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Infinite sum of squares converges

  1. Feb 25, 2008 #1
    1. The problem statement, all variables and given/known data
    So, L2 is defined to be the set of all infinite sequences of real numbers, s.t. the sum of their squares converges:
    L2 = {x=(x1,...,xn,...) | [tex]\Sigma[/tex]xi < [tex]\infty[/tex]}
    we have d(x,y) = [tex]\sqrt{\Sigma (xi-yi)^2}[/tex]

    I need to show that this is a metric, starting by showing that if [tex]xi,yi\in[/tex] then [tex]xi-yi\in[/tex]
    2. Relevant equations

    definition of a metric

    3. The attempt at a solution
    so my initial problem was with the first step:
    x = (x1,..., xn,....), y = (y1,...,yn,...), and then x - y = (x1 - y1, ...,xn-yn,...)
    to show [tex]xi,yi\in[/tex], we must show that [tex]\Sigma (xi -yi) < \infty[/tex].
    First, I broke up the squared quantity to get [tex]\Sigma (xi^2- 2xi*yi + yi^2) [/tex].
    Carrying the sumation through, [tex]\Sigma xi^2 < \infty [/tex] (as for yi^2).
    But where I get stuck is how to show or determine that [tex]\Sigma xi*yi < \infty [/tex].

    So then the first three parts of the definition of a metric ([tex]d(p,p) = 0, p\rightarrow d(p,q) > 0, [/tex] and [tex] d(p,q) = d(q,p)[/tex]) are easy enough to prove assuming [tex]xi-yi \in L2[/tex].
    I get caught up again in proving that d(p,q) < d(p,r) + d(r,q).
    I tried squaring both sides and distributing the summation, but found the resulting right hand side of the equation did not really simplify in a useful way. I'm not really sure where to go from there... any ideas for either of these roadblocks would be greatly appreciated. Thanks.
  2. jcsd
  3. Feb 25, 2008 #2


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    Science Advisor
    Homework Helper

    Your missing ingredients are various facets of the 'Cauchy-Schwarz inequality'. Look that up.
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