Theorem 1Let with , such that , where , if ; for all , if . If satisfiesthen .

# Category: Analysis

# Two fixed point theorems in Banach space

Here is a famous fixed point theorem in finite dimension by Brouwer:

Theorem 1 (Brouwer fixed point theorem)Let be a convex compact set, for any continuous function , there exists a point such that .

There are couple of ways to extend this theorem to Banach spaces. First recall that a mapping between two Banach spaces is called **compact**, if the mapping is continuous (not necessarily linear) and the images of bounded sets are pre-compact.

# Hausdorff Measure and Capacity

Definition 1Let . Definewhere .

Then is called d-dimensional Hausdorff measure in .

Remark 1Hausdorff measure is a Borel outer measure, it is not Radon for because is not finite with respect to . And it is a measure if restricted on Lebesgue measurable sets (by Caratheodory condition: We say a set satisfies Caratheodory condition, if for any set , ). Moreover, note that gives the volume of unit ball in dimension if is an integer, we naturally have (not trivially) , where is the n-dimensional Lebesgue measure.

# Fredholm Alternative and Riesz Schauder Theorem

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.

# Quotient norm of self-adjoint element can be achieved

TheoremLet 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.

# High order BV function has continuous representative

As known, bounded variation (BV) functions are not continuous, but evidently higher order BV functions on corresponding dimension have continuous representative. As a consequence, .

DefinitionA function , if , and the nth order distributional derivative is a finite Radon measure.

TheoremIf , 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 ,

Indeed,

This gives us the desire inequality. Then by density result we know that admits continuous representative for every element.

Corollary.

RemarkIf 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.

# Equivalence of compactness in metric space

LemmaIn 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 .

TheoremIn 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.