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Norms in sequence spaces.

  1. Mar 10, 2012 #1
    Im trying to understand the following. We have l_1(R)=( x=x_n in l(R): summation from n=1 to infinity for absolute value of x_n). It says that this summation converges, but converges to what?

    Also , its says (1) is not in l_1(R) but 1/n^2 is. Can some one explain how these are so.

    This is not homework, just trying 2 gain an understanding. Thanks.
  2. jcsd
  3. Mar 10, 2012 #2


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    Gold Member

    Please clarify. What is l_1(R)? What is (1)? Also your parenthetical expression ( x=x_n etc.) is not very clear.
  4. Mar 11, 2012 #3
    Ok, I will include a few additional lines previous to it. We are examining norms on sequences spaces.
    The notes say that ##|| ||_1## cannot be extended to the set of all real sequences
    ie ##x=(1)=(1,1,1...)## ##||x||_1=1+1+1+1## does not exist. What does this mean?

    So we define

    ##l_1(R)=\{x=(x_n) \in l(R) : Ʃ |x_n|\}## from n=1 to infinity.
    It says this summation converges, but to what?

    Then it says
    eg ##(1) \notin l_1(R)## and ##1/n^2 \in l_1(R)##
    How are these determined?

  5. Mar 11, 2012 #4


    Staff: Mentor

    You're being a little sloppy here.
    As you have defined x, ||x||1 = 1 + 1 + 1 + ... + 1 + ..., not 1 + 1 + 1 + 1, which equals 4.

    The expression "does not exist" here means that ||x||1 is not a finite number. I.e., the sum is not finite.
    Have you omitted something here? The usual way of saying a summation converges is to write it as Ʃ |xn| < ∞.
    Ʃ 1 = 1 + 1 + 1 + ..., which is divergent.
    Ʃ 1/n2 is a well known convergent series (a p-series, with p = 2).
  6. Mar 12, 2012 #5

    I am looking at some ones notes. Yes you are correct. It appears in another similar example.
    It seems against my intuition. Isn't it not approaching infinity? How would it converge to < infinity?

  7. Mar 12, 2012 #6


    Staff: Mentor

    I gave two examples, so I'm not sure which one you're referring to by "it."
    Ʃ1 diverges
    Ʃ1/n2 converges, so Ʃ1/n2 < ∞.
  8. Mar 12, 2012 #7
    Ok, I think I get you now. Thanks
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