Elements in a Commutative Banach Algebra Determining the Norm Topology

PROCEEDINGS OF THE AMERICAN MATHEMATICAL SOCIETY Volume 129, Number 4, Pages 1057–1064 S 0002-9939(00)05659-8 Article electronically published on November 8, 2000

ELEMENTS IN A COMMUTATIVE BANACH ALGEBRA DETERMINING THE NORM TOPOLOGY

A. R. VILLENA

(Communicated by Dale Alspach)

Abstract. For an element a of a commutative complex Banach algebra (A, k·k) we investigate the following property: every complete norm |·| on A making the multiplication by a from (A, |·|) to itself continuous is equivalent to k·k.

1. Introduction and statements of the theorems

Let (A, k·k) be a commutative complex Banach algebra and let ΦA denote the set of all nonzero multiplicative linear functionals on A. Itisabasicfactthatthese functionals are continuous. The carrier space of A is the set ΦA endowed with the Gelfand topology, which is the relative weak-star topology inherited from the 0 topological dual space A of A.ΦA is a locally compact Hausdorff space and the Gelfand representation of A is the mapping a 7→ aˆ from A to the Banach algebra C0(ΦA) defined bya ˆ(φ)=φ(a). The hull-kernel topology on ΦA is the topology on ΦA whose closed sets are of the form h(S)={φ ∈ ΦA : φ(S)=0} with some S ⊂ A. This is a T1-topology which is not necessarily Hausdorff and it is weaker than the Gelfand topology. In general the Gelfand topology and the hull-kernel topology are different and A is regular if both of them coincide. For further information, we refer to [1, Chapters 17 and 23] and [4, Chapter 3]. From the uniform boundedness theorem and the continuity of multiplicative linear functionals on a Banach algebra, it follows immediately that any complete norm on the underlying linear space of a semisimple commutative Banach algebra (A, k·k) under which all the multiplication operators are continuous is automatically equivalent to k·k. It is thus natural to ask whether or not the norm topology in the Banach algebra A may be determined by a given multiplication operator by itself. An element a in a commutative Banach algebra (A, k·k)issaidtodetermine the norm topology of A if every norm |·| on A for which it becomes a Banach space and which makes the multiplication operator by a from (A, |·|) to itself continuous is equivalent to k·k. The element a is said to almost determine the norm topology of A if the following property holds: for every complete norm on A making the multiplication by a from (A, |·|) to itself continuous there exists an idempotent e ∈ A such that dim eA < ∞ and the quotient norms of |·| and k·k on A/(eA)are equivalent.

Received by the editors May 19, 1999 and, in revised form, June 28, 1999. 2000 Mathematics Subject Classiﬁcation. Primary 43A20, 46E25, 46J10, 46J15.

c 2000 American Mathematical Society 1057

License or copyright restrictions may apply to redistribution; see https://www.ams.org/journal-terms-of-use 1058 A. R. VILLENA T The radical of A is the subset of A given by Rad(A)= ker φ provided φ∈ΦA that ΦA =6 ∅ and Rad(A)=A otherwise. A is semisimple if Rad(A) = 0 and it is radical if Rad(A)=A. In the present paper we prove the following two theorems. Theorem 1. Let A be a semisimple commutative complex Banach algebra and let a ∈ A.

(i) If, for every λ ∈ C,theset{φ ∈ ΦA :â(φ)=λ} has no interior points for the hull-kernel topology on ΦA, then a determines the norm topology of A. (ii) If, for every λ ∈ C,theset{φ ∈ ΦA :â(φ)=λ} has a finite number of interior points for the hull-kernel topology on ΦA,thena almost determines the norm topology of A. (iii) If there is λ ∈ C such that the set {φ ∈ ΦA :â(φ)=λ} has infinitely many interior points for the hull-kernel topology on ΦA,thena does not determine the norm topology of A.

