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Hindawi Publishing CorporationAbstract and Applied AnalysisVolume 2011, Article ID 874949, 6 pagesdoi:10.1155/2011/874949
Research ArticleApproximately Multiplicative Functionals onthe Spaces of Formal Power Series
F. Ershad and S. H. Petroudi
Department of Mathematics, Payame Noor University, P.O. Box 19395-4697, Tehran 19569, Iran
Correspondence should be addressed to F. Ershad, [email protected]
Received 20 February 2011; Accepted 5 May 2011
Academic Editor: John Rassias
Copyright q 2011 F. Ershad and S. H. Petroudi. This is an open access article distributed underthe Creative Commons Attribution License, which permits unrestricted use, distribution, andreproduction in any medium, provided the original work is properly cited.
We characterize the conditions under which approximately multiplicative functionals are nearmultiplicative functionals on weighted Hardy spaces.
1. Introduction
LetA be a commutative Banach algebra and A the set of all its characters, that is, the nonzeromultiplicative linear functionals onA. If ϕ is a linear functional onA, then define
ϕ(a, b) = ϕ(ab) − ϕ(a)ϕ(b) (1.1)
for all a, b ∈ A. We say that ϕ is δ-multiplicative if ‖ϕ‖ ≤ δ.For each ϕ ∈ A� define
d(
ϕ)
= inf{
∥
∥ϕ − ψ∥∥ : ψ ∈ A ∪ {0}}
. (1.2)
We say that A is an algebra in which approximately multiplicative functionals are near mul-tiplicative functionals or A is AMNM for short if, for each ε > 0, there is δ > 0 such thatd(ϕ) < ε whenever ϕ is a δ-multiplicative linear functional.
We deal with an algebra in which every approximately multiplicative functional isnear a multiplicative functional (AMNM algebra). The question whether an almost multi-plicative map is close to a multiplicative, constitutes an interesting problem. Johnson hasshown that various Banach algebras are AMNM and some of them fail to be AMNM [1–3].Also, this property is still unknown for some Banach algebras such asH∞, Douglas algebras,
2 Abstract and Applied Analysis
and R(K) where K is a compact subset of C. Here, we want to investigate conditions underwhich a weighted Hardy space is to be AMNM. For some sources on these topics one canrefer to [1–8].
Let {β(n)}∞n=0 be a sequence of positive numbers with β(0) = 1 and 1 < p < ∞. Weconsider the space of sequences f = { f(n)}∞n=0 such that
∥
∥f∥
∥
p =∥
∥f∥
∥
p
β =∞∑
n=0
∣
∣
∣
f(n)∣
∣
∣
pβp(n) <∞. (1.3)
The notation f(z) =∑∞
n=0f(n)zn will be used whether or not the series converges for any
value of z. These are called formal power series or weighted Hardy spaces. LetHp(β) denotethe space of all such formal power series. These are reflexive Banach spaces with norm ‖ · ‖β.Also, the dual of Hp(β) is Hq(βp/q), where 1/p + 1/q = 1 and βp/q = {β(n)p/q}∞n=0 (see [9]).Let fk(n) = δk(n). So fk(z) = zk, and then {fk}∞k=0 is a basis such that ‖fk‖ = β(k) for all k. Forsome sources one can see [9–21].
2. Main Results
In this section we investigate the AMNM property of the spaces of formal power series. Forthe proof of our main theorem we need the following lemma.
Lemma 2.1. Let 1 < p <∞ and 1/p+1/q = 1. Then,Hp(β)� = Hq(β−1), where β−1 = {β−1(n)}∞n=0.
Proof. Define L : Hq(βp/q) → Hq(β−1) by L(f) = F, where
f(z) =∞∑
n=0
f(n)zn,
F(z) =∞∑
n=0
f(n)βp(n)zn.
(2.1)
Then,
‖F‖qHq(β−1) =
∞∑
n=0
∣
∣
∣
f(n)∣
∣
∣
q(
βpq(n)βq(n)
)
=∞∑
n=0
∣
∣
∣
f(n)∣
∣
∣
qβp(n)
=∥
∥f∥
∥
q
Hp(βp/q ).
(2.2)
Thus, L is an isometry. It is also surjective because, if
F(z) =∞∑
n=0
F(n)zn ∈ Hq(
β−1)
, (2.3)
Abstract and Applied Analysis 3
then L(f) = F, where
f(z) =∞∑
n=0
(
F(n)βp(n)
)
zn. (2.4)
Hence, Hq(βp/q) and Hq(β−1) are norm isomorphic. Since Hp(β)� = Hq(βp/q), the proof iscomplete.
In the proof of the following theorem, our technique is similar to B. E. Johnson’s tech-nique in [2].
Theorem 2.2. Let lim inf β(n) > 1 and 1 < p <∞. Then,Hp(β) with multiplication
( ∞∑
n=0
f(n)zn)( ∞
∑
n=0
g(n)zn)
=∞∑
n=0
f(n)g(n)zn (2.5)
is a commutative Banach algebra that is AMNM.
