Finite element method for an eigenvalue optimization problem of the Schrödinger operator

1.   Introduction

Let $\mathcal{A}$ be an associative algebra. For $A, B\in\mathcal{A}$, denote by $[A, B] = AB-BA$ the Lie product of $A$ and $B$. An additive (a linear) map $\delta: \mathcal{A}\rightarrow \mathcal{A}$ is called a global Lie triple derivation if $\delta([[A, B], C]) = [[\delta(A), B], C]+[[A, \delta(B)], C]+[[A, B], \delta(C)]$ for all $A, B, C\in \mathcal{A}$. The study of global Lie triple derivations on various algebras has attracted several authors' attention, see for example ^{[2,11,16,17,20]}. Next, let $\delta: \mathcal{A}\rightarrow \mathcal{A}$ be a map (without the additivity (linearity) assumption). $\delta$ is called a global nonlinear Lie triple derivation if $\delta$ satisfies $\delta([[A, B], C]) = [[\delta(A), B], C]+[[A, \delta(B)], C]+[[A, B], \delta(C)]$ for all $A, B, C\in \mathcal{A}$. Ji, Liu and Zhao ^[4] gave the concrete form of global nonlinear Lie triple derivations on triangular algebras. Chen and Xiao ^[3] investigated global nonlinear Lie triple derivations on parabolic subalgebras of finite-dimensional simple Lie algebras. Very recently, Zhao and Hao ^[21] paid attention to non-global nonlinear Lie triple derivations. Let $F:\mathcal{A}\times \mathcal{A}\times \mathcal{A}\rightarrow \mathcal{A}$ be a map and $\mathcal{Q}$ be a proper subset of $\mathcal{A}$. $\delta$ is called a non-global nonlinear Lie triple derivation if $\delta$ satisfies $\delta([[A, B], C]) = [[\delta(A), B], C]+[[A, \delta(B)], C]+[[A, B], \delta(C)]$ for any $A, B, C\in \mathcal{A}$ with $F(A, B, C)\in\mathcal{Q}$. Let $\mathcal{M}$ be a finite von Neumann algebra with no central summands of type $I_{1}$. Zhao and Hao ^[21] proved that if $\delta: \mathcal{M}\rightarrow \mathcal{M}$ satisfies $\delta([[A, B], C]) = [[\delta(A), B], C]+[[A, \delta(B)], C]+[[A, B], \delta(C)]$ for any $A, B, C\in \mathcal{M}$ with $ABC = 0$, then $\delta = d+\tau$, where $d$ is a derivation from $\mathcal{M}$ into itself and $\tau$ is a nonlinear map from $\mathcal{M}$ into its center such that $\tau([[A, B], C]) = 0$ with $ABC = 0$.

Let $\mathcal{A}$ be an associative $\ast$-algebra. For $A, B\in\mathcal{A}$, denote by $[A, B]_{\ast} = AB-BA^{\ast}$ the skew Lie product of $A$ and $B$. The skew Lie product arose in representability of quadratic functionals by sesquilinear functionals ^[12,13]. In recent years, the study related to skew Lie product has attracted some authors' attention, see for example ^{[1,5,6,7,8,9,10,14,15,18,19,22]} and references therein. A map $\delta: \mathcal{A}\rightarrow \mathcal{A}$ (without the additivity (linearity) assumption) is called a global nonlinear skew Lie triple derivation if $\delta([[A, B]_{\ast}, C]_{\ast}) = [[\delta(A), B]_{\ast}, C]_{\ast}+[[A, \delta(B)]_{\ast}, C]_{\ast}+[[A, B]_{\ast}, \delta(C)]_{\ast}$ for all $A, B, C\in \mathcal{A}$. A map $\delta: \mathcal{A}\rightarrow \mathcal{A}$ is called an additive $\ast$-derivation if it is an additive derivation and satisfies $\delta(A^{\ast}) = \delta(A)^{\ast}$ for all $A\in \mathcal{A}$. Li, Zhao and Chen ^[5] proved that every global nonlinear skew Lie triple derivation on factor von Neumann algebras is an additive $\ast$-derivation. Taghavi, Nouri and Darvish ^[15] proved that every global nonlinear skew Lie triple derivation on prime $\ast$-algebras is additive. Similarly, let $F:\mathcal{A}\times \mathcal{A}\times \mathcal{A}\rightarrow \mathcal{A}$ be a map and $\mathcal{Q}$ be a proper subset of $\mathcal{A}$. If $\delta$ satisfies $\delta([[A, B]_{\ast}, C]_{\ast}) = [[\delta(A), B]_{\ast}, C]_{\ast}+[[A, \delta(B)]_{\ast}, C]_{\ast}+[[A, B]_{\ast}, \delta(C)]_{\ast}$ for any $A, B, C\in \mathcal{A}$ with $F(A, B, C)\in\mathcal{Q}$, then $\delta$ is called a non-global nonlinear skew Lie triple derivation.

