1. Introduction

Let \( {\mathcal{A}} \) be a \( * \)-algebra over the complex field \( {𝔺} \). Given \( A,B\in\mathcal{A} \), we call the product \( [A,B]_{\ast}=AB-BA^{\ast} \) the skew Lie product and \( A\bullet B=AB+BA^{\ast} \) the Jordan \( * \)-product. The two new products are fairly meaningful and important and have been studied by many authors (see [1,2,3,4,5,6,7,8,9,10,11,12,13,14,15]). Recall that an additive map \( \Phi:\mathcal{A}\rightarrow{\mathcal{A}} \) is said to be an additive derivation if

$$ \Phi(AB)=\Phi(A)B+A\Phi(B) $$

for all \( A,B\in{\mathcal{A}} \). Furthermore, \( \Phi \) is said to be an additive \( * \)-derivation if \( \Phi \) is an additive derivation and satisfies \( \Phi(A^{*})=\Phi(A)^{*} \) for all \( A\in{\mathcal{A}} \). A map (without the additivity assumption) \( \Phi:{\mathcal{A}}\rightarrow{\mathcal{A}} \) is said to be a nonlinear Jordan \( * \)-derivation or a nonlinear skew Lie derivation if \( \Phi(A\bullet B)=\Phi(A)\bullet B+A\bullet\Phi(B) \) or \( \Phi([A,B]_{\ast})=[\Phi(A),B]_{\ast}+[A,\Phi(B)]_{\ast} \) for all \( A,B\in{\mathcal{A}} \). Many authors have paid more attentions on the problem about Jordan \( * \)-derivations, skew Lie derivations and triple derivations, such as Jordan triple \( * \)-derivations and skew Lie triple derivations (see [15,16,17,18,19,20,21,22,23,24]). For example, Taghavi et al. [24] investigated a nonlinear \( \lambda \)-Jordan triple \( * \)-derivation on prime \( * \)-algebras; i.e., for all \( A,B,C\in{\mathcal{A}} \),

$$ \Phi(A\diamondsuit_{\lambda}B\diamondsuit_{\lambda}C)=\Phi(A)\diamondsuit_{\lambda}B\diamondsuit_{\lambda}C+A\diamondsuit_{\lambda}\Phi(B)\diamondsuit_{\lambda}C+A\diamondsuit_{\lambda}B\diamondsuit_{\lambda}\Phi(C) $$

where \( A\diamondsuit_{\lambda}B=AB+\lambda BA^{*} \) such that a complex scalar \( |\lambda|\neq 0,1 \), and \( \Phi \) is additive. Moreover, if \( \Phi(I) \) is self-adjoint, then \( \Phi \) is a \( * \)-derivation.

Recently, many authors have studied the isomorphisms and derivations corresponding to the new products of the mixture of Lie product and skew Lie product. For example, Yang and Zhang [25, 26] studied the nonlinear maps that preserve the mixed skew Lie triple product \( [[A,B]_{\ast},C] \) and \( [[A,B],C]_{\ast} \) on factor von Neumann algebras. Zhou, Yang, and Zhang [27] studied the structure of the nonlinear mixed Lie triple derivations on prime \( * \)-algebras. In this paper, we consider the derivations corresponding to the new product of the mixture of the skew Lie product and the Jordan \( * \)-product. A map \( \Phi:{\mathcal{A}}\rightarrow{\mathcal{A}} \) is said to be a nonlinear mixed Jordan triple \( * \)-derivation if

$$ \Phi([A\bullet B,C]_{*})=[\Phi(A)\bullet B,C]_{*}+[A\bullet\Phi(B),C]_{*}+[A\bullet B,\Phi(C)]_{*} $$

for all \( A,B,C\in{\mathcal{A}} \). Under some mild conditions on a \( * \)-algebra \( \mathcal{A} \), we prove that a map \( \Phi:{\mathcal{A}}\rightarrow{\mathcal{A}} \) is a nonlinear mixed Jordan triple \( * \)-derivation if and only if \( \Phi \) is an additive \( * \)-derivation. In particular, we apply the above result to prime \( \ast \)-algebras, von Neumann algebras with no central summands of type \( I_{1} \), factor von Neumann algebras, and standard operator algebras.

2. The Main Result and Its Proof

Our main result in this paper reads as follows:

Theorem 2.1

Let \( {\mathcal{A}} \) be a unital \( \ast \)-algebra with the unit \( I \). Assume that \( {\mathcal{A}} \) contains a nontrivial projection \( P \) that satisfies

$$ \kern 142.26378ptX{\mathcal{A}}P=0\quad\text{implies}\quad X=0\kern 199.169291pt(\spadesuit) $$

and

$$ \kern 142.26378ptX{\mathcal{A}}(I-P)=0\quad\text{implies}\quad X=0.\kern 170.716535pt(\clubsuit) $$

Then a map \( \Phi:{\mathcal{A}}\rightarrow{\mathcal{A}} \) satisfies

$$ \Phi([A\bullet B,C]_{*})=[\Phi(A)\bullet B,C]_{*}+[A\bullet\Phi(B),C]_{*}+[A\bullet B,\Phi(C)]_{*} $$

for all \( A,B,C\in{\mathcal{A}} \) if and only if \( \Phi \) is an additive \( * \)-derivation.

