Stability of a Generalized Trigonometric-Quadratic Functional Equation

Abstract

A stability result for the generalized trigonometric-quadratic functional equation ℱ(𝑥+𝑦)−𝒢(𝑥−𝑦)=2ℋ(𝑥)𝒦(𝑦)+ℒ(𝑥)+ℳ(𝑦) over the domain of a normed space and the range of a Banach space is derived. It states that under appropriate conditions on the function ℋ(𝑥)𝒦(𝑦), the four functions ℱ(𝑥), 𝒢(𝑥), ℒ(𝑥), and ℳ(𝑥) are approximately equal to combinations of certain quadratic and additive functions.

1. Introduction

A generalized trigonometric functional equation is a functional equation of the form 𝐹𝑥+𝑦−𝐺𝑥−𝑦=2𝐻(𝑥)𝐾𝑦, (1) which is so named because the two best known trigonometric functions, sine and cosine, are solutions of special cases of this equation. In [1], the authors investigated the stability of (1), where 𝐹, 𝐺, 𝐻, and 𝐾 are nonzero functions from an abelian group (𝐺,+) to the complex fieldℂ. Later in the same year, a generalized trigonometric functional equation with either the function 𝐻 or 𝐾 being bounded was investigated to complement the above earlier work where such function boundedness was not treated. A generalized quadratic functional equation is a functional equation of the form 𝑓₁𝑥+𝑦+𝑓₂𝑥−𝑦=𝑓₃(𝑥)+𝑓₄𝑦, (2) which is so named because the quadratic function 𝑥² is a solution of one of its particular cases, namely, 𝑓𝑥+𝑦+𝑓𝑥−𝑦=2𝑓(𝑥)+2𝑓𝑦, (3) and any solution of (3) is referred to as a quadratic function. The stability of the quadratic functional equation was first proved by Skof [2] in 1983 for functions from a normed space to a Banach space. In 1984, Cholewa [3] demonstrated that Skof’s theorem is also valid if a normed space is replaced by an abelian group. Later, Czerwik [4] extended Cholewa’s result by changing a control function and considering functions from a normed space to a Banach space. In another paper, [5], Czerwik investigated the stability of the “partially pexiderized” quadratic functional equation of the form 𝑓(𝑥+𝑦)+𝑓(𝑥−𝑦)=𝑔(𝑥)+𝑔(𝑦). In 2000, Jung [6] considered (2), where𝑓₁, 𝑓₂, 𝑓₃, and 𝑓₄ are functions from a normed space 𝐸₁ to a Banach space 𝐸₂. In the paper [7], the authors considered the stability of a combined generalized trigonometric-quadratic functional equation of the form ℱ𝑥+𝑦+𝒢𝑥−𝑦=2ℋ(𝑥)𝒦𝑦+ℒ(𝑥)+ℳ𝑦 (4) over the domain of an abelian group (𝐺,+) and the range ℂ, except the control function 𝜙 whose range is taken to be the nonnegative real numbers [0,∞). There, the functions ℋ and 𝒦 are explicitly determined under certain restrictions on the remaining functions ℱ, 𝒢, ℒ, and ℳ. It is natural to ask whether the restrictions on the functions ℱ, 𝒢, ℒ, and ℳ can be removed and/or altered to obtain some stability result of (4). We answer this question affirmatively using Jung’s techniques elaborated in [6, 8].

2. Notation

We collect in this section the notation and terminology that will be kept standard throughout. A normed space 𝐸₁ is the domain and a Banach space (𝐸₂,‖⋅‖) is the range of all functions involved. Let ℋ,𝒦∶𝐸₁→𝐸₂. Define 𝐻𝐾𝑥,𝑦∶=ℋ(𝑥)𝒦𝑦−ℋ(0)𝒦(0),𝐻𝐾_𝑒𝑥,𝑦∶=𝐻𝐾𝑥,𝑦+𝐻𝐾−𝑥,−𝑦2,𝐻𝐾_𝑜𝑥,𝑦∶=𝐻𝐾𝑥,𝑦−𝐻𝐾−𝑥,−𝑦2. (5)Let 𝜑∶𝐸₁×𝐸₁→[0,∞) be a given function satisfying the following three conditions (i) 𝜑(𝑦,𝑥)=𝜑(𝑥,𝑦), (ii) 𝜑(𝑥,−𝑦)=𝜑(𝑥,𝑦), and (iii) there is an integer 𝑘≥2 such that ^∞_𝑖=01𝑘^𝑖𝜑𝑘^𝑖𝑥,𝑘^𝑖𝑦<∞,or^∞_𝑖=0𝑘^𝑖𝜑𝑥𝑘^𝑖,𝑦𝑘^𝑖<∞𝑥,𝑦∈𝐸₁. (6) For 𝑥,𝑦∈𝐸₁, define 𝜑_𝑒𝑥,𝑦=2𝜑𝑥,𝑦+4𝜑(𝑥,0)+3𝜑𝑥+𝑦,0+3𝜑𝑥−𝑦,0+6𝜑𝑥+𝑦2,𝑥+𝑦2+2𝜑𝑥−𝑦2,𝑥−𝑦2(),𝜑+20𝜑0,0_𝑜()()(),𝜓𝑥,𝑦=2𝜑𝑥,𝑦+2𝜑𝑥,𝑥+2𝜑𝑦,𝑦+2𝜑2𝑥,0+2𝜑2𝑦,0+2𝜑𝑥+𝑦,0+2𝜑𝑥−𝑦,0+2𝜑0,0_𝑘𝑥,𝑦=^∞_𝑖=01𝑘^2𝑖𝜑𝑘^𝑖𝑥,𝑘^𝑖𝑦,𝜓_𝑘𝑥,𝑦=^∞_𝑖=0𝑘^2(𝑖+1)𝜑𝑥𝑘^𝑖+1,𝑦𝑘^𝑖+1,Φ_𝑘𝑥,𝑦=^∞_𝑖=01𝑘^{2𝑖𝑘−1}_𝑗=1𝑏_𝑗𝜑_𝑒𝑘^𝑖𝑘−𝑗𝑥,𝑘^𝑖𝑦,Φ_𝑘𝑥,𝑦=^∞_𝑖=0𝑘^{2(𝑖+1)𝑘−1}_𝑗=1𝑏_𝑗𝜑_𝑒𝑘−𝑗𝑥𝑘^𝑖+1,𝑦𝑘^𝑖+1,Φ^′_𝑘𝑥,𝑦=^∞_𝑖=01𝑘^𝑖𝑘−1_𝑗=1𝜑_𝑜𝑘^𝑖𝑗𝑥+𝑦2,𝑘^𝑖𝑗𝑥−𝑦2,Φ^′_𝑘𝑥,𝑦=^∞_𝑖=0𝑘^{𝑖+1𝑘−1}_𝑗=1𝜑_𝑜𝑗𝑥+𝑦2𝑘^𝑖+1,𝑗𝑥−𝑦2𝑘^𝑖+1,Φ^′′_𝑘𝑥,𝑦=^∞_𝑖=01𝑘^𝑖𝑘−1_𝑗=1𝜑_𝑜𝑘^𝑖𝑦𝑗𝑥+2,𝑘^𝑖𝑦2,Φ^′′_𝑘𝑥,𝑦=^∞_𝑖=0𝑘^{𝑖+1𝑘−1}_𝑗=1𝜑_𝑜2𝑗𝑥+𝑦2𝑘^𝑖+1,𝑦2𝑘^𝑖+1,𝜓_{𝐻𝐾,𝑘}=𝑥,𝑦_∞𝑖=01𝑘^2𝑖2𝐻𝐾_𝑒𝑘^𝑖𝑥,𝑘^𝑖𝑦+𝐻𝐾_𝑒𝑘^𝑖𝑥+𝑦2𝑘,−^𝑖𝑥+𝑦2−𝐻𝐾_𝑒𝑘^𝑖𝑥+𝑦2,𝑘^𝑖𝑥+𝑦2+𝐻𝐾_𝑒𝑘^𝑖𝑥−𝑦2,𝑘^𝑖𝑥−𝑦2−𝐻𝐾_𝑒𝑘^𝑖𝑥−𝑦2𝑘,−^𝑖𝑥−𝑦2+2𝐻𝐾_𝑒0,𝑘^𝑖𝑥−2𝐻𝐾_𝑒𝑘^𝑖𝑥,0−𝐻𝐾_𝑒0,𝑘^𝑖𝑥+𝑦−𝐻𝐾_𝑒0,𝑘^𝑖,𝜓𝑥−𝑦_{𝐻𝐾,𝑘}=𝑥,𝑦_∞𝑖=0𝑘^2(𝑖+1)2𝐻𝐾_𝑒𝑥𝑘^𝑖+1,𝑦𝑘^𝑖+1+𝐻𝐾_𝑒𝑥+𝑦2𝑘^𝑖+1,−𝑥+𝑦2𝑘^𝑖+1−𝐻𝐾_𝑒𝑥+𝑦2𝑘^𝑖+1,𝑥+𝑦2𝑘^𝑖+1+𝐻𝐾_𝑒𝑥−𝑦2𝑘^𝑖+1,𝑥−𝑦2𝑘^𝑖+1−𝐻𝐾_𝑒𝑥−𝑦2𝑘^𝑖+1,−𝑥−𝑦2𝑘^𝑖+1+2𝐻𝐾_𝑒𝑥0,𝑘^𝑖+1−2𝐻𝐾_𝑒𝑥𝑘^𝑖+1,0−𝐻𝐾_𝑒0,𝑥+𝑦𝑘^𝑖+1−𝐻𝐾_𝑒0,𝑥−𝑦𝑘^𝑖+1,Φ_{𝐻𝐾,𝑘}=2𝑥,𝑦𝑘^2∞_𝑖=01𝑘^{2𝑖𝑘−1}_𝑗=12𝐻𝐾_𝑒𝑘−𝑗𝑘^𝑖𝑥,𝑘^𝑖𝑦+𝐻𝐾_𝑒𝑘−𝑗𝑥+𝑦2𝑘^𝑖,−𝑘−𝑗𝑥+𝑦2𝑘^𝑖−𝐻𝐾_𝑒𝑘−𝑗𝑥+𝑦2𝑘^𝑖,𝑘−𝑗𝑥+𝑦2𝑘^𝑖+𝐻𝐾_𝑒𝑘−𝑗𝑥−𝑦2𝑘^𝑖,𝑘−𝑗𝑥−𝑦2𝑘^𝑖−𝐻𝐾_𝑒𝑘−𝑗𝑥−𝑦2𝑘^𝑖,−𝑘−𝑗𝑥−𝑦2𝑘^𝑖+2𝐻𝐾_𝑒0,𝑘−𝑗𝑘^𝑖𝑥−2𝐻𝐾_𝑒𝑘−𝑗𝑘^𝑖𝑥,0−𝐻𝐾_𝑒0,𝑘−𝑗𝑥+𝑦𝑘^𝑖−𝐻𝐾_𝑒0,𝑘−𝑗𝑥−𝑦𝑘^𝑖,Φ_{𝐻𝐾,𝑘}=2𝑥,𝑦𝑘^2∞_𝑖=01𝑘^{2𝑖𝑘−1}_𝑗=12𝐻𝐾_𝑒𝑗𝑘^𝑖𝑥,𝑘^𝑖𝑦+𝐻𝐾_𝑒𝑗𝑥+𝑦2𝑘^𝑖,−𝑗𝑥+𝑦2𝑘^𝑖−𝐻𝐾_𝑒𝑗𝑥+𝑦2𝑘^𝑖,𝑗𝑥+𝑦2𝑘^𝑖+𝐻𝐾_𝑒𝑗𝑥−𝑦2𝑘^𝑖,𝑗𝑥−𝑦2𝑘^𝑖−𝐻𝐾_𝑒𝑗𝑥−𝑦2𝑘^𝑖,−𝑗𝑥−𝑦2𝑘^𝑖+2𝐻𝐾_𝑒0,𝑗𝑘^𝑖𝑥−2𝐻𝐾_𝑒𝑗𝑘^𝑖𝑥,0−𝐻𝐾_𝑒0,𝑗𝑥+𝑦𝑘^𝑖−𝐻𝐾_𝑒0,𝑗𝑥−𝑦𝑘^𝑖,Φ^′_{𝐻𝐾,𝑘}=2𝑥,𝑦𝑘^∞_𝑖=01𝑘^𝑖𝑘−1_𝑗=1𝐻𝐾_𝑜𝑗𝑥+𝑦2𝑘^𝑖,𝑗𝑥−𝑦2𝑘^𝑖−𝐻𝐾_𝑜−𝑗𝑥+𝑦2𝑘^𝑖,𝑗𝑥−𝑦2𝑘^𝑖−𝐻𝐾_𝑜𝑗𝑥+𝑦2𝑘^𝑖,𝑗𝑥+𝑦2𝑘^𝑖+𝐻𝐾_𝑜−𝑗𝑥+𝑦2𝑘^𝑖,𝑗𝑥+𝑦2𝑘^𝑖+𝐻𝐾_𝑜𝑗𝑥+𝑦𝑘^𝑖,0−𝐻𝐾_𝑜𝑗𝑘^𝑖𝑥,0−𝐻𝐾_𝑜𝑘^𝑖,Φ𝑥,0_{′𝐻𝐾,𝑘}=2𝑥,𝑦𝑘^∞_𝑖=0𝑘^{𝑖+1𝑘−1}_𝑗=1𝐻𝐾_𝑜𝑗𝑥+𝑦2𝑘^𝑖+1,𝑗𝑥−𝑦2𝑘^𝑖+1−𝐻𝐾_𝑜−𝑗𝑥+𝑦2𝑘^𝑖+1,𝑗𝑥−𝑦2𝑘^𝑖+1−𝐻𝐾_𝑜𝑗𝑥+𝑦2𝑘^𝑖+1,𝑗𝑥+𝑦2𝑘^𝑖+1+𝐻𝐾_𝑜−𝑗𝑥+𝑦2𝑘^𝑖+1,𝑗𝑥+𝑦2𝑘^𝑖+1+𝐻𝐾_𝑜𝑗𝑥+𝑦𝑘^𝑖+1,0−𝐻𝐾_𝑜𝑗𝑥𝑘^𝑖+1,0−𝐻𝐾_𝑜𝑥𝑘^𝑖+1,0. (7)

