Borel $\sigma$- algebra on $\mathbb R^n$

Question

The $p$-norm on $\mathbb R^n$ is given by $\|x\|_{p}=\big(\sum_{i=1}^n |x_{i}|^p\big)^{1/p}$.

It's shown that $$\Vert x\Vert_q\leq \Vert x\Vert_p\leq n^{1/p-1/q}\Vert x\Vert_q ,\quad 1≤p≤q<\infty \tag{$*$}$$ my question is by using $(*)$ show that the Borel $\sigma$- algebra on $\mathbb R^n$, with respect to the $p$-norm is independent of $p$.

Any help would be appreciated. Thanks

Answer 1

1. If $A$ is a subset of $ \mathbb R^n$, then $A$ is open with respect to the $p$ - norm $\iff A$ is open with respect to the $q$ - norm . Hence all the above norms generate the same collection of open sets.

2. The Borel $ \sigma - $ algebra is generated by the open subsets of $ \mathbb R^n.$

Can you take it from here ?

Answer 2

Since the set $\{|x_i - y_i|\}_{i = 1}^n$ is finite, we may let $M =$ max $\{|x_i - x_j|\}$. So $ \frac{1}{M}(\sum^n_{i = 1} |x_i- y_i|^q)^\frac{1}{q} = (\frac{1}{M^q}\sum^n_{i = 1} |x_i- y_i|^q)^\frac{1}{q} = (\sum^n_{i = 1} \frac{1}{M^q}|x_i- y_i|^q)^\frac{1}{q} = (\sum^n_{i = 1} |\frac{x_i- y_i}{M}|^q)^\frac{1}{q} \leq (\sum^n_{i = 1} |\frac{x_i- y_i}{M}|^p)^\frac{1}{q} \leq (\sum^n_{i = 1} |\frac{x_i- y_i}{M}|^p)^\frac{1}{p} = (\sum^n_{i = 1} \frac{1}{M^p}|x_i- y_i|^p)^\frac{1}{p} = (\frac{1}{M^p}\sum^n_{i = 1} |x_i- y_i|^p)^\frac{1}{p} = \frac{1}{M}(\sum^n_{i = 1} |x_i- y_i|^p)^\frac{1}{p}$ where the $4^{th}$ inequality holds from the fact that $|\frac{x_i - y_i}{M}| \leq 1$ for all $i$, and the $5^{th}$ inequality holds from the fact that $\frac{1}{q} \leq \frac{1}{p}$ and $\sum^n_{i = 1} |\frac{x_i- y_i}{M}|^p \geq 1$ since at least one of the terms of the sum is equal to $1$, and all of the terms of the sum are non negative.

We use holders inequality that if $r > 1$, then $\sum_{i = 1}^n |a_i||b_i| \leq (\sum_{i = 1}^n |a_i|^r)^{\frac{1}{r}} (\sum_{i = 1}^n |b_i|^{\frac{r}{r - 1}})^{1 - \frac{1}{r}} $. We set $a_i = (x_i - y_i)^p$, $b_i = 1$ and $r = \frac{q}{p} > 1$, and take everything to the $\frac{1}{p}$. So

$[\sum_{i = 1}^n |x_i - y_i|^p]^{\frac{1}{p}} = [\sum_{i = 1}^n |(x_i - y_i)^p| \, |1|]^{\frac{1}{p}} \leq [(\sum_{i = 1}^n |(x_i - y_i)^p|^{\frac{q}{p}})^{\frac{1}{\frac{q}{p}}}(\sum_{i = 1}^n |1|^{\frac{\frac{q}{p}}{1 - \frac{q}{p}}})^{1 - \frac{1}{\frac{q}{p}}}]^{\frac{1}{p}}= [(\sum_{i = 1}^n |x_i - y_i|^{\frac{pq}{p}})^{\frac{p}{q}}(\sum_{i = 1}^n |1|^{\frac{\frac{q}{p}}{1 - \frac{q}{p}}})^{1 -\frac{p}{q}}]^{\frac{1}{p}} = [(\sum_{i = 1}^n |x_i - y_i|^{q})^{\frac{p}{q}}(\sum_{i = 1}^n |1|^{\frac{q}{p - q}})^{1 - \frac{p}{q}}]^{\frac{1}{p}} = [(\sum_{i = 1}^n |x_i - y_i|^{q})^{\frac{p}{q}}(\sum_{i = 1}^n |1|)^{1 - \frac{1}{q}}]^{\frac{1}{p}} = [(\sum_{i = 1}^n |x_i - y_i|^{q})^{\frac{p}{q}}(n)^{1 - \frac{p}{q}}]^{\frac{1}{p}}= [(\sum_{i = 1}^n |x_i - y_i|^{q})^{\frac{p}{q}}]^{\frac{1}{p}} [(n)^{1 - \frac{1}{q}}]^{\frac{1}{p}} = (\sum_{i = 1}^n |x_i - y_i|^{q})^{\frac{p}{pq}} (n)^{\frac{1}{p} - \frac{p}{pq}} = (\sum_{i = 1}^n |x_i - y_i|^{q})^{\frac{1}{q}} (n)^{\frac{1}{p} - \frac{1}{q}}$

To show that the Borel $\sigma$-algebra defined with respect to $d_p$ is the same as the Borel $\sigma$-algebra defined with respect to $d_q$, it suffices to show that the open stets with respect to $d_P$ are the same as the open sets with respect to $d_q$. Consider an arbitrary open set $E$ with respect to $d_p$. Consider an arbitrary $x \in E$. Since $E$ is open in $d_p$, there exists a ball $B_p(x,r)$ of radius $r$, such that $B_p(x,r) \subseteq E$. Since $B_q(x,r) \subseteq B_p(x,r) \subseteq E$, we have that $E$ is open with respect to $d_q$ as well. Suppose $U$ is open with respect to $d_q$. consider an arbitrary $x \in U$. Since $A$ is open in $d_q$, there exists a ball $B_q(x,r)$ of radius $r$, such that $B_q(x,r) \subseteq E$. Since $B_p(x,n^{-\frac{1}{p} + \frac{1}{q}}r) \subseteq B_q(x,r) \subseteq A$, we have that $A$ is open with respect to $d_p$ as well.