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Unformatted text preview: 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 . 95 LECTURE NOTES FOR 18.102, SPRING 2009 Lecture 16. Tuesday, April 7: partially reconstructed From last time Proposition 23. The invertible elements form an open subset GL( H ) B ( H ) . Proof. Recall that we showed using the convergence of the Neumann series that if B B ( H ) and B < 1 then Id B is invertible, meaning it has a twosided inverse in B ( H ) (which we know, from the open mapping Theorem to be equivalent to it being a bijection). So, suppose G GL( H ) , meaning it has a twosided (and unique) inverse G 1 B ( H ) : (16.1) G 1 G = GG 1 = Id . Then we wish to show that B ( G ; ) GL( H ) for some > . In fact we shall see that we can take = G 1 1 . The idea is that we wish to show that G + B is a bijection, and hence invertible. To do so set (16.2) E = G 1 B = G + B = G 1 (Id+ G 1 B ) . This is injective if Id + G 1 B is injective, and surjective if Id + G 1 B is surjective, since G 1 is a bijection. From last time we know that (16.3) G 1 B < 1 = Id+ G 1 B is invertible. Since G 1 B G 1 B this follows if B < G 1 1 as anticipated. Thus GL( H ) B ( H ) , the set of invertible elements, is open. It is also a group since the inverse of G 1 G 2 if G 1 , G 2 GL( H ) is G 1 G 1 . 2 1 This group of invertible elements has a smaller subgroup, U( H ) , the unitary group, defined by (16.4) U( H ) = { U GL( H ); U 1 = U } . The unitary group consists of the linear isometric isomorphisms of H onto itself thus (16.5) ( Uu,Uv ) = ( u,v ) , Uu = u u,v H , U U( H ) . This is an important object and we will use it a little bit later on. The unitary group on a separable Hilbert space may seem very similar to the familiar unitary group of n n matrices, U( n ) . It is, of course it is much bigger for one thing. In fact there are some other important differences which I will describe a little later on (or get you to do some of it in the problems). On important fact that you should know, even though I will not prove it, is that U( H ) is contractible as a metric space it has no significant topology. This is to be constrasted with the U( n ) which have a lot of topology, and not at all simple spaces especially for large n. One upshot of this is that U( H ) does not look much like the limit of the U( n ) as n ....
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