The Fixed Point Property of

Volume 2010, Article ID 362829,13pages doi:10.1155/2010/362829

Research Article

The Fixed Point Property of

Unital Abelian Banach Algebras

W. Fupinwong and S. Dhompongsa

Department of Mathematics, Faculty of Science, Chiang Mai University, Chiang Mai 50200, Thailand

Correspondence should be addressed to S. Dhompongsa,[email protected] Received 28 October 2009; Accepted 22 January 2010

Academic Editor: Anthony To Ming Lau

Copyrightq2010 W. Fupinwong and S. Dhompongsa. 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 give a general condition for infinite dimensional unital Abelian Banach algebras to fail the fixed point property. Examples of those algebras are given including the algebras of continuous functions on compact sets.

1. Introduction

LetXbe a Banach space. A mappingT :E⊂X → Xis nonexpansive if

Tx−Ty≤x−y 1.1

for eachx, y∈E.The fixed point set ofT is FixT {x∈E:Txx}.We say that the space X has the fixed point propertyor weak fixed point propertyif for every nonempty bounded closed convexor weakly compact convex, resp. subset E of X and every nonexpansive mappingT :E → Ewe have FixT/∅.One of the central goals in fixed point theory is to solve the problem: which Banach spaces have theweakfixed point property?

For weak fixed point property, Alspach1exhibited a weakly compact convex subset Eof the Lebesgue spaceL10,1and an isometryT :E → Ewithout a fixed point, proving that the spaceL₁0,1does not have the weak fixed point property. Lau et al.2proved the following results.

Theorem 1.1. LetXbe a locally compact Hausdorffspace. IfC0Xhas the weak fixed point property, then X is dispersed.

Corollary 1.2. LetGbe a locally compact group. Then theC^∗-algebraC0Ghas the weak fixed point property if and only ifGis discrete.

Corollary 1.3. A von Neumann algebraMhas the weak fixed point property if and only ifMis finite dimensional.

Continuing in this direction, Benavides and Pineda 3 developed the concept of ω-almost weak orthogonality in the Banach latticeCKand obtained the results.

Theorem 1.4. Let X be a ω-almost weakly orthogonal closed subspace of CK where K is a metrizable compact space. ThenXhas the weak fixed point property.

Theorem 1.5. LetKbe a metrizable compact space. Then, the following conditions are all equivalent:

1CKisω-almost weakly orthogonal, 2CKisω-weakly orthogonal, 3K^ω∅.

Corollary 1.6. LetKbe a compact set withK^ω∅.ThenCKhas the weak fixed point property.

As for the fixed point property, Dhompongsa et al.4showed that aC^∗-algebra has the fixed point property if and only if it is finite dimensional. In this paper, we approach the question on the fixed point property from the opposite direction by identifying unital abelian Banach algebras which fail to have the fixed point property. As consequences, we obtain results on the algebra of continuous functionsCS,whereSis a compact set, and there is a unital abelian subalgebra of the algebral_∞Nwhich does not have the fixed point property and does not contain the spacec0.

2. Preliminaries and Lemmas

The fields of real and complex numbers are denoted byRandC, respectively. The symbolF denotes a field that can be eitherRorC.The elements ofFare called scalars.

An elementxin a unital algebraXis said to be invertible if there is an elementyinX such that

xyyx1. 2.1

In this caseyis unique and writtenx⁻¹.

We define the spectrum of an elementxof a unital algebraXoverFto be the set σx {λ∈F:λ1−xis not invertible}. 2.2

The spectral radius ofxis defined to be

rx sup

λ∈σx|λ|. 2.3

We note that a subalgbra of a normed algebra is itself a normed algebra with the norm got by restriction. The closure of a subalgebra is a subalgebra. A closed subalgebra of a Banach algebra is a Banach algebra. IfBα_α∈Λis a family of subalgebras of an algebraX,then

α∈ΛBα

is a subalgebra also. Hence, for any subsetEofX,there is the smallest subalgebraAEofX containingE.This algebra is called the subalgebra ofX generated byE.IfEis the singleton {x},thenAEis the linear span of all powersxⁿofx.IfXis a normed algebra, the closed algebraBEgenerated by a setEis the smallest closed subalgebra containingE.We can see thatBE AE.

