Stability of 𝑛-Jordan Homomorphisms from a Normed Algebra to a Banach Algebra

Abstract

We establish the hyperstability of 𝑛-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.

1. Introduction

Let 𝐴,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(𝑎^𝑛)=(𝑔(𝑎))^𝑛 for all 𝑎∈𝐴 and an additive mapping ℎ∶𝐴→𝐵 is called an 𝑛-ring homomorphism if ℎ(∏^𝑛_𝑖=1𝑎_𝑖)=∏^𝑛_𝑖=1ℎ(𝑎_𝑖) for all 𝑎₁,𝑎₂,…,𝑎_𝑛∈𝐴.

In 2009, Gordji et al. [1] showed the following theorems.

Theorem 1. Let 𝑛∈{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. Let 𝑛∈{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: 𝑓(𝑎+𝑏)−𝑓(𝑎)−𝑓(𝑏)≤𝜀‖𝑎‖^𝑝+‖𝑏‖^𝑝,𝑓𝑎^𝑛−𝑓(𝑎)^𝑛≤𝛿‖𝑎‖^𝑛𝑞, (1) for all 𝑎,𝑏∈𝐴. Then, there exists a unique 𝑛-ring homomorphism ℎ∶𝐴→𝐵 such that 𝑓(𝑎)−ℎ(𝑎)≤2𝜀2−2^𝑝‖𝑎‖^𝑝, (2) 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 Theorems 1 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 Theorem 1

Lemma 3. Let 𝑛,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: ^𝑛−1_𝑖1^𝑖_=𝑘−1¹⁻¹_{𝑖2=𝑘−2}⋯^{𝑖𝑘−2−1}_{𝑖𝑘−1=1}𝑛𝑖₁𝑖₁𝑖₂⋯𝑖_𝑘−2𝑖_𝑘−1×𝑓𝑥^𝑛−𝑖1₁𝑥^{𝑖1−𝑖2}₂𝑥^{𝑖2−𝑖3}₃⋯𝑥^𝑖𝑘−1_𝑘=^𝑛−1_𝑖1^𝑖_=𝑘−1¹⁻¹_{𝑖2=𝑘−2}⋯^{𝑖𝑘−2−1}_{𝑖𝑘−1=1}𝑛𝑖₁𝑖₁𝑖₂⋯𝑖_𝑘−2𝑖_𝑘−1𝑓𝑥₁^𝑛−𝑖1×𝑓𝑥₂^{𝑖1−𝑖2}⋯𝑓𝑥_𝑘^𝑖𝑘−1, (3) for all 𝑥₁,𝑥₂,𝑥₃,…,𝑥_𝑘∈𝐴. Then one gets ^𝑛−1_𝑖1^𝑖_=𝑘¹⁻¹_{𝑖2=𝑘−1}⋯^{𝑖𝑘−1−1}_𝑖𝑘=1𝑛𝑖₁𝑖₁𝑖₂⋯𝑖_𝑘−1𝑖_𝑘×𝑓𝑥^𝑛−𝑖1₁𝑥^{𝑖1−𝑖2}₂⋯𝑥^𝑖𝑘_𝑘+1=^𝑛−1_𝑖1^𝑖_=𝑘¹⁻¹_{𝑖2=𝑘−1}⋯^{𝑖𝑘−1−1}_𝑖𝑘=1𝑛𝑖₁𝑖₁𝑖₂⋯𝑖_𝑘−1𝑖_𝑘𝑓𝑥₁^𝑛−𝑖1×𝑓𝑥₂^{𝑖1−𝑖2}⋯𝑓𝑥_𝑘+1^𝑖𝑘, (4) for all 𝑥₁,𝑥₂,𝑥₃,…,𝑥_𝑘+1∈𝐴.

