Corollary every separable hilbert space h is

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Corollary. Every separable Hilbert space H is isometrically isomorphic to ` 2 . Proof. It is a countable orthonormal set { e k } , and use the preceding theorem. Note. In fact, T : H ! ` 2 : x 7! ( h x, e k i ) 1 k =1 preserves inner products. Proof. h x, y i H = * 1 X k =1 h x, e k i e k , y + = 1 X k =1 h x, e k i h e k , y i = 1 X k =1 h x, e k i h y, e k i = h T ( x ) , T ( y ) i ` 2 Lecture 20: June 19 O ffi ce Hours Next Week. Monday 2-3; Tuesday 2:30-3:30; None on Wednesday, Thursday. Example. We consider L 2 [0 , 1] with h f, g i = R 1 0 f g . This is separable. A countable orthonormal basis is { e 2 inx } 1 n = -1 . We verify that this is an orthonormal set in L 2 [0 , 1]. e 2 inx , e 2 inx = Z 1 0 1 = 1 . Now if n 6 = m , e 2 inx , e 2 imx = Z 1 0 e 2 i ( n - m ) x dx = 1 2 i ( n - m ) h e 2 i ( n - m ) x i 1 0 = 0 . We will prove that it is a basis for L 2 [0 , 1] (eventually). Assuming that, this means for all f 2 L 2 , f = 1 X -1 f, e 2 inx e 2 inx . where f, e 2 inx = Z 1 0 f ( x ) e - 2 inx dx = ˆ f ( n ) 2 C is called the n th Fourier coe ffi cient of f . Then f = P 1 n = -1 ˆ f ( n ) e 2 inx . Then we have Parseval’s theorem. k f k L 2 = v u u t 1 X n = -1 | ˆ f ( n ) | 2 = k ( ˆ f ( n )) k ` 2 3.4 Brief Historical Introduction to Fourier Analysis Wave Equation – Vibrating String Problem. Model: Let u ( x, t ) be the vertical displacement at position x at time t . Assume the string has uniform density. It is known that u satisfies the wave equation: @ 2 u @ x 2 = c 2 @ 2 u @ t 2 (1)
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3 FOURIER ANALYSIS 31 for 0 x 1, and t 0, where c is determined by the string. To solve this, we need two initial conditions: u ( x, 0) = g ( x ) (2) @ u @ t ( x, 0) = v ( x ) . (3) This was solved by d’Alembert in 1740: where the solution is u ( x, t ) = 1 2 g x + t c + g x - t c + c 2 Z x + t c x - t c v ( s ) ds ! Bernoulli (1755) gave the following “proof” of uniqueness: Assume we can write g ( x ) = P 1 n = -1 a n e 2 inx (can we?). Then a n = ˆ g ( n ) = R 1 0 g ( x ) e - 2 inx dx . Similarly suppose that for each t that we can write u ( x, t ) = P 1 n = -1 c n ( t ) e 2 inx . Let’s di erentiate: @ u @ x = @ u @ x 1 X n = -1 c n ( t ) e 2 inx =? 1 X n = -1 c n ( t ) @ u @ x e 2 inx = 1 X n = -1 c n ( t )2 ine 2 inx @ 2 u @ x 2 =? 1 X n = -1 c n ( t )(2 in ) 2 e 2 inx Similarly @ 2 u @ t 2 =? 1 X n = -1 c 00 n ( t ) e 2 inx By (1), 1 X n = -1 c n ( t )(2 in ) 2 e 2 inx = c 2 1 X n = -1 c 00 n ( t ) e 2 inx ) 1 X n = -1 ( c n ( t )(2 in ) 2 - c 2 c 00 n ( t )) e 2 inx = 0 Thus c n ( t )(2 in ) 2 - c 2 c 00 n ( t ) =?0 ) c 00 n ( t ) = c n ( t ) (2 n ) 2 c 2 This has the solution c n ( t ) = A n e 2 i n c t + B n e - 2 i n c t for some constants A n , B n . Hence u ( x, t ) = 1 X n = -1 A n e 2 in n c t + B n e - 2 i n c t e 2 inx = 1 X n = -1 A n e 2 in ( x + t c ) + B n e 2 in ( x - t c ) From (2), we have u ( x, 0) = g ( x ) =? 1 X n = -1 ( A n + B n ) e 2 inx = 1 X n = -1 a n e 2 inx (do we have pointwise convergence?) so A n + B n = a n = ˆ g ( n ). Now using (3), @ u @ t =? 1 X n = -1 A N 2 i n c e 2 in ( x + t c ) + B n - 2 i n c e 2 in ( x - t c )
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3 FOURIER ANALYSIS 32 @ u @ t ( x, 0) =?
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