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Volume 2009, Article ID 153084,26pages doi:10.1155/2009/153084

Research Article

On the Stability of a Generalized

Quadratic and Quartic Type Functional Equation in Quasi-Banach Spaces

M. Eshaghi Gordji,

1

S. Abbaszadeh,

1

and Choonkil Park

2

1Department of Mathematics, Semnan University, P.O. Box 35195-363, Semnan, Iran

2Department of Mathematics, Research Institute for Natural Sciences, Hanyang University, Seoul 133-791, South Korea

Correspondence should be addressed to Choonkil Park,baak@hanyang.ac.kr Received 31 May 2009; Accepted 9 September 2009

Recommended by Nikolaos Papageorgiou

We establish the general solution of the functional equationfnxy fnxy n2fxy n2fxy 2fnx−n2fx−2n2−1fyfor fixed integersnwithn /0,±1 and investigate the generalized Hyers-Ulam stability of this equation in quasi-Banach spaces.

Copyrightq2009 M. Eshaghi Gordji et al. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

1. Introduction

The stability problem of functional equations originated from a question of Ulam1in 1940, concerning the stability of group homomorphisms. LetG1,·be a group and let G2,∗be a metric group with the metricd·,·.Givenε > 0, does there exist a δ > 0, such that if a mappingh: G1G2 satisfies the inequalitydhx·y, hxhy < δfor allx, yG1, then there exists a homomorphismH :G1G2 withdhx, Hx< εfor allxG1? In other words, under what condition does there exist a homomorphism near an approximate homomorphism? The concept of stability for functional equation arises when we replace the functional equation by an inequality which acts as a perturbation of the equation. In 1941, Hyers 2gave a first affirmative answer to the question of Ulam for Banach spaces. Let f:EEbe a mapping between Banach spaces such that

f xy

fxf

yδ 1.1

for allx, yE,and for someδ >0.Then there exists a unique additive mappingT :EE such that

fxTx≤δ 1.2

(2)

for allxE.Moreover, ifftxis continuous int∈Rfor each fixedxE,thenT isR-linear.

In 1978, Th. M. Rassias3 provided a generalization of Hyers’ theorem which allows the Cauchy difference to be unbounded. The functional equation

f xy

f xy

2fx 2f y

1.3

is related to a symmetric biadditive mapping4–7. It is natural that this functional equation is called a quadratic functional equation. In particular, every solution of the quadratic equation 1.3is said to be a quadratic mapping. It is well known that a mappingfbetween real vector spaces is quadratic if and only if there exists a unique symmetric biadditive mappingBsuch thatfx Bx, xfor allxsee4,7. The biadditive mappingBis given by

B x, y

1 4

f xy

f xy

. 1.4

The generalized Hyers-Ulam stability problem for the quadratic functional equation1.3was proved by Skof for mappingsf:AB, whereAis a normed space andBis a Banach space see8. Cholewa9noticed that the theorem of Skof is still true if relevant domainAis replaced by an abelian group. In10, Czerwik proved the generalized Hyers-Ulam stability of the functional equation1.3. Grabiec11has generalized these results mentioned above.

In12, Park and Bae considered the following quartic functional equation:

f x2y

f x−2y

4 f

xy f

xy 6f

y

−6fx. 1.5

In fact, they proved that a mappingf between two real vector spacesX andY is a solution of1.5if and only if there exists a unique symmetric multiadditive mappingD:X×X×X× XY such thatfx Dx, x, x, xfor allx. It is easy to show thatfx x4satisfies the functional equation1.5, which is called a quartic functional equationsee also13.

In addition, Kim14has obtained the generalized Hyers-Ulam stability for a mixed type of quartic and quadratic functional equation between two real linear Banach spaces.

Najati and Zamani Eskandani15have established the general solution and the generalized Hyers-Ulam stability for a mixed type of cubic and additive functional equation, wheneverf is a mapping between two quasi-Banach spacessee also16,17.