Proof

Let \( P_{1}=P \) and \( P_{2}=I-P \). Put \( {\mathcal{A}}_{jk}=P_{j}{\mathcal{A}}P_{k} \), \( j,k=1,2 \). Then

$$ \mathcal{A}=\sum\limits^{2}_{j,k=1}{\mathcal{A}}_{jk}. $$

In the sequel \( A_{jk} \) indicates that \( A_{jk}\in{\mathcal{A}}_{jk} \). Clearly, we only need to prove the necessity. We will complete the proof by several claims:

Claim 1

\( \Phi(0)=0 \).


Indeed,

$$ \Phi(0)=\Phi([0\bullet 0,0]_{*})=[\Phi(0)\bullet 0,0]_{*}+[0\bullet\Phi(0),0]_{*}+[0\bullet 0,\Phi(0)]_{*}=0. $$

Claim 2

\( \Phi \) is additive.


We will complete the proof of Claim 2 in several steps.

Step 2.1. Given \( A_{12}\in{\mathcal{A}}_{12} \) and \( B_{21}\in{\mathcal{A}}_{21} \), we have \( \Phi(A_{12}+B_{21})=\Phi(A_{12})+\Phi(B_{21}) \).

We only need show that \( T=\Phi(A_{12}+B_{21})-\Phi(A_{12})-\Phi(B_{21})=0 \). Since

$$ [I\bullet(i(P_{2}-P_{1})),A_{12}]_{*}=[I\bullet(i(P_{2}-P_{1})),B_{21}]_{*}=0, $$

where \( i \) is the imaginary unit; it follows from Claim 1 that

$$ \begin{gathered}\displaystyle[\Phi(I)\bullet(i(P_{2}-P_{1})),A_{12}+B_{21}]_{*}+[I\bullet\Phi(i(P_{2}-P_{1})),A_{12}+B_{21}]_{*}\\ \displaystyle+[I\bullet(i(P_{2}-P_{1})),\Phi(A_{12}+B_{21})]_{*}=\Phi([I\bullet(i(P_{2}-P_{1})),A_{12}+B_{21}]_{*})\\ \displaystyle=\Phi([I\bullet(i(P_{2}-P_{1})),A_{12}]_{*})+\Phi([I\bullet(i(P_{2}-P_{1})),B_{21}]_{*})\\ \displaystyle=[\Phi(I)\bullet(i(P_{2}-P_{1})),A_{12}+B_{21}]_{*}+[I\bullet\Phi(i(P_{2}-P_{1})),A_{12}+B_{21}]_{*}\\ \displaystyle+[I\bullet(i(P_{2}-P_{1})),\Phi(A_{12})+\Phi(B_{21})]_{*}.\end{gathered} $$

From this we get \( [I\bullet(i(P_{2}-P_{1})),T]_{*}=0 \). So \( T_{11}=T_{22}=0 \).

Since \( [I\bullet A_{12},P_{1}]_{*}=0 \), it follows that

$$ \begin{gathered}\displaystyle[\Phi(I)\bullet(A_{12}+B_{21}),P_{1}]_{*}+[I\bullet\Phi(A_{12}+B_{21}),P_{1}]_{*}+[I\bullet(A_{12}+B_{21}),\Phi(P_{1})]_{*}\\ \displaystyle=\Phi([I\bullet(A_{12}+B_{21}),P_{1}]_{*})=\Phi([I\bullet A_{12},P_{1}]_{*})+\Phi([I\bullet B_{21},P_{1}]_{*})\\ \displaystyle=[\Phi(I)\bullet(A_{12}+B_{21}),P_{1}]_{*}+[I\bullet(\Phi(A_{12})+\Phi(B_{21})),P_{1}]_{*}+[I\bullet(A_{12}+B_{21}),\Phi(P_{1})]_{*}.\end{gathered} $$

Hence \( [I\bullet T,P_{1}]_{*}=0 \), from which we get that \( T_{21}=0 \). Similarly, we can show that \( T_{12}=0 \), proving the step.

Step 2.2. For all \( A_{11}\in{\mathcal{A}}_{11},B_{12}\in{\mathcal{A}}_{12} \), \( C_{21}\in{\mathcal{A}}_{21} \), and \( D_{22}\in{\mathcal{A}}_{22} \) we have

$$ \Phi(A_{11}+B_{12}+C_{21})=\Phi(A_{11})+\Phi(B_{12})+\Phi(C_{21}) $$

and

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

Let \( T=\Phi(A_{11}+B_{12}+C_{21})-\Phi(A_{11})-\Phi(B_{12})-\Phi(C_{21}) \).

