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I have $d_p(x,y) = [\sum | x_i - y_i|^p]^{1/p}$ and $d_q(x,y) = [\sum | x_i - y_i|^q]^{1/q}$ metrics in $\mathbb{R}^n$ and I want to prove that they are equivalent. I already know that $d_{\infty}(x,y) \leq d_2(x,y) \leq d_1(x,y) \leq n d_{\infty}(x,y)$, so I thought of proving that if $p \leq q$ then $d_p(x,y) \geq d_q(x,y)$ or to make it simpler, $d_{p+1}(x,y) \leq d_p(x,y)$, but couldn't advance much

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    $\begingroup$ It will be enough to prove $d_\infty(x,y) \le d_p(x,y) \le n d_\infty(x,y)$. From this we can get $d_p(x,y) \le n d_q(x,y)$ and $d_q(x,y) \le n d_p(x,y)$ $\endgroup$ – GEdgar Feb 2 '20 at 16:06
  • $\begingroup$ I think the proposed duplicate is much more difficult than this question. $\endgroup$ – GEdgar Feb 2 '20 at 16:08
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    $\begingroup$ It is, but what @GEdgar proposed is quite simple and solves my problem. If it's ok with him, I'll write a complete answer in case someone falls here looking for it $\endgroup$ – Silkking Feb 2 '20 at 16:09
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    $\begingroup$ Go ahead and write your solution. $\endgroup$ – GEdgar Feb 2 '20 at 16:11
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I'll write an answer using @GEdgar's idea:

He proposed to prove that for any $p \in \mathbb{N}$, $d_{\infty}(x,y) \leq d_p(x,y) \leq n d_{\infty}(x,y)$ (after writing it, I found a better bound, but both are good). If this is true for every $p$, then $d_p(x,y) \leq n d_{\infty}(x,y) \leq nd_q(x,y) \leq n^2 d_{\infty}(x,y) \leq n^2d_p(x,y)$, so in particular $d_p(x,y) \leq nd_q(x,y) \leq n^2 d_p(x,y)$, so they would be equivalent.

$1)\text{ } d_{\infty}(x,y)= max(|x_i - y_i|)=max((|x_i - y_i|^p)^{1/p}) \leq (\sum |x_i - y_i|^p) ^{1/p} = d_p(x,y)$

$2) \text{ } d_p(x,y) = (\sum |x_i - y_i|^p) ^{1/p} \leq (\sum max(|x_i - y_i|)^p)^{1/p} = (max(|x_i - y_i|)^p \sum 1)^{1/p} = max(|x_i - y_i|) n^{1/p} = n^{1/p}d_{\infty}(x,y)$

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