Stability of the Drygas Functional Equation on ... - Springer Link

Results. Math. 68 (2015), 11–24 c 2014 The Author(s). This article is published with open access at Springerlink.com 1422-6383/15/010011-14 published online October 24, 2014 DOI 10.1007/s00025-014-0418-y

Results in Mathematics

Stability of the Drygas Functional Equation on Restricted Domain Magdalena Piszczek and Joanna Szczawi´ nska Abstract. We study the stability of the Drygas functional equation on a restricted domain. The main tool used in the proofs is the fixed point theorem for functional spaces. Mathematics Subject Classification. 39B82, 47H10, 47J20. Keywords. Drygas equation, stability, fixed point theorem.

1. Introduction We say that a function f : R → R satisfies the Drygas equation iff f (x + y) + f (x − y) = 2f (x) + f (y) + f (−y),

x, y ∈ R.

(1)

The above equation was introduced in [3] in order to obtain a characterization of the quasi-inner-product spaces. Ebanks, Kannappan and Sahoo in [4] have obtained the general solution of the Eq. (1) as f (x) = a(x) + q(x),

x ∈ R,

where a : R → R is an additive function and q : R → R is a quadratic function, i.e. q satisfies the quadratic functional equation q(x + y) + q(x − y) = 2q(x) + 2q(y),

x, y ∈ R.

A set-valued version of Eq. (1) was considered by Smajdor in [8]. The stability in the Hyers–Ulam sense of the Drygas equation has been investigated by Jung and Sahoo in [6]. They have proved that if a function f : X → Y , where X is a real vector space and Y is a Banach space satisfies the inequality f (x + y) + f (x − y) − 2f (x) − f (y) − f (−y) ≤ ε,

x, y ∈ X

(2)

12

M. Piszczek and J. Szczawi´ nska

Results. Math.

for same ε > 0, then there exists a unique additive function a : X → Y and and a unique quadratic function q : X → Y such that 25 ε, x ∈ X. 3 Their result was improved first by Yang in [9] and later by Sikorska in [7]. In the case when X is an Abelian group they obtained sharper bounds: 32 ε and ε respectively instead of 25 3 ε (cf. Proposition 1 in [9] and Theorem 3.2 in [7]). The stability and solution of the Drygas equation under some additional conditions was also studied by Forti and Sikorska in [5] in the case when X and Y are amenable groups. In the paper we present the stability results for the Drygas equation on restricted domain. Let X be a nonempty subset of a normed space and Y be a normed space. We say that a function f : X → Y satisfies the Drygas functional equation on X if f (x) − a(x) − q(x) ≤

f (x+y)+f (x−y) = 2f (x)+f (y)+f (−y),

x, y ∈ X, x+y, x−y ∈ X.

(3)

One of the method of the proof is based on a fixed point result that can be derived from [1] (Theorem 1). To present it we need the following three hypothesis: (H1) X is a nonempty set, Y is a Banach space, f1 , . . . , fk : X → X and L1 , . . . , Lk : X → R+ are given. (H2) T : Y X → Y X is an operator satisfying the inequality T ξ(x) − T μ(x) ≤

k

Li (x)ξ(fi (x)) − μ(fi (x))

i=1

for all ξ, μ ∈ Y X , x ∈ X. (H3) Λ : R+ X → R+ X is defined by Λδ(x) :=

k

Li (x)δ(fi (x)),

δ ∈ R+ X , x ∈ X.

i=1

Now we are in a position to present the above mentioned fixed point theorem. Theorem 1. Let hypotheses (H1)–(H3) be valid and functions ε : X → R+ and ϕ : X → Y fulfil the following two conditions T ϕ(x) − ϕ(x) ≤ ε(x), ∞ ε∗ (x) := Λn ε(x) < ∞,

x ∈ X, x ∈ X.

n=0

Then there exists a unique fixed point ψ of T with ϕ(x) − ψ(x) ≤ ε∗ (x),

x ∈ X.

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Stability of the Drygas Functional Equation on Restricted Domain

Moreover, ψ(x) := lim T n ϕ(x), n→∞

x ∈ X.

