Intuitionistic Fuzzy Stability of Functional Equations Associated with Inner Product Spaces

Abstract

In intuitionistic fuzzy normed spaces, we investigate some stability results for the functional equation ∑^𝑛_𝑖=1𝑓(𝑥_𝑖−(1/𝑛)∑^𝑛_𝑗=1𝑥_𝑗)=∑^𝑛_𝑖=1𝑓(𝑥_𝑖)−𝑛𝑓((1/𝑛)∑^𝑛_𝑖=1𝑥_𝑖) which is said to be a functional equation associated with inner products space.

1. Introduction and Preliminaries

The aim of this article is to prove an intuitionistic fuzzy version of the Hyers-Ulam-Rassias stability for the functional equation: ^𝑛_𝑖=1𝑓𝑥_𝑖−1𝑛^𝑛_𝑗=1𝑥_𝑗=^𝑛_𝑖=1𝑓𝑥_𝑖1−𝑛𝑓𝑛^𝑛_𝑖=1𝑥_𝑖, (1.1) which is said to be a functional equation associated with inner product spaces. It was shown by Rassias [1] that the norm defined over a real vector space 𝑋 is induced by an inner product if and only if for a fixed integer 𝑛≥2 it follows ^𝑛_𝑖=1𝑥_𝑖−1𝑛^𝑛_𝑗=1𝑥_𝑗²=^𝑛_𝑖=1𝑥_𝑖²1−𝑛𝑛^𝑛_𝑖=1𝑥_𝑖², (1.2) for all 𝑥_𝑖,…,𝑥_𝑛∈𝑋. Interesting new results concerning functional equations associated with inner product spaces have recently been obtained by Park et al. [2, 3] and Najati and Rassias [4] as well as for the fuzzy stability of a functional equation associated with inner product spaces [5].

Stability problem of a functional equation was first posed by Ulam [6] which was answered by Hyers [7] on approximately additive mappings and then generalized by Aoki [8] and Rassias [9] for additive mappings and linear mappings, respectively. Later there have been proved several new results on stability of various classes of functional equations in the Hyers-Ulam sense (cf. the following books and papers [10–18] and the references cited therein), as well as various fuzzy stability results concerning Cauchy, Jensen, quadratic and cubic functional equations (cf. [19–22]). Furthermore some stability results concerning Jensen, cubic, mixed-type additive and cubic functional equations were investigated (cf. [23–26]) in the spirit of intuitionistic fuzzy normed spaces, while the idea of intuitionistic fuzzy normed space was introduced in [27] and further studied in [28–35].

In this section, we recall some notations and basic definitions used in this paper as follows.

Definition 1.1 (cf. [36]). A binary operation ∗∶[0,1]×[0,1]→[0,1] is said to be a continuous 𝑡-norm if it satisfies the following conditions:(a)∗ is commutative and associative, (b) ∗ is continuous,(c)𝑎∗1=𝑎 for all 𝑎∈[0,1], (d)𝑎∗𝑏≤𝑐∗𝑑 whenever 𝑎≤𝑐 and 𝑏≤𝑑 for each 𝑎,𝑏,𝑐,𝑑∈[0,1].

Definition 1.2 (cf. [36]). A binary operation ⋄∶[0,1]×[0,1]→[0,1] is said to be a continuous 𝑡-conorm if it satisfies the following conditions:(a^’)⋄ is commutative and associative, (b^’) ⋄ is continuous,(c^’)𝑎⋄0=𝑎 for all 𝑎∈[0,1], (d^’)𝑎⋄𝑏≤𝑐⋄𝑑 whenever 𝑎≤𝑐 and 𝑏≤𝑑 for each 𝑎,𝑏,𝑐,𝑑∈[0,1].
Using the notions of continuous 𝑡-norm and continuous 𝑡-conorm, Saadati and Park [27] have recently introduced the concept of intuitionistic fuzzy normed spaces as follows.

Definition 1.3. The five-tuple (𝑋,𝜇,𝜈,∗,⋄) is said to be an intuitionistic fuzzy normed space (for short, IFNS) if 𝑋 is a vector space, ∗ is a continuous 𝑡-norm, ⋄ is a continuous 𝑡-conorm, and 𝜇,𝜈 are fuzzy sets on 𝑋×(0,∞) satisfying the following conditions. For every 𝑥,𝑦∈𝑋 and 𝑠,𝑡>0,(i)𝜇(𝑥,𝑡)+𝜈(𝑥,𝑡)≤1, (ii)𝜇(𝑥,𝑡)>0, (iii)𝜇(𝑥,𝑡)=1 if and only if 𝑥=0,(iv)𝜇(𝛼𝑥,𝑡)=𝜇(𝑥,𝑡/|𝛼|) for each 𝛼≠0, (v)𝜇(𝑥,𝑡)∗𝜇(𝑦,𝑠)≤𝜇(𝑥+𝑦,𝑡+𝑠), (vi)𝜇(𝑥,⋅)∶(0,∞)→[0,1] is continuous, (vii)lim_𝑡→∞𝜇(𝑥,𝑡)=1 and lim_𝑡→0𝜇(𝑥,𝑡)=0,(viii)𝜈(𝑥,𝑡)<1, (ix) 𝜈(𝑥,𝑡)=0 if and only if 𝑥=0, (x)𝜈(𝛼𝑥,𝑡)=𝜈(𝑥,𝑡/|𝛼|) for each 𝛼≠0, (xi) 𝜈(𝑥,𝑡)⋄𝜈(𝑦,𝑠)≥𝜈(𝑥+𝑦,𝑡+𝑠), (xii)𝜈(𝑥,⋅)∶(0,∞)→[0,1] is continuous, (xiii)lim_𝑡→∞𝜈(𝑥,𝑡)=0 and lim_𝑡→0𝜈(𝑥,𝑡)=1. In this case (𝜇,𝜈) is called an intuitionistic fuzzy norm.

Example 1.4 (cf. [37]). Let(𝑋,‖⋅‖)be a normed space, 𝑎∗𝑏=𝑎𝑏,and𝑎⋄𝑏=min(𝑎+𝑏,1)for all 𝑎,𝑏∈[0,1]. For all 𝑥∈𝑋and every𝑡>0and𝑘=1,2, consider 𝜇_𝑘()=𝑡𝑥,𝑡𝑡+𝑘‖𝑥‖𝜈,if𝑡>0,0,if𝑡≤0,^𝑘()=𝑘‖𝑥‖𝑥,𝑡𝑡+𝑘‖𝑥‖,if𝑡>0,0,if𝑡≤0. (1.3) Then (𝑋,𝜇,𝜈,∗,⋄) is an IFNS.

The concepts of convergence and Cauchy sequences in an intuitionistic fuzzy normed space are studied in [27].

Let (𝑋,𝜇,𝜈,∗,⋄) be an IFNS. Then, a sequence {𝑥_𝑘} is said to be intuitionistic fuzzy convergent to 𝑥∈𝑋 if, for every 𝜀>0 and 𝑡>0, there exists 𝑘₀∈ℕ such that 𝜇(𝑥_𝑘−𝑥,𝑡)>1−𝜀 and 𝜈(𝑥_𝑘−𝑥,𝑡)<𝜀 for all 𝑘≥𝑘₀. In this case we write (𝜇,𝜈)−lim𝑥_𝑘=𝑥. The sequence {𝑥_𝑘} is said to be intuitionistic fuzzy Cauchy sequence if, for every 𝜀>0 and 𝑡>0, there exists 𝑘₀∈ℕ such that 𝜇(𝑥_𝑘−𝑥_ℓ,𝑡)>1−𝜀 and 𝜈(𝑥_𝑘−𝑥_ℓ,𝑡)<𝜀 for all 𝑘,ℓ≥𝑘₀. (𝑋,𝜇,𝜈,∗,⋄) is said to be complete if every intuitionistic fuzzy Cauchy sequence in (𝑋,𝜇,𝜈,∗,⋄) is intuitionistic fuzzy convergent in (𝑋,𝜇,𝜈,∗,⋄).

2. Intuitionistic Fuzzy Stability

Throughout this section, assume that 𝑋, (𝑍,𝜇^′,𝜈^′), and (𝑌,𝜇,𝜈) are linear space, IFNS, and intuitionistic fuzzy Banach space, respectively. For convenience, we use the following abbreviation for a given function 𝑓∶𝑋→𝑌: Δ𝑓𝑥₁,…,𝑥_𝑛=^𝑛_𝑖=1𝑓𝑥_𝑖−1𝑛^𝑛_𝑗=1𝑥_𝑗−^𝑛_𝑖=1𝑓𝑥_𝑖1+𝑛𝑓𝑛^𝑛_𝑖=1𝑥_𝑖. (2.1)

We begin with the Hyers-Ulam-Rassias type theorem in IFNS for the functional (1.1) which is said to be a functional equation associated with inner product spaces.

