Banach algebras associated with linear operator pencils

ISSN 0001-4346, Mathematical Notes, 2009, Vol. 86, No. 3, pp. 361–367. © Pleiades Publishing, Ltd., 2009.Original Russian Text © I. V. Kurbatova, 2009, published in Matematicheskie Zametki, 2009, Vol. 86, No. 3, pp. 394–401.

Banach Algebras Associated with Linear Operator Pencils

I. V. Kurbatova*

Voronezh State UniversityReceived January 15, 2008; in final form, October 4, 2008

Abstract—A direct relationship between the theory of pseudoresolvents and the spectral theory oflinear operator pencils is established.

DOI: 10.1134/S0001434609090090

Key words: differential equation not solved with respect to derivatives, operator pencil,resolvent, spectrum, pseudoresolvent, maximal pseudoresolvent, Banach algebra.

Differential equations of the form Fx = Gx + f(t) not solved with respect to derivatives arise in manyapplications; see, e.g., [1]–[5]. Here F,G : X → Y are (possibly, unbounded) linear operators, and Xand Y are Banach spaces,

For a differential equation x = Gx + f(t) solved with respect to the derivative, finding a generalsolution reduces to constructing the operator exponential eGt; a natural method for constructing andstudying such an operator exponential is based on the spectral properties of the operator G.

The transfer of this approach to equations not solved with respect to the derivative was consideredby many authors [5]–[15]. The first step in its implementation consists in replacing the resolvent(λ1 − G)−1 of the operator G by the resolvent (λF − G)−1 of the pencil λ �→ λF − G. Unfortunately,as a rule, the domain Y of the pencil resolvent (λF − G)−1 does not coincide with its range X. For thisreason, consideration of the spectral properties of pencils usually begins by passing to the space X or Y(see, e.g., [5], [6], [11]).

The objective of this paper is to describe a commutative Banach algebra generated by the pencilλ �→ λF − G (see Theorem 8), which reduces the study of the properties of the resolvent (λF − G)−1 toa direct application of classical facts from spectral theory [16]–[18].

In Sec. 1, we recall the terminology of the theory of Banach algebras [16]–[18] and basic properties ofpseudoresolvents [18]. In Sec. 2, we define a Banach algebra B(F,G)(Y,X∗) under F-multiplication (seeTheorem 8). The resolvent of the pencil (λF − G)−1 turns out to be a maximal pseudoresolvent withvalues in this algebra (see Proposition 9). As an application, we prove Theorem 11, which describesfunctional calculus, and Theorem 12 on spectral decomposition. These theorems imply, in particular,Corollary 14 on the decomposition of X and Y into the direct sum generated by a partition of the extendedsingular set of the pencil resolvent into two parts, which is the best known spectral fact in the theory ofoperator pencils [3], [5], [6], [8], [9], [11], [14], [15].

1. PRELIMINARIES

All algebras considered below are assumed to be complex [16]–[18]. If an algebra B contains anelement 1 = 1B ∈ B such that A1 = 1A = A for any A ∈ B, then this element 1 is called a unit, andthe algebra B is said to be unital. An algebra can have at most one unit. An algebra is said to becommutative if AB = BA for any A and B.

If an algebra B is a Banach space and ‖AB‖ ≤ ‖A‖ · ‖B‖, then B is called a Banach algebra. If,in addition, B is unital and ‖1‖ = 1, then B is a unital Banach algebra. The simplest example of aBanach algebra is the algebra B(X) of bounded linear operators acting on a Banach space X.

*E-mail: [email protected]

361

362 KURBATOVA

Let B be a unital algebra. The inverse of A ∈ B is defined as an element A−1 ∈ B for whichAA−1 = A−1A = 1.

For a unital algebra B and its element A ∈ B, the set of λ ∈ C for which the element λ1 − A has noinverse is called the spectrum of A and denoted by σ(A) or σB(A). Its complement ρ(A) = C \ σ(A) isthe resolvent set of A. The function (the family)

Rλ = (λ1 − A)−1, λ ∈ ρ(A)is called the resolvent of A. The spectrum of any nonzero element of a unital Banach algebra is anonempty compact subset of C.

