# Approximately Multiplicative Functionals on the Spaces of Formal Power Series

Authors: F. Ershad; S. H. Petroudi
Journal: Abstract and Applied Analysis
 (2011)

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## Abstract

We characterize the conditions under which approximately multiplicative functionals are near multiplicative functionals on weighted Hardy spaces.


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## Body

## 1. Introduction

Let𝒜 be a commutative Banach algebra and 𝒜̂ the set of all its characters, that is, the nonzero multiplicative linear functionals on 𝒜. If φ is a linear functional on 𝒜, then define (1.1)φ̆(a,b)=φ(ab)-φ(a)φ(b)for alla,b∈𝒜. We say that φ is δ-multiplicative if ∥φ̆∥≤δ.For eachφ∈𝒜⋆ define (1.2)d(φ)=inf⁡{‖φ-ψ‖:ψ∈Â∪{0}}.We say that 𝒜 is an algebra in which approximately multiplicative functionals are near multiplicative functionals or 𝒜 is AMNM for short if, for each ε>0, there is δ>0 such that d(φ)<ε whenever φ is a δ-multiplicative linear functional.We deal with an algebra in which every approximately multiplicative functional is near a multiplicative functional (AMNM algebra). The question whether an almost multiplicative map is close to a multiplicative, constitutes an interesting problem. Johnson has shown 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 as H∞, Douglas algebras, and R(K) where K is a compact subset of 𝒞. Here, we want to investigate conditions under which a weighted Hardy space is to be AMNM. For some sources on these topics one can refer to [1–8].Let{β(n)}n=0∞ be a sequence of positive numbers with β(0)=1 and 1<p<∞. We consider the space of sequences f={f̂(n)}n=0∞ such that (1.3)‖f‖p=‖f‖βp=∑n=0∞|f̂(n)|pβp(n)<∞.
The notation f(z)=∑n=0∞f̂(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. Let Hp(β) denote the 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 f̂k(n)=δk(n). So fk(z)=zk, and then {fk}k=0∞ is a basis such that ∥fk∥=β(k) for all k. For some sources one can see [9–21].


