Left and right B-fredholm operators
UDC 517.9 We introduce the families of left and right B-Fredholm operators in Banach space, realize their stabilization with the help of finite-rank operators, and  prove a spectral mapping theorem for the left and right B-Fredholm operators.
Gespeichert in:
| Datum: | 2022 |
|---|---|
| Hauptverfasser: | , |
| Format: | Artikel |
| Sprache: | Englisch |
| Veröffentlicht: |
Institute of Mathematics, NAS of Ukraine
2022
|
| Online Zugang: | https://umj.imath.kiev.ua/index.php/umj/article/view/6331 |
| Tags: |
Tag hinzufügen
Keine Tags, Fügen Sie den ersten Tag hinzu!
|
| Назва журналу: | Ukrains’kyi Matematychnyi Zhurnal |
| Завантажити файл: |  |
Institution
Ukrains’kyi Matematychnyi Zhurnal| _version_ | 1860512333523058688 |
|---|---|
| author | Abdmouleh, F. Ben Lakhal, T. Abdmouleh, F. Ben Lakhal, T. |
| author_facet | Abdmouleh, F. Ben Lakhal, T. Abdmouleh, F. Ben Lakhal, T. |
| author_sort | Abdmouleh, F. |
| baseUrl_str | https://umj.imath.kiev.ua/index.php/umj/oai |
| collection | OJS |
| datestamp_date | 2023-01-07T13:45:40Z |
| description |
UDC 517.9
We introduce the families of left and right B-Fredholm operators in Banach space, realize their stabilization with the help of finite-rank operators, and  prove a spectral mapping theorem for the left and right B-Fredholm operators. |
| doi_str_mv | 10.37863/umzh.v74i10.6331 |
| first_indexed | 2026-03-24T03:27:07Z |
| format | Article |
| fulltext |
DOI: 10.37863/umzh.v74i10.6331
UDC 517.9
F. Abdmouleh1, T. Ben Lakhal (Univ. Sfax, Ins. Supér. Gestion Industrielle Sfax, Tunisia)
LEFT AND RIGHT B-FREDHOLM OPERATORS
ЛIВИЙ ТА ПРАВИЙ B-ФРЕДГОЛЬМОВI ОПЕРАТОРИ
We introduce the families of left and right B-Fredholm operators in Banach space, realize their stabilization with the help
of finite-rank operators, and prove a spectral mapping theorem for the left and right B-Fredholm operators.
Уведено сiм’ї лiвих i правих B-фредгольмових операторiв у банаховому просторi, проведено стабiлiзацiю за до-
помогою операторiв скiнченного рангу та доведено теорему про спектральне вiдображення для лiвих i правих
B-фредгольмових операторiв.
1. Introduction. M. Berkani in [1] studied the class of B-Fredholm operators on a Banach space.
This class is defined by: If we have T a bounded linear operator acting on the Banach space X
and for each integer n, then we define the restriction from T to R(Tn) denoted by Tn viewed as
Tn = T | R(Tn) : R(Tn) \rightarrow R(Tn) (for n = 0, T0 = T ). Now, we say that T is a B-Fredholm
operator if for some integer n the range space R(Tn) is closed and Tn is a Fredholm operator, in the
sense of having null space \scrN (Tn) of finite dimension \alpha (Tn) and range \scrR (Tn) of finite codimension
\beta (Tn), the difference \mathrm{i}\mathrm{n}\mathrm{d}(Tn) = \alpha (Tn) - \beta (Tn) = \mathrm{i}\mathrm{n}\mathrm{d}(T ) is known as the index of B-Fredholm
operator T (see [1]). M. Berkani and M. Sarih extended in [2] this notion and they given the class
of semi-B-Fredholm for which Tn is either upper or lower semi-Fredholm, in the sense that either
\scrN (Tn) is finite dimensional and \scrR (Tn) closed, or \scrR (Tn) is closed of finite codimension. In this
paper, we extend our research to ”left and right B-Fredholm operators”. We say that T is a left
Fredholm operator if \scrR (T ) is closed, \alpha (T ) < \infty and \scrR (T ) is a complemented subspace of X, and
we call T a right Fredholm operator if \beta (T ) < \infty and \scrN (T ) is a complemented subspace of X.
The notion of left and right Fredholm operators was introduced by the several mathematicians, for
example, in [3] A. A. Boichuk, A. M. Samoilenko studied this notion. We shall see that the left
B-Fredholm operator \scrB \scrF l(X) on a Banach space X in general properly contain the left Fredholm
operator \Phi l(X), and the right B-Fredholm operator \scrB \scrF r(X) on a Banach space X contain the
left Fredholm operator \Phi r(X). And we show that each a left B-Fredholm (resp., right B-Fredholm)
operator is a quasi-Fredholm operator in the sense of M. Mbekhta and V. Muller in [7]. Conversely,
a quasi-Fredholm operator such as there exists d such that \scrR (Tn) is a closed subspace of X for
each integer n \geq d and \scrR (T ) +\scrN (T d) is a closed subspace of X, is a left B-Fredholm (resp., right
B-Fredholm) operator.
In Theorem 2.1 and in the case of operators acting on a Hilbert space H we prove that T \in \scrL (H)
is a left B-Fredholm (resp., right B-Fredholm) operator if and only if T = Q \oplus F, where Q is a
nilpotent operator and F is a left Fredholm (resp., right Fredholm) operator. In Proposition 2.4, we
1 Corresponding author, e-mail: faical_abdmouleh@yahoo.fr.
c\bigcirc F. ABDMOULEH, T. BEN LAKHAL, 2022
ISSN 1027-3190. Укр. мат. журн., 2022, т. 74, № 10 1299
1300 F. ABDMOULEH, T. BEN LAKHAL
prove that if T is a left B-Fredholm (resp., right B-Fredholm) operator and if F is a finite dimensional
operator then T + F is also a left B-Fredholm (resp., right B-Fredholm) operator.
