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I'm confused by the different definitions of equivalent norms. The standard one I know is that two norms (on a vector space) are equivalent if they induce equivalent metrics; two metrics are equivalent if a sequence is Cauchy w.r.t. one metric iff it is Cauchy w.r.t. the other.

It can be proven that an equivalent definition of equivalence of norms is the following: we can find real constants $A$ and $B$ such that $A|x|_1 \leq |x|_2 \leq B|x|_1$ for all $x$ in our vector space. For a reference, see e.g. Theorem 2.1 in this document.

My problem is as follows. I want to prove that a norm on $\mathbb Q$ given by $||x|| = |x|^\alpha$, where $|\cdot|$ is the Euclidean norm and $0 < \alpha \leq 1$, is equivalent to the Euclidean norm. I can prove that the above definition of $||\cdot||$ induces a metric equivalent to the Euclidean metric. I can also prove that in this case, it is impossible to find constants $A$ and $B$ such that $A||x|| \leq |x| \leq B||x||$ for all $x \in \mathbb Q$. What's going on?

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  • $\begingroup$ I don’t have a direct answer for your question but let me say something related: Two norms are equivalent if they induce the same topology. Thus taken an open set S with respect to norm C. If the two norms are equivalent, this set S should also be open with respect to norm D. Then take an open set in norm D and show it is also open in norm C. This should show you intuitively why the two norms in your question are equivalent. $\endgroup$ – AnlamK Oct 6 '17 at 17:45
  • $\begingroup$ I'm guessing perhaps the issue is that the definition you give (which is also used in the document I linked to) is not equivalent to "inducing equivalent metrics" (although it looks equivalent at first sight). I'll try and see if I can (dis)prove this equivalence first. $\endgroup$ – user Oct 6 '17 at 17:58
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The problem comes from the fact that the results you are quoting only hold if you consider a norm over a vector space, as defined at en.wikipedia.org/wiki/Norm_(mathematics)#Definition (not a norm in a field, as you quoted in the answer I deleted since I had misunderstood your question: encyclopediaofmath.org/index.php/Norm_on_a_field).

Indeed, if you want to prove that:

(i) Two norms $N_{1}$ and $N_{2}$ (on a vector space) are equivalent according to the definitions in your first paragraph, i.e. that for every sequence $(x_{n})$, $lim_{m,n \rightarrow +\infty}N_{1}(x_{m}-x_{n})=0 \Leftrightarrow lim_{m,n \rightarrow +\infty}N_{2}(x_{m}-x_{n})=0$

$\Leftrightarrow$

(ii) $\exists A,B \in \mathbb{R}, \forall x, A.N_{1}(x) \leq N_{2}(x) \leq B.N_{1}(x)$

Then you need the absolute homogeneity of the norm (which does not hold for a norm in a field where you have some kind of sub-multiplicity instead).

Proof of the usual case (norm of a vector space) is rather easy by contradiction (assuming (ii) does not hold, you can build a sequence that is Cauchy for one norm but not the other). This proof also shows why you need that absolute homogeneity.

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  • $\begingroup$ Ah, I see - I got my definitions mixed up. I thought a norm on a field was always a norm on a vector space, too - simply because a field is a one-dimensional vector space over itself. I didn't realise the absolute homogeneity property explicitly assumes the underlying field is equipped with the absolute value norm. Thus, equivalence of norms on fields and equivalence of norms on vector spaces are entirely different things, even though the definition of equivalence is the same; it's because the norms are different in nature. Thank you very much for helping me see this! $\endgroup$ – user Oct 7 '17 at 15:50

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