Stability of a Bi-Additive Functional Equation in Banach Modules Over a C

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Volume 2012, Article ID 835893,12pages doi:10.1155/2012/835893

Research Article

Stability of a Bi-Additive Functional Equation in Banach Modules Over a C

-Algebra

Won-Gil Park

¹

and Jae-Hyeong Bae

²

1Department of Mathematics Education, College of Education, Mokwon University, Daejeon 302-729, Republic of Korea

2Graduate School of Education, Kyung Hee University, Yongin 446-701, Republic of Korea

Correspondence should be addressed to Jae-Hyeong Bae,[email protected] Received 6 April 2012; Accepted 30 May 2012

Academic Editor: Baodong Zheng

Copyrightq2012 W.-G. Park and J.-H. Bae. 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.

We solve the bi-additive functional equationfxy, z−w fx−y, zw 2fx, z−2fy, w and prove that every biadditive Borel function is bilinear. And we investigate the stability of a biadditive functional equation in Banach modules over a unitalC-algebra.

1. Introduction

In 1940, Ulam proposed the stability problemsee1.

Let G₁ be a group, and let G₂ be a metric group with the metric d·,·. Givenε >

0, does there exist a δ > 0 such that if a mapping h : G1 → G2 satisfies the inequality dhxy, hxhy< δfor allx, y ∈ G₁ then there is a homomorphismH :G₁ → G₂with dhx, Hx< εfor allx∈G₁?

In 1941, this problem was solved by Hyers2in the case of Banach space. Thereafter, many authors investigated solutions or stability of various functional equationssee3–21.

Let X and Y be real or complex vector spaces. In 1989, Acz´el and Dhombres 22 proved that a mappingg:X → Y satisfies the quadratic functional equation

g xy

g x−y

2gx 2g y

1.1 if and only if there exists a symmetric bi-additive mappingS :X×X → Y such thatgx Sx, x, where

S x, y

: 1 4

g xy

−g

x−y 1.2

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for allx, y∈X. For a mappingf:X×X → Y, consider the bi-additive functional equation:

f

xy, z−w f

x−y, zw

2fx, z−2f y, w

. 1.3

For a mappingg :X → Y satisfying1.1, the Acz´el’s bi-additive mappingS:X×X → Y given by1.2is a solution of1.3.

In this paper, we find out the general solution of the bi-additive functional equation 1.3 and investigate the linearity of bi-additive Borel functions. And we investigate the stability of1.3in Banach modules over a unitalC-algebra.

2. Solution of the bi-additive Functional Equation 1.3

The general solution of the bi-additive functional equation1.3is as follows.

Theorem 2.1. A mappingf:X×X → Y satisfies1.3if and only if the mappingfis bi-additive.

Proof. Assume that the mappingfsatisfies1.3. Lettingxyzw0 in1.3, we gain f0,0 0. Puttingwzin1.3, we get

f

xy,0 f

x−y,2z

2fx, z−2f y, z

2.1

for allx, y, z∈X. Settingyxin2.1, we have

fx,0 −f0, z 2.2

for allx, z∈X. Takingz0resp.,x0in the above equation, we obtain fx,0 0

resp., f0, z 0

2.3

for allx∈Xresp., for allz∈X. Lettingxw0 in1.3and using2.3, we gain

f

−y, z −f

y, z

2.4

for ally, z∈X. Puttingy0 in2.1and using2.3, we get

fx,2z 2fx, z 2.5

for allx, z ∈ X. Replacingyby−y in2.1and using 2.3,2.4, and2.5and the above equation, we see thatfxy, z fx, z fy, zfor allx, y, z∈X.

On the other hand, lettingyxin1.3and using2.3, we gain

f2x, z−w 2fx, z−2fx, w 2.6

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for allx, z, w∈X. Puttingyz0 in1.3and using2.3, we get

fx,−w −fx, w 2.7

for allx, w∈X. Settingw0 in2.6and using2.3, we have

f2x, z 2fx, z 2.8

for allx, z∈X. Replacingwby−win2.6and using2.7and2.8, we obtain thatfx, z w fx, z fx, wfor allx, z, w∈X.

The converse is trivial.

The bi-additive functional equation1.3is related to the quadratic functional equation 1.1.

