Stability of a Functional Equation Deriving from Quadratic and Additive Functions in Non-Archimedean Normed Spaces

Abstract

We obtain the general solution of the generalized mixed additive and quadratic functional equation 𝑓(𝑥+𝑚𝑦)+𝑓(𝑥−𝑚𝑦)=2𝑓(𝑥)−2𝑚²𝑓(𝑦)+𝑚²𝑓(2𝑦), 𝑚 is even; 𝑓(𝑥+𝑦)+𝑓(𝑥−𝑦)−2(𝑚²−1)𝑓(𝑦)+(𝑚²−1)𝑓(2𝑦), 𝑚 is odd, for a positive integer 𝑚. We establish the Hyers-Ulam stability for these functional equations in non-Archimedean normed spaces when 𝑚 is an even positive integer or 𝑚=3.

1. Introduction

The basic problem of the stability of functional equations was formulated by Ulam in 1940 in the following form. Suppose that a mapping 𝑓 satisfies the additive functional equation 𝑓(𝑥+𝑦)=𝑓(𝑥)+𝑓(𝑦) only approximately. Then does there exist an additive function which approximates 𝑓? (See also [1].) In 1941, Hyers [2] gave the following answer to this question for Banach spaces. The result of Hyers was generalized in 1950 by Aoki [3] for approximately additive mappings and in 1978 by Rassias [4] for approximately linear mappings, by considering the unbounded Cauchy differences. A further generalization was obtained by Găvruţa [5] in 1994, by replacing the Cauchy differences by a control function 𝜑 satisfying a very simple condition of convergence.

The Hyers-Ulam stability problem for the quadratic functional equation 𝑓𝑥+𝑦+𝑓𝑥−𝑦=2𝑓(𝑥)+2𝑓𝑦 (1) was first proved by Skof for a function 𝑓∶𝒳→𝒴, where 𝒳 is a normed space and 𝒴 is a Banach space [6]. One year later, Cholewa [7] demonstrated that Skof’s theorem is still true if relevant domain is replaced by an abelian group. After that, in [8], Czerwik proved the Hyers-Ulam stability of the quadratic functional equation (1) as a special case. In [9], it was shown that a mapping 𝑓∶𝒳→𝒴 is quadratic if and only if 𝑓(𝑘𝑥+𝑦)+𝑓(𝑘𝑥−𝑦)=2𝑘²𝑓(𝑥)+2𝑓(𝑦) for all 𝑥,𝑦∈𝒳. Also, 𝑓 is quadratic if and only if  2𝑓((𝑘𝑥+𝑘𝑦)/2)+2𝑓((𝑘𝑥−𝑘𝑦)/2)=𝑘²𝑓(𝑥)+𝑘²𝑓(𝑦) for all 𝑥,𝑦∈𝒳 [10]. Cădariu and Radu investigated the stability of the Cauchy functional equation [11] and for the quadratic functional equation [12]. Stability problems of miscellaneous functional equations have been investigated by several authors during the last decades (see, e.g., [13–15]).

In [16], Eskandani et al. determined the general solution of the following mixed additive and quadratic functional equation: 𝑓𝑥+2𝑦+𝑓𝑥−2𝑦+8𝑓𝑦=2𝑓(𝑥)+4𝑓2𝑦. (2)

They studied the Hyers-Ulam stability of (2) in non-Archimedean Banach modules over a unital Banach algebra. In [17], Najati and Moghimi established the general solution of the mixed type additive and quadratic functional equation 𝑓()(𝑥)2𝑥+𝑦+𝑓2𝑥−𝑦=𝑓𝑥+𝑦+𝑓𝑥−𝑦+2𝑓2𝑥−2𝑓 (3) and investigated the stability of this equation in quasi-Banach spaces. The stability of (3) in random normed spaces is proved in [18].

In this paper, we consider the following functional equations: 𝑓𝑥+𝑚𝑦+𝑓𝑥−𝑚𝑦=2𝑓(𝑥)−2𝑚²𝑓𝑦+𝑚²𝑓2𝑦, (4) where 𝑚 is an even positive integer and 𝑓𝑦.𝑥+3𝑦+𝑓𝑥−3𝑦=𝑓𝑥+𝑦+𝑓𝑥−𝑦−16𝑓+8𝑓2𝑦 (5) Indeed, (4) and (5) are different from (2) and (3). It is easily verified that the function 𝑓(𝑥)=𝛼𝑥²+𝛽𝑥 is a solution of the functional equations (4) and (5). We show that these functional equations are mixed additive and quadratic mappings. We also prove the Hyers-Ulam stability problem for these equations. As a corollary, the hyperstability of (4) and (5) under some conditions in non-Archimedean normed spaces is shown as well.

2. General Solution of (4) and (5)

To achieve our aim in this paper, we need the following lemma which is a fundamental tool.

Lemma 1. Let 𝒳 and 𝒴 be real vector spaces. (i)If an odd mapping 𝑓∶𝒳→𝒴 satisfies the functional equation (4), then 𝑓 is additive. (ii)If an odd mapping 𝑓∶𝒳→𝒴 satisfies the functional equation (5), then 𝑓 is additive. (iii)If an even mapping 𝑓∶𝒳→𝒴 satisfies the functional equation (4), then 𝑓 is quadratic. (iv)If an even mapping 𝑓∶𝒳→𝒴 satisfies the functional equation (5), then 𝑓 is quadratic.

