BANACH ALGEBRAS

BANACH ALGEBRAS

JOS ´E SA ´UL CAMPOS OROZCO AND ANTONI WAWRZY ´NCZYK Received 27 October 2004 and in revised form 7 July 2005

We introduce regularities in commutative Banach algebras in such a way that each regu- larity defines a joint spectrum on the algebra that satisfies the spectral mapping formula.

1. Introduction

LetBbe a complex commutative Banach algebra with unit element denoted bye. The space of linear continuous functionals onBis denoted byB.

We call regularity inBevery nontrivial open subsetR⊂Bwhich satisfies the following conditions:

ab∈R iffa∈R,b∈R, (1.1)

R=R^#, whereR^#=

b∈B| ∀ϕ∈Bϕ(b)=0=⇒0∈ϕ(R). (1.2) The setG(B) of invertible elements ofBis the main example of a regularity. As was proved in [4], the set of elements ofBwhich are not topological zero divisors is also a regularity.

In the present paper, we investigate a construction of joint spectra inBby means of regularities inB.

Letσ(a)= {µ∈C|a−µe∈G(B)}be the ordinary spectrum inB.

Recall that according to the terminology introduced by ˙Zelazko [6], a subspectrumτ inBis a mapping which associates to everyk-tuple (a1,...,ak)∈B^ka nonempty compact setτ(a1,...,ak) such that

(a)τ(a1,...,ak)⊂k i=1σ(ai),

(b)τ(p(a1,...,ak))=p(τ(a1,...,ak)) for every polynomial mappingp=(p1,...,pm) : C^k→C^m.

InTheorem 2.1, we prove that an arbitrary subspectrumτinBdefines a regularityRτ

by the formula

Rτ=

a∈B|0∈τ(a). (1.3)

Copyright©2005 Hindawi Publishing Corporation

International Journal of Mathematics and Mathematical Sciences 2005:15 (2005) 2399–2407 DOI:10.1155/IJMMS.2005.2399

Lemma 2.3used in the proof of this theorem permits us to obtain an elementary proof of a theorem belonging to ˙Zelazko which provides the complete description of all sub- spectra inB.

LetM(B) be the space of multiplicative functionals onBas usually identified with the space of maximal ideals inB.M(B) endowed with the Gelfand topology is a compact space. Fora∈B,ϕ∈M(B), we denote by ˆa(ϕ)=ϕ(a) the Gelfand transform ofa.

Theorem of ˙Zelazko [6] states that for every subspectrumτ inB, there is a unique compact subsetK⊂M(B) such that

τa1,...,ak

=ϕa1,...,ϕak

|ϕ∈K, (1.4)

for (a1,...,ak)∈B^k.

Our proof emphasizes the role played by the spectral mapping formula (b) while the original elegant proof in [6] involves more advanced methods.

The principal result of the paper isTheorem 4.1 which states that for an arbitrary regularityRthe formula

σR=

a1,...,ak

=

λ1,...,λk

∈C^k|IB

a1−λ1,...,ak−λk

∩R= ∅

(1.5) defines a subspectrum inB. ByIB(a1−λ1,...,ak−λk) the ideal generated inB by the elementsa1−λ1,...,ak−λkis denoted.

It follows that, given an arbitrary subspectrumτ, we can construct the regularityRτ

and then the subspectrumσRτ. Both subspectraτandσRτ, according to ˙Zelazko theorem, are uniquely determined by compact subsets ofM(B), sayKandK1, respectively.

We show that

K1=K=ϕ∈M(B)| ∀a∈B ϕ(a)=0 =⇒0∈a(Kˆ ). (1.6) The idea of describing spectra of single elements in a (noncommutative) Banach algebra by means of regularities appears in [1] by Kordula and M¨uller (see also [2]). The present paper is concerned with the case of a commutative Banach algebra and characterizes those regularities and corresponding spectra which admit an extension to a subspectrum.

2. Regularity corresponding to a subspectrum

Letτbe a subspectrum in a commutative unital Banach algebraBand letRτ= {a∈B| 0∈τ(a)}.

For the completness of the paper, we include the elementary proof of the basic fact in the following theorem.

Theorem2.1. Rτis a regularity.

Proof. By the property (a) of subspectra, we have∅ =τ(a)⊂σ(a) for an arbitrarya∈B.

In particular,∅ =τ(0)⊂σ(0)= {0}. Henceτ(0)= {0}and 0∈Rτ.

