Poles of Generalized Resolvent Operators

Abstract Linear relations, or multi-valued linear operators, have proven a useful tool in many branches of operator theory. They arise naturally in the context of linear operators, as their formal inverse, adjoint or closure. Although the general concept is known for over 75 years now (see for instance [7], [8], [1]), there is still only one book treating the subject, namely [3]. Therefore a general introduction to the theory of linear relations is given in the first part of the present work, focusing on algebraic properties and some generalizations of linear operator concepts. Also, the new notion of splitting a linear relation L into two linear operators S and T such that L = S −1 T is introduced and studied, as well as the linear relation T S −1 arising from the commutation of both factors S −1 and T The second part is concerned with the application of the formerly introduced ideas to the treatment of poles of a generalized resolvent λS −T for two operators S and T . In [2], the authors examine the same problem. The vector spaces they 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 linear relations. Zusammenfassung Das Konzept linearer Relationen, auch mehrwertige lineare Operatoren genannt, hat sich in vielen Bereichen der Operatortheorie als nützliches Hilfsmittel erwiesen. Lineare Relationen stellen eine natürliche Verallgemeinerung (einwertiger) linearer Operatoren dar. Relationale Inverse, Adjungierte und Abschlüsse der Graphen von linearen Operatoren sind oft selbst keine Operatoren, man kann sie aber stets als lineare Relationen auffassen. Obwohl das Konzept schon vor mindestens 75 Jahren eingeführt wurde (siehe [7], [8], [1]), ist die Fachliteratur zu dem Thema immer noch rar gesät. Das Buch [3] ist die einzige weitverbreitete Referenz. Aus diesem Grund befasst sich der erste Teil der vorliegenden Arbeit mit der Einführung in die Theorie der linearen Relationen, mit dem Hauptaugenmerk auf ihre algebraischen Eigenschaften und der Verallgemeinerung von Konzepten linearer Operatoren. Zudem wird eine neue Darstellungsmöglichkeit linearer Relationen eingeführt: Jede solche Relation L lässt sich als S −1 T schreiben, wobei S und T lineare Operatoren zwischen entsprechenden Vektorräumen sind. Die Eigenschaften solcher sogenannter „Splits“ werden untersucht, zusammen mit der natürlich auftretenden, neuen Relation T S −1 . Der zweite Teil befasst sich mit der Anwendung der allgemeinen Konzepte auf die Charakterisierung von Polen einer verallgemeinerten Resolvente λS − T über verallgemeinerte Auf- und Absteigeindizes, wobei T ein abgeschlossener und S ein T -beschränkter linearer Operator ist. H. Bart und D. C. Lay befassen sich in ihrem Artikel [2] mit derselben Fragestellung. Die in jenem Artikel eingeführten Vektorräume stellen sich als Kerne und Bildräume der linearen Relationen S −1 T und T S −1 heraus, und können mit in der im ersten Teil eingeführten Terminologie linearer Relationen strukturierter beschrieben und mit den entwickelten Mitteln untersucht werden.

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Contents 1 Preliminaries 2 1.1 Linear relations and related notions . . . . . . . . . . . . . . . . . . . . . 2 1.2 Ascent, descent and generalizations . . . . . . . . . . . . . . . . . . . . . 11 1.3 Splitting linear relations . . . . . . . . . . . . . . . . . . . . . . . . . . . 23 2 Poles of the generalized resolvent 37 2.1 Algebraic decomposition . . . . . . . . . . . . . . . . . . . . . . . . . . . 37 2.2 Poles of the generalized resolvent . . . . . . . . . . . . . . . . . . . . . . 43

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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) linear operator between two vector spaces. Some of the properties which we are about to show do not need the linear structure, but we will not concern ourselves with fine details. X, Y and Z will always denote F -vector spaces, where for now, F may be an arbitrary field. In particular, “subspace” will always mean “linear subspace”, without any 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 X to Y . In case of Y = X, we write R(X) := R(X, X). Furthermore, we define: • ker L := {x ∈ X | (x, 0) ∈ L} (kernel of L), • mul L := {y ∈ Y | (0, y) ∈ L} (multivalued part of L), • 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 single vector x ∈ X: Lx := L[{x}] = {y ∈ Y | (x, y) ∈ L} .1 • The inverse linear relation L−1 ⊆ Y × X is defined as L−1 := {(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 restriction of L to subspace U ⊆ X of its domain, so L|U := L(U,Y ) = {(x, y) ∈ U × Y | (x, y) ∈ L} .

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From now on, let L ⊆ X × Y be a linear relation. One can easily verify that kernel 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 to be defined as triple (X, Y, L), as in the case of functions. But in most cases, it will be clear from 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 of their definition regarding domain and range. For an operator T : X → Y , the inverse T −1 need not be an operator, but it always is a linear relation. Lemma 1.1 The following symmetries hold true: (i) ker L = mul L−1 , (ii) ran L = dom L−1 , (iii) (L−1 )−1 = L. Proof: An easy calculation shows: (x, 0) ∈ L ⇐⇒ (0, x) ∈ L−1 and (x, y) ∈ L ⇐⇒ (y, x) ∈ L−1 ⇐⇒ (x, y) ∈ (L−1 )−1 .  By simply adding or subtracting the vectors, one verifies that the following implications 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}.

1 We 2 In

will justify this notation in Definition 1.3 and Proposition 1.12. the rest of this work we shall not distinguish between a linear operator and its graph.

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Proof: If L is an operator, then its graph is a linear subspace of X × Y , and L0 = 0, or in our relational notation: mul L = L0 = {0}. On the other hand, if mul L = {0}, then L 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 ∈ N3 in the obvious way, where L0 := IX (or “I” for short) is the identity map on X. For m ∈ Z, m < 0, we set Lm := (L−1 )m . (iii) For a scalar α ∈ F we set αL := {(x, αy) ∈ 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 subb so spaces, we will explicitly use “ +”, b M := {(x1 + x2 , y1 + y2 ) ∈ X × Y | (x1 , y1 ) ∈ L, (x2 , y2 ) ∈ M } . L+

The usual sum of linear subspaces U, V ⊆ X which are no linear relations will be denoted by U + V . If the sum is direct, that is U ∩ V = {0}, then we express this by writing U ⊕ V . b LV , we call L completely reduced (v) Let U, V ⊆ X be subspaces. If L = LU ⊕ by 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, α ∈ F . Then KL, αL and L + M are linear relations. “◦” is associative; also, (R(X, Y ), +) 3 In

this work, 0 ∈ N.

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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 and U ⊆ V ⊆ X, W ⊆ Y be subsets. Then: S (i) L[U ] = x∈U Lx. (ii) If L is an operator, taking preimages is the same as taking images under the inverse relation: (L)−1 [W ] = L−1 [W ].

(iii) M [ker L] = ker LM −1 . (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 ∈ L−1 [W ]; thus: if L[U ] ⊆ W , then U ⊆ L−1 [W ]. (vii) (L + N )[U ] ⊆ L[U ] + N [U ]4 .

Proof: (i) If x ∈ U \ dom L, then Lx = ∅, so we can write [ [ L[U ] = Lx = Lx. x∈U∩dom L

x∈U

(ii) (L)−1 [W ] = {x ∈ X | ∃ y ∈ W : (x, y) ∈ L}  = x ∈ X ∃ y ∈ W ∩ dom L−1 : (y, x) ∈ L−1 = L−1 [W ].

(iii)

4 Note

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) ∈ M −1  = z ∈ Z (z, 0) ∈ LM −1 = ker LM −1 .

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

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(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) ∈ L−1 , which means x ∈ L−1 [W ]. Now the second part follows 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 conditions. 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) ∈ K and (y, z) ∈ L+M . There are vectors z ′ , z ′′ ∈ Z with (y, z ′ ) ∈ L, (y, z ′′ ) ∈ M and z = z ′ + z ′′ . This implies (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 of the operator-like sum). Let (x, z) = (x, z ′ + z ′′) ∈ LK + M K, 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.

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PRELIMINARIES • To show the second statement, let (x, z) ∈ LK + LJ, that means (x, z ′ ) ∈ LK and (x, z ′′ ) ∈ LJ for some z ′ , z ′′ ∈ Z with z = z ′ + z ′′ . Consequently, there are some 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 already shows (x, z) ∈ L(K + J). Now, suppose mul L ⊆ mul LK + LJ and ran K + ran J ⊆ dom L in order to show that equality holds. Let (x, z) ∈ L(K + J), that is, there is some y ∈ Y such that (x, y) ∈ K + J and (y, z) ∈ L. By the definition of the operatorlike sum, we find vectors y ′ , y ′′ ∈ Y with y = y ′ + y ′′ such that (x, y ′ ) ∈ K and (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 in X to Y do not form a vector space in general. As we have noted in Proposition 1.4, R(X, Y ) equipped with the addition from Definition 1.2, (iv), is a commutative monoid with identity element 0 := X × {0F }. Of course, the everywhere defined linear operators from X to Y form a submonoid of (R(X, Y ), +), which is even an abelian group. And more precisely: Proposition 1.7 L − L = 0 · L holds if and only if L is a linear operator. L is (additively) invertible in (R(X, Y ), +) if and only if it is an everywhere defined linear operator. Proof: We have L − L = 0 · L ⇐⇒ L − L = dom L × {0} , which is only true, if mul L = {0}: If y1 , y2 ∈ mul L with y1 − y2 6= 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, note that dom 0 = X and dom(L + K) = dom L ∩ dom K for all L, K ∈ R(X, Y ). Hence we must have dom L = dom 0 = X. Furthermore, for (x, 0) ∈ L + K to hold, it is necessary that (x, −z) ∈ K for some z ∈ Y with (x, z) ∈ L. If there was some y ∈ Y with y 6= z and (x, y) ∈ L, we would have (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 even expect (R(X, Y ), +, ·) to form an F -semimodule5 , where · denotes the scalar multiplication 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 ),

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which by Proposition 1.7 can only be true in trivial cases. But we have Proposition 1.8 Let G be a subsemifield of F such that α+β = 0 implies α = β = 0 for all α, β ∈ G6 . Then (R(X, Y ), +, ·) is a commutative G-semimodule with zero.

Proof: The fact that α(L + M ) = αL + αM , as well as α(βL) = (αβ)L, hold true for all α, β ∈ F and L, M ∈ R(X, Y ) is easy to see. Now, let α, β ∈ G. From Lemma 1.6, we know (α + β)L = (α + β)IL = (αI + βI)L ⊆ αIL + βIL = αL + βL. If α + β = 0, then α = β = 0. We note that 0 · L = dom L × {0} is additively idempotent, so (α + β)L = 0 · L = 0 · L + 0 · L = αL + βL. If α + β 6= 0, we consider (x, αy1 + βy2 ) ∈ (αL + βL), where (x, y1 ), (x, y2 ) ∈ L. Then (x, (α + β)y1 ), (x, (α + β)y2 ) ∈ (α + β)L, so



   β α x, αy1 , x, βy2 ∈ (α + β)L, α+β α+β

which eventually implies

(x, αy1 + βy2 ) ∈ (α + β)L.



