $是含单位元$ I $和非平凡投影$ P $的$ * $- 代数, 满足: (1) 若$ X{\cal A} P=0, $则$ X=0; $ (2) 若$ X{\cal A}(I-P)=0, $则$ X=0. $如果$ \phi $是$ {\cal A} $上的非线性$ \xi $-$ * $-Jordan- 型导子当且仅当$ \phi $是可加的$ * $- 导子, 且对所有的$ A\in\mathcal A, $有$ \phi(\xi A)=\xi\phi(A) $. 本文来证明这个猜想.

当$ \xi=1 $时文献[3] 已得到了结果. 当$ \xi=-1 $时, 本文定理2.1将给予证明. 当$ \xi\in{\Bbb C}\setminus\{0, \pm 1\} $时, 本文定理3.1和定理3.2将给予讨论.

本节中设$ \xi=-1. $

定理2.1  设$ {\cal A} $是复数域上含单位元$ I $和非平凡投影$ P $的$ * $- 代数, 满足: (1) 若$ X{\cal A} P=0, $则$ X=0; $ (2) 若$ X{\cal A}(I-P)=0, $则$ X=0. $如果映射$ \phi:{\cal A\rightarrow A} $满足对所有$ A_{n-1}, A_{n}\in{\cal A}, A_{1}=\mbox{i}I $或$ A_{1}=I, A_{2}=\ldots=A_{n-2}=I, n\geq 2, $有$ \phi(A_{1}\diamond A_{2}\diamond\ldots\diamond A_{n})=\Sigma_{k=1}^{n}A_{1}\diamond\ldots\diamond A_{k-1}\diamond\phi(A_{k})\diamond A_{k+1}\diamond\ldots\diamond A_{n}, $那么$ \phi $是可加的$ * $- 导子.

证   设$ P=P_{1}, 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}. $若$ n=2, $则$ A_{1}, A_{2} $是$ {\cal A} $中的任意元. 若$ n=3, $则$ A_{2}, A_{3} $是$ {\cal A} $中的任意元.

断言2.1   $ \phi(0)=0. $

易得$ \phi(0)=\phi(\mbox{i}I\diamond I\diamond\ldots\diamond I\diamond 0\diamond 0)=0. $

断言2.2   任取$ A_{12}\in{\cal A}_{12}, A_{21}\in{\cal A}_{21}, $则$ \phi(A_{12}+A_{21})=\phi(A_{12})+\phi(A_{21}). $

设$ T=\phi(A_{12}+A_{21})-\phi(A_{12})-\phi(A_{21}). $由$ \mbox{i}I\diamond\ldots\diamond I\diamond(P_{1}-P_{2})\diamond A_{12}=0 $和断言2.1得

$ \begin{eqnarray*} &&\phi(\mbox{i}I)\diamond\ldots\diamond I\diamond(P_{1}-P_{2})\diamond(A_{12}+A_{21})+\ldots+\mbox{i}I\diamond\ldots\diamond I\diamond\phi(P_{1}-P_{2})\diamond(A_{12}+A_{21})\\ &&+\mbox{i}I\diamond\ldots\diamond I\diamond(P_{1}-P_{2})\diamond\phi(A_{12}+A_{21})\\ &=&\phi(\mbox{i}I\diamond\ldots\diamond I\diamond(P_{1}-P_{2})\diamond(A_{12}+A_{21}))\\ &=&\phi(\mbox{i}I\diamond\ldots\diamond I\diamond(P_{1}-P_{2})\diamond A_{12})+\phi(\mbox{i}I\diamond\ldots\diamond I\diamond(P_{1}-P_{2})\diamond A_{21})\\ &=&\phi(\mbox{i}I)\diamond\ldots\diamond I\diamond(P_{1}-P_{2})\diamond(A_{12}+A_{21})+\ldots+\mbox{i}I\diamond\ldots\diamond I\diamond\phi(P_{1}-P_{2})\diamond(A_{12}+A_{21})\\ &&+\mbox{i}I\diamond\ldots\diamond I\diamond(P_{1}-P_{2})\diamond(\phi(A_{12})+\phi(A_{21})). \end{eqnarray*} $

由此可得$ \mbox{i}I\diamond\ldots\diamond I\diamond(P_{1}-P_{2})\diamond T=0, $所以$ T_{11}=T_{22}=0. $

又由断言2.1得

$ \begin{eqnarray*} &&\phi(\mbox{i}I)\diamond\ldots\diamond I\diamond(A_{12}+A_{21})\diamond P_{1}+\ldots+\mbox{i}I\diamond\ldots\diamond I\diamond\phi(A_{12}+A_{21})\diamond P_{1}\\ &&+\mbox{i}I\diamond\ldots\diamond I\diamond(A_{12}+A_{21})\diamond\phi(P_{1})\\ &=&\phi(\mbox{i}I\diamond\ldots\diamond I\diamond(A_{12}+A_{21})\diamond P_{1})\\ &=&\phi(\mbox{i}I\diamond\ldots\diamond I\diamond A_{12}\diamond P_{1})+\phi(\mbox{i}I\diamond\ldots\diamond I\diamond A_{21}\diamond P_{1})\\ &=&\phi(\mbox{i}I)\diamond\ldots\diamond I\diamond(A_{12}+A_{21})\diamond P_{1}+\ldots+\mbox{i}I\diamond\ldots\diamond I\diamond(\phi(A_{12})+\phi(A_{21}))\diamond P_{1}\\ &&+\mbox{i}I\diamond\ldots\diamond I\diamond(A_{12}+A_{21})\diamond\phi(P_{1}). \end{eqnarray*} $

由此可得$ \mbox{i}I\diamond\ldots\diamond I\diamond T\diamond P_{1}=0, $所以$ T_{21}=0. $类似可得$ T_{12}=0. $

断言2.3  任取$ A_{11}\in{\cal A}_{11}, A_{12}\in{\cal A}_{12}, A_{21}\in{\cal A}_{21}, A_{22}\in{\cal A}_{22}, $则$ \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}). $

令$ T=\phi(A_{11}+A_{12}+A_{21})-\phi(A_{11})-\phi(A_{12})-\phi(A_{21}). $由断言2.1和断言2.2得

$ \begin{eqnarray*} &&\phi(\mbox{i}I)\diamond\ldots\diamond I\diamond P_{2}\diamond(A_{11}+A_{12}+A_{21})+\ldots+\mbox{i}I\diamond\ldots\diamond I\diamond\phi(P_{2})\diamond(A_{11}+A_{12}+A_{21})\\ &&+\mbox{i}I\diamond\ldots\diamond I\diamond P_{2}\diamond\phi(A_{11}+A_{12}+A_{21})\\ &=&\phi(\mbox{i}I\diamond\ldots\diamond I\diamond P_{2}\diamond(A_{11}+A_{12}+A_{21}))\\ &=&\phi(\mbox{i}I\diamond\ldots\diamond I\diamond P_{2}\diamond A_{11})+\phi(\mbox{i}I\diamond\ldots\diamond I\diamond P_{2}\diamond (A_{12}+A_{21}))\\ &=&\phi(\mbox{i}I)\diamond\ldots\diamond I\diamond P_{2}\diamond(A_{11}+A_{12}+A_{21})+\ldots+\mbox{i}I\diamond\ldots\diamond I\diamond\phi(P_{2})\diamond(A_{11}+A_{12}+A_{21})\\ &&+\mbox{i}I\diamond\ldots\diamond I\diamond P_{2}\diamond(\phi(A_{11})+\phi(A_{12})+\phi(A_{21})). \end{eqnarray*} $

由此可得$ \mbox{i}I\diamond\ldots\diamond I\diamond P_{2}\diamond T=0, $所以$ T_{12}=T_{21}=T_{22}=0. $

