Contents lists available atScienceDirect
Journal
of
Functional
Analysis
www.elsevier.com/locate/jfa
Triple
variational
principles
for
self-adjoint
operator
functions
Matthias Langera, Michael Straussb,∗
aDepartmentofMathematicsandStatistics,UniversityofStrathclyde,
26 RichmondStreet,GlasgowG1 1XH,UnitedKingdom
b
DepartmentofMathematics,UniversityofSussex,FalmerCampus, Brighton BN1 9QH,UnitedKingdom
a r t i c l e i n f o a b s t r a c t
Articlehistory:
Received 13 June 2013 Accepted 3 September 2015 Available online 19 January 2016 Communicated by L. Gross
MSC:
primary 49R05
secondary 47A56, 47A10
Keywords:
Variational principles for eigenvalues Operator functions
Spectral decomposition
For a very general class of unbounded self-adjoint operator functions we prove upper bounds for eigenvalues which lie within arbitrary gaps of the essential spectrum. These upper bounds are given by triple variations. Furthermore, we find conditions which imply that a point is in the resolvent set. For norm resolvent continuous operator functions we show that the variational inequality becomes an equality.
© 2016 The Authors. Published by Elsevier Inc. This is an open access article under the CC BY license (http://creativecommons.org/licenses/by/4.0/).
1. Introduction
In many applications of operator and spectral theory eigenvalue problems appear which are nonlinear in the eigenvalue parameter, e.g.polynomially or rationally. Very oftensuchproblemscanbedealtwithbyintroducingafunctionofthespectralparameter
* Corresponding author.
E-mailaddresses:[email protected](M. Langer), [email protected](M. Strauss). http://dx.doi.org/10.1016/j.jfa.2015.09.004
whose valuesarelinearoperators inaHilbertspace. Tobemorespecific,let T(·) be an operatorfunctionthatisdefinedonsomeset Δ⊂Candwhose valuesareclosed linear operatorsinaHilbertspace(H, ·, ·);foreach λ ∈Δ thedomainoftheoperator T(λ) is denotedbydom(T(λ)).Anumber λ ∈Δ iscalledaneigenvalueoftheoperatorfunction
T if there exists an x ∈ dom(T(λ))\ {0} such that T(λ)x = 0, i.e. 0 is in the point spectrumoftheoperator T(λ).Thespectrum,essentialspectrum,discretespectrumand
resolvent set of T aredefinedasfollows:
σ(T) :=λ∈Δ : 0∈σ(T(λ)),
σess(T) :=
λ∈Δ : 0∈σess(T(λ))
=λ∈Δ :T(λ) is not Fredholm,
σdis(T) :=σ(T)\σess(T),
ρ(T) :=λ∈Δ : 0∈ρ(T(λ));
note that a closed operator is called Fredholm if the dimension of the kernel and the (algebraic)co-dimensionoftherangearefinite.Atrivialexampleofanoperatorfunction is givenby T(λ)=A −λI where A isaclosedoperator;inthis casethespectraof the operatorfunction T andtheoperator Aclearlycoincide.Morecomplicatedexamplesare operatorpolynomialsor Schurcomplementsof blockoperatormatrices;see,e.g.[25,31]
and thereferences therein;see alsothesurvey article[29]about numericalmethodsfor eigenvaluesofquadraticmatrixpolynomials.
It is our aim to show spectral enclosures and variational principles for eigenvalues of operator functions. In the 1950s R.J. Duffin [6] proved a variational principle for eigenvalues of certain quadratic matrixpolynomials, whichwas generalised to infinite-dimensional spaces and moregeneral operator functions inthe following decades; see, e.g.[27,32,14,36,1,25].Basically,thefollowing situationwasconsidered.Let T be a dif-ferentiable functiondefined onan interval [α, β] whose values arebounded self-adjoint operators inaHilbertspaceHsuchthat T(α)0 (i.e. T(α) isuniformlypositive)and
T(β)0.Moreover,forevery x ∈H\{0}thescalarfunction λ → T(λ)x, xhasexactly onezeroin(α, β),whichwedenoteby p(x),andtheinequalityT(p(x))x, x <0 holds. The mapping x → p(x) iscalled ageneralised Rayleigh functional.The eigenvaluesof
T belowtheessentialspectrumof T canaccumulateat mostat thebottomof σess(T); if they are denoted by λ1 ≤ λ2 ≤ · · ·, then they are characterised by the following
variationalprinciple:
λn= min L⊂H dimL=n
max
x∈L x=0
p(x) = max
L⊂H dimL=n−1
min
x∈H x⊥L, x=0
p(x); (1.1)
In [2] the assumption that T(α) is uniformly positive was relaxed and replaced by theassumptionthatthenegativespectrumof T(λ) consistsofonlyafinitenumber κof eigenvalues(counted with multiplicities), inwhich case n hasto be replaced by n +κ
in the variations over the subspaces; also the generalised Rayleigh quotient has to be slightlymodified(seeDefinition 2.1(i)below);cf.also[35,34].In[8]alsofunctionswhose valuesare unboundedoperators wereallowed;seealso[15].
Themain aimofourpaperistoremovetheassumptionofthefinitenessofthe nega-tivespectrumof T(α) andtoallowalsothecharacterisationofeigenvaluesingapsofthe essentialspectrum.Inordertodothis,athirdvariationisneeded;seeTheorem 5.1,the main resultofthepaper. This theoremgreatly sharpens andextends [7,Theorem 2.4], whereonlyaninequalitywasshownforoperatorfunctionsandwhereitwasassumedthat thevaluesareboundedoperators(forsomequadraticpolynomialsequalitywasproved). As part of the proof of Theorem 5.1 we also show such an inequality (Theorem 2.3) for aclass of operatorfunctions with lesscontinuity assumptions then needed in The-orem 5.1.To ourknowledgethefirsttriple variationalprincipleappearedin[26]where eigenvaluesofpositiveoperatorsinKreinspaceswerecharacterised;thiswasgeneralised in[28].
Oursecondmainresult,Theorem 2.2,isconnectedwiththeinequalityinTheorem 2.3
and gives a sufficient condition for points being in the resolvent set of an operator function.InTheorem 3.4thisisusedtoobtaintheexistenceofspectralgapsforperturbed self-adjointoperators.Inaforthcomingpaper[24]wewillalsoapplyTheorem 2.2toprove spectralinclusionsforcertainblockoperatormatrices.
Let us give a brief synopsis of the paper. In Section 2 we state and prove the re-sult about points in the resolvent set of an operator function (Theorem 2.2) and the variationalinequality(Theorem 2.3).Weshouldmentionthatalsoaninequalityforthe essentialspectrumisobtained.InSection3weconsiderself-adjointoperators,whichneed notbe semi-bounded,andproveavariationalprincipleforeigenvaluesinarbitrarygaps of theessential spectrum(Theorem 3.1). Moreover, the abovementioned perturbation resultforspectralgapsisprovedthere(Theorem 3.4).TheseresultsareappliedtoDirac operatorsandtoSchrödingeroperators withperturbedperiodicpotentials.InSection4
weproveadecompositionoftheHilbertspaceintoadirectsumofthreesubspaces,one being the span of theeigenvectors corresponding to eigenvaluesinan interval and the other two being spectral subspacesconnected with the operators at thetwo endpoints ofthe interval (Theorem 4.1). This isthe main ingredientintheproof ofthe other in-equalityofthevariationalprincipleinTheorem 5.1.Further,inSection4weprovethat eigenvalues cannot accumulate outside the essential spectrum of an analytic operator function(Proposition 4.2).Finally,inSection5weprovethetriplevariationalprinciple foreigenvaluesofnormresolventcontinuousoperatorfunctions.Theresultisillustrated withaquadratic operatorpolynomial.
