Poles of Generalized Resolvent Operators

8/9/2019 Poles of Generalized Resolvent Operators

1/57

Abstract

Linear relations, or multi-valued linear operators, have proven a useful tool inmany branches of operator theory. They arise naturally in the context of linear

operators, as their formal inverse, adjoint or closure. Although the general con-cept is known for over 75 years now (see for instance [7], [8], [1]), there is still onlyone book treating the subject, namely [3]. Therefore a general introduction tothe theory of linear relations is given in the first part of the present work, focus-ing on algebraic properties and some generalizations of linear operator concepts.Also, the new notion of splitting a linear relation L into two linear operatorsS and T such that L = S1T is introduced and studied, as well as the linearrelation T S1 arising from the commutation of both factors S1 and T

The second part is concerned with the application of the formerly introducedideas to the treatment of poles of a generalized resolvent ST for two operatorsS and T. In [2], the authors examine the same problem. The vector spacesthey introduce turn out to be kernels and ranges of powers of linear relations,objects similar to those used in the case of an ordinary resolvent. The theorems

and proofs in [2] are translated and adapted into the modern language of linearrelations.

Zusammenfassung

Das Konzept linearer Relationen, auch mehrwertige lineare Operatoren ge-nannt, hat sich in vielen Bereichen der Operatortheorie als ntzliches Hilfsmittelerwiesen. Lineare Relationen stellen eine natrliche Verallgemeinerung (einwerti-ger) linearer Operatoren dar. Relationale Inverse, Adjungierte und Abschlsse derGraphen von linearen Operatoren sind oft selbst keine Operatoren, man kann sieaber stets als lineare Relationen auffassen. Obwohl das Konzept schon vor min-destens 75 Jahren eingefhrt wurde (siehe [7], [8], [1]), ist die Fachliteratur zudem Thema immer noch rar gest. Das Buch [3] ist die einzige weitverbreitete Re-ferenz. Aus diesem Grund befasst sich der erste Teil der vorliegenden Arbeit mit

der Einfhrung in die Theorie der linearen Relationen, mit dem Hauptaugenmerkauf ihre algebraischen Eigenschaften und der Verallgemeinerung von Konzeptenlinearer Operatoren. Zudem wird eine neue Darstellungsmglichkeit linearer Re-lationen eingefhrt: Jede solche Relation L lsst sich als S1T schreiben, wobei Sund T lineare Operatoren zwischen entsprechenden Vektorrumen sind. Die Ei-genschaften solcher sogenannter Splits werden untersucht, zusammen mit dernatrlich auftretenden, neuen Relation T S1.

Der zweite Teil befasst sich mit der Anwendung der allgemeinen Konzepte aufdie Charakterisierung von Polen einer verallgemeinerten Resolvente ST berverallgemeinerte Auf- und Absteigeindizes, wobei T ein abgeschlossener und Sein T-beschrnkter linearer Operator ist. H. Bart und D. C. Lay befassen sich inihrem Artikel [2] mit derselben Fragestellung. Die in jenem Artikel eingefhrtenVektorrume stellen sich als Kerne und Bildrume der linearen Relationen S1Tund T S1 heraus, und knnen mit in der im ersten Teil eingefhrten Terminologielinearer Relationen strukturierter beschrieben und mit den entwickelten Mittelnuntersucht werden.

1


2/57

Contents

1 Preliminaries 2

1.1 Linear relations and related notions . . . . . . . . . . . . . . . . . . . . . 21.2 Ascent, descent and generalizations . . . . . . . . . . . . . . . . . . . . . 111.3 Splitting linear relations . . . . . . . . . . . . . . . . . . . . . . . . . . . 23

2 Poles of the generalized resolvent 37

2.1 Algebraic decomposition . . . . . . . . . . . . . . . . . . . . . . . . . . . 372.2 Poles of the generalized resolvent . . . . . . . . . . . . . . . . . . . . . . 43

1


3/57

1 PRELIMINARIES

1 Preliminaries

1.1 Linear relations and related notions

We will introduce notation for linear relations, which we will use throughout the work.Linear relations generalize the concept of a (not necessarily everywhere defined) linearoperator between two vector spaces. Some of the properties which we are about toshow do not need the linear structure, but we will not concern ourselves with finedetails.

X, Y and Z will always denote F-vector spaces, where for now, F may be anarbitrary field. In particular, subspace will always mean linear subspace, withoutany topological conditions.

Definition 1.1 A linear subspace L X Y is called linear relation in X to Y,or simply in X Y. Denote by

R(X, Y) := {L X Y | L linear relation}the set of all linear relations in Xto Y. In case ofY = X, we write R(X) := R(X, X).Furthermore, we define:

ker L := {x X | (x, 0) L} (kernel ofL),

mul L := {y Y | (0, y) L} (multivalued part ofL),

dom L := {x X| y Y : (x, y) L} (domain of L),

ran L := {y Y | x X : (x, y) L} (range of L).

For a subset M X we set

L[M] := {y Y | x M dom L : (x, y) L} ,

the so-called image of M under L. By abuse of notation, we set for a singlevector x X:

Lx := L[{x}] = {y Y | (x, y) L} .1

The inverse linear relation L1 Y X is defined as

L1 := {(y, x) Y X : (x, y) L} .

Given subspaces U X, V Y, we call

L(U,V) := L (U V) = {(x, y) U V | (x, y) L}

the compression of L to the spaces U and V. In case X = Y and U = V

we write LU := L(U,U). Another special kind of compression is the restrictionof L to subspace U X of its domain, so

L|U := L(U,Y) = {(x, y) U Y | (x, y) L} .

2


4/57

1 PRELIMINARIES

From now on, let L X Y be a linear relation. One can easily verify thatkernel and domain as well as multivalued part and range are subspaces of X and Y,respectively. Also, if M X is a linear subspace of X, then L[M] is linear subspace

of Y. Of course, compressions and hence restrictions are linear relations in X to Y,too.

Remark: If one wanted to be particularly exact, a linear relation L in X Y would have tobe defined as triple (X , Y , L), as in the case of functions. But in most cases, it will be clearfrom the context which spaces are involved, and we will freely consider a compression L(U,V)as linear relation in X Y as well as in U V.

One advantage of studying linear relations rather than operators is the symmetry oftheir definition regarding domain and range. For an operator T : X Y, the inverseT1 need not be an operator, but it always is a linear relation.

Lemma 1.1 The following symmetries hold true:

(i) ker L = mul L1,

(ii) ran L = dom L1,

(iii) (L1)1 = L.

Proof: An easy calculation shows: (x, 0) L (0, x) L1 and (x, y) L (y, x) L1 (x, y) (L1)1.

By simply adding or subtracting the vectors, one verifies that the following impli-cations hold true:

Lemma 1.2

(i) x ker L and y mul L (x, y) L,

(ii) (x, y) L and x ker L y mul L,

(iii) (x, y) L and y mul L x ker L.

As already stated, linear relations generalize the concept of linear operators. In fact:

Lemma 1.3 L is (graph of)2

an operator in X to Y mul L = {0}.

1We will justify this notation in Definition 1.3 and Proposition 1.12.2In the rest of this work we shall not distinguish between a linear operator and its graph.

3


5/57

1 PRELIMINARIES

Proof: IfL is an operator, then its graph is a linear subspace ofX Y, and L0 = 0, orin our relational notation: mul L = L0 = {0}. On the other hand, ifmul L = {0}, thenL is (the graph of) a well-defined function. It is linear, since L is a linear subspace.

We want to define L + M and K L to further generalize the operator case.

Definition 1.2 Let M X Y and K Y Z be additional linear relations.

(i) In case M L, M is called subrelation of L.

(ii) The product or composition of K and L is

KL := K L := K L := {(x, z) X Z| y Y : (x, y) L, (y, z) K} .

In the case X = Y, we define Lm for m N3in the obvious way, where L0 := IX(or I for short) is the identity map on X. For m Z, m < 0, we set

Lm := (L1)m.

(iii) For a scalar F we set

L := {(x,y) X Y | (x, y) L} .

We write L := (1)L.

(iv) The operator-like sum of two linear relations L and M is defined as

L + M := {(x, y + z) X Y | (x, y) L, (x, z) M} .

If we want to consider the sum of two linear relations regarded as linear sub-spaces, we will explicitly use +, so

L + M := {(x1 + x2, y1 + y2) X Y | (x1, y1) L, (x2, y2) M} .The usual sum of linear subspaces U, V X which are no linear relations willbe denoted by U+ V. If the sum is direct, that is U V = {0}, then we expressthis by writing U V.

(v) Let U, V X be subspaces. If L = LU LV, we call L completely reducedby the pair (U, V).

Verifying the next proposition is easy, but tedious.

Proposition 1.4 Let L, M X Y, K Y Z be linear relations, F.Then KL, L and L + M are linear relations. is associative; also, (R(X, Y), +)

3In this work, 0 N.

4


6/57

1 PRELIMINARIES

and (R(X), ) form monoids with identity element X {0} and I respectively.(R(X, Y), +) is even commutative.

Images of subsets under linear relations behave exactly as in the case of operators.

Lemma 1.5 Let N X Y, M X Z, K Y Z be linear relations andU V X, W Y be subsets. Then:

(i) L[U] =xULx.

(ii) If L is an operator, taking preimages is the same as taking images under theinverse relation: (L)1[W] = L1[W].

(iii) M[ker L] = ker LM1.

(iv) K[L[U]] = (KL)[U], in particular K[ran L] = ran KL.

(v) L[U] L[V].

(vi) For x dom L, Lx W implies x L1[W]; thus: if L[U] W, thenU L1[W].

(vii) (L + N)[U] L[U] + N[U]4.

Proof:(i) Ifx U \ dom L, then Lx = , so we can write

L[U] = xUdomL

Lx = xU

Lx.

(ii)

(L)1[W] = {x X | y W : (x, y) L}

=

x X y W dom L1 : (y, x) L1

= L1[W].

(iii)

M[ker L] = {z Y | x ker L dom M : (x, z) M}

= z Z x dom L dom M : (x, 0) L, (z, x) M1=

z Z (z, 0) LM1 = ker LM1.

4Note that the plus sign has a different meaning on each side.

5


7/57

1 PRELIMINARIES

(iv)

K[L[U]] = {z Z| y L[U] dom K : (y, z) K}

= {z Z| x U dom L, y dom K : (x, y) L, (y, z) K}= {z Z| x U dom KL : (x, z) KL} = (KL)[U].

For the second part, we note that ran L = L[X].(v) This is clear from the definition.

(vi) Suppose x dom L and Lx W, then there exists some y W ran L with(x, y) L, or (y, x) L1, which means x L1[W]. Now the second partfollows from (i).

(vii) Let y (L + N)[U], then there exists an x U dom(L + N) = U dom L dom N such that (x, y) (L+N). That means there further exist y1, y2 Y with(x, y1) L and (x, y2) N as well as y = y1 + y2. This shows y L[U] + N[U].

The addition and multiplication of linear relations is distributive under certain con-ditions.

Lemma 1.6 Let L, M Y Z and K, J X Y be linear relations. Then

(L + M)K LK+ M K,

where equality holds, if mul K ker M or mul K ker L; furthermore:

LK+ LJ L(K+ J),

where equality holds, if mul L mul LK+ LJ and ran K+ ran J dom L.

Proof:

Let (x, z) (L+M)K, so there is some y Y with (x, y) Kand (y, z) L+M.There are vectors z, z Z with (y, z) L, (y, z) M and z = z + z. Thisimplies (x, z) LK and (x, z) M K, so

(x, z) = (x, z + z) LK+ M K.

