A note on the spectrum of irreducible operators and semigroups
aa r X i v : . [ m a t h . F A ] F e b A NOTE ON THE SPECTRUM OF IRREDUCIBLE OPERATORSAND SEMIGROUPS
JOCHEN GL ¨UCK
Dedicated with great pleasure to Rainer Nagel on the occasion of his 80th birthday
Abstract.
Let T denote a positive operator with spectral radius 1 on, say, an L p -space. A classical result in infinite dimensional Perron–Frobenius theorysays that, if T is irreducible and power bounded, then its peripheral pointspectrum is either empty or a subgroup of the unit circle.In this note we show that the analogous assertion for the entire peripheralspectrum fails. More precisely, for every finite union U of finite subgroupsof the unit circle we construct an irreducible stochastic operator on ℓ whoseperipheral spectrum equals U .We also give a similar construction for the C -semigroup case. Introduction
Main result.
Consider a positive operator T with spectral radius 1 on an L p -space ( p ∈ [1 , ∞ ]) or, more generally, on a complex Banach lattice E . Assume that T is power-bounded (i.e., that sup n ∈ N k T n k < ∞ ), has spectral radius 1 and isirreducible (i.e., T does not leave any closed ideal invariant, except for { } and E ; see e.g. [13, Definition 4.2.1(1)]). Then the peripheral point spectrum of T –i.e., the set of all eigenvalues of T of modulus 1 – is either empty or a subgroupof the complex unit circle T . This result, which is a strong generalisation of thecorresponding Perron–Frobenius result in finite dimensions, was proved by Lotz[12, Theorem 5.2(2)] (for an English presentation of this result see, for instance,[17, Theorem V.5.2 and Lemma V.4.8]).For the peripheral spectrum of T – i.e., the set of all spectral values of modulus 1– no such result is known on general Banach lattices (but see Remark 1.2 for spacesof continuous functions). And indeed, the purpose of this note is to demonstratethat the peripheral spectrum of an irreducible operator need not be a subgroup ofthe unit circle, in general. More precisely, we show: Theorem 1.1.
Let U = ∅ be a finite union of finite subgroups of the complex unitcircle. Then there exists an irreducible stochastic operator T on ℓ with peripheralspectrum U . By stochastic we mean that, for each 0 ≤ f ∈ ℓ , we have T f ≥ k T f k = k f k . In order to prove the theorem, we explicitly construct an operator T with thedesired properties in Section 2. An analogous result for C -semigroups rather thanfor single operators is given in Section 3. Remark 1.2.
Complementary to Theorem 1.1, there is a positive result on spaces ofcontinuous functions: consider the space C ( K ) of continuous scalar-valued functionson a compact Hausdorff space K ; if T is a positive and irreducible operator on C ( K ) Date : February 9, 2021.2010
Mathematics Subject Classification.
Key words and phrases.
Irreducible operator; irreducible operator semigroup; peripheral spec-trum; cyclicity. and if the constant function is a fixed vector of T , then the peripheral spectrumof T is indeed a subgroup of the complex unit circle T . This was proved by Schaeferin [16, Theorem 7]. Related results and literature.
For an overview of classical Perron–Frobeniustype results on infinite-dimensional Banach lattices we refer to the book chapters[17, Sections V.4 and V.5] or [13, Chapter 4], or to the survey article [9].If one considers merely positive rather than irreducible operators, there is noreason to expect the peripheral (point) spectrum to be a subgroup of the unitcircle – consider, for instance, the direct sum of two cyclic permutation matricesin dimensions 2 and 3. However, there is a weaker notion of symmetry that issatisfied by the peripheral spectrum of many positive operators: we call a subsetof the unit circle T cyclic if it is a union of subgroups of T . It was shown byKrieger [11, Folgerungen 2.2.1(b) and 2.2.2(b)] and Lotz [12, Theoreme 4.7, 4.9and 4.10] that, under certain growth assumptions, the peripheral spectrum of apositive operator on a Banach lattices is cyclic (the results can be found in Englishin [17, Theorem V.4.9 and its corollary]). Whether the aforementioned growthassumptions can be dropped is a long open problem in spectral theory. A recentoverview of this problem, along with several partial results, can be found in [8].The cyclicity result of Krieger and Lotz also explains the assumed form of theset U in Theorem 1.1: the positivity of T (together with its contractivity) alreadyimplies the the peripheral spectrum must be a union of subgroups of T .The related question whether the peripheral point spectrum of a positive op-erator is cyclic, is very subtle, and its answer depends on the precise propertiesof the operator under consideration as well as on the geometric properties of theunderlying space. For various results and counterexamples, as well as for furtherreferences, we refer to [7, Sections 5 and 6]. Notation and terminology.