Motivated by the mentioned works, we will concentrate on characterizing a kind of non-global nonlinear skew Lie triple derivations $\delta$ on factor von Neumann algebras satisfying $\delta([[A, B]_{\ast}, C]_{\ast}) = [[\delta(A), B]_{\ast}, C]_{\ast}+[[A, \delta(B)]_{\ast}, C]_{\ast}+[[A, B]_{\ast}, \delta(C)]_{\ast}$ for any $A, B, C\in \mathcal{A}$ with $A^{\ast}B^{\ast}C = 0$.

As usual, $\mathbb{C}$ denotes the complex number field. Let $H$ be a complex Hilbert space and $B(H)$ be the algebra of all bounded linear operators on $H$. Let $\mathcal{A}\subseteq B(H)$ be a factor von Neumann algebra (i.e., the center of $\mathcal{A}$ is $\mathbb{C}I$, where $I$ is the identity of $\mathcal{A}$). Recall that $\mathcal{A}$ is prime (i.e., for any $A, B\in \mathcal{A}$, $A\mathcal{A}B = \{0\}$ implies $A = 0$ or $B = 0$).

2.   Main result

The main result is the following theorem.

Theorem 2.1. Let $\mathcal{A}$ be a factor von Neumann algebra acting on a complex Hilbert space $H$ with dim$\mathcal{A} > 1$. If a map $\delta: \mathcal{A}\rightarrow \mathcal{A}$ satisfies

for any $A, B, C\in \mathcal{A}$ with $A^{\ast}B^{\ast}C = 0$, then $\delta$ is an additive $\ast$-derivation.

Let $P_{1}\in \mathcal{A}$ be a nontrivial projection. Write $P_{2} = I-P_{1}$, ${\mathcal{A}}_{ij} = P_{i}{\mathcal{A}}P_{j}$ $(i, j = 1, 2)$. Then $\mathcal{A} = {\mathcal{A}}_{11}+{\mathcal{A}}_{12}+{\mathcal{A}}_{21}+{\mathcal{A}}_{22}$. For any $A\in \mathcal{A}$, $A = A_{11}+A_{12}+A_{21}+A_{22}$, $A_{ij}\in$ ${\mathcal A}_{ij}$ $(i, j = 1, 2)$.

Lemma 2.1. (a) $\delta(P_i)^{\ast} = \delta(P_i)$ $(i = 1, 2)$;

(b) $P_i\delta(P_i)P_j = -P_i\delta(P_j)P_j$ $(1\leq i\neq j\leq 2)$.

Proof. (a) It is clear that $\delta(0) = 0$. For any $X_{21}\in\mathcal{A}_{21}$, it follows from $P_1^{\ast}P_1^{\ast}X_{21} = 0$ and $[[P_1, P_1]_{\ast}, X_{21}]_{\ast} = 0$ that

Multiplying (2.1) by $P_2$ from the left and by $P_1$ from the right, we have $X_{21}(\delta(P_1)P_1-\delta(P_1)^{\ast}P_1) = 0$. Then by the primeness of $\mathcal{A}$, we get

By $P_1^{\ast} P_2^{\ast} P_2 = 0$ and $[[P_1, P_2]_{\ast}, P_2]_{\ast} = 0$, we have

Multiplying (2.3) by $P_2$ from both sides, we see that

From $P_1^{\ast} P_1^{\ast} P_2 = 0$ and $[[P_1, P_1]_{\ast}, P_2]_{\ast} = 0$, we have

Multiplying (2.5) by $P_1$ from the left and by $P_2$ from the right, then

Multiplying (2.5) by $P_2$ from the left and by $P_1$ from the right, then

It follows from (2.2), (2.4), (2.6) and (2.7) that $\delta(P_1)^{\ast} = \delta(P_1)$. Similarly, we can obtain that $\delta(P_2)^{\ast} = \delta(P_2)$.