In the third section, we prove the stability of a left and a right B-Fredholm operators, we show on
Theorems 3.1 and 3.2 that if T and S are two left B-Fredholm (resp., right B-Fredholm) operators
and the condition TA + ST = I is satisfy, then TS is a left B-Fredholm (resp., right B-Fredholm)
operators. Conversely, if TS is a left B-Fredholm (resp., right B-Fredholm) operator, then T and S
are left B-Fredholm (resp., right B-Fredholm) operators such that TA + ST = I. Also, we prove a
spectral mapping theorem for left and right B-Fredholm operators, more precisely in Theorem 3.3, for
T \in \scrL (X) and f an analytic function on the usual spectrum \sigma (T ) of T, we prove that f
\bigl(
\sigma \scrB \scrF l
(T )
\bigr)
=
= \sigma \scrB \scrF l
(f(T )), where \sigma \scrB \scrF l
(T ) = \{ \lambda \in \BbbC such that (T - \lambda I) /\in \scrB \scrF l(X)\} , and f(\sigma \scrB \scrF r(T )) =
= \sigma \scrB \scrF r(f(T )), where \sigma \scrB \scrF r(T ) = \{ \lambda \in \BbbC such that (T - \lambda I) /\in \scrB \scrF r(X)\} .
In the sequel if E and F are two vector spaces, the notation E \simeq F will mean that E and F are
isomorphic. If E and F are vector subspaces of the same vector space H we shall write E =e F
if there exist two finite dimensional vector subspaces G1 and G2 of H such that E \subset F +G1 and
F \subset E + G2. Next, if E \subset F then we denote the quotient space E modulo F by
E
F
(see [4],
Definition 1).
2. Definition and properties of left and right B-Fredholm operators.
Proposition 2.1. Let T \in \scrL (X). If there exists an integer n \in \BbbN such that \scrR (Tn) is closed
and the operator Tn is a left Fredholm operator, then \scrR (Tm) is closed and the operator Tm is a
left Fredholm operator and \mathrm{i}\mathrm{n}\mathrm{d}(Tm) = \mathrm{i}\mathrm{n}\mathrm{d}(Tn) for each m \geq n.
Proof. If Tn : \scrR (Tn) \rightarrow \scrR (Tn) is a left Fredholm operator, then Tn is upper semi-Fredholm
operator, so, for each m \geq n, the operator Tm - n
n : \scrR (Tn) \rightarrow \scrR (Tn) is also an upper semi-
Fredholm operator. Hence, \scrR (Tm - n
n ) = \scrR (Tm) is closed in \scrR (Tn). Since \scrR (Tn) is closed in
X, then \scrR (Tm) is closed in X. Consider now the operator Tm : \scrR (Tm) \rightarrow \scrR (Tm). We have
\scrN (Tm) = \scrN (T ) \cap \scrR (Tm) \subset \scrN (T ) \cap \scrR (Tn) = \scrN (Tn). So, \alpha (Tm) < \infty .
If the operator Tn is a left Fredholm operator, then \scrR (Tn) is a complemented subspace of \scrR (Tn).
Since \scrN (Tm - n
n ) is of finite dimension, then \scrR (Tn) +\scrN (Tm - n
n ) is also a complemented subspace
of \scrR (Tn).
This means that there exists a finite dimensional subspace F1 of \scrR (Tn) such that
\scrR (Tn) = F1 \oplus
\bigl(
\scrR (Tn) +\scrN (Tm - n
n )
\bigr)
.
Then \scrR (Tm) = Tm - n(F1) + Tm - n
\bigl(
\scrR (Tn)
\bigr)
.
First, it is known that the image of a closed subspace by an operator upper semi-Fredholm
operator is closed, then Tm - n(F1) is a closed subspace of \scrR (Tm). It remains to show that the sum
is direct: Let z \in Tm - n(F1) \cap Tm - n
\bigl(
\scrR (Tn)
\bigr)
. Then there exist x \in F1 and y \in \scrR (Tn) such that
z = Tm - n(x) = Tm - n(y). We obtain x - y \in \scrN (Tm - n
n ), therefore, x = y + (x - y) \in (\scrR (Tn) +
+ \scrN (Tm - n
n )) \cap F1 = \{ 0\} . Hence, x = 0 and therefore z = 0, whence \scrR (Tm) = Tm - n(F1) +
+\scrR (Tm).
Thus, \scrR (Tm) is a complemented subspace of \scrR (Tm). Consequently, Tm is a left Fredholm
operator.
Moreover, from [4] (Lemma 3.5), we have
\scrN (T ) \cap \scrR (Tn)
\scrN (T ) \cap \scrR (Tn+1)
\sim =
\scrN (Tn+1) +\scrR (T )
\scrN (Tn) +\scrR (T )
.
ISSN 1027-3190. Укр. мат. журн., 2022, т. 74, № 10
LEFT AND RIGHT B-FREDHOLM OPERATORS 1301
Also, from [4] (Lemma 3.2), we get
\scrR (Tn)
\scrR (Tn+1)
\sim =
X
\scrR (T ) +\scrN (Tn)
and
\scrR (Tn+1)
\scrR (Tn+2)
\sim =
X
\scrR (T ) +\scrN (Tn+1)
.
Hence, \alpha (Tn) - \alpha (Tn+1) = \beta (Tn) - \beta (Tn+1), which means that \mathrm{i}\mathrm{n}\mathrm{d}(Tn) = \mathrm{i}\mathrm{n}\mathrm{d}(Tn+1). It follows
then that \mathrm{i}\mathrm{n}\mathrm{d}(Tm) = \mathrm{i}\mathrm{n}\mathrm{d}(Tn) for each m \geq n.
