Multicellular model of angiogenesis

1.   Introduction

Let $\mathcal{A}$ be an associative $\ast$-algebra. For $A, B\in\mathcal{A}$, denote by $[A, B] = AB-BA$, $[A, B]_{\ast} = AB-BA^{\ast}$ and $A\bullet B = AB+BA^{\ast}$ the Lie product, skew Lie product and skew Jordan product of $A$ and $B$, respectively. A map $\delta: \mathcal{A}\rightarrow \mathcal{A}$ is called an additive derivation if $\delta(A+B) = \delta(A)+\delta(B)$ and $\delta(AB) = \delta(A)B+A\delta(B)$ for all $A, B\in \mathcal{A}$. Moreover, $\delta$ is called an additive $\ast$-derivation if it is an additive derivation and satisfies $\delta(A^{\ast}) = \delta(A)^{\ast}$ for all $A\in \mathcal{A}$. A map $\varphi: \mathcal{A}\rightarrow \mathcal{A}$ (without the additivity assumption) is called a nonlinear Lie derivation (resp. nonlinear Lie triple derivation) if $\varphi([A, B]) = [\varphi(A), B]+[A, \varphi(B)]$ for all $A, B\in \mathcal{A}$ (resp. $\varphi([[A, B], C]) = [[\varphi(A), B], C]+[[A, \varphi(B)], C]+[[A, B], \varphi(C)]$ for all $A, B, C\in \mathcal{A}$). In the past years, nonlinear Lie derivations and Lie triple derivations on various algebras have been studied. Chen and Zhang ^[1], Yu and Zhang ^[11], and Yang ^[9] gave the structure of nonlinear Lie derivations on upper triangular matrices, triangular algebras, and incidence algebras, respectively. Ji, Liu, and Zhao ^[4] proved that every nonlinear Lie triple derivation on triangular algebras can be expressed as the sum of an additive derivation and a center value map vanishing on Lie triple products. A map $\varphi: \mathcal{A}\rightarrow \mathcal{A}$ (without the additivity assumption) is called a nonlinear skew Jordan derivation (resp. nonlinear skew Jordan triple derivation) if $\varphi(A\bullet B) = \varphi(A)\bullet B+A\bullet \varphi(B)$ for all $A, B\in \mathcal{A}$ (resp. $\varphi(A\bullet B\bullet C) = \varphi(A)\bullet B\bullet C+A\bullet \varphi(B)\bullet C+A\bullet B\bullet \varphi(C)$ for all $A, B, C\in \mathcal{A}$). Nonlinear skew Jordan derivations and skew Jordan triple derivations on various algebras also have been studied by some authors. Taghavi, Rohi, and Darvish ^[8], and Zhang ^[12] proved every nonlinear skew Jordan derivation on factor von Neumann algebras is an additive $\ast$-derivation, respectively. Darvish et al.^[2] proved that every nonlinear skew Jordan triple derivation on prime $\ast$-algebras is an additive $\ast$-derivation under some assumption. Recently, derivations (nonlinear maps) corresponding to (preserving) the new products of the mixture of (skew) Lie product and skew Jordan product have attracted the attention of several authors. Li and Zhang ^[6] studied nonlinear mixed Jordan triple $\ast$-derivations on factor von Neumann algebras. Yang and Zhang ^[10] characterized nonlinear maps preserving the second mixed Lie triple products on factor von Neumann algebras. Zhou, Yang, and Zhang ^[15] proved that any map $\varphi$ from a unital prime $\ast$-algebra $\mathcal{A}$ to itself satisfying $\varphi([[A, B]_{\ast}, C]) = [[\varphi(A), B]_{\ast}, C]+[[A, \varphi(B)]_{\ast}, C]+[[A, B]_{\ast}, \varphi(C)]$ for all $A, B, C\in \mathcal{A}$ is an additive $\ast$-derivation. Let $\mathcal{A}$ and $\mathcal{B}$ be two factors with dim$\mathcal{A} > 4$. Zhao, Li, and Chen ^[14] characterized the concrete structure of any bijective map $\varphi: \mathcal{A}\rightarrow \mathcal{B}$ satisfying

for all $A, B, C\in \mathcal{A}$.

