Math. Log. Quart. 47 (2001) 3, 299 – 304

Mathematical LogicQuarterly

c© WILEY-VCH Verlag Berlin GmbH 2001

A Constructive Version of the Spectral Mapping Theorem

Douglas Bridges and Robin Havea1)

Department of Mathematics and Statistics, University of Canterbury,Private Bag 4800, Christchurch, New Zealand2)

Abstract. The spectral mapping theorem in a unital Banach algebra is examined for itsconstructive content.

Mathematics Subject Classification: 03F60, 46S30.

Keywords: Spectral mapping theorem, Constructive mathematics.

0 Introduction

Let B be a (complex) Banach algebra that is unital – that is, has an identity e. Leta be an element of B, Ra = λ ∈ C : a − λe has a two-sided inverse, the resolventset of a, and σa = λ ∈ C : (∀λ′ ∈ Ra) (λ = λ′), the spectrum of a. The ClassicalSpectral Mapping Theorem says that if p is a monic polynomial of degree ≥ 1, thenσp(a) = p(σa). The classical proof of this theorem hangs on the decomposition ofp(z) − λ, where λ ∈ C, into linear factors, and the observation that p(a) − λe hasno inverse if and only if at least one of its corresponding factors has no inverse.However, this proof is not valid in constructive mathematics – that is, mathematicsbased on intuitionistic logic (which leads to theorems valid in a wider range of models,including recursive and other computational ones, than those proved with classicallogic). Our main aims in this paper are (i) to find a constructive proof of the inclusionp(σa) ⊂ σp(a), and (ii) to recover constructively as much as is possible of the reverseinclusion. In order to fulfil these aims, we need relatively little technical backgroundin constructive mathematics. Much of what is needed will be stated in context below;for more detailed information on constructive mathematics, see [1, 2, 4, 7].

1 The first half of the spectral mapping theorem

Our proof of the inclusion p(σa) ⊂ σp(a) will be based on some elementary constructivesemigroup theory. First, we define an inequality on a set X to be a binary relation =on X with the following two properties:

x = y ⇒ ¬(x = y), x = y ⇒ y = x.

1)The authors wish to thank the Artington-DavyFund for providing the scholarship that supportedRobin Havea during the writing of the paper.

2)e-mail: d.bridges@math.canterbury.ac.nz

300 Douglas Bridges and Robin Havea

We call the inequality = discrete if also (∀x, y ∈ X) (x = y ∨ x = y). The complementof a subset Y of a set X with an inequality is the set

∼Y = x ∈ X : (∀y ∈ Y ) (x = y).The standard inequality on a normed space X is given by

x = y if and only if ‖x − y‖ > 0.

In this case, the statement x = y is equivalent to ¬(x = y) if and only if we postulateMarkov’s Principle (MP),

For each binary sequence (an), if ¬∀n (an = 0), then ∃n (an = 1),

a form of unbounded search that is independent of the Peano axioms plus intuitionisticlogic and so is not accepted in constructive mathematics. In the case where X = R,the standard inequality satisfies the special property

(x = y) ⇒ (∀z ∈ R) (x = z ∨ z = y);

see [2, p. 26, Cor. (2.17)].Let S be a semigroup with identity e, and let inv(S) be the set of invertible

elements of S. An inequality = on S is said to be quasi-discrete if

(∀x ∈ S) (x = e ∨ x ∈ inv(S)).

The discrete inequality on S is clearly quasi-discrete. A more interesting example oc-curs when S is the multiplicative semigroup of a Banach algebra B with an identity e.In that case, for each x ∈ B either 0 < ‖e− x‖ or ‖e− x‖ < 1, so either x = e or (bya well known argument) x has a two-sided inverse.

L emma 1. Let S be a semigroup with identity and a quasi-discrete inequality,and let a be an element of S such that ¬(a ∈ inv(S)) 3). Then

(∀x ∈ S) (ax = e ∨ e ∈ ∼Sa) and (∀x ∈ S) (xa = e ∨ e ∈ ∼aS).

P r o o f . Fix x ∈ S. Either ax = e or ax is invertible. In the latter case, a hasright inverse x(ax)−1. Moreover, for each s ∈ S, either e = sa or sa is invertible. Thesecond alternative implies that a has left inverse (sa)−1s; whence a has inverses oneither side and therefore a two-sided inverse, which contradicts the assumption thata /∈ inv(S). We conclude that if ax is invertible, then e ∈ ∼Sa.

