# Stability of a Generalized Trigonometric-Quadratic Functional Equation

Authors: Janyarak Tongsomporn; Vichian Laohakosol
Journal: Journal of Function Spaces and Applications
 (2013)

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## Abstract

A stability result for the generalized trigonometric-quadratic functional equationℱ(x+y)-𝒢(x-y)=2ℋ(x)𝒦(y)+ℒ(x)+ℳ(y) over the domain of a normed space and the range of a Banach space is derived. It states that under appropriate conditions on the function ℋ(x)𝒦(y), the four functions ℱ(x), 𝒢(x), ℒ(x), and ℳ(x) are approximately equal to combinations of certain quadratic and additive functions.


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## Body

## 1. Introduction

A generalized trigonometric functional equation is a functional equation of the form(1)F(x+y)-G(x-y)=2H(x)K(y),
which is so named because the two best known trigonometric functions, sine and cosine, are solutions of special cases of this equation. In [1], the authors investigated the stability of (1), where F, G, H, and K are nonzero functions from an abelian group (G,+) to the complex fieldℂ. Later in the same year, a generalized trigonometric functional equation with either the function H or K being bounded was investigated to complement the above earlier work where such function boundedness was not treated. A generalized quadratic functional equation is a functional equation of the form
(2)f1(x+y)+f2(x-y)=f3(x)+f4(y),
which is so named because the quadratic function x2 is a solution of one of its particular cases, namely,
(3)f(x+y)+f(x-y)=2f(x)+2f(y),
and any solution of (3) is referred to as a quadratic function. The stability of the quadratic functional equation was first proved by Skof [2] in 1983 for functions from a normed space to a Banach space. In 1984, Cholewa [3] demonstrated that Skof’s theorem is also valid if a normed space is replaced by an abelian group. Later, Czerwik [4] extended Cholewa’s result by changing a control function and considering functions from a normed space to a Banach space. In another paper, [5], Czerwik investigated the stability of the “partially pexiderized” quadratic functional equation of the form f(x+y)+f(x-y)=g(x)+g(y). In 2000, Jung [6] considered (2), wheref1, f2, f3, and f4 are functions from a normed space E1 to a Banach space E2. In the paper [7], the authors considered the stability of a combined generalized trigonometric-quadratic functional equation of the form
(4)ℱ(x+y)+𝒢(x-y)=2ℋ(x)𝒦(y)+ℒ(x)+ℳ(y)
over the domain of an abelian group (G,+) and the range ℂ, except the control function ϕ whose range is taken to be the nonnegative real numbers [0,∞). There, the functions ℋ and 𝒦 are explicitly determined under certain restrictions on the remaining functions ℱ, 𝒢, ℒ, and ℳ. It is natural to ask whether the restrictions on the functions ℱ, 𝒢, ℒ, and ℳ can be removed and/or altered to obtain some stability result of (4). We answer this question affirmatively using Jung’s techniques elaborated in [6, 8].


## 2. Notation

We collect in this section the notation and terminology that will be kept standard throughout. A normed spaceE1 is the domain and a Banach space (E2,∥·∥) is the range of all functions involved. Let ℋ,𝒦:E1→E2. Define
(5)HK(x,y)∶=ℋ(x)𝒦(y)-ℋ(0)𝒦(0),HKe(x,y)∶=HK(x,y)+HK(-x,-y)2,HKo(x,y)∶=HK(x,y)-HK(-x,-y)2.Let φ:E1×E1→[0,∞) be a given function satisfying the following three conditions (i) φ(y,x)=φ(x,y), (ii) φ(x,-y)=φ(x,y), and (iii) there is an integer k≥2 such that
(6)∑i=0∞1kiφ(kix,kiy)<∞,or∑i=0∞kiφ(xki,yki)<∞(x,y∈E1).
For x,y∈E1, define
