The following is exercise 3.3 from Terence Tao's "Solving Mathematical Problems." Emphasis added.
Suppose $f$ is a function on the positive integers which takes real values with the following properties:
(a) $f(2)=2$
(b) $f(mn)=f(m)f(n)$ for all positive integers $m$, $n$.
(c) $f(m)>f(n)$ if $m>n$.
Find $f(1983)$. (Hint: First try to prove that $f(3)=3$ by comparing $f(2^n)$ with $f(3^m)$ for various integers $m$, $n$.) For an additional challenge, replace $f(2)=2$ with $f(n)=n$ for at least one integer $n \geq 2$.
Tao demonstrates in the book that $f(n)=n$ is the only function with these properties when $f$ takes integer values, and the proof relies heavily on this restriction. If $f$ takes real values, then his proof does not work, although it's implied in the hint that the solution is the same. Despite the hint, I've been unable to prove even that $f(3)=3$, much less get to the "challenge" part of the problem. Any help would be appreciated.
$\endgroup$ 01 Answer
$\begingroup$Let $n \in \mathbf N$, then
$$ 3^n = 2^{n \log_2 3} $$
hence, as $f(2^k) = 2^k$ for $k \in \mathbf N$ by multiplicativity,
$$ 2^{\lfloor n \log_2 3\rfloor} \le f(3^n) = f(3)^n \le 2^{\lceil n \log_2 3\rceil}
$$
hence
$$ 2^{n^{-1}\lfloor n \log_2 3\rfloor} \le f(3)\le 2^{n^{-1}\lceil n \log_2 3\rceil}
$$
For $n \to \infty$, we get
$$ 2^{\log_2 3} \le f(3) \le 2^{\log_2 3} $$
Hence $f(3) = 3$. Along the same line of thought, we get $f(n) = n$ for all $n \in \mathbf N$.
