Ostrowski's Theorem

Theorem
Every nontrivial norm on $$\mathbb{Q}$$ is Cauchy equivalent to $|*|_p$ for some prime $$p \ $$ or the Euclidean metric.

Proof
Let $$||*||$$ be a norm.

Case 1: $$\exists n \in \mathbb{N} \text{ such that} ||n|| > 1$$
Let $$n_0 \ $$ be the least such integer. Since $$||n_0||>1, \exists \alpha \in \mathbb{R}_+$$ such that $$||n_0||=n_0^\alpha$$.

Now any positive integer $$n \ $$ can be written

$$n= a_0+a_1n_0+a_2n_0^2+...+a_sn_0^s$$, where $$0 \leq a_i < n_0$$ and $$a_s \neq 0$$.

Then

$$||n|| \leq ||a_0| + ||a_1n_0|| + ||a_2n_0^2|| + ... + ||A_sn_0^s||$$

$$= ||a_0||+||a_1||n_0^\alpha+||a_2||n_0^{2\alpha}+...+||a_s||n_0^{s\alpha}$$

Since all of the $$a_i<n_0$$, we have $$||a_i|| \leq 1$$ and hence

$$||n|| \leq 1+n_0^\alpha+n_0^{2\alpha}+...+n_0^{s\alpha}$$

$$n_0^{s\alpha}(1+n_0^{-\alpha}+n_0^{-2\alpha}+...+n_0^{-s\alpha})$$

$$\leq n^\alpha \left[ \sum_{i=0}^{\infty}(\frac{1}{n_0^\alpha})^i \right]$$

because $$n \geq n_0^s$$. The expression in brackets is a finite constant, call it $$C$$. Hence

$$||n|| \leq Cn^\alpha$$ for all positive integers. For any positive integer n and some large positive integer N, we can use this formula to obtain

$$||n|| \leq \sqrt[N]{C}n^\alpha$$. Letting $$N \to \infty$$ for fixed $$n \ $$ gives

$$||n|| \leq n^\alpha$$.

Now consider again the formulation of $$n \ $$ in base $$n_0 \ $$. We have $$n_0^{s+1}>n \geq n_0^s$$. Since $$||n_0^{s+1}|| = ||n+n_0^{s+1} -n|| \leq ||n|| + ||n_0^{s+1}-n||$$, we have

$$||n|| \geq ||n_0^{s+1}||-||n_0^{s+1}-n||$$

$$\geq n_0^{(s+1)\alpha} - (n_0^{s+1}-n)^\alpha$$

since $$||n_0^{s+1}|| = ||n_0||^{s+1}$$, and by the first inequality ($$||n|| \leq n^\alpha$$) on the term being subtracted. Thus

$$||n|| \geq n_0^{(s+1)\alpha} - (n_0^{s+1} - n_0^s)^\alpha$$

$$=n_0^{(s+1)\alpha} \left[1-(1-\frac{1}{n_0}^\alpha \right]$$

$$\geq C'n^\alpha$$

for some constant $$C' \ $$ which may depend on $$n_0 \ $$ and $$\alpha \ $$ but not on $$n / $$. As before, for very large $$N \ $$, use this inequality on $$n^N \ $$, take $$N \ $$th roots and let $$N \to \infty$$, to get

$$||n|| \geq n^\alpha$$

These two results imply $$||n||=n^\alpha \ $$. By the second property of norms, this result extends to all $$q \in \mathbb{Q}$$.

Suppose a series $$\left\{{x_1,x_2,...}\right|$$ is Cauchy on the Euclidean metric. We have $$||x_j-x_i|| \leq |x_j-x_i|$$, and so the series is Cauchy on $$||*||$$. Now suppose a series is Cauchy on $$||*||$$. Then for any $$N \ $$ such that $$\forall i,j >N, \text{log}_\alpha |x_j-x_i|<\epsilon, ||x_j-x_i||<\epsilon$$, so the series is Cauchy on the Euclidean metric.