On Higher {g_n, h_n}-derivations

TL;DR

This paper introduces and characterizes higher {g_n, h_n}-derivations and Jordan higher {g_n, h_n}-derivations, proving that on semiprime algebras, Jordan higher derivations are actual higher derivations.

Contribution

It defines new classes of higher derivations and provides a characterization linking them to existing {g, h}-derivations, with a key result on semiprime algebras.

Findings

Jordan higher {g_n, h_n}-derivations are actual higher {g_n, h_n}-derivations on semiprime algebras.

Characterization of higher {g_n, h_n}-derivations via {g, h}-derivations.

Introduction of higher {g_n, h_n}-derivation concepts.

Abstract

In this article, we introduce the concepts of higher {g_n, h_n}-derivation and Jordan higher {g_n, h_n}-derivation, and then we give a characterization of higher {g_n, h_n}-derivations in terms of {g, h}-derivations. Using this result, we prove that every Jordan higher {g_n, h_n}-derivation on a semiprime algebra is a higher {g_n, h_n}-derivation.

Equations

$$f_n(ab)=\sum _{k=0}^n{g}_{n-k}(a)h_k(b)=\sum _{k=0}^n{h}_{n-k}(a)g_k(b),$$

$$f_n(a\circ b)=\sum _{k=0}^n{g}_{n-k}(a)\circ h_k(b),$$

$$\left\{\begin{array}[]{c l}f_{n}=\sum_{i=1}^{n}\Bigg{(}\sum_{\sum_{j=1}^{i}r_{j}=n}\Big{(}\prod_{j=1}^{i}\frac{1}{r_{j}+...+r_{i}}\Big{)}F_{r_{1}}...F_{r_{i}}\Bigg{)},&\\ \\ g_{n}=\sum_{i=1}^{n}\Bigg{(}\sum_{\sum_{j=1}^{i}r_{j}=n}\Big{(}\prod_{j=1}^{i}\frac{1}{r_{j}+...+r_{i}}\Big{)}G_{r_{1}}...G_{r_{i}}\Bigg{)},&\\ \\ h_{n}=\sum_{i=1}^{n}\Bigg{(}\sum_{\sum_{j=1}^{i}r_{j}=n}\Big{(}\prod_{j=1}^{i}\frac{1}{r_{j}+...+r_{i}}\Big{)}H_{r_{1}}...H_{r_{i}}\Bigg{)},\end{array}\right.$$

$$\left\{\begin{array}[]{c l}(n+1)f_{n+1}=\sum_{k=0}^{n}F_{k+1}f_{n-k},&\\ \\ (n+1)g_{n+1}=\sum_{k=0}^{n}G_{k+1}g_{n-k},&\\ \\ (n+1)h_{n+1}=\sum_{k=0}^{n}H_{k+1}h_{n-k}\end{array}\right.$$

$$\left\{\begin{array}[]{c l}(r+1)f_{r+1}=\sum_{k=0}^{r}F_{k+1}f_{r-k},&\\ \\ (r+1)g_{r+1}=\sum_{k=0}^{r}G_{k+1}g_{r-k},&\\ \\ (r+1)h_{r+1}=\sum_{k=0}^{r}H_{k+1}h_{r-k}\end{array}\right.$$

$$F_{n+1}(ab)$$

$$\displaystyle=(n+1)\sum_{k=0}^{n+1}g_{k}(a)h_{n+1-k}(b)-\sum_{k=0}^{n-1}F_{k+1}\Big{(}\sum_{l=0}^{n-k}g_{l}(a)h_{n-k-l}(b)\Big{)}.$$

$$F_{n+1}(ab)$$

$$\displaystyle=\sum_{k=0}^{n+1}(k+n+1-k)g_{k}(a)h_{n+1-k}(b)-\sum_{k=0}^{n-1}F_{k+1}\Big{(}\sum_{l=0}^{n-k}g_{l}(a)h_{n-k-l}(b)\Big{)}.$$

$$F_{n+1}(ab)$$

$$\displaystyle-\sum_{k=0}^{n-1}\sum_{l=0}^{n-k}\Big{[}G_{k+1}\big{(}g_{l}(a)\big{)}h_{n-k-l}(b)+g_{l}(a)H_{k+1}\big{(}h_{n-k-l}(b)\big{)}\Big{]}.$$