2. Ageneralvariationalinequality
Inthissectionweconsiderarathergeneralclassofself-adjointoperatorfunctionsand provevariationalinequalitiesforeigenvalues.Moreover,wegivesufficientconditionsfor points tobelongtotheresolventsetof suchoperatorfunctions.
Let Abeaself-adjointoperatorinaHilbertspaceHandlet Ebeitsspectralmeasure. Wedefine thecorrespondingsesquilinearform aby
a[x, y] :=
R
μdE(μ)x, y (2.1)
for x, y∈dom(a):= dom|A|1/2.Moreover, weintroducethequadraticform
a[x] :=a[x, x], x∈dom(a).
Note that,for x ∈dom(A) and y ∈dom(a),wehave a[x, y]=Ax, y. If A isbounded from below,thenthisdefinitionclearlycoincideswiththedefinitionin[16,§IV.1.5].For moreinformationonnon-semi-boundedformssee, e.g.[10,13].
LetLbea(notnecessarilyclosed)subspaceofdom(a).WesaythatLis a-non-negative
if
a[x]≥0 for everyx∈L;
L is called maximal a-non-negative ifit cannot be extended to alarger subspacewith thesameproperty.
ThroughoutthepaperdenotebyLΔ(A) thespectralsubspacefor Acorrespondingto
aBorelsetΔ⊂R,i.e.LΔ(A)= ranE(Δ).
Assumptions(A1)–(A3).Let T beanoperatorfunctiondefinedonsomeintervalΔ⊂R whosevaluesareoperatorsinaHilbertspaceH.Weassumethatthefollowingconditions are satisfied:
(A1) T(λ) isself-adjointforevery λ ∈Δ withcorresponding quadraticform t(λ); (A2) dom(t(λ))= dom|T(λ)|1/2isindependentof λanddenotedbyD;
(A3) for each x ∈ D\ {0}, the function λ → t(λ)[x] is continuous and decreasing at value0,i.e.if t(λ0)[x]= 0 forsome λ0∈Δ,then
t(λ)[x]>0 forλ∈Δ such thatλ < λ0,
t(λ)[x]<0 forλ∈Δ such thatλ > λ0.
Occasionally — inparticular, when the essential spectrumis involved — we need the following condition,whichisnamedafterA. Virozub andV. Matsaev(see[33]andalso
(VM−) forevery u ∈D,thefunction t(·)[u] isdifferentiableonΔ and,foreverycompact subinterval I of Δ,there exist ε, δ > 0 such that,for all x ∈D with x= 1 andall λ ∈I,
t(λ)[x]≤ε =⇒ t(λ)[x]≤ −δ. (2.2)
Obviously,forfixed λ ∈I,thiscondition(i.e.(2.2)for x ∈Dwithx= 1)isequivalent totheconditionthat
t(λ)[x]≤εx2 =⇒ t(λ)[x]≤ −δx2 (2.3)
forall x ∈D.
In [33,22,20] the Virozub–Matsaev condition was studiedwith t(λ) ≥ δ instead of
t(λ)≤ −δ.Moreover,thedefinitionwasslightlydifferentbutequivalent toours(apart fromthedifferentsign)forthefunctionsconsideredin[20],whichwereassumedtohave boundedoperatorsasvaluesandtobecontinuouslydifferentiableintheoperatornorm, cf.[20, Lemma 3.6].
Next we define the notion of a generalised Rayleigh functional. First note that, by Assumption(A3),thefunction λ →t(λ)[x] hasatmostonezeroforagiven x ∈D\ {0}. Ifithasazero,wedefineageneralisedRayleighfunctional p(x) tobeequaltothiszero; otherwise,weassign avalueoutsideΔ.Moreprecisely,wedefineageneralisedRayleigh functionalas follows.
Definition 2.1. Let T be an operator function defined on Δ that satisfies Assump-tions (A1)–(A3)andlet t(λ) be thecorresponding forms.
(i) Afunctional p :D\ {0}→R∪ {±∞}iscalledageneralisedRayleigh functionalfor
T onΔ if, forall x ∈D\ {0},
p(x) =λ0 ift(λ0)[x] = 0,
p(x)< λ for allλ∈Δ ift(μ)[x]<0 for allμ∈Δ,
p(x)> λ for allλ∈Δ ift(μ)[x]>0 for allμ∈Δ.
(ii) For γ∈Δ set
M+γ :=M:Mis a maximalt(γ)-non-negative subspace ofD.
In [2,7,8] generalised Rayleigh functionals were defined such that p(x) = −∞ and
p(x)λ ⇐⇒ t(λ)[x]0. (2.4)
Moreover, if λ0 is an eigenvalue of T with eigenvector x0, i.e. T(λ0)x0 = 0, then p(x0) =λ0.
Thenexttwo theoremsarethemainresultsofthissection.Thefirstonecanbeused to show that some point is in the resolvent set of an operator function. The second one,whichisageneralisationof[7,Theorem 2.4],givestriplevariationalinequalitiesfor eigenvaluesandthebottomoftheessentialspectrumofanoperatorfunction.
Theorem 2.2. Assume that T satisfies (A1)–(A3) and (VM−). Let μ1, μ2 ∈ Δ with
μ1< μ2.If thereexistM∈M+
μ1 and a >0suchthat
t(μ2)[x]≥ax2 for all x∈M, (2.5)
then μ2∈ρ(T).
Theorem 2.3. Let Δ ⊂ R be an interval with right endpoint β ∈ R∪ {+∞} and let T
be anoperatorfunctiondefinedonΔ whichsatisfiesAssumptions (A1)–(A3).Moreover, let pbeageneralisedRayleigh functionalfor T on Δ,let γ∈ρ(T)with γ < β,andset
λe :=
infσess(T)∩(γ, β)
ifσess(T)∩(γ, β)=∅,
β otherwise. (2.6)
Let (λj)Nj=1, N ∈ N0∪ {∞},be afinite or infinite sequence of eigenvalues of T in the
interval(γ, λe)innon-decreasingordersuchthat,foreachsetof kcoincidingeigenvalues, say λi=λi+1=. . .=λi+k−1,onehas dim kerT(λi)≥k.Then
sup
M∈M+γ sup
L⊂M dimL=n−1
inf
x∈M\{0} x⊥L
p(x)≤λn, n∈N, n≤N. (2.7)
Moreover, if T satisfiesthecondition (VM−)and σess(T)∩(γ, β)=∅,then
sup
M∈M+γ sup
L⊂M dimL=n−1
inf
x∈M\{0} x⊥L
p(x)≤λe, n∈N. (2.8)
Remark2.4.
(i) The statement inTheorem 2.2 is false without the assumption (VM−) as can be seenfromthefollowingexample.Let Abeaself-adjointoperatorinaHilbertspace
(ii) Notethatthevariationsontheleft-handsidesof(2.7)and(2.8)areovernon-empty sets for those n considered there, i.e. there exists an M ∈ M+γ which is at least
n-dimensional;seethebeginningoftheproofofTheorem 2.3.
(iii) Under our assumptions one obtains in general only an inequality and not equal-ity as the following example shows. Consider the operator function T(λ) = diag(T1(λ), T2(λ), . . .), λ ∈ Δ = R, in the space H = 2 where the piece-wise linearfunctions Tk aredefinedas
Tk(λ) = ⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩
1, λ≤0,
1−kλ, 0< λ < 2k,
−1, λ≥ 2k.
Thespectrumof T consistsonlyofeigenvalues:
σ(T) =σp(T) =
1
k :k∈N
,
butthe variationsontheleft-hand sideof (2.7)areequalto 0 forall n ∈Nifone chooses,e.g. γ= 0.
(iv) Notethatinthelaststatementofthetheoremthecondition(VM−) isnecessaryas canbeseenfromtheexamplegivenin[8,Remark 2.10].
Beforeweprovethetheorems, weneedacoupleoflemmas.
