Approximation of eigenvalues using variational principles

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Approximation of Eigenvalues using

Variational Principles

Yacir Andr´

es Ram´ırez

ya.ramirez1763@uniandes.edu.co

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Chapter 1 Introduction

The objective of this text is to describe a method to find eigenvalues (or approximations for eigenvalues) of a linear operator, or more general, of an operator valued function, without having to calculate the corresponding eigenvectors explicitly. Note that given a selfadjoint operator T(H →H), it is in general not an easy task to find its eigenvalues. The idea of variational principles is to show that isolated eigenvalues are solutions to certain maxi-mization or minimaxi-mization problems. Roughly speaking, if T is a selfadjoint linear operator which is bounded from below, then the classical variational principle gives the following formula for the eigenvalues of T below its essen-tial spectrum:

λn= min L⊂D(T) dimL=n

max

x∈L

kxk=1

hT x, xi.

Even if this procedure does not immediately lead to an expression that can be evaluated easily numerically, it can be used to obtain at least bounds for the eigenvalues. According to [RS78] and [Dav95], these methods to approximate eigenvalues are very useful in quantum electrodynamics and the computation of eigenvalues of complicated N-body molecular Hamiltonians in quantum chemistry.

In this text we will show a generalized variational principle that applies to certain operator valued functions (also called pencils). Let us consider the pencil T : Ω ⊂ _R → C(H); here, Ω is a subset of _C and C(H) is the set of closed operators on a Hilbert space H. Probably the easiest non-trivial example is the following. Fix a closed linear operator T ∈C(H) and define

T(λ) :=T −λ. (1.1)

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Note that in this case

λ∈σp(T) ⇐⇒ 0∈σp(T(λ)).

So a somewhat natural definition of the spectrum of a pencil is to say that

λ∈Ω belongs to the spectrum of a pencil if and only if 0 is in the spectrum of T(λ). Moreover, for x ∈ D(T) with x 6= 0, let us define p(x) as the number λ where hT(λ)x, xi = 0. Then, if T is selfadjoint, it is easy to see that the classical variational principle of min-max (see [RS78]) in the case of the pencil (1.1) can be rewritten as

λn= min L⊂D dimL=n

max

x∈L x6=0

p(x). (1.2)

So it seems natural to look for a variational principle for a general pencil in terms of zeros of certain functionals. We need to assume that for any givenx

in a certain subspace ofH the functionλ 7→ hT(λ)x, xihas at most one zero so thatp(x) is well defined. Note that we allow that the operators T(λ) may be unbounded, and a priori their domains may be different for different values of λ, so some care has to be taken with respect to the domains. Actually, for this reason, it turns out to be easier to work with the sesquilinear forms associated toT. Moreover, in general the maximum in (1.2) has to be taken over a subspaceL of dimension n+k wherek is a fixed natural number. The aim of this paper is to present variational principles of this type. We will follow the paper [EL04] by Eschw´e and Langer.

This text is organized in the following way:

• In chapter 2 we briefly talk about a classical variational principle and we give examples of operators where it fails to give proper information of the spectrum of certain operators.

• Chapter 3 contains the preliminaries that will be needed to understand the proof of the generalized variational principle. First we give some useful definitions and notation of functional analysis, and then we prove lemma3.6, needed extensively along the text. Next, we announce the spectral theorem for selfadjoint operators and a few important corol-laries that follow from it. Then we will define symmetric sesquilinear form, as well as closed and closable forms. Finally, we will define the norm resolvent convergence, its relation to the generalized convergence (see [Kat95]), and properties that follow from this type of convergence.

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Chapter 1. Introduction 7

• In chapter 4 we prove the generalized variational principle. In order to do so, we will prove several lemmas first, and then the variational principle itself. It is necessary to point out that the main ideas in the proofs of the lemmas and the main theorem come from [EL04].

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Chapter 2 Classical Variational Principles

Here we cite a classical variational principle, the min-max principle, men-tioned in [RS78].

Theorem 2.1. (see [RS78], theorem XIII.1). Let T(H → H) be a semi-bounded selfadjoint operator with domain D(T). Define

µn= sup

L⊂H

dimL=n−1

inf

x⊥L, x∈D(T)

x6=0

hT x, xi. (2.1)

Then, for each fixed n, either:

(a) There are n eigenvalues below the essential spectrum of T, and µn is

the nth eigenvalue of T counting multiplicity. or

(b) The value µn is the infimum of the essential spectrum ofT, and in that

case µn=µn+1 =. . .and there are at most n−1eigenvalues (counting

multiplicity) below the essential spectrum of T.

Unfortunately, for operators like

T :D(T)⊂l2(Z)→l2(Z), (xn)n∈_Z 7→(nxn)n∈_Z

the min-max technique will not say much of the spectrum ofT. For example, computing µ1 we have that for alln inN

µ1 = inf

x∈D(T)

x6=0

hT x, xi ≤ hT e−n, e−ni=−n

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with en defined as the tuple with 1 in the nth position and zero elsewhere.

Therefore, µ1 = −∞. Moreover, for all n in N, µn = −∞. To see this, let n in _N, and letL be a subspace of H with dimL=n−1. Then, for each k

in _N with k > n−1 there exists xk in L⊥∩span{e−k, e−k−1, . . .} such that

xk6= 0. Therefore

inf

x⊥L, x∈D(T)

x6=0

hT x, xi ≤ hT xk, xki ≤ −k.

Since Lwas arbitrary we get that

µn= sup

L⊂H,

dimL=n−1

inf

x⊥L, x∈D(T)

x6=0

hT x, xi=−∞.

Another example where this technique does not provide much information about the spectrum is

S :l2(N)→l2(N),

(xn)n∈N 7→

1

nxn

n∈_N.

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In this case µ1 = 0. Moreover, by the min-max principle µn = 0 for all n

in _N. Therefore, the min-max principle only says what the infimum of the essential spectrum of S is. Nevertheless, if we apply the min-max principle to the operator S0 :=−S, we will get the spectrum of S that is above of 0, which is practically the whole spectrum of S except the point 0.

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Chapter 3 Preliminaries

3.1 Functional Analysis

In this paper, H will always be a Hilbert space, andh·,·iwill denote its inner product. Here we introduce some useful notation and definitions.

Notation 3.1. We use the notation T(H → H) to indicate that T is a densely defined linear operator from H to H. Additionally we use D(T) to identify the domain of T.

Definition 3.2. Given a selfadjoint operatorT(H →H), define

ρ(T) := {λ∈_C:T −λ is bijective}, σ(T) := {λ∈_C:T −λ is not bijective}, σp(T) := {λ∈C:T −λ is not injective},

σc(T) := {λ∈C:T −λ is injective but not surjective and

Rg(T −λ) =H},

σr(T) := {λ∈Ω :T −λ is injective but not surjective and

Rg(T −λ)6=H},

σd(T) := {λ∈σp(T) :λ has finite multiplicity and is not an accumulation point},

σess(T) := {λ∈σ(T) :λ is accumulation point of σ(T) or has infinite

multiplicity}.