Throughout the paper N0 denotes the set of all non-negative integers.

2. Stability Results Theorem 2. Let X be a subset with 0 of a normed space, Y be a Banach space and c ≥ 0. Assume that p > 0 and a function f : X → Y satisfies f (x + y) + f (x − y) − 2f (x) − f (y) − f (−y) ≤ c(xp + yp )

(4)

for all x, y ∈ X such that x + y, x − y ∈ X. (1) If p > 2 and −x, x2 ∈ X for all x ∈ X, then there exists a function g : X → Y satisfying the Drygas equation on X such that 2c xp , x ∈ X. f (x) − g(x) ≤ p 2 −4 (2) If 0 < p < 1 and −x, 2x ∈ X for all x ∈ X, then there exists a function g : X → Y satisfying the Drygas equation on X such that 2c xp , x ∈ X. f (x) − g(x) ≤ 2 − 2p (3) If 1 < p < 2 and −x, 12 x, 2x ∈ X for all x ∈ X, then there exists a function g : X → Y satisfying the Drygas equation on X such that 2c 2c + x ∈ X. f (x) − g(x) ≤ xp , 4 − 2p 2p − 2 Moreover, g is the unique solution of the Eq. (3) such that f (x) − g(x) ≤ M xp for all x ∈ X and some M ≥ 0. Proof. First observe that the inequality (4) clearly forces f (0) = 0. (1) Replacing x and y by x2 in (4) we obtain x x 2c −f − x ∈ X. f (x) − 3f ≤ xp , 2 2 2p Consider functions T : Y X → Y X and ε : X → R+ given as follows x x +ξ − , x ∈ X, ξ ∈ Y X T ξ(x) = 3ξ 2 2 and 2c x ∈ X. ε(x) = p xp , 2 The inequality (5) now becomes T f (x) − f (x) ≤ ε(x),

x ∈ X.

(5)

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Results. Math.

For every ξ, μ ∈ Y X and x ∈ X x x x x −μ −μ − T ξ(x) − T μ(x) ≤ 3ξ + ξ − , 2 2 2 2 so T satisfies the inequality (H2) with f1 (x) = x2 , f2 (x) = − x2 , L1 (x) = 3, X L2 (x) = 1, x ∈ X. By (H3), the operator Λ : RX + → R+ is given by x x Λη(x) = 3η +η − , x ∈ X, η ∈ RX +. 2 2 In particular Λε(x) = 4ε

x 2

=

4 ε(x), 2p

x ∈ X.

Since Λ is linear, we can prove by induction n 4 n ε(x), x ∈ X, n ∈ N0 . Λ ε(x) = 2p As p > 2 we have

4 2p

< 1. Consequently the series

every x ∈ X and ∗

ε (x) =

∞ n=0

n

Λ ε(x) =

∞ n=0

4 2p

n ε(x) =

∞

n=0

Λn ε(x) is convergent for

2c 2p ε(x) = p xp , −4 2 −4

2p

x ∈ X.

By Theorem 1, there exists a function g : X → Y such g(x) = lim T n f (x), x ∈ X, n→∞ x x g(x) = 3g +g − , x∈X 2 2 and f (x) − g(x) ≤

2c xp , −4

2p

x ∈ X.

Next we prove that g satisfies the Drygas equation. Observe first that if a function h : X → Y satisfies the inequality h(x + y) + h(x − y) − 2h(x) − h(y) − h(−y) ≤ M (xp + yp )

(6)

for all x, y ∈ X such that x + y, x − y ∈ X and some M > 0, then T h(x + y) + T h(x − y) − 2T h(x) − T h(y) − T h(−y) ≤

