Ostrowski's Theorem/Archimedean Norm/Lemma 1.1

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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:

\(\displaystyle \norm n\) \(\le\) \(\displaystyle \norm {a_0} + \norm {a_1 n_0} + \norm {a_2 n_0^2} + \cdots + \norm {a_s n_0^s}\) Norm axiom (N3) (Triangle Inequality)
\(\displaystyle \) \(=\) \(\displaystyle \norm {a_0} + \norm {a_1} \norm {n_0} + \norm {a_2} \norm {n_0}^2 + \cdots + \norm {a_s} \norm {n_0}^s\) Norm axiom (N2) (Multiplicativity)
\(\displaystyle \) \(=\) \(\displaystyle \norm {a_0} + \norm {a_1} n_0^\alpha + \norm {a_2} n_0^{2 \alpha} + \cdots + \norm {a_s} n_0^{s\alpha}\) as $\norm {n_0} = n_0^\alpha$
\(\displaystyle \) \(\le\) \(\displaystyle 1 + n_0^\alpha + n_0^{2 \alpha} + \cdots + n_0^{s \alpha}\) as $\forall a_i: \norm {a_i} \le 1$
\(\displaystyle \) \(=\) \(\displaystyle n_0^{s \alpha} \paren {1 + n_0^{-\alpha} + n_0^{-2 \alpha} + \cdots + n_0^{-s \alpha} }\)
\(\displaystyle \) \(\le\) \(\displaystyle n_0^{s \alpha} \paren {\sum_{i \mathop = 0}^\infty \paren{\frac 1 {n_0^\alpha} }^i}\)
\(\displaystyle \) \(=\) \(\displaystyle n_0^{s \alpha} \paren {\dfrac {n_0^\alpha} {n_0^\alpha - 1} }\) Sum of Infinite Geometric Progression (and as $n_0^\alpha > 1$)
\(\displaystyle \) \(\le\) \(\displaystyle n^\alpha \paren {\dfrac {n_0^\alpha} {n_0^\alpha - 1} }\) 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:

\(\displaystyle \norm {n^N} \le C \paren {n^N}^\alpha\) \(\leadsto\) \(\displaystyle \norm n^N \le C \paren {n^N}^\alpha\) Norm axiom (N2) (Multiplicativity)
\(\displaystyle \) \(\leadsto\) \(\displaystyle \norm n^N \le C \paren {n^\alpha}^N\)
\(\displaystyle \) \(\leadsto\) \(\displaystyle \norm n \le \sqrt [N] C n^\alpha\) 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.

$\blacksquare$


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