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Czechoslovak Mathematical Journal
Alireza Medghalchi; Taher Yazdanpanah
Problems concerning nweak amenability of a Banach algebra
Czechoslovak Mathematical Journal, Vol. 55 (2005), No. 4, 863876
Persistent URL: http://dml.cz/dmlcz/128029
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Czechoslovak Mathematical Journal, 55 (130) (2005), 863–876
PROBLEMS CONCERNING nWEAK AMENABILITY
OF A BANACH ALGEBRA
, Teheran, and , Boushehr
(Received October 24, 2002)
Abstract. In this paper we extend the notion of nweak amenability of a Banach algebra A
when n ∈ . Technical calculations show that when A is Arens regular or an ideal in A
∗∗
,
then A
∗
is an A
(2n)
module and this idea leads to a number of interesting results on
Banach algebras. We then extend the concept of nweak amenability to n ∈
.
Keywords: Banach algebra, weakly amenable, Arens regular, nweakly amenable
MSC 2000 : 46H20, 46H40
1. Introduction
Let A be a Banach algebra, X a Banach A bimodule. Then we denote by X
∗
the
topological dual space of X; the value of x
∗
∈ X
∗
at x ∈ X is denoted by hx, x
∗
i.
We recall that X
∗
is a Banach A bimodule under the actions
hx, ax
∗
i = hxa, x
∗
i, hx, x
∗
ai = hax, x
∗
i (a ∈ A , x ∈ X, x
∗
∈ X
∗
).
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
x
(a) = ax − xa is a derivation, which is called inner. The ﬁrst
cohomology group H
1
(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 H
1
(A , X) = {0} for every
Banach A bimodule X, A is called amenable if H
1
(A , X
∗
) = {0} for every Ba
nach A bimodule X, A is called nweakly amenable if H
1
(A , A
(n)
) = {0}, and
863
weakly amenable if A is 1weakly amenable. For the theory of amenable and weakly
amenable Banach algebras see [1], [2], [4], [6], [8] and [9] for example.
Let A be a Banach algebra. Given a
∗
∈ A
∗
and F ∈ A
∗∗
, then F a
∗
and a
∗
F are
deﬁned in A
∗
by the formulae
ha, F a
∗
i = ha
∗
a, F i, ha, a
∗
F i = haa
∗
, F i (a ∈ A ).
Next, for F, G ∈ A
∗∗
, F G and F 4 G are deﬁned in A
∗∗
by the formulae
ha
∗
, F Gi = hGa
∗
, F i, ha
∗
, F 4 Gi = ha
∗
F, Gi (a
∗
∈ A
∗
).
Then A
∗∗
is a Banach algebra with respect to either of the products and 4. These
products are called the ﬁrst and second Arens products on A
∗∗
, respectively. The
algebra A is called Arens regular if the two products and 4 coincide. For the
general theory of Arens products, see [5] and [10], for example.
Let A be a Banach algebra, n ∈ ∪ {0} and let P
n
: A
(n)
−→ A
(n+2)
be the
natural embedding, i.e., hϕ
n+1
, P
n
ϕ
n
i = hϕ
n
, ϕ
n+1
i (ϕ
n
∈ A
(n)
, ϕ
n+1
∈ A
(n+1)
),
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
α
) and (b
β
) are nets in A
with P
0
a
α
−→ F and P
0
b
β
−→ G in (A
∗∗
, σ), where σ = σ(A
∗∗
, A
∗
) is the weak
∗
topology on A
∗∗
. Then F G = lim
α
lim
β
P
0
(a
α
b
β
) and F 4 G = lim
β
lim
α
P
0
(a
α
b
β
)
in (A
∗∗
, σ). Also, for a ∈ A and F ∈ A
∗∗
, we have P
0
(a) 4 F = P
0
(a) F and
F 4 P
0
(a) = F P
0
(a).
By easy calculations we can obtain the following properties of the P
n
maps.
Lemma 1.1. Let m ∈
and n ∈ ∪ {0}. Then
(i) P
∗∗
n
P
n
= P
n+2
P
n
;
(ii) P
∗
n
P
n+1
= id;
(iii) P
(2m+1)
n
P
n+2m+1
. . . P
n+3
P
n+1
= P
n+2m−1
. . . P
n+3
P
n+1
;
(iv) P
(2m)
n
P
n+2m−2
= P
n+2m
P
(2m−2)
n
.
Lemma 1.2. Let A be a Banach algebra, n ∈
and let D : A −→ A
(n)
be a
derivation. Then P
∗
n−1
P
∗
n+1
. . . P
∗
n+2m−3
D
(2m)
P
2m−2
P
2m−4
. . . P
0
= D (m ∈ ).
. It is enough to show that P
∗
n+(2m−3)
D
(2m)
P
2m−2
= D
(2m−2)
for all
m ∈ . For ϕ ∈ A
(2m−2)
and ψ ∈ A
(n+2m−3)
we have
hψ, P
∗
n+2m−3
D
(2m)
P
2m−2
(ϕ)i = hD
(2m−1)
P
n+2m−3
(ψ), P
2m−2
(ϕ)i
hϕ, D
(2m−1)
P
n+2m−1
(ψ)i = hψ, D
(2m−2)
(ϕ)i,
and so P
∗
n+(2m−3)
D
(2m)
P
2m−2
= D
(2m−2)
.
