On a Rational Function in a Linear Relation

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.

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 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 [1,2,3, 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 Omladic̆ [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 $\mathbb {C}$. 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 $\mathcal {P}(X)\backslash \emptyset$ (the set of all non-empty subsets of X) such that

$$\begin{aligned} \alpha _1 A(x_1)+\alpha _2 A(x_2) \subseteq A(\alpha _1 x_1+\alpha _2 x_2), \end{aligned}$$

for all scalars $\alpha _1$ and $\alpha _2$ and $x_1, x_2 \in D(T)$. By LR(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

$$\begin{aligned} G(A) := \{(x,y): x \in D(A)~\text{ and }~y \in Ax\} \end{aligned}$$

of A is a linear subspace of $X\times X$. 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 $X\times X$, we say that A is closed and we shall denote CR(X) the set of all closed linear relations in X. The linear relation, denoted by $A^{-1}$, and defined as

$$\begin{aligned} A^{-1} := \{(y,x): (x,y) \in A\} \end{aligned}$$

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 $A, B \in LR(X)$ and $\lambda \in \mathbb {K}$. Define the linear relations $\lambda A$, $A+B$ and AB by

$$\begin{aligned} \lambda A= & {} \{(x, \lambda y) : (x, y) \in A \},\\ A+B= & {} \{(x,y + z): (x,y) \in A~\text{ and }~ (x,z) \in B\},\;\text{ and }\\ AB:= & {} \{(x,z) \in X\times X: (x,y) \in B, (y,z) \in A\quad \text{ for } \text{ some }~ y \in X\}, \end{aligned}$$

while $A-\lambda$ stands for $A -\lambda I$, where I is the identity operator on X. Since the product of linear relations is clearly associative, then the sequence of natural powers $(A^n )_{n\in \mathbb {N}}$ is defined recursively by $A^0 = I$ and $A^{n+1} := A^nA$. We can easily see that $(A^{-1})^n = (A^n)^{-1}, n\in \mathbb {Z}.$ A simple induction argument shows the validity of the law of exponents $A^{n+m} = A^nA^m$, for all $n,m \in \mathbb {N}$. In particular, $A^{n+1} = AA^n$, implying that the sequence of domains $D(A^n)$ is decreasing, i.e., $D(A^{n+1}) \subset D(A^n)$. Hence, $D(A^{n+m})\subset D(A^n)$, for all nonnegative integers n, m.

The kernels and the ranges of the iterates $A^n, n \in \mathbb {N}$, form two increasing and decreasing chains, respectively, i.e., the chain of kernels

$$\begin{aligned} N(A^0)= \{0\}\subset N(A)\subset N(A^2)\subset .... \end{aligned}$$

and the chain of ranges

$$\begin{aligned} R(A^0)= X\supset R(A)\supset R(A^2)\supset .... \end{aligned}$$

The singular chain manifold of A is defined by

$$\begin{aligned} \mathcal {R}_c(A) = \left( \displaystyle \mathop {\bigcup }_{n=1}^\infty A^n(0)\right) \cap \left( \displaystyle \mathop {\bigcup }_{n=1}^\infty N(A^n)\right) . \end{aligned}$$

Lemma 2.1

[18, Lemma 3.4] Let X be a linear space and $A \in LR(X)$.

• (i) If $N(A^k)= N(A^{k+1})$ for some $k \in \mathbb {N}$, then $N(A^n)= N(A^k)$ for all $n\ge k$

• (ii) If $R(A^k)= R(A^{k+1})$ for some $k \in \mathbb {N}$, then $R(A^n)= R(A^k)$ for all $n\ge k$.

The statement of Lemma 2.1 leads to define the ascent of A by

$$\begin{aligned} a(A):= inf\{n \in \mathbb {N}: N(A^n) = N(A^{n+1})\}, \end{aligned}$$

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

$$\begin{aligned} d(A) := \inf \{n \in \mathbb {N}: R(A^n)= R(A^{n+1}) \}, \end{aligned}$$

the infimum over the empty set is taken to be infinite.

Definition 2.2

Let $A \in LR(X)$. We define the ascent index and the descent index of A, respectively, by

$$\begin{aligned} \alpha _n(A) := dim N(A^{n+1})/N(A^{n}), n \in \mathbb {N}, \end{aligned}$$

and

$$\begin{aligned} \beta _n(A) := dim R(A^n)/R(A^{n+1}), n \in \mathbb {N}. \end{aligned}$$

Lemma 2.3

Let $A \in LR(X)$. Then

• (i) $\alpha _n(A) = dim [R(A^n)\cap N(A)]/ [A^n(0)\cap N(A)]$ for all $n \in \mathbb {N}$,

• (ii) $\beta _n(A) = dim D(A^n)/[R(A)+ N(A^n)]\cap D(A^n)$ for all $n \in \mathbb {N}$,

• (iii) $\beta _{n+1}(A)\le \beta _n(A)$ for all $n \in \mathbb {N}$. Moreover, if $\beta _k(A)<\infty$ for some $k \in \mathbb {N}$, then $\beta _m(A)<\infty$ for all $m\ge k$.

• (iv) $\alpha _{n+1}(A)\le \alpha _n(A)$ for all $n \in \mathbb {N}$. Moreover, if $\alpha _k(A)<\infty$ for some $k \in \mathbb {N}$, then $\alpha _m(A)<\infty$ for all $m\ge k$.

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 $A^k(0) \subset R(A^n)$ for all $k,n \in \mathbb {N}$, then

$$\begin{aligned} \alpha _{n+1}(A) =&~ dim [R(A^{n+1})\cap N(A)]/ [A^n(0)\cap N(A)]\\ \le&~ dim [R(A^{n})\cap N(A)]/ [A^n(0)\cap N(A)]\;(\text{ as }~R(A^{n+1})\subset R(A^{n}))\\ \le&~dim [R(A^{n})\cap N(A)]/ [A^{n-1}(0)\cap N(A)] \;(\text{ as }~ A^{n-1}(0)\subset A^{n}(0))\\ =&~ \alpha _{n}(A). \end{aligned}$$

On the other hand, suppose that $\alpha _k(A)<\infty$ for some $k \in \mathbb {N}$, then $\alpha _m(A)\le \alpha _k(A) <\infty$ for all $m \ge k.$$\square$

From Lemma 2.3, one can deduce that $\alpha _n(A)$ and $\beta _n(A)$ are decreasing sequences. This leads to define the essential ascent of A by

$$\begin{aligned} a_e(A) := \inf \{n \in \mathbb {N}: \alpha _n(A)<\infty \}, \end{aligned}$$

where the infimum over the empty set is taken to be infinite, is finite. Likewise, we define the essential descent of A by

$$\begin{aligned} d_e(A) := \inf \{n \in \mathbb {N}: \beta _n(A)<\infty \}, \end{aligned}$$

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 $a_e(A)=0$, and every linear relation with finite descent has a finite essential descent, more precisely, $d_e(A)= 0$ whenever d(A) is finite.

