© Wuhan University 2025

0 Introduction

Let $\mathcal{A}$ be a $*$-algebra over the complex field $\mathbb{C}$. For $A,B\in \mathcal{A}$, define the Jordan $*$-product of A and B by $A•B=AB+B{A}^{*}$ and the skew Lie product of A and B by ${[A,B]}_{*}=AB-B{A}^{*}$. Among some research topics, the Jordan $*$-product and the skew Lie product are of great significance^[1-9]. Recall that an additive map $\Phi :\mathcal{A}\to \mathcal{A}$ is an additive derivation if $\Phi (AB)=\Phi (A)B+A\Phi (B)$ for all $A,B\in \mathcal{A}$. In addition, $\Phi$ is said to be an additive $*$-derivation if $\Phi$ is an additive derivation and $\Phi (A^{*})=\Phi {(A)}^{*}$ for all $A\in \mathcal{A}$. Note that $\Phi$ is nonlinear if its additivity is removed.

Let $d:\mathcal{A}\to \mathcal{A}$ be a nonlinear map, $d$ is called

(i) a nonlinear Jordan triple $*$-derivation if $d(A•B•C)=d(A)•B•C+A•d(B)•C+A•B•d(C)$ for all $A,B,C\in \mathcal{A}$, where $A•B•C=(A•B)•C$.

(ii) a nonlinear skew Lie triple derivation if $d([{[A,B]}_{*}{,C]}_{*})=[[d{(A),B]}_{*}{,C]}_{*}+[[A,d{(B)]}_{*}{,C]}_{*}+[{[A,B]}_{*},d{(C)]}_{*}$ for all $A,B,C\in \mathcal{A}$. The characterization of Jordan triple $*$-derivations and skew Lie triple derivations have been studied by many authors^[10-19]. Recently the derivations corresponding to the new products of the mixture of various products have attracted the attentions of many researchers (see Refs. [20-22]). Li and Zhang^[23] considered the mixed Jordan triple $*$-product ${[A•B,C]}_{*}$ and investigated the nonlinear mixed Jordan triple $*$-derivation $d:\mathcal{A}\to \mathcal{A}$, i.e.

$d({[A•B,C]}_{*})=[d{(A)•B,C]}_{*}+[A•d{(B),C]}_{*}+[A•B,d{(C)]}_{*},$

and they proved a nonlinear mixed Jordan triple $*$-derivation is an additive $*$-derivation on $*$-algebras under some mild condition.

Let $\mathbb{N}$ be the set of non-negative integers, $\mathcal{D}=\{{\Phi }_n{\}}_{n\in \mathbb{N}}$ be a family of additive maps ${\Phi }_n:\mathcal{A}\to \mathcal{A}$ and ${\Phi }_{\mathrm{0}}$ be the identity map on $\mathcal{A}$. Then $\mathcal{D}=\{{\Phi }_n{\}}_{n\in \mathbb{N}}$ is called an additive higher derivation if for every $n\in \mathbb{N}$, ${\Phi }_n(AB)=\sum _{i+j=n}{\Phi }_i(A){\Phi }_j(B)$ for all $A,B\in \mathcal{A}$. $\mathcal{D}=\{{\Phi }_n{\}}_{n\in \mathbb{N}}$ is said to be an additive $*$-higher derivation if $\mathcal{D}=\{{\Phi }_n{\}}_{n\in \mathbb{N}}$ is an additive higher derivation with ${\Phi }_n(A^{*})={\Phi }_n{(A)}^{*}$ for all $A\in \mathcal{A}$ and $n\in \mathbb{N}$. In addition, $\mathcal{D}=\{{\Phi }_n{\}}_{n\in \mathbb{N}}$ is nonlinear if the additivity is removed. In recent years, many authors study different types of nonlinear higher derivations. Some researchers studied the nonlinear higher derivations on semi-prime rings and triangular algebras^[24-26].

The questions of characterizing Lie higher derivations and the relationship between different types of higher derivations have been studied by many authors (see Refs. [27-29]).