TheoremT 2. Let A be a commutative complex Banach algebra and let a ∈ A such ∞ n that n=1 a Rad(A)=0. Assume that A is an integral domain and that dim A = ∞.Thena determines the norm topology of A if and only if it is not a complex multiple of the identity.

Remarks 1. 1. Since the hull-kernel topology on ΦA is weaker than the Gelfand topology, assertions (i) and (ii) of Theorem 1 still hold if we replace the hull-kernel topology by the Gelfand topology. If A is regular, then these topologies are the same. 2. The Banach algebras under consideration need not have an identity. For a Banach algebra without identity the assertion of a 6∈ C1 in Theorem 2 means that a =0.6

2. Examples It should be noted that the class of semisimple commutative Banach algebras coincides with the class of Banach function algebras (regarded as being defined on the carrier space). We consider in the next corollary the suggestive form taken by Theorem 1 for regular Banach function algebras. For a locally compact Hausdorff space Ω, let C0(Ω) denote the space of continuous functions on Ω which vanish at infinity. With pointwise operations and the supremum norm k·k∞, C0(Ω) is a semisimple regular commutative Banach algebra. The nonzero multiplicative linear functionals on C0(Ω) are the evaluation maps εω (ω ∈ Ω) and the map ω 7→ εω is a homeomorphism from Ω onto the carrier space. A subalgebra A of C0(Ω) is a natural Banach function algebra on Ω if it has the following properties: (i) A separates the points of Ω and vanishes nowhere on Ω, (ii) A is a Banach algebra with respect to some norm, and (iii) the natural embedding ω 7→ εω from Ω into ΦA is onto. Corollary. Let A be a natural regular Banach function algebra on a locally compact Hausdorff space Ω and let f ∈ A. (i) If f is constant on no open subset of Ω,thenf determines the norm topology of A. (ii) If, for every λ ∈ C, f −1(λ) has a finite number of interior points, then a almost determines the norm topology of A.

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(iii) If f is constant on an inﬁnite open subset of Ω,thenf does not determine the norm topology of C0(Ω). Remarks 2. 1. Of course, if the topological space Ω has no isolated points, then the second assertion of the preceding result disappears and therefore f determines the norm topology of A if and only if it is constant on no open subset of Ω. This fact was proved in [7] for the algebra C(K) for a compact Hausdorﬀ space K without isolated points. 2. Let A be as in the Corollary and let e be an idempotent in A such that dim eA < ∞. Then there are ω1,...,ωm isolated points of Ω such that e is the characteristic function of the set {ω1,...,ωm}.WritingE =Ω\{ω1,...,ωm} we can assert that the quotient algebra A/(eA) is isomorphic to the restriction algebra AE = {f|E : f ∈ A}.

Examples 1. 1. Of course, for a locally compact Hausdorﬀ space Ω, C0(Ω) is a natural regular Banach function algebra on Ω. Cn([0, 1]), BV C([0, 1]), and AC([0, 1]) are natural regular Banach function algebras on [0, 1] (see [4, A.2.4/5]) and a function in any of these algebras determines the norm topology of the algebra if and only if it is constant on no open interval of [0, 1]. For a compact metric space K ≤ and 0 <α 1, Lipα(K) is a natural regular Banach function algebra on K ([5, Section 2]). 2. If G is a locally compact abelian group, then the group algebra L1(G)isa semisimple regular commutative Banach algebra. The carrier space of L1(G)may be identiﬁed, via the Fourier transform, with the dual group of G and the Gelfand and Fourier transforms for L1(G) coincide. Consequently, Theorem 1 is true if we replace the Gelfand transform of f ∈ L1(G) and the carrier space of L1(G)by the Fourier transform of f and the dual group of G, respectively. Since the dual group of Rn has no isolated points we conclude that f ∈ L1(Rn) determines the norm topology of L1(Rn) if and only if its Fourier transform is constant on no open n subset of R .Forρ>0letχ[−ρ,ρ] be the characteristic function of the interval − sin(2πρx) [ ρ, ρ]. The Fourier transform of χ[−ρ,ρ] is the function πx and therefore 1 χ[−ρ,ρ] determines the norm topology of L (R). In the next example we show that in the presence of isolated points in the carrier space even the most natural elements of the algebra can fail to determine the norm topology. Example 2. Let A = C([0, 1] ∪{2}) and consider the elements a and u in A given by ( ( t if t ∈ [0, 1], 0ift ∈ [0, 1], a(t)= and u(t)= 0ift =2, 1ift =2.

