# Jordan Higher Derivable Mappings on Rings

Authors: Mohammad Ashraf; Nazia Parveen
Journal: Algebra
 (2014)

---

## Abstract

LetR be a ring. We say that a family of maps D
=
{
d
n
}
n
∈
N is a Jordan higher derivable map (without assumption of additivity) on R if d
0
=
I
R (the identity map on R) and d
n
(
a
b
+
b
a
)
=
∑
p
+
q
=
n
‍
d
p
(
a
)
d
q
(
b
)
+
∑
p
+
q
=
n
‍
d
p
(
b
)
d
q
(
a
) hold for all a
,
b
∈
R and for each n
∈
N. In this paper, we show that every Jordan higher derivable map on a ring under certain assumptions becomes a higher derivation. As its application, we get that every Jordan higher derivable map on Banach algebra is an additive higher derivation.


---

## Body

## 1. Introduction

LetR be a ring. An additive mapping d
:
R
→
R is said to be a derivation (resp., Jordan derivation) if d
(
a
b
)
=
d
(
a
)
b
+
a
d
(
b
) (resp., d
(
a
2
)
=
d
(
a
)
a
+
a
d
(
a
)) holds for any a
,
b
∈
R. Note that, in case of 2-torsion free ring R, d
(
a
2
)
=
d
(
a
)
a
+
a
d
(
a
) is equivalent to d
(
a
b
+
b
a
)
=
d
(
a
)
b
+
a
d
(
b
)
+
d
(
b
)
a
+
b
d
(
a
) for all a
,
b
∈
R. A map d
:
R
→
R, (without assumption of additivity) is said to be derivable (resp., Jordan derivable) if d
(
a
b
)
=
d
(
a
)
b
+
a
d
(
b
) (resp., d
(
a
b
+
b
a
)
=
d
(
a
)
b
+
a
d
(
b
)
+
d
(
b
)
a
+
b
d
(
a
)), holds for all a
,
b
∈
R. Recall that a ring R is prime if a
R
b
=
{
0
} implies that either a
=
0 or b
=
0 and is semiprime if a
R
a
=
{
0
} implies a
=
0.Now suppose thatR is a ring with a nontrivial idempotent e
1. We write e
2
=
1
-
e
1. Note that R need not have identity element. Put e
i
R
e
j
=
R
i
j for any i
,
j
=
1,2. Then by Peirce decomposition of R, we have R
=
R
11
+
R
12
+
R
21
+
R
22. Throughout this paper, the notation a
i
j will denote an arbitrary element of R
i
j and any element a
∈
R can be expressed as a
=
a
11
+
a
12
+
a
21
+
a
22.Characterizing the interrelation between the multiplicative and the additive structures of a ring is an interesting topic and has received attention of many mathematicians. It is a well-known result due to Martindale III [1] that every multiplicative bijective map from a prime ring containing a nontrivial idempotent onto an arbitrary ring is necessarily additive. Long ago, Lu [2] proved that “each derivable map on a 2-torsion free unital prime ring containing a nontrivial idempotent is a derivation.”Motivated by the above result due to Lu [2], in the year 2011, Jing and Lu [3] characterized Jordan derivable maps to a larger class of rings and proved the following theorem.Theorem 1 (see [3, Theorem 1.2]).
LetR be a ring containing a nontrivial idempotent and satisfying the following conditions for i
,
j
,
k
∈
{
1,2
}.(i)
Ifa
i
j
x
j
k
=
0 for all x
j
k
∈
R
j
k, then a
i
j
=
0.
(ii)
Ifx
i
j
a
j
k
=
0 for all x
i
j
∈
R
i
j, then a
j
k
=
0.
(iii)
Ifa
i
i
x
i
i
+
x
i
i
a
i
i
=
0 for all x
i
i
∈
R
i
i, then a
i
i
=
0.
If a mappingδ
:
R
→
R satisfies
(1)
δ
(
a
b
+
b
a
)
=
δ
(
a
)
b
+
a
δ
(
b
)
+
δ
(
b
)
a
+
b
δ
(
a
)
for all a
,
b
∈
R, then δ is additive. In addition, if R is 2-torsion free, then δ is a Jordan derivation.The concept of derivation was extended to higher derivation by Hasse and Schmidt [4] (see [5, 6] for a historical account and applications). Let D
=
{
d
n
}
n
∈
N be a family of additive mappings d
n
:
R
→
R and let N be the set of nonnegative integers. Following Hasse and Schmidt [4], D is said to be a higher derivation (resp., Jordan higher derivation) on R if d
0
=
I
R (the identity map on R) and d
n
(
a
b
)
=
∑
‍
d
p
(
a
)
d
q
(
b
) (resp., d
n
(
a
2
)
=
∑
‍
d
p
(
a
)
d
q
(
a
)) for all a
,
b
∈
R and for each n
∈
N. In an attempt to generalize Herstein’s result for higher derivations, Haetinger [7] proved that, on a prime ring with c
h
a
r
(
R
)
≠
2, every Jordan higher derivation is a higher derivation (see [6, 8] for English versions). Further many results were obtained in this direction. In the present paper we introduce the notion of higher derivable (resp., Jordan higher derivable) map on R and provide a sufficient condition on a ring R under which a Jordan higher derivable map becomes a higher derivation.


## 2. Additivity of Jordan Higher Derivable Mappings

In order to illustrate the results in the literature focused on derivable maps and higher derivation we will introduce the concept of higher derivable maps (resp., Jordan higher derivable maps) on certain classes of rings.A familyD
=
{
d
n
}
n
∈
N of mappings d
n
:
R
→
R (without assumption of additivity) such that d
0
=
I
R, is said to be(i)
a higher derivable map ifd
n
(
a
b
)
=
∑
p
+
q
=
n
d
p
(
a
)
d
q
(
b
) holds for all a
,
b
∈
R and for each n
∈
N,
(ii)
a Jordan higher derivable map ifd
n
(
a
b
+
b
a
)
=
∑
p
+
q
=
n
(
d
p
(
a
)
d
q
(
b
)
+
d
p
(
b
)
d
q
(
a
)
) holds for all a
,
b
∈
R and for each n
∈
N.The main result of our paper states as follows.Theorem 2.
LetR be a ring containing a nontrivial idempotent satisfying the following conditions for i
,
j
,
k
∈
{
1,2
}.(
P
1)
Ifa
i
j
x
j
k
=
0 for all x
j
k
∈
R
j
k, then a
i
j
=
0.
