A pexider difference for a pexider functional equation

Abstract

We deal with a Pexider difference

$$f\left(2x+y\right)+f\left(2x-y\right)-g\left(x+y\right)-g\left(x-y\right)-2g\left(2x\right)+2g\left(x\right)$$

where f and g map be a given abelian group (G, +) into a sequentially complete Hausdorff topological vector space. We also investigate the Hyers-Ulam stability of the following Pexiderized functional equation

$$f\left(2x+y\right)+f\left(2x-y\right)=g\left(x+y\right)+g\left(x-y\right)+2g\left(2x\right)-2g\left(x\right)$$

in topological vector spaces.

Mathematics subject classification (2000): Primary 39B82; Secondary 34K20, 54A20.

1. Introduction and preliminaries

In 1940, Ulam [1] proposed the general stability problem: Let G₁ be a group, G₂ be a metric group with the metric d. Given ε > 0, does there exists δ > 0 such that if a function h: G₁→ G₂ satisfies the inequality

$$d\left(h\left(xy\right)-h\left(x\right)h\left(y\right)\right)<\delta ,\left(x,y\in G_1\right),$$

then there is a homomorphism H: G ₁ → G ₂ with

$$d\left(h\left(x\right),H\left(x\right)\right)<\epsilon ,\left(x\in G_1\right)?$$

Hyers [2] gave a partial affirmative answer to the question of Ulam in the context of Banach spaces. In 1950, Aoki [3] extended the theorem of Hyers by considering the unbounded cauchy difference inequality

$$∥f\left(x+y\right)-f\left(x\right)-f\left(y\right)∥\le \epsilon \left({∥x∥}^p+{∥y∥}^p\right)\left(\epsilon >0,p\epsilon \left[0,1\right)\right).$$

In 1978, Rassias [4] also generalized the Hyers' theorem for linear mappings under the assumption t ↦ f (tx) is continuous in t for each fixed x.

Recently, Adam and Czerwik [5] investigated the problem of the Hyers-Ulam stability of a generalized quadratic functional equation in linear topological spaces. Najati and Moghimi [6] investigated the Hyers-Ulam stability of the functional equation

$$f\left(2x+y\right)+f\left(2x-y\right)=f\left(x+y\right)+f\left(x-y\right)+2f\left(2x\right)-2f\left(x\right)$$

in quasi-Banach spaces. In this article, we prove that the Pexiderized functional equation

$$f\left(2x+y\right)+f\left(2x-y\right)=g\left(x+y\right)+g\left(x-y\right)+2g\left(2x\right)-2g\left(x\right)$$

is stable for functions f, g defined on an abelian group and taking values in a topological vector space.

Throughout this article, let G be an abelian group and X be a sequentially complete Hausdorff topological vector space over the field ℚ of rational numbers.

A mapping f: G → X is said to be quadratic if and only if it satisfies the following functional equation

$$f\left(x+y\right)+f\left(x-y\right)=2f\left(x\right)+2f\left(y\right)$$

for all x y ∈ G. A mapping f G → X is said to be additive if and only if it satisfies f (x + y) = f (x) + f (y) for all x y ε G. For a given f: G → X, we will use the following notation

$$Df\left(x,y\right):=f\left(2x+y\right)+f\left(2x-y\right)-f\left(x+y\right)-f\left(x-y\right)-2f\left(2x\right)+2f\left(x\right).$$

For given sets A B ⊆ X and a number k ∈ ℝ, we define the well known operations

$$A+B:=\left\{a+b:a\in A,b\in B\right\},kA:=\left\{ka:a\in A\right\}.$$

We denote the convex hull of a set U ⊆ X by conv(U) and by $\overline{U}$ the sequential closure of U. Moreover it is well know that:

• (1) If A ⊆ X are bounded sets, then conv(A) and $\overline{A}$ are bounded subsets of X.

• (2) If A, B ⊆ X and α β ∈ ℝ, then α conv(A) + β conv(B) = conv(αA + βB).

• (3) Let X₁ and X₂ be linear spaces over ℝ. If f: X₁→ X₂ is a additive (quadratic) function, then f (rx) = rf (x) (f (rx) = r²f (x)), for all x ∈ X₁ and all r ∈ ℚ.

2. Main results

We start with the following lemma.