We denote byC_FS the Banach algebra of continuous functions from a topological spaceStoF,with the sup-norm

f

∞sup

x∈S

fx. 2.4 The following theorems are known as the Stone-Weierstrass approximation theorem forC_RSandC_CS,respectively. For the details, the readers are referred to5.

Theorem 2.1. LetAbe a subalgebra ofC_RSsuch that 1^◦Aseparates the points ofS,

2^◦Aannihilates no point ofS.

ThenAis dense inC_RS.

Theorem 2.2. LetSbe a compact space,Aa subalgebra ofC_CSsuch that 1^◦Aseparates the points ofS,

2^◦Aannihilates no point ofS,

3^◦f ∈Aimplies that the conjugatefoffis inA.

ThenAis dense inC_CS.

A character on a unital algebraXoverFis a nonzero homomorphismτ :X → F.We denote byΩXthe set of characters onX.Note that ifXis a unital abelian complex Banach algebra, then

σx {τx:τ∈ΩX} 2.5

for eachx∈Xsee6.

Remark 2.3. It is unknown if2.5is valid wheneverΩX/∅.Equation2.5obviously does not hold for a spaceXwithΩX ∅as the following example shows.

Example 2.4. LetX Cbe considered as a real unital abelian Banach algebra under ordinary complex multiplication and whose norm is the absolute value. We haveΩX ∅.Indeed, assume to the contrary that there is a non-zero homomorphismτ0onXandτ0i λ∈R,so

τ₀1 ii τ₀i−1 λ−1,

τ₀1 iτ0i 1 λλλ λ². 2.6

Thusλ−1λ λ²; soλis not a real number, which is a contradiction.

SinceΩX ∅,so{τ1:τ∈ΩX}∅butσ1 {1}.

We consider throughout this paper on Banach algebras X for which ΩX/∅ and satisfy2.5.

IfX is a unital abelian Banach algebra, it follows fromProposition 2.5thatΩXis contained in the closed unit ball ofX^∗.We endowΩXwith the relative weak^∗topology and call the topological spaceΩXthe character space ofX.

Detailed proofs of the following propositions can be found in6.

Proposition 2.5. LetXbe a unital abelian Banach algebra. Ifτ ∈ΩX,thenτ1.

Proposition 2.6. IfXis a unital Banach algebra, thenΩXis compact.

IfXis a unital abelian Banach algebra, andx∈X,we define a continuous functionx by

x:ΩX−→F, τ −→τx. 2.7

We callxthe Gelfand transform ofx,and the homomorphism

ϕ:X −→C_FΩX, x→x 2.8

is called the Gelfand representation.

The following two lemmas, Lemmas 2.7 and 2.10, will be used to prove our main theorem.

Lemma 2.7. LetXbe a unital abelian real Banach algebra with

inf{rx:x∈X, x1}>0. 2.9

Then one has the following:

ithe Gelfand representationϕis a bounded isomorphism, iithe inverseϕ⁻¹is also a bounded isomorphism.

Proof. i ϕ is injective since inf{rx : x ∈ X, x 1} > 0 implies kerϕ 0. It is easily checked thatϕis a bounded homomorphism, andϕXis a subalgebra ofC_RΩX separating the points of ΩX, and having the property that for anyτ ∈ ΩXthere is an elementx ∈ X such that xτ /0. In order to use the Stone-Weierstrass theorem to show thatϕX C_RΩX,we shall show thatϕXis closed. We show thatϕXis closed by showing thatϕXis complete. Let{x_n}be a Cauchy sequence inϕX.First, we show that the sequence{xn} {ϕ⁻¹xn}is Cauchy. Assume on the contrary that{xn}is not Cauchy.

Thus there existsε₀>0 and subsequences{zn}and{z_n}of{xn}such that

zn−z_n≥ε0 2.10

for eachn∈N.Writeyn zn−z_n/ε0,thenyn ≥1,for eachn∈N.But{xn}is Cauchy, and so we havey_n → 0.Thus

0<inf{rx:x∈X, x1} ≤ inf

n∈Nr y_n

yn

inf

n∈N

y_n yn

∞

0, 2.11

which is a contradiction. Hence{xn}must be Cauchy and sox_n → x₀,for somex₀ ∈ X.

Sincex _∞ ϕx_∞ ≤ x,for each x ∈ X, so xn → x₀.Thus ϕXis complete. The Stone-Weierstrass theorem can be applied to conclude thatϕis surjective.

iifollows from the open mapping theorem.

