EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS
2025, Vol. 18, Issue 1, Article Number 5757
ISSN 1307-5543 – ejpam.com
Published by New York Business Global

Hyperstability of Functional Equation Deriving from
Quintic Mapping in Banach Spaces by Fixed Point

Method

Subramani Karthikeyan1, Siriluk Donganont2,∗, Choonkil Park3,
Kandhasamy Tamilvanan1, Yongqiao Wang4

1 Department of Mathematics, Faculty of Science & Humanities, R.M.K. Engineering
College, Kavaraipettai, Tiruvallur 601 206, Tamil Nadu, India
2 School of Science, University of Phayao, Phayao 56000, Thailand
3 Department of Mathematics, Research Institute for Convergence of Basic Science,
Hanyang University, Seoul 04763, Korea
4 School of Science, Dalian Maritime University, Dalian 116026, P. R. China

Abstract. In this work, we examine the hyperstability of the quintic functional equation ϕ(u +
3v)− 5ϕ(u+ 2v)− ϕ(u− 2v) + 10ϕ(u+ v) + 5ϕ(u− v)− 10ϕ(u)− 120ϕ(v) = 0, in Banach spaces
by means of Brzdȩk’s fixed point theorem.

2020 Mathematics Subject Classifications: 39B52, 39B72, 39B82, 47H10

Key Words and Phrases: Quintic functional equation, hyperstability, fixed point, stability

1. Introduction and preliminaries

One of the most important areas of mathematical research, which has its origins in
problems relating to applied mathematics, is the investigation of stability issues for func-
tional equations. Ulam [28] stated the following as the first query pertaining to the stability
of homomorphisms.

Let U be a group and V be a metric group with a metric d(·, ·). Given ϵ > 0, is there
a δ > 0 such that if a function ϕ : U → V fulfills

d(ϕ(uv), ϕ(u)ϕ(v)) < δ,

for all u, v ∈ U , then there is a homomorphism Φ : U → V with

d (ϕ(u),Φ(u)) < ϵ,

∗Corresponding author.
DOI: https://doi.org/10.29020/nybg.ejpam.v18i1.5757

Email addresses: karthik.sma204@yahoo.com (S. Karthikeyan),
siriluk.pa@up.ac.th (S. Donganont), baak@hanyang.ac.kr (C. Park),
tamiltamilk7@gmail.com (K. Tamilvanan), wangyq@dimu.edu.cn (Y. Wang)

https://www.ejpam.com 1 Copyright: © 2025 The Author(s). (CC BY-NC 4.0)



S. Karthikeyan et al. / Eur. J. Pure Appl. Math, 18 (1) (2025), 5757 2 of 18

for all u ∈ U?
Hyers provided the first partial answer to Ulam’s concern regarding the Cauchy equa-

tion in Banach spaces in [19]. Later, Aoki was the first to generalize Hyers’ findings and
not until much later by Rassias [24] and Găvruţa [18]. Since then, several functional equa-
tions’ stability issues have been thoroughly researched (see [7, 8, 20, 25]). If any function
f approximates (in some sense) the solution to the functional equation, then the func-
tional equation is said to be hyperstable. It appears that the first hyperstability finding,
which dealt with ring homomorphisms, was published in [6]. Hyperstability, however, is
mentioned for the first time in [21].

Brzdȩk investigated the hyperstability results for the Cauchy equation (see [9–11]). The
hyperstability of the parametric basic equation of information was addressed by Gselmann
in [17]. Bahyrycz and Piszczek reported the Jensen functional equation’s hyperstability in
[4]. For a certain class of complete metric spaces, Brzdȩk and Ciepliński [13] demonstrated
a simple fixed point theorem, namely, complete non-Archimedean metric spaces, p-adic
strings, and superstrings that are related to several quantum physics-related phenomena.
They also demonstrated that how effective and practical this theorem is for demonstrating
the Hyers-Ulam stability of a huge class of functional equations in a single variable.

The fixed point theorem [12, Theorem 1] was restated in 2-Banach spaces by El-Fassi
[15] in 2017, and a radical quartic functional equation was introduced and examined its
Ulam stability in 2-Banach spaces by fixed-point approache. In [5], Bounader examined
the hyperstability of the quartic functional equation in Banach spaces. In [2], Aribou et al.
presented the hyperstability results of a cubic- quartic functional equation in ultrametric
Banach spaces. And also, in 2020, Sayar and Bergam [27], examined stability and hyper-
stability for the quadratic functional equation in 2-Banach space by Brzdȩk fixed-point
theorem.

Motivated by the above results on the hyperstability of additive functional equations,
quadratic functional equations, cubic functional equations and quartic functional equa-
tions, in the current work, we try to examine the hyperstability of the following quintic
functional equation

ϕ(u+ 3v)− 5ϕ(u+ 2v)− ϕ(u− 2v) + 10ϕ(u+ v) + 5ϕ(u− v)− 10ϕ(u)− 120ϕ(v) = 0,

in Banach spaces by means of Brzdȩk’s fixed point approach.
A concept employed by Brzdȩk in [9–11] and later by Piszczek [23] served as the

inspiration for the manner of the primary results’ confirmation. Its foundation is a fixed-
point theorem for functional spaces discovered by Brzdȩk (see [12], Theorem 1]). Most
frequently, the superstability and hyperstability, which also accepts bounded functions-are
concerned. Numerous articles have been written on this topic and we refer to [1, 3, 4, 14,
16, 17, 21, 22, 26, 29].