Now we introduce the following functional equation for fixed integersnwithn /0,±1:

f nxy

f nxy

n2f xy

n2f xy

2fnx−2n2fx−2 n2−1

f y

1.6

in quasi-Banach spaces. It is easy to see that the function fx ax4 bx2 is a solution of the functional equation1.6. In the present paper we investigate the general solution of the functional equation1.6whenf is a mapping between vector spaces, and we establish the generalized Hyers-Ulam stability of this functional equation wheneverf is a mapping between two quasi-Banach spaces.

We recall some basic facts concerning quasi-Banach space and some preliminary results.

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Definition 1.1See18,19. LetXbe a real linear space. A quasinorm is a real-valued function onXsatisfying the following.

1x ≥0 for allxXandx0 if and only ifx0.

2λ·x|λ| · xfor allλ∈Rand allxX.

3There is a constantM≥1 such thatxy ≤Mxyfor allx, yX.

It follows from the condition3that

2m

i1

xi

Mm

2m i1

xi,

2m1

i1

xi

Mm1

2m1

i1

xi 1.7

for allm≥1 and allx1, x2, . . . , x2m1X.

The pairX,·is called a quasinormed space if·is a quasinorm onX. The smallest possibleMis called the modulus of concavity of · . A quasi-Banach space is a complete quasi-normed space.

A quasi-norm · is called ap-norm0< p≤1if

xypxpyp 1.8 for allx, yX. In this case, a quasi-Banach space is called ap-Banach space.

Given ap-norm, the formuladx, y:x−ypgives us a translation invariant metric on X. By the Aoki-Rolewicz theorem 19 see also 18, each quasi-norm is equivalent to some p-norm. Since it is much easier to work withp-norms, henceforth we restrict our attention mainly top-norms. In20, Tabor has investigated a version of Hyers-Rassias-Gajda theoremsee3,21in quasi-Banach spaces.

2. General Solution

Throughout this section, X and Y will be real vector spaces. We here present the general solution of1.6.

Lemma 2.1. If a mappingf :XY satisfies the functional equation1.6, thenfis a quadratic and quartic mapping.

Proof. Lettingxy0 in1.6, we getf0 0. Settingx0 in1.6, we getfy f−y for allyX. So the mappingfis even. Replacingxbyxyin1.6and thenxbyxyin 1.6, we get

f

nx n1y f

nx n−1y n2f

x2y

n2fx 2f

nxny

−2n2f xy

−2 n2−1

f y

, 2.1

f

nx−n−1y f

nx−n1y n2fx n2f

x−2y 2f

nxny

−2n2f xy

−2 n2−1

f

y 2.2

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for allx, yX. Interchangingxandyin1.6and using the evenness off, we get the relation

f xny

f xny n2f

xy n2f

xy 2f

ny

−2n2f y

−2 n2−1

fx 2.3

for allx, yX. Replacingybynyin1.6and then using2.3, we have

f

nxny f

nxny n4f

xy n4f

xy 2f

ny

2fnx−2n4fx−2n4f

y 2.4

for allx, yX. If we add2.1to2.2and use2.4, then we have

f

nx n1y f

nx−n1y f

nx n−1y f

nx−n−1y n2f

x2y n2f

x−2y 2n2

n2−1 f

xy 2n2

n2−1 f

xy 4f

ny

4fnx

−4n42n2

fx

−4n4−4n24 f

y

2.5

for allx, yX. Replacing ybyxy in1.6and then ybyxy in1.6and using the evenness off, we obtain that

f

n1xy f

n−1x−y n2f

2xy n2f

y

2fnx−2n2fx−2 n2−1

f xy

, 2.6

f

n1x−y f

n−1xy n2f

2x−y n2f

y

2fnx−2n2fx−2 n2−1

f

xy 2.7

for allx, yX. Interchangingxwithyin2.6and2.7and using the evenness off, we get the relations

f

x n1y f

x−n−1y n2f

x2y

n2fx 2f ny

−2n2f y

−2 n2−1

f xy

, 2.8

f

x−n1y f

x n−1y n2f

x−2y

n2fx 2f ny

−2n2f y

−2 n2−1

f

xy 2.9

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for allx, yX. Replacingybyn1yin1.6and thenybyn−1yin1.6, we have f

