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Linear functionals on a normed vector space

  1. Jul 23, 2011 #1
    I have a question: If [itex]x\in X[/itex] is a normed vector space, [itex]X^*[/itex] is the space of bounded linear functionals on [itex]X[/itex], and [itex]f(x) = 0[/itex] for every [itex]f\in X^*[/itex], is it true that [itex]x = 0[/itex]? I'm reasonably confident this has to be the case, but why? The converse is obviously true, but I don't see why there couldn't be an example of a normed space for which all the functionals in [itex]X^*[/itex] are zero at some other value of [itex]x[/itex]...
     
    Last edited: Jul 23, 2011
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  3. Jul 24, 2011 #2

    jambaugh

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    Start by considering expanding [itex]x[/itex] in a basis, dual some basis of [itex]X^*[/itex]. Clearly for the basis [itex]\{f_k\}[/itex] and dual basis [itex]\{e^k\}[/itex](if there is one) of [itex]X[/itex] you have [itex] x = x_k e^k [/itex] where [itex] x_k = f_k(x)[/itex].

    The question then becomes whether there exists such a dual basis. You don't have to worry about considering every element of [itex]X^*[/itex] but rather only a basis. Finally the fact that you have a metric space may shed some light on this question.
     
    Last edited: Jul 24, 2011
  4. Jul 24, 2011 #3

    micromass

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    You will need the Hahn-Banach theorem for this. It states exactly what you want:


    If V is a normed vector space and if z is in V, then there is a continuous function such that [itex]f(z)=\|z\|[/itex] and such that [itex]\|f\|\leq 1[/itex]. ​

    The correspond result for metric spaces may fail. For example

    [tex]L^{1/2}(\mathbb{R})=\{f:\mathbb{R}\rightarrow \mathbb{R}~\vert~\int_{-\infty}^{+\infty}{\sqrt{f}}<+\infty\}[/tex]

    with metric

    [tex]d(f,g)=\int_{-\infty}^{+\infty}{\sqrt{f-g}}[/tex]

    is a metric vector space such that 0 is its only continuous functional!
     
  5. Jul 25, 2011 #4
    Haha...yep, that's perfect! Thanks a lot!
     
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