Ostrowski's Theorem/Archimedean Norm/Lemma 1.1
Theorem
Let $\norm {\, \cdot \,}$ be a non-trivial Archimedean norm on the rational numbers $\Q$.
Let $n_0 = \min \set {n \in \N : \norm n > 1}$
Let $\alpha = \dfrac {\log \norm {n_0} } {\log n_0 }$
Then:
- $\forall n \in N: \norm n \le n^\alpha$
Proof
By the definition of $\alpha$ then:
- $\norm {n_0} = {n_0}^\alpha$
By the definition of $n_0$ then:
- ${n_0}^\alpha > 1$
Let $n \in \N$.
By Basis Representation Theorem then $n$ 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 $a_s \ne 0$
Since all of the $a_i < n_0$, by choice of $n_0$ then:
- $\forall a_i: \norm {a_i} \le 1$
Then:
\(\ds \norm n\) | \(\le\) | \(\ds \norm {a_0} + \norm {a_1 n_0} + \norm {a_2 {n_0}^2} + \cdots + \norm {a_s {n_0}^s}\) | Norm Axiom $\text N 3$: Triangle Inequality | |||||||||||
\(\ds \) | \(=\) | \(\ds \norm {a_0} + \norm {a_1} \norm {n_0} + \norm {a_2} \norm {n_0}^2 + \cdots + \norm {a_s} \norm {n_0}^s\) | Norm Axiom $\text N 2$: Multiplicativity | |||||||||||
\(\ds \) | \(=\) | \(\ds \norm {a_0} + \norm {a_1} n_0^\alpha + \norm {a_2} {n_0}^{2 \alpha} + \cdots + \norm {a_s} {n_0}^{s\alpha}\) | as $\norm {n_0} = {n_0}^\alpha$ | |||||||||||
\(\ds \) | \(\le\) | \(\ds 1 + {n_0}^\alpha + {n_0}^{2 \alpha} + \cdots + {n_0}^{s \alpha}\) | as $\forall a_i: \norm {a_i} \le 1$ | |||||||||||
\(\ds \) | \(=\) | \(\ds n_0^{s \alpha} \paren {1 + {n_0}^{-\alpha} + {n_0}^{-2 \alpha} + \cdots + {n_0}^{-s \alpha} }\) | ||||||||||||
\(\ds \) | \(\le\) | \(\ds {n_0}^{s \alpha} \paren {\sum_{i \mathop = 0}^\infty \paren {\frac 1 { {n_0}^\alpha} }^i}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds {n_0}^{s \alpha} \paren {\dfrac { {n_0}^\alpha} { {n_0}^\alpha - 1} }\) | Sum of Infinite Geometric Progression (and as ${n_0}^\alpha > 1$) | |||||||||||
\(\ds \) | \(\le\) | \(\ds n^\alpha \paren {\dfrac { {n_0}^\alpha} { {n_0}^\alpha - 1} }\) | as $n \ge {n_0}^s$ |
Let $C = \paren {\dfrac { {n_0}^\alpha} { {n_0}^\alpha - 1} }$
Hence:
- $\norm n \le C n^\alpha$
As $n \in \N$ was arbitrary:
- $\forall n \in N: \norm n \le C n^\alpha$
Let $n, N \in N$
Then:
- $\norm {n^N} \le C \paren {n^N}^\alpha$
Now:
\(\ds \norm {n^N} \le C \paren {n^N}^\alpha\) | \(\leadsto\) | \(\ds \norm n^N \le C \paren {n^N}^\alpha\) | Norm Axiom $\text N 2$: Multiplicativity | |||||||||||
\(\ds \) | \(\leadsto\) | \(\ds \norm n^N \le C \paren {n^\alpha}^N\) | ||||||||||||
\(\ds \) | \(\leadsto\) | \(\ds \norm n \le \sqrt [N] C n^\alpha\) | taking $N$th roots |
By Limit of Root of Positive Real Number:
- $\sqrt [N] C \to 1$ as $N \to \infty$
By the Multiple Rule for Real Sequences:
- $\sqrt [N] C n^\alpha \to n^\alpha$ as $N \to \infty$
By Inequality Rule for Real Sequences, letting $N \to \infty$ for fixed $n$:
- $\norm n \le n^\alpha$
The result follows.
$\blacksquare$
Sources
- 1997: Fernando Q. Gouvea: p-adic Numbers: An Introduction: $\S 3.1$ Absolute Values on $\Q$, Theorem $3.1.3$
- 2007: Svetlana Katok: p-adic Analysis Compared with Real: $\S 1.9$ Metrics and norms on the rational numbers. Ostrowski’s Theorem, Theorem $1.50$