X e k 8 k since e k is complete it follows that y x 0

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x ? e k 8 k . Since { e k } is complete, it follows that y - x = 0, so y = x . (b) Suppose x ? S . Then x ? y 8 y 2 sp( S ). Since H = sp( S ), then x = lim x N where x N 2 sp( S ). We know that h x, x N i = 0 8 N . But h x, x N i ! h x, x i , so h x, x i = 0 and thus x = 0, so S is complete. Corollary. Let { e k } be orthonormal. Then { e k } is complete i k x k 2 = P | h x, e k i | 2 8 x 2 H (i.e. Besel’s inequality holds with equality).
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3 FOURIER ANALYSIS 29 Proof. ( ) ) By part (a) of the theorem, x = P 1 k =1 h x, e k i e k . By the generalized Pythagorean Theorem, k x k 2 = 1 X k =1 k h x, e k i e k k 2 = 1 X k =1 | h e k , x i | 2 . ( ( ) Suppose { e k } is not complete. Get x 6 = 0 such that x ? e k 8 k . By assumption, k x k 2 = X | h x, e k i | 2 = X 0 = 0 ) x = 0 . Thus the result holds by contradiction. Corollary. The orthonormal set { e k } is complete i T : H ! ` 2 : x 7! ( h x, e k i ) 1 k =1 is 1-1. Proof. ( ) ) Suppose T ( x ) = T ( y ). Then h x, e k i = h y, e k i 8 k , so x - y ? e k 8 e k , so x = y by completeness. ( ( ) Suppose { e k } is not complete. Then 9 x 6 = 0 such that x ? e k 8 k . Then T ( x ) = 0 = T (0), so T is not 1-1. Lecture 19: June 17 Summary. The orthonormal set { e k } 1 k =1 is complete i any of the following are true: (0) h x, e k i = 0 8 k ) x = 0. (The definition.) (1) H = sp { e k } . (A basis for H .) (2) x = P 1 k =1 h x, e k i e k 8 x 2 H . (3) k x k 2 = P 1 k =1 | h x, e k i | 2 8 x 2 H (Bessel’s inequality holds with equality.) (4) T : H ! ` 2 : x 7! ( h x, e k i ) 1 k =1 is 1-1. We saw previously that T is onto: Given ( β k ) 2 ` 2 , x = P 1 k =1 β k e k 2 H and β k = h x, e k i . This combined with (3) gives k T ( x ) k 2 ` 2 = 1 X k =1 | h x, e k i | 2 = k x k 2 H . (5) T is an isometric (norm-preserving) isomorphism (1-1, onto, linear). Grand Theorem. Every Hilbert space H has a basis. If H is separable, then any orthonormal set is countable (or finite). Important Fact. L 2 [ a, b ] is separable because polynomials with rational coe ffi cients are dense. Proof of Grand Theorem. We first prove every Hilbert space H has a basis. Let S the the set of orthonormal sets in H . Define a partial order on S by inclusion: A B () A B . We need to show that every chain C has an upper bound. Take X = S C 2 C C . Certainly X C 8 C 2 C , so we must show X 2 S . If x 2 X , then x 2 C for some C 2 C , so k x k = 1. Let x 6 = y 2 X . Since C is a chain, there is some C 2 C such that x, y 2 C . But C is an orthonormal set, so x ? y . Thus X 2 S is an upper bound for C . By Zorn’s lemma, S has a maximal element A . If 9 x 6 = 0 such that h x, a i = 0 8 a 2 A , then A [ { x k x k } would be an orthonormal set that properly contains A , contradicting the maximality of A . Hence A is a complete orthonormal set, i.e. a basis for H . Now suppose H is separable, and we shall prove A is countable. Notice that if { e 1 , e 2 } is orthonormal, then k e 1 - e 2 k 2 = k e 1 k 2 + k - e 2 k 2 = 2. Say { e } is an uncountable orthonormal set. Let { f n } 1 1 be a countable dense subset of H . Take e and choose n such that e 2 B ( f n , 1 2 ). Since there are countably many such balls, and uncountably many e s, there exists n and e 6 = e β such that e , e β 2 B ( f n , 1 2 ). Then p 2 = k e - e β k  k e - f n k + k f n - e β | 1 ,
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3 FOURIER ANALYSIS 30 a contradiction.
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