It follows from Step 2.1 that

$$ \begin{gathered}\displaystyle[\Phi(I)\bullet(iP_{2}),A_{11}+B_{12}+C_{21}]_{*}+[I\bullet\Phi(iP_{2}),A_{11}+B_{12}+C_{21}]_{*}\\ \displaystyle+[I\bullet(iP_{2}),\Phi(A_{11}+B_{12}+C_{21})]_{*}=\Phi([I\bullet(iP_{2}),A_{11}+B_{12}+C_{21}]_{*})\\ \displaystyle=\Phi([I\bullet(iP_{2}),A_{11}]_{*})+\Phi([I\bullet(iP_{2}),B_{12}+C_{21}]_{*})\\ \displaystyle=\Phi([I\bullet(iP_{2}),A_{11}]_{*})+\Phi([I\bullet(iP_{2}),B_{12}]_{*})+\Phi([I\bullet(iP_{2}),C_{21}]_{*})\\ \displaystyle=[\Phi(I)\bullet(iP_{2}),A_{11}+B_{12}+C_{21}]_{*}+[I\bullet\Phi(iP_{2}),A_{11}+B_{12}+C_{21}]_{*}\\ \displaystyle+[I\bullet(iP_{2}),\Phi(A_{11})+\Phi(B_{12})+\Phi(C_{21})]_{*}.\end{gathered} $$

From this we get \( [I\bullet(iP_{2}),T]_{*}=0 \). So \( T_{12}=T_{21}=T_{22}=0 \).

Since

$$ [I\bullet(i(P_{2}-P_{1})),B_{12}]_{*}=[I\bullet(i(P_{2}-P_{1})),C_{21}]_{*}=0, $$

it follows that

$$ \begin{gathered}\displaystyle[\Phi(I)\bullet(i(P_{2}-P_{1})),A_{11}+B_{12}+C_{21}]_{*}+[I\bullet\Phi(i(P_{2}-P_{1})),A_{11}+B_{12}+C_{21}]_{*}\\ \displaystyle+[I\bullet(i(P_{2}-P_{1})),\Phi(A_{11}+B_{12}+C_{21})]_{*}=\Phi([I\bullet(i(P_{2}-P_{1})),A_{11}+B_{12}+C_{21}]_{*})\\ \displaystyle=\Phi([I\bullet(i(P_{2}-P_{1})),A_{11}]_{*})+\Phi([I\bullet(i(P_{2}-P_{1})),B_{12}]_{*})+\Phi([I\bullet(i(P_{2}-P_{1})),C_{21}]_{*})\\ \displaystyle=[\Phi(I)\bullet(i(P_{2}-P_{1})),A_{11}+B_{12}+C_{21}]_{*}+[I\bullet\Phi(i(P_{2}-P_{1})),A_{11}+B_{12}+C_{21}]_{*}\\ \displaystyle+[I\bullet(i(P_{2}-P_{1})),\Phi(A_{11})+\Phi(B_{12})+\Phi(C_{21})]_{*},\end{gathered} $$

from which we get \( [I\bullet(i(P_{2}-P_{1})),T]_{*}=0 \). So \( T_{11}=0 \), and then \( T=0 \). Similarly, \( \Phi(B_{12}+C_{21}+D_{22})=\Phi(B_{12})+\Phi(C_{21})+\Phi(D_{22}) \).

Step 2.3. For all \( A_{11}\in{\mathcal{A}}_{11} \), \( B_{12}\in{\mathcal{A}}_{12} \), \( C_{21}\in{\mathcal{A}}_{21} \), and \( D_{22}\in{\mathcal{A}}_{22} \), we have

$$ \Phi(A_{11}+B_{12}+C_{21}+D_{22})=\Phi(A_{11})+\Phi(B_{12})+\Phi(C_{21})+\Phi(D_{22}). $$

Let \( T=\Phi(A_{11}+B_{12}+C_{21}+D_{22})-\Phi(A_{11})-\Phi(B_{12})-\Phi(C_{21})-\Phi(D_{22}) \). It follows from Step 2.2 that

