1. INTRODUCTION

Let \(\mathcal{R}\) be a \(*\)-algebra. For \(A,B\in\mathcal{R}\), we write \(A\bullet B=AB+BA^{*}\) and \([A,B]_{*}=AB-BA^{*}\) for the \(\ast\)-Jordan product and \(\ast\)-Lie product, respectively. These products play an important role in some research topics, and their study has recently attracted the attention of many authors (for example, see [1]–[5]).

Recall that a map \(\Phi:\mathcal{R}\to\mathcal{R}\) is said to be an additive derivation if

$$\Phi(A+B)=\Phi(A)+\Phi(B) \qquad \text{and} \qquad \Phi(AB)=\Phi(A)B+A\Phi(B)$$

for all \(A,B\in\mathcal{R}\). A map \(\Phi\) is an additive \(\ast\)-derivation if it is an additive derivation and \(\Phi(A^{*})=\Phi(A)^{*}\). Derivations are very important maps both in theory and applications and have been studied intensively ([6]–[11]).

A von Neumann algebra\(\mathcal{A}\) is a self-adjoint subalgebra of \(B(H)\), the algebra of all bounded linear operators acting on a complex Hilbert space, which satisfies the double commutant property: \(\mathcal{A}^{''}=\mathcal{A}\) where \(\mathcal{A}^{'}=\{T\in B(H), TA=AT\}\) for all \(A\in\mathcal{A}\), and \(\mathcal{A}^{''}=\{\mathcal{A}^{'}\}^{'}\). We denote by \(\mathcal{Z}(\mathcal{A})=\mathcal{A}^{'}\cap \mathcal{A}\) the center of \(\mathcal{A}\). A von Neumann algebra \(\mathcal{A}\) is called a factor if its center is trivial, i.e., \(\mathcal{Z}(\mathcal{A})=\mathbb{C}I\). For \(A\in\mathcal{A}\), recall that the central carrier of \(A\), denoted by \(\overline{A}\), is the smallest central projection \(P\) such that \(PA=A\). It is not difficult to see that \(\overline{A}\) is the projection onto the closed subspace spanned by \(\{BAx : B\in \mathcal{A}, x\in H\}\). If \(A\) is self-adjoint, then the core of \(A\), denoted by \(\underline{A}\), is \(\sup\{S\in\mathcal{Z}(\mathcal{A}): S=S^{*}, S\leq A\}\). If \(A=P\) is a projection, it is clear that \(\underline{P}\) is the largest central projection \(Q\) satisfying \(Q\leq P\). A projection \(P\) is said to be core-free if \(\underline{P}=0\) (see [12]). It is easy to see that \(\underline{P}=0\) if and only if \(\overline{I-P}=I\), [13]-[14].

Recently, Yu and Zhang in [15] proved that every nonlinear \(\ast\)-Lie derivation from a factor von Neumann algebra into itself is an additive \(\ast\)-derivation. Also, in [16], Li, Lu, and Fang investigated nonlinear \(\lambda\)-Jordan \(\ast\)-derivations. They showed that if \(\mathcal{A}\subseteq\mathcal{B(H)}\) is a von Neumann algebra without central Abelian projections and \(\lambda\) is a nonzero scalar, then \(\Phi:\mathcal{A} \longrightarrow \mathcal{B(H)}\) is a nonlinear \(\lambda\)-Jordan \(\ast\)-derivation if and only if \(\Phi\) is an additive \(\ast\)-derivation.

On the other hand, many mathematicians have studied the \(\ast\)-Jordan product \(A\bullet B=AB+BA^{*}\). In [17], F. Zhang proved that every nonlinear \(\ast\)-Jordan derivation map \(\Phi:\mathcal{A}\to\mathcal{A}\) on a factor von Neumann algebra is an additive \(\ast\)-derivation.

In [18], we showed that \(\ast\)-Jordan derivation map on every factor von Neumann algebra \(\mathcal{A} \subseteq \mathcal{B(H)}\) is an additive \(\ast\)-derivation.

Quite recently, the authors of [19] discussed some bijective maps preserving the new product \(A^{*}B+B^{*}A\) between von Neumann algebras with no central Abelian projections. In other words, they considered the map \(\Phi\) that satisfies the following assumption:

$$\Phi(A^{*}B+B^{*}A)=\Phi(A)^{*}\Phi(B)+\Phi(B)^{*}\Phi(A).$$

They showed that such a map is the sum of a linear \(\ast\)-isomorphism and a conjugate linear \(\ast\)-isomorphism.

We say that \(\mathcal{A}\) is prime, i.e., if \(A\mathcal{A}B=\lbrace0\rbrace\) for \(A,B \in \mathcal{A}\), then \(A = 0\) or \(B = 0\).