We claim that aA =6 aA. By the Weierstrass approximation theorem, the function (√ t if t ∈ [0, 1], b(t)= 0ift =2

lies in aA.Ifthereexistedc ∈ A such that a(t)c(t)=b(t) for all t ∈ [0, 1], we would have c(t)=t−1/2 for all t ∈]0, 1], a contradiction. Since aA =6 aA and u 6∈ aA,it follows that there exists a discontinuous linear functional f on C([0, 1] ∪{2})such that f(aA)=0andf(u) = 1. The map |b| = k2b − f(b)uk deﬁnes a norm on A for

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which it becomes a Banach space and |·| is not equivalent to k·k. On the other hand, |ab| = k2ab − f(ab)uk = k2abk = ka(2b − f(b)u)k ≤kakk2b − f(b)uk = kak|b| for all b ∈ A, which shows that the multiplication by a from (A, |·|)toitselfis continuous. Examples 3. 1. Let C[[x]] denote the algebra of formal power series with complex coefficients in the indeterminate x. A Banach algebra of power series is a Banach space A which is a subalgebra of C[[x]] containing the indeterminate x and such that the coefficient functionals are continuous. The closed graph theorem shows that multiplication is separately continuous, and so A becomes a Banach algebra for an equivalent norm. Further A Tis an integral domain and for any nonconstant power ∞ n series a in A it is clear that n=1 a A = 0. Consequently, Theorem 2 shows that any nonconstant element of A determines the norm topology of A. For an example of Banach algebra of power series, let ω be an algebra weight on N0,thatis,ω is a + function from N0 to R satisfying ω(0) = 1 and ω(m + n) ≤ ω(m)ω(n) ∀m, n ∈ N. Let ( ) X∞ X∞ 1 n ` (ω)= f = αnx : kfk = |αn|ω(n) < ∞ . n=0 n=0 `1(ω) is a Banach algebra of power series. It is well known that `1(ω) is semisimple if lim(ω(n))1/n > 0 and Rad(`1(ω)) = {f ∈ `1(ω): f(0) = 0} if lim(ω(n))1/n =0. For further information we refer the reader to [2]. 2. Let ω be a positive measurable function on [0, +∞[ such that ω(s + t) ≤ ω(s)ω(t) for all s, t ∈ [0, +∞[; such a function is said to be a weight function on [0, +∞[. Let L1(ω) denote the space of (equivalence classes of) measurable functions R ∞ f on [0, +∞[forwhichkfk = |f(t)|ω(t)dt < ∞. L1(ω) becomes a complex ω 0 R k·k ∗ t − Banach algebra for the norm ω and the convolution (f g)(t)= 0 f(t s)g(s)ds as product. For a thorough treatment of these algebras we refer the reader to [3]. It 1 1/t is known that L (ω) is an integral domain. It is semisimple if limt→∞ ω(t) > 0, 1/t 1 and radical if limt→∞ ω(t) =0.Forf ∈ L (ω)\{0},let α(f)=sup{δ ≥ 0: f = 0 almost everywhere on [0,δ]}, 1 and set α(0) = ∞.Iff ∈ L (ω)satisfiesT α(f) =6 0, then Titchmarsh’s convolution ∞ n∗ 1 theorem [3, Theorem 3.10] shows that n=1 f L (ω) = 0. According to Theorem 2 we deduce that a nonzero function f ∈ L1(ω) determines the norm topology of 1 1/t L (ω) provided that either limt→∞ ω(t) > 0orα(f) =6 0. Therefore for any ∞ weight function ω and 0 <ρ1 <ρ2 < the function χ[ρ1,ρ2] determines the norm topology of L1(ω).