Proof. (i) Letting 𝑥=0 in (4), we get 𝑓(2𝑦)=2𝑓(𝑦) for all 𝑦∈𝒳. This equality implies that 𝑓𝑥+𝑚𝑦+𝑓𝑥−𝑚𝑦=2𝑓(𝑥) (6) for all 𝑥,𝑦∈𝒳. Replacing 𝑦 by 𝑦/𝑚 in (6), we have 𝑓𝑥+𝑦+𝑓𝑥−𝑦=2𝑓(𝑥) (7) for all 𝑥,𝑦∈𝒳. Substituting 𝑥,𝑦 by 𝑦,𝑥 in (7), respectively, we obtain 𝑓𝑥+𝑦−𝑓𝑥−𝑦=2𝑓𝑦 (8) for all 𝑥,𝑦∈𝒳. The equalities (7) and (8) show that 𝑓𝑥+𝑦=𝑓(𝑥)+𝑓𝑦𝑥,𝑦∈𝒳. (9)
(ii) Suppose that 𝑓 satisfies (5). Similar to the part (i), by the oddness of 𝑓, we have 𝑓(2𝑥)=2𝑓(𝑥) for all 𝑥∈𝒳. Thus 𝑓𝑥+3𝑦+𝑓𝑥−3𝑦=𝑓𝑥+𝑦+𝑓𝑥−𝑦 (10) for all 𝑥,𝑦∈𝒳. We substitute 𝑥 by 𝑥+𝑦 in (10) and then 𝑥 by 𝑥−𝑦 in (10); we get 𝑓(𝑥),𝑓(𝑥)𝑥+4𝑦+𝑓𝑥−2𝑦=𝑓𝑥+2𝑦+𝑓𝑥+2𝑦+𝑓𝑥−4𝑦=𝑓+𝑓𝑥−2𝑦 (11)(12) for all 𝑥,𝑦∈𝒳. Then, by adding (11) to (12), we lead to 𝑓𝑥+4𝑦+𝑓𝑥−4𝑦=2𝑓(𝑥) (13) for all 𝑥,𝑦∈𝒳. Similar to the part (i), we can show that 𝑓 is additive.
(iii) By the assumption, the equality 𝑓𝑥+𝑚𝑦+𝑓𝑥−𝑚𝑦=2𝑓(𝑥)−2𝑚²𝑓𝑦+𝑚²𝑓2𝑦 (14) holds for a fixed even positive integer 𝑚. Putting 𝑥=𝑦=0 in (14), we get 𝑓(0)=0. Once more, by letting 𝑥=0 in (14), we have 2𝑓𝑚𝑦=−2𝑚²𝑓𝑦+𝑚²𝑓2𝑦 (15) for all 𝑦∈𝒳. Interchanging 𝑥,𝑦 into 𝑚𝑥,𝑥 in (14), respectively, we deduce that 𝑓(2𝑚𝑥)=2𝑓(𝑚𝑥)−2𝑚²𝑓(𝑥)+𝑚²𝑓(2𝑥) (16) for all 𝑥∈𝒳. Plugging (15) into (16), we have 𝑓(2𝑚𝑥)=4𝑓(𝑚𝑥) for all 𝑥∈𝒳 and thus 𝑓(2𝑥)=4𝑓(𝑥) for all 𝑥∈𝒳. Using the last equality and (14), we have 𝑓𝑥+𝑚𝑦+𝑓𝑥−𝑚𝑦=2𝑓(𝑥)+2𝑚²𝑓𝑦 (17) for all 𝑥,𝑦∈𝒳. Setting 𝑥=0 in (17), we obtain 𝑓(𝑚𝑦)=𝑚²𝑓(𝑦) for all 𝑦∈𝒳. Applying this equality and putting 𝑥 by 𝑚𝑥 in (17), we get 𝑓𝑥+𝑦+𝑓𝑥−𝑦=2𝑓(𝑥)+2𝑓𝑦 (18) for all 𝑥,𝑦∈𝒳. This shows that 𝑓 is a quadratic mapping.
(iv) Suppose that 𝑓 satisfies (5). Replacing 𝑥 by 𝑥+𝑦 and 𝑥−𝑦 in (5), respectively, we have 𝑓(𝑥)𝑦,𝑓(𝑥)𝑦𝑥+4𝑦+𝑓𝑥−2𝑦=𝑓𝑥+2𝑦+𝑓−16𝑓+8𝑓2𝑦𝑥+2𝑦+𝑓𝑥−4𝑦=𝑓+𝑓𝑥−2𝑦−16𝑓+8𝑓2𝑦 (19)(20) for all 𝑥,𝑦∈𝒳. The equalities (19) and (20) imply that 𝑓𝑥+4𝑦+𝑓𝑥−4𝑦=2𝑓(𝑥)−32𝑓𝑦+16𝑓2𝑦 (21) for all 𝑥,𝑦∈𝒳. Now, the above equality is a special case of the part (iii) when 𝑚=4.

In the following theorem, we solve (4) in which 𝑚 is an even positive integer and 𝑓𝑚𝑥+𝑚𝑦+𝑓𝑥−𝑚𝑦=𝑓𝑥+𝑦+𝑓𝑥−𝑦−2²−1𝑓𝑦+𝑚²−1𝑓2𝑦, (22) where 𝑚 is an odd positive integer.

Theorem 2. Let 𝒳 and 𝒴 be real vector spaces. Then a mapping 𝑓∶𝒳→𝒴 satisfies the functional equation (2) if and only if it satisfies 𝑓=(𝑥)𝑥+𝑚𝑦+𝑓𝑥−𝑚𝑦2𝑓−2𝑚²𝑓𝑦+𝑚²𝑓𝑓𝑚2𝑦,𝑚𝑖𝑠𝑒𝑣𝑒𝑛,𝑥+𝑦+𝑓𝑥−𝑦−2²−1𝑓𝑦+𝑚²−1𝑓2𝑦,𝑚𝑖𝑠𝑜𝑑𝑑 (23) for all 𝑚≥𝑘, where 𝑘 is a fixed positive integer with 𝑘≥3.

