Linear functionals on a normed vector space

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Discussion Overview

The discussion revolves around the properties of bounded linear functionals in normed vector spaces, specifically addressing the question of whether a vector \( x \) in a normed vector space \( X \) must be zero if every functional \( f \) in the dual space \( X^* \) evaluates to zero at \( x \). The scope includes theoretical exploration and implications of the Hahn-Banach theorem.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant questions whether \( x = 0 \) is the only solution if \( f(x) = 0 \) for all \( f \in X^* \), expressing uncertainty about the existence of a normed space where this might not hold.
  • Another participant suggests considering the expansion of \( x \) in terms of a basis and its dual, raising the issue of whether a dual basis exists.
  • A third participant introduces the Hahn-Banach theorem, asserting that it provides a continuous functional that can demonstrate the property in question, but notes that similar results may not hold in metric spaces.
  • A later reply acknowledges the relevance of the Hahn-Banach theorem, indicating it aligns with the initial inquiry.

Areas of Agreement / Disagreement

Participants express differing views on the necessity of \( x \) being zero under the given conditions, with some supporting the assertion through the Hahn-Banach theorem while others remain skeptical about the generality of the claim.

Contextual Notes

The discussion highlights the dependency on the existence of a dual basis and the implications of the Hahn-Banach theorem, as well as the potential limitations when considering metric spaces.

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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]...
 
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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.
 
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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!
 
micromass said:
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].​

Haha...yep, that's perfect! Thanks a lot!
 

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