Advances in Pure Mathematics, 2012, 2, 124-127
http://dx.doi.org/10.4236/apm.2012.22018 Published Online March 2012 (http://www.SciRP.org/journal/apm)
The Index of Invariant Subspaces of Bounded below
Operators on Banach Spaces
George Chailos
Department of Computer Science, University of Nicosia, Nicosia, Cyprus
Email: chailos.g@unic.ac.cy
Received November 17, 2011; revised November 30, 2011; accepted December 8, 2011
ABSTRACT
For an operator on a Banach space S
X
, let
,LatS X
S
be the collection of all its invariant subspaces. We consider
the index function on and we show, amongst others, that if is a bounded below operator and if
,S X
Lat

,
i
M
LatS X, , then If in addition iI ii
iI
ind M




.
iI
in M

i
iI
indM
d
i
M
are index 1 invariant sub-
spaces of , with nonzero intersection, we show that
S
i
iI
M
<Mi
ind
iI
ind
. Furthermore, using the index
function, we provide an example where for some
,
i
M
LatS X=
ii
iI iI
, holds
M
M
.
Keywords: Index; Invariant Subs paces; Bounded below Operators; Banach Spaces
1. Introduction, the Index Function
If is an operator on a Banach space S
X
, then a
closed subspace
M
of
X
is called invariant for if
. The collection of all invariant subspaces of an
operator is denoted by
S
SM M
S
,LaXt S. It forms a com-
plete lattice with respect to intersections and closed spans.
One of the important notions in the general theory of
operators, such as bounded below operators, is the index
of an element in
,LatS X, which is defined as follows.
(This definition is taken from [1].)
Definition 1.1. The map


0:,indLatS X
is defined as

=indM

=0Mdim MSM=0dM and in if
and only if . We say that
M
has index if
. n
=indM n
The index function plays an essential role in the study
of invariant subspaces of Banach spaces. (For example,
see an extensive study in [2] for index 1 invariant sub-
spaces in Banach spaces of analytic functions.) In this
article we generalize and extend the results obtained in
[3], utilizing new proving techniques, deriving algebraic
properties of the index functions. Moreover, we provide
new results that are applied in Bergman space theory.
Amongst others, and as a corollary to our main result, we
show that if

,
i
M
LatS Xind , and
then ,
=1

ii
iI
nd M
i
indM

<
iI
M i


0
i
iI
M
where i
iI
M
i
M
, denotes the closed span of
iI
. (Equivalently, i
iI
M
i
iI
is the closure of
M
). An analogous result, but in not such a general
setting as the one presented here, was proved by Richter
([2], Corollary 3.12 ), using operato r theoretical tools and
results from analysis. Here we prove our general result
using only algebraic tools and a rather standard result
from functional analysis. Furthermore, we provide an
example where for some i

,
M
LatS X
=,
ii
iI iI
M
M
holds, and we present an application
of this result in Bergman space theory.
2. Algebraic Properties of the Index
Function—Main Results
In the sequel we denote with
I
an index subset of ,
and with
X
a Banach space.
Theorem 2.1. Let be a commutative ring with
identity and let i
R
,
A
A
i
be free unitary R-modules such
that
A
are free submodules of
A
, . Then iI

=.
iii
iI iI
iI
rankAArank AArank AA


 

 




,
Proof. We shall prove this theorem using mathe- ma-
tical induction. Henceforth at first we establish the fol-
lowing equation (which it is the initial step of mathe-
matical induction.) We supposes that
A
B

are free
unitary R-modules that are free submodules of
A
, then
C
opyright © 2012 SciRes. APM
G. CHAILOS 125
 




A AB

=
rankA ArankAB
rank AABrank


 ()
To prove (), consider the following sequence
 
0
fg
AA BAAABA
 
 0,A B


where


=,
f
yyy

,



,=
g
xy xy
and
denotes the equivalence class in the appropriate
quotient module. We claim that the sequence above is
exact.
[]
For its proof we first show that
f
and
g
are well-
defined homomorphisms. Letting

y
AA B

 and
x
AB

, we obtain that


=,=
,
f
yx yxyxy y. Hence,
f
is
well defined. Moreover,
f
is a homomorphism, since