Theorem 2.1. Let 𝜑∶𝑋→𝑍 be a function such that 𝜑(2𝑥)=𝛼𝜑(𝑥) for some real number 𝛼 with 0<|𝛼|<4. Suppose that an even function 𝑓∶𝑋→𝑌 with 𝑓(0)=0 satisfies the inequality 𝜇Δ𝑓𝑥₁,…,𝑥_𝑛,𝑡₁+⋯+𝑡_𝑛≥𝜇^′𝜑𝑥₁,𝑡₁∗⋯∗𝜇^′𝜑𝑥_𝑛,𝑡_𝑛,𝜈Δ𝑓𝑥₁,…,𝑥_𝑛,𝑡₁+⋯+𝑡_𝑛≤𝜈^′𝜑𝑥₁,𝑡₁⋄⋯⋄𝜈^′𝜑𝑥_𝑛,𝑡_𝑛, (2.2) for all 𝑥₁,…,𝑥_𝑛∈𝑋 and all 𝑡₁,…,𝑡_𝑛>0. Then there exists a unique quadratic function 𝑄∶𝑋→𝑌 such that 𝜇𝑄(𝑥)−𝑓(𝑥),𝑡≥𝜇₁^′′(|𝛼𝑡𝑥,4−|)8,𝜈𝑄(𝑥)−𝑓(𝑥),𝑡≤𝜈₁^′′(|𝛼𝑡𝑥,4−|)8, (2.3) for all 𝑥∈𝑋 and 𝑡>0, where 𝜇₁^′′(𝑥,𝑡)∶=𝜇^′𝜑(),8()𝑛𝑥𝑛−12𝑛²𝑡+9𝑛∗𝜇^′𝜑)𝑥),8()((𝑛−1𝑛−12𝑛²𝑡+9𝑛∗𝜇^′𝜑(𝑥),8(𝑛−1)2𝑛²𝑡+9𝑛∗𝜇^′𝜑(0),8(𝑛−1)2𝑛²𝑡,𝜈+9𝑛_1′′(𝑥,𝑡)∶=𝜈^′𝜑(),8()𝑛𝑥𝑛−12𝑛²𝑡+9𝑛⋄𝜈^′𝜑)𝑥),8()((𝑛−1𝑛−12𝑛²𝑡+9𝑛⋄𝜈^′𝜑(𝑥),8(𝑛−1)2𝑛²𝑡+9𝑛⋄𝜈^′𝜑(0),8(𝑛−1)2𝑛²𝑡.+9𝑛 (2.4)

