FOR A LOCALLY COMPACT TOPOLOGICAL SEMIGROUP

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MULTIPLIERS ON L(S) , L(S)

, AND LUC(S)

FOR A LOCALLY COMPACT TOPOLOGICAL SEMIGROUP

ALIREZA MEDGHALCHI Received 16 January 2001

We study compact and weakly compact multipliers onL(S),L(S)^∗∗, andLUC(S)^∗, where the latter is the dual of LUC(S). We show that a left cancellative semigroup S is left amenable if and only if there is a nonzero compact (or weakly compact) multiplier on L(S)^∗∗. We also prove thatSis left amenable if and only if there is a nonzero compact (or weakly compact) multiplier onLUC(S)^∗.

2000 Mathematics Subject Classification: 43A20.

1. Introduction. LetS be a locally compact, Hausdorff topological semigroup. Let M(S)be the space of all complex Borel measures onS. It is known thatM(S)=C0(S)^∗, therefore,M(S)is a Banach space and with convolutionµ∗ν(ψ)=

ψ(xy)dµ(x)dν(y) (µ, ν∈M(S), ψ∈C0(ψ)),M(S) is a Banach algebra. The subalgebraL(S)ofM(S) is defined by L(S)= {µ ∈M(S)|x → |µ| ∗δx, x →δx∗ |µ|from S to M(S) are norm continuous}[1]. A semigroupS is called foundation if S =

µ∈L(S)suppµ. A trivial example is a topological group and in this caseL(S)=L¹(G). LetCb(S)be the set of all bounded continuous function onS. LetLUC(S)= {f∈Cb(S)|x→lxf is norm continuous}, RUC(S)= {f |f ∈Cb(S), x→rxf is norm continuous}where lxf (y)=f (xy),rxf (y)=f (yx). WhenS is foundation, it is known thatL(S)has a bounded approximate identity [1], and therefore, the multiplier algebra ofL(S)is M(S)[4]. LetL(S)^∗ andL(S)^∗∗be the first and second duals ofL(S)and similarly, M(S)^∗ andM(S)^∗∗be the first and second duals ofM(S). We also use the notation LUC(S)^∗,RUC(S)^∗ for the duals ofLUC(S), andRUC(S), respectively. The subalge- brasLUC(S)andRUC(S)are BanachC^∗-subalgebras ofCb(S). With Arens product, L(S)^∗∗andM(S)^∗∗are Banach algebra. Also, with the same type productLUC(S)^∗is a Banach algebra. In this paper, among other things, we show that whenSis a left can- cellative foundation semigroup, thenSis left (right) amenable if and only if there is a nonzero left (right) compact or weakly compact multiplier onL(S)^∗∗(orLUC(S)^∗).

2. Preliminaries. For a Banach algebraA, we denote byA^∗ andA^∗∗the first and second dual ofA, respectively. OnA^∗∗we define the first Arens product by

mn, f = m, nf, nf , a = n, f a, f a, b =f (ab) (2.1) (m, n∈A^∗∗; f∈A^∗; a, b∈A). With this productA^∗∗is a Banach algebra. We can also define a similar product on LUC(S)^∗ such that mn, f = m, nf, nf (x)= n(lxf ),lxf (y)=f (xy) (m, n∈LUC(S)^∗; f∈LUC(S); x, y∈S). Clearly,LUC(S)^∗ is a Banach algebra. A linear map on a Banach algebra A is called a multiplier if

T (xy)=T (x)y=xT (y) (x, y∈A). The left (right) multiplier onL(S)^∗∗is defined by lm(n)=mn,(lm(n)=nm). In general,LUC(S)andRUC(S)are different subalgebras ofCb(S)andLUC(S)=RUC(S)if and only ifLUC(S)(resp.,RUC(S)) is right (resp., left) introverted, (see [2, Theorem 4.4.5]). For example, ifSis a compact semitopolog- ical semigroup or a totally bounded topological group, thenLUC(S)=RUC(S)[2].