Throughout the paper, N denote the set of all natural numbers, N0 = N ∪ {0}, R
denote the set of all real numbers, Nm denote the set of all natural numbers greater than
or equal to m, for every m ∈ N and R+ denote the set of all positive real numbers. We
use the notation U0 for the set U\{0}.



S. Karthikeyan et al. / Eur. J. Pure Appl. Math, 18 (1) (2025), 5757 3 of 18

Theorem 1. [13] Let U be a non-empty set, (V, d) be a complete metric space, and Υ :
VU → VU fulfill the hypothesis

lim
m→+∞

Υδm = 0,

for {δm}m∈N in VU with
lim

m→+∞
δm = 0.

Suppose that an operator Φ : VU → VU fulfills

d (Φψ(u),Φς(u)) ≤ Υ(∆(ψ, ς))), ψ, ς ∈ VU ,

for all u ∈ U , where a mapping ∆ : VU × VU → RU
+ is defined by

∆(ψ, ς) (u) := d(ψ(u), ς(u)), ψ, ς ∈ VU , u ∈ U .

If there is a mapping ϑ : U → R+ and ζ : U → V fulfilling

d (Φψ(u),Φς(u)) ≤ ψ(u)

and
ϑ∗(u) :=

∑
m∈N0

(Υmϑ) (u) <∞

for all u ∈ U , then the limit
lim

m→+∞
(Φmζ) (u)

exists for each u ∈ U . Furthermore, the mapping χ ∈ VU , defined by

χ(u) := lim
m→+∞

(Φmζ) (u)

is a fixed point of Φ with
d (ζ(u), χ(u)) ≤ ϑ∗(u)

for all u ∈ U .

The upcoming fixed point theorem, which corresponds to Theorem 1 in complete
normed space, is then discussed. This outcome is an important factor in the formula-
tion of stability findings.

Theorem 2. Let U be a nonempty set, (V, ∥ · ∥) be a Banach space and let ϕ1, ϕ2, · · · , ϕl :
U → U be mappings and L1, · · · , Ll : X → R+be functions. Suppose that Φ : VU → VU

and two operators Υ : RU×U
+ → RU×U

+ fulfill the conditions:

∥Φψ(u)− Φς(u)∥ ≤
l∑

i=1

Li(u) ∥ψ (ϕi(u))− ς (ϕi(u))∥



S. Karthikeyan et al. / Eur. J. Pure Appl. Math, 18 (1) (2025), 5757 4 of 18

for all ψ, ς ∈ VU , u ∈ U and

Υδ(u) :=
l∑

i=1

Li(u)δ (ϕi(u)) , δ ∈ RU×U
+ , u ∈ U .

If there exist mappings ϑ : U × U → R+ and ζ : U → V satisfying

∥Φζ(u)− ζ(u)∥ ≤ ϑ(u)

and

ϑ∗(u) :=

∞∑
m=0

(Υmϑ) (u) <∞

for all u ∈ U , then the limit
lim

m→+∞
(Φmζ) (u) (1)

exists for every u ∈ U . Furthermore, the mapping χ : U → V defined by

χ := lim
m→+∞

(Φmζ) (u)

is a fixed point of Φ with
∥ζ(u)− χ(u)∥ ≤ ϑ∗(u)

for all u ∈ U .

For our notational handiness, we use the abbreviation

Dϕ(u, v) = ϕ(u+ 3v)− 5ϕ(2v + u)− ϕ(u− 2v) + 10ϕ(v + u)

+5ϕ(u− v)− 10ϕ(u)− 120ϕ(v). (2)

2. Main results

In this section, we demonstrate various hyperstability and stability of (2) in Banach
spaces utilizing Theorem 2. Suppose that U is a normed space along with U0 = U\{0}
and (V, ∥ · ∥) is a Banach space.

Theorem 3. Let τ1, τ2 : U0 × U0 → R+ be two functions such that

W := {m ∈ N : αm < 1} = ∅,

where

αm :=
1

120
η1(m+ 3)η2(m+ 3) +

1

120
η1(m− 2)η2(m− 2) +

1

24
η1(m+ 2)η2(m+ 2)

+
1

12
η1(m+ 1)η2(m+ 1) +

1

12
η1(m)η2(m) +

1

24
η1(m− 1)η2(m− 1)



S. Karthikeyan et al. / Eur. J. Pure Appl. Math, 18 (1) (2025), 5757 5 of 18

and

ηi(m) := inf {l ∈ R+ : τi(mu) ≤ lτi(u), u ∈ U0}

for all m ∈ N, where i = 1, 2. Assume that ϕ : U → V fulfills

∥Dϕ(u, v)∥ ≤ τ1(u)τ2(v), u, v ∈ U0 (3)

such that 3v+ u ̸= 0, 2v+ u ̸= 0, u− 2v ̸= 0, u− v ̸= 0 and v+ u ̸= 0. Then there is only
one quintic mapping H : U → V fulfilling

∥ϕ(u)−H(u)∥ ≤ η0τ1(u)τ2(u)

for all u ∈ U0, where

η0 := inf
m∈W

{
η1(m)

120(1− αm)

}
.

