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Fuzzy approximation of an Euler-Lagrange quadratic mappings

Journal of Inequalities and Applications 2013, 2013:358 doi:10.1186/1029-242X-2013-358

Hark-Mahn Kim (hmkim@cnu.ac.kr)John M Rassias (jrassias@primedu.uoa.gr)

Juri Lee (annans@hanmail.net)

ISSN 1029-242X

Article type Research

Submission date 26 October 2012

Acceptance date 9 July 2013

Publication date 1 August 2013

Article URL http://www.journalofinequalitiesandapplications.com/content/2013/1/358

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Fuzzy approximation of Euler-Lagrange quadratic mappings

Hark-Mahn Kim1, John M. Rassias2and Juri Lee∗3

1Department of Mathematics, Chungnam National University, Deajeon, 305-764, Korea2Pedagogical Department E.E., National and Capodistrian University of Athens, 4 Agamemnonos Street, Aghia Paraskevi, Athens15342, Greece3Department of Mathematics, Chungnam National University, Deajeon, 305-764, Korea

Email: Hark-Mahn Kim - hmkim@cnu.ac.kr; John M. Rassias - jrassias@primedu.uoa.gr; Juri Lee∗- annans@hanmail.net;

∗Corresponding author

Abstract

In this article, we consider the Hyers-Ulam stability of the Euler-Lagrange quadratic functional equation

f(kx+ ly) + f(kx− ly) = kl[f(x+ y) + f(x− y)] + 2(k − l)[kf(x)− lf(y)]

in fuzzy Banach spaces, where k, l are nonzero rational numbers with k �= l.

1 Introduction

The theory of fuzzy spaces has much progressed as the theory of randomness has developed. Some math-

ematicians have defined fuzzy norms on a vector space from various points of view [1–5]. Following Cheng

and Mordeson [6], Bag and Samanta [1] gave the idea of fuzzy norm in such a manner that the correspond-

ing fuzzy metric is of Kramosil and Michalek type [7] and investigated some properties of fuzzy normed

spaces [8].

We use the definition of fuzzy normed spaces given in [1, 4, 9].

Definition 1.1 [1, 4, 9]. Let X be a real vector space. A function N : X ×R → [0, 1] is said to be a fuzzy

norm on X if for all x, y ∈ X and all s, t ∈ R,

(N1) N(x, t) = 0 for t ≤ 0;

(N2) x = 0 if and only if N(x, t) = 1 for all t > 0;

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(N3) N(cx, t) = N(x, t|c| ) for c �= 0;

(N4) N(x+ y, s+ t) ≥ min{N(x, s), N(y, t)};

(N5) N(x, ·) is a non-decreasing function on R and limt→∞ N(x, t) = 1;

(N6) for x �= 0, N(x, ·) is continuous on R.

The pair (X,N) is called a fuzzy normed vector space. The properties of fuzzy normed vector spaces

and examples of fuzzy norms are given in [4, 10].

Definition 1.2 [1, 4, 9]. Let (X,N) be a fuzzy normed vector space. A sequence {xn} in X is said to be

convergent or to converges to x if there exists an x ∈ X such that limn→∞ N(xn − x, t) = 1 for all t > 0. In

this case, x is called the limit of the sequence {xn}, and we denote it by N -limn→∞ xn = x.

Definition 1.3 [1,4,9]. Let (X,N) be a fuzzy normed vector space. A sequence {xn} in X is called Cauchy

if for each ε > 0 and each t > 0, there exists an n0 ∈ N such that for all n ≥ n0 and all p > 0, we have

N(xn+p − xn, t) > 1− ε.

It is wellknown that every convergent sequence in a fuzzy normed space is a Cauchy sequence. If each

Cauchy sequence is convergent, then the fuzzy norm is said to be complete and the fuzzy normed vector

space is called a fuzzy Banach space.

We say that a mapping f : X → Y between fuzzy normed spaces X and Y is continuous at x0 ∈ X if for

each sequence {xn} converging to each x0 ∈ X, the sequence {f(xn)} converges to f(x0). If f : X → Y is

continuous at each x ∈ X, then f : X → Y is said to be continuous on X (see [8]).

We recall the fixed point theorem from [12], which is needed in Section 4.

Let X be a set. A function d : X ×X → [0,∞] is called a generalized metric on X if d satisfies:

(1) d(x, y) = 0 if and only if x = y;

(2) d(x, y) = d(y, x) for all x, y ∈ X;

(3) d(x, z) ≤ d(x, y) + d(y, z) for all x, y, z ∈ X.

