Hindawi Publishing Corporation Journal of Function Spaces and Applications Volume 2013, Article ID 673810, 8 pages http://dx.doi.org/10.1155/2013/673810
Research Article Solution and Stability of a General Mixed Type Cubic and Quartic Functional Equation Xiaopeng Zhao,1,2 Xiuzhong Yang,3 and Chin-Tzong Pang4 1
Department of Mathematics, Zhejiang University, Hangzhou 310027, China Department of Applied Mathematics, National Sun Yat-sen University, Kaohsiung 80424, Taiwan 3 College of Mathematics and Information Science, Hebei Normal University, and Hebei Key Laboratory of Computational Mathematics and Applications, Shijiazhuang 050024, China 4 Department of Information Management, Yuan Ze University, Chung-Li 32003, Taiwan 2
Correspondence should be addressed to Chin-Tzong Pang;
[email protected] Received 23 August 2013; Accepted 5 September 2013 Academic Editor: Jinlu Li Copyright Β© 2013 Xiaopeng Zhao et al. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited. We consider the following mixed type cubic and quartic functional equation π[π(π₯ + ππ¦) + π(π₯ β ππ¦)] = π3 [π(π₯ + π¦) + π(π₯ β π¦)] β 2π3 (π + 1)π(π¦) β 2π(π2 β 1)π(π₯) + 2(π + 1)π(ππ¦), where π is a fixed integer. We establish the general solution of the functional equation when the integer π =ΜΈ 0, Β±1, and then, by using the fixed point alternative, we investigate the generalized Hyers-UlamRassias stability for this functional equation when the integer π β₯ 2.
1. Introduction In 1940, Ulam [1] asked the fundamental question for the stability for the group homomorphisms. Let (πΊ1 , β) be a group, and let (πΊ2 , ⬦, π) be a metric group with the metric π(β
, β
). Given π > 0, does there exist a πΏ(π) > 0 such that if a mapping β : πΊ1 β πΊ2 satisfies the inequality π (β (π₯ β π¦) , β (π₯) ⬦ β (π¦)) < πΏ
(1)
for all π₯, π¦ β πΊ1 , then there is a homomorphism π» : πΊ1 β πΊ2 with π (β (π₯) , π» (π₯)) < π
(2)
for all π₯ β πΊ1 ? In other words, under what conditions, does there exist a homomorphism near an approximately homomorphism? In the next year, Hyers [2] gave the first affirmative answer to the question of Ulam for Cauchy equation in the Banach spaces. Then, Rassias [3] generalized Hyersβ result by considering an unbounded Cauchy difference, and this stability phenomenon is known as generalized Hyers-Ulam-Rassias stability or Hyers-Ulam-Rassias stability. During the last three decades, the stability problem for several functional
equations has been extensively investigated by many mathematicians; see, for example, [4β9] and the references therein. We also refer the readers to the books [10β13]. In [14], Jun and Kim introduced the following functional equation π (2π₯ + π¦) + π (2π₯ β π¦) = 2π (π₯ + π¦)+2π (π₯ β π¦)+12π (π₯) . (3) It is easy to see that the function π(π₯) = π₯3 satisfies the functional equation (3). Thus, it is natural that (3) is called a cubic functional equation and every solution of the cubic functional equation is said to be a cubic function. In [14], Jun and Kim established the general solution and the generalized Hyers-Ulam-Rassias stability for (3). They proved that a function π : π β π between real vector spaces is a solution of the functional equation (3) if and only if there exists a function πΊ : π Γ π Γ π β π such that π(π₯) = πΊ(π₯, π₯, π₯) for all π₯ β π and πΊ is symmetric for each fixed one variable and is additive for fixed two variables. In [15], Lee et al. considered the following quartic functional equation π (2π₯ + π¦) + π (2π₯ β π¦) (4) = 4 [π (π₯ + π¦) + π (π₯ β π¦)] + 24π (π₯) β 6π (π¦) .
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Since the function π(π₯) = π₯4 satisfies the functional equation (4), the functional equation (4) is called a quartic functional equation and every solution of the quartic functional equation is said to be a quartic function. In [15], the authors solved the functional equation (4) and proved the stability for it. Actually, they obtained that a function π : π β π between real vector spaces satisfies the functional equation (4) if and only if there exists a symmetric biquadratic function π΅ : π Γ π β π such that π(π₯) = π΅(π₯, π₯) for all π₯ β π. A function π : π β π between real vector spaces is said to be quadratic if it satisfies the following functional equation π (π₯ + π¦) + π (π₯ β π¦) = 2π (π₯) + 2π (π¦)
(5)
for all π₯, π¦ β π, and a function π΅ : π Γ π β π is said to be bi-quadratic if π΅ is quadratic for each fixed one variable (see [8]). The following mixed type cubic and quartic functional equation was introduced by Eshaghi Gordji et al. [16]
Lemma 1. If an odd mapping π : π β π satisfies (7), then π is cubic. Proof. Note that, in view of the oddness of π, we have π(0) = 0 and π(βπ₯) = βπ(π₯) for all π₯ β π. Letting π₯ = 0 in (7), we get π (ππ¦) = π3 π (π¦)
(8)
for all π¦ β π. Applying (8) to (7), we obtain π (π₯ + ππ¦) + π (π₯ β ππ¦)
π (π₯ + 2π¦) + π (π₯ β 2π¦) = 4 [π (π₯ + π¦) + π (π₯ β π¦)]
be a cubic function and a quartic function, respectively. In this section, we investigate the general solution of the mixed type cubic and quartic functional equation (7). Throughout this section, let π and π be two real vector spaces, and we always assume that the integer π in the functional equation (7) is different from 0, β1, and 1. Before proving our main theorem, we first give the following two lemmas.