nx n1y f

nx−n1y n2f

x n1y n2f

x−n1y

2fnx−2n2fx−2 n2−1

f

n1y , 2.10 f

nx n−1y f

nx−n−1y n2f

x n−1y n2f

x−n−1y

2fnx−2n2fx−2 n2−1

f

n−1y 2.11

for allx, yX. Replacingxbyyin1.6, we obtain

f

n1y f

n−1y n2f

2y

−2

2n2−1 f

y 2f

ny

2.12

for allyX. Adding2.10to2.11and using2.8,2.9, and2.12, we get f

nx n1y f

nx−n1y f

nx n−1y f

nx−n−1y n4f

x2y n4f

x−2y

−2n2 n2−1

f xy

−2n2 n2−1

f xy 4f

ny

4fnx−2n2 n2−1

f 2y

2n4−4n2

fx

4n4−12n24 f

y 2.13

for allx, yX. By2.5and2.13, we obtain f

x2y f

x−2y 4f

xy 4f

xy 2f

2y

−8f y

−6fx 2.14

for allx, yX. Interchanging xand y in2.14and using the evenness of f, we get the relation

f 2xy

f 2x−y

4f xy

4f xy

2f2x−8fx−6f y

2.15

for allx, yX.

Now we show that2.15is a quadratic and quartic functional equation. To get this, we show that the mappingsg :XY, defined bygx f2x−16fx, andh:XY, defined byhx f2x−4fx, are quadratic and quartic, respectively.

Replacingyby 2yin2.15and using the evenness off, we have f

2x2y f

2x−2y 4f

2yx 4f

2y−x

2f2x−8fx−6f 2y

2.16

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for allx, yX. Interchangingxwith y in2.16 and then using2.15, we obtain by the evenness off

f

2x2y f

2x−2y 4f

2xy 4f

2x−y 2f

2y

−8f y

−6f2x 16f

xy 16f

xy

2f2x 2f 2y

−32fx−32f

y 2.17

for allx, yX. By2.17, we have f

2x2y

−16f

xy

f

2x−2y

−16f

x−y 2

f2x−16fx 2 f

2y

−16f y 2.18 for allx, yX. This means that

g xy

g xy

2gx 2g y

2.19

for allx, yX. Thus the mappingg:XY is quadratic.

To prove thath:XY is quartic, we have to show that h

2xy h

2x−y 4h

xy 4h

xy

24hx−6h y

2.20

for allx, yX. Replacingxandyby 2xand 2yin2.15, respectively, we get f

4x2y f

4x−2y 4f

2x2y 4f

2x−2y

2f4x−8f2x−6f 2y

2.21

for allx, yX. Sinceg2x 4gxfor allxXandg:XY is a quadratic mapping, we have

f4x 20f2x−64fx 2.22

for allxX. So it follows from2.15,2.21, and2.22that h

2xy h

2x−y

f

4x2y

−4f

2xy

f

4x−2y

−4f 2x−y 4

f

2x2y

−4f

xy 4 f

2x−2y

−4f xy 24

f2x−4fx −6 f

2y

−4f y 4h

xy 4h

xy

24hx−6h y

2.23

for allx, yX. Thush:XYis a quartic mapping.

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Theorem 2.2. A mappingf : XY satisfies1.6if and only if there exist a unique symmetric multiadditive mapping D : X ×X ×X×XY and a unique symmetric bi-additive mapping B:X×XY such that

fx Dx, x, x, x Bx, x 2.24

for allxX.

Proof. We first assume that the mappingf : XY satisfies1.6. Let g, h : XY be mappings defined by

gx:f2x−16fx, hx:f2x−4fx 2.25

for allxX.ByLemma 2.1, we achieve that the mappingsgandhare quadratic and quartic, respectively, and

fx: 1

12hx− 1

12gx 2.26

for allxX.Thus there exist a unique symmetric multiadditive mappingD:X×X×X×X → Y and a unique symmetric bi-additive mappingB : X×XY such thatDx, x, x, x 1/12hxandBx, x −1/12gxfor allxXsee citead, ki. So

fx Dx, x, x, x Bx, x 2.27

for allxX.