Iff :X×X → Y is a mapping satisfying1.3andg :X → Y is the mapping given bygx:fx, xfor allx∈X, then one can easily obtain thatgsatisfies1.1.

Leta ∈ Randg : X → Y be a mapping satisfying1.1. Iff : X×X → Y is the mapping given byfx, y: a/4gxy−gx−yfor allx, y ∈X, then one can easily prove thatfsatisfies1.3. Furthermore,gx fx, xholds for allx∈Xifa1.

The following is a result on bi-additive Borel functions.

Theorem 2.2. Letψ:R×R → Rbe a bi-additive Borel function; then it is bilinear, that is, it satisfies ψs, t stψ1,1for alls, t∈R.

Proof. Since the functionψis bi-additive, we gain

ψ pu, qv

pqψu, v 2.9

for allp, q∈Qand allu, v∈R. Lettingpv1 in equality2.9, we get

ψ u, q

qψu,1 2.10

for allq∈Qand allu∈R. Puttinguv1 in equality2.9again, we have

ψ p, q

pqψ1,1 2.11

for allp, q∈Q. Note that the functionv → ψu, vis measurable for each fixedu∈Rsee23, Proposition 2.34. Since the functionv → ψu, vis additive for each fixedu∈R, by24, it is continuous for each fixedu∈R. By the same reasoning, the functionu → ψu, vis also continuous for each fixedv∈R. Lets, t∈Rbe fixed. Sinceψis measurable, by25, Theorem 7.14.26, for everym∈Nthere is a closed setFm⊂s, s1such thatμs, s1\Fm<1/m and ψ|Fm×R is continuous. Since μFm → 1, one can chooseu_m ∈ F_m satisfyingu_m → s.

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Take a sequence{qn}inQconverging tot. For each fixedm∈N, take a sequence{pn}inQ converging tou_m. By equalities2.10and2.11, we see that

ψum, t ψ

u_m, lim

n→ ∞q_n

lim

n→ ∞ψ u_m, q_n

lim

n→ ∞q_nψum,1 tψum,1 tψ

nlim→ ∞p_n,1

tlim

n→ ∞ψ p_n,1

tlim

n→ ∞p_nψ1,1 tu_mψ1,1

2.12

for allm∈N. Hence we obtain that ψs, t ψ

mlim→ ∞um, t

lim

m→ ∞ψum, t lim

m→ ∞tumψ1,1 stψ1,1, 2.13 as desired.

3. Stability of the bi-additive Functional Equation 1.3

From now on, letXbe a normed space,Y a complete normed space, andr /2 a nonnegative real number. In this section, we investigate the stability of the bi-additive functional equation 1.3.

Lemma 3.1. Letf :X×X → Ybe a mapping such that f

xy, z−w f

x−y, zw

−2fx, z 2f y, w

≤ 4ε, r0,

ε

x^ry^rz^rw^r

, 0< r /2

3.1

for allx, y, z, w∈X. Then there exists a unique bi-additive mappingF:X×X → Y satisfying1.3 such that

f x, y

−F

x, y≤

⎧⎪

⎪⎪

⎨

⎪⎪

⎪⎩

2εf0,0, r0,

3ε 4−2^r

x^ry^r

, 0< r <2, 3ε

2^r−4

x^ry^r

, r >2

3.2

for allx, y∈X. The mappingFis given by

F x, y

:

⎧⎪

⎪⎨

⎪⎪

⎩

jlim→ ∞

1 4^jf

2^jx,2^jy

, 0≤r <2,

jlim→ ∞4^jf x

2^j, y 2^j

, r >2 3.3

for allx, y∈X.

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Proof. Consider the caser∈0,2. Lettingyxandw−zin3.1, we gain f2x,2z f0,0−2fx, z 2fx,−z≤2ε