Example 1.1 By what we just have shown, we see that the condition of Lemma 1.6 is not necessary. For all α, β ∈ F with α + β 6= 0 and all K ∈ R(X, Y ) we have (αI + βI)K = (α + β)IK = (α + β)K = αK + βK = (αI)K + (βI)K, 5 Let

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.

(S ∗ , ·)

It is called (nontrivial) semifield, if additionally is a group, where S ∗ := S \ {o}, if o is the zero of (S, +), and S ∗ := S otherwise. Let (S, +) be a semigroup, (R, +, ·) a semiring and · : R × S → S a mapping called scalar multiplication denoted by (α, s) 7→ α · s = αs. (R S, +, ·) is called a (left) R-semimodule, if α(a + b) = αa + αb, (α + β)a = αa + βa, α(βa) = (αβ)a

∀ α, β ∈ R, a, b ∈ S.

It is called unitary, if (R, +, ·) has an (multiplicative) identity ε and εa = 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 the zero in R. 6 Semifields satisfying that property are called zero-sum free, see [4] page 428.

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but ker(αI) = ker(βI) = {0}, while mul K may be arbitrary. On the other hand we have already seen in Proposition 1.7 that 0·K = (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 the inverse linear relation and inverse operator coincide. However, the product of a linear relation and its inverse is in general not the identity operator. Lemma 1.9 The following identities hold true: b ({0} × mul L). In particular, if L is a linear operator, we have • LL−1 = Iran L + −1 LL = Iran L . b (ker L × {0}), • L−1 L = Idom L + where Iran L := {(y, y) ∈ Y × Y : y ∈ ran L} is the identity operator on ran L and Idom L is defined analogously. Proof: Let (y, y ′ ) ∈ LL−1 , that is, there exists an element x ∈ X with (y, x) ∈ L−1 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 b ({0} × mul L). (y, y ′ ) ∈ I +

b Conversely, let (y, y +z) ∈ Iran L +({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) ∈ L−1 and (x, y + z) ∈ L, (y, y + z) ∈ LL−1 holds true. The second equation follows from passing to inverses: from the first equation we know b ({0} × mul L−1 ) = Idom L + b ({0} × ker L). L−1 L = L−1 (L−1 )−1 = Iran L−1 +

Inverting on both sides yields, since passing to inverses commutes with taking sums of subspaces, −1 −1 b b (ker L × {0}). L−1 L = Idom = Idom L + L + ({0} × ker L)



Though we have seen that the relational inverse L−1 and the operator-like negative −L are no inverses in an algebraic sense, they are obviously involutive, and they even commute. Lemma 1.10 We always have (L−1 )−1 = L and −(−L) = L, as well as (−L)−1 = −L−1 .

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Proof: We only proof the last claim: (x, y) ∈ L ⇐⇒ (y, x) ∈ L−1 ⇐⇒ (y, −x) ∈ −L−1 ⇐⇒ (−y, x) ∈ −L−1 , 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 = L−1 K −1 . In particular, if X = Y , we have L−n := (L−1 )n = (Ln )−1 ∀ n ∈ N. Proof: Let (z, x) ∈ (KL)−1 , so we have (x, z) ∈ KL, which means there exists some y ∈ Y such that (x, y) ∈ L and (y, z) ∈ K. But that implies (y, x) ∈ L−1 and (z, y) ∈ K −1 , and thus (z, x) ∈ L−1 K −1 . The converse inclusion follows analogously. Since Lx is a coset of mul L, we can associate a linear operator with a linear relation L by factoring Y by the multivalued part. More precisely Definition 1.3 Let QL : Y → Y / mul L be the canonical quotient map. Define the linear relation Lop := QL L ⊆ X × (Y / mul L), which we call linear operator associated with L. This name is justified in the following proposition. Proposition 1.12 The linear relation Lop is a well-defined linear operator with ran L Setdom Lop = dom L, ker Lop = ker L and ran Lop = QL [ran L] = mul L. theoretically we have Lop x = Lx ∀ x ∈ dom Lop . Lop and mul L determine L ⊆ X × Y uniquely, that is, if M ⊆ X × Y is another linear relation then Lop = Mop and mul L = mul M already imply L = M .

Proof: • We use Lemma 1.3: Let [y] ∈ mul QL L, 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 with the second statement, we get [y] = [0].

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PRELIMINARIES • Now for the mnemonic equality Lop x = Lx, let y ∈ Lx for some x ∈ X, which means (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 Lop x = 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 the converse inclusion, let x ∈ ker L. Now, Lemma 1.2 (ii) implies Lop x ⊆ 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 of X and {yi : i ∈ I} ⊆ Y , then L 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 = Q−1 L Lop . Proof: Let (x, y) ∈ L, so x ∈ dom L and y ∈ ran L. We have Lop x = y + mul L = QL y, −1 so (x, y) ∈ Q−1 L Lop . Now, let (x, y) ∈ QL Lop , then x ∈ dom L and y ∈ ran L with

Lx = Lop x = QL y = 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 relations L ⊆ X × X. In definition 1.2, we introduced powers of such relations. It is often convenient to characterize them as follows.

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Definition 1.4 Let n ∈ N, n ≥ 1 and x1 , . . . , xn ∈ X. The n-tuple (x1 , . . . , xn ) is said 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 ( ) [ m Ch(L) := (x1 , . . . , xn ) ∈ X n ∈ N, n ≥ 1 . m∈N

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 if x = y, so if and only if there is some – admittedly degenerated – L-chain (x1 ) with x = x1 = y. The case n ≥ 1 follows 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 6= 0 for at least one 1 < k < n. In that case, if (x, y) ∈ Lk+(n−k) , there exists some z ∈ X such that (x, z) ∈ Lk and (z, y) ∈ Ln−k . Since also (0, xk ) ∈ Lk , (xk , 0) ∈ Ln−k , we have (x, z + xk ) ∈ Lk , (z + xk , y) ∈ Ln−k , 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 6= 0, the singular chain is said to be non-trivial, otherwise it will be called trivial. As indicated beforehand, the existence of non-trivial singular chains has an easy characterization. Proposition 1.15 There exists an non-trivial singular L-chain of length n + 1 ≥ 3 if and only if there exists a 1 ≤ k < n with ker Ln−k ∩ mul Lk 6= {0}.

7 Note

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

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Proof: Suppose (x0 , . . . , xn ) ∈ Ch(L) is a non-trivial singular chain of length n + 1 ≥ 3 with xk 6= 0 for some 1 ≤ k < n. Then we have (0, xk ) ∈ Lk and (xk , 0) ∈ Ln−k , so 0 6= xk ∈ ker Lk ∩ mul Ln−k . – If on the other hand ker Ln−k ∩ mul Lk 6= {0} for some 1 ≤ k < n, choose 0 6= xk ∈ ker Ln−k ∩ 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) = ker Lm ∩ mul Lm , m∈N

m∈N

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 and mul 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 Lk ∩ ran Lm . = ker 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 lefthand side simplifies to ker Lk for the same reason; for the right-hand side we observe ran Lm = ran I = X and mul Lm = mul I = {0}, which means on the right we are also left 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

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whose kernel coincides with ker Lm . We define it as follows: Given x ∈ ker Lm+k , we find an L-chain (x, x1 , . . . , xm+k−1 , 0) using which we set Ax := xm + (ker Lk ∩ mul Lm ) = [xm ]. • A is indeed a map with stated codomain: (x, . . . , xm ) and (xm , . . . , xm+k−1 , 0) are L-chains, that is xm ∈ ran Lm and xm ∈ ker Lk , so xm ∈ ker Lk ∩ ran Lm . If we are given another L-chain (x, x′1 , . . . , x′m+k−1 , 0), then (0, x1 − x′1 , . . . , xm − x′m , . . . , xm+k−1 − x′m+k−1 , 0) is also an L-chain, likewise (0, x1 −x′1 , . . . , xm −x′m ) and (xm −x′m , . . . , xm+k−1 − x′m+k−1 , 0). That means we have xm − x′m ∈ mul Lm and xm − x′m ∈ ker Lk , so xm − x′m ∈ ker Lk ∩ mul Lm und hence [xm ] = [x′m ]. • The linearity of A is evident. k )∩ran(Lm ) • Concerning surjectivity: Let [y] ∈ ker(L ker Lk ∩mul Lm , which means there are Lchains (y, x1 , . . . , xk−1 , 0), (x′0 , x′1 , . . . , x′m−1 , y). Then (x′0 , x′1 , . . . , x′m−1 , y, x1 , . . . , xk−1 , 0) is an L-chain with length m+ k and y as m-th place. We conclude x′0 ∈ ker Lm+k and Ax′0 = [y]. • Concerning the kernel of A: Let x ∈ ker Lm , then there exists an L-chain (x, x1 , . . . , xm−1 , 0, . . . , 0 ), | {z } (k+1) times

which means Ax = [0] = 0. Now, let x ∈ ker Lm+k together with an L-chain (x, x1 , . . . , xm+k−1 , 0) and Ax = 0, which means we have xm ∈ ker Lk ∩ mul Ln . Hence there exist L-chains ′ (xm , y1 , . . . , yk−1 , 0), (0, y1′ , . . . , ym−1 , xm ).

It follows ′ (0, y1′ , . . . , ym−1 , xm , y1 , . . . , yk−1 , 0) ∈ Ch(L).

Subtracting this chain from the one belonging to x we get ′ (x, x1 − y1′ , . . . , xm−1 − ym−1 , xm − xm , xm+1 − y1 , . . . , xm+k−1 − yk−1 , 0). | {z } =0

In particular:

′ (x, x1 − y1′ , . . . , xm−1 − ym−1 , 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 simplified to Lemma 4.4 of [6].

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Corollary 1.18 Let L ⊆ X × X be a linear relation with no non-trivial singular chains, and k, m ∈ N. Then: ker Lm+k ∼ = ker Lk ∩ ran Lm . ker Lm Proof: The absence of non-trivial singular chains implies ker Lk ∩ mul Lm = {0}. Now apply the preceding lemma.  Analogously, we find a descriptive isomorphism for the ranges of a relation’s powers. The first part of the following lemma can be found in [6], numbered 4.1, the second one in [2] as Lemma 3.1. Lemma 1.19 Let L ⊆ X × X be a linear relation and k, m ∈ N, then: m k ran Lm ∼ dom Lm ∼ dom L + ran L . = = m+k m k m m k ran L (ker L + ran L ) ∩ dom L ker L + ran L

Proof: We show the first isomorphism: Firstly, we care about trivialities again: Is k = 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 the right using dom L0 = X:

X ran Lk

on the left, as well as on

X X . = (ker L0 + ran Lk ) ∩ 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 . Furthermore: Given another L-chain (x, x′1 , . . . , x′m ), then (0, x1 − x′1 , . . . , xm − x′m ) ∈ Ch(L),

15

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PRELIMINARIES hence ( 0, . . . , 0 , x1 − x′1 , . . . , xm − x′m ) ∈ Ch(L), | {z } (k+1) times

so we have xm − x′m ∈ ran Lm+k , which gives [xm ] = [x′m ]. • A is clearly linear. ran Lm , which means there is an L-chain • Concerning surjectivity: Let [y] ∈ ran Lm+k (x0 , x1 , . . . , xm−1 , y). Then x0 ∈ dom Lm and Ax0 = [y]. • Concerning the kernel of A: Firstly let x ∈ ker A, so because of Ax = [0] we have an L-chain (x, x1 , . . . , xm−1 , xm ) with xm ∈ ran Lm+k . This means on the other hand that there exists an L-chain (x′0 , . . . , x′m+k−1 , x′m+k = xm ). Using the last two chains, we get (x′k − x, x′k+1 − x1 , . . . , x′m+k−1 − xm−1 , 0) ∈ Ch(L). It follows x′k − x ∈ ker Lm . Together with x′k ∈ ran Lk , we get x = −(x′k − x) + x′k ∈ ker Lm + 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 some u ∈ ker Lm , v ∈ ran Lk with associated L-chains (u, u1 , . . . , um−1 , 0), (v0 , . . . , vk−1 , 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 , . . . , xm−1 − um−1 , xm ) ∈ Ch(L). Joining this chain and the one associated with v, we obtain (v0 , . . . , vk−1 v, x1 − u1 , . . . , xm−1 − um−1 , xm ) ∈ Ch(L), which means xm ∈ ran Lm+k und consequently Ax = [xm ] = 0. The second isomorphism is an easy consequence of the general result U ∼ U +V = U ∩V V for subspaces U, V of X.