Proof. Replacing 𝑥_𝑘 by 𝑥_𝑘+1 in (3), we obtain ^𝑛−1_{𝑖1=𝑘−1}⋯^{𝑖𝑘−2−1}_{𝑖𝑘−1=1}𝑛𝑖₁⋯𝑖_𝑘−2𝑖_𝑘−1𝑖_𝑘−1𝑖_𝑘−1×𝑓𝑥^𝑛−𝑖1₁⋯𝑥^{𝑖𝑘−2−𝑖𝑘−1}_𝑘−1𝑥^𝑖𝑘−1_𝑘+1=^𝑛−1_{𝑖1=𝑘−1}⋯^{𝑖𝑘−2−1}_{𝑖𝑘−1=1}𝑛𝑖₁⋯𝑖_𝑘−2𝑖_𝑘−1𝑖_𝑘−1𝑖_𝑘−1×𝑓𝑥₁^𝑛−𝑖1⋯𝑓𝑥_𝑘−1^{𝑖𝑘−2−𝑖𝑘−1}×𝑓𝑥_𝑘+1^𝑖𝑘−1, (5) for all 𝑥₁,𝑥₂,𝑥₃,…,𝑥_𝑘−1,𝑥_𝑘+1∈𝐴. In particular, the equality (3) implies that ^𝑛−1_{𝑖1=𝑘−1}⋯^{𝑖𝑘−2−1}_{𝑖𝑘−1=1}𝑛𝑖₁⋯𝑖_𝑘−2𝑖_𝑘−1𝑖_𝑘−10×𝑓𝑥^𝑛−𝑖1₁⋯𝑥^{𝑖𝑘−2−𝑖𝑘−1}_𝑘−1𝑥^𝑖𝑘−1_𝑘=^𝑛−1_{𝑖1=𝑘−1}⋯^{𝑖𝑘−2−1}_{𝑖𝑘−1=1}𝑛𝑖₁⋯𝑖_𝑘−2𝑖_𝑘−1𝑖_𝑘−10×𝑓𝑥₁^𝑛−𝑖1⋯𝑓𝑥_𝑘−1^{𝑖𝑘−2−𝑖𝑘−1}𝑓𝑥_𝑘^𝑖𝑘−1, (6) for all 𝑥₁,𝑥₂,𝑥₃,…,𝑥_𝑘∈𝐴. Recall that the equality, 𝑥_𝑘+𝑥_𝑘+1^𝑖𝑘−1=^𝑖𝑘−1_𝑖𝑘=0𝑖_𝑘−1𝑖_𝑘𝑥^{𝑖𝑘−1−𝑖𝑘}_𝑘𝑥^𝑖𝑘_𝑘+1, (7) holds for all 𝑥_𝑘,𝑥_𝑘+1∈𝐴. Replacing 𝑥_𝑘 by 𝑥_𝑘+𝑥_𝑘+1 in (3), we obtain ^𝑛−1_{𝑖1=𝑘−1}⋯^{𝑖𝑘−2−1}_𝑖𝑘−1^𝑖₌₁^𝑘−1_𝑖𝑘=0𝑛𝑖₁⋯𝑖_𝑘−2𝑖_𝑘−1𝑖_𝑘−1𝑖_𝑘×𝑓𝑥^𝑛−𝑖1₁⋯𝑥^{𝑖𝑘−2−𝑖𝑘−1}_𝑘−1𝑥^{𝑖𝑘−1−𝑖𝑘}_𝑘𝑥^𝑖𝑘_𝑘+1=^𝑛−1_𝑖1^𝑖_=𝑘−1¹⁻¹_{𝑖2=𝑘−2}⋯^{𝑖𝑘−2−1}_{𝑖𝑘−1=1}𝑛𝑖₁⋯𝑖_𝑘−2𝑖_𝑘−1×𝑓𝑥^𝑛−𝑖1₁⋯𝑥^{𝑖𝑘−2−𝑖𝑘−1}_𝑘−1×𝑥_𝑘+𝑥_𝑘+1^𝑖𝑘−1=^𝑛−1_𝑖1^𝑖_=𝑘−1¹⁻¹_{𝑖2=𝑘−2}⋯^{𝑖𝑘−2−1}_{𝑖𝑘−1=1}𝑛𝑖₁𝑖₁𝑖₂⋯𝑖_𝑘−2𝑖_𝑘−1×𝑓𝑥₁^𝑛−𝑖1⋯𝑓𝑥_𝑘+𝑥_𝑘+1^𝑖𝑘−1=^𝑛−1_𝑖1^𝑖_=𝑘−1¹⁻¹_{𝑖2=𝑘−2}⋯^{𝑖𝑘−2−1}_{𝑖𝑘−1=1}𝑛𝑖₁𝑖₁𝑖₂⋯𝑖_𝑘−2𝑖_𝑘−1×𝑓𝑥₁^𝑛−𝑖1⋯𝑓𝑥_𝑘+𝑓𝑥_𝑘+1^𝑖𝑘−1=^𝑛−1_{𝑖1=𝑘−1}⋯^{𝑖𝑘−2−1}_𝑖𝑘−1^𝑖₌₁^𝑘−1_𝑖𝑘=0𝑛𝑖₁⋯𝑖_𝑘−2𝑖_𝑘−1𝑖_𝑘−1𝑖_𝑘×𝑓𝑥₁^𝑛−𝑖1⋯𝑓𝑥_𝑘^{𝑖𝑘−1−𝑖𝑘}𝑓𝑥_𝑘+1^𝑖𝑘, (8) for