Now, suppose mul K ker M (the other case follows from the commutativity ofthe operator-like sum). Let (x, z) = (x, z + z) LK+ MK, such that (x, z) LK and (x, z) M K. Then there are some y, y Y with (x, y), (x, y) K,(y, z) L and (y, z) M. We have y y mul K ker M. Hence,

(y, z) = (y, z) + (y y, 0) M

and this shows (y, z) = (y, z + z) L + M. But then (x, z) (L + M)K.

6


8/57

1 PRELIMINARIES

To show the second statement, let (x, z) LK + LJ, that means (x, z) LKand (x, z) LJ for some z, z Z with z = z + z. Consequently, there aresome y, y Y with (x, y) K, (y, z) L, (x, y) J and (y, z) L. Now,

(x, y + y) K + J and (y + y, z + z) = (y + y, z) L. But this alreadyshows (x, z) L(K+ J).

Now, suppose mul L mul LK + LJ and ran K + ran J dom L in order toshow that equality holds. Let (x, z) L(K + J), that is, there is some y Ysuch that (x, y) K + J and (y, z) L. By the definition of the operator-like sum, we find vectors y, y Y with y = y + y such that (x, y) Kand (x, y) J. Choose z, z Z with (y, z), (y, z) L. This is possible,because y ran K dom L and y ran J dom L. This implies (y, z) +(y, z) = (y, z + z) L and (y, z) (y, z + z) = (0, z (z + z)) L. Thus,z = z + z + u with u mul L mul LK + LJ. But then (x, z) LK and(x, z) LJ, so (x, z + z) LK + LJ. Since (0, u) LK + LJ, we obtain(x, z + z) + (0, u) = (x, z) LK+ LJ. This shows the equality.

Although we have notions of adding and multiplying by scalars, linear relations inX to Y do not form a vector space in general. As we have noted in Proposition1.4, R(X, Y) equipped with the addition from Definition 1.2, (iv), is a commutativemonoid with identity element 0 := X {0F}. Of course, the everywhere defined linearoperators from X to Y form a submonoid of (R(X, Y), +), which is even an abeliangroup. And more precisely:

Proposition 1.7 L L = 0 L holds if and only if L is a linear operator. Lis (additively) invertible in (R(X, Y), +) if and only if it is an everywhere definedlinear operator.

Proof: We haveL L = 0 L L L = dom L {0} ,

which is only true, if mul L = {0}: If y1, y2 mul L with y1 y2 = 0, then (0, y1),(0, y2) L, so (0, y1 y2) L L, thus {0} span {y1 y2} ran(L L).

For the second statement, suppose L possesses an additive inverse K. Firstly, notethat dom 0 = X and dom(L + K) = dom L dom K for all L, K R(X, Y). Hencewe must have dom L = dom0 = X.

Furthermore, for (x, 0) L + K to hold, it is necessary that (x, z) K for somez Y with (x, z) L. If there was some y Y with y = z and (x, y) L, we wouldhave (x, y z) L + K, which contradicts L + K = 0. So mul L = {0}.

Except for trivial cases, (R(X, Y), +) is not an abelian group. We cannot evenexpect (R(X, Y), +, ) to form an F-semimodule5, where denotes the scalar multipli-cation from Definition 1.2, (iii). This is because otherwise we had

dom L {0} = 0 L = (1 1) L = 1 L 1 L = L L L R(X, Y),

7


9/57

1 PRELIMINARIES

which by Proposition 1.7 can only be true in trivial cases.But we have

Proposition 1.8 Let G be a subsemifield ofF such that + = 0 implies = = 0for all , G6. Then (R(X, Y), +, ) is a commutative G-semimodule with zero.

Proof: The fact that (L + M) = L + M, as well as (L) = ()L, hold true forall , F and L, M R(X, Y) is easy to see. Now, let , G. From Lemma 1.6,we know

( + )L = ( + )IL = (I + I)L IL + I L = L + L.

If + = 0, then = = 0. We note that 0 L = dom L {0} is additively

idempotent, so ( + )L = 0 L = 0 L + 0 L = L + L.

If + = 0, we consider (x,y1 + y2) (L + L), where (x, y1), (x, y2) L. Then

(x, ( + )y1), (x, ( + )y2) ( + )L,

so

+ x,y1

,

+ x,y2

( + )L,

which eventually implies

(x,y1 + y2) ( + )L.

Example 1.1 By what we just have shown, we see that the condition of Lemma 1.6is not necessary. For all , F with + = 0 and all K R(X, Y) we have

(I + I)K = ( + )IK = ( + )K = K+ K = (I)K + (I)K,

5Let S be a non-empty set equipped with two binary operations +, : S S S. Following [4] (p.427-428), (S, +, ) is called semiring, if (S, +) and (S, ) are semigroups such that

a(b + c) = ab + ac, (a + b)c = ac + bc a,b,c S.

It is called (nontrivial) semifield, if additionally (S, ) is a group, where S := S \ {o}, if o isthe zero of (S, +), and S := S otherwise.

Let (S, +) be a semigroup, (R, +, ) a semiring and : R S S a mapping called scalarmultiplication denoted by (, s) s = s. (RS, +, ) is called a (left) R-semimodule, if

(a + b) = a + b, ( + )a = a + a, (a) = ()a , R, a, b S.

It is called unitary, if (R, +, ) has an (multiplicative) identity and a = a holds for all a S.We differ from [4] (p. 444) in that we do not assume that 0 a = 0 for all a S, where 0 is thezero in R.

6Semifields satisfying that property are called zero-sum free, see [4] page 428.

8


10/57

1 PRELIMINARIES

but ker(I) = ker(I) = {0}, while mul K may be arbitrary.On the other hand we have already seen in Proposition 1.7 that 0K = (1+(1))K =

1 K + (1)K = K K can only hold, if K is a linear operator, so a strict inclusion

can occur in Lemma 1.6.

It is clear that kernel, domain and range of linear relations generalize the kernel,domain and range of linear operators. If the operator L is also injective, then theinverse linear relation and inverse operator coincide. However, the product of a linearrelation and its inverse is in general not the identity operator.

Lemma 1.9 The following identities hold true: LL1 = IranL + ({0} mul L). In particular, ifL is a linear operator, we have

LL1 = IranL. L1L = IdomL

+ (ker L {0}),

where IranL := {(y, y) Y Y : y ran L} is the identity operator on ran L and

IdomL is defined analogously.

Proof: Let (y, y) LL1, that is, there exists an element x X with (y, x) L1 and(x, y) L. So, (x, y), (x, y) L, which also implies y, y ran L and y y mul L.Now the decomposition (y, y) = (y, y) + (0, y y) shows

(y, y) I + ({0} mul L).Conversely, let (y, y +z) IranL +({0}mul L) with z mul L. There is an element

x dom L with (x, y) L and thus (x, y + z) L as well. Since (y, x) L1 and(x, y + z) L, (y, y + z) LL1 holds true.

The second equation follows from passing to inverses: from the first equation weknow

L1L = L1(L1)1 = IranL1 + ({0} mul L1) = IdomL + ({0} ker L).Inverting on both sides yields, since passing to inverses commutes with taking sums ofsubspaces,

L1L = I1domL + ({0} ker L)1 = IdomL + (ker L {0}). Though we have seen that the relational inverse L1 and the operator-like negative

L are no inverses in an algebraic sense, they are obviously involutive, and they evencommute.

Lemma 1.10 We always have (L1)1 = L and (L) = L, as well as (L)1 =L1.

9


11/57

1 PRELIMINARIES

Proof: We only proof the last claim:

(x, y) L (y, x) L1 (y, x) L1 (y, x) L1,

where we used linearity in the last step; on the other hand:

(x, y) L (x, y) L (y, x) (L)1.

We also have a group-like relation between taking products and taking inverses.

Lemma 1.11 Let K Y X, then (KL)1 = L1K1. In particular, if X = Y,we have

Ln := (L1)n = (Ln)1 n N.

Proof: Let (z, x) (KL)1

, so we have (x, z) KL, which means there exists somey Y such that (x, y) L and (y, z) K. But that implies (y, x) L1 and(z, y) K1, and thus (z, x) L1K1. The converse inclusion follows analogously.

Since Lx is a coset ofmul L, we can associate a linear operator with a linear relationL by factoring Y by the multivalued part. More precisely

Definition 1.3 Let QL : Y Y / mul L be the canonical quotient map. Definethe linear relation Lop := QLL X (Y / mul L), which we call linear operatorassociated with L.

This name is justified in the following proposition.

Proposition 1.12 The linear relation Lop is a well-defined linear operator withdom Lop = dom L, ker Lop = kerL and ran Lop = QL[ran L] = ranLmulL . Set-theoretically we have

Lopx = Lx x dom Lop.

Lop and mul L determine L X Y uniquely, that is, if M X Y is anotherlinear relation then Lop = Mop and mul L = mul M already imply L = M.

Proof:

We use Lemma 1.3: Let [y] mul QLL, which means there is some z Y with(0, z) L and (z, [y]) QL. But since the first statement means z mul L, so(z, [0]) by the very definition of QL, and since QL is an operator, together withthe second statement, we get [y] = [0].

10


12/57

1 PRELIMINARIES

Now for the mnemonic equality Lopx = Lx, let y Lx for some x X, whichmeans (x, y) L. Then we claim Lx = y + mul L: z Lx implies (x, z) L, so

L (x x, y z) = (0, y z),

which yields y z mul L, or equivalently z y + mul L. On the other hand,if z y + mul L, then we find some w mul L (so (0, w) L) with z = y + w.This implies (0, z y) L, and hence

L (0 + x, z y + y) = (x, z),

or equivalently z Lx. dom Lop = dom L and ran Lop = QL[ran L] are easily verfied equalities. Concerning the kernel: If Lopx = mul L, then there exists a y mul L such

that (x, y) L. Lemma 1.2 (iii) implies x ker L, so ker Lop ker L. For theconverse inclusion, let x ker L. Now, Lemma 1.2 (ii) implies Lopx mul L.

Since Lop is well-defined, we have shown that x ker Lop. To show that L = M, if Lop = Mop and mul L = mul M, let (x, y) L, that is

y + mul M = y + mul L = Lx = M x.

Then (x, y) M. Analogously, if (x, y) M, then (x, y) L.

This proposition shows that if{ei : i I} is a basis ofX and {yi : i I} Y, thenL is unambiguously defined by (ei, yi) L for all i I and its multivalued part.Furthermore, given Lop and mul L, we can reconstruct the relation L.

Lemma 1.13 We always have L = Q1L Lop.

Proof: Let (x, y) L, so x dom L and y ran L. We have

Lopx = y + mul L = QLy,

so (x, y) Q1L Lop. Now, let (x, y) Q1L Lop, then x dom L and y ran L with

Lx = Lopx = QLy = y + mul L,

so indeed we have (x, y) L.

1.2 Ascent, descent and generalizations

In this section, we want to examine some characteristical properties of linear relationsL X X. In definition 1.2, we introduced powers of such relations. It is oftenconvenient to characterize them as follows.

11


13/57

1 PRELIMINARIES

Definition 1.4 Let n N, n 1 and x1, . . . , xn X. The n-tuple (x1, . . . , xn) issaid to be an L-chain (of length n), if

(xk, xk+1) L k = 1, . . . , n 1.7

We will denote the set of all such tuples by

Ch(L) :=

(x1, . . . , xn)

mN

Xm

n N, n 1

.

Proposition 1.14 Let n 0. We have (x, y) Ln if and only if there exists an

L-chain (x1, . . . , xn+1) with x = x1 and xn+1 = y.