In [20], we assumed that \(\mathcal{A}\) is a prime \(\ast\)-algebra and the map \(\Phi:\mathcal{A}\to\mathcal{A}\) satisfies the following condition:

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

where \(A\diamond B = A^{*}B + B^{*}A\) for all \(A,B\in\mathcal{A}\). We proved that, in this case, \(\Phi\) is an additive \(\ast\)-derivation.

The authors of [21] introduced the concept of \(\ast\)-Lie triple derivations. A map \(\Phi:\mathcal{A}\to\mathcal{A}\) is a nonlinear\(\ast\)-Lie triple derivation if

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

for all \(A, B, C\in\mathcal{A}\), where \([A,B]_{\ast}=AB-BA^{*}\). They showed that if \(\Phi\) preserves the above characterization of factor von Neumann algebras, then \(\Phi\) is an additive \(\ast\)-derivation.

Motivated by the above results, we introduce the triple product \(A\diamond B\diamond C:=(A\diamond B) \diamond C\), where \(A\diamond B = A^{*}B + B^{*}A\). In this paper, let \(\mathcal{A}\) be a prime \(\ast\)-algebra, and let \(\Phi:\mathcal{A}\to\mathcal{A}\) satisfy the following equality:

$$\Phi(A\diamond B \diamond C)=\Phi(A)\diamond B \diamond C+A\diamond\Phi(B) \diamond C+A \diamond B \diamond \Phi(C)$$

for all \(A,B,C\in\mathcal{A}\). We prove that \(\Phi\) is an additive \(\ast\)-derivation.

2. MAIN RESULTS

Our main theorem is as follows.

Theorem 1.

Let \(\mathcal{A}\) be a prime \(\ast\) -algebra, and let \(\Phi:\mathcal{A}\to \mathcal{A}\) satisfy the condition

$$ \Phi(A\diamond B \diamond C)=\Phi(A)\diamond B \diamond C+A\diamond\Phi(B) \diamond C+A \diamond B \diamond \Phi(C)$$
(2.1)

for all \(A,B,C\in\mathcal{A}\) , then \(\Phi\) is an additive \(\ast\) -derivation.

Proof. Let \(P_{1}\) be a nontrivial projection in \(\mathcal{A}\), and let \(P_{2}=I_{\mathcal{A}}-P_{1}\). Denote \(\mathcal{A}_{ij}=P_{i}\mathcal{A}P_{j}\) for \(i,j=1,2\); then \(\mathcal{A}=\sum_{i,j=1}^{2}\mathcal{A}_{ij}\). For every \(A\in\mathcal{A}\), we can write \(A=A_{11}+A_{12}+A_{21}+A_{22}\). In what follows, when we write \(A_{ij}\), this will indicate that \(A_{ij}\in\mathcal{A}_{ij}\). In order to show additivity of \(\Phi\) on \(\mathcal{A}\), we apply the above partitions of \(\mathcal{A}\) and establish some claims that imply that \(\Phi\) is additive on each \(\mathcal{A}_{ij}\) for \(i,j=1,2\).

Thus, the above theorem is a consequence of the following claims.

Claim 1.

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

This claim is easy to prove.

Claim 2.

\(\Phi({I}/{2})=0\) , \(\Phi(-{I}/{2})=0\) , and \(\Phi(i{I}/{2})=0\) .

To show that \(\Phi({I}/{2})=0\), we write

$$\Phi\bigg(\frac{I}{2}\diamond \frac{I}{2} \diamond \frac{I}{2}\bigg)=\Phi\bigg(\frac{I}{2}\bigg)\diamond \frac{I}{2} \diamond \frac{I}{2}+\frac{I}{2}\diamond\Phi\bigg(\frac{I}{2}\bigg) \diamond \frac{I}{2}+\frac{I}{2} \diamond \frac{I}{2} \diamond \Phi\bigg(\frac{I}{2}\bigg).$$

Thus,

$$ \Phi\bigg(\frac{I}{2}\bigg)=\frac{3}{2}\bigg(\Phi\bigg(\frac{I}{2}\bigg)+\Phi\bigg(\frac{I }{2}\bigg)^{*}\bigg).$$
(2.2)

From (2.2), we deduce that \(\Phi({I}/{2})\) is self-adjoint. Therefore, we have the desired result.