Define a real sequence (𝑏_𝑖) by 𝑏₁=1,𝑏₂=2,𝑏_𝑖=2𝑏_𝑖−1+𝑏_𝑖−2(𝑖≥3), (8) and let Ψ_𝑘𝑥,𝑦=𝑏₁𝜓_𝑘(𝑘−1)𝑥,𝑦+𝑏₂𝜓_𝑘(𝑘−2)𝑥,𝑦+⋯+𝑏_𝑘−1𝜓_𝑘,Ψ𝑥,𝑦_𝑘𝑥,𝑦=𝑏₁𝜓_𝑘(𝑘−1)𝑥,𝑦+𝑏₂𝜓_𝑘(𝑘−2)𝑥,𝑦+⋯+𝑏_𝑘−1𝜓_𝑘,Ψ𝑥,𝑦_{𝐻𝐾,𝑘}(𝑥,𝑥)=𝑏₁𝜓_{𝐻𝐾,𝑘}(𝑘−1)𝑥,𝑦+𝑏₂𝜓_{𝐻𝐾,𝑘}(𝑘−2)𝑥,𝑦+⋯+𝑏_𝑘−1𝜓_{𝐻𝐾,𝑘},Ψ𝑥,𝑦_{𝐻𝐾,𝑘}(𝑥,𝑥)=𝑏₁𝜓_{𝐻𝐾,𝑘}(𝑘−1)𝑥,𝑦+𝑏₂𝜓_{𝐻𝐾,𝑘}(𝑘−2)𝑥,𝑦+⋯+𝑏_𝑘−1𝜓_{𝐻𝐾,𝑘}𝑥,𝑦. (9) Clearly, the functions 𝜑_𝑒, 𝜑_𝑜 satisfy the conditions (i), (ii), and (iii).

3. Auxiliary Results

We start with an important lemma.

Lemma 1. If 𝑓∶𝐸₁→𝐸₂, 𝑓(0)=0, satisfies 𝑓𝑥+𝑦+𝑓𝑥−𝑦−2𝑓(𝑥)−2𝑓𝑦≤𝜑_𝑒𝑥,𝑦+22𝐻𝐾_𝑒𝑥,𝑦+𝐻𝐾_𝑒𝑥+𝑦2,−𝑥+𝑦2−𝐻𝐾_𝑒𝑥+𝑦2,𝑥+𝑦2+𝐻𝐾_𝑒𝑥−𝑦2,𝑥−𝑦2−𝐻𝐾_𝑒𝑥−𝑦2,−𝑥−𝑦2+2𝐻𝐾_𝑒(0,𝑥)−2𝐻𝐾_𝑒(𝑥,0)−𝐻𝐾_𝑒0,𝑥+𝑦−𝐻𝐾_𝑒0,𝑥−𝑦, (10) then one has, for integer 𝑘≥2, 𝑥∈𝐸₁ and 𝑛∈ℕ, 𝑓𝑘^𝑛𝑥−𝑘^2𝑛𝑓(𝑥≤_{𝑛−1𝑖=0})𝑘^{2(𝑛−1−𝑖)𝑘−1}_𝑗=1𝑏_𝑗𝜑𝑘−𝑗𝑘^𝑖𝑥,𝑘^𝑖𝑥+2^𝑛−1_𝑖=0𝑘^{2(𝑛−1−𝑖)}×^𝑘−1_𝑗=1𝑏_𝑗2𝐻𝐾_𝑒𝑘−𝑗𝑘^𝑖𝑥,𝑘^𝑖𝑥+𝐻𝐾_𝑒𝑘−𝑗+1𝑘^𝑖𝑥2𝑘,−𝑘−𝑗+1^𝑖𝑥2−𝐻𝐾_𝑒𝑘−𝑗+1𝑘^𝑖𝑥2,𝑘−𝑗+1𝑘^𝑖𝑥2+𝐻𝐾_𝑒𝑘−𝑗−1𝑘^𝑖𝑥2,𝑘−𝑗−1𝑘^𝑖𝑥2−𝐻𝐾_𝑒𝑘−𝑗−1𝑘^𝑖𝑥2𝑘,−𝑘−𝑗−1^𝑖𝑥2+2𝐻𝐾_𝑒0,𝑘−𝑗𝑘^𝑖𝑥−2𝐻𝐾_𝑒𝑘−𝑗𝑘^𝑖𝑥,0−𝐻𝐾_𝑒0,𝑘−𝑗+1𝑘^𝑖𝑥−𝐻𝐾_𝑒0,𝑘−𝑗−1𝑘^𝑖𝑥. (11)

Proof. We prove the lemma by induction on 𝑛. To verify the case 𝑛=1, we use induction on 𝑘≥2. Putting 𝑦=𝑥 in (10), we obtain 𝑓(2𝑥)−2²𝑓(𝑥)≤𝜑_𝑒(𝑥,𝑥)+2𝐻𝐾_𝑒(𝑥,𝑥)+𝐻𝐾_𝑒(𝑥,−𝑥)+2𝐻𝐾_𝑒(0,𝑥)−2𝐻𝐾_𝑒(𝑥,0)−𝐻𝐾_𝑒(0,2𝑥); (12) that is, the assertion is true for 𝑘=2. Assume that the statement is true for all 𝑚∈{2,…,𝑘}. Replacing 𝑥 and 𝑦 in (10) by 𝑘𝑥 and 𝑥, we get 𝑓((𝑘+1)𝑥)+𝑓((𝑘−1)𝑥)−2𝑓(𝑘𝑥)−2𝑓(𝑥)≤𝜑_𝑒(𝑘𝑥,𝑥)+22𝐻𝐾_𝑒(𝑘𝑥,𝑥)+𝐻𝐾_𝑒(𝑘+1)𝑥2()𝑥,−𝑘+12−𝐻𝐾_𝑒(𝑘+1)𝑥2,(𝑘+1)𝑥2+𝐻𝐾_𝑒(𝑘−1)𝑥2,(𝑘−1)𝑥2−𝐻𝐾_𝑒(𝑘−1)𝑥2()𝑥,−𝑘−12+2𝐻𝐾_𝑒(0,𝑘𝑥)−2𝐻𝐾_𝑒(𝑘𝑥,0)−𝐻𝐾_𝑒(0,(𝑘+1)𝑥)−𝐻𝐾_𝑒(0,(𝑘−1)𝑥), (13) and so 𝑓((𝑘+1)𝑥)−(𝑘+1)²𝑓(𝑥≤𝑓)𝑥))𝑥)()(𝑥𝑓())((𝑘+1+𝑓((𝑘−1−2𝑓𝑘𝑥−2𝑓)+2𝑘𝑥−𝑘²𝑓(𝑥+𝑓)𝑥)−())((𝑘−1𝑘−1²𝑓(𝑥≤_𝑘𝑗=1)𝑏_𝑗𝜑_𝑒𝑘+1−𝑗𝑥,𝑥+2^𝑘_𝑗=1𝑏_𝑖2𝐻𝐾_𝑒𝑘+1−𝑗𝑥,𝑥+𝐻𝐾_𝑒𝑘+2−𝑗𝑥2𝑥,−𝑘+2−𝑗2−𝐻𝐾_𝑒𝑘+2−𝑗𝑥2,𝑘+2−𝑗𝑥2+𝐻𝐾_𝑒𝑘−𝑗𝑥2,𝑘−𝑗𝑥2−𝐻𝐾_𝑒𝑘−𝑗𝑥2𝑥,−𝑘−𝑗2+2𝐻𝐾_𝑒0,𝑘+1−𝑗𝑥−2𝐻𝐾_𝑒𝑘+1−𝑗𝑥,0−𝐻𝐾_𝑒0,𝑘+2−𝑗𝑥−𝐻𝐾_𝑒0,𝑘−𝑗𝑥, (14) and we are done for 𝑛=1. Assuming that the lemma is true for some 𝑛≥1, we have 𝑓𝑘^𝑛+1𝑥−𝑘^2(𝑛+1)𝑓(𝑥≤𝑓𝑘_𝑛+1)𝑥−𝑘²𝑓𝑘^𝑛𝑥+𝑘²𝑓𝑘^𝑛𝑥−𝑘^2𝑛𝑓(𝑥≤_𝑛𝑖=0)𝑘^{2(𝑛−𝑖)𝑘−1}_𝑗=1𝑏_𝑗𝜑𝑘−𝑗𝑘^𝑖𝑥,𝑘^𝑖𝑥+2^𝑛_𝑖=0𝑘^{2(𝑛−𝑖)}×^𝑘−1_𝑗=1𝑏_𝑗2𝐻𝐾_𝑒𝑘−𝑗𝑘^𝑖𝑥,𝑘^𝑖𝑥+𝐻𝐾_𝑒𝑘−𝑗+1𝑘^𝑖𝑥2𝑘,−𝑘−𝑗+1^𝑖𝑥2−𝐻𝐾_𝑒𝑘−𝑗+1𝑘^𝑖𝑥2,𝑘−𝑗+1𝑘^𝑖𝑥2+𝐻𝐾_𝑒𝑘−𝑗−1𝑘^𝑖𝑥2,𝑘−𝑗−1𝑘^𝑖𝑥2−𝐻𝐾_𝑒𝑘−𝑗−1𝑘^𝑖𝑥2𝑘,−𝑘−𝑗−1^𝑖𝑥2+2𝐻𝐾_𝑒0,𝑘−𝑗𝑘^𝑖𝑥−2𝐻𝐾_𝑒𝑘−𝑗𝑘^𝑖𝑥,0−𝐻𝐾_𝑒0,𝑘−𝑗+1𝑘^𝑖𝑥−𝐻𝐾_𝑒0,𝑘−𝑗−1𝑘^𝑖𝑥, (15) verifying the lemma in general.