(1.1) 

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For the case of linear operators in X, the notions of ascent and descent play an important role, for instance regarding the poles of the resolvent, which we seek to generalize, following [2]. We can recover their definitions for linear relations, but we will also introduce stronger versions. Definition 1.7 We  define: • α(L) := inf m ∈ N ker Lm = ker Lm+1 – ascent of L, • α b(L) := inf {m ∈ N | ker L ∩ ran Lm ⊆ mul Lm }, • α e(L) := inf {m ∈ N | ker L ∩ ran Lm = {0}} – generalized ascent of L, • δ(L) := inf m ∈ N ran Lm = ran Lm+1 – descent of L, b • δ(L) := inf {m ∈ N | ker Lm + ran L ⊇ dom Lm }, e • δ(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 proofs here 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: Assume ker Lm ⊇ ker Lm+k was already shown for some k ∈ N. Now, given x ∈ ker Lm+k+1 , we find some (x, 0) ∈ Lm+k+1 = Lm+k L, 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

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PRELIMINARIES (x, y) ∈ Lm+k = Lk Lm for some x ∈ X. But that means (x, xm ) ∈ Lm , (xm , y) ∈ Lk ,

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 we conclude (z, y) ∈ Lm+k+1 , which means y ∈ ran Lm+k+1 . (iii) Let ker L∩ran 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 (iv) Let ker Lm + ran L ⊇ dom Lm . dom Lm+k , we conclude

∀ k ∈ N.

Since ker Lm+k ⊇ ker Lm and dom Lm ⊇

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 ascent and descent. Proposition 1.21 Let m ∈ N. We have (i) ker L ∩ ran Lm = {0} ⇒ ker L ∩ ran Lm ⊆ mul Lm , so α b(L) ≤ α e(L), and

b e (ii) ker Lm + ran L = X ⇒ ker Lm + ran L ⊇ dom Lm , so δ(L) ≤ δ(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 6= {0}, which means α b(L) = 1. But Lm = L ∗ m holds for all m ∈ N , so we always have ker L ∩ ran L = span {e1 } = 6 {0}, which implies α e(L) = ∞. b Furthermore, we have ker L + ran L = span {e1 } = dom L 6= X, so δ(L) = 1. Since m ∗ e L = L holds for all m ∈ N , we also have δ(L) = ∞.

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

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Proposition 1.22 We always have α(L) = α b(L).

Proof: It suffices to show: ker L ∩ ran Lm ⊆ mul Lm ⇐⇒ ker Lm = ker Lm+1 . But this is trivial using Lemma 1.17, setting k = 1 there: ker Lm+1 ∼ ker L ∩ ran Lm . = ker Lm ker L ∩ mul Lm



b Proposition 1.23 We always have δ(L) = δ(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 ∼ dom Lm . = m+1 m ran L (ker L + ran L) ∩ dom Lm



Proposition 1.24 The ascent and generalized ascent have the property that the defining (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 for some k ∈ N. Let x ∈ ker Lk+1 ∩ ran Lm , then we have (x0 , x) ∈ Lm and (x, x1 ) ∈ L, (x1 , 0) ∈ Lk for some x1 ∈ X, so actually x1 ∈ ran Lm+1 and x1 ∈ ker Lk , which in view of ker L ∩ ran Lm+1 = {0} implies x1 = 0. But then we 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, using Lemma 1.17, we see {0} =

ker Lm+k ∼ ker Lk ∩ ran Lm = ker 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.

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Proposition 1.25 The characterizing inclusion of the descent remains true for the ranges 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} =

dom Lm ran Lm ∼ ⇒ ker Lm + ran Lk ⊇ dom Lm = ran Lm+k (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) α e(L) = p < ∞,

(ii) Rc (L) = {0} and α(L) = p < ∞, and (iii) either p = 0 or ker L ∩ ran Lp−1 6= {0}, together with ker Lk ∩ ran Lp = {0} for all k ∈ N. Proof: (i) ⇒ (ii) Let α e(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 α(L): Proposition 1.21 tells us α(L) ≤ p, so we only have to prove the converse inequality. But that one follows again from Lemma 1.17, or its Corollary 1.18: Suppose we had α(L) = q < p, then ker Lq+1 ∼ ker L ∩ ran Lq ∼ = = {0} , ker Lq so ker L ∩ ran Lq = {0}, which would imply α e(L) = q < p, a contradiction! (ii) ⇒ (iii) Let Rc (L) = {0} and α(L) = p. If p = 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, we see with ker Lp−1 ( ker Lp : ker Lp ∼ 6 {0} ker L ∩ ran Lp−1 ∼ = = ker Lp−1 and

ker Lk ∩ ran Lp ∼ ker Lp+1 ∼ ker Lk ∩ ran Lp ∼ = = {0} = ker Lk ∩ mul Lp ker Lp so ker Lk ∩ ran Lp = {0}.

20

∀ k ∈ N,

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(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 imply the absence of singular chains. Consider X = ℓ2 (C) and the linear relation L ⊆ X × X which arises from the left-shift with the multivalued part mul L = {λ ∈ C : (λ, 0, 0, . . .)}. Now, mul L ∩ ker L 6= {0}, but α e(L) = ∞ = α(L). To see that the abscence of singular chains alone does not imply that the generalized ascent is finite, consider any operator with infinite ascent, for example the left-shift in ℓ2 (C). Then, the generalized ascent has to be infinite as well. But operators do not have singular chains. The authors of [6] have established various connections between α(L) and δ(L), as well 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 are equal and coincide with (the usual) ascent and descent of L. In short form: e e α e(L), δ(L) < ∞ =⇒ α e(L) = α(L) = δ(L) = δ(L). Proof: Let 1.21

1.26 e p := α e(L) = α(L), q := δ(L) ≤ δ(L) =: r.

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

ker L ∩ ran Lq = ker L ∩ ran Lp = {0} , so α e(L) ≤ q < p = α e(L), an obvious contradiction, which means we have e α e(L) = α(L) = p ≤ q = δ(L) ≤ δ(L) =r

(1.2)

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

X = ker Lr + ran L = ker Lp + ran L, e so actually r = δ(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, we present its proof for the sake of completeness.

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Lemma 1.27 Let p ≥ 1. If dom L ⊆ ran Lp ⊕ ker Lp , then L is completely reduced by (ran Lp , ker Lp ).

Proof: Since ran L + ker Lp ⊇ ran Lp ⊕ ker Lp ⊇ dom L ⊇ dom Lp , we have δ(L) ≤ p. Abbreviate R := ran Lp and N := ker Lp . The intersection LN ∩ LR is trivial, because b LR ⊆ L is obvious. The N ∩ R = {0}. Also, by definition of compression, LN ⊕ converse inclusion remains to be shown. Let (x, y) ∈ L, then in particular x ∈ dom L, so x = x1 + x2 for some x1 ∈ R and x2 ∈ N . Now, there is a y2 ∈ X with (x2 , y2 ) ∈ L and (y2 , 0) ∈ Lp−1 . We see that (x1 , y −y2 ) = (x, y)−(x2 , y2 ) ∈ L. But x1 ∈ ran Lp , so y −y2 ∈ ran Lp+1 = ran Lp = R. Hence, (x1 , y − y2 ) ∈ LR . Moreover, (x2 , y2 ) ∈ LN , since x2 ∈ ker Lp = N and also y2 ∈ ker Lp−1 ⊆ 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 L−1 , we obtain dom Lm = dom Lm+1 ⇐⇒ ran(Lm )−1 = ran(Lm+1 )−1 ⇐⇒ ran(L−1 )m = ran(L−1 )m+1 ⇐⇒ ker(L−1 )m + ran L−1 ⊇ dom(L−1 )m ⇐⇒ mul Lm + dom L ⊇ ran Lm .



 Remark: We have shown that for the co-descent δc (L) := inf m ∈ N : dom Lm = dom Lm+1 the corresponding δbc (L) should be defined as δbc (L) := inf {m ∈ N : ran Lm ⊆ dom L + mul Lm } .

e Theorem 1.2 Let α e(L) = p < ∞, δ(L) = q < ∞ (by Theorem 1.1 p = q) and ran L ⊆ dom L + mul L. Set N := ker Lp and R := ran Lp . Then L is completely reduced by (N, R), LN is a nilpotent operator with nilpotence p and LR is surjective onto R. Proof: We first show that L is completely reduced by the pair (N, R). According to Lemma 1.27, we have to show dom L ⊆ N ⊕ R.

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Since α e(L) = p, the intersection ker L ∩ ran Lp = {0} is trivial. Proposition 1.24 implies ker Lp ∩ ran Lp = {0}, so the sum is direct. The sum ker Lp + ran L equals X, e because δ(L) = p. Now, this implies ker Lp +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 implies the absence of non-trivial singular chains. In particular, x = 0. Thus, LN is an operator. If (x, y) ∈ LpN , then y ∈ N ∩ ran LpN ⊆ ker Lp ∩ ran Lp = {0}. Proposition 1.26 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 e δ(L) ≤ δ(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 S −1 T for linear operators S, T as they are used in [2]. As we have seen in Lemma 1.13, we can write every linear relation L in such a way: L = Q−1 L Lop . First, we want to characterize relations of type S −1 T , 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 := S −1 T is a linear relation in X × Y , which can be characterized by (x, y) ∈ S −1 T ⇐⇒ x ∈ dom T ∧ y ∈ dom S ∧ T x = Sy.

Proof: We have  S −1 T = (x, y) ∈ X × Y ∃z ∈ Z : (x, z) ∈ T, (z, y) ∈ S −1

= {(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 = S −1 T with S, T as in the preceding proposition, T represents the “operator-like” behavior of L whereas S determines its multivalued part, though we have to be careful, since for L only those x ∈ X and y ∈ Y matter which are mapped into ran S ∩ ran T by T and S respectively. The first evidence for that intuition is the following lemma.

23

1

PRELIMINARIES

Lemma 1.30 Let S, T be linear operators as in Lemma 1.29 and let L := S −1 T . Then: (i) dom L = T −1 [ran T ∩ ran S] ⊆ dom T , (ii) ran L = S −1 [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 L−1 , we get mul L = ker L−1 = ker S, since L−1 = T −1 S.



