Ostrowski's Theorem

Theorem
Every non-trivial norm on $\Q$ is Cauchy equivalent to either:
 * the $p$-adic Metric $\left|{*}\right|_p$ for some prime $p$

or:
 * the Euclidean metric.

Proof
Let $\left \Vert {*}\right \Vert$ be a norm.

Case 1:
$\exists n \in \N$ such that $\left \Vert {n} \right \Vert > 1$:

Let $n_0$ be the least such number.

Since $\left \Vert {n_0}\right \Vert > 1$, it follows that $\exists \alpha \in \R_{>0}$ such that $\left \Vert {n_0}\right \Vert = n_0^\alpha$.

From the Basis Representation Theorem, any positive integer $n$ can be written:


 * $n = a_0 + a_1 n_0 + a_2 n_0^2 + \cdots + a_s n_0^s$, where $0 \le a_i < n_0$ and $a_s \ne 0$

Then:

Since all of the $a_i < n_0$, we have $\left \Vert {a_i}\right \Vert \le 1$.

Hence:

because $n \ge n_0^s$.

By Sum of Infinite Geometric Progression (and since $n_0^\alpha > 1$), the expression in brackets is a finite constant; call it $C$.

Hence $\left \Vert {n}\right \Vert \le C n^\alpha$ for all positive integers.

For any positive integers $n$ and $N$, we can use this inequality on $n^N$ and take $N$th roots to obtain:

$\left \Vert {n}\right \Vert \le \sqrt[N]{C} n^\alpha$.

Letting $N \to \infty$ for fixed $n$ gives $\left \Vert {n}\right \Vert \le n^\alpha$.

Now consider again the formulation of $n$ in base $n_0$.

We have $n_0^{s+1} > n \ge n_0^s$.

Since:


 * $\left \Vert {n_0^{s+1}}\right \Vert = \left \Vert {n+n_0^{s+1} - n}\right \Vert \le \left \Vert {n}\right \Vert + \left \Vert {n_0^{s+1} - n}\right \Vert$

we have:

since $\left \Vert {n_0^{s+1}}\right \Vert = \left \Vert {n_0}\right \Vert^{s+1}$, and by the first inequality ($\left \Vert {n}\right \Vert \le n^\alpha$) on the term being subtracted.

Thus:

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:


 * $\left \Vert {n}\right \Vert \ge n^\alpha$

These two results imply $\left \Vert {n}\right \Vert = n^\alpha$.

By the second property of norms (namely multiplicativity), this result extends to all $q \in \Q$.

Suppose a series $\left\{{x_1, x_2, \ldots}\right\}$ is Cauchy on the Euclidean metric.

We have $\left \Vert {x_j - x_i}\right \Vert \le \left \vert {x_j - x_i}\right \vert$, and so the series is Cauchy on $\left \Vert {*}\right \Vert $.

Now suppose a series is Cauchy on $\left \Vert {*}\right \Vert$.

Then for any $N$ such that $\forall i, j > N: \log_\alpha \left \vert{x_j - x_i}\right \vert < \epsilon, \left \Vert {x_j - x_i}\right \Vert < \epsilon$, so the series is Cauchy on the Euclidean metric.

Thus, $\left \Vert {*}\right \Vert$ is Cauchy equivalent to the Euclidean metric.

Case 2:
$\forall n \in \N: \left \Vert {n}\right \Vert \le 1$:

Let $n_0$ be the least positive integer such that $\left \Vert {n}\right \Vert < 1$.

Such a number exists because we have assumed $\left \Vert {*}\right \Vert$ is non-trivial.

If $n_0$ is composite, there are positive integers $n_1, n_2 < n_0$ such that $n_0 = n_1 n_2$.

Then:


 * $\left \Vert {n_1}\right \Vert = \left \Vert {n_2}\right \Vert = 1$

by the choice of $n_0$, and so:


 * $\left \Vert {n_0}\right \Vert = \left \Vert {n_1}\right \Vert \left \Vert {n_2}\right \Vert = 1$

which is a contradiction. So $n_0$ must be prime.

Let $n_0 = p$.

Claim: $\left \Vert {q}\right \Vert = 1$ if $q$ is a prime not equal to $p$.

Suppose not; then $\left \Vert {q}\right \Vert < 1$, and for some large $N$ we have:


 * $\left \Vert {q^N}\right \Vert = \left \Vert {q} \right \Vert ^N < \frac 1 2$

Also, for some large $M$ we have $\left \Vert {p}\right \Vert^M < \frac 1 2$.

Since $p^M, q^N$ are relatively prime, by Bézout's Identity we can find integers $n, m$ such that $m p^M + n q^N = 1$.

But then:

Now, $\left \Vert {k}\right \Vert \le 1$ for any integer $k$: If $k=0$, then $\left \Vert {k}\right \Vert=0$. If $k\ne 0$, then either $k$ or $-k$ is positive and since $\left \Vert {k}\right \Vert=\left \Vert {-k}\right \Vert$ for any norm, it follows by assumption.

Thus $\left \Vert {m}\right \Vert, \left \Vert {n}\right \Vert \le 1$, so that:


 * $1 \le \left \Vert {m}\right \Vert \left \Vert {p^M}\right \Vert + \left \Vert {n}\right \Vert \left \Vert {q^N}\right \Vert \le \left \Vert {p^M}\right \Vert + \left \Vert {q^N}\right \Vert < \frac 1 2 + \frac 1 2 = 1$

which is a contradiction.

Hence $\left \Vert {q}\right \Vert = 1$.

By the Fundamental Theorem of Arithmetic, any positive integer $a$ can be factored into prime divisors: $a = p_1^{b_1} p_2^{b_2} \dots p_r^{b_r}$.

Then:


 * $\left \Vert {a}\right \Vert = \left \Vert {p_1}\right \Vert ^{b_1} \left \Vert {p_2}\right \Vert ^{b_2} \dots \left \Vert {p_r}\right \Vert^{b_r}$

But $\left \Vert {p_i}\right \Vert$ will be different from $1$ only if $p_i = p$.

Its corresponding $b_i$ will be $\nu_p^\Z \left({a}\right)$, where $\nu_p^\Z$ is the $p$-adic valuation on $\Z$.

Hence, if we let $\rho = \left \Vert {p}\right \Vert < 1$, we have:


 * $\left \Vert {a}\right \Vert = \rho^{\nu_p^\Z \left({a}\right)}$

By the properties of norms, this same formula holds with any nonzero rational number in place of $a$.

Also see
In the same paper, published in 1918, Ostrowski also proved that, up to isomorhpism, $\R$ and $\C$ are the only fields that are complete with respect to an archimedean norm.

That result is also sometimes called Ostrowski's theorem.