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Prove that the dual norm is in fact a norm

  1. Jul 4, 2010 #1
    1. The problem statement, all variables and given/known data
    Let [tex]||\cdot |[/tex]| denote any norm on [tex]\mathbb{C}^m[/tex]. The corresponding dual norm [tex]||\cdot ||'[/tex] is defined by the formula [tex]||x||^=sup_{||y||=1}|y^*x|[/tex].
    Prove that [tex]||\cdot ||'[/tex] is a norm.

    2. Relevant equations
    I think the Hölder inequality is relevant: [tex]|x^*y|\leq ||x||_p ||y||_q, 1/p+1/q=1[/tex] with [tex]1\leq p, q\leq\infty[/tex]

    3. The attempt at a solution
    Since a norm is a function satisfying three properties, I need to show that they hold.

    (1) [tex]||x||'=0[/tex] if and only if [tex]x=0[/tex].
    (2) [tex]||\alpha x||'=|\alpha| ||x||^[/tex].
    (3) [tex]||x+z||'\leq ||x||^+||z||^[/tex].

    I manage to do (1) and (2) just fine, but the triangle inequality (3) is giving me problems.

    I use the Hölder inequality to get the following:

    [tex]||x+z||'=sup_{||y||=1}|y^*(x+z)|\leq ||y|| ||x+z||=||x+z||[/tex]

    [tex]||x||'=sup_{||y||=1}|y^*x|\leq ||y|| ||x|| =||x||[/tex]

    [tex]||z||'=sup_{||y||=1}|y^*z|\leq ||y|| ||z|| =||z||[/tex]

    (1) [tex]||x||'\leq ||x||[/tex]
    (2) [tex]||z||'\leq ||z||[/tex]
    (3) [tex]||x||'+||z||' \leq ||x||+||z||[/tex]

    I also know that
    (4) [tex]||x+z|| \leq ||x||+||z||[/tex]

    I am unable to show that [tex]||x+z||\leq ||x||'+||z||'[/tex] which I think I must if I am to prove the triangle inequality.

    Any help is appreciated.
     
  2. jcsd
  3. Jul 4, 2010 #2
    [tex]||x+z||'=sup_{||y||=1} |y^*(x+z)| = sup_{||y||=1} | y^*x + y^*z | \leq sup_{||y||=1} | y^*x| + sup_{||y||=1} |y^*z | = ||x||' + ||z||' [/tex]

    The supremum-norm is a norm.
     
    Last edited: Jul 4, 2010
  4. Jul 4, 2010 #3
    Ah, didn't think of it that way. Thank you very much.
     
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