We come now to our first main proposition.

Proposition 2. Assume that 𝑓∶𝐸₁→𝐸₂ satisfies 𝑓𝑥+𝑦+𝑓𝑥−𝑦−2𝑓(𝑥)−2𝑓𝑦≤𝜑_𝑒𝑥,𝑦+22𝐻𝐾_𝑒𝑥,𝑦+𝐻𝐾_𝑒𝑥+𝑦2,−𝑥+𝑦2−𝐻𝐾_𝑒𝑥+𝑦2,𝑥+𝑦2+𝐻𝐾_𝑒𝑥−𝑦2,𝑥−𝑦2−𝐻𝐾_𝑒𝑥−𝑦2,−𝑥−𝑦2+2𝐻𝐾_𝑒(0,𝑥)−2𝐻𝐾_𝑒(𝑥,0)−𝐻𝐾_𝑒0,𝑥+𝑦−𝐻𝐾_𝑒0,𝑥−𝑦. (16) When 𝜓_𝑘(𝑥,𝑦)<∞ for all 𝑥,𝑦∈𝐸₁, one further assumes that 𝑓(0)=0. Then there exists a unique quadratic function 𝑄∶𝐸₁→𝐸₂ satisfying 𝑄(𝑥)(𝑥≤1−𝑓)𝑘²Ψ_𝑘()+2𝑥,𝑥𝑘²Ψ_{𝐻𝐾,𝑘}(𝑥,𝑥) (17) provided that 𝜓_𝑘(𝑥,𝑦)<∞ and 𝜓_{𝐻𝐾,𝑘}(𝑥,𝑦)<∞ for all 𝑥,𝑦∈𝐸₁, or 𝑄(𝑥)(𝑥≤1−𝑓)𝑘²Ψ_𝑘()+2𝑥,𝑥𝑘²Ψ_{𝐻𝐾,𝑘}(𝑥,𝑥) (18) provided that 𝜓_𝑘(𝑥,𝑦)<∞ and 𝜓_{𝐻𝐾,𝑘}(𝑥,𝑥)<∞ for all 𝑥,𝑦∈𝐸₁. Moreover, if 𝑓 is measurable or 𝑓(𝑡𝑥) is continuous in 𝑡∈ℝ for every fixed 𝑥∈𝐸₁, then the function 𝑄 satisfies 𝑄(𝑡𝑥)=𝑡²𝑄(𝑥)𝑥∈𝐸₁,𝑡∈ℝ. (19)

Proof. From Lemma 1, we have 1𝑘^2𝑛𝑓𝑘^𝑛𝑥(𝑥≤1−𝑓)𝑘^2𝑛−1_𝑖=01𝑘^{2𝑖𝑘−1}_𝑗=1𝑏_𝑗𝜑𝑘−𝑗𝑘^𝑖𝑥,𝑘^𝑖𝑥+2𝑘^2𝑛−1_𝑖=01𝑘^2𝑖×^𝑘−1_𝑗=1𝑏_𝑗2𝐻𝐾_𝑒𝑘−𝑗𝑘^𝑖𝑥,𝑘^𝑖𝑥+𝐻𝐾_𝑒𝑘−𝑗+1𝑘^𝑖𝑥2𝑘,−𝑘−𝑗+1^𝑖𝑥2−𝐻𝐾_𝑒𝑘−𝑗+1𝑘^𝑖𝑥2,𝑘−𝑗+1𝑘^𝑖𝑥2+𝐻𝐾_𝑒𝑘−𝑗−1𝑘^𝑖𝑥2,𝑘−𝑗−1𝑘^𝑖𝑥2−𝐻𝐾_𝑒𝑘−𝑗−1𝑘^𝑖𝑥2𝑘,−𝑘−𝑗−1^𝑖𝑥2+2𝐻𝐾_𝑒0,𝑘−𝑗𝑘^𝑖𝑥−2𝐻𝐾_𝑒𝑘−𝑗𝑘^𝑖𝑥,0−𝐻𝐾_𝑒0,𝑘−𝑗+1𝑘^𝑖𝑥−𝐻𝐾_𝑒0,𝑘−𝑗−1𝑘^𝑖𝑥. (20) For 𝑚,𝑛∈ℕ with 𝑛>𝑚, from (20), we obtain 1𝑘^2𝑛𝑓𝑘^𝑛𝑥−1𝑘^2𝑚𝑓𝑘^𝑚𝑥=1𝑘^2𝑚1𝑘^{2(𝑛−𝑚)}𝑓𝑘^𝑛−𝑚𝑘^𝑚𝑥−𝑓𝑘^𝑚𝑥≤1𝑘^2𝑘−1_𝑗=1𝑏_𝑗^𝑛−1_𝑖=𝑚1𝑘^2𝑖𝜑𝑘−𝑗𝑘^𝑖𝑥,𝑘^𝑖𝑥+2𝑘^2𝑘−1_𝑗=1𝑏_𝑗×^𝑛−1_𝑖=𝑚1𝑘^2𝑖2𝐻𝐾_𝑒𝑘−𝑗𝑘^𝑖𝑥,𝑘^𝑖𝑥+𝐻𝐾_𝑒𝑘−𝑗+1𝑘^𝑖𝑥2𝑘,−𝑘−𝑗+1^𝑖𝑥2−𝐻𝐾_𝑒𝑘−𝑗+1𝑘^𝑖𝑥2,𝑘−𝑗+1𝑘^𝑖𝑥2+𝐻𝐾_𝑒𝑘−𝑗−1𝑘^𝑖𝑥2,𝑘−𝑗−1𝑘^𝑖𝑥2−𝐻𝐾_𝑒𝑘−𝑗−1𝑘^𝑖𝑥2𝑘,−𝑘−𝑗−1^𝑖𝑥2+2𝐻𝐾_𝑒0,𝑘−𝑗𝑘^𝑖𝑥−2𝐻𝐾_𝑒𝑘−𝑗𝑘^𝑖𝑥,0−𝐻𝐾_𝑒0,𝑘−𝑗+1𝑘^𝑖𝑥−𝐻𝐾_𝑒0,𝑘−𝑗−1𝑘^𝑖𝑥. (21) In (21), since the last term can be made as small as we wish by choosing 𝑚 sufficiently large, we deduce that {(1/𝑘^2𝑛)𝑓(𝑘^𝑛𝑥)} is a Cauchy sequence in 𝐸₂. Since 𝐸₂ is a Banach space, the limit 𝑄(𝑥)=lim_𝑛→∞1𝑘^2𝑛𝑓𝑘^𝑛𝑥𝑥∈𝐸₁ (22) is well defined. Using 𝑄 and (20), the validity of the inequality (17) is easily verified. Next, replacing 𝑥, 𝑦 in (16) by 𝑘^𝑛𝑥, 𝑘^𝑛𝑦, respectively, dividing the resulting inequality by 𝑘^2𝑛, and using 𝜓_𝑘(𝑥,𝑦)<∞, 𝜓_{𝐻𝐾,𝑘}(𝑥,𝑦)<∞, we get 1𝑘^2𝑛𝑓𝑘^𝑛+1𝑥+𝑦𝑘^2𝑛𝑓𝑘^𝑛−2𝑥−𝑦𝑘^2𝑛𝑓𝑘^𝑛𝑥−2𝑘^2𝑛𝑓𝑘^𝑛𝑦≤1𝑘^2𝑛𝜑𝑘^𝑛𝑥,𝑘^𝑛𝑦+2𝑘^2𝑛×2𝐻𝐾_𝑒𝑘^𝑛𝑥,𝑘^𝑛𝑦+𝐻𝐾_𝑒𝑘^𝑛𝑥+𝑦2𝑘,−^𝑛𝑥+𝑦2−𝐻𝐾_𝑒𝑘^𝑛𝑥+𝑦2,𝑘^𝑛𝑥+𝑦2+𝐻𝐾_𝑒𝑘^𝑛𝑥−𝑦2,𝑘^𝑛𝑥−𝑦2−𝐻𝐾_𝑒𝑘^𝑛𝑥−𝑦2𝑘,−^𝑛𝑥−𝑦2+2𝐻𝐾_𝑒0,𝑘^𝑛𝑥−2𝐻𝐾_𝑒𝑘^𝑛𝑥,0−𝐻𝐾_𝑒0,𝑘^𝑛𝑥+𝑦−𝐻𝐾_𝑒0,𝑘^𝑛𝑥−𝑦⟶0(𝑛⟶∞), (23) which implies that 𝑄 is a quadratic function. If 𝑄₁∶𝐸₁→𝐸₂ is another quadratic function satisfying the inequality (17), since 𝑄 and 𝑄₁ are quadratic functions, we have 𝑄𝑘^𝑛𝑥=𝑘^2𝑛𝑄(𝑥),𝑄₁𝑘^𝑛𝑥=𝑘^2𝑛𝑄₁(𝑥)(𝑛∈ℕ). (24) It thus follows from (24), (17), 𝜓_𝑘(𝑥,𝑦)<∞, and 𝜓_{𝐻𝐾,𝑘}(𝑥,𝑦)<∞ that 𝑄(𝑥)−𝑄₁(𝑥=1)𝑘^2𝑛𝑄𝑘^𝑛𝑥−𝑄₁𝑘^𝑛𝑥≤1𝑘^2𝑛2𝑘²Ψ_𝑘𝑘^𝑛𝑥,𝑘^𝑛𝑥+1𝑘^2𝑛2𝑘²Ψ_𝑘𝑘^𝑛𝑥,𝑘^𝑛𝑥⟶0(𝑛⟶∞); (25) that is, 𝑄(𝑥)=𝑄₁(𝑥).
The other provision 𝜓_𝑘(𝑥,𝑦)<∞ and 𝜓_{𝐻𝐾,𝑘}(𝑥,𝑥)<∞ is treated analogously by replacing 𝑥 by 𝑥/𝑘^𝑛 in (11) to obtain an upper bound for ‖𝑓(𝑥)−𝑘^2𝑛𝑓(𝑥/𝑘^𝑛)‖, which is then used to show that {𝑘^2𝑛𝑓(𝑥/𝑘^𝑛)} is a Cauchy sequence in 𝐸₂ so that the limit 𝑄(𝑥)=lim_𝑛→∞𝑘^2𝑛𝑓𝑥𝑘^𝑛𝑥∈𝐸₁ (26) is well defined and satisfies (18). Finally, the proof that (19) holds when 𝑓 is measurable or 𝑓(𝑡𝑥) is continuous is the same as that in [4].

We now proceed to our second main proposition.