For|µ|>a, the elementa−µis invertible. So 0∈σ(a−µ) and 0∈τ(a−µ) neither.

The setRτis not empty and not equal toB.

The particular case of the spectral mapping formula (b) is the addition formula τ(a+b)=

λ+µ|(λ,µ)∈τ(a,b), (2.1)

corresponding to the polynomialp(x,y)=x+y.

On the other hand, by (a), we have

τ(a,b)⊂σ(a)×σ(b)⊂σ(a)×D0,b

. (2.2)

If 0∈τ(a) andb<min{|λ| |λ∈τ(a)}, then 0∈τ(a+b). The setRτis open.

We apply the spectral mapping formula in the case ofp(x,y)=xy. We obtain τ(ab)=λµ|(λ,µ)∈τ(a,b). (2.3) Immediately, we conclude that 0∈τ(ab) if and only if 0∈τ(a) and 0∈τ(b).

The setRτhas property (1.1).

The proof of property (1.2) is based on the following two lemmas.

Lemma2.2. (1) If (µ1,...,µk)∈τ(a1,...,ak) andb1,...,bm∈B, then there existλ1,..., λm∈Csuch that

µ1,...,µk,λ1,...,λm

∈τa1,...,ak,b1,...,bm

. (2.4)

(2) If (0,..., 0)∈τ(a1,...,ak) andb1ⁱ,...,bⁱ_k∈B, 1≤i≤m, then (0,..., 0)∈τ

_k

j=1

ajb¹j,..., k j=1

ajb^mj . (2.5)

Proof. (1) The spectral mapping property (b) applied to the polynomial p(x1,...,xk, y1,...,ym)=(x1,...,xk) gives us the first formula.

(2) We can find inτ(a1,...,ak,b¹1,...,b¹_k,...,b^m1,...,b_k^m) an element of the form (0,..., 0, λ¹1,...,λ¹_k,...,λ^m1,...,λ^m_k) using the first part of the lemma. If we apply the spectral mapping property to the polynomial mapping

px1,...,xk,y₁¹,...,y_k¹,...,y₁^m,...,y^m_k)= _k

j=1

xjy¹_j,..., k j=1

xjy^m_j , (2.6)

we obtain the desired property.

Lemma2.3. Let(0,..., 0)∈τ(a1,...,ak)for somea1,...,ak∈B. Then there exists a max- imal idealJ∈M(B)such thatIB(a1,...,ak)⊂Jand(0,..., 0)∈τ(b1,...,bm)for arbitrary b1,...,bm∈J.

Proof. Ifb1,...,bm∈I0=IB(a1,...,ak), then (0,..., 0)∈τ(b1,...,bm) byLemma 2.2(2).

Denote by᏶the family of all idealsI inBwhich containI0and have the property that (0,..., 0)∈τ(b1,...,bm) for arbitraryb1,...,bm∈I. For every linearly ordered subfamily Iα,α∈Sof᏶, the setα∈SIα∈᏶. So by Kuratowski-Zorn lemma, the family᏶contains a maximal elementJ. It remains to prove thatJ∈M(B). Suppose thatJis not maximal.

There existsc∈Bsuch thatc+λ∈Jfor allλ∈C.

However, byLemma 2.2(1), for arbitraryc1,...,ck∈J, the set δc1,...,ck

=λ∈C|0∈τc1,...,ck,c−λ (2.7) is nonempty. It is a compact set as an intersection of the compact setτ(c1,...,ck,c) with a line.

By the spectral mapping property again, δc1,...,ck,b1,...,bm

⊂δc1,...,ck

∩δb1,...,bm

. (2.8)

The family of compact setsδ(c1,...,ck) has the finite intersection property, so there exists λ0∈Cwhich belongs toδ(c1,...,ck) for every (c1,...,ck)∈J^k.

ByLemma 2.2(2), the ideal generated byJ andc−λ0 also belongs to᏶, which is a

contradiction.Lemma 2.3is proved.

We return to the proof ofTheorem 2.1.

Takea∈Rτ. In order to prove thatR^#_τ=Rτ, we must find a functionalφ∈B such that φ(a)=0 and 0∈φ(Rτ).By definition 0∈τ(a) and byLemma 2.2(2), (0,..., 0)∈ τ(b1,...,bm) for allb1,...,bm∈IB(a).Lemma 2.3says that in particular,abelongs to some J∈M(B) that does not intersectRτ.Jbeing a maximal ideal, it is equal to the kernel of a

linear (multiplicative) functional. The proof follows.