Lemma2.5.Let Abeaself-adjoint operator, athecorresponding form,andassumethat
0∈ρ(A).LetMbeamaximal a-non-negativesubspaceofdom(a)= dom(|A|1/2),M an a-non-negative subspaceofdom(a)andL⊂M∩M.Then
dim(M/L)≥dim(M/L).
Proof. Since 0∈ ρ(A), K := dom(a) is aKrein space with inner product a[· , ·], i.e. it isadirectand orthogonalsumof theHilbertspaceK+=L(0,∞)(A)∩dom(a) and the
anti-HilbertspaceK−=L(−∞,0)(A)∩dom(a).Accordingto[18, Proposition I.1.1]and
itsfirstcorollary,Mand M haveangularoperatorrepresentations,i.e.,withrespect to thedecompositionK=K++˙ K−,theycanbewrittenas
M=
x CMx
:x∈K+
, M=
x CMx
:x∈dom(CM)
, (2.9)
where CM and CM are bounded operators from K+ to K− with dom(CM)=K+ and
dom(CM)⊂K+. This impliesthatM isisomorphic to K+ and M is isomorphicto a
Lemma2.6.Let abeaquadraticformwithdomaindom(a)correspondingtoaself-adjoint operator A.Letu ∈dom(a), v∈dom(A)and a, b, c >0suchthat
a[u]≥au2, Av ≤bv
and
ac > b(a+b+ 3c).
If u = 0 or v= 0,then
a[u+v] +cu+v2>0.
Proof. Wecanestimate
a[u+v] +cu+v2
=a[u] + 2 ReAv, u+Av, v+cu2+ 2cRev, u+cv2
≥a[u]−2Av u − Av v+cu2−2cu v+cv2
≥au2−2bv u −bv2+cu2−2cu v+cv2
= (a+c)u2−2(b+c)u v+ (c−b)v2. (2.10)
Since a +c >0,thequadraticform inuandvispositivedefiniteifandonlyif
(a+c)(c−b)−(b+c)2>0,
whichisequivalent to
ac > b(a+b+ 3c).
Asthis inequalityistruebyassumption,theexpressionin(2.10)ispositiveunlessboth
uandvarezero. 2
Inthenextlemmas T isanoperatorfunctiondefinedonaninterval Δ.
Lemma2.7.Assumethat T satisfies(A1)–(A3)and(VM−).Let μ1, μ2∈Δwith μ1< μ2
and letεand δ besuchthat (2.3)isvalidforall λ ∈[μ1, μ2]and x ∈D.Then
t(μ2)[x]≥ −εx2 =⇒ t(μ1)[x]≥min
εx2,t(μ2)[x] +δ(μ2−μ1)x2
Proof. Without loss of generality we may assume that x = 1. If t(λ0)[x] ≥ ε for some λ0 ∈ [μ1, μ2], then, clearly, t(λ)[x] ≥ ε for all λ ∈ [μ1, λ0]; if t(λ0)[x] ≤ −ε for some λ0 ∈ [μ1, μ2], then t(λ)[x] ≤ −ε for all λ ∈ [λ0, μ2]. Now, if t(μ1)[x] ≥ ε, then there is nothingto prove. Otherwise, t(λ)[x]∈[−ε, ε] forall λ ∈ [μ1, μ2] and therefore
t(λ)[x]≤ −δforsuch λ.Hence
t(μ1)[x] =t(μ2)[x]−
μ2
μ1
t(λ)[x] dλ≥t(μ2)[x] +δ(μ2−μ1),
whichshowstheassertion. 2
Lemma2.8.Assumethat T satisfies(A1)–(A3)and(VM−).Let μ1, μ2∈Δwith μ1< μ2
andlet εand δ be suchthat (2.3) isvalidfor allλ ∈[μ1, μ2] andx ∈D.Moreover, let
a, b >0,set c := min{ε, δ(μ2−μ1)}andsuppose that
ac > b(a+b+ 3c).
If u ∈D, v∈dom(T(μ2))aresuchthat
t(μ2)[u]≥au2, T(μ2)v ≤bv
and u = 0or v= 0,then
t(μ1)[u+v]>0.
Proof. Itfollows fromLemma 2.6appliedto a =t(μ2) that
t(μ2)[u+v] +cu+v2>0. (2.11)
Since c ≤ε, wehave t(μ2)[u +v]≥ −εu +v2.NowLemma 2.7impliesthat
t(μ1)[u+v]≥min
εu+v2,t(μ2)[u+v] +δ(μ2−μ1)u+v2
.
Thefirstexpressionintheminimumispositivebecause u +v= 0 by(2.11).Thesecond expressionintheminimum isalsopositive:
t(μ2)[u+v] +δ(μ2−μ1)u+v2>−cu+v2+δ(μ2−μ1)u+v2≥0
bythedefinitionof c,whichimpliestheassertion. 2
Lemma2.9. Assumethat T satisfies (A1)–(A3) and(VM−).Leta >0, μ1, μ2∈Δwith μ1< μ2,andlet Mbeasubspaceof D.Moreover, suppose that μ2∈σ(T)and that
Then thereexistsasubspaceM suchthatMM⊂D and
t(μ1)[x]>0 for allx∈M, x= 0.
Proof. Let ε and δ be such that (2.3) is valid for all λ ∈ [μ1, μ2] and x ∈ D. Set c := min{ε, δ(μ2−μ1)} and choose a positive number b such that b < a and ac > b(a +b + 3c).Since μ2∈σ(T),there exists a v0 ∈dom(T(μ2))⊂D suchthatv0= 1
and T(μ2)v0 ≤ b. Set M := M+ span{v0}. The space M is strictly larger than
M because b < a and (2.12) is satisfied. Now let u ∈ M and v ∈ span{v0}. Then
t(μ1)[u +v] >0 if u +v= 0 byLemma 2.8. 2
Theorem 2.2isnowanimmediateconsequenceofthepreviouslemma.
ProofofTheorem 2.2. If μ2∈σ(T),then,byLemma 2.9,thereexistsM ⊂D,MM such that t(μ1)[x] ≥ 0 for all x ∈ M, which contradicts the maximality of M as a
t(μ1)-non-negativesubspace. 2
BeforeweproveTheorem 2.3we needtwomorelemmas.
Lemma2.10.Assumethat T satisfies(A1)–(A3).Let λ1, . . . , λmbeeigenvaluesof T with eigenvectors u1, . . . , um andlet μ, ν ∈ Δ such that μ ≤λ1 ≤ · · · ≤λm≤ν.Moreover, let y∈D andc1, . . . , cm∈C.
(i) If t(ν)[y]≥0,then
t(μ)[y+c1u1+. . .+cmum]≥0.
(ii) If t(μ)[y]≤0,then
t(ν)[y+c1u1+. . .+cmum]≤0.
Proof. Weproveonlytheassertionin(i);thestatementin(ii)isprovedanalogously. Since t(ν)[y] ≥0 and T satisfies Assumption(A3), wehave t(λm)[y] ≥0. Using the factthat λmisaneigenvalueof Twitheigenvector um,i.e.that T(λm)um= 0,weobtain
t(λm)[y+cmum] =t(λm)[y] + 2 RecmT(λm)um, y+|cm|2T(λm)um, um
=t(λm)[y]≥0
andhence,againbyAssumption(A3), t(λm−1)[y+cmum]≥0.Repeatingthisargument
we get
t(λ1)[y+c1u1+· · ·+cmum]≥0.
Lemma2.11.Assumethat T satisfies (A1)–(A3).Let λ1≤λ2≤ · · · ≤λm beeigenvalues of T such that,for each set of k coinciding eigenvalues, say λi =λi+1 =. . .=λi+k−1,
onehas dim kerT(λi)≥k.Thenthereexistlinearlyindependentvectors u1, . . . , umsuch that uj isaneigenvectorof T corresponding to λj, j= 1, . . . , m.