The second part of the lemma is proved similarly to the first part.

P r o po s i t i o n 2. Let S be a semigroup with identity and a quasi-discrete inequal-ity, and let a be an element of S such that a /∈ inv(S). Then aS ∩ Sa ⊂ ∼ inv(S).

P r o o f . Let x ∈ aS ∩ Sa and y ∈ ∼ inv(S). Choose b, c ∈ S with x = ab = ca.By Lemma 1, either y−1x = (y−1c)a = e and therefore4) x = y, or else, as we mayassume, e ∈ aS. Then e = a(by−1) = (ab)y−1 = xy−1, and so y = ey = (xy−1)y = x.

Co r o l l a r y 3. If S is a semigroup with identity and a quasi-discrete inequality,and a is an element of S such that a /∈ inv(S), then a ∈ ∼ inv(S).

P r o o f . Apply Proposition 2, noting that a ∈ aS ∩ Sa.

3)This means a /∈ inv(S).4)We are assuming that the operations of multiplication on the left and on the right are strongly

extensional, in the sense that, for example, if ax = bx, then a = b.

A Constructive Version of the Spectral Mapping Theorem 301

We can now dispose of the first half of the spectral mapping theorem.Th e o r em 4. Let a be an element of a unital Banach algebra, and p a nonconstant

monic polynomial over C. Then p(σa) ⊂ σp(a).P r o o f . Arguing as in the first half of the proof on [6, p. 182], we see that for

each λ ∈ C, p(a)− p(λ)e ∈ (a − λe)B ∩ B(a − λe). If λ ∈ σa, it follows immediatelyfrom Proposition 2 that p(a) − p(λ)e ∈ ∼ inv(B), whence p(λ) ∈ σp(a).

The classical proof of Theorem 4 uses the observation that if λ ∈ σa, then it isimpossible that e belong to both the ideals (a − λe)B and B(a − λe), so at least oneof these ideals is proper. Constructively, we cannot be sure of telling which of theideals is proper.

2 The other half of the spectral mapping theorem

What about the other half of the spectral mapping theorem ? The classical proofon [6, p. 182] depends on the result that if a product xy in a Banach algebra is notinvertible, then at least one of its factors is not invertible. Constructively, we need todecide which of the factors is not invertible. There is no guarantee that we will beable to make this decision: this is a simple consequence of the fact that, given tworeal numbers x, y whose product is 0, we may not be able to decide which of x, yequals 0 (see [3, p. 44]). Nevertheless, as we show in this section, the other half of thespectral mapping theorem holds constructively under some additional hypotheses onthe element a of B, hypotheses that are trivially provable with classical logic.

We say that a subset S of a metric space (X, ) is· located if the distance (x, S) = inf(x, s) : s ∈ S exists for each x ∈ X;· compact if it is complete and totally bounded;· locally totally bounded if each bounded subset is contained in a totally bounded

set in X.A compact set is locally totally bounded, and a locally totally bounded set is located;moreover, a located subset of a locally totally bounded set is locally totally bounded([4, p. 33, (4.11)]).

When we write an inequality of the form (x, S) < r, we mean that there existss ∈ S such that (x, s) < r; we do not require that (x, S) exist as a real num-ber. Likewise, we use the inequalities (x, S) ≤ r and (x, S) > 0 as shorthand for(∀ε > 0)(∃s ∈ S) ((x, s) < r + ε) and (∃r > 0)(∀s ∈ S) ((x, s) ≥ r), respectively.With such interpretations in mind, we define the metric complement of S in X to beX − S = x ∈ X : (x, S) > 0. Bishop’s Principle states that if S is a completelocated subset of X, then for each x ∈ X there is s ∈ S such that x = s implies thatx ∈ X − S ([2, p. 92, Lemma (3.8)]).

Let Ω be a subset of the metric space X. We say that a subset K of Ω is wellcontained in Ω, and we write K⊂⊂Ω, if there exists r > 0 such that x ∈ Ω whenever(x,K) ≤ r. We say that

· K approximates Ω internally to within ε > 0 if K⊂⊂Ω and (x, ∂Ω) < ε for eachx ∈ Ω−K (where, as usual, ∂Ω denotes the boundary Ω ∩ ∼Ω of Ω);

· Ω is approximated internally by sets of type T if for each ε > 0 it can be approx-imated internally by a set of type T .