Proof. Replacing (u, v) by (nu, u) in (3), we obtain∥∥∥∥ 1

120
ϕ((3 + n)u)− 1

24
ϕ((2 + n)u)− 1

120
ϕ((n− 2)u) +

1

12
ϕ((1 + n)u)

+
1

24
ϕ((n− 1)u)− 1

12
ϕ(nu)− ϕ(u)

∥∥∥∥ ≤ 1

120
τ1(nu)τ2(u) (4)

for all u ∈ U0 and all n ∈ N. For any n ∈ N, we define the operator Φn : VU0 → VU0 by

Φnψ(u) :=
1

120
ψ((3 + n)u)− 1

24
ψ((2 + n)u)− 1

120
ψ((n− 2)u)

+
1

12
ψ((1 + n)u) +

1

24
ψ((n− 1)u)− 1

12
ψ(nu) (5)

for all ψ ∈ VU0 and all u ∈ U0. Moreover, putting

ϑn(u) :=
1

120
τ1(nu)τ2(u) (6)

for all u ∈ U0, and observe that

ϑn(u) =
1

120
τ1(nu)τ2(u) ≤

1

120
η1(n)τ1(u)τ2(u) (7)

for all u ∈ U0 and all n ∈ N. Using the conditions (5) and (7) in (4), we get

∥ϕ(u)− Φnϕ(u)∥ ≤ ϑn(u)

for all u ∈ U0. Moreover, for any u ∈ U0 and every ψ, ς ∈ VU0 , we obtain

∥Φnψ(u)− Φnς(u)∥ =

∥∥∥∥ 1

120
ψ((3 + n)u)− 1

24
ψ((2 + n)u)− 1

120
ψ((n− 2)u)



S. Karthikeyan et al. / Eur. J. Pure Appl. Math, 18 (1) (2025), 5757 6 of 18

+
1

12
ψ((1 + n)u) +

1

24
ψ((n− 1)u)− 1

12
ψ(nu)

− 1

120
ς((3 + n)u) +

1

24
ς((2 + n)u) +

1

120
ς((n− 2)u)

− 1

12
ς((1 + n)u)− 1

24
ς((n− 1)u) +

1

12
ς(nu)

∥∥∥∥
≤ 1

120

∥∥∥∥(ψ − ς)((3 + n)u)

∥∥∥∥+
1

24

∥∥∥∥(ψ − ς)((2 + n)u)

∥∥∥∥
+

1

120

∥∥∥∥(ψ − ς)((n− 2)u)

∥∥∥∥+
1

12

∥∥∥∥(ψ − ς)((1 + n)u)

∥∥∥∥
+

1

24

∥∥∥∥(ψ − ς)((n− 1)u)

∥∥∥∥+
1

12

∥∥∥∥(ψ − ς)(nu)

∥∥∥∥.
This brings us to define the operator Υn : RU0×U0

+ → RU0×U0
+ by

Υnδ(u) :=
1

120
δ((3 + n)u) +

1

24
δ((2 + n)u) +

1

120
δ((n− 2)u)

+
1

12
δ((1 + n)u) +

1

24
δ((n− 1)u) +

1

12
δ(nu)

for all u ∈ U0 and all δ ∈ RU0×U0
+ . For every n ∈ N, the operator previously defined has the

form specified in (1) with ϕ1(u) = (3 + n)u, L1(u) =
1

120 ; ϕ2(u) = (2 + n)u, L2(u) =
1
24 ;

ϕ3(u) = (n−2)u, L3(u) =
1

120 ; ϕ4(u) = (1+n)u, L4(u) =
1
12 ; ϕ5(u) = (n−1)u, L5(u) =

1
24 ;

ϕ6(u) = nu, L6(u) =
1
12 for all u ∈ U0.

By induction, we will prove that for all u ∈ U0, m ∈ N0, and n ∈W , we have

(Υm
n ϑn) (u) ≤

1

120
η1(n)α

m
n τ1(u)τ2(u). (8)

We may deduce the inequality (8) holds for m = 0 from (6) and (7). Following that,
we suppose that (8) is true for m = k, where k ∈ N. Then(
Υk+1

n ϑn

)
(u) = Υn

((
Υk

nϑn

)
(u)

)
=

1

120
(Υn

nϑn) ((3 + n)x) +
1

24
(Υn

nϑn) ((2 + n)u) +
1

120
(Υn

nϑn) ((n− 2)u)

+
1

12
(Υn

nϑn) ((1 + n)u) +
1

24
(Υn

nϑn) ((n− 1)u) +
1

12
(Υn

nϑn) (nu)

≤ 1

120

(
1

120
η1(n)α

k
nτ1((3 + n)u)τ2((3 + n)u)

)
+

1

24

(
1

120
η1(n)α

k
nτ1((2 + n)u)τ2((2 + n)u)

)
+

1

120

(
1

120
η1(n)α

k
nτ1((n− 2)u)τ2((n− 2)u)

)



