数学杂志  2025, Vol. 45 Issue (3): 271-282   PDF    
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张婷
余志强
胥兵
齐霄霏
$\mathcal{B}(H)$上可加导子的新刻画
张婷, 余志强, 胥兵, 齐霄霏    
山西大学数学与统计学院, 山西 太原 030006
摘要:令$H$是复数域$\mathbb{C}$上的Hilbert空间, $\mathcal{B}(H)$表示$H$上所有有界线性算子构成的代数, $C, D\in\mathcal{B}(H)$是任意两个固定算子. 本文研究了$\mathcal{B}(H)$上可加导子的局部刻画问题. 利用算子分解的方法, 证明了可加映射$\phi:\mathcal{B}(H)\to\mathcal{B}(H)$满足条件对任意$A, B\in\mathcal B(H)$, $AB=C$蕴涵$\phi(A)B+A\phi(B)=D$成立当且仅当存在$\mathcal{B}(H)$上的可加导子$\delta$与数$\lambda\in\mathbb{C}$使得$\phi(C)+\lambda C=D$$\phi(A)=\delta(A)+\lambda A$对所有$A\in\mathcal{B}(H)$成立. 该结果推广改进了已有的一些相关成果.
关键词可加导子    局部导子    有界线性算子    Hilbert空间    
A NEW CHARACTERIZATION OF ADDITIVE DERIVATIONS ON $\mathcal{B}(H)$
ZHANG Ting, YU Zhi-qiang, XU Bing, QI Xiao-fei    
School of Mathematics and Statistics, Shanxi University, Taiyuan 030006, China
Abstract: Let $H$ be any complex Hilbert space and $\mathcal{B}(H)$ the algebra of all bounded linear operators on $H$, and let $C, D\in\mathcal{B}(H)$ be any two fixed operators. In this paper, we study the question of local characterization of additive derivations on $\mathcal{B}(H)$. By using operator block methods, we show that an additive map $\delta:\mathcal B(H)\rightarrow\mathcal{B}(H)$ satisfies that, for any $A, B\in\mathcal B(H)$, $AB=C$ implies $\phi(A)B+A\phi(B)=D$, if and only if there exists some additive derivation $\delta$ on $\mathcal B(H)$ and $\lambda\in\mathbb{C}$ such that $\phi(C)+\lambda C=D$ and $\phi(A)=\delta(A)+\lambda A$ for all $A\in\mathcal{B}(H)$. Our result generalizes and improves some known related ones.
Keywords: Additive derivations     local derivations     bounded linear operators     Hilbert space    
1 引言

导子是环与代数上的重要映射之一, 其与上同调理论有着密切的联系. 令$ \mathcal{R} $是环或代数. 假设$ \delta $$ \mathcal{R} $上的可加或线性映射. 回忆, 若$ \delta $满足$ \delta(AB)=\delta(A)B+A\delta(B) $对所有$ A, B\in\mathcal{R} $成立, 则称$ \delta $是导子; 特别地, 若存在$ T\in\mathcal R $使得$ \delta(A)=AT-TA $对所有$ A\in\mathcal{R} $成立, 则称$ \delta $是内导子. 显然, 内导子一定是导子, 反之不一定成立. 关于不同环与代数上导子与内导子关系的刻画, 吸引了国内外许多学者对其展开研究. 譬如, Herstein在文献[1]中证明了满足一定条件的素环上的可加导子一定是内导子. Kadison在文献[2]中证明了von Neumann代数上的线性导子也是内导子. 令$ \mathcal{A} $$ C^* $-代数, $ \mathcal{M} $$ \mathcal{A} $上的Hilbert $ C^* $-模, $ {\rm End}_{\mathcal{A}}\mathcal{M} $$ \mathcal{M} $上的有界线性$ \mathcal{A} $-同态全体构成的Banach代数. 在一些条件的假设下, Moghadam和Miri等人在文献[3]中得到$ {\rm End}_{\mathcal{A}}\mathcal{M} $上的线性导子也是内导子. 关于其它相关成果, 可参见文献[46]及其里面的参考文献.