This justifies setting α e(L) := inf {n ∈ N | ker L ∩ ran Ln = {0}},

compare the definition of the ascent of T relative to S in [2] in section 1, page 149:  α(T : S) := inf n ∈ N ker T ∩ ran(S −1 T )n = {0} .

Now, let L ⊆ X × Y be a linear relation. We examine to which extend L can be decomposed into a product S −1 T 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, if L = S −1 T . The triple (Z, S, T ) will be called split of L. In case dom T = dom L and 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 (Zmin , Smin , Tmin) by setting Zmin := Z, Smin := S|ran L and Tmin := T |dom L . Proof: This follows immediately from Lemma 1.29. A trivial, yet useful observation is

24

split



1

PRELIMINARIES

Proposition 1.32 If (Z, S, T ) is a (minimal) split of L, then (Z, T, S) is a (minimal) split of L−1 .

Lemma 1.33 Every linear relation L is splittable. More precisely: ran L ( mul L , QL , Lop ) is a minimal split of L, so L can be written as −1 L = Q−1 L Lop (= QL QL L),

where Lop : X ⊇ dom L →

ran L mul L

is the associated linear operator and QL : Y ⊇ ran L →

ran L mul L

is the canonical quotient map. Y Y Proof: This was shown in Lemma 1.13 for QL : Y → mul L and Lop : dom L → mul L , ran L ran L which we can obviously compress to ran L → mul L and dom L → mul L 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 operators can always be achieved by modifying the operators involved marginally (in algebraic terms). Applying the lemma to L−1 yields

Corollary 1.34 L

−1

minimally.



dom L ker L , QL−1 ,

L−1


 op

with QL−1 : X ⊇ dom L →

dom L ker L

splits

Now applying Proposition 1.32 to the corollary, we get

Corollary 1.35



dom L ker L ,

L−1




, QL−1 op



splits L.

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

25

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 = JS1 and T2 = JT1 will be called a morphism (Z1 , S1 , T1 ) → (Z2 , S2 , T2 ). We will also say J induces such a morphism.

Proposition 1.36 If (Z1 , S1 , T1 ) is a split of L and J : Z1 → Z2 is an injective linear map, then (Z2 , S2 , T1 ) with T2 = J ◦ T1 and S2 = J ◦ S1 is a split of L. Proof: We have S2−1 T2 = (J ◦ S1 )−1 (J ◦ T1 ) = S1−1 J −1 JT1 = S1−1 T1 = L, because J −1 J = 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 = S −1 T 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 given by 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 commutes: dom L QL−1 dom L ker L where J :

ran L mul L

ran L mul L

J ∼ = L−1

QL


ran L

op

−1 ) ^ is the canonical linear isomorphism (L op associated g is the canonical linear isomorphism Lop associated with Lop .

dom L ker L −1

→

with (L−1 )op and J

Lop

26

1

PRELIMINARIES

g Proof: The only thing we have to prove is J −1 = L op : let x ∈ dom L, then ^ −1 ) L g g JL op (x + ker L) = (L op op (x + ker L) ^ −1 ) L x = (L op op

^ −1 ) (y + mul L) = (L op for some y ∈ ran L such that (x, y) ∈ L. Furthermore:
. . . = L−1 op y = z + ker L

for some z ∈ dom L such that (z, y) ∈ L. But then (x−z, 0) ∈ L, so z+ker L = x+ker L. This shows g JL op = I dom L ker L

Exchanging the roles of L and L−1 yields the other relation g L op J = I ran L . mul L



Corollary 1.38 The map J from the preceding proposition induces an isomorpism     ran L dom L −1 , QL , Lop → , (L )op , QL−1 . mul L ker L 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 ). If T is surjective, then there exists an isomorphism  
dom L , L−1 op , QL−1 → (Z, S, T ), ker L 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 Te :

dom L ker T

→ Z be the linear isomorphism associated   with T .
By Lemma L dom L −1 e −1 , Q 1.30, we have ker T = ker L, so actually T : ker L → Z. Since dom L ker L , L op is a split it suffices to show
S = Te L−1 op and T = TeQL−1 . 27

1

PRELIMINARIES

The second equality is the very definition of Te, so we only have to prove the
first equality. It is more convenient to show the equivalent equation Te−1 S = L−1 op : we have
−1 , Te−1 S = QL−1 T −1 S = QL−1 L−1 = QL−1 Q−1 L−1 L op where the last equality is an immediate consequence of Corollary 1.38. Now, since QL−1 is an operator, by Lemma 1.9 it holds true that QL−1 Q−1 L . But that L−1 = I dom ker L means

Te−1 S = I dom L L−1 op = L−1 op .  ker L

Corollary 1.39 For every minimal split (Z, S, T ) of L we have: T is surjective ⇐⇒ S is surjective. ran L Proof: Consider “⇒”: This is obvious for the split ( mul L , QL , Lop ), and for every other ran L minimal split (Z, S, T ) of L we have an isomorphism J : mul L → Z such that S = JQL , which immediately implies the surjectivity of S. – The other implication follows from duality. 

Definition 1.10 A minimal split (Z, S, T ) with surjective S or T will itself henceforth be called surjective. For algebraic considerations, it suffices to consider surjective minimal splits, since every 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, which coincides with the range of T . Lemma 1.40 Let (Z, S, T ) be a minimal split of L, then ran S = ran T =: Z1 . The maps S1 : Y ⊇ ran L → Z2 , T1 : X ⊇ dom L → Z2 split L in Z2 , and the embedding Z2 ֒→ 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 for some y ∈ ran L. But since L = S −1 T , 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 L−1 . The rest of the lemma is trivial now.  Remark: At this point it becomes clear that all results of [2] using only numbers and spaces concerning S −1 T with operators S, T can be applied to arbitrary relations.

28

1

PRELIMINARIES

Suppose we have two linear relations L, K ⊆ X × Y with L ⊆ K, and let K be split by (Z, S, T ). One could surmise that now (Z1 , S1 , T1 ) splits L, where Z1 := Z, S1 := S|ran L and T1 := T |dom L . But, unfortunately, this is not true in general. For instance, Lemma 1.30 tells us that ker L = ker T1 = ker T ∩ dom L = ker K ∩ dom L and analogously mul L = mul K ∩ ran L are necessary conditions, though in general, we only have “⊆” between these spaces. Example 1.4 Let X = Y = R2 , and consider K := span {(e1 , 0), (0, e1 ), (0, e2 )} and L := span {(e1 , e2 )}. Then L ⊆ K, ker L = {0}, but ker K ∩ dom L = span {e1 } ∩ span {e1 } = span {e1 } ) {0} . It can be immediately verified that K = S −1 T , where S : R2 → R2 , S = 0 and T : span {e1 } → R2 , T = 0. But L is already an operator, so (R2 , IR2 , L) splits L. Obviously, it is impossible to construct a split isomorphic to that formed by S and T . But if we have ker L = ker K ∩ dom L or mul L = mul K ∩ ran L, then we have both equalities, and (Z1 , S1 , T1 ), as defined above, splits L. Lemma 1.41 Let L, K ⊆ X × Y be linear relations with L ⊆ K, and let K be split by (Z, S, T ). If additionally ker L = ker K ∩ dom L

(1.3)

mul L = mul K ∩ ran L,

(1.4)

or then both equalities hold and (Z1 , S1 , T1 ) splits L, where Z1 := Z, S1 := S|ran L and T1 := T |dom L ). Proof: We have dom T |dom L = dom T ∩ dom L = dom L, since dom T ⊇ dom K ⊇ dom L, and dom S|ran L = ran L analogously. Furthermore, (x, y) ∈ L =⇒ x ∈ dom L ∧ y ∈ ran L ∧ (x, y) ∈ K = S −1 T ⇐⇒ x ∈ dom L ∧ y ∈ ran L ∧ T x = Sy ⇐⇒ x ∈ dom T |dom L ∧ y ∈ dom S|ran L ∧ T |dom L (x) = S|ran L (y) ⇐⇒ (x, y) ∈ S1−1 T1 . This shows L ⊆ S1−1 T1 . Now, suppose (1.3) holds true. We will first show that equality (1.4) also holds. Let y ∈ mul K ∩ ran L, so (0, y) ∈ K and there is some x ∈ dom L such that (x, y) ∈ L.

29

1

PRELIMINARIES

Then (x, 0) ∈ K, so x ∈ ker K ∩ dom L = ker L, which means (x, 0) ∈ L. Together with (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 L−1 = ker K −1 ∩ dom L−1 , which by what we just showed implies mul L−1 = mul K −1 ∩ ran L−1 , or equivalently ker L = ker K ∩ dom L. Again, assume (1.3), in order to show S1−1 T1 ⊆ L. Using the equivalences we have proved in the beginning, we examine (x, y) ∈ (dom L) × (ran L) with (x, y) ∈ K. We must have some x′ ∈ dom L such that (x′ , y) ∈ L. But then (x − x′ , 0) ∈ K, which means 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 S −1 for operators S, T . In the following paragraphs, 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 always consider a linear relation L ⊆ X × X. Definition 1.11 Let (Y, S, T ) be a split of L. The relation Lc := T S −1 = {(y, y ′ ) ∈ Y × Y | ∃ x ∈ dom S ∩ dom T : Sx = y ∧ T x = y ′ } = {(Sx, T x) | 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 of splits, so we just call it “L commutated”. Right from the definition, we see that Lc does only depend on the behavior of S and T on dom S ∩ dom T . Proposition 1.42 Let S, T be linear operators in X to Y . Then T S −1 = T |dom S∩dom T (S|dom S∩dom T )−1 . Again, it is easy to see, but useful, that commutating a relation commutes with inverting.

30

1

PRELIMINARIES

Proposition 1.43 We always have Lc−1 = (L−1 )c . −1 −1 Proof: L−1 ) = ST −1 = (L−1 )c , since L−1 is split by (Z, T, S). c = (T S



The usual linear subspaces ker, ran, dom associated with the powers of S −1 T and T S −1 are the basis of all considerations in [2], cf. section 1, page 149 ibid. We will partly 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 those with 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 if dom 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 Lc−1 = 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 = T −1 [ran T ] ⊆ T −1 [ran Lc ] = T −1 [T [dom S]]. So, let x ∈ dom T , then T x = T v for some v ∈ dom S, which menas x − v ∈ ker T and thus 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

1

PRELIMINARIES

Concerning ker Lc : ker Lc = {y ∈ Y | ∃ x ∈ dom S ∩ dom T : Sx = y, T x = 0 = S0} = S[{x ∈ dom S ∩ dom T | (x, 0) ∈ L}] = S[ker L] = S[ker T ]. Also: −1 mul Lc = ker(Lc )−1 = ker L−1 ] = T [mul L] = T [ker S]. c = T [ker L



This justifies setting e c ) := inf {n ∈ N | ran Lc + ker Ln = Y }, δ(L c

compare the definition of the descent of T relative to S in [2], section 1, page 149:  δ(T : S) := inf n ∈ N ran T + ker(T S −1 )n = Y .