$$\displaystyle G=\sum_{k=0}^{n+1}kg_{k}(a)h_{n+1-k}(b)-\sum_{k=0}^{n-1}\sum_{l=0}^{n-k}G_{k+1}\big{(}g_{l}(a)\big{)}h_{n-k-l}(b),$$

$$\displaystyle H=\sum_{k=0}^{n+1}g_{k}(a)(n+1-k)h_{n+1-k}(b)-\sum_{k=0}^{n-1}\sum_{l=0}^{n-k}g_{l}(a)H_{k+l}\big{(}h_{n-k-l}(b)\big{)},$$

$$G$$

$$\displaystyle=\sum_{k=0}^{n+1}kg_{k}(a)h_{n+1-k}(b)-\sum_{r=0}^{n-1}\sum_{k=0}^{r}G_{k+1}\big{(}g_{r-k}(a)\big{)}h_{n-r}(b)-\sum_{k=0}^{n-1}G_{k+1}\big{(}g_{n-k}(a)\big{)}b.$$

$$G+\sum_{k=0}^{n-1}G_{k+1}\big{(}g_{n-k}(a)\big{)}b=\sum_{k=0}^{n+1}kg_{k}(a)h_{n+1-k}(b)-\sum_{r=0}^{n-1}\sum_{k=0}^{r}G_{k+1}\big{(}g_{r-k}(a)\big{)}h_{n-r}(b).$$

$$G$$

$$\displaystyle=\sum_{r=0}^{n}(r+1)g_{r+1}(a)h_{n-r}(b)-\sum_{r=0}^{n-1}\sum_{k=0}^{r}G_{k+1}\big{(}g_{r-k}(a)\big{)}h_{n-r}(b)$$

$$\displaystyle=\sum_{r=0}^{n-1}\Big{[}(r+1)g_{r+1}(a)-\sum_{k=0}^{r}G_{k+1}\big{(}g_{r-k}(a)\big{)}\Big{]}h_{n-r}(b)+(n+1)g_{n+1}(a)b.$$

$$\displaystyle G=\Big{[}(n+1)g_{n+1}(a)-\sum_{k=0}^{n-1}G_{k+1}\big{(}g_{n-k}(a)\big{)}\Big{]}b=G_{n+1}(a)b.$$

$$\displaystyle H=a\Big{[}(n+1)h_{n+1}(b)-\sum_{k=0}^{n-1}H_{k+1}\big{(}h_{n-k}(b)\big{)}\Big{]}=aH_{n+1}(b).$$

$$F_{n+1}(ab)$$

$$\displaystyle=(n+1)\sum_{k=0}^{n+1}h_{k}(a)g_{n+1-k}(b)-\sum_{k=0}^{n-1}F_{k+1}\Big{(}\sum_{l=0}^{n-k}h_{l}(a)g_{n-k-l}(b)\Big{)}.$$

$$F_{n+1}(ab)$$

$$\displaystyle=\sum_{k=0}^{n+1}(k+n+1-k)h_{k}(a)g_{n+1-k}(b)-\sum_{k=0}^{n-1}F_{k+1}\Big{(}\sum_{l=0}^{n-k}h_{l}(a)g_{n-k-l}(b)\Big{)}.$$

$$F_{n+1}(ab)$$

$$\displaystyle-\sum_{k=0}^{n-1}\sum_{l=0}^{n-k}\Big{[}H_{k+1}\big{(}h_{l}(a)\big{)}g_{n-k-l}(b)+h_{l}(a)G_{k+1}\big{(}g_{n-k-l}(b)\big{)}\Big{]}.$$

$$\displaystyle\mathfrak{G}=\sum_{k=0}^{n+1}kh_{k}(a)g_{n+1-k}(b)-\sum_{k=0}^{n-1}\sum_{l=0}^{n-k}H_{k+1}\big{(}h_{l}(a)\big{)}g_{n-k-l}(b),$$

$$\displaystyle\mathfrak{H}=\sum_{k=0}^{n+1}h_{k}(a)(n+1-k)g_{n+1-k}(b)-\sum_{k=0}^{n-1}\sum_{l=0}^{n-k}h_{l}(a)G_{k+l}\big{(}g_{n-k-l}(b)\big{)},$$

$$\mathfrak{G}$$

$$\displaystyle=\sum_{k=0}^{n+1}kh_{k}(a)g_{n+1-k}(b)-\sum_{r=0}^{n-1}\sum_{k=0}^{r}H_{k+1}\big{(}h_{r-k}(a)\big{)}g_{n-r}(b)-\sum_{k=0}^{n-1}H_{k+1}\big{(}h_{n-k}(a)\big{)}b.$$

