# Hyperstability of the Drygas Functional Equation

Authors: Magdalena Piszczek; Joanna Szczawińska
Journal: Journal of Function Spaces and Applications
 (2013)

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## Abstract

We use the fixed point theorem for functional spaces to obtain the hyperstability result for the Drygas functional equation on a restricted domain. Namely, we show that a function satisfying the Drygas equation approximately must be exactly the solution of it.


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## Body

## 1. Introduction

We say that a functionf:ℝ→ℝsatisfies the Drygas equation if
(1)f(x+y)+f(x-y)=2f(x)+f(y)+f(-y),mmmmmmmmmmmmmmmmmmmllix,y∈ℝ.
The above equation was introduced in [1] in order to obtain a characterization of the quasi-inner-product spaces. The general solution of (1), obtained by Ebanks et al. in [2] (see also [3]), is of the following form:
(2)f(x)=a(x)+q(x),x∈ℝ,
where a:ℝ→ℝis an additive function and q:ℝ→ℝis a quadratic function; that is, q satisfies the quadratic functional equation
(3)q(x+y)+q(x-y)=2q(x)+2q(y),x,y∈ℝ.
A set-valued version of (1) was considered by Smajdor in [4].The stability in the Hyers-Ulam sense of the Drygas equation has been investigated, for example, in [5–8]. For more information and numerous references on the stability of functional equations in general, we refer to [9, 10].The term hyperstability was used for the first time probably in [11]; however, it seems that the first hyperstability result was published in [12] and concerned the ring homomorphisms. The hyperstability results for the Cauchy equation were investigated by Brzdȩk in [13, 14]. Gselmann in [15] studied the hyperstability of the parametric fundamental equation of information.In the paper, we prove the hyperstability of the Drygas equation on a restricted domain. LetX be a nonempty subset of an Abelian group and Y a semigroup. We say that a function f:X→Y satisfies the Drygas functional equation on X if
(4)f(x+y)+f(x-y)=2f(x)+f(y)+f(-y)
for all x,y∈X such that -y,x+y,x-y∈X. We will show that (4) is hyperstable; that is, each function f:X→Y (under some additional assumptions on X) satisfying the inequality
(5)∥f(x+y)+f(x-y)-2f(x)-f(y)-f(-y)∥≤c(∥x∥p+∥y∥p)
for all x,y∈X such that -y,x+y,x-y∈X with p<0 must satisfy the Drygas equation (4).The method of the proof of the main theorem is motivated by an idea used by Brzdȩk in [13] and further by Piszczek in [16]. It is based on a fixed point theorem for functional spaces obtained by Brzdęk et al. (see [17, Theorem  1]). Some generalizations of their result were proved by Cădariu et al. in [18]. The case of fixed point theorem for non-Archimedean metric spaces was also studied by Brzdęk and Ciepliński in [19]. It is worth mentioning that using fixed point theorems is now one of the most popular methods of investigating the stability of functional equations in single as well as in several variables. Let us recall a few recent approaches of Jung in [20], Lee and Jung in [21], or Cădariu and Radu in [22, 23]. More information on the application of the fixed point method was collected by Ciepliński in [24].First, we take the following three hypotheses (all notations come from [17]).(H1)
X is a nonempty set, Y is a Banach space, and f1,…,fk:X→X and L1,…,Lk:X→ℝ+ are given.(H2)
𝒯:YX→YX is an operator satisfying the inequality
(6)∥𝒯ξ(x)-𝒯μ(x)∥≤∑i=1k‍Li(x)∥ξ(fi(x))-μ(fi(x))∥
for all ξ,μ∈YX, x∈X.(H3)
Λ:ℝ+X→ℝ+X is defined by
(7)Λδ(x):=∑i=1k‍Li(x)δ(fi(x)),δ∈ℝ+X,x∈X.The mentioned fixed point theorem is stated as follows.Theorem 1.
Let hypotheses (H1)–(H3) be valid and functionsε:X→ℝ+ and let φ:X→Y fulfil the following two conditions:
(8)∥𝒯φ(x)-φ(x)∥≤ε(x),x∈X,ε*(x):=∑n=0∞‍Λnε(x)<∞,x∈X.
Then, there exists a unique fixed point ψ of 𝒯 with
(9)∥φ(x)-ψ(x)∥≤ε*(x),x∈X.
Moreover,
(10)ψ(x):=limn→∞𝒯nφ(x),x∈X.Throughout the paper,ℕ, ℕ0, and ℕm0 denote the set of all positive integers, the set of all nonnegative integers, and the set of all integers greater than or equal to m0, respectively.