(7)φe(x,y)=2φ(x,y)+4φ(x,0)+3φ(x+y,0)+3φ(x-y,0)+6φ(x+y2,x+y2)+2φ(x-y2,x-y2)+20φ(0,0),φo(x,y)=2φ(x,y)+2φ(x,x)+2φ(y,y)+2φ(2x,0)+2φ(2y,0)+2φ(x+y,0)+2φ(x-y,0)+2φ(0,0),ψk(x,y)=∑i=0∞1k2iφ(kix,kiy),ψ~k(x,y)=∑i=0∞k2(i+1)φ(xki+1,yki+1),Φk(x,y)=∑i=0∞1k2i∑j=1k-1bjφe(ki(k-j)x,kiy),Φ~k(x,y)=∑i=0∞k2(i+1)∑j=1k-1bjφe((k-j)xki+1,yki+1),Φk′(x,y)=∑i=0∞1ki∑j=1k-1φo(kijx+y2,kijx-y2),Φ~k′(x,y)=∑i=0∞ki+1∑j=1k-1φo(jx+y2ki+1,jx-y2ki+1),Φk′′(x,y)=∑i=0∞1ki∑j=1k-1φo(ki(jx+y2),kiy2),Φ~k′′(x,y)=∑i=0∞ki+1∑j=1k-1φo(2jx+y2ki+1,y2ki+1),ψHK,k(x,y)=∑i=0∞‍1k2i∥2HKe(kix,kiy)+HKe(ki(x+y)2,-ki(x+y)2)-HKe(ki(x+y)2,ki(x+y)2)+HKe(ki(x-y)2,ki(x-y)2)-HKe(ki(x-y)2,-ki(x-y)2)+2HKe(0,kix)-2HKe(kix,0)-HKe(0,ki(x+y))-HKe(0,ki(x-y))∥,ψ~HK,k(x,y)=∑i=0∞k2(i+1)∥2HKe(xki+1,yki+1)+HKe(x+y2ki+1,-x+y2ki+1)-HKe(x+y2ki+1,x+y2ki+1)+HKe(x-y2ki+1,x-y2ki+1)-HKe(x-y2ki+1,-x-y2ki+1)+2HKe(0,xki+1)-2HKe(xki+1,0)-HKe(0,x+yki+1)-HKe(0,x-yki+1)∥,ΦHK,k(x,y)=2k2∑i=0∞1k2i∑j=1k-1∥2HKe((k-j)kix,kiy)+HKe(((k-j)x+y2)ki,-((k-j)x+y2)ki)-HKe(((k-j)x+y2)ki,((k-j)x+y2)ki)+HKe(((k-j)x-y2)ki,((k-j)x-y2)ki)-HKe(((k-j)x-y2)ki,-((k-j)x-y2)ki)+2HKe(0,(k-j)kix)-2HKe((k-j)kix,0)-HKe(0,((k-j)x+y)ki)-HKe(0,((k-j)x-y)ki)∥,Φ~HK,k(x,y)=2k2∑i=0∞1k2i∑j=1k-1∥2HKe(jkix,kiy)+HKe((jx+y2)ki,-(jx+y2)ki)-HKe((jx+y2)ki,(jx+y2)ki)+HKe((jx-y2)ki,(jx-y2)ki)-HKe((jx-y2)ki,-(jx-y2)ki)+2HKe(0,jkix)-2HKe(jkix,0)-HKe(0,(jx+y)ki)-HKe(0,(jx-y)ki)∥,ΦHK,k′(x,y)=2k∑i=0∞1ki∑j=1k-1∥HKo((jx+y2)ki,(jx-y2)ki)-HKo(-(jx+y2)ki,(jx-y2)ki)-HKo((jx+y2)ki,(jx+y2)ki)+HKo(-(jx+y2)ki,(jx+y2)ki)+HKo((jx+y)ki,0)-HKo(jkix,0)-HKo(kix,0)HKo((jx+y2)ki,(jx-y2)ki)∥,Φ~HK,k′(x,y)=2k∑i=0∞ki+1∑j=1k-1∥HKo(jx+y2ki+1,jx-y2ki+1)-HKo(-jx+y2ki+1,jx-y2ki+1)-HKo(jx+y2ki+1,jx+y2ki+1)+HKo(-jx+y2ki+1,jx+y2ki+1)+HKo(jx+yki+1,0)-HKo(jxki+1,0)-HKo(xki+1,0)∥.Define a real sequence(bi) by
(8)b1=1,b2=2,bi=2bi-1+bi-2(i≥3),
and let
(9)Ψk(x,y)=b1ψk((k-1)x,y)+b2ψk((k-2)x,y)+⋯+bk-1ψk(x,y),Ψ~k(x,y)=b1ψ~k((k-1)x,y)+b2ψ~k((k-2)x,y)+⋯+bk-1ψ~k(x,y),ΨHK,k(x,x)=b1ψHK,k((k-1)x,y)+b2ψHK,k((k-2)x,y)+⋯+bk-1ψHK,k(x,y),Ψ~HK,k(x,x)=b1ψ~HK,k((k-1)x,y)+b2ψ~HK,k((k-2)x,y)+⋯+bk-1ψ~HK,k(x,y).
Clearly, the functions φe, φo satisfy the conditions (i), (ii), and (iii).


## 3. Auxiliary Results

We start with an important lemma.Lemma 1.
Iff:E1→E2, f(0)=0, satisfies
(10)∥f(x+y)+f(x-y)-2f(x)-2f(y)∥≤φe(x,y)+2∥2HKe(x,y)+HKe(x+y2,-x+y2)-HKe(x+y2,x+y2)+HKe(x-y2,x-y2)-HKe(x-y2,-x-y2)+2HKe(0,x)-2HKe(x,0)-HKe(0,x+y)-HKe(0,x-y)∥,
then one has, for integer k≥2, x∈E1 and n∈ℕ,
(11)∥f(knx)-k2nf(x)∥≤∑i=0n-1‍k2(n-1-i)∑j=1k-1‍bjφ((k-j)kix,kix)+2∑i=0n-1‍k2(n-1-i)×∑j=1k-1‍bj∥2HKe((k-j)kix,kix)+HKe((k-j+1)kix2,-(k-j+1)kix2)-HKe((k-j+1)kix2,(k-j+1)kix2)+HKe((k-j-1)kix2,(k-j-1)kix2)-HKe((k-j-1)kix2,-(k-j-1)kix2)+2HKe(0,(k-j)kix)-2HKe((k-j)kix,0)-HKe(0,(k-j+1)kix)-HKe(0,(k-j-1)kix)((k-j)kix,kix)∥.Proof.
We prove the lemma by induction onn. To verify the case n=1, we use induction on k≥2. Putting y=x in (10), we obtain
(12)∥f(2x)-22f(x)∥≤φe(x,x)+2∥HKe(x,x)+HKe(x,-x)+2HKe(0,x)-2HKe(x,0)-HKe(0,2x)∥;
that is, the assertion is true for k=2. Assume that the statement is true for all m∈{2,…,k}. Replacing x and y in (10) by kx and x, we get
(13)∥f((k+1)x)+f((k-1)x)-2f(kx)-2f(x)∥≤φe(kx,x)+2∥2HKe(kx,x)+HKe((k+1)x2,-(k+1)x2)-HKe((k+1)x2,(k+1)x2)+HKe((k-1)x2,(k-1)x2)-HKe((k-1)x2,-(k-1)x2)+2HKe(0,kx)-2HKe(kx,0)-HKe(0,(k+1)x)-HKe(0,(k-1)x)HKe((k+1)x2,-(k+1)x2)∥,
and so
(14)∥f((k+1)x)-(k+1)2f(x)∥≤∥f((k+1)x)+f((k-1)x)-2f(kx)-2f(x)∥+2∥f(kx)-k2f(x)∥+∥f((k-1)x)-(k-1)2f(x)∥≤∑j=1kbjφe((k+1-j)x,x)+2∑j=1kbi∥2HKe((k+1-j)x,x)+HKe((k+2-j)x2,-(k+2-j)x2)-HKe((k+2-j)x2,(k+2-j)x2)+HKe((k-j)x2,(k-j)x2)-HKe((k-j)x2,-(k-j)x2)+2HKe(0,(k+1-j)x)-2HKe((k+1-j)x,0)-HKe(0,(k+2-j)x)-HKe(0,(k-j)x)∥,
and we are done for n=1. Assuming that the lemma is true for some n≥1, we have
(15)∥f(kn+1x)-k2(n+1)f(x)∥≤∥f(kn+1x)-k2f(knx)∥+k2∥f(knx)-k2nf(x)∥≤∑i=0nk2(n-i)∑j=1k-1bjφ((k-j)kix,kix)+2∑i=0nk2(n-i)×∑j=1k-1bj∥2HKe((k-j)kix,kix)+HKe((k-j+1)kix2,-(k-j+1)kix2)-HKe((k-j+1)kix2,(k-j+1)kix2)+HKe((k-j-1)kix2,(k-j-1)kix2)-HKe((k-j-1)kix2,-(k-j-1)kix2)+2HKe(0,(k-j)kix)-2HKe((k-j)kix,0)-HKe(0,(k-j+1)kix)-HKe(0,(k-j-1)kix)2HKe((k-j)kix,kix)∥,
verifying the lemma in general.We come now to our first main proposition.Proposition 2.