(6)
β 24π (π¦) β 6π (π₯) + 3π (2π¦) . It can be verified that the function π(π₯) = π₯3 + π₯4 satisfies the functional equation (6). In [16], the authors obtained the general solution and the generalized Hyers-Ulam-Rassias stability of (6) in quasi-Banach space. The literature on the stability of the mixed type functional equations is very rich; see [17β22]. In the present paper, we extend (6) and consider the following functional equation π [π (π₯ + ππ¦) + π (π₯ β ππ¦)] = π3 [π (π₯ + π¦) + π (π₯ β π¦)] β 2π3 (π + 1) π (π¦) (7) 2
β 2π (π β 1) π (π₯) + 2 (π + 1) π (ππ¦) , where π is a fixed integer. One can see that the functional equation (6) is a special case of (7) when we take the integer π = 2. In 2003, Radu [23] noticed that the fixed point theorem, which was established by Diaz and Margolis [24], plays an important part in solving the stability problem of functional equations. Subsequently, this method has been successfully used by many mathematicians to investigate the stability of several functional equations; see, for example, [21, 25, 26] and the references therein. In this paper, we first establish the general solution of functional equation (7) when the integer π =ΜΈ 0, Β±1 and π is a mapping between vector spaces. Then, by using the fixed point method, we prove the generalized Hyers-Ulam-Rassias stability of the functional equation (7) when the integer π β₯ 2 and π is a mapping from the normed space to the Banach space.
2. Solution of the Functional Equation (7) Recall form [14, 15] that every solution of the cubic functional equation (3) and the quartic functional equation (4) is said to
= π2 [π (π₯ + π¦) + π (π₯ β π¦)] β 2 (π2 β 1) π (π₯)
(9)
for all π₯, π¦ β π. Replacing π¦ by π₯ + π¦ in (9), we get π ((π + 1) π₯ + ππ¦) β π ((π β 1) π₯ + ππ¦) = π2 [π (2π₯ + π¦) β π (π¦)] β 2 (π2 β 1) π (π₯)
(10)
for all π₯, π¦ β π. Now, if we replace π₯ by ππ₯ + π¦ in (9) and use (8), we see that π (ππ₯ + (π + 1) π¦) + π (ππ₯ β (π β 1) π¦) = π2 [π (ππ₯ + 2π¦) + π3 π (π₯)] β 2 (π2 β 1) π (ππ₯ + π¦) (11) for all π₯, π¦ β π. Interchanging π₯ with π¦ in (11) and using the oddness of π, we get the relation π ((π + 1) π₯ + ππ¦) β π ((π β 1) π₯ β ππ¦) = π2 [π (2π₯ + ππ¦) + π3 π (π¦)] β 2 (π2 β 1) π (π₯ + ππ¦) (12) for all π₯, π¦ β π. Then, subtracting (12) from (10), one has π ((π β 1) π₯ β ππ¦) β π ((π β 1) π₯ + ππ¦) = π2 [π (2π₯ + π¦) β π (2π₯ + ππ¦)] β (π2 + π5 ) π (π¦) + 2 (π2 β 1) (π (π₯ + ππ¦) β π (π₯)) (13) for all π₯, π¦ β π. Now, replacing π¦ by βπ¦ in (13) and noting that π is odd, we have π ((π β 1) π₯ + ππ¦) β π ((π β 1) π₯ β ππ¦) = π2 [π (2π₯ β π¦) β π (2π₯ β ππ¦)] + (π2 + π5 ) π (π¦) + 2 (π2 β 1) (π (π₯ β ππ¦) β π (π₯)) (14)
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for all π₯, π¦ β π. Adding (13) to (14) gives