Proof. Put 𝑥₁=𝑛𝑥₁,𝑥_𝑖=𝑛𝑥₂(𝑖=2,…,𝑛),𝑡_𝑖=𝑡(𝑖=1,…,𝑛) in (2.2), and, using the evenness of 𝑓, we obtain 𝜇𝑛𝑓𝑥₁+(𝑛−1)𝑥₂+𝑓(𝑛−1)𝑥₁−𝑥₂+(𝑛−1)𝑓𝑥₁−𝑥₂−𝑓𝑛𝑥₁−(𝑛−1)𝑓𝑛𝑥₂,𝑛𝑡≥𝜇^′𝜑𝑛𝑥₁,𝑡∗𝜇^′𝜑𝑛𝑥₂,𝜈𝑥,𝑡𝑛𝑓₁+(𝑛−1)𝑥₂+𝑓(𝑛−1)𝑥₁−𝑥₂+(𝑛−1)𝑓𝑥₁−𝑥₂−𝑓𝑛𝑥₁−(𝑛−1)𝑓𝑛𝑥₂,𝑛𝑡≤𝜈^′𝜑𝑛𝑥₁,𝑡⋄𝜈^′𝜑𝑛𝑥₂,𝑡, (2.5) for all 𝑥₁,𝑥₂∈𝑋 and 𝑡>0. Interchanging 𝑥₁ with 𝑥₂ in (2.5) and using the evenness of 𝑓, we get 𝜇𝑛𝑓(𝑛−1)𝑥₁+𝑥₂+𝑓(𝑛−1)𝑥₁−𝑥₂+(𝑛−1)𝑓𝑥₁−𝑥₂−(𝑛−1)𝑓𝑛𝑥₁−𝑓𝑛𝑥₂,𝑛𝑡≥𝜇^′𝜑𝑛𝑥₁,𝑡∗𝜇^′𝜑𝑛𝑥₂,𝜈()𝑥,𝑡𝑛𝑓𝑛−1₁+𝑥₂+𝑓(𝑛−1)𝑥₁−𝑥₂+(𝑛−1)𝑓𝑥₁−𝑥₂−(𝑛−1)𝑓𝑛𝑥₁−𝑓𝑛𝑥₂,𝑛𝑡≤𝜈^′𝜑𝑛𝑥₁,𝑡⋄𝜈^′𝜑𝑛𝑥₂,𝑡, (2.6) for all 𝑥₁,𝑥₂∈𝑋 and 𝑡>0. It follows from (2.5) and (2.6) that 𝜇𝑛𝑓(𝑛−1)𝑥₁+𝑥₂+𝑛𝑓𝑥₁+(𝑛−1)𝑥₂+2𝑓(𝑛−1)𝑥₁−𝑥₂+2(𝑛−1)𝑓𝑥₁−𝑥₂−𝑛𝑓𝑛𝑥₁−𝑛𝑓𝑛𝑥₂,2𝑛𝑡≥𝜇^′𝜑𝑛𝑥₁,𝑡∗𝜇^′𝜑𝑛𝑥₂,𝜈)𝑥,𝑡𝑛𝑓(𝑛−1₁+𝑥₂+𝑛𝑓𝑥₁+(𝑛−1)𝑥₂+2𝑓(𝑛−1)𝑥₁−𝑥₂+2(𝑛−1)𝑓𝑥₁−𝑥₂−𝑛𝑓𝑛𝑥₁−𝑛𝑓𝑛𝑥₂,2𝑛𝑡≤𝜈^′𝜑n𝑥₁,𝑡⋄𝜈^′𝜑𝑛𝑥₂,𝑡, (2.7) for all 𝑥₁,𝑥₂∈𝑋 and 𝑡>0. Putting 𝑥₁=𝑛𝑥₁,𝑥₂=−𝑛𝑥₂,𝑥_𝑖=0(𝑖=3,…,𝑛),𝑡_𝑖=𝑡(𝑖=1,…,𝑛) in (2.2) and using the evenness of 𝑓, we obtain 𝜇𝑓(𝑛−1)𝑥₁+𝑥₂+𝑓𝑥₁+(𝑛−1)𝑥₂+2(𝑛−1)𝑓𝑥₁−𝑥₂−𝑓𝑛𝑥₁−𝑓𝑛𝑥₂,𝑛𝑡≥𝜇^′𝜑𝑛𝑥₁,𝑡∗𝜇^′𝜑−𝑛𝑥₂,𝑡∗𝜇^′𝜑(0),𝜈𝑓()𝑥,𝑡𝑛−1₁+𝑥₂+𝑓𝑥₁+(𝑛−1)𝑥₂+2(𝑛−1)𝑓𝑥₁−𝑥₂−𝑓𝑛𝑥₁−𝑓𝑛𝑥₂,𝑛𝑡≤𝜈^′𝜑𝑛𝑥₁,𝑡⋄𝜈^′𝜑−𝑛𝑥₂,𝑡⋄𝜈^′𝜑(0),𝑡, (2.8) for all 𝑥₁,𝑥₂∈𝑋 and 𝑡>0. Hence, we obtain from (2.7) and (2.8) that 𝜇𝑓(𝑛−1)𝑥₁−𝑥₂−(𝑛−1)²𝑓𝑥₁−𝑥₂,𝑛²+2𝑛2𝑡≥𝜇^′𝜑𝑛𝑥₁,𝑡∗𝜇^′𝜑𝑛𝑥₂,𝑡∗𝜇^′𝜑−𝑛𝑥₂,𝑡∗𝜇^′𝜑(0),𝜈𝑓(𝑥,𝑡𝑛−1)₁−𝑥₂−(𝑛−1)²𝑓𝑥₁−𝑥₂,𝑛²+2𝑛2𝑡≤𝜈^′𝜑𝑛𝑥₁,𝑡⋄𝜈^′𝜑𝑛𝑥₂,𝑡⋄𝜈^′𝜑−𝑛𝑥₂,𝑡⋄𝜈^′𝜑(0),𝑡, (2.9) for all 𝑥₁,𝑥₂∈𝑋 and 𝑡>0. So 𝜇𝑓((𝑛−1)𝑥)−(𝑛−1)²𝑓(𝑥),𝑛²+2𝑛2𝑡≥𝜇^′𝜑(𝑛𝑥),𝑡∗𝜇^′𝜑(0),𝜈𝑓)𝑥)−(),𝑡((𝑛−1𝑛−1²𝑓(𝑥),𝑛²+2𝑛2𝑡≤𝜈^′𝜑(𝑛𝑥),𝑡⋄𝜈^′𝜑(0),𝑡, (2.10) for all 𝑥∈𝑋 and 𝑡>0. Putting 𝑥₁=𝑛𝑥,𝑥_𝑖=0(𝑖=2,…,𝑛),𝑡_𝑖=𝑡(𝑖=1,…,𝑛) in (2.2), we obtain 𝜇𝑓(𝑛𝑥)−𝑓((𝑛−1)𝑥)−(2𝑛−1)𝑓(𝑥),𝑛𝑡≥𝜇^′𝜑(𝑛𝑥),𝑡∗𝜇^′𝜑(0),𝜈𝑓())𝑥)−()𝑓(𝑥),𝑡𝑛𝑥−𝑓((𝑛−12𝑛−1,𝑛𝑡≤𝜈^′𝜑(𝑛𝑥),𝑡⋄𝜈^′𝜑(0),𝑡, (2.11) for all 𝑥∈𝑋 and 𝑡>0. It follows from (2.10) and (2.11) that 𝜇𝑓(𝑛𝑥)−𝑛²𝑓(𝑥),𝑛²+4𝑛2𝑡≥𝜇^′𝜑(𝑛𝑥),𝑡∗𝜇^′𝜑(0),𝜈𝑓(),𝑡𝑛𝑥−𝑛²𝑓(𝑥),𝑛²+4𝑛2𝑡≤𝜈^′𝜑(𝑛𝑥),𝑡⋄𝜈^′𝜑(0),𝑡, (2.12) for all 𝑥∈𝑋 and 𝑡>0. Letting 𝑥₂=−(𝑛−1)𝑥₁ in (2.8) and replacing 𝑥₁ by 𝑥/𝑛 in the obtained inequality, we get 𝜇𝑓((𝑛−1)𝑥)−𝑓((𝑛−2)𝑥)−(2𝑛−3)𝑓(𝑥),𝑛𝑡≥𝜇^′𝜑(𝑥),𝑡∗𝜇^′𝜑((𝑛−1)𝑥),𝑡∗𝜇^′𝜑(0),𝜈𝑓)𝑥))𝑥)−()𝑓(𝑥),𝑡((𝑛−1−𝑓((𝑛−22𝑛−3,𝑛𝑡≤𝜈^′𝜑(𝑥),𝑡⋄𝜈^′𝜑((𝑛−1)𝑥),𝑡⋄𝜈^′𝜑(0),𝑡, (2.13) for all 𝑥∈𝑋 and 𝑡>0. It follows from (2.10), (2.11), (2.12), and (2.13) that 𝜇𝑓((𝑛−2)𝑥)−(𝑛−1)²𝑓(𝑥),𝑛²+4𝑛2𝑡≥𝜇^′𝜑(𝑛𝑥),𝑡∗𝜇^′𝜑((𝑛−1)𝑥),𝑡∗𝜇^′𝜑(𝑥),𝑡∗𝜇^′𝜑(0),𝜈𝑓)𝑥)−(),𝑡((𝑛−2𝑛−1²𝑓(𝑥),𝑛²+4𝑛2𝑡≤𝜈^′𝜑(𝑛𝑥),𝑡⋄𝜈^′𝜑((𝑛−1)𝑥),𝑡⋄𝜈^′𝜑(𝑥),𝑡⋄𝜈^′𝜑(0),𝑡, (2.14) for all 𝑥∈𝑋 and 𝑡>0. Applying (2.12) and (2.14), we obtain 𝜇𝑓(𝑛𝑥)−𝑓((𝑛−2)𝑥)−4(𝑛−1)𝑓(𝑥),𝑛²+4𝑛𝑡≥𝜇^′𝜑(𝑛𝑥),𝑡∗𝜇^′𝜑((𝑛−1)𝑥),𝑡∗𝜇^′𝜑(𝑥),𝑡∗𝜇^′𝜑(0),𝜈𝑓())𝑥)()𝑓(𝑥),𝑛,𝑡𝑛𝑥−𝑓((𝑛−2−4𝑛−1²+4𝑛𝑡≤𝜈^′𝜑(𝑛𝑥),𝑡⋄𝜈^′𝜑((𝑛−1)𝑥),𝑡⋄𝜈^′𝜑(𝑥),𝑡⋄𝜈^′𝜑(0),𝑡, (2.15) for all 𝑥∈𝑋 and 𝑡>0. Setting 𝑥₁=𝑥₂=𝑛𝑥,𝑥_𝑖=0(𝑖=3,…,𝑛),𝑡_𝑖=𝑡(𝑖=1,…,𝑛) in (2.2), we obtain 𝜇𝑓)𝑥)+()𝑓()(),𝑛((𝑛−2𝑛−12𝑥−𝑓𝑛𝑥2𝑡≥𝜇^′𝜑(𝑛𝑥),𝑡∗𝜇^′𝜑(0),𝜈𝑓)𝑥)+()𝑓()(),𝑛,𝑡((𝑛−2𝑛−12𝑥−𝑓𝑛𝑥2𝑡≤𝜈^′𝜑(𝑛𝑥),𝑡⋄𝜈^′𝜑(0),𝑡, (2.16) for all 𝑥∈𝑋 and 𝑡>0. It follows from (2.15) and (2.16) that 𝜇𝑓(2𝑥)−4𝑓(𝑥),2𝑛²+9𝑛𝑡2𝑛−2≥𝜇^′𝜑(𝑛𝑥),𝑡∗𝜇^′𝜑((𝑛−1)𝑥),𝑡∗𝜇^′𝜑(𝑥),𝑡∗𝜇^′𝜑(0),𝜈𝑓()(𝑥),,𝑡2𝑥−4𝑓2𝑛²+9𝑛𝑡2𝑛−2≤𝜈^′𝜑(𝑛𝑥),𝑡⋄𝜈^′𝜑((𝑛−1)𝑥),𝑡⋄𝜈^′𝜑(𝑥),𝑡⋄𝜈^′𝜑(0),𝑡. (2.17) It follows that 𝜇𝑓(𝑥)−4⁻¹𝑓(2𝑥),𝑡≥𝜇^′𝜑(),8()𝑛𝑥𝑛−12𝑛²𝑡+9𝑛∗𝜇^′𝜑)𝑥),8()((𝑛−1𝑛−12𝑛²𝑡+9𝑛∗𝜇^′𝜑(𝑥),8(𝑛−1)2𝑛²𝑡+9𝑛∗𝜇^′𝜑(0),8(𝑛−1)2𝑛²𝑡,𝜈𝑓(𝑥)+9𝑛−4⁻¹𝑓(2𝑥),𝑡≤𝜈^′𝜑(),8()𝑛𝑥𝑛−12𝑛²𝑡+9𝑛⋄𝜈^′𝜑)𝑥),8()((𝑛−1𝑛−12𝑛²𝑡+9𝑛⋄𝜈^′𝜑(𝑥),8(𝑛−1)2𝑛²𝑡+9𝑛⋄𝜈^′𝜑(0),8(𝑛−1)2𝑛²𝑡.+9𝑛 (2.18) Define 𝜇₁^′′(𝑥,𝑡)∶=𝜇^′𝜑(),8()𝑛𝑥𝑛−12𝑛²𝑡+9𝑛∗𝜇^′𝜑)𝑥),8()((𝑛−1𝑛−12𝑛²𝑡+9𝑛∗𝜇^′𝜑(𝑥),8(𝑛−1)2𝑛²𝑡+9𝑛∗𝜇^′𝜑(0),8(𝑛−1)2𝑛²𝑡,𝜈+9𝑛_1′′(𝑥,𝑡)∶=𝜈^′𝜑(),8()𝑛𝑥𝑛−12𝑛²𝑡+9𝑛⋄𝜈^′𝜑)𝑥),8()((𝑛−1𝑛−12𝑛²𝑡+9𝑛⋄𝜈^′𝜑(𝑥),8(𝑛−1)2𝑛²𝑡+9𝑛⋄𝜈^′𝜑(0),8(𝑛−1)2𝑛²𝑡.