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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 . 59 LECTURE NOTES FOR 18.102, SPRING 2009 Lecture 10. Tuesday, Mar 10 All of this is easy to find in the various reference notes and/or books so I will keep these notes very brief. (1) Bessel’s inequality If in a preHilbert space H, e i , i = 1 ,...,N are orthonormal – so ( e i ,e j ) = δ ij then for any element u ∈ H, set N v = ( u,e i ) e i then i =1 N (10.1) 2 2 2 v =  ( u,e i )  ≤ u H H , i =1 ( u − v ) ⊥ e i , i = 1 ,...,N. The last statement follows immediately by computing ( u,e j ) = ( v,e j ) and similarly v 2 can be computed directly. Then the inequality, which is Bessel’s inequality, follows from Cauchy’s inequality since from the last statement (10.2) v 2 = ( v,v ) = ( v,u ) + ( v,v − u ) = ( v,u ) =  ( v,u )  ≤ v u shows v ≤ u . (2) Orthonormal bases: Since in the inequality in (10.1) the right side is independent of N it follows that if { e i } ∞ is a countable orthonormal set then i =1 ∞ (10.3)  ( u,e i )  2 ≤ u 2 H . i =1 From this it follows that the sequence n (10.4) v n = ( u,e i ) e i i =1 is Cauchy since if m > n, ∞ (10.5) v n − v m 2 = ( u,e j )  2 ≤ ( u,e j )  2 n<j ≤ m j = n +1 and the right side is small if n is large, independent of m. Lemma 5. If H is a Hilbert space – so now we assume completeness – and { e i } ∞ is an orthonormal sequence then for each u ∈ H, i =1 ∞ (10.6) v = ( u,e j ) e j ∈ H j =1 converges and ( u − v ) ⊥ e j for all j. Proof. The limit exists since the sequence is Cauchy and the space is com plete. The orthogonality follows from the fact that ( u − v n ,e j ) = as soon as n ≥ j and (10.7) ( u − v,e j ) = lim ( u − v n ,e j ) = 0 n →∞ 60 LECTURE NOTES FOR 18.102, SPRING 2009 by continuity of the inner product (which follows from Cauchy’s inequality). Now, we say an orthonormal sequence is complete, or is and orthonormal basis of H if u ⊥ e j = for all j implies u = 0 . Then we see: Proposition 15. If { e j ∞ is an orthonormal basis in a Hilbert space H j =1 then ∞ (10.8) u = ( u,e j ) e j ∀ u ∈ H. j =1 Proof. From the lemma the series converges to v and ( u − v ) ⊥ e j for all j so by the assumed completeness, u = v which is (10.8). (3) GramSchmidt Theorem 6. Every separable Hilbert space has an orthonormal basis. Proof. Take a countable dense subset – which can be arranged as a se quence { v j } and the existence of which is the definition of separability – and orthonormalize it. Thus if v 1 = 0 set e i = v 1 / v 1 . Proceeding by induction we can suppose to have found for a given integer n elements e i , i = 1 ,...,m, where m ≤ n, which are orthonormal and such that the linear span (10.9) sp( e 1 ,...,e m ) = sp( v 1 ,...,v n ) ....
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 Linear Algebra, CN, Hilbert space, Orthonormal basis, cj ej

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