Proposition 3. If 𝑓∶𝐸₁→𝐸₂ satisfies 𝑓𝑥+𝑦−𝑓(𝑥)−𝑓𝑦≤𝜑_𝑜𝑥,𝑦+2𝐻𝐾_𝑜𝑥+𝑦2,𝑥−𝑦2−𝐻𝐾_𝑜−𝑥+𝑦2,𝑥−𝑦2−𝐻𝐾_𝑜𝑥+𝑦2,𝑥+𝑦2+𝐻𝐾_𝑜−𝑥+𝑦2,𝑥+𝑦2+𝐻𝐾_𝑜𝑥+𝑦,0−𝐻𝐾_𝑜(𝑥,0)−𝐻𝐾_𝑜𝑦,0, (27) then there exists a unique additive function 𝐴∶𝐸₁→𝐸₂ such that either 𝐴(𝑥)(𝑥≤1−𝑓)𝑘Φ^′_𝑘(𝑥,𝑥)+Φ^′_{𝐻𝐾,𝑘}(𝑥,𝑥) (28) provided that for 𝑥,𝑦∈𝐸₁, one has ∑^∞_𝑖=0(1/𝑘^𝑖)𝜑(𝑘^𝑖𝑥,𝑘^𝑖𝑦)<∞ and ^∞_𝑖=0𝐻𝐾_𝑜𝑘^𝑖𝑥+𝑦,𝑘^𝑖𝑥−𝑦−𝐻𝐾_𝑜−𝑘^𝑖𝑥+𝑦,𝑘^𝑖𝑥−𝑦−𝐻𝐾_𝑜𝑘^𝑖𝑥+𝑦,𝑘^𝑖𝑥+𝑦+𝐻𝐾_𝑜−𝑘^𝑖𝑥+𝑦,𝑘^𝑖𝑥+𝑦+𝐻𝐾_𝑜𝑘^𝑖𝑥+𝑦,0−𝐻𝐾_𝑜2𝑘^𝑖𝑥,0−𝐻𝐾_𝑜2𝑘^𝑖𝑦,0<∞ (29) or 𝐴(𝑥)(𝑥≤1−𝑓)𝑘Φ^′_𝑘(𝑥,𝑥)+Φ^′_{𝐻𝐾,𝑘}(𝑥,𝑥) (30) provided that for 𝑥,𝑦∈𝐸₁, one has ∑^∞_𝑖=0𝑘^𝑖𝜑(𝑥/𝑘^𝑖,𝑦/𝑘^𝑖)<∞ and ^∞_𝑖=0𝑘^𝑖+1𝐻𝐾_𝑜𝑥+𝑦𝑘^𝑖+1,𝑥−𝑦𝑘^𝑖+1−𝐻𝐾_𝑜−𝑥+𝑦𝑘^𝑖+1,𝑥−𝑦𝑘^𝑖+1−𝐻𝐾_𝑜𝑥+𝑦𝑘^𝑖+1,𝑥+𝑦𝑘^𝑖+1+𝐻𝐾_𝑜−𝑥+𝑦𝑘^𝑖+1,𝑥+𝑦𝑘^𝑖+1+𝐻𝐾_𝑜2𝑥+𝑦𝑘^𝑖+1,0−𝐻𝐾_𝑜2𝑥𝑘^𝑖+1,0−𝐻𝐾_𝑜2𝑦𝑘^𝑖+1,0<∞. (31) Moreover, if 𝑓(𝑡𝑥) is continuous in 𝑡∈ℝ for every fixed 𝑥∈𝐸₁, then 𝐴 is linear.