Proposition 2.2. Let T \in \scrL (X). If there exists an integer n \in \BbbN such that \scrR (Tn) is closed
and the operator Tn is a right Fredholm operator, then \scrR (Tm) is closed, the operator Tm is a right
Fredholm operator and \mathrm{i}\mathrm{n}\mathrm{d}(Tm) = \mathrm{i}\mathrm{n}\mathrm{d}(Tn) for each m \geq n.
Proof. In the same way as the previous proposition we show that if \scrR (Tn) is closed, then
\scrR (Tm) is closed. For n \in \BbbN , suppose that Tn is a right Fredholm operator. We shall show that
Tn+1 is a right Fredholm operator.
Tn is a right Fredholm operator, then \mathrm{c}\mathrm{o}\mathrm{d}\mathrm{i}\mathrm{m}(\scrR (Tn)) < \infty in \scrR (Tn). So, there exists F a finite
dimensional subspace of \scrR (Tn) such that \scrR (Tn) = F \oplus \scrR (Tn) = F \oplus \scrR (Tn+1). This implies that
the injection in : \scrR (Tn+1) \rightarrow \scrR (Tn) and the projection pn : \scrR (Tn) \rightarrow \scrR (Tn+1) are both Fredholm
operators. We can easily check that Tn+1 = pn \circ T \circ in. Hence, if Tn is a right Fredholm operator,
then Tn+1 is also right Fredholm operator. Consequently, if Tn is a right Fredholm operator, then
Tm is likewise right Fredholm operator. We get the equality of the index by the same way as in the
proof of the previous proposition.
Definition 2.1. Let T \in \scrL (X).
(i) If , for some integer n \in \BbbN , \scrR (Tn) is closed and the operator Tn is a left Fredholm operator,
then T is called a left B-Fredholm operator.
(ii) If , for some integer n \in \BbbN , \scrR (Tn) is closed and the operator Tn is a right Fredholm
operator, then T is called a right B-Fredholm operator.
Observe from the definition of left and right B-Fredholm operators all nilpotent operators and
all bounded linear projections are left and right B-Fredholm operators. Hence the class \scrB \scrF l(X)
(resp., \scrB \scrF r(X)) of left B-Fredholm (resp., right B-Fredholm) operators contains the class of left
Fredholm (resp., right Fredholm) operators as a proper subclass. Note also that obviously every B-
Fredholm operator is a left B-Fredholm (resp., right B-Fredholm) and every left B-Fredholm (resp.,
right B-Fredholm) operator is upper semi B-Fredholm (resp., lower semi-B-Fredholm).
Definition 2.2. Let T \in \scrL (X) be a left (resp., right) B-Fredholm operator and n any integer
such that \scrR (Tn) is closed and Tn is a left (resp., right) Fredholm operator. Then we define the
index of T denote by \mathrm{i}\mathrm{n}\mathrm{d}(T ) as the index of the left (resp., right) Fredholm operator Tn. From
Propositions 2.1 and 2.2, this definition is independent of the choice of the integer n. Furthermore,
if T is a Fredholm operator, this reduces to the usual definition of the index.
Definition 2.3 [5]. Let T \in \scrL (X) and
\Delta (T ) =
\bigl\{
n \in \BbbN ; \forall m \in \BbbN , m \geq n \Rightarrow (\scrR (Tn) \cap \scrN (T )) \subset (\scrR (Tm) \cap \scrN (T ))
\bigr\}
.
Then the degree of stable iteration \mathrm{d}\mathrm{i}\mathrm{s}(T ) of T is defined as \mathrm{d}\mathrm{i}\mathrm{s}(T ) = \mathrm{i}\mathrm{n}\mathrm{f}(\Delta (T )). If \Delta (T ) = \varnothing
then \mathrm{d}\mathrm{i}\mathrm{s}(T ) = \infty .
Definition 2.4. Let T \in \scrL (X). Then T is called a quasi-Fredholm operator of degree d if there
is an integer d \in \BbbN such that:
(i) \mathrm{d}\mathrm{i}\mathrm{s}(T ) = d,
(ii) \scrR (Tn) is a closed subspace of X for each integer n \geq d,
ISSN 1027-3190. Укр. мат. журн., 2022, т. 74, № 10
1302 F. ABDMOULEH, T. BEN LAKHAL
(iii) \scrR (T ) +\scrN (T d) is a closed subspace of X.
In the sequel \scrQ \scrF (d) will denote the set of quasi-Fredholm operators of degree d.
Proposition 2.3. Let T \in \scrL (X). Then T is a left (resp., right) B-Fredholm operator if and only
if there exists an integer d \in \BbbN such that T \in \scrQ \scrF (d) and:
(i) \mathrm{d}\mathrm{i}\mathrm{m}(\scrR (T d) \cap \scrN (T )) < \infty (resp., \mathrm{c}\mathrm{o}\mathrm{d}\mathrm{i}\mathrm{m}(\scrR (T ) +\scrN (T d)) < \infty ),
(ii) \scrR (T ) +\scrN (T d) (resp., \scrR (T d) \cap \scrN (T )) is a complemented subspace of \scrR (T d).
Proof. Suppose that T \in \scrB \scrF l(X). Then, there exists n \in \BbbN such that \scrR (Tn) is closed and Tn
is a left Fredholm operator in \scrR (Tn). Then \mathrm{d}\mathrm{i}\mathrm{m}(\scrR (Tn) \cap \scrN (T )) < \infty and \scrR (T ) + \scrN (Tn) is a
complemented subspace of \scrR (Tn).