all 𝑥₁,𝑥₂,𝑥₃,…,𝑥_𝑘,𝑥_𝑘+1∈𝐴. From (5), (6), and the above equality, we get the desired equality: ^𝑛−1_𝑖1^𝑖_=𝑘¹⁻¹_{𝑖2=𝑘−1}⋯^{𝑖𝑘−1−1}_𝑖𝑘=1𝑛𝑖₁𝑖₁𝑖₂⋯𝑖_𝑘−1𝑖_𝑘×𝑓𝑥^𝑛−𝑖1₁𝑥^{𝑖1−𝑖2}₂⋯𝑥^𝑖𝑘_𝑘+1=^𝑛−1_{𝑖1=𝑘−1}⋯^{𝑖𝑘−2−1}_𝑖𝑘−1^𝑖₌₁^𝑘−1_𝑖𝑘=0𝑛𝑖₁⋯𝑖_𝑘−2𝑖_𝑘−1𝑖_𝑘−1𝑖_𝑘×𝑓𝑥^𝑛−𝑖1₁⋯𝑥^{𝑖𝑘−2−𝑖𝑘−1}_𝑘−1×𝑥^{𝑖𝑘−1−𝑖𝑘}_𝑘𝑥^𝑖𝑘_𝑘+1−^𝑛−1_{𝑖1=𝑘−1}⋯^{𝑖𝑘−2−1}_{𝑖𝑘−1=1}𝑛𝑖₁⋯𝑖_𝑘−2𝑖_𝑘−1𝑖_𝑘−1𝑖_𝑘−1×𝑓𝑥^𝑛−𝑖1₁⋯𝑥^{𝑖𝑘−2−𝑖𝑘−1}_𝑘−1𝑥^𝑖𝑘−1_𝑘+1−^𝑛−1_{𝑖1=𝑘−1}⋯^{𝑖𝑘−2−1}_{𝑖𝑘−1=1}𝑛𝑖₁⋯𝑖_𝑘−2𝑖_𝑘−1𝑖_𝑘−10×𝑓𝑥^𝑛−𝑖1₁⋯𝑥^{𝑖𝑘−2−𝑖𝑘−1}_𝑘−1𝑥^𝑖𝑘−1_𝑘=^𝑛−1_{𝑖1=𝑘−1}⋯^{𝑖𝑘−2−1}_𝑖𝑘−1^𝑖₌₁^𝑘−1_𝑖𝑘=0𝑛𝑖₁⋯𝑖_𝑘−2𝑖_𝑘−1𝑖_𝑘−1𝑖_𝑘×𝑓𝑥₁^𝑛−𝑖1⋯𝑓𝑥_𝑘^{𝑖𝑘−1−𝑖𝑘}×𝑓𝑥_𝑘+1^𝑖𝑘−^𝑛−1_{𝑖1=𝑘−1}⋯^{𝑖𝑘−2−1}_{𝑖𝑘−1=1}𝑛𝑖₁⋯𝑖_𝑘−2𝑖_𝑘−1𝑖_𝑘−1𝑖_𝑘−1×𝑓𝑥₁^𝑛−𝑖1⋯𝑓𝑥_𝑘−1^{𝑖𝑘−2−𝑖𝑘−1}×𝑓𝑥_𝑘+1^𝑖𝑘−1−^𝑛−1_{𝑖1=𝑘−1}⋯^{𝑖𝑘−2−1}_{𝑖𝑘−1=1}𝑛𝑖₁⋯𝑖_𝑘−2𝑖_𝑘−1𝑖_𝑘−10×𝑓𝑥₁^𝑛−𝑖1⋯𝑓𝑥_𝑘−1^{𝑖𝑘−2−𝑖𝑘−1}×𝑓𝑥_𝑘^𝑖𝑘−1=^𝑛−1_𝑖1^𝑖_=𝑘¹⁻¹_{𝑖2=𝑘−1}⋯^{𝑖𝑘−1−1}_𝑖𝑘=1𝑛𝑖₁𝑖₁𝑖₂⋯𝑖_𝑘−1𝑖_𝑘×𝑓𝑥₁^𝑛−𝑖1𝑓𝑥₂^{𝑖1−𝑖2}⋯𝑓𝑥_𝑘+1^𝑖𝑘, (9) for all 𝑥₁,𝑥₂,𝑥₃,…,𝑥_𝑘,𝑥_𝑘+1∈𝐴.

