Let R be a ring. We say that a family of maps D={dn}n∈N is a Jordan higher derivable map (without assumption of additivity) on R if d0=IR (the identity map on R) and dn(ab+ba)=∑p+q=ndp(a)dq(b)+∑p+q=ndp(b)dq(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.
1. Introduction
Let R be a ring. An additive mapping d:R→R is said to be a derivation (resp., Jordan derivation) if d(ab)=d(a)b+ad(b) (resp., d(a2)=d(a)a+ad(a)) holds for any a,b∈R. Note that, in case of 2-torsion free ring R, d(a2)=d(a)a+ad(a) is equivalent to d(ab+ba)=d(a)b+ad(b)+d(b)a+bd(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(ab)=d(a)b+ad(b) (resp., d(ab+ba)=d(a)b+ad(b)+d(b)a+bd(a)), holds for all a,b∈R. Recall that a ring R is prime if aRb={0} implies that either a=0 or b=0 and is semiprime if aRa={0} implies a=0.
Now suppose that R is a ring with a nontrivial idempotent e1. We write e2=1-e1. Note that R need not have identity element. Put eiRej=Rij for any i,j=1,2. Then by Peirce decomposition of R, we have R=R11+R12+R21+R22. Throughout this paper, the notation aij will denote an arbitrary element of Rij and any element a∈R can be expressed as a=a11+a12+a21+a22.
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]).
Let R be a ring containing a nontrivial idempotent and satisfying the following conditions for i,j,k∈{1,2}.
If aijxjk=0 for all xjk∈Rjk, then aij=0.
If xijajk=0 for all xij∈Rij, then ajk=0.
If aiixii+xiiaii=0 for all xii∈Rii, then aii=0.
If a mapping δ:R→R satisfies
(1)δ(ab+ba)=δ(a)b+aδ(b)+δ(b)a+bδ(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={dn}n∈N be a family of additive mappings dn: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 d0=IR (the identity map on R) and dn(ab)=∑dp(a)dq(b) (resp., dn(a2)=∑dp(a)dq(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 char(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 family D={dn}n∈N of mappings dn:R→R (without assumption of additivity) such that d0=IR, is said to be
a higher derivable map if dn(ab)=∑p+q=ndp(a)dq(b) holds for all a,b∈R and for each n∈N,
a Jordan higher derivable map if dn(ab+ba)=∑p+q=n(dp(a)dq(b)+dp(b)dq(a)) holds for all a,b∈R and for each n∈N.
The main result of our paper states as follows.
Theorem 2.
Let R be a ring containing a nontrivial idempotent satisfying the following conditions for i,j,k∈{1,2}.
If aijxjk=0 for all xjk∈Rjk, then aij=0.
If xijajk=0 for all xij∈Rij, then ajk=0.
If aiixii+xiiaii=0 for all xii∈Rii, then aii=0.
If the family D={dn}n∈N of mappings dn:R→R such that d0=IR satisfy
(2)dn(ab+ba)=∑p+q=n(dp(a)dq(b)+dp(b)dq(a))
for all a,b∈R and for each n∈N, then D={dn}n∈N is additive. In addition, if R is 2-torsion free, then D={dn}n∈N is a Jordan higher derivation.
Throughout the paper, we always assume that R is a ring with a nontrivial idempotent e1 and satisfies conditions (P1), (P2), and (P3). We also assume that D={dn}n∈N is the family of Jordan higher derivable maps dn:R→R; that is,
(3)dnab+ba=∑p+q=ndpadqb+∑p+q=ndpbdqahhhhhhhhhhhhhhhhhhhhihhh∀a,b∈R,n∈N.
We may see that, for n=1, (3) reduces to d1(ab+ba)=d1(a)b+ad1(b)+d1(b)a+bd1(a) for all a,b∈R. In view of Theorem 1, it is clear that d1 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.
Let D={dn}n∈N be the family of Jordan higher derivable mappings dn:R→R. Then for each n∈N,
dn(a11+b12)=dn(a11)+dn(b12),
dn(a11+b21)=dn(a11)+dn(b21),
dn(a22+b12)=dn(a22)+dn(b12),
dn(a22+b21)=dn(a22)+dn(b21).