Let $d_{\mathrm{0}}$ be the identity map on $\mathcal{A}$, and let $d_n:\mathcal{A}\to \mathcal{A}$ be a nonlinear map for all $n\ge \mathrm{1}$. We say that $\mathcal{D}=\{d_n{\}}_{n\in \mathbb{N}}$ is

(i) a nonlinear Jordan triple $*$-higher derivation if $d_n(A•B•C)=\sum _{i+j+k=n}{d}_i(A)•d_j(B)•d_k(C).$

(ii) a nonlinear skew Lie triple higher derivation if $d_n([{[A,B]}_{*}{,C]}_{*})=\sum _{i+j+k=n}[[d_i(A),d_j{(B)]}_{*},d_k(C)]$ for all $A,B,C\in \mathcal{A}$ and $n\in \mathbb{N}$. Some reasearchers proved that a nonlinear Jordan triple $*$-higher derivation or skew Lie triple higher derivation is an additive $*$-higher derivation on standard operator algebras^[30-31]. Motivated by the related works^{[23, 30-31]}, we consider nonlinear mixed Jordan triple $*$-higher derivations $\mathcal{D}=\{d_n{\}}_{n\in \mathbb{N}}$, that is, $d_n({[A•B,C]}_{*})=\sum _{i+j+k=n}[d_i(A)•d_j(B),d_k{(C)]}_{*}$ for all $A,B,C\in \mathcal{A}$ and $n\in \mathbb{N}$. It is clear that $d_{\mathrm{1}}$ is a nonlinear mixed Jordan triple $*$-derivation on $\mathcal{A}$ if $\mathcal{D}=\{d_n{\}}_{n\in \mathbb{N}}$ is a nonlinear mixed Jordan triple $*$-higher derivation of $\mathcal{A}$, then we will study the nonlinear mixed Jordan triple $*$-higher derivations on $\mathcal{A}$.

1 Nonlinear Mixed Jordan Triple $\mathit{*}$-Derivation

Theorem 1   Let $\mathcal{A}$ be a $*$-algebra with the unit $I$. Assume that $\mathcal{A}$ contains a nontrivial projection $P$ which satisfies for $A\in \mathcal{A}$ ,

$XAP=\mathrm{0}\mathrm{i}\mathrm{m}\mathrm{p}\mathrm{l}\mathrm{i}\mathrm{e}\mathrm{s}X=\mathrm{0}$(1)

and

$XA\left(I-P\right)=\mathrm{0}\mathrm{i}\mathrm{m}\mathrm{p}\mathrm{l}\mathrm{i}\mathrm{e}\mathrm{s}X=\mathrm{0}$(2)

Then every nonlinear mixed Jordan triple $*$-derivation $d:\mathcal{A}\to \mathcal{A}$, which satisfies

$d({[A•B,C]}_{*})=[d{(A)•B,C]}_{*}+[A•d{(B),C]}_{*}+[A•B,d{(C)]}_{*}$

for all $A,B,C\in \mathcal{A}$, is an additive $\mathrm{*}$-derivation.

Proof   Let $P_{\mathrm{1}}=P$ and $P_{\mathrm{2}}=I-P$. We write ${\mathcal{A}}_{jk}=P_j\mathcal{A}{P}_k$ for $j,k=\mathrm{1,2}$. Then $\mathcal{A}={\displaystyle \sum _{j,k=\mathrm{1}}^{\mathrm{2}}}{\mathcal{A}}_{jk}.$

In the sequel ${\mathcal{A}}_{jk}$ indicates that $A_{jk}\in {\mathcal{A}}_{jk}$. We set $\mathcal{S}=\{A\in \mathcal{A},A=A^{*}\}$. We complete the proof by several Claims.

Claim 1 $d_n\left(\mathrm{0}\right)=\mathrm{0}$ for all $n\in \mathbb{N}$.

Proof   It follows from ${\left[\mathrm{0}•\mathrm{0,0}\right]}_{*}=\mathrm{0}$ that

$d\left(\mathrm{0}\right)=d({[\mathrm{0}•\mathrm{0,0}]}_{*})=[d{(\mathrm{0})•\mathrm{0,0}]}_{*}+[\mathrm{0}•d{(\mathrm{0}),\mathrm{0}]}_{*}+[\mathrm{0}•\mathrm{0},d{(\mathrm{0})]}_{*}=\mathrm{0}.$

Claim 2 $d$ has the following properties: (a) For any $\lambda \in \mathbb{R}$, $d\left(\lambda I\right)\in \mathcal{Z}(\mathcal{A})$ and $d\left(\lambda I\right)=d{\left(\lambda I\right)}^{*}$; (b) For any $A\in \mathcal{S}$, $d\left(A\right)=d\left(A^{*}\right)=d{\left(A\right)}^{*}$; (c) For any $\lambda \in \mathbb{C}$, $d\left(\lambda I\right)\in \mathcal{Z}(\mathcal{A})$.