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Definition 3.3. Given a closed operator T(H→H) and λ in ρ(T), define

R(λ, T) := (T −λ)−1.

It is well defined by definition of ρ(T).

Remark 3.4. Since T is closed, we have by the closed graph theorem (see [Win15b]) thatR(λ, T) is bounded.

Definition 3.5. The numerical range of a linear operator T(H → H) is denoted byW(T) and defined by

W(T) := {hT x, xi:x∈ D(T), kxk= 1}.

The following lemma from functional analysis will be used frequently through-out this text. We strongly recommend the reader to get familiarized with it and its contrapositive.

Lemma 3.6. Given closed subspaces A and B of H, if A⊥∩B ={0} then

dim(B)≤dim(A).

Proof. Consider the function fromB to A defined as

x7→PA(x)

where PA is the orthogonal projection over A. It is a linear injection from A to B, since PA is linear and if PA(x) = 0 then x = 0. The last assertion

follows from the fact that kerPA = A⊥ and A⊥ ∩ B = {0}. Therefore

dim(B)≤dim(A).

Notation 3.7. Given an orthogonal projectionP, we set dimP = dim Rg(P).

3.2 Spectral Theorem for Selfadjoint

Opera-tors

The idea of this section is to announce the spectral theorem for selfadjoint operators and some corollaries that follow from it which will be useful later. First of all, some definitions:

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Chapter 3. Preliminaries 13

Definition 3.8. (See [Win15b] and [Tes09]). A spectral resolution (Eλ)λ∈R

is defined as follows:

• Eλ is an orthogonal projection for all λ∈R.

• EλEµ =EµEλ =Eµ for all µ≤λ.

• Eµx→Eλx if µ&λ (Strong-right continuity).

• Eµx→xfor µ→ ∞.

• Eµx→0 forµ→ −∞.

Definition 3.9. A projection-valued measure E(·) (see [Bel14] and [Tes09]) is a function from the Borel sigma-algebra of _R to the set of orthogonal projections on H, with the following properties:

• E(∅) = 0 and E(_R) = idH.

• Let (Bn)n∈_Nbe a subset of the Borel sigma-algebra ofRwhose members are pairwise disjoint. Then

E [ n≥1 Bn ! =X n≥1

E(Bn), (3.1)

where P

n≥1

E(Bn) refers to the strong limit of P

1≤n≤N

E(Bn)

!

N∈N

.

Remark 3.10. We use the same notation E(·) both for the spectral reso-lution and the projection-valued measure, since we can define a projection valued measure by setting E((−∞, λ]) equal to Eλ and extending to the

whole sigma-algebra, or define a spectral resolution by setting Eλ equal to E((−∞, λ]).

Definition 3.11. (see [Win15b], Definition 2.18). Given a spectral resolu-tion (Eλ)λ∈Rand a step functionf on [a, b], we define an integral with respect

to Eλ as follows

b

Z

a

f dEλ := n

X

j=1

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Before moving on, we recommend the reader to check Section 2.2 of Chapter 2 in [Win15b] and Section 3.1 of Chapter 3 in [Tes09] in order to become familiar with the properties of the integral defined above. The following theorem of [Win15b] will be useful to understand the spectral theorem.

Theorem 3.12. (see [Win15b], Theorem 2.43). Let (Eλ)λ∈R be a spectral

resolution and f in C(_R,_C). For x in H the following is equivalent:

• ∞ R

−∞

f(λ) dEλx:= lim a→−∞blim→∞

b

R

a

f(λ) dEλx exists.

• f belongs to L2(R, dhEλx, xi), that is,

∞ R

−∞

|f(λ)|2 _d_h_E

λx, xi exists.

Now we state the spectral theorem for selfadjoint operators, taken from [Win15b] and [Tes09].

Theorem 3.13. (see [Win15b,Tes09], Theorems 2.46 and 3.7 respectively).

Given a selfadjoint operator A(H → H), there exists a spectral resolution

(Eλ)λ∈R such that

∞ Z

−∞

λ dhEλx, xi=hAx, xi, x∈ D(A).

The integral R(·)dhEλx, xi is defined with respect to the finite variation given

by the function λ7→ hEλx, xi.

Notation 3.14. Let A(H → H) be a selfadjoint operator with spectral resolution Eλ and projection valued measure E(·). If B is a Borel set of R,

we set LB(A) := Rg(E(B)).

We give some useful corollaries of the spectral theorem. From now on the operator A(H → H) will be selfadjoint and (Eλ)λ∈R will be its spectral

resolution.

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R we have that Eλx is in D(A). Moreover

D(A) =  



x∈H : ∞ Z

−∞

λ2 dhEλx, xi<∞

 



and for all x in D(A)

Ax= ∞ Z

−∞

λ dEλx.

Proof. See Theorem 2.47 of [Win15b].

Corollary 3.16. For all Borel set B of _R, and all x in D(A) we have that

Z

B

λ dEλx = E(B)Ax=AE(B)x.

Proof. See [Win15b].

The following corollary is a consequence of corollaries 3.15 and 3.16.

Corollary 3.17. If B is a bounded Borel set of _R, then LB(A)⊂ D(A).

Corollary 3.18. If B 6=∅is a Borel set of_R, thenD(A)is dense inLB(A). Proof. IfB is bounded then the assertion follows from the previous corollary. Otherwise, set a:= infB and b:= supB. By equation (3.1) the strong limit of E(B ∩[−n, n]) as n tends to infinity, is E(B). Therefore the assertion follows since L[−n,n](A)⊂ D(A) by the previous corollary.

Remark 3.19. The previous corollary shows that the operator A(LB(A)→

LB(A)) is well defined.

Theorem 3.20. (see [Win15b], Theorem 2.52). The following is equivalent

• Spectrum:

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– For all >0, dimE((λ−, λ+))6= 0.

• Point spectrum:

– λ∈σd(A).

– dimE({λ})6= 0.

• Essential spectrum:

– λ∈σd(A).

– For all >0, dimE((λ−, λ+)) =∞.

• Discrete spectrum:

– λ∈σd(A).

– 0<dimE({λ})<∞.

Proof. See Theorem 2.52 of [Win15b] and Section 6.4 of [Tes09].

Remark 3.21. The previous theorem also tells us that the “measure” of the resolvent set of A under E(·) is zero.

Corollary 3.22. Let a < b in _R∪ {−∞,∞}. If dim(E((a, b))) < ∞, then

(a, b)∩σ(A)⊂σd(A), |(a, b)∩σ(A)|<∞ and L(a,b)(A)⊂ D(A).

Proof. The claim follows from the previous theorem and corollary 3.17. Corollary 3.23. Let a < b in _R∪ {−∞,∞}. If dim(E((a, b))) < ∞ or both a and b are real numbers, then the operator A(L(a,b)(A)→L(a,b)(A)) is

bounded.