These notions permit to define the following algebraic spectra

$$\begin{aligned} \sigma _{asc}(A):= & {} \{\lambda \in \mathbb {C}: a(A-\lambda )= \infty \},\quad \sigma _{asc}^e(A) := \{\lambda \in \mathbb {C}: a_e(A-\lambda )= \infty \},\\ \sigma _{dsc}(A):= & {} \{\lambda \in \mathbb {C}: d(A-\lambda )= \infty \},\quad \sigma _{dsc}^e(A) := \{\lambda \in \mathbb {C}: d_e(A-\lambda )= \infty \}. \end{aligned}$$

Combining the algebraic conditions defining the spectra $\sigma _i(A)$, $i \in \{asc, asc^e,dsc, dsc^e \}$ with a topological condition, we can consider the following topological spectra

$$\begin{aligned} \sigma _{ta}(A):= & {} \mathbb {C}\backslash \rho _{ta}(A) \quad \text{ where }\quad \rho _{ta}(A) := \{\lambda \in \mathbb {C}: a(A-\lambda )<\infty ,\\&R((A-\lambda )^{a(A-\lambda )+1}) \quad \text{ is } \text{ closed }~\},\\ \sigma _{ta}^e(A):= & {} \mathbb {C}\backslash \rho _{ta}^e(A) \quad \text{ where }\quad \rho _{ta}^e(A) := \{\lambda \in \mathbb {C}: a_e(A-\lambda )<\infty ,\\&R((T-\lambda )^{a_e(T-\lambda )+1}) \quad \text{ is } \text{ closed }~\},\\ \sigma _{td}(A):= & {} \mathbb {C}\backslash \rho _{td}(A) \quad \text{ where }\quad \rho _{td}(A) := \{\lambda \in \mathbb {C}: d(A-\lambda )<\infty ,\\&R((A-\lambda )^{d(A-\lambda )}) \quad \text{ is } \text{ closed }~\} \end{aligned}$$

and

$$\begin{aligned} \sigma _{td}^e(A):= & {} \mathbb {C}\backslash \rho _{td}^e(A) \quad \text{ where }\quad \rho _{td}^e(A) := \{\lambda \in \mathbb {C}: d_e(A-\lambda )<\infty , \\&R((A-\lambda )^{d_e(A-\lambda )}) \quad \text{ is } \text{ closed }~\}. \end{aligned}$$

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 $\sigma _i(p(A)) = p(\sigma _i(A))$ where $i \in \{asc,asc^e,dsc,dsc^e,ta,ta^e,td,td^e\}$. 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.

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 $\mathbb {C}$ of complex numbers. Let $p(z) = {\prod }_{i=1}^k(z-\lambda _i)^{m_i}$ be a polynomial and $\mathrm {p} = :{\sum }_{i=1}^km_i$ its degree, where k and $m_i, 1\le i \le k$, are positive integers and $\lambda _i, 1\le i\le k$, are distinct constants. Then the polynomial p in T given by

$$\begin{aligned} p(T) := \displaystyle \mathop {\prod }_{i=1}^k(T-\lambda _i)^{m_i} \end{aligned}$$

(3.1)

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 $k \in \mathbb {N}, \lambda _i \in \mathbb {C}, m_i \in \mathbb {N}, 1 \le i \le k.$ Assume that $\lambda _i, \quad 1 \le i\le k$ are distinct and let p(T) be as in (3.1). Then

(i):

$D(p(T))= D(T^{\mathrm {p}})$.

(ii):

$R(p(T))= \displaystyle \mathop {\bigcap }_{i=1}^kR(T-\lambda _i)^{m_i}$.

(iii):

$N(p(T))= \displaystyle \mathop {\sum }_{i=1}^kN(T-\lambda _i)^{m_i}$.

(iv):

$p(T)(0)= T^{\mathrm {p}}(0)$.

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 $T-\lambda _i$ is bijective, for all $1\le i \le k$.

Let S and T be linear relations in a linear space X. Assume that S and Tcommute, i.e., $ST = TS$ in the sense of the product of linear relations, then (see [19, (2.9)])

$$\begin{aligned} (S-\lambda )^n(T-\mu )^m = (T-\mu )^m(S-\lambda )^n \end{aligned}$$

holds true for all $n,m \in \mathbb {N}$ and $\lambda ,\mu \in \mathbb {C}$. 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

$$\begin{aligned} (pq)(T)= p(T)q(T) = q(T)p(T) \end{aligned}$$

for all polynomials p and q in T. Furthermore, the following results are straightforward;

$$\begin{aligned} p(T)^{-1} = \displaystyle \mathop {\prod }_{i=1}^k(T-\lambda _i)^{-m_i} \end{aligned}$$

and

$$\begin{aligned} p(T)^{-1} q(T)^{-1}= q(T)^{-1} p(T)^{-1}. \end{aligned}$$

Proposition 3.3

Let T be a linear relation in a linear space X and $p, q \in \mathbb {C}[z]$. Consider the relation    $r(T)= p(T)q(T)^{-1}$ and let $\mathrm {p} = deg(P)$ and $\mathrm {q} = deg(q)$. Then the following statements hold:

(i):

$D(r(T))= \left\{ \begin{array}{ll} D(T^{\mathrm {p}-\mathrm {q}})\cap R(q(T)) + T^\mathrm {q}(0) &{}\quad \mathrm{if} \mathrm {p} \ge \mathrm {q} \\ R(q(T)) &{}\quad \text{ else }. \end{array}\right.$ In particular, if T is everywhere defined, then $D(r(T))= R(q(T))$.

(ii):

$R(r(T)) =$$\left\{ \begin{array}{ll} D(T^{\mathrm {q}-\mathrm {p}})\cap R(p(T)) + T^\mathrm {p}(0) &{}\quad \mathrm{if} \mathrm {p} \le \mathrm {q} \\ R(p(T)) &{}\quad \text{ else }. \end{array}\right.$ In particular, if T is everywhere defined, then $R(r(T))= R(p(T))$.

Proof

(i):

First suppose that $\mathrm {p} \ge \mathrm {q}$. Then, by using [7, I.1.3 (2)] and Proposition 3.1 we obtain

$$\begin{aligned} D(r(T)) =&\, D(p(T)q(T)^{-1}) \\ =&\,q(T)(D(p(T)) \\ =&\,q(T)(D(T^\mathrm {p})) \\ =&\,q(T)(D(T^{\mathrm {p}-\mathrm {q}+\mathrm {q}}))\\ =&\,q(T)(T^{\mathrm {q}-\mathrm {p}}(D(T^q))) \\ =&\,q(T)(T^{\mathrm {q}-\mathrm {p}}(D(q(T)))) \\ =&\,q(T)(D(q(T)T^{\mathrm {p}-\mathrm {q}})) \\ =&\,q(T)(D(T^{\mathrm {p}-\mathrm {q}}q(T))) \\ =&\,q(T)q(T)^{-1}(D(T^{\mathrm {p}-\mathrm {q}})) \\ =&\,D(T^{\mathrm {p}-\mathrm {q}})\cap R(q(T)) + q(T)(0) \\ =&\,D(T^{\mathrm {p}-\mathrm {q}})\cap R(q(T)) + T^\mathrm {q}(0). \end{aligned}$$

Now suppose that $\mathrm {p}< \mathrm {q}$. In this case $D(q(T)) = D(T^\mathrm {q}) \subset D(T^\mathrm {p}) = D(p(T))$ which implies that

$$\begin{aligned} D(r(T)) = q(T) (D(p(T)) \supset q(T)( D(q(T)))= R(q(T)). \end{aligned}$$

The inclusion $q(T)(D(p(T)) \subset R(q(T))$ is trivial. From this one can deduce that

$$\begin{aligned} D(r(T)) = R(q(T)). \end{aligned}$$

(ii):

Follows immediately from part (i) and the fact that $R(r(T)) = D(q(T)p(T)^{-1})$. $\square$

Example 3.4

Let $X = l^p, 1\le p<\infty$ be the Banach space of all complex sequences $x = (x_0,x_1,...)$ such that ${\sum }_{n=0}^\infty |x_n|^p<\infty$. We define the operator A by

$$\begin{aligned} A:(x_0,x_1,...,x_n,...) \in X \mapsto (0,x_1,x_2,...,x_n,...) \in X\quad \text{ and }\quad T:= A^{-1}. \end{aligned}$$

Clearly A is a bounded operator in X, so T is a closed linear relation in X with $D(T)= R(A) = span\{e_1,e_2,....\}$, $N(T)= A(0) = \{0\}$, $R(T) = D(A)= X$ and $T(0)= N(A) = span\{e_0\}$, where $e_i = (\displaystyle \mathop {\underbrace{0,0,0,...}}_{i~ times} ,1,0,0,0...)$ for $i\ge 0$. Let us consider the polynomials $p(z) := (z-1)^3$ and $q(z)= z^2$ in $\mathbb {C}[z]$. Since q(T) is clearly bijective (Lemma 3.2) and $T^2(0) = T(T(0)) = T(T(0)\cap D(T))= T(0)$, then

$$\begin{aligned} D(r(T)) = D(T) + T^2(0) = D(T) + T(0) = span\{e_1,e_2,....\} + span\{e_0\} = X \end{aligned}$$

and

$$\begin{aligned} R(r(T)) = R(p(T)) = R(T-I)^3 = R(T-I) = span\{e_0\}. \end{aligned}$$

