Problems concerning $ n $-weak amenability of a Banach algebra

15 Pages·2016·0.3 MB·English

Czechoslovak Mathematical Journal
Alireza Medghalchi; Taher Yazdanpanah
n
Problems concerning -weak amenability of a Banach algebra
CzechoslovakMathematicalJournal,Vol.55(2005),No.4,863–876
PersistentURL:http://dml.cz/dmlcz/128029
Terms of use:
©InstituteofMathematicsASCR,2005
InstituteofMathematicsoftheCzechAcademyofSciencesprovidesaccesstodigitizeddocuments
strictlyforpersonaluse.EachcopyofanypartofthisdocumentmustcontaintheseTermsofuse.
Thisdocumenthasbeendigitized,optimizedforelectronicdeliveryand
stampedwithdigitalsignaturewithintheprojectDML-CZ:TheCzechDigital
MathematicsLibraryhttp://dml.cz
CzechoslovakMathematical Journal,55(130)(2005),863–876
PROBLEMS CONCERNING n-WEAK AMENABILITY
OF A BANACH ALGEBRA
(cid:0) (cid:1)(cid:3)(cid:2)(cid:4) (cid:5)(cid:7)(cid:6)(cid:9)(cid:8)(cid:11)(cid:10)(cid:12)(cid:5)(cid:14)(cid:13) (cid:15) (cid:16) (cid:8)(cid:17)(cid:1) (cid:18) (cid:16) (cid:2),Teheran, and (cid:8) (cid:16) (cid:5) (cid:4) (cid:20)(cid:21)(cid:8)(cid:22)(cid:6)(cid:9)(cid:13)(cid:23)(cid:8) (cid:24) (cid:25)(cid:9)(cid:8) (cid:24) (cid:8) (cid:16) ,Boushehr
(cid:19)
(Received October 24, 2002)
Abstract. Inthispaperweextendthenotionofn-weakamenabilityofaBanachalgebraA
wthheennAn∗∈i(cid:26)s.aTnecAhn(2icna)l-mcaolcduullaetiaonndssthhoiswitdheaatlweahdesntAo iasAnurmenbserregouflainrtoerreasntinidgearlesinultAs∗o∗n,
Banach algebras. We thenextendthe concept of n-weak amenability ton .
∈ (cid:27)
Keywords: Banach algebra, weakly amenable, Arensregular, n-weakly amenable
MSC 2000: 46H20, 46H40
1. Introduction
LetA beaBanachalgebra,X aBanachA-bimodule. ThenwedenotebyX the
∗
topological dual space of X; the value of x X at x X is denoted by x,x .
∗ ∗ ∗
∈ ∈ h i
We recall that X is a Banach A-bimodule under the actions
∗
x,ax = xa,x , x,x a = ax,x (a A, x X, x X ).
∗ ∗ ∗ ∗ ∗ ∗
h i h i h i h i ∈ ∈ ∈
A derivation D: A X is a (bounded) linear map such that
−→
D(ab)=D(a)b+aD(b) (a,b A).
∈
For each x X, δ (a) = ax xa is a derivation, which is called inner. The first
x
∈ −
cohomology group H1(A,X) is the quotient of the space of derivations by the in-
ner derivations, and in many situations triviality of this space is of considerable
importance. In particular, A is called contractible if H1(A,X) = 0 for every
{ }
Banach A-bimodule X, A is called amenable if H1(A,X ) = 0 for every Ba-
∗
{ }
nach A-bimodule X, is called n-weakly amenable if H1(A,A(n)) = 0 , and
A { }
863
weaklyamenableifA is1-weaklyamenable. Forthetheoryofamenableandweakly
amenable Banach algebras see [1], [2], [4], [6], [8] and [9] for example.
LetA beaBanachalgebra. Givena A andF A , then Fa anda F are
∗ ∗ ∗∗ ∗ ∗
∈ ∈
defined in A by the formulae
∗
a,Fa∗ = a∗a,F , a,a∗F = aa∗,F (a A).
h i h i h i h i ∈
Next, for F,G A∗∗, F (cid:3)G and F G are defined in A∗∗ by the formulae
∈ 4
a∗,F (cid:3)G = Ga∗,F , a∗,F G = a∗F,G (a∗ A∗).
h i h i h 4 i h i ∈
ThenA∗∗ isaBanachalgebrawithrespecttoeitheroftheproducts(cid:3)and . These
4
products are called the first and second Arens products on A , respectively. The
∗∗
algebra A is called Arens regular if the two products (cid:3) and coincide. For the
4
general theory of Arens products, see [5] and [10], for example.
