rre Selfadjoint operators with an

Selfadjoint operators with an inner singularity and Pontryagin

spaces

Chris Zeinstra

CQv' ^1Qt

^IiW...L

rre

Vakgroep

Wiskunde

F' 'r'!3,t Groningen ::)eok

t'11:..j0 'InforrnatJcqp,a..1

Landleven 5

Master thesis

Selfadjoint operators with an inner singularity and Pontryagin

spaces

Chris Zeinstra

Rijksuniversiteit Groningen Vakgroep Wiskunde

Postbus 800

9700 AV Groningen August 1996

Voorwoord

and when at last the work is done

don't sit down, it's time to dig another one..

Voor u ligt een scriptie die een ruim jaar nodig heeft gehad om deze vorm en omvang te krijgen. Goede dingen moeten blijkbaar rijpen.

1k wil een aantal mensen danken voor hun medewerking en luisterend oor. Ten eerste natuurlijk Prof.Dr.Ir. Aad Dijksma die mij duidelijk heeft gemaakt waar Lineaire Analyse, in het bijzonder Operatoren Theorie, over gaat. ^Verder gaat mijn dank uit naar familie en vrienden, in het bijzonder mijn ouders, Abe Douwe, Matthijs, Colinda en Fleur, for giving it a kick without knowing it somewhere in February.

Chris Zeinstra Groningen, augustus 1996

Contents

1

Introduction

²

1.1 Motivation of the problem ²

1.2 Some remarks on definitions and notation ⁴

2 The theorectical construction of Ht and the corresponding Straus extension Tt(z) associated with a restriction of the original operator A

⁶

2.1 The concept of rigging ⁶

2.2 The rigging applied to an unbounded selfadjoint operator ⁸

2.3 The spaces 1m and Tim ¹²

2.4

The generalized resolvent R(z) and the H relation

¹⁶

2.5 Selfadjoint extensions of a restriction of the H°°-relation ²⁰

2.6 The Straus-extension and Om function associated with So ²²

2.7 Calculation of nonreal eigenvalues of Ht 27

2.8 Looking back at the construction: Krein's formula ²⁸

3 The iD operator on L2(R) perturbed by a delta functional

³⁴

3.1 The transition from theory to A = iD ³⁴

3.2 The non real eigenvalues of Ht for m = 1 and A = i 38

3.3

The real spectrum ofHt form=1 and A=i

⁴⁰

A The selection of ge,, for i + j <0

A.1 The compatability condition and the first algorithm ⁴⁴ A.2 The selection of g... and

gm,k, k = 1,. ..

^{, m 46}

B The calculation of Om(z)

⁵³

B.1 Solution of the induced linear system ⁵³

B.2 Three lemmas ⁵⁸

1

Chapter 1 Introduction

1.1 Motivation of the problem

In 1991 the article "Scattering from generalized point interaction using selfadjoint extensions in Pontryagin spaces" written by J.F.Van Diejen and A.Tip appeared in Journal of Mathe- matical Physics, Volume 82, No. 8, p680-641. In this paper they aim to give (general) point interactions, which are frequently used in quantum mechanics to enable more convenient calculations, an operator theorectical basis.

Consider an unbounded selfadjoint operator A in a filbert space (fl0, (•,-)o) and give a rigourous meaning to the expression

A=A+s(,)o, s€R

^(1.1)

For example, A denotes the Hamiltonian and x is some 'local' functional, that is, a delta funtional or a derivative of the delta functional. The abstract setting of this problem is formulated as follows. Define the Hubert spaces (uk, (, ),), k

= 1,... by

=

domAk and _{(u, v)k} =

>(Ai

^{A'v)o. (1.2)}

One can define the space ILk as the dual space of uk, the space of continuous antilinear functionals, with respect to the inner product (., .)°. This process is called rigging of Hilbert spaces. In Section 2.1 and 2.2 we will show how the actual construction is made. Now fix a integer m and select an element x e li—rn—i \ li—rn. ^First suppose m —1, then one could interpreted (1.1) as follows. We first restrict A to those admissible functions çô E ¹¹¹

such that A0 =

_AI(,x,o=o and we calculate all selfadjoint extensions of this operator. It is convenient to describe all selfadjoint extensions in terms of their resolvent using Krein's formula that relates the resolvent of A to the resolvents R(z) = ^{(Ht —}

z)'

^{of the one}

parameter family of extensions Ht:

Rt(z) = (A —

z)

— Q(z)±Y(z)' ^(1.3)

with 1', = R(z), the function Q

^will be explained in Section 2.8 and T is a real function that depends on t.

In the remainder of this thesis we assume that m 1 and we would like to use (1.3). The main problem is that the defect elements R(z) lie outside the original space 11o• One ^could now interpret (1.1) as an operator on flm+1 with values in fl—rn-i and (A, t/)o =

(, A)o.

However, this point of view is not taken in [DT]. In this thesis we will adapt a procedure de- scribed in [DT] and which is due to Shondin [S1,S2]. In order to 'restore' the Kreln formula, we introduce the inner product space P, consistingof elements of the form m+>Z',JCiXi,

with ,

^R(A)m+Xj, A E C C p(A). Note that this space is invariant under the resolvent R(z). The next step is the completion of Pm into a Pontryagin space urn = lb ^{® Ctm}

by means of an isometric operator r : Pm '1m with a dense range. The additional space C' is equipped with the indefinite inner product

(c (e d)'),J)1 G)d)' ^{(eV(Om Im\(C\ (.)}

14

where G is a certain m x m hermitian matrix.

It can be shown that the matrix on the

righthandside of (1.4) has m positive and m negative eigenvalues, hence Hm is indeed a Pon- tryagin space with negative index m. The transition from Pm to H has two mathematical advantages: 11m has a natural norm topology and it contains the original space fib. We prove

in Section 2.4 that the resolvent R(z) on 7',, can be copied to an operator that admits a bounded closure R°°(z) on flm. R00(z) ^inheritsthe resolvent properties of R(z) and thus one can define a selfadjoint relation H°° = {{R°°(A)f, (I + AR°°(A))f}If C flm}. In Section 2.5 we proceed as above, we restrict H°° and construct an one parameterfamily of selfadjoint operators (P. In Section 2.8 we give a Krein formula for this particular extension. ^One could conclude that the family Ht is an interpretation of (1.1). However, in Section 2.6 we formulate a slightly different approach. In this section we define a symmetric restriction So of the original operator A, namely

50 = {{f,g}

AI(f,xj)o = 0}

^(1.5)

and we can characterize T(z), the traus projection of Ht onto fib, or

traus extension of

S0, with aid of this S0 operator:

T(z)

^{—3' So+span{{o} Aem(z)xo}}, Im(z) >0

t '16

— 'j So + span{{xo — em()o, Ao — Xem()o}}, Im(z) <0,

where em ^belongs to Sm(C+), the space of generalized Schur functions. The advantage of introducing the Straus extension is that the operator T(z) acts on the original space 110 and the information of the extension to H is contained in em(z). ^Moreover, the Straus extension can be used to find a relative simple formula in terms of e(z) ^forcalculating any nonreal eigenvalue of Ht. This is done in Section 2.7.