又由断言2.1得

$ \begin{eqnarray*} &&\phi(\mbox{i}I)\diamond\ldots\diamond(P_{1}-P_{2})\diamond(A_{11}+A_{12}+A_{21})+\ldots \\ &&+\mbox{i}I\diamond\ldots\diamond\phi(P_{1}-P_{2})\diamond(A_{11}+A_{12}+A_{21}) +\mbox{i}I\diamond\ldots\diamond(P_{1}-P_{2})\diamond\phi(A_{11}+A_{12}+A_{21})\\ &=&\phi(\mbox{i}I\diamond\ldots\diamond(P_{1}-P_{2})\diamond(A_{11}+A_{12}+A_{21}))\\ &=&\phi(\mbox{i}I\diamond\ldots\diamond(P_{1}-P_{2})\diamond A_{11})\\ &&+\phi(\mbox{i}I\diamond\ldots\diamond(P_{1}-P_{2})\diamond A_{12})+\phi(\mbox{i}I\diamond\ldots\diamond(P_{1}-P_{2})\diamond A_{21})\\ &=&\phi(\mbox{i}I)\diamond\ldots\diamond(P_{1}-P_{2})\diamond(A_{11}+A_{12}+A_{21})+\ldots \\ &&+\mbox{i}I\diamond\ldots\diamond\phi(P_{1}-P_{2})\diamond(A_{11}+A_{12}+A_{21})\\ &&+\mbox{i}I\diamond\ldots\diamond(P_{1}-P_{2})\diamond(\phi(A_{11})+\phi(A_{12})+\phi(A_{21})). \end{eqnarray*} $

由此可得$ \mbox{i}I\diamond\ldots\diamond(P_{1}-P_{2})\diamond T=0, $所以$ T_{11}=0. $类似可以证明第二种情况成立.

断言2.4  任取$ A_{11}\in{\cal A}_{11}, A_{12}\in{\cal A}_{12}, A_{21}\in{\cal A}_{21}, A_{22}\in{\cal A}_{22}, $则$ \phi(A_{11}+A_{12}+A_{21}+A_{22})=\phi(A_{11})+\phi(A_{12})+\phi(A_{21})+\phi(A_{22}) $

设$ T=\phi(A_{11}+A_{12}+A_{21}+A_{22})-\phi(A_{11})-\phi(A_{12})-\phi(A_{21})-\phi(A_{22}). $由断言2.1和断言2.3得

$ \begin{eqnarray*} &&\phi(\mbox{i}I)\diamond\ldots\diamond I\diamond P_{2}\diamond(A_{11}+A_{12}+A_{21}+A_{22})+\ldots+\mbox{i}I\diamond\ldots\diamond I\diamond\phi(P_{2})\\ &&\diamond(A_{11}+A_{12}+A_{21}+A_{22})+\mbox{i}I\diamond\ldots\diamond I\diamond P_{2}\diamond\phi(A_{11}+A_{12}+A_{21}+A_{22})\\ &=&\phi(\mbox{i}I\diamond\ldots\diamond I\diamond P_{2}\diamond(A_{11}+A_{12}+A_{21}+A_{22}))\\ &=&\phi(\mbox{i}I\diamond\ldots\diamond I\diamond P_{2}\diamond A_{11})+\phi(\mbox{i}I\diamond\ldots\diamond I\diamond P_{2}\diamond(A_{12}+A_{21}+A_{22}))\\ &=&\phi(\mbox{i}I)\diamond\ldots\diamond I\diamond P_{2}\diamond(A_{11}+A_{12}+A_{21}+A_{22})+\ldots\\ &&+\mbox{i}I\diamond\ldots\diamond I\diamond\phi(P_{2}) \diamond(A_{11}+A_{12}+A_{21}+A_{22})\\ &&+\mbox{i}I\diamond\ldots\diamond I\diamond P_{2}\diamond(\phi(A_{11})+\phi(A_{12})+\phi(A_{21})+\phi(A_{22})). \end{eqnarray*} $

由此可得$ \mbox{i}I\diamond\ldots\diamond I\diamond P_{2}\diamond T=0, $所以$ T_{12}=T_{21}=T_{22}=0. $ 类似可得$ T_{11}=0. $

断言2.5  设$ i, j\in\{1, 2\}, $若$ i\neq j, A_{ij}, B_{ij}\in{\cal A}_{ij}, $则有$ \phi(A_{ij}+B_{ij})=\phi(A_{ij})+\phi(B_{ij}). $

由断言2.4得

$ \begin{eqnarray*} &&\phi(A_{ij}+B_{ij})+\phi(A_{ij}^{*})+\phi(B_{ij}A_{ij}^{*})=\phi(\mbox{i}I\diamond\ldots\diamond I\diamond\frac{P_{i}+A_{ij}}{2^{n-2}}\diamond(-\mbox{i}(P_{j}+B_{ij})))\\ &=&\phi(\mbox{i}I)\diamond\ldots\diamond I\diamond\frac{P_{i}+A_{ij}}{2^{n-2}}\diamond(-\mbox{i}(P_{j}+B_{ij}))+\ldots\\ &&+\mbox{i}I\diamond\ldots\diamond I\diamond(\phi(\frac{P_{i}}{2^{n-2}})+\phi(\frac{A_{ij}}{2^{n-2}}))\diamond(-\mbox{i}(P_{j}+B_{ij}))\\ &&+\mbox{i}I\diamond\ldots\diamond I\diamond\frac{P_{i}+A_{ij}}{2^{n-2}}\diamond(\phi(-\mbox{i}P_{j})+\phi(-\mbox{i}B_{ij}))\\ &=&\phi(\mbox{i}I\diamond\ldots\diamond I\diamond\frac{P_{i}}{2^{n-2}}\diamond -\mbox{i}P_{j})+\phi(\mbox{i}I\diamond\ldots\diamond I\diamond\frac{P_{i}}{2^{n-2}}\diamond -\mbox{i}B_{ij})\\ &&+\phi(\mbox{i}I\diamond\ldots\diamond I\diamond\frac{A_{ij}}{2^{n-2}}\diamond -\mbox{i}P_{j})+\phi(\mbox{i}I\diamond\ldots\diamond I\diamond\frac{A_{ij}}{2^{n-2}}\diamond -\mbox{i}B_{ij})\\ &=&\phi(B_{ij})+\phi(A_{ij}+A_{ij}^{*})+\phi(B_{ij}A_{ij}^{*})=\phi(B_{ij})+\phi(A_{ij})+\phi(A_{ij}^{*})+\phi(B_{ij}A_{ij}^{*}), \end{eqnarray*} $

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

断言2.6  设$ i\in\{1, 2\}, $若$ A_{ii}, B_{ii}\in{\cal A}_{ii}, $则有$ \phi(A_{ii}+B_{ii})=\phi(A_{ii})+\phi(B_{ii}). $

设$ T=\phi(A_{ii}+B_{ii})-\phi(A_{ii})-\phi(B_{ii}). $令$ i\neq j, $由断言2.1得

$ \begin{eqnarray*} &&\phi(\mbox{i}I)\diamond\ldots\diamond I\diamond P_{j}\diamond(A_{ii}+B_{ii})+\ldots+\mbox{i}I\diamond\ldots\diamond I\diamond\phi(P_{j})\diamond(A_{ii}+B_{ii})\\ &&+\mbox{i}I\diamond\ldots\diamond I\diamond P_{j}\diamond\phi(A_{ii}+B_{ii})\\ &=&\phi(\mbox{i}I\diamond\ldots\diamond I\diamond P_{j}\diamond(A_{ii}+B_{ii}))\\ &=&\phi(\mbox{i}I\diamond\ldots\diamond I\diamond P_{j}\diamond A_{ii})+\phi(\mbox{i}I\diamond\ldots\diamond I\diamond P_{j}\diamond B_{ii})\\ &=&\phi(\mbox{i}I)\diamond\ldots\diamond I\diamond P_{j}\diamond(A_{ii}+B_{ii})+\ldots+\mbox{i}I\diamond\ldots\diamond I\diamond\phi(P_{j})\diamond(A_{ii}+B_{ii})\\ &&+\mbox{i}I\diamond\ldots\diamond I\diamond P_{j}\diamond(\phi(A_{ii})+\phi(B_{ii})), \end{eqnarray*} $