(b) From $P_2^{\ast} P_1^{\ast} P_2 = 0$ and $[[P_2, P_1]_{\ast}, P_2]_{\ast} = 0$, we have

Multiplying (2.8) by $P_1$ from the left and by $P_2$ from the right, we have $P_1\delta(P_1)P_2 = -P_1\delta(P_2)^{\ast}P_2.$ Then $P_1\delta(P_1)P_2 = -P_1\delta(P_2)P_2$ by (a). Similarly, we can obtain that $P_2\delta(P_2)P_1 = -P_2\delta(P_1)P_1$.

Lemma 2.2. For any $A_{ij}\in\mathcal{A}_{ij}$ $(1\leq i\neq j\leq 2)$, we have

Proof. Let $A_{12}\in\mathcal{A}_{12}$. For any $X_{12}\in\mathcal{A}_{12}$, since $A_{12}^{\ast}X_{12}^{\ast}P_2 = 0$ and $[[A_{12}, X_{12}]_{\ast}, P_{2}]_{\ast} = 0$, we have

Multiplying (2.9) by $P_2$ from both sides, we have

Replacing $X_{12}$ with $iX_{12}$ in (2.10) yields that

Combining (2.10) and (2.11), we see that $P_2\delta(A_{12})X_{12} = 0$. Then $P_2\delta(A_{12})P_1 = 0$ by the primeness of $\mathcal{A}$. Similarly, we can obtain that $P_1\delta(A_{21})P_2 = 0$.

Lemma 2.3. For any $A_{12}\in\mathcal{A}_{12}, B_{21}\in\mathcal{A}_{21}$, there exist $G_{A_{12}, B_{21}} \in\mathcal{A}_{11}, K_{A_{12}, B_{21}}\in\mathcal{A}_{22}$ such that

Proof. Let $T = \delta(A_{12}+B_{21})-\delta(A_{12})-\delta(B_{21})$. From $P_2^{\ast} (A_{12}+B_{21})^{\ast} P_{2} = P_2^{\ast} A_{12}^{\ast} P_{2} = P_2^{\ast} B_{21}^{\ast} P_{2} = 0$ and $[[P_2, B_{21}]_{\ast}, P_{2}]_{\ast} = 0$, we have

which implies

Multiplying (2.12) by $P_1$ from the left, we get $T_{12} = 0$. Similarly, $T_{21} = 0$. Let

Then $G_{A_{12}, B_{21}} \in\mathcal{A}_{11}, K_{A_{12}, B_{21}}\in\mathcal{A}_{22}$, and so $\delta(A_{12}+B_{21}) = \delta(A_{12})+\delta(B_{21})+G_{A_{12}, B_{21}}+K_{A_{12}, B_{21}}.$

Lemma 2.4. (a) $P_j\delta(P_i)P_j = 0$ $(1\leq i\neq j\leq 2)$;

(b) $P_i\delta(P_i)P_i = 0$ $(i = 1, 2)$.

Proof. (a) For any $X_{12}\in\mathcal{A}_{12}$, since $P_1^{\ast} X_{12}^{\ast} P_{1} = 0$ and $[[P_1, X_{12}]_{\ast}, P_1]_{\ast} = 0$, we have

Multiplying (2.13) by $P_1$ from the left and by $P_2$ from the right, we have

It follows from Lemma 2.2 that $X_{12}\delta(P_1)P_2 = -(P_2\delta(X_{12})P_1)^{\ast} = 0.$ Then $P_2\delta(P_1)P_2 = 0$. Similarly, $P_1\delta(P_2)P_1 = 0$.

(b) For any $X_{21}\in\mathcal{A}_{21}$, from $(iX_{21})^{\ast} P_1^{\ast} P_{1} = 0$, $[[iX_{21}, P_1]_{\ast}, P_1]_{\ast} = iX_{21}^{\ast}+iX_{21}$, Lemma 2.1(a) and Lemma 2.3, there exist $G_{iX_{21}^{\ast}, iX_{21}} \in\mathcal{A}_{11}, K_{iX_{21}^{\ast}, iX_{21}}\in\mathcal{A}_{22}$ such that

Multiplying (2.14) by $P_2$ from the left and by $P_1$ from the right, we have

By (2.15), Lemma 2.2 and the fact that $P_2\delta(P_1)P_2 = 0$, we obtain $X_{21}\delta(P_1)P_1 = 0$. Then $P_1\delta(P_1)P_1 = 0$. Similarly, $P_2\delta(P_2)P_2 = 0$.