S. Karthikeyan et al. / Eur. J. Pure Appl. Math, 18 (1) (2025), 5757 7 of 18

+
1

12

(
1

120
η1(n)α

k
nτ1((1 + n)u)τ2((1 + n)u)

)
+

1

24

(
1

120
η1(n)α

k
nτ1((n− 1)u)τ2((n− 1)u)

)
+

1

12

(
1

120
η1(n)α

k
nτ1(nu)τ2(nu)

)
≤ 1

120
η1(n)α

k+1
n τ1(u)τ2(u)

for all u ∈ U0 and all n ∈ W . Thus, for m = k + 1, the inequality (8) holds. Since the
inequality (8) holds for all m ∈ N0, we may obtain this conclusion. Thus we obtain

ϑ∗n(u) =
∞∑

m=0

(Υm
n ϑn) (u)

≤
∑

0≤n≤∞

1

120
η1(n)α

m
n τ1(u)τ2(u)

≤ η1(n)

120(1− αn)
τ1(u)τ2(u) <∞

for all u ∈ U0 and n ∈W . Therefore, according to Theorem 2, we obtain the limit mapping

Hn(u) := lim
m→+∞

(Φm
n ϕ) (u)

exists for each u ∈ U0 and n ∈W , and

∥ϕ(u)−Hn(u)∥ ≤ η1(n)τ1(u)τ2(u)

120(1− αn)
(9)

for all u ∈ U0 and n ∈W .
Now, we will prove that Hn fulfills (2). It is enough to prove the following inequality

∥D(Φm
n ϕ)(u, v)∥ ≤ αm

n τ1(u)τ2(v), (10)

for all u, v ∈ U0 and n ∈ W . Consider k ∈ N and suppose that (10) holds for m = k.
Then, for each u, v ∈ U0 and n ∈W , we get

∥∥∥D(Φk+1
n ϕ)(u, v)

∥∥∥ ≤ 1

120
αk
nτ1((3 + n)u)τ2((3 + n)v) +

1

24
αk
nτ1((2 + n)u)τ2((2 + n)v)

+
1

120
αk
nτ1((n− 2)u)τ2((n− 2)v) +

1

12
αk
nτ1((1 + n)u)τ2((1 + n)v)

+
1

24
αk
nτ1((n− 1)u)τ2((n− 1)v) +

1

12
αk
nτ1((n)u)τ2((n)v)

≤ αk+1
n τ1(u)τ2(v).



S. Karthikeyan et al. / Eur. J. Pure Appl. Math, 18 (1) (2025), 5757 8 of 18

By induction, we need to prove that (10) holds for all u, v ∈ U0, m ∈ N0, and n ∈ W .
Taking the limit m→ ∞ in (10), we get

Hn(u+ 3v)− 5Hn(u+ 2v)−Hn(u− 2v) + 10Hn(u+ v)

+5Hn(u− v)− 10Hn(u)− 120Hn(v) = 0

for all u, v ∈ U0 such that 3v+u ̸= 0, 2v+u ̸= 0, u−2v ̸= 0, u−v ̸= 0, v+u ̸= 0, m ∈ N0,
and n ∈W . This implies that we can be defined H : U → V which fulfills

H(u) :=
1

120
H((n+ 3)u)− 1

120
H((n− 2)u)− 1

24
H((n+ 2)u)

+
1

12
H((n+ 1)u)− 1

12
H(nu) +

1

24
H((n− 1)u), (11)

for all u ∈ U0 and all n ∈W .
Next, we need to show that each quintic mapping H : U → V fulfills the inequality

∥ϕ(u)−H(u)∥ ≤ Lτ1(u)τ2(u) (12)

for all u ∈ U0, with 0 < L, is equal to Hn for every n ∈ W . As a result, we set n0 ∈ W
and H : U → V fulfilling (12). From (9), for every u ∈ U0, we have

∥H(u)−Hn0(u)∥ ≤ ∥H(u)− ϕ(u)∥+ ∥ϕ(u)−Hn0(u)∥
≤ Lτ1(u)τ2(u) + ϑ∗n0

(u)

≤ L0τ1(u)τ2(u)

∞∑
m=0

αm
n0
, (13)

where L0 := (1−αn0)L+
1

120η1(n0) > 0 and we exclude the case that τ1(u) ≡ 0 or τ2(u) ≡ 0
which is trivial. From the observation, the functions H and Hn0 are the solutions to the
functional equation (11) for every n ∈W .

Next, we prove that, for every j ∈ N0, we obtain

∥H(u)−Hn0(u)∥ ≤ L0τ1(u)τ2(u)

∞∑
m=j

α∞
n0

(14)

for all u ∈ U0. The inequality (13) is valid for the case j = 0. The next step is to correct
k ∈ N and suppose that (14) is true for j = k. In the sense of (13), for every u ∈ U0, we
obtain

∥H(u)−Hn0(u)∥ ≤ 1

120
L0τ1((3 + n0)u)τ2((3 + n0)u))

∞∑
m=k

αm
n0

+
1

24
L0τ1((2 + n0)u)τ2((2 + n0)u))

∞∑
m=k

αm
n0



S. Karthikeyan et al. / Eur. J. Pure Appl. Math, 18 (1) (2025), 5757 9 of 18

+
1

120
L0τ1((−2 + n0)u)τ2((−2 + n0)u))

∞∑
m=k

αm
n0

+
1

12
L0τ1((1 + n0)u)τ2((1 + n0)u))

∞∑
m=k

αm
n0

+
1

24
L0τ1((−1 + n0)u)τ2((−1 + n0)u))

∞∑
m=k

αm
n0

+
1

12
L0τ1((n0)u)τ2((n0)u))

∞∑
m=k

αm
n0

≤ L0αn0τ1(u)τ2(u)
∞∑

m=k

αm
n0

≤ L0τ1(u)τ2(u)
∞∑

m=k+1

αm
n0
.