The conditions under which linear (nonlinear) Lie (triple) derivations of operator algebras can be completely determined by the action on some proper subsets of these operator algebras were considered. Let $\mathcal{M}$ be a factor with dim$\mathcal{M} > 1$. Liu ^[7] investigated any linear map $\varphi: \mathcal{M}\rightarrow \mathcal{M}$ satisfying $\varphi([[A, B], C]) = [[\varphi(A), B], C]+[[A, \varphi(B)], C]+[[A, B], \varphi(C)]$ for any $A, B, C\in \mathcal{A}$ with $AB = 0$ (resp. $AB = P$, where $P$ is a fixed non-trivial projection of $\mathcal{M}$). Zhao and Hao ^[13] proved that if a nonlinear map $\varphi$ from a finite von Neumann algebra $\mathcal{M}$ with no central summands of type $I_{1}$ to itself 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 an additive 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 $ABC = 0$. In sequel, we introduce the notion of non-global nonlinear mixed skew Jordan Lie triple derivation. 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 $\varphi$ satisfies

for any $A, B, C\in \mathcal{A}$ with $F(A, B, C)\in\mathcal{Q}$, then $\varphi$ is called a non-global nonlinear mixed skew Jordan Lie triple derivation.

A ring $\mathcal{R}$ is called a $\ast$-ring if there is an additive map $\ast: \mathcal{R}\rightarrow \mathcal{R}$ satisfying $(AB)^{\ast} = B^{\ast}A^{\ast}$ and $(A^{\ast})^{\ast} = A$ for all $A, B\in \mathcal{R}$. $\mathcal{R}$ is called prime when for $A, B\in \mathcal{R}$, if $A\mathcal{R}B = \{0\}$, then $A = 0$ or $B = 0$. Let $Z(\mathcal {R})$ be the center of $\mathcal {R}$ and $Z_{c} (\mathcal {R}) = \{A\in Z(\mathcal {R}): A^\ast = -A\}$ be the anti-symmetric center of $\mathcal{R}$. Motivated by the above-mentioned works, we will study the concrete structure of a kind of non-global nonlinear mixed skew Jordan Lie triple derivations $\varphi$ on prime $\ast$-rings $\mathcal{R}$ satisfying $\varphi([A\bullet B, C]) = [\varphi(A)\bullet B, C]+[A\bullet\varphi(B), C]+[A\bullet B, \varphi(C)]$ for any $A, B, C\in \mathcal{R}$ with $ABC^{\ast} = 0$.

2.   Main results

The main result is the following theorem:

Theorem 2.1. Let $\mathcal{R}$ be a 2-torsion free unital prime $\ast$-ring containing a nontrivial symmetric idempotent. If a map $\varphi:\mathcal{R}\rightarrow \mathcal{R}$ satisfies

for any $A, B, C\in \mathcal{R}$ with $ABC^{\ast} = 0$, then there exists an additive $\ast$-derivation $\Theta$ of $\mathcal{R}$ and a nonlinear map $g:\mathcal{R}\rightarrow Z_{c}(\mathcal {R})$ such that

for any $A\in \mathcal{R}$.

Let $P\in\mathcal{R}$ be the nontrivial symmetric idempotent. Write $P_{1} = P$, $P_{2} = I-P_{1}$, ${\mathcal R}_{ij} = P_{i}{\mathcal{R}}P_{j}$ $(i, j = 1, 2)$, then $\mathcal{R}$ = ${\mathcal R}_{11}$+${\mathcal R}_{12}$+${\mathcal R}_{21}$+${\mathcal R}_{22}$. For every $A\in \mathcal{R}$, $A = A_{11}+A_{12}+A_{21}+A_{22}$, where $A_{ij}\in$ ${\mathcal R}_{ij}$ $(i, j = 1, 2)$.