Assume thatf:E1→E2 satisfies
(16)∥f(x+y)+f(x-y)-2f(x)-2f(y)∥≤φe(x,y)+2∥2HKe(x,y)+HKe(x+y2,-x+y2)-HKe(x+y2,x+y2)+HKe(x-y2,x-y2)-HKe(x-y2,-x-y2)+2HKe(0,x)-2HKe(x,0)-HKe(0,x+y)-HKe(0,x-y)2HKe(x,y)+HKe(x+y2,-x+y2)∥.
When ψ~k(x,y)<∞ for all x,y∈E1, one further assumes that f(0)=0. Then there exists a unique quadratic function Q:E1→E2 satisfying
(17)∥Q(x)-f(x)∥≤1k2Ψk(x,x)+2k2ΨHK,k(x,x)
provided that ψk(x,y)<∞ and ψHK,k(x,y)<∞ for all x,y∈E1, or
(18)∥Q(x)-f(x)∥≤1k2Ψ~k(x,x)+2k2Ψ~HK,k(x,x)
provided that ψ~k(x,y)<∞ and ψ~HK,k(x,x)<∞ for all x,y∈E1. Moreover, if f is measurable or f(tx) is continuous in t∈ℝ for every fixed x∈E1, then the function Q satisfies
(19)Q(tx)=t2Q(x)(x∈E1,t∈ℝ).Proof.
From Lemma1, we have
(20)∥1k2nf(knx)-f(x)∥≤1k2∑i=0n-11k2i∑j=1k-1bjφ((k-j)kix,kix)+2k2∑i=0n-11k2i×∑j=1k-1bj∥2HKe((k-j)kix,kix)+HKe((k-j+1)kix2,-(k-j+1)kix2)-HKe((k-j+1)kix2,(k-j+1)kix2)+HKe((k-j-1)kix2,(k-j-1)kix2)-HKe((k-j-1)kix2,-(k-j-1)kix2)+2HKe(0,(k-j)kix)-2HKe((k-j)kix,0)-HKe(0,(k-j+1)kix)-HKe(0,(k-j-1)kix)((k-j)kix,kix)∥.
For m,n∈ℕ with n>m, from (20), we obtain
(21)∥1k2nf(knx)-1k2mf(kmx)∥=1k2m∥1k2(n-m)f(kn-mkmx)-f(kmx)∥≤1k2∑j=1k-1bj∑i=mn-11k2iφ((k-j)kix,kix)+2k2∑j=1k-1bj×∑i=mn-11k2i∥2HKe((k-j)kix,kix)+HKe((k-j+1)kix2,-(k-j+1)kix2)-HKe((k-j+1)kix2,(k-j+1)kix2)+HKe((k-j-1)kix2,(k-j-1)kix2)-HKe((k-j-1)kix2,-(k-j-1)kix2)+2HKe(0,(k-j)kix)-2HKe((k-j)kix,0)-HKe(0,(k-j+1)kix)-HKe(0,(k-j-1)kix)((k-j)kix,kix)∥.
In (21), since the last term can be made as small as we wish by choosing m sufficiently large, we deduce that {(1/k2n)f(knx)} is a Cauchy sequence in E2. Since E2 is a Banach space, the limit
(22)Q(x)=limn→∞1k2nf(knx)(x∈E1)
is well defined. Using Q and (20), the validity of the inequality (17) is easily verified. Next, replacing x, y in (16) by knx, kny, respectively, dividing the resulting inequality by k2n, and using ψk(x,y)<∞, ψHK,k(x,y)<∞, we get
(23)∥1k2nf(kn(x+y))+1k2nf(kn(x-y))-2k2nf(knx)-2k2nf(kny)∥≤1k2nφ(knx,kny)+2k2n×∥2HKe(knx,kny)+HKe(kn(x+y)2,-kn(x+y)2)-HKe(kn(x+y)2,kn(x+y)2)+HKe(kn(x-y)2,kn(x-y)2)-HKe(kn(x-y)2,-kn(x-y)2)+2HKe(0,knx)-2HKe(knx,0)-HKe(0,kn(x+y))-HKe(0,kn(x-y))∥⟶0(n⟶∞),
which implies that Q is a quadratic function. If Q1:E1→E2 is another quadratic function satisfying the inequality (17), since Q and Q1 are quadratic functions, we have
(24)Q(knx)=k2nQ(x),Q1(knx)=k2nQ1(x)(n∈ℕ).
It thus follows from (24), (17), ψk(x,y)<∞, and ψHK,k(x,y)<∞ that
(25)∥Q(x)-Q1(x)∥=1k2n∥Q(knx)-Q1(knx)∥≤1k2n2k2Ψk(knx,knx)+1k2n2k2Ψ~k(knx,knx)⟶0(n⟶∞);
that is, Q(x)=Q1(x).
The other provisionψ~k(x,y)<∞ and ψ~HK,k(x,x)<∞ is treated analogously by replacing x by x/kn in (11) to obtain an upper bound for ∥f(x)-k2nf(x/kn)∥, which is then used to show that {k2nf(x/kn)} is a Cauchy sequence in E2 so that the limit
(26)Q(x)=limn→∞k2nf(xkn)(x∈E1)
is well defined and satisfies (18). Finally, the proof that (19) holds when f is measurable or f(tx) is continuous is the same as that in [4].We now proceed to our second main proposition.Proposition 3.