Remark 2.1. Let $S = P_1\delta(P_1)P_2-P_2\delta(P_1)P_1$. Then $S^{\ast} = -S$ by Lemma 2.1. We define a map $\Delta: \mathcal{A}\rightarrow \mathcal{A}$ by

for any $X\in\mathcal{A}$. It is easy to verify that $\Delta$ is a map satisfying

for any $A, B, C\in \mathcal{A}$ with $A^{\ast}B^{\ast}C = 0$. By Lemmas 2.1–2.4, it follows that

${\rm{ (a) }}$ $\Delta(P_i) = 0$ $(i = 1, 2)$;

${\rm{ (b) }}$ For any $A_{ij}\in\mathcal{A}_{ij}$ $(1\leq i\neq j\leq 2)$, we have $P_j\Delta(A_{ij})P_i = 0$;

${\rm{(c)}}$ For any $A_{12}\in\mathcal{A}_{12}, B_{21}\in\mathcal{A}_{21}$, there exist $U_{A_{12}, B_{21}} \in\mathcal{A}_{11}, V_{A_{12}, B_{21}}\in\mathcal{A}_{22}$ such that

Lemma 2.5. $\Delta(\mathcal{A}_{ii})\subseteq\mathcal{A}_{ii}$ $(i = 1, 2)$.

Proof. Let $A_{11}\in\mathcal{A}_{11}$. From $A_{11}^{\ast}P_2^{\ast}P_2 = 0$, $[[A_{11}, P_2]_{\ast}, P_2]_{\ast} = 0$ and $\Delta(P_2) = 0$, we have

Multiplying (2.16) by $P_1$ from the left, we get $P_1\Delta(A_{11})P_2 = 0$. Since $P_2^{\ast}A_{11}^{\ast}P_1 = 0$, $[[P_2, A_{11}]_{\ast}, P_1]_{\ast} = 0$ and $\Delta(P_1) = \Delta(P_2) = 0$, we have

Multiplying (2.17) by $P_2$ from the left, we get $P_2\Delta(A_{11})P_1 = 0$. For any $X_{12}\in\mathcal{A}_{12}$, from $X_{12}^{\ast}A_{11}^{\ast}P_2 = 0$, $[[X_{12}, A_{11}]_{\ast}, P_{2}]_{\ast} = 0$ and $\Delta(P_{2}) = 0$, we have

Multiplying (2.18) by $P_1$ from the left, we get $-A_{11}\Delta(X_{12})^{\ast}P_{2}+X_{12}\Delta(A_{11})P_{2} = 0$. It follows from Remark 2.1(b) that $X_{12}\Delta(A_{11})P_{2} = A_{11}(P_{2}\Delta(X_{12})P_{1})^{\ast} = 0.$ Then $P_2\Delta(A_{11})P_{2} = 0$. Hence $\Delta(\mathcal{A}_{11})\subseteq\mathcal{A}_{11}$. Similarly, $\Delta(\mathcal{A}_{22})\subseteq\mathcal{A}_{22}$.

Lemma 2.6. $\Delta(\mathcal{A}_{ij})\subseteq\mathcal{A}_{ij}$ $(1\leq i\neq j\leq 2)$.

Proof. Let $A_{12}\in\mathcal{A}_{12}$. Then $P_2\Delta(A_{12})P_{1} = 0$ by Remark 2.1(b). For any $X_{12}\in\mathcal{A}_{12}$, from $X_{12}^{\ast}A_{12}^{\ast}P_1 = 0$ and $\Delta(P_{1}) = 0$, we have

Multiplying (2.19) by $P_2$ from the left and by $P_1$ from the right, then by Lemma 2.5, we get $P_2\Delta(A_{12})X_{12}^{\ast} = 0$. Hence $P_2\Delta(A_{12})P_{2} = 0$. Since $A_{12}^{\ast}X_{12}^{\ast}P_2 = 0$, $[[A_{12}, X_{12}]_{\ast}, P_{2}]_{\ast} = 0$ and $\Delta(P_{2}) = 0$, we have

Multiplying (2.20) by $P_1$ from the left and by $P_2$ from the right, then by $P_2\Delta(A_{12})P_{2} = P_2\Delta(X_{12})P_{2} = 0$, we have $P_{1}\Delta(A_{12})X_{12} = 0$. It follows that $P_1\Delta(A_{12})P_{1} = 0$. Therefore $\Delta(\mathcal{A}_{12})\subseteq\mathcal{A}_{12}$. Similarly, $\Delta(\mathcal{A}_{21})\subseteq\mathcal{A}_{21}$.