Lemma 2.1. For every $A_{ii}\in\mathcal{R}_{ii}, B_{ij}\in\mathcal{R}_{ij}, C_{ji}\in\mathcal{R}_{ji}$ $(1\leq i\neq j\leq 2)$, we have

Proof: Clearly, $\varphi(0) = 0$. Let

Next, we show that $T = 0$. For any $X_{ij}\in\mathcal{R}_{ij}$ with $1\leq i\neq j\leq2$, since

and

We have

This implies that

Multiplying Eq (2.1) by $P_{j}$ from the right, we have $X_{ij}TP_{j} = 0$. Hence, $T_{jj} = 0$ by the primeness of $\mathcal{R}$.

From

and

We have

It follows that

Multiplying Eq (2.2) by $P_{j}$ from the right and by the fact that $T_{jj} = 0$, we have $P_{i}TX_{ij} = 0$. Thus, $T_{ii} = 0$.

Since

and

We have

Then,

Multiplying Eq (2.3) by $P_{i}$ from the left, we obtain $T_{ij} = 0$.

From

and

We have

It follows that

Multiplying Eq (2.4) by $P_{j}$ from the left, we obtain $T_{ji} = 0$. Therefore, $T = 0$.

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

Proof: Since $(P_{i}-A_{ij})(P_{j}-B_{ij})P_{i}^{\ast} = 0$,

and

We have from Lemma 2.1 that

Then, $\varphi(A_{ij}+B_{ij}) = \varphi(A_{ij})+\varphi(B_{ij})$.

Lemma 2.3. For every $A_{ii}, B_{ii}\in\mathcal{R}_{ii}$ $(i = 1, 2)$, we have

Proof: Let

Since

and

We have

This gives that

Multiplying Eq (2.5) by $P_{i}$ from the left, by $P_{i}$ from the right, respectively, we obtain $T_{ij} = 0$, $T_{ji} = 0$, respectively.

For any $X_{ij}\in \mathcal{R}_{ij}$ with $1\leq i\neq j\leq 2$, since

and

We have

It follows that

Multiplying Eq (2.6) by $P_{i}$ from the left, we have $X_{ij}TP_{j} = 0$, and so $T_{jj} = 0$.

For any $X_{ji}\in \mathcal{R}_{ji}$ with $1\leq i\neq j\leq2$, since

We have from Lemmas 2.1 and 2.2 that

which gives that

Multiplying Eq (2.7) by $P_{i}$ from the right, we have $X_{ji}TP_{i} = 0$. Hence, $T_{ii} = 0$.

Lemma 2.4. (a) $\varphi(P_i)^{\ast} = \varphi(P_i)$ $(i = 1, 2)$;

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

Proof: (a) Let $1\leq i\neq j\leq 2$. Since $P_{j}P_{i}P_{j}^{\ast} = 0$ and $[P_{j}\bullet P_{i}, P_{j}] = 0$, we have

Multiplying Eq (2.8) by $P_{i}$ from the left and by $P_{j}$ from the right, we obtain

It follows that

Multiplying Eq (2.8) by $P_{j}$ from the left and by $P_{i}$ from the right, we obtain

Comparing Eqs (2.10) and (2.11), we obtain

Since $P_{j}P_{j}P_{i}^{\ast} = 0$ and $[P_{j}\bullet P_{j}, P_{i}] = 0$, we have

Multiplying Eq (2.13) by $P_{i}$ from the left and by $P_{j}$ from the right, we have

Since $\mathcal{R}$ is 2-torsion free, we have from Eq (2.14) that

From $P_{i}P_{i}P_{j}^{\ast} = 0$ and $[P_{i}\bullet P_{i}, P_{j}] = 0$, we have

Multiplying Eq (2.16) by $P_{i}$ from the left and by $P_{j}$ from the right, then by Eq. (2.15), we have

For any $X_{ij}\in\mathcal{R}_{ij}$ with $1\leq i\neq j\leq 2$, we have from $X_{ij}P_{i}P_{i}^{\ast} = 0$ and $[X_{ij}\bullet P_{i}, P_{i}] = 0$ that

