The following representation in terms of the spectral family is of interest in itself.
Lemma 2.1.
Let
be a self-adjoint operator in the Hilbert space
with the spectrum
for some real numbers
and let
be its spectral family. If
is a continuous function on
with
, then one has the representation
Proof.
We observe
which is an equality of interest in itself.
Since
are projections, we have
for any
and then we can write
Integrating by parts in the Riemann-Stieltjes integral and utilizing the spectral representation (1.1), we have
which together with (2.3) and (2.2) produce the desired result (2.1).
The following vector version may be stated as well.
Corollary 2.2.
With the assumptions of Lemma 2.1 one has the equality
for any
.
The following result that provides some bounds for continuous functions of bounded variation may be stated as well.
Theorem 2.3.
Let
be a self-adjoint operator in the Hilbert space
with the spectrum
for some real numbers
, and let
be its spectral family. If
is a continuous function of bounded variation on
with
, then we have the inequality
for any
.
Proof.
It is well known that if
is a bounded function,
is of bounded variation, and the Riemann-Stieltjes integral
exists, then the following inequality holds:
where
denotes the total variation of
on
.
Utilising this property and the representation (2.5), we have by the Schwarz inequality in Hilbert space
that
for any
.
Since
are projections, in this case we have
then from (2.8), we deduce the first part of (2.6).
Now, by the same property (2.7) for vector-valued functions
with values in Hilbert spaces, we also have
for any
and
.
Since
in the operator order, then
which gives that
, that is,
for any
, which implies that
for any
. Therefore,
which together with (2.10) prove the last part of (2.6).
The case of Lipschitzian functions is as follows.
Theorem 2.4.
Let
be a self-adjoint operator in the Hilbert space
with the spectrum
for some real numbers
, and let
be its spectral family. If
is a Lipschitzian function with the constant
on
and with
, then one has the inequality
for any
.
Proof.
Recall that if
is a Riemann integrable function and
is Lipschitzian with the constant
, that is,
then the Riemann-Stieltjes integral
exists and the following inequality holds:
Now, on applying this property of the Riemann-Stieltjes integral, then we have from the representation (2.5) that
for any
and the first inequality in (2.11) is proved.
Further, observe that
for any
.
If we use the vector-valued version of the property (2.13), then we have
for any
and the second part of (2.11) is proved.
Further on, by applying the double-integral version of the Cauchy-Buniakowski-Schwarz inequality, we have
for any
.
Now, by utilizing the fact that
are projections for each
, then we have
for any
.
If we integrate by parts and use the spectral representation (1.2), then we get
and by (2.18), we then obtain the following equality of interest:
for any
.
On making use of (2.20) and (2.17), we then deduce the third part of (2.11).
Finally, by utilizing the elementary inequality in inner product spaces
we also have that
for any
, which proves the last inequality in (2.11).
The case of nondecreasing monotonic functions is as follows.
Theorem 2.5.
Let
be a self-adjoint operator in the Hilbert space
with the spectrum
for some real numbers
, and let
be its spectral family. If
is a monotonic nondecreasing function on
, then one has the inequality
for any
.
Proof.
From the theory of Riemann-Stieltjes integral, it is also well known that if
is of bounded variation and
is continuous and monotonic nondecreasing, then the Riemann-Stieltjes integrals
and
exist and
Now, on applying this property of the Riemann-Stieltjes integral, we have from the representation (2.5) that
for any
, which proves the first inequality in (2.23).
On utilizing the Cauchy-Buniakowski-Schwarz-type inequality for the Riemann-Stieltjes integral of monotonic nondecreasing integrators, we have
for any
.
Observe that
and, integrating by parts in the Riemann-Stieltjes integral, we have
for any
.
On making use of the equalities (2.28), we have
for any
.
Therefore, we obtain the following equality of interest in itself as well:
for any 
On making use of the inequality (2.26), we deduce the second inequality in (2.23).
The last part follows by (2.21), and the details are omitted.