First case

Suppose that there exists a positive integer n such that ${\displaystyle |n|_{*}>1.}$ Let k be a non-negative integer and b be a positive integer greater than ${\displaystyle 1}$. We express ${\displaystyle n^{k}}$ in base b: there exist a positive integer m and integers ${\displaystyle (c_{i})_{0\leq i<m}}$ such that for all i, ${\displaystyle 0\leq c_{i}<b}$ and ${\displaystyle n^{k}=\sum _{i<m}c_{i}b^{i}}$. In particular, ${\displaystyle n^{k}\geq b^{m-1}}$ so ${\displaystyle m\leq 1+k\log _{b}n}$.

Each term ${\displaystyle |c_{i}b^{i}|_{*}}$ is smaller than ${\displaystyle (b-1)|b|_{*}^{i}}$. (By the multiplicative property, ${\displaystyle |c_{i}b^{i}|_{*}=|c_{i}|_{*}|b|_{*}^{i}}$, then using the fact that ${\displaystyle c_{i}}$ is a digit, write ${\displaystyle c_{i}=1+1+\cdots +1}$ so by the triangle inequality, ${\displaystyle |c_{i}|\leq |1|+|1|+\cdots +|1|=c_{i}\leq b-1}$.) Besides, ${\displaystyle |b|_{*}^{i}}$ is smaller than ${\displaystyle \max\{1,|b|_{*}^{m-1}\}}$. By the triangle inequality and the above bound on m, it follows:

${\displaystyle {\begin{aligned}|n|_{*}^{k}&\leq m\max _{i<m}\{|c_{i}b^{i}|_{*}\}\\&\leq m(b-1)\max\{1,|b|_{*}^{m-1}\}\\&\leq (1+k\log _{b}n)(b-1)\max\{1,|b|_{*}^{k\log _{b}n}\}.\end{aligned}}}$

Therefore, raising both sides to the power ${\displaystyle 1/k}$, we obtain

${\displaystyle |n|_{*}\leq \left((1+k\log _{b}n)(b-1)\right)^{1/k}\max\{1,|b|_{*}^{\log _{b}n}\}.}$

Finally, taking the limit as k tends to infinity shows that

${\displaystyle |n|_{*}\leq \max\{1,|b|_{*}^{\log _{b}n}\}.}$

Together with the condition ${\displaystyle |n|_{*}>1,}$ the above argument leads to ${\displaystyle |b|_{*}>1}$ regardless of the choice of b (otherwise ${\displaystyle |b|_{*}^{\log _{b}n}\leq 1}$ implies ${\displaystyle |n|_{*}\leq 1}$). As a result, all integers greater than one have an absolute value strictly greater than one. Thus generalizing the above, for any choice of integers n and b greater than or equal to 2, we get

${\displaystyle |n|_{*}\leq |b|_{*}^{\log _{b}n},}$

i.e.

${\displaystyle \log _{n}|n|_{*}\leq \log _{b}|b|_{*}.}$

By symmetry, this inequality is an equality. In particular, for all ${\displaystyle n\geq 2}$, ${\displaystyle \log _{n}|n|_{*}=\log _{2}|2|_{*}=\lambda }$, i.e. ${\displaystyle |n|_{*}=n^{\lambda }=|n|_{\infty }^{\lambda }}$. Because the triangle inequality implies that for all positive integers n we have ${\displaystyle |n|_{*}\leq n}$, in this case we obtain more precisely that ${\displaystyle 0<\lambda \leq 1}$.

As per the above result on the determination of an absolute value by its values on the prime numbers, we easily see that ${\displaystyle |r|_{*}=|r|_{\infty }^{\lambda }}$ for all rational r, thus demonstrating equivalence to the real absolute value.

Second case

Suppose that for all integer n, one has ${\displaystyle |n|_{*}\leq 1}$. As our absolute value is non-trivial, there must exist a positive integer n for which ${\displaystyle |n|_{*}<1.}$ Decomposing ${\displaystyle |n|_{*}}$ on the prime numbers shows that there exists ${\displaystyle p\in \mathbb {P} }$ such that ${\displaystyle |p|_{*}<1}$. We claim that in fact this is so for one prime number only.

Suppose per contra that p and q are two distinct primes with absolute value strictly less than 1. Let k be a positive integer such that ${\displaystyle |p|_{*}^{k}}$ and ${\displaystyle |q|_{*}^{k}}$ are smaller than ${\displaystyle 1/2}$. By Bezout's identity, since ${\displaystyle p^{k}}$ and ${\displaystyle q^{k}}$ are coprime, there exist two integers a and b such that ${\displaystyle ap^{k}+bq^{k}=1.}$ This yields a contradiction, as

${\displaystyle 1=|1|_{*}\leq |a|_{*}|p|_{*}^{k}+|b|_{*}|q|_{*}^{k}<{\frac {|a|_{*}+|b|_{*}}{2}}\leq 1.}$

This means that there exists a unique prime p such that ${\displaystyle |p|_{*}<1}$ and that for all other prime q, one has ${\displaystyle |q|_{*}=1}$ (from the hypothesis of this second case). Let ${\displaystyle \lambda =-\log _{p}|p|_{*}}$. From ${\displaystyle |p|_{*}<1}$, we infer that ${\displaystyle 0<\lambda <\infty }$. (And indeed in this case, all positive ${\displaystyle \lambda }$ give absolute values equivalent to the p-adic one.)

We finally verify that ${\displaystyle |p|_{p}^{\lambda }=p^{-\lambda }=|p|_{*}}$ and that for all other prime q, ${\displaystyle |q|_{p}^{\lambda }=1=|q|_{*}}$. As per the above result on the determination of an absolute value by its values on the prime numbers, we conclude that ${\displaystyle |r|_{*}=|r|_{p}^{\lambda }}$ for all rational r, implying that this absolute value is equivalent to the p-adic one. ${\displaystyle \blacksquare }$