注意到, 在许多情形, 我们只能得到关于映射的部分信息. 这样, 一个自然的问题是, 如何从这些部分信息中获得映射的完整信息. 近三十多年来, 国内外学者致力于研究在何种条件下, 环与代数上的导子可以完全由这类映射的某种局部行为来决定. Kadison与Larson, Sourour分别在文献[7]和[8]中独立地引入了局部导子的概念, 即称$ \delta $是局部导子, 若$ \delta $满足对任意$ A\in\mathcal{R} $, 总存在可加或线性导子$ \theta_A:\mathcal{R}\to\mathcal{R} $使得$ \delta(A)=\theta_A(A) $成立. Kadison在文献[7]中得到了从任意von Neumann代数到其对偶Banach双模的连续线性局部导子一定是导子. Larson和Sourour在文献[8]中证明了$ \mathcal B(X) $ (Banach空间$ X $上所有有界线性算子构成的代数)上的线性局部导子是导子. Johnson在文献[9]中证明了任意$ C^* $-代数到其Banach双模的连续线性局部导子也是导子. 上世纪九十年代末到本世纪初, 国内外学者从另一角度展开对导子的研究, 并取得了丰富的研究成果. 譬如, Brešar在文献[10]中证明了含非平凡幂等元的素环$ \mathcal{R} $上的可加映射$ \delta $若满足条件对任意元$ A, B\in \mathcal{R} $, $ AB=0 $蕴涵$ \delta(A)B+A\delta(B)=0 $成立, 则$ \delta(A)=d(A)+\lambda A $对所有$ A\in \mathcal{R} $成立, 其中$ d:\mathcal{R}\to\mathcal{R} $是可加导子, $ \lambda $$ \mathcal{R} $的扩展中心中的元. 令$ X $是维数大于2的实或复Banach空间, $ \mathcal{L} $$ X $上的$ \mathcal{J} $-子空间格且$ \rm {Alg}\mathcal{L} $$ \mathcal{J} $-子空间格代数. Hou和Qi在文献[11]中证明了$ \rm {Alg}\mathcal{L} $上的线性映射$ \delta $满足对任意$ A, B\in \rm {Alg}\mathcal{L} $, $ AB=I $(单位算子)蕴涵$ \delta(A)B+A\delta(B)=\delta(I) $成立当且仅当$ \delta $是导子. 令$ \mathcal{A} $是含单位元和非平凡幂等元$ P $的Banach代数且$ \mathcal{M} $是含单位元的$ \mathcal{A} $双模. 在一些条件的假设下, Ghahramani在文献[12]中证明了可加映射$ \delta:\mathcal{A}\to\mathcal{M} $若满足对任意$ A, B\in \mathcal A $, $ AB=P $蕴涵$ \delta(A)B+A\delta(B)=\delta(P) $, 那么$ \delta $是导子. Zhu和Xiong等人在文献[13]中证明了$ \mathcal B(H) $ ($ H $为任意Hilbert空间)上的线性映射$ \delta $若满足对任意$ A, B\in \mathcal B(H) $, $ AB=C $蕴涵$ \delta(A)B+A\delta(B)=\delta(C) $成立(其中$ C\in\mathcal B(H) $为任意固定算子), 则$ \delta $是导子当且仅当$ C\ne 0 $. 令$ \mathcal{A} $是任意$ C^* $- 代数, $ X $是本性Banach $ \mathcal{A} $双模. 近来, Garcés和Khrypchenko在文献[14]中证明了有界线性映射$ \delta:\mathcal{A}\to\mathcal{A} $满足条件对任意$ A, B\in\mathcal A $, $ AB=0 $蕴涵$ \delta(A)B+A\delta(B)=0 $成立当且仅当存在线性导子$ d:\mathcal{A}\to X^{**} $$ \xi\in X^{**} $ ($ X^{**} $表示$ X $的二次共轭空间)使得$ a\xi=\xi a $$ \delta(a)=d(a)+\xi a $对所有$ a\in\mathcal{A} $成立. 对于其它相关研究成果, 可参见文献[1522]及其里面的参考文献.

2018年, Catalano在文献[23]中将上述研究导子局部行为的问题进行了更一般化的讨论. 假设$ \mathcal R $是特征不为2的除环, $ C, D\in \mathcal{R} $是任意两个固定元且$ C\ne0 $. Catalano证明了, $ \mathcal R $上的可加映射$ \delta $若满足条件对任意$ A, B\in \mathcal{R} $, $ AB=C $蕴涵$ \delta(A)B+A\delta(B)=D $成立, 那么$ \delta(A)=d(A)+ZA $对所有$ A\in \mathcal{R} $成立, 其中$ d:\mathcal{R}\to\mathcal{R} $是可加导子, $ Z\in \mathcal{R} $是中心元.

受以上工作的启发, 本文将继续研究$ \mathcal B(H) $上导子的局部行为与导子之间的关系. 下面是本文的主要结果.

定理1    令$ H $是复数域$ \mathbb{C} $上的Hilbert空间, $ C, D\in\mathcal{B}(H) $是任意两个固定算子. 假设$ \phi:\mathcal{B}(H)\to\mathcal{B}(H) $是可加映射. 则$ \phi $满足条件对任意$ A, B\in\mathcal{B}(H) $,

$ AB=C\ \mbox{ 蕴涵}\ \phi(A)B+A\phi(B)=D\ \mbox{成立}, $

当且仅当存在可加导子$ \delta:\mathcal{B}(H)\to\mathcal{B}(H) $$ \lambda\in\mathbb{C} $使得$ \phi(C)+\lambda C=D $$ \phi(A)=\delta(A)+\lambda A $对所有$ A\in\mathcal{B}(H) $成立.

注意到, 当$ H $是无限维Hilbert空间时, $ \mathcal{B}(H) $上的可加导子一定是内导子[24]. 利用定理1, 立得下面的结论.

定理2    令$ H $是复数域$ \mathbb{C} $上的无限维Hilbert空间, $ C, D\in\mathcal{B}(H) $是任意两个固定算子. 假设$ \phi:\mathcal{B}(H)\to\mathcal{B}(H) $是可加映射. 则$ \phi $满足条件对任意$ A, B\in\mathcal{B}(H) $,

$ AB=C\ \mbox{ 蕴涵}\ \phi(A)B+A\phi(B)=D\ \mbox{成立}, $

当且仅当存在$ T\in\mathcal{B}(H) $$ \lambda\in\mathbb{C} $使得$ \phi(C)+\lambda C=D $$ \phi(A)=AT-TA+\lambda A $对所有$ A\in\mathcal{B}(H) $成立.

$ H $是复数域$ \mathbb{C} $上的有限维Hilbert空间时, 不妨假设$ \dim H=n $, 则通过固定$ H $的一组基, $ \mathcal{B}(H) $同构于$ \mathbb{C} $上的矩阵代数$ \mathcal{M}_n(\mathbb{C}) $. 根据文献[24]对$ \mathcal{M}_n(\mathbb{C}) $上可加导子刻画的结论, 定理1可表述为:

定理3    假设$ \phi:\mathcal{M}_n(\mathbb{C})\to\mathcal{M}_n(\mathbb{C}) $是可加映射且$ C, D\in\mathcal{M}_n(\mathbb{C}) $是任意两个固定矩阵. 则$ \phi $满足条件对任意$ A, B\in\mathcal{M}_n(\mathbb{C}) $,

$ AB=C\ \mbox{蕴涵}\ \phi(A)B+A\phi(B)=D\ \mbox{成立}, $

当且仅当存在$ T\in\mathcal{M}_n(\mathbb{C}) $, 可加导子$ f:\mathbb{C}\to\mathbb{C} $$ \lambda\in\mathbb{C} $使得$ \phi(C)+\lambda C=D $$ \phi(A)=AT-TA+(f(a_{ij}))+\lambda A $对所有$ A\in\mathcal{M}_n(\mathbb{C}) $成立, 其中$ A=(a_{ij}) $.