Multiplying Eq (2.18) by $P_{i}$ from the left and by $P_{j}$ from the right, we have

Multiplying Eq (2.18) by $P_{j}$ from the left and by $P_{i}$ from the right, we have $P_{j}\varphi(X_{ij})P_{i}+P_{j}\varphi(P_{i})X_{ij}^{\ast} = 0$. Then,

Comparing Eqs (2.19) and (2.20), we obtain $X_{ij}(P_{j}\varphi(P_{i})P_{j}-P_{j}\varphi(P_{i})^{\ast}P_{j}) = 0$. It follows that

From $P_{i}P_{i}X_{ij}^{\ast} = 0$ and $[P_{i}\bullet P_{i}, X_{ij}] = 2X_{ij}$, we have

Multiplying Eq (2.22) by $P_{i}$ from the left and by $P_{j}$ from the right, we have

It follows from Eq (2.23) and Lemma 2.2 that $3P_{i}\varphi(P_{i})X_{ij}+P_{i}\varphi(P_{i})^{\ast}X_{ij} = 0$, and so

For any $X_{ji}\in\mathcal{R}_{ji}$ with $1\leq i\neq j\leq 2$, since $P_{i}X_{ji}P_{i}^{\ast} = 0$ and $[P_{i}\bullet X_{ji}, P_{i}] = X_{ji}$, we have

Multiplying Eq (2.25) by $P_{j}$ from the left and by $P_{i}$ from the right, we have $X_{ji}(P_{i}\varphi(P_{i})^{\ast}P_{i}+P_{i}\varphi(P_{i})P_{i}) = 0$. Then,

By Eqs (2.24) and (2.26), we obtain that $2P_{i}\varphi(P_{i})P_{i} = 0$. Then by $\mathcal{R}$ is 2-torsion free, we obtain

From Eqs (2.12), (2.17), (2.21), and (2.27), we can see that $\varphi(P_{i})^{\ast} = \varphi(P_{i})$.

(b) It follows from Eq (2.9) and (a) that (b) holds.

Lemma 2.5. $P_{j}\varphi(P_{i})P_{j} = 0$ $(1\leq i\neq j\leq 2)$.

Proof: For any $X_{ij}\in\mathcal{R}_{ij}$ with $1\leq i\neq j\leq 2$, since $P_{j}X_{ij}P_{j}^{\ast} = 0$ and $[P_{j}\bullet X_{ij}, P_{j}] = X_{ij}$, we have

Multiplying Eq (2.28) by $P_{j}$ from the left and by $P_{i}$ from the right, we obtain $2P_{j}\varphi(X_{ij})P_{i} = 0$, and so,

Combining Eqs (2.19) and (2.29), we have $X_{ij}\varphi(P_{i})P_{j} = 0$. It follows that $P_{j}\varphi(P_{i})P_{j} = 0$.

Lemma 2.6. $P_{i}\varphi(P_{i})P_{i} = 0$ $(i = 1, 2)$.

Proof: For any $X_{ij}\in\mathcal{R}_{ij}$ with $1\leq i\neq j\leq 2$, since

we have from Lemma 2.4 that

Multiplying Eq (2.30) by $P_{i}$ from the left and by $P_{j}$ from the right, we have

It follows from Eq (2.31) and Lemma 2.2 that $2P_{i}\varphi(P_{i})X_{ij} = 0$, and so $P_{i}\varphi(P_{i})X_{ij} = 0$. Hence, $P_{i}\varphi(P_{i})P_{i} = 0$.

Let $T = P_{1}\varphi(P_{1})P_{2}-P_{2}\varphi(P_{1})P_{1}$. Then, $T^{\ast} = -T$ by Lemma 2.4. We define a map $\Phi: \mathcal{R}\rightarrow \mathcal{R}$ by

for all $A\in \mathcal{R}$.

