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Linear forms and complete metric space

  1. Nov 11, 2007 #1

    quasar987

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    [SOLVED] Linear forms and complete metric space

    1. The problem statement, all variables and given/known data
    Question:

    Let L be a linear functional/form on a real Banach space X and let {x_k} be a sequence of vectors such that L(x_k) converges. Can I conclude that {x_k} has a limit in X?

    It would help me greatly in solving a certain problem if I knew the answer to that question.

    3. The attempt at a solution

    The natural approach is to try to show that {x_k} is Cauchy.

    Since the sequence of real numbers {L(x_k)} converges, then it is Cauchy, so for n,k large enough,

    [tex]|L(x_k)-L(x_n)|=|L(x_k - x_n)|<\epsilon[/tex]

    Now what??
     
  2. jcsd
  3. Nov 11, 2007 #2
    If L is not invertible what can happen?
     
  4. Nov 11, 2007 #3

    morphism

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    edit: removing my too explicit hint. :tongue2:
     
    Last edited: Nov 11, 2007
  5. Nov 11, 2007 #4
    Boo. I'm sure he could have figured it out on his own.

    What is the particular quandary?

    Edit: Suppose L is invertible! What can you say then?
     
    Last edited: Nov 11, 2007
  6. Nov 11, 2007 #5

    quasar987

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    Too late morphism! :D
     
  7. Nov 11, 2007 #6

    quasar987

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    Anyway, guys, in the problem I'm working on, I must show that in a particular situation, the x_k do have a limit in X.

    Are you willing to help with the general problem?

    If so, I will type it out. Not very long, but a little complicated notation-wise.
     
  8. Nov 11, 2007 #7

    quasar987

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    It's at the core, a problem on measure theory.

    Let [tex]\mathcal{L}^2(\mathbb{R})[/tex] denote the space of square-integrable functions f on R (with respect to the Lebesgue sigma-algebra and Lebesgue measure [itex]\lambda[/itex]) and let [tex]\mathcal{L}^2_{\lambda}(\mathbb{R})[/tex] denote the space of their equivalence classes f where f=g if f=g almost everywhere. [f]+[g]=[f+g] and c[f]=[cf] are well defined.

    Now with the norm

    [tex]||\mathbf{f}||_2=\int_{-\infty}^{+\infty}|f(x)|^2dx[/tex]

    [tex](\mathcal{L}^2_{\lambda}(\mathbb{R}),||||_2)[/tex] is a real Banach space.

    A continuous linear form on [tex](\mathcal{L}^2_{\lambda}(\mathbb{R}),||||_2)[/tex] is a linear form [tex]L:\mathcal{L}^2_{\lambda}(\mathbb{R})\rightarrow \mathbb{R}[/tex] such that

    [tex]||L||=\sup\left\{\frac{|L(\mathbf{f})|}{||\mathbf{f}||_2}: \mathbf{f}\neq \mathbf{0}\right\}=\sup\left\{|L(\mathbf{f})|:||\mathbf{f}||_2=1\right\}<+\infty[/tex]

    Now consider a sequence [tex]\mathbf{g}_k\in \mathcal{L}^2_{\lambda}(\mathbb{R})[/tex] be such that

    [tex]||\mathbf{g}_k||_2=1[/tex] and [tex]\lim_{k\rightarrow\infty}L(\mathbf{g}_k)=||L||[/tex]

    Show that [tex]\mathbf{g}_k[/tex] has a limit in [tex]\mathcal{L}^2_{\lambda}(\mathbb{R})[/tex].
     
    Last edited: Nov 11, 2007
  9. Nov 11, 2007 #8

    morphism

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    Is L a fixed functional? If so, I don't think this is true. Try constructing a counterexample using the zero functional.
     
  10. Nov 11, 2007 #9

    quasar987

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    Apologies!

    There is an additional hypothese! L is a non-identically vanishing functional!
     
  11. Nov 11, 2007 #10

    morphism

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    I'm tempted to use the Riesz representation theorem: we know there is a nonzero f in L^2 such that L(g)=<f,g> for all g in L^2, and ||L||=||f||_2.

    Now consider ||f - g_n||[itex]_2^2[/itex]. (Hint: apply the polarization identity, and use the fact that <f,g_n> -> ||f||.) Try to see if you can guess what (g_n) converges to using this.
     
  12. Nov 12, 2007 #11

    quasar987

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    We are not to use this theorem in this problem, because in a sense, the whole problem sheet comes down to showing explicitely that the Riesz representation theorem hold in the case of L². No prior knowledge of functional analysis should be needed to do this problem.

    Someone told me he succeeded in answering this question by effectively proving that the sequence g_k was Cauchy!
     
  13. Nov 13, 2007 #12

    morphism

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    Did you manage to do it without Riesz?
     
  14. Nov 13, 2007 #13

    quasar987

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    Yes, with the help of aforementioned person. :)

    With the parallelogram identity, we reduce the problem to showing [tex]||\mathbf{g}_k+\mathbf{g}_{k+p}||_2^2 \rightarrow 4[/tex]

    Then notice that because ||L|| is the sup,

    [tex]\frac{L(\mathbf{g}_k+\mathbf{g}_{k+p})}{||\mathbf{g}_k+\mathbf{g}_{k+p}||_2}\leq ||L|| \ \ \ \ \ \ (*)[/tex]

    On the other hand, write out the facts that L(g_k)-->||L|| and L(g_{k+p})-->||L|| and add the inequalities to obtain

    [tex]L(\mathbf{g}_k+\mathbf{g}_{k+p})>2(||L||-\epsilon')[/tex]

    Combine with equation (*) to obtain an inequality involving [tex]||\mathbf{g}_k+\mathbf{g}_{k+p}||_2[/tex] and [tex]\epsilon'[/tex]. Show that to any [tex]\epsilon>0[/tex], you can find an [tex]\epsilon'(\epsilon)[/tex] such that [tex]4-||\mathbf{g}_k+\mathbf{g}_{k+p}||_2^2<\epsilon^2[/tex] for k large enough.
     
    Last edited: Nov 13, 2007
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