$$\mathfrak{G}+\sum_{k=0}^{n-1}H_{k+1}\big{(}h_{n-k}(a)\big{)}b=\sum_{k=0}^{n+1}kh_{k}(a)g_{n+1-k}(b)-\sum_{r=0}^{n-1}\sum_{k=0}^{r}H_{k+1}\big{(}h_{r-k}(a)\big{)}g_{n-r}(b).$$

$$\mathfrak{G}$$

$$\displaystyle=\sum_{r=0}^{n}(r+1)h_{r+1}(a)g_{n-r}(b)-\sum_{r=0}^{n-1}\sum_{k=0}^{r}H_{k+1}\big{(}h_{r-k}(a)\big{)}g_{n-r}(b)$$

$$\displaystyle=\sum_{r=0}^{n-1}\Big{[}(r+1)h_{r+1}(a)-\sum_{k=0}^{r}H_{k+1}\big{(}h_{r-k}(a)\big{)}\Big{]}g_{n-r}(b)+(n+1)h_{n+1}(a)b.$$

$$\displaystyle\mathfrak{G}=\Big{[}(n+1)h_{n+1}(a)-\sum_{k=0}^{n-1}H_{k+1}\big{(}h_{n-k}(a)\big{)}\Big{]}b=H_{n+1}(a)b.$$

$$\displaystyle\mathfrak{H}=a\Big{[}(n+1)g_{n+1}(b)-\sum_{k=0}^{n-1}G_{k+1}\big{(}g_{n-k}(b)\big{)}\Big{]}=aG_{n+1}(b).$$

Full text

1. Introduction and preliminaries

Recently, in 2016 Bre$\check{s}$ar [1] introduced the notion of $\{g,h\}$-derivation. Let $\mathcal{A}$ be an algebra over a field $\mathbb{F}$ with $char(\mathbb{F})\neq 2$, and let $f,g,h:\mathcal{A}\rightarrow\mathcal{A}$ be linear maps. We say that $f$ is a $\{g,h\}$-derivation if $f(ab)=g(a)b+ah(b)=h(a)b+ag(b)$ for all $a,b\in\mathcal{A}$, and $f$ is called a Jordan $\{g,h\}$-derivation if $f(a\circ b)=g(a)\circ b+a\circ h(b)$ for all $a,b\in\mathcal{A}$, where $a\circ b=ab+ba$. We call $a\circ b$ the Jordan product of $a$ and $b$. It is evident that $a\circ b=b\circ a$ for all $a,b\in\mathcal{A}$. The notion of a Jordan $\{g,h\}$-derivation is a generalization of what is called a Jordan generalized derivation in [8]. Recall that a linear mapping $f:\mathcal{A}\rightarrow\mathcal{A}$ is called a Jordan generalized derivation if there exists a linear mapping $d:\mathcal{A}\rightarrow\mathcal{A}$ such that $f(a\circ b)=f(a)\circ b+a\circ d(b)$ for all $a,b\in\mathcal{A},$ where $d$ is called an associated linear map of $f$. It is clear that $f(a\circ b)=d(a)\circ b+a\circ f(b)$ for all $a,b\in\mathcal{A}.$ Obviously, the definition of a Jordan generalized derivation is generally not equivalent to the ordinary Jordan case of generalized derivations. For more details in this regard, see [1, 8] and references therein.

As an important result, Bre$\check{s}$ar [1, Theorem 4.3] established that every Jordan $\{g,h\}$-derivation of a semiprime algebra $\mathcal{A}$ is a $\{g,h\}$-derivation. He also showed that every Jordan $\{g,h\}$-derivation of the tensor product of a semiprime and a commutative algebra is a $\{g,h\}$-derivation. Obviously, every $\{g,h\}$-derivation is a Jordan $\{g,h\}$-derivation, but the converse is in general not true. For instance in this regard see [1, Example 2.1].

