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How to find the distance between these elements?

  1. Apr 4, 2009 #1
    I'm studying the proof that the l2 metric space is complete and I just don't get something about this proof.

    What I know:

    1. l2 is the metric space whose elements are sequences that converge when you square each term and sum them. In other words, a sequence, f is in l2 if SUM( |f(k)|2) < infinity.
    2. The norm on this metric space is: (SUM( |f(k)|2))0.5 = || f ||2
    3. A sequence in the l2 space is a sequence of sequences (confusing isn't it?)
    4. To show that the space is complete I need to show that every cauchy sequence in l2 converges to a sequence in l2.


    My book says:

    Suppose that for e > o there exists n0 such that:

    ||fn - fm||2 < e when m, n >= n0.

    I don't understand how to think about ||fn - fm||2.

    Suppose fn = .1, .01, .001, ...
    Suppose fm = .2, .02, .002, ...

    How would I find ||fn - fm||2 computationally?

    Do I do:

    ||fn - fm||2 = (|(.12 + .012, .0012 + ...) - (.22 + .022 + .0022 + ...)|)0.5
    ||fn - fm||2 = (|(.01 + .001, .0001 + ...) - (.04+ .004 + .0004 + ...)|)0.5
    ||fn - fm||2 = (|(.01 + .001, .0001 + ...) - (.04 + .004 + .0004 + ...)|)0.5
    ||fn - fm||2 = ( |1/90 - 4/90|)0.5
    ||fn - fm||2 = (3/90)0.5

    or

    [(.1 - .2)2 + (.01-.02)2, + ...]0.5 ?

    I think it's the first way... but for some reason I'm confused. Deeply.
     
  2. jcsd
  3. Apr 4, 2009 #2

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    It's the second way. First find the difference between the sequences, then calculate the norm. Your first calculation is | ||fn|| - ||fm|| |.
     
  4. Apr 4, 2009 #3
    Thanks dick... OK... So...

    || fn - fm||2 = [(.1 - .2)^2 + (.01-.02)^2 + ...]^0.5
    || fn - fm||2 = [(.1)^2 + (.01)^2 + ...]^0.5
    || fn - fm||2 = [.01 + .001 + ...]^0.5
    || fn - fm||2 = [1/90]^0.5

    1 / sqrt(90) is about 0.105409255


    Grrr... Ok so:

    |fn(k) - fm(k)| <= || fn - fm||2

    |fn(k) - fm(k)| is the absolute difference between the kth terms in fn and fm ... sequences that are "close enough together" so that || fn - fm||2 < e.

    Suppose

    fn = .1, .01, .001, ...
    fm = .2, .02, .002, ...
    are that close together. This is saying that

    .1 < = || fn - fm||2

    .1 < = 0.105409255

    Yay! it worked! I don't know why but it did!

    Thanks!

    God... but why is |fn(k) - fm(k)| <= || fn - fm||2?

    I'm not quite seeing that...
     
    Last edited: Apr 4, 2009
  5. Apr 4, 2009 #4
    Oh wait I think I see that now... it's like:

    x-y <= sqrt((x-y)^2 + positive stuff or zero)


    I wish I knew what this was all about... It seems very disconnected from the other things in the course... I just don't know where it's going ...
     
  6. Apr 4, 2009 #5

    Dick

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    Science Advisor
    Homework Helper

    You mean |x-y|. Yes, that's what it's like. BTW I get 1/sqrt(99) for your first example for ||fn-fm||. The common ratio in the geometric series is wrong. It's 1/100, not 1/10.
     
  7. Apr 4, 2009 #6
    ah yes.. see that now. OK... this proof is starting to make a little more sense to me now. Thank you!
     
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