Proof: For n = 0 we have (x, y) L0 = IX if and only ifx = y, so if and only if thereis some admittedly degenerated L-chain (x1) with x = x1 = y. The case n 1follows easily from the the definition of Ln.

Linear relations show degenerated behavior if there exist L-chains (x1, . . . , xn) (n 3) with leading and trailing zero, so x1 = 0 = xn and xk = 0 for at least one 1 < k < n.In that case, if (x, y) Lk+(nk), there exists some z X such that (x, z) Lk and(z, y) Lnk. Since also (0, xk) Lk, (xk , 0) Lnk, we have

(x, z + xk) Lk, (z + xk, y) L

nk,

so z is not unique.

Definition 1.5 An L-chain (x1, . . . , xn) is called singular chain, if x1 = 0 = xn.If there exists a 1 < k < n such that xk = 0, the singular chain is said to benon-trivial, otherwise it will be called trivial.

As indicated beforehand, the existence of non-trivial singular chains has an easycharacterization.

Proposition 1.15 There exists an non-trivial singular L-chain of length n + 1 3if and only if there exists a 1 k < n with ker Lnk mul Lk = {0}.

7Note that this is no condition for the case n = 1.

12


14/57

1 PRELIMINARIES

Proof: Suppose (x0, . . . , xn) Ch(L) is a non-trivial singular chain of length n + 1 3with xk = 0 for some 1 k < n. Then we have (0, xk) Lk and (xk, 0) Lnk,so 0 = xk ker Lk mul Lnk. If on the other hand ker Lnk mul Lk = {0} for

some 1 k < n, choose 0 = xk ker Lnk mul Lk. Then there exist L-chains(x0, . . . xk), (xk, . . . , xn) of length k + 1 and n k respectively with x0 = 0 = xn, so(x0, . . . , xk, . . . , xn) is a non-trivial singular chain.

Definition 1.6 This justifies defining the singular component manifold

Rc(L) =mN

ker Lm mN

mul Lm,

which coincides with the same notion in [6], subsection 3.2.

It is an easy exercise to prove the following proposition.

Proposition 1.16 For all k, m N we have

ker Lk ker Lk+1 dom Lm+1 dom Lm

andmul Lk mul Lk+1 ran Lm+1 ran Lm.

Now we can examine the relationships between range, kernel etc. of powers of L.We generalize Lemma 2.1 from [2].

Lemma 1.17 Let L X X be a linear relation and k, m N. Then:

ker Lm+k

ker Lm=

ker Lk ran Lm

ker Lk mul Lm.

Proof: We begin by eliminating trivialities: For k = 0 the left-hand side is trivial,as is the right-hand side because of ker L0 = ker I = {0}. For m = 0 the left-hand side simplifies to ker Lk for the same reason; for the right-hand side we observeran Lm = ran I = X and mul Lm = mul I = {0}, which means on the right we are alsoleft with ker Lk. In the rest of the proof we are concerned with k,m > 0.

It suffices to find an onto linear map

A : ker Lm+k ker Lk ran Lm

ker Lk mul Lm,

13


15/57

1 PRELIMINARIES

whose kernel coincides with ker Lm. We define it as follows: Given x ker Lm+k, wefind an L-chain (x, x1, . . . , xm+k1, 0) using which we set

Ax := xm + (ker L

k

mul L

m

) = [xm]. A is indeed a map with stated codomain: (x , . . . , xm) and (xm, . . . , xm+k1, 0)

are L-chains, that is xm ran Lm and xm ker Lk, so xm ker Lk ran Lm. Ifwe are given another L-chain (x, x1, . . . , x

m+k1, 0), then

(0, x1 x1, . . . , xm x

m, . . . , xm+k1 x

m+k1, 0)

is also an L-chain, likewise (0, x1x1, . . . , xmxm) and (xmx

m, . . . , xm+k1

xm+k1, 0). That means we have xm xm mul L

m and xm xm ker Lk, so

xm xm ker L

k mul Lm und hence [xm] = [xm]. The linearity ofA is evident.

Concerning surjectivity: Let [y] ker(Lk)ran(Lm)

kerLkmulLm , which means there are L-chains

(y, x1, . . . , xk1, 0), (x0, x1, . . . , xm1, y).

Then(x0, x

1, . . . , x

m1, y , x1, . . . , xk1, 0)

is an L-chain with length m+ k and y as m-th place. We conclude x0 ker Lm+k

and Ax0 = [y]. Concerning the kernel ofA: Let x ker Lm, then there exists an L-chain

(x, x1, . . . , xm1, 0, . . . , 0 (k+1) times

),

which means Ax = [0] = 0.Now, let x ker Lm+k together with an L-chain (x, x1, . . . , xm+k1, 0) and Ax =0, which means we have xm ker Lk mul Ln. Hence there exist L-chains

(xm, y1, . . . , yk1, 0), (0, y1, . . . , y

m1, xm).

It follows(0, y1, . . . , y

m1, xm, y1, . . . , yk1, 0) Ch(L).

Subtracting this chain from the one belonging to x we get

(x, x1 y1, . . . , xm1 y

m1, xm xm

=0

, xm+1 y1, . . . , xm+k1 yk1, 0).

In particular:(x, x1 y

1, . . . , xm1 y

m1, 0) Ch(L),

so x ker Lm. This shows the inclusion ker A ker Lm.

In case of the absence of non-trivial singular chains, the formula can be simplifiedto Lemma 4.4 of [6].

14


16/57

1 PRELIMINARIES

Corollary 1.18 Let L X X be a linear relation with no non-trivial singularchains, and k, m N. Then:

ker Lm+k

ker Lm= ker Lk ran Lm.

Proof: The absence of non-trivial singular chains implies ker Lk mul Lm = {0}. Nowapply the preceding lemma.

Analogously, we find a descriptive isomorphism for the ranges of a relations powers.The first part of the following lemma can be found in [6], numbered 4.1, the secondone in [2] as Lemma 3.1.

Lemma 1.19 Let L X X be a linear relation and k, m N, then:

ran Lm

ran Lm+k=

dom Lm

(ker Lm + ran Lk) dom Lm=

dom Lm + ran Lk

ker Lm + ran Lk.

Proof: We show the first isomorphism: Firstly, we care about trivialities again: Isk = 0, then the left-hand side is trivial, and since ran L0 = X and hence

(ker Lm + ran L0) dom Lm = dom Lm,

the right-hand side is also trivial. For m = 0 we get XranLk

on the left, as well as onthe right using dom L0 = X:

X

(ker L0 + ran Lk) X=

X

ran Lk.

Now let k, m 1. Again, it suffices to construct an onto linear map

A : dom Lm ran Lm

ran Lm+k

with ker A = (ker Lm +ran Lk) dom Lm. We define it as follows: Given x dom Lm,we find an L-chain (x, x1, . . . , xm) using which we set

Ax := xm + ran Lm+k = [xm].

A is a function with stated codomain: Evidently we have xm ran Lm. Further-more: Given another L-chain (x, x1, . . . , x

m), then

(0, x1 x1, . . . , xm x

m) Ch(L),

15


17/57

1 PRELIMINARIES

hence( 0, . . . , 0

(k+1) times, x1 x

1, . . . , xm x

m) Ch(L),

so we have xm xm ran Lm+k, which gives [xm] = [xm].

A is clearly linear. Concerning surjectivity: Let [y] ranL

m

ranLm+k, which means there is an L-chain

(x0, x1, . . . , xm1, y).

Then x0 dom Lm and Ax0 = [y]. Concerning the kernel ofA: Firstly let x ker A, so because ofAx = [0] we have

an L-chain(x, x1, . . . , xm1, xm)

with xm ran Lm+k. This means on the other hand that there exists an L-chain

(x

0, . . . , x

m+k1, x

m+k = xm).

Using the last two chains, we get

(xk x, xk+1 x1, . . . , x

m+k1 xm1, 0) Ch(L).

It follows xk x ker Lm. Together with xk ran L

k, we get

x = (xk x) + xk ker L

m + ran Lk.

This shows ker A (ker Lm + ran Lk) dom Lm.Conversely let x (ker Lm + ran Lk) dom Lm, which means there are someu ker Lm, v ran Lk with associated L-chains

(u, u1, . . . , um1, 0), (v0, . . . , vk1, v) respectively,

and x = u + v. Furthermore, there exists an L-chain

(x = u + v, x1, . . . , xm).

But then,

(u + v u = v, x1 u1, . . . , xm1 um1, xm) Ch(L).

Joining this chain and the one associated with v, we obtain

(v0, . . . , vk1v, x1 u1, . . . , xm1 um1, xm) Ch(L),

which means xm ran Lm+k und consequently Ax = [xm] = 0.

The second isomorphism is an easy consequence of the general resultU

U V=

U + V

V(1.1)

for subspaces U, V ofX.

16


18/57

1 PRELIMINARIES

For the case of linear operators in X, the notions of ascent and descent play animportant role, for instance regarding the poles of the resolvent, which we seek togeneralize, following [2]. We can recover their definitions for linear relations, but we

will also introduce stronger versions.

Definition 1.7 We define: (L) := inf

m N

ker Lm = ker Lm+1 ascent of L, (L) := inf{m N | ker L ran Lm mul Lm}, (L) := inf{m N | ker L ran Lm = {0}} generalized ascent of L, (L) := inf

m N

ran Lm = ran Lm+1 descent of L, (L) := inf{m N | ker Lm + ran L dom Lm}, (L) := inf{m N | ker Lm + ran L = X} generalized descent of L.

Of course, inf := , as usual.

Firstly we check that taking infima is reasonable.

Proposition 1.20 For all m, k N we have:

(i) ker Lm = ker Lm+1 ker Lm = ker Lm+k,

(ii) ran Lm = ran Lm+1 ran Lm = ran Lm+k,

(iii) ker L ran Lm mul Lm ker L ran Lm+k mul Lm+k,

(iv) ker Lm + ran L dom Lm ker Lm+k + ran L dom Lm+k,

(v) ker L ran Lm = {0} ker L ran Lm+k = {0} and

(vi) ker Lm + ran L = X ker Lm+k + ran L = X.

The first two points are Lemma 3.4 and Lemma 3.5 in [6]. We repeat their proofshere for the sake of completeness.

Proof: Let m N.(i) Let ker Lm = ker Lm+1. ker Lm ker Lm+k follows inductively from Lemma

1.16 for all k N. We show the other inclusion by another induction: As-sume ker Lm ker Lm+k was already shown for some k N. Now, givenx ker Lm+k+1, we find some (x, 0) Lm+k+1 = Lm+kL, which means (x, y) L, (y, 0) Lm+k, so y ker Lm+k ker Lm for some y X. But then

(x, 0) Lm+1 = Lm, so x ker Lm.(ii) Let ran Lm = ran Lm+1. Again, ran Lm ran Lm+k follows from Lemma 1.16

for all k N. Concerning the other inclusion: Assuming ran Lm ran Lm+k

is true for some k N, we argue: given y ran Lm ran Lm+k, we find

17


19/57

1 PRELIMINARIES

(x, y) Lm+k = LkLm for some x X. But that means

(x, xm) Lm, (xm, y) L

k,

so xm ran Lm = ran Lm+1 by assumption, for some xm X. xm ran Lm+1

implies (z, xm) Lm+1 for some z X, so together with (xm, y) Lk weconclude (z, y) Lm+k+1, which means y ran Lm+k+1.