$$ \begin{gathered}\displaystyle[\Phi(I)\bullet(iP_{2}),A_{11}+B_{12}+C_{21}+D_{22}]_{*}+[I\bullet\Phi(iP_{2}),A_{11}+B_{12}+C_{21}+D_{22}]_{*}\\ \displaystyle+[I\bullet(iP_{2}),\Phi(A_{11}+B_{12}+C_{21}+D_{22})]_{*}=\Phi([I\bullet(iP_{2}),A_{11}+B_{12}+C_{21}+D_{22}]_{*})\\ \displaystyle=\Phi([I\bullet(iP_{2}),A_{11}]_{*})+\Phi([I\bullet(iP_{2}),B_{12}+C_{21}+D_{22})\\ \displaystyle=\Phi([I\bullet(iP_{2}),A_{11}]_{*})+\Phi([I\bullet(iP_{2}),B_{12}]_{*})+\Phi([I\bullet(iP_{2}),C_{21}]_{*})+\Phi([I\bullet(iP_{2}),D_{22}]_{*})\\ \displaystyle=[\Phi(I)\bullet(iP_{2}),A_{11}+B_{12}+C_{21}++D_{22}]_{*}+[I\bullet\Phi(iP_{2}),A_{11}+B_{12}+C_{21}+D_{22}]_{*}\\ \displaystyle+[I\bullet(iP_{2}),\Phi(A_{11})+\Phi(B_{12})+\Phi(C_{21})+\Phi(D_{22})]_{*}.\end{gathered} $$

From this we get \( [I\bullet(iP_{2}),T]_{*}=0 \). So \( T_{12}=T_{21}=T_{22}=0 \). Similarly, we can show that \( T_{11}=0 \), proving Step 2.3.

Step 2.4. Given \( A_{jk},B_{jk}\in{\mathcal{A}}_{jk} \), with \( 1\leq j\neq k\leq 2 \), we have \( \Phi(A_{jk}+B_{jk})=\Phi(A_{jk})+\Phi(B_{jk}) \).

Since

$$ \bigg{[}\frac{I}{2}\bullet(P_{j}+A_{jk}),P_{k}+B_{jk}\bigg{]}_{*}=(A_{jk}+B_{jk})-A^{\ast}_{jk}-B_{jk}A^{\ast}_{jk}, $$

we get from Step 2.3 that

$$ \begin{gathered}\displaystyle\Phi(A_{jk}+B_{jk})+\Phi(-A^{\ast}_{jk})+\Phi(-B_{jk}A^{\ast}_{jk})=\Phi\bigg{(}\bigg{[}\frac{I}{2}\bullet(P_{j}+A_{jk}),P_{k}+B_{jk}\bigg{]}_{*}\bigg{)}\\ \displaystyle=\bigg{[}\Phi(\frac{I}{2})\bullet(P_{j}+A_{jk}),P_{k}+B_{jk}\bigg{]}_{*}+\bigg{[}\frac{I}{2}\bullet\Phi(P_{j}+A_{jk}),P_{k}+B_{jk}\bigg{]}_{*}\\ \displaystyle+\bigg{[}\frac{I}{2}\bullet(P_{j}+A_{jk}),\Phi(P_{k}+B_{jk})\bigg{]}_{*}=\bigg{[}\Phi(\frac{I}{2})\bullet(P_{j}+A_{jk}),P_{k}+B_{jk}\bigg{]}_{*}\\ \displaystyle+\bigg{[}\frac{I}{2}\bullet\bigg{(}\Phi(P_{j})+\Phi(A_{jk})\bigg{)},P_{k}+B_{jk}\bigg{]}_{*}+\bigg{[}\frac{I}{2}\bullet(P_{j}+A_{jk}),(\Phi(P_{k})+\Phi(B_{jk}))\bigg{]}_{*}\\ \displaystyle=\Phi\bigg{(}\bigg{[}\frac{I}{2}\bullet P_{j},P_{k}\bigg{]}_{*}\bigg{)}+\Phi\bigg{(}\bigg{[}\frac{I}{2}\bullet P_{j},B_{jk}\bigg{]}_{*}\bigg{)}+\Phi\bigg{(}\bigg{[}\frac{I}{2}\bullet A_{jk},P_{k}\bigg{]}_{*}\bigg{)}+\Phi\bigg{(}\bigg{[}\frac{I}{2}\bullet A_{jk},B_{jk}\bigg{]}_{*}\bigg{)}\\ \displaystyle=\Phi(B_{jk})+\Phi(A_{jk}-A^{\ast}_{jk})+\Phi(-B_{jk}A^{\ast}_{jk})\\ \displaystyle=\Phi(B_{jk})+\Phi(A_{jk})+\Phi(-A^{\ast}_{jk})+\Phi(-B_{jk}A^{\ast}_{jk}).\end{gathered} $$

Then \( \Phi(A_{jk}+B_{jk})=\Phi(A_{jk})+\Phi(B_{jk}) \).