由此可得$ \mbox{i}I\diamond\ldots\diamond I\diamond P_{j}\diamond T=0, $所以$ T_{ij}=T_{ji}=T_{jj}=0. $

任取$ C_{ij}\in{\cal A}_{ij}, i\neq j, $由断言2.5得

$ \begin{eqnarray*} &&\phi(\mbox{i}I)\diamond\ldots\diamond I\diamond(A_{ii}+B_{ii})\diamond C_{ij}+\ldots+\mbox{i}I\diamond\ldots\diamond I\diamond\phi(A_{ii}+B_{ii})\diamond C_{ij}\\ &&+\mbox{i}I\diamond\ldots\diamond I\diamond(A_{ii}+B_{ii})\diamond\phi(C_{ij})\\ &=&\phi(\mbox{i}I\diamond\ldots\diamond I\diamond(A_{ii}+B_{ii})\diamond C_{ij})\\ &=&\phi(\mbox{i}I\diamond\ldots\diamond I\diamond A_{ii}\diamond C_{ij})+\phi(\mbox{i}I\diamond\ldots\diamond I\diamond B_{ii}\diamond C_{ij})\\ &=&\phi(\mbox{i}I)\diamond\ldots\diamond I\diamond(A_{ii}+B_{ii})\diamond C_{ij}+\ldots+\mbox{i}I\diamond\ldots\diamond I\diamond(\phi(A_{ii})+\phi(B_{ii}))\diamond C_{ij}\\ &&+\mbox{i}I\diamond\ldots\diamond I\diamond(A_{ii}+B_{ii})\diamond\phi(C_{ij}). \end{eqnarray*} $

由此可得$ \mbox{i}I\diamond\ldots\diamond I\diamond T_{ii}\diamond C_{ij}=0, $则$ T_{ii}C_{ij}=0, $即对所有$ C\in{\cal A}, $有$ T_{ii}CP_{j}=0, $由已知得$ T_{ii}=0. $

断言2.7  $ \phi $是可加的.

由断言2.4–2.6得$ \phi $是可加的.

断言2.8  $ \phi(I)=0, \phi(\mbox{i}I)=0. $

由$ 0=\phi(I\diamond\ldots\diamond I\diamond I)=\phi(I)\diamond\ldots\diamond I\diamond I=2^{n-2}(\phi(I)-\phi(I)^{*}) $得$ \phi(I)=\phi(I)^{*}. $另一方面, 由断言2.7得

$ \begin{eqnarray*} 2^{n-1}\phi(\mbox{i}I)&=&\phi(\mbox{i}I\diamond\ldots\diamond I\diamond I)\\ &=&\phi(\mbox{i}I)\diamond\ldots\diamond I\diamond I+\ldots+\mbox{i}I\diamond\ldots\diamond\phi(I)\diamond I +\mbox{i}I\diamond\ldots\diamond I\diamond\phi(I)\\ &=&2^{n-2}(\phi(\mbox{i}I)-\phi(\mbox{i}I)^{*})+(n-1)2^{n-1}\mbox{i}\phi(I), \end{eqnarray*} $

即$ \phi(\mbox{i}I)=-\phi(\mbox{i}I)^{*}+2(n-1)\mbox{i}\phi(I), $进而$ \phi(\mbox{i}I)^{*}=-\phi(\mbox{i}I)-2(n-1)\mbox{i}\phi(I)=\phi(\mbox{i}I)^{*}-4(n-1)\mbox{i}\phi(I), $所以$ \phi(I)=0, \phi(\mbox{i}I)^{*}=-\phi(\mbox{i}I). $由

$ \begin{eqnarray*} 0&=&-2^{n-1}\phi(I)=\phi(\mbox{i}I\diamond I\diamond\ldots\diamond I\diamond\mbox{i}I)\\ &=&\phi(\mbox{i}I)\diamond I\diamond\ldots\diamond I\diamond\mbox{i}I+\mbox{i}I\diamond I\diamond\ldots\diamond I\diamond\phi(\mbox{i}I)=2^{n}\mbox{i}\phi(\mbox{i}I) \end{eqnarray*} $

得$ \phi(\mbox{i}I)=0. $

断言2.9  $ \phi(P_{i})=P_{i}\phi(P_{i})P_{j}+P_{j}\phi(P_{i})P_{i}, i, j=1, 2. $

任取$ A\in{\cal A}, $由$ 2^{n-1}\phi(\mbox{i}A)=\phi(\mbox{i}I\diamond I\diamond\ldots\diamond I\diamond A)=\mbox{i}I\diamond I\diamond\ldots\diamond I\diamond\phi(A)=2^{n-1}\mbox{i}\phi(A) $得

(2.1) $ \begin{align} \phi(\mbox{i}A)=\mbox{i}\phi(A). \end{align} $

任取$ A, B\in{\cal A}, $由(2.1) 式和断言2.8得

$ \begin{eqnarray*} 2^{n-2}\mbox{i}\phi(A\bullet B)&=&2^{n-2}\phi(\mbox{i}A\diamond B)=\phi(\mbox{i}I\diamond I\diamond\ldots\diamond I\diamond A\diamond B)\\ &=&\mbox{i}I\diamond I\diamond\ldots\diamond I\diamond \phi(A)\diamond B+\mbox{i}I\diamond I\diamond\ldots\diamond I\diamond A\diamond \phi(B)\\ &=&2^{n-2}\mbox{i}(\phi(A)\bullet B+A\bullet\phi(B)), \end{eqnarray*} $

所以

(2.2) $ \begin{align} \phi(A\bullet B)=\phi(A)\bullet B+A\bullet\phi(B). \end{align} $

任取$ A\in{\cal A}, $由(2.2) 式和断言2.8得$ \phi(A)+\phi(A^{*})=\phi(A\bullet I)=\phi(A)\bullet I=\phi(A)+\phi(A)^{*}, $所以

(2.3) $ \begin{align} \phi(A^{*})=\phi(A)^{*}. \end{align} $

(2.2) 式和(2.3) 式得$ 2\phi(P_{i})=\phi(P_{i}\bullet P_{i})=2(\phi(P_{i})P_{i}+P_{i}\phi(P_{i})), $由此可得$ \phi(P_{i})=P_{i}\phi(P_{i})P_{j}+P_{j}\phi(P_{i})P_{i}. $

设$ T=P_{i}\phi(P_{i})P_{j}-P_{j}\phi(P_{i})P_{i}, $则$ T^{*}=-T. $对所有$ A\in{\cal A}, $定义映射$ \varphi:{\cal A\rightarrow A} $为$ \varphi(A)=\phi(A)-(AT-TA). $易得$ \varphi $有下面性质.

断言2.10   (1) 对所有$ A, B\in{\cal A}, $有$ \varphi(A\bullet B)=\varphi(A)\bullet B+A\bullet\varphi(B). $

(2) $ \varphi(P_{i})=0, i=1, 2. $

(3) $ \varphi $是可加的.

(4) 对所有$ A\in{\cal A}, $有$ \varphi(A^{*})=\varphi(A)^{*}. $

(5) $ \varphi $是可加的导子当且仅当$ \phi $是可加的导子.