In this study, we introduce the concept of a higher $\{g_{n},h_{n}\}$-derivation, and then we characterize it on algebras. Throughout this paper, $\mathcal{A}$ denotes an algebra over a field of characteristic zero, and $I$ denotes the identity mapping on $\mathcal{A}$. Let $f$ be a $\{g,h\}$-derivation on an algebra $\mathcal{A}$. An easy induction argument implies that f^{n}(ab)=\sum_{k=0}^{n}\Big{(}_{k}^{n}\Big{)}g^{n-k}(a)h^{k}(b)=\sum_{k=0}^{n}\Big{(}_{k}^{n}\Big{)}h^{n-k}(a)g^{k}(b) (Leibniz rule) for each $a,b\in\mathcal{A}$ and each non-negative integer $n$, where $f^{0}=g^{0}=h^{0}=I$. If we define the sequences $\{f_{n}\}$, $\{g_{n}\}$ and $\{h_{n}\}$ of linear mappings on $\mathcal{A}$ by $f_{0}=g_{0}=h_{0}=I$, and $f_{n}=\frac{f^{n}}{n!}$, $g_{n}=\frac{g^{n}}{n!}$ and $h_{n}=\frac{h^{n}}{n!}$, then it follows from the Leibniz rule that $f_{n}$’s, $g_{n}$’s and $h_{n}$’s satisfy

[TABLE]

for each $a,b\in\mathcal{A}$ and each non-negative integer $n$. This is our motivation to consider the sequences $\{f_{n}\}$, $\{g_{n}\}$ and $\{h_{n}\}$ of linear mappings on an algebra $\mathcal{A}$ satisfying (1.1). A sequence $\{f_{n}\}$ of linear mappings on $\mathcal{A}$ is called a higher $\{g_{n},h_{n}\}$-derivation if there exist two sequences $\{g_{n}\}$ and $\{h_{n}\}$ of linear mappings on $\mathcal{A}$ satisfying $f_{n}(ab)=\sum_{k=0}^{n}g_{n-k}(a)h_{k}(b)=\sum_{k=0}^{n}h_{n-k}(a)g_{k}(b)$ for any $a,b\in\mathcal{A}$ and any non-negative integer $n$. Additionally, a sequence $\{f_{n}\}$ of linear mappings on $\mathcal{A}$ is called a Jordan higher $\{g_{n},h_{n}\}$-derivation if

[TABLE]

for each $a,b\in\mathcal{A}$ and each non-negative integer $n$. Notice that if $\{f_{n}\}$ is a higher $\{f_{n},f_{n}\}$-derivation (resp. Jordan higher $\{f_{n},f_{n}\}$-derivation), then it is an ordinary higher derivation (resp. Jordan higher derivation). We know that if $f$ is a $\{g,h\}$-derivation, then $\{f_{n}=\frac{f^{n}}{n!}\}$ is a higher $\{g_{n},h_{n}\}$-derivation, where $g_{n}=\frac{g^{n}}{n!}$, $h_{n}=\frac{h^{n}}{n!}$ and $f_{0}=g_{0}=h_{0}=I$. We call this kind of higher $\{g_{n},h_{n}\}$-derivation an ordinary higher $\{g_{n},h_{n}\}$-derivation, but this is not the only example of a higher $\{g_{n},h_{n}\}$-derivation.

In 2010, Miravaziri [9] characterized all higher derivations on an algebra $\mathcal{A}$ in terms of the derivations on $\mathcal{A}$. In this article, by getting idea from [9], our aim is to characterize all higher $\{g_{n},h_{n}\}$-derivations on an algebra $\mathcal{A}$ in terms of the $\{g,h\}$-derivations on $\mathcal{A}$. Indeed, we show that each higher $\{g_{n},h_{n}\}$-derivation is a combination of compositions of $\{g,h\}$-derivations. As the main result of this article, we prove that if $\{f_{n}\}$ is a higher $\{g_{n},h_{n}\}$-derivation on an algebra $\mathcal{A}$ with $f_{0}=g_{0}=h_{0}=I$, then there is a sequence $\{F_{n}\}$ of $\{G_{n},H_{n}\}$-derivations on $\mathcal{A}$ such that

[TABLE]