## 2. Main Results

In this section we investigate theAMNM property of the spaces of formal power series. For the proof of our main theorem we need the following lemma.Lemma 2.1.
Let1<p<∞ and 1/p+1/q=1. Then, Hp(β)⋆=Hq(β-1), where β-1={β-1(n)}n=0∞.Proof.
DefineL:Hq(βp/q)→Hq(β-1) by L(f)=F, where
(2.1)f(z)=∑n=0∞f̂(n)zn,F(z)=∑n=0∞f̂(n)βp(n)zn.
Then,
(2.2)‖F‖Hq(β-1)q=∑n=0∞|f̂(n)|q(βpq(n)βq(n))=∑n=0∞|f̂(n)|qβp(n)=‖f‖Hp(βp/q)q.
Thus, L is an isometry. It is also surjective because, if
(2.3)F(z)=∑n=0∞F̂(n)zn∈Hq(β-1),
then L(f)=F, where
(2.4)f(z)=∑n=0∞(F̂(n)βp(n))zn.
Hence, Hq(βp/q) and Hq(β-1) are norm isomorphic. Since Hp(β)⋆=Hq(βp/q), the proof is complete.In the proof of the following theorem, our technique is similar to B. E. Johnson’s technique in [2].Theorem 2.2.
Letliminfβ(n)>1 and 1<p<∞. Then, Hp(β) with multiplication
(2.5)(∑n=0∞f̂(n)zn)(∑n=0∞ĝ(n)zn)=∑n=0∞f̂(n)ĝ(n)zn
is a commutative Banach algebra that is AMNM.Proof.
First note that clearlyHp(β) is a commutative Banach algebra. To prove that it is AMNM, let 0<ε<1 and put δ=ε2/16. Suppose that φ∈Hq(β-1) and ∥φ̆∥≤δ, where 1/p+1/q=1. It is sufficient to show that d(φ)<ε. Since d(φ)≤∥φ∥, if ∥φ∥<ε, then d(φ)≤∥φ∥. So suppose that ∥ϕ∥≥ε. For each subset E of ℕ0(=ℕ∪{0}), let
(2.6)nφ(E)=(∑j∈E|φ̂(j)|qβ-q(j))1/q,where
(2.7)φ(z)=∑j=0∞φ̂(j)zj.
For any subsets E1 and E2 of ℕ0 we have that
(2.8)nφq(E1∪E2)=∑j∈E1∪E2|φ̂(j)|qβ-q(j)≤∑j∈E1|φ̂(j)|qβ-q(j)+∑j∈E2|φ̂(j)|qβ-q(j)=nφq(E1)+nφq(E2)≤(nφ(E1)+nφ(E2))q.
Hence,
(2.9)nφ(E1∪E2)≤nφ(E1)+nφ(E2)
for all E1,E2⊆ℕ0. Also if E1∩E2=∅, then, by considering f,g with support, respectively, in E1 and E2, we get that fg=0 and so
(2.10)|φ(f)||φ(g)|=|φ̆(f,g)|≤δ‖f‖‖g‖.
By taking supremum over all such f and g with norm one, we see that
(2.11)nφ(E1)⋅nφ(E2)≤δ.
So either nφ(E1)≤ε/4 or nφ(E2)≤ε/4 whenever E1∩E2=∅.
For allE⊆ℕ0 we have that
(2.12)ε≤‖φ‖=nφ(N0)≤nφ(E)+nφ(N0∖E).
Thus, we get that
(2.13)nφ(N0∖E)≥ε-nφ(E).
Since(ℕ0∖E)∩E=∅, as we saw earlier, it should be nφ(E)≤ε/4 or nφ(ℕ0∖E)≤ε/4 and equivalently it should be nφ(E)≤ε/4 or nφ(E)≥3ε/4 for all E⊆ℕ0.
Note that, ifE1,E2⊆ℕ0 with nφ(Ei)≤ε/4 for i=1,2, then
(2.14)nφ(E1∪E2)≤nφ(E1)+nφ(E2)≤ε2.
Thus, the relation nφ(E1∪E2)≥3ε/4 is not true and so it should be
(2.15)nφ(E1∪E2)≤ε4.
Since ∥φ∥>ε, clearly there exists a positive integer n0 such that nφ(Sj)>ε for all j≥n0, where
(2.16)Sj={i∈N0:i≤j}
for all j∈ℕ0. Now, let nφ({i})≤ε/4 for i=0,1,2,…,n0. Since nφ(S0)≤ε/4 and nφ({1})≤ε/4, nφ(S1)≤ε/4. By continuing this manner we get that nφ(Sn0)≤ε/4, which is a contradiction. Hence there exists m0∈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.
Remember thatfj(z)=zj for all j∈ℕ0. Now we have that
(2.17)|φ̆(fm0,fm0)|=|φ(fm02)-φ(fm0)φ(fm0)|=|φ(fm0)-φ2(fm0)|=|φ(fm0)||1-φ(fm0)|=|φ̂(fm0)||1-φ̂(fm0)|≤δβ2(m0).
Therefore,
(2.18)|φ̂(m0)|β-1(m0)(β-1(m0)|1-φ̂(m0)|)≤ε216,
and so
(2.19)nφ({m0})(β-1(m0)|1-φ̂(m0)|)≤ε216.
But nφ({m0})≥3ε/4, and thus
(2.20)β-1(m0)|1-φ̂(m0)|≤ε12.Define ψ(z)=zm0. Then ψ∈Ĥp(β), and we have that
(2.21)‖φ-ψ‖=‖∑n≠m0φ̂(n)zn+(φ̂(m0)-1)zm0‖=(∑n≠m0|φ̂(n)|qβ-q(n))1/q+β-1(m0)|1-φ̂(m0)|=nφ(N0∖{m0})+β-1(m0)|1-φ̂(m0)|≤ε4+ε12<ε.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 first author.



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*Source: 874949-2011-07-03.xml*