for all π₯, π¦ β π. Interchanging the roles of π₯ and π¦ in (21), we obtain
π2 [π (2π₯ + π¦) + π (2π₯ β π¦)]
π ((π + 1) π₯ + ππ¦) + π ((π β 1) π₯ β ππ¦)
2
+ 2 (π β 1) [π (π₯ + ππ¦) + π (π₯ β ππ¦)] 2
= π2 [π (2π₯ + ππ¦) + π4 π (π¦)]
2
= π [π (2π₯ + ππ¦) + π (2π₯ β ππ¦)] + 4 (π β 1) π (π₯) (15) for all π₯, π¦ β π. Replacing π₯ by 2π₯ in (9) and using (8), we get
= π2 [π (2π₯ + π¦) + π (2π₯ β π¦)] β 16 (π2 β 1) π (π₯) . (16)
(17)
for all π₯, π¦ β π. Thus, the mapping π : π β π is cubic. This completes the proof. Lemma 2. If an even mapping π : π β π satisfies (7) for all π₯, π¦ β π, then π is quartic. Proof. In view of the evenness of π, we have π(βπ₯) = π(π₯) for all π₯ β π. Putting π₯ = π¦ = 0 in (7), we get π(0) = 0. Then, let π₯ = 0 in (7), we obtain π (ππ¦) = π4 π (π¦)
(18)
for all π¦ β π. Combing (7) and (18) implies the following equation
(19)
for all π₯, π¦ β π. Replacing π¦ by π₯ + π¦ in (19) and note that π is even, we get
(20)
+ 2π2 (π2 β 1) π (π₯ + π¦) for all π₯, π¦ β π. Replacing π₯ by ππ₯ + π¦ in (19) and using (18), we get
π ((π β 1) π₯ β ππ¦) β π ((π β 1) π₯ + ππ¦) = π2 [π (2π₯ β π¦) β π (2π₯ β ππ¦)] + 2 (π2 β 1) π (π₯ β ππ¦) + (π2 β π6 ) π (π¦) + 2π2 (π2 β 1) π (π₯ β π¦) β 2 (π2 + 1) (π2 β 1) π (π₯) (24) for all π₯, π¦ β π. If we add (23) to (24), we have π2 [π (2π₯ + π¦) + π (2π₯ β π¦)] β π2 [π (2π₯ + ππ¦) + π (2π₯ β ππ¦)] (25)
+ 2π2 (π2 β 1) [π (π₯ + π¦) + π (π₯ β π¦)]
for all π₯, π¦ β π. Replacing π₯ by 2π₯ in (19) and using (18), we get
= π2 [π (2π₯ + π¦) + π (2π₯ β π¦)]
(21)
(26)
β 32 (π2 β 1) π (π₯) + 2π2 (π2 β 1) π (π¦) for all π₯, π¦ β π. Applying (19) and (26) to (25), we obtain that π (2π₯ + π¦) + π (2π₯ β π¦) = 4 [π (π₯ + π¦) + π (π₯ β π¦)] + 24π (π₯) β 6π (π¦)
π (ππ₯ + (π + 1) π¦) + π (ππ₯ β (π β 1) π¦)
β 2 (π2 β 1) π (ππ₯ + π¦) + 2π2 (π2 β 1) π (π¦)
for all π₯, π¦ β π. Replacing π¦ by βπ¦ in (23), we get
π (2π₯ + ππ¦) + π (2π₯ β ππ¦)
π ((π + 1) π₯ + ππ¦) + π ((π β 1) π₯ + ππ¦)
= π2 [π (ππ₯ + 2π¦) + π4 π (π₯)]
+ 2 (π2 β 1) π (π₯ + ππ¦) + (π2 β π6 ) π (π¦)
= 4 (π2 + 1) (π2 β 1) π (π₯) + 2 (π6 β π2 ) π (π¦)
β 2 (π2 β 1) π (π₯) + 2π2 (π2 β 1) π (π¦)
= π2 [π (2π₯ + π¦) + π (π¦)] β 2 (π2 β 1) π (π₯)
= π2 [π (2π₯ + π¦) β π (2π₯ + ππ¦)]
+ 2 (π2 β 1) [π (π₯ + ππ¦) + π (π₯ β ππ¦)]
π (π₯ + ππ¦) + π (π₯ β ππ¦) = π2 [π (π₯ + π¦) + π (π₯ β π¦)]
for all π₯, π¦ β π. If we subtract (22) from (20), we obtain
+ 2π2 (π2 β 1) π (π₯ + π¦) β 2 (π2 + 1) (π2 β 1) π (π₯) (23)
Applying (9) and (16) to (15) yields that
= 2 [π (π₯ + π¦) + π (π₯ β π¦)] + 12π (π₯)
β 2 (π2 β 1) π (π₯ + ππ¦) + 2π2 (π2 β 1) π (π₯)
π ((π β 1) π₯ + ππ¦) β π ((π β 1) π₯ β ππ¦)
π (2π₯ + ππ¦) + π (2π₯ β ππ¦)
π (2π₯ + π¦) + π (2π₯ β π¦)
(22)
(27)
for all π₯, π¦ β π. Therefore, the mapping π : π β π is quartic and the proof is complete. Now, we are ready to find out the general solution of (7).