Remark 3.5

(i):

In Example 3.4 and according to Proposition 3.1, we have

$$\begin{aligned} r(T)(0) = p(T)q(T)^{-1}(0) = p(T)(0) = T^3(0) = T(0) = span\{e_0\}. \end{aligned}$$

Define the linear relation $r_1(T):= q(T)^{-1}p(T)$. Then

$$\begin{aligned} r_1(T)(0)= & {} q(T)^{-1}p(T)(0)= q(T)^{-1}T(0) \\= & {} T^{-2}T(0) = T(0)\cap D(T) + T^{-1}(0) = T^{-1}(0) = \{0\}. \end{aligned}$$

This implies that $r_1(T)$ is an operator, so that r(T) and $r_1(T)$ do not define the same linear relation. Furthermore,

$$\begin{aligned} D(r_1(T)) = p(T)^{-1}(R(q(T))) = p(T)^{-1}(X) = D(p(T)) \subset D(r(T))= X. \end{aligned}$$

(ii):

If T is single valued and q(T) is bijective (that is $q^{-1}(0)\cap \sigma (T)= \emptyset$), 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 ?

Example 3.6

Let $X = l^p, 1<p<\infty$ be the Banach space of all complex sequences $x = (x_0,x_1,...)$ such that ${\sum }_{n=0}^\infty |x_n|^p<\infty$. We define the left shift operator L in X by

$$\begin{aligned} L:(x_0,x_1,...) \in X \mapsto (x_1,x_2,...) \in X. \end{aligned}$$

Let $T := L^{-1} \in LR(X)$ and the polynomials $p(z) = z-1$, $q(z) = z$, $p_1(z)= (z-1)z$ and $q_1(z)= z^2$. Then

$$\begin{aligned} p(T)q(T)^{-1}(0) =&~ (T-I)T^{-1}(0) \\ =&~ (T-I)L(0) \\ =&~ T(0) \\ =&~ span(e_0)\quad \text{ where }~ e_0 = (1,0,0,...). \end{aligned}$$

and

$$\begin{aligned} p_1(T)q_1(T)(0)=&~ (T-I)TT^{-2}(0) \\ =&~ (T-I)T(0) \\ =&~ span(e_0,e_1,...) \quad \text{ where }~ e_1 = (0,1,0,...). \end{aligned}$$

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

$$\begin{aligned} \mathcal {R}(z) := \{ p/q: p, q \in \mathbb {C}[z]~\text{ and }~ p/q~{is~irreducible}\} \end{aligned}$$

the set of all irreducible rational functions.

Let $T \in LR(X)$. In the sequel of this paper we define a rational function in T by

$$\begin{aligned} r(T) := p(T)q(T)^{-1},\quad \text{ for }~~ p/q \in \mathcal {R}(z). \end{aligned}$$

(3.2)

Lemma 3.8

Let X be a linear space, $T \in LR(X)$ and $\lambda \in \mathbb {C}\backslash \{0\}$. Then

$$\begin{aligned} N(T-\lambda )^k \subset T^n(N(T-\lambda )^k), \quad \text{ for } \text{ all }~~k, n \in \mathbb {N}. \end{aligned}$$

(3.3)

Proof

If either $k = 0$ or $n = 0$, the desired inclusion is trivial. Assume that $n \ge 1$ and $k \ge 1.$ The proof will be given by induction on $n \in \mathbb {N}$.

For $n=1$ let be consider two cases:

Case 1: $k=1$. Let $x_0 \in N(T-\lambda )$. Then $\lambda x_0 \in Tx_0 \subset T(N(T-\lambda ))$ which implies that $x_0 \in T(N(T-\lambda ))$ (as $\lambda \ne 0$).

Case 2: $k\ge 2$. Let $x_0 \in N(T-\lambda )^k$. Then there exist elements $x_1,x_2,...,x_{k-1}$ such that

$$\begin{aligned} (x_0,x_1),(x_1,x_2),...,(x_{k-1},0) \in T-\lambda . \end{aligned}$$

Define $x_k :=0$. It follows that

$$\begin{aligned} (x_i,x_{i+1}) \in T-\lambda , \quad \text{ for }~~ 0\le i \le k, \end{aligned}$$

and hence

$$\begin{aligned} \displaystyle \mathop \sum _{i=0}^{k-1}(-\lambda )^{k-i-1}(x_i,x_{i+1}+\lambda x_i) = \left( \displaystyle \mathop \sum _{i=0}^{k-1}(-\lambda )^{k-i-1}x_i, -(-\lambda )^kx_0\right) \in T. \end{aligned}$$

This shows that

$$\begin{aligned} x_0 \in T\left( \displaystyle \mathop \sum _{i=0}^{k-1}(-\lambda )^{-i-2}x_i\right) . \end{aligned}$$

On the other hand, since $x_0 \in N(T-\lambda )^k$, then $x_i \in N(T-\lambda )^{k-i}$, for  $0\le i \le k-1$. It follows that

$$\begin{aligned} x_0 \in T\left( \displaystyle \mathop \sum _{i=0}^{k-1}(-\lambda )^{-i-2}x_i\right) \subset T\left( \displaystyle \mathop \sum _{i=0}^{k-1}N(T-\lambda )^{k-i}\right) = T(N(T-\lambda )^k). \end{aligned}$$

Now assume that (3.3) is satisfied for some $n\ge 1$ and for all $k \in \mathbb {N}$. Then

$$\begin{aligned} N(T-\lambda )^k \subset T^n\left( N(T-\lambda )^k\right) \subset T^n\left[ T(N(T-\lambda )^k)\right] = T^{n+1}(N(T-\lambda )^k). \end{aligned}$$

This completes the proof. $\square$

Lemma 3.9

Let T be a linear relation in a linear space X and let $\alpha , \beta \in \mathbb {C}$ such that $\alpha \ne \beta$. Then

$$\begin{aligned} (T-\alpha )^s (T-\beta )^{-k}(0) = (T-\alpha )^s(0) + (T-\beta )^{-k}(0). \end{aligned}$$

(3.4)

for all $s, k \in \mathbb {N}$.

Proof

Clearly $(T-\alpha )^s(0)\subset (T-\alpha )^s (T-\beta )^{-k}(0)$ and by using (3.3) we have

$$\begin{aligned} (T-\beta )^{-k}(0) \subset (T-\alpha )^s (T-\beta )^{-k}(0). \end{aligned}$$

This means that

$$\begin{aligned} (T-\alpha )^s(0) + (T-\beta )^{-k}(0) \subset (T-\alpha )^s (T-\beta )^{-k}(0). \end{aligned}$$

(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 $s \ge 1$ and $k \ge 1.$

For $s=1$. Let $k\ge 1$, then

$$\begin{aligned} (T-\alpha )(T-\beta )^{-k}(0) =&~ (T-\beta +\beta -\alpha )(T-\beta )^{-k}(0) \\ \subset&~ (T-\beta )(T-\beta )^{-k}(0) + (T-\beta )^{-k}(0) \ \ \ (\text{ by }\quad [7, \hbox {I.}4.2]) \\ =&~ (T-\beta )(0) + (T-\beta )^{-k+1}(0)\cap R(T-\beta ) \\&+ (T-\beta )^{-k}(0)~ (\text{ by }\quad [7, \hbox {I.}3.1\hbox { (d)}]) \\ \subset&~(T-\beta )(0) + (T-\beta )^{-k+1}(0)+ (T-\beta )^{-k}(0) \\ =&~(T-\alpha )(0) + (T-\beta )^{-k}(0) \\&\quad (\text{ as }~(T-\beta )(0)= T(0)= (T-\alpha )(0)). \end{aligned}$$