Proof. • Case 1: dim(E((a, b))) < ∞. Set m := min(a, b)∩σ(A) and

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Now note that for all xin D(A)∩L(a,b)(A) we have

kAxk2 = ∞ Z

−∞

λ2 dhEλx, xi= M

Z

m

λ2 dhEλx, xi

≤

M

Z

m

max{m2, M2} dhEλx, xi

= max{m2, M2}kxk2_.

The second equality follows from remark 3.21.

• Case 2: a and b are real numbers. The proof is the same as in the previous case setting m:=a and M :=b.

Corollary 3.24. Consider A(H →H) a selfadjoint operator, k in _N∪ {∞}

and (a, b) such that −∞ ≤ a < b ≤ ∞ and 0 ∈/ (a, b). If dimL(a,b)(A) =k

then dim(A(L(a,b)(A)∩ D(A))) =k.

Proof. The proof will be split into several cases. • Case 1: k <∞.

Notice that dim ker(A−λ) = dimA(ker(A−λ)) for allλinσd(A)\ {0}.

Additionally, by corollary 3.22 we know that σ(A)∩(a, b) ⊆ σd(A) is

finite. Therefore

dim(A(L(a,b)(A)∩ D(A))) = dim(A(

M

λ∈σ(A)∩(a,b)

ker(A−λ)))

= X

λ∈σ(A)∩(a,b)

dim(A(ker(A−λ)))

= X

λ∈σ(A)∩(a,b)

dim(ker(A−λ))

= dim( M

λ∈σ(A)∩(a,b)

ker(A−λ))

= dimL(a,b)(A) = k.

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• Case 2: k =∞ and there exists λ ∈(a, b) such that dim(E{λ}) =∞.

Choose (vj)j∈N ⊆ker(A−λ) such that they are orthonormal andAvj 6=

0 for allj in_N. Therefore

dim(A(L(a,b)(A))) ≥dim(A(span{vj})) =∞.

• Case 3: k =∞ and there is no λ ∈(a, b) such that dim(E{λ}) =∞.

The hypothesis implies that there has to be an accumulation point of the spectrum in (a, b). Therefore we can find a discrete sequence (λj)j∈_N and j >0 such that Bj(λj)∩Bi(λi) =∅ for all j 6=i and 0 is not in

Bj(λj). For each j in N choose uj ∈Rg(E(λj−j,λj+j)) with kujk= 1. SoAuj belongs to Rg(E(λj−j,λj+j)) for allj inN, and Auj 6= 0. Then the Auj are pairwise orthogonal and

dim(A(L(a,b)(A)))≥dim(span{(Auj)j∈N}) = ∞.

This finishes the proof.

3.3 Sesquilinear Symmetric Forms

Definition 3.25. Let H be a complex Hilbert space and D(q) a subspace of H. Asesquilinear symmetric form q is a function from D(q)× D(q) to _C with the following properties:

• q is linear in the first component and anti-linear in the second compo-nent. This property is calledsesquilinearity.

• qhas the property that for allx, yinD(q) we have thatq[x, y] =q[y, x]. This property is called symmetry.

Notation 3.26. To simplify notation, we setq[x] := q[x, x] for allxinD(q).

Example 3.27. The inner product h·,·i of H is a symmetric sesquilinear form.

Definition 3.28. The numerical range of a form q is denoted as W(q) and it is defined as

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Closed and Closable Forms

Now we announce the definitions of sectorial, closed and closable forms and a few corollaries associated with these properties.

Definition 3.29. A symmetric sesquilinear form q is said to be sectorial

with angle ϑ in 0,π₂ and shift γ in_R if and only if W(q) is a subset of the sector

{η∈_C : |arg(η−γ)| ≤ϑ}.

Example 3.30. Define the form

q :l2(N)×l2(N)→C, (x, y)7→ hSx, yi

where S is the operator defined in (2.2). The form q is sectorial since S is bounded and selfadjoint. In this case, the sector with both shift and angle equal to 0 contains W(q).

Definition 3.31. A sectorial symmetric sesquilinear form q with domain D(q) is said to be closed if and only if it satisfies the following property: If a sequence vn converges to some element v of H and q[vn−vm] converges

to zero as n and m tend to infinity, then v belongs to D(q) and q[v −vn]

tends to zero as n tends to infinity.

Definition 3.32. A sectorial sesquilinear form q with domain D(q) is said to be closable if and only if it satisfies the following property:

if a sequence vn converges to zero and q[vn−vm] converges to zero as n and m tend to infinity, then q[vn] tends to zero as n tends to infinity.

Note that this is equivalent to the fact that q admits a closed extension.

Definition 3.33. The closure of a closable form q will be denoted asq. It is defined as follows:

D(q) := {x∈H : ∃(xn)n∈N ⊂H, xn →xand lim_n_→∞ m→∞

q[xn−xm] = 0},

and for x, y in D(q) we set

q[x, y] := lim

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Remark 3.34. By Theorem 1.12, Chapter VI of [Kat95], the form q is well defined. Moreoverq is closed, and if a formwis a closed extension of q, then

w is an extension ofq (see theorem 1.17, Chapter VI of [Kat95]).

Example 3.35. (see [KD14], Example 29.18). Let H=L2([−1,1]), D(q) =

C([−1,1]) andq[f, g] :=f(0)g(0). The formq is not closable.

Proof. Let fn(x) = (1 +x2)−n, then fn → 0 in L2([−1,1]) by the

domi-nated convergence theorem (see [Els05]). Moreover, q[fn −fm] = (1−n −

1−m)(1−n₋₁−m_{) = 0 for all}_{n, m}_∈

N, butq[fn] = 1 for alln∈N. Therefore q is not closable.

The next lemmas will be useful later. From now on, q will be a symmetric sesquilinear form.

Lemma 3.36. (see [Kat95], Chapter VI, Theorem 1.18) If q is a closable form, then W(q) is dense in W(q)

Proof. Letq[y] in W(q) wherey belongs to D(q). Therefore, by definition of

q there exists (xn)n∈_N ⊂ D(q) such that xn → y and q[y] = lim

n→∞q[xn]. This finishes the proof since (q[xn])n∈N⊂W(q).

Lemma 3.37. LetH1 be a closed subspace of H. If q is closable andD(q)is

dense in H1, then the form q1 := q D(q)∩H1×D(q)∩H1 is closable. Notice that

we consider q1 as a form on H1.

Proof. Let (xn)n∈N ⊂ D(q)∩H1 such that xn → 0 and q1[xn−xm] → 0 as

n, m tend to infinity. Therefore, since q is closable, q1[xn] = q[xn]→ 0 as n

tends to infinity.

Lemma 3.38. Let A(H → H) be a selfadjoint operator and set a[x, y] := hAx, yi. Additionally, letA1 :=AL[0,∞)(A), let a1 be its corresponding closed

form and assume that k−:= dimL(−∞,0)(A) is finite. Then

D(a1) =D(a)∩L[0,∞)(A). (3.2)

Proof. Notice thatA1 is well defined by remark 3.19, and that a1 is closable by lemma 3.37. The proof of3.2 will follow by double inclusion.