+9𝑛 (2.19) Then, by our assumption, 𝜇₁^′′(2𝑥,𝑡)=𝜇₁^′′𝑡𝑥,𝛼,𝜈₁^′′(2𝑥,𝑡)=𝜈₁^′′𝑡𝑥,𝛼. (2.20) Replacing 𝑥 by 2^𝑛𝑥 in (2.18) and applying (2.20), we get 𝜇𝑓2^𝑛𝑥4^𝑛−𝑓2^𝑛+1𝑥4^𝑛+1,𝛼^𝑛𝑡4^𝑛=𝜇𝑓2^𝑛𝑥−𝑓2^𝑛+1𝑥4,𝛼^𝑛𝑡≥𝜇₁^′′2^𝑛𝑥,𝛼^𝑛𝑡=𝜇₁^′′(),𝜈𝑓2𝑥,𝑡_𝑛𝑥4^𝑛−𝑓2^𝑛+1𝑥4^𝑛+1,𝛼^𝑛𝑡4^𝑛=𝜈𝑓2^𝑛𝑥−𝑓2^𝑛+1𝑥4,𝛼^𝑛𝑡≤𝜈₁^′′2^𝑛𝑥,𝛼^𝑛𝑡=𝜈₁^′′(𝑥,𝑡). (2.21) Thus for each 𝑛>𝑚, we have 𝜇𝑓2^𝑚𝑥4^𝑚−𝑓2^𝑛𝑥4^𝑛,^𝑛−1_𝑘=𝑚𝛼^𝑘𝑡4^𝑘=𝜇^𝑛−1_𝑘=𝑚𝑓2^𝑘𝑥4^𝑘−𝑓2^𝑘+1𝑥4^𝑘+1,^𝑛−1_𝑘=𝑚𝛼^𝑘𝑡4^𝑘≥^𝑛−1_𝑘=𝑚𝜇𝑓2^𝑘𝑥4^𝑘−𝑓2^𝑘+1𝑥4^𝑘+1,𝛼^𝑘𝑡4^𝑘≥𝜇₁^′′(),𝜈𝑓2𝑥,𝑡_𝑚𝑥4^𝑚−𝑓2^𝑛𝑥4^𝑛,^𝑛−1_𝑘=𝑚𝛼^𝑘𝑡4^𝑘=𝜈^𝑛−1_𝑘=𝑚𝑓2^𝑘𝑥4^𝑘−𝑓2^𝑘+1𝑥4^𝑘+1,^𝑛−1_𝑘=𝑚𝛼^𝑘𝑡4^𝑘≤^𝑛−1_𝑘=𝑚𝜈𝑓2^𝑘𝑥4^𝑘−𝑓2^𝑘+1𝑥4^𝑘+1,𝛼^𝑘𝑡4^𝑘≤𝜈₁^′′(𝑥,𝑡), (2.22) where ∏^𝑛_𝑗=1𝑎_𝑗=𝑎₁∗𝑎₂∗⋯∗𝑎_𝑛,∐^𝑛_𝑗=1𝑎_𝑗=𝑎₁⋄𝑎₂⋄⋯⋄𝑎_𝑛. Let 𝜀>0 and 𝛿>0 be given. Since lim_𝑡→∞𝜇₁^′′(𝑥,𝑡)=1 and lim_𝑡→∞𝜈₁^′′(𝑥,𝑡)=0, there exists some 𝑡₀>0 such that 𝜇₁^′′(𝑥,𝑡₀)>1−𝜀 and 𝜈₁^′′(𝑥,𝑡₀)<𝜀. Since ∑^∞_𝑘=0(𝛼^𝑘𝑡₀/4^𝑘)<∞, there is some 𝑛₀∈ℕ such that ∑^𝑛−1_𝑘=𝑚(𝛼^𝑘𝑡₀/4^𝑘)<𝛿 for each 𝑛>𝑚≥𝑛₀. It follows that 𝜇𝑓2^𝑚𝑥4^𝑚−𝑓2^𝑛𝑥4^𝑛𝑓2,𝛿≥𝜇^𝑚𝑥4^𝑚−𝑓2^𝑛𝑥4^𝑛,^𝑛−1_𝑘=𝑚𝛼^𝑘𝑡₀4^𝑘≥𝜇₁^′′𝑥,𝑡₀𝜈𝑓2>1−𝜀,_𝑚𝑥4^𝑚−𝑓2^𝑛𝑥4^𝑛𝑓2,𝛿≤𝜈^𝑚𝑥4^𝑚−𝑓2^𝑛𝑥4^𝑛,^𝑛−1_𝑘=𝑚𝛼^𝑘𝑡₀4^𝑘≤𝜈₁^′′𝑥,𝑡₀<𝜀, (2.23) for all 𝑡>𝑡₀. This shows that the sequence {𝑓(2^𝑛𝑥)/4^𝑛} is Cauchy in (𝑌,𝜇,𝜈). Since (𝑌,𝜇,𝜈) is intuitionistic fuzzy Banach space, {𝑓(2^𝑛𝑥)/4^𝑛} converges to some point 𝑄(𝑥)∈𝑌. Thus, we can define a mapping 𝑄(𝑥)∶𝑋→𝑌 such that 𝑄(𝑥)∶=(𝜇,𝜈)−lim_𝑛→∞(𝑓(2^𝑛𝑥)/4^𝑛). Moreover, if we put 𝑚=0 in (2.22), we get 𝜇𝑓2^𝑛𝑥4^𝑛−𝑓(𝑥),^𝑛−1_𝑘=0𝛼^𝑘𝑡4^𝑘≥𝜇₁^′′()𝑓2𝑥,𝑡,𝜈^𝑛𝑥4^𝑛−𝑓(𝑥),^𝑛−1_𝑘=0𝛼^𝑘𝑡4^𝑘≤𝜈₁^′′(𝑥,𝑡). (2.24) Thus, 𝜇𝑓2^𝑛𝑥4^𝑛−𝑓(𝑥),𝑡≥𝜇₁^′′𝑡𝑥,∑^𝑛−1_𝑘=0(𝛼/4)^𝑘,𝜈𝑓2^𝑛𝑥4^𝑛−𝑓(𝑥),𝑡≤𝜈₁^′′𝑡𝑥,∑^𝑛−1_𝑘=0(𝛼/4)^𝑘. (2.25)
Now, we will show that 𝑄 is quadratic. Setting 𝑥_𝑖=2^𝑚𝑥_𝑖(𝑖=1,…,𝑛) and 𝑡_𝑖=(𝑡/𝑛)(𝑖=1,…,𝑛) in (2.2), we obtain 𝜇Δ𝑓2^𝑚𝑥₁,…,2^𝑚𝑥_𝑛4^𝑚,𝑡≥𝜇^′𝜑2^𝑚𝑥₁,4^𝑚𝑡𝑛∗⋯∗𝜇^′𝜑2^𝑚𝑥_𝑛,4^𝑚𝑡𝑛=𝜇^′𝜑𝑥₁,4𝛼^𝑚𝑡𝑛∗⋯∗𝜇^′𝜑𝑥_𝑛,4𝛼^𝑚𝑡𝑛,𝜈Δ𝑓2^𝑚𝑥₁,…,2^𝑚𝑥_𝑛4^𝑚,𝑡≤𝜈^′𝜑2^𝑚𝑥₁,4^𝑚𝑡𝑛⋄⋯⋄𝜈^′𝜑2^𝑚𝑥_𝑛,4^𝑚𝑡𝑛=𝜈^′𝜑𝑥₁,4𝛼^𝑚𝑡𝑛⋄⋯⋄𝜈^′𝜑𝑥_𝑛,4𝛼^𝑚𝑡𝑛, (2.26) for all 𝑥₁,…,𝑥_𝑛∈𝑋 and 𝑡>0. Letting 𝑛→∞ in (2.26), we obtain 𝜇Δ𝑄𝑥₁,…,𝑥_𝑛,𝑡=1,𝜈Δ𝑄𝑥₁,…,𝑥_𝑛,𝑡=0, (2.27) for all 𝑥₁,…,𝑥_𝑛∈𝑋 and all 𝑡>0. This means that 𝑄 satisfies the functional (1.1) and so it is quadratic (see Lemma 2.2 of [4]).
Next, we approximate the difference between 𝑓 and 𝑄 in intuitionistic fuzzy sense. By (2.25), we have 𝜇𝑄(𝑥)(𝑥)𝑄(𝑥)−𝑓2−𝑓,𝑡≥𝜇^𝑛𝑥4^𝑛,𝑡2𝑓2∗𝜇^𝑛𝑥4^𝑛(𝑥),𝑡−𝑓2≥𝜇₁^′′𝑡𝑥,2∑^∞_𝑘=0(𝛼/4)^𝑘=𝜇₁^′′()𝑡𝑥,4−𝛼8,𝜈𝑄(𝑥)(𝑥)𝑄(𝑥)−𝑓2−𝑓,𝑡≤𝜈^𝑛𝑥4^𝑛,𝑡2𝑓2⋄𝜈^𝑛𝑥4^𝑛(𝑥),𝑡−𝑓2≤𝜈₁^′′𝑡𝑥,2∑^∞_𝑘=0(𝛼/4)^𝑘=𝜈₁^′′()𝑡𝑥,4−𝛼8, (2.28) for every 𝑥∈𝑋,𝑡>0 and large enough 𝑛. To prove the uniqueness of 𝑄, assume that 𝑄^′ is another quadratic mapping from 𝑋 to 𝑌, which satisfies the required inequality. Then, for each 𝑥∈𝑋 and 𝑡>0, 𝜇𝑄(𝑥)−𝑄^′(𝑥)𝑄(𝑥)(𝑥),𝑡,𝑡≥𝜇−𝑓2∗𝜇𝑄^′(𝑥)(𝑥),𝑡−𝑓2≥𝜇₁^′′()𝑡𝑥,4−𝛼,𝜈𝑄(𝑥)16−𝑄^′(𝑥)𝑄(𝑥)(𝑥),𝑡,𝑡≤𝜈−𝑓2⋄𝜈𝑄^′(𝑥)(𝑥),𝑡−𝑓2≤𝜈₁^′′()𝑡𝑥,4−𝛼.16 (2.29) Since 𝑄 and 𝑄^′ are quadratic, we have 𝜇𝑄(𝑥)−𝑄^′(𝑥),𝑡=𝜇𝑄2^𝑛𝑥−𝑄^′2^𝑛𝑥,4^𝑛𝑡≥𝜇₁^′′()𝑥,4/𝛼^𝑛(4−𝛼)𝑡,𝜈𝑄(𝑥)16−𝑄^′(𝑥),𝑡=𝜈𝑄2^𝑛𝑥−𝑄^′2^𝑛𝑥,4^𝑛𝑡≤𝜈₁^′′()𝑥,4/𝛼^𝑛(4−𝛼)𝑡,16 (2.30) for all 𝑥∈𝑋,𝑡>0 and 𝑛∈ℕ. Since 0<𝛼<4 and lim_𝑛→∞(4/𝛼)^𝑛=∞, wegetlim_𝑛→∞𝜇₁^′′()𝑥,4/𝛼^𝑛(4−𝛼)𝑡16=1,lim_𝑛→∞𝜈₁^′′()𝑥,4/𝛼^𝑛(4−𝛼)𝑡16=0. (2.31) Therefore 𝜇(𝑄(𝑥)−𝑄^′(𝑥),𝑡)=1 and 𝜈(𝑄(𝑥)−𝑄^′(𝑥),𝑡)=0 for all 𝑥∈𝑋 and 𝑡>0. Hence, 𝑄(𝑥)=𝑄^′(𝑥) for all 𝑥∈𝑋. This completes the proof of the theorem.

