This preview shows pages 1–3. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: ANALYSIS TOOLS WITH APPLICATIONS 579 35. Compact and Fredholm Operators and the Spectral Theorem In this section H and B will be Hilbert spaces. Typically H and B will be separable, but we will not assume this until it is needed later. 35.1. Compact Operators. Proposition 35.1. Let M be a f nite dimensional subspace of a Hilbert space H then (1) M is complete (hence closed). (2) Closed bounded subsets of M are compact. Proof. Using the GramSchmidt procedure, we may choose an orthonormal basis { 1 , . . . , n } of M. De f ne U : M C n to be the unique unitary map such that U i = e i where e i is the i th standard basis vector in C n . It now follows that M is complete and that closed bounded subsets of M are compact since the same is true for C n . De f nition 35.2. A bounded operator K : H B is compact if K maps bounded sets into precompact sets, i.e. K ( U ) is compact in B, where U := { x H : k x k < 1 } is the unit ball in H. Equivalently, for all bounded sequences { x n } n =1 H, the sequence { Kx n } n =1 has a convergent subsequence in B. Notice that if dim( H ) = and T : H B is invertible, then T is not compact. De f nition 35.3. K : H B is said to have f nite rank if Ran ( K ) B is f nite dimensional. Corollary 35.4. If K : H B is a f nite rank operator, then K is compact. In particular if either dim( H ) < or dim( B ) < then any bounded operator K : H B is f nite rank and hence compact. Example 35.5. Let ( X, ) be a measure space, H = L 2 ( X, ) and k ( x, y ) n X i =1 f i ( x ) g i ( y ) where f i , g i L 2 ( X, ) for i = 1 , . . . , n. De f ne ( Kf )( x ) = R X k ( x, y ) f ( y ) d ( y ) , then K : L 2 ( X, ) L 2 ( X, ) is a f nite rank operator and hence compact. Lemma 35.6. Let K := K ( H, B ) denote the compact operators from H to B. Then K ( H, B ) is a norm closed subspace of L ( H, B ) . Proof. The fact that K is a vector subspace of L ( H, B ) will be left to the reader. Now let K n : H B be compact operators and K : H B be a bounded operator such that lim n k K n K k op = 0 . We will now show K is compact. First Proof. Given > , choose N = N ( ) such that k K N K k < . Using the fact that K N U is precompact, choose a f nite subset U such that min x k y K N x k < for all y K N ( U ) . Then for z = Kx K ( U ) and x , k z Kx k = k ( K K N ) x + K N ( x x ) + ( K N K ) x k 2 + k K N x K N x k . 580 BRUCE K. DRIVER Therefore min x k z K N x k < 3 , which shows K ( U ) is 3 bounded for all > , K ( U ) is totally bounded and hence precompact....
View
Full
Document
This note was uploaded on 10/11/2010 for the course MATH 11 taught by Professor Smith during the Three '10 term at ADFA.
 Three '10
 Smith

Click to edit the document details