Let m \geq n, then \scrR (Tm) \cap \scrN (T ) \subset \scrR (Tn) \cap \scrN (T ). Since \mathrm{d}\mathrm{i}\mathrm{m}(\scrR (Tn) \cap \scrN (T )) < \infty , the
sequence (\scrR (T p) \cap \scrN (T ))p is a stationary sequence for p large enough. Therefore,
d = \mathrm{d}\mathrm{i}\mathrm{s}(T ) \in \BbbN and \mathrm{d}\mathrm{i}\mathrm{m}
\bigl(
\scrR (T d) \cap \scrN (T )
\bigr)
< \infty .
If \scrR (T ) +\scrN (Tn) is a complemented subspace of \scrR (Tn), then there exists F \in \scrR (Tn) such that
\scrR (Tn) = F \oplus \scrR (T ) +\scrN (Tn).
We have, for each n \geq d, \scrR (T ) \subset \scrN (T d) + \scrR (T ) such that \scrR (T ) + \scrN (Tn) \subset \scrR (T ) +
+\scrN (T d). Since \scrN (T d) \subset \scrN (Tn), then \scrR (T )+\scrN (T d) \subset \scrR (T )+\scrN (Tn). This shows that \scrR (T )+
+\scrN (Tn) = \scrR (T ) +\scrN (T d). If \scrR (Tn) \subset \scrR (T d), then there exists F \in \scrR (T d) such that \scrR (T d) =
= F \oplus \scrR (T ) +\scrN (T d). Therefore, \scrR (T ) +\scrN (T d) is a complemented subspace of \scrR (T d).
As though, \scrR (Tm) is closed for each m \geq n, we deduced that \scrR (Tm) is closed for each
m \geq d. Moreover, we have \scrR (T ) + \scrN (T d) = (T d) - 1
\bigl(
\scrR (T d+1)
\bigr)
. Hence, \scrR (T ) + \scrN (T d) is
a closed subspace of X. Consequently, T \in \scrQ \scrF (d) such that \mathrm{d}\mathrm{i}\mathrm{m}(\scrR (T d) \cap \scrN (T )) < \infty and
\scrR (T ) +\scrN (T d) is a complemented subspace of \scrR (T d).
Conversely, suppose that T \in \scrQ \scrF (d) such that \mathrm{d}\mathrm{i}\mathrm{m}(\scrR (T d) \cap \scrN (T )) < \infty and \scrR (T ) +\scrN (T d)
is a complemented subspace of \scrR (T d). Thus, \scrR (Tn) is closed for each n \geq d, since, \mathrm{d}\mathrm{i}\mathrm{m}(\scrR (T d)\cap
\cap \scrN (T )) < \infty and \scrR (T ) + \scrN (T d) is a complemented subspace of \scrR (T d). Hence, Td is a left
Fredholm operator. Finally, T \in \scrB \scrF l(X).
Suppose that T \in \scrB \scrF r(X). Then there exists n \in \BbbN such that \scrR (Tn) is closed and Tn is a
right Fredholm operator of \scrR (Tn). Then \mathrm{c}\mathrm{o}\mathrm{d}\mathrm{i}\mathrm{m}(\scrR (T ) + \scrN (Tn)) < \infty and \scrR (T ) \cap \scrN (Tn) is a
complemented subspace of \scrR (Tn).
Let m \geq n, then \scrR (T )+\scrN (Tn) \subset \scrR (T )+\scrN (Tm). Since \mathrm{c}\mathrm{o}\mathrm{d}\mathrm{i}\mathrm{m}(\scrR (T )+\scrN (Tn)) < \infty , thus,
the sequence (\scrR (T ) +\scrN (T p))p is a stationary sequence for p large enough. This shows that
d = \mathrm{d}\mathrm{i}\mathrm{s}(T ) \in \BbbN and \mathrm{c}\mathrm{o}\mathrm{d}\mathrm{i}\mathrm{m}
\bigl(
\scrR (T ) +\scrN (T d)
\bigr)
< \infty .
If \scrR (T ) \cap \scrN (Tn) is a complemented subspace of \scrR (Tn), then there exists F \in \scrR (Tn) such that
\scrR (Tn) = F \oplus \scrR (T ) \cap \scrN (Tn). Hence \scrR (T ) \cap \scrN (T d) is a complemented subspace of \scrR (T d).
We have \mathrm{c}\mathrm{o}\mathrm{d}\mathrm{i}\mathrm{m}(\scrR (T ) + \scrN (Tn)) < \infty . Thus, \scrR (Tm) is closed for each m \geq n, and then
\scrR (Tm) is a closed for each m \geq d. Therefore, \scrR (T ) +\scrN (T d) is a closed subspace of X.
Hence, T \in \scrQ \scrF (d) such that \mathrm{c}\mathrm{o}\mathrm{d}\mathrm{i}\mathrm{m}(\scrR (T ) + \scrN (T d)) < \infty and \scrR (T ) \cap \scrN (T d) is a comple-
mented subspace of \scrR (T d).
Conversely, we suppose that T \in \scrQ \scrF (d) such that \mathrm{c}\mathrm{o}\mathrm{d}\mathrm{i}\mathrm{m}(\scrR (T ) +\scrN (T d)) < \infty and \scrR (T ) \cap
\cap \scrN (T d) is a complemented subspace of \scrR (T d). Thus, \scrR (Tn) is closed for each n \geq d, as though,
\mathrm{c}\mathrm{o}\mathrm{d}\mathrm{i}\mathrm{m}(\scrR (T )+\scrN (T d)) < \infty and \scrR (T )\cap \scrN (T d) is a complemented subspace of \scrR (T d). Therefore,
Td is a right Fredholm operator. Consequently, T \in \scrB \scrF r(X).
