1 Product measures
Let be a -finite measure space. Then with the product -algebra and the product measure on , is itself a -finite measure space.
Write and . For any measurable space , it is a fact that if is measurable then is measurable for each and is measurable for each .11 1 Heinz Bauer, Measure and Integration Theory, p. 138, Lemma 23.5.
Suppose that , . Fubini’s theorem tells us the following.22 2 Heinz Bauer, Measure and Integration Theory, p. 139, Corollary 23.7. 33 3 Suppose that is measurable. Tonelli’s theorem, Heinz Bauer, Measure and Integration Theory, p. 138, Theorem 23.6, tells us that the functions are measurable , and that There are sets with and such that if then and if then . Define
and , and
2 Integral operators in L2
Let and let . By Fubini’s theorem, there is a set with such that if then . For , by the Cauchy-Schwarz inequality,
Since is -finite, there are , , with . For each , the function belongs to and hence, by the Cauchy-Schwarz inequality, belongs to . Applying Fubini’s theorem, there is a set with such that if then belongs to , and the function defined by
belongs to .
Let , for which
For , define by
which shows that is measurable . For any , for sufficiently large we have , thus pointwise, which implies that is measurable.44 4 Charalambos D. Aliprantis and Kim C. Border, Infinite Dimensional Analysis: A Hitchhiker’s Guide, third ed., p. 142, Lemma 4.29.
Using the Cauchy-Schwarz inequality and then Fubini’s theorem,
This shows that , with
Recapitulating, for there is some with such that for , , and such that defined by (1) belongs to . If is any set satisfying these conditions, then for ,
and . Therefore, for it makes sense to define by .
If and in , check that in . We thus define for as
Let be a -finite measure space. For , it makes sense to define by
is a bounded linear operator with .
3 Integrals of functions
Suppose that is a function, which we do not ask to be measurable, and that , , , satisfy . We have
Therefore if there is some with and , it makes sense to define
However, only if is itself measurable do we write .
4 Self-adjoint operators
Let be a -finite measure space. For satisfying , is self-adjoint.
It follows that is self-adjoint. ∎
5 Hilbert-Schmidt operators
Let be a measure space and let . It is a fact that if is -finite and is countably generated, then the Banach space is separable.55 5 Donald L. Cohn, Measure Theory, second ed., p. 102, Proposition 3.4.5.
Let be a -finite countably generated measure space. For , is a Hilbert-Schmidt operator with
is separable, so there is an orthonormal basis for . Using Parseval’s formula and then Fubini’s theorem,
This shows that
If is a compact linear operator on , then is a positive compact operator on . Then is a positive compact operator.66 6 See Anton Deitmar and Siegfried Echterhoff, Principles of Harmonic Analysis, second ed., p. 109, Theorem 5.1.3 Let be the nonzero eigenvalues of repeated according to geometric multiplicity, with , , called the singular values of . By the spectral theorem, there is an orthonormal basis for for such that for each . Then
6 Trace class operators
A compact operator on is called trace class if , where
For a trace class operator it makes sense to define
which does not depend on the orthonormal basis of .
Let be a locally compact Hausdorff space and let be the Borel -algebra of . A Borel measure on is a measure on . We say that a Borel measure on is locally finite if for each there is an open set with and . A Radon measure on is a locally finite Borel measure on such that for each and for any there is an open set with and
and for each open set and for any there is a compact set with and
By definition, if is a Radon measure then can be approximated by for compact sets contained in . We prove that this holds for if .77 7 Anton Deitmar and Siegfried Echterhoff, Principles of Harmonic Analysis, second ed., p. 291, Lemma B.2.1.
Let be a locally compact Hausdorff space and let be a Radon measure on . If with , there for any there is a compact set , , such that
If is a compact set, , and , let . For there is an open set , , such that . Let , and because is Hausdorff, is closed and hence is closed and therefore compact. Now, as ,
We have proved that if is a compact set and is a Borel set contained in , then for any then there is a compact set with and
Now let be an open set with and , say . Let be a compact set with and
Let . Because is a Borel set contained in a compact set , there is a compact set contained in such that
As and ,
Let be a locally compact Hausdorff space and let be a Radon measure on . An admissible kernel is a function for which there is some such that for all . We call an admissible integral operator if there is an admissible kernel such that
The following gives conditions under which we can calculate the trace of an integral operator.88 8 Anton Deitmar and Siegfried Echterhoff, Principles of Harmonic Analysis, second ed., p. 172, Proposition 9.3.1.
Let be a first-countable locally compact Hausdorff space and let be a Radon measure on . Let and let
If there are admissible integral operators and such that , then is of trace class and
The following is Mercer’s theorem.99 9 E. Brian Davies, Linear Operators and their Spectra, p. 156, Proposition 5.6.9.
Theorem 6 (Mercer’s theorem).
If and is a positive operator, then