To prove that \(\Phi({I}/{2})=0\), we write

$$\Phi\bigg(\frac{I}{2} \diamond \frac{I}{2}\diamond -\frac{I}{2}\bigg)=\frac{I}{2}\diamond \frac{I}{2} \diamond \Phi\bigg(-\frac{I}{2}\bigg).$$

It follows that

$$ \Phi\bigg(-\frac{I}{2}\bigg)=\frac{1}{2}\bigg(\Phi\bigg(-\frac{I}{2}\bigg) +\Phi\bigg(-\frac{I}{2}\bigg)^{*}\bigg).$$
(2.3)

Then

$$ \Phi\bigg(-\frac{I}{2}\bigg)^{*}=\Phi\bigg(-\frac{I}{2}\bigg).$$
(2.4)

On the other hand, we have

$$\Phi\bigg(\frac{I}{2} \diamond -\frac{I}{2}\diamond -\frac{I}{2}\bigg)=\frac{I}{2}\diamond \Phi\bigg(-\frac{I}{2}\bigg) \diamond \bigg(-\frac{I}{2}\bigg)+\frac{I}{2} \diamond -\frac{I}{2} \diamond \Phi\biggl(-\frac{I}{2}\biggr).$$

It follows that

$$ \Phi\bigg(-\frac{I}{2}\bigg)^{*}=-\Phi\bigg(\frac{-I}{2}\bigg).$$
(2.5)

Then, from (2.4) and (2.5), we obtain \(\Phi(-\frac{I}{2})=0\). To show that \(\Phi(i\frac{I}{2})=0\), we write

$$\Phi\bigg(i\frac{I}{2}\diamond i\frac{I}{2} \diamond \frac{I}{2}\bigg)=\Phi\bigg(i\frac{I}{2}\bigg)\diamond i\frac{I}{2} \diamond \frac{I}{2}+i\frac{I}{2} \diamond \Phi\bigg(i\frac{I}{2} \bigg) \diamond \frac{I}{2}.$$

Thus,

$$ \Phi\bigg(i\frac{I}{2}\bigg)^{*}-\Phi\bigg(i\frac{I}{2}\bigg)=0.$$
(2.6)

Also, we have

$$\Phi\bigg(\frac{I}{2} \diamond \frac{I}{2}\diamond i\frac{I}{2}\bigg)=\frac{I}{2}\diamond \frac{I}{2} \diamond \Phi\bigg(i\frac{I}{2}\bigg).$$

Thus,

$$ \Phi\bigg(i\frac{I}{2}\bigg)^{*}+\Phi\bigg(i\frac{I}{2}\bigg)=0.$$
(2.7)

From (2.6) and (2.7), we obtain \(\Phi(i\frac{I}{2})=0\).

Claim 3.

Suppose that, for each \(A \in \mathcal{A}\) ,

  1. 1.

    \(\Phi(-iA)=-i\Phi(A).\)

  2. 2.

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

It is easy to see that

$$\Phi\bigg(-iA\diamond \frac{I}{2} \diamond \frac{I}{2}\bigg)=\Phi\bigg(A\diamond i\frac{I}{2} \diamond \frac{I}{2}\bigg).$$

Thus,

$$\Phi(-iA)\diamond\frac{I}{2} \diamond \frac{I}{2}=\Phi(A)\diamond i\frac{I}{2} \diamond \frac{I}{2}.$$

It follows that

$$ \Phi(-iA)^{*}+\Phi(-iA)=i\Phi(A)^{*}-i\Phi(A).$$
(2.8)

On the other hand, one can check that

$$\Phi\bigg(-iA\diamond i\frac{I}{2} \diamond \frac{I}{2}\bigg)=\Phi\bigg(-\frac{I}{2}\diamond A \diamond \frac{I}{2}\bigg).$$

Thus,

$$\Phi(-iA)\diamond i\frac{I}{2} \diamond \frac{I}{2}=-\frac{I}{2}\diamond\Phi(A) \diamond \frac{I}{2}.$$

It follows that

$$ i\Phi(-iA)^{*}-i\Phi(-iA)=-\Phi(A)-\Phi(A)^{*}.$$
(2.9)

Equivalently, we have

$$ -\Phi(-iA)^{*}+\Phi(-iA)=-i\Phi(A)-i\Phi(A)^{*}.$$
(2.10)

By adding equeations (2.8) and (2.10), we obtain

$$\Phi(-iA)=-i\Phi(A).$$

Similarly, we can show that \(\Phi(iA)=i\Phi(A)\).

Claim 4.