(P
2)
Ifx
i
j
a
j
k
=
0 for all x
i
j
∈
R
i
j, then a
j
k
=
0.
(P
3)
Ifa
i
i
x
i
i
+
x
i
i
a
i
i
=
0 for all x
i
i
∈
R
i
i, then a
i
i
=
0.
If the familyD
=
{
d
n
}
n
∈
N of mappings d
n
:
R
→
R such that d
0
=
I
R satisfy
(2)
d
n
(
a
b
+
b
a
)
=
∑
p
+
q
=
n
‍
(
d
p
(
a
)
d
q
(
b
)
+
d
p
(
b
)
d
q
(
a
)
)
for all a
,
b
∈
R and for each n
∈
N, then D
=
{
d
n
}
n
∈
N is additive. In addition, if R is 2-torsion free, then D
=
{
d
n
}
n
∈
N is a Jordan higher derivation.Throughout the paper, we always assume thatR is a ring with a nontrivial idempotent e
1 and satisfies conditions (
P
1
), (
P
2
), and (
P
3
). We also assume that D
=
{
d
n
}
n
∈
N is the family of Jordan higher derivable maps d
n
:
R
→
R; that is,
(3)
d
n
a
b
+
b
a
=
∑
p
+
q
=
n
‍
d
p
a
d
q
b
+
∑
p
+
q
=
n
‍
d
p
b
d
q
a
hhhhhhhhhhhhhhhhhhhhihhh
∀
a
,
b
∈
R
,
n
∈
N
.We may see that, forn
=
1, (3) reduces to d
1
(
a
b
+
b
a
)
=
d
1
(
a
)
b
+
a
d
1
(
b
)
+
d
1
(
b
)
a
+
b
d
1
(
a
) for all a
,
b
∈
R. In view of Theorem 1, it is clear that d
1 is additive. We will use this result throughout the paper whenever needed without specific mention.We facilitate our discussion with the following lemmas.Lemma 3.
LetD
=
{
d
n
}
n
∈
N be the family of Jordan higher derivable mappings d
n
:
R
→
R. Then for each n
∈
N, (i)
d
n
(
a
11
+
b
12
)
=
d
n
(
a
11
)
+
d
n
(
b
12
),
(ii)
d
n
(
a
11
+
b
21
)
=
d
n
(
a
11
)
+
d
n
(
b
21
),
(iii)
d
n
(
a
22
+
b
12
)
=
d
n
(
a
22
)
+
d
n
(
b
12
),
(iv)
d
n
(
a
22
+
b
21
)
=
d
n
(
a
22
)
+
d
n
(
b
21
).Proof.
We only prove (i). The rest of the proofs follow similarly. Sinced
1 is additive, we have d
1
(
a
11
+
b
12
)
=
d
1
(
a
11
)
+
d
1
(
b
12
). Now by induction hypothesis let the result hold for all positive integer m
<
n. For any x
22
∈
R
22, we compute
(4)
d
n
a
11
+
b
12
x
22
+
x
22
a
11
+
b
12
=
∑
p
+
q
=
n
‍
d
p
a
11
+
b
12
d
q
x
22
+
∑
p
+
q
=
n
‍
d
p
x
22
d
q
a
11
+
b
12
=
d
n
a
11
+
b
12
x
22
+
a
11
+
b
12
d
n
x
22
+
∑
p
+
q
=
n
0
<
p
,
q
≤
n
-
1
‍
d
p
a
11
+
b
12
d
q
x
22
+
d
n
x
22
a
11
+
b
12
+
x
22
d
n
a
11
+
b
12
+
∑
p
+
q
=
n
0
<
p
,
q
≤
n
-
1
‍
d
p
(
x
22
)
d
q
(
a
11
+
b
12
)
.
On the other hand, we have
(5)
d
n
a
11
+
b
12
x
22
+
x
22
a
11
+
b
12
=
d
n
b
12
x
22
=
d
n
a
11
x
22
+
x
22
a
11
+
d
n
b
12
x
22
+
x
22
b
12
=
∑
p
+
q
=
n
‍
d
p
a
11
d
q
x
22
+
∑
p
+
q
=
n
‍
d
p
x
22
d
q
a
11
+
∑
p
+
q
=
n
‍
d
p
b
12
d
q
x
22
+
∑
p
+
q
=
n
‍
d
p
x
22
d
q
b
12
=
d
n
a
11
x
22
+
a
11
d
n
x
22
+
∑
p
+
q
=
n
0
<
p
,
q
≤
n
-
1
‍
d
p
a
11
d
q
x
22
+
d
n
x
22
a
11
+
x
22
d
n
a
11
+
∑
p
+
q
=
n
0
<
p
,
q
≤
n
-
1
‍
d
p
x
22
d
q
a
11
+
d
n
b
12
x
22
+
b
12
d
n
x
22
+
∑
p
+
q
=
n
0
<
p
,
q
≤
n
-
1
‍
d
p
b
12
d
q
x
22
+
d
n
x
22
b
12
+
x
22
d
n
b
12
+
∑
p
+
q
=
n
0
<
p
,
q
≤
n
-
1
‍
d
p
(
x
22
)
d
q
(
b
12
)
.
Since the result holds for all m
<
n, on comparing the above two equalities we obtain
(6)
d
n
a
11
+
b
12
-
d
n
a
11
-
d
n
b
12
x
22
+
x
22
[
d
n
(
a
11
+
b
12
)
-
d
n
(
a
11
)
-
d
n
(
b
12
)
]
=
0
.
On using (
P
1
), (
P
2
), and (
P
3
), respectively, we have
(7)
d
n
a
11
+
b
12
-
d
n
a
11
-
d
n
b
12
12
=
0
,
[
d
n
(
a
11
+
b
12
)
-
d
n
(
a
11
)
-
d
n
(
b
12
)
]
21
=
0
,
[
d
n
(
a
11
+
b
12
)
-
d
n
(
a
11
)
-
d
n
(
b
12
)
]
22
=
0
.
In order to complete the proof, we have to show that [
d
n
(
a
11
+
b
12
)
-
d
n
(
a
11
)
+
d
n
(
b
12
)
]
11
=
0.
For anyx
12
∈
R
12, we have
(8)
d
n
a
11
+
b
12
x
12
+
x
12
a
11
+
b
12
=
d
n
a
11
x
12
=
d
n
(
a
11
x
12
+
x
12
a
11
)
+
d
n
(
b
12
x
12
+
x
12
b
12
)
.