Remark 2.1. It is easy to check that $\Phi$ also satisfies

for any $A, B, C\in \mathcal{R}$ with $ABC^{\ast} = 0$. By Lemmas 2.1–2.6, it follows that

(a) For every $A_{ii}\in\mathcal{R}_{ii}, B_{ij}\in\mathcal{R}_{ij}, C_{ji}\in\mathcal{R}_{ji}$ $(1\leq i\neq j\leq 2)$, we have

(b) For every $A_{ij}, B_{ij}\in\mathcal{R}_{ij}$ $(1\leq i\neq j\leq 2)$, we have

(c) For every $A_{ii}, B_{ii}\in\mathcal{R}_{ii}$ $(i = 1, 2)$, we have

(d) $\Phi(P_i) = 0$ $(i = 1, 2)$.

Lemma 2.7. $\ \Phi(\mathcal{R}_{ij})\subseteq\mathcal{R}_{ij}$ $(1\leq i\neq j\leq 2)$.

Proof: Let $A_{ij}\in\mathcal{R}_{ij}$ with $1\leq i\neq j\leq 2$. From $A_{ij}P_{i}P_{i}^{\ast} = 0, [A_{ij}\bullet P_{i}, P_{i}] = 0$ and $\Phi(P_i) = 0$, we have

Multiplying Eq (2.32) by $P_{j}$ from the left, we obtain $P_{j}\Phi(A_{ij})P_{i} = 0$.

Since $P_{j}(-A_{ij})P_{i}^{\ast} = 0, [P_j\bullet(-A_{ij}), P_i] = A_{ij}$ and $\Phi(P_i) = \Phi(P_{j}) = 0$, we have

Multiplying Eq (2.33) by $P_{i}$ from both sides, by $P_{j}$ from both sides, respectively, we have $P_{i}\Phi(A_{ij})P_{i} = 0$, $P_{j}\Phi(A_{ij})P_{j} = 0$, respectively. Therefore, $\Phi(\mathcal{R}_{ij})\subseteq\mathcal{R}_{ij}$.

Lemma 2.8. $\Phi(\mathcal{R}_{ii})\subseteq\mathcal{R}_{ii}$ $(i = 1, 2)$.

Proof: It follows from $P_{i}A_{ii}P_{j}^{\ast} = 0, [P_i\bullet A_{ii}, P_j] = 0$ and $\Phi(P_i) = \Phi(P_{j}) = 0$ that

Multiplying Eq (2.34) by $P_{i}$ from the left, by $P_{j}$ from the left, respectively, we have $P_{i}\Phi(A_{ii})P_{j} = 0$, $P_{j}\Phi(A_{ii})P_{i} = 0$, respectively.

For any $X_{ij}\in\mathcal{R}_{ij}$ with $1\leq i\neq j\leq 2$, since $X_{ij}A_{ii}P_{j}^{\ast} = 0, [X_{ij}\bullet A_{ii}, P_{j}] = 0$ and $\Phi(P_{j}) = 0$, we have

Multiplying Eq (2.35) by $P_{i}$ from the left, and by Lemma 2.7, we have

Then, $P_{j}\Phi(A_{ii})P_{j} = 0$. Hence, $\Phi(\mathcal{R}_{ii})\subseteq\mathcal{R}_{ii}$.

Lemma 2.9. For every $A_{ii}\in\mathcal{R}_{ii}, B_{ij}\in\mathcal{R}_{ij}, C_{ji}\in\mathcal{R}_{ji}, D_{jj}\in\mathcal{R}_{jj}$ $(1\leq i\neq j\leq 2)$, we have

(a) $\Phi(A_{ii}+B_{ij}+D_{jj}) = \Phi(A_{ii})+\Phi(B_{ij})+\Phi(D_{jj})$;

(b) $\Phi(A_{ii}+C_{ji}+D_{jj}) = \Phi(A_{ii})+\Phi(C_{ji})+\Phi(D_{jj})$.

Proof: (a) Let

Since

and

We have

This implies that

Multiplying Eq (2.36) by $P_{i}$ from the left, by $P_{j}$ from the left, respectively, we have $T_{ij} = 0$, $T_{ji} = 0$, respectively.

