AAAAbstract and Applied Analysis1687-04091085-3375Hindawi Publishing Corporation87494910.1155/2011/874949874949Research ArticleApproximately Multiplicative Functionals on the Spaces of Formal Power SeriesErshadF.1PetroudiS. H.1RassiasJohn1Department of MathematicsPayame Noor UniversityP.O. Box 19395-4697, Tehran 19569Iranpnu.ac.ir2011862011201120022011050520112011Copyright © 2011 F. Ershad and S. H. Petroudi.This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

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

1. Introduction

Let 𝒜 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 φ̆(a,b)=φ(ab)-φ(a)φ(b)

for all a,b𝒜. We say that φ is δ-multiplicative if φ̆δ.

For each φ𝒜 define d(φ)=inf{φ-ψ:ψÂ{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 . 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 .

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 fp=fβp=n=0|f̂(n)|pβp(n)<. The notation f(z)=n=0f̂(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 ). 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 .

2. Main Results

In this section we investigate the AMNM property of the spaces of formal power series. For the 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=0f̂(n)zn,F(z)=n=0f̂(n)βp(n)zn. Then, FHq(β-1)q=n=0|f̂(n)|q(βpq(n)βq(n))=n=0|f̂(n)|qβp(n)=fHp(βp/q  )q. Thus, L is an isometry. It is also surjective because, if F(z)=n=0F̂(n)znHq(β-1), then L(f)=F, where 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 .

Theorem 2.2.

Let liminfβ(n)>1 and 1<p<. Then, Hp(β) with multiplication (n=0f̂(n)zn)(n=0ĝ(n)zn)=n=0f̂(n)ĝ(n)zn is a commutative Banach algebra that is AMNM.

Proof.

First note that clearly Hp(β) is a commutative Banach algebra. To prove that it is AMNM, let 0<ε<1 and put δ=ε2/16. Suppose that φHq(β-1) and φ̆δ, 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 ϕε. For each subset E of 0(={0}), let nφ(E)=(jE|φ̂(j)|qβ-q(j))1/q,where φ(z)=j=0φ̂(j)zj. For any subsets E1 and E2 of 0 we have that nφq(E1E2)=jE1E2|φ̂(j)|qβ-q(j)jE1|φ̂(j)|qβ-q(j)+jE2|φ̂(j)|qβ-q(j)=nφq(E1)+nφq(E2)(nφ(E1)+nφ(E2))q. Hence, nφ(E1E2)nφ(E1)+nφ(E2) for all E1,E20. Also if E1E2=, then, by considering f,g with support, respectively, in E1 and E2, we get that fg=0 and so |φ(f)||φ(g)|=|φ̆(f,g)|δfg. By taking supremum over all such f and g with norm one, we see that nφ(E1)nφ(E2)δ. So either nφ(E1)ε/4 or nφ(E2)ε/4 whenever E1E2=.

For all E0 we have that εφ=nφ(N0)nφ(E)+nφ(N0E). Thus, we get that nφ(N0E)ε-nφ(E).

Since (0E)E=, as we saw earlier, it should be nφ(E)ε/4 or nφ(0  E)ε/4 and equivalently it should be nφ(E)ε/4 or nφ(E)3ε/4 for all E0.

Note that, if E1,E20 with nφ(Ei)ε/4 for i=1,2, then nφ(E1E2)nφ(E1)+nφ(E2)ε2. Thus, the relation nφ(E1E2)3ε/4 is not true and so it should be nφ(E1E2)ε4. Since φ>ε, clearly there exists a positive integer n0 such that nφ(Sj)>ε for all jn0, where Sj={iN0:ij} for all j0. 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 m0Sn0 such that nφ({m0})3ε/4. On the other hand, since (0{m0}){m0}=, nφ({m0})ε/4 or nφ(0{m0})ε/4. But nφ({m0})3ε/4, and so it should be nφ(0{m0})ε/4.

Remember that fj(z)=zj for all j0. Now we have that |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, |φ̂(m0)|β-1(m0)(β-1(m0)|1-φ̂(m0)|)ε216, and so nφ({m0})(β-1(m0)|1-φ̂(m0)|)ε216. But nφ({m0})3ε/4, and thus β-1(m0)|1-φ̂(m0)|ε12.Define ψ(z)=zm0. Then ψĤp(β), and we have that φ-ψ=nm0φ̂(n)zn+(φ̂(m0)-1)zm0=(nm0|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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