The semigroupSis called left amenable if there is a positive functionalmonLUC(S) such thatm(laf )=m(f ),m =1 for allf∈LUC(S),a∈S. Suchmis called a left invariant mean onLUC(S)[7].

LetAbe a Banach algebra andBa closed subalgebra ofAandi:B→Athe inclusion mapping, thenπ:A^∗→B^∗is the restriction mapping which is norm decreasing and onto (by the Hahn-Banach theorem). Following Ghahramani and Lau [3], we have the following lemma (see [3, Lemmas 1.1, 1.2, 1.4, Proposition 1.3]).

Lemma2.1. (a)Letf∈A^∗,b∈B. Thenbπ (f )=π (i(b)f ).

(b)The mappingπ^∗:B^∗∗→A^∗∗is a homeomorphism wheneverB^∗∗has the weak^∗- topology andπ^∗(B^∗∗)has the relative weak^∗-topology.

Lemma2.2. LetBbe a closed ideal inA,n∈A^∗∗. If(aα)is a bounded net inAsuch thataα→n, theni(b)aα

ω^∗

→π^∗(b)n (b∈B).

Proposition2.3. LetBbe a right (or left) ideal ofA. Thenπ^∗(B^∗∗)is a right (resp., left) ideal ofA^∗∗.

Lemma2.4. LetAbe a commutative Banach algebra. Then any weak^∗-closed right ideal inA^∗∗is an ideal. IfX=specA, thenh(n)= n, δxis a multiplicative onA^∗∗, whereδx(ψ)= x, ψ.

3. Multipliers onLUC(S)^∗andL(S)^∗∗. First we prove a theorem which is new even for topological groups.

Theorem3.1. LetS be a right cancellative topological semigroup with identitye.

Then the following are equivalent:

(a)Sis left amenable.

(b)There is a nonzero compact (or weakly compact) right multiplier on LUC(S)^∗. Proof. (a)⇒(b). LetSbe left amenable andmbe a left invariant mean onLUC(S).

Then nm, f = n, mf, mf (x)=m(lxf )=m(f ) (f ∈LUC(S)^∗, f ∈LUC(S)).

Therefore,nm, f = n, m(f ) =m(f )n,1, that is,nm= n,1m. Thuslm(n)= n,1mis a rank one operator and hence compact.

(b)⇒(a). Let T be a nonzero weakly compact right multiplier on LUC(S)^∗. Then T (m)=T (mδe)=mT (δe)=lT (δ_e)m. So, T =ln wheren=T (δe). Note that δe∈ LUC(S)^∗andδe(f )=f (e) (f∈LUC(S)). Now, letA= {δxn|x∈S} = {δxT (δe)|x∈ S} = {T (δx)|x∈S}which is weakly compact. By Krein-Smulian’s theoremK=co^ωA is weakly compact [2]. Now, we show that ifk≠k∈K, thenδxk1 ≤ k1. On the other hand, if we define

g(y)=





f (t), y=tx,

0, otherwise, (3.1)

thengis well defined and belongs toβ(S)(the space of bounded functions onS), then δxg(t)=δx(ltg)=g(tx)=rxg(t)=f (t). Let ¯k1be the extension ofk1toβ(S)(by the Hahn-Banach theorem). Then

k1=¯k≤sup ¯k1, ff∈β(S)

=sup ¯k1, δxgg∈β(S)

=sup δxk¯1, gg∈β(S)

=δx¯k1

=δxk1.

(3.2)

It follows thatδxk1 = k1≠0. Now, we show that ifk, k∈co(A), andk≠k, then a similar argument shows thatδx(k−k)≠0. Finally, we show that 0∈ {δx(k−k)| x∈S}since, by a completely similar argument, we haveδx_α(k−k) = k−k≠0.

Therefore, 0∈ {δx(k−k)|x∈S}⁻. Hence, by Ryll-Nardzewski fixed point theorem [2], there exists a pointq ∈K such thatδxq =q. It follows that δx|q| = |δxq| =

|q|, and q = n≠0. Now, if we take m= |q|/q, then clearly δxm=m, so, m(f )=δxm(f )=δx(mf )=mf (x)=m(xf ). Therefore,mis a left invariant mean onLUC(S), that is,Sis left amenable.