Chapter 3 is devoted to an application of the theory of Chapter 2, we will study Ht in the case A = iD on L2(R), m = 1, > = ^{5(') and A} = i. For example, we will investigate what kind of eigenvalues Ht attains as t runs through several values. In Appendix A we give a complete algorithm to determine the entries of the matrix G appearing in the righthandside of (1.4) and Appendix B deals with the algebraic tools needed to derive the formula for em(z).

3

1.2 Some remarks on definitions and notation

In the sequel we will frequently use familiar concepts like inner products etc. In this section we describe some of these concepts in more detail.

Definition 1.2.1 Let V be a vectorspace over C.

An inner product on V ss a function

V x V —÷ C which satisfies the following properties:

(with f,g,h E

V and A E C)

• (f + g,

^h)

₌

(1h) + (g, h),

• (Af,g) =

A(f,g),

• (f,g)=(g,f),

• (f,g)=O VgeV='f=O.

^-

The first three properties are also called the sesquilinearity of the inner product, the last property is its non-degenerateness. The third item implies (f, f) is real, f E V. In addition, if the inner product satisfies

(f,f) >

^O,f

E V and (f,f) =

⁰

f ⁼

^{0 (1.7)}

we call the inner product positive definite. Note that such an inner productdefines a norm:

iii=vT.

An inner product space is a vectorspace equipped with an inner product. In this paper we will only use two types of (complete) inner product spaces. The first type is the Hubert space, having a positive definite inner product and the second type is the Pontryagin space.

We adopt the convention that this is a KreTn space with finite negative index. (Much more details can be found in, for instance, [Bog].)

When dealing with an inner product space we will mostly use the name or an index in its name as a subscript of the inner product and corresponding norm.

Let IL and A be Hubert spaces.

Notation 1.2.2 The space of continuous antilinear functionals on IL is denoted by 1-C.

Observe that this notation is the same as for the dual space of IL, since these two spaces can always be identified.

Notation 1.2.3 By IL .-+ K; we mean:

• As sets we have IL c ^K;

• The_embedding E : IL — ^ftC defined by Ex = x has dense range and is contractive:

ranE = ^K; and IIEXIIK IIXIIn

Note that c_+ defines a partial ordering (a reflexive, anti-symmetric and transitive relation) on any collection of filbert spaces. We write A C B to denote a strict inclusion.

Recall that we write (Cj)1, m 1 to denote the column vector

(

(1.8)

m)

If the second delimiter is smaller than the first one, the running order will reverse, thus for

m1

Notation 1.2.4

(c)". ₌ (

_I ^(1.9)

\C_mJ

This notation is also used in matrices. However, we do not use this convention in the evaluation of (finite) sums. The starting value of the dummy variable has to be equal or smaller than the last value, otherwise we define the sum to be zero, for example i

=

⁰

and i

=

^—1.

5

Chapter 2

The theorectical construction of ^Ht and the corresponding Straus

extension Tt(z) associated with a

restriction of the original operator A

2.1 The concept of rigging

In this first section we present the basic idea behind rigging of Hubert spaces. A much more thorough treatment of this subject can be found in [Ben], LBer2] and [GG]. Let hf and hi0 be Hubert spaces such that 1-Lb —+ It0. The definition of c_ implies

kerE =

(ranE)'

=

{O}, so the following definition does make sense:

Definition 2.1.1 IL.. is the Hubert space completion of 110 with respect to the inner product:

(f,g)_ = (Ef,Etg)÷,f,h

^E _ho. ^(2.1)

We would like to have hiJ = IL.. It turns out that this is indeed true, but before we arrive at this conclusion a few remarks and definitions are in order.

Note that Definition 2.1.1 leads to a partial ordering of the three Hilbert-spaces ho

fl

since the embeddings are contractive and have dense range:

lifli— = IIE*fII+ Ill IIo,f E hi0. ^(2.2)

The inequality holds since E: 1-1÷ —÷

hi

is a contraction and 11E11 = hEll.

The definition of (., .).... implies

that E :

hi0 C IL. ^—÷

hi

is isometric and since hio is densely contained in IL.. it follows that there exists an isometric extension E : ^{It.... —+}

1i.

But there is more to be said about this operator:

Theorem 2.1.2 The extension E : IL.. —+ hi... is unitary.

Proof It suffices to show that E is surjective.

Let u E li+ 0 ranE C i+ 0 ranE C 14, then there holds for ^all h 14:

0

= (u,Eh)..

(Eu,h)o = (n,h)o, ^(2.3)

thus u = 0. Since ranE is closed, we have ranE = lLf. I

We

want to define how an element of IL works on ?L. To do this

unambiguously, the following inequality is needed:

Theorem 2.1.3 There holds I(f,Eu)oI S IIfIIIIuII+,f E fl0,u E

'lL.

Proof Choose

f

Iio(C IL). Then

Ill II- ^{IIEf 11+ =} Efl+

I(Ef,u)÷I

^IIuII÷

=

u€fl ^I(f7E)oIIIuII+ ^(2.4) The second equality is a result of the Cauchy-Schwarz inequality.

I

Definition 2.1.4 For a C IL, ^{a sequence}liD 3 f, ^a and u C li+ we set:

(a, u)o = urn (I,,, Eu)o. ^(2.5)

n—,00

Furthermore, (u, &)o (a, u)0.

This definition does make sense according to Theorem 2.1.3, the limit exists and does not depend on the chosen sequence. Also, if a E 14 then Definition 2.1.4 reduces to the inner product (., .)° by virtue of the continuity of (., .)°.

We can reformulate the definition above in terms of the inner product of Ii+.

Lemma 2.1.5 We have (a,u)o = (Ea,u)+,a IL.,u

Ii.

Proof

(a, u)o = lim (f,, Eu)o = lim (E*f, u)+

n-400 ^n—400

=

lim(Ef,u) = (Ea,t4.

^(2.6)

I

Finally, we prove what we wanted to have:

Theorem 2.1.6 There holds IL.. = 1i, the space of continuous antilinear functionals on

Proof The form u c

Ii

^{i-4 (a,} u)0 is antilinear in u and is bounded by Theorem 2.1.3:

I(a,

u)oI = urn I(f

Eu)oI 1m IIfIIiIuII+ = IIaII_IIuII+. ^(2.7)

n—+oo n—Ioo

7

From this we conclude that ?L. c

fl.

For the reverse inclusion, assume £ E

fl..

By the Riesz representation lemma there exists a

e n such that

£(u) =

(v,u),u fl.

^(2.8)

Theorem 2.1.2 implies that a = (E)—'v E fl.... is well defined and according to Lemma 2.1.5

£(u) = ^(v,u)÷ =

(Ea,

u)÷ = (a,U)O. (2.9)

U

So what have we obtained? We started with two Hubert spaces: fl and fl÷, the space of

te3t functions. We used the inner product on 1to to construct IL., which turned out to be equal to Ii. The elements of IL. are also called generalized elements. This process is called

a rigging of the Hilbert spaces fl÷ -+ Ito c

IL.