特别地, 由以上结果可立得下面的推论, 即文献[13]中的结论.

推论4    令$ H $是复数域$ \mathbb{C} $上的Hilbert空间, $ C\in\mathcal{B}(H) $是任意固定算子. 假设$ \phi:\mathcal{B}(H)\to\mathcal{B}(H) $是可加映射. 则$ \phi $满足条件对任意$ A, B\in\mathcal{B}(H) $,

$ AB=C\ \mbox{ 蕴涵}\ \phi(A)B+A\phi(B)=\phi(C)\ \mbox{成立}, $

当且仅当下列表述成立:

(1)若$ \dim H=\infty $, 则存在$ T\in\mathcal{B}(H) $$ \lambda\in\mathbb{C} $使得$ \lambda C=0 $$ \phi(A)=AT-TA+\lambda A $对所有$ A\in\mathcal{B}(H) $成立.

(2)若$ \dim H=n<\infty $, 则存在$ T\in{\mathcal B}(H)=\mathcal{M}_n(\mathbb{C}) $, 可加导子$ f:\mathbb{C}\to\mathbb{C} $$ \lambda\in\mathbb{C} $使得$ \lambda C=0 $$ \phi(A)=AT-TA+(f(a_{ij}))+\lambda A $对所有$ A=(a_{ij})\in\mathcal{M}_n(\mathbb{C}) $成立.

    由定理1的结论可以看出, $ C $$ D $之间存在某种联系, 且$ C=0 $定蕴涵$ D=0 $; 但反过来, $ D=0 $并不一定蕴涵$ C=0 $. 因此, 我们的结论是对相关已有结果的进一步推广与改进.

2 主要结果的证明

本节将主要给出本文定理1的证明.

为证明定理1, 首先引入一个命题.

命题2.1    ([23, 命题2.0])令$ \mathcal{F} $是域且$ |\mathcal{F}|\ge3 $, $ \mathcal{V} $$ \mathcal{F} $上的向量空间. 对任意固定元$ u, v, w\in\mathcal{V} $, 定义$ p(t)=ut^2+vt+w, t\in\mathcal{F}. $如果$ p(t)=0 $$ \mathcal{F} $中有三个不同的根, 则$ u=v=w=0. $

定理1的证明    定理的充分性显然. 对于必要性, 我们分两种情形来证明之.

情形1    $ \overline{{\rm ran} C}\ne H $$ \overline{{\rm ran} C}\ne 0. $

$ P $$ C $的值域投影. 那么$ P $是非平凡的, 且满足$ C=PC. $$ Q=I-P $, $ \mathcal{B}_{11}=P\mathcal{B}(H)P $, $ \mathcal{B}_{12}=P\mathcal{B}(H)Q $, $ \mathcal{B}_{21}=Q\mathcal{B}(H)P $, $ \mathcal{B}_{22}=Q\mathcal{B}(H)Q $. 则$ \mathcal{B}(H)=\mathcal{B}_{11}+\mathcal{B}_{12}+\mathcal{B}_{21}+\mathcal{B}_{22}. $显然, 对任意算子$ A\in{\mathcal B}(H) $均可以表示为$ A=\sum\limits_{i, j=1, 2}A_{ij} $, 其中$ A_{ij}\in{\mathcal B}_{ij} $.

$ C=C_{11}+C_{12}+C_{21}+C_{22} $. 利用条件$ C=PC $可得$ C=C_{11}+C_{12}. $

任取可逆元$ A_{11}\in\mathcal{B}_{11} $, 任取$ A_{12}\in\mathcal{B}_{12} $, $ A_{22}\in\mathcal{B}_{22} $以及非零有理数$ t $. 直接计算可得

$ (A_{11}+tA_{11}A_{12})(A_{11}^{-1}C-A_{12}A_{22}+t^{-1}A_{22})=C. $

根据假设条件, 我们有$ \phi(A_{11}+tA_{11}A_{12})(A_{11}^{-1}C-A_{12}A_{22}+t^{-1}A_{22}) +(A_{11}+tA_{11}A_{12})\phi(A_{11}^{-1}C-A_{12}A_{22}+t^{-1}A_{22})=D. $利用$ \phi $的可加性, 上式整理得到

$ \begin{array}{rl} D=&\phi(A_{11})A_{11}^{-1}C-\phi(A_{11})A_{12}A_{22}+A_{11}\phi(A_{11}^{-1}C)\\ &-A_{11}\phi(A_{12}A_{22})+\phi(A_{11}A_{12})A_{22}+A_{11}A_{12}\phi(A_{22})\\ &+t[\phi(A_{11}A_{12})A_{11}^{-1}C-\phi(A_{11}A_{12})A_{12}A_{22}+A_{11}A_{12}\phi(A_{11}^{-1}C)-A_{11}A_{12}\phi(A_{12}A_{22})]\\ &+t^{-1}[\phi(A_{11})A_{22}+A_{11}\phi(A_{22})].\end{array} $

根据命题2.1, 可得

$ \begin{equation} \begin{array}{rl} D=&\phi(A_{11})A_{11}^{-1}C-\phi(A_{11})A_{12}A_{22}+A_{11}\phi(A_{11}^{-1}C)\\ &-A_{11}\phi(A_{12}A_{22})+\phi(A_{11}A_{12})A_{22}+A_{11}A_{12}\phi(A_{22}), \end{array} \end{equation} $ (2.1)
$ \begin{equation} \phi(A_{11}A_{12})A_{11}^{-1}C-\phi(A_{11}A_{12})A_{12}A_{22}+A_{11}A_{12}\phi(A_{11}^{-1}C)-A_{11}A_{12}\phi(A_{12}A_{22})=0 \end{equation} $ (2.2)