Lemma 2.7. For any $A_{ii}\in\mathcal{A}_{ii}, B_{ij}\in\mathcal{A}_{ij}, B_{ji}\in\mathcal{A}_{ji}$ $(1\leq i\neq j\leq 2)$, we have

(a) $\Delta(A_{ii}+B_{ij}) = \Delta(A_{ii})+\Delta(B_{ij})$;

(b) $\Delta(A_{ii}+B_{ji}) = \Delta(A_{ii})+\Delta(B_{ji})$.

Proof. (a) Let $T = \Delta(A_{ii}+B_{ij})-\Delta(A_{ii})-\Delta(B_{ij})$. Since $(iP_{j})^{\ast}I^{\ast}(A_{ii}+B_{ij}) = (iP_{j})^{\ast}I^{\ast}A_{ii} = (iP_{j})^{\ast}I^{\ast}B_{ij} = 0$ and $[[iP_{j}, I]_{\ast}, A_{ii}]_{\ast} = 0$, we have

which implies

Multiplying (2.21) by $P_i$ from the left, by $P_i$ from the right, by $P_j$ from both sides, respectively, we get $T_{ij} = T_{ji} = T_{jj} = 0$. Hence

For any $X_{ij}\in\mathcal{A}_{ij}$, from $(A_{ii}+B_{ij})^{\ast}X_{ij}^{\ast}P_j = A_{ii}^{\ast}X_{ij}^{\ast}P_j = B_{ij}^{\ast}X_{ij}^{\ast}P_j = 0$, $[[B_{ij}, X_{ij}]_{\ast}, P_{j}]_{\ast} = 0$ and (2.22), we have

This implies

Multiplying (2.23) by $P_j$ from the right, we see that $T_{ii}X_{ij} = 0$. Hence $T_{ii} = 0$, and so we obtain (a).

Similarly, we can show that (b) holds.

Lemma 2.8. For any $A_{ij}, B_{ij}\in\mathcal{A}_{ij}$ $(1\leq i\neq j\leq 2)$, we have

Proof. For any $A_{12}, B_{12}\in\mathcal{A}_{12}$, it follows that

Then by (2.24) and Remark 2.1(c), there exist $U_{A_{12}+B_{12}, -A_{12}^{\ast}-B_{12}^{\ast}} \in\mathcal{A}_{11}$, $V_{A_{12}+B_{12}, -A_{12}^{\ast}-B_{12}^{\ast}}$ $\in\mathcal{A}_{22}$ such that

From $(P_1+A_{12})^{\ast}(P_2+B_{12})^{\ast}P_2 = 0$, $\Delta(P_1) = \Delta(P_2) = 0$, (2.25), Lemmas 2.6 and 2.7, we have

Multiplying (2.26) by $P_1$ from the left and by $P_2$ from the right, then by Lemma 2.6 and the fact that $U_{A_{12}+B_{12}, -A_{12}^{\ast}-B_{12}^{\ast}} \in\mathcal{A}_{11}, V_{A_{12}+B_{12}, -A_{12}^{\ast}-B_{12}^{\ast}}\in\mathcal{A}_{22}$, we see that $\Delta(A_{12}+B_{12}) = \Delta(A_{12})+\Delta(B_{12})$. Similarly, we can show that $\Delta(A_{21}+B_{21}) = \Delta(A_{21})+\Delta(B_{21})$.

Lemma 2.9. For any $A_{ii}, B_{ii}\in\mathcal{A}_{ii}$ $(i = 1, 2)$, we have

Proof. For any $A_{11}, B_{11}\in\mathcal{A}_{11}, B_{12}\in\mathcal{A}_{12}$, from $A_{11}^{\ast}B_{12}^{\ast}P_2 = 0$, $\Delta(P_2) = 0$, $[[A_{11}, B_{12}]_{\ast}, P_2]_{\ast} = A_{11}B_{12}-B_{12}^{\ast}A_{11}^{\ast}$, Lemmas 2.5, 2.6 and 2.8, we have

Multiplying (2.27) by $P_1$ from the left and by $P_2$ from the right, we have

Similarly, we can show that

For any $X_{12}\in\mathcal{A}_{12}$, it follows from Lemma 2.8 and (2.28) that

It follows that $(\Delta(A_{11}+B_{11})-\Delta(A_{11})-\Delta(B_{11}))X_{12} = 0$. Then $\Delta(A_{11}+B_{11}) = \Delta(A_{11})+\Delta(B_{11})$. Similarly, we can show that $\Delta(A_{22}+B_{22}) = \Delta(A_{22})+\Delta(B_{22})$.

