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Unformatted text preview: PDE LECTURE NOTES, MATH 237AB 157 11. Introduction to the Spectral Theorem The following spectral theorem is a minor variant of the usual spectral theo rem for matrices. This reformulation has the virtue of carrying over to general (unbounded) self adjoint operators on in f nite dimensional Hilbert spaces. Theorem 11.1. Suppose A is an n n complex self adjoint matrix, i.e. A = A or equivalently A ji = A ij and let be counting measure on { 1 , 2 , . . . , n } . Then there exists a unitary map U : C n L 2 ( { 1 , 2 , . . . , n } , d ) and a real function : { 1 , 2 , . . . , n } R such that UA = U for all C n . We summarize this equation by writing UAU 1 = M where M : L 2 ( { 1 , 2 , . . . , n } , d ) L 2 ( { 1 , 2 , . . . , n } , d ) is the linear operator, g L 2 ( { 1 , 2 , . . . , n } , d ) g L 2 ( { 1 , 2 , . . . , n } , d ) . Proof. By the usual form of the spectral theorem for selfadjoint matrices, there exists an orthonormal basis { e i } n i =1 of eigenvectors of A, say Ae i = i e i with i R . De f ne U : C n L 2 ( { 1 , 2 , . . . , n } , d ) to be the unique (unitary) map determined by Ue i = i where i ( j ) = 1 if i = j if i 6 = j and let : { 1 , 2 , . . . , n } R be de f ned by ( i ) := i . De f nition 11.2. Let A : H H be a possibly unbounded operator on H. We let D ( A ) = { y H : z H 3 ( Ax, y ) = ( x, z ) x D ( A ) } and for y D ( A ) set A y = z . De f nition 11.3. If A = A the A is self adjoint. Proposition 11.4. Let ( X, ) be f nite measure space, H = L 2 ( X, d ) and f : X C be a measurable function. Set Ag = fg = M f g for all g D ( M f ) = { g H : fg H } . Then D ( M f ) is a dense subspace of H and M f = M f . Proof. For any g H = L 2 ( X, d ) and m N , let g m := g 1  f  m . Since  fg m  m  g  it follows that fg m H and hence g m D ( M f ) . By the dominated convergence theorem, it follows that g m g in H as m , hence D ( M f ) is dense in H. Suppose h D ( M f ) then there exists k L 2 such that ( M f g, h ) = ( g, k ) for all g D ( M f ) , i.e. Z X fg h d = Z X g k d for all g D ( M f ) or equivalently (11.1) Z X g ( fh k ) d = 0 for all g D ( M f ) . Choose X n X such that X n X and ( X n ) < for all n. It is easily checked that g n := 1 X n fh k fh k 1  f  n 158 BRUCE K. DRIVER is in D ( M f ) and putting this function into Eq. (11.1) shows Z X fh k 1  f  n d = 0 for all n. Using the monotone convergence theorem, we may let n in this equation to f nd R X fh k d = 0 and hence that fh = k L 2 . This shows h D ( M f ) and M f h = fh....
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