The Local spectral theory and surjective spectrum of linear relations
UDC 517.98 This paper initiates a study of local spectral theory for linear relations. At the beginning, we define the local spectrum and study its properties. Then we obtain results related to the correlation analytic core $K\prime (T)$ and quasinilpotent part $H_0(T)$ of a linear relation $T$ in a...
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| author | Mnif, M. Ouled-Hmed, A.-A. Mnif, Maher Ouled-Hmed, A.-A. Mnif, M. Ouled-Hmed, A.-A. |
| author_facet | Mnif, M. Ouled-Hmed, A.-A. Mnif, Maher Ouled-Hmed, A.-A. Mnif, M. Ouled-Hmed, A.-A. |
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This paper initiates a study of local spectral theory for linear relations. At the beginning, we define the local spectrum and study its properties. Then we obtain results related to the correlation analytic core $K\prime (T)$ and quasinilpotent part $H_0(T)$ of a linear relation $T$ in a Banach space $X$. As an application, we give a characterization of the surjective spectrum $\sigma_{su}(T)$ in terms of the local spectrum and show that if $X = H_0(\lambda I - T) + K\prime (\lambda I - T)$, then $\sigma_{su}(T)$ does not cluster at $\lambda$. |
| doi_str_mv | 10.37863/umzh.v73i2.81 |
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DOI: 10.37863/umzh.v73i2.81
UDC 517.98
M. Mnif, A.-A. Ouled-Hmed (Univ. Sfax, Tunisia)
LOCAL SPECTRAL THEORY AND SURJECTIVE SPECTRUM
OF LINEAR RELATIONS
ТЕОРIЯ ЛОКАЛЬНОГО СПЕКТРА ТА СЮР’ЄКТИВНИЙ СПЕКТР
ЛIНIЙНИХ ВIДНОШЕНЬ
This paper initiates a study of local spectral theory for linear relations. At the beginning, we define the local spectrum and
study its properties. Then we obtain results related to the correlation analytic core K\prime (T ) and quasinilpotent part H0(T ) of
a linear relation T in a Banach space X. As an application, we give a characterization of the surjective spectrum \sigma su(T )
in terms of the local spectrum and show that if X = H0(\lambda I - T ) +K\prime (\lambda I - T ), then \sigma su(T ) does not cluster at \lambda .
Ця робота започатковує вивчення теорiї локального спектра для лiнiйних вiдношень. Спочатку наведено означення
та властивостi локального спектра. Пiсля цього отримано деякi результати, що вiдносяться до кореляцiйного аналi-
тичного ядра K\prime (T ) та квазiнiльпотентної частини H0(T ) лiнiйного вiдношення T у банаховому просторi X. Як
застосування наведено характеризацiю сюр’єктивного спектра \sigma su(T ) у термiнах локального спектра та доведено,
що якщо X = H0(\lambda I - T ) +K\prime (\lambda I - T ), то \sigma su(T ) не кластеризується для \lambda .
1. Introduction. Notice that throughout this paper (X, | | | | ) denotes a complex Banach space.
A linear relation T is any mapping having domain D(T ) a subspace of X, and taking values in
\scrP (X)\setminus \varnothing (the collection of nonempty subsets of X ) such that T (\alpha x1 + \beta x2) = \alpha T (x1) + \beta T (x2)
for all non zero scalars \alpha , \beta \in \BbbK and x1, x2 \in D(T ). If x /\in D(T ) then Tx = \varnothing . With this
convention we have D(T ) = \{ u \in X : T (u) \not = \varnothing \} . The set of all linear relations in X is denoted
by LR(X). A linear relation T is uniquely defined by its graph G(T ) = \{ (u, v) \in X \times X :
u \in D(T ), v \in T (u)\} . The inverse of T is the relation T - 1 given by
G(T - 1) = \{ (v, u) \in X \times X : (u, v) \in G(T )\} .
We denotes by
R(T ) :=
\bigcup
x\in D(T )
Tx, \mathrm{k}\mathrm{e}\mathrm{r}(T ) := \{ x \in X : (x, 0) \in G(T )\} , T (0) := \{ x \in X : (0, x) \in G(T )\}
the range, the kernel and the multivalued part of T, respectively. The generalized range of T is
defined by R\infty (T ) =
\bigcap
n\in \BbbN
R(Tn). Note that if T (0) = \{ 0\} , then T is said to be an operator or
single-valued. If G(T ) is closed, then T is said to be closed. Let QT denotes the quotient map from
X onto X/T (0). We can easy show that QTT is single valued and so we can define the norm of
T by \| T\| := \| QTT\| . T is said to be continuous if \| T\| < \infty , bounded if T is continuous and
everywhere defined. We denote by CR(X), BR(X), and BCR(X) the sets of closed, bounded,
and bounded closed linear relations, respectively.
Let U and V be two nonempty subsets of X. We define the distance between U and V by the
formula
\mathrm{d}\mathrm{i}\mathrm{s}(U, V ) = \mathrm{i}\mathrm{n}\mathrm{f} \{ \| u - v\| , u \in U and v \in V \} .
We shall also write \mathrm{d}\mathrm{i}\mathrm{s}(x, V ) for the distance between \{ x\} and V.
c\bigcirc M. MNIF, A.-A. OULED-HMED, 2021
222 ISSN 1027-3190. Укр. мат. журн., 2021, т. 73, № 2
LOCAL SPECTRAL THEORY AND SURJECTIVE SPECTRUM . . . 223
A linear relation T is called open if T - 1 is continuous equivalently if \gamma (T ) > 0 where
\gamma (T ) =
\left\{ +\infty , \mathrm{i}\mathrm{f}D(T ) \subset \mathrm{k}\mathrm{e}\mathrm{r}(T ),
\mathrm{i}\mathrm{n}\mathrm{f}
\biggl\{
\| Tx\|
\mathrm{d}\mathrm{i}\mathrm{s}(x, \mathrm{k}\mathrm{e}\mathrm{r}(T ))
;x \in D(T )\setminus \mathrm{k}\mathrm{e}\mathrm{r}(T )
\biggr\}
\mathrm{o}\mathrm{t}\mathrm{h}\mathrm{e}\mathrm{r}\mathrm{w}\mathrm{i}\mathrm{s}\mathrm{e}.
A linear operator A is called a selection of T if T = A+ T - T and D(A) = D(T ). The resolvent
set of a closed linear relation T is the set
\rho (T ) = \{ \lambda \in \BbbC such that (\lambda I - T ) - 1 is everywhere defined and single valued\} ,
and the spectrum of T is defined by
\sigma (T ) = \BbbC \setminus \rho (T ).
For a survey of result related to linear relations, the reader may see [5].
The concept of local spectral theory, for bounded operators in Banach space, was firstly appeared
in Dunford [6, 7]. It’s studied also by Kjeld B. Laursen and M. M. Neumann in [8] where they
mentioned that this theory played a very naturel role in commutative harmonic analysis. More
recently, P. Aiena was attached by this notion in [1, 2] and he was developed many results based
on the single valued extension property. Our objective is to generalize the concept of local spectral
theory to the general setting of linear relations and we give some of their properties. More precisely,
let T \in BR(X) and x \in X. The local resolvent of T at x, denoted by \rho T (x), is defined as the set
of all \lambda \in \BbbC for which there exist an open neighborhood U\lambda of \lambda and an analytic function f\lambda ,x :
U\lambda \rightarrow X such that the equation (\mu I - T )f\lambda ,x(\mu ) = x+T (0) holds for all \mu \in U\lambda . The complement
of \rho T (x) in \BbbC is called the local spectrum of T at x and denoted by \sigma T (x).
This paper is organized as follows. In Section 2, we introduce some auxiliary results which are
important for the following sections. We introduce the notions of quasinilpotent part and correlation
analytic core of a linear relation and we give some of their basic properties. In Section 3, we introduce
the definition of the local spectrum of linear relation and we give some of their properties. Then
we give the definitions of the local and glocal spectral subspaces and their applications in Sections 4
and 5, respectively. In particular, we investigate some connections between local spectral subspace
and the correlation analytic core on the one hand and between the glocal spectral subspace and the
quasinilpotent part on the other hand. We finish by giving, in Section 6 as application of the local
spectral theory, an important result concerning the surjective spectrum of a linear relation.
2. Preliminary results. In this section, we will introduce some auxiliary results which are
important for the next parts of this paper. We begin by giving the definitions of the algebraic core
C(T ), the correlation analytic core K \prime (T ) and the quasinilpotent part H0(T ) of a linear relation T
defined in a Banach space X.