For each \(A_{11} \in \mathcal{A}_{11}\) , \(A_{12} \in \mathcal{A}_{12}\) , the following equality holds:

$$\Phi(A_{11}+A_{12})=\Phi(A_{11})+\Phi(A_{12}).$$

Setting \(T=\Phi(A_{11}+A_{12})-\Phi(A_{11})-\Phi(A_{12})\) let us prove that \(T=0\). We have

$$\begin{aligned} \, &\Phi(A_{11}+A_{12})\diamond C_{21} \diamond I+(A_{11}+A_{12}) \diamond \Phi(C_{21}) \diamond I+(A_{11}+A_{12}) \diamond C_{21} \diamond \Phi(I) \\ &\qquad =\Phi(A_{11}+A_{12} \diamond C_{21} \diamond I) \\ &\qquad =\Phi(A_{11} \diamond C_{21} \diamond I)+\Phi(A_{12} \diamond C_{21} \diamond I) \\ &\qquad =\Phi(A_{11}) \diamond C_{21} \diamond I+A_{11} \diamond \Phi(C_{21}) \diamond I+A_{11} \diamond C_{21} \diamond \Phi(I)+\Phi(A_{12}) \diamond C_{21} \diamond I \\ &\qquad\qquad +A_{12} \diamond \Phi(C_{21}) \diamond I+A_{12} \diamond C_{21} \diamond \Phi(I) \\ &\qquad =(\Phi(A_{11})+\Phi(A_{12})) \diamond C_{21} \diamond I+(A_{11}+A_{12}) \diamond \Phi(C_{21}) \diamond I +(A_{11}+A_{12}) \diamond C_{21} \diamond \Phi(I). \end{aligned}$$

Since \(T_{11}+T_{12}+T_{21}+T_{22}\), it follows that

$$T_{22}^{*}C_{21}+T_{21}^{*}C_{21}+C_{21}^{*}T_{22}+C_{21}^{*}T_{21}=0.$$

So \(T_{22}=T_{21}=0.\) Similarly, we have

$$\begin{aligned} \, &\Phi(A_{11}+A_{12}) \diamond C_{12} \diamond P_{1}+(A_{11}+A_{12}) \diamond \Phi(C_{12}) \diamond P_{1}+(A_{11}+A_{12}) \diamond C_{12} \diamond \Phi(P_{1})\\ &\qquad=\Phi((A_{11}+A_{12}) \diamond C_{12} \diamond P_{1}) \\ &\qquad=\Phi(A_{11} \diamond C_{12} \diamond P_{1})+\Phi(A_{12} \diamond C_{12} \diamond P_{1})\\ &\qquad =(\Phi(A_{11})+\Phi(A_{12}))\diamond C_{12} \diamond P_{1}+(A_{11}+A_{12}) \diamond \Phi(C_{12}) \diamond P_{1} +(A_{11}+A_{12}) \diamond C_{12} \diamond \Phi(P_{1}). \end{aligned}$$

Therefore, \(T\diamond C_{12} \diamond P_{1}=0\). So \(T_{11}^{*}C_{12}+C_{12}^{*}T_{11}=0\). It follows that \(T_{11}^{*}C_{12}=0\). Hence, for all \(C \in \mathcal{A}\), we have \(T_{11}^{*}C P_{2}=0\). Since \(\mathcal{A}\) is prime, it follows that \(T_{11}=0\). Similarly, we can show that \(T_{12}=0\) by applying \(P_{2}\) instead of \(P_{1}\) in the above.

Claim 5.

For each \(A_{11}\in\mathcal{A}_{11}, A_{12}\in\mathcal{A}_{12}, A_{21}\in\mathcal{A}_{21}\) , and \(A_{22}\in\mathcal{A}_{22}\) ,

  1. 1.

    \(\Phi(A_{11}+A_{12}+A_{21})=\Phi(A_{11})+\Phi(A_{12})+\Phi(A_{21}).\)

  2. 2.

    \(\Phi(A_{12}+A_{21}+A_{22})=\Phi(A_{12})+\Phi(A_{21})+\Phi(A_{22}).\)

Then

$$T=\Phi(A_{11}+A_{12}+A_{21})-\Phi(A_{11})-\Phi(A_{12})-\Phi(A_{21})=0.$$

From Claim 4, we obtain

$$\begin{aligned} \, &\Phi(A_{11}+A_{12}+A_{21})\diamond C_{21}\diamond I+(A_{11}+A_{12}+A_{21})\diamond \Phi(C_{21}) \diamond I+(A_{11}+A_{12}+A_{21}) \diamond C_{21} \diamond \Phi(I) \\ &\qquad =\Phi(A_{11}+A_{21}+A_{12} \diamond C_{21} \diamond I) \\ &\qquad =\Phi(A_{11}\diamond C_{21} \diamond I)+\Phi(A_{21}\diamond C_{21} \diamond I)+\Phi(A_{12}\diamond C_{21} \diamond I) \\ &\qquad =(\Phi(A_{11})+\Phi(A_{12})+\Phi(A_{21})) \diamond C_{21} \diamond I +(A_{11}+A_{12}+A_{21})\diamond \Phi(C_{21}) \diamond I \\ &\qquad\qquad +(A_{11}+A_{12}+A_{21}) \diamond C_{21}\diamond \Phi(I). \end{aligned}$$

