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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?

Tsangares
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    How do you define the equivalence of two norms? E.g. for me, $1$ is actually the definition of equivalence –  Nov 11 '16 at 00:14
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    I think it is unfortunate that you've put a bounty on this question without answering the comment. Unfortunately, it is not possible to give you an honest answer without your definition of equivalence of norms. – davidlowryduda Dec 06 '16 at 03:28
  • I suppose one can explain the motivation for why equivalence of norms is defined that way (i.e. equivalent norms induce the same topology on the vector space), though that's not the question as stated. –  Dec 06 '16 at 03:45
  • $\lim_{x\rightarrow \infty}| x_1 - x | 1 = \lim{x\rightarrow \infty}|_2 - x | _2$ – Tsangares Dec 06 '16 at 22:55
  • It must be $\alpha |{x}|_1 \le |{x}|_2 \le \beta |{x}|_1.$ The strict inequality doesn't satisfy if $x=0.$ Anyway, as asked in other comments, can you say what do you understand by equivalence of norms? – mfl Dec 07 '16 at 20:13
  • This question deserves "unclear what you are asking" closing if not for the bounty. – Moishe Kohan Dec 09 '16 at 09:29

1 Answers1

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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.

Skullgreymon
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