Proof   (a) It follows from ${\left[I•\lambda I,I\right]}_{*}=\mathrm{0}$ that

$\mathrm{0}=d\left({\left[I•\lambda I,I\right]}_{*}\right)={\left[d\left(I\right)•\lambda I,I\right]}_{*}+{\left[I•d\left(\lambda I\right),I\right]}_{*}+{\left[I•\lambda I,d\left(I\right)\right]}_{*}={\left[I•d\left(\lambda I\right),I\right]}_{*}=\mathrm{2}d\left(\lambda I\right)-\mathrm{2}d{\left(\lambda I\right)}^{*}.$

From this we get $d\left(\lambda I\right)=d{\left(\lambda I\right)}^{*}.$ Then for all $A\in \mathcal{A}$, it follows from ${\left[\lambda I•\lambda I,A\right]}_{*}=\mathrm{0}$ that

$\begin{array}{l}\mathrm{0}=d\left({\left[\lambda I•\lambda I,A\right]}_{*}\right)={\left[d\left(\lambda I\right)•\lambda I,A\right]}_{*}+{\left[\lambda I•d\left(\lambda I\right),A\right]}_{*}+{\left[\lambda I•\lambda I,d\left(A\right)\right]}_{*}\\ ={\left[d\left(\lambda I\right)•\lambda I,A\right]}_{*}+{\left[\lambda I•d\left(\lambda I\right),A\right]}_{*}=\mathrm{4}\lambda d\left(\lambda I\right)A-\mathrm{4}\lambda Ad{\left(\lambda I\right)}^{*}.\end{array}$

Then we get $d\left(\lambda I\right)A=Ad\left(\lambda I\right),$ for all $A\in \mathcal{A}$, which implies $d\left(\lambda I\right)\in \mathcal{Z}(\mathcal{A}).$

(b) For any $A\in \mathcal{S},$ we see from Claim 2(a) that

$\mathrm{0}=d\left({\left[I•A,I\right]}_{*}\right)={\left[d\left(I\right)•A,I\right]}_{*}+{\left[I•d\left(A\right),I\right]}_{*}+{\left[I•A,d\left(I\right)\right]}_{*}={\left[I•d\left(A\right),I\right]}_{*}=\mathrm{2}d\left(A\right)-\mathrm{2}d{\left(A\right)}^{*}.$

(c) For any $\lambda \in \mathbb{C}$ and $A\in \mathcal{S},$ we see from Claim 2(a) and Claim 2(b) that

$\mathrm{0}=d\left({\left[A•I,\lambda I\right]}_{*}\right)={\left[d\left(A\right)•I,\lambda I\right]}_{*}+{\left[A•d\left(I\right),\lambda I\right]}_{*}+{\left[A•I,d\left(\lambda I\right)\right]}_{*}={\left[A•I,d\left(\lambda I\right)\right]}_{*}=\mathrm{2}Ad\left(\lambda I\right)-\mathrm{2}d\left(\lambda I\right)A.$

Thus $Ad\left(\lambda I\right)=d\left(\lambda I\right)A,$ for all $A\in \mathcal{S},$ which implies $d\left(\lambda I\right)\in \mathcal{Z}(\mathcal{A}).$

Claim 3 For any $A\in \mathcal{A},$ $d\left(\frac{\mathrm{1}}{\mathrm{2}}I\right)=d\left(\frac{\mathrm{1}}{\mathrm{2}}\mathrm{i}I\right)=\mathrm{0},d\left(\mathrm{i}A\right)=\mathrm{i}d\left(A\right).$

Proof   It follows from ${\left[\frac{\mathrm{1}}{\mathrm{2}}I•\frac{\mathrm{1}}{\mathrm{2}}\mathrm{i}I,\frac{\mathrm{1}}{\mathrm{2}}\mathrm{i}I\right]}_{*}=-\frac{\mathrm{1}}{\mathrm{2}}I$ , where $\mathrm{i}$ is the imaginary unit. We see from Claim 2 that

