Functional Analysis/Special topics

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

Fredholm theory[edit]

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


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 groups[edit]

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