$ \begin{equation} \phi(A_{11})A_{22}+A_{11}\phi(A_{22})=0. \end{equation} $ (2.3)

特别地, 在等式(2.3)中取$ A_{11}=P, A_{22}=Q $, 得到

$ \begin{equation} P\phi(Q)P=0. \end{equation} $ (2.4)

另一方面, 由于$ A_{11}(A_{11}^{-1}C)=C $, 由题设条件可知$ D=\phi(A_{11})A_{11}^{-1}C+A_{11}\phi(A_{11}^{-1}C). $结合该式与等式(2.1), 可得

$ \begin{equation} \phi(A_{11}A_{12})A_{22}+A_{11}A_{12}\phi(A_{22})=\phi(A_{11})A_{12}A_{22}+A_{11}\phi(A_{12}A_{22}). \end{equation} $ (2.5)

在等式(2.2)中, 利用$ 2A_{11} $来代替$ A_{11} $, 有

$ \phi(A_{11}A_{12})A_{11}^{-1}C-2\phi(A_{11}A_{12})A_{12}A_{22}+A_{11}A_{12}\phi(A_{11}^{-1}C)-2A_{11}A_{12}\phi(A_{12}A_{22})=0. $

比较上式和等式(2.2), 即得

$ \phi(A_{11}A_{12})A_{11}^{-1}C+A_{11}A_{12}\phi(A_{11}^{-1}C)=0 $

$ \begin{equation} \phi(A_{11}A_{12})A_{12}A_{22}+A_{11}A_{12}\phi(A_{12}A_{22})=0. \end{equation} $ (2.6)

现令$ M=Q\phi(Q)P-P\phi(Q)Q $, 并定义映射$ \tau:{\mathcal B}(H)\rightarrow {\mathcal B}(H) $$ \tau(A)=\phi(A)-(AM-MA)\ \ \mbox{对任意}\ \ A\in\mathcal{B}(H)\ \ \mbox{成立}. $易证$ \tau $是可加的; 满足条件对任意$ A, B\in{\mathcal B}(H) $,

$ AB=C \Rightarrow \tau(A)B+A\tau(B)=D', \ \ \mbox{其中}\ \ D'=D-(CM-MC); $

且对任意可逆元$ A_{11}\in\mathcal{B}_{11} $以及任意元$ A_{12}\in\mathcal{B}_{12} $$ A_{22}\in\mathcal{B}_{22} $, 等式(2.2)–(2.3), 等式(2.5)–(2.6)对映射$ \tau $也成立. 此外, 等式(2.4)蕴涵

$ \begin{equation} \tau(Q)=\phi(Q)-(P\phi(Q)Q+Q\phi(Q)P)=Q\phi(Q)Q\in\mathcal{B}_{22}. \end{equation} $ (2.7)

以下分几个断言证明$ \tau $是广义导子, 即$ \tau $满足$ \tau(AB)=\tau(A)B+A\tau(B)-\lambda AB $对所有$ A, B\in\mathcal B(H) $成立.

断言1    $ \tau(\mathcal B_{22})\subseteq\mathcal{B}_{22} $.

在等式(2.3)中取$ A_{22}=Q $, 得到$ \tau(A_{11})Q+A_{11}\tau(Q)=0 $对任意可逆元$ A_{11}\in\mathcal{B}_{11} $成立. 该式结合等式(2.7), 可得$ \tau(A_{11})Q=A_{11}\tau(Q)=0\ \mbox{对任意可逆元}\ A_{11}\in\mathcal{B}_{11} \ \mbox{成立}. $上式与等式(2.3)又蕴涵

$ \begin{equation} \tau(A_{11})A_{22}=A_{11}\tau(A_{22})=0\ \ \mbox{对任意可逆元}\ A_{11}\in\mathcal{B}_{11}\ \mbox{与任意}\ A_{22}\in\mathcal{B}_{22}\ \mbox{成立}. \end{equation} $ (2.8)

特别地, 我们有

$ \begin{equation} \tau(P)Q=0 \end{equation} $ (2.9)

$ \begin{equation} P\tau(A_{22})=0\ \ \mbox{对任意元}\ A_{22}\in\mathcal{B}_{22}\ \mbox{成立}. \end{equation} $ (2.10)

在等式(2.5)中取$ A_{11}=P $, 则有

$ \begin{equation} P\tau(A_{12}A_{22})=\tau(A_{12})A_{22}+A_{12}\tau(A_{22})-\tau(P)A_{12}A_{22} \end{equation} $ (2.11)

对所有元$ \ A_{12}\in\mathcal{B}_{12} $$ A_{22}\in\mathcal{B}_{22} $成立. 在等式(2.11)的右边乘以$ P $, 可得

$ P\tau(A_{12}A_{22})P=A_{12}\tau(A_{22})P. $

特别地, 在上式中取$ A_{22}=Q $, 并结合等式(2.7), 得到

$ \begin{equation} P\tau(A_{12})P=A_{12}\tau(Q)P=0\ \mbox{对任意元}\ A_{12}\in\mathcal{B}_{12}\ \mbox{成立}. \end{equation} $ (2.12)

因此, 对任意$ A_{12}\in\mathcal{B}_{12} $$ A_{22}\in\mathcal{B}_{22} $, 我们有

$ A_{12}\tau(A_{22})P=P\tau(A_{12}A_{22})P=A_{12}A_{22}\tau(Q)P=0, $

$ PAQ\tau(A_{22})P=0\ \ \mbox{对任意的}\ \ A\in\mathcal B(H)\ \ \mbox{成立}. $

注意到$ \mathcal{B}(H) $是素代数. 上式蕴涵$ Q\tau(A_{22})P=0 $. 结合该式与等式(2.10), 即得$ \tau(A_{22})\in\mathcal{B}_{22} $.