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 (A
4
, ) = ((A
∗∗
, )
∗∗
, ), (A
4
, 4) =
((A
∗∗
, 4)
∗∗
, ), (A
4
, 4) = ((A
∗∗
, )
∗∗
, 4), (A
4
, 44) = ((A
∗∗
, 4)
∗∗
, 4). For
a ∈ A and ϕ ∈ A
(4)
it is easy to check that
P
2
P
0
(a) ϕ = P
2
P
0
(a) 4 ϕ = P
2
P
0
(a) 4 ϕ = P
2
P
0
(a) 44 ϕ,
ϕ P
2
P
0
(a) = ϕ 4 P
2
P
0
(a) = ϕ 4 P
2
P
0
(a) = ϕ 44 P
2
P
0
(a).
Let A be a Banach algebra and n ∈ . Consider the maps (a
∗
, ϕ
2n
) 7→ a
∗
· ϕ
2n
and
(a
∗
, ϕ
2n
) 7→ ϕ
2n
· a
∗
from A
∗
× A
(2n)
into A
∗
deﬁned by
ha, a
∗
· ϕ
2n
i = hP
2n−3
. . . P
3
P
1
(aa
∗
), ϕ
2n
i,
ha, ϕ
2n
· a
∗
i = hP
2n−3
. . . P
3
P
1
(a
∗
a), ϕ
2n
i (a ∈ A ).
Then a
∗
· ϕ
2n
= a
∗
P
∗
1
P
∗
3
. . . P
∗
2n−3
(ϕ
2n
) and ϕ
2n
· a
∗
= P
∗
1
P
∗
3
. . . P
∗
2n−3
(ϕ
2n
)a
∗
.
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 bilinear maps from A
(m)
× A
(2n)
into A
(m)
for every m ∈
.
For example, for F ∈ A
∗∗
and ϕ
2n
∈ A
(2n)
we have
ha
∗
, F · ϕ
2n
i = hϕ
2n
· a
∗
, F i
= hP
∗
1
P
∗
3
. . . P
∗
2n−3
(ϕ
2n
)a
∗
, F i
= ha
∗
, F P
∗
1
P
∗
3
. . . P
∗
2n−3
(ϕ
2n
)i (a
∗
∈ A
∗
),
and so F ·ϕ
2n
= F P
∗
1
P
∗
3
. . . P
∗
2n−3
(ϕ
2n
). Similarly, ϕ
2n
·F = P
∗
1
P
∗
3
. . . P
∗
2n−3
(ϕ
2n
)4
F . From now on we regard these actions as A
(2n)
actions on A
(m)
induced from A
∗
.
Now consider the maps (F, ϕ
2n
) 7→ F ·ϕ
2n
and (F, ϕ
2n
) 7→ ϕ
2n
·F from A
∗∗
×A
(2n)
into A
∗∗
deﬁned by
ha
∗
, F · ϕ
2n
i = hP
2n−3
. . . P
3
P
1
(a
∗
F ), ϕ
2n
i,
ha
∗
, ϕ
2n
· F i = hP
2n−3
. . . P
3
P
1
(F a
∗
), ϕ
2n
i (a
∗
∈ A
∗
).
Clearly these are continuous bilinear maps, F · ϕ
2n
= F 4 P
∗
1
P
∗
3
. . . P
∗
2n−3
(ϕ
2n
) and
similarly ϕ
2n
· F = P
∗
1
P
∗
3
. . . P
∗
2n−3
(ϕ
2n
) F . Note that these actions are diﬀerent
from the actions induced from A
∗
. Again by dualizing these actions we have con
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 ∈ be such that n > 2k. Set B =
(A
(2k)
, ·), where · is one of the 2
k
Arens products on A
(2k)
. Then B is a Banach
algebra and B
∗
is a Banach Bmodule. By a similar argument we have continuous
bilinear maps from B
∗
× A
(2n)
into B
∗
and from B
∗∗
× A
(2n)
into B
∗∗
. Therefore
for every m > 2k + 1 we have A
(2n)
actions on A
(m)
induced from B
∗
and for every
m > 2k + 2 we have A
(2n)
actions on A
(m)
induced from B
∗∗
.
Proposition 2.1. Let A be an Arens regular Banach algebra and n ∈ . Then
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
∗
.