For any $X_{ij}\in\mathcal{R}_{ij}$ with $1\leq i\neq j\leq 2$, from

and

We have

which implies that

Multiplying Eq (2.37) by $P_{j}$ from the right, we obtain $X_{ij}TP_{j} = 0$, and so $T_{jj} = 0$.

For any $X_{ji}\in\mathcal{R}_{ji}$ with $1\leq i\neq j\leq 2$, since

and

We have from Remark 2.1 (b) that

It follows that

Multiplying Eq (2.38) by $P_{i}$ from the right and by the fact that $T_{jj} = 0$, we obtain $X_{ji}T^{\ast}P_{i} = 0$ and so $T_{ii} = 0$. Hence, $T = 0$.

(b) Similarly, we can show that (b) holds.

Lemma 2.10. For $A\in\mathcal{R}$, there exists a map $g:\mathcal{R}\rightarrow Z_{c} (\mathcal {R})$ such that

Proof: For $A\in\mathcal{R}$, write $A = \sum_{i, j = 1}^{2}A_{ij}$. Let

For any $X_{ij}\in\mathcal{R}_{ij}$ with $1\leq i\neq j\leq 2$, since

and

We have from Lemma 2.9 and Remark 2.1 (b) that

It follows that

Multiplying Eq (2.40) by $P_{i}$ from the right, we obtain $X_{ij}TP_{i} = 0$. Then, $T_{ji} = 0$. Multiplying Eq (2.40) by $P_{i}$ from the left and by $P_{j}$ from the right, we have $P_{i}TX_{ij} = X_{ij}TP_{j}$. It follows from [5, Lemma 1.6] that $T_{ii}+T_{jj}\in Z(\mathcal {R})$. Similarly, we can show that $T_{ij} = 0$.

Since $(A_{ii}+A_{ij}+A_{ji}+A_{jj})P_{i}X_{ij}^{\ast} = 0$,

and

We have from Lemma 2.9 that

which implies that

Multiplying Eq (2.41) by $P_{i}$ from the left and by $P_{j}$ from the right, we have $P_{i}TX_{ij}+P_{i}T^{\ast}X_{ij} = 0$, and so $T_{ii} = -T_{ii}^{\ast}$. Similarly, $T_{jj} = -T_{jj}^{\ast}$. It follows that $T_{ii}+T_{jj}\in Z_{c} (\mathcal {R})$ by $T_{ii}+T_{jj}\in Z(\mathcal {R})$. Define a map $g:\mathcal{R}\rightarrow Z_{c} (\mathcal {R})$ by

Combining Eqs (2.39) and (2.42), we can obtain the desired result.

Lemma 2.11. For every $A_{ii}, B_{ii}\in\mathcal{R}_{ii}, A_{ij}, B_{ij}\in\mathcal{R}_{ij}, B_{ji}\in\mathcal{R}_{ji}, B_{jj}\in\mathcal{R}_{jj}$ $(1\leq i\neq j\leq 2)$, we have

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

(b) $\Phi(A_{ii}B_{ii}) = \Phi(A_{ii})B_{ii}+A_{ii}\Phi(B_{ii})$;

(c) $\Phi(A_{ij}B_{ji}) = \Phi(A_{ij})B_{ji}+A_{ij}\Phi(B_{ji})$;

(d) $\Phi(A_{ij}B_{jj}) = \Phi(A_{ij})B_{jj}+A_{ij}\Phi(B_{jj})$.

Proof: (a) Since

we have from Remark 2.1 (c), (d), and Lemmas 2.7 and 2.8 that

(b) For any $X_{ij}\in\mathcal{R}_{ij}$ with $1\leq i\neq j\leq 2$, it follows from (a) that

This yields that $(\Phi(A_{ii}B_{ii})-\Phi(A_{ii})B_{ii}-A_{ii}\Phi(B_{ii}))X_{ij} = 0$. Hence, (b) holds by Lemma 2.8.

