Bull. Malays. Math. Sci. Soc. https://doi.org/10.1007/s40840-018-0643-8

On a Rational Function in a Linear Relation Ezzeddine Chafai1,2

Received: 15 January 2018 / Revised: 24 May 2018 © Malaysian Mathematical Sciences Society and Penerbit Universiti Sains Malaysia 2018

Abstract The behavior of the domain, the range, the kernel and the multi-valued part of a rational function in a linear relation is analyzed, respectively. We give some basic properties of such linear relations, and we prove that the rational form of the spectral mapping theorem holds in terms of ascent, essential ascent, descent and essential descent. Keywords Linear relation · Ascent · Descent · Spectral mapping theorem · Rational function Mathematics Subject Classification 47A06 · 47A53 · 47A10

1 Introduction In the operator theory, it is always required that the operators are single valued, and further are densely defined. But in many problems, such as when considering the minimal and maximal operators corresponding to a linear discrete Hamiltonian system or a linear symmetric difference equation [15,16], it was found that such operators in the discrete and time scales cases are multi-valued or non-densely defined. So, the classical operator theory (single-valued operator theory) seems inadequate and insufficient. Moreover, for a given multi-valued operator T from a linear space X to a

Communicated by Fuad Kittaneh.

B

Ezzeddine Chafai [email protected]

1

Department of Mathematics, Prince Sattam bin Abdulaziz University, Al kharj, Saudi Arabia

2

Department of Mathematics, Faculty of sciences of Sfax, University of Sfax, Sfax, Tunisia

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linear space Y , which are also called a linear relation, one can introduce concepts of the inverse and the adjoint under no conditions on T . This flexibility of linear relations motivated by necessity of taking adjoints of operators with a non-dense domain used in applications to the theory of generalized differential equations [6,23] and the need of considering the inverse of certain operators, used, for example, in the study of some Cauchy problems associated to parabolic type equations in Banach spaces [8,10], explains the importance of the theory of linear relations. More details related to this theory and its applications can be found in [1–3,6,7,10,11] and elsewhere. In this paper and following the work [3] we focus on the study of the ascent, the essential ascent, the descent and the essential descent spectra of a rational function of a linear relation defined in linear spaces. In particular, we prove that the above spectra satisfy the rational form of the spectral mapping property. The notion of ascent and descent extends the classical theory for single operators (see for instance [13,21, 22]) and the work of Grunenfelder and Omladi˘c [12] which deals with commuting tuples. Riesz [17] introduced the concept of ascent and descent for a linear operator in connection with his investigations of compact linear operators. Recently, Sandovici et al. [18] have extended the study of ascent and descent for linear relations in linear spaces. The rest of this paper is organized as follows. In Sect. 2, we recall some notations, basic concepts of ascent and descent, and fundamental known results about the theory of linear relations that we will need to prove the main theorems. Section 3 is devoted to investigate the concept of a rational function of a given linear relation A in a linear space X . We define and present some basic properties of the relation r (A) = p(A)q(A)−1 , and we establish that r (A) is independent of the particular choice of p and q but depend only of the rational function p/q. Finally, in Sect. 4 we prove that the descent spectrum, the essential descent spectrum, the ascent spectrum and essential ascent spectrum verify the rational function version of the spectral mapping theorem. As an application we prove that the topological descent, essential descent, ascent and essential ascent spectrums of a closed linear relation in a Banach space satisfy, under some conditions, the rational form of the spectral mapping property.

2 Preliminaries Let X be a linear space over C. A linear relation or multi-valued linear operator A in X is a mapping from a subspace D(A) of X called the domain of A into P(X )\∅ (the set of all non-empty subsets of X ) such that α1 A(x1 ) + α2 A(x2 ) ⊆ A(α1 x1 + α2 x2 ), for all scalars α1 and α2 and x1 , x2 ∈ D(T ). By L R(X ) denote the set of all linear relations in X . If A maps the points of its domain to singletons, then A is said to be single valued or simply an operator. The graph G(A) := {(x, y) : x ∈ D(A) and y ∈ Ax}

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of A is a linear subspace of X × X . It is easy to see that A is uniquely determined by its graph. For convenience, A will always be identified with G(A). If G(A) is a closed subspace of X × X , we say that A is closed and we shall denote C R(X ) the set of all closed linear relations in X . The linear relation, denoted by A−1 , and defined as A−1 := {(y, x) : (x, y) ∈ A} is often called the inverse of A. It is evident that A closed if and only if A−1 is closed. If A−1 is single valued, then A is called injective, that is, A is injective if and only if its the null space (or the kernel) N (A) := A−1 (0) = {0}, and A is said to be surjective if its range R(A) := A(D(A)) = X . When A is both injective and surjective we say that A is bijective. Let A, B ∈ L R(X ) and λ ∈ K. Define the linear relations λA, A + B and AB by λA = {(x, λy) : (x, y) ∈ A}, A + B = {(x, y + z) : (x, y) ∈ A and (x, z) ∈ B}, and AB := {(x, z) ∈ X × X : (x, y) ∈ B, (y, z) ∈ A for some y ∈ X }, while A − λ stands for A − λI , where I is the identity operator on X. Since the product of linear relations is clearly associative, then the sequence of natural powers (An )n∈N is defined recursively by A0 = I and An+1 := An A. We can easily see that (A−1 )n = (An )−1 , n ∈ Z. A simple induction argument shows the validity of the law of exponents An+m = An Am , for all n, m ∈ N. In particular, An+1 = A An , implying that the sequence of domains D(An ) is decreasing, i.e., D(An+1 ) ⊂ D(An ). Hence, D(An+m ) ⊂ D(An ), for all nonnegative integers n, m. The kernels and the ranges of the iterates An , n ∈ N, form two increasing and decreasing chains, respectively, i.e., the chain of kernels N (A0 ) = {0} ⊂ N (A) ⊂ N (A2 ) ⊂ .... and the chain of ranges R(A0 ) = X ⊃ R(A) ⊃ R(A2 ) ⊃ .... The singular chain manifold of A is defined by  Rc (A) =

∞ 

n=1

 A (0) ∩ n



∞ 

 N (A ) . n

n=1

Lemma 2.1 [18, Lemma 3.4] Let X be a linear space and A ∈ L R(X ). (i) If N (Ak ) = N (Ak+1 ) for some k ∈ N, then N (An ) = N (Ak ) for all n ≥ k (ii) If R(Ak ) = R(Ak+1 ) for some k ∈ N, then R(An ) = R(Ak ) for all n ≥ k.

