设环${{\cal R}}$,称可加映射$\delta:{{\cal R}}\rightarrow{{\cal R}}$是导子,若$\delta(AB)=\delta(A)B+A\delta(B), \ \forall A, B\in{{\cal R}};$称$\delta$是Lie导子,若$\delta([A, B])=[\delta(A), B]+[A, \delta(B)], \ \forall A, B\in {{\cal R}},$其中$[A, B]=AB-BA$为$A, B$的Lie积.显然,导子一定是Lie导子,反之不一定成立.在文献[3]中, Bresar证明了特征值不是2的素环${{\cal R}}$上的Lie导子可以被分解为$\delta+\zeta$,其中$\delta$是一个从${\mathcal R}$到其中心闭包的导子, $\zeta$为一个从${{\cal R}}$到其中心扩张(extended centroid)的可加映射.在文献[7]中,作者证明了C$^\ast$-代数上每一个Lie导子可以被分解为一个导子与一个中心值的迹(center-valued trace)的和.导子和Lie导子在理论和实际中有着重要应用,许多学者对它们进行了研究(见文献[1-4, 6-7, 9-10]及其参考文献).近些年来,许多学者探讨可加映射成为导子或Lie导子的条件,其中研究热点之一是可导映射和Lie可导映射.称可加映射$\delta: {\cal R}\rightarrow {\cal R}$在$\Omega\in{\cal R}$可导,若$\delta(AB)=\delta(A)B+A\delta(B), \ \forall A, B\in {\cal R}, AB=\Omega;$在$\Omega\in{\cal R}$ Lie可导,若$\delta([A, B])=[\delta(A), B]+[A, \delta(B)], \ \forall A, B\in {\cal R}, AB=\Omega.$在文献[2]中,作者给出了CSL代数到自身的线性映射在任意点可导的充分必要条件,进而证明了套代数到自身在任意非零点可导的可加映射是导子.在文献[6]中,作者证明了若可加映射$\delta: {\cal B}(X)\rightarrow {\cal B}(X)$在0 (非平凡幂等元$P$) Lie可导,则

其中$\tau:{\cal B}(X)\rightarrow {\cal B}(X)$是导子, $f:{\cal B}(X)\rightarrow {\Bbb C}$为一个可加映射且对满足$AB=0$ ($AB=P$)的$A, B\in{\cal B}(X)$有$f([A, B])=0$.随后,在文献[4]中, Ji和Qi刻画了三角代数上在零点或标准幂等元Lie可导的可加映射.本文刻画von Neumann代数${\cal A}$上的Lie可导映射.设投影$P\in{\cal A}$使得$\underline{P}=0, \overline{P}=I$, $\Omega\in{\cal A}$满足$P\Omega=\Omega$,我们证明可加映射$\delta: {\mathcal A}\rightarrow {\cal A}$在$\Omega$ Lie可导当且仅当存在导子$\tau:{\cal A} \rightarrow {\cal A}$和可加映射$f: {\cal A}\rightarrow {\cal Z}({\cal A})$使得

其中$f([A, B])=0, \forall A, B\in {\cal A}, AB=\Omega$.特别地,若${\cal A}$是因子von Neumann代数, $\Omega\in {\cal A}$满足$\mbox{ker}(\Omega)\neq {0}$或$\overline{\mbox{ran}(\Omega)}\neq H$,则可加映射$\delta: {\cal A}\rightarrow {\cal A}$在$\Omega$ Lie可导当且仅当$\delta$有上述形式.

本文设$H$是复数域上的维数至少是2的Hilbert空间, ${\mathcal B}(H)$是$H$上有界线性算子全体构成的代数.设$A\in{\cal B}(H)$, $\mbox{ker}(A), \ \mbox{ran}(A)$分别为$A$的零空间和值域. ${{\cal B}}(H)$的子代数${{\cal A}}$是von Neumann代数,若${{\cal A}}^{\prime\prime}={\cal A},$其中${{\cal A}}^\prime =\{T\in{{\cal B}}(H), \ TA=AT, \ \forall A\in{{\cal A}}\}$, ${{\cal A}}^{\prime\prime}=\{{{\cal A}}^\prime\}^\prime$.令${{\cal Z}}({{\cal A}})={{\cal A}}\cap{{\cal A}}^\prime$为${\cal A}$的中心.若${{\cal Z}}({{\cal A}})=\Bbb{CI}$,则称${{\cal A}}$为因子von Neumann代数.对$A\in{\cal A}$, $A$的中心覆盖$\overline{A}$是满足$PA=A$的最小中心投影.不难证明$\overline{A}$是由$\{BAx, \ B\in{\cal A}, \ x\in H\}$张成的闭子空间上的投影(见文献[5,定理5.5.2]).若$A^*=A\in{\mathcal A}$,记

若$P\in{\cal A}$是一个投影,则那么$\underline{P}$是小于等于$P$的最大中心投影.称$P\in{\cal A}$是core-free的,若$\underline{P}=0$,容易验证$\underline{P}=0$当且仅当$\overline{I-P}=I$.