Proof.
We only prove (i). The rest of the proofs follow similarly. Since d1 is additive, we have d1(a11+b12)=d1(a11)+d1(b12). Now by induction hypothesis let the result hold for all positive integer m<n. For any x22∈R22, we compute
(4)dna11+b12x22+x22a11+b12=∑p+q=ndpa11+b12dqx22+∑p+q=ndpx22dqa11+b12=dna11+b12x22+a11+b12dnx22+∑p+q=n0<p,q≤n-1dpa11+b12dqx22+dnx22a11+b12+x22dna11+b12+∑p+q=n0<p,q≤n-1dp(x22)dq(a11+b12).
On the other hand, we have
(5)dna11+b12x22+x22a11+b12=dnb12x22=dna11x22+x22a11+dnb12x22+x22b12=∑p+q=ndpa11dqx22+∑p+q=ndpx22dqa11+∑p+q=ndpb12dqx22+∑p+q=ndpx22dqb12=dna11x22+a11dnx22+∑p+q=n0<p,q≤n-1dpa11dqx22+dnx22a11+x22dna11+∑p+q=n0<p,q≤n-1dpx22dqa11+dnb12x22+b12dnx22+∑p+q=n0<p,q≤n-1dpb12dqx22+dnx22b12+x22dnb12+∑p+q=n0<p,q≤n-1dp(x22)dq(b12).
Since the result holds for all m<n, on comparing the above two equalities we obtain
(6)dna11+b12-dna11-dnb12x22+x22[dn(a11+b12)-dn(a11)-dn(b12)]=0.
On using (P1), (P2), and (P3), respectively, we have
(7)dna11+b12-dna11-dnb1212=0,[dn(a11+b12)-dn(a11)-dn(b12)]21=0,[dn(a11+b12)-dn(a11)-dn(b12)]22=0.
In order to complete the proof, we have to show that [dn(a11+b12)-dn(a11)+dn(b12)]11=0.
For any x12∈R12, we have
(8)dna11+b12x12+x12a11+b12=dna11x12=dn(a11x12+x12a11)+dn(b12x12+x12b12).
Application of (3) yields that
(9)∑p+q=ndpa11+b12dqx12+∑p+q=ndpx12dqa11+b12=∑p+q=ndpa11dqx12+∑p+q=ndpx12dqa11+∑p+q=ndp(b12)dq(x12)+∑p+q=ndp(x12)dq(b12).
Also the above equation can be rewritten as
(10)dna11+b12x12+a11+b12dnx12+∑p+q=n0<p,q≤n-1dpa11+b12dqx12+dnx12a11+b12+x12dna11+b12+∑p+q=n0<p,q≤n-1dpx12dqa11+b12=dna11x12+a11dnx12+∑p+q=n0<p,q≤n-1dpa11dqx12+dnx12a11+x12dna11+∑p+q=n0<p,q≤n-1dpx12dqa11+dnb12x12+b12dnx12+∑p+q=n0<p,q≤n-1dpb12dqx12+dnx12b12+x12dnb12+∑p+q=n0<p,q≤n-1dp(x12)dq(b12).
Using hypothesis, we find that
(11)dna11+b12-dna11-dnb12x12+x12[dn(a11+b12)-dn(a11)-dn(b12)]=0.
Consequently,
(12)dna11+b12-dna11-dnb1211x12=0.
Therefore, on application of (P1), we obtain, [dn(a11+b12)-dn(a11)-dn(b12)]11=0. This completes the proof.
Lemma 4.
Let D={dn}n∈N be the family of Jordan higher derivable mappings dn:R→R. Then for each n∈N,
dn(a12+b12c22)=dn(a12)+dn(b12c22),
dn(a21+b22c21)=dn(a21)+dn(b22c21).
Proof.