Step 2.5. Given \( A_{jj},B_{jj}\in{\mathcal{A}}_{jj} \), with \( 1\leq j\leq 2 \), we have

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

Let \( T=\Phi(A_{jj}+B_{jj})-\Phi(A_{jj})-\Phi(B_{jj}) \). For \( 1\leq j\neq k\leq 2 \), it follows that

$$ \begin{gathered}\displaystyle[\Phi(I)\bullet(iP_{k}),A_{jj}+B_{jj}]_{*}+[I\bullet\Phi(iP_{k}),A_{jj}+B_{jj}]_{*}\\ \displaystyle+[I\bullet(iP_{k}),\Phi(A_{jj}+B_{jj})]_{*}=\Phi([I\bullet(iP_{k}),A_{jj}+B_{jj}]_{*})=\Phi([I\bullet(iP_{k}),A_{jj}]_{*})+\Phi([I\bullet(iP_{k}),B_{jj}]_{*})\\ \displaystyle=[\Phi(I)\bullet(iP_{k}),A_{jj}+B_{jj}]_{*}+[I\bullet\Phi(iP_{k}),A_{jj}+B_{jj}]_{*}\\ \displaystyle+[I\bullet(iP_{k}),\Phi(A_{jj})+\Phi(B_{jj})]_{*}.\end{gathered} $$

From this we get \( [I\bullet(iP_{k}),T]_{*}=0 \). So \( T_{jk}=T_{kj}=T_{kk}=0 \). Now \( T=T_{jj} \).

For all \( C_{jk}\in{\mathcal{A}}_{jk} \), \( j\neq k \), it follows from Step 2.4 that

$$ \begin{gathered}\displaystyle[\Phi(I)\bullet(A_{jj}+B_{jj}),C_{jk}]_{*}+[I\bullet\Phi(A_{jj}+B_{jj}),C_{jk}]_{*}\\ \displaystyle+[I\bullet(A_{jj}+B_{jj}),\Phi(C_{jk})]_{*}=\Phi([I\bullet(A_{jj}+B_{jj}),C_{jk}]_{*})\\ \displaystyle=\Phi([I\bullet A_{jj},C_{jk}]_{*})+\Phi([I\bullet B_{jj},C_{jk}]_{*})\\ \displaystyle=[\Phi(I)\bullet(A_{jj}+B_{jj}),C_{jk}]_{*}+[I\bullet\Phi(A_{jj})+\Phi(B_{jj}),C_{jk}]_{*}\\ \displaystyle+[I\bullet(A_{jj}+B_{jj}),\Phi(C_{jk})]_{*}.\end{gathered} $$

Hence \( [I\bullet T_{jj},C_{jk}]_{*}=0 \) for all \( C_{jk}\in{\mathcal{A}}_{jk} \); i.e., \( T_{jj}CP_{k}=0 \) for all \( C\in{\mathcal{A}} \). It follows from \( (\spadesuit) \) and \( (\clubsuit) \) that \( T=T_{jj}=0 \), proving the step.

Now, it follows from Steps 2.3–2.5 that \( \Phi \) is additive, proving Claim 2.

Claim 3

\( \Phi(I) \) is a self-adjoint central element in \( \mathcal{A} \).


On the one hand,

$$ \begin{gathered}\displaystyle 0=\Phi([I\bullet I,I]_{*})=[\Phi(I)\bullet I,I]_{*}+[I\bullet\Phi(I),I]_{*}+[I\bullet I,\Phi(I)]_{*}\\ \displaystyle=[2\Phi(I),I]_{*}=2\Phi(I)-2\Phi(I)^{*},\end{gathered} $$

which implies that \( \Phi(I) \) is a self-adjoint element in \( {\mathcal{A}} \).

On the other hand, for all \( A\in{\mathcal{A}} \) we get

$$ \begin{gathered}\displaystyle 0=\Phi([I\bullet I,A]_{*})=[\Phi(I)\bullet I,A]_{*}+[I\bullet\Phi(I),A]_{*}+[I\bullet I,\Phi(A)]_{*}\\ \displaystyle=2[2\Phi(I),A]_{*}=4(\Phi(I)A-A\Phi(I)),\end{gathered} $$

which implies that \( \Phi(I) \) is a central element in \( {\mathcal{A}} \).

Claim 4

\( P_{1}\Phi(P_{1})P_{2}=-P_{1}\Phi(P_{2})P_{2} \), \( P_{2}\Phi(P_{1})P_{1}=-P_{2}\Phi(P_{2})P_{1} \), and \( P_{1}\Phi(P_{2})P_{1}=P_{2}\Phi(P_{1})P_{2}=0 \).