根据文献[3] 的断言14–16得定理2.1成立. 证毕.

定理3.1    设$ {\cal A} $是含单位元$ I $和非平凡投影$ P $的$ * $- 代数, $ \xi\in{\Bbb C}\setminus\{0\}, $满足: (1) 若$ X{\cal A} P=0, $则$ X=0; $ (2) 若$ X{\cal A}(I-P)=0, $则$ X=0. $如果映射$ \phi:{\cal A\rightarrow A} $满足对所有$ A_{n-1}, A_{n}\in{\cal A}, A_{1}=I, A_{2}=\ldots=A_{n-2}=\frac{1}{1+\xi}I, n\geq 2, $有$ \phi(A_{1}\diamond A_{2}\diamond\ldots\diamond A_{n})=\Sigma_{k=1}^{n}A_{1}\diamond\ldots\diamond A_{k-1}\diamond\phi(A_{k})\diamond A_{k+1}\diamond\ldots\diamond A_{n}, $那么$ \phi $是可加的.

证   当$ n=2 $时, 按照文献[5] 的证明可得定理3.1成立, 下面的证明中假设$ n\geq 3. $当$ \xi=1 $时文献[3] 已得到了结果. 当$ \xi=-1 $时定理2.1已给予证明. 下面的证明中假设$ \xi\neq -1. $

断言3.1   $ \phi(0)=0. $

$ \begin{eqnarray*} \phi(0)=\phi(I\diamond\frac{1}{1+\xi}I\diamond\ldots\diamond\frac{1}{1+\xi}I\diamond 0\diamond 0)=0. \end{eqnarray*} $

断言3.2  任取$ A_{12}\in {\cal A}_{12}, A_{21}\in {\cal A}_{21}, $则$ \phi(A_{12}+A_{21})=\phi(A_{12})+\phi(A_{21}). $

设$ T=\phi(A_{12}+A_{21})-\phi(A_{12})-\phi(A_{21}). $由$ (P_{1}-\frac{1}{\overline{\xi}}P_{2})\diamond A_{12}=0 $和断言3.1得

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

则$ I\diamond \frac{1}{1+\xi}I\diamond\ldots\diamond \frac{1}{1+\xi}I\diamond\frac{P_{1}-\frac{1}{\overline{\xi}}P_{2}}{1+\xi}\diamond 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. $由于$ \xi\neq 0, -1, $所以$ T_{11}=T_{22}=0. $

由$ A_{12}\diamond P_{1}=0 $得

$ \begin{eqnarray*} &&\phi(I)\diamond \frac{1}{1+\xi}I\diamond\ldots\diamond \frac{1}{1+\xi}I\diamond (A_{12}+A_{21})\diamond P_{1}+\ldots \\ &&+I\diamond \frac{1}{1+\xi}I\diamond\ldots\diamond \frac{1}{1+\xi}I\diamond\phi(A_{12}+A_{21})\diamond P_{1}\\ &&+I\diamond \frac{1}{1+\xi}I\diamond\ldots\diamond \frac{1}{1+\xi}I\diamond(A_{12}+A_{21})\diamond\phi(P_{1})\\ &=&\phi(I\diamond \frac{1}{1+\xi}I\diamond\ldots\diamond \frac{1}{1+\xi}I\diamond(A_{12}+A_{21})\diamond P_{1})\\ &=&\phi(I\diamond \frac{1}{1+\xi}I\diamond\ldots \diamond \frac{1}{1+\xi}I \diamond A_{12}\diamond P_{1})\\ &&+\phi(I\diamond \frac{1}{1+\xi}I\diamond\ldots\diamond \frac{1}{1+\xi}I\diamond A_{21}\diamond P_{1})\\ &=&\phi(I)\diamond \frac{1}{1+\xi}I\diamond\ldots\diamond \frac{1}{1+\xi}I\diamond (A_{12}+A_{21})\diamond P_{1}+\ldots \\ &&+I\diamond \frac{1}{1+\xi}I\diamond\ldots\diamond \frac{1}{1+\xi}I\diamond(\phi(A_{12})+\phi(A_{21}))\diamond P_{1}\\ &&+I\diamond \frac{1}{1+\xi}I\diamond\ldots\diamond \frac{1}{1+\xi}I\diamond (A_{12}+A_{21})\diamond \phi(P_{1}), \end{eqnarray*} $

即$ I\diamond \frac{1}{1+\xi}I\diamond\ldots \diamond \frac{1}{1+\xi}I\diamond T\diamond P_{1}=0, $由此可得$ (1+\xi)T_{21}+(\xi+|\xi|^{2})T_{21}^{*}=0, $所以$ T_{21}=0. $类似可得$ T_{12}=0. $

断言3.3  设$ i, j, k\in\{1, 2\}, i\neq j. $对所有$ A_{kk}\in {\cal A}_{kk}, A_{ij}\in {\cal A}_{ij}, $有$ \phi(A_{kk}+A_{ij})=\phi(A_{kk})+\phi(A_{ij}). $

下面只证明情形$ i=k=1, j=2, $其他情形类似可证. 设$ T=\phi(A_{11}+A_{12})-\phi(A_{11})-\phi(A_{12}). $由$ P_{2}\diamond A_{11}=0 $和断言3.1得

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

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

另一方面, 由$ (P_{1}-\frac{1}{\overline{\xi}}P_{2})\diamond A_{12}=0 $得

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

即$ I\diamond \frac{1}{1+\xi}I\diamond\ldots\diamond \frac{1}{1+\xi}I\diamond\frac{P_{1}-\frac{1}{\overline{\xi}}P_{2}}{1+\xi}\diamond T=0, $由此可得$ T_{11}=0. $

断言3.4  任取$ A_{11}\in {\cal A}_{11}, A_{12}\in {\cal A}_{12}, A_{21}\in {\cal A}_{21}, A_{22}\in {\cal A}_{22}, $则$ \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}). $

设$ T=\phi(A_{11}+A_{12}+A_{21})-\phi(A_{11})-\phi(A_{12})-\phi(A_{21}). $由断言3.1和断言3.2得

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

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

另一方面, 由$ (-\frac{1}{\overline{\xi}}P_{1}+P_{2})\diamond A_{21}=0 $和断言3.3得

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

即$ I\diamond \frac{1}{1+\xi}I\diamond\ldots\diamond \frac{1}{1+\xi}I\diamond\frac{-\frac{1}{\overline{\xi}}P_{1}+P_{2}}{1+\xi}\diamond T=0, $由此可得$ T_{11}=0, $所以$ \phi(A_{11}+A_{12}+A_{21})=\phi(A_{11})+\phi(A_{12})+\phi(A_{21}). $第二种情况类似可证.