Definition 2.1. Let T \in LR(X). The algebraic core C(T ) of T is defined to be the greatest
subspace M of X for which T (M) = M. It’s clear that C(T ) \subseteq R\infty (T ).
Definition 2.2. Let T \in LR(X). The correlation analytic core K \prime (T ) of T is defined by
K \prime (T ) =
\Bigl\{
x \in X such that there exist a > 0 and a sequence (un) \subset X, which verify
x = u0, un \in Tun+1, and \mathrm{d}\mathrm{i}\mathrm{s}(un, T (0) \cap \mathrm{k}\mathrm{e}\mathrm{r}(T )) \leq an\mathrm{d}\mathrm{i}\mathrm{s}(x, T (0) \cap \mathrm{k}\mathrm{e}\mathrm{r}(T )) \forall n \in \BbbN
\Bigr\}
.
ISSN 1027-3190. Укр. мат. журн., 2021, т. 73, № 2
224 M. MNIF, A.-A. OULED-HMED
Definition 2.3. Let T \in LR(X). Then, we define its quasinilpotent part by H0(T ) =
\Bigl\{
x \in
\in X such that there exists a sequence (xn) \subset X, which verify x = x0, xn+1 \in Txn and
\mathrm{l}\mathrm{i}\mathrm{m}n\mapsto \rightarrow \infty \| xn\|
1
n = 0
\Bigr\}
.
Proposition 2.1. Let T \in BCR(X). Then we have:
1) T (K \prime (T )) = K \prime (T );
2) if F be a closed subspace of X such that T (F ) = F, then F \subseteq K \prime (T ).
Proof. 1. Let x \in K \prime (T ). Then there exist a > 0 and (un) \subset X such that x = u0, un \in Tun+1
and \mathrm{d}\mathrm{i}\mathrm{s}(un, T (0) \cap \mathrm{k}\mathrm{e}\mathrm{r}(T )) \leq an\mathrm{d}\mathrm{i}\mathrm{s}(x, T (0) \cap \mathrm{k}\mathrm{e}\mathrm{r}(T )) \forall n \in \BbbN . So it suffices to prove that u1 \in
\in K \prime (T ). If u1 \in T (0) \cap \mathrm{k}\mathrm{e}\mathrm{r}(T ), then there is nothing to prove. If not, we consider the sequence
(wn) defined by
w0 := u1 and wn := un+1.
Then, for all n \in \BbbN , wn = un+1 \in Tun+2 = Twn+1 and we have
\mathrm{d}\mathrm{i}\mathrm{s}(wn, T (0) \cap \mathrm{k}\mathrm{e}\mathrm{r}(T )) = \mathrm{d}\mathrm{i}\mathrm{s}(un+1, T (0) \cap \mathrm{k}\mathrm{e}\mathrm{r}(T )) \leq
\leq bn \mathrm{d}\mathrm{i}\mathrm{s}(u1, T (0) \cap \mathrm{k}\mathrm{e}\mathrm{r}(T ))
with b > a2
\mathrm{d}\mathrm{i}\mathrm{s}(x, T (0) \cap \mathrm{k}\mathrm{e}\mathrm{r}(T ))
\mathrm{d}\mathrm{i}\mathrm{s}(u1, T (0) \cap \mathrm{k}\mathrm{e}\mathrm{r}(T ))
. Hence, u1 \in K \prime (T ) which permits us to deduce that K \prime (T ) \subset
\subset T (K \prime (T )).
Moving to the reverse inclusion. Let y \in T (K \prime (T )). Then y \in Tx for some x \in K \prime (T ).
Let a > 0 and (un) \subset X be such that x = u0, un \in Tun+1 and \mathrm{d}\mathrm{i}\mathrm{s}(un, T (0) \cap \mathrm{k}\mathrm{e}\mathrm{r}(T )) \leq
\leq an\mathrm{d}\mathrm{i}\mathrm{s}(x, T (0) \cap \mathrm{k}\mathrm{e}\mathrm{r}(T )) \forall n \in \BbbN .
We want to show that y \in K \prime (T ). If y \in T (0) \cap \mathrm{k}\mathrm{e}\mathrm{r}(T ), then there is nothing to prove. If not,
let consider the sequence (wn) defined by
w0 := y and wn := un - 1.
Then, for all n \in \BbbN \ast , we have wn = Twn+1 and
\mathrm{d}\mathrm{i}\mathrm{s}(wn, T (0) \cap \mathrm{k}\mathrm{e}\mathrm{r}(T )) = \mathrm{d}\mathrm{i}\mathrm{s}(un - 1, T (0) \cap \mathrm{k}\mathrm{e}\mathrm{r}(T )) \leq
\leq bn \mathrm{d}\mathrm{i}\mathrm{s}(y, T (0) \cap \mathrm{k}\mathrm{e}\mathrm{r}(T ))
with b > \mathrm{m}\mathrm{a}\mathrm{x}
\biggl(
a,
\mathrm{d}\mathrm{i}\mathrm{s}(x, T (0) \cap \mathrm{k}\mathrm{e}\mathrm{r}(T ))
\mathrm{d}\mathrm{i}\mathrm{s}(y, T (0) \cap \mathrm{k}\mathrm{e}\mathrm{r}(T ))
\biggr)
. Thus, y \in K \prime (T ) and, therefore, T (K \prime (T )) \subseteq K \prime (T ).
2. Let F be a closed subspace of X such that T (F ) = F. Then the restriction T0 : F - \rightarrow F is
an open map (see [5], Theorem III.4.2). Therefore, by [5] (Proposition II.3.2) we see that \gamma (T0) > 0.
Now let x \in F. So, there exists u \in F such that x \in Tu.
On the other hand, we have
\mathrm{d}\mathrm{i}\mathrm{s}(u, \mathrm{k}\mathrm{e}\mathrm{r}(T0)) \leq
1
\gamma (T0)
\| T0u \| .
Let \delta >
1
\gamma (T0)
. Then there exists y \in \mathrm{k}\mathrm{e}\mathrm{r}(T ) \cap F such that
\| u - y \| \leq \delta \mathrm{d}\mathrm{i}\mathrm{s}(x, T (0)) \leq \delta \mathrm{d}\mathrm{i}\mathrm{s}(x, T (0) \cap \mathrm{k}\mathrm{e}\mathrm{r}(T )).
ISSN 1027-3190. Укр. мат. журн., 2021, т. 73, № 2
LOCAL SPECTRAL THEORY AND SURJECTIVE SPECTRUM . . . 225
Take u1 = u - y. We have x \in Tu1 and
\mathrm{d}\mathrm{i}\mathrm{s}(u1, T (0) \cap \mathrm{k}\mathrm{e}\mathrm{r}(T )) \leq \| u1 \| \leq \delta \mathrm{d}\mathrm{i}\mathrm{s}(x, T (0) \cap \mathrm{k}\mathrm{e}\mathrm{r}(T )).
By repeating this process we find a sequence (un) \subset X such that x = u0, un \in Tun+1, and
\mathrm{d}\mathrm{i}\mathrm{s}(un, T (0) \cap \mathrm{k}\mathrm{e}\mathrm{r}(T )) \leq \delta n\mathrm{d}\mathrm{i}\mathrm{s}(x, T (0) \cap \mathrm{k}\mathrm{e}\mathrm{r}(T )) \forall n \in \BbbN . This implies that x \in K \prime (T ) and,
therefore, F \subseteq K \prime (T ).
Proposition 2.1 is proved.
The next theorem gives the connection between algebraic and correlation analytic core when
C(T ) is closed.
Theorem 2.1. Let T \in BCR(X). If C(T ) is closed, then we have
K \prime (T ) = C(T ).
Proof. From the definition of C(T ), by using the first assertion of Proposition 2.1, we observe
that K \prime (T ) \subseteq C(T ). Now, since C(T ) is closed and T (C(T )) = C(T ), then from the second
assertion of Proposition 2.1 we get C(T ) \subseteq K \prime (T ). Thus, K \prime (T ) = C(T ).
3. The local spectrum of a linear relation. In this section, we will define the local spectrum for
linear relations and we want to give some of their properties. It’s well-known that if \mu \in \rho (T ), then
the resolvent function R(\mu , T ) := (\mu I - T ) - 1 is everywhere defined and single valued. Moreover,
for any x \in X, the function fx : \rho (T ) \rightarrow X, defined by fx(\mu ) := (\mu I - T ) - 1x, is an analytic
function which verify
(\mu I - T )fx(\mu ) = x+ T (0) for all \mu \in \rho (T ).