## 2. The Main Result

Theorem 2.
Assume thatX is a nonempty, symmetric with respect to 0 subset of a normed space such that 0∉X and there exists n0∈ℕwith nx∈X for x∈X, n∈ℕn0. Let Y be a Banach space, c≥0, and p<0. If f:X→Y satisfies
(11)∥f(x+y)+f(x-y)-2f(x)-f(y)-f(-y)∥≤c(∥x∥p+∥y∥p)
for all x,y∈X such that x+y,x-y∈X, then f satisfies the Drygas equation on X.Proof.
First observe that there existsm0∈ℕn0 such that
(12)2(m+1)p+2mp+(2m+1)p<1form≥m0.
Let us fix m∈ℕm0. Replacing x with (m+1)x and y with mx in (11), we get
(13)∥2f((m+1)x)+f(mx)+f(-mx)-f((2m+1)x)-f(x)∥≤c((m+1)p+mp)∥x∥p,x∈X.
Further put
(14)𝒯ξ(x)≔2ξ((m+1)x)+ξ(mx)+ξ(-mx)-ξ((2m+1)x),x∈X,ξ∈YX,ε(x):=c((m+1)p+mp)∥x∥p,x∈X.
Inequality (13) now takes the form
(15)∥𝒯f(x)-f(x)∥≤ε(x),x∈X.
The operator
(16)Λη(x)≔2η((m+1)x)+η(mx)+η(-mx)+η((2m+1)x),η∈ℝ+X,x∈X
has the form described in (H3) with k=4 and f1(x)=(m+1)x, f2(x)=mx, f3(x)=-mx, f4(x)=(2m+1)x, L1(x)=2, and L2(x)=L3(x)=L4(x)=1 for x∈X. Moreover, for every ξ,μ∈YX,  x∈X,
(17)∥𝒯ξ(x)-𝒯μ(x)∥=∥2ξ((m+1)x)+ξ(mx)+ξ(-mx)-ξ((2m+1)x)-2μ((m+1)x)-μ(mx)-μ(-mx)+μ((2m+1)x)∥≤2∥(ξ-μ)((m+1)x)∥+∥(ξ-μ)(mx)∥+∥(ξ-μ)(-mx)∥+∥(ξ-μ)((2m+1)x)∥=∑i=14‍Li(x)∥(ξ-μ)(fi(x))∥.
As 2(m+1)p+2mp+(2m+1)p<1, we have
(18)ε*(x):=∑n=0∞‍Λnε(x)=∑n=0∞‍c((m+1)p+mp)(2(m+1)p+2mpmmmmmmmmmmmmm+(2m+1)p)n∥x∥p=c((m+1)p+mp)∥x∥p1-2(m+1)p-2mp-(2m+1)p.
Thus, according to Theorem 1 there exists a unique solution F:X→Y of the equation
(19)F(x)=2F((m+1)x)+F(mx)+F(-mx)-F((2m+1)x)
such that
(20)∥f(x)-F(x)∥≤c((m+1)p+mp)∥x∥p1-2(m+1)p-2mp-(2m+1)p,mmmmmmmmmmmmmmmmmmmmmmllx∈X.
Moreover,
(21)F(x)=limn→∞𝒯nf(x).
To prove thatF satisfies the Drygas equation on X, observe that
(22)∥𝒯nf(x+y)+𝒯nf(x-y)-2𝒯nf(x)-𝒯nf(y)-𝒯nf(-y)∥≤c(2(m+1)p+2mp+(2m+1)p)n(∥x∥p+∥y∥p)
for every x,y∈X,  x+y,x-y∈X, and n∈ℕ0.
Indeed, ifn=0, then (22) is simply (11). So, fix n∈ℕ0 and suppose that (22) holds for n and x,y∈X such that x+y,x-y∈X. Then