Remark: Without the additional assumption dom T ⊆ dom S, we would have to consider ′ the relation (T S −1 )|ran T , if we wanted to reconstruct exactly the same spaces Rm from [2]. Indeed, we have ′ R1′ = T X = ran T, Rm = ran((T S −1 )|ran T )m−1

∀ m ≥ 2.

But we see, that the case m = 1 seems to be an exception, and the relation which we have to consider for m ≥ 2 is rather cumbersome. A closer examination can improve the situation: It is easy to see that ((T S −1 )|ran T )m = ((T S −1 )m )|ran T for all m ≥ 1. Concerning the first flaw, we observe ran((T S −1 )0 )|ran T = ran Iran T = ran T. So we have ′ Rm = ran((T S −1 )m )|ran T ∀ 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. e So unless dom S 6= X, δ(L) < ∞ is impossible.

The “pointwise characterization” of Lc is by far not as handy as it was in the case of S −1 T . But powers of Lc are closely related to those of L. Proposition 1.45 Let n ∈ N, n ≥ 1, then Ln = S −1 Lcn−1 T and Lnc = T Ln−1S −1 , or in terms of pairs: (x1 , xn+1 ) ∈ Ln if and only if ∃ (y1 , yn ) ∈ Lcn−1 : T x1 = y1 ∧ yn = Sxn+1 , so if and only if (T x1 , Sxn+1 ) ∈ Lcn−1 . Respectively: (y1 , yn+1 ) ∈ Lnc if and only if ∃ x1 ∈ dom S, xn ∈ dom T : (x1 , xn ) ∈ Ln−1 ∧ y1 = Sx1 ∧ T xn = yn+1 .

32

1

PRELIMINARIES

Proof: Follows easily by induction.



Corollary 1.46 Let n ∈ N, n ≥ 1, then (i) ker Ln = T −1 [ker Lcn−1 ], (ii) ker Lnc = S[ker Ln ]. If additionally dom T ⊆ dom S, we have for n ≥ 1: (iii) ran Ln = S −1 [ran Lnc ] and (iv) ran Lnc = T [ran Ln−1 ]. Proof: For n ≥ 1 argue using the preceding proposition: x ∈ ker Ln ⇐⇒ (x, 0) ∈ Ln ⇐⇒ (T x, S0) = (T x, 0) ∈ Lcn−1 ⇐⇒ x ∈ T −1 [ker Lcn−1 ]; 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 ) ∈ Lcn−1 : T z = y1 , yn = S0 = 0, Sz = x

⇐⇒ ∃ z ∈ dom S ∩ dom T

∃ (y1 , 0) ∈ Lcn−1 : T z = y1 , Sz = x

⇐⇒ ∃ y1 ∈ Y : (x, y1 ) ∈ Lc , (y1 , 0) ∈ Lcn−1 ⇐⇒ (x, 0) ∈ Lnc ⇐⇒ x ∈ ker Lnc . Concerning (iii): x ∈ ran Ln ⇐⇒ ∃ x1 ∈ dom T : (x1 , x) ∈ Ln ⇐⇒ ∃ x1 ∈ dom T : (T x1 , Sx) ∈ Lcn−1 , where the last step is possible, because x ∈ ran Ln ⊆ ran L ⊆ dom S. Since we assumed dom 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) ∈ Lcn−1 ⇐⇒ ∃ x1 ∈ dom S ∩ dom T : (Sx1 , Sx) ∈ Lnc ⇐⇒ Sx ∈ ran Lnc ⇐⇒ x ∈ S −1 [ran Lnc ].

33

1

PRELIMINARIES

And for (iv): y ∈ ran Lnc ⇐⇒ ∃ z ∈ Y : (z, y) ∈ Lnc ⇐⇒ ∃ z ∈ Y, x1 ∈ dom S, xn ∈ dom T : (x1 , xn ) ∈ Ln−1 , z = Sx1 , T xn = y ⇐⇒ ∃ x1 ∈ dom S, xn ∈ dom T : (x1 , xn ) ∈ Ln−1 , T xn = y. (x1 , xn ) ∈ Ln−1 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 in fact equivalent to ∃ x1 ∈ dom T, xn ∈ dom T : (x1 , xn ) ∈ Ln−1 , T xn = y ⇐⇒ y ∈ T [dom T ∩ ran Ln−1 ] = T [ran Ln−1 ].



Remark: Note that (iv) is in general not true for n = 0. We only have ran L0 = X ⊇ dom S = S −1 [Y ] = S −1 [ran L0c ].

The following lemma condenses Lemma 1.3 and Lemma 1.4 from [2]. We will need it later on to prove a pertubation result. Lemma 1.47 Let λ 6= 0 and m ≥ 1. Then −1 (i) ker Lm [ker Lm−1 ] = (λS − T )[ker Lm ], c = (λS − T )T c

(ii) ker Lm = T −1 (λS − T )[ker Lm−1 ], (iii) ran Lm ∩ dom T = (λS − T )−1 T [ran Lm−1 ] = (λS − T )−1 [ran Lm c ] and −1 (iv) ran Lm [ran Lm−1 ]. c = T (λS − T ) c

Proof: (i) Using Corollary 1.46 as well as Lemmas 1.6, 1.9 and 1.5, we get (λS − T )[ker Lm ]

1.46

(λS − T )T −1 [ker Lm−1 ] c

=

1.6 + 1.9

⊆

1.5

(λST −1 − Iran T )[ker Lm−1 ] c

⊆

m−1 L−1 ] + ker Lm−1 c [ker Lc c

1.5

m−1 = ker Lm ker Lm c . c + ker Lc

=

1.5

m For the converse inclusion, suppose y ∈ ker Lm c = S[ker L ], so there exists some m m x0 ∈ ker L such that Sx0 = y. x0 ∈ ker L means we find an L-chain

(x0 , x1 , . . . , xn−1 , 0).

34

1

PRELIMINARIES In particular: xn−1 ∈ ker L = ker T . Set u :=

n−1 X

λ−(k+1) xk .

k=0

Obviously, u ∈ ker Lm ⊆ dom T ⊆ dom S. We have Tu =

=

n−2 X

k=0 n−2 X

λ−(k+1) T xk λ−(k+1) Sxk+1

k=0

= (λS)

n−1 X

λ−(k+1) Sxk+1

k=1

= (λS)u − Sx0 = (λS)u − y, or equivalently y = (λS − T )u ∈ (λS − T )[ker Lm ]. (ii) By the first point, we have ker Lm = T −1 [ker Lm−1 ] = T −1 [(λS − T )[ker Lm−1 ]]. c (iii) For “⊆”, we note that (λS − T )[ran Lm ∩ dom T ] = S[ran Lm ∩ dom T ] + T [ran Lm ] m+1 ⊆ ran Lm c + ran Lc m−1 ⊆ ran Lm ], c = T [ran L

so by Lemma 1.5, we have ran Lm ∩dom T ⊆ (λS − T )−1 T [ran Lm−1 ]. – We show the converse inclusion by induction. For m = 1 let x ∈ (λS − T )−1 T [ran Lm−1 ] = (λS − T )−1 T [Y ] so there exists some u ∈ dom T such that (λS − T )x = T u, or equivalently Sx = T (λ−1 (x + u)), which means (λ−1 (x + u), x) ∈ L. So we have x ∈ ran L ∩ dom T . The induction step is very similar: Let x ∈ (λS − T )−1 T [ran Lm ] ⊆ (λS − T )−1 T [ran Lm−1 ] = ran Lm ∩ dom T, so there exists some u ∈ ran Lm ∩ dom T such that (λS − T )x = T u, or (λ−1 (x + u), x) ∈ L. Since λ−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 (λS − T )[ran Lm−1 ] = T (λS − T )−1 [X] = T [dom S ∩ dom T ] = ran T = ran L1c , c

35

1

PRELIMINARIES where the last step follows from Lemma 1.44. For all other natural numbers, we use (iii): ran Lm+1 = T [ran Lm ∩ dom T ] c = T [(λS − T )−1 [ran Lm c ]] = T (λS − T )−1 [ran Lm c ] ∀ m ≥ 1.

36



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 decompose the domain of our relation L into a direct sum of a kernel part and a range part, in our case depending on a given split (Y, S, T ) of L, in such a way that the restrictions of 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 following, let L ⊆ X × X be a linear relation, (Y, S, T ) a split of L and let Lc be L commutated with respect to (Y, S, T ), so Lc = T S −1 . The following lemma can also be found on page 149 of [2]. It establishes a relatione e c ) respectively. ship between α e(L) and α e(Lc ), δ(L) and δ(L 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 k m (i) ker Lkc ∩ ran Lm c = S[ker L ∩ ran L ] and

(ii) (ker Lk + ran Lm ) ∩ dom S = S −1 [ker Lkc + ran Lm c ]. Proof: (i) For k = 0 we have k m ker Lkc ∩ ran Lm c = {0} = S[{0}] = S[ker L ∩ ran L ].

For m = 0 we have k k m k ker Lkc ∩ ran Lm c = ker Lc = S[ker L ] = S[ker L ∩ ran L ].

Now let k, m ∈ N, k ≥ 1, m ≥ 1. Then k m−1 y ∈ ker Lkc ∩ ran Lm ] c = S[ker L ] ∩ T [ran L

⇐⇒ ∃ (x1 , 0) ∈ Lk , (x′1 , x′m ) ∈ Lm−1 : Sx1 = y = T (x′m ) ⇐⇒ ∃ (x1 , 0) ∈ Lk , (x′1 , x1 ) ∈ Lm : Sx1 = y ⇐⇒ y ∈ S[ker Lk ∩ ran Lm ]. (ii) For m = 0 we have (ker Lk +ran L0 )∩dom S = X ∩dom S = dom S = S −1 [Y ] = S −1 [ker Lkc +ran L0c ].

37

2

POLES OF THE GENERALIZED RESOLVENT 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 ∈ k n S −1 [ran Lm c ]. But also Sx ∈ S[ker L ] = ker Lc , so Sz = S(x + y) = Sx + Sy ∈ ker Lkc + ran Lm c , as desired. – If on the other the other hand z ∈ S −1 [ker Lkc + ran Lm c ] ⊆ dom S, m−1 we find x ∈ ker Lkc = S[ker Lk ], y ∈ ran Lm = T [ran L ] with Sz = x + y. We c c thus find v ∈ dom S ∩ ker Lk and w ∈ dom T ∩ ran Lm−1 such that Sv = x and c T 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 α e(Lc ) ≤ α e(L).

Proof: Let p := α e(L). The case p = ∞ is trivial, so let p < ∞, then

ker L ∩ ran Lp = {0} ⇒ ker Lc ∩ ran Lpc = S[ker L ∩ ran Lp ] = {0} ⇔ α e(Lc ) ≤ p.



The situation is more complex for the generalized descent. For the traditional descent we had Proposition 1.25, which stated that the definition of the descent remains true for ran Lk instead of ran L. We can establish a similar result for the generalized descent of Lc under the condition dom L ⊆ ran L. In [2], this is Proposition 3.3. e c ) = p < ∞, then for all k ∈ N, k ≥ 1: Proposition 2.3 Let dom L ⊆ ran L. If δ(L (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 = S −1 [ker Lpc + ran Lkc ], so suppose we have proved (ii), we can conclude (ker Lp + ran Lk ) ∩ dom S = S −1 [ker Lpc + ran Lkc ] = dom S. That means ker Lp + ran Lk ⊇ dom S.