(i) Using Lemma 3, we obtain
(13)dna12+b12c22=dne1+b12a12+c22+a12+c22e1+b12=∑p+q=ndpe1+b12dqa12+c22+∑p+q=ndpa12+c22dqe1+b12=∑p+q=ndpe1+dpb12dqa12+dqc22+∑p+q=ndpa12+dpc22dqe1+dqb12=dna12e1+e1a12+dnb12c22+c22b12+dnc22e1+e1c22+dnb12a12+a12b12=dn(a12)+dn(b12c22).
(ii) Since (a21+b22c21)=(e1+c21)(a21+b22)+(a21+b22)(e1+c21). On using the same technique used in (i), we obtain the result.
Lemma 5.
Let D={dn}n∈N be the family of Jordan higher derivable mappings dn:R→R. Then for each n∈N,
dn(a12+b12)=dn(a12)+dn(b12),
dn(a21+b21)=dn(a21)+dn(b21).
Proof.
We prove (i); proof of (ii) follows similarly. Since d1 is additive, we have d1(a12+b12)=d1(a12)+d1(b12). Now by induction hypothesis let the result hold for all positive integer m<n. For any x22∈R22, we calculate dn[(a12+b12)x22+x22(a12+b12)] in two ways.
On one hand,
(14)dna12+b12x22+x22a12+b12=∑p+q=ndpa12+b12dqx22+∑p+q=ndpx22dqa12+b12=dna12+b12x22+a12+b12dnx22+∑p+q=n0<p,q≤n-1dpa12+b12dqx22+dnx22a12+b12+x22dna12+b12+∑p+q=n0<p,q≤n-1dp(x22)dq(a12+b12).
On the other hand, by Lemma 4, we have
(15)dna12+b12x22+x22a12+b12=dna12x22+b12x22=dna12x22+dnb12x22=dna12x22+x22a12+dnb12x22+x22b12=∑p+q=ndpa12dqx22+∑p+q=ndpx22dqa12+∑p+q=ndpb12dqx22+∑p+q=ndpx22dqb12=dna12x22+a12dnx22+∑p+q=n0<p,q≤n-1dpa12dqx22+dnx22a12+x22dna12+∑p+q=n0<p,q≤n-1dpx22dqa12+dnb12x22+b12dnx22+∑p+q=n0<p,q≤n-1dpb12dqx22+dnx22b12+x22dnb12+∑p+q=n0<p,q≤n-1dp(x22)dq(b12).
On comparing the above two equalities and using the hypothesis that the result holds for all m<n, we get
(16)dna12+b12-dna12-dnb12x22+x22[dn(a12+b12)-dn(a12)-dn(b12)]=0,
for all x22∈R22. Therefore,
(17)dna12+b12-dna12-dnb1212x22=0,x22dna12+b12-dna12-dnb1221=0
and [dn(a12+b12)-dn(a12)-dn(b12)]22x22+x22[dn(a12+b12)-dn(a12)-dn(b12)]22=0. By (P1), (P2), and (P3), we can deduce that
(18)dna12+b12-dna12-dnb1212=0,dna12+b12-dna12-dnb1221=0,dna12+b12-dna12-dnb1222=0.
We now complete the proof by showing that [dn(a12+b12)-dn(a12)-dn(b12)]11=0. Now for any x12∈R12, we observe that
(19)0=dna12+b12x12+x12a12+b12=∑p+q=ndpa12+b12dqx12+∑p+q=ndpx12dqa12+b12=dna12+b12x12+a12+b12dnx12+∑p+q=n0<p,q≤n-1dpa12+b12dqx12+dnx12a12+b12+x12dna12+b12+∑p+q=n0<p,q≤n-1dp(x12)dq(a12+b12).
Also,
(20)0=dna12x12+x12a12+dnb12x12+x12b12=∑p+q=ndpa12dqx12+∑p+q=ndpx12dqa12+∑p+q=ndpb12dqx12+∑p+q=ndqx12dqb12=dna12x12+a12dnx12+∑p+q=n0<p,q≤n-1dpa12dqx12+dnx12a12+x12dna12+∑p+q=n0<p,q≤n-1dpx12dqa12+dnb12x12+b12dnx12+∑p+q=n0<p,q≤n-1dpb12dqx12+dnx12b12+x12dnb12+∑p+q=n0<p,q≤n-1dq(x12)dq(b12).