Thus the condition (14) is valid for j = 1+ k. As a result, we may say that the inequality
(14) is true for all j ∈ N0. Taking the limit j → ∞ in inequality (14), we obtain

H = Hn0 . (15)

Also, in view of (9), we have

∥ϕ(u)−Hn0(u)∥ ≤ η1(n)τ1(u)τ2(u)

120(1− αn)
,

for all u ∈ U0 and all n ∈W . This implies the condition (3) with H = Hn0 and (15) shows
the uniqueness of H.

Theorem 4. Let τ : U0 × U0 → R+ be a function such that

W := {m ∈ N : αm < 1} = ∅,

where

αm :=
1

120
η(3 +m) +

1

24
η(2 +m) +

1

120
η(−2 +m)

+
1

12
η(1 +m) +

1

24
η(−1 +m) +

1

12
η(m)

and

η(m) := inf {l ∈ R+ : τ(mu) ≤ lτ(u), u ∈ U0} , (16)

for every m ∈ N. Assume that ϕ : U → V fulfills

∥Dϕ(u, v)∥ ≤ τ(u) + τ(v) (17)



S. Karthikeyan et al. / Eur. J. Pure Appl. Math, 18 (1) (2025), 5757 10 of 18

for all u, v ∈ U0, such that 3v + u ̸= 0, 2v + u ̸= 0, u− 2v ̸= 0, u− v ̸= 0 and v + u ̸= 0.
Then there is only one quintic mapping H : U → V satisfying

∥H(u)− ϕ(u)∥ ≤ η0τ(u), (18)

for all u ∈ U0, where

η0 := inf
m∈W

{
1 + η(m)

120(1− αm)

}
.

Proof. Replacing (u, v) by (nu, u) in (17), we have∥∥∥∥ 1

120
ϕ((3 + n)u)− 1

24
ϕ((2 + n)u)− 1

120
ϕ((−2 + n)u) +

1

12
ϕ((1 + n)u)

+
1

24
ϕ((−1 + n)u)− 1

12
ϕ(nu)− ϕ(u)

∥∥∥∥ ≤ 1

120
(τ1(nu) + τ2(u)) (19)

for all u ∈ U0 and all n ∈ N. For any n ∈ N, we define the operator Φn : VU0 → VU0 by

Φnψ(u) :=
1

120
ψ((n+ 3)u)− 1

24
ψ((n+ 2)u)− 1

120
ψ((n− 2)u)

+
1

12
ψ((n+ 1)u) +

1

24
ψ((n− 1)u)− 1

12
ψ(nu)

for all ψ ∈ VU0 and all u ∈ U0. Moreover, put

ϑn(u) :=
1

120
(τ(nu) + τ(u)) , (20)

for all u ∈ U0, and observe that

ϑn(u) =
1

120
(τ(nu) + τ(u)) ≤ 1

120
(η(n) + 1) τ(u) (21)

for all u ∈ U0 and all n ∈ N. The inequality (19), then has the following form

∥ϕ(u)− Φnϕ(u)∥ ≤ ϑn(u)

for all u ∈ U0. Moreover, for any u ∈ U0 and every ψ, ς ∈ VU0 , we obtain

∥Φnψ(u)− Φnς(u)∥ ≤ 1

120

∥∥∥∥(ψ − ς)((n+ 3)u)

∥∥∥∥+
1

24

∥∥∥∥(ψ − ς)((n+ 2)u)

∥∥∥∥
+

1

120

∥∥∥∥(ψ − ς)((n− 2)u)

∥∥∥∥+
1

12

∥∥∥∥(ψ − ς)((n+ 1)u)

∥∥∥∥
+

1

24

∥∥∥∥(ψ − ς)((n− 1)u)

∥∥∥∥+
1

12

∥∥∥∥(ψ − ς)(nu)

∥∥∥∥.
This brings us to define the operator Υn : RU0×U0

+ → RU0×U0
+ by

Υnδ(u) :=
1

120
δ((n+ 3)u) +

1

24
δ((n+ 2)u) +

1

120
δ((n− 2)u)



S. Karthikeyan et al. / Eur. J. Pure Appl. Math, 18 (1) (2025), 5757 11 of 18

+
1

12
δ((n+ 1)u) +

1

24
δ((n− 1)u) +

1

12
δ(nu)

for all u ∈ U0 and all δ ∈ RU0×U0
+ . For every n ∈ N, the form of the operator previously

specified is given in (1) with

ϕ1(u) = (n+ 3)u, L1(u) =
1

120
,

ϕ2(u) = (n+ 2)u, L2(u) =
1

24
,

ϕ3(u) = (n− 2)u, L3(u) =
1

120
,

ϕ4(u) = (n+ 1)u, L4(u) =
1

12
,

ϕ5(u) = (n− 1)u, L5(u) =
1

24
,

ϕ6(u) = nu, L6(u) =
1

12

for all u ∈ U0. By induction, we will verify that for every u ∈ U0, m ∈ N0, and n ∈W , we
have