Proof. For every λj choose an eigenvector uj such that for coinciding eigenvalues the
eigenvectorsarelinearlyindependent.Assumethatthereexistnumbers α1, . . . , αm∈C,
notallequalto0,suchthat
α1u1+. . .+αmum= 0.
Let αn be thelastnon-zerocoefficient,i.e. αn= 0 and
α1u1+. . .+αnun= 0.
Becausethe ujarechosentobelinearlyindependentforcoincidingeigenvalues,wehave
λ1< λn.Let k besuchthat
λk < λk+1=. . .=λn.
Since uk+1, . . . , un arelinearlyindependentand αn= 0, itfollowsthat
α1u1+. . .+αkuk =−(αk+1uk+1+. . .+αnun)= 0.
FromLemma 2.10(ii)with y= 0 we obtainthat
t(λk)[α1u1+. . .+αkuk]≤0. (2.13)
Thefactthat αk+1uk+1+. . .+αnun isaneigenvectortotheeigenvalue λn impliesthat
t(λn)[αk+1uk+1+. . .+αnun]= 0.Hence
0 =t(λn)[αk+1uk+1+. . .+αnun] =t(λn)[α1u1+. . .+αkuk]
<t(λk)[α1u1+. . .+αkuk]
by(A3),whichisacontradictionto(2.13). 2
Notethat,withoutAssumption(A3),thestatementofthepreviouslemmaisfalsein general;see,e.g.theexamplein[22,Remark 7.7].
NowwecanturntotheproofofTheorem 2.3.
ProofofTheorem 2.3. FirstweshowRemark 2.4(ii).Let n ∈Nandassumethat Thasat least neigenvaluesin(γ, λe) countedwithmultiplicities.ItfollowsfromLemma 2.11that
ByLemma 2.10(i)thespacespan{u1, . . . , un}is t(γ)-non-negativeandcanbeextended to amaximal t(γ)-non-negative subspacebyZorn’s lemma, whichshows thestatement concerning (2.7). For the analogous statement for (2.8) let μ2 ∈ σess(T)∩(γ, β) and
a, b, c as inLemma 2.8 where ε, δ aresuch that(2.3)is valid on[γ, μ2]. If n ∈ N, then
thereexistsan n-dimensionalsubspaceMofdom(T(μ2)) suchthatT(μ2)v≤bvfor v ∈M. It follows from Lemma 2.8with u = 0 that thespace M is t(γ)-non-negative. Againwecanextendthisspacetoamaximal t(γ)-non-negative subspace.
Supposethattheinequalityin(2.7)isfalseforsome n.Thenthereexist anM∈M+ γ
and asubspaceL⊂MwithdimL=n −1 suchthat
inf
x∈M\{0} x⊥L
p(x)> λn.
Hence
t(λn)[y]>0, y∈ML, y= 0. (2.14)
By Lemma 2.11there exist linearlyindependent eigenvectors u1, . . . , un corresponding to theeigenvalues λ1, . . . , λn.AccordingtoLemma 2.10(i)wehave
t(γ)[y+c1u1+· · ·+cnun]≥0
forall y∈MLand c1, . . . , cn∈C.Thisimpliesthat
M := (ML) + span{u
1, . . . , un} (2.15)
is a t(γ)-non-negative subspaceofD.Thesumin(2.15)is directbecauseof(2.14)and
t(λn)[x]≤0, x∈span{u1, . . . , un},
whichistruebyLemma 2.10(ii).Hence
dimM/(ML)=n, dimM/(ML)=n−1.
Lemma 2.5showsthatthiscontradictsthemaximalityofM.
Forthesecondpartassumethattheinequalityin(2.8)isfalseforsome n ∈N.There exist anM∈M+γ andasubspaceL⊂MwithdimL=n −1 suchthat
μ2:= inf x∈M\{0}
x⊥L
p(x)> λe. (2.16)
Accordingtothedefinitionof λethereexistsanumber μ1∈σess(T) sothat λe≤μ1< μ2.
t(μ1)[x]≥ax2, x∈ML, (2.17)
where a := min{ε, δ(μ2−μ1)}and ε, δaresuchthat(2.3)isvalid forall λ ∈[γ, μ2] and x ∈D.Set c := min{ε, δ(μ1−γ)}andchoose b >0 suchthat b < aand ac > b(a +b +3c). Since μ1 ∈ σess(T), i.e. 0 ∈ σess(T(μ1)), there exists an n-dimensional subspace V of dom(T(μ1))⊂D suchthat
T(μ1)v ≤bv for allv∈V. (2.18)
Set M := (ML)+˙ V; the sum is directbecause of (2.17), (2.18) and the inequality
b < a.It followsfrom(2.17),(2.18)andLemma 2.8that
t(γ)[y]≥0 for ally∈M, (2.19)
i.e.M is t(γ)-non-negative. Since
dimM/(ML)=n, dimM/(ML)=n−1,
thisisacontradictiontothemaximalityofMaccordingto Lemma 2.5. 2
3. Self-adjointoperators
Let Abeaself-adjointoperatorand athecorrespondingquadraticformwithdomain
D := dom(a)= dom|A|1/2. We introduce theoperator function T(λ)=A −λI and
the associated form t(λ)[x, y] = a[x, y]−λx, y, where λ ∈ R and x, y ∈ D. Note that T satisfiesAssumptions (A1)–(A3)andthecondition(VM−) onany intervalsince
t(λ)[x]=−x2. AsinDefinition 2.1(ii)set
M+γ :=M:Mis a maximal (a−γ)-non-negative subspace ofD
for γ∈R.
In thefollowing theorem eigenvalues ina gap of the essential spectrum are charac-terisedbyatriplevariationalprinciple.Thisresultisageneralisationof[7,Theorem 3.1]
tounboundedoperators.Forothertypesofvariationalprinciplesforeigenvaluesof self-adjointoperators in gapsof the essentialspectrumsee, e.g. [5,12,17,21], where agiven decompositionof the space isused. Note that Theorem 3.1is not acorollary of Theo-rem 5.1belowsincethereweassumethatthevaluesoftheoperatorfunctionareoperators thatarebounded frombelow,whichisnotassumed inTheorem 3.1.
Theorem3.1. Letγ∈ρ(A)∩Randset
Moreover, let (λj)Nj=1, N ∈ N0∪ {∞}, be the finite or infinite sequence of
eigenval-ues in (γ, λe) in non-decreasing order and counted according to their multiplicities:
λ1≤λ2≤ · · ·.Then
λn= sup M∈M+
γ
sup
L⊂M dimL=n−1
inf
x∈M\{0} x⊥L
a[x]
x2, n∈N, n≤N. (3.1)
Moreover, if N isfiniteand σess(A)∩(γ, ∞)=∅,then
minσess(A)∩(γ,∞)
= sup
M∈M+
γ
sup
L⊂M dimL=n−1
inf
x∈M\{0} x⊥L
a[x]
x2, n > N. (3.2)
Proof. Theinequalities‘≤’in(3.1)and(3.2)followfrom Theorem 2.3since
p(x) = a[x]
x2
is ageneralisedRayleighfunctional fortheoperatorfunction T on(γ, ∞).Toshow the reverse inequalities,setM:=L(γ,∞)(A)∩DwhereL(γ,∞)(A) denotesthespectral sub-spacefor Acorrespondingtotheinterval(γ, ∞).Clearly,Mismaximal t(γ)-non-negative because
D=L(−∞,γ)(A)∩D
˙
+L(γ,∞)(A)∩D
.
The operator A|dom(A)∩L(γ,∞)(A) is self-adjoint in H := L(γ,∞)(A) and bounded from
below.Astandardvariationalprincipleyields that
sup
L⊂M dimL=n−1
inf
x∈M\{0} x⊥L
a[x]
x2 =λn, n∈N, n≤N,
sup
L⊂M dimL=n−1
inf
x∈M\{0} x⊥L
a[x]
x2 =λe, n > N,
whichshows theinequalities‘≥’in(3.1)and (3.2). 2
Remark 3.2. Let us denote by M++
γ the set of M∈ M+γ on which a −γ is uniformly
positive,i.e.