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• “D(a1)⊆ D(a)∩L[0,∞)(A)”

Letx∈ D(a1). Then there exists a sequence (xn)n∈N⊂ D(A1)⊂ D(A)

such that xn converges to x and a1[xn −xm] converges to 0. So by

definition of a, x ∈ D(a) and x∈ L[0,∞)(A), since (xn)n∈N ⊂ D(A1)⊂

L[0,∞)(A) and L[0,∞)(A) is closed. • “D(a)∩L[0,∞)(A)⊆ D(a1)”

The idea of the proof of this inclusion is that given a fixedx inD(a)∩

L[0,∞)(A), there exists a sequence (yn)n∈N⊂ D(a)∩L[0,∞)(A) such that

yn→ x and lim_n_→∞ m→∞

a1[yn−ym]→0. So, by definition of D(a1), x would

belong to D(a1).

Now let x ∈ D(a) ∩ L[0,∞)(A) fixed. Then there exists a sequence (xn)n∈N⊂ D(A) such that xn converges to x and a[xn−xm] converges

to 0 as n, m tend to infinity. Now set

H+:=L[0,∞)(A),

H−:=L(−∞,0)(A).

Notice thatH+andH−are Hilbert spaces with dimH−=k− <∞and

H+ ⊥ H−. So for all n ∈N, xn can be written as xn =x+n +x

−

n with x+_n in H+ and x−n in H−. Additionally, H− ⊂ D(A) by corollary 3.22. So x−_n belongs to D(A)⊂ D(a) for all n∈_N. Therefore x+

n belongs to

D(A)⊂ D(a) for all n ∈_N.

Now we want to prove that x+_n → x and lim

n→∞

m→∞

a1[x+n −x+m] → 0. By

definition of D(a1),x belongs toD(a1). Since x∈L[0,∞)(A), we have that

kx−xnk2 =kx+n −xk

2

+kx−_nk2.

By the equality lim

n→∞kx−xnk

2 _{= 0, it follows that lim}

n→∞kx +

n −xk2 = 0

and lim

n→∞kx −

nk2 = 0. Therefore it only remains to prove that

lim

n→∞

m→∞

a1[x+n −x

+

m] = 0.

For that, notice

a[xn−xm] = hA(xn−xm), xn−xmi

= hA(x−_n −x_m−), x−_n −x−_mi+hA(x+_n −x_m+), x+_n −x+_mi +hA(x+_n −x_m+), x−_n −x−_mi+hA(x−_n −x−_m), x+_n −x+_mi.

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But

hA(x+_n −x+_m), x−_n −x−_mi= 0,

hA(x−_n −x−_m), x+_n −x+_mi= 0

since bothA(x+_n−x+_m) andx+_n−x+_m belong toH+and both A(x−n−x−m)

and x−_n −x−_m belong to H− by corollary 3.16.

Additionally,A H− is bounded by corollary 3.23. Therefore

lim

n→∞|hA(x −

n −x

−

m), x

−

n −x

−

mi| ≤_nlim_→∞kAH− k kx

−

n −x

−

mk

2

≤ kAH− k lim

n→∞ kx −

n −x

−

mk

2

= 0.

So

0 = lim

n→∞a[xn−xm] = lim

n→∞hA(x +

n −x

+

m), x

+

n −x

+

mi

= lim

n→∞a1[xn−xm].

This finishes the proof.

3.4 Norm Resolvent Topology

The idea of this section is to introduce both norm resolvent convergence and generalized convergence and the fact that under certain circumstances they are equivalent.

Notation 3.39. Let H be a Hilbert space. The set of all closed linear operators on H is denoted by C(H).

Definition 3.40. The generalized convergence (see page 197 of [Kat95]) is given by the following metric onC(H). LetN andM be two closed subspaces of H×H and let SM be the unit sphere of M. Then we define

δ(M, N) := sup

u∈SM

dist(u, N),

b

δ(M, N) := max{δ(N, M), δ(M, N)}.

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To show that bδ(·,·) is a metric we will show that

ˆ

δ(M, N) = kid−PM −(id−PN)k.

Proof. First set ρm,n := sup{kid − PN(x)k : x ∈ Rg(PM), kxk ≤ 1} = δ(M, N) and ρn,m analogously. Then, by definition of the operator norm we

have that

kid−PM −(id−PN)k ≥ max{ρm,n, ρn,m}

= max{δ(M, N), δ(N, M)}

= ˆδ(M, N).

Now we only have to prove that kid−PM −(id−PN)k ≤max{ρm,n, ρn,m}. For this notice that for all x inH

k(id−PM −(id−PN))xk2 =k(id−PM)PNx−PM(id−PN)xk2

=k(id−PM)PNxk2+kPM(id−PN)xk2 ≤ k(id−PM)PNk2kPNxk2

+kPM(id−PN)k2k(id−PN)xk2

≤ ρ2_n,mkPNxk2

+k[PM(id−PN)]∗k2k(id−PN)xk2

=ρ2_n,mkPNxk2+k(id−PN)PMk2k(id−PN)xk2

≤ ρ2_n,mkPNxk2+ρ2m,nk(id−PN)xk2

≤ max{ρ2_n,m, ρ2_m,n}(kPNxk2+k(id−PN)xk2)

≤ max{ρ2_n,m, ρ2_m,n}kxk2_.

This finishes the poof.

Notation 3.41. LetT be inC(H), and set G(T) := {(x, y)∈H×H : x∈ D(T), y =T x}.

Notation 3.42. Let T and S in C(H). We set δ(T, S) := δ(G(T), G(S)) and bδ(T, S) := bδ(G(T), G(S)). Note thatG(T) is the graph of T.

Notation 3.43. Let H be a Hilbert space. The set of all bounded linear operators on H is denoted as B(H).

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Now we mention a couple of very nice lemmas that will be helpful to under-stand the main theorems of this section.

Lemma 3.44. (see [Kat95], Chapter IV, Lemma 2.11 ). LetT be an element of B(H). If S is in C(H) and δ(S, T)<(1 +kTk2₎−1

2, then S is bounded.

Proof. See ChapterIV, Lemma 2.11 of [Kat95].

Theorem 3.45. (see [Kat95], Chapter IV, Theorem 2.13). Let T be in

B(H). If S is in C(H) andδb(T, S)<(1 +kTk2)−

1

2, thenS belongs to B(H).

Proof. See ChapterIV, Lemma 2.13 of [Kat95].

Remark 3.46. A nice consequence of the previous theorem is that the set B(H) is open in C(H) with the metric bδ(·,·).

Theorem 3.47. (see [Kat95], Chapter IV, Theorem 2.20). If T and S in

C(H) are invertible, then δ(S−1, T−1) = δ(S, T) and bδ(S−1, T−1) = bδ(S, T).

Proof. See Chapter IV, Theorem 2.20 of [Kat95].

Theorem 3.48. (see [Kat95], Chapter IV, theorem 2.21). Let T in C(H)be invertible withT−1 in B(H). If S is inC(H) withbδ(T, S)<(1 +kT−1k2)−

1 2,

then S is invertible and S−1 is in B(H).

