Why does $\alpha \|{x}\|_1 < \|{x}\|_2 < \beta \|{x}\|_1$ show equivalence of norms?

Question

I am reading a book on Hilbert space (“Introduction to Hilbert Space by Debnath and Mikusinski”) and I came across a proof for equivalence of norms that I do not understand. $E$ is a vector space and a non empty set, elements of $E$ are vectors. $\alpha,\beta$ are positive numbers and $x\in E$. Then: $$\alpha \|{x}\|_1 < \|{x}\|_2 < \beta \|{x}\|_1. \tag{1}\label{first}$$ The proof starts off saying that this “clearly” implies the equivalence of norms $\|{\cdot}\|_1$ and $\|{\cdot}\|_2$: Question 1: Why does $\ref{first}$ imply that $\|{\cdot}\|_1$ and $\|{\cdot}\|_2$ are equivalent?

Answer 1

I hope that this can help you:

For two norms $||·||_{1},||·||_{2}$ defined in a vectorial space $X$ the following two statements are equivalent:

i) $\exists \rho\in \mathbb{R}$ (constant) such that $||x||_{2}\le \rho ||x||_{1}$ $\forall x\in X$.

ii) The topology of $||·||_{2}$ is included of the topology of $||·||_{1}$.

Notation: $B_{i}(x,r)$ is the open ball of center $x\in X$ and radius $r>0$ with the norm $||· ||_{i}$, $i=1,2$

$i)\Rightarrow ii)$ If $U$ is an open set with the norm $||·||_{2}$ , for each $x\in U$ $\exists \epsilon>0$ such that $B_{2}(x,\epsilon)\subset U.$ From i) we deduce that $B_{1}(x,\epsilon / \rho) \subset B_{2}(x,\epsilon) \subset U$, then $U$ is an open set with the norm $||·||_{1}$ as we want.

$ii)\Rightarrow i)$ As $B_{2}(0,1)$ is open with $||·||_{2}$, it will be open with $||·||_{1}$ too, then exists $\delta >0$ such that $B_{1}(0,\delta)\subset B_{2}(0,1)$. Taking $\rho=1/\delta>0$ we have the inequality. Indeed, if $x\in X$ verify that $||x||_{2}>\rho ||x||_{1}$, taking $y=x/||x||_{2}$ we have $$||y||_{1}=\frac{||x||_{1}}{||x||_{2}}<\frac{1}{\rho}=\delta$$ Where $||y||_{2}<1$ , which is a contradiction, $||y||_{2}=1$ . So $||x||_{2}\le \rho ||x||_{1}$ $\forall x\in X$.

Then in the equivalence of norms you have both inequalities and both inclusions at the topologies.