(iii) Let ker Lran Lm mul Lm. Since ran Lm+k ran Lm and mul Lm mul Lm+k,we conclude

ker L ran Lm+k ker L ran Lm mul Lm mul Lm+k k N.

(iv) Let ker Lm + ran L dom Lm. Since ker Lm+k ker Lm and dom Lm dom Lm+k, we conclude

ker Lm+k + ran L ker Lm + ran L dom Lm dom Lm+k k N.

(v) and (vi) are trivial.

The following proposition allows us to compare traditional and generalized ascentand descent.

Proposition 1.21 Let m N. We have

(i) ker L ran Lm = {0} ker L ran Lm mul Lm, so (L) (L), and(ii) ker Lm + ran L = X ker Lm + ran L dom Lm, so (L) (L).The converse implications (and consequently the converse inequalities) do not hold

in general.

Example 1.2 Consider for instance X = R2 and L := span {(0, e1), (e1, 0)}, then

ker L = ran L = dom L = mul L = span {e1} .

So ker L ran L = span {e1} = mul L = {0}, which means (L) = 1. But Lm = Lholds for all m N, so we always have ker L ran Lm = span {e1} = {0}, whichimplies (L) = .

Furthermore, we have ker L + ran L = span {e1} = dom L = X, so (L) = 1. SinceLm = L holds for all m N, we also have

(L) = .

It turns out that the numbers and coincide with usual definitions of ascent and descent respectively.

18


20/57

1 PRELIMINARIES

Proposition 1.22 We always have (L) =

(L).

Proof: It suffices to show: ker L ran Lm mul Lm ker Lm = ker Lm+1. Butthis is trivial using Lemma 1.17, setting k = 1 there:

ker Lm+1

ker Lm=

ker L ran Lm

ker L mul Lm.

Proposition 1.23 We always have (L) = (L).Proof: We have to show: ran Lm = ran Lm+1 ker Lm+ran L dom Lm. Again,

this is trivial using Lemma 1.19, setting k = 1:ran Lm

ran Lm+1=

dom Lm

(ker Lm + ran L) dom Lm.

Proposition 1.24 The ascent and generalized ascent have the property that thedefining (characterizing) inclusions hold true even for kernels of any powers of L.

(i) ker L ran Lm = {0} ker Lk ran Lm = {0} for all k N,

(ii) ker L ran Lm mul Lm ker Lk ran Lm mul Lm for all k N

Proof:

1. The statement is obviously true for k = 1. Suppose it has been shown forsome k N. Let x ker Lk+1 ran Lm, then we have (x0, x) Lm and(x, x1) L, (x1, 0) L

k for some x1 X, so actually x1 ran Lm+1 andx1 ker L

k, which in view of ker L ran Lm+1 = {0} implies x1 = 0. But thenwe have x ker Lk ran Lm = {0}, so x = 0.

2. The inclusion ker L ran Lm mul Lm implies ker Lm = ker Lm+k. So, usingLemma 1.17, we see

{0} =ker Lm+k

ker Lm=

ker Lk ran Lm

ker Lk mul Lm

Hence, ker Lk ran Lm ker Lk mul Lm mul Lm.

We obtain a (partially) dual result from Lemma 1.19.

19


21/57

1 PRELIMINARIES

Proposition 1.25 The characterizing inclusion of the descent remains true for theranges of powers of L:

ker Lm + ran L dom Lm ker Lm + ran Lk dom Lm for k N.

Proof: Let k N. The inclusion ker Lm+ran L dom Lm implies ran Lm = ran Lm+k,by Proposition 1.23. Using Lemma 1.19, we obtain:

{0} =ran Lm

ran Lm+k=

dom Lm

(ker Lm + ran Lk) dom Lm ker Lm + ran Lk dom Lm

The following lemma condenses Lemma 3.3 and Lemma 5.5 from [6].

Lemma 1.26 The following conditions are equivalent:

(i) (L) = p < ,(ii) Rc(L) = {0} and (L) = p < , and

(iii) either p = 0 or ker L ran Lp1 = {0}, together with ker Lk ran Lp = {0} forall k N.

Proof:(i) (ii) Let (L) = p < , so ker L ran Lm = {0} for all m p. By Proposition

1.24 i, we have ker Lk ran Lm = {0} for all k N. Since ker Lk mul Lm ker Lk ran Lm = {0}, there are no non-trivial singular chains by Lemma 1.15.

Concerning (L): Proposition 1.21 tells us (L) p, so we only have to provethe converse inequality. But that one follows again from Lemma 1.17, or itsCorollary 1.18: Suppose we had (L) = q < p, then

ker L ran Lq =ker Lq+1

ker Lq= {0} ,

so ker L ran Lq = {0}, which would imply (L) = q < p, a contradiction!(ii) (iii) Let Rc(L) = {0} and (L) = p. Ifp = 0, then {0} = ker L = ker Lk for all

k N, so ker Lk ran Lp = {0}. Now, let p > 0. Using Corollary 1.18 again, wesee with ker Lp1 ker Lp:

ker L ran Lp1 =ker Lp

ker Lp1= {0}

and

ker Lk ran Lp =ker Lk ran Lp

ker Lk mul Lp=

ker Lp+1

ker Lp= {0} k N,

so ker Lk ran Lp = {0}.

20


22/57

1 PRELIMINARIES

(iii) (i) is trivial.

We have proved: If the generalized ascent is finite, it coincides with the traditional

ascent and there are no singular chains.Example 1.3 In general, the equality of ascent and generalized ascent does not im-ply the absence of singular chains. Consider X = 2(C) and the linear relationL X X which arises from the left-shift with the multivalued part mul L ={ C : (, 0, 0, . . .)}. Now, mul L ker L = {0}, but (L) = = (L).

To see that the abscence of singular chains alone does not imply that the generalizedascent is finite, consider any operator with infinite ascent, for example the left-shift in2(C). Then, the generalized ascent has to be infinite as well. But operators do nothave singular chains.

The authors of [6] have established various connections between (L) and (L), aswell as our generalized numbers, without explicitly introducing them. We express their

(more sophisticated) Theorem 5.7 with our notions.

Theorem 1.1 If generalized ascent and generalized descent of L are finite, they areequal and coincide with (the usual) ascent and descent of L. In short form:

(L), (L) < = (L) = (L) = (L) = (L).Proof: Let

p :=

(L)

1.26= (L), q := (L)

1.21

(L) =: r.

Assume p > q , then ran Lp = ran Lq, in particular

ker L ran Lq = ker L ran Lp = {0} ,

so (L) q < p = (L), an obvious contradiction, which means we have(L) = (L) = p q = (L) (L) = r (1.2)

Now (L) = p r implies ker Lr = ker Lp, in particular

X = ker Lr + ran L = ker Lp + ran L,

so actually r =

(L) p. This shows: (1.2) holds with equality everywhere.

The following lemma is (in a similar form) part of Theorem 8.3 in [ 6]. Again, wepresent its proof for the sake of completeness.

21


23/57

1 PRELIMINARIES

Lemma 1.27 Let p 1. If dom L ran Lp ker Lp, then L is completely reducedby (ran Lp, ker Lp).

Proof: Since ran L + ker Lp ran Lp ker Lp dom L dom Lp, we have (L) p.Abbreviate R := ran Lp and N := ker Lp. The intersection LN LR is trivial, becauseN R = {0}. Also, by definition of compression, LN LR L is obvious. Theconverse inclusion remains to be shown.

Let (x, y) L, then in particular x dom L, so x = x1 + x2 for some x1 R andx2 N. Now, there is a y2 X with (x2, y2) L and (y2, 0) Lp1. We see that(x1, y y2) = (x, y)(x2, y2) L. But x1 ran Lp, so yy2 ran Lp+1 = ran Lp = R.Hence, (x1, y y2) LR.

Moreover, (x2, y2) LN, since x2 ker Lp = N and also y2 ker Lp1 ker Lp =N. Hence, (x, y) = (x1, y y2) + (x2, y2) LR LN.

Lemma 1.28 dom Lm = dom Lm+1 ran Lm dom L + mul Lm

Proof: We will use Proposition 1.23, which states

ran Lm = ran Lm+1 ker Lm + ran L dom Lm.

Now, replacing L with L1, we obtain

dom Lm = dom Lm+1 ran(Lm)1 = ran(Lm+1)1

ran(L1)m = ran(L1)m+1 ker(L1)m + ran L1 dom(L1)m

mul Lm + dom L ran Lm.

Remark: We have shown that for the co-descent c(L) := inf

m N : dom Lm = dom Lm+1

the corresponding c(L) should be defined asc(L) := inf{m N : ran Lm dom L + mul Lm} .

Theorem 1.2 Let (L) = p < , (L) = q < (by Theorem 1.1 p = q) andran L dom L + mul L. Set N := ker Lp and R := ran Lp. Then L is completelyreduced by (N, R), LN is a nilpotent operator with nilpotence p and LR is surjectiveonto R.

Proof: We first show that L is completely reduced by the pair (N, R). According toLemma 1.27, we have to show dom L N R.

22


24/57

1 PRELIMINARIES

Since (L) = p, the intersection ker L ran Lp = {0} is trivial. Proposition 1.24implies ker Lp ran Lp = {0}, so the sum is direct. The sum ker Lp +ran L equals X,because (L) = p. Now, this implies ker L

p+ran Lp dom Lp by Proposition 1.25. The

inclusion ran L dom L + mul L shows dom L = dom L2 and thus dom L = dom Lp.So we also have ker Lp + ran Lp dom L.

We verify mul LN = {0}:

x mul LN ((0, x) L and x N) x mul L ker Lp.

But this intersection is trivial, because the finiteness of the generalized ascent impliesthe absence of non-trivial singular chains. In particular, x = 0. Thus, LN is anoperator. If (x, y) LpN, then y N ran L

pN ker L

p ran Lp = {0}. Proposition1.26 implies that LN is of nilpotence p.

Let y R = ran Lp = ran Lp+1, where the last equality is due to the fact that(L)

(L). But this already means that there is an element x R = ran Lp with

(x, y) L, so (x, y) LR holds true as well.

1.3 Splitting linear relations

This section is motivated by linear relations of type S1T for linear operators S, T asthey are used in [2]. As we have seen in Lemma 1.13, we can write every linear relationL in such a way: L = Q1L Lop.

First, we want to characterize relations of type S1T, where S and T are operators.

Lemma 1.29 For two linear operators S : Y dom S Z, T : X dom T Z,the product L := S1T is a linear relation in X Y, which can be characterized by

(x, y) S1T x dom T y dom S T x = Sy.

Proof: We have

S1T =

(x, y) X Y z Z : (x, z) T, (z, y) S1

= {(x, y) X Y | z Z : (x, z) T, (y, z) S}

= {(x, y) X Y | x dom T, y dom S z Z : T x = z,Sy = z}

= {(x, y) X Y | x dom T y dom S T x = Sy} .

Intuitively, in a relation L = S1T with S, T as in the preceding proposition, Trepresents the operator-like behavior ofL whereas S determines its multivalued part,though we have to be careful, since for L only those x X and y Y matter whichare mapped into ran S ran T by T and S respectively. The first evidence for thatintuition is the following lemma.

23


25/57

1 PRELIMINARIES

Lemma 1.30 Let S, T be linear operators as in Lemma 1.29 and let L := S1T.Then:

(i) dom L = T1[ran T ran S] dom T,

(ii) ran L = S1[ran T ran S] dom S,

(iii) ker L = ker T and

(iv) mul L = ker S.

Proof: The first two points follow immediately from Lemma 1.29. We have

ker L = {x X | (x, 0) L}1.29= {x X | x dom T 0 ran S T x = S0 = 0}

= {x dom T | T x = 0} = ker T.