Conversely assume that

fx Dx, x, x, x Bx, x 2.28

for allxX,where the mappingD :X×X×X×XY is symmetric multi-additive and B:X×XY is bi-additive. By a simple computation, one can show that the mappingsD andBsatisfy the functional equation1.6, so the mapping f satisfies1.6.

3. Generalized Hyers-Ulam Stability of 1.6

From now on, letX andY be a quasi-Banach space with quasi-norm · X and ap-Banach space withp-norm · Y, respectively. LetM be the modulus of concavity of · Y. In this section, using an idea of G˘avruta22, we prove the stability of1.6in the spirit of Hyers, Ulam, and Rassias. For convenience we use the following abbreviation for a given mapping

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f:XY:

Δf x, y

f nxy

f nxy

n2f xy

n2f xy

−2fnx 2n2fx 2 n2−1

f

y 3.1

for allx, yX. Letϕpqx, y: ϕqx, yp. We will use the following lemma in this section.

Lemma 3.1see15. Let 0< p1 and letx1, x2, . . . , xnbe nonnegative real numbers. Then n

i1

xi

p

n

i1

xip. 3.2

Theorem 3.2. Letϕq :X×X → 0,∞be a function such that

mlim→ ∞4mϕq

x 2m, y

2m

0 3.3

for allx, yXand

i1

4piϕqp

x 2i,y

2i

<∞ 3.4

for allxX and ally∈ {x,2x,3x, nx,n1x,n−1x,n2x,n−2x,n−3x}.Suppose that a mappingf:XYwithf0 0 satisfies the inequality

Δf x, y

Yϕq x, y

3.5

for allx, yX.Then the limit

Qx: lim

m→ ∞4m

f x

2m−1

−16f x 2m

3.6

exists for allxXandQ:XY is a unique quadratic mapping satisfying

f2x−16fx−Qx

YM11 4

ψqx 1/p 3.7

(9)

for allxX,where

ψqx:

i1

4pi

1 n2pn2−1p

ϕpq

x

2i,n2x 2i

ϕpq

x

2i,n−2x 2i

4pϕpq

x

2i,n1x 2i

4pϕpq x

2i,n−1x 2i

4pϕpq

x 2i,nx

2i

ϕpq 2x

2i ,2x 2i

4pϕpq

2x 2i,x

2i

n2pϕpq x

2i,3x 2i

2p

3n2−1p ϕpq

x 2i,2x

2i

17n2−8p ϕpq

x 2i,x

2i

n2p n2−1p

ϕpq

0,xn1x 2i

ϕpq

0,n−3x 2i

10pϕpq

0,n−1x 2i

4pϕpq

0,nx 2i

4pϕpq

0,n−2x 2i

n41p

n2−1pϕpq

0,2x 2i

2

3n4n22p n2−1p ϕpq

0,x

2i

.

3.8 Proof. Settingx0 in3.5and then interchangingxandy, we get

n2−1

fxn2−1

f−xϕq0, x 3.9

for allxX. Replacingybyx, 2x,nx,n1xandn−1xin3.5, respectively, we get fn1x fn−1x−n2f2x−2fnx

4n2−2

fxϕqx, x, 3.10 fn2x fn−2x−n2f3xn2f−x−2fnx 2n2fx

2 n2−1

f2x≤ϕqx,2x,

3.11 f2nx−n2fn1x−n2f1nx 2

n2−2

fnx 2n2fxϕqx, nx, 3.12 f2n1x f−xn2fn2x−n2f−nx−2fnx 2n2fx

2 n2−1

fn1x≤ϕqx,n1x,

3.13 f2n−1x fxn2f2nx

n22

fnx 2n2fx 2

n2−1

fn−1x≤ϕqx,n−1x, 3.14

f2n1x f−2xn2fn3x−n2f−n1x−2fnx 2n2fx 2

n2−1

fn2x≤ϕqx,n2x, 3.15

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f2n−1x f2xn2fn−1x−n2f−n−3x−2fnx 2n2fx 2