2.2 The rigging applied to an unbounded selfadjoint operator

In this section we will apply the concepts of the previous section to an unbounded selfadjoint operator in a Hilbert space flu. Let A be such an operator and choose A E C+. Define R(z) (A —

z)'

and construct inner product spaces II+k:

fl+k = ranR(A)' = domAk (2.10)

with inner product.:

(u, V)+k = (Au,Av)o. ^(2.11)

Propostion 2.2.1 The space (fik, (, )+k) is a Hubert space.

Proof First of all we have to check that (•, •) is indeed an inner product: the sesquilineair properties of (., •)+k are quite obvious. If we set

u =

^0, the definition of (., .)o and the linearity of A show that the inner product becomes zero. Conversely, if (u, U)k = ⁰ and taking j = 0 in its definition one sees this implies that u =^0. It remains to show that Ilk is complete.

Pick a Cauchy-sequence (u)flEN in ltk• The definition of (., )k shows that A'u, j = ^{0,.. . ,k} are Cauchy-sequences in fl0 and its completeness implies convergence:

u,, -4 u

Au —4

^V1

(2.12)

k Ito

A

u —*

^Vk

Since A is a closed operator, we have u E domA and v1 = Au. Next, proceed by induction:

suppose u

domA1,

^v3_1

= A'u, j =

^{2,. . .,}

r < k. This range of j and (ufl)EN fl

domAk C domK imply domA ^Ar_lUg * Vr_1 = A'-1u. The closedness of A shows

Ar_lu E domA and A'u = AA''u AAu = Ku. Thus, u E '1tr and V,. = Ku. From

this we conclude that u C ?-Lk and

v, = A'u, j

⁼

1,..

^.,k.

Choose E >0 and fix N EN, j

=

0,...,k such that IIA3(u

—u)II < for all n ^N3.

Set N = max{No,.^{. . ,}N,}, then for all n N

iiun —

uII

= k

IIA(u — u)II

2,

_(2.13)

proving the actual convergence.

I

The correspondering norms on 1tk are related in an obvious way.

Lemma 2.2.2

The norms . Ik k

C N sat zsfy

Ilulik lIUIIk+1, u C Nk+l. ^(2.14)

Proof

k k-i-i

IIuII =

(Au, Au)o < (A'u, A'u)o = IIuII+i.

^(2.15)

3=0 j=0

I Furthermore, it is clear from the definition that Nk+i Ilk

c

No holds for all

k>1.

Lemma 2.2.3 The spaces ILk with k 1 are dense in Ito

Proof Ni is dense in No since At = A is an operator. It suffices to prove that fl2÷i is

dense in fl2, n 0. Suppose Ii2

v J. fl2+i, then the following equality holds ^{for all}

U C 1-12fl+1:

0 = (u,v)j2 = (Au,Av)o = (>2A23u,v)o.

^(2.16)

The last equality is a consequence of the selfadjointness of A in Ho. The polynomial does not possess any real root and we obtain the factorisation E-0 =

with ) C C \ It

This implies that for any w C N0 there exists a u E N2n+i such that

w = E0

^A2'zz,just take u = R(A1)R(T) . ^.. R(A2)R(A2)w, and we conclude that v = 0.

I

Theorem 2.2.1, Lemma 2.2.2 and 2.2.3 enable us to write down the chain of filbert spaces

Nm-i-i Nm c_ ... ^c—p _flo. (2.17)

If we introduce the negative spaces, what is the relationship between these spaces? This is the contents of the next lemma.

9

Lemma 2.2.4 Suppose 1(2 fl-f I ^{hO ,} then

fl+2 fl+i ho

fl-i

^{11-2. (2.18)}

Proof Since

fl2

^fl+i fib implies 1•L2 '—+ hO we are able to construct IL.1 and 11—2.

Choose &

IL1 and u

11+2 C IL,4. Then

I(a

^u)oI IfrII_iIIuII+i IIaII_lIIuhI+2 ^(2.19) which shows that u '-4 ^(cr,u)0 is a continuous antilinear functional on fl2 ^and IIaII_2 hall-.1. ^So fl_i ç 11—2 ^and the embedding is contractive. Since It0 C fl C fl_2 and 110 is dense in fl_2, the conclusion is that the embedding has a dense range.

I

This lemma guarantees the existence of the chain

4

^c_

^... 4 fib 4 4 fl-rn + fl-rn-1

^(2.20)

The Hilbert space fl...,, j > 1 is the dual space of 11, obtained by rigging with respect to the inner product of ho• The domain of A cannot be ho; since A is closed the closed graph theorem implies that if the domain of A is closed, the operator will be bounded. (We are merely restating the Flellinger-Toeplitz theorem.) This implies that flk+I C 11k, k > 0, since

=

{xlx E domAkAAcx e domA} = flkfl{xlAkx E domA} c

flkflfll c _flkflflo = 14.

(Note that 'ltk {0} for all k.) It follows from the construction of a negative space that fl_k ^C fl_k_I, ^k 0 and thus there exists a x E fl_rn_i \ fl_rn. We know how R)t) works

on the positive spaces and we want to define its action on x Definition 2.2.5 Define R(A)x 11_rn by

(R(A),u)o = (x,RC't)u)o, U ^{flm. (2.21)}

Theorem 2.2.6 R(A) is well defined, that is: if u E him, then

R())u

flm+i, ^and

R(A) is

bonnded on flm R(.A) fl-rn-

Before we prove this we give some properties concerning

the action of R(p), p

p(A) on elements of flm—i and 11m in the following lemma.

Lemma 2.2.7 For all p E C there holds (a) domA'

=

dom(A ^—

Now suppose p p(A), then

(b) E flrn-i * U = R(p)u E

flm

(c)

If

U E flm then there exists a u e flrn...i such that U = R(p)u.

(d)

If

^U E fim then A'R(j4U = R(p)A'U, j = 0,. ^{. ., m.}

Proof of part (a): The equality domA = dom(A_—p) is obvious and we proceed by induction.

Suppose domAk = dom(A — ^{p)k and} notice that

domA

^{{x domAAx domAk} = {x domAdlAkx domA}. (2.22)}

Choose x E domA'4', then x E domA = dom(A — p) and Ax e domA' = dom(A —

)k.

According to the second set we even have x E ^{domAc = dom(A — ,j)C and} thus —tx E

dom(A —i)'. Summing ^{up: x E dom(A —}

i)

^{and (A — p)x C dom(A — i)k,} from which the inclusion domAkl C dom(A _—i)'1 follows. The reverse inclusion is similarly proven.

Proof of part (b): Using part (a) in the first step, we find U E 7lm = domAtm = ^{dom(A —}

U E dom(A —ii) = domA A (A — it)u C dom(A —

U ranR(t) A (A — = u E dom(A — )m.1

= Ii

Proof of part (c): As in part (b) we have U E domA and (A — i)U ^{C domAm_l} = Define u = (A — jt)U, then R(p)u = U.

Proof of part (d): Notice

that AU, j

= 0,.^{. . , m} is properly defined since ñ C 'Urn. The case j = 0 is trivial and j = 1 is straightforward:

= (A — u

+ )R()U = (I

+ 1tR(1i))U (2.23) and

R(p)AU = R(t)(A — j.t+ p)U = (I + iR(p))U. ^(2.24) The general case is easily proved by induction. Suppose A'R(L)U = R(t)AU with 1 i <

m, then A1R(ji)U = AA3R(i)U =

AR(fl)AU = R(iz)AAU = R()A'U.