Remark 2.8. i Lemma 2.7 tells us that if X is a unital abelian real Banach algebra with property

inf{rx:x∈X,x1}>0, 2.12

thenXandC_RΩXare homeomorphic and isomorphic underϕ.Hence if we would like to consider the convergence of a sequence{xn}inX,we could look at the convergence of the corresponding sequence{xn}.

iiProperty2.12clearly implies the semisimplicity propertyrx⇔x0but the following example shows that it is stronger.

Example 2.9. Let l₁Zdenote the Banach algebra of complex-valued absolutely summable functions on the group of integersZunder convolution regarded as a real Banach algebra and letX be the real subalgebra ofl₁Z consisting of those functions that satisfyf−n fn, n∈Z.Then the maximal ideal space ofXequalsT R/Zand the Gelfand transform is precisely the Fourier transform which maps X into the real Banach algebra C_RT of continuous real-valued functions onC_RT under pointwise multiplication and maximum norm. Although the image of the Fourier transform is dense, it is clearly not all ofC_RT since it is simply the real-valued functions in the Wiener space which consists of complex- valued functions whose Fourier series are absolutely summable. ThereforeLemma 2.7shows thatXdoes not have Property2.12.

Lemma 2.10. LetXbe an infinite dimensional unital abelian real Banach algebra with

inf{rx:x∈X, x1}>0. 2.13

Then one has the following:

i ΩXis an infinite set,

iiif there exists a bounded sequence{xn}inX which contains no convergent subsequences and such that{τxn:τ∈ΩX}is finite for eachn∈N,then there is an elementx0∈X with

{τx0:τ ∈ΩX} 1,1 2,2

3,3 4, . . .

, 2.14

iiithere is an elementx0∈Xsuch that{τx0:τ ∈ΩX}is an infinite set,

ivthere exists a sequence{xn}inXsuch that{τxn:τ ∈ΩX} ⊂0,1,for eachn∈N, and{xn⁻¹{1}}is a sequence of nonempty pairwise disjoint subsets ofΩX.

Proof. LetXbe an infinite dimensional unital abelian real Banach algebra with

inf{rx:x∈X,x1}>0. 2.15

iIf suffices to show that ifΩXis a finite set, for then the closed unit ballBX ofX is compact, and this will lead to us a contradiction. Let ΩXbe a finite set, say {τ1, τ₂, . . . , τ_m}, and let{xn}be a sequence inB_X.

Since the sequences {τpxn}, p 1,2, . . . , m, are bounded, we can choose a subsequence{xnm}of{xn}such thatτpxnm → λp,for eachp1,2, . . . , m.

Defineψ :ΩX → Rbyψτp λ_p.Thus there existsx ∈ X such thatψ x, and consequently,x_nm → xsince

xnm−x _∞ sup

τ∈ΩX|xnmτ−xτ| max

1≤p≤mxnm

τp

−x

τp, 2.16

andτ_pxnm → xτ p,for eachp1,2, . . . , m.

So{xnm}is a subsequence of{xn}such thatxnm → x. ByRemark 2.8,xnm → x,where xϕ⁻¹x. ThusBXis compact.

iiLet{xn}be a bounded sequence inXwhich has no convergent subsequences and suppose that the set {τxn : τ ∈ ΩX}is finite for each n ∈ N.By Remark 2.8, we will consider{xn}as a sequence of Gelfand transforms{xn}.

First, we show that we can write ΩX

n∈N

G_n

∪F, 2.17

whereFis closed,Gn are all closed and open, and{F, G1, G₂, . . .}is a partition ofΩX.For eachn∈N,write

{τx_n:τ ∈ΩX}

λ_n,i:i1,2, . . . , m_n , Ln

x_n⁻¹ λ_n,i

:i1,2, . . . , m_n

. 2.18

Define

L

k∈N

A_k:A_k∈ Lk

\ {∅}. 2.19

Note thatxn⁻¹{λn,i}are all closed and open. SinceLn is a partition ofΩXfor eachn∈ N,Lis a partition ofΩX.There are two cases to be considered.

