Variation of the Spectrum of Operators in Infinite Dimensional Spaces ()
1. Introduction
Let
be an infinite dimensional Banach space. We denote by
an arbitrary bounded operator on
and by
the identity operator on
. Let
be the closed unit disc of the complex plane
. The restriction on
of the spectrum of an operator
, denoted by
, is the unit spectrum defined as follows:

Essential spectra of some matrix operators on Banach spaces (see [1]) and spectra of some block operator matrices (see [2]) were investigated, with applications to differential and transport operators. This paper investigates the variations of the unit spectrum
as
varies over the space
of all bounded operator on the Banach space
. First, we introduce the sets and the topologies required for this study.
Definition 1
•
the set of all compact subsets of the closed unit disc
of the complex plane
;
•
the spectrum function defined from
into
that maps an operator
to its unit spectrum
.
The set
is endowed with the Hausdorff topology generated by the families of all subsets in one of the following forms

and

for
an open subset of
. Therefore,
is a Polish space, i.e., a separable metrizable complete space, since
is Polish (see [3-5]). It is shown below that we can reduce the families that generate the above Hausdorff topology.
Proposition 1 Let
be the set of compact subsets of the closed unit disc
. Then
equipped with the Hausdorff topology is a Polish space; where the Borel structure is generated by one of the following two families

and

Proof 1 Let
be an open subset of
. There exists a decreasing sequence of open subsets
; for example
such that

We have

On the other hand,

Indeed, if for all
, there exists
, then there exists a subsequence
of
that converges to
, and since
is decreasing, we have

2. Norm Operator Topology and the Spectrum Function
We equip
with the canonical norm of operators defined by

Note that the map
is not continuous when
is endowed with its canonical norm.
Indeed, the operators
converge to the identity
while
and
. However, we have the following result.
Proposition 2 Let X be a Banach space,
the space of bounded operators equipped with the norm of operators, and
the set of compact subsets of the unit disc
equipped with the Hausdorff topology. Then the spectrum map


is upper-semi continuous.
Proof 2 Let
be an open subset of
. By proposition 1, it is only needed to show that the set

is
-open in
. Let
be fixed in
. Since
, then for all
• The operator
is invertible;
• And the map
is continuous (see [6]).
It follows that

since
is compact. Put

Let
such that
.
For any
we have

Thus,
is invertible and hence
. In other terms,
for all
with 
. Therefore
is an open subset of
.
3. Strong Operator Topology and the Spectrum Function
Consider now
equipped with the strong operator topology
(see [6]). In general,
equipped with the strong operator topology is not a polish space (since it is not a Baire space). However, if
is separable, then
is a standard Borel space. Indeed, it is Borel-isomorph to a Borel subset of the Polish space
equipped with the norm product topology via the map


where
is a dense
-vector space in
.
The next result shows how this topology on
affects the spectrum function.
Theorem 1 For any separable infinite dimensional Banach
, the map


which maps a bounded operator to its unit spectrum, is Borel when
is endowed with the strong operator topology
and
with the Hausddorf topology.
Proof 3 As
is equipped with the Hausdorff topology, it follows from the proposition 1, that it is enough to show that for any open subset
of the disc
, the following subset
is Borel in 

Let
be a fixed open subset of
. We have

where
stands for the canonical projection of
onto
, and

By a descriptive set theory result from ([7]), to show that
is a Borel set it suffices to show that
is a Borel set with
vertical sections.
For
, the vertical section of the set
along the direction
is given by

Thus, it is indeed a
of
.
Now, we need to prove that
is a Borel set. Put

Therefore

Hence, to finish the proof, it is enough to prove the following claim.
Claim:
is a Borel set of
.
First, note that
with


Indeed, if
is an isomorphism onto its range, then
is a closed subspace that will be strict if
, and thus not dense in
. On the other hand, since
is separable, there exists a countable and dense subset
in the sphere
of
, and there exists a dense sequence
in
.
Now, we will show that
and
are Borel sets. Let
. From the definition of
, We have
if and only if

In other terms, this is equivalent to

By choosing the subsequence
instead of
, the previous statement is equivalent to

or again,

Therefore,

with

Since
is equipped with the the strong operator convergence
, it follows that the sets
are open. Hence,
is a Borel set.
On the other hand, “
is not dense in
” is equivalent to

or again,

Therefore

with

Similarly to
, it is not difficult to see that the sets
are Borel sets. Hence
is also a Borel set. This proves the claim and ends the proof of the theorem 1.
4. Conclusions
The variation of the unit spectrum of operators in infinite dimensional Banach spaces is investigated. The unit spectrum of an operator
, denoted by
, is defined as the restriction on the closed unit disc
of the complex plane
of the spectrum of
given by
.
First, the paper presents a simplified characterization of the Borel structure making the set
of compact subsets of the closed unit disc
a Polish space. It is also shown that for a Banach space
, the map
that for an operator associates its unit spectrum
is upper-semi continuous when
is endowed with the norm of operators. On the other hand, when
is endowed with the strong operator topology, it is shown that first
needed to be a separable infinite dimensional Banach to guarantee a standard Borel structure on
, then it is shown that the that the map
is Borel in this case. Therefore, this topology is making the spectrum function more rigourous, and as a consequence the variations of the spectrum following changes in an operator or a sequence of operators.