x^rz^r

3.4

for allx, z∈X. Puttingxz0 in3.4, we getf0,0 0. Puttingxz0 in3.1, we get f

y,−w f

−y, w 2f

y, w≤εy^rw^r

3.5

for ally, w∈X. Replacingybyxandwbyzin the above inequality, we have

fx,−z f−x, z 2fx, z ≤εx^r z^r 3.6

for allx, z∈X. Settingy−xandwzin3.1, we obtain f2x,2z−2fx, z 2f−x, z≤2ε

x^rz^r

3.7

for allx, z∈X. By3.4and3.6, we gain

f2x,2z−4fx, z fx,−z−f−x, z≤3ε

x^rz^r

3.8

for allx, z∈X. By3.4and3.7, we get

fx,−z−f−x, z≤2ε

x^rz^r

3.9

for allx, z∈X. By3.4,3.6, and3.7, we have f2x,2z−4fx, z≤3ε

x^rz^r

3.10

for allx, z∈X. Replacingxby 2^jxandzby 2^jzand dividing 4^j1, we obtain that 1

4^jf

2^jx,2^jz

− 1 4^j1f

2^j1x,2^j1z

≤ 3ε·2^rj 4^j1

x^rz^r

3.11

for allx, z∈Xand allj0,1,2, . . .. For given integersl, m0≤l < m, we obtain that 1

4^lf

2^lx,2^lz

− 1

4^mf2^mx,2^mz ≤^m−1

jl

3ε·2^rj 4^j1

x^rz^r

3.12

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for allx, z∈X. By3.12, the sequence{1/4^jf2^jx,2^jy}is a Cauchy sequence for allx, y∈ X. SinceY is complete, the sequence{1/4^jf2^jx,2^jy}converges for allx, y ∈ X. Define F:X×X → Y byFx, y:lim_j_{→ ∞}1/4^jf2^jx,2^jyfor allx, y∈X. By3.1, we have

1 4^jf

2^j xy

,2^jz−w 1

4^jf 2^j

x−y

,2^jzw

− 2 4^jf

2^jx,2^jz 2

4^jf

2^jy,2^jw

≤ε2^rj 4^j

x^ry^r z^rw^r

3.13 for allx, y, z, w ∈Xand allj 0,1,2, . . .. Lettingj → ∞in the above inequality, we see that Fsatisfies1.3. Settingl0 and takingm → ∞in3.12, one can obtain inequality3.2. If G:X×X → Yis another mapping satisfying1.3and3.2, byTheorem 2.1, we obtain that

F x, y

−G

x, y 1 4ⁿF

2ⁿx,2ⁿy

−G

2ⁿx,2ⁿy

≤ 1 4ⁿF

2ⁿx,2ⁿy

−f

2ⁿx,2ⁿy 1 4ⁿf

2ⁿx,2ⁿy

−G

2ⁿx,2ⁿy

≤ 6ε·2^nr−2 4−2^r

x^ry^r

−→0 asn−→ ∞

3.14 for allx, y∈X. Hence the mappingFis the unique bi-additive mapping satisfying1.3, as desired.

The proof of the caser∈ {0} ∪2,∞is similar to that of the caser ∈0,2.

From now on, letAbe a unitalC-algebra with a norm| · |, and letAMandANbe left BanachA-modules with norms|| · ||and · , respectively. PutA₁:{a∈A| |a|1}.

A bi-additive mappingF : _AM × AM → ANsatisfying1.3is calledA-quadratic if Fax, ay a²Fx, yfor alla∈Aand allx, y∈AM.

Theorem 3.2. Letf : _AM ×AM → ANbe a mapping such that f

axay, az−aw f

ax−ay, azaw

−2a²fx, z 2a²f y, w

≤ 4ε, r0,

ε

x^ry^rz^rw^r

, 0< r /2

3.15

for alla∈A₁and allx, y, z, w∈AM. Ifftx, tyis continuous int∈Rfor each fixedx, y∈AM, then there exists a unique bi-additiveA-quadratic mappingF : AM ×AM → ANsatisfying1.3 and inequality3.2.

Proof. Consider the case r ∈ 0,2. By Lemma 3.1, it follows from the inequality of the statement fora 1 that there exists a unique bi-additive mappingF : AM × AM → AN satisfying1.3and inequality3.2. Letx₀, y₀ ∈ AMbe fixed. And letL: _AN → Rbe any

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real continuous linear functional, that is,Lis an arbitrary real functional element of the dual space of_ANrestricted to the scalar fieldR. Forn ∈ N, consider the functionsψ_n : R → R defined byψ_nt: 1/4ⁿLf2ⁿtx₀,2ⁿty₀for allt∈R. By the assumption thatftx, tyis continuous int∈Rfor each fixedx, y∈AM, the functionψnis continuous for alln∈N. Note thatψnt 1/4ⁿLf2ⁿtx0,2ⁿty0 L1/4ⁿf2ⁿtx0,2ⁿty0for alln ∈Nand allt ∈R.

