# $d_p$ and $d_\infty$ in $\mathbb{C}^n$ are uniformly equivalent

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 $$u=x-y$$.
• I think that $|| u ||_p \leq (n || u ||_{\infty}^{p})^{\frac{1}{p}}$. You wrote it backwards. – Thiago Alexandre Mar 5 at 14:07
• 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 $a||u||_{\infty} \leq ||u||_{p}$. – Thiago Alexandre Mar 5 at 15:48
• 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. – Thiago Alexandre Mar 5 at 16:08