GLEASON-KAHANE- ˙ZELAZKO THEOREM FOR SPECTRALLY BOUNDED ALGEBRA

We prove by elementary methods the following generalization of a theorem due to Gleason, Kahane, and Żelazko. Let A be a real algebra with unit 1 such that the spectrum of every element in A is bounded and let φ:A→ℂ be a linear map such that φ(1)=1 and (φ(a))2


Introduction
Let A be a real algebra with unit 1 and let φ : A → C be a linear transformation with φ(1) = 1. When is φ multiplicative? That is, when is φ(ab) = φ(a)φ(b) for all a, b in A? This question was first answered for the case of a complex Banach algebra by Gleason [3], Kahane andŻelazko [6]. Their result, now known as the Gleason-Kahane-Żelazko theorem, states that, if φ(a) = 0 for every invertible element a in A (or equivalently φ(a) lies in the spectrum of a for every a in A), then φ is multiplicative. Subsequently several generalizations of this result were published by many authors. These include (i) real Banach algebra-Kulkarni [7], (ii) complex spectrally bounded algebra-Roitman and Sternfeld [10]. The articles by Jarosz [4,5] and Sourour [11] contain surveys of many of these results.
The aim of the present article is two-fold. First we extend this result to a real spectrally bounded algebra (Theorem 2.9), that is, the algebra in which the spectrum of each element is bounded (Definition 2.6). The result says φ is multiplicative if and only if (φ(a)) 2 + (φ(b)) 2 = 0 for all a,b in A such that ab = ba and a 2 + b 2 is invertible. The class of real spectrally bounded algebras includes all the above-mentioned algebras. All these characterizations including the ones to be discussed in this paper are mainly in terms of the spectrum.
Our second aim is to give simple proofs. The classical proofs make use of the tools from the complex function theory, in particular Hadamard's theorem. Our proof uses the elementary properties of polynomials, namely, relations between roots and coefficients. The essential ideas are in Lemma 2.5. Similar ideas were used by Roitman and Sternfeld in [10] (see also [8,Theorem 2.4.3]).
In Sections 3 and 4, we attempt to relate these ideas to Ransford's generalized spectrum [9]. In Section 3, it is proved that if for each x in a complex algebra, φ(x) lies in the generalized spectrum of x, then φ is multiplicative. A statement of this theorem was published by Catalin Badea in [1], where it was mentioned that the proof will be published elsewhere, but the proof was not published anywhere. Here is the first instance where a proof is given for that theorem.
In Section 4, the result in Section 3 is extended to a real algebra E in terms of Ransford's spectrum. We have also extended the concept of Ransford's spectrum to the real case. It is shown that if (φ(a)) 2 + (φ(b)) 2 = 0 for all a,b in E such that ab = ba and a 2 + b 2 in Ω R , then φ is multiplicative (Theorem 4.8). Examples are given to show that this condition is not necessary.
In the last section, using the sufficient conditions obtained in Sections 3 and 4, we give a sufficient condition for a linear transformation between spectrally bounded, (complex or real) algebras, to be multiplicative.

Notation. Let
A be an algebra with the unit 1. An algebra element λ · 1 (product of λ and one), where λ ∈ C, will be denoted just as λ. Let Inv(A) and Sing(A) denote the set of invertible and singular (noninvertible) elements in A, respectively. For an element a in A the spectrum is denoted by Sp(a,A). If A is a complex algebra, Sp(a,A) := λ ∈ C : λ − a ∈ Sing(A) . (2.1) If A is a real algebra,

Complexification.
Complexification of a real algebra A, denoted by A C , is the set A × A with addition, scalar multiplication, and multiplication are defined in the following way. For every (a,b), (c,d) in A C and α + iβ in C, With these operations A C becomes a complex algebra. Let us recall some results in [2]. These results will be used to prove a lemma. Proof. (2.4) The following lemma will be used repeatedly. Proof.
=⇒ (a,b) ∈ Sing A C (using Proposition 2.2). (2.7) Lemma 2.5. Let A be a complex algebra with unit 1, let ψ : A → C be a complex linear functional with ψ(1) = 1. Fix a ∈ A and define P : C → C by Let λ j , j = 1,...,n, be the roots of the polynomial P. Then (2.9) Proof. As λ j , j = 1,...,n, are the roots, On the other hand by expanding P, This means for every a in A, there exist M a > 0 such that |λ| ≤ M a whenever λ ∈ Sp(a,A). In other words, if is the spectral radius, then, r(a) ≤ M a . This is a property which we will be using to establish the result.
Definition 2.7 (spectral algebra). A norm which dominates the spectral radius is called a spectral norm. A spectral algebra is an algebra on which a spectral norm can be defined.
In view of the spectral radius formula, every Banach algebra is a spectral algebra. See [8] for examples of spectral algebras that are not Banach algebras. Also, every spectral algebra is a spectrally bounded algebra. The next example shows that the converse is not true.

S. H. Kulkarni and D. Sukumar 2451
Example 2.8. Let C(z) denote the set of all complex rational functions. Consider the algebra C ⊕ C(z). Then for an element (λ, f ) in the algebra, (2.17) Hence the algebra is spectrally bounded. But it is not a spectral algebra because in a commutative spectral algebra the spectral radius is subadditive and submultiplicative by [8,Theorem 2.4.11]. Here the spectral radius is neither subadditive nor submultiplicative by the following inequalities: . Let A be a real unital algebra. Let φ : A → C be linear and unital (i.e., φ(1) = 1). The first four conditions below are equivalent and imply the last two conditions. If A is a spectrally bounded algebra, then all six conditions are equivalent: The implication is trivial. This shows that the first four conditions are equivalent.
(4) obviously implies (1). This establishes equivalence of the first four statements. (a −1 ). This shows that φ(a) can not be zero. (6)⇒ (2) The implication holds when A is spectrally bounded algebra. Let a,b ∈ A be such that ab = ba and a 2 + b 2 is invertible. Then, since a 2 + b 2 = (a + ib)(a − ib), both a + ib and a − ib are invertible. Now, by (6) . Now the conclusion follows from Theorem 2.9.
The following example shows that the condition (1) of Corollary 2.10 does not imply condition (2) when A is a real algebra.

