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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 twosided 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 CauchySchwarz 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.
 Spring '10
 SERGIUKLAINERMANN
 Differential Equations, Equations, Partial Differential Equations, Fourier Series, Vector Space

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