Proof. Suppose that 𝑓∶𝒳→𝒴 satisfies the functional equation (2). Putting 𝑥=𝑦=0 in (2), we get 𝑓(0)=0. Replacing 𝑥 by 𝑥+𝑦 and 𝑥−𝑦 in (2), respectively, we have 𝑓𝑦.𝑥+3𝑦+𝑓𝑥−3𝑦=𝑓𝑥+𝑦+𝑓𝑥−𝑦−16𝑓+8𝑓2𝑦 (24) Similar to the above, we get 𝑓𝑥+4𝑦+𝑓𝑥−4𝑦=2𝑓(𝑥)−32𝑓𝑦+16𝑓2𝑦. (25) Using the above method, we can deduce that 𝑓=(𝑥)𝑥+𝑚𝑦+𝑓𝑥−𝑚𝑦2𝑓−2𝑝_𝑚𝑓𝑦+𝑝_𝑚𝑓𝑓2𝑦,𝑚iseven,𝑥+𝑦+𝑓𝑥−𝑦−2𝑝_𝑚𝑓𝑦+𝑝_𝑚𝑓2𝑦,𝑚isodd (26) for all 𝑥,𝑦∈𝒳 for which 𝑝₂=4, 𝑝₃=8 and 𝑝_𝑚=2𝑝_𝑚−1−𝑝_𝑚−2+4𝑚iseven,2𝑝_𝑚−1−𝑝_𝑚−2𝑚isodd. (27) Solving the above recurrence equations, we get 𝑝_𝑚=𝑚²𝑚𝑚iseven,₂−1𝑚isodd (28) for all positive integers 𝑚≥2.
Conversely, assume that 𝑓∶𝒳→𝒴 satisfies the functional equations (4) and (22) for each 𝑘≥𝑚. Firstly, we assume that 𝑚 is even. For 𝑘=𝑚(𝑚−1) and for each 𝑥,𝑦∈𝒳, we have 𝑓()𝑦()𝑦(𝑥)𝑥+𝑚𝑚−1+𝑓𝑥−𝑚𝑚−1=2𝑓−2𝑚²𝑓(𝑚−1)𝑦+𝑚²𝑓2(𝑚−1)𝑦 (29) for all 𝑥,𝑦∈𝒳. On the other hand, 𝑓𝑥+𝑚²−𝑚𝑦+𝑓𝑥−𝑚²𝑦(𝑥)𝑚−𝑚=2𝑓−2²−𝑚²𝑓𝑦+𝑚²−𝑚²𝑓2𝑦 (30) for all 𝑥,𝑦∈𝒳. It follows from (29) and (30) that ()𝑦2()𝑦()−2𝑓𝑚−1+𝑓𝑚−1=−2𝑚−1²𝑓𝑦+(𝑚−1)²𝑓2𝑦 (31) for all 𝑥,𝑦∈𝒳. Since (𝑚+1)(𝑚−1) is an odd number, we have 𝑓()𝑦()𝑦()𝑦()𝑦()𝑥+𝑚+1)(𝑚−1+𝑓𝑥−𝑚+1)(𝑚−1=𝑓𝑥+𝑚−1+𝑓𝑥−𝑚−1−2𝑚+1²𝑓()𝑦+()−1𝑚−1𝑚+1²−1𝑓2(𝑚−1)𝑦 (32) for all 𝑥,𝑦∈𝒳. Also, 𝑓𝑥+𝑚²−1𝑦+𝑓𝑥−𝑚²𝑦𝑚−1=𝑓𝑥+𝑦+𝑓𝑥−𝑦−2²−1²𝑓𝑦+𝑚−1₂−1²−1𝑓2𝑦 (33) for all 𝑥,𝑦∈𝒳. Plugging (32) into (33) and using (31), we get 𝑓()𝑦()𝑦()𝑥+𝑚−1+𝑓𝑥−𝑚−1=𝑓𝑥+𝑦+𝑓𝑥−𝑦−2𝑚−1²𝑓𝑦+()−1𝑚−1²−1𝑓2𝑦. (34) For the odd case 𝑚, we have 𝑓()𝑦()𝑦(𝑥)()𝑥+𝑚+1)(𝑚−1+𝑓𝑥−𝑚+1)(𝑚−1=2𝑓−2𝑚+1²𝑓()𝑦+()𝑚−1𝑚+1²𝑓2(𝑚−1)𝑦 (35) for all 𝑥,𝑦∈𝒳. Also 𝑓𝑥+𝑚²−1𝑦+𝑓𝑥−𝑚²𝑦(𝑥)𝑚−1=2𝑓−2²−1²𝑓𝑦+𝑚²−1²𝑓2𝑦 (36) for all 𝑥,𝑦∈𝒳. The comparison of (35) and (36) shows that ()𝑦2()𝑦()−2𝑓𝑚−1+𝑓𝑚−1=−2𝑚−1²𝑓𝑦+(𝑚−1)²𝑓2𝑦 (37) for all 𝑥,𝑦∈𝒳. For 𝑘=𝑚(𝑚−1) and for each 𝑥,𝑦∈𝒳, we have 𝑓()𝑦()𝑦()𝑦()𝑦𝑚𝑥+𝑚𝑚−1+𝑓𝑥−𝑚𝑚−1=𝑓𝑥+𝑚−1+𝑓𝑥−𝑚−1−2²−1𝑓(𝑚−1)𝑦+𝑚²−1𝑓2(𝑚−1)𝑦 (38) for all 𝑥,𝑦∈𝒳. On the other hand, 𝑓𝑥+𝑚²−𝑚𝑦+𝑓𝑥−𝑚²𝑦(𝑥)𝑚−𝑚=2𝑓−2²−𝑚²𝑓𝑦+𝑚²−𝑚²𝑓2𝑦 (39) for all 𝑥,𝑦∈𝒳. Now, by comparing (38) with (39) and applying (37), we obtain 𝑓()𝑦()𝑦(𝑥)()𝑥+𝑚−1+𝑓𝑥−𝑚−1=2𝑓−2𝑚−1²𝑓𝑦+(𝑚−1)²𝑓2𝑦 (40) for all 𝑥,𝑦∈𝒳. This completes the proof.

Theorem 3. Let 𝒳 and 𝒴 be real vector spaces. A mapping 𝑓∶𝒳→𝒴 satisfies either (4) or (5) if and only if there exist a symmetric biadditive mapping 𝐵∶𝒳×𝒳→𝒴 and an additive mapping 𝐴∶𝒳→𝒴 such that 𝑓(𝑥)=𝐵(𝑥,𝑥)+𝐴(𝑥) for all 𝑥∈𝒳.

Proof. Assume that there exist a symmetric biadditive mapping 𝐵∶𝒳×𝒳→𝒴 and an additive mapping 𝐴∶𝒳→𝒴 such that 𝑓(𝑥)=𝐵(𝑥,𝑥)+𝐴(𝑥) for all 𝑥∈𝒳. A simple computation shows that the mappings 𝐴 and 𝐵^′∶𝒳→𝒴 given by 𝐵^′(𝑥)=𝐵(𝑥,𝑥) satisfy the functional equations (4) and (5). Therefore the mapping 𝑓 satisfies (4) and (5).
Conversely, we decompose 𝑓 into the even part and odd part by setting 𝑓_𝑒(𝑥)=𝑓(𝑥)+𝑓(−𝑥)2,𝑓_𝑜(𝑥)=𝑓(𝑥)−𝑓(−𝑥)2 (41) for all 𝑥∈𝒳. Obviously, 𝑓(𝑥)=𝑓_𝑒(𝑥)+𝑓_𝑜(𝑥) for all 𝑥∈𝒳. One can easily check that the mappings 𝑓_𝑒 and 𝑓_𝑜 satisfy (4) and (5). It follows from Lemma 1 that the mappings 𝑓_𝑒 and 𝑓_𝑜 are quadratic and additive, respectively. Since 𝑓_𝑒 is quadratic, by [19], there exists a symmetric biadditive mapping 𝐵∶𝒳×𝒳→𝒴 such that 𝑓_𝑒(𝑥)=𝐵(𝑥,𝑥) for all 𝑥∈𝒳. Thus 𝑓(𝑥)=𝐵(𝑥,𝑥)+𝐴(𝑥) for all 𝑥∈𝒳, where 𝐴(𝑥)=𝑓_𝑜(𝑥) for all 𝑥∈𝒳.