38

2

POLES OF THE GENERALIZED RESOLVENT

On the other hand ker Lp + ran Lk ⊆ dom L + ran L ⊆ dom T + dom S = dom S. e c ) = p, we have the equality for k = 1. Concerning (ii): “⊆” is trivial. Since δ(L 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 Lk−1 ], so there is some w ∈ dom T ∩ ran Lk−1 with y = T w. Now, also w ∈ dom S, so from the induction assumption and Lemma 2.1 we get w =u+v with u ∈ ker Lp and v ∈ ran Lk . Hence, 1.46

y = T w ∈ T [ker Lp ] + T [ran Lk ] = T [T −1 [ker Lpc ]] + ran Lk+1 ⊆ ker Lpc + ran Lk+1 , c c as desired.



The following corollary is numbered 3.4 in [2]. b c ) < ∞, then dom S = dom Lm +ran Lk and Y = S[dom Lm ]+ Corollary 2.4 If δ(L m ran Lc for all k, m ∈ N. b c ) < ∞, then by the preceding proposition dom S = ker Lp +ran Lk Proof: Let p := δ(L p and Y = ker Lc + ran Lkc for all k ∈ N. By Proposition 1.16, we have ker Lp ⊆ dom Lm , and subsequently S[ker Lp ] = ker Lpc ⊆ S[dom Lm ] 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 Lkc ⊆ S[dom Lm ] + ran Lk ⊆ Y.



Before we can proof connections between δ and δe for L and Lc similar to Lemma 1.26, we need an isomorphism-type result similar to 1.19, which makes use of Lc . The following lemma is the second part of Lemma 3.2 in [2]. We give the proof omitted there. Lemma 2.5 Let dom L ⊆ ran L. For all k, m ≥ 1 we have ran Lm ∼ S[dom Lm ] + ran Lkc . = k ran Lm+k ker Lm c + ran Lc Proof: From Lemma 1.19 we know ran Lm ∼ dom Lm + ran Lk = ran Lm+k ker Lm + ran Lk

39

∀ k, m ≥ 1.

2

POLES OF THE GENERALIZED RESOLVENT

Let k, m ≥ 1. We define a linear map A : (dom Lm + ran Lk ) →

S[dom Lm ] + ran Lkc , x 7→ [Sx]. k ker Lm c + ran Lc

It suffices to prove that A is well-defined, surjective and ker A = ker Lm + ran Lk . The proof 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 if x ∈ dom Lm and y ∈ ran Lk = S −1 [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 . Then k m k Sx ∈ ker Lm c and Sy ∈ ran Lc , so S(x + y) ∈ ker Lc + ran Lc , which means m k −1 [S(x + y)] = 0. – Now, let x ∈ dom L , y ∈ ran L = S [ran Lkc ] such that k m k [S(x + y)] = 0, which means S(x + y) ∈ ker Lm c + ran Lc = S[ker L ] + ran Lc , m k so we find x1 ∈ ker Lc and y1 ∈ ran Lc such that Sx + Sy = Sx1 + y1 ⇐⇒ S(x − x1 ) = y1 − Sy ∈ ran Lkc = T [ran Lk−1 ]. Hence, there exists some y2 ∈ ran Lk−1 with S(x − x1 ) = T y2 , which in turn means (y2 , x − x1 ) ∈ L. We get x − x1 ∈ ran Lk , so there is a y3 ∈ ran Lk such that 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]. e c ) = 0, then dom S = ran Lk and Y = ran Lk for k ≥ 1, in Proposition 2.6 If δ(L c particular δ(Lc ) = 0, and ( 0, dom S = X δ(L) = 1, dom S 6= X. e c ) =: p < ∞, then δ(L) = δ(Lc ) = p. If 0 < δ(L

e c ) = 0, then ran Lc = Y and subsequently ran Lk = Y and Proof: Firstly, suppose δ(L c k −1 thus ran L = S [ran Lkc ] = dom S for k ≥ 1. We note that δ(Lc ) = 0 can also be deduced from Proposition 1.21. e c ) = p < ∞, which means ker Lp + Now we examine the situation in case of 0 < δ(L c p−1 ran L = Y and ker Lc + ran L 6= Y . 40

2

POLES OF THE GENERALIZED RESOLVENT e c ) = p. • From Proposition 1.21 we already know δ(Lc ) ≤ δ(L

• Concerning δ(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 gives ran Lp ∼ dom Lp + ran L = {0} , = ran Lp+1 ker Lp + ran L so ran Lp ⊆ ran Lp+1 , which shows δ(L) ≤ p. • Suppose δ(L) < p, then ran Lp−1 = ran Lp . Now, Lemma 2.5 gives {0} =

ran Lp−1 ∼ S[dom Lp−1 ] + ran Lc ∼ Y , = = p−1 p−1 ran Lp−1+1 ker Lc + ran Lc ker Lc + ran Lc

e c ) = p. So ran Lp−1 6= ran Lp , so ker Lp−1 + ran Lc = Y , which contradicts δ(L c p−1 −1 p−1 and since S is an operator, Lc = S [ran L ] 6= S −1 [ran Lp ] = ran Lpc , which implies δ(Lc ) ≥ p.  Now, we are able to show analogons of Theorem 1.1 and Theorem 1.2 for the case e c ) are finite. Note again that due to our assumption dom T ⊆ where both α e(L) and δ(L e dom S we must have δ(L) = ∞, so the conditions of the previously mentioned theorems cannot be satisfied. Proposition 2.7 If generalized ascent of L and generalized descent of Lc are both finite, then they are equal and coincide with the ascent of L. In short form: e c ) < ∞ =⇒ α e c ). α e(L), δ(L e(L) = α(L) = δ(L

e e In case δ(L) = 0 as well as dom S = X, or δ(L) > 0, the numbers also coincide with the descent of L. This is Proposition 4.1 in [2]. Proof: The proof is very similar to the one of Theorem 1.1: Let 1.26 e c ) =: r. p := α e(L) = α(L), q := δ(L), δ(L

e First we examine the case of r = δ(L) = 0. Then dom S ⊆ ran Lm for all m ∈ N by Proposition 2.6, and because of ker L = ker T ⊆ dom T ⊆ dom S ⊆ ran Lp ,

we have ker L ∩ ran L0 = ker L = ker L ∩ ran Lp = {0} .

41

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POLES OF THE GENERALIZED RESOLVENT

e c ). If additionally dom S = X, we can continue this equation Thus, α e(L) = 0 = δ(L with . . . = δ(L). e c ) = r by Proposition 2.6. Assume p > q, then Now, let r > 0, then q = δ(L) = δ(L p q ran L = ran L , in particular ker L ∩ ran Lq = ker L ∩ ran Lp = {0} ,

so α e(L) ≤ q < p = α e(L), an obvious contradiction, which means we have e c ) = r. α e(L) = α(L) = p ≤ q = δ(L) = δ(L

(2.5)

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

e c ) ≤ p. This shows: (2.5) holds with equality everywhere. so actually r = δ(L



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

e c ) =: p < ∞, then Theorem 2.1 If 0 < α e(L) = δ(L

dom S = ker Lp ⊕ ran Lp , Y = ker Lpc ⊕ ran Lpc .

(2.6)

e c ) ≤ p. If on the other hand (2.6) holds for some p > 0, we have α e(L) = δ(L

Proof: By Proposition 2.3, we have ker Lp + ran Lp = dom S and ker Lpc + ran Lpc = Y . Since α e(L) = p, the intersection ker L ∩ ran Lp = {0} is trivial. Then, using Lemma 2.1, we get ker Lc ∩ ran Lpc = S[{0}] = {0}. We already know from Proposition 1.24 that 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 Lpc = Y, e c ) ≤ p, and by the preceding proposition, they coincide. so α e(L), δ(L



Theorem 2.2 L is completely reduced by (ker Lp , ran Lp ), and we have S = S1 ⊕ S2 and T = T1 ⊕ T2 , where S1 , T1 : ker Lp → ker Lpc and S2 , T2 : ran Lp → ran Lpc are defined by S1 := S|ker Lp , T1 := T |ker Lp , S2 := S|ran Lp and T2 := T |ran Lp . We have (i) T2 is bijective, (ii) S1 is bijective and Lker Lp is a nilpotent operator whose nilpotence index is p.

42

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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 ; p

T [ker L ] ⊆

ker Lcn−1

⊆

ker Lnc ,

p

T [ran L ] =

ran Lp+1 c

(2.7) ⊆

ran Lpc .

(2.8)

So S1 , S2 , T1 , T2 are well-defined, and by the preceding theorem we have S = S1 ⊕ S2 and T = T1 ⊕ T2 . From the calculations in (2.7) and (2.8) we also see that T2 and S1 are surjective. Also: ran Lp ∩ ker T = ran Lp ∩ ker L = {0} and p

ker L ∩ ker S = ker Lp ∩ mul L ⊆ 8 ker Lp ∩ ran Lp = {0} .

(2.9)

since α e(L) = p, so T2 is also injective. (2.9) also shows that Lker Lp is an operator. Obviously, Lpker Lp = 0ker Lp , so Lker Lp is nilpotent with nilpotence index less or equal than p. Suppose it was strictly smaller than p, so Lp−1 ker Lp = 0ker Lp . That would mean ker Lp ⊆ ker Lp−1 , so α(L) ≤ p − 1, which contradicts α(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 denote F-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 ρS (T ) := {λ ∈ F | λS − T : X ⊇ dom T → Y is bijective} is called S-resolvent set. For λ ∈ ρS (T ) we call Rλ (S, T ) : Y → X, Rλ (S, T )y = (λS − T )−1 y, S-resolvent of T in λ. We differ from [2] in that we first assume S and T to be bounded, everywhere defined operators, and relax that assumption later on.

8 Proposition

1.16

43

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POLES OF THE GENERALIZED RESOLVENT

Lemma 2.8 ρS (T ) ∋ λ 7→ Rλ (S, T ) ∈ B(Y, X) is well-defined and analytic on the open set ρS (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 such that A−1 ∈ with dom B ⊆ dom A and BA−1 ∈ B(X)

B(X)

and B a linear operator −1 −1

such that BA < 1. Then (A − B) ∈ B(X) with −1

(A − B)

−1

=A

∞ X


k BA−1 .

k=0

(2.10)

Proof: We set C := BA−1 ∈ B(X). Then ∞ ∞ X

k X k

C ≤ kCk < ∞,

k=0

k=0

P∞ since kCk < 1. Thus, because X is a Banach space, k=0 C k converges and C k → 0. We observe ! n n X X k n+1 k (I − C) C =I −C = C (I − C), k=0

k=0

so (I − C) is invertible with

(I − C)−1 =

∞ X

Ck.

k=0

Furthermore: A−1

∞ P

C k = A−1 (I − C)−1

k=0

= A−1 (AA−1 − BA−1 )−1 = ((AA−1 − BA−1 )A)−1 1.6

= ((A − B)A−1 A)−1

1.9

= (A − B)−1 .

