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Homework Help: Equivalent Norms

  1. Feb 21, 2010 #1
    1. The problem statement, all variables and given/known data
    In [tex]l_1[/tex] (each element is a sequence of numbers such that the series converges absolutely) show that no pair of the norms [tex]||\cdot||_1, \, ||\cdot||_2, \, ||\cdot||_{\infty} [/tex] are equivalent norms.


    2. Relevant equations
    [tex]||x||_1=\sum_{i=1}^{\infty}|x_i|[/tex]
    [tex]||x||_2=\left(\sum_{i=1}^{\infty}|x_i|^2\right)^{\frac{1}{2}}[/tex]
    [tex]||x||_{\infty}=\mbox{ max}\{|x_i|:i=1,2,\ldots\}[/tex]


    3. The attempt at a solution
    I realize I need to show:
    [tex]\dfrac{||x||_1}{||x||_2}=\infty \text{ or } 0\quad \dfrac{||x||_1}{||x||_{\infty}}=\infty \text{ or } 0\quad \dfrac{||x||_2}{||x||_{\infty}}=\infty \text{ or } 0[/tex]
    but I am having problems finding such a sequence and showing this is true.
     
  2. jcsd
  3. Feb 21, 2010 #2

    Dick

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    You can't find a SINGLE sequence that will prove they aren't equivalent. You need to find a sequence of sequences. For example, can you find a sequence of sequences x_n where ||x_n||_infinity remains constant as n increases but ||x_n||_1 goes to infinity? That would prove ||_infinity and ||_1 were not equivalent norms, right?
     
    Last edited: Feb 21, 2010
  4. Feb 21, 2010 #3
    How about [tex]x_n=n^{-1/2}[/tex]

    Then: [tex]||x_n||_1[/tex] diverges while [tex]||x_n||_{\infty}=1[/tex]

    But, this sequence does not converge absolutely so it wouldn't be in the space, right?
     
  5. Feb 21, 2010 #4

    Dick

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    That's still a single sequence. You want a sequence of sequences. Define x_n to be the sequence where the first n terms are 1 and the rest are zero. What are ||x_n||_infinity and ||x_n||_1?
     
  6. Feb 21, 2010 #5
    In that case,
    [tex] ||x_n||_1=\infty[/tex]
    and
    [tex] ||x_n||_{\infty}=1[/tex]

    But again, this is not a convergent sequence, so it wouldn't be in the space in the first place right?
     
    Last edited: Feb 21, 2010
  7. Feb 21, 2010 #6

    Dick

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    You aren't quite getting this. x_n isn't a single sequence. It's a different sequence for each value of n. x_1=(1,0,0,0...), x_2=(1,1,0,0,0..), x_3=(1,1,1,0,0...) etc etc. All of those are in l_1. And, yes, ||x_n||_infinity=1. But ||x_n||_1=n, right? What does that tell you about the possibility of the norms being equivalent?
     
  8. Feb 21, 2010 #7
    You're right, I was being dense (my steady state as of late).

    So,

    [tex]\dfrac{||x_n||_1}{||x_n||_{\infty}}=n [/tex]

    Then,
    [tex]\dfrac{||x_n||_1}{||x_n||_2}=\dfrac{n}{\sqrt{n}}=\sqrt{n} [/tex]

    Lastly,
    [tex]\dfrac{||x_n||_2}{||x_n||_{\infty}}=\sqrt{n} [/tex]

    Or, am I still lost?
     
  9. Feb 21, 2010 #8

    Dick

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    No, I think you are getting it. Since all of those diverge as n->infinity, that means you can't bound any of your norms by a constant times the other norm, right?
     
  10. Feb 21, 2010 #9
    Right, Thanks Dick. I appreciate it. I guess I kept getting stuck because I wasn't thinking about a sequence of sequences correctly and I kept thinking of [tex]l_1[/tex] incorrectly.
     
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