Now comparing the above two expressions, we find that
(21)dna12+b12-dna12-dnb12x12+x12dna12+b12-dna12-dnb12=0.
Therefore, we have [dn(a12+b12)-dn(a12)-dn(b12)]11x12=0, for all x12∈R12. By (P1) we can infer that [dn(a12+b12)-dn(a12)-dn(b12)]11=0. This completes the proof.
Lemma 6.
Let D={dn}n∈N be the family of Jordan higher derivable mappings dn:R→R. Then for each n∈N,
dn(a11+b11)=dn(a11)+dn(b11),
dn(a22+b22)=dn(a22)+dn(b22).
Proof.
We will prove (i). The proof of (ii) follows similarly. Since d1 is additive, we have d1(a11+b11)=d1(a11)+d1(b11). Now by induction hypothesis let the result hold for all positive integer m<n. For any x22∈R22, we have
(22)0=dna11+b11x22+x22a11+b11=∑p+q=ndpa11+b11dqx22+∑p+q=ndpx22dqa11+b11=dna11+b11x22+a11+b11dnx22+∑p+q=n0<p,q≤n-1dpa11+b11dqx22+dnx22a11+b11+x22dna11+b11+∑p+q=n0<p,q≤n-1dpx22dqa11+b11.
Also,
(23)0=dna11x22+x22a11+dnb11x22+x22b11=∑p+q=ndpa11dqx22+∑p+q=ndpx22dqa11+∑p+q=ndpb11dqx22+∑p+q=ndqx22dqb11=dna11x22+a11dnx22+∑p+q=n0<p,q≤n-1dpa11dqx22+dnx22a11+x22dna11+∑p+q=n0<p,q≤n-1dpx22dqa11+dnb11x22+b11dnx22+∑p+q=n0<p,q≤n-1dpb11dqx22+dnx22b11+x22dnb11+∑p+q=n0<p,q≤n-1dq(x22)dq(b11).
On comparing the above two equalities and using the hypothesis that the result holds for all m<n, we get
(24)dna11+b11-dna11-dnb11x22+x22dna11+b11-dna11-dnb11=0.
This implies that
(25)[dn(a11+b11)-dna11-dn(b11)]12=0,dn(a11+b11)-dna11-dnb1121=0,dna11+b11-dna11-dnb1122=0.
Similarly, by considering (a11+b11)x12+x12(a11+b11) and using Lemma 5 we can find that [dn(a11+b11)-dn(a11)-dn(b11)]11=0.
Lemma 7.
Let D={dn}n∈N be the family of Jordan higher derivable mappings dn:R→R. Then for each n∈N, dn(a12+b21)=dn(a12)+dn(b21).
Proof.
Since d1 is additive, we have d1(a12+b21)=d1(a12)+d1(b21). Now by induction hypothesis let the result hold for all positive integer m<n. Consider a12+b21=(a12+b21)e1+e1(a12+b21). We have
(26)dna12+b21=dna12+b21e1+e1a12+b21=∑p+q=ndpa12+b21dqe1+∑p+q=ndpe1dqa12+b21=dna12+b21e1+a12+b21dne1+∑p+q=n0<p,q≤n-1dpa12+b21dqe1+dne1a12+b21+e1dna12+b21+∑p+q=n0<p,q≤n-1dp(e1)dq(a12+b21).
On multiplying right hand side by e1 and using the fact that result holds good for all m<n in the above relation, we obtain
(27)0=a12+b21dne1e1+∑p+q=n0<p,q≤n-1dpa12+dpb21dqe1e1+dne1b21+e1dna12+b21e1+∑p+q=n0<p,q≤n-1dp(e1)(dq(a12)+dq(b21))e1.
Also on using the same technique, we compute dn(a12)=dn(a12e1+e1a12) and dn(b21)=dn(b21e1+e1b21) to obtain
(28)0=a12dne1e1+∑p+q=n0<p,q≤n-1dpa12dqe1e1+e1dna12e1+∑p+q=n0<p,q≤n-1dpe1dqa12e1,(29)0=b21dne1e1+∑p+q=n0<p,q≤n-1dpb21dqe1e1+dne1b21+e1dnb21e1+∑p+q=n0<p,q≤n-1dp(e1)dq(b21)e1.
