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.