$d\left(-\frac{\mathrm{1}}{\mathrm{2}}I\right)=d\left({\left[\frac{\mathrm{1}}{\mathrm{2}}I•\frac{\mathrm{1}}{\mathrm{2}}\mathrm{i}I,\frac{\mathrm{1}}{\mathrm{2}}\mathrm{i}I\right]}_{*}\right)={\left[d\left(\frac{\mathrm{1}}{\mathrm{2}}I\right)•\frac{\mathrm{1}}{\mathrm{2}}\mathrm{i}I,\frac{\mathrm{1}}{\mathrm{2}}\mathrm{i}I\right]}_{*}+{\left[\frac{\mathrm{1}}{\mathrm{2}}I•d\left(\frac{\mathrm{1}}{\mathrm{2}}\mathrm{i}I\right),\frac{\mathrm{1}}{\mathrm{2}}\mathrm{i}I\right]}_{*}+{\left[\frac{\mathrm{1}}{\mathrm{2}}I•\frac{\mathrm{1}}{\mathrm{2}}\mathrm{i}I,d\left(\frac{\mathrm{1}}{\mathrm{2}}\mathrm{i}I\right)\right]}_{*}$

$=-d\left(\frac{\mathrm{1}}{\mathrm{2}}I\right)+\frac{\mathrm{1}}{\mathrm{2}}\mathrm{i}\left(d\left(\frac{\mathrm{1}}{\mathrm{2}}\mathrm{i}I\right)-d{\left(\frac{\mathrm{1}}{\mathrm{2}}\mathrm{i}I\right)}^{*}\right)+\mathrm{i}d\left(\frac{\mathrm{1}}{\mathrm{2}}\mathrm{i}I\right).$

Since $d(\frac{\mathrm{1}}{\mathrm{2}}I),d\left(-\frac{\mathrm{1}}{\mathrm{2}}I\right),\mathrm{i}\left(d\left(\frac{\mathrm{1}}{\mathrm{2}}\mathrm{i}I\right)-d{\left(\frac{\mathrm{1}}{\mathrm{2}}\mathrm{i}I\right)}^{*}\right)$ are self-adjoint, $\mathrm{i}d\left(\frac{\mathrm{1}}{\mathrm{2}}\mathrm{i}I\right)$ is also self-adjoint. Hence

$d\left(-\frac{\mathrm{1}}{\mathrm{2}}I\right)=-d\left(\frac{\mathrm{1}}{\mathrm{2}}I\right)+\mathrm{2}\mathrm{i}d\left(\frac{\mathrm{1}}{\mathrm{2}}\mathrm{i}I\right).$(3)

Similarly, we can also obtain from the fact that $-\frac{\mathrm{1}}{\mathrm{2}}I={\left[\frac{\mathrm{1}}{\mathrm{2}}I•\left(-\frac{\mathrm{1}}{\mathrm{2}}\mathrm{i}I\right),-\frac{\mathrm{1}}{\mathrm{2}}\mathrm{i}I\right]}_{*}$, that

$d\left(-\frac{\mathrm{1}}{\mathrm{2}}\mathrm{i}I\right)=-d{\left(-\frac{\mathrm{1}}{\mathrm{2}}\mathrm{i}I\right)}^{\mathrm{*}}$(4)

and

$d\left(-\frac{\mathrm{1}}{\mathrm{2}}I\right)=-d\left(\frac{\mathrm{1}}{\mathrm{2}}I\right)-\mathrm{2}\mathrm{i}d\left(-\frac{\mathrm{1}}{\mathrm{2}}\mathrm{i}I\right).$(5)

Now by (3) and (5) we have

$d\left(-\frac{\mathrm{1}}{\mathrm{2}}\mathrm{i}I\right)=-d\left(\frac{\mathrm{1}}{\mathrm{2}}\mathrm{i}I\right).$(6)

It follows from (4) and (6), we compute

$d\left(\frac{\mathrm{1}}{\mathrm{2}}\mathrm{i}I\right)=d\left({\left[\frac{\mathrm{1}}{\mathrm{2}}I•\left(-\frac{\mathrm{1}}{\mathrm{2}}\mathrm{i}I\right),-\frac{\mathrm{1}}{\mathrm{2}}I\right]}_{*}\right)$

$={\left[d\left(\frac{\mathrm{1}}{\mathrm{2}}I\right)•\left(-\frac{\mathrm{1}}{\mathrm{2}}\mathrm{i}I\right),-\frac{\mathrm{1}}{\mathrm{2}}I\right]}_{*}+{\left[\frac{\mathrm{1}}{\mathrm{2}}I•d\left(-\frac{\mathrm{1}}{\mathrm{2}}\mathrm{i}I\right),-\frac{\mathrm{1}}{\mathrm{2}}I\right]}_{*}+{\left[\frac{\mathrm{1}}{\mathrm{2}}I•\left(-\frac{\mathrm{1}}{\mathrm{2}}\mathrm{i}I\right),d\left(-\frac{\mathrm{1}}{\mathrm{2}}I\right)\right]}_{*}$

