Let H be a separable Hilbert space and let B(H) be the set of all bounded operators acting on H. Given T∈ B(H), we show that T has a proper invariant subspace, i.e., there exists a proper Hilbert subspace L⊂H such that T(L)⊆L. This problem has only been solved for special cases so far and in this article we try to offer a solution that can take care of the most general cases.
The preliminary definitions and notations can be found in the thesis of Jonathan Noel “The invariant subspace problem”, Department of Mathematics and Statistic, Thompson River University, 2011.
Given T as in the above, without loss generality, we can assume that is one to one and onto, otherwise T will have an proper invariant subspace. So if
is any vector in
then the subspace
which is spanned by the elements of the form
will be equal to
, because
is invariant under the action of
. Thus any element
can be written as a norm convergent sum,
, where
is a complex number. Before we propose the general solution, we consider the following special case. Suppose we have
, defined by
with
Lemma 1.1. If for some vector
, then
has a proper invariant subspace.
Proof. Let be the roots of the polynomial
Hence we have
Then at least one of the factors, say
, is not invertible. This implies that either
has a proper invariant subspace, or there exits an integer
with
which implies the existence of a one dimensional invariant subspace for
The case of , with
, is much more complex and Lemma 1 is not working for this case. At this point we assume that
, and
. Next, let us pick an element
, with
. Now it is well known that a subset
of
is compact if every sequence in
has a bounded subsequence with its limits included in
. Next, consider the following set,
Subsequently, define the following set
to be the subset of the closure of convex hall of
consisting of all elements of the form
, with
.
It is easy using the above comment to check that is a convex and compact subset of
Hence by the Brower fixed point theorem there exists an element
in
for which
If
is different from zero then we have a non trivial invariant subspace and we are done. Otherwise, we have to consider the case where
This means we have
But assuming that
does not have a non trivial invariant subspace implies that for every
where, by the above, each
and
.
Next, let us denote . Then continue the above argument for each
If for any integer
the subspace generated by the set
, will not span
, then this implies that
has a non trivial invariant subspace, otherwise the Brower fixed point acting over the compact convex hall of the above set implies the existence of a fixed point under the action of
, in which case we have the following infinite dimensional polynomial
with
.
On the other hand, if we have a nontrivial fixed point then the proof is complete. As we mentioned in the above if does not have nontrivial invariant subspace we get that the polynomial
converges in norm topology to zero. But considering the fact that
and
therefore the sets
have a following limit point
with
and this will implies contradiction since as
tends to infinity this will force
to be equal to zero. This proves the final theorem which is the main focus of this article.
Theorem 1.2. Every Bounded Operator acting on a separable Hilbert space has a proper invariant subspace.
[1] | R. Abraham, J.E. Mardsen, T.Ratiu. “Manifolds, Tensor Analysis and applications”. | ||
In article | |||
[2] | Jonathan Noel, “The invariant subspace problem”, Thesis, Department of Mathematics and Statistic, Thompson River University, 2011. | ||
In article | |||
Published with license by Science and Education Publishing, Copyright © 2022 Bahman Mashood
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[1] | R. Abraham, J.E. Mardsen, T.Ratiu. “Manifolds, Tensor Analysis and applications”. | ||
In article | |||
[2] | Jonathan Noel, “The invariant subspace problem”, Thesis, Department of Mathematics and Statistic, Thompson River University, 2011. | ||
In article | |||