断言2    存在数$ \lambda\in\mathbb C $使得$ \tau(P)=\lambda P $$ \tau(Q)=\lambda Q $.

对任意的$ A_{12}\in\mathcal B_{12} $, 等式(2.11)成立. 在该式中取$ A_{22}=Q $, 并左边乘以$ P $且右边乘以$ Q $, 得到

$ \begin{align*} P\tau(A_{12})Q=P\tau(A_{12})Q+A_{12}\tau(Q)Q-P\tau(P)A_{12}, \end{align*} $

即得

$ \begin{equation} P\tau(P)A_{12}=A_{12}\tau(Q)Q\ \ \mbox{对任意的}\ \ A_{12}\in\mathcal B_{12}\ \ \mbox{成立}. \end{equation} $ (2.13)

进而, 对任意$ A_{22}\in\mathcal{B}_{22} $, 等式(2.13)蕴涵

$ A_{12}\tau(Q)A_{22}=P\tau(P)A_{12}A_{22}=A_{12}A_{22}\tau(Q)Q, $

$ PA(Q\tau(Q)A_{22}-A_{22}\tau(Q)Q)=0 \ \mbox{对任意的}\ \ A\in\mathcal B(H) \ \mbox{成立}. $

利用$ \mathcal{B}(H) $的素性, $ A_{22}\in\mathcal{B}_{22} $的任意性以及断言1, 知存在数$ \lambda\in\mathbb C $使得$ \tau(Q)=\lambda Q $. 该式与等式(2.13)结合, 可得$ P\tau(P)P=\lambda P $.

断言3    对任意的$ A_{22}, B_{22}\in\mathcal{B}_{22} $, 我们有$ \tau(A_{22}B_{22})=\tau(A_{22})B_{22}+A_{22}\tau(B_{22})-\lambda A_{22}B_{22} $.

任取$ A_{22}, B_{22}\in\mathcal{B}_{22} $. 对任意的$ A_{12}\in\mathcal{B}_{12} $, 利用等式(2.11), 有

$ \begin{align} P\tau(A_{12}A_{22}B_{22})=\tau(A_{12})A_{22}B_{22}+A_{12}\tau(A_{22}B_{22})-\tau(P)A_{12}A_{22}B_{22} \end{align} $ (2.14)

$ \begin{align} &P\tau(A_{12}A_{22}B_{22}) \\ =&\tau(A_{12}A_{22})B_{22}+A_{12}A_{22}\tau(B_{22})-\tau(P)A_{12}A_{22}B_{22}\\ =&P\tau(A_{12}A_{22})B_{22}+A_{12}A_{22}\tau(B_{22})-\tau(P)A_{12}A_{22}B_{22} \\ =&[\tau(A_{12})A_{22}+A_{12}\tau(A_{22})-\tau(P)A_{12}A_{22}]B_{22}+A_{12}A_{22}\tau(B_{22})-\tau(P)A_{12}A_{22}B_{22} . \end{align} $ (2.15)

比较等式(2.14)和等式(2.15), 并利用断言2, 可得

$ A_{12}(\tau(A_{22}B_{22})-\tau(A_{22})B_{22}-A_{22}\tau(B_{22})+\lambda A_{22}B_{22})=0 $

对所有$ A_{12}\in\mathcal{B}_{12} $成立. 现再次利用$ \mathcal{B}(H) $的素性, 即得

$ \tau(A_{22}B_{22})=\tau(A_{22})B_{22}+A_{22}\tau(B_{22})-\lambda A_{22}B_{22}. $

断言4    对任意的$ A_{11}, B_{11}\in\mathcal{B}_{11} $, 我们有

$ \tau(A_{11}B_{11})=\tau(A_{11})B_{11}+A_{11}\tau(B_{11})-\lambda A_{11}B_{11}. $

任取可逆元$ A_{11}\in\mathcal{B}_{11} $. 对任意的$ A_{12}\in\mathcal{B}_{12} $, 在等式(2.5)中令$ A_{22}=Q $, 并结合断言2, 知

$ \begin{align*} \tau(A_{11}A_{12})Q=\tau(A_{11})A_{12}+A_{11}\tau(A_{12})-\lambda A_{11}A_{12}. \end{align*} $

注意到$ \mathcal{B}_{11} $中的每个元$ A_{11} $, 均存在正整数$ n $使得$ nP-A_{11} $$ \mathcal{B}_{11} $中是可逆的. 利用这一事实以及$ \tau $的可加性, 上式蕴涵

$ \begin{equation} \tau(A_{11}A_{12})Q=\tau(A_{11})A_{12}+A_{11}\tau(A_{12})-\lambda A_{11}A_{12} \end{equation} $ (2.16)

对所有$ A_{11}\in\mathcal{B}_{11} $$ A_{12}\in\mathcal{B}_{12} $成立. 进而, 对任意元$ A_{11}, B_{11}\in\mathcal{B}_{11} $与任意元$ A_{12}\in\mathcal{B}_{12} $, 根据等式(2.12)和等式(2.16), 可得

$ \begin{align*} A_{11}\tau(B_{11}A_{12}) &=A_{11}\tau(B_{11}A_{12})Q\\ &=A_{11}\tau(B_{11})A_{12}+A_{11}B_{11}\tau(A_{12})-\lambda A_{11}B_{11}A_{12}. \end{align*} $

结合上式与等式(2.16), 我们有

$ \begin{align} \tau(A_{11}B_{11}A_{12})Q &=\tau(A_{11}B_{11})A_{12}+A_{11}B_{11}\tau(A_{12})-\lambda A_{11}B_{11}A_{12} \end{align} $ (2.17)