(Υm
n ϑn) (u) ≤

1

120
(η(n) + 1)αm

n τ(u). (22)

The condition (22) for m = 0 is derived from (20) and (21).
Suppose that (22) holds for m = k. Then(

Υk+1
n ϑn

)
(u) = Υn

((
Υk

nϑn

)
(u)

)
=

1

120
(Υn

nϑn) ((n+ 3)x) +
1

24
(Υn

nϑn) ((n+ 2)u) +
1

120
(Υn

nϑn) ((n− 2)u)

+
1

12
(Υn

nϑn) ((n+ 1)u) +
1

24
(Υn

nϑn) ((n− 1)u) +
1

12
(Υn

nϑn) (nu)

≤ 1

120
(η(n) + 1)αk+1

n τ(u),

for all u ∈ U0 and every n ∈W . Therefore, for m = k+1, the inequality (22) holds. As a
result, we may say that the inequality (22) holds for all m ∈ N0. Thus we have

ϑ∗n(u) =

∞∑
m=0

(Υm
n ϑn) (u)

≤
∞∑
n=0

1

120
(η(n) + 1)αm

n τ(u)

≤ (η(n) + 1)

120(1− αn)
τ(u) <∞



S. Karthikeyan et al. / Eur. J. Pure Appl. Math, 18 (1) (2025), 5757 12 of 18

for all u ∈ U0 and n ∈W . Therefore, according to Theorem 2, we obtain the limit function

Hn(u) := lim
m→+∞

(Φm
n ϕ) (u)

exists for every u ∈ U0 and every n ∈W , and

∥ϕ(u)−Hn(u)∥ ≤ (η(n) + 1)τ(u)

120(1− αn)
(23)

for all u ∈ U0 and n ∈W . Now, we want to prove that Hn fulfills (2), it is enough to show
the following inequality

∥D(Φm
n ϕ)(u, v)∥ ≤ αm

n (τ(u) + τ(v)) (24)

for all u, v ∈ U0, m ∈ N0, and n ∈ W . Since the condition (17) is all that is required in
the situation m = 0, assume that k ∈ N and suppose that (24) holds for every m = k and
u, v ∈ U0 and n ∈W . Then, for each u, v ∈ U0 and n ∈W , we have∥∥∥D(Φk+1

n ϕ)(u, v)
∥∥∥ ≤ αk+1

n (τ(u) + τ2(v)) .

By induction, we need to prove that (24) holds for every u, v ∈ U0, m ∈ N0, and n ∈ W .
Taking the limit m→ ∞ in (24), we obtain

DHn(u, v) = 0

for all u, v ∈ U0, m ∈ N0, and n ∈W . This implies that the mapping H : U → V satisfies

H(u) :=
1

120
H((n+ 3)u)− 1

24
H((n+ 2)u)− 1

120
H((n− 2)u)

+
1

12
H((n+ 1)u) +

1

24
H((n− 1)u)− 1

12
H(nu) (25)

for all u ∈ U0 and all n ∈W .
Next, we need to show that each quintic mapping H : U → V fulfills the inequality

∥ϕ(u)−H(u)∥ ≤ Lτ(u) (26)

for all u ∈ U0, with some 0 < L, is equal to Hn for every n ∈ W . As a result, we fix
n0 ∈W and H : U → V fulfills (26). From (16), for every u ∈ U0, we get

∥H(u)−Hn0(u)∥ ≤ ∥H(u)− ϕ(u)∥+ ∥ϕ(u)−Hn0(u)∥
≤ Lτ(u) + ϑ∗n0

(u)

≤ L0τ(u)

∞∑
m=0

αm
n0
, (27)



S. Karthikeyan et al. / Eur. J. Pure Appl. Math, 18 (1) (2025), 5757 13 of 18

where L0 := (1 − αn0)L + 1
120η(n0) > 0 and we exclude the case that τ(u) ≡ 0 which is

trivial. From the observation, the functions H and Hn0 are the solutions to the equations
(25) for every n ∈W . Next, we prove that, for every j ∈ N0, we obtain

∥H(u)−Hn0(u)∥ ≤ L0τ(u)
∞∑

m=j

α∞
n0
, (28)

for all u ∈ U0. The inequality (27) is valid for the case j = 0.
Next, we set k ∈ N and assume (28) is true for j = k. In view of (27), for every u ∈ U0,

we have

∥H(u)−Hn0(u)∥ ≤ 1

120
L0τ((n0 + 3)u)

∞∑
m=k

αm
n0

+
1

24
L0τ((n0 + 2)u)

∞∑
m=k

αm
n0

+
1

120
L0τ((n0 − 2)u)

∞∑
m=k

αm
n0

+
1

12
L0τ((n0 + 1)u)

∞∑
m=k

αm
n0

+
1

24
L0τ((n0 − 1)u)

∞∑
m=k

αm
n0

+
1

12
L0τ((n0)u)

∞∑
m=k

αm
n0

≤ L0αn0τ(u)(u)
∞∑

m=k

αm
n0
.

