Proposition 4

Let $f$ be a mapping of $\mathbf{Q}$ to $\mathbf{R}_+$ belonging to $\mathscr{V}(\mathbf{Q})$. Then:

(i) Either $f$ is the improper absolute value on $\mathbf{Q}$.

(ii) Or there exists a real number $a$ and a prime number $p$ such that $0 < a < 1$ and $f = a^{v_p}$, where $v_p$ is the $p$-adic valuation.

(iii) Or there exists $s > 0$ such that $f(x) = |x|^s$ for all $x \in \mathbf{Q}$.

In care (iii) for $f$ to be an absolute value on $\mathbf{Q}$, it is necessary and sufficient that $0 < s \leq 1$.

Suppose first that $f(n) \leq 1$ for every integer $n > 0$. By Proposition 3 of no. 2 there exist a real number $b$ and a valuation $v$ on $\mathbf{Q}$ such that $0 < b < 1$ and $f = b^v$. Now, we know (§ 3, no. 4, Example 4) that the only valuations on $\mathbf{Q}$ are (up to equivalence) the improper valuation and the $p$-adic valuations $v_p$; we therefore have either case (i) or case (ii).

Suppose from now on that there exists an integer $h > 0$ such that $f(h) > 1$; by no. 1, Corollary 2 to Proposition 2, there exists a number $\rho > 0$ such that $f^\rho$ is an absolute value; let us write

for every rational number $x \neq 0$. Let $a, b$ be two integers $\geq 2$; for every integer $n \geq 2$ let $q(n)$ denote the integral part of $n \log a/\log b$, in other words the least integer $m$ such that $a^n < b^{m+1}$; the expansion of $a^n$ to base $b$ is therefore

where $0 \leq c_i < b$ for $0 \leq i \leq q(n)$. As $f^\rho$ is an absolute value, $f^\rho(c_i) \leq c_i \leq b$ and we therefore deduce from (2) that

Taking logarithms on both sides of this inequality and dividing by $n \log a$, we obtain

Note now that as $n$ tends to $+\infty$, $q(n)/n$ tends to $\log a/\log b$; therefore $q(n)$ tends to $+\infty$ and

tends to 0 (Functions of a Real Variable, Chapter III, § 2, no. 1). Taking the limit in (3), we obtain

(4)

But $f(h) > 1$, whence $g(h) > 0$; if $a$ is replaced by $h$ in (4), we obtain $\sup(0, g(b)) > 0$ and hence

Then, for any integers $a, b$ at least equal to 2, $g(a) \leq g(b)$ and therefore $g(a) = g(b)$, exchanging the roles of $a$ and $b$. In other words, there exists a constant $\lambda$ such that $g(a) = A$ for every integer $a \geq 2$; if we write $s = \lambda / \rho$, then $f(a) = |a|^s$ for every integer $a \geq 2$. As $f(xy) = f(x)f(y)$ and $f(-x) = f(x)$, $f(x) = |x|^s$ for all $x \in Q$. Finally, if $0 < s \leq 1$, we know that $x \mapsto |x|^s$ is an absolute value (General Topology, Chapter IX, § 3, no. 2); conversely, if $s$ is such that $x \mapsto |x|^s$ is an absolute value on $Q$, then $(1 + 1)^s \leq 1^s + 1^s$, that is $2^s \leq 2$, whence $s \leq 1$.