By replacing L with L1, we get mul L = ker L1 = ker S, since L1 = T1S.

This justifies setting

(L) := inf{n N | ker L ran Ln = {0}},compare the definition of the ascent ofT relative to S in [2] in section 1, page 149:

(T : S) := inf

n N ker T ran(S1T)n = {0}.

Now, let L X Y be a linear relation. We examine to which extend L can bedecomposed into a product S1T of operators S, T : X Y.

Definition 1.8 We say a vector space Z and two linear operators T : X dom T Z and S : Y dom S Z split L with respect to the auxiliary space Z, ifL = S1T. The triple (Z,S,T) will be called split of L. In case dom T = dom Land dom S = ran L, we call (Z,S,T) a minimal split of L with respect to Z.

Proposition 1.31 Every split (Z,S,T) of L induces a minimal split(Zmin, Smin, Tmin) by setting Zmin := Z, Smin := S|ranL and Tmin := T|domL.

Proof: This follows immediately from Lemma 1.29.

A trivial, yet useful observation is

24


26/57

1 PRELIMINARIES

Proposition 1.32 If(Z,S,T) is a (minimal) split ofL, then (Z , T , S ) is a (minimal)split ofL1.

Lemma 1.33 Every linear relation L is splittable. More precisely:( ranLmulL

, QL, Lop) is a minimal split ofL, so L can be written as

L = Q1L Lop(= Q1L QLL),

where

Lop : X dom L ran L

mul L

is the associated linear operator and

QL : Y ran L ran Lmul L

is the canonical quotient map.

Proof: This was shown in Lemma 1.13 for QL : Y YmulL and Lop : dom L Y

mulL,

which we can obviously compress to ran L ranLmulL and dom L ranLmulL respectively.

This way we guarantee the minimality of the split and Lop and QL being surjective.

We will see that minimality of the split and surjectivity of the involved operatorscan always be achieved by modifying the operators involved marginally (in algebraicterms).

Applying the lemma to L1 yields

Corollary 1.34domLkerL , QL1 ,

L1

op

with QL1 : X dom L

domLkerL splits

L1 minimally.

Now applying Proposition 1.32 to the corollary, we get

Corollary 1.35

domLkerL

,

L1

op

, QL1

splits L.

But the two splits ofL we constructed are actually isomorphic, in an adequate sense.

25


27/57

1 PRELIMINARIES

Definition 1.9 Let (Z1, S1, T1) and (Z2, S2, T2) be splits of L. A linear map J :Z1 Z2 such that

S2 = J S1 and T2 = JT1

will be called a morphism (Z1, S1, T1) (Z2, S2, T2). We will also say J inducessuch a morphism.

Proposition 1.36 If (Z1, S1, T1) is a split of L and J : Z1 Z2 is an injectivelinear map, then (Z2, S2, T1) with T2 = J T1 and S2 = J S1 is a split ofL.

Proof: We have

S12 T2 = (J S1)1(J T1) = S

11 J

1J T1 = S11 T1 = L,

because J1J = IZ1 , see Lemma 1.9.

Remark: A straightforward calculation verifies that SplitL is a category given by the class

Ob(SplitL) :=

(Z, S, T) | Z F-vector space, S : Y dom L Z, T : X dom T Z

linear operators such that L = S1T

of splits, together with morphisms

MorSplitL((Z1, S1, T1), (Z2, S2, T2)) := {J : Z1 Z2 linear map such that

S2 = JS1, T2 = JS2}

for two given splits (Z1, S1, T1), (Z2, S2, T2), where the composition and an identity are givenby the composition of linear maps and identity on the vector space respectively. That justifies

the term morphism defined above.

Proposition 1.37 The following diagram of vector spaces and linear operators com-mutes:

dom L ran Lmul L

dom Lker L ran L

Lop

L1

opQL1 QL

J=

where J : ranLmulL

domLkerL

is the canonical linear isomorphism (L1)op associated

with (L1)op and J1 is the canonical linear isomorphism Lop associated with Lop.

26


28/57

1 PRELIMINARIES

Proof: The only thing we have to prove is J1 = Lop: let x dom L, thenJLop(x + ker L) =

(L1)opLop(x + ker L)= (L1)opLopx= (L1)op(y + mul L)

for some y ran L such that (x, y) L. Furthermore:

. . . =

L1op

y

= z + ker L

for some z dom L such that (z, y) L. But then (xz, 0) L, so z+ker L = x+ker L.This shows

J

Lop = IdomL

kerL

Exchanging the roles of L and L1 yields the other relationLopJ = IranLmulL

.

Corollary 1.38 The map J from the preceding proposition induces an isomorpismran L

mul L, QL, Lop

dom L

ker L, (L1)op, QL1

.

Using Lemma 1.30, we get the following import result about minimal splits (Z,S,T)with surjective T or surjective S: they are actually all isomorphic.

Theorem 1.3 Let L X Y be a linear relation minimally split by (Z,S,T). IfT is surjective, then there exists an isomorphism

dom L

ker L,

L1op

, QL1

(Z,S,T),

which is induced by the linear isomorphism associated with T : X dom L Z.As consequence all splits (Z,S,T) with surjective T are isomorphic.

The dual statement for splits (Z,S,T) with surjective S also holds true.

Proof: Let T : domLkerT Z be the linear isomorphism associated with T. By Lemma1.30, we have ker T = ker L, so actually T : domLkerL Z. Since domLkerL , L1op , QL1is a split it suffices to show

S = TL1op

and T = T QL1 .27


29/57

1 PRELIMINARIES

The second equality is the very definition of T, so we only have to prove the firstequality. It is more convenient to show the equivalent equation

T1S =

L1

op

: wehave T1S = QL1T1S = QL1L1 = QL1Q1L1 L1op ,where the last equality is an immediate consequence of Corollary 1.38. Now, sinceQL1 is an operator, by Lemma 1.9 it holds true that QL1Q

1L1

= IdomLkerL

. But thatmeans T1S = IdomL

kerL

L1

op

=

L1op

.

Corollary 1.39 For every minimal split (Z,S,T) ofL we have: T is surjective S is surjective.

Proof: Consider : This is obvious for the split ( ranLmulL , QL, Lop), and for every otherminimal split (Z,S,T) ofL we have an isomorphism J : ranL

mulL Zsuch that S = J QL,

which immediately implies the surjectivity of S. The other implication follows fromduality.

Definition 1.10 A minimal split (Z,S,T) with surjective S or T will itself hence-forth be called surjective.

For algebraic considerations, it suffices to consider surjective minimal splits, sinceevery split (Z,S,T) can be made minimal by restricting the domains of S and T, and

it can be made surjective by replacing the auxiliary space Z by the range of S, whichcoincides with the range ofT.

Lemma 1.40 Let (Z,S,T) be a minimal split ofL, then ran S = ran T =: Z1. Themaps S1 : Y ran L Z2, T1 : X dom L Z2 split L in Z2, and the embeddingZ2 Z1 is a morphism (Z2, S1, T1) (Z,S,T).

Proof: Let z ran T, then there exists an x dom L with T x = z and (x, y) L forsome y ran L. But since L = S1T, the latter means Sy = T x = z, so z ran S.The converse inclusion follows from using what we just showed on the split (Z , T , S )of L1. The rest of the lemma is trivial now.

Remark: At this point it becomes clear that all results of [2] using only numbers and spacesconcerning S1T with operators S, T can be applied to arbitrary relations.

28


30/57


31/57

1 PRELIMINARIES

Then (x, 0) K, so x ker K dom L = ker L, which means (x, 0) L. Togetherwith (x, y) L this implies (0, y) L, so y mul L. The converse inclusion is trivial. If conversely (1.4) holds true, we have

ker L1 = ker K1 dom L1,

which by what we just showed implies mul L1 = mul K1 ran L1, or equivalently

ker L = ker K dom L.

Again, assume (1.3), in order to show S11 T1 L. Using the equivalences we haveproved in the beginning, we examine (x, y) (dom L) (ran L) with (x, y) K. Wemust have some x dom L such that (x, y) L. But then (x x, 0) K, whichmeans x x ker K dom L = ker L. So actually (x x, 0) L, which implies

(x, y) = (x x, 0) + (x, y) L.

In [2], the authors also make use of T S1 for operators S, T. In the following para-graphs, we want to concern ourselves with this commutation of the split components.This is, of course, only possible for the case X = Y, so in the following we will alwaysconsider a linear relation L X X.

Definition 1.11 Let (Y,S,T) be a split of L. The relation

Lc := T S1 = {(y, y) Y Y | x dom S dom T : Sx = y T x = y}

= {(Sx,Tx) | x dom S dom T}

is called L commutated with respect to (Y,S,T).

In case of a surjective minimal split (Y,S,T), Lc is unique up to an isomorphism ofsplits, so we just call it L commutated.

Right from the definition, we see that Lc does only depend on the behavior ofS andT on dom S dom T.

Proposition 1.42 Let S, T be linear operators in X to Y. Then

T S1 = T|domSdomT(S|domSdomT)1.

Again, it is easy to see, but useful, that commutating a relation commutes withinverting.

30


32/57

1 PRELIMINARIES

Proposition 1.43 We always have L1c = (L1)c.

Proof: L1c = (T S1)1 = ST1 = (L1)c, since L1 is split by (Z , T , S ).

The usual linear subspaces ker, ran, dom associated with the powers of S1T andT S1 are the basis of all considerations in [2], cf. section 1, page 149 ibid. We willpartly recover their notions in the more general context of linear relations here.

Following [2], later we will only consider relations L and splits (Z,S,T) of thosewith dom T dom S. Dual to Lemma 1.30 we have

Lemma 1.44 Let L be splitted by (Y,S,T), then we have

(i) dom Lc = S[dom T], and dom Lc = ran S if and only ifdom S dom T+ker S,

(ii) ran Lc = T[dom S], and ran Lc = ran T if and only if dom T dom S+ ker T,

(iii) ker Lc = S[ker L] = S[ker T] and

(iv) mul Lc = T[mul L] = T[ker T].

Proof: We have

dom Lc = {y Y | z Y : (y, z) Lc}

= {y Y | z Y x dom S dom T : Sx = y T x = z}

= {y Y | x dom S dom T : Sx = y} = S[dom T].

Passing to the inverse, we get the following by what we have just shown:

ran Lc = dom L1c = dom(L

1)c = T[dom S]( ran T).

If dom T dom S+ ker T, then

ran T = T[dom T] T[dom S+ ker T] = T[dom S] = ran Lc.

If on the other hand ran T ran Lc, then

dom T = T1[ran T] T1[ran Lc] = T1[T[dom S]].

So, let x dom T, then T x = T v for some v dom S, which menas x v ker T andthus

x = v + (x v) dom S+ ker T.

Again, exchanging the roles of S and T via inverting Lc yields

dom Lc = ran S dom S dom T + ker S.

31


33/57

1 PRELIMINARIES

Concerning ker Lc:

ker Lc = {y Y | x dom S dom T : Sx = y, Tx = 0 = S0}

= S[{x dom S dom T | (x, 0) L}] = S[ker L] = S[ker T].Also:

mul Lc = ker(Lc)1 = ker L1c = T[ker L

1] = T[mul L] = T[ker S].

This justifies setting(Lc) := inf{n N | ran Lc + ker Lnc = Y},compare the definition of the descent of T relative to S in [2], section 1, page 149:

(T : S) := inf

n N ran T + ker(T S1)n = Y.