The following theorem is the generalization of Theorem 1.

Theorem 4. Let 𝐴,𝐵 be two commutative algebras, and let 𝑓∶𝐴→𝐵 be an 𝑛-Jordan homomorphism. Then 𝑓 is an 𝑛-ring homomorphism.

Proof. Since 𝑓 is an 𝑛-Jordan homomorphism, together with the additivity of 𝑓, we get ^𝑛_𝑖=0𝑛𝑖𝑓𝑥₁^𝑛−𝑖𝑥^𝑖₂=𝑓𝑥₁+𝑥₂^𝑛=𝑓𝑥₁+𝑥₂^𝑛=𝑓𝑥₁+𝑓𝑥₂^𝑛=^𝑛_𝑖=0𝑛𝑖𝑓𝑥₁^𝑛−𝑖𝑓𝑥₂^𝑖, (10) for all 𝑥₁,𝑥₂∈𝐴. It is clear that 𝑓(𝑥^𝑛₁)=𝑓(𝑥₁)^𝑛 and 𝑓(𝑥^𝑛₂)=𝑓(𝑥₂)^𝑛, so we obtain ^𝑛−1_𝑖=1𝑛𝑖𝑓𝑥₁^𝑛−𝑖𝑥^𝑖₂=^𝑛−1_𝑖=1𝑛𝑖𝑓𝑥₁^𝑛−𝑖𝑓𝑥₂^𝑖, (11) 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, ^𝑛−1_𝑖1^𝑖_=𝑛−1¹⁻¹_{𝑖2=𝑛−2}⋯^{𝑖𝑛−2−1}_{𝑖𝑛−1=1}𝑛𝑖₁𝑖₁𝑖₂⋯𝑖_𝑛−2𝑖_𝑛−1×𝑓𝑥^𝑛−𝑖1₁𝑥^{𝑖1−𝑖2}₂𝑥^{𝑖2−𝑖3}₃⋯𝑥^𝑖𝑛−1_𝑛=^𝑛−1_𝑖1^𝑖_=𝑛−1¹⁻¹_{𝑖2=𝑛−2}⋯^{𝑖𝑛−2−1}_{𝑖𝑛−1=1}𝑛𝑖₁𝑖₁𝑖₂⋯𝑖_𝑛−2𝑖_𝑛−1×𝑓𝑥₁^𝑛−𝑖1𝑓𝑥₂^{𝑖1−𝑖2}⋯𝑓𝑥_𝑛^𝑖𝑛−1, (12) holds for all 𝑥₁,𝑥₂,𝑥₃,…,𝑥_𝑛∈𝐴. Notice that 𝑛−1>𝑖₁>𝑖₂>⋯>𝑖_𝑛−2>𝑖_𝑛−1≥1 (13) implies 𝑖₁=𝑛−1, 𝑖₂=𝑛−2, …, 𝑖_𝑛−2=2, 𝑖_𝑛−1=1 and so 𝑛−𝑖₁,𝑖₁−𝑖₂,…,𝑖_𝑛−1−𝑖_𝑛−2,𝑖_𝑛−1=1. (14) Therefore we get the desired equality: 𝑓𝑥₁𝑥₂𝑥₃⋯𝑥_𝑛=𝑓𝑥₁𝑓𝑥₂𝑓𝑥₃⋯𝑓𝑥_𝑛, (15) for all 𝑥₁,𝑥₂,𝑥₃,…,𝑥_𝑛∈𝐴.

3. Generalization of Theorem 2

We need the following lemmas to prove the generalization of Theorem 2.

Lemma 5 (see [9, Corollaries 2.5 and 3.5]). Let 𝑉 be a normed space, and let 𝑊 be a Banach space. Assume that 𝑓,𝑔,ℎ∶𝑉→𝑊 are mappings such that 𝑓𝑥+𝑦−𝑔(𝑥)−ℎ𝑦≤𝜀‖𝑥‖^𝑝+𝑦^𝑝, (16) for all 𝑥,𝑦∈𝑉⧵{0}, where 𝑝≠1 and 𝜀≥0. Then there exists a unique additive mapping 𝑇∶𝑉→𝑊 such that 𝑓(𝑥)(𝑥)(0≤4−𝑇−𝑓)3+3^𝑝𝜀2^𝑝3−3^𝑝‖𝑥‖^𝑝, (17) for all 𝑥∈𝑉⧵{0}. In particular, 𝑇 is given by 𝑇(𝑥)=lim_𝑚→∞𝑓3^𝑠𝑚𝑥−𝑓(0)3^𝑠𝑚, (18) for all 𝑥∈𝑉⧵{0}, where 𝑠∶=−sgn(𝑝−1).