44



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POLES OF THE GENERALIZED RESOLVENT

Corollary 2.10 If A has an inverse A−1 ∈ B(X) and B ∈ B(X) such that kBk <

−1 −1

A , then (A − B) is invertible and (2.10) holds. In particular, if A, A−1 , B ∈

−1 B(X) and kA − Bk < A−1 , then B is invertible with B −1 = A−1

∞ X

((A − B)A−1 )k

(2.11)

k=0



Proof: We have BA−1 ∈ B(X) such that BA−1 ≤ kBk A−1 < 1. The claim thus follows immediately from the preceding proposition. 

Proof (of Lemma 2.8): By our assumption, X and Y are Banach spaces and S, T : X → Y are bounded linear operators. Thus, Rλ (S, T ) = (λS − T )−1 is bounded by the open mapping theorem for all λ ∈ ρS (T ). Let µ ∈ F, then µ→λ

kλS − T − µS + T k = |λ − µ| kSk −→ 0, so we find some r > 0 such that µS − T is invertible for all µ ∈ F with |λ − µ| < r using Corollary 2.10. So ρS (T ) is indeed open and we have, by (2.11), (µS − T )−1 = (λS − T )−1

∞ X

((λS − T − µS + T )(λS − T )−1 )k

k=0

= (λS − T )−1

∞ X

(λ − µ)k (S(λS − T )−1 )k ,

k=0

which shows that λ 7→ Rλ (S, T ) is analytic.



The general strategy of the main result’s proof is similar to the one used in the single e c ) = p < ∞, then we will show: operator-case: Let α e(L) = δ(L 1. X can be decomposed into ker Lp ⊕ ran Lp using Theorem 2.1. 2. λS − T is bijective on a deleted neighborhood of 0. 3. ker Lp and ran Lp are closed. 4. λ 7→ λS − T restricted to ker Lp has a pole of order p at 0, and on the other hand, if we restrict it to ran Lp , it is analytic. The converse direction uses the identity theorem for Laurent series expansions and e c ) are finite. algebraic considerations to show that α e(L) and δ(L In order to show point 2 of the strategy, we have to show a pertubation result to e c ) and α guarantee surjectivity and injectivity using the finiteness of δ(L e(L), respectively. We will be using a standard theorem for that, which requires us to control kernel 45

2

POLES OF THE GENERALIZED RESOLVENT

and range of the operators which we examine. It is natural to restrict these operatores, e c ) respectively) are so that they become injective (respectively, surjective), if α e(L) (δ(L finite. The pertubation result will also require the involved normed vector spaces to be complete, which might fail, if we compress our operators. Since we are mainly interested in the stability of algebraic properties, we are free to choose norms on our spaces that guarantee their completeness. The following is called Remark 5.1 in [2]. Proposition 2.11 Let (E, k·kE ) and (F, k·k)F be Banach spaces, A, B ∈ B(E, F ). Set F0 := ran A, E0 := B −1 [F0 ]. Then A0 := A|E0 , B0 := B|E0 : E0 → F0 are well-defined operators and there exist norms k·kE0 on E0 and k·kF0 on F0 such that (i) k·kE0 and k·kF0 are stronger than k·kE and k·kF respectively. (ii) (E0 , k·kE0 ) and (F0 , k·kF0 ) are complete. (iii) A0 and B0 are bounded linear operators from (E0 , k·kE0 ) to (F0 , k·kF0 ). e : E/ ker A → F be the injective operator induced by A. A e is bounded, Proof: Let A −1 e so it is closed, and its inverse A : ran A → E/ ker A is also closed. Equip F0 = ran A with the graph norm

e−1 kykF0 := kykAe−1 = kykF + A y ∀ y ∈ F0 E/ ker A

e−1 , which is stronger than k·k and with respect to which F0 is complete. induced by A F Equip E0 = B −1 [F0 ] ⊆ dom B with the norm

e−1 kxkE0 := kxkE + kBxkF0 = kxkE + kBxkF + A Bx ∀ x ∈ E0 . E/ ker A

induced by B|E0 : (E0 , k·kE ) → (F0 , k·kF0 ), which is stronger than the graph norm k·kB induced by B, which in turn is stronger than k·kE . Hence, B0 : (E0 , k·kE0 ) → (F0 , k·kF0 ) is obviously bounded, and B|E0 : (E0 , k·kE ) → (F0 , k·kF0 ) is closed: let (xn )n∈N ⊆ E0 satisfy k·kF

k·k

E xn −→ x, Bxn −→0 y,

so, by the definition of k·kE0 , k·kF

k·kE

xn −→0 x, B0 xn −→0 y, which means, since B0 is closed, x ∈ E0 and Bx = B0 x = y. We conclude that (E0 , k·kE0 ) is complete. Thus, if we can show that A0 is closed, it is bounded by the closed graph theorem. But since k·kE0 and k·kF0 are stronger norms than k·kE and k·kF respectively, the same is true for the associated product norms on E × F , which means that A0 is necessarily closed, because its graph coincides with that of the closed operator A. 

46

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POLES OF THE GENERALIZED RESOLVENT

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 operators from (X, k·k) to (Y, k·k). Lemma 2.12 Let m ≥ 0, set Sm := S|ran Lm , Tm := T |ran Lm : ran Lm → ran Lm c . There exist stronger norms on ran Lm and ran Lm c such that these spaces are complete and Tm and Sm are bounded with respect to those norms. Proof: For m = 0 there is nothing to do, we set k·k0 := k·kX as norm on ran L0 = X and |k·k|0 := k·kY as norm on ran L0c = Y . – The rest follows by induction: Suppose we have proved the claim for some m ≥ 0, then S m , T m are bounded operators between the Banach spaces (ran Lm ∩ dom T, k·km ) and (ran Lm c , |k·k|m ). Also: ran Lm+1 = T [ran Lm ] c = ran Tm , ] ∩ dom T ran Lm+1 ∩ dom T = S −1 [ran Lm+1 c −1 = Sm [ran Lm+1 ]. c

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 of A) 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 there exists an ε > 0 such that for all B ∈ B(E, F ) with kBk < ε, we have: A + B is also a semi-Fredholm operator and n(A + B) ≤ n(A), d(A + B) ≤ d(A) as well as n(A + B) − d(A + B) = n(A) − d(A).

9 In

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

47

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POLES OF THE GENERALIZED RESOLVENT

All we further need is a purely algebraic result, which relates kernel and range of the compressions from Lemma 2.12 to those of S and T . The following is a reformulation of Lemma 1.5 in [2]. Lemma 2.13 Let m ≥ 0 and Sm , Tm as in Lemma 2.12. (i) If λ 6= 0, then ker(λSm − Tm ) = ker(λS − T ). (ii) If m ≥ 1 and S[dom Lm ] + ran Lm c = Y , and if λ 6= 0, then Y ran Lm c ∼ . = ran(λSm − Tm ) ran(λS − T )

Proof: (i) Let λ 6= 0. For m = 0 the equality is trivial, so let m ≥ 1. ker(λSm − Tm ) = ker(λS − T ) ∩ ran Lm ⊆ ker(λS − T ) is obvious. Suppose x ∈ ker(λS − T ), so T x = λSx = S(λx), which means (x, λx) ∈ L. Since λ 6= 0, (λ−1 x, x) ∈ L. A trivial induction shows (λ−m x, x) ∈ Lm , so x ∈ ran Lm . (ii) Firstly, we show ran(λS − T ) ∩ ran Lm c = ran(λSm − Tm ).

(2.12)

“⊇” is obvious. Concerning “⊆”: Suppose x ∈ dom T and (λS − T )x ∈ ran Lm c = T [ran Lm−1 ]. Then there exists some u ∈ ran Lm−1 ∩dom T such that λSx−T x = T u, so (λ−1 (x + u), x) ∈ L = L1 , in particular x ∈ ran L1 . Also: v := λ−1 (x + u) ∈ dom T and (λS − T )u = λSu − T u ∈ S[ran Lm−1 ] + T [ran Lm−1 ] m−1 ⊆ ran Lm−1 + ran Lm . c c = ran Lc

So we have (λS − T )v ⊆ ran Lm−1 . c We can apply the same procedure as above to v instead of x, and thus a finite induction shows x ∈ ran Lm , so (λS − T )x = (λSm − Tm )x ∈ ran(λSm − Tm ). Equation 2.12 implies (1.1) ran Lm + ran(λS − T ) ran Lm ran Lm c c c ∼ = . = ran(λSm − Tm ) ran(λS − T ) ∩ ran Lm ran(λS − T ) c

48

2

POLES OF THE GENERALIZED RESOLVENT By our hypothesis S[dom Lm ] + ran Lm = Y , it suffices to show S[dom Lm ] ⊆ m ran Lm c + ran(λS − T ). To see this, let x ∈ dom L ∩ dom S. Then there exists an L-chain (x0 , . . . , xm ) with x0 = x, so T xi = Sxi+1

(2.13)

∀ 0 ≤ i < m.

We claim: Sx =

k X

λ−(i+1) (λS − T )xi + λ−(k+1) T xk

∀ 0 ≤ k ≤ m − 1.

i=0

This is true for k = 0: Sx = λ−1 (λSx − T x) + λ−1 T x = λ−1 (λS − T )x0 + λ−1 T x0 .

(2.14)

Suppose it is true for some 0 ≤ k < m − 1. Then: Sx = 2.13

=

k P

i=0 k P

λ−(i+1) (λS − T )xi + λ−(k+1) T xk λ−(i+1) (λS − T )xi + λ−(k+1) Sxk+1 .

i=0

Replacing x by xk+1 in (2.14) and replacing Sxk+1 with the result in (ii) yields the desired induction step.  Finally, we are ready to prove two pertubation results (Proposition 5.3, Corollary 5.4 and Proposition 5.5 in [2]) for S and T , using the better behaved operators Sm and Tm introduced before. e c ) = p < ∞. Then there exists ε > 0 such that Proposition 2.14 Suppose δ(L λS − T is surjective and dim ker(λS − T ) = dim(ker T ∩ ran Lp ),

(2.15)

for all 0 < |λ| < ε. Proof: If p = 0, then Y = ker L0 + ran T = ran T , so T is surjective, which means d(T ) = 0. And since it is also bounded, T is (lower) semi-Fredholm. By Theorem 2.3, there exists an ε > 0 such that if |λ| < ε, λS − T is also semi-Fredholm, since S is bounded. Also, the thereom gives d(λS − T ) ≤ d(T ) = 0 and n(λS − T ) = n(λS − T ) − d(λS − T ) = n(T ) − d(T ) = n(T ).

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POLES OF THE GENERALIZED RESOLVENT

That means λS − T is surjective and we have dim ker(λS − T ) = dim ker T = dim(ker T ∩ ran L0 ). e c ) = p, Now, suppose p > 0. Let Sp and Tp be as in Lemma 2.12. Since δ(Lc ) ≤ δ(L we have ran Tp = T [ran Lp ] = ran Lp+1 = ran Lpc , c so Tp is surjective. Again, Sp is bounded, so Theorem 2.3 guarantees the existence of some ε > 0 such that for all |λ| < ε, λSp − Tp is also semi-Fredholm, surjective and satisfies dim ker(λSp − Tp ) = dim ker Tp = dim(ker T ∩ ran Lp ). By Lemma 2.13 (i), we have for λ 6= 0: dim ker(λS − T ) = dim ker(λSp − 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 way we get ran Lpc Y ∼ ∼ = = {0} , ran(λS − T ) ran(λSp − Tp ) since ran(λSp − Tp ) = ran Lpc . So λS − T is indeed surjective.



e c ) =: p < ∞, and let ε > 0 as in the preceding propoCorollary 2.15 Suppose δ(L sition. (i) If α e(L) < ∞, then λS − T is bijective for 0 < |λ| < ε.