Comparing (27), (28), and (29) we see that
(30)dna12+b21-dna12-dnb2111=e1dna12+b21-dna12-dnb21e1=0.
Now, for any x12∈R12, we have
(31)dna12+b21x12+a12+b21dnx12+∑p+q=n0<p,q≤n-1dpa12+b21dqx12+dnx12a12+b21+x12dna12+b21+∑p+q=n0<p,q≤n-1dpx12dqa12+b21=∑p+q=ndpa12+b21dqx12+∑p+q=ndpx12dqa12+b21=dn(a12+b21)x12+x12(a12+b21)=dnb21x12+x12b21=dnb21x12+x12b21+dna12x12+x12a12=∑p+q=ndpb21dqx12+∑p+q=ndpx12dqb21+∑p+q=ndpa12dqx12+∑p+q=ndpx12dqa12=dnb21x12+b21dnx12+∑p+q=n0<p,q≤n-1dpb21dqx12+dnx12b21+x12dnb21+∑p+q=n0<p,q≤n-1dpx12dqb21+dna12x12+a12dnx12+∑p+q=n0<p,q≤n-1dpa12dqx12+dnx12a12+x12dna12+∑p+q=n0<p,q≤n-1dpx12dqa12.
This further leads to
(32)dna12+b21-dna12-dnb21x12+x12dna12+b21-dna12-dnb21=0.
Since [dn(a12+b21)-dn(a12)-dn(b21)]11=0, we see that
(33)dna12+b21-dna12-dnb2121x12=0,x12dna12+b21-dna12-dnb2122=0.
Using (P1) and (P2), we find that
(34)dna12+b21-dna12-dnb2121=dna12+b21-dna12-dnb2122=0.
Similarly, by considering dn[(a12+b21)x21+x21(a12+b21)] for all x21∈R21, we get [dn(a12+b12)-dn(a12)-dn(b21)]x21+x21[dn(a12+b12)-dn(a12)-dn(b21)]x12=0. Consequently, [dn(a12+b12)-dn(a12)-dn(b21)]12=0.
Lemma 8.
Let D={dn}n∈N be the family of Jordan higher derivable mappings dn:R→R. Then for each n∈N,
dn(a11+b12+c21)=dn(a11)+dn(b12)+dn(c21),
dn(a12+b21+c22)=dn(a12)+dn(b21)+dn(c22).
Proof.
We prove (i). The proof of (ii) follows similarly. Since d1 is additive, we have d1(a11+b12+c21)=d1(a11)+d1(b12)+d1(c21). Now by induction hypothesis let the result hold for all positive integer m<n. For any x22∈R22, we have
(35)dna11+b12+c21x22+x22a11+b12+c21=∑p+q=ndpa11+b12+c21dqx22+∑p+q=ndpx22dqa11+b12+c21=dna11+b12+c21x22+a11+b12+c21dnx22+∑p+q=n0<p,q≤n-1dpa11+b12+c21dqx22+dnx22a11+b12+c21+x22dna11+b12+c21+∑p+q=n0<p,q≤n-1dp(x22)dq(a11+b12+c21).
On the other hand, by Lemma 7, we also have
(36)dna11+b12+c21x22+x22a11+b12+c21=dnb12x22+x22c21=dnb12x22+dnx22c21=dna11x22+x22a11+dnb12x22+x22b12+dnc21x22+x22c21=∑p+q=ndpa11dqx22+∑p+q=ndpx22dqa11+∑p+q=ndpb12dqx22+∑p+q=ndpx22dqb12+∑p+q=ndpc21dqx22+∑p+q=ndpx22dqc21=dna11x22+a11dnx22+∑p+q=n0<p,q≤n-1dpa11dqx22+dnx22a11+x22dna11+∑p+q=n0<p,q≤n-1dpx22dqa11+dnb12x22+b12dnx22+∑p+q=n0<p,q≤n-1dpb12dqx22+dpx22b12+x22dnb12+∑p+q=n0<p,q≤n-1dpx22dqb12+dnc21x22+c21dnx22+∑p+q=n0<p,q≤n-1dpc21dqx22+dnx22c21+x22dnc21+∑p+q=n0<p,q≤n-1dp(x22)dq(c21).