Let A be a Banach algebra, n 0 and let P : A(n) A(n+2) be the
∈ (cid:28) ∪{ } n −→
natural embedding, i.e., ϕ ,P ϕ = ϕ ,ϕ (ϕ A(n), ϕ A(n+1)),
n+1 n n n n+1 n n+1
h i h i ∈ ∈
where A(0) = A and A(n) is the nth dual of A. We shall require the following
standard properties of the Arens products. Suppose (a ) and (b ) are nets in A
α β
with P a F and P b G in (A ,σ), where σ =σ(A ,A ) is the weak
0 α 0 β ∗∗ ∗∗ ∗ ∗
−→ −→
topology on A∗∗. Then F (cid:3)G = limlimP0(aαbβ) and F G = limlimP0(aαbβ)
α β 4 β α
in (A∗∗,σ). Also, for a A and F A∗∗, we have P0(a) F = P0(a)(cid:3)F and
∈ ∈ 4
F P0(a)=F (cid:3)P0(a).
4
By easy calculations we can obtain the following properties of the P maps.
n
Lemma 1.1. Let m and n 0 . Then
∈ (cid:28) ∈ (cid:28) ∪{ }
(i) P P =P P ;
n∗∗ n n+2 n
(ii) P P =id;
n∗ n+1
(iii) P(2m+1)P ...P P =P ...P P ;
n n+2m+1 n+3 n+1 n+2m 1 n+3 n+1
(iv) Pn(2m)Pn+2m 2 =Pn+2mPn(2m−2). −
−
Lemma 1.2. Let A be a Banach algebra, n and let D: A A(n) be a
∈ (cid:28) −→
derivation. Then P P ...P D(2m)P P ...P =D (m ).
n∗−1 n∗+1 n∗+2m−3 2m−2 2m−4 0 ∈ (cid:28)
(cid:29)(cid:31)(cid:30)! " $# . It is enough to show that P D(2m)P = D(2m 2) for all
n∗+(2m 3) 2m 2 −
m . For ϕ A(2m 2) and ψ A(n+2m 3) we−have −
∈ (cid:28) ∈ − ∈ −
ψ,P D(2m)P (ϕ) = D(2m 1)P (ψ),P (ϕ)
h n∗+2m−3 2m−2 i h − n+2m−3 2m−2 i
ϕ,D(2m 1)P (ψ) = ψ,D(2m 2)(ϕ) ,
− n+2m 1 −
h − i h i
and so Pn∗+(2m 3)D(2m)P2m 2 =D(2m−2). (cid:3)
− −
864
2. When A(m) is an A(2n)-module?
Let A be a Banach algebra. Clearly A(4) is a Banach algebra with four Arens
products. We denote these algebras by (A4,(cid:3)(cid:3)) = ((A∗∗,(cid:3))∗∗,(cid:3)), (A4, (cid:3)) =
4
((A∗∗, )∗∗,(cid:3)), (A4,(cid:3) ) = ((A∗∗,(cid:3))∗∗, ), (A4, ) = ((A∗∗, )∗∗, ). For
4 4 4 44 4 4
a A and ϕ A(4) it is easy to check that
∈ ∈
P2P0(a)(cid:3)(cid:3)ϕ=P2P0(a)(cid:3) ϕ=P2P0(a) (cid:3)ϕ=P2P0(a) ϕ,
4 4 44
ϕ(cid:3)(cid:3)P2P0(a)=ϕ(cid:3) P2P0(a)=ϕ (cid:3)P2P0(a)=ϕ P2P0(a).
4 4 44
Let A be a Banachalgebra and n . Consider the maps (a ,ϕ ) a ϕ and
∈ (cid:28) ∗ 2n 7→ ∗· 2n
(a ,ϕ ) ϕ a from A A(2n) into A defined by
∗ 2n 2n ∗ ∗ ∗
7→ · ×
a,a ϕ = P ...P P (aa ),ϕ ,
∗ 2n 2n 3 3 1 ∗ 2n
h · i h − i
a,ϕ2n a∗ = P2n 3...P3P1(a∗a),ϕ2n (a A).
h · i h − i ∈
Then a ϕ = a P P ...P (ϕ ) and ϕ a = P P ...P (ϕ )a .
