# Stability ofn-Jordan Homomorphisms from a Normed Algebra to a Banach Algebra

Authors: Yang-Hi Lee
Journal: Abstract and Applied Analysis
 (2013)

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## Abstract

We establish the hyperstability ofn-Jordan homomorphisms from a normed algebra to a Banach algebra, and also we show that an n-Jordan homomorphism between two commutative Banach algebras is an n-ring homomorphism.


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## Body

## 1. Introduction

LetA,B be two rings (algebras) and n a positive integer greater than 1. An additive mapping g:A→B is called an n-Jordan homomorphism if g(an)=(g(a))n for all a∈A and an additive mapping h:A→B is called an n-ring homomorphism if h(∏i=1nai)=∏i=1n‍h(ai) for all a1,a2,…,an∈A.In 2009, Gordji et al. [1] showed the following theorems.Theorem 1.
Letn∈{2,3,4,5} be fixed. Suppose that A,B are two commutative algebras. Let h:A→B be an n-Jordan homomorphism. Then h is an n-ring homomorphism.Theorem 2.
Letn∈{2,3,4,5} be fixed. Suppose that A,B are commutative Banach algebras. Let δ and ε be nonnegative real numbers, and let p,q be real numbers such that (p-1)(q-1)>0, q≥0 or (p-1)(q-1)>0, q<0, and f(0)=0. Assume that f:A→B satisfies the system of functional inequalities:
(1)∥f(a+b)-f(a)-f(b)∥≤ε(∥a∥p+∥b∥p),∥f(an)-f(a)n∥≤δ∥a∥nq,
for all a,b∈A. Then, there exists a unique n-ring homomorphism h:A→B such that
(2)∥f(a)-h(a)∥≤2ε|2-2p|∥a∥p,
for all a∈A.The stability problem of group homomorphisms was formulated by Ulam [2] in 1940. Bourgin [3] and Badora [4] solved the stability problem of ring homomorphisms (see [5]). The term hyperstability was used for the first time in [6]. Some recent results on hyperstability of Cauchy or linear equation can be founded in [5, 7, 8].In this paper, we improve Theorems1 and 2 into Theorems 4 and 8, respectively. In particular, we prove the hyperstability of n-Jordan homomorphisms between two commutative Banach algebras.