Definition 3.1. Let X be a Banach space, T \in LR(X), and x \in X. Let \rho T (x) denote the
set of all \lambda \in \BbbC for which there exist an open neighborhood \scrU \lambda and an analytic function f\lambda ,x :
\scrU \lambda \rightarrow X such that the equation
(\mu I - T )f\lambda ,x(\mu ) = x+ T (0)
holds for all \mu \in \scrU \lambda .
\rho T (x) is called the local resolvent set of T at the point x. The local spectrum of T at the point
x is then defined by
\sigma T (x) := \BbbC \setminus \rho T (x).
Evidently, by Definition 3.1, we have \rho T (x) :=
\bigcup
\lambda \in \rho
T
(x)
\scrU \lambda and, hence, it is an open subset of \BbbC .
Moreover, for all x \in X,
\rho (T ) \subseteq \rho T (x) and \sigma T (x) \subseteq \sigma (T ).
In the following three propositions we gather some elementary properties of \sigma T (x).
Proposition 3.1. Let T \in LR(X). Then:
1) for all x \in T (0), we have \sigma T (x) = \varnothing ;
2) \sigma T (\alpha x+ \beta y) \subseteq \sigma T (x) \cup \sigma T (y) for all x, y \in X and \alpha , \beta \in \BbbC ;
3) \sigma - T (x) = \{ - \lambda , \lambda \in \sigma T (x)\} for all x \in X;
4) for every F \subseteq \BbbC , \sigma \lambda I+T (x) \subseteq F if and only if \sigma T (x) \subseteq F - \{ \lambda \} ; in particular, \sigma \lambda I - T (x) \subseteq
\subseteq \{ 0\} if and only if \sigma T (x) \subseteq \{ \lambda \} .
ISSN 1027-3190. Укр. мат. журн., 2021, т. 73, № 2
226 M. MNIF, A.-A. OULED-HMED
Proof. 1. It suffices to prove that for all x \in T (0) we have \rho T (x) = \BbbC . Let \lambda \in \BbbC . The null
analytic function f \equiv 0 defined on \BbbC verify that (\mu I - T )f(\mu ) = x + T (0) for all x \in T (0) and
for all \mu \in \BbbC . Then \lambda \in \rho T (x).
2. It is equivalent to prove that \rho T (x) \cap \rho T (y) \subseteq \rho T (\alpha x + \beta y) for all x, y \in X and for
all \alpha , \beta \in \BbbC . If \alpha = \beta = 0, then there is nothing to prove. If not, let \delta \in \rho T (x) \cap \rho T (y).
Then there exist two open neighborhoods U\delta , V\delta and two analytic functions f\delta ,x : U\delta \rightarrow X, g\delta ,y :
V\delta \rightarrow X such that
(\mu I - T )f\delta ,x(\nu ) = x+ T (0) \forall \mu \in U\delta ,
(\mu I - T )g\delta ,y(\mu ) = y + T (0) \forall \mu \in V\delta .
Now, let W\delta = U\delta \cap V\delta and let h : W\delta \rightarrow X be the analytic function defined by h = \alpha f\delta ,x + \beta g\delta ,y.
Then we have
(\mu I - T )h(\mu ) = (\alpha x+ \beta y) + T (0) \forall \mu \in W\delta .
So, \delta \in \rho T (\alpha x+ \beta y).
3. Let \lambda \in \rho - T (x). Then there exist an open neighborhood U\lambda and an analytic function
f : U\lambda \rightarrow X such that (\mu I - ( - T ))f(\mu ) = x + T (0) for all \mu \in U\lambda . If we take the change of
variables \nu = - \mu , then we find (\nu I - T )g(\nu ) = x+ T (0) \forall \nu \in U - \lambda , where g(\nu ) = - f( - \nu ) is an
analytic function defined on an open neighborhood U - \lambda of - \lambda . Therefore, - \lambda \in \rho T (x).
4. It is similar to prove that \BbbC \setminus F \subseteq \rho \lambda I+T (x) if and only if \BbbC \setminus (F - \{ \lambda \} ) \subseteq \rho T (x). For the only
if part, let \lambda 0 \in \BbbC \setminus (F - \{ \lambda \} ). So, \lambda 0+\lambda \in \BbbC \setminus F \subseteq \rho \lambda I+T (x). Then there exist an open neighborhood
U\lambda 0+\lambda and an analytic function f\lambda 0+\lambda ,x : U\lambda 0+\lambda \rightarrow X such that (\mu I - (\lambda I + T ))f\lambda 0+\lambda ,x(\mu ) =
= x+ T (0) for all \mu \in U\lambda 0+\lambda . If we take the change of variables \nu = \mu - \lambda , then we get
(\nu I - T )g(\nu ) = x+ T (0) \forall \nu \in U\lambda 0 ,
where g(\nu ) = f(\nu + \lambda ) is an analytic function defined on an open neighborhood U\lambda 0 of \lambda 0. Thus,
\lambda 0 \in \rho T (x).
In particular, let \lambda 0 \in \BbbC \setminus F. Then \lambda 0 - \lambda \in \BbbC \setminus (F - \{ \lambda \} ) \subseteq \rho T (x). So, there exist an open
neighborhood U\lambda 0 - \lambda and an analytic function f\lambda 0 - \lambda ,x : U\lambda 0 - \lambda \rightarrow X such that
((\mu + \lambda )I - (\lambda I + T ))f\lambda 0 - \lambda ,x(\mu ) = x+ T (0) \forall \mu \in U\lambda 0 - \lambda .
If we take the change of variables \nu = \mu + \lambda , we get
(\nu I - (\lambda I + T ))g(\nu ) = x+ T (0) \forall \nu \in U\lambda 0 ,
where g(\nu ) = f(\nu - \lambda ) is an analytic function defined on an open neighborhood U\lambda 0 of \lambda 0. Thus,
\lambda 0 \in \rho \lambda I+T (x).
Now, if we take F = \{ 0\} and we replace T by - T, we can easily find the particular result.
Proposition 3.1 is proved.
Proposition 3.2. Let T \in BR(X), x \in X and \scrU an open subset of \BbbC . If there exists an analytic
function fx : \scrU \rightarrow X such that (\mu I - T )f(\mu ) = x+ T (0) for all \mu \in \scrU , then \scrU \subseteq \rho T (f(\lambda )) for all
\lambda \in \scrU . If moreover T has a continuous selection A such that T (0) \subset \mathrm{k}\mathrm{e}\mathrm{r}(A), then
\sigma T (x) = \sigma T (f(\lambda )) for all \lambda \in \scrU .
ISSN 1027-3190. Укр. мат. журн., 2021, т. 73, № 2
LOCAL SPECTRAL THEORY AND SURJECTIVE SPECTRUM . . . 227
Proof. Let \lambda \in \scrU . Define
h(\mu ) =
\left\{
f(\lambda ) - f(\mu )
\mu - \lambda
, if \mu \not = \lambda ,
- f \prime (\lambda ), if \mu = \lambda .
Then h is an analytic function on \scrU and we have, for all \mu \in \scrU \setminus \{ \lambda \} ,
(\mu I - T )h(\mu ) =
1
\mu - \lambda
\bigl[
((\mu - \lambda )I + (\lambda I - T ))(f(\lambda )) - x+ T (0)
\bigr]
= f(\lambda ) + T (0).
Therefore (\mu I - T )h(\mu ) = f(\lambda ) + T (0) holds for all \mu \in \scrU \setminus \{ \lambda \} . On the other hand, we have
(\mu I - T )f(\mu ) = x + T (0). Then \mu QT (f(\mu )) - QTTf(\mu ) = QTx. Now, as QT is a bounded
operator, by derivation on \mu we get QT (f(\mu )) + \mu QT (f
\prime (\mu )) - QTTf(\mu ) = 0. Thus, for \mu = \lambda ,
we have - (\lambda I - T )f \prime (\lambda ) = f(\lambda ) + T (0). Hence, the equality (\mu I - T )h(\mu ) = f(\lambda ) + T (0) holds
for all \mu \in \scrU . Therefore, \scrU \subseteq \rho T (f(\lambda )).
To show the equality \sigma T (x) = \sigma T (f(\lambda )) we begin by proving the inclusion \rho T (x) \subseteq \rho T (f(\lambda )).
Let w \in \rho T (x). If w \in \scrU then w \in \rho T (f(\lambda )) for all \lambda \in \scrU , by the first part of the proof. If
w \in \rho T (x)\setminus \scrU , then there exist an open neighborhood W which not contains \lambda and an analytic
function g : W \rightarrow X such that (\mu I - T )g(\mu ) = x+ T (0) for all \mu \in W. Define
K(\mu ) =
g(\lambda ) - g(\mu )
\mu - \lambda
for all \mu \in W.