We use the convention N := { , , . . . } ; the com-plex unit circle is denoted by T .We call a bounded linear operator T on a complex Banach lattice E positive if T f ≥ ≤ f ∈ E . If λ ∈ C is in the resolvent set of a linear operator T ,we denote the resolvent of T at λ by R ( λ, T ) := ( λ − T ) − .2. Proof of the main result
Let G , . . . , G n ⊆ T be finite subgroups such that U = G ∪ · · · ∪ G n . We denotethe cardinality of G k by d k and we set d := d + · · · + d n . It is very easy to constructa finite (column) stochastic matrix with spectrum U : for each k ∈ { , . . . , n } let P k ∈ R d k × d k denote a cyclic permutation matrix. Then the spectrum of P k is G k ,so the permutation matrix P := P ⊕ · · · ⊕ P n ∈ R d × d has spectrum U . The point here is, of course, that P is not irreducible. In order toget an irreducible operator, we now take a direct sum of infinitely many copies of P and slightly perturb it. The main difficulty is then to check that the perturbedoperator does not have spectral values in T \ U . Here is the detailed construction: Construction 2.1.
The space:
Endow C d with the 1-norm k · k and consider thespace E := { f = ( f , f , f , . . . ) : f ∈ C , f , f , · · · ∈ C d and k f k E < ∞} , where k f k E := | f | + ∞ X n =1 k f n k . HE SPECTRUM OF IRREDUCIBLE OPERATORS 3
Obviously, E is isometrically lattice isomorphic to ℓ . The operator:
Choose a sequence of numbers ( q n ) n ∈ N in (0 , ] such that ∞ X n =1 dq n = 1 . Clearly, we have q n → n → ∞ .We use the symbol ∈ R d to denote the vector in R d whose entries are all equalto 1, and we consider the vectors e = (1 , , , . . . ) ∈ E and q = (0 , q , q , . . . ) ∈ E Both vectors e and q have norm 1.Let us define the operator T : E → E by the formula T f f f ... = P ∞ n =1 q n h , f n i (1 − q ) P f + f q (1 − q ) P f + f q ... . Note that we can write T in the form T = S + e ⊗ q + q ⊗ e , where the operator S : E → E has the block diagonal form S = − q ) P (1 − q ) P . . . , and where the rank-1 operators e ⊗ q and q ⊗ e are given by( e ⊗ q ) f := h e, f i q and ( q ⊗ e ) f := h q, f i e for all f ∈ E (note that this makes sense since both sequences e and q can be seenas elements of E ≃ ℓ and as elements of E ′ ≃ ℓ ∞ ).Now that we have constructed our operator T on E , let us check that it satisfiesall the desired properties. Obviously, T is positive, and since all q n are >
0, onereadily sees that T is irreducible, too. The fact that T is stochastic follows from theequality P ∞ n =1 dq n = 1 and from the fact that the matrix P is (column) stochastic.It is also not difficult to see that each number in U is a spectral value of T : Proposition 2.2.