Ransford spectrum in a complex algebra
Ransford extended the concept of spectrum for a general complex normed linear space in [9] by replacing the set of all invertible elements with a set, denoted as Ω, satisfying some properties as follows. Let X be a complex linear space and 1 a fixed nonzero element in X. Let Ω be a subset of X such that (1) 0 / ∈ Ω, Then, for every x ∈ X, Ransford's Ω spectrum of X is given by It is proved in [9] that if X is a normed linear space and Ω an open subset of X, then 2454 Gleason-Kahane-Żelazko theorem Sp Ω (x) is bounded for every x ∈ X. That is, if λ ∈ Sp Ω (x) then |λ| ≤ M x for some M x > 0. In fact, it is proved in [9] that Sp Ω (x) is a nonempty compact subset of C for every x in X. He also proved an analog of the spectral radius formula using a property called pseudoconvexity. When X is an algebra, we assume another property for the set Ω in terms of multiplication as follows.
(4) There is an increasing sequence {n j } (i.e., n 1 < n 2 < n 3 ··· where n j ∈ N for j = 1,2,3,...) such that holds true for all j ∈ N. The statement of the following theorem, with slight modifications, was given in [1] but the proof is not published anywhere.
Next we prove that (1) implies (3). Fix x ∈ X and n j ∈ N. Define P : C → C as follows: Consider the roots λ i for 1 ≤ i ≤ n j of the polynomial P. These roots satisfy the equation In view of (1), this implies that [λ i − x] nj is not in Ω. Hence (λ i − x) is also not in Ω. So λ i ∈ Sp Ω (x) for 1 ≤ i ≤ n j by definition of spectrum. Also by Lemma 2.5 we get Since n j ∈ N is arbitrary and spectrum is bounded, allowing n j → ∞ and noting | Now φ is multiplicative by Theorem 2.9.
The following example shows that the third condition in the above theorem does not imply any of the first two equivalent conditions. Example 3.2. Consider X = C 2 with coordinatewise multiplication, then (1,1) is the unit element. Let then Ω is an open set satisfying the conditions of hypothesis. Define φ :

Ransford spectrum in a real algebra
In this section, we extend the ideas in Section 3 to the case of a real algebra. For this, first we need to define Ransford's spectrum in this case.
Definition 4.1. Let E be a real algebra with unit 1. Let Ω R be a subset of E that satisfies For every x ∈ E, spectrum of x is defined as (4) for a certain increasing sequence n 1 ,n 2 ,n 3 ,... (i.e., n 1 < n 2 < n 3 ··· ) where n j ∈ N for j = 1,2,3,..., Example 4.2. In R with usual multiplication, the set R * satisfies all conditions with a sequence 1,2,3,... . Consider the complexification E C of E and a subset Ω C of E C defined by Then Ω C satisfies the following conditions: (4) x ∈ Ω C ⇒ x nj ∈ Ω C for the same increasing sequence n 1 < n 2 < n 3 ··· where n j ∈ N for j = 1,2,3,... . The following propositions and lemma are general in the sense that Inv(A) and Inv(A C ) in Propositions 2.1, 2.2, and 2.3, and Lemma 2.4 are replaced by Ω R , Ω C . But proofs are similar.
Then F is complex linear. Assume for all a,b in E, satisfying ab = ba and a 2 + b 2 in Ω R , for all a,b ∈ E such that ab = ba and a 2 + b 2 ∈ Ω R , then φ is multiplicative.
Proof. Fix a ∈ E and n j ∈ N. Define P : C → C as follows: Consider the roots λ i for 1 ≤ i ≤ n j of the polynomial P. The equation that is, implies that [λ i (1,0) − (a,0)] nj / ∈ Ω C by Lemma 4.7. Hence λ i (1,0) − (a,0) is also not in Ω C by property (5) in definition. That is, λ i ∈ Sp ΩC (a,0), which is equivalent to λ i ∈ Sp ΩR (a) for 1 ≤ i ≤ n j by Proposition 4.6. Also by Lemma 2.5, we get Since n j ∈ N is arbitrary and Sp ΩR (a) is bounded for every a in E, letting n j → ∞ and Now the conclusion follows by Theorem 2.9.

Operators
In this section, we give sufficient conditions for a linear transformation, between spectrally bounded algebras, to be multiplicative. Let X and Ω be as in Theorem 3.1. Ransford defined Ω-radical, in [9], as Rad Ω (X) := {a ∈ X : a + Ω = Ω}.
In other words φ • T ∈ M ΩA . Hence by Theorem 3.1, φ • T is multiplicative. Thus, as φ is multiplicative. Hence Since every real Banach algebra is a spectrally bounded real algebra, we show in the next corollary that [7, Theorem 7] follows from the above theorem using [7, Theorem 2]. Proof. (1) implies (2) and (2) implies (3)  The assumption, commutativity, on B in Theorem 5.3 is necessary by [7, Example 10]. Here we give an example which shows that semisimple condition on B is necessary, in Theorem 5.2, to get T as multiplicative operator.
Example 5.5. Let X and Ω be as in Example 3.2. Then X is semisimple by the explanation in Example 5.1. Now define T : X → X as T(z 1 ,z 2 ) = (z 1 ,(z 1 + z 2 )/2). Clearly T satisfies hypothesis of Theorem 5.2 but is not multiplicative.