$ \begin{align} &\tau(A_{11}B_{11}A_{12})Q \\ =&\tau(A_{11})B_{11}A_{12}+A_{11}\tau(B_{11}A_{12})-A_{11}B_{11}A_{12}\tau(Q) \\ =&\tau(A_{11})B_{11}A_{12}+A_{11}\tau(B_{11})A_{12}+A_{11}B_{11}\tau(A_{12})-2\lambda A_{11}B_{11}A_{12} . \end{align} $ (2.18)

比较等式(2.17)和等式(2.18), 得到

$ \tau(A_{11}B_{11})A_{12} =\tau(A_{11})B_{11}A_{12}+A_{11}\tau(B_{11})A_{12}-\lambda A_{11}B_{11}A_{12}, $

进而得到

$ \begin{align*} \tau(A_{11}B_{11})=A_{11}\tau(B_{11})+\tau(A_{11})B_{11}-\lambda A_{11}B_{11} \end{align*} $

对所有元$ A_{11}, B_{11}\in\mathcal{B}_{11} $成立. 该断言成立.

断言5    $ \tau(\mathcal B_{11})\subseteq\mathcal{B}_{11} $.

由断言4知,

$ \tau(A_{11}B_{11})=\tau(A_{11})B_{11}+A_{11}\tau(B_{11})-\lambda A_{11}B_{11} $

对所有$ A_{11}, B_{11}\in\mathcal{B}_{11} $成立. 现在上式中左乘$ Q $并令$ A_{11}=P $, 利用断言2, 可得$ Q\tau(B_{11})=0 $对所有$ B_{11}\in\mathcal{B}_{11} $成立; 在上式中右乘$ Q $并令$ B_{11}=P $, 再次利用断言2, 可得$ \tau(A_{11})Q=0 $对所有$ A_{11}\in\mathcal{B}_{11} $成立. 因此$ \tau(\mathcal B_{11})\subseteq\mathcal{B}_{11} $.

断言6    $ \tau(\mathcal B_{12})\subseteq\mathcal{B}_{12} $; 且对任意元$ A_{12}\in\mathcal{B}_{12} $$ A_{22}\in\mathcal{B}_{22} $, 我们有

$ \tau(A_{12}A_{22})=\tau(A_{12})A_{22}+A_{12}\tau(A_{22})-\lambda A_{12}A_{22} $

$ \tau(A_{11}A_{12})=\tau(A_{11})A_{12}+A_{11}\tau(A_{12})-\lambda A_{11}A_{12}. $

任取$ A_{12}\in\mathcal{B}_{12} $. 一方面, 注意到等式(2.16)成立. 那么, 在该式左边乘以$ Q $, 并取$ A_{11}=P $, 利用断言2, 我们有

$ \begin{equation} Q\tau(A_{12})Q=0. \end{equation} $ (2.19)

另一方面, 由于等式(2.6)对映射$ \tau $亦成立, 在该等式中取$ A_{11}=P $$ A_{22}=Q $, 可得

$ \tau(A_{12})A_{12}+A_{12}\tau(A_{12})=0. $

在上式左边乘以$ Q $, 并利用等式(2.12), 即得$ \tau(A_{12})A_{12}=A_{12}\tau(A_{12})=0 $对所有$ A_{12}\in\mathcal{B}_{12} $成立. 进而, 对任意$ A_{12}, B_{12}\in\mathcal{B}_{12} $, 我们有

$ \begin{align*} \tau(A_{12})B_{12}+\tau(B_{12})A_{12} &=\tau(A_{12})B_{12}+\tau(B_{12})A_{12}+\tau(A_{12})A_{12}+\tau(B_{12})B_{12}\notag\\ &=\tau(A_{12}+B_{12})(A_{12}+B_{12})=0. \end{align*} $

在上式右边乘以$ \tau(A_{12}) $, 得到

$ \tau(A_{12})B_{12}\tau(A_{12})=0\ \mbox{对所有}\ \ A_{12}, \ B_{12}\in\mathcal{B}_{12}\ \mbox{成立}, $

即有

$ Q\tau(A_{12})PBQ\tau(A_{12})P=0\ \ \mbox{对所有}\ \ B\in{\mathcal B}(H), A_{12}\in\mathcal{B}_{12}\ \ \mbox{成立}. $

利用$ \mathcal{B}(H) $的素性, 知$ Q\tau(A_{12})P=0 $对所有$ A_{12}\in\mathcal{B}_{12} $成立. 现结合等式(2.12)和等式(2.19), 可得$ \tau(A_{12})\in\mathcal{B}_{12} $.

最后, 利用等式(2.11), 等式(2.16)与上面所证事实, 即知$ \tau(A_{12}A_{22})=\tau(A_{12})A_{22}+A_{12}\tau(A_{22})-\lambda A_{12}A_{22} $$ \tau(A_{11}A_{12})=\tau(A_{11})A_{12}+A_{11}\tau(A_{12})-\lambda A_{11}A_{12} $对任意元$ A_{11}\in\mathcal{B}_{11} $, $ A_{12}\in\mathcal{B}_{12} $$ A_{22}\in\mathcal{B}_{22} $成立.

该断言成立.