3. Proofs of the theorems Although the Banach algebras under consideration need not have an identity, given a and b in the algebra A and λ ∈ C,wewrite(a + λ)b instead of ab + λb. If X is a linear space, then we measure the equivalence of two norms |·| and k·k on X for which X becomes a Banach space by considering the separating subspace of the identity map from (X, |·|)onto(X, k·k). This is deﬁned as the subspace S of those x ∈ X for which there exists a sequence {xn} in X such that {xn} converges

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to zero for the norm |·|and converges to x for the norm k·k. The closed graph theorem shows that |·|and k·kare equivalent if, and only if, S =0. Lemma 1. Let X be a linear space and let |·| and k·k be two norms on X for which it becomes a Banach space.

(i) If Fn is a sequence of linear operators from X to itself which are |·|-|·|- continuous and k·k-k·k-continuous, then there exists N ∈ N such that k·k k·k F1 ···FN (S) = F1 ···Fn(S) for all n ≥ N. (ii) If F is a linear operator form X to itself which is |·|-|·|-continuous and k·k-k·k- k·k T∞ k·k ∈ N N n continuous, then there exists N such that F (S) = n=1 F (S) . Proof. The ﬁrst assertion follows immediately from [6, Lemma 1.6]. k·k k·k From (i) we see that there exists N ∈ N such that F n(S) = F N (S) for all n ≥ N. Therefore F is a continuous linear operator from the Banach space k·k (F N (S) , k·k) onto a dense linear subspace of itself. According to Mittag-Leﬄer theorem [2, Theorem 5.3], we have k·k ∞ k·k \ k·k F N (S) = F n(F N (S) ) . n=1 k·k It is immediate that F (S) ⊂ S and that S is closed in (X, k·k), and so F N (S) ⊂ T∞ k·k T∞ k·k n N ⊂ n ⊂ B S. Hence n=1 F (F (S) ) n=1 F (S) F (S) , and our assertion follows. Lemma 2. Let (A, k·k) be a commutative complex Banach algebra and let a ∈ A. Assume that |·| is a norm on A for which A is a Banach space and the multiplication operator by a from (A, |·|) to itself is continuous and let S be the separating subspace of the identity map from (A, |·|) onto (A, k·k). Then there are λ1,...,λm ∈ C such that (a − λ1) ···(a − λm)S ⊂ Rad(A).

Proof. We claim that the set {φ(a): φ ∈ ΦA\h(S)} is ﬁnite, where h(S)={φ ∈ ΦA : φ(S)=0}. Suppose the claim were false. Then we could ﬁnd a sequence {φn} in ΦA\h(S) such that φn(a) =6 φm(a)ifn =6 m. For every n ∈ N the linear operator b 7→ (a − φn(a))b from A to itself is |·|-|·|-continuous and k·k-k·k-continuous. From Lemma 1(i) we deduce that there exists N ∈ N such that k·k k·k (a − φ1(a)) ···(a − φN (a))S = (a − φ1(a)) ···(a − φn(a))S for all n ≥ N.Consequently, k·k (a − φ1(a)) ···(a − φN (a))S ⊂ (a − φ1(a)) ···(a − φN+1(a))S .