Theorem 2.2. Let 𝜑∶𝑋→𝑍 be a function such that 𝜑(2𝑥)=𝛼𝜑(𝑥) for some real number 𝛼 with 0<|𝛼|<2. Suppose that an odd function 𝑓∶𝑋→𝑌 satisfies the inequality 𝜇Δ𝑓𝑥₁,…,𝑥_𝑛,𝑡₁+⋯+𝑡_𝑛≥𝜇^′𝜑𝑥₁,𝑡₁∗⋯∗𝜇^′𝜑𝑥_𝑛,𝑡_𝑛,𝜈Δ𝑓𝑥₁,…,𝑥_𝑛,𝑡₁+⋯+𝑡_𝑛≤𝜈^′𝜑𝑥₁,𝑡₁⋄⋯⋄𝜈^′𝜑𝑥_𝑛,𝑡_𝑛, (2.32) for all 𝑥₁,…,𝑥_𝑛∈𝑋 and all 𝑡₁,…,𝑡_𝑛>0. Then there exists a unique additive function 𝐴∶𝑋→𝑌 such that 𝜇𝐴(𝑥)−𝑓(𝑥),𝑡≥𝜇₂^′′(|𝛼𝑡𝑥,2−|)4,𝜈𝐴(𝑥)−𝑓(𝑥),𝑡≤𝜈₂^′′(|𝛼𝑡𝑥,2−|)4, (2.33) for all 𝑥∈𝑋 and 𝑡>0, where 𝜇₂^′′(𝑥,𝑡)∶=𝜇^′𝜑(),42𝑥𝑛²𝑡+4𝑛∗𝜇^′𝜑(𝑥),4𝑛²𝑡+4𝑛∗𝜇^′𝜑(),4−𝑥𝑛²𝑡+4𝑛∗𝜇^′𝜑(0),4𝑛²𝑡,𝜈+4𝑛_2′′(𝑥,𝑡)∶=𝜈^′𝜑(),42𝑥𝑛²𝑡+4𝑛⋄𝜈^′𝜑(𝑥),4𝑛²𝑡+4𝑛⋄𝜈^′𝜑(),4−𝑥𝑛²𝑡+4𝑛⋄𝜈^′𝜑(0),4𝑛²𝑡.+4𝑛 (2.34)