. When n = 1, one can immediately see that A
∗
is a left Banach
(A
∗∗
, )module and a right Banach (A
∗∗
, 4)module. Since A is Arens regular,
A
∗
is a left and right Banach A
∗∗
module. For a ∈ A , a
∗
∈ A and F, G ∈ A
∗∗
we
have
ha, (F a
∗
)Gi = h(aF )a
∗
, Gi = ha
∗
, G (aF )i
= ha
∗
G, P
0
(a) F i = hF (a
∗
G), P
0
(a)i
= ha, F (a
∗
G)i,
and so (F a
∗
)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)
)
∗∗
, ). Let
a ∈ A , a
∗
∈ A
∗
, ϕ, ψ ∈ A
(2n+2)
and let (ϕ
α
) , (ψ
β
) be nets in A
(2n)
such that
P
2n
(ϕ
α
) −→ ϕ and P
2n
(ψ
β
) −→ ψ in the weak
∗
topology. Then
ha, a
∗
· (ϕ ψ)i = lim
α
lim
β
hP
2n−3
. . . P
3
P
1
(aa
∗
), ϕ
α
ψ
β
i
= lim
α
lim
β
ha, a
∗
· (ϕ
α
ψ
β
)i
= lim
α
lim
β
ha, (a
∗
· ϕ
α
) · ψ
β
i
= lim
α
lim
β
ha, a
∗
P
∗
1
P
∗
3
. . . P
∗
2n−3
(ϕ
α
)P
∗
1
P
∗
3
. . . P
∗
2n−3
(ψ
β
)i
= lim
α
lim
β
hP
2n−3
. . . P
3
P
1
(aa
∗
P
∗
1
P
∗
3
. . . P
∗
2n−3
(ϕ
α
)), ψ
β
i
= lim
α
haa
∗
P
∗
1
P
∗
3
. . . P
∗
2n−3
(ϕ
α
), P
∗
1
P
∗
3
. . . P
∗
2n−1
(ψ)i
= lim
α
hP
∗
1
. . . P
∗
2n−1
(ψ)aa
∗
, P
∗
1
P
∗
3
. . . P
∗
2n−3
(ϕ
α
)i
= ha, a
∗
P
∗
1
. . . P
∗
2n−1
(ϕ)P
∗
1
. . . P
∗
2n−1
(ψ)i
= ha, (a
∗
· ϕ) · ψi,
866
and so a
∗
· (ϕ ψ) = (a
∗
· ϕ) · ψ. Similarly (ϕ ψ) · a
∗
= ϕ · (ψ · a
∗
). On the other
hand,
(ϕ · a
∗
) · ψ = (P
∗
1
. . . P
∗
2n−1
(ϕ)a
∗
)P
∗
1
. . . P
∗
2n−1
(ψ)
= P
∗
1
. . . P
∗
2n−1
(ϕ)(a
∗
P
∗
1
. . . P
∗
2n−1
(ψ))
= ϕ · (a
∗
· ψ).
Hence A
∗
is a Banach A
(2n+2)
bimodule. So we are done by induction.
Proposition 2.2. Let A be an Arens regular Banach algebra and n ∈ . Then
with any of the Arens products on A
(2n)
, the A
(2n)
actions on A
∗∗
induced from A
∗
and A
∗∗
coincide. In particular, A
∗∗
is a Banach A
(2n)
bimodule with any of these
actions.
is straightforward.
Let A be a Banach algebra and m, n ∈ . Then A
(2m)
is a Banach algebra
with one of the 2
m
Arens products. We recall that every closed subalgebra of an
Arens regular Banach algebra is Arens regular. In particular, when A
(2m)
is Arens
regular for a Banach algebra A and m ∈ , then A
∗∗
, A
(4)
, . . . , A
(2m−2)
are Arens
regular, and these algebras have only one Arens product. The following proposition
is a generalization of Proposition 2.1 and Proposition 2.2.
Proposition 2.3. Let A be a Banach algebra and let n, m ∈ be such
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.
It is well known that every C
∗
algebra is Arens regular and the second dual of a
C
∗
algebra is a C
∗
algebra. Therefore, every C
∗
algebra is completely Arens regular.
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)
.
. A direct consequence of Proposition 2.3.
867
Lemma 2.6. Let A be a Banach algebra and P
0
(A ) a left (right) ideal in A
∗∗
.
Then A
∗
is a Banach (A
∗∗
, )module((A
∗∗
, 4))module).
. For a
∗
∈ A
∗
, a ∈ A and F, G ∈ A
∗∗
we have
ha, a
∗
(F G)i = hG(aa
∗
), F i = hG 4 P
0
(a)a
∗
, F i
= ha
∗
, F 4 G 4 P
0
(a)i = h(a
∗
F )G, P
0
(a)i
= ha, (a
∗
F )Gi
and
ha, F (a
∗
G)i = ha
∗
G P
0
(a), F i = ha
∗
, G P
0
(a) F i
= hF (a
∗
G), P
0
(a)i = ha, F (a
∗
G)i.
Therefore A
∗
is a Banach (A
∗∗
, )module.
Lemma 2.7. Let A be a Banach algebra. Then P
0
(A ) is an ideal in A
∗∗
with
any of the Arens products if and only if P
2
P
0
(A ) is an ideal in A
(4)
with any of the
Arens products.
. Let P
0
(A ) be an ideal in A
∗∗
. For a ∈ A and ϕ ∈ A
(4)
, one can
immediately see that
P
2
P
0
(a) ϕ = P
2
(P
0
(a) P
∗
1
(ϕ)) and ϕ P
2
P
0
(a) = P
2
(P
∗
1
(ϕ) P
0
(a)).