3. Hyers-Ulam Stability of (4) and (5)

We recall some basic facts concerning non-Archimedean spaces and some preliminary results.

By a non-Archimedean field, we mean a field 𝕂 equipped with a function (valuation) |⋅| from 𝕂 into [0,∞) such that |𝑟|=0 if and only if 𝑟=0,  |𝑟𝑠|=|𝑟||𝑠|, and |𝑟+𝑠|≤max{|𝑟|,|𝑠|} for all 𝑟,𝑠∈𝕂. Clearly |1|=|−1|=1 and |𝑛|≤1 for all 𝑛∈ℕ.

Let 𝒳 be a vector space over a scalar field 𝕂 with a non-Archimedean nontrivial valuation |⋅|. A function ‖⋅‖∶𝒳→ℝ is a non-Archimedean norm (valuation) if it satisfies the following conditions: (i)‖𝑥‖=0 if and only if 𝑥=0; (ii)‖𝑟𝑥‖=|𝑟|‖𝑥‖, (𝑥∈𝒳,𝑟∈𝕂); (iii)the strong triangle inequality (ultrametric), namely, 𝑥+𝑦≤max‖𝑥‖,𝑦𝑥,𝑦∈𝒳. (42)Then (𝒳,‖⋅‖) is called a non-Archimedean normed space. Due to the fact that 𝑥_𝑛−𝑥_𝑚≤max𝑥_𝑗+1−𝑥_𝑗;𝑚≤𝑗≤𝑛−1(𝑛≥𝑚), (43) a sequence {𝑥_𝑛} is Cauchy if and only if {𝑥_𝑛+1−𝑥_𝑛} converges to zero in a non-Archimedean normed space 𝒳. By a complete non-Archimedean normed space, we mean one in which every Cauchy sequence is convergent.

In [20], Hensel discovered the 𝑝-adic numbers as a number theoretical analogue of power series in complex analysis. The most interesting example of non-Archimedean normed spaces is 𝑝-adic numbers. A key property of 𝑝-adic numbers is that they do not satisfy the Archimedean axiom; for all 𝑥,𝑦>0, there exists an integer 𝑛 such that 𝑥<𝑛𝑦.

Let 𝑝 be a prime number. For any nonzero rational number 𝑥=𝑝^𝑟(𝑚/𝑛) in which 𝑚 and 𝑛 are coprime to the prime number 𝑝. Consider the 𝑝-adic absolute value |𝑥|_𝑝=𝑝^−𝑟 on ℚ. It is easy to check that |⋅| is a non-Archimedean norm on ℚ. The completion of ℚ with respect to |⋅| which is denoted by ℚ_𝑝 is said to be the 𝑝-adic number field. One should remember that if 𝑝>2, then |2^𝑛|=1 for all integers 𝑛. The stability of some functional equations in non-Archimedean spaces was investigated, for instance, in [21–24] (see also [13, 25]).

Let 𝑚 be an even positive integer. We use the abbreviation for the given mapping 𝑓∶𝒳→𝒴 as follows: 𝒟_𝑚𝑓(𝑥)𝑥,𝑦∶=𝑓𝑥+𝑚𝑦+𝑓𝑥−𝑚𝑦−2𝑓+2𝑚²𝑓𝑦−𝑚²𝑓,𝒟2𝑦₃𝑓𝑦.𝑥,𝑦∶=𝑓𝑥+3𝑦+𝑓𝑥−3𝑦−𝑓𝑥+𝑦−𝑓𝑥−𝑦+16𝑓−8𝑓2𝑦 (44)

From now on, we assume that 𝒳 is a real vector space and 𝒴 is a complete non-Archimedean space unless otherwise stated explicitly. In the upcoming theorem, we prove the stability of functional equations (4) and (5).

Theorem 4. Let 𝜙∶𝒳×𝒳→[0,∞) be a function such that lim_𝑘→∞1|2|^𝑘𝜙2^𝑘𝑥,2^𝑘𝑦=0 (45) for all 𝑥,𝑦∈𝒳. Suppose that 𝑓∶𝒳→𝒴 is an odd mapping satisfying the inequality 𝒟_𝑚𝑓𝑥,𝑦≤𝜙𝑥,𝑦 (46) for all 𝑥,𝑦∈𝒳, where 𝑚 is an even positive integer or 𝑚=3. Then there exists a unique additive mapping 𝐴∶𝒳→𝒴 such that 𝑓(𝑥)(𝑥≤1−𝐴)2𝑚²𝜙(𝑥)1,𝑚𝑖𝑠𝑒𝑣𝑒𝑛,|8|𝜙(𝑥),𝑚=3 (47) for all 𝑥∈𝒳, where 𝜙(𝑥)=sup{(𝜙(0,2^𝑗𝑥)/|2|^𝑗)∶𝑗∈ℕ∪{0}}.

Proof. We prove the result when 𝑚 is even; another case is similar. Putting 𝑥=0 in (46), we have 𝑦≤12𝑓−𝑓2𝑦|𝑚|²𝜙0,𝑦 (48) for all 𝑦∈𝒳. Replacing 𝑦 by 2^𝑛𝑥 in (48) and then dividing both sides by |2|^𝑛+1, we get 12^𝑛𝑓2^𝑛𝑥−12^𝑛+1𝑓2^𝑛+1𝑥≤1|𝑚|²|2|^𝑛+1𝜙0,2^𝑛𝑥 (49) for all 𝑥∈𝒳 and all nonnegative integers 𝑛. Thus the sequence {𝑓(2^𝑛𝑥)/2^𝑛} is Cauchy by (45) and (49). Completeness of the non-Archimedean space 𝒴 allows us to assume that there exists a mapping 𝐴, so that lim_𝑛→∞𝑓2^𝑛𝑥2^𝑛=𝐴(𝑥). (50) For each 𝑥∈𝒳 and nonnegative integers 𝑛, we have 𝑓2^𝑛𝑥2^𝑛(𝑥=−𝑓)_{𝑛−1𝑗=0}𝑓2^𝑗+1𝑥2^𝑗+1−𝑓2^𝑗𝑥2^𝑗𝑓2≤max^𝑗+1𝑥2^𝑗+1−𝑓2^𝑗𝑥2^𝑗≤1∶0≤𝑗<𝑛2𝑚²𝜙max0,2^𝑗𝑥|2|^𝑗∶0≤𝑗<𝑛. (51) Taking that 𝑛 tends to approach infinity in (51) and applying (50), we can see that inequality (47) holds when 𝑚 is even. It follows from (45), (46), and (50) that for all 𝑥,𝑦∈𝒳, 𝒟_𝑚𝐴𝑥,𝑦=lim_𝑛→∞1|2|^𝑛𝒟_𝑚𝑓2^𝑛𝑥,2^𝑛𝑦≤lim_𝑛→∞1|2|^𝑛𝜙2^𝑛𝑥,2^𝑛𝑦=0. (52) Hence, the mapping 𝐴 satisfies (4). Part (i) of Lemma 1 shows that the mapping 𝐴 is additive. Now, let 𝐴^′∶𝒳→𝒴 be another additive mapping satisfying (47). Then we have 𝐴(𝑥)−𝐴^′(𝑥)=lim_𝑘→∞1|2|^𝑘𝐴2^𝑘𝑥−𝐴^′2^𝑘𝑥≤lim_𝑘→∞1|2|^𝑘max𝐴2^𝑘𝑥−𝑓2^𝑘𝑥,𝑓2^𝑘𝑥−𝐴^′2^𝑘𝑥≤12𝑚²lim_𝑘→∞lim_𝑛→∞𝜙max0,2^𝑗𝑥|2|^𝑗=1∶𝑘≤𝑗<𝑛+𝑘2𝑚²lim_𝑘→∞𝜙sup0,2^𝑗𝑥|2|^𝑗∶𝑘≤𝑗<∞=0 (53) for all 𝑥∈𝒳. This shows the uniqueness of 𝐴.