Assume that $(T-\alpha )^s (T-\beta )^{-k}(0) \subset (T-\alpha )^s(0) + (T-\beta )^{-k}(0)$ for some $s\ge 1$ and all $k \in \mathbb {N}$. Then

$$\begin{aligned}&(T-\alpha )^{s+1}(T-\beta )^{-k}(0) = (T-\alpha )(T-\alpha )^{s}(T-\beta )^{-k}(0) \\&\quad \subset (T-\alpha )[(T-\alpha )^s(0) + (T-\beta )^{-k}(0)] \\&\quad = (T-\alpha )^{s+1}(0) + (T-\alpha )(T-\beta )^{-k}(0)\quad (\text{ by }~[7, \hbox {I.}3.1\hbox {(c)}]) \\&\quad \subset (T-\alpha )^{s+1}(0) + (T-\alpha )(0) + (T-\beta )^{-k}(0)\quad (\text{ by } \text{ hypothesis }) \\&\quad = (T-\alpha )^{s+1}(0) + (T-\beta )^{-k}(0). \end{aligned}$$

Hence

$$\begin{aligned} (T-\alpha )^{s}(T-\beta )^{-k}(0) \subset (T-\alpha )^{s}(0) + (T-\beta )^{-k}(0) \quad \text{ for } \text{ all } s, k \in \mathbb {N}. \end{aligned}$$

(3.6)

Now, a combination of (3.5) and (3.6) leads to deduce that (3.4) holds for all $s, k \in \mathbb {N}$. $\square$

Proposition 3.10

Let X be a linear space, $T\in LR(X)$ and r(T) be defined as in (3.2). Then

(i):

$r(T)(0) = p(T)(0) + N(q(T)).$

(ii):

$N(r(T)) = N(p(T)) + q(T)(0).$

Proof

Let $p(T):= {\prod }_{i=1}^\mathrm {p}(T-\alpha _i)^{s_i}$ and $q(T):= {\prod }_{j=1}^\mathrm {q}(T-\beta _j)^{k_j}$, where $\mathrm {p}$ and $\mathrm {q}$ are positive integers, $s_i \in \mathbb {N}, 1\le i\le \mathrm {p}$, $k_j \in \mathbb {N}, 1\le j\le \mathrm {q}$, and $\alpha _i, 1\le i\le \mathrm {p}$, $\beta _j, 1\le j\le \mathrm {q}$ are distinct constants. Put $s := \sum _{i=1}^\mathrm {p}s_i$. Then

$$\begin{aligned}&(T-\beta _j)^{-k_j}(0)\subset (T-\beta _j)^{-k_j-s}(0) \subset D(T-\beta _j)^{k_j+s} \\&\quad = D(T^{k_j+s}) \subset D(T^{s})= D(p(T)). \end{aligned}$$

According to Proposition 3.1 and [7, I.3.1 (c)], it follows that

$$\begin{aligned} r(T)(0)= & {} p(T)q(T)^{-1}(0) = p(T)\left[ \displaystyle \mathop {\sum }_{j=1}^\mathrm {q} (T-\beta _j)^{-k_j}(0)\right] \\= & {} \displaystyle \mathop {\sum }_{j=1}^\mathrm {q} p(T)(T-\beta _j)^{-k_j}(0). \end{aligned}$$

By combining (3.4), [7, I.3.1 (c)] and the fact that $(T-\beta _j)^{-k_j}(0) \subset D(T-\alpha _i)^{s_i}$ for all $1\le i\le \mathrm {p}$ and $1\le j\le \mathrm {q}$, one can deduce that

$$\begin{aligned} p(T)(T-\beta _j)^{-k_j}(0)= & {} \displaystyle \mathop {\prod }_{i=1}^\mathrm {p}(T-\alpha _i)^{s_i}(0) + (T-\beta _j)^{-k_j}(0)\\= & {} p(T)(0)+(T-\beta _j)^{-k_j}(0) \quad \text{ for }~ 1\le j \le q. \end{aligned}$$

Hence

$$\begin{aligned} r(T)(0)= & {} \displaystyle \mathop {\sum }_{j=1}^\mathrm {q} p(T)(T-\beta _j)^{-k_j}(0) = \displaystyle \mathop {\sum }_{j=1}^\mathrm {q} \left[ p(T)(0) +(T-\beta _j)^{-k_j}(0)\right] \\= & {} p(T)(0) + N(q(T)). \end{aligned}$$

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. $\square$

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 $\lambda \in \mathbb {C}$

$$\begin{aligned} r(T)-\lambda \subset (p(T)-\lambda q(T))q(T)^{-1} \end{aligned}$$

with equality if $deg(p) > deg(q)$ or if T is an operator.

Proof

The result is trivial for $\lambda =0$, so we can suppose that $\lambda \ne 0$. By $\mathrm {p}$ and $\mathrm {q}$ we denote the degrees of p and q, respectively. If $\mathrm {p}\le \mathrm {q}$ then $deg(p-\lambda q) \le max(\mathrm {p},\mathrm {q})= \mathrm {q}$. By using Proposition 3.3 we obtain

$$\begin{aligned} D(p(T)-\lambda q(T))q(T)^{-1} = R(q(T)) = D(r(T)) = D(r(T)-\lambda ). \end{aligned}$$

Let $x \in D(r(T))\subset R(q(T))$. It follows that

$$\begin{aligned} (p(T)-\lambda q(T))q(T)^{-1}x =&~ p(T)q(T)^{-1}x - \lambda q(T)q(T)^{-1}x \ \ \ ([4, \hbox {Lemma }4.1]) \\ =&~ p(T))q(T)^{-1}x - \lambda x + q(T)(0) \ \ \ ([7, \hbox {I.}3.1 \hbox { (d)}]) \\ =&~ r(T)x - \lambda x + q(T)(0) \\ \supset&~r(T)x - \lambda x, \end{aligned}$$

which means that $r(T) - \lambda \subset (p(T) - \lambda q(T))q(T)^{-1}$.

Now suppose that $\mathrm {p} > \mathrm {q}$ or T is an operator then $deg(p-\lambda q) = \mathrm {p}$ and hence the use of Proposition 3.3 gives

$$\begin{aligned} D(p(T)-\lambda q(T))q(T)^{-1}= & {} D(T^{\mathrm {p}- \mathrm {q}})\cap R(q(T)) \\&+ T^\mathrm {q}(0)= D(r(T))= D(r(T)-\lambda ). \end{aligned}$$

Moreover $q(T)(0) \subset r(T)(0)$, so that

$$\begin{aligned} r(T)x - \lambda x =&~ r(T)x + q(T)(0) - \lambda x \\ =&~ r(T)x -\lambda q(T)q(T)^{-1}x \\ =&~ (p(T)-\lambda q(T))q(T)^{-1}x \end{aligned}$$

for all $x \in D(r(T)-\lambda ) = D(p(T)-\lambda q(T))q(T)^{-1})$. Consequently $r(T) - \lambda = (p(T)-\lambda q(T))q(T)^{-1}$. $\square$

Remark 3.12

Let X be a linear space and $T \in LR(X)$ be injective (not single valued). Let $p(z) =1, q(z) = z$ be polynomials and $r(T) = p(T)q(T)^{-1}$. Then $(r(T) - \lambda )(0) = r(T)(0) = T^{-1}(0) = 0 \varsubsetneq (p(T)- \lambda q(T))q(T)^{-1}(0) = r(T)(0) + q(T)(0) = T^{-1}(0)+ T(0) = T(0)$. This shows that the inclusion in Proposition 3.11 is, in general, strictly whenever $deg(p) \le deg(q)$.