On comparing the above two equalities, we get
(37)dna11+b12+c21-dna11-dnb12-dnc21x22+x22[dn(a11+b12+c21)-dn(a11)-dn(b12)-dn(c21)]=0.
Hence we obtain that
(38)[dn(a11+b12+c21)-dn(a11)-dn(b12)-dn(c21)]12=0,[dn(a11+b12+c21)-dn(a11)-dn(b12)-dn(c21)]21=0,[dn(a11+b12+c21)-dn(a11)-dn(b12)-dn(c21)]22=0.
From
(39)dna11+b12+c21x12+x12a11+b12+c21=dna11+c21x12+x12a11+c21+dnb12x12+x12b12=0
and using Lemma 3, one can easily obtain that
(40)dna11+b12+c21-dna11-dnb12-dnc21x12+x12[dn(a11+b12+c21)-dn(a11)-dn(b12)-dn(c21)]=0.
It follows that [dn(a11+b12+c21)-dn(a11)-dn(b12)-dn(c21)]11x12=0, and therefore we get [dn(a11+b12+c21)-dn(a11)-dn(b12)-dn(c21)]11=0.
Lemma 9.
Let D={dn}n∈N be the family of Jordan higher derivable mappings dn:R→R. Then for each n∈N, dn(a11+b12+c21+u22)=dn(a11)+dn(b12)+dn(c21)+dn(u22).
Proof.
Since d1 is additive, we have d1(a11+b12+c21)=d1(a11)+d1(b12)+d1(c21). Now by induction hypothesis let the result hold for all positive integer m<n. For any x11∈R11, we have
(41)dna11+b12+c21+u22x11+a11+b12+c21+u22dnx11+∑p+q=n0<p,q≤n-1dpa11+b12+c21+u22dqx11+dnx11a11+b12+c21+u22+x11dna11+b12+c21+u22+∑p+q=n0<p,q≤n-1dpx11dqa11+b12+c21+u22=∑p+q=ndpa11+b12+c21+u22dqx11+∑p+q=ndpx11dqa11+b12+c21+u22=dna11+b12+c21+u22x11+x11a11+b12+c21+u22=dn(a11x11+c21x11+x11a11+x11b12)=dna11x11+x11a11+dnx11b12+dnc21x11=dna11x11+x11a11+dnb12x11+b12x11+dnc21x11+x11c21+dnu22x11+x11u22=∑p+q=ndpa11dqx11+∑p+q=ndpx11dqa11+∑p+q=ndpb12dqx11+∑p+q=ndpx11dqb12+∑p+q=ndpc21dqx11+∑p+q=ndpx11dqc21+∑p+q=ndp(u22)dq(x11)+∑p+q=ndp(x11)dq(u22)=dna11x11+a11dnx11+∑p+q=n0<p,q≤n-1dpa11dqx11+dnx11c21+x11dnc21+∑p+q=n0<p,q≤n-1dpx11dqc21+dnb12x11+b12dnx11+∑p+q=n0<p,q≤n-1dpb12dqx11+dnx11b12+x11dnb12+∑p+q=n0<p,q≤n-1dpx11dqb12+dnc21x11+c21dnx11+∑p+q=n0<p,q≤n-1dpc21dqx11+dnx11c21+x11dnc21+∑p+q=n0<p,q≤n-1dpx11dqc21+dnu22x11+u22dnx11+∑p+q=n0<p,q≤n-1dpu22dqx11+dnx11u22+x11dnu22+∑p+q=n0<p,q≤n-1dp(x11)dq(u22).
This gives us
(42)dn(a11+b12+c21+u22)-dn(a11)-dn(b12)-dn(c21)-dn(u22)x11+x11dn(a11+b12+c21+u22)-dn(a11)-dn(b12)-dn(c21)-dn(u22)=0.