## 2. Generalization of Theorem1

Lemma 3.
Letn,k be fixed natural numbers with n>k≥2. Let A,B be two commutative algebras, and let f:A→B be an additive mapping. Assume that f satisfies the following equality:
(3)∑i1=k-1n-1∑i2=k-2i1-1⋯∑ik-1=1ik-2-1(ni1)(i1i2)⋯(ik-2ik-1)×f(x1n-i1x2i1-i2x3i2-i3⋯xkik-1)=∑i1=k-1n-1∑i2=k-2i1-1⋯∑ik-1=1ik-2-1(ni1)(i1i2)⋯(ik-2ik-1)f(x1)n-i1×f(x2)i1-i2⋯f(xk)ik-1,
for all x1,x2,x3,…,xk∈A. Then one gets
(4)∑i1=kn-1∑i2=k-1i1-1⋯∑ik=1ik-1-1(ni1)(i1i2)⋯(ik-1ik)×f(x1n-i1x2i1-i2⋯xk+1ik)=∑i1=kn-1∑i2=k-1i1-1⋯∑ik=1ik-1-1(ni1)(i1i2)⋯(ik-1ik)f(x1)n-i1×f(x2)i1-i2⋯f(xk+1)ik,
for all x1,x2,x3,…,xk+1∈A.Proof.
Replacingxk by xk+1 in (3), we obtain
(5)∑i1=k-1n-1⋯∑ik-1=1ik-2-1(ni1)⋯(ik-2ik-1)(ik-1ik-1)×f(x1n-i1⋯xk-1ik-2-ik-1xk+1ik-1)=∑i1=k-1n-1⋯∑ik-1=1ik-2-1(ni1)⋯(ik-2ik-1)(ik-1ik-1)×f(x1)n-i1⋯f(xk-1)ik-2-ik-1×f(xk+1)ik-1,
for all x1,x2,x3,…,xk-1,xk+1∈A. In particular, the equality (3) implies that
(6)∑i1=k-1n-1⋯∑ik-1=1ik-2-1(ni1)⋯(ik-2ik-1)(ik-10)×f(x1n-i1⋯xk-1ik-2-ik-1xkik-1)=∑i1=k-1n-1⋯∑ik-1=1ik-2-1(ni1)⋯(ik-2ik-1)(ik-10)×f(x1)n-i1⋯f(xk-1)ik-2-ik-1f(xk)ik-1,
for all x1,x2,x3,…,xk∈A. Recall that the equality,
(7)(xk+xk+1)ik-1=∑ik=0ik-1(ik-1ik)xkik-1-ikxk+1ik,
holds for all xk,xk+1∈A. Replacing xk by xk+xk+1 in (3), we obtain
(8)∑i1=k-1n-1⋯∑ik-1=1ik-2-1∑ik=0ik-1(ni1)⋯(ik-2ik-1)(ik-1ik)×f(x1n-i1⋯xk-1ik-2-ik-1xkik-1-ikxk+1ik)=∑i1=k-1n-1∑i2=k-2i1-1⋯∑ik-1=1ik-2-1(ni1)⋯(ik-2ik-1)×f((xk+xk+1)ik-1x1n-i1⋯xk-1ik-2-ik-1×(xk+xk+1)ik-1x1n-i1⋯xk-1ik-2-ik-1)=∑i1=k-1n-1∑i2=k-2i1-1⋯∑ik-1=1ik-2-1(ni1)(i1i2)⋯(ik-2ik-1)×f(x1)n-i1⋯f(xk+xk+1)ik-1=∑i1=k-1n-1∑i2=k-2i1-1⋯∑ik-1=1ik-2-1(ni1)(i1i2)⋯(ik-2ik-1)×f(x1)n-i1⋯(f(xk)+f(xk+1))ik-1=∑i1=k-1n-1⋯∑ik-1=1ik-2-1∑ik=0ik-1(ni1)⋯(ik-2ik-1)(ik-1ik)×f(x1)n-i1⋯f(xk)ik-1-ikf(xk+1)ik,
for all x1,x2,x3,…,xk,xk+1∈A. From (5), (6), and the above equality, we get the desired equality:
(9)∑i1=kn-1∑i2=k-1i1-1⋯∑ik=1ik-1-1(ni1)(i1i2)⋯(ik-1ik)×f(x1n-i1x2i1-i2⋯xk+1ik)=∑i1=k-1n-1‍⋯∑ik-1=1ik-2-1‍∑ik=0ik-1(ni1)⋯(ik-2ik-1)(ik-1ik)×f(x1n-i1⋯xk-1ik-2-ik-1×xkik-1-ikxk+1ik)-∑i1=k-1n-1⋯∑ik-1=1ik-2-1(ni1)⋯(ik-2ik-1)(ik-1ik-1)×f(x1n-i1⋯xk-1ik-2-ik-1xk+1ik-1)-∑i1=k-1n-1⋯∑ik-1=1ik-2-1(ni1)⋯(ik-2ik-1)(ik-10)×f(x1n-i1⋯xk-1ik-2-ik-1xkik-1)=∑i1=k-1n-1⋯∑ik-1=1ik-2-1∑ik=0ik-1‍(ni1)⋯(ik-2ik-1)(ik-1ik)×f(x1)n-i1⋯f(xk)ik-1-ik×f(xk+1)ik-∑i1=k-1n-1⋯∑ik-1=1ik-2-1(ni1)⋯(ik-2ik-1)(ik-1ik-1)×f(x1)n-i1⋯f(xk-1)ik-2-ik-1×f(xk+1)ik-1-∑i1=k-1n-1⋯∑ik-1=1ik-2-1(ni1)⋯(ik-2ik-1)(ik-10)×f(x1)n-i1⋯f(xk-1)ik-2-ik-1×f(xk)ik-1=∑i1=kn-1∑i2=k-1i1-1⋯∑ik=1ik-1-1(ni1)(i1i2)⋯(ik-1ik)×f(x1)n-i1f(x2)i1-i2⋯f(xk+1)ik,
for all x1,x2,x3,…,xk,xk+1∈A.The following theorem is the generalization of Theorem1.Theorem 4.
LetA,B be two commutative algebras, and let f:A→B be an n-Jordan homomorphism. Then f is an n-ring homomorphism.Proof.
Sincef is an n-Jordan homomorphism, together with the additivity of f, we get
(10)∑i=0n(ni)f(x1n-ix2i)=f((x1+x2)n)=f(x1+x2)n=(f(x1)+f(x2))n=∑i=0n(ni)f(x1)n-if(x2)i,
for all x1,x2∈A. It is clear that f(x1n)=f(x1)n and f(x2n)=f(x2)n, so we obtain
(11)∑i=1n-1(ni)f(x1n-ix2i)=∑i=1n-1(ni)f(x1)n-if(x2)i,
for all x1,x2∈A. If n=2, then by (11) we have f(x1x2)=f(x1)f(x2). Now let n>2. Together with Lemma 3 and (11), we can say that the equality (4) holds for k=n-1; that is,
(12)∑i1=n-1n-1∑i2=n-2i1-1⋯∑in-1=1in-2-1(ni1)(i1i2)⋯(in-2in-1)×f(x1n-i1x2i1-i2x3i2-i3⋯xnin-1)=∑i1=n-1n-1∑i2=n-2i1-1⋯∑in-1=1in-2-1(ni1)(i1i2)⋯(in-2in-1)×f(x1)n-i1f(x2)i1-i2⋯f(xn)in-1,
holds for all x1,x2,x3,…,xn∈A. Notice that
(13)n-1>i1>i2>⋯>in-2>in-1≥1
implies i1=n-1, i2=n-2, …, in-2=2, in-1=1 and so
(14)n-i1,i1-i2,…,in-1-in-2,in-1=1.
Therefore we get the desired equality:
(15)f(x1x2x3⋯xn)=f(x1)f(x2)f(x3)⋯f(xn),
for all x1,x2,x3,…,xn∈A.