where the inner summation is taken over all positive integers $r_{j}$with $\sum_{j=1}^{i}r_{j}=n$. The importance of this result is to transfer the problems such as the characterization of Jordan higher $\{g_{n},h_{n}\}$-derivations on semiprime algebras and automatic continuity of higher $\{g_{n},h_{n}\}$-derivations into the same problems concerning $\{g,h\}$-derivations. Let $\mathcal{A}$ be an algebra, $D$ be the set of all higher derivations $\{d_{n}\}_{n=0,1,...}$ on $\mathcal{A}$ with $d_{0}=I$ and $\Delta$ be the set of all sequences $\{\delta_{n}\}_{n=0,1,...}$ of derivations on $\mathcal{A}$ with $\delta_{0}=0$. As an application of the main result of this article, we investigate Jordan higher $\{g_{n},h_{n}\}$-derivations on algebras. It is a classical question in which algebras (and rings) a Jordan derivation is necessarily a derivation. Let us give a brief background in this issue. In 1957, Herstein [7] achieved a result which asserts any Jordan derivation on a prime ring of characteristic different from two is a derivation. A brief proof of Herstein’s result can be found in [3]. In 1975, Cusack [5] generalized Herstein’s result to 2-torsion free semiprime rings (see also [2] for an alternative proof). Moreover, Vukman [12] investigated generalized Jordan derivations on semiprime rings and he proved that every generalized Jordan derivation of a 2-torsion free semiprime ring is a generalized derivation. Recently, the first name author along with Ajda Fo$\check{s}$ner [6] have studied the same problem for $(\sigma,\tau$)-derivations from a $C^{\ast}$-algebra $\mathcal{A}$ into a Banach $\mathcal{A}$-module $\mathcal{M}$. In this paper, we show that if $\{f_{n}\}$ is a Jordan higher $\{g_{n},h_{n}\}$-derivation of a semiprime algebra $\mathcal{A}$ with $f_{0}=g_{0}=h_{0}=I$, then it is a higher $\{g_{n},h_{n}\}$-derivation.

2. characterization of higher $\{g_{n},h_{n}\}$-derivations on algebras

Throughout the article, $\mathcal{A}$ denotes an algebra over a field of characteristic zero, and $I$ is the identity mapping on $\mathcal{A}$. Let $f,g,h:\mathcal{A}\rightarrow\mathcal{A}$ be linear maps. We say that $f$ is a $\{g,h\}$-derivation if $f(ab)=g(a)b+ah(b)=h(a)b+ag(b)$ for all $a,b\in\mathcal{A}$, and $f$ is called a Jordan $\{g,h\}$-derivation if $f(a\circ b)=g(a)\circ b+a\circ h(b)$ for all $a,b\in\mathcal{A}$, where $a\circ b=ab+ba$.

We begin with the following definition.

Definition 2.1**.**

A sequence $\{f_{n}\}$ of linear mappings on $\mathcal{A}$ is called a higher $\{g_{n},h_{n}\}$-derivation if there are two sequences $\{g_{n}\}$ and $\{h_{n}\}$ of linear mappings on $\mathcal{A}$ such that $f_{n}(ab)=\sum_{k=0}^{n}g_{n-k}(a)h_{k}(b)=\sum_{k=0}^{n}h_{n-k}(a)g_{k}(b)$ for each $a,b\in\mathcal{A}$ and each non-negative integer $n$.

We begin our results with the following lemma which will be used extensively to prove the main theorem of this article. The following lemma has been motivated by [9].

Lemma 2.2**.**

Let $\{f_{n}\}$ be a higher $\{g_{n},h_{n}\}$-derivation on an algebra $\mathcal{A}$ with $f_{0}=g_{0}=h_{0}=I$. Then there is a sequence $\{F_{n}\}$ of $\{G_{n},H_{n}\}$-derivations on $\mathcal{A}$ such that

[TABLE]

for each non-negative integer $n$.

Proof.

Using induction on $n$, we prove this lemma. Let $n=0$. We know that $f_{1}(ab)=g_{1}(a)b+ah_{1}(b)=h_{1}(a)b+ag_{1}(b)$ for all $a,b\in\mathcal{A}$. Thus, if $F_{1}=f_{1}$, $G_{1}=g_{1}$ and $H_{1}=h_{1}$, then $F_{1}$ is a $\{G_{1},H_{1}\}$-derivation on $\mathcal{A}$ and further, $(0+1)f_{0+1}=\sum_{k=0}^{0}F_{k+1}f_{0-k}$, $(0+1)g_{0+1}=\sum_{k=0}^{0}G_{k+1}g_{0-k}$ and $(0+1)h_{0+1}=\sum_{k=0}^{0}H_{k+1}h_{0-k}$. As induction assumption, suppose that $F_{k}$ is a $\{G_{k},H_{k}\}$-derivation for any $k\leq n$ and further

[TABLE]

for $r=0,1,...,n-1$. Put $F_{n+1}=(n+1)f_{n+1}-\sum_{k=0}^{n-1}F_{k+1}f_{n-k}$, $G_{n+1}=(n+1)g_{n+1}-\sum_{k=0}^{n-1}G_{k+1}g_{n-k}$ and $H_{n+1}=(n+1)h_{n+1}-\sum_{k=0}^{n-1}H_{k+1}h_{n-k}$. Our next task is to show that $F_{n+1}$ is a $\{G_{n+1},H_{n+1}\}$-derivation on $\mathcal{A}$. For $a,b\in\mathcal{A}$, we have

