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Sequence in N.L.S

  1. Mar 9, 2012 #1
    1. The problem statement, all variables and given/known data

    Consider ##R^2## with the sup norm ##|| ||_∞## defined by ##||x||_\infty=sup(|x_1|,|x_2|)## for ##x = (x1, x2)##.

    Show that a sequence

    ##x^{n} \in (R^2, || ||_\infty)## where

    ##x^{n} =(x^ {n}_1, x^{n}_2) ## converges to

    ##x = (x_1, x_2) \in R^2## (with the sup norm) if and only if

    ## x^{n}_1 \to x_1##, and ##x^{n}_2→ x_2 \in R##.




    3. The attempt at a solution

    We need to show ##x^n_1 \to x_1## and ##x^n_2 \to x_2 \in R##

    Assume that ##x^n_1 \to x_1 \in (R^2, || ||_\infty)##, we know


    ## \exists n_0 in N s.t ||x^n-x||_\infty < \epsilon \forall n>n_0## ie

    ##||(x^n_1,x^n_2)-(x_1,x_2)||_\infty =sup|(x^n_1-x_1, x^n_2-x_2)| < \epsilon ##
    ##= max|(x^n_1-x_1, x^n_2-x_2)|##

    We have that

    ##|x^n_1-x_1| \le max (x^n_1-x_1, x^n_2-x_2) < \epsilon \forall n>n_0##

    ##|x^n_1-x_1| < \epsilon##

    This shows ##x^n_1 \to x_1 as n \to \infty##
    Similarly for ##|x^n_2-x_2|##.....?
     
  2. jcsd
  3. Mar 9, 2012 #2

    LCKurtz

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    I would reverse the order of these last two sentences, putting the assumption before the statement of what we need to show.
    Be more careful with your notation here. You have absolute value signs around an ordered pair. Don't write ##|(a,b)|## when you mean ##(|a|,|b|)##.
    Yes, you have the idea. But you realize that that completes only half the problem, right? You now have to show that if the two components converge then the sequence converges in the ##\|\cdot \|_\infty## norm.

    I would also suggest a slightly less cumbersome notation. You might call your point ##x \in R^2## and denote its components by ##(x(1),x(2))## so you can just think of ##x## as a function on the first two integers. Using this notation avoids using superscripts which could be confused with exponents. So what you have done so far is to show that if ##x_n \rightarrow x## in ##\|\cdot \|_\infty## then ##x_n(1)\rightarrow x(1)##. When you do ##x(2)## you will have finished the first half. You have used the fact that for any ##x##, ##|x(1)|\le \| x\|_\infty## and ##|x(2)|\le \| x\|_\infty##

    Now you have to do the converse. You might want to think about how ##\| x\|_\infty## compares with ##|x(1)|+|x(2)|##.
     
    Last edited: Mar 9, 2012
  4. Mar 10, 2012 #3
    Ok, Im starting from scratch because I was a little confused. Using your notation:

    Assume ##x_n \to x##. We need to show ##x_n(1) \to x(1)## and ##x_n(2) \to x(2)##
    ie, we need to find ##n_0 \in N## s.t ##|x_n(1)-x(1)| < \epsilon \forall n > n_0##

    Since we know ##x_n \to x \in R^2, || ||_\infty## we know ##\exists n_0 \in N## s.t

    ##||x_n-x||_\infty < \epsilon \forall n \ge n_0## ie

    ##||(x_n(1)-x(1), x_n(2)-x(2))||_\infty < \epsilon##

    We have that ##|x_n(1)-x(1)| \le max (|x_n(1)-x(1)|, |x_n(2)-x(2)|) < \epsilon##
    This shows ##x_n(1) \to x(1)## as ##n \to \infty##

    Similarly

    We have that ##|x_n(2)-x(2)| \le max (|x_n(1)-x(1)|, |x_n(2)-x(2)|) < \epsilon##
    This shows ##x_n(2) \to x(2)## as ##n \to \infty##

    Now, conversely assume ##x_n(1) \to x(1)## and ##x_n(2) \to x(2)##
    Need to show that ##x_n \to x in R^2, || ||_\infty##. Need to find
    ##n_0 \in N## s.t. ##||x_n-x||_\infty \forall n \ge n_0##

    but we know that ##x_n(1) \to x(1), \exists n \in N## s.t.

    ##|x_n(1)-x(1)| < \epsilon/2 \forall n \ge n_1## and
    ##|x_n(2)-x(2)| < \epsilon/2 \forall n \ge n_2##

    Take ##n_0=max(n_1,n_2)##.

    Then ##\forall n > n_0## we have ##||x_n-x||_\infty=max(|x_n(1)-x(1)|,|x_n(2)-x(2)|) < \epsilon/2 + \epsilon/2 < \epsilon##

    Therefore
    ##x_n \to x##.........?
     
  5. Mar 10, 2012 #4

    LCKurtz

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    OK, you have the idea. You just need to be a little more careful in the writeup.

    You haven't told us what ##\epsilon## is so I inserted it. And I see you did both below but you should still state both above in what you need to prove.

    Insert ## < \epsilon## there
    Yes. I think you have it with those minor corrections.
     
  6. Mar 12, 2012 #5
    Thank you LCKurtz, its appreciated.
     
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