We have the following result which is analogous to Theorem 4 for the functional equations (4) and (5). We include the proof for (4). The proof of (5) is similar.

Theorem 5. Let 𝜙∶𝒳×𝒳→[0,∞) be a function such that lim_𝑘→∞|2|^𝑘𝜙𝑥2^𝑘,𝑦2^𝑘=0 (54) for all 𝑥,𝑦∈𝒳. Suppose that 𝑓∶𝒳→𝒴 is an odd mapping satisfying the inequality 𝒟_𝑚𝑓𝑥,𝑦_𝒴≤𝜙𝑥,𝑦 (55) for all 𝑥,𝑦∈𝒳, where 𝑚 is an even positive integer or 𝑚=3. Then there exists a unique additive mapping 𝐴∶𝒳→𝒴 such that 𝑓(𝑥)(𝑥≤1−𝐴)|𝑚|²𝜙(𝑥)1,𝑚𝑖𝑠𝑒𝑣𝑒𝑛,|8|𝜙(𝑥),𝑚=3 (56) for all 𝑥∈𝒳, where 𝜙(𝑥)=sup{|2|^𝑗𝜙(0,(𝑥/2^𝑗+1))∶𝑗∈ℕ∪{0}}.

Proof. We only obtain the result for the even integers. Similar to the proof of Theorem 4, we have 𝑦≤12𝑓−𝑓2𝑦|𝑚|²𝜙0,𝑦 (57) for all 𝑦∈𝒳. If we replace 𝑦 by 𝑥/2^𝑛+1 in inequality (57) and then multiply both sides of the result to |2|^𝑛, we get 2^𝑛+1𝑓𝑥2^𝑛+1−2^𝑛𝑓𝑥2^𝑛≤|2|^𝑛|𝑚|²𝜙𝑥0,2^𝑛+1 (58) for all 𝑥∈𝒳 and all nonnegative integers 𝑛. Thus, we conclude from (54) and (58) that the sequence {2^𝑛𝑓(𝑥/2^𝑛)} is Cauchy. Since the non-Archimedean normed space 𝒴 is complete, this sequence converges in 𝒴 to the mapping 𝐴. Indeed, 𝐴(𝑥)=lim_𝑛→∞2^𝑛𝑓𝑥2^𝑛(𝑥∈𝒳). (59) Using induction and (57), one can show that 2^𝑛𝑓𝑥2^𝑛(𝑥≤1−𝑓)|𝑚|²max|2|^𝑗𝜙𝑥0,2^𝑗+1∶0≤𝑗<𝑛 (60) for all 𝑥∈𝒳 and nonnegative integers 𝑛. Since the right-hand side of inequality (60) tends to be 0 as 𝑛 to approach infinity, by applying (59), we deduce inequality (56). Now, similar to the proof of Theorem 4, we can complete the rest of the proof.

Corollary 6. Let 𝛼,𝑟, and 𝑠 be positive real numbers such that 𝑟,𝑠≠1 and |2|<1. Suppose that 𝒳 is a non-Archimedean normed space and 𝑓∶𝒳→𝒴 is an odd mapping fulfilling 𝒟_𝑚𝑓𝑥,𝑦≤𝛼‖𝑥‖^𝑟+𝑦^𝑠 (61) for all 𝑥,𝑦∈𝒳, where 𝑚 is an even positive integer or 𝑚=3. Then there exists a unique additive mapping 𝐴∶𝒳→𝒴 such that 𝑓(𝑥)(𝑥≤𝛼‖𝑥‖−𝐴)^𝑠2𝑚²𝛼‖𝑥‖𝑚𝑖𝑠𝑒𝑣𝑒𝑛,𝑟,𝑠>1,_𝑠|8|𝛼‖𝑥‖𝑚=3,𝑟,𝑠>1,_𝑠|𝑚|²|2|^𝑠𝛼‖𝑥‖𝑚𝑖𝑠𝑒𝑣𝑒𝑛,𝑟,𝑠<1,_𝑠|8||2|^𝑠𝑚=3,𝑟,𝑠<1 (62) for all 𝑥∈𝒳.

Proof. The result follows from Theorems 4 and 5 by letting 𝜙(𝑥,𝑦)=𝛼(‖𝑥‖^𝑟+‖𝑦‖^𝑠).

In the next result, we prove the hyperstability of the functional equations (4) and (5) under some conditions. Recall that a functional equation is called hyperstable if every approximate solution is an exact one (see, e.g., [13, 26–29]).

Corollary 7. Let 𝛼,𝑟, and 𝑠 be positive real numbers such that 𝑟+𝑠≠1 and |2|<1. Suppose that 𝒳 is a non-Archimedean normed space and 𝑓∶𝒳→𝒴 is an odd mapping fulfilling 𝒟_𝑚𝑓𝑥,𝑦≤𝛼‖𝑥‖^𝑟𝑦^𝑠 (63) for all 𝑥,𝑦∈𝒳, where 𝑚 is an even positive integer or 𝑚=3. Then 𝑓 is an additive mapping.

Proof. Taking 𝜙(𝑥,𝑦)=𝛼‖𝑥‖^𝑟‖𝑦‖^𝑠 in Theorems 4 and 5, we can obtain the desired result.

