1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    User Avatar
    Science Advisor
    Homework Helper

    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

    User Avatar
    Science Advisor
    Homework Helper

    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

    User Avatar
    Science Advisor
    Homework Helper

    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

    User Avatar
    Science Advisor
    Homework Helper

    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.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Equivalent Norms
  1. Equivalent norms (Replies: 2)

  2. Equivalent norms? (Replies: 4)

  3. Equivalent norms (Replies: 11)

  4. Equivalent Norms =( (Replies: 5)

Loading...