On the one hand, for \( 1\leq j\neq k\leq 2 \), it follows from Claim 3 that

$$ \begin{gathered}\displaystyle 0=\Phi([I\bullet P_{j},P_{k}]_{*})=[\Phi(I)\bullet P_{j},P_{k}]_{*}+[I\bullet\Phi(P_{j}),P_{k}]_{*}+[I\bullet P_{j},\Phi(P_{k})]_{*}\\ \displaystyle=[2\Phi(P_{j}),P_{k}]_{*}+[2P_{j},\Phi(P_{k})]_{*}=2\Phi(P_{j})P_{k}-2P_{k}\Phi(P_{j})^{*}+2P_{j}\Phi(P_{k})-2\Phi(P_{k})P_{j}.\end{gathered} $$

Multiplying both sides of the above equation by \( P_{j} \) and \( P_{k} \) from the left and right, respectively, we infer that \( P_{1}\Phi(P_{1})P_{2}=-P_{1}\Phi(P_{2})P_{2} \) and \( P_{2}\Phi(P_{1})P_{1}=-P_{2}\Phi(P_{2})P_{1} \).

On the other hand, we get

$$ \begin{gathered}\displaystyle 0=\Phi([I\bullet(iP_{j}),P_{k}]_{*})=[\Phi(I)\bullet(iP_{j}),P_{k}]_{*}+[I\bullet\Phi(iP_{j}),P_{k}]_{*}+[I\bullet(iP_{j}),\Phi(P_{k})]_{*}\\ \displaystyle=[2\Phi(iP_{j}),P_{k}]_{*}+[2iP_{j},\Phi(P_{k})]_{*}\\ \displaystyle=2\Phi(iP_{j})P_{k}-2P_{k}\Phi(iP_{j})^{*}+2i(P_{j}\Phi(P_{k})+\Phi(P_{k})P_{j}).\end{gathered} $$

Multiplying both sides of the above equation by \( P_{j} \) from the left and right, respectively, we obtain that \( P_{1}\Phi(P_{2})P_{1}=P_{2}\Phi(P_{1})P_{2}=0 \).

Claim 5

\( P_{1}\Phi(P_{1})P_{1}=P_{2}\Phi(P_{2})P_{2}=0 \).


For every \( A_{12}\in{\mathcal{A}}_{12} \), on the one hand, it follows from Claims 2 and 3 that

$$ \begin{gathered}\displaystyle 2\Phi(A_{12})=\Phi([I\bullet P_{1},A_{12}]_{*})=[\Phi(I)\bullet P_{1},A_{12}]_{*}+[I\bullet\Phi(P_{1}),A_{12}]_{*}+[I\bullet P_{1},\Phi(A_{12})]_{*}\\ \displaystyle=[2\Phi(I)P_{1},A_{12}]_{*}+[2\Phi(P_{1}),A_{12}]_{*}+[2P_{1},\Phi(A_{12})]_{*}\\ \displaystyle=2\Phi(I)A_{12}+2\Phi(P_{1})A_{12}-2A_{12}\Phi(P_{1})^{*}+2P_{1}\Phi(A_{12})-2\Phi(A_{12})P_{1}.\end{gathered} $$

Multiplying both sides of the above equation by \( P_{1} \) and \( P_{2} \) from the left and right, respectively, by Claim 4, we get that

$$ P_{1}\Phi(P_{1})A_{12}+\Phi(I)A_{12}=0. $$
(2.1)

On the other hand, we have

$$ \begin{gathered}\displaystyle 2\Phi(A_{12})=\Phi([P_{1}\bullet P_{1},A_{12}]_{*})\\ \displaystyle=[\Phi(P_{1})\bullet P_{1},A_{12}]_{*}+[P_{1}\bullet\Phi(P_{1}),A_{12}]_{*}+[P_{1}\bullet P_{1},\Phi(A_{12})]_{*}\\ \displaystyle=[\Phi(P_{1})P_{1}+P_{1}\Phi(P_{1})^{*},A_{12}]_{*}+[P_{1}\Phi(P_{1})+\Phi(P_{1})P_{1},A_{12}]_{*}+[2P_{1},\Phi(A_{12})]_{*}\\ \displaystyle=\Phi(P_{1})A_{12}+P_{1}\Phi(P_{1})^{*}A_{12}-A_{12}\Phi(P_{1})P_{1}+P_{1}\Phi(P_{1})A_{12}\\ \displaystyle+\Phi(P_{1})A_{12}-A_{12}\Phi(P_{1})^{*}P_{1}+2P_{1}\Phi(A_{12})-2\Phi(A_{12})P_{1}.\end{gathered} $$

Multiplying both sides of the above equation by \( P_{1} \) and \( P_{2} \) from the left and right, respectively, we get that

$$ 3P_{1}\Phi(P_{1})A_{12}+P_{1}\Phi(P_{1})^{*}A_{12}=0. $$
(2.2)