断言3.5  任取$ A_{11}\in {\cal A}_{11}, A_{12}\in {\cal A}_{12}, A_{21}\in {\cal A}_{21}, A_{22}\in {\cal A}_{22}, $则$ \phi(A_{11}+A_{12}+A_{21}+A_{22})=\phi(A_{11})+\phi(A_{12})+\phi(A_{21})+\phi(A_{22}). $

设$ T=\phi(A_{11}+A_{12}+A_{21}+A_{22})-\phi(A_{11})-\phi(A_{12})-\phi(A_{21})-\phi(A_{22}). $由断言3.1和断言3.4得

$ \begin{eqnarray*} &&\phi(I)\diamond \frac{1}{1+\xi}I\diamond\ldots\diamond \frac{1}{1+\xi}I\diamond\frac{P_{1}}{1+\xi}\diamond(A_{11}+A_{12}+A_{21}+A_{22})+\ldots\\ &&+I\diamond \frac{1}{1+\xi}I\diamond\ldots\diamond \frac{1}{1+\xi}I\diamond\phi(\frac{P_{1}}{1+\xi})\diamond(A_{11}+A_{12}+A_{21}+A_{22})\\ &&+I\diamond \frac{1}{1+\xi}I\diamond\ldots\diamond \frac{1}{1+\xi}I\diamond\frac{P_{1}}{1+\xi}\diamond\phi(A_{11}+A_{12}+A_{21}+A_{22})\\ &=&\phi(I\diamond \frac{1}{1+\xi}I\diamond\ldots\diamond \frac{1}{1+\xi}I\diamond\frac{P_{1}}{1+\xi}\diamond(A_{11}+A_{12}+A_{21}+A_{22}))\\ &=&\phi(I\diamond \frac{1}{1+\xi}I\diamond\ldots\diamond \frac{1}{1+\xi}I\diamond\frac{P_{1}}{1+\xi}\diamond(A_{11}+A_{12}+A_{21})) \\ &&+\phi(I\diamond \frac{1}{1+\xi}I\diamond\ldots\diamond \frac{1}{1+\xi}I\diamond\frac{P_{1}}{1+\xi}\diamond A_{22})\\ &=&\phi(I)\diamond \frac{1}{1+\xi}I\diamond\ldots\diamond \frac{1}{1+\xi}I\diamond\frac{P_{1}}{1+\xi}\diamond(A_{11}+A_{12}+A_{21}+A_{22})+\ldots\\ &&+I\diamond \frac{1}{1+\xi}I\diamond\ldots\diamond \frac{1}{1+\xi}I\diamond\phi(\frac{P_{1}}{1+\xi})\diamond(A_{11}+A_{12}+A_{21}+A_{22})\\ &&+I\diamond \frac{1}{1+\xi}I\diamond\ldots\diamond \frac{1}{1+\xi}I\diamond\frac{P_{1}}{1+\xi}\diamond(\phi(A_{11})+\phi(A_{12})+\phi(A_{21})+\phi(A_{22})), \end{eqnarray*} $

即$ I\diamond \frac{1}{1+\xi}I\diamond\ldots\diamond \frac{1}{1+\xi}I\diamond\frac{P_{1}}{1+\xi}\diamond T=0, $由此可得$ T_{11}=T_{12}=T_{21}=0. $类似可证$ T_{22}=0. $

断言3.6  设$ i, j\in\{1, 2\}, i\neq j. $对所有$ A_{ij}, B_{ij}\in {\cal A}_{ij}, $有$ \phi(A_{ij}+B_{ij})=\phi(A_{ij})+\phi(B_{ij}). $

由$ (P_{i}+A_{ij})\diamond(P_{j}+B_{ij})=A_{ij}+B_{ij}+\xi A_{ij}^{*}+\xi B_{ij}A_{ij}^{*} $和断言3.5得

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

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

断言3.7  任取$ A_{ii}, B_{ii}\in {\cal A}_{ii}, 1\leq i\leq 2, $则$ \phi(A_{ii}+B_{ii})=\phi(A_{ii})+\phi(B_{ii}). $

下面证明$ T=\phi(A_{ii}+B_{ii})-\phi(A_{ii})-\phi(B_{ii})=0. $设$ i\neq j, $由

$ \begin{eqnarray*} &&\phi(I)\diamond \frac{1}{1+\xi}I\diamond\ldots\diamond \frac{1}{1+\xi}I\diamond\frac{P_{j}}{1+\xi}\diamond(A_{ii}+B_{ii})+\ldots\\ &&+I\diamond \frac{1}{1+\xi}I\diamond\ldots\diamond \frac{1}{1+\xi}I\diamond\phi(\frac{P_{j}}{1+\xi})\diamond(A_{ii}+B_{ii})\\ &&+I\diamond \frac{1}{1+\xi}I\diamond\ldots\diamond \frac{1}{1+\xi}I\diamond\frac{P_{j}}{1+\xi}\diamond\phi(A_{ii}+B_{ii})\\ &=&\phi(I\diamond \frac{1}{1+\xi}I\diamond\ldots\diamond \frac{1}{1+\xi}I\diamond\frac{P_{j}}{1+\xi}\diamond(A_{ii}+B_{ii}))\\ &=&\phi(I\diamond \frac{1}{1+\xi}I\diamond\ldots\diamond \frac{1}{1+\xi}I\diamond\frac{P_{j}}{1+\xi}\diamond A_{ii})+\phi(I\diamond \frac{1}{1+\xi}I\diamond\ldots\diamond \frac{1}{1+\xi}I\diamond\frac{P_{j}}{1+\xi}\diamond B_{ii})\\ &=&\phi(I)\diamond \frac{1}{1+\xi}I\diamond\ldots\diamond \frac{1}{1+\xi}I\diamond\frac{P_{j}}{1+\xi}\diamond(A_{ii}+B_{ii})+\ldots\\ &&+I\diamond \frac{1}{1+\xi}I\diamond\ldots\diamond \frac{1}{1+\xi}I\diamond\phi(\frac{P_{j}}{1+\xi})\diamond(A_{ii}+B_{ii})\\ &&+I\diamond \frac{1}{1+\xi}I\diamond\ldots\diamond \frac{1}{1+\xi}I\diamond\frac{P_{j}}{1+\xi}\diamond(\phi(A_{ii})+\phi(B_{ii})) \end{eqnarray*} $

得$ I\diamond \frac{1}{1+\xi}I\diamond\ldots\diamond \frac{1}{1+\xi}I\diamond\frac{P_{j}}{1+\xi}\diamond T=0, $则$ T_{ij}=T_{ji}=T_{jj}=0. $

任取$ C_{ij}\in{\cal A}_{ij}, i\neq j, $由断言3.6得

$ \begin{eqnarray*} &&\phi(I)\diamond \frac{1}{1+\xi}I\diamond\ldots\diamond \frac{1}{1+\xi}I\diamond(A_{ii}+B_{ii})\diamond C_{ij}+\ldots\\ &&+I\diamond \frac{1}{1+\xi}I\diamond\ldots\diamond \frac{1}{1+\xi}I\diamond\phi(A_{ii}+B_{ii})\diamond C_{ij}\\ &&+I\diamond \frac{1}{1+\xi}I\diamond\ldots\diamond \frac{1}{1+\xi}I\diamond(A_{ii}+B_{ii})\diamond\phi(C_{ij})\\ &=&\phi(I\diamond \frac{1}{1+\xi}I\diamond\ldots\diamond \frac{1}{1+\xi}I\diamond(A_{ii}+B_{ii})\diamond C_{ij})\\ &=&\phi((1+\xi)A_{ii}C_{ij}+(1+\xi)B_{ii}C_{ij})=\phi((1+\xi)A_{ii}C_{ij})+\phi((1+\xi)B_{ii}C_{ij})\\ &=&\phi(I\diamond \frac{1}{1+\xi}I\diamond\ldots\diamond \frac{1}{1+\xi}I\diamond A_{ii}\diamond C_{ij})+\phi(I\diamond \frac{1}{1+\xi}I\diamond\ldots\diamond \frac{1}{1+\xi}I\diamond B_{ii}\diamond C_{ij})\\ &=&\phi(I)\diamond \frac{1}{1+\xi}I\diamond\ldots\diamond \frac{1}{1+\xi}I\diamond(A_{ii}+B_{ii})\diamond C_{ij}+\ldots\\ &&+I\diamond \frac{1}{1+\xi}I\diamond\ldots\diamond \frac{1}{1+\xi}I\diamond(\phi(A_{ii})+\phi(B_{ii}))\diamond C_{ij}\\ &&+I\diamond \frac{1}{1+\xi}I\diamond\ldots\diamond \frac{1}{1+\xi}I\diamond(A_{ii}+B_{ii})\diamond\phi(C_{ij}), \end{eqnarray*} $

即$ I\diamond \frac{1}{1+\xi}I\diamond\ldots\diamond \frac{1}{1+\xi}I\diamond T\diamond C_{ij}=0, $则$ (1+\xi)T_{ii}C_{ij}=0, $即对所有$ C\in{\cal A}, $有$ T_{ii}CP_{j}=0. $由已知得$ T_{ii}=0, $所以$ \phi(A_{ii}+B_{ii})=\phi(A_{ii})+\phi(B_{ii}). $

由断言3.5–3.7得$ \phi $是可加的. 定理3.1证毕.