Theorem 1.4 [11, 12]. Let (X, d) be a complete generalized metric space and let J : X → X be a strictly

contractive mapping with the Lipschitz constant L < 1. Then, for each given element x ∈ X, either

d(Jnx, Jn+1x) = ∞

for all nonnegative integers n or there exists a positive integer n0 such that

(1) d(Jnx, Jn+1x) < ∞, ∀n ≥ n0;

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(2) the sequence {Jnx} converges to a fixed point y∗ of J ;

(3) y∗ is the unique fixed point of J in the set Y = {y ∈ X|d(Jn0x, y) < ∞};

(4) d(y, y∗) ≤ 11−Ld(y, Jy) for all y ∈ Y .

In 1996, Isac and Th.M Rassias [13] were the first to provide a new application of fixed point theorems to

prove the of stability theory of functional equations. By using fixed point methods, the stability problems of

several functional equations have been extensively investigated by a number of authors (see [10–12,14–17]).

The stability problem of functional equations originated from a question of Ulam [18] concerning the

stability of group homomorphisms. Hyers [19] gave the first affirmative partial answer to the question of

Ulam for additive mappings on Banach spaces. Hyers’s theorem was generalized by Aoki [20] for additive

mappings and by Th.M. Rassias [21] for linear mappings by considering an unbounded Cauchy difference. A

generalization of the Th.M. Rassias theorem was obtain by Gavruta [22] by replacing the unbounded Cauchy

difference by a general control function in the spirit of Th.M. Rassias’s approach.

The functional equation

f(x+ y) + f(x− y) = 2f(x) + 2f(y)

is called a quadratic functional equation. In particular, every solution of the quadratic functional equation

is said to be a quadratic function. A Hyers-Ulam stability problem for the quadratic functional equation

was proved by Skof [23] for mappings f : X → Y , where X is a normed space and Y is a Banach space.

Cholewa [24] noticed that the theorem of Skof is still true if the relevant domain X is replaced by an Abelian

group. Czerwik [25] proved the Hyers-Ulam stability of the quadratic functional equation. In particular,

J.M. Rassias investigated the Hyers-Ulam stability for the relative Euler-Lagrange functional equation

f(ax+ by) + f(bx− ay) = (a2 + b2)[f(x) + f(y)]

in [26, 27]. The stability problems of several functional equations have been extensively investigated by a

number of authors, and there are many interesting results concerning this problem (see [28–32]).

In [33], Kim, Lee and Son have investigated the Hyers-Ulam stability of the quadratic functional equation

f(kx+ y) + f(kx− y) = kf(x+ y) + kf(x− y) + 2(k − 1)[kf(x)− f(y)] (1)

for any fixed integer k with k �= 0, 1.

In this paper, we prove the generalized Hyers-Ulam stability of the Euler-Lagrange quadratic functional

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equation

f(kx+ ly) + f(kx− ly) = kl[f(x+ y) + f(x− y)] + 2(k − l)[kf(x)− lf(y)] (2)

in fuzzy Banach spaces, where k, l are nonzero rational numbers with k �= l.

2 General solution of (2)

Lemma 2.1. A mapping f : X → Y between linear spaces satisfies the functional equation

f(kx+ y) + f(kx− y) = kf(x+ y) + kf(x− y) + 2(k − 1)[kf(x)− f(y)]

for any fixed rational number k with k �= 0, 1 if and only if f is quadratic.

Proof. Let f be a solution of equation (1). Letting x = y = 0 in (1), we have f(0) = 0. Putting y = 0

in (1), we get f(kx) = k2f(x). Putting x = 0 in (1), we get f(−y) = f(y). Thus, the mapping f is even.

Therefore, it suffices to prove that if a mapping f satisfies equation (1) for any fixed rational k with k �= 0, 1,

then f is quadratic. Now, replacing y by x+ y in (1), we have

f((k + 1)x+ y) + f((k − 1)x− y)

= kf(2x+ y) + kf(y) + 2k(k − 1)f(x)− 2(k − 1)f(x+ y) (3)

for all x, y ∈ X. Replacing y by −y in (3), we obtain

f((k + 1)x− y) + f((k − 1)x+ y)

= kf(2x− y) + kf(y) + 2k(k − 1)f(x)− 2(k − 1)f(x− y) (4)

for all x, y ∈ X. Adding (3) to (4), we get

f((k + 1)x+ y) + f((k + 1)x− y) + f((k − 1)x+ y) + f((k − 1)x− y)

= k[f(2x+ y) + f(2x− y)]− 2(k − 1)[f(x+ y) + f(x− y)]