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Theorem 3. A mapping π : π β π satisfies (7) for all π₯, π¦ β π if and only if there exist a symmetric multiadditive mapping πΊ : π Γ π Γ π β π and a symmetric bi-quadratic mapping π΅ : π Γ π β π such that π(π₯) = πΊ(π₯, π₯, π₯) + π΅(π₯, π₯) for all π₯ β π. Proof. First, we assume that there exist a symmetric multiadditive mapping πΊ : π Γ π Γ π β π and a symmetric bi-quadratic mapping π΅ : π Γ π β π such that π(π₯) = πΊ(π₯, π₯, π₯) + π΅(π₯, π₯) for all π₯ β π. We need to show that the function π satisfies (7). By a simple computation, one can obtain that the function π₯ σ³¨β πΊ(π₯, π₯, π₯) satisfies (7). Thus, to show that the function π satisfies (7), we only need to show that the function π₯ σ³¨β π΅(π₯, π₯) also satisfies (7); namely, π [π΅ (π₯ + ππ¦, π₯ + ππ¦) + π΅ (π₯ β ππ¦, π₯ β ππ¦)] (28)
for all π₯, π¦ β π. Since π΅ : π Γ π β π is a bi-quadratic mapping, it can be verified that (29)
for all integer π and all π₯, π¦ β π. Then, (28) becomes π΅ (π₯ + ππ¦, π₯ + ππ¦) + π΅ (π₯ β ππ¦, π₯ β ππ¦) = π2 [π΅ (π₯ + π¦, π₯ + π¦) + π΅ (π₯ β π¦, π₯ β π¦)]
β 4 [π΅ (π₯, π₯) + π΅ (π¦, π₯)] + 4 [π΅ (π¦, π₯) + π΅ (π¦, 2π¦)] β [2π΅ (π₯, π₯) + 8π΅ (π₯, π¦)] = 4 [π΅ (π₯ + π¦, π₯ + π¦) + π΅ (π₯ β π¦, π₯ β π¦)] + 4 [π΅ (π₯ + π¦, π¦) + π΅ (π₯ β π¦, π¦)] β 6π΅ (π₯, π₯) β 8π΅ (π₯, π¦) + 16π΅ (π¦, π¦) = 4 [π΅ (π₯ + π¦, π₯ + π¦) + π΅ (π₯ β π¦, π₯ β π¦)]
β 6π΅ (π₯, π₯) β 8π΅ (π₯, π¦) + 16π΅ (π¦, π¦) = 4 [π΅ (π₯ + π¦, π₯ + π¦) + π΅ (π₯ β π¦, π₯ β π¦)]
β 2π (π2 β 1) π΅ (π₯, π₯) + 2 (π + 1) π΅ (ππ¦, ππ¦)
π΅ (ππ₯, ππ¦) = π4 π΅ (π₯, π¦)
+ 4 [π΅ (π₯ β π¦, π₯ β π¦) + π΅ (π₯ β π¦, π¦)]
+ 8 [π΅ (π₯, π¦) + π΅ (π¦, π¦)]
= π3 [π΅ (π₯ + π¦, π₯ + π¦) + π΅ (π₯ β π¦, π₯ β π¦)] β 2π3 (π + 1) π΅ (π¦, π¦)
= 4 [π΅ (π₯ + π¦, π₯ + π¦) + π΅ (π₯ + π¦, π¦)]
(30)
β 2 (π2 β 1) π΅ (π₯, π₯) + 2π2 (π2 β 1) π΅ (π¦, π¦) . To establish (28), it suffices to show (30). Note that if (30) holds for some integer π, then so does βπ. Thus, in the following, we will show that (30) holds for all positive integers π with π β₯ 2 and all π₯, π¦ β π. To do this, we use induction on π. Fix any π₯, π¦ β π. In the case when π = 2, we have π΅ (π₯ + 2π¦, π₯ + 2π¦) + π΅ (π₯ β 2π¦, π₯ β 2π¦) = [π΅ (π₯ + π¦ + π¦, π₯ + 2π¦) + π΅ (π₯ + π¦ β π¦, π₯ + 2π¦)] + [π΅ (π₯ β π¦ + π¦, π₯ β 2π¦) + π΅ (π₯ β π¦ β π¦, π₯ β 2π¦)] β [π΅ (π₯, π₯ + 2π¦) + π΅ (π₯, π₯ β 2π¦)] = 2 [π΅ (π₯ + π¦, π₯ + 2π¦) + π΅ (π¦, π₯ + 2π¦)] + 2 [π΅ (π₯ β π¦, π₯ β 2π¦) + π΅ (π¦, π₯ β 2π¦)] β [2π΅ (π₯, π₯) + 2π΅ (π₯, 2π¦)] = 2 [π΅ (π₯ + π¦, π₯ + π¦ + π¦) + π΅ (π₯ + π¦, π₯ + π¦ β π¦)] + 2 [π΅ (π₯ β π¦, π₯ β π¦ β π¦) + π΅ (π₯ β π¦, π₯ β π¦ + π¦)]