∗ · 2n ∗ 1∗ 3∗ 2∗n−3 2n 2n · ∗ 1∗ 3∗ 2∗n−3 2n ∗
Clearly these maps are continuous and bilinear. Note that with respect to these
actions A is not necessarily a Banach A(2n)-module. By dualizing these actions
∗
we obtain continuous bilinearmaps from A(m) A(2n) into A(m) for everym .
× ∈ (cid:28)
For example, for F A and ϕ A(2n) we have
∗∗ 2n
∈ ∈
a ,F ϕ = ϕ a ,F
∗ 2n 2n ∗
h · i h · i
= P P ...P (ϕ )a ,F
h 1∗ 3∗ 2∗n−3 2n ∗ i
=ha∗,F (cid:3)P1∗P3∗...P2∗n−3(ϕ2n)i (a∗ ∈A∗),
aFn.dFsroomF·nϕo2wn =onFw(cid:3)ePre1∗gPa3r∗d..th.Pes2∗en−ac3t(iϕo2nns)a.sSAim(i2lna)r-layc,tϕio2nns·Fon=AP(1∗mP)3∗in.d..uPce2∗nd−fr3o(ϕm2nA)4.
∗
Nowconsiderthemaps(F,ϕ ) F ϕ and(F,ϕ ) ϕ F fromA A(2n)
2n 2n 2n 2n ∗∗
7→ · 7→ · ×
into A defined by
∗∗
a ,F ϕ = P ...P P (a F),ϕ ,
∗ 2n 2n 3 3 1 ∗ 2n
h · i h − i
a∗,ϕ2n F = P2n 3...P3P1(Fa∗),ϕ2n (a∗ A∗).
h · i h − i ∈
Clearly these are continuous bilinear maps, F ϕ =F P P ...P (ϕ ) and
· 2n 4 1∗ 3∗ 2∗n−3 2n
sfrimomilatrhlye ϕac2tnio·nFs i=ndPu1∗cPed3∗f.r.o.mP2∗nA−3.(ϕA2ng)ai(cid:3)nFby. dNuoatleiztinhgatththeeseseaacctitoionnsswaerehadviffeecroennt-
∗
tinuous bilinear maps from A(m) A(2n) into A(m) for every m > 2. So we have
×
A(2n)-actions on A(m) (m>2) induced from A∗∗.
865
Let A be a Banach algebra and let n,k ∈ (cid:28) be such that n > 2k. Set B =
(A(2k), ), where is one of the 2k Arens products on A(2k). Then B is a Banach
· ·
algebra and B is a Banach B-module. By a similar argument we have continuous
∗
bilinear maps from B A(2n) into B and from A(2n) into B . Therefore
∗ ∗ ∗∗ ∗∗
× B ×
foreverym>2k+1wehaveA(2n)-actionsonA(m) inducedfromB∗ andforevery
m>2k+2 we have A(2n)-actions on A(m) induced from B∗∗.
Proposition 2.1. Let A be an Arens regularBanachalgebra and n . Then
∈ (cid:28)
A is a Banach A(2n)-bimodule with actions induced from A and any of Arens
∗ ∗
products on A(2n). In particular, A(m) is a Banach A(2n)-bimodule by actions
induced from A .