Iff:E1→E2 satisfies
(27)∥f(x+y)-f(x)-f(y)∥≤φo(x,y)+2∥HKo(x+y2,x-y2)-HKo(-x+y2,x-y2)-HKo(x+y2,x+y2)+HKo(-x+y2,x+y2)+HKo(x+y,0)-HKo(x,0)-HKo(y,0)HKo(x+y2,x-y2)∥,
then there exists a unique additive function A:E1→E2 such that either
(28)∥A(x)-f(x)∥≤1kΦk′(x,x)+ΦHK,k′(x,x)
provided that for x,y∈E1, one has ∑i=0∞‍(1/ki)φ(kix,kiy)<∞ and
(29)∑i=0∞∥HKo(ki(x+y),ki(x-y))-HKo(-ki(x+y),ki(x-y))-HKo(ki(x+y),ki(x+y))+HKo(-ki(x+y),ki(x+y))+HKo(ki(x+y),0)-HKo(2kix,0)-HKo(2kiy,0)HKo(ki(x+y),ki(x-y))∥<∞
or
(30)∥A(x)-f(x)∥≤1kΦ~k′(x,x)+Φ~HK,k′(x,x)
provided that for x,y∈E1, one has ∑i=0∞‍kiφ(x/ki,y/ki)<∞ and
(31)∑i=0∞‍ki+1∥HKo(x+yki+1,x-yki+1)-HKo(-x+yki+1,x-yki+1)-HKo(x+yki+1,x+yki+1)+HKo(-x+yki+1,x+yki+1)+HKo(2(x+y)ki+1,0)-HKo(2xki+1,0)-HKo(2yki+1,0)∥<∞.
Moreover, if f(tx) is continuous in t∈ℝ for every fixed x∈E1, then A is linear.Proof.
First, we show that(32)∥f(kx)-kf(x)∥≤∑j=1k-1φo((j+12)x,(j-12)x)+2∑j=1k-1∥HKo((j+1)x2,(j-1)x2)-HKo(-(j+1)x2,(j-1)x2)-HKo((j+1)x2,(j+1)x2)+HKo(-(j+1)x2,(j+1)x2)+HKo((j+1)x)-HKo(jx,0)-HKo(x,0)HKo((j+1)x2,(j-1)x2)∥.
Putting y=x in (27), we obtain
(33)∥f(2x)-2f(x)∥≤φo(x,0)+2∥-HKo(x,x)+HKo(-x,x)+HKo(2x,0)∥.
This shows that (32) holds for k=2. Assume that (32) is valid for k=m; that is,
(34)∥f(mx)-mf(x)∥≤∑j=1m-1φo((j+12)x,(j-12)x)+2∑j=1m-1∥HKo((j+1)x2,(j-1)x2)-HKo(-(j+1)x2,(j-1)x2)-HKo((j+1)x2,(j+1)x2)+HKo(-(j+1)x2,(j+1)x2)+HKo((j+1)x)-HKo(jx,0)-HKo(x,0)HKo((j+1)x2,(j-1)x2)∥.
Replacing y by mx in (27) and using (34), we have
(35)∥f((m+1)x)-(m+1)f(x)∥≤∥f((m+1)x)-f(mx)-f(x)∥+∥f(mx)-mf(x)∥≤∑j=1m‍φo((j+12)x,(j-12)x)+2∑j=1m‍∥HKo((j+1)x2,(j-1)x2)-HKo(-(j+1)x2,(j-1)x2)-HKo((j+1)x2,(j+1)x2)+HKo(-(j+1)x2,(j+1)x2)+HKo((j+1)x)-HKo(jx,0)-HKo(x,0)HKo((j+1)x2,(j-1)x2)∥,
verifying the case m+1, and so the assertion (32) holds for all n>1 and all x∈E1.
Next, we use induction onn to show that
(36)∥f(knx)-knf(x)∥≤∑i=1nkn-i∑j=1k-1φo((j+12)ki-1x,(j-12)ki-1x)+2∑i=1nkn-i×∑j=1k-1∥HKo((j+12)ki-1x,(j-12)ki-1x)-HKo(-(j+12)ki-1x,(j-12)ki-1x)-HKo((j+12)ki-1x,(j+12)ki-1x)+HKo(-(j+12)ki-1x,(j+12)ki-1x)+HKo((j+1)ki-1x,0)-HKo(jki-1x,0)-HKo(ki-1x,0)HKo((j+12)ki-1x,(j-12)ki-1x)∥.
The case n=1 follows from (32). Assuming that this holds for n=m, putting n=1, and then substituting kmx for x in (36), we obtain
(37)∥f(km+1x)-km+1f(x)∥≤∥f(k·kmx)-kf(kmx)∥+k∥f(kmx)-kmf(x)∥≤∑i=0mkm-i∑j=1k-1φo((j+12)kix,(j-12)kix)+2∑i=0mkm-i×∑j=1k-1∥HKo((j+12)kix,(j-12)kix)-HKo(-(j+12)kix,(j-12)kix)-HKo((j+12)kix,(j+12)kix)+HKo(-(j+12)kix,(j+12)kix)+HKo((j+1)kix,0)-HKo(jkix,0)-HKo(kix,0)HKo((j+12)kix,(j-12)kix)∥,
verifying (36) generally. Rewriting (36), we get
(38)∥1knf(knx)-f(x)∥≤∑i=1n1ki∑j=1k-1φo((j+12)ki-1x,(j-12)ki-1x)+2∑i=1n1ki×∑j=1k-1∥HKo((j+12)ki-1x,(j-12)ki-1x)-HKo(-(j+12)ki-1x,(j-12)ki-1x)-HKo((j+12)ki-1x,(j+12)ki-1x)+HKo(-(j+12)ki-1x,(j+12)ki-1x)+HKo((j+1)ki-1x,0)-HKo(jki-1x,0)-HKo(ki-1x,0)HKo((j+12)ki-1x,(j-12)ki-1x)∥.