Remark: Without the additional assumption dom T dom S, we would have to considerthe relation (T S1)|ranT, if we wanted to reconstruct exactly the same spaces R

m from [2].Indeed, we have

R

1 = T X = ran T, R

m = ran((T S1)|ranT)

m1 m 2.

But we see, that the case m = 1 seems to be an exception, and the relation which we have toconsider for m 2 is rather cumbersome. A closer examination can improve the situation:It is easy to see that ((T S1)|ranT)

m = ((T S1)m)|ranT for all m 1. Concerning the firstflaw, we observe

ran((T S1)0)|ranT = ran IranT = ran T.

So we haveR

m = ran((T S1)m)|ranT m 1,

which is compact enough, but we would no longer deal with ranges of powers of a relation.On the other hand, note that dom T dom S implies

ker Lp + ran L dom L + ran L dom T + dom S dom S.

So unless dom S= X, (L) < is impossible.The pointwise characterization of Lc is by far not as handy as it was in the case

of S1T. But powers of Lc are closely related to those of L.

Proposition 1.45 Let n N, n 1, then

Ln = S1Ln1c T and Lnc = T L

n1S1,

or in terms of pairs: (x1, xn+1) Ln if and only if

(y1, yn) Ln1c : T x1 = y1 yn = Sxn+1, so if and only if

(T x1, Sxn+1) Ln1c .

Respectively: (y1, yn+1) Lnc if and only if

x1 dom S, xn dom T : (x1, xn) Ln1 y1 = Sx1 T xn = yn+1.

32


34/57

1 PRELIMINARIES

Proof: Follows easily by induction.

Corollary 1.46 Let n N, n 1, then(i) ker Ln = T1[ker Ln1c ],

(ii) ker Lnc = S[ker Ln].

If additionally dom T dom S, we have for n 1:

(iii) ran Ln = S1[ran Lnc ] and

(iv) ran Lnc = T[ran Ln1].

Proof: For n 1 argue using the preceding proposition:

x ker Ln (x, 0) Ln

(T x , S 0) = (T x, 0) Ln1c

x T1[ker Ln1c ];

also:

x S[ker Ln]

z dom S ker Ln : Sz = x

z dom S dom T : (z, 0) Ln Sz = x

z dom S dom T (y1, yn) Ln1c : T z = y1, yn = S0 = 0, Sz = x

z dom S dom T (y1, 0) Ln1

c : T z = y1, Sz = x y1 Y : (x, y1) Lc, (y1, 0) L

n1c

(x, 0) Lnc x ker Lnc .

Concerning (iii):

x ran Ln x1 dom T : (x1, x) Ln

x1 dom T : (T x1, Sx) Ln1c ,

where the last step is possible, because x ran Ln ran L dom S. Since we assumeddom T dom S, we can also apply S to x1, so that the last line is equivalent to

x1 dom T dom S : (Sx1, T x1) Lc, (T x1, Sx) Ln1c

x1 dom S dom T : (Sx1, Sx) Lnc

Sx ran Lnc x S1[ran Lnc ].

33


35/57

1 PRELIMINARIES

And for (iv):

y ran Lnc

z Y : (z, y) Lnc

z Y, x1 dom S, xn dom T : (x1, xn) Ln1, z = Sx1, T xn = y

x1 dom S, xn dom T : (x1, xn) Ln1, T xn = y.

(x1, xn) Ln1 already implies x1 dom T, and on the other hand, because of the

additional condition dom T dom S, the last line in the chain of equivalences is infact equivalent to

x1 dom T, xn dom T : (x1, xn) Ln1, T xn = y

y T[dom T ran Ln1] = T[ran Ln1].

Remark: Note that (iv) is in general not true for n = 0. We only have

ran L0 = X dom S = S1[Y] = S1[ran L0c ].

The following lemma condenses Lemma 1.3 and Lemma 1.4 from [2]. We will needit later on to prove a pertubation result.

Lemma 1.47 Let = 0 and m 1. Then

(i) ker Lmc = (S T)T1[ker Lm1c ] = (S T)[ker L

m],

(ii) ker Lm = T1(S T)[ker Lm1],

(iii) ran Lm dom T = (S T)1T[ran Lm1] = (S T)1[ran Lmc ] and

(iv) ran Lmc = T(S T)1[ran Lm1c ].

Proof:(i) Using Corollary 1.46 as well as Lemmas 1.6, 1.9 and 1.5, we get

(S T)[ker Lm]1.46= (S T)T1[ker Lm1c ]

1.6 + 1.9 (ST1 IranT)[ker L

m1c ]

1.5 L1c [ker L

m1c ] + ker L

m1c

1.5= ker Lmc + ker L

m1c

1.5= ker Lmc .

For the converse inclusion, suppose y ker Lmc = S[ker Lm], so there exists some

x0 ker Lm such that Sx0 = y. x0 ker Lm means we find an L-chain

(x0, x1, . . . , xn1, 0).

34


36/57

1 PRELIMINARIES

In particular: xn1 ker L = ker T. Set

u :=

n1

k=0 (k+1)xk.Obviously, u ker Lm dom T dom S. We have

T u =n2k=0

(k+1)T xk

=n2k=0

(k+1)Sxk+1

= (S)n1

k=1(k+1)Sxk+1

= (S)u Sx0 = (S)u y,

or equivalently y = (S T)u (S T)[ker Lm].(ii) By the first point, we have

ker Lm = T1[ker Lm1c ] = T1[(S T)[ker Lm1]].

(iii) For , we note that

(S T)[ran Lm dom T] = S[ran Lm dom T] + T[ran Lm]

ran Lmc + ran Lm+1c

ran Lmc = T[ran Lm1],

so by Lemma 1.5, we have ran Lmdom T (ST)1T[ran Lm1]. We showthe converse inclusion by induction. For m = 1 let

x (S T)1T[ran Lm1] = (S T)1T[Y]

so there exists some u dom T such that (S T)x = T u, or equivalently Sx =T(1(x + u)), which means (1(x + u), x) L. So we have x ran L dom T.The induction step is very similar: Let

x (S T)1T[ran Lm] (S T)1T[ran Lm1] = ran Lm dom T,

so there exists some u ran Lm dom T such that (S T)x = T u, or (1(x +

u), x) L. Since 1

(x + u) ran Lm

dom T, we have x ran Lm+1

dom T.The second equality follows from Corollary 1.46.(iv) For m = 1 we have

T(S T)[ran Lm1c ] = T(S T)1[X] = T[dom S dom T] = ran T = ran L1c,

35


37/57

1 PRELIMINARIES

where the last step follows from Lemma 1.44. For all other natural numbers, weuse (iii):

ran Lm+1c = T[ran L

m

dom T]= T[(S T)1[ran Lmc ]]

= T(S T)1[ran Lmc ] m 1.

36


38/57

2 POLES OF THE GENERALIZED RESOLVENT

2 Poles of the generalized resolvent

2.1 Algebraic decomposition

As in the treatment of this subject in the single operator case, we seek out to decomposethe domain of our relation L into a direct sum of a kernel part and a range part, inour case depending on a given split (Y,S,T) of L, in such a way that the restrictionsof S and T yield a bijective and a nilpotent part of L.

We have to further investigate the properties of commutated splits. So in the fol-lowing, let L X X be a linear relation, (Y,S,T) a split of L and let Lc be Lcommutated with respect to (Y,S,T), so Lc = T S1.

The following lemma can also be found on page 149 of [2]. It establishes a relation-ship between (L) and (Lc), (L) and (Lc) respectively.From now on we will only consider splits (Z,S,T) of L with dom T dom S.

Lemma 2.1 Let k, m N, then

(i) ker Lkc ran Lmc = S[ker L

k ran Lm] and

(ii) (ker Lk + ran Lm) dom S = S1[ker Lkc + ran Lmc ].

Proof:

(i) For k = 0 we have

ker Lkc ran Lmc = {0} = S[{0}] = S[ker L

k ran Lm].

For m = 0 we have

ker Lkc ran Lmc = ker L

kc = S[ker L

k] = S[ker Lk ran Lm].

Now let k, m N, k 1, m 1. Then

y ker Lkc ran Lmc = S[ker L

k] T[ran Lm1]

(x1, 0) Lk, (x1, x

m) L

m1 : Sx1 = y = T(xm)

(x1, 0) Lk, (x1, x1) L

m : Sx1 = y

y S[ker Lk ran Lm].

(ii) For m = 0 we have

(ker Lk+ran L0)dom S = Xdom S = dom S = S1[Y] = S1[ker Lkc +ran L0c].

37


39/57


Now let k, m N, m 1. If z (ker Lk + ran Lm) dom S, there exist x ker Lk, y ran Lm such that z = x + y. Using Corollary 1.46 we have y S1[ran Lmc ]. But also Sx S[ker L

k] = ker Lnc , so

Sz = S(x + y) = Sx + Sy ker Lkc + ran Lmc ,

as desired. If on the other the other hand z S1[ker Lkc + ran Lmc ] dom S,

we find x ker Lkc = S[ker Lk], y ran Lmc = T[ran L

m1c ] with Sz = x + y. We

thus find v dom S ker Lk and w dom T ran Lm1c such that Sv = x andT w = y. We conclude

Sz = Sv + T w,

or equivalently S(z v) = T w, so (w, z v) L, which means z v ran Lm.So

z = v + (z v) ker Lk + ran Lm.

Corollary 2.2 We always have (Lc) (L).Proof: Let p := (L). The case p = is trivial, so let p < , then

ker L ran Lp = {0} ker Lc ran Lpc = S[ker L ran L

p] = {0} (Lc) p. The situation is more complex for the generalized descent. For the traditional de-

scent we had Proposition 1.25, which stated that the definition of the descent remainstrue for ran Lk instead of ran L. We can establish a similar result for the generalizeddescent of Lc under the condition dom L ran L. In [2], this is Proposition 3.3.

Proposition 2.3 Let dom L ran L. If(Lc) = p < , then for all k N, k 1:(i) ker Lp + ran Lk = dom S and

(ii) ker Lpc + ran Lkc = Y.

Proof: It suffices to prove (ii): From Lemma 2.1 we know

(ker Lp + ran Lk) dom S = S1[ker Lpc + ran Lkc ],

so suppose we have proved (ii), we can conclude

(ker Lp + ran Lk) dom S = S1[ker Lpc + ran Lkc ] = dom S.

That meansker Lp + ran Lk dom S.

38


40/57


On the other hand ker Lp + ran Lk dom L + ran L dom T + dom S = dom S.Concerning (ii): is trivial. Since

(Lc) = p, we have the equality for k = 1.

Now, suppose it was shown for some k 1 and let z Y. Then we have some

x ker Lpc , y ran Lkc such that z = x + y. Corollary 1.46 gives y T[ran Lk1], sothere is some w dom T ran Lk1 with y = T w. Now, also w dom S, so from theinduction assumption and Lemma 2.1 we get

w = u + v

with u ker Lp and v ran Lk. Hence,

y = T w T[ker Lp] + T[ran Lk]1.46= T[T1[ker Lpc]] + ran L

k+1c ker L

pc + ran L

k+1c ,

as desired.

The following corollary is numbered 3.4 in [2].

Corollary 2.4 If(Lc) < , then dom S = dom Lm+ran Lk and Y = S[dom Lm]+ran Lmc for all k, m N.

Proof: Let p := (Lc) < , then by the preceding proposition dom S = ker Lp+ran Lkand Y = ker Lpc +ran L

kc for all k N. By Proposition 1.16, we have ker L

p dom Lm,and subsequently S[ker Lp] = ker Lpc S[dom L

m] for all m N. We get

dom S = ker Lp + ran Lk dom Lm + ran Lk dom T + dom S dom S

and

Y = ker Lpc + ran L

kc S[dom L

m

] + ran Lk

Y.