本文刻画不含交换中心投影的von Neumann代数上的Lie可导映射.下列引理是文献[8]中的引理4.

引理2.1  设${\cal A}$是不含交换中心投影的von Neumann代数.则对任意的投影$Q\in{\mathcal Z}({\cal A})$,存在非零投影$P\in{\cal A}$使得$\underline{P}=0$, $\overline{P}=Q$.

由引理2.1知若${\cal A}$是不含交换中心投影的von Neumann代数,则存在非零投影$P\in{\cal A}$使得$\underline{P}=0$, $\overline{P}=I$.下面是本文的主要结论.

定理2.1  设${\cal A}$是不含交换中心投影的von Neumann代数,投影$P\in{\cal A}$使得$\underline{P}=0,$$\overline{P}=I$.若$\Omega\in{\cal A}$使得$P\Omega=\Omega$,则可加映射$\delta:{\cal A}\rightarrow {\cal A}$在$\Omega$ Lie可导,即

当且仅当存在导子$\tau:{\cal A}\rightarrow {\cal A}$和可加映射$f: {\cal A}\rightarrow \mathcal Z({\cal A})$使得$\delta(A)=\tau(A)+f(A), \forall A\in {\cal A},$$f([A, B])=0, \forall A, B\in {\cal A}, AB=\Omega.$

为证明定理2.1,我们需要下列引理.

引理2.2  设${\cal A}$是von Neumann代数,投影$P\in{\cal A}$使得$\underline{P}=0, \overline{P}=I$.

(1)对$A\in{\cal A}$,若$A(I-P)BP=0$, $\forall B\in{\cal A}$,则$A(I-P)=0$.

(2)对$A\in{\cal A}$,若$PB(I-P)A=0$, $\forall B\in{\cal A}$,则$(I-P)A=0$.

(3)对$A\in{\cal A}$,若$APB(I-P)=0$, $\forall B\in{\cal A}$,则$AP=0$.

(4)对$A\in{\cal A}$,若$(I-P)BPA=0$, $\forall B\in{\cal A}$,则$PA=0$.

证  由core-free和中心覆盖的定义知$\underline{I-P}=0, \overline{I-P}=I$.因此只证明(1), (2)即可.

由$\overline{P}=I$知$\{BPx:\ B\in{\cal A}, \ x\in H\}$在$H$中稠密,因此若$A(I-P)BP=0$, $\forall B\in{\cal A}$,则$A(I-P)=0$.由$PB(I-P)A=0$得$A^*(I-P)B^*P=0$, $\forall B\in{\cal A}$.因此由(1)知$A^*(I-P)=0$, $(I-P)A=0$.结论(1), (2)成立.

定理2.1的证明  充分性.假设存在导子$\tau: \mathcal A\rightarrow {\cal A}$,可加映射$f: {\cal A}\rightarrow {\cal Z}({\cal A})$使得$\delta(A)=\tau(A)+f(A), \forall A\in {\cal A},$$f([A, B])=0, \ \forall A, B\in {\cal A}, AB=\Omega.$则对任意的$A, B \in{\cal A}$, $AB=\Omega$, $\delta([A, B])=\tau([A, B])+f([A, B])=\tau([A, B]).$另一方面

因此, $\delta([A, B]) = [\delta(A), B] + [A, \delta(B)],$$\forall A, B\in{\cal A}$, $AB=\Omega,$$\delta在\Omega$ Lie可导.

下证必要性.分几个断言证之.设投影$P\in{\cal A}$使得$\underline{P}=0, \overline{P}=I$.令$P_{1}=P, \ P_{2}=I-P,$${\cal A}_{ij}={\cal P}_{i}{\cal A}{\cal P}_{j}$,则${\cal A}={\cal A}_{11}+{\cal A}_{12}+{\cal A}_{21}+{\cal A}_{22}$.对任意的$A\in{\cal A}$, $A_{ij}=P_iAP_j\in{\cal A}_{ij}$, $1\leq i, j\leq2.$

断言1  若$A_{ii}\in{\cal A}_{ii}, \ i=1, 2$使得$A_{ii}T_{ij}=T_{ij}A_{jj}, \ \forall T_{ij}\in{\mathcal A}_{ij}, \ 1\leq i\neq j\leq2$,则$A_{ii}+A_{jj}\in{\cal Z}({\mathcal A})$.