n2−1

fn−2x≤ϕqx,n−2x, 3.16

fn3x fn−3x−n2f4xn2f−2x−2fnx 2n2fx 2

n2−1

f3x≤ϕqx,3x 3.17

for all xX. Combining 3.9 and 3.11–3.17, respectively, yields the following ine- qualities:

fn2x fn−2x−n2f3xn2fx−2fnx 2n2fx 2 n2−1

f2x

ϕqx,2x n2

n2−1ϕq0, x,

3.18 f2nx−n2fn1x−n2fn−1x 2

n2−2

fnx 2n2fx

ϕqx, nx n2

n2−1ϕq0,n−1x,

3.19 f2n1x fxn2fn2x−n2fnx−2fnx 2n2fx 2

n2−1

fn1x

ϕqx,n1x n2

n2−1ϕq0, nx 1

n2−1ϕq0, x,

3.20 f2n−1x fxn2fn−2x−

n22

fnx 2n2fx 2 n2−1

fn−1x

ϕqx,n−1x n2

n2−1ϕq0,n−2x,

3.21 f2n1xf2x−n2fn3x−n2fn1x−2fnx2n2fx2

n2−1

fn2x

ϕqx,n2x n2

n2−1ϕq0,n1x ϕq0,2x, 3.22

f2n−1xf2x−n2fn−1x−n2fn−3x−2fnx2n2fx2 n2−1

fn−2x

ϕqx,n−2x n2

n2−1ϕq0,n−3x, 3.23

fn3x fn−3x−n2f4xn2f2x−2fnx 2n2fx 2 n2−1

f3x

ϕqx,3x n2

n2−1ϕq0,2x

3.24 for allxX.

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Replacingxandyby 2xandxin3.5, respectively, we obtain f2n1x f2n−1x−n2f3x−2f2nx 2n2f2x

n2−2

fxϕq2x, x 3.25

for allxX. Putting 2xand 2yinstead ofxandyin3.5, respectively, we have f2n1x f2n−1x−n2f4x−2f2nx 2

2n2−1

f2xϕq2x,2x 3.26

for allxX. It follows from3.10,3.18,3.19,3.20,3.21, and3.25that f3x−6f2x 15fx

M5 n2n2−1

ϕqx,n1x ϕqx,n−1x ϕq2x, x 2ϕqx, nx n2ϕqx,2x

4n2−2

ϕqx, x n2 n2−1

q0,n−1x ϕq0, nx ϕq0,n−2x

n41

n2−1ϕq0, x

3.27

for allxX. Also, from3.10,3.18,3.19,3.22,3.23,3.24, and3.26, we conclude f4x−4f3x 4f2x 4fx

M6 n2n2−1

ϕqx,n2x ϕqx,n−2x ϕq2x,2x 2ϕqx, nx n2

ϕqx,3x ϕqx, x 2

n2−1

ϕqx,2x n2

n2−1

q0,n−1x ϕq0,n−3x ϕq0,n1x

n41

n2−1ϕq0,2x 2n2ϕq0, x

3.28

(12)

for allxX. Finally, combining3.27and3.28yields f4x−24f2x 64fx

M8 n2n2−1

ϕqx,n2x ϕqx,n−2x 4ϕqx,n1x 4ϕqx,n−1x 10ϕqx, nx ϕq2x,2x 4ϕq2x, x n2ϕqx,3x 2

3n2−1

ϕqx,2x

17n2−8

ϕqx, x n2

n2−1

ϕq0,n1x ϕq0,n−3x 10ϕq0,n−1x

q0, nx4ϕq0,n−2x

n41

n2−1ϕq0,2x 2

3n4n22

n2−1 ϕq0, x

3.29

for allxX. Let

ψqx: 1 n2n2−1

ϕqx,n2x ϕqx,n−2x 4ϕqx,n1x

qx,n−1x 10ϕqx, nx ϕq2x,2x 4ϕq2x, x n2ϕqx,3x 2

3n2−1

ϕqx,2x

17n2−8

ϕqx, x n2

n2−1

ϕq0,n1xϕq0,n−3x10ϕq0,n−1x4ϕq0, nx

q0,n−2x

n41

n2−1ϕq0,2x2

3n4n22

n2−1 ϕq0, x

. 3.30

Then the inequality3.29implies that

f4x−20f2x 64fx≤M8ψqx 3.31

for allxX.