∗
(cid:29)(cid:31)(cid:30)! " $# . When n = 1, one can immediately see that A is a left Banach
∗
(A∗∗,(cid:3))-module and a right Banach (A∗∗, )-module. Since A is Arens regular,
4
A is a left and right Banach A -module. For a A, a A and F,G A we
∗ ∗∗ ∗ ∗∗
∈ ∈ ∈
have
a,(Fa∗)G = (aF)a∗,G = a∗,G(cid:3)(aF)
h i h i h i
= a∗G,P0(a)(cid:3)F = F(a∗G),P0(a)
h i h i
= a,F(a G) ,
∗
h i
and so (Fa )G=F(a G). Hence A is a Banach A -bimodule. Now suppose the
∗ ∗ ∗ ∗∗
result has been proved for n. We may assume that A(2n+2) = ((A(2n))∗∗,(cid:3)). Let
a A, a A , ϕ,ψ A(2n+2) and let (ϕ ) , (ψ ) be nets in A(2n) such that
∗ ∗ α β
∈ ∈ ∈
P (ϕ ) ϕ and P (ψ ) ψ in the weak topology. Then
2n α 2n β ∗
−→ −→
a,a∗ (ϕ(cid:3)ψ) = limlim P2n 3...P3P1(aa∗),ϕαψβ
h · i α β h − i
= limlim a,a∗ (ϕαψβ)
α β h · i
= limlim a,(a ϕ ) ψ
∗ α β
α β h · · i
= limlim a,a P P ...P (ϕ )P P ...P (ψ )
α β h ∗ 1∗ 3∗ 2∗n−3 α 1∗ 3∗ 2∗n−3 β i
= limlim P ...P P (aa P P ...P (ϕ )),ψ
α β h 2n−3 3 1 ∗ 1∗ 3∗ 2∗n−3 α βi
= liαmhaa∗P1∗P3∗...P2∗n−3(ϕα),P1∗P3∗...P2∗n−1(ψ)i
= lim P ...P (ψ)aa ,P P ...P (ϕ )
α h 1∗ 2∗n−1 ∗ 1∗ 3∗ 2∗n−3 α i
= a,a P ...P (ϕ)P ...P (ψ)
h ∗ 1∗ 2∗n−1 1∗ 2∗n−1 i
= a,(a ϕ) ψ ,
∗
h · · i
866
and so a∗ (ϕ(cid:3)ψ)=(a∗ ϕ) ψ. Similarly (ϕ(cid:3)ψ) a∗ =ϕ (ψ a∗). On the other
· · · · · ·
hand,
(ϕ·a∗)·ψ =(P1∗...P2∗n−1(ϕ)a∗)P1∗...P2∗n−1(ψ)
=P ...P (ϕ)(a P ...P (ψ))
1∗ 2∗n 1 ∗ 1∗ 2∗n 1
− −
=ϕ (a ψ).
∗
· ·
Hence A∗ is a Banach A(2n+2)-bimodule. So we are done by induction. (cid:3)
Proposition 2.2. Let A be an Arens regularBanachalgebra and n . Then
∈ (cid:28)
withanyoftheArensproductsonA(2n),theA(2n)-actionsonA inducedfromA
∗∗ ∗
andA coincide. Inparticular,A isaBanachA(2n)-bimodulewith anyofthese
∗∗ ∗∗
actions.
(cid:29)(cid:31)(cid:30)! " $# is straightforward. (cid:3)
Let A be a Banach algebra and m,n . Then A(2m) is a Banach algebra
∈ (cid:28)
with one of the 2m Arens products. We recall that every closed subalgebra of an
Arens regular Banach algebra is Arens regular. In particular, when A(2m) is Arens
regularforaBanachalgebraA andm , then A ,A(4),...,A(2m 2) areArens
∈ (cid:28) ∗∗ −
regular, and these algebras have only one Arens product. The following proposition
is a generalizationof Proposition 2.1 and Proposition 2.2.
Proposition 2.3. Let A be a Banach algebra and let n,m be such
∈ (cid:28)
that n > 2m. If A(2m) is Arens regular, then A(2m+1) and A(2m+2) are Banach
A(2n)-bimodules with actions induced from A(2m+1). Moreover, the A(2n)-actions
on A(2m+2) induced from A(2m+1) and A(2m+2) coincide.
Definition 2.4. Let A be a Banach algebra. A is called completely Arens
regular, if for every n , A(2n) is Arens regular.
∈ (cid:28)
It is well known that every C -algebra is Arens regular and the second dual of a
∗
C -algebraisaC -algebra. Therefore,everyC -algebraiscompletelyArensregular.
∗ ∗ ∗
Proposition 2.5. Let A be a completely Arens regular Banach algebra. Then
A(m) is a Banach A(2n)-module with actions induced A(m).
(cid:29)(cid:31)(cid:30)! " $# . A direct consequence of Proposition 2.3. (cid:3)
867
Lemma 2.6. LetA be aBanachalgebraand P (A) aleft (right)idealinA .
0 ∗∗
Then A∗ is a Banach (A∗∗,(cid:3))-module((A∗∗, ))-module).