Case 1Lis infinite. Thus there existsi1such that

x₁⁻¹ λ_1,i1

∩

k

A_k

:A_k∈ Lk, k≥2

2.20

is an infinite set. Similarly, there existsi₂such that

x₁⁻¹ λ_1,i1

∩x₂⁻¹ λ_2,i2

∩

k

A_k

:A_k∈ Lk, k≥3

2.21

is an infinite set. Continuing in this process we obtain a sequence of the setsxn⁻¹{λn,in} ∈ Lnsuch that

⎧⎨

⎩ n j1

xj₋₁ λ_j,ij

∩

k

Ak

:Ak∈ Lk, k≥n 1

⎫⎬

⎭ 2.22

is an infinite set, for eachn∈N.

Write

H₁

i /i1

x₁⁻¹ λ_1,i

, H₂ x₁⁻¹

λ_1,i1

∩

i /i2

x₂⁻¹ λ_2,i

,

H₃ x₁⁻¹ λ_1,i1

∩x₂⁻¹ λ_2,i2

∩

i /i3

x₃⁻¹ λ_3,i

, . . . .

2.23

ThusH_nare all closed and open, and

ΩX

n∈N

Hn

∪

ΩX\

n∈N

Hn

, 2.24

where ΩX\

n∈NH_n is a nonempty closed set sinceΩX is compact. And sinceLhas infinite elements, we can see that there exists a subsequence{Gn}of{Hn}such that

n∈NGn

n∈NHnandGn/∅,for eachn∈N.

Hence we have

ΩX

n∈N

Gn

∪

ΩX\

n∈N

Gn

, 2.25

and{ΩX\

n∈NG_n, G1, G₂, . . .}is a partition ofΩX.

Case 2L {Li : i 1,2, . . . , m}. To show that this case leads to a contradiction, we first observe that ifτ, τare in the sameL_i∈ L,then

τxn τxn 2.26

for each n ∈ N. Write α_n,i τxn, if τ ∈ Li.There exists a subsequence {αn,1, α_n,2, . . . , α_n,m}_n∈Nof{α_n,1, α_n,2, . . . , α_n,m}_n∈Nsuch that for eachi1,2, . . . , m,

α_n,i−→αi 2.27

for someα1, α₂, . . . , αm ∈R^m.Define a Gelfand transformx :ΩX → Rbyxτ αi,if τ ∈Li.Since

x_n−x _∞ sup

τ∈ΩX|x_nτ−xτ| max

1≤i≤mα_n,i−α_i, 2.28

sox_n → x, which is a contradiction.

Now we conclude that

ΩX

n∈N

Gn

∪F, 2.29

whereF is closed,G_n is closed and open for eachn∈N,and{F, G1, G₂, . . .}is a partition of ΩX.Define a mapψ:ΩX → Rby

ψτ

⎧⎪

⎨

⎪⎩ n

n 1 ifτ∈Gn,

1 ifτ∈F. 2.30

We can check that the inverse image of each closed set inψΩXis closed. Therefore,ϕ⁻¹ψ is an element inX,sayx0,with

{τx0:τ ∈ΩX} 1,1 2,2

3,3 4, . . .

. 2.31

iiiAssume to the contrary that{τx :τ ∈ΩX}is finite for eachx ∈ X.SinceX is infinite dimensional, so, asBX is noncompact, there exists a bounded sequence{xn}inX which has no convergent subsequences. Hence{τxn:τ ∈ΩX}is finite for eachn∈N.It follows fromiithat

{τx0:τ∈ΩX} 1,1 2,2

3,3 4, . . .

2.32 for somex₀∈X,which is the contradiction.

ivFromiii, there is an elementx1 ∈X such that{τx1 :τ ∈ΩX}is an infinite set. We can choosex₁so that there exists a strictly decreasing sequence{an}such that

{an} ⊂x₁ΩX⊂0,1, a1<1, 2.33 andτx1 1 for someτ ∈ ΩX.Define a continuous functiong1 : 0,1 → 0,1to be linear on0, a1and ona1,1joining the points0,0anda1,1,andg1⊂g1a2,1.Put

x₂g₁◦x₁,for somex₂∈X,and define a continuous functiong₂ :0,1 → 0,1similar to the way we constructg1.The left part ofg2is the line joining the point0,0andg1a2,1 andg₂1∈g2g1a2,1.Then putx₃g₂◦x₂,for somex₃∈X.Continuing in this process we obtain a sequence of points{xn}such that{τxn:τ∈ΩX} ⊂0,1,for eachn∈N,and {xn⁻¹{1}}is a sequence of nonempty pairwise disjoint subsets ofΩX.We then obtain the required result.