断言7    $ \tau(\mathcal B_{21})\subseteq\mathcal{B}_{21} $; 且对任意元$ A_{12}\in\mathcal{B}_{12} $$ A_{21}\in\mathcal{B}_{21} $, 我们有

$ \tau(A_{12}A_{21})=\tau(A_{12})A_{21}+A_{12}\tau(A_{21})-\lambda A_{12}A_{21}. $

任取$ A_{21}\in\mathcal{B}_{21} $. 由于$ P(C+A_{21})=PC=C $, 所以

$ \tau(P)(C+A_{21})+P\tau(C+A_{21})=D=\tau(P)C+P\tau(C). $

化简可得

$ \tau(P)A_{21}+P\tau(A_{21})=0. $

利用断言2, 得到

$ \begin{equation} P\tau(A_{21})=0. \end{equation} $ (2.20)

对任意元$ A_{12}\in\mathcal{B}_{12} $, 由于$ (P+A_{12})(C-A_{12}A_{21}-A_{12}+A_{21}+Q)=PC=C $, 所以

$ \begin{array}{rl}&\tau(P+A_{12})(C-A_{12}A_{21}-A_{12}+A_{21}+Q)\\ &+(P+A_{12})\tau(C-A_{12}A_{21}-A_{12}+A_{21}+Q)=\tau(P)C+P\tau(C).\end{array} $

利用断言2, 断言5-6, 等式(2.13)和等式(2.20), 上式可化简为

$ \begin{equation} \tau(A_{12}A_{21})=\tau(A_{12})A_{21}+A_{12}\tau(A_{21})-\lambda A_{12}A_{21}. \end{equation} $ (2.21)

在等式(2.21)右边乘以$ Q $, 并结合断言5, 可知

$ A_{12}\tau(A_{21})Q=0 \ \mbox{对所有}\ A_{12}\in\mathcal{B}_{12}, \ A_{21}\in\mathcal{B}_{21}\ \mbox{均成立}. $

这蕴涵$ Q\tau(A_{21})Q=0 $. 该式结合等式(2.20), 可得$ \tau(A_{21})\in\mathcal{B}_{21} $. 该断言成立.

断言8    对任意元$ A_{11}\in{\mathcal B}_{11} $$ A_{21}\in{\mathcal B}_{21} $, 我们有

$ \tau(A_{21}A_{11})=\tau(A_{21})A_{11}+A_{21}\tau(A_{11})-\lambda A_{21}A_{11}. $

任取$ A_{11}\in{\mathcal B}_{11} $, $ A_{12}\in{\mathcal B}_{12} $$ A_{21}\in{\mathcal B}_{21} $. 利用断言4与断言7, 我们有

$ \begin{align*} \tau(A_{12}A_{21}A_{11}) =&\tau(A_{12})A_{21}A_{11}+A_{12}\tau(A_{21}A_{11})-\lambda A_{12}A_{21}A_{11} \end{align*} $

$ \begin{align*} \tau(A_{12}A_{21}A_{11}) =&\tau(A_{12}A_{21})A_{11}+A_{12}A_{21}\tau(A_{11})-\lambda A_{12}A_{21}A_{11}\notag \\ =&[\tau(A_{12})A_{21}+A_{12}\tau(A_{21})-\lambda A_{12}A_{21}]A_{11}+A_{12}A_{21}\tau(A_{11})-\lambda A_{12}A_{21}A_{11}. \end{align*} $

比较上面两个等式, 则有

$ A_{12}\tau(A_{21}A_{11})=A_{12}\tau(A_{21})A_{11}+A_{12}A_{21}\tau(A_{11})-\lambda A_{12}A_{21}A_{11}. $

由于$ A_{12} $的任意性, 上式蕴涵

$ \tau(A_{21}A_{11})=\tau(A_{21})A_{11}+A_{21}\tau(A_{11})-\lambda A_{21}A_{11}. $

断言9    对任意元$ A_{12}\in{\mathcal B}_{12} $$ A_{21}\in{\mathcal B}_{21} $, 我们有

$ \tau(A_{21}A_{12})=\tau(A_{21})A_{12}+A_{21}\tau(A_{12})-\lambda A_{21}A_{12}. $

对任意元$ A_{12}, B_{12}\in{\mathcal B}_{12} $$ A_{21}\in{\mathcal B}_{21} $, 利用断言6-7, 得到

$ \begin{align*} \tau(B_{12}A_{21}A_{12}) =\tau(B_{12})A_{21}A_{12}+B_{12}\tau(A_{21}A_{12})-\lambda B_{12}A_{21}A_{12} \end{align*} $

$ \begin{align*} \tau(B_{12}A_{21}A_{12})=&\tau(B_{12}A_{21})A_{12}+B_{12}A_{21}\tau(A_{12})-\lambda B_{12}A_{21}A_{12}\\ =&[\tau(B_{12})A_{21}+B_{12}\tau(A_{21})-\lambda B_{12}A_{21}]A_{12}+B_{12}A_{21}\tau(A_{12})-\lambda B_{12}A_{21}A_{12}. \end{align*} $

比较上面两个等式, 则有

$ B_{12}\tau(A_{21}A_{12})=B_{12}\tau(A_{21})A_{12}+B_{12}A_{21}\tau(A_{12})-\lambda B_{12}A_{21}A_{12}. $

由于$ B_{12} $是任意的, 上式蕴涵

$ \tau(A_{21}A_{12})=\tau(A_{21})A_{12}+A_{21}\tau(A_{12})-\lambda A_{21}A_{12}. $

类似地, 利用断言6-7通过两种方式计算$ \tau(A_{12}A_{22}A_{21}) $, 可证下面断言成立.

断言10    对任意元$ A_{21}\in{\mathcal B}_{21} $$ A_{22}\in{\mathcal B}_{22} $, 我们有

$ \tau(A_{22}A_{21})=\tau(A_{22})A_{21}+A_{22}\tau(A_{21})-\lambda A_{22}A_{21}. $

断言11    对任意元$ A, B\in{\mathcal B}(H) $, 我们有

$ \tau(AB)=\tau(A)B+A\tau(B)-\lambda AB. $

对任意元$ A=\sum_{ij}A_{ij}, B=\sum_{ij}B_{ij}\in{\mathcal B}(H) $, 结合断言3-10与$ \tau $的可加性, 容易验证$ \tau(AB)=\tau(A)B+A\tau(B)-\lambda AB $成立.