β 6π΅ (π₯, π₯) + 24π΅ (π¦, π¦) . (31) So (30) is true for π = 2. Here, we have used the bi-quadratic property of the function π΅ and (29). Now, assume that (30) is true for all positive integers that are less than or equal to some integer π(> 2). Then, π΅ (π₯ + (π + 1) π¦, π₯ + (π + 1) π¦) + π΅ (π₯ β (π + 1) π¦, π₯ β (π + 1) π¦) = [π΅ (π₯ + π¦ + ππ¦, π₯ + π¦ + ππ¦) +π΅ (π₯ + π¦ β ππ¦, π₯ + π¦ β ππ¦)] + [π΅ (π₯ β π¦ + ππ¦, π₯ β π¦ + ππ¦) +π΅ (π₯ β π¦ β ππ¦, π₯ β π¦ β ππ¦)] β [π΅ (π₯ + (π β 1) π¦, π₯ + (π β 1) π¦) +π΅ (π₯ β (π β 1) π¦, π₯ β (π β 1) π¦)] = π2 [π΅ (π₯ + 2π¦, π₯ + 2π¦) + π΅ (π₯, π₯)] β 2 (π2 β 1) π΅ (π₯ + π¦, π₯ + π¦) + 2π2 (π2 β 1) π΅ (π¦, π¦) + π2 [π΅ (π₯, π₯) + π΅ (π₯ β 2π¦, π₯ β 2π¦)] β 2 (π2 β 1) π΅ (π₯ β π¦, π₯ β π¦) + 2π2 (π2 β 1) π΅ (π¦, π¦) β {(π β 1)2 [π΅ (π₯ + π¦, π₯ + π¦) + π΅ (π₯ β π¦, π₯ β π¦)]
β 2 [π΅ (π₯ + π¦, π₯) + π΅ (π₯ β π¦, π₯)]
β 2 ((π β 1)2 β 1) π΅ (π₯, π₯)
+ 2 [π΅ (π¦, π₯ + 2π¦) + π΅ (π¦, π₯ β 2π¦)]
+2(π β 1)2 ((π β 1)2 β 1) π΅ (π¦, π¦)} .
β [2π΅ (π₯, π₯) + 8π΅ (π₯, π¦)]
(32)
Journal of Function Spaces and Applications
5 Then, for each fixed element π₯0 β πΈ, either
Applying (31) to (32), one can obtain that π΅ (π₯ + (π + 1) π¦, π₯ + (π + 1) π¦)
π (π½π π₯0 , π½π+1 π₯0 ) = +β
+ π΅ (π₯ β (π + 1) π¦, π₯ β (π + 1) π¦)
for all nonnegative integers π or there exists a non-negative integer π0 such that
= (π + 1)2 [π΅ (π₯ + π¦, π₯ + π¦) + π΅ (π₯ β π¦, π₯ β π¦)]
(a) π(π½π π₯0 , π½π+1 π₯0 ) < +β for all π β₯ π0 ;
β 2 ((π + 1)2 β 1) π΅ (π₯, π₯)
(b) the sequence {π½π π₯0 } converges to a fixed point π₯β of π½;
+ 2(π + 1)2 ((π + 1)2 β 1) π΅ (π¦, π¦) . (33) This means that (30) is true for π+1, and we have showed that the function π₯ σ³¨β π΅(π₯, π₯) satisfies (7). Therefore, the mapping π : π β π satisfies (7). Conversely, we decompose π into the odd part and the even part by putting π (π₯) β π (βπ₯) , ππ (π₯) = 2
π (π₯) + π (βπ₯) ππ (π₯) = 2
(34)
for all π₯ β π. Then, π(π₯) = ππ (π₯) + ππ (π₯) for all π₯ β π. It is easy to show that the mappings ππ and ππ satisfy (7). Hence, it follows from Lemmas 1 and 2 that the function ππ is cubic and ππ is quartic, respectively. Therefore, there exist a symmetric multi-additive mapping πΊ : π Γ π Γ π β π such that ππ (π₯) = πΊ(π₯, π₯, π₯) for all π₯ β π (see [14, Theorem 2.1]) and a symmetric bi-quadratic mapping π΅ : π Γ π β π such that ππ (π₯) = π΅(π₯, π₯) for all π₯ β π (see [15, Theorem 2.1]). Hence, we get π(π₯) = πΊ(π₯, π₯, π₯) + π΅(π₯, π₯) for all π₯ β π. The proof is complete.
