Are the Metrics \rho^{(p)} and \rho^{(q)} Equivalent on \Re^n?

[jjou]

Show equivalence of family of metrics on $$\Re^n$$: $$\rho^{(p)}:(x,y)\rightleftharpoons(\sum_{i=1}^{n}|x_i-y_i|^p)^{\frac{1}{p}}$$ for $$p\geq1$$
The attempt at a solution
Want to show for $$p,q\geq1$$, $$p\neq q$$, that $$\rho^{(p)}$$ and $$\rho^{(q)}$$ generate the same topology. I tried two methods:

1. Show that a set is open in $$(\Re^n,\rho^{(p)})$$ iff it is open in $$(\Re^n,\rho^{(q)})$$.

2. Show that the bases are equivalent (an element of the base of the topology generated by $$\rho^{(p)}$$ is a union of elements of the base of the topology generated by $$\rho^{(q)}$$.

Both reduced down to considering, for a fixed $$x_0\in\Re^n$$ and $$r_p$$, the set $$B_{r_p}(x_0)=\{x\in\Re^n|\rho^{(p)}(x,x_0)<r_p\}$$. Then fix a point $$y\in B_{r_p}(x_0)$$ and show there exists a $$r_q$$ such that:

for any u satisfying $$\rho^{(q)}(u,y)<r_q$$, u also satisfies $$\rho^{(p)}(u,y)<r_p-\rho^{(p)}(y,x_0)$$. Ie. that any point in $$B_{r_q}(y)$$ is also in $$B_{r_p}(x_0)$$.

Also tried to get ideas by simplifying problem to $$\Re^2$$ with p=1 and p=2. Was not very illuminating since I was still stuck working with the same expressions.

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Another method is to show that for any $$x,y\in\Re^n$$ there exists $$c,C>0$$ such that:
$$c\rho^{(p)}(x,y)\leq\rho^{(q)}(x,y)\leq C\rho^{(p)}(x,y)$$
Ie. that:
$$c(\sum_{i=1}^{n}|x_i-y_i|^p)^{\frac{1}{p}}\leq(\sum_{i=1}^{n}|x_i-y_i|^q)^{\frac{1}{q}}\leq C(\sum_{i=1}^{n}|x_i-y_i|^p)^{\frac{1}{p}}$$

I didn't know whether to attempt to show both inequalities at once or separately, but either way, I'm not sure how to manipulate all the exponents and absolute values...

Please help!

 

[morphism]

Each of these metrics comes from a norm, and as such, they are all equivalent because R^n is finite dimensional.

If you don't know this, then I suggest you prove each of the metrics is equivalent to $\rho^{(1)}$. Holder's inequality will be useful here, because we can write (for p>1):

$$\sum |x_i - y_i| \leq \left(\sum |x_i - y_i|^p\right)^{1/p} \left(\sum 1^{p/(p-1)}\right)^{(p-1)/p}$$

 

[jjou]

Thank you so much! Very simple solution using Holder's, if I'm not wrong...

Using $$c(\sum_{i=1}^{n}|x_i-y_i|^p)^{\frac{1}{p}}\leq(\sum_{i=1}^{n}|x_i-y_i|^q)^{\frac{1}{q}}\leq C(\sum_{i=1}^{n}|x_i-y_i|^p)^{\frac{1}{p}}$$, take
c=1 and $$C=\left(\sum 1^{p/(p-1)}\right)^{(p-1)/p}=n^{(p-1)/p}$$.

The first inequality holds iff $$c^p(\sum_{i=1}^{n}|x_i-y_i|^p)\leq(\sum_{i=1}^{n}|x_i-y_i|)^p$$ which is obviously true by expansion.

The second inequality fulfills Holder's.

Yes?

Again, thanks so much! :)

 

[morphism]

Looks good (as long as that 'q' is actually a '1'!).