Lemma 2.1. Let G be a 2-divisible abelian group and B ⊆ X be a nonempty set. If the functions f, g: G → X satisfy

$$f\left(2x+y\right)+f\left(2x-y\right)-g\left(x+y\right)-g\left(x-y\right)-2g\left(2x\right)+2g\left(x\right)\in B$$

(2.1)

for all x, y ∈ G, then

$$Df\left(x,y\right)\in 2\mathsf{\text{ conv}}\left(B-B\right),$$

(2.2)

$$Dg\left(x,y\right)\in \mathsf{\text{conv}}\left(B-B\right)$$

(2.3)

for all x, y ∈ G.

Proof. Putting y = 0 in (2.1), we get

$$2f\left(2x\right)-2g\left(2x\right)\in B$$

(2.4)

for all x ∈ G. If we replace x by $\frac{1}{2}x$ in (2.4), then we have

$$f\left(x\right)-g\left(x\right)\in \frac{1}{2}B$$

(2.5)

for all x ∈ G. It follows from (2.5) and (2.1) that

$$\begin{array}{ll}\hfill Df\left(x,y\right)& =f\left(2x+y\right)+f\left(2x-y\right)-g\left(x+y\right)-g\left(x-y\right)-2g\left(2x\right)+2g\left(x\right)\\ -\left[f\left(x+y\right)-g\left(x+y\right)\right]-\left[f\left(x-y\right)-g\left(x-y\right)\right]\\ -\left[2f\left(2x\right)-2g\left(2x\right)\right]+\left[2f\left(x\right)-2g\left(x\right)\right]\\ \in 2\mathsf{\text{ conv}}\left(B-B\right).\end{array}$$

Moreover, we have

$$\begin{array}{ll}\hfill Dg\left(x,y\right)& =f\left(2x+y\right)+f\left(2x-y\right)-g\left(x+y\right)-g\left(x-y\right)-2g\left(2x\right)+2g\left(x\right)\\ -\left[f\left(2x+y\right)-g\left(2x+y\right)\right]-\left[f\left(2x-y\right)-g\left(2x-y\right)\right]\\ \in \mathsf{\text{conv}}\left(B-B\right).\end{array}$$

Theorem 2.2. Let G be a 2-divisible abelian group and B ⊆ X be a bounded set. Suppose that the odd functions f, g: G → X satisfy (2 1) for all x, y ∈ G. Then there exists exactly one additive function $\mathcal{A}:G\to X$ such that

$$\mathcal{A}\left(x\right)-f\left(x\right)\in 4\overline{\mathsf{\text{conv}}\left(B-B\right)}\mathsf{\text{,}}\mathcal{A}\left(x\right)-g\left(x\right)\in 2\overline{\mathsf{\text{conv}}\left(B-B\right)}$$

(2.6)

for all x ∈ G. Moreover the function $\mathcal{A}$ is given by

$$\mathcal{A}\left(x\right)=\underset{n\to \infty }{\text{lim}}\frac{1}{2^n}f\left(2^{n}x\right)=\underset{n\to \infty }{\text{lim}}\frac{1}{2^n}g\left(2^{n}x\right)$$

for all x ∈ G. Moreover, the convergence of the sequences are uniform on G.

Proof. By Lemma 2.1, we get (2.2). Setting y = x, y = 3x and y = 4x in (2.2), we get

$$f\left(3x\right)-3f\left(2x\right)+3f\left(x\right)\in 2\mathsf{\text{ conv}}\left(B-B\right),$$

(2.7)

$$f\left(5x\right)-f\left(4x\right)-f\left(2x\right)+f\left(x\right)\in 2\mathsf{\text{conv}}\left(B-B\right),$$

(2.8)

$$f\left(6x\right)-f\left(5x\right)+f\left(3x\right)-3f\left(2x\right)+2f\left(x\right)\in 2\mathsf{\text{conv}}\left(B-B\right)$$

(2.9)

for all x ∈ G. It follows from (2.7), (2.8), and (2.9) that

$$f\left(6x\right)-f\left(4x\right)-f\left(2x\right)\in 6\mathsf{\text{conv}}\left(B-B\right)$$

for all x ∈ G. So

$$f\left(3x\right)-f\left(2x\right)-f\left(x\right)\in 6\mathsf{\text{conv}}\left(B-B\right)$$

