# Functional Analysis/Special topics

This chapter collect some materials that didn't quite fit in the main development of the theory.

## Fredholm theoryEdit

We recall that the closed unit ball of a Banach space is compact if and only if the space is finite-dimensional. This is a special case of the next lemma:

**7 Lemma** *Let be a closed densely defined operator. Then the following are equivalent.*

*(i) and the range of*T*is closed.**(ii) Every bounded sequence has a convergent subsequence when*is convergent.

Proof: We may assume *T* has dense range. (i) (ii): Suppose is a bounded sequence such that is convergent. In view of the Hahn-Banach theorem, *X* is a direct sum of the kernel of and some other subspace, say, . Thus, we can write:

By the closed graph theorem, the inverse of is continuous. Since , the continuity implies that is convergent. Since contains a convergent subsequence by the paragraph preceding the theorem, has a convergent subsequence then. (ii) (i): (ii) implies the first condition of (i), again by the preceding paragraph. For the second, suppose is convergent. Then by (ii) has a subsequence converging to, say, . Since the graph of *T* is closed, converges to .

A bounded linear operator between Hilbert spaces is said to be *Fredholm* if *T* and both satisfy the condition of (i) in the lemma. The definition is equivalent to requiring that the kernel of *T* and the quotient are finite-dimensional. In fact, if is finite-dimensional, then is a complemented subspace; thus, closed. That has closed range implies that has closed range. For a Fredholm operator at least, it thus makes sense to define:

- .

Because of the first isomorphism theorem, the index is actually independent of any operator *T* when *T* is a map between finite-dimensional spaces. This is no longer the case for operators acting on infinite-dimensional spaces.

**7 Lemma** *Let and . If and are Fredholm operators, then is a Fredholm operator with*

- .

Conversely, if , and both and are Fredholm operators, then is a Fredholm operator.
Proof:
Since

- , and ,

we see that is Fredholm. Next, using the identity

we compute:

For the conversely, let be a bounded sequence such that is convergent. Then is convergent and so has a convergent sequence when is Fredholm. Thus, and has closed range. That is a Fredholm operator shows that this is also true for and we conclude that is Fredholm.

**7 Theorem** *The mapping*

*is a locally constant function on the set of Fredhold operators .*

Proof: By the Hahn-Banach theorem, we have decompositions:

- .

With respect to these, we represent *T* by a block matrix:

where . By the above lemma, is invariant under row and column operations. Thus, for any , we have:

- ,

since is invertible when is small. *A* depends on *S* but the point is that is a linear operator between finite-dimensional spaces. Hence, the index of *A* is independent of *A*; thus, of *S*.

**7 Corollary** *If is a Fredholm operator and *K* is a compact operator, then is a Fredholm operator with*

Proof: Let be a bounded sequence such that is convergent. By compactness, has a convergent subsequence such that is convergent. is then convergent and so contains a convergent subsequence. Since is compact, the same argument applies to . The invariance of the index follows from the preceding theorem since is Fredholm for any complex number , and the index of is constant.

The next result, known as Fredholm alternative, is now easy but is very important in application.

**7 Corollary** *If is a compact, then*

- and

have the same (finite) dimension for any nonzero complex number , and consists of eigenvalues of *K*.

Proof: The first assertion follows from:

- ,

and the second is the immediate consequence.

**7 Theorem** *Let . Then is a Fredholm operator if and only if and are finite-rank operators for some . Moreover, when and are of trace class (e.g., of finite-rank),*

Proof: Since the identities are Fredholm operators (in fact, any invertible operator) and since

and are Fredholm operators, which implies *T* is a Fredholm operator. Conversely, suppose *T* is a Fredholm operator. Then, as before, we can write:

where is invertible. If we set, for example, , then has required properties. Next, suppose *S* is given arbitrary: . Then

- .

Similarly, we compute:

- .

Now, since , and is invertible, we have:

- .

## Representations of compact groupsEdit

**Theorem** *Every irreducible unitary representation of a compact group is finite-dimensional.*