We assume everywhere that, on a Hilbert complex space
H, endowed with the norm
, a self-adjoint unbounded linear operator
A with the dense domain
is given. By the spectral theorem, the operator
A and its positive powers have the following spectral expansions:
such that
for all
(see, e.g., [
7]), where
is a unique projection-valued measure determined on its spectrum
with values in the Banach space of bounded linear operators
that can be extended on
as zero. For any Borel set
, the spectral subspace is defined to be the range
Let
and
. The following mappings
are well defined. Hence, for any
, we can assign the linear subspace
which by definition contains all elements
such that the scalar function
has the finite norm
This definition is correct, since, for any
and
, the following inequality holds:
for any
. The case
is similar, which proves the linearity of
.
We will apply a quadratic modified real interpolation method. Given a couple of normed spaces
,
where elements
of the algebraic sum
are such that
, we assign the quadratic
K-functional with
(see [
2] (Definition 3.3), [
8] (p. 318)). Note that this couple of normed spaces with a fixed
and an operator
A can be considered as a subspace in
endowed with the quasi-norm
which guarantees its compatibility ([
1] (3.11)).
For any pair indexes
or
with the help of a quadratic
K-functional, we define the interpolation space
endowed with the norm
that is determined in the case
using the Haar measure
on the multiplicative group
. For this interpolation space, we will briefly denote
In particular, for any
and
, the Lorentz-type subspace in
H can be defined with the help of the following linear isomorphism (see, e.g., [
1] (p. 109), [
9] (Proposition 2))
In accordance with this definition, the Lorentz subspace
contains all scalar functions
in the variable
belonging to
, i.e., such that
where the non-increasing rearrangement
of the function
,
is defined via the Lebesgue measure
on
. In other words, the function
should be
-integrable.
Proof. (a) The inequality immediately yields the contractive embedding .
For
, the following inequality with arbitrary
holds:
By Lagrange’s mean value theorem, for any
, there exists
such that
This yields the contractive embedding
for any
.
Consider the case of a vector index
with
. Let
. By the known interpolation property of
K-functionals (see [
1] (p. 81) or [
8] (Theorem B.2)), the contractive inclusions
with both indexes
imply that the inclusion
is contractive. In particular, it holds for the Lorentz-type spaces. Thus, as a summary result, we obtain the inequality
For any
and
, we have
. Hence,
yields the contractive embedding
. Likewise,
Thus,
for
is also contractive.
Similarly, for the subspaces
, we obtain the inequality
The inequality (
6) and (
7) together yield all inclusions (
3).
(b) Since
for all
, we get
Applying the already mentioned interpolation property of
K-functionals (see, [
1] (p. 81) or [
8] (Theorem B.2)), we obtain the inequality
and any
with
. It at once follows (
4).
(c) Let
be a fundamental sequence in
. For every
, there exists
such that
for all
. By the inequality (
5), for any
and
, we find
In particular, this inequality hold for the indexes
. Consequently, by the known interpolation property (see, e.g., [
9] (Theorem 4)), we obtain
It follows that and , for every , are fundamental sequences in H. By the completeness of H, there exist such that and by the norm in H. The graph of the operator is a closed subspace in ; therefore, and . Since, it holds for any , we have . Hence, is convergent by the norm in H for any .
Furthermore, we may apply a standard reasoning. For every
, there exists the following limits
and
for all
such that
for all
. From
, it follows that
By integration with the weight
, we find
Hence,
, which is the same in the case
. Moreover, by integration with the weight
, the inequality
and we find that
for all
. Thus,
is complete. The case
is fully similar.
An alternative reasoning can also be used. Since both spaces
and
are complete, the interpolation space
is also complete in accordance with ([
1] (Theorem 3.4.2 & Lemma 3.10.2)), [
6].
(d) Let
and
. In accordance with the spectral theorem, the restriction
is a bounded operator on the spectral subspace
. Hence, using the inequality
, we get
for any
, where
Thus, every spectral subspace
is contained in some
with a large enough
.
For
, we similarly obtain
for all
, i.e.,
for a large enough
. □