Let λ ∈ U . Then λ is an approximate eigenvalue of T .Proof. Clearly, λ is a spectral value of the matrix P . Let z ∈ C d be a correspondingeigenvector of norm k z k = 1.Now we can easily construct an approximate eigenvector for T : for each n ∈ N ,let z ( n ) ∈ E be the vector that is equal to z at the n -th component, and 0 everywhereelse. This vector has norm 1 in E .Consider the vector T z ( n ) : its 0-th component is q n h , z i , its n -th component is λ (1 − q n ) z , and all other components are 0. Thus, ( T − λ ) z ( n ) → E as n → ∞ .This proves that λ is an approximate eigenvalue of T . (cid:3) It only remains to show that no other unimodular number is in the spectrum of T . This is a bit more involved since it is, in general, not easy to show that a givencomplex number is not in the spectrum of a given operator. What will save us inour concrete situation is the formula T = S + e ⊗ q + q ⊗ e, which says that T is an additive perturbation of the very simple operator S bytwo rank-1 operators; for rank-1 perturbations, there is a variant of the famous JOCHEN GL¨UCK
Sherman–Morrison formula which will allow us to retrieve very precise informationabout the spectrum of T . But first, we analyse the spectrum of the simpler operator S : Proposition 2.3.
Let λ ∈ T \ U . Then λ is in the resolvent set of S Proof.
Clearly, λ is in the resolvent set of (1 − q n ) P for each n . Due to the blockdiagonal structure of S , it suffices to prove that the resolvents R (cid:0) λ, (1 − q n ) P (cid:1) areuniformly bounded as n varies.Fortunately, this is easy: since λ is not in the spectrum of P , there exists aconstant M ≥ kR ( rλ, P ) k ≤ M for all r ∈ [1 , q n are in the interval (0 , ], this yields (cid:13)(cid:13) R (cid:0) λ, (1 − q n ) P (cid:1)(cid:13)(cid:13) = 11 − q n (cid:13)(cid:13)(cid:13)(cid:13) R (cid:16) λ − q n , P (cid:17)(cid:13)(cid:13)(cid:13)(cid:13) ≤ M for all n ∈ N . (cid:3) In order to show that the perturbation by e ⊗ q and q ⊗ e does not destroythe spectral properties on the unit circle, we now use the following version of theSherman–Morrison formula: Proposition 2.4.
Let A : X ⊇ dom A → X be a closed linear operator on acomplex Banach space X , let w ∈ X and ϕ ∈ X ′ . Let λ ∈ C be in the resolvent setof A .Then λ is in the resolvent set of A + ϕ ⊗ w if and only if h ϕ, R ( λ, A ) w i 6 = 1 ; inthis case, the resolvent of A + ϕ ⊗ w at λ is given by the formula R ( λ, A + ϕ ⊗ w ) = R ( λ, A ) + 11 − h ϕ, R ( λ, A ) w i R ( λ, A )( ϕ ⊗ w ) R ( λ, A ) . (2.1)The reason why we formulated the proposition for closed rather than merely forbounded operators is that we will also employ this result for semigroup generatorsin the subsequent section.In finite dimensions (and for λ = 0) Proposition 2.4 is a classical result in matrixanalysis; for an historical overview we refer to [10, Section 1]. In infinite dimensionsthe proposition can, for bounded operators A , be found in [6, Theorem 1.1]; for thecase of unbounded A we refer to [1, Lemma 1.1] or [4, Proposition A.1].We will now use Proposition 2.4 to show that T has no spectral values in T \ U .Since T is a rank-2 perturbation of S , we have to employ the proposition twice.To perform all the necessary computations, the following simple observation aboutcomplex numbers is useful: Proposition 2.5.
Let p ∈ [0 , . For every complex λ ∈ T \ { } we have p < | λ − p | . Proof.
It is easy to see this geometrically: the circle T − p (which is centered at − p ) is located outside the circle p T , and they only intersect in the point p . (cid:3) Now, finally, we can conclude the proof of our main result, Theorem 1.1, byshowing the following proposition:
Proposition 2.6.