Application of (3) yields that
(9)
∑
p
+
q
=
n
‍
d
p
a
11
+
b
12
d
q
x
12
+
∑
p
+
q
=
n
‍
d
p
x
12
d
q
a
11
+
b
12
=
∑
p
+
q
=
n
‍
d
p
a
11
d
q
x
12
+
∑
p
+
q
=
n
‍
d
p
x
12
d
q
a
11
+
∑
p
+
q
=
n
‍
d
p
(
b
12
)
d
q
(
x
12
)
+
∑
p
+
q
=
n
‍
d
p
(
x
12
)
d
q
(
b
12
)
.
Also the above equation can be rewritten as
(10)
d
n
a
11
+
b
12
x
12
+
a
11
+
b
12
d
n
x
12
+
∑
p
+
q
=
n
0
<
p
,
q
≤
n
-
1
‍
d
p
a
11
+
b
12
d
q
x
12
+
d
n
x
12
a
11
+
b
12
+
x
12
d
n
a
11
+
b
12
+
∑
p
+
q
=
n
0
<
p
,
q
≤
n
-
1
‍
d
p
x
12
d
q
a
11
+
b
12
=
d
n
a
11
x
12
+
a
11
d
n
x
12
+
∑
p
+
q
=
n
0
<
p
,
q
≤
n
-
1
‍
d
p
a
11
d
q
x
12
+
d
n
x
12
a
11
+
x
12
d
n
a
11
+
∑
p
+
q
=
n
0
<
p
,
q
≤
n
-
1
‍
d
p
x
12
d
q
a
11
+
d
n
b
12
x
12
+
b
12
d
n
x
12
+
∑
p
+
q
=
n
0
<
p
,
q
≤
n
-
1
‍
d
p
b
12
d
q
x
12
+
d
n
x
12
b
12
+
x
12
d
n
b
12
+
∑
p
+
q
=
n
0
<
p
,
q
≤
n
-
1
‍
d
p
(
x
12
)
d
q
(
b
12
)
.
Using hypothesis, we find that
(11)
d
n
a
11
+
b
12
-
d
n
a
11
-
d
n
b
12
x
12
+
x
12
[
d
n
(
a
11
+
b
12
)
-
d
n
(
a
11
)
-
d
n
(
b
12
)
]
=
0
.
Consequently,
(12)
d
n
a
11
+
b
12
-
d
n
a
11
-
d
n
b
12
11
x
12
=
0
.
Therefore, on application of (
P
1
), we obtain, [
d
n
(
a
11
+
b
12
)
-
d
n
(
a
11
)
-
d
n
(
b
12
)
]
11
=
0. This completes the proof.Lemma 4.
LetD
=
{
d
n
}
n
∈
N be the family of Jordan higher derivable mappings d
n
:
R
→
R. Then for each n
∈
N, (i)
d
n
(
a
12
+
b
12
c
22
)
=
d
n
(
a
12
)
+
d
n
(
b
12
c
22
),
(ii)
d
n
(
a
21
+
b
22
c
21
)
=
d
n
(
a
21
)
+
d
n
(
b
22
c
21
).Proof.
(i) Using Lemma3, we obtain
(13)
d
n
a
12
+
b
12
c
22
=
d
n
e
1
+
b
12
a
12
+
c
22
+
a
12
+
c
22
e
1
+
b
12
=
∑
p
+
q
=
n
‍
d
p
e
1
+
b
12
d
q
a
12
+
c
22
+
∑
p
+
q
=
n
‍
d
p
a
12
+
c
22
d
q
e
1
+
b
12
=
∑
p
+
q
=
n
‍
d
p
e
1
+
d
p
b
12
d
q
a
12
+
d
q
c
22
+
∑
p
+
q
=
n
‍
d
p
a
12
+
d
p
c
22
d
q
e
1
+
d
q
b
12
=
d
n
a
12
e
1
+
e
1
a
12
+
d
n
b
12
c
22
+
c
22
b
12
+
d
n
c
22
e
1
+
e
1
c
22
+
d
n
b
12
a
12
+
a
12
b
12
=
d
n
(
a
12
)
+
d
n
(
b
12
c
22
)
.
(ii) Since (
a
21
+
b
22
c
21
)
=
(
e
1
+
c
21
)
(
a
21
+
b
22
)
+
(
a
21
+
b
22
)
(
e
1
+
c
21
). On using the same technique used in (
i
), we obtain the result.Lemma 5.
LetD
=
{
d
n
}
n
∈
N be the family of Jordan higher derivable mappings d
n
:
R
→
R. Then for each n
∈
N, (i)
d
n
(
a
12
+
b
12
)
=
d
n
(
a
12
)
+
d
n
(
b
12
),
(ii)
d
n
(
a
21
+
b
21
)
=
d
n
(
a
21
)
+
d
n
(
b
21
).Proof.
We prove (i); proof of (ii) follows similarly. Sinced
1 is additive, we have d
1
(
a
12
+
b
12
)
=
d
1
(
a
12
)
+
d
1
(
b
12
). Now by induction hypothesis let the result hold for all positive integer m
<
n. For any x
22
∈
R
22, we calculate d
n
[
(
a
12
+
b
12
)
x
22
+
x
22
(
a
12
+
b
12
)
] in two ways.
On one hand,(14)
d
n
a
12
+
b
12
x
22
+
x
22
a
12
+
b
12
=
∑
p
+
q
=
n
‍
d
p
a
12
+
b
12
d
q
x
22
+
∑
p
+
q
=
n
‍
d
p
x
22
d
q
a
12
+
b
12
=
d
n
a
12
+
b
12
x
22
+
a
12
+
b
12
d
n
x
22
+
∑
p
+
q
=
n
0
<
p
,
q
≤
n
-
1
‍
d
p
a
12
+
b
12
d
q
x
22
+
d
n
x
22
a
12
+
b
12
+
x
22
d
n
a
12
+
b
12
+
∑
p
+
q
=
n
0
<
p
,
q
≤
n
-
1
‍
d
p
(
x
22
)
d
q
(
a
12
+
b
12
)
.