M++γ :=M∈Mγ+:∃c >0 such that a[x]−γx2≥cx2 for x∈M.
Onecanreplacethefirstsupremumin(3.1)and(3.2)bysupM∈M++
γ because M
++ γ ⊂M+γ
Assumefortherestofthisremarkthat γ= 0,whichiswithoutloss ofgenerality.Let
M∈M++0 andlet CMbeasin(2.9).ThenCM <1,andhencetheform aM:=a|Misa closedpositiveformintheHilbertspaceM.Let AMbetherepresentingoperatorof aMin thesenseof[16,Theorem VI.2.1],andlet λ1(AM)≤λ2(AM)≤ · · · betheeigenvaluesof AMbelowtheessentialspectrum;ifthereisonlyafinitenumber,say NM,ofeigenvalues, then set λn(AM):= minσess(AM) forn > NM. Thestandard variationalprinciple for semi-boundedoperators yields
λn(AM) = sup L⊂M dimL=n−1
inf
x∈M\{0} x⊥L
a[x]
x2, n∈N.
Hencerelation(3.1)withM+0 replacedbyM++0 turnsinto
λn= sup
M∈M++0
λn(AM), n∈N, n≤N, (3.3)
andasimilarrelationholdsfor λe if N isfinite.
Example3.3.ConsidertheDiracoperator
D:=
3
j=1
αj
1
i∂j+β+V
inthespace L2(R3;C4) whereα
j andβarethe4×4 complexmatrices
αj=
0 σj
σj 0
, β=
I 0 0 −I
withσj being thePaulimatrices,
σ1=
0 1 1 0
, σ2=
0 −i i 0
, σ3=
1 0 0 −1
,
andwhere V istheelectrostaticpotential.Assumethat V issuchthat Disaself-adjoint operatorwithform domain D= dom(|D|1/2) and form d as in(2.1).If γ∈ρ(D), then onecancharacteriseeigenvaluesof D in(γ, min(σess(D)∩(γ, ∞))) with(3.1).
Let D0 be thefree Diracoperator,i.e. theoperatorfrom abovewith V ≡0 andset M0:=L(0,∞)(D0)∩D.Assumethatthereexists c >0 such that
d[x]≥cx2 for x∈M0,
These conditionsaresatisfied,e.g.when
−|μ
x| ≤V(x)≤0 (3.5)
with 0 ≤μ ≤2/π 2 +
2 π
; see [30, Theorem 1]. If (3.4) is satisfied, then M0 ∈ M++0 ,
where M++0 is as in Remark 3.2. Then d|M0 defines apositive self-adjoint operatorB
in the Hilbert space L(0,∞)(D0), which was called Brown–Ravenhall operator in the literature; see, e.g. [4,9,30,12]. Let λn(B) be the eigenvalues of B below its essential spectrum;if B hasonlyafinite number,say NB,ofsucheigenvalues,thenset λn(B):= minσess(B) for n > NB. Moreover, let λn(D) be the eigenvalues of D in the interval (0, min(σess(D)∩(0, ∞)));againifthereareonlyfinitelymanysucheigenvalues,say ND,
then set λn(D)= min(σess(D)∩(0, ∞)) for n > ND.Relation (3.3)impliesthat
λn(B)≤λn(D), n∈N.
This inequality was proved for V satisfying (3.5) with μ < √3/2 in [12, Theorem 6]. Therelationin(3.3)alsoshowsthattheeigenvaluesoftheDiracoperatorareobtained bymaximisingtheeigenvaluesofoperators thatareobtainedinasimilarway as B but with arbitraryM∈M++0 .
The following theorem shows that for a non-negative perturbation of aself-adjoint operatoraspectralgapclosesonlyfromoneside.
Theorem 3.4. Let A be a self-adjoint operatorwith corresponding quadratic form a and
α, β ∈ R such that (α, β) ⊂ ρ(A). Moreover, let b be a non-negative quadratic form with dom(b)⊃dom(a) such that a +b with domain dom(a) isthe quadratic form of a self-adjoint operator C,andassume that
b[x]≤ax2+ba[x], x∈dom(a), (3.6)
with some a, b ≥0.If α < βˆ where
ˆ
α:=α+a+bα, (3.7)
then ( ˆα, β)⊂ρ(C).
Proof. Considertheoperatorfunction T(λ):=C−λI withcorrespondingforms t(λ)=
a +b −λwithdomains D= dom(a) andthesubspace
M:=L(α,∞)(A)∩D=L[β,∞)(A)∩D.
Let μ ∈( ˆα, β) andchoosesome μ1∈( ˆα, μ).For x ∈Mand λ ∈(α, β) wehave
Inparticular,thisshowsthatMisa t(μ1)-non-negativesubspaceofD.AssumethatMis notmaximal t(μ1)-non-negative.Thenthereexistsanon-zeroelement x0inL(−∞,α](A)∩
Dsuchthat t(μ1)[x0]≥0.However,
t(μ1)[x0] =a[x0] +b[x0]−μ1x02
≤a[x0] +ax02+ba[x0]−μ1x02
≤(1 +b)α+a−μ1
x02= ( ˆα−μ1)x02<0,
which is a contradiction. Hence M ∈ M+
μ1. Since β > μ, it follows from (3.8) with λ
replacedby μandTheorem 2.2 that μ ∈ρ(T)=ρ(C). 2
If Aisboundedfrombelowand bisanon-negativeformwithdom(b)⊃dom(a),then
a +b with domain dom(a) is aclosed form that is bounded from below (see, e.g. [16, Theorem VI.1.31]).Hencethereexistsaself-adjointoperator Cthatrepresentstheform
a +bby[16,Theorem VI.2.1].ThereforeTheorem 3.4canbeappliedif(3.6)issatisfied. If B is abounded non-negative operator,then αˆ=a +BinTheorem 3.4; see [3, Section 9.4]forrelatedconsiderations.
Example3.5.ConsideraSchrödingeroperator H0 inRn withpotential V0 suchthat H0
isboundedfrom belowandhasagap(α, β) in thespectrum.Forinstance,V0 canbea
periodicpotential.Moreover, let V1 be non-negative perturbation of V0. Let h0 and v1
bethequadraticformscorrespondingto H0andthemultiplicationoperatorwith V1and assumethatdom(h0)⊂dom(v1) andthatthereexist a, b ≥0 suchthat
Rn
V1(x)|u(x)|2dx≤a
Rn
|u(x)|2dx+b
Rn
|∇u(x)|2+V0(x)|u(x)|2dx
for u ∈dom(h0).Let H betheoperatorcorrespondingtotheform h :=h0+v1.Ifα < βˆ withαˆ definedas in(3.7),then( ˆα, β)⊂ρ(H).
4. Aspectraldecomposition
InProposition 4.2 weprovethat,forholomorphicfunctionsof type (B),no accumu-lation ofeigenvalues outsidethe essentialspectrumcan occur,so thatthe discreteness assumption ofTheorem 4.1is automaticallysatisfiedoutsidetheessentialspectrum.