$=\mathrm{i}d\left(\frac{\mathrm{1}}{\mathrm{2}}I\right)-\frac{\mathrm{1}}{\mathrm{2}}\left(d\left(-\frac{\mathrm{1}}{\mathrm{2}}\mathrm{i}I\right)-d{\left(-\frac{\mathrm{1}}{\mathrm{2}}\mathrm{i}I\right)}^{*}\right)-\mathrm{i}d\left(-\frac{\mathrm{1}}{\mathrm{2}}I\right).$

Then we can get

$d\left(\frac{\mathrm{1}}{\mathrm{2}}I\right)=d\left(-\frac{\mathrm{1}}{\mathrm{2}}I\right).$(7)

Now we compute

$d(-\frac{\mathrm{1}}{\mathrm{2}}\mathrm{i}I)=d\left({\left[-\frac{\mathrm{1}}{\mathrm{2}}I•\left(-\frac{\mathrm{1}}{\mathrm{2}}\mathrm{i}I\right),-\frac{\mathrm{1}}{\mathrm{2}}I\right]}_{\mathrm{*}}\right)$

$={\left[d\left(-\frac{\mathrm{1}}{\mathrm{2}}I\right)•\left(-\frac{\mathrm{1}}{\mathrm{2}}\mathrm{i}I\right),-\frac{\mathrm{1}}{\mathrm{2}}I\right]}_{\mathrm{*}}+{\left[-\frac{\mathrm{1}}{\mathrm{2}}I•d\left(-\frac{\mathrm{1}}{\mathrm{2}}\mathrm{i}I\right),-\frac{\mathrm{1}}{\mathrm{2}}I\right]}_{\mathrm{*}}+{\left[-\frac{\mathrm{1}}{\mathrm{2}}I•\left(-\frac{\mathrm{1}}{\mathrm{2}}\mathrm{i}I\right),d\left(-\frac{\mathrm{1}}{\mathrm{2}}I\right)\right]}_{*}$

$=\mathrm{2}\mathrm{i}d\left(-\frac{\mathrm{1}}{\mathrm{2}}I\right)+\frac{\mathrm{1}}{\mathrm{2}}\left(d\left(-\frac{\mathrm{1}}{\mathrm{2}}\mathrm{i}I\right)-d{\left(-\frac{\mathrm{1}}{\mathrm{2}}\mathrm{i}I\right)}^{*}\right).$

By using equation (4) we get $d\left(-\frac{\mathrm{1}}{\mathrm{2}}I\right)=\mathrm{0},$ then by equations (3), (6) and (7), we have $d\left(-\frac{\mathrm{1}}{\mathrm{2}}I\right)=d\left(\frac{\mathrm{1}}{\mathrm{2}}I\right)=d\left(-\frac{\mathrm{1}}{\mathrm{2}}\mathrm{i}I\right)=d\left(\frac{\mathrm{1}}{\mathrm{2}}\mathrm{i}I\right)=\mathrm{0}.$

For any $A\in \mathcal{A}$ we have

$d(\mathrm{i}A)=d\left({\left[\frac{\mathrm{1}}{\mathrm{2}}I•\frac{\mathrm{1}}{\mathrm{2}}\mathrm{i}I,A\right]}_{\mathrm{*}}\right)={\left[d\left(\frac{\mathrm{1}}{\mathrm{2}}I\right)•\frac{\mathrm{1}}{\mathrm{2}}\mathrm{i}I,A\right]}_{\mathrm{*}}+{\left[\frac{\mathrm{1}}{\mathrm{2}}I•d\left(\frac{\mathrm{1}}{\mathrm{2}}\mathrm{i}I\right),A\right]}_{\mathrm{*}}+{\left[\frac{\mathrm{1}}{\mathrm{2}}I•\frac{\mathrm{1}}{\mathrm{2}}\mathrm{i}I,d\left(A\right)\right]}_{\mathrm{*}}={\left[\frac{\mathrm{1}}{\mathrm{2}}I•\frac{\mathrm{1}}{\mathrm{2}}\mathrm{i}I,d\left(A\right)\right]}_{*}=\mathrm{i}d(A).$

Claim 4^[23] $d$ is additive.