It follows that \(T\diamond C_{21} \diamond I=0\). Since \(T=T_{11}+T_{12}+T_{21}+T_{22}\), we have

$$T_{22}^{*}C_{21}+T_{21}^{*}C_{21}+C_{21}^{*}T_{22}+C_{21}^{*}T_{21}=0.$$

Therefore, \(T_{22}=T_{21}=0\). From Claim 4, we obtain

$$\begin{aligned} \, &\Phi(A_{11}+A_{12}+A_{21})\diamond P_{1} \diamond P_{1}+(A_{11}+A_{12}+A_{21})\diamond \Phi(P_{1}) \diamond P_{1}+(A_{11}+A_{12}+A_{21})\diamond P_{1} \diamond \Phi(P_{1}) \\ &\qquad =\Phi((A_{11}+A_{12}+A_{21})\diamond P_{1} \diamond P_{1}) \\ &\qquad =\Phi(A_{11}\diamond P_{1} \diamond P_{1})+\Phi(A_{12}\diamond P_{1} \diamond P_{1})+\Phi(A_{21}\diamond P_{1} \diamond P_{1}) \\ &\qquad =(\Phi(A_{11})+\Phi(A_{12})+\Phi(A_{21}))\diamond P_{1} \diamond P_{1}+(A_{11}+A_{12}+A_{21})\diamond \Phi(P_{1}) \diamond P_{1} \\ &\qquad\qquad +(A_{11}+A_{12}+A_{21}) \diamond P_{1} \diamond \Phi(P_{1}). \end{aligned}$$

So \(T \diamond P_{1} \diamond P_{1}=0\) Then \(2T_{11}+2T_{11}^{*}+T_{12}+T_{12}^{*}=0\). Therefore,

$$ T_{12}=0,T_{11}+T_{11}^{*}=0.$$
(2.11)

Using Claim 3 and Claim 4, we obtain

$$\begin{aligned} \, &\Phi(A_{11}+A_{12}+A_{21}) \diamond iP_{1} \diamond I+(A_{11}+A_{12}+A_{21}) \diamond \Phi(iP_{1}) \diamond I+(A_{11}+A_{12}+A_{21}) \diamond iP_{1} \diamond \Phi(I)\\ &\qquad=\Phi(A_{11}+A_{12} \diamond iP_{1} \diamond I)+\Phi(A_{21} \diamond iP_{1} \diamond I)\\ &\qquad=\Phi(A_{11} \diamond iP_{1} \diamond I)+\Phi(A_{12} \diamond iP_{1} \diamond I)+\Phi(A_{21} \diamond iP_{1} \diamond I)\\ &\qquad=\Phi(A_{11}) \diamond iP_{1} \diamond I+A_{11} \diamond \Phi(iP_{1}) \diamond I+A_{11} \diamond iP_{1} \diamond \Phi(I) \\ &\qquad\qquad +\Phi(A_{12}) \diamond iP_{1} \diamond I+A_{12} \diamond \Phi(iP_{1}) \diamond I+A_{12} \diamond iP_{1} \diamond \Phi(I) \\ &\qquad\qquad +\Phi(A_{21}) \diamond iP_{1} \diamond I+A_{21} \diamond \Phi(iP_{1}) \diamond I+A_{21} \diamond iP_{1} \diamond \Phi(I) \\ &\qquad =(\Phi(A_{11})+\Phi(A_{12})+\Phi(A_{21})) \diamond iP_{1} \diamond I +(A_{11}+A_{12}+A_{21})\diamond \Phi(iP_{1}) \diamond I \\ &\qquad\qquad +(A_{11}+A_{12}+A_{21}) \diamond iP_{1} \diamond \Phi(I). \end{aligned}$$

Thus, \(T \diamond iP_{1} \diamond I =0\). We obtain

$$ T_{11}-T_{11}^{*}=0.$$
(2.12)

Relations (2.11) and (2.12) imply \(T_{11}=0\). Similarly, we can show that

$$\Phi(A_{12}+A_{21}+A_{22})=\Phi(A_{12})+\Phi(A_{21})+\Phi(A_{22}).$$

Claim 6.