Lemma 6. Let 𝑉,𝑊,𝑓,𝑔,ℎ,𝜀 be as in Lemma 5. If 𝑝<0 and 𝑓(0)=0, then 𝑓 is an additive mapping.

Proof. Let 𝑇∶𝑉→𝑊 be the additive mapping satisfying (17). Then we have (𝑥)(𝑥≤𝑓(2()𝑥)(2()𝑥+𝑓()(+𝑓(𝑥))𝑥)(+𝑓(𝑥)())𝑥+𝑓(2()𝑥))𝑥))𝑥+𝑓()()(≤42𝑓−2𝑇)𝑛+1−𝑇𝑛+1)−2𝑛𝑥−𝑇−2𝑛𝑥)−𝑔((𝑛+1−ℎ−𝑛𝑥)−𝑔−𝑛𝑥−ℎ((𝑛+1)𝑛+1−𝑔((𝑛+1−ℎ((𝑛+1)−2𝑛𝑥−𝑔−𝑛𝑥−ℎ−𝑛𝑥)3+3^𝑝2^𝑝3−3^𝑝+4(𝑛+1)^𝑝𝜀‖𝑥‖^𝑝+43+3^𝑝2^𝑝3−3^𝑝+4𝑛^𝑝𝜀‖𝑥‖^𝑝, (19) 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. Let 𝑉,𝑊,𝜀 be as in Lemma 5 and 𝑝<0. If 𝑓∶𝑉→𝑊 is a mapping such that 𝑓𝑥+𝑦−𝑓(𝑥)−𝑓𝑦≤𝜀‖𝑥‖^𝑝+𝑦^𝑝∀𝑥,𝑦∈𝑉⧵{0}, (20) then 𝑓 is an additive mapping.

Proof. By Lemma 5, we can take an additive mapping 𝑇∶𝑉→𝑊 satisfying (17). Observe that 𝑓(0≤𝑓()()(0+𝑓()()(0+𝑓(0)()(≤8)𝑛𝑥−𝑇𝑛𝑥−𝑓)−𝑛𝑥−𝑇−𝑛𝑥−𝑓)−𝑓𝑛𝑥−𝑓−𝑛𝑥)3+3^𝑝2^𝑝3−3^𝑝+2𝑛^𝑝𝜀‖𝑥‖^𝑝, (21) 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 Theorem 2.

Theorem 8. Let 𝐴 be a commutative normed algebra and 𝐵 a commutative Banach algebra. Assume that 𝑓,𝑔,ℎ∶𝐴→𝐵 satisfy (16) and 𝑓𝑥^𝑛−𝑓(𝑥)^𝑛≤𝛿‖𝑥‖^𝑛𝑞, (22) 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 Lemma 5, there exists a unique additive mapping  𝑇 satisfying (17). By Theorem 4, it suffices to show that 𝑇(𝑥^𝑛)=𝑇(𝑥)^𝑛. Put 𝑠∶=−sgn(𝑞−1). From the equality below (17) in Lemma 5, we have 𝑇(𝑥)=lim_𝑚→∞𝑓3^𝑠𝑚𝑥3^𝑠𝑚, (23) for all 𝑥∈𝐴⧵{0}. It follows from (22) that 𝑇𝑥^𝑛−𝑇(𝑥)^𝑛=lim_𝑚→∞13^𝑠𝑚𝑛𝑓3^𝑠𝑚𝑥^𝑛−𝑓3^𝑠𝑚𝑥^𝑛≤lim_𝑚→∞𝛿3^𝑠𝑚𝑛3^𝑠𝑚𝑥^𝑛𝑞=lim_𝑚→∞3^{𝑠𝑚𝑛(𝑞−1)}𝛿‖𝑥‖^𝑛𝑞=0, (24) 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 of 𝑛-ring homomorphisms between Banach algebras.

Corollary 9. Let 𝐴,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. Let 𝑇 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. Let 𝐴,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 Corollary 9, with Lemma 6 replaced by Lemma 7.

Acknowledgments

The author thanks the reviewers and the editor for their very helpful comments that improve the paper. This research was supported by Basic Science Research Program through the National Research Foundation of Korea (NRF) funded by the Ministry of Education, Science and Technology (no. 2012R1A1A4A01002971).