# A Functional Equation Originating from Elliptic Curves

Authors: Won-Gil Park; Jae-Hyeong Bae
Journal: Abstract and Applied Analysis
 (2008)

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## Abstract

We obtain the general solution and the stability of the functional
equationf(x+y+z,u+v+w)+f(x+y−z,u+v+w)+2f(x,u−w)+2f(y,v−w)=f(x+y,u+w)+f(x+y,v+w)+f(x+z,u+w)+f(x−z,u+v−w)+f(y+z,v+w)+f(y−z,u+v−w).
The function f(x,y)=x3+ax+b−y2 having level curves as elliptic curves is a solution of the above functional equation.


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## Body

## 1. Introduction

A graph of an equation of the formy2=x3+ax+b is called an elliptic curve [1],  where a and b are constants. Since f(x,y)=x3+ax+b−y2 has level curves as elliptic curves,
functional equations having the mapping f(x,y)=x3+ax+b−y2 as a solution are helpful to study
cryptography and their applications.Recently, Jun and Kim [2] solved the cubic functional
equation(1.1)g(x+y+z)+g(x+y−z)+2g(x)+2g(y)=2g(x+y)+g(x+z)+g(x−z)+g(y+z)+g(y−z).We consider the quadratic functional
equation(1.2)g(x+y+z)+g(x−z)+g(y−z)=g(x+y−z)+g(x+z)+g(y+z),and the 2-dimensional vector variable functional
equation(1.3)f(x+y+z,u+v+w)+f(x+y−z,u+v+w)+2f(x,u−w)+2f(y,v−w)=f(x+y,u+w)+f(x+y,v+w)+f(x+z,u+w)+f(x−z,u+v−w)+f(y+z,v+w)+f(y−z,u+v−w).The functiong:ℝ→ℝ given by g(x):=ax3+bx+c is a particular solution of (1.1), and the
function g:ℝ→ℝ given by g(x):=ax2+b is a particular solution of (1.2). The function f:ℝ×ℝ→ℝ given by f(x,y):=x3+ax+b−y2 is a particular solution of (1.3). The
functional (1.3) is a mixed type of (1.1) and (1.2).Problems on solutions or stability of various
functional equations have been extensively investigated by a number of
mathematicians  [3–12].In this paper, letX, Y, and Z be real vector spaces. We find out the general
solution and investigate the stability of (1.2) and (1.3).