302 Douglas Bridges and Robin Havea

For us the most important types are located and compact.We call Ω coherent if Ω = −∼Ω.The next four results are proved in [5].P r o po s i t i o n 5. Let Ω be a subset of R

N whose interior is the metric com-plement of a located set L ⊂ ∼Ω. Then Ω is approximated internally by locatedsubsets.

P r o po s i t i o n 6. Let Ω be a subset of RN that is approximated internally bylocated subsets. Then ∼Ω is located, and Ω is coherent.

P r o po s i t i o n 7. A located subset K of a Banach space has the boundary crossingproperty, i. e. if x ∈ K and y ∈ ∼K, then for each ε > 0 there exists a point z ∈ ∂Kwhose distance from the segment joining x and y is less than ε.

P r o po s i t i o n 8. If Ω is a coherent subset of a Banach space such that both ∼Ωand ∂Ω are located, then Ω is located.

We now apply these results to the resolvent and spectrum of an element of ourunital Banach algebra.

P r o po s i t i o n 9. The following are equivalent conditions on an element a of aBanach algebra B with an identity.(i) σa is compact and Ra is coherent.(ii) Ra is approximated internally by located sets.

If either, and hence each, of these conditions holds, then a necessary and sufficientcondition for Ra to be located is that ∂σa be located.

P r o o f . To prove the equivalence of (i) and (ii), take Ω = Ra in Propositions 5and 6, and note that if σa is located, then, being closed in C, it is compact. Nowsuppose that (i) holds. Then as σa is located, σa ∪ −σa – that is, ∼Ra ∪ Ra – isdense in C. Since ∂σa = ∂Ra, Proposition 8 immediately shows that Ra is located ifand only if ∂σa is located.

We denote by B(z, r) (respectively, B(z, r)) the open (respectively, closed) ballwith centre z and radius r in C. By a border for a compact subset K of C wemean a totally bounded subset B of K such that B(z, (z, B)) ⊂ K for each z ∈ K.The importance of borders for us lies in the following result, in which, for example,m(f,K) = inf|f(z)| : z ∈ K.

P r o po s i t i o n 10. Let K ⊂ C be compact, and let B be a border for K. Let f bea differentiable function on K such that m(f, B) > m(f,K) = 0. Then there existsz ∈ K such that f(z) = 0. (See [2, p. 156, (5.8)].)

P r o po s i t i o n 11. Let Ω be an open subset of C such that ∼Ω and ∂Ω arecompact. Then ∂Ω is a border for ∼Ω.

P r o o f . Let K = ∼Ω, fix z0 ∈ K, and write r = (z0, ∂Ω). Given z ∈ B(z0, r),for each t ∈ [0, 1] write zt = tz0 + (1 − t)z. Assume that z ∈ Ω, and choose δ > 0such that B(z, 3δ) ⊂ Ω; then r ≥ 3δ. Applying Proposition 7, compute ζ ∈ ∂Ω andt ∈ [0, 1] such that |ζ − zt| < δ. Since ζ ∈ K = K, we have

3δ ≤ (z,K) ≤ |ζ − z| ≤ |ζ − zt|+ |z − zt| < δ + |z − zt|.Hence |z − zt| > 2δ and therefore

|z0 − ζ| ≤ |z0 − zt|+ |ζ − zt| = |z − z0| − |z − zt|+ |ζ − zt|<r − 2δ + δ = r − δ.

A Constructive Version of the Spectral Mapping Theorem 303

But this is absurd, since |z0 − ζ| ≥ (z0 , ∂Ω) = r. It follows that ¬(z ∈ Ω). Since Ωis open, we conclude that z ∈ ∼Ω.

Co r o l l a r y 12. Let a be an element of a unital Banach algebra B such that boththe spectrum of a and its boundary are compact. Then ∂σa is a border for σa.

P r o o f . Take Ω = Ra in Proposition 11.

Here is our constructive version of the second part of the spectral mapping theo-rem.

Th e o r em 13. Let a be an element of a unital Banach algebra B such that Ra isapproximated internally by located sets and ∂σa is compact. Let p be a nonconstantmonic polynomial over C, and let λ be an element of σp(a) ∩∼p(∂σa). Then λ ∈ p(σa).