K is an analytic function on W, and we have (\mu I - T )K(\mu ) = f(\lambda ) + T (0) for all \mu \in W. Thus,
w \in \rho T (f(\lambda )). So, \rho T (x) \subseteq \rho T (f(\lambda )).
It remains to prove the reverse inclusion. Let \nu \in \rho T (f(\lambda )). Then there exist an open neigh-
borhood \scrV of \nu and an analytic function h : \scrV \rightarrow X such that (\mu I - T )h(\mu ) = f(\lambda ) + T (0) is
satisfied for all \mu \in \scrV . By hypothesis A is a continuous selection of T then the function hA defined
on \scrV by hA(\mu ) = (\lambda I - A)h(\mu ) is analytic on \scrV . Besides, since T (0) \subset \mathrm{k}\mathrm{e}\mathrm{r}(A), then we have, for
all \mu \in \scrV ,
(\mu I - T )hA(\mu ) = (\lambda I - A)(\mu I - T )h(\mu ) =
= (\lambda I - A+ T - T )(f(\lambda )) + T (0) - AT (0) =
= x+ T (0).
Thus, \nu \in \rho T (x) and, hence, \rho T (f(\lambda )) \subseteq \rho T (x) which ends the proof.
Proposition 3.3. Let T, S \in BR(X) be such that ST = TS, S(0) \subset \mathrm{k}\mathrm{e}\mathrm{r}(T ), and S have a
linear continuous selection S1. Then, for all y \in Sx, we have
\sigma T (y) \subset \sigma T (x).
Proof. We shall prove that \rho T (x) \subset \rho T (y) for all y \in Sx. Let y \in Sx and \lambda \in \rho T (x). Then
there exist an open neighborhood U\lambda and an analytic function f : U\lambda \rightarrow X such that (\nu I - T )f(\nu ) =
= x+ T (0) for all \nu \in U\lambda . Then we have
S(\nu I - T )f(\nu ) = S(x+ T (0)).
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228 M. MNIF, A.-A. OULED-HMED
By using the fact that TS = ST, S(0) \subset \mathrm{k}\mathrm{e}\mathrm{r}(T ) and [3] (Lemma 2.4), we get (\nu I - T )Sf(\nu ) =
= Sx+ ST (0). Thus, (\nu I - T )(S1f(\nu ) + S(0)) = Sx+ ST (0). So, (\nu I - T )S1f(\nu ) + ST (0) =
= y + S(0) + ST (0). Hence, (\nu I - T )S1f(\nu ) = y + T (0). Then \lambda \in \rho T (y) which ends the
proof.
Let T \in LR(X). We say that T verifies the stabilization criteria if T (0) = T 2(0). We denote
by \scrS \scrT \scrR 1(X) the set of all linear relations satisfying the stabilization criteria.
Lemma 3.1. Let R,S \in BR(X). Then the following equivalence holds:
S(0) \subset \mathrm{k}\mathrm{e}\mathrm{r}(SR) and R(0) \subset \mathrm{k}\mathrm{e}\mathrm{r}(RS) if and only if
SR and RS are both in \scrS \scrT \scrR 1(X).
Proof. For the only if part, we have by hypothesis SRS(0) = SR(0) and RSR(0) = RS(0).
Then SRSR(0) = SR(0). Thus, SR \in \scrS \scrT \scrR 1(X). In a similar way we can prove that RS \in
\in \scrS \scrT \scrR 1(X).
For the if part, we have SRSR(0) = SR(0). So, SRS(0) \subset SR(0). Then, for all y \in S(0),
we have SR(y) \subset SR(0). So, SR(y) = SR(0). Thus, SRS(0) = SR(0) and we conclude that
S(0) \subset \mathrm{k}\mathrm{e}\mathrm{r}(SR). By the same way we can prove that R(0) \subset \mathrm{k}\mathrm{e}\mathrm{r}(RS).
Lemma 3.1 is proved.
In the next theorem we study the relation between the local spectrums of the linear relations SR
and RS.
Theorem 3.1. Let S,R \in BR(X) be such that SR and RS are both in \scrS \scrT \scrR 1(X). If S and
R have continuous linear selection S1 and R1, respectively, then, for all x \in X and y \in Sx, we
have
\sigma SR(y) \subseteq \sigma RS(x) \subseteq \sigma SR(y) \cup \{ 0\} .
Proof. Let beginning by showing the first inclusion. It is equivalent to prove that \rho RS(x) \subseteq
\subseteq \rho SR(y). Let \lambda \in \rho RS(x). Then there exist an open neighborhood U\lambda and an analytic function
f : U\lambda \rightarrow X such that (\nu I - RS)f(\nu ) = x + RS(0) \forall \nu \in U\lambda . Then we have S(\nu I - RS)f(\nu ) =
= Sx + SRS(0). By using Lemma 3.1 and [3] (Lemma 2.4), we get (\nu I - SR)Sf(\nu ) = Sx +
+SRS(0). Thus, (\nu I - SR)(S1f(\nu )+S(0)) = Sx+SRS(0). Since (\nu I - SR)S(0) \subseteq (\nu I - SR)(0),
then we obtain (\nu I - SR)S1f(\nu ) = y + SR(0) for all \nu \in U\lambda . So we conclude that \lambda \in \rho SR(y)
and, therefore, \rho RS(x) \subseteq \rho SR(y).
In order to show the second inclusion, it suffices to prove that \rho SR(y) \cap \BbbC \ast \subset \rho RS(x). Let
\lambda \in \rho SR(y) \cap \BbbC \ast . So, there exist an open neighborhood V\lambda \subset \BbbC \ast and an analytic function
f : V\lambda \rightarrow X such that (\mu I - SR)f(\mu ) = y + SR(0) \forall \mu \in V\lambda . Let h be the function defined
by
h(\mu ) =
1
\mu
(x+R1f(\mu )).
So, we have
(\mu I - RS)h(\mu ) = x - 1
\mu
RS(x) +
1
\mu
(\mu R1 - RSR1)f(\mu ) =
= x - 1
\mu
RS(x) +
1
\mu
R[(\mu I - SR1)f(\mu )] \subset x - 1
\mu
R(y) +
1
\mu
R(y) +RS(0) \subset x+RS(0).
Then (\mu I - RS)h(\mu ) = x + RS(0) for all \mu \in V\lambda . Therefore, \lambda \in \rho RS(x), and we obtain the
desired inclusion.
Theorem 3.1 is proved.
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LOCAL SPECTRAL THEORY AND SURJECTIVE SPECTRUM . . . 229
4. The local spectral subspace. For every subset F of \BbbC the local spectral subspace of T
associated with F is the set
XT (F ) := \{ x \in X : \sigma T (x) \subseteq F\} .
Obviously, if F1 \subseteq F2 then XT (F1) \subseteq XT (F2).
We begin with the following proposition, which gives some properties of the local spectral
subspace.
Proposition 4.1. Let T \in BR(X) and F be a subset of \BbbC . Then:
1) XT (F ) is a subspace of X;
2) XT (F ) is a linear hyperinvariant subspace for T, i.e., for every bounded operator S such
that TS = ST we have S(XT (F )) \subseteq XT (F );
3) if moreover T \in \scrS \scrT \scrR 1(X) and have a continuous selection, then
T (XT (F )) \subseteq XT (F ).
Proof. 1. By the first part of Proposition 3.1 we have T (0) \subset XT (F ). Then XT (F ) \not = \varnothing .
Let x, y \in XT (F ). We have \sigma T (x) \subset F and \sigma T (y) \subset F. Then, by using the second part of
Proposition 3.1, we get, for all \alpha , \beta \in \BbbC ,
\sigma T (\alpha x+ \beta y) \subseteq \sigma T (x) \cup \sigma T (y) \subseteq F.
Thus, \alpha x+ \beta y \in XT (F ) and, therefore, XT (F ) is a subspace of X.
2. Let S \in B(X) be such that ST = TS. Let x \in XT (F ) and let \lambda /\in \sigma T (x). Then there exist
an open neighborhood U\lambda of \lambda and an analytic function f : U\lambda \rightarrow X which satisfies (\mu I - T )f(\mu ) =
= x+ T (0) for all \mu \in U\lambda . Thus, S(\mu I - T )f(\mu ) = Sx+ ST (0) for all \mu \in U\lambda . Since ST = TS
and, by using [3] (Lemma 2.4), we get, for all \mu \in U\lambda ,
(\mu I - T )Sf(\mu ) = Sx+ T (0).