4
(cid:29)(cid:31)(cid:30)! " $# . For a A , a A and F,G A we have
∗ ∗ ∗∗
∈ ∈ ∈
a,a∗(F (cid:3)G) = G(aa∗),F = G P0(a)a∗,F
h i h i h 4 i
= a ,F G P (a) = (a F)G,P (a)
∗ 0 ∗ 0
h 4 4 i h i
= a,(a∗F)G
h i
and
a,F(a∗G) = a∗G(cid:3)P0(a),F = a∗,G(cid:3)P0(a)(cid:3)F
h i h i h i
= F(a G),P (a) = a,F(a G) .
∗ 0 ∗
h i h i
Therefore A∗ is a Banach (A∗∗,(cid:3))-module. (cid:3)
Lemma 2.7. Let A be a Banach algebra. Then P (A) is an ideal in A with
0 ∗∗
anyof the Arensproductsif andonlyif P P (A)isanidealinA(4) with anyof the
2 0
Arens products.
(cid:29)(cid:31)(cid:30)! " $# . Let P (A) be an ideal in A . For a A and ϕ A(4), one can
0 ∗∗
∈ ∈
immediately see that
P2P0(a)(cid:3)(cid:3)ϕ=P2(P0(a)(cid:3)P1∗(ϕ)) and ϕ(cid:3)(cid:3)P2P0(a)=P2(P1∗(ϕ)(cid:3)P0(a)).
Therefore P P (A) is an ideal in A(4) with any of the Arens products on A(4).
2 0
Conversely, let P P (A) be an ideal in A(4). Take a A and F A . It is easy
2 0 ∗∗
∈ ∈
toseethatP2(P0(a)(cid:3)F)=P2P0(a)(cid:3)(cid:3)P2(F) P2P0(A). HenceP0(a)(cid:3)F P0(A)
∈ ∈
and similarly F (cid:3)P0(a) P0(A). Therefore P0(A) is an ideal in A∗∗. (cid:3)
∈
Proposition2.8. LetA beaBanachalgebraandP (A)anidealinA . Then
0 ∗∗
A is a Banach A(2n)-module with any of the Arens products on A(2n) (n ).
∗ ∈ (cid:28)
(cid:29)(cid:31)(cid:30)! " $# . Whenn=1theresultistruebyLemma2.6. Nowsuppose,inductively,
theresulthasbeenprovedforn 1. WemayassumethatA(2n+2) =((A(2n))∗∗,(cid:3)).
−
Let a A, a A and ϕ,ψ A(2n+2), and let (ϕ ), (ψ ) be nets in A(2n) such
∗ ∗ α β
∈ ∈ ∈
that P (ϕ ) ϕ and P (ψ ) ψ in the weak topology. Now we have
2n α 2n β ∗
−→ −→
a,a∗ (ϕ(cid:3)ψ) = P2n 1...P3P1(aa∗),ϕ(cid:3)ψ
h · i h − i
= limlim a ,P P ...P (ϕ ) P P ...P (ψ ) P (a)
α β h ∗ 1∗ 3∗ 2∗n−3 α 4 1∗ 3∗ 2∗n−3 β 4 0 i
= lim aa P P ...P (ϕ ),P P ...P (ψ)
α h ∗ 1∗ 3∗ 2∗n−3 α 1∗ 3∗ 2∗n−1 i
868
= liαmha∗,P1∗P3∗...P2∗n−3(ϕα)(cid:3)P1∗P3∗...P2∗n−1(ψ)(cid:3)P0(a)i
=haa∗,P1∗P3∗...P2∗n−1(ϕ)(cid:3)P1∗P3∗...P2∗n−1(ψ)i
= a,(a ϕ) ψ ,
∗
h · · i
and so a∗ (ϕ(cid:3)ψ)=(a∗ ϕ) ψ. Similarly (ϕ(cid:3)ψ) a∗ =ϕ (ψ a∗). Since A∗ is a
· · · · · ·
Banach A -module,
∗∗
(ϕ a ) ψ =(P P ...P (ϕ)a )P P ...P (ψ)
· ∗ · 1∗ 3∗ 2∗n−1 ∗ 1∗ 3∗ 2∗n−1
=P1∗P3∗...P2∗n 1(ϕ)(a∗P1∗P3∗...P2∗n 1(ψ))
− −
=ϕ (a ψ).