假设$A_{ii}T_{ij}=T_{ij}A_{jj}, \ \forall T_{ij}\in{\cal A}_{ij}$.对任意的$T_{ii}\in{\cal A}_{ii}$,用$T_{ii}T_{ij}$代替$T_{ij}$得

由引理2.2得$A_{ii}T_{ii}=T_{ii}A_{ii}$.因此由${\cal A}$是von Neumann代数知存在$Z_i\in{\cal Z}({\cal A})$使得$A_{ii}=Z_i P_i$.同理$A_{jj}=Z_j P_j$, $Z_j\in{\cal Z}({\cal A})$.因此$A_{ii}T_{ij}=T_{ij}A_{jj}$可重写为

由引理2.2得

因此$Z_i=Z_j$, $A_{ii}+A_{jj}=Z_i(P_i+P_j)=Z_i\in{\cal Z}({\mathcal A})$.

令$T=P_{1}\delta(P_{1})P_{2}-P_{2}\delta(P_{1})P_{1}.$定义$\Delta(A)=\delta(A)-(AT-TA), \ \forall A\in{\cal A}$,则$\Delta$在$\Omega$ Lie可导,且

断言2   $\Delta(P_{1})\in {\cal Z}({\cal A}).$

对$\forall A_{21}\in {{\cal A}}_{21},$由$P_1(\Omega+A_{21})=\Omega$及(2.1)式有

用$tA_{21}$代替$A_{21}$, $t\in\Bbb Q$得

(2.2)式左乘$P_{2}$,右乘$P_{1}$得$P_{2}\Delta(P_{1})A_{21}=A_{21}\Delta(P_{1})P_{1},$由$\Delta(P_{1})\in{\cal A}_{11}+{\cal A}_{22}$及断言1知$\Delta (P_{1})\in {\cal Z}({\cal A}).$

断言3   $\Delta(I)\in {\cal Z}({\cal A}), \Delta(P_{2})\in {\cal Z}({\cal A})$.

对可逆的$A_{11}\in {\cal A}_{11}$,任意的$A_{12}\in {\cal A}_{12}, C_{22}\in {\cal A}_{22}.$令

则$AB=\Omega$且$BA=(A_{11}^{-1}\Omega-A_{12}C_{22}+t^{-1}C_{22}) (A_{11}+tA_{11}A_{12}).$因此

在(2.5)式中令$A_{11}=P_1$, $C_{22}=P_2$得

此式左乘$P_1$,右乘$P_2$且由断言2得

在(2.3)式中令$A_{11}=P_{1}, C_{22}=P_{2}$得

用$tA_{12}$代替$A_{12}$, $t\in\Bbb Q$得

因此

此式左乘$P_2$得$P_2\Delta(I)A_{12}=0$,因此由引理2.2得$P_2\Delta(I)P_1=0$.注意到$P_1\Delta(I)P_2=P_1\Delta(P_1)P_2+P_1\Delta(P_2)P_2=0$.因此由断言1得$\Delta(I)\in {\cal Z}({\cal A})$,由断言2得$\Delta(P_{2})=\Delta(I-P_{1})=\Delta(I)-\Delta(P_{1})\in \mathcal Z({\cal A})$.

断言4   $\Delta(A_{12})\in {\cal A}_{12}, \forall A_{12}\in {\cal A}_{12}.$

由$P_{1}\Omega=\Omega$及(2.1)式有

由$(P_{1}+A_{12})(\Omega-A_{12}+P_{2})=\Omega$及(2.1)有

两式相减得

用$tA_{12}$代替$A_{12}$, $t\in\Bbb Q$,由$\Delta(P_{2})\in {\cal Z}({\cal A})$得

(2.6)式两边分别同乘$P_{1}$, $P_2$得$P_{1}\Delta(A_{12})P_{1}=P_{2}\Delta(A_{12})P_{2}=0.$ (2.6)式两边左乘$P_{2}$,右乘$P_{1}$得$P_{2}\Delta(A_{12})P_{1}=0.$因此$\Delta(A_{12})=P_{1}\Delta(A_{12})P_{2}\in {\cal A}_{12}, \ \forall A_{12}\in {\cal A}_{12}$.