For a foundation semigroupS, leti:LUC(S)→L(S)^∗be such thati(f ), µ = µ, f (f∈LUC(S), µ∈L(S))is an embedding andπ=i^∗:L(S)^∗∗→LUC(S)^∗is onto. It is clear from the proof of [3, Lemma 2.2] for topological groups thatπ (E)=δewhereE is a right identity,πis a homomorphism andF G=F π (G). Also we have the following proposition which is similar to [6, Theorem 2.3].

We prove the following proposition for foundation semigroups with identitye.

Proposition3.2. LetE be a right identity inL(S)^∗∗. Thenπ is an isometric iso- morphism ofEL(S)^∗∗onto LUC(S)^∗.

Proof. LetIbe the identity operator onL(S)^∗∗. Then

L(S)^∗∗=EL(S)^∗∗+(I−E)L(S)^∗∗. (3.3) Now, if m∈L(S)^∗∗, then π ((I−E)m)=π (m)−π (E)π (m)=π (m)−δeπ (m)= π (m)−π (m)=0. Thus (I−E)m∈kerπ. On the other hand, if m∈kerπ, then Em=Eπ (m)=0. Som=m−Em=(I−E)m∈(I−E)L(S)^∗∗. Thus,

kerπ=(I−E)L(S)^∗∗. (3.4)

So, we have

L(S)^∗∗=EL(S)^∗∗+kerπ . (3.5) It follows thatπis injective fromEL(S)^∗∗ontoL(S)^∗∗/kerπ, thereforeπis injec- tive fromEL(S)^∗∗ontoLUC(S)^∗, and soπ is an algebra isomorphism. We also have Em = Eπ (m) ≤ Eπ (m) = π (m) ≤ m, sinceπ is a quotient map. Thus π (Em) ≤ πEm ≤ Em ≤ π (m). Soπ (Em) = π (m) = Em, that is, πis an isometry.

Now, we have another main theorem.

Theorem3.3. LetS be a right cancellative locally compact foundation semigroup with identitye. Then the following are equivalent:

(a)Sis left amenable.

(b)There is a nonzero compact (or weakly compact) right multiplier onL(S)^∗∗. Proof. (a)⇒(b). The proof of this part exactly reads the same line of the proof of (a)⇒(b) ofTheorem 3.1, so it is omitted.

(b)⇒(a). LetT be a nonzero weakly compact right multiplier onL(S)^∗∗, soT =ln

for somen∈L(S)^∗∗. NowlEn is also a nonzero right multiplier onEL(S)^∗∗where E is a right identity of L(S)^∗∗ with norm 1, sincelEn(Em)=EmEn=Emn. Now byProposition 3.2,π (EL(S)^∗∗)=(LUC(S))^∗ isometrically isomorphic. If we define l_n=lEn◦π, thenl_n is a nonzero right multiplier onLUC(S)^∗. Therefore, S is left amenable.

In [3, Theorem 2.1] it was also shown that a locally compact groupGis amenable if and only if there is a nonzero compact (weakly compact) right multiplier onM(G)^∗∗. But we are not able to extend this result toM(S)^∗∗.

Proposition3.4. A right multiplierln(m)=mn (m∈LUC(S)^∗)is compact if and only if the restriction oflntoM(S)is compact.

Note3.5. It is clear thatM(S)⊆LUC(S)^∗ since, if µ∈M(S), then we can take µ, f =

Sf dµ (f∈LUC(S)).

Proof. Letlnbe compact, then clearly the restriction ofln toM(S)is compact.

Conversely, letln:M(S)→LUC(S)^∗be compact, whereln(µ)=µn (µ∈M(S)). Let m∈LUC(S)^∗withm ≤1. Since, the linear span ofδx’s is weak^∗-dense inLUC(S)^∗, there is a netµα=nα

i=1λα,iδx_α,isuch thatµα→min weak^∗-topology. By compactness ofln, there is a subnet(µα(β))such that(µα(β)n)converges in norm.