Therefore P
2
P
0
(A ) is an ideal in A
(4)
with any of the Arens products on A
(4)
.
Conversely, let P
2
P
0
(A ) be an ideal in A
(4)
. Take a ∈ A and F ∈ A
∗∗
. It is easy
to see that P
2
(P
0
(a)F ) = P
2
P
0
(a)P
2
(F ) ∈ P
2
P
0
(A ). Hence P
0
(a)F ∈ P
0
(A )
and similarly F P
0
(a) ∈ P
0
(A ). Therefore P
0
(A ) is an ideal in A
∗∗
.
Proposition 2.8. Let A be a Banach algebra and P
0
(A ) an ideal in A
∗∗
. Then
A
∗
is a Banach A
(2n)
module with any of the Arens products on A
(2n)
(n ∈ ).
. When n = 1 the result is true by Lemma 2.6. Now suppose, inductively,
the result has been proved for n−1. We may assume that A
(2n+2)
= ((A
(2n)
)
∗∗
, ).
Let a ∈ A , a
∗
∈ A
∗
and ϕ, ψ ∈ A
(2n+2)
, and let (ϕ
α
), (ψ
β
) be nets in A
(2n)
such
that P
2n
(ϕ
α
) −→ ϕ and P
2n
(ψ
β
) −→ ψ in the weak
∗
topology. Now we have
ha, a
∗
· (ϕ ψ)i = hP
2n−1
. . . P
3
P
1
(aa
∗
), ϕ ψi
= lim
α
lim
β
ha
∗
, P
∗
1
P
∗
3
. . . P
∗
2n−3
(ϕ
α
) 4 P
∗
1
P
∗
3
. . . P
∗
2n−3
(ψ
β
) 4 P
0
(a)i
= lim
α
haa
∗
P
∗
1
P
∗
3
. . . P
∗
2n−3
(ϕ
α
), P
∗
1
P
∗
3
. . . P
∗
2n−1
(ψ)i
868
= lim
α
ha
∗
, P
∗
1
P
∗
3
. . . P
∗
2n−3
(ϕ
α
) P
∗
1
P
∗
3
. . . P
∗
2n−1
(ψ) P
0
(a)i
= haa
∗
, P
∗
1
P
∗
3
. . . P
∗
2n−1
(ϕ) P
∗
1
P
∗
3
. . . P
∗
2n−1
(ψ)i
= ha, (a
∗
· ϕ) · ψi,
and so a
∗
· (ϕ ψ) = (a
∗
· ϕ) · ψ. Similarly (ϕ ψ) · a
∗
= ϕ · (ψ · a
∗
). Since A
∗
is a
Banach A
∗∗
module,
(ϕ · a
∗
) · ψ = (P
∗
1
P
∗
3
. . . P
∗
2n−1
(ϕ)a
∗
)P
∗
1
P
∗
3
. . . P
∗
2n−1
(ψ)
= P
∗
1
P
∗
3
. . . P
∗
2n−1
(ϕ)(a
∗
P
∗
1
P
∗
3
. . . P
∗
2n−1
(ψ))
= ϕ · (a
∗
· ψ).
Consequently, A
∗
is a Banach A
(2n)
module.
Proposition 2.9. Let A be a Banach algebra. Then P
2
((A
∗∗
, )) is a left (right,
twosided) ideal in (A
(4)
, ) if and only if P
∗
2
is an A
(4)
module homomorphism
between left (right, twosided) Banach A
(4)
modules.
. Let P
2
((A
∗∗
, )) be a left ideal in (A
(4)
, ). For F ∈ A
∗∗
, ϕ
4
∈ A
(4)
and ϕ
5
∈ A
(5)
we have
hF, P
∗
2
(ϕ
4
ϕ
5
)i = hP
2
(F ), ϕ
4
ϕ
5
i = hP
2
(F ) ϕ
4
, ϕ
5
i
= hP
∗
2
(ϕ
5
), P
2
(F ) ϕ
4
i = hF, ϕ
4
P
∗
2
(ϕ
5
)i.
Hence P
∗
2
is an A
(4)
module homomorphism between left Banach A
(4)
modules.
Conversely, for F ∈ A
∗∗
, ϕ
4
∈ A
(4)
it is easy to see that
P
2
(F ) ϕ
4
= P
2
(P
∗
1
(P
2
(F ) ϕ
4
)),
so P
2
((A
∗∗
, )) is a left ideal in (A
(4)
, ).
3. N weak amenability for N ∈ Z
Lemma 3.1. Let A be a Banach algebra and D : A −→ A
∗
a derivation. Then
(i) D
∗∗
: (A
∗∗
, ) −→ (A
∗∗
)
∗
is satisﬁed in
D
∗∗
(F G) = D
∗∗
(F )G + P
∗∗
0
(F )D
∗∗
(G) (F, G ∈ A
∗∗
),
(ii) D
∗∗
: (A
∗∗
, 4) −→ (A
∗∗
)
∗
is satisﬁed in
D
∗∗
(F 4 G) = D
∗∗
(F )P
∗∗
0
(G) + F D
∗∗
(G) (F, G ∈ A
∗∗
).