^U

Proof of Theorem 2.2.6 The first conjecture is just part (b) of the previous lemma, since

A p(A) A p(A). Let us prove the second conjecture, that is R(A) C li_rn.

If

u C flrn then according to Lemma 2.2.7 (d)

A'R(A)u = R(A)A3u, j = 0,.^{. ., m (2.25)} and thus, by taking j =

m we find, with R(A)u C fli

Am+lR()u

_{= (Atm}

+R()Am)u.

(2.26)

Using these identities we obtain

IIARG)uII

=

IIRG)AuII

(ImA))2

j =0,...,m

^(2.27)

and

IIA'1R()uII

^{= IIAmu}

+R()AmuII

(1 +

1))2IlAmutI.

(2.28)

Hence, by defining C =

())2

_{+ (1 +}

jJ))

we infer that IIRG)uII÷i

=

IIARG)uII <C>J IIAuII = CIIuIl,

^(2.29)

proving the estimate

I(R(.\)x,u)oI IIXII_m_iIIRG)uuIIm+i V'IIXII_rn_1IIUIIm. ^(2.30) The

last inequality implies R(A) E fl =

fl-rn.

The choice of x ^leads to an essential property of R(A):

Theorem 2.2.8 We have R(A) E fl_rn \ fl_rn+i.

Proof Suppose R(A) C '7t_m+l, ^then for all u C flm-.i ^{we have the following estimate,} with c C

RR(A)x,u)oI CIUIm_i ^{= cII(A} )i)R\)UIIm_i. ^(2.31) Looking at the definition of (.,•)+k we obtain

IIARG)u

R()UIIm_i II4R()ttIIm_i

+ IAIIIR(A)UIIm_i

IIR()ttIIm + IIIIR(A)UIIm. ^(2.32) Thus

I(R(A)x,u)oI c(1 + II)IIR(A)UIIm. ^(2.33)

We know that each element in firn can be written as R(A)u, with u E flm....i, see Lemma 2.2.7(c). Furthermore, since (R(.A), u)0 = (x, ^Rc7t)u)o the estimate found leads to the con- clusion that x C fl-rn, ^which is impossible since we assumed x ^C fl—rn—i

\

^fl—rn.

I

Definition 2.2.9 For i = —m — 1,...,m ^— 1 we define the functionals Xi = R(A)mX.

As in the case for x—m =

R(.\) we have

Theorem 2.2.10 There holds xi

li1 \ fl,, i = —m —

1,. ^{. ., rn —1.}

ProofTotally analogous to the proof of Theorem 2.2.8. I

2.3 The spaces Pm and Urn

In this section we will consider two inner product spaces, namely Pm, ^a linear space con- sisting of the functionals we constructed in the previous section, and a Pontryagin space 11mrn We will concoct an isometry 'r : Pm 11m with a dense range. Loosely speaking, this enables us to consider Pm as a pre-Pontryagin space and flm as its completion. In Section 2.5 we will use r to define a generalized resolvent R°°(z) acting on 11m

Definition 2.3.1 Pm i3 the linear space of elements:

=

ço

+ >.CiXi,

^{ccm C firn and c C C. (2.34)}

Writing c1 = + EIXI, it is equipped with the inner product:

rn—i rn-i rn-i

(' )pm

^{= (cOrn,ç0m) +} >(ci(Xi, çOm)⁺ Ci(çOm, jo) + (2.35)

The scalar ge,, is defined by g2,j = _(Xi,Xi)O, this inner product is well defined if and only if

i +

j ^0.

In Appendix A we will describe an algorithm that determines gtj for i + j

<0

in a consistent way. This means that for example 9i,j = still holds for i + j <0. Let us turn to the second inner product space.

Definition 2.3.2 Let the inner product space llm be II.OC' and define its inner product

by:

1nrn)(C).

_(2.36)

Here, °m,rn ^is the m x m zero-matrix, 'rn,rn ^the m x in identity-matrix and G is

(

^{g—1,_i 9-1,-rn}

-. ^:

.

^(2.37)

\9_rn,_i

^g_,,,_,,,

Theorem 2.3.3 rn is a Pontryagin space with negative index m.

Proof It suffices to show that ( ^rn,m

)

^has m negative and in positive elgenval-

ues. Its completeness follows directly from the completeness of 1L0 and

Cl. We prove it

by induction. Since

? = —1 for any g E R, we infer that this matrix has one negative and one positive eigenvalue. In the remainder of the proof we use an index in the name of a matrix to denote the dimension of the space on which it acts, for example Gr,r E crx..

The matrix °m,n is the zero matrix and 'rn,n is the identity matrix, that is 'rn,n= (k,I)ji

with k,1 = ¹ if k = I and zero otherwise. 'rn,a ^denotes the matrix whose entries are given by 'm,n = ^(ik,I)1 with k,1 = 1 if k + 1 m + 1 and zero otherwise. Now suppose that

(

^{0r,r 'r,r ") has} r positive and r negative eigenvalues for all hermitian matrices Gr,r. ^Let

\

^r,r

r,r/

G+i,+i

be an arbitrary hermitian matrix, then there exists an unitary matrix Ur+i,r+i such that Ur+i,r+iGr+i,r+iU,i,r+i ^Ar+i,r+i =

diag\i,.

^. . , )tT+i), with E R, hence

(

^{Uri,r+i °r+I,r+i}

'\ (

^{°r+i,r+i 'r+i,r+i '\}

(

^{Ur+i,r+i °r+1,r+i}

\

\. °r+i,r+i ^Ur+i,r+i

I ' 'r+l,r+i

Gr+i,r+i ) ' °r+i,r+i Ur+i,r+i I

=

( Ur,ri

°r-p1,r--i ' ( 0r+i,r+i ^'r+l,r-t-i ( ^°r+i,rl-i (2.38)

\

^°r+i,r+i Ur+i,r+i I \ ^'r+i,r+i Gr+i,r+i ) \ °r+i,r-f-i

U,i,r+i

I

— ( °r+i,r+i ^'r+1,r-i-i

—

'r-i-i,r+i Ar+i,r+i

Define the unitary (permutation) matrix

p r+1,r+1( ci

^—

^{( 'r+i,i}

^{'r+I,r °r+1,r+Ir ( )}

\ '-'r4-1,r+l 'r+l,r+I

and it is directly verified that

°r+l,r+I

°r,l

'r,r

(

^Or+i,r+i 'r+l,r+l '\ ^{—1 — '1,1 O1,r}

Pr+i,r+i 1 A I 1'r+i,r+i — r ^. (2.40)

\

^'r+l,r+i "r+I,r+i / ^11,1

'r,r

°r,i

^Ar+i,r+i

From this we infer that the operator T whose matrix representation is ( ^°r+I,r+1 'r+l,r+l

)

'r+i,r+l Gr+i,r+i with respect to the standard basis has a two-dimensional invariant subspace E2 such that TIE2 has one positive and one negative eigenvalue and a 2r-dimensional invariant subspace E2 such that TIE2,, has r positive and r negative eigenvalues due to the induction hypothesis.