Proof. Put 𝑥₁=𝑛𝑥₁,𝑥_𝑖=𝑛𝑥^′₁(𝑖=2,…,𝑛),𝑡_𝑖=𝑡(𝑖=1,…,𝑛) in (2.32) and using the oddness of 𝑓, we obtain 𝜇𝑛𝑓𝑥₁+(𝑛−1)𝑥^′₁+𝑓(𝑛−1)𝑥₁−𝑥^′₁−(𝑛−1)𝑓𝑥₁−𝑥^′₁−𝑓𝑛𝑥₁−(𝑛−1)𝑓𝑛𝑥^′₁,𝑛𝑡≥𝜇^′𝜑𝑛𝑥₁,𝑡∗𝜇^′𝜑𝑛𝑥^′₁,𝜈𝑥,𝑡𝑛𝑓₁+(𝑛−1)𝑥^′₁+𝑓(𝑛−1)𝑥₁−𝑥^′₁−(𝑛−1)𝑓𝑥₁−𝑥^′₁−𝑓𝑛𝑥₁−(𝑛−1)𝑓𝑛𝑥^′₁,𝑛𝑡≤𝜈^′𝜑𝑛𝑥₁,𝑡⋄𝜈^′𝜑𝑛𝑥^′₁,𝑡, (2.35) for all 𝑥₁,𝑥^′₁∈𝑋 and 𝑡>0. Interchanging 𝑥₁ with 𝑥^′₁ in (2.35) and using the oddness of 𝑓, we get 𝜇𝑛𝑓(𝑛−1)𝑥₁+𝑥^′₁−𝑓(𝑛−1)𝑥₁−𝑥^′₁+(𝑛−1)𝑓𝑥₁−𝑥^′₁−(𝑛−1)𝑓𝑛𝑥₁−𝑓𝑛𝑥^′₁,𝑛𝑡≥𝜇^′𝜑𝑛𝑥₁,𝑡∗𝜇^′𝜑𝑛𝑥^′₁,𝜈()𝑥,𝑡𝑛𝑓𝑛−1₁+𝑥^′₁−𝑓(𝑛−1)𝑥₁−𝑥^′₁+(𝑛−1)𝑓𝑥₁−𝑥^′₁−(𝑛−1)𝑓𝑛𝑥₁−𝑓𝑛𝑥^′₁,𝑛𝑡≤𝜈^′𝜑𝑛𝑥₁,𝑡⋄𝜈^′𝜑𝑛𝑥^′₁,𝑡, (2.36) for all 𝑥₁,𝑥^′₁∈𝑋 and 𝑡>0. It follows from (2.35) and (2.36) that 𝜇𝑛𝑓𝑥₁+(𝑛−1)𝑥^′₁−𝑛𝑓(𝑛−1)𝑥₁+𝑥^′₁+2𝑓(𝑛−1)𝑥₁−𝑥^′₁−2(𝑛−1)𝑓𝑥₁−𝑥^′₁+(𝑛−2)𝑓𝑛𝑥₁−(𝑛−2)𝑓𝑛𝑥^′₁,2𝑛𝑡≥𝜇^′𝜑𝑛𝑥₁,𝑡∗𝜇^′𝜑𝑛𝑥^′₁,𝜈𝑥,𝑡𝑛𝑓₁+(𝑛−1)𝑥^′₁−𝑛𝑓(𝑛−1)𝑥₁+𝑥^′₁+2𝑓(𝑛−1)𝑥₁−𝑥^′₁−2(𝑛−1)𝑓𝑥₁−𝑥^′₁+(𝑛−2)𝑓𝑛𝑥₁−(𝑛−2)𝑓𝑛𝑥^′₁,2𝑛𝑡≤𝜈^′𝜑𝑛𝑥₁,𝑡⋄𝜈^′𝜑𝑛𝑥^′₁,𝑡, (2.37) for all 𝑥₁,𝑥^′₁∈𝑋 and 𝑡>0. Setting 𝑥₁=𝑛𝑥₁,𝑥₂=−𝑛𝑥^′₁,𝑥_𝑖=0(𝑖=3,…,𝑛),𝑡_𝑖=𝑡(𝑖=1,…,𝑛) in (2.32) and using the oddness of 𝑓, we get 𝜇𝑓(𝑛−1)𝑥₁+𝑥^′₁−𝑓𝑥₁+(𝑛−1)𝑥^′₁+2𝑓𝑥₁−𝑥^′₁−𝑓𝑛𝑥₁+𝑓𝑛𝑥^′₁,𝑛𝑡≥𝜇^′𝜑𝑛𝑥₁,𝑡∗𝜇^′𝜑−𝑛𝑥^′₁,𝑡∗𝜇^′𝜑(0),𝜈𝑓()𝑥,𝑡𝑛−1₁+𝑥^′₁−𝑓𝑥₁+(𝑛−1)𝑥^′₁+2𝑓𝑥₁−𝑥^′₁−𝑓𝑛𝑥₁+𝑓𝑛𝑥^′₁,𝑛𝑡≤𝜈^′𝜑𝑛𝑥₁,𝑡⋄𝜈^′𝜑−𝑛𝑥^′₁,𝑡⋄𝜈^′𝜑(0),𝑡, (2.38) for all 𝑥₁,𝑥^′₁∈𝑋 and 𝑡>0. It follows from (2.37) and (2.38) that 𝜇𝑓(𝑛−1)𝑥₁−𝑥^′₁+𝑓𝑥₁−𝑥^′₁−𝑓𝑛𝑥₁+𝑓𝑛𝑥^′₁,𝑛²+2𝑛2𝑡≥𝜇^′𝜑𝑛𝑥₁,𝑡∗𝜇^′𝜑𝑛𝑥^′₁,𝑡∗𝜇^′𝜑−𝑛𝑥^′₁,𝑡∗𝜇^′𝜑(0),𝜈𝑓()𝑥,𝑡𝑛−1₁−𝑥^′₁+𝑓𝑥₁−𝑥^′₁−𝑓𝑛𝑥₁+𝑓𝑛𝑥^′₁,𝑛²+2𝑛2𝑡≤𝜈^′𝜑𝑛𝑥₁,𝑡⋄𝜈^′𝜑𝑛𝑥^′₁,𝑡⋄𝜈^′𝜑−𝑛𝑥^′₁,𝑡⋄𝜈^′𝜑(0),𝑡, (2.39) for all 𝑥₁,𝑥^′₁∈𝑋 and 𝑡>0. Putting 𝑥₁=𝑛(𝑥₁−𝑥^′₁),𝑥_𝑖=0(𝑖=2,…,𝑛),𝑡_𝑖=𝑡(𝑖=1,…,𝑛) in (2.32), we obtain 𝜇𝑓𝑛𝑥₁−𝑥^′₁−𝑓(𝑛−1)𝑥₁−𝑥^′₁−𝑓𝑥₁−𝑥^′₁,𝑛𝑡≥𝜇^′𝜑𝑛𝑥₁−𝑥^′₁,𝑡∗𝜇^′𝜑(0),𝜈𝑓𝑛𝑥,𝑡₁−𝑥^′₁−𝑓(𝑛−1)𝑥₁−𝑥^′₁−𝑓𝑥₁−𝑥^′₁,𝑛𝑡≤𝜈^′𝜑𝑛𝑥₁−𝑥^′₁,𝑡⋄𝜈^′𝜑(0),𝑡, (2.40) for all 𝑥₁,𝑥^′₁∈𝑋 and 𝑡>0. It follows from (2.39) and (2.40) that 𝜇𝑓𝑛𝑥₁−𝑥^′₁−𝑓𝑛𝑥₁+𝑓𝑛𝑥^′₁,𝑛²+4𝑛2𝑡≥𝜇^′𝜑𝑛𝑥₁−𝑥^′₁,𝑡∗𝜇^′𝜑𝑛𝑥₁,𝑡∗𝜇^′𝜑𝑛𝑥^′₁,𝑡∗𝜇^′𝜑−𝑛𝑥^′₁,𝑡∗𝜇^′𝜑(0),𝜈𝑓𝑛𝑥,𝑡₁−𝑥^′₁−𝑓𝑛𝑥₁+𝑓𝑛𝑥^′₁,𝑛²+4𝑛2𝑡≤𝜈^′𝜑𝑛𝑥₁−𝑥^′₁,𝑡⋄𝜈^′𝜑𝑛𝑥₁,𝑡⋄𝜈^′𝜑𝑛𝑥^′₁,𝑡⋄𝜈^′𝜑−𝑛𝑥^′₁,𝑡⋄𝜈^′𝜑(0),𝑡, (2.41) for all 𝑥₁,𝑥^′₁∈𝑋 and 𝑡>0. Replacing 𝑥₁ and 𝑥^′₁ by 𝑥/𝑛 and −𝑥/𝑛 in (2.41); respectively, we have 𝜇𝑓()(𝑥),𝑛2𝑥−2𝑓²+4𝑛2𝑡≥𝜇^′𝜑(2𝑥),𝑡∗𝜇^′𝜑(𝑥),𝑡∗𝜇^′𝜑(−𝑥),𝑡∗𝜇^′𝜑(0),𝜈𝑓()(𝑥),𝑛,𝑡2𝑥−2𝑓²+4𝑛2𝑡≤𝜈^′𝜑(2𝑥),𝑡⋄𝜈^′𝜑(𝑥),𝑡⋄𝜈^′𝜑(−𝑥),𝑡⋄𝜈^′𝜑(0),𝑡. (2.42) It follows that 𝜇𝑓(𝑥)−2⁻¹𝑓(2𝑥),𝑡≥𝜇^′𝜑(),42𝑥𝑛²𝑡+4𝑛∗𝜇^′𝜑(𝑥),4𝑛²𝑡+4𝑛∗𝜇^′𝜑(),4−𝑥𝑛²𝑡+4𝑛∗𝜇^′𝜑(0),4𝑛²𝑡,𝜈𝑓(𝑥)+4𝑛−2⁻¹𝑓(2𝑥),𝑡≤𝜈^′𝜑(),42𝑥𝑛²𝑡+4𝑛⋄𝜈^′𝜑(𝑥),4𝑛²𝑡+4𝑛⋄𝜈^′𝜑(),4−𝑥𝑛²𝑡+4𝑛⋄𝜈^′𝜑(0),4𝑛²𝑡.+4𝑛 (2.43) Define 𝜇₂^′′(𝑥,𝑡)∶=𝜇^′𝜑(),42𝑥𝑛²𝑡+4𝑛∗𝜇^′𝜑(𝑥),4𝑛²𝑡+4𝑛∗𝜇^′𝜑(),4−𝑥𝑛²𝑡+4𝑛∗𝜇^′𝜑(0),4𝑛²𝑡,𝜈+4𝑛_2′′(𝑥,𝑡)∶=𝜈^′𝜑(),42𝑥𝑛²𝑡+4𝑛⋄𝜈^′𝜑(𝑥),4𝑛²𝑡+4𝑛⋄𝜈^′𝜑(),4−𝑥𝑛²𝑡+4𝑛⋄𝜈^′𝜑(0),4𝑛²𝑡.+4𝑛 (2.44) Then by the assumption 𝜇₂^′′(2𝑥,𝑡)=𝜇₂^′′𝑡𝑥,𝛼,𝜈₂^′′(2𝑥,𝑡)=𝜈₂^′′𝑡𝑥,𝛼. (2.45) Replacing 𝑥 by 2^𝑛𝑥 in (2.43) and using (2.45), we obtain 𝜇𝑓2^𝑛𝑥2^𝑛−𝑓2^𝑛+1𝑥2^𝑛+1,𝛼^𝑛t2^𝑛=𝜇𝑓2^𝑛𝑥−𝑓2^𝑛+1𝑥2,𝛼^𝑛𝑡≥𝜇₂^′′2^𝑛𝑥,𝛼^𝑛𝑡=𝜇₂^′′(),𝜈𝑓2𝑥,𝑡_𝑛𝑥2^𝑛−𝑓2^𝑛+1𝑥2^𝑛+1,𝛼^𝑛𝑡2^𝑛=𝜈𝑓2^𝑛𝑥−𝑓2^𝑛+1𝑥2,𝛼^𝑛𝑡≤𝜈₂^′′2^𝑛𝑥,𝛼^𝑛𝑡=𝜈₂^′′(𝑥,𝑡). (2.46) Thus, for each 𝑛>𝑚, we have 𝜇𝑓2^𝑚𝑥2^𝑚−𝑓2^𝑛𝑥2^𝑛,^𝑛−1_𝑘=𝑚𝛼^𝑘𝑡2^𝑘=𝜇^𝑛−1_𝑘=𝑚𝑓2^𝑘𝑥2^𝑘−𝑓2^𝑘+1𝑥2^𝑘+1,^𝑛−1_𝑘=𝑚𝛼^𝑘𝑡2^𝑘≥^𝑛−1_𝑘=𝑚𝜇𝑓2^𝑘𝑥2^𝑘−𝑓2^𝑘+1𝑥2^𝑘+1,𝛼^𝑘𝑡2^𝑘≥𝜇₂^′′(),𝜈𝑓2𝑥,𝑡_𝑚𝑥2^𝑚−𝑓2^𝑛𝑥2^𝑛,^𝑛−1_𝑘=𝑚𝛼^𝑘𝑡2^𝑘=𝜈^𝑛−1_k=𝑚𝑓2^𝑘𝑥2^𝑘−𝑓2^𝑘+1𝑥2^𝑘+1,^𝑛−1_𝑘=𝑚𝛼^𝑘𝑡2^𝑘≤^𝑛−1_𝑘=𝑚𝜈𝑓2^𝑘𝑥2^𝑘−𝑓2^𝑘+1𝑥2^𝑘+1,𝛼^𝑘𝑡2^𝑘≤𝜈₂^′′(𝑥,𝑡), (2.47) where ∏^𝑛_𝑗=1𝑎_𝑗=𝑎₁∗𝑎₂∗⋯∗𝑎_𝑛, ∐^𝑛_𝑗=1𝑎_𝑗=𝑎₁⋄𝑎₂⋄⋯⋄𝑎_𝑛. Let 𝜀>0 and 𝛿>0 be given. Since lim_𝑡→∞𝜇₂^′′(𝑥,𝑡)=1 and lim_𝑡→∞𝜈₂^′′(𝑥,𝑡)=0, there exists some 𝑡₀>0 such that 𝜇₂^′′(𝑥,𝑡₀)>1−𝜀 and 𝜈₂^′′(𝑥,𝑡₀)<𝜀. Since ∑^∞_𝑘=0𝛼^𝑘𝑡₀/2^𝑘<∞, there is some 𝑛₀∈ℕ such that ∑^𝑛−1_𝑘=𝑚𝛼^𝑘𝑡₀/2^𝑘<𝛿 for each 𝑛>𝑚≥𝑛₀. It follows that 𝜇𝑓2^𝑚𝑥2^𝑚−𝑓2^𝑛𝑥2^𝑛𝑓2,𝛿≥𝜇^𝑚𝑥2^𝑚−𝑓2^𝑛𝑥2^𝑛,^𝑛−1_𝑘=𝑚𝛼^𝑘𝑡₀2^𝑘≥𝜇₂^′′𝑥,𝑡₀𝜈𝑓2>1−𝜀,_𝑚𝑥2^𝑚−𝑓2^𝑛𝑥2^𝑛𝑓2,𝛿≤𝜈^𝑚𝑥2^𝑚−𝑓2^𝑛𝑥2^𝑛,^𝑛−1_𝑘=𝑚𝛼^𝑘𝑡₀2^𝑘≤𝜈₂^′′𝑥,𝑡₀<𝜀, (2.48) for all 𝑡>𝑡₀. This shows that the sequence {𝑓(2^𝑛𝑥)/2^𝑛} is Cauchy in (𝑌,𝜇,𝜈). Since (𝑌,𝜇,𝜈) is intuitionistic fuzzy Banach space, {𝑓(2^𝑛𝑥)/4^𝑛} converges to some point 𝐴(𝑥)∈𝑌. Thus, we can define a mapping 𝐴(𝑥)∶𝑋→𝑌 such that 𝐴(𝑥)∶=(𝜇,𝜈)−lim_𝑛→∞𝑓(2^𝑛𝑥)/2^𝑛. Moreover, if we put 𝑚=0 in (2.47), we get 𝜇𝑓2^𝑛𝑥2^𝑛−𝑓(𝑥),^𝑛−1_𝑘=0𝛼^𝑘𝑡2^𝑘≥𝜇₂^′′()𝑓2𝑥,𝑡,𝜈^𝑛𝑥2^𝑛−𝑓(𝑥),^𝑛−1_𝑘=0𝛼^𝑘𝑡2^𝑘≤𝜈₂^′′(𝑥,𝑡). (2.49) Thus, 𝜇𝑓2^𝑛𝑥2^𝑛−𝑓(𝑥),𝑡≥𝜇₂^′′𝑡𝑥,∑^𝑛−1_𝑘=0(𝛼/2)^𝑘𝑓2,𝜈^𝑛𝑥2^𝑛−𝑓(𝑥),𝑡≤𝜈₂^′′𝑡𝑥,∑^𝑛−1_𝑘=0(𝛼/2)^𝑘. (2.50) Next we will show that 𝐴 is additive. Putting 𝑥_𝑖=2^𝑚𝑥_𝑖(𝑖=1,…,𝑛) and 𝑡_𝑖=(𝑡/𝑛)(𝑖=1,…,𝑛) in (2.32), we obtain 𝜇Δ𝑓2^𝑚𝑥₁,…,2^𝑚𝑥_𝑛2^𝑚,𝑡≥𝜇^′𝜑2^𝑚𝑥₁,2^𝑚𝑡𝑛∗⋯∗𝜇^′𝜑2^𝑚𝑥_𝑛,2^𝑚𝑡𝑛=𝜇^′𝜑𝑥₂,2𝛼^𝑚𝑡𝑛∗⋯∗𝜇^′𝜑𝑥_𝑛,2𝛼^𝑚𝑡𝑛,𝜈Δ𝑓2^𝑚𝑥₁,…,2^𝑚𝑥_𝑛2^𝑚,𝑡≤𝜈^′𝜑2^𝑚𝑥₁,2^𝑚𝑡𝑛⋄⋯⋄𝜈^′𝜑2^𝑚𝑥_𝑛,2^𝑚𝑡𝑛=𝜈^′𝜑𝑥₂,2𝛼^𝑚𝑡𝑛⋄⋯⋄𝜈^′𝜑𝑥_𝑛,2𝛼^𝑚𝑡𝑛, (2.51) for all 𝑥₁,…,𝑥_𝑛∈𝑋 and 𝑡>0. Letting 𝑛→∞ in (2.51), we obtain 𝜇Δ𝐴𝑥₁,…,𝑥_𝑛,𝑡=1,𝜈Δ𝐴𝑥₁,…,𝑥_𝑛,𝑡=0, (2.52) for all 𝑥₁,…,𝑥_𝑛∈𝑋 and all 𝑡>0. This means that 𝐴 satisfies the functional (1.1), and so it is additive (see Lemma  2.1 of [4]).
Now, we approximate the difference between 𝑓 and 𝐴 in intuitionistic fuzzy sense. For every 𝑥∈𝑋,𝑡>0, and sufficiently large 𝑛, by (2.50), we have 𝜇𝐴(𝑥)(𝑥)𝐴(𝑥)−𝑓2−𝑓,𝑡≥𝜇^𝑛𝑥2^𝑛,𝑡2𝑓2∗𝜇^𝑛𝑥2^𝑛(𝑥),𝑡−𝑓2≥𝜇₂^′′𝑡𝑥,2∑^∞_𝑘=0(𝛼/2)^𝑘=𝜇₂^′′()𝑡𝑥,2−𝛼4,𝜈𝐴(𝑥)(𝑥)𝐴(𝑥)−𝑓2−𝑓,𝑡≤𝜈^𝑛𝑥2^𝑛,𝑡2𝑓2⋄𝜈^𝑛𝑥2^𝑛(𝑥),𝑡−𝑓2≤𝜈₂^′′𝑡𝑥,2∑^∞_𝑘=0(𝛼/2)^𝑘=𝜈₂^′′()𝑡𝑥,2−𝛼4. (2.53) To prove the uniqueness of 𝐴, assume that 𝐴^′ is another additive mapping from 𝑋 to 𝑌, which satisfies the required inequality. Then, for each 𝑥∈𝑋 and 𝑡>0, 𝜇𝐴(𝑥)−𝐴^′(𝑥)𝐴(𝑥)(𝑥),𝑡,𝑡≥𝜇−𝑓2∗𝜇𝐴^′(𝑥)(𝑥),𝑡−𝑓2≥𝜇₂^′′()𝑡𝑥,2−𝛼8,𝜈𝐴(𝑥)−𝐴^′(𝑥)𝐴(𝑥)(𝑥),𝑡,𝑡≤𝜈−𝑓2⋄𝜈𝐴^′(𝑥)(𝑥),𝑡−𝑓2≤𝜈₂^′′()𝑡𝑥,2−𝛼8. (2.54) Therefore, by the additivity of 𝐴 and 𝐴^′, we have 𝜇𝐴(𝑥)−𝐴^′(𝑥),𝑡=𝜇𝐴2^𝑛𝑥−𝐴^′2^𝑛𝑥,2^𝑛𝑡≥𝜇₂^′′()𝑥,2/𝛼^𝑛(2−𝛼)𝑡8,𝜈𝐴(𝑥)−𝐴^′(𝑥),𝑡=𝜈𝐴2^𝑛𝑥−𝐴^′2^𝑛𝑥,2^𝑛𝑡≤𝜈₂^′′()𝑥,2/𝛼^𝑛(2−𝛼)𝑡8, (2.55) for all 𝑥∈𝑋,𝑡>0 and 𝑛∈ℕ. Since 0<𝛼<2 and lim_𝑛→∞(2/𝛼)^𝑛=∞, we get lim_𝑛→∞𝜇₂^′′()𝑥,2/𝛼^𝑛(2−𝛼)𝑡8=1,lim_𝑛→∞𝜈₂^′′()𝑥,4/𝛼^𝑛(2−𝛼)𝑡8=0. (2.56) Therefore, 𝜇(𝐴(𝑥)−𝐴^′(𝑥),𝑡)=1 and 𝜈(𝐴(𝑥)−𝐴^′(𝑥),𝑡)=0 for all 𝑥∈𝑋 and 𝑡>0. Hence, 𝐴(𝑥)=𝐴^′(𝑥) for all 𝑥∈𝑋. This completes the proof of the theorem.