[TABLE]

So, we have

[TABLE]

Since $F_{k}$ is a $\{G_{k},H_{k}\}$-derivation for each $k=1,2,...,n$,

[TABLE]

Letting

[TABLE]

we have $F_{n+1}(ab)=G+H$. Here, we compute $G$ and $H$. In the summation $\sum_{k=0}^{n-1}\sum_{l=0}^{n-k}$, we have $0\leq k+l\leq n$ and $k\neq n$. Thus if we put $r=k+l$, then we can write it as the form $\sum_{r=0}^{n}\sum_{k+l=r,k\neq n}$. Putting $l=r-k$, we find that

[TABLE]

It means that

[TABLE]

Putting $r+1$ instead of $k$ in the first summation of above, we have

[TABLE]

According to the induction hypothesis, (r+1)g_{r+1}(a)=\sum_{k=0}^{r}G_{k+1}\big{(}g_{r-k}(a)\big{)} for $r=0,...,n-1$. So, it is obtained that

[TABLE]

Like above, we achieve that

[TABLE]

Therefore, we have $F_{n+1}(ab)=G+H=G_{n+1}(a)b+aH_{n+1}(b)$. In the next step, we will show that $F_{n+1}(ab)=H_{n+1}(a)b+aG_{n+1}(b)$. We have

[TABLE]

So,

[TABLE]

Since $F_{k}$ is a $\{G_{k},H_{k}\}$-derivation for each $k=1,2,...,n$,

[TABLE]

Letting

[TABLE]

we have $F_{n+1}(ab)=\mathfrak{G}+\mathfrak{H}$. The next step is to compute $\mathfrak{G}$ and $\mathfrak{H}$. In the summation $\sum_{k=0}^{n-1}\sum_{l=0}^{n-k}$, we have $0\leq k+l\leq n$ and $k\neq n$. Thus if we put $r=k+l$, then we can write it as the form $\sum_{r=0}^{n}\sum_{k+l=r,k\neq n}$. Putting $l=r-k$, we find that

[TABLE]

It means that

[TABLE]

Putting $r+1$ instead of $k$ in the first summation of above, we have

[TABLE]

According to the induction hypothesis, (r+1)h_{r+1}(a)=\sum_{k=0}^{r}H_{k+1}\big{(}h_{r-k}(a)\big{)} for $r=0,...,n-1$. So, it is obtained that

[TABLE]

By a reasoning like above, we get that

[TABLE]

Therefore, we have $F_{n+1}(ab)=\mathfrak{G}+\mathfrak{H}=H_{n+1}(a)b+aG_{n+1}(b)$. Consequently, $F_{n+1}$ is a $\{G_{n+1},H_{n+1}\}$-derivation and the proof is complete. ∎

Example 2.3**.**

Using Lemma 2.2, the first five terms of a higher $\{g_{n},h_{n}\}$-derivation $\{f_{n}\}$ are

[TABLE]

Now the main result of this paper reads as follows:

Theorem 2.4**.**

Let $\{f_{n}\}$ be a higher $\{g_{n},h_{n}\}$-derivation on an algebra $\mathcal{A}$ with $f_{0}=g_{0}=h_{0}=I$. Then there is a sequence $\{F_{n}\}$ of $\{G_{n},H_{n}\}$-derivations on $\mathcal{A}$ such that

[TABLE]

where the inner summation is taken over all positive integers $r_{j}$with $\sum_{j=1}^{i}r_{j}=n$.

Proof.

First, we show that if $f_{n}$, $g_{n}$ and $h_{n}$ are of the above form, then they satisfy the recursive relations of Lemma 2.2. Simplifying the notation, we put $a_{r_{1},...,r_{i}}=\prod_{j=1}^{i}\frac{1}{r_{j}+...+r_{i}}$. Note that if $r_{1}+...+r_{i}=n+1$, then $(n+1)a_{r_{1},...,r_{i}}=a_{r_{2},...,r_{i}}$. Furthermore, $a_{n+1}=\frac{1}{n+1}$. According to the aforementioned assumptions, we have

[TABLE]

So,

[TABLE]