Proof. IfS is invertible, then the boundedness of S−1 follows from theorem

3.45and the previous theorem. So we only have to prove thatS is invertible.

Suppose by contradiction that S is not invertible. Then there exists u in D(S), such that kuk= 1 and Su = 0. Therefore, (u,0) belongs to the unit sphere ofG(S), so there is a (v, T v) inG(T) such thatku−vk2₊_k_{T v}_k_{< β}2_,

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where β is a number such that bδ(S, T)< β <(1 +kT−1k2)−

1

2. Hence

1 = ku2k

≤ (ku−vk+kvk)2

≤ (ku−vk+kT−1kkT vk)2

≤ ku−vk2+ (kT−1kkT vk)2 + 2ku−vkkT−1kkT vk ≤ ku−vk2_{+ (k}_T−1_kk

T vk)2 _{+ (k}_T−1_kk

u−vk)2 ₊_k_{T v}_k2 ≤ (1 +kT−1k2_)(k_u₋_v_k2₊_k_{T v}_k2₎

≤ (1 +kT−1k2₎_β2

< 1,

which is a contradiction.

Theorem 3.49. (see [Kat95], Chapter IV, Theorem 2.23). Let T, Tn in

C(H) with n in _N.

• Assume that T is in B(H). Then Tn → T in the generalized sense if and only if Tn∈B(H) for sufficiently large n and kTn−Tk →0.

• If T−1 exists and belongs to B(H), then Tn → T in the generalized

sense if and only if T_n−1 exists and belongs to B(H) for sufficiently large n and kT_n−1−T−1k →0.

• IfTn→T in the generalized sense and A∈B(H), thenTn+A→T+A

in the generalized sense.

Now, by the previous theorem we have that under certain conditions the norm resolvent convergence and the generalized convergence are the same. This will be useful, since in this text the conditions of the following theorem will be fulfilled, and therefore we can use all the results on generalized convergence.

Theorem 3.50. (see [Kat95], Chapter IV, Theorem 2.25). Let T ∈ C(H)

have nonempty resolvent set ρ(T). In order that a sequence Tn ∈ C(H)

converges to T in the generalized sense, it is necessary that each η ∈ ρ(T)

belongs to ρ(Tn) for sufficiently large n andkR(η, Tn)−R(η, T)k →0, while

it is sufficient that this is true for some η ∈ρ(T).

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Now we introduce several useful lemmas and theorems.

Lemma 3.51. Let T and (Tn)n∈_N be selfadjoint operators and λ in R. If

Tn →T in the norm resolvent sense and for all n in N, λ is in σ(Tn), then λ is in σ(T).

Proof. Suppose by contradiction that λis in ρ(T)). Then (T −λ)−1 belongs to B(H). Therefore, by theorem 3.49, we have that for sufficiently large n, the operator (Tn−λ)−1 belongs toB(H). This contradicts thatλis inσ(Tn)

for all n in _N.

Theorem 3.52. (see [Kat95], Chapter IV, Theorem 3.16). LetT in C(H)be selfadjoint and let σ(T) be separated into two parts σ1(T), σ2(T) by a simple

rectifiable closed curve Γ. Let H = Lσ₁(T)(T)LLσ2(T)(T) be the associated

decomposition of H due to the spectral resolution of T. Then there exists a

δ >0, depending onT andΓ, with the following properties. Any selfadjointS

inC(H)withbδ(S, T)< δ, has spectrumσ(S)likewise separated byΓ into two

partsσ1(S), σ2(S) (Γrunning inρ(S)). In the associated decompositionH =

Lσ1(S)(S)

LLσ

2(S)(S), Lσ1(S)(S) and Lσ2(S)(S) are respectively isomorphic

with Lσ₁(T)(T) and Lσ2(T)(T). In particular dimLσ1(S)(S) = dimLσ1(T)(T)

and dimLσ₂(S)(S) = dimLσ2(T)(T).

Proof. See page 212 of [Kat95].

Lemma 3.53. Let T and (Tn)n∈N be selfadjoint operators and λ in R. If

Tn → T in the norm resolvent sense and if λ is in σess(Tn) for all n in N,

then λ is in σess(T).

Proof. By lemma 3.51 we have that λ is in σ(T). Assume by contradiction that λ is in σd(T). Additionally, Let Γ be a simple rectifiable closed curve

that separates λ from σ(T)\ {λ}. By the previous theorem

dimL{λ}(T) = dimLσλ(Tn)(Tn),

for sufficiently largen, where σλ(Tn) is the part of the spectrum of Tn inside

Γ. Moreover, dimLσλ(Tn)(Tn) = ∞ since λ is in both σλ(Tn) and σess(Tn).

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Chapter 4 Generalized Variational

Principle

Before passing to the main theorem, we will introduce some useful notation and definitions.

Notation 4.1. ∆⊆_Rwill be an open, half-open or closed interval with end points α and β, such that −∞ ≤α < β≤ ∞.

Notation 4.2. The set of closed operators with domain and range in H is denoted as C(H).

Definition 4.3. A function T : Ω⊂_C→C(H) is called a pencil.

Definition 4.4. Given a pencil T, we define t0(λ)[x, y] := hT(λ)x, yi and

t0(λ)[x] := hT(λ)x, xi, where x, y belong to D(T(λ)) and λ belong to Ω. If

t0(λ)[·,·] is closable we denote its closure by t(λ)[·,·].

Definition 4.5. The spectrum, essential spectrum and discrete spectrum of a pencil T is defined respectively as

σ(T) :={λ ∈Ω : 0 ∈σ(T(λ))}, σess(T) :={λ ∈Ω : 0 ∈σess(T(λ))},

σd(T) :={λ ∈Ω : 0 ∈σd(T(λ))}.

Notation 4.6. From now on T is a pencil with domain ∆ ⊂ _R and such

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that T(λ) is selfadjoint for all λ in ∆.

Notation 4.7. For all λin ∆, we denote the dimension ofL(−∞,0)(T(λ)) by

k−(λ).

Now we will introduce the main hypotheses that will be needed to demon-strate the lemmas of the next section and the main theorem.

(A1) Either D(T(λ))≡ D is independent of λ

or the form t0(λ)[·] is closable and there exists a dense subspace of H such that

D(T(λ))⊂ D ⊆ D(t(λ)), ∀λ∈∆.

(A2) The pencil T is continuous in the norm resolvent topology. Moreover, if the first case of (A1) happens, then the function λ 7→ t0(λ)[x], is continuous in ∆, for every x∈ D fixed. If the second one occurs, the same would be true for the function λ7→t(λ)[x].

(A3) For all x ∈ D with x 6= 0, the function t0(·)[x], or t(·)[x] if the second case of(A1)occurs, is decreasing at value zero in ∆, i.e., ift(λ0)[x] = 0 for someλ0 ∈∆, this implies

t0(λ)[x]>0 for λ < λ0,

t0(λ)[x]<0 for λ > λ0.

(A4) There exists γ ∈∆ such that dimL(−∞,0)(T(γ))<∞.

In lemma4.24 an additional assumption is used.