Using (38), ∑i=0∞‍kiφ(x/ki,y/ki)<∞, and the inequality
(39)∑i=0∞∥HKo(ki(x+y),ki(x-y))-HKo(-ki(x+y),ki(x-y))-HKo(ki(x+y),ki(x+y))+HKo(-ki(x+y),ki(x+y))+HKo(ki(x+y),0)-HKo(2kix,0)-HKo(2kiy,0)HKo(ki(x+y),ki(x-y))∥<∞
and proceeding as in the proof of Proposition 2, we deduce that {k-nf(knx)} is a Cauchy sequence. Thus, the limit
(40)A(x)=limn→∞1knf(knx)(x∈E1)
is well defined. To show that A is additive, we substitute knx and kny for x and y in (27), respectively, to get
(41)∥f(knx+kny)-f(knx)-f(kny)∥≤φo(knx,kny)+2∥HKo(kn(x+y)2,kn(x-y)2)-HKo(-kn(x+y)2,kn(x-y)2)-HKo(kn(x+y)2,kn(x+y)2)+HKo(-kn(x+y)2,kn(x+y)2)+HKo(knx+kny,0)-HKo(knx,0)-HKo(kny,0)HKo(kn(x+y)2,kn(x-y)2)∥.
From ∑i=0∞‍(1/ki)φ(kix,kiy)<∞ and
(42)∑i=0∞∥HKo(ki(x+y),ki(x-y))-HKo(-ki(x+y),ki(x-y))-HKo(ki(x+y),ki(x+y))+HKo(-ki(x+y),ki(x+y))+HKo(ki(x+y),0)-HKo(2kix,0)-HKo(2kiy,0)∥<∞,
it follows that
(43)1kn{HKo(kn(x+y)2,kn(x-y)2)φo(knx,kny)+2∥HKo(kn(x+y)2,kn(x-y)2)-HKo(-kn(x+y)2,kn(x-y)2)-HKo(kn(x+y)2,kn(x+y)2)+HKo(-kn(x+y)2,kn(x+y)2)+HKo(knx+kny,0)-HKo(knx,0)-HKo(kny,0)HKo(kn(x+y)2,kn(x-y)2)∥}⟶0(n⟶∞).
Using this last inequality, dividing both sides of (41) by kn and letting n→∞, we obtain A(x+y)=A(x)+A(y).
According to (38), we have
(44)∥f(x)-A(x)∥≤∑i=1∞1ki∑j=1k-1φo((j+12)ki-1x,(j-12)ki-1x)+2∑i=1∞1ki×∑j=1k-1∥HKo((j+12)ki-1x,(j-12)ki-1x)-HKo(-(j+12)ki-1x,(j-12)ki-1x)-HKo((j+12)ki-1x,(j+12)ki-1x)+HKo(-(j+12)ki-1x,(j+12)ki-1x)+HKo((j+1)ki-1x,0)-HKo(jki-1x,0)-HKo(ki-1x,0)HKo((j+12)ki-1x,(j-12)ki-1x)∥.
To show that A is unique, let A1:E1→E2 be another additive mapping satisfying (44). Thus,
(45)∥A(x)-A1(x)∥≤1kn∥A(knx)-f(knx)∥+1kn∥f(knx)-A1(knx)∥≤2∑i=n+1∞‍1ki∑j=1k-1‍φo((j+12)ki-1x,(j-12)ki-1x)+4∑i=n+1∞‍1ki×∑j=1k-1‍∥HKo((j+12)ki-1x,(j-12)ki-1x)-HKo(-(j+12)ki-1x,(j-12)ki-1x)-HKo((j+12)ki-1x,(j+12)ki-1x)+HKo(-(j+12)ki-1x,(j+12)ki-1x)+HKo((j+1)ki-1x,0)-HKo(jki-1x,0)-HKo(ki-1x,0)HKo((j+12)ki-1x,(j-12)ki-1x)∥.
Using ∑i=0∞‍(1/ki)φ(kix,kiy)<∞ and
(46)∑i=0∞∥HKo(ki(x+y),ki(x-y))-HKo(-ki(x+y),ki(x-y))-HKo(ki(x+y),ki(x+y))+HKo(-ki(x+y),ki(x+y))+HKo(ki(x+y),0)-HKo(2kix,0)-HKo(2kiy,0)∥<∞,
we see that the last term becomes arbitrarily small for sufficiently large n, implying A=A1. Finally, the same proof as in [9] shows that A is linear if E1 is a Banach space and f(tx) is continuous in t for every fixed x∈E1.
The other provision∑i=0∞‍kiφ(x/ki,y/ki)<∞ and
(47)∑i=0∞ki+1∥HKo(x+yki+1,x-yki+1)-HKo(-x+yki+1,x-yki+1)-HKo(x+yki+1,x+yki+1)+HKo(-x+yki+1,x+yki+1)+HKo(2(x+y)ki+1,0)-HKo(2xki+1,0)-HKo(2yki+1,0)∥<∞
is treated analogously by replacing x by x/kn in (36) to obtain
(48)∥f(x)-knf(xkn)∥≤1k∑i=0n-1ki+1∑j=1k-1φo((j+1)x2ki+1,(j-12)xki+1)+2k∑i=0n-1ki+1×∑j=1k-1∥HKo((j+1)x2ki+1,(j-1)x2ki+1)-HKo(-(j+1)x2ki+1,(j-1)x2ki+1)-HKo((j+1)x2ki+1,(j+1)x2ki+1)+HKo(-(j+1)x2ki+1,(j+1)x2ki+1)+HKo((j+1)xki+1,0)-HKo(jxki+1,0)-HKo(xki+1,0)HKo((j+1)x2ki+1,(j-1)x2ki+1)∥,
which is then used to show that {knf(x/kn)} is a Cauchy sequence in E2 for all x∈E1, so that the limit
(49)A(x)=limn→∞knf(xkn)(x∈E1)
is a well-defined additive function. Using
(50)∥A(x)-f(x)∥≤1k∑i=0∞ki+1∑j=1k-1φo((j+1)x2ki+1,(j-1)x2ki+1)+2k∑i=0∞ki+1×∑j=1k-1∥HKo((j+1)x2ki+1,(j-1)x2ki+1)-HKo(-(j+1)x2ki+1,(j-1)x2ki+1)-HKo((j+1)x2ki+1,(j+1)x2ki+1)+HKo(-(j+1)x2ki+1,(j+1)x2ki+1)+HKo((j+1)xki+1,0)-HKo(jxki+1,0)-HKo(xki+1,0)HKo((j+1)x2ki+1,(j-1)x2ki+1)∥,
we can show that A is unique. Finally, that A is linear if E1 is a Banach space and f(tx) is continuous in t for every fixed x∈E1 is proved by the same technique as that in [9].