ISSN 1027-3190. Укр. мат. журн., 2022, т. 74, № 10
LEFT AND RIGHT B-FREDHOLM OPERATORS 1303
Theorem 2.1. Let X be an Hilbert space and T \in \scrL (X). Then T is a left B-Fredholm (resp.,
right B-Fredholm) operator if and only if there exist two closed subspaces M and N of X and an
integer d \in \BbbN such that:
(i) X = M \oplus N,
(ii) T (N) \subset N and T | N is a nilpotent operator,
(iii) T (M) \subset M and T | M is a left Fredholm (resp., right Fredholm) operator.
Proof. Since H is a Hilbert space, then each subspace of H admits a complemented. So
according to [2] (Theorem 2.6) we obtain the result.
Proposition 2.4. Let T \in \scrL (X) be a left B-Fredholm (resp., right B-Fredholm) operator and
F \in \scrL (X) be a finite rank operator. Then T +F is also a left B-Fredholm (resp., right B-Fredholm)
operator.
Proof. If T is a left B-Fredholm (resp., right B-Fredholm) operator, then T is an upper semi-
B-Fredholm (resp., lower semi-B-Fredholm) operator. Hence, from [2] (Proposition 2.7) we obtain
that T + F is an upper semi-B-Fredholm.
Moreover, we have \scrR
\bigl(
(T + F )n
\bigr)
= \scrR ((T + F )n+1). Since \scrR ((T + F )n+1) =e \scrR (Tn+1) =
= \scrR (Tn) and \scrR ((T + F )n) =e \scrR (Tn).
If T is left B-Fredholm, then \scrR (Tn) is a complemented subspace of \scrR (Tn) for some n \in \BbbN .
Thus, \scrR
\bigl(
(T + F )n
\bigr)
is a complemented subspace of \scrR ((T + F )n). Consequently, T + F is a left
B-Fredholm operator. Now suppose that T is a right B-Fredholm.
Let us show that \scrN
\bigl(
(T + F )n
\bigr)
is a complemented subspace of \scrR ((T + F )n). We have
\scrN
\bigl(
(T + F )n
\bigr)
= \scrN ((T + F )) \cap \scrR ((T + F )n) =e \scrN (T ) \cap \scrR (Tn) = \scrN (Tn).
As T is a right B-Fredholm operator, then \scrN (Tn) is a complemented subspace of \scrR (Tn). Hence,
\scrN
\bigl(
(T +F )n
\bigr)
is a complemented subspace of \scrR ((T +F )n). Therefore, T +F is a right B-Fredholm
operator.
Proposition 2.5. Let T \in \scrL (X). The following properties are equivalent:
(i) T \in \scrB \scrF l(X),
(ii) Tm \in \scrB \scrF l(X) for each m > 0,
(iii) Tm \in \scrB \scrF l(X) for some m > 0.
Proof. (i)\Rightarrow (ii). Suppose that T \in \scrB \scrF l(X) and let d = \mathrm{d}\mathrm{i}\mathrm{s}(T ). From Proposition 2.1 we obtain
that \scrR (Tmd) is a closed subspace of X and Tmd is a left Fredholm operator. Since (Tmd)
m =
= (Tm)d, then the operator (Tm)d is a left Fredholm operator. Consequently, Tm is a left B-
Fredholm operator.
(ii)\Rightarrow (iii). This is obvious.
(iii)\Rightarrow (i). Suppose that Tm is a left B-Fredholm for some m > 0. Then there exists an integer
n such that \scrR (Tmn) is a closed subspace of X and (Tm)n is a left Fredholm operator. Since
(Tmn)
m = (Tm)n, hence, (Tmn)
m is a left Fredholm operator. Therefore, if the operator (Tmn) is a
left Fredholm operator, then T is a left B-Fredholm operator.
Proposition 2.6. Let T \in \scrL (X). The following properties are equivalent:
(i) T \in \scrB \scrF r(X),
(ii) Tm \in \scrB \scrF r(X) for each m > 0,
(iii) Tm \in \scrB \scrF r(X) for some m > 0.
ISSN 1027-3190. Укр. мат. журн., 2022, т. 74, № 10
1304 F. ABDMOULEH, T. BEN LAKHAL
Proof. (i)\Rightarrow (ii). Suppose that T \in \scrB \scrF r(X) and let d = \mathrm{d}\mathrm{i}\mathrm{s}(T ). From Proposition 2.2 we obtain
that \scrR (Tmd) is a closed subspace of X and Tmd is a right Fredholm operator. Since (Tmd)
m =
= (Tm)d, then the operator (Tm)d is a right Fredholm operator. Consequently, Tm is a right
B-Fredholm operator.
(ii)\Rightarrow (iii). This is obvious.
(iii)\Rightarrow (i). Suppose that Tm is a right B-Fredholm for some m > 0. Then there exists an integer
n such that \scrR (Tmn) is a closed subspace of X and (Tm)n is a right Fredholm operator. Since
(Tmn)
m = (Tm)n, then (Tmn)
m is a right Fredholm operator. Therefore, the operator (Tmn) is a
right Fredholm operator, which means that T is a right B-Fredholm operator.
3. A spectral mapping theorem for left and right B-Fredholm operators.
Definition 3.1. (i) Let T \in \scrL (X). We call right B-Fredholm resolving set of T and we write
\rho \scrB \scrF r(T ) the set \rho \scrB \scrF r(T ) =
\bigl\{
\lambda \in \BbbC , (T - \lambda I) \in \scrB \scrF r(X)
\bigr\}
. We call right B-Fredholm spectrum of
T, denoted \sigma \scrB \scrF r(T ), the set \sigma \scrB \scrF r(T ) =
\bigl\{
\lambda \in \BbbC , (T - \lambda I) /\in \scrB \scrF r(X)
\bigr\}
.
