# Stability of the Pexiderized Lobacevski Equation

Authors: Gwang Hui Kim
Journal: Journal of Applied Mathematics
 (2011)

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## Abstract

The aim of this paper is to investigate the solution and the superstability of the Pexiderized Lobacevski equationf((x+y)/2)2=g(x)h(y), where f, g, h : G2→ℂ are unknown functions on an Abelian semigroup (G,+). The obtained result is a generalization of Gǎvruţa's result in 1994 and Kim's result in 2010.


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## Body

## 1. Introduction

The stability problem of the functional equation was conjectured by Ulam [1] during the conference in the University of Wisconsin in 1940. In the next year, it was solved by Hyers [2] in the case of additive mapping, which is called the Hyers-Ulam stability. Thereafter, this problem was improved by Bourgin [3], Aoki [4], Rassias [5],  Ger [6], and  Gǎvruţa et al. [7, 8] in which Rassias’ result is called the Hyers-Ulam-Rassias stability.In 1979, Baker et al. [9] developed the superstability, which is that if f is a function from a vector space to ℝ satisfying
(1.1)|f(x+y)-f(x)f(y)|≤ε
for some fixed ε>0, then either f is bounded or satisfies the exponential functional equation
(E)f(x+y)=f(x)f(y).In 1983, the superstability bounded by a constant for the sine functional equation(S)f(x+y2)2-f(x-y2)2=f(x)f(y)
was investigated by Cholewa [10] and was improved by  Badora and Ger [11]. Recently, the superstability bounded by some function for the Pexider type sine functional equation
(1.2)f(x+y2)2-f(x-y2)2=g(x)h(y)
has been investigated by Kim [12, 13].In 1994, Gǎvruţa [14] proved the superstability of the Lobacevski equation
(L)f(x+y2)2=f(x)f(y)
under the condition bounded by a constant.Kim [15] improved his result under the condition bounded by an unknown function. In there, author conjectured through an example that the Lobacevski equation (L) will have a solution as an exponential function. Namely, for a simple example of this equation, we can find the functional equation (e(x+y)/2)2=exey.The aim of this paper is to investigate the solution and the superstability of the Pexiderized Lobacevski equation(PL)f(x+y2)2=g(x)h(y)
under the condition bounded by a function. Namely, this has improved in the Pexider type for the results of Gǎvruţa and Kim.Furthermore, the range of the function in all results is expanded to the Banach space.The solution of (PL) will be represented as an exponential, namely, for a simple example of this equation, it will be considered as a geometric mean
(1.3)f(x)=αβex=(αex)(βex)=g(x)h(x),whereα,β>0.In this paper, let(G,+) be a uniquely 2-divisible Abelian semigroup (i.e., for each x∈G, there exists a unique y∈G such that y+y=x: such y will be denoted by x/2), ℂ is the field of complex numbers, ℝ the field of real numbers, and ℝ+ the set of positive reals. We assume that f,g,h:G→ℂ are nonzero and nonconstant functions, ε is a nonnegative real constant, and φ:G→ℝ+ is a mapping.