Since S is a bounded operator, then Sf is analytic on U\lambda . So, we conclude that \lambda /\in \sigma T (Sx). Thus,
\sigma T (Sx) \subseteq \sigma T (x) \subseteq F and, therefore, Sx \in XT (F ) which implies that S(XT (F )) \subseteq XT (F ).
3. Since T admits a continuous linear selection and T (0) \subseteq \mathrm{k}\mathrm{e}\mathrm{r}(T ), then by using Proposition 3.3
the result follows immediately.
We prove the following auxiliary assertion.
Lemma 4.1. Let T be a bounded linear relation in X and (xn) be a sequence of X such that
xn \in Txn+1 for all n \in \BbbN . Let R denote the convergence radius of the entire series
\sum
n\geq 1
\lambda n - 1xn.
Then, for all scalar \lambda such that | \lambda | < R, we have
T
\left( \sum
n\geq 1
\lambda n - 1xn
\right) =
\sum
n\geq 1
\lambda n - 1xn - 1 + T (0).
Proof. Let \lambda such that | \lambda | < R. We get
T
\left( \sum
n\geq 1
\lambda n - 1xn
\right) =
N\sum
n=1
\lambda n - 1Txn + T
\left( \sum
n\geq N+1
\lambda n - 1xn
\right) =
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230 M. MNIF, A.-A. OULED-HMED
=
\sum
n\geq 1
\lambda n - 1xn - 1 -
\sum
n\geq N+1
\lambda n - 1xn - 1 + T
\left( \sum
n\geq N+1
\lambda n - 1xn
\right) .
Now, since the operators QT and QTT are bounded, we have
QT
\left( T
\left( \sum
n\geq 1
\lambda n - 1xn
\right) \right) = QT
\left[ \sum
n\geq 1
\lambda n - 1xn - 1 -
\sum
n\geq N+1
\lambda n - 1xn - 1 + T
\left( \sum
n\geq N+1
\lambda n - 1xn
\right) \right] =
= QT
\left( \sum
n\geq 1
\lambda n - 1xn - 1
\right) .
Then QT
\Bigl(
T
\Bigl( \sum
n\geq 1
\lambda n - 1xn
\Bigr)
-
\sum
n\geq 1
\lambda n - 1xn - 1
\Bigr)
= 0, which implies that
T
\left( \sum
n\geq 1
\lambda n - 1xn
\right) -
\sum
n\geq 1
\lambda n - 1xn - 1 \subseteq T (0).
Thus, T
\Bigl( \sum
n\geq 1
\lambda n - 1xn
\Bigr)
=
\sum
n\geq 1
\lambda n - 1xn - 1 + T (0).
Lemma 4.1 is proved.
We now establish the relationship between the correlation analytic core and the local spectral
subspace of a relation T.
Theorem 4.1. Let X be a Banach space and T \in BCR(X). We have
K \prime (T ) = XT (\BbbC \setminus \{ 0\} ) = \{ x \in X such that 0 \in \rho T (x)\} .
Proof. Let prove the first inclusion K \prime (T ) \subseteq XT (\BbbC \setminus \{ 0\} ). Let x \in K \prime (T ). So, there exist a
sequence (xn) \subset X and a positive scalar a such that x = x0, xn \in Txn+1 and
\mathrm{d}\mathrm{i}\mathrm{s}(xn, T (0) \cap \mathrm{k}\mathrm{e}\mathrm{r}(T )) \leq an\mathrm{d}\mathrm{i}\mathrm{s}(x, T (0) \cap \mathrm{k}\mathrm{e}\mathrm{r}(T )) \forall n \in \BbbN .
Let b > a. From the last inequality, we deduce that for all n \geq 1 there exists \alpha n \in T (0)\cap \mathrm{k}\mathrm{e}\mathrm{r}(T )
such that \| xn - \alpha n\| \leq bn\| x\| . Let (yn) be the sequence defined by y0 = x, and, for all n \geq 1,
yn = xn - \alpha n. Then we have \| yn\| \leq bn\| x\| . Let f be the analytic function f : B
\Bigl(
0,
1
b
\Bigr)
\rightarrow X
defined by f(\lambda ) = -
\sum
n\geq 1
\lambda n - 1yn. We can easily verify that yn \in Tyn+1 for all n \in \BbbN . Moreover,
by Lemma 4.1, for all \lambda \in B
\Bigl(
0,
1
b
\Bigr)
, we have
(\lambda I - T )f(\lambda ) =
\sum
n\geq 1
\lambda n - 1yn - 1 + T (0) - \lambda
\sum
n\geq 1
\lambda n - 1yn =
= y0 + T (0) = x+ T (0).
Thus, 0 \in \rho T (x) and then x \in XT (\BbbC \setminus \{ 0\} ), which provide the desired inclusion.
Moving to the reverse inclusion. Let x \in XT (\BbbC \setminus \{ 0\} ). So, 0 \in \rho T (x) which implies that
there exist an open disc \BbbD (0, \varepsilon ) and an analytic function f : \BbbD (0, \varepsilon ) \rightarrow X such that the equation
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LOCAL SPECTRAL THEORY AND SURJECTIVE SPECTRUM . . . 231
(\mu I - T )f(\mu ) = x+T (0) holds for all \mu \in \BbbD (0, \varepsilon ). Since f is an analytic function, then there exists
a sequence (un)n\geq 1 \subset X such that
f(\lambda ) = -
\sum
n\geq 1
\lambda n - 1un for all \lambda \in \BbbD (0, \varepsilon ).
Clearly, f(0) = - u1. Then T (u1) = x + T (0) and x \in Tu1. Take u0 = x and let prove that
un \in Tun+1 for all n \in \BbbN . The proof is given by induction. For n = 0, we have u0 \in Tu1.
Suppose that this property is true until the order n, and let prove it for the order n + 1. For all
\lambda \in \BbbD (0, \varepsilon ), we get
(\lambda I - T )f(\lambda ) = -
\sum
k\geq 1
\lambda kuk + T
\left( \sum
k\geq 1
\lambda k - 1uk
\right) =
= -
n+1\sum
k=1
\lambda kuk -
\sum
k\geq n+2
\lambda kuk + T
\Biggl(
n+2\sum
k=1
\lambda k - 1uk
\Biggr)
+ T
\left( \sum
k\geq n+3
\lambda k - 1uk
\right) .
So,
Tu1 + \lambda n+1(Tun+2 - un+1) -
\sum
k\geq n+2
\lambda kuk + T
\left( \sum
k\geq n+3
\lambda k - 1uk
\right) = Tu1.
Since
\lambda n+1T
\left( \sum
k\geq n+3
\lambda k - (n+2)uk
\right) \subset T
\left( \lambda n+1
\sum
k\geq n+3
\lambda k - (n+2)uk
\right) ,
then
(Tun+2 - un+1) -
\sum
k\geq n+2
\lambda k - (n+1)uk + T
\left( \sum
k\geq n+3
\lambda k - (n+2)uk
\right) \subset T (0).
If we take \lambda = 0, then we find that Tun+2 - un+1 \subset T (0) and, hence, un+1 \in Tun+2. Consequently,
un \in Tun+1 for all n \in \BbbN .
It remains to prove that there exists a positive scalar which verify the second condition of K \prime (T ).
If x \in T (0) \cap \mathrm{k}\mathrm{e}\mathrm{r}(T ) then there is nothing to prove. If not, since the series
\Bigl(
-
\sum
n\geq 1
\lambda n - 1un
\Bigr)
converges then | \lambda | n - 1\| un\| \rightarrow 0 as n \rightarrow \infty for all | \lambda | < \varepsilon . In particular,
1
\mu n - 1
\| un\| \rightarrow 0 as n \rightarrow \infty
for \mu >
1
\varepsilon
. Take \mu 0 >
1
\varepsilon
. So, there exists c > 0 such that \| un\| \leq c\mu n - 1
0 . Hence, we obtain
\| un\| \leq
\biggl(
\mu 0 +
c
\mathrm{d}\mathrm{i}\mathrm{s}(x, T (0) \cap \mathrm{k}\mathrm{e}\mathrm{r}(T ))
\biggr) n
\mathrm{d}\mathrm{i}\mathrm{s}(x, T (0) \cap \mathrm{k}\mathrm{e}\mathrm{r}(T )).
Thus, we have
\mathrm{d}\mathrm{i}\mathrm{s}(un, T (0) \cap \mathrm{k}\mathrm{e}\mathrm{r}(T )) \leq \| un\| \leq bn\mathrm{d}\mathrm{i}\mathrm{s}(x, T (0) \cap \mathrm{k}\mathrm{e}\mathrm{r}(T ))
with b = \mu 0 +
c
\mathrm{d}\mathrm{i}\mathrm{s}(x, T (0) \cap \mathrm{k}\mathrm{e}\mathrm{r}(T ))
. So, x \in K \prime (T ) and, therefore, XT (\BbbC \setminus \{ 0\} ) \subseteq K \prime (T ).