On the other hand, by Lemma 4, we have
(15)
d
n
a
12
+
b
12
x
22
+
x
22
a
12
+
b
12
=
d
n
a
12
x
22
+
b
12
x
22
=
d
n
a
12
x
22
+
d
n
b
12
x
22
=
d
n
a
12
x
22
+
x
22
a
12
+
d
n
b
12
x
22
+
x
22
b
12
=
∑
p
+
q
=
n
‍
d
p
a
12
d
q
x
22
+
∑
p
+
q
=
n
‍
d
p
x
22
d
q
a
12
+
∑
p
+
q
=
n
‍
d
p
b
12
d
q
x
22
+
∑
p
+
q
=
n
‍
d
p
x
22
d
q
b
12
=
d
n
a
12
x
22
+
a
12
d
n
x
22
+
∑
p
+
q
=
n
0
<
p
,
q
≤
n
-
1
‍
d
p
a
12
d
q
x
22
+
d
n
x
22
a
12
+
x
22
d
n
a
12
+
∑
p
+
q
=
n
0
<
p
,
q
≤
n
-
1
‍
d
p
x
22
d
q
a
12
+
d
n
b
12
x
22
+
b
12
d
n
x
22
+
∑
p
+
q
=
n
0
<
p
,
q
≤
n
-
1
‍
d
p
b
12
d
q
x
22
+
d
n
x
22
b
12
+
x
22
d
n
b
12
+
∑
p
+
q
=
n
0
<
p
,
q
≤
n
-
1
‍
d
p
(
x
22
)
d
q
(
b
12
)
.
On comparing the above two equalities and using the hypothesis that the result holds for all m
<
n, we get
(16)
d
n
a
12
+
b
12
-
d
n
a
12
-
d
n
b
12
x
22
+
x
22
[
d
n
(
a
12
+
b
12
)
-
d
n
(
a
12
)
-
d
n
(
b
12
)
]
=
0
,
for all x
22
∈
R
22. Therefore,
(17)
d
n
a
12
+
b
12
-
d
n
a
12
-
d
n
b
12
12
x
22
=
0
,
x
22
d
n
a
12
+
b
12
-
d
n
a
12
-
d
n
b
12
21
=
0
and [
d
n
(
a
12
+
b
12
)
-
d
n
(
a
12
)
-
d
n
(
b
12
)
]
22
x
22
+
x
22
[
d
n
(
a
12
+
b
12
)
-
d
n
(
a
12
)
-
d
n
(
b
12
)
]
22
=
0. By (
P
1
), (
P
2
), and (
P
3
), we can deduce that
(18)
d
n
a
12
+
b
12
-
d
n
a
12
-
d
n
b
12
12
=
0
,
d
n
a
12
+
b
12
-
d
n
a
12
-
d
n
b
12
21
=
0
,
d
n
a
12
+
b
12
-
d
n
a
12
-
d
n
b
12
22
=
0
.
We now complete the proof by showing that [
d
n
(
a
12
+
b
12
)
-
d
n
(
a
12
)
-
d
n
(
b
12
)
]
11
=
0. Now for any x
12
∈
R
12, we observe that
(19)
0
=
d
n
a
12
+
b
12
x
12
+
x
12
a
12
+
b
12
=
∑
p
+
q
=
n
‍
d
p
a
12
+
b
12
d
q
x
12
+
∑
p
+
q
=
n
‍
d
p
x
12
d
q
a
12
+
b
12
=
d
n
a
12
+
b
12
x
12
+
a
12
+
b
12
d
n
x
12
+
∑
p
+
q
=
n
0
<
p
,
q
≤
n
-
1
d
p
a
12
+
b
12
d
q
x
12
+
d
n
x
12
a
12
+
b
12
+
x
12
d
n
a
12
+
b
12
+
∑
p
+
q
=
n
0
<
p
,
q
≤
n
-
1
‍
d
p
(
x
12
)
d
q
(
a
12
+
b
12
)
.
Also,
(20)
0
=
d
n
a
12
x
12
+
x
12
a
12
+
d
n
b
12
x
12
+
x
12
b
12
=
∑
p
+
q
=
n
‍
d
p
a
12
d
q
x
12
+
∑
p
+
q
=
n
‍
d
p
x
12
d
q
a
12
+
∑
p
+
q
=
n
‍
d
p
b
12
d
q
x
12
+
∑
p
+
q
=
n
‍
d
q
x
12
d
q
b
12
=
d
n
a
12
x
12
+
a
12
d
n
x
12
+
∑
p
+
q
=
n
0
<
p
,
q
≤
n
-
1
‍
d
p
a
12
d
q
x
12
+
d
n
x
12
a
12
+
x
12
d
n
a
12
+
∑
p
+
q
=
n
0
<
p
,
q
≤
n
-
1
‍
d
p
x
12
d
q
a
12
+
d
n
b
12
x
12
+
b
12
d
n
x
12
+
∑
p
+
q
=
n
0
<
p
,
q
≤
n
-
1
‍
d
p
b
12
d
q
x
12
+
d
n
x
12
b
12
+
x
12
d
n
b
12
+
∑
p
+
q
=
n
0
<
p
,
q
≤
n
-
1
‍
d
q
(
x
12
)
d
q
(
b
12
)
.
Now comparing the above two expressions, we find that
(21)
d
n
a
12
+
b
12
-
d
n
a
12
-
d
n
b
12
x
12
+
x
12
d
n
a
12
+
b
12
-
d
n
a
12
-
d
n
b
12
=
0
.
Therefore, we have [
d
n
(
a
12
+
b
12
)
-
d
n
(
a
12
)
-
d
n
(
b
12
)
]
11
x
12
=
0, for all x
12
∈
R
12. By (
P
1
) we can infer that [
d
n
(
a
12
+
b
12
)
-
d
n
(
a
12
)
-
d
n
(
b
12
)
]
11
=
0. This completes the proof.Lemma 6.
LetD
=
{
d
n
}
n
∈
N be the family of Jordan higher derivable mappings d
n
:
R
→
R. Then for each n
∈
N, (i)
d
n
(
a
11
+
b
11
)
=
d
n
(
a
11
)
+
d
n
(
b
11
),
(ii)
d
n
(
a
22
+
b
22
)
=
d
n
(
a
22
)
+
d
n
(
b
22
).Proof.
We will prove (i). The proof of (ii) follows similarly. Sinced
1 is additive, we have d
1
(
a
11
+
b
11
)
=
d
1
(
a
11
)
+
d
1
(
b
11
). Now by induction hypothesis let the result hold for all positive integer m
<
n. For any x
22
∈
R
22, we have
(22)
0
=
d
n
a
11
+
b
11
x
22
+
x
22
a
11
+
b
11
=
∑
p
+
q
=
n
‍
d
p
a
11
+
b
11
d
q
x
22
+
∑
p
+
q
=
n
‍
d
p
x
22
d
q
a
11
+
b
11
=
d
n
a
11
+
b
11
x
22
+
a
11
+
b
11
d
n
x
22
+
∑
p
+
q
=
n
0
<
p
,
q
≤
n
-
1
‍
d
p
a
11
+
b
11
d
q
x
22
+
d
n
x
22
a
11
+
b
11
+
x
22
d
n
a
11
+
b
11
+
∑
p
+
q
=
n
0
<
p
,
q
≤
n
-
1
‍
d
p
x
22
d
q
a
11
+
b
11
.