Lemma 2.10. For any $A_{12}\in\mathcal{A}_{12}, B_{21}\in\mathcal{A}_{21}$, we have

Proof. For any $X_{12}\in\mathcal{A}_{12}$, by $X_{12}^{\ast}(A_{12}+B_{21})^{\ast}P_1 = X_{12}^{\ast}A_{12}^{\ast}P_1 = X_{12}^{\ast}B_{21}^{\ast}P_1 = 0$,

Remark 2.1(c) and Lemma 2.9, there exist $U_{A_{12}, B_{21}} \in\mathcal{A}_{11}, V_{A_{12}, B_{21}}\in\mathcal{A}_{22}$ such that

Then

Multiplying (2.30) by $P_1$ from the right, we get $V_{A_{12}, B_{21}}X_{12}^{\ast} = 0$. Hence $V_{A_{12}, B_{21}} = 0$. Then by Remark 2.1(c), we get

For any $X_{21}\in\mathcal{A}_{21}$, from $X_{21}^{\ast}(A_{12}+B_{21})^{\ast}P_2 = X_{21}^{\ast}A_{12}^{\ast}P_2 = X_{21}^{\ast}B_{21}^{\ast}P_2 = 0$,

Lemma 2.9 and (2.31), we have

which implies

Multiplying (2.32) by $P_2$ from the right, we obtain $U_{A_{12}, B_{21}}X_{21}^{\ast} = 0$. Then $U_{A_{12}, B_{21}} = 0$. Hence we obtain the desired result.

Lemma 2.11. For any $A_{11}\in\mathcal{A}_{11}, B_{12}\in\mathcal{A}_{12}, C_{21}\in\mathcal{A}_{21}, D_{22}\in\mathcal{A}_{22}$, we have

(a) $\Delta(A_{11}+B_{12}+ C_{21}) = \Delta(A_{11})+\Delta(B_{12})+\Delta(C_{21})$;

(b) $\Delta(B_{12}+ C_{21}+D_{22}) = \Delta(B_{12})+\Delta(C_{21})+\Delta(D_{22})$.

Proof. (a) Let $T = \Delta(A_{11}+B_{12}+ C_{21})-\Delta(A_{11})-\Delta(B_{12})-\Delta(C_{21})$. From $P_2^{\ast}(A_{11}+B_{12}+ C_{21})^{\ast}P_2 = P_2^{\ast}A_{11}^{\ast}P_2 = P_2^{\ast}B_{12}^{\ast}P_2 = P_2^{\ast}C_{21}^{\ast}P_2 = 0$ and $[[P_2, A_{11}]_{\ast}, P_2]_{\ast} = [[P_2, C_{21}]_{\ast}, P_2]_{\ast} = 0$, we have

This implies

Multiplying (2.33) by $P_1$ from the left, we have $T_{12} = 0$. For any $X_{12}\in\mathcal{A}_{12}$, from $P_1^{\ast}X_{12}^{\ast}(A_{11}+B_{12}+ C_{21}) =$$P_1^{\ast}X_{12}^{\ast}A_{11} = P_1^{\ast}X_{12}^{\ast}B_{12} = P_1^{\ast}X_{12}^{\ast}C_{21} = 0$, $[[P_1, X_{12}]_{\ast}, A_{11}+B_{12}+ C_{21}]_{\ast} = X_{12}C_{21}-B_{12}X_{12}^{\ast}$ and Lemma 2.9, we have

which implies

Multiplying (2.34) by $P_1$ from both sides, we obtain $X_{12}TP_1-P_1TX_{12}^{\ast} = 0$. Then $X_{12}TP_1 = 0$ by $T_{12} = 0$. Hence $T_{21} = 0$. Multiplying (2.34) by $P_2$ from the right, we have $X_{12}TP_2 = 0$ and so $T_{22} = 0$. Let $S_{A_{11}, B_{12}, C_{21}} = T_{11}$. Then $S_{A_{11}, B_{12}, C_{21}}\in \mathcal{A}_{11}$ and

Similarly, there exists a $R_{B_{12}, C_{21}, D_{22}}\in \mathcal{A}_{22}$ such that