P r o o f . By Proposition 9, σa is compact and Ra = −σa. By Corollary 12, ∂σa isa compact border for σa. If deg(p) = n, then by [2, p. 156, (5.10)], there exist complexnumbers λ1, . . . , λn such that p(z)−λ = (z−λ1) · · · (z−λn) (z ∈ C). For each z ∈ ∂σa,since p(z) = λ, we have λk = z (1 ≤ k ≤ n). By Bishop’s Principle, λk ∈ C − ∂σa foreach k; whence m(p−λ, ∂σa) > 0. On the other hand, if inf1≤k≤n (λk, σa) > 0, thenfor each k, λk ∈ −σa = Ra and so a− λke is invertible; whence p(a)−λe is invertible– a contradiction. Thus inf1≤k≤n (λk , σa) = 0 and therefore m(p−λ, σa) = 0. It nowfollows from Proposition 10 that there exists ζ ∈ σa such that p(ζ) − λ = 0. Henceλ ∈ p(σa).

If λ ∈ p(∂σa), then trivially λ ∈ p(σa). So if Ra is approximated internally bylocated sets and ∂σa is compact, then λ ∈ p(σa) for each λ in the dense subsetσp(a) ∩ (p(∂σa) ∪ ∼p(∂σa)) of σp(a). Perhaps surprisingly, this is the best result thatwe can hope for in the constructive setting. To justify this remark, let B be a unitalBanach algebra containing an element a whose resolvent set is the exterior of theclosed unit disc D in C and whose spectrum is that disc. Let ζ ∈ [−1

2 ,12 ], and define

p(z) = (z+1− ζ)(z−1− ζ). Suppose that p(a) is invertible. Then both a− (−1+ ζ)eand a−(1+ζ)e are invertible, so −1+ζ ∈ −D and 1+ζ ∈ −D; this implies that ζ < 0and ζ > 0, which is absurd. Hence p(a) is not invertible, and therefore ¬(0 ∈ Rp(a)).Since Rp(a) is open, we conclude that 0 ∈ ∼Rp(a) = σp(a).

Now suppose that p(ζ0) = 0 for some ζ0 ∈ D. Either ζ0 > −12 or ζ0 < 1

2 . In thefirst case, if ζ0 = 1+ ζ, then the quadratic equation p(z) = 0 has three distinct roots,which is absurd; so 1 + ζ = ζ0 ∈ [−1, 1] and therefore ζ ≤ 0. Similarly, in the secondcase, −1 + ζ = ζ0 ∈ [−1, 1] and so ζ ≥ 0. Thus the proposition,

If a is an element of a unital Banach algebra such that Ra is approximatedinternally by located sets and ∂σa is located (and hence compact), and if pis a nonconstant monic polynomial over C, then σp(a) ⊂ p(σa)

entails(∀x ∈ R) (x ≥ 0 ∨ x ≤ 0),

which is known to be essentially nonconstructive; indeed, it is false in the recursivemodel (see [4, pp. 4 and 53–54]).

References

[1] Beeson, M. J., Foundations of Constructive Mathematics. Springer-Verlag, Berlin-Heidelberg-New York 1985.

304 Douglas Bridges and Robin Havea

[2] Bridges, D., and E. Bishop, Constructive Analysis. Springer-Verlag, Berlin-Heidel-berg-New York 1985.

[3] Bridges, D., A constructive look at the real number line. In: Real Numbers, Gener-alizations of the Reals and Theories of Continua (P. Ehrlich, ed.), Kluwer AcademicPublishers, Amsterdam 1994, pp. 29 – 92.

[4] Bridges, D., and F. Richman, Varieties of Constructive Mathematics. London Math.Soc. Lecture Notes 97, Cambridge Univ. Press, London 1987.

[5] Bridges, D., F. Richman, and Wang Yuchuan, Sets, complements and boundaries.Indag. Math. N.S. 7(4) (1996), 425–445.

[6] Kadison, R. V., and J. R. Ringrose, Fundamentals of the Theory of Operator Alge-bras (Vol. 1). Academic Press, New York 1983.

[7] Troelstra, A. S., and D. van Dalen, Constructivism in Mathematics: An Introduc-tion (two volumes). North Holland Publ. Comp., Amsterdam 1988.

(Received: March 18, 2000; Revised: May 31, 2000)