(23)∥𝒯n+1f(x+y)+𝒯n+1f(x-y)-2𝒯n+1f(x)-𝒯n+1f(y)-𝒯n+1f(-y)∥=∥2𝒯nf((m+1)(x+y))+𝒯nf(m(x+y))+𝒯nf(-m(x+y))-𝒯nf((2m+1)(x+y))+2𝒯nf((m+1)(x-y))+𝒯nf(m(x-y))+𝒯nf(-m(x-y))-𝒯nf((2m+1)(x-y))-2(2𝒯nf((m+1)x)+𝒯nf(mx)+𝒯nf(-mx)-𝒯nf((2m+1)x))-2𝒯nf((m+1)y)-𝒯nf(my)-𝒯nf(-my)+𝒯nf((2m+1)y)-2𝒯nf((m+1)(-y))-𝒯nf(m(-y))-𝒯nf(-m(-y))+𝒯nf((2m+1)(-y))∥≤c(2(m+1)p+2mp+(2m+1)p)n×[2(∥(m+1)x∥p+∥(m+1)y∥p)+∥mx∥p+∥my∥p+∥-mx∥p+∥-my∥p+∥(2m+1)x∥p+∥(2m+1)y∥p]=c(2(m+1)p+2mp+(2m+1)p)n+1×(∥x∥p+∥y∥p).
By induction, we have shown that (22) holds for all x,y∈X,  x+y,x-y∈X, and n∈ℕ0. Letting n→∞ in (22), we obtain
(24)F(x+y)+F(x-y)=2F(x)+F(y)+F(-y),mmmmmmmmiimx,y∈X,x+y,x-y∈X.
Thus, we have proved that for every m∈ℕm0 there exists a function Fm:X→Y such that Fm is a solution of the Drygas equation on X and
(25)∥f(x)-Fm(x)∥≤c((m+1)p+mp)∥x∥p1-2(m+1)p-2mp-(2m+1)p,mmmmmmmmmmmmmmmmmmmmmmmmlx∈X.
Since p<0, the sequence ((c((m+1)p+mp)∥x∥p)/(1-2(m+1)p-2mp-(2m+1)p))m∈ℕm0 tends to zero. Consequently, f satisfies the Drygas equation on X as the pointwise limit of (Fm)m∈ℕm0.Remark 3.
IfX is a normed space and f:X→Y satisfies (11) for x,y∈X∖{0}, with p<0, then by Theorem 2 we know that f satisfies the Drygas equation on X∖{0}. It is easy to see that if f(0)=0, then f satisfies the Drygas equation on the whole X.We end with two examples. First notice that the assumption of unboundedness of the setX is necessary.Example 4.
LetX=[-1,1]∖{0} and let f:X→ℝbe defined by f(x)=|x|, x∈X. Then for all x,y∈X such that x+y,x-y∈X,
(26)|f(x+y)+f(x-y)-2f(x)-f(y)-f(-y)|≤|x|p+|y|p,
with p<0, but f does not satisfy the Drygas equation on X.In the casep≥0, the considered Drygas equation is not hyperstable.Example 5.
LetX=ℝ∖[-1,1] and let f:X→ℝbe a constant function f(x)=c, x∈X for some c>0. Then f satisfies the inequality
(27)|f(x+y)+f(x-y)-2f(x)-f(y)-f(-y)|≤c(|x|p+|y|p),
for all x,y∈X such that x+y,x-y∈X, with p≥0, but is not a solution of the Drygas equation on X.



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*Source: 912718-2013-07-29.xml*