Thus

∥H(u)−Hn0(u)∥ ≤ L0τ(u)

∞∑
m=k+1

αm
n0
.

So the condition (28) is valid for j = k + 1. Hence we may infer that for any j ∈ N0, the
inequality (28) holds. Taking the limit j → ∞ in (28), we obtain

H = Hn0 . (29)

Also, in view of (23), we have

∥ϕ(u)−Hn0(u)∥ ≤ (η(n) + 1)τ(u)

120(1− αn)

for all u ∈ U0 and all n ∈ W . This implies the condition (18) with H = Hn0 and (29)
confirms the uniqueness of H.

3. Hyperstability

The φ-hyperstability of (2) in Banach spaces is the subject of the following theorem.
In particular, we take into account functions ϕ : U → V fulfilling (2), i.e.,

∥Dϕ(u, v)∥ ≤ φ(u, v)



S. Karthikeyan et al. / Eur. J. Pure Appl. Math, 18 (1) (2025), 5757 14 of 18

for all u, v ∈ U0 such that u+ 3v ̸= 0, u+ 2v ̸= 0, u− 2v ̸= 0, u− v ̸= 0, u+ v ̸= 0 with a
given mapping φ : U0 × U0 → R+.

Next, we find a unique quintic mapping H : U → V which is near to ϕ. After that,
assuming some further φ assumptions, we demonstrate that the conditional functional
equation (2) belongs to the family of functions ϕ : U → V and is φ-hyperstable.

Theorem 5. Let τ1, τ2, αm and W be as in Theorem 3. Assume that
limm→+∞ η1(m) = 0,

limm→+∞ η1(m)η2(m) = 0,

limm→+∞ η1(m− 2)η2(m− 2) = 0.

Then every mapping ϕ : U → V fulfilling (3) is a solution of (2) on U0.

Proof. Suppose that ϕ : U → V fulfills (3). By Theorem 3, there is a mapping
H : U → V fulfilling (2) and

∥ϕ(u)−H(u)∥ ≤ η0τ1(u)τ2(u)

for all u ∈ U0, where

η0 := inf
m∈W

{
η1(m)

120(1− αm)

}
.

So, in view of (5), η0 = 0. This means that ϕ(u) = H(u) for all u ∈ U0. So

Dϕ(u, v) = 0

for all u, v ∈ U0 such that u+ 3v ̸= 0, u+ 2v ̸= 0, u− 2v ̸= 0, u− v ̸= 0, u+ v ̸= 0 which
gives that ϕ fulfills (2) on U0.

Corollary 1. Let ϵ ≥ 0, c1, c2 ∈ R with c1 + c2 < 0. Assume that a mapping ϕ : U → V
fulfills ϕ(0) = 0 and

∥Dϕ(u, v)∥ ≤ ϵ∥u∥c1∥v∥c2 (30)

for all u, v ∈ U0 such that u+3v ̸= 0, u+2v ̸= 0, u− 2v ̸= 0, u− v ̸= 0, u+ v ̸= 0. Then
ϕ is quintic on U0.

Proof. Theorem 3 is supported by the proof, which defines

τ1, τ2 : U0 × U0 → R+ by τ1(u) = ϵ1∥u∥c1 , τ2(v) = ϵ2∥v∥c2

and τ1(0) = τ2(0) = 0 with ϵ1, ϵ2 ∈ R+ and c1, c2 ∈ R such that ϵ1ϵ2 = ϵ and c1 + c2 < 0.
For each m ∈ N, we obtain

η1(m) = inf{l ∈ R+ : τ1(mx) ≤ τ1(u), u ∈ U0}
= inf{l ∈ R+ : ϵ1∥mu∥c1 ≤ lϵ1∥u∥c1 , u ∈ U0}



S. Karthikeyan et al. / Eur. J. Pure Appl. Math, 18 (1) (2025), 5757 15 of 18

= mc1 .

Similarly, we obtain η2(m) = mc2 for all m ∈ N.
Now, we can find m0 ∈ N such that

αm =
1

120
(m+ 3)c1+c2 +

1

24
(m+ 2)c1+c2 +

1

120
(m− 2)c1+c2

+
1

12
(m+ 1)c1+c2 +

1

24
(m− 1)c1+c2 +

1

12
(m)c1+c2 < 1

for all m ≥ m0. According to Theorem 3, there is only one quintic mapping H : U → V
satisfying

∥ϕ(u)−H(u)∥ ≤ ϵη0τ1(u)τ2(u)

for all m ∈ U0. Since c1 + c2 < 0, one of c1 and c2 must be negative. Assume that c2 < 0.
Then 

limm→+∞ η1(m) = limm→+∞mc1 = 0,

limm→+∞ η1(m)η2(m) = limm→+∞mc1+c2 = 0,

limm→+∞ η1(m− 2)η2(m− 2) = limm→+∞(m− 2)c1+c2 = 0.

The expected outcomes are thus obtained by Theorem 5.