∗
· ·
Consequently, A∗ is a Banach A(2n)-module. (cid:3)
Proposition2.9. LetA beaBanachalgebra. ThenP2((A∗∗,(cid:3)))isaleft(right,
two-sided) ideal in (A(4),(cid:3)(cid:3)) if and only if P2∗ is an A(4)-module homomorphism
between left (right, two-sided) Banach A(4)-modules.
(cid:29)(cid:31)(cid:30)! " $# . LetP2((A∗∗,(cid:3)))bealeftidealin(A(4),(cid:3)(cid:3)). ForF A∗∗,ϕ4 A(4)
∈ ∈
and ϕ A(5) we have
5
∈
hF,P2∗(ϕ4ϕ5)i=hP2(F),ϕ4ϕ5i=hP2(F)(cid:3)(cid:3)ϕ4,ϕ5i
=hP2∗(ϕ5),P2(F)(cid:3)(cid:3)ϕ4i=hF,ϕ4P2∗(ϕ5)i.
Hence P is an A(4)-module homomorphism between left Banach A(4)-modules.
2∗
Conversely, for F A , ϕ A(4) it is easy to see that
∗∗ 4
∈ ∈
P2(F)(cid:3)(cid:3)ϕ4 =P2(P1∗(P2(F)(cid:3)(cid:3)ϕ4)),
so P2((A∗∗,(cid:3))) is a left ideal in (A(4),(cid:3)(cid:3)). (cid:3)
3. N-weak amenability for N Z
∈
Lemma 3.1. LetA beaBanachalgebraandD: A A aderivation. Then
∗
−→
(i) D∗∗: (A∗∗,(cid:3)) ( ∗∗)∗ is satisfied in
−→ A
D∗∗(F (cid:3)G)=D∗∗(F)G+P0∗∗(F)D∗∗(G) (F,G∈A∗∗),
(ii) D : (A , ) (A ) is satisfied in
∗∗ ∗∗ ∗∗ ∗
4 −→
D (F G)=D (F)P (G)+FD (G) (F,G A ).
∗∗ 4 ∗∗ 0∗∗ ∗∗ ∈ ∗∗
869
(cid:29)(cid:31)(cid:30)! " $# . (i) Let F,G A and let (a ), (b ) be nets in A such that
∗∗ α β
∈
P (a ) F and P (b ) G in the weak topology. We have
0 α 0 β ∗
−→ −→
H,D∗∗(F (cid:3)G) = D∗(H),F (cid:3)G
h i h i
= limlim D(a )b +a D(b ),H
α β α β
α β h i
= limlim b ,HD(a )+D (Ha )
β α ∗ α
α β h i
= liαmhaα,D∗(G(cid:3)H)+P0∗(D∗∗(G)H)i
= H,D (F)G+P (F)D (G)
h ∗∗ 0∗∗ ∗∗ i
and so D∗∗(F (cid:3)G)=D∗∗(F)G+P0∗∗(F)D∗∗(G).
(ii) The proof is similar to (i). (cid:3)
Corollary 3.2. Let A be a Banach algebra and D: A A a derivation.
∗
−→
Then
(i) D∗∗: (A∗∗,(cid:3)) −→ (A∗∗)∗ is a derivation if and only if P0∗∗(F)D∗∗(G) =
FD (G) for F,G A ;
∗∗ ∗∗
∈
(ii) D : (A , ) (A ) is a derivation if and only if D (F)P (G) =
∗∗ ∗∗ 4 −→ ∗∗ ∗ ∗∗ 0∗∗
D (F)G for F,G A .
∗∗ ∗∗
∈
Definition3.3. LetA beaBanachalgebram,n∈ (cid:28) , and16m<2n. TheBa-
nachalgebraA(2n) iscalled( m)-weaklyamenable,ifA(2n m) isaBanachA(2n)-
−
−
bimodule with actions induced from A(2n m) and H1(A(2n),A(2n m))= 0 .
− −
{ }
Theorem3.4. LetA beaBanachalgebraandP (A)aleft(right)idealinA .
0 ∗∗
If (A∗∗,(cid:3))((A∗∗, )) is ( 1)-weakly amenable, then A is weakly amenable.
