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Limit in a metric space

  1. Oct 9, 2011 #1
    1. The problem statement, all variables and given/known data
    Prove that [tex]\rho_{0}(x,y)=max_{1 \leq k \leq n}|x_{k} - y_{k}|=lim_{p\rightarrow\infty}(\sum^{n}_{k=1}|x_{k}-y_{k}|^{p})^{\frac{1}{p}}[/tex]


    2. Relevant equations



    3. The attempt at a solution
    My approach was to define [itex]a_{m}=max_{1 \leq k \leq n}|x_{k} - y_{k}|[/itex] and [itex]a_{k}=|x_{k} - y_{k}|[/itex]. Then since [itex]a_{m} \geq a_{k} \geq 0[/itex] we can replace [itex]lim_{p\rightarrow\infty}(\sum^{n}_{k=1}|x_{k}-y_{k}|^{p})^{\frac{1}{p}}[/itex] with [itex]lim_{p\rightarrow\infty}(\sum^{n}_{k=1}({a_{m} \frac{a_{k}}{a_{m}})^{p}})^{\frac{1}{p}}[/itex].

    When trying to break this down I get stuck at [tex]a_m * lim_{p\rightarrow\infty}(\sum^{n}_{k=1}{(\frac{a_{k}}{a_{m}})^{p}})^{\frac{1}{p}} [/tex]

    The limit here should be 1 since [itex] 0 \leq \frac{a_{k}}{a_{m}} \leq 1 [/itex]. However I need to be careful of the case where there are multiple k such that [itex]a_k = a_m[/itex]. Does anyone have suggestions for how to proceed? Thanks.
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Oct 9, 2011 #2

    Dick

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    Science Advisor
    Homework Helper

    The worst case would be ALL of the a_k=a_m, right? Would that change your limit?
     
  4. Oct 9, 2011 #3
    No it wouldn't. Good point. Many thanks.
     
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