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Convergence with L2 norm functions

  1. Oct 21, 2007 #1
    1. The problem statement, all variables and given/known data
    (I'm posting this because my proofs seem to be lousy. I want to see if I'm missing anything.)
    Show that if [tex] f_n \in L^2(a,b) [/tex] and [tex] f_n \rightarrow f [/tex] in norm, then [tex] <f_n,g> \rightarrow <f,g> [/tex] for all [tex] g \in L^2(a,b) [/tex]

    2. Relevant equations
    [tex] L^2(a,b) [/tex] is the space of square-integrable functions,
    [tex] {f_n} [/tex] is a finite sequence of piecewise continuous functions, and
    < , > is the inner product


    3. The attempt at a solution

    I started with a linear combination of inner products and applied the Cauchy Schwarz inequality:
    [tex] \vert <f_n - f,g> \vert \leq \Vert f_n - f \Vert \Vert g \Vert [/tex]
    By the definition of norm convergence, I have
    [tex] \Vert f_n - f \Vert \rightarrow 0 [/tex], which means that
    [tex] \vert <f_n - f,g> \vert \rightarrow 0 [/tex]
    Since this is absolutely convergent, that means that
    [tex] <f_n,g> - <f,g> [/tex] is also convergent to 0. So therefore,
    [tex] <f_n,g> -<f,g> \rightarrow 0[/tex]
    [tex] <f_n,g> \rightarrow <f,g> [/tex]

    Is that reasonable? Or am I missing something?
     
  2. jcsd
  3. Oct 21, 2007 #2

    EnumaElish

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    I suggest going from <fn - f , g> ---> 0 to <fn , g> - <f , g> ---> 0 more explicitly.
     
  4. Oct 22, 2007 #3
    As in:
    <fn-f,g> --> 0 (absolutely convergent series are convergent)
    <fn,g> - <f,g> --> 0 (linearity property wrt the first variable for inner products)
     
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