minkowski distance at the infinity












1












$begingroup$


Given that the Minkowski distance between points $X = (x_1,dots,x_n)$ and $Y = (y_1,dots ,y_n)$ is
$$
d(X, Y)
= left(sum_{i=1}^n|x_i−y_i|^pright)^{1/p},
$$
I want to show that
$$
lim_{ptoinfty}d(X, Y) = max_{i=1, dots, n} |x_i-y_i|.
$$

How can I prove that as $p$ approaches infinity, the Minkowski distance approaches maximum difference of coordinates?






geometry limits norm normed-spaces







share|cite|improve this question











$endgroup$

















    1












    $begingroup$


    Given that the Minkowski distance between points $X = (x_1,dots,x_n)$ and $Y = (y_1,dots ,y_n)$ is
    $$
    d(X, Y)
    = left(sum_{i=1}^n|x_i−y_i|^pright)^{1/p},
    $$
    I want to show that
    $$
    lim_{ptoinfty}d(X, Y) = max_{i=1, dots, n} |x_i-y_i|.
    $$

    How can I prove that as $p$ approaches infinity, the Minkowski distance approaches maximum difference of coordinates?






    geometry limits norm normed-spaces







    share|cite|improve this question











    $endgroup$















      1












      1








      1


      0



      $begingroup$


      Given that the Minkowski distance between points $X = (x_1,dots,x_n)$ and $Y = (y_1,dots ,y_n)$ is
      $$
      d(X, Y)
      = left(sum_{i=1}^n|x_i−y_i|^pright)^{1/p},
      $$
      I want to show that
      $$
      lim_{ptoinfty}d(X, Y) = max_{i=1, dots, n} |x_i-y_i|.
      $$

      How can I prove that as $p$ approaches infinity, the Minkowski distance approaches maximum difference of coordinates?






      geometry limits norm normed-spaces







      share|cite|improve this question











      $endgroup$




      Given that the Minkowski distance between points $X = (x_1,dots,x_n)$ and $Y = (y_1,dots ,y_n)$ is
      $$
      d(X, Y)
      = left(sum_{i=1}^n|x_i−y_i|^pright)^{1/p},
      $$
      I want to show that
      $$
      lim_{ptoinfty}d(X, Y) = max_{i=1, dots, n} |x_i-y_i|.
      $$

      How can I prove that as $p$ approaches infinity, the Minkowski distance approaches maximum difference of coordinates?






      geometry limits norm normed-spaces




      geometry limits norm normed-spaces






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Oct 29 '16 at 13:30









      Andrew D. Hwang

      53.3k449115




      53.3k449115










      asked Oct 29 '16 at 10:27









      Muaa2404Muaa2404

      4316




      4316






















          1 Answer
          1






          active

          oldest

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          0












          $begingroup$

          Fix $X, Y$ And assume that $$ |x_1-y_1|geq |x_i-y_i| ast$$ Then begin{align*}
          |x_1-y_1|&leq d(X,Y) \&=|x_1-y_1| bigg(1+ sum_{i=2}^n
          frac{|x_i-y_i|^p}{|x_1-y_1|^p} bigg)^frac{1}{p} \&leq |x_1-y_1|
          n^frac{1}{p} end{align*}



          Here second equality is followed from




          $$sum_i |x_i-y_i|^p=|x_1-y_1|^p bigg{ 1+
          sum_{i=2}^n |x_i-y_i|^p/|x_1-y_1|^pbigg} $$




          And last inequality is followed from $ast$



          And




          $$lim_{prightarrow infty} n^frac{1}{p}=1$$




          so that we complete the proof






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Can you write your answer more clearly? I did not understand the last two line. Thank you very much.
            $endgroup$
            – Muaa2404
            Oct 31 '16 at 20:21














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          1 Answer
          1






          active

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          active

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          active

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          0












          $begingroup$

          Fix $X, Y$ And assume that $$ |x_1-y_1|geq |x_i-y_i| ast$$ Then begin{align*}
          |x_1-y_1|&leq d(X,Y) \&=|x_1-y_1| bigg(1+ sum_{i=2}^n
          frac{|x_i-y_i|^p}{|x_1-y_1|^p} bigg)^frac{1}{p} \&leq |x_1-y_1|
          n^frac{1}{p} end{align*}



