Lecture 17 june 12 note an inner product always gives

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Lecture 17: June 12 Note. An inner product always gives a norm defined by k f k = h f, f i 1 / 2 . It can be shown that this satisfies the norm definitions: k f k = h f, f i 1 / 2 = ( h f, f i ) = | | k f k and from Cauchy Schwarz ( | h f, g i |  k f kk g k ) we get k f + g k 2 = h f + g, f + g i = h f, f i + h f, g i + h g, f i + h g, g i = k f k 2 +2Re h f, g i + k g k 2  k f k 2 +2 k f kk g k + k g k 2 = ( k f k + k g k ) 2 Example. (1) C n with k ( x 1 , . . . , x n ) k = (P n i =1 | x i | 2 ) 1 / 2 . (2) ` 2 with k ( x n ) k ` 2 = (P 1 i =1 | x i | 2 ) 1 / 2 . (3) L 2 ( E ) with k f k L 2 ( E ) = (R E | f | 2 ) 1 / 2 . (4) C [0 , 1] with k f k = k f k L 2 [0 , 1] . (Note this is not the usual sup-norm.)
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3 FOURIER ANALYSIS 26 Note. k f k = h f, f i 1 / 2 = h f, f i 1 h f, f i 1 / 2 = h f, f i 1 k f k = f, f k f k = h f, g i where g = f k f k and k g k = f k f k = 1. Definition. A Hilbert space is a complete inner product space, i.e. a Banach space where the norm comes from an inner product. Example. 1. C n is a Hilbert space. 2. ` 2 is a Hilbert space. 3. L 2 ( E ) is a Hilbert space, by Reisz-Fisher theorem. 4. C [0 , 1] is not a Hilbert space, because it is dense in L 2 with k · k L 2 . (It is complete with respect to the sup-norm, but the sup-norm does not come from an inner product.) 3.2 Orthogonality Let H be a Hilbert space. Definition. x, y 2 H are orthogonal (denoted x ? y ) are orthogonal if h x, y i = 0. Example. 1. (1 , 0 , 0) ? (0 , 1 , 0) in C 3 . 2. χ E ? χ E C in L 2 ( R ). 3. x ? x i x = 0. Definition. We say S H is an orthogonal set if x ? y whenever x, y 2 S and x 6 = y . Moreover, S is orthonormal if it is orthogonal and k x k = 1 8 x 2 S . Note. In an orthonormal set S , h x, y i = ( 0 x 6 = y 1 x = y . Example. Consider H = ` 2 . Put e n have a 1 in the n th position and a 0 everywhere else. S = { e n } n is orthonormal. Pythagorean Theorem. If x ? y , then k x + y k 2 = k x k 2 + k y k 2 . Proof. k x + y k 2 = h x + y, x + y i = h x, x i + h x, y i | {z } 0 + h y, x i | {z } 0 + h y, y i = k x k 2 + k y k 2 . Corollary. More generally, if { x k k N 1 is orthogonal, then P N k =1 x k 2 = P N k =1 k x k k 2 . Theorem. If S is orthogonal and consists of nonzero vectors, then S is a linearly independent set. Proof. Suppose P N i =1 a i x i = 0, where x i 2 S and a i 2 C . Then 0 = N sup i =1 a i x i 2 = X k a i x i k 2 = X | a i | 2 k x i k 2 ) | a i | 2 k x i k 2 = 0 8 i and thus x i = 0 8 i .
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3 FOURIER ANALYSIS 27 Example. Let x = P N i =1 a n e n 2 ` 2 . Then k x k 2 ` 2 = N X i =1 k a n e n k 2 = N X i =1 | a n | 2 . Let ( a n ) C with P 1 i =1 | a n | 2 < 1 . We claim that x = P n i =1 a n e n 2 ` 2 . Let x N = P N n =1 a n e n 2 ` 2 . We check that this is a Cauchy sequence in ` 2 . Suppose J < K . Then k x J - x K k 2 = K X n = J +1 a n e n 2 = K X n = J +1 k a n e n k 2 = K X n = J +1 | a n | 2 . Hence ( x N ) converges in ` 2 , i.e. lim N !1 P N n =1 a n e n converges. Proposition. If { x k } k is orthogonal in H , then P 1 k =1 x k converges (in H ) i P n k =1 k x k k 2 H converges (in R ). Proof. Let S n = P N k =1 x k 2 H . Suppose M < N . Then k S N - S M k 2 = N X k = M +1 x k 2 = N X k = M +1 k x k k 2 = N X k =1 k x k k 2 - M X k =1 k x k k 2 Hence ( S N ) is Cauchy in H i P N k =1 k x k k 2 is Cauchy in R .
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