断言5   $\Delta(A_{21})\in {\cal A}_{21}, \forall A_{21}\in {\cal A}_{21}.$

对任意的$A_{21}\in {\cal A}_{21},$由(2.2)式及$\Delta(P_{1})\in {\cal Z}({\cal A})$有

此式两边分别乘以$P_{1}$, $P_2$得$P_{1}\Delta(A_{21})P_{1}=0,$$P_{2}\Delta(A_{21})P_{2}=0.$左乘$P_{1}$,右乘$P_{2}$得$P_{1}\Delta(A_{21})P_{2}=0,$因此$\Delta(A_{21})\in {\cal A}_{21}, \ \forall A_{21}\in {\cal A}_{21}.$

断言6  存在可加映射$f_i:{\cal A}_{ii}\rightarrow{\cal Z}({\cal A})$使得$\Delta(A_{ii})-f_i(A_{ii})\in {\cal A}_{ii}$, $\forall A_{ii}\in{\cal A}_{ii}$, $i=1, 2$.

对任意可逆的$A_{11}\in {\cal A}_{11}$及任意的$C_{22}\in {\cal A}_{22}$,由(2.5)式得

在(2.7)式中令$C_{22}=P_{2}$,由$\Delta(P_{2})\in {\cal Z}({\mathcal A})$得

因此$P_{1}\Delta(A_{11})P_{2}=0,$$P_{2}\Delta(A_{11})P_{1}=0.$ (2.7)式两边同乘$P_{2}$得

因此存在可加映射$f_1:{\mathcal A}_{11}\rightarrow{\cal Z}({\cal A})$使得$P_{2}\Delta(A_{11})P_{2}=f_1(A_{11})P_2$对任意可逆的$A_{11}\in {\cal A}_{11}$成立.若$A_{11}\in {\cal A}_{11}$不可逆,则存在$n\in{\mathbb N}$使得$nP_1-A_{11}在{\cal A}_{11}$中可逆, $P_{2}\Delta(nP_1-A_{11})C_{22}=C_{22}\Delta(nP_1-A_{11})P_{2}.$由$\Delta(P_1)\in{\cal Z}({\mathcal A})$知$P_{2}\Delta(A_{11})C_{22}=C_{22}\Delta(A_{11})P_{2}, \forall A_{11}\in{\cal A}_{11}, C_{22}\in{\cal A}_{22}$.因此

同理存在可加映射$f_2:{\mathcal A}_{22}\rightarrow{\cal Z}({\cal A})$使得$\Delta(A_{22})-f_2(A_{22})\in {\cal A}_{22}, \ \forall A_{22}\in{\cal A}_{22}$.

对任意的$A=\sum\limits_{i, j=1}^{2}A_{ij}\in {\cal A},$定义可加映射$f:{\cal A}\rightarrow {\cal Z}({\cal A}), \ f(A)=f_1(A_{11})+f_{2}(A_{22})$, $d(A)=\sum\limits_{i, j=1}^{2}\Delta(A_{ij})-f(A).$则$f(A_{ii})\in{\mathcal Z}({\cal A})$, $i=1, 2,$$f(A_{ij})=0$, $1\leq i\neq j\leq 2.$$d(A_{ij})\in {\cal A}_{ij},$$1\leq i, j\leq 2$,且$i \neq j$时$d(A_{ij})=\Delta( A_{ij}).$

断言7   $d$是一个导子.

分以下几个步骤证明.

步骤1   $d(A_{11}A_{12})=d(A_{11})A_{12}+A_{11}d(A_{12}),$$d(A_{12}A_{22})=d(A_{12})A_{22}+A_{12}d(A_{22}),$$\forall A_{11}\in{{\cal A}}_{11},$$A_{12}\in{{\cal A}}_{12}, \ \forall A_{22}\in{{\cal A}}_{22}.$

对可逆的$A_{11}\in{{\cal A}}_{11}$及任意的$A_{12}\in{\mathcal A}_{12}, C_{22}\in{\cal A}_{22}$,由(2.3)及$A_{12}$的任意性得

