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'!).