= −2(k − 1)[f(x+ y) + f(x− y)] + 4k(k − 1)f(x) + 2kf(y) (5)

for all x, y ∈ X. From the substitution y := kx+ y in (1), we have

f(2kx+ y) + f(y) = k[f((k + 1)x+ y) + f((k − 1)x+ y)]

+2k(k − 1)f(x)− 2(k − 1)f(kx+ y) (6)

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for all x, y ∈ X. Replacing y by −y in (6), we get

f(2kx− y) + f(y) = k[f((k + 1)x− y) + f((k − 1)x− y)]

+2k(k − 1)f(x)− 2(k − 1)f(kx− y) (7)

for all x, y ∈ X. Adding (6) to (7), we get

f(2kx+ y) + f(2kx− y) = k[f((k + 1)x+ y) + f((k + 1)x− y)]

+k[f((k − 1)x+ y) + f((k − 1)x− y)]

−2(k − 1)[f(kx+ y) + f(kx− y)]

+4k(k − 1)f(x)− 2f(y) (8)

for all x, y ∈ X. It follows from (8), by using (1) and (5), that

f(2kx+ y) + f(2kx− y) = k2[f(2x+ y) + f(2x− y)]

−4k(k − 1)[f(x+ y) + f(x− y)]

+8k(k − 1)f(x) + 2(k − 1)(3k − 1)f(y) (9)

for all x, y ∈ X. If we replace x by 2x in (1), then we obtain that

f(2kx+ y) + f(2kx− y) = k[f(2x+ y) + f(2x− y)]

+2k(k − 1)f(2x)− 2(k − 1)f(y) (10)

for all x, y ∈ X. Associating (9) with (10), we conclude that the mapping f satisfies the equation

f(2x+ y) + f(2x− y) = 4[f(x+ y) + f(x− y)] + 2[f(2x)− 4f(x)]− 6f(y)

for all x, y ∈ X. Then f(x) = Q1(x) +Q2(x) for all x ∈ X, where Q1 is quadratic and Q2 is quartic by the

papers [34–36]. Therefore, f is quadratic because of the property f(kx) = k2f(x).

Conversely, if a mapping f is quadratic, then it is obvious that f satisfies (1).

Theorem 2.2. A mapping f : X → Y with f(0) = 0 between linear spaces satisfies the functional equation

(2) if and only if f is quadratic.

Proof. Let f be a solution of equation (2) and f(0) = 0. Putting y = 0 in (2), we get f(kx) = k2f(x) for

all x ∈ X. Putting x = 0 and replacing y by x, we have

f(lx) + f(−lx) = l(2l − k)f(x) + klf(−x) (11)

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for all x ∈ X. Replacing x by −x in (11), one gets

f(−lx) + f(lx) = l(2l − k)f(−x) + klf(x) (12)

for all x ∈ X. Subtracting equation (11) from (12), we find f(−x) = f(x) and so f(lx) = l2f(x) for all

x ∈ X. Thus equation (2) can be rewritten as

f(ax+ y) + f(ax− y) = af(x+ y) + af(x− y) + 2a(a− 1)f(x)− 2(a− 1)f(y)

where a := kl �= 0, 1 for all x, y ∈ X. Therefore, it follows from Lemma 2.1 that f is quadratic.

Conversely, if a mapping f is quadratic, then it is obvious that f satisfies equation (2).

3 Stability of equation (2) by direct method

Throughout this paper, we assume that X is a linear space, (Y,N) is a fuzzy Banach space and (Z,N ′) is a

fuzzy normed space.

For notational convenience, given a mapping f : X → Y , we define the difference operatorDklf : X2 → Y

of equation (2) by

Dklf(x, y) := f(kx+ ly) + f(kx− ly)− kl[f(x+ y) + f(x− y)]− 2(k − l)[kf(x)− lf(y)]

for all x, y ∈ X.

Theorem 3.1. Assume that a mapping f : X → Y with f(0) = 0 satisfies the inequality

N(Dklf(x, y), t) ≥ N ′(ϕ(x, y), t) (13)

and ϕ : X2 → Z is a mapping for which there is a constant s ∈ R satisfying 0 < |s| < k2 such that

N ′(ϕ(kx, ky), t) ≥ N ′(sϕ(x, y), t) (14)

for all x ∈ X and all t > 0. Then we can find a unique quadratic mapping Q : X → Y satisfying the equation

DklQ(x, y) = 0 and the inequality

N(f(x)−Q(x), t) ≥ N ′( ϕ(x, 0)

2(k2 − |s|) , t), t > 0 (15)

for all x ∈ X.