 

=,
=,,
=,=
fy zy zy
yy z
fryryry r





,,.
z
z
yy r
R
Similarly, if

,
x
yAAAB

 1
, and
x
A
,
2
x
B
, then



111
11
,=gxx yyxxy

1
==
,
y
x
yx
 
 y xy




since 11
x
yA
B
. Thus,
g
is well defined.
Moreover,
g
is a homomorphism, since





  



,,
=( ,)
=,=
==
gxy xy
gxx yy


g
xx yyxx
xyx yxy




 yy
x y
 



 
and




,= ,
=,
=
.
g
rxyg rxry
rx y
rx ry
r

R
It remains to show that ker
g
= im
f
. For this let

,
x
yAAAB

be such that

,=0.gx y
Then
=0,xy and thus
x
yA
 B
. This im-
plies that =
x
Ay

B, i.e.,
/
=
AA
x
/AB
y
where-
fore
AA
i
//AB
, m
x
yfker
 im, and hence
g
f.
Conversely, if

im,
x
yf then =
x
AyB

=
and hence
x
AB

 yAB

. It follows that

,=gx yxy=0 so that im ker
f
g

. The proof
of the claim is complete.
Since
A
AB

is a free module, it is in particular
projective, and hence the above exact sequence splits
(see [4]). Therefore

=.
A
AABAAB
 

AAB
 

The above equation immediately implies (). Now
since finite intersection and finite sum of free submo-
dules of
A
, are also free submodules of
A
, a standard
use of Mathematical Induction concludes the proof of the
theorem.
As every vector space is fr ee over its ground field, the
following is an immediate consequence of the above
theorem.
Corollary 2.1. If
X
is a Banach space and an
operat o r o n S

,
i
X
, for all
M
LatS XiI
=
ii i
iI iI
iI
indM indMindM






S

,
,
In the case where is a bounded below operator,
like the shift operator on Banach spaces of analytic func-
tions, the following, which is the fundamental lemma of
this article, holds.
Lemma 2.1. Suppose iLatS XiI,
M
, where
is a bo u n ded below operator on a Ba nach space
X
. S1) We have
.
iii
iI iI iI
indMindMindM




 
 

2) If
,LatS X
,2mm

,,niI
contains an invariant subspace of
index and i

=nm , with
i
iI
, then there are invariant subspace ,
i
Ni I
,
such that
=, ,and=.
iii i
iI iI
ind NniIindNind N



Proof. 1) Once more, we shall make use of mathe-
matical induction to prove this corollary. We assume that
,,
M
NLatSX and we show that
indMindNindMNindMind N
indMind N<indM
<ind N
(as the initial step of mathematical induction.) If either
or is infinite, then there is nothing to
prove. So we may assume that and
. Thus there are finite-dimensional subspaces
1
M
and of
1
N
M
and , respectively, such that N
1
=
M
SM M
, 1
=NSNN
, where
and . We find that 1=dimM indM
1=dimNind N

 
11
11
11
=
=
.
MNSMM SNN
SM NMN
SM NMN
MN




S
Since is a bounded below operator, its range is
closed (see, e.g., [5], Proposition 6.4, chapter VII), and
henceforth the second to last expression as the sum of a
closed and a finite-dimensional subspace, it is closed.
Since
M
N
M
is dense in N we obtain that the
last inclusion in above relations is actually an equality.
Copyright © 2012 SciRes. APM
G. CHAILOS
126
From this it follows that

11
=
.
indMNdim MN
indMind N
 


ind MN

,LatSX

,LatS X
Now, since the closed span of a finite number of
elements in is an element of , the
proof of (1) follows by mathematical induction.
2) To prove that the equality in 1) can actually occur
let us assume that
2mm, and that con-
tains an invariant subspace
M
with index . Without
loss of generality let m

2 ,=1,,3,
I
l
i
iI
mn
i
n
NL=,Nn iI
1
,
. At first we assume that
123
<<<<nl
nnn
=, . We shall construct