4M (xp + yp ), 2p

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15

for x, y ∈ X satisfying x + y, x − y ∈ X. Indeed, fix h : X → Y and assume (6). Then T h(x + y) + T h(x − y) − 2T h(x) − T h(y) − T h(−y) x−y x y y x+y +h − 2h −h −h − =3 h 2 2 2 2 2 x+y y x−y x y + h − +h − −2h − −h − −h 2 2 2 2 2 for all x, y ∈ X, x + y, x − y ∈ X. Hence T h(x + y) + T h(x − y) − 2T h(x) − T h(y) − T h(−y) x+y x−y x y y ≤ 3h +h − 2h −h −h − 2 2 2 2 2 x+y y x−y x y + h − +h − −2h − −h − −h 2 2 2 2 2 p p x y x p y p ≤ 3M + + M − + − 2 2 2 2 4M (xp + yp ). 2p Consequently, proceeding by induction we get that if a function h : X → Y satisfies the inequality (6), then =

T n h(x + y) + T n h(x − y) − 2T n h(x) − T n h(y) − T n h(−y) n 4 (xp + yp ) ≤M 2p for all n ∈ N0 and x, y ∈ X, x + y, x − y ∈ X. On account of the above observation and (4) T n f (x + y) + T n f (x − y) − 2T n f (x) − T n f (y) − T n f (−y) n 4 (xp + yp ) ≤c p 2 for every n ∈ N0 and x, y ∈ X such that x + y, x − y ∈ X. Letting n → ∞ we get g(x + y) + g(x − y) = 2g(x) + g(y) + g(−y),

x, y ∈ X, x + y, x − y ∈ X.

(2) The idea of the proof is the same as before so we only give a sketch. Replacing y by x in (4) we obtain 2c 1 1 x ∈ X. (7) f (2x) − f (−x) − f (x) ≤ xp , 3 3 3

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Results. Math.

Let functions T : Y X → Y X and ε : X → R+ be define by formulas 1 1 x ∈ X, ξ ∈ Y X T ξ(x) = ξ(2x) − ξ(−x), 3 3 and 2c ε(x) = xp , x ∈ X. 3 The inequality (7) takes now the form T f (x) − f (x) ≤ ε(x),

x ∈ X.

Obviously T satisfies the inequality (H2) with f1 (x) = 2x, f2 (x) = −x, L1 (x) = X L2 (x) = 13 , x ∈ X. The operator Λ : RX + → R+ is given by Λη(x) =

1 1 η(2x) + η(−x), 3 3

x ∈ X, η ∈ RX +.

In particular 1 1 2p + 1 ε(2x) + ε(−x) = ε(x), x ∈ X. 3 3 3 Proceeding by induction, we obtain n p + 1 2 Λn ε(x) = ε(x), x ∈ X, n ∈ N0 . 3 Λε(x) =

Since p < 1, 2

p

+1 3

< 1, so the series

and ε∗ (x) =

∞

Λn ε(x) =

n=0

∞

n=0

Λn ε(x) is convergent for every x ∈ X

3 2c ε(x) = xp , 2 − 2p 2 − 2p

x ∈ X.

By Theorem 1, there exists a function g : X → Y such g(x) = lim T n f (x), n→∞

g(x) =

1 1 g(2x) − g(−x), 3 3

x ∈ X, x∈X

and f (x) − g(x) ≤

2c xp , 2 − 2p

x ∈ X.

A trivial verification shows that T n f (x + y) + T n f (x − y) − 2T n f (x) − T n f (y) − T n f (−y) n 2p + 1 c(xp + yp ), ≤ 3 for every n ∈ N0 and x, y ∈ X satisfying x + y, x − y ∈ X. Hence, letting n → ∞ we obtain g(x + y) + g(x − y) = 2g(x) + g(y) + g(−y),

x, y ∈ X, x + y, x − y ∈ X.

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17

(3) In this case let fe : X → Y and fo : X → Y be the even and the odd (−x) part of the function f , respectively. That means fe (x) = f (x)+f , fo (x) = 2 f (x)−f (−x) for x ∈ X and f = fe + fo . It is easy to see that f (0) = fe (0) = 2 fo (0) = 0. It follows that fe (x + y) + fe (x − y) − 2fe (x) − fe (y) − fe (−y) 1 = f (x + y) + f (−x − y) + f (x − y) + f (−x + y) 2 − 2f (x) − 2f (−x) − f (y) − f (−y) − f (−y) − f (y) 1 ≤ (f (x + y) + f (x − y) − 2f (x) − f (y) − f (−y) 2 + f (−x − y) + f (−x + y) − 2f (−x) − f (−y) − f (y)) ≤ c(xp + yp ) and analogously fo (x + y) + fo (x − y) − 2fo (x) − fo (y) − fo (−y) ≤ c(xp + yp ) for every x, y ∈ X such that x + y, x − y ∈ X. Hence fe , fo satisfy the inequality (4). Replace y by x in (4). By the evenness of fe , fe (2x) − 4fe (x) ≤ 2cxp , which gives