Using a reasoning like above, we get that

[TABLE]

for each non-negative integer $n$. Putting $n+1=m$, we find that $mf_{m}=\sum_{k=0}^{m-1}F_{k+1}f_{m-1-k}=\sum_{k=0}^{m-2}F_{k+1}f_{m-1-k}+F_{m}$, and consequently

[TABLE]

Similarly, we have

[TABLE]

Therefore, we can define $F_{n},G_{n},H_{n}:\mathcal{A}\rightarrow\mathcal{A}$ by $F_{0}=G_{0}=H_{0}=0$ and

[TABLE]

for each positive integer $n$. It follows from Lemma 2.2 that $\{F_{n}\}$ is a sequence of $\{G_{n},H_{n}\}$-derivations. In addition, we prove that if $f_{n}$, $g_{n}$ and $h_{n}$ are of the form

[TABLE]

where $\{F_{n}\}$ is a sequence of $\{G_{n},H_{n}\}$-derivations, then $\{f_{n}\}$ is a higher $\{g_{n},h_{n}\}$-derivation on $\mathcal{A}$ with $f_{0}=g_{0}=h_{0}=I$. To see this, we use induction on $n$. For $n=0$, we have $f_{0}(ab)=ab=g_{0}(a)h_{0}(b)=h_{0}(a)g_{0}(b)$. As the inductive hypothesis, assume that

[TABLE]

Therefore, we have

[TABLE]

According to the above-mentioned recursive relations, we continue the previous expressions as follows:

[TABLE]

which means that $f_{n+1}(ab)=\sum_{i=0}^{n+1}g_{i}(a)h_{n+1-i}(b)$. By an argument like above, we can obtain that $f_{n+1}(ab)=\sum_{i=0}^{n+1}h_{i}(a)g_{n+1-i}(b)$. Thus, $\{f_{n}\}$ is a higher $\{g_{n},h_{n}\}$-derivation on $\mathcal{A}$ which is characterized by the sequence $\{F_{n}\}$ of $\{G_{n},H_{n}\}$-derivations. This completes the proof. ∎

In the next example, using the above theorem, we characterize term $f_{4}$ of a higher $\{g_{n},h_{n}\}$-derivation $\{f_{n}\}$.

Example 2.5**.**

We compute the coefficients $a_{r_{1},...,r_{i}}$ for the case $n=4$. First, note that $4=1+3=3+1=2+2=1+1+2=1+2+1=2+1+1=1+1+1+1.$ Based on the definition of $a_{r_{1},...,r_{i}}$ we have

[TABLE]

Therefore, $f_{4}$ is characterized as follows:

[TABLE]

In the following there are some immediate consequences of the above theorem. Before it, recall that a sequence $\{f_{n}\}$ of linear mappings on $\mathcal{A}$ is called a Jordan higher $\{g_{n},h_{n}\}$-derivation if there exist two sequences $\{g_{n}\}$ and $\{h_{n}\}$ of linear mappings on $\mathcal{A}$ such that $f_{n}(a\circ b)=\sum_{k=0}^{n}g_{n-k}(a)\circ h_{k}(b)$ holds for each $a,b\in\mathcal{A}$ and each non-negative integer $n$. Since the Jordan product is commutative, we have

[TABLE]

So, it is observed that if $\{f_{n}\}$ is a Jordan higher $\{g_{n},h_{n}\}$-derivation, then

[TABLE]

for all $a,b\in\mathcal{A}$.

Corollary 2.6**.**

Let $\{f_{n}\}$ be a Jordan higher $\{g_{n},h_{n}\}$-derivation on a semiprime algebra $\mathcal{A}$ with $f_{0}=g_{0}=h_{0}=I$. Then $\{f_{n}\}$ is a higher $\{g_{n},h_{n}\}$-derivation.

Proof.

Using the proof of Theorem 2.4, we can show that if $\{f_{n}\}$ is a Jordan higher $\{g_{n},h_{n}\}$-derivation on an algebra $\mathcal{A}$ with $f_{0}=g_{0}=h_{0}=I$, then there exists a sequence $\{F_{n}\}$ of Jordan $\{G_{n},H_{n}\}$-derivations on $\mathcal{A}$ such that

[TABLE]

where the inner summation is taken over all positive integers $r_{j}$ with $\sum_{j=1}^{i}r_{j}=n$. Since $\mathcal{A}$ is a semiprime algebra, [1, Theorem 4.3] proves the corollary. ∎