在(2.8)式中令$C_{22}=P_2$得

因为$\Delta(P_2)\in{\cal Z}({\cal A}), \ \Delta(A_{12})\in{\mathcal A}_{12}, \ \Delta(A_{11})=d(A_{11})+f_1(A_{11}), \ f_1(A_{11})\in{\cal Z}({\cal A})$.因此$d(A_{11}A_{12})=d(A_{11})A_{12}+A_{11}d(A_{12}).$对任意的$A_{11}\in{{\cal A}}_{11},$存在正整数$n$使得$nP_{1}-A_{11}$为${{\cal A}}_{11}$中的可逆元,因此$d((nP_{1}-A_{11})A_{12})=d(nP_{1}-A_{11})A_{12}+(nP_{1}-A_{11})d(A_{12}).$又$d(P_{1}A_{12})=d(P_{1})A_{12}+P_{1}d(A_{12}),$因此$d(A_{11}A_{12})=d(A_{11})A_{12}+A_{11}d(A_{12}),$$\forall A_{11}\in{{\cal A}}_{11}, A_{12}\in{{\cal A}}_{12}.$

类似可证$d(A_{12}A_{22})=d(A_{12})A_{22}+A_{12}d(A_{22}), \ \forall A_{12}\in{{\cal A}}_{12}, A_{22}\in{{\cal A}}_{22}.$

步骤2   $d(A_{ii}B_{ii})=d(A_{ii})B_{ii}+A_{ii}d(B_{ii}), \ \forall A_{ii}, B_{ii}\in{{\cal A}}_{ii},$$i=1, 2$.

对任意的$A_{11}, B_{11}\in{{\cal A}}_{11},$$A_{12}\in{\mathcal A}_{12},$一方面,由步骤1得

另一方面

比较这两个等式,可得

因此由引理2.2有$d(A_{11}B_{11})=d(A_{11})B_{11}+A_{11}d(B_{11}), \ \forall A_{11}, B_{11}\in{{\cal A}}_{11}.$类似地,对任意的$A_{22}, B_{22}\in{{\cal A}}_{22}$有$d(A_{22}B_{22})=d(A_{22})B_{22}+A_{22}d(B_{22}).$

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

对任意的$A_{12}\in{{\cal A}}_{12}, A_{21}\in{{\cal A}}_{21}.$由$(P_{1}+A_{12})(\Omega-A_{12}A_{21}-A_{12}+A_{21}+P_{2})=P_1\Omega=\Omega$及$\Delta(P_{1})\in{\cal Z}({\cal A})$有

将$A_{12}$以及$A_{21}$替换成$sA_{12}$, $tA_{21}, \ s, t\in{\Bbb Q}$,由$d(A_{12})=\Delta(A_{12}), \ d(A_{21})=\Delta(A_{21})$得

从而有

上式左、右乘$A_{12}$得

两式相加得

另一方面由步骤1-2得

由(2.9), (2.10)式知$2Z A_{12}=A_{12}Z=0,$$\forall A_{12}\in{\mathcal A}_{12}.$因此由引理2.2得$ZP_1=0$, $P_2Z=ZP_2=0$, $Z=Z(P_1+P_2)=0$.从而,对任意的$A_{12}\in{{\cal A}}_{12}$以及$A_{21}\in{\mathcal A}_{21},$有

步骤4  $d(A_{21}A_{11})=d(A_{21})A_{11}+A_{21}d(A_{11}), \ \forall A_{11}\in{{\cal A}}_{11}, \ \forall A_{21}\in{\mathcal A}_{21}.$

对任意的$A_{11}\in{{\cal A}}_{11}, \ A_{21}\in{{\cal A}}_{21}, \ A_{12}\in{\cal A}_{12},$一方面

另一方面

所以

因此由断言5-6和引理2.2得

类似于文献[6]定理2.1可证$d$是一个导子.

断言8  存在导子$\tau:{\cal A}\rightarrow\mathcal A$使得$\delta(A)=\tau(A)+f(A)$, $f([A, B])=0, \forall A, B\in{\cal A}, AB=\Omega.$

注意到$\Delta(A)=\delta(A)-(AT-TA)$, $d(A)=\Delta(A)-f(A)且d$是导子,令$\tau(A)=d(A)+AT-TA, \forall A\in{\cal A}$,则$\tau$是导子.对任意的$A, B\in{\cal A}$, $AB=\Omega,$由$f(A)\in{\cal Z}({\cal A}),$得

因此,存在导子$\tau:{\cal A}\rightarrow{\cal A}$使得$\delta(A)=\tau(A)+f(A)$, $f([A, B])=0, \forall A, B\in{\cal A}, AB=\Omega.$定理得证.