Now, we havemn=ω^∗−limµα(β)n. Thusmn=limµα(β)nwith norm topology. It follows that

mn| m ≤1

⊆

µn|µ∈L(S), µ ≤1

. (3.6)

Thus,lnis compact.

Theorem3.6. LetSbe a right cancellative semigroup with identityeandlna right multiplier on LUC(S)^∗. Thenlncan be written as a linear combination of four compact right multiplierln_i (i=1,2,3,4),ni≥0,ni∈LUC(S)^∗.

Proof. Letebe the identity ofS. ThenT (m)=T (mδe)=mT (δe)lT (δe)(m). So, T=ln(n=T (δe)∈LUC(S)^∗). Letn=n⁺₁−n⁻₁+i(n⁺₂−n⁻₂)wheren⁺_k, n⁻_k (k=1,2) are Hermitian. So, it suffices to show thatl_n+

k andl_n−

k are compact. ByProposition 3.4 it suffices to prove that the restrictions of these operators toM(S)are compact. Now sincelnis compact onLUC(S)^∗,{δxn|x∈S}⁻is compact. So{δxnx∈S}⁻is com- pact. Since,n⁺ ≤ n,{(δxn)⁺|x∈S}is compact. It follows that{δxn⁺|x∈S}⁻ is compact. Since the linear span ofδx,sis weak^∗dense inLUC(S)^∗,{µn⁺|µ∈M(S), µ ≤1}⁻is compact. Therefore,ln⁺is compact. This completes the proof.

We denote byβS the space of all multiplicative linear functional onLUC(S). We have another main theorem.

Theorem 3.7. LetS be a finite topological semigroup. Then there exists n∈βS such thatlnis compact. Conversely, ifSis a subsemigroup of a topological group with identity, and there existsn∈βSsuch thatlnis compact, thenS is finite.

Proof. LetSbe finite, then by [2, Corollary 4.1.8],AP (S)=C(S). Also, by [2, Propo- sition 4.4.8],AP (S)=LUC(S)=RUC(S). Therefore,LUC(S)=C(S). SoβSis topologi- cally isomorphic toS. On the other hand, sincelsS⊆Sis compact,l^∗_sC(S)is compact.

Hence,lnis compact.

Conversely, letln be compact for somen∈βS, byTheorem 3.6, we may assume thatnis positive, thenTn(f )=nf (f∈LUC(S))is compact. Now, letF=rangeTn. ClearlyTn is an algebra homomorphism, since,Tn(f g)=n(f g)(x)= n, lxf g = n((lxf )(lxg))=n(lxf )n(lxg)=Tn(f )Tn(g). AlsoTnpreserves conjugation. So, by [8, Theorem 5.3],Txf ≥ f(f∈LUC(S)). So by open mapping theorem,Tnis a homeomorphism. SinceTnis compact,Fis closed. Also,{Tnf|f∈LUC(S),Tnf ≤ 1} ⊆ {Tnf | f ∈LUC(S), f ≤ 1}, so {Tnf |f ∈LUC(S), Tnf ≤ 1} is com- pact. ThereforeF is reflexive. It follows thatF is finite dimensional (see [8, Exercise 2]). Let{m1, m2, . . . , mk}be the spectrum ofF and we can assume thatmi is posi- tive. If we definem(f )=(1/k)k

i=1mi(Tnf ), then clearly,m≥0,m(1)=1. Also, sinceSis left cancellative,l^∗_x{m1, . . . , mk} = {m1, . . . , mk}. Therefore,mi, Tnlx(f ) = mi, lxTn(f ) = l^∗_xmi, Tn(f ) = mj, T n(f ), for some 1 ≤j ≤ k. It follows that m(lxf )=m(f ), that is,mis a left-invariant mean onLUC(S), so by [5, Theorem 3]

Sis finite.

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Alireza Medghalchi: Department of Mathematics, Teacher Training University,566 Taleghani Avenue,13614Tehran, Iran

E-mail address:[email protected]