Also,
(23)
0
=
d
n
a
11
x
22
+
x
22
a
11
+
d
n
b
11
x
22
+
x
22
b
11
=
∑
p
+
q
=
n
‍
d
p
a
11
d
q
x
22
+
∑
p
+
q
=
n
‍
d
p
x
22
d
q
a
11
+
∑
p
+
q
=
n
‍
d
p
b
11
d
q
x
22
+
∑
p
+
q
=
n
‍
d
q
x
22
d
q
b
11
=
d
n
a
11
x
22
+
a
11
d
n
x
22
+
∑
p
+
q
=
n
0
<
p
,
q
≤
n
-
1
‍
d
p
a
11
d
q
x
22
+
d
n
x
22
a
11
+
x
22
d
n
a
11
+
∑
p
+
q
=
n
0
<
p
,
q
≤
n
-
1
‍
d
p
x
22
d
q
a
11
+
d
n
b
11
x
22
+
b
11
d
n
x
22
+
∑
p
+
q
=
n
0
<
p
,
q
≤
n
-
1
‍
d
p
b
11
d
q
x
22
+
d
n
x
22
b
11
+
x
22
d
n
b
11
+
∑
p
+
q
=
n
0
<
p
,
q
≤
n
-
1
‍
d
q
(
x
22
)
d
q
(
b
11
)
.
On comparing the above two equalities and using the hypothesis that the result holds for all m
<
n, we get
(24)
d
n
a
11
+
b
11
-
d
n
a
11
-
d
n
b
11
x
22
+
x
22
d
n
a
11
+
b
11
-
d
n
a
11
-
d
n
b
11
=
0
.
This implies that
(25)
[
d
n
(
a
11
+
b
11
)
-
d
n
a
11
-
d
n
(
b
11
)
]
12
=
0
,
d
n
(
a
11
+
b
11
)
-
d
n
a
11
-
d
n
b
11
21
=
0
,
d
n
a
11
+
b
11
-
d
n
a
11
-
d
n
b
11
22
=
0
.
Similarly, by considering (
a
11
+
b
11
)
x
12
+
x
12
(
a
11
+
b
11
) and using Lemma 5 we can find that [
d
n
(
a
11
+
b
11
)
-
d
n
(
a
11
)
-
d
n
(
b
11
)
]
11
=
0.Lemma 7.
LetD
=
{
d
n
}
n
∈
N be the family of Jordan higher derivable mappings d
n
:
R
→
R. Then for each n
∈
N, d
n
(
a
12
+
b
21
)
=
d
n
(
a
12
)
+
d
n
(
b
21
).Proof.
Sinced
1 is additive, we have d
1
(
a
12
+
b
21
)
=
d
1
(
a
12
)
+
d
1
(
b
21
). Now by induction hypothesis let the result hold for all positive integer m
<
n. Consider a
12
+
b
21
=
(
a
12
+
b
21
)
e
1
+
e
1
(
a
12
+
b
21
). We have
(26)
d
n
a
12
+
b
21
=
d
n
a
12
+
b
21
e
1
+
e
1
a
12
+
b
21
=
∑
p
+
q
=
n
‍
d
p
a
12
+
b
21
d
q
e
1
+
∑
p
+
q
=
n
‍
d
p
e
1
d
q
a
12
+
b
21
=
d
n
a
12
+
b
21
e
1
+
a
12
+
b
21
d
n
e
1
+
∑
p
+
q
=
n
0
<
p
,
q
≤
n
-
1
‍
d
p
a
12
+
b
21
d
q
e
1
+
d
n
e
1
a
12
+
b
21
+
e
1
d
n
a
12
+
b
21
+
∑
p
+
q
=
n
0
<
p
,
q
≤
n
-
1
‍
d
p
(
e
1
)
d
q
(
a
12
+
b
21
)
.
On multiplying right hand side by e
1 and using the fact that result holds good for all m
<
n in the above relation, we obtain
(27)
0
=
a
12
+
b
21
d
n
e
1
e
1
+
∑
p
+
q
=
n
0
<
p
,
q
≤
n
-
1
‍
d
p
a
12
+
d
p
b
21
d
q
e
1
e
1
+
d
n
e
1
b
21
+
e
1
d
n
a
12
+
b
21
e
1
+
∑
p
+
q
=
n
0
<
p
,
q
≤
n
-
1
‍
d
p
(
e
1
)
(
d
q
(
a
12
)
+
d
q
(
b
21
)
)
e
1
.
Also on using the same technique, we compute d
n
(
a
12
)
=
d
n
(
a
12
e
1
+
e
1
a
12
) and d
n
(
b
21
)
=
d
n
(
b
21
e
1
+
e
1
b
21
) to obtain
(28)
0
=
a
12
d
n
e
1
e
1
+
∑
p
+
q
=
n
0
<
p
,
q
≤
n
-
1
‍
d
p
a
12
d
q
e
1
e
1
+
e
1
d
n
a
12
e
1
+
∑
p
+
q
=
n
0
<
p
,
q
≤
n
-
1
‍
d
p
e
1
d
q
a
12
e
1
,
(29)
0
=
b
21
d
n
e
1
e
1
+
∑
p
+
q
=
n
0
<
p
,
q
≤
n
-
1
‍
d
p
b
21
d
q
e
1
e
1
+
d
n
e
1
b
21
+
e
1
d
n
b
21
e
1
+
∑
p
+
q
=
n
0
<
p
,
q
≤
n
-
1
‍
d
p
(
e
1
)
d
q
(
b
21
)
e
1
.
Comparing (27), (28), and (29) we see that
(30)
d
n
a
12
+
b
21
-
d
n
a
12
-
d
n
b
21
11
=
e
1
d
n
a
12
+
b
21
-
d
n
a
12
-
d
n
b
21
e
1
=
0
.