(ii) Let T \in \scrL (X). We call left B-Fredholm resolving set of T and we write \rho \scrB \scrF l
(T ) the
set \rho \scrB \scrF l
(T ) =
\bigl\{
\lambda \in \BbbC , (T - \lambda I) \in \scrB \scrF l(X)
\bigr\}
. We call left B-Fredholm spectrum of T, denoted
\sigma \scrB \scrF l(T ), the set \sigma \scrB \scrF l
(T ) =
\bigl\{
\lambda \in \BbbC , (T - \lambda I) /\in \scrB \scrF l(X)
\bigr\}
.
Proposition 3.1. Let T \in \scrL (X), then the left B-Fredholm (resp., right B-Fredholm) spectrum
of T is a closed subset of \BbbC such that
\sigma \scrB \scrF l
(T ) \subset \sigma (T ) and \sigma \scrB \scrF r(T ) \subset \sigma (T ).
Proof. If \lambda /\in \sigma (T ) then T - \lambda I \in \scrB \scrF (X). Thus, T - \lambda I is a left B-Fredholm (resp., right
B-Fredholm ) operator. So, \sigma \scrB \scrF l
(T ) \subset \sigma (T ) and \sigma \scrB \scrF r(T ) \subset \sigma (T ). If \alpha /\in \sigma \scrB \scrF l
(T ) then S =
= T - \alpha I \in \scrB \scrF l(X). If \epsilon is small and not equal to zero, by [7, p. 144] (Table 2) S - \epsilon I is a
quasi-Fredholm operator. From [2] (Theorem 3.1) we have \mathrm{d}\mathrm{i}\mathrm{m}(\scrN (S - \epsilon I)n) = \mathrm{d}\mathrm{i}\mathrm{m}(\scrN (Sn)) < \infty .
Furthermore, \scrR (Sn) is a complemented subspace of \scrR (Sn). So, there exists F \subset \scrR (Sn) such that
\scrR (Sn) = F \oplus \scrR (Sn). Hence, \scrR ((S - \epsilon I)n) =e \scrR (Sn) = F \oplus \scrR (Sn) =e F \oplus \scrR ((S - \epsilon I)n). Then
S - \epsilon I is a left B-Fredholm operator. So, \rho \scrB \scrF l
(T ) is open in \BbbC . Consequently, \sigma \scrB \scrF l
(T ) is a closed
subset of \BbbC .
On the same way, let \beta /\in \sigma \scrB \scrF r(T ), then A = T - \alpha I \in \scrB \scrF r(X). If \epsilon is small and not equal
to zero, by [7, p. 144] (Table 2), A - \epsilon I is a quasi-Fredholm operator. From [2] (Theorem 3.1),
\mathrm{c}\mathrm{o}\mathrm{d}\mathrm{i}\mathrm{m}(\scrR (S - \epsilon I)n) = \mathrm{c}\mathrm{o}\mathrm{d}\mathrm{i}\mathrm{m}(\scrR (Sn)) < \infty . Furthermore, \scrN (Sn) is a complemented subspace of
\scrR (Sn). So, there exists F1 \subset \scrR (Sn) such that \scrR (Sn) = F1\oplus \scrN (Sn). If \scrR
\bigl(
(S - \epsilon I)n
\bigr)
=e \scrR (Sn),
then, \scrR ((S - \epsilon I)n) = F \oplus \scrN (Sn) =e F \oplus \scrN
\bigl(
(S - \epsilon I)n
\bigr)
. Then S - \epsilon I is a right B-Fredholm
operator. So, \rho \scrB \scrF r(T ) is open in \BbbC . Finally, \sigma \scrB \scrF r(T ) is a closed subset of \BbbC .
Proposition 3.2. Let T \in \scrL (X), then the right B-Fredholm spectrum \sigma \scrB \scrF r(T ) of T is a closed
subset of \BbbC contained in the usual spectrum \sigma (T ) of T.
Proof. If \lambda /\in \sigma (T ) then T - \lambda I \in \scrB \scrF (X). Thus, T - \lambda I \in \scrB \scrF r(X) and \lambda /\in \sigma \scrB \scrF r(T ). So,
\sigma \scrB \scrF r(T ) \subset \sigma (T ).
Theorem 3.1. Let S, T, A, B be mutually commuting operators in \scrL (X), satisfying TA +
+BS = I . T, S \in \scrB \scrF l(X) if and only if TS \in \scrB \scrF l(X).
Proof. Suppose that T and S are a left B-Fredholm operator. Then there exist An \in \scrL (\scrR (Tn)),
Bn \in \scrL (\scrR (Sn)) and Kn \in \scrK (\scrR (Tn)), Cn \in \scrK (\scrR (Sn)) such that
AnTn = I +Kn, BnSn = I + Cn.
ISSN 1027-3190. Укр. мат. журн., 2022, т. 74, № 10
LEFT AND RIGHT B-FREDHOLM OPERATORS 1305
Let
Gn : \scrR ((TS)n) - \rightarrow \scrR ((TS)n),
x \mapsto - \rightarrow (BnAn +AnBn)x.
If TA + BS = I, then, according to [6] (Lemma 2.6), we obtain \scrR ((TS)n) = \scrR (Tn) \cap \scrR (Sn).
Then Gn becomes
Gn : \scrR (Tn) \cap \scrR (Sn) - \rightarrow \scrR (Tn) \cap \scrR (Sn),
x \mapsto - \rightarrow (BnAn +AnBn)x.
If Tn(\scrR (Sn)) \subset \scrR (Sn), in fact, if x \in Tn
\bigl(
\scrR (Sn)
\bigr)
, then there exists x\prime \in X such that Sn(x\prime ) = x.
Thus, (ST )nx = (TS)nx = Tnx \subset \scrR (Sn). Therefore, the operator Gn is well defined. Since
GnTnSn = (BnAn +AnBn)TnSn =
= BnAnTnSn +AnBnTnSn =
= Bn(I +Kn)Sn +An(I + Cn)Tn =
= BnSn +BnKnSn +AnTn +AnCnTn =
= BnSn +AnTn + (BnKnSn +AnCnTn) = I + \~K.