869
. (i) Let F, G ∈ A
∗∗
and let (a
α
), (b
β
) be nets in A such that
P
0
(a
α
) −→ F and P
0
(b
β
) −→ G in the weak
∗
topology. We have
hH, D
∗∗
(F G)i = hD
∗
(H), F Gi
= lim
α
lim
β
hD(a
α
)b
β
+ a
α
D(b
β
), Hi
= lim
α
lim
β
hb
β
, HD(a
α
) + D
∗
(Ha
α
)i
= lim
α
ha
α
, D
∗
(G H) + P
∗
0
(D
∗∗
(G)H)i
= hH, D
∗∗
(F )G + P
∗∗
0
(F )D
∗∗
(G)i
and so D
∗∗
(F G) = D
∗∗
(F )G + P
∗∗
0
(F )D
∗∗
(G).
(ii) The proof is similar to (i).
Corollary 3.2. Let A be a Banach algebra and D : A −→ A
∗
a derivation.
Then
(i) D
∗∗
: (A
∗∗
, ) −→ (A
∗∗
)
∗
is a derivation if and only if P
∗∗
0
(F )D
∗∗
(G) =
F D
∗∗
(G) for F, G ∈ A
∗∗
;
(ii) D
∗∗
: (A
∗∗
, 4) −→ (A
∗∗
)
∗
is a derivation if and only if D
∗∗
(F )P
∗∗
0
(G) =
D
∗∗
(F )G for F, G ∈ A
∗∗
.
Deﬁnition 3.3. Let A be a Banach algebra m, n ∈ , and 1 6 m < 2n. The Ba
nach algebra A
(2n)
is called (−m)weakly amenable, if A
(2n−m)
is a Banach A
(2n)

bimodule with actions induced from A
(2n−m)
and H
1
(A
(2n)
, A
(2n−m)
) = {0}.
Theorem 3.4. Let A be a Banach algebra and P
0
(A ) a left (right) ideal in A
∗∗
.
If (A
∗∗
, )((A
∗∗
, 4)) is (−1)weakly amenable, then A is weakly amenable.
. By Lemma 2.6, A
∗
is a Banach (A
∗∗
, )module. Let D : A −→ A
∗
be a derivation. Put d = P
∗
0
D
∗∗
: (A
∗∗
, ) −→ A
∗
. For F, G ∈ A
∗∗
, a ∈ A we
have
ha, P
∗
0
(D
∗∗
(F )G)i = hG P
0
(a), D
∗∗
(F )i = hd(F ), G 4 P
0
(a)i = ha, d(F )Gi
and
ha, P
∗
0
(P
∗∗
0
(F )D
∗∗
(G))i = hP
∗
0
(D
∗∗
(G)P
0
(a)), F i = hd(G)P
0
(a), F i = ha, F d(G)i.
Therefore, by Lemma 3.1, d is a derivation. Since H
1
((A
∗∗
, ), A
∗
) = {0}, there
exists a
∗
∈ A
∗
such that d = δ
a
∗
. Using Lemma 1.2 we obtain
aa
∗
− a
∗
a = P
0
(a)a
∗
− a
∗
P
0
(a) = dP
0
(a)
= P
∗
0
D
∗∗
P
0
(a) = D(a) (a ∈ A ).
Hence D = δ
a
∗
is an inner derivation.
870
Theorem 3.5. Let A be a Banach algebra. If P
0
(A ) is an ideal in A
∗∗
and the
Banach algebra A
(2n)
(n ∈ ) with one of 2
n
Arens products is (−2n + 1)weakly
amenable, then A is weakly amenable.
. Let D : A −→ A
∗
be a derivation. We claim that
d
n
= P
∗
0
P
∗∗∗
0
. . . P
(2n−1)
0
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 proved for n. We may suppose that A
(2n+2)
=
((A
(2n)
)
∗∗
, ). For a ∈ A , a
∗
∈ A
∗
, ϕ, ψ ∈ A
(2n+2)
, let (ϕ
α
) and (ψ
β
) be nets
in A
(2n)
such that P
2n
(ϕ
α
) −→ ϕ and P
2n
(ψ
β
) −→ ψ in the weak
∗
topology. Then
we have
ha, d
n+1
(ϕ ψ)i = lim
α
lim
β
ha, d
n
(ϕ
α
)ψ
β
+ ϕ
α
d
n
(ψ
β
)i
= lim
α
lim
β
hP
2n−3
. . . P
3
P
1
(ad
n
(ϕ
α
)), ψ
β
i
+ lim
α
lim
β
hψ
β
, D
(2n+1)
P
(2n)
0
. . . P
∗∗
0
(aP
∗
1
. . . P
∗
2n−3
(ϕ
α
))i
= lim
α
had
n
(ϕ
α
), P
∗
1
. . . P
∗
2n−1
(ψ)i
+ lim
α
hd
n+1
(ψ)a, P
∗
1
. . . P
∗
2n−3
(ϕ
α
)i
= ha, d
n+1
(ϕ) · ψ + ϕ · d
n+1
(ψ)i,
so d
n+1
is a derivation. Since H
1
(A
(2n)
, A
∗
) = {0}, there exists a
∗
∈ A
∗
such that
d
n
= δ
a
∗
. Using Lemma 1.1 (iv) and Lemma 1.2, we conclude that
aa
∗
− a
∗
a = P
2n−2
. . . P
2
P
0
(a) · a
∗
− a
∗
· P
2n−2
. . . P
2
P
0
(a)
= d
n
P
2n−2
. . . P
2
P
0
(a) = D(a) (a ∈ A ).
