compact

# compact - ANALYSIS TOOLS WITH APPLICATIONS 579 35. Compact...

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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 Gram-Schmidt 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....
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## This note was uploaded on 10/11/2010 for the course MATH 11 taught by Professor Smith during the Three '10 term at ADFA.

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compact - ANALYSIS TOOLS WITH APPLICATIONS 579 35. Compact...

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