Theorem 4.1 is proved.
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232 M. MNIF, A.-A. OULED-HMED
5. The glocal spectral subspace. Let F \subset \BbbC be a closed subset and let T \in LR(X). We define
the glocal spectral subspace \scrX T (F ) as the set of all x \in X such that there exists an analytic function
f : \BbbC \setminus F \rightarrow X checking
(\lambda I - T )f(\lambda ) = x+ T (0) for all \lambda \in \BbbC \setminus F.
Some basic properties of the glocal spectral subspace are gathered in the following proposition.
Proposition 5.1. Let T \in BR(X) and F \subset \BbbC be a closed subset. Then:
1) \scrX T (F ) is a subspace of X and \scrX T (F ) \subset XT (F );
2) \scrX T (F ) is a linear hyperinvariant subspace for T, i.e., for every bounded operator S such
that TS = ST we have S(\scrX T (F )) \subseteq \scrX T (F );
3) if moreover T \in \scrS \scrT \scrR 1(X) and have a continuous selection, then
T (\scrX T (F )) \subseteq \scrX T (F ).
Proof. Proceeding as in the proof of Proposition 4.1 we obtain the desired results.
In the sequel we need the two following elementary lemmas.
Lemma 5.1. Let T belongs to BCR(X) and (xn) \subset X be such that xn \in Txn - 1 for all
n \in \BbbN \ast . Let R be the convergence radius of the power series
\sum
n\geq 1
\lambda nxn - 1. Then, for all scalar \lambda
such that | \lambda | > 1
R
, we have
T
\left( \sum
n\geq 1
\lambda - nxn - 1
\right) =
\sum
n\geq 1
\lambda - nxn + T (0).
Proof. For a scalar \lambda such that | \lambda | > 1
R
, we obtain
T
\left( \sum
n\geq 1
\lambda - nxn - 1
\right) =
N\sum
n=1
\lambda - nTxn - 1 + T
\left( \sum
n\geq N+1
\lambda - nxn - 1
\right) =
=
\sum
n\geq 1
\lambda - nxn -
\sum
n\geq N+1
\lambda - nxn + T
\left( \sum
n\geq N+1
\lambda - nxn - 1
\right) .
Now, since the operators QT and QTT are bounded, then we can easily seen that
QT
\left( T
\left( \sum
n\geq 1
\lambda - nxn - 1
\right) \right) = QT
\left( \sum
n\geq 1
\lambda - nxn
\right) .
Then
T
\left( \sum
n\geq 1
\lambda - nxn - 1
\right) -
\sum
n\geq 1
\lambda - nxn \subseteq T (0).
Thus,
T
\left( \sum
n\geq 1
\lambda - nxn - 1
\right) =
\sum
n\geq 1
\lambda - nxn + T (0).
Lemma 5.1 is proved.
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LOCAL SPECTRAL THEORY AND SURJECTIVE SPECTRUM . . . 233
Lemma 5.2. Let T belongs to BCR(X) and (xn) \subset X be such that \mathrm{l}\mathrm{i}\mathrm{m} \mathrm{s}\mathrm{u}\mathrm{p}
n\rightarrow \infty
\| xn\|
1
n \leq \varepsilon . Then,
for all n \in \BbbN , there exists \alpha n+1 \in Txn such that for all | \lambda | < 1
\varepsilon
we have
T
\left( \sum
n\geq 0
\lambda nxn
\right) =
\sum
n\geq 0
\lambda n\alpha n+1 + T (0).
Proof. Let \alpha \prime
n+1 \in Txn. Then we have \mathrm{d}\mathrm{i}\mathrm{s}(\alpha \prime
n+1, T (0)) = \| Txn\| \leq \| T\| \| xn\| . Hence, for a
fixed \gamma > 0, there exists \beta n+1 \in T (0) such that
\| \alpha \prime
n+1 - \beta n+1\| \leq (\| T\| + \gamma )\| xn\| .
Take \alpha n+1 = \alpha \prime
n+1 - \beta n+1. Then \alpha n+1 \in Txn and the series
\sum
n\geq 0
\lambda n\alpha n+1 is absolutely conver-
gent for all \lambda such that | \lambda | < 1
\varepsilon
. Thus, we obtain
QTT
\left( \sum
n\geq 0
\lambda nxn
\right) = QT
\left( \sum
n\geq 0
\lambda n\alpha n+1
\right) .
Hence,
T
\left( \sum
n\geq 0
\lambda nxn
\right) =
\sum
n\geq 0
\lambda n\alpha n+1 + T (0).
Lemma 5.2 is proved.
Theorem 5.1. Let T \in BCR(X). Then
\scrX T (\BbbD (0, \varepsilon )) \supset H\varepsilon (T ) + T (0),
where
H\varepsilon (T ) :=
\biggl\{
x \in X : \exists (xn) \subset X such that x = x0, xn+1 \in Txn and \mathrm{l}\mathrm{i}\mathrm{m} \mathrm{s}\mathrm{u}\mathrm{p}
n\rightarrow \infty
\| xn\|
1
n \leq \varepsilon
\biggr\}
.
If moreover \sigma (T ) is bounded, then the equality holds.
Proof. Let x \in H\varepsilon (T ). Then there exists (xn) \subset X such that x = x0, xn+1 \in Txn, and
\mathrm{l}\mathrm{i}\mathrm{m} \mathrm{s}\mathrm{u}\mathrm{p}n\rightarrow \infty \| xn\| 1/n \leq \varepsilon . Thus, the series defined by f(\lambda ) :=
\sum
n\geq 1
\lambda - nxn - 1 converges uni-
formly on \BbbC \setminus \BbbD (0, \varepsilon ). So, f is analytic on \BbbC \setminus \BbbD (0, \varepsilon ). Besides, using Lemma 5.1, we get, for all \lambda
such that | \lambda | > \varepsilon ,
(\lambda I - T )f(\lambda ) =
\sum
n\geq 0
\lambda - nxn -
\sum
n\geq 1
\lambda - nxn - T (0) = x+ T (0).
Then x \in \scrX T (\BbbD (0, \varepsilon )). On the other hand, since T (0) \subset \scrX T (\BbbD (0, \varepsilon )) and \scrX T (\BbbD (0, \varepsilon )) is a subspace
of X, then the inclusion H\varepsilon (T ) + T (0) \subseteq \scrX T (\BbbD (0, \varepsilon )) is satisfied.
Conversely, let x \in \scrX T (\BbbD (0, \varepsilon )). There exists an analytic function f : \BbbC \setminus \BbbD (0, \varepsilon ) \rightarrow X such that
(\lambda I - T )f(\lambda ) = x+T (0) holds for every \lambda \in \BbbC \setminus \BbbD (0, \varepsilon ). By assumption, we have \sigma (T ) is bounded,
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234 M. MNIF, A.-A. OULED-HMED
then the set V :=
\bigl(
\BbbC \setminus \BbbD (0, \varepsilon )
\bigr)
\cap \rho (T ) is not empty and open and the function f is analytic on V.
Besides, for all \lambda \in V, we have f(\lambda ) = (\lambda I - T ) - 1x. Indeed, we get (\lambda I - T )f(\lambda ) = x + T (0)
so, f(\lambda ) + \mathrm{k}\mathrm{e}\mathrm{r}(\lambda I - T ) = (\lambda I - T ) - 1x+\mathrm{k}\mathrm{e}\mathrm{r}(\lambda I - T ). Then f(\lambda ) = (\lambda I - T ) - 1x, and we get the
result. Now, let \lambda \in V. By using [5] (Proposition VI.3.2), we have \mathrm{l}\mathrm{i}\mathrm{m}| \lambda | \mapsto \rightarrow \infty f(\lambda ) = 0. Let consider
the analytic function g defined by
g(\mu ) :=
\left\{ f
\biggl(
1
\mu
\biggr)
, if 0 \not = \mu \in \BbbD (0, 1/\varepsilon ),
0, if \mu = 0.
Since g is analytic on \BbbD (0, 1/\varepsilon ) and g(0) = 0, then there exists a sequence (xn) \subset X such that
x0 = 0 and g(\mu ) =
\sum
n\geq 0
\mu nxn holds even for all \BbbD (0, 1/\varepsilon ). This shows that the radius of
convergence of the power series representing g(\mu ) is greater then 1/\varepsilon . Hence, \mathrm{l}\mathrm{i}\mathrm{m} \mathrm{s}\mathrm{u}\mathrm{p}
n\rightarrow \infty
\| xn\| 1/n \leq \varepsilon .