引理3.1   设$ |\xi|\neq 0, 1, $则$ I\diamond I\diamond\ldots\diamond I\diamond I\neq 0. $

证   用数学归纳法. 当$ n=2 $时, 则$ I\diamond I=I+\xi I\neq 0. $当$ n = k $时, 假设$ \alpha I=I\diamond I\diamond\ldots\diamond I $和$ \alpha\neq 0. $当$ n=k+1 $时, 则$ I\diamond I\diamond\ldots\diamond I\diamond I=(\alpha+\xi \alpha^{*})I, $若$ \alpha+\xi \alpha^{*}=0, $则$ \xi=-\frac{\alpha}{\alpha^{*}}, $进而$ |\xi|^{2}=\xi\overline{\xi}=1, $与已知矛盾, 所以$ I\diamond I\diamond\ldots\diamond I\diamond I\neq 0. $证毕.

定理3.2   设$ {\cal A} $是含单位元$ I $和非平凡投影$ P $的$ * $- 代数, $ \xi\in{\Bbb C}\setminus\{0\}, $满足: (1) 若$ X{\cal A} P=0, $则$ X=0; $ (2) 若$ X{\cal A}(I-P)=0, $则$ X=0. $如果映射$ \phi:{\cal A\rightarrow A} $满足对所有$ A_{n-2}, A_{n-1}, A_{n}\in{\cal A}, A_{1}=I $或$ \mbox{i}I, A_{2}=\ldots=A_{n-3}=I, $有$ \phi(A_{1}\diamond A_{2}\diamond\ldots\diamond A_{n})=\Sigma_{k=1}^{n}A_{1}\diamond\ldots\diamond A_{k-1}\diamond\phi(A_{k})\diamond A_{k+1}\diamond\ldots\diamond A_{n}, $那么$ \phi $是可加的$ * $- 导子, 且对所有的$ A\in{\cal A}, $有$ \phi(\xi A)=\xi\phi(A) $.

证   当$ n=2 $时, 按照文献[5] 的证明可得定理3.2成立, 下面的证明中假设$ n\geq 3. $

当$ \xi=1, -1 $时, 已有结果, 下面证明中假设$ \xi\neq\pm 1. $分两种情况进行讨论.

情形1  $ |\xi|=1 $且$ \xi\neq \pm 1 $.

由

$ \begin{eqnarray*} 0&=&\phi(I\diamond \ldots \diamond I\diamond\mbox{i}I\diamond\mbox{i}I)\\ &=&\phi(I)\diamond \ldots \diamond I\diamond\mbox{i}I\diamond\mbox{i}I+\ldots +I\diamond \ldots \diamond \phi(I)\diamond\mbox{i}I\diamond\mbox{i}I\\ &&+I\diamond \ldots \diamond I\diamond\phi(\mbox{i}I)\diamond\mbox{i}I+I\diamond \ldots \diamond I\diamond\mbox{i}I\diamond\phi(\mbox{i}I)\\ &=&2^{n-3}\mbox{i}(1+\xi)(\phi(\mbox{i}I)+\phi(\mbox{i}I)^{*}) \end{eqnarray*} $

得$ \phi(\mbox{i}I)^{*}=-\phi(\mbox{i}I). $一方面, 由

$ \begin{eqnarray*} 2^{n-2}\phi(\mbox{i}(1-\xi)I)&=&\phi(2^{n-2}\mbox{i}(1-\xi)I)=\phi(\mbox{i}I\diamond I\diamond \ldots \diamond I)\\ &=&\phi(\mbox{i}I)\diamond I\diamond \ldots \diamond I+\mbox{i}I\diamond\phi(I)\diamond \ldots \diamond I+\ldots+\mbox{i}I\diamond I\diamond \ldots \diamond\phi(I)\\ &=&2^{n-2}(1-\xi)\phi(\mbox{i}I)+2^{n-3}(n-2)\mbox{i}(1-\xi)\phi(I)^{*}+2^{n-3}n\mbox{i}(1-\xi)\phi(I) \end{eqnarray*} $

得

(3.1) $ \begin{align} 2\phi(\mbox{i}(1-\xi)I)=2(1-\xi)\phi(\mbox{i}I)+(n-2)\mbox{i}(1-\xi)\phi(I)^{*}+n\mbox{i}(1-\xi)\phi(I). \end{align} $

另一方面, 由

$ \begin{eqnarray*} 2^{n-2}\phi(\mbox{i}(1+\xi)I)&=&\phi(2^{n-2}\mbox{i}(1+\xi)I)=\phi(I\diamond I\diamond \ldots \diamond I\diamond\mbox{i}I)\\ &=&\phi(I)\diamond I\diamond \ldots \diamond I\diamond\mbox{i}I+I\diamond\phi(I)\diamond \ldots \diamond I\diamond\mbox{i}I+\ldots+I\diamond I\diamond \ldots \diamond I\diamond\phi(\mbox{i}I)\\ &=&2^{n-3}\mbox{i}(n+(n-2)\xi)\phi(I)+2^{n-3}\mbox{i}(n-2+n\xi)\phi(I)^{*}+2^{n-2}(1+\xi)\phi(\mbox{i}I) \end{eqnarray*} $

得

(3.2) $ \begin{align} 2\phi(\mbox{i}(1+\xi)I)=2(1+\xi)\phi(\mbox{i}I)+\mbox{i}(n-2+n\xi)\phi(I)^{*}+\mbox{i}(n+(n-2)\xi)\phi(I). \end{align} $

注意到$ \xi+n-2\neq 0, $由(3.1) 式和(3.2) 式得

(3.3) $ \begin{align} \phi(I)^{*}=\frac{\xi-n}{\xi+n-2}\phi(I). \end{align} $

(3.3) 式取伴随得

$ \phi(I)=\frac{\overline{\xi}-n}{\overline{\xi}+n-2}\phi(I)^{*}=\frac{\overline{\xi}-n}{\overline{\xi}+n-2}\cdot\frac{\xi-n}{\xi+n-2}\phi(I). $

若$ \phi(I)\neq 0, $则$ \xi=1, $矛盾. 所以

(3.4) $ \begin{align} \phi(I)=0. \end{align} $

任取$ A\in{\cal A}, $由

$ \begin{eqnarray*} 2^{n-2}\phi((1+\xi)A)&=&\phi(2^{n-2}(1+\xi)A)=\phi(I\diamond I\diamond \ldots \diamond I\diamond A)\\ &=&I\diamond I\diamond \ldots \diamond I\diamond\phi(A)=2^{n-2}(1+\xi)\phi(A) \end{eqnarray*} $

得

(3.5) $ \begin{align} \phi((1+\xi)A)=(1+\xi)\phi(A). \end{align} $

对所有$ A, B\in{\cal A}, $有

$ \begin{eqnarray*} 2^{n-3}\phi((1+\xi)(AB+BA^{*}))&=&\phi(I\diamond I\diamond \ldots \diamond I\diamond A\diamond B)\\ &=&I\diamond I\diamond \ldots \diamond I\diamond\phi(A)\diamond B+I\diamond I\diamond \ldots \diamond I\diamond A\diamond\phi(B)\\ &=&2^{n-3}(1+\xi)(\phi(A)B+B\phi(A)^{*}+A\phi(B)+\phi(B)A^{*}). \end{eqnarray*} $