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Proof. We observe from (14) that

N ′(ϕ(knx, kny), t) ≥ N ′(snϕ(x, y), t)

= N ′(ϕ(x, y),

t

|s|n), t > 0

N ′(ϕ(knx, kny), |s|nt) ≥ N ′(ϕ(x, y), t), t > 0 (16)

for all x, y ∈ X. Putting y := 0 in (13), we obtain

N(2f(kx)− 2k2f(x), t) ≥ N ′(ϕ(x, 0), t),

or, N(f(x)− f(kx)

k2,

t

2k2

)≥ N ′(ϕ(x, 0), t) (17)

for all x ∈ X. Therefore it follows from (16), (17) that

N(f(knx)

k2n− f(kn+1x)

k2(n+1),

|s|nt2k2(n+1)

)≥ N ′(ϕ(knx, 0), |s|nt) ≥ N ′(ϕ(x, 0), t)

for all x ∈ X and any integer n ≥ 0. So,

N(f(x)− f(knx)

k2n,

n−1∑i=0

|s|it2k2(i+1)

)= N

(n−1∑i=0

(f(kix)k2i

− f(ki+1x)

k2(i+1)

),

n−1∑i=0

|s|it2k2(i+1)

)

≥ min0≤i≤n−1

{N(f(kix)

k2i− f(ki+1x)

k2(i+1),

|s|it2k2(i+1)

)}

≥ N ′(ϕ(x, 0), t), t > 0, (18)

which yields

N(f(kmx)

k2m− f(km+px)

k2(m+p),

m+p−1∑i=m

|s|it2k2(i+1)

)

= N(m+p−1∑

i=m

(f(kix)k2i

− f(ki+1x)

k2(i+1)

),

m+p−1∑i=m

|s|it2k2(i+1)

)

≥ minm≤i≤m+p−1

{N(f(kix)

k2i− f(ki+1x)

k2(n+1),

|s|it2k2(i+1)

)}

≥ N ′(ϕ(x, 0), t), t > 0,

for all x ∈ X and any integers p > 0,m ≥ 0. Hence, one obtains

N(f(kmx)

k2m− f(km+px)

k2(m+p), t)≥ N ′

(ϕ(x, 0),

t∑m+p−1i=m

|s|i2k2(i+1)

)(19)

for all x ∈ X and any integers p > 0,m ≥ 0, t > 0. Since∑m+p−1

i=m|s|ik2i is a convergent series, we see, by

taking the limit m → ∞ in the last inequality, that the sequence { f(knx)k2n } is Cauchy in the fuzzy Banach

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JIA_517_edited [07/23 10:35] 7/19

space (Y,N) and so it converges in Y . Therefore a mapping Q : X → Y defined by

Q(x) := N - limn→∞

f(knx)

k2n

is well defined for all x ∈ X. It means that limn→∞ N( f(knx)

k2n − Q(x), t) = 1, t > 0, for all x ∈ X. In

addition, we see from (18) that

N(f(x)− f(knx)

k2n, t)≥ N ′

(ϕ(x, 0),

t∑n−1i=0

|s|i2k2(i+1)

)(20)

and so

N(f(x)−Q(x), t) ≥ min

{N(f(x)− f(knx)

k2n, (1− ε)t

), N

(f(knx)k2n

−Q(x), εt)}

≥ N ′

(ϕ(x, 0),

(1− ε)t∑n−1i=0

|s|i2k2(i+1)

)

≥ N ′(ϕ(x, 0), 2(1− ε)(k2 − |s|)t), 0 < ε < 1, (21)

for sufficiently large n and for all x ∈ X and all t > 0. Since ε is arbitrary and N ′ is left continuous, we

obtain

N(f(x)−Q(x), t) ≥ N ′(ϕ(x, 0), 2(k2 − |s|)t), t > 0,

for all x ∈ X, which yields the approximation (15).

In addition, it is clear from (13) and (N5) that the relation

N(Dklf(k

nx, kny)

k2n, t)

≥ N ′(ϕ(knx, kny), k2nt)

≥ N ′(ϕ(x, y),

k2n

|s|n t)

→ 1 as n → ∞

holds for all x, y ∈ X and all t > 0. Therefore, we obtain by use of limn→∞ N( f(knx)

k2n −Q(x), t) = 1 (t > 0)

that

N(DklQ(x, y), t) ≥ min

{N(DklQ(x, y)− Dklf(k

nx, kny)

k2n,t

2

), N

(Dklf(knx, kny)

k2n,t

2

)}

= N(Dklf(k

nx, kny)

k2n,t

2

)(for sufficiently large n)

≥ N ′(ϕ(x, y),

k2n

2|s|n t), t > 0

→ 1 as n → ∞,

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JIA_517_edited [07/23 10:35] 8/19

which implies DklQ(x, y) = 0 by (N2). Thus we find that Q is a quadratic mapping satisfying equation (2)

and inequality (15) near the approximate quadratic mapping f : X → Y .