Besides, we obtain f(\lambda ) = g
\Bigl( 1
\lambda
\Bigr)
=
\sum
n\geq 0
\lambda - nxn. From Lemma 5.2, for all n \in \BbbN , there exists
\alpha n+1 \in Txn such that for all \lambda with | \lambda | > \varepsilon we get
(\lambda I - T )f(\lambda ) =
\sum
n\geq 0
\lambda - n+1xn -
\sum
n\geq 0
\lambda - n\alpha n+1 + T (0) =
=
\sum
n\geq 0
\lambda - n(xn+1 - \alpha n+1) + T (0).
Thus, we have QT (x) =
\sum
n\geq 0
\lambda - nQT (xn+1 - \alpha n+1). Hence,
QT (x1 - \alpha 1) = QT (x),
QT (xn+1 - \alpha n+1) = 0, n \geq 1,
and, so,
x = x1 + \alpha with \alpha \in T (0),
xn+1 \in Txn, n \geq 1.
Now, let prove that x1 \in H\varepsilon (T ). Let consider the sequence (yn) defined by yn := xn+1. We
have x1 = y0 and yn+1 = xn+2 \in Txn+1 = Tyn. Besides, we have
\mathrm{l}\mathrm{i}\mathrm{m} \mathrm{s}\mathrm{u}\mathrm{p}
n\rightarrow \infty
\| yn\| 1/n \leq \mathrm{l}\mathrm{i}\mathrm{m} \mathrm{s}\mathrm{u}\mathrm{p}
n\rightarrow \infty
\| xn+1\| 1/n \leq \varepsilon .
So, x1 \in H\varepsilon (T ) and, therefore, \scrX T (\BbbD (0, \varepsilon )) \subseteq H\varepsilon (T ) + T (0).
Theorem 5.1 is proved.
As a consequence of Theorem 5.1, we prove that the quasinilpotent part of a linear relation may
be characterized in terms of the glocal spectral subspace as follows.
Corollary 5.1. For every T \in BCR(X), we have
\scrX T (\{ 0\} ) \supset H0(T ) + T (0).
If moreover \sigma (T ) is bounded, then the equality holds.
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LOCAL SPECTRAL THEORY AND SURJECTIVE SPECTRUM . . . 235
6. Some properties of the surjective spectrum of a linear relation. Let T be a linear relation
in BCR(X). Recall that the surjective spectrum of T is defined by
\sigma su(T ) :=
\bigl\{
\lambda \in \BbbC : \lambda I - T is not surjective
\bigr\}
.
The next lemma is a consequence of [3] (Corollary 1.4.3).
Lemma 6.1. Let T \in BCR(X). Then \sigma su(T ) is a closed subset in \BbbC .
The surjective spectrum may be characterized by means of the local spectrum as follows.
Theorem 6.1. Let T \in BCR(X). Then we have:
1) \sigma su(T ) =
\bigcup
x\in X
\sigma T (x);
2) the set \{ x \in X such that \sigma T (x) = \sigma su(T )\} is of the second category in X.
Proof. 1. First observe that the desired result is equivalent to \rho su(T ) =
\bigcap
x\in X
\rho T (x). For the
opposite inclusion, let \lambda \in
\bigcap
x\in X
\rho T (x). Then \lambda \in \rho T (x) for all x \in X. Let x \in X. There exists
an open neighborhood U\lambda of \lambda and an analytic function fx : U\lambda \rightarrow X such that
(\mu I - T )fx(\mu ) = x+ T (0) \forall \mu \in U\lambda .
This implies that x \in (\lambda I - T )fx(\lambda ) \subset R(\lambda I - T ) for all x \in X. Then (\lambda I - T ) is surjective.
Thus, \lambda \in \rho su(T ).
For the direct inclusion, let \lambda \in \rho su(T ). Then (\lambda I - T ) is surjective. So, by Proposition 2.1,
we have K \prime (\lambda I - T ) = X and, from Theorem 4.1, we obtain that 0 \in \rho \lambda I - T (x). So, there exist an
open neighborhood U0 of 0 and an analytic function f : U0 \rightarrow X such that
((\mu - \lambda )I + T )f(\mu ) = x+ T (0) \forall \mu \in U0.
Therefore,
(\delta I - T )g(\delta ) = x+ T (0) \forall \delta \in U\lambda ,
where U\lambda be the open neighborhood of \lambda given by U\lambda = \lambda - U0 and g be the analytic function
defined on U\lambda by g(\delta ) = - f(\lambda - \delta ). Therefore, \lambda \in \rho T (x) for all x \in X, which ends the proof.
2. Let E be a dense countable subset of \sigma su(T ). Then, for each \lambda \in E, we have (\lambda I - T )X \not = X.
We claim that (\lambda I - T )X is of the first category in X. Indeed, let us suppose that (\lambda I - T )X is of
the second category in X. We show that (\lambda I - T )X = X, which is absurd. To do this, it suffices
to prove that (\lambda I - T ) is an open mapping. Let U be the open ball in X with center 0 and radius
r > 0. We prove that (\lambda I - T )(U) contains a neighborhood of 0 in X. Define
Un := \{ x \in X : \| x\| < 2 - nr\} , n = 0, 1, 2, . . . .
We note that U1 \supset U2 - U2. So, (\lambda I - T )U1 \supset (\lambda I - T )U2 - (\lambda I - T )U2. Hence,
(\lambda I - T )U1 \supset (\lambda I - T )U2 - (\lambda I - T )U2 \supset (\lambda I - T )U2 - (\lambda I - T )U2. (6.1)
On the other hand, we have
(\lambda I - T )X =
\infty \bigcup
k=1
k(\lambda I - T )(U2).
ISSN 1027-3190. Укр. мат. журн., 2021, т. 73, № 2
236 M. MNIF, A.-A. OULED-HMED
As, the union of countably many first category sets is first category and as (\lambda I - T )X is of the second
category, then at least one k(\lambda I - T )(U2) is of the second category of X and, so, (\lambda I - T )(U2) is
of the second category. Hence, \mathrm{i}\mathrm{n}\mathrm{t}((\lambda I - T )(U2)) \not = \varnothing . Thus, by (6.1), there is some neighborhood
W of 0 in X such that
W \subset (\lambda I - T )(U1).
We shall now show that (\lambda I - T )(U1) \subset (\lambda I - T )(U). Fix y1 \in (\lambda I - T )(U1). As what just proved
for U1 holds by the same way for U2, so, (\lambda I - T )(U2) contains a neighborhood of 0. Thus,\bigl(
y1 - (\lambda I - T )(U2)
\bigr)
\cap (\lambda I - T )(U1) \not = \varnothing .
Hence, there exists \alpha 1 \in (\lambda I - T )(x1) with x1 \in U1 such that y2 = y1 - \alpha 1 \in (\lambda I - T )(U2).
Proceeding by induction we can construct the sequences (\alpha n)n\geq 1, (xn)n\geq 1 and (yn)n\geq 1 such that
for all n \geq 1, xn \in Un,
\alpha n \in (\lambda I - T )(xn) and yn+1 = yn - \alpha n \in (\lambda I - T )(Un+1).
Since, \| xn\| <
r
2n
, then
\sum
n\geq 1
xn converges. Let x =
\sum \infty
n=1
xn. Then \| x\| < r and, so, x \in U.
By the construction of the sequences (yn)n\geq 1 and (\alpha n)n\geq 1, we have
m\sum
n=1
QT\alpha n =
m\sum
n=1
QT (yn - yn+1) = QT y1 - QT ym+1.
So,
\sum m
n=1
QT (\lambda I - T )xn = QT y1 - QT ym+1. Thus,
QT (\lambda I - T )
\Biggl(
m\sum
n=1
xn
\Biggr)
= QT y1 - QT ym+1.
Now, we claim that QT ym \rightarrow 0 as m \rightarrow \infty . Indeed, let \alpha \in (\lambda I - T )(U1). Then there exists
\beta \in U1 such that \alpha \in (\lambda I - T )(\beta ). Hence, \mathrm{d}\mathrm{i}\mathrm{s}(\alpha , T (0)) = \| (\lambda I - T )(\beta )\| \leq \| (\lambda I - T )\| \| \beta \| .
Thus, we obtain
(\lambda I - T )(U1) \subset
\Bigl\{
\alpha \in X : \mathrm{d}\mathrm{i}\mathrm{s}(\alpha , T (0)) \leq \| (\lambda I - T )\| r
2n
\Bigr\}
.