Claim 5 For any $A\in \mathcal{A},d(A^{*})=d{(A)}^{*}.$

Proof   For any $A\in \mathcal{A}$, where $A=A_{\mathrm{1}}+\mathrm{i}{A}_{\mathrm{2}}$, $A_{\mathrm{1}},A_{\mathrm{2}}\in \mathcal{S}$, it follows from Claim 2, Claim 3 and Claim 4 that

$d(A^{\mathrm{*}})=d(A_{\mathrm{1}}-\mathrm{i}{A}_{\mathrm{2}})=d(A_{\mathrm{1}})-\mathrm{i}d(A_{\mathrm{2}})=d(A_{\mathrm{1}}{)}^{*}+{\left(\mathrm{i}d(A_{\mathrm{2}})\right)}^{*}=d(A_{\mathrm{1}}+\mathrm{i}{A}_{\mathrm{2}}{)}^{*}=d{(A)}^{*}.$

Claim 6 $d$ is an additive $\mathrm{*}$-derivation on $\mathcal{A}$.

Proof   To complete the proof, we only need to show that $d$ is a derivation on $\mathcal{A}$. Since $d$ is additive. ${[A,B]}_{\mathrm{*}}=AB-B{A}^{\mathrm{*}}$ and ${[\mathrm{i}A,\mathrm{i}B]}_{\mathrm{*}}=-AB-B{A}^{\mathrm{*}}$ for all $A,B\in \mathcal{A}$. It follows from Claim 3 that

$d(AB)-d(B{A}^{\mathrm{*}})=d({[A,B]}_{\mathrm{*}})=d\left({\left[\frac{\mathrm{1}}{\mathrm{2}}I•A,B\right]}_{\mathrm{*}}\right)={\left[d\left(\frac{\mathrm{1}}{\mathrm{2}}I\right)•A,B\right]}_{*}+{\left[\frac{\mathrm{1}}{\mathrm{2}}I•d\left(A\right),B\right]}_{*}+{\left[\frac{\mathrm{1}}{\mathrm{2}}I•A,d\left(B\right)\right]}_{*}$

$={\left[\frac{\mathrm{1}}{\mathrm{2}}I•d\left(A\right),B\right]}_{*}+{\left[\frac{\mathrm{1}}{\mathrm{2}}I•A,d\left(B\right)\right]}_{*}=d(A)B-Bd{(A)}^{*}+Ad(B)-d(B)A^{*}.$

And

$-d(AB)-d(B{A}^{\mathrm{*}})=d({[\mathrm{i}A,\mathrm{i}B]}_{\mathrm{*}})=d\left({\left[\frac{\mathrm{1}}{\mathrm{2}}I•\mathrm{i}A,\mathrm{i}B\right]}_{\mathrm{*}}\right)={\left[d\left(\frac{\mathrm{1}}{\mathrm{2}}I\right)•\mathrm{i}A,\mathrm{i}B\right]}_{*}+{\left[\frac{\mathrm{1}}{\mathrm{2}}I•d\left(\mathrm{i}A\right),\mathrm{i}B\right]}_{*}+{\left[\frac{\mathrm{1}}{\mathrm{2}}I•\mathrm{i}A,d\left(\mathrm{i}B\right)\right]}_{*}$

$={\left[\frac{\mathrm{1}}{\mathrm{2}}I•d\left(\mathrm{i}A\right),\mathrm{i}B\right]}_{*}+{\left[\frac{\mathrm{1}}{\mathrm{2}}I•\mathrm{i}A,d\left(\mathrm{i}B\right)\right]}_{*}=-d(A)B-Bd{(A)}^{*}-Ad(B)-d(B)A^{*}.$

Using the two equations above, we obtain $d(AB)=d(A)B+Ad(B)$, and Claim 6 is proved.

It follows from Claims 4-6 that $d$ is an additive $\mathrm{*}$-derivation on $\mathcal{A}$, which completes the proof of Theorem 1.

2 Nonlinear Mixed Jordan Triple $\mathit{*}$-Higher Derivation

Theorem 2   Let $\mathcal{A}$ be a $*$-algebra with the unit $I$, and let $\mathbb{N}$ be the set of non-negative integers. Assume that $\mathcal{A}$ contains a nontrivial projection $P$ which satisfies for $A\in \mathcal{A}$, $XAP=\mathrm{0}\mathrm{i}\mathrm{m}\mathrm{p}\mathrm{l}\mathrm{i}\mathrm{e}\mathrm{s}X=\mathrm{0}$, and $XA\left(I-P\right)=\mathrm{0}\mathrm{i}\mathrm{m}\mathrm{p}\mathrm{l}\mathrm{i}\mathrm{e}\mathrm{s}X=\mathrm{0}.$

Then every nonlinear mixed Jordan triple $\mathit{*}$-higher derivation $\mathcal{D}=\{d_n{\}}_{n\in \mathbf{\mathbb{N}}}$ on $\mathcal{A}$ which satisfies

$d_n({[A•B,C]}_{*})=\sum _{i+j+k=n}[d_i(A)•d_j(B),d_k{(C)]}_{*}$

for all $A,B,C\in \mathcal{A}$ and $n\in \mathbb{N}$, is an additive $\mathrm{*}$-higher derivation.