(ii) If α e(L) = ∞, then λS − T is surjective, but not injective for 0 < |λ| < ε.

Proof:

(i) If α e(L) < ∞, by Proposition 2.7, it equals p, so we have ker L ∩ ran Lp = {0} ,

which in view of Proposition 2.14 yields for all 0 < |λ| < ε: dim ker(λS − T ) = dim ker L ∩ ran Lp = 0, so λS − T is injective. Also, by that proposition, λS − T is surjective. (ii) Follows from the same arguments.



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

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Theorem 2.4 Let X, Y be Banach spaces and let S, T : X → Y be bounded linear operators. Let p ≥ 1, then: e c ) if and only if R• (S, T ) has a pole of order p at 0. α e(L) = p = δ(L

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

e c ). By the algebraic decomposition Theorem Proof: Firstly, suppose α e(L) = p = δ(L 2.1, 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 ε > 0 such that λS − T is bijective for all 0 < |λ| < ε. Fix such a λ and let A := (λS − T )−1 T, B := T (λS − T ). Using Lemma 1.47 and a straightforward induction, we see that ker Lm = (A−1 )m [ker L0 ] = ker Am , −1 m ) [ker L0c ] = ker B m , ker Lm c = (B

ran Lm = ran Lm ∩ dom T = Am [ran L0 ] = Am [X] = ran Am , m 0 m m ran Lm c = B [ran Lc ] = B [Y ] = ran B .

Since A and B are bounded, ker Lm and ker Lm c are closed, and because of our decomposition (2.16), ran Lp and ran Lpc must also be closed, which follows from a well-known result, see for instance [5], Satz 55.3. By Theorem 2.2, we can decompose S and T as X = ker Lp ⊕ ran Lp S T Y

S1 T 1

S2 T 2

= ker Lpc ⊕ ran Lpc

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 show that λ 7→ (λS2 − T2 )−1 is an analytic function on a neighborhood Bε (0) of 0. We also know that S1 is bijective and that Lker Lp = S1−1 T1 is a nilpotent operator with index p. The latter means that ρ(Lker Lp ) = F \ {0}, where ρ denotes the usual resolvent set. Thus: λS1 − T1 = S1 (λI − S1−1 T1 ),

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is bijective for λ 6= 0, and for those we have the following series expansion (λS1 − T1 )−1 = (λS1 (I − λ−1 Lker Lp ))−1
−1 −1 S1 = λ−1 I − λ−1 Lker Lp ! ∞ X = λ−1 λ−k Lkker Lp S1−1 k=0

=

=

p−1 X

λ−(k+1) Lkker Lp S1−1

k=0 −1 X

−(k+1)

λk Lker Lp S1−1 .

k=−p

That shows: λ 7→ (λS1 − T1 )−1 has a pole of order p at 0. It is easy to verify that (λS − T )−1 = (λS1 − T1 )−1 ⊕ (λS2 − T2 )−1 , for λ ∈ Bε (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 ∞ X Rλ (S, T ) = λk Ck k=−p

on a punctured neighborhood Bε (0) of 0, where Ck ∈ B(Y, X) for all −p ≤ k and C−p 6= 0. Let y ∈ Y and set xk = Ck y for all k ≥ −p. Then λ0 y = y = (λS − T )Rλ (S, T )y ∞ X = (λS − T ) λk xk k=−p

∞ X

= =

λk+1 Sxk −

k=−p ∞ X

∞ X

λk T xk

k=−p ∞ X

λk Sxk−1 −

k=−p+1

λk T xk

k=−p

= λ−p T x−p +

∞ X

λk (Sxk−1 − T xk ).

k=−p+1

Comparing both sides, we get T x−p = 0 Sxk−1 − T xk = 0

∀ − p + 1 ≤ k ≤ −1 or 1 ≤ k

(2.17) (2.18) (2.19)

Sx−1 − T x0 = y.

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POLES OF THE GENERALIZED RESOLVENT

From (2.18) we get (xk , xk−1 ) ∈ L for all −p + 1 ≤ k ≤ −1, and from (2.17) we get (x−p , 0) ∈ L. That means (x−1 , . . . , x−p , 0) ∈ Ch(L), hence x−1 ∈ ker Lp and Sx−1 ∈ S[ker Lp ] = ker Lpc . But then by (2.19) y = Sx−1 − T x0 ∈ ker Lpc + ran T = ker Lpc + ran Lc . e c ) ≤ p < ∞. By Corollary 2.15 and our This proves Y = ker Lpc + ran Lc , so δ(L assumption that R• (S, T ) is defined on a punctured neighborhood of 0, we get α e(L) < e ∞. Using Proposition 2.7, we have q := α e(L) = δ(Lc ) ≤ p. Suppose by contradiction that q < p, then, by the first part of the proof, R• (S, T ) had a pole of order q < p in e c ). 0, which contradicts C−p 6= 0. So indeed, we have p = α e(L) = δ(L 

We will now weaken the topological assumptions on the involved operators. As in [2], page 159, in the following, we assume S and T to be (not necessarily everywhere defined) linear operators in X to Y . Furthermore, T shall be closed and S shall be T -bounded. Definition 2.2 S is called T -bounded, if dom T ⊆ dom S and S is bounded with respect to the graph norm induced by T , that is: there exists some C > 0 such that kSxk ≤ C(kxk + kT xk) ∀ x ∈ dom T. We are interested in the behavior of the function ρS (T ) ∋ λ 7→ Rλ (S, T ) ∈ L(Y, X). A priori, it is not clear, whether Rλ (S, T ) is bounded, and also, dom T need not be complete. But we can remedy that by equipping dom T with the graph-norm k·kT induced by T : Set S0 , T0 : (dom T, k·kT ) → (Y, k·k). Note that (dom T, k·kT ) and (Y, k·k) are Banach spaces and S0 , T0 are everywhere defined, bounded linear operators. It is clear that ρS (T ) coincides with ρS0 (T0 ), and Rλ (S, T ) has the same graph as Rλ (S0 , T0 ) for all λ ∈ ρS (T ). Lemma 2.16 ρS (T ) ∋ λ 7→ Rλ (S, T ) ∈ B(Y, X) is well-defined and analytic on the open set ρS (T ). Furthermore, it has a pole of order p at λ0 ∈ C if and only if that is the case for ρS0 (T0 ) ∋ λ 7→ Rλ (S0 , T0 ) ∈ B(Y, dom T ). Proof: The canonical inclusion map J : (dom T, k·kT ) → (X, k·k) is obviously bounded, so for λ ∈ ρS (T ) we get Rλ (S, T ) = J ◦Rλ (S0 , T0 ) ∈ B(Y, X). We immediately see that R• (S, T ) is analytic, and if R• (S0 , T0 ) has a pole of order p at λ0 ∈ F, then R• (S, T )

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POLES OF THE GENERALIZED RESOLVENT

has a pole of order p at λ0 . Conversely, let R• (S, T ) have pole of order p at λ0 , that is: there are Qk ∈ B(Y, X) such that Rλ (S, T ) =

∞ X

(λ0 − λ)k Qk

k=−p

for all λ ∈ ρS (T ) close enough to λ0 . But we also have operators Pk ∈ B(Y, dom T ) such that ∞ X Rλ (S0 , T0 ) = (λ0 − λ)k Pk k=−∞

for the λ above, since ρS0 (T0 ) = ρS (T ). Because of Rλ (S, T ) = JRλ (S0 , T0 ), we have JPk = Qk = 0

∀ k < −p,

so Pk = 0 for all k < −p, since J is injective.



Eventually, we only have to show that the algebraic quantities used in the main theorem are the same for S, T and S0 , T0 . e c) = Lemma 2.17 Set L0 := S0−1 T0 , then dom L0 = dom L, α e(L) = α e(L0 ) and δ(L e δ((L0 )c ).

Proof: We have dom L0 = dom L, since dom L ⊆ dom T . Furthermore: • It is clear that ker L0 = ker L. Also, ran L00 = dom T = X ∩ dom T = ran L0 ∩ dom T,

m as well as ran L0 = ran L ∩ dom T . Suppose we have shown ran Lm 0 = ran L ∩ m+1 m+1 dom T for some m ≥ 1. Then ran L0 ⊆ ran L ∩ dom T is trivial; for the converse inclusion, we let y ∈ ran Lm+1 ∩ dom T , so there exists some x ∈ ran Lm such that (x, y) ∈ L. In particular, x ∈ dom L ⊆ dom T , so (x, y) ∈ L0 and x ∈ m+1 ran Lm ∩ dom T = ran Lm . 0 by induction assumption, which yields y ∈ ran L0 This gives ker L ∩ ran Lm = ker L0 ∩ ran Lm ∩ dom T = ker L0 ∩ ran Lm for all m ≥ 0, and thus α e(L) = α e(L0 ). • By Proposition 1.42, we know that Lc does only depend on the behavior of S and T on dom S ∩ dom T = dom T . Thus (L0 )c = Lc , and in particular e 0 )c ) = δ(L e c ). δ((L 

Theorem 2.5 Let X, Y be Banach spaces and let S, T be linear operators in X to Y such that T is closed and S is T -bounded. Let p ≥ 1, then: e c ) if and only if R• (S, T ) has a pole of order p at 0. α e(L) = p = δ(L 54

2

POLES OF THE GENERALIZED RESOLVENT

e c ) if and only if α e 0 )c ). Proof: By Lemma 2.17, we have α e(L) = p = δ(L e(L0 ) = p = δ((L Since S0 , T0 are bounded linear operators in Banach spaces, by Theorem 2.4, we have e 0 )c ) if and only if R• (S0 , T0 ) has a pole of order p at 0, which, by Lemma α e(L0 ) = δ((L 2.16, is necessary and sufficent for a pole of the function R• (S, T ) of order p at 0. 

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References

References [1] Richard Arens. Operational calculus of linear relations. Pacific J. Math, 11(1):9– 23, 1961. [2] Harm Bart and David C. Lay. Poles of a generalised resolvent operator. Proceedings of the Royal Irish Academy. Section A: Mathematical and Physical Sciences, 74:147–168, 1974. [3] Ronald Cross. Multivalued Linear Operators. Marcel Dekker, Inc., 2. edition, 1998. [4] Michiel Hazewinkel, editor. Handbook of Algebra, volume 1. North Holland, 1996. [5] Harro Heuser. Funktionalanalysis. Teubner, 4. edition, 2007. [6] Adrian Sandovici, Henk de Snoo, and Henrik Winkler. Ascent, descent, nullity, defect, and related notions for linear relations in linear spaces. Linear Algebra and its Applications, 423(2-3):456–497, 2007. [7] John von Neumann. Über Adjungierte Funktionaloperatoren. The Annals of Mathematics, 33(2,):294–310, 1932. [8] John von Neumann. Functional Operators, Volume 2: The Geometry of Orthogonal Spaces. Princeton University Press, 1950.

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