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MIT18_102s09_lec19

# MIT18_102s09_lec19 - MIT OpenCourseWare http/ocw.mit.edu...

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MIT OpenCourseWare http://ocw.mit.edu 18.102 Introduction to Functional Analysis Spring 2009 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms .

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108 LECTURE NOTES FOR 18.102, SPRING 2009 Lecture 19. Thursday, April 16 I am heading towards the spectral theory of self-adjoint compact operators. This is rather similar to the spectral theory of self-adjoint matrices and has many useful applications. There is a very effective spectral theory of general bounded but self- adjoint operators but I do not expect to have time to do this. There is also a pretty satisfactory spectral theory of non-selfadjoint compact operators, which it is more likely I will get to. There is no satisfactory spectral theory for general non-compact and non-self-adjoint operators as you can easily see from examples (such as the shift operator). In some sense compact operators are ‘small’ and rather like finite rank operators. If you accept this, then you will want to say that an operator such as (19.1) Id K, K ∈ K ( H ) is ‘big’. We are quite interested in this operator because of spectral theory. To say that λ C is an eigenvalue of K is to say that there is a non-trivial solution of (19.2) Ku λu = 0 where non-trivial means other than than the solution u = 0 which always exists. If λ = 0 we can divide by λ and we are looking for solutions of (19.3) (Id λ 1 K ) u = 0 which is just (19.1) for another compact operator, namely λ 1 K. What are properties of Id K which migh show it to be ‘big? Here are three: Proposition 26. If K ∈ K ( H ) is a compact operator on a separable Hilbert space then null(Id K ) = { u ∈ H ; (Id K ) u = 0 } is finite dimensional (19.4) Ran(Id K ) = { v ∈ H ; u ∈ H , v = (Id K ) u } is closed and Ran(Id K ) = { w ∈ H ; ( w, Ku ) = 0 u H} is finite dimensional and moreover (19.5) dim (null(Id K )) = dim Ran(Id K ) . Definition 9 . A bounded operator F ∈ B ( H ) on a Hilbert space is said to be Fredholm if it has the three properties in (19.4) its null space is finite dimensional, its range is closed and the orthocomplement of its range is finite dimensional. For general Fredholm operators the row-rank=colum-rank result (19.5) does not hold. Indeed the difference of these two integers (19.6) ind( F ) = dim (null(Id K )) dim Ran(Id K ) is a very important number with lots of interesting properties and uses. Notice that the last two conditions are really independent since the orthocom- plement of a subspace is the same as the orthocomplement of its closure. There are for instance bounded opertors on a separable Hilbert space with trivial null space and dense range which is not closed. How could this be? Think for instance of the operator on L 2 (0 , 1) which is multiplication by the function x. This is assuredly bounded and an element of the null space would have to satisfy xu ( x ) = 0 almost everywhere, and hence vanish almost everywhere. Moreover the density of the L 2
109 LECTURE NOTES FOR 18.102, SPRING 2009 functions vanishing in x < � for some (non-fixed) � > 0 shows that the range is dense.

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