,atSX ind

=
i
iI
N m
i
in
m
with ii , and
. As in the proof of part 1) there is an
dimensional subspace
d
-
M
of
M
such that
1
=
M
SM M. Let
12 1
,,,
n
g
gg1
be a basis for
M
and define to be the smallest invariant subspace of
1
N
S12
,, which contains

1
,
n
g
gk
N
12
11
,,
g
S
. Define to be
the smallest invariant subspace of which contains

,
n
k
nn
kk
g
g


=2,3,,1kl
N
S
12
11
,,,
m
ll
g, where , and si-
milarly define to be the smallest invariant subspace
l
nn
of which contains
g
gg


N
S
. It is easy
to observe that i
iI is the smallest invariant sub-
space of which contains
12
,,,
m
g
gg
ind
1
,,,
n
.
Claim:
11
Proof of Claim:
=N n
Let be the linear span of

12
g
gg. Then
1
M
1
SN SM
11
==m n
. We have , thus
implies that . Furthermore,
11
is closed, since 1 is closed an d is
finite dimensional. We note that 1 is invariant
for , thus by definitio n of 1 we have
. This implies that
1=0MSM

1=0 SN
Nind
SN
N di
SN
N
=N
SN
S
SN
11
Similarly we see that for
and that =
ii
d Nn
=nm
in =2,3,,il
=
j
n

=
i
iI
N m
=m
ind
Finally if and i
iI
, then there is at
least one index such that j
n. If jI=
for
some , then set j. If also
i then set i. In this case we are
done because iI
j

j
I
=
=
=NM
=NM,\Ini
M
M
. So suppose that that there is
some such that j. Since
Inj =indM
there
is an
j
n-dimensional subspace
j
M
of
M
such that
j. Define

=0SM M
j
N to be the smallest inva-
riant subspace which contains all of
j
M
. As in the arg-
ument given above it follows that =
j
j. Clearly,
j, thus ind Nn
N
M=iI i
M
N
and the proof of part (2) is
now complete.
Corollary 2.2. Under the hypothesis of Lemma 2.1 part
(2), for the family of
,tXS iI
=
ii
iI iI
NN
i
NLa, , it holds:
Proof. To see this, observe that by applying the con-
clusion of Lemma 2.1 part (1) to the equation of Cor-
ollary 2.1 we obtain
=0ind N
i
iI. Henc e
=0N
=
ii
iI iI
NN
S
i
iI. Therefore by the definition of direct sum
(of vector spaces), we get
.
If is a bounded below operator on
X
the fol-
lowing is true:

,
i
Corollary 2.3. Suppose that LatS XiI,
M
,
.
=1
i
indM
If
0M
i
iI, then strict inequality holds in Lem-
ma 2.1 (a), that is,

<
ii
iI iI
indM indM
.
0
i
iI
M

0ind M, then i
iI
Proof. If
=1
i
indM iI , and
since ,
, then by [2] Theorem 3.16,
=1ind M
i
iI. Thus from Lemma 2.1 (a) and the
equation in Corollary 2.1 , we obtain,
<indMindM
ii
iI iI
.
The next theorem, which is our main result, follows
immediately from Corollary 2.1 and Lemma 2.1, part (1).
Theorem 2.2. if
X
is a Banach space and is a
bounded below operator, then for S
,
i
M
LatS X
iI ,
.
iii
iI
iI iI
indMind MindM

 



 

S
=2m
S
,
Remark 2.1. We would like to note that [2], Pro-
position (2.16), Richter proved a special case of our
Lemma 2.1 whe n is the shift operator on any Banach
space of analytic functions on an open and con-
nected subset of the complex plane and .
Example 2.1. It is well known ([6], Corollary 6.5) that
when is the shift operator on a weighted Bergman
space on the unit disk, then for all
m
Nm
=
mm
NN
there
are invariant subspaces m, of index . Thus, for this
operator Corollary 2.2 applies and we have
.
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G. CHAILOS
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