Since φN+1 is continuous on (A, k·k), we see that

(φN+1(a) − φ1(a)) ···(φN+1(a) − φN (a))φN+1(S)

= φN+1((a − φ1(a)) ···(a − φN (a))S)

⊂ φN+1((a − φ1(a)) ···(a − φN+1(a))S)=0,

which clearly forces φN+1(S) = 0, a contradiction. Let φ1,...,φm ∈ ΦA\h(S)be such that {φ(a): φ ∈ ΦA\h(S)} = {φ1(a),...,φm(a)}. We proceed to show that φ((a−φ1(a)) ···(a−φm(a))S) = 0 for all φ ∈ ΦA. This equality is easily seen to be

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true whenever φ lies in h(S). On the other hand, if φ 6∈ h(S), then φ(a)=φk(a) for some k ∈{1,...,m} and therefore the equality holds.

Lemma 3. Let X be a topological space, let C1,...,CS n be closed subsets of X, ⊂ n and let D be an open subset of X such that D k=1 Ck.S Assume that, for each ∈{ } ◦ ⊂ n ◦ k 1,...,n , the interior Ck of Ck is closed. Then D k=1 Ck . Proof of Theorem 1. Let k·k be the norm of A and let |·| be a norm on A for which A is a Banach space and the multiplication operator by a from (A, |·|)toitselfis continuous. Let S be the separating subspace of the identity map from (A, |·|) k·k ∈ C − ··· − onto (A, ). Lemma 2 givesS λ1,...,λm such that (a λ1) (a λm)S =0. \ ⊂ m { ∈ } Consequently, ΦA h(S) k=1 φ ΦA : φ(a)=λk . Let us suppose that, for every λ ∈ C,theset{φ ∈ ΦA :â(φ)=λ} has a finite number of interior points for the hull-kernel topology on ΦA and let us consider the topological space X =ΦA endowed with the hull-kernel topology. For each k ∈{1,...,m} the set Ck = {φ ∈ } ◦ ΦA : φ(a)=λk is closed in X ([4, Lemma 3.1.16])S and its interior Ck is finite and \ ⊂ m ◦ so it is closed. Lemma 3 now yields ΦA h(S) k=1 Ck . Consequently, if the requirement in the first assertion is fulfilled, then we can assert that ΦA\h(S)= ∅.ThusφA\h(S)=∅.Thusφ(S) = 0 for all φ ∈ ΦA and therefore S =0, which shows that |·|and k·kare equivalent. We now turn to the case traced in our second assertion. From what has already been proved, it may be concluded that ΦA\h(S)={φ1,...,φm} with φ1,...,φm ∈ ΦA and note that, for each k ∈ {1,...,m},theset{φk} is open. Let k ∈{1,...,m}.SinceΦA\{φk} is closed and φk 6∈ ΦA\{φk}, [1, Proposition 23.4] gives ek ∈ A such that φk(ek)=1and 6 2 − ∈ φ(ek) = 0 for all φ = φk. It is immediate that φ(ek ek) = 0 for all φ ΦA, 2 ∈{ } 6 and so ek = ek.Furthermore,ifi, j 1,...,m with i = j,thenφ(eiej)=0 for all φ ∈ ΦA, which implies eiej =0.Thuse = e1 + ···+ em is an idempotent, φk(e)=1fork =1,...,m,andφ(e) = 0 for all φ ∈ ΦA\{φ1,...,φm}.The m map eb 7→ (φ1(b),...,φm(b)) from eA to C is easily seen to be injective and therefore dim eA < ∞. It follows immediately that φ((1 − e)S) = 0 for all φ ∈ ΦA and therefore (1 − e)S = 0. The closed graph theorem now shows that the map b 7→ b − eb from (A, |·|)to(A, k·k) is continuous. Let {an} be a sequence in A such that {an + eA} converges to zero in the quotient Banach space (A/(eA), |·|). Then there exists a sequence {bn} in A such that {an + ebn} converges to zero in (A, |·|) and hence an − ean =(1− e)(an + ebn) converges to zero in (A, k·k), which shows that the sequence {an + eA} converges to zero in the quotient Banach space (A/(eA), k·k). From this we conclude that the quotient norms of |·|and k·kon A/(eA)areequivalent. We finally assume that the set {φ ∈ ΦA :â(φ)=λ} has an infinitely many interior, say D, for the hull-kernel topology and some λ ∈ C.Let{φn} be a sequence of pairwise different elements of D.SinceΦA\D is closed and φ1 6∈ ΦA\D, [1, Proposition 23.4(i)] gives a1 ∈ A such that φ1(a1)=1andφ(a1) = 0 for all φ ∈ ΦA\D.Foreachn ∈ N we can choose an+1 ∈ A such that φk(an+1)=0 for k =1,...,n, φn+1(an+1) =0,and6 kan+1k = 1 (see [1, Lemma 18.17]). It { ∈ N} is a simple matterT to show that the set an : n is linearly independent and { ∈ N} the intersection of φ∈D ker φ with the linear subspace spanned by an : n is zero.T Consequently, there is a linear functional f on A such that f(b)=0for ∈ ∈ N all b φ∈D ker φ and f(an)=n for all n .Ofcoursef is discontinuous and therefore the map |b| = k2b − f(b)a1k defines a norm on A for which it becomes a Banach space and |·|is not equivalent to k·k. The proof is completed by showing