Letg :XY be a mapping defined bygx : f2x−16fxfor allxX.From 3.31, we conclude that

g2x−4gx≤M8ψqx 3.32

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for allxX.If we replacexin3.32byx/2m1and multiply both sides of3.32by 4m,then we get

4m1g x

2m1

−4mg x 2m

Y

M84mψq x

2m1

3.33

for allxXand all nonnegative integersm. SinceYis a p-Banach space, the inequality3.33 gives

4m1g x

2m1

−4kg x

2k p

Y

m

ik

4i1g x

2i1

−4ig x

2i p

Y

M8p m ik

4ipψqp x

2i1

3.34

for all nonnegative integersmandkwithmkand allxX.Since 0< p≤1, byLemma 3.1 and3.30, we conclude that

ψqpx≤ 1 n2pn2−1p

ϕpqx,n2x ϕpqx,n−2x 4pϕpqx,n1x 4pϕpqx,n−1x10pϕpqx, nxϕpq2x,2x 4pϕpq2x, x n2pϕpqx,3x2p

3n2−1p

ϕpqx,2x

17n2−8p

ϕpqx, x n2p n2−1p

×

ϕpq0,n1xϕpq0,n−3x10pϕpq0,n−1x4pϕpq0, nx 4pϕpq0,n−2x

n41p

n2−1pϕpq0,2x 2

3n4n22p

n2−1p ϕpq0, x

3.35

for allxX.Therefore, it follows from3.4and3.35that

i1

4ipψqp

x 2i

<∞ 3.36

for allxX.It follows from3.34and3.36that the sequence{4mgx/2m}is Cauchy for allxX.SinceY is complete, the sequence{4mgx/2m}converges for allxX.So one can define the mappingQ:XYby

Qx lim

m→ ∞4mg x 2m

3.37

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for allxX.Lettingk0 and passing the limitm → ∞in3.34, we get gxQxp

YM8p

i0

4ipψqp x

2i1

M8p 4p

i1

4ipψqp x

2i

3.38

for allxX.Thus3.7follows from3.4and3.38.

Now we show thatQis quadratic. It follows from3.3,3.33and3.37that

Q2x−4QxY lim

m→ ∞

4mg x

2m−1

−4m1gx 2m

Y

4 lim

m→ ∞

4m−1g x

4m−1

−4mg x 2m

Y

M11 lim

m→ ∞4mψq x 2m

0

3.39

for allxX.So

Q2x 4Qx 3.40

for allxX.On the other hand, it follows from3.3,3.5,3.6and3.37that ΔQx, y

Y lim

m→ ∞4mΔgx 2m, y

2m

Y

lim

m→ ∞4m Δf

x 2m−1, y

2m−1

−16Δfx 2m, y

2m

Y

Mlim

m→ ∞4m Δf

x 2m−1, y

2m−1

Y

16Δf x 2m, y

2m

Y

Mlim

m→ ∞4m

ϕq

x 2m−1, y

2m−1

16ϕq

x 2m, y

2m 0

3.41

for allx, yX.Hence the mappingQ satisfies1.6. ByLemma 2.1, the mappingQ2x− 4Qxis quadratic. Hence3.40implies that the mappingQis quadratic.