By the proof ofLemma 3.1, the sequenceψ_ntis a Cauchy sequence for allt ∈ R. Define a functionψ :R → Rbyψt:limn→ ∞ψntfor allt∈R. Note thatψt LFtx0, ty0for allt∈R. SinceFis bi-additive, we get

ψst ψs−t L F

stx0,sty0

L F

s−tx0,s−ty0

L

F

stx0,sty0

F

s−tx0,s−ty0

L

F

sx0tx0, sy0ty0

F

sx0−tx0, sy0−ty0

L

2F

sx₀, sy₀ 2F

tx₀, ty₀ 2L

F sx0, sy0

2L F

tx0, ty0

2ψs 2ψt

3.16

for alls, t ∈R. Sinceψ is the pointwise limit of continuous functions, it is a Borel function.

Thus the functionψ as a measurable quadratic function is continuoussee26so has the formψt t²ψ1for allt∈R. Hence we have

L F

tx0, ty0

ψt t²ψ1 t²L F

x0, y0

L t²F

x0, y0

3.17

for allt∈R. SinceLis any continuous linear functional, the bi-additive mappingF : AM ×

AM → ANsatisfiesFtx0, ty₀ t²Fx0, y₀for allt∈R. Therefore we obtain F

tx, ty t²F

x, y

3.18 for allt∈Rand allx, y∈AM. Letjbe an arbitrary positive integer. Replacingxandzby 2^jx and 2^jz, respectively, and lettingyw0 in inequality3.15, we gain

f

2^jax,2^jaz

−a²f

2^jx,2^jz

a²f0,0≤2^rj−1ε

x^rz^r

3.19

for alla∈A₁and allx, z∈AM. Note that there is a constantK >0 such that the condition

av ≤K|a|v 3.20

for eacha∈Aand eachv∈ANsee27, Definition 12. For alla∈A1 and allx, y ∈AM, we get

1 4^j

f

2^jax,2^jay

−a²f

2^jx,2^jy≤2^r−2j−1ε

x^rz^r

K|a|²

4^j f0,0−→0 3.21

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asj → ∞. Hence we have

F ax, ay

lim

j→ ∞

1 4^jf

2^jax,2^jay

a²lim

j→ ∞

1 4^jf

2^jx,2^jy a²F

x, y

3.22

for alla∈A1and allx, y ∈ AM. SinceFax, ay a²Fx, yfor eacha∈A1, by3.18, we obtain

F ax, ay

F

|a|a

|a|x,|a|a

|a|y

|a|²F a

|a|x, a

|a|y

a²F x, y

3.23

for all nonzeroa∈Aand allx, y∈ AM. By3.18, we getF0x,0y 0²Fx, yfor allx, y∈

AM. Therefore the bi-additive mappingFis the uniqueA-quadratic mapping satisfying the inequality3.2.

We obtain the Hyers-Ulam stability of1.3as a corollary ofTheorem 3.2.

Corollary 3.3. LetEbe a complex normed space andf :E×E → Ca function such that f

λxλy, λz−λw f

λx−λy, λzλw

−2λ²fx, z 2λ²f

y, w≤ε 3.24 for allλ∈T:{λ∈C:|λ|1}and allx, y, z, w∈E. Ifftx, tyis continuous int∈Rfor each fixedx, y∈E, then there exists a unique bi-additiveC-quadratic mappingF:E×E → Csatisfying 1.3such thatfx, y−Fx, y ≤ε/2f0,0for allx, y∈E.

PutAin : {a ∈ A | ais invertible inA},Asa : {a ∈ A | a a},A : {a ∈ Asa | Spa⊂0,∞}, andA₁ :A₁∩A.

A unitalC-algebraAis said to have real rank 0see28if the invertible self-adjoint elements are dense inAsa.

For any elementa∈A,a a₁ia₂, wherea₁ : aa/2 anda₂ : a−a/2iare self-adjoint elements, furthermore,aa₁−a⁻₁ia₂−ia⁻₂, wherea₁, a⁻₁, a₂, anda⁻₂are positive elementssee27, Lemma 38.8.