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The statement of Lemma 2.1 leads to define the ascent of A by a(A) := in f {n ∈ N : N (An ) = N (An+1 )}, whenever these minima exist. If no such numbers exists the ascent of A is defined to be infinite. Likewise, this statement leads to the introduction of the descent of A by d(A) := inf{n ∈ N : R(An ) = R(An+1 )}, the infimum over the empty set is taken to be infinite. Definition 2.2 Let A ∈ L R(X ). We define the ascent index and the descent index of A, respectively, by αn (A) := dim N (An+1 )/N (An ), n ∈ N, and βn (A) := dim R(An )/R(An+1 ), n ∈ N. Lemma 2.3 Let A ∈ L R(X ). Then (i) αn (A) = dim[R(An ) ∩ N (A)]/[An (0) ∩ N (A)] for all n ∈ N, (ii) βn (A) = dim D(An )/[R(A) + N (An )] ∩ D(An ) for all n ∈ N, (iii) βn+1 (A) ≤ βn (A) for all n ∈ N. Moreover, if βk (A) < ∞ for some k ∈ N, then βm (A) < ∞ for all m ≥ k. (iv) αn+1 (A) ≤ αn (A) for all n ∈ N. Moreover, if αk (A) < ∞ for some k ∈ N, then αm (A) < ∞ for all m ≥ k. Proof We only prove the statement (iv) which extends Lemma 2.8 of [3]. The proofs of the statements (i), (ii) and (iii) can be found in [20, Lemma 3.2], [18, Lemma 4.1]) and [3, Lemma 2.7], respectively. Since Ak (0) ⊂ R(An ) for all k, n ∈ N, then αn+1 (A) = dim[R(An+1 ) ∩ N (A)]/[An (0) ∩ N (A)] ≤ dim[R(An ) ∩ N (A)]/[An (0) ∩ N (A)] (as R(An+1 ) ⊂ R(An )) ≤ dim[R(An ) ∩ N (A)]/[An−1 (0) ∩ N (A)] (as An−1 (0) ⊂ An (0)) = αn (A). On the other hand, suppose that αk (A) < ∞ for some k ∈ N, then αm (A) ≤ αk (A) < ∞ for all m ≥ k.  From Lemma 2.3, one can deduce that αn (A) and βn (A) are decreasing sequences. This leads to define the essential ascent of A by ae (A) := inf{n ∈ N : αn (A) < ∞},

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where the infimum over the empty set is taken to be infinite, is finite. Likewise, we define the essential descent of A by de (A) := inf{n ∈ N : βn (A) < ∞}, where the infimum over the empty set is taken to be infinite. Clearly, every linear relation with finite ascent has a finite essential ascent, more precisely, if a(A) is finite, then ae (A) = 0, and every linear relation with finite descent has a finite essential descent, more precisely, de (A) = 0 whenever d(A) is finite. These notions permit to define the following algebraic spectra e σasc (A) := {λ ∈ C : a(A − λ) = ∞}, σasc (A) := {λ ∈ C : ae (A − λ) = ∞}, e σdsc (A) := {λ ∈ C : d(A − λ) = ∞}, σdsc (A) := {λ ∈ C : de (A − λ) = ∞}.

Combining the algebraic conditions defining the spectra σi (A), i ∈ {asc, asce , dsc, dsce } with a topological condition, we can consider the following topological spectra σta (A) := C\ρta (A) where ρta (A) := {λ ∈ C : a(A − λ) < ∞, e σta (A)

R((A − λ)a(A−λ)+1 ) is closed }, e e := C\ρta (A) where ρta (A) := {λ ∈ C : ae (A − λ) < ∞, R((T − λ)ae (T −λ)+1 ) is closed },

σtd (A) := C\ρtd (A) where ρtd (A) := {λ ∈ C : d(A − λ) < ∞, R((A − λ)d(A−λ) ) is closed } and e e e σtd (A) := C\ρtd (A) where ρtd (A) := {λ ∈ C : de (A − λ) < ∞,

R((A − λ)de (A−λ) ) is closed }. In a recent paper [5], Chafai and Àlvarez prove that, for a given complex polynomial P and a linear relation A in a linear space X , we have σi ( p(A)) = p(σi (A)) where i ∈ {asc, asce , dsc, dsce , ta, ta e , td, td e }. In the present paper we continue the investigation initiated in [5] in terms of a rational function of a linear relation, and thus we establish that the rational form of the spectral mapping property is valid for the algebraic spectra and the topological spectra of a linear relation in X under additional conditions.

3 Rational Functions in a Linear Relation The concept of a rational function of a linear relation extend that of a polynomial of a linear relation which is introduced and studied in several papers (see for examples [14] and [19]). This section contains some basic properties related to them.

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We start by recalling some properties of polynomials of linear relations due to Sandovici. Let T be a linear relation in a linear space X over the field C of complex numbers. Let k k (z − λi )m i be a polynomial and p =: i=1 m i its degree, where k and p(z) = i=1 m i , 1 ≤ i ≤ k, are positive integers and λi , 1 ≤ i ≤ k, are distinct constants. Then the polynomial p in T given by k  p(T ) := (T − λi )m i

(3.1)

i=1

is a linear relation in X too. The behavior of the domain, the range, the null space and the multi-valued part of p(T ) is described in the next proposition. Proposition 3.1 [19, 3.2, 3.3, 3.4, 3.5 and 3.6] Let T be a linear relation in a linear space X , let k ∈ N, λi ∈ C, m i ∈ N, 1 ≤ i ≤ k. Assume that λi , 1 ≤ i ≤ k are distinct and let p(T ) be as in (3.1). Then (i) D( p(T )) = D(T p ). k  R(T − λi )m i . (ii) R( p(T )) = (iii) N ( p(T )) =

i=1 k

N (T − λi )m i .

i=1

(iv) p(T )(0) = T p (0). The next lemma is an immediate consequence of Proposition 3.1. Lemma 3.2 Let T be a linear relation in a linear space X and let p(T ) be a polynomial in T defined as in (3.1). Then p(T ) is bijective if and only if T − λi is bijective, for all 1 ≤ i ≤ k. Let S and T be linear relations in a linear space X . Assume that S and T commute, i.e., ST = T S in the sense of the product of linear relations, then (see [19, (2.9)]) (S − λ)n (T − μ)m = (T − μ)m (S − λ)n holds true for all n, m ∈ N and λ, μ ∈ C. In particular, if S commutes with T then S commutes with p(T ) for any polynomial p in T and by associativity of the product of linear relations, we have the following consequence ( pq)(T ) = p(T )q(T ) = q(T ) p(T ) for all polynomials p and q in T . Furthermore, the following results are straightforward; k  p(T )−1 = (T − λi )−m i i=1

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and

p(T )−1 q(T )−1 = q(T )−1 p(T )−1 .