Proof. First, we show that 𝑓()(𝑥≤𝑘𝑥−𝑘𝑓)_{𝑘−1𝑗=1}𝜑_𝑜𝑗+12𝑥,𝑗−12𝑥+2^𝑘−1_𝑗=1𝐻𝐾_𝑜𝑗+1𝑥2,𝑗−1𝑥2−𝐻𝐾_𝑜−𝑗+1𝑥2,𝑗−1𝑥2−𝐻𝐾_𝑜𝑗+1𝑥2,𝑗+1𝑥2+𝐻𝐾_𝑜−𝑗+1𝑥2,𝑗+1𝑥2+𝐻𝐾_𝑜𝑗+1𝑥−𝐻𝐾_𝑜𝑗𝑥,0−𝐻𝐾_𝑜(𝑥,0). (32) Putting 𝑦=𝑥 in (27), we obtain 𝑓(2𝑥)−2𝑓(𝑥)≤𝜑_𝑜(𝑥,0)+2−𝐻𝐾_𝑜(𝑥,𝑥)+𝐻𝐾_𝑜(−𝑥,𝑥)+𝐻𝐾_𝑜(2𝑥,0). (33) This shows that (32) holds for 𝑘=2. Assume that (32) is valid for 𝑘=𝑚; that is, 𝑓()(𝑥≤𝑚𝑥−𝑚𝑓)_{𝑚−1𝑗=1}𝜑_𝑜𝑗+12𝑥,𝑗−12𝑥+2^𝑚−1_𝑗=1𝐻𝐾_𝑜𝑗+1𝑥2,𝑗−1𝑥2−𝐻𝐾_𝑜−𝑗+1𝑥2,𝑗−1𝑥2−𝐻𝐾_𝑜𝑗+1𝑥2,𝑗+1𝑥2+𝐻𝐾_𝑜−𝑗+1𝑥2,𝑗+1𝑥2+𝐻𝐾_𝑜𝑗+1𝑥−𝐻𝐾_𝑜𝑗𝑥,0−𝐻𝐾_𝑜(𝑥,0). (34) Replacing 𝑦 by 𝑚𝑥 in (27) and using (34), we have 𝑓)𝑥)−()𝑓(𝑥≤𝑓)𝑥)()(𝑥+𝑓()(𝑥≤((𝑚+1𝑚+1)((𝑚+1−𝑓𝑚𝑥−𝑓)𝑚𝑥−𝑚𝑓)_𝑚𝑗=1𝜑_𝑜𝑗+12𝑥,𝑗−12𝑥+2^𝑚_𝑗=1𝐻𝐾_𝑜𝑗+1𝑥2,𝑗−1𝑥2−𝐻𝐾_𝑜−𝑗+1𝑥2,𝑗−1𝑥2−𝐻𝐾_𝑜𝑗+1𝑥2,𝑗+1𝑥2+𝐻𝐾_𝑜−𝑗+1𝑥2,𝑗+1𝑥2+𝐻𝐾_𝑜𝑗+1𝑥−𝐻𝐾_𝑜𝑗𝑥,0−𝐻𝐾_𝑜(𝑥,0), (35) verifying the case 𝑚+1, and so the assertion (32) holds for all 𝑛>1 and all 𝑥∈𝐸₁.
Next, we use induction on 𝑛 to show that 𝑓𝑘^𝑛𝑥−𝑘^𝑛𝑓(𝑥≤_𝑛𝑖=1)𝑘^{𝑛−𝑖𝑘−1}_𝑗=1𝜑_𝑜𝑗+12𝑘^𝑖−1𝑥,𝑗−12𝑘^𝑖−1𝑥+2^𝑛_𝑖=1𝑘^𝑛−𝑖×^𝑘−1_𝑗=1𝐻𝐾_𝑜𝑗+12𝑘^𝑖−1𝑥,𝑗−12𝑘^𝑖−1𝑥−𝐻𝐾_𝑜−𝑗+12𝑘^𝑖−1𝑥,𝑗−12𝑘^𝑖−1𝑥−𝐻𝐾_𝑜𝑗+12𝑘^𝑖−1𝑥,𝑗+12𝑘^𝑖−1𝑥+𝐻𝐾_𝑜−𝑗+12𝑘^𝑖−1𝑥,𝑗+12𝑘^𝑖−1𝑥+𝐻𝐾_𝑜𝑗+1𝑘^𝑖−1𝑥,0−𝐻𝐾_𝑜𝑗𝑘^𝑖−1𝑥,0−𝐻𝐾_𝑜𝑘^𝑖−1𝑥,0. (36) The case 𝑛=1 follows from (32). Assuming that this holds for 𝑛=𝑚, putting 𝑛=1, and then substituting 𝑘^𝑚𝑥 for 𝑥 in (36), we obtain 𝑓𝑘^𝑚+1𝑥−𝑘^𝑚+1𝑓(𝑥≤𝑓)𝑘⋅𝑘^𝑚𝑥−𝑘𝑓𝑘^𝑚𝑥+𝑘𝑓𝑘^𝑚𝑥−𝑘^𝑚𝑓(𝑥≤_𝑚𝑖=0)𝑘^{𝑚−𝑖𝑘−1}_𝑗=1𝜑_𝑜𝑗+12𝑘^𝑖𝑥,𝑗−12𝑘^𝑖𝑥+2^𝑚_𝑖=0𝑘^𝑚−𝑖×^𝑘−1_𝑗=1𝐻𝐾_𝑜𝑗+12𝑘^𝑖𝑥,𝑗−12𝑘^𝑖𝑥−𝐻𝐾_𝑜−𝑗+12𝑘^𝑖𝑥,𝑗−12𝑘^𝑖𝑥−𝐻𝐾_𝑜𝑗+12𝑘^𝑖𝑥,𝑗+12𝑘^𝑖𝑥+𝐻𝐾_𝑜−𝑗+12𝑘^𝑖𝑥,𝑗+12𝑘^𝑖𝑥+𝐻𝐾_𝑜𝑗+1𝑘^𝑖𝑥,0−𝐻𝐾_𝑜𝑗𝑘^𝑖𝑥,0−𝐻𝐾_𝑜𝑘^𝑖𝑥,0, (37) verifying (36) generally. Rewriting (36), we get 1𝑘^𝑛𝑓𝑘^𝑛𝑥(𝑥≤−𝑓)_𝑛𝑖=11𝑘^𝑖𝑘−1_𝑗=1𝜑_𝑜𝑗+12𝑘^𝑖−1𝑥,𝑗−12𝑘^𝑖−1𝑥+2^𝑛_𝑖=11𝑘^𝑖×^𝑘−1_𝑗=1𝐻𝐾_𝑜𝑗+12𝑘^𝑖−1𝑥,𝑗−12𝑘^𝑖−1𝑥−𝐻𝐾_𝑜−𝑗+12𝑘^𝑖−1𝑥,𝑗−12𝑘^𝑖−1𝑥−𝐻𝐾_𝑜𝑗+12𝑘^𝑖−1𝑥,𝑗+12𝑘^𝑖−1𝑥+𝐻𝐾_𝑜−𝑗+12𝑘^𝑖−1𝑥,𝑗+12𝑘^𝑖−1𝑥+𝐻𝐾_𝑜𝑗+1𝑘^𝑖−1𝑥,0−𝐻𝐾_𝑜𝑗𝑘^𝑖−1𝑥,0−𝐻𝐾_𝑜𝑘^𝑖−1𝑥,0. (38) Using (38), ∑^∞_𝑖=0𝑘^𝑖𝜑(𝑥/𝑘^𝑖,𝑦/𝑘^𝑖)<∞, and the inequality ^∞_𝑖=0𝐻𝐾_𝑜𝑘^𝑖𝑥+𝑦,𝑘^𝑖𝑥−𝑦−𝐻𝐾_𝑜−𝑘^𝑖𝑥+𝑦,𝑘^𝑖𝑥−𝑦−𝐻𝐾_𝑜𝑘^𝑖𝑥+𝑦,𝑘^𝑖𝑥+𝑦+𝐻𝐾_𝑜−𝑘^𝑖𝑥+𝑦,𝑘^𝑖𝑥+𝑦+𝐻𝐾_𝑜𝑘^𝑖𝑥+𝑦,0−𝐻𝐾_𝑜2𝑘^𝑖𝑥,0−𝐻𝐾_𝑜2𝑘^𝑖𝑦,0<∞ (39) and proceeding as in the proof of Proposition 2, we deduce that {𝑘^−𝑛𝑓(𝑘^𝑛𝑥)} is a Cauchy sequence. Thus, the limit 𝐴(𝑥)=lim_𝑛→∞1𝑘^𝑛𝑓𝑘^𝑛𝑥𝑥∈𝐸₁ (40) is well defined. To show that 𝐴 is additive, we substitute 𝑘^𝑛𝑥 and 𝑘^𝑛𝑦 for 𝑥 and 𝑦 in (27), respectively, to get 𝑓𝑘^𝑛𝑥+𝑘^𝑛𝑦−𝑓𝑘^𝑛𝑥−𝑓𝑘^𝑛𝑦≤𝜑_𝑜𝑘^𝑛𝑥,𝑘^𝑛𝑦+2𝐻𝐾_𝑜𝑘^𝑛𝑥+𝑦2,𝑘^𝑛𝑥−𝑦2−𝐻𝐾_𝑜−𝑘^𝑛𝑥+𝑦2,𝑘^𝑛𝑥−𝑦2−𝐻𝐾_𝑜𝑘^𝑛𝑥+𝑦2,𝑘^𝑛𝑥+𝑦2+𝐻𝐾_𝑜−𝑘^𝑛𝑥+𝑦2,𝑘^𝑛𝑥+𝑦2+𝐻𝐾_𝑜𝑘^𝑛𝑥+𝑘^𝑛𝑦,0−𝐻𝐾_𝑜𝑘^𝑛𝑥,0−𝐻𝐾_𝑜𝑘^𝑛𝑦,0. (41) From ∑^∞_𝑖=0(1/𝑘^𝑖)𝜑(𝑘^𝑖𝑥,𝑘^𝑖𝑦)<∞ and ^∞_𝑖=0𝐻𝐾_𝑜𝑘^𝑖𝑥+𝑦,𝑘^𝑖𝑥−𝑦−𝐻𝐾_𝑜−𝑘^𝑖𝑥+𝑦,𝑘^𝑖𝑥−𝑦−𝐻𝐾_𝑜𝑘^𝑖𝑥+𝑦,𝑘^𝑖𝑥+𝑦+𝐻𝐾_𝑜−𝑘^𝑖𝑥+𝑦,𝑘^𝑖𝑥+𝑦+𝐻𝐾_𝑜𝑘^𝑖𝑥+𝑦,0−𝐻𝐾_𝑜2𝑘^𝑖𝑥,0−𝐻𝐾_𝑜2𝑘^𝑖𝑦,0<∞, (42) it follows that 1𝑘^𝑛𝜑_𝑜𝑘^𝑛𝑥,𝑘^𝑛𝑦+2𝐻𝐾_𝑜𝑘^𝑛𝑥+𝑦2,𝑘^𝑛𝑥−𝑦2−𝐻𝐾_𝑜−𝑘^𝑛𝑥+𝑦2,𝑘^𝑛𝑥−𝑦2−𝐻𝐾_𝑜𝑘^𝑛𝑥+𝑦2,𝑘^𝑛𝑥+𝑦2+𝐻𝐾_𝑜−𝑘^𝑛𝑥+𝑦2,𝑘^𝑛𝑥+𝑦2+𝐻𝐾_𝑜𝑘^𝑛𝑥+𝑘^𝑛𝑦,0−𝐻𝐾_𝑜𝑘^𝑛𝑥,0−𝐻𝐾_𝑜𝑘^𝑛𝑦,0⟶0(𝑛⟶∞). (43) Using this last inequality, dividing both sides of (41) by 𝑘^𝑛 and letting 𝑛→∞, we obtain 𝐴(𝑥+𝑦)=𝐴(𝑥)+𝐴(𝑦).
According to (38), we have 𝑓(𝑥)(𝑥≤−𝐴)_∞𝑖=11𝑘^𝑖𝑘−1_𝑗=1𝜑_𝑜𝑗+12𝑘^𝑖−1𝑥,𝑗−12𝑘^𝑖−1𝑥+2^∞_𝑖=11𝑘^𝑖×^𝑘−1_𝑗=1𝐻𝐾_𝑜𝑗+12𝑘^𝑖−1𝑥,𝑗−12𝑘^𝑖−1𝑥−𝐻𝐾_𝑜−𝑗+12𝑘^𝑖−1𝑥,𝑗−12𝑘^𝑖−1𝑥−𝐻𝐾_𝑜𝑗+12𝑘^𝑖−1𝑥,𝑗+12𝑘^𝑖−1𝑥+𝐻𝐾_𝑜−𝑗+12𝑘^𝑖−1𝑥,𝑗+12𝑘^𝑖−1𝑥+𝐻𝐾_𝑜𝑗+1𝑘^𝑖−1𝑥,0−𝐻𝐾_𝑜𝑗𝑘^𝑖−1𝑥,0−𝐻𝐾_𝑜𝑘^𝑖−1𝑥,0. (44) To show that 𝐴 is unique, let 𝐴₁∶𝐸₁→𝐸₂ be another additive mapping satisfying (44). Thus, 𝐴(𝑥)−𝐴₁(𝑥≤1)𝑘^𝑛𝐴𝑘^𝑛𝑥−𝑓𝑘^𝑛𝑥+1𝑘^𝑛𝑓𝑘^𝑛𝑥−𝐴₁𝑘^𝑛𝑥≤2^∞_𝑖=𝑛+11𝑘^𝑖𝑘−1_𝑗=1𝜑_𝑜𝑗+12𝑘^𝑖−1𝑥,𝑗−12𝑘^𝑖−1𝑥+4^∞_𝑖=𝑛+11𝑘^𝑖×^𝑘−1_𝑗=1𝐻𝐾_𝑜𝑗+12𝑘^𝑖−1𝑥,𝑗−12𝑘^𝑖−1𝑥−𝐻𝐾_𝑜−𝑗+12𝑘^𝑖−1𝑥,𝑗−12𝑘^𝑖−1𝑥−𝐻𝐾_𝑜𝑗+12𝑘^𝑖−1𝑥,𝑗+12𝑘^𝑖−1𝑥+𝐻𝐾_𝑜−𝑗+12𝑘^𝑖−1𝑥,𝑗+12𝑘^𝑖−1𝑥+𝐻𝐾_𝑜𝑗+1𝑘^𝑖−1𝑥,0−𝐻𝐾_𝑜𝑗𝑘^𝑖−1𝑥,0−𝐻𝐾_𝑜𝑘^𝑖−1𝑥,0. (45) Using ∑^∞_𝑖=0(1/𝑘^𝑖)𝜑(𝑘^𝑖𝑥,𝑘^𝑖𝑦)<∞ and ^∞_𝑖=0𝐻𝐾_𝑜𝑘^𝑖𝑥+𝑦,𝑘^𝑖𝑥−𝑦−𝐻𝐾_𝑜−𝑘^𝑖𝑥+𝑦,𝑘^𝑖𝑥−𝑦−𝐻𝐾_𝑜𝑘^𝑖𝑥+𝑦,𝑘^𝑖𝑥+𝑦+𝐻𝐾_𝑜−𝑘^𝑖𝑥+𝑦,𝑘^𝑖𝑥+𝑦+𝐻𝐾_𝑜𝑘^𝑖𝑥+𝑦,0−𝐻𝐾_𝑜2𝑘^𝑖𝑥,0−𝐻𝐾_𝑜2𝑘^𝑖𝑦,0<∞, (46) we see that the last term becomes arbitrarily small for sufficiently large 𝑛, implying 𝐴=𝐴₁. Finally, the same proof as in [9] shows that 𝐴 is linear if 𝐸₁ is a Banach space and 𝑓(𝑡𝑥) is continuous in 𝑡 for every fixed 𝑥∈𝐸₁.
The other provision ∑^∞_𝑖=0𝑘^𝑖𝜑(𝑥/𝑘^𝑖,𝑦/𝑘^𝑖)<∞ and ^∞_𝑖=0𝑘^𝑖+1𝐻𝐾_𝑜𝑥+𝑦𝑘^𝑖+1,𝑥−𝑦𝑘^𝑖+1−𝐻𝐾_𝑜−𝑥+𝑦𝑘^𝑖+1,𝑥−𝑦𝑘^𝑖+1−𝐻𝐾_𝑜𝑥+𝑦𝑘^𝑖+1,𝑥+𝑦𝑘^𝑖+1+𝐻𝐾_𝑜−𝑥+𝑦𝑘^𝑖+1,𝑥+𝑦𝑘^𝑖+1+𝐻𝐾_𝑜2𝑥+𝑦𝑘^𝑖+1,0−𝐻𝐾_𝑜2𝑥𝑘^𝑖+1,0−𝐻𝐾_𝑜2𝑦𝑘^𝑖+1,0<∞ (47) is treated analogously by replacing 𝑥 by 𝑥/𝑘^𝑛 in (36) to obtain 𝑓(𝑥)−𝑘^𝑛𝑓𝑥𝑘^𝑛≤1𝑘^𝑛−1_𝑖=0𝑘^{𝑖+1𝑘−1}_𝑗=1𝜑_𝑜𝑗+1𝑥2𝑘^𝑖+1,𝑗−12𝑥𝑘^𝑖+1+2𝑘^𝑛−1_𝑖=0𝑘^𝑖+1×^𝑘−1_𝑗=1𝐻𝐾_𝑜𝑗+1𝑥2𝑘^𝑖+1,𝑗−1𝑥2𝑘^𝑖+1−𝐻𝐾_𝑜−𝑗+1𝑥2𝑘^𝑖+1,𝑗−1𝑥2𝑘^𝑖+1−𝐻𝐾_𝑜𝑗+1𝑥2𝑘^𝑖+1,𝑗+1𝑥2𝑘^𝑖+1+𝐻𝐾_𝑜−𝑗+1𝑥2𝑘^𝑖+1,𝑗+1𝑥2𝑘^𝑖+1+𝐻𝐾_𝑜𝑥𝑗+1𝑘^𝑖+1,0−𝐻𝐾_𝑜𝑗𝑥𝑘^𝑖+1,0−𝐻𝐾_𝑜𝑥𝑘^𝑖+1,0, (48) which is then used to show that {𝑘^𝑛𝑓(𝑥/𝑘^𝑛)} is a Cauchy sequence in 𝐸₂ for all 𝑥∈𝐸₁, so that the limit 𝐴(𝑥)=lim_𝑛→∞𝑘^𝑛𝑓𝑥𝑘^𝑛𝑥∈𝐸₁ (49) is a well-defined additive function. Using 𝐴(𝑥)(𝑥≤1−𝑓)𝑘^∞_𝑖=0𝑘^{𝑖+1𝑘−1}_𝑗=1𝜑_𝑜𝑗+1𝑥2𝑘^𝑖+1,𝑗−1𝑥2𝑘^𝑖+1+2𝑘^∞_𝑖=0𝑘^𝑖+1×^𝑘−1_𝑗=1𝐻𝐾_𝑜𝑗+1𝑥2𝑘^𝑖+1,𝑗−1𝑥2𝑘^𝑖+1−𝐻𝐾_𝑜−𝑗+1𝑥2𝑘^𝑖+1,𝑗−1𝑥2𝑘^𝑖+1−𝐻𝐾_𝑜𝑗+1𝑥2𝑘^𝑖+1,𝑗+1𝑥2𝑘^𝑖+1+𝐻𝐾_𝑜−𝑗+1𝑥2𝑘^𝑖+1,𝑗+1𝑥2𝑘^𝑖+1+𝐻𝐾_𝑜𝑗+1𝑥𝑘^𝑖+1,0−𝐻𝐾_𝑜𝑗𝑥𝑘^𝑖+1,0−𝐻𝐾_𝑜𝑥𝑘^𝑖+1,0, (50) we can show that 𝐴 is unique. Finally, that 𝐴 is linear if 𝐸₁ is a Banach space and 𝑓(𝑡𝑥) is continuous in 𝑡 for every fixed 𝑥∈𝐸₁ is proved by the same technique as that in [9].