To prove the aforementioned uniqueness, we assume now that there is another quadratic mapping Q′ :

X → Y which satisfies inequality (15). Then one establishes by the equality Q′(knx) = k2nQ(x) and (15)

that

N(Q(x)−Q′(x), t) = N(Q(knx)

k2n− Q′(knx)

k2n, t)

≥ min

{N(Q(knx)

k2n− f(knx)

k2n,t

2

), N

(f(knx)k2n

− Q′(knx)

k2n,t

2

)}

≥ N ′(ϕ(knx, 0), (k2 − |s|)k2nt)

≥ N ′(ϕ(x, 0),

(k2 − |s|)k2nt|s|n

), t > 0,

for all n ∈ N, which tends to 1 as n → ∞ by (N5). Therefore one obtains Q(x) = Q′(x) for all x ∈ X,

completing the proof of uniqueness.

We remark that if k = 1 in Theorem 3.1, then N ′(ϕ(x, y), t) ≥ N ′(ϕ(x, y), t|s|n

)→ 1 as n → ∞, and so

ϕ(x, y) = 0 for all x, y ∈ X. Hence Dklf(x, y) = 0 for all x, y ∈ X and f is itself a quadratic mapping.

Theorem 3.2. Assume that a mapping f : X → Y with f(0) = 0 satisfies the inequality

N(Dklf(x, y), t) ≥ N ′(ϕ(x, y), t) (22)

and that ϕ : X2 → Z is a mapping for which there is a constant s ∈ R satisfying |s| > k2 such that

N ′(ϕ(xk,y

k

), t)≥ N ′

(1sϕ(x, y), t

), t > 0, (23)

for all x ∈ X and all t > 0. Then we can find a unique quadratic mapping Q : X → Y satisfying the equation

DklQ(x, y) = 0 and the inequality

N(f(x)−Q(x), t) ≥ N ′( ϕ(x, 0)

2(|s| − k2), t), t > 0, (24)

for all x ∈ X.

Proof. It follows from (17) and (23) that

N(f(x)− k2f

(xk

),

t

2|s|

)≥ N ′(ϕ(x, 0), t), t > 0,

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JIA_517_edited [07/23 10:35] 9/19

for all x ∈ X. Therefore it follows that

N

(f(x)− k2nf

( x

kn

),

n−1∑i=0

k2i

2|s|i+1t

)≥ N ′(ϕ(x, 0), t), t > 0,

for all x ∈ X and any integer n > 0. Thus we see from the last inequality that

N(f(x)− k2nf

( x

kn

), t)

≥ N ′

(ϕ(x, 0),

t∑n−1i=0

k2i

2|s|i+1

)

≥ N ′(ϕ(x, 0), 2(|s| − k2)t), t > 0.

The remaining assertion goes in a similar way as the corresponding part of Theorem 3.1.

We also observe that if k = 1 in Theorem 3.2, then N ′(ϕ(x, y), t) ≥ N ′(ϕ(x, y), |s|nt) → 1 as n → ∞,

and so ϕ(x, y) = 0 for all x, y ∈ X. Hence, Dklf = 0 and f is itself a quadratic mapping.

Corollary 3.3. Let X be a normed space and (R, N ′) be a fuzzy normed space. Assume that there exist

real numbers θ1 ≥ 0, θ2 ≥ 0 and that p is a real number such that either 0 < p < 2 or p > 2. If a mapping

f : X → Y with f(0) = 0 satisfies the inequality

N(Dklf(x, y), t) ≥ N ′(θ1‖x‖p + θ2‖y‖p, t)

for all x, y ∈ X and all t > 0, then we can find a unique quadratic mapping Q : X → Y satisfying the

equation DklQ(x, y) = 0 and the inequality

N(f(x)−Q(x), t) ≥{

N ′( θ1‖x‖p

2(k2−|k|p) , t), if 0 < p < 2, |k| > 1, (p > 2, |k| < 1)

N ′( θ1‖x‖p

2(|k|p−k2) , t), if p > 2, |k| > 1, (0 < p < 2, |k| < 1)

for all x ∈ X and all t > 0.

Proof. Taking ϕ(x, y) = θ1‖x‖p + θ2‖y‖p and applying Theorems 3.1 and 3.2, we obtain the desired

approximation, respectively.