Now using (P1) and (P3), the above equation reduces to
(43)dn(a11+b12+c21+u22)-dn(a11)-dnb12-dnc21-dnu2211=0,dn(a11+b12+c21+u22)-dn(a11)-dnb12-dnc21-dnu2212=0,dn(a11+b12+c21+u22)-dn(a11)-dnb12-dnc21-dnu2221=0.
Similarly one can get
(44)dn(a11+b12+c21+u22)-dn(a11)-dn(b12)-dn(c21)-dn(u22)22=0.
This completes the proof.
We are now well equipped to prove our main theorem.
Proof of Theorem 2.
For any a,b∈R, we write a=a11+a12+a21+a22 and b=b11+b12+b21+b22. Applying Lemmas 3–9, we have
(45)dna+b=dna11+a12+a21+a22+b11+b12+b21+b22=dna11+b11+a12+b12+a21+b21+a22+b22=dna11+a12+a21+a22+dnb11+b12+b21+b22=dn(a)+dn(b).
Hence we conclude that D={dn}n∈N is additive.
In addition, if R is 2-torsion free, then for any a∈R, we have
(46)2dn(a2)=dn(2a2)=dn(aa+aa)=2∑p+q=ndp(a)dq(a).
Therefore, D={dn}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.
Let R be a 2-torsion free semiprime ring with a nontrivial idempotent satisfying the following conditions for i,j∈{1,2}.
If aiixij=0 for all xij∈Rij, (i≠j), then aii=0.
If xjiaii=0 for all xji∈Rji, (i≠j), then aii=0.
If the family D={dn}n∈N of mappings dn:R→R such that d0=IR satisfy
(47)dn(ab+ba)=∑p+q=n(dp(a)dq(b)+dp(b)dq(a))
for all a,b∈R, then D={dn}n∈N is additive. Moreover, D={dn}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.
Let R be a prime ring of characteristic different from two with a nontrivial idempotent and let a family D={dn}n∈N of mappings dn:R→R such that d0=IR satisfy
(48)dn(ab+ba)=∑p+q=n(dp(a)dq(b)+dp(b)dq(a))
for all a,b∈R; then D={dn}n∈N is additive. Moreover, D={dn}n∈N is a higher derivation.
Since every standard operator algebra is prime, we can easily have the following.
Corollary 12.
Let A be standard operator algebra in a Banach space X whose dimension is greater than 1. If the family D={dn}n∈N of mappings dn:A→A such that d0=IA satisfy
(49)dn(AB+BA)=∑p+q=n(dp(A)dq(B)+dp(B)dq(A))
for all A,B∈A, then D={dn}n∈N is additive. Moreover, D={dn}n∈N is an additive higher derivation.
Conflict of Interests
The authors declare that there is no conflict of interests regarding the publication of this paper.
Acknowledgment
The authors are indebted to the referees for their valuable comments and helpful suggestions.
MartindaleW. S.IIIWhen are multiplicative mappings additive?19692169569810.1090/S0002-9939-1969-0240129-7MR0240129LuF.Jordan derivable maps of prime rings201038124430444010.1080/00927870903366884MR2764829ZBL1218.160082-s2.0-84858035114JingW.LuF.Additivity of Jordan (triple) derivations on rings20124082700271910.1080/00927872.2011.584927MR29689062-s2.0-84865270908HasseF.SchmidtF. K.Noch eine Begründung der Theorie der höheren DiKerentialquotienten einem algebraischen Funktionenköroer einer Unbestimmten1937177215237AshrafM.AliS.HaetingerC.On derivations in rings and their applications200625279107MR2537802FerreroM.HaetingerC.Higher derivations of semiprime rings200230523212333MR1904640ZBL1010.1602810.1081/AGB-1200034712-s2.0-0035999233HaetingerC.2000Porto Alegre, BrazilUFRGSFerreroM.HaetingerC.Higher derivations and a theorem by Herstein200225224925710.2989/16073600209486012MR1916335