(2.10)

for all x ∈ G. Using (2.7) and (2.10), we obtain

$$\frac{\mathsf{\text{1}}}{\mathsf{\text{2}}}f\left(2x\right)-f\left(x\right)\in 2\mathsf{\text{conv}}\left(B-B\right)$$

for all x ∈ G. Therefore

$$\begin{array}{ll}\hfill \frac{1}{2^n}f\left(2^{n}x\right)-\frac{1}{2^m}f\left(2^{m}x\right)& =\sum _{k=m}^{n-1}\left[\frac{1}{2^{k+1}}f\left(2^{k+1}x\right)-\frac{1}{2^k}f\left(2^{k}x\right)\right]\\ \in \sum _{k=m}^{n-1}\frac{2}{2^k}\mathsf{\text{conv}}\left(B-B\right)\\ \subseteq \frac{4}{2^m}\mathsf{\text{conv}}\left(B-B\right)\end{array}$$

(2.11)

for all x ∈ G and all integers n > m ≥ 0. Since B is bounded, we conclude that conv(B - B) is bounded. It follows from (2.11) and boundedness of the set conv(B - B) that the sequence $\left\{\frac{1}{2^n}f\left(2^{n}x\right)\right\}$ is (uniformly) Cauchy in X for all x ∈ G. Since X is a sequential complete topological vector space, the sequence $\left\{\frac{1}{2^n}f\left(2^{n}x\right)\right\}$ is convergent for all x ∈ G, and the convergence is uniform on G. Define

$${\mathcal{A}}_1:G\to X,{\mathcal{A}}_1\left(x\right):=\underset{n\to \infty }{\text{lim}}\frac{1}{2^n}f\left(2^{n}x\right).$$

Since conv(B - B) is bounded, it follows from (2.2) that

$$D{\mathcal{A}}_1\left(x,y\right)=\underset{n\to \infty }{\text{lim}}\frac{1}{2^n}Df\left(2^{n}x,2^{n}y\right)=0$$

for all x y ∈ G. So ${\mathcal{A}}_1$ is additive (see [6]). Letting m = 0 and n →∞ in (2.11), we get

$${\mathcal{A}}_1\left(x\right)-f\left(x\right)\in 4\overline{\mathsf{\text{conv}}\left(B-B\right)}$$

(2.12)

for all x ∈ G. Similarly as before applying (2.3) we have an additive mapping ${\mathcal{A}}_2:G\to X$ defined by ${\mathcal{A}}_2\left(x\right):=\underset{n\to \infty }{\text{lim}}\frac{1}{2^n}g\left(2^{n}x\right)$ which is satisfying

$${\mathcal{A}}_2\left(x\right)-g\left(x\right)\in 2\overline{\mathsf{\text{conv}}\left(B-B\right)}$$

(2.13)

for all x ∈ G. Since B is bounded, it follows from (2.5) that ${\mathcal{A}}_1={\mathcal{A}}_2$. Letting $\mathcal{A}:={\mathcal{A}}_1$, we obtain (2.6) from (2.12) and (2.13).

To prove the uniqueness of $\mathcal{A}$, suppose that there exists another additive function ${\mathcal{A}}^{\prime }$: G → X satisfying (2.6). So

$${\mathcal{A}}^{\prime }\left(x\right)-\mathcal{A}\left(x\right)=\left[{\mathcal{A}}^{\prime }\left(x\right)-f\left(x\right)\right]+\left[f\left(x\right)-\mathcal{A}\left(x\right)\right]\in 8\overline{\mathsf{\text{conv}}\left(B-B\right)}$$

for all x ∈ G. Since ${\mathcal{A}}^{\prime }$ and $\mathcal{A}$ are additive, replacing x by 2ⁿx implies that

$${\mathcal{A}}^{\prime }\left(x\right)-\mathcal{A}\left(x\right)\in \frac{8}{2^n}\overline{\mathsf{\text{conv}}\left(B-B\right)}$$

for all x ∈ G and all integers n. Since $\overline{\mathsf{\text{conv}}\left(B-B\right)}$ is bounded, we infer ${\mathcal{A}}^{\prime }=\mathcal{A}$. This completes the proof of theorem.

