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

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

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

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

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

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

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

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

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

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

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

任取$A, B\in{\cal A},$由(2.1) 式和断言2.8得

所以

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

(2.2) 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   $\phi(0)=0.$

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

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

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

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

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

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

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

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

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

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

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

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

得

另一方面, 由

得

注意到$\xi+n-2\neq 0,$由(3.1) 式和(3.2) 式得

(3.3) 式取伴随得

若$\phi(I)\neq 0,$则$\xi=1,$矛盾. 所以

任取$A\in{\cal A},$由

得

对所有$A, B\in{\cal A},$有

联合(3.5) 式得

(3.6) 式中取$B=I$得

由(3.4)–(3.5) 式和

得

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

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

任取$A, B\in{\cal A},$由(3.6) 式得

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

计算得

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

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

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

由(3.9) 式和(3.10) 式得

由(3.11) 式得

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

由(3.10) 式得

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

由引理3.1得

另一方面

对所有$A^{*}=A\in{\cal A},$由(3.15) 式得

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

对所有$A\in{\cal A},$有

由(3.15) 式和(3.17) 式得

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

任取$A, B\in{\cal A},$有

另一方面, 由(3.14) 式得

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.