Proposition 3.3 Let T be a linear relation in a linear space X and p, q ∈ C[z]. Consider the relation r (T ) = p(T )q(T )−1 and let p = deg(P) and q = deg(q). Then the following statements hold:

D(T p−q ) ∩ R(q(T )) + T q (0) ifp ≥ q (i) D(r (T )) = R(q(T )) else. In particular, everywhere defined, then D(r (T )) = R(q(T )).

if T is D(T q−p ) ∩ R( p(T )) + T p (0) ifp ≤ q (ii) R(r (T )) = R( p(T )) else. In particular, if T is everywhere defined, then R(r (T )) = R( p(T )). Proof (i) First suppose that p ≥ q. Then, by using [7, I.1.3 (2)] and Proposition 3.1 we obtain D(r (T )) = D( p(T )q(T )−1 ) = q(T )(D( p(T )) = q(T )(D(T p )) = q(T )(D(T p−q+q )) = q(T )(T q−p (D(T q ))) = q(T )(T q−p (D(q(T )))) = q(T )(D(q(T )T p−q )) = q(T )(D(T p−q q(T ))) = q(T )q(T )−1 (D(T p−q )) = D(T p−q ) ∩ R(q(T )) + q(T )(0) = D(T p−q ) ∩ R(q(T )) + T q (0). Now suppose that p < q. In this case D(q(T )) = D(T q ) ⊂ D(T p ) = D( p(T )) which implies that D(r (T )) = q(T )(D( p(T )) ⊃ q(T )(D(q(T ))) = R(q(T )). The inclusion q(T )(D( p(T )) ⊂ R(q(T )) is trivial. From this one can deduce that D(r (T )) = R(q(T )). (ii) Follows immediately from part (i) and the fact that R(r (T )) = D(q(T ) p(T )−1 ).  Example 3.4 Let X = l p , 1 ≤ p < ∞ be the Banach space of all complex sequences p x = (x0 , x1 , ...) such that ∞ n=0 |x n | < ∞. We define the operator A by A : (x0 , x1 , ..., xn , ...) ∈ X → (0, x1 , x2 , ..., xn , ...) ∈ X and T := A−1 .

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Clearly A is a bounded operator in X , so T is a closed linear relation in X with D(T ) = R(A) = span{e1 , e2 , ....}, N (T ) = A(0) = {0}, R(T ) = D(A) = X and T (0) = N (A) = span{e0 }, where ei = (0, 0, 0, ..., 1, 0, 0, 0...) for i ≥ 0. Let us   i times

consider the polynomials p(z) := (z − 1)3 and q(z) = z 2 in C[z]. Since q(T ) is clearly bijective (Lemma 3.2) and T 2 (0) = T (T (0)) = T (T (0) ∩ D(T )) = T (0), then D(r (T )) = D(T ) + T 2 (0) = D(T ) + T (0) = span{e1 , e2 , ....} + span{e0 } = X and R(r (T )) = R( p(T )) = R(T − I )3 = R(T − I ) = span{e0 }. Remark 3.5 (i) In Example 3.4 and according to Proposition 3.1, we have r (T )(0) = p(T )q(T )−1 (0) = p(T )(0) = T 3 (0) = T (0) = span{e0 }. Define the linear relation r1 (T ) := q(T )−1 p(T ). Then r1 (T )(0) = q(T )−1 p(T )(0) = q(T )−1 T (0) = T −2 T (0) = T (0) ∩ D(T ) + T −1 (0) = T −1 (0) = {0}. This implies that r1 (T ) is an operator, so that r (T ) and r1 (T ) do not define the same linear relation. Furthermore, D(r1 (T )) = p(T )−1 (R(q(T ))) = p(T )−1 (X ) = D( p(T )) ⊂ D(r (T )) = X. (ii) If T is single valued and q(T ) is bijective (that is q −1 (0) ∩ σ (T ) = ∅), one could have equally defined r (T ) = q(T )−1 p(T ). This is essentially the same operator but with a smaller domain. Question: Is the linear relation p(T )q(T )−1 independent of the particular choice of p and q ? 1 < p < ∞ be the Banach space of all complex sequences Example 3.6 Let X = l p ,  p x = (x0 , x1 , ...) such that ∞ n=0 |x n | < ∞. We define the left shift operator L in X by L : (x0 , x1 , ...) ∈ X → (x1 , x2 , ...) ∈ X.

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Let T := L −1 ∈ L R(X ) and the polynomials p(z) = z−1, q(z) = z, p1 (z) = (z−1)z and q1 (z) = z 2 . Then p(T )q(T )−1 (0) = (T − I )T −1 (0) = (T − I )L(0) = T (0) = span(e0 ) where e0 = (1, 0, 0, ...). and p1 (T )q1 (T )(0) = (T − I )T T −2 (0) = (T − I )T (0) = span(e0 , e1 , ...) where e1 = (0, 1, 0, ...). From this example one can deduce that the linear relation r (T ) = p(T )q(T )−1 depend of the special choice of p and q. The following definition helps to understand the definition of a rational function r (T ) of a linear relation T (see (3.2) blow). Definition 3.7 Let p and q be two complex polynomials. We say that the rational function p/q is irreducible if p and q have no common divisors. We denote by R(z) := { p/q : p, q ∈ C[z] and p/q is irr educible} the set of all irreducible rational functions. Let T ∈ L R(X ). In the sequel of this paper we define a rational function in T by r (T ) := p(T )q(T )−1 , for p/q ∈ R(z).

(3.2)

Lemma 3.8 Let X be a linear space, T ∈ L R(X ) and λ ∈ C\{0}. Then N (T − λ)k ⊂ T n (N (T − λ)k ), for all k, n ∈ N.