But, the map \alpha \mapsto \rightarrow \mathrm{d}\mathrm{i}\mathrm{s}(\alpha , T (0)) is continuous on X. Then the set\Bigl\{
\alpha \in X : \mathrm{d}\mathrm{i}\mathrm{s}(\alpha , T (0)) \leq \| (\lambda I - T )\| r
2n
\Bigr\}
is closed and we get
yn \in (\lambda I - T )(Un) \subset
\Bigl\{
\alpha \in X : \mathrm{d}\mathrm{i}\mathrm{s}(\alpha , T (0)) \leq \| (\lambda I - T )\| r
2n
\Bigr\}
.
So, \mathrm{d}\mathrm{i}\mathrm{s}(yn, T (0)) \leq \| (\lambda I - T )\| r
2n
. Therefore \| QT yn\| \rightarrow 0 as n \rightarrow \infty . Then we conclude that
QT (\lambda I - T )(x) = QT (y1) and, hence, y1 \in (\lambda I - T )(x), which confirm our claim.
Now, since, for all \lambda \in E, we have (\lambda I - T )X is of the first category, then F =
\bigcup
\lambda \in E
(\lambda I - T )X
is also of the first category. Thus, X\setminus F is of the second category. We claim that, for all x \in X\setminus F,
ISSN 1027-3190. Укр. мат. журн., 2021, т. 73, № 2
LOCAL SPECTRAL THEORY AND SURJECTIVE SPECTRUM . . . 237
\sigma su(T ) \subset \sigma T (x). Indeed, let x \in X\setminus F then E \subset \sigma T (x). So,
\sigma su(T ) = E \subset \sigma T (x) = \sigma T (x).
Hence, \sigma su(T ) = \sigma T (x). Therefore, \{ x \in X such that \sigma T (x) = \sigma su(T )\} is of the second category
in X.
Theorem 6.1 is proved.
We can now state the main result of this section, which is a sufficient condition to ensure that the
surjective spectrum \sigma su(T ) is not cluster at a point \lambda .
Theorem 6.2. Let T \in BCR(X). If X = H0(\lambda I - T ) +K \prime (\lambda I - T ), then \sigma su(T ) does not
cluster at \lambda .
Proof. Without loss of generality we can suppose that \lambda = 0. Suppose that 0 \in \sigma su(T ) and
X = H0(T ) +K \prime (T ). Then every x \in X may be written as x = x1 + x2, where x1 \in H0(T ) and
x2 \in K \prime (T ). From Corollary 5.1 we have \sigma T (x1) \subseteq \{ 0\} . Therefore,
\sigma T (x) \subseteq \sigma T (x1) \cup \sigma T (x2) \subseteq \{ 0\} \cup \sigma T (x2).
Now, by Theorem 4.1 and since \sigma T (x2) is closed, we conclude that 0 is isolated in \sigma T (x) for any
x \in X. By using Theorem 6.1, we conclude that there exists x0 \in X such that \sigma T (x0) = \sigma su(T ).
Hence, 0 is isolated in \sigma su(T ).
Theorem 6.2 is proved.
We close this section by stating the next corollary which is related to Theorems 6.1 and 6.2.
Corollary 6.1. Let T \in BCR(X). Then, if X = H0(\lambda I - T )+K \prime (\lambda I - T ), \sigma T (x) does not
cluster at \lambda for every x \in X.
References
1. P. Aiena, Fredholm and local spectral theory, with applications to multipliers, Kluwer Acad. Publ. (2004).
2. P. Aiena, Fredholm theory and localized SVEP, Funct. Anal., Approxim. and Comput., 7, № 2, 9 – 58 (2015).
3. E. Chafai, Ascent, descent and some perturbation results for linear relation: Doctor. Thesis, Univ. Sfax (2013).
4. E. Chafai, M. Mnif, Perturbation of normally solvable linear relations in paracomplete space, Linear Algebra and
Appl., 439, 1875 – 1885 (2013).
5. R. W. Cross, Multivalued linear operators, Marcel Dekker, New York (1998).
6. N. Dunford, Spectral theory. II. Resolution of the identity, Pacif. J. Math., 2, 559 – 614 (1952).
7. N. Dunford, Spectral operators, Pacif. J. Math., 4, 321 – 354 (1954).
8. K. B. Laursen, M. M. Neuman, An introduction to local spectral theory, London Math. Soc. Monogr., 20, Clarendon
Press, Oxford (2000).
Received 09.05.18
ISSN 1027-3190. Укр. мат. журн., 2021, т. 73, № 2
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| id | umjimathkievua-article-81 |
| institution | Ukrains’kyi Matematychnyi Zhurnal |
| keywords_txt_mv | keywords |
| language | English |
| last_indexed | 2026-03-24T02:04:08Z |
| publishDate | 2021 |
| publisher | Institute of Mathematics, NAS of Ukraine |
| record_format | ojs |
| resource_txt_mv | umjimathkievua/6e/4c8236b2a758c597b6acb6563328f86e.pdf |
| spelling | umjimathkievua-article-812025-03-31T08:48:28Z The Local spectral theory and surjective spectrum of linear relations Local spectral theory and surjective spectrum of linear relations Local spectral theory and surjective spectrum of linear relations Mnif, M. Ouled-Hmed, A.-A. Mnif, Maher Ouled-Hmed, A.-A. Mnif, M. Ouled-Hmed, A.-A. Linear relations local spectrum surjective spectrum correlation analytic core local and glocal spectral subspaces Linear relations local spectrum surjective spectrum correlation analytic core local and glocal spectral subspaces UDC 517.98 This paper initiates a study of local spectral theory for linear relations. At the beginning, we define the local spectrum and study its properties. Then we obtain results related to the correlation analytic core $K\prime (T)$ and quasinilpotent part $H_0(T)$ of a linear relation $T$ in a Banach space $X$. As an application, we give a characterization of the surjective spectrum $\sigma_{su}(T)$ in terms of the local spectrum and show that if $X = H_0(\lambda I - T) + K\prime (\lambda I - T)$, then $\sigma_{su}(T)$ does not cluster at $\lambda$. УДК 517.98 Теорія локального спектра та сюр'єктивний спектр лінійних відношень Ця робота починає вивчення теорiї локального спектру для лiнiйних вiдношень. Спочатку дається визначення та вивчаються властивостi локального спектру. Пiсля цього отримано деякi результати, що вiдносяться до кореляцiйного аналiтичного ядра $K\prime (T)$ та квазiнiльпотентної частини $H_0(T)$ лiнiйного вiдношення $T$ у банаховому просторi $X$.Як застосування дається характеризацiя сюр’єктивного спектру $\sigma_{su}(T)$ у термiнах локального спектру та доведено, що якщо $X = H_0(\lambda I - T) + K\prime (\lambda I - T)$, то $\sigma_{su}(T)$ не кластеризується для $\lambda$. Institute of Mathematics, NAS of Ukraine 2021-02-22 Article Article application/pdf https://umj.imath.kiev.ua/index.php/umj/article/view/81 10.37863/umzh.v73i2.81 Ukrains’kyi Matematychnyi Zhurnal; Vol. 73 No. 2 (2021); 222 - 237 Український математичний журнал; Том 73 № 2 (2021); 222 - 237 1027-3190 en https://umj.imath.kiev.ua/index.php/umj/article/view/81/8940 Copyright (c) 2021 Maher Mnif, Aman-Allah Ouled-Hmed |
| spellingShingle | Mnif, M. Ouled-Hmed, A.-A. Mnif, Maher Ouled-Hmed, A.-A. Mnif, M. Ouled-Hmed, A.-A. The Local spectral theory and surjective spectrum of linear relations |
| title | The Local spectral theory and surjective spectrum of linear relations |
| title_alt | Local spectral theory and surjective spectrum of linear relations Local spectral theory and surjective spectrum of linear relations |
| title_full | The Local spectral theory and surjective spectrum of linear relations |
| title_fullStr | The Local spectral theory and surjective spectrum of linear relations |
| title_full_unstemmed | The Local spectral theory and surjective spectrum of linear relations |
| title_short | The Local spectral theory and surjective spectrum of linear relations |
| title_sort | local spectral theory and surjective spectrum of linear relations |
| topic_facet | Linear relations local spectrum surjective spectrum correlation analytic core local and glocal spectral subspaces Linear relations local spectrum surjective spectrum correlation analytic core local and glocal spectral subspaces |
| url | https://umj.imath.kiev.ua/index.php/umj/article/view/81 |
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