All norms of $\mathbb R^n$ are equivalent

1) First, all $\|x\|_p=\sqrt[p]{|x_1|^p+...+|x_n|^p}$ are equivalent. In my course, they used Hölder inequality, i.e. that $$\|x\cdot y\|_1\leq \|x\|_p\|x\|_q$$ for $p,q\geq 1$ s.t. $\frac{1}{p}+\frac{1}{q}=1$. But can I do as follow :

Let $1\leq p<\infty $. I denote $\|x\|_\infty =\max\{|x_1|,...,|x_n|\}$. Then, $$\|x\|_\infty ^p\leq |x_1|^p+...+|x_n|^p\leq n\|x\|_\infty^p ,$$

and thus $$\|x\|_\infty \leq \|x\|_p\leq n^{1/p}\|x\|_\infty .$$ Therefore, if $1\leq r,p <\infty $, then $$\|x\|_p\leq C\|x\|_\infty \leq C\|x\|_r\leq Cn^{1/r}\|x\|_\infty \leq Cn^{1/r}\|x\|_p.$$

Therefore, $\|\cdot \|_p$ and $\|\cdot \|_r$ are equivalent.

Question 1: Does it work ? (then no need Hölder?)

2) All norms of $\mathbb R^n$ are equivalent. Let $N$ a norm.

Question 2: By what I did previously, I just need to prove that $N$ is equivalent to $\|\cdot \|_\infty $ to conclude that all norm are equivalent, true ?

To do it I tried as follow : $$N(x)=N(x_1e_1+...+x_ne_n)\leq |x_1|N(e_1)+...+|x_n|N(e_n)\leq \|x\|_\infty (N(e_1)+...+N(e_n)),$$ where $e_i=(0,...,0,1,0,...,0)$ where $1$ is at the $i-$th position.

Question 3: How can I show that $ N(x)\geq C\|x\|_\infty$ ?

Solution 1:

For your first question, I don't think you need Minkowski. I think the best strategy is what you allude to later: prove that everything is equivalent to a single norm, and then since norm-equivalence is an equivalence relation, they are all equivalent to each other. The $\infty$-norm works pretty well: note that $$\|x\|_\infty \le \|x\|_p \le n^{1/p} \|x\|_\infty $$ for all $p$ which shows that all $p$-norms are equivalent to the $\infty$-norm and thus all $p$-norms are equivalent to each other.

For question 2, you're reasoning is exactly correct, and putting $D = N(e_1) + \cdots + N(e_n)$, you've shown that $N(x) \le D \|x\|_\infty$.

You've shown that $N(x) \le D \|x \|_\infty$. Then by the reverse triangle inequality, you have $$\lvert N(x) - N(y) \rvert \le N(x-y) \le D\|x-y\|_\infty$$ which shows that $N$ is continuous in the $\infty$-norm topology. The unit $\infty$-norm ball $S = \{x\in \mathbb R^n \, : \, \|x\|_\infty = 1\}$ is compact and continuous functions meet their minimum on compact set so there is some $x_m \in S$ such that $$0 < m := N(x_m) \le N(x) \,\,\,\, \text{ for all } x \in S.$$ Take any $x \in \mathbb R^n \setminus\{0\}$. Then $x /\|x\|_\infty \in S$, and so $$m \le N(x/\|x\|_\infty),$$ but then by homogeneity, you have $$m\|x\|_\infty \le N(x)$$ and of course, the same inequality holds for $x = 0$, so this holds for all $x \in \mathbb R^n$.

Thus $$m \|x\|_\infty \le N(x) \le D\|x\|_\infty$$ shows that any norm is equaivalent to the $\infty$-norm and thus any norm is equivalent to any other norm.