(3.3)

Proof If either k = 0 or n = 0, the desired inclusion is trivial. Assume that n ≥ 1 and k ≥ 1. The proof will be given by induction on n ∈ N. For n = 1 let be consider two cases: Case 1: k = 1. Let x0 ∈ N (T − λ). Then λx0 ∈ T x0 ⊂ T (N (T − λ)) which implies that x0 ∈ T (N (T − λ)) (as λ = 0). Case 2: k ≥ 2. Let x0 ∈ N (T − λ)k . Then there exist elements x1 , x2 , ..., xk−1 such that (x0 , x1 ), (x1 , x2 ), ..., (xk−1 , 0) ∈ T − λ.

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Define xk := 0. It follows that (xi , xi+1 ) ∈ T − λ, for 0 ≤ i ≤ k, and hence k−1  k−1 (−λ)k−i−1 (xi , xi+1 + λxi ) = (−λ)k−i−1 xi , −(−λ)k x0 ∈ T. i=0

i=0

This shows that k−1  −i−2 x0 ∈ T (−λ) xi . i=0

On the other hand, since x0 ∈ N (T − λ)k , then xi ∈ N (T − λ)k−i , for 0 ≤ i ≤ k − 1. It follows that k−1  k−1  −i−2 k−i (−λ) xi ⊂ T N (T − λ) x0 ∈ T = T (N (T − λ)k ). i=0

i=0

Now assume that (3.3) is satisfied for some n ≥ 1 and for all k ∈ N. Then     N (T − λ)k ⊂ T n N (T − λ)k ⊂ T n T (N (T − λ)k ) = T n+1 (N (T − λ)k ). 

This completes the proof.

Lemma 3.9 Let T be a linear relation in a linear space X and let α, β ∈ C such that α = β. Then (T − α)s (T − β)−k (0) = (T − α)s (0) + (T − β)−k (0).

(3.4)

for all s, k ∈ N. Proof Clearly (T − α)s (0) ⊂ (T − α)s (T − β)−k (0) and by using (3.3) we have (T − β)−k (0) ⊂ (T − α)s (T − β)−k (0). This means that (T − α)s (0) + (T − β)−k (0) ⊂ (T − α)s (T − β)−k (0).

(3.5)

Now, we claim to show the converse inclusion by induction on s. If either s = 0 or k = 0, the desired inclusion is trivial. Assume that s ≥ 1 and k ≥ 1.

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For s = 1. Let k ≥ 1, then (T − α)(T − β)−k (0) = (T − β + β − α)(T − β)−k (0) ⊂ (T − β)(T − β)−k (0) + (T − β)−k (0) (by [7, I.4.2]) = (T − β)(0) + (T − β)−k+1 (0) ∩ R(T − β) + (T − β)−k (0) (by [7, I.3.1 (d)]) ⊂ (T − β)(0) + (T − β)−k+1 (0) + (T − β)−k (0) = (T − α)(0) + (T − β)−k (0) (as (T − β)(0) = T (0) = (T − α)(0)). Assume that (T − α)s (T − β)−k (0) ⊂ (T − α)s (0) + (T − β)−k (0) for some s ≥ 1 and all k ∈ N. Then (T − α)s+1 (T − β)−k (0) = (T − α)(T − α)s (T − β)−k (0) ⊂ (T − α)[(T − α)s (0) + (T − β)−k (0)] = (T − α)s+1 (0) + (T − α)(T − β)−k (0) (by [7, I.3.1(c)]) ⊂ (T − α)s+1 (0) + (T − α)(0) + (T − β)−k (0) (by hypothesis) = (T − α)s+1 (0) + (T − β)−k (0). Hence (T − α)s (T − β)−k (0) ⊂ (T − α)s (0) + (T − β)−k (0) for all s, k ∈ N.

(3.6)

Now, a combination of (3.5) and (3.6) leads to deduce that (3.4) holds for all s, k ∈ N.  Proposition 3.10 Let X be a linear space, T ∈ L R(X ) and r (T ) be defined as in (3.2). Then (i) r (T )(0) = p(T )(0) + N (q(T )). (ii) N (r (T )) = N ( p(T )) + q(T )(0). p q Proof Let p(T ) := i=1 (T − αi )si and q(T ) := j=1 (T − β j )k j , where p and q 1 ≤ j ≤ q, and αi , 1 ≤ i ≤ p, are positive integers, si ∈ N, 1 ≤ i ≤ p, k j ∈ N, p β j , 1 ≤ j ≤ q are distinct constants. Put s := i=1 si . Then (T − β j )−k j (0) ⊂ (T − β j )−k j −s (0) ⊂ D(T − β j )k j +s = D(T k j +s ) ⊂ D(T s ) = D( p(T )).

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According to Proposition 3.1 and [7, I.3.1 (c)], it follows that ⎡ r (T )(0) = p(T )q(T )−1 (0) = p(T ) ⎣

q

⎤ (T − β j )−k j (0)⎦

j=1

q

=

p(T )(T − β j )−k j (0).

j=1

By combining (3.4), [7, I.3.1 (c)] and the fact that (T − β j )−k j (0) ⊂ D(T − αi )si for all 1 ≤ i ≤ p and 1 ≤ j ≤ q, one can deduce that p(T )(T − β j )−k j (0) =

p  (T − αi )si (0) + (T − β j )−k j (0) i=1

= p(T )(0) + (T − β j )−k j (0) for 1 ≤ j ≤ q. Hence r (T )(0) =

q

p(T )(T − β j )

−k j

(0) =

j=1

q 

p(T )(0) + (T − β j )−k j (0)



j=1

= p(T )(0) + N (q(T )). To prove the part (ii) we proceed in the same way by taking q(T ) and p(T ) instead of p(T ) and q(T ), respectively.  Proposition 3.11 Let T be a linear relation in a linear space X and let r (T ) be defined as in (3.2). Then for all λ ∈ C r (T ) − λ ⊂ ( p(T ) − λq(T ))q(T )−1 with equality if deg( p) > deg(q) or if T is an operator. Proof The result is trivial for λ = 0, so we can suppose that λ = 0. By p and q we denote the degrees of p and q, respectively. If p ≤ q then deg( p −λq) ≤ max(p, q) = q. By using Proposition 3.3 we obtain D( p(T ) − λq(T ))q(T )−1 = R(q(T )) = D(r (T )) = D(r (T ) − λ). Let x ∈ D(r (T )) ⊂ R(q(T )). It follows that ( p(T ) − λq(T ))q(T )−1 x = p(T )q(T )−1 x − λq(T )q(T )−1 x ([4, Lemma 4.1]) = p(T ))q(T )−1 x − λx + q(T )(0) ([7, I.3.1 (d)]) = r (T )x − λx + q(T )(0) ⊃ r (T )x − λx,