3. Main Theorem

Now we prove our main theorem.

Theorem 3.1. LetXbe an infinite dimensional unital abelian real Banach algebra satisfying each of the following:

iifx, y∈Xis such that|τx| ≤ |τy|,for eachτ ∈ΩX,then x ≤ y, iiinf{rx:x∈X, x1}>0.

ThenXdoes not have fixed point property.

Proof. LetXbe an infinite dimensional unital abelian real Banach algebra satisfyingiand ii. Assume to the contrary that X has fixed point property. From Lemma 2.10iv, there exists a sequence{xn}inXsuch that

{τx_n:τ∈ΩX} ⊂0,1 3.1

for eachn∈N,and{xn⁻¹{1}}is a sequence of nonempty pairwise disjoint subsets ofΩX.

WriteAn xn⁻¹{1},and defineTn :En → Enby

x−→xnx, 3.2

where

E_n{x∈X: 0≤τx≤1 for each τ∈ΩX,and τx 1 ifτ ∈A_n}. 3.3 It follows fromithatTn : En → En is a nonexpansive mapping on the bounded closed convex setE_n,for eachn∈N.Indeed,E_nis bounded since

0<inf{rx:x∈X, x1} ≤r x x

! sup

τ∈ΩX

τ x x

! 1 x sup

τ∈ΩX|τx| 3.4

for eachx∈ X.It follows thatTn has a fixed point, sayyn,for eachn∈ N.Sinceyn xnyn, thusy_n x_ny_n,and then

ynτ

⎧⎨

⎩

0 ifτ is not inAn,

1 ifτ is inA_n, 3.5

for eachn∈N.SinceA₁, A₂, A₃, . . .are pairwise disjoint, soy_m−y_n1,ifm /n.Hence{y_n} has no convergent subsequences. FromLemma 2.7,{yn}has no convergent subsequences too.

It follows from the existence of{yn}andLemma 2.10iithat there exists an elementx0inX with

{τx0:τ ∈ΩX} 1,1 2,2

3,3 4, . . .

. 3.6

WriteA₀ x₀⁻¹{1}.DefineT₀:E₀ → E₀by

x−→x0x, 3.7

where

E₀{x∈X: 0≤τx≤1 for eachτ ∈ΩX,andτx 1 ifτ∈A₀}. 3.8

By i,T₀ is a nonexpansive mapping on the bounded closed convex setE₀.ThusT₀ has a fixed point, sayy₀,that is,y₀x₀y₀.Thusy₀x₀y₀.Consequently,

y₀τ

⎧⎨

⎩

0, if τ is not inx₀⁻¹{1},

1, if τ is inx₀⁻¹{1}. 3.9

The set x₀⁻¹{1} y₀⁻¹{1} is open in ΩX, since y₀ is continuous. Also the set x0⁻¹{n/n 1}is open inΩXfor eachn∈N,sincex0is continuous. Thus,

x₀⁻¹ n n 1

:n∈N

∪

x₀⁻¹{1}

3.10

is an open covering ofΩX.This leads to a contradiction, sinceΩXis compact.

From the above theorem we have the following.

Corollary 3.2. LetSbe a compact Hausdorfftopological space. IfC_RSis infinite dimensional, then C_RSfails to have the fixed point property.

Proof. C_RS satisfiesi,iiin Theorem 3.1. Indeed, if x, y ∈ C_RS is such that|τx| ≤

|τy|,for eachτ ∈Ω C_RS,then|xs| ≤ |ys|,for eachs∈S.Hencex ≤ y.And since rx sup

λ∈σx|λ|sup

s∈S|xs|x, 3.11

so inf{rx:x∈X, x1}1>0.

Let_∞Ndenote the Banach algebra of all real bounded sequences with the sup-norm.

The following two propositions tell us that there is a subalgebra of_∞Nwhich does not containc₀but fails to have the fixed point property.

Proposition 3.3. IfE is a subset of _∞N which contains an infinite bounded sequence and the identity, then the Banach subalgebra BE of_∞N generated by E fails to have the fixed point property.

