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I am not able to wrap my head around this result.

$\lim_{p\to\infty}(x_1^p + x_2^p + x_3^p .. + x_n^p)^{1/p} = \max [x_1, x_2, .. x_n]$ ?

Davide Giraudo
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user2340452
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    The hint I gave here does generalize to $n$ components. – JimmyK4542 Aug 29 '14 at 06:08
  • @JimmyK4542 what if I have some of these elements equal to the maximum value? all of those fractions will be one !! so if I have half of the set equal to maximum value what I get is 0.5 * n * maximum value !!!! – user2340452 Aug 29 '14 at 06:47

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Reading JimmyK4542's comment made me want to delete this question as this is very much covered in that thread. But then again some poor soul like me might find this useful in future so here goes.

let $x_1$ be the maximum of the lot. And I take it out of the expression.

$x_1(1 + (x_2/x_1)^p + (x_2/x_1)^p + (x_3/x_1)^p .. + (x_n/x_1)^p)^{1/p}$ ---- (1)

Since $x_1$ is the largest all of those fractions and when I take the limit all of them will become 0. So the bracket after $x_1$ will be equal to 1. Hence what I will be left with is $x_1$ which is the maximum of the given terms.

Edit: This works for the case when exactly one of the $x's$ equal to the maximum value.

Say some q terms are equal to the maximum value, then (1) will be $x_1(1 + q)^{1/p}$

user2340452
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  • You still need to cover the case where $|x_i| = |x_1|$ for one or more values of $i \neq 1$. To do this, note that $(1+|x_2/x_1|+\cdots+|x_n/x_1|)^{1/p} \le (1+1+\cdots+1)^{1/p} = n^{1/p}$. – JimmyK4542 Aug 29 '14 at 06:49
  • @JimmyK4542 Haha !! Read my comment to this question :) – user2340452 Aug 29 '14 at 06:51