Definition 3.13

We define $\mathbb {C}_{\infty } := \mathbb {C}\cup \{\infty \}$ to be the one-point compactification of $\mathbb {C}$ (Riemann sphere) with the usual conventions for computation. Let $\lambda \in \mathbb {C}_{\infty }$ and T be a linear relation in a linear space X. The eigenspace of T at $\lambda$ is defined as

$$\begin{aligned} N(T,\lambda ) := \left\{ \begin{array}{ll} N(T-\lambda ) &{}\quad \text{ if }\, \lambda \in \mathbb {C}\\ T(0) &{}\quad \text{ if }\, \lambda = \infty \end{array}\right. \end{aligned}$$

Similarly, we define the range space of T at $\lambda$ as

$$\begin{aligned} R(T,\lambda ) := \left\{ \begin{array}{ll} R(T-\lambda ) &{} \text{ if } \lambda \in \mathbb {C}\\ D(T) &{}\quad \text{ if } \lambda = \infty \end{array}\right. \end{aligned}$$

Using this notation we define the extended point spectrum, the surjectivity spectrum and the extended spectrum of T by

$$\begin{aligned} \widetilde{\sigma }_p(T):= & {} \{\lambda \in \mathbb {C}_{\infty }: N(T,\lambda )\ne 0\},\quad \widetilde{\sigma }_s(T)\\:= & {} \{\lambda \in \mathbb {C}_{\infty }: R(T,\lambda )\ne X\}~\text{ and }~ \widetilde{\sigma }(T):= \widetilde{\sigma }_p(T)\cup \widetilde{\sigma }_s(T). \end{aligned}$$

Lemma 3.14

Let X be a linear space and $T\in LR(X)$ be bijective and let $\lambda \ne 0$. Then

(i):

$N(T-\lambda )^n = N(T^{-1}-\lambda ^{-1})^n$, for all $n\in \mathbb {N}$,

(ii):

$R(T-\lambda )^n = R(T^{-1}-\lambda ^{-1})^n$, for all $n\in \mathbb {N}$.

Proof

First observe that

$$\begin{aligned} T-\lambda = -\lambda (T^{-1}-\lambda ^{-1})T = -\lambda T(T^{-1}-\lambda ^{-1}). \end{aligned}$$

(3.7)

The equality $T-\lambda = -\lambda (T^{-1}-\lambda ^{-1})T$ follows immediately from [7, VI.4.2]. On the other hand

$$\begin{aligned} T^{-1}T \subset TT^{-1}. \end{aligned}$$

(3.8)

Indeed, let $(x,y)\in T^{-1}T$ then $y \in T^{-1}Tx = x+ T^{-1}(0) = x$ (as T is injective). Now, let $z \in T^{-1}x \ne \emptyset$ (as T is surjective). Then $x \in Tz \subset TT^{-1}x$ which implies that $(x,y) \in TT^{-1}$. Hence (3.8) holds. It follows that

$$\begin{aligned} (T^{-1}-\lambda ^{-1})T =&~ T^{-1}T -\lambda ^{-1}T \ \ \ [4, \hbox {Lemma }4.1]\end{aligned}$$

(3.9)

$$\begin{aligned} \subset&~ TT^{-1} -\lambda ^{-1}T \ \ \ \text{(by } ~3.8)\end{aligned}$$

(3.10)

$$\begin{aligned} \subset&~ T(T^{-1}-\lambda ^{-1}) \ \ \ [7, \hbox {I.}4.2\hbox {(e)}]. \end{aligned}$$

(3.11)

Now, we shall prove that

$$\begin{aligned} (T^{-1}-\lambda ^{-1})T(0)= T(T^{-1}-\lambda ^{-1})(0) \end{aligned}$$

(3.12)

and

$$\begin{aligned} D[(T^{-1}-\lambda ^{-1})T] = D[T(T^{-1}-\lambda ^{-1})]. \end{aligned}$$

(3.13)

By using [4, Lemma 4.1] and the fact that T is injective, it follows that

$$\begin{aligned} (T^{-1}-\lambda ^{-1})T(0)= & {} T^{-1}T(0) -\lambda ^{-1}T(0)= T^{-1}(0) \\&+T(0)= T(0) = T(T^{-1}-\lambda ^{-1})(0) . \end{aligned}$$

On the other hand, since $T-\lambda \subset T(T^{-1}-\lambda ^{-1})$ and $T-\lambda = -\lambda (T^{-1}-\lambda ^{-1})T$, then

$$\begin{aligned} D[(T^{-1}-\lambda ^{-1})T]= D(T -\lambda ) \subset D[T(T^{-1}-\lambda ^{-1})]. \end{aligned}$$

Conversely, let $x \in D[T(T^{-1}-\lambda ^{-1})]$ then

$$\begin{aligned} \emptyset \ne T(T^{-1}-\lambda ^{-1})x =&~ T(T^{-1}x-\lambda ^{-1}x) \\ =&~ TT^{-1}x-\lambda ^{-1}Tx \ \ \ ([7, \hbox {I.}3.1\hbox {(c)}])\\ =&~ x + T(0) -\lambda ^{-1}Tx ~~ \ \ \ ([7, \hbox {I.}3.1\hbox {(c)}] \text{ and }~ T~ \text{ is } \text{ surjective })\\ =&~ x -\lambda ^{-1}Tx. \end{aligned}$$

It follows that $Tx \ne \emptyset$, so that $x \in D(T)$. Thus $D[T(T^{-1}-\lambda ^{-1})] \subset D(T)= D[(T^{-1}-\lambda ^{-1})T]$ 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 $n \in \mathbb {N}$.

• (i) For $n=0$ the result is evident and for $n=1$ the result follows immediately from [7, VI.2.3]. Now suppose $N(T-\lambda )^n = N(T^{-1}-\lambda ^{-1})^n$ for some $n \ge 1$. Then

  $$\begin{aligned} N(T-\lambda )^{n+1} =&~ (T-\lambda )^{-1}(N(T-\lambda )^{n}) \\ =&~ (T-\lambda )^{-1}(N(T-\lambda )^{n}) \\ =&~ (-\lambda )^{-1}[T(T^{-1}-\lambda ^{-1})]^{-1}(N(T-\lambda )^{n})~~ \ \ (\text{ by }~3.7)\\ =&~ (-\lambda )^{-1}(T^{-1}-\lambda ^{-1})^{-1}T^{-1}(T-\lambda )^{-n}(0) \\ =&~ (T^{-1}-\lambda ^{-1})^{-1}N((T-\lambda )T)\\ =&~ (T^{-1}-\lambda ^{-1})^{-1}N(T(T-\lambda )^n)~~ \ \ \ \ \ \ \ \ \ \ \ \ ([19, (2.9)])\\ =&~ (T^{-1}-\lambda ^{-1})^{-1}(T-\lambda )^{-n}T^{-1}(0)\\ =&~ (T^{-1}-\lambda ^{-1})^{-1}(T-\lambda )^{-n}(0)~~ \ \ \ \ \ \ \ \ \ \ \ \ \ (\text{ as } T~\text{ is } \text{ injective })\\ =&~ (T^{-1}-\lambda ^{-1})^{-1}N(T^{-1}-\lambda ^{-1})^n \\ =&~ N(T^{-1}-\lambda ^{-1})^{n+1}. \end{aligned}$$

• (ii) For $n=0$ the result is trivial. For $n=1$, by using (3.7) and the fact that T is surjective, we obtain $R(T-\lambda ) = R[(T^{-1}-\lambda ^{-1})T] = (T^{-1}-\lambda ^{-1})(R(T)) = (T^{-1}-\lambda ^{-1})(X) = R(T^{-1}-\lambda ^{-1})$.