现在, 定义映射$ d:\mathcal B(H)\rightarrow \mathcal B(H) $为:

$ d(A)=\tau(A)-\lambda A, \ \ \forall A\in\mathcal B(H). $

易验证$ d $是可加导子. 再由$ \tau $的定义, 可令$ \delta(A)=d(A)+AM-MA $对任意$ A\in{\mathcal B}(H) $成立. 显然, $ \delta $也是$ \mathcal B(H) $上的可加导子, 且$ \phi(A)=\delta(A)+\lambda A $对任意$ A\in{\mathcal B}(H) $成立.

情形2    $ \overline{\operatorname{ran}C}=\{0\} $$ \overline{\operatorname{ran}C}=H $.

如果$ \overline{\operatorname{ran}C}=\{0\} $, 则$ C=0 $. 任取$ A\in\mathcal{B}(H) $. 由于$ 0 \cdot A=C $, 所以$ \phi(0)\cdot A+0\cdot\phi(A)=D $. 根据$ \phi $的可加性, 可知$ \phi(0)=0 $. 因而$ D=0 $. 现利用文献[25, 定理2.2], 可知此时结论成立.

如果$ \overline{\operatorname{ran}C}={H} $, 则$ C $是稠值域算子.

首先, 对任意可逆元$ A\in\mathcal{B}(H) $, 因为$ A^{-1}AC=C $, 所以$ \phi(A^{-1})AC+A^{-1}\phi(AC)=D $, 即

$ \phi(AC)-AD=-A\phi(A^{-1})AC. $

接着, 定义映射$ \Phi:\mathcal{B}(H)\to\mathcal{B}(H) $

$ \Phi(A)=\phi(AC)-AD\ \mbox{对任意}\ A\in\mathcal B(H)\ \mbox{成立}. $

显然, $ \Phi $是可加的, 且满足

$ \begin{equation} \Phi(A)=-A\phi(A^{-1})AC\ \mbox{对任意可逆元}\ A\in\mathcal B(H)\ \mbox{成立}. \end{equation} $ (2.22)

再次, 对任意可逆元$ A\in\mathcal B(H) $, 选择适当的整数$ n $使得$ A-nI $是可逆元. 注意到

$ \begin{equation*} (A-n I)^{-1}-A^{-1}=n(A-nI)^{-1}A^{-1}. \end{equation*} $

上式作用$ \Phi $, 并利用$ \Phi $的可加性, 可得

$ n\Phi((A-n I)^{-1}A^{-1})=\Phi((A-nI)^{-1})-\Phi(A^{-1}). $

对上式利用等式(2.22), 有

$ \begin{align*} &-n(A-nI)^{-1}A^{-1}\phi(A^2-n A)(A-n I)^{-1}A^{-1}C\\ =&-(A-nI)^{-1}\phi(A-nI)(A-nI)^{-1}C+A^{-1}\phi(A)A^{-1}C. \end{align*} $

又因为$ C $$ \mathcal{B}(H) $中的稠值域算子, 上式蕴涵

$ \begin{align*} &n(A-nI)^{-1}A^{-1}\phi(A^2-n A)(A-n I)^{-1}A^{-1}\\ =&(A-nI)^{-1}\phi(A-nI)(A-nI)^{-1}-A^{-1}\phi(A)A^{-1}. \end{align*} $

在上式两边同乘$ (A-nI)A=A(A-nI) $, 可得

$ n\phi(A^2-nA)=A\phi(A-nI)A-(A-n I)\phi(A)(A-n I). $

利用$ \phi $的可加性, 上式化简得到

$ \phi(A^2)=\phi(A)A+A\phi(A)-A\phi(I)A\ \mbox{对任意可逆元}\ A\in\mathcal B(H)\ \mbox{成立}. $

注意到$ \mathcal{B}(H) $中每一个元都可以表示为$ \mathcal{B}(H) $中两个可逆元的和. 对上式利用该事实, 易知$ \phi(A^2)=\phi(A)A+A\phi(A)-A\phi(I)A\ \mbox{对任意元}\ A\in\mathcal B(H)\ \mbox{成立}, $即有

$ \phi(A^2)-\phi(I)A^2=\phi(A)A+A\phi(A)-A\phi(I)A-\phi(I)A^2=(\phi(A)-\phi(I)A)A+A(\phi(A)-\phi(I)A) $

对任意元$ A\in\mathcal B(H) $成立.

对任意元$ A\in\mathcal B(H) $, 令$ \delta(A)=\phi(A)-\phi(I)A. $表明$ \delta $$ \mathcal B(H) $上的可加Jordan导子, 即$ \delta(A^2)=\delta(A)A+A\delta(A) $对所有$ A\in\mathcal B(H) $成立. 根据文献[26], $ \delta $是导子.

最后, 仍需验证$ \phi(I)\in\mathbb{C}I $. 为此, 任取$ A, B\in\mathcal B(H) $. 若$ AB=C $, 则$ \phi(A)B+A\phi(B)=D $, 即

$ \begin{align*} D=&\delta(A)B+\phi(I)AB+A\delta(B)+A\phi(I)B\\ =&\delta(AB)+\phi(I)AB+A\phi(I)B\\ =&\phi(AB)+A\phi(I)B=\phi(C)+A\phi(I)B. \end{align*} $

因为$ D $$ \mathcal{B}(H) $中的任意固定元, 上式蕴涵$ \mbox{当}\ AB=C \ \mbox{时}, \ \ A\phi(I)B\ \mbox{也是}\ \mathcal{B}(H)\ \mbox{中的固定元}. $特别地, 取$ A $为任意可逆元. 那么$ A^{-1}AC=IC=C $. 因此, 由上式得到$ A^{-1}\phi(I)AC=\phi(I)C $, 即$ \phi(I)AC=A\phi(I)C\ \mbox{对所有可逆元}\ A\in\mathcal B(H)\ \mbox{成立}. $现在, 对上式利用$ C $的稠值域性可知, $ \phi(I)A=A\phi(I) $对所有$ A\in\mathcal{B}(H) $成立, 进而得到$ \phi(I)\in\mathbb{C}I $.

定理得证.

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