联合(3.5) 式得

(3.6) $ \begin{align} \phi(AB+BA^{*})=\phi(A)B+B\phi(A)^{*}+A\phi(B)+\phi(B)A^{*}. \end{align} $

(3.6) 式中取$ B=I $得

(3.7) $ \begin{align} \phi(A^{*})=\phi(A)^{*}. \end{align} $

由(3.4)–(3.5) 式和

$ \begin{eqnarray*} 2^{n-2}\phi((\xi-1)I)&=&\phi(\mbox{i}I\diamond I\diamond \ldots \diamond I\diamond \mbox{i}I)\\ &=&\phi(\mbox{i}I)\diamond I\diamond \ldots \diamond I\diamond \mbox{i}I+\mbox{i}I\diamond I\diamond \ldots \diamond I\diamond \phi(\mbox{i}I)=2^{n-1}\mbox{i}(1-\xi)\phi(\mbox{i}I) \end{eqnarray*} $

得

$ \phi(\mbox{i}I)=0. $

任取$ A=A^{*}\in{\cal A}, $由(3.6) 式和(3.7) 式得

$ 2\phi(\mbox{i}A)=\phi(A(\mbox{i}I)+(\mbox{i}I)A^{*})=\phi(A)\mbox{i}I+\mbox{i}I\phi(A)^{*}=2\mbox{i}\phi(A), $

即$ \phi(\mbox{i} A)=\mbox{i} \phi(A). $对所有$ A\in{\cal A}, $由$ A=A_{1}+\mbox{i}A_{2}, A_{1}=\frac{A+A^{*}}{2}, A_{2}=\frac{A-A^{*}}{2\mbox{i}} $得

$ \begin{eqnarray*} \phi(\mbox{i}A)&=&\phi(\mbox{i}(A_{1}+\mbox{i}A_{2}))=\phi(\mbox{i}A_{1}-A_{2})=\mbox{i} \phi(A_{1})-\phi(A_{2})=\mbox{i} \phi(A_{1})+\mbox{i}^{2}\phi(A_{2})\\ &=&\mbox{i} \phi(A_{1})+\mbox{i}\phi(\mbox{i}A_{2})=\mbox{i}\phi(A_{1}+\mbox{i}A_{2})=\mbox{i}\phi(A). \end{eqnarray*} $

任取$ A, B\in{\cal A}, $由(3.6) 式得

(3.8) $ \begin{align} \begin{array}[b]{rcl} \phi(-AB+BA^{*})&=&\phi((\mbox{i}A)(\mbox{i}B)+(\mbox{i}B)(\mbox{i}A)^{*})\\ &=&\phi(\mbox{i}A)(\mbox{i}B)+(\mbox{i}B)\phi(\mbox{i}A)^{*}+(\mbox{i}A)\phi(\mbox{i}B)+\phi(\mbox{i}B)(\mbox{i}A)^{*}\\ &=&-\phi(A)B+B\phi(A)^{*}-A\phi(B)+\phi(B)A^{*}. \end{array} \end{align} $

由(3.6) 式和(3.8) 式得$ \phi(AB)=\phi(A)B+A\phi(B). $所以$ \phi $是$ * $- 导子. 由(3.5) 式得对所有$ A\in{\cal A}, $有$ \phi(\xi A)=\xi\phi(A). $

情形2  $ |\xi|\neq 1. $

下面的证明中, 设$ \alpha_{n}I=\underbrace{I\diamond I\diamond\ldots\diamond I}\limits_{n}. $

由$ \phi(I\diamond\ldots\diamond I\diamond I\diamond\mbox{i}I\diamond\mbox{i}I)=\phi(I\diamond\ldots\diamond I\diamond\mbox{i}I\diamond\mbox{i}I\diamond I) $得

$ \begin{eqnarray*} &&\alpha_{n-3}\diamond\phi(I)\diamond\mbox{i}I\diamond\mbox{i}I+\alpha_{n-3}\diamond I\diamond\phi(\mbox{i}I)\diamond\mbox{i}I+\alpha_{n-3}\diamond I\diamond\mbox{i}I\diamond\phi(\mbox{i}I)\\ &=&\alpha_{n-3}\diamond\phi(\mbox{i}I)\diamond\mbox{i}I\diamond I+\alpha_{n-3}\diamond\mbox{i}I\diamond\phi(\mbox{i}I)\diamond I+\alpha_{n-3}\diamond\mbox{i}I\diamond\mbox{i}I\diamond\phi(I), \end{eqnarray*} $

计算得

$ \begin{eqnarray*} &&\mbox{i}(2\alpha_{n-3}+3\alpha_{n-3}^{*}\xi-\alpha_{n-3}^{*}\xi|\xi|^{2})\phi(\mbox{i}I)+\mbox{i}(\alpha_{n-3}^{*}\xi +2\alpha_{n-3}|\xi|^{2} +\alpha_{n-3}^{*}\xi|\xi|^{2})\phi(\mbox{i}I)^{*} \\ &&+(\alpha_{n-3}+\alpha_{n-3}^{*}\xi)(|\xi|^{2}-1)\phi(I)\\ &=&\mbox{i}(2\alpha_{n-3}+\alpha_{n-3}^{*}\xi-\alpha_{n-3}^{*}\xi|\xi|^{2}-2\alpha_{n-3}|\xi|^{2})\phi(\mbox{i}I)\\ &&-\mbox{i}(\alpha_{n-3}^{*}\xi-\alpha_{n-3}^{*}\xi|\xi|^{2})\phi(\mbox{i}I)^{*}+(\alpha_{n-3}+\alpha_{n-3}^{*}\xi)(|\xi|^{2}-1)\phi(I), \end{eqnarray*} $

则$ \xi(\alpha_{n-3}^{*}+\alpha_{n-3}\overline{\xi})(\phi(\mbox{i}I)+\phi(\mbox{i}I)^{*})=0. $由引理3.1得$ (\alpha_{n-3}^{*}+\alpha_{n-3}\overline{\xi})^{*}=\alpha_{n-3}+\alpha_{n-3}^{*}\xi=\alpha_{n-2}\neq 0, $所以

(3.9) $ \begin{align} \phi(\mbox{i}I)^{*}=-\phi(\mbox{i}I). \end{align} $

另一方面, 由$ \phi(I\diamond\ldots\diamond I\diamond\mbox{i}I\diamond\mbox{i}I\diamond\mbox{i}I)=\phi(I\diamond\ldots\diamond I\diamond -I\diamond\mbox{i}I\diamond I) $得

$ \begin{eqnarray*} &&\alpha_{n-3}\diamond\phi(\mbox{i}I)\diamond\mbox{i}I\diamond\mbox{i}I+\alpha_{n-3}\diamond \mbox{i}I\diamond\phi(\mbox{i}I)\diamond\mbox{i}I+\alpha_{n-3}\diamond \mbox{i}I\diamond\mbox{i}I\diamond\phi(\mbox{i}I)\\ &=&\alpha_{n-3}\diamond\phi(-I)\diamond\mbox{i}I\diamond I+\alpha_{n-3}\diamond -I\diamond\phi(\mbox{i}I)\diamond I+\alpha_{n-3}\diamond-I\diamond\mbox{i}I\diamond\phi(I), \end{eqnarray*} $

结合(3.9) 式得$ (\alpha_{n-3}+\alpha_{n-3}^{*}\xi)(1-|\xi|^{2})\phi(\mbox{i}I)=(\alpha_{n-3}+\alpha_{n-3}^{*}\xi)(1-|\xi|^{2})\mbox{i}\phi(I), $由引理3.1得