4. Main Result

Theorem 4. Adopting the notation of Section 2, if the functions ℱ,𝒢,ℋ,𝒦,ℒ,ℳ∶𝐸₁→𝐸₂ satisfy ℱ(𝑥)𝒦𝑦(𝑥)𝑦𝑥+𝑦+𝒢𝑥−𝑦−2ℋ−ℒ−ℳ≤𝜑𝑥,𝑦𝑥,𝑦∈𝐸₁, (51) then there exist a quadratic function 𝑄∶𝐸₁→𝐸₂ and two additive functions 𝐴₁,𝐴₂∶𝐸₁→𝐸₂ such that, for all 𝑥∈𝐸₁, one has ℱ(𝑥)−𝑄(𝑥)−𝐴₁(𝑥)−𝐴₂(𝑥)(0≤1−ℱ)2𝑘²Φ_𝑘()+1𝑥,𝑥Φ2𝑘^′𝑘()+1𝑥,𝑥Φ2𝑘^′′𝑘()+1𝑥,𝑥2𝑘²Φ_{𝐻𝐾,𝑘}()+1𝑥,𝑥Φ2𝑘^{′𝐻𝐾,𝑘}()+1𝑥,𝑥Φ2𝑘^{′′𝐻𝐾,𝑘}(𝑥,𝑥)+𝐻𝐾_𝑜(+𝑥,0)𝐻𝐾_𝑜(0,𝑥)+𝐻𝐾_𝑒(𝑥𝑥,0)+3𝜑2,𝑥2+52𝜑(𝑥,0)+112𝜑(),𝒢(𝑥)(𝑥)0,0−𝑄−𝐴₁(𝑥)−𝐴₂(𝑥)(0≤1−𝒢)2𝑘²Φ_𝑘()+1𝑥,𝑥Φ2𝑘^′𝑘()+1𝑥,𝑥Φ2𝑘^′′𝑘()+1𝑥,𝑥2𝑘²Φ_{𝐻𝐾,𝑘}()+1𝑥,𝑥Φ2𝑘^{′𝐻𝐾,𝑘}()+1𝑥,𝑥Φ2𝑘^{′′𝐻𝐾,𝑘}(𝑥,𝑥)+𝐻𝐾_𝑜(+𝑥,0)𝐻𝐾_𝑜(0,𝑥)+𝐻𝐾_𝑒(𝑥𝑥,0)+𝜑2,𝑥2+52𝜑()+7𝑥,02𝜑(),ℒ(𝑥)(𝑥)0,0−2𝑄−2𝐴₁(𝑥)(0≤1−ℒ)𝑘²Φ_𝑘()+1𝑥,𝑥𝑘Φ^′_𝑘()()()+1𝑥,𝑥+2𝜑𝑥,0+2𝜑0,0𝑘²Φ_{𝐻𝐾,𝑘}()+1𝑥,𝑥𝑘Φ^′_{𝐻𝐾,𝑘}(),ℳ(𝑥)(𝑥)𝑥,𝑥−2𝑄−2𝐴₂(𝑥)(0≤1−ℳ)𝑘²Φ_𝑘()+1𝑥,𝑥𝑘Φ^′′_𝑘()+1𝑥,𝑥𝑘²Φ_{𝐻𝐾,𝑘}()+1𝑥,𝑥𝑘Φ^′′_{𝐻𝐾,𝑘}(𝑥,𝑥), (52) where Φ_𝑘=Φ𝑥,𝑦_𝑘()Φ𝑥,𝑦𝑖𝑓𝑡ℎ𝑒𝑓𝑖𝑟𝑠𝑡𝑐𝑜𝑛𝑑𝑖𝑡𝑖𝑜𝑛𝑖𝑛𝑑𝑒𝑓𝑖𝑛𝑖𝑡𝑖𝑜𝑛𝑖𝑖𝑖𝑖𝑠𝑡𝑟𝑢𝑒,_𝑘()Φ𝑥,𝑦𝑖𝑓𝑡ℎ𝑒𝑠𝑒𝑐𝑜𝑛𝑑𝑐𝑜𝑛𝑑𝑖𝑡𝑖𝑜𝑛𝑖𝑛𝑑𝑒𝑓𝑖𝑛𝑖𝑡𝑖𝑜𝑛𝑖𝑖𝑖𝑖𝑠𝑡𝑟𝑢𝑒,_′𝑘=Φ𝑥,𝑦_′𝑘()Φ𝑥,𝑦𝑖𝑓𝑡ℎ𝑒𝑓𝑖𝑟𝑠𝑡𝑐𝑜𝑛𝑑𝑖𝑡𝑖𝑜𝑛𝑖𝑛𝑑𝑒𝑓𝑖𝑛𝑖𝑡𝑖𝑜𝑛𝑖𝑖𝑖𝑖𝑠𝑡𝑟𝑢𝑒,_′𝑘()Φ𝑥,𝑦𝑖𝑓𝑡ℎ𝑒𝑠𝑒𝑐𝑜𝑛𝑑𝑐𝑜𝑛𝑑𝑖𝑡𝑖𝑜𝑛𝑖𝑛𝑑𝑒𝑓𝑖𝑛𝑖𝑡𝑖𝑜𝑛𝑖𝑖𝑖𝑖𝑠𝑡𝑟𝑢𝑒,_′′𝑘=Φ𝑥,𝑦_′′𝑘()Φ𝑥,𝑦𝑖𝑓𝑡ℎ𝑒𝑓𝑖𝑟𝑠𝑡𝑐𝑜𝑛𝑑𝑖𝑡𝑖𝑜𝑛𝑖𝑛𝑑𝑒𝑓𝑖𝑛𝑖𝑡𝑖𝑜𝑛𝑖𝑖𝑖𝑖𝑠𝑡𝑟𝑢𝑒,_′′𝑘()Φ𝑥,𝑦𝑖𝑓𝑡ℎ𝑒𝑠𝑒𝑐𝑜𝑛𝑑𝑐𝑜𝑛𝑑𝑖𝑡𝑖𝑜𝑛𝑖𝑛𝑑𝑒𝑓𝑖𝑛𝑖𝑡𝑖𝑜𝑛𝑖𝑖𝑖𝑖𝑠𝑡𝑟𝑢𝑒,_{𝐻𝐾,𝑘}=Φ𝑥,𝑦_{𝐻𝐾,𝑘}()Φ𝑥,𝑦𝑖𝑓𝑡ℎ𝑒𝑓𝑖𝑟𝑠𝑡𝑐𝑜𝑛𝑑𝑖𝑡𝑖𝑜𝑛𝑖𝑛𝑑𝑒𝑓𝑖𝑛𝑖𝑡𝑖𝑜𝑛𝑖𝑖𝑖𝑖𝑠𝑡𝑟𝑢𝑒,_{𝐻𝐾,𝑘}()Φ𝑥,𝑦𝑖𝑓𝑡ℎ𝑒𝑠𝑒𝑐𝑜𝑛𝑑𝑐𝑜𝑛𝑑𝑖𝑡𝑖𝑜𝑛𝑖𝑛𝑑𝑒𝑓𝑖𝑛𝑖𝑡𝑖𝑜𝑛𝑖𝑖𝑖𝑖𝑠𝑡𝑟𝑢𝑒,_{′𝐻𝐾,𝑘}=Φ𝑥,𝑦_{′𝐻𝐾,𝑘}()Φ𝑥,𝑦𝑖𝑓𝑡ℎ𝑒𝑓𝑖𝑟𝑠𝑡𝑐𝑜𝑛𝑑𝑖𝑡𝑖𝑜𝑛𝑖𝑛𝑑𝑒𝑓𝑖𝑛𝑖𝑡𝑖𝑜𝑛𝑖𝑖𝑖𝑖𝑠𝑡𝑟𝑢𝑒,_{′𝐻𝐾,𝑘}()Φ𝑥,𝑦𝑖𝑓𝑡ℎ𝑒𝑠𝑒𝑐𝑜𝑛𝑑𝑐𝑜𝑛𝑑𝑖𝑡𝑖𝑜𝑛𝑖𝑛𝑑𝑒𝑓𝑖𝑛𝑖𝑡𝑖𝑜𝑛𝑖𝑖𝑖𝑖𝑠𝑡𝑟𝑢𝑒,_{′′𝐻𝐾,𝑘}=Φ𝑥,𝑦_{′′𝐻𝐾,𝑘}()Φ𝑥,𝑦𝑖𝑓𝑡ℎ𝑒𝑓𝑖𝑟𝑠𝑡𝑐𝑜𝑛𝑑𝑖𝑡𝑖𝑜𝑛𝑖𝑛𝑑𝑒𝑓𝑖𝑛𝑖𝑡𝑖𝑜𝑛𝑖𝑖𝑖𝑖𝑠𝑡𝑟𝑢𝑒,_{′′𝐻𝐾,𝑘}()𝑥,𝑦𝑖𝑓𝑡ℎ𝑒𝑠𝑒𝑐𝑜𝑛𝑑𝑐𝑜𝑛𝑑𝑖𝑡𝑖𝑜𝑛𝑖𝑛𝑑𝑒𝑓𝑖𝑛𝑖𝑡𝑖𝑜𝑛𝑖𝑖𝑖𝑖𝑠𝑡𝑟𝑢𝑒. (53) Moreover, if ℒ(𝑡𝑥) and ℳ(𝑡𝑥) are continuous in 𝑡∈ℝ for each 𝑥∈𝐸₁, then the two additive functions 𝐴₁, 𝐴₂ are linear, while the quadratic function 𝑄 satisfies 𝑄(𝑡𝑥)=𝑡²𝑄(𝑥). (54)