Proof. LetEbe a subset of_∞Nwhich contains an infinite bounded sequence{zn}and the identity. It follows thatBEis unital and abelian.BEis infinite dimensional, since the set {{zn}ⁿ:n∈N}is a linearly independent subset ofBE.Next, we show thatBEsatisfies iandiiinTheorem 3.1.

Leta{a1, a₂, a₃, . . .}, b{b1, b₂, b₃, . . .} ∈BEbe such thata /band|τa| ≤ |τb|, for eachτ ∈ΩX.Defineτ_n:BE → Rby

τn{x1, x2, x3, . . .} xn 3.12

for eachn∈N.Henceτn ∈Ω BEfor eachn∈N,and thus

|an||τna| ≤ |τnb||bn| 3.13 for eachn∈N.Clearly,a ≤ b.

Since for eachx∈BEwe have x ≥rx sup

λ∈σx|λ| sup

τ∈ΩBE|τx| ≥sup

n∈N|τnx|x, 3.14

so inf{rx :x ∈ X, x 1} 1 > 0.Now it follows fromTheorem 3.1thatBEdoesn’t have the fixed point property.

Proposition 3.4. Letz{1/p,1/p²,1/p³, . . .}withp >1.Then the Banach subalgebraB{1, z}

of_∞Ngenerated by the identity andzdoes not contain the spacec0. Proof. We have

A{1, z}

"_n

i0

αizⁱ:αi ∈R, n∈N

. 3.15

Ifa{a1, a2, a3, . . .} ∈A{1, z},thena#_N

i0αizⁱ,for someN∈Nandαi ∈R.It follows thata_n#_N

i0α_i1/pⁿⁱ,for eachn∈N.WriteMmax_i1,...,N|αi|.Hence α0−M 1

pⁿ−1

!

≤an≤α0 M 1 pⁿ−1

!

3.16

for each n ∈ N. From the above inequality, and since a is arbitrary, we can see that the sequence{1,1/2,1/3, . . .}does not lie inA{1, z} B{1, z}.

4. Results on Complex Banach Algebras

LetXbe a unital abelian complex Banach algebra. Consider the following condition.

(A) For eachx∈X,there exists an elementy∈Xsuch thatτy τx,for eachτ ∈ΩX.

IfXsatisfies conditionA, thenϕXis a subspace ofC_CΩXwhich is closed under the complex conjugation. By using the Stone-Weierstrass theorem for the complex Banach algebraC_CSand following the proof ofLemma 2.7, we obtain the following result.

Lemma 4.1. LetXbe a unital abelian complex Banach algebra satisfying (A) and

inf{rx:x∈X, x1}>0. 4.1

Then one has the following:

ithe Gelfand representationϕis a bounded isomorphism, iithe inverseϕ⁻¹is also a bounded isomorphism.

UsingLemma 4.1we obtain the complex counterpart ofLemma 2.10.

Lemma 4.2. LetXbe an infinite dimensional unital abelian complex Banach algebra satisfying (A) and

inf{rx:x∈X, x1}>0. 4.2

Then one has the following:

i ΩXis an infinite set,

iiif there exists a bounded sequence{xn}inX which contains no convergent subsequences and such that{τxn:τ∈ΩX}is finite for eachn∈N,then there is an elementx0∈X with

{τx0:τ ∈ΩX} 1,1 2,2

3,3 4, . . .

, 4.3

iiithere is an elementx₀∈Xsuch that{τx0:τ ∈ΩX}is an infinite set,

ivthere exists a sequence{xn}inXsuch that{τxn:τ ∈ΩX} ⊂0,1,for eachn∈N, and{xn⁻¹{1}}is a sequence of nonempty pairwise disjoint subsets ofΩX.

By using Lemmas4.1and4.2, and by following the proof ofTheorem 3.1, we get the following theorem.

Theorem 4.3. LetXbe an infinite dimensional unital abelian complex Banach algebra satisfying (A) and each of the following:

iifx, y∈Xis such that|τx| ≤ |τy|,for eachτ ∈ΩX,thenx ≤ y, iiinf{rx:x∈X, x1}>0.

ThenXdoes not have the fixed point property.

Acknowledgments

The authors would like to express their thanks to the referees for valuable comments, especially, to whom that provides them Remark 2.8iiand Example 2.9 for completeness.

This work was supported by the Thailand Research Fund, grant BRG50800016.

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