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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 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 = 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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 Spring '10
 SERGIUKLAINERMANN
 Differential Equations, Equations, Partial Differential Equations, Fourier Series, Vector Space

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