Hence D = δ
a
∗
is inner.
Lemma 3.6. Let A be a Banach algebra, n ∈
and let D : A −→ A
(2n)
be a
derivation. Then for every F, G ∈ A
∗∗
(i) D
∗∗
: (A
∗∗
, ) −→ ((A
(2n)
)
∗∗
, ) holds in
D
∗∗
(F G) = D
∗∗
(F ) P
∗∗
2n−2
. . . P
∗∗
2
P
∗∗
0
(G) + P
∗∗
2n−2
. . . P
∗∗
2
P
∗∗
0
(F ) D
∗∗
(G).
(ii) D
∗∗
: (A
∗∗
, 4) −→ ((A
(2n)
)
∗∗
, 4) holds in
D
∗∗
(F 4G) = D
∗∗
(F )4P
∗∗
2n−2
. . . P
∗∗
2
P
∗∗
0
(G)+P
∗∗
2n−2
. . . P
∗∗
2
P
∗∗
0
(F )4D
∗∗
(G).
is straightforward.
871
Proposition 3.7. Let A be a Banach algebra, n ∈ and let D : A −→ A
(2n)
be a derivation. If A
(2n)
is Arens regular and
D
∗∗
(A
∗∗
)A
(2n+1)
∪ A
(2n+1)
D
∗∗
(A
∗∗
) ⊆ P
2n−1
. . . P
3
P
1
(A
∗
),
then D
∗∗
: A
∗∗
−→ (A
(2n)
)
∗∗
is a derivation.
. Since A
(2n)
is Arens regular, A is Arens regular. For ϕ
2n+1
∈ A
(2n+1)
,
F, G ∈ A
∗∗
, there exists a
∗
∈ A
∗
such that
ϕ
2n+1
· D
∗∗
(F ) = P
2n−1
. . . P
1
(a
∗
).
By Lemma 1.1 we have
hϕ
2n+1
, D
∗∗
(F ) P
∗∗
2n−2
. . . P
∗∗
0
(G)i = hP
2n−1
. . . P
1
(a
∗
), P
∗∗
2n−2
. . . P
∗∗
0
(G)i
= hP
2n−2
. . . P
2
(G), ϕ
2n+1
D
∗∗
(F )i
= hϕ
2n+1
, D
∗∗
(F )Gi.
Similarly, P
∗∗
2n−2
. . . P
∗∗
0
(F ) D
∗∗
(G) = F D
∗∗
(G). Hence D
∗∗
is a derivation by
Lemma 3.6.
Lemma 3.8. Let A be a Banach algebra and D : A −→ A
(2n+1)
(n ∈
) a
derivation. Then D
∗∗
: (A
∗∗
, ) −→ ((A
(2n)
)
∗∗
, )
∗
is valid in
D
∗∗
(F G) = D
∗∗
(F )P
∗∗
2n−2
. . . P
∗∗
0
(G) + P
∗∗
2n
. . . P
∗∗
0
(F )D
∗∗
(G) (F, G ∈ A
∗∗
).
is straightforward.
Proposition 3.9. Let A be a Banach algebra and let D : A −→ A
(2n+1)
(n ∈ ) be a derivation. If
D
∗∗
(A
∗∗
) · A
(2n+2)
∪ A
(2n+2)
· D
∗∗
(A
∗∗
) ⊆ P
2n+1
. . . P
1
(A
∗
),
then D
∗∗
: (A
∗∗
, ) −→ (A
∗∗
)
(2n+1)
is a derivation.
. By Lemma 3.8, it is clear.
872
Lemma 3.10. Let A be a Banach algebra and D : A −→ A
∗
a derivation.
Then for every ϕ and ψ in A
(4)
(i) D
(4)
: (A
(4)
, ) −→ (A
(4)
, )
∗
holds in D
(4)
(ϕψ) = D
(4)
(ϕ)ψ+P
(4)
0
(ϕ)
D
(4)
(ψ);
(ii) D
(4)
: (A
(4)
, 44) −→ (A
(4)
, 44)
∗
holds in D
(4)
(ϕ44ψ) = D
(4)
(ϕ)P
(4)
0
(ψ)+
ϕD
(4)
(ψ).
. (i) Let ξ, ϕ, ψ ∈ A
(4)
and let (F
α
), (G
β
) be nets in A
∗∗
such that
P
2
(F
α
) −→ ϕ and P
2
(G
β
) −→ ψ in the weak
∗
topology. By Lemma 3.1 we have
hξ, D
(4)
(ϕ ψ)i = lim
α
lim
β
hD
∗∗
(F
α
G
β
), ξi
= lim
α
lim
β
hD
∗∗
(F
α
)G
β
+ P
∗∗
0
(F
α
)D
∗∗
(G
β
), ξi
= lim
α
hξD
∗∗
(F
α
) + D
(3)
(ξ P
∗∗
0
(F
α
)), ψi
= hD
(3)
(ψ ξ) + P
(3)
0
(D
(4)
(ψ)ξ), ϕi
= hξ, D
(4)
(ϕ)ψ + P
(4)
0
(ϕ)D
(4)
(ψ)i.