Proof. Putting 𝑥=𝑦=0 in (51), we obtain ‖ℱ(0)+𝒢(0)−2ℋ(0)𝒦(0)−ℒ(0)−ℳ(0)‖≤𝜑(0,0). (55) For 𝑥,𝑦∈𝐸₁, setting 𝐹(𝑥)(𝑥)(0)(𝑥)(𝑥)(0),𝐿(𝑥)(𝑥)(0)(𝑥)(𝑥)(0),(𝑥)𝒦𝑦(0)𝒦(0),∶=ℱ−ℱ,𝐺∶=𝒢−𝒢∶=ℒ−ℒ,𝑀∶=ℳ−ℳ𝐻𝐾𝑥,𝑦∶=ℋ−ℋ (56) we get 𝐹(0)=𝐺(0)=𝐻𝐾(0,0)=𝐿(0)=𝑀(0)=0. Subtracting (51) and (55), we have 𝐹(𝑥)𝑦().𝑥+𝑦+𝐺𝑥−𝑦−2𝐻𝐾𝑥,𝑦−𝐿−𝑀≤𝜑𝑥,𝑦+𝜑0,0 (57) Substituting −𝑥 for 𝑥 and −𝑦 for 𝑦 in (57), we get 𝐹−−()().𝑥+𝑦+𝐺𝑥−𝑦−2𝐻𝐾−𝑥,−𝑦−𝐿−𝑥−𝑀−𝑦≤𝜑𝑥,𝑦+𝜑0,0 (58) Combining (57) and (58), we obtain 𝐹−+𝐺−−{𝐿(𝑥)(−𝑀𝑦().𝑥+𝑦+𝐹𝑥+𝑦𝑥−𝑦+𝐺𝑥−𝑦−2𝐻𝐾𝑥,𝑦+𝐻𝐾−𝑥,−𝑦+𝐿−𝑥)}+𝑀−𝑦≤2𝜑𝑥,𝑦+2𝜑0,0 (59) Let 𝐹_𝑒(𝑥)𝐹(𝑥)()∶=+𝐹−𝑥2,𝐺_𝑒(𝑥)𝐺(𝑥)()∶=+𝐺−𝑥2,𝐿_𝑒(𝑥)𝐿(𝑥)()∶=+𝐿−𝑥2,𝑀_𝑒(𝑥)𝑀(𝑥)()∶=+𝑀−𝑥2,𝐻𝐾_𝑒𝑥,𝑦∶=𝐻𝐾𝑥,𝑦+𝐻𝐾−𝑥,−𝑦2𝑥,𝑦∈𝐸₁. (60) Note that 𝐹_𝑒, 𝐺_𝑒, 𝐿_𝑒, 𝑀_𝑒 are the even parts of the functions 𝐹, 𝐺, 𝐿, 𝑀, respectively, and that 𝐻𝐾_𝑒−𝑥,−𝑦=𝐻𝐾_𝑒𝑥,𝑦,𝐻𝐾_𝑒𝑥,−𝑦=𝐻𝐾_𝑒−𝑥,𝑦,𝐻𝐾_𝑒(0,−𝑥)=𝐻𝐾_𝑒(0,𝑥),𝐻𝐾_𝑒(−𝑥,0)=𝐻𝐾_𝑒(𝑥,0). (61) From (59), we obtain 𝐹_𝑒𝑥+𝑦+𝐺_𝑒𝑥−𝑦−2𝐻𝐾_𝑒𝑥,𝑦−𝐿_𝑒(𝑥)−𝑀_𝑒𝑦≤𝜑𝑥,𝑦+𝜑(0,0). (62) Substituting 𝑦=0 in (62), we get 𝐹_𝑒(𝑥)+𝐺_𝑒(𝑥)−2𝐻𝐾_𝑒(𝑥,0)−𝐿_𝑒(𝑥)≤𝜑(𝑥,0)+𝜑(0,0). (63) Putting 𝑥=0 in (62) and replacing 𝑦 by 𝑥, we get 𝐹_𝑒(𝑥)+𝐺_𝑒(𝑥)−2𝐻𝐾_𝑒(0,𝑥)−𝑀_𝑒(𝑥()().)≤𝜑𝑥,0+𝜑0,0 (64) Putting 𝑦=𝑥 in (62), we have 𝐹_𝑒(2𝑥)−2𝐻𝐾_𝑒(𝑥,𝑥)−𝐿_𝑒(𝑥)−𝑀_𝑒(𝑥()().)≤𝜑𝑥,𝑥+𝜑0,0 (65) Substituting 𝑦=−𝑥 in (62), we get 𝐺_𝑒(2𝑥)−2𝐻𝐾_𝑒(𝑥,−𝑥)−𝐿_𝑒(𝑥)−𝑀_𝑒(𝑥()().)≤𝜑𝑥,𝑥+𝜑0,0 (66) From (64) and (63), we have 𝐿_𝑒(𝑥)−𝑀_𝑒(𝑥)−2𝐻𝐾_𝑒(0,𝑥)−𝐻𝐾_𝑒(()().𝑥,0)≤2𝜑𝑥,0+2𝜑0,0 (67) From (65) and (66) and replacing 𝑥 by 𝑥/2, we obtain 𝐹_𝑒(𝑥)−𝐺_𝑒(𝑥)−2𝐻𝐾_𝑒𝑥2,𝑥2−𝐻𝐾_𝑒𝑥2,−𝑥2𝑥≤2𝜑2,𝑥2+2𝜑(0,0). (68) From (62), (67), and (68), we get 𝐺_𝑒𝑥+𝑦+𝐺_𝑒𝑥−𝑦−𝑀_𝑒(𝑥)−𝑀_𝑒𝑦−2𝐻𝐾_𝑒𝑥,𝑦+𝐻𝐾_𝑒𝑥+𝑦2,−𝑥+𝑦2−𝐻𝐾_𝑒𝑥+𝑦2,𝑥+𝑦2+𝐻𝐾_𝑒(0,𝑥)−𝐻𝐾_𝑒(()𝑥,0)≤𝜑𝑥,𝑦+2𝜑𝑥,0+2𝜑𝑥+𝑦2,𝑥+𝑦2+5𝜑(0,0). (69) Substituting 𝑦=0 in (69), we get 2𝐺_𝑒(𝑥)−𝑀_𝑒(𝑥)−2𝐻𝐾_𝑒𝑥2𝑥,−2−𝐻𝐾_𝑒𝑥2,𝑥2+𝐻𝐾_𝑒(()𝑥0,𝑥)≤3𝜑𝑥,0+2𝜑2,𝑥2+5𝜑(0,0). (70) From (69) and (70), we obtain 𝑀_𝑒𝑥+𝑦+𝑀_𝑒𝑥−𝑦−2𝑀_𝑒(𝑥)−2𝑀_𝑒𝑦−22𝐻𝐾_𝑒𝑥,𝑦+𝐻𝐾_𝑒𝑥+𝑦2,−𝑥+𝑦2−𝐻𝐾_𝑒𝑥+𝑦2,𝑥+𝑦2+𝐻𝐾_𝑒𝑥−𝑦2,𝑥−𝑦2−𝐻𝐾_𝑒𝑥−𝑦2,−𝑥−𝑦2+2𝐻𝐾_𝑒(0,𝑥)−2𝐻𝐾_𝑒(𝑥,0)−𝐻𝐾_𝑒0,𝑥+𝑦−𝐻𝐾_𝑒0,𝑥−𝑦≤2𝜑𝑥,𝑦+6𝜑𝑥+𝑦2,𝑥+𝑦2+2𝜑𝑥−𝑦2,𝑥−𝑦2()()+3𝜑𝑥+𝑦,0+3𝜑𝑥−𝑦,0+4𝜑𝑥,0+20𝜑0,0=𝜑_𝑒𝑥,𝑦, (71) and so 𝑀_𝑒𝑥+𝑦+𝑀_𝑒𝑥−𝑦−2𝑀_𝑒(𝑥)−2𝑀_𝑒𝑦≤𝜑_𝑒𝑥,𝑦+22𝐻𝐾_𝑒𝑥,𝑦+𝐻𝐾_𝑒𝑥+𝑦2,−𝑥+𝑦2−𝐻𝐾_𝑒𝑥+𝑦2,𝑥+𝑦2+𝐻𝐾_𝑒𝑥−𝑦2,𝑥−𝑦2−𝐻𝐾_𝑒𝑥−𝑦2,−𝑥−𝑦2+2𝐻𝐾_𝑒(0,𝑥)−2𝐻𝐾_𝑒(𝑥,0)−𝐻𝐾_𝑒0,𝑥+𝑦−𝐻𝐾_𝑒0,𝑥−𝑦. (72) By Proposition 2, there is a unique quadratic function 𝑄∶𝐸₁→𝐸₂ such that 𝑀_𝑒(𝑥)(𝑥≤1−2𝑄)𝑘²Φ_𝑘()+1𝑥,𝑥𝑘²Φ_{𝐻𝐾,𝑘}(𝑥,𝑥). (73) If ℳ(𝑡𝑥) is continuous in 𝑡∈ℝ for each 𝑥∈𝐸₁, then the quadratic function 𝑄 satisfies 𝑄(𝑡𝑥)=𝑡²𝑄(𝑥). From (68), (70), and (73), we have 𝐹_𝑒(𝑥)(𝑥≤𝐹−𝑄)_𝑒(𝑥)−𝐺_𝑒(𝑥+𝐺_𝑒)(𝑥)−12𝑀_𝑒(𝑥)−𝐻𝐾_𝑒(+1𝑥,0)2𝑀_𝑒(𝑥)(𝑥≤−𝑄)𝐻𝐾_𝑒(𝑥𝑥,0)+3𝜑2,𝑥2+32𝜑()+9𝑥,02𝜑()+10,02𝑘²Φ_𝑘()+1𝑥,𝑥2𝑘²Φ_{𝐻𝐾,𝑘}(𝑥,𝑥). (74) From (70) and (73), we get 𝐺_𝑒(𝑥)(𝑥≤𝐺−𝑄)_𝑒(𝑥)−12𝑀_𝑒(𝑥)−𝐻𝐾_𝑒(+1𝑥,0)2𝑀_𝑒(𝑥)(𝑥≤−𝑄)𝐻𝐾_𝑒(𝑥𝑥,0)+𝜑2,𝑥2+32𝜑()+5𝑥,02𝜑()+10,02𝑘²Φ_𝑘()+1𝑥,𝑥2𝑘²Φ_{𝐻𝐾,𝑘}(𝑥,𝑥). (75) From (67) and (73), we obtain 𝐿_𝑒(𝑥)(𝑥≤𝐿−2𝑄)_𝑒(𝑥)−𝑀_𝑒(𝑥)+𝑀_𝑒(𝑥)(𝑥()()+1−2𝑄)≤2𝜑𝑥,0+2𝜑0,02𝑘²Φ_𝑘()+1𝑥,𝑥2𝑘²Φ_{𝐻𝐾,𝑘}(𝑥,𝑥). (76) Next, we consider the odd part of functions. Subtracting (57) and (58), we obtain 𝐹−+𝐺−−{𝐿(𝑥)(−𝑀𝑦().