For any $X_{21}\in\mathcal{A}_{21}$, by $[[P_2, X_{21}]_{\ast}, A_{11}+B_{12}+ C_{21}]_{\ast} =$$-A_{11}X_{21}^{\ast}+X_{21}A_{11}+X_{21}B_{12}-C_{21}X_{21}^{\ast}$ and (2.35), there exist a $R_{-A_{11}X_{21}^{\ast}, X_{21}A_{11}, X_{21}B_{12}-C_{21}X_{21}^{\ast}}\in \mathcal{A}_{22}$ such that

From $P_2^{\ast}X_{21}^{\ast}(A_{11}+B_{12}+ C_{21}) = P_2^{\ast}X_{21}^{\ast}A_{11} =$$P_2^{\ast}X_{21}^{\ast}B_{12} = P_2^{\ast}X_{21}^{\ast}C_{21} = 0$, (2.36), Lemmas 2.9 and 2.10, we have

It follows that

Multiplying (2.37) by $P_1$ from the right, then by $R_{-A_{11}X_{21}^{\ast}, X_{21}A_{11}, X_{21}B_{12}-C_{21}X_{21}^{\ast}}\in \mathcal{A}_{22}$, we obtain $X_{21}TP_1 = 0$. Hence $S_{A_{11}, B_{12}, C_{21}} = T_{11} = 0$, and so $\Delta(A_{11}+B_{12}+ C_{21}) = \Delta(A_{11})+\Delta(B_{12})+\Delta(C_{21})$.

Similarly, we can show that (b) holds.

Lemma 2.12. For any $A_{11}\in\mathcal{A}_{11}, B_{12}\in\mathcal{A}_{12}, C_{21}\in\mathcal{A}_{21}, D_{22}\in\mathcal{A}_{22}$, we have

Proof. Let $T = \Delta(A_{11}+B_{12}+ C_{21}+D_{22})-$$\Delta(A_{11})-\Delta(B_{12})-\Delta(C_{21})-\Delta(D_{22})$. From $(A_{11}+B_{12}+ C_{21}+D_{22})^{\ast}P_1^{\ast}P_2 = A_{11}^{\ast}P_1^{\ast}P_2 =$$B_{12}^{\ast}P_1^{\ast}P_2 = C_{21}^{\ast}P_1^{\ast}P_2 =$$D_{22}^{\ast}P_1^{\ast}P_2 = 0$ and $[[A_{11}+B_{12}+ C_{21}+D_{22}, P_1]_{\ast}, P_2]_{\ast} =$$-C_{21}^{\ast}+C_{21}$, we have

This implies

Multiplying (2.38) by $P_2$ from the left, we have $T_{21} = 0$. Similarly, $T_{12} = 0$. For any $X_{12}\in\mathcal{A}_{12}$, from $P_1^{\ast}X_{12}^{\ast}(A_{11}+B_{12}+ C_{21}+D_{22}) =$$P_1^{\ast}X_{12}^{\ast}A_{11} =$$P_1^{\ast}X_{12}^{\ast}B_{12} =$$P_1^{\ast}X_{12}^{\ast}C_{21} =$$P_1^{\ast}X_{12}^{\ast}D_{22} = 0$, $[[P_1, X_{12}]_{\ast}, A_{11}+B_{12}+ C_{21}+D_{22}]_{\ast} =$$X_{12}C_{21}-B_{12}X_{12}^{\ast}+X_{12}D_{22}-D_{22}X_{12}^{\ast},$ Lemmas 2.9 and 2.12, we have

This implies

Multiplying (2.39) by $P_2$ from the right, we obtain $X_{12}TP_2 = 0$. Then $T_{22} = 0$. Similarly, $T_{11} = 0$. Hence we obtain the desired result.

Lemma 2.13. For any $A_{ii}, B_{ii}\in\mathcal{A}_{ii}, A_{ij}, B_{ij}\in\mathcal{A}_{ij}, B_{ji}\in\mathcal{A}_{ji}, B_{jj}\in\mathcal{A}_{jj}$ $(1\leq i\neq j\leq 2)$, we have

(a) $\Delta(A_{ii}B_{ij}) = \Delta(A_{ii})B_{ij}+A_{ii}\Delta(B_{ij})$;

(b) $\Delta(A_{ij}B_{jj}) = \Delta(A_{ij})B_{jj}+A_{ij}\Delta(B_{jj})$;

(c) $\Delta(A_{ii}B_{ii}) = \Delta(A_{ii})B_{ii}+A_{ii}\Delta(B_{ii})$;

(d) $\Delta(A_{ij}B_{ji}) = \Delta(A_{ij})B_{ji}+A_{ij}\Delta(B_{ji})$.