I

Definition 2.3.4 The mapping T Pm

4

urn is given by:

S0rn+j01CiXz

(('Pm,xi)o +

j' cjgjj)"1

)

. (2.41)

(Cj)1_1

This mapping satisfies several properties described in the following two theorems.

Theorem 2.3.5 The mapping r: Pm 4 Hm iS isometric:

(TI,i)flm

=

(I',)pm.

^(2.42)

ProofExpanding

(r, T4)rim

^according to Definition 2.3.1 and Definition 2.3.4, using (Xi, Xi)o = g,3 we find

(corn + Cr1,_'Pm

+ Xi)O + ((SOm,

Xi)o

+

+ jejgj,.)c +

rn—i rn—i rn—I

= ^(cOrn,Prn)o + >cigijcj + >Ci(Xi, corn)o + Ci(çOm,x)o ^(2.43)

and this equals ('I', )pm according to Definition ^2.3.2.

I

Theorem 2.3.6 The mapping r Pm 11m iS injective and has a dense range : = fl,n.

Proof Set

rn—I

/

I ^{CiXi j 0}

r('I') = ⁽ ((SOm, Xi)O +

'J cjgjj)..1

) = ( (0) (2.44)

(c!l /

^{\ (0)}

and it is immediate that Cj = 0 fori = —m, . . ., —1. Since 1Lm ^cOrn and E 1 \?L,

i

= 0,...,m— 1 are independent we also see that 0m = ⁰ and Cj = 0, ^{i =}

1,...,m— 1, thus r

is injective.

It is sufficient to show that r(P;), ^with

'P

=

{4II E

^{Pm ,co = = C,,,l =} 0}, is dense in Fim.

The image of G P is:

I

^cOin

.

^((corn,

x)o)2).

^(2.45)

(c)_1 By taking com = 0 we get

(0, (0,.^{. .,}0), Cm)T C iiW. (2.46)

The functionals X—rn+i,• •,X—l are continuous on ?iml and X—m is not continuous on Iim...I.

Since ?-t 4 flm—l, there exists 1tm ^tt',

'

⁰ satisfying:

• 1'n ^0,

• (1/.'n,Xi)o+O, i=—1,...,—m+1,

• (/'n,X—m)O4i.

The first two properties hold for any sequence lim 0 and Lemma 2.3.7 guarantees the existence of a sequence satisfying (ib, X—m)O —÷ 1. We conclude:

TtJ) —÷ e_m ^{(0, (0,.. . ,}0, ^{1), (0,.. .}, 0))' _{E 15iW.} (2.47) This argument can be repeated: the functionals X—rn+2,• . , X—i are continuous on 1Im—2, X—m+i is not continuous on flm—2 and 1trn is dense in flm—2• Again, there exists a sequence

flrn 0 such that

•

• (I)n,Xi)OO, i=—1,...,—-m+2,

• (t/'n,X_m+i)O —+ ^1.

(Again, we refer to Lemma 2.3.7). So r — (, X_m)oe_m converges to

t

^(0,

^(0,..

^.,0, 1,0), (0, ^{.. .,}0))" E i5?i. ^(2.48) Repeating the preceding method several times one obtains

(0,Cm,(O,...,0))TçFiw.

^(2.49)

Remember we also had

(2.50)

and this implies, taking I = so ^'Km

(fl,(0,.. ^.,0),(0,..

^.,0))T

ciiw.

^(2.51)

Summing up: äiW = ¹

The following lemma deals with 'Km L4 'Km_i, but

it holds more generally for 1(m 4

'K,, i =0,...,m— 1, since 'Km is dense in 'K,.

Lemma 2.3.7 There exists ^{a sequence} 'Km J!n 0 with ( Wn,X_rn)O ^1.

Proof Consider the space flm, being the space 'Km equipped with the urn-i topology.

Suppose that (X—m,) is a continuous functional on 'Km. Since 'Km = 'Km—i, the closure is

of course with respect to the 11m—1 topology, we can extend X—m continuously to 'Km—i, contradicting Theorem ^2.2.10.

I 2.4 The generalized resolvent R°°(z) and the H°° rela-

tion

In this section we will show that the operator TR(Z)T', with

1'; = {çorn+>'_cjXjIm

^E

'Km and c5 C C} and z

p(A) is bounded on r(P;) C Hm• Note that r is well defined

according to Theorem 2.3.6. In the proof of this theorem it is mentioned that r(P) ^isdense i m, and we conclude that rR(z)r' admits a bounded closure on 11m, being denoted by R°°(z). This operator inherits the resolvent properties of R(z), which enables us to define the selfadjoint relation

H°°

=

{

{iro()

()

^,

^{i +} ( ^{) } I ()}

^E

^m}.

^(2.52)

Hopefully this diagram will elucidate the action of the various operators.

r(P)

R(z)j jTR(z)T1 jRoo(z)

Pm ^Hm

T

Figure 1: The spaces on which the operators TR(z)T', R°°(z) and the relation !I°° act.

Theorem 2.4.1 The bounded closure of rR(z)'r1 is denoted by R°°(z) and is given by:

f \ f

R(z)+po(z)R(z)x_i

R(z) ( (d)1

)

⁼

(L(Z

^—

)i''d ^{+ qj(z))1}

)

, (2.53)

(cj)=_1 (Pk(z))k_1

the polynomiaLs pk(z) of degree m + k — ¹ are

pk(z) =

cj(z —

(2.54)

and the functions q.(z) are

(z —

)''

(R(z)cp, x—i)o + (po(z)R(z)_i, xi)o. ^(2.55)

Proof The action of rR(z)r' on T(P)

^{is given by}

TR(Z)T' (((OmXi)o)i) f ^=rR(z)W,

^(2.56)

(cj)_1

with W E 1';. Recall we have:

R(z) = (I + (z — )t)R(z))R(A) and R(A)2, i = —m,.. ^., m — 2, (2.57) and thus

R(z)W = R(z)m + c_iR(z)_i + cj[X+i + (z — )R(z)+iJ.

^(2.58)

A repetitive use of the resolvent identity until x—i is the only functional left on which R(z) acts yields

R(z)W = R(Z)Wm + ^—

1'R(z)_i ^{+ (z —}

^(2.59)

After the substitution k = i + j

+ 1 and the introduction of the polynomials pk(z) =

— ))lc_1_1, k = —m + 1,..., —1 we write (2.59) more compactly R(z)W = R(z)çøm +po(z)R(z)_i +

k,n+1

(2.60)

Looking at this expansion for R(z)W we see that we cannot directly apply r since ^the

representation is not of the form corn + Zm c,,,

with corn E flm.ReMThite R(z)_i as

01(z

^{— )t)kxk + (z —}))mR(z)Xm_i, yielding

R(z)W = R(z)ço + po(z)(z — A)mR(Z)m_i

—

)kx

^(2.61)

This is the desired representation, since R(z)x_1 E 'Km. Thus rR(z)W = R(Z)çm + po(z)(z — A)mR(Z)Xm_i