(c) π₯β is the unique fixed point of π½ in the set F := {π₯ β πΈ | π(π½π0 π₯0 , π₯) < +β}; (d) π(π₯, π₯β ) β€ (1/(1 β πΏ))π(π₯, π½π₯) for all π₯ β F. Lemma 5. Let π : π β π be an odd function for which there exists a function π : π Γ π β [0, +β) such that σ΅© σ΅©σ΅© σ΅©σ΅©π·π (π₯, π¦)σ΅©σ΅©σ΅© β€ π (π₯, π¦)
π (ππ₯, ππ¦) β€ π3 πΏπ (π₯, π¦)
πΏ σ΅© σ΅©σ΅© π (π₯) σ΅©σ΅©π (π₯) β πΆ (π₯)σ΅©σ΅©σ΅© β€ 1βπΏ
Proof. It follows from (39) that
σ΅©σ΅© 3 σ΅© σ΅©σ΅©π π (π₯) β π (ππ₯)σ΅©σ΅©σ΅© β€ π (ππ₯) σ΅© σ΅©
β π3 [π (π₯ + π¦) + π (π₯ β π¦)]
Proposition 4 (see [24]). Let (πΈ, π) be a complete generalized metric space (i.e., one for which π may assume infinite value), and let π½ : πΈ β πΈ be a strictly contractive mapping with Lipschitz constant πΏ < 1; that is, π (π½π₯, π½π¦) β€ πΏπ (π₯, π¦)
βπ₯, π¦ β π.
(36)
(42)
for all π₯ β π. By (39), we have π(ππ₯) β€ π3 πΏπ(π₯) for all π₯ β π. This, together with (42), implies that σ΅©σ΅© σ΅© σ΅©σ΅©π (π₯) β 1 π (ππ₯)σ΅©σ΅©σ΅© β€ πΏπ (π₯) σ΅©σ΅© σ΅©σ΅© 3 π σ΅© σ΅©
for all π₯, π¦ β π. Let us recall the following result by Diaz and Margolis.
(41)
for all π₯, π¦ β π. Letting π₯ = 0 in (38) and replacing π¦ by π₯, we have
π·π (π₯, π¦) := π [π (π₯ + ππ¦) + π (π₯ β ππ¦)]
β 2 (π + 1) π (ππ¦)
(40)
for all π₯ β π, where π(π₯) := (1/2(π + 1))π(0, π₯/π).
π (ππ π₯, ππ π¦) =0 πββ π3π
(35)
(39)
for all π₯, π¦ β π, then there exists a unique cubic mapping πΆ : π β π such that
lim
In this section, we will investigate the stability of the functional equation (7) by using the fixed point alternative. Throughout this section, let π be a real normed space and π be a real Banach space, and we always assume that the integer π used in the section is greater than or equal to 2. For convenience, we use the following abbreviation for a given function π : π β π:
(38)
for all π₯, π¦ β π. If there exists a constant 0 < πΏ < 1 such that
3. Generalized Hyers-Ulam-Rassias Stability of the Functional Equation (7)
+ 2π3 (π + 1) π (π¦) + 2π (π2 β 1) π (π₯)
(37)
(43)
for all π₯ β π. Let Ξ© be the set of all odd mappings π : π β π. We introduce the generalized metric on Ξ©: π (π, β) σ΅© σ΅© := inf {πΎ β [0, β) | σ΅©σ΅©σ΅©π (π₯) β β (π₯)σ΅©σ΅©σ΅© β€ πΎπ (π₯) , π₯ β π} . (44) It is easy to show that (Ξ©, π) is complete. Now, we define a function π : Ξ© β Ξ© by ππ (π₯) :=
1 π (ππ₯) π3
(45)
6
Journal of Function Spaces and Applications
for all π β Ξ© and all π₯ β π. Note that, for all π, β β Ξ©,
Proof. It follows from (52) that
π (π, β) β€ πΎ σ΅© σ΅© σ³¨β σ΅©σ΅©σ΅©π (π₯) β β (π₯)σ΅©σ΅©σ΅© β€ πΎπ (π₯) , σ΅©σ΅© 1 σ³¨β σ΅©σ΅©σ΅©σ΅© 3 π (ππ₯) β σ΅©π σ΅©σ΅© 1 σ³¨β σ΅©σ΅©σ΅©σ΅© 3 π (ππ₯) β σ΅©π
lim π3π π (
πββ
π₯βπ
1 σ΅©σ΅© 1 β (ππ₯)σ΅©σ΅©σ΅©σ΅© β€ 3 πΎπ (ππ₯) , π₯ β π 3 π σ΅© π σ΅©σ΅© 1 β (ππ₯)σ΅©σ΅©σ΅©σ΅© β€ πΏπΎπ (π₯) , π₯ β π 3 π σ΅©
σ³¨β π (ππ, πβ) β€ πΏπΎ. (46) Hence, we obtain that π (ππ, πβ) β€ πΏπ (π, β)
πΆ : π σ³¨β π,
π (ππ π₯) π β β π3π (48)
πΆ (π₯) = lim (ππ π) (π₯) = lim πββ