Since the way fromLemma 2.3to ˙Zelazko theorem is short, we include the complete proof of this important theorem.

Theorem 2.4 [6]. For every subspectrumτ on a commutative algebra B, there exists a unique compact setK⊂M(B)such that

τa1,...,ak

= ϕa1

,...,ϕak

|ϕ∈K. (2.9)

Proof. We defineKas the set of those multiplicative functionalsϕonBfor which (0,..., 0)∈τb1,...,bm

for arbitraryb1,...,bm∈kerϕ. (2.10) If (µ1,...,µk)∈τ(a1,...,ak), then (0,..., 0)∈τ(a1−µ1,...,ak−µk) and byLemma 2.3, the ideal generated bya1−µ1,...,ak−µk is contained in the kernel of a multiplicative functionalϕsuch that condition (2.10) is satisfied.

This proves thatKis nonempty and τa1,...,ak

⊂ ϕa1

,...,ϕak

|ϕ∈K. (2.11)

Now suppose thatϕ∈K anda1,...,ak∈B. Obviously,a1−ϕ(a1),...,ak−ϕ(ak)∈ kerϕand (0,..., 0)∈τ(a1−ϕ(a1),...,ak−ϕ(ak)) that implies that (ϕ(a1),...,ϕ(ak))∈ τ(a1,...,ak).

It remains to prove thatK is compact. Letφ∈K. There existb1,...,bm∈kerφsuch that (φ(b1),...,φ(bm))∈τ(b1,...,bm). By the definition of the Gelfand topology and the compactness ofτ(b1,...,bm), the property (ψ(b1),...,ψ(bm))∈τ(b1,...,bm) holds forψ in some neighborhood ofφ. The setK^cis open andKis compact.

3.Ᏺ-rationally convex sets and regularities

LetXbe a topological Hausdorffspace andᏲa family of continuous functions onX. For an arbitrary setC⊂X, we define theᏲ-rationally convex hull ofCas follows:

C=

x∈X| ∀f ∈Ᏺ f(x)=0=⇒0∈ f(C). (3.1) The termᏲ-rationally convex hull is justified at least whenCis compact andᏲis a vector space that contains constant functions.

The case is beingx∈Cif and only if f

g

(x)≤sup

y∈C

f g

(y) (3.2)

for every f,g∈Ᏺwith 0∈g(C).

A subsetC⊂XisᏲ-rationally convex ifC=C.

The hullR^#that appears in the definition of a regularity is just theB-rationally convex hull of a setR⊂B. Condition (1.2) means that every regularity isB-rationally convex.

We observe some basic properties of regularities.

Proposition3.1. Let∅ =R⊂B.

(1)IfR⊂Bsatisfies (1.1), then it contains the setG(B)of all invertible elements inB, (2)ifRis a regularity, then

R^c=

I∈M(B),I∩R=∅

{I}. (3.3)

Proof. (1) Letb∈R. Thenb=be∈R. By condition(1.1),e∈R. Ifa∈G(B), thenaa⁻¹= e∈Rand again by (1.1), we obtain thata∈R.

(2) By the definition,_{I∈M(B),I∩R=∅}{I} ⊂R^c.

Leta∈R. By condition (1.2), there existsφ∈Bsuch thatφ(a)=0 and 0∈φ(R). In particular, (kerφ)∩G(B)= ∅. By Gleason-Kahane- ˙Zelazko theorem,φ∈M(B) (see [5, page 81]) and

a∈kerφ⊂

I∈M(B),I∩R=∅

{I}. (3.4)

Proposition3.2. A nontrivial open subsetR⊂Bis a regularity if and only ifG(B)⊂R andR^#=R.

Proof. We show that the right-hand side condition implies the property (1.1). By con- dition (1.2) and Gleason-Kahane- ˙Zelazko theorem,ab∈Rif and only if ϕ(ab)=0 for someϕ∈M(B) with kerϕ∩R= ∅.This holds if and only ifϕ(a)=0 orϕ(b)=0.The

proof follows.

In general, condition (1.1) does not imply (1.2). The simplest counterexample is the setQ=B\ {0}, whereBis an integral domain.

We us observe the following hereditary property.