Let λ ∈ T \ U . Then λ is in the resolvent set of T .Proof. From Proposition 2.3 we know that λ is in the resolvent set of S . Now, letus first show that λ is in the resolvent set of S + e ⊗ q , and let us also compute theresolvent of this operator at λ . HE SPECTRUM OF IRREDUCIBLE OPERATORS 5
The resolvent of S at λ can be written down in block diagonal form, and fromthis, we obtain h e, R ( λ, S ) q i = 0. Thus, by Proposition 2.4, λ is indeed in theresolvent set of S + e ⊗ q , and we have R ( λ, S + e ⊗ q ) = R ( λ, S ) + R ( λ, S )( e ⊗ q ) R ( λ, S ) . Next, we perturb S + e ⊗ q by q ⊗ e , and we apply Proposition 2.4 a second timein order to see that λ is in the resolvent set of T = S + e ⊗ q + q ⊗ e : we have h q, R ( λ, S + e ⊗ q ) e i = h q, R ( λ, S ) e i + h q, R ( λ, S )( e ⊗ q ) R ( λ, S ) e i = h e, R ( λ, S ) e i · h q, R ( λ, S ) q i = 1 λ · h q, R ( λ, S ) q i ;for the second equality we used that h q, R ( λ, S ) e i = 0.We only need to show that the result of the preceding computation cannot be 1,and to this end, it suffices to show that the modulus of h q, R ( λ, S ) q i is strictly lessthan 1. To see this, we once again use the block diagonal representation of R ( λ, S ),together with the fact that R ( λ, (1 − q n ) P ) = λ − q n since P = . Thus, weobtain |h q, R ( λ, S ) q i| = (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ∞ X n =1 (cid:10) q n , R ( λ, (1 − q n ) P ) q n (cid:11)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) = (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ∞ X n =1 q n dλ − q n (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≤ ∞ X n =1 q n d q n | λ − q n | . The numbers q n d sum up to 1, and the numbers q n | λ − q n | are all strictly lessthan 1 according to Proposition 2.5. Hence, |h q, R ( λ, S ) q i| <
1, which shows that h q, R ( λ, S + e ⊗ q ) e i 6 = 1. Therefore, λ is in the resolvent set of T . (cid:3) A C -semigroup version In this section we present an analogous construction for the case of C -semigroups.Throughout the section we freely make use of C -semigroup theory; standard ref-erences for this topic include [15, 5]. Perron–Frobenius type results for positive C -semigroups can, for instance, be found in [14, Chapters B-III and C-III] or in[3, Chapters 12 and 14]. As a motivation, we first observe the following property: Proposition 3.1.
Let ( T ( t )) t ∈ [0 , ∞ ) be a bounded, positive and irreducible C -semi-group on a Banach lattice E . Then the set σ pnt ( A ) ∩ i R is either empty or an (additive) subgroup of i R . For the single operator case, we quoted an analogous result from [17, Theo-rem V.5.2 and Lemma V.4.8] in the previous section. For C -semigroups, we foundthis result only under slightly different assumptions in the literature, so we includea brief argument here: Proof of Proposition 3.1.
Assume that σ pnt ( A ) ∩ i R is non-empty and let iβ be apoint in this set; let z ∈ E denote a corresponding eigenvector. Then, by thepositivity of the semigroup, we have | z | = | T ( t ) z | ≤ T ( t ) | z | for each time t ≥ | z | is a non-zero super fixed vector of the semigroup and hence, we can use thesame argument as in [17, Lemma V.4.8] to conclude that the dual semigroup has anon-zero positive fixed vector.Under this assumption, the subgroup property of σ pnt ( A ) ∩ i R is proved in [14,Theorem C-III-3.8(a)] or in [3, Proposition 14.15(c)]. (cid:3) JOCHEN GL¨UCK
Similarly as in the previous section, we now show that the assertion of Proposi-tion 3.1 is not true, in general, for the spectrum instead of the point spectrum. Inorder to keep the technicalities as simple as possible, we will not prove the resultin the same generality as Theorem 1.1; instead, we will restrict ourselves to theconstruction of a single semigroup with a certain spectral property:
Theorem 3.2.
There exists an L -space E over a σ -finite measure space and anirreducible and stochastic C -semigroup on E such that the spectrum σ ( A ) of itsgenerator A satisfies σ ( A ) ∩ i R = i ( −∞ , − ∪ { } ∪ i [1 , ∞ ) . This theorem is – in a negative sense – relevant for the analysis of the long-termbehaviour of positive C -semigroups: Remark 3.3.