## 2. Solutions

We find out the general solution of the functional (1.2)
as follows.Lemma 2.1.
A mappingg:X→Y satisfies (1.2) if and only if there exists a symmetric
biadditive mapping S:X×X→Y such that(2.1)g(x)=S(x,x)+g(0)for all x∈X.Proof.
We first assume thatg is a solution of (1.2). Let h:X→Y be a mapping given by h(x):=g(x)−g(0) for all x∈X.
Then, h(0)=0 and(2.2)h(x+y+z)+h(x−z)+h(y−z)=h(x+y−z)+h(x+z)+h(y+z)for all x,y,z∈X.
Putting x=y=0 in (2.2), h is even. Replacing z by x+y in (2.2),(2.3)h(2x+2y)+h(y)+h(x)=h(2x+y)+h(x+2y)for all x,y∈X.
Taking z=x in (2.2), we get that(2.4)h(2x+y)+h(x−y)=h(y)+h(2x)+h(x+y)for all x,y∈X.
Interchanging x and y in (2.4), we see that(2.5)h(x+2y)+h(x−y)=h(x)+h(2y)+h(x+y)for all x,y∈X.
By (2.3), (2.4), and (2.5), we obtain that (2.6)h(2x+2y)=h(2x+y)−h(y)+h(x+2y)−h(x)=h(2x)+h(x+y)−h(x−y)+h(2y)+h(x+y)−h(x−y)=h(2x)+h(2y)+2h(x+y)−2h(x−y)  for all x,y∈X.
Letting y=x in (2.6), we get that(2.7)h(2x)=4h(x)for all x∈X.
By (2.6) and (2.7), we obtain that(2.8)h(x+y)+h(x−y)=2h(x)+2h(y)for all x,y∈X.
By [13], there exists a
symmetric biadditive mapping S:X×X→Y such that h(x)=S(x,x) for all x∈X.
Conversely, we assume that there exists a symmetric
biadditive mappingS:X×X→Y such that g(x)=S(x,x)+g(0) for all x∈X.
Since S is biadditive,(2.9)g(x+y+z)+g(x−z)+g(y−z)=S(x+y+z,x+y+z)+S(x−z,x−z)+S(y−z,y−z)+3g(0)=2[S(x,x)+S(x,y)+S(y,y)]+3S(z,z)+3g(0)=S(x+y−z,x+y−z)+S(x+z,x+z)+S(y+z,y+z)+3g(0)=g(x+y−z)+g(x+z)+g(y+z)for all x,y,z∈X.Example 2.2.
Letr/s be a rational number, where r and s are integers with gcd⁡(r,s)=1.
Define H(r/s):=max⁡{|r|,|s|} and g(r/s):=log⁡H(r/s).
Let ℚ be the set of rational numbers and let  E(ℚ):={∞}∪{(x,y)∈ℚ×ℚ∣y2=x3+ax+b}
be an elliptic
curve over ℚ.
The addition in E(ℚ) is given in [1]. The height function h:E(ℚ)→[0,∞) is defined by
(2.10)h(x,y)=g(x),h(∞)=0for all (x,y)∈E(ℚ).
The canonical height function h^:E(ℚ)→[0,∞) given by(2.11)h^(P)=12lim⁡n→∞14nh(2nP)is a solution of (1.2).We find out the general solution of the functional (1.3)
as follows.Theorem 2.3.
A mappingf:X×Y→Z satisfies (1.3) if and only if there exist a symmetric
multiadditive mapping S1:X×X×X→Z,
an additive mapping A:X→Z, and a symmetric biadditive mapping S2:Y×Y→Z such that(2.12)f(x,y)=S1(x,x,x)+A(x)+f(0,0)+S2(y,y)for all x∈X and all y∈Y.Proof.
We first assume thatf satisfies (1.3). Define f1:X→Z and f2:Y→Z by f1(x):=f(x,0) for all x∈X and f2(y):=f(0,y) for all y∈Y.
Then, f1 satisfies (1.1) and f2 satisfies (1.2). By [2], there exist a symmetric
multiadditive mapping S1:X×X×X→Z and an additive mapping A:X→Z such that(2.13)f1(x)=S1(x,x,x)+A(x)+f1(0)for all x∈X.