Theorem4.1.Let T beanoperatorfunctiondefinedontheinterval[α, β],where α, β ∈R,
α < β, which satisfies Assumptions (A1)–(A3), is continuous in the norm resolvent sense, and T(λ)isboundedfrombelowforeach λ ∈[α, β].Assumethat α, β∈ρ(T)and that
σ(T)∩(α, β) ={λ1, . . . , λn} ⊂σdis(T)
where λ1 ≤ · · · ≤ λn are repeated according to their multiplicities. Moreover, let u1, . . . , unbecorrespondinglinearlyindependenteigenvectors,whichexistbyLemma 2.11. Then
H=L(−∞,0)T(α)span{u1, . . . , un}L(0,∞)
T(β). (4.1)
The next proposition gives a sufficient condition for σ(T) having no accumulation point on an interval. Note that, without any further continuity assumption, functions satisfying (A1)–(A3) mayhave asequence ofeigenvalues thataccumulatesoutside the essential spectrum; see, e.g. the example in Remark 2.4(iii). Recall that an operator function T definedon adomain U ⊂Cis saidtobe holomorphicoftype(B) if T(λ) is
m-sectorialforevery λ ∈U,thedomainofthecorresponding closedquadraticform t(λ) is independent of λ: dom(t(λ)) ≡D, and t(·)[x] isholomorphic on U for every x ∈ D. Instead of(A3)weassumetheslightlystrongerassumption:
(A3) if t(λ0)[x]= 0 forsome x ∈D\ {0}and λ0∈Δ,then t(λ0)[x] <0.
In[20]thisconditionwiththereverseinequalityforthederivativewascalled(vm).Note that,withoutanyassumptionoftype(A3)or(A3),theresultwouldbeincorrectasthe zerofunctiononafinite-dimensionalspaceshows.
Proposition 4.2.LetU be adomain inCandlet T beaholomorphicfamilyof operators of type (B) defined on U. Moreover, let α, β ∈R with α < β be such that (α, β)⊂U,
T satisfies Assumptions (A1), (A2), (A3) on (α, β) and σess(T)∩(α, β) = ∅. Then σ(T)∩(α, β)has no accumulationpointin (α, β).
We first prove Theorem 4.1. The main idea is to add eigenvalues successively (see
Lemma 4.3. LetL1,L2 be closed subspacesof H andassume that L1∩L2 ={0}.The
sum L1L2 isnot closed ifandonly ifthere existxn ∈L1, yn∈L2 suchthat
xn= 1 for all n∈N, xn+yn →0 as n→ ∞. (4.2)
Proof. IfL1L2isnotclosed,then,by[11,Theorem 2.1.1],thereexist xn ∈L1, yn∈L2
suchthat
xn+yn<
1
n
xn+yn
.
Clearly, xn = 0 for all n ∈ N. Without loss of generality we can choose xn such that
xn= 1.Therelation
1
n
xn+yn
>xn+yn ≥ yn − xn
impliesthatyn≤ n+1n−1 for n ≥2 andhencethatxn+yn→0,whichis(4.2). Conversely,assumethatthereexist xn∈L1, yn∈L2thatsatisfy(4.2).Then,clearly,
thereexists no K suchthat
xn+yn ≥Kxn+yn for alln∈N.
Hence,by[11, Theorem 2.1.1],thesumL1L2 isnotclosed. 2
InLemmas 4.4–4.6weassumethattheassumptionsofTheorem 4.1 aresatisfied.
Lemma 4.4. Leta, b ∈[α, β] with a < b and letH1⊂D be a closed subspace such that t(a)[x]≤0forall x ∈H1.Assumethat (0, δ)⊂ρ(T(b))forsome δ >0.Thenthesums
H1+L(0,∞)T(b), H1+L[0,∞)T(b) (4.3)
aredirect andclosed.
Proof. ThecaseH1={0}istrivial;sointhefollowingweassumethatH1={0}.First
observethat t(b)[x] <0 forevery x ∈H1\ {0}byAssumption (A3).Hencethefirstsum
in(4.3)isdirect.Assumethatitisnotclosed.Then,byLemma 4.3,thereexist xn ∈H1
and yn ∈L(0,∞)(T(b)), n ∈N,suchthat
xn= 1 and xn+yn →0 as n→ ∞.
Set M0:= minσ(T(b)),whichisnegativebecause t(b)[x] <0 for x ∈H1\ {0}.Let Ebe
0>t(b)[xn] =
0
M0
λdE(λ)xn, xn+
∞
δ
λdE(λ)xn, xn
≥M0E
(−∞,0)xn 2
+δE(0,∞)xn 2
. (4.4)
Since
E
(−∞,0)xn≤E(−∞,0)yn+E
(−∞,0)(xn+yn)
=E(−∞,0)(xn+yn)
≤ xn+yn →0 asn→ ∞,
E(0,∞)xn≥E(0,∞)yn−E(0,∞)(xn+yn)
≥ yn − xn+yn →1 asn→ ∞,
it follows thatthe right-hand side of (4.4) is positive for all sufficiently large n. This contradiction showsthatthefirstsumin(4.3)isclosed.
Since thesecond sumcanbe writtenas
H1+L(0,∞)
T(b)+ kerT(b),
itisclosedbythefirstpartoftheproofandthefactthatker(T(b)) isfinite-dimensional; see, e.g. [11, Corollary 2.1.1]. Assume that the sum is not direct. Then there exist
u ∈ H1, v ∈ L(0,∞)(T(b)), w ∈ ker(T(b)) such that u +v+w = 0 and v+w = 0.
Clearly, t(b)[v+w] ≥ 0. Assumption (A3)implies that t(a)[v+w] >0, which contra-dicts t(a)[u]≤0.Hencealsothesecond sumin(4.3)isdirectandclosed. 2
Lemma4.5.Let a, b ∈[α, β]be suchthat a < b and(a, b)⊂ρ(T).Moreover,letH1⊂D
be aclosed subspacesuchthat t(a)[x]≤0forallx ∈H1.Assumethat
H=H1L(0,∞)
T(μ) for all μ∈(a, b). (4.5)
Then
H=H1L[0,∞)T(b). (4.6)
Proof. Since b ∈ σdis(T)∪ρ(T), the sum in (4.6) is direct by Lemma 4.4 and there
exists δ >0 suchthat[−δ, 0)⊂ρ(T(b)).Itfollowsfrom[16,Theorem VI.5.10]thatthere exists ε > 0 such that −δ, −δ3 ⊂ ρ(T(μ)) for all μ ∈ [b −ε, b]. For such μ we have
−δ 3,
δ 3
⊂ρT(μ)+2δ3and
L(−2δ
3,∞)(T(μ)) =L(0,∞)
T(μ) +2δ 3
=L(0,∞)
T(μ) +2δ 3
By[16,Theorem VII.4.2]theoperators T(μ)+2δ3 areuniformlyboundedfrombelowon [b −ε, b],say T(μ)+2δ3 M.Then
σ
T(μ) +2δ 3
−1
⊂−∞, 1 M
∪0,3 δ
.
IfΓ isacirclepassingthrough M1 and 3δ,then
L(0,∞)
T(μ) +2δ 3
−1
= ranP(μ)
where
P(μ) :=− 1 2πi
Γ
T(μ) +2δ 3
−1
−z
−1
dz.
Since T iscontinuousinnormresolventsense,thefamilyofspectralprojections P(μ) is uniformlycontinuousontheinterval[b −ε, b].
Now let x0 ∈ H. We show that x0 is contained in the set on the right-hand side of(4.6).Tothisend,let bn∈(b −ε, b) for n ∈Nwith bn→b.By(4.5)wecanwrite
x0=xn+yn with xn∈H1, yn∈L(0,∞)T(bn).
Supposethatynis notbounded. Without loss ofgenerality assumethatyn→ ∞, which implies that xn → ∞. Clearly, P(bn)yn = yn since yn ∈ L(0,∞)(T(bn)) ⊂
L(−2δ
3,∞)(T(bn)).Setyˆn :=P(b)yn ∈L[0,∞)(T(b)).Since δn :=P(bn)−P(b)→0 as n → ∞,wehave
yn
xn−
ˆ
yn
xn
= 1
xnP(bn)−P(b)
yn
≤δn yn
xn≤ δn
x0+xn
xn →
0 asn→ ∞.