If a C -semigroup is bounded and the spectrum of its generator A intersects i R in at most countably many points, this implies – under a certainergodicity assumption – that the semigroup is asymptotically almost periodic; thisis a version of the ABLV theorem, see for instance [2, Theorem 5.5.5]. So it isnatural to ask for sufficient conditions of σ ( A ) ∩ i R to be at most countable.If irreducibility of a positive C -semigroup implied that σ ( A ) ∩ i R is a groupthen, for such a C -semigroup, it would suffice to know that at least one number in i R is not a spectral value in order to conclude that σ ( A ) ∩ i R is at most countable(since all closed non-trivial subgroups of i R are countable).Theorem 3.2, though, shows that such an argument cannot work, in general.Our construction of the semigroup generator A in Theorem 3.2 is quite similarto what we did in the previous section. Here are the details: Construction 3.4.
The space:
Endow the complex unit circle T with the Haarmeasure that assigns the measure 2 π to the whole space T . We denote the normon L ( T ) by k · k , and we set E := { f = ( f , f , f , . . . ) : f ∈ C , f , f , · · · ∈ L ( T ) and k f k E < ∞} , where k f k E := | f | + ∞ X n =1 k f n k . Clearly, E is isometrically lattice isomorphic to the L -space over the σ -finite mea-sure space { } ˙ ∪ T ˙ ∪ T ˙ ∪ . . . . The generator:
We specify our semigroup by defining its generator A . Let D : L ( T ) ⊇ dom D → L ( T ) denote the generator of the shift semigroup on L ( T ); then the spectrum of D equals i Z .Choose a sequence ( q n ) n ∈ N ⊆ (0 ,
1] such that P ∞ n =1 πq n = 1; clearly, q n con-verges to 0 as n → ∞ . In addition, let ( ω n ) n ∈ N ⊆ [1 , ∩ Q be a sequence whichcontains each rational number in [1 ,
2] infinitely often. We first define a blockdiagonal operator B : E ⊇ dom B → E as B = − ω D − q ω D − q . . . ; HE SPECTRUM OF IRREDUCIBLE OPERATORS 7 its domain is given bydom B := (cid:8) f = ( f , f , f , . . . ) : f ∈ C , f , f , · · · ∈ dom D and ∞ X n =1 k Df n k < ∞ (cid:9) . By standard perturbation theory, B is the generator of a positive semigroup on E .Now, let ∈ L ( T ) denote the constant function with value 1 (it has norm 2 π ) andconsider the vectors e = (1 , , , . . . ) and q = (0 , q , q , . . . );we can interpret both of them as vectors in E and as vectors in the dual space E ′ .Similarly as in Construction 2.1 we define the operator A : E ⊇ dom A → E by A := B + e ⊗ q + q ⊗ e, with domain dom A := dom B .Clearly, A generates a positive C -semigroup on E , and a straightforward com-putation shows that the vector (1 , , , . . . ) ∈ E ′ is in the kernel of the dual operator A ′ (one merely has to use that ∈ ker D ′ ).Hence, the semigroup generated by A is stochastic.Moreover, one readily checks that the rank-2 operator e ⊗ q + q ⊗ e is irreducible.Thus, the semigroup generated by A is irreducible, too [14, Proposition C-III-3.3].So in order to obtain Theorem 3.2, it only remains to compute the peripheralspectrum of A . We start with the easier of both inclusions: Proposition 3.5.
We have σ ( A ) ∩ i R ⊇ i ( −∞ , − ∪ { } ∪ i [1 , ∞ ) . Proof.