Theorem 2.3. Let 𝜑∶𝑋→𝑍 be a function such that 𝜑(2𝑥)=𝛼𝜑(𝑥) for some real number 𝛼 with 0<|𝛼|<2. Suppose that a function 𝑓∶𝑋→𝑌 with 𝑓(0)=0 satisfies the inequality 𝜇Δ𝑓𝑥₁,…,𝑥_𝑛,𝑡₁+⋯+𝑡_𝑛≥𝜇^′𝜑𝑥₁,𝑡₁∗⋯∗𝜇^′𝜑𝑥_𝑛,𝑡_𝑛,𝜈Δ𝑓𝑥₁,…,𝑥_𝑛,𝑡₁+⋯+𝑡_𝑛≤𝜈^′𝜑𝑥₁,𝑡₁⋄⋯⋄𝜈^′𝜑𝑥_𝑛,𝑡_𝑛, (2.57) for all 𝑥₁,…,𝑥_𝑛∈𝑋 and all 𝑡₁,…,𝑡_𝑛>0. Then there exists a unique quadratic function 𝑄∶𝑋→𝑌 and a unique additive function 𝐴∶𝑋→𝑌 such that 𝜇𝑄(𝑥)−𝐴(𝑥)−𝑓(𝑥),𝑡≥𝑀₁^′′(|𝛼𝑡𝑥,4−|)∗𝑀16^1′′(|𝛼𝑡𝑥,2−|)8,𝜈𝑄(𝑥)−𝐴(𝑥)−𝑓(𝑥),𝑡≤𝑀₂^′′(|𝛼𝑡𝑥,4−|)⋄𝑀16^2′′(|𝛼𝑡𝑥,2−|)8, (2.58) for all 𝑥∈𝑋 and 𝑡>0, where 𝑀₁^′′(𝑥,𝑡)∶=𝜇^′𝜑(),8()𝑛𝑥𝑛−12𝑛²𝑡+9𝑛∗𝜇^′𝜑)𝑥),8()((𝑛−1𝑛−12𝑛²𝑡+9𝑛∗𝜇^′𝜑(𝑥),8(𝑛−1)2𝑛²𝑡+9𝑛∗𝜇^′𝜑(),8()−𝑛𝑥𝑛−12𝑛²𝑡+9𝑛∗𝜇^′𝜑(−()𝑥),8()𝑛−1𝑛−12𝑛²𝑡+9𝑛∗𝜇^′𝜑(),8()−𝑥𝑛−12𝑛²𝑡+9𝑛∗𝜇^′𝜑(0),8(𝑛−1)2𝑛²𝑡,𝑀+9𝑛_1′′(𝑥,𝑡)∶=𝜇^′𝜑(),42𝑥𝑛²𝑡+4𝑛∗𝜇^′𝜑(𝑥),4𝑛²𝑡+4𝑛∗𝜇^′𝜑(),4−𝑥𝑛²𝑡+4𝑛∗𝜇^′𝜑(),4−2𝑥𝑛²𝑡+4𝑛∗𝜇^′𝜑(0),4𝑛²𝑡,𝑀+4𝑛_2′′(𝑥,𝑡)∶=𝜈^′𝜑(),8()𝑛𝑥𝑛−12𝑛²𝑡+9𝑛⋄𝜈^′𝜑)𝑥),8()((𝑛−1𝑛−12𝑛²𝑡+9𝑛⋄𝜈^′𝜑(𝑥),8(𝑛−1)2𝑛²𝑡+9𝑛⋄𝜈^′𝜑(),8()−𝑛𝑥𝑛−12𝑛²𝑡+9𝑛⋄𝜈^′𝜑(−()𝑥),8()𝑛−1𝑛−12𝑛²𝑡+9𝑛⋄𝜈^′𝜑(),8()−𝑥𝑛−12𝑛²𝑡+9𝑛⋄𝜈^′𝜑(0),8(𝑛−1)2𝑛²𝑡,𝑀+9𝑛_2′′(𝑥,𝑡)∶=𝜈^′𝜑(),42𝑥𝑛²𝑡+4𝑛⋄𝜈^′𝜑(𝑥),4𝑛²𝑡+4𝑛⋄𝜈^′𝜑(),4−𝑥𝑛²𝑡+4𝑛⋄𝜈^′𝜑(),4−2𝑥𝑛²𝑡+4𝑛⋄s𝜈^′𝜑(0),4𝑛²𝑡.+4𝑛 (2.59)