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which means that r (T ) − λ ⊂ ( p(T ) − λq(T ))q(T )−1 . Now suppose that p > q or T is an operator then deg( p − λq) = p and hence the use of Proposition 3.3 gives D( p(T ) − λq(T ))q(T )−1 = D(T p−q ) ∩ R(q(T )) +T q (0) = D(r (T )) = D(r (T ) − λ). Moreover q(T )(0) ⊂ r (T )(0), so that r (T )x − λx = r (T )x + q(T )(0) − λx = r (T )x − λq(T )q(T )−1 x = ( p(T ) − λq(T ))q(T )−1 x for all x ∈ D(r (T ) − λ) = D( p(T ) − λq(T ))q(T )−1 ). Consequently r (T ) − λ =  ( p(T ) − λq(T ))q(T )−1 . Remark 3.12 Let X be a linear space and T ∈ L R(X ) be injective (not single valued). Let p(z) = 1, q(z) = z be polynomials and r (T ) = p(T )q(T )−1 . Then (r (T ) − λ)(0) = r (T )(0) = T −1 (0) = 0  ( p(T ) − λq(T ))q(T )−1 (0) = r (T )(0) + q(T )(0) = T −1 (0) + T (0) = T (0). This shows that the inclusion in Proposition 3.11 is, in general, strictly whenever deg( p) ≤ deg(q). Definition 3.13 We define C∞ := C ∪ {∞} to be the one-point compactification of C (Riemann sphere) with the usual conventions for computation. Let λ ∈ C∞ and T be a linear relation in a linear space X . The eigenspace of T at λ is defined as

N (T, λ) :=

N (T − λ) T (0)

if λ ∈ C if λ = ∞

Similarly, we define the range space of T at λ as

R(T, λ) :=

R(T − λ) ifλ ∈ C D(T ) ifλ = ∞

Using this notation we define the extended point spectrum, the surjectivity spectrum and the extended spectrum of T by σs (T )  σ p (T ) := {λ ∈ C∞ : N (T, λ) = 0},  := {λ ∈ C∞ : R(T, λ) = X } and  σ (T ) :=  σ p (T ) ∪  σs (T ). Lemma 3.14 Let X be a linear space and T ∈ L R(X ) be bijective and let λ = 0. Then (i) N (T − λ)n = N (T −1 − λ−1 )n , for all n ∈ N, (ii) R(T − λ)n = R(T −1 − λ−1 )n , for all n ∈ N.

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Proof First observe that T − λ = −λ(T −1 − λ−1 )T = −λT (T −1 − λ−1 ).

(3.7)

The equality T − λ = −λ(T −1 − λ−1 )T follows immediately from [7, VI.4.2]. On the other hand (3.8) T −1 T ⊂ T T −1 . Indeed, let (x, y) ∈ T −1 T then y ∈ T −1 T x = x + T −1 (0) = x (as T is injective). Now, let z ∈ T −1 x = ∅ (as T is surjective). Then x ∈ T z ⊂ T T −1 x which implies that (x, y) ∈ T T −1 . Hence (3.8) holds. It follows that (T −1 − λ−1 )T = T −1 T − λ−1 T [4, Lemma 4.1] ⊂ TT

−1

⊂ T (T

−λ

−1

−1

−λ

T (by (3.8))

−1

) [7, I.4.2(e)].

(3.9) (3.10) (3.11)

Now, we shall prove that

and

(T −1 − λ−1 )T (0) = T (T −1 − λ−1 )(0)

(3.12)

D[(T −1 − λ−1 )T ] = D[T (T −1 − λ−1 )].

(3.13)

By using [4, Lemma 4.1] and the fact that T is injective, it follows that (T −1 − λ−1 )T (0) = T −1 T (0) − λ−1 T (0) = T −1 (0) +T (0) = T (0) = T (T −1 − λ−1 )(0). On the other hand, since T − λ ⊂ T (T −1 − λ−1 ) and T − λ = −λ(T −1 − λ−1 )T , then D[(T −1 − λ−1 )T ] = D(T − λ) ⊂ D[T (T −1 − λ−1 )]. Conversely, let x ∈ D[T (T −1 − λ−1 )] then ∅ = T (T −1 − λ−1 )x = T (T −1 x − λ−1 x) = T T −1 x − λ−1 T x ([7, I.3.1(c)]) = x + T (0) − λ−1 T x =x −λ

−1

([7, I.3.1(c)]and T is surjective)

T x.

It follows that T x = ∅, so that x ∈ D(T ). Thus D[T (T −1 − λ−1 )] ⊂ D(T ) = D[(T −1 − λ−1 )T ] and consequently (3.13) holds. By combining (3.9), (3.12), (3.13) and [7, I.2.14(b)] one can deduce that (3.7) is satisfied. Now we prove the parts (i) and (ii) by induction on n ∈ N.

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(i) For n = 0 the result is evident and for n = 1 the result follows immediately from [7, VI.2.3]. Now suppose N (T − λ)n = N (T −1 − λ−1 )n for some n ≥ 1. Then N (T − λ)n+1 = (T − λ)−1 (N (T − λ)n ) = (T − λ)−1 (N (T − λ)n ) = (−λ)−1 [T (T −1 − λ−1 )]−1 (N (T − λ)n )

(by 3.7)

= (−λ)−1 (T −1 − λ−1 )−1 T −1 (T − λ)−n (0) = (T −1 − λ−1 )−1 N ((T − λ)T ) = (T −1 − λ−1 )−1 N (T (T − λ)n ) = (T

−1

= (T

−1

= (T

−1

−λ

−1 −1

−λ

−1 −1

−λ

−1 −1

) ) )

−n

(T − λ)

−n

(T − λ) N (T

−1

T

−1

([19, (2.9)]) (0)

(0)

−λ

(as T is injective)

−1 n

)

= N (T −1 − λ−1 )n+1 . (ii) For n = 0 the result is trivial. For n = 1, by using (3.7) and the fact that T is surjective, we obtain R(T − λ) = R[(T −1 − λ−1 )T ] = (T −1 − λ−1 )(R(T )) = (T −1 − λ−1 )(X ) = R(T −1 − λ−1 ). Now, suppose that R(T − λ)n = R(T −1 − λ−1 )n for some n ∈ N. Then R(T − λ)n+1 = (T − λ)R(T − λ)n = (−λ)(T −1 − λ−1 )T R(T − λ)n = (T

−1

= (T

−1

= (T

−1

= (T

−1

(by 3.7)

−λ

−1

)R(T (T − λ) ))

−λ

−1

)R((T − λ)n )T )

([19, (2.9)])

−λ

−1

)R((T − λ) ))

(as T is surjective)

−λ

−1

n

n

)R(T

−1

−λ

−1 n

)

= R(T −1 − λ−1 )n+1 .