(A5) For allλ ∈∆ and >0 such thatλ+ < β, there existsδ=δ(λ, )>0, such that if 0< t0(λ)[x]< δforx∈ Dwith kxk= 1, thent0(λ+)[x]≤ 0. If the second case of (A1)occurs, then assume the above for t(·)[·].

Remark 4.8. An important observation that will be frequently used is that under assumptions (A1)-(A3), λ1 < λ2 with λ1 and λ2 in ∆ and x in D we have that: if t(λ1)[x] ≤ 0 then t(λ2)[x] < 0. Otherwise there would exist

λ > λ1 such that t(λ)[x] = 0 due to the continuity of t(·)[x], given by (A2), and this is a direct contradiction with (A3).

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Chapter 4. Generalized Variational Principle 29

4.1 Auxiliary Lemmas

In order to prove the main theorem we will use the following lemmas.

Lemma 4.9. Let A be a selfadjoint operator in H. Assume that the form

a[x, y] := hAx, yi is closable and denote its closure also by a. Let D(A) ⊂ D ⊂ D(a), where D(A) and D(a) respectively denote the domains of A and

a. Then k0 := dimL(−∞,0)(A) is equal to the dimension of every maximal

subspace of the set

N :={x∈ D:a[x]<0} ∪ {0}.

Proof. Let Lbe a maximal subspace of N. First we will show that dimL≥

k0. For this assume by contradiction that dimL < k0. By corollary 3.24, we have that dim Rg(A _L(−∞_,₀₎(A)∩D(A)) = k0. So, by lemma 3.6 we have that there exists x inL(−∞,0)(A)∩ D(A) with x6= 0, such that hAx, yi= 0 for all

y in L. This implies

L+ span{x} ⊂ N,

and x is linearly independent of L. This contradicts the fact thatLis maxi-mal in N.

Now we will show that dimL ≤ k0. For this assume by contradiction that dimL > k0. Therefore, again by lemma 3.6, there exists x ∈ L∩L[0,∞)(A) withx6= 0. LetA1 =AL[0,∞)(A)and leta1be the corresponding closed form

of A1. Notice that A1 is well defined by remark 3.19 and that a1 is closable by lemma 3.37.

Next, by lemma3.38we have thatD(a1) = D(a)∩L[0,∞)(A). So according to lemma3.36,W(A1), the numerical range ofA1, is dense inW(a1) and there-fore W(a1) =W(A1) ⊂[0,∞). This implies that a1[x]≥0, but x∈L ⊂ N and that contradicts the definition of N.

Before we pass to next lemma we will need the following definition.

Definition 4.10. For all λ in ∆ define

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Lemma 4.11. Let T be an operator pencil satisfying the conditions (A1)

-(A4) and let [α1, α2]⊂∆. Assume that k−(α1)<∞ and that

σess(T)∩[α1, α2] =∅.

Then forλ∈[α1, α2], the function which assignsλtok−(λ)is non-decreasing,

left-continuous and constant between eigenvalues ofT. The eigenvalues ofT

cannot accumulate in [α1, α2]. Moreover lim

λ&µk−(λ)−k−(µ) = dim kerT(µ) (4.1)

for µ∈[α1, α2).

Proof. Denote

k(λ) := dim kerT(λ)

and notice that k(λ) is finite for every λ∈[α1, α2], since

σess(T)∩[α1, α2] =∅ =⇒ ∀λ∈[α1, α2] λ6∈σess(T(λ)).

=⇒ 06∈σess(T(λ))

=⇒ 0∈σd(T(λ))∪ρ(T(λ)).

The next two arguments will complete the proof:

(i) If λ1 < λ2 then NT(λ1)⊕kerT(λ1) is a subspace of NT(λ2).

(ii) There is an interval (δ1, δ2) around every elementλin [α1, α2] such that

k−(·) is constant on (δ1, λ] and on (λ, δ2). The difference between the valuek−(·) on (δ1, λ] and the value of k−(·) on (λ, δ2) is k(λ).

Let us check that the assertions follow from these arguments. Letλ1 < λ2 in ∆. Then, by lemma4.9, both k−(λ1) and k−(λ2) are equal to the dimension of any maximal subspace ofNT(λ1) andNT(λ2)respectively. This implies that

k−(λ1) +k(λ1) = dimNT(λ1)⊕kerT(λ1)

≤dimNT(λ2)

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Note that the inequality in the second line follows from argument (i). Equa-tion (4.1) and the left-continuity ofk−(·) follow directly from argument (ii). The fact that k−(·) is constant between eigenvalues ofT and that the eigen-values of T cannot accumulate in [α1, α2] follows from argument(ii) and the fact that [α1, α2] is compact.

Now we just have to show that (i)and (ii) are true. To prove (i)first notice that, for all x in D if t(λ1)[x] is negative or zero, then we have that t(λ)[x] is negative for all λ bigger thanλ1. This follows from remark 4.8. Therefore

t(λ2)[x]<0 for all x inNT(λ1)⊕kerT(λ1), which implies that

NT(λ1)⊕kerT(λ1)⊂ NT(λ2).

Notice that NT(λ1) ⊕ kerT(λ1) is indeed a direct sum because NT(λ1) ∩

kerT(λ1) = {0} by definition of NT(λ1).

Now we prove argument (ii). Take µ in [α1, α2] and let > 0 such that [−,0)∪(0, ] ⊂ ρ(T(µ)). This can be done since 0 ∈ σd(T(µ))∪ρ(T(µ))

and the resolvent set is open. Now separate σ(T(µ)) ∩ (−∞,−) from

σ(T(µ))∩(−,∞) with a simple rectifiable closed curve Γ. This can be done since σ(T(µ))∩(−∞,−) is finite by corollary 3.22 . Therefore by lemma

3.52and(A2)there existsδ1 >0 such that for allλin (µ−δ1, µ+δ1)∩[α1, α2]

dimL(−∞,−)(T(µ)) = dimL(−∞,−)(T(λ)).

This implies that

k−(λ)≥dimL(−∞,−)(T(λ)) =k−(µ).

Therefore for all λ in (µ−δ1, µ)∩[α1, α2]

k−(λ) +k(λ)≤k−(µ)≤k−(λ).

So k−(λ) =k−(µ) and k(λ) = 0.

Now let Γ be a rectifiable closed simple curve that separates σ(T(µ))∩ (−∞, ) from σ(T(µ))∩(,∞). Using lemma 3.52 and (A2)we know again that there exists δ2 >0 such that for all λ in (µ−δ2, µ+δ2)∩[α1, α2]

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Therefore for allλ in (µ, µ+δ2)∩[α1, α2]

k−(λ) +k(λ)≤dimL(−∞,)(T(λ)) =k−(µ) +k(µ),

and

k−(λ) +k(λ)≤k−(µ) +k(µ)≤k−(λ).

So we have thatk−(λ) =k−(µ) +k(µ). This finishes the proof of (ii) and of the lemma.

Before moving on, we will need some definitions.

Definition 4.12. For n≤N

m(n) := min{k ∈_N:λk =λn}, M(n) := max{k ∈_N:λk=λn}.

