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For an arbitrary number of dimensions, I know that the mean minimizes the distance using the $L_2$ norm and that the geometric median minimizes the distance function using the $L_1$ norm (though I have yet to find a good proof of this). So what minimizes the $L_{\infty}$ (or Chebyshev) norm?

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Rodrigo de Azevedo
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larry
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The mid-range minimizes the $\:L_{\hspace{.02 in}\infty}\:$ norm.

  • So how do I know that this is correct? – larry Mar 12 '14 at 02:41
  • Look at the distance from points above the mid-range to the smallest value, then look at the distance from points below the mid-range to the largest value. $;$ –  Mar 12 '14 at 02:55
  • @RicyDemer Not sure what you are trying to show or what you are getting at. – larry Mar 12 '14 at 02:57
  • Observe that all those distances are at least half the range, so no point can make the $:L_{\hspace{.02 in}\infty}:$ norm smaller than half the range. $;;$ Since the mid-range makes the $:L_{\hspace{.02 in}\infty}:$ norm equal to half the range, the mid-range minimizes the $:L_{\hspace{.02 in}\infty}:$ norm. $;;;;$ –  Mar 12 '14 at 02:59
  • Can you state in this is a non-confusing way? – larry Mar 12 '14 at 03:01
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    No. ${}{}{}{};$ –  Mar 12 '14 at 03:05