4 −
(cid:29)(cid:31)(cid:30)! " $# . By Lemma 2.6, A∗ is a Banach (A∗∗,(cid:3))-module. Let D: A A∗
−→
be a derivation. Put d = P0∗D∗∗: (A∗∗,(cid:3)) −→ A∗. For F,G ∈ A∗∗, a ∈ A we
have
ha,P0∗(D∗∗(F)G)i=hG(cid:3)P0(a),D∗∗(F)i=hd(F),G4P0(a)i=ha,d(F)Gi
and
ha,P0∗(P0∗∗(F)D∗∗(G))i=hP0∗(D∗∗(G)P0(a)),Fi=hd(G)P0(a),Fi=ha,Fd(G)i.
Therefore, by Lemma 3.1, d is a derivation. Since H1((A∗∗,(cid:3)),A∗) = 0 , there
{ }
exists a A such that d=δ . Using Lemma 1.2 we obtain
∗ ∗ a
∈ ∗
aa a a=P (a)a a P (a)=dP (a)
∗ ∗ 0 ∗ ∗ 0 0
− −
=P D P (a)=D(a) (a A).
0∗ ∗∗ 0 ∈
Hence D=δa is an inner derivation. (cid:3)
∗
870
Theorem 3.5. LetA beaBanachalgebra. IfP (A)isanidealinA andthe
0 ∗∗
Banach algebra A(2n) (n ) with one of 2n Arens products is ( 2n+1)-weakly
∈ (cid:28) −
amenable, then A is weakly amenable.
(cid:29)(cid:31)(cid:30)! " $# . Let D: A A be a derivation. We claim that
∗
−→
dn =P0∗P0∗∗∗...P0(2n−1)D(2n): A(2n) −→A∗
is a derivation. By Proposition 3.4, the result is true for n = 1. Now suppose,
inductively, that the result has been provedfor n. We may suppose that A(2n+2) =
((A(2n))∗∗,(cid:3)). For a A, a∗ A∗, ϕ,ψ A(2n+2), let (ϕα) and (ψβ) be nets
∈ ∈ ∈
in A(2n) such that P (ϕ ) ϕ and P (ψ ) ψ in the weak topology. Then
2n α 2n β ∗
−→ −→
we have
a,dn+1(ϕ(cid:3)ψ) = limlim a,dn(ϕα)ψβ +ϕαdn(ψβ)
h i α β h i
= limlim P ...P P (ad (ϕ )),ψ
2n 3 3 1 n α β
α β h − i
+limlim ψ ,D(2n+1)P(2n)...P (aP ...P (ϕ ))
α β h β 0 0∗∗ 1∗ 2∗n−3 α i
= liαmhadn(ϕα),P1∗...P2∗n−1(ψ)i
+lim d (ψ)a,P ...P (ϕ )
α h n+1 1∗ 2∗n−3 α i
= a,d (ϕ) ψ+ϕ d (ψ) ,
n+1 n+1
h · · i
so d is a derivation. Since H1(A(2n), )= 0 , there exists a A such that
n+1 ∗ ∗ ∗
A { } ∈
d =δ . Using Lemma 1.1(iv) and Lemma 1.2, we conclude that
n a
∗
aa∗ a∗a=P2n 2...P2P0(a) a∗ a∗ P2n 2...P2P0(a)
− − · − · −
=d P ...P P (a)=D(a) (a A).
n 2n 2 2 0
− ∈
Hence D=δa is inner. (cid:3)
∗
Lemma 3.6. Let A be a Banach algebra, n and let D: A A(2n) be a
∈ (cid:28) −→
derivation. Then for every F,G A
∗∗
∈
(i) D∗∗: (A∗∗,(cid:3)) ((A(2n))∗∗,(cid:3)) holds in
−→
D∗∗(F(cid:3)G)=D∗∗(F)(cid:3)P2∗n∗ 2...P2∗∗P0∗∗(G)+P2∗n∗ 2...P2∗∗P0∗∗(F)(cid:3)D∗∗(G).
− −
(ii) D : (A , ) ((A(2n)) , ) holds in
∗∗ ∗∗ ∗∗
4 −→ 4
D (F G)=D (F) P ...P P (G)+P ...P P (F) D (G).
∗∗ 4 ∗∗ 4 2∗n∗−2 2∗∗ 0∗∗ 2∗n∗−2 2∗∗ 0∗∗ 4 ∗∗
(cid:29)(cid:31)(cid:30)! " $# is straightforward. (cid:3)
871