Proposition 1 ([18, Theorem 4.1.8]). If B is a unital algebra, then the resolvent Rλ of any elementA ∈ B satisfies Hilbert’s identity

Rλ − Rμ = −(λ − μ)RλRμ, λ, μ ∈ ρ(A). (1)

Let B be an algebra without unit. Obviously, the set ˜B = C ⊕ B under the coordinatewise linearoperations and the multiplication (α,A)(β,B) = (αβ, αB + βA + AB) is an algebra with the unit1 = (1, 0). The element (α,A) is denoted by α1 + A. The algebra ˜B is referred to as the algebra Bwith adjoint unit. If the algebra B is Banach, then we set ‖α1 + A‖ = |α| + ‖A‖; obviously, in thiscase, the algebra ˜B is Banach as well. The spectrum (resolvent) of an element in a nonunital algebrais defined as the spectrum (resolvent) of this element in the algebra with adjoint unit. If B is unital, thenby the algebra ˜B with adjoint unit we understand the algebra B itself.

Let A and B be algebras. A map ϕ : A → B is called an algebra morphism [16] if

ϕ(A + B) = ϕ(A) + ϕ(B), ϕ(αA) = αϕ(A), ϕ(AB) = ϕ(A)ϕ(B).If A and B are unital and ϕ(1A) = 1B, then ϕ is said to be a unital algebra morphism.

Suppose that A and B are algebras and A has no unit. If ϕ : A → B is an algebra morphism, then,obviously, the unique extension of ϕ to a unital algebra morphism ϕ : ˜A → ˜B is the map

ϕ(α1 + A) = α1 + ϕ(A).

Let B be a Banach algebra, and let G ⊆ C be a nonempty set. A pseudoresolvent (on G withvalues in B) is a function (family) λ �→ Rλ defined on G, taking values in B, and satisfying Hilbert’sidentity (1) [18, Chap. 5, Sec. 2, p. 201 (Russian transl.)]. A pseudoresolvent is said to be maximal [6] ifit has no extension to a larger set satisfying identity (1). Any pseudoresolvent can be uniquely extendedto a maximal pseudoresolvent (see Theorem 3). The domain ρ(R( · )) of this maximal pseudoresolventis called the regular set of the initial pseudoresolvent, and the complement σ(R( · )) to ρ(R( · )) is itssingular set.

Hilbert’s identity (1) can be written in the equivalent form

Rλ + (λ − μ)RλRμ = Rμ

or in the formRλ(1 + (λ − μ)Rμ) = Rμ

(if the given algebra is nonunital, then the algebra with adjoint unit is considered).

Proposition 2 ([18, Corollary 1 of Theorem 5.8.4]). If two commuting elements Rλ, Rμ ∈ B sat-isfy (1), then the element

1 + (λ − μ)Rμ ∈ ˜B

is invertible.

Theorem 3 ([18, Theorem 5.8.6]). Any pseudoresolvent admits a unique extension to a maximalpseudoresolvent. The domain of this maximal pseudoresolvent is the set of all λ ∈ C for whichthe element 1 + (λ − μ)Rμ is invertible in ˜B. The extension has the form

Rλ = Rμ(1 + (λ − μ)Rμ)−1.

Corollary 4 ([18, Theorem 5.8.2], [13, Chap. 6, Sec. 1]). The domain of any maximal pseudoresol-vent is an open set, and such a pseudoresolvent itself is an analytic function with values in B.

MATHEMATICAL NOTES Vol. 86 No. 3 2009

BANACH ALGEBRAS ASSOCIATED WITH LINEAR OPERATOR PENCILS 363

2. F-ALGEBRASGiven complex linear spaces X and Y , linear subspaces XF ,XG ⊆ X, and linear operators

F : XF → Y and G : XG → Y , the (operator) pencil is defined as a function

λ �→ λF − G : X∗ → Y, λ ∈ C,

where X∗ = XF ∩ XG [5], [6], [10]. In particular, 0F − G denotes the operator −G : X∗ → Y . It followsfrom Proposition 7 below that the case of bounded operators F,G : X∗ → Y can be considered general.

The resolvent set of such a pencil is the set ρ(F,G) consisting of those λ ∈ C for which the operatorλF − G : X∗ → Y is invertible, and its resolvent is the function (the family)

Rλ = (λF − G)−1 : Y → X∗, λ ∈ ρ(F,G).

The complement σ(F,G) to ρ(F,G) is called the spectrum of the pencil.

