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I have two statements in mind that taken without further caution could seem contradictory:

  • all norms are equivalent in finite dimension
  • there are infinitely many non-equivalent norm over the rationals (Ostrowski)

So I should be missing a point. Is it that the first statement is only valid for the reals or complexes? (however I have the impression that the proof still holds over the rationals)

Or is it rather than the two notions of equivalence (one with bounds, continuity of the identity; the other with equality up to a certain power) are different? Case in which: why are these two natural, what motivates one in some cases and the other one in others?

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    $\begingroup$ The first is true over a complete field. The rationals are not complete in the natural norm you'd want to use. You end up doing some sort of limiting procedure (effectively) in the proof and the rationals are not complete so you can leave the space. I suspect that the proof actually helps in building the infinitely many examples. $\endgroup$ – Cameron Williams Jun 22 '18 at 17:18
  • $\begingroup$ @CameromWilliams However I do not get where the completeness is so critical. Indeed, I though that the proof mainly relied on the compactness of the unit sphere, what is still true for p-adics isn’t it? Has it something to do with the Heine-Borel failure for non-complete spaces? $\endgroup$ – Gory Jun 23 '18 at 8:31
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A norm on a field is not the same as a norm on a field as a vector space over itself.

The definitions are:

Let $F$ be a field. A function $|\cdot|: F \times F \to F$ is a norm on $F$ if for all $x,y \in F$ we have

  • $\left|x\right| \ge 0 \text{ and } \left|x\right| = 0 \iff x = 0$
  • $\left|xy\right| = \left|x\right|\left|y\right|$
  • $\left|x+y\right| \le \left|x\right| + \left|y\right|$

$|\cdot|$ is more commonly called a valuation on $F$ and $(F, \left|\,\cdot\,\right|)$ is a valued field.

Let $V$ be a vector space over $F$ with a fixed valuation $|\cdot|$. A function $\|\cdot\| : V \times V \to F$ is a norm on $V$ if for all $x,y \in V$ and scalars $\alpha \in F$ holds:

  • $\|x\| \ge 0 \text{ and } \|x\| = 0 \iff x = 0$
  • $\|\alpha x\| = \left|\alpha\right|\|x\|$
  • $\|x+y\| \le \|x\| + \|y\|$

For example, the $p$-adic absolute value $\left|\,\cdot\,\right|_p$ is a valuation on $\mathbb{Q}$ but it is not a norm on the vector space $\mathbb{Q}$ over the valued field $(\mathbb{Q}, |\cdot|)$ where $|\cdot|$ is the standard absolute value on $\mathbb{Q}$.

The claim that "all norms on a finite-dimensional space are equivalent" holds only for norms on a vector space, not valuations (as Ostrowski's theorem shows).

Furthermore, all norms on a finite-dimensional vector space over a valued field $(F, \left|\,\cdot\,\right|)$ do not have to be equivalent if $(F, \left|\,\cdot\,\right|)$ is not complete as a metric space with the metric $(x,y) \mapsto \left|x-y\right|$.

Consider the vector space $\mathbb{Q}^2$ over over the valued field $(\mathbb{Q}, |\cdot|)$ where $|\cdot|$ is the standard absolute value.

Define two norms on the vector space $\mathbb{Q}^2$ as $\|(x,y)\|_1 = \left|x\right| + \left|y\right|$ and $\|(x,y)\|_2 = |x + \sqrt{2}y|$.

They are not equivalent. Namely, let $(x_n)_n$ be a sequence of rational numbers such that $x_n \to \sqrt{2}$. Then

$$\|(x_n,-1)\|_2 = |x_n - \sqrt{2}| \xrightarrow{n\to\infty} 0$$ but $\|(x_n,-1)\|_1 \ge 1$.

This example is taken from this question.

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