and πΆ(ππ₯) = π3 πΆ(π₯) for all π₯ β π. Also πΆ is the unique fixed point of π in the set Ξ = {π β Ξ© | π(π, π) < β} and π (π, πΆ) β€
1 πΏ π (ππ, π) β€ , 1βπΏ 1βπΏ
(49)
which yields the inequality (40). It follows from the definition of πΆ, (38), and (41) that σ΅©σ΅© σ΅© π (ππ π₯, ππ π¦) σ΅©σ΅©π·π (ππ π₯, ππ π¦)σ΅©σ΅©σ΅© σ΅©σ΅© σ΅©σ΅© β€ lim =0 σ΅©σ΅©π·πΆ (π₯, π¦)σ΅©σ΅© = πlim ββ πββ π3π π3π (50) for all π₯, π¦ β π; that is, the mapping πΆ : π β π satisfies (7). Since π is odd, πΆ is odd. Therefore, Lemma 1 guarantees that πΆ is cubic. Finally, it remains to prove the uniqueness of πΆ. Let π : π β π be another cubic function satisfying (40). Since π(π, π) β€ πΏ/(1 β πΏ) and π is cubic, we get π β Ξ and ππ(π₯) = (1/π3 )π(ππ₯) = π(π₯) for all π₯ β π; that is, π is a fixed point of π. Since πΆ is the unique fixed point of π in Ξ, it follows that π = πΆ. Lemma 6. Let π : π β π be an odd function for which there exists a function π : π Γ π β [0, +β) such that σ΅© σ΅©σ΅© (51) σ΅©σ΅©π·π (π₯, π¦)σ΅©σ΅©σ΅© β€ π (π₯, π¦) for all π₯, π¦ β π. If there exists a constant 0 < πΏ < 1 such that π3 π (π₯, π¦) β€ πΏπ (ππ₯, ππ¦)
(52)
for all π₯, π¦ β π, then there exists a unique cubic mapping πΆ : π β π such that 1 σ΅© σ΅©σ΅© (53) π (π₯) σ΅©σ΅©π (π₯) β πΆ (π₯)σ΅©σ΅©σ΅© β€ 1βπΏ for all π₯ β π, where π(π₯) := (1/2(π + 1))π(0, π₯/π).
(54)
for all π₯, π¦ β π. Letting π₯ = 0 in (51) and replacing π¦ by π₯, we have σ΅© σ΅©σ΅© 3 σ΅©σ΅©π π ( π₯ ) β π (π₯)σ΅©σ΅©σ΅© β€ π (π₯) (55) σ΅©σ΅© σ΅©σ΅© π σ΅© σ΅© for all π₯ β π. We introduce the same definitions for Ξ© and π as in the proof of Lemma 5 (by replacing π by π) such that (Ξ©, π) becomes a generalized complete metric space. Let π : Ξ© β Ξ© be the mapping defined by π₯ ππ (π₯) := π3 π ( ) π
(47)
for all π, β β Ξ©; that is, π is a strictly contractive mapping of Ξ© with Lipschitz constant πΏ. It follows from (43) that π(ππ, π) β€ πΏ. Therefore, according to Proposition 4, the sequence {ππ π} converges to a fixed point πΆ of π; that is,
π₯ π¦ , )=0 ππ ππ
(56)
for all π β Ξ© and all π₯ β π. One can show that π(ππ, πβ) β€ πΏπ(π, β) for all π, β β Ξ©. It follows from (55) that π(ππ, π) β€ 1. Due to Proposition 4, the sequence {ππ π} converges to a fixed point πΆ of π; that is, πΆ : π σ³¨β π,
πββ
π₯ ) ππ (57)
1 1 π (ππ, π) β€ , 1βπΏ 1βπΏ
(58)
πΆ (π₯) = lim (ππ π) (π₯) = lim π3π π ( πββ
and πΆ(ππ₯) = π3 πΆ(π₯) for all π₯ β π. Also, π (π, πΆ) β€
which yields the inequality (53). The rest of the proof is similar to the proof of Lemma 5, and we omit the details. Similarly, we can prove the following two lemmas on even functions. Lemma 7. Let π : π β π be an even function with π(0) = 0 for which there exists a function π : π Γ π β [0, +β) such that σ΅© σ΅©σ΅© σ΅©σ΅©π·π (π₯, π¦)σ΅©σ΅©σ΅© β€ π (π₯, π¦)
(59)
for all π₯, π¦ β π. If there exists a constant 0 < πΏ < 1 such that π (ππ₯, ππ¦) β€ π4 πΏπ (π₯, π¦)
(60)
for all π₯, π¦ β π, then there exists a unique quartic mapping π : π β π such that πΏ σ΅© σ΅©σ΅© π (π₯) σ΅©σ΅©π (π₯) β π (π₯)σ΅©σ΅©σ΅© β€ 1βπΏ
(61)