Definition 4.13. Recall that according to definition 4.1, β is the right end point of ∆ andα is the left end point, so define

Nβ :={x∈ D :∃λ ∈∆ such that t(λ)[x]<0} ∪ {0}.

Remark 4.14. Observe that Nβ = S λ∈∆

Nλ.

Definition 4.15.

λe :=

(

infσess(T) if σess(T)6=∅,

β if σess(T) =∅.

Definition 4.16. Assumption (A3) implies that for all x in D \ {0} the function t(·)[x] has at most one zero in ∆. If a zero exists, that zero will be denoted byp(x). Otherwise we define

p(x) :=  



−∞ if t(λ)[x]<0 ∀λ ∈∆,

∞ if t(λ)[x]>0 ∀λ ∈∆.

Remark 4.17. Notice that by definition of p(x)

p(x)> λ ⇔ t(λ)[x]>0, p(x)< λ ⇔ t(λ)[x]<0.

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Definition 4.18.

µn:= inf L⊂D dimL=n

sup

x∈L x6=0

p(x), (4.2)

µ0_n:= sup

L⊂H

dimL=n−1 inf

x∈D,x6=0

x⊥L

p(x). (4.3)

Definition 4.19. Set

∆0 :={λ∈∆ : λ <infσess(T)}.

Remark 4.20. By lemma 4.11 we have that for all > 0 the number of eigenvalues of T in [α+, λe−] is finite. Moreover, the eigenvalues of T in

[α+, λe−] is a discrete set. Therefore ∆0 would be countable and discrete since it is the countable union of all the eigenvalues of T in sets of the form [α+_n1, λe− _n1] for n in N.

Lemma 4.21. If λ1 ≤λ2 ≤. . . are the eigenvalues of T in ∆

0

, then

dimL(−∞,0)(T(λn)) = k+m(n)−1,

dimL(−∞,0](T(λn)) = k+M(n).

Proof. By Lemma 4.11 we have that dimL(−∞,0)(T(λn)) = k−(λn),

= k+ number of eigenvalues of T

less than λn counted with multiplicity,

= k+m(n)−1,

and

dimL(−∞,0](T(λn)) = k−(λn) + kerT(λn)

= k+m(n)−1 + (M(n)−m(n) + 1).

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Lemma 4.22. IfLis a finite dimensional subspace of D, then the supremum

sup

x∈L x6=0

p(x) is attained.

Proof. If there is anxinLsuch thatp(x) = ∞, then the claim is clear. Oth-erwise denote the supremum of thep(x) by µ. Then there exists a sequence (xk)k∈N ⊂ L such that p(xk) → µ. Notice that we can take this sequence

such that kxkk= 1 by remark 4.17. Since dimL < ∞, the unit sphere of L

is compact and therefore we can assume that xk → y for some y in L with

kyk= 1. We will show that the supremum is attained in y.

By definition ofµwe have that p(y)≤µ. Now assume by contradiction that

p(y)< µ. Then there is aλ0 with p(y)< λ0 < µ. So there exists k0 ∈Nsuch that p(xk) ≥ λ0 for every k ≥ k0, or equivalently t(λ0)[xk] ≥ 0 by remark 4.17. But we also have that t(λ0)[y] < 0 which is a contradiction to the continuity of the formt(λ0)[·] in the finite dimensional subspace L.

Remark 4.23. By the previous lemma notice that in definition (4.2) we can change sup by max.

Lemma 4.24. Assume that T satisfies the assumptions (A1)-(A5). If λe < β, then k−(λ) =∞ for every λ > λe.

Proof. By definition of λe there exists (λn)n∈N ⊂σess(T) such that λn &λ.

Therefore, T(λn) → T(λe) in the norm resolvent sense and 0 belongs to σess(T(λn)) for all n in N. Therefore, 0 belongs to σess(T(λe)) by the

corollary3.53. Hence, dimL(−η,η)(T(λe)) = ∞ for allη > 0. Which implies

that either dimL(−η,0)(T(λe)) = ∞ or dimL[0,η)(T(λe)) = ∞. Therefore

either the value k−(λe) = ∞ or dimL[0,η)(T(λe)) =∞.

Now let λ in ∆ such that λe < λ < β. The proof of this lemma will

fol-low by the cases befol-low.

• Case 1: k−(λe) = ∞.

By remark 4.8 we know that Nλe ⊂ Nλ. Hence, by lemma 4.9 this implies that

dimL(−∞,0](T(λe))≤dimL(−∞,0](T(λ)).

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• Case 2: for all η >0 dimL[0,η)(T(λe)) =∞.

The idea of this case is to prove that there exists η >0 such that

L[0,η)(T(λe))⊂ Nλ. (4.4)

By lemma 4.9 and the assumption of this case we have

∞= dimL(−∞,0](T(λe))≤k−(λ).

So the only thing missing is the proof of formula (4.4) for some η >0. For this we will need assumption (A5)and we recall it here:

– For all λ0 ∈∆ and >0 such that λ0+ < β, there exists δ >0 depending on λ0 and , with the following property:

if 0< t(λ0)[x]< δ for x∈ D, kxk= 1 then t(λ0+)[x]≤0.

Now considering λ0 = λe and = ₂1(λ−λe), we take η equal to the δ

given by(A5). So we just have to check equation (4.4). Notice that for all x in L[0,δ)(T(λe))⊂ D, we get t(λe)[x] < δ. Therefore, by (A5) we

have that for all x inL[0,δ)(T(λe))⊂ D

0≥t λe+1₂(λ−λe)

[x] =t 1₂(λ+λe)

[x].

By remark 4.8 we obtain t(λ)[x]<0, as desired.

4.2 Main Theorem

Theorem 4.25. Assume that T satisfies the assumptions (A1)-(A4) and that ∆0 is not empty. If ∆ is closed at the left end point, set k := k−(α).

Otherwise there exists an α0 ∈ ∆0 such that (α, α0) ⊂ ρ(T) and we set

k :=k−(α0). In both cases, k is a finite number.

Then σ(T)∩∆0 consists only of a finite or infinite sequence of isolated eigen-values λ1 ≤λ2 ≤λ3 ≤ · · · ≤λN, N ∈N0 ∪ {∞}, counted with their

multi-plicity, and for n= 1, . . . , N :

λn=µk+n = min L⊂D dimL=k+n

max

x∈L x6=0

p(x), (4.5)

λn=µ0k+n = max_L_⊂_H

dimL=k+n−1 inf

x∈D,x6=0

x⊥L

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Moreover

N =

(

k−(β)−k+ dim kerT(β) if β ∈∆ and σess(T) =∅,

k−(λe)−k otherwise. (4.7)

(i) If N =∞, then lim

n→∞λn =λe.

(ii) If N <∞ and σess(T) = ∅, then µn=∞ for n > k+N.

(iii) If N < ∞ and λe < β (which implies σess(T) 6= ∅), and assumption

(A5)is fulfilled, then µn=µ0n =λe for n > k+N.

