# Homework Help: Difficult Functional Equation

1. Oct 11, 2009

### snipez90

1. The problem statement, all variables and given/known data
Suppose we define an absolute value on the rationals to be a function f: Q -> Q satisfying:
$a(x) \geq 0$ for all x in Q and $a(x) = 0 \Leftrightarrow x = 0$
$a(xy) = a(x)a(y)$ for all x,y in Q
$a(x + y) \leq a(x) + a(y)$ for all x,y in Q

Determine all such functions and prove they are the only ones that exist.

2. Relevant equations

3. The attempt at a solution
All right this is by far the hardest problem on my analysis pset. Obviously the usual absolute value (typical metric used on the reals) satisfies this and so does a trivial one where you set a(x) = 1 for x =/= 0 and a(0) = 0. I had suspected that the p-adic absolute value also satisfies this mainly because I've seen the metric before when I studied basic topology.

I think I figured out the proof for the case that forces the function described in the problem statement to be the p-adic absolute value:

Suppose $a(n) \leq 1$ for each natural number n. If our absolute value is not the trivial one, we can stipulate the existence of some n such that a(n) < 1. By well-ordering, choose the smallest such n and call it p. Now p must be prime, since otherwise p = xy and a(p) = a(x)a(y) < 1 which implies a(x) < 1 and a(y) < 1, contrary to our choice of p as the least natural number n satisfying a(n) < 1. If q is a prime distinct from p, then we claim a(q) = 1. Suppose a(q) < 1. Then for sufficiently large N, $a(q^N) = [a(q)]^N < 1/2$ and similarly for sufficiently large M, $a(p^M) < 1/2.$ Since gcd(q^N, p^M) = 1, there exist integers s and t such that $sp^M + tq^N = 1.$ But then

$$1 = a(sp^M + tq^N) \leq a(sp^M) + a(tq^N) \leq a(s)a(p^M) + a(t)a(q^N) < 1/2 + 1/2 = 1,$$

which is a contradiction (we assumed a(1) = 1, but this has to follow from a(xy) = a(x)(ay) because otherwise a(1*n) = a(1)a(n) = 0 for each natural number n). Thus a(q) = 1, and due to uniqueness of prime factorization, we can write a(n) = a(p)^M.

Unfortunately, I'm having trouble showing that the usual absolute value is also implied (presumably by the case a(n) > 1 for all natural n). Does anyone have any ideas? Thanks in advance.

2. Oct 11, 2009

### LCKurtz

I haven't really looked at your work but just a quick note, you can't have a(n) > 1 for all n because a(1) = a(1)*a(1) implies a(1) = 1 or 0, But a(1) > 0 so you know a(1) = 1. Also 1 = a(1) = a(-1*-1) = a(-1)*a(-1) implies a(-1) = 1 by a similar argument.