By Lemma 2.1, there exists a symmetric biadditive mapping S2:Y×Y→Z such that(2.14)f2(y)=S2(y,y)+f2(0)for all y∈Y.
Note that f1(0)=f2(0)=f(0,0) and that f(0,0)≠0 in general. Let k:X×Y→Z be a mapping given by k(x,y):=f(x,y)−f(0,0) for all x∈X and all y∈Y.
Then, k(0,0)=0 and(2.15)k(x+y+z,u+v+w)+k(x+y−z,u+v+w)+2k(x,u−w)+2k(y,v−w)=k(x+y,u+w)+k(x+y,v+w)+k(x+z,u+w)+k(x−z,u+v−w)+k(y+z,v+w)+k(y−z,u+v−w)for all x,y,z∈X and all u,v,w∈Y.
Sincek(0,u)=S2(u,u) for all u∈Y, k(0,−u)=k(0,u) and k(0,2u)=4k(0,u) for all u∈Y.
Letting y=z=u=v=0 in (2.15), we get(2.16)k(x,w)=k(x,−w)for all x∈X and all w∈Y.
Putting y=z=0 in (2.15), we get(2.17)2k(x,u+v+w)+2k(x,u−w)+2k(0,v−w)=2k(x,u+w)+k(x,v+w)+k(0,v+w)+k(x,u+v−w)+k(0,u+v−w)for all x∈X and all u,v,w∈Y.
Setting v=w=−u in (2.17) and using (2.16),(2.18)k(x,u)+k(x,2u)=2k(x,0)+5k(0,u)for all x∈X and all u∈Y.
Taking v=−w=−u in (2.17) and using (2.16),(2.19)k(x,u)+k(x,0)+7k(0,u)=2k(x,2u)for all x∈X and all u∈Y.
By (2.18) and (2.19), k(x,u)=k(x,0)+k(0,u) for all x∈X and all u∈Y.
Hence, we obtain that(2.20)f(x,y)=k(x,y)+f(0,0)=k(x,0)+k(0,y)+f(0,0)=f1(x)−f1(0)+f2(y)−f2(0)+f(0,0)=S1(x,x,x)+A(x)+f(0,0)+S2(y,y)for all x∈X and all y∈Y.
Conversely, we assume that there exist a symmetric
multiadditive mappingS1:X×X×X→Z,
an additive mapping A:X→Z, and a symmetric biadditive mapping S2:Y×Y→Z such that(2.21)f(x,y)=S1(x,x,x)+A(x)+f(0,0)+S2(y,y)for all x∈X and all y∈Y.
Since S1 is symmetric multiadditive and S2 is symmetric biadditive,(2.22)f(x+y+z,u+v+w)+f(x+y−z,u+v+w)+2f(x,u−w)+2f(y,v−w)=S1(x+y+z,x+y+z,x+y+z)+S1(x+y−z,x+y−z,x+y−z)+2S1(x,x,x)+2S1(y,y,y)+A(x+y+z)+A(x+y−z)+2A(x)+2A(y)+2S2(u+v+w,u+v+w)+2S2(u−w,u−w)+2S2(v−w,v−w)+6f(0,0)=4S1(x,x,x)+4S1(y,y,y)+6S1(x,x,y)+6S1(x,y,y)+6S1(x,z,z)+6S1(y,z,z)+4A(x)+4A(y)+4S2(u,u)+4S2(u,v)+4S2(v,v)+6S2(w,w)+6f(0,0)=2S1(x+y,x+y,x+y)+S1(x+z,x+z,x+z)+S1(x−z,x−z,x−z)+S1(y+z,y+z,y+z)+S1(y−z,y−z,y−z)+2A(x+y)+A(x+z)+A(x−z)+A(y+z)+A(y−z)+2S2(u+w,u+w)+2S2(v+w,v+w)+2S2(u+v−w,u+v−w)+6f(0,0)=f(x+y,u+w)+f(x+y,v+w)+f(x+z,u+w)+f(x−z,u+v−w)+f(y+z,v+w)+f(y−z,u+v−w)for all x,y,z∈X and all u,v,w∈Y.Example 2.4.
LetX and Y be the vector spaces M3(ℂ) and M2(ℂ),
respectively. Consider a function f:X×Y→ℂ given by f(x,y):=det⁡(x)+det⁡(y) for all x∈X and all y∈Y.
One can easily verify that f satisfies (1.3).
By Theorem 2.3, there exist a symmetric multiadditive mapping S1:X×X×X→ℂ,
an additive mapping A:X→ℂ, and a symmetric biadditive mapping S2:Y×Y→ℂ such that
(2.23)f(x,y)=S1(x,x,x)+A(x)+f(0,0)+S2(y,y)for all x∈X and all y∈Y.
In fact,(2.24)S1(x,y,z)=124[det⁡(x+y+z)+det⁡(x−y−z)−det⁡(x+y−z)−det⁡(x−y+z)],A(x)=f(0,0)=0,S2(u,v)=14[det⁡(u+v)−det⁡(u−v)]for all x,y,z∈X and all u,v∈Y