Before we can proof connections between and for L and Lc similar to Lemma1.26, we need an isomorphism-type result similar to 1.19, which makes use of Lc. Thefollowing lemma is the second part of Lemma 3.2 in [2]. We give the proof omittedthere.

Lemma 2.5 Let dom L ran L. For all k, m 1 we have

ran Lm

ran Lm+k=

S[dom Lm] + ran Lkcker Lmc + ran L

kc

.

Proof: From Lemma 1.19 we know

ran Lm

ran Lm+k=

dom Lm + ran Lk

ker Lm + ran Lk k, m 1.

39


41/57


Let k, m 1. We define a linear map

A : (dom Lm + ran Lk) S[dom Lm] + ran Lkc

ker Lmc + ran L

kc

, x [Sx].

It suffices to prove that A is well-defined, surjective and ker A = ker Lm +ran Lk. Theproof relies heavily on Corollary 1.46, which we will use without further notice.

A is well-defined: Firstly, dom Lm + ran Lk dom T + dom S dom S. And ifx dom Lm and y ran Lk = S1[ran Lkc ], we have

S(x + y) = Sx + Sy S[dom Lm] + ran Lkc .

We show ker A = ker Lm + ran Lk. So let x ker Lm, y ran Lk. ThenSx ker Lmc and Sy ran L

kc , so S(x + y) ker L

mc + ran L

kc , which means

[S(x + y)] = 0. Now, let x dom Lm, y ran Lk = S1[ran Lkc ] such that[S(x + y)] = 0, which means S(x + y) ker Lmc + ran L

kc = S[ker L

m] + ran Lkc ,

so we find x1 ker Lm

c and y1 ran Lk

c such thatSx + Sy = Sx1 + y1 S(x x1) = y1 Sy ran L

kc = T[ran L

k1].

Hence, there exists some y2 ran Lk1 with S(x x1) = T y2, which in turnmeans (y2, x x1) L. We get x x1 ran Lk, so there is a y3 ran Lk suchthat x = x1 + y3. Eventually, we get

x + y = x1 + (y3 + y) ker Lm + ran Lk.

Concerning surjectivity: Let x dom Lm, y ran Lkc . We have

A(x + 0) = [Sx] = [Sx + y].

Using this result, we give a shorter proof of Proposition 3.5 in [2].

Proposition 2.6 If(Lc) = 0, then dom S = ran Lk and Y = ran Lkc for k 1, inparticular (Lc) = 0, and

(L) =

0, dom S = X

1, dom S = X.

If 0 < (Lc) =: p < , then (L) = (Lc) = p.Proof: Firstly, suppose (Lc) = 0, then ran Lc = Y and subsequently ran Lkc = Y andthus ran Lk = S1[ran Lkc ] = dom S for k 1. We note that (Lc) = 0 can also bededuced from Proposition 1.21.

Now we examine the situation in case of 0 < (Lc) = p < , which means ker Lpc +ran L = Y and ker Lp1c + ran L = Y.

40


42/57


From Proposition 1.21 we already know (Lc) (Lc) = p. Concerning (L) p: By Proposition 2.3 and its Corollary 2.4 we know ker Lp +

ran L = dom S = dom Lp

+ ran L, and Lemma 1.19 givesran Lp

ran Lp+1=

dom Lp + ran L

ker Lp + ran L= {0} ,

so ran Lp ran Lp+1, which shows (L) p.

Suppose (L) < p, then ran Lp1 = ran Lp. Now, Lemma 2.5 gives

{0} =ran Lp1

ran Lp1+1=

S[dom Lp1] + ran Lc

ker Lp1c + ran Lc=

Y

ker Lp1c + ran Lc,

so ker Lp1c + ran Lc = Y, which contradicts

(Lc) = p. So ran Lp1 = ran Lp,

and since S is an operator, Lp1c = S1[ran Lp1] = S1[ran Lp] = ran Lpc , which

implies (Lc) p.

Now, we are able to show analogons of Theorem 1.1 and Theorem 1.2 for the casewhere both (L) and (Lc) are finite. Note again that due to our assumption dom T dom Swe must have (L) = , so the conditions of the previously mentioned theoremscannot be satisfied.

Proposition 2.7 If generalized ascent ofL and generalized descent of Lc are bothfinite, then they are equal and coincide with the ascent of L. In short form:

(L),

(Lc) < =

(L) = (L) =

(Lc).

In case (L) = 0 as well as dom S = X, or (L) > 0, the numbers also coincide withthe descent of L.

This is Proposition 4.1 in [2].

Proof: The proof is very similar to the one of Theorem 1.1: Let

p := (L) 1.26= (L), q := (L), (Lc) =: r.First we examine the case of r = (L) = 0. Then dom S ran Lm for all m N byProposition 2.6, and because of

ker L = ker T dom T dom S ran Lp,

we haveker L ran L0 = ker L = ker L ran Lp = {0} .

41


43/57


Thus, (L) = 0 = (Lc). If additionally dom S = X, we can continue this equationwith . . . = (L).

Now, let r > 0, then q = (L) = (Lc) = r by Proposition 2.6. Assume p > q, thenran Lp = ran Lq, in particularker L ran Lq = ker L ran Lp = {0} ,

so (L) q < p = (L), an obvious contradiction, which means we have(L) = (L) = p q = (L) = (Lc) = r. (2.5)

It remains to show that r p. By Corollary 2.2, we know (Lc) (L) = p < .Lemma 1.26 thus gives (Lc) = (Lc) p r, so ker Lrc = ker Lpc , in particular

Y = ker Lrc + ran Lc = ker Lpc + ran Lc,

so actually r =

(Lc) p. This shows: (2.5) holds with equality everywhere.

The following two theorems are Theorem 4.2 and Theorem 4.3 respectively in [2].

Theorem 2.1 If 0 < (L) = (Lc) =: p < , thendom S = ker Lp ran Lp, Y = ker Lpc ran L

pc . (2.6)

If on the other hand (2.6) holds for some p > 0, we have (L) = (Lc) p.Proof: By Proposition 2.3, we have ker Lp +ran Lp = dom S and ker Lpc +ran L

pc = Y.

Since (L) = p, the intersection ker L ran Lp = {0} is trivial. Then, using Lemma2.1, we get ker Lc ran Lpc = S[{0}] = {0}. We already know from Proposition 1.24that in that case

ker Lp ran Lp = {0} and ker Lpc ran Lpc = {0} .

(2.6) follows. If now (2.6) holds for 0 < p < , we have

ker L ran Lp ker Lp ran Lp = {0} , ker Lpc + ran Lc ker Lpc + ran L

pc = Y,

so (L), (Lc) p, and by the preceding proposition, they coincide. Theorem 2.2 L is completely reduced by (ker Lp, ran Lp), and we have S = S1 S2and T = T1 T2, where S1, T1 : kerLp ker Lpc and S2, T2 : ran L

p ran Lpc aredefined by

S1 := S|kerLp, T1 := T|kerLp, S2 := S|ranLp and T2 := T|ranLp .

We have(i) T2 is bijective,

(ii) S1 is bijective and LkerLp is a nilpotent operator whose nilpotence index is p.

42


44/57


Proof: L is completely reduced by (ker Lp, ran Lp), by Lemma 1.27, since

dom L dom T dom S ker Lp ran Lp.

We have, using Corollary 1.46,

S[ker Lp] = ker Lpc , S[ran Lp] ran Lpc ; (2.7)

T[ker Lp] ker Ln1c ker Lnc , T[ran L

p] = ran Lp+1c ran Lpc . (2.8)

So S1, S2, T1, T2 are well-defined, and by the preceding theorem we have S = S1 S2and T = T1 T2. From the calculations in (2.7) and (2.8) we also see that T2 and S1are surjective. Also:

ran Lp ker T = ran Lp ker L = {0} and

ker Lp ker S = ker Lp mul L 8 ker Lp ran Lp = {0} . (2.9)

since (L) = p, so T2 is also injective. (2.9) also shows that LkerLp is an operator.Obviously, LpkerLp = 0kerLp , so LkerLp is nilpotent with nilpotence index less or equalthan p. Suppose it was strictly smaller than p, so Lp1kerLp = 0kerLp . That would mean

ker Lp ker Lp1,

so (L) p 1, which contradicts (L) = p.

2.2 Poles of the generalized resolvent

We turn on to the main part of this work. In the following, X and Y will denoteF-Banach spaces, where F {R,C}, and S, T will denote linear operators in X to Y.

First, we generalized the notions of resolvent sets and resolvents.

Definition 2.1 The set

S(T) := { F | S T : X dom T Y is bijective}

is called S-resolvent set. For S(T) we call

R(S, T) : Y X, R(S, T)y = (S T)1y,

S-resolvent of T in .

We differ from [2] in that we first assume Sand T to be bounded, everywhere definedoperators, and relax that assumption later on.

8Proposition 1.16

43


45/57


Lemma 2.8 S(T) R(S, T) B(Y, X) is well-defined and analytic on theopen set S(T).

Before we prove that lemma, we recall two standard results.

Proposition 2.9 Let X be a Banach space, A X X be a linear relation suchthat A1 B (X) and B a linear operator with dom B dom A and BA1 B(X)such that

BA1 < 1. Then (A B)1 B(X) with(A B)1 = A1

k=0

BA1

k. (2.10)

Proof: We set C := BA1 B (X). Then

k=0

Ck k=0

Ck < ,

since C < 1. Thus, because X is a Banach space,

k=0 Ck converges and Ck 0.

We observe

(I C)n

k=0

Ck = I Cn+1 =

n

k=0

Ck

(I C),

so (I C) is invertible with

(I C)1 =k=0

Ck.

Furthermore:

A1k=0

Ck = A1(I C)1

= A1(AA1 BA1)1

= ((AA1 BA1)A)1

1.6= ((A B)A1A)1

1.9= (A B)1.

44


46/57


Corollary 2.10 If A has an inverse A1 B(X) and B B(X) such that B 0 such that S T is invertible for all F with | | < rusing Corollary 2.10. So S(T) is indeed open and we have, by (2.11),

(S T)1 = (S T)1k=0

((S T S+ T)(S T)1)k

= (S T)1k=0

( )k(S(S T)1)k,

which shows that R(S, T) is analytic.

The general strategy of the main results proof is similar to the one used in the singleoperator-case: Let (L) = (Lc) = p < , then we will show:

1. X can be decomposed into ker Lp ran Lp using Theorem 2.1.

2. S T is bijective on a deleted neighborhood of0.

3. ker Lp and ran Lp are closed.

4. S T restricted to ker Lp has a pole of order p at 0, and on the otherhand, if we restrict it to ran Lp, it is analytic.

The converse direction uses the identity theorem for Laurent series expansions andalgebraic considerations to show that (L) and (Lc) are finite.In order to show point 2 of the strategy, we have to show a pertubation result to

guarantee surjectivity and injectivity using the finiteness of (Lc) and (L), respec-tively. We will be using a standard theorem for that, which requires us to control kernel

45


47/57


and range of the operators which we examine. It is natural to restrict these operatores,so that they become injective (respectively, surjective), if

(L) (

(Lc) respectively) are

finite.

The pertubation result will also require the involved normed vector spaces to becomplete, which might fail, if we compress our operators. Since we are mainly inter-ested in the stability of algebraic properties, we are free to choose norms on our spacesthat guarantee their completeness. The following is called Remark 5.1 in [2].