Proof   Let $P_{\mathrm{1}}=P$ and $P_{\mathrm{2}}=I-P$. We write ${\mathcal{A}}_{jk}=P_j\mathcal{A}{P}_k$ for $j,k=\mathrm{1,2}$. Then $\mathcal{A}={\displaystyle \sum _{j,k=\mathrm{1}}^{\mathrm{2}}}{\mathcal{A}}_{jk}.$

In the sequel ${\mathcal{A}}_{jk}$ indicates that $A_{jk}\in {\mathcal{A}}_{jk}$. We set $\mathcal{S}=\{A\in \mathcal{A},A=A^{*}\}$. We complete the proof by several Claims.

Claim 7 $d_n\left(\mathrm{0}\right)=\mathrm{0}$ for all $n\in \mathbb{N}$.

Proof   It follows from Claim 1 that $d_{\mathrm{1}}(\mathrm{0})=\mathrm{0}$. Now assume that $d_k\left(\mathrm{0}\right)=\mathrm{0}$ for $k<n$. Then we compute

$d_n(\mathrm{0})=d_n\left({\left[\mathrm{0}•\mathrm{0,0}\right]}_{*}\right)=\sum _{m+l+j=n}{\left[d_m\left(\mathrm{0}\right)•d_l\left(\mathrm{0}\right),d_j\left(\mathrm{0}\right)\right]}_{\mathrm{*}}=\sum _{\begin{array}{c}m+l+j=n\\ j<n\end{array}}{\left[d_m\left(\mathrm{0}\right)•d_l\left(\mathrm{0}\right),d_j\left(\mathrm{0}\right)\right]}_{*}+{\left[d_{\mathrm{0}}\left(\mathrm{0}\right)•d_{\mathrm{0}}\left(\mathrm{0}\right),d_n(\mathrm{0})\right]}_{*}=\mathrm{0}.$

Claim 8 For any $n\in \mathbb{N}$, $d_n$ has the following properties: (a) For any $\lambda \in \mathbb{R}$, $d_n\left(\lambda I\right)\in \mathcal{Z}(\mathcal{A})$ and $d_n\left(\lambda I\right)=d_n{\left(\lambda I\right)}^{*}$; (b) For any $A\in \mathcal{S}$, $d_n\left(A\right)=d_n\left(A^{*}\right)=d_n{\left(A\right)}^{*}$; (c) For any $\lambda \in \mathbb{C}$, $d_n\left(\lambda I\right)\in \mathcal{Z}(\mathcal{A})$.

Proof   It follows from Claim 2 and Claim 6 that $d_{\mathrm{1}}(\lambda I)=\mathrm{0}$ for $\lambda \in \mathbb{C}$ and $d_{\mathrm{1}}\left(A^{*}\right)=d_{\mathrm{1}}{\left(A\right)}^{*}$ for $A\in \mathcal{A}$.

(a) For any λ ∈$\mathbb{R}$, we assume that $d_k\left(\lambda I\right)=d_k{\left(\lambda I\right)}^{*}\in \mathcal{Z}(\mathcal{A})$ for k<n. Then

$\mathrm{0}=d_n\left({\left[I•\lambda I,I\right]}_{*}\right)=\sum _{m+l+j=n}[d_m(I)•d_l(\lambda I),d_j{(I)]}_{*}$

$={\left[d_n\left(I\right)•\lambda I,I\right]}_{*}+{\left[I•d_n\left(\lambda I\right),I\right]}_{*}+{\left[I•\lambda I,d_n\left(I\right)\right]}_{*}+{\displaystyle \sum _{\begin{array}{c}m+l+j=n\\ \mathrm{0}\le m,l,j\le n-\mathrm{1}\end{array}}^{}}{\left[d_m\left(I\right)•d_l\left(\lambda I\right),d_j\left(I\right)\right]}_{*}$

$={\left[d_n\left(I\right)•\lambda I,I\right]}_{*}+{\left[I•d_n\left(\lambda I\right),I\right]}_{*}+{\left[I•\lambda I,d_n\left(I\right)\right]}_{*}=\mathrm{2}{d}_n\left(\lambda I\right)-\mathrm{2}{d}_n{\left(\lambda I\right)}^{*}.$