for all π₯ β π, where π(π₯) := (1/2)π(0, π₯/π). Lemma 8. Let π : π β π be an even function with π(0) = 0 for which there exists a function π : π Γ π β [0, +β) such that σ΅© σ΅©σ΅© σ΅©σ΅©π·π (π₯, π¦)σ΅©σ΅©σ΅© β€ π (π₯, π¦)
(62)
Journal of Function Spaces and Applications
7
for all π₯, π¦ β π. If there exists a constant 0 < πΏ < 1 such that π4 π (π₯, π¦) β€ πΏπ (ππ₯, ππ¦)
(63)
for all π₯, π¦ β π, then there exists a unique quartic mapping π : π β π such that 1 σ΅© σ΅©σ΅© π (π₯) σ΅©σ΅©π (π₯) β π (π₯)σ΅©σ΅©σ΅© β€ 1βπΏ
Theorem 9. Let π : π β π be a function with π(0) = 0 for which there exists a function π : π Γ π β [0, +β) such that (65)
for all π₯, π¦ β π. If there exists a constant 0 < πΏ < 1 such that
(67)
π₯ 1 Ξ¦ (0, ) , 2 (π + 1) π
1 π₯ ππ (π₯) := Ξ¦ (0, ) , 2 π
1 Ξ¦ (π₯, π¦) := [π (π₯, π¦) + π (βπ₯, βπ¦)] . 2
(68)
Proof. Let ππ and ππ denote the odd and the even part of π, respectively. Then, it can be verified from (65) that σ΅© σ΅©σ΅© σ΅©σ΅©π·ππ (π₯, π¦)σ΅©σ΅©σ΅© β€ Ξ¦ (π₯, π¦) ,
σ΅©σ΅© σ΅© σ΅©σ΅©π·ππ (π₯, π¦)σ΅©σ΅©σ΅© β€ Ξ¦ (π₯, π¦) (69)
for all π₯, π¦ β π. Moreover, by (66), it is easy to compute that Ξ¦ (ππ₯, ππ¦) β€ π3 πΏΞ¦ (π₯, π¦)
π4 π (π₯, π¦) β€ πΏπ (ππ₯, ππ¦)
(74)
σ΅© σ΅©σ΅© σ΅©σ΅©π (π₯) β πΆ (π₯) β π (π₯)σ΅©σ΅©σ΅© β€
1 [π (π₯) + ππ (π₯)] 1βπΏ π
(75)
for all π₯ β π, where ππ (π₯) :=
1 π₯ Ξ¨ (0, ) , 2 (π + 1) π Ξ¨ (π₯, π¦) :=
1 π₯ ππ (π₯) := Ξ¨ (0, ) , 2 π
1 [π (π₯, π¦) + π (βπ₯, βπ¦)] . 2
(76)
Proof. Similar to the proof of Theorem 9, the result follows from Lemmas 6 and 8.
Conflict of Interests The authors declare that there is no conflict of interests regarding the publication of this paper.
for all π₯ β π, where ππ (π₯) :=
for all π₯, π¦ β π. If there exists a constant 0 < πΏ < 1 such that
(66)
for all π₯, π¦ β π, then there exist a unique cubic mapping πΆ : π β π and a unique quartic mapping π : π β π such that πΏ σ΅© σ΅©σ΅© [π (π₯) + ππ (π₯)] σ΅©σ΅©π (π₯) β πΆ (π₯) β π (π₯)σ΅©σ΅©σ΅© β€ 1βπΏ π
(73)
for all π₯, π¦ β π, then there exist a unique cubic mapping πΆ : π β π and a unique quartic mapping π : π β π such that
Now, we are ready to give our main theorems in this section.
π (ππ₯, ππ¦) β€ π3 πΏπ (π₯, π¦)
σ΅© σ΅©σ΅© σ΅©σ΅©π·π (π₯, π¦)σ΅©σ΅©σ΅© β€ π (π₯, π¦)
(64)
for all π₯ β π, where π(π₯) := (1/2)π(0, π₯/π).
σ΅© σ΅©σ΅© σ΅©σ΅©π·π (π₯, π¦)σ΅©σ΅©σ΅© β€ π (π₯, π¦)
Theorem 10. Let π : π β π be a function with π(0) = 0 for which there exists a function π : π Γ π β [0, +β) such that
(70)
for all π₯, π¦ β π. Thus, by applying Lemmas 5 and 7, one can obtain that there exist a unique cubic mapping πΆ : π β π and a unique quartic mapping π : π β π such that πΏ σ΅© σ΅©σ΅© π (π₯) , σ΅©σ΅©ππ (π₯) β πΆ (π₯)σ΅©σ΅©σ΅© β€ 1βπΏ π
(71)
πΏ σ΅© σ΅©σ΅© π (π₯) σ΅©σ΅©ππ (π₯) β π (π₯)σ΅©σ΅©σ΅© β€ 1βπΏ π
(72)
for all π₯ β π, where ππ (π₯) := (1/2(π + 1))Ξ¦(0, π₯/π) and ππ (π₯) := (1/2)Ξ¦(0, π₯/π). Moreover, combining (71) and (72) yields the inequality (67). The proof is complete.
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