For each \(A_{11}\in\mathcal{A}_{11}, A_{12}\in\mathcal{A}_{12}, A_{21}\in\mathcal{A}_{21}\) , and \(A_{22}\in\mathcal{A}_{22}\) ,

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

Then

$$T=\Phi(A_{11}+A_{12}+A_{21}+A_{22})-\Phi(A_{11})-\Phi(A_{12})-\Phi(A_{21})-\Phi(A_{22})= 0.$$

From Claim 5, we obtain

$$\begin{aligned} \, &\Phi(A_{11}+A_{12}+A_{21}+A_{22})\diamond C_{12} \diamond I+(A_{11}+A_{12}+A_{21}+A_{22})\diamond\Phi(C_{12}) \diamond I\\ &\qquad\qquad +(A_{11}+A_{12}+A_{21}+A_{22})\diamond C_{12} \diamond \Phi(I)\\ &\qquad=\Phi((A_{11}+A_{12}+A_{21}+A_{22})\diamond C_{12} \diamond I)\\ &\qquad=\Phi((A_{11}+A_{12}+A_{21})\diamond C_{12} \diamond I)+\Phi(A_{22}\diamond C_{12}\diamond I)\\ &\qquad=\Phi(A_{11}\diamond C_{12} \diamond I)+\Phi(A_{12}\diamond C_{12} \diamond I)+\Phi(A_{21}\diamond C_{12} \diamond I)+\Phi(A_{22}\diamond C_{12} \diamond I) \\ &\qquad=(\Phi(A_{11})+\Phi(A_{12})+\Phi(A_{21})+\Phi(A_{22}))\diamond C_{12} \diamond I \\ &\qquad\qquad +(A_{11}+A_{12}+A_{21}+A_{22})\diamond \Phi(C_{12}) \diamond I \\ &\qquad\qquad +(A_{11}+A_{12}+A_{21}+A_{22}) \diamond C_{12} \diamond \Phi(I). \end{aligned}$$

Thus, \(T\diamond C_{12} \diamond I=0\). It follows that

$$C_{12}^{*}T_{11}+C_{12}^{*}T_{12}+T_{11}^{*}C_{12}+T_{12}^{*}C_{12}=0.$$

Therefore, \(T_{11}=T_{12}=0\). Similarly, by applying \(C_{21}\) instead of \(C_{12}\) in the above, we obtain \(T_{21}=T_{22}=0\).

Claim 7.

For each \(A_{ij},B_{ij} \in \mathcal{A}_{i}\) such that \(i\neq j\) , the following equality holds:

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

It is easy to show that

$$(P_{i}+A_{ij}^{*})\diamond (P_{j}+B_{ij}) \diamond \frac{I}{2}=A_{ij}+B_{ij}+A_{ij}^{*}+B_{ij}^{*}.$$

Thus, we can write

$$\begin{aligned} \, \Phi(A_{ij}+B_{ij})+\Phi(A_{ij}^{*}+B_{ij}^{*}) &=\Phi\bigg((P_{i}+A_{ij}^{*})\diamond (P_{j}+B_{ij}) \diamond \frac{I}{2}\bigg) \\ &=\Phi(P_{i}+A_{ij}^{*})\diamond (P_{j}+B_{ij}) \diamond \frac{I}{2}+(P_{i}+A_{ij}^{*})\diamond \Phi(P_{j}+B_{ij}) \diamond \frac{I}{2}\\ &\qquad +(P_{i}+A_{ij}^{*})\diamond (P_{j}+B_{ij}) \diamond \Phi\bigg(\frac{I}{2}\bigg)\\ &=(\Phi(P_{i})+\Phi(A_{ij}^{*}))\diamond (P_{j}+B_{ij}) \diamond \frac{I}{2}+(P_{i}+A_{ij}^{*})\diamond (\Phi(P_{j})\\ &\qquad +\Phi(B_{ij})) \diamond \frac{I}{2}+(P_{i}+A_{ij}^{*})\diamond (P_{j}+B_{ij})\diamond \Phi\bigg(\frac{I}{2}\bigg)\\ &=\Phi\bigg(P_{i}\diamond B_{ij} \diamond \frac{I}{2}\bigg)+\Phi\bigg(A_{ij}^{*} \diamond P_{j} \diamond \frac{I}{2}\bigg)\\ &=\Phi(B_{ij})+\Phi(B_{ij}^{*})+\Phi(A_{ij})+\Phi(A_{ij}^{*}). \end{aligned}$$

Thus, we have shown that

$$ \Phi(A_{ij}+B_{ij})+\Phi(A_{ij}^{*}+B_{ij}^{*})=\Phi(A_{ij})+\Phi(B_{ij})+\Phi(A_{ij}^{*}) +\Phi(B_{ij}^{*}).$$
(2.13)

By an easy computation, we obtain

$$(P_{i}+A_{ij}^{*}) \diamond (iP_{j}+iB_{ij}) \diamond \frac{I}{2}=iA_{ij}+iB_{ij}-iA_{ij}^{*}-iB_{ij}^{*}.$$