Now, for any x
12
∈
R
12, we have
(31)
d
n
a
12
+
b
21
x
12
+
a
12
+
b
21
d
n
x
12
+
∑
p
+
q
=
n
0
<
p
,
q
≤
n
-
1
‍
d
p
a
12
+
b
21
d
q
x
12
+
d
n
x
12
a
12
+
b
21
+
x
12
d
n
a
12
+
b
21
+
∑
p
+
q
=
n
0
<
p
,
q
≤
n
-
1
‍
d
p
x
12
d
q
a
12
+
b
21
=
∑
p
+
q
=
n
‍
d
p
a
12
+
b
21
d
q
x
12
+
∑
p
+
q
=
n
‍
d
p
x
12
d
q
a
12
+
b
21
=
d
n
(
a
12
+
b
21
)
x
12
+
x
12
(
a
12
+
b
21
)
=
d
n
b
21
x
12
+
x
12
b
21
=
d
n
b
21
x
12
+
x
12
b
21
+
d
n
a
12
x
12
+
x
12
a
12
=
∑
p
+
q
=
n
‍
d
p
b
21
d
q
x
12
+
∑
p
+
q
=
n
‍
d
p
x
12
d
q
b
21
+
∑
p
+
q
=
n
‍
d
p
a
12
d
q
x
12
+
∑
p
+
q
=
n
‍
d
p
x
12
d
q
a
12
=
d
n
b
21
x
12
+
b
21
d
n
x
12
+
∑
p
+
q
=
n
0
<
p
,
q
≤
n
-
1
‍
d
p
b
21
d
q
x
12
+
d
n
x
12
b
21
+
x
12
d
n
b
21
+
∑
p
+
q
=
n
0
<
p
,
q
≤
n
-
1
‍
d
p
x
12
d
q
b
21
+
d
n
a
12
x
12
+
a
12
d
n
x
12
+
∑
p
+
q
=
n
0
<
p
,
q
≤
n
-
1
‍
d
p
a
12
d
q
x
12
+
d
n
x
12
a
12
+
x
12
d
n
a
12
+
∑
p
+
q
=
n
0
<
p
,
q
≤
n
-
1
‍
d
p
x
12
d
q
a
12
.
This further leads to
(32)
d
n
a
12
+
b
21
-
d
n
a
12
-
d
n
b
21
x
12
+
x
12
d
n
a
12
+
b
21
-
d
n
a
12
-
d
n
b
21
=
0
.
Since [
d
n
(
a
12
+
b
21
)
-
d
n
(
a
12
)
-
d
n
(
b
21
)
]
11
=
0, we see that
(33)
d
n
a
12
+
b
21
-
d
n
a
12
-
d
n
b
21
21
x
12
=
0
,
x
12
d
n
a
12
+
b
21
-
d
n
a
12
-
d
n
b
21
22
=
0
.
Using (
P
1
) and (
P
2
), we find that
(34)
d
n
a
12
+
b
21
-
d
n
a
12
-
d
n
b
21
21
=
d
n
a
12
+
b
21
-
d
n
a
12
-
d
n
b
21
22
=
0
.
Similarly, by considering d
n
[
(
a
12
+
b
21
)
x
21
+
x
21
(
a
12
+
b
21
)
] for all x
21
∈
R
21, we get [
d
n
(
a
12
+
b
12
)
-
d
n
(
a
12
)
-
d
n
(
b
21
)
]
x
21
+
x
21
[
d
n
(
a
12
+
b
12
)
-
d
n
(
a
12
)
-
d
n
(
b
21
)
]
x
12
=
0. Consequently, [
d
n
(
a
12
+
b
12
)
-
d
n
(
a
12
)
-
d
n
(
b
21
)
]
12
=
0.Lemma 8.
LetD
=
{
d
n
}
n
∈
N be the family of Jordan higher derivable mappings d
n
:
R
→
R. Then for each n
∈
N, (i)
d
n
(
a
11
+
b
12
+
c
21
)
=
d
n
(
a
11
)
+
d
n
(
b
12
)
+
d
n
(
c
21
),
(ii)
d
n
(
a
12
+
b
21
+
c
22
)
=
d
n
(
a
12
)
+
d
n
(
b
21
)
+
d
n
(
c
22
).Proof.
We prove (i). The proof of (ii) follows similarly. Sinced
1 is additive, we have d
1
(
a
11
+
b
12
+
c
21
)
=
d
1
(
a
11
)
+
d
1
(
b
12
)
+
d
1
(
c
21
). Now by induction hypothesis let the result hold for all positive integer m
<
n. For any x
22
∈
R
22, we have
(35)
d
n
a
11
+
b
12
+
c
21
x
22
+
x
22
a
11
+
b
12
+
c
21
=
∑
p
+
q
=
n
‍
d
p
a
11
+
b
12
+
c
21
d
q
x
22
+
∑
p
+
q
=
n
‍
d
p
x
22
d
q
a
11
+
b
12
+
c
21
=
d
n
a
11
+
b
12
+
c
21
x
22
+
a
11
+
b
12
+
c
21
d
n
x
22
+
∑
p
+
q
=
n
0
<
p
,
q
≤
n
-
1
‍
d
p
a
11
+
b
12
+
c
21
d
q
x
22
+
d
n
x
22
a
11
+
b
12
+
c
21
+
x
22
d
n
a
11
+
b
12
+
c
21
+
∑
p
+
q
=
n
0
<
p
,
q
≤
n
-
1
‍
d
p
(
x
22
)
d
q
(
a
11
+
b
12
+
c
21
)
.
On the other hand, by Lemma 7, we also have
(36)
d
n
a
11
+
b
12
+
c
21
x
22
+
x
22
a
11
+
b
12
+
c
21
=
d
n
b
12
x
22
+
x
22
c
21
=
d
n
b
12
x
22
+
d
n
x
22
c
21
=
d
n
a
11
x
22
+
x
22
a
11
+
d
n
b
12
x
22
+
x
22
b
12
+
d
n
c
21
x
22
+
x
22
c
21
=
∑
p
+
q
=
n
‍
d
p
a
11
d
q
x
22
+
∑
p
+
q
=
n
‍
d
p
x
22
d
q
a
11
+
∑
p
+
q
=
n
‍
d
p
b
12
d
q
x
22
+
∑
p
+
q
=
n
‍
d
p
x
22
d
q
b
12
+
∑
p
+
q
=
n
‍
d
p
c
21
d
q
x
22
+
∑
p
+
q
=
n
‍
d
p
x
22
d
q
c
21
=
d
n
a
11
x
22
+
a
11
d
n
x
22
+
∑
p
+
q
=
n
0
<
p
,
q
≤
n
-
1
‍
d
p
a
11
d
q
x
22
+
d
n
x
22
a
11
+
x
22
d
n
a
11
+
∑
p
+
q
=
n
0
<
p
,
q
≤
n
-
1
‍
d
p
x
22
d
q
a
11
+
d
n
b
12
x
22
+
b
12
d
n
x
22
+
∑
p
+
q
=
n
0
<
p
,
q
≤
n
-
1
‍
d
p
b
12
d
q
x
22
+
d
p
x
22
b
12
+
x
22
d
n
b
12
+
∑
p
+
q
=
n
0
<
p
,
q
≤
n
-
1
‍
d
p
x
22
d
q
b
12
+
d
n
c
21
x
22
+
c
21
d
n
x
22
+
∑
p
+
q
=
n
0
<
p
,
q
≤
n
-
1
‍
d
p
c
21
d
q
x
22
+
d
n
x
22
c
21
+
x
22
d
n
c
21
+
∑
p
+
q
=
n
0
<
p
,
q
≤
n
-
1
‍
d
p
(
x
22
)
d
q
(
c
21
)
.
