Theorem (Analytic Fredholm Alternative) Let be a connected open subset of and be a separable Hilbert space. Suppose that is an analytic operator-valued function such that is compact . Then either
- does not exist .
- exists , where is discrete.
Theorem Let be a closed ideal of a C* algebra , then for any self-adjoint element , there exists an such that
Proof: Since is a closed ideal, is also a C* algebra and one can define a natural projection homomorphism , with . Fix an element , define a continuous function as follow:
Let , then is identical on . By continuous functional calculus, . And since can be approximated by polynomials, we know . The spectral radius , so we know is identical on the spectrum of , which implies . Therefore is in the kernel of , and hence in the ideal . Since and by the Gelfand Naimark isomorphism theorem, we know and hence . So is the we are looking for.
Note: this is a homework exercise given by Dr. Carlen.
As known, bounded variation (BV) functions are not continuous, but evidently higher order BV functions on corresponding dimension have continuous representative. As a consequence, .
Definition A function , if , and the nth order distributional derivative is a finite Radon measure.
Theorem If , then has a continuous representative.
Proof: Note that is dense in , i.e. for any , there exists a sequence such that in , and . (See for example, book by Evans and Gariepy). So it suffice to show for all ,
This gives us the desire inequality. Then by density result we know that admits continuous representative for every element.
Remark If the dimension is greater than the order of BV function, then the above theorem will fail. For example, let be defined in a neighborhood around 0, with smooth boundary in . and . So . Then we can extend on , since is smooth (See for example, PDE book by Evans). But then as , which doesn’t admit a continuous representative.
Lemma In a metric space , if is a sequence of compact sets such that for and diam. Then , where is a point in .
Proof: First note that cannot contain more than 1 points. If , then . Since diam=0, it forces .
It remains to show is nonempty. Assume it is empty, then . Since metric space is Hausdorff, is closed for all and is an open cover for . Since is compact, there exists finite subcover such that , which contradicts .
Theorem In a metric space (no matter it has countable basis or not), let be a subset in . The following statement are equivalent by assuming axiom of choice:
- is compact;
- is sequentially compact;
- is complete and totally bounded.
I will show an version of Arzela-Ascoli theorem, which is called Kolmogorov-Riesz theorem, by using a fact in topology as following.
Definition In a metric space, a set is totally bounded, if for any fixed , the set can be covered by finitely many open balls of radius .
Lemma In a metric space, a set is compact if and only if is complete and totally bounded.
Proof of the lemma can be find here.
Theorem 1 (Kolmogorov Riesz Theorem (finite measure domain)) Let be a bounded set of functions, Assume that , there exists a such that
whenever , for all . Then has a compact closure in for any finite measure set .
Definition Let be a banach space, be the space of bounded linear operator from . If , then say a complex number is in the resolvent set of if has a bounded inverse. Then we define the resolvent of at to be . If , then is said to be in the spectrum of T.
We will learn some properties of the resolvent and the spectrum.
Definition are Banach spaces, we say is embedded in , denoted by , if and .
Here are some embedding results.