Proposition3.3. LetRbe a regularity inB. LetAbe a commutative unital Banach algebra andφ:A→Ba continuous homomorphism of algebras. ThenQ=φ⁻¹(R)is a regularity in A.

Proof. The setQis obviously open inA. Moreover,

G(A)⊂φ⁻¹G(B)⊂φ⁻¹(R)=Q. (3.5)

ByProposition 3.2, it is sufficent to prove thatQ^#=Q. To this end, givena∈Q, we must findϕ∈Asuch thatϕ(a)=0 and kerϕ∩Q= ∅. Sinceφ(a)∈R, there existsψ∈Bsuch thatψ(φ(a))=0 and kerψ∩R= ∅. Hence,ϕ=ψ◦φhas the desired properties.

We denote byBthe set of all Gelfand transforms of elements ofB.

Theorem3.4. LetRbe a regularity inBand let

K=

ϕ∈M(B)|0∈ϕ(R)=

ϕ∈M(B)|kerϕ∩R= ∅

. (3.6)

ThenKis a nonempty, compact,B-rationally convex set.

Proof. As we know byProposition 3.1(2),R^cis a union of a nonempty family of maximal ideals ofBwhich are precisely kernels of eachϕ∈K. HenceKis nonempty.

Ifϕ∈K^c, then ˆa(ϕ)=0 for somea∈R. If at the same timeϕ∈K, we obtain 0∈a(Kˆ ).

Hence,ϕ0(a)=0 for someϕ0∈K. This contradics the definition ofK, and soK\K= ∅. Take againϕ∈K^c anda∈R such that ϕ(a)=0. Since Ris open, there exists δ >

0 such thata−b< δ implies thatb∈R. The setV = {ψ∈M(B)| |a(ψ)ˆ |< δ}is a neighborhood ofϕinM(B). Forψ∈V, we have thata−ψ(a)∈Randψ(a−ψ(a))=0.

It follows thatV⊂K^c. SoK^cis open,Kis closed, and hence compact.

4. Subspectrum associated to a regularity

LetRbe a regularity inB. For (a1,...,ak)∈B^k, denote

σRa1,...,ak

=λ1,...,λk

∈C^k|IBa1−λ1,...,ak−λk

∩R= ∅. (4.1)

Theorem4.1. For an arbitrary regularityRin a commutative unital Banach algebra,σRis a subspectrum. IfK= {ϕ∈M(B)|0∈ϕ(R)}, then

σR

a1,...,ak

= ϕa1

,...,ϕak

|ϕ∈K. (4.2)

Proof. The condition (a) defining subspectrum is obviously satisfied becauseG(B)⊂R.

We introduce the operatorT:B→C(K) by the formula

T(a)=aˆ|K. (4.3)

The operatorT is a continuous homomorphism of algebras and its imageAis a unital subalgebra ofC(K). Ifa∈R, thenT(a) nowhere vanishes onK, hence it is invertible in C(K). Conversely, if a∈R, then by the property R^#=R and Gleason-Kahane- ˙Zelazko theorem, there existsϕ∈K such thatϕ(a)=0. So ˆavanishes atϕ∈K andT(a) is not invertible inC(K). It follows thatT(R)=G(C(K))∩A.

Theorem 3.1 in [3] states that the mapping τf1,...,fk

=λ1,...,λk

∈C^k|IAf1−λ1,...,fk−λk

∩GC(K)= ∅

(4.4) is a subspectrum onA. We extendT onA^k in a natural way:T(a1,...,ak)=(T(a1),..., T(ak)).

Notice that σR

a1,...,ak

=τTa1

,...,Tak

=τTa1,...,ak

. (4.5)

Then for an arbitrary polynomial mappingp:C^k→C^m, we have pσR

a1,...,ak

=pτTa1

,...,Tak

=τpTa1

,...,Tak

=τTpa1,...,ak

=σRpa1,...,ak. (4.6) Thus the spectral mapping formula (b) holds forσR.

For everyϕ∈Kanda1,...,ak∈B, we have IB

a1−ϕa1

,...,ak−ϕak

⊂kerϕ. (4.7)

The kernel ofϕdoes not intersectR, so (ϕ(a1),...,ϕ(ak))∈σR(a1,...,ak).

Now suppose that (µ1,...,µk)∈σR(a1,...,ak), which implies that (0,..., 0)∈σR(a1− µ1,...,ak−µk). ByLemma 2.3, we know that the idealIB(a1−µ1,...,ak−µk) is contained in the kernel of someϕ∈M(B) and 0∈σR(b) for allb∈kerϕ. It follows thatϕ∈Kand

(µ1,...,µk)=(ϕ(a1),...,ϕ(ak)).