+ p0(z) 01(z —

^A)kxk

( (R(Z)(pm+ po(z)(z — .A)mR(Z)Xm_i,Xi)O +... ^-m )i=_1 (Pk(z))k_1

f

^{R(z)prn +}po(z)R(z)x_i

= ((R(Z)Wm+PD(Z)R(Z)X_i,Xi)O)11

)

. (2.62)

(Pk(z))k=_1

Note that p_m(Z) = ^0. The first term in the second entry of (2.62) can be rewritten, again by using the resolvent identity:

(R(z)W, Xi)O = ^{(z —}

)'(Wm,

^{Xj)O + (z}

'(R(Z)Wm, x-i)o

^(2.63)

j=i+1

Furthermore, define (i = —m,. ^{. . ,} —1)

q(z) =

^(z

—)'(R(z)W,X_i)o + (po(z)R(z)_i,)o,

^(2.64)

I

and

we finally obtain rR(z)T ((, ^{xjo)' =}

(c)1

f

R(z)ço+po(z)R(z)X_i

(1+

^—

X)''(Wm,xj)o + q2(z))1

)

. (2.65)

(Pk (z) )k=—1

Suppose we have a sequence in 'Km and a sequence (c')1 in Cin such that

I ⁿ I

I I ^Win

T('Prn)

((Wxi)o)i) (d)1

^E

^r(1').

^(2.66)

(c,)1=_1

(c),_1

A consequence of the convergence in Hm is that ço, Om. Let p(z) be the polynomial

and q'(z) the function associated with (c)1 and .

^Since any polynomial is a con- tinuous function of its coefficients, it is clear that p(z) —+ pk(z) = ^{cj(z — ))k_i_1.}

Furthermore, we have the estimate

IIR(z)(W

co)II MIIW

— WmIIO, M SU IIR(z)c0IIi ^(2.67) II'IIo'

which implies

— Wm), x—i)oI IIR(2)(Wz — Wm)IIiIIX—iII—i

< Mço — comIIoIIx—i Il_i ^{0. (2.68)}

Together with p(z) —+ po(z) this results in q'(z) — q(z). Looking more carefully at the entries of

f

R(z)co,+p(z)R(z)x_i

\

(E+1(z

^—

+ q(z))1

)

(2.69) (Pk (z))k_1

we conclude that its convergence in 11rn is forced by the convergence of

/

^,fl

I

ji n

^%.

I \'Pm,Xi/O)i=—i In\—tfl

in 11m so rR(z)T' admits a bounded closure R°°(z) on

I

Theorem 2.4.2 The operator R°°(z) satisfies, with z, w e p(A):

• R°°(z) -

R°°(w)

=

(z —w)R°°(z)R°°(w)

• Roo(z)* = R°°()

• kerR°°(z) = {(O,(O,. ..,O, d_m), (O,.. .,O))TId_m E C}.

Proof The first equality follows from the resolvent formula applied to R(z). In the second equality the adjoint is taken with respect tothe indefinite inner product of flm• It is proved by noting that

(R(z), I')im =

(4', R()'F)pm implies (R°°(z)rF, T4')flm (r'.I', R°°()T'I')flm and again extending this equality continuously.

Setting pk(z) = ^0, k = —m,. ^{.. ,}—1 implies C_rn = ... = c_2

=

^0. Furthermore, since

E fl

and X—i E fl—i \ 7-b (see Theorem 2.2.10) are independent we also have p0(z) = c_1

=

^0.

Hence q,(z) = 0,i = —m,...,—1 and this implies d

^0,

i = —m + 1,...,—1, so

kerR°°(z) = {(0,(0,...,0,d_m),(0,...,0))TId_m C}. ^U

Theorem 2.4.3 Define the relation H°°(z) = {{R°°(z)f, (I + zR°°(z))f}If E flm, Z E p(A)}, then

• H(z)*=H(), zEp(A),

• H°°(z) =

H°°(w), z,w p(A).

Hence, the relation H°° = H°°(A) is selfadjoint and is given by H°°

= {

^{{ R°°(A)}

( )

^{, +}

^{00 () }}

E 1i0, c, d E

III'h+c_ixo\ f Ah+Ac_io \) ^c,d€Cm,

}

=

(C)m

_{) , (} (ds + AC1)T1 _J I C_rn_I = 0,

1. I. (Cj_) ) \ (ci + Ac_1)1 ,/ J

^{h domA}

and the vector (C)1 equals

((A —

)_i_l(h, _x-')o

_{+ c_1g1,0 +} (A — (2.71)

j=i+1

Proof From kerR°°(z) = (H°°(z))(O) and Theorem 2.4.2 it is inferred that H°°(z) is indeed a relation. Furthermore, it is easily seen that R°°(z) ^is the resolvent of H°°(z). We now show

that H°°(z) = H°°().

According to the definition of H°°(z)t there holds

{a, /3} E Hoo(z)* (/3, R°°(z)g)nm — (a,g)n,, — (a,zR°°(z)g)ri = 0, Vg E urn (f3—a,R°°(Z)g)nm — (a,g)nm = 0, Vg E Hm

R00(z) = R°°(i)

There exists f E flm ^such

that a = R°°()f

and /3 —

a

=

f

{a,/3}EH°°().

The equality on the third line follows from Theorem 2.4.2. It remains to prove that II°°(z) = H°°(w).

Suppose {a, /3} E H°°(z), then there exists f C rn such that a =

^{R°°(z)f and}

/3 =

(I

+ zR°°(z))f. We have to show that there exists g E llm

such that a =

R°°(w)g and /9 = (I+wR°°(w))g. ^{Define g =} (I+(z—w)R°°(z))f, then by using the resolvent properties of R°°(z) listed in Theorem 2.4.2 we find

R°°(w)g =

R°°(w)f

+ (z -

w)R°°(w)R°°(z)f

=

R°°(w)f

+ R°°(z)f -

R°°(w)f

R°°(z)f = a ^(2.72)

and

(I + wR°°(w))g = (I

+ wR°°(w))(I + (z — w)R°°(z))f

=

^{(z —}

w)R(z)f

+ f + w(z — w)R°°(w)R°°(z)f + wR°°(w)f

(I+zR°°(z))f=/3.

^(2.73)

From this we conclude that H°° is selfadjoint.

I

2.5 Selfadjoint extensions of a restriction of the H°°- relation

We now choose a (necessarily symmetric) restriction S of H°° such that the defect space ker(S — A)

=

^{ran(S — ))(I1} has dimension one and is spanned by the vector TX.m =

(0,(0,. ^{. .,0),}(0,. ^{.. ,}0, i))". This leads to the following definition of S.

Definition 2.5.1 The symmetric restriction S of H°° is defined by

S = H°° fl {TX_m, )tTX_m}t. (2.74)

We describe S more explicitly by using the entries of H°°.