Example 1. Let U be a normed space and V be a Banach space. Let ϕ : U → V be a
mapping such that Dϕ(u0, v0) ̸= 0 for some u0, v0 ∈ U and

∥Dϕ(u, v)∥ ≤ c∥u∥c1∥v∥c2

for all u, v ∈ U0 such that u + 3v ̸= 0, u + 2v ̸= 0, u − 2v ̸= 0, u − v ̸= 0, u + v ̸= 0,
where ϵ > 0 and c1, c2 ∈ R. Assume that the numbers c1, c2 satisfy c1 + c2 < 0. Then the
functional equation

ϕ(u+ 3v)− 5ϕ(u+ 2v)− ϕ(u− 2v) + 10ϕ(u+ v) + 5ϕ(u− v)− 10ϕ(u)− 120ϕ(v)

= 0, ∀u, v ∈ U0, (31)

has no solution in the class of functions ϕ : U → V.

Proof. Suppose that ϕ : U → V is a solution of (31). Then (30) holds, and consequently,
according to Corollary 1, ϕ is a quintic mapping on U0, which means that Dϕ(u0, v0) = 0.
This is a contradiction.

Corollary 2. Let ϵ ≥ 0, c ∈ R with c < 0. If a mapping ϕ : U → V fulfills ϕ(0) = 0 and

∥Dϕ(u, v)∥ ≤ ϵ (∥u∥c + ∥v∥c) (32)

for all u, v ∈ U0 such that u+3v ̸= 0, u+2v ̸= 0, u− 2v ̸= 0, u− v ̸= 0, u+ v ̸= 0. Then
ϕ is quintic on U0.



S. Karthikeyan et al. / Eur. J. Pure Appl. Math, 18 (1) (2025), 5757 16 of 18

Example 2. Let U0 be a normed space and V be a Banach space. Let ϕ : U0 → V be a
mapping such that Dϕ(u0, v0) ̸= 0 for some u0, v0 ∈ U0 and

∥Dϕ(u, v)∥ ≤ c∥u∥c∥v∥c

for all u, v ∈ U0 such that u+3v ̸= 0, u+2v ̸= 0, u− 2v ̸= 0, u− v ̸= 0, u+ v ̸= 0, where
ϵ > 0 and c ∈ R. Assume that the number c satisfies c < 0. Then the functional equation

ϕ(u+ 3v)− 5ϕ(u+ 2v)− ϕ(u− 2v) + 10ϕ(u+ v) + 5ϕ(u− v)− 10ϕ(u)− 120ϕ(v)

= 0, ∀u, v ∈ U0, (33)

has no solution in the class of functions ϕ : U → V.

Proof. Suppose that ϕ : U → V is a solution of (33). Then (32) holds, and consequently,
according to Corollary 2, ϕ is a quintic mapping on U0, which means that Dϕ(u0, v0) = 0.
This is a contradiction.

The findings of hyperstability for inhomogeneous quintic functional equations are
demonstrated by the following corollary.

Corollary 3. Let ϵ, c1, c2 ∈ R with ϵ ≥ 0 and c1 + c2 < 0. Assume that mappings
G : U2 → V and ϕ : U → V fulfill ϕ(0) = 0 and

∥Dϕ(u, v)−G(u, v)∥ ≤ ϵ∥u∥c1∥v∥c2 (34)

for all u, v ∈ U0 such that u + 3v ̸= 0, u + 2v ̸= 0, u − 2v ̸= 0, u − v ̸= 0, u + v ̸= 0. If
the functional equation

Dϕ(u, v) = G(u, v)

for all u, v ∈ U0, has a solution ϕ0 : U → V on U0, then ϕ satisfies (34) on U0.

Proof. From (34), we obtain the mapping f : U → V defined by f := ϕ − ϕ0 fulfills
(32).

The equation (2) on U0 is therefore implied by Corollary 1, which states that f is a
solution. Thus

Dϕ(u, v)−G(u, v) = D(f + ϕ0)(u, v)−G(u, v)

= 0

for all u, v ∈ U0 such that u+ 3v ̸= 0, u+ 2v ̸= 0, u− 2v ̸= 0, u− v ̸= 0, u+ v ̸= 0, which
means that ϕ is a solution to (3) on U0.

4. Conclusion

We proved the hyperstability of the quintic functional equation ϕ(u + 3v) − 5ϕ(u +
2v) − ϕ(u − 2v) + 10ϕ(u + v) + 5ϕ(u − v) − 10ϕ(u) − 120ϕ(v) = 0, in Banach spaces by
means of Brzdȩk’s fixed point theorem.



S. Karthikeyan et al. / Eur. J. Pure Appl. Math, 18 (1) (2025), 5757 17 of 18

Acknowledgements

The authors are thankful to the editors and the anonymous reviewers for many valuable
suggestions to improve this paper.

Fundings

S. Donganont was supported by the University of Phayao and Thailand Science Research
and Innovation Fund (Fundamental Fund 2025, Grant No. 5020/2567).

Declarations

Availablity of data and materials
Not applicable.
Human and animal rights
We would like to mention that this article does not contain any studies with animals and
does not involve any studies over human being.
Conflict of interest
The authors declare that they have no competing interests.

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