  Now, suppose that $R(T-\lambda )^n = R(T^{-1}-\lambda ^{-1})^n$ for some $n \in \mathbb {N}$. Then

  $$\begin{aligned} R(T-\lambda )^{n+1} =&~ (T-\lambda )R(T-\lambda )^n \\ =&~ (-\lambda )(T^{-1}-\lambda ^{-1})TR(T-\lambda )^n~~ \ \ \ (\text{ by }~3.7)\\ =&~ (T^{-1}-\lambda ^{-1})R(T(T-\lambda )^n))~~\\ =&~ (T^{-1}-\lambda ^{-1})R((T-\lambda )^n)T)~~ \ \ \ \ \ \ ([19, (2.9)])\\ =&~ (T^{-1}-\lambda ^{-1})R((T-\lambda )^n))~~ \ \ \ \ \ \ \ \ (\text{ as } T~ \text{ is } \text{ surjective })\\ =&~ (T^{-1}-\lambda ^{-1})R(T^{-1}-\lambda ^{-1})^n \\ =&~ R(T^{-1}-\lambda ^{-1})^{n+1}. \end{aligned}$$

$\square$

Remark 3.15

From Definition 3.13 we have $N(T,0) = N(T) = T^{-1}(0) = N(T^{-1},\infty )$ and $R(T,0)= R(T)= D(T^{-1})= R(T^{-1},\infty )$. Then Lemma 3.14 holds for all $\lambda \in \mathbb {C}_\infty$, that is

$$\begin{aligned} N(T,\lambda ) = N(T^{-1},\lambda ^{-1})~~\text{ and }~~ R(T,\lambda ) = R(T^{-1},\lambda ^{-1})~~ \text{ for } \text{ all }~ \lambda \in \mathbb {C}_\infty . \end{aligned}$$

(3.14)

Proposition 3.16

Let X be a linear space and $T \in LR(X)$ be bijective. Then

$$\begin{aligned} \widetilde{\sigma }(T^{-1}) = [\widetilde{\sigma }(T)]^{-1}. \end{aligned}$$

Proof

Let $\lambda \in \widetilde{\sigma }(T^{-1})$. Then $N(T^{-1}-\lambda )\ne \{0\}$ or $R(T^{-1}-\lambda )\ne X$. By using (3.14) it follows that $N(T - \lambda ^{-1})\ne \{0\}$ or $R(T - \lambda ^{-1})\ne X$ so that $\lambda ^{-1} \in \widetilde{\sigma }(T)$. Consequently $\lambda \in [\widetilde{\sigma }(T)]^{-1}$. Conversely, let $\lambda \in [\widetilde{\sigma }(T)]^{-1}$, then $\lambda ^{-1} \in \widetilde{\sigma }(T)$. This means that $N(T - \lambda ^{-1})\ne \{0\}$ or $R(T - \lambda ^{-1})\ne X$. Again from (3.14) one can deduce that $N(T^{-1}-\lambda )\ne \{0\}$ or $R(T^{-1}-\lambda )\ne X$. Thus $\lambda \in \widetilde{\sigma }(T^{-1})$. $\square$

Corollary 3.17

Let T be a linear relation in a linear space X and let p be a polynomial in T. Suppose that $p^{-1}(0)\cap \sigma (T)= \emptyset$. Then

$$\begin{aligned} \widetilde{\sigma }(p(T)^{-1})= [p(\widetilde{\sigma }(T))]^{-1}. \end{aligned}$$

The next result is the rational form of the spectral mapping property.

Theorem 3.18

Let X be a linear space, $T \in LR(X)$ be everywhere defined and r(T) defined as in (3.2). Then

(i):

$\widetilde{\sigma _s}(r(T)) \supset r(\widetilde{\sigma _s}(T)),$ with equality if $deg(p)>deg(q)$,

(ii):

$\widetilde{\sigma }_p(r(T)) \supset r(\widetilde{\sigma }_p(T)),$ with equality if T is an operator.

Proof

(i):

Let $\infty \ne \lambda \in r(\widetilde{\sigma }_s(T))$ and put $h(T) := p(T)-\lambda q(T):= {\prod }_{i=1}^k(T-\lambda _i)^{m_i}$. Then $\lambda = r(\alpha )$ for some $\alpha \in \widetilde{\sigma }_s(T)$, so that $\alpha = \lambda _j$ for some $1\le j \le k$, say $j=1$. Since $T-\lambda _1$ is not surjective, then so is h(T) (Proposition 3.1 (ii)). It follows that $X \ne R(r(T)-\lambda )) = R(h(T))$ and consequently $\lambda \in \widetilde{\sigma }_s(r(T))$.

If $\lambda = \infty \in r(\widetilde{\sigma }_s(T))$, then $\lambda = r(\alpha )$, for some $\alpha \in \widetilde{\sigma }_s(T)$, so that $q(\alpha )=0$. It follows that q(T) is not surjective, so $D(r(T)) = R(q(T)) \ne X$ (Propositions 3.3 (i)). This means that $\infty \in \widetilde{\sigma }_s(r(T))$. Consequently $r(\widetilde{\sigma _s}(T))\subset \widetilde{\sigma _s}(r(T))$.

Now, suppose moreover that $deg(p)>deg(q)$ and let $\infty \ne \lambda \in \widetilde{\sigma }_s(r(T))$. Put $h(T) := p(T) - \lambda q(T) := {\prod }_{i=1}^k(T-\lambda _i)^{m_i}$. Since p and q are irreducible then so are $h = p -\lambda q$ and q. Since $deg(h) = deg(p)>deg(q)$, then $r(T) - \lambda = h(T)q(T)^{-1}$ (Proposition 3.11) and $R(r(T)-\lambda ) = R(h(T))$ (Proposition 3.10). It follows that $R(T-\lambda _j)\ne X$, for some $1\le j \le k$ (since $r(T)-\lambda$ is not surjective), say $j=1$. This means that $\lambda _1 \in \widetilde{\sigma }_s(T)$. But $\lambda = r(\lambda _1)$, which implies that $\lambda \in r(\widetilde{\sigma }_s(T))$.

If $\lambda = \infty \in \widetilde{\sigma }_s(r(T))$, then $R(q(T))= D(r(T)) \ne X$. This implies that q(T) is not surjective. If follows that $T-\alpha$ is also not surjective for some $\alpha$ such that $q(\alpha )=0$ (Proposition 3.1 (ii)). Therefore $\infty = r(\alpha ) \in r(\widetilde{\sigma }_s(T))$. Consequently, $\widetilde{\sigma _s}(r(T)) \subset r(\widetilde{\sigma _s}(T)).$

(ii):

Let $\lambda \in \mathbb {C}_\infty$. If $\lambda \ne \infty$, then $N(r(T),\lambda )= N(p(T)-\lambda q(T)) + q(T)(0)$. Moreover $N(r(T)),\infty ) = r(T)(0) = p(T)(0) + q(T)(0) + N(q(T))$. Suppose that $T(0)\ne 0$, then clearly $N(r(T),\lambda )\ne \{0\}$ for all $\lambda \in \mathbb {C}_\infty$. It follows that $\widetilde{\sigma }_p(r(T)) = \mathbb {C}_\infty \supset r(\widetilde{\sigma }_p(T))$.

Now, suppose that T is an operator, that is $T(0) =0$ and let $\infty \ne \lambda \in \widetilde{\sigma }_p(r(T))$. Then $r(T) -\lambda = (p(T)-\lambda q(T))q(T)^{-1}$ (Proposition 3.11) and $\{0\}\ne N(r(T)-\lambda )) = N(p(T)-\lambda q(T))$ (Proposition 3.10). Put $p(T)-\lambda q(T) = {\prod }_{i=1}^k(T-\lambda _i)^{m_i}$. The use of Proposition 3.1 ensures that $N(T-\lambda _i)\ne \{0\}$ for some $1\le i\le k$, say $i=1$. It follows that $\lambda _1 \in \widetilde{\sigma }_p(T)$. Since $r(\lambda _1)= 0$, then $\lambda \in r(\widetilde{\sigma }_p(T)$.

On the other hand suppose that $\infty \in \widetilde{\sigma }_p(r(T))$, then $r(T)(0)\ne \{0\}$, that is $N(q(T))\ne \{0\}$ (Proposition 3.10). Put $q(T)= {\prod }_{i=1}^l(T-\beta _i)^{n_i}$. Then $T-\beta _i$ is not injective for some $1\le i \le l$, say $i=1$. It follows that $\beta _1 \in \widetilde{\sigma }_p(T)$. Since $q(\beta _1)= 0$ and $p/q \in \mathcal {R}[z]$, then $r(\beta _1)= \infty$. Consequently $\infty \in r(\widetilde{\sigma }_p(T))$.