Theorem 2.5.2 We have

S =

^{{u,v} C H°°I(A —)rn(h, x-i)o

+ A —

+ c_s(g_m,_, + (A — 5)g—m,—i+i) = 0). (2.75)

Proof Note that we have by definition S = {{u,v} E H°°I(v ^— AU,TX_rn) Elm = O} and by looking at Theorem 2.4.3 one sees that

f Ah—Ah+A—.A)c_io

\

v—Au= ₍

(d+(A—A)C)1

₎ ^(2.76)

\ (cj+(A—A)cj_i)1 /

holds, so by using Definition 2.3.2 we derive the following restriction on {u, v} e H°°, with

Ah—Ah+(A—A)c_jo \

1 ⁰

\

(v —

u,

TX_m)Q =

(d +

^{(A —}

)C)1

J 1 (0) _{I )Hm}

(c+(A—)i)' / \e_/

= d_m + (A — )C_m + ^{+ (A —} = 0. (2.77)

By using (2.71), (2.77) is equivalent to

d_rn + (A — A)m(h, x-i)o ^{+ (A —} .A)c_ig_m,o

+ ^{— 5)rn_id_1}

+

^{+ (A —})C_j_i)g_rn,_i

(A )rn(h, X-i)o +

A 3)rn_d_

+ >jj c_i(g_m,_t + (A —)g—rn,—+i) = 0, ^(2.78)

the first equality holds since C_rn_i = 0.

I

From Definition 2.5.1 we conclude that

S = H°° + span{{TX_m,

ATX_m}}

= S + span{{TX_rn, ATX_m}} +span{{(0, (0),^(o))T, (0,e_rn, (o))T}}, _(2.79) or equivalently

X =

Y = (0,(0,. ..,0,y),(0,...,0,XA))T,

M = {{X,Y}Ix,y e

^{C) (2.80)}

and S = S + M. Note that (2.79) is a direct sum decomposition. Set w = ^{(0, e_rn,(o))T} 11mrn S is an operator, this follows from the definition: suppose S contains an element of the form {0,/3w} E H°°, then

({0,/3W},{TX_m,ATX_m}) = 0

s

I3(W,Tx_t)iim ⁰

s /3=0.

^(2.81)

Moreover, {TX_m,ATX_m} H°°, SO {TX_m, ATX_m} S. Hence, {0, w} S and it is clear that span{{rX_rn, ,\TX_rn}} flspan{{0,w}} = ^{O,O}. From this we conclude that (2.79) is a direct sum decomposition.

Let us now consider a selfadjoint extension of S:

H =

S

+ {{J, c'}

^E

M(X, ')im — çc, )m

^{= 0}. (2.82)}

The restriction on {X, '} ^can be translated into an implicit relation between x and y. We have by definition of the inner product of urn

=

(0, (0,. ^.. ,0,0), (0,.. . ,0,X))T)iIm. (2.83) Since

(, Y)n,,, = (i" X)iim the implicit relation between x and y is y — ix

+ 1x12(A —

)')9_rn,—rn = 0. Thus, if x = ⁰ set t = ^oo and if x 0 we can rewrite this relation as t — t = ⁰

with t = + Ag_m,_m. So t is real and we define the selfadjoint operators Ht as follows.

Definition 2.5.3 The selfadjoint eztensions Ht, t E R, of S are defined by the direct sum

Ht = S+{{(0,(0,...,O,O),(0,...,0,C_m_i)T,

(0,(0,.. .,O,C_m_i(t —Ag_mi_rn)), (0,.. .,0,C_rn_1A))T}IC_rn_i E C}. (2.84) Hence, an element of Ht looks like, writing out S:

h+c_io ^Ah+Ac_io

(C)1 ₍ ^(d+AC)''

₂₈₅

d_ + AC_, + C_rn_i(t — Ag_m,_m) )

(Cj_j)

^{((Ci +}

)ci_i)i)

with

CE C"', d E C", hE domA, the equality (2.71) and

—

x—°

^{+ —}

)m-d ₊

c_1(_,,,,_1 + (A —)g-m,_i+i) ^{= 0. (2.86)} Note that if we replace C_rn_i b C_rn_lit and let t —p^oo the H°° -relation emerges.

In the forthcoming section we will define a symmetric restriction S0 of A and study the Straus-extension associated with S0. This gives a different interpretation of (1.1).

2.6 The Straus-extension and êm ^{function associated}

with S0

As we indicated in previous section and the introduction we will study a certain restriction S0 of the operator A. The Straus-extension associated with So C Ht gives a 'different' in- terpretation of the action of (1.1). The novelty is that the action now takes place in the

original space fl0 and that the information of the extension to the larger space 11m ^{is now} contained in the em-function.

In this section we frequently identify operators with their graphs, so that, for example, {h, k} e A means Ii domA and k = Ah.

Definition 2.6.1 The symmetric restriction So of A is defined by So =

An

{Xo,)Xo}.

Lemma 2.6.2 There holds So = AI(h,x_1)0=o, h E domA.

Proof

h S0 (Ah, Xo)o^{— (h,} )¼Xo)o = 0 ((A —

)h, R(A)_i)o =

⁰

h E AI(h,x_l)0o ^(2.87)

I

It follows from Definition 2.6.1 that S = A+span{{o, Ao}}. We also have a Von Neumann formula.

Theorem 2.6.3 There exists a o C 1jo such that

S

^{= S0 +}span{{o,)o}} +span{{xo,Axo}}. ^(2.88)

Proof Suppose {f,Af} E S, then according to the decomposition S = A+span{{o, )'Xo}}

there exist {h,k} C A and 11 ^E C such that {f,3f} = {h,k} + v{Xo,)tXo}. It suffices to consider the case 1' = 1, since we are only looking for an element that spans the subspace

{f, Af} of S. Set {o, )o} = {h, k} + {xo, Ao} with {h,

k} E A and we find:

Ao—Ao=O Ah+.Axo—k—Axo=O

*

h=(A—A)R)o,

^(2.89)

and thus

oh+xo =

xo+(3—A)R()xo

=

(I+(3— X)R(3))R(A)_i

= R()_i C 11g.

^(2.90)

There holds o domS0, since the selection procedure described in Appendix A guarantees that

= (R)_i,_i)o ⁼

^{g0,_1 0.} (Notice that Theorem 2.2.10 with a minor modification implies that

= H(A)_i C lb \

^{li1, so} domS0 C fl1.) I

This functional o =

R(A)x_i has a nice property.

Lemma 2.6.4

We have (o,o)o =

(Xo,Xo)o.

Proof

(go,

o)o = (R()_i,

R(A)x_i)o =

(R(A)R)_i, _x-i)o

R(A) ^—R(A) R(A) —

R)

=

^—

)x-i x_')o =

(( ^—

x-i,

= (R()R(A)_i,

X—i)o =

(R\)_i,

^R(A)x_i)o

=

(Xo,Xo)o• (2.91)

I

This lemma implies that every unitary U : span{{Xo}} — span{{o}} can be written as

U =

span{{o,0o}} with

=

^1. Moreover, Xo spans ker(S — A) = ran(So ^— A)1, and

a similar statement holds for .

^Introduce

the Cayley-transform C of So, C1 =

{{g — Af,g^— A!

}{f,g}

So} and we can draw the setting as follows:

x

— :

o

ran(So—A) ^ran(So—A)

Figure 2: The theoretical setting of the unitary mappings U ^: span{{Xo}} —+ span{{o}} and the isometric Cayley-transform C1 of S0.