## 3. Generalization of Theorem2

We need the following lemmas to prove the generalization of Theorem2.Lemma 5 (see [9, Corollaries 2.5 and 3.5]).
LetV be a normed space, and let W be a Banach space. Assume that f,g,h:V→W are mappings such that
(16)∥f(x+y)-g(x)-h(y)∥≤ε(∥x∥p+∥y∥p),
for all x,y∈V∖{0}, where p≠1 and ε≥0. Then there exists a unique additive mapping T:V→W such that
(17)∥f(x)-T(x)-f(0)∥≤|4(3+3p)|ε|2p(3-3p)|∥x∥p,
for all x∈V∖{0}. In particular, T is given by
(18)T(x)=limm→∞f(3smx)-f(0)3sm,
for all x∈V∖{0}, where s:=-sgn(p-1).Lemma 6.
LetV,W,f,g,h,ε be as in Lemma 5. If p<0 and f(0)=0, then f is an additive mapping.Proof.
LetT:V→W be the additive mapping satisfying (17). Then we have
(19)∥2f(x)-2T(x)∥≤∥f(2(n+1)x)-T(2(n+1)x)∥+∥f(-2nx)-T(-2nx)∥+∥f(x)-g((n+1)x)-h(-nx)∥+∥f(x)-g(-nx)-h((n+1)x)∥+∥f(2(n+1)x)-g((n+1)x)-h((n+1)x)∥+∥f(-2nx)-g(-nx)-h(-nx)∥≤(4(3+3p)2p(3-3p)+4)(n+1)pε∥x∥p+(4(3+3p)2p(3-3p)+4)npε∥x∥p,
for all x∈V∖{0} and n∈ℕ. Taking the limit as n→∞, we get f(x)=T(x) as desired.The following result has already been proved in [7] (see also [8]). We show that it can also be derived from Lemma 6.Lemma 7.
LetV,W,ε be as in Lemma 5 and p<0. If f:V→W is a mapping such that
(20)∥f(x+y)-f(x)-f(y)∥≤ε(∥x∥p+∥y∥p)∀x,y∈V∖{0},
then f is an additive mapping.Proof.
By Lemma5, we can take an additive mapping T:V→W satisfying (17). Observe that
(21)∥f(0)∥≤∥f(nx)-T(nx)-f(0)∥+∥f(-nx)-T(-nx)-f(0)∥+∥f(0)-f(nx)-f(-nx)∥≤(8(3+3p)2p(3-3p)+2)npε∥x∥p,
for all x∈V∖{0} and for all n∈ℕ. Taking the limit as n→∞, we get f(0)=0. By Lemma 6, f is an additive mapping.Now we can prove the following theorem which is the generalization of Theorem2.Theorem 8.
LetA be a commutative normed algebra and B a commutative Banach algebra. Assume that f,g,h:A→B satisfy (16) and
(22)∥f(xn)-f(x)n∥≤δ∥x∥nq,
for all x∈A∖{0}, where δ≥0 and (p-1)(q-1)>0. If f(0)=0, then there exists a unique n-ring homomorphism T:A→B satisfying (17).Proof.
By Lemma5, there exists a unique additive mapping  T satisfying (17). By Theorem 4, it suffices to show that T(xn)=T(x)n. Put s:=-sgn(q-1). From the equality below (17) in Lemma 5, we have
(23)T(x)=limm→∞f(3smx)3sm,
for all x∈A∖{0}. It follows from (22) that
(24)∥T(xn)-T(x)n∥=limm→∞13smn{∥f((3smx)n)-(f(3smx))n∥}≤limm→∞δ3smn∥3smx∥nq=limm→∞(3smn(q-1))δ∥x∥nq=0,
for all x∈A∖{0}. Hence T is an n-Jordan homomorphism. By Theorem 4, T is an n-ring homomorphism.The following two corollaries give results on the hyperstability ofn-ring homomorphisms between Banach algebras.Corollary 9.
LetA,B,q,δ,f,g,h be as in Theorem 8. If f(0)=0 and p<0, then f is an n-ring homomorphism.Proof.
LetT be the unique n-ring homomorphism satisfying (17) in Theorem 8. By Lemma 6, f is the unique additive mapping satisfying (17). So f is the unique n-ring homomorphism.Corollary 10.
LetA,B,p,q be as in Corollary 9. Assume that f:A→B satisfies the system of functional inequalities (20) and (22) for all x∈A∖{0}. Then f is an n-ring homomorphism.Proof.
The proof is analogous as for Corollary9, with Lemma 6 replaced by Lemma 7.



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