Finally,

$$ \begin{gathered}\displaystyle 2\Phi(A_{12})=\Phi([P_{1}\bullet I,A_{12}]_{*})\\ \displaystyle=[\Phi(P_{1})\bullet I,A_{12}]_{*}+[P_{1}\bullet\Phi(I),A_{12}]_{*}+[P_{1}\bullet I,\Phi(A_{12})]_{*}\\ \displaystyle=[\Phi(P_{1})+\Phi(P_{1})^{*},A_{12}]_{*}+[2P_{1}\Phi(I),A_{12}]_{*}+[2P_{1},\Phi(A_{12})]_{*}\\ \displaystyle=(\Phi(P_{1})+\Phi(P_{1})^{*})A_{12}-A_{12}(\Phi(P_{1})+\Phi(P_{1})^{*})+2\Phi(I)A_{12}+2P_{1}\Phi(A_{12})-2\Phi(A_{12})P_{1}.\end{gathered} $$

Multiplying both sides of the above equation by \( P_{1} \) and \( P_{2} \) from the left and right, respectively, by Claim 4, we get that

$$ P_{1}\Phi(P_{1})A_{12}+P_{1}\Phi(P_{1})^{*}A_{12}+2\Phi(I)A_{12}=0. $$
(2.3)

It follows from (2.2) and (2.3) that

$$ P_{1}\Phi(P_{1})A_{12}-\Phi(I)A_{12}=0. $$
(2.4)

Now, by (2.1) and (2.4), we have \( P_{1}\Phi(P_{1})A_{12}=0 \); i.e., \( P_{1}\Phi(P_{1})P_{1}AP_{2}=0 \) for all \( A\in{\mathcal{A}} \). It follows from \( (\clubsuit) \) that \( P_{1}\Phi(P_{1})P_{1}=0 \). Similarly, we can prove that \( P_{2}\Phi(P_{2})P_{2}=0 \).

Claim 6

\( \Phi(I)=0 \).


By Claims 2, 4, and 5, we can get that

$$ \Phi(I)=\Phi(P_{1})+\Phi(P_{2})=P_{1}\Phi(P_{1})P_{2}+P_{2}\Phi(P_{1})P_{1}+P_{1}\Phi(P_{2})P_{2}+P_{2}\Phi(P_{2})P_{1}=0. $$

Claim 7

\( \Phi([A,B]_{*})=[\Phi(A),B]_{*}+[A,\Phi(B)]_{*} \) for all \( A,B\in{\mathcal{A}} \).


It follows from Claims 2 and 6 that

$$ \begin{gathered}\displaystyle 2\Phi([A,B]_{*})=\Phi([I\bullet A,B]_{*})=[\Phi(I)\bullet A,B]_{*}+[I\bullet\Phi(A),B]_{*}+[I\bullet A,\Phi(B)]_{*}\\ \displaystyle=[2\Phi(A),B]_{*}+[2A,\Phi(B)]_{*}=2([\Phi(A),B]_{*}+[A,\Phi(B)]_{*}),\end{gathered} $$

which implies that \( \Phi([A,B]_{*})=[\Phi(A),B]_{*}+[A,\Phi(B)]_{*} \).

Claim 8

\( \Phi(A^{*})=\Phi(A)^{*} \) for all \( A\in{\mathcal{A}} \).


For every \( A\in{\mathcal{A}} \), by Claims 2, 6, and 7, we have

$$ \Phi(A)-\Phi(A^{*})=\Phi([A,I]_{*})=[\Phi(A),I]_{*}=\Phi(A)-\Phi(A)^{*}. $$

Hence \( \Phi(A^{*})=\Phi(A)^{*} \).

Claim 9

\( \Phi(iI)=0 \).


By Claims 2 and 8, we can get \( \Phi(iI)^{*}=-\Phi(iI) \). So

$$ 0=-2\Phi(I)=\Phi([iI,iI]_{*})=[\Phi(iI),iI]_{*}+[iI,\Phi(iI)]_{*}=4i\Phi(iI), $$

which implies that \( \Phi(iI)=0 \).

Claim 10

\( \Phi(iA)=i\Phi(A) \) for all \( A\in{\mathcal{A}} \).


It follows from Claims 2 and 9 that

$$ 2\Phi(iA)=\Phi(2iA)=\Phi([iI,A]_{*})=[\Phi(iI),A]_{*}+[iI,\Phi(A)]_{*}=2i\Phi(A), $$

and then \( \Phi(iA)=i\Phi(A) \).

Claim 11

\( \Phi \) is a derivation.


On the one hand, by Claim 7, we have

$$ \begin{gathered}\displaystyle\Phi(AB)-\Phi(BA^{*})=\Phi(AB-BA^{*})=\Phi([A,B]_{*})=[\Phi(A),B]_{*}+[A,\Phi(B)]_{*}\\ \displaystyle=\Phi(A)B-B\Phi(A)^{*}+A\Phi(B)-\Phi(B)A^{*}.\end{gathered} $$
(2.5)

On the other hand, by Claims 2, 7, and 10, we also have

$$ \begin{gathered}\displaystyle-\Phi(AB)-\Phi(BA^{*})=\Phi([iA,iB]_{*})=[i\Phi(A),iB]_{*}+[iA,i\Phi(B)]_{*}\\ \displaystyle=-\Phi(A)B-B\Phi(A)^{*}-A\Phi(B)-\Phi(B)A^{*}.\end{gathered} $$
(2.6)

From (2.5) and (2.6) we obtain \( \Phi(AB)=\Phi(A)B+A\Phi(B) \).

