330 HW 03

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

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Unformatted text preview: MAT 330: Fourier Series and Partial Differential Equations Professor Sergiu Klainerman Problem Set 3 Rik Sengupta [email protected] 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 ω = sgn( h x , y i ), and observe that ¯ ω 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- tω y k 2 = h x- tω y , x + tω y i = X n ∈ Z ( x n- tωy n ) ( x n- tωy n ) . We wish to show that P ( t ) = 0 for some t , which will show that every term in the sequence x- t ω y is zero, or that x = ( t ω ) y , as desired. We will expand, noting that z + ¯ z = 2Re( z ) and | ω | 2 = 1. P ( t ) = X n ∈ Z ( | x n | 2- x n tωy n- tωy n x n + t 2 | y n | 2 ) = X n ∈ Z ( | x n | 2- 2Re( x n tωy n ) + t 2 | y n | 2 ) = X n ∈ Z | x n | 2- 2 t X n ∈ Z Re( x n ωy n ) + t 2 X n ∈ Z s | y n | 2 = k x k 2- 2 t Re ¯ ω X n ∈ Z x n y n !...
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This note was uploaded on 05/07/2010 for the course MAT 330 taught by Professor Sergiuklainermann during the Spring '10 term at Princeton.

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

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