## 3. Stability

From now on, letY be a Banach space, and let φ:X3→[0,∞) be a function satisfying(3.1)φ˜(x,y,z):=∑j=0∞14j+1φ(2jx,2jy,2jz)<∞for all x,y,z∈X.Lemma 3.1.
Letg:X→Y be a mapping such that
(3.2)∥g(x+y+z)+g(x−z)+g(y−z)−g(x+y−z)−g(x+z)−g(y+z)∥≤φ(x,y,z)
for all x,y,z∈X.
Then, there exists a unique quadratic mapping G:X→Y satisfying (1.2) such that(3.3)∥g(x)−G(x)∥≤φ˜(x2,x2,x)+2φ˜(x2,x2,x2)+2φ˜(0,0,x2)+13[φ(0,0,0)+∥g(0)∥]for all x∈X.
The mapping G is given by G(x):=lim⁡j→∞14jg(2jx) for all x∈X.Proof.
Lettingx=y=0 in (3.2), we have(3.4)∥g(−z)−g(z)∥≤φ(0,0,z)for all z∈X.
Replacingz by x+y in (3.2), we get
(3.5)∥g(2x+2y)+g(−y)+g(−x)−g(0)−g(2x+y)−g(x+2y)∥≤φ(x,y,x+y)for all x,y∈X.
Taking z=x in (3.2), we see that(3.6)∥g(2x+y)+g(0)+g(y−x)−g(y)−g(2x)−g(x+y)∥≤φ(x,y,x)for all x,y∈X.
Interchanging x and y in (3.6), we see that(3.7)∥g(x+2y)+g(0)+g(x−y)−g(x)−g(2y)−g(x+y)∥≤φ(y,x,x)for all x,y∈X.
By (3.5), (3.6), and (3.7),(3.8)∥g(2x+2y)−2g(x+y)−g(2x)−g(2y)∥≤φ(x,y,x+y)+φ(x,y,x)+φ(y,x,x)+∥g(−x)+g(x)∥+∥g(−y)−g(y)∥+∥g(y−x)−g(x−y)∥+∥g(0)∥≤φ(x,y,x+y)+φ(x,y,x)+φ(y,x,x)+φ(0,0,x)+φ(0,0,y)+φ(0,0,x−y)+∥g(0)∥for all x,y∈X.
Putting y=x in the above inequality, we
get(3.9)∥g(4x)−4g(2x)∥≤φ(x,x,2x)+2φ(x,x,x)+2φ(0,0,x)+φ(0,0,0)+∥g(0)∥for all x∈X.
So,(3.10)∥g(x)−14g(2x)∥≤14φ*(x),where(3.11)φ*(x):=φ(x2,x2,x)+2φ(x2,x2,x2)+2φ(0,0,x2)+φ(0,0,0)+∥g(0)∥for all x∈X.
Thus, we obtain(3.12)∥14jg(2jx)−14j+1g(2j+1x)∥≤14j+1φ*(2jx)for all x∈X and all j.
For given integers l,m⁢(0≤l<m),
we get(3.13)∥14lg(2lx)−14mg(2mx)∥≤∑j=lm−114j+1φ*(2jx)for all x∈X.
By (3.13), the sequence {(1/4j)g(2jx)} is a Cauchy sequence for all x∈X.
Since Y is complete, the sequence {(1/4j)g(2jx)} converges for all x∈X.
Define G:X→Y by(3.14)G(x):=lim⁡j→∞14jg(2jx)for all x∈X.
By (3.2), we have(3.15)∥14j[g(2j(x+y+z))+g(2j(x−z))+g(2j(y−z))−g(2j(x+y−z))−g(2j(x+z))−g(2j(y+z))]∥≤14jφ(2jx,2jy,2jz)for all x,y,z∈X and all j.
Letting j→∞ and using (3.1), we see that G satisfies (1.2). Setting l=0 and taking m→∞ in (3.13), one can obtain the inequality (3.3).
If H:X→Y is another quadratic mapping satisfying (1.2)
and (3.3), we obtain(3.16)∥G(x)−H(x)∥=14n∥G(2nx)−H(2nx)∥≤14n∥G(2nx)−g(2nx)∥+14n∥g(2nx)−H(2nx)∥≤24n[φ˜(2n−1x,2n−1x,2nx)+2φ˜(2n−1x,2n−1x,2n−1x)+2φ˜(0,0,2n−1x)+13[φ(0,0,0)+∥g(0)∥]]⟶0asn⟶∞for all x∈X.
Hence, G is a unique quadratic mapping, as desired.Example 3.2.
If the height functionh in Example 2.2 satisfies
(3.17)∥h(P+Q+R)+h(P−R)+h(Q−R)−h(P+Q−R)−h(P+R)−h(Q+R)∥≤cfor all P,Q,R∈E(ℚ),
