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Normed vector space: convex set

  1. Apr 12, 2010 #1
    1. The problem statement, all variables and given/known data
    Show that the closed unit ball {x E V:||x||≤1} of a normed vector space, (V,||.||), is convex, meaning that if ||x||≤1 and ||y||≤1, then every point on the line segment between x and y has norm at most 1.
    (hint: describe the line segment algebraically in terms of x and y and a parameter t.)

    2. Relevant equations
    3. The attempt at a solution
    For the line segment between x and y, is it described by x+t(y-x), 0≤t≤1? I'm pretty sure this is correct in [tex]R^n[/tex], but is it still true in a general normed vector space? Does it still describe a straight line? Why or why not?

    Also, how can we prove that the norm of it is ≤1?

    Thanks!
     
  2. jcsd
  3. Apr 12, 2010 #2
    Yes, that is the equation for a line segment in Euclidean space (see the last part of http://planetmath.org/encyclopedia/ConvexSet.html [Broken])
    I have not thought about this much but what have you tried in showing the norm is less than 1? Have you used the triangle inequality? That was the first thing that came to my mind
     
    Last edited by a moderator: May 4, 2017
  4. Apr 12, 2010 #3
    Yes, that is the equation for a line segment in Euclidean space Rn, but is it still true in a GENERAL normed vector space? If so, why??
    In a general space, I think the norms are all screwed up and skewed, so does it even describe a straight line anymore?

    About showing the norm is ≤1, I tried the triangle inequality, but I don't think it works.
    ||x+t(y-x)|| ≤ ||x|| + t||y-x|| ≤ 1 + t||y-x||
     
  5. Apr 12, 2010 #4
    That is the definition of a convex set in a vector space I've always used. It is also in Royden's Real Analysis, in the section on Banach Spaces.
     
  6. Apr 12, 2010 #5

    LCKurtz

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    Try writing it as (1-t)x + ty first.
     
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