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Proof that a given subspace of C[−1,1] with L2 norm is closed

  1. Feb 28, 2013 #1
    1. The problem statement, all variables and given/known data

    Let H= C[-1,1] with L^2 norm and consider G={f belongs to H| f(1) = 0}. Show that G is a closed subspace of H.


    2. Relevant equations

    L^2 inner product: [tex]<f,g>\to \int_{-1}^{1}f(t)\overline{g(t)} dt[/tex]

    3. The attempt at a solution

    I've been trying to prove this for a while but i can't establish that given [tex]||f_{n}-f||< \epsilon[/tex] (where the norm is the L^2 one) we have uniform convergence for the sequence [tex](f_{n})[/tex]If I could prove this the result would follow easily given that G is contained in its closure and if [tex](f_{n})[/tex] converged uniformly we would have [tex]f(1)=\lim_{x \to 1} \lim_{n \to \infty} f_n(x)=\lim_{n \to \infty} \lim_{x \to 1} f_n=0[/tex] and thus that f belongs G
     
    Last edited: Feb 28, 2013
  2. jcsd
  3. Feb 28, 2013 #2

    micromass

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    You won't be able to show uniform convergence.

    Try to prove it by contradiction. Assume that [itex]f(1)\neq 0[/itex].
     
  4. Feb 28, 2013 #3
    Hmm, if [tex]f(1)\neq 0[/tex] then [tex]f(t)-f_n(t) \neq 0[/tex] in a open interval around one.
    But shouldn't it be possible to have a [tex]f_n[/tex] as close as possible to f(t) until 1-1/n say and then decrease linearly to zero? I know this function in particular is discontinuous in the n to infinity limit but can you give me a hint of how i prove it must me discontinuous for all possible [tex]f_n[/tex]
     
  5. Feb 28, 2013 #4

    micromass

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    This function [itex]f_n[/itex] that you mention converges to [itex]f[/itex]. Remember that you're dealing with the integral norm. The integral norm behaves very strangely.

    But let's proceed naively. We know [itex]f_n\rightarrow f[/itex]. We assumed [itex]f(1)\neq 0[/itex]. And we know that [itex]f_n(1)=0[/itex].

    Can you use continuity to find a point [itex]x_n[/itex] such that [itex]f(x_n)[/itex] is "far away" from 0. But [itex]f[/itex] and [itex]f_n[/itex] are closed in the L2-norm. Can you use this to say anything about the [itex]x_n[/itex]?? It seems reasonably that [itex]x_n[/itex] is rather close to 1. Can you give a formula for how close?
     
  6. Feb 28, 2013 #5
    Hmm, i see it now. I was confusing norms. Thanks for the help!
     
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