Theorem 8. Let 𝜙∶𝒳×𝒳→[0,∞) be a function such that lim_𝑘→∞1|4|^𝑘𝜙2^𝑘𝑥,2^𝑘𝑦=0 (64) for all 𝑥,𝑦∈𝒳. Suppose that 𝑓∶𝒳→𝒴 is an even mapping satisfying the inequality 𝒟_𝑚𝑓𝑥,𝑦≤𝜙𝑥,𝑦 (65) for all 𝑥,𝑦∈𝒳, where 𝑚 is an even positive integer. Then there exists a unique quadratic mapping 𝑄∶𝒳→𝒴 such that 𝑓(𝑥)(𝑥≤1−𝑄)|4|𝜙(𝑥) (66) for all 𝑥∈𝒳, where 𝜙(𝑥)=sup{(𝜙(0,(2^𝑗𝑥/𝑚)/|4|^𝑗),(𝜙(2^𝑗𝑥,(2^𝑗𝑥/𝑚))/|4|^𝑗)∶𝑗∈ℕ∪{0}}.

Proof. It follows from (64) that 𝜙(0,0)=0. Thus (65) implies that 𝑓(0)=0. Putting 𝑥=0 in (65) and interchanging 𝑦 into 𝑥, we have 2𝑓(𝑚𝑥)+2𝑚²𝑓(𝑥)−𝑚²𝑓(2𝑥)≤𝜙(0,𝑥) (67) for all 𝑥∈𝒳. Substituting 𝑥,𝑦 by 𝑚𝑥,𝑥 in (65), respectively, we get 𝑓(2𝑚𝑥)−2𝑓(𝑚𝑥)+2𝑚²𝑓(𝑥)−𝑚²𝑓(2𝑥)≤𝜙(𝑚𝑥,𝑥) (68) for all 𝑥∈𝒳. It follows from (67) and (68) that 𝑓(2𝑚𝑥)−4𝑓(𝑚𝑥)≤max𝜙(0,𝑥),𝜙(𝑚𝑥,𝑥) (69) for all 𝑥∈𝒳. Thus we have 𝑓()(𝑥𝜙𝑥2𝑥−4𝑓)≤max0,𝑚𝑥,𝜙𝑥,𝑚 (70) for all 𝑥∈𝒳. Replacing 𝑥 by 2^𝑛𝑥 in (70) and then dividing both sides by |4|^𝑛+1, we get 14^𝑛+1𝑓2^𝑛+1𝑥−14^𝑛𝑓2^𝑛𝑥≤1|4|𝜙2max0,^𝑛𝑥/𝑚|4|^𝑛,𝜙2^𝑛𝑥,2^𝑛𝑥/𝑚|4|^𝑛 (71) for all 𝑥∈𝒳 and all nonnegative integers 𝑛. Thus the sequence {𝑓(2^𝑛𝑥)/4^𝑛} is Cauchy by (64) and (71). Since 𝒴 is complete, the sequence {𝑓(2^𝑛𝑥)/4^𝑛} converges in 𝒴 for all 𝑥∈𝒳. So one can define the mapping 𝑄∶𝒳→𝒴 by 𝑄(𝑥)∶=lim_𝑛→∞𝑓2^𝑛𝑥4^𝑛(𝑥∈𝒳). (72) For each 𝑥∈𝒳 and nonnegative integers 𝑛, we have 𝑓2^𝑛𝑥4^𝑛(𝑥=−𝑓)_{𝑛−1𝑗=0}𝑓2^𝑗+1𝑥4^𝑗+1−𝑓2^𝑗𝑥4^𝑗𝑓2≤max^𝑗+1𝑥4^𝑗+1−𝑓2^𝑗𝑥4^𝑗≤1∶0≤𝑗<𝑛|4|𝜙2max0,^𝑗𝑥/𝑚|4|^𝑗,𝜙2^𝑗𝑥,2^𝑗𝑥/𝑚|4|^𝑗∶0≤𝑗<𝑛. (73) Taking 𝑛 to approach infinity in (73) and using (64) and (72), we find (66). Employing (64), (65), and (72), we obtain 𝒟_𝑚𝑄𝑥,𝑦=lim_𝑛→∞1|4|^𝑛𝒟_𝑚𝑓2^𝑛𝑥,2^𝑛𝑦≤lim_𝑛→∞1|4|^𝑛𝜙2^𝑛𝑥,2^𝑛𝑦=0. (74) Hence, the mapping 𝑄 satisfies (4). It follows from part (iii) of Lemma 1 that the mapping 𝑄 is quadratic. If 𝑄^′∶𝒳→𝒴 is another quadratic mapping satisfying (66), then 𝑄(𝑥)−𝑄^′(𝑥)=lim_𝑘→∞1|4|^𝑘𝑄2^𝑘𝑥−𝑄^′2^𝑘𝑥≤lim_𝑘→∞1|4|^𝑘max𝑄2^𝑘𝑥−𝑓2^𝑘𝑥,𝑓2^𝑘𝑥−𝑄^′2^𝑘𝑥≤1|4|lim_𝑘→∞lim_𝑛→∞𝜙2max0,^𝑗𝑥/𝑚|4|^𝑗,𝜙2^𝑗𝑥,2^𝑗𝑥/𝑚|4|^𝑗=1∶𝑘≤𝑗<𝑛+𝑘|4|lim_𝑘→∞𝜙2sup0,^𝑗𝑥/𝑚|4|^𝑗,𝜙2^𝑗𝑥,2^𝑗𝑥/𝑚|4|^𝑗∶𝑘≤𝑗<∞=0 (75) for all 𝑥∈𝒳. Therefore 𝑄=𝑄^′. This completes the proof of the uniqueness of 𝑄.

Theorem 9. Let 𝜙∶𝒳×𝒳→[0,∞) be a function such that lim_𝑘→∞|4|^𝑘𝜙𝑥2^𝑘,𝑦2^𝑘=0 (76) for all 𝑥,𝑦∈𝒳. Suppose that 𝑓∶𝒳→𝒴 is an even mapping satisfying 𝑓(0)=0 and the inequality 𝒟_𝑚𝑓𝑥,𝑦≤𝜙𝑥,𝑦 (77) for all 𝑥,𝑦∈𝒳, where 𝑚 is an even positive integer. Then there exists a unique quadratic mapping 𝑄∶𝒳→𝒴 such that 𝑓(𝑥)(𝑥≤1−𝑄)|4|𝜙(𝑥) (78) for all 𝑥∈𝒳, where 𝜙(𝑥)=sup{|4|^𝑛𝜙(0,(𝑥/2^𝑛𝑚)),|4|^𝑛𝜙((𝑥/2^𝑛),(𝑥/2^𝑛𝑚))∶𝑛∈ℕ}.

