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Help with continuous functions in metric spaces

  1. Nov 28, 2011 #1
    hi guys,
    I have a question I would like assistance with:

    let (v,||.||) be a norm space over ℝ, and let f:v→ℝ be a linear functional.
    if f is continuous on 0 (by the metric induced by the norm), prove that there is k>0 such that for each u in v, |f(u)| ≤ k*||u||.

    thanks :)
  2. jcsd
  3. Nov 28, 2011 #2


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    I am a little confused here, do you consider ℝ also as a normed space, or is f an element of the dual V*? If f is an element of the dual, what notion of continuity do you have?

    If f is a linear operator between normed spaces, then continuity at a point implies global continuity (by translation; given T(x), we can get T(y)=T(y-x+x) , and linearity does the rest), and, for linear operators on finite-dimensional space, continuity implies boundedness; express any vector v in terms of a finite basis, and then use inequalities to find a bound for ||T(v)||. Is that the question?
  4. Nov 29, 2011 #3
    Thank you for replying (I am new to this forum and I just realized that I am not supposed to ask homework-style questions here...)

    R is also considered as a normed space, but v is not necessarily a finite dimensional space, therefore I can't use boundedness...
  5. Nov 29, 2011 #4


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    Continuous operators are bounded at the origin even for the infinite dimensional case, but that's essentially what you're proving here so if it hasn't been brought up in class I would be hesitant to use it.

    Suppose that the claim is not true. Then for each integer n, there is some un such that |f(u)| > n||u||. Prove f is not continuous at the origin
  6. Nov 30, 2011 #5

    We have some δ st for all p with ||p||<δ we have |f(p)-f(0)| = |f(p)| < ε for any ε. Then for any unit vector v in our space

    |f(v)| = 1/δ |f(δ v )| < ε||v||/δ =ε/δ since ||δ v || = δ. But the norm of the opearator is


    so we are done if you just pick for ε any number n .
    Last edited: Nov 30, 2011
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