Thisrelationtogetherwith xn→ ∞ yields
xn
xn+
ˆ
yn
xn = x0
xn−
yn
xn−
ˆ
yn
xn
→0.
It follows from Lemma 4.3 that H1 L[0,∞)(T(b)) is not closed, which contradicts
Lemma 4.4. Hence the sequences (xn) and (yn) are uniformly bounded and therefore
yn−ynˆ →0.Setting
weobtain x0−x0(n)=P(bn)−P(b)yn→0.Thisimpliesthat x0∈H1L[0,∞)(T(b))
sincethelatterspaceisclosedbyLemma 4.4. 2
Lemma4.6.Let a, μ0∈[α, β)with a < μ0andletH1⊂D beaclosedsubspacesuchthat t(a)[x]≤0forallx ∈H1.Assumethat
H=H1L[0,∞)T(μ0). (4.7)
Then thereexistsan ε >0suchthat
H=H1ker
T(μ0)L(0,∞)T(μ) for allμ∈[μ0, μ0+ε). (4.8)
Proof. Firstweprovethatthe sumontheright-hand sideof (4.8)isdirectand closed for allμ ∈[μ0, β].Itfollows from Lemma 4.4thatthesumH1+L(0,∞)(T(μ)) isdirect
and closed. Since ker(T(μ0)) is finite-dimensional, the sum on the right-hand side of
(4.8) is closed; see [11, Corollary 2.1.1]. Assume thatit is notdirect. Then there exist
u ∈ H1, v ∈ ker(T(μ0)), w ∈ L(0,∞)(T(μ)) such that u +v+w = 0 and w = 0. By
Lemma 2.10(ii)wehave t(μ)[u +v]≤0,whichcontradicts w∈L(0,∞)(T(μ)).Hencethe
sumontheright-handsideof(4.8)isdirectand closed. Nextweshowthatthereexists a K >0 suchthat
x∈H1, y∈kerT(μ0), w∈L(0,∞)T(μ0), x+y+w= 1
=⇒ w ≤K.
(4.9)
Assume that this is not true. Then there exist xn ∈ H1, yn ∈ ker(T(μ0)), wn ∈ L(0,∞)(T(μ0)) such that xn +yn +wn = 1 and wn → ∞. In this case also
yn+wn→ ∞ andhencexn→ ∞.Since
xn
xn+
yn+wn
xn =
xn+yn+wn
xn →0,
Lemma 4.3 implies that the sum H1L[0,∞)(T(μ0)) is not closed, which contradicts
(4.7).Hencea K >0 withthedesiredpropertyexists.
Let P(μ) betheorthogonalprojectionontoL(0,∞)(T(μ)) for μ ∈[μ0, β].Similarlyas
intheproofofthepreviouslemmaoneshowsthat dμ:=P(μ)−P(μ0)→0 as μ μ0.
Hencethereexistsan ε >0 suchthat δμK <1 forall μ ∈[μ0, μ0+ε).Weshowthat(4.8)
holds for all suchμ. Assumethatthis is notthe case. Then, forsome μ ∈[μ0, μ0+ε) thereexistsan x0∈Hwithx0= 1 whichisorthogonaltotheright-handsideof(4.8). Since (4.7)is truebyassumption,wecanwrite
x0=u+v+w with u∈H1, v∈kerT(μ0), w∈L(0,∞)T(μ0).
x0−y=w−P(μ)w=P(μ0)−P(μ)w≤δμK <1,
whichisacontradiction tothefactsthat x0⊥y andx0= 1. 2
ProofofTheorem 4.1. Letλ1ˆ <· · ·<λmˆ betheeigenvaluesof T intheinterval(α, β)
not countedwithmultiplicitiesandset ˆλ0:=α.For γ∈[α, β] considerthestatement
H=L(−∞,0)T(α)kerT(ˆλ1). . .kerT(ˆλk)L[0,∞)T(γ)
where kis such that ˆλ1, . . . ,λkˆ are the eigenvalues ofT in (α, γ).
(4.10)
If(α, γ) containsnoeigenvalues,then k= 0.Forγ=αthestatementis certainlytrue. We provethat(4.10) holds for all γ ∈ [α, β]. Assume thatthis is not thecase and let
γ0:= infγ∈[α, β]:(4.10)does not hold.Set
H1:=L(−∞,0)T(α)kerT(ˆλ1). . .kerT(ˆλk)
where k is such that λ1, . . . , ˆ λkˆ are the eigenvalues of T in the interval (α, γ0).
Lemma 2.10(ii)impliesthat t(ˆλk)[x]≤0 forall x ∈H1.
Itfollows fromLemma 4.5with a = ˆλk and b =γ0 that(4.10)holds alsofor γ=γ0. Now,if γ0< β,then Lemma 4.6yieldsacontradiction withthedefinitionof γ0.Hence
(4.10)holdsforall γ∈[α, β].For γ=β thisisexactlytheassertionofthetheorem. 2
InordertoproveProposition 4.2, wefirstneedthefollowinglemma.
Lemma4.7.Let T beaholomorphicfamilyofoperatorsoftype(B)definedonthecomplex domain U ⊂C with closed forms t such that dom(t(λ))=D for all λ ∈U.Moreover, let x(λ)∈D for λ ∈U suchthat x(·)isholomorphic.
(i) Assume that t(λ)[x(λ)] is locally bounded and let y0 ∈ D. Then t(λ)[x(λ), y0] is holomorphicin λ, x(λ)∈D and
d dλ
t(λ)[x(λ), y0]
=t(λ)[x(λ), y0] +t(λ)[x(λ), y0] (4.11)
forallλ ∈U.
(ii) Assume that T(λ)x(λ)=ν(λ)x(λ)where ν is ascalarholomorphic functionon U. Further, let λ0 ∈ U and assume that there exists a y0 ∈ dom(T(λ0)∗) such that
T(λ0)∗y0=ν(λ0)y0 andx(λ0), y0= 0.Then
ν(λ0) = t(λ0)
x(λ0), y0
x(λ0), y0
. (4.12)
Proof. (i)Fix λ0∈U andchoose M∈RsuchthatRet(λ0)+M0.Accordingto[16, (VII.4.4)]theform tcanbewrittenas
t(λ)[u, v] =T0(λ)Gu, Gv−Mu, v, u, v∈D,
where T0 is a holomorphic operatorfunction whose values are bounded operators and
G := (ReT(λ0)+M)1/2.
Nowset y(λ):=Gx(λ) for λ ∈U.Itfollowsfrom[16,(VII.4.7)]that,foreachcompact subset U0of U with λ0∈U0there exists C >0 suchthat
y(λ)2=Gx(λ), Gx(λ)= Ret(λ0)[x(λ)] +Mx(λ)2
≤Ct(λ)[x(λ)]+Mx(λ)2
forall λ ∈U0.Sincethelastexpressionisboundedon U0byassumption,itfollowsthat y(λ) islocallybounded. For u ∈D,thescalarfunction
y(λ), u=Gx(λ), u=x(λ), Gu
is holomorphic in λ. Hence y(λ) is strongly holomorphic in λ; see, e.g. [16, §VII.1.1]. Moreover, y(λ), u =x(λ), Gu for all u ∈ D = domG, which implies thatx(λ)∈ domG =D and y(λ)=Gx(λ).
Weconcludethatthefunction
t(λ)[x(λ), y0] =
T0(λ)y(λ), Gy0
−Mx(λ), y0
is holomorphicand that
d dλ
t(λ)[x(λ), y0]= d dλ
T0(λ)y(λ), Gy0−Mx(λ), y0
=T0(λ)y(λ), Gy0+T0(λ)y(λ), Gy0−Mx(λ), y0
=T0(λ)Gx(λ), Gy0
+T0(λ)Gx(λ), Gy0
−Mx(λ), y0
=t(λ)[x(λ), y0] +t(λ)[x(λ), y0],
whichshows (4.11).