The inverse Cayley-transform of U is Cj'(U) = span{{xo — 0o, A, —

0o}},

^{thus ev-}

ery selfadjoint extension A9 of So is described by

A0 = S0 + C'(U)

=

S0 + span{{Xo —

0,

^{Axo — (2.92)}

with

=

^1. Since A itself is an selfadjoint extension of S0, there must exist a 0 E T such

that A =

A0.

Lemma 2.6.5 A = A1, thus A can be written as the direct sum

A = S0 +span{{Xo —

o,Ao o}}.

^(2.93)

Proof It suffices to show that {Xo—

o, Ao

—

o}

E A and (Xo —

o,

x-i)o ^0, so (2.93) is a direct sum. We have

— = (R(A) —

R()))_i ⁼

^{(A —} )t)R(A)R(A)_i E fl1 =^{domA, (2.94)}

F

and we calculate

A(o—o) = (A—))AR(A)R(A)_i

= (A —

— +

= AR(A)_i

^— AR(A)_i ^{+ )t(A —}

= AR(A)_i

^—

R(A)_i ^{+ R(A)_i —}

= AR(A)_i

^—

AR\)_i

= ^{AXO — A, (2.95)} and thus {Xo —

o,

^{Axo —}Xo} E A. Moreover,

(xo —

o,

^{x—i)o = (A —}

)(R(A)R(A)_i,

x—i)o = (A — A)go,o 0, (2.96)

proving that {Xo _—Xo,AXo —

5} So• ^I

This characterisation of A yields a simple formula for (h, x—i)o, h e domA.

Lemma 2.6.6 Suppose {f, g} c S0 and A ^{{h, k}}

= {f,

^{g} + v{Xo — Xo,}

^Ao

^—

o}

^then

(h, x-..i)o = (A —)vgo,o.

Proof It is somewhat trivial.

(h, X—i)o = (1 + ^zi(xD

o)

x—i)o = "((Xo— o) x—i)o = v(A — )go,o. ^(2.97) The second equality holds since (f, X—i>o = 0 and the last equality can be found in the proof

of the previous lemma. I

This formula is quite useful when we are dealing with

lit.

Theorem 2.6.7 The selfadjoint operator Ht is decomposable into a direct sum, Ht = So + Wt,

with Wt=

c_1Xo+ v(xo —

o)

^Ac_io

^{+ v(Ao}

^—

\—rn f_i

%.. I)i=—1

Uj + /

_{iJj=—1 (2}

d_m + AC_rn + C_m..i(t ^{— Ag_rn,_m)} 98

(Cj + ACii)1

and

= ^{(A —} + c_igj,o +

(A — )1'd,

i = —m,. . ., —1. (2.99) j=i+1

Note In terms of graphs we identify S0 in an obvious way with

(

^,

⁽⁽⁰⁾

^{1 501 E flrn. (2.100)}

t,(°)) \(°))J

^J

Proof Looking at Definition 2.5.3 and using Lemma 2.6.5 we find, with {f,g} So and h e domA

h+cio = 1

+v(Xo—o)+c_iXo, ^(2.101)

and

Ah + Xc_iXo = g+ z4.\ —))+

\c_iXo, ^(2.102) from which the desired decomposition follows. (Note thato fl1 so we indeed have a direct sum decomposition.) The substitution (h, x—i)o= CA — ))vgo,o in the original definition of

C1, see (2.71), yields the formula (2.99). ^U

Definition 2.6.8 The traus extension Tt(z) associated with S0 C Ht 13 defined by:

Tt(z) = {{P0H,P0K}I{H,K} ^E

Ht, K

^— zH E ?L0}. (2.103) where P0 denotes the orthogonal projection from flm onto flo.

Note We

have T() C T(z) and equality holds if z c

^p(Ht). More details can be found in for example [DLS2].

The decomposition of Ht helps us to find the em-function emerging in Tt(z).

Theorem 2.6.9 There holds

T ( ₎ ^— J So + span{{o — Om(z)Xo,)Zo — )tem(z)xo}}, ^Im(z) > 0,

2 104

1 lm(z) <0,

with = Pvn(Z) ^— t

(2.105) Pm(Z) — t

Pm(Z) = [(z

^{—)9_m,_j(Z —}

^)mi

^{+ (z — )m+lg •0(z —}

— ))(z ^{— ,\)m} + zg_m,_m (2.106)

and

_____

(z)

= Pm()• ^(2.107)

Note Actually, one can also write

T1(z) =S0 + span{{o — em(z)xo,xo ^— #\em(z)xo}} ^(2.108) for all z E C \ ^]R, since no restriction appears on z during the calculation of êm(Z), except at most 2rn points where em(z) has a pole. We have written it this way to emphasize

em E Sm(C),

^being the class of generalized Schur functions.

Proof (partially) The first entry of domWt and ranWt can be put into the right form by observing that

c_iXo + V(Xo —

o) =

^{v((1 +} c_i)(Xo —Xo) (2.109) and

c_iXo + vXo —X0)

⁼

vCA(!

+ 1)xo ^(2.110)

We write Ht = ^So

+ W,

wt—

f( ((1+)(Xo—o)

'\

( v(A(!+1)xoo) _'\1

₍₂₁₁₁

(H1) (K1)

)J'

with (H1), (K1) E C2m. Applying the definition of T(z) to this form of Ht one finds the required formula for T(z), except for ern(z). In Appendix B we will derive the formula for

ern(z)=+1.

2.7 Calculation of nonreal eigenvalues of ^Ht

In this (short) section we prove an explicit formula for finding a nonreal eigenvalue of Ht.

In the next chapter we will look at a more concrete operator, namely A = iD and choose

x =

^5(rn) and calculate some elgenvalues in the case m = ^1. But first of all, we state and prove the next theorem.

Theorem 2.7.1 The relation H does not have any nonreal eigenvalue.

ProofLookingat Definition 2.4.3 we see that the followingsystem of equations is non-trivially satisfied if z C C \ ]R ^is an eigenvalue.

Ah+Ac_io—zh—zc_io =

⁰

dj+(A—z)C1 = 0,i=—m,...,—1

Cj + (A —

z)c'_i = 0,

i = —m,. ^.., —1 (2.112) with

C =

^(A

—3)1(h,x_i) +c_1g,0+ (A—5)'d.

^(2.113)

j=i+1

and C_rn_I = 0. It is easily seen that C_i = ... = C_rn ^C_rn_I = ^0. This implies that (A—z)h = ^0, which results in Ii = 0, since z e C\R C p(A). hence C_1 = ⁰and thus d_1 = 0,

repeating this argument m times we see that d_1 = ... = d_m

= C_1 = ...

= C_rn = 0. I

Theorem 2.7.2 All nonreal cigenvalues of Ht (if any) are given by the nonreal solutions

of

em(z) = (z A) (go,o + (z ))(R(z)o,Xo)o) _(2.114)

(z — ^A) (go,o + (z — ))(R(z)xo,xo)o)

Note In the next chapter we will show that in the case A = iD

and x =

^any nonreal solution can occur.

Proof Note that the following holds, P0 denotes the orthogonal projection on H ker(Ht — z) ^{K —} zH

=

0 and {H, K} C ^Ht

K—zH=0, {H,K}EHtandK—zHEflo

PoK — zP0H

=

^0, {H, K}

lit

_{and K — zH C}

PH C ker(T(z) — z).