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that the multiplication operator by a from (A,T|·|) to itself is continuous. For ∈ − ∈ every b A, we see at once that (a λ)b φ∈D ker φ and therefore we have 0=f((a − λ)b)=f(ab) − λf(b). Since φ(a)=λ for all φ ∈ D and φ(a1)=0for all φ ∈ ΦA\D, we conclude that φ((a − λ)a1) = 0 for all φ ∈ ΦA and hence that aa1 = λa1.Consequently,

|ab| = k2ab − f(ab)a1k = k2ab − λf(b)a1k

= k2ab − f(b)aa1k≤kakk2b − f(b)a1k = kak|b| for all b ∈ A, which establishes the desired conclusion.

Proof of Theorem 2. If a = λ1forsomeλ ∈ C, then for any discontinuous linear functional f on A and any element u ∈ A such that f(u)=1themap|b| = k2b − f(b)uk deﬁnes a complete norm on A which clearly makes the multiplication by a continuous and it is not equivalent to the norm of A. Assume that a is not a complex multiple of the identity. Let k·k be the norm of A and let |·| be a norm on A for which A is a Banach space and the multiplication operator by a from (A, |·|) to itself is continuous. Let S be the separating subspace of the identity map from (A, |·|)onto(A, k·k). By considering the complex numbers λ1,...,λm given in Lemma 2 we have

(a − λ1) ···(a − λm)S ⊂ Rad(A). From Lemma 1(ii) it follows that there exists N ∈ N such that

∞ k·k k·k \ N N N n a (a − λ1) ···(a − λm) S = [a(a − λ1) ···(a − λm)] S , n=1 which gives k·k \∞ N N N n n a (a − λ1) ···(a − λm) S ⊂ a [(a − λ1) ···(a − λm)] S n=1 k·k \∞ ⊂ an Rad(a) =0. n=1 We claim that S = 0. On the contrary, suppose that S =0.Since6 A is n an integral domain, it follows that [a(a − λ1) ···(a − λm)] = 0 and therefore a(a − λ1) ···(a − λm) = 0. Since a[(a − λ1)A] ···[(a − λm)A]=0itmaybecon- cluded that (a − λk)A =0forsomek ∈{1,...,m}. Clearly Ak =6 0, and hence −1 ∈ (λk a)b = b for all b A, a contradiction.

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6. A. M. Sinclair, Automatic continuity of linear operators, Cambridge University Press, 1976. MR 58:7011 7. A. R. Villena, Operators determining the complete norm topology of C(K), Studia Math. 124 (1997), 155–160. MR 98h:46026

Departamento de Analisis Matematico, Facultad de Ciencias, Universidad de Granada, 18071 Granada, Spain E-mail address: [email protected]

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