Remark 2.7*.*

We know that the notion of a Jordan $\{g,h\}$-derivation is a generalization of a Jordan generalized derivation (see Introduction). A sequence $\{f_{n}\}$ of linear mappings on an algebra $\mathcal{A}$ is called a Jordan higher generalized derivation if there exists a sequence $\{d_{n}\}$ of linear mappings on $\mathcal{A}$ such that $f_{n}(a\circ b)=\sum_{k=0}^{n}f_{n-k}(a)\circ d_{k}(b)$ for all $a,b\in\mathcal{A}.$ Obviously, if $\{f_{n}\}$ is a Jordan higher generalized derivation associated with a sequence $\{d_{n}\}$ of linear mappings on $\mathcal{A}$, then it is a Jordan higher $\{f_{n},d_{n}\}$-derivation. So, Corollary 2.6 is also valid for Jordan higher generalized derivations.

Theorem 2.8**.**

If $\{f_{n}\}_{n=0,1,...}$ is a higher $\{g_{n},h_{n}\}$-derivation on $\mathcal{A}$ with $f_{0}=g_{0}=h_{0}=I$, then there is a sequence $\{F_{n}\}_{n=0,1,...}$ of $\{G_{n},H_{n}\}$-derivations with $F_{0}=G_{0}=H_{0}=0$ characterizing the higher $\{g_{n},h_{n}\}$-derivation $\{f_{n}\}_{n=0,1,...}$. As well as if $\{F_{n}\}_{n=0,1,...}$ is a sequence of $\{G_{n},H_{n}\}$-derivations with $F_{0}=G_{0}=H_{0}=0$, then there exists a higher $\{g_{n},h_{n}\}$-derivation $\{f_{n}\}$ with $f_{0}=g_{0}=h_{0}=I$ which is characterized by the sequence $\{F_{n}\}_{n=0,1,...}$.

Proof.

Let $\{f_{n}\}$ be a higher $\{g_{n},h_{n}\}$-derivation on $\mathcal{A}$ with $f_{0}=g_{0}=h_{0}=I$. We are going to obtain a sequence $\{F_{n}\}_{n=0,1,...}$ of $\{G_{n},H_{n}\}$-derivations with $F_{0}=G_{0}=H_{0}=0$ that characterizes the higher $\{g_{n},h_{n}\}$-derivation $\{f_{n}\}$. Define $F_{n},G_{n},H_{n}:\mathcal{A}\rightarrow\mathcal{A}$ by $F_{0}=G_{0}=H_{0}=0$ and

[TABLE]

for each positive integer $n$. Then it follows from Lemma 2.2 that $\{F_{n}\}$ is a sequence of $\{G_{n},H_{n}\}$-derivations characterizing the higher $\{g_{n},h_{n}\}$-derivation $\{f_{n}\}$. Now suppose that $\{F_{n}\}_{n=0,1,...}$ is a sequence of $\{G_{n},H_{n}\}$-derivations with $F_{0}=G_{0}=H_{0}=0$. We will show that there exists a higher $\{g_{n},h_{n}\}$-derivation $\{f_{n}\}$ with $f_{0}=g_{0}=h_{0}=I$ which is characterized by the sequence $\{F_{n}\}_{n=0,1,...}$. We define $f_{n},g_{n},h_{n}:\mathcal{A}\rightarrow\mathcal{A}$ by $f_{0}=g_{0}=h_{0}=I$ and

[TABLE]

By Theorem 2.4, $f_{n}$, $g_{n}$ and $h_{n}$ satisfy the following recursive relations:

[TABLE]

Based on the last part of the proof of Theorem 2.4, $\{f_{n}\}$ is a higher $\{g_{n},h_{n}\}$-derivation on $\mathcal{A}$ with $f_{0}=g_{0}=h_{0}=I$. This yields the desired result. ∎

An immediate corollary of the precede theorem reads as follows:

Corollary 2.9**.**

If $\{d_{n}\}_{n=0,1,...}$ with $d_{0}=I$ is a higher derivation on $\mathcal{A}$, then there is a sequence $\{\delta_{n}\}_{n=0,1,...}$ of derivations with $\delta_{0}=0$ characterizing the higher derivation $\{d_{n}\}_{n=0,1,...}$. As well as if $\{\delta_{n}\}_{n=0,1,...}$ is a sequence of derivations with $\delta_{0}=0$, then there exists a higher derivation $\{d_{n}\}$ with $d_{0}=I$ which is characterized by the sequence $\{\delta_{n}\}_{n=0,1,...}$.