2. Notation and Complementary Results
A bounded operator
on a complex infinite dimensional Banach space
X is said to have
the single valued extension property at
. In short,
T has the SVEP at
, if for every open disc
centered at
the only analytic function
which satisfies the equation
is the constant function
.
T is said to have the SVEP if T has the SVEP for every .
To facilitate the reader, we remember that the SVEP is a typical tool of the local spectral theory. If
denote the local resolvent set of
T at the point
, defined as the union of all open subsets
of
for which there exists an analytic function
that satisfies
then the local spectrum
of
T at
x is defined by
and, obviously,
, where
denotes the spectrum of
T.
Remark 1. Let and denotes an open neighborhood of λ. If satisfies the equation on , then for all (see [7], Lemma 1.2.14). Moreover, if and only if . Theorem 1. Let , X a Banach space. Then, T has SVEP if and only if every the local spectrum is non-empty.
Proof. See ([
7], Proposition 1.2.16). □
The SVEP has a decisive role in local spectral theory it has a certain interest to find conditions for which an operator has the SVEP.
Definition 1. Let T is a linear operator on a vector space X. The hyperrange
of T is the subspace Generally,
, thus we are interested in finding conditions for which
. For every linear operator
T on a vector space
X, there corresponds the two chains:
and
The ascent of T is the smallest positive integer , whenever it exists, such that = . If such p does not exist, we let . Analogously, the descent of T is defined to be the smallest integer , whenever it exists, such that = . If such q does not exist, we let .
It is possible to prove that, if and are both finite, then Note that means that T is injective, and that T is surjective.
Theorem 2. If and X is a Banach space, then As noted in [
1] (Lemma 1.1), the local spectrum of
and
x may differ only at
i.e.,
For every
and
, we have
Moreover, if
T is injective, then
For every subset
the
analytic spectral subspace of
T associated with
F is the set
For every subset
the
global spectral subspace consists of all
for which there exists an analytic function
that satisfies
In general,
for every closed sets
The identity
holds for all closed sets
whenever
T has SVEP, precisely.
T has SVEP if and only if
holds for all closed sets
Definition 2. The analytical core of is the set The analytic core of an operator
T is an invariant subspace, which, in general, is not closed [
8].
Definition 3. An operator is said to be upper semi-Fredholm, if T(X) is closed and the kernel is finite-dimensional. An operator is said to be lower semi-Fredholm, if the range has finite codimension.
Definition 4. An operator is said to be Drazin invertible if there exist such that
- 1.
for some integer
- 2.
; and
- 3.
In this case, C is called Drazin inverse of T and the smallest in (4) is called the index of T.
3. Operator Equation
As mentioned in the Introduction, in this section, we show some results concerning the transmission of some local spectral properties from R to
We study the relationship between the local spectral properties of an operator R and the local spectral properties S, if this exists. In particular, we study a reciprocal relationship, analogous to that of (2). We also show that many local spectral properties, such as SVEP and Dunford property (C), are transferred from operator R to S somehow through a bond. While these properties are, in general, not preserved under sums and products of commuting operators, we obtain positive results in the case of our perturbations.
We suppose that
satisfy
for some integers
The case
and
is studied in [
1,
9,
10]; if
, the operators
A and
B are relatively regular.
Moreover, if
is
Drazin invertible operator with
, then, by (4),
Therefore, in this case,
and
Lemma 1. For every , we haveMoreover, Proof. Suppose that
; then, there exists an open neighborhood
if
and an analytic function
such that
From this, it then follows that
for all
Hence,
; thus,
To show the first inclusion (
11), let
; then, there exists an open neighborhood
of
and an analytic function
such that
Consequently,
for all
and since
is analytic, we obtain
Hence, this shows the first inclusion of (
11). To show the second inclusion, let
; then, there exists an open neighborhood
of
and an analytic function
such that
Consequently, the argument is similar to that first part. □
Theorem 3. Suppose that is a closed subset of and Then, is closed if and only if is closed.
Proof. Suppose that
is closed and let
be a sequence of
which converges to
. Then, for every
, we have
By (
10), we have
Since
, by (
6) where
, we have
Therefore,
i.e.,
By [
9] (Lemma 2.3),
and by assumption
is closed. We then have
, i.e.,
By (
11),
Then,
by [
9] (Lemma 2.3)
thus
is closed. Conversely, suppose that
is closed and let
be a sequence of
which converges to
; then,
for every
By (
11),
, and then
By [
9] (Lemma 2.3)
therefore
Hence,
Since by (
10)
for all
then, if
, we have
. By (
6), we have
i.e.,
Hence,
i.e.,
By [
9] (Lemma 2.3)
The following result is inspired by [
1] and ([
11], Theorem 2.1). □
Lemma 2. Let be such that for same integers If has SVEP, then and have SVEP.