Proposition 5. The resolvent of a pencil satisfies Hilbert’s F -identity

Rλ − Rμ = −(λ − μ)RλFRμ, λ, μ ∈ ρ(F,G). (2)

Proof. Indeed, for λ, μ ∈ ρ(F,G), we have

Rλ = (λF − G)−1 = (λF − G)−1(μF − G)(μF − G)−1

= (λF − G)−1(

λF − G + (μ − λ)F)

(μF − G)−1

= (λF − G)−1(λF − G)(μF − G)−1 + (λF − G)−1(μ − λ)F (μF − G)−1

= (μF − G)−1 + (μ − λ)(λF − G)−1F (μF − G)−1 = Rμ − (λ − μ)RλFRμ.

Suppose that the following hypothesis holds.

Conjecture. The space Y is Banach (on X, no norm is defined in advance), and ρ(F,G) containsat least two points λ = μ for which the operators

(λF − G)(μF − G)−1 : Y → Y, (μF − G)(λF − G)−1 : Y → Y

are bounded.

Proposition 6. Under the above conjecture, the isomorphisms

λF − G, μF − G : X∗ → Y

generate equivalent norms on X∗.

In what follows, we assume X∗ to be endowed with one of the equivalent norms generated by theisomorphisms λF − G and μF − G.

Proposition 7. If λ, μ ∈ ρ(F,G), λ = μ, and the operators

(λF − G)(μF − G)−1 and (μF − G)(λF − G)−1

are bounded, then the operators F : X∗ → Y and G : X∗ → Y are bounded as well. Thus, forany ν ∈ ρ(F,G), the operator νF − G : X∗ → Y generates a norm on X∗ equivalent to the normsgenerated by λF − G and μF − G.

Proof. By assumption, the operators (λF − G) : X∗ → Y and (μF − G) : X∗ → Y are bounded.Therefore, so is the operator

(λF − G) − (μF − G) = λF − μF.

Thus, the operator F : X∗ → Y is bounded. But this implies the boundedness of the operator

G = λF − (λF − G) : X∗ → Y

and of the operator νF − G : X∗ → Y . The second assertion follows from Banach’s inverse operatortheorem.


364 KURBATOVA

We emphasize that the space X may be Banach, and the operators F : XF → Y and G : XG → Ymay be unbounded. According to Proposition 7, under the assumption made above, even unboundedoperators F : XF → Y and G : XG → Y generate bounded operators F,G : X∗ → Y . This reduces thecase of unbounded operators to that of bounded operators.

Let B(Y,X∗) denote the Banach space of all bounded linear operators A : Y → X∗, and letB(F,G)(Y,X∗) be the closure in B(Y,X∗) of the linear span of the operators Rλ with λ ∈ ρ(F,G). OnB(F,G)(Y,X∗), we define the F -multiplication by setting

A � B = AFB.

Theorem 8. The space B(F,G)(Y,X∗) is a commutative Banach algebra under F -multiplication.This algebra is unital if and only if the operator F : X∗ → Y is invertible; in this case, the unitis F−1. If the algebra B(F,G)(Y,X∗) is unital, then an algebraic norm (i.e., a norm satisfying thecondition ‖A � B‖ ≤ ‖A‖ · ‖B‖) on this algebra can be defined as

‖A‖F = sup{

‖A � B‖ : ‖B‖ ≤ 1, B ∈ B(F,G)(Y,X∗)}

.

If the algebra B(F,G)(Y,X∗) is nonunital, then the function

‖A‖˜F = sup

{

‖βA + A � B‖ : |β| + ‖B‖ ≤ 1, β ∈ C, B ∈ B(F,G)(Y,X∗)}

can be taken for the algebraic norm. Both these norms are equivalent to the initial norm on thespace B(Y,X∗).

Proof. Let us check that B(F,G)(Y,X∗) is closed under F-multiplication. It follows from Hilbert’sidentity (2) that the product RλRμ with λ = μ belongs to the linear span of the family of Rλ withλ ∈ ρ(F,G). By continuity, the elements RλRμ with λ = μ belong to the closure of the linear span ofthe family of Rλ with λ ∈ ρ(F,G). Obviously, F-multiplication is continuous with respect to the initialnorm on the space B(Y,X∗). It follows that the closure of the linear span of Rλ with λ ∈ ρ(F,G) formsan algebra.