Proof. First note that clearly Hp(β) is a commutative Banach algebra. To prove that it isAMNM, let 0 < ε < 1 and put δ = ε2/16. Suppose that ϕ ∈ Hq(β−1) and ‖ϕ‖ ≤ δ, where1/p + 1/q = 1. It is sufficient to show that d(ϕ) < ε. Since d(ϕ) ≤ ‖ϕ‖, if ‖ϕ‖ < ε, thend(ϕ) ≤ ‖ϕ‖. So suppose that ‖φ‖ ≥ ε. For each subset E of N0(= N ∪ {0}), let
nϕ(E) =
⎛
⎝
∑
j∈E
∣
∣ϕ(
j)∣
∣
qβ−q(
j)
⎞
⎠
1/q
, (2.6)
where
ϕ(z) =∞∑
j=0
ϕ(
j)
zj . (2.7)
For any subsets E1 and E2 of N0 we have that
nqϕ(E1 ∪ E2) =
∑
j∈E1∪E2
∣
∣ϕ(
j)∣
∣
qβ−q(
j)
≤∑
j∈E1
∣
∣ϕ(
j)∣
∣
qβ−q(
j)
+∑
j∈E2
∣
∣ϕ(
j)∣
∣
qβ−q(
j)
= nqϕ(E1) + nqϕ(E2)
≤ (nϕ(E1) + nϕ(E2))q.
(2.8)
Hence,
nϕ(E1 ∪ E2) ≤ nϕ(E1) + nϕ(E2) (2.9)
4 Abstract and Applied Analysis
for all E1, E2 ⊆ N0. Also if E1 ∩ E2 = ∅, then, by considering f, g with support, respectively, inE1 and E2, we get that fg = 0 and so
∣
∣ϕ(
f)∣
∣
∣
∣ϕ(
g)∣
∣ =∣
∣ϕ(
f, g)∣
∣ ≤ δ∥∥f∥∥∥∥g∥∥. (2.10)
By taking supremum over all such f and g with norm one, we see that
nϕ(E1) · nϕ(E2) ≤ δ. (2.11)
So either nϕ(E1) ≤ ε/4 or nϕ(E2) ≤ ε/4 whenever E1 ∩ E2 = ∅.For all E ⊆ N0 we have that
ε ≤ ∥∥ϕ∥∥ = nϕ(N0) ≤ nϕ(E) + nϕ(N0 \ E). (2.12)
Thus, we get that
nϕ(N0 \ E) ≥ ε − nϕ(E). (2.13)
Since (N0 \ E) ∩ E = ∅, as we saw earlier, it should be nϕ(E) ≤ ε/4 or nϕ(N0 \ E) ≤ ε/4and equivalently it should be nϕ(E) ≤ ε/4 or nϕ(E) ≥ 3ε/4 for all E ⊆ N0.
Note that, if E1, E2 ⊆ N0 with nϕ(Ei) ≤ ε/4 for i = 1, 2, then
nϕ(E1 ∪ E2) ≤ nϕ(E1) + nϕ(E2) ≤ ε
2. (2.14)
Thus, the relation nϕ(E1 ∪ E2) ≥ 3ε/4 is not true and so it should be
nϕ(E1 ∪ E2) ≤ ε
4. (2.15)
Since ‖ϕ‖ > ε, clearly there exists a positive integer n0 such that nϕ(Sj) > ε for all j ≥ n0,where
Sj ={
i ∈ N0 : i ≤ j}
(2.16)
for all j ∈ N0. Now, let nϕ({i}) ≤ ε/4 for i = 0, 1, 2, . . . , n0. Since nϕ(S0) ≤ ε/4 andnϕ({1}) ≤ ε/4, nϕ(S1) ≤ ε/4. By continuing this manner we get that nϕ(Sn0) ≤ ε/4, whichis a contradiction. Hence there existsm0 ∈ Sn0 such that nϕ({m0}) ≥ 3ε/4. On the other hand,since (N0 \ {m0}) ∩ {m0} = ∅, nϕ({m0}) ≤ ε/4 or nϕ(N0 \ {m0}) ≤ ε/4. But nϕ({m0}) ≥ 3ε/4,and so it should be nϕ(N0 \ {m0}) ≤ ε/4.
Abstract and Applied Analysis 5
Remember that fj(z) = zj for all j ∈ N0. Now we have that
∣
∣ϕ(
fm0 , fm0
)∣
∣ =∣
∣
∣ϕ(
f2m0
)
− ϕ(fm0
)
ϕ(
fm0
)
∣
∣
∣
=∣
∣
∣ϕ(
fm0
) − ϕ2(fm0
)
∣
∣
∣
=∣
∣ϕ(
fm0
)∣
∣
∣
∣1 − ϕ(fm0
)∣
∣
=∣
∣ϕ(
fm0
)∣
∣
∣
∣1 − ϕ(fm0
)∣
∣
≤ δβ2(m0).
(2.17)
Therefore,
∣
∣ϕ(m0)∣
∣β−1(m0)(
β−1(m0)∣
∣1 − ϕ(m0)∣
∣
)
≤ ε2
16, (2.18)
and so
nϕ({m0})(
β−1(m0)∣
∣1 − ϕ(m0)∣
∣
)
≤ ε2
16. (2.19)
But nϕ({m0}) ≥ 3ε/4, and thus
β−1(m0)∣
∣1 − ϕ(m0)∣
∣ ≤ ε
12. (2.20)
Define ψ(z) = zm0 . Then ψ ∈ Hp(β), and we have that
∥
∥ϕ − ψ∥∥ =
∥
∥
∥
∥
∥
∑
n/=m0
ϕ(n)zn +(
ϕ(m0) − 1)
zm0
∥
∥
∥
∥
∥
=
(
∑
n/=m0
∣
∣ϕ(n)∣
∣
qβ−q(n)
)1/q
+ β−1(m0)∣
∣1 − ϕ(m0)∣
∣
= nϕ(N0 \ {m0}) + β−1(m0)∣
∣1 − ϕ(m0)∣
∣
≤ ε
4+ε
12< ε.
(2.21)
Thus, indeed d(ϕ) ≤ ε, and so the proof is complete.
Disclosure
This is a part of the second author’s Doctoral thesis written under the direction of the firstauthor.
6 Abstract and Applied Analysis
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