(adsbygoogle = window.adsbygoogle || []).push({}); Show equivalence of family of metrics on [tex]\Re^n[/tex]: [tex]\rho^{(p)}:(x,y)\rightleftharpoons(\sum_{i=1}^{n}|x_i-y_i|^p)^{\frac{1}{p}}[/tex] for [tex]p\geq1[/tex]

The attempt at a solution

Want to show for [tex]p,q\geq1[/tex], [tex]p\neq q[/tex], that [tex]\rho^{(p)}[/tex] and [tex]\rho^{(q)}[/tex] generate the same topology. I tried two methods:

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

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

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

for anyusatisfying [tex]\rho^{(q)}(u,y)<r_q[/tex],ualso satisfies [tex]\rho^{(p)}(u,y)<r_p-\rho^{(p)}(y,x_0)[/tex]. Ie. that any point in [tex]B_{r_q}(y)[/tex] is also in [tex]B_{r_p}(x_0)[/tex].

Also tried to get ideas by simplifying problem to [tex]\Re^2[/tex] 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 [tex]x,y\in\Re^n[/tex] there exists [tex]c,C>0[/tex] such that:

[tex]c\rho^{(p)}(x,y)\leq\rho^{(q)}(x,y)\leq C\rho^{(p)}(x,y)[/tex]

Ie. that:

[tex]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}}[/tex]

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!

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# Homework Help: Equivalence of Metrics

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