Proof. By ([
12], Proposition 2.1),
has SVEP if and only if
has Svep. Suppose that
has SVEP at
and let
be an analytic function for which
for all
Then,
The SVEP of
at
implies that
and hence
Thus, if
, then
for
and by continuity
Therefore,
has SVEP at
□
We now consider the case where
Theorem 4. Let be a closed subset of such that Suppose that satisfy for some integers and has SVEP. If is closed, then is closed.
Proof. Let
; by assumption,
is closed. By (3),
is closed. By (2),
has SVEP; therefore, by ([
9], Lemma 1.4),
is closed. □
Definition 5. An operator is said to have Dunford’s property (abbreviated property ) if is closed for every closed set
It is known that Dunford property (C) entails SVEP for T.
Theorem 5. Let be such that for some integers If has the property (C), then and have the property (C).
Proof. Suppose that is a closed set and has property (C); then, has SVEP. If , by (3) and by assumptions is closed, it follows that is closed. Similarly, if , then by (4) we have that is closed. Therefore, has property (C). □
We prove that somehow there exists a bond, i.e., SR and RS share Dunford’s property (C) when for same integers
Definition 6. An operator is said to have property if the quasi-nilpotent part
of defined by is closed for every
It is known that
and moreover for operator
T we have
Every multiplier of a semi-simple commutative Banach algebra has property
, see ([
13], Theorem 1.8), in particular every convolution operator
,
, on the group algebra
has property
, but there are convolution operators which do not enjoy property
(see [
7], Chapter 4).
Observe that, if
T has property
and
f is an injective analytic function defined on an open neighborhood
U of
, then
also has property
. To see this, recall first that the equality
holds for every closed subset of
and every analytic function
f on an open neighborhood
U of
, see ([
7], Theorem 3.3.6). Now, to show that
has property
amd
f is injective, we have to prove that
is closed for every
. If
, then
, while, if
, then
where
, and, consequently,
is closed. In particular, considering the function
, we see that, if
T is invertible and has property
, then its inverse has property
. Furthermore, property
for
T implies property
for
, for every
.
Theorem 6. Let be such that for some integers If has the property , then has the property .
Proof. Suppose that
has property
. Then,
has SVEP, hence by Lemma 2
has SVEP. Therefore, by (
13) and by assumption,
is closed for every
By (
13) and (3),
is closed. Following the procedure of [
1], let
; by ([
7], Proposition 3.3.1, part (f)) we have
Since
is upper semi-Fredholm, the SVEP at 0 implies that
is finite-dimensional (see [
8], Theorem 3.18). Then,
is closed. By Theorem 5, we then have
is closed, therefore
has property
□
Following the procedure of [
1] (Theorem 3), it is possible to prove the following:
Theorem 7. Let be such that for same integers
- 1.
(i) If , then is closed if and only is closed, or equivalently is closed.
- 2.
(ii) If is injective, then is closed if and only is closed, or equivalently is closed for all
Corollary 1. Suppose for some integers and Then, the following statements are equivalent:
- 1.
is closed.
- 2.
is closed.
- 3.
is closed.
- 4.
is closed.
When R is injective, the equivalence also holds for
Proof. The equivalence of (3) and (4) follows from Theorem 3. Since, the injectivity of R is equivalent to the injectivity of S, the equivalence of (1) and (4) also holds for . □
We show now that property is also transmitted between operators R into Let be such that for some integers If R has the property and has the property , then has the property , therefore has the property , thus S has the property .
4. Example: Drazin Invertible Operators
In this section, we give an example that plays a crucial role for the theory, of operators that satisfy the equation for some integers
In the literature, the concept of invertibility admits several generalizations. Another generalization of the notion of invertibility, which satisfies the relationships of “reciprocity” observed above, is provided by the concept of Drazin invertibility.
The concept of Drazin invertibility has been introduced in a more abstract setting than operator theory [
14]. In the case of the Banach algebra
,
is said to be
Drazin invertible (with a finite index) if there exists an operator
and
such that
The smallest nonnegative integer
such that (
15) holds is called the index
of
R. In this case, the operator
S is called
Drazin inverse of
R.
Clearly, any invertible operator or a nilpotent operator R is Drazin invertible.