Proof. (a) It follows from (2.28) and (2.29) that (a) holds.

(b) Let $A_{12}\in\mathcal{A}_{12}, B_{22}\in\mathcal{A}_{22}$. From $A_{12}^{\ast}B_{22}^{\ast}P_2 = 0$, $\Delta(P_2) = 0$, $[[A_{12}, B_{22}]_{\ast}, P_2]_{\ast} = A_{12}B_{22}-B_{22}^{\ast}A_{12}^{\ast}$, Lemmas 2.5, 2.6 and 2.12, we have

Multiplying (2.40) by $P_1$ from the left and by $P_2$ from the right, we have $\Delta(A_{12}B_{22}) = \Delta(A_{12})B_{22}+A_{12}\Delta(B_{22}).$ Similarly, $\Delta(A_{21}B_{11}) = \Delta(A_{21})B_{11}+A_{21}\Delta(B_{11}).$

(c) Let $A_{11}, B_{11}\in\mathcal{A}_{11}, X_{12}\in\mathcal{A}_{12}$. It follows from (a) that

It follows that $(\Delta(A_{11}B_{11})-\Delta(A_{11})B_{11}-A_{11}\Delta(B_{11}))X_{12} = 0$. Hence $\Delta(A_{11}B_{11}) = \Delta(A_{11})B_{11}+A_{11}\Delta(B_{11})$. Similarly, $\Delta(A_{22}B_{22}) = \Delta(A_{22})B_{22}+A_{22}\Delta(B_{22})$.

(d) Let $A_{12}\in\mathcal{A}_{12}, B_{21}\in\mathcal{A}_{21}$. From $B_{21}^{\ast}P_{1}^{\ast}A_{12} = 0$, $\Delta(P_1) = 0$, $[[B_{21}, P_{1}]_{\ast}, A_{12}]_{\ast} = B_{21}A_{12}+A_{12}B_{21}$, Lemmas 2.6 and 2.12, we have

Multiplying (2.41) by $P_1$ from both sides, we have $\Delta(A_{12}B_{21}) = \Delta(A_{12})B_{21}+A_{12}\Delta(B_{21}).$ Similarly, $\Delta(A_{21}B_{12}) = \Delta(A_{21})B_{12}+A_{21}\Delta(B_{12}).$

Now, we give the proof of Theorem 2.1 in the following.

Proof of Theorem 2.1. By Lemmas 2.5, 2.6, 2.8, 2.9, 2.12 and 2.13, it is easy to verify that $\Delta$ is an additive derivation on $\mathcal{A}$. Let $A_{ij}\in\mathcal{A}_{ij}$ $(1\leq i\neq j\leq 2)$. By $A_{ij}^{\ast}P_{j}^{\ast}P_{j} = 0$, $\Delta(P_j) = 0$ and Lemma 2.6, we have

It follows that

Let $A_{ii}\in\mathcal{A}_{ii}$, $X_{ji}\in\mathcal{A}_{ji}$ $(1\leq i\neq j\leq 2)$. Since $A_{ii}^{\ast}P_{i}^{\ast}X_{ji} = 0$, $\Delta(P_i) = 0$, $[[A_{ii}, P_{i}]_{\ast}, X_{ji}]_{\ast} = X_{ji}A_{ii}-X_{ji}A_{ii}^{\ast}$, Lemmas 2.5, 2.6 and 2.13(b), we have

It follows that $X_{ji}(\Delta(A_{ii}^{\ast})-\Delta(A_{ii})^{\ast}) = 0$. Then

For any $A\in\mathcal{A}$, we have $A = \sum_{i, j = 1}^{2}A_{ij}$. By (2.42), (2.43) and the additivity of $\Delta$ on $\mathcal{A}$, it follows that

Hence $\Delta$ is an additive $\ast$-derivation. Therefore, $\delta$ is an additive $\ast$-derivation on $\mathcal{A}$ by Remark 2.1.

3.   Conclusions

In this paper, we gave the characterization of a kind of non-global nonlinear skew Lie triple derivations on factor von Neumann algebras.

Acknowledgments

This research is supported by Scientific Research Project of Shangluo University (21SKY104).

Conflict of interest

The authors declare that there are no conflicts of interest.