TITLE: Find all continuous functions such that $f(x)f(2x)\dots f(nx) \le an^k$
QUESTION [7 upvotes]: Find all continuous functions $f : \mathbb{R} \rightarrow [1,\infty)$ for which there exist $a \in \mathbb{R}$ and $k$ a positive integer such that 
$$f(x)f(2x)\dots f(nx) ≤ an^k,$$
for every real number $x$ and positive integer $n$.
EDIT: A trivial solution I could find was the constant function equal to one.

REPLY [3 votes]: If $f$ is not constant, by continuity there is some $m>1$ and an interval $[b,c]$ such that $f\geqslant m$ on $[b,c]$.
We can assume without loss of generality that $0<b$ and $b=\frac{p_1}{q}$ while $c=\frac{p_2}{q}$ for some natural numbers $p_1,p_2,q$ (making the interval smaller if necessary). Taking $p=p_1q$, we can actually assume that $b=\frac{p}{q^2}$ and $c=\frac{p+q}{q^2}$.
For $x=q^{-2l}$ and $n=(p+q)q^{2l-2}$ we have
$$ f(x)f(2x)\dots f(nx) \geqslant f(pq^{2l-2}q^{-2k})\cdot f((pq^{2l-2}+1)q^{-2l})\cdots f((p+q)q^{2l-2}q^{-2l})\geqslant m^{q\cdot q^{2l-2}}=m^{q^{2l-1}}$$
Now $n$, as well as any $an^k=a(p+q)^kq^{2lk-2k}$, grows only exponentially in $l$, while this expression grows superexponentially, so it cannot be bounded (for all $l$) by any $an^k$.