As the semigroup generated by A is stochastic, 0 is a spectral value of A .Now, let r be a rational number in [1 , ∞ ). It suffices to prove that ir ∈ σ ( A )(since the spectrum is closed and invariant under complex conjugation). Let k ∈ N denote the largest integer that is smaller than r . Then there exists a rationalnumber ω ∈ [1 ,
2) such that ωk = r . As the number ik is an eigenvalue value of thedifferential operator D , it follows that iωk = ir is an eigenvalue of ωD .Since the operator ωD occurs, up to the perturbations − q n , infinitely manyoften in the definition of B , one can use the same argument as in the proof ofProposition 2.2 to show that ir is an approximate eigenvalue of A . (cid:3) We still have to show the converse inclusion for the peripheral spectrum, andto this end we proceed similarly as in the single operator case: we start with theoperator B , and then we use the Sherman–Morrison formula. Proposition 3.6.
Let β ∈ ( − , \ { } . Then iβ is in the resolvent set of B .Proof. Since the spectrum of D equals i Z , we can find a number M > kR ( λ, D ) k ≤ M for all λ in the rectangle [0 ,
1] + iβ [ , kR ( iβ, ω n D − q n ) k = 1 ω n (cid:13)(cid:13)(cid:13)(cid:13) R (cid:0) iβ + q n ω n , D (cid:1)(cid:13)(cid:13)(cid:13)(cid:13) ≤ M for all n ∈ N (as we assumed the numbers ω n to be in [1 ,
2] and the numbers q n to be in (0 , R ( iβ, ω n D − q n ) are uniformly bounded as n varies. Due to the block diagonal form of B this implies that iβ is in the resolventset of B . (cid:3) JOCHEN GL¨UCK
Now we can finally prove that the spectrum of A is the claimed set, again byemploying the Sherman–Morrison type result from Proposition 2.4 twice: Proposition 3.7.
Let β ∈ ( − , \ { } . Then iβ is in the resolvent set of A .Proof. We already know from Proposition 3.6 that iβ is in the resolvent set of B .Let us first note that it is also in the resolvent set of B + e ⊗ q : indeed, we have h e, R ( iβ, B ) q i = 0, so Proposition 2.4 tells us that iβ is in the resolvent set of B + e ⊗ q , and that the resolvent of this operator at iβ is given by R ( iβ, B + e ⊗ q ) = R ( iβ, B ) + R ( iβ, B )( e ⊗ q ) R ( iβ, B ) . In order to prove that iβ is also in the resolvent set of A = B + e ⊗ q + q ⊗ e , wehave to show, according to Proposition 2.4, that h q, R ( iβ, B + e ⊗ q ) e i 6 = 1. So letus compute h q, R ( iβ, B + e ⊗ q ) e i = h q, R ( iβ, B ) e i + h q, R ( iβ, B )( e ⊗ q ) R ( iβ, B ) e i = h e, R ( iβ, B ) e i · h q, R ( iβ, B ) q i = 1 iβ + 1 · h q, R ( iβ, B ) q i ;for the second equality we used that h q, R ( iβ, B ) e i = 0. In order to further simplifythe expression, we note that D = 0, so R ( iβ, ω n D − q n ) = iβ + q n for each n ∈ N . Therefore, h q, R ( iβ, B ) q i = ∞ X n =1 πq n q n iβ + q n . The numbers 2 πq n sum up to 1 and the modulus of q n iβ + q n is strictly less than 1 foreach n . Hence, |h q, R ( iβ, B ) q i| < iβ +1 is also strictly less than 1, so |h q, R ( iβ, B + e ⊗ q ) e i| = (cid:12)(cid:12)(cid:12)(cid:12) iβ + 1 (cid:12)(cid:12)(cid:12)(cid:12) · |h q, R ( iβ, B ) q i| < . In particular, the number h q, R ( iβ, B + e ⊗ q ) e i is not 1, so iβ is indeed in theresolvent set of A . (cid:3) Acknowledgements.
It is my pleasure to thank Wolfgang Arendt for a very in-teresting discussion that motivated the results presented in this note, and to thankUlrich Groh for pointing out Schaefer’s result in [16, Theorem 7] to me.
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Jochen Gl¨uck, Universit¨at Passau, Fakult¨at f¨ur Informatik und Mathematik, 94032Passau, Germany
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