Then, we have

$$\begin{aligned} \, \Phi(iA_{ij}+iB_{ij})+\Phi(-iA_{ij}^{*}-iB_{ij}^{*})&=\Phi\bigg((P_{i}+A_{ij}^{*})\diamond (iP_{j}+iB_{ij}) \diamond \frac{I}{2}\bigg) \\ & =\Phi(P_{i}+A_{ij}^{*})\diamond (iP_{j}+iB_{ij}) \diamond \frac{I}{2}+(P_{i}+A_{ij}^{*})\diamond \Phi(iP_{j}+iB_{ij}) \diamond \frac{I}{2} \\ &\qquad +(P_{i}+A_{ij}^{*}) \diamond (iP_{j}+iB_{ij}) \diamond \Phi\bigg(\frac{I}{2}\bigg) \\ & =(\Phi(P_{i})+\Phi(A_{ij}^{*}))\diamond(iP_{j}+iB_{ij}) \diamond \frac{I}{2}+(P_{i}+A_{ij}^{*})\diamond (\Phi(iP_{j}) \\ &\qquad +\Phi(iB_{ij}))\diamond \frac{I}{2} +(P_{i}+A_{ij}^{*})\diamond (iP_{j}+iB_{ij})\diamond \Phi\bigg(\frac{I}{2}\bigg) \\ & =\Phi\bigg(P_{i}\diamond iB_{ij}\diamond \frac{I}{2}\bigg)+\Phi\bigg(A_{ij}^{*}\diamond iP_{j} \diamond \frac{I}{2}\bigg) \\ & =\Phi(iB_{ij})+\Phi(-iB_{ij}^{*})+\Phi(iA_{ij})+\Phi(-iA_{ij}^{*}). \end{aligned}$$

We have shown that

$$\Phi(iA_{ij}+iB_{ij})+\Phi(-iA_{ij}^{*}-iB_{ij}^{*})=\Phi(iA_{ij})+\Phi(iB_{ij})+\Phi(-i A_{ij}^{*})+\Phi(-iB_{ij}^{*}).$$

From Claim 3 and the above equation, we have

$$ \Phi(A_{ij}+B_{ij})-\Phi(A_{ij}^{*}+B_{ij}^{*})=\Phi(B_{ij})-\Phi(B_{ij}^{*})+\Phi(A_{ij}) -\Phi(A_{ij}^{*}).$$
(2.14)

By adding equations (2.13) and (2.14), we obtain

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

Claim 8.

For each \(A_{ii},B_{ii} \in \mathcal{A}_{ii}\) such that \(1\leq i \leq 2\) , the following equality holds:

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

Let us show that

$$T=\Phi(A_{ii}+B_{ii})-\Phi(A_{ii})-\Phi(B_{ii})=0.$$

We have

$$\begin{aligned} \, &\Phi(A_{ii}+B_{ii})\diamond P_{j} \diamond I+(A_{ii}+B_{ii})\diamond \Phi(P_{j})\diamond I+(A_{ii}+B_{ii})\diamond P_{j} \diamond \Phi(I)\\ &\qquad=\Phi((A_{ii}+B_{ii}) \diamond P_{j} \diamond I)\\ &\qquad=\Phi(A_{ii} \diamond P_{j} \diamond I)+\Phi(B_{ii} \diamond P_{j} \diamond I)\\ &\qquad=\Phi(A_{ii}) \diamond P_{j} \diamond I+A_{ii} \diamond \Phi(P_{j}) \diamond I+A_{ii} \diamond P_{j} \diamond \Phi(I)+\Phi(B_{ii}) \diamond P_{j} \diamond I \\ &\qquad\qquad +B_{ii} \diamond \Phi(P_{j}) \diamond I+B_{ii} \diamond P_{j} \diamond \Phi(I)\\ &\qquad=(\Phi(A_{ii})+\Phi(B_{ii}))\diamond P_{j} \diamond I+(A_{ii}+B_{ii}) \diamond \Phi(P_{j}) \diamond I+(A_{ii}+B_{ii})\diamond P_{j} \diamond \Phi(I). \end{aligned}$$

Therefore,

$$T\diamond P_{j} \diamond I=0.$$

Thus, \(T_{ij}=T_{ji}=T_{jj}=0\).

On the other hand, for every \(C_{ij}\in\mathcal{A}_{ij}\), we have

$$\begin{aligned} \, &\Phi(A_{ii}+B_{ii})\diamond C_{ij} \diamond I+(A_{ii}+B_{ii})\diamond \Phi(C_{ij}) \diamond I+(A_{ii}+B_{ii})\diamond C_{ij} \diamond \Phi(I)\\ &\qquad=\Phi((A_{ii}+B_{ii})\diamond C_{ij} \diamond I) \\ &\qquad=\Phi(A_{ii}\diamond C_{ij} \diamond I)+\Phi(B_{ii}\diamond C_{ij} \diamond I)\\ &\qquad=(\Phi(A_{ii})+\Phi(B_{ii})) \diamond C_{ij} \diamond I+(A_{ii}+B_{ii}) \diamond \Phi(C_{ij}) \diamond I \\ &\qquad\qquad +(A_{ii}+B_{ii}) \diamond C_{ij} \diamond \Phi(I). \end{aligned}$$

