# Equivalence of Metrics

1. Jan 27, 2008

### 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.

----

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...

2. Jan 27, 2008

### 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):

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

3. Jan 27, 2008

### 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! :)

4. Jan 27, 2008

### morphism

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