Conversely, let $\lambda \in r(\widetilde{\sigma }_p(T))$. Then $\lambda = r(\beta )$, for some $\beta \in \widetilde{\sigma }_p(T)$, that is $N(T,\beta )\ne \{0\}$. Since $T(0)= 0$, then $\beta \ne \infty$. Suppose that $\lambda \ne \infty$ and put $r(T)-\lambda :={\prod }_{i=1}^s(T-\mu _i)^{r_i}$. From Proposition 3.1, one can deduce that $T-\mu _i$ is not injective, for some $1\le i \le s$, say $i=1$. It follows that $\mu _1 \in \widetilde{\sigma }_p(T)$ and since $\lambda = r(\mu _1)$, then $\lambda \in r(\widetilde{\sigma }_p(T))$.

Now, suppose that $\lambda =\infty$, so necessarily $q(\beta )= 0$. Let $q(T) := {\prod }_{i=1}^l(T-\beta _i)^{n_i}$. Then $\beta = \beta _i$ for some $1\le i \le l$, say $i=1$. Since $\beta _1 \in \widetilde{\sigma }_p(T)$, then $T-\beta _1$ is not injective and so is q(T) (Proposition 3.1). It follows that $r(T)(0) \ne 0$ (Proposition 3.10). Finally, since $r(\beta _1)= \infty$ (as $q(\beta _1) = 0)$, then $\infty \in r(\widetilde{\sigma }_p(T))$. $\square$

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

(i):

$\dfrac{M+N}{N}\simeq \dfrac{M}{M\cap N}$,

(ii):

if $M\subset W$, then $(M+N)\cap W = M+N\cap W.$

Lemma 4.2

[19, Lemma 6.1] Let T be a linear relation in a linear space X and let $n, m \in \mathbb {N}$. Then if $\rho (T)\ne \emptyset$, then $N(T^n)\cap T^m(0) = \{0\}$, in particular $\mathcal {R}_c(T)=\{0\}$.

The following proposition describes the behavior of the kernels and ranges of the iterates $(r(T)^n), n \in \mathbb {N}$ 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 $n \in \mathbb {N}$.

(i):

$r(T)^n(0) = \left\{ \begin{array}{lll} T^\mathrm {p}(0) +N(q(T)^n) &{}\quad ~\text{ if }~~ &{} \mathrm {p}<\mathrm {q} \\ T^{np-(n-1)q}(0) +N(q(T)^n) &{}\quad ~\text{ if }~~ &{} p\ge q \end{array}\right.$

(ii):

$N(r(T)^n) = \left\{ \begin{array}{lll} T^\mathrm {q}(0) +N(p(T)^n) &{}\quad ~\text{ if }~~ &{} \mathrm {p}\ge \mathrm {q} \\ T^{n\mathrm {q}-(n-1)\mathrm {p}}(0) +N(p(T)^n) &{}\quad ~\text{ if }~~ &{} \mathrm {p}< \mathrm {q} \end{array}\right.$

(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 $n\ge 1$ and suppose that (i) is satisfied for n. If $\mathrm {p}<\mathrm {q}$ then q(T) can be written as $q(T) = h(T)k(T)$ where h(z) and k(z) are two complex polynomials with degrees $\mathrm {q}-\mathrm {p}$ and $\mathrm {p}$, respectively (obviously h(T) and k(T) commute). If follows that

$$\begin{aligned} r(T)^{n+1}(0) =&~ p(T)q(T)^{-1}[T^\mathrm {p}(0) +N(q(T)^n)] \\ =&~~ p(T)q(T)^{-1}T^\mathrm {p}(0) + p(T)q(T)^{-1}N(q(T)^n) \\ =&~~ p(T)h(T)^{-1}k(T)^{-1}k(T)(0) + p(T)q(T)^{-n-1}(0) \\ =&~~ p(T)h(T)^{-1}k(T)^{-1}(0) + p(T)q(T)^{-n-1}(0) \\ =&~~ p(T)q(T)^{-1}(0) + p(T)q(T)^{-n-1}(0) \\ =&~~ p(T)(0) + q(T)^{-1}(0) + p(T)(0) + q(T)^{-n-1}(0)\\ =&~~ T^\mathrm {p}(0) + q(T)^{-n-1}(0). \end{aligned}$$

Now assume that $\mathrm {p}\ge \mathrm {q}$ and $r(T)^n(0)= T^{n\mathrm {p}-(n-1)\mathrm {q}}(0) +N(q(T)^n)$ for some $n\ge 1$. Then

$$\begin{aligned} r(T)^{n+1}(0) =&~ p(T)q(T)^{-1}[T^{n\mathrm {p}-(n-1)\mathrm {q}}(0) +N(q(T)^n)] \\ =&~~ p(T)q(T)^{-1}T^{n\mathrm {p}-(n-1)\mathrm {q}}(0) + p(T)q(T)^{-1}N(q(T)^n)\\ =&~~ p(T)q(T)^{-1}T^{n\mathrm {p}-(n-1)\mathrm {q}}(0) + p(T)q(T)^{-n-1}(0) \\ =&~~ p(T)q(T)^{-1}T^{n\mathrm {p}-n\mathrm {q}}q(T)(0) + p(T)q(T)^{-n-1}(0) \\ =&~~ p(T)q(T)^{-1}q(T)T^{n\mathrm {p}-n\mathrm {q}}(0) + p(T)q(T)^{-n-1}(0) \\ =&~~ p(T)[T^{n\mathrm {p}-n\mathrm {q}}(0)\cap R(q(T))+ q(T)^{-1}(0)] + p(T)q(T)^{-n-1}(0) \\ =&~~ p(T)[T^{n\mathrm {p}-n\mathrm {q}}(0)+ q(T)^{-1}(0)] + p(T)q(T)^{-n-1}(0) \\ =&~~ T^{(n+1)\mathrm {p}-n\mathrm {q}}(0)+ p(T)q(T)^{-1}(0) + p(T)q(T)^{-n-1}(0) \\ =&~~ T^{(n+1)\mathrm {p}-n\mathrm {q}}(0)+ p(T)(0)+q(T)^{-1}(0) + p(T)(0)+ q(T)^{-n-1}(0) \\ =&~~ T^{(n+1)\mathrm {p}-n\mathrm {q}}(0)+ q(T)^{-n-1}(0). \end{aligned}$$

(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 $n\ge 1$. Since T and $p(T)^n$ commute then $T(R(p(T)^n)) \subset R(p(T)^n)$. Let $T_n :R(p(T)^n) \rightarrow R(p(T)^n)$ defined by $T_nx = Tx$ for all $x\in R(p(T)^n)$. From Proposition 3.3 it follows that $R(r(T)^{n+1}) = r(R(r(T)^n)= r(R(p(T)^n) = R(r(T_n)) = R(p(T_n)) = p(T_n)(R(p(T)^n)) = p(T)(R(p(T)^n)) = R(p(T)^{n+1})$. $\square$

Recall that the resolvent set of a linear relation T is the set $\rho (T)$ defined by $\rho (T) := \{\lambda \in \mathbb {C}: T-\lambda ~~\text{ is } \text{ bijective }\}$ and the spectrum of T is the set $\sigma (T) := \mathbb {C}\backslash \rho (T)$.

Lemma 4.4

Let T be a linear relation everywhere defined in a linear space X such that $dim(T(0))<\infty$ and $\rho (T)\ne \emptyset$. Let r(T) be defined as in (3.2). The following statements hold.

(i):

If $a_e(r(T))<\infty$, then $a_e(p(T))<\infty$, with equivalence if $\mathrm {p}\ge \mathrm {q}$,

(ii):

if $a(r(T))<\infty$, then $a(p(T))<\infty$, with equivalence if $\mathrm {p}\ge \mathrm {q}$,

(iii):

$d_e(r(T))<\infty$ if and only if $d_e(p(T))<\infty$,

(iv):

$d(r(T))<\infty$ if and only if $d(p(T))<\infty$.