The following is a simple example that the quadratic functional equation Dklf(x, y) = 0, k ≥ 2, k > l ≥ 1

is not stable for p = 2 in Corollary 3.3. This is a counterexample for the singular case p = 2 in a real space

with a fuzzy norm N(x, t) = N ′(x, t) = tt+‖x‖ .

Example 3.4. Let φ : R → R be defined by

φ(x) =

{μx2 if |x| < 1,μ otherwise,

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JIA_517_edited [07/23 10:35] 10/19

where μ > 0 is a positive constant, and define f : R → R by

f(x) =∞∑i=0

φ(kix)

k2ifor all x ∈ R.

Then f satisfies the functional inequality

|f(kx+ ly) + f(kx− ly)− kl[f(x+ y) + f(x− y)] (25)

−2k(k − l)[kf(x)− lf(y)]|

≤ 2k6μ(k2 + kl − l2 + 1)

k2 − 1(|x|2 + |y|2)

for all x, y ∈ R, but there do not exist a quadratic function Q : R → R and a constant β > 0 such that

|f(x)−Q(x)| ≤ β|x|2 for all x ∈ R. (26)

Proof. It is easy to see that φ is bounded by μ and f is bounded by k2μk2−1 on R. First, if |x|2 + |y|2 ≥ 1

k4 or

0, then

|Dklf(x, y)| ≤k2μ

k2 − 1{2(k2 + kl − l2 + 1)} ≤ 2k6μ(k2 + kl − l2 + 1)

k2 − 1(|x|2 + |y|2)

and thus (25) is true. Now suppose that 0 < |x|2 + |y|2 < 1k4 . Then there exists a positive integer i0 such

that

1

k2i0+3≤ |x|2 + |y|2 <

1

k2i0+2, (27)

so that k2i0 |x|2 < 1k2 , k

2i0 |y|2 < 1k2 and ki0−1(kx± ly), ki0−1(x±y), ki0−1x, ki0−1y all belong to the interval

(−1, 1). Hence, for i = 0, 1, · · · , i0 − 1,

Dklφ(kix, kiy) = 0

since each term of Dklφ(kix, kiy) is defined by μx2. Therefore, it follows from the definition of f and

inequality (27) that

|Dklf(x, y)| ≤∞∑i=0

1

k2i|Dklφ(k

ix, kiy)| ≤∞∑

i=i0

1

k2i|Dklφ(k

ix, kiy)| (28)

≤∞∑

i=i0

1

k2i{2(k2 + kl − l2 + 1)}μ =

2k2μ(k2 + kl − l2 + 1)

k2i0(k2 − 1)

≤ 2k5μ(k2 + kl − l2 + 1)

k2 − 1(|x|2 + |y|2)

for all x, y ∈ R with 0 < |x|2 + |y|2 < 1k4 . Thus f satisfies inequality (25) for all x, y ∈ R.

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We claim that the quadratic functional equation Dklf(x, y) = 0 is not stable for p = 2 in Corollary 3.3.

Suppose on the contrary that there exist a quadratic mapping Q : R → R and a constant β > 0 satisfying

(26). Since f is bounded and continuous for all x ∈ R, Q is bounded on any open interval containing the

origin and continuous at the origin. Therefore, Q must have the form Q(x) = ηx2 for any x in R. Thus we

obtain that

|f(x)| ≤ (β + |η|)|x|2 for all x ∈ R. (29)

However, we can choose a positive integer p with pμ > β + |η|. Then if x ∈ (0, 1kp−1 ), then kix ∈ (0, 1)

for all i = 0, 1, · · · , p− 1, and so for this x we get

f(x) =

∞∑i=0

φ(kix)

k2i≥

p−1∑i=0

μ(kix)2

k2i= pμx2 > (β + |η|)x2,

which contradicts (29). Therefore the quadratic functional equation Dklf(x, y) = 0 is not stable if p = 2 = q

is assumed in Corollary 3.3.

Corollary 3.5. Assume that for k �= 1, there exists a real number θ ≥ 0 such that the mapping f : X → Y

with f(0) = 0 satisfies the inequality

N(Dklf(x, y), t) ≥ N ′(θ, t)

for all x, y ∈ X and all t > 0. Then we can find a unique quadratic mapping Q : X → Y satisfying the

equation DklQ(x, y) = 0 and the inequality

N(f(x)−Q(x), t) ≥ N ′( θ

2|k2 − 1| , t)

for all x ∈ X and all t > 0.

We remark that if θ = 0, then N(Dklf(x, y), t) ≥ N ′(0, t) = 1, and so Dklf(x, y) = 0. Thus we get f = Q

is itself a quadratic mapping.