then there exists a unique quadratic function h˜:E(ℚ)→[0,∞) satisfying (1.2) such that(3.18)∥h(P)−h˜(P)∥≤2cfor all P∈E(ℚ).Letψ:X3×Y3→[0,∞) be a function satisfying (3.19)ψ˜(x,y,z,u,v,w):=∑j=0∞[12j+1ψ1(2jx,2jy,2jz)+14j+1ψ2(2ju,2jv,2jw)]<∞,  where(3.20)ψ1(x,y,z):=ψ(x,y,z,0,0,0),ψ2(u,v,w):=ψ(0,0,0,u,v,w)for all x,y,z∈X and all u,v,w∈Y.
Also, let ψ˜1:X3→[0,∞), and
ψ˜2:Y3→[0,∞) be two functions given by(3.21)ψ˜1(x,y,z):=ψ˜(x,y,z,0,0,0),ψ˜2(u,v,w):=ψ˜(0,0,0,u,v,w)for all x,y,z∈X and all u,v,w∈Y.Theorem 3.3.
Letf:X×Y→Z be a mapping such that(3.22)∥f(x+y+z,u+v+w)+f(x+y−z,u+v+w)+2f(x,u−w)+2f(y,v−w)−f(x+y,u+w)−f(x+y,v+w)−f(x+z,u+w)−f(x−z,u+v−w)−f(y+z,v+w)−f(y−z,u+v−w)∥≤ψ(x,y,z,u,v,w)  for all x,y,z∈X and all u,v,w∈Y.
Then, there exists a mapping F:X×Y→Z satisfying (1.3) such that (3.23)∥f(x,y)−F(x,y)∥≤ψ(x,0,0,y,0,0)+ψ˜1(x2,x2,x2)+13ψ˜1(x,x2,x2)+ψ˜2(y2,y2,y)+2ψ˜2(y2,y2,y2)+2ψ˜2(0,0,y2)+13[ψ2(0,0,0)+∥f(0,0)∥] for all x∈X and all y∈Y.
The mapping F is given by(3.24)F(x,y):=lim⁡j→∞[4j+1−13⋅8jf(2jx,0)−4j−16⋅8jf(2j+1x,0)+14jf(0,2jy)]−f(0,0)for all x∈X and all y∈Y.Proof.
Lettingy=z=v=w=0 in (3.22), we have(3.25)∥f(x,u)−f(x,0)−f(0,u)+f(0,0)∥≤ψ(x,0,0,u,0,0)for all x∈X and all u∈Y.
Definef1,f2:X→Y by f1(x):=f(x,0) and f2(u):=f(0,u) for all x∈X and all u∈Y.
Putting u=v=w=0 in (3.22), we have(3.26)∥f1(x+y+z)+f1(x+y−z)+2f1(x)+2f1(y)−2f1(x+y)−f1(x+z)−f1(x−z)−f1(y+z)−f1(y−z)∥≤ψ1(x,y,z)for all x,y,z∈X.
Setting x=y=z=0 in (3.22), we have(3.27)∥f2(u+v+w)+f2(u−w)+f2(v−w)−f2(u+v−w)−f2(u+w)−f2(v+w)∥≤ψ2(u,v,w)for all u,v,w∈Y.
By [2], there exists a cubic mapping F1:X→Z satisfying (1.1) such that(3.28)∥f1(x)−F1(x)∥≤ψ˜1(x2,x2,x2)+13ψ˜1(x,x2,x2)for all x∈X.
By Lemma 3.1, there exists a quadratic mapping F2:Y→Z satisfying (1.2) such that(3.29)∥f2(y)−F2(y)∥≤ψ˜2(y2,y2,y)+2ψ˜2(y2,y2,y2)+2ψ˜2(0,0,y2)+13[ψ2(0,0,0)+∥f(0,0)∥]for all y∈Y.
If we define(3.30)F(x,y):=F1(x)+F2(y)−f(0,0)for all x∈X and all y∈Y,
we conclude that(3.31)∥f(x,y)−F(x,y)∥≤ψ(x,0,0,y,0,0)+ψ˜1(x2,x2,x2)+13ψ˜1(x,x2,x2)+ψ˜2(y2,y2,y)+2ψ˜2(y2,y2,y2)+2ψ˜2(0,0,y2)+13[ψ2(0,0,0)+∥f(0,0)∥]
for allx∈X and all y∈Y.Letφ′:X3→[0,∞) be a function satisfying(3.32)∑j=0∞4j+1φ′(2−jx,2−jy,2−jz)<∞for all x,y,z∈X.
For the function φ′,
we can obtain similar results to Lemma 3.1 and
Theorem 3.3.As a corollary, one can obtain a result when the
control mapping is a summation of terms∥⋅∥p,
in which p is a suitable constant by referring to
[14].



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*Source: 135237-2008-04-24.xml*