Proof

(i):

Arguing exactly as in the proof of Theorem VI.5.4 in [7] we obtain $\rho (p(T))\ne \emptyset$ and hence $\mathcal {R}_c(p(T))= \{0\}$ (Lemma 4.2). If $\mathrm {p}\ge \mathrm {q}$ or T is an operator, then the use of Lemmas 2.3, 4.1, 4.2 together with Proposition 4.3 leads to

$$\begin{aligned} \dfrac{N(r(T)^{n+1})}{N(r(T)^n)}&= \dfrac{T^{\mathrm {q}}(0)+ N(p(T)^{n+1})}{T^{\mathrm {q}}(0)+ N(p(T)^{n})} \\&\simeq \dfrac{N(p(T)^{n+1})}{N(p(T)^{n})+ T^\mathrm {q}(0)\cap N(p(T)^{n+1})} \\&= \dfrac{N(p(T)^{n+1})}{N(p(T)^n)}. \end{aligned}$$

for all $n\in \mathbb {N}$. Consequently, $a_e(r(T)) <\infty$ (resp. $a(r(T)) <\infty$) if and only if $a_e(r(T)) <\infty$ (resp. $a(r(T)) <\infty$). Now suppose that $T(0) \ne 0$ (that is T is not single valued) and $\mathrm {p}<\mathrm {q}$. Arguing as above we obtain

$$\begin{aligned} \dfrac{N(r(T)^{n+1})}{N(r(T)^n)}&= \dfrac{T^{(n+1)\mathrm {q}-n\mathrm {p}}(0)+ N(p(T)^{n+1})}{T^{n\mathrm {q}-(n-1)\mathrm {p}}(0)+ N(p(T)^{n})} \\&\supset \dfrac{T^{n\mathrm {q}-(n-1)\mathrm {p}}(0)+ N(p(T)^{n+1})}{T^{n\mathrm {q}-(n-1)\mathrm {p}}(0)+ N(p(T)^{n})}\\&= \dfrac{N(p(T)^{n+1})}{N(p(T)^n)}. \end{aligned}$$

Thus $a_e(p(T))<\infty$ (resp. $a(p(T))<\infty$) whenever $a_e(r(T))<\infty$ (resp. $a(r(T))<\infty$).

(iii):

For $n \in \mathbb {N}$, put $m = \mathrm {q}$ if $\mathrm {p}\ge \mathrm {q}$ and $m = n\mathrm {q} -(n-1)\mathrm {p}$ if $\mathrm {p}< \mathrm {q}$. Then by combining Proposition 3.3, Lemma 2.3 and Proposition 4.3 we obtain

$$\begin{aligned} \dfrac{R(r(T)^n)}{R(r(T)^{n+1})}&\simeq \dfrac{X}{N(r(T)^n) +R(r(T))} \\&= \dfrac{X}{T^m(0) + N(p(T)^n) + R(p(T))} \\&= \dfrac{X}{N(p(T)^n) + R(p(T))} \\&\simeq \dfrac{R(p(T)^n)}{R(p(T)^{n+1})}. \end{aligned}$$

The results of (ii) and (iv) follow immediately. $\square$

Theorem 4.5

Let T be a linear relation everywhere defined in a linear space X with $\rho (T)\ne \emptyset$, and let r(T) be defined as in (3.2). Then

(i):

$\sigma _{asc}^e(r(T)) \subset r(\sigma _{asc}^e(T))$,

(ii):

$\sigma _{asc}(r(T)) \subset r(\sigma _{asc}(T))$,

(iii):

$\sigma _{dsc}^e(r(T)) \subset r(\sigma _{dsc}^e(T))$,

(iv):

$\sigma _{dsc}(r(T)) \subset r(\sigma _{dsc}(T))$.

Proof

We only prove (i), the proofs of statements (ii), (iii) and (iv) are similar. For $\lambda \in \mathbb {C}$, let $h(T) := p(T) - \lambda q(T)= {\prod }_{i=1}^\mathrm {p}(T-\lambda _i)^{m_i}$, where $\mathrm {p}$ and $m_i, 1\le i \le \mathrm {p}$ are positive integers and $\lambda _i \in \mathbb {C}$, $1\le i \le \mathrm {p}$ are distinct constants. Then $r(T) - \lambda \subset h(T)q(T)^{-1}$ (Proposition 3.11). Let $\lambda \in \sigma _{asc}^e(r(T)))$. Then $a_e(r(T)-\lambda ) = \infty$, so that $a_e(h(T))=\infty$ by virtue of Lemma 4.4. It follows from [5, Proposition 3.1] that $a_e(T-\lambda _j) = \infty$ for some $\lambda _j$, and since $r(\lambda ) = \lambda _j$ one can deduce that $\lambda \in r(\sigma _{asc}^e(T))$. $\square$

It is more interesting from the point of view of the operator theory to combine the algebraic conditions defining the resolvent sets $\rho _{dsc}(.), \rho _{dsc}^e(.), \rho _{asc}(.)$ and $\rho _{asc}^(.)$ 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 $q^{-1}(0)\cap \sigma (T) = \emptyset$. Then r(T) is closed.

Proof

Since T is closed and $\rho (T)\ne \emptyset$ 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. $\square$

Lemma 4.7

Let T be a closed linear relation in a Banach space X.

(i):

If $a(T)<\infty$ and $R(T^k)$ is closed for some $k>a(T)$, then $R(T^n)$ is closed for all $n\ge a(T)$,

(ii):

if $a_e(T)<\infty$ and $R(T^k)$ is closed for some $k>a_e(T)$, then $R(T^n)$ is closed for all $n\ge a_e(T)$,

(iii):

if $d(T)<\infty$ and $R(T^k)$ is closed for some $k\ge d(T)$, then $R(T^n)$ is closed for all $n\ge d(T)$,

(iv):

if $d_e(T)<\infty$ and $R(T^k)$ is closed for some $k\ge d_e(T)$, then $R(T^n)$ is closed for all $n\ge d_e(T)$.

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. $\square$

Corollary 4.8

Let X be a Banach space and $T \in CR(X)$ be everywhere defined. Let r(T) be defined as in (3.2) and suppose that $\{\lambda : q(\lambda )= 0\}\subset \rho (T)$. If $deg(p) \ge deg(q)$ or T is single valued, then $\sigma _i(r(T)) \subset r(\sigma _i(T))\, \text{ for }\,i \in \{ta, ta^e, td, td^e\}.$

Proof

Let $\lambda \in \sigma _{ta}(r(T))$. From Proposition 3.11 we have $r(T)-\lambda = (p(T)-\lambda q(T))q(T)^{-1} := h(T)q(T)^{-1}:= {\prod }_{i=1}^\mathrm {p}(T-\lambda _i)^{m_i}$, where $\lambda _i, 1\le i \le \mathrm {p}$ are distinct scalars and $m_i, 1\le i\le \mathrm {p}$ are positive integers. If $a(r(T)-\lambda ) = \infty$ then Theorem 4.5 ensures that $\lambda \in r(\sigma _{ta}(T))$. Now suppose that $a = a(r(T)-\lambda ) < \infty$ and that $R(r(T)-\lambda )^{d+1})$ is not closed. It follows from Proposition 4.3 that $R(h(T)^{a+1})$ is not closed so that $R(T-\lambda _j)^{am_j+m_j}$ for some $1\le j \le p$ [5, Proposition 3.1]. The use of Lemma 4.7 ensures that $R(T-\lambda _j)^{a(T)+1}$ is not closed which implies that $\lambda _j \in \sigma _{ta}(T)$. Since $r(\lambda )= \lambda _j$, then $\lambda \in r(\sigma _{ta}(T))$. Proceeding as in above we show similarly that $\sigma _i(r(T)) \subset r(\sigma _i(T))$ for $i= ta^e, td, td^e$. $\square$

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

$$\begin{aligned}{}[\sigma _i(q(T)^{-1})] = [q(\sigma _i(T))]^{-1} \quad \text{ for }\quad i \in \{asc, asc^e, dsc, dsc^e, ta, ta^e, td, td^e\}, \end{aligned}$$

whenever $T \in CR(X)$ everywhere defined with $\{\lambda : Q(\lambda )= 0\}\subset \rho (T)$. In particular if $0 \in \rho (T)$, then

$$\begin{aligned} \sigma _i(T^{-1})= [\sigma _i(T)]^{-1}\quad \text{ for }\quad i \in \{asc, asc^e, dsc, dsc^e, ta, ta^e, td, td^e\}. \end{aligned}$$