由引理2.2知$P$和$I-P$是对称的,因此由定理2.1有

注2.1  设${\cal A}$是不含交换中心投影的von Neumann代数,投影$P\in{\cal A}$使得$\underline{P}=0, \overline{P}=I$.若$\Omega\in{\cal A}$满足下列条件之一: (1) $P\Omega=\Omega$; (2) $\Omega P=\Omega$; (3) $(I-P)\Omega=\Omega$; (4) $\Omega(I-P)=\Omega$,则可加映射$\delta: {\cal A}\rightarrow {\cal A}$在$\Omega$ Lie可导当且仅当存在导子$\tau: {\cal A}\rightarrow {\cal A}$和可加映射$f: {\cal A}\rightarrow {\cal Z}({\cal A})$使得$\delta(A)=\tau(A)+f(A), \forall A\in {\cal A}, \ f([A, B])=0, \ \forall A, B\in {\cal A}, AB=\Omega.$

显然因子von Neumann代数不含交换中心投影.且若$P$是因子von Neumann代数中的非平凡投影,则$\underline{P}=0, \overline{P}=I$.因此由注2.1有

推论2.1  设${\cal A}$是因子von Neumann代数, $\Omega\in{\cal A}$.若存在非平凡投影$P\in{\cal A}$使得$\Omega$满足注2.1中条件(1)-(4)之一,则可加映射$\delta:{\cal A}\rightarrow {\cal A}$在$\Omega$ Lie可导当且仅当存在导子$\tau:{\cal A}\rightarrow {\cal A}$和可加映射$f: {\cal A}\rightarrow {\mathbb C}$使得$\delta(A)=\tau(A)+f(A)I, \forall A\in {\cal A},$$f([A, B])=0, \forall A, B\in {\cal A}, AB=\Omega.$

更一般地,有

定理2.2  设${\cal A}$是因子von Neumann代数, $\Omega\in {\cal A}$.若$\overline {\mbox{ran}(\Omega)}\neq H$或${\mbox{ker}(\Omega)}\neq 0$,则$\delta: {\cal A}\rightarrow {\cal A}$在$\Omega$ Lie可导当且仅当存在导子$\tau:{\cal A}\rightarrow {\cal A}$和可加映射$f: {\cal A}\rightarrow \mathbb C$使得$\delta(A)=\tau(A)+f(A)I, \forall A\in {\cal A},$$f([A, B])=0, \forall A, B\in {\cal A}, AB=\Omega.$

证  若$\Omega=0$,由文献[6]中定理2.1知结论成立.下面假设$\Omega\neq 0$,

情形1  $\overline{ran(\Omega)}\neq H$.

此时存在非平凡投影$P\in {\cal A}$使得$(I-P)\Omega=0$.事实上设$\Omega=V|\Omega^{*}\Omega|^{\frac{1}{2}}$是$\Omega$的极分解,那么$V^{*}\in {\cal A}, VV^{*}=P|_{\overline{ran(\Omega)}}\neq 0$是${\cal A}$中的投影. $\overline{ran(\Omega)}\neq H$表明$P|_{\overline{ran(\Omega)}}\neq I$, $P|_{\overline{ran(\Omega)}}$是非平凡投影.因此由$(I-P|_{\overline{ran(\Omega)}})\Omega=0$和推论2.1知结论成立.

情形2   ${\mbox{ker}(\Omega)}\neq 0$.

此时$\overline{ran(\Omega^{*})}\neq H$.定义可加映射$\delta^{*}(A^{*})=\delta(A)^{*}$,则$\delta^*$在$\Omega^{*}$ Lie可导.事实上,对任意的$A, B\in {\cal A}, \ B^*A^*=\Omega^{*}$,则$AB=\Omega$,

即$\delta^*$在$\Omega^{*}$ Lie可导.由情形1知存在导子$\tau^*: {\cal A}\rightarrow {\cal A}$和可加映射$f^*: {\cal A}\rightarrow \Bbb C,$使得$\delta^*(A^*)=\tau^*(A^*)+f^*(A^*)I, \forall A\in {\cal A}, \ f^*([B^*, A^*])=f^*([A, B]^*)=0, \ \forall A, B\in {\mathcal A}, \ B^*A^*=\Omega^*$.定义可加映射$\tau(A)=(\tau^{*}(A^{*}))^*,$$f(A)=(f^{*}(A^{*}))^*,$则$\tau: {\cal A}\rightarrow {\mathcal A}$是导子, $f: {\cal A}\rightarrow \Bbb C$使得$f([A, B])=0, \ \forall A, B\in {\cal A}, AB=\Omega,$且$\delta(A)=\tau(A)+f(A)I, \ \forall A\in{\cal A}$.证毕.