From this we get $d_n\left(\lambda I\right)=d_n{\left(\lambda I\right)}^{*}$. Then for all $A\in \mathcal{A}$,

$\mathrm{0}=d_n\left({\left[\lambda I•\lambda I,A\right]}_{*}\right)=\sum _{m+l+j=n}{\left[d_m\left(\lambda I\right)•d_l\left(\lambda I\right),d_j\left(A\right)\right]}_{*}$

$={\left[d_n\left(\lambda I\right)•\lambda I,A\right]}_{*}+{\left[\lambda I•d_n\left(\lambda I\right),A\right]}_{*}+{\left[\lambda I•\lambda I,d_n\left(A\right)\right]}_{*}+\sum _{\begin{array}{c}m+l+j=n\\ \mathrm{0}\le m,l,j\le n-\mathrm{1}\end{array}}{\left[d_m\left(\lambda I\right)•d_{\mathrm{l}}\left(\lambda I\right),d_j\left(A\right)\right]}_{\mathrm{*}}$

$={\left[d_n\left(\lambda I\right)•\lambda I,A\right]}_{*}+{\left[\lambda I•d_n\left(\lambda I\right),A\right]}_{*}+{\left[\lambda I•\lambda I,d_n\left(A\right)\right]}_{*}=\mathrm{4}\lambda d_n\left(\lambda I\right)A-\mathrm{4}\lambda A{d}_n{\left(\lambda I\right)}^{*}.$

So we get $d_n\left(\lambda I\right)A=A{d}_n\left(\lambda I\right)$ for all $A\in \mathcal{A}$, which implies $d_n\left(\lambda I\right)\in \mathcal{Z}(\mathcal{A})$.

(b) For any $A\in \mathcal{S}$, we assume that $d_k\left(A\right)=d_k\left(A^{*}\right)=d_k{\left(A\right)}^{*}$ for $k<n$. Then we can get from Claim 8(a) that

$\mathrm{0}=d_n\left({\left[I•A,I\right]}_{*}\right)=\sum _{m+l+j=n}{\left[d_m\left(I\right)•d_l\left(A\right),d_j\left(I\right)\right]}_{*}$

$={\left[d_n\left(I\right)•A,I\right]}_{*}+{\left[I•d_n\left(A\right),I\right]}_{*}+{\left[I•A,d_n\left(I\right)\right]}_{*}+\sum _{\begin{array}{c}m+l+j=n\\ \mathrm{0}\le m,l,j\le n-\mathrm{1}\end{array}}{\left[d_m\left(I\right)•d_l\left(A\right),d_j\left(I\right)\right]}_{\mathrm{*}}$

$={\left[d_n\left(I\right)•A,I\right]}_{*}+{\left[I•d_n\left(A\right),I\right]}_{*}+{\left[I•A,d_n\left(I\right)\right]}_{*}=\mathrm{2}{d}_n\left(A\right)-\mathrm{2}{d}_n{\left(A\right)}^{*}.$

(c) For any $\lambda \in \mathbb{C}$ and $A\in \mathcal{S}$, assume that $d_k\left(\lambda I\right)\in \mathcal{Z}(\mathcal{A})$ for $k<n$. Then we see from Claim 8(a) and 8(b) that

$\mathrm{0}=d_n\left({\left[A•I,\lambda I\right]}_{*}\right)=\sum _{m+l+j=n}{\left[d_m\left(A\right)•d_l\left(I\right),d_j\left(\lambda I\right)\right]}_{*}$

$={\left[d_n\left(A\right)•I,\lambda I\right]}_{*}+{\left[A•d_n\left(I\right),\lambda I\right]}_{*}+{\left[A•I,d_n\left(\lambda I\right)\right]}_{*}+{\displaystyle \sum _{\begin{array}{c}m+l+j=n\\ \mathrm{0}\le m,l,j\le n-\mathrm{1}\end{array}}^{}}{\left[d_m\left(A\right)•d_l\left(I\right),d_j\left(\lambda I\right)\right]}_{\mathrm{*}}$

$={\left[d_n\left(A\right)•I,\lambda I\right]}_{*}+{\left[A•d_n\left(I\right),\lambda I\right]}_{*}+{\left[A•I,d_n\left(\lambda I\right)\right]}_{*}=\mathrm{2}A{d}_n\left(\lambda I\right)-\mathrm{2}{d}_n\left(\lambda I\right)A.$