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Let

$\eqalign{ & d\left( {x,y} \right) = \mathop {\max }\limits_{1 \leqslant i \leqslant n} \left\{ {\left| {{x_i} - {y_i}} \right|} \right\} \cr & d'\left( {x,y} \right) = \sqrt {\sum\limits_{i = 1}^n {{{\left( {{x_i} - {y_i}} \right)}^2}} } \cr & d''\left( {x,y} \right) = \sum\limits_{i = 1}^n {\left| {{x_i} - {y_i}} \right|} \cr} $

for any two points $x,y \in \Bbb R^n$.

How to prove the following holds?

$\eqalign{ & d\left( {x,y} \right) \leqslant d'\left( {x,y} \right) \leqslant \sqrt n \cdot d\left( {x,y} \right) \cr & d\left( {x,y} \right) \leqslant d''\left( {x,y} \right) \leqslant n \cdot d\left( {x,y} \right) \cr} $

I think I got the second one:

It is trivial that

$\mathop {\max }\limits_{1 \leqslant i \leqslant n} \left\{ {\left| {{x_i} - {y_i}} \right|} \right\} < \sum\limits_{i = 1}^n {\left| {{x_i} - {y_i}} \right|} $

Now let $k$ be the integer such that

$d\left( {x,y} \right) = \left| {{x_k} - {y_k}} \right|$

Then for each $1 \leq i \leq n$ we have that $\left| {{x_i} - {y_i}} \right|\leq \left| {{x_k} - {y_k}} \right|$

So summing from $1$ to $n$ one gets:

$d''\left( {x,y} \right) \leqslant n \cdot d\left( {x,y} \right)$

  • 0
    But I think that generalising to$n$dimensions doesn't change too much - just replaces 2 by $n$, or $\sqrt{2}$ with $\sqrt{n}$, from memory.2012-07-10

1 Answers 1

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For the second result, use: $ \sum_{i=1}^n |x_i-y_i|\le \sum_{i=1}^n \max_i |x_i-y_i|=n\max_i|x_i-y_i| $ and $ \sum_{i=1}^n |x_i-y_i|\ge |x_j-y_j|, \ \text{for each}\ j $

Similar inequalities will establish your first result:

$ \sum_{i=1}^n |x_i-y_i|^2\le \sum_{i=1}^n \max_i |x_i-y_i|^2=n(\max_i |x_i-y_i|)^2 $ and $ \sum_{i=1}^n |x_i-y_i|^2\ge |x_j-y_j|^2, \ \text{for each}\ j $

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    I should have tried doing it myself before asking! Wasn't really challenging. Thanks anyways.2012-07-10