Proof. Notice that the assertion about σ(T)∩∆0 follows from remark4.20. Now we will show (4.5) and (4.6). First we will prove that λn ≤ µk+n. Let Lbe a subspace of H with dimL=k+n. By lemma 4.21 we have

dimL(−∞,0)(T(λn)) =k−(λn) = k+m(n)−1<dimL.

Moreover, by lemma 3.6 there exists x0 ∈ L∩L[0,∞)(T(λn)) with x0 6= 0. This implies that t(λn)[x0] ≥ 0 and therefore p(x0) ≥ λn. Since L was an

arbitrary subspace of D of dimension n+k we have that max

x∈L x6=0

p(x) ≥ λn.

Hence

inf

L⊂D dimL=k+n

max

x∈L x6=0

p(x)≥λn.

Now we will prove that λn ≥µ0k+n. Let L be a subspace ofH with dimL= n+k−1. Using lemma 4.21 we see

dimL(−∞,0](T(λn)) =k+M(n)> k+n−1 = dimL.

So by lemma 3.6 there exists x0 6= 0 such that x0 ∈ L⊥ ∩L(−∞,0](T(λn)).

Notice that t(λn)[x0] is well defined since x0 is in D, which follows from corollaries 3.22 and 3.17. Sincex0 is in L(−∞,0](T(λn)) we gett(λn)[x0]≤0, which implies that p(x0)≤λn. Therefore λn ≥ inf

x∈D,x6=0

x⊥L

p(x) and

λn≥ sup

L⊂H

dimL=k+n−1 inf

x∈D,x6=0

x⊥L

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Now we only need to prove the remaining inequalities, which are λn≥µk+n

and λn ≤ µ0k+n. To do this let xm(n), . . . , xM(n) be an orthonormal basis of ker(T(λn)) and set

L(₁n) :=L(−∞,0)(T(λn)) + span{xm(n), . . . , xn},

L(₂n) :=L(−∞,0)(T(λn)) + span{xm(n), . . . , xn−1}.

Then dimL(₁n) = k+n and dimL₂(n) = k +n−1. Note that if n = m(n), then {xm(n), . . . , xn−1} is empty. In order to prove λn ≥µk+n notice that

L(₁n) ⊆L(−∞,0](T(λn))⊂ D.

Therefore,xbelongs toD(T(λn)) andt(λn)[x]≤0 for allxinL

(n)

1 . Moreover, by definition of xn we get t(λn)[xn] = 0, so

λn= max x∈L(₁n)

x6=0

p(x).

This proves λn ≥ µk+n and the fact that the infimum in the definition of µk+n is attained.

Now we will prove that λn≤µ0k+n. By definition of L

(n)

2 we have

L⊥₂ ⊂L[0,∞)(T(λn))

which implies that for all x in (L(₂n))⊥∩ D, we get t(λn)[x] ≥ 0. Moreover, p(x)≥λnby remark4.17. By definition ofxnwe also have thatt(λn)[xn] = 0,

so p(x) =λn and therefore

λn = min x∈D x6=0

x⊥L2

p(x).

This shows λn ≤ µ0k+n and that the supremum in the definition of µ

0

k+n is

attained. Now we will show (4.7). The proof is divided in several parts.

• Case 1: β ∈∆and σess(T) = ∅.

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• Case 2: β /∈∆ and σess(T)6=∅.

By lemma4.9, we only have to prove that the dimension of any maximal subspace ofNλe is equal tok+N. LetL⊂ Nλe be a maximal subspace with dimensionM. First we will prove that M ≤k+N, and then, by contradiction we will show that the inequality cannot be strict.

– Subcase 1: M is finite.

Then according to lemma 4.22 we can define

µ:= max

x∈L, x6=0

p(x),

and by definition ofNλe we haveµ < λe. Notice that by definition of µ and remark 4.8 we have that t(µ)[x] ≤ 0 for all x in L. Therefore

n ≤dimL(−∞,0](T(µ))≤k+N.

– Subcase 2: M is infinite.

If n is infinite, the same reasoning as in the previous subcase will apply for every finite dimensional subspace of L, so the assertion in this case follows.

Now assume by contradiction that M < k+N. Therefore M would be finite and there exists a µ as before. Next notice that there exist at least n−k+ 1 eigenvalues, since n < k +N. Choose λ in ∆ with

λe> λ >max{µ, λn−k+1 }. Therefore we have L⊂ Nλ due to remark 4.8. Moreover, by the same remark, we have that

L⊂ Nλ ⊂ Nλe.

Therefore every maximal subspace ofNλe has dimension at least

k−(λ)≥k−(λn−k+1) + ker(T(λn−k+1)) = n−k+ 1 +k =n+ 1.

But L is a maximal subspace of Nλe with dimL =n. This gives us a contradiction.

Finally, we will prove the claims(i),(ii), and (iii) about N.

(i): if N =∞, then lim

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We have that the sequence (λn)n∈N is non-decreasing, cannot have an

accu-mulation point below λe and is bounded above by λe. Therefore

lim

n→∞λn =λe.

(ii): if N <∞ and σess(T) =∅, then µn =∞ for n > k+N.

This assertion will be proved by contradiction. Let n in _N with n > k+N

and assume that there exists a subspace L of D with dimension n such that

max

x∈L x6=0

p(x)<∞.

Then there existsµin ∆ such thatp(x)≤µfor allx∈L, which is equivalent to t(µ)[x]≤0 for all x∈L. So by lemma 4.9,

k−(µ)≥dimL=n > k+N.

But by lemma4.21this would imply that there are more thanN eigenvalues, and this gives as contradiction.

(iii): if N <∞, λe< β and assumption (A5) is fulfilled, then µn =µ0n =λe

for n > k+N.

Given n > k+N we have that µn ≥λe and µ0n ≥λe. Otherwise either µn or µ0_n would be an eigenvalue of T and this contradicts the fact that there are only N eigenvalues. Now we just need to prove that µn ≤ λe and µ0n ≤λe.

For µn ≤ λe let λ in (λe, β). By lemma 4.24 we can find a subspace L of L(−∞,0)(T(λ))∩ D such that dimL = n. Therefore for all x in L we have that p(x)≤λ. So

max

x∈L x6=0

p(x)≤λ.

This implies thatµn≤λ by definition ofµn. Since λwas arbitrary in (λe, β)

the relation µn≤λe follows.

For µ0_n ≤ λe let λ in (λe, β) and let L be a subspace of H such that

dimL=n−1. By lemma4.24there exists a subspaceL0ofL(−∞,0)(T(λ))∩D such that dimL0 > n−1. Therefore by lemma 3.6 there exists x0 6= 0 in

L0 ∩L⊥. Moreover, t(λ)[x0] < 0 since L0 is a subspace of L(−∞,0)(T(λ)). This implies that p(x0)≤λ and

inf

x∈D,x6=0

x⊥L

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So µ0_n ≤λ by definition of µ0_n. Since λ was arbitrary in (λe, β) the relation µ0_n ≤λe follows. This finishes the proof.

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Approximation of eigenvalues using variational principles