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spectral_theorem

# spectral_theorem - P DE LECTURE NOTES M ATH 237A-B 157 11...

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PDE LECTURE NOTES, MATH 237A-B 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 fi 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 µ be counting measure on { 1 , 2 , . . . , n } . Then there exists a unitary map U : C n L 2 ( { 1 , 2 , . . . , n } , dµ ) and a real function λ : { 1 , 2 , . . . , n } R such that UAξ = λ · for all ξ C n . We summarize this equation by writing UAU 1 = M λ where M λ : L 2 ( { 1 , 2 , . . . , n } , dµ ) L 2 ( { 1 , 2 , . . . , n } , dµ ) is the linear operator, g L 2 ( { 1 , 2 , . . . , n } , dµ ) λ · g L 2 ( { 1 , 2 , . . . , n } , dµ ) . Proof. By the usual form of the spectral theorem for self-adjoint matrices, there exists an orthonormal basis { e i } n i =1 of eigenvectors of A, say Ae i = λ i e i with λ i R . De fi ne U : C n L 2 ( { 1 , 2 , . . . , n } , dµ ) to be the unique (unitary) map determined by Ue i = δ i where δ i ( j ) = ½ 1 if i = j 0 if i 6 = j and let λ : { 1 , 2 , . . . , n } R be de fi ned by λ ( i ) := λ i . De fi 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 fi nition 11.3. If A = A the A is self adjoint. Proposition 11.4. Let ( X, µ ) be σ fi nite measure space, H = L 2 ( X, dµ ) 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, dµ ) 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 dµ = Z X g k dµ for all g D ( M f ) or equivalently (11.1) Z X g ( fh k ) = 0 for all g D ( M f ) . Choose X n X such that X n X and µ ( X n ) < for all n. It is easily checked that g n := 1 X n fh k ¯ ¯ fh k ¯ ¯ 1 | f | n

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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 = 0 for all n. Using the monotone convergence theorem, we may let n → ∞ in this equation to fi nd R X ¯ ¯ fh k ¯ ¯ = 0 and hence that ¯ fh = k L 2 . This shows h D ( M ¯ f ) and M f h = fh. Theorem 11.5 (Spectral Theorem) . Suppose A = A then there exists ( X, µ ) a σ fi nite measure space, f : X R measurable, and U : H L 2 ( x, µ ) unitary such that UAU 1 = M f . Note this is a statement about domains as well, i.e. UD ( M f ) = D ( A ) . I would like to give some examples of computing A and Theorem 11.5 as well. We will consider here the case of constant coe cient di ff erential operators on L 2 ( R n ) . First we need the following de fi nition.
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