fe (2x) 1 fe (x) − ≤ cxp , 4 2

x∈X

x ∈ X.

Let ξ(2x) , 4 δ(2x) , Λδ(x) = 4

T ξ(x) =

ξ ∈ Y X , x ∈ X, δ ∈ RX +, x ∈ X

and ε(x) = 12 cxp , x ∈ X. By Theorem 1, there exists a function ge : X → Y such that ge (x) = lim T n fe (x), n→∞

ge (x) =

ge (2x) , 4

x ∈ X,

x∈X

and fe (x) − ge (x) ≤

2c xp , 4 − 2p

x ∈ X.

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Results. Math.

Moreover, T n fe (x + y) + T n fe (x − y) − 2T n fe (x) − T n fe (y) − T n fe (−y) n 2p c(xp + yp ), ≤ 4 for every n ∈ N0 and x, y ∈ X satisfying x + y, x − y ∈ X. Hence ge satisfies the Drygas equation. In the same way, replacing y by x in (4) and using the oddness of fo , we obtain fo (2x) − 2fo (x) ≤ 2cxp , which with x replacing by x2 yields 2 x fo (x) − 2fo ≤ p cxp , 2 2 Define now

and ε(x) = such that

x T ξ(x) = 2ξ , 2 x , Λδ(x) = 2δ 2 2 p 2p cx , x

x∈X

x ∈ X.

ξ ∈ Y X , x ∈ X, δ ∈ RX +, x ∈ X

∈ X. By Theorem 1, there exists a function go : X → Y go (x) = lim T n fo (x), x ∈ X, n→∞ x go (x) = 2go , x∈X 2

and fo (x) − go (x) ≤

2c xp , 2p − 2

x ∈ X.

By T n fo (x + y) + T n fo (x − y) − 2T n fo (x) − T n fo (y) − T n fo (−y) n 2 c(xp + yp ), n ∈ N0 , x, y ∈ X, x + y, x − y ∈ X, ≤ 2p go satisfies the Drygas equation. Thus g = ge + go also satisfies the Drygas equation and f (x) − g(x) ≤

2c 2c xp , xp + p 4 − 2p 2 −2

x ∈ X.

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19

It remains to prove the uniqueness of the function g. We show the case p > 2 in details. Let us assume that functions g1 , g2 : X → Y fulfill the Drygas equation on X and f (x) − gi (x) ≤ Mi xp ,

x∈X

for some Mi ≥ 0, i = 1, 2. Hence g1 (x) − g2 (x) ≤ (M1 + M2 )xp , x ∈ X. Since g1 , g2 satisfy the Drygas equation, x x gi (x) = 3gi + gi − , x ∈ X, i = 1, 2. 2 2 Thus

x x x x g1 (x) − g2 (x) ≤ 3g1 − g2 − g2 − + g1 − 2 2 2 2

4 (M1 + M2 )xp 2p for all x ∈ X. It is easy to check that, n 4 g1 (x) − g2 (x) ≤ (M1 + M2 )xp , 2p ≤

x ∈ X, n ∈ N0 .

Letting n to ∞ we obtain g1 (x) = g2 (x),

x ∈ X.

The proofs of the other cases runs as before.