(3.10) $ \begin{align} \phi(\mbox{i}I)=\mbox{i}\phi(I). \end{align} $

由(3.9) 式和(3.10) 式得

(3.11) $ \begin{align} \phi(I)^{*}=\phi(I). \end{align} $

由(3.11) 式得

$ \begin{eqnarray*} \phi(\alpha_{n}I)=\phi(I\diamond \ldots \diamond I\diamond I)=\phi(I)\diamond\ldots\diamond I\diamond I+\ldots+I\diamond \ldots\diamond I\diamond\phi(I)=n\alpha_{n}\phi(I) \end{eqnarray*} $

对所有的$ |\xi|\neq 1 $成立. 若$ \xi $为有理数, 则$ \alpha_{n}\phi(I)=n\alpha_{n}\phi(I). $如果$ \phi(I)\neq 0, $由引理3.1得$ n=1, $与题设矛盾. 所以

(3.12) $ \begin{align} \phi(I)=0. \end{align} $

由(3.10) 式得

(3.13) $ \begin{align} \phi(\mbox{i}I)=0. \end{align} $

对所有$ A\in{\cal A}, $因为$ \phi(I\diamond \ldots \diamond I\diamond A\diamond\mbox{i}I\diamond\mbox{i}I)=I\diamond \ldots \diamond I\diamond \phi(A)\diamond\mbox{i}I\diamond\mbox{i}I, $由此可得$ \phi(\alpha_{n-2}(|\xi|^{2}-1)A)=\alpha_{n-2}(|\xi|^{2}-1)\phi(A), $进而

$ \begin{eqnarray*} \alpha_{n-2}(1-|\xi|^{2})\phi(\mbox{i}A)&=&\phi(\alpha_{n-2}(1-|\xi|^{2})\mbox{i}A) =\phi(I\diamond\ldots\diamond I\diamond I\diamond \mbox{i}I\diamond A)\\ &=&I\diamond\ldots\diamond I\diamond I\diamond \mbox{i}I\diamond \phi(A)=\mbox{i}\alpha_{n-2}(1-|\xi|^{2})\phi(A). \end{eqnarray*} $

由引理3.1得

(3.14) $ \begin{align} \phi(\mbox{i}A)=\mbox{i}\phi(A). \end{align} $

另一方面

(3.15) $ \begin{align} \begin{array}[b]{rcl} \phi((-\alpha_{n-1}+\xi\alpha_{n-1}^{*}) A)&=&\phi(I\diamond \ldots \diamond I\diamond\mbox{i}I\diamond \mbox{i}A) =I\diamond \ldots \diamond I\diamond\mbox{i}I\diamond \phi(\mbox{i}A) \\ &=&(-\alpha_{n-1}+\xi\alpha_{n-1}^{*}) \phi(A). \end{array} \end{align} $

对所有$ A^{*}=A\in{\cal A}, $由(3.15) 式得

$ \begin{eqnarray*} (-\alpha_{n-1}+\xi\alpha_{n-1}^{*})\phi(A) &=&\phi((-\alpha_{n-1}+\xi\alpha_{n-1}^{*}) A)=\phi(I\diamond \ldots \diamond I\diamond\mbox{i}A\diamond \mbox{i}I)\\ &=&I\diamond\ldots\diamond \phi(\mbox{i}A)\diamond \mbox{i}I=-\alpha_{n-1}\phi(A)+\xi\alpha_{n-1}^{*}\phi(A)^{*}. \end{eqnarray*} $

由此可得$ \xi\alpha_{n-1}^{*}(\phi(A)-\phi(A)^{*})=0, $又由于$ \xi\alpha_{n-1}^{*}\neq 0, $所以$ \phi(A)=\phi(A)^{*} $对所有$ A^{*}=A\in{\cal A} $成立, 结合(3.14) 式, 对所有$ A\in{\cal A}, $由$ A=A_{1}+\mbox{i}A_{2}, A_{1}=\frac{A+A^{*}}{2}, A_{2}=\frac{A-A^{*}}{2\mbox{i}} $得

(3.16) $ \begin{align} \begin{array}[b]{rcl} \phi(A^{*})&=&\phi(A_{1}-\mbox{i}A_{2})=\phi(A_{1})-\phi(\mbox{i}A_{2})=\phi(A_{1})^{*}-\mbox{i}\phi(A_{2})^{*}\\ &=&\phi(A_{1})^{*}+(\mbox{i}\phi(A_{2}))^{*}=\phi(A_{1})^{*}+\phi(\mbox{i}A_{2})^{*}=\phi(A)^{*}. \end{array} \end{align} $

对所有$ A\in{\cal A}, $有

(3.17) $ \begin{align} \phi((\alpha_{n-1}+\xi\alpha_{n-1}^{*}) A)=\phi(I\diamond \ldots \diamond I\diamond I\diamond A)=(\alpha_{n-1}+\xi\alpha_{n-1}^{*}) \phi(A). \end{align} $

由(3.15) 式和(3.17) 式得

(3.18) $ \begin{align} \phi(\alpha_{n-1} A)=\alpha_{n-1}\phi(A) \end{align} $

和$ \phi(\xi\alpha_{n-1}^{*}A)=\xi\alpha_{n-1}^{*}\phi(A). $联合(3.16) 式得$ \xi\alpha_{n-1}^{*}\phi(A)=\phi(\xi\alpha_{n-1}^{*}A)=\phi((\alpha_{n-1}\overline{\xi} A^{*})^{*})=\phi(\alpha_{n-1}\overline{\xi} A^{*})^{*}=(\alpha_{n-1}\phi(\overline{\xi} A^{*}))^{*}=\alpha_{n-1}^{*}\phi(\xi A), $所以

(3.19) $ \begin{align} \phi(\xi A)=\xi\phi(A). \end{align} $

任取$ A, B\in{\cal A}, $有

(3.20) $ \begin{align} \begin{array}[b]{rcl} \phi(\alpha_{n-1}AB+\xi\alpha_{n-1}^{*}BA^{*}) &=&\phi(I\diamond \ldots \diamond I\diamond I\diamond A\diamond B)\\ &=&I\diamond\ldots\diamond I\diamond I\diamond\phi(A)\diamond B+I\diamond\ldots\diamond I\diamond I\diamond A\diamond \phi(B)\\ &=&\alpha_{n-1}(\phi(A)B+A\phi(B))+\xi\alpha_{n-1}^{*}(B\phi(A)^{*}+\phi(B)A^{*}). \end{array} \end{align} $

另一方面, 由(3.14) 式得

(3.21) $ \begin{align} \begin{array}[b]{rcl} \phi(-\alpha_{n-1}AB+\xi\alpha_{n-1}^{*}BA^{*}) &=&\phi(I\diamond \ldots \diamond I\diamond I\diamond \mbox{i}A\diamond \mbox{i}B)\\ &=&I\diamond\ldots\diamond I\diamond I\diamond\phi(\mbox{i}A)\diamond \mbox{i}B+I\diamond\ldots\diamond I\diamond I\diamond \mbox{i}A\diamond \phi(\mbox{i}B)\\ &=&-\alpha_{n-1}(\phi(A)B+A\phi(B))+\xi\alpha_{n-1}^{*}(B\phi(A)^{*}+\phi(B)A^{*}). \end{array} \end{align} $

由(3.20) 式和(3.21) 式得$ \phi(\alpha_{n-1}AB)=\alpha_{n-1}(\phi(A)B+A\phi(B)). $由(3.18) 式得$ \alpha_{n-1}\phi(AB)=\alpha_{n-1}(\phi(A)B+A\phi(B)). $由引理3.1得$ \alpha_{n-1}\neq 0, $所以$ \phi(AB)=\phi(A)B+A\phi(B). $证毕.