## 4. Main Result

Theorem 4.
Adopting the notation of Section2, if the functions ℱ,𝒢,ℋ,𝒦,ℒ,ℳ:E1→E2 satisfy
(51)∥ℱ(x+y)+𝒢(x-y)-2ℋ(x)𝒦(y)-ℒ(x)-ℳ(y)∥≤φ(x,y)(x,y∈E1),
then there exist a quadratic function Q:E1→E2 and two additive functions A1,A2:E1→E2 such that, for all x∈E1, one has
(52)∥ℱ(x)-Q(x)-A1(x)-A2(x)-ℱ(0)∥≤12k2Φ^k(x,x)+12kΦ^k′(x,x)+12kΦ^k′′(x,x)+12k2Φ^HK,k(x,x)+12kΦ^HK,k′(x,x)+12kΦ^HK,k′′(x,x)+∥HKo(x,0)∥+∥HKo(0,x)∥+∥HKe(x,0)∥+3φ(x2,x2)+52φ(x,0)+112φ(0,0),∥𝒢(x)-Q(x)-A1(x)-A2(x)-𝒢(0)∥≤12k2Φ^k(x,x)+12kΦ^k′(x,x)+12kΦ^k′′(x,x)+12k2Φ^HK,k(x,x)+12kΦ^HK,k′(x,x)+12kΦ^HK,k′′(x,x)+∥HKo(x,0)∥+∥HKo(0,x)∥+∥HKe(x,0)∥+φ(x2,x2)+52φ(x,0)+72φ(0,0),∥ℒ(x)-2Q(x)-2A1(x)-ℒ(0)∥≤1k2Φ^k(x,x)+1kΦ^k′(x,x)+2φ(x,0)+2φ(0,0)+1k2Φ^HK,k(x,x)+1kΦ^HK,k′(x,x),∥ℳ(x)-2Q(x)-2A2(x)-ℳ(0)∥≤1k2Φ^k(x,x)+1kΦ^k′′(x,x)+1k2Φ^HK,k(x,x)+1kΦ^HK,k′′(x,x),
where
(53)Φ^k(x,y)={Φk(x,y)ifthefirstconditionindefinition(iii)istrue,Φ~k(x,y)ifthesecondconditionindefinition(iii)istrue,Φ^k′(x,y)={Φk′(x,y)ifthefirstconditionindefinition(iii)istrue,Φ~k′(x,y)ifthesecondconditionindefinition(iii)istrue,Φ^k′′(x,y)={Φk′′(x,y)ifthefirstconditionindefinition(iii)istrue,Φ~k′′(x,y)ifthesecondconditionindefinition(iii)istrue,Φ^HK,k(x,y)={ΦHK,k(x,y)ifthefirstconditionindefinition(iii)istrue,Φ~HK,k(x,y)ifthesecondconditionindefinition(iii)istrue,Φ^HK,k′(x,y)={ΦHK,k′(x,y)ifthefirstconditionindefinition(iii)istrue,Φ~HK,k′(x,y)ifthesecondconditionindefinition(iii)istrue,Φ^HK,k′′(x,y)={ΦHK,k′′(x,y)ifthefirstconditionindefinition(iii)istrue,Φ~HK,k′′(x,y)ifthesecondconditionindefinition(iii)istrue.
Moreover, if ℒ(tx) and ℳ(tx) are continuous in t∈ℝ for each x∈E1, then the two additive functions A1, A2 are linear, while the quadratic function Q satisfies
(54)Q(tx)=t2Q(x).Proof.
Puttingx=y=0 in (51), we obtain
(55)∥ℱ(0)+𝒢(0)-2ℋ(0)𝒦(0)-ℒ(0)-ℳ(0)∥≤φ(0,0).
For x,y∈E1, setting
(56)F(x)∶=ℱ(x)-ℱ(0),G(x)∶=𝒢(x)-𝒢(0),L(x)∶=ℒ(x)-ℒ(0),M(x)∶=ℳ(x)-ℳ(0),HK(x,y)∶=ℋ(x)𝒦(y)-ℋ(0)𝒦(0),
we get F(0)=G(0)=HK(0,0)=L(0)=M(0)=0. Subtracting (51) and (55), we have
(57)∥F(x+y)+G(x-y)-2HK(x,y)-L(x)-M(y)∥≤φ(x,y)+φ(0,0).
Substituting -x for x and -y for y in (57), we get
(58)∥F(-(x+y))+G(-(x-y))-2HK(-x,-y)-L(-x)-M(-y)∥≤φ(x,y)+φ(0,0).
Combining (57) and (58), we obtain
(59)∥{F(x+y)+F(-(x+y))}+{G(x-y)+G(-(x-y))}-2{HK(x,y)+HK(-x,-y)}-{L(x)+L(-x)}-{M(y)+M(-y)}∥≤2φ(x,y)+2φ(0,0).
Let
(60)Fe(x)∶=F(x)+F(-x)2,Ge(x)∶=G(x)+G(-x)2,Le(x)∶=L(x)+L(-x)2,Me(x)∶=M(x)+M(-x)2,HKe(x,y)∶=HK(x,y)+HK(-x,-y)2(x,y∈E1).
Note that Fe, Ge, Le, Me are the even parts of the functions F, G, L, M, respectively, and that
(61)HKe(-x,-y)=HKe(x,y),HKe(x,-y)=HKe(-x,y),HKe(0,-x)=HKe(0,x),HKe(-x,0)=HKe(x,0).