It is clear that BnKnSn +AnCnTn \in \scrK (\scrR (Tn) \cap \scrR (Sn)). Consequently, TS \in \scrB \scrF l(X).
Conversely, if TS \in \scrB \scrF l(X), then (TS)n and (ST )n are a left Fredholm operators.
Let \~T = T | \scrR ((TS)n) and \~S = S| \scrR ((TS)n) . Then (TS)n = \~T \~S and (ST )n = \~S \~T . Thus, \~T \~S and
\~S \~T are a left Fredholm operators. Therefore, \~T and \~S are a left Fredholm operators in \scrR ((TS)n). If
TA+BS = I, then, according to [6] (Lemma 2.6), we obtain \scrR ((TS)n) = \scrR (Tn)\cap \scrR (Sn). Since
Sn and Tn are a left Fredholm operators. Consequently, T and S are a left B-Fredholm operators.
Theorem 3.2. Let S, T, A, B be mutually commuting operators in \scrL (X), satisfying TA +
+BS = I . T, S \in \scrB \scrF r(X) if and only if TS \in \scrB \scrF r(X).
Proof. Suppose that T and S are a right B-Fredholm operators. Then there exist An \in
\in \scrL (\scrR (Tn)), Bn \in \scrL (\scrR (Sn)) and Kn \in \scrK (\scrR (Tn)), Cn \in \scrK (\scrR (Sn)) such that
TnAn = I +Kn and SnBn = I + Cn.
Let
Gn : \scrR ((TS)n) - \rightarrow \scrR ((TS)n),
x \mapsto - \rightarrow (BnA
n +AnB
n)x.
If Sn(\scrR (Tn)) \subset \scrR (Tn), in fact, if x \in Sn(\scrR (Tn)), then there exists x\prime \in X such that Tn(x\prime ) = x.
Thus, (TS)nx = (ST )nx = Snx \subset \scrR (Tn). Therefore, the operator Gn is well defined. Since
TnSnGn = TnSn(BnA
n +AnB
n) =
= TnSnBnA
n + TnSnAnB
n =
ISSN 1027-3190. Укр. мат. журн., 2022, т. 74, № 10
1306 F. ABDMOULEH, T. BEN LAKHAL
= Tn(I + Cn)A
n + Sn(I +Kn)B
n =
= TnA
n + TnCnA
n + SnB
n + SnKnBn =
= TnA
n + SnB
n + (TnCnA
n + SnKnBn) = I + \~C.
It is clear that TnCnA
n + SnKnBn \in \scrK (\scrR (Tn) \cap \scrR (Sn)). Consequently, TS \in \scrB \scrF r(X).
Conversely, if TS \in \scrB \scrF r(X), then (TS)n and (ST )n are a right Fredholm operators.
Let \~T = T | \scrR ((TS)n) and \~S = S| \scrR ((TS)n) . Then (TS)n = \~T \~S and (ST )n = \~S \~T , thus, \~T \~S and
\~S \~T are a right Fredholm operators. Therefore, \~T and \~S are a right Fredholm operators in \scrR ((TS)n).
If TA + BS = I, then, according to [6] (Lemma 2.6), we obtain \scrR ((TS)n) = \scrR (Tn) \cap \scrR (Sn).
Since, Sn and Tn are a right Fredholm operators. Consequently, T and S are a right B-Fredholm
operators.
Corollary 3.1. Let P (X) = (X - \lambda 1I)
m1 . . . (X - \lambda nI)
mn be a polynomial with complex coef-
ficients. Then P (T ) = (T - \lambda 1I)
m1 . . . (T - \lambda nI)
mn is a left B-Fredholm (resp., right B-Fredholm)
operator if and only if , for some 1 \leq i \leq n, (T - \lambda iI) is a left B-Fredholm (resp., right B-Fredholm)
operator.
Theorem 3.3. Let T \in \scrL (X) and f an analytic function in a neighborhood of \sigma (T ) of T. Then
f(\sigma \scrB \scrF l
(T )) = \sigma \scrB \scrF l
(f(T )) and f(\sigma \scrB \scrF r(T )) = \sigma \scrB \scrF r(f(T )).
Proof. Let \mu \in \sigma \scrB \scrF l
(T ) and f an analytic function in a neighborhood of \sigma (T ). If \sigma (T ) is a
compact subset of \BbbC , then the function f(z) - f(\mu ) possesses at most a finite number of zeros in
\sigma (T ). So,
f(z) - f(\mu ) = (z - \mu )m0(z - \lambda 1)
m1 . . . (z - \lambda n)
mng(z),
where g(z) is a non-vanishing analytic function on \sigma (T ). Thus,
f(T ) - f(\mu )I = (T - \mu I)m0(T - \lambda 1I)
m1 . . . (T - \lambda nI)
mng(T ),
where g(T ) an invertible operator. So, [g(T )] - 1 is a B-Fredholm operator, then [g(T )] - 1 is a
left B-Fredholm operator. If f(T ) - f(\mu )I \in \scrB \scrF l(X), then by Theorem 3.1 we obtain (f(T ) -
- f(\mu )I)[g(T )] - 1 \in \scrB \scrF l(X). Thus, (T - \mu I)m0(T - \lambda 1I)
m1 . . . (T - \lambda nI)
mn \in \scrB \scrF l(X). So,
from Corollary 3.1, we have (T - \mu I) \in \scrB \scrF l(X), a fact which contradicts our assumption. So,
\mu \in \sigma \scrB \scrF l
(T ). Hence, f(\mu ) \in \sigma \scrB \scrF l
(f(T )). Consequently, f(\sigma \scrB \scrF l
(T )) \subset \sigma \scrB \scrF l
(f(T )).