Hilbert’s identity (2) also implies the permutability of Rλ and Rμ for λ, μ ∈ ρ(F,G). By continuity,this implies the commutativity of the algebra B(F,G)(Y,X∗).

Let us check the existence of a unit. If F is invertible, then, obviously, we can take F−1 for 1F .Conversely, suppose that the algebra has a unit 1F . Substituting a surely invertible operator (e.g., oneof the resolvent values) for A into the equality

A � 1F = 1F � A = A or AF1F = A and 1F FA = A

and multiplying the result by A−1 on suitable sides, we obtain the equalities F1F = 1 and 1F F = 1,which mean that 1F = F−1.

A direct verification shows that the functions ‖ · ‖F and ‖ · ‖˜F

are seminorms on B(F,G)(Y,X∗)equivalent to the initial norm on B(Y,X∗). The equality ‖1F ‖F = ‖F−1‖F = 1 is obvious.

Let us show that these seminorms are algebraic. The algebraicity of the seminorm ‖ · ‖F is a specialcase of [17, Theorem 10.2]. Clearly, for the seminorm ‖ · ‖

˜F, we have ‖A‖F ≤ ‖A‖

˜F. Therefore,

‖A � B‖˜F

= sup{‖γ(A � B) + (A � B) � C‖ : |γ| + ‖C‖ ≤ 1}= sup{‖A � (γB + B � C)‖ : |γ| + ‖C‖ ≤ 1}≤ ‖A‖F · sup{‖γB + B � C‖ : |γ| + ‖C‖ ≤ 1}≤ ‖A‖F · ‖B‖

˜F≤ ‖A‖

˜F· ‖B‖

˜F.

Proposition 9. The resolvent of a pencil is a maximal F -pseudoresolvent; i.e., it cannot beextended over a set larger than ρ(F,G) to a pseudoresolvent satisfying Hilbert’s F -identity (2).



Proof. Suppose that, on the contrary, the resolvent of a pencil can be extended to a point λ /∈ ρ(F,G)so that the extension satisfies Hilbert’s F-identity. Take any auxiliary point μ ∈ ρ(F,G). For λ and μ,Hilbert’s F-identity holds. Applying the transformations

Rλ − Rμ = −(λ − μ)RλFRμ, Rλ + (λ − μ)RλFRμ = Rμ,

Rλ(1X∗ + (λ − μ)FRμ) = Rμ, Rλ((μF − G)Rμ + (λ − μ)FRμ) = Rμ,

Rλ((μF − G) + (λ − μ)F)

Rμ = Rμ, Rλ((μF − G) + (λ − μ)F ) = 1X∗ ,

Rλ(λF − G) = 1X∗ ,

we see that Rλ is a left inverse of (λF − G). Similarly, (λF − G)Rλ = 1Y , i.e., Rλ is a right inverseof (λF − G). Thus, the point λ belongs to the resolvent set.

The extended resolvent set of a pencil λ �→ λF − G is defined as a subset ρ(F,G) of the extendedcomplex plane C consisting of ρ(F,G) and, possibly, ∞ [6, p. 31]. The point λ = ∞ is included in ρ(F,G)if the resolvent λ �→ (λF − G)−1 is defined in a punctured neighborhood of λ = ∞, the operator F isinvertible, and limλ→∞ λRλ = F−1 in the norm of B(Y,X∗). Otherwise, the point λ = ∞ is included inthe extended spectrum σ(F,G).

Proposition 10. The point ∞ belongs to ρ(F,G) if and only if the operator F is invertible.

By Theorem 8, in the most interesting case in which F has no inverse, the algebra B(F,G)(Y,X∗) has

no unit. In this case, let ˜B(F,G)(Y,X∗) denote the algebra B(F,G)(Y,X∗) with adjoint unit I. In the case

of invertible F , by ˜B(F,G)(Y,X∗) we mean the algebra B(F,G)(Y,X∗) itself and by I, the operator F−1.

Let K ⊆ C be a closed subset of the extended complex plane C, and let O(K) denote the set of allanalytic functions f : U → C defined on open neighborhoods U of the set K (for different functions f ,the neighborhoods U may be different). We say that two functions f1 : U1 → C and f2 : U2 → C areequivalent if the set K has an open neighborhood U ⊆ U1 ∩ U2 on which f1 and f2 coincide (that is,f1(λ) = f2(λ) for all λ ∈ U ). It is easy to show that this is indeed an equivalence relation. Thus, strictlyspeaking, the elements of O(K) are classes of equivalent functions. Obviously, O(K) is an algebra(without a norm) with unit u(λ) = 1.