Remark 3.15 From Definition 3.13 we have N (T, 0) = N (T ) = T −1 (0) = N (T −1 , ∞) and R(T, 0) = R(T ) = D(T −1 ) = R(T −1 , ∞). Then Lemma 3.14 holds for all λ ∈ C∞ , that is N (T, λ) = N (T −1 , λ−1 ) and R(T, λ) = R(T −1 , λ−1 ) for all λ ∈ C∞ .

(3.14)

Proposition 3.16 Let X be a linear space and T ∈ L R(X ) be bijective. Then  σ (T −1 ) = [ σ (T )]−1 . Proof Let λ ∈  σ (T −1 ). Then N (T −1 − λ) = {0} or R(T −1 − λ) = X . By using σ (T ). (3.14) it follows that N (T − λ−1 ) = {0} or R(T − λ−1 ) = X so that λ−1 ∈ 

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Consequently λ ∈ [ σ (T )]−1 . Conversely, let λ ∈ [ σ (T )]−1 , then λ−1 ∈  σ (T ). This −1 −1 means that N (T −λ ) = {0} or R(T −λ ) = X . Again from (3.14) one can deduce σ (T −1 ).  that N (T −1 − λ) = {0} or R(T −1 − λ) = X . Thus λ ∈  Corollary 3.17 Let T be a linear relation in a linear space X and let p be a polynomial in T . Suppose that p −1 (0) ∩ σ (T ) = ∅. Then σ (T ))]−1 .  σ ( p(T )−1 ) = [ p( The next result is the rational form of the spectral mapping property. Theorem 3.18 Let X be a linear space, T ∈ L R(X ) be everywhere defined and r (T ) defined as in (3.2). Then (i) σs (r (T )) ⊃ r ( σs (T )), with equality if deg( p) > deg(q), σ p (T )), with equality if T is an operator. (ii)  σ p (r (T )) ⊃ r (

k Proof (i) Let ∞ = λ ∈ r ( σs (T )) and put h(T ) := p(T ) − λq(T ) := i=1 (T − σs (T ), so that α = λ j for some 1 ≤ j ≤ k, λi )m i . Then λ = r (α) for some α ∈  say j = 1. Since T − λ1 is not surjective, then so is h(T ) (Proposition 3.1 (ii)). It follows that X = R(r (T ) − λ)) = R(h(T )) and consequently λ ∈  σs (r (T )). σs (T ), so that q(α) = 0. If λ = ∞ ∈ r ( σs (T )), then λ = r (α), for some α ∈  It follows that q(T ) is not surjective, so D(r (T )) = R(q(T )) = X (Propositions σs (T )) ⊂ σs (r (T )). 3.3 (i)). This means that ∞ ∈  σs (r (T )). Consequently r ( Now, suppose moreover that deg( p) > deg(q) and let ∞ = λ ∈  σs (r (T )). Put k (T − λi )m i . Since p and q are irreducible then h(T ) := p(T ) − λq(T ) := i=1 so are h = p − λq and q. Since deg(h) = deg( p) > deg(q), then r (T ) − λ = h(T )q(T )−1 (Proposition 3.11) and R(r (T ) − λ) = R(h(T )) (Proposition 3.10). It follows that R(T − λ j ) = X , for some 1 ≤ j ≤ k (since r (T ) − λ is not σs (T ). But λ = r (λ1 ), which implies surjective), say j = 1. This means that λ1 ∈  that λ ∈ r ( σs (T )). If λ = ∞ ∈  σs (r (T )), then R(q(T )) = D(r (T )) = X . This implies that q(T ) is not surjective. If follows that T − α is also not surjective for some α such that q(α) = 0 (Proposition 3.1 (ii)). Therefore ∞ = r (α) ∈ r ( σs (T )). Consequently, σs (T )). σs (r (T )) ⊂ r ( (ii) Let λ ∈ C∞ . If λ = ∞, then N (r (T ), λ) = N ( p(T ) − λq(T )) + q(T )(0). Moreover N (r (T )), ∞) = r (T )(0) = p(T )(0) + q(T )(0) + N (q(T )). Suppose that T (0) = 0, then clearly N (r (T ), λ) = {0} for all λ ∈ C∞ . It follows that σ p (T )).  σ p (r (T )) = C∞ ⊃ r ( Now, suppose that T is an operator, that is T (0) = 0 and let ∞ = λ ∈  σ p (r (T )).  = Then r (T )−λ = ( p(T )−λq(T ))q(T )−1 (Proposition 3.11) and {0} k N (r (T )− (T −λi )m i . λ)) = N ( p(T )−λq(T )) (Proposition 3.10). Put p(T )−λq(T ) = i=1 The use of Proposition 3.1 ensures that N (T − λi ) = {0} for some 1 ≤ i ≤ k, say σ p (T ). Since r (λ1 ) = 0, then λ ∈ r ( σ p (T ). i = 1. It follows that λ1 ∈  On the other hand suppose that ∞ ∈  σ p (r (T )), then r (T )(0) = {0}, that is  N (q(T )) = {0} (Proposition 3.10). Put q(T ) = li=1 (T − βi )n i . Then T − βi is not injective for some 1 ≤ i ≤ l, say i = 1. It follows that β1 ∈  σ p (T ). Since σ p (T )). q(β1 ) = 0 and p/q ∈ R[z], then r (β1 ) = ∞. Consequently ∞ ∈ r (

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On a Rational Function in a Linear Relation

Conversely, let λ ∈ r ( σ p (T )). Then λ = r (β), for some β ∈  σ p (T ), that is N (T, β) = {0}. Since T (0) = 0, then β  = ∞. Suppose that λ  = ∞ and put s (T − μi )ri . From Proposition 3.1, one can deduce that T − μi r (T ) − λ := i=1 is not injective, for some 1 ≤ i ≤ s, say i = 1. It follows that μ1 ∈  σ p (T ) and σ p (T )). since λ = r (μ1 ), then λ ∈ r (  Now, suppose that λ = ∞, so necessarily q(β) = 0. Let q(T ) := li=1 (T − n σ p (T ), then βi ) i . Then β = βi for some 1 ≤ i ≤ l, say i = 1. Since β1 ∈  T −β1 is not injective and so is q(T ) (Proposition 3.1). It follows that r (T )(0) = 0 (Proposition 3.10). Finally, since r (β1 ) = ∞ (as q(β1 ) = 0), then ∞ ∈ r ( σ p (T )). 

