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Metrics Spaces Inequalities

  1. Oct 26, 2011 #1
    1. The problem statement, all variables and given/known data

    (a) Show that for [itex]x,y\in \mathbb{R}^N[/itex]:

    (i) [itex]d_{\infty} (x,y) \leq d_1 (x,y) \leq N d_{\infty} (x,y)[/itex] ;

    (ii) [itex]d_{\infty} (x,y) \leq d_2 (x,y) \leq \sqrt{N} d_{\infty} (x,y)[/itex].

    (b) Find constants [itex]A[/itex] and [itex]B[/itex] such that [itex]A d_1 (x,y) \leq d_2 (x,y) \leq B d_1 (x,y)[/itex].

    2. Relevant equations

    Sum metric: [itex]\displaystyle d_1 (x,y) = \sum^N_{j=1} |x_j - y_j|[/itex]

    Euclidean metric: [itex]\displaystyle d_2 (x,y) = \left( \sum^N_{j=1} (x_j - y_j)^2 \right)^{1/2}[/itex]

    Maximum metric: [itex]\displaystyle d_{\infty} (x,y) = \text{max}^N_{j=1} |x_j - y_j |[/itex]

    3. The attempt at a solution

    For any [itex]a_1,\cdots , a_N \geq 0[/itex] and [itex]p\geq 1[/itex]:

    [itex] \sum_j (a_j)^p \geq (a_j)^p [/itex] for all [itex]j \in \mathbb{N}[/itex]. Taking the maximum over [itex]j[/itex] of both sides yields the lower bound.

    Notice that [itex](a_j)^p \leq (\text{max}_ja_j)^p [/itex] for all [itex]j\in\mathbb{N}[/itex]. Summing both sides over [itex]j[/itex] yields the upper bound.

    So these lead to the inequalities [tex] \sum_j (a_j)^p \geq (\text{max}_j a_j )^p[/tex] and [tex]\displaystyle \sum_j (a_j)^p \leq N( \text{max}_j a_j )^p[/tex] which leads to (a)(i) and (a)(ii) for [itex]p=1,2[/itex].

    For part (b) is the following correct:

    By (a)(i), [itex]d_1(x,y) \leq Nd_{\infty}(x,y) \iff N^{-1} d_1(x,y)\leq d_{\infty}(x,y)[/itex] (noting that [itex]N\in\mathbb{N}[/itex] so non-zero and non-negative).

    But by (a)(ii), [itex]d_{\infty}(x,y) \leq d_2(x,y) \leq N^{1/2}d_{\infty}(x,y)[/itex].

    So we have [itex]N^{-1}d_1(x,y) \leq d_2(x,y) \leq N^{1/2} d_1(x,y) \iff N^{-3/2}d_1(x,y) \leq d_2(x,y) \leq d_1(x,y)[/itex].

    Therefore [itex]A=N^{-3/2}[/itex] and [itex]B=1[/itex] are 2 constants satisfying part (b).
     
    Last edited: Oct 26, 2011
  2. jcsd
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