𝑥+𝑦−𝐹𝑥+𝑦𝑥−𝑦−𝐺𝑥−𝑦−2𝐻𝐾𝑥,𝑦−𝐻𝐾−𝑥,−𝑦−𝐿−𝑥)}−𝑀−𝑦≤2𝜑𝑥,𝑦+2𝜑0,0 (77) For 𝑥,𝑦∈𝐸₁, let 𝐹_𝑜(𝑥)𝐹(𝑥)()∶=−𝐹−𝑥2,𝐺_𝑜(𝑥)𝐺(𝑥)()∶=−𝐺−𝑥2,𝐿_𝑜(𝑥)𝐿(𝑥)()∶=−𝐿−𝑥2,𝑀_𝑜(𝑥)𝑀(𝑥)()∶=−𝑀−𝑥2,𝐻𝐾_𝑜𝑥,𝑦∶=𝐻𝐾𝑥,𝑦−𝐻𝐾−𝑥,−𝑦2. (78) Clearly, 𝐹_𝑜, 𝐺_𝑜, 𝐿_𝑜, 𝑀_𝑜 are the odd parts of 𝐹, 𝐺, 𝐿, 𝑀, respectively. Note that 𝐻𝐾_𝑜−𝑥,−𝑦=−𝐻𝐾_𝑜𝑥,𝑦,𝐻𝐾_𝑜𝑥,−𝑦=−𝐻𝐾_𝑜−𝑥,𝑦,𝐻𝐾_𝑜(0,−𝑥)=−𝐻𝐾_𝑜(0,𝑥),𝐻𝐾_𝑜(−𝑥,0)=−𝐻𝐾_𝑜(𝑥,0). (79) From (77), we get 𝐹_𝑜𝑥+𝑦+𝐺_𝑜𝑥−𝑦−2𝐻𝐾_𝑜𝑥,𝑦−𝐿_𝑜(𝑥)−𝑀_𝑜𝑦≤𝜑𝑥,𝑦+𝜑(0,0). (80) Substituting 𝑦=0 in (80), we get 𝐹_𝑜(𝑥)+𝐺_𝑜(𝑥)−2𝐻𝐾_𝑜(𝑥,0)−𝐿_𝑜(𝑥()().)≤𝜑𝑥,0+𝜑0,0 (81) Putting 𝑥=0 in (80) and replacing 𝑦 by 𝑥, we get 𝐹_𝑜(𝑥)−𝐺_𝑜(𝑥)−2𝐻𝐾_𝑜(0,𝑥)−𝑀_𝑜(𝑥()().)≤𝜑𝑥,0+𝜑0,0 (82) Letting 𝑦=𝑥 in (80), we have 𝐹_𝑜(2𝑥)−2𝐻𝐾_𝑜(𝑥,𝑥)−𝐿_𝑜(𝑥)−𝑀_𝑜(𝑥()().)≤𝜑𝑥,𝑥+𝜑0,0 (83) Substituting 𝑦=−𝑥 in (80), we get 𝐺_𝑜(2𝑥)−2𝐻𝐾_𝑜(𝑥,−𝑥)−𝐿_𝑜(𝑥)+𝑀_𝑜(𝑥()().)≤𝜑𝑥,𝑥+𝜑0,0 (84) From (81) and (82), we obtain 2𝐹_𝑜(𝑥)−𝐿_𝑜(𝑥)−𝑀_𝑜(𝑥)−2𝐻𝐾_𝑜(𝑥,0)+𝐻𝐾_𝑜(()(),0,𝑥)≤2𝜑𝑥,0+2𝜑0,02𝐺_𝑜(𝑥)−𝐿_𝑜(𝑥)+𝑀_𝑜(𝑥)−2𝐻𝐾_𝑜(𝑥,0)−𝐻𝐾_𝑜(()().0,𝑥)≤2𝜑𝑥,0+2𝜑0,0 (85)(86) From (83) and (85), we have 𝐿_𝑜(2𝑥)−2𝐿_𝑜(𝑥)+𝑀_𝑜(2𝑥)−2𝑀_𝑜(𝑥)−22𝐻𝐾_𝑜(𝑥,𝑥)−𝐻𝐾_𝑜(2𝑥,0)−𝐻𝐾_𝑜(()()().0,2𝑥)≤2𝜑𝑥,𝑥+2𝜑2𝑥,0+4𝜑0,0 (87) From (84) and (86), we get 𝐿_𝑜(2𝑥)−2𝐿_𝑜(𝑥)−𝑀_𝑜(2𝑥)+2𝑀_𝑜(𝑥)+22𝐻𝐾_𝑜(−𝑥,𝑥)+𝐻𝐾_𝑜(2𝑥,0)−𝐻𝐾_𝑜(()()().0,2𝑥)≤2𝜑𝑥,𝑥+2𝜑2𝑥,0+4𝜑0,0 (88) From (87) and (88), we have 𝐿_𝑜(2𝑥)−2𝐿_𝑜(𝑥)−2𝐻𝐾_𝑜(𝑥,𝑥)−𝐻𝐾_𝑜(−𝑥,𝑥)−𝐻𝐾_𝑜(()()(),𝑀2𝑥,0)≤2𝜑𝑥,𝑥+2𝜑2𝑥,0+4𝜑0,0_𝑜(2𝑥)−2𝑀_𝑜(𝑥)−2𝐻𝐾_𝑜(𝑥,𝑥)+𝐻𝐾_𝑜(−𝑥,𝑥)−𝐻𝐾_𝑜(()()().0,2𝑥)≤2𝜑𝑥,𝑥+2𝜑2𝑥,0+4𝜑0,0 (89) In view of (80), (85), (86), and (89), we have 𝐿_𝑜𝑥+𝑦+𝑀_𝑜𝑥+𝑦+𝐿_𝑜𝑥−𝑦−𝑀_𝑜𝑥−𝑦−𝐿_𝑜(2𝑥)−𝑀_𝑜2𝑦−22𝐻𝐾_𝑜𝑥,𝑦−𝐻𝐾_𝑜(𝑥,𝑥)+𝐻𝐾_𝑜(−𝑥,𝑥)−𝐻𝐾_𝑜𝑦,𝑦−𝐻𝐾_𝑜−𝑦,𝑦+𝐻𝐾_𝑜(2𝑥,0)+𝐻𝐾_𝑜0,2𝑦−𝐻𝐾_𝑜𝑥+𝑦,0−𝐻𝐾_𝑜0,𝑥+𝑦−𝐻𝐾_𝑜𝑥−𝑦,0+𝐻𝐾_𝑜()()()0,𝑥−𝑦≤2𝜑𝑥,𝑦+2𝜑𝑥,𝑥+2𝜑𝑦,𝑦+2𝜑2𝑥,0+2𝜑2𝑦,0+2𝜑𝑥+𝑦,0+2𝜑𝑥−𝑦,0+14𝜑0,0=𝜑₀𝑥,𝑦. (90) Substituting 𝑦 by −𝑦 in (90), we obtain 𝐿_𝑜𝑥−𝑦+𝑀_𝑜𝑥−𝑦+𝐿_𝑜𝑥+𝑦−𝑀_𝑜𝑥+𝑦−𝐿_𝑜(2𝑥)+𝑀_𝑜2𝑦−2−2𝐻𝐾_𝑜−𝑥,𝑦−𝐻𝐾_𝑜(𝑥,𝑥)+𝐻𝐾_𝑜(−𝑥,𝑥)+𝐻𝐾_𝑜𝑦,𝑦+𝐻𝐾_𝑜−𝑦,𝑦+𝐻𝐾_𝑜(2𝑥,0)−𝐻𝐾_𝑜0,2𝑦−𝐻𝐾_𝑜𝑥+𝑦,0+𝐻𝐾_𝑜0,𝑥+𝑦−𝐻𝐾_𝑜𝑥−𝑦,0−𝐻𝐾_𝑜0,𝑥−𝑦≤𝜑₀𝑥,𝑦. (91) From (90) and (91), we have 𝐿_𝑜𝑥+𝑦+𝐿_𝑜𝑥−𝑦−𝐿_𝑜(2𝑥)−2𝐻𝐾_𝑜𝑥,𝑦−𝐻𝐾_𝑜−𝑥,𝑦−𝐻𝐾_𝑜(𝑥,𝑥)+𝐻𝐾_𝑜(−𝑥,𝑥)+𝐻𝐾_𝑜(2𝑥,0)−𝐻𝐾_𝑜𝑥+𝑦,0−𝐻𝐾_𝑜𝑥−𝑦,0≤𝜑₀,𝑀𝑥,𝑦_𝑜𝑥+𝑦−𝑀_𝑜𝑥−𝑦−𝑀_𝑜2𝑦−2𝐻𝐾_𝑜𝑥,𝑦+𝐻𝐾_𝑜−𝑥,𝑦−𝐻𝐾_𝑜𝑦,𝑦−𝐻𝐾_𝑜−𝑦,𝑦+𝐻𝐾_𝑜0,2𝑦−𝐻𝐾_𝑜0,𝑥+𝑦+𝐻𝐾_𝑜0,𝑥−𝑦≤𝜑₀𝑥,𝑦. (92)(93) Putting 𝑢=𝑥+𝑦 and 𝑣=𝑥−𝑦 in (92), we obtain 𝐿_𝑜(𝑢)+𝐿_𝑜(𝑣)−𝐿_𝑜(𝑢+𝑣)−2𝐻𝐾_𝑜𝑢+𝑣2,𝑢−𝑣2−𝐻𝐾_𝑜−𝑢+𝑣2,𝑢−𝑣2−𝐻𝐾_𝑜𝑢+𝑣2,𝑢+𝑣2+𝐻𝐾_𝑜−𝑢+𝑣2,𝑢+𝑣2+𝐻𝐾_𝑜(𝑢+𝑣,0)−𝐻𝐾_𝑜(𝑢,0)−𝐻𝐾_𝑜(𝑣,0)≤𝜑₀𝑢+𝑣2,𝑢−𝑣2. (94) By Proposition 3, there exists a unique additive function 𝐴₁∶𝐸₁→𝐸₂ such that 𝐿_𝑜(𝑥)−2𝐴₁(𝑥≤1)𝑘Φ^′_𝑘()+1𝑥,𝑥𝑘Φ^′_{𝐻𝐾,𝑘}(𝑥,𝑥)𝑥∈𝐸₁. (95) Moreover, if ℒ(𝑡𝑥) is continuous in 𝑡∈ℝ for every fixed 𝑥∈𝐸₁, we have that 𝐴₁ is a linear function. Putting 𝑢=𝑥−𝑦 and 𝑣=2𝑦 in (93), we get 𝑀_𝑜(𝑢+𝑣)−𝑀_𝑜(𝑢)−𝑀_𝑜(𝑣)−2𝐻𝐾_𝑜𝑣𝑢+2,𝑣2+𝐻𝐾_𝑜−𝑣𝑢+2,𝑣2−𝐻𝐾_𝑜𝑣2,𝑣2−𝐻𝐾_𝑜−𝑣2,𝑣2+𝐻𝐾_𝑜(0,𝑣)−𝐻𝐾_𝑜(0,𝑢+𝑣)+𝐻𝐾_𝑜(0,𝑢)≤𝜑_𝑜𝑣𝑢+2,𝑣2𝑢,𝑣∈𝐸₁. (96) Again by Proposition 3, there exists a unique additive function 𝐴₂∶𝐸₁→𝐸₂ such that 𝑀_𝑜(𝑥)−2𝐴₂(𝑥≤1)𝑘Φ^′′_𝑘()+1𝑥,𝑥𝑘Φ^′′_{𝐻𝐾,𝑘}(𝑥,𝑥)𝑥∈𝐸₁. (97) Furthermore, if ℳ(𝑡𝑥) is continuous in 𝑡∈ℝ for all 𝑥∈𝐸₁, then 𝐴₂ is also linear. From (85), (95), and (97), it follows that 𝐹_𝑜(𝑥)−𝐴₁(𝑥)−𝐴₂(𝑥≤1)Φ2𝑘^′𝑘()+1𝑥,𝑥Φ2𝑘^′′𝑘()+1𝑥,𝑥Φ2𝑘^{′𝐻𝐾,𝑘}()+1𝑥,𝑥Φ2𝑘^{′′𝐻𝐾,𝑘}()()()+𝑥,𝑥+𝜑𝑥,0+𝜑0,0𝐻𝐾_𝑜(+𝑥,0)𝐻𝐾_𝑜(,𝐺0,𝑥)_𝑜(𝑥)−𝐴₁(𝑥)−𝐴₂(𝑥≤1)Φ2𝑘^′𝑘()+1𝑥,𝑥Φ2𝑘^′′𝑘()+1𝑥,𝑥Φ2𝑘^{′𝐻𝐾,𝑘}()+1𝑥,𝑥Φ2𝑘^{′′𝐻𝐾,𝑘}()()()+𝑥,𝑥+𝜑𝑥,0+𝜑0,0𝐻𝐾_𝑜(+𝑥,0)𝐻𝐾_𝑜(0,𝑥). (98) The theorem now follows from the inequalities (73), (74), (75), (76), (95), (97), and (98).