4 Rational Spectral Mapping Theorem for Ascent, Essential Ascent, Descent and Essential Descent Many results in the theory of normal operators are based on the spectral theorem. In this section we extend the polynomial form of the spectral mapping theorem for ascent, essential ascent, descent and essential descent spectra studied in [3] to the rational form. We begin with some auxiliary results. Firstly, we recall the following elementary lemma. Lemma 4.1 Let M, N and W subspaces of a linear space X . Then M M+N  , N M∩N (ii) if M ⊂ W , then (M + N ) ∩ W = M + N ∩ W. (i)

Lemma 4.2 [19, Lemma 6.1] Let T be a linear relation in a linear space X and let n, m ∈ N. Then if ρ(T ) = ∅, then N (T n ) ∩ T m (0) = {0}, in particular Rc (T ) = {0}. The following proposition describes the behavior of the kernels and ranges of the iterates (r (T )n ), n ∈ N which is used to prove our main theorem of this section. Proposition 4.3 Let T be a linear relation everywhere defined in a linear space X and r (T ) be defined as in (3.2). The following assertions hold for all n ∈ N.

p if p < q T (0) + N (q(T )n ) n (i) r (T ) (0) = np−(n−1)q n (0) + N (q(T ) ) if p ≥ q T

q if p ≥ q T (0) + N ( p(T )n ) n (ii) N (r (T ) ) = T nq−(n−1)p (0) + N ( p(T )n ) if p < q (iii) R(r (T )n ) = R( p(T )n . Proof (i) We shall proceed by induction on n. The case n = 1 is proved in Proposition 3.10. Let us consider n ≥ 1 and suppose that (i) is satisfied for n. If p < q then q(T ) can be written as q(T ) = h(T )k(T ) where h(z) and k(z) are two complex

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polynomials with degrees q − p and p, respectively (obviously h(T ) and k(T ) commute). If follows that r (T )n+1 (0) = p(T )q(T )−1 [T p (0) + N (q(T )n )] = p(T )q(T )−1 T p (0) + p(T )q(T )−1 N (q(T )n ) = p(T )h(T )−1 k(T )−1 k(T )(0) + p(T )q(T )−n−1 (0) = p(T )h(T )−1 k(T )−1 (0) + p(T )q(T )−n−1 (0) = p(T )q(T )−1 (0) + p(T )q(T )−n−1 (0) = p(T )(0) + q(T )−1 (0) + p(T )(0) + q(T )−n−1 (0) = T p (0) + q(T )−n−1 (0). Now assume that p ≥ q and r (T )n (0) = T np−(n−1)q (0) + N (q(T )n ) for some n ≥ 1. Then r (T )n+1 (0) = p(T )q(T )−1 [T np−(n−1)q (0) + N (q(T )n )] = p(T )q(T )−1 T np−(n−1)q (0) + p(T )q(T )−1 N (q(T )n ) = p(T )q(T )−1 T np−(n−1)q (0) + p(T )q(T )−n−1 (0) = p(T )q(T )−1 T np−nq q(T )(0) + p(T )q(T )−n−1 (0) = p(T )q(T )−1 q(T )T np−nq (0) + p(T )q(T )−n−1 (0) = p(T )[T np−nq (0) ∩ R(q(T )) + q(T )−1 (0)] + p(T )q(T )−n−1 (0) = p(T )[T np−nq (0) + q(T )−1 (0)] + p(T )q(T )−n−1 (0) = T (n+1)p−nq (0) + p(T )q(T )−1 (0) + p(T )q(T )−n−1 (0) = T (n+1)p−nq (0) + p(T )(0) + q(T )−1 (0) + p(T )(0) + q(T )−n−1 (0) = T (n+1)p−nq (0) + q(T )−n−1 (0).

(ii) It suffices to interchange p(T ) and q(T ) in (i). (iii) For n = 1 the result is proved in Proposition 3.3 (ii). Suppose that R(r (T )n ) = R( p(T )n for some n ≥ 1. Since T and p(T )n commute then T (R( p(T )n )) ⊂ R( p(T )n ). Let Tn : R( p(T )n ) → R( p(T )n ) defined by Tn x = T x for all x ∈ R( p(T )n ). From Proposition 3.3 it follows that R(r (T )n+1 ) = r (R(r (T )n ) = r (R( p(T )n ) = R(r (Tn )) = R( p(Tn )) = p(Tn )(R( p(T )n )) =  p(T )(R( p(T )n )) = R( p(T )n+1 ). Recall that the resolvent set of a linear relation T is the set ρ(T ) defined by ρ(T ) := {λ ∈ C : T − λ is bijective} and the spectrum of T is the set σ (T ) := C\ρ(T ). Lemma 4.4 Let T be a linear relation everywhere defined in a linear space X such that dim(T (0)) < ∞ and ρ(T ) = ∅. Let r (T ) be defined as in (3.2). The following statements hold. (i) If ae (r (T )) < ∞, then ae ( p(T )) < ∞, with equivalence if p ≥ q,

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On a Rational Function in a Linear Relation