On comparing the above two equalities, we get
(37)
d
n
a
11
+
b
12
+
c
21
-
d
n
a
11
-
d
n
b
12
-
d
n
c
21
x
22
+
x
22
[
d
n
(
a
11
+
b
12
+
c
21
)
-
d
n
(
a
11
)
-
d
n
(
b
12
)
-
d
n
(
c
21
)
]
=
0
.
Hence we obtain that
(38)
[
d
n
(
a
11
+
b
12
+
c
21
)
-
d
n
(
a
11
)
-
d
n
(
b
12
)
-
d
n
(
c
21
)
]
12
=
0
,
[
d
n
(
a
11
+
b
12
+
c
21
)
-
d
n
(
a
11
)
-
d
n
(
b
12
)
-
d
n
(
c
21
)
]
21
=
0
,
[
d
n
(
a
11
+
b
12
+
c
21
)
-
d
n
(
a
11
)
-
d
n
(
b
12
)
-
d
n
(
c
21
)
]
22
=
0
.
From
(39)
d
n
a
11
+
b
12
+
c
21
x
12
+
x
12
a
11
+
b
12
+
c
21
=
d
n
a
11
+
c
21
x
12
+
x
12
a
11
+
c
21
+
d
n
b
12
x
12
+
x
12
b
12
=
0
and using Lemma 3, one can easily obtain that
(40)
d
n
a
11
+
b
12
+
c
21
-
d
n
a
11
-
d
n
b
12
-
d
n
c
21
x
12
+
x
12
[
d
n
(
a
11
+
b
12
+
c
21
)
-
d
n
(
a
11
)
-
d
n
(
b
12
)
-
d
n
(
c
21
)
]
=
0
.
It follows that [
d
n
(
a
11
+
b
12
+
c
21
)
-
d
n
(
a
11
)
-
d
n
(
b
12
)
-
d
n
(
c
21
)
]
11
x
12
=
0, and therefore we get [
d
n
(
a
11
+
b
12
+
c
21
)
-
d
n
(
a
11
)
-
d
n
(
b
12
)
-
d
n
(
c
21
)
]
11
=
0.Lemma 9.
LetD
=
{
d
n
}
n
∈
N be the family of Jordan higher derivable mappings d
n
:
R
→
R. Then for each n
∈
N, d
n
(
a
11
+
b
12
+
c
21
+
u
22
)
=
d
n
(
a
11
)
+
d
n
(
b
12
)
+
d
n
(
c
21
)
+
d
n
(
u
22
).Proof.
Sinced
1 is additive, we have d
1
(
a
11
+
b
12
+
c
21
)
=
d
1
(
a
11
)
+
d
1
(
b
12
)
+
d
1
(
c
21
). Now by induction hypothesis let the result hold for all positive integer m
<
n. For any x
11
∈
R
11, we have
(41)
d
n
a
11
+
b
12
+
c
21
+
u
22
x
11
+
a
11
+
b
12
+
c
21
+
u
22
d
n
x
11
+
∑
p
+
q
=
n
0
<
p
,
q
≤
n
-
1
‍
d
p
a
11
+
b
12
+
c
21
+
u
22
d
q
x
11
+
d
n
x
11
a
11
+
b
12
+
c
21
+
u
22
+
x
11
d
n
a
11
+
b
12
+
c
21
+
u
22
+
∑
p
+
q
=
n
0
<
p
,
q
≤
n
-
1
‍
d
p
x
11
d
q
a
11
+
b
12
+
c
21
+
u
22
=
∑
p
+
q
=
n
‍
d
p
a
11
+
b
12
+
c
21
+
u
22
d
q
x
11
+
∑
p
+
q
=
n
‍
d
p
x
11
d
q
a
11
+
b
12
+
c
21
+
u
22
=
d
n
a
11
+
b
12
+
c
21
+
u
22
x
11
+
x
11
a
11
+
b
12
+
c
21
+
u
22
=
d
n
(
a
11
x
11
+
c
21
x
11
+
x
11
a
11
+
x
11
b
12
)
=
d
n
a
11
x
11
+
x
11
a
11
+
d
n
x
11
b
12
+
d
n
c
21
x
11
=
d
n
a
11
x
11
+
x
11
a
11
+
d
n
b
12
x
11
+
b
12
x
11
+
d
n
c
21
x
11
+
x
11
c
21
+
d
n
u
22
x
11
+
x
11
u
22
=
∑
p
+
q
=
n
‍
d
p
a
11
d
q
x
11
+
∑
p
+
q
=
n
‍
d
p
x
11
d
q
a
11
+
∑
p
+
q
=
n
‍
d
p
b
12
d
q
x
11
+
∑
p
+
q
=
n
‍
d
p
x
11
d
q
b
12
+
∑
p
+
q
=
n
‍
d
p
c
21
d
q
x
11
+
∑
p
+
q
=
n
‍
d
p
x
11
d
q
c
21
+
∑
p
+
q
=
n
‍
d
p
(
u
22
)
d
q
(
x
11
)
+
∑
p
+
q
=
n
‍
d
p
(
x
11
)
d
q
(
u
22
)
=
d
n
a
11
x
11
+
a
11
d
n
x
11
+
∑
p
+
q
=
n
0
<
p
,
q
≤
n
-
1
‍
d
p
a
11
d
q
x
11
+
d
n
x
11
c
21
+
x
11
d
n
c
21
+
∑
p
+
q
=
n
0
<
p
,
q
≤
n
-
1
‍
d
p
x
11
d
q
c
21
+
d
n
b
12
x
11
+
b
12
d
n
x
11
+
∑
p
+
q
=
n
0
<
p
,
q
≤
n
-
1
‍
d
p
b
12
d
q
x
11
+
d
n
x
11
b
12
+
x
11
d
n
b
12
+
∑
p
+
q
=
n
0
<
p
,
q
≤
n
-
1
‍
d
p
x
11
d
q
b
12
+
d
n
c
21
x
11
+
c
21
d
n
x
11
+
∑
p
+
q
=
n
0
<
p
,
q
≤
n
-
1
‍
d
p
c
21
d
q
x
11
+
d
n
x
11
c
21
+
x
11
d
n
c
21
+
∑
p
+
q
=
n
0
<
p
,
q
≤
n
-
1
‍
d
p
x
11
d
q
c
21
+
d
n
u
22
x
11
+
u
22
d
n
x
11
+
∑
p
+
q
=
n
0
<
p
,
q
≤
n
-
1
‍
d
p
u
22
d
q
x
11
+
d
n
x
11
u
22
+
x
11
d
n
u
22
+
∑
p
+
q
=
n
0
<
p
,
q
≤
n
-
1
‍
d
p
(
x
11
)
d
q
(
u
22
)
.