Conversely, let \alpha \in \sigma \scrB \scrF l
(f(T )), then \alpha \in \sigma (f(T )). Hence, there exists \mu \in \sigma (T ) such that
f(\mu ) = \alpha . We have
f(z) - f(\mu ) = (z - \mu )m0(z - \mu 1)
m1 . . . (z - \mu n)
mng(z),
where g(z) is a non-vanishing analytic function on \sigma (T ). Thus,
f(T ) - f(\mu )I = (T - \mu I)m0(T - \mu 1I)
m1 . . . (T - \mu nI)
mng(T ) = f(T ) - \alpha I,
where g(T ) is an invertible operator. If f(T ) - \alpha I /\in \scrB \scrF l(X), then (f(T ) - \alpha I)[g(T )] - 1 /\in
/\in \scrB \scrF l(X). From Corollary 3.1, there exists \beta = \{ \mu , \mu 1, . . . , \mu n\} such that T - \beta I /\in \scrB \scrF l(X).
Hence, \beta \in \sigma \scrB \scrF l
(T ) and f(\beta ) = \alpha . Thus, \alpha = f(\beta ) \in f(\sigma \scrB \scrF l
(T )). Consequently, \sigma \scrB \scrF l
(f(T )) \subset
\subset f(\sigma \scrB \scrF l
(T )).
We do the same steps that we applied for the first equality and get \sigma \scrB \scrF r(f(T )) \subset f(\sigma \scrB \scrF r(T )).
ISSN 1027-3190. Укр. мат. журн., 2022, т. 74, № 10
LEFT AND RIGHT B-FREDHOLM OPERATORS 1307
References
1. M. Berkani, On a class of quasi-Fredholm operators, Int. Equat. Oper. Theory, 34, 244 – 249 (1999).
2. M. Berkani, M. Sarih, On semi B-Fredholm operators, Glasgow Math. J., 43, 457 – 465 (2001).
3. A. A. Boichuk, A. M. Samoilenko, Generalized inverse operators and Fredholm boundary value problems, 2nd ed.,
Inverse and Ill-Posed Probl. Ser., 59, De Gruyter, Berlin (2016).
4. M. A. Kaashoek, Ascent, descent, nullity and defect, a note on a paper by A. E. Taylor, Math. Ann., 172, 105 – 115
(1967).
5. J. P. Labrousse, Les Opérateurs quasi-Fredholm: Une généralisation des opérateurs semi-Fredholm, Rend. Circ. Mat.
Palermo (2), 29, 161 – 258 (1980).
6. V. Kordula, V. Muller, On the axiomatic theory of the spectrum, Stud. Math., 119, № 2, 109 – 128 (1996).
7. M. Mbekhta, V. Muller, On the axiomatic theory of the spectrum, II, Stud. Math., 119, № 2, 129 – 147 (1996).
Received 14.10.20
ISSN 1027-3190. Укр. мат. журн., 2022, т. 74, № 10
|
| id | umjimathkievua-article-6331 |
| institution | Ukrains’kyi Matematychnyi Zhurnal |
| keywords_txt_mv | keywords |
| language | English |
| last_indexed | 2026-03-24T03:27:07Z |
| publishDate | 2022 |
| publisher | Institute of Mathematics, NAS of Ukraine |
| record_format | ojs |
| resource_txt_mv | umjimathkievua/d9/376043eae05be634a1e2c0b3523917d9.pdf |
| spelling | umjimathkievua-article-63312023-01-07T13:45:40Z Left and right B-fredholm operators Left and right B-fredholm operators Abdmouleh, F. Ben Lakhal, T. Abdmouleh, F. Ben Lakhal, T. Fredholm operators, B-Fredholm operators, finite rank operators. UDC 517.9 We introduce the families of left and right B-Fredholm operators in Banach space, realize their stabilization with the help of finite-rank operators, and&nbsp; prove a spectral mapping theorem for the left and right B-Fredholm operators. УДК 517.9 Лівий та правий B-фредгольмові оператори Уведено сім'ї лівих і правих B-фредгольмових операторів у банаховому просторі, проведено стабілізацію за допомогою операторів скінченного рангу та доведено теорему про спектральне відображення для лівих і правих B-фредгольмових операторів. Institute of Mathematics, NAS of Ukraine 2022-11-27 Article Article application/pdf https://umj.imath.kiev.ua/index.php/umj/article/view/6331 10.37863/umzh.v74i10.6331 Ukrains’kyi Matematychnyi Zhurnal; Vol. 74 No. 10 (2022); 1299 - 1307 Український математичний журнал; Том 74 № 10 (2022); 1299 - 1307 1027-3190 en https://umj.imath.kiev.ua/index.php/umj/article/view/6331/9305 Copyright (c) 2022 Faiçal Abdmouleh |
| spellingShingle | Abdmouleh, F. Ben Lakhal, T. Abdmouleh, F. Ben Lakhal, T. Left and right B-fredholm operators |
| title | Left and right B-fredholm operators |
| title_alt | Left and right B-fredholm operators |
| title_full | Left and right B-fredholm operators |
| title_fullStr | Left and right B-fredholm operators |
| title_full_unstemmed | Left and right B-fredholm operators |
| title_short | Left and right B-fredholm operators |
| title_sort | left and right b-fredholm operators |
| topic_facet | Fredholm operators B-Fredholm operators finite rank operators. |
| url | https://umj.imath.kiev.ua/index.php/umj/article/view/6331 |
| work_keys_str_mv | AT abdmoulehf leftandrightbfredholmoperators AT benlakhalt leftandrightbfredholmoperators AT abdmoulehf leftandrightbfredholmoperators AT benlakhalt leftandrightbfredholmoperators |