The setK is exactly the compact set which describes the subspectrumσRin the sense of ˙Zelazko theorem (Theorem 2.4).

InSection 3, we have studied the regularity associated with a given subspectrum. Ac- cording to the definition, the regularity associated withσR is the set R1= {a∈B|0∈ σR(a)}. Obviously, R⊂R1. Ifa∈R1, thenIB(a)∩R= ∅. There existsb∈B such that ab∈R. Hencea∈Rby property (1.1). We conclude thatR1=R.

It is well known that different subspectra can lead to the same set of regular elements.

Letτbe the approximate point spectrum. The corresponding regularityRτ is the set of all elements ofBwhich are not topological zero divisors while the setKτdefiningτvia formula (2.9) is the set of maximal ideals which consists of joint topological zero divisors.

The spectrumσRτ was studied in [4] and it corresponds toK equal to the set of all maximal ideals consisting of topological zero divisors, which in general differs fromKτ.

IfK⊂M(B) is compact andτis the subspectrum defined by formula (2.9), then the regularityRτcan be described as

a∈B|0∈a(K)ˆ . (4.8)

Proposition4.2. LetK1,K2⊂M(B)and let Ri=

a∈B|0∈aˆKi

, (4.9)

i=1, 2. ThenR1=R2if and only ifK1=K2.

Proof. Suppose thatR1=R2. It means that fora∈B, the Gelfand transform ˆavanishes onK1if and only if it vanishes onK2. If ˆa(ϕ)=0, then ˆa(K1) contains zero if and only if a(Kˆ 2) does. HenceK1=K2.

Now suppose thatK1=K2and thata∈R1. It follows that ˆa(ϕ)=0 for someϕ∈K1⊂ K2. We obtain 0∈a(Kˆ 2). Soa∈R2. This shows thatR^c1⊂R^c2, andR2⊂R1. Similarly, we

can prove the opposite. ThenR1=R2.

For a given regularityRinB, the subspectrumσRis the largest subspectrum havingR as the corresponding regularity.

Proposition4.3. LetRbe a regularity and letτbe a subspectrum such thatRτ=R. Then for everyk-tuple(a1,...,ak)∈B^k,

τa1,...,ak

⊂σR

a1,...,ak

. (4.10)

Proof. IfRis a regularity, then according toTheorem 4.1, σR

a1,...,ak

= ϕa1

,...,ϕak

|ϕ∈K, (4.11)

whereK=KasTheorem 3.4asserts.

Ifτis a subspectrum of the form τa1,...,ak

= ϕa1

,...,ϕak

|ϕ∈K1

(4.12)

andRτ=R, thenK1=K=KbyProposition 4.2. In particular,K1⊂Kand τa1,...,ak

⊂σRa1,...,ak. (4.13)

References

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[3] A. Vel´azquez Gonz´alez and A. Wawrzy ´nczyk,The projection property of a family of ideals in subalgebras of Banach algebras, Bol. Soc. Mat. Mexicana (3)8(2002), no. 2, 155–160.

[4] A. Wawrzy ´nczyk,On ideals consisting of topological zero divisors, Studia Math.142(2000), no. 3, 245–251.

[5] W. ˙Zelazko,Banach Algebras, Elsevier, Amsterdam; PWN—Polish Scientific Publishers, War- saw, 1973.

[6] ,An axiomatic approach to joint spectra. I, Studia Math.64(1979), no. 3, 249–261.

Jos´e Sa ´uL Campos Orozco: Departamento de Matem´aticas, Divisi ´on de Ciencias B´asicas e Ingenier´ıa, Universidad Aut ´onoma Metropolitana - Unidad Iztapalapa, 186 Avenida San Rafael Atlixco, Col. Vicentina, C.P. 09 340, Distrito Federal, AP 55-534, M´exico

E-mail address:[email protected]

Antoni Wawrzy ´nczyk: Departamento de Matem´aticas, Divisi ´on de Ciencias B´asicas e Ingenier´ıa, Universidad Aut ´onoma Metropolitana - Unidad Iztapalapa, 186 Avenida San Rafael Atlixco, Col. Vi- centina, C.P. 09 340, Distrito Federal, AP 55-534, M´exico

E-mail address:[email protected]