From (59), we obtain
(62)∥Fe(x+y)+Ge(x-y)-2HKe(x,y)-Le(x)-Me(y)∥≤φ(x,y)+φ(0,0).
Substituting y=0 in (62), we get
(63)∥Fe(x)+Ge(x)-2HKe(x,0)-Le(x)∥≤φ(x,0)+φ(0,0).
Putting x=0 in (62) and replacing y by x, we get
(64)∥Fe(x)+Ge(x)-2HKe(0,x)-Me(x)∥≤φ(x,0)+φ(0,0).
Putting y=x in (62), we have
(65)∥Fe(2x)-2HKe(x,x)-Le(x)-Me(x)∥≤φ(x,x)+φ(0,0).
Substituting y=-x in (62), we get
(66)∥Ge(2x)-2HKe(x,-x)-Le(x)-Me(x)∥≤φ(x,x)+φ(0,0).
From (64) and (63), we have
(67)∥Le(x)-Me(x)-2[HKe(0,x)-HKe(x,0)]∥≤2φ(x,0)+2φ(0,0).
From (65) and (66) and replacing x by x/2, we obtain
(68)∥Fe(x)-Ge(x)-2{HKe(x2,x2)-HKe(x2,-x2)}∥≤2φ(x2,x2)+2φ(0,0).
From (62), (67), and (68), we get
(69)∥HKe(x,y)+HKe(x+y2,-x+y2)Ge(x+y)+Ge(x-y)-Me(x)-Me(y)-2{HKe(x,y)+HKe(x+y2,-x+y2)-HKe(x+y2,x+y2)+HKe(0,x)-HKe(x,0)HKe(x,y)+HKe(x+y2,-x+y2)}∥≤φ(x,y)+2φ(x,0)+2φ(x+y2,x+y2)+5φ(0,0).
Substituting y=0 in (69), we get
(70)∥(x2,x2)2Ge(x)-Me(x)-2{HKe(x2,-x2)-HKe(x2,x2)+HKe(0,x)}∥≤3φ(x,0)+2φ(x2,x2)+5φ(0,0).
From (69) and (70), we obtain
(71)∥2HKe(x,y)+HKe(x+y2,-x+y2)Me(x+y)+Me(x-y)-2Me(x)-2Me(y)-2{2HKe(x,y)+HKe(x+y2,-x+y2)-HKe(x+y2,x+y2)+HKe(x-y2,x-y2)-HKe(x-y2,-x-y2)+2HKe(0,x)-2HKe(x,0)-HKe(0,x+y)-HKe(0,x-y)2HKe(x,y)+HKe(x+y2,-x+y2)}∥≤2φ(x,y)+6φ(x+y2,x+y2)+2φ(x-y2,x-y2)+3φ(x+y,0)+3φ(x-y,0)+4φ(x,0)+20φ(0,0)=φe(x,y),
and so
(72)∥Me(x+y)+Me(x-y)-2Me(x)-2Me(y)∥≤φe(x,y)+2∥{2HKe(x,y)+HKe(x+y2,-x+y2)-HKe(x+y2,x+y2)+HKe(x-y2,x-y2)-HKe(x-y2,-x-y2)+2HKe(0,x)-2HKe(x,0)-HKe(0,x+y)-HKe(0,x-y)2HKe(x,y)+HKe(x+y2,-x+y2)}∥.
By Proposition 2, there is a unique quadratic function Q:E1→E2 such that
(73)∥Me(x)-2Q(x)∥≤1k2Φ^k(x,x)+1k2Φ^HK,k(x,x).
If ℳ(tx) is continuous in t∈ℝ for each x∈E1, then the quadratic function Q satisfies Q(tx)=t2Q(x). From (68), (70), and (73), we have
(74)∥Fe(x)-Q(x)∥≤∥Fe(x)-Ge(x)∥+∥Ge(x)-12Me(x)-HKe(x,0)∥+∥12Me(x)-Q(x)∥≤∥HKe(x,0)∥+3φ(x2,x2)+32φ(x,0)+92φ(0,0)+12k2Φ^k(x,x)+12k2Φ^HK,k(x,x).
From (70) and (73), we get
(75)∥Ge(x)-Q(x)∥≤∥Ge(x)-12Me(x)-HKe(x,0)∥+∥12Me(x)-Q(x)∥≤∥HKe(x,0)∥+φ(x2,x2)+32φ(x,0)+52φ(0,0)+12k2Φ^k(x,x)+12k2Φ^HK,k(x,x).
From (67) and (73), we obtain
(76)∥Le(x)-2Q(x)∥≤∥Le(x)-Me(x)∥+∥Me(x)-2Q(x)∥≤2φ(x,0)+2φ(0,0)+12k2Φ^k(x,x)+12k2Φ^HK,k(x,x).
Next, we consider the odd part of functions. Subtracting (57) and (58), we obtain
(77)∥{F(x+y)-F(-(x+y))}+{G(x-y)-G(-(x-y))}-2{HK(x,y)-HK(-x,-y)}-{L(x)-L(-x)}-{M(y)-M(-y)}∥≤2φ(x,y)+2φ(0,0).
For x,y∈E1, let
(78)Fo(x)∶=F(x)-F(-x)2,Go(x)∶=G(x)-G(-x)2,Lo(x)∶=L(x)-L(-x)2,Mo(x)∶=M(x)-M(-x)2,HKo(x,y)∶=HK(x,y)-HK(-x,-y)2.
Clearly, Fo, Go, Lo, Mo are the odd parts of F, G, L, M, respectively. Note that
(79)HKo(-x,-y)=-HKo(x,y),HKo(x,-y)=-HKo(-x,y),HKo(0,-x)=-HKo(0,x),HKo(-x,0)=-HKo(x,0).
From (77), we get
(80)∥Fo(x+y)+Go(x-y)-2HKo(x,y)-Lo(x)-Mo(y)∥≤φ(x,y)+φ(0,0).
Substituting y=0 in (80), we get
(81)∥Fo(x)+Go(x)-2HKo(x,0)-Lo(x)∥≤φ(x,0)+φ(0,0).
Putting x=0 in (80) and replacing y by x, we get
(82)∥Fo(x)-Go(x)-2HKo(0,x)-Mo(x)∥≤φ(x,0)+φ(0,0).