It remains to show that Q is unique. Suppose that there exists another quadratic mappingQ : XY which satisfies1.6and 3.7. Since Qx/2m 1/4mQxand Qx/2m 1/4mQxfor allxX, we conclude from3.7that

QxQxp

Y lim

m→ ∞4mpgx 2m

Q x 2m

Y

pM8p 4p lim

m→ ∞4mpψq

x 2m

3.42

(15)

for allxX.On the other hand, since

mlim→ ∞4mp i1

4ipϕqp x

2mi, y 2mi

lim

m→ ∞

im1

4ipϕqp x

2i,y 2i

0 3.43

for allxXand ally∈ {x,2x,3x, nx,n1x,n−1x,n2x,n−2x,n−3x},then

mlim→ ∞4mpψqx 2m

0 3.44

for allxX. Using3.44and3.42, we getQQ,as desired.

Theorem 3.3. Letϕq :X×X → 0,∞be a function such that

mlim→ ∞

1 4mϕq

2mx,2my

0 3.45

for allx, yXand

i0

1 4piϕqp

2ix,2iy

<∞ 3.46

for allxX and ally∈ {x,2x,3x, nx,n1x,n−1x,n2x,n−2x,n−3x}.Suppose that a mappingf:XYwithf0 0 satisfies the inequality

Δf x, y

Yϕq

x, y

3.47

for allx, yX.Then the limit

Qx: lim

m→ ∞

1 4m

f

2m1x

−16f2mx

3.48

exists for allxXandQ:XY is a unique quadratic mapping satisfying

f2x−16fx−Qx

YM8 4

ψqx 1/p 3.49

(16)

for allxX, where

ψqx:

i0

1 4pi

1 n2pn2−1p

ϕpq

2ix,2in2x ϕpq

2ix,2in−2x 4pϕpq

2ix,2in1x

4pϕpq

2ix,2in−1x 10pϕpq

2ix,2inx ϕpq

2i2x,2i2x 4pϕpq

2i2x,2ix

n2pϕpq

2ix,2i3x 2p

3n2−1p ϕpq

2ix,2i2x

17n2−8p ϕpq

2ix,2ix

n2p n2−1p

ϕpq

0,2in1x ϕpq

0,2in−3x 10pϕpq

0,2in−1x

4pϕpq

0,2inx

4pϕpq

0,2in−2x

n41p

n2−1pϕpq 0,2i2x

2

3n4n22p

n2−1p ϕpq

0,2ix .

3.50

Proof. The proof is similar to the proof ofTheorem 3.2.

Corollary 3.4. Letθ, r, sbe nonnegative real numbers such thatr, s > 2 ors < 2. Suppose that a mappingf:XYwithf0 0 satisfies the inequality

Δfx, y

Y

⎧⎪

⎪⎪

⎪⎪

⎪⎪

⎪⎪

⎪⎪

⎪⎪

⎪⎪

⎪⎪

⎪⎩

θ, r s0,

θxrX, r >0, s0, θys

X, r 0, s >0, θ

xrXys

X

, r, s >0

3.51

for allx, yX.Then there exists a unique quadratic mappingQ:XY satisfying

f2x−16fx−Qx

YM8θ n2n2−1

⎧⎪

⎪⎪

⎪⎪

⎪⎪

⎪⎪

⎪⎪

⎪⎪

⎪⎪

⎪⎪

⎪⎩

δq, rs0,

αqx, r >0, s0, βqx, r0, s >0,

αpqx βpqx1/p

, r, s >0

3.52

(17)

for allxX,where

δq

1 4p−1n2−1p

6n2−2p

n2−1p

17n2−8p

n2−1p

6n4−2n24p

n2p210p2∗4p

n41p n2p

n2−1p

3∗4p

n2−1p

10p

n2−1p 3

n2−1p1/p ,

αqx

4p22rp 10p

6n2−2p

17n2−8p

2rpn2p

|4p−2rp|

1/p xrX,

βqx

1

n2−1p|4p−2sp|

2sp

6n2−2p

n2−1p

17n2−8pn2−1p

6n4−2n24p n2p

n1sp n−3sp10pn−1sp4pnsp4pn−2sp 2sp

n41p

3spn2pn2−1p4pn2−1pn2sp

n2−1p

n−2sp

n2−1p

4pn1spn2−1p4pn−1sp

n2−1p

10pnsp

n2−1p1/p xsX.