This gives us
(42)
d
n
(
a
11
+
b
12
+
c
21
+
u
22
)
-
d
n
(
a
11
)
-
d
n
(
b
12
)
-
d
n
(
c
21
)
-
d
n
(
u
22
)
x
11
+
x
11
d
n
(
a
11
+
b
12
+
c
21
+
u
22
)
-
d
n
(
a
11
)
-
d
n
(
b
12
)
-
d
n
(
c
21
)
-
d
n
(
u
22
)
=
0
.
Now using (
P
1
) and (
P
3
), the above equation reduces to
(43)
d
n
(
a
11
+
b
12
+
c
21
+
u
22
)
-
d
n
(
a
11
)
-
d
n
b
12
-
d
n
c
21
-
d
n
u
22
11
=
0
,
d
n
(
a
11
+
b
12
+
c
21
+
u
22
)
-
d
n
(
a
11
)
-
d
n
b
12
-
d
n
c
21
-
d
n
u
22
12
=
0
,
d
n
(
a
11
+
b
12
+
c
21
+
u
22
)
-
d
n
(
a
11
)
-
d
n
b
12
-
d
n
c
21
-
d
n
u
22
21
=
0
.
Similarly one can get
(44)
d
n
(
a
11
+
b
12
+
c
21
+
u
22
)
-
d
n
(
a
11
)
-
d
n
(
b
12
)
-
d
n
(
c
21
)
-
d
n
(
u
22
)
22
=
0
.
This completes the proof.We are now well equipped to prove our main theorem.Proof of Theorem2.
For anya
,
b
∈
R, we write a
=
a
11
+
a
12
+
a
21
+
a
22 and b
=
b
11
+
b
12
+
b
21
+
b
22. Applying Lemmas 3–9, we have
(45)
d
n
a
+
b
=
d
n
a
11
+
a
12
+
a
21
+
a
22
+
b
11
+
b
12
+
b
21
+
b
22
=
d
n
a
11
+
b
11
+
a
12
+
b
12
+
a
21
+
b
21
+
a
22
+
b
22
=
d
n
a
11
+
a
12
+
a
21
+
a
22
+
d
n
b
11
+
b
12
+
b
21
+
b
22
=
d
n
(
a
)
+
d
n
(
b
)
.
Hence we conclude that D
=
{
d
n
}
n
∈
N is additive.
In addition, ifR is 2-torsion free, then for any a
∈
R, we have
(46)
2
d
n
(
a
2
)
=
d
n
(
2
a
2
)
=
d
n
(
a
a
+
a
a
)
=
2
∑
p
+
q
=
n
‍
d
p
(
a
)
d
q
(
a
)
.
Therefore, D
=
{
d
n
}
n
∈
N is a Jordan higher derivation.If the underlying ring is semiprime, applying Lemma 1.5 of [3] and the fact that every Jordan higher derivation on a 2-torsion free semiprime ring is a higher derivation (see [6]), we have the following.Corollary 10.
LetR be a 2-torsion free semiprime ring with a nontrivial idempotent satisfying the following conditions for i
,
j
∈
{
1,2
}. (C
1)
Ifa
i
i
x
i
j
=
0 for all x
i
j
∈
R
i
j, (
i
≠
j
), then a
i
i
=
0.
(C
2)
Ifx
j
i
a
i
i
=
0 for all x
j
i
∈
R
j
i, (
i
≠
j
), then a
i
i
=
0.
If the familyD
=
{
d
n
}
n
∈
N of mappings d
n
:
R
→
R such that d
0
=
I
R satisfy
(47)
d
n
(
a
b
+
b
a
)
=
∑
p
+
q
=
n
‍
(
d
p
(
a
)
d
q
(
b
)
+
d
p
(
b
)
d
q
(
a
)
)
for all a
,
b
∈
R, then D
=
{
d
n
}
n
∈
N is additive. Moreover, D
=
{
d
n
}
n
∈
N is a higher derivation.Applying Lemma 1.6 of [3] and the fact that every Jordan higher derivation on a prime ring of characteristic different from two is a higher derivation (see the main theorem of [8]), we find the following corollary.Corollary 11.
LetR be a prime ring of characteristic different from two with a nontrivial idempotent and let a family D
=
{
d
n
}
n
∈
N of mappings d
n
:
R
→
R such that d
0
=
I
R satisfy
(48)
d
n
(
a
b
+
b
a
)
=
∑
p
+
q
=
n
‍
(
d
p
(
a
)
d
q
(
b
)
+
d
p
(
b
)
d
q
(
a
)
)
for all a
,
b
∈
R; then D
=
{
d
n
}
n
∈
N is additive. Moreover, D
=
{
d
n
}
n
∈
N is a higher derivation.Since every standard operator algebra is prime, we can easily have the following.Corollary 12.
LetA be standard operator algebra in a Banach space X whose dimension is greater than 1. If the family D
=
{
d
n
}
n
∈
N of mappings d
n
:
A
→
A such that d
0
=
I
A satisfy
(49)
d
n
(
A
B
+
B
A
)
=
∑
p
+
q
=
n
‍
(
d
p
(
A
)
d
q
(
B
)
+
d
p
(
B
)
d
q
(
A
)
)
for all A
,
B
∈
A, then D
=
{
d
n
}
n
∈
N is additive. Moreover, D
=
{
d
n
}
n
∈
N is an additive higher derivation.



---

*Source: 672387-2014-11-19.xml*