# How to prove that the following metrics are topologically equivalent [duplicate]

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

• 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 $d_q(x,y) \le n d_p(x,y)$ – GEdgar Feb 2 '20 at 16:06
• I think the proposed duplicate is much more difficult than this question. – GEdgar Feb 2 '20 at 16:08
• 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 – Silkking Feb 2 '20 at 16:09
• Go ahead and write your solution. – GEdgar Feb 2 '20 at 16:11

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