Theorem 3.4. LetAbe of real rank 0, and letf: _AM × AM → ANbe a mapping such that f

axay, bz−bw f

ax−ay, bzbw

−2abfx, z 2ab y, w

≤ 4ε, r0,

ε

x^ry^rz^rw^r

, 0< r /2

3.25

for alla, b∈A₁ ∩Ain∪ {i}and allx, y, z, w ∈ AM. For each fixedx, y ∈ AM, let the sequence {1/4^jf2^jax,2^jby}converge uniformly onA₁×A₁. Iffax, byis continuous ina, b∈A1∪ R²for each fixedx, y∈ AM, then there exists a unique bi-additiveA-quadratic mappingF: _AM ×

AM → ANsatisfying1.3and inequality 3.2such thatFax, by abFx, yfor alla, b ∈ A₁∪ {i}and allx, y∈ AM.

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Proof. Consider the caser ∈0,2. ByLemma 3.1, there exists a unique bi-additive mapping F : _AM × AM → ANsatisfying1.3and inequality3.2on_AM × AM. Letx₀, y₀ ∈ AM be fixed. And letLbe an arbitrary real functional element of the dual space of_ANrestricted to the scalar fieldR. Forn∈N, consider the functionsψn :R×R → Rdefined byψns, t: 1/4ⁿLf2ⁿsx0,2ⁿty0for all s, t ∈ R. By the assumption thatfax, byis continuous in a, b ∈ A1 ∪R² for each fixedx, y ∈ AM, the functionψ_n is continuous for all n ∈ N.

Note thatψns, t 1/4ⁿLf2ⁿsx0,2ⁿty0 L1/4ⁿf2ⁿsx0,2ⁿty0for alln∈Nand all s, t∈R. By the proof ofLemma 3.1, the sequenceψns, tis a Cauchy sequence for alls, t∈R.

Define a functionψ : R×R → R byψs, t : lim_n_{→ ∞}ψ_ns, t for alls, t ∈ R. Note that ψs, t LFsx0, ty0for alls, t∈R. Since the mappingFis bi-additive, we have

ψs1s₂, t₁−t₂ ψs1−s₂, t₁t₂ L

F

s1s2x0,t1−t2y0

L F

s1−s2x0,t1t2y0

L

F

s1s₂x0,t1−t₂y0

F

s1−s₂x0,t1t₂y0

L

F

s₁x₀s₂x₀, t₁y₀−t₂y₀ F

s₁x₀−s₂x₀, t₁y₀t₂y₀ L

2F

s1x0, t1y0

−2F

s2x0, t2y0

2L F

s1x0, t1y0

−2L F

s2x0, t2y0

2ψs1, t₁−2ψs2, t₂

3.26

for alls₁, s₂, t₁, t₂ ∈ R. Since ψ is the pointwise limit of continuous functions, it is a Borel function. ByTheorem 2.2, we gainψs, t stψ1,1for alls, t∈R. Hence we get

L F

sx0, ty0

ψs, t stψ1,1 stL F

x0, y0

L stF

x0, y0

3.27

for alls, t∈R. SinceLis any continuous linear functional, the bi-additive mappingF: _AM ×

AM →ANsatisfiesFsx0, ty₀ stFx0, y₀for alls, t∈R. Therefore we obtain F

sx, ty stF

x, y

3.28 for alls, t∈Rand allx, y∈ AM. Letjbe an arbitrary positive integer. Replacingxandzby 2^jxand 2^jz, respectively, and lettingyw0 in inequality3.25, we get

f

2^jax,2^jbz

−abf

2^jx,2^jz

abf0,0≤2^rj−1ε

x^rz^r

3.29

for alla, b∈A₁∩Ain∪ {i}and allx, z∈ AM. By inequality3.20and the above inequality, for alla, b∈A₁ ∩A_in∪ {i}and allx, z∈ AM, we have

1 4^j

f

2^jax,2^jbz

−abf

2^jx,2^jz

≤2^r−2j−1ε

x^r z^r

K|a||b|

4^j f0,0−→0 asj −→ ∞.