Theorem 2.3 Let G be a 2, 3-divisible abelian group and B ⊆ X be a bounded set. Suppose that the even functions f, g: G → X satisfy (2 1) for all x, y ∈ G. Then there exists exactly one quadratic function $\mathcal{Q}:G\to X$such that

$$\mathcal{Q}\left(x\right)-f\left(x\right)+f\left(0\right)\in 4\overline{\mathsf{\text{conv}}\left(B-B\right)},\mathcal{Q}\left(x\right)-g\left(x\right)+g\left(0\right)\in 2\overline{\mathsf{\text{conv}}\left(B-B\right)}$$

for all x ∈ G. Moreover, the function $\mathcal{Q}$ is given by

$$\mathcal{Q}\left(x\right)=\underset{n\to \infty }{\text{lim}}\frac{1}{4^n}f\left(2^{n}x\right)=\underset{n\to \infty }{\text{lim}}\frac{1}{4^n}g\left(2^{n}x\right)$$

for all x ∈ G. Moreover, the convergence of the sequences are uniform on G.

Proof. By replacing y by x + y in (2.2), we get

$$\begin{array}{ll}\hfill f\left(3x+y\right)+f\left(x-y\right)& -f\left(2x+y\right)-f\left(y\right)\\ -2f\left(2x\right)+2f\left(x\right)\in 2conv\left(B-B\right)\end{array}$$

(2.14)

for all x, y ∈ G. Replacing y by - y in (2.14), we get

$$\begin{array}{ll}\hfill f\left(3x-y\right)+f\left(x+y\right)& -f\left(2x-y\right)-f\left(y\right)\\ -2f\left(2x\right)+2f\left(x\right)\in 2\mathsf{\text{conv}}\left(B-B\right)\end{array}$$

(2.15)

for all x, y ∈ G. It follows from (2.2), (2.14), and (2.15) that

$$\begin{array}{ll}\hfill f\left(3x+y\right)& +f\left(3x-y\right)-2f\left(y\right)\\ -6f\left(2x\right)+6f\left(x\right)\in 6\mathsf{\text{conv}}\left(B-B\right)\end{array}$$

(2.16)

for all x, y ∈ G. By letting y = 0 and y = 3x in (2.16), we get

$$2f\left(3x\right)-6f\left(2x\right)+6f\left(x\right)-2f\left(0\right)\in 6\mathsf{\text{conv}}\left(B-B\right),$$

(2.17)

$$f\left(6x\right)-2f\left(3x\right)-6f\left(2x\right)+6f\left(x\right)+f\left(0\right)\in 6\mathsf{\text{conv}}\left(B-B\right)$$

(2.18)

for all x ∈ G. Using (2.17) and (2.18), we obtain

$$f\left(6x\right)-4f\left(3x\right)+3f\left(0\right)\in 12\mathsf{\text{conv}}\left(B-B\right)$$

(2.19)

for all x ∈ G. If we replace x by $\frac{1}{3}x$ in (2.19), then

$$f\left(2x\right)-4f\left(x\right)+3f\left(0\right)\in 12\mathsf{\text{conv}}\left(B-B\right)$$

for all x ∈ G. Therefore

$$\frac{1}{4^{n+1}}f\left(2^{n+1}x\right)-\frac{1}{4^n}f\left(2^{n}x\right)+\frac{3}{4^{n+1}}f\left(0\right)\in \frac{3}{4^n}\mathsf{\text{conv}}\left(B-B\right)$$

(2.20)

for all x ∈ G and all integers n. So

$$\begin{array}{l}\frac{1}{4^n}f\left(2^{n}x\right)-\frac{1}{4^m}f\left(2^{m}x\right)\\ =\sum _{k=m}^{n-1}\frac{1}{4^{k+1}}f\left(2^{k+1}x\right)-\frac{1}{4^k}f\left(2^{k}x\right)\\ \in -\sum _{k=m}^{n-1}\frac{3}{4^{k+1}}f\left(0\right)+\sum _{k=m}^{n-1}\frac{3}{4^k}\mathsf{\text{conv}}\left(B-B\right)\\ \subseteq -\sum _{k=m}^{n-1}\frac{3}{4^{k+1}}f\left(0\right)+\frac{1}{4^{m-1}}\mathsf{\text{conv}}\left(B-B\right)\end{array}$$

(2.21)

for all x ∈ G and all integers n >m ≥ 0. It follows from (2.21) and boundedness of the set conv(B - B) that the sequence $\left\{\frac{1}{4^n}f\left(2^{n}x\right)\right\}$ is (uniformly) Cauchy in X for all x ∈ G. The rest of the proof is similar to proof of of Theorem 2.2.

Remark 2.4. If the functions f, g: G → X satisfy (2.1), where f is even (odd) and g is odd (even), then it is easy to show that f and g are bounded.