Now, by Claims 2, 8, and 11, we have proved that \( \Phi \) is an additive \( * \)-derivation. This completes the proof of Theorem 2.1.  ☐

3. Corollaries

In this section, we present some corollaries of the main result. An algebra \( {\mathcal{A}} \) is called prime if \( A\mathcal{A}B=\{0\} \) for \( A,B\in{\mathcal{A}} \) implies either \( A=0 \) or \( B=0 \). It is easy to see that prime \( * \)-algebras satisfy \( (\spadesuit) \) and \( (\clubsuit) \). So we have the following corollary.

Corollary 3.1

Let \( {\mathcal{A}} \) be a prime \( * \)-algebra with unit \( I \) and let \( P \) be a nontrivial projection in \( {\mathcal{A}} \). Then \( \Phi \) is a nonlinear mixed Jordan triple \( * \)-derivation on \( {\mathcal{A}} \) if and only if \( \Phi \) is an additive \( * \)-derivation.


We denote by \( B({\mathcal{H}}) \) the algebra of all bounded linear operators on a complex Hilbert space \( {\mathcal{H}} \) and by \( {\mathcal{F}}(H)\subseteq B({\mathcal{H}}) \), the subalgebra of all bounded finite rank operators. A subalgebra \( {\mathcal{A}}\subseteq B({\mathcal{H}}) \) is called a standard operator algebra if \( {\mathcal{A}} \) includes \( {\mathcal{F}}(H) \). Now we have the following corollary.

Corollary 3.2

Let \( {\mathcal{A}} \) be a standard operator algebra on an infinite-dimensional complex Hilbert space \( {\mathcal{H}} \) containing the identity operator \( I \). Suppose that \( {\mathcal{A}} \) is closed under the adjoint operation. Then \( \Phi:{\mathcal{A}}\rightarrow{\mathcal{A}} \) is a nonlinear mixed Jordan triple \( * \)-derivation if and only if \( \Phi \) is a linear \( * \)-derivation. Moreover, there exists an operator \( T\in B({\mathcal{H}}) \) satisfying \( T+T^{*}=0 \) such that \( \Phi(A)=AT-TA \) for all \( A\in A \), i.e., \( \Phi \) is inner.

Proof

Since \( {\mathcal{A}} \) is prime, we have that \( \Phi \) is an additive \( * \)-derivation. It follows from [28] that \( \Phi \) is a linear inner derivation, i.e., there exists an operator \( S\in B({\mathcal{H}}) \) such that \( \Phi(A)=AS-SA \). Since \( \Phi(A^{*})=\Phi(A)^{*} \), we have

$$ A^{*}S-SA^{*}=\Phi(A^{*})=\Phi(A)^{*}=-A^{*}S^{*}+S^{*}A^{*} $$

for all \( A\in A \). Hence

$$ A^{*}(S+S^{*})=(S+S^{*})A^{*}, $$

and then \( S+S^{*}=\lambda I \) for some \( \lambda\in{𝕉} \). Let

$$ T=S-\frac{1}{2}\lambda I. $$

It is easy to see that \( T+T^{*}=0 \) such that \( \Phi(A)=AT-TA \).  ☐

A von Neumann algebra \( {\mathcal{M}} \) is a weakly closed self-adjoint algebra of operators on a Hilbert space \( {\mathcal{H}} \) containing the identity operator \( I \). Note that \( {\mathcal{M}} \) is a factor von Neumann algebra if its center only contains the scalar operators. It is well known that a factor von Neumann algebra is prime. So we have the following corollary:

Corollary 3.3

Let \( {\mathcal{M}} \) be a factor von Neumann algebra with \( \dim{\mathcal{M}}\geq 2 \). Then \( \Phi:{\mathcal{M}}\rightarrow{\mathcal{M}} \) is a nonlinear mixed Jordan triple \( * \)-derivation if and only if \( \Phi \) is an additive \( * \)-derivation.


It is shown in [2] and [18] that if a von Neumann algebra has no central summands of type \( I_{1} \), then \( {\mathcal{M}} \) satisfies \( (\spadesuit) \) and \( (\clubsuit) \). Now we have the following corollary:

Corollary 3.4

Let \( {\mathcal{M}} \) be a von Neumann algebra with no central summands of type \( I_{1} \). Then \( \Phi:{\mathcal{M}}\rightarrow{\mathcal{M}} \) is a nonlinear mixed Jordan triple \( * \)-derivation if and only if \( \Phi \) is an additive \( * \)-derivation.