Letting y=x in (80), we have
(83)∥Fo(2x)-2HKo(x,x)-Lo(x)-Mo(x)∥≤φ(x,x)+φ(0,0).
Substituting y=-x in (80), we get
(84)∥Go(2x)-2HKo(x,-x)-Lo(x)+Mo(x)∥≤φ(x,x)+φ(0,0).
From (81) and (82), we obtain
(85)∥2Fo(x)-Lo(x)-Mo(x)-2[HKo(x,0)+HKo(0,x)]∥≤2φ(x,0)+2φ(0,0),(86)∥2Go(x)-Lo(x)+Mo(x)-2[HKo(x,0)-HKo(0,x)]∥≤2φ(x,0)+2φ(0,0).
From (83) and (85), we have
(87)∥Lo(2x)-2Lo(x)+Mo(2x)-2Mo(x)-2[2HKo(x,x)-HKo(2x,0)-HKo(0,2x)]∥≤2φ(x,x)+2φ(2x,0)+4φ(0,0).
From (84) and (86), we get
(88)∥Lo(2x)-2Lo(x)-Mo(2x)+2Mo(x)+2[2HKo(-x,x)+HKo(2x,0)-HKo(0,2x)]∥≤2φ(x,x)+2φ(2x,0)+4φ(0,0).
From (87) and (88), we have
(89)∥Lo(2x)-2Lo(x)-2[HKo(x,x)-HKo(-x,x)-HKo(2x,0)]∥≤2φ(x,x)+2φ(2x,0)+4φ(0,0),∥Mo(2x)-2Mo(x)-2[HKo(x,x)+HKo(-x,x)-HKo(0,2x)]∥≤2φ(x,x)+2φ(2x,0)+4φ(0,0).
In view of (80), (85), (86), and (89), we have
(90)∥Lo(x+y)+Mo(x+y)+Lo(x-y)-Mo(x-y)-Lo(2x)-Mo(2y)-2[2HKo(x,y)-HKo(x,x)+HKo(-x,x)-HKo(y,y)-HKo(-y,y)+HKo(2x,0)+HKo(0,2y)-HKo(x+y,0)-HKo(0,x+y)-HKo(x-y,0)+HKo(0,x-y)]∥≤2φ(x,y)+2φ(x,x)+2φ(y,y)+2φ(2x,0)+2φ(2y,0)+2φ(x+y,0)+2φ(x-y,0)+14φ(0,0)=φ0(x,y).
Substituting y by -y in (90), we obtain
(91)∥Lo(x-y)+Mo(x-y)+Lo(x+y)-Mo(x+y)-Lo(2x)+Mo(2y)-2[-2HKo(-x,y)-HKo(x,x)+HKo(-x,x)+HKo(y,y)+HKo(-y,y)+HKo(2x,0)-HKo(0,2y)-HKo(x+y,0)+HKo(0,x+y)-HKo(x-y,0)-HKo(0,x-y)]∥≤φ0(x,y).
From (90) and (91), we have
(92)∥Lo(x+y)+Lo(x-y)-Lo(2x)-2[HKo(x,y)-HKo(-x,y)-HKo(x,x)+HKo(-x,x)+HKo(2x,0)-HKo(x+y,0)-HKo(x-y,0)]∥≤φ0(x,y),(93)∥Mo(x+y)-Mo(x-y)-Mo(2y)-2[HKo(x,y)+HKo(-x,y)-HKo(y,y)-HKo(-y,y)+HKo(0,2y)-HKo(0,x+y)+HKo(0,x-y)]∥≤φ0(x,y).
Putting u=x+y and v=x-y in (92), we obtain
(94)∥+HKo(u+v,0)-HKo(u,0)-HKo(v,0)HKo(u+v2,u-v2)}Lo(u)+Lo(v)-Lo(u+v)-2{HKo(u+v2,u-v2)-HKo(-u+v2,u-v2)-HKo(u+v2,u+v2)+HKo(-u+v2,u+v2)+HKo(u+v,0)-HKo(u,0)-HKo(v,0)HKo(u+v2,u-v2)}∥≤φ0(u+v2,u-v2).
By Proposition 3, there exists a unique additive function A1:E1→E2 such that
(95)∥Lo(x)-2A1(x)∥≤1kΦ^k′(x,x)+1kΦ^HK,k′(x,x)(x∈E1).
Moreover, if ℒ(tx) is continuous in t∈ℝ for every fixed x∈E1, we have that A1 is a linear function. Putting u=x-y and v=2y in (93), we get
(96)∥HKo(u+v2,v2)Mo(u+v)-Mo(u)-Mo(v)-2[HKo(u+v2,v2)+HKo(-(u+v2),v2)-HKo(v2,v2)-HKo(-v2,v2)+HKo(0,v)-HKo(0,u+v)+HKo(0,u)HKo(u+v2,v2)]∥≤φo(u+v2,v2)(u,v∈E1).
Again by Proposition 3, there exists a unique additive function A2:E1→E2 such that
(97)∥Mo(x)-2A2(x)∥≤1kΦ^k′′(x,x)+1kΦ^HK,k′′(x,x)(x∈E1).
Furthermore, if ℳ(tx) is continuous in t∈ℝ for all x∈E1, then A2 is also linear. From (85), (95), and (97), it follows that
(98)∥Fo(x)-A1(x)-A2(x)∥≤12kΦ^k′(x,x)+12kΦ^k′′(x,x)+12kΦ^HK,k′(x,x)+12kΦ^HK,k′′(x,x)+φ(x,0)+φ(0,0)+∥HKo(x,0)∥+∥HKo(0,x)∥,∥Go(x)-A1(x)-A2(x)∥≤12kΦ^k′(x,x)+12kΦ^k′′(x,x)+12kΦ^HK,k′(x,x)+12kΦ^HK,k′′(x,x)+φ(x,0)+φ(0,0)+∥HKo(x,0)∥+∥HKo(0,x)∥.
The theorem now follows from the inequalities (73), (74), (75), (76), (95), (97), and (98).



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*Source: 309098-2013-11-28.xml*