          Here second equality is followed from




          $$sum_i |x_i-y_i|^p=|x_1-y_1|^p bigg{ 1+
          sum_{i=2}^n |x_i-y_i|^p/|x_1-y_1|^pbigg} $$




          And last inequality is followed from $ast$



          And




          $$lim_{prightarrow infty} n^frac{1}{p}=1$$




          so that we complete the proof






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Can you write your answer more clearly? I did not understand the last two line. Thank you very much.
            $endgroup$
            – Muaa2404
            Oct 31 '16 at 20:21


















          0












          $begingroup$

          Fix $X, Y$ And assume that $$ |x_1-y_1|geq |x_i-y_i| ast$$ Then begin{align*}
          |x_1-y_1|&leq d(X,Y) \&=|x_1-y_1| bigg(1+ sum_{i=2}^n
          frac{|x_i-y_i|^p}{|x_1-y_1|^p} bigg)^frac{1}{p} \&leq |x_1-y_1|
          n^frac{1}{p} end{align*}



          Here second equality is followed from




          $$sum_i |x_i-y_i|^p=|x_1-y_1|^p bigg{ 1+
          sum_{i=2}^n |x_i-y_i|^p/|x_1-y_1|^pbigg} $$




          And last inequality is followed from $ast$



          And




          $$lim_{prightarrow infty} n^frac{1}{p}=1$$




          so that we complete the proof






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Can you write your answer more clearly? I did not understand the last two line. Thank you very much.
            $endgroup$
            – Muaa2404
            Oct 31 '16 at 20:21
















          0












          0








          0





          $begingroup$

          Fix $X, Y$ And assume that $$ |x_1-y_1|geq |x_i-y_i| ast$$ Then begin{align*}
          |x_1-y_1|&leq d(X,Y) \&=|x_1-y_1| bigg(1+ sum_{i=2}^n
          frac{|x_i-y_i|^p}{|x_1-y_1|^p} bigg)^frac{1}{p} \&leq |x_1-y_1|
          n^frac{1}{p} end{align*}



          Here second equality is followed from




          $$sum_i |x_i-y_i|^p=|x_1-y_1|^p bigg{ 1+
          sum_{i=2}^n |x_i-y_i|^p/|x_1-y_1|^pbigg} $$




          And last inequality is followed from $ast$



          And




          $$lim_{prightarrow infty} n^frac{1}{p}=1$$




          so that we complete the proof






          share|cite|improve this answer











          $endgroup$



          Fix $X, Y$ And assume that $$ |x_1-y_1|geq |x_i-y_i| ast$$ Then begin{align*}
          |x_1-y_1|&leq d(X,Y) \&=|x_1-y_1| bigg(1+ sum_{i=2}^n
          frac{|x_i-y_i|^p}{|x_1-y_1|^p} bigg)^frac{1}{p} \&leq |x_1-y_1|
          n^frac{1}{p} end{align*}



          Here second equality is followed from




          $$sum_i |x_i-y_i|^p=|x_1-y_1|^p bigg{ 1+
          sum_{i=2}^n |x_i-y_i|^p/|x_1-y_1|^pbigg} $$




          And last inequality is followed from $ast$



          And




          $$lim_{prightarrow infty} n^frac{1}{p}=1$$




          so that we complete the proof







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Oct 31 '16 at 20:51

























          answered Oct 31 '16 at 10:18









          HK LeeHK Lee

          14.1k52361




          14.1k52361












          • $begingroup$
            Can you write your answer more clearly? I did not understand the last two line. Thank you very much.
            $endgroup$
            – Muaa2404
            Oct 31 '16 at 20:21




















          • $begingroup$
            Can you write your answer more clearly? I did not understand the last two line. Thank you very much.
            $endgroup$
            – Muaa2404
            Oct 31 '16 at 20:21


















          $begingroup$
          Can you write your answer more clearly? I did not understand the last two line. Thank you very much.
          $endgroup$
          – Muaa2404
          Oct 31 '16 at 20:21






          $begingroup$
          Can you write your answer more clearly? I did not understand the last two line. Thank you very much.
          $endgroup$
          – Muaa2404
          Oct 31 '16 at 20:21




















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