4 Stability of equation (2) by fixed point method

Now, in the next theorem, we are going to consider a stability problem concerning the stability of equation

(2) by using a fixed point theorem of the alternative for contraction mappings on generalized complete metric

spaces due to Margolis and Diaz [12].

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JIA_517_edited [07/23 10:35] 12/19

Theorem 4.1. Assume that there exists a constant s ∈ R with |s| �= 1 and q > 0 satisfying 0 < |s|1q < k2

such that a mapping f : X → Y with f(0) = 0 satisfies the inequality

N(Dklf(x, y), t1 + t2) ≥ min{N ′(ϕ(x), tq1), N′(ϕ(y), tq2)} (30)

for all x, y ∈ X, ti > 0 (i = 1, 2) and that ϕ : X → Z is a mapping satisfying

N ′(ϕ(kx), t) ≥ N ′(sϕ(x), t) (31)

for all x ∈ X and all t > 0. Then there exists a unique quadratic mapping Q : X → Y satisfying the equation

DklQ(x, y) = 0 and the inequality

N(f(x)−Q(x), t) ≥ min

{N ′

(ϕ(x)

(k2 − |s|1q )q

, tq

), N ′

(ϕ(0)

(k2 − |s|1q )q

, tq

)}(32)

for all x ∈ X and all t > 0.

Proof. We consider the set of functions

Ω := {g : X → Y |g(0) = 0}

and define a generalized metric on Ω as follows:

dΩ(g, h) := inf{K ∈ [0,∞] : N(g(x)− h(x),Kt) ≥ min{N ′(ϕ(x), tq), N ′(ϕ(0), tq)},

∀x ∈ X, ∀t > 0}.

Then one can easily see that (Ω, dΩ) is a complete generalized metric space [37, 38].

Now, we define an operator J : Ω → Ω as

Jg(x) =g(kx)

k2

for all g ∈ Ω, x ∈ X.

We first prove that J is strictly contractive on Ω. For any g, h ∈ Ω, let ε ∈ [0,∞) be any constant with

dΩ(g, h) ≤ ε. Then we deduce from the use of (31) and the definition of dΩ(g, h) that

N(g(x)− h(x), εt) ≥ min{N ′(ϕ(x), tq), N ′(ϕ(0), tq)}, ∀x ∈ X, t > 0

⇒ N

(g(kx)

k2− h(kx)

k2,|s|

1q εt

k2

)≥ min{N ′(ϕ(kx), |s|tq), N ′(ϕ(0), |s|tq)}

⇒ N

(Jg(x)− Jh(x),

|s|1q εt

k2

)≥ min{N ′(ϕ(x), tq), N ′(ϕ(0), tq)}, ∀x ∈ X, t > 0,

⇒ dΩ(Jg, Jh) ≤|s|

1q ε

k2.

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JIA_517_edited [07/23 10:35] 13/19

Since ε is an arbitrary constant with dΩ(g, h) ≤ ε, we see that for any g, h ∈ Ω,

dΩ(Jg, Jh) ≤|s|

1q

k2dΩ(g, h),

which implies J is strictly contractive with the constant |s|1q

k2 < 1 on Ω.

We now want to show that d(f, Jf) < ∞. If we put y := 0, ti := t (i = 1, 2) in (30), then we arrive at

N(f(x)− f(kx)

k2,t

k2

)≥ min{N ′(ϕ(x), tq), N ′(ϕ(0), tq)},

which yields dΩ(f, Jf) ≤ 1k2 and so dΩ(J

nf, Jn+1f) ≤ dΩ(f, Jf) ≤ 1k2 for all n ∈ N.

Using the fixed point theorem of the alternative for contractions on generalized complete metric spaces

due to Margolis and Diaz [12], we see the following (i), (ii) and (iii):

(i) There is a mapping Q : X → Y with Q(0) = 0 such that

dΩ(f,Q) ≤ 1

1− |s|1q

k2

dΩ(f, Jf) ≤1

k2 − |s|1q

and Q is a fixed point of the operator J , that is, 1k2Q(kx) = JQ(x) = Q(x) for all x ∈ X. Thus we can get

N

(f(x)−Q(x),

t

k2 − |s|1q

)≥ min{N ′(ϕ(x), tq), N ′(ϕ(0), tq)},

N(f(x)−Q(x), t) ≥ min{N ′

(ϕ(x), (k2 − |s|

1q

)q

tq), N ′

(ϕ(0), (k2 − |s|

1q

)q

tq)}

for all t > 0 and all x ∈ X.