The following examples show that the assumptions putting on the set X can not be omitted. Example 3. Let p > 2, X = (−∞, −1] ∪ {0} ∪ [1, ∞) and f (x) = |x|, x ∈ X. Then |f (x + y) + f (x − y) − 2f (x) − f (y) − f (−y)| ≤ 2(|x|p + |y|p ) for all x, y ∈ X such that x + y, x − y ∈ X. Consider functions ga : X → R given by ga (x) = ax2 , x ∈ X, where a is any real constant. The functions ga obviously satisfy the Drygas equation and |f (x) − ga (x)| ≤ |x|p ,

x∈X

for all a ∈ [0, 1]. Example 4. Let p ∈ (0, 1), X = [−1, 1] and f (x) = x3 , x ∈ X. Then for all x, y ∈ X such that x + y, x − y ∈ X |f (x + y) + f (x − y) − 2f (x) − f (y) − f (−y)| ≤ 6(|x|p + |y|p ). Every function ga : X → R given by ga (x) = ax, x ∈ X with a ∈ R satisfies the Drygas equation and |f (x) − ga (x)| ≤ |x|p , for all a ∈ [0, 1].

x∈X

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Results. Math.

Example 5. Let 1 < p < 2, X = (−∞, −1] ∪ {0} ∪ [1, ∞) and f (x) = 12 (|x| + x), x ∈ X. Then |f (x + y) + f (x − y) − 2f (x) − f (y) − f (−y)| ≤ |x|p + |y|p for all x, y ∈ X such that x + y, x − y ∈ X. Functions ga : X → R given by ga (x) = ax, x ∈ X are solutions of the Drygas equation and |f (x) − ga (x)| ≤ |x|p ,

x∈X

for all a ∈ [0, 1]. By the same method, we can also obtain the stability result for p = 0, but in order to obtain the best known bound we have to make more technical substitutions. The idea is adapted from [7]. Theorem 6. Let X be such a subset of an Abelian group that 0, −x, 2x, 3x ∈ X for all x ∈ X, Y a Banach space and c ≥ 0. If a function f : X → Y satisfies f (x + y) + f (x − y) − 2f (x) − f (y) − f (−y) ≤ c

(8)

for all x, y ∈ X with x + y, x − y ∈ X, then there exists a function g : X → Y satisfying the Drygas equation on X such that f (x) − g(x) ≤ c,

x ∈ X.

Moreover, g is the unique function satisfying equation (3), such that f (x) − g(x) ≤ M, x ∈ X for some M ≥ 0. Proof. Replace (x, y) in (8) by (2x, x), next by (x, 2x), (−x, −2x) and (x, x) (cf. the proof of Theorem 3.2 in [7]). Then f (3x) − 2f (2x) − f (−x) ≤ c, f (3x) + f (−x) − 2f (x) − f (2x) − f (−2x) ≤ c, − f (−3x) − f (x) + 2f (−x) + f (−2x) + f (2x) ≤ c, f (2x) + f (0) − 3f (x) − f (−x) ≤ c, for x ∈ X. Which with f (0) ≤

c 2

give

2f (3x) − f (−3x) − 9f (x) ≤ 6c, whence

2 2 1 f (x) − f (3x) + f (−3x) ≤ c, 3 9 9

x ∈ X, x ∈ X.

Let functions T : Y X → Y X and ε : X → R+ be defined as follows T ξ(x) =

1 2 ξ(3x) − ξ(−3x), 9 9

x ∈ X, ξ ∈ Y X

and ε(x) =

2 c, 3

x ∈ X.

(9)

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The inequality (9) now takes the form T f (x) − f (x) ≤ ε(x),

x ∈ X.

As before, using Theorem 1, there exists a function g : X → Y such that g(x) = lim T n (x), n→∞

g(x) =

x ∈ X,

2 1 g(3x) − g(−3x), 9 9

x∈X

and f (x) − g(x) ≤ c,

x ∈ X.

In the same manner as in the proofs of Theorem 2 we show that g satisfies the Drygas equation and g is unique.

