MATH501_HW6 - Serge Ballif MATH 501 Homework 6 November 9,...

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Serge Ballif MATH 501 Homework 6 November 9, 2007 1. Let ( V, , ·i ) be an inner product space. Prove that if |h x,y i| = k x k·k y k and y 6 = 0 , then x = ay for some a C . We follow the proof outlined in Folland. Define a = h x,y i |h x,y i| and z = ay . Claim 1. h x,z i = h z,x i = |h x,y i| and k z k = k y k . Proof. We use the fact that for α C , α α = | α | 2 . h x,z i = h x,ay i = a h x,y i = h y,x i |h x,y i| h x,y i = |h x,y i| 2 |h x,y i| = |h x,y i| . h z,x i = h ay,x i = a h y,x i = h x,y i |h x,y i| h y,x i = |h x,y i| 2 |h x,y i| = |h x,y i| . h z,z i = h ay,ay i = a a h y,y i = h x,y i |h x,y i| h y,x i |h x,y i| h y,y i = h y,y i . Now we can use these facts to show that for t R 0 ≤ h x - tz,x - tz i = k x k 2 + 2 Re h x, - tz i + k - tz k 2 = k x k 2 - 2 t |h x,z i| + t 2 k z k 2 = k x k 2 - 2 t |h x,y i| + t 2 k y k 2 . Since any quadratic function ax 2 + bx + c has its extremum at x = - b 2 a , our quadratic function in t has an absolute minimum at t = |h x,y i| k y k 2 . Substituting this value in for t yields 0 ≤ k x - tz k 2 = k x k 2 - 2 t h x,y i + t 2 k y k 2 = k x k 2 - 2 |h x,y i| k y k 2 |h x,y i| + |h x,y i| k y k 2 2 k y k 2 = k x k 2 - |h x,y i| k y k 2 . In this case |h x,y i| ≤ k x kk y k . We get equality in this statement iff x - tz = x - tay = 0 iff x and y are linearly dependent. 2. (a) Let y be a nonzero element of a Hilbert space H . Prove that, for every x H , d ( x, { y } ) = |h x,y i| k y k . (b) Consider the space L 2 ([0 , 1]) and E = { f L 2 ([0 , 1]) | R [0 , 1] f ( x ) dm = 0 } . Determine E and find the distance of the function e x to E . Page 1 of 5
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Serge Ballif MATH 501 Homework 6 November 9, 2007 (a) We know that there exists an element t of { y } such that d ( x,t ) = d ( x, { y } ). We want to show that
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MATH501_HW6 - Serge Ballif MATH 501 Homework 6 November 9,...

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