P-adic Norm is Norm/Proof 2
Theorem
The $p$-adic norm forms a norm on the rational numbers $\Q$.
Proof
Recall that the $p$-adic norm is defined as:
- $\forall q \in \Q: \norm r_p := \begin {cases} 0 & : r = 0 \\ p^{- k} & : r \ne 0 \end {cases}$
where:
- $r = p^k \dfrac m n$
and:
- $k, n \in \Z, m \in \Z_{\ne 0} : p \nmid m, n$
where $\nmid$ stands for "does not divide".
We must show that the norm axioms for all $r_1$, $r_2 \in \Q$:
\((\text N 1)\) | $:$ | \(\ds \forall r \in \Q:\) | \(\ds \norm r_p = 0 \) | \(\ds \iff \) | \(\ds x = 0 \) | ||||
\((\text N 2)\) | $:$ | \(\ds \forall r_1, r_2 \in \Q:\) | \(\ds \norm {r_1 r_2} \) | \(\ds = \) | \(\ds \norm {r_1}_p \times \norm {r_2}_p \) | ||||
\((\text N 3)\) | $:$ | \(\ds \forall r_1, r_2 \in \Q:\) | \(\ds \norm {r_1 + r_2}_p \) | \(\ds \le \) | \(\ds \norm {r_1}_p + \norm {r_2}_p \) |
Norm Axiom $\text N 1$: Positive Definiteness
Let $r \in \Q : r \ne 0$.
Let $k, m\in \Z, n \in \Z_{\ne 0} : p \nmid m, n$.
Suppose $r = 0$.
By definition:
- $\norm {r}_p = 0$
Suppose $r = p^k \dfrac m n \ne 0$
By definition:
- $\norm r_p = \dfrac 1 {p^k} > 0$
Suppose $\norm r_p = 0$.
By definition:
- $r = 0$
$\Box$
Norm Axiom $\text N 2$: Multiplicativity
Suppose $r_1 = 0$ or $r_2 = 0$.
From Norm Axiom $\text N 1$: Positive Definiteness, $\norm {r_1}_p = 0$ or $\norm {r_2}_p = 0$.
Suppose $r_1 \ne 0 \ne r_2$.
Let $k_1, k_2, m_1, m_2 \in \Z, n_1, n_2 \in \Z_{\ne 0} : p \nmid n_1, n_2, m_1, m_2$
Let $r_1 = p^{k_1} \dfrac {m_1} {n_1}, r_2 = p^{k_2} \frac {m_2} {n_2}$
Then:
- $r_1 r_2 = p^{k_1 + k_2} \dfrac {m_1 m_2}{n_1 n_2}$
We have that $p \nmid m_1$, $p \nmid m_2$.
Since $p$ is prime:
- $p \nmid m_1 m_2$.
Similarly:
- $p \nmid n_1 n_2$.
Therefore:
\(\ds \norm {r_1 r_2}_p\) | \(=\) | \(\ds p^{-\paren {k_1 + k_2} }\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds p^{-k_1} p^{-k_2}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \norm {r_1}_p \norm {r_2}_p\) |
$\Box$
Norm Axiom $\text N 3$: Triangle Inequality
Suppose one of the following is true:
- $r_1 = 0$
- $r_2 = 0$
- $r_1 + r_2 = 0$
Then the result is straightforward.
Suppose $r_1 \ne 0$, $r_2 \ne 0$, $r_1 + r_2 \ne 0$.
Let $r_1 = p^{k_1} \dfrac {m_1} {n_1}, r_2 = p^{k_2} \dfrac {m_2} {n_2}$ where:
- $k_1, k_2, m_1, m_2 \in \Z, n_1, n_2 \in \Z_{\ne 0} : p \nmid m_1, m_2, n_1, n_2$
Then:
\(\ds r_1 + r_2\) | \(=\) | \(\ds \frac {p^{k_1} m_1 n_2 + p^{k_2} m_2 n_1} {n_1 n_2}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds p^{\map \min {k_1, k_2} } \frac {p^{k_1 \mathop - \map \min {k_1, k_2} } m_1 n_2 + p^{k_2 \mathop - \map \min {k_1, k_2} } n_1 m_2} {n_1 n_2}\) | Definition of Min Operation | |||||||||||
\(\ds \) | \(=\) | \(\ds p^{\map \min {k_1, k_2} } \frac {\tilde m} {n_1 n_2}\) | $\tilde m := p^{k_1 \mathop - \map \min {k_1, k_2} } m_1 n_2 + p^{k_2 \mathop - \map \min {k_1, k_2} } n_1 m_2$ |
By Fundamental Theorem of Arithmetic:
- $\exists ! \tilde k \in \Z_{\ge 0} : \exists m \in \Z : p \nmid m : \tilde m = p^{\tilde k} m$
Obviously, $p \nmid n_1 n_2$
Hence:
\(\ds \norm {r_1 + r_2}_p\) | \(=\) | \(\ds \frac 1 {p^{\tilde k + \map \min {k_1, k_2} } }\) | ||||||||||||
\(\ds \) | \(\le\) | \(\ds \frac 1 {p^{\map \min {k_1, k_2} } }\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \map \max {p^{-k_1}, p^{-k_2} }\) | Definition of Max Operation | |||||||||||
\(\ds \) | \(=\) | \(\ds \map \max {\norm {r_1}_p, \norm {r_1}_p}\) | ||||||||||||
\(\ds \) | \(\le\) | \(\ds \map \max {\norm {r_1}_p, \norm {r_2}_p} + \map \min {\norm {r_1}_p, \norm {r_2}_p}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \norm {r_1}_p + \norm {r_2}_p\) |
$\Box$
All norm axioms are seen to be satisfied.
Hence the result.
$\blacksquare$
Sources
- 2017: Amol Sasane: A Friendly Approach to Functional Analysis ... (previous) ... (next): $\S 1.2$: Normed and Banach spaces. Normed spaces