3. Nonstability Results In this section we show that for p ∈ {1, 2} the Drygas equation is not stable. The idea of the construction of the examples comes from the paper [2]. Example 7. Let φ : R → R be defined as ⎧ x ≤ −1, ⎨ −α, −1 < x < 1, φ(x) = αx, ⎩ α, 1 ≤ x, where α > 0. The function f : R → R given by f (x) =

∞ φ(2n x) , 2n n=0

x∈R

satisfies |f (x + y) + f (x − y) − 2f (x) − f (y) − f (−y)| ≤ 8α(|x| + |y|),

(10)

but there exist no pair (g, k) of a function g : R → R satisfying the Drygas equation and a constant k ≥ 0 such that |f (x) − g(x)| ≤ k|x|,

x ∈ R.

Proof. We observe that f is odd and bounded by 2α. Now, we show that (10) holds. For x = y = 0 and x, y ∈ R such that |x|+|y| ≥ 1 it is obvious. Consider the case 0 < |x| + |y| < 1. There exists N ∈ N such that 1 1 ≤ |x| + |y| < N −1 . 2N 2 Then |2N −1 x| < 1, |2N −1 y| < 1, |2N −1 (x + y)| < 1, |2N −1 (x − y)| < 1. Hence 2n x, 2n y, 2n (x + y), 2n (x − y) ∈ (−1, 1)

for n = 0, 1, . . . , N − 1.

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Results. Math.

By the definition of f , |f (x + y) + f (x − y) − 2f (x) − f (y) − f (−y)| ∞ ∞ ∞ φ(2n (x + y)) φ(2n (x − y)) φ(2n x) = + −2 2n 2n 2n n=0 n=0 n=0 ≤ 4α

∞ 1 1 = 8α N n 2 2

n=N

≤ 8α(|x| + |y|). Assume that there exist a function g : R → R satisfying the Drygas equation and a constant k ≥ 0 such that |f (x) − g(x)| ≤ k|x|,

x ∈ R.

Since g fulfills the Drygas equation, there exist an additive function h : R → R and a quadratic function q : R → R such that g(x) = h(x) + q(x), x ∈ R. Whence, as f is bounded by 2α, we have |h(x) + q(x)| ≤ k|x| + 2α,

x ∈ R.

In particular, |h(nx) + q(nx)| ≤ k|nx| + 2α,

x ∈ R, n ∈ N.

The function q satisfies the quadratic functional equation, which implies 1 x ∈ R, n ∈ N, |h(x) + nq(x)| ≤ k|x| + 2α, n Hence q(x) = 0, x ∈ R and |h(x)| ≤ k|x| + 2α,

x ∈ R.

It follows that, the additive function h is bounded in the neighborhood of 0, and consequently h(x) = ax, x ∈ R for some constant a ∈ R. Thus |f (x) − ax| ≤ k|x|, which gives

f (x) ≤ k + |a|, x

x∈R x ∈ R \ {0}.

(11)

Let N be such that N α > k + |a| and take an x ∈ (0, 2N1−1 ). Then 2n x ∈ (0, 1) for n = 0, 1, . . . , N − 1 and f (x) =

N −1 n=0

∞ φ(2n x) φ(2n x) + > N xα 2n 2n n=N

so f (x) > N α > k + |a|, x which is contrary to (11).

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23

For p = 2 we have the same example like in the case of the quadratic equation (see [2]). Example 8. Let φ : R → R be defined as α, x ∈ (−∞, −1] ∪ [1, +∞), φ(x) = αx2 , x ∈ (−1, 1), where α > 0. Put f (x) =

∞ φ(2n x) , 4n n=0

x ∈ R.

Then f satisfies |f (x + y) + f (x − y) − 2f (x) − f (y) − f (−y)| ≤ 32α(|x|2 + |y|2 ) and there do not exist a function g : R → R satisfying the Drygas equation and a constant k ≥ 0 such that |f (x) − g(x)| ≤ k|x|2 ,

x ∈ R.

Open Access. This article is distributed under the terms of the Creative Commons Attribution License which permits any use, distribution, and reproduction in any medium, provided the original author(s) and the source are credited.

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Magdalena Piszczek and Joanna Szczawi´ nska Institute of Mathematics Pedagogical University Podchor¸az˙ ych 2 30-084 Krak´ ow, Poland e-mail: [email protected]; [email protected] Received: July 1, 2013. Accepted: October 13, 2014.

Results. Math.