3.30

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Hence we obtain that

F ax, by

lim

j→ ∞

1 4^jf

2^jax,2^jby

ablim

j→ ∞

1 4^jf

2^jx,2^jy

abF x, y

3.31

for alla, b ∈A₁ ∩Ain∪ {i}and allx, y∈ AM. Letc, d∈A₁\Ain. SinceAin∩Asais dense inA_sa, there exists two sequences{cj}and{dj}inA_in∩A_sasuch thatc_j → candd_j → d asj → ∞. Putp_j : 1/|cj|cj andq_j : 1/|dj|djfor allj ∈N. Thenp_j → candq_j → das j → ∞. Setaj :

pjpjandbj :

qjqjfor allj ∈N. Thenaj → candbj → dasj → ∞ anda_j, b_j ∈A₁ ∩A_in. Since{1/4^jf2^jax,2^jby}is uniformly converges onA₁×A₁for each x, y∈ AMandfax, byis continuous ina, b∈A₁for eachx, y∈ AM, we see thatFax, by is also continuous ina, b∈A1for eachx, y∈ AM. In fact, we gain

a,blim→c,dF ax, by

lim

a,b→c,dlim

j→ ∞

1 4^jf

2^jax,2^jby

lim

j→ ∞ lim

a,b→c,d

1 4^jf

2^jax,2^jby lim

j→ ∞

1 4^jf

2^jcx,2^jdy F

cx, dy 3.32

for allx, y∈ AM. Thus we get

jlim→ ∞F

ajx, bjy F

jlim→ ∞ajx,lim

j→ ∞bjy

F cx, dy

3.33

for allx, y∈ AM. By equality3.31, we have F

a_jx, b_jy

−cdF

x, ya_jb_jF x, y

−cdF x, y

−→cdF x, y

−cdF

x, y0 3.34

asj → ∞for allx, y∈ AM. By equality3.33and the above convergence, we see that F

cx, dy

−cdF

x, y≤F cx, dy

−F

ajx, bjyF

ajx, bjy

−cdF

x, y−→0 3.35

asj → ∞for allx, y∈ AM. By equality3.31and the above convergence, we obtain

F ax, by

abF x, y

3.36

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for alla, b∈A₁ ∪ {i}and allx, y∈ AM. Since the mappingFis bi-additive, we see that

F ax, ay

F

a₁x−a⁻₁xia₂x−ia⁻₂x, a₁y−a⁻₁yia₂y−ia⁻₂y F

a₁x, a₁y

−F

a₁x, a⁻₁y F

a₁x, ia₂y

−F

a₁x, ia⁻₂y

−F

a⁻₁x, a₁y F

a⁻₁x, a⁻₁y

−F

a⁻₁x, ia₂y F

a⁻₁x, ia⁻₂y F

ia₂x, a₁y

−F

ia₂x, a⁻₁y F

ia₂x, ia₂y

−F

ia₂x, ia⁻₂y

−F

ia⁻₂x, a₁y F

ia⁻₂x, a⁻₁y

−F

ia⁻₂x, ia₂y F

ia⁻₂x, ia⁻₂y

3.37

for alla∈Aand allx, y∈ AM. By3.28and equality3.36, we have

F px, qy

F p p

px,q q qy

pqF p

px, q qy

pqF

x, y

3.38

for allp, q ∈ {a₁, a⁻₁, a₂, a⁻₂}and allx, y ∈A M. Note thata₁a⁻₁ a⁻₁a₁ a₂a⁻₂ a⁻₂a₂ 0.

Hence we obtain that F

ax, ay

a₂₂ F

x, y

ia₁a₂F x, y

−ia₁a⁻₂F x, y

a⁻₁₂

F x, y

−ia⁻₁a₂F x, y

ia⁻₁a⁻₂F x, y

ia₂a₁F x, y

−ia₂a⁻₁F x, y

− a₂₂

F x, y

−ia⁻₂a₁F x, y

ia⁻₂a⁻₁F x, y

− a⁻₂₂

F x, y

a₁2ia₁a₂ −ia₁a⁻₂

a⁻₁2−ia⁻₁a₂ ia⁻₁a⁻₂ ia₂a₁−ia₂a⁻₁ −

a₂2−ia⁻₂a₁ia⁻₂a⁻₁− a⁻₂2

F x, y

a₁ −a⁻₁ia₂−ia⁻₂2

F x, y

a²F x, y

3.39

for alla∈Aand allx, y∈ AM.

Acknowledgment

This research was supported by Basic Science Research Program through the National Research Foundation of Korea NRF funded by the Ministry of Education, Science and TechnologyGrant no. 2012003499.

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