## 2. Stability of the Pexiderized Lobacevski Equation (PL)

We will investigate the solution and the superstability of the Pexiderized Lobacevski equation (PL).Theorem 2.1.
Suppose thatf,g,h:G→ℝ satisfy the inequality
(2.1)|f(x+y2)2-g(x)h(y)|≤ε
for all x,y∈G.
Then, either there existC1,C2,C3>0 such that
(2.2)|g(x)|≤C1,|h(x)|≤C2,|f(x)|≤C3,
for all x∈G, or else each function g and h satisfies (L). Here g and h are represented by
(2.3)g(x)=g(0)e(x),h(x)=h(0)e(x),(2.4)|f(x)2-g(0)h(0)⋅e(x)2|≤ε,
where e(x) is an exponential function.Proof.
Replacingx with y in (2.1), and then subtracting them and using triangle inequality, we have
(2.5)|g(x)h(y)-g(y)h(x)|≤2ε∀x,y∈G.
It follows from the inequality (2.5) that there exist constants c1,c2,d1,d2≥0 such that
(2.6)|g(x)|≤c1|h(x)|+d1,(2.7)|h(x)|≤c2|g(x)|+d2
for all x∈G. It follows from (2.6) and (2.7) that g is bounded if and only if h is bounded. If either of g or h is bounded, then we obtain (2.2) from (2.1).
Now ifh(x) is unbounded, then we can choose (yn)∈G so that |h(yn)|→∞ as n→∞. Letting y=yn in (2.1), dividing by |h(yn)|, and letting n→∞, we have
(2.8)g(x)=lim⁡n→∞f((x+yn)/2)2h(yn),∀x∈G.
It follows from (2.1) and (2.8) that
(2.9)g(x+y)g(z)=lim⁡n→∞f((x+y+yn)/2)2g(z)h(yn)=lim⁡n→∞g(x)h(y+yn)g(z)+R1h(yn)=lim⁡n→∞g(x)f((y+z+yn)/2)2+R1+R2h(yn)=g(x)g(y+z)+lim⁡n→∞R1+R2h(yn),
where |R1|≤ε|g(z)|,|R2|≤ε|g(x)|, which implies
(2.10)g(x+y)g(z)=g(x)g(y+z)
for all x,y,z∈G.
Lettingz=0 in (2.10), we get
(2.11)g(x+y)g(0)=g(x)g(y)
for all x,y∈G, which implies that
(2.12)g(x)=g(0)e1(x),
where g(0)≠0 (since g(x) is a nonzero and nonconstant function), and e1 is an exponential.
Exchanging the roles ofg and h, by the same proceeding, we have
(2.13)h(x)=h(0)e2(x),
where h(0)≠0, and e2 is an exponential.
Putting (2.12) and (2.13) in (2.5), it implies
(2.14)|e1(x)e2(y)-e1(y)e2(x)|≤2εg(0)h(0)=M∀x,y∈G.
Letx=0 in (2.14). Since e1 and e2 are exponentials, this implies that |e1(y)-e2(y)|≤M for all y∈G. Hence, from this and (2.14), we have
(2.15)e1(y)|e1(x)-e2(x)|=|e1(x)[e1(y)-e2(y)]+e1(x)e2(y)-e1(y)e2(x)|≤e1(x)M+M,
which is
(2.16)|e1(x)-e2(x)|≤e1(x)M+Me1(y),
for all x,y∈G.
Sinceg is unbounded from (2.2), we can choose (yn)∈G so that g(yn)=g(0)e1(yn)→∞ as n→∞. Letting y=yn in (2.16), we get that e1(x)=e2(x). Let it be denoted by e(x). Then (2.12) and (2.13) state nothing but (2.3). Putting (2.3) with x=y in (2.1), we get the inequality (2.4).
Finally, it is immediate thatg and h in (2.3) satisfy (L), respectively.Corollary 2.2.