Proof. Passing to the even part 𝑓_𝑒 and odd part 𝑓_𝑜 of 𝑓, we deduce from (2.57) that 𝜇Δ𝑓_𝑒𝑥₁,…,𝑥_𝑛,𝑡₁+⋯+𝑡_𝑛≥𝜇^′𝜑𝑥₁,𝑡₁∗𝜇^′𝜑−𝑥₁,𝑡₁∗⋯∗𝜇^′𝜑𝑥_𝑛,𝑡_𝑛∗𝜇^′𝜑−𝑥_𝑛,𝑡_𝑛,𝜈Δ𝑓_𝑒𝑥₁,…,𝑥_𝑛,𝑡₁+⋯+𝑡_𝑛≤𝜈^′𝜑𝑥₁,𝑡₁⋄𝜈^′𝜑−𝑥₁,𝑡₁⋄⋯⋄𝜈^′𝜑𝑥_𝑛,𝑡_𝑛⋄𝜈^′𝜑−𝑥_𝑛,𝑡_𝑛. (2.60) On the other hand, 𝜇Δ𝑓_𝑜𝑥₁,…,𝑥_𝑛,𝑡₁+⋯+𝑡_𝑛≥𝜇^′𝜑𝑥₁,𝑡₁∗𝜇^′𝜑−𝑥₁,𝑡₁∗⋯∗𝜇^′𝜑𝑥_𝑛,𝑡_𝑛∗𝜇^′𝜑−𝑥_𝑛,𝑡_𝑛,𝜈Δ𝑓_𝑜𝑥₁,…,𝑥_𝑛,𝑡₁+⋯+𝑡_𝑛≤𝜈^′𝜑𝑥₁,𝑡₁⋄𝜈^′𝜑−𝑥₁,𝑡₁⋄⋯⋄𝜈^′𝜑𝑥_𝑛,𝑡_𝑛⋄𝜈^′𝜑−𝑥_𝑛,𝑡_𝑛. (2.61) Applying the proofs of Theorems 2.1 and 2.2, we get a unique quadratic function 𝑄 and a unique additive function 𝐴 satisfying 𝜇𝑄(𝑥)−𝑓_𝑒(𝑥),𝑡≥𝑀₁^′′(|𝛼𝑡𝑥,4−|)8,𝜈𝑄(𝑥)−𝑓_𝑒(𝑥),𝑡≤𝑀₂^′′(|𝛼𝑡𝑥,4−|)8. (2.62) Also, 𝜇𝐴(𝑥)−𝑓_𝑜(𝑥),𝑡≥𝑀₁^′′(|𝛼𝑡𝑥,2−|)4,𝜈𝐴(𝑥)−𝑓_𝑜(𝑥),𝑡≤𝑀₂^′′(|𝛼𝑡𝑥,2−|)4. (2.63) Therefore, 𝜇𝑄(𝑥)−𝐴(𝑥)−𝑓(𝑥),𝑡≥𝑀₁^′′(|𝛼𝑡𝑥,4−|)∗𝑀16^1′′(|𝛼𝑡𝑥,2−|)8,𝜈𝑄(𝑥)−𝐴(𝑥)−𝑓(𝑥),𝑡≤𝑀₂^′′(|𝛼𝑡𝑥,4−|)⋄𝑀16^2′′(|𝛼𝑡𝑥,2−|)8. (2.64) This completes the proof of the theorem.

Acknowledgment

The authors are very grateful to the referees for their helpful comments and suggestions.