(ii) The proof is similar to (i).
Proposition 3.11. Let A be a Banach algebra and D : A −→ A
∗
a derivation.
(i) If D
(4)
(A
(4)
) · A
(4)
⊆ P
3
P
1
(A
∗
), then D
(4)
: (A
(4)
, ) −→ (A
(4)
)
∗
is a
derivation.
(ii) If A
(4)
· D
(4)
(A
(4)
) ⊆ P
3
P
1
(A
∗
), then D
(4)
: (A
(4)
, 44) −→ (A
(4)
)
∗
is a
derivation.
. (i) Let ξ, ϕ, ψ ∈ A
(4)
, there exists a
∗
∈ A
∗
such that D
(4)
(ϕ) · ξ =
P
3
P
1
(a
∗
). By Lemma 1.1 (iii) we have
hξ, P
(4)
0
(ϕ)D
(4)
(ψ)i = hP
(3)
0
P
3
P
1
(a
∗
), ϕi = hϕ, D
(4)
(ψ)ξi = hξ, ϕD
(4)
(ψ)i.
Therefore D
(4)
is a derivation by Lemma 3.10.
(ii) The proof is similar to (i).
We recall that an operator T : X −→ Y between Banach spaces is weakly compact
if and only if T
∗∗
X
∗∗
⊂ Y (considered as a subspace of Y
∗∗
) if and only if T
∗
is
weakly compact.
873
Lemma 3.12. Let A be a Banach algebra and D : A −→ A
∗
a weakly compact
operator. Then D
(2n)
(A
(2n)
) ⊆ P
2n−1
. . . P
3
P
1
(A
∗
) (n ∈ ).
. When n = 1, clearly the result is true. Now suppose, inductively, that
the result has been proved for n. Let ϕ, ξ ∈ A
(2n+2)
and let (ϕ
α
) be a net in A
(2n)
such that P
2n
(ϕ
α
) −→ ϕ in the weak
∗
topology. Then
hξ, D
(2n+2)
(ϕ)i = lim
α
hD
(2n)
(ϕ
α
), ξi = lim
α
hP
∗
2n−1
(ξ), D
(2n)
(ϕ
α
)i
= hD
(2n−1)
P
∗
2n−1
(ξ), P
∗
2n−1
(ϕ)i = hP
∗
2n−1
(ξ), D
(2n)
P
∗
2n−1
(ϕ)i
= hD
(2n)
P
∗
2n−1
(ϕ), ξi = hξ, P
2n+1
D
(2n)
P
∗
2n−1
(ϕ)i.
Consequently, D
(2n+2)
(ϕ) = P
2n+1
D
(2n)
P
∗
2n−1
(ϕ) ⊆ P
2n+1
. . . P
3
P
1
(A
∗
).
Dales, RodriguesPalacios and Velasco in [3] proved the following theorem.
Theorem 3.13. Let A be an Arens regular Banach algebra and D : A −→ A
∗
a weakly compact derivation. Then D
(∗∗)
: A
(∗∗)
−→ (A
∗∗
)
∗
is a derivation.
Now we have the same result for A
(4)
.
Theorem 3.14. Let A be an Arens regular Banach algebra and D : A −→
A
∗
a weakly compact derivation. Then D
(4)
: (A
(4)
, ) −→ (A
(4)
)
∗
and D
(4)
:
(A
(4)
, 44) −→ (A
(4)
)
∗
are derivations.
. Let ξ, ϕ, ψ ∈ A
(4)
and (F
α
), (G
β
), (H
γ
) be nets in A
∗∗
such that
P
2
(F
α
) −→ ϕ, P
2
(G
β
) −→ ψ and P
2
(H
γ
) −→ ξ in the weak
∗
topology, let a
∗
∈ A
∗
and let a
∗
α
be a net in A
∗
such that P
1
(a
∗
α
) = D
∗∗
(F
α
) and P
1
(a
∗
) = D
∗∗
P
∗
1
(ϕ). We
have
hξ, D
(4)
(ϕ)ψi = hψ ξ, D
(4)
(ϕ)i = lim
α
lim
β
lim
γ
ha
∗
α
, G
β
H
γ
i
= lim
α
lim
β
ha
∗
α
G
β
, P
∗
1
(ξ)i = lim
α
hP
∗
1
(ψ) P
∗
1
(ξ), D
∗∗
P
∗∗
1
(ϕ)i
= hξ, P
3
(P
1
(a
∗
)P
∗
1
(ψ))i = hξ, P
3
P
1
(a
∗
P
∗
1
(ψ))i,
and so D
(4)
(A
(4)
)A
(4)
⊆ P
3
P
1
(A
∗
) and by Proposition 3.11, D
(4)
: (A
(4)
, ) −→
(A
(4)
)
∗
is a derivation. The other part is similar.
