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Consider the question

Let $f:\mathbb N \to \mathbb R$ satisfy $f(mr)<f(r)$ for all $m,r\in \mathbb N$, $m>1$. Is $f(k)$ decreasing for all $k>k_0$ for some $k_0$?

The answer is clearly no as we have the following counterexample $$f(n)=\begin{cases} 1/n & \text{when $n$ is even}\\ 2/n & \text{when $n$ is odd} \end{cases}$$ where $f$ is increasing at all even values of $n$.

Now, my question is whether there are any extra (minimal) assumptions that we can put on $f(k)$ to make it decreasing for large $k$? Or rather what non-trivial conditions can be imposed on $f$ to make it decreasing?

I feel like I may have seen a similar problem before (in some Olympiad question probably) but I really don't remember where. And, a google search doesn't yield anything useful. Please let me know if anyone has seen this problem before or has any idea how to handle it.

I understand that given this setup, $f$ does not have any constraints at the primes except the upper bound $f(p)<f(1)$. But, I must point out that this does not imply we can set any values at the primes since we also need to make sure that $f$ at the multiples of these primes don't clash.

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    $\begingroup$ Is $m$ required to be $> 1$? Otherwise, we have $f(1 \times r) < f(r)$ $\endgroup$
    – Vezen BU
    Commented Mar 8 at 5:37
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    $\begingroup$ Your condition does not restrict values at primes. $\endgroup$
    – coffeemath
    Commented Mar 8 at 5:44
  • $\begingroup$ @VezenBU thanks, I made the edit! $\endgroup$
    – Dumbest person on earth
    Commented Mar 8 at 6:06
  • $\begingroup$ I think @coffeemath's point is very important! For any increasing sequence of primes $p_1, p_2, \ldots$, the values $f(p_i)$ have no constraints at all. $\endgroup$
    – Vezen BU
    Commented Mar 8 at 6:14
  • $\begingroup$ @VezenBU please check my edit $\endgroup$
    – Dumbest person on earth
    Commented Mar 11 at 17:16

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