(c) For any $X_{ij}\in\mathcal{R}_{ij}$ with $1\leq i\neq j\leq 2$, since $A_{ij}B_{ji}X_{ij}^{\ast} = 0$ and $[A_{ij}\bullet B_{ji}, X_{ij}] = A_{ij}B_{ji}X_{ij}$, we have from (a) and Lemma 2.7 that

Then, $(\Phi(A_{ij}B_{ji})-\Phi(A_{ij})B_{ji}-A_{ij}\Phi(B_{ji}))X_{ij} = 0$, and so (c) holds by Lemmas 2.7 and 2.8.

(d) For any $X_{ji}\in\mathcal{R}_{ji}$ with $1\leq i\neq j\leq 2$, we have from (a) and (c) that

It follows that $(\Phi(A_{ij}B_{jj})-\Phi(A_{ij})B_{jj}-A_{ij}\Phi(B_{jj}))X_{ji} = 0$. Then, (d) holds by Lemmas 2.7 and 2.8.

Lemma 2.12. For every $A_{ij}\in\mathcal{R}_{ij}$ $(i, j = 1, 2)$, we have

Proof: Let $1\leq i\neq j\leq 2$. Since $A_{ij}P_{j}(P_{i})^{\ast} = 0, [A_{ij}\bullet P_{j}, P_i] = A_{ij}^{\ast}-A_{ij}$, we have from Remark 2.1 (a), (b), (d), and Lemma 2.7 that

Hence, $\Phi(A_{ij}^{\ast}) = \Phi(A_{ij})^{\ast}$.

For any $X_{ij}\in\mathcal{R}_{ij}$ with $1\leq i\neq j\leq 2$, since $A_{ii}P_{i}X_{ij}^{\ast} = 0, [A_{ii}\bullet P_i, X_{ij}] = A_{ii}X_{ij}+A_{ii}^{\ast}X_{ij}$, we have from Remark 2.1 (a), (b), and Lemmas 2.7, 2.8, and 2.11 that

It follows that $(\Phi(A_{ii}^{\ast})-\Phi(A_{ii})^{\ast})X_{ij} = 0$. Then, $\Phi(A_{ii}^{\ast}) = \Phi(A_{ii})^{\ast}$ by Lemma 2.8.

Proof of Theorem 2.1: We define a map $\Psi:\mathcal{R}\rightarrow \mathcal{R}$ by

for all $A\in\mathcal{R}$. For every $A, B\in\mathcal{R}$, let $A = \sum_{i, j = 1}^{2}A_{ij}$, $B = \sum_{i, j = 1}^{2}B_{ij}$. From Remark 2.1 (b), (c), and Lemma 2.10, we have

Hence, $\Psi$ is additive on $\mathcal{R}$. Next, it follows from Remark 2.1 (b), (c), and Lemmas 2.7, 2.8, 2.10, and 2.11 that

It follows that $\Psi$ is an additive derivation on $\mathcal{R}$. Moreover, we have from Lemmas 2.10 and 2.12 that

Consequently, $\Psi$ is an additive $\ast$-derivation on $\mathcal{R}$. By the definitions of $\Phi$ and $\Psi$, we can see that $\varphi(A) = \Theta(A)+g(A)$, where $\Theta(A) = \Psi(A)+[A, T]$ is an additive $\ast$-derivation.

As applications of Theorem 2.1, we have the following corollaries.

Corollary 2.1. Let $\mathcal{A}$ be a factor von Neumann algebra acting on a complex Hilbert space $\mathcal{H}$ with dim$\mathcal {A}$ > 1. If a map $\varphi:\mathcal{A}\rightarrow \mathcal{A}$ satisfies $\varphi([A\bullet B, C]) = [\varphi(A)\bullet B, C]+[A\bullet\varphi(B), C] +[A\bullet B, \varphi(C)]$ for any $A, B, C\in \mathcal{A}$ with $ABC^{\ast} = 0$, then there exists an additive $\ast$-derivation $\Theta$ of $\mathcal{A}$ and a nonlinear map $g:\mathcal{A}\rightarrow \mathrm{i}\mathbb{R}I$ such that $\varphi(A) = \Theta(A)+g(A)$ for any $A\in \mathcal{A}$.