(ii) if a(r (T )) < ∞, then a( p(T )) < ∞, with equivalence if p ≥ q, (iii) de (r (T )) < ∞ if and only if de ( p(T )) < ∞, (iv) d(r (T )) < ∞ if and only if d( p(T )) < ∞. Proof (i) Arguing exactly as in the proof of Theorem VI.5.4 in [7] we obtain ρ( p(T )) = ∅ and hence Rc ( p(T )) = {0} (Lemma 4.2). If p ≥ q or T is an operator, then the use of Lemmas 2.3, 4.1, 4.2 together with Proposition 4.3 leads to T q (0) + N ( p(T )n+1 ) N (r (T )n+1 ) = n N (r (T ) ) T q (0) + N ( p(T )n ) N ( p(T )n+1 )  N ( p(T )n ) + T q (0) ∩ N ( p(T )n+1 ) N ( p(T )n+1 ) . = N ( p(T )n ) for all n ∈ N. Consequently, ae (r (T )) < ∞ (resp. a(r (T )) < ∞) if and only if ae (r (T )) < ∞ (resp. a(r (T )) < ∞). Now suppose that T (0) = 0 (that is T is not single valued) and p < q. Arguing as above we obtain T (n+1)q−np (0) + N ( p(T )n+1 ) N (r (T )n+1 ) = n N (r (T ) ) T nq−(n−1)p (0) + N ( p(T )n ) T nq−(n−1)p (0) + N ( p(T )n+1 ) ⊃ T nq−(n−1)p (0) + N ( p(T )n ) N ( p(T )n+1 ) . = N ( p(T )n ) Thus ae ( p(T )) < ∞ (resp. a( p(T )) < ∞) whenever ae (r (T )) < ∞ (resp. a(r (T )) < ∞). (iii) For n ∈ N, put m = q if p ≥ q and m = nq − (n − 1)p if p < q. Then by combining Proposition 3.3, Lemma 2.3 and Proposition 4.3 we obtain X R(r (T )n )  n+1 n R(r (T ) ) N (r (T ) ) + R(r (T )) X = m T (0) + N ( p(T )n ) + R( p(T )) X = N ( p(T )n ) + R( p(T )) R( p(T )n )  . R( p(T )n+1 ) The results of (ii) and (iv) follow immediately.



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Theorem 4.5 Let T be a linear relation everywhere defined in a linear space X with ρ(T ) = ∅, and let r (T ) be defined as in (3.2). Then (i) (ii) (iii) (iv)

e (r (T )) ⊂ r (σ e (T )), σasc asc σasc (r (T )) ⊂ r (σasc (T )), e (r (T )) ⊂ r (σ e (T )), σdsc dsc σdsc (r (T )) ⊂ r (σdsc (T )).

Proof We only prove (i), the proofs of  statements (ii), (iii) and (iv) are similar. For p λ ∈ C, let h(T ) := p(T ) − λq(T ) = i=1 (T − λi )m i , where p and m i , 1 ≤ i ≤ p are positive integers and λi ∈ C, 1 ≤ i ≤ p are distinct constants. Then r (T ) − λ ⊂ e (r (T ))). Then a (r (T ) − λ) = ∞, so h(T )q(T )−1 (Proposition 3.11). Let λ ∈ σasc e that ae (h(T )) = ∞ by virtue of Lemma 4.4. It follows from [5, Proposition 3.1] that e (T )). ae (T −λ j ) = ∞ for some λ j , and since r (λ) = λ j one can deduce that λ ∈ r (σasc  It is more interesting from the point of view of the operator theory to combine ( e (.), ρ the algebraic conditions defining the resolvent sets ρdsc (.), ρdsc asc (.) and ρasc .) with a topological condition. We shall use the following result which gives sufficient conditions for r (T ) to be closed. Proposition 4.6 Let T be a closed linear relation in a Banach space X and let r (T ) be defined as in (3.2). Suppose that q −1 (0) ∩ σ (T ) = ∅. Then r (T ) is closed. Proof Since T is closed and ρ(T ) = ∅ then, by [9, Lemma 3.1], p(T ) is closed. On the other q(T )−1 is a bounded operator hence by using [7, II.5.18] it follows that r (T ) is closed.  Lemma 4.7 Let T be a closed linear relation in a Banach space X . (i) If a(T ) < ∞ and R(T k ) is closed for some k > a(T ), then R(T n ) is closed for all n ≥ a(T ), (ii) if ae (T ) < ∞ and R(T k ) is closed for some k > ae (T ), then R(T n ) is closed for all n ≥ ae (T ), (iii) if d(T ) < ∞ and R(T k ) is closed for some k ≥ d(T ), then R(T n ) is closed for all n ≥ d(T ), (iv) if de (T ) < ∞ and R(T k ) is closed for some k ≥ de (T ), then R(T n ) is closed for all n ≥ de (T ). Proof The statement (ii) is proved in [3, Lemma 4.3] and (i) can be proved arguing in the same way as in the proof of (ii). The proofs of the statements (iii) and (iv) are trivial.  Corollary 4.8 Let X be a Banach space and T ∈ C R(X ) be everywhere defined. Let r (T ) be defined as in (3.2) and suppose that {λ : q(λ) = 0} ⊂ ρ(T ). If deg( p) ≥ deg(q) or T is single valued, then σi (r (T )) ⊂ r (σi (T )) for i ∈ {ta, ta e , td, td e }. 3.11 we have r (T ) − λ = ( p(T ) − Proof Let λ ∈ σta (r (T )). From Proposition p λq(T ))q(T )−1 := h(T )q(T )−1 := i=1 (T − λi )m i , where λi , 1 ≤ i ≤ p are

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On a Rational Function in a Linear Relation

distinct scalars and m i , 1 ≤ i ≤ p are positive integers. If a(r (T ) − λ) = ∞ then Theorem 4.5 ensures that λ ∈ r (σta (T )). Now suppose that a = a(r (T )−λ) < ∞ and that R(r (T ) − λ)d+1 ) is not closed. It follows from Proposition 4.3 that R(h(T )a+1 ) is not closed so that R(T − λ j )am j +m j for some 1 ≤ j ≤ p [5, Proposition 3.1]. The use of Lemma 4.7 ensures that R(T − λ j )a(T )+1 is not closed which implies that λ j ∈ σta (T ). Since r (λ) = λ j , then λ ∈ r (σta (T )). Proceeding as in above we show  similarly that σi (r (T )) ⊂ r (σi (T )) for i = ta e , td, td e . Remark 4.9 The case where p is a constant polynomial. In this case r (T ) = p(T )−1 . Combining Lemma 3.14 together with Lemma 4.7 and [5, Theorem 4.1] one can easily deduce that [σi (q(T )−1 )] = [q(σi (T ))]−1 for i ∈ {asc, asce , dsc, dsce , ta, ta e , td, td e }, whenever T ∈ C R(X ) everywhere defined with {λ : Q(λ) = 0} ⊂ ρ(T ). In particular if 0 ∈ ρ(T ), then σi (T −1 ) = [σi (T )]−1 for i ∈ {asc, asce , dsc, dsce , ta, ta e , td, td e }.

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