Thus, \(T\diamond C_{ij} \diamond I=0\); then \(T_{ii} \diamond C_{ij} \diamond I=0\). We have \(T_{ii}^{*}C_{ij}+C_{ij}^{*}T_{ii}=0\). We know that if \(\mathcal{A}\) is prime, then \(T_{ii}=0\). Hence the additivity of \(\Phi\) follows from the above claims. □

In the rest of this paper, we show that \(\Phi\) is a \(\ast\)-derivation.

Claim 9.

\(\Phi\) preserves stars.

Since \(\Phi({I}/{2})=0\), we have

$$\Phi\bigg(\frac{I}{2} \diamond \frac{I}{2}\diamond A\bigg)=\frac{I}{2} \diamond \frac{I}{2}\diamond \Phi(A).$$

Therefore,

$$\Phi(A+A^{*})=\Phi(A)+\Phi(A)^{*}.$$

Thus, we have shown that \(\Phi\) preserves stars.

Claim 10.

\(\Phi\) is a derivation.

For every \(A,B\in\mathcal{A}\), we have

$$\begin{aligned} \, \Phi(AB+A^{*}B+B^{*}A+B^{*}A^{*})&=\Phi(I \diamond A \diamond B)\\ &=I \diamond \Phi(A) \diamond B+I \diamond A \diamond \Phi(B)\\ &=(\Phi(A)+\Phi(A)^{*}) \diamond B+(A+A^{*}) \diamond \Phi(B)\\ &=\Phi(A)B+\Phi(A)^{*}B+B^{*}\Phi(A)+B^{*}\Phi(A)^{*}\\ &\qquad +A\Phi(B)+A^{*}\Phi(B) +\Phi(B)^{*}A+\Phi(B)^{*}A^{*}. \end{aligned}$$

Therefore,

$$\begin{aligned} \, \Phi(AB+A^{*}B+B^{*}A+B^{*}A^{*})&=\Phi(A)B+\Phi(A)^{*}B+B^{*}\Phi(A)+B^{*}\Phi(A)^{*}+ A\Phi(B)\nonumber\\ &\qquad +A^{*}\Phi(B)+\Phi(B)^{*}A+\Phi(B)^{*}A^{*}. \end{aligned}$$
(2.15)

Also

$$\begin{aligned} \, \Phi(AB-A^{*}B-B^{*}A+B^{*}A^{*})&=\Phi(I \diamond (-iA) \diamond iB)\\ &=I \diamond \Phi(-iA) \diamond iB+I \diamond (-iA) \diamond \Phi(iB)\\ &=\Phi(A)B-\Phi(A)^{*}B-B^{*}\Phi(A)+B^{*}\Phi(A)^{*}\\ &\qquad +A\Phi(B)-A^{*}\Phi(B)-\Phi(B)^{*}A+\Phi(B)^{*}A^{*}. \end{aligned}$$

So we have

$$\begin{aligned} \, \Phi(AB-A^{*}B-B^{*}A+B^{*}A^{*})&=\Phi(A)B-\Phi(A)^{*}B-B^{*}\Phi(A)\nonumber\\ &\qquad +B^{*}\Phi(A)^{*}+A\Phi(B)-A^{*}\Phi(B)\nonumber\\ &\qquad -\Phi(B)^{*}A+\Phi(B)^{*}A^{*}. \end{aligned}$$
(2.16)

By adding equations (2.15) and (2.16), we obtain

$$\Phi(AB+B^{*}A^{*})=\Phi(A)B+A\Phi(B)+\Phi(A)^{*}B^{*}+A^{*}\Phi(B)^{*}.$$
(2.17)

From (2.17), Claims 3 and 9, it follows that

$$\begin{aligned} \, \Phi(AB-B^{*}A^{*})&=i\Phi(A(-iB)+(-iB)^{*}A^{*})\\ &=i(\Phi(A)(-iB)+A\Phi(-iB)+\Phi(A)^{*}(-iB)^{*}+A^{*}\Phi(-iB)^{*})\\ &=\Phi(A)B+A\Phi(B)-\Phi(A)^{*}B^{*}-A^{*}\Phi(B)^{*}. \end{aligned}$$

Therefore,

$$\Phi(AB-B^{*}A^{*})=\Phi(A)B+A\Phi(B)-\Phi(A)^{*}B^{*}-A^{*}\Phi(B)^{*}.$$
(2.18)

From (2.17) and (2.18), we obtain

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

This completes the proof. □