Suppose thatf,g:G→ℝ satisfy the inequality
(2.17)|f(x+y2)2-g(x)g(y)|≤ε
for all x,y∈G.
Then, eitherg is bounded or g satisfies (L). In particular, g is represented by
(2.18)g(x)=g(0)e(x),
where e is exponential.Corollary 2.3.
Suppose thatf,g:G→ℝ satisfy the inequality
(2.19)|f(x+y2)2-g(x)f(y)|≤ε
for all x,y∈G.
Then either there existC1,C2>0 such that
(2.20)|g(x)|≤C1,|f(x)|≤C2
for all x∈G, or else each function f and g satisfies (L). In particular, f and g are represented by
(2.21)f(x)=f(0)e(x),g(x)=g(0)e(x),
where e:G→ℝ is exponential.Corollary 2.4.
Suppose thatf:G→ℝ satisfy the inequality
(2.22)|f(x+y2)2-f(x)f(y)|≤ε
for all x,y∈G.
Then eitherf is bounded or f satisfies (L). In particular, f is represented by
(2.23)f(x)=f(0)e(x),
where e:G→ℝ is exponential.In Corollary2.4, it is founded in papers [14, 15] that f satisfies (L).Theorem 2.5.
Suppose thatf,g,h:G→ℝ satisfy the inequality
(2.24)|f(x+y2)2-g(x)h(y)|≤φ(x)
for all x,y∈G.
Then, eitherh is bounded or g is an exponential by the multiplying of a scalar g(0) and satisfies (L).Proof.
Suppose thath(x) is unbounded. Then we can choose (yn)∈G such that |h(yn)|→∞ as n→∞. Letting y=yn in (2.24), dividing by |h(yn)|, and letting n→∞, we have
(2.25)g(x)=lim⁡n→∞f(x+yn)h(yn)∀x∈G.
Thus, it follows from (2.24) and (2.25) that
(2.26)g(x+y)g(z)=lim⁡n→∞f((x+y+yn)/2)2g(z)h(yn)=lim⁡n→∞g(x)h(y+yn)g(z)+R1h(yn)=lim⁡n→∞g(x)f((y+z+yn)/2)2+R1+R2h(yn)=g(x)g(y+z)+lim⁡n→∞R1+R2h(yn),
where |R1|≤ε|g(z)|,|R2|≤ε|g(x)|, which implies
(2.27)g(x+y)g(z)=g(x)g(y+z)
for all x,y,z∈G.
Lettingz=0 in (2.27), we get
(2.28)g(x+y)g(0)=g(x)g(y)
for all x,y,z∈G. Namely, it means that g is an exponential function by the multiplying of a scalar g(0) and satisfies (L).Theorem 2.6.
Suppose thatf,g,h:G→ℝ satisfy the inequality
(2.29)|f(x+y2)2-g(x)h(y)|≤φ(y)
for all x,y∈G.
Then, eitherg is bounded or h is an exponential by the multiplying of a scalar h(0) and satisfies (L).Proof.
The proof runs along a slight change in the step-by-step procedure in Theorem2.5.Remark 2.7.
(i) As Corollaries2.2–2.4 of Theorem 2.1, by replacing g and h with f in Theorems 2.5 and 2.6, we can obtain more corollaries for the following functional equations:
(2.30)f(x+y2)2=g(x)g(y),f(x+y2)2=f(x)g(y),f(x+y2)2=f(x)f(y),
in which the case of (2.30) is found in paper [15].
(ii) For the results obtained from each equation of the above (i), by applyingφ(y)=φ(x)=ε, we can obtain the same number of corollaries.