Proof. Similar to the proof of Theorem 8, we have 𝑓()(𝑥𝜙𝑥2𝑥−4𝑓)≤max0,𝑚𝑥,𝜙𝑥,𝑚 (79) for all 𝑥∈𝒳. Then we get 𝑓(𝑥)𝑥−4𝑓2𝜙𝑥≤max0,𝑥2𝑚,𝜙2,𝑥2𝑚 (80) for all 𝑥∈𝒳. Replacing 𝑥 by 𝑥/2^𝑛 in (80) and multiplying both sides to |4|^𝑛, we get 4^𝑛𝑓𝑥2^𝑛−4^𝑛+1𝑓𝑥2^𝑛+1≤1|4|max|4|^𝑛+1𝜙𝑥0,2^𝑛+1𝑚,|4|^𝑛+1𝜙𝑥2^𝑛+1,𝑥2^𝑛+1𝑚 (81) for all 𝑥∈𝒳 and all nonnegative integers 𝑛. Thus the sequence {4^𝑛𝑓(𝑥/2^𝑛)} is Cauchy by (76). The completeness of 𝒴 implies that the mentioned sequence is convergent. So we consider the mapping 𝑄∶𝒳→𝒴 by 𝑄(𝑥)∶=lim_𝑛→∞4^𝑛𝑓𝑥2^𝑛(𝑥∈𝒳). (82) For each 𝑥∈𝒳 and nonnegative integers 𝑛, we have 𝑓(𝑥)−4^𝑛𝑓𝑥2^𝑛≤1|4|max|4|^𝑗𝜙𝑥0,2^𝑗𝑚,|4|^𝑗𝜙𝑥2^𝑗,𝑥2^𝑗𝑚∶1≤𝑗≤𝑛. (83) Letting 𝑛 approach infinity in (83) and using (76) and (82), we can see that (78) holds. The rest of the proof is similar to the proof of Theorem 8.

Corollary 10. Let 𝛼,𝑟, and 𝑠 be nonnegative real numbers and |2|<1. Suppose that 𝒳 is a non-Archimedean normed space and 𝑓∶𝒳→𝒴 is an even mapping fulfilling 𝒟_𝑚𝑓≤𝛼‖𝑥‖𝑥,𝑦^𝑟𝛼‖𝑥‖𝑟≠2,_𝑟+𝑦^𝑠𝑟,𝑠≠2 (84) for all 𝑥,𝑦∈𝒳, where 𝑚 is an even positive integer. Then there exists a unique quadratic mapping 𝑄∶𝒳→𝒴 such that 𝑓(𝑥)(𝑥≤𝛼‖𝑥‖−𝑄)^𝑟|4|𝛼𝑟>2,𝑠=0,|4|‖𝑥‖^𝑟+𝑥𝑚^𝑠𝛼‖𝑥‖𝑟,𝑠>2,_𝑟|2|^𝑟𝛼‖𝑥‖𝑟<2,𝑠=0,_𝑟|2|^𝑟+𝑥2𝑚^𝑠𝑟,𝑠<2 (85) for all 𝑥∈𝒳.

Theorem 11. Let 𝜙∶𝒳×𝒳→[0,∞) be a function such that lim_𝑘→∞1|4|^𝑘𝜙2^𝑘𝑥,2^𝑘𝑦=0 (86) for all 𝑥,𝑦∈𝒳. Suppose that 𝑓∶𝒳→𝒴 is an even mapping satisfying the inequality 𝒟₃𝑓𝑥,𝑦≤𝜙𝑥,𝑦 (87) for all 𝑥,𝑦∈𝒳. Then there exists a unique quadratic mapping 𝑄∶𝒳→𝒴 such that 𝑓(𝑥)(𝑥≤1−𝑄)|4|𝜙(𝑥) (88) for all 𝑥∈𝒳, where 𝜙(𝑥)=sup{|2|(𝜙((2^𝑗𝑥/4),(2^𝑗𝑥/4))/|4|^𝑗),(𝜙((5/4)2^𝑗𝑥,(2^𝑗𝑥/4))/|4|^𝑗),(𝜙((3/4)2^𝑗𝑥,(2^𝑗𝑥/4))/|4|^𝑗)∶𝑗∈ℕ∪{0}}.

Proof. Similar to the proof of Theorem 8, we can show that 𝑓(0)=0. Replacing 𝑥 by 𝑥+𝑦 and 𝑥−𝑦 in (87), respectively, we get 𝑓(𝑥)𝑦,𝑓(𝑥)𝑦𝑥+4𝑦+𝑓𝑥−2𝑦−𝑓𝑥+2𝑦−𝑓+16𝑓−8𝑓2𝑦≤𝜙𝑥+𝑦,𝑦𝑥+2𝑦+𝑓𝑥−4𝑦−𝑓𝑥−2𝑦−𝑓+16𝑓−8𝑓2𝑦≤𝜙𝑥−𝑦,𝑦 (89)(90) for all 𝑥,𝑦∈𝒳. Inequalities (89) and (90) imply that 𝑓(𝑥)𝑦𝜙𝑥+4𝑦+𝑓𝑥−4𝑦−2𝑓+32𝑓−16𝑓2𝑦≤max𝑥+𝑦,𝑦,𝜙𝑥−𝑦,𝑦 (91) for all 𝑥,𝑦∈𝒳. Interchanging 𝑥,𝑦 into 4𝑥,𝑥 in (91), respectively, we obtain 𝑓()()(𝑥)(𝜙()(8𝑥−2𝑓4𝑥+32𝑓−16𝑓2𝑥)≤max5𝑥,𝑥,𝜙3𝑥,𝑥) (92) for all 𝑥∈𝒳. On the other hand, by putting 𝑥=𝑦 in (87), we can deduce that 𝑓(4𝑥)−8𝑓(2𝑥)+16𝑓(𝑥)≤𝜙(𝑥,𝑥) (93) for all 𝑥∈𝒳. It follows from (92) and (93) that 𝑓(8𝑥)−4𝑓(4𝑥)≤max|2|𝜙(𝑥,𝑥),𝜙(5𝑥,𝑥),𝜙(3𝑥,𝑥) (94) for all 𝑥∈𝒳. Thus 14𝑓()(𝑥≤12𝑥−𝑓)|4|2|𝜙𝑥max|4,𝑥45,𝜙4𝑥𝑥,43,𝜙4𝑥𝑥,4 (95) for all 𝑥∈𝒳. Substituting 𝑥 by 2^𝑛𝑥 in (95) and then dividing both sides by |4|^𝑛, we obtain 14^𝑛+1𝑓2^𝑛+1𝑥−14^𝑛𝑓2^𝑛𝑥≤1|4|2|𝜙2max|^𝑛𝑥/4,2^𝑛𝑥/4|4|^𝑛,𝜙(5/4)2^𝑛𝑥,2^𝑛𝑥/4|4|^𝑛,𝜙(3/4)2^𝑛𝑥,2^𝑛𝑥/4|4|^𝑛 (96) for all 𝑥∈𝒳 and all nonnegative integers 𝑛. Thus the sequence {𝑓(2^𝑛𝑥)/4^𝑛} is Cauchy by (86) and (96). The completeness of 𝒴 implies that the sequence {𝑓(2^𝑛𝑥)/4^𝑛} is convergent. Define the mapping 𝑄∶𝒳→𝒴 via 𝑄(𝑥)∶=lim_𝑛→∞𝑓2^𝑛𝑥4^𝑛(𝑥∈𝒳). (97) By a simple computation, one can show that 𝑓2^𝑛𝑥4^𝑛(𝑥≤1−𝑓)|4|2|𝜙2max|^𝑗𝑥/4,2^𝑗𝑥/4|4|^𝑗,𝜙(5/4)2^𝑗𝑥,2^𝑗𝑥/4|4|^𝑗,𝜙(3/4)2^𝑗𝑥,2^𝑗𝑥/4|4|^𝑗∶0≤𝑗<𝑛 (98) for all 𝑥∈𝒳 and for all 𝑛≥0. Taking 𝑛 to approach infinity in (98) and applying (86) and (97), we find (88). By (86), (87), and (97), we have 𝒟₃𝑄𝑥,𝑦=lim_𝑛→∞1|4|^𝑛𝒟₃𝑓2^𝑛𝑥,2^𝑛𝑦≤lim_𝑛→∞1|4|^𝑛𝜙2^𝑛𝑥,2^𝑛𝑦=0. (99) Hence, the mapping 𝑄 satisfies (5). It follows from part (iv) of Lemma 1 that the mapping 𝑄 is quadratic. Similar to the proof of Theorem 8, one can show that 𝑄 is unique.

