# Hyers-Ulam Stability of Polynomial Equations

Authors: M. Bidkham; H. A. Soleiman Mezerji; M. Eshaghi Gordji
Journal: Abstract and Applied Analysis
 (2010)

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## Abstract

We prove the Hyers-Ulam stability of the polynomial equationanxn+an−1xn−1+⋯+a1x+a0=0. We give an affirmative answer to a problem posed by Li and Hua (2009).


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## Body

## 1. Introduction and Preliminaries

A classical question in the theory of functional equations is that “when is it true that a function which approximately satisfies a functional equationℰ must be somehow close to an exact solution of ℰ''. Such a problem was formulated by Ulam [1] in 1940 and solved in the next year for the Cauchy functional equation by Hyers [2]. It gave rise to the stability theory for functional equations. The result of Hyers was generalized by Rassias [3]. The topic of the Hyers-Ulam stability of functional equations and its applications has been studied by a number of mathematicians; see [3–40] and references therein.Recently, Li and Hua [41] discussed and proved the Hyers-Ulam stability of the polynomial equation(1.1)xn+αx+β=0,
where x∈[-1,1] and proved the following.Theorem 1.1.
If|α|>n,|β|<|α|-1, and y∈[-1,1] satisfy the inequality
(1.2)|yn+αy+β|≤ε,
then there exists a solution v∈[-1,1] of (1.1) such that
(1.3)|y-v|≤Kε,
where K>0 is constant.They also asked an open problem whether the real polynomial equation(1.4)anxn+an-1xn-1+⋯+a1x+a0=0
has the Hyers-Ulam stability for the case that this real polynomial equation has some solutions in [a,b]. The aim of this paper is to give a positive answer to this problem. First of all, we give the definition of the Hyers-Ulam stability.Definition 1.2.
One says that (1.4) has the Hyers-Ulam stability if there exists a constant K>0 with the following property:
for everyε>0,y∈[-1,1], if
(1.5)|anyn+an-1yn-1+⋯+a1y+a0|≤ε
then there exists some z∈[-1,1] satisfying
(1.6)anzn+an-1zn-1+⋯+a1z+a0=0
such that |y-z|≤Kε. One calls such K a Hyers-Ulam stability constant for (1.4). For the complex polynomial equation, [-1,1] is replaced by closed unit disc
(1.7)D={z∈ℂ;|z|≤1}.


## 2. Main Results

The aim of this work is to investigate the Hyers-Ulam stability for (1.4).Theorem 2.1.
If(2.1)|a0|<|a1|-(|a2|+|a3|+⋯+|an|),(2.2)|a1|>2|a2|+3|a3|+⋯+(n-1)|an-1|+n|an|,
then there exists an exact solution v∈[-1,1] of (1.4).Proof.
If we set(2.3)g(x)=1a1(-a0-a2x2-a3x3-⋯-an-1xn-1-anxn),
for x∈[-1,1], then we have
(2.4)|g(x)|=1|a1||-a0-a2x2-⋯-an-1xn-1-anxn|≤1|a1|(|a0|+|a2|+⋯+|an-1|+|an|)≤1
by (2.1).
LetX=[-1,1] and d(x,y)=|x-y|. Then (X,d) is a complete metric space and g maps X to X. Now, we will show that g is a contraction from X to X. For any x,y∈X, we have
(2.5)d(g(x),g(y))=|1a1(-a0-a2x2-⋯-anxn)-1a1(-a0-⋯-anyn)|≤1|a1||x-y|{|a2||x+y|+⋯+|an||xn-1+⋯+yn-1|}≤1|a1||x-y|{2|a2|+3|a3|+⋯+(n-1)|an-1|+n|an|}.
For x,y∈[-1,1],x≠y, from (2.2), we obtain
(2.6)d(g(x),g(y))≤λd(x,y).
Here
(2.7)λ=2|a2|+3|a3|+⋯+(n-1)|an-1|+n|an||a1|<1.
Thus g is a contraction from X to X. By the Banach contraction mapping theorem, there exists a unique v∈X such that
(2.8)g(v)=v.
Hence (1.4) has a solution on [-1,1].As an application of Rouche's theorem, we prove the following theorem for complex polynomial equation(2.9)anzn+an-1zn-1+⋯+a1z+a0=0,
which is much better than the above result. In fact, we prove the following theorem.Theorem 2.2.
If(2.10)|a0|<|a1|-(|a2|+|a3|+⋯+|an|),
then there exists an exact solution in open unit disc for (2.9).Proof.
If we set(2.11)g(z)=1a1(-a0-a2z2-a3z3-⋯-an-1zn-1-anzn),
then we have
(2.12)|g(z)|=1|a1||-a0-a2z2-⋯-an-1zn-1-anzn|≤1|a1|(|a0|+|a2|+⋯+|an-1|+|an|),for|z|≤1<1
by (2.10).
Since|g(z)|<1 for |z|=1, then |g(z)|<|-z|=1 and by Rouche's theorem, we observe that g(z)-z has exactly one zero in |z|<1 which implies that g has a unique fixed point in |z|<1.Theorem 2.3.
If the conditions of Theorem2.1 hold and y∈[-1,1] satisfies the inequality
(2.13)|anyn+an-1yn-1+⋯+a1y+a0|≤ε,
then (1.4) has the Hyers-Ulam stability.Proof.
Letε>0 and y∈[-1,1] such that
(2.14)|anyn+an-1yn-1+⋯+a1y+a0|≤ε.
We will show that there exists a constant K independent of ε and v such that
(2.15)|y-v|≤Kε
for some v∈[-1,1] satisfying (1.4).
Let us introduce the abbreviationK=1/|a1|(1-λ). Then
(2.16)|y-v|=|y-g(y)+g(y)-g(v)|≤|y-g(y)|+|g(y)-g(v)|≤|y-1a1(-a0-a2y2-⋯-anyn)|+λ|y-v|=1|a1||anyn+an-1yn-1+⋯+a1y+a0|+λ|y-v|.
Thus, we have
(2.17)|y-v|≤1|a1|(1-λ)|anyn+an-1yn-1+⋯+a1y+a0|≤Kε
by (2.13) and so the result follows.Corollary 2.4.
In Theorem2.2, if there exists y∈D satisfying the inequality (2.13), then (2.9) has the Hyers-Ulam stability.Remark 2.5.
Foran=1, ai=0, for 2≤i≤n-1, combining Theorems 2.1 and 2.3 gives Theorem 1.1.Remark 2.6.
By the similar way, one can easily prove the Hyers-Ulam stability of (1.4) on any finite interval [a,b].Remark 2.7.
Letf be any complex function such that f is analytic in
(2.18)Δ={z∈ℂ:|z|〈R,R〉0}.
It is an interesting open problem whether f has the Hyers-Ulam stability for the case that f has some zeros in Δ.We note that there is an error in the proof of Theorem2.2 of [41], when Li and Hua stated that if (X,d) is a complete metric linear space then metric d is invariant, more precisely (2.19)d(x,y)=d(x-y,0)
for all x,y∈X. We give a counterexample for this case. Suppose that X=ℝ, and we define metric d on X as follows: (2.20)d(x,y)=|x+[x]-(y+[y])|,
for all x,y∈X(X,d) is a complete metric linear space, and d is not an invariant metric on X, that is, there are x,y∈X such that (2.21)d(x,y)≠d(x-y,0).



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*Source: 754120-2010-10-26.xml*