I need to prove this

Show with the metrics $d_p$ and $d_{\infty}$ in $\mathbb{C}^n$ are uniformly equivalents, with $p \in [1, \infty)$.

So, I have in my book two definitions about equivalence metrics in a metric space $(X,d)$. 1) Two metrics $d_1$ and $d_2$ are uniformly equivalents if there are constants $a, b >0$ such that $ad_1(x,y) \leq d_2(x,y) \leq bd_1(x,y), \forall x, y \in \mathbb{C}^n$ or equivalently, if $a \leq \dfrac{d_2(x,y)}{d_1(x,y)}\leq b $ for all $x \neq y$. 2) The second definition is about topology equivalence. We say that two metrics $d_1$ and $d_2$ are topology equivalents if any sequence convergent in space $X$ with the metric $d_1$ also converge in the metric $d_2$ and for the same limit point.

I have already proved two facts:

a) If two metrics $d_1$ and $d_2$ are uniformly equivalent in $X$, then a subset $M$ of $X$ is bounded with respect to the metric $d_1$ if, and only if, $M$ is bounded with respect to the metric $d_2$.

b) If two metrics are uniformly equivalent, then they are topologically equivalent.

But my problem above continues. I could this: By definition we know that $$ d_p (x,y) = \left( \sum_{i=1}^{n} \ |x_i -y_i|^{p} \right)^{1/p} \mbox{e}\;\; d_\infty (x,y) = \sup_{i=1,..., n}{ |x_i - y_i| }. $$ So, by Minkowski's inequality, we have

$$ d_p (x,y) = \left( \sum_{i=1}^{n} \ |x_i -y_i|^{p} \right)^{1/p} \leq \left( \sum_{i=1}^{n} \ |x_i|^{p} \right)^{1/p} + \left( \sum_{i=1}^{n} \ |y_i|^{p} \right)^{1/p} \leq M_1 + M_2 = M $$ and $$ d_\infty = \sup_{i=1,..., n}{ |x_i - y_i| } \leq N. $$ How $0 < d_p(x,y)$ and $0 < d_\infty (x,y)$ for all $x \neq y$, we have with statements above that $$ 0 \leq \dfrac{d_p(x,y)}{d_{\infty}(x,y)} \leq \dfrac{M}{N}=b, b>0. $$ My problem here is how can I to prove with there is a constant positive $a$ such that $a \leq \dfrac{d_p(x,y)}{d_{\infty}(x,y)}$. Thanks.


For any $p\ge 1$ and any $u=(u_1,u_2,\dots,u_n)\in\mathbb{R}^n$, $$\|u\|_p = \left(\sum_{k=1}^n |u_k|^p\right)^{1/p} \le \left(\sum_{k=1}^n (\sup_k|u_k|)^p\right)^{1/p} = \left(n\|u\|_{\infty}\right)^{1/p}=n^{1/p}\|u\|_{\infty}$$ Estimate each term in the sum by the largest one. It's that simple. For the distance, apply this to $u=x-y$.

  • $\begingroup$ thanks so much. I understand it. $\endgroup$ – Thiago Alexandre Mar 5 at 13:20
  • $\begingroup$ I think that $|| u ||_p \leq (n || u ||_{\infty}^{p})^{\frac{1}{p}}$. You wrote it backwards. $\endgroup$ – Thiago Alexandre Mar 5 at 14:07
  • $\begingroup$ Sorry for my confusion. But I think you've helped me show that $||u||_{p} \leq b||u||_{\infty}$. Now, my problem is actually finding that $ a||u||_{\infty} \leq ||u||_{p}$. $\endgroup$ – Thiago Alexandre Mar 5 at 15:48
  • $\begingroup$ Still I was thinking and I thought this way (I think this solve my problem). I can prove that $|| u ||_{\infty} \leq || u ||_{p} \leq n^{\frac{1}{p}} ||u||_{\infty}$. This prove the result. If this is correct, tell me please. I apreciate your tips my friend. Thanks so much. $\endgroup$ – Thiago Alexandre Mar 5 at 16:08
  • $\begingroup$ Inequality sign fixed. And yes, that's the final version of the inequality; the other direction comes from the fact that we have one big term. $\endgroup$ – jmerry Mar 5 at 20:09

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