Theorem 12. Let 𝜙∶𝒳×𝒳→[0,∞) be a function such that lim_𝑘→∞|4|^𝑘𝜙𝑥2^𝑘,𝑦2^𝑘=0 (100) for all 𝑥,𝑦∈𝒳. Suppose that 𝑓∶𝒳→𝒴 is an even mapping satisfying the inequality 𝒟₃𝑓𝑥,𝑦≤𝜙𝑥,𝑦 (101) for all 𝑥,𝑦∈𝒳. Then there exists a unique quadratic mapping 𝑄∶𝒳→𝒴 such that 𝑓(𝑥)(𝑥≤1−𝑄)|4|𝜙(𝑥)(𝑥∈𝒳), (102) where 𝜙(𝑥)=sup{|2||4|^𝑗𝜙((𝑥/2^𝑗+3),(𝑥/2^𝑗+3)),|4|^𝑗𝜙(5(𝑥/2^𝑗+3),(𝑥/2^𝑗+3)),|4|^𝑗𝜙(3(𝑥/2^𝑗+3),(𝑥/2^𝑗+3))∶𝑗∈ℕ∪{0}}.

Proof. Similar to the proof of Theorem 8, one can obtain 𝑓(8𝑥)−4𝑓(4𝑥)≤max|2|𝜙(𝑥,𝑥),𝜙(5𝑥,𝑥),𝜙(3𝑥,𝑥) (103) for all 𝑥∈𝒳. Thus 𝑓(𝑥)𝑥−4𝑓22|𝜙𝑥≤max|8,𝑥85,𝜙8𝑥𝑥,83,𝜙8𝑥𝑥,8 (104) for all 𝑥∈𝒳. Substituting 𝑥 by 𝑥/2^𝑛 in (104) and then multiplying both sides to |4|^𝑛, we have 4^𝑛𝑓𝑥2^𝑛−4^𝑛+1𝑓𝑥2^𝑛+1≤max|2||4|^𝑛𝜙𝑥2^𝑛+3,𝑥2^𝑛+3,|4|^𝑛𝜙5𝑥2^𝑛+3,𝑥2^𝑛+3,|4|^𝑛𝜙3𝑥2^𝑛+3,𝑥2^𝑛+3 (105) for all 𝑥∈𝒳 and all nonnegative integers 𝑛. The last inequality and (100) imply that the sequence {𝑓(2^𝑛𝑥)/4^𝑛} is Cauchy. So, this sequence converges to the mapping 𝑄∶𝒳→𝒴. In other words, 𝑄(𝑥)∶=lim_𝑛→∞𝑓2^𝑛𝑥4^𝑛(𝑥∈𝒳). (106) We can also show that 𝑓(𝑥)−4^𝑛𝑓𝑥2^𝑛≤1|4|max|2||4|^𝑗𝜙𝑥2^𝑗+3,𝑥2^𝑗+3,|4|^𝑗𝜙5𝑥2^𝑗+3,𝑥2^𝑗+3,|4|^𝑗𝜙3𝑥2^𝑗+3,𝑥2^𝑗+3∶0≤𝑗<𝑛 (107) for all 𝑥∈𝒳 and for all 𝑛≥0. Taking 𝑛 to approach infinity in (107) and applying (100) and (101), we see that inequality (102) holds. Now, similar to the proof of Theorem 8, one can show that 𝑄 is a unique quadratic mapping.

Corollary 13. Let 𝛼 and 𝑟 be positive real numbers such that 𝑟≠2 and |2|<1. Suppose that 𝒳 is a non-Archimedean normed space and 𝑓∶𝒳→𝒴 is an even mapping fulfilling 𝒟₃𝑓𝑥,𝑦≤𝛼𝑦^𝑟 (108) for all 𝑥,𝑦∈𝒳. Then there exists a unique quadratic mapping 𝑄∶𝒳→𝒴 such that 𝑓(𝑥)(𝑥≤|2𝑥‖−𝑄)|‖^𝑟|4|^𝑟+1‖𝑥‖𝛼𝑟>2,_𝑟|2||8|^𝑟𝛼𝑟<2 (109) for all 𝑥∈𝒳.

Proof. Letting 𝜙(𝑥,𝑦)=𝛼‖𝑦‖^𝑟 in Theorems 11 and 12, one can obtain the required result.

Conflict of Interests

The authors declare that there is no conflict of interests regarding the publication of this paper.

Acknowledgments

The authors express their sincere thanks to the reviewers for the careful and detailed reading of the paper and very helpful suggestions that improved the paper substantially.