330 HW 03 - MAT 330: Fourier Series and Partial...

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Unformatted text preview: MAT 330: Fourier Series and Partial Differential Equations Professor Sergiu Klainerman Problem Set 3 Rik Sengupta rsengupt@princeton.edu February 24, 2010 1. Stein and Shakarchi, p. 88, problem 2 Prove that the vector space 2 ( Z ) is complete. Solution. Recall that the vector space 2 ( Z ) over C is the set of all two-sided infinite sequences of complex numbers ( ...,a- n ,...,a- 1 ,a ,a 1 ,...,a n ,... ) such that X n Z | a n | 2 < . First we will prove the Cauchy-Schwarz inequality in this space. First of all, note that the product h x , y i makes sense for all x , y 2 ( Z ). To see this, it is enough to note that the series n Z x n y n converges absolutely. But this is clearly true, because | x n y n | = | x n || y n | = | x n || y n | , and we get X n Z x n y n X n Z | x n y n | X n Z | x n | 2 + | y n | 2 2 < , from the way the space is defined. If at least one of x or y is zero, then both sides of the inequality are zero, so the inequality holds, and we have equality. Then one of x and y that is zero is simply the scalar 0 times the other one. So the complete result holds here. If not in the case just discussed, neither of k x k and k y k can be zero. Consider the special case k x k = k y k = 1. The right hand side of the inequality is then 1 = k x kk y k , which shows that the inequality holds still. If neither x nor y is zero, we can define x = x / k x k , and y = y / k y k , and then, since k x k = k y k = 1, we get |h x , y i| = x k x k , y k y k 1 . When we now multiply both sides of the second inequality by k x kk y k , we have the desired inequality in all cases. It only remains to prove the statement about equality in the case when neither x nor y is . Thus, we suppose |h x , y i| = k x kk y k 6 = 0. Define the sign of a complex number z C to be 1 sgn( z ) = z/ | z | if z 6 = 0, and define sgn(0) = 0. Except at z = 0, this sign is a complex number with absolute value 1, a fact we will exploit. Furthermore, note that for all z C , we have | z | = z sgn( z ). Denote = sgn( h x , y i ), and observe that h x , y i = |h x , y i| = k x kk y k in our case under study. Next, we define a function P ( t ) for t R by P ( t ) = k x- t y k 2 = h x- t y , x + t y i = X n Z ( x n- ty n ) ( x n- ty n ) . We wish to show that P ( t ) = 0 for some t , which will show that every term in the sequence x- t y is zero, or that x = ( t ) y , as desired. We will expand, noting that z + z = 2Re( z ) and | | 2 = 1. P ( t ) = X n Z ( | x n | 2- x n ty n- ty n x n + t 2 | y n | 2 ) = X n Z ( | x n | 2- 2Re( x n ty n ) + t 2 | y n | 2 ) = X n Z | x n | 2- 2 t X n Z Re( x n y n ) + t 2 X n Z s | y n | 2 = k x k 2- 2 t Re X n Z x n y n !...
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330 HW 03 - MAT 330: Fourier Series and Partial...

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