## 3. Extension to Banach Algebra

All obtained results can be extended to the stability on the Banach algebras. We will illustrate only for the case of Theorem2.1 among them.Theorem 3.1.
Let(E,∥·∥) be a semisimple commutative Banach algebra. Assume that  f,g,h:G→E satisfy the inequality
(3.1)‖f(x+y2)2-g(x)h(y)‖≤ε
for all x,y∈G.
Then, for an arbitrary linear multiplicative functionalx*∈E*, either there exist C1,C2,C3>0 such that
(3.2)|(x*∘g)(x)|≤C1,|(x*∘h)(x)|≤C2,|(x*∘f)(x)|≤C3
for all x∈G, or else each function g and h satisfies (L). Here g and h are represented by
(3.3)g(x)=g(0)e(x),h(x)=h(0)e(x),(3.4)|f(x)2-g(0)h(0)⋅e(x)2|≤ε,
where e(x) is an exponential function.Proof.
Assume that (3.1) holds, and fix arbitrarily a linear multiplicative functional x*∈E. As well known, we have ∥x*∥=1, hence, for every x,y∈G, we have
(3.5)ε≥‖f(x+y2)2-g(x)h(y)‖=sup⁡∥y*∥=1|y*(f(x+y2)2-g(x)h(y))|≥|x*(f(x+y2))-x*(g(x))x*(h(y))|,
which states that the superpositions x*∘f, x*∘g, and x*∘h satisfy the inequality (2.1) of Theorem 2.1. Due to the same processing as from (2.5) to (2.7), to fix arbitrarily a linear multiplicative functional x*∈E, indeed, we have
(3.6)|(x*∘g)(x)(x*∘h)(y)-(x*∘g)(y)(x*∘h)(x)|≤2ε∀x,y∈G.
It follows from the inequality (3.6) that there exist constants c1,c2,d1,d2≥0 such that
(3.7)|(x*∘g)(x)|≤c1|(x*∘h)(x)|+d1,|(x*∘h)(x)|≤c2|(x*∘g)(x)|+d2
for all x∈G. Since x* is an arbitrarily linear multiplicative functional, it follows from (3.7) that g is bounded if and only if h is bounded. Assume that one of g or h is bounded. From (3.1), we arrive at (3.2).
By the assumption (3.2), an appeal to Theorem 2.1 shows that
(3.8)(x*∘g)(x)=(x*∘g(0)e1)(x),(3.9)(x*∘h)(x)=(x*∘h(0)e2)(x),(3.10)|(x*∘f)(x)-(x*∘g(0)h(0)e3)(x)|≤ε,
where e1,e2,e3:G→ℝ are exponentials. In other words, bearing the linear multiplicativity of x* in mind, for all x∈G, each difference derived from (3.8) and (3.9)
(3.11)D(3.8)(x)∶=g(x)-(g(0)e1)(x),D(3.9)(x)∶=h(x)-(h(0)e2)(x),
falls into the kernel of x*. Therefore, in view of the unrestricted choice of x*, we infer that
(3.12)D(3.8)(x),D(3.9)(x)∈⋂{ker⁡x*:x*isamultiplicativememberofE*}
for all x∈G. Since the algebra E has been assumed to be semisimple, the last term of the previous formula coincides with the singleton {0}, that is,
(3.13)g(x)-g(0)e1(x)=0,h(x)-h(0)e2(x)=0,x∈G.
Putting (3.13) in (3.6), following the same proceeding as after (2.13) in Theorem 2.1, then we arrive that e1(x)=e2(x). Indeed, we have
(3.14)|(x*∘g(0)e1)(x)(x*∘h(0)e2)(y)-(x*∘g(0)e1)(y)(x*∘h(0)e2)(x)|≤2ε
for all x,y∈G. This implies that
(3.15)|(x*∘e1)(x)(x*∘e2)(y)-(x*∘e1)(y)(x*∘e2)(x)|≤2εg(0)h(0)=M∀x,y∈G.
Lettingx=0 in (3.15), it implies |(x*∘e2)(y)-(x*∘e1)(y)|≤M/x*(1)=M′ for all y∈G. Thus, from this and (3.15), we have
(3.16)|(x*∘e1)(y)||(x*∘e1)(x)-(x*∘e2)(x)|=|(x*∘e1)(x)[(x*∘e1)(y)-(x*∘e2)(y)]+(x*∘e1)(x)(x*∘e2)(y)-(x*∘e1)(y)(x*∘e2)(x)|≤|(x*∘e1)(x)|M′+M,
which is
(3.17)|(x*∘e1)(x)-(x*∘e2)(x)|≤|(x*∘e1)(x)|M′+M|(x*∘e1)(y)|,
for all x,y∈G.
Sincex*∘g is unbounded from (3.2), we can choose (yn)∈G so that |(x*∘g)(yn)|=|g(0)(x*∘e1)(yn)|→∞ as n→∞. Letting y=yn in (3.17), which arrive that
(3.18)(x*∘e1)(x)=(x*∘e2)(x).
Using the same logic as before, that is, bearing the linear multiplicativity ofx* in mind, the difference derived from (3.18), D(3.18)(x):=e1(x)-e2(x), falls into the kernel of x*. Then, the semisimplicity of E implies that e1(x)=e2(x). Let it be denoted by e(x), which arrive the claimed (3.3) and (3.4).
Sincee(x):G→ℝ is exponential, it is immediate from (3.3) that each function g and h satisfies (L).Remark 3.2.
All results of Section2 containing Remark 2.7 can be extended to the Banach space as Theorem 3.1.



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