Proposition 2.11 Let (E, E) and (F, )F be Banach spaces, A, B B(E, F).Set F0 := ran A, E0 := B1[F0]. Then A0 := A|E0 , B0 := B|E0 : E0 F0 arewell-defined operators and there exist norms E0 on E0 and F0 on F0 such that

(i) E0 and F0 are stronger than E and F respectively.(ii) (E0, E0) and (F0, F0) are complete.

(iii) A0 and B0 are bounded linear operators from (E0, E0) to (F0, F0).

Proof: Let A : E/ ker A F be the injective operator induced by A. A is bounded,so it is closed, and its inverse A1 : ran A E/ ker A is also closed. Equip F0 = ran Awith the graph norm

yF0 := y A1 = yF + A1y

E/ kerA y F0

induced by A1, which is stronger than F and with respect to which F0 is complete.Equip E0 = B1[F0] dom B with the norm

xE0 := xE + BxF0 = xE + BxF +

A1Bx

E/ kerA x E0.

induced by B|E0 : (E0, E) (F0, F0), which is stronger than the graph normB induced by B, which in turn is stronger than E . Hence, B0 : (E0, E0) (F0, F0) is obviously bounded, and B|E0 : (E0, E) (F0, F0) is closed: let(xn)nN E0 satisfy

xn

E x, Bxn

F0 y,

so, by the definition of E0 ,

xn

E0 x, B0xn

F0 y,

which means, since B0 is closed, x E0 and Bx = B0x = y. We conclude that(E

0,

E0) is complete. Thus, if we can show that A

0is closed, it is bounded by the

closed graph theorem. But since E0 and F0 are stronger norms than E andF respectively, the same is true for the associated product norms on E F, whichmeans that A0 is necessarily closed, because its graph coincides with that of the closedoperator A.

46


48/57


Remark: The proof of the proposition shows that it suffices to assume A as closed.

Now, we move onto the aforementioned compressions, which is the content of Lemma

5.2 in [2]. Remember, for now, S, T are bounded, everywhere-defined linear operatorsfrom (X, ) to (Y, ).

Lemma 2.12 Let m 0, set

Sm := S|ranLm , Tm := T|ranLm : ran Lm ran Lmc .

There exist stronger norms on ran Lm and ran Lmc such that these spaces are completeand Tm and Sm are bounded with respect to those norms.

Proof: For m = 0 there is nothing to do, we set 0 := X as norm on ran L0 = X

and ||0 := Y as norm on ran L0c = Y. The rest follows by induction: Supposewe have proved the claim for some m 0, then Sm, Tm are bounded operators between

the Banach spaces (ran Lm dom T, m) and (ran Lmc , ||m). Also:

ran Lm+1c = T[ran Lm]

= ran Tm,

ran Lm+1 dom T = S1[ran Lm+1c ] dom T

= S1m [ran Lm+1c ].

Now, Proposition 2.11 gives the desired norms.

As already mentioned, we will need a theorem from standard pertubation theory for

semi-Fredholm operators. It can, for instance, be found in [5], Satz 82.4.

Theorem 2.3 Let E, F be Banach spaces, A B (E, F) a semi-Fredholm operator,that is: n(A) := dim ker A < (called nullity ofA) and ran A is closed, or d(A) :=dim F/ ran A < (called defect of A; ran A is closed in that case)9. Then thereexists an > 0 such that for all B B(E, F) with B < , we have: A + B is alsoa semi-Fredholm operator and

n(A + B) n(A), d(A + B) d(A)

as well asn(A + B) d(A + B) = n(A) d(A).

9In the first case, A is called upper semi-Fredholm, and in the second case it is called lowersemi-Fredholm.

47


49/57


All we further need is a purely algebraic result, which relates kernel and range of thecompressions from Lemma 2.12 to those of S and T. The following is a reformulationof Lemma 1.5 in [2].

Lemma 2.13 Let m 0 and Sm, Tm as in Lemma 2.12.

(i) If = 0, then ker(Sm Tm) = ker(S T).

(ii) Ifm 1 and S[dom Lm] + ran Lmc = Y, and if = 0, then

ran Lmcran(Sm Tm)

=Y

ran(S T).

Proof:

(i) Let = 0. For m = 0 the equality is trivial, so let m 1.

ker(Sm Tm) = ker(S T) ran Lm ker(S T)

is obvious. Suppose x ker(S T), so T x = Sx = S(x), which means(x,x) L. Since = 0, (1x, x) L. A trivial induction shows (mx, x) Lm, so x ran Lm.

(ii) Firstly, we show

ran(S T) ran Lmc = ran(Sm Tm). (2.12)

is obvious. Concerning : Suppose x dom T and (S T)x ran Lmc =T[ran Lm1]. Then there exists some u ran Lm1dom T such that SxT x =

T u, so (1(x + u), x) L = L1,

in particular x ran L1. Also: v := 1(x + u) dom T and

(S T)u = Su T u S[ran Lm1] + T[ran Lm1]

ran Lm1c + ran Lmc = ran L

m1c .

So we have(S T)v ran Lm1c .

We can apply the same procedure as above to v instead of x, and thus a finiteinduction shows x ran Lm, so

(S T)x = (Sm Tm)x ran(Sm Tm).

Equation 2.12 implies

ran Lmcran(Sm Tm)

=ran Lmc

ran(S T) ran Lmc

(1.1)=

ran Lmc + ran(S T)

ran(S T).

48


50/57


By our hypothesis S[dom Lm] + ran Lm = Y, it suffices to show S[dom Lm] ran Lmc + ran(S T). To see this, let x dom L

m dom S. Then there existsan L-chain (x0, . . . , xm) with x0 = x, so

T xi = Sxi+1 0 i < m. (2.13)

We claim:

Sx =ki=0

(i+1)(S T)xi + (k+1)T xk 0 k m 1.

This is true for k = 0:

Sx = 1(Sx T x) + 1T x = 1(S T)x0 + 1T x0. (2.14)

Suppose it is true for some 0 k < m 1. Then:

Sx =ki=0

(i+1)(S T)xi + (k+1)T xk

2.13=

ki=0

(i+1)(S T)xi + (k+1)Sxk+1.

Replacing x by xk+1 in (2.14) and replacing Sxk+1 with the result in (ii) yieldsthe desired induction step.

Finally, we are ready to prove two pertubation results (Proposition 5.3, Corollary5.4 and Proposition 5.5 in [2]) for S and T, using the better behaved operators Smand Tm introduced before.

Proposition 2.14 Suppose (Lc) = p < . Then there exists > 0 such thatS T is surjective and

dim ker(S T) = dim(ker T ran Lp), (2.15)

for all 0 < || < .

Proof: If p = 0, then Y = ker L0 + ran T = ran T, so T is surjective, which meansd(T) = 0. And since it is also bounded, T is (lower) semi-Fredholm. By Theorem 2.3,there exists an > 0 such that if || < , S T is also semi-Fredholm, since S isbounded. Also, the thereom gives

d(S T) d(T) = 0

andn(S T) = n(S T) d(S T) = n(T) d(T) = n(T).

49


51/57


That means S T is surjective and we have

dim ker(S T) = dim ker T = dim(ker T ran L0).

Now, suppose p > 0. Let Sp and Tp be as in Lemma 2.12. Since (Lc) (Lc) = p,we have

ran Tp = T[ran Lp] = ran Lp+1c = ran L

pc ,

so Tp is surjective. Again, Sp is bounded, so Theorem 2.3 guarantees the existence ofsome > 0 such that for all || < , Sp Tp is also semi-Fredholm, surjective andsatisfies

dim ker(Sp Tp) = dim ker Tp = dim(ker T ran Lp).

By Lemma 2.13 (i), we have for = 0:

dim ker(S T) = dim ker(Sp Tp) = dim(ker T ran Lp).

We can also use 2.13 (ii), since Y = S[dom Lp] + ran Lpc by Corollary 2.4. This waywe get

Y

ran(S T)=

ran Lpcran(Sp Tp)

= {0} ,

since ran(Sp Tp) = ran Lpc . So S T is indeed surjective.

Corollary 2.15 Suppose (Lc) =: p < , and let > 0 as in the preceding propo-sition.

(i) If

(L) < , then S T is bijective for 0 < || < .

(ii) If(L) = , then S T is surjective, but not injective for 0 < || < .Proof:

(i) If(L) < , by Proposition 2.7, it equals p, so we haveker L ran Lp = {0} ,

which in view of Proposition 2.14 yields for all 0 < || < :

dim ker(S T) = dim ker L ran Lp = 0,

so S T is injective. Also, by that proposition, S T is surjective.

(ii) Follows from the same arguments.

Eventually, we are able to prove the main theorem of this work.

50


52/57


Theorem 2.4 Let X, Y be Banach spaces and let S, T : X Y be bounded linearoperators. Let p 1, then:(L) = p = (Lc) if and only if R(S, T) has a pole of order p at 0.

This is Theorem 5.6 in [2], except that their assumptions on S and T are less strict.

Proof: Firstly, suppose (L) = p = (Lc). By the algebraic decomposition Theorem2.1, we know

X = ker Lp ran Lp, Y = ker Lpc ran Lpc . (2.16)

These decompositions are also topological: By Corollary 2.15, there exists some > 0such that S T is bijective for all 0 < || < . Fix such a and let

A := (S T)1T, B := T(S T).

Using Lemma 1.47 and a straightforward induction, we see that

ker Lm = (A1)m[ker L0] = ker Am,

ker Lmc = (B1)m[ker L0c ] = ker B

m,

ran Lm = ran Lm dom T = Am[ran L0] = Am[X] = ran Am,

ran Lmc = Bm[ran L0c ] = B

m[Y] = ran Bm.

Since A and B are bounded, ker Lm and ker Lmc are closed, and because of our decom-position (2.16), ran Lp and ran Lpc must also be closed, which follows from a well-knownresult, see for instance [5], Satz 55.3.

By Theorem 2.2, we can decompose S and T as

X = ker Lp ran Lp

Y = ker Lpc ran Lpc

S T S1 T1 S2 T2

Since the decomposition of X in (2.16) is topological, S1, S2 and T1, T2 are bounded.We know that T2 is bijective, so similar arguments as in the proof of Lemma 2.8 showthat (S2 T2)1 is an analytic function on a neighborhood B(0) of 0. We alsoknow that S1 is bijective and that LkerLp = S

11 T1 is a nilpotent operator with index

p. The latter means that (LkerLp) = F \ {0}, where denotes the usual resolvent set.Thus:

S1 T1 = S1(I S11 T1),

51


53/57


is bijective for = 0, and for those we have the following series expansion

(S1 T1)1 = (S1(I

1LkerLp))1

= 1 I 1LkerLp1 S11= 1

k=0

kLkkerLp

S11

=

p1k=0

(k+1)LkkerLpS11

=

1k=p

kL(k+1)kerLp S

11 .

That shows: (S1 T1)1 has a pole of order p at 0. It is easy to verify that

(S T)1 = (S1 T1)1 (S2 T2)1,

for B(0) \ {0}, so R(S, T) has a pole of order p at 0, indeed.Now conversely assume that R(S, T) has a pole of order p at 0, so we have a series

expansion of the form

R(S, T) =

k=p

kCk

on a punctured neighborhood B(0) of 0, where Ck B(Y, X) for all p k andCp = 0. Let y Y and set xk = Cky for all k p. Then

0y = y = (S T)R(S, T)y

= (S T)

k=p

kxk

=

k=p

k+1Sxk

Poles of Generalized Resolvent Operators

Documents

Transcript of Poles of Generalized Resolvent Operators