(ii)Theexpression t(λ)[x(λ)]=ν(λ)x(λ)2islocallyboundedin λ.Hencewecan
ap-plyitem(i)ofthislemmatothederivativeoftheequality t(λ)[x(λ), y0]=ν(λ)x(λ), y0,
which, for λ =λ0,yields
t(λ0)x(λ0), y0+t(λ0)x(λ0), y0=ν(λ0)x(λ0), y0+ν(λ0)x(λ0), y0. (4.13)
x(λ0), T(λ0)∗y0=x(λ0), ν(λ0)y0=ν(λ0)x(λ0), y0.
Hence(4.13) yieldsthedesiredresult. 2
Proofof Proposition 4.2. Let μ ∈(α, β). Theassumption σess(T)∩(α, β)=∅implies
thatthereexists an ε >0 suchthat σ(T(μ))∩(−ε, ε)⊂ {0}and n := dim ker(T(μ)) is finite.Further,thereexistsa δε>0 suchthat,forall λwith|λ −μ| < δεtheintersection σ(T(λ))∩(ε, ε) consistsonlyofeigenvaluesoffinitemultiplicitywithtotalmultiplicityn. Theseeigenvaluescanbeenumeratedsuchthattheyareanalyticfunctionsof λ.Let ν(λ) be suchan eigenvaluecurve withazero at λ0 (i.e. λ0 is aneigenvalueof T),extend it
alsotoacomplexneighbourhoodof λ0andlet x(λ) becorrespondingeigenvectors,which canbe chosentodependanalyticallyon λ;see[16,§§VII.6.2 andII.6.2].
NowwecanapplyLemma 4.7(ii)with y0=x(λ0),whichyields
ν(λ0) = t
(λ
0)[x(λ0)]
x(λ0)2 .
Since,byAssumption(A3),thisexpressionisnegative, eigenvaluecurvescancrossthe
λ-axisonlyinonedirection.Therefore T hasatmost N eigenvaluesin[μ −δ, μ +δ],and hencetheeigenvaluescannotaccumulateat μ. 2
5. Variationalprinciplesfornormresolvent continuousoperatorfunctions
Inthissection weprovethatunderstrongercontinuity assumptionsontheoperator functionwehaveequalityinthevariationalprinciplefrom Theorem 2.3.
Theorem 5.1. Let Δ ⊂ R be an interval with right endpoint β ∈ R∪ {∞} and let T
be an operator function defined on Δ which satisfies Assumptions (A1)–(A3) on Δ, is continuousin thenormresolvent sense on Δ,and T(λ)isbounded frombelowforeach
λ ∈Δ.Moreover,let pbeageneralisedRayleighfunctionalfor T onΔ,let γ∈ρ(T)∩Δ
with γ < β,let M+γ bedefined asin Definition 2.1andlet λe beas in(2.6).
Assume that the spectrum of T in (γ, λe) has no accumulation point in [γ, λe), i.e. σ(T)∩[γ, λe) is empty or consists of a finite or infinite non-decreasing sequence of eigenvalues (λn)Nn=1 with N ∈N∪ {∞},counted accordingtotheir multiplicities,which
canaccumulate atmost atλe. If σ(T)∩(γ, λe)=∅,then
λn= sup M∈M+γ
sup
L⊂M dimL=n−1
inf
x∈M\{0} x⊥L
Moreover, if, in addition, T satisfies the condition (VM−), N is finite and σess(T)∩ (γ, β)=∅,then
λe= sup
M∈M+γ sup
L⊂M dimL=n−1
inf
x∈M\{0} x⊥L
p(x), n > N. (5.2)
Remark 5.2. If T is a holomorphic family of type (B) in a neighbourhood of Δ and Assumption (A3) is satisfied, then onedoes not have to assume that the eigenvalues cannot accumulate in [γ, λe), but this follows from Proposition 4.2. Theorem 5.1 and
Proposition 4.2 canbe applied,e.g.to operatorpolynomialsand Schurcomplementsof certain blockoperatormatrices;forthelattersee[24].
Proof. Theinequalities ‘≥’ in(5.1) and (5.2)follow from Theorem 2.3. Wefirst prove ‘≤’ in(5.1).Let1≤n ≤N.Itis sufficienttofindasubspaceM∈M+
γ andasubspace L⊂MwithdimL=n −1 suchthat
inf
x∈M\{0} x⊥L
p(x)≥λn. (5.3)
Let m = max{k∈N:λk =λn}and choose μ > λn suchthat(λn, μ]⊂ρ(T).Moreover, let u1, . . . , umbelinearlyindependenteigenvectorsof T correspondingtotheeigenvalues
λ1, . . . λm (seeLemma 2.11).Considerthesubspace
M:= span{u1, . . . , um}+
L(0,∞)(T(μ))∩D
.
FromLemma 2.10(i)weobtainthat t(γ)[x]≥0 forall x ∈M.SinceL(−∞,0)(T(γ))⊂D and uk ∈D, k = 1, . . . , m, Theorem 4.1implies thatthefollowing decompositionof D is valid:
D=L(−∞,0)
T(γ)span{u1, . . . , um}
L(0,∞)(T(μ))∩D
. (5.4)
It follows from thisdecomposition thatM ismaximal t(γ)-non-negative, i.e.M∈M+ γ.
Let P be theorthogonalprojectioninHonto
K:= span{un, . . . , um}+L(0,∞)T(μ)
and set
L:= (I−P)M= (I−P) span{u1, . . . , un−1}.
Since
and
(I−P)uk=uk−P uk∈M+K, k= 1, . . . , n−1,
we have L ⊂ D ∩ (M +K) = M. We show that the mapping I − P is injec-tive on span{u1, . . . , un−1}. Assume that this is not the case. Then there exists a u ∈ span{u1, . . . , un−1}, u = 0, such that u ∈ ker(I −P) = K, i.e. there exist α1, . . . , αm∈Cand x ∈L(0,∞)(T(μ))∩Dsuchthat
u=α1u1+. . .+αn−1un−1=αnun+. . .+αmum+x.
Sincethesumin(5.4)isdirect,wehave x = 0,andtherefore α1=. . .=αm= 0 because ofthelinearindependenceof u1, . . . , um.Hence I−P isinjectiveonspan{u1, . . . , un−1},
whichshowsthatdimL=n −1.
Now let x ∈ M\ {0}such that x ⊥L = (I−P)M. Then x ∈ D, and the relation
x =P x + (I−P)ximpliesthat
x2=P x, x+(I−P)x, x=P x2,
whichshowsthat x ∈ranP=K.ItfollowsfromLemma 2.10(i)that t(λn)[x]≥0,which proves(5.3).
In order to prove (5.2), assume that N is finite and let n > N. Moreover, let
μ ∈(λN, λe) bearbitraryand P be theorthogonal projectioninH ontoL(0,∞)(T(μ)). Similarlytothefirstpartoftheproofwe canchoose
M:= span{u1, . . . , uN}+
L(0,∞)(T(μ))∩D
,
whichisin M+γ. ThespaceL := (I−P)Mis anN-dimensionalsubspaceof M, which canbe seenas above.Extend L toan(n −1)-dimensionalsubspaceLofM.Then
inf
x∈M\{0} x⊥L
p(x)≥μ,
whichshows(5.2)since μ ∈(λN, λe) wasarbitrary. 2
Example5.3.Let Cbeaself-adjointoperatorinaHilbertspaceHthatisboundedfrom below,let cbe thecorresponding quadraticform,andassumethat0∈ρ(C).Moreover, let bbeasymmetricnon-positivequadraticformthatis c-boundedwithrelative bound 0,i.e. dom(c)⊂dom(b) andforeach b >0 thereexistsan a ≥0 suchthat
b[x]≤ax2+bc[x], x∈dom(c).
Forinstance, bcanbeaform correspondingto a C-compactoperator.Thentheform