Theorem 11. The map ϕ : O(σ(F,G)) → ˜B(F,G)(Y,X∗) defined by

ϕ(f) =

⎧

⎪

⎪

⎨

⎪

⎪

⎩

12πi

∫

Γf(λ)(λF − G)−1 dλ if Γ does not enclose ∞,

12πi

∫

Γf(λ)(λF − G)−1 dλ + f(∞)I if Γ encloses ∞,

where the contour Γ is the oriented envelope [18, p. 183 (Russian transl.)] of the extended spectrumσ(F,G) with respect to the complement of the domain of f , is a unital algebra morphism; inparticular, ϕ(fg) = ϕ(f) � ϕ(g), and the function u(λ) = 1 is taken by ϕ to the unit I of thealgebra ˜B(F,G)(Y,X∗). The morphism ϕ takes the function

rλ0(λ) =1

λ0 − λ, where λ0 ∈ ρ(F,G),

to Rλ0 = (λ0F − G)−1, and if F is invertible, then ϕ takes v(λ) = λ to the operator F−1GF−1.For any function f ∈ O(σ(F,G)),

σ˜B(F,G)(Y,X∗)

(ϕ(f)) = {f(λ) : λ ∈ σ(F,G)}.

Proof. The theorem is proved by standard arguments used in functional calculus [16, Chap. 1, Sec. 4,Theorem 3], [17, Theorem 10.27], [18, Theorems 5.2.5 and 5.11.2] and in spectral mapping theorems [18,Theorem 5.12.1].


366 KURBATOVA

Theorem 12. If the set σ(F,G) decomposes into the disjoint union of two closed subsets σ0, σ1 ⊆C and σ0 is bounded, then there exists an idempotent Π0 ∈ B(F,G)(Y,X∗) for which

A = Π0 � A � Π0 + Π1 � A � Π1, A ∈ ˜B(F,G)(Y,X∗), where Π1 = I − Π0,

the extended singular set of the pseudoresolvent Π0 � R( · ) � Π0 in the algebra

Π0 � B(F,G)(Y,X∗) � Π0

coincides with σ0, and the (extended) singular set of the pseudoresolvent Π1 � R(·) � Π1 in thealgebra Π1 � B(F,G)(Y,X∗) � Π1 coincides with σ1 = σ1 \ {∞} (with σ1).

Proof. The proof of the theorem repeats that of spectral decomposition theorems (see, e.g., [16, Chap. 1,Sec. 4, Subsec. 4], [18, Theorem 5.13.1].

Proposition 13. The map A �→ AF is a morphism from the algebra B(F,G)(Y,X∗) to the algebraB(X∗), and the map A �→ FA is a morphism from the algebra B(F,G)(Y,X∗) to the algebra B(Y ).

Corollary 14 stated below is the best known spectral fact in pencil theory (see, e.g., [3], [5], [6,Theorem 6.3], [8], [9], [11, Theorem 1.1.1], [14, Sec. 4.1], [15]). It readily follows from the propertiesof the algebra ˜B(F,G)(Y,X∗) described above and Proposition 13.

Corollary 14. Under the assumptions of Theorem 12, the pairs of operators

P0 = Π0F , P1 = 1− P0 and Q0 = FΠ0, Q1 = 1 − Q0

are mutually complementary projectors and, therefore, they generate the direct sum decomposi-tions

X∗ = Im P0 ⊕ Im P1, Y = Im Q0 ⊕ Im Q1.

For any A ∈ ˜B(F,G)(Y,X∗), the relations P0A = AQ0 and P1A = AQ1 hold, which imply

FP0 = Q0F, FP1 = P1F,

GP0 = Q0G, GP1 = P1G.

This means that the operators F and G have diagonal matrices with respect to the above directsum decompositions. The extended spectrum of the pencil

λ �→ λF − G : Im P0 → Im Q0

is equal to σ0, and the extended spectrum of the pencil

λ �→ λF − G : Im P1 → Im Q1

is equal to σ1.

ACKNOWLEDGMENTS

This work was supported by the Russian Foundation for Basic Research (grant no. 07-01-00131).

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Banach algebras associated with linear operator pencils