3.53 Proof. InTheorem 3.2, puttingϕqx, y:θxrXysXfor allx, yX, we get the desired result.

Corollary 3.5. Letθ0 andr, s >0 be nonnegative real numbers such thatλ:rs /2. Suppose that a mapingf:XY withf0 0 satisfies the inequality

Δf x, y

YθxrXys

X 3.54

for allx, yX.Then there exists a unique quadratic mappingQ:XY satisfying f2x−16fx−Qx

Y

M8θ n2n2−1

1 4p−2λp

n2sp n−2sp4pn1sp

4pn−1sp10pnsp2rsp4p2rpn2p3sp 2sp

6n2−2p

17n2−8p1/p xλX

3.55

for allxX.

(18)

Proof. InTheorem 3.2, puttingϕqx, y : θxrXysX for allx, yX, we get the desired result.

Theorem 3.6. Letϕt:X×X → 0,∞be a function such that

m→ ∞lim16mϕt

x 2m, y

2m

0 3.56

for allx, yXand

i1

16piϕtp

x 2i,y

2i

<∞ 3.57

for allxX and ally∈ {x,2x,3x, nx,n1x,n−1x,n2x,n−2x,n−3x}.Suppose that a mappingf:XYwithf0 0 satisfies the inequality

Δf x, y

Yϕt

x, y

3.58 for allx, yX.Then the limit

Tx: lim

m→ ∞16m

f x

2m−1

−4f x 2m

3.59

exists for allxXandT :XY is a unique quartic mapping satisfying f2x−4fx−Tx

YM8 16

ψtx 1/p 3.60

for allxX,where

ψtx:

i1

16pi

1 n2pn2−1p

ϕpt

x

2i,n2x 2i

ϕpt

x

2i,n−2x 2i

4pϕpt

x

2i,n1x 2i

4pϕpt x

2i,n−1x 2i

10pϕpt

x 2i,nx

2i

ϕpt 2x

2i,2x 2i

4pϕpt

2x 2i, x

2i

n2pϕpt x

2i,3x 2i

2p

3n2−1p

ϕpt x

2i,2x 2i

17n2−8p

ϕpt x

2i,x 2i

n2p n2−1p

ϕpt

0,xn1x 2i

ϕpt

0,n−3x 2i

10pϕpt

0,n−1x 2i

4pϕpt

0,nx 2i

4pϕpt

0,n−2x 2i

n41p n2−1pϕpt

0,2x

2i

2

3n4n22p

n2−1p ϕpt

0,x 2i

.

3.61

(19)

Proof. Similar to the proofTheorem 3.2, we have

f4x−20f2x 64fx≤M8ψtx 3.62 for allxX,where

ψtx 1 n2n2−1

ϕtx,n2x ϕtx,n−2x 4ϕtx,n1x

tx,n−1x 10ϕtx, nx ϕt2x,2x 4ϕt2x, x n2ϕtx,3x 2

3n2−1

ϕtx,2x

17n2−8 ϕtx, x n2

n2−1

ϕt0,n1x ϕt0,n−3x 10ϕt0,n−1x 4ϕt0, nx

t0,n−2x

n41

n2−1ϕt0,2x 2

3n4n22 n2−1 ϕt0, x

. 3.63

Leth:XY be a mapping defined byhx:f2x−4fx. Then we conclude that

h2x−16hx ≤M8ψtx 3.64

for allxX.If we replacexin3.65byx/2m1 and multiply both sides of3.65by 16m, then we get

16m1h x

2m1

−16mh x 2m

Y

M816mψt

x 2m1

3.65

for allxXand all nonnegative integersm. SinceYis ap-Banach space, the inequality3.66 gives

16m1h x

2m1

−16kh x

2k p

Y

m

ik

16i1h x

2i1

−16ih x

2i p

Y

M8p m ik

16piψtp x

2i1

3.66

参照

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