874
Corollary 3.15. Let A be an Arens regular Banach algebra such that
(A
(4)
, ) or (A
(4)
, 44) is weakly amenable and each derivation from A to A
∗
is
weakly compact. Then A is weakly amenable.
. Let D : A −→ A
∗
be a derivation. We may suppose that (A
(4)
, ) is
weakly amenable. By Theorem 3.14, D
(4)
: (A
(4)
, ) −→ (A
(4)
)
∗
is a derivation.
So there exists ϕ
5
∈ (A
(4)
)
∗
such that D
(4)
= δ
ϕ
5
. Set a
∗
= P
∗
0
P
∗
2
(ϕ
5
). Then by
Lemma 1.2 we have
aa
∗
− a
∗
a = P
∗
0
P
∗
2
(P
2
P
0
(a)ϕ
5
− ϕ
5
P
2
P
0
(a))
= P
∗
0
P
∗
2
D
(4)
P
2
P
0
(a) = D(a) (a ∈ A ).
Therefore D = δ
a
∗
is inner. Hence A is weakly amenable.
Proposition 3.16. Let A be a Banach algebra, D : A −→ A
∗
a derivation
and A
(2n)
= ((. . . ((A
∗∗
, )
∗∗
, ) . . .)
∗∗
, ) (n ∈
). Then
(i) D
(2n)
: A
(2n)
−→ (A
(2n)
)
∗
holds in
D
(2n)
(ϕ ψ) = D
(2n)
(ϕ)ψ + P
(2n)
0
(ϕ)D
(2n)
(ψ) (ϕ, ψ ∈ A
(2n)
).
(ii) If D
(2n)
(A
(2n)
) · A
(2n)
⊆ P
2n−1
. . . P
3
P
1
(A
∗
), then D
(2n)
is a derivation.
(iii) If A
(2n−2)
is Arens regular and D is weakly compact, then D
(2n)
is a derivation.
Corollary 3.17. Let A be a completely regular Banach algebra such that
A
(2n)
is weakly amenable for some n ∈ , and each derivation from A to A
∗
is
weakly compact. Then A is weakly amenable.
Lemma 3.18. Let A be an Arens regular Banach algebra such that (A
(4)
, )
or (A
(4)
, 44) is (−2)weakly amenable. Then A is 2weakly amenable.
. Let D : A −→ A
∗∗
be a derivation, and let (A
(4)
, ) be (−2)weakly
amenable. Set d = P
∗
1
D
∗∗
P
∗
1
: (A
(4)
, ) −→ A
∗∗
. For a
∗
∈ A
∗
, ϕ, ψ ∈ A
(4)
let
(F
α
), (G
β
) be nets in A
∗∗
such that P
2
(F
α
) −→ ϕ and P
2
(G
β
) −→ ψ in the weak
∗
topology. Then
ha
∗
, d(ϕ ψ)i = hP
1
D
∗
P
1
(a
∗
), ϕ ψi
= lim
α
lim
β
hP
1
(a
∗
), D
∗∗
(F
α
G
β
)i
= lim
α
lim
β
ha
∗
, P
∗
1
D
∗∗
(F
α
)G
β
+ F
α
P
∗
1
D
∗∗
(G
β
)i
= lim
α
ha
∗
P
∗
1
D
∗∗
(F
α
), P
∗
1
(ψ)i + hD
∗
P
1
(a
∗
F
α
), P
∗
1
(ψ)i
= hP
∗
1
(ψ)a
∗
, d(ϕ)i + hd(ψ)a
∗
, P
∗
1
(ϕ)i
= ha
∗
, d(ϕ) · ψ + ϕ · d(ψ)i.
875
Therefore d is a derivation. Since H
1
(A
(4)
, A
∗∗
) = {0}, there exists F ∈ A
∗∗
such
that d = δ
F
. It is easy to see that D = δ
F
. So A is 2weakly amenable.
Proposition 3.19. Let A be an Arens regular Banach algebra such that
A
(2n+2)
(n ∈ ) with one of Arens products is (−2n)weakly amenable. Then
A is 2weakly amenable.
. Let D : A −→ A
∗∗
be a derivation. By Lemma 3.18 and by in
duction, d = P
∗
1
D
∗∗
P
∗
1
P
∗
3
. . . P
∗
2n−1
: A
(2n+2)
−→ A
∗∗
is a derivation. Since
H
1
(A
(2n+2)
, A
∗∗
) = {0}, there exists F ∈ A
∗∗
such that d = δ
F
. It is easy to see
that D = δ
F
is inner.
Acknowledgement. We would like to thank the referee for carefully reading the
paper and giving some interesting and fruitful suggestions.
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Authors’ address:
, Faculty of Mathematical Science, Teacher Train
ing University, 599, Taleghani Avenue, Tehran, 15614, Iran, email: medghalchi@saba.tmu.
ac.ir; , Department of Mathematics, Persian Gulf University, 75168
Boushehr, Iran, email: yazdanpanah@pgu.ac.ir.
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