(ii) dΩ(Jnf,Q) → 0 as n → ∞. Thus we obtain

N(f(knx)

k2n−Q(x), t

)= N(f(knx)−Q(knx), k2nt)

≥ min

{N ′

(ϕ(knx)

(k2 − |s|1q )q

, k2nqtq

), N ′

(ϕ(0)

(k2 − |s|1q )q

, k2nqtq

)}

= min

{N ′

(ϕ(x)

(k2 − |s|1q )q

,

(k2q

|s|

)n

tq

), N ′

(ϕ(0)

(k2 − |s|1q )q

,

(k2q

|s|

)n

tq

)}

→ 1 as n → ∞(k2q

|s| > 1

)

for all t > 0 and all x ∈ X, that is, the mapping Q : X → Y given by

N - limn→∞

f(knx)

k2n= Q(x) (33)

is well defined for all x ∈ X. In addition, it follows from the conditions (30), (31) and (N4) that

N(Dklf(k

nx, kny)

k2n, t)

≥ min

{N ′

(ϕ(knx),

k2nqtq

2q

), N ′

(ϕ(kny),

k2nqtq

2q

)}

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JIA_517_edited [07/23 10:35] 14/19

= min

{N ′

(|s|nϕ(x), k

2nqtq

2q

), N ′

(|s|nϕ(y), k

2nqtq

2q

)}

= min

{N ′

(ϕ(x),

(k2q

|s|

)ntq

2q

), N ′

(ϕ(y),

(k2q

|s|

)ntq

2q

)}

→ 1 as n → ∞, t > 0, (34)

for all x ∈ X. Therefore we obtain, by use of (N4), (33) and (34),

N(DklQ(x, y), t) ≥ min

{N(DklQ(x, y)− Dklf(k

nx, kny)

k2n,t

2

), N

(Dklf(knx, kny)

k2n,t

2

)}

= N(Dklf(k

nx, kny)

k2n,t

2

)(for sufficiently large n)

≥ min

{N ′

(ϕ(x),

(k2q

|s|

)ntq

4q

), N ′

(ϕ(y),

(k2q

|s|

)ntq

4q

)}

→ 1 as n → ∞, t > 0,

which implies DklQ(x, y) = 0 by (N2), and so the mapping Q is quadratic satisfying equation (2).

(iii) The mapping Q is a unique fixed point of the operator J in the set Δ = {g ∈ Ω|dΩ(f, g) < ∞}.

Thus if we assume that there exists another Euler-Lagrange type quadratic mapping Q′ : X → Y satisfying

inequality (32), then

Q′(x) =Q′(kx)

k2= JQ′(x), dΩ(f,Q

′) ≤ 1

(k2 − |s|1q )

< ∞,

and so Q′ is a fixed point of the operator J and Q′ ∈ Δ = {g ∈ Ω|dΩ(f, g) < ∞}. By the uniqueness of

the fixed point of J in Δ, we find that Q = Q′, which proves the uniqueness of Q satisfying inequality (32).

This ends the proof of the theorem.

We observe that if 0 < |s| < 1 in Theorem 4.1, then min{N ′(ϕ(x), tq), N ′(ϕ(0), tq)

}= N ′(ϕ(x), tq) for

all x ∈ X and all t > 0 since N ′(ϕ(x), tq) ≥ N ′(ϕ(0), tq

|s|n

)→ 1 as n → ∞ by (31).

Theorem 4.2. Assume that there exists a constant s ∈ R with |s| �= 1 and q > 0 satisfying |s|1q > k2 such

that a mapping f : X → Y with f(0) = 0 satisfies the inequality

N(Dklf(x, y), t1 + t2) ≥ min{N ′(ϕ(x), tq1), N′(ϕ(y), tq2)}

for all x, y ∈ X, ti > 0 (i = 1, 2) and that ϕ : X → Z is a mapping satisfying

N ′(ϕ(xk

), t)≥ N ′

(1sϕ(x), t

)

15

JIA_517_edited [07/23 10:35] 15/19

for all x ∈ X. Then there exists a unique quadratic mapping Q : X → Y satisfying the equationDklQ(x, y) =

0 and the inequality

N(f(x)−Q(x), t) ≥ min

{N ′

(ϕ(x)

(|s|1q − k2)q

, tq

), N ′

(ϕ(0)

(|s|1q − k2)q

, tq

)}, t > 0,

for all x ∈ X.

Proof. The proof of this theorem is similar to that of Theorem 4.1.

Authors’ contributions

All authors carried out the proof. All authors conceived of the study, and participated in its design and

coordination. All authors read and approved the final manuscript.

Competing interests

The authors declare that they have no competing interests.

Acknowledgements

The authors would like to thank the referees and the editors for carefully reading this article and for their

valuable comments. This work was supported by research fund of Chungnam National University.

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