Corollary 2.2. Let $\mathcal{H}$ be an infinite dimensional complex Hilbert space and $\mathcal{B(H)}$ be the algebra of all linear bounded operators on $\mathcal{H}$. If a map $\varphi:\mathcal{B(H)}\rightarrow \mathcal{B(H)}$ satisfies $\varphi([A\bullet B, C]) = [\varphi(A)\bullet B, C]+[A\bullet\varphi(B), C] +[A\bullet B, \varphi(C)]$ for any $A, B, C\in \mathcal{B(H)}$ with $ABC^{\ast} = 0$, then there exists $T\in\mathcal{B(H)}$ satisfying $T+T^{\ast} = 0$ and a nonlinear map $g:\mathcal{B(H)}\rightarrow \mathrm{i}\mathbb{R}I$ such that

for any $A\in \mathcal{B(H)}$.

Proof: It follows from Theorem 2.1 that there exists an additive $\ast$-derivation $\Theta$ of $\mathcal{B(H)}$ and a nonlinear map $g:\mathcal{B(H)}\rightarrow \mathrm{i}\mathbb{R}I$ such that $\varphi(A) = \Theta(A)+g(A)$ for any $A\in \mathcal{B(H)}$. By the result of ^[3], $\Theta$ is linear and, so it is inner. Hence, there exists $S\in \mathcal{B(H)}$ such that $\Theta(A) = AS-SA$ for any $A\in \mathcal{B(H)}$. Therefore,

for all $A\in \mathcal{B(H)}$. From this, we can see that $S+S^{\ast} = \lambda I$ for some $\lambda \in \mathbb{R}$. Write $T = S-\frac{1}{2}\lambda I$ and so $T+T^{\ast} = 0$. Hence $\varphi(A) = AT-TA+g(A)$ for any $A\in \mathcal{B(H)}$.

Corollary 2.3. Let $\mathcal{A}$ be a standard operator algebra on an infinite dimensional complex Hilbert space $\mathcal{H}$ with dim$\mathcal {A}$ > 1, which is closed under the adjoint operation and contains a nontrivial projection. If a map $\varphi:\mathcal{A}\rightarrow \mathcal{A}$ satisfies $\varphi([A\bullet B, C]) = [\varphi(A)\bullet B, C]+[A\bullet\varphi(B), C] +[A\bullet B, \varphi(C)]$ for any $A, B, C\in \mathcal{A}$ with $ABC^{\ast} = 0$, then there exists an additive $\ast$-derivation $\Theta$ of $\mathcal{A}$ and a nonlinear map $g:\mathcal{A}\rightarrow \mathrm{i}\mathbb{R}I$ such that $\varphi(A) = \Theta(A)+g(A)$ for any $A\in \mathcal{A}$.

3.   Conclusions

In this paper, we characterized the structure of a specific type of non-global nonlinear mixed skew Jordan Lie triple derivations on prime $\ast$-rings. Moreover, we applied the above result to factor von Neumann algebras and standard operator algebras.

Author contributions

Fenhong Li: Writing–original draft, writing–review & editing; Liang Kong: Writing–original draft, writing–review & editing, funding acquisition; Chao Li: Writing–original draft, writing–review & editing. All authors are contributed equally. All authors have read and approved the final version of the manuscript for publication.

Use of Generative-AI tools declaration

The authors declare that they have not used Artificial Intelligence (AI) tools in the creation of this article

Acknowledgments

The authors wish to thank the anonymous referees for their valuable comments and suggestions that improved the presentation of the paper. This research is supported by the Scientific Research Project of Shangluo University (21SKY104, 22KYPY05), and the Shaanxi Provincial Natural Science Basic Research Program (2024JC-YBMS-062).

Conflict of interest

The authors declare that there are no conflicts of interest.